Jan 01, 2013 College Algebra, Ninth Edition. I am proud to present to you this new edition. As with all editions, I have been able to incorporate many useful comments from you, our user. And while much has changed in this revision, you will still find what you expect—a pedagogically sound, mathematically precise, and comprehensive textbook. MyLab Math with Pearson eText - Standalone Access Card - for Algebra for College Students (9th Edition) by Margaret L. Lial, John Hornsby, et al. Jan 19, 2019.
College Algebra 7th Edition Textbook
College Algebra Textbooks Pdf
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College Algebra
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NINTH EDITION
College Algebra Raymond A. Barnett Merritt College
Michael R. Ziegler Marquette University
Karl E. Byleen Marquette University
Dave Sobecki Miami University Hamilton
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COLLEGE ALGEBRA, NINTH EDITION Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © 2011 by The McGraw-Hill Companies, Inc. All rights reserved. Previous editions © 2008, 2001, and 1999. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 DOW/DOW 1 0 9 8 7 6 5 4 3 2 1 0 ISBN 978–0–07–351949–4 MHID 0–07–351949–9 ISBN 978–0–07–729713–8 (Annotated Instructor’s Edition) MHID 0–07–729713–X Editorial Director: Stewart K. Mattson Sponsoring Editor: John R. Osgood Director of Development: Kristine Tibbetts Developmental Editor: Christina A. Lane Marketing Manager: Kevin M. Ernzen Lead Project Manager: Sheila M. Frank Senior Production Supervisor: Kara Kudronowicz Senior Media Project Manager: Sandra M. Schnee Designer: Tara McDermott Cover/Interior Designer: Ellen Pettergell (USE) Cover Image: © Bob Pool/Getty Images Senior Photo Research Coordinator: Lori Hancock Supplement Producer: Mary Jane Lampe Compositor: Aptara®, Inc. Typeface: 10/12 Times Roman Printer: R. R. Donnelley All credits appearing on page or at the end of the book are considered to be an extension of the copyright page. Chapter R Opener: © Corbis RF; p. 31: © The McGraw-Hill Companies, Inc./John Thoeming photographer. Chapter 1 Opener: © Corbis RF; p. 56: © Vol. 71/Getty RF; p. 92: © Getty RF. Chapter 2 Opener: © Vol. 88/Getty RF; p. 142: © Big Stock Photo; p. 147: © Corbis RF; p. 151: © Vol. 112/Getty RF. Chapter 3 Opener: © Getty RF; p. 170: © Getty RF; p. 187: © Vol. 88/Getty RF; p. 220: © Corbis RF; p. 250: © The McGraw-Hill Companies, Inc./Andrew Resek photographer. Chapter 4 Opener: © Corbis RF; p. 271: © Corbis RF; p. 272: © Vol. 4/Getty RF. Chapter 5 Opener: © Getty RF; p. 333: © Vol. 68/Getty RF; p. 345: © Corbis RF. Chapter 6 Opener: © Brand X/SuperStock RF; p. 401: © California Institute of Technology; Chapter 7 Opener: © Corbis RF; p. 435: Courtesy of Bill Tapenning, USDA; p. 439: © Vol. 5/Getty RF; p. 456: © Vol. 48/Getty RF; p. 460: © Getty RF. Chapter 8 Opener: © Vol.6/Getty RF; p. 531: © ThinkStock/PictureQuest RF; p. 543: © Corbis RF. Library of Congress Cataloging-in-Publication Data College algebra / Raymond A. Barnett ... [et al.]. — 9th ed. p. cm. Rev. ed. of: College algebra. 8th ed. / Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen. Includes index. ISBN 978-0-07-351949-4 — ISBN 0-07-351949-9 (hard copy : alk. paper) 1. Algebra–Textbooks. I. Barnett, Raymond A. QA154.3.B365 2011 512.9–dc22 2009019471
www.mhhe.com
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The Barnett, Ziegler, Byleen, and Sobecki Precalculus Series College Algebra, Ninth Edition This book is the same as Precalculus without the three chapters on trigonometry. ISBN 0-07-351949-9, ISBN 978-0-07-351-949-4
Precalculus, Seventh Edition This book is the same as College Algebra with three chapters of trigonometry added. The trigonometry functions are introduced by a unit circle approach. ISBN 0-07-351951-0, ISBN 978-0-07-351-951-7
College Algebra with Trigonometry, Ninth Edition This book differs from Precalculus in that College Algebra with Trigonometry uses right triangle trigonometric to introduce the trigonometric functions. ISBN 0-07-735010-3, ISBN 978-0-07-735010-9
College Algebra: Graphs and Models, Third Edition This book is the same as Precalculus: Graphs and Models without the three chapters on trigonometry. This text assumes the use of a graphing calculator. ISBN 0-07-305195-0, ISBN 978-0-07-305195-6
Precalculus: Graphs and Models, Third Edition This book is the same as College Algebra: Graphs and Models with three additional chapters on trigonometry. The trigonometric functions are introduced by a unit circle approach. This text assumes the use of a graphing calculator. ISBN 0-07-305196-9, ISBN 978-0-07-305-196-3
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About the Authors
Raymond A. Barnett, a native of and educated in California, received his B.A. in mathematical statistics from the University of California at Berkeley and his M.A. in mathematics from the University of Southern California. He has been a member of the Merritt College Mathematics Department and was chairman of the department for four years. Associated with four different publishers, Raymond Barnett has authored or co-authored 18 textbooks in mathematics, most of which are still in use. In addition to international English editions, a number of the books have been translated into Spanish. Co-authors include Michael Ziegler, Marquette University; Thomas Kearns, Northern Kentucky University; Charles Burke, City College of San Francisco; John Fujii, Merritt College; Karl Byleen, Marquette University; and Dave Sobecki, Miami University Hamilton. Michael R. Ziegler received his B.S. from Shippensburg State College and his M.S. and Ph.D. from the University of Delaware. After completing postdoctoral work at the University of Kentucky, he was appointed to the faculty of Marquette University where he held the rank of Professor in the Department of Mathematics, Statistics, and Computer Science. Dr. Ziegler published more than a dozen research articles in complex analysis and co-authored more than a dozen undergraduate mathematics textbooks with Raymond Barnett and Karl Byleen before passing away unexpectedly in 2008. Karl E. Byleen received his B.S., M.A., and Ph.D. degrees in mathematics from the University of Nebraska. He is currently an Associate Professor in the Department of Mathematics, Statistics, and Computer Science of Marquette University. He has published a dozen research articles on the algebraic theory of semigroups and co-authored more than a dozen undergraduate mathematics textbooks with Raymond Barnett and Michael Ziegler. Dave Sobecki earned a B.A. in math education from Bowling Green State University, then went on to earn an M.A. and a Ph.D. in mathematics from Bowling Green. He is an associate professor in the Department of Mathematics at Miami University in Hamilton, Ohio. He has written or co-authored five journal articles, eleven books and five interactive CD-ROMs. Dave lives in Fairfield, Ohio with his wife (Cat) and dogs (Macleod and Tessa). His passions include Ohio State football, Cleveland Indians baseball, heavy metal music, travel, and home improvement projects.
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Dedicated to the memory of Michael R. Ziegler, trusted author, colleague, and friend.
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Brief Contents
Preface xiv Features xvii Application Index xxviii
R CHAPTER 1 CHAPTER 2 CHAPTER 3 CHAPTER 4 CHAPTER 5 CHAPTER 6 CHAPTER 7 CHAPTER 8
CHAPTER
Basic Algebraic Operations 1 Equations and Inequalities 43 Graphs 109 Functions 161 Polynomial and Rational Functions 259 Exponential and Logarithmic Functions 327 Additional Topics in Analytic Geometry 385 Systems of Equations and Matrices 423 Sequences, Induction, and Probability 503 Appendix A Cumulative Review Exercises A-1 Appendix B Special Topics A-13 Appendix C Geometric Formulas A-29 Student Answers SA-1 Subject Index I-1
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SECTION 11–1
Systems of Linear Equations in Two Variables
xii
Contents Preface xiv Features xvii Applications Index xxviii
CHAPTER R-1 R-2 R-3 R-4
1-1 1-2 1-3 1-4 1-5 1-6
xii
3
Functions 161 Functions 162 Graphing Functions 175 Transformations of Functions 188 Quadratic Functions 203 Operations on Functions; Composition 223
Inverse Functions 235 Chapter 3 Review 250 Chapter 3 Review Exercises 252 Chapter 3 Group Activity: Mathematical Modeling: Choosing a Cell Phone Plan 257
CHAPTER 4-1 4-2 4-3 4-4 4-5
5-1 5-2 5-3 5-4 5-5
5
Exponential and Logarithmic Functions 327 Exponential Functions 328 Exponential Models 340 Logarithmic Functions 354 Logarithmic Models 365 Exponential and Logarithmic Equations 372 Chapter 5 Review 379 Chapter 5 Review Exercises 380 Chapter 5 Group Activity: Comparing Regression Models 383
CHAPTER 6-1 6-2 6-3
4
Polynomial and Rational Functions 259 Polynomial Functions, Division, and Models 260 Real Zeros and Polynomial Inequalities 278 Complex Zeros and Rational Zeros of Polynomials 288 Rational Functions and Inequalities 298 Variation and Modeling 315 Chapter 4 Review 321 Chapter 4 Review Exercises 323 Chapter 4 Group Activity: Interpolating Polynomials 326
CHAPTER
2
Graphs 109 Cartesian Coordinate Systems 110 Distance in the Plane 122 Equation of a Line 132 Linear Equations and Models 147 Chapter 2 Review 157 Chapter 2 Review Exercises 158 Chapter 2 Group Activity: Average Speed 160
CHAPTER 3-1 3-2 3-3 3-4 3-5
1
Equations and Inequalities 43 Linear Equations and Applications 44 Linear Inequalities 56 Absolute Value in Equations and Inequalities 65 Complex Numbers 74 Quadratic Equations and Applications 84 Additional Equation-Solving Techniques 97 Chapter 1 Review 104 Chapter 1 Review Exercises 106 Chapter 1 Group Activity: Solving a Cubic Equation 108
CHAPTER 2-1 2-2 2-3 2-4
R
Basic Algebraic Operations 1 Algebra and Real Numbers 2 Exponents and Radicals 11 Polynomials: Basic Operations and Factoring 21 Rational Expressions: Basic Operations 32 Chapter R Review 39 Chapter R Review Exercises 40
CHAPTER
3-6
6
Additional Topics in Analytic Geometry 385 Conic Sections; Parabola 386 Ellipse 395 Hyperbola 405 Chapter 6 Review 418 Chapter 6 Review Exercises 421 Chapter 6 Group Activity: Focal Chords 422
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SECTION 11–1
CHAPTER 7-1 7-2 7-3 7-4 7-5
7
Systems of Equations and Matrices 423 Systems of Linear Equations 424 Solving Systems of Linear Equations 441 Matrix Operations 457 Solving Systems of Linear Equations Using Matrix Inverse Methods 470 Determinants and Cramer’s Rule 487
Additional Topics Available Online: (Visit www.mhhe.com/barnett) 7-6 Systems of Nonlinear Equations 7-7 Systems of Linear Inequalities in Two Variables 7-8 Linear Programming Chapter 7 Review 496 Chapter 7 Review Exercises 498 Chapter 7 Group Activity: Modeling with Systems of Linear Equations 501
Systems of Linear Equations in Two Variables
CHAPTER 8-1 8-2 8-3 8-4 8-5 8-6
xiii
8
Sequences, Induction, and Probability 503 Sequences and Series 504 Mathematical Induction 511 Arithmetic and Geometric Sequences 520 Multiplication Principle, Permutations, and Combinations 531 Sample Spaces and Probability 543 The Binomial Formula 558 Chapter 8 Review 564 Chapter 8 Review Exercises 566 Chapter 8 Group Activity: Sequences Specified by Recursion Formulas 568
Appendix A Cumulative Review Exercises A-1 Appendix B Special Topics A-13 Appendix C Geometric Formulas A-29 Student Answers SA-1 Subject Index I-1
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Preface Enhancing a Tradition of Success The ninth edition of College Algebra represents a substantial step forward in student accessibility. Every aspect of the revision of this classic text focuses on making the text more accessible to students while retaining the precise presentation of the mathematics for which the Barnett name is renowned. Extensive work has been done to enhance the clarity of the exposition, improving the overall presentation of the content. This in turn has decreased the length of the text. Specifically, we concentrated on the areas of writing, exercises, worked examples, design, and technology. Based on numerous reviews, advice from expert consultants, and direct correspondence with the many users of previous editions, this edition is more relevant and accessible than ever before. Writing Without sacrificing breadth or depth or coverage, we have rewritten explanations to make them clearer and more direct. As in previous editions, the text emphasizes computational skills, essential ideas, and problem solving rather than theory. Exercises Over twenty percent of the exercises in the ninth edition are new. These exercises encompass both a variety of skill levels as well as increased content coverage, ensuring a gradual increase in difficulty level throughout. In addition, brand new writing exercises have been included at the beginning of each exercise set in order to encourage a more thorough understanding of key concepts for students. Examples Color annotations accompany many examples, encouraging the learning process for students by explaining the solution steps in words. Each example is then followed by a similar matched problem for the student to solve. Answers to the matched problems are located at the end of each section for easy reference. This active involvement in learning while reading helps students develop a more thorough understanding of concepts and processes. Technology Instructors who use technology to teach college algebra, whether it be exploring mathematics with a graphing calculator or assigning homework and quizzes online, will find the ninth edition to be much improved. Refined “Technology Connections” boxes included at appropriate points in the text illustrate how problems previously introduced in an algebraic context may be solved using a graphing calculator. Exercise sets include calculator-based exercises marked with a calculator icon. Note, however, that the use of graphing technology is completely optional with this text. We understand that at many colleges a single text must serve the purposes of teachers with widely divergent views on the proper use of graphing and scientific calculators in college algebra, and this text remains flexible regarding the degree of calculator integration. Additionally, McGraw-Hill’s MathZone offers a complete online homework system for mathematics and statistics. Instructors can assign textbook-specific content as well as customize the level of feedback students receive, including the ability to have students show their work for any given exercise. Assignable content for the ninth edition of College Algebra includes an array of videos and other multimedia along with algorithmic exercises, providing study tools for students with many different learning styles.
A Central Theme In the Barnett series, the function concept serves as a unifying theme. A brief look at the table of contents reveals this emphasis. A major objective of this book is the development of a library of elementary functions, including their important properties and uses. Employing this library as a basic working tool, students will be able to proceed through this book with greater confidence and understanding. xiv
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Preface
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Reflecting trends in the way college algebra is taught, the ninth edition emphasizes functions modeled in the real world more strongly than previous editions. In some cases, data are provided and the student is asked to produce an approximate corresponding function using regression on a graphing calculator. However, as with previous editions, the use of a graphing calculator remains completely optional and any such examples or exercises can be easily omitted without loss of continuity.
Key Features The revised full-color design gives the book a more contemporary feel and will appeal to students who are accustomed to high production values in books, magazines, and nonprint media. The rich color palette, streamlined calculator explorations, and use of color to signify important steps in problem material work in conjunction to create a more visually appealing experience for students. An emphasis on mathematical modeling is evident in section titles such as “Linear Equations and Models” and “Exponential Models.” These titles reflect a focus on the relationship between functions and real-world phenomena, especially in examples and exercises. Modeling problems vary from those where only the function model is given (e.g., when the model is a physical law such as F ⫽ ma), through problems where a table of data and the function are provided, to cases where the student is asked to approximate a function from data using the regression function of a calculator or computer. Matched problems following worked examples encourage students to practice problem solving immediately after reading through a solution. Answers to the matched problems are located at the end of each section for easy reference. Interspersed throughout each section, Explore-Discuss boxes foster conceptual understanding by asking students to think about a relationship or process before a result is stated. Verbalization of mathematical concepts, results, and processes is strongly encouraged in these explanations and activities. Many Explore-Discuss boxes are appropriate for group work. Refined Technology Connections boxes employ graphing calculators to show graphical and numerical alternatives to pencil-and-paper symbolic methods for problem solving—but the algebraic methods are not omitted. Screen shots are from the TI-84 Plus calculator, but the Technology Connections will interest users of any automated graphing utility. Think boxes (color dashed boxes) are used to enclose steps that, with some experience, many students will be able to perform mentally. Balanced exercise sets give instructors maximum flexibility in assigning homework. A wide variety of easy, moderate, and difficult level exercises presented in a range of problem types help to ensure a gradual increase in difficulty level throughout each exercise set. The division of exercise sets into A (routine, easy mechanics), B (more difficult mechanics), and C (difficult mechanics and some theory) is explicitly presented only in the Annotated Instructor’s Edition. This is due to our attempt to avoid fueling students’ anxiety about challenging exercises. This book gives the student substantial experience in modeling and solving applied problems. Over 500 application exercises help convince even the most skeptical student that mathematics is relevant to life outside the classroom. An Applications Index is included following the Guided Tour to help locate particular applications. Most exercise sets include calculator-based exercises that are clearly marked with a calculator icon. These exercises may use real or realistic data, making them computationally heavy, or they may employ the calculator to explore mathematics in a way that would be impractical with paper and pencil. As many students will use this book to prepare for a calculus course, examples and exercises that are especially pertinent to calculus are marked with an icon. A Group Activity is located at the end of each chapter and involves many of the concepts discussed in that chapter. These activities require students to discuss and write about mathematical concepts in a complex, real-world context.
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Preface
Changes to this Edition A more modernized, casual, and student-friendly writing style has been infused throughout the chapters without radically changing the tone of the text overall. This directly works toward a goal of increasing motivation for students to actively engage with their textbooks, resulting in higher degrees of retention. A significant revision to the exercise sets in the new edition has produced a variety of important changes for both students and instructors. As a result, over twenty percent of the exercises are new. These exercises encompass both a variety of skill levels as well as increased content coverage, ensuring a gradual increase in difficulty level throughout. In addition, brand new writing exercises have been included at the beginning of each exercise set in order to encourage a more thorough understanding of key concepts for students. Specific changes include: • The addition of hundreds of new writing exercises to the beginning of each exercise set. These exercises encourage students to think about the key concepts of the sections before attempting the computational and application exercises, ensuring a more thorough understanding of the material. • An update to the data in many application exercises to reflect more current statistics in topics that are both familiar and highly relevant to today’s students. • A significant increase the amount of moderate skill level problems throughout the text in response to the growing need expressed by instructors. The number of colored annotations that guide students through worked examples has been increased throughout the text to add clarity and guidance for students who are learning critical concepts. New instructional videos on graphing calculator operations posted on MathZone help students master the most essential calculator skills used in the college algebra course. The videos are closed-captioned for the hearing impaired, subtitled in Spanish, and meet the Americans with Disabilities Act Standards for Accessible Design. Though these are an entirely optional ancillary, instructors may use them as resources in a learning center, for online courses, and to provide extra help to students who require extra practice. Chapter R, “Basic Algebraic Operations,” has been extensively rewritten based upon feedback from reviewers to provide a streamlined review of basic algebra in four sections rather than six. Exponents and radicals are now covered in a single section (R-2), and the section covering operations on polynomials (R-3) now includes factoring. Chapter 7, “Systems of Equations and Matrices,” has been reorganized to focus on systems of linear equations, rather than on systems of inequalities or nonlinear systems. A section on determinants and Cramer’s rule (10-5) has been added. Three additional sections on systems of nonlinear equations, systems of linear inequalities, and linear programming are also available online.
Design: A Refined Look with Your Students in Mind The McGraw-Hill Mathematics Team has gathered a great deal of information about how to create a student-friendly textbook in recent years by going directly to the source—your students. As a result, two significant changes have been made to the design of the ninth edition based upon this feedback. First, example headings have been pulled directly out into the margins, making them easy for students to find. Additionally, we have modified the design of one of our existing features—the caution box—to create a more powerful tool for your students. Described by students as one of the most useful features in a math text, these boxes now demand attention with bold red headings pulled out into the margin, alerting students to avoid making a common mistake. These fundamental changes have been made entirely with the success of your students in mind and we are confident that they will improve your students’ overall reaction to and enjoyment of the course.
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Features Examples and Matched Problems Integrated throughout the text, completely worked examples and practice problems are used to introduce concepts and demonstrate problem-solving techniques—algebraic, graphical, and numerical. Each example is followed by a similar Matched Problem for the student to work through while reading the material. Answers to EXAMPLE the matched problems are located at the end of each section for easy reference. This active involvement in the learning process helps students develop a more thorough understanding of algebraic concepts and processes.
2
Using the Distance Formula Find the distance between the points (00033, 5) and (00032, 00038).*
SOLUTION
Let (x1, y1) 0002 (ⴚ3, 5) and (x2, y2) 0002 (ⴚ2, ⴚ8). Then, d 0002 2[(ⴚ2) 0003 (ⴚ3)] 2 0004 [(ⴚ8) 0003 5 ] 2 0002 2(00032 0004 3)2 0004 (00038 0003 5)2 0002 212 0004 (000313)2 0002 21 0004 169 0002 2170 Notice that if we choose (x1, y1) 0002 (00032, 00038) and (x2, y2) 0002 (00033, 5), then d 0002 2 [(00033) 0003 (00032)] 2 0004 [5 0003 (00038) ] 2 0002 21 0004 169 0002 2170 so it doesn’t matter which point we designate as P1 or P2.
MATCHED PROBLEM 2
Find the distance between the points (6, 00033) and (00037, 00035).
Z Midpoint of a Line Segment The midpoint of a line segment is the point that is equidistant from each of the endp A formula for finding the midpoint is given in Theorem 2. The proof is discussed i exercises.
Exploration and Discussion Would you like to incorporate more discovery learning in your course? Interspersed at appropriate places in every section, Explore-Discuss boxes encourage students to think critically about mathematics and to explore key concepts in more detail. Verbalization of mathematical concepts, results, and processes is ZZZ EXPLORE-DISCUSS 1 encouraged in these Explore-Discuss boxes, as well as in some matched problems, and in problems marked with color numerals in almost every exercise set. Explore-Discuss material can be used in class or in an out-of-class activity.
To graph the equation y ⫽ ⫺x3 ⫹ 2x, we use point-by-point plotting to obtain the graph in Figure 5. (A) Do you think this is the correct graph of the equation? If so, why? If not, why? (B) Add points on the graph for x ⫽ ⫺2, ⫺0.5, 0.5, and 2. (C) Now, what do you think the graph looks like? Sketch your version of the graph, adding more points as necessary. (D) Write a short statement explaining any conclusions you might draw from parts A, B, and C.
y 5
x
y
⫺1 ⫺1 0 0 1 1
⫺5
5
x
⫺5
Z Figure 5
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Applications One of the primary objectives of this book is to give the student substantial experience in modeling and solving real-world problems. Over 500 application exercises help convince even the most skeptical student that mathematics is relevant to everyday life. An Applications 15. 2 2 0002 0.426 16. 3 3 Index is included following the features to help In Problems 17–26, solve exactly. locate particular applications. 3 0005x
4 0005x
4.23
17. log5 x 0002 2
0002 0.089
55. I 0002
18. log3 y 0002 4
x 0002 25
19. log (t 0005 4) 0002 00051
t 0002 41 10
21. log 5 0003 log x 0002 2
20
y 0002 81
20. ln (2x 0003 3) 0002 0
x 0002 00051
22. log x 0005 log 8 0002 1
23. log x 0003 log (x 0005 3) 0002 1
54. L 0002 8.8 0003 5.1 log D for D (astronomy)
6.20
80
5
24. log (x 0005 9) 0003 log 100x 0002 3
10
25. log (x 0003 1) 0005 log (x 0005 1) 0002 1
11 9
26. log (2x 0003 1) 0002 1 0003 log (x 0005 2)
21 8
(1 0003 i)n 0005 1 56. S 0002 R for n (annuity) i
27. 2 0002 1.05x
28. 3 0002 1.06x
14.2
29. e00051.4x 0003 5 0002 0 No solution
31. 123 0002 500e00050.12x 0005x 2
33. e
0002 0.23
00071.21
11.7
32. 438 0002 200e0.25x x2
34. e 0002 125
00072.20
B In Problems 35–48, solve exactly. 35. log (5 0005 2x) 0002 log (3x 0003 1) 36. log (x 0003 3) 0002 log (6 0003 4x)
59. y 0002 No solution
e x 0005 e0005x e x 0003 e0005x
3.14
5
38. log (6x 0003 5) 0005 log 3 0002 log 2 0005 log x
2 0003 13
40. ln (x 0003 1) 0002 ln (3x 0003 1) 0005 ln x
1 0003 12 1 0003 189 4
42. 1 0005 log (x 0005 2) 0002 log (3x 0003 1)
3
n0002
ln(1 0003 i )
60. y 0002
x 0002 12 ln
e x 0005 e0005x 2
e x 0003 e0005x e x 0005 e0005x
1 y00031 x 0002 ln 2 y00051
10005y
In Problems 61–68, use a graphing calculator to approximate to two decimal places any solutions of the equation in the interval 0 0004 x 0004 1. None of these equations can be solved exactly using any step-by-step algebraic process. 0.38
62. 30005x 0005 3x 0002 0 64. xe2x 0005 1 0002 0
0.57
x 0002 0.25 x 0002 0.43
65. ln x 0003 2x 0002 0
0.43
66. ln x 0003 x2 0002 0
67. ln x 0003 e x 0002 0
0.27
68. ln x 0003 x 0002 0
x 0002 0.65 x 0002 0.57
2 3
39. ln x 0002 ln (2x 0005 1) 0005 ln (x 0005 2) 41. log (2x 0003 1) 0002 1 0005 log (x 0005 1)
RI L ln al 0005 b R E
ln(Si R 0003 1)
x 0002 ln [y 0003 2y 2 0003 1]
10003y
63. e0005x 0005 x 0002 0
00051
No solution
44. 1 0003 ln (x 0003 1) 0002 ln (x 0005 1)
58. y 0002
61. 20005x 0005 2x 0002 0 4 5
37. log x 0005 log 5 0002 log 2 0005 log (x 0005 3)
43. ln (x 0003 1) 0002 ln (3x 0003 3)
e x 0003 e0005x 2
x 0002 ln (y 0007 2y 2 0005 1)
18.9
30. e0.32x 0003 0.47 0002 0
D 0002 10(L00058.8)00065.1
t00020005
The following combinations of exponential functions define four of six hyperbolic functions, a useful class of functions in calculus and higher mathematics. Solve Problems 57–60 for x in terms of y. The results are used to define inverse hyperbolic functions, another useful class of functions in calculus and higher mathematics. 57. y 0002
In Problems 27–34, solve to three significant digits.
E (1 0005 e0005Rt0006L) for t (circuitry) R
No solution
APPLICATIONS 69. COMPOUND INTEREST How many years, to the nearest year, will it take a sum of money to double if it is invested at 7% compounded annually? 10 years 70. COMPOUND INTEREST How many years, to the nearest year, will it take money to quadruple if it is invested at 6% compounded annually? 24 years
Technology Connections Technology Connections Technology Connections boxes integrated at appropriate points in the text illustrate how conFigure 1 shows the details of constructing the logarithmic model of Example 5 on a graphing calculator. cepts previously introduced in an algebraic context may be approached using a graphing calculator. Students always learn the algebraic methods first so that they develop a solid grasp of these methods and do not become calculatordependent. The exercise sets contain calculatorZ Figure 1 based exercises that are clearly marked with a calculator icon. The use of technology is 62. g(x) 0002 4e 0003 7; f (x) 0002 e completely optional with this text. All technology 63. g(x) 0002 3 0003 4e ; f (x) 0002 e features and exercises may be omitted without sacrificing 64. g(x) 0002 00032 0003 5e ; f (x) 0002 e content coverage. 100
0
120
0
(a) Entering the data
(b) Finding the model
(c) Graphing the data and the model
x00041
0002
x
20003x
40003x
x
x
In Problems 65–68, simplify. 65.
00032x3e00032x 0003 3x2e00032x x6
66.
5x4e5x 0003 4x3e5x x8
67. (e x 0004 e0003x )2 0004 (e x 0003 e0003x )2
2e2x 0004 2e00032x
68. e x(e0003x 0004 1) 0003 e0003x(e x 0004 1)
ex 0003 e0003x
In Problems 69–76, use a graphing calculator to find local extrema, y intercepts, and x intercepts. Investigate the behavior as x S 0007 and as x 00030007 and identify any horizontal asymptotes. Round any approximate values to two decimal places. 69. f (x) 0002 2 0004 e x00032
70. g(x) 0002 00033 0004 e10004x
71. s(x) 0002 e0003x
72. r(x) 0002 e x
2
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Group Activities A Group Activity is located at the end of each chapter and involves many of the concepts discussed in that chapter. These activities strongly encourage the verbalization of mathematical concepts, results, and processes. All of these special activities are highlighted to emphasize their importance.
CHAPTER
ZZZ
5
GROUP ACTIVITY Comparing Regression Models
We have used polynomial, exponential, and logarithmic regression models to fit curves to data sets. How can we determine which equation provides the best fit for a given set of data? There are two principal ways to select models. The first is to use information about the type of data to help make a choice. For example, we expect the weight of a fish to be related to the cube of its length. And we expect most populations to grow exponentially, at least over the short term. The second method for choosing among equations involves developing a measure of how closely an equation fits a given data set. This is best introduced through an example. Consider the data set in Figure 1, where L1 represents the x coordinates and L2 represents the y coordinates. The graph of this data set is shown in Figure 2. Suppose we arbitrarily choose the equation y1 ⫽ 0.6x ⫹ 1 to model these data (Fig. 3).
Each of these differences is called a residual. Note that three of the residuals are positive and one is negative (three of the points lie above the line, one lies below). The most commonly accepted measure of the fit provided by a given model is the sum of the squares of the residuals (SSR). When squared, each residual (whether positive or negative or zero) makes a nonnegative contribution to the SSR. SSR ⫽ (4 ⫺ 2.2)2 ⫹ (5 ⫺ 3.4)2 ⫹ (3 ⫺ 4.6)2 ⫹ (7 ⫺ 5.8)2 ⫽ 9.8 (A) A linear regression model for the data in Figure 1 is given by y2 ⫽ 0.35x ⫹ 3 Compute the SSR for the data and y2, and compare it to the one we computed for y1.
10
0
10
0
Z Figure 1
Z Figure 2 10
0
It turns out that among all possible linear polynomials, the linear regression model minimizes the sum of the squares of the residuals. For this reason, the linear regression model is often called the least-squares line. A similar statement can be made for polynomials of any fixed degree. That is, the quadratic regression model minimizes the SSR over all quadratic polynomials, the cubic regression model minimizes the SSR over all cubic polynomials, and so on. The same statement cannot be made for exponential or logarithmic regression models. Nevertheless, the SSR can still be used to compare exponential, logarithmic, and polynomial models. (B) Find the exponential and logarithmic regression models for the data in Figure 1, compute their SSRs, and compare with the linear model. (C) National annual advertising expenditures for selected years since 1950 are shown in Table 1 where x is years since 1950 and y is total expenditures in billions of dollars. Which regression model would fit this data best: a quadratic model, a cubic model, or an exponential model? Use the SSRs to sup-
10
0
Z Figure 3 y1 ⫽ 0.6x ⫹ 1.
Foundation for Calculus As many students will use this book to prepare for a calculus course, examples and exercises that are especially pertinent to calculus are marked with an icon. EXAMPLE
6
Evaluating and Simplifying a Difference Quotient For f(x) ⫽ x2 ⫹ 4x ⫹ 5, find and simplify: (A) f(x ⫹ h)
SOLUTIONS
(B) f(x ⫹ h) ⫺ f(x)
(C)
f(x ⫹ h) ⫺ f(x) ,h⫽0 h
(A) To find f(x ⫹ h), we replace x with x ⫹ h everywhere it appears in the equation that defines f and simplify: f(x ⴙ h) ⫽ (x ⴙ h)2 ⫹ 4(x ⴙ h) ⫹ 5 ⫽ x2 ⫹ 2xh ⫹ h2 ⫹ 4x ⫹ 4h ⫹ 5 (B) Using the result of part A, we get f(x ⴙ h) ⫺ f(x) ⫽ x2 ⴙ 2xh ⴙ h2 ⴙ 4x ⴙ 4h ⴙ 5 ⫺ (x2 ⴙ 4x ⴙ 5) ⫽ x2 ⫹ 2xh ⫹ h2 ⫹ 4x ⫹ 4h ⫹ 5 ⫺ x2 ⫺ 4x ⫺ 5 ⫽ 2xh ⫹ h2 ⫹ 4h (C)
f(x ⫹ h) ⫺ f(x) 2xh ⫹ h2 ⫹ 4h ⫽ h h ⫽ 2x ⫹ h ⫹ 4
⫽
h(2x ⫹ h ⫹ 4) h
Divide numerator and denominator by h ⴝ 0.
0002
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Student Aids
The domain of f is all x values except ⫺52, or (⫺⬁, ⫺52 ) 傼 (⫺52, ⬁). The value of a fraction is 0 if and only if the numerator is zero:
Annotation of examples and explanations, in small colored type, is found throughout the text to help students through critical stages. Think Boxes are dashed boxes used to enclose steps that students may be encouraged to perform mentally.
Screen Boxes are used to highlight important definitions, theorems, results, and step-by-step processes.
4 ⫺ 3x ⫽ 0
Subtract 4 from both sides.
⫺3x ⫽ ⫺4 x⫽
Divide both sides by ⴚ3.
4 3
The x intercept of f is 43. The y intercept is f(0) ⫽
4 ⫺ 3(0) 2(0) ⫹ 5
4 ⫽ . 5
0002
Z COMPOUND INTEREST If a principal P is invested at an annual rate r compounded m times a year, then the amount A in the account at the end of n compounding periods is given by A 0002 P a1 0003
r n b m
Note that the annual rate r must be expressed in decimal form, and that n 0002 mt, where t is years.
Z DEFINITION 1 Increasing, Decreasing, and Constant Functions Let I be an interval in the domain of function f. Then, 1. f is increasing on I and the graph of f is rising on I if f(x1) 6 f(x2) whenever x1 6 x2 in I. 2. f is decreasing on I and the graph of f is falling on I if f(x1) 7 f(x2) whenever x1 6 x2 in I. 3. f is constant on I and the graph of f is horizontal on I if f(x1) 0002 f (x2) whenever x1 6 x2 in I.
Z THEOREM 1 Tests for Symmetry
Caution Boxes appear throughout the text to indicate where student errors often occur.
Symmetry with respect to the:
An equivalent equation results if:
y axis
x is replaced with ⫺x
x axis
y is replaced with ⫺y
Origin
x and y are replaced with ⫺x and ⫺y
ZZZ CAUTION ZZZ
A very common error occurs about now—students tend to confuse algebraic expressions involving fractions with algebraic equations involving fractions. Consider these two problems: (A) Solve:
x x ⫹ ⫽ 10 2 3
(B) Add:
x x ⫹ ⫹ 10 2 3
The problems look very much alike but are actually very different. To solve the equation in (A) we multiply both sides by 6 (the LCD) to clear the fractions. This works so well for equations that students want to do the same thing for problems like (B). The only catch is that (B) is not an equation, and the multiplication property of equality does not apply. If we multiply (B) by 6, we simply obtain an expression 6 times as large as the original! Compare these correct solutions: x x ⫹ ⫽ 10 2 3
(A) 6ⴢ
x x ⫹ 6 ⴢ ⫽ 6 ⴢ 10 2 3 3x ⫹ 2x ⫽ 60 5x ⫽ 60 x ⫽ 12
xx
(B)
x x ⫹ ⫹ 10 2 3 ⫽
3ⴢx 2ⴢx 6 ⴢ 10 ⫹ ⫹ 3ⴢ2 2ⴢ3 6ⴢ1
3x 2x 60 ⫹ ⫹ 6 6 6 5x ⫹ 60 ⫽ 6
⫽
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Chapter Review sections are provided at the end of each chapter and include a thorough review of all the important terms and symbols. This recap is followed by a comprehensive set of review exercises.
j
CHAPTER
5-1
5
Review
Exponential Functions
The equation f (x) 0003 bx, b 0004 0, b 0005 1, defines an exponential function with base b. The domain of f is (00060007, 0007) and the range is (0, 0007). The graph of f is a continuous curve that has no sharp corners; passes through (0, 1); lies above the x axis, which is a horizontal asymptote; increases as x increases if b 0004 1; decreases as x increases if b 1; and intersects any horizontal line at most once. The function f is one-to-one and has an inverse. We often use the following exponential function properties: 1. a xa y 0003 a x y
(a x) y 0003 a xy
a x ax a b 0003 x b b
(ab)x 0003 a xb x
ax 0003 a x0006y ay
2. a x 0003 a y if and only if x 0003 y. 3. For x 0005 0, a x 0003 b x if and only if a 0003 b. As x approaches 0007, the expression [1 (1兾x)]x approaches the irrational number e ⬇ 2.718 281 828 459. The function f (x) 0003 e x is called the exponential function with base e. The growth of money in an account paying compound interest is described by A 0003 P(1 r兾m)n, where P is the principal, r is the annual rate, m is the number of compounding periods in 1 year, and A is the amount in the account after n compounding periods. If the account pays continuous compound interest, the amount A in the account after t years is given by A 0003 Pert.
5-2
1. Population growth can be modeled by using the doubling time growth model A 0003 A02t d, where A is the population at time t, A0 is the population at time t 0003 0, and d is the doubling time—
CHAPTERS
1–3
3. Limited growth—the growth of a company or proficiency at learning a skill, for example—can often be modeled by the equation y 0003 A(1 0006 e0006kt ), where A and k are positive constants. Logistic growth is another limited growth model that is useful for modeling phenomena like the spread of an epidemic, or sales of a new product. The logistic model is A 0003 M/(1 ce0006kt ), where c, k, and M are positive constants. A good comparison of these different exponential models can be found in Table 3 at the end of Section 5-2. Exponential regression can be used to fit a function of the form y 0003 ab x to a set of data points. Logistic regression can be used to find a function of the form y 0003 c (1 ae0006bx ).
Logarithmic Functions
The logarithmic function with base b is defined to be the inverse of the exponential function with base b and is denoted by y 0003 logb x. So y 0003 logb x if and only if x 0003 b y, b 0004 0, b 0005 1. The domain of a logarithmic function is (0, 0007) and the range is (00060007, 0007). The graph of a logarithmic function is a continuous curve that always passes
Cumulative Review Exercises
*Additional answers can be found in the Instructor Answer Appendix.
Work through all the problems in this cumulative review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. 1. Solve for x:
2. Radioactive decay can be modeled by using the half-life decay model A 0003 A0(12)t h 0003 A020006t h, where A is the amount at time t, A0 is the amount at time t 0003 0, and h is the half-life—the time it takes for half the material to decay. Another model of radioactive decay, A 0003 A0e0006kt , where A0 is the amount at time zero and k is a positive constant, uses the exponential function with base e. This model can be used for other types of quantities that exhibit negative exponential growth as well.
5-3
Exponential Models
Exponential functions are used to model various types of growth:
Cumulative Review Exercise sets are provided in Appendix A for additional reinforcement of key concepts.
the time it takes for the population to double. Another model of population growth, A 0003 A0ekt, where A0 is the population at time zero and k is a positive constant called the relative growth rate, uses the exponential function with base e. This model is used for many other types of quantities that exhibit exponential growth as well.
7x 3 0003 2x x 0002 10 0002 0007 00032 5 2 3
x 0007 52
Problems 16–18 refer to the function f given by the graph: f(x) 5
(1-1)
00025
5
x
In Problems 2–4, solve and graph the inequality. 2. 2(3 0002 y) 0003 4 0004 5 0002 y
00025
3. 冟x 0002 2冟 0005 7
16. Find the domain and range of f. Express answers in interval notation. Domain: [00022, 3]; range: [00021, 2] (3-2)
4. x2 0003 3x 0006 10 5. Perform the indicated operations and write the answer in standard form: (A) (2 0002 3i) 0002 (00025 0003 7i) (B) (1 0003 4i)(3 0002 5i) 50003i (C) (A) 7 0002 10i (B) 23 0003 7i (C) 1 0002 i (1-4) 2 0003 3i In Problems 6–9, solve the equation. 7. 4x2 0002 20 0007 0
8. x2 0002 6x 0003 2 0007 0
9. x 0002 112 0002 x 0007 0
x 0007 000215, 15 (1-5)
x 0007 3 (1-6)
x 0007 3 17 (1-5)
10. Given the points A 0007 (3, 2) and B 0007 (5, 6), find: (A) Distance between A and B. (B) Slope of the line through A and B. (C) Slope of a line perpendicular to the line through A and B. (A) 215
(B) 2
(C)
000212
(2-2, 2-3)
11. Find the equation of the circle with radius 12 and center: (A) (0, 0) (B) (00023, 1) (A) x2 0003 y2 0007 2
Neither (3-3)
18. Use the graph of f to sketch a graph of the following: (A) y 0007 0002f(x 0003 1) (B) y 0007 2f (x) 0002 2 In Problems 19–21, solve the equation. 19.
6. 3x2 0007 000212x
x 0007 00024, 0 (1-5)
17. Is f an even function, an odd function, or neither? Explain.
(B) (x 0003 3)2 0003 (y 0002 1)2 0007 2 (2-2)
12. Graph 2x 0002 3y 0007 6 and indicate its slope and intercepts. 13. Indicate whether each set defines a function. Find the domain and range of each function. (A) {(1, 1), (2, 1), (3, 1)}
x00033 5x 0003 2 5 0003 0007 2x 0003 2 3x 0003 3 6
No solution
20.
21. 2x 0003 1 0007 312x 0002 1 x 0007 1, 52
3 6 1 0007 0002 x x00031 x00021
x 0007 12 , 3 (1-1)
(1-1)
(1-6)
In Problems 22–24, solve and graph the inequality. 22. 冟4x 0002 9冟 7 3
23. 2(3m 0002 4)2 0004 2
x00031 24. 7 x00022 2 25. For what real values of x does the following expression represent a real number? 1x 0002 2 x00024 26 P f
th i di t d
ti
d
it th fi l
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Experience Student Success! ALEKS is a unique online math tool that uses adaptive questioning and artificial intelligence to correctly place, prepare, and remediate students . . . all in one product! Institutional case studies have shown that ALEKS has improved pass rates by over 20% versus traditional online homework and by over 30% compared to using a text alone. By offering each student an individualized learning path, ALEKS directs students to work on the math topics that they are ready to learn. Also, to help students keep pace in their course, instructors can correlate ALEKS to their textbook or syllabus in seconds. To learn more about how ALEKS can be used to boost student performance, please visit www.aleks.com/highered/math or contact your McGraw-Hill representative.
ALEKS Pie
Easy Graphing Utility!
Each student is given their own individualized learning path.
Students can answer graphing problems with ease!
Course Calendar Instructors can schedule assignments and reminders for students.
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New ALEKS Instructor Module Enhanced Functionality and Streamlined Interface Help to Save Instructor Time The new ALEKS Instructor Module features enhanced functionality and streamlined interface based on research with ALEKS instructors and homework management instructors. Paired with powerful assignment driven features, textbook integration, and extensive content flexibility, the new ALEKS Instructor Module simplifies administrative tasks and makes ALEKS more powerful than ever.
New Gradebook! Instructors can seamlessly track student scores on automatically graded assignments. They can also easily adjust the weighting and grading scale of each assignment.
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Track Student Progress Through Detailed Reporting Instructors can track student progress through automated reports and robust reporting features.
Automatically Graded Assignments Instructors can easily assign homework, quizzes, tests, and assessments to all or select students. Deadline extensions can also be created for select students.
Learn more about ALEKS by visiting www.aleks.com/highered/math or contact your McGraw-Hill representative. Select topics for each assignment xxiii
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Preface
Supplements ALEKS (Assessment and Learning in Knowledge Spaces) is a dynamic online learning system for mathematics education, available over the Web 24/7. ALEKS assesses students, accurately determines their knowledge, and then guides them to the material that they are most ready to learn. With a variety of reports, Textbook Integration Plus, quizzes, and homework assignment capabilities, ALEKS offers flexibility and ease of use for instructors. • ALEKS uses artificial intelligence to determine exactly what each student knows and is ready to learn. ALEKS remediates student gaps and provides highly efficient learning and improved learning outcomes • ALEKS is a comprehensive curriculum that aligns with syllabi or specified textbooks. Used in conjunction with McGraw-Hill texts, students also receive links to text-specific videos, multimedia tutorials, and textbook pages. • ALEKS offers a dynamic classroom management system that enables instructors to monitor and direct student progress towards mastery of course objectives. ALEKS Prep/Remediation: • Helps instructors meet the challenge of remediating under prepared or improperly placed students. • Assesses students on their pre-requisite knowledge needed for the course they are entering (i.e. Calculus students are tested on Precalculus knowledge) and prescribes unique and efficient learning paths specific to their strengths and weaknesses. • Students can address pre-requisite knowledge gaps outside of class freeing the instructor to use class time pursuing course outcomes.
McGraw-Hill’s MathZone is a complete online homework system for mathematics and statistics. Instructors can assign textbook-specific content from over 40 McGraw-Hill titles as well as customize the level of feedback students receive, including the ability to have students show their work for any given exercise. Assignable content includes an array of videos and other multimedia along with algorithmic exercises, providing study tools for students with many different learning styles. MathZone also helps ensure consistent assignment delivery across several sections through a course administration function and makes sharing courses with other instructors easy. In addition, instructors can also take advantage of a virtual whiteboard by setting up a Live Classroom for online office hours or a review session with students. For more information, visit the book’s website (www.mhhe.com/barnett) or contact your local McGraw-Hill sales representative (www.mhhe.com/rep).
Tegrity Campus is a service that makes class time available all the time by automatically capturing every lecture in a searchable format for students to review when they study and complete assignments. With a simple one-click start and stop process, you capture all computer screens and corresponding audio. Students replay any part of any class with easy-touse browser-based viewing on a PC or Mac.
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Preface
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Educators know that the more students can see, hear, and experience class resources, the better they learn. With Tegrity Campus, students quickly recall key moments by using Tegrity Campus’s unique search feature. This search helps students efficiently find what they need, when they need it across an entire semester of class recordings. Help turn all your students’ study time into learning moments immediately supported by your lecture. To learn more about Tegrity watch a 2 minute Flash demo at http://tegritycampus.mhhe.com. Instructor Solutions Manual Prepared by Fred Safier of City College of San Francisco, this supplement provides detailed solutions to exercises in the text. The methods used to solve the problems in the manual are the same as those used to solve the examples in the textbook.
Student Solutions Manual Prepared by Fred Safier of City College of San Francisco, the Student’s Solutions Manual provides complete worked-out solutions to odd-numbered exercises from the text. The procedures followed in the solutions in the manual match exactly those shown in worked examples in the text. Lecture and Exercise Videos The video series is based on exercises from the textbook. J. D. Herdlick of St. Louis Community College-Meramec introduces essential definitions, theorems, formulas, and problem-solving procedures. Professor Herdlick then works through selected problems from the textbook, following the solution methodologies employed by the authors. The video series is available on DVD or online as part of MathZone. The DVDs are closed-captioned for the hearing impaired, subtitled in Spanish, and meet the Americans with Disabilities Act Standards for Accessible Design. NetTutor Available through MathZone, NetTutor is a revolutionary system that enables students to interact with a live tutor over the web. NetTutor’s web-based, graphical chat capabilities enable students and tutors to use mathematical notation and even to draw graphs as they work through a problem together. Students can also submit questions and receive answers, browse previously answered questions, and view previous sessions. Tutors are familiar with the textbook’s objectives and problem-solving styles. Computerized Test Bank (CTB) Online Available through the book’s website, this computerized test bank, utilizing Brownstone Diploma® algorithm-based testing software, enables users to create customized exams quickly. This user-friendly program enables instructors to search for questions by topic, format, or difficulty level; to edit existing questions or to add new ones; and to scramble questions and answer keys for multiple versions of the same test. Hundreds of text-specific open-ended and multiple-choice questions are included in the question bank. Sample chapter tests and a sample final exam in Microsoft Word® and PDF formats are also provided.
Acknowledgments In addition to the authors, many others are involved in the successful publication of a book. We wish to thank personally the following people who reviewed the text and offered invaluable advice for improvements: Marwan Abu-Sawwa, Florida Community College at Jacksonville Gerardo Aladro, Florida International University Eugene Allevato, Woodbury University Joy Becker, University of Wisconsin–Stout Susan Bradley, Angelina College Ellen Brook, Cuyahoga Community College, Eastern Campus Kelly Brooks, Pierce College Denise Brown, Collin County Community College Cheryl Davids, Central Carolina Technical College
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Preface
Timothy Delworth, Purdue University Marcial Echenique, Broward Community College Gay Ellis, Missouri State University Jackie English, Northern Oklahoma College Enid Mike Everett, Santa Ana College Nicki Feldman, Pulaski Technical College James Fightmaster, Virginia Western Community College Perry Gillespie, University of North Carolina at Fayetteville Vanetta Grier-Felix, Seminole Community College David Gurney, Southeastern Louisiana University Celeste Hernandez, Richland College Fredrick Hoffman, Florida Atlantic University Syed Hossain, University of Nebraska at Kearney Glenn Jablonski, Triton College Sarah Jackson, Pratt Community College Charles Johnson, South Georgia College Larry Johnson, Metropolitan State College of Denver Cheryl Kane, University of Nebraska Lincoln Raja Khoury, Collin County Community College Betty Larson, South Dakota State University Owen Mertens, Southwest Missouri State University Dana Nimic, Southeast Community College, Lincoln Campus Lyn Noble, Florida Community College at Jacksonville Luke Papademas, DeVry University, DeVry Chicago Campus David Phillips, Georgia State University Margaret Rosen, Mercer County Community College Patty Schovanec, Texas Tech University Eleanor Storey, Front Range Community College, Westminster Campus Linda Sundbye, Metropolitan State College of Denver Cynthia Woodburn, Pittsburg State University Martha Zimmerman, University of Louisville Bob Martin, Tarrant County College Susan Walker, Montana Tech of the University of Montana Lynn Cleaveland, University of Arkansas Michael Wodzak, Viterbo University Ryan Kasha, Valencia Community College Frank Juric, Brevard Community College Jerry Mayfield, North Lake College Andrew Shiers, Dakota State University Richard Avery, Dakota State University Mike Everett, Santa Ana College Greg Boyd, Murray State College Sarah Cook, Washburn University Nga Wai Liu, Bowling Green State University Donald Bennett, Murray State University Sharon Suess, Asheville-Buncombe Technical Community College Dale Rohm, University of Wisconsin at Stevens Point George Anastassiou, The University of Memphis Bill White, University of South Carolina Upstate Linda Sundbye, Metropolitan State College of Denver Khaled Hussein, University of Wisconsin Diane Cook, Okaloosa Walton College Celeste Hernandez, Richland College Thomas Riedel, University of Louisville Thomas English, College of the Mainland Hayward Allan Edwards, West Virginia University at Parkersburg
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Preface
xxvii
Debra Lehman, State Fair Community College Nancy Ressler, Oakton Community College Marwan Zabdawi, Gordon College Ianna West, Nicholls State University Tzu-Yi Alan Yang, Columbus State Community College Patricia Jones, Methodist University Kay Geving, Belmont University Linda Horner, Columbia State Community College Martha Zimmerman, University of Louisville Faye Childress, Central Piedmont Community College Bradley Thiessen, Saint Ambrose University Pamela Lasher, Edinboro University of Pennsylvania We also wish to thank Carrie Green for providing a careful and thorough check of all the mathematical calculations in the book (a tedious but extremely important job). Fred Safier for developing the supplemental manuals that are so important to the success of a text. Mitchel Levy for scrutinizing our exercises in the manuscript and making recommendations that helped us to build balanced exercise sets. Tony Palermino for providing excellent guidance in making the writing more direct and accessible to students. Pat Steele for carefully editing and correcting the manuscript. Christina Lane for editorial guidance throughout the revision process. Sheila Frank for guiding the book smoothly through all publication details. All the people at McGraw-Hill who contributed their efforts to the production of this book. Producing this new edition with the help of all these extremely competent people has been a most satisfying experience.
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APPLICATION INDEX Advertising, 326, 378 Aeronautical engineering, 404 AIDS epidemics, 348–349, 352 Airfreight, 469 Air search, 95–96 Airspeed, 434–435, 439 Air temperature, 146, 530 Alcohol consumption, 95, 221 Anthropology, 320 Approximation, 74 Architecture, 96, 132, 160, 174–175, 325, 417, 572 Astronomy, 371, 394, 530, 579 Atmospheric pressure, 373–374, 531 Automobile rental, 187 Average cost, 315 Bacterial growth, 341–342, 351, 530 Biology, 321 Body surface area, 148–149, 160 Body weight, 155, 249 Boiling point of water, 146 Braking distance, 20 Break-even analysis, 107, 213–214, 222, 440, 576 Business, 55, 64, 74, 121, 439, 500, 530, 576 Business markup policy, 149 Buying, 455 Carbon-14 dating, 343–344, 378, 381 Card hands, 540, 551 Cell division, 530 Cell phone cost, 174 Cell phone subscribers, 382 Chemistry, 55, 61–62, 74, 107, 371, 455–456, 576 Circuit analysis, 486–487 City planning, 147 Code words, 534 Coin problem, 31 Coin toss, 532–533, 548–549, 553, 557–558 Committee selection, 551–552 Communications, 421 Compound interest, 333–335, 373, 377, 567, 579 Computer design, 351 Computer-generated tests, 533 Computer science, 183, 187, 257, 576 Construction, 31, 42, 95, 97, 104, 132, 170–171, 175, 220, 256–257, 277, 285, 288, 298, 315, 325, 371, 576 Continuous compound interest, 335–336 Cost analysis, 107, 142, 146, 155, 160 Cost functions, 174, 202 Cost of high speed internet, 174 Counting card hands, 540 Counting code words, 534 Counting serial numbers, 540–541 Court design, 102 xxviii
Crime statistics, 326 Cryptography, 482–483, 486, 501 Data analysis, 160, 346 Delivery charges, 187, 456 Demographics, 147 Depreciation, 155–156, 159, 353, 576 Design, 104, 107, 404 Diamond prices, 152–153 Dice roles, 546, 554, 557 Diet, 500, 572 Distance-rate-time problems, 50–51, 92 Divorce, 277 Earthquakes, 366–368, 371, 376, 379, 382, 579 Earth science, 55, 64, 352, 440 Ecology, 371 Economics, 20, 42, 55, 64, 530, 567, 572 Economy stimulation, 527–528 Electric circuit, 42 Electricity, 320 Employee training, 314 Energy, 64 Engineering, 96, 132, 220–221, 321, 394, 404, 421, 530, 572 Epidemics, 345–346 Evaporation, 203, 234 Explosive energy, 371 Fabrication, 298 Falling object, 220, 256 Finance, 339, 440, 530, 572, 576 Flight conditions, 156 Flight navigation, 156 Fluid flow, 203, 234 Food chain, 530 Forestry, 155, 160 Gaming, 351 Gas mileage, 220 Genealogy, 530 Genetics, 321 Geometry, 31, 55, 103, 287, 321, 456, 487, 531, 572, 579 Health care, 277 Heat conduction, 501 History of technology, 351 Home ownership rates, 369 Hydroelectric power, 272–273 Illumination, 320 Immigration, 377 Income, 256
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APPLICATION INDEX
Income tax, 181–182 Infectious diseases, 347 Insecticides, 351 Insurance company data, 555–556 Internet access, 371–372 Inventory value, 468–469 Investment, 441 Investment allocation, 480–481 Labor costs, 464–465, 468, 500–501 Learning curve, 344–345 Learning theory, 315 Licensed drivers, 156 Life science, 55 Linear depreciation, 159 Loan repayment, 567 Logistic growth, 345–346 Manufacturing, 103, 174, 277, 287–288, 325, 572 Marine biology, 352, 378, 382 Market analysis, 558, 567–568 Market research, 234, 256 Markup policy, 156, 250, 256, 468 Marriage, 277 Maximizing revenue, 222, 277 Maximum area, 210–211 Medical research, 378 Medicare, 382 Medicinal lithotripsy, 401 Medicine, 352, 382 Mixing antifreeze, 150 Mixture problems, 52–53, 150 Money growth, 339, 382 Movie industry revenue, 220 Music, 56, 321, 531 Naval architecture, 404–405 Navigation, 95, 412–414 Newton’s law of cooling, 352, 378 Nuclear power, 353, 417–418 Numbers, 107, 576 Nutrition, 456, 469 Oceanography, 146–147 Officer selection, 537 Olympic games, 157 Ozone levels, 113 Packaging, 31, 298 Parachutes, 156 Pendulum, 21 Petty crime, 455 Photography, 321, 352, 378, 531 Physics, 122, 146, 320–321, 326, 510, 530, 576, 580 Physiology, 314–315 Player ranking, 470 Political science, 572 Politics, 107, 469
xxix
Pollution, 234 Population growth, 340–341, 351–352, 378, 381, 530, 579 Present value, 339, 382 Price and demand, 93, 95, 121, 249–250, 256, 576–577 Price and supply, 121, 250 Prize money, 524 Production costs, 202, 468 Production rates, 440 Production scheduling, 436–437, 440–441, 456, 462, 486 Profit analysis, 213–214, 220, 221–222, 230–231 Profit and loss analysis, 213–214, 576 Projectile flight, 220 Projectile motion, 211, 594–595, 596 Psychology, 56, 64, 321 Purchasing, 452–453, 572 Puzzle, 501, 530–531 Quality control, 568 Quantity-rate-time problems, 51–52 Radioactive decay, 342–343 Radioactive tracers, 351 Rate of descent, 156 Rate problems, 174 Rate-time, 576 Rate-time problems, 55–56, 107, 439 Regression, 346 Relativistic mass, 21 Replacement time, 315 Resource allocation, 456, 486, 500 Retention, 315 Revenue, 242–243, 277 Revenue analysis, 496 Rocket flight, 368–369 Safety research, 203 Salary increment, 510 Sales and commissions, 187, 460–461 Serial numbers, 540–541 Service charges, 187 Shipping, 288, 576, 579 Signal light, 394 Significant digits, 74 Simple interest, 326 Smoking statistics, 155 Sociology, 456 Sound, 365–366, 371, 382, 579 Space science, 321, 352, 394, 418, 421 Space vehicles, 371 Speed of sound, 155 Sports, 131–132 Sports medicine, 107, 160 State income tax, 187, 257 Statistics, 74 Stopping distance, 214–215, 221, 250, 256, 576 Storage, 298 Subcommittee selection, 539 Supply and demand, 157, 435–436, 440
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Telephone charges, 187 Telephone expenditures, 153–154 Temperature, 122 Test averaging, 572 Thumbtack toss, 552 Timber harvesting, 202–203 Tournament seeding, 469–470 Traffic flow, 501–502 Training, 353 Transportation, 96, 567
Underwater pressure, 151 Weather, 175 Weather balloon, 234 Weight of fish, 271 Well depth, 103 Wildlife management, 353, 382 Work, 326 Zeno’s paradox, 531
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College Algebra
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CHAPTER
Basic Algebraic Operations
R
C
OUTLINE
ALGEBRA is “generalized arithmetic.” In arithmetic we add, subtract,
multiply, and divide specific numbers. In algebra we use all that we know about arithmetic, but, in addition, we work with symbols that represent one or more numbers. In this chapter we review some important basic algebraic operations usually studied in earlier courses.
R-1
Algebra and Real Numbers
R-2
Exponents and Radicals
R-3
Polynomials: Basic Operations and Factoring
R-4
Rational Expressions: Basic Operations Chapter R Review
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Algebra and Real Numbers Z The Set of Real Numbers Z The Real Number Line Z Addition and Multiplication of Real Numbers Z Further Operations and Properties
3 The numbers 14, 00033, 0, 73, 12, and 1 6 are examples of real numbers. Because the symbols we use in algebra often stand for real numbers, we will discuss important properties of the real number system.
Z The Set of Real Numbers Informally, a real number is any number that has a decimal representation. So the real numbers are the numbers you have used for most of your life. The set of real numbers, denoted by R, is the collection of all real numbers. The notation 12 僆 R (read “ 12 is an element of R”) expresses the fact that 12 is a real number. The set Z 0002 {. . . , 00032, 00031, 0, 1, 2, . . .} of the natural numbers, along with their negatives and zero, is called the set of integers. We write Z ( R (read “Z is a subset of R”) to express the fact that every element of Z is an element of R; that is, that every integer is a real number. Table 1 describes the set of real numbers and some of its important subsets. Study Table 1 and note in particular that N ( Z ( Q ( R. No real number is both rational and irrational, so the intersection (overlap) of the sets Q and I is the empty set (or null set), denoted by 0004. The empty set contains no elements,
Table 1 The Set of Real Numbers Symbol
Name
Description
Examples
N
Natural numbers
Counting numbers (also called positive integers)
1, 2, 3, . . .
Z
Integers
Natural numbers, their negatives, and 0 (also called whole numbers)
. . . , 00032, 00031, 0, 1, 2, . . .
Q
Rational numbers
Numbers that can be represented as a兾b, where a and b are integers and b 0006 0; decimal representations are repeating or terminating
00034, 0, 1, 25, 000335, 23, 3.67, 00030.333,* 5.272727
I
Irrational numbers
Numbers that can be represented as nonrepeating and nonterminating decimal numbers
3 12, 0005, 1 7, 1.414213 . . . ,† 2.71828182 . . .†
R
Real numbers
Rational numbers and irrational numbers
*The overbar indicates that the number (or block of numbers) repeats indefinitely. †Note that the ellipsis does not indicate that a number (or block of numbers) repeats indefinitely.
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SECTION R–1
3
Algebra and Real Numbers
so it is true that every element of the empty set is an element of any given set. In other words, the empty set is a subset of every set. Two sets are equal if they have exactly the same elements. The order in which the elements of a set are listed does not matter. For example, {1, 2, 3, 4} 0002 {3, 1, 4, 2}
Z The Real Number Line A one-to-one correspondence exists between the set of real numbers and the set of points on a line. That is, each real number corresponds to exactly one point, and each point to exactly one real number. A line with a real number associated with each point, and vice versa, as in Figure 1, is called a real number line, or simply a real line. Each number associated with a point is called the coordinate of the point. The point with coordinate 0 is called the origin. The arrow on the right end of the line indicates a positive direction. The coordinates of all points to the right of the origin are called positive real numbers, and those to the left of the origin are called negative real numbers. The real number 0 is neither positive nor negative.
4 0003 3 Origin
0003兹27 000310
00035
0
7.64
5
10
Z Figure 1 A real number line.
Z Addition and Multiplication of Real Numbers How do you add or multiply two real numbers that have nonrepeating and nonterminating decimal expansions? The answer to this difficult question relies on a solid understanding of the arithmetic of rational numbers. The rational numbers are numbers that can be written in the form a兾b, where a and b are integers and b 0006 0 (see Table 1 on page 2). The numbers 7兾5 and 00032兾3 are rational, and any integer a is rational because it can be written in the form a兾1. Two rational numbers a兾b and c兾d are equal if ad 0002 bc; for example, 35兾10 0002 7兾2. Recall how the sum and product of rational numbers are defined:
Z DEFINITION 1 Addition and Multiplication of Rationals For rational numbers a兾b and c兾d, where a, b, c, and d are integers and b 0006 0, d 0006 0: Addition: Multiplication:
a 0007 b a ⴢ b
c ad 0007 bc 0002 d bd c ac 0002 d bd
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Addition and multiplication of rational numbers are commutative; changing the order in which two numbers are added or multiplied does not change the result. 3 0007 2 3 ⴢ 2
5 5 3 0002 0007 7 7 2 5 5 3 0002 ⴢ 7 7 2
Addition is commutative.
Multiplication is commutative.
Addition and multiplication of rational numbers is also associative; changing the grouping of three numbers that are added or multiplied does not change the result: 3 5 0007a 0007 2 7 3 5 ⴢa ⴢ 2 7
9 3 5 9 b0002a 0007 b0007 4 2 7 4 9 3 5 9 b0002a ⴢ bⴢ 4 2 7 4
Addition is associative.
Multiplication is associative.
Furthermore, the operations of addition and multiplication are related in that multiplication distributes over addition: 3 5 9 ⴢa 0007 b0002 2 7 4 9 3 5 a 0007 bⴢ 0002 7 4 2
3 5 3 9 ⴢ 0007 ⴢ 2 7 2 4 5 3 9 3 ⴢ 0007 ⴢ 7 2 4 2
Left distributive law
Right distributive law
The rational number 0 is an additive identity; adding 0 to a number does not change it. The rational number 1 is a multiplicative identity; multiplying a number by 1 does not change it. Every rational number r has an additive inverse, denoted 0003r; the additive inverse of 4兾5 is 00034兾5, and the additive inverse of 00033兾2 is 3兾2. The sum of a number and its additive inverse is 0. Every nonzero rational number r has a multiplicative inverse, denoted r00031; the multiplicative inverse of 4兾5 is 5兾4, and the multiplicative inverse of 00033兾2 is 00032兾3. The product of a number and its multiplicative inverse is 1. The rational number 0 has no multiplicative inverse.
EXAMPLE
1
Arithmetic of Rational Numbers Perform the indicated operations. (A)
1 6 0007 3 5
(C) (0003179)00031 SOLUTIONS
(B)
8 5 ⴢ 3 4
(D) (00036 0007 92)00031
(A)
1 6 5 0007 18 23 0007 0002 0002 3 5 15 15
(B)
8 5 40 10 ⴢ 0002 0002 3 4 12 3
40 10 ⴝ 12 3
because
40 ⴢ 3 ⴝ 12 ⴢ 10
(C) (0003179)00031 0002 0003917 (D) (00036 0007 92)00031 0002 a
00036 9 00031 000312 0007 9 00031 00033 00031 2 0007 b 0002a b 0002a b 00020003 1 2 2 2 3
0002
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MATCHED PROBLEM 1*
Algebra and Real Numbers
5
Perform the indicated operations. (A) 0003(52 0007 73) (C)
21 15 ⴢ 20 14
00031 (B) 0003(817)
(D) 5 ⴢ (12 0007 13) 0002
Rational numbers have decimal expansions that are repeating or terminating. For example, using long division, 2 0002 0.666 3 22 0002 3.142857 7 13 0002 1.625 8
The number 6 repeats indefinitely.
The block 142857 repeats indefinitely.
Terminating expansion
Conversely, any decimal expansion that is repeating or terminating represents a rational number (see Problems 49 and 50 in Exercise R-1). The number 12 is irrational because it cannot be written in the form a兾b, where a and b are integers, b 0006 0 (for an explanation, see Problem 89 in Section R-3). Similarly, 13 is irrational. But 14, which is equal to 2, is a rational number. In fact, if n is a positive integer, then 1n is irrational unless n belongs to the sequence of perfect squares 1, 4, 9, 16, 25, . . . (see Problem 90 in Section R-3). We now return to our original question: how do you add or multiply two real numbers that have nonrepeating and nonterminating decimal expansions? Although we will not give a detailed answer to this question, the key idea is that every real number can be approximated to any desired precision by rational numbers. For example, the irrational number 12 ⬇ 1.414 213 562 . . . is approximated by the rational numbers 14 10 141 100 1,414 1,000 14,142 10,000 141,421 100,000
0002 1.4 0002 1.41 0002 1.414 0002 1.4142 0002 1.41421 .. .
Using the idea of approximation by rational numbers, we can extend the definitions of rational number operations to include real number operations. The following box summarizes the basic properties of real number operations. *Answers to matched problems in a given section are found near the end of the section, before the exercise set.
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Z BASIC PROPERTIES OF THE SET OF REAL NUMBERS Let R be the set of real numbers, and let x, y, and z be arbitrary elements of R. Addition Properties Closure:
x 0007 y is a unique element in R.
Associative:
(x 0007 y) 0007 z 0002 x 0007 (y 0007 z)
Commutative:
x0007y0002y0007x
Identity:
0 is the additive identity; that is, 0 0007 x 0002 x 0007 0 0002 x for all x in R, and 0 is the only element in R with this property.
Inverse:
For each x in R, 0003x is its unique additive inverse; that is, x 0007 (0003x) 0002 (0003x) 0007 x 0002 0, and 0003x is the only element in R relative to x with this property.
Multiplication Properties Closure:
xy is a unique element in R.
Associative:
(xy)z 0002 x( yz)
Commutative:
xy 0002 yx
Identity:
1 is the multiplicative identity; that is, for all x in R, (1)x 0002 x(1) 0002 x, and 1 is the only element in R with this property.
Inverse:
For each x in R, x 0006 0, x00031 is its unique multiplicative inverse; that is, xx00031 0002 x00031x 0002 1, and x00031 is the only element in R relative to x with this property.
Combined Property Distributive:
EXAMPLE
2
x(y 0007 z) 0002 xy 0007 xz
(x 0007 y)z 0002 xz 0007 yz
Using Real Number Properties Which real number property justifies the indicated statement?
SOLUTIONS
(A) (B) (C) (D) (E)
(7x)y 0002 7(xy) a(b 0007 c) 0002 (b 0007 c)a (2x 0007 3y) 0007 5y 0002 2x 0007 (3y 0007 5y) (x 0007 y)(a 0007 b) 0002 (x 0007 y)a 0007 (x 0007 y)b If a 0007 b 0002 0, then b 0002 0003a.
(A) (B) (C) (D) (E)
Associative (ⴢ) Commutative (ⴢ) Associative (0007) Distributive Inverse (0007)
0002
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MATCHED PROBLEM 2
Algebra and Real Numbers
7
Which real number property justifies the indicated statement? (A) 4 0007 (2 0007 x) 0002 (4 0007 2) 0007 x (C) 3x 0007 7x 0002 (3 0007 7)x (E) If ab 0002 1, then b 0002 1a.
(B) (a 0007 b) 0007 c 0002 c 0007 (a 0007 b) (D) (2x 0007 3y) 0007 0 0002 2x 0007 3y 0002
Z Further Operations and Properties Subtraction of real numbers can be defined in terms of addition and the additive inverse. If a and b are real numbers, then a 0003 b is defined to be a 0007 (0003b). Similarly, division can be defined in terms of multiplication and the multiplicative inverse. If a and b are real numbers and b 0006 0, then a b (also denoted a兾b) is defined to be a ⴢ b00031.
Z DEFINITION 2 Subtraction and Division of Real Numbers For all real numbers a and b: Subtraction:
a 0003 b 0002 a 0007 (0003b)
Division:
a b 0002 a ⴢ b00031
5 ⴚ 3 ⴝ 5 ⴙ (ⴚ3) ⴝ 2
b00060
3 ⴜ 2 ⴝ 3 ⴢ 2ⴚ1 ⴝ 3 ⴢ
1 ⴝ 1.5 2
It is important to remember that Division by 0 is never allowed.
ZZZ EXPLORE-DISCUSS 1
(A) Give an example that shows that subtraction of real numbers is not commutative. (B) Give an example that shows that division of real numbers is not commutative.
The basic properties of the set of real numbers, together with the definitions of subtraction and division, lead to the following properties of negatives and zero.
Z THEOREM 1 Properties of Negatives For all real numbers a and b: 0003(0003a) 0002 a (0003a)b 0002 0003(ab) 0002 a(0003b) 0002 0003ab (0003a)(0003b) 0002 ab (00031)a 0002 0003a a a 0003a 5. 00020003 0002 b00060 b b 0003b 0003a 0003a a a 6. b00060 00020003 00020003 0002 0003b b 0003b b 1. 2. 3. 4.
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Z THEOREM 2 Zero Properties For all real numbers a and b: 1. a ⴢ 0 0002 0 ⴢ a 0002 0 2. ab 0002 0 if and only if*
a 0002 0 or b 0002 0 or both
Note that if b 0006 0, then 0b 0002 0 ⴢ b00031 0002 0 by Theorem 2. In particular, 03 0002 0; but the expressions 30 and 00 are undefined.
EXAMPLE
3
Using Negative and Zero Properties Which real number property or definition justifies each statement? (A) 3 0003 (00032) 0002 3 0007 [0003(00032)] 0002 5 (B) 0003(00032) 0002 2 00033 3 0002 (C) 0003 2 2 5 5 00020003 (D) 00032 2 (E) If (x 0003 3)(x 0007 5) 0002 0, then either x 0003 3 0002 0 or x 0007 5 0002 0.
SOLUTIONS
MATCHED PROBLEM 3
(A) (B) (C) (D) (E)
Subtraction (Definition 1 and Theorem 1, part 1) Negatives (Theorem 1, part 1) Negatives (Theorem 1, part 6) Negatives (Theorem 1, part 5) Zero (Theorem 2, part 2)
0002
Which real number property or definition justifies each statement? (A)
3 1 0002 3a b 5 5
(B) (00035)(2) 0002 0003(5 ⴢ 2)
(D)
7 00037 00020003 9 9
(E) If x 0007 5 0002 0, then (x 0003 3)(x 0007 5) 0002 0.
(C) (00031)3 0002 00033
0002
ZZZ EXPLORE-DISCUSS 2
A set of numbers is closed under an operation if performing the operation on numbers of the set always produces another number in the set. For example, the set of odd integers is closed under multiplication, but is not closed under addition. (A) Give an example that shows that the set of irrational numbers is not closed under addition. (B) Explain why the set of irrational numbers is closed under taking multiplicative inverses.
*Given statements P and Q, “P if and only if Q” stands for both “if P, then Q” and “if Q, then P.”
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Algebra and Real Numbers
9
If a and b are real numbers, b 0006 0, the quotient a b, when written in the form a兾b, is called a fraction. The number a is the numerator, and b is the denominator. It can be shown that fractions satisfy the following properties. (Note that some of these properties, under the restriction that numerators and denominators are integers, were used earlier to define arithmetic operations on the rationals.)
Z THEOREM 3 Fraction Properties For all real numbers a, b, c, d, and k (division by 0 excluded): 1.
a c 0002 b d 4 6 ⴝ 6 9
2.
since
4ⴢ9ⴝ6ⴢ6
ka a 0002 kb b
3.
7ⴢ3 3 ⴝ 7ⴢ5 5
5.
ad 0002 bc
if and only if
c a0007c a 0007 0002 b b b 3 4 3ⴙ4 7 ⴙ ⴝ ⴝ 6 6 6 6
a c ac ⴢ 0002 b d bd
4.
3 7 3ⴢ7 21 ⴢ ⴝ ⴝ 5 8 5ⴢ8 40
6.
c a0003c a 0003 0002 b b b 7 2 7ⴚ2 5 ⴚ ⴝ ⴝ 8 8 8 8
a c a d 0002 ⴢ b d b c 2 5 2 7 14 ⴜ ⴝ ⴢ ⴝ 3 7 3 5 15
7.
c ad 0007 bc a 0007 0002 b d bd 2 1 2ⴢ5ⴙ3ⴢ1 13 ⴙ ⴝ ⴝ 3 5 3ⴢ5 15
ANSWERS TO MATCHED PROBLEMS 1. (A) 0003296 (B) 000317 8 (C) 9 8 (D) 256 2. (A) Associative (0007) (B) Commutative (0007) (C) Distributive (D) Identity (0007) (E) Inverse (ⴢ) 3. (A) Division (Definition 1) (B) Negatives (Theorem 1, part 2) (C) Negatives (Theorem 1, part 4) (D) Negatives (Theorem 1, part 5) (E) Zero (Theorem 2, part 1)
R-1
Exercises
In Problems 1–16, perform the indicated operations, if defined. If the result is not an integer, express it in the form a/b, where a and b are integers. 1 1 1. 0007 3 5
1 1 2. 0007 2 7
3.
3 4 0003 4 3
4.
8 4 0003 9 5
5.
2 4 ⴢ 3 7
6. a0003
1 3 bⴢ 10 8
7.
11 1 5 3
9. 100 0 3 5 11. a0003 b a0003 b 5 3 13.
17 2 ⴢ 8 7
3 00031 15. a b 0007 200031 8
8.
7 2 9 5
10. 0 0 12.
6 4 a3 0003 b 7 2
2 5 14. a0003 b a0003 b 3 6 16. 0003(400031 0007 3)
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In Problems 17–28, each statement illustrates the use of one of the following properties or definitions. Indicate which one. Commutative (0007) Commutative (ⴢ) Associative (0007) Associative (ⴢ) Distributive Identity (0007) Identity (ⴢ)
Inverse (0007) Inverse (ⴢ) Subtraction Division Negatives (Theorem 1) Zero (Theorem 2)
17. x 0007 ym 0002 x 0007 my 1 21. (00032)(00032 )00021
18. 7(3m) 0002 (7 ⴢ 3)m u u 20. 0003 0002 0003v v 22. 8 0003 12 0002 8 0007 (000312)
23. w 0007 (0003w) 0002 0
1 24. 5 (00036) 0002 5(00036 )
19. 7u 0007 9u 0002 (7 0007 9)u
40. Indicate true (T) or false (F), and for each false statement find real number replacements for a, b, and c that will provide a counterexample. For all real numbers a, b, and c: (A) (a 0007 b) 0007 c 0002 a 0007 (b 0007 c) (B) (a 0003 b) 0003 c 0002 a 0003 (b 0003 c) (C) a(bc) 0002 (ab)c (D) (a b) c 0002 a (b c) In Problems 41–48, indicate true (T) or false (F), and for each false statement give a specific counterexample.
25. 3(xy 0007 z) 0007 0 0002 3(xy 0007 z) 26. ab(c 0007 d ) 0002 abc 0007 abd 0003x x 27. 0002 0003y y
39. Indicate true (T) or false (F), and for each false statement find real number replacements for a and b that will provide a counterexample. For all real numbers a and b: (A) a 0007 b 0002 b 0007 a (B) a 0003 b 0002 b 0003 a (C) ab 0002 ba (D) a b 0002 b a
28. (x 0007 y) ⴢ 0 0002 0
41. The difference of any two natural numbers is a natural number. 42. The quotient of any two nonzero integers is an integer. 43. The sum of any two rational numbers is a rational number.
29. If ab 0002 0, does either a or b have to be 0?
44. The sum of any two irrational numbers is an irrational number.
30. If ab 0002 1, does either a or b have to be 1?
45. The product of any two irrational numbers is an irrational number.
31. Indicate which of the following are true: (A) All natural numbers are integers. (B) All real numbers are irrational. (C) All rational numbers are real numbers. 32. Indicate which of the following are true: (A) All integers are natural numbers. (B) All rational numbers are real numbers. (C) All natural numbers are rational numbers. 33. Give an example of a rational number that is not an integer. 34. Give an example of a real number that is not a rational number. In Problems 35 and 36, list the subset of S consisting of (A) natural numbers, (B) integers, (C) rational numbers, and (D) irrational numbers. 35. S 0002 500033, 000323, 0, 1, 13, 95, 11446
36. S 0002 50003 15, 00031, 000312, 2, 17, 6, 16259, 00056
46. The product of any two rational numbers is a rational number. 47. The multiplicative inverse of any irrational number is an irrational number. 48. The multiplicative inverse of any nonzero rational number is a rational number. 49. If c 0002 0.151515 . . . , then 100c 0002 15.1515 . . . and 100c 0003 c 0002 15.1515 . . . 0003 0.151515 . . . 99c 0002 15 5 c 0002 15 99 0002 33
Proceeding similarly, convert the repeating decimal 0.090909 . . . into a fraction. (All repeating decimals are rational numbers, and all rational numbers have repeating decimal representations.) 50. Repeat Problem 49 for 0.181818. . . .
In Problems 37 and 38, use a calculator* to express each number in decimal form. Classify each decimal number as terminating, repeating, or nonrepeating and nonterminating. Identify the pattern of repeated digits in any repeating decimal numbers. 37. (A) 98
(B) 113
38. (A) 136
(B) 121
(C) 15 (C) 167
(D) 118 29 (D) 111
*Later in the book you will encounter optional exercises that require a graphing calculator. If you have such a calculator, you can certainly use it here. Otherwise, any scientific calculator will be sufficient for the problems in this chapter.
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SECTION R–2
R-2
Exponents and Radicals
11
Exponents and Radicals Z Integer Exponents Z Scientific Notation Z Roots of Real Numbers Z Rational Exponents and Radicals Z Simplifying Radicals
The French philosopher/mathematician René Descartes (1596–1650) is generally credited with the introduction of the very useful exponent notation “x n.” This notation as well as other improvements in algebra may be found in his Geometry, published in 1637. If n is a natural number, x n denotes the product of n factors, each equal to x. In this section, the meaning of x n will be expanded to allow the exponent n to be any rational number. Each of the following expressions will then represent a unique real number: 500034
75
3.140
612
14000353
Z Integer Exponents If a is a real number, then a6 0002 a ⴢ a ⴢ a ⴢ a ⴢ a ⴢ a
6 factors of a
In the expression a6, 6 is called an exponent and a is called the base. Recall that a00031, for a 0006 0, denotes the multiplicative inverse of a (that is, 1 a). To generalize exponent notation to include negative integer exponents and 0, we define a00036 to be the multiplicative inverse of a6, and we define a0 to be 1. n Z DEFINITION 1 a , n an Integer and a a Real Number
For n a positive integer and a a real number: an 0002 a ⴢ a ⴢ . . . ⴢ a 1 a0003n 0002 n a a0 0002 1
EXAMPLE
1
n factors of a
(a 0006 0) (a 0006 0)
Using the Definition of Integer Exponents Write parts (A) and (B) in decimal form and parts (C) and (D) using positive exponents. Assume all variables represent nonzero real numbers. (A) (u3v2)0
(B) 1000033
(C) x00038
(D)
x00033 y00035
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SOLUTIONS
(A) (u3v2)0 0002 1 (C) x00038 0002
MATCHED PROBLEM 1
1 x8
(B) 1000033 0002
1 1 0002 0.001 3 0002 1,000 10
x00033 y00035
x00033 1 1 y5 ⴢ 00035 0002 3 ⴢ 1 1 y x
(D)
0002
* 0002
y5 x3
0002
Write parts (A) and (B) in decimal form and parts (C) and (D) using positive exponents. Assume all variables represent nonzero real numbers. (B) 1000035
(A) (x2)0
(C)
1
u00037 v00033
(D)
00034
x
0002 To calculate with exponents, it is helpful to remember Definition 1. For example: 23 ⴢ 24 0002 (2 ⴢ 2 ⴢ 2)(2 ⴢ 2 ⴢ 2 ⴢ 2) 0002 2300074 0002 27 (23)4 0002 (2 ⴢ 2 ⴢ 2)4 0002 (2 ⴢ 2 ⴢ 2)(2 ⴢ 2 ⴢ 2)(2 ⴢ 2 ⴢ 2)(2 ⴢ 2 ⴢ 2) 0002 23ⴢ4 0002 212 These are instances of Properties 1 and 2 of Theorem 1. Z THEOREM 1 Properties of Integer Exponents For n and m integers and a and b real numbers: 1. aman 0002 am0007n
a5a ⴚ7
ⴝ a5ⴙ(ⴚ7)
2. (an)m 0002 amn 3. (ab)m 0002 ambm a m am 4. a b 0002 m b b
(a3)ⴚ2
ⴝ a(ⴚ2)3
再
am0003n a 1 5. n 0002 a an0003m m
EXAMPLE
2
ⴝ aⴚ2 ⴝ aⴚ6
(ab)3 ⴝ a3b3 a 4 a4 a b ⴝ 4 b b
b00060
a3 aⴚ2
a00060
a3 aⴚ2
ⴝ a3ⴚ(ⴚ2) ⴝ a5 ⴝ
1 aⴚ2ⴚ3
ⴝ
1 aⴚ5
Using Exponent Properties Simplify using exponent properties, and express answers using positive exponents only.†
SOLUTIONS
6x00032 8x00035
(A) (3a5)(2a00033)
(B)
(C) 00034y3 0003 (00034y)3
(D) (2a00033b2)00032
(A) (3a5)(2a00033)
(B)
6x00032 8x00035
0002
0002 (3 ⴢ 2)(a5a00033)
3x000320003(00035) 4
0002
0002 6a2
3x3 4
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally. †
By “simplify” we mean eliminate common factors from numerators and denominators and reduce to a minimum the number of times a given constant or variable appears in an expression. We ask that answers be expressed using positive exponents only in order to have a definite form for an answer. Later (in this section and elsewhere) we will encounter situations where we will want negative exponents in a final answer.
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(C) 00034y3 0003 (00034y)3 0002 00034y3 0003 (00034)3y3
Exponents and Radicals
13
0002 00034y3 0003 (000364)y3
0002 00034y3 0007 64y3 0002 60y3 (D) (2a00033b2)00032 0002 200032a6b00034 0002
MATCHED PROBLEM 2
a6 4b4
0002
Simplify using exponent properties, and express answers using positive exponents only. (A) (5x00033)(3x4)
(B)
9y00037 6y00034
(C) 2x4 0003 (00032x)4
(D) (3x4y00033)00032 0002
Z Scientific Notation Scientific work often involves the use of very large numbers or very small numbers. For example, the average cell contains about 200,000,000,000,000 molecules, and the diameter of an electron is about 0.000 000 000 0004 centimeter. It is generally troublesome to write and work with numbers of this type in standard decimal form. The two numbers written here cannot even be entered into most calculators as they are written. However, each can be expressed as the product of a number between 1 and 10 and an integer power of 10: 200,000,000,000,000 0002 2 1014 0.000 000 000 0004 0002 4 10000313 In fact, any positive number written in decimal form can be expressed in scientific notation, that is, in the form a 10n
EXAMPLE
3
1 a 6 10, n an integer, a in decimal form
Scientific Notation (A) Write each number in scientific notation: 6,430; 5,350,000; 0.08; 0.000 32 (B) Write in standard decimal form: 2.7 102; 9.15 104; 5 1000033; 8.4 1000035
SOLUTIONS
MATCHED PROBLEM 3
(A) 6,430 0002 6.43 103; 5,350,000 0002 5.35 106; 0.08 0002 8 1000032; 0.000 32 0002 3.2 1000034 (B) 270; 91,500; 0.005; 0.000 084
0002
(A) Write each number in scientific notation: 23,000; 345,000,000; 0.0031; 0.000 000 683 (B) Write in standard decimal form: 4 103; 5.3 105; 2.53 1000032; 7.42 1000036
0002
Most calculators express very large and very small numbers in scientific notation. Consult the manual for your calculator to see how numbers in scientific notation are entered in your calculator. Some common methods for displaying scientific notation on a calculator are shown here. Number Represented
Typical Scientific Calculator Display
Typical Graphing Calculator Display
5.427 493 10000317
5.427493 – 17
5.427493E – 17
2.359 779 1012
2.359779 12
2.359779E12
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4
Using Scientific Notation on a Calculator 325,100,000,000 by writing each number in scientific notation and then 0.000 000 000 000 0871 using your calculator. (Refer to the user’s manual accompanying your calculator for the procedure.) Express the answer to three significant digits* in scientific notation. Calculate
SOLUTION
325,100,000,000 3.251 1011 0002 0.000 000 000 000 0871 8.71 10000314 0002 3.732491389E24 24
0002 3.73 10
Z Figure 1
MATCHED PROBLEM 4
Calculator display To three significant digits
Figure 1 shows two solutions to this problem on a graphing calculator. In the first solution we entered the numbers in scientific notation, and in the second we used standard decimal notation. Although the multiple-line screen display on a graphing calculator enables us to enter very long standard decimals, scientific notation is usually more efficient and less prone to errors in data entry. Furthermore, as Figure 1 shows, the calculator uses scientific notation to display the answer, regardless of the manner in which the numbers are entered. 0002 Repeat Example 4 for: 0.000 000 006 932 62,600,000,000 0002
Z Roots of Real Numbers The solutions of the equation x2 0002 64 are called square roots of 64 and the solutions of x3 0002 64 are the cube roots of 64. So there are two real square roots of 64 (00038 and 8) and one real cube root of 64 (4 is a cube root, but 00034 is not). Note that 000364 has no real square root (x2 0002 000364 has no real solution because the square of a real number can’t be negative), but 00034 is a cube root of 000364 because (00034)3 0002 000364. In general:
Z DEFINITION 2 Definition of an nth Root For a natural number n and a and b real numbers: a is an nth root of b if an 0002 b
3 is a fourth root of 81, since 34 ⴝ 81.
The number of real nth roots of a real number b is either 0, 1, or 2, depending on whether b is positive or negative, and whether n is even or odd. Theorem 2 gives the details, which are summarized in Table 1.
*For those not familiar with the meaning of significant digits, see Appendix A for a brief discussion of this concept.
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Table 1 Number of Real nth Roots of b n even
n odd
b 7 0
2
1
b00020
1
1
b 6 0
0
1
Exponents and Radicals
15
Z THEOREM 2 Number of Real nth Roots of a Real Number b Let n be a natural number and let b be a real number: 1. b 7 0: If n is even, then b has two real nth roots, each the negative of the other; if n is odd, then b has one real nth root. 2. b 0002 0: 0 is the only nth root of b 0002 0. 3. b 6 0: If n is even, then b has no real nth root; if n is odd, then b has one real nth root.
Z Rational Exponents and Radicals To denote nth roots, we can use rational exponents or we can use radicals. For example, the square root of a number b can be denoted by b12 or 1b. To avoid ambiguity, both expressions denote the positive square root when there are two real square roots. Furthermore, both expressions are undefined when there is no real square root. In general:
Z DEFINITION 3 Principal nth Root For n a natural number and b a real number, the principal nth root of b, n denoted by b1n or 1b, is: 1. The real nth root of b if there is only one. 2. The positive nth root of b if there are two real nth roots. 3. Undefined if b has no real nth root.
n
In the notation 1b, the symbol 1 is called a radical, n is called the index, and b is the 2 radicand. If n 0002 2, we write 1b in place of 1 b.
EXAMPLE
5
Principal nth Roots Evaluate each expression: (A) 912
SOLUTIONS
MATCHED PROBLEM 5
(B) 1121
(A) 912 0002 3 3 (C) 10003125 0002 00035 (E) 2713 0002 3
3 (C) 1 0003125
(D) (000316)14
(E) 2713
5 (F) 1 32
(B) 1121 0002 11 (D) (000316)14 is undefined (not a real number). 5 (F) 132 0002 2
0002
Evaluate each expression: (A) 813
(B) 100034
4 (C) 110,000
(D) (00031)15
3
(E) 1000327
(F) 018 0002
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How should a symbol such as 723 be defined? If the properties of exponents are to hold for rational exponents, then 723 0002 (713)2; that is, 723 must represent the square of the cube root of 7. This leads to the following general definition: m兾n and bⴚm兾n, Rational Number Exponent Z DEFINITION 4 b
For m and n natural numbers and b any real number (except b cannot be negative when n is even): bmn 0002 (b1n)m 432 ⴝ (412)3 ⴝ 23 ⴝ 8
b0003mn 0002
and 4ⴚ32 ⴝ
1
ⴝ
432
1 8
1 bmn
(ⴚ4)32 is not real
(ⴚ32)35 ⴝ [(ⴚ32)15 ] 3 ⴝ (ⴚ2)3 ⴝ ⴚ8
We have now discussed bmn for all rational numbers mn and real numbers b. It can be shown, though we will not do so, that all five properties of exponents listed in Theorem 1 continue to hold for rational exponents as long as we avoid even roots of negative numbers. With the latter restriction in effect, the following useful relationship is an immediate consequence of the exponent properties: Z THEOREM 3 Rational Exponent/Radical Property For m and n natural numbers and b any real number (except b cannot be negative when n is even): (b1n)m 0002 (bm)1n
ZZZ EXPLORE-DISCUSS 1
and
n
n
(1b)m 0002 2bm
Find the contradiction in the following chain of equations: 00031 0002 (00031)22 0002 3(00031)2 4 12 0002 112 0002 1
(1)
Where did we try to use Theorem 3? Why was this not correct?
EXAMPLE
6
Using Rational Exponents and Radicals Simplify and express answers using positive exponents only. All letters represent positive real numbers. (A) 823
SOLUTIONS
4 (B) 2312
3 (C) (3 1 x)(2 1x)
(A) 823 0002 (813)2 0002 22 0002 4 or 4 12 12 1/4 3 (B) 23 0002 (3 ) 0002 3 0002 27
(D) a
4x13 12 b x12
823 0002 (82)13 0002 6413 0002 4
3 (C) (3 1x)(2 1x) 0002 (3x13)(2x12) 0002 6x13000712 0002 6x56
(D) a
2 2 4x13 12 412x16 b 0002 0002 14000316 0002 112 12 14 x x x x
0002
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MATCHED PROBLEM 6
Exponents and Radicals
17
Simplify and express answers using positive exponents only. All letters represent positive real numbers. (A) (⫺8)5Ⲑ3
5 (B) 2 324
(D) a
4 3 3 (C) (5 2 y )(2 1 y)
8x1Ⲑ2 1Ⲑ3 b x2Ⲑ3
0002
Z Simplifying Radicals The exponent properties considered earlier lead to the following properties of radicals.
Z THEOREM 4 Properties of Radicals For n a natural number greater than 1, and x and y positive real numbers: n
1. 2xn ⫽ x n n n 2. 2xy ⫽ 2x2y 3.
n
x 2x ⫽ n By 2y n
3 3 2 x ⫽x 5 5 5 2 xy ⫽ 2 x2 y 4 x 1 x ⫽ 4 Ay 1y 4
An algebraic expression that contains radicals is said to be in simplified form if all four of the conditions listed in the following definition are satisfied.
Z DEFINITION 5 Simplified (Radical) Form 1. No radicand (the expression within the radical sign) contains a factor to a power greater than or equal to the index of the radical. For example, 2x5 violates this condition.
2. No power of the radicand and the index of the radical have a common factor other than 1. 6 4 For example, 2 x violates this condition.
3. No radical appears in a denominator. For example, y/ 1x violates this condition.
4. No fraction appears within a radical. For example, 235 violates this condition.
EXAMPLE
7
Finding Simplified Form Write in simplified radical form. (A) 212x5y2
6 (B) 216x4y2
(C)
6 12x
(D)
8x4 B y 3
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SOLUTIONS
(A) Condition 1 is violated. First we convert to rational exponent form. 212x5y2 0002 (12x5y2)12 0002 1212x52y 0002 (4 ⴢ 3)12x2x12y 0002 213 x2 1x y 0002 2x2y13x
Use (ab)m ⴝ ambm and (an)m ⴝ amn. 12 ⴝ 4 ⴢ 3, x52 0002 x2x12 Write in radical form. Use commutative property and radical property 2.
(B) Condition 2 is violated. First we convert to rational exponent form. 6 2 16x4y2 0002 (16x4y2)16 0002 1616x23y13 0002 223x23y13 3 0002 2 4x2y
Use (ab)m ⴝ ambm and (an)m ⴝ amn. 16 ⴝ 24 Write in radical form.
(C) Condition 3 is violated. We multiply numerator and denominator by 12x; the effect is to multiply the expression by 1, so its value is unchanged, but the denominator is left free of radicals. 6 6 12x 612x 312x 0002 ⴢ 0002 0002 x 2x 12x 12x 12x (D) Condition 4 is violated. First we convert to rational exponent form. 8x4 813x43 0002 B y y13
y23
3
Multiply by
y23
ⴝ 1.
0002
2x43y23 y
x 43 ⴝ xx 13
0002
2xx13y23 y
Write in radical form.
0002
2x 2xy2 y
3
MATCHED PROBLEM 7
0002
Write in simplified radical form. (A) 218x4y3
9 (B) 2 8x6y3
(C)
30 1 16x 4
(D)
5x3 B y 0002
Eliminating a radical from a denominator [as in Example 7(C)] is called rationalizing the denominator. To rationalize the denominator, we multiply the numerator and denominator by a suitable factor that will leave the denominator free of radicals. This factor is called a rationalizing factor. If the denominator is of the form 1a 0007 1b, then 1a 0003 1b is a rationalizing factor because (1a 0007 1b)(1a 0003 1b) 0002 a 0003 b Similarly, if the denominator is of the form 1a 0003 1b, then 1a 0007 1b is a rationalizing factor.
EXAMPLE
8
Rationalizing Denominators Rationalize the denominator and write the answer in simplified radical form. (A)
8 16 0007 15
(B)
1x 0007 1y 1x 0003 1y
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SOLUTIONS
Exponents and Radicals
19
(A) Multiply numerator and denominator by the rationalizing factor 16 ⫺ 15. 8 8 16 ⫺ 15 ⫽ ⴢ 16 ⫹ 15 16 ⫹ 15 16 ⫺ 15 ⫽
8(16 ⫺ 15) 6⫺5
(1a ⴙ 1b) ( 1a ⴚ 1b) ⴝ a ⴚ b
Simplify.
⫽ 8( 16 ⫺ 15) (B) Multiply numerator and denominator by the rationalizing factor 1x ⫹ 1y. 1x ⫹ 1y 1x ⫹ 1y 1x ⫹ 1y ⫽ ⴢ 1x ⫺ 1y 1x ⫺ 1y 1x ⫹ 1y
MATCHED PROBLEM 8
⫽
x ⫹ 1x1y ⫹ 1y1x ⫹ y x⫺y
⫽
x ⫹ 21xy ⫹ y x⫺y
Expand numerator and denominator.
Combine like terms.
0002
Rationalize the denominator and write the answer in simplified radical form. (A)
6 1 ⫺ 13
(B)
21x ⫺ 31y 1x ⫹ 1y
0002
ANSWERS TO MATCHED PROBLEMS 1. (A) 1 (B) 0.000 01 (C) x4 (D) v3Ⲑu7 3 4 2. (A) 15x (B) 3Ⲑ(2y ) (C) ⫺14x (D) y6 Ⲑ(9x8) 4 8 ⫺3 3. (A) 2.3 ⫻ 10 ; 3.45 ⫻ 10 ; 3.1 ⫻ 10 ; 6.83 ⫻ 10⫺7 (B) 4,000; 530,000; 0.0253; 0.000 007 42 4. 1.11 ⫻ 10⫺19 5. (A) 2 (B) Not real (C) 10 (D) ⫺1 (E) ⫺3 6. (A) ⫺32 (B) 16 (C) 10y13Ⲑ12 (D) 2 Ⲑx1Ⲑ18 4 3 x 15xy 152 x 3 7. (A) 3x2y 12y (B) 2 (C) (D) 2x2y x y 2x ⫺ 51xy ⫹ 3y 8. (A) ⫺3 ⫺ 313 (B) x⫺y
R-2
(F) 0
Exercises
All variables represent positive real numbers and are restricted to prevent division by 0. In Problems 1–14, evaluate each expression. If the answer is not an integer, write it in fraction form. 1 8 1. 37 2. 56 3. a b 2 3 3 4. a b 5. 6⫺3 6. 2⫺6 5
7. (⫺5)4
8. (⫺4)5
10. (⫺7)⫺2
11. ⫺7⫺2
1 0 13. a b 3
14. a
1 ⫺1 b 10
9. (⫺3)⫺1 12. ⫺100
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In Problems 15–20, write the numbers in scientific notation. 15. 58,620,000
75.
16. 4,390
1 215
76.
2
17. 0.027
18. 0.11
19. 0.000 000 064
20. 0.000 0325
78.
In Problems 21–26, write each number in standard decimal form. 21. 4 1000033
22. 5 1000036
23. 2.99 105
24. 7.75 1011
25. 3.1 1000037
26. 8.167 1000034
12y
16y
5 6 7 11 81. x 2 3xy
84.
31y 21y 0003 3
79.
1
77.
3 1 7
4 16 0003 2
3 82. 2a2 8a8b13
85.
215 0007 312 515 0007 212
3 3
254
80.
12 16 0007 2
83.
12m15 120m
86.
312 0003 213 313 0003 212
2
In Problems 27–32, change to radical form. Do not simplify. 27. 3215
28. 62534
29. 4x000312
30. 32y000325
31. x13 0003 y13
32. (x 0003 y)13
87. What is the result of entering 23 on a calculator? 2
88. Refer to Problem 87. What is the difference between 2(3 ) and 3 2 32 (2 ) ? Which agrees with the value of 2 obtained with a calculator?
APPLICATIONS In Problems 33–38, change to rational exponent form. Do not simplify. 33. 1361
3 34. 2172
35. 4x 2y3
5
4 36. 27x3y2
3 37. 2x2 0007 y2
3 3 38. 2x2 0007 2y2
In Problems 39–50, evaluate each expression that represents a real number. 39. 10012
40. 16912
41. 1121
42. 1361
13
44. 27
3 45. 1000327
3 46. 164
N 0002 10x34y14
4
6
48. 100031
000332
50. 64000343
47. 1000316 49. 9
Estimate how many units of a finished product will be produced using 256 units of labor and 81 units of capital.
In Problems 51–64, simplify and express answers using positive exponents only. 51. x5x00032
52. y6y00038
53. (2y)(3y2)(5y4)
54. (6x3)(4x7)(x00035)
55. (a2b3)5
56. (2c4d00032)00033
13 53
57. u
000315 65
u
58. v
v
60. (49a4b00032)12
61. a
m00032n3 2 b m4n00031
w4 000312 b 9x00032
64. a
8a00034b3 13 b 27a2b00033
63. a
00033 16
59. (x ) 62. a
6mn00032 00033 b 3m00031n2
In Problems 65–86, write in simplified radical form. 65. 0003 1128
66. 00031125
67. 127 0003 5 13 3
3
68. 2 18 0007 118
69. 25 0003 225 0007 2625
3 3 70. 220 0007 240 0003 25
3 3 71. 225 210
72. 16114
4 8
73. 216m y
3
90. ECONOMICS If in the United States in 2007 the gross domestic product (GDP) was about $14,074,000,000,000 and the population was about 301,000,000, estimate to three significant digits the GDP per person. Write your answer in scientific notation and in standard decimal form. 91. ECONOMICS The number of units N of a finished product produced from the use of x units of labor and y units of capital for a particular Third World country is approximated by
23
43. 125
89. ECONOMICS If in the United States in 2007 the national debt was about $8,868,000,000,000 and the population was about 301,000,000, estimate to three significant digits each individual’s share of the national debt. Write your answer in scientific notation and in standard decimal form.
4 74. 216m4n8
92. ECONOMICS The number of units N of a finished product produced by a particular automobile company where x units of labor and y units of capital are used is approximated by N 0002 50x12y12 Estimate how many units will be produced using 256 units of labor and 144 units of capital. 93. BRAKING DISTANCE R. A. Moyer of Iowa State College found, in comprehensive tests carried out on 41 wet pavements, that the braking distance d (in feet) for a particular automobile traveling at v miles per hour was given approximately by d 0002 0.0212v73 Approximate the braking distance to the nearest foot for the car traveling on wet pavement at 70 miles per hour. 94. BRAKING DISTANCE Approximately how many feet would it take the car in Problem 93 to stop on wet pavement if it were traveling at 50 miles per hour? (Compute answer to the nearest foot.)
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Polynomials: Basic Operations and Factoring
and back again is called the period T and is given by
95. PHYSICS—RELATIVISTIC MASS The mass M of an object moving at a velocity v is given by
L Ag
T 0002 20005
M0
M0002 A
10003
v2 c2
where g is the gravitational constant. Show that T can be written in the form
where M0 0002 mass at rest and c 0002 velocity of light. The mass of an object increases with velocity and tends to infinity as the velocity approaches the speed of light. Show that M can be written in the form M0002
21
T0002
200051gL g
M0c2c2 0003 v2 c2 0003 v2
96. PHYSICS—PENDULUM A simple pendulum is formed by hanging a bob of mass M on a string of length L from a fixed support (see the figure). The time it takes the bob to swing from right to left
R-3
Polynomials: Basic Operations and Factoring Z Polynomials Z Addition and Subtraction Z Multiplication Z Factoring
In this section, we review the basic operations on polynomials. Polynomials are expressions such as x4 0003 5x2 0007 1 or 3xy 0003 2x 0007 5y 0007 6 that are built from constants and variables using only addition, subtraction, and multiplication (the power x4 is the product x ⴢ x ⴢ x ⴢ x). Polynomials are used throughout mathematics to describe and approximate mathematical relationships.
Z Polynomials Algebraic expressions are formed by using constants and variables and the algebraic operations of addition, subtraction, multiplication, division, raising to powers, and taking roots. Some examples are 3 3 2 x 00075 x00035 x2 0007 2x 0003 5
5x4 0007 2x2 0003 7 1 10007 1 10007 x
An algebraic expression involving only the operations of addition, subtraction, multiplication, and raising to natural number powers is called a polynomial. (Note that raising to a natural number power is repeated multiplication.) Some examples are 2x 0003 3 x 0003 2y
4x2 0003 3x 0007 7 x3 0003 3x2y 0007 xy2 0007 2y7
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In a polynomial, a variable cannot appear in a denominator, as an exponent, or within a radical. Accordingly, a polynomial in one variable x is constructed by adding or subtracting constants and terms of the form axn, where a is a real number and n is a natural number. A polynomial in two variables x and y is constructed by adding and subtracting constants and terms of the form axmyn, where a is a real number and m and n are natural numbers. Polynomials in three or more variables are defined in a similar manner. Polynomials can be classified according to their degree. If a term in a polynomial has only one variable as a factor, then the degree of that term is the power of the variable. If two or more variables are present in a term as factors, then the degree of the term is the sum of the powers of the variables. The degree of a polynomial is the degree of the nonzero term with the highest degree in the polynomial. Any nonzero constant is defined to be a polynomial of degree 0. The number 0 is also a polynomial but is not assigned a degree.
EXAMPLE
1
Polynomials and Nonpolynomials (A) Which of the following are polynomials? 2x 0002 5 0003
1 x
x2 0003 3x 0002 2
2x3 0003 4x 0002 1
x4 0002 12
(B) Given the polynomial 2x3 0003 x6 0002 7, what is the degree of the first term? The third term? The whole polynomial? (C) Given the polynomial x3y2 0002 2x2y 0002 1, what is the degree of the first term? The second term? The whole polynomial? SOLUTIONS
MATCHED PROBLEM 1
(A) x2 0003 3x 0002 2 and x4 0002 12 are polynomials. (The others are not polynomials since a variable appears in a denominator or within a radical.) (B) The first term has degree 3, the third term has degree 0, and the whole polynomial has degree 6. (C) The first term has degree 5, the second term has degree 3, and the whole polynomial has degree 5. 0002 (A) Which of the following are polynomials? 3x2 0003 2x 0002 1
1x 0003 3
x2 0003 2xy 0002 y2
x00031 x2 0002 2
(B) Given the polynomial 3x5 0003 6x3 0002 5, what is the degree of the first term? The second term? The whole polynomial? (C) Given the polynomial 6x4y2 0003 3xy3, what is the degree of the first term? The second term? The whole polynomial? 0002 In addition to classifying polynomials by degree, we also call a single-term polynomial a monomial, a two-term polynomial a binomial, and a three-term polynomial a trinomial. 5 2 3 2x y 3
x 0002 4.7 x4 0003 12x2 0002 9
Monomial Binomial Trinomial
A constant in a term of a polynomial, including the sign that precedes it, is called the numerical coefficient, or simply, the coefficient, of the term. If a constant doesn’t appear, or
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only a 0002 sign appears, the coefficient is understood to be 1. If only a 0003 sign appears, the coefficient is understood to be 00031. So given the polynomial 2x4 0003 4x3 0002 x2 0003 x 0002 5
2x4 ⴙ (ⴚ4)x3 ⴙ 1x2 ⴙ (ⴚ1)x ⴙ 5
the coefficient of the first term is 2, the coefficient of the second term is 00034, the coefficient of the third term is 1, the coefficient of the fourth term is 00031, and the coefficient of the last term is 5. Two terms in a polynomial are called like terms if they have exactly the same variable factors to the same powers. The numerical coefficients may or may not be the same. Since constant terms involve no variables, all constant terms are like terms. If a polynomial contains two or more like terms, these terms can be combined into a single term by making use of distributive properties. Consider the following example: 5x3y 0003 2xy 0003 x3y 0003 2x3y
0004 5x3y 0003 x3y 0003 2x3y 0003 2xy 0004 (5x3y 0003 x3y 0003 2 x3y) 0003 2xy 0004 (5 0003 1 0003 2) x3y 0003 2xy
Group like terms. Use the distributive property Simplify.
0004 2x3y 0003 2xy It should be clear that free use has been made of the real number properties discussed earlier. The steps done in the dashed box are usually done mentally, and the process is quickly done as follows: Like terms in a polynomial are combined by adding their numerical coefficients.
Z Addition and Subtraction Addition and subtraction of polynomials can be thought of in terms of removing parentheses and combining like terms. Horizontal and vertical arrangements are illustrated in the next two examples. You should be able to work either way, letting the situation dictate the choice.
EXAMPLE
2
Adding Polynomials Add: x4 0003 3x3 0002 x2,
SOLUTION
0003x3 0003 2x2 0002 3x,
and
3x2 0003 4x 0003 5
Add horizontally: (x4 0003 3x3 0002 x2) 0002 (0003x3 0003 2x2 0002 3x) 0002 (3x2 0003 4x 0003 5) 0004 x4 0003 3x3 0002 x2 0003 x3 0003 2x2 0002 3x 0002 3x2 0003 4x 0003 5 0004 x4 0003 4x3 0002 2x2 0003 x 0003 5
Remove parentheses. Combine like terms.
Or vertically, by lining up like terms and adding their coefficients: x4 0003 3x3 0002 x2 0003 x3 0003 2x2 0002 3x 3x2 0003 4x 0003 5 4 3 x 0003 4x 0002 2x2 0003 x 0003 5 MATCHED PROBLEM 2
0002
Add horizontally and vertically: 3x4 0003 2x3 0003 4x2,
x3 0003 2x2 0003 5x,
and
x2 0002 7x 0003 2
0002
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Subtracting Polynomials Subtract:
SOLUTION
MATCHED PROBLEM 3
ZZZ
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4x2 ⫺ 3x ⫹ 5
(x2 ⫺ 8) ⫺ (4x2 ⫺ 3x ⫹ 5) ⫽ x2 ⫺ 8 ⫺ 4x2 ⫹ 3x ⫺ 5 ⫽ ⫺3x2 ⫹ 3x ⫺ 13
from
x2 ⫺ 8 2 ⫺4x ⫹ 3x ⫺ 5 ⫺3x2 ⫹ 3x ⫺ 13
or
2x2 ⫺ 5x ⫹ 4
Subtract:
x2 ⫺ 8
from
d Change signs and add.
5x2 ⫺ 6
0002
0002
When you use a horizontal arrangement to subtract a polynomial with more than one term, you must enclose the polynomial in parentheses. For example, to subtract 2x ⫹ 5 from 4x ⫺ 11, you must write
CAUTION ZZZ
4x ⫺ 11 ⫺ (2x ⫹ 5)
and not
4x ⫺ 11 ⫺ 2x ⫹ 5
Z Multiplication Multiplication of algebraic expressions involves extensive use of distributive properties for real numbers, as well as other real number properties.
EXAMPLE
4
Multiplying Polynomials (2x ⫺ 3)(3x2 ⫺ 2x ⫹ 3)
Multiply: (2x ⫺ 3)(3x2 ⫺ 2x ⫹ 3)
SOLUTION
⫽ 2x(3x2 ⫺ 2x ⫹ 3) ⫺ 3(3x2 ⫺ 2x ⫹ 3) ⫽ 6x3 ⫺ 4x2 ⫹ 6x ⫺ 9x2 ⫹ 6x ⫺ 9 ⫽ 6x3 ⫺ 13x2 ⫹ 12x ⫺ 9
Distribute, multiply out parentheses.
Combine like terms.
Or, using a vertical arrangement, 3x2 ⫺ 2x ⫹ 3 2x ⫺ 3 6x3 ⫺ 4x2 ⫹ 6x ⫺ 9x2 ⫹ 6x ⫺ 9 6x3 ⫺ 13x2 ⫹ 12x ⫺ 9 MATCHED PROBLEM 4
0002
Multiply: (2x ⫺ 3)(2x2 ⫹ 3x ⫺ 2)
0002
To multiply two polynomials, multiply each term of one by each term of the other, and combine like terms.
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Z Factoring A factor of a number is one of two or more numbers whose product is the given number. Similarly, a factor of an algebraic expression is one of two or more algebraic expressions whose product is the given algebraic expression. For example, 30 0004 2 ⴢ 3 ⴢ 5 x2 0003 4 0004 (x 0003 2)(x 0002 2)
2, 3, and 5 are each factors of 30. (x ⴚ 2) and (x ⴙ 2) are each factors of x2 ⴚ 4.
The process of writing a number or algebraic expression as the product of other numbers or algebraic expressions is called factoring. We start our discussion of factoring with the positive integers. An integer such as 30 can be represented in a factored form in many ways. The products 6ⴢ5
(12)(10)(6)
15 ⴢ 2
2ⴢ3ⴢ5
all yield 30. A particularly useful way of factoring positive integers greater than 1 is in terms of prime numbers. An integer greater than 1 is prime if its only positive integer factors are itself and 1. So 2, 3, 5, and 7 are prime, but 4, 6, 8, and 9 are not prime. An integer greater than 1 that is not prime is called a composite number. The integer 1 is neither prime nor composite. A composite number is said to be factored completely if it is represented as a product of prime factors. The only factoring of 30 that meets this condition, except for the order of the factors, is 30 0004 2 ⴢ 3 ⴢ 5. This illustrates an important property of integers.
Z THEOREM 1 The Fundamental Theorem of Arithmetic Each integer greater than 1 is either prime or can be expressed uniquely, except for the order of factors, as a product of prime factors.
We can also write polynomials in completely factored form. A polynomial such as 2x2 0003 x 0003 6 can be written in factored form in many ways. The products (2x 0002 3)(x 0003 2)
2(x2 0003 12x 0003 3)
2(x 0002 32)(x 0003 2)
all yield 2x2 0003 x 0003 6. A particularly useful way of factoring polynomials is in terms of prime polynomials.
Z DEFINITION 1 Prime Polynomials A polynomial of degree greater than 0 is said to be prime relative to a given set of numbers if: (1) all of its coefficients are from that set of numbers; and (2) it cannot be written as a product of two polynomials (excluding constant polynomials that are factors of 1) having coefficients from that set of numbers. Relative to the set of integers: x2 ⴚ 2 is prime x2 ⴚ 9 is not prime, since x2 ⴚ 9 ⴝ (x ⴚ 3)(x ⴙ 3)
[Note: The set of numbers most frequently used in factoring polynomials is the set of integers.]
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A nonprime polynomial is said to be factored completely relative to a given set of numbers if it is written as a product of prime polynomials relative to that set of numbers. In Examples 5 and 6 we review some of the standard factoring techniques for polynomials with integer coefficients.
EXAMPLE
5
Factoring Out Common Factors Factor out, relative to the integers, all factors common to all terms: (A) 2x3y ⫺ 8x2y2 ⫺ 6xy3
SOLUTIONS
(A) 2x3y ⫺ 8x2y2 ⫺ 6xy3
(B) 2x(3x ⫺ 2) ⫺ 7(3x ⫺ 2) ⫽ (2xy)x2 ⫺ (2xy)4xy ⫺ (2xy)3y2
Factor out 2xy.
⫽ 2xy(x2 ⫺ 4xy ⫺ 3y2) (B) 2x(3x ⫺ 2) ⫺ 7(3x ⫺ 2)
⫽ 2x(3x ⫺ 2) ⫺ 7(3x ⫺ 2)
Factor out 3x ⴚ 2.
⫽ (2x ⫺ 7)(3x ⫺ 2) MATCHED PROBLEM 5
0002
Factor out, relative to the integers, all factors common to all terms: (A) 3x3y ⫺ 6x2y2 ⫺ 3xy3
(B) 3y(2y ⫹ 5) ⫹ 2(2y ⫹ 5) 0002
The polynomials in Example 6 can be factored by first grouping terms to find a common factor.
EXAMPLE
6
Factoring by Grouping Factor completely, relative to the integers, by grouping: (A) 3x2 ⫺ 6x ⫹ 4x ⫺ 8 (C) 3ac ⫹ bd ⫺ 3ad ⫺ bc
SOLUTIONS
(B) wy ⫹ wz ⫺ 2xy ⫺ 2xz
(A) 3x2 ⫺ 6x ⫹ 4x ⫺ 8 Group the first two and last two terms. ⫽ (3x2 ⫺ 6x) ⫹ (4x ⫺ 8) Remove common factors from each group. ⫽ 3x(x ⫺ 2) ⫹ 4(x ⫺ 2) Factor out the common factor (x ⴚ 2). ⫽ (3x ⫹ 4)(x ⫺ 2) (B) wy ⫹ wz ⫺ 2xy ⫺ 2xz Group the first two and last two terms—be careful of signs. ⫽ (wy ⫹ wz) ⫺ (2xy ⫹ 2xz) Remove common factors from each group. ⫽ w( y ⫹ z) ⫺ 2x(y ⫹ z) Factor out the common factor ( y ⴙ z). ⫽ (w ⫺ 2x)(y ⫹ z) (C) 3ac ⫹ bd ⫺ 3ad ⫺ bc In parts (A) and (B) the polynomials are arranged in such a way that grouping the first two terms and the last two terms leads to common factors. In this problem neither the first two terms nor the last two terms have a common factor. Sometimes rearranging terms will lead to a factoring by grouping. In this case, we interchange
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the second and fourth terms to obtain a problem comparable to part (B), which can be factored as follows: 3ac ⫺ bc ⫺ 3ad ⫹ bd ⫽ (3ac ⫺ bc) ⫺ (3ad ⫺ bd) ⫽ c(3a ⫺ b) ⫺ d(3a ⫺ b)
Factor out c, d. Factor out 3a ⴚ b.
⫽ (c ⫺ d)(3a ⫺ b) MATCHED PROBLEM 6
0002
Factor completely, relative to the integers, by grouping: (A) 2x2 ⫹ 6x ⫹ 5x ⫹ 15 (C) 6wy ⫺ xz ⫺ 2xy ⫹ 3wz
(B) 2pr ⫹ ps ⫺ 6qr ⫺ 3qs 0002
Example 7 illustrates an approach to factoring a second-degree polynomial of the form 2x2 ⫺ 5x ⫺ 3
2x2 ⫹ 3xy ⫺ 2y2
or
into the product of two first-degree polynomials with integer coefficients.
EXAMPLE
7
Factoring Second-Degree Polynomials Factor each polynomial, if possible, using integer coefficients: (A) 2x2 ⫹ 3xy ⫺ 2y2
SOLUTIONS
(B) x2 ⫺ 3x ⫹ 4
(A) 2x2 ⫹ 3xy ⫺ 2y2 ⫽ (2x ⫹ y)(x ⫺ y) c
c
?
?
(C) 6x2 ⫹ 5xy ⫺ 4y2 Put in what we know. Signs must be opposite. (We can reverse this choice if we get ⴚ3xy instead of ⴙ3xy for the middle term.)
Now, what are the factors of 2 (the coefficient of y2)? 2 1ⴢ2 2ⴢ1
(2x ⴙ y)(x ⴚ 2y) ⴝ 2x2 ⴚ 3xy ⴚ 2y2 (2x ⴙ 2y)(x ⴚ y) ⴝ 2x2 ⴚ 2y2
The first choice gives us ⫺3xy for the middle term—close, but not there—so we reverse our choice of signs to obtain 2x2 ⫹ 3xy ⫺ 2y2 ⫽ (2x ⫺ y)(x ⫹ 2y) (B) x2 ⫺ 3x ⫹ 4 ⫽ (x ⫺ )(x ⫺ ) 4 2ⴢ2 1ⴢ4 4ⴢ1
Signs must be the same because the third term is positive and must be negative because the middle term is negative.
(x ⴚ 2)(x ⴚ 2) ⴝ x2 ⴚ 4x ⴙ 4 (x ⴚ 1)(x ⴚ 4) ⴝ x2 ⴚ 5x ⴙ 4 (x ⴚ 4)(x ⴚ 1) ⴝ x2 ⴚ 5x ⴙ 4
No choice produces the middle term; so x2 ⫺ 3x ⫹ 4 is not factorable using integer coefficients. (C) 6x2 ⫹ 5xy ⫺ 4y2 ⫽ ( x ⫹ y)( x ⫺ y) c
c
c
c
?
?
?
?
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The signs must be opposite in the factors, because the third term is negative. We can reverse our choice of signs later if necessary. We now write all factors of 6 and of 4: 6 2ⴢ3 3ⴢ2 1ⴢ6 6ⴢ1
4 2ⴢ2 1ⴢ4 4ⴢ1
and try each choice on the left with each on the right—a total of 12 combinations that give us the first and last terms in the polynomial 6x2 0002 5xy 0003 4y2. The question is: Does any combination also give us the middle term, 5xy? After trial and error and, perhaps, some educated guessing among the choices, we find that 3 ⴢ 2 matched with 4 ⴢ 1 gives us the correct middle term. 6x2 0002 5xy 0003 4y2 0004 (3x 0002 4y)(2x 0003 y) If none of the 24 combinations (including reversing our sign choice) had produced the middle term, then we would conclude that the polynomial is not factorable using integer coefficients. 0002 MATCHED PROBLEM 7
Factor each polynomial, if possible, using integer coefficients: (A) x2 0003 8x 0002 12
(B) x2 0002 2x 0002 5
(C) 2x2 0002 7xy 0003 4y2
(D) 4x2 0003 15xy 0003 4y2
0002
The special factoring formulas listed here will enable us to factor certain polynomial forms that occur frequently.
Z SPECIAL FACTORING FORMULAS 1. u2 0002 2uv 0002 v2 0004 (u 0002 v)2 2
2
Perfect Square
2
2. u 0003 2uv 0002 v 0004 (u 0003 v)
Perfect Square
2
2
3
3
2
2
Difference of Cubes
3
3
2
2
Sum of Cubes
3. u 0003 v 0004 (u 0003 v)(u 0002 v)
Difference of Squares
4. u 0003 v 0004 (u 0003 v)(u 0002 uv 0002 v ) 5. u 0002 v 0004 (u 0002 v)(u 0003 uv 0002 v )
The formulas in the box can be established by multiplying the factors on the right.
ZZZ EXPLORE-DISCUSS 1
Explain why there is no formula for factoring a sum of squares u2 0002 v2 into the product of two first-degree polynomials with real coefficients.
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EXAMPLE
8
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29
Using Special Factoring Formulas Factor completely relative to the integers: (A) x2 0002 6xy 0002 9y2
SOLUTIONS
(A) x2 0002 6xy 0002 9y2
(B) 9x2 0003 4y2
(C) 8m3 0003 1
0004 x2 0002 2(x)(3y) 0002 (3y)2
(D) x3 0002 y3z3
0004 (x 0002 3y)2
(B) 9x2 0003 4y2
0004 (3x)2 0003 (2y)2
(C) 8m3 0003 1
0004 (2m)3 0003 13 0004 (2m 0003 1) 冤(2m)2 0002 (2m)(1) 0002 12冥
0004 (3x 0003 2y)(3x 0002 2y)
Perfect square
Difference of squares
Difference of cubes Simplify.
0004 (2m 0003 1)(4m2 0002 2m 0002 1) (D) x3 0002 y3z3
0004 x3 0002 ( yz)3
Sum of cubes
0004 (x 0002 yz)(x2 0003 xyz 0002 y2z2) MATCHED PROBLEM 8
0002
Factor completely relative to the integers: (A) 4m2 0003 12mn 0002 9n2
(B) x2 0003 16y2
(C) z3 0003 1
(D) m3 0002 n3
0002
ANSWERS TO MATCHED PROBLEMS (A) 3x2 0003 2x 0002 1, x2 0003 2xy 0002 y2 (B) 5, 3, 5 (C) 6, 4, 6 3. 3x2 0002 5x 0003 10 4. 4x3 0003 13x 0002 6 3x4 0003 x3 0003 5x2 0002 2x 0003 2 (A) 3xy(x2 0003 2xy 0003 y2) (B) (3y 0002 2)(2y 0002 5) (A) (2x 0002 5)(x 0002 3) (B) (p 0003 3q)(2r 0002 s) (C) (3w 0003 x)(2y 0002 z) (A) (x 0003 2)(x 0003 6) (B) Not factorable using integers (C) (2x 0003 y)(x 0002 4y) (D) (4x 0002 y)(x 0003 4y) 8. (A) (2m 0003 3n)2 (B) (x 0003 4y)(x 0002 4y) (C) (z 0003 1)(z2 0002 z 0002 1) 2 2 (D) (m 0002 n)(m 0003 mn 0002 n ) 1. 2. 5. 6. 7.
R-3
Exercises
Problems 1–8 refer to the polynomials (a) x2 0002 1 and (b) x4 0003 2x 0002 1. 1. What is the degree of (a)?
In Problems 9–14, is the algebraic expression a polynomial? If so, give its degree. 9. 4 0003 x2
10. x3 0003 5x6 0002 1
2. What is the degree of (b)?
11. x3 0003 7x 0002 81x
12. x4 0002 3x 0003 15
3. What is the degree of the sum of (a) and (b)?
13. x5 0003 4x2 0002 600032
14. 3x4 0003 2x00031 0003 10
4. What is the degree of the product of (a) and (b)? 5. Multiply (a) and (b). 6. Add (a) and (b). 7. Subtract (b) from (a). 8. Subtract (a) from (b).
In Problems 15–22, perform the indicated operations and simplify. 15. 2(x 0003 1) 0002 3(2x 0003 3) 0003 (4x 0003 5) 16. 2y 0003 3y [4 0003 2( y 0003 1)] 17. (m 0003 n)(m 0002 n)
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18. (5y ⫺ 1)(3 ⫺ 2y)
19. (3x ⫹ 2y)(x ⫺ 3y)
65. 2(x ⫹ h)2 ⫺ 3(x ⫹ h) ⫺ (2x2 ⫺ 3x)
20. (4x ⫺ y)2
21. (a ⫹ b)(a2 ⫺ ab ⫹ b2)
66. ⫺4(x ⫹ h)2 ⫹ 6(x ⫹ h) ⫺ (⫺4x2 ⫹ 6x)
2
2
22. (a ⫺ b)(a ⫹ ab ⫹ b )
67. (x ⫹ h)3 ⫺ 2(x ⫹ h)2 ⫺ (x3 ⫺ 2x2)
In Problems 23–28, factor out, relative to the integers, all factors common to all terms.
68. (x ⫹ h)3 ⫹ 3(x ⫹ h) ⫺ (x3 ⫹ 3x)
23. 6x4 ⫺ 8x3 ⫺ 2x2
24. 3x5 ⫹ 6x3 ⫹ 9x
Problems 69–74 are calculus-related. Factor completely, relative to the integers.
25. x2y ⫹ 2xy2 ⫹ x2y2
26. 8u3v ⫺ 6u2v2 ⫹ 4uv3
69. 2x(x ⫹ 1)4 ⫹ 4x2(x ⫹ 1)3
27. 2w( y ⫺ 2z) ⫺ x( y ⫺ 2z)
70. (x ⫺ 1)3 ⫹ 3x(x ⫺ 1)2
28. 2x(u ⫺ 3v) ⫹ 5y(u ⫺ 3v)
71. 6(3x ⫺ 5)(2x ⫺ 3)2 ⫹ 4(3x ⫺ 5)2(2x ⫺ 3)
In Problems 29–34, factor completely, relative to the integers. 2
2
29. x ⫹ 4x ⫹ x ⫹ 4 2
30. 2y ⫺ 6y ⫹ 5y ⫺ 15 2
31. x ⫺ xy ⫹ 3xy ⫺ 3y
32. 3a2 ⫺ 12ab ⫺ 2ab ⫹ 8b2
33. 8ac ⫹ 3bd ⫺ 6bc ⫺ 4ad
In Problems 35–42, perform the indicated operations and simplify. 35. 2x ⫺ 35x ⫹ 2 3x ⫺ (x ⫹ 5) 4 ⫹ 16
78. 15ac ⫺ 20ad ⫹ 3bc ⫺ 4bd
38. (x2 ⫺ 3xy ⫹ y2)(x2 ⫹ 3xy ⫹ y2)
79. 3x2 ⫺ 2xy ⫺ 4y2
2
39. (3u ⫺ 2v) ⫺ (2u ⫺ 3v)(2u ⫹ 3v)
80. 5u2 ⫹ 4uv ⫺ v2
40. (2a ⫺ b)2 ⫺ (a ⫹ 2b)2 3
42. (3a ⫹ 2b)
In Problems 43–62, factor completely, relative to the integers. If a polynomial is prime relative to the integers, say so. 43. 2x2 ⫹ x ⫺ 3
44. 3y2 ⫺ 8y ⫺ 3
45. x2 ⫹ 5xy ⫺ 14y2
46. x2 ⫹ 4y2
47. 4x2 ⫺ 20x ⫹ 25
48. a2b2 ⫺ c2
2
49. a b ⫹ c 2
2
50. 9x ⫺ 4
51. 4x ⫹ 9
52. 16x2 ⫺ 25
53. 6x2 ⫹ 48x ⫹ 72
54. 3z2 ⫺ 28z ⫹ 48
55. 2x4 ⫺ 24x3 ⫹ 40x2
56. 16x2y ⫺ 8xy ⫹ y
57. 6m2 ⫺ mn ⫺ 12n2
58. 4u3v ⫺ uv3
59. 3m3 ⫺ 6m2 ⫹ 15m
60. 2x3 ⫺ 2x2 ⫹ 8x
61. m3 ⫹ n3
62. 8x3 ⫺ 125
Problems 63–68 are calculus-related. Perform the indicated operations and simplify. 63. 3(x ⫹ h) ⫺ 7 ⫺ (3x ⫺ 7) 64. (x ⫹ h)2 ⫺ x2
75. (a ⫺ b)2 ⫺ 4(c ⫺ d )2 77. 2am ⫺ 3an ⫹ 2bm ⫺ 3bn
37. (2x2 ⫺ 3x ⫹ 1)(x2 ⫹ x ⫺ 2)
2 2
74. 3x4(x ⫺ 7)2 ⫹ 4x3(x ⫺ 7)3
76. (x ⫹ 2)2 ⫹ 9
36. m ⫺ 5m ⫺ 3m ⫺ (m ⫺ 1)4 6
41. (2m ⫺ n)
73. 5x4(9 ⫺ x)4 ⫺ 4x5(9 ⫺ x)3
In Problems 75–86, factor completely, relative to the integers. In polynomials involving more than three terms, try grouping the terms in various combinations as a first step. If a polynomial is prime relative to the integers, say so.
34. 3ux ⫺ 4vy ⫹ 3vx ⫺ 4uy
3
72. 2(x ⫺ 3)(4x ⫹ 7)2 ⫹ 8(x ⫺ 3)2(4x ⫹ 7)
81. x3 ⫺ 3x2 ⫺ 9x ⫹ 27
82. t3 ⫺ 2t 2 ⫹ t ⫺ 2
83. 4(A ⫹ B)2 ⫺ 5(A ⫹ B) ⫺ 5 84. x4 ⫹ 6x2 ⫹ 8
85. m4 ⫺ n4
86. y4 ⫺ 3y2 ⫺ 4 87. Show by example that, in general, (a ⫹ b)2 ⫽ a2 ⫹ b2. Discuss possible conditions on a and b that would make this a valid equation. 88. Show by example that, in general, (a ⫺ b)2 ⫽ a2 ⫺ b2. Discuss possible conditions on a and b that would make this a valid equation. 89. To show that 12 is an irrational number, explain how the assumption that 12 is rational leads to a contradiction of Theorem 1, the fundamental theorem of arithmetic, by the following steps: (A) Suppose that 12 ⫽ aⲐb, where a and b are positive integers, b ⫽ 0. Explain why a2 ⫽ 2b2. (B) Explain why the prime number 2 appears an even number of times (possibly 0 times) as a factor in the prime factorization of a2. (C) Explain why the prime number 2 appears an odd number of times as a factor in the prime factorization of 2b2. (D) Explain why parts (B) and (C) contradict the fundamental theorem of arithmetic.
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90. To show that 1n is an irrational number unless n is a perfect square, explain how the assumption that 1n is rational leads to a contradiction of the fundamental theorem of arithmetic by the following steps: (A) Assume that n is not a perfect square, that is, does not belong to the sequence 1, 4, 9, 16, 25, . . . . Explain why some prime number p appears an odd number of times as a factor in the prime factorization of n. (B) Suppose that 1n 0004 a0006b, where a and b are positive integers, b 0005 0. Explain why a2 0004 nb2. (C) Explain why the prime number p appears an even number of times (possibly 0 times) as a factor in the prime factorization of a2. (D) Explain why the prime number p appears an odd number of times as a factor in the prime factorization of nb2. (E) Explain why parts (C) and (D) contradict the fundamental theorem of arithmetic.
APPLICATIONS 91. GEOMETRY The width of a rectangle is 5 centimeters less than its length. If x represents the length, write an algebraic expression in terms of x that represents the perimeter of the rectangle. Simplify the expression. 92. GEOMETRY The length of a rectangle is 8 meters more than its width. If x represents the width of the rectangle, write an algebraic expression in terms of x that represents its area. Change the expression to a form without parentheses. 93. COIN PROBLEM A parking meter contains nickels, dimes, and quarters. There are 5 fewer dimes than nickels, and 2 more quarters than dimes. If x represents the number of nickels, write an algebraic expression in terms of x that represents the value of all the coins in the meter in cents. Simplify the expression.
Polynomials: Basic Operations and Factoring
31
0.3 centimeters thick, write an algebraic expression in terms of x that represents the volume of the plastic used to construct the container. Simplify the expression. [Recall: The volume V of a sphere of radius r is given by V 0004 430007r3.] 96. PACKAGING A cubical container for shipping computer components is formed by coating a metal mold with polystyrene. If the metal mold is a cube with sides x centimeters long and the polystyrene coating is 2 centimeters thick, write an algebraic expression in terms of x that represents the volume of the polystyrene used to construct the container. Simplify the expression. [Recall: The volume V of a cube with sides of length t is given by V 0004 t3.] 97. CONSTRUCTION A rectangular open-topped box is to be constructed out of 20-inch-square sheets of thin cardboard by cutting x-inch squares out of each corner and bending the sides up as indicated in the figure. Express each of the following quantities as a polynomial in both factored and expanded form. (A) The area of cardboard after the corners have been removed. (B) The volume of the box.
20 inches x
x
x
x
20 inches
x
x x
x
94. COIN PROBLEM A vending machine contains dimes and quarters only. There are 4 more dimes than quarters. If x represents the number of quarters, write an algebraic expression in terms of x that represents the value of all the coins in the vending machine in cents. Simplify the expression. 95. PACKAGING A spherical plastic container for designer wristwatches has an inner radius of x centimeters (see the figure). If the plastic shell is
0.3 cm x cm
Figure for 95
98. CONSTRUCTION A rectangular open-topped box is to be constructed out of 9- by 16-inch sheets of thin cardboard by cutting x-inch squares out of each corner and bending the sides up. Express each of the following quantities as a polynomial in both factored and expanded form. (A) The area of cardboard after the corners have been removed. (B) The volume of the box.
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Rational Expressions: Basic Operations Z Reducing to Lowest Terms Z Multiplication and Division Z Addition and Subtraction Z Compound Fractions
A quotient of two algebraic expressions, division by 0 excluded, is called a fractional expression. If both the numerator and denominator of a fractional expression are polynomials, the fractional expression is called a rational expression. Some examples of rational expressions are the following (recall that a nonzero constant is a polynomial of degree 0): x⫺2 2 2x ⫺ 3x ⫹ 5
1 4 x ⫺1
3 x
x2 ⫹ 3x ⫺ 5 1
In this section, we discuss basic operations on rational expressions, including multiplication, division, addition, and subtraction. Since variables represent real numbers in the rational expressions we are going to consider, the properties of real number fractions summarized in Section R-1 play a central role in much of the work that we will do. Even though not always explicitly stated, we always assume that variables are restricted so that division by 0 is excluded.
Z Reducing to Lowest Terms We start this discussion by restating the fundamental property of fractions (from Theorem 3 in Section R-1):
Z FUNDAMENTAL PROPERTY OF FRACTIONS If a, b, and k are real numbers with b, k ⫽ 0, then ka a ⫽ kb b
2ⴢ3 3 ⴝ 2ⴢ4 4
(x ⴚ 3)2
2 ⴝ (x ⴚ 3)x x x ⴝ 0, x ⴝ 3
Using this property from left to right to eliminate all common factors from the numerator and the denominator of a given fraction is referred to as reducing a fraction to lowest terms. We are actually dividing the numerator and denominator by the same nonzero common factor. Using the property from right to left—that is, multiplying the numerator and the denominator by the same nonzero factor—is referred to as raising a fraction to higher terms. We will use the property in both directions in the material that follows. We say that a rational expression is reduced to lowest terms if the numerator and denominator do not have any factors in common. Unless stated to the contrary, factors will be relative to the integers.
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EXAMPLE
1
Rational Expressions: Basic Operations
33
Reducing Rational Expressions Reduce each rational expression to lowest terms. (A)
SOLUTIONS
(A)
x2 ⫺ 6x ⫹ 9 x2 ⫺ 9
(B)
x3 ⫺ 1 x2 ⫺ 1
(x ⫺ 3)2 x2 ⫺ 6x ⫹ 9 ⫽ (x ⫺ 3)(x ⫹ 3) x2 ⫺ 9 x⫺3 ⫽ x⫹3
Factor numerator and denominator completely. Divide numerator and denominator by (x ⫺ 3); this is a valid operation as long as x ⴝ 3.
1
(x ⫺ 1)(x2 ⫹ x ⫹ 1) x3 ⫺ 1 (B) 2 ⫽ (x ⫺ 1)(x ⫹ 1) x ⫺1
Dividing numerator and denominator by (x ⴚ 1) can be indicated by drawing lines through both (x ⴚ 1)’s and writing the resulting quotients, 1’s.
1
x2 ⫹ x ⫹ 1 ⫽ x⫹1 MATCHED PROBLEM 1
CAUTION ZZZ
0002
Reduce each rational expression to lowest terms. (A)
ZZZ
x ⴝ ⴚ1 and x ⴝ 1
6x2 ⫹ x ⫺ 2 2x2 ⫹ x ⫺ 1
(B)
x4 ⫺ 8x 3x ⫺ 2x2 ⫺ 8x 3
0002
Remember to always factor the numerator and denominator first, then divide out any common factors. Do not indiscriminately eliminate terms that appear in both the numerator and the denominator. For example, 1
2x3 ⫹ y2 2x3 ⫹ y2 ⫽ 2 y y2 1
2x3 ⫹ y2 ⫽ 2x3 ⫹ 1 y2 Since the term y2 is not a factor of the numerator, it cannot be eliminated. In fact, (2x3 ⫹ y2)Ⲑy2 is already reduced to lowest terms.
Z Multiplication and Division Since we are restricting variable replacements to real numbers, multiplication and division of rational expressions follow the rules for multiplying and dividing real number fractions (Theorem 3 in Section R-1).
Z MULTIPLICATION AND DIVISION If a, b, c, and d are real numbers with b, d ⫽ 0, then: 1.
a c ac ⴢ ⫽ b d bd
2.
a c a d ⫼ ⫽ ⴢ b d b c
2 x 2x ⴢ ⴝ 3 xⴚ1 3(x ⴚ 1)
c⫽0
2 x 2 xⴚ1 ⴜ ⴝ ⴢ 3 xⴚ1 3 x
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2
Multiplying and Dividing Rational Expressions Perform the indicated operations and reduce to lowest terms. (A)
10x3y x2 ⫺ 9 ⴢ 2 3xy ⫹ 9y 4x ⫺ 12x
(C)
x3 ⫹ y3 2x3 ⫺ 2x2y ⫹ 2xy2 ⫼ 2 3 3 x y ⫺ xy x ⫹ 2xy ⫹ y2 5x2
SOLUTIONS
4 ⫺ 2x ⫼ (x ⫺ 2) 4
(B)
1ⴢ1
10x3y 10x3y (x ⫺ 3)(x ⫹ 3) x2 ⫺ 9 (A) ⴢ 2 ⫽ ⴢ 3xy ⫹ 9y 4x ⫺ 12x 3y(x ⫹ 3) 4x(x ⫺ 3) 3ⴢ1
2ⴢ1
Factor numerators and denominators; then divide any numerator and any denominator with a like common factor.
2
⫽
5x 6
1
2(2 ⫺ x) 4 ⫺ 2x 1 ⫼ (x ⫺ 2) ⫽ ⴢ (B) 4 4 x⫺2
x ⴚ 2 is the same as
xⴚ2 . 1
2
⫺1
⫺(x ⫺ 2) 2⫺x ⫽ ⫽ 2(x ⫺ 2) 2(x ⫺ 2)
b ⴚ a ⴝ ⴚ(a ⴚ b), a useful change in some problems.
1
⫽⫺ (C)
1 2
2x3 ⫺ 2x2y ⫹ 2xy2 x3 ⫹ y3 ⫼ x3y ⫺ xy3 x2 ⫹ 2xy ⫹ y2 2
1
a c a d ⴜ ⴝ ⴢ b d b c
1
2x(x2 ⫺ xy ⫹ y2) (x ⫹ y)2 ⫽ ⴢ xy(x ⫹ y)(x ⫺ y) (x ⫹ y)(x2 ⫺ xy ⫹ y2) y
1
1
Divide out common factors.
1
2 ⫽ y(x ⫺ y)
MATCHED PROBLEM 2
0002
Perform the indicated operations and reduce to lowest terms. (A)
12x2y3 y2 ⫹ 6y ⫹ 9 ⴢ 2xy2 ⫹ 6xy 3y3 ⫹ 9y2
(C)
m3n ⫺ m2n2 ⫹ mn3 m3 ⫹ n3 ⫼ 2m2 ⫹ mn ⫺ n2 2m3n2 ⫺ m2n3
(B) (4 ⫺ x) ⫼
x2 ⫺ 16 5
0002
Z Addition and Subtraction Again, because we are restricting variable replacements to real numbers, addition and subtraction of rational expressions follow the rules for adding and subtracting real number fractions (Theorem 3 in Section R-1).
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35
Z ADDITION AND SUBTRACTION For a, b, and c real numbers with b ⫽ 0: 1.
a c a⫹c ⫹ ⫽ b b b
2.
a c a⫺c ⫺ ⫽ b b b
x 2 xⴙ2 ⴙ ⴝ xⴚ3 xⴚ3 xⴚ3 x 2xy
2
ⴚ
xⴚ4 2xy
2
ⴝ
x ⴚ (x ⴚ 4) 2xy 2
So we add rational expressions with the same denominators by adding or subtracting their numerators and placing the result over the common denominator. If the denominators are not the same, we raise the fractions to higher terms, using the fundamental property of fractions to obtain common denominators, and then proceed as described. Even though any common denominator will do, our work will be simplified if the least common denominator (LCD) is used. Often, the LCD is obvious, but if it is not, the steps in the box describe how to find it.
Z THE LEAST COMMON DENOMINATOR (LCD) The LCD of two or more rational expressions is found as follows: 1. Factor each denominator completely. 2. Identify each different prime factor from all the denominators. 3. Form a product using each different factor to the highest power that occurs in any one denominator. This product is the LCD.
EXAMPLE
3
Adding and Subtracting Rational Expressions Combine into a single fraction and reduce to lowest terms. (A)
SOLUTIONS
3 5 11 ⫹ ⫺ 10 6 45
(B)
4 5x ⫺ 2⫹1 9x 6y
(C)
x⫹3 x⫹2 5 ⫺ 2 ⫺ 3⫺x x ⫺ 6x ⫹ 9 x ⫺9 2
(A) To find the LCD, factor each denominator completely:
冎
10 ⫽ 2 ⴢ 5 6 ⫽ 2 ⴢ 3 LCD ⫽ 2 ⴢ 32 ⴢ 5 ⫽ 90 45 ⫽ 32 ⴢ 5 Now use the fundamental property of fractions to make each denominator 90: 3 5 11 9ⴢ3 15 ⴢ 5 2 ⴢ 11 ⫹ ⫺ ⫽ ⫹ ⫺ 10 6 45 9 ⴢ 10 15 ⴢ 6 2 ⴢ 45 ⫽
27 75 22 ⫹ ⫺ 90 90 90
⫽
27 ⫹ 75 ⫺ 22 80 8 ⫽ ⫽ 90 90 9
Multiply.
Combine into a single fraction.
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(B)
冎
9x ⫽ 32x LCD ⫽ 2 ⴢ 32xy2 ⫽ 18xy2 6y2 ⫽ 2 ⴢ 3y2 2y2 ⴢ 4 18xy2 4 5x 3x ⴢ 5x ⫺ 2⫹1⫽ 2 ⫹ ⫺ 9x 6y 2y ⴢ 9x 3x ⴢ 6y2 18xy2 ⫽
(C)
Multiply, combine.
8y2 ⫺ 15x2 ⫹ 18xy2 18xy2
x⫹3 x⫹2 5 x⫹3 x⫹2 5 ⫺ 2 ⫺ ⫽ ⫺ ⫹ 3⫺x (x ⫺ 3)(x ⫹ 3) x⫺3 x2 ⫺ 6x ⫹ 9 x ⫺9 (x ⫺ 3)2 Note: ⫺
5 5 5 ⫽⫺ ⫽ 3⫺x ⫺(x ⫺ 3) x⫺3
We have again used the fact that a ⴚ b ⴝ ⴚ(b ⴚ a).
The LCD ⫽ (x ⫺ 3)2(x ⫹ 3). (x ⫹ 3)2 (x ⫺ 3)(x ⫹ 2) 5(x ⫺ 3)(x ⫹ 3) ⫺ ⫹ 2 2 (x ⫺ 3) (x ⫹ 3) (x ⫺ 3) (x ⫹ 3) (x ⫺ 3)2(x ⫹ 3)
MATCHED PROBLEM 3
Expand numerators.
⫽
(x2 ⫹ 6x ⫹ 9) ⫺ (x2 ⫺ x ⫺ 6) ⫹ 5(x2 ⫺ 9) (x ⫺ 3)2(x ⫹ 3)
Be careful of sign errors here.
⫽
x2 ⫹ 6x ⫹ 9 ⫺ x2 ⫹ x ⫹ 6 ⫹ 5x2 ⫺ 45 (x ⫺ 3)2(x ⫹ 3)
Combine like terms.
⫽
5x2 ⫹ 7x ⫺ 30 (x ⫺ 3)2(x ⫹ 3)
0002
Combine into a single fraction and reduce to lowest terms. (A)
5 1 6 ⫺ ⫹ 28 10 35
(C)
y⫹2 y⫺3 2 ⫺ 2 ⫺ 2⫺y y2 ⫺ 4 y ⫺ 4y ⫹ 4
(B)
1 2x ⫹ 1 3 ⫹ 2 ⫺ 3 12x 4x 3x
0002 ZZZ EXPLORE-DISCUSS 1
What is the result of entering 16 ⫼ 4 ⫼ 2 on a calculator? What is the difference between 16 ⫼ (4 ⫼ 2) and (16 ⫼ 4) ⫼ 2? How could you use fraction bars to distinguish between these two cases when 16 writing 4 ? 2
Z Compound Fractions A fractional expression with fractions in its numerator, denominator, or both is called a compound fraction. It is often necessary to represent a compound fraction as a simple fraction—that is (in all cases we will consider), as the quotient of two polynomials. The process does not involve any new concepts. It is a matter of applying old concepts and processes in the right sequence. We will illustrate two approaches to the problem, each with its own merits, depending on the particular problem under consideration.
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EXAMPLE
4
Rational Expressions: Basic Operations
37
Simplifying Compound Fractions Express as a simple fraction reduced to lowest terms: 2 ⫺1 x 4 ⫺1 x2
SOLUTION
Method 1. Multiply the numerator and denominator by the LCD of all fractions in the numerator and denominator—in this case, x2. (We are multiplying by 1 ⫽ x2Ⲑx2.) 2 x2a ⫺ 1b x x2a
4 ⫺ 1b x2
2 x2 ⫺ x2 x
⫽
x2
4 ⫺ x2 x2
1
x(2 ⫺ x) 2x ⫺ x2 ⫽ 2 ⫽ (2 ⫹ x)(2 ⫺ x) 4⫺x 1
⫽
x 2⫹x
Method 2. Write the numerator and denominator as single fractions. Then treat as a quotient. 2 ⫺1 x
2⫺x x
1
x
2⫺x 4 ⫺ x2 2⫺x x2 ⫽ ⫽ ⫼ ⫽ ⴢ x x 4 4 ⫺ x2 (2 ⫺ x)(2 ⫹ x) x2 1 1 2 ⫺ 1 2 x x ⫽
MATCHED PROBLEM 4
x 2⫹x
0002
Express as a simple fraction reduced to lowest terms. Use the two methods described in Example 4. 1⫹
1 x
x⫺
1 x 0002
ANSWERS TO MATCHED PROBLEMS 3x ⫹ 2 x2 ⫹ 2x ⫹ 4 ⫺5 (B) 2. (A) 2x (B) (C) mn x⫹1 3x ⫹ 4 x⫹4 2 2 2y ⫺ 9y ⫺ 6 1 3x ⫺ 5x ⫺ 4 1 3. (A) (B) (C) 4. 3 2 4 x⫺1 12x ( y ⫺ 2) ( y ⫹ 2)
1. (A)
R-4
Exercises
In Problems 1–10, reduce each rational expression to lowest terms. 1.
17 85
2.
91 26
3.
360 288
4.
63 105
5.
x⫹1 x ⫹ 3x ⫹ 2 2
6.
x2 ⫺ 2x ⫺ 24 x⫺6
7.
x2 ⫺ 9 x ⫹ 3x ⫺ 18 2
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3x2y3
10.
x4y
2a2b4c6 6a5b3c
In Problems 11–36, perform the indicated operations and reduce answers to lowest terms. Represent any compound fractions as simple fractions reduced to lowest terms. 7 19 12. ⫹ 10 25
5 11 11. ⫹ 6 15
41. 42.
⫺2x(x ⫹ 4)3 ⫺ 3(3 ⫺ x2)(x ⫹ 4)2 (x ⫹ 4)6 3x2(x ⫹ 1)3 ⫺ 3(x3 ⫹ 4)(x ⫹ 1)2 (x ⫹ 1)6
In Problems 43–54, perform the indicated operations and reduce answers to lowest terms. Represent any compound fractions as simple fractions reduced to lowest terms. y
2 1 ⫹ 2 y2 ⫺ 5y ⫹ 4 y ⫹y⫺2
13.
1 1 ⫺ 8 9
14.
9 8 ⫺ 8 9
43.
15.
1 1 ⫺ n m
16.
m n ⫺ n m
44.
x⫺8 x x⫹4 ⫹ ⫹ x ⫺ 6 x ⫺3 x ⫺ 9x ⫹ 18
17.
5 3 ⫼ 12 4
18.
10 5 ⫼ 3 2
45.
16 ⫺ m2 m⫺1 ⴢ m ⫹ 3m ⫺ 4 m ⫺ 4
y2 ⫺ 2y ⫺ 8
⫺
2
2
19. a
25 5 4 ⫼ bⴢ 8 16 15
20.
25 5 4 ⫼a ⴢ b 8 16 15
46.
x⫹1 x2 ⫺ 2x ⫹ 1 ⴢ x(1 ⫺ x) x2 ⫺ 1
21. a
b2 b a ⫼ 2b ⴢ 2a 3b a
22.
b2 b a ⫼a 2ⴢ b 2a a 3b
47.
y⫹9 x⫹7 ⫹ ax ⫺ bx by ⫺ ay
23.
x2 ⫺ 1 x⫹1 ⫼ 2 x⫹2 x ⫺4
24.
x⫺3 x2 ⫺ 9 ⫼ 2 x⫺1 x ⫺1
48.
c⫺2 c c⫹2 ⫺ ⫹ 5c ⫺ 5 3c ⫺ 3 1⫺c
25.
1 1 1 ⫹ ⫹ c a b
26.
1 1 1 ⫹ ⫹ ac bc ab
49.
x2 ⫺ 13x ⫹ 36 x2 ⫺ 16 ⫼ 2x ⫹ 10x ⫹ 8 x3 ⫹ 1
2a ⫺ b 2a ⫹ 3b ⫺ 2 27. 2 2 a ⫺b a ⫹ 2ab ⫹ b2 28.
x⫺2 x⫹2 ⫺ x2 ⫺ 1 (x ⫺ 1)2
29. m ⫹ 2 ⫺
m⫺2 m⫺1
31.
3 2 ⫺ x⫺2 2⫺x
33.
4y 3 2 ⫹ ⫺ 2 y⫹2 y⫺2 y ⫺4
34.
4x 3 2 ⫹ ⫺ 2 2 x ⫺ y x ⫹ y x ⫺y
x2 ⫺1 y2 35. x ⫹1 y
38. 39. 40.
50. a
x3 ⫺ y3
51. a
1 x 4 ⫺ b⫼ x⫹4 x⫹4 x2 ⫺ 16 3 1 x⫹4 ⫺ b⫼ x⫺2 x⫹1 x⫺2
y3
ⴢ
y x2 ⫹ xy ⫹ y2 b⫼ x⫺y y2
30.
x⫹1 ⫹x x⫺1
52. a
32.
1 2 ⫺ a⫺3 3⫺a
2 15 ⫺ 2 x x 53. 4 5 1⫹ ⫺ 2 x x 1⫹
y x ⫺2⫹ y x 54. y x ⫺ y x
Problems 55–58 are calculus-related. Perform the indicated operations and reduce answers to lowest terms. Represent any compound fractions as simple fractions reduced to lowest terms. 4 ⫺x x 36. 2 ⫺1 x
Problems 37–42 are calculus-related. Reduce each fraction to lowest terms. 37.
2
1 1 ⫺ x x⫹h 55. h
1 1 ⫺ 2 (x ⫹ h)2 x 56. h
(x ⫹ h)2 x2 ⫺ x⫹h⫹2 x⫹2 57. h
2x ⫹ 2h ⫹ 3 2x ⫹ 3 ⫺ x x⫹h 58. h
6x3(x2 ⫹ 2)2 ⫺ 2x(x2 ⫹ 2)3 x4 4x4(x2 ⫹ 3) ⫺ 3x2(x2 ⫹ 3)2 x6 2x(1 ⫺ 3x)3 ⫹ 9x2(1 ⫺ 3x)2 (1 ⫺ 3x)6 2x(2x ⫹ 3)4 ⫺ 8x2(2x ⫹ 3)3 (2x ⫹ 3)8
In Problems 59–62, perform the indicated operations and reduce answers to lowest terms. Represent any compound fractions as simple fractions reduced to lowest terms. y2 y⫺x 59. x2 1⫹ 2 y ⫺ x2 y⫺
s2 ⫺s s⫺t 60. 2 t ⫹t s⫺t
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Review
1
61. 2 ⫺
1
62. 1 ⫺
2 1⫺ a⫹2
Discuss possible conditions of a and b that would make this a valid equation.
1
1⫺
1⫺
1 x
64. Show by example that, in general, a2 ⫹ b2 ⫽a⫹b a⫹b
63. Show by example that, in general, a⫹b ⫽a⫹1 b
CHAPTER
R-1
R
Review 6.
A real number is any number that has a decimal representation. There is a one-to-one correspondence between the set of real numbers and the set of points on a line. Important subsets of the real numbers include the natural numbers, integers, and rational numbers. A rational number can be written in the form aⲐb, where a and b are integers and b ⫽ 0. A real number can be approximated to any desired precision by rational numbers. Consequently, arithmetic operations on rational numbers can be extended to operations on real numbers. These operations satisfy basic real number properties, including associative properties: x ⫹ ( y ⫹ z) ⫽ (x ⫹ y) ⫹ z and x( yz) ⫽ (xy)z; commutative properties: x ⫹ y ⫽ y ⫹ x and xy ⫽ yx; identities: 0 ⫹ x ⫽ x ⫹ 0 ⫽ x and (1)x ⫽ x(1) ⫽ x; inverses: ⫺x is the additive inverse of x and, if x ⫽ 0, x⫺1 is the multiplicative inverse of x; and distributive property: x( y ⫹ z) ⫽ xy ⫹ xz. Subtraction is defined by a ⫺ b ⫽ a ⫹ (⫺b) and division by aⲐb ⫽ ab⫺1. Division by 0 is never allowed. Additional properties include properties of negatives:
2. (⫺a)b ⫽ ⫺(ab) ⫽ a(⫺b) ⫽ ⫺ab 3. (⫺a)(⫺b) ⫽ ab 4. (⫺1)a ⫽ ⫺a
6.
⫺a a a ⫺a ⫽⫺ ⫽⫺ ⫽ ⫺b b ⫺b b
b⫽0 b⫽0
zero properties: 1. a ⴢ 0 ⫽ 0 2. ab ⫽ 0
if and only if
a⫽0
or
b⫽0
or both.
and fraction properties (division by 0 excluded): 1.
a c ⫽ b d
2.
ka a ⫽ kb b
3.
a c ac ⴢ ⫽ b d bd
4.
a c a d ⫼ ⫽ ⴢ b d b c
5.
a c a⫹c ⫹ ⫽ b b b
a c a⫺c ⫺ ⫽ b b b
R-2
7.
a c ad ⫹ bc ⫹ ⫽ b d bd
Exponents and Radicals
The notation an, in which the exponent n is an integer, is defined as follows. For n a positive integer and a a real number: an ⫽ a ⴢ a ⴢ . . . ⴢ a (n factors of a) a⫺n ⫽
1 an
a0 ⫽ 1
(a ⫽ 0) (a ⫽ 0)
Properties of integer exponents (division by 0 excluded): 1. aman ⫽ am⫹n
2. (an)m ⫽ amn
3. (ab)m ⫽ ambm
a m am 4. a b ⫽ m b b
5.
1. ⫺(⫺a) ⫽ a
⫺a a a ⫽⫺ ⫽ b b ⫺b
(assume a ⫽ ⫺b)
Discuss possible conditions of a and b that would make this a valid equation.
(assume b ⫽ 0)
Algebra and Real Numbers
5.
39
1 am ⫽ am⫺n ⫽ n⫺m an a
Any positive number written in decimal form can be expressed in scientific notation, that is, in the form a ⫻ 10n 1 ⱕ a 6 10, n an integer, a in decimal form. For n a natural number, a and b real numbers: a is an nth root of b if an ⫽ b. The number of real nth roots of a real number b is either 0, 1, or 2, depending on whether b is positive or negative, and whether n is even or odd. The principal nth root of b, denoted by n b1/n or 1b, is the real nth root of b if there is only one, and the posn itive nth root of b if there are two real nth roots. In the notation 1b, the symbol 1 is called a radical, n is called the index, and b is the 2 radicand. If n ⫽ 2 we write 1b in place of 1 b. We extend exponent notation so that exponents can be rational numbers, not just integers, as follows. For m and n natural numbers and b any real number (except b can't be negative when n is even), bmⲐn ⫽ (b1Ⲑn)m and b⫺mⲐn ⫽
if and only if ad ⫽ bc
1 bmⲐn
Rational exponent/radical property: (b1Ⲑn)m ⫽ (bm)1Ⲑn
and
n
n
(1b)m ⫽ 2bm
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Properties of radicals (x 7 0, y 7 0): n
1. 2xn ⫽ x
n
n
n
2. 2xy ⫽ 2x2y
A radical is in simplified form if:
3.
n
x 1x ⫽ n y A 1y n
1. No radicand contains a factor to a power greater than or equal to the index of the radical. 2. No power of the radicand and the index of the radical have a common factor other than 1. 3. No radical appears in a denominator.
number otherwise. Each composite number can be factored uniquely into a product of prime numbers. A polynomial is prime relative to a given set of numbers (usually the set of integers) if (1) all its coefficients are from that set of numbers, and (2) it cannot be written as a product of two polynomials of positive degree having coefficients from that set of numbers. A nonprime polynomial is factored completely relative to a given set of numbers if it is written as a product of prime polynomials relative to that set of numbers. Common factors can be factored out by applying the distributive properties. Grouping can be used to identify common factors. Second-degree polynomials can be factored by trial and error. The following special factoring formulas are useful:
4. No fraction appears within a radical.
1. u2 ⫹ 2uv ⫹ v2 ⫽ (u ⫹ v)2
Perfect Square
Eliminating a radical from a denominator is called rationalizing the denominator. To rationalize the denominator, we multiply the numerator and denominator by a suitable factor that will leave the denominator free of radicals. This factor is called a rationalizing factor. For example, if the denominator is of the form 1a ⫹ 1b, then 1a ⫺ 1b is a rationalizing factor.
2. u2 ⫺ 2uv ⫹ v2 ⫽ (u ⫺ v)2
Perfect Square
R-3
Polynomials: Basic Operations and Factoring
An algebraic expression is formed by using constants and variables and the operations of addition, subtraction, multiplication, division, raising to powers, and taking roots. A polynomial is an algebraic expression formed by adding and subtracting constants and terms of the form axn (one variable), axnym (two variables), and so on. The degree of a term is the sum of the powers of all variables in the term, and the degree of a polynomial is the degree of the nonzero term with highest degree in the polynomial. Polynomials with one, two, or three terms are called monomials, binomials, and trinomials, respectively. Like terms have exactly the same variable factors to the same powers and can be combined by adding their coefficients. Polynomials can be added, subtracted, and multiplied by repeatedly applying the distributive property and combining like terms. A number or algebraic expression is factored if it is expressed as a product of other numbers or algebraic expressions, which are called factors. An integer greater than 1 is a prime number if its only positive integer factors are itself and 1, and a composite
CHAPTER
R
⫺1 ⫺1
3. 7 9 5.
5 1 ⫼ a ⫺ 3⫺1 b 7 3
2
3
3
2
2
Difference of Cubes
3
3
2
2
Sum of Cubes
3. u ⫺ v ⫽ (u ⫺ v)(u ⫹ v)
Difference of Squares
4. u ⫺ v ⫽ (u ⫺ v)(u ⫹ uv ⫹ v ) 5. u ⫹ v ⫽ (u ⫹ v)(u ⫺ uv ⫹ v )
There is no factoring formula relative to the real numbers for u2 ⫹ v2.
R-4
Rational Expressions: Basic Operations
A fractional expression is the ratio of two algebraic expressions, and a rational expression is the ratio of two polynomials. The rules for adding, subtracting, multiplying, and dividing real number fractions (see Section R-1 in this review) all extend to fractional expressions with the understanding that variables are always restricted to exclude division by zero. Fractions can be reduced to lowest terms or raised to higher terms by using the fundamental property of fractions: ka a ⫽ kb b
with b, k ⫽ 0
A rational expression is reduced to lowest terms if the numerator and denominator do not have any factors in common relative to the integers. The least common denominator (LCD) is useful for adding and subtracting fractions with different denominators and for reducing compound fractions to simple fractions.
Review Exercises
In Problems 1–6, perform the indicated operations, if defined. If the result is not an integer, express it in the form a/b, where a and b are integers. 5 3 1. ⫹ 6 4
2
4 2 2. ⫺ 3 9 6 10 4. a⫺ b a⫺ b 3 5 6.
11 3 ⫼ a⫺ b 12 4
Problems 7–12 refer to the polynomials (a) x4 ⫹ 3x2 ⫹ 1 and (b) 4 ⫺ x4. 7. What is the degree of (a)? 8. What is the degree of (b)? 9. What is the degree of the sum of (a) and (b)? 10. What is the degree of the product of (a) and (b)? 11. Multiply (a) and (b). 12. Add (a) and (b).
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Review Exercises
In Problems 13–18, evaluate each expression that results in a rational number. 13. 2891Ⲑ2
14. 2161Ⲑ3
15. 8⫺2Ⲑ3
16. (⫺64)5Ⲑ3
9 ⫺1Ⲑ2 17. a b 16
1Ⲑ2
18. (121
⫹ 25
)
20. (3m ⫺ 5n)(3m ⫹ 5n)
21. (2x ⫹ y)(3x ⫺ 4y)
22. (2a ⫺ 3b)2
In Problems 23–25, write each polynomial in a completely factored form relative to the integers. If the polynomial is prime relative to the integers, say so. 24. t 2 ⫺ 4t ⫺ 6
25. 6n3 ⫺ 9n2 ⫺ 15n In Problems 26–29, perform the indicated operations and reduce to lowest terms. Represent all compound fractions as simple fractions reduced to lowest terms. 26.
28.
2 4 1 ⫺ 3⫺ 2 2 5b 3a 6a b y⫺2 2
y ⫺ 4y ⫹ 4
27.
⫼
1 3x ⫹ 6x 3x2 ⫺ 12x
1 u 29. 1 1⫺ 2 u u⫺
2
y ⫹ 2y 2
y ⫹ 4y ⫹ 4
Simplify Problems 30–35, and write answers using positive exponents only. All variables represent positive real numbers. 8 6
9u v 3u4v8
30. 6(xy3)5
31.
32. (2 ⫻ 105)(3 ⫻ 10⫺3)
33. (x⫺3y2)⫺2
5Ⲑ3 2Ⲑ3
34. u
35. (9a b ) 2Ⲑ5
36. Change to radical form: 3x
3 37. Change to rational exponent form: ⫺3 2 (xy)2
40.
6ab 13a
39. 22x2y5 218x3y2 41.
15 3 ⫺ 15
a a ⫽⫺ ⫺(b ⫺ c) b⫺c
48. 3xy ⫹ 0 ⫽ 3xy
49. Indicate true (T) or false (F): (A) An integer is a rational number and a real number. (B) An irrational number has a repeating decimal representation. 50. Give an example of an integer that is not a natural number. 51. Given the algebraic expressions: (a) 2x2 ⫺ 3x ⫹ 5 (b) x2 ⫺ 1x ⫺ 3 ⫺3 ⫺2 ⫺1 (c) x ⫹ x ⫺ 3x (d) x2 ⫺ 3xy ⫺ y2 (A) Identify all second-degree polynomials. (B) Identify all third-degree polynomials. In Problems 52–55, perform the indicated operations and simplify. 52. (2x ⫺ y)(2x ⫹ y) ⫺ (2x ⫺ y)2 53. (m2 ⫹ 2mn ⫺ n2)(m2 ⫺ 2mn ⫺ n2) 54. 5(x ⫹ h)2 ⫺ 7(x ⫹ h) ⫺ (5x2 ⫺ 7x)
55. ⫺2x5(x2 ⫹ 2)(x ⫺ 3) ⫺ x [x ⫺ x(3 ⫺ x)] 6 In Problems 56–61, write in a completely factored form relative to the integers. 56. (4x ⫺ y)2 ⫺ 9x2
57. 2x2 ⫹ 4xy ⫺ 5y2
58. 6x3y ⫹ 12x2y2 ⫺ 15xy3
59. (y ⫺ b)2 ⫺ y ⫹ b
60. y3 ⫹ 2y2 ⫺ 4y ⫺ 8
61. 2x(x ⫺ 4)3 ⫹ 3x2(x ⫺ 4)2
62.
Simplify Problems 38–42, and express answers in simplified form. All variables represent positive real numbers. 3 5 4 38. 3x 2 xy
47.
In Problems 62–65, perform the indicated operations and reduce to lowest terms. Represent all compound fractions as simple fractions reduced to lowest terms.
4 ⫺2 1Ⲑ2
u
44. 3y ⫹ (2x ⫹ 5) ⫽ (2x ⫹ 5) ⫹ 3y 46. 3 ⴢ (5x) ⫽ (3 ⴢ 5)x
1Ⲑ2 ⫺3Ⲑ4
19. 5x ⫺ 3x[4 ⫺ 3(x ⫺ 2) ]
23. 9x2 ⫺ 12x ⫹ 4
43. (⫺3) ⫺ (⫺2) ⫽ (⫺3) ⫹ [⫺(⫺2)] 45. (2x ⫹ 3)(3x ⫹ 5) ⫽ (2x ⫹ 3)3x ⫹ (2x ⫹ 3)5
In Problems 19–22, perform the indicated operations and simplify. 2
41
63.
64. 8 42. 2y6
In Problems 43–48, each statement illustrates the use of one of the following real number properties or definitions. Indicate which one. Commutative (⫹) Identity (⫹) Commutative (ⴢ) Identity (ⴢ) Division Associative (⫹) Inverse (⫹) Associative (ⴢ) Inverse (ⴢ) Zero Distributive Subtraction Negatives
3x2(x ⫹ 2)2 ⫺ 2x(x ⫹ 2)3 x4 m⫹3 2 m⫺1 ⫹ 2 ⫹ 2⫺m m ⫺ 4m ⫹ 4 m ⫺4 2
y x2
⫼a
1⫺
x3y ⫺ x2y x2 ⫹ 3x ⫼ b 2x2 ⫹ 5x ⫺ 3 2x2 ⫺ 3x ⫹ 1 1
1⫹ 65. 1⫺
x y
1 1⫺
x y
66. Convert to scientific notation and simplify: 0.000 000 000 52 (1,300)(0.000 002)
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BASIC ALGEBRAIC OPERATIONS
In Problems 67–75, perform the indicated operations and express answers in simplified form. All radicands represent positive real numbers. 5
6 7 11
2x2
67. ⫺2x 23 x y
68.
9 70. 2 8x6y12
3 71. 21 4x4
69.
3 2 4x
5
3y2
B 8x2
(C) What is the effect on production of doubling the units of labor and capital at any production level? 79. ELECTRIC CIRCUIT If three electric resistors with resistances R1, R2, and R3 are connected in parallel, then the total resistance R for the circuit shown in the figure is given by R⫽
72. (2 1x ⫺ 5 1y)( 1x ⫹ 1y) 73.
75.
3 1x 2 1x ⫺ 1y
74.
2 1u ⫺ 3 1v 2 1u ⫹ 3 1v
1 1 1 1 ⫹ ⫹ R1 R2 R3
Represent this compound fraction as a simple fraction.
y2
R1
2y2 ⫹ 4 ⫺ 2
R2 R3
APPLICATIONS 76. CONSTRUCTION A circular fountain in a park includes a concrete wall that is 3 ft high and 2 ft thick (see the figure). If the inner radius of the wall is x feet, write an algebraic expression in terms of x that represents the volume of the concrete used to construct the wall. Simplify the expression.
2 feet
x feet
80. CONSTRUCTION A box with a hinged lid is to be made out of a piece of cardboard that measures 16 by 30 inches. Six squares, x inches on a side, will be cut from each corner and the middle, and then the ends and sides will be folded up to form the box and its lid (see the figure). Express each of the following quantities as a polynomial in both factored and expanded form. (A) The area of cardboard after the corners have been removed. (B) The volume of the box.
3 feet
30 in. x
77. ECONOMICS If in the United States in 2007 the total personal income was about $11,580,000,000,000 and the population was about 301,000,000, estimate to three significant digits the average personal income. Write your answer in scientific notation and in standard decimal form. 78. ECONOMICS The number of units N produced by a petroleum company from the use of x units of capital and y units of labor is approximated by N ⫽ 20x1Ⲑ2y1Ⲑ2 (A) Estimate the number of units produced by using 1,600 units of capital and 900 units of labor. (B) What is the effect on production if the number of units of capital and labor are doubled to 3,200 units and 1,800 units, respectively?
16 in.
x
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CHAPTER
Equations and Inequalities
1
C
OUTLINE
SOLVING equations and inequalities is one of the most important
skills in algebra because it can be applied to solving a boundless supply of real-world problems. In this chapter, we will begin with a look at techniques for solving linear equations and inequalities. After a study of complex numbers, we’ll return to equations, learning how to solve a variety of nonlinear equations. For each type of equation and inequality we solve, we will look at some real-world problems that can be solved using those solution techniques. This doesn’t close the book on solving equations, though—we will learn how to solve new types of equations in many of the remaining chapters.
1-1
Linear Equations and Applications
1-2
Linear Inequalities
1-3
Absolute Value in Equations and Inequalities
1-4
Complex Numbers
1-5
Quadratic Equations and Applications
1-6
Additional Equation-Solving Techniques Chapter 1 Review Chapter 1 Group Activity: Solving a Cubic Equation
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EQUATIONS AND INEQUALITIES
Linear Equations and Applications Z Understanding Basic Terms Z Solving Linear Equations Z Solving Number and Geometric Problems Z Solving Rate–Time Problems Z Solving Mixture Problems
We begin this section with a quick look at what an equation is and what it means to solve one. After solving some linear equations, we move on to the main topic: using linear equations to solve word problems.
Z Understanding Basic Terms An algebraic equation is a mathematical statement that two algebraic expressions are equal. Some examples of equations with variable x are 3x 0002 2 0004 7 2x2 0002 3x 0003 5 0004 0
1 x 0004 10003x x00022 1x 0003 4 0004 x 0002 1
The replacement set, or domain, for a variable is defined to be the set of numbers that are permitted to replace the variable.
Z ASSUMPTION On Domains of Variables Unless stated to the contrary, we assume that the domain for a variable in an algebraic expression or equation is the set of those real numbers for which the algebraic expressions involving the variable are real numbers.
For example, the domain for the variable x in the expression 2x 0002 4 is R, the set of all real numbers, since 2x 0002 4 represents a real number for all replacements of x by real numbers. The domain of x in the equation 1 2 0004 x x00023 is the set of all real numbers except 0 and 3. These values are excluded because the expression on the left is not defined for x 0004 0 and the expression on the right is not defined for x 0004 3. Both expressions represent real numbers for all other replacements of x by real numbers. The solution set for an equation is defined to be the set of all elements in the domain of the variable that make the equation true. Each element of the solution set is called a solution, or root, of the equation. To solve an equation is to find the solution set for the equation.
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SECTION 1–1
Linear Equations and Applications
45
An equation is called an identity if the equation is true for all elements from the domain of the variable. An equation is called a conditional equation if it is true for certain domain values and false for others. For example, 2x 0002 4 0004 2(x 0002 2)
and
5 5 0004 x(x 0002 3) x2 0002 3x
are identities, since both equations are true for all elements from the respective domains of their variables. On the other hand, the equations 3x 0002 2 0004 5
and
1 2 0004 x x00021
are conditional equations, since, for example, neither equation is true for the domain value 2. Knowing what we mean by the solution set of an equation is one thing; finding it is another. We introduce the idea of equivalent equations to help us find solutions. We will call two equations equivalent if they both have the same solution set. To solve an equation, we perform operations on the equation to produce simpler equivalent equations. We stop when we find an equation whose solution is obvious. Then we check this obvious solution in the original equation. Any of the properties of equality given in Theorem 1 can be used to produce equivalent equations.
Z THEOREM 1 Properties of Equality For a, b, and c any real numbers: 1. If a 0004 b, then a 0003 c 0004 b 0003 c. 2. If a 0004 b, then a 0002 c 0004 b 0002 c. 3. If a 0004 b and c 0005 0, then ca 0004 cb. b a 4. If a 0004 b and c 0005 0, then 0004 . c c
Addition Property Subtraction Property Multiplication Property
5. If a 0004 b, then either may replace the other in any statement without changing the truth or falsity of the statement.
Substitution Property
Division Property
Z Solving Linear Equations We now turn our attention to methods of solving first-degree, or linear, equations in one variable.
Z DEFINITION 1 Linear Equation in One Variable Any equation that can be written in the form ax ⴙ b ⴝ 0
aⴝ0
Standard Form
where a and b are real constants and x is a variable, is called a linear, or firstdegree, equation in one variable. 5x ⴚ 1 ⴝ 2(x ⴙ 3) is a linear equation because after simplifying, it can be written in the standard form 3x ⴚ 7 ⴝ 0.
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EXAMPLE
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1
Solving a Linear Equation Solve 5x ⫺ 9 ⫽ 3x ⫹ 7 and check.
SOLUTION
We will use the properties of equality to transform the given equation into an equivalent equation whose solution is obvious. 5x ⫺ 9 ⫽ 3x ⫹ 7 5x ⫺ 9 ⫹ 9 ⫽ 3x ⫹ 7 ⫹ 9 5x ⫽ 3x ⫹ 16 5x ⫺ 3x ⫽ 3x ⫹ 16 ⫺ 3x 2x ⫽ 16 2x 16 ⫽ 2 2 x⫽8
Add 9 to both sides. Combine like terms. Subtract 3x from both sides. Combine like terms. Divide both sides by 2. Simplify.
The solution set for this last equation is obvious: Solution set: {8} And since the equation x ⫽ 8 is equivalent to all the preceding equations in our solution, {8} is also the solution set for all these equations, including the original equation. [Note: If an equation has only one element in its solution set, we generally use the last equation (in this case, x ⫽ 8) rather than set notation to represent the solution.] CHECK
MATCHED PROBLEM 1
A ⫽ lw
5x ⫺ 9 ⫽ 3x ⫹ 7 ? 5(8) ⫺ 9 ⫽ 3(8) ⫹ 7 ? 40 ⫺ 9 ⫽ 24 ⫹ 7 ✓ 31 ⫽ 31
Simplify each side.
A true statement
Solve and check: 7x ⫺ 10 ⫽ 4x ⫹ 5
0002
0002
We often encounter equations involving more than one variable. For example, if l and w are the length and width of a rectangle, respectively, the area of the rectangle is given by A ⫽ lw (see Fig. 1). Depending on the situation, we may want to solve this equation for l or w. To solve for w, we simply consider A and l to be constants and w to be a variable. Then the equation A ⫽ lw becomes a linear equation in w that can be solved easily by dividing both sides by l:
w
l
Z Figure 1 Area of a rectangle.
w⫽
EXAMPLE
Substitute x ⴝ 8.
2
A l
l⫽0
Solving an Equation with More Than One Variable Solve for P in terms of the other variables: A ⫽ P ⫹ Prt
SOLUTION
A ⫽ P ⫹ Prt A ⫽ P(1 ⫹ rt)
Factor to isolate P. Divide both sides by 1 ⴙ rt.
A ⫽P 1 ⫹ rt P⫽
A 1 ⫹ rt
Restriction: 1 ⴙ rt ⴝ 0
0002
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SECTION 1–1
MATCHED PROBLEM 2
Linear Equations and Applications
47
Solve for F in terms of C: C 0004 59(F 0002 32) 0002 A great many practical problems can be solved using algebraic techniques—so many, in fact, that there is no one method of attack that will work for all. However, we can put together a strategy that will help you organize your approach.
Z STRATEGY FOR SOLVING WORD PROBLEMS 1. Read the problem slowly and carefully, more than once if necessary. Write down information as you read the problem the first time to help you get started. Identify what it is that you are asked to find. 2. Use a variable to represent an unknown quantity in the problem, usually what you are asked to find. Then try to represent any other unknown quantities in terms of that variable. It’s pretty much impossible to solve a word problem without this step. 3. If it helps to visualize a situation, draw a diagram and label known and unknown parts. 4. Write an equation relating the quantities in the problem. Often, you can accomplish this by finding a formula that connects those quantities. Try to write the equation in words first, then translate to symbols. 5. Solve the equation, then answer the question in a sentence by rephrasing the question. Make sure that you’re answering all of the questions asked. 6. Check to see if your answers make sense in the original problem, not just the equation you wrote.
ZZZ EXPLORE-DISCUSS 1
Translate each of the following sentences involving two numbers into an equation. (A) The first number is 10 more than the second number. (B) The first number is 15 less than the second number. (C) The first number is half the second number. (D) The first number is three times the second number. (E) Ten times the first number is 15 more than the second number.
The remaining examples in this section contain solutions to a variety of word problems illustrating both the process of setting up word problems and the techniques used to solve the resulting equations. As you read an example, try covering up the solution and working the problem yourself. If you need a hint, uncover just part of the solution and try to work out the rest. After you successfully solve an example problem, try the matched problem. If you work through the remainder of the section in this way, you will already have experience with a wide variety of word problems.
Z Solving Number and Geometric Problems Example 3 introduces the process of setting up and solving word problems in a simple mathematical context. Examples 4–8 are more realistic.
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3
Setting Up and Solving a Word Problem Find four consecutive even integers so that the sum of the first three is 8 more than the fourth.
SOLUTION
Let x ⫽ the first even integer; then x
x⫹2
x⫹4
and
x⫹6
represent four consecutive even integers starting with the even integer x. (Remember, even integers are separated by 2.) The phrase “the sum of the first three is 8 more than the fourth” translates into an equation: Sum of the first three ⫽ Fourth ⫹ 8 x ⫹ (x ⫹ 2) ⫹ (x ⫹ 4) ⫽ (x ⫹ 6) ⫹ 8 3x ⫹ 6 ⫽ x ⫹ 14 2x ⫽ 8 x⫽4
Combine like terms. Subtract 6 and x from both sides. Divide both sides by 2.
The first even integer is 4, so the four consecutive integers are 4, 6, 8, and 10. CHECK
MATCHED PROBLEM 3
ZZZ EXPLORE-DISCUSS 2
4 ⫹ 6 ⫹ 8 ⫽ 18 ⫽ 10 ⫹ 8
Sum of first three is 8 more than the fourth.
0002
Find three consecutive odd integers so that 3 times their sum is 5 more than 8 times the middle one. 0002
According to Part 3 of Theorem 1, multiplying both sides of an equation by a nonzero number always produces an equivalent equation. By what number would you choose to multiply both sides of the following equation to eliminate all the fractions? x⫹1 x 1 ⫺ ⫽ 3 4 2 If you did not choose 12, the LCD of all the fractions in this equation, you could still solve the resulting equation, but with more effort. (For a discussion of LCDs and how to find them, see Section R-4.)
EXAMPLE
4
Using a Diagram in the Solution of a Word Problem A landscape designer plans a series of small triangular gardens outside a new office building. Her plans call for one side to be one-third of the perimeter, and another side to be onefifth of the perimeter. The space allotted for each will allow the third side to be 7 meters. Find the perimeter of the triangle.
SOLUTION p 5
p 3 7 meters
Z Figure 2
Draw a triangle, and label one side 7 meters. Let p ⫽ the perimeter: then the remaining sides are one-third p, or p Ⲑ3, and one-fifth p, or pⲐ 5 (see Fig. 2). Perimeter ⫽ Sum of the side lengths p⫽
p p ⫹ ⫹7 3 5
Multiply both sides by 15, the LCD. Make sure to multiply every term by 15!
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SECTION 1–1
p p 15 ⴢ p 0004 15 ⴢ a 0003 0003 7b 3 5 p p 15p 0004 15 ⴢ 0003 15 ⴢ 0003 15 ⴢ 7 3 5 15p 0004 5p 0003 3p 0003 105 15p 0004 8p 0003 105 7p 0004 105 p 0004 15
Linear Equations and Applications
49
*
Combine like terms. Subtract 8p from both sides. Divide both sides by 7.
The perimeter is 15 meters. p 15 0004 00045 3 3 p 15 0003 0004 00043 5 5 0003 7 15 meters
CHECK
MATCHED PROBLEM 4
ZZZ
CAUTION ZZZ
Side 1
Side 2 Side 3 Perimeter
0002
If one side of a triangle is one-fourth the perimeter, the second side is 7 centimeters, and the third side is two-fifths the perimeter, what is the perimeter? 0002
A very common error occurs about now—students tend to confuse algebraic expressions involving fractions with algebraic equations involving fractions. Consider these two problems: (A) Solve:
x x 0003 0004 10 2 3
(B) Add:
x x 0003 0003 10 2 3
The problems look very much alike but are actually very different. To solve the equation in (A) we multiply both sides by 6 (the LCD) to clear the fractions. This works so well for equations that students want to do the same thing for problems like (B). The only catch is that (B) is not an equation, and the multiplication property of equality does not apply. If we multiply (B) by 6, we simply obtain an expression 6 times as large as the original! Compare these correct solutions: x x 0003 0004 10 2 3
(A) 6ⴢ
x x 0003 6 ⴢ 0004 6 ⴢ 10 2 3 3x 0003 2x 0004 60 5x 0004 60 x 0004 12
(B)
x x 0003 0003 10 2 3 0004
3ⴢx 2ⴢx 6 ⴢ 10 0003 0003 3ⴢ2 2ⴢ3 6ⴢ1
0004
3x 2x 60 0003 0003 6 6 6
0004
5x 0003 60 6
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
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There are many problems in which a rate plays a key role. For example, if you’re losing weight at the rate of 2 lb per week, you can use that rate to find a total weight loss for some period of time. Rate problems can often be solved using the following basic formula: Z QUANTITY-RATE-TIME FORMULA The change in a quantity is the rate at which it changes times the time passed: Quantity 0004 Rate 0007 Time, or Q 0004 RT. If the quantity is distance, then D 0004 RT. The formulas can be solved for R or T to get a related formula to find the rate or the time. [Note: R is an average or uniform rate.]
ZZZ EXPLORE-DISCUSS 3
EXAMPLE
5
(A) If you drive at an average rate of 65 miles per hour, how far do you go in 3 hours? (B) If you make $750 for 2 weeks of part-time work, what is your weekly rate of pay? (C) If you eat at the rate of 1,900 calories per day, how long will it take you to eat 7,600 calories?
A Distance–Rate–Time Problem The distance along a shipping route between San Francisco and Honolulu is 2,100 nautical miles. If one ship leaves San Francisco at the same time another leaves Honolulu, and if the former travels at 15 knots* and the latter at 20 knots, how long will it take the two ships to rendezvous? How far will they be from Honolulu and San Francisco at that time?
SOLUTION
Let T 0004 number of hours until both ships meet. Draw a diagram and label known and unknown parts. Both ships will have traveled the same amount of time when they meet.
San Francisco 2,100
H
miles
D1 0004 20T
D2 0004 15T
20 knots
15 knots Meeting
D 0004 RT D1 0004 20 knots ⴢ Time D2 0004 15 knots ⴢ Time
SF
Distance ship 1 Distance ship 2 from Honolulu from San Francisco ± ≤ 0003 ± ≤ travels to travels to meeting point meeting point D1 0003 D2 20T 0003 15T 35T T
Honolulu
Total distance 0004 ° from Honolulu ¢ to San Francisco 0004 0004 0004 0004
2,100 2,100 2,100 60
Therefore, it takes 60 hours, or 2.5 days, for the ships to meet. Distance from Honolulu 0004 20 ⴢ 60 0004 1,200 nautical miles Distance from San Francisco 0004 15 ⴢ 60 0004 900 nautical miles CHECK
1,200 0003 900 0004 2,100 nautical miles
0002
*15 knots means 15 nautical miles per hour. There are 6,076.1 feet in 1 nautical mile, and 5,280 feet in 1 statute mile.
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MATCHED PROBLEM 5
51
Linear Equations and Applications
An old piece of equipment can print, stuff, and label 38 mailing pieces per minute. A newer model can handle 82 per minute. How long will it take for both pieces of equipment to prepare a mailing of 6,000 pieces? [Hint: Use Quantity 0004 Rate 0007 Time for each machine.] 0002 Some equations involving variables in a denominator can be transformed into linear equations. We can proceed in essentially the same way as in Example 5; however, we need to exclude any value of the variable that will make a denominator 0. With these values excluded, we can multiply through by the LCD even though it contains a variable, and, according to Theorem 1, the new equation will be equivalent to the old.
EXAMPLE
6
A Distance–Rate–Time Problem An excursion boat takes 1.5 times as long to go 360 miles up a river as to return. If the boat cruises at 15 miles per hour in still water, what is the rate of the current?
SOLUTION
360 miles
Let x 0004 Rate of current (in miles per hour) 15 0002 x 0004 Rate of boat upstream 15 0003 x 0004 Rate of boat downstream Time upstream 0004 (1.5)(Time downstream) Distance upstream Distance downstream 0004 (1.5) Rate upstream Rate downstream 360 360 0004 (1.5) 15 0002 x 15 0003 x 360 540 0004 15 0002 x 15 0003 x 360(15 0003 x) 0004 540(15 0002 x) 5,400 0003 360x 0004 8,100 0002 540x 5,400 0003 900x 0004 8,100 900x 0004 2,700 x00043
What we were asked to find.
Faster downstream.
Because D 0004 RT, T ⴝ
x cannot be 15 or 000215 Multiply both sides by the LCD, (15 ⴚ x)(15 ⴙ x). Multiply out parentheses. Add 540x to both sides. Subtract 5,400 from both sides. Divide both sides by 900.
The rate of the current is 3 miles per hour. The check is left to the reader. MATCHED PROBLEM 6
EXAMPLE
7
D R
0002
A jetliner takes 1.2 times as long to fly from Paris to New York (3,600 miles) as to return. If the jet cruises at 550 miles per hour in still air, what is the average rate of the wind blowing in the direction of Paris from New York? 0002
A Quantity–Rate–Time Problem An advertising firm has an old computer that can prepare a whole mailing in 6 hours. With the help of a newer model the job is complete in 2 hours. How long would it take the newer model to do the job alone?
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SOLUTION
Let x 0004 time (in hours) for the newer model to do the whole job alone. a
Part of job completed b 0004 (Rate)(Time) in a given length of time 1 job per hour 6 1 Rate of new model 0004 job per hour x Rate of old model 0004
Part of job completed Part of job completed ° by old model ¢ 0003 ° by new model ¢ 0004 1 whole job in 2 hours in 2 hours Rate of Time of Rate of Time of Recall: Q ⴝ RT a ba b0003a ba b00041 old model old model new model new model 1 (2) 6 1 3
0003
1 (2) x
00041
0003
2 x
00041
1 3x a b 3
0003
2 3x a b x
0004 3x
x
0003
6
0004 3x
6
0004 2x
3
0004x
x cannot be zero.
Multiply both sides by 3x, the LCD.
Subtract x from both sides. Divide both sides by 2.
Therefore, the new computer could do the job alone in 3 hours. CHECK
MATCHED PROBLEM 7
Part of job completed by old model in 2 hours 0004 2(16) 0004 13 0003 Part of job completed by new model in 2 hours 0004 2(13) 0004 23 Part of job completed by both models in 2 hours 0004 1
0002
Two pumps are used to fill a water storage tank at a resort. One pump can fill the tank by itself in 9 hours, and the other can fill it in 6 hours. How long will it take both pumps operating together to fill the tank? 0002
Z Solving Mixture Problems A variety of applications can be classified as mixture problems. Even though the problems come from different areas, their mathematical treatment is essentially the same.
EXAMPLE
8
A Mixture Problem How many liters of a mixture containing 80% alcohol should be added to 5 liters of a 20% solution to yield a 30% solution?
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SOLUTION
Linear Equations and Applications
53
Let x 0004 amount of 80% solution used. BEFORE MIXING 80% solution
AFTER MIXING 30% solution
20% solution
0003
0004 5 liters
x liters
(x 0003 5) liters
Amount of Amount of Amount of ° alcohol in ¢ 0003 ° alcohol in ¢ 0004 ° alcohol in ¢ first solution second solution mixture 0.8x 0003 0.2(5) 0004 0.3(x 0003 5) 0.8x 0003 1 0004 0.3x 0003 1.5 0.5x 0004 0.5 x00041 Add 1 liter of the 80% solution. CHECK First solution Second solution Mixture
Liters of solution
Liters of alcohol 0.8(1) 0004 0.8 0003 0.2(5) 0004 1 1.8
1 00035 6
Percent alcohol 80 or 0.8兾1 20 or 1兾5 1.8兾6 0004 0.3, or 30%
0002 MATCHED PROBLEM 8
A chemical storeroom has a 90% acid solution and a 40% acid solution. How many centiliters of the 90% solution should be added to 50 centiliters of the 40% solution to yield a 50% solution? 0002 ANSWERS TO MATCHED PROBLEMS 1. x 0004 5 2. F 0004 95C 0003 32 3. 3, 5, 7 4. 20 centimeters 5. 50 minutes 6. 50 miles per hour 7. 3.6 hours 8. 12.5 centiliters
1-1
Exercises
1. What does it mean to solve an equation? 2. Describe the difference between an equation and an expression. 3. How can you tell if an equation is linear?
4. In one or two sentences, describe what parts 1– 4 in Theorem 1 say about working with equations. 5. How can you check your solution to an equation? 6. How do you check your solution to a word problem?
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7. Explain why the following does not make sense: Solve the equation P 0002 2l 0003 2w.
41.
1 1 1 0003 for f (simple lens formula) 0002 f d1 d2
8. Explain why the following does not make sense: Solve y y 0004 0003 1. 4 5
42.
1 1 1 0003 0002 for R1 (electric circuit) R R1 R2
44. A 0002 2ab 0003 2ac 0003 2bc for c
In Problems 9–34, solve each equation. 9. 10 x 0004 7 0002 4x 0004 25
10. 11 0003 3y 0002 5y 0004 5
11. 3(x 0003 2) 0002 5(x 0004 6)
12. 3(y 0004 4) 0003 2y 0002 18
13. 5 0003 4(t 0004 2) 0002 2(t 0003 7) 0003 1 14. 4 0004 3(t 0003 2) 0003 t 0002 5(t 0004 1) 0004 7t 15. 5 0004 17.
3a 0004 4 7 0004 2a 0002 5 2
16. 5 0004
x00033 x00044 3 0004 0002 4 2 8
18.
43. A 0002 2ab 0003 2ac 0003 2bc for a (surface area of a rectangular solid)
2x 0004 1 x00032 0002 4 3
x 3x 0004 1 6x 0003 5 0003 0002 5 2 4
45. y 0002
21. 0.35(s 0003 0.34) 0003 0.15s 0002 0.2s 0004 1.66 22. 0.35(u 0003 0.34) 0004 0.15u 0002 0.2u 0004 1.66 2 2 5 23. 0003 0002 4 0004 y 2 3y 25. 27.
z 1 0002 00032 z00041 z00041
30003w 1 4 0002 0003 24. 6w 2w 3 26.
y y 0004 10 2y 0004 2 0003 0002 00043 3 5 4
29. 1 0004
x00043 2x 0004 3 0002 x00042 x00042
6 5 000310002 31. y00034 2y 0003 8
t 2 0002 00032 t00041 t00041 28.
30.
z00034 z z00038 0003 0002 00035 7 6 3
2x 0004 3 3x 0004 1 000220004 x00031 x00031
4y 12 000350002 32. y00043 y00043
33.
3 3 3a 0004 1 0004 2 0002 a a2 0003 4a 0003 4 a 0003 2a
34.
1 1 10 0002 0004 2 b00045 b 0003 5 b 0004 5b 0003 25
46. x 0002
47.
x 2x 0004 3 000340002 x00043 x00043 x 0003 4x 0004 12 0002 2x 0004 3 x00023
1 x 00023 49. 1 10003 x
x0004
x0004
x000310004 51. 10004
2 x
1 x
50.
2.34 5.67x 0003 5.67 0002 x x00034
x2 0003 4x 0004 3 x2 0003 1 0002 x00041 x00041 x2 0003 1 0002 x2 0003 4x 0004 3 x00021
00021
0002x00032
3 y x 0002a b 10004y 10004x
a
53. Solve for x in terms of y: y 0002 10003
b x0003c
54. Let m and n be real numbers with m larger than n. Then there exists a positive real number p such that m 0002 n 0003 p. Find the fallacy in the following argument: m0002n0003p (m 0004 n)m 0002 (m 0004 n)(n 0003 p) m2 0004 mn 0002 mn 0003 mp 0004 n2 0004 np 2
m 0004 mn 0004 mp 0002 mn 0004 n2 0004 np m(m 0004 n 0004 p) 0002 n(m 0004 n 0004 p) m0002n
36. 1.73y 0003 0.279(y 0004 3) 0002 2.66y 38.
1 x
2 x000310004 x
52. Solve for y in terms of x:
35. 3.142x 0004 0.4835(x 0004 4) 0002 6.795
2.32x 3.76 0004 0002 2.32 x x00042
48.
In Problems 49–51, solve the equation.
In Problems 35–38, use a calculator to solve each equation to three significant digits.*
37.
3y 0003 2 for y y00043
In Problems 47 and 48, imagine that the indicated “solutions” were given to you by a student whom you were tutoring in this class. Is the solution right or wrong? If the solution is wrong, explain what is wrong and show a correct solution.
19. 0.1(t 0003 0.5) 0003 0.2t 0002 0.3(t 0004 0.4) 20. 0.1(w 0003 0.5) 0003 0.2w 0002 0.2(w 0004 0.4)
2x 0004 3 for x 3x 0003 5
APPLICATIONS These problems are grouped according to subject area.
In Problems 39–46, solve for the indicated variable in terms of the other variables. 39. an 0002 a1 0003 (n 0004 1)d for d (arithmetic progressions) 40. F 0002 95C 0003 32 for C (temperature scale) *Appendix A contains a brief discussion of significant digits.
Numbers 55. Find a number so that 10 less than two-thirds the number is one-fourth the number. 56. Find a number so that 6 more than one-half the number is twothirds the number.
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57. Find four consecutive even integers so that the sum of the first three is 2 more than twice the fourth. 58. Find three consecutive even integers so that the first plus twice the second is twice the third. Geometry 59. Find the perimeter of a triangle if one side is 16 feet, another side is two-sevenths the perimeter, and the third side is one-third the perimeter. 60. Find the perimeter of a triangle if one side is 11 centimeters, another side is two-fifths the perimeter, and the third side is onehalf the perimeter. 61. A new game show requires a playing field with a perimeter of 54 yards and length 3 yards less than twice the width. What are the dimensions? 62. A celebrity couple wants to have a rectangular pool put in the backyard of their vacation home. They want it to be 24 meters long, and they insist that it have at least as much area as the neighbor’s pool, which is a square 12 meters on a side. Find the dimensions of the smallest pool that meets these criteria. Business and Economics 63. The sale price of an MP3 player after a 30% discount was $140. What was the original price? 64. A sporting goods store marks up each item it sells 60% above wholesale price. What is the wholesale price on a snowboard that sells for $144? 65. One employee of a computer store is paid a base salary of $2,150 a month plus an 8% commission on all sales over $7,000 during the month. How much must the employee sell in 1 month to earn a total of $3,170 for the month? 66. A second employee of the computer store in Problem 65 is paid a base salary of $1,175 a month plus a 5% commission on all sales during the month. (A) How much must this employee sell in 1 month to earn a total of $3,170 for the month? (B) Determine the sales level where both employees receive the same monthly income. If employees can select either of these payment methods, how would you advise an employee to make this selection? Earth Science 67. In 1970, Russian geologists began drilling a very deep borehole in the Kola Peninsula. Their goal was to reach a depth of 15 kilometers, but high temperatures in the borehole forced them to stop in 1994 after reaching a depth of 12 kilometers. They found that below 3 kilometers the temperature T increased 2.5°C for each additional 100 meters of depth. (A) If the temperature at 3 kilometers is 30°C and x is the depth of the hole in kilometers, write an equation using x that will give the temperature T in the hole at any depth beyond 3 kilometers. (B) What would the temperature be at 12 kilometers? (C) At what depth (in kilometers) would they reach a temperature of 200°C?
Linear Equations and Applications
55
68. An earthquake emits a primary wave and a secondary wave. Near the surface of the Earth the primary wave travels at about 5 miles per second, and the secondary wave travels at about 3 miles per second. From the time lag between the two waves arriving at a given seismic station, it is possible to estimate the distance to the quake. Suppose a station measures a time difference of 12 seconds between the arrival of the two waves. How far is the earthquake from the station? (The epicenter can be located by obtaining distance bearings at three or more stations.) Life Science 69. The kangaroo rat is an endangered species native to California. In order to keep track of their population size in a state nature preserve, a conservation biologist trapped, tagged, and released 80 individuals from the population. After waiting 2 weeks for the animals to mix back in with the general population, she again caught 80 individuals and found that 22 of them were tagged. Assuming that the ratio of tagged animals to total animals in the second sample is the same as the ratio of all tagged animals to the total population in the preserve, estimate the total number of kangaroo rats in the preserve. 70. Repeat Problem 69 with a first (marked) sample of 70 and a second sample of 30 with only 11 marked animals. Chemistry 71. How many gallons of distilled water must be mixed with 50 gallons of 30% alcohol solution to obtain a 25% solution? 72. How many gallons of hydrochloric acid must be added to 12 gallons of a 30% solution to obtain a 40% solution? 73. A chemist mixes distilled water with a 90% solution of sulfuric acid to produce a 50% solution. If 5 liters of distilled water are used, how much 50% solution is produced? 74. A fuel oil distributor has 120,000 gallons of fuel with 0.9% sulfur content, which exceeds pollution control standards of 0.8% sulfur content. How many gallons of fuel oil with a 0.3% sulfur content must be added to the 120,000 gallons to obtain fuel oil that will comply with the pollution control standards? Rate–Time 75. An old computer can do the weekly payroll in 5 hours. A newer computer can do the same payroll in 3 hours. The old computer starts on the payroll, and after 1 hour the newer computer is brought on-line to work with the older computer until the job is finished. How long will it take both computers working together to finish the job? (Assume the computers operate independently.) 76. One pump can fill a gasoline storage tank in 8 hours. With a second pump working simultaneously, the tank can be filled in 3 hours. How long would it take the second pump to fill the tank operating alone? 77. The cruising speed of an airplane is 150 miles per hour (relative to the ground). You plan to hire the plane for a 3-hour sightseeing trip. You instruct the pilot to fly north as far as she can and still return to the airport at the end of the allotted time. (A) How far north should the pilot fly if the wind is blowing from the north at 30 miles per hour? (B) How far north should the pilot fly if there is no wind?
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78. Suppose you are at a river resort and rent a motor boat for 5 hours starting at 7 A.M. You are told that the boat will travel at 8 miles per hour upstream and 12 miles per hour returning. You decide that you would like to go as far up the river as you can and still be back at noon. At what time should you turn back, and how far from the resort will you be at that time?
82. A minor chord is composed of notes whose frequencies are in the ratio 10:12:15. If the first note of a minor chord is A, with a frequency of 220 hertz, what are the frequencies of the other two notes?
79. A two-woman rowing team can row 1,200 meters with the current in a river in the same amount of time it takes them to row 1,000 meters against that same current. In each case, their average rowing speed without the effect of the current is 3 meters per second. Find the speed of the current.
83. In an experiment on motivation, Professor Brown trained a group of rats to run down a narrow passage in a cage to receive food in a goal box. He then put a harness on each rat and connected it to an overhead wire attached to a scale. In this way, he could place the rat different distances from the food and measure the pull (in grams) of the rat toward the food. He found that the relationship between motivation (pull) and position was given approximately by the equation
80. The winners of the men’s 1,000-meter double sculls event in the 2008 Olympics rowed at an average of 11.3 miles per hour. If this team were to row this speed for a half mile with a current in 80% of the time they were able to row that same distance against the current, what would be the speed of the current? Music 81. A major chord in music is composed of notes whose frequencies are in the ratio 4:5:6. If the first note of a chord has a frequency of 264 hertz (middle C on the piano), find the frequencies of the other two notes. [Hint: Set up two proportions using 4:5 and 4:6.]
Psychology
p 0004 000215d 0003 70
30 d 170
where pull p is measured in grams and distance d in centimeters. When the pull registered was 40 grams, how far was the rat from the goal box? 84. Professor Brown performed the same kind of experiment as described in Problem 83, except that he replaced the food in the goal box with a mild electric shock. With the same kind of apparatus, he was able to measure the avoidance strength relative to the distance from the object to be avoided. He found that the avoidance strength a (measured in grams) was related to the distance d that the rat was from the shock (measured in centimeters) approximately by the equation a 0004 000243d 0003 230
30 d 170
If the same rat were trained as described in this problem and in Problem 83, at what distance (to one decimal place) from the goal box would the approach and avoidance strengths be the same? (What do you think the rat would do at this point?)
1-2
Linear Inequalities Z Understanding Inequality and Interval Notation Z Solving Linear Inequalities Z Applying Linear Inequalities
An equation is a statement that two expressions are equal. Sometimes it is useful to find when one expression is more or less than another, so in this section we turn our attention to linear inequalities in one variable, like 3x 0003 5 7 x 0002 10
and
00024 6 3 0002 2x 6 7
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57
Z Understanding Inequality and Interval Notation The preceding mathematical statements use the inequality, or order, relations, more commonly known as “greater than” and “less than.” Just as we use the symbol “0004” to replace the words “is equal to,” we use the inequality symbols and to replace “is less than” and “is greater than,” respectively. You probably have a natural understanding of how to compare numbers using these symbols, but to be precise about using inequality symbols, we should have a clear definition of what they mean.
Z DEFINITION 1 a < b and b > a For two real numbers a and b, we say that a is less than b, and write a b, if there is a positive real number p so that a 0003 p 0004 b. The statement b a, read b is greater than a, means exactly the same as a b.
This definition basically says that if you add a positive number to any number, the sum is larger than the original number. When we write a b we mean a 6 b or a 0004 b and say a is less than or equal to b. When we write a b we mean a 7 b or a 0004 b and say a is greater than or equal to b. The inequality symbols 6 and 7 have a very clear geometric interpretation on the real number line. If a 6 b, then a is to the left of b; if c 7 d, then c is to the right of d (Fig. 1). This is called a line graph.
a
d
b
c
Z Figure 1 a b, c d.
If we want to state that some number x is between a and b, we could use two inequalities: x a and x b. Instead, we will write one double inequality, a x b. For example, the inequality 00022 x 5 indicates that x is between 00022 and 5, and could be equal to 5, but not 00022. The set of all real numbers that satisfy this inequality is called an interval, and is commonly represented by (00022, 5]. In general, (a, b] 0004 5x ƒ a 6 x b6*
The number a is called the left endpoint of the interval, and the symbol “(” indicates that a is not included in the interval. The number b is called the right endpoint of the interval, and the symbol “]” indicates that b is included in the interval. An interval is closed if it contains its endpoint(s) and open if it does not contain any endpoint. Other types of intervals of real numbers are shown in Table 1. Note that the symbol “ ,” read “infinity,” used in Table 1 is not a numeral. When we write [b, ), we are simply referring to the interval starting at b and continuing indefinitely to the right. We would never write [b, ] or b x , because cannot be used as an endpoint of an interval. The interval (0002 , ) represents the set of real numbers R, since its graph is the entire real number line. *In general, 5x ƒ P(x)6 represents the set of all x such that statement P(x) is true. To express this set verbally, just read the vertical bar as “such that.”
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Table 1 Interval Notation Interval notation
Inequality notation
[a, b]
aⱕxⱕb
[a, b)
aⱕx⬍b
Line graph
(a, b]
a⬍xⱕb
(a, b)
a⬍x⬍b
[b, ⬁)
xⱖb
(b, ⬁)
x⬎b
(⫺⬁, a]
xⱕa
(⫺⬁, a)
x⬍a
Type
[
]
x
Closed
)
x
Half-open
]
x
Half-open
)
x
Open
[
x
Closed*
(
x
Open
]
x
Closed*
)
x
Open
a
b
[
a
b
(
a
b
(
a
b b b a a
*These intervals are closed because they contain all of their endpoints; they have only one endpoint.
ZZZ
It is important to note that
CAUTION ZZZ
5 7 x ⱖ ⫺3
is equivalent to [⫺3, 5) and not to (5, ⫺3]
In interval notation, the smaller number is always written to the left. It may be useful to rewrite the inequality as ⫺3 ⱕ x 6 5 before rewriting it in interval notation. The symbol (5, ⫺3] is meaningless.
EXAMPLE
1
Graphing Intervals and Inequalities Write each of the following in inequality notation and graph on a real number line: (A) [⫺2, 3)
SOLUTIONS
(C) [ ⫺2, ⬁)
(B) (⫺4, 2)
(A) ⫺2 ⱕ x 6 3 (B) ⫺4 6 x 6 2 (C) x ⱖ ⫺2 (D) x 6 3
[
⫺5
⫺2
(
⫺5 ⫺4 ⫺5
[
⫺5
MATCHED PROBLEM 1
3
x
5
)
2
x
5
0
x
5 0
) 3
5
0002
x
Write each of the following in interval notation and graph on a real number line: (A) ⫺3 6 x ⱕ 3
ZZZ EXPLORE-DISCUSS 1
)
0 0
⫺2
(D) (⫺⬁, 3)
(B) 2 ⱖ x ⱖ ⫺1
(C) x 7 1
(D) x ⱕ 2
Example 1C shows the graph of the inequality x ⱖ ⫺2. What is the graph of x 6 ⫺2? What is the corresponding interval? Describe the relationship between these sets.
0002
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59
Since intervals are sets of real numbers, the set operations of union and intersection are often useful when working with intervals. The union of sets A and B, denoted by A 傼 B, is the set formed by combining all the elements of A and all the elements of B. The intersection of sets A and B, denoted by A 傽 B, is the set of elements of A that are also in B. Symbolically: Z DEFINITION 2 Union and Intersection A 傼 B 0004 5x ƒ x is in A or x is in B6
Union:
{1, 2, 3} ´ {2, 3, 4, 5} ⴝ {1, 2, 3, 4, 5}
Intersection: A 傽 B 0004 5x ƒ x is in A and x is in B6 {1, 2, 3} 傽 {2, 3, 4, 5} ⴝ {2, 3}
EXAMPLE
2
Graphing the Union and Intersection of Intervals If A 0004 (00022, 5] and B 0004 (1, ), graph the sets A 傼 B and A 傽 B and write them in interval notation. )
00022
1
00022
1
5
1
5
)
[
)
00022
ZZZ EXPLORE-DISCUSS 2
5
)
00022
MATCHED PROBLEM 2
[
SOLUTION
1
5
x
A 0004 (00022, 5]
x
B 0004 (1, )
x
A 傼 B 0004 (00022, )
x
A 傽 B 0004 (1, 5]
0002
If C 0004 [ 00024, 3) and D 0004 (0002 , 00021] , graph the sets C 傼 D and C 傽 D and write them in interval notation. 0002 Replace ? with 6 or 7 in each of the following. (A) 00021 ? 3
and
2(00021) ? 2(3)
(B) 00021 ? 3
and
00022(00021) ? 00022(3)
(C) 12 ? 00028
and
12 00028 ? 4 4
(D) 12 ? 00028
and
12 00028 ? 00024 00024
Based on your results, describe verbally the effect of multiplying or dividing both sides of an inequality by a number.
Z Solving Linear Inequalities We now turn to the problem of solving linear inequalities in one variable, such as 2(2x 0003 3) 6 6(x 0002 2) 0003 10
and
00023 6 2x 0003 3 9
The solution set for an inequality is the set of all values of the variable that make the inequality a true statement. Each element of the solution set is called a solution of the inequality. To solve an inequality is to find its solution set. Two inequalities are equivalent
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if they have the same solution set. Just as with equations, we perform operations on inequalities that produce simpler equivalent inequalities, and continue the process until an inequality is reached whose solution is obvious. The properties of inequalities given in Theorem 1 can be used to produce equivalent inequalities. Z THEOREM 1 Inequality Properties An equivalent inequality will result and the sense (or direction) will remain the same if each side of the original inequality • Has the same real number added to or subtracted from it • Is multiplied or divided by the same positive number An equivalent inequality will result and the sense (or direction) will reverse if each side of the original inequality • Is multiplied or divided by the same negative number Note: Multiplication by 0 and division by 0 are not permitted.
Theorem 1 tells us that we can perform essentially the same operations on inequalities that we perform on equations, with the exception that the sense (or direction) of the inequality reverses if we multiply or divide both sides by a negative number: Otherwise the sense of the inequality does not change. Now let’s see how the inequality properties are used to solve linear inequalities. Examples 3, 4, and 5 will illustrate the process.
EXAMPLE
3
Solving a Linear Inequality Solve and graph: 2(2x 0003 3) 0002 10 6 6(x 0002 2)
SOLUTION
2(2x 0003 3) 0002 10 6 6(x 0002 2) 4x 0003 6 0002 10 6 6x 0002 12
Multiply out parentheses. Combine like terms.
4x 0002 4 6 6x 0002 12
Add 4 to both sides.
4x 0002 4 0003 4 6 6x 0002 12 0003 4 4x 6 6x 0002 8
Subtract 6x from both sides.
4x 0002 6x 6 6x 0002 8 0002 6x 00022x 6 00028
Divide both sides by ⴚ2. Note that direction reverses because ⴚ2 is negative.
00022x 00028 7 00022 00022 x 7 4 2
3
( 4
5
6
7
8
(4, )
or 9
x
Graph of solution set
MATCHED PROBLEM 3
Solution set
0002
Solve and graph: 3(x 0002 1) 5(x 0003 2) 0002 5 0002
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EXAMPLE
4
61
Solving a Linear Inequality Involving Fractions Solve and graph:
4x 2x 0002 3 00036 20003 4 3
2x 0002 3 4x 00036 20003 4 3
SOLUTION
12 ⴢ
Multiply both sides by 12, the LCD.
2x 0002 3 4x 0003 12 ⴢ 6 12 ⴢ 2 0003 12 ⴢ 4 3
Direction doesn’t change: we multiplied by a positive number.
3(2x 0002 3) 0003 72 24 0003 4(4x) 6x 0002 9 0003 72 24 0003 16x 6x 0003 63 24 0003 16x 6x 000239 0003 16x 000210x 000239 x 3.9 ]
3.9
MATCHED PROBLEM 4
Linear Inequalities
Solve and graph:
or
Multiply out parentheses. Combine like terms. Subtract 63 from both sides. Subtract 16x from both sides.
(0002 , 3.9]
Order reverses when both sides are divided by ⴚ10, a negative number.
x
0002
Graph of solution set
3x 4x 0002 3 00038 6 60003 3 2 0002
EXAMPLE
5
Solving a Double Inequality Solve and graph: 00023 4 0002 7x 6 18
SOLUTION
We proceed as before, except we try to isolate x in the middle with a coefficient of 1, being sure to perform operations on all three parts of the inequality. 00023 4 0002 7x 6 18
Subtract 4 from each member.
00023 0002 4 4 0002 7x 0002 4 6 18 0002 4 00027 00027x 6 14
Divide each member by ⴚ7 and reverse each inequality.
00027x 14 00027 7 00027 00027 00027 1 x 7 00022 (
00022
MATCHED PROBLEM 5
00022 6 x 1
or ]
1
x
or
Graph of solution set
(00022, 1] 0002
Solve and graph: 00023 6 7 0002 2x 7 0002
Z Applying Linear Inequalities to Chemistry EXAMPLE
6
Chemistry In a chemistry experiment, a solution of hydrochloric acid is to be kept between 30°C and 35°C—that is, 30 C 35. What is the range in temperature in degrees Fahrenheit if the Celsius/Fahrenheit conversion formula is C 0004 59 (F 0002 32)?
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30 C 35 5 30 (F 0002 32) 35 9
SOLUTION
Replace C with
5 (F ⴚ 32). 9
9 Multiply each member by 5 to clear fractions.
9 9 5 9 ⴢ 30 ⴢ (F 0002 32) ⴢ 35 5 5 9 5 54 F 0002 32 63
Add 32 to each member.
54 0003 32 F 0002 32 0003 32 63 0003 32 86 F 95 The range of the temperature is from 86°F to 95°F, inclusive. MATCHED PROBLEM 6
0002
A film developer is to be kept between 68°F and 77°F—that is, 68 F 77. What is the range in temperature in degrees Celsius if the Celsius/Fahrenheit conversion formula is F 0004 95C 0003 32? 0002
ANSWERS TO MATCHED PROBLEMS
(B) [ 00021, 2]
[
] 2
( ]
) 3
00021
3
3. x 00024 or (0002 , 00024] 5
2
x x x
5
0
00021
5 5
0 1
00025
4. x 7 6 or (6, )
3
00021 0
[
[
00024
0
00025
(D) (0002 , 2]
00024
00023
00025
(C) (1, )
2.
)
00025
[
1. (A) (00023, 3]
5
x
x
C 傼 D 0004 (0002 , 3)
x
C 傽 D 0004 [ 00024, 00021] ]
00027
00024
( 6
0 12
5. 5 7 x 0 or 0 x 6 5 or [0, 5)
00021
[ 0
x
x
) 5
6
x
6. 20 C 25: the range in temperature is from 20°C to 25°C
1-2
Exercises
1. Explain in your own words what it means to solve an inequality.
3. What is the main difference between the procedures for solving linear equations and linear inequalities?
2. Explain why the “interval” [5, 00023) is meaningless.
4. Describe how to graph the solution set of an inequality.
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45. [ ⫺1, 4) 傽 (2, 6]
46. [⫺1, 4) 傼 (2, 6]
47. (⫺⬁, 1) 傼 (⫺2, ⬁)
48. (⫺⬁, 1) 傽 (2, ⬁)
49. (⫺⬁, ⫺1) 傼 [3, 7)
50. (1, 6] ´ [9, ⬁)
51. [2, 3] 傼 (1, 5)
52. [2, 3] 傽 (1, 5)
In Problems 11–16, rewrite in interval notation and graph on a real number line.
53. (⫺⬁, 4) 傼 (⫺1, 6]
54. (⫺3, 2) 傼 [0, ⬁)
11. ⫺2 6 x ⱕ 6
12. ⫺5 ⱕ x ⱕ 5
13. ⫺7 6 x 6 8
In Problems 55–70, solve and graph.
14. ⫺4 ⱕ x 6 5
15. x ⱕ ⫺2
16. x 7 3
55.
q q⫺4 ⫺3 7 ⫹1 7 3
57.
2x 1 3 2x ⫺ (x ⫺ 3) ⱕ ⫺ (x ⫹ 2) 5 2 3 10
In Problems 5–10, rewrite in inequality notation and graph on a real number line. 5. [ ⫺8, 7 ]
6. (⫺4, 8)
7. [ ⫺6, 6)
8. (⫺3, 3 ]
9. [⫺6, ⬁)
10. (⫺⬁, 7)
In Problems 17–20, write in interval and inequality notation. 17. 18. 19. 20.
⫺10 ⫺10
[
[
⫺5
⫺10
⫺10
⫺5
0 0
5 5
]
⫺5
⫺5
)
0
0
(
] 5
5
x
10 10
x x
10
10
x
In Problems 21–28, replace each ? with ⬎ or ⬍ to make the resulting statement true. and
12 ⫹ 5 ? 6 ⫹ 5
22. ⫺4 ? ⫺2
and
⫺4 ⫺ 7 ? ⫺2 ⫺ 7
23. ⫺6 ? ⫺8
and
⫺6 ⫺ 3 ? ⫺8 ⫺ 3
21.
12 ? 6
56.
2 62. 24 ⱕ (x ⫺ 5) 6 36 3 63. 16 6 7 ⫺ 3x ⱕ 31 64. 19 ⱕ 7 ⫺ 6x 6 49 1 65. ⫺8 ⱕ ⫺ (2 ⫺ x) ⫹ 3 6 10 4
4?9
and
4⫹2?9⫹2
25.
2 ? ⫺1
and
⫺2(2) ? ⫺2(⫺1)
66. 0 6
and
4(⫺3) ? 4(2)
67. 0.1(x ⫺ 7) 6 0.8 ⫺ 0.05x
27.
2?6
28. ⫺10 ? ⫺15
and
2 6 ? 2 2
and
⫺10 ⫺15 ? 5 5
In Problems 29–42, solve and graph. 29. 7x ⫺ 8 6 4x ⫹ 7
30. 5x ⫺ 21 ⱖ 3x ⫹ 5
31. 12 ⫺ y ⱖ 2(9 ⫺ 2y)
32. 4(y ⫹ 1) ⫺ 7 6 ⫺9 ⫺ 2y
N 33. 7 4 ⫺2
Z 34. ⱕ3 ⫺10
35. ⫺5t 6 ⫺10
36. ⫺20m ⱖ 100
37. 3 ⫺ m 6 4(m ⫺ 3)
38. 6(5 ⫺ 2k) ⱖ 6 ⫺ 8k
39. ⫺2 ⫺
B 1⫹B ⱕ 4 3
41. ⫺4 6 5t ⫹ 6 ⱕ 21
40.
t t⫺2 ⫹2 7 5 3
42. ⫺2 ⱕ 4r ⫺ 14 6 2
In Problems 43–54, graph the indicated set and write as a single interval, if possible. 43. (⫺5, 5) 傼 [ 4, 7 ]
44. (⫺5, 5) 傽 [4, 7]
p⫺2 p p ⫺ ⱕ ⫺4 3 2 4
x 1 x 2 58. (x ⫹ 7) ⫺ 7 (3 ⫺ x) ⫹ 3 4 2 6 9 4 59. ⫺4 ⱕ x ⫹ 32 ⱕ 68 60. 2 ⱕ z ⫹ 6 6 18 5 5 5 61. ⫺20 6 (4 ⫺ x) 6 ⫺5 2
24.
26. ⫺3 ? 2
63
1 (4 ⫺ x) ⫺ 10 ⱕ 16 3
68. 0.4(x ⫹ 5) 7 0.3x ⫹ 17 69. 0.3x ⫺ 2.04 ⱖ 0.04(x ⫹ 1) 70. 0.02x ⫺ 5.32 ⱕ 0.5(x ⫺ 2) Problems 71–76 are calculus-related. For what real number(s) x does each expression represent a real number? 71. 11 ⫺ x
72. 1x ⫹ 5
73. 13x ⫹ 5
74. 17 ⫺ 2x
75.
1
76.
4
22x ⫹ 3
1 4
25 ⫺ 6x
77. What can be said about the signs of the numbers a and b in each case? (A) ab 7 0 (B) ab 6 0 a a (C) 7 0 (D) 6 0 b b 78. What can be said about the signs of the numbers a, b, and c in each case? ab (A) abc 7 0 (B) 6 0 c (C)
a 7 0 bc
(D)
a2 6 0 bc
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79. Replace each question mark with 6 or 7, as appropriate: (A) If a 0002 b 0004 1, then a ? b. (B) If u 0002 v 0004 00022, then u ? v. 80. For what p and q is p 0003 q 6 p 0002 q? 81. If both a and b are negative numbers and b0006a is greater than 1, then is a 0002 b positive or negative? 82. If both a and b are positive numbers and b0006a is greater than 1, then is a 0002 b positive or negative? 83. Indicate true (T) or false (F): (A) If p 7 q and m 7 0, then mp 6 mq. (B) If p 6 q and m 6 0, then mp 7 mq. (C) If p 7 0 and q 6 0, then p 0003 q 7 q. 84. Assume that m 7 n 7 0; then mn 7 n2 mn 0002 m2 7 n2 0002 m2 m(n 0002 m) 7 (n 0003 m)(n 0002 m) m 7 n0003m 0 7 n But it was assumed that n 7 0. Find the error. Prove each inequality property in Problems 85–88, given a, b, and c are arbitrary real numbers. 85. If a 6 b, then a 0003 c 6 b 0003 c. 86. If a 6 b, then a 0002 c 6 b 0002 c. 87. (A) If a 6 b and c is positive, then ca 6 cb. (B) If a 6 b and c is negative, then ca 7 cb. b a 6 . c c a b (B) If a 6 b and c is negative, then 7 . c c
88. (A) If a 6 b and c is positive, then
APPLICATIONS Write all your answers using inequality notation. 89. EARTH SCIENCE In 1970, Russian geologists began drilling a very deep borehole in the Kola Peninsula. Their goal was to reach a depth of 15 kilometers, but high temperatures in the borehole forced them to stop in 1994 after reaching a depth of 12 kilometers. They found that the approximate temperature x kilometers below the surface of the Earth is given by T 0004 30 0003 25(x 0002 3)
3 x 12
where T is temperature in degrees Celsius. At what depth is the temperature between 150°C and 250°C, inclusive? 90. EARTH SCIENCE As dry air moves upward it expands, and in so doing it cools at a rate of about 5.5°F for each 1,000-foot rise up to about 40,000 feet. If the ground temperature is 70°F, then the temperature T at height h is given approximately by T 0004 70 0002 0.0055h.
For what range in altitude will the temperature be between 26°F and 000240°F, inclusive? 91. BUSINESS AND ECONOMICS An electronics firm is planning to market a new graphing calculator. The fixed costs are $650,000 and the variable costs are $47 per calculator. The wholesale price of the calculator will be $63. For the company to make a profit, it is clear that revenues must be greater than costs. (A) How many calculators must be sold for the company to make a profit? (B) How many calculators must be sold for the company to break even? (C) Discuss the relationship between the results in parts A and B. 92. BUSINESS AND ECONOMICS A video game manufacturer is planning to market a handheld version of its game machine. The fixed costs are $550,000 and the variable costs are $120 per machine. The wholesale price of the machine will be $140. (A) How many game machines must be sold for the company to make a profit? (B) How many game machines must be sold for the company to break even? (C) Discuss the relationship between the results in parts A and B. 93. BUSINESS AND ECONOMICS The electronics firm in Problem 91 finds that rising prices for parts increases the variable costs to $50.50 per calculator. (A) Discuss possible strategies the company might use to deal with this increase in costs. (B) If the company continues to sell the calculators for $63, how many must they sell now to make a profit? (C) If the company wants to start making a profit at the same production level as before the cost increase, how much should they increase the wholesale price? 94. BUSINESS AND ECONOMICS The video game manufacturer in Problem 92 finds that unexpected programming problems increases the fixed costs to $660,000. (A) Discuss possible strategies the company might use to deal with this increase in costs. (B) If the company continues to sell the game machines for $140, how many must they sell now to make a profit? (C) If the company wants to start making a profit at the same production level as before the cost increase, how much should they increase the wholesale price? 95. ENERGY If the power demands in a 110-volt electric circuit in a home vary between 220 and 2,750 watts, what is the range of current flowing through the circuit? (W 0004 EI, where W 0004 Power in watts, E 0004 Pressure in volts, and I 0004 Current in amperes.) 96. PSYCHOLOGY A person’s IQ is given by the formula IQ 0004
MA 100 CA
where MA is mental age and CA is chronological age. If 80 IQ 140 for a group of 12-year-old children, find the range of their mental ages.
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Absolute Value in Equations and Inequalities
65
Absolute Value in Equations and Inequalities Z Relating Absolute Value and Distance Z Solving Absolute Value Equations and Inequalities Z Using Absolute Value to Solve Radical Inequalities
We can express the distance between two points on a number line using the concept of absolute value. As a result, absolute values often appear in equations and inequalities that are associated with distance. In this section, we define absolute value and we show how to solve equations and inequalities that involve absolute value.
Z Relating Absolute Value and Distance We start with a geometric definition of absolute value. If a is the coordinate of a point on a real number line, then the distance from the origin to a is represented by |a| and is referred to as the absolute value of a. So |5| 0002 5, since the point with coordinate 5 is five units from the origin, and 冟 00036 冟 0002 6, since the point with coordinate 00036 is six units from the origin (Fig. 1).
兩00036兩 0002 6 00036
兩5兩 0002 5 0
5
x
Z Figure 1 Absolute value.
We can use symbols to write a formal definition of absolute value:
Z DEFINITION 1 Absolute Value 冟x冟 0002
再
0003x x
if x 6 0 if x 0004 0
For example, 冟00033冟
0002 0003(00033)
00023
For example, 冟4冟 0002 4
[Note: 0003x is positive if x is negative.]
Both the geometric and algebraic definitions of absolute value are useful, as will be seen in the material that follows. Remember: The absolute value of a number is never negative.
EXAMPLE
1
Finding Absolute Value Write without the absolute value sign: (A) 冟 0005 0003 3 冟
(B) 冟 3 0003 0005 冟
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(A) 冟 0005 0003 3 冟 0002 0005 0003 3
Because 0005 ⬇ 3.14, 0005 0003 3 is positive.
(B) 冟 3 0003 0005 冟 0002 0003(3 0003 0005) 0002 0005 0003 3
Because 3 0003 0005 is negative.
0002
Write without the absolute value sign: (A) 冟 8 冟
3 (B) 冟 29 0003 2 冟
(C) 冟 000312 冟
3 (D) 冟 2 0003 29 冟
0002
Notice that the solution in both parts of Example 1 was the same. This suggests Theorem 1, which will be proved in Problem 81.
Z THEOREM 1 For All Real Numbers a and b, 冟b 0003 a冟 0002 冟a 0003 b冟
To find the distance between two numbers, we subtract, larger minus smaller. But if we don’t know which is larger, we can use absolute value; Theorem 1 tells us that the order is immaterial.
Z DEFINITION 2 Distance Between Points A and B Let A and B be two points on a real number line with coordinates a and b, respectively. The distance between A and B is given by d(A, B) 0002 冟 b 0003 a 冟 This distance is also called the length of the line segment joining A and B.
It will come in very handy to observe that an expression like 冟 b 0003 a 冟 can always be interpreted as the distance between two numbers a and b, and that the order of the subtraction doesn’t matter.
Z Solving Absolute Value Equations and Inequalities The connection between algebra and geometry is an important tool when working with equations and inequalities involving absolute value. For example, the algebraic statement 冟x 0003 1冟 0002 2 can be interpreted geometrically as stating that the distance from x to 1 is 2.
ZZZ EXPLORE-DISCUSS 1
Write geometric interpretations of the following algebraic statements: (A) 冟 x 0003 1 冟 6 2
(B) 0 6 冟 x 0003 1 冟 6 2
(C) 冟 x 0003 1 冟 7 2
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EXAMPLE
2
Absolute Value in Equations and Inequalities
67
Solving Absolute Value Problems Geometrically Interpret geometrically, solve, and graph. For inequalities, write solutions in both inequality and interval notation.
SOLUTIONS
(A) 冟 x 0003 3 冟 0002 5
(B) 冟 x 0003 3 冟 6 5
(C) 0 6 冟 x 0003 3 冟 6 5
(D) 冟 x 0003 3 冟 7 5
(A) The expression |x 0003 3| represents the distance between x and 3, so the solutions to |x 0003 3| 0002 5 are all numbers that are exactly 5 units away from 3 on a number line. x 0002 3 5 0002 00032 or 8 The solution set is {00032, 8}. 5
This is not interval notation.
5
00032
3
x
8
(B) Solutions to |x 0003 3| 0006 5 are all numbers whose distance from 3 is less than 5. These are the numbers between 00032 and 8: 00032 6 x 6 8 The solution set is (00032, 8). (
00032
This is interval notation.
)
3
x
8
(C) Expressions like 0 0006 |x 0003 3| 0006 5 are important in calculus. The solutions are all numbers whose distance from 3 is less than 5, and is not zero. This excludes 3 itself from the solution set: 00032 6 x 6 8
x3
or
(00032, 3) 0004 (3, 8)
Hole
(
00032
)
3
x
8
(D) The solutions to 冟 x 0003 3 冟 7 5 are all numbers whose distance from 3 is greater than 5; that is, x 6 00032 )
00032
ZZZ
CAUTION ZZZ
3
or
x 7 8
(00030007, 00032) 0004 (8, 0007)
(
8
The pair of inequalities 00032 0006 x and x 0006 8 can be written as a double inequality: 00032 6 x 6 8 or in interval notation (00032, 8) But the pair x 0006 00032 or x 8 from Example 2(D) cannot be written as a double inequality, or as a single interval: no number is both less than 00032 and greater than 8.
0002
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MATCHED PROBLEM 2
Interpret geometrically, solve, and graph. For inequalities, write solutions in both inequality and interval notation. [Hint: |x ⫹ 2| ⫽ |x ⫺ (⫺2)|.] (A) 冟 x ⫹ 2 冟 ⫽ 6
(B) 冟 x ⫹ 2 冟 6 6
(C) 0 6 冟 x ⫹ 2 冟 6 6
(D) 冟 x ⫹ 2 冟 7 6
0002
The preceding results are summarized in Table 1. Table 1 Geometric Interpretation of Absolute Value Equations and Inequalities Form (d ⬎ 0)
冟x ⫺ c冟 ⫽ d
Geometric interpretation
Solution
Graph
c⫺d
冟x ⫺ c冟 6 d
Distance between x and c is less than d.
(c ⫺ d, c ⫹ d )
0 6 冟x ⫺ c冟 6 d
Distance between x and c is less than d, but x ⫽ c.
(c ⫺ d, c) 傼 (c, c ⫹ d )
Distance between x and c is greater than d.
(⫺⬁, c ⫺ d ) 傼 (c ⫹ d, ⬁)
冟x ⫺ c冟 7 d
EXAMPLE
3
d
5c ⫺ d, c ⫹ d6
Distance between x and c is equal to d.
(
c⫺d
(
c⫺d
)
c⫺d
d c
c⫹d
c
c⫹d
c
c⫹d
c
c⫹d
x
)
x
)
x
(
x
Interpreting Verbal Statements Algebraically Express each verbal statement as an absolute value equation or inequality. (A) x is 4 units from 2. (B) y is less than 3 units from ⫺5. (C) t is no more than 5 units from 7. (D) w is no less than 2 units from ⫺1.
SOLUTIONS
MATCHED PROBLEM 3
⫽
|x ⫺ 2| ⫽ 4
The distance from x to 2 is 4.
(B)
d( y, ⫺5) ⫽
| y ⫹ 5| 6 3
The distance from y to 00025 is 0003 3.
(C)
d(t, 7)
⫽
|t ⫺ 7| ⱕ 5
The distance from t to 7 is 0004 5.
(D) d(w, ⫺1) ⫽
|w ⫹ 1| ⱖ 2
The distance from w to 00021 is 0004 2.
(A) d(x, 2)
0002
Express each verbal statement as an absolute value equation or inequality. (A) x is 6 units from 5. (B) y is less than 7 units from ⫺6. (C) w is no less than 3 units from ⫺2. (D) t is no more than 4 units from 3. 0002
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ZZZ EXPLORE-DISCUSS 2
Absolute Value in Equations and Inequalities
69
Describe the set of numbers that satisfies each of the following: (A) 2 7 x 7 1
(B) 2 7 x 6 1
(C) 2 6 x 7 1
(D) 2 6 x 6 1
Explain why we never write double inequalities with inequality symbols pointing in different directions.
The results of Example 2 can be generalized as Theorem 2. [Note: |x| 0002 |x 0003 0|.]
Z THEOREM 2 Properties of Equations and Inequalities Involving 冟x冟 For p 0:
p has to be positive!
1. 冟 x 冟 0002 p is equivalent to
x0002p
or
The distance from x to zero is p.
x 0002 0003p.
0003p
2. 冟 x 冟 6 p is equivalent to 0003p 6 x 6 p.
(
0003p
The distance from x to zero is less than p.
3. 冟 x 冟 7 p is equivalent to
x 6 0003p
or
x 7 p.
)
0003p
The distance from x to zero is greater than p.
0
p
0
p
0
p
x
)
x
(
x
If we replace x in Theorem 2 with ax b, we obtain the more general Theorem 3.
Z THEOREM 3 Properties of Equations and Inequalities Involving |ax b| For p 0:
EXAMPLE
4
p has to be positive!
1. 冟 ax b 冟 0002 p
is equivalent to
ax b 0002 p
2. 冟 ax b 冟 6 p
is equivalent to
0003p 6 ax b 6 p.
3. 冟 ax b 冟 7 p
is equivalent to
ax b 6 0003p
ax b 0002 0003p.*
or or
ax b 7 p.
Solving Absolute Value Problems Solve each equation or inequality. For inequalities, write solutions in both inequality and interval notation. (A) 冟 3x 5 冟 0002 4
(B) 冟 x 冟 6 5
(C) 冟 2x 0003 1 冟 6 3
(D) 冟 7 0003 3x 冟 2
*This can be more concisely written as ax b 0002 ; p.
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SOLUTIONS
(A) 冟 3x 5 冟 0002 4 Use Theorem 3, part 1 3x 5 0002 4 3x 0002 00035 4 00035 4 x0002 3 x 0002 00033, 000313
(B) 冟 x 冟 6 5 Use Theorem 2, part 2 00035 6 x 6 5 or (00035, 5)
or 500033, 000313 6 (C) 冟 2x 0003 1 冟 6 3 Use Theorem 3, part 2 00033 6 2x 0003 1 6 3 00032 6 2x 6 4 00031 6 x 6 2 or (00031, 2)
MATCHED PROBLEM 4
5
0002
Solve each equation or inequality. For inequalities, write solutions in both inequality and interval notation. (A) 冟 2x 0003 1 冟 0002 8
EXAMPLE
(D) 冟 7 0003 3x 冟 2 Use Theorem 3, part 2 00032 7 0003 3x 2 00039 00033x 00035 3 0004 x 0004 53 5 3 x 3 or [ 53, 3]
(B) 冟 x 冟 7
(C) 冟 3x 3 冟 9
(D) 冟 5 0003 2x 冟 6 9
0002
Solving Absolute Value Inequalities Solve, and write solutions in both inequality and interval notation. (A) 冟 x 冟 7 3
SOLUTIONS
(B) 冟 2x 0003 1 冟 0004 3
(A) 冟 x 冟 7 3 or x 7 3 x 6 00033 (00030007, 00033) 0004 (3, 0007) (B) 冟 2x 0003 1 冟 0004 3 2x 0003 1 00033 2x 00032 x 00031 (00030007, 00031] 0004 (C) 冟 7 0003 3x 冟 7 2 7 0003 3x 6 00032 00033x 6 00039
Use Theorem 2, part 3. Solution in inequality notation Solution in interval notation Use Theorem 3, part 3.
or 2x 0003 1 0004 3 or 2x 0004 4 or x00042 [2, 0007)
Add 1 to both sides. Divide both sides by 2. Solution in inequality notation Solution in interval notation Use Theorem 3, part 3.
or or
7 0003 3x 7 2 00033x 7 00035
x 7 3 or (00030007, 53) 0004 (3, 0007)
MATCHED PROBLEM 5
(C) 冟 7 0003 3x 冟 7 2
x 6
5 3
Subtract 7 from both sides. Divide both sides by 00033 and reverse the direction of the inequality. Solution in inequality notation Solution in interval notation
0002
Solve, and write solutions in both inequality and interval notation. (A) 冟 x 冟 0004 5
(B) 冟 4x 0003 3 冟 7 5
(C) 冟 6 0003 5x 冟 7 16
0002
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EXAMPLE
6
Absolute Value in Equations and Inequalities
71
An Absolute Value Problem with Two Cases Solve: 冟 x 4 冟 0002 3x 0003 8
SOLUTION
We can’t use Theorem 3 directly, because we don’t know that 3x 0003 8 is positive. However, we can use the definition of absolute value and two cases: x 4 0004 0 and x 4 0006 0. Case 1. x 4 0004 0 (in which case, x 0004 00034) For this case, the only acceptable values of x are greater than or equal to 00034. 冟 x 4 冟 0002 3x 0003 8 x 4 0002 3x 0003 8 00032x 0002 000312 x00026
If x 0005 4 is positive, |x 0005 4| 0002 x 0005 4. Subtract 3x and 4 from both sides. Divide both sides by 00032. A solution, because 6 is among the acceptable values of x (6 0004 00024).
Case 2. x 4 0006 0 (in which case, x 0006 00034) In this case, the only acceptable values of x are less than 00034. 冟 x 4 冟 0002 3x 0003 8 0003(x 4) 0002 3x 0003 8 0003x 0003 4 0002 3x 0003 8 00034x 0002 00034 x00021
If x 0005 4 is negative, |x 0005 4| 0002 0003(x 0005 4). Distribute 00021. Subtract 3x and add 4 to both sides. Divide both sides by 00034. Not a solution, since 1 is not among the acceptable values of x (1 0005 00024).
Combining both cases, we see that the only solution is x 0002 6. As a final check, we substitute x 0002 6 and x 0002 1 in the original equation. 冟 x 4 冟 0002 3x 0003 8 ? 冟 6 4 冟 0002 3(6) 0003 8 ✓ 10 000210 MATCHED PROBLEM 6
冟 x 4 冟 0002 3x 0003 8 ? 冟 1 4 冟 0002 3(1) 0003 8 5 00035
Solve: 冟 3x 0003 4 冟 0002 x 5
0002
0002
Z Using Absolute Value to Solve Radical Inequalities In Section R-2, we found that if x is positive or zero, 2x2 0002 x. But what if x is negative? Let’s look at an example: 2(00032)2 0002 14 0002 2 We see that for negative x, 2x2 0002 0003x. So for any real number, 2x2 0002
0003x x
冦
if x 6 0 if x 0004 0
But this is exactly how we defined 冟 x 冟 at the beginning of this section (see Definition 1). So for any real number x, 2x2 0002 冟 x 冟
(1)
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7
Solving a Radical Inequality Solve the inequality. Write your answer in both inequality and interval notation. 2(x 0003 2)2 5 2(x 0003 2)2 5 冟x 0003 2冟 5
SOLUTION
Use equation (1): 2(x 0003 2)2 0002 冟x 0003 2冟 Use Theorem 3, part 2.
00035 x 0003 2 5 00033 x 7 or [00033, 7] MATCHED PROBLEM 7
Add 2 to each member. Solution in inequality notation
0002
Solution in interval notation
Solve the inequality. Write your answers in both inequality and interval notation. 2(x 2)2 6 3 0002
ANSWERS TO MATCHED PROBLEMS 3 3 1. (A) 8 (B) 2 (C) 12 (D) 2 900032 900032 2. (A) x is a number whose distance from 00032 is 6. x 0002 00038, 4 or 500038, 46
00038
00032
x
4
(B) x is a number whose distance from 00032 is less than 6. 00038 6 x 6 4 or (00038, 4) ( ) 00038
00032
4
x
(C) x is a number whose distance from 00032 is less than 6, but x cannot equal 00032. 00038 6 x 6 4, x 00032, or (00038, 00032) 0004 (00032, 4) ( ) (D) x is a number whose distance from 00032 is greater than 6. x 6 00038 or x 7 4, or (00030007, 00038) 0004 (4, 0007) ) 00038
00038
00032
00032
( 4
4
x
x
3. (A) 冟 x 0003 5 冟 0002 6 (B) 冟 y 6 冟 6 7 (C) 冟 w 2 冟 0004 3 (D) 冟 t 0003 3 冟 4 4. (A) x 0002 000372, 92 or 5000372, 92 6 (B) 00037 x 7 or [ 00037, 7] (C) 00034 x 2 or [ 00034, 2] (D) 00032 6 x 6 7 or (00032, 7) 5. (A) x 00035 or x 0004 5, or (00030007, 00035] 0004 [5, 0007) (B) x 6 000312 or x 7 2, or (00030007, 000312) 0004 (2, 0007) 22 22 (C) x 6 00032 or x 7 5 , or (00030007, 00032) 0004 ( 5 , 0007) 6. x 0002 000314, 92 or 5000314, 92 6 7. 00035 6 x 6 1 or (00035, 1)
1-3
Exercises
1. Describe how to find the absolute value of a number, then explain how your description matches Definition 1.
4. Repeat Problem 3 for the inequalities |x 0003 5| 0006 10 and |x 0003 5| 10.
2. Explain what the expression |x 0003 5| represents geometrically, and why.
5. Explain why it is incorrect to say that 2x2 0002 x.
3. Describe the equation |x 0003 5| 0002 10 in terms of your answer to Problem 2, then explain how that helps you to solve it.
6. Why can’t the following be a legitimate solution to an inequality? x 0006 1 and x 5.
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In Problems 7–14, simplify, and write without absolute value signs. Do not replace radicals with decimal approximations. 8. 冟 ⫺34 冟
7. 冟 15 冟 9. 冟 (⫺6) ⫺ (⫺2) 冟
10. 冟 (⫺2) ⫺ (⫺6) 冟
11. 冟 5 ⫺ 15 冟
12. 冟 17 ⫺ 2 冟
13. 冟 15 ⫺ 5 冟
14. 冟 2 ⫺ 17 冟
Absolute Value in Equations and Inequalities
55. 2x2 6 2
56. 2m2 7 3
57. 2(1 ⫺ 3t)2 ⱕ 2
58. 2(3 ⫺ 2x)2 6 5
59. 2(2t ⫺ 3)2 7 3
60. 2(3m ⫹ 5)2 ⱖ 4
73
In Problems 61–64, solve and write answers in inequality notation. Round decimals to three significant digits. 61. 冟 2.25 ⫺ 1.02x 冟 ⱕ 1.64
In Problems 15–20, use the number line shown to find the indicated distances. A
B
⫺10
⫺5
O
C
0
5
D 10
15. d(B, O)
16. d(A, B)
17. d(O, B)
18. d(B, A)
19. d(B, C)
20. d(D, C)
62. 冟 0.962 ⫺ 0.292x 冟 ⱕ 2.52 63. 冟 21.7 ⫺ 11.3x 冟 ⫽ 15.2 x
Write each of the statements in Problems 21–30 as an absolute value equation or inequality. 21. x is 4 units from 3. 22. y is 3 units from 1.
64. 冟 195 ⫺ 55.5x 冟 ⫽ 315 Problems 65⫺68 involve expressions that are important in the study of limits in calculus. First, provide a verbal translation of the inequality. Then solve and graph, writing your solution in interval notation. 65. 0 6 冟 x ⫺ 3 冟 6 0.1 1 67. 0 6 冟 x ⫺ a 冟 6 10
66. 0 6 冟 x ⫹ 5 冟 6 0.5 68. 0 6 冟 x ⫺ 8 冟 6 d
23. m is 5 units from ⫺2.
In Problems 69–76, for what values of x does each hold?
24. n is 7 units from ⫺5.
69. 冟 x ⫺ 2 冟 ⫽ 2x ⫺ 7
70. 冟 x ⫹ 4 冟 ⫽ 3x ⫹ 8
25. x is less than 5 units from 3.
71. 冟 3x ⫹ 5 冟 ⫽ 2x ⫹ 6
72. 冟 7 ⫺ 2x 冟 ⫽ 5 ⫺ x
26. z is less than 8 units from ⫺2.
73. 冟 x 冟 ⫹ 冟 x ⫹ 3 冟 ⫽ 3
74. 冟 x 冟 ⫺ 冟 x ⫺ 5 冟 ⫽ 5
27. p is more than 6 units from ⫺2.
75. 冟 3 ⫺ x 冟 ⫽ 2 (4 ⫹ x)
28. c is no greater than 7 units from ⫺3.
76. 冟 5 ⫺ 2x 冟 ⫽ 4(x ⫺ 5)
29. q is no less than 2 units from 1.
77. What are the possible values of
x ? 冟x冟
78. What are the possible values of
冟x ⫺ 1冟 ? x⫺1
30. d is no more than 4 units from 5. In Problems 31–42, solve, interpret geometrically, and graph. When applicable, write answers using both inequality notation and interval notation.
79. Explain why 冟 ax ⫹ b 冟 6 ⫺3 has no solution for any values of a and b.
31. 冟 y ⫺ 5 冟 ⫽ 3
32. 冟 t ⫺ 3 冟 ⫽ 4
33. 冟 y ⫺ 5 冟 6 3
34. 冟 t ⫺ 3 冟 6 4
35. 冟 y ⫺ 5 冟 7 3
36. 冟 t ⫺ 3 冟 7 4
80. Explain why 冟 ax ⫹ b 冟 7 ⫺3 has solution all real numbers for any values of a and b.
37. 冟 u ⫹ 8 冟 ⫽ 3
38. 冟 x ⫹ 1 冟 ⫽ 5
39. 冟 u ⫹ 8 冟 ⱕ 3
81. Prove that 冟 b ⫺ a 冟 ⫽ 冟 a ⫺ b 冟 for all real numbers a and b. [Hint: Apply Definition 1 and use cases.]
40. 冟 x ⫹ 1 冟 ⱕ 5
41. 冟 u ⫹ 8 冟 ⱖ 3
42. 冟 x ⫹ 1 冟 ⱖ 5
82. Prove that 冟 x 冟2 ⫽ x2 for all real numbers x.
In Problems 43–60, solve the equation or inequality. Write solutions to inequalities using both inequality and interval notation. 43. 冟 2x ⫺ 11 冟 ⱕ 13
44. 冟 5x ⫹ 20 冟 ⱖ 5
45. 冟 100 ⫺ 40t 冟 7 60
46. 冟 150 ⫺ 20y 冟 6 10
47. 冟 4x ⫺ 7 冟 ⫽ 13
48. 冟 ⫺8x ⫹ 3 冟 ⱕ 91
49. 冟 12w
50. 冟 13z ⫹ 56 冟 ⫽ 1
⫺
3 4冟
6 2
51. 冟 0.2u ⫹ 1.7 冟 ⱖ 0.5 53. 冟 95 C
⫹ 32 冟 6 31
52. 冟 0.5v ⫺ 2.5 冟 7 1.6 54.
冟 59 (F
⫺ 32) 冟 6 40
83. Prove that the average of two numbers is between the two numbers; that is, if m 6 n, then m 6
m⫹n 6 n 2
84. Prove that for m 6 n, d am,
m⫹n m⫹n b ⫽ da , nb 2 2
85. Prove that 冟 ⫺m 冟 ⫽ 冟 m 冟. 86. Prove that 冟 m 冟 ⫽ 冟 n 冟 if and only if m ⫽ n or m ⫽ ⫺n.
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87. Prove that for n 0, `
冟m冟 m `0002 冟n冟 n
88. Prove that 冟 mn 冟 0002 冟 m 冟冟 n 冟.
94. CHEMISTRY In order to manufacture a polymer for soft drink containers, a chemical reaction must take place within 10°C of 200°C. Write this temperature restriction as an absolute value inequality, then solve to find the acceptable temperatures.
89. Prove that 0003冟 m 冟 m 冟 m 冟. 90. Prove the triangle inequality: 冟m n冟 冟m冟 冟n冟 Hint: Use Problem 89 to show that 0003冟 m 冟 0003 冟 n 冟 m n 冟 m 冟 冟 n 冟
APPLICATIONS 91. STATISTICS Inequalities of the form `
x0003m ` 6 n s
occur frequently in statistics. If m 0002 45.4, s 0002 3.2, and n 0002 1, solve for x. 92. STATISTICS Repeat Problem 91 for m 0002 28.6, s 0002 6.5, and n 0002 2.
1-4
93. BUSINESS The daily production P in an automobile assembly plant is always within 20 units of 500 units. Write the daily production as an absolute value inequality, then solve to find the range of daily productions possible.
;
95. APPROXIMATION The area A of a region is approximately equal to 12.436. The error in this approximation is less than 0.001. Describe the possible values of this area both with an absolute value inequality and with interval notation. 96. APPROXIMATION The volume V of a solid is approximately equal to 6.94. The error in this approximation is less than 0.02. Describe the possible values of this volume both with an absolute value inequality and with interval notation. 97. SIGNIFICANT DIGITS If N 0002 2.37 represents a measurement, then we assume an accuracy of 2.37 0.005. Express the accuracy assumption using an absolute value inequality. 98. SIGNIFICANT DIGITS If N 0002 3.65 000e 1000033 is a number from a measurement, then we assume an accuracy of 3.65 000e 1000033 5 000e 1000036. Express the accuracy assumption using an absolute value inequality.
Complex Numbers Z Understanding Complex Number Terminology Z Performing Operations with Complex Numbers Z Relating Complex Numbers and Radicals Z Solving Equations Involving Complex Numbers
The idea of inventing new numbers might seem odd to you, but think about this example: A group of mathematicians known as the Pythagoreans proved over 2,000 years ago that the equation x2 0002 2 has no solutions that are rational numbers. You may be thinking that the solutions are 12 and 000312, but at the time, those numbers had not been defined, so the Pythagoreans invented a new kind of number—irrational numbers, like 12 and 000312. Now consider the similar equation x2 0002 00031. To be a solution, a number has to result in 00031 when squared. But we know that the square of any real number cannot be negative, so again a new type of number is invented—a number whose square is 00031. The concept of square roots of negative numbers had been kicked around for a few centuries, but in 1748, the Swiss mathematician Leonhard Euler (pronounced “Oiler”) used the letter i to represent a square root of 00031. From this simple beginning, it is possible to build a new system of numbers called the complex number system.
Z Understanding Complex Number Terminology The number i, whose square is 00031, is called the imaginary unit. Complex numbers are defined in terms of the imaginary unit.
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Complex Numbers
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Z DEFINITION 1 Complex Number A complex number is a number of the form a bi, where a and b are real numbers, and i is the imaginary unit (a square root of 00031). A complex number written this way is said to be in standard form. The real number a is called the real part, and bi is called the imaginary part.
Some examples of complex numbers are 3 0003 2i 0 3i
1 2
2 0003 13i 0 0i
5i 5 0i
The notation 3 0003 2i is shorthand for 3 (00032)i. Particular kinds of complex numbers are given special names as follows:
Z DEFINITION 2 Special Terms a 0005 bi b0 0 0005 bi 0002 bi b0 a 0005 0i 0002 a 0 0002 0 0005 0i a 0003 bi
EXAMPLE
1
Imaginary Number Pure Imaginary Number Real Number Zero Conjugate of a 0005 bi
Complex Numbers Identify the real part, the imaginary part, and the conjugate of each of the following numbers: (A) 3 0003 2i
SOLUTIONS
(B) 2 5i
(C) 7i
(D) 6
(A) Real part: 3; imaginary part: 00032i; conjugate: 3 2i (B) Real part: 2; imaginary part: 5i; conjugate: 2 0003 5i (C) Real part: 0; imaginary part: 7i; conjugate: 00037i (D) Real part: 6; imaginary part: 0; conjugate: 6
MATCHED PROBLEM 1
0002
Identify the real part, the imaginary part, and the conjugate of each of the following numbers: (A) 6 7i
(B) 00033 0003 8i
(C) 00034i
(D) 00039 0002
We will identify a complex number of the form a 0i with the real number a, a complex number of the form 0 bi, b 0, with the pure imaginary number bi, and the complex number 0 0i with the real number 0. So a real number is also a complex number, just as a rational number is also a real number. Any complex number that is not a real number is called an imaginary number. If we combine the set of all real numbers with the set of all imaginary numbers, we obtain C, the set of complex numbers. The relationship of the complex number system to the other number systems we have studied is shown in Figure 1.
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Z Figure 1 Natural numbers (N ) Zero Negative integers
N0006Z0006Q0006R0006C Integers (Z ) Noninteger rational numbers
Rational numbers (Q) Irrational numbers (I )
Real numbers (R ) Imaginary numbers
Complex numbers (C)
Z Performing Operations with Complex Numbers To work with complex numbers, we will need to know how to add, subtract, multiply, and divide them. We start by defining equality, addition, and multiplication.
Z DEFINITION 3 Equality and Basic Operations 1. Equality: a bi 0002 c di if and only if a 0002 c and b 0002 d 2. Addition: (a bi) (c di) 0002 (a c) (b d )i 3. Multiplication: (a bi)(c di) 0002 (ac 0003 bd) (ad bc)i
In Section R-1 we listed the basic properties of the real number system. Using Definition 3, it can be shown that the complex number system possesses the same properties. That is, 1. 2. 3. 4. 5.
Addition and multiplication of complex numbers are commutative and associative operations. There is an additive identity and a multiplicative identity for complex numbers. Every complex number has an additive inverse or negative. Every nonzero complex number has a multiplicative inverse or reciprocal. Multiplication distributes over addition.
This is actually really good news: it tells us that we don’t have to memorize the formulas for adding and multiplying complex numbers in Definition 3. Instead: We can treat complex numbers of the form a 0005 bi exactly as we treat algebraic expressions of the form a 0005 bx. We just need to remember that in this case, i stands for the imaginary unit; it is not a variable that represents a real number. The first two arithmetic operations we consider are addition and subtraction.
EXAMPLE
2
Addition and Subtraction of Complex Numbers Carry out each operation and express the answer in standard form: (A) (2 0003 3i) (6 2i) (C) (7 0003 3i) 0003 (6 2i)
(B) (00035 4i) (0 0i) (D) (00032 7i) (2 0003 7i)
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SOLUTIONS
Complex Numbers
77
(A) We could apply the definition of addition directly, but it is easier to use complex number properties. (2 0003 3i) (6 2i) 0002 2 0003 3i 6 2i
Use the commutative property.
0002 (2 6) (00033 2)i
Combine like terms.
000280003i (B) (00035 4i) (0 0i) 0002 00035 4i 0 0i 0002 00035 4i (C) (7 0003 3i) 0003 (6 2i) 0002 7 0003 3i 0003 6 0003 2i 0002 1 0003 5i
Make sure you distribute the negative sign!
(D) (00032 7i) (2 0003 7i) 0002 00032 7i 2 0003 7i 0002 0
MATCHED PROBLEM 2
0002
Carry out each operation and express the answer in standard form: (A) (3 2i) (6 0003 4i)
(B) (0 0i) (7 0003 5i)
(C) (3 0003 5i) 0003 (1 0003 3i)
(D) (00034 9i) (4 0003 9i)
0002
Example 2, part B, illustrates the following general property: For any complex number a bi, (a bi) (0 0i) 0002 a bi and
(0 0i) (a bi) 0002 a bi
That is, 0 0i is the additive identity or zero for the complex numbers. This is why we identify 0 0i with the real number zero in Definition 2. Example 2, part D, illustrates a different result: In general, the additive inverse or negative of a bi is 0003a 0003 bi because (a bi) (0003a 0003 bi) 0002 0 and (0003a 0003 bi) (a bi) 0002 0 Now we turn our attention to multiplication. Just like addition and subtraction, multiplication of complex numbers can be carried out by treating a bi in the same way we treat the algebraic expression a bx. The key difference is that we replace i2 with 00031 each time it occurs.
EXAMPLE
3
Multiplying Complex Numbers Carry out each operation and express the answer in standard form:
SOLUTIONS
(A) (2 0003 3i)(6 2i)
(B) 1(3 0003 5i)
(C) i(1 i)
(D) (3 4i)(3 0003 4i)
(A) (2 0003 3i)(6 2i)
(B) 1(3 0003 5i)
0002 12 4i 0003 18i 0003 6i2 0002 12 0003 14i 0003 6(00031) 0002 18 0003 14i
0002 1 0007 3 0003 1 0007 5i
Replace i 2 with 00031. 00036(00031) 0002 6; combine like terms.
0002 3 0003 5i
(C) i(1 i) 0002 i i2 0002 i 0003 1 0002 00031 i (D) (3 4i)(3 0003 4i) 0002 9 0003 12i 12i 0003 16i2 0002 9 16 0002 25
Answer in standard form. 000316i 2 0002 000316(00031) 0002 16
0002
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MATCHED PROBLEM 3
Carry out each operation and express the answer in standard form: (A) (5 2i)(4 0003 3i)
(B) 3(00032 6i)
(C) i(2 0003 3i)
(D) (2 3i)(2 0003 3i) 0002
For any complex number a bi, 1(a bi) 0002 a bi and (a bi)1 0002 a bi (see Example 3, part B). This indicates that 1 is the multiplicative identity for complex numbers, just as it is for real numbers. Earlier we stated that every nonzero complex number has a multiplicative inverse or reciprocal. We will denote this as a fraction, just as we do with real numbers: 1 a bi
a bi
is the reciprocal of
a bi 0
The following important property of the conjugate of a complex number is used to express reciprocals and quotients in standard form. (See Example 3, part D)
Z THEOREM 1 Product of a Complex Number and Its Conjugate (a bi)(a 0003 bi) 0002 a2 b2
A real number
To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator.
EXAMPLE
4
Reciprocals and Quotients Write each expression in standard form: (A)
SOLUTIONS
1 2 3i
(B)
7 0003 3i 1000bi
(A) Multiply numerator and denominator by the conjugate of the denominator: 1 1 2 0003 3i 0002 0007 2 3i 2 3i 2 0003 3i 0002
0002
2 3 2 0003 3i 0002 0003 i 13 13 13
2 0003 3i 2 0003 3i 0002 4000b9 4 0003 9i2 Answer in standard form.
This answer can be checked by multiplication: CHECK
(2 3i) a
2 3 4 6 6 9 2 0003 ib 0002 0003 i i0003 i 13 13 13 13 13 13 4 9 0002 00021 ✓ 13 13
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(B)
7 ⫺ 3i 7 ⫺ 3i 1 0003 i ⫽ 0002 1⫹i 1⫹i 10003i ⫽
CHECK
MATCHED PROBLEM 4
⫽
7 ⫺ 7i ⫺ 3i ⫹ 3i2 1 ⫺ i2
4 ⫺ 10i ⫽ 2 ⫺ 5i 2
Complex Numbers
79
3i 2 0004 00033
Answer in standard form.
(1 ⫹ i)(2 ⫺ 5i) ⫽ 2 ⫺ 5i ⫹ 2i ⫺ 5i2 ⫽ 7 ⫺ 3i
0002
Carry out each operation and express the answer in standard form: (A)
1 4 ⫹ 2i
(B)
6 ⫹ 7i 2⫺i 0002
EXAMPLE
5
Combined Operations Carry out the indicated operations and write each answer in standard form: (A) (3 ⫺ 2i)2 ⫺ 6(3 ⫺ 2i) ⫹ 13
SOLUTIONS
2 ⫺ 3i 2i
(B)
(A) (3 ⫺ 2i)2 ⫺ 6(3 ⫺ 2i) ⫹ 13 ⫽ 9 ⫺ 12i ⫹ 4i2 ⫺ 18 ⫹ 12i ⫹ 13 ⫽ 9 ⫺ 12i ⫺ 4 ⫺ 18 ⫹ 12i ⫹ 13 ⫽0 (B) If a complex number is divided by a pure imaginary number, we can make the denominator real by multiplying numerator and denominator by i. (We could also multiply by the conjugate of 2i, which is ⫺2i.) 2i ⫺ 3i2 2i ⫹ 3 3 2 ⫺ 3i i 0002 ⫽ ⫽ ⫽⫺ ⫺i 2 2i i ⫺2 2 2i
MATCHED PROBLEM 5
0002
Carry out the indicated operations and write each answer in standard form: (A) (3 ⫹ 2i)2 ⫺ 6(3 ⫹ 2i) ⫹ 13
4⫺i 3i
(B)
0002
ZZZ EXPLORE-DISCUSS 1
Natural number powers of i take on particularly simple forms: i i 2 ⫽ ⫺1 i 3 ⫽ i 2 0002 i ⫽ (⫺1)i ⫽ ⫺i i 4 ⫽ i 2 0002 i 2 ⫽ (⫺1)(⫺1) ⫽ 1
i5 i6 i7 i8
⫽ ⫽ ⫽ ⫽
i4 i4 i4 i4
0002 0002 0002 0002
i ⫽ (1)i ⫽ i i 2 ⫽ 1(⫺1) ⫽ ⫺1 i 3 ⫽ 1(⫺i) ⫽ ⫺i i4 ⫽ 1 0002 1 ⫽ 1
In general, what are the possible values for i n, n a natural number? Explain how you could easily evaluate i n for any natural number n. Then evaluate each of the following: (A) i 17
(B) i 24
(C) i 38
(D) i 47
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Z Relating Complex Numbers and Radicals Recall that we say that a is a square root of b if a2 0002 b. If x is a positive real number, then x has two square roots, the principal square root, denoted by 1x, and its negative, 00031x (Section R-2). If x is a negative real number, then x still has two square roots, but now these square roots are imaginary numbers.
Z DEFINITION 4 Principal Square Root of a Negative Real Number The principal square root of a negative real number, denoted by 10003a, where a is positive, is defined by 10003a 0002 i 1a
For example 100033 0002 i 13; 100039 0002 i 19 0002 3i
The other square root of 0003a, a 7 0, is 0003 10003a 0002 0003i1a.
Note in Definition 4 that we wrote i1a and i 13 in place of the standard forms 1ai and 13i. We follow this convention to avoid confusion over whether the i should or should not be under the radical. (Notice that 13i and 13i look a lot alike, but are not the same number.)
EXAMPLE
6
Complex Numbers and Radicals Write in standard form: (A) 100034 (C)
SOLUTIONS
(B) 4 100035
00033 0003 100035 2
(D)
1 1 0003 100039
(A) 100034 0002 i 14 0002 2i (B) 4 100035 0002 4 i 15 (C)
00033 0003 100035 00033 0003 i 15 3 15 0002 00020003 0003 i 2 2 2 2
(D)
1 0007 (1 0005 3i ) 1 1 0002 0002 1 0003 3i (1 0003 3i) 0007 (1 0005 3i ) 1 0003 100039 0002
MATCHED PROBLEM 6
1 3i 1 3 1 3i 0002 0002 i 10 10 10 1 0003 9i2
Answer in standard form.
Standard form
0002
Write in standard form: (A) 1000316 (C)
00035 0003 100032 2
(B) 5 100037 (D)
1 3 0003 100034 0002
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ZZZ EXPLORE-DISCUSS 2
81
Complex Numbers
From Theorem 4 in Section R-2, we know that if a and b are positive real numbers, then 1a 1b 0002 1ab
(1)
So we can evaluate expressions like 19 14 two ways: 19 14 0002 1(9)(4) 0002 136 0002 6
and
19 14 0002 (3)(2) 0002 6
Evaluate each of the following two ways. Is equation (1) a valid property to use in all cases? (A) 19 100034
ZZZ
(B) 100039 14
(C) 100039 100034
Note that in Example 6, part D, we wrote 1 0003 100039 0002 1 0003 3i before proceeding with the simplification. This is a necessary step because some of the properties of radicals that are true for real numbers turn out not to be true for complex numbers. In particular, for positive real numbers a and b,
CAUTION ZZZ
1a 1b 0002 1ab
but
10003a 10003b 1(0003a)(0003b)
(See Explore-Discuss 2.) To avoid having to worry about this, always convert expressions of the form 10003a to the equivalent form in terms of i before performing any operations.
Z Solving Equations Involving Complex Numbers EXAMPLE
7
Equations Involving Complex Numbers (A) Solve for real numbers x and y: (3x 2) (2y 0003 4)i 0002 00034 6i (B) Solve for complex number z: (3 2i)z 0003 3 6i 0002 8 0003 4i
SOLUTIONS
(A) This equation is really a statement that two complex numbers are equal: (3x 2) (2y 0003 4)i, and 00034 6i. In order for these numbers to be equal, the real parts must be the same, and the imaginary parts must be the same as well. 3x 2 0002 00034 3x 0002 00036 x 0002 00032
and
2y 0003 4 0002 6 2y 0002 10 y00025
(B) Solve for z, then write the answer in standard form. (3 2i)z 0003 3 6i 0002 8 0003 4i (3 2i)z 0002 11 0003 10i 11 0003 10i 3 2i (11 0003 10i)(3 0003 2i) 0002 (3 2i)(3 0003 2i)
z0002
0002
13 0003 52i 13
Add 3 and subtract 6i from both sides. Divide both sides by 3 0005 2i. Multiply numerator and denominator by the conjugate of the denominator.
Simplify.
Write in standard form.
0002 1 0003 4i A check is left to the reader.
0002
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(A) Solve for real numbers x and y: (2y 0003 7) (3x 4)i 0002 1 i (B) Solve for complex number z: (1 3i)z 4 0003 5i 0002 3 2i 0002 The truth is that the numbers we studied in this section weren’t received very well when they were invented, as you can guess from the names they were given: complex and imaginary. These names are not exactly ringing endorsements. Still, complex numbers eventually came into widespread use in areas like electrical engineering, physics, chemistry, statistics, and aeronautical engineering. Our first application of complex numbers will be in solving second-degree equations in Section 1-5.
ANSWERS TO MATCHED PROBLEMS 1. (A) Real part: 6; imaginary part: 7i; conjugate: 6 0003 7i (B) Real part: 00033; imaginary part: 00038i; conjugate: 00033 8i (C) Real part: 0; imaginary part: 00034i; conjugate: 4i (D) Real part: 00039; imaginary part: 0; conjugate: 00039 2. (A) 9 0003 2i (B) 7 0003 5i (C) 2 0003 2i (D) 0 3. (A) 26 0003 7i (B) 00036 18i (C) 3 2i (D) 13 4. (A) 15 0003 101 i (B) 1 4i 5. (A) 0 (B) 000313 0003 43 i 6. (A) 4i (B) 5 i17 (C) 000352 0003 (12 2)i (D) 133 132 i 7. (A) x 0002 00031, y 0002 4 (B) z 0002 2 i
1-4
Exercises 4 3 i 5 5
1. Do negative real numbers have square roots? Explain.
10. 4.2 0003 9.7i
11. 6.5 2.1i
12.
2. Arrange the following sets of numbers so that each one contains the one that comes before it in the list: rational numbers, complex numbers, integers, real numbers, natural numbers.
13. i0005
14. 60005
15. 40005
16. 000320005i
17. 00035 i12
18. 4 0003 i 17
3. Is it possible to square an imaginary number and get a real number? Explain. 4. What is the conjugate of a complex number? How do we use conjugates?
In Problems 19–44, perform the indicated operations and write each answer in standard form. 19. (3 5i) (2 4i)
20. (4 i) (5 3i)
5. Which statement is false, and which is true? Justify your response. (A) Every real number is a complex number. (B) Every complex number is a real number.
21. (8 0003 3i) (00035 6i)
22. (00031 2i) (4 0003 7i)
23. (9 5i) 0003 (6 2i)
24. (3 7i) 0003 (2 5i)
25. (3 0003 4i) 0003 (00035 6i)
26. (00034 0003 2i) 0003 (1 i)
6. Is it possible to add a real number and an imaginary number? If so, what kind of number is the result?
27. 2 (3i 5)
28. (2i 7) 0003 4i
29. (2i)(4i)
31. 00032i(4 0003 6i)
For each number in Problems 7–18, find the (A) real part, (B) imaginary part, and (C) conjugate. 7. 2 0003 9i
8. 00036i 4
3 5 9. 0003 i 2 6
30. (3i)(5i)
32. (00034i)(2 0003 3i)
33. (1 2i)(3 0003 4i)
34. (2 0003 i)(00035 6i)
35. (3 0003 i)(4 i)
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36. (5 2i)(4 0003 3i)
37. (2 9i)(2 0003 9i)
38. (3 8i)(3 0003 8i)
39.
40.
i 3000bi
43.
7000bi 2000bi
41.
4 3i 1 2i
1 2 4i
42.
3 0003 5i 20003i
44.
00035 10i 3 4i
Complex Numbers
77.
(1 x) ( y 0003 2)i 000220003i 1000bi
78.
(2 x) ( y 3)i 0002 00033 i 10003i
83
In Problems 79–82, solve for z and write your answer in standard form. 79. (10 0003 2i)z (5 i) 0002 2i
In Problems 45–52, evaluate and express results in standard form.
80. (3 0003 2i)z (4i 6) 0002 8i
45. 12 18
46. 13 112
81. (4 2i)z (7 0003 2i) 0002 (4 0003 i)z (3 5i)
47. 12 100038
48. 100033 112
82. (00032 3i) (4 5i)z 0002 (1 i) 0003 (00034 2i)z
49. 100032 18
50. 13 1000312
83. Show that 2 0003 i and 00032 i are square roots of 3 0003 4i.
51. 100032 100038
52. 100033 1000312
84. Show that 00033 2i and 3 0003 2i are square roots of 5 0003 12i.
In Problems 53–62, convert imaginary numbers to standard form, perform the indicated operations, and express answers in standard form.
85. Explain what is wrong with the following “proof ” that 00031 0002 1: 00031 0002 i2 0002 100031 100031 0002 1(00031)(00031) 0002 11 0002 1 86. Explain what is wrong with the following “proof ” that 1 i 0002 i. What is the correct value of 1 i?
53. (2 0003 100034) (5 0003 100039) 54. (3 0003 100034) (00038 1000325)
1 1 11 1 0002 0002 0002 0002 100031 0002 i i 100031 100031 A 00031
55. (9 0003 100039) 0003 (12 0003 1000325) 56. (00032 0003 1000336) 0003 (4 1000349)
87. Show that i 4k 0002 1, k a natural number
57. (3 0003 100034)(00032 1000349)
88. Show that i 4k000b1 0002 i, k a natural number
58. (2 0003 100031)(5 100039) 59.
5 0003 100034 7
60.
6 0003 1000364 2
Supply the reasons in the proofs for the theorems stated in Problems 89 and 90.
61.
1 2 0003 100039
62.
1 3 0003 1000316
89. Theorem: The complex numbers are commutative under addition. Proof: Let a bi and c di be two arbitrary complex numbers; then:
In Problems 63–68, write the complex number in standard form. 63. 0003
5 i
64.
65. (2i)2 0003 5(2i) 6
1 10i
66. (i13)4 2(i13)2 15
67. (5 2i)2 0003 4(5 2i) 0003 1 68. (7 0003 3i)2 8(7 0003 3i) 0003 30 69. Evaluate x2 0003 2x 2 for x 0002 1 0003 i. 70. Evaluate x2 0003 2x 2 for x 0002 1 i. In Problems 71–74, for what real values of x does each expression represent an imaginary number? 71. 13 0003 x
72. 15 x
73. 12 0003 3x
74. 13 2x
In Problems 75–78, solve for x and y. 75. (2x 0003 1) (3y 2)i 0002 5 0003 4i 76. 3x ( y 0003 2)i 0002 (5 0003 2x) (3y 0003 8)i
Statement 1. (a bi) (c di) 0002 (a c) (b d )i 2. 0002 (c a) (d b)i 3. 0002 (c di) (a bi) Reason 1. 2. 3. 90. Theorem: The complex numbers are commutative under multiplication. Proof: Let a bi and c di be two arbitrary complex numbers; then: Statement 1. (a bi) 0007 (c di) 0002 (ac 0003 bd ) (ad bc)i 0002 (ca 0003 db) (da cb)i 2. 0002 (c di)(a bi) 3. Reason 1. 2. 3.
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Letters z and w are often used as complex variables, where z 0002 x yi, w 0002 u vi, and x, y, u, v are real numbers. The conjugates of z and w, denoted by z and w, respectively, are given by z 0002 x 0003 yi and w 0002 u 0003 vi. In Problems 91–98, express each property of conjugates verbally and then prove the property. 91. zz is a real number.
1-5
93. z 0002 z if and only if z is real.
94. z 0002 z
95. z w 0002 z w
96. z 0003 w 0002 z 0003 w
97. zw 0002 z 0007 w
98. z w 0002 z w
92. z z is a real number.
Quadratic Equations and Applications Z Using Factoring to Solve Quadratic Equations Z Using the Square Root Property to Solve Quadratic Equations Z Using Completing the Square to Solve Quadratic Equations Z Using the Quadratic Formula to Solve Quadratic Equations Z Solving Applications Involving Quadratic Equations
The next class of equations we consider are the second-degree polynomial equations in one variable, called quadratic equations.
Z DEFINITION 1 Quadratic Equation A quadratic equation in one variable is any equation that can be written in the form ax2 0005 bx 0005 c 0002 0
a0
Standard Form
where x is a variable and a, b, and c are constants.
Now that we have discussed the complex number system, we can use complex numbers when solving equations. Recall that a solution of an equation is also called a root of the equation. A real number solution of an equation is called a real root, and an imaginary number solution is called an imaginary root. In this section, we develop methods for finding all real and imaginary roots of a quadratic equation.
Z Using Factoring to Solve Quadratic Equations There is one single reason why factoring is so important in solving equations. It’s called the zero product property.
ZZZ EXPLORE-DISCUSS 1
(A) Write down a pair of numbers whose product is zero. Is one of them zero? Can you think of two nonzero numbers whose product is zero? (B) Choose any number other than zero and call it a. Write down a pair of numbers whose product is a. Is one of them a? Can you think of a pair, neither of which is a, whose product is a?
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85
Z ZERO PRODUCT PROPERTY If m and n are complex numbers, then m0002n00020
m 0002 0 or n 0002 0 (or both)
if and only if
It is very helpful to think about what this says in words: If the product of two factors is zero, then at least one of those factors has to be zero. It’s also helpful to observe that zero is the only number for which this is true.
EXAMPLE
1
Solving Quadratic Equations by Factoring Solve by factoring: (B) 6x2 0003 19x 0003 7 0002 0 (D) 2x2 0002 3x
(A) (x 0003 5)(x 0004 3) 0002 0 (C) x2 0003 6x 0004 5 0002 00034 SOLUTIONS
(A) The product of two factors is zero, so by the zero product property, one of the two must be zero. This enables us to write two easier equations to solve. (x 0003 5)(x 0004 3) 0002 0 x0003500020 x00025 (B)
6x2 0003 19x 0003 7 0002 0 (2x 0003 7)(3x 0004 1) 0002 0 2x 0003 7 0002 0 x0002
(C)
Solution set: {00033, 5}.
Factor the left side. Use the zero product property.
or
3x 0004 1 0002 0
7 2
x2 0003 6x 0004 5 0002 00034 x2 0003 6x 0004 9 0002 0 (x 0003 3)(x 0003 3) 0002 0 x0003300020 x00023
x0004300020 x 0002 00033
or
x 0002 000313
Solution set: 5000313 , 72 6.
Add 4 to both sides. Factor left side. Use the zero product property.
Solution set: {3}.
The equation has one root, 3. But because it came from two factors, we call 3 a double root or a root of multiplicity 2. (D)
2x2 0002 3x 2x 0003 3x 0002 0 x(2x 0003 3) 0002 0 x00020 2
MATCHED PROBLEM 1
Subtract 3x from both sides. Factor the left side. Use the zero product property.
or
2x 0003 3 0002 0 x 0002 32
Solution set: 50, 32 6
0002
Solve by factoring: (A) (2x 0004 4)(x 0003 7) 0002 0 (C) 4x2 0004 12x 0004 9 0002 0
(B) 3x2 0004 7x 0003 20 0002 0 (D) 4x2 0002 5x 0002
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1. One side of an equation must be 0 before the zero product property can be applied. So
CAUTION ZZZ
x2 0003 6x 0004 5 0002 00034 (x 0003 1)(x 0003 5) 0002 00034 does not mean that x 0003 1 0002 00034 or x 0003 5 0002 00034. See Example 1, part C, for the correct solution of this equation. 2. The equations 2x2 0002 3x
2x 0002 3
and
are not equivalent. The first has solution set 50, 32 6, but the second has solution set 5 32 6. The root x 0002 0 is lost when each member of the first equation is divided by the variable x. See Example 1, part D, for the correct solution of this equation.
Never divide both sides of an equation by an expression containing the variable for which you are solving. You may be dividing by 0, which of course is not allowed.
Z Using the Square Root Property to Solve Quadratic Equations We now turn our attention to quadratic equations that do not have the first-degree term— that is, equations of the special form ax2 0004 c 0005 0
a00060
The method of solution of this special form makes direct use of the square root property:
Z SQUARE ROOT PROPERTY If A2 0002 C, then A 0002 00051C.
The use of the square root property is illustrated in Example 2.
EXAMPLE
2
Using the Square Root Property Solve using the square root property: (A) 9x2 0003 7 0002 0
SOLUTIONS
(B) 3x2 0004 27 0002 0
(A) 9x2 0003 7 0002 0 9x2 0002 7 7 x2 0002 9 x00020005
(C) (x 0004 12)2 0002 54
Add 7 to both sides. Divide both sides by 9. Apply the square root property; don’t forget the 0007!
7 17 00020005 B9 3
Solution set: e
17 000317 , f 3 3
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(B) 3x2 0004 27 0002 0 x2 0002 00039 x 0002 0005 100039 000200053i (C) (x
MATCHED PROBLEM 2
0004 12)2 x 0004 12
5 4
0002 0002 0005 254 1 15 x00020003 0005 2 2 00031 0005 15 0002 2
Quadratic Equations and Applications
87
Solve for x2. Apply the square root property. Solution set: 500033i, 3i6
Apply the square root property. Subtract
1 2
from both sides, and simplify 254 .
Combine fractions with common denominators.
0002
Solve using the square root property: (A) 9x2 0003 5 0002 0
(B) 2x2 0004 8 0002 0
(C) (x 0004 13)2 0002 29 0002
Note: It is common practice to represent solutions of quadratic equations informally by the last equation (Example 2, part C) rather than by writing a solution set using set notation (Example 2, parts A and B). From now on, we will follow this practice unless we need to make a special point.
Z Using Completing the Square to Solve Quadratic Equations The methods of square root property and factoring are generally fast when they apply; however, there are equations, such as x2 0004 6x 0003 2 0002 0, that cannot be solved directly by these methods. A more general procedure must be developed to take care of this type of equation. One is called the method of completing the square.* This method is based on the process of transforming the standard quadratic equation ax2 0004 bx 0004 c 0002 0 into the form (x 0004 A)2 0002 B where A and B are constants. Equations of this form can easily be solved by using the square root property. But how do we transform the first equation into the second? We will need to find a way to make the left side factor as a perfect square.
ZZZ EXPLORE-DISCUSS 2
Replace ? in each of the following with a number that makes the equation valid. (A) (x 0004 1)2 0002 x2 0004 2x 0004 ?
(B) (x 0004 2)2 0002 x2 0004 4x 0004 ?
(C) (x 0004 3)2 0002 x2 0004 6x 0004 ?
(D) (x 0004 4)2 0002 x2 0004 8x 0004 ?
Replace ? in each of the following with a number that makes the expression a perfect square of the form (x 0004 h)2. (E) x2 0004 10x 0004 ?
(F) x2 0004 12x 0004 ?
(G) x2 0004 bx 0004 ?
Given the quadratic expression x2 0004 bx *We will find many other uses for this important method.
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what number should be added to this expression to make it a perfect square? To find out, consider the square of the following expression:
{
{
(x 0004 m)2 0002 x2 0004 2mx 0004 m2
m2 is the square of one-half the coefficient of x.
We see that the third term on the right side of the equation is the square of one-half the coefficient of x in the second term on the right; that is, m2 is the square of 12(2m). This observation leads to the following rule:
Z COMPLETING THE SQUARE To complete the square of a quadratic expression of the form x2 0004 bx, add the square of one-half the coefficient of x; that is, add (b兾2)2, or b2兾4. The resulting expression factors as a perfect square,
EXAMPLE
3
x2 0004 bx
For example, x2 0004 5x
b 2 b 2 x2 0004 bx 0004 a b 0002 ax 0004 b 2 2
5 2 5 2 x2 0004 5x 0004 a b 0005 ax 0004 b 2 2
Completing the Square Complete the square for each of the following: (A) x2 0003 3x
SOLUTIONS
00033 2 9 b ; that is, and factor. 2 4
(A) x2 0003 3x x2 0003 3x 0004 94 0002 (x 0003 32)2
Add a
(B) x2 0003 bx
Add a
x2 0003 bx 0004
MATCHED PROBLEM 3
(B) x2 0003 bx
b 2 b2 0002 ax 0003 b 4 2
0003b 2 b2 b ; that is, and factor. 2 4
0002
Complete the square for each of the following: (A) x2 0003 5x
(B) x2 0004 mx
0002
You should note that the rule for completing the square applies only if the coefficient of the second-degree term is 1. This causes little trouble, however, as you will see. To solve equations by completing the square, we will add b2兾4 to both sides after moving the constant term to the right side.
EXAMPLE
4
Solution by Completing the Square Solve by completing the square: (A) x2 0004 6x 0003 2 0002 0
(B) 2x2 0003 4x 0004 3 0002 0
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SOLUTIONS
(A) x2 0004 6x 0003 2 0002 0 x2 0004 6x 0002 2
Complete the square on the left side and add (2b )2 0005 (62 )2 0005 9 to both sides. Factor the left side; add on the right. Use the square root property. Don’t forget the ⫾! Add 00033 to both sides.
(B) 2x2 0003 4x 0004 3 0002 0 x2 0003 2x 0004 32 0002 0
Make the leading coefficient 1 by dividing both sides by 2. Subtract 000332 from both sides. 2 Complete the square on the left side and add (2b )2 0005 (00032 2 ) 0005 1 to both sides.
x2 0003 2x 0002 000332 x2 0003 2x 0004 1 0002 000332 0004 1 (x 0003 1)2 0002 000312 x 0003 1 0002 0005 2000312 x 0002 1 0005 i212
MATCHED PROBLEM 4
89
Add 2 to both sides to obtain the form x2 0004 bx on the left side.
x2 0004 6x 0004 9 0002 2 0004 9 (x 0004 3)2 0002 11 x 0004 3 0002 0005 111 x 0002 00033 0005 111
000210005
Quadratic Equations and Applications
Factor the left side; add on the right. Use the square root property. Add 1 to both sides and simplify 2000312 .
12 i 2
Answer in a 0004 bi form.
0002
Solve by completing the square: (A) x2 0004 8x 0003 3 0002 0
(B) 3x2 0003 12x 0004 13 0002 0
0002
Z Using the Quadratic Formula to Solve Quadratic Equations If we solve a generic quadratic equation using the method of completing the square, the result will be a formula for solving any quadratic equation. ax2 0004 bx 0004 c 0002 0 b c x2 0004 x 0004 0002 0 a a
a00060
b c x2 0004 x 0002 0003 a a x2 0004
b b2 b2 c x0004 20002 20003 a a 4a 4a
b 2 b2 0003 4ac b 0002 2a 4a2 b b2 0003 4ac x0004 00020005 2a B 4a2 b 2b2 0003 4ac x00020003 0005 2a 2a 0003b 0005 2b2 0003 4ac x0002 2a
ax 0004
Make the leading coefficient 1 by dividing by a. Subtract
c from both sides. a
Complete the square on the left side and add b2 b 2 a b 0005 to both sides. 2a 4a2 Factor the left side and combine terms on the right side, getting a common denominator. Use the square root property. b2 0003 4ac b to both sides and simplify B 4a2 2a (see Problem 75 in Exercises 1-5). Add 0003
Combine terms on the right side.
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The result is known as the quadratic formula: Z THEOREM 1 Quadratic Formula If ax2 0004 bx 0004 c 0002 0, a 0006 0, then 0003b 0007 2b2 0003 4ac 2a
x0005
The quadratic formula should be memorized and used to solve quadratic equations when other methods fail, or are more difficult to apply.
EXAMPLE
5
Using the Quadratic Formula Solve 2x 0004 32 0002 x2 using the quadratic formula. Leave the answer in simplest radical form. 2x 0004 32 0002 x2 4x 0004 3 0002 2x2
SOLUTION
Multiply both sides by 2. Write in standard form.
Identify a, b, and c and use the quadratic 2x2 0003 4x 0003 3 0002 0 formula: a 0005 2, b 0005 00034, c 0005 00033 2 0003b 0005 2b 0003 4ac x0002 2a 0003(00034) 0005 2(00034)2 0003 4(2)(00033) 0002 2(2) 4 0005 2110 2 0005 110 4 0005 140 0002 0002 0002 4 4 2
ZZZ
1. 000342 0006 (00034)2
CAUTION ZZZ
2. 2 0004 3.
MATCHED PROBLEM 5
110 2 0004 110 0006 2 2
4 0005 2110 0006 00052110 4
0002
000342 0005 000316 and (00034)2 0005 16 20004
110 4 0004 110 0005 2 2
2(2 0007 110) 4 0007 2 110 2 0007 110 0005 0005 4 4 2
Solve x2 0003 52 0002 00033x by use of the quadratic formula. Leave the answer in simplest radical form. 0002 The expression under the square root in the quadratic formula, b2 0003 4ac, is called the discriminant. It gives us useful information about the corresponding roots, as shown in Table 1. Table 1 Discriminant and Roots Discriminant b2 0003 4ac
Roots of ax2 0004 bx 0004 c 0005 0 a, b, and c real numbers, a 0006 0
Positive
Two distinct real roots
0
One real root (a double root)
Negative
Two imaginary roots, one the conjugate of the other
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EXAMPLE
6
Quadratic Equations and Applications
91
Using the Discriminant Find the number of real roots of each quadratic equation. (A) 2x2 0002 4x 0003 1 0004 0
SOLUTIONS
(B) 2x2 0002 4x 0003 2 0004 0
(C) 2x2 0002 4x 0003 3 0004 0
(A) b2 0002 4ac 0004 (00024)2 0002 4(2)(1) 0004 8 7 0; two real roots (B) b2 0002 4ac 0004 (00024)2 0002 4(2)(2) 0004 0; one real (double) root (C) b2 0002 4ac 0004 (00024)2 0002 4(2)(3) 0004 00028 6 0; no real roots (two imaginary roots)
MATCHED PROBLEM 6
0002
Find the number of real roots of each quadratic equation. (A) 3x2 0002 6x 0003 5 0004 0
(B) 3x2 0002 6x 0003 1 0004 0
(C) 3x2 0002 6x 0003 3 0004 0
0002
Z Solving Applications Involving Quadratic Equations Now that we’re good at solving quadratic equations, we can use them to solve many applied problems. It would be a good idea to review the problem-solving strategy on page 47 before beginning.
EXAMPLE
7
Setting Up and Solving a Word Problem The sum of a number and its reciprocal is
SOLUTION
Find all such numbers.
Let x 0004 the number we’re asked to find; then its reciprocal is 1x . x0003
13 1 0004 x 6
Multiply both sides by the LCD, 6x. [Note: x cannot be zero.]
1 13 (6x)x 0003 (6x) 0004 (6x) x 6 6x2 0003 6 0004 13x 6x 0002 13x 0003 6 0004 0 (2x 0002 3)(3x 0002 2) 0004 0 2x 0002 3 0004 0 x 0004 32
Make sure to multiply every term by 6x.
Subtract 13x from both sides.
2
These are two such numbers: CHECK
MATCHED PROBLEM 7
13 6.
3 2
Factor the left side. Use the zero product property.
3x 0002 2 0004 0 x 0004 23
or
3 2
Solve each equation for x.
and 23.
0003 23 0004 136
2 3
0003 32 0004 136
0002
The sum of two numbers is 23 and their product is 132. Find the two numbers. [Hint: If one number is x, then the other number is 23 0002 x.] 0002
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8
A Distance–Rate–Time Problem A casino boat takes 1.6 hours longer to go 36 miles up a river than to return. If the rate of the current is 4 miles per hour, what is the speed of the boat in still water?
SOLUTION
Let x 0002 Speed of boat in still water x 0004 4 0002 Speed downstream x 0003 4 0002 Speed upstream
a
Time Time b0003a b 0002 1.6 upstream downstream 36 x00034
0003
36(x 0004 4) 0003
36 x00044
Use Time 0005
Multiply both sides by (x 0003 4)(x 0004 4), the LCD.
0002 1.6
36(x 0003 4)
Distance . Rate
0002 1.6(x 0003 4)(x 0004 4)
Multiply out parentheses. Combine like terms and isolate 1.6x 2 on one side of the equation.
2
36x 0004 144 0003 36x 0004 144 0002 1.6x 0003 25.6 1.6x2 0002 313.6
Divide both sides by 1.6.
2
x 0002 196
Use the square root property.
x 0002 0005 1196 0002 000514 The speed in still water is 14 miles per hour. (The negative answer is thrown out, because it doesn’t make sense in the problem to have a negative speed.) CHECK
Time upstream 0002 0003 Time downstream 0002
D 36 0002 0002 3.6 R 14 0003 4 36 D 0002 00022 R 14 0004 4 1.6
MATCHED PROBLEM 8
Difference of times
0002
Two boats travel at right angles to each other after leaving a dock at the same time. One hour later they are 25 miles apart. If one boat travels 5 miles per hour faster than the other, what is the rate of each? [Hint: Use the Pythagorean theorem,* remembering that distance equals rate times time.] 0002 In Example 9, we introduce some concepts from economics that will be used throughout this book. The quantity of a product that people are willing to buy during some period
*Pythagorean theorem: In a right triangle, the square of the length of the longest side is equal to the sum of the squares of the lengths of the two shorter sides.
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93
of time is called the demand for that product. The price p of a product and the demand q for that product are often related by a price–demand equation of the following form: q 0002 a 0003 bp
q is the number of items that can be sold at $p per item.
The constants a and b in a price–demand equation are usually determined by using historical data and statistical analysis. The amount of money received from the sale of q items at $p per item is called the revenue and is given by R 0002 (Number of items sold) 0007 (Price per item) 0002 qp 0002 (a 0003 bp)p
EXAMPLE
9
Using the price-demand equation
Price and Demand The daily price–demand equation for whole milk in a chain of supermarkets is q 0002 5,600 0003 800p where p is the price per gallon and q is the number of gallons sold per day. Find the price(s) that will produce a revenue of $9,500. Round answer(s) to two decimal places.
SOLUTION
The revenue equation is R 0002 qp 0002 (5,600 0003 800p)p 0002 5,600p 0003 800p2 To get a revenue of $9,500, we substitute 9,500 for R: 5,600p 0003 800p2 0002 9,500 00039,500 0004 5,600p 0003 800p2 0002 0 p2 0003 7p 0004 11.875 0002 0 7 0005 11.5 2 0002 2.89, 4.11
Subtract 9,500 from both sides. Divide both sides by 0003800. Use the quadratic formula with a 0005 1, b 0005 00037, and c 0005 11.875.
p0002
Selling whole milk for either $2.89 per gallon or $4.11 per gallon will produce a revenue of $9,500. 0002 MATCHED PROBLEM 9
If the price–demand equation for milk is q 0002 4,800 0003 600p, find the price that will produce revenues of (A) $9,300
(B) $10,500
ANSWERS TO MATCHED PROBLEMS 1. (A) x 0002 00032, 7 (B) x 0002 00034, 53 (C) x 0002 000332 (a double root) (D) x 0002 0, 54 2. (A) x 0002 0005 15 2 (B) x 0002 00052i (C) x 0002 (00031 0005 12)3 3. (A) x2 0003 5x 0004 254 0002 (x 0003 52)2 (B) x2 0004 mx 0004 (m24) 0002 [x 0004 (m2)] 2 4. (A) x 0002 000340005 119 (B) x 0002 (6 0005 i13)3 or 20005 (133)i 5. x 0002 (000330005 119)2 6. (A) No real roots (two imaginary roots) (B) Two real roots (C) One real (double) root 7. 11 and 12 8. 15 and 20 miles per hour 9. (A) $3.29 or $4.71 (B) Not possible
0002
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Exercises
Leave all answers involving radicals in simplified radical form unless otherwise stated.
In Problems 41–56, solve by any method. 41. 12x2 ⫹ 7x ⫽ 10
42. 9x2 ⫹ 9x ⫽ 4
1. How can you tell when an equation is quadratic?
43. (2y ⫺ 3)2 ⫽ 5
44. (3m ⫹ 2)2 ⫽ ⫺4
2. What do a, b, and c in the quadratic formula stand for?
45. x2 ⫽ 3x ⫹ 1
46. x2 ⫹ 2x ⫽ 2
3. Explain what the zero product property says in your own words.
47. 7n2 ⫽ ⫺4n
48. 8u2 ⫹ 3u ⫽ 0
4. Explain what the square root property says in your own words. 5. If you could only use one of factoring, completing the square, and quadratic formula on an important test featuring a variety of quadratic equations, which would you choose, and why?
49. 1 ⫹
9. ⫺8 ⫽ 22t ⫺ 6t2 11. 3w2 ⫹ 13w ⫽ 10
10. 25z2 ⫽ ⫺10z 12. 36x2 ⫽ ⫺12x ⫺ 1
In Problems 13–24, solve by using the square root property.
2 3 ⫽ 2⫹1 u u
53.
4 1 2 ⫽ ⫺ x⫺2 x⫺3 x⫹1
24 24 ⫹1⫽ 10 ⫹ m 10 ⫺ m
52.
1.2 1.2 ⫽1 ⫹ y y⫺1
54.
2 4 3 ⫺ ⫽ x⫺1 x⫹3 x⫺2
55.
x⫺1 x⫹2 x2 ⫽1⫺ ⫺ 2 x⫹3 3⫺x x ⫺9
56.
x⫹3 2x ⫺ 3 11 ⫹ ⫽ 2⫺x x⫹2 x2 ⫺ 4
In Problems 7–12, solve by factoring. 8. 3y2 ⫽ y ⫹ 10
50.
51.
6. Does every quadratic equation have two solutions? Explain.
7. 2x2 ⫽ 8x
8 4 ⫽ x x2
13. m2 ⫺ 25 ⫽ 0
14. n2 ⫹ 16 ⫽ 0
15. c2 ⫹ 9 ⫽ 0
16. d 2 ⫺ 36 ⫽ 0
In Problems 57–60, solve for the indicated variable in terms of the other variables. Use positive square roots only.
17. 4y2 ⫹ 9 ⫽ 0
18. 9x2 ⫺ 25 ⫽ 0
57. s ⫽ 12gt 2
2
2
19. 25z ⫺ 32 ⫽ 0 2
21. (2k ⫺ 5) ⫽ 16 2
23. (n ⫺ 3) ⫽ ⫺4
20. 16w ⫹ 27 ⫽ 0 2
22. (t ⫺ 2) ⫽ ⫺3
26. y2 ⫺ 4y ⫹ 7 ⫽ 0
27. x2 ⫺ 2x ⫹ 3 ⫽ 0
28. y2 ⫺ 4y ⫹ 1 ⫽ 0
2
30. 9s ⫹ 2 ⫽ 12s
2
32. 9s2 ⫹ 7 ⫽ 12s
31. 2t ⫹ 1 ⫽ 6t
2
In Problems 33–40, solve by completing the square. 2
33. x ⫺ 4x ⫺ 1 ⫽ 0
2
34. y ⫹ 4y ⫺ 3 ⫽ 0
2
36. 2s ⫺ 6s ⫹ 7 ⫽ 0
2
38. 4v2 ⫹ 16v ⫹ 23 ⫽ 0
35. 2r ⫹ 10r ⫹ 11 ⫽ 0 37. 4u ⫹ 8u ⫹ 15 ⫽ 0 2
39. 3w ⫹ 4w ⫹ 3 ⫽ 0
2
2
40. 3z ⫺ 8z ⫹ 1 ⫽ 0
60. A ⫽ P(1 ⫹ r)2 for r
61. Consider the quadratic equation
24. (5m ⫺ 6) ⫽ 7
25. x2 ⫺ 2x ⫺ 1 ⫽ 0 29. 2t ⫹ 8 ⫽ 6t
59. P ⫽ EI ⫺ RI 2 for I
2
In Problems 25–32, use the discriminant to determine the number of real roots of each equation and then solve each equation using the quadratic formula.
58. a2 ⫹ b2 ⫽ c2 for a
for t
x2 ⫹ 4x ⫹ c ⫽ 0 where c is a real number. Discuss the relationship between the values of c and the three types of roots listed in Table 1. 62. Consider the quadratic equation x2 ⫺ 2x ⫹ c ⫽ 0 where c is a real number. Discuss the relationship between the values of c and the three types of roots listed in Table 1. Solve the equation in Problems 63–66 and leave answers in simplified radical form (i is the imaginary unit). 63. x2 ⫹ 3ix ⫺ 2 ⫽ 0
64. x2 ⫺ 7ix ⫺ 10 ⫽ 0
65. x2 ⫹ 2ix ⫽ 3
66. x2 ⫽ 2ix ⫺ 3
In Problems 67 and 68, find all solutions. 67. x3 ⫺ 1 ⫽ 0
68. x4 ⫺ 1 ⫽ 0
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69. Prove that when the discriminant of a quadratic equation with real coefficients is negative, the equation has two imaginary solutions. 70. Prove that when the discriminant of a quadratic equation with real coefficients is zero, the equation has one real solution.
Quadratic Equations and Applications
83. CONSTRUCTION A gardener has a 30 foot by 20 foot rectangular plot of ground. She wants to build a brick walkway of uniform width on the border of the plot (see the figure). If the gardener wants to have 400 square feet of ground left for planting, how wide (to two decimal places) should she build the walkway? x
71. Can a quadratic equation with rational coefficients have one rational root and one irrational root? Explain. 72. Can a quadratic equation with real coefficients have one real root and one imaginary root? Explain. 73. Show that if r1 and r2 are the two roots of ax2 0004 bx 0004 c 0002 0, then r1r2 0002 ca. 74. For r1 and r2 in Problem 73, show that r1 0004 r2 0002 0003ba. 75. In one stage of the derivation of the quadratic formula, we replaced the expression 00052(b2 0003 4ac)4a2 00052b2 0003 4ac2a What justifies using 2a in place of 冟 2a 冟? 76. Find the error in the following “proof ” that two arbitrary numbers are equal to each other: Let a and b be arbitrary numbers such that a 0006 b. Then 2
2
20 feet 30 feet
84. CONSTRUCTION Refer to Problem 83. The gardener buys enough bricks to build 160 square feet of walkway. Is this sufficient to build the walkway determined in Problem 83? If not, how wide (to two decimal places) can she build the walkway with these bricks? 85. CONSTRUCTION A 1,200 square foot rectangular garden is enclosed with 150 feet of fencing. Find the dimensions of the garden to the nearest tenth of a foot.
with
2
95
2
2
(a 0003 b) 0002 a 0003 2ab 0004 b 0002 b 0003 2ab 0004 a
86. CONSTRUCTION The intramural fields at a small college will cover a total area of 140,000 square feet, and the administration has budgeted for 1,600 feet of fence to enclose the rectangular field. Find the dimensions of the field. 87. PRICE AND DEMAND The daily price–demand equation for hamburgers at a fast-food restaurant is
(a 0003 b)2 0002 (b 0003 a)2
q 0002 1,600 0003 200p
a0003b0002b0003a
where q is the number of hamburgers sold daily and p is the price of one hamburger (in dollars). Find the demand and the revenue when the price of a hamburger is $3.
2a 0002 2b a0002b 77. Find two numbers such that their sum is 21 and their product is 104. 78. Find all numbers with the property that when the number is added to itself the sum is the same as when the number is multiplied by itself. 79. Find two consecutive positive even integers whose product is 168. 80. The sum of a number and its reciprocal is
10 3.
Find the number.
APPLICATIONS 81. ALCOHOL CONSUMPTION The beer consumption by Americans for the years 1960–2005 can be modeled by the equation y 0002 00030.0665x2 0004 3.58x 0004 122, where x is the number of years after 1960, and y is the number of ounces of beer consumed per person in that year. Find the per person consumption in 1960, then find in what year the model predicts that consumption will return to that level. 82. ALCOHOL CONSUMPTION The wine consumption by Americans for the years 1985–2005 can be modeled by the equation y 0002 0.0951x2 0003 2.06x 0004 49.0, where x is the number of years after 1985, and y is the number of ounces of wine consumed per person in that year. In what year does the model predict that consumption will reach the 1960 level of beer consumption (see Problem 81)?
88. PRICE AND DEMAND The weekly price–demand equation for medium pepperoni pizzas at a fast-food restaurant is q 0002 8,000 0003 400p where q is the number of pizzas sold weekly and p is the price of one medium pepperoni pizza (in dollars). Find the demand and the revenue when the price is $8. 89. PRICE AND DEMAND Refer to Problem 87. Find the price p that will produce each of the following revenues. Round answers to two decimal places. (A) $2,800 (B) $3,200 (C) $3,400 90. PRICE AND DEMAND Refer to Problem 88. Find the price p that will produce each of the following revenues. Round answers to two decimal places. (A) $38,000 (B) $40,000 (C) $42,000 91. NAVIGATION Two planes travel at right angles to each other after leaving the same airport at the same time. One hour later they are 260 miles apart. If one travels 140 miles per hour faster than the other, what is the rate of each? 92. NAVIGATION A speedboat takes 1 hour longer to go 24 miles up a river than to return. If the boat cruises at 10 miles per hour in still water, what is the rate of the current? 93. AIR SEARCH A search plane takes off from an airport at 6:00 A.M. and travels due north at 200 miles per hour. A second plane leaves that airport at the same time and travels due east at 170 miles
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per hour. The planes carry radios with a maximum range of 500 miles. When (to the nearest minute) will these planes no longer be able to communicate with each other? 94. AIR SEARCH If the second plane in Problem 93 leaves at 6:30 A.M. instead of 6 A.M., when (to the nearest minute) will the planes lose communication with each other? 95. ENGINEERING One pipe can fill a tank in 5 hours less than another. Together they can fill the tank in 5 hours. How long would it take each alone to fill the tank? Compute the answer to two decimal places.
(B) A potential buyer for the building needs to have a floor area of 25,000 square feet. Can the builder accommodate them? 100. ARCHITECTURE An architect is designing a small A-frame cottage for a resort area. A cross section of the cottage is an isosceles triangle with an area of 98 square feet. The front wall of the cottage must accommodate a sliding door that is 6 feet wide and 8 feet high (see the figure). Find the width and height of the cross section of the cottage. [Recall: The area of a triangle with base b and altitude h is bh兾2.]
96. ENGINEERING Two gears rotate so that one completes 1 more revolution per minute than the other. If it takes the smaller gear 1 second less than the larger gear to complete 15 revolution, how many revolutions does each gear make in 1 minute? 97. PHYSICS—ENGINEERING For a car traveling at a speed of v miles per hour, under the best possible conditions the shortest distance d necessary to stop it (including reaction time) is given by the formula d 0002 0.044v2 0004 1.1v, where d is measured in feet. Estimate the speed of a car that requires 165 feet to stop in an emergency. 98. PHYSICS—ENGINEERING If a projectile is shot vertically into the air (from the ground) with an initial velocity of 176 feet per second, its distance y (in feet) above the ground t seconds after it is shot is given by y 0002 176t 0003 16t 2 (neglecting air resistance). (A) Find the times when y is 0, and interpret the results physically. (B) Find the times when the projectile is 16 feet off the ground. Compute answers to two decimal places.
REBEKAH DRIVE
200 feet
99. ARCHITECTURE A developer wants to erect a rectangular building on a triangular-shaped piece of property that is 200 feet wide and 400 feet long (see the figure).
Property A
8 feet
6 feet
101. TRANSPORTATION A delivery truck leaves a warehouse and travels north to factory A. From factory A the truck travels east to factory B and then returns directly to the warehouse (see the figure). The driver recorded the truck’s odometer reading at the warehouse at both the beginning and the end of the trip and also at factory B, but forgot to record it at factory A (see the table). The driver does recall that it was farther from the warehouse to factory A than it was from factory A to factory B. Since delivery charges are based on distance from the warehouse, the driver needs to know how far factory A is from the warehouse. Find this distance.
Property Line
l Proposed Building
Factory A
Factory B
w
FIRST STREET 400 feet
(A) Building codes require that industrial buildings on lots that size have a floor area of at least 15,000 square feet. Find the dimensions that will yield the smallest building that meets code. [Hint: Use Euclid’s theorem* to find a relationship between the length and width of the building.]
*Euclid’s theorem: If two triangles are similar, their corresponding sides are proportional: c
a b
a
c b
b c a 0002 0002 a¿ b¿ c¿
Warehouse
Odometer readings Warehouse
5 2 8 4 6
Factory A
5 2 ? ? ?
Factory B
5 2 9 3 7
Warehouse
5 3 0 0 2
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102. CONSTRUCTION A 14-mile track for racing stock cars consists of two semicircles connected by parallel straightaways (see the figure). In order to provide sufficient room for pit crews, emergency vehicles, and spectator parking, the track must enclose an area of 100,000 square feet. Find the length of the straightaways and the diameter of the semicircles to the nearest foot. [Recall: The area A and circumference C of a circle of diameter d are given by A 0002 d 24 and c 0002 d. ]
1-6
97
100,000 square feet
Additional Equation-Solving Techniques Z Solving Equations Involving Radicals Z Revisiting Equations Involving Absolute Value Z Solving Equations of Quadratic Type
In this section, we’ll study equations that are not quadratic but can be transformed into quadratic equations. We can then solve the quadratic equation, and with a little bit of interpretation, use the solutions to solve the original equation.
Z Solving Equations Involving Radicals In solving an equation involving a radical, like x 0002 1x 0004 2 it seems reasonable that we can remove the radical by squaring each side and then proceed to solve the resulting quadratic equation. Let’s give it a try: Square both sides. x 0002 1x 0004 2 2 2 Recall that ( 1a)2 0005 a if a 0. x 0002 ( 1x 0004 2) 2 Subtract x 0004 2 from both sides. x 0002x00042 2 Factor the left side. x 0003x0003200020 Use the zero product property. (x 0003 2)(x 0004 1) 0002 0 or x0003200020 x0004100020 or x00022 x 0002 00031
Now we check these results in the original equation. Check: x 0002 2
Check: x 0002 00031
x 0002 1x 0004 2 ? 2 0002 12 0004 2 ? 2 0002 14 ✓ 200022
x0002 ? 00031 0002 ? 00031 0002 00031 0006
1x 0004 2 100031 0004 2 11 1
That’s interesting: 2 is a solution, but 00031 is not. These results are a special case of Theorem 1.
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Z THEOREM 1 Squaring Operation on Equations If both sides of an equation are squared, then the solution set of the original equation is a subset of the solution set of the new equation. Equation x00053 x2 0005 9
Solution Set {3} {00033, 3}
This theorem provides us with a method of solving some equations involving radicals. It is important to remember that any new equation obtained by raising both sides of an equation to the same power may have solutions, called extraneous solutions, that are not solutions of the original equation. Fortunately though, any solution of the original equation must be among those of the new equation. When raising both sides of an equation to a power, checking solutions is not just a good idea—it is essential to identify any extraneous solutions.
Squaring both sides of the equations x 0002 1x and x 0002 0003 1x produces the new equation x2 0002 x. Find the solutions to the new equation and then check for extraneous solutions in each of the original equations.
ZZZ EXPLORE-DISCUSS 1
EXAMPLE
1
Solving Equations Involving Radicals Solve: (A) x 0004 1x 0003 4 0002 4
SOLUTIONS
(A)
(B) 12x 0004 3 0003 1x 0003 2 0002 2
x 0004 1x 0003 4 0002 4 1x 0003 4 0002 4 0003 x
Isolate radical on one side. Square both sides.
(1x 0003 4)2 0002 (4 0003 x)2
See the upcoming caution on squaring the right side.
x 0003 4 0002 16 0003 8x 0004 x2 x 0003 9x 0004 20 0002 0 (x 0003 5)(x 0003 4) 0002 0 2
CHECK
Write in standard form. Factor left side. Use the zero product property.
x0003500020
or
x0003400020
x00025
or
x00024
x00025 x 0004 1x 0003 4 0002 4 ? 5 0004 15 0003 4 0002 4 600064
x00024 x 0004 1x 0003 4 0002 4 ? 4 0004 14 0003 4 0002 4 ✓ 400024
This shows that 4 is a solution to the original equation and 5 is extraneous. The only solution is x 0002 4.
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99
(B) To solve an equation that contains more than one radical, isolate one radical at a time and square both sides to eliminate the isolated radical. Repeat this process until all the radicals are eliminated. 12x 0004 3 0003 1x 0003 2 0002 2 12x 0004 3 0002 1x 0003 2 0004 2
Isolate one of the radicals. Square both sides. See the upcoming caution on squaring the right side.
(12x 0004 3)2 0002 ( 1x 0003 2 0004 2)2 2x 0004 3 0002 x 0003 2 0004 41x 0003 2 0004 4 x 0004 1 0002 41x 0003 2
Isolate the remaining radical. Square both sides.
(x 0004 1)2 0002 (4 1x 0003 2)2 x2 0004 2x 0004 1 0002 16(x 0003 2) x2 0003 14x 0004 33 0002 0 (x 0003 3)(x 0003 11) 0002 0
CHECK
x0003300020
or
x00023
or
x00023 12x 0004 3 0003 1x 0003 2 0002 2 ? 12(3) 0004 3 0003 13 0003 2 0002 2 ✓ 200022
Write in standard form. Factor left side. Use the zero property.
x 0003 11 0002 0 x 0002 11 x 0002 11 12x 0004 3 0003 1x 0003 2 0002 2 ? 12(11) 0004 3 0003 111 0003 2 0002 2 ✓ 2 00022
Both solutions check, so there are two solutions: x 0002 3, 11. MATCHED PROBLEM 1
Solve: (A) x 0003 5 0002 1x 0003 3
ZZZ
0002
(B) 12x 0004 5 0004 1x 0004 2 0002 5 0002
1. When squaring both sides, it is very important to isolate the radical first. 2. Be sure to square binomials like (4 0003 x) by first writing as (4 0003 x)(4 0003 x) and then multiplying. Remember: (4 0003 x)2 0006 42 0003 x2.
CAUTION ZZZ
Z Revisiting Equations Involving Absolute Value Squaring both sides of an equation can be a useful operation even if the equation does not involve any radicals. Because |x|2 0002 x2 for any x, squaring can be helpful in some absolute value equations.
EXAMPLE
2
Absolute Value Equations Revisited Solve the following equation by squaring both sides: |x 0004 4| 0002 3x 0003 8
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|x 0004 4| 0002 3x 0003 8
SOLUTION
Square both sides.
|x 0004 4|2 0002 (3x 0003 8)2
Use |x 0004 4|2 0005 (x 0004 4)2 and expand each side.
x2 0004 8x 0004 16 0002 9x2 0003 48x 0004 64 8x2 0003 56x 0004 48 0002 0 x2 0003 7x 0004 6 0002 0 (x 0003 1)(x 0003 6) 0002 0
Divide both sides by 8. Factor the left side. Use the zero product property.
x0003100020
or
x0003600020
x00021
or
x00026
x00021 |x 0004 4| 0002 3x 0003 8 ? |1 0004 4| 0002 3(1) 0003 8 ? |5| 0002 00035 5 0006 00035
CHECK
Write in standard form.
x00026 冟 x 0004 4 冟 0002 3x 0003 8 ? |6 0004 4| 0002 3(6) 0003 8 ? 冟 10 冟 0002 10 ✓ 10 0002 10
The only solution is x 0002 6. Compare this solution with the solution of Example 6, Section 1-3. Squaring both sides eliminates the need to consider two separate cases. 0002 MATCHED PROBLEM 2
Solve the following equation by squaring both sides: 冟 3x 0003 4 冟 0002 x 0004 4 0002
Z Solving Equations of Quadratic Type Quadratic equations in standard form have two terms with the variable; one has power 2, the other power 1. When equations have two variable terms where the larger power is twice the smaller, we can use quadratic solving techniques.
EXAMPLE
3
Solving an Equation of Quadratic Type Solve x23 0003 x13 0003 6 0002 0.
SOLUTIONS
Method I. Direct solution: Note that the larger power (23) is twice the smaller. Using the properties of exponents from basic algebra, we can write x23 as (x13)2 and solve by factoring. (x13)2 0003 x13 0003 6 0002 0 (x13 0003 3)(x13 0004 2) 0002 0 or x13 0002 3 x13 0002 00032 13 3
(x
3
) 00023
x 0002 27 The solution is x 0002 27, 00038
13 3
(x
Factor left side. Use the zero product property. Cube both sides. 3
) 0002 (00032) x 0002 00038
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101
Method II. Using substitution: Replace x100053 (the smaller power) with a new variable u. Then the larger power x200053 is u2. This gives us a quadratic equation with variable u. u2 0004 u 0004 6 0003 0 (u 0004 3)(u 0002 2) 0003 0 u 0003 3, 00042
Factor. Use the zero product property.
This is not the solution! We still need to find the values of x that correspond to u 0003 3 and u 0003 00042. Replacing u with x100053, we obtain x100053 0003 3 x 0003 27
or
x100053 0003 00042 x 0003 00048
Cube both sides.
The solution is x 0003 27, 00048. MATCHED PROBLEM 3
0002
Solve algebraically using both Method I and Method II: x100052 0004 5x100054 0002 6 0003 0.
0002
In general, if an equation that is not quadratic can be transformed to the form au2 0002 bu 0002 c 0003 0 where u is an expression in some other variable, then the equation is called an equation of quadratic type. Equations of quadratic type often can be solved using quadratic methods.
ZZZ EXPLORE-DISCUSS 2
Which of the following can be transformed into a quadratic equation by making a substitution of the form u 0003 xn? What is the resulting quadratic equation? (A) 3x00044 0002 2x00042 0002 7 0003 0
(B) 7x5 0004 3x2 0002 3 0003 0
(C) 2x5 0002 4x2 1x 0004 6 0003 0
(D) 8x00042 1x 0004 5x00041 1x 0004 2 0003 0
In general, if a, b, c, m, and n are nonzero real numbers, when can an equation of the form axm 0002 bxn 0002 c 0003 0 be transformed into an equation of quadratic type?
EXAMPLE
4
Solving an Equation of Quadratic Type Solve: 3x4 0004 5x2 0002 1 0003 0
SOLUTION
If we let u 0003 x2, then u2 0003 x4, and the equation becomes 3u2 0004 5u 0002 1 0003 0 5 0006 113 u0003 6 x2 0003
5 0006 113 6
x00030006
MATCHED PROBLEM 4
5 0006 113 B 6
Use the quadratic formula with a ⴝ 3, b ⴝ ⴚ5, c ⴝ 1. Substitute x2 back in for u.
Use the square root property to solve for x.
There are four solutions.
0002
Solve: 2x4 0002 3x2 0004 4 0003 0 0002
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Many applied problems result in equations that can be solved using the techniques in this section.
EXAMPLE
5
An Application: Court Design A hardcourt version of the game broomball becomes popular on college campuses because it enables people to hit each other with a stick. The court is a rectangle with diagonal 30 feet and area 400 square feet. Find the dimensions to one decimal place.
SOLUTION
t
30
fee
Draw a rectangle and label the dimensions as shown in Figure 1. The area is given by A 0002 xy. Also, x2 0004 y2 0002 302 (Pythagorean theorem), and we can solve for y to get y 0002 2900 0003 x2. Now substitute in for y in our area equation, then set area equal to 400 and solve.
y
x 2900 0003 x2 0002 400 x2(900 0003 x2) 0002 160,000 2
4
900x 0003 x 0002 160,000
x
2 2
2
(x ) 0003 900x 0004 160,000 0002 0
Z Figure 1
Square both sides. Multiply out parentheses. Write in standard quadratic form.
Use quadratic formula with a 0005 1, b 0005 0003900, and c 0005 160,000.
900 0005 2(0003900)2 0003 4(1)(160,000) x2 0002 2 0002
900 0005 1170,000 2
Simplify inside the square root.
Use a calculator.
x2 ⬇ 656.2, 243.8 x 0002 1656.2 ⬇ 25.6 or 1243 ⬇ 15.6
Use square root property; discard negative solutions.
If x 0002 25.6, then y 0002 2900 0003 25.62 ⬇ 15.6. If x 0002 15.6, then y 0002 2900 0003 15.62 ⬇ 25.6. In either case, the dimensions are 25.6 feet by 15.6 feet. CHECK Area: 25.6 0007 15.6 0002 399.36 ⬇ 400
Diagonal: 225.62 0004 15.62 ⬇ 30 MATCHED PROBLEM 5
0002
If the area of a right triangle is 24 square inches and the hypotenuse is 12 inches, find the lengths of the legs of the triangle correct to one decimal place. 0002 ANSWERS TO MATCHED PROBLEMS 1. (A) x 0002 7 (B) x 0002 2 2. x 0002 0, 4 3. x 0002 16, 81 0005200033 0005 141 4. x 0002 5. 11.2 inches by 4.3 inches 2
1-6
Exercises
1. What is meant by the term “extraneous solution”? 2. When is it necessary to check for extraneous solutions? 3. How can squaring both sides help in solving absolute value equations?
4. How can you recognize when an equation is of quadratic type? In Problems 5–12, determine the validity of each statement. If a statement is false, explain why. 5. If x2 0002 5, then x 0002 0005 15.
6. 125 0002 00055
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7. (x 0004 5)2 0002 x2 0004 25 9. (1x 0003 1 0004 1)2 0002 x 11. If x3 0002 2, then x 0002 8.
8. (2x 0003 1)2 0002 4x2 0003 1 10. (1x 0003 1)2 0004 1 0002 x 12. If x1兾3 0002 8, then x 0002 2.
Additional Equation-Solving Techniques
59. t 0003 111t 0004 18 0002 0
103
60. x 0002 15 0003 21x
In Problems 61–68, solve the equation. 61. 17 0003 2x 0003 1x 0004 2 0002 1x 0004 5 62. 11 0004 3x 0003 12x 0003 1 0002 1x 0004 2
In Problems 13–26, solve the equation. 13. 1x 0004 2 0002 4
14. 1x 0003 4 0002 2
63. 3 0004 x00034 0002 5x00032
64. 2 0004 4x00034 0002 7x00032
15. 13y 0003 5 0004 10 0002 0
16. 14 0003 x 0004 5 0002 0
65. 21x 0004 5 0002 0.01x 0004 2.04
17. 13y 0003 2 0002 y 0003 2
18. 14y 0004 1 0002 5 0003 y
66. 3 1x 0003 1 0002 0.05x 0004 2.9
19. 15w 0004 6 0003 w 0002 2
20. 12w 0003 3 0004 w 0002 1
67. 2x000325 0003 5x000315 0004 1 0002 0
21. 冟 2x 0004 1 冟 0002 x 0004 2
22. 冟 2x 0004 2 冟 0002 5 0003 x
68. x000325 0003 3x000315 0004 1 0002 0
23. 冟 x 0003 5 冟 0002 7 0003 2x
24. 冟 x 0004 7 冟 0002 1 0003 2x
69. Explain why the following “solution” is incorrect:
25. 冟 3x 0003 4 冟 0002 2x 0003 5
26. 冟 3x 0003 1 冟 0002 x 0003 1
1x 0004 3 0004 5 0002 12 x 0004 3 0004 25 0002 144
In Problems 27–32, transform each equation of quadratic type into a quadratic equation in u and state the substitution used in the transformation. If the equation is not an equation of quadratic type, say so. 4 3 6 27. 2x00036 0003 4x00033 0002 0 28. 0003 0004 2 0002 0 x 7 x 29. 3x3 0003 4x 0004 9 0002 0
30. 7x00031 0004 3x00031/2 0004 2 0002 0
10 4 7 0004 20003 400020 9 x x
32. 3x3/2 0003 5x1/2 0004 12 0002 0
31.
In Problems 33–56, solve the equation. 33. 13t 0003 2 0002 1 0003 2 1t
34. 15t 0004 4 0003 21t 0002 1
35. m4 0004 2m2 0003 15 0002 0
36. m4 0004 4m2 0003 12 0002 0
37. 3x 0002 2x2 0003 2
38. x 0002 25x2 0004 9
23
39. 2y
13
0004 5y
0003 12 0002 0
2
2
2
2
2
2
23
40. 3y
13
0004 2y
0003800020
41. (m 0003 2m) 0004 2(m 0003 2m) 0002 15 42. (m 0004 2m) 0003 6(m 0004 2m) 0002 16 43. 12t 0004 3 0004 2 0002 1t 0003 2
70. Explain why the following “solution” is incorrect. 2x2 0003 16 0002 2x 0004 3 x 0003 4 0002 2x 0004 3 00037 0002 x
APPLICATIONS 71. PHYSICS—WELL DEPTH When a stone is dropped into a deep well, the number of seconds until the sound of a splash is heard is x 1x given by the formula t 0002 0004 , where x is the depth of the 4 1,100 well in feet. For one particular well, the splash is heard 14 seconds after the stone is released. How deep (to the nearest foot) is the well? 72. PHYSICS—WELL DEPTH Refer to Problem 71. For a different well, the sound of the splash is heard 2 seconds after the stone is released. How deep (to the nearest foot) is the well? 73. GEOMETRY The diagonal of a rectangle is 10 inches and the area is 45 square inches. Find the dimensions of the rectangle, correct to one decimal place. 74. GEOMETRY The hypotenuse of a right triangle is 12 inches and the area is 24 square inches. Find the dimensions of the triangle, correct to one decimal place.
44. 12x 0003 1 0003 1x 0003 5 0002 3 45. 1w 0004 3 0004 12 0003 w 0002 3
75. MANUFACTURING A lumber mill cuts rectangular beams from circular logs (see the figure). If the diameter of the log is 16 inches and the cross-sectional area of the beam is 120 square inches, find the dimensions of the cross section of the beam correct to one decimal place.
46. 1w 0004 7 0002 2 0004 13 0003 w 47. 18 0003 z 0002 1 0004 1z 0004 5 48. 13z 0004 1 0004 2 0002 1z 0003 1 49. 24x2 0004 12x 0004 1 0003 6x 0002 9 50. 6x 0003 24x2 0003 20x 0004 17 0002 15 51. y00032 0003 2y00031 0004 3 0002 0
52. y00032 0003 3y00031 0004 4 0002 0
53. 2t00034 0003 5t00032 0004 2 0002 0
54. 15t00034 0003 23t00032 0004 4 0002 0
55. 3z00031 0003 3z00031/2 0004 1 0002 0
56. 2z00031 0003 3z00031/2 0004 2 0002 0
Solve Problems 57–60 two ways: by squaring and by substitution. 57. m 0003 7 1m 0004 12 0002 0
x 0002 116
58. y 0003 6 0004 1y 0002 0
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76. DESIGN A food-processing company packages an assortment of their products in circular metal tins 12 inches in diameter. Four identically sized rectangular boxes are used to divide the tin into eight compartments (see the figure). If the cross-sectional area of each box is 15 square inches, find the dimensions of the boxes correct to one decimal place.
78. DESIGN A paper drinking cup in the shape of a right circular cone is constructed from 125 square centimeters of paper (see the figure). If the height of the cone is 10 centimeters, find the radius correct to two decimal places. r
h
77. CONSTRUCTION A water trough is constructed by bending a 4- by 6-foot rectangular sheet of metal down the middle and attaching triangular ends (see the figure). If the volume of the trough is 9 cubic feet, find the width correct to two decimal places.
Lateral surface area: S ⫽ r 兹r 2 ⫹ h 2
6 feet
2 feet
CHAPTER
1-1
1
Review
Linear Equations and Applications
Solving an equation is the process of finding all values of the variable that make the equation a true statement. An equation that is true for some values of the variable is called a conditional equation. An equation that is true for all permissible values of the variable is called an identity. An equation that is false for all permissible values of the variable is called a contradiction, and has no solution. An equation that can be written in the standard form ax ⫹ b ⫽ 0, a ⫽ 0, is a linear or first-degree equation. Linear
equations are solved by performing algebraic steps that result in equivalent equations until the result is an equation whose solution is obvious. When an equation has fractions, begin by multiplying both sides by the least common denominator of all the fractions. The formula Quantity ⫽ Rate ⫻ Time is useful in modeling problems that involve a rate of change, like speed.
Z STRATEGY FOR SOLVING WORD PROBLEMS 1. Read the problem slowly and carefully, more than once if
4. Write an equation relating the quantities in the problem.
necessary. Write down information as you read the problem the first time to help you get started. Identify what it is that you are asked to find.
Often, you can accomplish this by finding a formula that connects those quantities. Try to write the equation in words first, then translate to symbols.
2. Use a variable to represent an unknown quantity in the
5. Solve the equation, then answer the question in a sentence
problem, usually what you are asked to find. Then try to represent any other unknown quantities in terms of that variable. It’s pretty much impossible to solve a word problem without this step.
by rephrasing the question. Make sure that you’re answering all of the questions asked.
3. If it helps to visualize a situation, draw a diagram and label known and unknown parts.
6. Check to see if your answers make sense in the original problem, not just the equation you wrote.
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1-2
Linear Inequalities
The inequality symbols , , 000e, are used to express inequality relations. Line graphs, interval notation, and the set operations of union and intersection are used to describe inequality relations. A solution of a linear inequality in one variable is a value of the variable that makes the inequality a true statement. Two inequalities are equivalent if they have the same solution set. Linear inequalities can be solved using the same basic procedure as linear equations, with one important difference: the direction of an inequality reverses if we multiply or divide both sides by a negative number.
1-3
Absolute Value in Equations and Inequalities
The absolute value of a number x is the distance on a real number line from the origin to the point with coordinate x and is given by 冟x冟 0002
0003x x
冦
if x 6 0 if x 0
The distance between points A and B with coordinates a and b, respectively, is d(A, B) 0002 冟 b 0003 a 冟, which has the following geometric interpretations: 冟 x 0003 c 冟 0002 d Distance between x and c is equal to d.
Because complex numbers obey the same commutative, associative, and distributive properties as real numbers, most operations with complex numbers are performed by using these properties in the same way that algebraic operations are performed on the expression a 0004 bx. Keep in mind that i2 0002 00031. The property of conjugates, (a 0004 bi)(a 0003 bi) 0002 a2 0004 b2 can be used to find reciprocals and quotients. To divide by a complex number, we multiply the numerator and denominator by the conjugate of the denominator. This enables us to write the result in a 0004 bi form. If a 7 0, then the principal square root of the negative real number 0003a is 10003a 0002 i1a. To solve equations involving complex numbers, set the real and imaginary parts equal to each other and solve.
1-5
冟 x 0003 c 冟 7 d Distance between x and c is greater than d.
ax2 0004 bx 0004 c 0002 0
1. Factoring and using the zero product property: m0002n00020
m 0002 0 or n 0002 0 (or both)
If A2 0002 C, then A 0002 00051C 3. Completing the square: b 2 b 2 x2 0004 bx 0004 a b 0002 ax 0004 b 2 2 4. Using the quadratic formula: x0002
3. 冟 x 冟 7 p is equivalent to x 6 0003p or x 7 p.
1-4
if and only if
2. Using the square root property:
2. 冟 x 冟 6 p is equivalent to 0003p 6 x 6 p. These relationships also hold if x is replaced with ax 0004 b. For x any real number, 2x2 0002 冟 x 冟.
a00060
where x is a variable and a, b, and c are constants. Methods of solution include:
Equations and inequalities involving absolute values are solved using the following relationships for p 0: 1. 冟 x 冟 0002 p is equivalent to x 0002 p or x 0002 0003p.
Quadratic Equations and Applications
A quadratic equation is an equation that can be written in the standard form
冟 x 0003 c 冟 6 d Distance between x and c is less than d. 0 6 冟 x 0003 c 冟 6 d Distance between x and c is less than d, but x 0006 c.
105
0003b 0005 2b2 0003 4ac 2a
If the discriminant b2 0003 4ac is positive, the equation has two distinct real roots; if the discriminant is 0, the equation has one real double root; and if the discriminant is negative, the equation has two imaginary roots, each the conjugate of the other.
Complex Numbers
A complex number in standard form is a number in the form a 0004 bi where a and b are real numbers and i denotes a square root of 00031. The number i is known as the imaginary unit. For a complex number a 0004 bi, a is the real part and bi is the imaginary part. If b 0006 0 then a 0004 bi is also called an imaginary number. If a 0002 0 then 0 0004 bi 0002 bi is also called a pure imaginary number. If b 0002 0 then a 0004 0i 0002 a is a real number. The complex zero is 0 0004 0i 0002 0. The conjugate of a 0004 bi is a 0003 bi. Equality, addition, and multiplication are defined as follows:
1. a 0004 bi 0002 c 0004 di if and only if a 0002 c and b 0002 d 2. (a 0004 bi) 0004 (c 0004 di) 0002 (a 0004 c) 0004 (b 0004 d)i 3. (a 0004 bi)(c 0004 di) 0002 (ac 0003 bd) 0004 (ad 0004 bc)i
1-6
Additional Equation-Solving Techniques
A square root radical can be eliminated from an equation by isolating the radical on one side of the equation and squaring both sides of the equation. The new equation formed by squaring both sides may have extraneous solutions. Consequently, every solution of the new equation must be checked in the original equation to eliminate extraneous solutions. If an equation contains more than one radical, then the process of isolating a radical and squaring both sides can be repeated until all radicals are eliminated. If a substitution transforms an equation into the form au2 0004 bu 0004 c 0002 0, where u is an expression in some other variable, then the equation is an equation of quadratic type that can be solved by quadratic methods.
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Review Exercises
Work through all the problems in this chapter review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text.
1. 8x 0004 10 0002 4x 0003 30 2. 4 0003 3(x 0004 2) 0002 5x 0003 7(4 0003 x) y 0004 10 y00041 1 1 0003 0002 0003 15 5 6 10
Solve the equation in Problems 25–30. 25. ay 0004
Solve and graph the inequality in Problems 4–6. 4. 3(2 0003 x) 0003 2 000e 2x 0003 1
5. 冟 y 0004 9 冟 6 5
26. 1 0004
6. 冟 3 0003 2x 冟 000e 5 7. Find the real part, the imaginary part, and the conjugate: (A) 9 0003 4i (B) 5i (C) 000310
27.
8. Perform the indicated operations and write the answer in standard form. (A) (4 0004 7i) 0004 (00032 0003 3i) (B) (00033 0004 5i) 0003 (4 0003 8i) (C) (1 0003 2i)(3 0004 4i) 21 0004 9i (D) 5 0003 2i
12. 2x2 0002 7x 0003 3
13. m2 0004 m 0004 1 0002 0
14. y2 0002 32 ( y 0004 1)
2 3 0002 u u2
2 x 0003 0002 3 28. 2x23 0003 5x13 0003 12 0002 0 x00033 x 0003x00036 2
29. m4 0004 5m2 0003 36 0002 0
30. 1y 0003 2 0003 15y 0004 1 0002 00033
Solve the equation or inequality in Problems 31–35, and round answers to three significant digits if necessary. 31. 2.15x 0003 3.73(x 0003 0.930) 0002 6.11x 32. 00031.52 000e 0.770 0003 2.04x 000e 5.33
4 1 8 34. 2 0003 t2 000e 3 5 2
10. 5x2 0004 20 0002 0
11. 2x2 0002 4x
11 2 b 0002 20 3
33. 冟 9.71 0003 3.62x 冟 7 5.48
Solve the equation in Problems 9–15. 9. 2x2 0003 7 0002 0
23. Perform the indicated operations and write the final answers in standard form: (A) (3 0004 i)2 0003 2(3 0004 i) 0004 3 (B) i 27 24. Convert to a 0004 bi forms, perform the indicated operations, and write the final answers in standard form: (A) (2 0003 100034) 0003 (3 0003 100039) 4 0004 1000325 2 0003 100031 (B) (C) 3 0004 100034 100034
In Problems 1–3, solve the equation.
3.
22. If points A, B, and C have coordinates on a number line of 5, 20, and 00038 respectively, find (A) d(A, B) (B) d(A, C) (C) d(B, C)
35. 6.09x2 0004 4.57x 0003 8.86 0002 0
15. 15x 0003 6 0003 x 0002 0 16. For what values of x does the expression 115 0004 6x represent a real number?
Solve the equation in Problems 36–38 for the indicated variable in terms of the other variables. 36. P 0002 M 0003 Mdt for M (mathematics of finance) 37. P 0002 EI 0003 RI 2 for I (electrical engineering)
Solve the equation in Problems 17 and 18. 7 10 0003 4x 17. 0002 2 20003x x 0004 3x 0003 10
38. x 0002
u00033 1 10003u 18. 0002 0003 2u 0003 2 6 3u 0003 3
x00043 20003x 000e50003 8 3
21. 2(1 0003 2m)2 000e 3
for y
39. Find the error in the following “solution” and then find the correct solution. 3 4 0002 2 x2 0003 4x 0004 3 x 0003 3x 0004 2
Solve and graph the inequality in Problems 19–21. 19.
4y 0004 5 2y 0004 1
20. 冟 3x 0003 8 冟 7 2 [
00031
[
2
m
4x2 0003 12x 0004 8 0002 3x2 0003 12x 0004 9 x2 0002 1 or x00021 x 0002 00031
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40. Consider the quadratic equation x2 ⫺ 8x ⫹ c ⫽ 0 , where c is a real number. Describe the number and type of solutions for c ⫽ ⫺16, 16, and 32. Use your result to make a general statement about the number and type of solutions for certain values of c, then use an inequality to prove your statement. 41. For what values of a and b is the inequality a ⫹ b 6 b ⫺ a true? 42. If a and b are negative numbers and a 7 b, then is aⲐb greater than 1 or less than 1? 1
43. Solve for x in terms of y: y ⫽ 1⫺
1 1⫺x
44. Solve and graph: 0 6 冟 x ⫺ 6 冟 6 d Solve the equation in Problems 45–47. 45. 2x2 ⫽ 13x ⫺ 12 46. 4 ⫽ 8x⫺2 ⫺ x⫺4 47. 2ix2 ⫹ 3ix ⫺ 5i ⫽ 0 48. Evaluate: (a ⫹ bi) a
a b ⫺ 2 ib, a, b ⫽ 0 a2 ⫹ b2 a ⫹ b2
(C) If the crew wants to increase their still-water speed to 18 km/h, how fast must they make the round-trip? Express answer in minutes rounded to one decimal place. 54. COST ANALYSIS Cost equations for manufacturing companies are often quadratic in nature. If the cost equation for manufacturing inexpensive calculators is C ⫽ x2 ⫺ 10x ⫹ 31, where C is the cost of manufacturing x units per week (both in thousands), find: (A) The output for a $15 thousand weekly cost (B) The output for a $6 thousand weekly cost 55. BREAK-EVEN ANALYSIS The manufacturing company in Problem 54 sells its calculators to wholesalers for $3 each. So its revenue equation is R ⫽ 3x, where R is revenue and x is the number of units sold per week (both in thousands). Find the break-even point(s) for the company—that is, the output at which revenue equals cost. 56. POLITICS Before the 2008 presidential election, one news outlet estimated that the percentage of voters casting a vote for Barack Obama would be within 1.2% of 54%. Express this range as an absolute value inequality, then solve the inequality. 57. DESIGN The pages of a textbook have uniform margins of 2 centimeters on all four sides (see the figure). If the area of the entire page is 480 square centimeters and the area of the printed portion is 320 square centimeters, find the dimensions of the page.
APPLICATIONS 49. NUMBERS Find a number such that subtracting its reciprocal 16 from the number gives 15 . 50. SPORTS MEDICINE The following quotation was found in a sports medicine handout: “The idea is to raise and sustain your heart rate to 70% of its maximum safe rate for your age. One way to determine this is to subtract your age from 220 and multiply by 0.7.” (A) If H is the maximum safe sustained heart rate (in beats per minute) for a person of age A (in years), write a formula relating H and A. (B) What is the maximum safe sustained heart rate for a 20-yearold? (C) If the maximum safe sustained heart rate for a person is 126 beats per minute, how old is the person? 51. CHEMISTRY A chemical storeroom has an 80% alcohol solution and a 30% alcohol solution. How many milliliters of each should be used to obtain 50 milliliters of a 60% solution? 52. RATE–TIME A student group flies to Cancun for spring break, a distance of 1,200 miles. The plane used for both trips has an average cruising speed of 300 miles per hour in still air. The trip down is with the prevailing winds and takes 112 hours less than the trip back, against the same strength wind. What is the wind speed? 53. RATE–TIME A crew of four practices by rowing up a river for a fixed distance and then returning to their starting point. The river has a current of 3 km/h. (A) Currently the crew can row 15 km/h in still water. If it takes them 25 minutes to make the round-trip, how far upstream did they row? (B) After some additional practice the crew cuts the round-trip time to 23 minutes. What is their still-water speed now? Round answers to one decimal place.
107
2
2
2
2
2
2 2
2
Figure for 57.
58. DESIGN A landscape designer uses 8-foot timbers to form a pattern of isosceles triangles along the wall of a building (see the figure). If the area of each triangle is 24 square feet, find the base correct to two decimal places.
8 feet
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1
GROUP ACTIVITY Solving a Cubic Equation
If a, b, and c are real numbers with a 0006 0, then the quadratic equation ax2 0004 bx 0004 c 0002 0 can be solved by a variety of methods, including the quadratic formula. How can we solve the cubic equation 3
2
ax 0004 bx 0004 cx 0004 d 0002 0,
a00060
(1)
and is there a formula for the roots of this equation? The first published solution of equation (1) is generally attributed to the Italian mathematician Girolamo Cardano (1501–1576) in 1545. His work led to a complicated formula for the roots of equation (1) that involves topics that are discussed later in this text. For now, we will use Cardano’s method to find a real solution in special cases of equation (1). Note that because a is nonzero, we can always multiply both sides of (1) by 1 a to make the coefficient of x3 equal to 1.
Let x3 0004 bx2 0004 cx 0004 d 0002 0 Example problem: x3 0003 6x2 0004 6x 0003 5 0005 0. Steps will be in red.
Step 1. Substitute x 0002 y 0003 b兾3 to obtain the reduced cubic y3 0004 my 0002 n. 00036 x0005y0003 or x 0005 y 0004 2. The equation becomes 3 (y 0004 2)3 0003 6(y 0004 2) 0004 6(y 0004 2) 0003 5 0005 0,
3
Step 2. Define u and v by m 0002 3uv and n 0002 u 0003 v . Use v 0002 write
m 3 b 3u
Step 3. Using either of the solutions found in step 2, x0002y0003
b b 0002u0003v0003 3 3
is a solution to x3 0004 bx2 0004 cx 0004 d 0002 0 For u 0005 2, v 0005 00031, x 0005 2 0003 (00031) 0003
00036 0005 5 3
(Solution)
x3 0003 6x2 0003 3x 0003 8 0002 0
m 3u
to
(2)
Use a calculator to find a decimal approximation of your solution and check your answer by substituting this approximate value in equation (2). (C) Use Cardano’s method to solve x3 0003 6x2 0004 9x 0003 6 0002 0
which simplifies to y3 0003 6y 0005 9: m 0005 00036, n 0005 9.
n 0002 u3 0003 a
00036 2 2 3 8 or v 0005 0003 ; 9 0005 u3 0003 a0003 b 0005 u3 0004 Multiply both sides by 3u u u u3 3 6 3 u to obtain u 0003 9u 0004 8 0005 0; solve by factoring to get u 0005 2 (in which case v 0005 00031) or u 0005 1 (in which case v 0005 00032). v0005
(A) The key to Cardano’s method is to recognize that if u and v are defined as in step 2, then y 0002 u 0003 v is a solution of the reduced cubic. Verify this by substituting y 0002 u 0003 v, m 0002 3uv, and n 0002 u3 0003 y3 in y3 0004 my 0002 n and show that the result is an identity. (B) Use Cardano’s method to solve
CARDANO’S METHOD FOR SOLVING A CUBIC EQUATION
3
Multiply both sides by u3 to obtain an equation quadratic in u3. Solve for u3 by factoring or by using the quadratic formula. Then solve for u, and find the associated value of v.
(3)
Use a calculator to find a decimal approximation of your solution and check your answer by substituting this approximate value in equation (3). (D) In step 2 of Cardano’s method, show that u3 is real if n 2 0003m 3 a b a b. 2 3
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CHAPTER
Graphs C EQUATIONS and inequalities are algebraic objects. A graph, on the other hand, is a geometric object such as a line, circle, or parabola. The idea of visualizing an equation or inequality by means of a graph was crucial to the development of analytic geometry, a subject that combines algebra and geometry. In this chapter, we study the fundamentals of analytic geometry: The Cartesian coordinate system, named after the French mathematician and philosopher René Descartes (1596–1650); the calculation of distances in the plane; and equations of lines and circles. We conclude the chapter by applying linear models to solve real-world problems.
2 OUTLINE 2-1
Cartesian Coordinate Systems
2-2
Distance in the Plane
2-3
Equations of a Line
2-4
Linear Equations and Models Chapter 2 Review Chapter 2 Group Activity: Average Speed
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2-1
Cartesian Coordinate Systems Z Reviewing Cartesian Coordinate Systems Z Graphing: Point by Point Z Using Symmetry as an Aid in Graphing
In Chapter 1, we discussed algebraic methods for solving equations. In this section we show how to find a geometric representation ( graph) of an equation. Examining the graph of an equation often results in additional insight into the nature of the equation’s solutions.
Z Reviewing Cartesian Coordinate Systems y 10
II
I
000210
x
10
III
IV
000210
Z Figure 1 Cartesian coordinate system.
y 10
R 0003 (5, 10)
Q 0003 (000210, 5) a 000210
Origin b (0, 0)
10
P 0003 (a, b)
000210
Z Figure 2 Coordinates in a plane.
x
Just as a real number line is formed by establishing a one-to-one correspondence between the points on a line and the elements in the set of real numbers, we can form a real plane by establishing a one-to-one correspondence between the points in a plane and elements in the set of all ordered pairs of real numbers. This can be done by means of a Cartesian coordinate system. To form a Cartesian or rectangular coordinate system, we select two real number lines, one horizontal and one vertical, and let them cross through their origins, as indicated in Figure 1. Up and to the right are the usual choices for the positive directions. These two number lines are called the horizontal axis and the vertical axis, or, together, the coordinate axes. The horizontal axis is usually referred to as the x axis and the vertical axis as the y axis, and each is labeled accordingly. Other labels may be used in certain situations. The coordinate axes divide the plane into four parts called quadrants, which are numbered counterclockwise from I to IV (see Fig. 1). Given an arbitrary point P in the plane, pass horizontal and vertical lines through the point (Fig. 2). The vertical line will intersect the horizontal axis at a point with coordinate a, and the horizontal line will intersect the vertical axis at a point with coordinate b. These two numbers written as the ordered pair* (a, b) form the coordinates of the point P. The first coordinate a is called the abscissa of P; the second coordinate b is called the ordinate of P. The abscissa of Q in Figure 2 is 000210, and the ordinate of Q is 5. The coordinates of a point can also be referenced in terms of the axis labels. The x coordinate of R in Figure 2 is 5, and the y coordinate of R is 10. The point with coordinates (0, 0) is called the origin. The procedure we have just described assigns to each point P in the plane a unique pair of real numbers (a, b). Conversely, if we are given an ordered pair of real numbers (a, b), then, reversing this procedure, we can determine a unique point P in the plane. There is a one-to-one correspondence between the points in a plane and the elements in the set of all ordered pairs of real numbers. This correspondence is often referred to as the fundamental theorem of analytic geometry. Because of this correspondence, we regularly speak of the point (a, b) when we are referring to the point with coordinates (a, b). We also write P 0003 (a, b) to identify the coordinates of the point P. In Figure 2, referring to Q as the point (000210, 5) and writing R 0003 (5, 10) are both acceptable statements.
*An ordered pair of real numbers is a pair of numbers in which the order is specified. We now use (a, b) as the coordinates of a point in a plane. In Chapter 1, we used (a, b) to represent an interval on a real number line. These concepts are not the same. You must always interpret the symbol (a, b) in terms of the context in which it is used.
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Z Graphing: Point by Point Given any set of ordered pairs of real numbers S, the graph of S is the set of points in the plane corresponding to the ordered pairs in S. The fundamental theorem of analytic geometry enables us to look at an algebraic object (a set of ordered pairs) geometrically and to look at a geometric object (a graph) algebraically. We begin by considering an equation in two variables: y 0003 x2 0002 4
(1)
A solution to equation (1) is an ordered pair of real numbers (a, b) such that b 0003 a2 0002 4. The solution set of equation (1) is the set of all its solutions. To find a solution to equation (1) we simply replace one of the variables with a number and solve for the other variable. For example, if x 0003 2, then y 0003 22 0002 4 0003 0, and the ordered pair (2, 0) is a solution. Similarly, if y 0003 5, then 5 0003 x2 0002 4, x2 0003 9, x 0003 00043, and the ordered pairs (3, 5) and (00023, 5) are solutions. Sometimes replacing one variable with a number and solving for the other variable will introduce imaginary numbers. For example, if y 0003 00025 in equation (1), then 00025 0003 x2 0002 4 x2 0003 00021 x 0003 0004 100021 0003 0004i So (0002i, 00025) and (i, 00025) are solutions to y 0003 x2 0002 4. However, the coordinates of a point in a rectangular coordinate system must be real numbers. For that reason, when graphing an equation, we consider only those values of the variables that produce real solutions to the equation. The graph of an equation in two variables is the graph of its solution set. In equation (1), we find that its solution set will have infinitely many elements and its graph will extend off any paper we might choose, no matter how large. To sketch the graph of an equation, we include enough points from its solution set so that the total graph is apparent. This process is called point-by-point plotting.
EXAMPLE
1
Graphing an Equation Using Point-by-Point Plotting Sketch a graph of y 0003 x2 0002 4.
SOLUTION y
y 0003 x2 0002 4
We make a table of solutions—ordered pairs of real numbers that satisfy the given equation.
15
(00024, 12)
(4, 12) 10
(00023, 5)
5
(00022, 0)
(3, 5)
(2, 0)
00025
5
(00021, 00023) 00025
Z Figure 3
(1, 00023) (0, 00024)
x
x
00024
00023
00022
00021
0
1
2
3
4
y
12
5
0
00023
00024
00023
0
5
12
After plotting these solutions, if there are any portions of the graph that are unclear, we plot additional points until the shape of the graph is apparent. Then we join all these plotted points with a smooth curve, as shown in Figure 3. Arrowheads are used to indicate that the graph continues beyond the portion shown here with no significant changes in shape. The resulting figure is called a parabola. Notice that if we fold the paper along the y axis, the right side will match the left side. We say that the graph is symmetric with respect to the y axis and call the y axis the axis of the parabola. More will be said about parabolas later in the text. 0002
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MATCHED PROBLEM 1
Sketch a graph of y2 0003 x. 0002 This book contains a number of activities that use a graphing calculator or computer with appropriate software. All of these activities are clearly marked and easily omitted if no such device is available.
Technology Connections To graph the equation in Example 1 on a graphing calculator, we first enter the equation in the calculator’s equation editor* [Fig. 4(a)]. Using Figure 3 as a guide, we next enter values for the window variables [Fig. 4(b)], and then we graph the equation [Fig. 4(c)]. The values of the window variables, shown in red in Figure 4(c), are not displayed on the calculator screen. We add them to give you additional insight into the graph.
Compare the graphs in Figure 3 and Figure 4(c). They are similar in shape, but they are not identical. The discrepancy is due to the difference in the axes scales. In Figure 3, one unit on the x axis is equal to one unit on the y axes. In Figure 4(c), one unit on the x axis is equal to about three units on the y axis. We will have more to say about axes scales later in this section. 15
00025
5
00025
Enter the equation. (a)
Enter the window variables. (b)
Graph the equation. (c)
Z Figure 4
*See the Technology Index for a list of graphing calculator terms used in this book.
ZZZ EXPLORE-DISCUSS 1
To graph the equation y 0003 0002x3 0005 2x, we use point-by-point plotting to obtain the graph in Figure 5. (A) Do you think this is the correct graph of the equation? If so, why? If not, why? (B) Add points on the graph for x 0003 00022, 00020.5, 0.5, and 2. (C) Now, what do you think the graph looks like? Sketch your version of the graph, adding more points as necessary. (D) Write a short statement explaining any conclusions you might draw from parts A, B, and C.
y 5
x
y
00021 00021 0 0 1 1
00025
5
00025
Z Figure 5
x
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Graphs illustrate the relationship between two quantities, one represented by x coordinates and the other by y coordinates. If no equation for the graph is available, we can find specific examples of this relationship by estimating coordinates of points on the graph. Example 2 illustrates this process.
EXAMPLE
2
Ozone Levels The ozone level during a 12-hour period in a suburb of Milwaukee, Wisconsin, on a particular summer day is given in Figure 6, where L is ozone in parts per billion and t is time in hours. Use this graph to estimate the following ozone levels to the nearest integer and times to the nearest quarter hour. (A) The ozone level at 6 P.M. (B) The highest ozone level and the time when it occurs. (C) The time(s) when the ozone level is 90 ppb. L 120
Parts per billion (ppb)
100
80
60
40
20
0 Noon 1
2
3
4
5
6
7
8
9
10
11
12
t
Z Figure 6 Ozone level. SOLUTIONS
MATCHED PROBLEM 2
(A) The L coordinate of the point on the graph with t coordinate 6 is approximately 97 ppb. (B) The highest ozone level is approximately 109 ppb at 3 P.M. (C) The ozone level is 90 ppb at about 12:30 P.M. and again at 10 P.M. 0002 Use Figure 6 to estimate the following ozone levels to the nearest integer and times to the nearest quarter hour. (A) The ozone level at 7 P.M. (B) The time(s) when the ozone level is 100 ppb. 0002
Z Using Symmetry as an Aid in Graphing We noticed that the graph of y 0003 x2 0002 4 in Example 1 is symmetric with respect to the y axis; that is, the two parts of the graph coincide if the paper is folded along the y axis. Similarly, we say that a graph is symmetric with respect to the x axis if the parts above and
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below the x axis coincide when the paper is folded along the x axis. To make the intuitive idea of folding a graph along a line more concrete, we introduce two related concepts— reflection and symmetry.
Z DEFINITION 1 Reflection 1. The reflection through the y axis of the point (a, b) is the point (0002a, b). 2. The reflection through the x axis of the point (a, b) is the point (a, 0002b). 3. The reflection through the origin of the point (a, b) is the point (0002a, 0002b). 4. To reflect a graph just reflect each point on the graph.
EXAMPLE
3
Reflections In a Cartesian coordinate system, plot the point P 0003 (4, 00022) along with its reflection through (A) the y axis, (B) the x axis, (C) and the origin. y
SOLUTION 5
C 0003 (00024, 2)
B 0003 (4, 2)
00025
5
A 0003 (00024, 00022)
x
P 0003 (4, 00022) 00025
MATCHED PROBLEM 3
0002
In a Cartesian coordinate system, plot the point P 0003 (00023, 5) along with its reflection through (A) the x axis, (B) the y axis, and (C) the origin. 0002
Z DEFINITION 2 Symmetry A graph is symmetric with respect to 1. The x axis if (a, 0002b) is on the graph whenever (a, b) is on the graph— reflecting the graph through the x axis does not change the graph. 2. The y axis if (0002a, b) is on the graph whenever (a, b) is on the graph— reflecting the graph through the y axis does not change the graph. 3. The origin if (0002a, 0002b) is on the graph whenever (a, b) is on the graph—reflecting the graph through the origin does not change the graph.
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Figure 7 illustrates these three types of symmetry. y (0002a, b)
y
y (a, b)
Symmetry with respect to x axis (b)
Symmetry with respect to y axis (a)
(a, b)
x
x (a, 0002b)
(0002a, b)
(a, b)
(a, b) x
y
x (0002a, 0002b)
(0002a, 0002b)
(a, 0002b)
Symmetry with respect to y axis, x axis, and origin (d)
Symmetry with respect to origin (c)
Z Figure 7 Symmetry.
ZZZ EXPLORE-DISCUSS 2
If a graph possesses two of the three types of symmetry in Definition 1, must it also possess the third? Explain.
Given an equation, if we can determine the symmetry properties of its graph ahead of time, we can save a lot of time and energy in sketching the graph. For example, we know that the graph of y 0003 x2 0002 4 in Example 1 is symmetric with respect to the y axis, so we can carefully sketch only the right side of the graph; then reflect the result through the y axis to obtain the whole sketch—the point-by-point plotting is cut in half! The tests for symmetry are given in Theorem 1. These tests are easily applied and are very helpful aids to graphing. Recall, two equations are equivalent if they have the same solution set.
Z THEOREM 1 Tests for Symmetry
EXAMPLE
4
Symmetry with respect to the:
An equivalent equation results if:
y axis
x is replaced with 0002x
x axis
y is replaced with 0002y
Origin
x and y are replaced with 0002x and 0002y
Using Symmetry as an Aid to Graphing Test the equation y 0003 x3 for symmetry and sketch its graph.
SOLUTION
Test y Axis Replace x with 0002x:
Test x Axis Replace y with 0002y:
Test Origin Replace x with 0002x and y with 0002y:
y 0003 (0002x)3 y 0003 0002x3
0002y 0003 x3 y 0003 0002x3
0002y 0003 (0002x)3 0002y 0003 0002x3 y 0003 x3
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The only test that produces an equivalent equation is replacing x with 0002x and y with 0002y. So the only symmetry property for the graph of y 0003 x3 is symmetry with respect to the origin. Note that positive values of x produce positive values for y and negative values of x produce negative values for y. So the graph is in the first and third quadrants. First we make a careful sketch in the first quadrant [Fig. 8(a)]. It is easier to perform a reflection through the origin if you first reflect through one axis [Fig. 8(b)] and then through the other axis [Fig. 8(c)].
x
0
1
2
y
0
1
8
y
y
y 0003 x3
10
10
(2, 8) (1, 1)
00025
5
y
x
10
(2, 8) (1, 1)
00025
(1, 00021)
5
(1, 1)
x
00025
(00021, 00021)
(b)
EXAMPLE
5
x
(c)
0002
Z Figure 8
MATCHED PROBLEM 4
5
000210
000210
(a)
(2, 8)
(00022, 00028)
(2, 00028) 000210
y 0003 x3
Test the equation y 0003 x for symmetry and sketch its graph.
0002
Using Symmetry as an Aid to Graphing Test the equation y 0003 冟x冟 for symmetry and sketch its graph.
SOLUTION
Test y Axis Replace x with 0002x:
Test x Axis Replace y with 0002y:
Test Origin Replace x with 0002x and y with 0002y:
y 0003 冟 0002x 冟 y 0003 冟x冟
0002y 0003 冟 x 冟 y 0003 0002冟 x 冟
0002y 0003 冟 0002x 冟 0002y 0003 冟 x 冟 y 0003 0002冟 x 冟
The only symmetry property for the graph of y 0003 冟x冟 is symmetry with respect to the y axis. Since 冟x冟 is never negative, this graph is in the first and second quadrants. We make a careful sketch in the first quadrant; then reflect this graph through the y axis to obtain the complete sketch shown in Figure 9.
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y 5
y 0003 兩x兩 00025
x
0
2
4
y
0
2
4
5
x
00025
0002
Z Figure 9
MATCHED PROBLEM 5
EXAMPLE
6
Test the equation y 0003 0002冟 x 冟 for symmetry and sketch its graph.
0002
Using Symmetry as an Aid to Graphing Test the equation y2 0002 x2 0003 4 for symmetry and sketch its graph.
SOLUTION
Since (0002x)2 0003 x2 and (0002y)2 0003 y2, the equation y2 0002 x2 0003 4 will be unchanged if x is replaced with 0002x or if y is replaced with 0002y. So the graph is symmetric with respect to the y axis, the x axis, and the origin. We need to make a careful sketch in only the first quadrant, reflect this graph through the y axis, and then reflect everything through the x axis. To find quadrant I solutions, we solve the equation for either y in terms of x or x in terms of y. We choose to solve for y. y2 0002 x2 0003 4 y2 0003 x2 0005 4 y 0003 0004 2x2 0005 4 To obtain the quadrant I portion of the graph, we sketch y 0003 2x2 0005 4 for x 0003 0, 1, 2, . . . . The final graph is shown in Figure 10.
x
0
1
2
3
4
y
2
15 ⬇ 2.2
18 ⬇ 2.8
113 ⬇ 3.6
120 ⬇ 4.5
y 5
y 2 0002 x2 0003 4
(3, √13)
(4, √20)
(2, √8) (0, 2) (1, √5) 00025
5
x
00025
Z Figure 10
MATCHED PROBLEM 6
0002
Test the equation 2y2 0002 x2 0003 2 for symmetry and sketch its graph. 0002
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Technology Connections 5
To graph y2 0002 x2 0003 4 on a graphing calculator, we enter both 2x2 0005 4 and 00022x2 0005 4 in the equation editor [Fig. 11(a)] and graph. 00025
5
00025
(a)
(b)
Z Figure 11
ANSWERS TO MATCHED PROBLEMS y
1.
2. (A) 96 ppb 3.
5
(1, 1) (0, 0)
5
00025
(1, 00021)
(4, 00022)
00025
P 0003 (00023, 5)
(9, 3)
(4, 2)
10
(B) 1:45 P.M. and 5 P.M. y 5
x 00025
5
00025
4. Symmetric with respect to the origin
y
5
5
x
00025
00025
5
00025
6. Symmetric with respect to the x axis, the y axis, and the origin y 5
00025
5
00025
C 0003 (3, 00025)
5. Symmetric with respect to the y axis
y
5
x
(9, 00023) A 0003 (00023, 00025)
00025
B 0003 (3, 5)
x
x
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119
Exercises
1. Describe the one-to-one correspondence between points in the plane and ordered pairs of real numbers.
20. Reflect A, B, C, and D through the x axis. y
2. Explain how to graph an equation in two variables using pointby-point plotting.
5
A
3. Explain how to sketch the reflection of a graph through the y axis. 00025
4. How can you tell whether the graph of an equation is symmetric with respect to the origin?
5
5. 5(x, y) ƒ x 0003 06
7. 5(x, y) ƒ x 6 0, y 6 06 9. 5(x, y) ƒ x 7 0, y 6 06
11. 5(x, y) ƒ x 7 0, y 0006 06 13. 5(x, y) ƒ xy 6 06
6. 5(x, y) ƒ x 7 0, y 7 06
x
D
B
In Problems 5–14, give a verbal description of the indicated subset of the plane in terms of quadrants and axes.
C
00025
21. Reflect A, B, C, and D through the origin. y
8. 5(x, y) ƒ y 0003 06
5
10. 5(x, y) ƒ y 6 0, x 0006 06
B
12. 5(x, y) ƒ x 6 0, y 7 06
C
14. 5(x, y) ƒ xy 7 06
00025
D
[Hint: In Problems 13 and 14, consider two cases.]
5
x
A 00025
In Problems 15–18, plot the given points in a rectangular coordinate system. 15. (5, 0), (3, 00022), (00024, 2), (4, 4)
22. Reflect A, B, C, and D through the x axis and then through the y axis. y
16. (0, 4), (00023, 2), (5, 00021), (00022, 00024)
5
17. (0, 00022), (00021, 00023), (4, 00025), (00022, 1)
C
18. (00022, 0), (3, 2), (1, 00024), (00023, 5)
A D
00025
In Problems 19–22, find the coordinates of points A, B, C, and D and the coordinates of the indicated reflections.
00025
y
Test each equation in Problems 23–30 for symmetry with respect to the x axis, y axis, and the origin. Sketch the graph of the equation.
5
00025
A
C
B
00025
x
B
19. Reflect A, B, C, and D through the y axis.
D
5
5
x
23. y 0003 2x 0002 4
24. y 0003 12x 0005 1
25. y 0003 12x
26. y 0003 2x
27. 冟 y 冟 0003 x
28. 冟 y 冟 0003 0002x
29. 冟 x 冟 0003 冟 y 冟
30. y 0003 0002x
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In Problems 31–34, use the graph to estimate to the nearest integer the missing coordinates of the indicated points. (Be sure you find all possible answers.) 31. (A) (8, ?) (D) (?, 6)
(B) (00025, ?) (E) (?, 00025)
(C) (0, ?) (F) (?, 0) y
The figures in Problems 35 and 36 show a portion of a graph. Extend the given graph to one that exhibits the indicated type of symmetry.
10
x
10
(C) (0, ?) (F) (?, 0)
x
5
00025
y
36. (A) x axis only (B) y axis only (C) origin only (D) x axis, y axis, and origin
000210
(B) (00025, ?) (E) (?, 00024)
5
00025
000210
32. (A) (3, ?) (D) (?, 3)
y
35. (A) x axis only (B) y axis only (C) origin only (D) x axis, y axis, and origin
5
00025
y
5
x
10 00025
000210
10
x
Test each equation in Problems 37–46 for symmetry with respect to the x axis, the y axis, and the origin. Do not sketch the graph. 37. 2x 0005 7y 0003 0 38. x2 0005 6y 0005 y2 0003 25
000210
33. (A) (1, ?) (D) (?, 00026)
(B) (00028, ?) (E) (?, 4)
39. x2 0002 4xy2 0003 3
(C) (0, ?) (F) (?, 0)
40. 3x 0002 5y 0003 2 41. x4 0002 5x2y 0005 y4 0003 1
y
42. x4 0002 y4 0003 16
10
43. x3 0002 y3 0003 8 000210
10
x
44. x2 0005 2xy 0005 3y2 0003 12 45. x4 0002 4x2y2 0005 y4 0003 81 46. x3 0002 4y2 0003 1
000210
34. (A) (6, ?) (D) (?, 00022)
(B) (00026, ?) (E) (?, 1)
Test each equation in Problems 47–58 for symmetry with respect to the x axis, the y axis, and the origin. Sketch the graph of the equation.
(C) (0, ?) (F) (?, 0)
y 10
000210
10
000210
x
47. y2 0003 x 0005 2
48. y2 0003 x 0002 2
49. y 0003 x2 0005 1
50. y 0005 2 0003 x2
51. 4y2 0002 x2 0003 1
52. 4x2 0002 y2 0003 1
53. y3 0003 x
54. y 0003 x4
55. y 0003 0.6x2 0002 4.5
56. x 0003 0.8y2 0002 3.5
57. y 0003 x2兾3
58. y2兾3 0003 x
59. (A) Graph the triangle with vertices A 0003 (1, 1), B 0003 (7, 2), and C 0003 (4, 6). (B) Now graph the triangle with vertices A0007 0003 (1, 00021), B0007 0003 (7, 00022), and C0007 0003 (4, 00026) in the same coordinate system. (C) How are these two triangles related? How would you describe the effect of changing the sign of the y coordinate of all the points on a graph?
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61. (A) Graph the triangle with vertices A 0002 (1, 1), B 0002 (7, 2), and C 0002 (4, 6). (B) Now graph the triangle with vertices A0003 0002 (00041, 00041), B0003 0002 (00047, 00042), and C0003 0002 (00044, 00046) in the same coordinate system. (C) How are these two triangles related? How would you describe the effect of changing the signs of the x and y coordinates of all the points on a graph? 62. (A) Graph the triangle with vertices A 0002 (1, 2), B 0002 (1, 4), and C 0002 (3, 4). (B) Now graph y 0002 x and the triangle obtained by reversing the coordinates for each vertex of the original triangle: A0003 0002 (2, 1), B0003 0002 (4, 1), B0003 0002 (4, 3). (C) How are these two triangles related? How would you describe the effect of reversing the coordinates of each point on a graph? In Problems 63–66, solve for y, producing two equations, and then graph both of these equations in the same viewing window. 63. 2x 0005 y2 0002 3
64. x3 0005 y2 0002 8
65. x 2 0004 ( y 0005 1)2 0002 4
66. ( y 0004 2)2 0004 x 2 0002 9
Test each equation in Problems 67–76 for symmetry with respect to the x axis, the y axis, and the origin. Sketch the graph of the equation. 68. 冟 y 冟 0002 x3
67. y3 0002 冟 x 冟
69. xy 0002 1 2
70. xy 0002 00041
71. y 0002 6x 0004 x
72. y 0002 x2 0004 6x
73. y2 0002 冟 x 冟 0005 1
74. y2 0002 4冟 x 冟 0005 1
75. 冟 xy 冟 0005 2冟 y 冟 0002 6
76. 冟 xy 冟 0005 冟 y 冟 0002 4
77. If a graph is symmetric with respect to the x axis and to the origin, must it be symmetric with respect to the y axis? Explain. 78. If a graph is symmetric with respect to the y axis and to the origin, must it be symmetric with respect to the x axis? Explain.
R 0002 np 0002 (10 0004 p)p
82. BUSINESS Repeat Problem 81 for the demand equation n000280004p
81. BUSINESS After extensive surveys, the marketing research department of a producer of popular compact discs developed the demand equation n 0002 10 0004 p
5 0006 p 0006 10
40006p00068
83. PRICE AND DEMAND The quantity of a product that consumers are willing to buy during some period of time depends on its price. The price p and corresponding weekly demand q for a particular brand of diet soda in a city are shown in the figure. Use this graph to estimate the following demands to the nearest 100 cases. (A) What is the demand when the price is $6.00 per case? (B) Does the demand increase or decrease if the price is increased from $6.00 to $6.30 per case? By how much? (C) Does the demand increase or decrease if the price is decreased from $6.00 to $5.70? By how much? (D) Write a brief description of the relationship between price and demand illustrated by this graph. p $7
$6
$5 2,000
3,000
4,000
q
Number of cases
84. PRICE AND SUPPLY The quantity of a product that suppliers are willing to sell during some period of time depends on its price. The price p and corresponding weekly supply q for a particular brand of diet soda in a city are shown in the figure. Use this graph to estimate the following supplies to the nearest 100 cases. (A) What is the supply when the price is $5.60 per case? (B) Does the supply increase or decrease if the price is increased from $5.60 to $5.80 per case? By how much? (C) Does the supply increase or decrease if the price is decreased from $5.60 to $5.40 per case? By how much? (D) Write a brief description of the relationship between price and supply illustrated by this graph. p $7
Price per case
APPLICATIONS
5 0006 p 0006 10
Graph the revenue equation for the indicated values of p.
79. If a graph is symmetric with respect to the origin, must it be symmetric with respect to the x axis? Explain. 80. If a graph is symmetric with respect to the origin, must it be symmetric with respect to the y axis? Explain.
121
where n is the number of units (in thousands) retailers are willing to buy per day at $p per disc. The company’s daily revenue R (in thousands of dollars) is given by
Price per case
60. (A) Graph the triangle with vertices A 0002 (1, 1), B 0002 (7, 2), and C 0002 (4, 6). (B) Now graph the triangle with vertices A0003 0002 (00041, 1), B0003 0002 (00047, 2), and C0003 0002 (00044, 6) in the same coordinate system. (C) How are these two triangles related? How would you describe the effect of changing the sign of the x coordinate of all the points on a graph?
Cartesian Coordinate Systems
$6
$5 2,000
3,000
4,000
Number of cases
q
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85. TEMPERATURE The temperature during a spring day in the Midwest is given in the figure. Use this graph to estimate the following temperatures to the nearest degree and times to the nearest hour. (A) The temperature at 9:00 A.M. (B) The highest temperature and the time when it occurs. (C) The time(s) when the temperature is 49°F. 70
x
(A) Graph v for 0 x 2. (B) Describe the relationship between this graph and the physical behavior of the ball as it swings back and forth.
60
88. PHYSICS The speed (in meters per second) of a ball oscillating at the end of a spring is given by 50
v 0003 4 225 0002 x2
40 Midnight
where x is the vertical displacement (in centimeters) of the ball from its position at rest (positive displacement measured downward—see the figure). 6 AM
Noon
6 PM
Midnight
86. TEMPERATURE Use the graph in Problem 85 to estimate the following temperatures to the nearest degree and times to the nearest half hour. (A) The temperature at 7:00 P.M. (B) The lowest temperature and the time when it occurs. (C) The time(s) when the temperature is 52°F.
x000b0
87. PHYSICS The speed (in meters per second) of a ball swinging at the end of a pendulum is given by v 0003 0.5 12 0002 x where x is the vertical displacement (in centimeters) of the ball from its position at rest (see the figure).
2-2
x 0
(A) Graph v for 00025 x 5. (B) Describe the relationship between this graph and the physical behavior of the ball as it oscillates up and down.
Distance in the Plane Z Distance Between Two Points Z Midpoint of a Line Segment Z Circles
Two basic problems studied in analytic geometry are 1. 2.
Given an equation, find its graph. Given a figure (line, circle, parabola, ellipse, etc.) in a coordinate system, find its equation.
The first problem was discussed in Section 2-1. In this section, we introduce some tools that are useful when studying the second problem.
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Z Distance Between Two Points Given two points P1 and P2 in a rectangular coordinate system, we denote the distance between P1 and P2 by d(P1, P2). We begin with an example.
EXAMPLE
1
Distance Between Two Points Find the distance between the points P1 ⫽ (1, 2) and P2 ⫽ (4, 6).
SOLUTION
Connecting the points P1, P2, and P3 ⫽ (4, 2) with straight line segments forms a right triangle (Fig. 1). y
P1 ⫽ (1, 2)
P
兩6 ⫺ 2兩 ⫽ 4
d(
P
1,
5
2)
P2 ⫽ (4, 6)
P3 ⫽ (4, 2) 兩4 ⫺ 1兩 ⫽ 3 5
10
x
Z Figure 1
From the figure, we see that the lengths of the legs of the triangle are d(P1, P3) ⫽ 冟 4 ⫺ 1 冟 ⫽ 3 and d(P3, P2) ⫽ 冟 6 ⫺ 2 冟 ⫽ 4 The length of the hypotenuse is d(P1, P2), the distance we are seeking. Applying the Pythagorean theorem (see Appendix B), we get [d(P1, P2)] 2 ⫽ ⫽ ⫽ ⫽
[ d(P1, P3)] 2 ⫹ [ d(P3, P2)] 2 32 ⫹ 42 9 ⫹ 16 25
Therefore, d(P1, P2) ⫽ 125 ⫽ 5
MATCHED PROBLEM 1
0002
Find the distance between the points P1 ⫽ (1, 2) and P2 ⫽ (13, 7).
0002
The ideas used in Example 1 can be applied to any two distinct points in the plane. If P1 ⫽ (x1, y1) and P2 ⫽ (x2, y2 ) are two points in a rectangular coordinate system (Fig. 2), then [d(P1, P2)] 2 ⫽ 冟 x2 ⫺ x1 冟2 ⫹ 冟 y2 ⫺ y1 冟2 ⫽ (x2 ⫺ x1)2 ⫹ ( y2 ⫺ y1)2 Taking square roots gives the distance formula.
Because 冟N冟2 ⴝ N2
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Z Figure 2 Distance between two points.
P1 0003 (x1, y1)
,
1
兩x2 0002 x1兩 x1
P2 0003 (x2, y2) y2 兩y2 0002 y1兩
d(P
P 2)
x
y1 (x2, y1) x2
Z THEOREM 1 Distance Formula The distance between P1 0003 (x1, y1) and P2 0003 (x2, y2) is d(P1, P2) 0003 2(x2 0002 x1)2 0005 ( y2 0002 y1)2
EXAMPLE
2
Using the Distance Formula Find the distance between the points (00023, 5) and (00022, 00028).*
SOLUTION
Let (x1, y1) 0003 (ⴚ3, 5) and (x2, y2) 0003 (ⴚ2, ⴚ8). Then, d 0003 2 [(ⴚ2) 0002 (ⴚ3)] 2 0005 [(ⴚ8) 0002 5] 2 0003 2(00022 0005 3)2 0005 (00028 0002 5)2 0003 212 0005 (000213)2 0003 21 0005 169 0003 2170 Notice that if we choose (x1, y1) 0003 (00022, 00028) and (x2, y2) 0003 (00023, 5), then d 0003 2[(00023) 0002 (00022)] 2 0005 [5 0002 (00028)] 2 0003 21 0005 169 0003 2170 so it doesn’t matter which point we designate as P1 or P2.
MATCHED PROBLEM 2
Find the distance between the points (6, 00023) and (00027, 00025).
0002
0002
Z Midpoint of a Line Segment The midpoint of a line segment is the point that is equidistant from each of the endpoints. A formula for finding the midpoint is given in Theorem 2. The proof is discussed in the exercises. Z THEOREM 2 Midpoint Formula The midpoint of the line segment joining P1 0003 (x1, y1) and P2 0003 (x2, y2) is M0003a
x1 0005 x2 y1 0005 y2 , b 2 2
The point M is the unique point satisfying 1 d(P1, M ) 0003 d(M, P2) 0003 d(P1, P2) 2
*We often speak of the point (a, b) when we are referring to the point with coordinates (a, b). This shorthand, though not technically accurate, causes little trouble, and we will continue the practice.
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Note that the coordinates of the midpoint are simply the averages of the respective coordinates of the two given points.
EXAMPLE
3
Using the Midpoint Formula Find the midpoint M of the line segment joining A 0003 (00023, 2) and B 0003 (4, 00025). Plot A, B, and M and verify that d(A, M ) 0003 d(M, B) 0003 12d(A, B).
SOLUTION
We use the midpoint formula with (x1, y1) 0003 (00023, 2) and (x2, y2) 0003 (4, 00025) to obtain the coordinates of the midpoint M. M0003a
x1 0005 x2 y1 0005 y2 , b 2 2
00023 0005 4 2 0005 (00025) , b 2 2 1 00023 0003a , b 2 2 0003a
Substitute x1 ⴝ ⴚ3, y1 ⴝ 2, x2 ⴝ 4, and y2 ⴝ ⴚ5.
Simplify.
0003 (0.5, 00021.5) We plot the three points (Fig. 3) and compute the distances d(A, M ), d(M, B), and d(A, B):
y 5
d(A, M ) 0003 2(00023 0002 0.5)2 0005 [2 0002 (00021.5)] 2 0003 212.25 0005 12.25 0003 224.5
A 0003 (00023, 2) 00025
5
x
M 0003 冢2 , 0002 2 冣 1
00025
3
d(A, B) 0003 2(00023 0002 4)2 0005 [2 0002 (00025)] 2 0003 249 0005 49 0003 298 1 1 98 d(A, B) 0003 198 0003 0003 124.5 0003 d(A, M ) 0003 d(M, B) 2 2 B4
B 0003 (4, 00025)
Z Figure 3
This verifies that M is the midpoint of the line segment joining A and B.
MATCHED PROBLEM 3
EXAMPLE
d(M, B) 0003 2(0.5 0002 4)2 0005 [ 00021.5 0002 (00025)] 2 0003 212.25 0005 12.25 0003 224.5
4
0002
Find the midpoint M of the line segment joining A 0003 (4, 1) and B 0003 (00023, 00025). Plot A, B, and M and verify that d(A, M ) 0003 d(M, B) 0003 12 d(A, B). 0002
Using the Midpoint Formula If M 0003 (1, 1) is the midpoint of the line segment joining A 0003 (00023, 00021) and B 0003 (x, y), find the coordinates of B.
SOLUTION
From the midpoint formula, we have M 0003 (1, 1) 0003 a
00023 0005 x 00021 0005 y , b 2 2
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We equate the corresponding coordinates and solve the resulting equations for x and y: 00021 0005 y 2 2 0003 00021 0005 y
00023 0005 x 2 2 0003 00023 0005 x 10003
10003
2 0005 3 0003 00023 0005 x 0005 3
*
2 0005 1 0003 00021 0005 y 0005 1
50003x
30003y
Therefore, B 0003 (5, 3). MATCHED PROBLEM 4
0002
If M 0003 (1, 00021) is the midpoint of the line segment joining A 0003 (00021, 00025) and B 0003 (x, y), find the coordinates of B. 0002
Z Circles The distance formula would be helpful if its only use were to find actual distances between points, such as in Example 2. However, its more important use is in finding equations of figures in a rectangular coordinate system. We start with an example.
EXAMPLE
5
Equations and Graphs of Circles Write an equation for the set of all points that are 5 units from the origin. Graph your equation.
SOLUTION
The distance between a point (x, y) and the origin is d 0003 2(x 0002 0)2 0005 ( y 0002 0)2 0003 2x2 0005 y2 So, an equation for the set of points that are 5 units from the origin is 2x2 0005 y2 0003 5 We square both sides of this equation to obtain an equation that does not contain any radicals. x2 0005 y2 0003 25 Because (0002x)2 0003 x2 and (0002y)2 0003 y2, the graph will be symmetric with respect to the x axis, y axis, and origin. We make up a table of solutions, sketch the curve in the first quadrant, and use symmetry properties to produce a familiar geometric object—a circle (Fig. 4). x
y
0
5
3
4
4
3
5
0
y (00023, 4) (00024, 3) (00025, 0)
(00024, 00023) (00023, 00024)
(0, 5)
(3, 4) (4, 3) (5, 0) x
(4, 00023) (3, 00024) (0, 00025)
Z Figure 4
MATCHED PROBLEM 5
0002
Write an equation for the set of all points that are three units from the origin. Graph your equation. 0002 *Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
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127
Technology Connections Refer to Example 5. To graph this circle on a graphing calculator, first we solve x 2 ⫹ y 2 ⫽ 25 for y:
x2 ⴙ y2 ⴝ 25 y2 ⴝ 25 ⴚ x2 y ⴝ ⴞ225 ⴚ x2 Next we enter y ⴝ 225 ⴚ x2 and y ⴝ ⴚ 225 ⴚ x2 in the equation editor of a graphing calculator [Fig. 5(a)], enter appropriate window variables [Fig. 5(b)], and graph [Fig. 5(c)].
The graph in Figure 5(c) doesn’t look like a circle. (A circle is as wide as it is tall.) This distortion is caused by the difference between axes scales. One unit on the x axis appears to be longer than one unit on the y axis. Most graphing calculators have an option called ZSquare under the zoom menu [Fig. 6(a)] that automatically adjusts the x axis scale [Fig. 6(b)] to produce a squared viewing window. The graph of a circle in a squared viewing window is not distorted [Fig. 6(c)]. 5
00025
5
00025
(a)
(b)
(c)
Z Figure 5 5
00027.6
7.6
00025
(a)
(b)
(c)
Z Figure 6
In Example 5, we began with a verbal description of a set of points, produced an algebraic equation that these points must satisfy, constructed a numerical table listing some of these points, and then drew a graphical representation of this set of points. The interplay between verbal, algebraic, numerical, and graphical concepts is one of the central themes of this book. Now we generalize the ideas introduced in Example 5. Z DEFINITION 1 Circle y
A circle is the set of all points in a plane equidistant from a fixed point. The fixed distance is called the radius, and the fixed point is called the center. r C 0003 (h, k)
Z Figure 7 Circle.
P 0003 (x, y)
x
Let’s find the equation of a circle with radius r (r 0) and center C at (h, k) in a rectangular coordinate system (Fig. 7). The circle consists of all points P 0003 (x, y) satisfying d(P, C ) 0003 r; that is, all points satisfying 2(x 0002 h)2 0005 ( y 0002 k)2 0003 r
r 7 0
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or, equivalently, (x ⫺ h)2 ⫹ ( y ⫺ k)2 ⫽ r 2 r 7 0
Z THEOREM 3 Standard Form of the Equation of a Circle The standard form of a circle with radius r and center at (h, k) is: (x ⫺ h)2 ⫹ ( y ⫺ k)2 ⫽ r2 r 7 0
EXAMPLE
6
Equations and Graphs of Circles Find the equation of a circle with radius 4 and center at C ⫽ (⫺3, 6). Graph the equation. C ⫽ (h, k) ⫽ (⫺3, 6) and r ⫽ 4 (x ⫺ h)2 ⫹ ( y ⫺ k)2 ⫽ r2 Substitute h ⴝ ⴚ3, k ⴝ 6 [x ⫺ (⫺3)] 2 ⫹ ( y ⫺ 6)2 ⫽ 42 Simplify 2 2 (x ⫹ 3) ⫹ ( y ⫺ 6) ⫽ 16
SOLUTION
To graph the equation, plot the center and a few points on the circle (the easiest points to plot are those located 4 units from the center in either the horizontal or vertical direction), then draw a circle of radius 4 (Fig. 8). y (⫺3, 10)
10
C ⫽ (⫺3, 6)
(⫺7, 6)
r⫽4
5
(1, 6)
(⫺3, 2) ⫺5
x
(x ⫹ 3)2 ⫹ (y ⫺ 6)2 ⫽ 16
Z Figure 8
MATCHED PROBLEM 6
ZZZ EXPLORE-DISCUSS 1
0002
Find the equation of a circle with radius 3 and center at C ⫽ (3, ⫺2). Graph the equation. 0002
Explain how to find the equation of the circle with diameter AB, if A ⫽ (3, 8) and B ⫽ (11, 12).
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EXAMPLE
7
Distance in the Plane
129
Finding the Center and Radius of a Circle Find the center and radius of the circle with equation x2 0005 y2 0005 6x 0002 4y 0003 23.
SOLUTION
We transform the equation into the form (x 0002 h)2 0005 (y 0002 k)2 0003 r2 by completing the square relative to x and relative to y (see Section 1-5). From this standard form we can determine the center and radius. x2 0005 y2 0005 6x 0002 4y 0003 23
Group together the terms involving x and those involving y.
(x2 0005 6x ) 0005 ( y2 0002 4y ) 0003 23 2 2 (x 0005 6x 0005 9) 0005 ( y 0002 4y 0005 4) 0003 23 0005 9 0005 4 (x 0005 3)2 0005 ( y 0002 2)2 0003 36 [ x 0002 (00023)] 2 0005 ( y 0002 2)2 0003 62 (h, k) 0003 (00023, 2) Center: Radius: r 0003 136 0003 6 MATCHED PROBLEM 7
Complete the squares. Factor each trinomial. Write ⴙ3 as ⴚ(ⴚ3) to identify h.
0002
Find the center and radius of the circle with equation x2 0005 y2 0002 8x 0005 10y 0003 000225.
ANSWERS TO MATCHED PROBLEMS 1. 13 2. 1173 3. M 0003 (12, 00022) 0003 (0.5, 00022); d(A, B) 0003 185; d(A, M ) 0003 121.25 0003 d(M, B) 0003 12 d(A, B) y 5
A 0003 (4, 1) 00025
x
5
00025
B 0003 (00023, 00025)
4. B 0003 (3, 3) 5. x2 + y2 0003 9
6. (x 0002 3)2 0005 ( y 0005 2)2 0003 9 y
y 5
(0, 3) (00023, 0)
(3, 1) (3, 0)
00025
5
(0, 00023)
x
(0, 00022) 00025
00025
7. (x 0002 4)2 0005 ( y 0005 5)2 0003 16; radius: 4, center: (4, 00025)
C (3, 00022) (3, 00025)
5
x (6, 00022)
0002
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Exercises
1. State the Pythagorean theorem.
29. Find x such that (x, 7) is 10 units from (00024, 1).
2. Explain how to calculate the distance between two points in the plane if you know their coordinates.
30. Find x such that (x, 2) is 4 units from (3, 00023). 31. Find y such that (2, y) is 3 units from (00021, 4).
3. Explain how to calculate the midpoint of a line segment if you know the coordinates of the endpoints.
32. Find y such that (3, y) is 13 units from (00029, 2).
4. Explain how to find the standard form of the equation of the circle with center (1, 5) and radius 12.
In Problems 33–36, write a verbal description of the graph and then write an equation that would produce the graph.
In Problems 5–12, find the distance between each pair of points and the midpoint of the line segment joining the points. Leave distance in radical form, if applicable. 5. (1, 0), (4, 4)
6. (0, 1), (3, 5)
7. (0, 00022), (5, 10)
8. (3, 0), (00022, 00023)
9. (00026, 00024), (3, 4)
10. (00025, 4), (6, 00021)
11. (00026, 00023), (00022, 00021)
y
33. 5
00025
13. C 0003 (0, 0), r 0003 7
14. C 0003 (0, 0), r 0003 5
15. C 0003 (2, 3), r 0003 6
16. C 0003 (5, 6), r 0003 2
17. C 0003 (00024, 1), r 0003 17
18. C 0003 (00025, 6), r 0003 111
19. C 0003 (00023, 00024), r 0003 12
20. C 0003 (4, 00021), r 0003 15
x
00025
12. (00025, 00022), (00021, 2)
In Problems 13–20, write the equation of a circle with the indicated center and radius.
5
y
34. 5
00025
5
x
00025
In Problems 21–26, write an equation for the given set of points. Graph your equation.
y
35. 5
21. The set of all points that are two units from the origin. 22. The set of all points that are four units from the origin. 23. The set of all points that are one unit from (1, 0).
00025
5
x
24. The set of all points that are one unit from (0, 00021). 25. The set of all points that are three units from (00022, 1).
00025
26. The set of all points that are two units from (3, 00022). 27. Let M be the midpoint of A and B, where
y
36.
A 0003 (a1, a2), B 0003 (1, 3), and M 0003 (00022, 6).
5
(A) Use the fact that 00022 is the average of a1 and 1 to find a1. (B) Use the fact that 6 is the average of a2 and 3 to find a2. (C) Find d(A, M ) and d(M, B). 28. Let M be the midpoint of A and B, where
00025
5
A 0003 (00023, 5), B 0003 (b1, b2), and M 0003 (4, 00022). (A) Use the fact that 4 is the average of 00023 and b1 to find b1. (B) Use the fact that 00022 is the average of 5 and b2 to find b2. (C) Find d(A, M ) and d(M, B).
00025
x
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SECTION 2–2
In Problems 37–42, M is the midpoint of A and B. Find the indicated point. Verify that d(A, M) 0003 d(M, B) 0003 12d(A, B). 37. A 0003 (00024.3, 5.2), B 0003 (9.6, 00021.7), M 0003 ? 38. A 0003 (2.8, 00023.5), B 0003 (00024.1, 7.6), M 0003 ?
131
Distance in the Plane
62. A parallelogram ABCD is shown in the figure. (A) Find the midpoint of the line segment joining A and C. (B) Find the midpoint of the line segment joining B and D. (C) What can you conclude about the diagonals of the parallelogram?
39. A 0003 (25, 10), M 0003 (00025, 00022), B 0003 ?
y
40. M 0003 (2.5, 3.5), B 0003 (12, 10), A 0003 ?
B 0003 (a, b)
41. M 0003 (00028, 00026), B 0003 (2, 4), A 0003 ? 42. A 0003 (00024, 00022), M 0003 (00021.5, 00024.5), B 0003 ?
A 0003 (0, 0)
C 0003 (a 0005 c, b)
x D 0003 (c, 0)
In Problems 43–52, find the center and radius of the circle with the given equation. Graph the equation. 43. x2 0005 ( y 0005 2)2 0003 9 44. (x 0002 5)2 0005 y2 0003 16 45. (x 0005 4)2 0005 (y 0002 2)2 0003 7 46. (x 0002 5)2 0005 (y 0005 7)2 0003 15 47. x2 0005 6x 0005 y2 0003 16 48. x2 0005 y2 0002 8y 0003 9 49. x2 0005 y2 0002 6x 0002 4y 0003 36 50. x2 0005 y2 0002 2x 0002 10y 0003 55 51. 3x2 0005 3y2 0005 24x 0002 18y 0005 24 0003 0 2
2
52. 2x 0005 2y 0005 8x 0005 20y 0005 30 0003 0
In Problems 63–68, find the standard form of the equation of the circle that has a diameter with the given endpoints. 63. (00024, 3), (6, 3) 64. (5, 00021), (5, 7) 65. (4, 0), (0, 10) 66. (00026, 0), (0, 00028) 67. (11, 00022), (3, 00024) 68. (00028, 9), (12, 15) In Problems 69–72, find the standard form of the equation of the circle with the given center that passes through the given point. 69. Center: (0, 5); point on circle: (2, 00024)
In Problems 53–56, solve for y, producing two equations, and then graph both of these equations in the same viewing window. 53. x2 0005 y2 0003 3 54. x2 0005 y2 0003 5
70. Center: (00023, 0); point on circle: (6, 1) 71. Center: (00022, 9); point on circle: (8, 00027) 72. Center: (7, 000212); point on circle: (13, 8)
55. (x 0005 3)2 0005 (y 0005 1)2 0003 2
APPLICATIONS
56. (x 0002 2)2 0005 (y 0002 1)2 0003 3
73. SPORTS A singles court for lawn tennis is a rectangle 27 feet wide and 78 feet long (see the figure). Points B and F are the midpoints of the end lines of the court.
In Problems 57 and 58, show that the given points are the vertices of a right triangle (see the Pythagorean theorem in Appendix B). Find the length of the line segment from the midpoint of the hypotenuse to the opposite vertex. 57. (00023, 2), (1, 00022), (8, 5)
18 feet B
C
A
18 feet
58. (00021, 3), (3, 5), (5, 1)
D 78 feet
Find the perimeter (to two decimal places) of the triangle with the vertices indicated in Problems 59 and 60. 59. (00023, 1), (1, 00022), (4, 3) 60. (00022, 4), (3, 1), (00023, 00022) x1 0005 x2 y1 0005 y2 , b, 2 2 show that d(P1, M ) 0003 d(M, P2) 0003 12d(P1, P2). (This is one step in the proof of Theorem 2.)
61. If P1 0003 (x1, y1), P2 0003 (x2, y2) and M 0003 a
27 feet
E F
G
(A) Sketch a graph of the court with A at the origin of your coordinate system, C on the positive y axis, and G on the positive x axis. Find the coordinates of points A through G. (B) Find d(B, D) and d(F, C ) to the nearest foot.
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74. SPORTS Refer to Problem 73. Find d(A, D) and d(C, G) to the nearest foot. 75. ARCHITECTURE An arched doorway is formed by placing a circular arc on top of a rectangle (see the figure). If the doorway is 4 feet wide and the height of the arc above its ends is 1 foot, what is the radius of the circle containing the arc? [Hint: Note that (2, r 0002 1) must satisfy x2 0005 y2 0003 r 2.] y
77. CONSTRUCTION Town B is located 36 miles east and 15 miles north of town A (see the figure). A local telephone company wants to position a relay tower so that the distance from the tower to town B is twice the distance from the tower to town A. (A) Show that the tower must lie on a circle, find the center and radius of this circle, and graph. (B) If the company decides to position the tower on this circle at a point directly east of town A, how far from town A should they place the tower? Compute answer to one decimal place.
(2, r 0002 1)
y
r
25
x
Town B
Tower
(36, 15)
(x, y) 4 feet
Town A
Arched doorway
76. ENGINEERING The cross section of a rivet has a top that is an arc of a circle (see the figure). If the ends of the arc are 12 millimeters apart and the top is 4 millimeters above the ends, what is the radius of the circle containing the arc?
25
x
78. CONSTRUCTION Repeat Problem 77 if the distance from the tower to town A is twice the distance from the tower to town B.
Rivet
2-3
Equations of a Line Z Graphing Lines Z Finding the Slope of a Line Z Determining Special Forms of the Equation of a Line Z Finding Slopes of Parallel or Perpendicular Lines
In this section, we consider one of the most basic geometric figures—a line. When we use the term line in this book, we mean straight line. We will learn how to recognize and graph a line and how to use information concerning a line to find its equation.
Z Graphing Lines With your past experience in graphing equations in two variables, you probably remember that first-degree equations in two variables, such as y 0003 00023x 0005 5
3x 0002 4y 0003 9
y 0003 000223 x
have graphs that are lines. This fact is stated in Theorem 1.
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Z THEOREM 1 The Equation of a Line If A, B, and C are constants, with A and B not both 0, and x and y are variables, then the graph of the equation Ax ⴙ By ⴝ C
Standard Form
(1)
is a line. Any line in a rectangular coordinate system has an equation of this form.
Also, the graph of any equation of the form y ⴝ mx ⴙ b
(2)
where m and b are constants, is a line. Equation (2), which we will discuss in detail later, is simply a special case of equation (1) for B 0002 0. This can be seen by solving equation (1) for y in terms of x: C A y00030004 x0005 B B
B00020
To graph either equation (1) or (2), we plot any two points from the solution set and use a straightedge to draw a line through these two points. The points where the line crosses the axes are convenient to use and easy to find. The y intercept* is the y coordinate of the point where the graph crosses the y axis, and the x intercept is the x coordinate of the point where the graph crosses the x axis. To find the y intercept, let x = 0 and solve for y; to find the x intercept, let y = 0 and solve for x. It is often advisable to find a third point as a checkpoint. All three points must lie on the same line or a mistake has been made.
EXAMPLE
1
Using Intercepts to Graph a Line Graph the equation 3x 0004 4y 0003 12.
SOLUTION
Find intercepts, a third checkpoint (optional), and draw a line through the two (three) points (Fig. 1). y 5
(8, 3) (4, 0)
00045
5
y intercept is 00043
MATCHED PROBLEM 1
x
0
4
8
y
00043
0
3
Graph the equation 4x 0005 3y 0003 12.
Checkpoint 10
x
x intercept is 4 (0, 00043)
00045
Z Figure 1
0002
0002
*If the x intercept is a and the y intercept is b, then the graph of the line passes through the points (a, 0) and (0, b). It is common practice to refer to both the numbers a and b and the points (a, 0) and (0, b) as the x and y intercepts of the line.
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Technology Connections To solve Example 1 on a graphing calculator, we first solve the equation for y:
3x ⴚ 4y ⴝ 12 ⴚ4y ⴝ ⴚ3x ⴙ 12 y ⴝ 0.75x ⴚ 3 To find the y intercept of this line, we graph the preceding equation, press TRACE, and then enter 0 for x [Fig. 2(a)]. The displayed y value is the y intercept.
The x intercept can be found by using the zero option on the CALC menu. After selecting the zero option, you will be asked to provide three x values: a left bound (a number less than the zero), a right bound (a number greater than the zero), and a guess (a number between the left and right bounds). You can enter the three values from the keypad, but most find it easier to use the cursor. The zero or x intercept is displayed at the bottom of the screen [Fig. 2(b)].
5
00045
5
10
00045
10
00045
00045
(a) y intercept
(b) x intercept
Z Figure 2
Z Finding the Slope of a Line If we take two different points P1 0003 (x1, y1) and P2 0003 (x2, y2) on a line, then the ratio of the change in y to the change in x as we move from point P1 to point P2 is called the slope of the line. Roughly speaking, slope is a measure of the “steepness” of a line. Sometimes the change in x is called the run and the change in y is called the rise. Z DEFINITION 1 Slope of a Line If a line passes through two distinct points P1 0003 (x1, y1) and P2 0003 (x2, y2), then its slope m is given by the formula m0003 0003
y2 0004 y1 x2 0004 x1
y
x1 0002 x2
P2 0003 (x2, y2)
Vertical change (rise) Horizontal change (run)
y2 0004 y1 Rise x
P1 0003 (x1, y1) x2 0004 x1 Run
(x2, y1)
For a horizontal line, y doesn’t change as x changes, so its slope is 0. On the other hand, for a vertical line, x doesn’t change as y changes, so its slope is not defined: y2 0004 y1 y2 0004 y1 0003 x2 0004 x1 0
For a vertical line, slope is not defined.
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Equations of a Line
In general, the slope of a line may be positive, negative, 0, or not defined. Each of these cases is interpreted geometrically as shown in Table 1. Table 1 Geometric Interpretation of Slope Line
Slope
Example y
Rising as x moves from left to right y values are increasing
Positive
x
y
Falling as x moves from left to right y values are decreasing
Negative
x
y
Horizontal y values are constant
0
x
y
Vertical x values are constant
Not defined
x
In using the formula to find the slope of the line through two points, it doesn’t matter which point is labeled P1 or P2, because changing the labeling will change the sign in both the numerator and denominator of the slope formula: y2 0004 y1 y1 0004 y2 0003 x2 0004 x1 x1 0004 x2
b0006 a0006
For example, the slope of the line through the points (3, 2) and (7, 5) is
b
500042 3 00043 200045 0003 0003 0003 700043 4 00044 300047
a m0003
b0006 b 0003 a0006 a
In addition, it is important to note that the definition of slope doesn’t depend on the two points chosen on the line as long as they are distinct. This follows from the fact that the ratios of corresponding sides of similar triangles are equal (Fig. 3).
Z Figure 3
EXAMPLE
2
Finding Slopes For each line in Figure 4, find the run, the rise, and the slope. (All the horizontal and vertical line segments have integer lengths.) y
y
5
5
00045
5
00045
00045
5
00045
(a)
Z Figure 4
x
(b)
x
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SOLUTION
MATCHED PROBLEM 2
In Figure 4(a), the run is 3, the rise is 6 and the slope is 63 0003 2. In Figure 4(b), the run is 6, 2 the rise is 00044 and the slope is 00044 0002 6 0003 00043 . For each line in Figure 5, find the run, the rise, and the slope. (All the horizontal and vertical line segments have integer lengths.) y
y
5
5
00045
5
x
00045
5
00045
x
00045
(a)
(b)
Z Figure 5
0002
EXAMPLE
3
Finding Slopes Sketch a line through each pair of points and find the slope of each line.
SOLUTIONS
(A) (00043, 00044), (3, 2)
(B) (00042, 3), (1, 00043)
(C) (00044, 2), (3, 2)
(D) (2, 4), (2, 00043) y
(A)
y
(B)
5
5
(00042, 3)
(3, 2) 00045
x
5
00045
5
(1, 00043)
(00043, 00044) 00045
m0003
x
00045
2 0004 (00044) 6 0003 00031 3 0004 (00043) 6
m0003
y
(C)
y
(D)
5
(00044, 2)
00043 0004 3 00046 0003 0003 00042 1 0004 (00042) 3
5
(2, 4)
(3, 2)
00045
5
x
00045
5
x
(2, 00043) 00045
m0003
200042 0 0003 00030 3 0004 (00044) 7
00045
00043 0004 4 00047 0003 ; 200042 0 slope is not defined m0003
0002
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MATCHED PROBLEM 3
137
Equations of a Line
Find the slope of the line through each pair of points. Do not graph. (A) (00043, 00043), (2, 00043)
(B) (00042, 00041), (1, 2)
(C) (0, 4), (2, 00044)
(D) (00043, 2), (00043, 00041)
0002
Z Determining Special Forms of the Equation of a Line We start by investigating why y 0003 mx 0005 b is called the slope–intercept form for a line.
ZZZ EXPLORE-DISCUSS 1
(A) Graph y 0003 x 0005 b for b 0003 00045, 00043, 0, 3, and 5 simultaneously in the same coordinate system. Verbally describe the geometric significance of b. (B) Graph y 0003 mx 0004 1 for m 0003 00042, 00041, 0, 1, and 2 simultaneously in the same coordinate system. Verbally describe the geometric significance of m.
As you see from the preceding exploration, constants m and b in y = mx 0005 b have special geometric significance. If we let x = 0, then y = b and the graph of y = mx 0005 b crosses the y axis at (0, b). So the constant b is the y intercept. For example, the y intercept of the graph of y = 2x – 7 is 00047. We have already seen that the point (0, b) is on the graph of y = mx 0005 b. If we let x = 1, then it follows that the point (1, m 0005 b) is also on the graph (Fig. 6). Because the graph of y = mx 0005 b is a line, we can use these two points to compute the slope:
f (x)
(0, b)
(1, m 0005 b) x
Slope 0003
Z Figure 6
y2 0004 y1 (m 0005 b) 0004 b 0003 0003m x2 0004 x1 100040
(x1, y1) ⴝ (0, b) (x2, y2) ⴝ (1, m ⴙ b)
So m is the slope of the line with equation y = mx 0005 b.
Z THEOREM 2 Slope–Intercept Form An equation of the line with slope m and y intercept b is y 0003 mx 0005 b
y y 0003 mx 0005 b
which is called the slope–intercept form.
EXAMPLE
4
m0003
Rise
y intercept b
Run
x
Using the Slope–Intercept Form (A) Write the slope–intercept form of a line with slope (B) Find the slope and y intercept, and graph y 0003
3 4x
2 3
and y intercept 00045.
0004 1.
Rise Run
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SOLUTIONS
(A) Substitute m 0003 23 and b 0003 00045 in y = mx 0005 b to obtain y 0003 23x 0004 5. (B) The y intercept of y 0003 34 x 0004 1 is 00041 and the slope is 34. If we start at the point (0, 00041) and move four units to the right (run), then the y coordinate of a point on the line must move up three units (rise) to the point (4, 2). Drawing a line through these two points produces the graph shown in Figure 7. y 5
3 00045
5
x
4 00045
0002
Z Figure 7
MATCHED PROBLEM 4
Write the slope–intercept form of the line with slope equation.
5 4
and y intercept 00042. Graph the 0002
y (x, y) x (x1, y 1)
Z Figure 8
In Example 4 we found the equation of a line with a given slope and y intercept. It is also possible to find the equation of a line passing through a given point with a given slope or to find the equation of a line containing two given points. Suppose a line has slope m and passes through the point (x1, y1). If (x, y) is any other point on the line (Fig. 8), then y 0004 y1 0003m x 0004 x1
(x, y 1)
that is, y 0004 y1 0003 m(x 0004 x1)
(3)
Because the point (x1, y1) also satisfies equation (3), we can conclude that equation (3) is an equation of a line with slope m that passes through (x1, y1).
Z THEOREM 3 Point–Slope Form An equation of the line with slope m that passes through (x1, y1) is y 0004 y1 0003 m(x 0004 x1) which is called the point–slope form.
If we are given the coordinates of two points on a line, we can use the given coordinates to find the slope and then use the point–slope form with either of the given points to find the equation of the line.
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EXAMPLE
5
Equations of a Line
139
Point–Slope Form (A) Find an equation for the line that has slope 23 and passes through the point (00042, 1). Write the final answer in the form Ax 0005 By 0003 C. (B) Find an equation for the line that passes through the two points (4, 00041) and (00048, 5). Write the final answer in the form y 0003 mx 0005 b.
SOLUTIONS
(A) If m 0003 23 and (x1, y1) 0003 (00042, 1), then y 0004 y1 0003 m(x 0004 x1) y000410003
Substitute y1 ⴝ 1, x1 ⴝ ⴚ2, and m ⴝ 23 .
2 [x 0004 (00042)] 3
Multiply both sides by 3.
3( y 0004 1) 0003 2(x 0005 2) 3y 0004 3 0003 2x 0005 4 or 00042x 0005 3y 0003 7
Distribute. Write in standard form.
2x 0004 3y 0003 00047
(B) First use the slope formula to find the slope of the line: m0003
y2 0004 y1 5 0004 (00041) 6 1 0003 0003 00030004 x2 0004 x1 00048 0004 4 000412 2
Substitute x1 ⴝ 4, y1 ⴝ ⴚ1, x2 ⴝ ⴚ8, and y2 ⴝ 5 in the slope formula.
Now we choose (x1, y1) 0003 (4, 00041) and proceed as in part A: y 0004 y1 0003 m(x 0004 x1) 1 y 0004 (00041) 0003 0004 (x 0004 4) 2 1 y0005100030004 x00052 2 1 y00030004 x00051 2
1 Substitute x1 ⴝ 4, y1 ⴝ ⴚ1, and m ⴝ ⴚ . 2 y ⴚ (ⴚ1) ⴝ y ⴙ 1; Distribute on right side.
Subtract 1 from both sides.
You may want to verify that choosing (x1, y1) = (00048, 5), the other given point, produces the same equation. 0002 MATCHED PROBLEM 5
(A) Find an equation for the line that has slope 000425 and passes through the point (3, 00042). Write the final answer in the form Ax 0005 By 0003 C. (B) Find an equation for the line that passes through the two points (00043, 1) and (7, 00043). Write the final answer in the form y 0003 mx 0005 b. 0002 The simplest equations of lines are those for horizontal and vertical lines. Consider the following two equations: x 0005 0y 0003 a 0x 0005 y 0003 b
or or
x0003a y0003b
(4) (5)
In equation (4), y can be any number as long as x 0003 a. So the graph of x 0003 a is a vertical line crossing the x axis at (a, 0). In equation (5), x can be any number as long as y 0003 b.
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So the graph of y 0003 b is a horizontal line crossing the y axis at (0, b). We summarize these results as follows:
Z THEOREM 4 Vertical and Horizontal Lines Equation x0003a (short for x 0005 0y 0003 a) y0003b
Graph Vertical line through (a, 0) (Slope is undefined.) Horizontal line through (0, b) (Slope is 0.)
(short for 0x 0005 y 0003 b) y
x0003a
y0003b
b a
EXAMPLE
6
x
Graphing Horizontal and Vertical Lines Graph the line x 0003 00042 and the line y 0003 3. y
SOLUTION 5
y00033
00045
5
x
x 0003 00042 00045
MATCHED PROBLEM 6
0002
Graph the line x 0003 4 and the line y 0003 00042. 0002 The various forms of the equation of a line that we have discussed are summarized in Table 2 for convenient reference. Table 2 Equations of a Line Standard form
Ax 0005 By 0003 C
A and B not both 0
Slope–intercept form
y 0003 mx 0005 b
Slope: m; y intercept: b
Point–slope form
y 0004 y1 0003 m(x 0004 x1)
Slope: m; Point: (x1, y1)
Horizontal line
y0003b
Slope: 0
Vertical line
x0003a
Slope: Undefined
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Z Finding Slopes of Parallel or Perpendicular Lines From geometry, we know that two vertical lines are parallel to each other and that a horizontal line and a vertical line are perpendicular to each other. How can we tell when two nonvertical lines are parallel or perpendicular to each other? Theorem 5, which we state without proof, provides a convenient test.
Z THEOREM 5 Parallel and Perpendicular Lines Given two nonvertical lines L1 and L2 with slopes m1 and m2, respectively, then L1 储 L2 L1 ⬜ L2
if and only if if and only if
m1 0003 m2 m1m2 0003 00041
The symbols 储 and ⬜ mean, respectively, “is parallel to” and “is perpendicular to.” In the case of perpendicularity, the condition m1m2 = 00041 also can be written as m2 0003 0004
1 m1
or
m1 0003 0004
1 m2
Therefore, Two nonvertical lines are perpendicular if and only if their slopes are the negative reciprocals of each other.
EXAMPLE
7
Parallel and Perpendicular Lines Given the line L: 3x 0004 2y = 5 and the point P 0003 (00043, 5), find an equation of a line through P that is (A) Parallel to L
(B) Perpendicular to L
Write the final answers in the slope–intercept form y = mx 0005 b. SOLUTIONS
First, find the slope of L by writing 3x 0004 2y = 5 in the equivalent slope–intercept form y = mx 0005 b: 3x 0004 2y 0003 5 00042y 0003 00043x 0005 5 y 0003 32 x 0004 52 So the slope of L is 32. The slope of a line parallel to L is the same, 32, and the slope of a line perpendicular to L is 000423. We now can find the equations of the two lines in parts A and B using the point–slope form. (A) Parallel (m 0003 32): y 0004 y1 0003 m(x 0004 x1) y 0004 5 0003 32 (x 0005 3) y 0004 5 0003 32 x 0005 92 y 0003 32 x 0005 192
(B) Perpendicular (m 0003 000423): y 0004 y1 0003 m(x 0004 x1) y 0004 5 0003 000423 (x 0005 3) y 0004 5 0003 000423x 0004 2 y 0003 000423x 0005 3
Substitute for x1, y1, and m. Distribute. Add 5 to both sides.
0002
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MATCHED PROBLEM 7
Given the line L: 4x 0005 2y = 3 and the point P = (2, 00043), find an equation of a line through P that is (A) Parallel to L
(B) Perpendicular to L
Write the final answers in the slope–intercept form y = mx 0005 b.
EXAMPLE
8
0002
Cost Analysis A hot dog vendor pays $25 per day to rent a pushcart and $1.25 for the ingredients in one hot dog. (A) Find the cost of selling x hot dogs in 1 day. (B) What is the cost of selling 200 hot dogs in 1 day? (C) If the daily cost is $355, how many hot dogs were sold that day?
SOLUTIONS
(A) The rental charge of $25 is the vendor’s fixed cost—a cost that is accrued every day and does not depend on the number of hot dogs sold. The cost of the ingredients for x hot dogs is $1.25x. This is the vendor’s variable cost—a cost that depends on the number of hot dogs sold. The total cost for selling x hot dogs is C(x) 0003 1.25x 0005 25
Total Cost ⴝ Variable Cost ⴙ Fixed Cost
(B) The cost of selling 200 hot dogs in 1 day is C(200) 0003 1.25(200) 0005 25 0003 $275 (C) The number of hot dogs that can be sold for $355 is the solution of the equation 1.25x 0005 25 0003 355 1.25x 0003 330 330 x0003 1.25 0003 264 hot dogs MATCHED PROBLEM 8
Subtract 25 from each side. Divide both sides by 1.25. Simplify.
0002
It costs a pretzel vendor $20 per day to rent a cart and $0.75 for each pretzel. (A) Find the cost of selling x pretzels in 1 day. (B) What is the cost of selling 150 pretzels in 1 day? (C) If the daily cost is $275, how many pretzels were sold that day? 0002
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143
Technology Connections A graphing calculator can be used to solve equations like 1.25x ⴙ 25 ⴝ 355 (see Example 8). First enter both sides of the equation in the equation editor [Fig. 9(a)] and choose window variables [Fig. 9(b)] so that the graphs of both equations appear on the screen. There is no “right” choice for the window variables. Any choice that displays the intersection point will do. (Here is how we chose our window variables: We chose Ymax ⴝ 600 to place the graph of the horizontal
line below the top of the window. We chose Ymin ⴝ ⴚ200 to place the graph of the x axis above the text displayed at the bottom of the screen. Since x cannot be negative, we chose Xmin ⴝ 0. We used trial and error to determine a reasonable choice for Xmax.) Now choose intersect on the CALC menu, and respond to the prompts from the calculator. The coordinates of the intersection point of the two graphs are shown at the bottom of the screen [Fig. 9(c)]. 600
0
(a)
400
0004200
(b)
(c)
Z Figure 9
ANSWERS TO MATCHED PROBLEMS 2. (A) Run 0003 5, rise 0003 4, slope 0003 45 (B) Run 0003 3, rise 0003 00046, slope 0003 00046 3 0003 00042 3. (A) m 0003 0 (B) m 0003 1 (C) m 0003 00044 (D) m is not defined
y
1.
00045
5
x
00045
4. y 0003 54 x 0004 2
5. (A) 2x 0005 5y 0003 00044 y
y
6.
5
5
5 5
00045
x
4 00045
7. (A) y 0003 00042x 0005 1 (B) y 0003 12 x 0004 4 8. (A) C(x) 0003 0.75x 0005 20 (B) $132.50
(B) y 0003 000425 x 0004 15
x00034
00045
5
y 0003 00042 00045
(C) 340 pretzels
x
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Exercises
1. Explain how to find the x and y intercepts of a line if its equation is written in standard form.
y
10. 6
2. Given the graph of a line, explain how to determine whether the slope is negative. 3. Explain why y ⫽ mx ⫹ b is called the slope–intercept form. ⫺6
4. Explain why y ⫺ y1 ⫽ m(x ⫺ x1) is called the point–slope form. 5. Given the equations of two lines in standard form, explain how to determine whether the lines are parallel. 6. Given the equations of two lines in standard form, explain how to determine whether the lines are perpendicular.
6
x
⫺6
y
11. 5
In Problems 7–12, use the graph of each line to find the rise, run, and slope. Write the equation of each line in the standard form Ax ⫹ By ⫽ C, A ⱖ 0. (All the horizontal and vertical line segments have integer lengths.)
⫺5
5
x
y
7.
⫺5
5
y
12. ⫺5
5
5
x
⫺5
⫺5
5
x
y
8. 5
⫺5
⫺5
5
x
In Problems 13–18, use the graph of each line to find the x intercept, y intercept, and slope, if they exist. Write the equation of each line, using the slope–intercept form whenever possible. y
13.
⫺5
5
y
9. 5
⫺5
⫺5
5
⫺5
x
5
⫺5
x
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SECTION 2–3 y
14.
25. 4x 0004 5y 0003 000424
5
27.
00045
5
x
145
26. 6x 0004 7y 0003 000449 28.
y x 0004 00031 6 5
29. x 0003 00043
30. y 0003 00042
31. y 0003 3.5
32. x 0003 2.5
In Problems 33–38, find an equation of the line with the indicated slope and y intercept, and write it in the form Ax 0005 By 0003 C, A 0007 0, where A, B, and C are integers.
00045
y
15.
y x 0004 00031 8 4
Equations of a Line
33. Slope 0003 00043; y intercept 0003 7
5
34. Slope 0003 4; y intercept 0003 000410 35. Slope 0003 72; y intercept 0003 000413
00045
5
x
36. Slope 0003 000454; y intercept 0003 115 37. Slope 0003 0; y intercept 0003 23 38. Slope 0003 0; y intercept 0003 0
00045
In Problems 39–44, find the equation of the line passing through the given point with the given slope. Write the final answer in the slope–intercept form y 0003 mx 0005 b.
y
16. 5
39. (0, 3); m 0003 00042 00045
5
x
43. (00042, 00043); m 0003 000412
y
45. (0, 4); m 0003 00043
5
47. (00045, 4); m 0003 00045
5
x
00045
y
18. 5
00045
40. (4, 0); m 0003 3 42. (2, 00043); m 0003 000445 44. (2, 1); m 0003
4 3
In Problem 45–58, write the equation of the line that contains the indicated point(s), and/or has the given slope or intercepts; use either the slope–intercept form y 0003 mx 0005 b, or the form x 0003 c.
00045
17.
41. (00045, 4); m 0003
3 2
5
x
000425
46. (2, 0); m 0003 2 48. (00044, 00042); m 0003 12
49. (1, 6); (5, 00042)
50. (00043, 4); (6, 1)
51. (00044, 8); (2, 0)
52. (2, 00041); (10, 5)
53. (00043, 4); (5, 4)
54. (0, 00042); (4, 00042)
55. (4, 6); (4, 00043)
56. (00043, 1); (00043, 00044)
57. x intercept 00044; y intercept 3
58. x intercept 00044; y intercept 00045
In Problems 59–66, write an equation of the line that contains the indicated point and meets the indicated condition(s). Write the final answer in the standard form Ax 0005 By 0003 C, A 0007 0. 59. (00043, 4); parallel to y 0003 3x 0004 5 60. (00044, 0); parallel to y 0003 00042x 0005 1
00045
61. (2, 00043); perpendicular to y 0003 000413 x
Graph each equation in Problems 19–32, and indicate the slope, if it exists.
62. (00042, 00044); perpendicular to y 0003 23 x 0004 5 63. (5, 0); parallel to 3x 0004 2y 0003 4
19. y 0003 000435 x 0005 4
20. y 0003 000432 x 0005 6
64. (3, 5); parallel to 3x 0005 4y 0003 8
21. y 0003 000434 x
22. y 0003 23 x 0004 3
65. (0, 00044); perpendicular to x 0005 3y 0003 9
23. 4x 0005 2y 0003 0
24. 6x 0004 2y 0003 0
66. (00042, 4); perpendicular to 4x 0005 5y 0003 0
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Problems 67–72 refer to the quadrilateral with vertices A 0003 (0, 2), B 0003 (4, 00041), C 0003 (1, 00045), and D 0003 (00043, 00042).
(A) Complete Table 4.
Table 4
67. Show that AB 储 DC.
68. Show that DA 储 CB.
69. Show that AB ⬜ BC.
70. Show that AD ⬜ DC.
x
71. Find an equation of the perpendicular bisector* of AD.
A
72. Find an equation of the perpendicular bisector of AB. 73. Prove that if a line L has x intercept (a, 0) and y intercept (0, b), then the equation of L can be written in the intercept form y x 0005 00031 a b
a, b 0002 0
74. Prove that if a line L passes through P1 0003 (x1, y1) and P2 0003 (x2, y2), then the equation of L can be written in the twopoint form ( y 0004 y1)(x2 0004 x1) 0003 ( y2 0004 y1)(x 0004 x1)
75. x2 0005 y2 0003 25, (3, 4)
76. x2 0005 y2 0003 100, (00048, 6)
77. x2 0005 y2 0003 50, (5, 00045)
78. x2 0005 y2 0003 80, (00044, 00048)
79. (x 0004 3)2 0005 ( y 0005 4)2 0003 169, (8, 000416) 80. (x 0005 5)2 0005 ( y 0004 9)2 0003 289, (000413, 00046)
APPLICATIONS 81. BOILING POINT OF WATER At sea level, water boils when it reaches a temperature of 212°F. At higher altitudes, the atmospheric pressure is lower and so is the temperature at which water boils. The boiling point B in degrees Fahrenheit at an altitude of x feet is given approximately by B 0003 212 0004 0.0018x (A) Complete Table 3.
Table 3 0
5,000
10,000
15,000
20,000
1
2
3
4
5
(B) Based on the information in the table, write a brief verbal description of the relationship between altitude and air temperature. 83. COST ANALYSIS A doughnut shop has a fixed cost of $124 per day and a variable cost of $0.12 per doughnut. Find the total daily cost of producing x doughnuts. How many doughnuts can be produced for a total daily cost of $250? 84. COST ANALYSIS A small company manufactures picnic tables. The weekly fixed cost is $1,200 and the variable cost is $45 per table. Find the total weekly cost of producing x picnic tables. How many picnic tables can be produced for a total weekly cost of $4,800? 85. PHYSICS Hooke’s law states that the relationship between the stretch s of a spring and the weight w causing the stretch is linear (a principle upon which all spring scales are constructed). For a particular spring, a 5-pound weight causes a stretch of 2 inches, while with no weight the stretch of the spring is 0. (A) Find a linear equation that expresses s in terms of w. (B) What is the stretch for a weight of 20 pounds? (C) What weight will cause a stretch of 3.6 inches?
Problems 75–80 are calculus related. Recall that a line tangent to a circle at a point is perpendicular to the radius drawn to that point (see the figure). Find the equation of the line tangent to the circle at the indicated point. Write the final answer in the standard form Ax 0005 By 0003 C, A 0007 0. Graph the circle and the tangent line on the same coordinate system.
x
0
25,000
30,000
B (B) Based on the information in the table, write a brief verbal description of the relationship between altitude and the boiling point of water. 82. AIR TEMPERATURE As dry air moves upward, it expands and cools. The air temperature A in degrees Celsius at an altitude of x kilometers is given approximately by A 0003 25 0004 9x *The perpendicular bisector of a line segment is a line perpendicular to the segment and passing through its midpoint.
86. PHYSICS The distance d between a fixed spring and the floor is a linear function of the weight w attached to the bottom of the spring. The bottom of the spring is 18 inches from the floor when the weight is 3 pounds and 10 inches from the floor when the weight is 5 pounds. (A) Find a linear equation that expresses d in terms of w. (B) Find the distance from the bottom of the spring to the floor if no weight is attached. (C) Find the smallest weight that will make the bottom of the spring touch the floor. (Ignore the height of the weight.) 87. PHYSICS The two most widespread temperature scales are Fahrenheit* (F) and Celsius† (C). It is known that water freezes at 32°F or 0°C and boils at 212°F or 100°C. (A) Find a linear equation that expresses F in terms of C. (B) If a European family sets its house thermostat at 20°C, what is the setting in degrees Fahrenheit? If the outside temperature in Milwaukee is 86°F, what is the temperature in degrees Celsius? 88. PHYSICS Two other temperature scales, used primarily by scientists, are Kelvin‡ (K) and Rankine** (R). Water freezes at 273 K or 492°R and boils at 373 K or 672°R. Find a linear equation that expresses R in terms of K. 89. OCEANOGRAPHY After about 9 hours of a steady wind, the height of waves in the ocean is approximately linearly related to *Invented in 1724 by Daniel Gabriel Fahrenheit (1686–1736), a German physicist. † Invented in 1742 by Anders Celsius (1701–1744), a Swedish astronomer. ‡ Invented in 1848 by Lord William Thompson Kelvin (1824–1907), a Scottish mathematician and physicist. Note that the degree symbol “ ° ” is not used with degrees Kelvin. **Invented in 1859 by John Maquorn Rankine (1820–1872), a Scottish engineer and physicist.
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Express all calculated quantities to three significant digits. 90. OCEANOGRAPHY Refer to Problem 89. A steady 25-knot wind produces a wave 7 feet high after 9 hours and 11 feet high after 25 hours. (A) Write a linear equation that expresses height h in terms of time t. (B) How long will the wind have been blowing for the waves to be 20 feet high? 91. DEMOGRAPHICS Life expectancy in the United States has increased from about 49.2 years in 1900 to about 77.3 years in 2000. The growth in life expectancy is approximately linear with respect to time. (A) If L represents life expectancy and t represents the number of years since 1900, write a linear equation that expresses L in terms of t. (B) What is the predicted life expectancy in the year 2020? Express all calculated quantities to three significant digits.
147
92. DEMOGRAPHICS The average number of persons per household in the United States has been shrinking steadily for as long as statistics have been kept and is approximately linear with respect to time. In 1900, there were about 4.76 persons per household and in 2000, about 2.59. (A) If N represents the average number of persons per household and t represents the number of years since 1900, write a linear equation that expresses N in terms of t. (B) What is the predicted household size in the year 2025? Express all calculated quantities to three significant digits. 93. CITY PLANNING The design of a new subdivision calls for three parallel streets connecting First Street with Main Street (see the figure). Find the distance d1 (to the nearest foot) from Avenue A to Avenue B. First Street Avenue A
Distance in feet
the duration of time the wind has been blowing. During a storm with 50-knot winds, the wave height after 9 hours was found to be 23 feet, and after 24 hours it was 40 feet. (A) If t is time after the 50-knot wind started to blow and h is the wave height in feet, write a linear equation that expresses height h in terms of time t. (B) How long will the wind have been blowing for the waves to be 50 feet high?
Linear Equations and Models
5,000
Avenue B
Avenue C
d2
d1
0
5,000
Main Street
Distance in feet
94. CITY PLANNING Refer to Problem 93. Find the distance d2 (to the nearest foot) from Avenue B to Avenue C.
2-4
Linear Equations and Models Z Slope as a Rate of Change Z Linear Models Z Linear Regression
Mathematical modeling is the process of using mathematics to solve real-world problems. This process can be broken down into three steps (Fig. 1): Step 1. Construct the mathematical model, a mathematics problem that, when solved, will provide information about the real-world problem.
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Step 2. Solve the mathematical model. Step 3. Interpret the solution to the mathematical model in terms of the original real-world problem.
Real-world problem
3. I
t uc str
nt er p
on
re t
C 1.
Mathematical solution
2. Solve
Mathematical model
Z Figure 1
In more complex problems, this cycle may have to be repeated several times to obtain the required information about the real-world problem. In this section, we discuss one of the simplest mathematical models, a linear equation. With the aid of a graphing calculator, we also learn how to analyze a linear model based on real-world data.
Z Slope as a Rate of Change If x and y are related by the equation y 0003 mx 0005 b, where m and b are constants with m 0002 0, then x and y are linearly related. If (x1, y1) and (x2, y2) are two distinct points on this line, then the slope of the line is m0003
Change in y y2 0004 y1 0003 x2 0004 x1 Change in x
(1)
In applications, ratio (1) is called the rate of change of y with respect to x. Since the slope of a line is unique, the rate of change of two linearly related variables is constant. Here are some examples of familiar rates of change: miles per hour, revolutions per minute, price per pound, passengers per plane, etc. If y is distance and x is time, then the rate of change is also referred to as speed or velocity. If the relationship between x and y is not linear, ratio (1) is called the average rate of change of y with respect to x.
EXAMPLE
1
Estimating Body Surface Area Appropriate doses of medicine for both animals and humans are often based on body surface area (BSA). Since weight is much easier to determine than BSA, veterinarians use the weight of an animal to estimate BSA. The following linear equation expresses BSA for canines in terms of weight*: a 0003 16.21w 0005 375.6 where a is BSA in square inches and w is weight in pounds. (A) Interpret the slope of the BSA equation. (B) What is the effect of a 1-pound increase in weight?
SOLUTIONS
(A) The rate of change BSA with respect to weight is 16.21 square inches per pound. (B) Since slope is the ratio of rise to run, increasing w by 1 pound (run) increases a by 16.21 square inches (rise). 0002 *Based on data from Veterinary Oncology Consultants, PTY LTD.
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MATCHED PROBLEM 1
Linear Equations and Models
149
The following linear equation expresses BSA for felines in terms of weight: a ⫽ 28.55w ⫹ 118.7 where a is BSA in square inches and w is weight in pounds. (A) Interpret the slope of the BSA equation. (B) What is the effect of a 1-pound increase in weight? 0002
Z Linear Models We can use our experience with lines in Section 2-3 to construct linear models for applications involving linearly related quantities. This process is best illustrated through examples.
EXAMPLE
2
Business Markup Policy A sporting goods store sells a fishing rod that cost $60 for $82 and a pair of cross-country ski boots that cost $80 for $106. (A) If the markup policy of the store for items that cost more than $30 is assumed to be linear, find a linear model that express the retail price P in terms of the wholesale cost C. (B) What is the effect on the price of a $1 increase in cost for any item costing over $30? (C) Use the model to find the retail price for a pair of running shoes that cost $40.
SOLUTIONS
(A) If price P is linearly related to cost C, then we are looking for the equation of a line whose graph passes through (C1, P1) ⫽ (60, 82) and (C2, P2) ⫽ (80, 106). We find the slope, and then use the point–slope form to find the equation. m⫽
P2 ⫺ P1 106 ⫺ 82 24 ⫽ ⫽ ⫽ 1.2 C2 ⫺ C1 80 ⫺ 60 20
Substitute P1 ⴝ 82, C1 ⴝ 60, and m ⴝ 1.2 into the point–slope formula.
P ⫺ P1 ⫽ m(C ⫺ C1) P ⫺ 82 ⫽ 1.2(C ⫺ 60) P ⫺ 82 ⫽ 1.2C ⫺ 72 P ⫽ 1.2C ⫹ 10
Substitute C1 ⴝ 60, P1 ⴝ 82, C2 ⴝ 80, and P2 ⴝ 106 into the slope formula.
Distribute Add 82 to both sides.
C 7 30
Linear model
(B) If the cost is increased by $1, then the price will increase by 1.2(1) ⫽ $1.20. (C) P ⫽ 1.2(40) ⫹ 10 ⫽ $58. MATCHED PROBLEM 2
ZZZ EXPLORE-DISCUSS 1
0002
The sporting goods store in Example 2 is celebrating its twentieth anniversary with a 20% off sale. The sale price of a mountain bike is $380. What was the presale price of the bike? How much did the bike cost the store? 0002 The wholesale supplier for the sporting goods store in Example 2 offers the store a 15% discount on all items. The store decides to pass on the savings from this discount to the consumer. Which of the following markup policies is better for the consumer? 1. Apply the store’s markup policy to the discounted cost. 2. Apply the store’s markup policy to the original cost and then reduce this price by 15%. Support your choice with examples.
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3
Mixing Antifreeze Ethylene glycol and propylene glycol are liquids used in antifreeze and deicing solutions. Ethylene glycol is listed as a hazardous chemical by the Environmental Protection Agency, while propylene glycol is generally regarded as safe. Table 1 lists solution concentration percentages and the corresponding freezing points for each chemical. Table 1 Concentration
Ethylene Glycol
Propylene Glycol
20%
15°F
17°F
50%
000436°F
000428°F
(A) Assume that the concentration and the freezing point for ethylene glycol are linearly related. Construct a linear model for the freezing point. (B) Interpret the slope in part (A). (C) What percentage (to one decimal place) of ethylene glycol will result in a freezing point of 000410°F? SOLUTIONS
(A) We begin by defining appropriate variables: Let p 0003 percentage of ethylene glycol in the antifreeze solution f 0003 freezing point of the antifreeze solution From Table 1, we see that (20, 15) and (50, ⴚ36) are two points on the line relating p and f. The slope of this line is m0003
f2 0004 f1 15 0004 (ⴚ36) 51 0003 0003 0003 00041.7 p2 0004 p1 20 0004 50 000430
and its equation is f 0004 15 0003 00041.7( p 0004 20) f 0003 00041.7p 0005 49
Linear model
(B) The rate of change of the freezing point with respect to the percentage of ethylene glycol in the antifreeze solution is 00041.7 degrees per percentage of ethylene glycol. Increasing the amount of ethylene glycol by 1% will lower the freezing point by 1.7°F. (C) We must find p when f is 000410°. f 0003 00041.7p 0005 49 000410 0003 00041.7p 0005 49 1.7p 0003 59 59 p0003 0003 34.7% 1.7
MATCHED PROBLEM 3
Add 10 ⴙ 1.7p to both sides. Divide both sides by 1.7.
0002
Refer to Table 1. (A) Assume that the concentration and the freezing point for propylene glycol are linearly related. Construct a linear model for the freezing point. (B) Interpret the slope in part (A). (C) What percentage (to one decimal place) of propylene glycol will result in a freezing point of 000415°F? 0002
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EXAMPLE
4
Linear Equations and Models
151
Underwater Pressure The pressure at sea level is 14.7 pounds per square inch. As you descend into the ocean, the pressure increases linearly at a rate of about 0.445 pounds per square foot. (A) Find the pressure p at a depth of d feet. (B) If a diver’s equipment is rated to be safe up to a pressure of 40 pounds per square foot, how deep (to the nearest foot) is it safe to use this equipment?
SOLUTIONS
(A) Let p ⫽ md ⫹ b. At the surface, d ⫽ 0 and p ⫽ 14.7, so b ⫽ 14.7. The slope m is the given rate of change, m ⫽ 0.445. So the pressure at a depth of d feet is p ⫽ 0.445d ⫹ 14.7 (B) The safe depth is the solution of the equation 0.445d ⫹ 14.7 ⫽ 40 0.445d ⫽ 25.3 25.3 d⫽ 0.445 ⬇ 57 feet
MATCHED PROBLEM 4
Subtract 14.7 from each side. Divide both sides by 0.445. Simplify.
0002
The rate of change of pressure in fresh water is 0.432 pounds per square foot. Repeat Example 4 for a body of fresh water. 0002
Technology Connections 80
Figure 2 shows the solution of Example 4(B) on a graphing calculator. 0
100
⫺20
Z Figure 2 y1 ⫽ 0.445x ⫹ 14.7, y2 ⫽ 40
Z Linear Regression In real-world applications we often encounter numerical data in the form of a table. The very powerful mathematical tool, regression analysis, can be used to analyze numerical data. In general, regression analysis is a process for finding an equation that provides a useful model for a set of data points. Graphs of equations are often called curves and regression analysis is also referred to as curve fitting. In Example 5, we use a linear model obtained by using linear regression on a graphing calculator.
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GRAPHS
5
Table 2 Round-Shaped Diamond Prices Weight (Carats)
Price
0.5
$1,340
0.6
$1,760
0.7
$2,540
0.8
$3,350
0.9
$4,130
1.0
$4,920
Source: www.tradeshop.com
SOLUTIONS
Diamond Prices Prices for round-shaped diamonds taken from an online trader are given in Table 2. (A) A linear model for the data in Table 2 is given by p ⫽ 7,380c ⫺ 2,530
(2)
where p is the price of a diamond weighing c carats. (We will discuss the source of models like this later in this section.) Plot the points in Table 2 on a Cartesian coordinate system, producing a scatter plot, and graph the model on the same axes. (B) Interpret the slope of the model in equation (2). (C) Use the model to estimate the cost of a 0.85-carat diamond and the cost of a 1.2-carat diamond. Round answers to the nearest dollar. (D) Use the model to estimate the weight of a diamond that sells for $3,000. Round the answer to two significant digits. (A) A scatter plot is simply a plot of the points in Table 2 [Fig. 3(a)]. To add the graph of the model to the scatter plot, we find any two points that satisfy equation (2) [we choose (0.4, 422) and (1.4, 7,802)]. Plotting these points and drawing a line through them gives us Figure 3(b). p
p
$8,000
$8,000
Price
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$4,000
0.5
1
1.5
$4,000
c
0.5
Carats
1
1.5
c
Carats (b) Linear model
(a) Scatter plot
Z Figure 3
(B) The rate of change of the price of a diamond with respect to its weight is 7,380. Increasing the weight by 1 carat will increase the price by about $7,380. (C) The graph of the model [Fig. 3(b)] does not pass through any of the points in the scatter plot, but it comes close to all of them. [Verify this by evaluating equation (2) at c ⫽ 0.5, 0.6, . . . , 1.] So we can use equation (2) to approximate points not in Table 2. c ⫽ 0.85 p ⫽ 7,380(0.85) ⫺ 2,530 ⫽ $3,743
c ⫽ 1.2 p ⫽ 7,380(1.2) ⫺ 2,530 ⫽ $6,326
A 0.85-carat diamond will cost about $3,743 and a 1.2-carat diamond will cost about $6,326. (D) To find the weight of a $3,000 diamond, we solve the following equation for c: 7,380c ⫺ 2,530 ⫽ 3,000 7,380c ⫽ 3,000 ⫹ 2,530 ⫽ 5,530 5,530 ⫽ 0.75 c⫽ 7,380
To two significant digits
A $3,000 diamond will weigh about 0.75 carats.
0002
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MATCHED PROBLEM 5 Table 3 Emerald-Shaped Diamond Prices Weight (Carats)
Price
0.5
$1,350
0.6
$1,740
0.7
$2,610
0.8
$3,320
0.9
$4,150
1.0
$4,850
Source: www.tradeshop.com
Linear Equations and Models
153
Prices for emerald-shaped diamonds taken from an online trader are given in Table 3. Repeat Example 5 for this data with the linear model p ⫽ 7,270c ⫺ 2,450 where p is the price of an emerald-shaped diamond weighing c carats. 0002 The model we used in Example 5 was obtained by using a technique called linear regression and the model is called the regression line. This technique produces a line that is the best fit for a given data set. We will not discuss the theory behind this technique, nor the meaning of “best fit.” Although you can find a linear regression line by hand, we prefer to leave the calculations to a graphing calculator or a computer. Don’t be concerned if you don’t have either of these electronic devices. We will supply the regression model in the applications we discuss, as we did in Example 5.
Technology Connections If you want to use a graphing calculator to construct regression lines, you should consult your user’s manual.* The process varies from one calculator to another. Figure 4
shows three of the screens related to the construction of the model in Example 5 on a Texas Instruments TI-84 Plus. 8,000
0
1.5
⫺1,000
(a) Entering the data.
(b) Finding the model.
(c) Graphing the data and the model.
Z Figure 4 *User’s manuals for the most popular graphing calculators are readily available on the Internet.
In Example 5, we used the regression line to approximate points that were not given in Table 2, but would fit between points in the table. This process is called interpolation. In the next example we use a regression model to approximate points outside the given data set. This process is called extrapolation and the approximations are often referred to as predictions.
EXAMPLE
6
Telephone Expenditures Table 4 gives information about expenditures for residential and cellular phone service. The linear regression model for residential service is r ⫽ 722 ⫺ 33.1t where r is the average annual expenditure (in dollars per consumer unit) on residential service and t is time in years with t ⫽ 0 corresponding to 2000. (A) Interpret the slope of the regression line as a rate of change. (B) Use the regression line to predict expenditures for residential service in 2018.
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Table 4 Average Annual Telephone Expenditures (dollars per consumer unit) 2001
2003
2005
2007
Residential
686
620
570
482
Cellular
210
316
455
608
Source: Bureau of Labor Statistics
SOLUTIONS
(A) The slope m 0003 000433.1 is the rate of change of expenditures with respect to time. Because the slope is negative, the expenditures for residential service are decreasing at a rate of $33.10 per year. (B) If t 0003 18, then r 0003 722 0004 33.1(18) 0003 $126 So the model predicts that expenditures for residential phone service will be approximately $126 in 2018. Repeat Example 6 using the following linear regression model for cellular service: c 0003 66.7t 0005 131 where c is the average annual expenditure (in dollars per consumer unit) on cellular service and t is time in years with t = 0 corresponding to 2000. 0002
ANSWERS TO MATCHED PROBLEMS 1. (A) The rate of change of BSA with respect to weight is 28.55 square inches per pound. (B) Increasing w by 1 pound increases a by 28.55 square inches. 2. Presale price is $475. Cost is $387.50 3. (A) f 0003 00041.5p 0005 47 (B) The rate of change of the freezing point with respect to the percentage of propylene glycol in the antifreeze solution is 00041.5. Increasing the percentage of propylene glycol by 1% will lower the freezing point by 1.5°F. (C) 41.3% 4. (A) p 0003 0.432d 0005 14.7 (B) 59 ft p 5. (A) $8,000
Price
MATCHED PROBLEM 6
0002
$4,000
0.5
1
1.5
c
Carats
(B) The rate of change of the price of a diamond with respect to the size is 7,270. Increasing the size by 1 carat will increase the price by about $7,270. (C) $3,730; $6,274 (D) 0.75 carats 6. (A) The expenditures for cellular service are increasing at a rate of $66.70 per year. (B) $1,332.
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2-4
Linear Equations and Models
155
Exercises
1. Explain the steps that are involved in the process of mathematical modeling.
9. Dr. J. D. Robinson and Dr. D. R. Miller published the following models for estimating the weight of a woman:
2. If two variables x and y are linearly related, explain how to calculate the rate of change.
Robinson: w 0003 108 0005 3.7h
3. If two variables x and y are not linearly related, explain how to calculate the average rate of change from x 0003 x1 to x 0003 x2.
where w is weight (in pounds) and h is height over 5 feet (in inches). (A) Interpret the slope of each model. (B) If a woman is 500066 tall, what does each model predict her weight to be? (C) If a woman weighs 140 pounds, what does each model predict her height to be?
4. Explain the difference between interpolation and extrapolation in the context of regression analysis.
APPLICATIONS 5. COST ANALYSIS A plant can manufacture 80 golf clubs per day for a total daily cost of $8,147 and 100 golf clubs per day for a total daily cost of $9,647. (A) Assuming that daily cost and production are linearly related, find the total daily cost of producing x golf clubs. (B) Interpret the slope of this cost equation. (C) What is the effect of a 1 unit increase in production? 6. COST ANALYSIS A plant can manufacture 50 tennis rackets per day for a total daily cost of $4,174 and 60 tennis rackets per day for a total daily cost of $4,634. (A) Assuming that daily cost and production are linearly related, find the total daily cost of producing x tennis rackets. (B) Interpret the slope of this cost equation. (C) What is the effect of a 1 unit increase in production? 7. FORESTRY Forest rangers estimate the height of a tree by measuring the tree’s diameter at breast height (DBH) and then using a model constructed for a particular species.* A model for white spruce trees is h 0003 4.06d 0005 24.1 where d is the DBH in inches and h is the tree height in feet. (A) Interpret the slope of this model. (B) What is the effect of a 1-inch increase in DBH? (C) How tall is a white spruce with a DBH of 12 inches? Round answer to the nearest foot. (D) What is the DBH of a white spruce that is 100 feet tall? Round answer to the nearest inch. 8. FORESTRY A model for black spruce trees is h 0003 2.27d 0005 33.1 where d is the DBH in inches and h is the tree height in feet. (A) Interpret the slope of this model. (B) What is the effect of a 1-inch increase in DBH? (C) How tall is a black spruce with a DBH of 12 inches? Round answer to the nearest foot. (D) What is the DBH of a black spruce that is 100 feet tall? Round answer to the nearest inch. *Models in Problems 7 and 8 are based on data found at http://flash.lakeheadu.ca/~fluckai/htdbh04.xls
Miller: w 0003 117 0005 3.0h
10. Dr. J. D. Robinson and Dr. D. R. Miller also published the following models for estimating the weight of a man: Robinson: w 0003 115 0005 4.2h Miller: w 0003 124 0005 3.1h where w is weight (in pounds) and h is height over 5 feet (in inches). (A) Interpret the slope of each model. (B) If a man is 5000610 tall, what does each model predict his weight to be? (C) If a man weighs 160 pounds, what does each model predict his height to be? 11. SPEED OF SOUND The speed of sound through the air near sea level is linearly related to the temperature of the air. If sound travels at 741 mph at 32°F and at 771 mph at 72°F, construct a linear model relating the speed of sound (s) and the air temperature (t). Interpret the slope of this model. 12. SPEED OF SOUND The speed of sound through the air near sea level is linearly related to the temperature of the air. If sound travels at 337 mps (meters per second) at 10°C and at 343 mps at 20°C, construct a linear model relating the speed of sound (s) and the air temperature (t). Interpret the slope of this model. 13. SMOKING STATISTICS The percentage of male cigarette smokers in the United States declined from 25.7% in 2000 to 23.9% in 2006. Find a linear model relating the percentage m of male smokers to years t since 2000. Use the model to predict the first year for which the percentage of male smokers will be less than or equal to 18%. 14. SMOKING STATISTICS The percentage of female cigarette smokers in the United States declined from 21.0% in 2000 to 18.0% in 2006. Find a linear model relating the percentage f of female smokers to years t since 2000. Use the model to predict the first year for which the percentage of female smokers will be less than or equal to 10%. 15. BUSINESS—DEPRECIATION A farmer buys a new tractor for $142,000 and assumes that it will have a trade-in value of $67,000 after 10 years. The farmer uses a constant rate of depreciation (commonly called straight-line depreciation—one of several methods permitted by the IRS) to determine the annual value of the tractor. (A) Find a linear model for the depreciated value V of the tractor t years after it was purchased.
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(B) Interpret the slope of this model. (C) What is the depreciated value of the tractor after 6 years? 16. BUSINESS—DEPRECIATION A charter fishing company buys a new boat for $154,900 and assumes that it will have a trade-in value of $46,100 after 16 years. (A) Use straight-line depreciation (See Problem 15) to find a linear model for the depreciated value V of the boat t years after it was purchased. (B) Interpret the slope of this model. (C) In which year will the depreciated value of the boat fall below $100,000?
23. LICENSED DRIVERS Table 5 contains the state population and the number of licensed drivers in the state (both in millions) for the states with population under 1 million. The regression model for this data is y 0003 0.72x 0005 0.03 where x is the state population and y is the number of licensed drivers in the state.
Table 5 Licensed Drivers in 2006 State
Population
Licensed Drivers
17. BUSINESS—MARKUP POLICY A drugstore sells a drug costing $85 for $112 and a drug costing $175 for $238. (A) If the markup policy of the drugstore is assumed to be linear, write an equation that expresses retail price R in terms of cost C (wholesale price). (B) What is the slope of the graph of the equation found in part A? Interpret verbally. (C) What does a store pay (to the nearest dollar) for a drug that retails for $185?
Alaska
0.67
0.49
Delaware
0.85
0.62
Montana
0.94
0.72
North Dakota
0.64
0.47
South Dakota
0.78
0.58
Vermont
0.62
0.53
18. BUSINESS—MARKUP POLICY A clothing store sells a shirt costing $20 for $33 and a jacket costing $60 for $93. (A) If the markup policy of the store for items costing over $10 is assumed to be linear, write an equation that expresses retail price R in terms of cost C (wholesale price). (B) What is the slope of the equation found in part A? Interpret verbally. (C) What does a store pay for a suit that retails for $240?
Wyoming
0.52
0.39
19. FLIGHT CONDITIONS In stable air, the air temperature drops about 5 F for each 1,000-foot rise in altitude. (A) If the temperature at sea level is 70°F and a commercial pilot reports a temperature of 000420 F at 18,000 feet, write a linear equation that expresses temperature T in terms of altitude A (in thousands of feet). (B) How high is the aircraft if the temperature is 0 F? 20. FLIGHT NAVIGATION An airspeed indicator on some aircraft is affected by the changes in atmospheric pressure at different altitudes. A pilot can estimate the true airspeed by observing the indicated airspeed and adding to it about 2% for every 1,000 feet of altitude. (A) If a pilot maintains a constant reading of 200 miles per hour on the airspeed indicator as the aircraft climbs from sea level to an altitude of 10,000 feet, write a linear equation that expresses true airspeed T (miles per hour) in terms of altitude A (thousands of feet). (B) What would be the true airspeed of the aircraft at 6,500 feet? 21. RATE OF DESCENT—PARACHUTES At low altitudes, the altitude of a parachutist and time in the air are linearly related. A jump at 2,880 ft using the U.S. Army’s T-10 parachute system lasts 120 seconds. (A) Find a linear model relating altitude a (in feet) and time in the air t (in seconds). (B) The rate of descent is the speed at which the jumper falls. What is the rate of descent for a T-10 system? 22. RATE OF DESCENT—PARACHUTES The U.S. Army is considering a new parachute, the ATPS system. A jump at 2,880 ft using the ATPS system lasts 180 seconds. (A) Find a linear model relating altitude a (in feet) and time in the air t (in seconds). (B) What is the rate of descent for an ATPS system parachute?
Source: Bureau of Transportation Statistics
(A) Plot the data in Table 5 and the model on the same axes. (B) If the population of New Hampshire in 2006 was about 1.3 million, use the model to estimate the number of licensed drivers in New Hampshire. (C) If the population of Nebraska in 2006 was about 1.8 million, use the model to estimate the number of licensed drivers in Nebraska. 24. LICENSED DRIVERS Table 6 contains the state population and the number of licensed drivers in the state (both in millions) for several states with population over 10 million. The regression model for this data is y 0003 0.60x 0005 1.15 where x is the state population and y is the number of licensed drivers in the state.
Table 6 Licensed Drivers in 2006 State
Population
Licensed Drivers
California
36
23
Florida
18
14
Illinois
13
8
Michigan
10
7
New York
19
11
Ohio
11
8
Pennsylvania
12
9
Texas
24
15
Source: Bureau of Transportation Statistics
(A) Plot the data in Table 6 and the model on the same axes. (B) If the population of Georgia in 2006 was about 9.4 million, use the model to estimate the number of licensed drivers in Georgia. (C) If the population of New Jersey in 2006 was about 8.7 million, use the model to estimate the number of licensed drivers in New Jersey.
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Review
Problems 25–28 require a graphing calculator or a computer that can calculate the linear regression line for a given data set. 25. OLYMPIC GAMES Find a linear regression model for the men’s 100-meter freestyle data given in Table 7, where x is years since 1968 and y is winning time (in seconds). Do the same for the women’s 100-meter freestyle data. (Round regression coefficients to four significant digits.) Do these models indicate that the women will eventually catch up with the men?
Table 7 Winning Times in Olympic Swimming Events 100-Meter Freestyle
200-Meter Backstroke
Men
Women
Men
Women
1968
52.20
60.0
2:09.60
2:24.80
1976
49.99
55.65
1:59.19
2:13.43
1984
49.80
55.92
2:00.23
2:12.38
1992
49.02
54.65
1:58.47
2:07.06
2000
48.30
53.83
1:56.76
2:08.16
2008
47.21
53.12
1:53.94
2:05.24
26. OLYMPIC GAMES Find a linear regression model for the men’s 200-meter backstroke data given in Table 7 where x is years since 1968 and y is winning time (in seconds). Do the same for the women’s 200-meter backstroke data. (Round regression coefficients to five significant digits.) Do these models indicate that the women will eventually catch up with the men? 27. SUPPLY AND DEMAND Table 8 contains price–supply data and price–demand data for corn. Find a linear regression model for the price–supply data where x is supply (in billions of bushels) and y is price (in dollars). Do the same for the price–demand data. (Round regression coefficients to three significant digits.) Find the price at which supply and demand are equal. (In economics, this price is referred to as the equilibrium price.)
2-1
2
Price ($/bu.)
Supply (Billion bu.)
Price ($/bu.)
Demand (Billion bu.)
2.15
6.29
2.07
9.78
2.29
7.27
2.15
9.35
2.36
7.53
2.22
8.47
2.48
7.93
2.34
8.12
2.47
8.12
2.39
7.76
2.55
8.24
2.47
6.98
Source: www.usda.gov/nass/pubs/histdata.htm
28. SUPPLY AND DEMAND Table 9 contains price–supply data and price–demand data for soybeans. Find a linear regression model for the price–supply data where x is supply (in billions of bushels) and y is price (in dollars). Do the same for the price–demand data. (Round regression coefficients to three significant digits.) Find the equilibrium price for soybeans.
Table 9 Supply and Demand for U.S. Soybeans
Source: www.infoplease.com
CHAPTER
Table 8 Supply and Demand for U.S. Corn
Price ($/bu.)
Supply (Billion bu.)
Price ($/bu.)
Demand (Billion bu.)
5.15
1.55
4.93
2.60
5.79
1.86
5.48
2.40
5.88
1.94
5.71
2.18
6.07
2.08
6.07
2.05
6.15
2.15
6.40
1.95
6.25
2.27
6.66
1.85
Source: www.usda.gov/nass/pubs/histdata.htm
Review
Cartesian Coordinate System
A Cartesian or rectangular coordinate system is formed by the intersection of a horizontal real number line and a vertical real number line at their origins. These lines are called the coordinate axes. The horizontal axis is often referred to as the x axis and the vertical axis as the y axis. These axes divide the plane into four quadrants. Each point in the plane corresponds to its coordinates— an ordered pair (a, b) determined by passing horizontal and vertical lines through the point. The abscissa or x coordinate a is the coordinate of the intersection of the vertical line with the horizontal axis, and the ordinate or y coordinate b is the coordinate of the intersection of the horizontal line with the vertical axis. The point (0, 0) is
called the origin. A solution of an equation in two variables is an ordered pair of real numbers that makes the equation a true statement. The solution set of an equation is the set of all its solutions. The graph of an equation in two variables is the graph of its solution set formed using point-by-point plotting or with the aid of a graphing calculator. The reflection of the point (a, b) through the y axis is the point (0004a, b), through the x axis is the point (a, 0004b), and through the origin is the point (0004a, 0004b). The reflection of a graph is the reflection of each point on the graph. If reflecting a graph through the y axis, x axis, or origin does not change its shape, the graph is said to be symmetric with respect to the y axis, x axis, or origin, respectively. To test an equation for symmetry, determine if the equation is unchanged when y is replaced with 0004y (x axis symmetry), x is replaced
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with 0004x ( y axis symmetry), or both x and y are replaced with 0004x and 0004y (origin symmetry).
The slope is not defined for a vertical line where x1 0003 x2. Two lines with slopes m1 and m2 are parallel if and only if m1 0003 m2 and perpendicular if and only if m1m2 0003 00041.
2-2
Equations of a Line Standard form Ax 0005 By 0003 C
Distance in the Plane
The distance between the two points P1 0003 (x1, y1) and P2 0003 (x2, y2) is d(P1, P2) 0003 2(x2 0004 x1)2 0005 ( y2 0004 y1)2 and the midpoint of the line segment joining P1 0003 (x1, y1) and P2 0003 (x2, y2) is M0003a
x1 0005 x2 y1 0005 y2 , b 2 2
The standard form for the equation of a circle with radius r and center at (h, k) is (x 0004 h)2 0005 ( y 0004 k)2 0003 r2,
2-3
r 7 0
Equations of a Line
The standard form for the equation of a line is Ax 0005 By 0003 C, where A, B, and C are constants, A and B not both 0. The y intercept is the y coordinate of the point where the graph crosses the y axis, and the x intercept is the x coordinate of the point where the graph crosses the x axis. The slope of the line through the points (x1, y1) and (x2, y2) is m0003
CHAPTER
y2 0004 y1 x2 0004 x1
2
if x1 0002 x2
Slope–intercept form y 0003 mx 0005 b
A and B not both 0 Slope: m; y intercept: b
Point–slope form
y 0004 y1 0003 m(x 0004 x1) Slope: m; Point: (x1, y1)
Horizontal line
y0003b
Slope: 0
Vertical line
x0003a
Slope: Undefined
2-4
Linear Equations and Models
A mathematical model is a mathematics problem that, when solved, will provide information about a real-world problem. If y 0003 mx 0005 b, then the variables x and y are linearly related and the rate of change of y with respect to x is the constant m. If x and y are not linearly related, the ratio ( y2 0004 y1)兾(x2 0004 x1) is called the average rate of change of y with respect to x. Regression analysis produces an equation whose graph is a curve that fits (approximates) a set of data points. A scatter plot is the graph of the points in a data set. Linear regression produces a regression line that is the best fit for a given data set. Graphing calculators or other electronic devices are frequently used to find regression lines.
Review Exercises
Work through all the problems in this chapter review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text.
y 5
00045
5
x
1. Plot A 0003 (00044, 1), B 0003 (2, 00043), and C 0003 (00041, 00042) in a rectangular coordinate system. 2. Refer to Problem 1. Plot the reflection of A through the x axis, the reflection of B through the y axis, and the reflection of C through the origin. 3. Test each equation for symmetry with respect to the x axis, y axis, and origin and sketch its graph. (A) y 0003 2x (B) y 0003 2x 0004 1 (C) y 0003 2|x| (D) | y| 0003 2x 4. Use the following graph to estimate to the nearest integer the missing coordinates of the indicated points. (Be sure you find all possible answers.) (A) (0, ?) (B) (?, 0) (C) (?, 4)
00045
5. Given the points A 0003 (00042, 3) and B 0003 (4, 0), find: (A) Distance between A and B (B) Slope of the line through A and B (C) Slope of a line perpendicular to the line through A and B 6. Write the equation of a circle with radius 17 and center: (A) (0, 0) (B) (3, 00042) 7. Find the center and radius of the circle given by (x 0005 3)2 0005 ( y 0004 2)2 0003 5
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8. Let M be the midpoint of A and B, where A 0003 (a1, a2), B 0003 (2, 00045), and M 0003 (00044, 3). (A) Use the fact that 00044 is the average of a1 and 2 to find a1. (B) Use the fact that 3 is the average of a2 and 00045 to find a2. (C) Find d(A, M ) and d(M, B). 9. (A) Graph the triangle with vertices A 0003 (00041, 00042), B 0003 (4, 3), and C 0003 (1, 4). (B) Find the perimeter to two decimal places. (C) Use the Pythagorean theorem to determine if the triangle is a right triangle. (D) Find the midpoint of each side of the triangle. 10. Use the graph of the linear function in the figure to find the rise, run, and slope. Write the equation of the line in the form Ax 0005 By 0003 C, where A, B, and C are integers with A 0. (The horizontal and vertical line segments have integer lengths.) y
159
21. Write the slope–intercept form of the equation of the line that passes through the point (00042, 1) and is (A) parallel to the line 6x 0005 3y 0003 5 (B) perpendicular to the line 6x 0005 3y 0003 5 22. Find the equation of a circle that passes through the point (00041, 4) with center at (3, 0). 23. Find the center and radius of the circle given by x2 0005 y2 0005 4x 0004 6y 0003 3 24. Find the equation of the set of points equidistant from (3, 3) and (6, 0). What is the name of the geometric figure formed by this set? 25. Are the graphs of mx 0004 y 0003 b and x 0005 my 0003 b parallel, perpendicular, or neither? Justify your answer. 26. Use completing the square to find the center and radius of the circle with equation: x2 0004 4x 0005 y2 0004 2y 0004 3 0003 0
5
00045
5
x
27. Refer to Problem 26. Find the equation of the line tangent to the circle at the point (4, 3). Graph the circle and the line on the same coordinate system. 28. Find the equation of a circle with center (4, 00043) whose graph passes through the point (1, 2).
00045
11. Graph 3x 0005 2y 0003 9 and indicate its slope. 12. Write an equation of a line with x intercept 6 and y intercept 4. Write the final answer in the standard form Ax 0005 By 0003 C, where A, B, and C are integers.
29. Extend the following graph to one that exhibits the indicated symmetry: (A) x axis only (B) y axis only (C) origin only (D) x axis, y axis, and origin y 5
13. Write the slope–intercept form of the equation of the line with slope 000423 and y intercept 2. 14. Write the equations of the vertical and horizontal lines passing through the point (00043, 4). What is the slope of each?
00045
5
00045
Test each equation in Problems 15–18 for symmetry with respect to the x axis, y axis, and the origin. Sketch the graph of the equation. 15. y 0003 x2 0004 2
16. y2 0003 x 0004 2
17. 9y2 0005 4x2 0003 36
18. 9y2 0004 4x2 0003 36
x
Problems 30 and 31 refer to a triangle with base b and height h (see the figure). Write a mathematical expression in terms of b and h for each of the verbal statements in Problems 30 and 31.
19. Write a verbal description of the graph shown in the figure and then write an equation that would produce the graph.
h
y
b
30. The base is five times the height.
5
31. The height is one-fourth of the base. 00045
5
x
00045
20. (A) Find an equation of the line through P 0003 (00044, 3) and Q 0003 (0, 00043). Write the final answer in the standard form Ax 0005 By 0003 C, where A, B, and C are integers with A 0. (B) Find d(P, Q).
APPLICATIONS 32. LINEAR DEPRECIATION A computer system was purchased by a small company for $12,000 and is assumed to have a depreciated value of $2,000 after 8 years. If the value is depreciated linearly from $12,000 to $2,000: (A) Find the linear equation that relates value V (in dollars) to time t (in years). (B) What would be the depreciated value of the system after 5 years?
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33. COST ANALYSIS A video production company is planning to produce an instructional CD. The producer estimates that it will cost $24,900 to produce the CD and $5 per unit to copy and distribute the CD. The budget for this project is $62,000. How many CDs can be produced without exceeding the budget? 34. FORESTRY Forest rangers estimate the height of a tree by measuring the tree’s diameter at breast height (DBH) and then using a model constructed for a particular species. A model for sugar maples is h 0003 2.9d 0005 30.2 where d is the DBH in inches and h is the tree height in feet. (A) Interpret the slope of this model. (B) What is the effect of a 1-inch increase in DBH? (C) How tall is a sugar maple with a DBH of 3 inches? Round answer to the nearest foot. (D) What is the DBH of a sugar maple that is 45 feet tall? Round answer to the nearest inch. 35. ESTIMATING BODY SURFACE AREA An important criterion for determining drug dosage for children is the patient’s body surface area (BSA). John D. Current published the following useful model for estimating BSA*: BSA 0003 1,321 0005 0.3433 Wt where BSA is given in square centimeters and Wt in grams. (A) Interpret the slope of this model. (B) What is the effect of a 100-gram increase in weight? (C) What is the BSA for a child that weighs 15 kilograms?
*“Body Surface Area in Infants and Children,” The Internet Journal of Anesthesiology, 1998, Volume 2, Number 2.
CHAPTER
ZZZ
36. ARCHITECTURE A circular arc forms the top of an entryway with 6-foot vertical sides 8 feet apart. If the top of the arc is 2 feet above the ends, what is the radius of the arc? 37. SPORTS MEDICINE The following quotation was found in a sports medicine handout: “The idea is to raise and sustain your heart rate to 70% of its maximum safe rate for your age. One way to determine this is to subtract your age from 220 and multiply by 0.7.” (A) If H is the maximum safe sustained heart rate (in beats per minute) for a person of age A (in years), write a formula relating H and A. (B) What is the maximum safe sustained heart rate for a 20-year-old? (C) If the maximum safe sustained heart rate for a person is 126 beats per minute, how old is the person? 38. DATA ANALYSIS Winning times in the men’s Olympic 400-meter freestyle event in minutes for selected years are given in Table 1. A mathematical model for these data is y 0003 00040.021x 0005 5.57 where x is years since 1900. (A) Compare the model and the data graphically and numerically. (B) Estimate (to three decimal places) the winning time in 2024.
Table 1 Year
Time
1912
5.41
1932
4.81
1952
4.51
1972
4.00
1992
3.75
2
GROUP ACTIVITY Average Speed
If you score 40 on the first exam and 80 on the second, then your average score for the two exams is (40 0005 80) 2 0003 60. The number 60 is the arithmetic average of 40 and 80. On the other hand, if you drive 100 miles at a speed of 40 mph, and then drive an additional 100 miles at 80 mph, your average speed for the entire trip is not 60 mph. Average speed is defined to be the constant speed at which you could drive the same distance in the same length of time. So to calculate average speed, total distance (200 miles) must be divided by total time: The time t1 it takes to drive 100 miles at 40 mph is t1 0003 (100 miles) (40 mph) 0003 2.5 hours. Similarly, the time t2 it takes to drive 100 miles at 80 mph is t2 0003 (100 miles) (80 mph) 0003 1.25 hours. Therefore, your average speed is 200 200 200 miles 0003 0003 0003 53.3 mph t1 0005 t2 2.5 0005 1.25 3.75
(A) You bicycle 15 miles at 21 mph, then 20 miles at 18 mph, and finally 30 miles at 12 mph. Find the average speed. (B) You bicycle for 2 hours at 18 mph, then 2 more hours at 12 mph. Find the average speed. (C) You run a 10-mile race by running at a pace of 8 minutes per mile for 1 hour, and after that at a pace of 9 minutes per mile. Define average pace, find it (to the nearest second) for the 10-mile race, and discuss the connection between average pace (in minutes per mile) and average speed (in miles per hour).
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CHAPTER
Functions
3
C
OUTLINE
THE function concept is one of the most important ideas in mathe-
matics. To study math beyond the elementary level, you absolutely need to have a solid understanding of functions and their graphs. In this chapter, you’ll learn the fundamentals of what functions are all about, and how to apply them. As you work through this and subsequent chapters, this will pay off as you study specific types of functions in depth. Everything you learn in this chapter will increase your chance of success in this course, and in almost any other course you may take that involves mathematics.
3-1
Functions
3-2
Graphing Functions
3-3
Transformations of Functions
3-4
Quadratic Functions
3-5
Operations on Functions; Composition
3-6
Inverse Functions Chapter 3 Review Chapter 3 Group Activity: Mathematical Modeling: Choosing a Cell Phone Plan
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Functions Z Definition of Function Z Defining Functions by Equations Z Using Function Notation Z Application
The idea of correspondence plays a really important role in understanding the concept of functions, which is easily one of the most important ideas in this book. The good news is that you have already had years of experience with correspondences in everyday life. For example, For For For For For
every every every every every
person, there is a corresponding age. item in a store, there is a corresponding price. football season, there is a corresponding Super Bowl champion. circle, there is a corresponding area. number, there is a corresponding cube.
One of the most basic and important ways that math can be applied to other areas of study is the establishment of correspondence among various types of phenomena. In many cases, once a correspondence is known, it can be used to make important decisions and predictions. An engineer can use a formula to predict the weight capacity of a stadium grandstand. A political operative decides how many resources to allocate to a race given current polling results. A computer scientist can use formulas to compare the efficiency of algorithms for sorting data stored on a computer. An economist would like to be able to predict interest rates, given the rate of change of the money supply. And the list goes on and on.
Z Definition of a Function What do all of the preceding examples have in common? Each describes the matching of elements from one set with elements from a second set. Consider the correspondences in Tables 1 and 2. Table 1 Top Four Weekly Average Primetime Network Viewers for the 2007–2008 Season
Table 2 Top Four Best Selling Automobiles in the United States for 2008
Network
Manufacturer
Model
Viewers (Millions)
Fox
10.9
Toyota
Camry
CBS
10.1
Honda
Accord
ABC
8.9
Toyota
Corolla
NBC
7.8
Honda
Civic
Source: tvbythenumbers.com
Source: www.2-speed.com
Table 1 specifies a function, but Table 2 does not. Why not? The definition of function will explain.
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Z DEFINITION 1 Definition of Function A function is a correspondence between two sets of elements such that to each element in the first set there corresponds one and only one element in the second set. The first set is called the domain and the set of all corresponding elements in the second set is called the range.
Table 1 specifies a function with domain {Fox, CBS, ABC, NBC} and range {10.9, 10.1, 8.9, 7.8} because every network in the first set corresponds with exactly one number in the second set. Table 2 does not specify a function, because each manufacturer in the first set corresponds to two different models in the second set. Functions can also be specified by using ordered pairs of elements, where the first component represents an element from the domain, and the second component represents the corresponding element from the range. The function in Table 1 can be written as F 0002 {(Fox, 10.9), (CBS, 10.1), (ABC, 8.9), (NBC, 7.8)} Notice that no two ordered pairs have the same first component and different second component. On the other hand, if we list the set H of ordered pairs determined by Table 2, we get H 0002 {(Toyota, Camry), (Honda, Accord), (Toyota, Corolla), (Honda, Civic)} In this case, there are ordered pairs with the same first component but different second components. This means that H does not specify a function. This ordered pair approach leads to a second (but equivalent) way to define a function.
Z DEFINITION 2 Set Form of the Definition of Function A function is a set of ordered pairs with the property that no two ordered pairs have the same first component and different second components. The set of all first components in a function is called the domain of the function, and the set of all second components is called the range.
EXAMPLE
1
Functions Specified as Sets of Ordered Pairs Determine whether each set specifies a function. If it does, then state the domain and range. (A) S 0002 5(1, 4), (2, 3), (3, 2), (4, 3), (5, 4)6 (B) T 0002 5(1, 4), (2, 3), (3, 2), (2, 4), (1, 5)6
SOLUTIONS
(A) Because all the ordered pairs in S have distinct first components, this set specifies a function. The domain and range are Domain 0002 51, 2, 3, 4, 56 Range 0002 52, 3, 46
Set of first components Set of second components written with no repeats
(B) Because there are ordered pairs in T with the same first component [for example, (1, 4) and (1, 5)], this set does not specify a function.
0002
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Determine whether each set defines a function. If it does, then state the domain and range. (A) S 0002 5(00042, 1), (00041, 2), (0, 0), (00041, 1), (00042, 2)6 (B) T 0002 5(00042, 1), (00041, 2), (0, 0), (1, 2), (2, 1)6
0002
Z Defining Functions by Equations So far, we have described a particular function in various ways: (1) by a verbal description, (2) by a table, and (3) by a set of ordered pairs. We will see that if the domain and range are sets of numbers, we can also define a function by an equation, or by a graph. If the domain of a function is a large or infinite set, it may be impractical or impossible to actually list all of the ordered pairs that belong to the function, or to display the function in a table. Such a function can often be defined by a verbal description of the “rule of correspondence” that clearly specifies the element of the range that corresponds to each element of the domain. One example is “to each real number corresponds its square.” When the domain and range are sets of numbers, the algebraic and graphical analogs of the verbal description are the equation and graph, respectively. We will find it valuable to be able to view a particular function from multiple perspectives—algebraic (in terms of an equation), graphical (in terms of a graph), and numeric (in terms of a table or ordered pairs). Both versions of our definition of function are very general. The objects in the domain and range can be pretty much anything, and there is no restriction on the number of elements in each. In this text, we are primarily interested, however, in functions with real number domains and ranges. Unless otherwise indicated, the domain and range of a function will be sets of real numbers. For such a function we often use an equation with two variables to specify both the rule of correspondence and the set of ordered pairs. Consider the equation y 0002 x2 0003 2x
x any real number
(1)
This equation assigns to each domain value x exactly one range value y. For example, If x 0002 4, If x 0002 000413,
then then
y 0002 (4)2 0003 2(4) 0002 24 y 0002 (000413)2 0003 2(000413) 0002 000459
We can view equation (1) as a function with rule of correspondence y 0002 x2 0003 2x
any x corresponds to x 2 ⴙ 2x
The variable x is called an independent variable, indicating that values can be assigned “independently” to x from the domain. The variable y is called a dependent variable, indicating that the value of y “depends” on the value assigned to x and on the given equation. In general, any variable used as a placeholder for domain values is called an independent variable; any variable used as a placeholder for range values is called a dependent variable. We often refer to a value of the independent variable as the input of the function, and the corresponding value of the dependent variable as the associated output. In this regard, a function can be thought of as a process that accepts an input from the domain and outputs an appropriate range element. We next address the question of which equations can be used to define functions.
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165
Z FUNCTIONS DEFINED BY EQUATIONS In an equation with two variables, if to each value of the independent variable there corresponds exactly one value of the dependent variable, then the equation defines a function. If there is any value of the independent variable to which there corresponds more than one value of the dependent variable, then the equation does not define a function.
Since an equation is just one way to represent a function, we will say “an equation defines a function” rather than “an equation is a function.”
EXAMPLE
2
Determining if an Equation Defines a Function Determine if each equation defines a function with independent variable x. (A) y ⫽ x2 ⫺ 4
SOLUTIONS
(B) x2 ⫹ y2 ⫽ 16
(A) For any real number x, the square of x is a unique real number. When you subtract 4, the result is again unique. So for any input x, there is exactly one output y, and the equation defines a function. (B) In this case, it will be helpful to solve the equation for the dependent variable. x2 ⫹ y2 ⫽ 16 y2 ⫽ 16 ⫺ x2 y ⫽ ⫾ 216 ⫺ x2
Subtract x2 from both sides. Take the square root of both sides.
For any x that provides an output (when 16 ⫺ x2 ⱖ 0), there are two choices for y, one positive and one negative. The equation has more than one output for some inputs, so does not define a function. 0002 MATCHED PROBLEM 2
Determine if each equation defines a function with independent variable x. (A) y2 ⫹ x4 ⫽ 4
(B) y3 ⫺ x3 ⫽ 3
0002
It is very easy to determine whether an equation defines a function if you have the graph of the equation. The two equations we considered in Example 2 are graphed next in Figure 1. y
Z Figure 1 Graphs of equations and the vertical line test.
y
5
5
y ⫽ x2 ⫺ 4
(2, 2兹3) x 2 ⫹ y 2 ⫽ 16
⫺5
5
x
⫺5
5
(1, ⫺3) ⫺5
(2, ⫺2兹3) ⫺5
(a)
x
(b)
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In Figure 1(a), any vertical line will intersect the graph of y 0002 x2 0004 4 exactly once. This shows that every value of the independent variable x corresponds to exactly one value of the dependent variable y, and confirms our conclusion that y 0002 x2 0004 4 defines a function. But in Figure 1(b), there are many vertical lines that intersect the graph of x2 0003 y2 0002 16 in two points. This shows that there are values of the independent variable x that correspond to two different values of the dependent variable y, which confirms our conclusion that x2 0003 y2 0002 16 does not define a function. These observations lead to Theorem 1.
Z THEOREM 1 Vertical Line Test for a Function An equation defines a function if each vertical line in a rectangular coordinate system passes through at most one point on the graph of the equation. If any vertical line passes through two or more points on the graph of an equation, then the equation does not define a function.
ZZZ EXPLORE-DISCUSS 1
The definition of a function specifies that to each element in the domain there corresponds one and only one element in the range. (A) Give an example of a function such that to each element of the range there correspond exactly two elements of the domain. (B) Give an example of a function such that to each element of the range there corresponds exactly one element of the domain.
Sometimes when a function is defined by an equation, a domain is specified, as in f (x) 0002 2x2 0003 5, x 7 0 The “x 7 0” tells us that the domain is all positive real numbers. More often, a function is defined by an equation with no domain specified. Unless a domain is specified, we will use the following convention regarding domains and ranges for functions defined by equations.
Z AGREEMENT ON DOMAINS AND RANGES If a function is defined by an equation and the domain is not stated explicitly, then we assume that the implied domain is the set of all real number replacements of the independent variable that produce real values for the dependent variable. The range is the set of all values of the dependent variable corresponding to the domain values.
EXAMPLE
3
Finding the Domain of a Function Find the domain of the function defined by the equation y 0002 1x 0004 3, assuming x is the independent variable.
SOLUTION
For y to be real, x 0004 3 must be greater than or equal to 0. That is, x0004300050
The domain is 5x ƒ x 0005 36, or [3, 0007).
or
x00053 0002
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MATCHED PROBLEM 3
Functions
167
Find the domain of the function defined by the equation y 0002 1x 0003 5, assuming x is the independent variable. 0002
Z Using Function Notation We will use letters to name functions and to provide a very important and convenient notation for defining functions. For example, if f is the name of the function defined by the equation y 0002 2x 0003 1, we could use the formal representations f : y 0002 2x 0003 1
Rule of correspondence
or f :5(x, y) | y 0002 2x 0003 16
Set of ordered pairs
But instead, we will simply write f (x) 0002 2x 0003 1
Function notation
The symbol f (x) is read “f of x,” “f at x,” or “the value of f at x” and represents the number in the range of the function f (the output) that is paired with the domain value x (the input).
ZZZ
CAUTION ZZZ
The symbol “f (x)” should never be read as “f times x.” The notation does not represent a product. It tells us that the function named f has independent variable x. f (x) is the value of the function f at x. 2(x) 0002 2x is algebraic multiplication.
Using function notation, f (3) is the output for the function f associated with the input 3. We find this range value by replacing x with 3 wherever x occurs in the function definition f(x) 0002 2x 0003 1
f x
f (x)
and evaluating the right side, f (3) 0002 2 ⴢ 3 0003 1 0002 6 0003 1 0002 7
DOMAIN
RANGE
The function f “maps” the domain value x into the range value f (x).
Z Figure 2 Function notation.
The statement f(3) 0002 7 indicates in a concise way that the function f assigns the range value 7 to the domain value 3 or, equivalently, that the ordered pair (3, 7) belongs to f. The symbol f : x S f(x), read “f maps x into f (x),” is also used to denote the relationship between the domain value x and the range value f (x) (Fig. 2). Letters other than f and x can be used to represent functions and independent variables. For example, g(t) 0002 t 2 0004 3t 0003 7 defines g as a function of the independent variable t. To find g(00042), we replace t by 00042 wherever t occurs in the equation g(t) 0002 t 2 0004 3t 0003 7 and evaluate the right side: g(ⴚ2) 0002 (ⴚ2)2 0004 3(ⴚ2) 0003 7 000240003600037 0002 17 The function g assigns the range value 17 (output) to the domain value 00042 (input); the ordered pair (00042, 17) belongs to g.
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It is important to understand and remember the definition of the symbol f(x): Z DEFINITION 3 The Symbol f(x) The symbol f(x), read “f of x,” represents the real number in the range of the function f corresponding to the domain value x. The symbol f (x) is also called the value of the function f at x. The ordered pair (x, f (x)) belongs to the function f. If x is a real number that is not in the domain of f, then f is undefined at x and f (x) does not exist.
EXAMPLE
4
Evaluating Functions (A) Find f(6), f(a), and f(6 0003 a) for f (x) 0002
15 . x00043
(B) Find g(7), g(h), and g(7 0003 h) for g(x) 0002 16 0003 3x 0004 x2. (C) Find k(9), 4k(a), and k(4a) for k(x) 0002
SOLUTIONS
(A)
f (6)
0002
15 600043
* 0002
15 00025 3
Substitute 6 for x.
15 a00043 15 15 f (6 ⴙ a) 0002 0002 (6 ⴙ a) 0004 3 30003a f (a) 0002
(B)
0002 16 0003 3(7) 0004 (7)2
g(7)
2 . 1x 0004 2
Substitute a for x.
Substitute (6 ⫹ a) for x and simplify.
0002 16 0003 21 0004 49 0002 000412
g(h) 0002 16 0003 3h 0004 h2 g(7 ⴙ h) 0002 16 0003 3(7 ⴙ h) 0004 (7 ⴙ h)2 0002 16 0003 21 0003 3h 0004 (49 0003 14h 0003 h2) 0002 37 0003 3h 0004 49 0004 14h 0004 h2 0002 000412 0004 11h 0004 h2 (C) k (9)
0002
2 19 0004 2
0002
2 00022 300042
2 8 0002 1a 0004 2 1a 0004 2 2 k(4a) 0002 14a 0004 2 2 0002 21a 0004 2 1 0002 1a 0004 1
Multiply out the first set of parentheses and square (7 ⫹ h). Combine like terms and distribute the negative through the parentheses. Combine like terms.
19 ⴝ 3, not ⴞ3.
4k(a) 0002 4
14a ⴝ 141a ⴝ 2 1a.
Divide numerator and denominator by 2.
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
0002
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MATCHED PROBLEM 4
Functions
169
4 . 20004x (B) Find G(3), G(h), and G(3 0003 h) for G(x) 0002 x2 0003 5x 0004 2. 6 (C) Find K(4), K(9x), and 9K(x) for K(x) 0002 . 3 0004 1x (A) Find F(4), F(4 0003 h), and F(4) 0003 F(h) for F(x) 0002
0002
EXAMPLE
5
Finding Domains of Functions Find the domain of each of the following functions. Express the answer in both set notation and inequality notation.* (A) f (x) 0002
SOLUTIONS
15 x00043
(B) g(x) 0002 16 0003 3x 0004 x2
(C) k(x) 0002
2 1x 0004 2
(A) The rational expression 15兾(x 0004 3) represents a real number for all replacements of x by real numbers except x 0002 3, since division by 0 is not defined. So f(3) does not exist, and the domain of f is 5x ƒ x 36
or
R
or
(00040007, 3) 傼 (3, 0007)
(B) Since 16 0003 3x 0004 x2 represents a real number for all replacements of x by real numbers, the domain of g is (00040007, 0007)
(C) Since 1x is not a real number for negative real numbers x, x must be a nonnegative real number. Because division by 0 is not defined, we must exclude any values of x that make the denominator 0. Set the denominator equal to zero and solve: 2 0004 1x 0002 0 2 0002 1x 40002x
Add 1x to both sides. Square both sides.
The domain of f is all nonnegative real numbers except 4. This can be written as 5x ƒ x 0005 0, x 46
MATCHED PROBLEM 5
[0, 4) 傼 (4, 0007)
0002
Find the domain of each of the following functions. Express the answer in both set notation and inequality notation. (A) F(x) 0002
ZZZ EXPLORE-DISCUSS 2
or
4 20004x
(B) G(x) 0002 x2 0003 5x 0004 2
(C) K(x) 0002
6 3 0004 1x
Let x and h be real numbers. (A) If f(x) 0002 4x 0003 3, which of the following is true: (1) f (x 0003 h) 0002 4x 0003 3 0003 h (2) f(x 0003 h) 0002 4x 0003 4h 0003 3 (3) f (x 0003 h) 0002 4x 0003 4h 0003 6 (B) If g(x) 0002 x2, which of the following is true: (1) g(x 0003 h) 0002 x2 0003 h (2) g(x 0003 h) 0002 x2 0003 h2 (3) g(x 0003 h) 0002 x2 0003 2hx 0003 h2 (C) If M(x) 0002 x2 0003 4x 0003 3, describe the operations that must be performed to evaluate M(x 0003 h). *A review of Table 1 in Section 1-2 might prove to be helpful at this point.
0002
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In addition to evaluating functions at specific numbers, it is useful to be able to evaluate functions at expressions that involve one or more variables. For example, the difference quotient f (x 0003 h) 0004 f (x) h
x and x 0003 h in the domain of f, h 0
is very important in calculus courses.
EXAMPLE
6
Evaluating and Simplifying a Difference Quotient For f(x) 0002 x2 0003 4x 0003 5, find and simplify: (A) f(x 0003 h)
SOLUTIONS
(B) f(x 0003 h) 0004 f(x)
(C)
f (x 0003 h) 0004 f (x) ,h0 h
(A) To find f(x 0003 h), we replace x with x 0003 h everywhere it appears in the equation that defines f and simplify: f (x ⴙ h) 0002 (x ⴙ h)2 0003 4(x ⴙ h) 0003 5 0002 x2 0003 2xh 0003 h2 0003 4x 0003 4h 0003 5 (B) Using the result of part A, we get f (x ⴙ h) 0004 f (x) 0002 x2 ⴙ 2xh ⴙ h2 ⴙ 4x ⴙ 4h ⴙ 5 0004 (x2 ⴙ 4x ⴙ 5) 0002 x2 0003 2xh 0003 h2 0003 4x 0003 4h 0003 5 0004 x2 0004 4x 0004 5 0002 2xh 0003 h2 0003 4h (C)
f (x 0003 h) 0004 f (x) 2xh 0003 h2 0003 4h 0002 h h
0002
h(2x 0003 h 0003 4) h
Divide numerator and denominator by h ⴝ 0.
0002 2x 0003 h 0003 4 MATCHED PROBLEM 6
ZZZ
Repeat Example 6 for f (x) 0002 x2 0003 3x 0003 7.
1. Remember, f(x 0003 h) is not a multiplication! 2. In general, f(x 0003 h) is not equal to f(x) 0003 f(h), nor is it equal to f(x) 0003 h.
CAUTION ZZZ
Z Application EXAMPLE
7
Construction A rectangular feeding pen for cattle is to be made with 100 meters of fencing. (A) If x represents the width of the pen, express its area A in terms of x. (B) What is the domain of the function A (determined by the physical restrictions)?
0002
0002
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171
Functions
(A) Draw a figure and label the sides.
x ( Width)
Perimeter ⴝ 100 meters of fencing. Half the perimeter ⴝ 50. If x ⴝ Width, then 50 ⴚ x ⴝ Length.
50 0004 x (Length)
A 0002 (Width)(Length) 0002 x(50 0004 x) (B) To have a pen, x must be positive, but x must also be less than 50 (or the length will not exist). So the domain is 5x ƒ 0 6 x 6 506
(0, 50) MATCHED PROBLEM 7
Inequality notation
0002
Interval notation
Rework Example 7 with the added assumption that a large barn is to be used as one of the sides that run the length of the pen. 0002 ANSWERS TO MATCHED PROBLEMS 1. (A) S does not define a function. (B) T defines a function with domain {00042, 00041, 0, 1, 2} and range {0, 1, 2}. 2. (A) Does not define a function (B) Defines a function 3. 5x ƒ x 0005 000456 or [ 00045, 0007) 4 2h 4. (A) F(4) 0002 00042, F(4 0003 h) 0002 0004 , F(4) 0003 F(h) 0002 20003h 20004h (B) G(3) 0002 22, G(h) 0002 h2 0003 5h 0004 2, G(3 0003 h) 0002 22 0003 11h 0003 h2 2 54 (C) K(4) 0002 6, K(9x) 0002 , 9K(x) 0002 1 0004 1x 3 0004 1x 5. (A) 5x ƒ x 26 or (00040007, 2) ´ (2, 0007) (B) R or (00040007, 0007) (C) 5x ƒ x 0005 0, x 96 or [0, 9) 傼 (9, 0007) 6. (A) x2 0003 2xh 0003 h2 0003 3x 0003 3h 0003 7 2 (B) 2xh 0003 h 0003 3h (C) 2x 0003 h 0003 3 7. (A) A 0002 x(100 0004 2x) (B) Domain: 5x ƒ 0 6 x 6 506 or (0, 50)
3-1
Exercises
1. Is every correspondence between two sets a function? Why or why not? 2. Describe four different ways that we represented functions in this section. 3. Explain what the domain and range of a function are. Don’t just think about functions defined by equations.
6. Describe how to determine if an equation defines a function by looking at the graph of the equation. Indicate whether each table in Problems 7–12 defines a function. 7. Domain
Range
8. Domain
Range
4. What do the terms “input” and “output” refer to when working with functions?
00041
1
2
1
0
2
4
3
5. If 2(x 0003 h) 0002 2x 0003 2h, why doesn’t f (x 0003 h) 0002 f (x) 0003 f (h), where f is a function?
1
3
6
5
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9. Domain
Range
1
3
00041
0
3
5
00042
5
7
00043
8
5
10. Domain
Range
y
21. 10
000410
10
x
9
11. Domain
Range
12. Domain
000410
Range
English
A
Auburn
Tigers
Math
B
Memphis
Tigers
Sociology
A
Georgia
Bulldogs
Chemistry
B
Fresno State
Bulldogs
y
22. 10
000410
Indicate whether each set in Problems 13–18 defines a function. Find the domain and range of each function. 13. {(2, 4), (3, 6), (4, 8), (5, 10)}
10
x
000410
y
23.
14. {(00041, 4), (0, 3), (1, 2), (2, 1)}
10
15. {(10, 000410), (5, 00045), (0, 0), (5, 5), (10, 10)} 16. {(0, 1), (1, 1), (2, 1), (3, 2), (4, 2), (5, 2)} 000410
17. {(Ohio, Obama), (Alabama, McCain), (West Virginia, McCain), (California, Obama)} 18. {(Democrat, Obama), (Republican, Bush), (Democrat, Clinton), (Republican, Reagan)}
10
000410
y
24. Indicate whether each graph in Problems 19–24 is the graph of a function. 19.
x
10
y 10
000410
000410
10
x
y 10
000410
10
000410
x
000410
In Problems 25 and 26, which of the indicated correspondences define functions? Explain.
000410
20.
10
x
25. Let F be the set of all faculty teaching Math 125 at Enormous State University, and let S be the set of all students taking that course. (A) Students from set S correspond to their Math 125 instructors. (B) Faculty from set F correspond to the students in their Math 125 class. 26. Let A be the set of floor advisors in Hoffmann Hall, a dorm at Enormous State. Assume that each floor has one floor advisor. Let R be the set of residents of that dorm. (A) Floor advisors from set A correspond to the residents on their floor. (B) Students from set R correspond to their floor advisor.
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27. Let f (x) 0002 3x 0004 5. Find (A) f(3) (B) f (h) (C) f(3) 0003 f(h) (D) f (3 0003 h)
173
In Problems 47–62, find the domain of the indicated function. Express answers in both interval notation and inequality notation.
28. Let g(y) 0002 7 0004 2y. Find (A) g(4) (B) g(h) (C) g(4) 0003 g(h) (D) g(4 0003 h) 29. Let F(w) 0002 0004w2 0003 2w. Find (A) F(4) (B) F(00044) (C) F(4 0003 a) (D) F(2 0004 a)
47. f(x) 0002 4 0004 9x 0003 3x2
48. g(t) 0002 1 0003 7t 0004 2t2
49. L(u) 0002 23u2 0003 4
50. M(w) 0002
2 40004z
51. h(z) 0002
30. Let G(t) 0002 5t 0004 t2. Find (A) G(8) (B) G(00048) (C) G(00041 0003 h) (D) G(6 0004 t) 2
31. Let f(t) 0002 2 0004 3t . Find (A) f(00042) (B) f(0004t) (C) 0004f(t) (D) 0004f(0004t) 32. Let k(z) 0002 40 0003 20z2. Find (A) k(00042) (B) k(0004z) (C) 0004k(z) (D) 0004k(0004z)
52. k(z) 0002
w00045 23 0003 2w2
z z00043
53. g(t) 0002 1t 0004 4
54. h(t) 0002 16 0004 t
55. k(w) 0002 17 0003 3w
56. j(w) 0002 19 0003 4w
57. H(u) 0002
u u2 0003 4
58. G(u) 0002
u u2 0004 4
59. M(x) 0002
1x 0003 4 x00041
60. N(x) 0002
1x 0004 3 x00032
61. s(t) 0002
33. Let F(u) 0002 u2 0004 u 0004 1. Find (A) F(10) (B) F(u2) (C) F(5u) (D) 5F(u)
1 3 0004 1t
62. r(t) 0002
1 1t 0004 4
The verbal statement “function f multiplies the square of the domain element by 3 and then subtracts 7 from the result” and the algebraic statement “f(x) 0002 3x2 0004 7” define the same function. In Problems 63–66, translate each verbal definition of a function into an algebraic definition.
34. Let G(u) 0002 4 0004 3u 0004 u2. Find (A) G(00048) (B) G(u2) (C) G(00042u) (D) 00042G(u) Problems 35–36 refer to the following graph of a function f.
63. Function g subtracts 5 from twice the cube of the domain element. 64. Function f multiplies the square of the domain element by 10 then adds 1,000 to the result.
f (x) y 0002 f (x)
65. Function F multiplies the square root of the domain element by 8, then subtracts the product of 4 and the sum of the domain element and two.
10
000410
Functions
10
x
000410
66. Function G divides the sum of the domain element and 7 by the cube root of the domain element. In Problems 67–70, translate each algebraic definition of the function into a verbal definition.
35. (A) Find f (00042) to the nearest integer. (B) Find all values of x, to the nearest integer, so that f (x) 0002 00044.
67. f(x) 0002 2x2 0003 5
36. (A) Find f(4) to the nearest integer. (B) Find all values of x, to the nearest integer, so that f (x) 0002 0.
69. z(x) 0002
Determine which of the equations in Problems 37–46 define a function with independent variable x. For those that do, find the domain. For those that do not, find a value of x to which there corresponds more than one value of y. 37. y 0004 x2 0002 1
38. y2 0004 x 0002 1
39. 2x3 0003 y2 0002 4
40. 3x2 0003 y3 0002 8
41. x3 0004 y 0002 2
42. x3 0003 冟 y 冟 0002 6
43. 2x 0003 冟 y 冟 0002 7
44. y 0004 2冟 x 冟 0002 3
45. 3y 0003 2|x| 0002 12
46. x| y| 0002 x 0003 1
68. g(x) 0002 00042x 0003 7
4x 0003 5 1x
70. M(t) 0002 5t 0004 21t
71. If F(s) 0002 3s 0003 15, find:
F(2 0003 h) 0004 F(2) h
72. If K(r) 0002 7 0004 4r, find:
K(1 0003 h) 0004 K(1) h
73. If g(x) 0002 2 0004 x2, find:
g(3 0003 h) 0004 g(3) h
74. If P(m) 0002 2m2 0003 3, find:
P(2 0003 h) 0004 P(2) h
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In Problems 75–84, find and simplify: (A)
f (x 0004 h) 0003 f (x) h
(B)
75. f (x) 0002 4x 0003 7
f (x) 0003 f (a) x0003a 76. f (x) 0002 00035x 0004 2
2
77. f(x) 0002 2x 0003 4
78. f(x) 0002 5 0003 3x2
79. f (x) 0002 00034x2 0004 3x 0003 2
80. f (x) 0002 3x2 0003 5x 0003 9
81. f (x) 0002 1x 0004 2
82. f (x) 0002 1x 0003 1
83. f (x) 0002
4 x
84. f (x) 0002
s(2 0004 h) 0003 s(2) . h (C) Evaluate the expression in part (B) for h 0002 00051, 00050.1, 00050.01, 00050.001. (D) What happens in part (C) as h gets closer and closer to 0? Interpret physically. (B) Find and simplify
3 x00042
85. The area of a rectangle is 64 square inches. Express the perimeter P as a function of the width w and state the domain. 86. The perimeter of a rectangle is 50 inches. Express the area A as a function of the width w and state the domain. 87. The altitude of a right triangle is 5 meters. Express the hypotenuse h as a function of the base b and state the domain. 88. The altitude of a right triangle is 4 meters. Express the base b as a function of the hypotenuse h and state the domain.
94. PHYSICS—RATE An automobile starts from rest and travels along a straight and level road. The distance in feet traveled by the automobile is given by s(t) 0002 10t2, where t is time in seconds. (A) Find: s(8), s(9), s(10), and s(11). s(11 0004 h) 0003 s(11) (B) Find and simplify . h (C) Evaluate the expression in part (B) for h 0002 00051, 00050.1, 00050.01, 00050.001. (D) What happens in part (C) as h gets closer and closer to 0? Interpret physically. 95. MANUFACTURING A candy box is to be made out of a piece of cardboard that measures 8 by 12 inches. Squares, x inches on a side, will be cut from each corner, and then the ends and sides will be folded down (see the figure). Find a formula for the volume of the box V in terms of x. What is the domain of the function V that makes sense in this problem?
APPLICATIONS Most of the applications in this section are calculus-related. That is, similar problems will appear in a calculus course, but additional analysis of the functions will be performed. 89. COST FUNCTION The fixed costs per day for a doughnut shop are $300, and the variable costs are $1.75 per dozen doughnuts produced. If x dozen doughnuts are produced daily, express the daily cost C(x) as a function of x. 90. COST FUNCTION A manufacturer of MP3 players has fixed daily costs of 15,700 Chinese yuan, and it costs 178 yuan to produce one MP3 player. If the manufacturer produces x players daily, express the daily cost C in yuan as a function of x. 91. CELL PHONE COST Since Don usually borrows his roommate’s cell phone for long-distance calls, he chooses an inexpensive plan for his own phone with a monthly access charge, and a variable charge for each hour of calls used. The function C(h) 0002 17 0004 2.40h is used to calculate Don’s monthly bill, where C is the cost in dollars and h is hours of airtime used. Translate this equation into a verbal statement that you could use to explain Don’s monthly charge. 92. COST OF HIGH SPEED INTERNET A college offers highspeed Internet in dorm rooms. The monthly access fee in dollars is calculated using the function
x CITR
US D
x
ELIG
HTS
x
x
CITRUS DELIGHTS
CITRUS DELIGHTS
bar19499_ch03_161-258.qxd
x
x
x x
96. CONSTRUCTION A rancher has 20 miles of fencing to fence a rectangular piece of grazing land along a straight river. If no fence is required along the river and the sides perpendicular to the river are x miles long, find a formula for the area A of the rectangle in terms of x. What is the domain of the function A that makes sense in this problem? 97. CONSTRUCTION The manager of an animal clinic wants to construct a kennel with four identical pens, as indicated in the figure. State law requires that each pen have a gate 3 feet wide and an area of 50 square feet. If x is the width of one pen, express the total amount of fencing F (excluding the gates) required for the construction of the kennel as a function of x. Complete the following table (round values of F to one decimal place): 4
x
5
6
7
F x
A(m) 0002 15 0004 0.02m where m is the number of minutes spent online. Translate this equation into a verbal statement that can be used to explain the monthly charges to an incoming freshman. 93. PHYSICS—RATE The distance in feet that an object falls (ignoring air resistance) is given by s(t) 0002 16t2, where t is time in seconds. (A) Find: s(0), s(1), s(2), and s(3).
3 feet
98. ARCHITECTURE An architect wants to design a window with an area of 24 square feet in the shape of a rectangle with a
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semicircle on top, as indicated in the figure. If x is the width of the window, express the perimeter P of the window as a function of x. Complete the following table (round each value of P to one decimal place): 5
6
Island Lake 8 miles
4
x
175
Graphing Functions
7
Pipe
Freshwater source
Land
P
x
20 0004 x 20 miles
Figure for 99
100. WEATHER An observation balloon is released at a point 10 miles from the station that receives its signal and rises vertically, as indicated in the figure. Express the distance d between the balloon and the receiving station as a function of the altitude h of the balloon.
x
99. CONSTRUCTION A freshwater pipeline is to be run from a source on the edge of a lake to a small resort community on an island 8 miles offshore, as indicated in the figure. It costs $10,000 per mile to lay the pipe on land and $15,000 per mile to lay the pipe in the lake. Express the total cost C of constructing the pipeline as a function of x. From practical considerations, what is the domain of the function C ?
d
h
10 miles Figure for 100
3-2
Graphing Functions Z Basic Concepts Z Linear Functions Z Piecewise-Defined Functions
One of the ways we represented functions in Section 3-1 was with sets of ordered pairs. If these ordered pairs reminded you of points on a graph, you already understand the most important idea in this section—that graphs are a natural fit for functions because a graph matches up a pair of numbers in exactly the same way a function matches up a pair of objects. y or f (x) y intercept
(x, y) or (x, f (x))
Z Basic Concepts
f
y or f (x) x x intercept
Z Figure 1 Graph of a function.
When we graph a function whose domain and range are both sets of numbers, we are drawing a visual representation of the pairs of numbers matched up by that function. We will associate domain values with the horizontal axis, and range values with the vertical axis. The graph of a function f (x) is the set of all points whose first coordinate is an element of the domain of f, and whose second coordinate is the associated element of the range. We can use the symbol y or f (x) to represent the dependent variable. See Figure 1. Since it is
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typical to use the variables x and y for the independent and dependent variables, respectively, we usually refer to the first coordinate of a point as the x coordinate, and the second coordinate as the y coordinate. The x coordinate of a point where the graph of a function intersects the x axis is called an x intercept or zero of the function. An x intercept is also a real solution or root of the equation f (x) 0002 0. The y coordinate of a point where the graph of a function crosses the y axis is called the y intercept of the function. The y intercept is given by f (0), provided 0 is in the domain of f. Note that a function can have more than one x intercept but can never have more than one y intercept—a consequence of the vertical line test from Section 3-1.
EXAMPLE
1
Finding the Domain and Intercepts of a Function Find the domain, x intercept, and y intercept of f (x) 0002
SOLUTION
4 0004 3x . 2x 0003 5
The rational expression (4 0004 3x)兾(2x 0003 5) is defined for every x except those that make the denominator zero: 2x 0003 5 0002 0 2x 0002 00045 x 0002 000452
Subtract 5 from both sides. Divide both sides by 2.
The domain of f is all x values except 000452, or (00040007, 000452 ) 傼 (000452, 0007). The value of a fraction is 0 if and only if the numerator is zero: 4 0004 3x 0002 0 00043x 0002 00044 x 0002 43
Subtract 4 from both sides. Divide both sides by ⴚ3.
The x intercept of f is 43. The y intercept is f (0) 0002
MATCHED PROBLEM 1
4 0004 3(0) 2(0) 0003 5
4 0002 . 5
Find the domain, x intercept, and y intercept of f (x) 0002
0002 4x 0003 5 . 3x 0004 2 0002
The domain of a function is the set of all the x coordinates of points on the graph of the function and the range is the set of all the y coordinates. It is very useful to view the domain and range as subsets of the coordinate axes as in Figure 2 on the next page. Note the effective use of interval notation in describing the domain and range of the functions in this figure. In Figure 2(a) a solid dot is used to indicate that a point is on the graph of the function and in Figure 2(b) an open dot is used to indicate that a point is not on the graph of the function. An open or solid dot at the end of a graph indicates that the graph terminates there, whereas an arrowhead indicates that the graph continues indefinitely beyond the portion shown with no significant changes of direction [see Fig. 2(b) and note that the arrowhead indicates that the domain extends infinitely far to the right, and the range extends infinitely far downward].
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]
(
d
a
[
]
x
b
[
177
f (x)
f (x) d
Graphing Functions
(
x
a
c Domain f 0002 (a, 0007) Range f 0002 (00040007, d )
Domain f 0002 [a, b] Range f 0002 [c, d ] (a)
(b)
Z Figure 2 Domain and range.
EXAMPLE
2
Finding the Domain and Range from a Graph (A) Find the domain and range of the function f whose graph is shown in Figure 3. (B) Find f(1), f (3), and f(5). y or f (x) 4
1 00043
3
5
x
y 0002 f (x) 00044 00045
Z Figure 3 SOLUTIONS y or f (x)
3
Domain: 00043 6 x 6 0007
y 0002 f (x) 1
5
00044
00044
Z Figure 4
MATCHED PROBLEM 2
(A) The dot at the left end of the graph indicates that the graph terminates at that point, while the arrowhead on the right end indicates that the graph continues infinitely far to the right. So the x coordinates on the graph go from 00043 to 0007. The open dot at (00043, 4) indicates that 00043 is not in the domain of f.
x
or
(00043, 0007)
The least y coordinate on the graph is 00045, and there is no greatest y coordinate. (The arrowhead tells us that the graph continues infinitely far upward.) The closed dot at (3, 00045) indicates that 00045 is in the range of f. Range: 00045 y 6 0007
or
[00045, 0007)
(B) The point on the graph with x coordinate 1 is (1, 00044), so f(1) 0002 00044. Likewise, (3, 00045) and (5, 00044) are on the graph, so f (3) 0002 00045 and f (5) 0002 00044.
0002
(A) Find the domain and range of the function f given by the graph in Figure 4. (B) Find f(–4), f (0), and f(2). 0002
ZZZ
CAUTION ZZZ
When using interval notation to describe domain and range, make sure that you always write the least number first! You should find the domain by working left to right along the x axis, and find the range by working bottom to top along the y axis.
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Z Identifying Increasing and Decreasing Functions We will now take a look at increasing and decreasing properties of functions. Informally, a function is increasing over an interval if its graph rises as the x coordinate increases (moves from left to right) over that interval. A function is decreasing over an interval if its graph falls as the x coordinate increases over that interval. A function is constant on an interval if its graph is horizontal (i.e., the height doesn’t change) over that interval (Fig. 5). g(x)
f (x)
5
5
f (x) 0002 0004x 3
g(x) 0002 2x 0003 2 00045
5
x
00045
00045
5
x
00045
(a) Increasing on (ⴚⴥ, ⴥ)
(b) Decreasing on (ⴚⴥ, ⴥ)
h(x)
p (x)
5
5
h(x) 0002 2 00045
5
x
00045
(c) Constant on (ⴚⴥ, ⴥ)
p (x) 0002 x 2 0004 1 x
00045
5
00045
(d) Decreasing on (ⴚⴥ, 0 ] Increasing on [0, ⴥ)
Z Figure 5 Increasing, decreasing, and constant functions.
More formally, we define increasing, decreasing, and constant functions as follows:
Z DEFINITION 1 Increasing, Decreasing, and Constant Functions Let I be an interval in the domain of function f. Then, 1. f is increasing on I and the graph of f is rising on I if f(x1) 6 f(x2) whenever x1 6 x2 in I. 2. f is decreasing on I and the graph of f is falling on I if f(x1) 7 f(x2) whenever x1 6 x2 in I. 3. f is constant on I and the graph of f is horizontal on I if f(x1) 0002 f(x2) whenever x1 6 x2 in I.
Z Linear Functions In Section 2-3, we studied the slope–intercept form of the equation of a line: y 0002 mx 0003 b, where m is the slope, and b is the y intercept. We can carry over what we learned to the study of linear functions.
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Graphing Functions
Z DEFINITION 2 Linear Function A function of the form f (x) 0002 mx 0003 b is called a linear function. If m 0002 0, the result is f(x) 0002 b, which is called a constant function. If m 0002 1 and b 0002 0, then the result is f(x) 0002 x, which is called the identity function. The domain of any linear function is all real numbers. If m 0, then the range is also all real numbers. If m 0002 0, the function is constant and the range is {b}.
Z GRAPH PROPERTIES OF f(x) ⴝ mx ⴙ b The graph of a linear function is a line with slope m and y intercept b. f (x)
b
f (x)
f(x)
b
b
x
mⴝ0 Constant on (ⴚ0007, 0007) Domain: (ⴚ0007, 0007) Range: {b}
m⬍0 Decreasing on (ⴚ0007, 0007) Domain: (ⴚ0007, 0007) Range: (ⴚ0007, 0007)
ZZZ EXPLORE-DISCUSS 1
EXAMPLE
3
x
x
m⬎0 Increasing on (ⴚ0007, 0007) Domain: (ⴚ0007, 0007) Range: (ⴚ0007, 0007)
(A) Is it possible for a linear function to have two x intercepts? No x intercepts? If either of your answers is yes, give an example. (B) Is it possible for a linear function to have two y intercepts? No y intercept? If either of your answers is yes, give an example.
Graphing a Linear Function Find the slope and intercepts, and then sketch the graph of the linear function defined by f (x) 0002 000423 x 0003 4
SOLUTION f(x)
The y intercept is f(0) 0002 4, and the slope is 000423. To find the x intercept, we solve the equation f(x) 0002 0 for x: 000423 x
5
5
x
f (x) 0002 0 0003400020 000423 x 0002 00044 x0002
00044 0002 (000432)(00044) 0002 6 000423
The graph of f is shown in Figure 6. Z Figure 6
Substitute ⴚ23 x ⴙ 4 for f(x). Subtract 4 from both sides Divide both sides by ⴚ23 . x intercept
0002
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MATCHED PROBLEM 3
Find the slope and intercepts, and then sketch the graph of the linear function defined by f (x) 0002 32 x 0004 6 0002
Z Piecewise-Defined Functions The absolute value function can be defined using the definition of absolute value from Section 1-3: f (x) 0002 冟 x 冟 0002
再
0004x x
if x 6 0 if x 0005 0
Notice that this function is defined by different expressions for different parts of its domain. Functions whose definitions involve more than one expression are called piecewise-defined functions. Example 4 will show you how to work with a piecewise-defined function.
EXAMPLE
4
Analyzing a Piecewise-Defined Function The function f is defined by
再
4x 0003 11 f (x) 0002 3 000412 x 0003 72
if x 6 00042 if 00042 x 1 if x 7 1
(A) Find f (00043), f (00042), f (1), and f(3). (B) Graph f. (C) Find the domain, range, and intervals where f is increasing, decreasing, or constant. SOLUTIONS
(A) Since 00043 is an x value less than 00042, we use the formula 4x 0003 11 to calculate f(00043). f (ⴚ3) 0002 4(ⴚ3) 0003 11 0002 000412 0003 11 0002 00041 Since both 00042 and 1 are in the interval 00042 x 1, the output is 3 for both. f (ⴚ2) 0002 3
f(1) 0002 3
and
Since 3 is an x value greater than 1, we use the formula 000412 x 0003 72 to calculate f (3). f (3) 0002 000412 (3) 0003 72 0002 000432 0003 72 0002 42 0002 2 (B) To graph f, we graph each expression in the definition of f over the appropriate interval. That is, we graph y 0002 4x 0003 11 y00023 y 0002 000412 x 0003 72
for x 6 00042 for 00042 x 1 for x 7 1
y y00023
5
(1, 3)
(00042, 3)
1
y 0002 00042x 0003
00045
5
y 0002 4x 0003 11 00045
x
7 2
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181
We used a solid dot at the point (00042, 3) to indicate that y 0002 4x 0003 11 and y 0002 3 agree at x 0002 00042. The solid dot at the point (1, 3) indicates that y 0002 3 and y 0002 000412 x 0003 72 agree at x 0002 1. (C) The domain of a piecewise-defined function is the union of the intervals used in its definition: Domain of f: (00040007, 00042) ´ [00042, 1] ´ (1, 0007) 0002 (00040007, 0007) The graph of f shows that the range of f is (00040007, 3]. The function f is increasing on (00040007, 00042), constant on [00042, 1], and decreasing on (1, 0007). 0002 MATCHED PROBLEM 4
The function f is defined by
再
000413 x 0004 73 f (x) 0002 00042 5x 0004 17
if x 00041 if 00041 6 x 6 3 if x 0005 3
(A) Find f(00044), f(00041), f(3), and f(4). (B) Graph f. (C) Find the domain, range, and intervals where f is increasing, decreasing, or constant. 0002 Notice that the graph of f in Example 4 contains no breaks. Informally, a graph (or portion of a graph) is said to be continuous if it contains no breaks or gaps. (A formal presentation of continuity can be found in calculus texts.) Piecewise-defined functions occur naturally in many applications, especially ones involving money. A very useful example is income tax.
EXAMPLE
5
Income Tax Table 1 contains a recent tax rate chart for a single filer in the state of Oregon. If T(x) is the tax on an income of $x, write a piecewise definition for T. Find the tax on each of the following incomes: $2,000, $5,000, and $9,000. Table 1 2009 Tax Rate Chart for Persons Filing Single, or Married Filing Separately If the taxable income is:
The tax is:
Not over $3,050
5% of taxable income
Over $3,050 but not over $7,600
$153 plus 7% of the excess over $3,050
Over $7,600
$471 plus 9% of the excess over $7,600
Source: Oregon Department of Revenue
SOLUTION
Since taxes are computed differently on [0, 3,050], (3,050, 7,600] and (7,600, 0007), we must find an expression for the tax on incomes in each of these intervals. [0, 3,050]: Tax is 0.05x. (3,050, 7,600]: Tax is $153 0003 0.07(x 0004 3,050) 0002 0.07x – 61* (7,600, 0007): Tax is $471 0003 0.09(x 0004 7,600) 0002 0.09x 0004 213
*In the Oregon tax rate chart, dollar amounts ending with 0.50 were rounded up to the next dollar. We will do the same.
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Combining the three intervals with the preceding linear expressions, we can write
再
if 0 x 3,050 if 3,050 6 x 7,600 if x 7 7,600
0.05x T(x) 0002 0.07x 0004 61 0.09x 0004 213 Using the piecewise definition of T, we have
T(2,000) 0002 0.05(2,000) 0002 $100 T(5,000) 0002 0.07(5,000) 0004 61 0002 $289 T(9,000) 0002 0.09(9,000) 0004 213 0002 $597 MATCHED PROBLEM 5
0002
Table 2 contains a recent tax rate chart for persons filing a joint return in the state of Oregon. If T(x) is the tax on an income of $x, write a piecewise definition for T. Find the tax on each of the following incomes: $4,000, $10,000, and $18,000. Table 2 2009 Tax Rate Chart for Persons Filing Jointly If the taxable income is:
The tax is:
Not over $6,100
5% of taxable income
Over $6,100 but not over $15,200
$305 plus 7% of the excess over $6,100
Over $15,200
$942 plus 9% of the excess over $15,200
0002 We will conclude the section with a look at a particular piecewise function that is especially useful in computer science. It is called the greatest integer function. The greatest integer for a real number x, denoted by 冀x冁, is the integer n such that n x n 0003 1; that is, 冀x冁 is the largest integer less than or equal to x. For example, 冀3.45 冁 0002 3 冀7 冁 0002 7
5
f(x) 0002 冚x 军 00045
5
00045
Z Figure 7 Greatest integer function.
x
冀5.99冁 0002 5 冀0冁 0002 0
冀00042.13 冁 0002 00043 冀00048 冁 0002 00048 冀00043.79 冁 0002 00044
Not ⴚ2
The greatest integer function f is defined by the equation f(x) 0002 冀x冁. A piecewise definition of f for 00042 x 3 is shown below, and a sketch of the graph of f for 00045 x 5 is shown in Figure 7. Since the domain of f is all real numbers, the piecewise definition continues indefinitely in both directions, as does the stairstep pattern in the figure. So the range of f is the set of all integers.
f (x) 0002 冀x冁 0002
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭
f (x)
o 00042 00041 0 1 2 o
if 00042 if 00041 if 0 if 1 if 2
x x x x x
6 00041 6 0 6 1 6 2 6 3
Notice in Figure 7 that at each integer value of x there is a break in the graph, and between integer values of x there is no break. In other words, the greatest integer function is discontinuous at each integer n and continuous on each interval of the form [n, n 0003 1).
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Graphing Functions
183
Technology Connections graph and Figure 7. If your graphing calculator supports both a connected mode and a dot mode for graphing functions (consult your manual), which mode is preferable for this graph?
Most graphing calculators denote the greatest integer function as int (x), although not all define it the same way we have here. Graph y ⴝ int (x) for ⴚ5 ⱕ x ⱕ 5 and ⴚ5 ⱕ y ⱕ 5 and discuss any differences between your
EXAMPLE
6
Computer Science Let f (x) 0002
冀10x 0003 0.5冁 10
Find: (A) f(6)
(B) f (1.8)
(C) f (3.24)
(D) f(4.582)
(E) f(00042.68)
What operation does this function perform? SOLUTIONS
Table 3 x
f(x)
6
6
1.8
1.8
3.24
3.2
4.582
4.6
00042.68
00042.7
MATCHED PROBLEM 6
(A) f (6) 0002
冀60.5冁 60 0002 00026 10 10
(C) f (3.24) 0002
(B) f (1.8) 0002
冀32.9冁 32 0002 0002 3.2 10 10
(E) f (00042.68) 0002
冀18.5冁 18 0002 0002 1.8 10 10
(D) f (4.582) 0002
冀46.32冁 46 0002 0002 4.6 10 10
冀000426.3冁 000427 0002 0002 00042.7 10 10
Comparing the values of x and f (x) in Table 3 in the margin, we conclude that this function rounds decimal fractions to the nearest tenth. The greatest integer function is used in programming (spreadsheets, for example) to round numbers to a specified accuracy. 0002 Let f(x) 0002 冀x 0003 0.5冁. Find: (A) f(6)
(B) f (1.8)
(C) f(3.24)
(D) f(00044.3)
(E) f(00042.69)
What operation does this function perform? 0002 ANSWERS TO MATCHED PROBLEMS 1. Domain: (00040007, 23) 傼 (23, 0007); x intercept: 000454; y intercept: f (0) 0002 000452 2. (A) Domain: (00044, 5); range: (00044, 3] (B) f (00044) 0002 1, f (0) 0002 3, f (2) 0002 2 3. y intercept: f(0) 0002 00046 y x intercept: 4 3 Slope: 2 5
00045
x
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4. (A) f (00044) 0002 00041; f (00041) 0002 00042; f (3) 0002 00042; f (4) 0002 3 (B) (C) Domain: (00040007, 0007); y range: [00042, 0007); 5 increasing: [3, 0007); y 0002 5x 0003 17 decreasing: (00040007, 00041]; constant: (00041, 3) 1
y 0002 00043x 0004
7 3
00045
y 0002 00042
(00041, 00042)
再
x
5
(3, 00042)
00045
0.05x x 6,100 5. T(x) 0002 0.07x 0004 122 6,100 6 x 15,200 0.09x 0004 426 x 7 15,200 T(4,000) 0002 $200; T(10,000) 0002 $578; T(18,000) 0002 $1,194 T(4,000) 0002 $200; T(10,000) 0002 $594; T(18,000) 0002 $1,248 6. (A) 6 (B) 2 (C) 3 (D) 00044 (E) 00043; f rounds decimal fractions to the nearest integer.
3-2
Exercises
1. Describe in your own words what the graph of a function is.
13. Repeat Problem 9 for the function p.
2. Explain how to find the domain and range of a function from its graph.
14. Repeat Problem 9 for the function q.
3. How many y intercepts can a function have? What about x intercepts? Explain. 4. True or false: On any interval in its domain, every function is either increasing or decreasing. Explain. 5. Explain in your own words what it means to say that a function is increasing on an interval.
f (x)
g(x)
5
5
00045
6. Explain in your own words what it means to say that a function is decreasing on an interval.
5
x
00045
00045
00045
h (x)
7. What does it mean for a function to be defined piecewise?
x
5
k (x)
5
5
8. Explain how the output of the greatest integer function is calculated for any real number input. Problems 9–20 refer to functions f, g, h, k, p, and q given by the following graphs.
00045
9. For the function f, find: (A) Domain (B) Range (C) x intercepts (D) y intercept (E) Intervals over which f is increasing (F) Intervals over which f is decreasing (G) Intervals over which f is constant (H) Any points of discontinuity 10. Repeat Problem 9 for the function g.
5
x
00045
00045
5
00045
p (x)
q(x)
5
5
00045
5
x
00045
5
11. Repeat Problem 9 for the function h. 12. Repeat Problem 9 for the function k.
00045
x
00045
x
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SECTION 3–2
15. Find f (00044), f (0), and f (4).
Graphing Functions
185
17. Find h(00043), h(0), and h(2).
In Problems 47–58, (A) find the indicated values of f; (B) graph f and label the points from part A, if they exist; and (C) find the domain, range, and the values of x in the domain of f at which f is discontinuous.
18. Find k (0), k(2), and k(4).
47. f(00041), f(0), f(1)
16. Find g (00045), g(0), and g(5).
f (x) 0002 e
19. Find p(00042), p(2), and p (5). 20. Find q(00044), q(00043), and q (1).
x00031 0004x 0003 1
if 00041 x 6 0 if 0 x 1
48. f(00042), f(1), f(2) Problems 21–26 describe the graph of a continuous function f over the interval [00045, 5]. Sketch the graph of a function that is consistent with the given information. 21. The function f is increasing on [00045, 00042], constant on [00042, 2], and decreasing on [2, 5]. 22. The function f is decreasing on [00045, 00042], constant on [00042, 2], and increasing on [2, 5]. 23. The function f is decreasing on [00045, 00042], constant on [00042, 2], and decreasing on [2, 5]. 24. The function f is increasing on [00045, 00042], constant on [00042, 2], and increasing on [2, 5]. 25. The function f is decreasing on [00045, 00042], increasing on [00042, 2], and decreasing on [2, 5]. 26. The function f is increasing on [00045, 00042], decreasing on [00042, 2], and increasing on [2, 5]. In Problems 27–32, find the slope and intercepts, and then sketch the graph. 27. f(x) 0002 2x 0003 4
28. f (x) 0002 3x 0004 3
29. f (x) 0002 000412 x 0004 53
30. f (x) 0002 000434 x 0003 65
31. f (x) 0002 00042.3x 0003 7.1
32. f (x) 0002 5.2x 0004 3.4
In Problems 33–36, find a linear function f satisfying the given conditions. 33. f (00042) 0002 2 and f(0) 0002 10
f (x) 0002 e
if 00042 x 6 1 if 1 x 2
x 0004x 0003 2
49. f (00043), f(00041), f(2) f (x) 0002 e
00042 4
if 00043 x 6 00041 if 00041 6 x 2
50. f (00042), f(2), f(5) f (x) 0002 e
if 00042 x 6 2 if 2 6 x 5
1 00043
51. f(00042), f(00041), f(0) f (x) 0002 e
x00032 x00042
if x 6 00041 if x 7 00041
52. f(0), f(2), f(4) f (x) 0002 e
00041 0004 x 50004x
if x 6 2 if x 7 2
53. f(00043), f(00042), f(0), f(3), f(4)
再 再 再 再 再 再
00042x 0004 6 f (x) 0002 00042 6x 0004 20
if x 6 00042 if 00042 x 6 3 if x 0005 3
54. f(00042), f(00041), f(0), f(2), f(3) 2 3x
0003 113
f (x) 0002 3 000432 x 0003 6
if x 00041 if 00041 6 x 2 if x 7 2
55. f(00043), f(00042), f(0), f(3), f(4)
34. f(4) 0002 00047 and f(0) 0002 5
5 2x
00036 f (x) 0002 1 3 7 2x 0004 2
35. f (00042) 0002 7 and f (4) 0002 00042 36. f (00043) 0002 00042 and f(5) 0002 4
if x 6 00042 if 00042 x 3 if x 7 3
56. f(00043), f(00042), f(0), f(1), f(2) In Problems 37–46, find the domain, x intercept, and y intercept. 3x 0004 12 37. f (x) 0002 2x 0003 4 39. f (x) 0002
3x 0004 2 4x 0004 5
4x 41. f (x) 0002 (x 0004 2)2 43. f (x) 0002
x2 0004 16 x2 0004 9
2x 0003 9 38. f (x) 0002 x00043 40. f (x) 0002
2x 42. f (x) 0002 (x 0003 1)2 44. f (x) 0002
2
45. f (x) 0002
x 00037 x2 0004 25
2x 0003 7 5x 0003 8
x2 0004 4 x2 0003 10 2
46. f (x) 0002
x 0003 11 x2 0003 5
3 f (x) 0002 000413 x 0003 73 00043x 0003 5
if x 00042 if 00042 6 x 6 1 if x 0005 1
57. f(00041), f(0), f(1), f(2), f(3) f (x) 0002
2 3x 000412 x 000412 x
00034 00033
if x 6 0 if 0 6 x 6 2 if x 7 2
58. f(00043), f(00042), f(0), f(2), f(3) 000432 x 0004 2 3 1 f (x) 0002 4x 0004 2 3 5 4x 0004 2
if x 6 00042 if 00042 6 x 6 2 if x 7 2
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FUNCTIONS
In Problems 59–64, use the graph of f to find a piecewise definition for f.
f (x)
64.
5
f (x)
59.
(2, 4)
5
(2, 0) 00045
(00042, 3) 00045
x
5
(0, 00041)
In Problems 65–68, find a piecewise definition of f that does not involve the absolute value function. (Hint: Use the definition of absolute value on page 180 to consider cases.) Sketch the graph of f, and find the domain, range, and the values of x at which f is discontinuous.
00045
f (x) 5
(0, 2)
(2, 2)
00045
x
5
65. f (x) 0002 1 0003 冟 x 冟
66. f (x) 0002 2 0004 冟 x 冟
67. f (x) 0002 冟 x 0004 2 冟
68. f (x) 0002 冟 x 0003 1 冟
69. The function f is continuous and increasing on the interval [1, 9] with f (1) 0002 00045 and f(9) 0002 4. (A) Sketch a graph of f that is consistent with the given information. (B) How many times does your graph cross the x axis? Could the graph cross more times? Fewer times? Support your conclusions with additional sketches and/or verbal arguments.
(0, 00042) 00045
61.
x
(00044, 00043) (00043, 00043) 00045
(2, 00043)
(00042, 00042)
5
(00043, 00041)
(0, 1)
60.
(4, 4)
f (x) 5
(00044, 3) (00041, 3)
70. Repeat Problem 69 if the function is not continuous. 00045
x
5
71. The function f is continuous on the interval [00045, 5] with f(00045) 0002 00044, f (1) 0002 3, and f (5) 0002 00042. (A) Sketch a graph of f that is consistent with the given information. (B) How many times does your graph cross the x axis? Could the graph cross more times? Fewer times? Support your conclusions with additional sketches and/or verbal arguments.
(1, 00043) 00045
f (x)
62.
5
72. Repeat Problem 71 if f is continuous on [00048, 8] with f (00048) 0002 00046, f(00044) 0002 3, f(3) 0002 00042, and f (8) 0002 5. (4, 1)
00045
x
5
In Problems 73–78, first graph functions f and g in the same viewing window, then graph m(x) and n(x) in their own viewing windows:
(00044, 00042) (00042, 00042) 00045
m(x) 0002 0.5[ f (x) 0003 g(x) 0003 冟 f (x) 0004 g(x) 冟 ] n(x) 0002 0.5[ f (x) 0003 g(x) 0004 冟 f (x) 0004 g(x) 冟 ]
f (x)
63.
Problems 73–80 require the use of a graphing calculator.
5
73. f(x) 0002 00042x, g(x) 0002 0.5x
(00044, 3) (00042, 3)
74. f(x) 0002 3x 0003 1, g(x) 0002 00040.5x 0004 4
(00042, 2) 00045
5
x
(1, 00041) (4, 00041) 00045
(1, 00044)
75. f(x) 0002 5 0004 0.2x2, g(x) 0002 0.3x2 0004 4 76. f (x) 0002 0.15x2 0004 5, g(x) 0002 5 0004 1.5冟 x 冟 77. f (x) 0002 0.2x2 0004 0.4x 0004 5, g(x) 0002 0.3x 0004 3 78. f (x) 0002 8 0003 1.5x 0004 0.4x2, g(x) 0002 00040.2x 0003 5 79. How would you characterize the relationship between f, g, and m in Problems 73–78? [Hint: Use the trace feature on the calculator and the up/down arrows to examine all 3 graphs at several points.]
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SECTION 3–2
80. How would you characterize the relationship between f, g, and n in Problems 73–78? [Hint: Use the trace feature on the calculator and the up/down arrows to examine all 3 graphs at several points.]
Graphing Functions
187
(B) Can the function f defined by f (x) 0002 15 0003 3冀x冁 be used to compute the delivery charges for all x, 0 x 6? Justify your answer.
APPLICATIONS Table 4 contains daily automobile rental rates from a New Jersey firm.
Table 4 Vehicle Type
Daily Charge
Included Miles
Mileage Charge*
Compact
$32.00
100/Day
$0.16/mile
Midsize
$41.00
200/Day
$0.18/mile
*Mileage charge does not apply to included miles.
81. AUTOMOBILE RENTAL Use the data in Table 4 to construct a piecewise-defined model for the daily rental charge for a compact automobile that is driven x miles. 82. AUTOMOBILE RENTAL Use the data in Table 4 to construct a piecewise-defined model for the daily rental charge for a midsize automobile that is driven x miles. 83. SALES COMMISSIONS A high-volume website pays salespeople to solicit advertisements for placement on their site. The sales staff each gets $200 per week in salary, and a commission of 4% on all sales over $3,000 for the week. In addition, if the weekly sales are $8,000 or more, the salesperson gets a $100 bonus. Find a piecewise definition for the weekly earnings E (in dollars) in terms of the weekly sales x (in dollars). Graph this function and find the values of x at which the function is discontinuous. Find the weekly earnings for sales of $5,750 and of $9,200. 84. SERVICE CHARGES On weekends and holidays, an emergency plumbing repair service charges $2.00 per minute for the first 30 minutes of a service call and $1.00 per minute for each additional minute. Express the total service charge S (in dollars) as a piecewise-defined function of the duration of a service call x (in minutes). Graph this function and find the values of x at which the function is discontinuous. Find the charge for a 25-minute service call and for a 45-minute service call. 85. COMPUTER SCIENCE Let f (x) 0002 10 冀 0.5 0003 x 10冁 . Evaluate f at 4, 00044, 6, 00046, 24, 25, 247, 0004243, 0004245, and 0004246. What operation does this function perform? 86. COMPUTER SCIENCE Let f (x) 0002 100 冀 0.5 0003 x 100冁 . Evaluate f at 40, 000440, 60, 000460, 740, 750, 7,551, 0004601, 0004649, and 0004651. What operation does this function perform? 87. COMPUTER SCIENCE Use the greatest integer function to define a function f that rounds real numbers to the nearest hundredth. 88. COMPUTER SCIENCE Use the greatest integer function to define a function f that rounds real numbers to the nearest thousandth. 89. DELIVERY CHARGES A nationwide package delivery service charges $15 for overnight delivery of packages weighing 1 pound or less. Each additional pound (or fraction thereof ) costs an additional $3. Let C be the charge for overnight delivery of a package weighing x pounds. (A) Write a piecewise definition of C for 0 x 6, and sketch the graph of C.
90. TELEPHONE CHARGES Calls to 900 numbers are charged to the caller. A 900 number hot line for gambling advice on college football games charges $4 for the first minute of the call and $2 for each additional minute (or fraction thereof). Let C be the charge for a call lasting x minutes. (A) Write a piecewise definition of C for 0 x 6, and sketch the graph of C. (B) Can the function f defined by f(x) 0002 4 0003 2冀x冁 be used to compute the charges for all x, 0 x 6? Justify your answer. 91. STATE INCOME TAX The Connecticut state income taxes for an individual filing a single return are 3% for the first $10,000 of taxable income and 5% on the taxable income in excess of $10,000. Find a piecewise-defined function for the taxes owed by a single filer with an income of x dollars and graph this function. 92. STATE INCOME TAX The Connecticut state income taxes for an individual filing a head of household return are 3% for the first $16,000 of taxable income and 5% on the taxable income in excess of $16,000. Find a piecewise-defined function for the taxes owed by a head of household filer with an income of x dollars and graph this function. Table 5 contains income tax rates for Minnesota in a recent year. Table 5
Status
Taxable Income Over
But Not Over
Tax Is
Of the Amount Over
Single
$0
$19,890
5.35%
$0
19,890
65,330
$1,064 0003 7.05%
19,890
65,330
...
4,268 0003 7.85%
65,330
0
29,070
5.35%
0
29,070
115,510
1,555 0003 7.05%
29,070
115,510
...
7,649 0003 7.85%
115,510
Married
93. STATE INCOME TAX Use the schedule in Table 5 to construct a piecewise-defined model for the taxes due for a single taxpayer with a taxable income of x dollars. Find the tax on the following incomes: $10,000, $30,000, $100,000. 94. STATE INCOME TAX Use the schedule in Table 5 to construct a piecewise-defined model for the taxes due for a married taxpayer with a taxable income of x dollars. Find the tax on the following incomes: $20,000, $60,000, $200,000.
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FUNCTIONS
Transformations of Functions Z A Library of Elementary Graphs Z Shifting Graphs Horizontally and Vertically Z Reflecting Graphs Z Stretching and Shrinking Graphs Z Even and Odd Functions
We have seen that the graph of a function can provide valuable insight into the information provided by that function. But there is a seemingly endless variety of functions out there, and it seems like an insurmountable task to learn about so many different graphs. In this section, we will see that relationships between the formulas for certain functions lead to relationships between their graphs as well. For example, the functions g(x) 0002 x2 0003 2
h(x) 0002 (x 0003 2)2
k(x) 0002 2x2
can be expressed in terms of the function f(x) 0002 x2 as follows: g(x) 0002 f (x) 0003 2
h(x) 0002 f (x 0003 2)
k(x) 0002 2f(x)
We will see that the graphs of functions g, h, and k are closely related to the graph of function f. Once we understand these relationships, knowing the graph of a very simple function like f (x) 0002 x2 will enable us to learn about the graphs of many related functions.
Z A Library of Elementary Graphs As you progress through this book, you will encounter a number of basic functions that you will want to add to your library of elementary functions. Figure 1 shows six basic functions that you will encounter frequently. You should know the definition, domain, and range of each of these functions, and be able to draw their graphs.
f (x)
g(x)
h(x)
5
5
5
00045
5
x
00045
5
x
00045
00045
(a) Identity function f(x) 0002 x Domain: R Range: R
(b) Absolute value function g(x) 0002 |x| Domain: R Range: [0, 0003)
(c) Square function h(x) 0002 x2 Domain: R Range: [0, 0003)
5
x
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SECTION 3–3 m (x)
n (x)
5
p(x)
5
00045
5
x
5
x
5
00045
00045
5
00045
(d) Cube function m(x) 0002 x3 Domain: R Range: R
189
Transformations of Functions
x
00045
(f) Cube root function 3 p(x) 0002 1x Domain: R Range: R
(e) Square root function n(x) 0002 1x Domain: [0, 0003) Range: [0, 0003)
Z Figure 1 Some basic functions and their graphs. [Note: Letters used to designate these functions may vary from context to context; R represents the set of all real numbers.]
Z Shifting Graphs Vertically and Horizontally If a new function is formed by performing an operation on a given function, then the graph of the new function is called a transformation of the graph of the original function. For example, if we add a constant k to f (x), then the graph of y 0002 f (x) is transformed into the graph of y 0002 f (x) 0003 k.
ZZZ EXPLORE-DISCUSS 1
The following activities refer to the graph of f shown in Figure 2 and the corresponding points on the graph shown in Table 1. (A) Use the points in Table 1 to construct a similar table and then sketch a graph for each of the following functions: y 0002 f (x) 0003 2, y 0002 f(x) 0004 3. Describe the relationship between the graph of y 0002 f (x) and the graph of y 0002 f(x) 0003 k for k any real number. (B) Use the points in Table 1 to construct a similar table and then sketch a graph for each of the following functions: y 0002 f (x 0003 2), y 0002 f(x 0004 3). [Hint: Choose values of x so that x 0003 2 or x 0004 3 is in Table 1.] Describe the relationship between the graph of y 0002 f (x) and the graph of y 0002 f (x 0003 h) for h any real number. y
Table 1
5
x
B
00045
E
C
A
5
x
y 0002 f (x) D 00045
Z Figure 2
f(x)
A
00044
0
B
00042
3
C
0
0
D
2
00043
E
4
0
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1
Vertical and Horizontal Shifts (A) How are the graphs of y 0002 x2 0003 2 and y 0002 x2 0004 3 related to the graph of y 0002 x2? Confirm your answer by graphing all three functions in the same coordinate system. (B) How are the graphs of y 0002 (x 0003 2)2 and y 0002 (x 0004 3)2 related to the graph of y 0002 x2? Confirm your answer by graphing all three functions in the same coordinate system.
SOLUTIONS
(A) Note that the output of y 0002 x2 0003 2 is always exactly two more than the output of y 0002 x2. Consequently, the graph of y 0002 x2 0003 2 is the same as the graph of y 0002 x2 shifted upward two units, and the graph of y 0002 x2 0004 3 is the same as the graph of y 0002 x2 shifted downward three units. Figure 3 confirms these conclusions. (It appears that the graph of y 0002 f(x) 0003 k is the graph of y 0002 f (x) shifted up if k is positive and down if k is negative.) y 5
y 0002 x2 0003 2 y 0002 x2 x
00045
5
y 0002 x2 0004 3 00045
Z Figure 3 Vertical shifts.
(B) Note that the output of y 0002 (x 0003 2)2 is zero for x 0002 00042, while the output of y 0002 x2 is zero for x 0002 0. This suggests that the graph of y 0002 (x 0003 2)2 is the same as the graph of y 0002 x2 shifted to the left two units, and the graph of y 0002 (x 0004 3)2 is the same as the graph of y 0002 x2 shifted to the right three units. Figure 4 confirms these conclusions. It appears that the graph of y 0002 f(x 0003 h) is the graph of y 0002 f(x) shifted right if h is negative and left if h is positive. y
y 0002 x2
y 0002 (x 0004 3)2
5
y 0002 (x 0003 2)2 00045
Z Figure 4 Horizontal shifts.
MATCHED PROBLEM 1
5
x
0002
(A) How are the graphs of y 0002 1x 0003 3 and y 0002 1x 0004 1 related to the graph of y 0002 1x? Confirm your answer by graphing all three functions in the same coordinate system. (B) How are the graphs of y 0002 1x 0003 3 and y 0002 1x 0004 1 related to the graph of y 0002 1x? Confirm your answer by graphing all three functions in the same coordinate system. 0002 To summarize our experiences in Explore-Discuss 1 and Example 1: We can graph y 0002 f(x) 0003 k by vertically shifting the graph of y 0002 f(x) upward k units if k is positive
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SECTION 3–3
Transformations of Functions
191
and downward 冟 k 冟 units if k is negative. We can graph y 0002 f(x 0003 h) by horizontally shifting the graph of y 0002 f (x) left h units if h is positive and right 冟 h 冟 units if h is negative.
EXAMPLE
2
Vertical and Horizontal Shifts The graphs in Figure 5 are either horizontal or vertical shifts of the graph of f(x) 0002 |x|. Write appropriate equations for functions H, G, M, and N in terms of f. y
G
y
f
5
f
H
00045
5
M
5
N
x 00045
x
5
00045
Z Figure 5 Vertical and horizontal shifts. SOLUTION
The graphs of functions H and G are 3 units lower and 1 unit higher, respectively, than the graph of f, so H and G are vertical shifts given by H(x) 0002 冟 x 冟 0004 3
G(x) 0002 冟 x 冟 0003 1
The graphs of functions M and N are 2 units to the left and 3 units to the right, respectively, of the graph of f, so M and N are horizontal shifts given by M(x) 0002 冟 x 0003 2 冟 MATCHED PROBLEM 2
N(x) 0002 冟 x 0004 3 冟
0002
The graphs in Figure 6 are either horizontal or vertical shifts of the graph of f (x) 0002 x3. Write appropriate equations for functions H, G, M, and N in terms of f. G y
y MfN
f H 5
5
00045
5
x
00045
5
x
Z Figure 6 Vertical and horizontal shifts.
0002
Z Reflecting Graphs In Section 2-1, we discussed reflections of graphs and developed symmetry properties that we used as an aid in graphing equations. Now we will consider reflection as an operation that transforms the graph of a function.
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ZZZ EXPLORE-DISCUSS 2
The following activities refer to the graph of f shown in Figure 7 and the corresponding points on the graph shown in Table 2. (A) Construct a similar table for y 0002 0004f (x) and then sketch the graph of y 0002 0004f(x). Describe the relationship between the graph of y 0002 f(x) and the graph of y 0002 0004f(x) in terms of reflections. (B) Construct a similar table for y 0002 f(0004x) and then sketch the graph of y 0002 f(0004x). [Hint: Choose x values so that 0004x is in Table 2.] Describe the relationship between the graph of y 0002 f (x) and the graph of y 0002 f (0004x) in terms of reflections. (C) Construct a similar table for y 0002 0004f(0004x) and then sketch the graph of y 0002 0004f(0004x). [Hint: Choose x values so that 0004x is in Table 2.] Describe the relationship between the graph of y 0002 f(x) and the graph of y 0002 0004f(0004x) in terms of reflections. y
Table 2
5
A
E
x
y 0002 f (x) 00045
D
B
00045
5
x
C
Z Figure 7
EXAMPLE
3
f(x)
A
00042
5
B
00041
0
C
1
00044
D
3
0
E
4
5
Reflecting the Graph of a Function Let f(x) 0002 (x 0004 1)2. (A) How are the graphs of y 0002 f (x) and y 0002 0004f(x) related? Confirm your answer by graphing both functions in the same coordinate system. (B) How are the graphs of y 0002 f (x) and y 0002 f (0004x) related? Confirm your answer by graphing both functions in the same coordinate system. (C) How are the graphs of y 0002 f (x) and y 0002 0004f(0004x) related? Confirm your answer by graphing both functions in the same coordinate system.
SOLUTIONS
Refer to Definition 1 in Section 2-1. (A) The graph of y 0002 0004f (x) can be obtained from the graph of y 0002 f (x) by changing the sign of each y coordinate. This has the effect of moving every point to the opposite side of the x axis. So the graph of y 0002 0004f (x) is the reflection through the x axis of the graph of y 0002 f (x) [Fig. 8(a)]. (B) The graph of y 0002 f (0004x) can be obtained from the graph of y 0002 f (x) by changing the sign of each x coordinate. This has the effect of moving every point to the opposite side of the y axis. So the graph of y 0002 f (0004x) is the reflection through the y axis of the graph of y 0002 f (x) [Fig. 8(b)]. (C) The graph of y 0002 0004f (0004x) can be obtained from the graph of y 0002 f (x) by changing the sign of each x and y coordinate. So the graph of y 0002 0004f (0004x) is the reflection through the origin of the graph of y 0002 f (x) [Fig. 8(c)].
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SECTION 3–3 y 5
00045
5
x
5
5
(00043, 4)
y 0002 f(x)
y 0002 0004f(x)
y
y (3, 4)
(3, 4) y 0002 f(x)
y 0002 f(0004x) 00045
5
y 0002 f(x) x
00045
(00043, 00044) 00045
(a) y 0002 f(x) and y 0002 0004f(x); reflection through the x axis
(3, 4)
5
x
y 0002 0004f(0004x)
(3, 00044)
00045
193
Transformations of Functions
(b) y 0002 f(x) and y 0002 f(0004x); reflection through the y axis
00045
(c) y 0002 f(x) and y 0002 0004f(0004x); reflection through the origin
0002
Z Figure 8
MATCHED PROBLEM 3
Repeat Example 3 for f (x) 0002 |x 0003 2|.
0002
Z Stretching and Shrinking Graphs Horizontal shifts, vertical shifts, and reflections are called rigid transformations because they do not change the shape of a graph, only its location. Now we consider some nonrigid transformations that change the shape by stretching or shrinking a graph.
ZZZ EXPLORE-DISCUSS 3 y
(A) Use the points in Table 3 to construct a similar table and sketch a graph for each of the following functions: y 0002 2f(x) and y 0002 12 f (x). If A 0005 1, does multiplying f by A stretch or shrink the graph of y 0002 f(x) in the vertical direction? What happens if 0 0006 A 0006 1?
8
A
E B
D C
00043
The following activities refer to the graph of f shown in Figure 9 and the corresponding points on the graph shown in Table 3.
x
7
(B) Use the points in Table 3 to complete the following tables and then sketch a graph of y 0002 f (2x) and of y 0002 f (12x):
00042
x
2x
f (2x)
x
Z Figure 9
Table 3 x
f(x)
A
00042
5
B
0
3
C
2
1
D
4
3
E
6
5
00041
00044
0
0
1
4
2
8
3
12
1 2x
f (12x)
If A 0005 1, is the graph of y 0002 f(Ax) a horizontal stretch or a horizontal shrink of the graph of y 0002 f(x)? What if 0 0006 A 0006 1?
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In general, the graph of y 0002 Af(x) can be obtained from the graph of y 0002 f(x) by multiplying the y coordinate of each point on the graph f by A. This vertically stretches the graph of y 0002 f(x) if A 0005 1 and vertically shrinks the graph if 0 0006 A 0006 1. The graph of y 0002 f(Ax) can be obtained from the graph of y 0002 f(x) by multiplying the x coordinate of each point on the graph f by 1兾A. This horizontally stretches the graph of y 0002 f(x) if 0 0006 A 0006 1 and horizontally shrinks the graph if A 0005 1. Another common name for a stretch is an expansion and another common name for a shrink is a contraction.
EXAMPLE
4
Stretching or Shrinking a Graph Let f(x) 0002 1 0003 x2. (A) How are the graphs of y 0002 2f(x) and y 0002 12 f (x) related to the graph of y 0002 f(x)? Confirm your answer by graphing all three functions in the same coordinate system. (B) How are the graphs of y 0002 f(2x) and y 0002 f (12 x) related to the graph of y 0002 f(x)? Confirm your answer by graphing all three functions in the same coordinate system.
SOLUTIONS
(A) The graph of y 0002 2f(x) 0002 2 0003 2x2 can be obtained from the graph of f by multiplying each y value by 2. This stretches the graph of f vertically (away from the x axis) by a factor of 2. The graph of y 0002 12 f (x) 0002 12 0003 12 x2 can be obtained from the graph of f by multiplying each y value by 12. This shrinks the graph of f vertically (toward the x axis) by a factor of 12 [Fig. 10(a)]. (B) The graph of y 0002 f(2x) 0002 1 0003 4x2 can be obtained from the graph of f by multiplying each x value by 12. This shrinks the graph of f horizontally (toward the y axis) by a factor of 12. The graph of y 0002 f (12 x) 0002 1 0003 14 x2 can be obtained from the graph of f by multiplying each x value by 2. This stretches the graph of f horizontally (away from the y axis) by a factor of 2 [Fig. 10(b)].
y 7
y 0002 2 0003 2x 2 y 0002 1 0003 x2 1 1 y 0002 2 0003 2 x2
y 7
(1, 4)
5
x
00043
(a) Vertical stretching and shrinking
y 0002 1 0003 x2 1 y 0002 1 0003 4 x2 (4, 5) (2, 5) (1, 5) x
(1, 2) (1, 1) 00045
y 0002 1 0003 4x 2
00045
5
00043
(b) Horizontal stretching and shrinking
Z Figure 10
MATCHED PROBLEM 4
0002
Let f(x) 0002 4 0004 x2. (A) How are the graphs of y 0002 2f(x) and y 0002 12 f (x) related to the graph of y 0002 f(x)? Confirm your answer by graphing all three functions in the same coordinate system. (B) How are the graphs of y 0002 f(2x) and y 0002 f (12 x) related to the graph of y 0002 f(x)? Confirm your answer by graphing all three functions in the same coordinate system. 0002 Plotting points with the same x coordinate will help you recognize vertical stretches and shrinks [Fig. 10(a)]. And plotting points with the same y coordinate will help you recognize horizontal stretches and shrinks [Fig. 10(b)].
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Transformations of Functions
195
Note that for some functions, a horizontal stretch or shrink can also be interpreted as a vertical stretch or shrink. For example, if y 0002 f(x) 0002 x2, then y 0002 4f (x) 0002 4x2 0002 (2x)2 0002 f (2x) So the graph of y 0002 4x2 is both a vertical stretch and a horizontal shrink of the graph of y 0002 x2. The transformations we’ve studied are summarized next for easy reference.
Z GRAPH TRANSFORMATIONS (SUMMARY) Z Figure 11 Graph transformations.
Vertical Shift [Fig. 11(a)]:
y 5
g
00045
5
再
y 0002 f (x) 0003 k
Shift graph of y 0002 f (x) up k units Shift graph of y 0002 f (x) down 冟 k 冟 units
k 7 0 k 6 0
f
Horizontal Shift [Fig. 11(b)]:
x h
y 0002 f (x 0003 h)
再
h 7 0 h 6 0
Shift graph of y 0002 f (x) left h units Shift graph of y 0002 f (x) right 冟 h 冟 units
Vertical Stretch and Shrink [Fig. 11(c)]:
00045
(a) Vertical translation g(x) 0002 f(x) 0005 2 h(x) 0002 f(x) 0004 3
g y
f
再 再
y 0002 Af (x)
A 7 1
Vertically stretch the graph of y 0002 f (x) by multiplying each y value by A
0 6 A 6 1
Vertically shrink the graph of y 0002 f (x) by multiplying each y value by A
Horizontal Stretch and Shrink [Fig. 11(d)]:
h
A 7 1
5
y 0002 f (Ax) 00045
0 6 A 6 1
x
5
Horizontally shrink the graph of y 0002 f (x) by multiplying each x value by A1 Horizontally stretch the graph of y 0002 f (x) by multiplying each x value by A1
Reflection [Fig. 11(e)]: y 0002 0004f (x) y 0002 f (0004x) y 0002 0004f(0004x)
00045
(b) Horizontal translation g(x) 0002 f(x 0005 3) h(x) 0002 f(x 0004 2)
y
y g f
g
Reflect the graph of y 0002 f(x) through the x axis Reflect the graph of y 0002 f(x) through the y axis Reflect the graph of y 0002 f(x) through the origin
y
g h
f
5
5
f
5
h 00045
5
00045
5
x
00045
5
x 00045 00045
(c) Vertical expansion and contraction g(x) 0002 2f(x) h(x) 0002 12 f (x)
(d) Horizontal expansion and contraction g(x) 0002 f(2x) h(x) 0002 f(12 x)
k
h (e) Reflection g(x) 0002 f(0004x) h(x) 0002 0004f(x) k(x) 0002 0004f(0004x)
x
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EXAMPLE
5
Combining Graph Transformations The graph of y 0002 g(x) in Figure 12 is a transformation of the graph of y 0002 x2. Find an equation for the function g. y 5
y 0002 g(x) 00045
5
x
Z Figure 12 SOLUTION
To transform the graph of y 0002 x2 [Fig. 13(a)] into the graph of y 0002 g(x), we first reflect the graph of y 0002 x2 through the x axis [Fig. 13(b)], then shift it to the right two units [Fig. 13(c)]. An equation for the function g is g(x) 0002 0004(x 0004 2)2 y
y 5
5
5
x
00045
5
x
00045
y 0002 0004(x 0004 2)2 x
(b) y 0002 0004x2
(c) y 0002 0004(x 0004 2)2
Z Figure 13
MATCHED PROBLEM 5 y 5
5
00045
Z Figure 14
0002
The graph of y 0002 h(x) in Figure 14 is a transformation of the graph of y 0002 x3. Find an equation for the function h. 0002
Z Even and Odd Functions
y 0002 h(x)
00045
5
00045
00045
(a) y 0002 x2
5
y 0002 0004x 2
y 0002 x2 00045
y
x
Certain transformations leave the graphs of some functions unchanged. For example, reflecting the graph of y 0002 x2 through the y axis does not change the graph. Functions with this property are called even functions. Similarly, reflecting the graph of y 0002 x3 through the origin does not change the graph. Functions with this property are called odd functions. More formally, we have the following definitions. Z EVEN AND ODD FUNCTIONS If f(x) 0002 f (0004x) for all x in the domain of f, then f is an even function. If f(0004x) 0002 0004f(x) for all x in the domain of f, then f is an odd function.
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The graph of an even function is symmetric with respect to the y axis and the graph of an odd function is symmetric with respect to the origin (Fig. 15). f (x)
f (x) f
f
f (x)
f (⫺x) ⫽ f (x) ⫺x
x
f (x)
⫺x
x
f (⫺x) ⫽ ⫺f (x)
x
x
Even function (symmetric with respect to y axis)
Odd function (symmetric with respect to origin)
Z Figure 15 Even and odd functions.
EXAMPLE
6
Testing for Even and Odd Functions Determine whether the functions f, g, and h are even, odd, or neither. (A) f(x) ⫽ x4 ⫹ 1
SOLUTIONS
(B) g(x) ⫽ x3 ⫹ 1
(C) h(x) ⫽ x5 ⫹ x
It will be useful to note the following: if n is an even integer, then (⫺x)n ⫽ (⫺1)n xn ⫽ xn because (⫺1)n ⫽ 1 if n is even. But if n is an odd integer, (⫺x)n ⫽ (⫺1)n xn ⫽ ⫺xn because (⫺1)n ⫽ ⫺1 when n is odd. (A)
f (x) ⫽ x4 ⫹ 1 f (⫺x) ⫽ (⫺x)4 ⫹ 1 ⫽ x4 ⫹ 1 ⫽ f(x)
(⫺x)4 ⫽ x4 because 4 is even.
This shows that f is even. (B)
g(x) ⫽ x3 ⫹ 1 g(⫺x) ⫽ (⫺x)3 ⫹ 1 ⫽ ⫺x3 ⫹ 1 ⫺g(x) ⫽ ⫺(x3 ⫹ 1) ⫽ ⫺x3 ⫺ 1
(⫺x)3 ⫽ ⫺x3 because 3 is odd.
Distribute the negative.
The function g(⫺x) is neither g(x) nor ⫺g(x), so g is neither even nor odd. (C)
h(x) ⫽ x5 ⫹ x h(⫺x) ⫽ (⫺x)5 ⫹ (⫺x) ⫽ ⫺x5 ⫺ x ⫺h(x) ⫽ ⫺(x5 ⫹ x) ⫽ ⫺x5 ⫺ x
(⫺x)5 ⫽ ⫺x5 because 5 is odd.
Distribute the negative.
Since h(⫺x) ⫽ ⫺h(x), h is odd. MATCHED PROBLEM 6
0002
Determine whether the functions F, G, and H are even, odd, or neither: (A) F(x) ⫽ x3 ⫺ 2x
(B) G(x) ⫽ x2 ⫹ 1
(C) H(x) ⫽ 2x ⫹ 4
0002
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ANSWERS TO MATCHED PROBLEMS 1. (A) The graph of y 0002 1x 0003 3 is the same as the graph of y 0002 1x shifted upward 3 units, and the graph of y 0002 1x 0004 1 is the same as the graph of y 0002 1x shifted downward 1 unit. The figure confirms these conclusions. (B) The graph of y 0002 1x 0003 3 is the same as the graph of y 0002 1x shifted to the left 3 units, and the graph of y 0002 1x 0004 1 is the same as the graph of y 0002 1x shifted to the right 1 unit. The figure confirms these conclusions. y
y y 0002 兹x 0003 3
5
00045
y 0002 兹x y 0002 兹x 0004 1 x
5
5
y 0002 兹x 0003 3 y 0002 兹x y 0002 兹x 0004 1
00045
00045
5
x
00045
2. G(x) 0002 (x 0003 3)3, H(x) 0002 (x 0004 1)3, M(x) 0002 x3 0003 3, N(x) 0002 x3 0004 4 (B) The graph of y 0002 f(0004x) is the 3. (A) The graph of y 0002 0004f(x) is the reflection through the y axis of the reflection through the x axis of graph of y 0002 f(x). the graph of y 0002 f(x). y
y 5
5
(00042, 4)
(2, 4) y 0002 f(x) 00045
5
y 0002 0004f(x)
x
(2, 4) y 0002 f(0004x)
y 0002 f(x)
5
00045
x
(2, 00044) 00045
00045
(C) The graph of y 0002 0004f (0004x) is the reflection through the origin of the graph of y 0002 f(x). y 5
y 0002 f(x) (2, 4)
5
00045
(00042, 00044) 00045
x
y 0002 0004f(0004x)
4. (A) The graph of y 0002 2f (x) is a vertical stretch of the graph of y 0002 f (x) by a factor of 2. The graph of y 0002 12 f (x) is a vertical shrink of the graph of y 0002 f (x) by a factor of 12.
(B) The graph of y 0002 f(2x) is a horizontal shrink of the graph of y 0002 f(x) by a factor of 12. The graph of y 0002 f (12 x) is a horizontal stretch of the graph of y 0002 f(x) by a factor of 2.
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199
Transformations of Functions
y
10
10
(1, 6)
y 0002 8 0003 2x 2 y 0002 4 0003 x2 1 y 0002 2 0003 2 x2
(1, 3) (1, 32 )
00035
5
(1, 0) (2, 0) (4, 0)
1
y 0002 4 0003 4 x2 x
00035
5
x
y 0002 4 0003 x2 000310
000310
y 0002 4 0003 4x 2
5. The graph of function h is a reflection through the x axis and a horizontal translation of three units to the left of the graph of y 0002 x3. An equation for h is h(x) 0002 0003(x 0004 3)3. 6. (A) Odd (B) Even (C) Neither
3-3
Exercises
1. Explain why the graph of y 0002 f (x) 0004 k is the same as the graph of y 0002 f (x) moved upward k units when k is positive. 2. Explain why the graph of y 0002 Af (x) is a vertical stretch of the graph of y 0002 f (x) when A 0005 1, and a vertical shrink when A 0006 1.
18. h(x) 0002 f(x 0003 1)
g (x)
19. h(x) 0002 0003f(x)
5
20. h(x) 0002 0003g(x)
3. Explain why the graph of y 0002 0003f (x) is a reflection of the graph of y 0002 f (x) about the x axis, and why the graph of y 0002 f (0003x) is a reflection about the y axis.
21. h(x) 0002 2g(x)
4. Is every function either even or odd? Explain your answer.
23. h(x) 0002 g(2x)
In Problems 5–10, without looking back in the text, indicate the domain and range of each of the following functions. (Making rough sketches on scratch paper may help.) 3
5. h(x) 0002 0003 1x
00035
5
x
00035
1 24. h(x) 0002 f a xb 2 25. h(x) 0002 f (0003x) 26. h(x) 0002 0003g(0003x)
6. m(x) 0002 0003 1 x
7. g(x) 0002 00032x2
1 22. h(x) 0002 f (x) 2
8. f (x) 0002 00030.5|x| 3
9. F(x) 0002 00030.5x
Indicate whether each function in Problems 27–36 is even, odd, or neither.
10. G(x) 0002 4x3
Problems 11–26 refer to the functions f and g given by the graphs below. The domain of each function is [00032, 2], the range of f is [00032, 2], and the range of g is [00031, 1]. Use the graph of f or g, as required, to graph the function h and state the domain and range of h. 11. h(x) 0002 f (x) 0004 2
f (x)
12. h(x) 0002 g(x) 0003 1
5
13. h(x) 0002 g(x) 0004 2 14. h(x) 0002 f(x) 0003 1 15. h(x) 0002 f(x 0003 2)
00035
5
16. h(x) 0002 g(x 0003 1) 17. h(x) 0002 g(x 0004 2)
00035
x
27. g(x) 0002 x3 0004 x
28. f(x) 0002 x5 0003 x
29. m(x) 0002 x4 0004 3x2
30. h(x) 0002 x4 0003 x2
31. F(x) 0002 x5 0004 1
32. f (x) 0002 x5 0003 3
33. G(x) 0002 x4 0004 2
34. P(x) 0002 x4 0003 4
35. q(x) 0002 x2 0004 x 0003 3
36. n(x) 0002 2x 0003 3
In Problems 37–44, the graph of the function g is formed by applying the indicated sequence of transformations to the given function f. Find an equation for the function g. Check your work by graphing f and g in a standard viewing window. 3 37. The graph of f (x) 0002 1x is shifted four units to the left and five units down.
38. The graph of f (x) 0002 x3 is shifted five units to the right and four units up.
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39. The graph of f (x) 0002 1x is shifted six units up, reflected in the x axis, and vertically shrunk by a factor of 0.5.
y
65. 5
40. The graph of f (x) 0002 1x is shifted two units down, reflected in the x axis, and vertically stretched by a factor of 4. 41. The graph of f (x) 0002 x2 is reflected in the x axis, vertically stretched by a factor of 2, shifted four units to the left, and shifted two units down. 42. The graph of f (x) 0002 冟 x 冟 is reflected in the x axis, vertically shrunk by a factor of 0.5, shifted three units to the right, and shifted four units up.
00045
y
66. 5
3 44. The graph of f (x) 0002 1x is horizontally shrunk by a factor of 2, shifted three units up, and reflected in the y axis.
45. f (x) 0002 4x
1 46. g(x) 0002 1x 3
47. h(x) 0002 |x 0003 2|
48. k(x) 0002 |x 0004 4|
49. m(x) 0002 0004|4x 0004 8|
50. n(x) 0002 0004|9 0003 3x|
51. p(x) 0002 3 0004 1x
52. q(x) 0002 00042 0003 1x 0003 3
53. r(x) 0002 3 1x 0004 1 0003 2
54. s(x) 0002 1x 0004 1 0003 2
2
2
55. h(x) 0002 x 0003 3
56. h(x) 0002 4 0004 x2
57. k(x) 0002 2x3 0003 1
58. h(x) 0002 3x3 0004 1
59. n(x) 0002 (x 0003 2)2
60. m(x) 0002 (x 0004 4)2
61. q(x) 0002 4 0004
1 (x 0004 2)2 2
x
00045
43. The graph of f (x) 0002 1x is horizontally stretched by a factor of 0.5, reflected in the y axis, and shifted two units to the left.
Use graph transformations to sketch the graph of each function in Problems 45–62.
5
00045
5
x
00045
y
67. 5
00045
5
x
00045
y
68. 5
2 62. p(x) 0002 5 0004 (x 0003 3)2 3
00045
5
x
Each graph in Problems 63–78 is a transformation of one of the six basic functions in Figure 1. Find an equation for the given graph. 00045
y
63.
y
69.
5
5
00045
5
x 00045
5
x
00045 00045
y
64.
y
70.
5
5
00045
5
x 00045
5
00045 00045
x
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SECTION 3–3 y
71.
5
5
x
00045
y
5
00045
5
x
00045
00045
5
00045
5
x
00045
5
00045
5
x
00045
y 5
5
00045
00045
5
x
00045
y
00045
x
y
78.
5
75.
5
00045
y
74.
x
y
77.
5
73.
5
00045
00045
72.
201
y
76.
5
00045
Transformations of Functions
x
3 3 79. Consider the graphs of f (x) 0002 18x and g (x) 0002 2 1x . 3 (A) Describe each as a stretch or shrink of y 0002 1x. (B) Graph both functions in the same viewing window on a graphing calculator. What do you notice? (C) Rewrite the formula for f algebraically to show that f and g are in fact the same function. (This shows that for some functions, a horizontal stretch or shrink can also be interpreted as a vertical stretch or shrink.)
80. Consider the graphs of f (x) 0002 (3x)3 and g(x) 0002 27x3. (A) Describe each as a stretch or shrink of y 0002 x3. (B) Graph both functions in the same viewing window on a graphing calculator. What do you notice? (C) Rewrite the formula for f algebraically to show that f and g are in fact the same function. (This shows that for some functions, a horizontal stretch or shrink can also be interpreted as a vertical stretch or shrink.) 81. (A) Starting with the graph of y 0002 x2, apply the following transformations. (i) Shift downward 5 units, then reflect in the x axis. (ii) Reflect in the x axis, then shift downward 5 units. What do your results indicate about the significance of order when combining transformations? (B) Write a formula for the function corresponding to each of the above transformations. Discuss the results of part A in terms of order of operations.
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82. (A) Starting with the graph of y 0002 冟 x 冟, apply the following transformations. (i) Stretch vertically by a factor of 2, then shift upward 4 units. (ii) Shift upward 4 units, then stretch vertically by a factor of 2. What do your results indicate about the significance of order when combining transformations? (B) Write a formula for the function corresponding to each of the above transformations. Discuss the results of part A in terms of order of operations.
95. Let f be any function with the property that 0004x is in the domain of f whenever x is in the domain of f, and let E and O be the functions defined by
83. Based on the graphs of the six elementary functions in Figure 1, which are odd, which are even, and which are neither? Use the definitions of odd and even functions to prove your answers.
96. Let f be any function with the property that –x is in the domain of f whenever x is in the domain of f, and let g(x) 0002 xf(x). (A) If f is even, is g even, odd, or neither? (B) If f is odd, is g even, odd, or neither?
84. Based on the results of Example 6, why do you think the terms “even” and “odd” are used to describe functions with particular symmetry properties? Changing the order in a sequence of transformations may change the final result. Investigate each pair of transformations in Problems 85–90 to determine if reversing their order can produce a different result. Support your conclusions with specific examples and/or mathematical arguments.
E(x) 0002 12 [ f (x) 0003 f (0004x)] and O(x) 0002 12 [ f (x) 0004 f (0004x)] (A) Show that E is always even. (B) Show that O is always odd. (C) Show that f (x) 0002 E(x) 0003 O(x). What is your conclusion?
APPLICATIONS 97. PRODUCTION COSTS Total production costs for a product can be broken down into fixed costs, which do not depend on the number of units produced, and variable costs, which do depend on the number of units produced. So, the total cost of producing x units of the product can be expressed in the form C(x) 0002 K 0003 f(x)
85. Vertical shift, horizontal shift
where K is a constant that represents the fixed costs and f (x) is a function that represents the variable costs. Use the graph of the variable-cost function f(x) shown in the figure to graph the total cost function if the fixed costs are $30,000.
86. Vertical shift, reflection in y axis 87. Vertical shift, reflection in x axis 88. Vertical shift, expansion
f (x)
89. Horizontal shift, reflection in x axis
150,000
Problems 91–94 refer to two functions f and g with domain [00045, 5] and partial graphs as shown here. f (x)
g (x)
5
5
00045
5
00045
x
00045
5
x
00045
91. Complete the graph of f over the interval [00045, 0], given that f is an even function. 92. Complete the graph of f over the interval [00045, 0], given that f is an odd function. 93. Complete the graph of g over the interval [00045, 0], given that g is an odd function. 94. Complete the graph of g over the interval [00045, 0], given that g is an even function.
Variable production costs
90. Horizontal shift, contraction
100,000
50,000
500
1,000
x
Units produced
98. COST FUNCTIONS Refer to the variable-cost function f (x) in Problem 97. Suppose construction of a new production facility results in a 25% decrease in the variable cost at all levels of output. If F is the new variable-cost function, use the graph of f to graph y 0002 F(x), then graph the total cost function for fixed costs of $30,000. 99. TIMBER HARVESTING To determine when a forest should be harvested, forest managers often use formulas to estimate the number of board feet a tree will produce. A board foot equals 1 square foot of wood, 1 inch thick. Suppose that the number of board feet y yielded by a tree can be estimated by y 0002 f (x) 0002 C 0003 0.004(x 0004 10)3
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where x is the diameter of the tree in inches measured at a height of 4 feet above the ground and C is a constant that depends on the species being harvested. Graph y 0002 f(x) for C 0002 10, 15, and 20 simultaneously in the viewing window with Xmin 0002 10, Xmax 0002 25, Ymin 0002 10, and Ymax 0002 35. Write a brief verbal description of this collection of functions.
Quadratic Functions
203
a brief verbal description of this collection of functions. Based on the graphs, do larger values of C correspond to a larger or smaller opening? 4 feet
100. SAFETY RESEARCH If a person driving a vehicle slams on the brakes and skids to a stop, the speed v in miles per hour at the time the brakes are applied is given approximately by
4 feet
4 feet
v 0002 f (x) 0002 C 1x where x is the length of the skid marks and C is a constant that depends on the road conditions and the weight of the vehicle. The table lists values of C for a midsize automobile and various road conditions. Graph v 0002 f (x) for the values of C in the table simultaneously in the viewing window with Xmin 0002 0, Xmax 0002 100, Ymin 0002 0, and Ymax 0002 60. Write a brief verbal description of this collection of functions. Road Condition
C
Wet (concrete)
3.5
Wet (asphalt)
4
Dry (concrete)
5
Dry (asphalt)
5.5
101. FLUID FLOW A cubic tank is 4 feet on a side and is initially full of water. Water flows out an opening in the bottom of the tank at a rate proportional to the square root of the depth (see the figure). Using advanced concepts from mathematics and physics, it can be shown that the volume of the water in the tank t minutes after the water begins to flow is given by 64 V(t) 0002 2 (C 0004 t)2 C
Figure for 101
102. EVAPORATION A water trough with triangular ends is 9 feet long, 4 feet wide, and 2 feet deep (see the figure). Initially, the trough is full of water, but due to evaporation, the volume of the water in the trough decreases at a rate proportional to the square root of the volume. Using advanced concepts from mathematics and physics, it can be shown that the volume after t hours is given by V(t) 0002
0 t 6 |C|
where C is a constant. Sketch by hand the graphs of y 0002 V(t) for C 0002 00044, 00045, and 00046. Write a brief verbal description of this collection of functions. Based on the graphs, do values of C with a larger absolute value correspond to faster or slower evaporation? 4 feet 9 feet
0tC
where C is a constant that depends on the size of the opening. Sketch by hand the graphs of y 0002 V(t) for C 0002 1, 2, 4, and 8. Write
3-4
1 (t 0003 6C)2 C2
2 feet
Quadratic Functions Z Graphing Quadratic Functions Z Modeling with Quadratic Functions Z Solving Quadratic Inequalities Z Modeling with Quadratic Regression
The graph of the squaring function h(x) 0002 x2 is shown in Figure 1 on page 204. Notice that h is an even function; that is, the graph of h is symmetric with respect to the y axis. Also, the lowest point on the graph is (0, 0). Let’s explore the effect of applying a sequence of basic transformations to the graph of h.
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Indicate how the graph of each function is related to the graph of h(x) 0002 x2. Discuss the symmetry of the graphs and find the highest or lowest point, whichever exists, on each graph.
ZZZ EXPLORE-DISCUSS 1
(A) f (x) 0002 (x 0004 3)2 0004 7 0002 x2 0004 6x 0003 2 (B) g(x) 0002 0.5(x 0003 2)2 0003 3 0002 0.5x2 0003 2x 0003 5 (C) m(x) 0002 0004(x 0004 4)2 0003 8 0002 0004x2 0003 8x 0004 8 (D) n(x) 0002 00043(x 0003 1)2 0004 1 0002 00043x2 0004 6x 0004 4
00045
h(x)
Z Graphing Quadratic Functions
5
Graphing the functions in Explore-Discuss 1 produces figures similar in shape to the graph of the squaring function in Figure 1. These figures are called parabolas. The functions that produced these parabolas are examples of the important class of quadratic functions, which we will now define. 5
x
Z DEFINITION 1 Quadratic Functions Z Figure 1 Squaring function
If a, b, and c are real numbers with a 0, then the function
h(x) 0002 x2.
f (x) 0002 ax2 0003 bx 0003 c is called a quadratic function and its graph is called a parabola. This is known as the general form of a quadratic function.
Because the expression ax2 0003 bx 0003 c represents a real number no matter what number we substitute for x, the domain of a quadratic function is the set of all real numbers. We will discuss methods for determining the range of a quadratic function later in this section. Typical graphs of quadratic functions are illustrated in Figure 2. y
Z Figure 2 Graphs of quadratic functions.
10
00045
f(x) 0002 x2 0004 4 (a)
10
10
5
000410
y
y
x
00045
5
x
000410
g(x) 0002 3x 2 0004 12x 0005 14 (b)
00045
5
x
000410
h(x) 0002 3 0004 2x 0004 x2 (c)
We will begin our detailed study of quadratic functions by examining some in a special form, which we will call the vertex form:* f (x) 0002 a(x 0004 h)2 0003 k *In Problem 75 of Exercises 3-4, you will be asked to show that any function of this form fits the definition of quadratic function in Definition 1.
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We’ll see where the name comes from in a bit. For now, refer to Explore-Discuss 1. Any function of this form is a transformation of the basic squaring function g(x) ⫽ x2, so we can use transformations to analyze the graph.
EXAMPLE
1
The Graph of a Quadratic Function Use transformations of g(x) ⫽ x2 to graph the function f (x) ⫽ 2(x ⫺ 3)2 ⫹ 4. Use your graph to determine the graphical significance of the constants 2, 3, and 4 in this function.
SOLUTION
Multiplying by 2 vertically stretches the graph by a factor of 2. Subtracting 3 inside the square moves the graph 3 units to the right. Adding 4 outside the square moves the graph 4 units up. The graph of f is shown in Figure 3, along with the graph of g(x) ⫽ x2.
y ⫽ x2
y 10
y ⫽ 2(x ⫺ 3)2 ⫹ 4
5
(3, 4)
⫺5
5
x
Z Figure 3
The lowest point on the graph of f is (3, 4), so h ⫽ 3 and k ⫽ 4 determine the key point where the graph changes direction. The constant a ⫽ 2 affects the width of the parabola. 0002 MATCHED PROBLEM 1
Use transformations of g(x) ⫽ x2 to graph the function f (x) ⫽ ⫺12(x ⫺ 2)2 ⫹ 5. Use your graph to determine the significance of the constants ⫺12, 2, and 5 in this function. 0002 Every parabola has a point where the graph reaches a maximum or minimum and changes direction. We will call that point the vertex of the parabola. Finding the vertex is key to many of the things we’ll do with parabolas. Example 1 and Explore-Discuss 1 demonstrate that if a quadratic function is in the form f (x) ⴝ a(x ⴚ h)2 ⴙ k, then the vertex is the point (h, k). Next, notice that the graph of h(x) ⫽ x2 is symmetric about the y axis. As a result, the transformation f (x) ⫽ 2(x ⫺ 3)2 ⫹ 4 is symmetric about the vertical line x ⫽ 3 (which runs through the vertex). We will call this vertical line of symmetry the axis, or axis of symmetry of a parabola. If the page containing the graph of f is folded along the line x ⫽ 3, the two halves of the graph will match exactly. Finally, Explore-Discuss 1 illustrates the significance of the constant a in f (x) ⫽ a(x ⫺ h)2 ⫹ k. If a is positive, the graph has a minimum and opens upward. But if a is negative, the graph will be a vertical reflection of h(x) ⫽ x2 and will have a maximum and open downward. The size of a determines the width of the parabola: if 冟 a 冟 7 1, the graph is narrower than h(x) ⫽ x2, and if 冟 a 冟 6 1, it is wider. These properties of a quadratic function in vertex form are summarized next.
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Z PROPERTIES OF A QUADRATIC FUNCTION AND ITS GRAPH Given a quadratic function in vertex form f (x) 0002 a(x 0004 h)2 0003 k
a 0
we summarize general properties as follows: 1. The graph of f is a parabola: f (x)
f (x)
Axis x0002h
Axis x0002h Vertex (h, k)
k
Max f(x)
Vertex (h, k) k
Min f (x) h
x
a00050 Opens upward
h
x
a00060 Opens downward
2. Vertex: (h, k) (parabola rises on one side of the vertex and falls on the other). 3. Axis (of symmetry): x 0002 h (parallel to y axis). 4. f (h) 0002 k is the minimum if a 7 0 and the maximum if a 6 0. 5. Domain: all real numbers; range: (00040007, k] if a 6 0 or [ k, 0007) if a 7 0. 6. The graph of f is the graph of g(x) 0002 ax2 translated horizontally h units and vertically k units.
Now that we can recognize the key properties of quadratic functions in vertex form, the obvious question is “What if a quadratic function is not in vertex form?” More often than not, the quadratic functions we will encounter will be in the form f(x) 0002 ax2 0003 bx 0003 c. The method of completing the square, which we studied in Section 1-5, can be used to find the vertex form in this case.
EXAMPLE
2
Finding the Vertex Form of a Parabola Find the vertex form of f(x) 0002 2x2 0004 8x 0003 4 by completing the square, then write the vertex and the axis.
SOLUTION
We will begin by separating the first two terms with parentheses; then we will complete the square to factor part of f as a perfect square. f (x) 0002 2x2 0004 8x 0003 4 0002 (2x2 0004 8x) 0003 4 0002 2(x2 0004 4x) 0003 4 0002 2(x2 0004 4x 0003 ?) 0003 4 0002 2(x2 0004 4x 0003 4) 0003 4 0004 8 0002 2(x 0004 2)2 0004 4
Group first two terms. Factor out 2. (b a)2 0002 (00042)2 0002 4 Add 4 inside parentheses; because of the 2 in front, we really added 8, so subtract 8 as well. Factor inside parentheses; simplify 4 0004 8.
The vertex form is f (x) 0002 2(x 0004 2)2 0004 4; the vertex is (2, 00044) and the axis is x 0002 2.
0002
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MATCHED PROBLEM 2
EXAMPLE
3
Quadratic Functions
207
Find the vertex form of g(x) 0002 3x2 0004 18x 0003 2 by completing the square, then write the vertex and axis. 0002
Graphing a Quadratic Function Let f(x) 0002 00040.5x2 0004 x 0003 2. (A) Use completing the square to find the vertex form of f. State the vertex and the axis of symmetry. (B) Graph f and find the maximum or minimum of f (x), the domain, the range, and the intervals where f is increasing or decreasing.
SOLUTIONS
(A) Complete the square: f (x) 0002 00040.5x2 0004 x 0003 2 0002 (00040.5x2 0004 x) 0003 2
Group first two terms Factor out 00040.5 Add 1 inside the parentheses to complete the square and 0.5 outside the parentheses.
0002 00040.5(x2 0003 2x 0003 ?) 0003 2 0002 00040.5(x2 0003 2x 0003 1) 0003 2 0003 0.5 0002 00040.5(x 0003 1)2 0003 2.5
Factor the trinomial and combine like terms.
From this last form we see that h 0002 00041 and k 0002 2.5, so the vertex is (00041, 2.5) and the axis of symmetry is x 0002 00041. (B) To graph f, locate the axis and vertex; then plot several points on either side of the axis Axis x 0002 00041
y 5
00045
Vertex (00041, 2.5)
5
00045
x
x
f(x)
00044
00042
00042
2
00041
2.5
0
2
2
00042
The domain of f is (00040007, 0007). From the graph we see that the maximum value is f(00041) 0002 2.5 and that f is increasing on (00040007, 00041] and decreasing on [00041, 0007). Also, y 0002 f(x) can be any number less than or equal to 2.5; the range of f is y 2.5 or (00040007, 2.5]. 0002 MATCHED PROBLEM 3
Let f (x) 0002 0004x2 0003 4x 0003 2. (A) Use completing the square to find the vertex form of f. State the vertex and the axis of symmetry. (B) Graph f and find the maximum or minimum of f (x), the domain, the range, and the intervals where f is increasing or decreasing. 0002
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We can develop a simple formula for finding the vertex of a parabola if we apply completing the square to f (x) ⫽ ax2 ⫹ bx ⫹ c. f (x) ⫽ ax2 ⫹ bx ⫹ c ⫽ a ax2 ⫹
Factor a out of the first two terms. Add a
b ⫹ ?b ⫹ c a
b 2 b inside the parentheses and 2a
subtract a a
b2 b b2 ⫹ 2b ⫹ c ⫺ a 4a 4a 2 2 b b ⫽ a ax ⫹ b ⫹ c ⫺ 2a 4a
⫽ a ax2 ⫹
b2 b 2 outside the parentheses. b ⴝ 4a 2a
Factor the trinomial.
This is in vertex form, and the x coordinate of the vertex is ⫺bⲐ2a. Z FINDING THE VERTEX OF A PARABOLA When a quadratic function is written in the form f (x) ⫽ ax2 ⫹ bx ⫹ c, the first coordinate of the vertex can be found using the formula x⫽⫺
b 2a
The second coordinate can then be found by evaluating f at the first coordinate.
EXAMPLE
4
Graphing a Quadratic Function Let f (x) ⫽ x2 ⫺ 6x ⫹ 4. (A) Use the vertex formula to find the vertex and the axis of symmetry of f. (B) Graph f and find the maximum or minimum of f (x), the domain, the range, and the intervals where f is increasing or decreasing.
SOLUTIONS
(A) Using a ⫽ 1 and b ⫽ ⫺6 in the vertex formula, x⫽⫺
b ⫺6 ⫽⫺ ⫽ 3; f (3) ⫽ 32 ⫺ 6(3) ⫹ 4 ⫽ ⫺5 2a 2
The vertex is (3, ⫺5) and the axis of symmetry is x ⫽ 3. (B) Locate the axis of symmetry, the vertex, and several points on either side of the axis of symmetry, and graph f. y
x
f (x)
0
4
2
⫺4
3
⫺5
4
⫺4
6
4
x⫽3 9
⫺2
8
⫺5
Vertex (3, ⫺5)
x
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The minimum of f(x) is 00045, the domain is (00040007, 0007), the range is [00045, 0007), f is decreasing on (00040007, 3] and increasing on [3, 0007).
0002
Let f (x) 0002 14 x2 0003 12x 0004 5.
MATCHED PROBLEM 4
(A) Use the vertex formula to find the vertex and the axis of symmetry of f. (B) Graph f and find the maximum or minimum of f (x), the domain, the range, and the intervals where f is increasing or decreasing. 0002
EXAMPLE
5
Finding the Equation of a Parabola Find the equation of the parabola with vertex (3, 00042) and x intercept 4. Since the vertex is (3, 00042), the vertex form for the equation is
SOLUTION
f (x) 0002 a(x 0004 3)2 0004 2
h 0002 3, k 0002 00042 in a(x 0004 h)2 0005 k
Since 4 is an x intercept, f (4) 0002 0. Substituting x 0002 4 and f (x) 0002 0 into the vertex formula, we have f (4) 0002 a(4 0004 3)2 0004 2 0002 0 a00022
Add 2 to both sides.
The equation of this parabola is f (x) 0002 2(x 0004 3)2 0004 2 0002 2x2 0004 12x 0003 16 Find the equation of the parabola with vertex (4, 00042) and y intercept 2.
MATCHED PROBLEM 5
0002
0002
We have presented two methods for locating the vertex of a parabola: completing the square and evaluating the vertex formula. You may prefer to use the completing the square process or to remember the formula. Unless directed otherwise, we will leave this choice to you. If you have a graphing calculator, there is a third approach.
Technology Connections The maximum and minimum options on the CALC menu of a graphing calculator can be used to find the vertex of a parabola. After selecting the appropriate option (maximum or minimum), you will be asked to provide three x values: a
left bound, a right bound, and a guess. The maximum or minimum is displayed at the bottom of the screen. Figure 4(a) locates the vertex of the parabola in Example 1 and Figure 4(b) locates the vertex of the parabola in Example 4. 10
5
00045
5
8
000410
00045
(a) f(x) 0002 00040.5x 0004 x 0005 2 2
Z Figure 4
00042
(b) f(x) 0002 x 2 0004 6x 0005 4
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Z Modeling with Quadratic Functions We will now look at some applications that can be modeled using quadratic functions.
EXAMPLE
6
Maximum Area A dairy farm has a barn that is 150 feet long and 75 feet wide. The owner has 240 feet of fencing and plans to use all of it in the construction of two identical adjacent outdoor pens, with part of the long side of the barn as one side of the pens, and a common fence between the two (Fig. 5). The owner wants the pens to be as large as possible.
150 feet
x x 75 feet
y
x
Z Figure 5
(A) Construct a mathematical model for the combined area of both pens in the form of a function A(x) (see Fig. 5) and state the domain of A. (B) Find the value of x that produces the maximum combined area. (C) Find the dimensions and the area of each pen. SOLUTIONS
(A) The combined area of the two pens is A 0002 xy Adding up the lengths of all four segments of fence, we find that building the pens will require 3x 0003 y feet of fencing. We have 240 feet of fence to use, so 3x 0003 y 0002 240 y 0002 240 0004 3x Because the distances x and y must be nonnegative, x and y must satisfy x 0 and y 0002 240 0004 3x 0. It follows that 0 x 80. Substituting for y in the combined area equation, we have the following model for this problem: A(x) 0002 x(240 0004 3x) 0002 240x 0004 3x2
0 x 80
(B) The function A(x) 0002 240x 0004 3x2 is a parabola that opens downward, so the maximum value of area will occur at the vertex. b 240 00020004 0002 40; 2a 2(00043) A(40) 0002 240(40) 0004 3(40)2 0002 4,800 x00020004
A value of x 0002 40 gives a maximum area of 4,800 square feet.
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(C) When x 0002 40, y 0002 240 0004 3(40) 0002 120. Each pen is x by y 2, or 40 feet by 60 feet. The area of each pen is 40 feet 60 feet 0002 2,400 square feet. 0002 MATCHED PROBLEM 6
Repeat Example 6 with the owner constructing three identical adjacent pens instead of two. 0002 The great sixteenth-century astronomer and physicist Galileo was the first to discover that the distance an object falls is proportional to the square of the time it has been falling. This makes quadratic functions a natural fit for modeling falling objects. Neglecting air resistance, the quadratic function h(t) 0002 h0 0004 16t2 represents the height of an object t seconds after it is dropped from an initial height of h0 feet. The constant 000416 is related to the force of gravity and is dependent on the units used. That is, 000416 only works for distances measured in feet and time measured in seconds. If the object is thrown either upward or downward, the quadratic model will also have a term involving t. (See Problems 93 and 94 in Exercises 3-4.)
EXAMPLE
7
Projectile Motion As a publicity stunt, a late-night talk show host drops a pumpkin from a rooftop that is 200 feet high. When will the pumpkin hit the ground? Round your answer to two decimal places.
SOLUTION
Because the initial height is 200 feet, the quadratic model for the height of the pumpkin is h(t) 0002 200 0004 16t2 Because h(t) 0002 0 when the pumpkin hits the ground, we must solve this equation for t. h(t) 0002 200 0004 16t2 0002 0 Add 16t 2 to both sides. 2 16t 0002 200 Divide both sides by 16. 200 t2 0002 0002 12.5 Take the square root of both sides. 16 t 0002 112.5 Only the positive solution is relevant. ⬇ 3.54 seconds
MATCHED PROBLEM 7
0002
A watermelon is dropped from a rooftop that is 300 feet high. When will the melon hit the ground? Round your answer to two decimal places. 0002
Z Solving Quadratic Inequalities Given a quadratic function f (x) 0002 ax2 0003 bx 0003 c, a 0, the zeros of f are the solutions of the quadratic equation ax2 0003 bx 0003 c 0002 0
(1)
(see Section 1-5). If the equal sign in equation (1) is replaced with 0006, 0005, , or , the result is a quadratic inequality in standard form. Just as was the case with linear inequalities (see Section 1-2), the solution set for a quadratic inequality is the subset of the real number line that makes the inequality a true statement. We can identify this subset by examining the graph of a quadratic function. We begin with a specific example and then generalize the results. The graph of f (x) 0002 x2 0004 2x 0004 3 0002 (x 0004 3)(x 0003 1)
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is shown in Figure 6. Information obtained from the graph is listed in Table 1. y
Table 1
5
f(x) > 0 f(x) > 0 (00040007, 00041) (00041, 3) (3, 0007) x )( )(
00045
5
00045
f(x) < 0
Z Figure 6
x
f (x)
00040007 0006 x 0006 00041
Positive
x 0002 00041
Zero
00041 0006 x 0006 3
Negative
x00023
Zero
30006x00060007
Positive
y 0002 f(x) 0002 x 0004 2x 0004 3 0002 (x 0004 3)(x 0003 1) 2
Because we now know where the output of f is positive, negative, and zero, we can use the graph or the table to solve a number of related inequalities involving f. For example, f (x) 7 0 on (00040007, 00041) 0006 (3, 0007)
and
f (x) 0 on [ 00041, 3]
The key steps in the preceding process are summarized in the box. Z SOLVING A QUADRATIC INEQUALITY 1. Write the inequality in standard form (a form where one side of the inequality defines a quadratic function f and the other side is 0). 2. Find the zeros of f. 3. Graph f and plot its zeros. 4. Use the graph to identify the intervals on the x axis that satisfy the original inequality.
EXAMPLE
8
Solving a Quadratic Inequality Solve: x2 0004 4x 14
SOLUTION
Step 1. Write in standard form. x2 0004 4x 14 x2 0004 4x 0004 14 0 f (x) 0002 x2 0004 4x 0004 14 0
y f(x) 0002 x2 0004 4x 0004 14
2
Step 2. Solve: f(x) 0002 x 0004 4x 0004 14 0002 0
10
Subtract 14 from both sides. Write using function notation. Standard form
Use the quadratic formula with a 0002 1, b 0002 00044, and c 0002 000414.
2
x0002 2 0004 3√2 000410
2 0003 3√2 x 10
(0, 000414) 000420
Z Figure 7
(4, 000414) (2, 000418)
0004b 2b 0004 4ac 2a
0004(00044) 2(00044)2 0004 4(1)(000414) 2(1) 4 172 4 612 0002 0002 2 2 0002 2 312
0002
Divide both terms in numerator by 2.
The zeros of f are 2 0004 312 ⬇ 00042.24 and 2 0003 312 ⬇ 6.24. Step 3. Plot these zeros, along with a few other points, and graph f (Figure 7).
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213
Step 4. We need to identify intervals where f (x) 0. From the graph we see that f(x) 0 for x 2 0004 312 and for x 2 0003 312. Returning to the original inequality, the solution to x2 0004 4x 14 MATCHED PROBLEM 8
EXAMPLE
9
(00040007, 2 0004 312] 0006 [2 0003 312, 0007)
is
Solve: x2 0003 6x 6
0002 0002
Break-Even, Profit, and Loss
Table 2 Price–Demand Data Table 2 contains price–demand data for a paint manufacturer. A linear regression model for this data is Weekly Sales (in gallons)
Price per Gallon
1,400
$43.00
2,550
$37.25
3,475
$32.60
4,856
$25.72
5,625
$21.88
6,900
$15.50
SOLUTIONS
p 0002 50 0004 0.005x
Price–demand equation
where x is the weekly sales (in gallons) and $p is the price per gallon. The manufacturer has weekly fixed costs of $58,500 and variable costs of $3.50 per gallon produced. (A) Find the weekly revenue function R and weekly cost function C as functions of the sales x. What is the domain of each function? (B) Graph R and C on the same coordinate axes and find the level of sales for which the company will break even. (C) Describe verbally and graphically the sales levels that result in a profit and those that result in a loss. (A) If x gallons of paint are sold weekly at a price of $p per gallon, then the weekly revenue is R 0002 xp 0002 x(50 0004 0.005x) 0002 50x 0004 0.005x2 Since the sales x and the price p cannot be negative, x must satisfy x000b0
and
p 0002 50 0004 0.005x 0
Subtract 50 from both sides.
0004 0.005x 000450 000450 x 0002 10,000 00040.005
Divide both sides by 00040.005 and reverse the inequality. Simplify.
The revenue function and its domain are R(x) 0002 50x 0004 0.005x2
0 x 10,000
The cost of producing x gallons of paint weekly is C(x) 0002 58,500 0003 3.5x
x000b0
Fixed costs 0005 $3.50 times number of gallons
(B) The graph of C is a line and the graph of R is a parabola opening downward. Using the vertex formula, b 50 00020004 0002 5,000 2a 2(00040.005) R(5,000) 0002 50(5,000) 0004 0.005(5,000)2 0002 125,000 x00020004
The vertex is (5,000, 125,000).
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After plotting a few points (Table 3), we sketch the graphs of R and C (Fig. 8). y
Table 3 x
R(x)
C(x)
150,000
0
0
58,500
5,000
125,000
76,000
10,000
0
93,500
y ⫽ R(x) ⫽ 50x ⫺ 0.005x2
100,000
y ⫽ C(x) ⫽ 58,500 ⫹ 3.5x 50,000
Break-even points (
Loss
Profit
)( 1,500
Loss )( ) 7,800 10,000
x
Z Figure 8 Profit when R ⬎ C; loss when R ⬍ C
The company breaks even if cost equals revenue: C(x) ⫽ R(x) 58,500 ⫹ 3.5x ⫽ 50x ⫺ 0.005x2 0.005x2 ⫺ 46.5x ⫹ 58,000 ⫽ 0
x⫽
46.5 ⫾ 246.52 ⫺ 4(0.005)(58,500) 2(0.005)
⫽
Use the quadratic formula with a ⴝ 0.005, b ⴝ ⴚ46.5, and c ⴝ 58,000.
46.5 ⫾ 1992.25 46.5 ⫾ 31.5 ⫽ 0.01 0.01
⫽ 1,500 or 7,800 Now we find the corresponding points on the graph: C(1,500) ⫽ R(1,500) ⫽ $63,750 C(7,800) ⫽ R(7,800) ⫽ $85,800 The graphs of C and R intersect at the points (1,500, 63,750) and (7,800, 85,800) (see Figure 8). These intersection points are called the break-even points. (C) If the company produces and sells between 1,500 and 7,800 gallons of paint weekly, then R ⬎ C and the company will make a profit. These sales levels are shown in blue in Figure 8. If it produces and sells between 0 and 1,500 gallons or between 7,800 and 10,000 gallons of paint, then R ⬍ C and the company will lose money. These sales levels are shown in red in Figure 8. 0002 MATCHED PROBLEM 9
Refer to Example 9. (A) Find the profit function P and state its domain. (B) Find the sales levels for which P(x) ⬎ 0 and those for which P(x) ⬍ 0. (C) Find the maximum profit and the sales level at which it occurs. 0002
Z Modeling with Quadratic Regression We obtained the linear model for the price–demand data in Example 9 by applying linear regression to the data in Table 2. Regression is not limited to just linear functions. In Example 10 we will use a quadratic model obtained by applying quadratic regression to a data set.
EXAMPLE
10
Stopping Distance Automobile accident investigators often use the length of skid marks to approximate the speed of vehicles involved in an accident. The skid mark length depends on a number of factors, including the make and weight of the vehicle, the road surface, and the road
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Table 4 Length of Skid Marks (in feet) Speed (mph)
Wet Asphalt
Dry Concrete
20
22
16
30
49
33
40
84
61
50
137
94
60
197
133
Quadratic Functions
215
conditions at the time of the accident. Investigators conduct tests to determine skid mark length for various vehicles under varying conditions. Some of the test results for a particular vehicle are listed in Table 4. Using the quadratic regression feature on a graphing calculator, (see the Technology Connections following this example) we find a model for the skid mark length on wet asphalt: L(x) 0002 0.06x2 0003 0.42x 0004 6.6 where x is speed in miles per hour. (A) Graph y 0002 L(x) and the data for skid mark length on wet asphalt on the same axes. (B) How fast (to the nearest mile) was the vehicle traveling if it left skid marks 100 feet long? y L(x) 0002 0.06x2 0003 0.42x 0004 6.6
(A) Skid mark length (feet)
SOLUTIONS
300
(60, 197) (50, 137) (40, 84) (30, 49) (20, 22)
50 10
80
x
Speed (mph)
(B) To approximate the speed from the skid mark length, we solve L(x) 0002 100 0.06x 0003 0.42x 0004 6.6 0002 100 0.06x2 0003 0.42x 0003 93.4 0002 0 2
x0002 0002
Use the quadratic formula.
0003(00030.42) 0005 2(00030.42)2 0003 4(0.06)(000393.4) 2(0.06) 0.42 0005 122.5924 0.12
x ⬇ 43 mph
MATCHED PROBLEM 10
Subtract 100 from both sides.
The negative root was discarded.
0002
A model for the skid mark length on dry concrete in Table 4 is M(x) 0002 0.035x2 0004 0.15x 0003 1.6 where x is speed in miles per hour. (A) Graph y 0002 L(x) and the data for skid mark length on dry concrete on the same axes. (B) How fast (to the nearest mile) was the vehicle traveling if it left skid marks 100 feet long? 0002
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Technology Connections we expand our library of functions, we will see that regression can be used to construct models involving these new functions.
Figure 9 shows three of the screens related to the construction of the quadratic model in Example 10 on a Texas Instruments TI-84 Plus. The use of regression to construct mathematical models is not limited to just linear and quadratic models. As
240
0
(a) Enter the data.
(b) Use the QuadReg option on a calculator.
80
0
(c) Graph the data and the model.
Z Figure 9
ANSWERS TO MATCHED PROBLEMS 1.
y
y 0002 x2 10
(2, 5) 000410
10
000410
x
1
y 0002 0004 2 (x 0004 2)2 0003 5
The 000412 makes the graph open downward and vertically shrinks it by a factor of 12, the 2 moves it 2 units right, and the 5 moves it 5 units up. 2. g(x) 0002 3(x 0004 3)2 0004 25; vertex: (3, 000425); axis: x 0002 3 3. (A) Vertex form: f (x) 0002 0004(x 0004 2)2 0003 6; vertex: (2, 6); axis of symmetry: x 0002 2. y (B) 7
00043
x00022
Vertex (2, 6)
7
x
00043
Max f (x) 0002 f (2) 0002 6; domain: (00040007, 0007); range: (00040007, 6]; increasing on (00040007, 2]; decreasing on [2, 0007)
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217
Quadratic Functions
4. (A) Vertex: (00041, 0004214); axis of symmetry: x 0002 00041 y (B) 10
000410
x
10
000410
Vertex (00041, 000421/4)
x 0002 00041
Skid mark length (feet)
Min f (x) 0002 f (00041) 0002 0004214; domain: (00040007, 0007); range: [ 0004214, 0007); decreasing on (00040007, 00041]; increasing on [00041, 0007) 5. y 0002 14(x 0004 4)2 0004 2 0002 0.25x2 0004 2x 0003 2 6. (A) A(x) 0002 (240 0004 4x)x, 0 x 60 (B) The maximum combined area of 3,600 ft.2 occurs at x 0002 30 feet. (C) Each pen is 30 feet by 40 feet with area 1,200 ft.2 7. 4.33 seconds 8. [00043 0004 115, 00043 0003 115] 9. (A) P(x) 0002 46.5x 0004 0.005x2 0004 58,500, 0 x 10,000 (B) Profit is positive for sales between 1,500 and 7,800 gallons per week and negative for sales less than 1,500 or for sales between 7,800 and 10,000. (C) The maximum profit is $49,612.50 at a sales level of 4,650 gallons. y M(x) 0002 0.035x2 0003 0.15x 0004 1.6 10. (A) 200
(60, 133) (50, 94) (40, 61)
50
(30, 33) (20, 16) 10
80
x
Speed (mph)
(B) 52 mph
3-4
Exercises
1. Describe the graph of any quadratic function. 2. How can you tell from a quadratic function whether its graph opens up or down?
In Problems 7–12, find the vertex and axis of the parabola, then draw the graph. 7. f (x) 0002 (x 0003 3)2 0004 4
3. True or False: Every quadratic function has a maximum. Explain.
3 2 9. f (x) 0002 0004ax 0004 b 0004 5 2
4. Using transformations, explain why the vertex of f (x) 0002 a(x 0004 h)2 0003 k is (h, k).
11. f (x) 0002 2(x 0003 10)2 0003 20
5. What information does the constant a provide about the graph of a function of the form f (x) 0002 ax2 0003 bx 0003 c? 6. Explain how to find the maximum or minimum value of a quadratic function.
8. f (x) 0002 (x 0003 2)2 0004 2 10. f (x) 0002 0004ax 0004
11 2 b 00033 2
1 12. f (x) 0002 0004 (x 0003 8)2 0003 12 2
In Problems 13–18, write a brief verbal description of the relationship between the graph of the indicated function and the graph of y 0002 x2. 13. f (x) 0002 (x 0004 2)2 0003 1
14. g(x) 0002 0004(x 0003 1)2 0004 2
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15. h(x) 0002 0004(x 0003 1)2
16. k(x) 0002 (x 0004 2)2
17. m(x) 0002 (x 0004 2)2 0004 3
18. n(x) 0002 0004(x 0003 1)2 0003 4
In Problems 19–24, match each graph with one of the functions in Problems 13–18.
y
24. 5
00045
5
x
y
19. 5
00045
00045
5
x
00045
y
20. 5
00045
5
25. f (x) 0002 x2 0004 4x 0003 5
26. g(x) 0002 x2 0004 6x 0003 1
27. h(x) 0002 0004x2 0004 2x 0004 3
28. k(x) 0002 0004x2 0004 10x 0003 3
29. m(x) 0002 2x2 0004 12x 0003 22 1 7 31. f (x) 0002 x2 0003 3x 0004 2 2 33. f (x) 0002 2x2 0004 24x 0003 90
30. n(x) 0002 3x2 0003 6x 0004 2 3 11 32. g(x) 0002 0004 x2 0003 9x 0003 2 2 34. g(x) 0002 3x2 0003 24x 0003 30
x
In Problems 35–46, use the formula x 0002 0004b 2a to find the vertex. Then write a description of the graph using all of the following words: axis, increases, decreases, range, and maximum or minimum. Finally, draw the graph.
00045
35. f (x) 0002 x2 0003 8x 0003 8
y
21.
In Problems 25–34, complete the square and find the vertex form of each quadratic function, then write the vertex and the axis and draw the graph.
36. f (x) 0002 x2 0003 10x 0003 10
5
37. f (x) 0002 0004x2 0004 7x 0003 4 00045
5
x
38. f (x) 0002 0004x2 0004 11x 0003 1 39. f (x) 0002 4x2 0004 18x 0003 25 40. f (x) 0002 5x2 0003 30x 0004 17 41. f (x) 0002 000410x2 0003 50x 0003 12
00045
42. f (x) 0002 00048x2 0004 24x 0003 16
y
22.
43. f(x) 0002 x2 0003 3x
5
44. f (x) 0002 4x 0004 x2 00045
5
x
y 5
00045
5
00045
46. f(x) 0002 0.4x2 0003 4x 0003 4 In Problems 47–60, solve and write the answer using interval notation.
00045
23.
45. f(x) 0002 0.5x2 0004 2x 0004 7
x
47. x2 0006 10 0004 3x
48. x2 0003 x 0006 12
49. x2 0003 21 0005 10x
50. x2 0003 7x 0003 10 0005 0
51. x2 8x
52. x2 0003 6x 0
53. x2 0003 5x 0
54. x2 4
55. x2 0003 1 0006 2x
56. x2 0003 25 0006 10x
57. x2 0006 3x 0004 3
58. x2 0003 3 0005 2x
59. x2 0004 1 4x
60. 2x 0003 2 0005 x2
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In Problems 61–68, find the standard form of the equation for the quadratic function whose graph is shown.
Quadratic Functions
219
y
66. 9
y
61. 5
(00041, 4)
(0, 5)
(3, 4)
00045
(5, 0) x
(00041, 0)
x
5
00045
(1, 00044)
5
00041
y
67.
00045 5
62.
y
(0, 2.5)
5
(5, 0)
(00041, 0) 00043
00045
(00043, 00041)
x
5
(00041, 00041)
00045
y
68.
(00042, 00044)
x
7
00045 5
y
63.
(0, 2.5)
5
(00041, 4)
(00045, 0)
(00041, 0)
00048
00045
x
5
00045
00045
In Problems 69–74, find the equation of a quadratic function whose graph satisfies the given conditions.
y 5
x
(1, 2)
(00043, 2)
64.
2
69. Vertex: (4, 8); x intercept: 6 70. Vertex: (00042, 000412); x intercept: 00044
(3, 3) (6, 0)
(0, 0) 00042
8
71. Vertex: (00044, 12); y intercept: 4 x
72. Vertex: (5, 8); y intercept: 00042 73. Vertex: (00045, 000425); additional point on graph: (00042, 20) 74. Vertex: (6, 000440); additional point on graph: (3, 50)
00045
75. For f (x) 0002 a(x 0004 h)2 0003 k, expand the parentheses and simplify to write in the form f (x) 0002 ax2 0003 bx 0003 c. This proves that any function in vertex form is a quadratic function as defined in Definition 1.
y
65. 5
(3, 0)
(00041, 0) 00045
5
(0, 00043) 00045
x
76. Find a general formula for the constant term c when expanding f (x) 0002 a(x 0004 h)2 0003 k into the form f (x) 0002 ax2 0003 bx 0003 c. 77. Let g(x) 0002 x2 0003 kx 0003 1. Graph g for several different values of k and discuss the relationship between these graphs. 78. Confirm your conclusions in Problem 77 by finding the vertex form for g.
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79. Let f(x) 0002 (x 0004 1)2 0003 k. Discuss the relationship between the values of k and the number of x intercepts for the graph of f. Generalize your comments to any function of the form f (x) 0002 a(x 0004 h)2 0003 k, a 7 0 80. Let f (x) 0002 0004(x 0004 2)2 0003 k. Discuss the relationship between the values of k and the number of x intercepts for the graph of f. Generalize your comments to any function of the form f (x) 0002 a(x 0004 h)2 0003 k, a 6 0
Horse barn 50 feet x Corral y
81. Find the minimum product of two numbers whose difference is 30. Is there a maximum product? Explain. 82. Find the maximum product of two numbers whose sum is 60. Is there a minimum product? Explain.
APPLICATIONS 83. PROFIT ANALYSIS A consultant hired by a small manufacturing company informs the company owner that their annual profit can be modeled by the function P(x) 0002 00041.2x2 0003 62.5x 0004 491, where x represents the number of employees and P is profit in thousands of dollars. How many employees should the company have to maximize annual profit? What is the maximum annual profit they can expect in that case? 84. PROFIT ANALYSIS The annual profits (in thousands of dollars) from 2000 to 2009 for the company in Problem 83 can be modeled by the function P(t) 0002 6.8t2 0004 80.5t 0003 427.3, 0 t 9, where t is years after 2000. How much profit did the company make in their worst year? 85. MOVIE INDUSTRY REVENUE The annual U.S. box office revenue in billions of dollars for a span of years beginning in 2002 can be modeled by the function B(x) 0002 00040.19x2 0003 1.2x 0003 7.6, 0 x 7, where x is years after 2002. (A) In what year was box office revenue at its highest in that time span? (B) Explain why you should not use the exact vertex in answering part A in this problem. 86. GAS MILEAGE The speed at which a car is driven can have a big effect on gas mileage. Based on EPA statistics for compact cars, the function m(x) 0002 00040.025x2 0003 2.45x 0004 30, 30 x 65, models the average miles per gallon for compact cars in terms of the speed driven x (in miles per hour). (A) At what speed should the owner of a compact car drive to maximize miles per gallon? (B) If one compact car has a 14-gallon gas tank, how much farther could you drive it on one tank of gas driving at the speed you found in part A than if you drove at 65 miles per hour? 87. CONSTRUCTION A horse breeder plans to construct a corral next to a horse barn that is 50 feet long, using all of the barn as one side of the corral (see the figure). He has 250 feet of fencing available and wants to use all of it. (A) Express the area A(x) of the corral as a function of x and indicate its domain. (B) Find the value of x that produces the maximum area. (C) What are the dimensions of the corral with the maximum area?
88. CONSTRUCTION Repeat Problem 87 if the horse breeder has only 140 feet of fencing available for the corral. Does the maximum value of the area function still occur at the vertex? Explain. Problems 89–92 use the falling object function described on page 211. 89. FALLING OBJECT A sandbag is dropped off a high-altitude balloon at an altitude of 10,000 ft. When will the sandbag hit the ground? 90. FALLING OBJECT A prankster drops a water balloon off the top of a 144-ft.-high building. When will the balloon hit the ground? 91. FALLING OBJECT A cliff diver hits the water 2.5 seconds after diving off the cliff. How high is the cliff? 92. FALLING OBJECT A forest ranger drops a coffee cup off a fire watchtower. If the cup hits the ground 1.5 seconds later, how high is the tower? 93. PROJECTILE FLIGHT An arrow shot vertically into the air reaches a maximum height of 484 feet after 5.5 seconds of flight. Let the quadratic function d(t) represent the distance above ground (in feet) t seconds after the arrow is released. (If air resistance is neglected, a quadratic model provides a good approximation for the flight of a projectile.) (A) Find d(t) and state its domain. (B) At what times (to two decimal places) will the arrow be 250 feet above the ground?
94. PROJECTILE FLIGHT Repeat Problem 93 if the arrow reaches a maximum height of 324 feet after 4.5 seconds of flight. 95. ENGINEERING The arch of a bridge is in the shape of a parabola 14 feet high at the center and 20 feet wide at the base (see the figure).
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h(x)
221
(B) Graph y ⫽ L(x) and the data for skid mark length on the same axes. (C) How fast (to the nearest mile per hour) was the car traveling if it left skid marks 150 feet long?
14 ft
x 20 ft
(A) Express the height of the arch h(x) in terms of x and state its domain. (B) Can a truck that is 8 feet wide and 12 feet high pass through the arch? (C) What is the tallest 8-ft.-wide truck that can pass through the arch? (D) What (to two decimal places) is the widest 12-ft.-high truck that can pass through the arch? 96. ENGINEERING The roadbed of one section of a suspension bridge is hanging from a large cable suspended between two towers that are 200 feet apart (see the figure). The cable forms a parabola that is 60 feet above the roadbed at the towers and 10 feet above the roadbed at the lowest point. 200 feet d(x)
Quadratic Functions
98. STOPPING DISTANCE (A) Use the quadratic regression feature on a graphing calculator to find a quadratic model M(x) for the skid mark length for Car B, where x is speed in miles per hour. (Round to two significant digits.) (B) Graph y ⫽ M(x) and the data for skid mark length on the same axes. (C) How fast (to the nearest mile) was the car traveling if it left skid marks 100 feet long? 99. ALCOHOL CONSUMPTION Table 6 contains data related to the per capita ethanol consumption in the United States from 1960 to 2000 (Source: NIAAA). A quadratic regression model for the per capita beer consumption is B(x) ⫽ ⫺0.0006x2 ⫹ 0.03x ⫹ 1 (A) If beer consumption continues to follow the trend exhibited in Table 6, when (to the nearest year) would the consumption return to the 1960 level? (B) What does this model predict for beer consumption in the year 2005? Use the Internet or a library to compare the predicted results with the actual results.
60 feet
Table 6 Per Capita Alcohol Consumption (in gallons) x feet
(A) Express the vertical distance d(x) (in feet) from the roadbed to the suspension cable in terms of x and state the domain of d. (B) The roadbed is supported by seven equally spaced vertical cables (see the figure). Find the combined total length of these supporting cables. 97. STOPPING DISTANCE Table 5 contains data related to the length of the skid marks left by two different cars when making emergency stops.
Year
Beer
Wine
1960
0.99
0.22
1970
1.14
0.27
1980
1.38
0.34
1990
1.34
0.33
2000
1.22
0.31
100. ALCOHOL CONSUMPTION Refer to Table 6. A quadratic regression model for the per capita wine consumption is W(x) ⫽ ⫺0.00016x2 ⫹ 0.009x ⫹ 0.2
Table 5 Length of Skid Marks (in feet)
Speed (mph)
Car A
Car B
20
26
38
30
45
62
40
73
102
50
118
158
60
171
230
(A) If wine consumption continues to follow the trend exhibited in Table 6, when (to the nearest year) would the consumption return to the 1960 level? (B) What does this model predict for wine consumption in the year 2005? Use the Internet or a library to compare the predicted results with the actual results. 101. PROFIT ANALYSIS A screen printer produces custom silkscreen apparel. The cost C(x) of printing x custom T-shirts and the revenue R(x) from the sale of x T-shirts (both in dollars) are given by C(x) ⫽ 245 ⫹ 1.6x R(x) ⫽ 10x ⫺ 0.04x2
(A) Use the quadratic regression feature on a graphing calculator to find a quadratic model L(x) for the skid mark length for Car A, where x is speed in miles per hour. (Round to two significant digits.)
Find the break-even points and determine the sales levels x (to the nearest integer) that will result in the printer showing a profit.
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102. PROFIT ANALYSIS Refer to Problem 101. Determine the sales levels x (to the nearest integer) that will result in the printer showing a profit of at least $60. 103. MAXIMIZING REVENUE A company that manufactures beer mugs has collected the price–demand data in Table 7. A linear regression model for this data is p 0002 d(x) 0002 9.3 0004 0.15x where x is the number of mugs (in thousands) that the company can sell at a price of $p. Find the price that maximizes the company’s revenue from the sale of beer mugs.
where x is the number of gallons of orange juice that can be sold at a price of $p. (A) Find the revenue and cost functions as functions of the sales x. What is the domain of each function? (B) Graph R and C on the same coordinate axes and find the sales levels for which the company will break even. (C) Describe verbally and graphically the sales levels that result in a profit and those that result in a loss. (D) Find the sales and the price that will produce the maximum profit. Find the maximum profit.
Table 9 Orange Juice
Table 7 Demand
Price
45,800
$2.43
40,500
$3.23
37,900
$3.67
34,700
$4.10
30,400
$4.74
28,900
$4.97
25,400
$5.49
Demand
Price
21,800
$1.97
24,300
$1.80
26,700
$1.63
28,900
$1.48
29,700
$1.42
33,700
$1.14
34,800
$1.06
104. MAXIMIZING REVENUE A company that manufactures inexpensive flash drives has collected the price–demand data in Table 8. A linear regression model for this data is
106. BREAK-EVEN ANALYSIS Table 10 contains weekly price– demand data for grapefruit juice for a fruit-juice producer. The producer has weekly fixed cost of $4,500 and variable cost of $0.15 per gallon of grapefruit juice produced. A linear regression model for the data in Table 10 is
p 0002 d(x) 0002 12.3 0004 0.15x
p 0002 d(x) 0002 3 0004 0.0003x
where x is the number of drives (in thousands) that the company can sell at a price of $p. Find the price that maximizes the company’s revenue from the sale of flash drives.
where x is the number of gallons of grapefruit juice that can be sold at a price of $p. (A) Find the revenue and cost functions as functions of the sales x. What is the domain of each function? (B) Graph R and C on the same coordinate axes and find the sales levels for which the company will break even. (C) Describe verbally and graphically the sales levels that result in a profit and those that result in a loss. (D) Find the sales and the price that will produce the maximum profit. Find the maximum profit.
Table 8 Demand
Price
47,800
$5.13
45,600
$5.46
42,700
$5.90
39,600
$6.36
34,700
$7.10
31,600
$7.56
27,800
$8.13
105. BREAK-EVEN ANALYSIS Table 9 contains weekly price– demand data for orange juice for a fruit-juice producer. The producer has weekly fixed cost of $24,500 and variable cost of $0.35 per gallon of orange juice produced. A linear regression model for the data in Table 9 is p 0002 d(x) 0002 3.5 0004 0.00007x
Table 10 Grapefruit Juice Demand
Price
2,130
$2.36
2,480
$2.26
2,610
$2.22
2,890
$2.13
3,170
$2.05
3,640
$1.91
4,350
$1.70
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223
Operations on Functions; Composition Z Performing Operations on Functions Z Composition Z Mathematical Modeling
Perhaps the most basic thing you’ve done in math classes is operations on numbers: things like addition, subtraction, multiplication, and division. In this section, we will explore the concept of operations on functions. In many cases, combining functions will enable us to model more complex and useful situations. If two functions f and g are both defined at some real number x, then f (x) and g(x) are both real numbers, so it makes sense to perform the four basic arithmetic operations with f(x) and g(x). Furthermore, if g(x) is a number in the domain of f, then it is also possible to evaluate f at g(x). We will see that operations on the outputs of the functions can be used to define operations on the functions themselves.
Z Performing Operations on Functions The functions f and g given by f (x) 0002 2x 0003 3 and g(x) 0002 x2 0004 4 are both defined for all real numbers. Note that f(3) 0002 9 and g(3) 0002 5, so it would seem reasonable to assign the value 9 0003 5, or 14, to a new function ( f 0003 g)(x). Based on this idea, for any real x we can perform the operation f(x) 0003 g(x) 0002 (2x 0003 3) 0003 (x2 0004 4) 0002 x2 0003 2x 0004 1 Similarly, we can define other operations on functions: f (x) 0004 g(x) 0002 (2x 0003 3) 0004 (x2 0004 4) 0002 0004x2 0003 2x 0003 7 f (x)g(x) 0002 (2x 0003 3)(x2 0004 4) 0002 2x3 0003 3x2 0004 8x 0004 12 For x 0005 00062 (to avoid zero in the denominator) we can also form the quotient f (x) 2x 0003 3 0002 2 g(x) x 00044
x 0005 00062
Notice that the result of each operation is a new function. So, we have ( f 0003 g)(x) 0002 f(x) 0003 g(x) 0002 x2 0003 2x 0004 1 ( f 0004 g)(x) 0002 f(x) 0004 g(x) 0002 0004x2 0003 2x 0003 7 ( fg)(x) 0002 f(x)g(x) 0002 2x3 0003 3x2 0004 8x 0004 12 f (x) f 2x 0003 3 0002 2 a b(x) 0002 g g(x) x 00044
x 0005 00062
Sum Difference Product
Quotient
The sum, difference, and product functions are defined for all values of x, as were the original functions f and g, but the domain of the quotient function must be restricted to exclude those values where g(x) 0002 0.
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Z DEFINITION 1 Operations on Functions The sum, difference, product, and quotient of the functions f and g are the functions defined by ( f ⫹ g)(x) ⫽ f (x) ⫹ g(x) ( f ⫺ g)(x) ⫽ f(x) ⫺ g(x) ( fg)(x) ⫽ f(x)g(x) f(x) f a b(x) ⫽ g g(x)
Sum function Difference function Product function
g(x) ⫽ 0
Quotient function
The domain of each function consists of all elements in the domains of both f and g, with the exception that the values of x where g(x) ⫽ 0 must be excluded from the domain of the quotient function.
ZZZ EXPLORE-DISCUSS 1
The following activities refer to the graphs of f and g shown in Figure 1 and the corresponding points on the graph shown in Table 1. Table 1
y
x
10
y ⫽ f (x)
y ⫽ g(x)
10
x
Z Figure 1
f(x)
g(x)
0
8
0
2
7
2
4
6
3
6
5
3
8
4
2
10
3
0
For each of the following functions, construct a table of values, sketch a graph, and state the domain and range. (A) ( f ⫹ g)(x)
EXAMPLE
1
(B) ( f ⫺ g)(x)
(C) ( fg)(x)
f (D) a b(x) g
Finding the Sum, Difference, Product, and Quotient Functions Let f (x) ⫽ 14 ⫺ x and g(x) ⫽ 13 ⫹ x. Find the functions f ⫹ g, f ⫺ g, fg, and f Ⲑg, and find their domains.
SOLUTION
( f ⫹ g)(x) ⫽ f (x) ⫹ g(x) ⫽ ( f ⫺ g)(x) ⫽ f (x) ⫺ g(x) ⫽ ( fg)(x) ⫽ f (x)g(x) ⫽ ⫽ ⫽
14 ⫺ x ⫹ 13 ⫹ x 14 ⫺ x ⫺ 13 ⫹ x 14 ⫺ x 13 ⫹ x 1(4 ⫺ x)(3 ⫹ x) 212 ⫹ x ⫺ x2
f (x) f 14 ⫺ x 4⫺x ⫽ a b(x) ⫽ ⫽ g g(x) 13 ⫹ x A 3 ⫹ x
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SECTION 3–5 Domain of f 00043
4
x
Domain of f: x 4 or (0004, 4] [Fig. 2(a)] Domain of g: x 00043 or [00043, ) [Fig. 2(b)]
(a)
Domain of g
[
4
0
x
The intersection of these domains is shown in Figure 2(c): (0004, 4] 傽 [00043, ) 0002 [00043, 4]
(b)
Domain of f 0003 g, f 0004 g, and fg
[
00043
225
The domains of f and g are [
0
00043
Operations on Functions; Composition
[ 4
0
x
This is the domain of the functions f 0003 g, f 0004 g, and fg. Since g(00043) 0002 0, x 0002 00043 must be excluded from the domain of the quotient function, and
(c)
Domain of
Z Figure 2
MATCHED PROBLEM 1
f : (00043, 4] g
0002
Let f (x) 0002 1x and g(x) 0002 110 0004 x. Find the functions f 0003 g, f 0004 g, fg, and f兾g, and find their domains. 0002
Technology Connections A graphing calculator can be used to check the domains in the solution of Example 1. To check the domain of f ⴙ g, we enter y1 ⴝ 14 ⴚ x, y2 ⴝ 13 ⴙ x, and y3 ⴝ y1 ⴙ y2 in the equation editor of a graphing calculator and graph y3 (Fig. 3).
5
00045
5
00045
5
Z Figure 5 00045
5
Figures 6 and 7 indicate that y3 is not defined for x 0002 4. This confirms that the domain of y3 ⴝ f ⴙ g is [ⴚ3, 4]. 5
00045
Z Figure 3 00045
Next we press TRACE and enter ⴚ3 (Fig. 4). Pressing the left cursor indicates that y3 is not defined for x 0003 ⴚ3 (Fig. 5).
5
00045
Z Figure 6 5
5
00045
5
00045
00045
00045
Z Figure 4
5
Z Figure 7
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2
Finding the Quotient of Two Functions Let f (x) 0002
SOLUTION
f x00044 x . Find the function and find its domain. and g(x) 0002 g x00041 x00033
Because division by 0 must be excluded, the domain of f is all x except x 0002 1 and the domain of g is all x except x 0002 00043. Now we find f兾g. x f (x) f x00041 a b(x) 0002 0002 g g (x) x00044 x00033 x x00033 0002 ⴢ x00041 x00044 x(x 0003 3) 0002 (x 0004 1)(x 0004 4)
(1)
The fraction in equation (1) indicates that 1 and 4 must be excluded from the domain of f兾g to avoid division by 0. But equation (1) does not indicate that 00043 must be excluded also. Although the fraction in equation (1) is defined at x 0002 00043, 00043 was excluded from the domain of g, so it must be excluded from the domain of f兾g also. The domain of f兾g is all real numbers x except 00043, 1, and 4. 0002 MATCHED PROBLEM 2
Let f (x) 0002
f 1 x00045 . Find the function and find its domain. and g (x) 0002 x g x00032
0002
Z Composition Consider the functions f and g given by f (x) 0002 1x and
g(x) 0002 4 0004 2x
Note that g(0) 0002 4 0004 2(0) 0002 4 and f(4) 0002 14 0002 2. So if we apply these two functions consecutively, we get f (g(0)) 0002 f (4) 0002 2 In a diagram, this would look like
x00020
g(x)
4
f (x)
2
When two functions are applied consecutively, we call the result the composition of functions. We will use the symbol f 0002 g to represent the composition of f and g, which we formally define now.
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227
Z DEFINITION 2 Composition of Functions The composition of a function f with another function g is denoted by f 0002 g (read “f composed with g”) and is defined by ( f 0002 g)(x) 0002 f(g(x))
EXAMPLE
3
Computing Composition From a Table Functions f and g are defined by Table 2. Find ( f 0002 g)(2), ( f 0002 g)(5), and ( f 0002 g)(00043). Table 2
SOLUTION
x
f(x)
g(x)
00045
00048
11
00043
00046
2
0
00041
00046
2
5
00043
5
12
0
We will use the formula provided by Definition 2. (f 0002 g)(2) 0002 f (g(2)) 0002 f(00043) 0002 00046 ( f 0002 g)(5) 0002 f (g(5)) 0002 f (0) 0002 00041 ( f 0002 g)(00043) 0002 f(g(00043)) 0002 f(2) 0002 5
MATCHED PROBLEM 3
0002
Functions h and k are defined by Table 3. Find (h 0002 k)(10), (h 0002 k)(00048), and (h 0002 k)(0). Table 3 x
h(x)
k(x)
00048
12
0
00044
18
22
0
40
00044
10
52
00048
20
70
000430
0002
ZZZ
CAUTION ZZZ
When computing f 0002 g, it’s important to keep in mind that the first function that appears in the notation ( f, in this case) is actually the second function that is applied. For this reason, some people read f 0002 g as “f following g.”
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ZZZ EXPLORE-DISCUSS 2
Refer to the functions f and g on page 226, and let h(x) ⫽ (f ⴰ g)(x). Complete Table 4 and graph h. Table 4 x 0
g(x)
h(x) ⴝ f(g(x))
g(0) ⫽ 4 h(0) ⫽ f (g(0)) ⫽ f (4) ⫽ 2
1 2 3 4
The domain of f is {x ƒ x ⱖ 0} and the domain of g is the set of all real numbers. What is the domain of h?
So far, we have looked at composition on a point-by-point basis. Using algebra, we can find a formula for the composition of two functions.
EXAMPLE
4
Finding the Composition of Two Functions Find ( f ⴰ g)(x) for f(x) ⫽ x2 ⫺ x and g(x) ⫽ 3 ⫹ 2x.
SOLUTION
We again use the formula in Definition 2. (f ⴰ g)(x) ⫽ f(g(x)) ⫽ f(3 ⫹ 2x) ⫽ (3 ⫹ 2x)2 ⫺ (3 ⫹ 2x) ⫽ 9 ⫹ 12x ⫹ 4x2 ⫺ 3 ⫺ 2x ⫽ 4x2 ⫹ 10x ⫹ 6
MATCHED PROBLEM 4
0002
Find (h ⴰ k)(x) for h(x) ⫽ 11 ⫹ x2 and k(x) ⫽ 4x ⫺ 1. 0002
ZZZ EXPLORE-DISCUSS 3
(A) For f (x) ⫽ x ⫺ 10 and g(x) ⫽ 3 ⫹ 7x, find ( f ⴰ g)(x) and (g ⴰ f )(x). Based on this result, what do you think is the relationship between f ⴰ g and g ⴰ f in general? x⫺1 . Does this change your thoughts 2 on the relationship between f ⴰ g and g ⴰ f ? (B) Repeat for f (x) ⫽ 2x ⫹ 1 and g(x) ⫽
Explore-Discuss 3 tells us that order is important in composition. Sometimes f ⴰ g and g ⴰ f are equal, but more often they are not. Finding the domain of a composition of functions can sometimes be a bit tricky. Based on the definition ( f ⴰ g)(x) ⫽ f (g(x)), we can see that for an x value to be in the domain of f ⴰ g, two things must occur. First, x must be in the domain of g so that g(x) is defined. Second, g(x) must be in the domain of f, so that f (g(x)) is defined.
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EXAMPLE
5
Operations on Functions; Composition
229
Finding the Composition of Two Functions Find ( f ⴰ g)(x) and (g ⴰ f )(x) and their domains for f (x) ⫽ x10 and g(x) ⫽ 3x4 ⫺ 1.
SOLUTION
( f ⴰ g)(x) ⫽ f (g(x)) ⫽ f(3x4 ⫺ 1) ⫽ (3x4 ⫺ 1)10 (g ⴰ f )(x) ⫽ g( f (x)) ⫽ g(x10) ⫽ 3(x10)4 ⫺ 1 ⫽ 3x40 ⫺ 1 Note that the functions f and g are both defined for all real numbers. If x is any real number, then x is in the domain of g, so g(x) is a real number. This then tells us that g(x) is in the domain of f, which means that f (g(x)) is a real number. In other words, every real number is in the domain of f ⴰ g. Using similar reasoning, we can conclude that the domain of g ⴰ f is also the set of all real numbers. 0002
MATCHED PROBLEM 5
3 Find ( f ⴰ g)(x) and (g ⴰ f )(x) and their domains for f (x) ⫽ 1 x and g(x) ⫽ 7x ⫹ 5.
0002 The line of reasoning used in Example 5 can be used to deduce the following fact: If two functions are both defined for all real numbers, then so is their composition. If either function in a composition is not defined for some real numbers, then, as Example 6 illustrates, the domain of the composition may not be what you first think it should be.
EXAMPLE
6
Finding the Composition of Two Functions Find ( f ⴰ g)(x) for f(x) ⫽ 24 ⫺ x2 and g(x) ⫽ 13 ⫺ x, then find the domain of f ⴰ g.
SOLUTION
We begin by stating the domains of f and g, which is a good idea in any composition problem: Domain f : ⫺2 ⱕ x ⱕ 2 Domain g: x ⱕ 3 or
or [⫺2, 2] (⫺⬁, 3]
Next we find the composition: ( f ⴰ g)(x) ⫽ f(g(x)) ⫽ f (13 ⫺ x) ⫽ 24 ⫺ (13 ⫺ x)2 ⫽ 24 ⫺ (3 ⫺ x) ⫽ 11 ⫹ x
Substitute 13 ⴚ x for g(x). Square: (1t)2 ⴝ t as long as t 0002 0. Subtract.
Although 11 ⫹ x is defined for all x ⱖ ⫺1, we must restrict the domain of f ⴰ g to those values that also are in the domain of g. Domain f ⴰ g: x ⱖ ⫺1 and x ⱕ 3
MATCHED PROBLEM 6
or
[⫺1, 3]
Find f ⴰ g for f (x) ⫽ 29 ⫺ x2 and g(x) ⫽ 1x ⫺ 1, then find the domain of f ⴰ g.
0002
0002
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The domain of f 0002 g cannot always be determined simply by examining the final form of ( f 0002 g)(x). Any numbers that are excluded from the domain of g must also be excluded from the domain of f 0002 g.
CAUTION ZZZ
In calculus, it is not only important to be able to find the composition of two functions, but also to recognize when a given function is the composition of simpler functions.
EXAMPLE
7
Recognizing Composition Forms Express h as a composition of two simpler functions for h(x) 0002 21 0003 3x4
SOLUTION
MATCHED PROBLEM 7
If we were to evaluate this function for some x value, say, x 0002 1, we would do so in two stages. First, we would find the value of 1 0003 3(1)4, which is 4. Then we would apply the square root to get 2. This shows that h can be thought of as two consecutive functions: First, g(x) 0002 1 0003 3x4, then f(x) 0002 1x. So h(x) 0002 f (g(x)), and we have written h as f 0002 g. 0002 Express h as the composition of two simpler functions for h(x) 0002 (4x3 0004 7)4.
0002
The answers to Example 7 and Matched Problem 7 are not unique. For example, if f(x) 0002 11 0003 3x and g(x) 0002 x4, then f (g(x)) 0002 21 0003 3g(x) 0002 21 0003 3x4 0002 h(x)
Z Mathematical Modeling The operations discussed in this section can be applied in many different situations. Example 8 shows how they are used to construct a model in economics.
EXAMPLE
8
Modeling Profit The research department for an electronics firm estimates that the weekly demand for a certain brand of headphones is given by x 0002 f( p) 0002 20,000 0004 1,000p
0 p 20
Demand function
This function describes the number x of pairs of headphones retailers are likely to buy per week at p dollars per pair. The research department also has determined that the total cost (in dollars) of producing x pairs per week is given by C(x) 0002 25,000 0003 3x
Cost function
and the total weekly revenue (in dollars) obtained from the sale of these headphones is given by R(x) 0002 20x 0004 0.001x2
Revenue function
Express the firm’s weekly profit as a function of the price p and find the price that produces the largest profit. What is the largest possible profit?
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SOLUTION
Operations on Functions; Composition
231
The basic economic principle we are using is that profit is revenue minus cost. So the profit function P is the difference of the revenue function R and the cost function C. P(x) ⫽ (R ⫺ C)(x) ⫽ R(x) ⫺ C(x) ⫽ (20x ⫺ 0.001x2) ⫺ (25,000 ⫹ 3x) ⫽ 17x ⫺ 0.001x2 ⫺ 25,000 This is a function of the demand x. We were asked to find the profit P as a function of the price p; we can accomplish this using composition, because x ⫽ f( p). (P ⴰ f )( p) ⫽ P( f ( p)) ⫽ P(20,000 ⫺ 1,000p) ⫽ 17(20,000 ⫺ 1,000p) ⫺ 0.001(20,000 ⫺ 1,000p)2 ⫺ 25,000 ⫽ 340,000 ⫺ 17,000p ⫺ 400,000 ⫹ 40,000p ⫺ 1,000p2 ⫺ 25,000 ⫽ ⫺85,000 ⫹ 23,000p ⫺ 1,000p2 Technically, P ⴰ f and P are different functions, because the first has independent variable p and the second has independent variable x. However, because both functions represent the same quantity (the profit), it is customary to use the same symbol to name each function. So P( p) ⫽ ⫺85,000 ⫹ 23,000p ⫺ 1,000p2 expresses the weekly profit P as a function of price p. Now we can use the vertex formula to find the maximum. p⫽⫺
23,000 b ⫽⫺ ⫽ 11.5 2a ⫺2,000
P(11.5) ⫽ ⫺85,000 ⫹ 23,000(11.5) ⫺ 1,000(11.5)2 ⫽ 47,250 Since a ⬍ 0, the parabola opens downward, and the maximum value of P occurs at the vertex. So the largest profit is $47,250 and it will occur when the price of the headphones is $11.50. 0002
MATCHED PROBLEM 8
Repeat Example 8 for the functions x ⫽ f( p) ⫽ 10,000 ⫺ 1,000p 0 ⱕ p ⱕ 10 C(x) ⫽ 10,000 ⫹ 2x R(x) ⫽ 10x ⫺ 0.001x2
ANSWERS TO MATCHED PROBLEMS 1. ( f ⫹ g)(x) ⫽ 1x ⫹ 110 ⫺ x, ( f ⫺ g)(x) ⫽ 1x ⫺ 110 ⫺ x, ( fg)(x) ⫽ 210x ⫺ x2, ( fⲐg)(x) ⫽ 1xⲐ(10 ⫺ x); the functions f ⫹ g, f ⫺ g, and fg have domain:[0, 10] , the domain of fⲐg is [0, 10) f x 2. a b(x) ⫽ ; domain: all real numbers x except ⫺2, 0, and 5 g (x ⫹ 2)(x ⫺ 5) 3. (h ⴰ k)(10) ⫽ 12; (h ⴰ k)(⫺8) ⫽ 40; (h ⴰ k)(0) ⫽ 18 4. (h ⴰ k)(x) ⫽ 16x2 ⫺ 8x ⫹ 12 3 5. ( f ⴰ g)(x) ⫽ 1 7x ⫹ 5, domain: (⫺⬁, ⬁) 3 (g ⴰ f )(x) ⫽ 71 x ⫹ 5, domain: (⫺⬁, ⬁) 6. ( f ⴰ g)(x) ⫽ 110 ⫺ x; domain: x ⱖ 1 and x ⱕ 10 or [1, 10] 7. h(x) ⫽ ( f ⴰ g)(x) where f (x) ⫽ x4 and g(x) ⫽ 4x3 ⫺ 7 8. P( p) ⫽ ⫺30,000 ⫹ 12,000p ⫺ 1,000p2. The largest profit is $6,000 and occurs when the price is $6.
0002
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Exercises 27. Functions f and g are defined by Table 5. Find ( f ° g)(00027), ( f ° g)(0), and ( f ° g)(4).
1. Explain how to find the sum of two functions. 2. Explain how to find the product of two functions. 3. Describe in your own words what the composition of two functions means. Don’t focus on how to find composition, but rather on what it really means. 4. Is the domain of f0005g always the same as the intersection of the domains of f and g? Explain.
28. Functions h and k are defined by Table 6. Find ( h ° k)(000215), ( h ° k)(000210), and ( h ° k)(15).
Table 5 x
Table 6 f (x)
g(x)
x
h (x)
k(x)
5. When composing two functions, why can’t you always find the domain by simply looking at the simplified form of the composition?
00027
5
4
000220
0002100
30
6. Describe a real-world situation where the composition of two functions would have significance.
00022
9
10
000215
0002200
5
0
0
00022
000210
0002300
15
4
3
6
5
0002150
8
6
000210
00023
15
000290
000210
Problems 7–18 refer to functions f and g whose graphs are shown below. f(x)
g (x)
5
5
00025
5
x
00025
In Problems 29–42, for the indicated functions f and g, find the functions f 0003 g, f 0002 g, fg, and f兾g, and find their domains.
5
x
29. f (x) 0004 4x;
g(x) 0004 x 0003 1
30. f (x) 0004 3x;
g(x) 0004 x 0002 2
2
31. f (x) 0004 2x ; 00025
00025
In Problems 7–10 use the graphs of f and g to construct a table of values and sketch the graph of the indicated function. 7. ( f 0003 g)(x) 9. ( fg)(x)
8. (g 0002 f )(x) 10. ( f 0002 g)(x)
32. f(x) 0004 3x;
g(x) 0004 x2 0003 1 g(x) 0004 x2 0003 4
33. f(x) 0004 3x 0003 5;
g(x) 0004 x2 0002 1
34. f(x) 0004 2x 0002 7;
g(x) 0004 9 0002 x2
35. f (x) 0004 12 0002 x; g(x) 0004 1x 0003 3 36. f (x) 0004 1x 0003 4; g(x) 0004 13 0002 x 37. f (x) 0004 1x 0003 2; g(x) 0004 1x 0002 4
In Problems 11–18, use the graphs of f and g to find each of the following: 11. ( f ° g)(00021)
12. ( f ° g)(2)
13. ( g ° f )(00022)
14. ( g ° f )(3)
15. f (g(1))
16. f(g(0))
17. g( f (2))
18. g( f (00023))
38. f (x) 0004 1 0002 1x; g(x) 0004 2 0002 1x 39. f (x) 0004 2x2 0003 x 0002 6; g(x) 0004 27 0003 6x 0002 x2 40. f (x) 0004 28 0003 2x 0002 x2; g(x) 0004 2x2 0002 7x 0003 10 1 1 41. f (x) 0004 x 0003 ; g(x) 0004 x 0002 x x 42. f (x) 0004 x 0002 1; g(x) 0004 x 0002
In Problems 19–26, find the indicated function value, if it exists, given f(x) 0004 2 0002 x and g(x) 0004 13 0002 x. 19. ( f 0003 g)(00023)
20. (g 0002 f )(00025)
23. ( f ° g)(00022)
f 22. a b(3) g 24. ( f ° g)(1)
25. (g ° f )(1)
26. (g ° g)(00027)
21. ( fg)(00021)
6 x00021
In Problems 43–60, for the indicated functions f and g, find the functions f ° g, and g ° f, and find their domains. 43. f(x) 0004 x3;
g(x) 0004 x2 0002 x 0003 1
44. f(x) 0004 x2;
g(x) 0004 x3 0003 2x 0003 4
45. f(x) 0004 |x 0003 1|;
g(x) 0004 2x 0003 3
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46. f (x) 0002 |x 0004 4|;
y
g(x) 0002 3x 0003 2
100073
47. f (x) 0002 x
3
; g(x) 0002 2x 0003 4
200073
48. f (x) 0002 x
y
5
5
3
; g(x) 0002 8 0004 x
49. f (x) 0002 1x; g(x) 0002 x 0004 4
00045
50. f (x) 0002 1x; g(x) 0002 2x 0003 5 51. f (x) 0002 x 0003 2; g(x) 0002
1 x
52. f (x) 0002 x 0004 3; g(x) 0002
1 x2
5
x
00045
00045
x
(d)
In Problems 65–72, find f ° g and g ° f. Graph f, g, f ° g, and g ° f in the same coordinate system and describe any apparent symmetry between these graphs.
54. f (x) 0002 1x 0004 1; g(x) 0002 x2 55. f (x) 0002
x00035 ; x
56. f (x) 0002
x 2x 0004 4 ; g(x) 0002 x x00041
66. f (x) 0002 3x 0003 2; g(x) 0002 13 x 0004 23
57. f (x) 0002
2x 0003 1 1 ; g(x) 0002 x x00042
68. f (x) 0002 00042x 0003 3; g(x) 0002 000412 x 0003 32
2 2 0004 3x ; g(x) 0002 x x00033
69. f (x) 0002
g(x) 0002
5
00045
(c)
53. f (x) 0002 14 0004 x; g(x) 0002 x2
58. f (x) 0002
233
Operations on Functions; Composition
x x00042
65. f (x) 0002 12 x 0003 1; g(x) 0002 2x 0004 2 67. f (x) 0002 000423 x 0004 53; g(x) 0002 000432 x 0004 52
3 g(x) 0002 22 x
3 70. f (x) 0002 3 2x; g(x) 0002
59. f (x) 0002 225 0004 x2; g(x) 0002 29 0003 x2 60. f (x) 0002 2x2 0004 9; g(x) 0002 2x2 0003 25
x3 27
3 71. f (x) 0002 2x 0004 2; g(x) 0002 x3 0003 2
Use the graphs of functions f and g shown below to match each function in Problems 61–64 with one of graphs (a)–( d). y 0002 f (x) y 5
x3 ; 8
3 72. f (x) 0002 x3 0004 3; g(x) 0002 2x 0003 3
In Problems 73–80, express h as a composition of two simpler functions f and g. 73. h(x) 0002 (2x 0004 7)4
y 0002 g (x)
74. h(x) 0002 (3 0004 5x)7 00045
5
75. h(x) 0002 14 0003 2x
x
76. h(x) 0002 13x 0004 11 77. h(x) 0002 3x7 0004 5
00045
78. h(x) 0002 5x6 0003 3
61. ( f 0003 g)(x)
62. ( f 0004 g)(x)
63. ( g 0004 f )(x)
64. ( fg)(x)
y
79. h(x) 0002
80. h(x) 0002 0004 y
5
5
00045
x
82. Are the functions f ° g and g ° f identical? Justify your answer.
00045
5
x
83. Is there a function g that satisfies f ° g 0002 g ° f 0002 f for all functions f ? If so, what is it? 84. Is there a function g that satisfies fg 0002 gf 0002 f for all functions f ? If so, what is it?
00045
(a)
2 00031 1x
81. Are the functions fg and gf identical? Justify your answer.
5
00045
4 00033 1x
(b)
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In Problems 85–88, for the indicated functions f and g, find the functions f 0003 g, f 0004 g, fg, and f0007g, and find their domains. 1 1 85. f (x) 0002 x 0003 ; g(x) 0002 x 0004 x x 86. f (x) 0002 x 0004 1; g(x) 0002 x 0004 87. f (x) 0002 1 0004
6 x00041
x x ; g(x) 0002 1 0003 冟x冟 冟x冟
92. WEATHER BALLOON A weather balloon is rising vertically. An observer is standing on the ground 100 meters from the point where the weather balloon was released. (A) Express the distance d between the balloon and the observer as a function of the balloon’s distance h above the ground. (B) If the balloon’s distance above ground after t seconds is given by h 0002 5t, express the distance d between the balloon and the observer as a function of t. 93. FLUID FLOW A conical paper cup with diameter 4 inches and height 4 inches is initially full of water. A small hole is made in the bottom of the cup and the water begins to flow out of the cup. Let h and r be the height and radius, respectively, of the water in the cup t minutes after the water begins to flow.
88. f (x) 0002 x 0003 冟 x 冟; g(x) 0002 x 0004 冟 x 冟
APPLICATIONS
4 inches
89. MARKET RESEARCH The demand x and the price p (in dollars) for new release CDs for a large online retailer are related by x 0002 f ( p) 0002 4,000 0004 200p
0 p 20
The revenue (in dollars) from the sale of x units is given by R(x) 0002 20x 0004
r
1 2 x 200
4 inches h
and the cost (in dollars) of producing x units is given by C(x) 0002 2x 0003 8,000 Express the profit as a function of the price p and find the price that produces the largest profit. 90. MARKET RESEARCH The demand x and the price p (in dollars) for portable iPod speakers at a national electronics store are related by x 0002 f(p) 0002 5,000 0004 100p
0 p 50
The revenue (in dollars) from the sale of x units and the cost (in dollars) of producing x units are given, respectively, by R(x) 0002 50x 0004
1 2 x 100
C(x) 0002 20x 0003 40,000
and
Express the profit as a function of the price p and find the price that produces the largest profit.
1
V 0002 3 r 2h
(A) Express r as a function of h. (B) Express the volume V as a function of h. (C) If the height of the water after t minutes is given by h(t) 0002 4 0004 0.51t express V as a function of t. 94. EVAPORATION A water trough with triangular ends is 6 feet long, 4 feet wide, and 2 feet deep. Initially, the trough is full of water, but due to evaporation, the volume of the water is decreasing. Let h and w be the height and width, respectively, of the water in the tank t hours after it began to evaporate.
91. POLLUTION An oil tanker aground on a reef is leaking oil that forms a circular oil slick about 0.1 foot thick (see the figure). The radius of the slick (in feet) t minutes after the leak first occurred is given by r(t) 0002 0.4t100073 Express the volume of the oil slick as a function of t.
r
4 feet 6 feet
2 feet
w h
V 0002 3wh
(A) Express w as a function of h. (B) Express V as a function of h. (C) If the height of the water after t hours is given by h(t) 0002 2 0004 0.21t express V as a function of t.
A 0002 r 2 V 0002 0.1A
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3-6
Inverse Functions
235
Inverse Functions Z One-to-One Functions Z Finding the Inverse of a Function Z Mathematical Modeling Z Graphing Inverse Functions
We have seen that many important mathematical relationships can be expressed in terms of functions. For example, C 0002 d
The circumference of a circle is a function of the diameter d.
V 0002 s3 d 0002 1,000 0004 100p 9 F 0002 C 0003 32 5
The volume of a cube is a function of length s of the edges. The demand for a product is a function of the price p. Temperature measured in °F is a function of temperature in °C.
In many cases, we are interested in reversing the correspondence determined by a function. For our examples, C 3 s0002 1 V
d0002
p 0002 10 0004
The diameter of a circle is a function of the circumference C. The length of the edge of a cube is a function of the volume V.
1 d 100
5 C 0002 (F 0004 32) 9
The price of a product is a function of the demand d.
Temperature measured in °C is a function of temperature in °F.
As these examples illustrate, reversing the correspondence between two quantities often produces a new function. This new function is called the inverse of the original function. Later in this text we will see that many important functions are actually defined as the inverses of other functions. In this section, we develop techniques for determining whether the inverse of a function exists, some general properties of inverse functions, and methods for finding the rule of correspondence that defines the inverse function. A review of function basics in Section 3-1 would be very helpful at this point.
Z One-to-One Functions Recall the set form of the definition of function: A function is a set of ordered pairs with the property that no two ordered pairs have the same first component and different second components. However, it is possible that two ordered pairs in a function could have the same second component and different first components. If this does not happen, then we call the function a one-to-one function. In other words, a function is one-to-one if there are no duplicates among the second components.
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Z DEFINITION 1 One-to-One Function A function is one-to-one if no two ordered pairs in the function have the same second component and different first components.
ZZZ EXPLORE-DISCUSS 1
Given the following sets of ordered pairs: f 0002 5(0, 1), (0, 2), (1, 1), (1, 2)6 g 0002 5(0, 1), (1, 1), (2, 2), (3, 2)6 h 0002 5(0, 1), (1, 2), (2, 3), (3, 0)6 (A) Which of these sets represent functions? (B) Which of the functions are one-to-one functions? (C) For each set that is a function, form a new set by reversing each ordered pair in the set. Which of these new sets represent functions? (D) What do these results tell you about the result of reversing the ordered pairs for functions that are one-to-one, and for functions that are not one-to-one?
Explore-Discuss 1 illustrates an important idea that we will examine later: Only oneto-one functions have inverses.
EXAMPLE
1
Determining Whether a Function Is One-to-One Determine whether f is a one-to-one function for (A) f (x) 0002 x2
SOLUTIONS
(B) f (x) 0002 2x 0004 1
(A) To show that a function is not one-to-one, all we have to do is find two different ordered pairs in the function with the same second component and different first components. Because f (2) 0002 22 0002 4
and
f (00042) 0002 (00042)2 0002 4
the ordered pairs (2, 4) and (00042, 4) both belong to f, and f is not one-to-one. (Note that there’s nothing special about 2 and 00042 here: Any real number and its negative can be used in the same way.) (B) To show that a function is one-to-one, we have to show that no two ordered pairs have the same second component and different first components. To do this, we’ll show that if any two ordered pairs (a, f (a)) and (b, f(b)) in f have the same second components, then the first components must also be the same. That is, we show that f (a) 0002 f (b) implies a 0002 b. We proceed as follows: f (a) 0002 f (b) 2a 0004 1 0002 2b 0004 1 2a 0002 2b a0002b
Assume second components are equal. Evaluate f(a) and f(b).
Simplify. Conclusion: f is one-to-one.
By Definition 1, f is a one-to-one function.
0002
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MATCHED PROBLEM 1
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237
Determine whether f is a one-to-one function for (A) f (x) 0002 4 0004 x2
(B) f (x) 0002 4 0004 2x
0002
The methods used in the solution of Example 1 can be stated as a theorem.
Z THEOREM 1 One-to-One Functions 1. If f (a) 0002 f (b) for at least one pair of domain values a and b, a 0005 b, then f is not one-to-one. 2. If the assumption f (a) 0002 f (b) always implies that the domain values a and b are equal, then f is one-to-one.
Applying Theorem 1 is not always easy—try testing f (x) 0002 x 3 0003 2x 0003 3, for example. (Good luck!) However, the graph of a function can help us develop a simple procedure for determining if a function is one-to-one. If any horizontal line intersects the graph in more than one point [as shown in Fig. 1(a)], then there is a second component (height) that corresponds to two different first components (x values). This shows that the function is not one-to-one. On the other hand, if every horizontal line intersects the graph in just one point or not at all [as shown in Fig. 1(b)], the function is one-to-one. These observations form the basis of the horizontal line test. y
y
y 0002 f (x) (a, f (a))
(b, f (b))
(a, f (a))
y 0002 f (x) a
b
f(a) ⴝ f(b) for a 0005 b f is not one-to-one (a)
x
a
x
Only one point has second component f (a); f is one-to-one (b)
Z Figure 1 Intersections of graphs and horizontal lines.
Z THEOREM 2 Horizontal Line Test A function is one-to-one if and only if every horizontal line intersects the graph of the function in at most one point.
The graphs of the functions considered in Example 1 are shown in Figure 2 on page 238. Applying the horizontal line test to each graph confirms the results we obtained in Example 1. A function that is increasing throughout its domain or decreasing throughout its domain will always pass the horizontal line test [Figs. 3(a) and 3(b)]. This gives us the following theorem.
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FUNCTIONS y
y 5
5
(00042, 4)
(2, 4)
00045
00045
5
5
x
x 00045
f(x) ⴝ 2x 0006 1 passes the horizontal line test; f is one-to-one (b)
f(x) ⴝ x2 does not pass the horizontal line test; f is not one-to-one (a)
Z Figure 2 Applying the horizontal line test.
Z THEOREM 3 Increasing and Decreasing Functions If a function f is increasing throughout its domain or decreasing throughout its domain, then f is a one-to-one function.
y
y
y
x
x
An increasing function is always one-to-one (a)
A decreasing function is always one-to-one (b)
x
A one-to-one function is not always increasing or decreasing (c)
Z Figure 3 Increasing, decreasing, and one-to-one functions.
Figure 3(c) shows that a function can still be one-to-one even if it is neither increasing nor decreasing. The function illustrated is increasing on [ 0004, 0] and decreasing on (0, ).
Z Finding the Inverse of a Function Now we will demonstrate how we can form a new function by reversing the correspondence determined by a given function. Let g be the function defined as follows: g 0002 5(00043, 9), (0, 0), (3, 9)6
g is not one-to-one.
Notice that g is not one-to-one because the domain elements 00043 and 3 both correspond to the range element 9. We can reverse the correspondence determined by function g simply by reversing the components in each ordered pair in g, producing the following set: G 0002 5(9, 00043), (0, 0), (9, 3)6
G is not a function.
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But the result is not a function because the domain element 9 corresponds to two different range elements, ⫺3 and 3. On the other hand, if we reverse the ordered pairs in the function f ⫽ 5(1, 2), (2, 4), (3, 9)6
f is one-to-one; all second components are distinct.
we obtain F ⫽ 5(2, 1), (4, 2), (9, 3)6
F is a function.
This time f is a one-to-one function, and the set F turns out to be a function also. This new function F, formed by reversing all the ordered pairs in f, is called the inverse of f and is usually denoted by f ⫺1 (this is read as “inverse f ” or “the inverse of f ”): f ⫺1 ⫽ 5(2, 1), (4, 2), (9, 3)6
The inverse of f
Notice that f ⫺1 is also a one-to-one function and that the following relationships hold: Domain of f ⫺1 ⫽ 52, 4, 96 ⫽ Range of f Range of f ⫺1 ⫽ 51, 2, 36 ⫽ Domain of f We conclude that reversing all the ordered pairs in a one-to-one function forms a new one-to-one function and reverses the domain and range in the process. We are now ready to present a formal definition of the inverse of a function.
Z DEFINITION 2 Inverse of a Function If f is a one-to-one function, then the inverse of f, denoted f ⫺1, is the function formed by reversing all the ordered pairs in f. That is, f ⫺1 ⫽ 5( y, x) | (x, y) is in f } If f is not one-to-one, then f does not have an inverse and f ⫺1 does not exist.
ZZZ
CAUTION ZZZ
Be careful not to confuse inverse notation and reciprocal notation. For numbers, a 1 superscript of ⫺1 means reciprocal: 2⫺1 ⫽ . For functions, a superscript of ⫺1 2 1 means inverse: f ⫺1(x) is the inverse of f (x), which is not the same as . f (x)
The following properties of inverse functions follow directly from the definition.
Z THEOREM 4 Properties of Inverse Functions For a given function f, if f ⫺1 exists, then 1. f ⫺1 is a one-to-one function. 2. The domain of f ⫺1 is the range of f. 3. The range of f ⫺1 is the domain of f.
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ZZZ EXPLORE-DISCUSS 2
(A) For the function f 0002 5(3, 5), (7, 11), (11, 17)6, find f 00041.
(B) What do you think would be the result of composing f with f 00041? Justify your answer using Definition 2. (C) Check your conjecture from part B by finding both f 0002 f 00041 and f 00041 0002 f. Were you correct?
Explore-Discuss 2 brings up an important point: If you apply a function to any number in its domain, then apply the inverse of that function to the result, you’ll get right back where you started. This leads to the following theorem. Z THEOREM 5 Inverse Functions and Composition If f 00041 exists, then 1. f( f 00041(x)) 0002 x for all x in the domain of f 00041. 2. f 00041( f (x)) 0002 x for all x in the domain of f. If f and g are one-to-one functions satisfying f(g(x)) 0002 x for all x in the domain of g and g( f (x)) 0002 x for all x in the domain of f then f and g are inverses of one another.
We can use Theorem 5 to see if two functions defined by equations are inverses.
EXAMPLE
2
Deciding If Two Functions Are Inverses Use Theorem 5 to decide if these two functions are inverses. f (x) 0002 3x 0004 7
SOLUTION
g(x) 0002
x00037 3
The domain of both functions is all real numbers. For any x, f (g(x)) 0002 f a 0002 3a
x00037 b 3
Substitute into f(x).
x00037 b00047 3
Multiply.
0002x0003700047 0002x g( f(x)) 0002 g(3x 0004 7) 3x 0004 7 0003 7 0002 3 0002
3x 3
Add.
Substitute into g(x). Add.
Simplify.
0002x By Theorem 5, f and g are inverses.
0002
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MATCHED PROBLEM 2
Inverse Functions
241
Use Theorem 5 to decide if these two functions are inverses. 2 f(x) 0002 (11 0004 x) 5
5 g(x) 0002 0004 x 0003 11 2 0002
There is one obvious question that remains: when a function is defined by an equation, how can we find the inverse? Given a function y 0002 f(x), the first coordinates of points on the graph are represented by x, and the second coordinates are represented by y. Finding the inverse by reversing the order of the coordinates would then correspond to switching the variables x and y. This leads us to the following procedure, which can be applied whenever it is possible to solve y 0002 f (x) for x in terms of y. Z FINDING THE INVERSE OF A FUNCTION f Step 1. Find the domain of f and verify that f is one-to-one. If f is not one-to-one, then stop, because f 00041 does not exist. Step 2. If the function is written with function notation, like f (x), replace the function symbol with the letter y. Then interchange x and y. Step 3. Solve the resulting equation for y. The result is f 00041(x). Step 4. Find the domain of f 00041. Remember, the domain of f 00041 must be the same as the range of f. You can check your work using Theorem 5.
EXAMPLE
3
Finding the Inverse of a Function Find f 00041 for f(x) 0002 1x 0004 1.
SOLUTION
y 5
y 0002 1x 0004 1 x 0002 1y 0004 1
y 0002 f (x) 5
00045
f(x) 0002 兹x 0004 1, x 1
Z Figure 4
Step 1. Find the domain of f and verify that f is one-to-one. Since 1x 0004 1 is defined only for x 0004 1 0, the domain of f is [1, ). The graph of f in Figure 4 shows that f is one-to-one, so f 00041 exists. Step 2. Replace f (x) with y, then interchange x and y.
x
Interchange x and y.
Step 3. Solve the equation for y. x 0002 1y 0004 1 x2 0002 y 0004 1 2 x 000310002y
Square both sides. Add 1 to each side.
The inverse is f 00041(x) 0002 x2 0003 1. Step 4. Find the domain of f 00041. The equation we found for f 00041 is defined for all x, but the domain should be the range of f. From Figure 4, we see that the range of f is [0, ) so that is the domain of f 00041. Therefore, f 00041(x) 0002 x2 0003 1
x 0
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Find the composition of f with the alleged inverse (in both orders!). For x in [1, ), the domain of f, we have
CHECK
f 00041( f (x)) 0002 f 00041(1x 0004 1) 0002 ( 1x 0004 1)2 0003 1 0002x0004100031 ✓ 0002x
Substitute 1x 0004 1 into f ⴚ1. Square 1x 0004 1. Add.
For x in [0, ), the domain of f 00041, we have f( f 00041(x)) 0002 f (x2 0003 1) 0002 2(x2 0003 1) 0004 1 0002 2x2 0002 冟x冟
Substitute x2 ⴙ 1 into f. Add. 2x2 ⴝ 円x円 for any real number x. 円x円 ⴝ x for x 0.
✓
0002x MATCHED PROBLEM 3
0002
Find f 00041 for f (x) 0002 1x 0003 2.
0002
The technique of finding an inverse by interchanging x and y leads to the following property of inverses that comes in very handy later in the course.
Z THEOREM 6 A Property of Inverses If f 00041 exists, then x 0002 f 00041( y) if and only if y 0002 f (x).
Z Mathematical Modeling Example 4 shows how an inverse function is used in constructing a revenue model. It is based on Example 8 in Section 3-5.
EXAMPLE
4
Modeling Revenue The research department for an electronics firm estimates that the weekly demand for a certain brand of headphones is given by x 0002 f ( p) 0002 20,000 0004 1,000p
Demand function
where x is the number of pairs retailers are likely to buy per week at p dollars per pair. Express the revenue as a function of the demand x and state its domain. SOLUTION
If x pairs of headphones are sold at p dollars each, the total revenue is Revenue 0002 (Number of pairs)(price of each pair) R 0002 xp To express the revenue as a function of the demand x, we need to express the price in terms of x. That is, we must find the inverse of the demand function.
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Step 1. Find the domain of f and verify that f is one-to-one. Price and demand are never negative, so p 0002 0 and x 0005 20,000 0004 1,000p 0005 1,000(20 0004 p) 0002 0 20 0004 p 0002 0 20 0002 p
Factor. Divide both sides by 1,000. Add p to both sides.
p 0003 20
or
Since p must satisfy both p 0002 0 and p 0003 20, the domain of f is [0, 20]. The graph of f (Fig. 5) shows that f is one-to-one. x 20,000
x 0005 20,000 0004 1,000p
0
p
20
Z Figure 5
Step 2. Since x and p have specific meaning in the context of this problem, interchanging them does not apply here. Step 3. Solve the equation x 0005 20,000 0004 1,000p for p. x 0005 20,000 0004 1,000p x 0004 20,000 0005 00041,000p 00040.001x 0006 20 0005 p
Subtract 20,000 from both sides. Divide both sides by ⴚ1,000.
The inverse of the demand function is p 0005 f 00041(x) 0005 20 0004 0.001x Step 4. From Figure 5, we see that the range of f is [0, 20,000], so this must also be the domain of f 00041. p 0005 f 00041(x) 0005 20 0004 0.001x We should check that f ( f the reader.
00041
(x)) 0005 x and f
0 0003 x 0003 20,000 00041
( f ( p)) 0005 p, but we will leave that to
The revenue R is given by R 0005 xp R(x) 0005 x(20 0004 0.001x) 0005 20x 0004 0.001x2 and the domain of R is [0, 20,000]. MATCHED PROBLEM 4
0002
Repeat Example 3 for the demand function x 0005 f ( p) 0005 10,000 0004 1,000p
0 0003 p 0003 10 0002
The demand function in Example 4 was defined with independent variable p and dependent variable x. When we found the inverse function, we did not rewrite it with independent variable p. Because p represents price and x represents number of players, to interchange these variables would be confusing. In most applications, the variables have specific meaning and should not be interchanged as part of the inverse process.
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Z Graphing Inverse Functions The following activities refer to the graph of f in Figure 6 and Tables 1 and 2.
ZZZ EXPLORE-DISCUSS 3
y 0002 f (x)
Table 1
Table 2
x
5
f(x)
f ⴚ1(x)
x
00044 00045
5
00042
x
0 2
00045
Z Figure 6
(A) Complete the second column in Table 1. (B) Reverse the ordered pairs in Table 1 and list the results in Table 2. (C) Add the points in Table 2 to Figure 6 (or a copy of the figure) and sketch the graph of f 00041. (D) Discuss any symmetry you observe between the graphs of f and f 00041.
Explore-Discuss 3 is based on an important relationship between the graph of any function and its inverse. In a rectangular coordinate system, the points (a, b) and (b, a) are symmetric with respect to the line y 0002 x [Fig. 7(a)]. Theorem 6 is an immediate consequence of this observation.
y
Z Figure 7 Symmetry with respect to the line y 0002 x.
5
y
y0002x (1, 4)
y 0002 f (x)
y
y0002x
5
y 0002 f 00041(x)
y 0002 f 00041(x)
y0002x
10
(00043, 2) (4, 1) x
00045
5
00045
5
x
y 0002 f(x)
(00045, 00042) (2, 00043) 00045
(00042, 00045)
(a, b) and (b, a) are symmetric with respect to the line y ⴝ x (a)
00045
10
f(x) ⴝ 2x ⴚ 1 f ⴚ1(x) ⴝ 12 x ⴙ 12
f (x) ⴝ 1x ⴚ 1 f ⴚ1(x) ⴝ x 2 ⴙ 1, x 0004 0
(b)
(c)
00041 Z THEOREM 7 Symmetry Property for the Graphs of f and f
The graphs of y 0002 f (x) and y 0002 f 00041(x) are symmetric with respect to the line y 0002 x.
x
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245
Inverse Functions
Knowledge of this symmetry property allows us to graph f 00041 if the graph of f is known, and vice versa. Figures 7(b) and 7(c) illustrate this property for the two inverse functions we found earlier. If a function is not one-to-one, we can usually restrict the domain of the function to produce a new function that is one-to-one. Then we can find an inverse for the restricted function. Suppose we start with f (x) 0002 x2 0004 4. Because f is not one-to-one, f 00041 does not exist [Fig. 8(a)]. But there are many ways the domain of f can be restricted to obtain a oneto-one function. Figures 8(b) and 8(c) illustrate two such restrictions. In essence, we are “forcing” the function to be one-to-one by throwing out a portion of the graph that would make it fail the horizontal line test. y
Z Figure 8 Restricting the domain of a function.
y
y 0002 f (x)
5
y
y 0002 h(x)
y0002x
5
y0002x
5
y 0002 g00041(x) 00045
5
x
00045
00045
5
00045
f(x) ⴝ x2 ⴚ 4 f ⴚ1 does not exist (a)
x
00045
5
y 0002 g(x)
00045
x
y 0002 h 00041(x)
h(x) ⴝ x2 ⴚ 4, x 0007 0 hⴚ1(x) ⴝ ⴚ1x ⴙ 4, x 0004 ⴚ4 (c)
g(x) ⴝ x 2 ⴚ 4, x 0004 0 g ⴚ1(x) ⴝ 1x ⴙ 4, x 0004 ⴚ4 (b)
Recall from Theorem 3 that increasing and decreasing functions are always one-to-one. This provides the basis for a convenient method of restricting the domain of a function: If the domain of a function f is restricted to an interval on the x axis over which f is increasing (or decreasing), then the new function determined by this restriction is one-to-one and has an inverse. We used this method to form the functions g and h in Figure 8.
EXAMPLE
5
Finding the Inverse of a Function Find the inverse of f(x) 0002 4x 0004 x2, x 2. Graph f, f 00041, and the line y 0002 x in the same coordinate system.
SOLUTION
Step 1. Find the domain of f and verify that f is one-to-one. We are given that the domain of f is (0004, 2]. The graph of y 0002 4x 0004 x2 is a parabola opening downward with vertex (2, 4) (Fig. 9). The graph of f is the left side of this parabola (Fig. 10). From the graph of f, we see that f is increasing and one-to-one on (0004, 2]. y 5
00045
y 0002 4x 0004 x2
5
00045
Z Figure 9
y 5
x
00045
y 0002 f(x)
5
00045
Z Figure 10
x
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Step 2. Replace f (x) with y, then interchange x and y. y 0002 4x 0004 x2 x 0002 4y 0004 y2 Step 3. Solve the equation for y. x 0002 4y 0004 y2 y2 0004 4y 0002 0004x y2 0004 4x 0003 4 0002 0004x 0003 4 (y 0004 2)2 0002 4 0004 x y 0004 2 0002 0006 14 0004 x y 0002 2 000614 0004 x
Rewrite so that the coefficient of y 2 is ⴙ1. Add 4 to both sides to complete the square. Factor the left side. Take the square root of both sides. Add 2 to both sides.
Now we have two possible solutions. The domain of f was (–, 2], and this should be the range of f 00041. In other words, the output of the inverse is never greater than 2. But y 0002 2 0003 14 0004 x would always be greater than or equal to 2, so we must instead choose y 0002 2 0004 14 0004 x. y 5
y0002x y 0002 f (x)
5
y0002
f 00041(x)
f 00041(x) 0002 2 0004 14 0004 x Step 4. The domain of f 00041 is the range of f. We can see from Figure 10 that this is (0004, 4]. Notice that the equation we found for f 00041(x) is defined for these values. Our final answer is f 00041(x) 0002 2 0004 14 0004 x
x
x 4
The check is again left for the reader. The graphs of f, f 00041, and y 0002 x are shown in Figure 11. To aid in graphing f 00041, we plotted several points on the graph of f and then reflected these points in the line y 0002 x. 0002
Z Figure 11
MATCHED PROBLEM 5
Find the inverse of f(x) 0002 4x 0004 x2, x 2. Graph f, f 00041, and y 0002 x in the same coordinate system. 0002
Technology Connections To reproduce Figure 11 on a graphing calculator, first enter
y1 ⴝ (4x ⴚ x2)0007(x 0007 2) in the equation editor (Fig. 12) and graph (Fig. 13). (For graphs involving both f and f 00061 it is best to use a squared viewing window.) The Boolean expression (x 0007 2) is
assigned the value 1 if the inequality is true and 0 if it is false. The calculator recognizes that division by 0 is an undefined operation and no graph is drawn for x 0002 2. Now enter
y2 ⴝ 2 ⴚ 14 ⴚ x
5
7.6
00047.6
00045
Z Figure 12
Z Figure 13
y3 ⴝ x
in the equation editor and graph (Fig. 14).
5
00047.6
and
7.6
00045
Z Figure 14
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ANSWERS TO MATCHED PROBLEMS 1. (A) Not one-to-one (B) One-to-one 2. They are inverses. 3. f ⫺1(x) ⫽ x2 ⫺ 2, x ⱖ 0 4. R(x) ⫽ 10x ⫺ 0.001x2
5. f ⫺1(x) ⫽ 2 ⫹ 14 ⫺ x, x ⱕ 4 y ⫽ f ⫺1(x)
y
y⫽x
5
⫺5
5
⫺5
3-6
x
y ⫽ f (x)
Exercises
1. When a function is defined by ordered pairs, how can you tell if it is one-to-one?
In Problems 13–30, determine if the function is one-to-one. 13. Domain
2. When you have the graph of a function, how can you tell if it is one-to-one?
Range
14. Domain
⫺2
⫺4
⫺2
3. Why does a function fail to have an inverse if it is not one-toone? Give an example using ordered pairs to illustrate your answer.
⫺1
⫺2
⫺1
0
0
0
4. True or False: Any function whose graph changes direction is not one-to-one. Explain.
1
1
1
2
5
2
5. What is the result of composing a function with its inverse? Why does this make sense? 6. What is the relationship between the graphs of two functions that are inverses?
15. Domain
For each set of ordered pairs in Problems 7–12, determine if the set is a function, a one-to-one function, or neither. Reverse all the ordered pairs in each set and determine if this new set is a function, a one-to-one function, or neither.
Range
17.
9
Range 5
2
2
3
3
1
4
4
2
5
5
4
7
f (x)
10. {(5, 4), (4, 3), (3, 2), (2, 1)} 11. 5(1, 2), (1, 4), (⫺3, 2), (⫺3, 4)6 12. 5(0, 5), (⫺4, 5), (⫺4, 2), (0, 2)6
7
1
8. {(⫺1, 0), (0, 1), (1, ⫺1), (2, 1)6 9. {(5, 4), (4, 3), (3, 3), (2, 4)}
⫺3
1
3
7. {(1, 2), (2, 1), (3, 4), (4, 3)}
16. Domain
Range
x
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g (x)
r (x)
23.
x
19.
x
s(x)
24.
h(x)
x
x
20.
k(x)
x
21.
m(x)
x
25. F(x) 0002 12x 0003 2
26. G(x) 0002 000413x 0003 1
27. H(x) 0002 4 0004 x2
28. K(x) 0002 14 0004 x
29. M(x) 0002 1x 0003 1
30. N(x) 0002 x2 0004 1
In Problems 31–40, determine if g is the inverse of f. 31. f (x) 0002 3x 0003 5;
g(x) 0002 13x 0004 53
32. f (x) 0002 2x 0004 4;
g(x) 0002 12x 0004 2
33. f (x) 0002 2 0004 (x 0003 1)3;
3 g(x) 0002 2 30004x00041
34. f (x) 0002 (x 0004 3)3 0003 4;
3 g(x) 0002 2 x0004400033
35. f (x) 0002
2x 0004 3 ; x00034
g(x) 0002
3 0003 4x 20004x
36. f (x) 0002
x00031 ; 2x 0004 3
g(x) 0002
3x 0003 1 2x 0003 1
37. f (x) 0002 4 0003 x2, x 0; 38. f (x) 0002 1x 0003 2;
g(x) 0002 1x 0004 4
g(x) 0002 x2 0004 2, x 0
39. f (x) 0002 1 0004 x2, x 0; 40. f (x) 0002 0004 1x 0004 2; 22.
g(x) 0002 0004 11 0004 x
g(x) 0002 x2 0003 2, x 0
n(x)
In Problems 41–44, find the domain and range of f, sketch the graph of f 00041, and find the domain and range of f 00041. y
41. x
y0002x
5
y 0002 f (x) 00045
5
00045
x
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42.
Inverse Functions
249
67. f (x) ⫽ x2 ⫹ 2x ⫺ 2, x ⱕ ⫺1
y⫽x
68. f (x) ⫽ x2 ⫹ 8x ⫹ 7, x ⱖ ⫺4
5
69. f (x) ⫽ ⫺ 29 ⫺ x2, 0 ⱕ x ⱕ 3 ⫺5
5
70. f (x) ⫽ 29 ⫺ x2, 0 ⱕ x ⱕ 3
x
71. f (x) ⫽ 29 ⫺ x2, ⫺3 ⱕ x ⱕ 0
y ⫽ f (x)
72. f (x) ⫽ ⫺ 29 ⫺ x2, ⫺3 ⱕ x ⱕ 0
⫺5
73. f (x) ⫽ 1 ⫺ 21 ⫺ x2, ⫺1 ⱕ x ⱕ 0 y
43.
74. f (x) ⫽ 1 ⫹ 21 ⫺ x2, ⫺1 ⱕ x ⱕ 0
y⫽x
5
The functions in Problems 75–84 are one-to-one. Find f ⫺1. 75. f (x) ⫽ 3 ⫺
⫺5
y ⫽ f (x)
5
x
y⫽x
78. f (x) ⫽
2x ⫹ 5 3x ⫺ 4
82. f (x) ⫽
5 83. f (x) ⫽ 4 ⫺ 2x ⫹ 2
5
4 x
3 x⫹4 4x 80. f (x) ⫽ 2⫺x
2 x⫺1 2x 79. f (x) ⫽ x⫹1 81. f (x) ⫽
y
76. f (x) ⫽ 5 ⫹
77. f (x) ⫽
⫺5
44.
2 x
5 ⫺ 3x 7 ⫺ 4x
3 84. f (x) ⫽ 2x ⫹ 3 ⫺ 2
85. How are the x and y intercepts of a function and its inverse related? ⫺5
5
x
86. Does a constant function have an inverse? Explain.
y ⫽ f (x)
87. Are the functions f (x) ⫽ x2 and g(x) ⫽ 1x inverses? Why or why not?
⫺5
In Problems 45–74, graph f and verify that f is a one-to-one function. Find f ⫺1and add the graph of f ⫺1 and the line y ⫽ x to the graph of f. State the domain and range of f and the domain and range of f ⫺1. 45. f (x) ⫽ 3x 47. f (x) ⫽ 4x ⫺ 3
1 46. f (x) ⫽ x 2 1 5 48. f (x) ⫽ ⫺ x ⫹ 3 3
49. f (x) ⫽ 0.2x ⫹ 0.4
50. f (x) ⫽ 0.25x ⫹ 2.25
51. f (x) ⫽ 1x ⫹ 3
52. f (x) ⫽ 2 ⫺ 1x
53. f (x) ⫽
1 116 ⫺ x 2
54. f (x) ⫽
1 136 ⫺ x 3
55. f (x) ⫽ 3 ⫺ 1x ⫺ 1
56. f (x) ⫽ 2 ⫹ 15 ⫺ x
57. f (x) ⫽ x2 ⫹ 5, x ⱖ 0
58. f (x) ⫽ x2 ⫹ 5, x ⱕ 0
59. f (x) ⫽ 4 ⫺ x2, x ⱕ 0
60. f (x) ⫽ 4 ⫺ x2, x ⱖ 0
61. f (x) ⫽ x2 ⫹ 8x, x ⱖ ⫺4 62. f (x) ⫽ x2 ⫹ 8x, x ⱕ ⫺4 63. f (x) ⫽ (2 ⫺ x)2, x ⱕ 2 64. f (x) ⫽ (2 ⫺ x)2, x ⱖ 2 65. f (x) ⫽ (x ⫺ 1)2 ⫹ 2, x ⱖ 1 66. f (x) ⫽ 3 ⫺ (x ⫺ 2)2, x ⱕ 2
3 x inverses? Why or 88. Are the functions f (x) ⫽ x3 and g(x) ⫽ 1 why not?
In Problems 89–92, the given function is not one-to-one. Find a way to restrict the domain so that the function is one-to-one, then find the inverse of the function with that domain. 89. f (x) ⫽ (2 ⫺ x)2
90. f (x) ⫽ (1 ⫹ x)2
91. f (x) ⫽ 24x ⫺ x2
92. f (x) ⫽ 26x ⫺ x2
APPLICATIONS 93. BODY WEIGHT Two formulas for estimating body weight as a function of height that are commonly used are Women: p ⫽ W(h) ⫽ 100 ⫹ 5h Men: p ⫽ M(h) ⫽ 110 ⫹ 5h where p is weight in pounds and h is height over 5 feet (in inches). Find h ⫽ W ⫺1(p) and state its domain. 94. BODY WEIGHT Refer to Problem 93. Find h ⫽ M ⫺1( p) and state its domain. 95. PRICE AND DEMAND The number q of CD players consumers are willing to buy per week from a retail chain at a price of $p is given approximately by (see the figure) q ⫽ d(p) ⫽
3,000 0.2p ⫹ 1
10 ⱕ p ⱕ 70
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(A) Find the range of d. (B) Find p ⫽ d⫺1(q), and find its domain and range. q 1,000
q ⫽ d(p) q ⫽ s(p) 70
p
Figure for 95–96
96. PRICE AND SUPPLY The number q of CD players a retail chain is willing to supply at a price of $p is given approximately by (see the figure) q ⫽ s( p) ⫽
900p p ⫹ 20
10 ⱕ p ⱕ 70
(A) Find the range of s. (B) Find p ⫽ s⫺1(q), and find its domain and range.
97. BUSINESS—MARKUP POLICY A bookstore sells a book with a wholesale price of $6 for $10.50 and one with a wholesale price of $10 for $15.50. (A) If the markup policy for the store is assumed to be linear, find a function r ⫽ m(w) that expresses the retail price r as a function of the wholesale price w and find its domain and range. (B) Find w ⫽ m⫺1(r) and find its domain and range. 98. BUSINESS—MARKUP POLICY Repeat Problem 97 if the second book has a wholesale price of $11 and sells for $18.50. Problems 99 and 100 are related to Problems 97 and 98 in Exercises 3-4. 99. STOPPING DISTANCE A model for the length L (in feet) of the skid marks left by a particular automobile when making an emergency stop is L ⫽ f (s) ⫽ 0.06s2 ⫺ 1.2s ⫹ 26, s ⱖ 10 where s is speed in miles per hour. Find s ⫽ f ⫺1(L) and find its domain and range. 100. STOPPING DISTANCE A model for the length L (in feet) of the skid marks left by a second automobile when making an emergency stop is L ⫽ f (s) ⫽ 0.08s2 ⫺ 1.6s ⫹ 38, s ⱖ 10 where s is speed in miles per hour. Find s ⫽ f ⫺1(L) and find its domain and range.
CHAPTER
3-1
3
Review
Functions
A function is a correspondence between two sets of elements such that to each element in the first set there corresponds one and only one element in the second set. The first set is called the domain and the set of all corresponding elements in the second set is called the range. Equivalently, a function is a set of ordered pairs with the property that no two ordered pairs have the same first component and different second components. The domain is the set of all first components, and the range is the set of all second components. An equation in two variables defines a function if to each value of the independent variable, the placeholder for domain values, there corresponds exactly one value of the dependent variable, the placeholder for range values. The vertical line test states that a vertical line will intersect the graph of a function in at most one point. Unless otherwise specified, the implied domain of a function defined by an equation is assumed to be the set of all real number replacements for the independent variable that produce real values for the dependent variable. The symbol f(x) represents the real number in the range of the function f corresponding to the domain value x. Equivalently, the ordered pair (x, f (x)) belongs to the function f.
3-2
Graphing Functions
The graph of a function f is the set of all points (x, f(x)), where x is in the domain of f and f(x) is the associated output. This is also the same as the graph of the equation y ⫽ f (x). The first coordinate of a point where the graph of a function intersects the x axis is called an x intercept or real zero of the function. The x intercept is also a real solution or root of the equation f (x) ⫽ 0. The second coordinate of a point where the graph of a function crosses the y axis is called the y intercept of the function. The y intercept is given by f(0), provided 0 is in the domain of f. A solid dot on a graph of a function indicates a point that belongs to the graph and an open dot indicates a point that does not belong to the graph. Dots are also used to indicate that a graph terminates at a point, and arrows are used to indicate that the graph continues indefinitely with no significant changes in direction. Let I be an interval in the domain of a function f. Then, 1. f is increasing on I and the graph of f is rising on I if f (x1) 6 f (x2) whenever x1 6 x2 in I. 2. f is decreasing on I and the graph of f is falling on I if f (x1) 7 f (x2) whenever x1 6 x2 in I.
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Review
3. f is constant on I and the graph of f is horizontal on I if f (x1) ⫽ f (x2) whenever x1 6 x2 in I. A function of the form f(x) ⫽ mx ⫹ b, where m and b are constants, is a linear function. If m ⫽ 0, then f(x) ⫽ b is a constant function, and if m ⫽ 1 and b ⫽ 0, then f(x) ⫽ x is the identity function. A piecewise-defined function is a function whose definition involves more than one formula. The absolute value function is a piecewise-defined function. The graph of a function is continuous if it has no holes or breaks and discontinuous at any point where it has a hole or break. Intuitively, the graph of a continuous function can be sketched without lifting a pen from the paper. The greatest integer for a real number x, denoted by 冀x冁 , is the largest integer less than or equal to x; that is, 冀x 冁 ⫽ n, where n is an integer, n ⱕ x 6 n ⫹ 1. The greatest integer function f is defined by the equation f (x) ⫽ 冀x 冁.
3-3
Transformations of Functions
The first six basic functions in a library of elementary functions are defined by f (x) ⫽ x (identity function), g(x) ⫽ 冟 x 冟 (absolute value function), h(x) ⫽ x2 (square function), m(x) ⫽ x3 (cube function), 3 n(x) ⫽ 1x (square root function), and p(x) ⫽ 2 x (cube root function) (see Figure 1, Section 3-3). Performing an operation on a function produces a transformation of the graph of the function. The basic transformations are the following:
251
A function f is called an even function if f (x) ⫽ f (⫺x) for all x in the domain of f and an odd function if f (⫺x) ⫽ ⫺f (x) for all x in the domain of f. The graph of an even function is said to be symmetric with respect to the y axis and the graph of an odd function is said to be symmetric with respect to the origin.
3-4
Quadratic Functions
If a, b, and c are real numbers with a ⫽ 0, then the function f (x) ⫽ ax2 ⫹ bx ⫹ c is a quadratic function and its graph is a parabola. Completing the square of the quadratic expression x2 ⫹ bx produces a perfect square: b 2 b 2 x2 ⫹ bx ⫹ a b ⫽ ax ⫹ b 2 2 Completing the square for f (x) ⫽ ax2 ⫹ bx ⫹ c produces the vertex form f (x) ⫽ a(x ⫺ h)2 ⫹ k and gives the following properties:
1. The graph of f is a parabola: f (x)
Axis of symmetry x⫽h
Vertex (h, k)
Vertical Translation: k 7 0 Shift graph of y ⫽ f(x) up k units y ⫽ f(x) ⫹ k e k 6 0 Shift graph of y ⫽ f(x) down 冟 k 冟 units
e
Min f(x) h
h 7 0 Shift graph of y ⫽ f (x) left h units h 6 0 Shift graph of y ⫽ f (x) right 冟 h 冟 units
f (x)
Reflection: y ⫽ ⫺f (x) y ⫽ f (⫺x) y ⫽ ⫺f (⫺x)
x
a⬎0 Opens upward
Horizontal Translation: y ⫽ f(x ⫹ h)
k
Reflect the graph of y ⫽ f (x) through the x axis Reflect the graph of y ⫽ f (x) through the y axis Reflect the graph of y ⫽ f (x) through the origin
Axis of symmetry x⫽h Vertex (h, k)
k
Max f(x)
Vertical Stretch and Shrink: A 7 1 y ⫽ Af (x) f
0 6 A 6 1
Vertically stretch the graph of y ⫽ f (x) by multiplying each y value by A Vertically shrink the graph of y ⫽ f (x) by multiplying each y value by A
Horizontal Stretch and Shrink: A 7 1
y ⫽ f (Ax) h
0 6 A 6 1
Horizontally shrink the graph of y ⫽ f (x) by multiplying 1 each x value by A Horizontally stretch the graph of y ⫽ f (x) by multiplying 1 each x value by A
h
x
a⬍0 Opens downward
2. Vertex: (h, k) (Parabola increases on one side of the vertex and decreases on the other.)
3. Axis (of symmetry): x ⫽ h (parallel to y axis) 4. f (h) ⫽ k is the minimum if a 7 0 and the maximum if a 6 0. 5. Domain: All real numbers
Range: (⫺⬁, k] if a 6 0 or [k, ⬁) if a 7 0
6. The graph of f is the graph of g(x) ⫽ ax2 translated horizontally h units and vertically k units.
The first coordinate of the vertex of a parabola in standard form can be located using the formula x ⫽ ⫺b/2a. This can then be substituted into the function to find the second coordinate. The vertex
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form of a parabola can be used to find the equation when the vertex and one other point on the graph are known. Replacing the equal sign in a quadratic equation with 6, 7 ,
, or produces a quadratic inequality. The set of all values of the variable that make the inequality a true statement is the solution set.
3-5
Combining Functions; Composition
The sum, difference, product, and quotient of the functions f and g are defined by ( f 0003 g)(x) 0002 f (x) 0003 g (x)
( f 0004 g)(x) 0002 f (x) 0004 g (x)
( fg)(x) 0002 f (x)g (x)
f (x) f a b (x) 0002 g g (x)
3
Inverse Functions
A function is one-to-one if no two ordered pairs in the function have the same second component and different first components. According to the horizontal line test, a horizontal line will intersect the graph of a one-to-one function in at most one point. A function that is increasing (or decreasing) throughout its domain is one-to-one. The inverse of the one-to-one function f is the function f 00041 formed by reversing all the ordered pairs in f. If f is a one-to-one function, then: 1. f 00041 is one-to-one. 2. Domain of f 00041 0002 Range of f.
g(x) 0005 0
The domain of each function is the intersection of the domains of f and g, with the exception that values of x where g(x) 0002 0 must be excluded from the domain of f兾g. The composition of functions f and g is defined by ( f ° g)(x) 0002 f ( g (x)). The domain of f ° g is the set of all real numbers x in the domain of g such that g(x) is in the domain of f. The domain of f ° g is always a subset of the domain of g.
CHAPTER
3-6
3. Range of f 00041 0002 Domain of f. 4. x 0002 f 00041( y) if and only if y 0002 f (x). 5. f 00041 ( f (x)) 0002 x for all x in the domain of f. 6. f ( f 00041(x)) 0002 x for all x in the domain of f 00041. 7. To find f 00041, solve the equation y 0002 f (x) for x. Interchanging x and y at this point is an option. 8. The graphs of y 0002 f (x) and y 0002 f 00041 (x) are symmetric with respect to the line y 0002 x.
Review Exercises
Work through all the problems in this review and check answers in the back of the book. Answers to most review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. 1. Indicate whether each table defines a function. (A) Domain
Range
(B) Domain
1
4
7
3
6
8
5
8
9
Range
0
years during which a Super Bowl was played. If each team corresponds to the year or years in which they won the Super Bowl, does this correspondence define a function? Explain your answer. 4. Indicate whether each graph specifies a function: (A)
(C) Domain
Range
5
1
10
2
y
x
20
2. Indicate whether each set defines a function. Indicate whether any of the functions are one-to-one. Find the domain and range of each function. Find the inverse of any one-to-one functions. Find the domain and range of any inverse functions. (A) {(1, 1), (2, 4), (3, 9)} (B) {(1, 1), (1, 00041), (2, 2), (2, 00042)} (C) {(Albany, New York), (Utica, New York), (Akron, Ohio), (Dayton, Ohio)} (D) {(Albany, New York),(Akron, Ohio), (Tucson, Arizona), (Atlanta, Georgia), (Muncie, Indiana)} 3. Let T be the set of teams in the National Football League that have won at least one Super Bowl, and let Y be the set of
(B)
y
x
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Review Exercises y
(C)
17. Find f(00044), f(0), f(3), and f(5). 18. Find all values of x for which f (x) 0002 00042. 19. Find the domain and range of f. 20. Find the intervals over which f is increasing and decreasing.
x
21. Find any points of discontinuity. Problems 22–29 refer to the graphs of f and g shown here. y
(D)
x
f (x)
g(x)
5
5
00045
5
x
00045
5
00045
5. Which of the following equations define functions? (A) y 0002 x (B) y2 0002 x 3 (C) y 0002 x (D) 冟 y 冟 0002 x Problems 6–15 refer to the functions f, g, k, and m given by: g(x) 0002 4 0004 x2
f (x) 0002 3x 0003 5
k(x) 0002 5 m(x) 0002 2冟 x 冟 0004 1
8.
f (2 0003 h) 0004 f (2) h
m(00042) 0003 1 g (2) 0003 4
9.
g (a 0003 h) 0004 g (a) h
11. ( f 0004 g)(x)
12. ( fg)(x)
f 13. a b (x) g
14. ( f ° g)(x)
15. (g ° f )(x)
In Problems 24–27, use the graphs of f and g to find: 24. ( f ° g)(00041)
25. (g ° f )(00042)
26. f [g(1)]
27. g[ f(00043)]
Problems 31–36 refer to the graph of the function f used in Problems 17–21. Sketch the graph of each of the following.
16. For f (x) 0002 x 0004 2x, find (B) f (00044)
23. Construct a table of values of ( fg)(x) for x 0002 00043, 00042, 00041, 0, 1, 2, and 3, and sketch the graph of fg.
30. Indicate whether each function is even, odd, or neither: (A) f (x) 0002 x5 0003 6x (B) g(t) 0002 t 4 0003 3t 2 (C) h(z) 0002 z5 0003 4z2
2
(A) f(1)
22. Construct a table of values of ( f 0004 g)(x) for x 0002 00043, 00042, 00041, 0, 1, 2, and 3, and sketch the graph of f 0004 g.
29. Is g a one-to-one function?
7.
10. ( f 0003 g)(x)
00045
28. Is f a one-to-one function?
Find the indicated quantities or expressions. 6. f (2) 0003 g (00042) 0003 k (0)
(C) f (2) ⴢ f (00041)
(D)
f (0) f (3)
Problems 17–21 refer to the function f given by the following graph.
31. f (x) 0003 1
32. f (x 0003 1)
33. 0004f (x)
34. 0.5f (x)
35. f (2x)
36. 0004f (0004x)
37. Match each equation with a graph of one of the functions f, g, m, or n in the figure. Each graph is a graph of one of the equations. (B) y 0002 0004(x 0003 2)2 0003 4 (A) y 0002 (x 0004 2)2 0004 4 2 (C) y 0002 0004(x 0004 2) 0003 4 (D) y 0002 (x 0003 2)2 0004 4 y
f
g
5
f(x) 5
00045
5
00045
x
00045
5
m
n
x
x
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38. Referring to the graph of function f in the figure for Problem 37 and using known properties of quadratic functions, find each of the following to the nearest integer: (A) Intercepts (B) Vertex (C) Maximum or minimum (D) Range (E) Interval of increase (F) Interval of decrease 39. Let f (x) ⫽ x 2 ⫺ 4 and g(x) ⫽ x ⫹ 3. Find each of the following functions and find their domains. (A) f兾g (B) g兾f (C) f ° g (D) g ° f 40. For each function, find the maximum or minimum value without graphing. Then write the coordinates of the vertex. (A) f (x) ⫽ ⫺2(x ⫹ 4) 2 ⫺ 10 (B) f (x) ⫽ x 2 ⫺ 6x ⫹ 11 41. Complete the square to write the quadratic function in vertex form: q (x) ⫽ 2x 2 ⫺ 14x ⫹ 3
In Problems 52–57, find the domain, y intercept (if it exists), and any x intercepts. 52. m(x) ⫽ x2 ⫺ 4x ⫹ 5
53. r(x) ⫽ 2 ⫹ 31x
2
54. p(x) ⫽
1⫺x x3
55. f (x) ⫽
x 13 ⫺ x
56. g(x) ⫽
2x ⫹ 3 x2 ⫺ 4
57. h(x) ⫽
1 4 ⫺ 1x
58. Let f (x) ⫽ 0.5x2 ⫺ 4x ⫹ 5. (A) Sketch the graph of f and label the axis and the vertex. (B) Where is f increasing? Decreasing? What is the range? (Express answers in interval notation.) 59. Find the equations of the linear function g and the quadratic function f whose graphs are shown in the figure. This line is called the tangent line to the graph of f at the point (⫺1, 0).
42. How are the graphs of the following related to the graph of y ⫽ x2? (A) y ⫽ ⫺x2 (B) y ⫽ x2 ⫺ 3 (C) y ⫽ (x ⫹ 3)2
y 5
y ⫽ g (x)
y ⫽ f (x) ⫺5
5
x
Problems 43–49 refer to the function q given by the following graph. ⫺5
q(x)
60. Let
5
f (x) ⫽ e ⫺5
5
x
⫺5
43. Find y to the nearest integer: (A) y ⫽ q(0) (B) y ⫽ q(1) (C) y ⫽ q(2) (D) y ⫽ q(⫺2) 44. Find x to the nearest integer: (A) q(x) ⫽ 0 (B) q(x) ⫽ 1 (C) q(x) ⫽ ⫺3 (D) q(x) ⫽ 3 45. Find the domain and range of q. 46. Find the intervals over which q is increasing, decreasing, and constant. 47. Identify any points of discontinuity. 48. The function f multiplies the cube of the domain element by 4 and then subtracts the square root of the domain element. Write an algebraic definition of f. 49. Write a verbal description of the function f(x) ⫽ 3x2 ⫹ 4x ⫺ 6. In Problems 50 and 51, determine if the indicated equation defines a function. Justify your answer. 50. x ⫹ 2y ⫽ 10
51. x ⫹ 2y2 ⫽ 10
⫺x ⫺ 5 0.2x2
for ⫺4 ⱕ x 6 0 for 0 ⱕ x ⱕ 5
(A) Find f (⫺4), f(⫺2), f(0), f(2), and f(5). (B) Sketch the graph of y ⫽ f (x). (C) Find the domain and range. (D) Find any points of discontinuity. (E) Find the intervals over which f is increasing, decreasing, and constant. 61. Given f (x) ⫽ 1x ⫺ 8 and g(x) ⫽ 冟 x 冟: (A) Find f ° g and g ° f. (B) Find the domains of f ° g and g ° f. 62. Which of the following functions are one-to-one? (A) f(x) ⫽ x3 (B) g(x) ⫽ (x ⫺ 2)2 (C) h(x) ⫽ 2x ⫺ 3 (D) F(x) ⫽ (x ⫹ 3)2, x ⱖ ⫺3 63. Is u(x) ⫽ 4x ⫺ 8 the inverse of v(x) ⫽ 0.25x ⫹ 2? 64. The function f(x) ⫽ 2(x ⫺ 3)2 is not one-to-one. (A) Graph f using transformations of y ⫽ x2. (B) Restrict the domain of f to make it a one-to-one function. (C) Find the inverse of the one-to-one function. 65. Given f (x) ⫽ 3x ⫺ 7: (A) Find f ⫺1(x). (B) Find f ⫺1(5). (C) Find f ⫺1 [f (x)]. (D) Is f increasing, decreasing, or constant on (⫺⬁, ⬁)?
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66. The following graph is the result of applying a sequence of transformations to the graph of y 0002 x2. Describe the transformations verbally and write an equation for the given graph.
255
y
(A) 5
Check by graphing your equation on a graphing calculator. y
00045
5
x
5
00045 00045
5
x y
(B) 5
00045
67. The graph of f (x) 0002 冟 x 冟 is vertically stretched by a factor of 3, reflected through the x axis, and shifted 2 units to the right and 5 units up to form the graph of the function g. Find an equation for the function g and graph g. 68. Write an equation for the following graph in the form y 0002 a(x 0004 h)2 0003 k, where a is either 00041 or 00031 and h and k are integers. Check by graphing your equation on a graphing calculator. y
00045
5
x
00045
72. The graph of f (x) 0002 冟 x 冟 is stretched vertically by a factor of 3, reflected through the x axis, shifted four units to the right and eight units up to form the graph of the function g. Find an equation for the function g and graph g. 73. The graph of m(x) 0002 x2 is stretched horizontally by a factor of 2, shifted two units to the left and four units down to form the graph of the function t. Find an equation for the function t and graph t.
5
00045
5
x
00045
69. The following graph is the result of applying a sequence of 3 transformations to the graph of y 0002 1 x. Describe the transformations verbally, and write an equation for the given graph. y
Use graph transformations to sketch the graph of each equation in Problems 74–81: 74. y 0002 冟 x 0003 1 冟
3 75. y 0002 1 0003 1 10004x
76. y 0002 冟 x 冟 0004 2
77. y 0002 9 0004 3 1x
78. y 0002 12 冟 x 冟
3 79. y 0002 1 4 0004 0.5x
80. y 0002 2 0004 3(x 0004 1)3
81. y 0002 0004冟 x 0003 1 冟 0004 1
Solve Problems 82 and 83. Express answers in interval notation.
5
82. x2 0003 x 6 20
83. x2 7 4x 0003 12
84. Find the domain of f (x) 0002 225 0004 x2. 00045
5
x
00045
85. Given f (x) 0002 x 2 and g(x) 0002 11 0004 x, find each function and its domain. (A) fg (B) f兾g (C) f ° g (D) g ° f 86. For the one-to-one function f given by
Check by graphing your equation on a graphing calculator. 70. How is the graph of f(x) 0002 0004(x 0004 2)2 0004 1 related to the graph of g(x) 0002 x2? 71. Each of the following graphs is the result of applying one or more transformations to the graph of one of the six basic functions in Figure 1, Section 3-3. Find an equation for the graph. Check by graphing the equation on a graphing calculator.
f (x) 0002 (A) Find f 00041(x). (B) Find f 00041(3). (C) Find f 00041 [ f (x)].
x00032 x00043
87. Given f (x) 0002 1x 0004 1: (A) Find f 00041(x). (B) Find the domain and range of f and f 00041. (C) Graph f, f 00041, and y 0002 x on the same coordinate system.
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Check by graphing f, f 00041, and y 0002 x in a squared window on a graphing calculator. 88. Given f(x) 0002 x2 0004 1, x 0: (A) Find the domain and range of f and f 00041. (B) Find f 00041(x). (C) Find f 00041(3). (D) Find f 00041[ f (4)]. (E) Find f 00041[ f (x)].
94. STOPPING DISTANCE Table 1 contains data related to the length of the skid marks left by an automobile when making an emergency stop. A model for the skid mark length L (in feet) of the auto is L 0002 f (s) 0002 0.06s2 0004 2.4s 0003 50, s 20 where s is speed in miles per hour.
Table 1
Check by graphing f, f 00041, and y 0002 x in a squared window on a graphing calculator. 89. A partial graph of the function f is shown in the figure. Complete the graph of f over the interval [0, 5] given that: (A) f is symmetric with respect to the y axis. (B) f is symmetric with respect to the origin. y 5
00045
5
x
00045
90. The function f is decreasing on [00045, 5] with f(00045) 0002 4 and f(5) 0002 00043. (A) If f is continuous on [00045, 5], how many times can the graph of f cross the x axis? Support your conclusion with examples and/or verbal arguments. (B) Repeat part A if the function does not have to be continuous.
APPLICATIONS 91. INCOME Megan works 20 hours per week at an electronics store to help pay for tuition and rent. She gets a base salary of $6 per hour, a commission of 10% on all sales over $2,000 for the week, and a bonus of $250 if her weekly sales are over $5,000. (A) Write a function that describes Megan’s weekly earnings, where x represents her weekly sales. (B) Find Megan’s weekly earnings if her sales are $2,000, $4,000, and $6,000. (C) If Megan needs to average at least $400 per week to cover her tuition and rent, how much does she need to sell on average each week? 92. On the set of a movie, a stuntman will be jumping from a helicopter that is hovering at a height of 120 feet, and landing in a moving truck full of chicken feathers. How many seconds after he jumps does the truck need to be in position? 93. BUSINESS—MARKUP POLICY A sporting goods store sells tennis shorts that cost $30 for $48 and sunglasses that cost $20 for $32. (A) If the markup policy of the store for items that cost over $10 is assumed to be linear and is reflected in the pricing of these two items, find a function r 0002 f(c) that expresses retail price r as a function of cost c. (B) What should be the retail price of a pair of skis that cost $105? (C) Find c 0002 f 00041(r) and find its domain and range. (D) What is the cost of a box of golf balls that retail for $39.99?
Speed (mph)
Length of Skid Marks (feet)
20
26
30
32
40
49
50
80
60
122
70
176
80
242
(A) Graph L 0002 f(s) and the data for skid mark length on the same axes. (B) Find s 0002 f 00041(L) and find its domain and range. (C) How fast (to the nearest mile) was the auto traveling if it left skid marks 200 feet long? 95. PRICE AND DEMAND The price $p per hot dog at which q hot dogs can be sold during a baseball game is given approximately by 9 p 0002 g(q) 0002 1,000 q 4,000 1 0003 0.002q (A) Find the range of g. (B) Find q 0002 g00041( p) and find its domain and range. (C) Express the revenue as a function of p. (D) Express the revenue as a function of q. 96. MARKET RESEARCH A market research firm is hired to study demand for a new blanket that looks an awful lot like a bathrobe worn backwards. They determine that if x units are produced each week and sold at a price of $p per unit, then the weekly demand, revenue, and cost equations are, respectively x 0002 500 0004 10p R(x) 0002 50x 0004 0.1x2 C(x) 0002 10x 0003 1,500 Express the weekly profit as a function of the price p and find the price that produces the largest profit. 97. CONSTRUCTION A farmer has 120 feet of fencing to be used in the construction of two identical rectangular pens sharing a common side (see the figure).
x
y y
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Group Activity
(A) Express the total area A(x) enclosed by both pens as a function of the width x. (B) From physical considerations, what is the domain of the function A? (C) Find the dimensions of the pens that will make the total enclosed area maximum. 98. COMPUTER SCIENCE In computer programming, it is often necessary to check numbers for certain properties (even, odd, perfect square, etc.). The greatest integer function provides a convenient method for determining some of these properties. Consider the function f (x) ⫽ x ⫺ ( 冀 1x冁)2 (A) Evaluate f for x ⫽ 1, 2, . . . , 16. (B) Find f (n2), where n is a positive integer. (C) What property of x does this function determine?
CHAPTER
ZZZ GROUP
99. Use the schedule in Table 2 to construct a piecewise-defined model for the taxes due for a single taxpayer in Virginia with a taxable income of x dollars. Find the tax on the following incomes: $2,000, $4,000, $10,000, $30,000.
Table 2 Virginia Tax Rate Schedule
Status Single
Taxable Income Over $
But Not Over
Tax Is
Of the Amount Over
0
$ 3,000
2%
$ 3,000
$ 5,000
$ 60 ⫹ 3%
$
$ 3,000
$ 5,000
$17,000
$120 ⫹ 5%
$ 5,000
$17,000
—
$720 ⫹ 5.75%
$17,000
3 ACTIVITY Mathematical Modeling: Choosing a Cell Phone Plan
The number of companies offering cellular telephone service has grown rapidly in recent years. The plans they offer vary greatly and it can be difficult to select the plan that is best for you. Here are five typical plans: Plan 1: A flat fee of $50 per month for unlimited calls. Plan 2: A $30 per month fee for a total of 30 hours of calls and an additional charge of $0.01 per minute for all minutes over 30 hours. Plan 3: A $5 per month fee and a charge of $0.04 per minute for all calls. Plan 4: A $2 per month fee and a charge of $0.045 per minute for all calls; the fee is waived if the charge for calls is $20 or more. Plan 5: A charge of $0.05 per minute for all calls; there are no additional fees.
(A) Construct a mathematical model for each plan that gives the total monthly cost in terms of the total number of minutes of calls placed in a month. (B) Compare plans 1 and 2. Determine how many minutes per month would make plan 1 cheaper and how many would make plan 2 cheaper. (C) Repeat part (B) for plans 1 and 3; plans 1 and 4; plans 1 and 5. (D) Repeat part (B) for plans 2 and 3; plans 2 and 4; plans 2 and 5. (E) Repeat part (B) for plans 3 and 4; plans 3 and 5. (F) Repeat part (B) for plans 4 and 5. (G) Is there one plan that is always better than all the others? Based on your personal calling history, which plan would you choose and why?
0
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CHAPTER
Polynomial and Rational Functions
4
C IN Chapters 2 and 3, we used lines and parabolas to model a variety of situations. But the graph of a line doesn't change direction, and the graph of a parabola has just one turning point. So to model more complicated phenomena, we will study the more general class of polynomial functions in Chapter 4. A polynomial function can have many turning points. We will investigate the graphs and zeros of polynomials and apply that knowledge to study functions that can be written as quotients of polynomials, that is, the rational functions. Finally, we will use the language of variation to describe a wide range of mathematical models used in engineering and the physical, social, and health sciences.
OUTLINE 4-1
Polynomial Functions, Division, and Models
4-2
Real Zeros and Polynomial Inequalities
4-3
Complex Zeros and Rational Zeros of Polynomials
4-4
Rational Functions and Inequalities
4-5
Variation and Modeling Chapter 4 Review Chapter 4 Group Activity: Interpolating Polynomials
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Polynomial Functions, Division, and Models Z Graphs of Polynomial Functions Z Polynomial Division Z Remainder and Factor Theorems Z Mathematical Modeling and Data Analysis
In this section, we will study polynomial functions, a class that includes the linear and quadratic functions of Chapter 3. Graphs of polynomials exhibit much greater variety than just lines and parabolas. We will examine the properties of the graphs of polynomial functions, and we will use tools from algebra (division and factorization) to understand those properties. We also will show how polynomials are used to model data for which linear and quadratic functions are unsuitable.
Z Graphs of Polynomial Functions In Chapter 3 we introduced linear and quadratic functions and their graphs (Fig. 1): f (x) 0002 ax 0003 b, a00040 f (x) 0002 ax2 0003 bx 0003 c, a00040
Linear function Quadratic function
10
10
000510
10
000510
10
000510
000510
Z Figure 1 Graphs of linear and quadratic functions.
A function such as g(x) 0002 7x4 0005 5x3 0003 (2 0003 9i)x2 0003 3x 0005 1.95 which is the sum of a finite number of terms, each of the form axk, where a is a number and k is a nonnegative integer, is called a polynomial function. The polynomial function g(x) is said to have degree 4 because x4 is the highest power of x that appears among the terms of g(x). Therefore, linear and quadratic functions are polynomial functions of degrees 1 and 2, respectively. The two functions h(x) 0002 x00051 and k(x) 0002 x100022, however, are not polynomial functions (the exponents 00051 and 12 are not nonnegative integers). Z DEFINITION 1 Polynomial Function If n is a nonnegative integer, a function that can be written in the form P(x) 0002 an xn 0003 an00051xn00051 0003 . . . 0003 a1x 0003 a0,
an 0004 0
is called a polynomial function of degree n. The numbers an, an00051, . . ., a1, a0 are called the coefficients of P(x).
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261
We will assume that the coefficients of a polynomial function are complex numbers, or real numbers, or rational numbers, or integers, depending on our interest. Similarly, the domain of a polynomial function can be the set of complex numbers, the set of real numbers, or an appropriate subset of either, depending on the situation. According to Definition 1, a nonzero constant function like f (x) 0002 5 has degree 0 (it can be written as f (x) 0002 5x0). The constant function with value 0 is considered to be a polynomial but is not assigned a degree.
Z DEFINITION 2 Zeros or Roots A number r is said to be a zero or root of a function P(x) if P(r) 0002 0.
The zeros of P(x) are the solutions of the equation P(x) 0002 0. So if the coefficients of a polynomial P(x) are real numbers, then the real zeros of P(x) are just the x intercepts of the graph of P(x). For example, the real zeros of the polynomial P(x) 0002 x2 0005 4 are 2 and 00052, the x intercepts of the graph of P(x) [Fig. 2(a)]. However, a polynomial may have zeros that are not x intercepts. Q(x) 0002 x2 0003 4, for example, has zeros 2i and 00052i, but its graph has no x intercepts [Fig. 2(b)]. 10
10
000510
10
000510
10
000510
000510
(a)
(b)
Z Figure 2 Real zeros are x intercepts.
EXAMPLE
1
Zeros and x Intercepts (A) Figure 3 shows the graph of a polynomial function of degree 5. List its real zeros. 200
00055
5
0005200
Z Figure 3
(B) List all zeros of the polynomial function P(x) 0002 (x 0005 4)(x 0003 7)3(x2 0003 9)(x2 0005 2x 0003 2) Which zeros of P(x) are x intercepts?
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(A) The real zeros are the x intercepts: 00054, 00052, 0, and 3. (B) Note first that P(x) is a polynomial because it can be written in the form of Definition 1 (it is not necessary to actually multiply out P(x) to find that form). The zeros of P(x) are the solutions to the equation P(x) 0002 0. Because a product equals 0 if and only if one of the factors equals 0, we can find the zeros by solving each of the following equations (the last was solved using the quadratic formula): x0005400020 x00024
(x 0003 7)3 0002 0 x 0002 00057
x2 0003 9 0002 0 x 0002 00063i
x2 0005 2x 0003 2 0002 0 x000210006i
Therefore, the zeros of P(x), are 4, 00057, 3i, 00053i, 1 0003 i, and 1 0005 i. Only two of the six zeros are real numbers and therefore x intercepts: 4 and 00057. 0002 MATCHED PROBLEM 1
(A) Figure 4 shows the graph of a polynomial function of degree 4. List its real zeros. 5
00055
5
00055
Z Figure 4
(B) List all zeros of the polynomial function P(x) 0002 (x 0003 5)(x2 0005 4)(x2 0003 4)(x2 0003 2x 0003 5) Which zeros of P(x) are x intercepts? 0002
A point on a continuous graph that separates an increasing portion from a decreasing portion, or vice versa, is called a turning point. The vertex of a parabola, for example, is a turning point. Linear functions with real coefficients have exactly one real zero and no turning points; quadratic functions with real coefficients have at most two real zeros and exactly one turning point.
ZZZ EXPLORE-DISCUSS 1
Examine Figures 2(a), 2(b), 3, and 4, which show the graphs of polynomial functions of degree 2, 2, 5, and 4, respectively. In each figure, all real zeros and all turning points of the function appear in the given viewing window. (A) Is the number of real zeros ever less than the degree? Equal to the degree? Greater than the degree? How is the number of real zeros of a polynomial related to its degree? (B) Is the number of turning points ever less than the degree? Equal to the degree? Greater than the degree? How is the number of turning points of a polynomial related to its degree?
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263
Polynomial Functions, Division, and Models
Explore-Discuss 1 suggests that graphs of polynomial functions with real coefficients have the properties listed in Theorem 1, which we accept now without proof. Property 3 is proved later in this section. The other properties are established in calculus.
Z THEOREM 1 Properties of Graphs of Polynomial Functions Let P(x) be a polynomial of degree n 0 with real coefficients. Then the graph of P(x): 1. 2. 3. 4. 5.
Is continuous for all real numbers Has no sharp corners Has at most n real zeros Has at most n 0005 1 turning points Increases or decreases without bound as x → 0007 and as x → 00050007*
Figure 5 shows graphs of representative polynomial functions of degrees 1 through 6, illustrating the five properties of Theorem 1.
y
y 5
5
00055
5
x
5
00055
x
5
00055
00055
00055
(c) h(x) ⴝ x5 ⴚ 6x3 ⴙ 8x ⴙ 1
y
y
y
5
5
5
x
00055
(d) F(x) ⴝ x2 ⴚ x ⴙ 1
5
5
x
00055
00055
x
5
00055
(b) g(x) ⴝ x3 ⴙ 5x
(a) f(x) ⴝ x ⴚ 2
00055
y
(e) G(x) ⴝ 2x4 ⴚ 7x2 ⴙ x ⴙ 3
00055
5
x
00055
(f) H(x) ⴝ x6 ⴚ 7x4 ⴙ 12 x 2 ⴚ x ⴚ 2
Z Figure 5 Graphs of polynomial functions.
*Remember that 0007 and 00050007 are not real numbers. The statement the graph of P(x) increases without bound as x → 00050007 means that for any horizontal line y 0002 b there is some interval (00050007, a] 0002 {x0003 x a} on which the graph of P(x) is above the horizontal line.
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2
Properties of Graphs of Polynomials Explain why each graph is not the graph of a polynomial function by listing the properties of Theorem 1 that it fails to satisfy. (A)
(B)
y
5
5
00055
5
x
MATCHED PROBLEM 2
y 5
00055
5
x
00055
00055
00055
SOLUTIONS
(C)
y
5
x
00055
(A) The graph has a sharp corner when x 0002 0. Property 2 fails. (B) There are no points on the graph with x coordinate less then or equal to 0, so properties 1 and 5 fail. (C) There are an infinite number of zeros and an infinite number of turning points, so properties 3 and 4 fail. Furthermore, the graph is bounded by the horizontal lines y 0002 00061, so property 5 fails.
0002
Explain why each graph is not the graph of a polynomial function by listing the properties of Theorem 1 that it fails to satisfy. (A)
(B)
y 5
(C)
y 5
00055
5
x
5
00055
00055
y
5
x
00055
00055
5
x
00055
0002 The shape of the graph of a polynomial function with real coefficients is similar to the shape of the graph of the leading term, that is, the term of highest degree. Figure 6 compares the graph of the polynomial h(x) 0002 x5 0005 6x3 0003 8x 0003 1 from Figure 5 with the graph of its leading term p(x) 0002 x5. The graphs are dissimilar near the origin, but as we zoom out, the shapes of the two graphs become quite similar. The leading term in the polynomial dominates all other terms combined. Because the graph of p(x) increases without bound as x → 0007, the same is true of the graph of h(x). And because the graph of p(x) decreases without bound as x → 00050007, the same is true of the graph of h(x). y ph
5 Z Figure 6 p(x) 0002 x ,
h(x) 0002 x 0005 6x 0003 8x 0003 1. 5
3
y
5
ph
500
ZOOM OUT 00055
5
00055
x
00055
5
0005500
x
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The left and right behavior of a polynomial function with real coefficients is determined by the left and right behavior of its leading term (see Fig. 6). Property 5 of Theorem 1 can therefore be refined. The various possibilities are summarized in Theorem 2.
Z THEOREM 2 Left and Right Behavior of Polynomial Functions Let P(x) 0004 an xn 0005 an00031xn00031 0005 . . . 0005 a1x 0005 a0 be a polynomial function with real coefficients, an 0006 0, n 0007 0. 1. an > 0, n even: The graph of P(x) increases without bound as x S 0002 and increases without bound as x S 00030002 (like the graphs of x2, x4, x6, etc.). 2. an > 0, n odd: The graph of P(x) increases without bound as x S 0002 and decreases without bound as x S 00030002 (like the graphs of x, x3, x5, etc.). 3. an < 0, n even: The graph of P(x) decreases without bound as x S 0002 and decreases without bound as x S 00030002 (like the graphs of 0003x2, 0003x4, 0003x6, etc.). 4. an < 0, n odd: The graph of P(x) decreases without bound as x S 0002 and increases without bound as x S 00030002 (like the graphs of 0003x, 0003x3, 0003x5, etc.). y
y
y
Case 1
x
x
x
Case 2
y
Case 3
x
Case 4
It is convenient to write P(x) → 0002 as an abbreviation for the phrase the graph of P(x) increases without bound. Using this notation, the left and right behavior in Case 4 of Theorem 2, for example, is P(x) → 00030002 as x → 0002 and P(x) → 0002 as x → 00030002.
EXAMPLE
3
Left and Right Behavior of Polynomials Determine the left and right behavior of each polynomial. (A) The degree of P(x) 0004 3 0003 x2 0005 4x3 0003 x4 0003 2x6 (B) The degree of Q(x) 0004 4x5 0005 8x3 0005 5x 0003 1
SOLUTIONS
(A) The degree P(x) is 6 (even) and the coefficient a6 is 00032 (negative), so the left and right behavior is the same as that of 0003x6 (Case 3 of Theorem 2): P(x) → 00030002 as x → 0002 and P(x) → 00030002 as x → 00030002. (B) The degree Q(x) is 5 (odd) and the coefficient a5 is 4 (positive), so the left and right behavior is the same as that of x5 (Case 2 of Theorem 2): P(x) → 0002 as x → 0002 and P(x) → 00030002 as x → 00030002. 0002
MATCHED PROBLEM 3
Determine the left and right behavior of each polynomial. (A) P(x) 0004 4x9 0003 3x11 0005 5 (B) Q(x) 0004 1 0003 2x50 0005 x100
0002
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EXAMPLE
4
Graphing a Polynomial Graph the polynomial P(x) 0002 x3 0005 12x 0005 16, 00055 x 5. List the real zeros and turning points.
SOLUTION
First we construct a table of values by calculating P(x) for each integer x, 00055 x 5. For example, P(00055) 0002 (00055)3 0005 12(00055) 0005 16 0002 000581
y 100
00055
5
x
0005100 3 Z Figure 7 P(x) 0002 x 0005 12x 0005 16.
MATCHED PROBLEM 4
ZZZ
x
P(x)
x
P(x)
00055
000581
1
000527
00054
000532
2
000532
00053
00057
3
000525
00052
0
4
0
00051
00055
5
49
0
000516
Then we plot the points in the table and join them with a smooth curve (Fig. 7). The zeros are 00052 and 4. The turning points are (00052, 0) and (2, 000532). Note that P(x) has the maximum number of turning points for a polynomial of degree 3, but one fewer than the maximum number of real zeros. 0002 Graph P(x) 0002 x4 0005 6x2 0005 8x 0005 3, 00054 x 4. List the real zeros and turning points. 0002 Finding the real zeros and turning points of a polynomial is usually more difficult than suggested by Example 4. In Example 4, how did we know that the real zeros were between 00055 and 5 rather than between, say, 95 and 105? Could there be another real zero just to the left or right of 00052? How do we know that (00052, 0) and (2, 000532), rather than nearby points having noninteger coordinates, are the turning points? To answer such questions we must view polynomials from an algebraic perspective. Polynomials can be factored. So next we will study the division and factorization of polynomials.
CAUTION ZZZ
Z Polynomial Division We can find quotients of polynomials by a long-division process similar to the one used in arithmetic. Example 5 will illustrate the process.
EXAMPLE
5
Polynomial Long Division Divide P(x) 0002 3x3 0005 5 0003 2x4 0005 x by 2 0003 x.
SOLUTION
First, rewrite the dividend P(x) in descending powers of x, inserting 0 as the coefficient for any missing terms of degree less than 4: P(x) 0002 2x4 0003 3x3 0003 0x2 0005 x 0005 5
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Similarly, rewrite the divisor 2 0002 x in the form x 0002 2. Then divide the first term x of the divisor into the first term 2x4 of the dividend. Multiply the result, 2x3, by the divisor, obtaining 2x4 0002 4x3. Line up like terms, subtract as in arithmetic, and bring down 0x2. Repeat the process until the degree of the remainder is less than the degree of the divisor. 2x3 0003 x2 0002 2x 0003 5 x 0002 2 0002 2x4 0002 3x3 0002 0x2 0003 x 0003 5 2x4 0002 4x3 0003x3 0002 0x2 0003x3 0003 2x2 2x2 0003 x
Divisor
Quotient Dividend Subtract
Subtract
2x2 0002 4x 00035x 0003 5 00035x 0003 10 5
Subtract
Subtract Remainder
Therefore, 5 2x4 0002 3x3 0003 x 0003 5 0004 2x3 0003 x2 0002 2x 0003 5 0002 x00022 x00022 CHECK
You can always check division using multiplication: 5 d x00022 0004 (x 0002 2)(2x3 0003 x2 0002 2x 0003 5) 0002 5 0004 2x4 0002 3x3 0003 x 0003 5
(x 0002 2) c 2x3 0003 x2 0002 2x 0003 5 0002
MATCHED PROBLEM 5
Multiply and collect like terms
0002
Divide 6x2 0003 30 0002 9x3 by x 0003 2. 0002 The procedure illustrated in Example 5 is called the division algorithm. The concluding equation of Example 5 (before the check) may be multiplied by the divisor x 0002 2 to give the following form: Dividend 4
ⴝ
3
Divisor ⴢ Quotient 3
ⴙ Remainder 2
2x 0002 3x 0003 x 0003 5 0004 (x 0002 2)(2x 0003 x 0002 2x 0003 5) 0002 5 This last equation is an identity: it is true for all replacements of x by real or complex numbers including x 0004 00032. Theorem 3, which we state without proof, gives the general result of applying the division algorithm when the divisor has the form x 0003 r.
Z THEOREM 3 Division Algorithm For each polynomial P(x) of degree greater than 0 and each number r, there exists a unique polynomial Q(x) of degree 1 less than P(x) and a unique number R such that P(x) 0004 (x 0003 r)Q(x) 0002 R The polynomial Q(x) is called the quotient, x 0003 r is the divisor, and R is the remainder. Note that R may be 0.
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There is a shortcut called synthetic division for the long division of Example 5. First write the coefficients of the dividend and the negative of the constant term of the divisor in the format shown below at the left. Bring down the 2 as indicated next on the right, multiply by 00052, and record the product 00054. Add 3 and 00054, bringing down their sum 00051. Repeat the process until the coefficients of the quotient and the remainder are obtained. Dividend coefficients
Dividend coefficients
2
3
0
00051
00055
00052 Negative of constant term of divisor
2
3
0
00051
00055
00052 2
00054 00051
2 2
00054 00055
10 5
Quotient coefficients
Remainder
Compare the preceding synthetic division to the long division shown below, in which the essential numerals appear in color, to convince yourself that synthetic division produces the correct quotient and remainder. (In synthetic division we use the negative of the constant term of the divisor so we can add rather than subtract.)
Divisor
2x3 ⴚ 1x2 ⴙ 2x ⴚ 5 x ⴙ 20004 2x4 0003 3x3 0003 0x2 ⴚ 1x ⴚ 5 2x4 0003 4x3 ⴚ1x3 0003 0x2 00051x3 ⴚ 2x2 2x2 0005 1x 2x2 ⴙ 4x ⴚ5x 0005 5 00055x ⴚ 10 5
Quotient Dividend
Remainder
Z KEY STEPS IN THE SYNTHETIC DIVISION PROCESS To divide the polynomial P(x) by x 0005 r: Step 1. Arrange the coefficients of P(x) in order of descending powers of x. Write 0 as the coefficient for each missing power. Step 2. After writing the divisor in the form x 0005 r, use r to generate the second and third rows of numbers as follows. Bring down the first coefficient of the dividend and multiply it by r; then add the product to the second coefficient of the dividend. Multiply this sum by r, and add the product to the third coefficient of the dividend. Repeat the process until a product is added to the constant term of P(x). Step 3. The last number to the right in the third row of numbers is the remainder. The other numbers in the third row are the coefficients of the quotient, which is of degree 1 less than P(x).
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Synthetic Division Use synthetic division to divide P(x) 0004 4x5 0003 30x3 0003 50x 0003 2 by x 0002 3. Find the quotient and remainder. Write the conclusion in the form P(x) 0004 (x 0003 r)Q(x) 0002 R of Theorem 3.
SOLUTION
Because x 0002 3 0004 x 0003 (00033), we have r 0004 00033, and 4 00033 4
4
0 000312 000312
000330 36 6
0 000318 000318
000350 54 4
00032 000312 000314
3
The quotient is 4x 0003 12x 0002 6x2 0003 18x 0002 4 with a remainder of 000314. So 4x5 0003 30x3 0003 50x 0003 2 0004 (x 0002 3)(4x4 0003 12x3 0002 6x2 0003 18x 0002 4) 0003 14 MATCHED PROBLEM 6
0002
Repeat Example 6 with P(x) 0004 3x4 0003 11x3 0003 18x 0002 8 and divisor x 0003 4. 0002
Z Remainder and Factor Theorems ZZZ EXPLORE-DISCUSS 2
Let P(x) 0004 x3 0003 3x2 0003 2x 0002 8. (A) Evaluate P(x) for (i) x 0004 00032 (ii) x 0004 1
(iii) x 0004 3
(B) Use synthetic division to find the remainder when P(x) is divided by (i) x 0002 2 (ii) x 0003 1 (iii) x 0003 3 What conclusion does a comparison of the results in parts A and B suggest? Explore-Discuss 2 suggests that when a polynomial P(x) is divided by x 0003 r, the remainder is equal to P(r), the value of the polynomial P(x) at x 0004 r. In Problem 87 of Exercises 4-1, you are asked to complete a proof of this fact, which is called the remainder theorem. Z THEOREM 4 Remainder Theorem If R is the remainder after dividing the polynomial P(x) by x 0003 r, then P(r) 0004 R
EXAMPLE
7
Two Methods for Evaluating Polynomials If P(x) 0004 4x4 0002 10x3 0002 19x 0002 5, find P(00033) by (A) Using the remainder theorem and synthetic division (B) Evaluating P(00033) directly
SOLUTIONS
(A) Use synthetic division to divide P(x) by x 0003 (00033). 4 00033
4
10 000312 00032
0 6 6
19 000318 1
5 00033 2 0004 R 0004 P(00033)
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(B) P(00053) 0002 4(00053)4 0003 10(00053)3 0003 19(00053) 0003 5 00022
MATCHED PROBLEM 7
0002
Repeat Example 7 for P(x) 0002 3x4 0005 16x2 0005 3x 0005 7 and x 0002 00052. 0002 You might think the remainder theorem is not a very effective tool for evaluating polynomials. But let’s consider the number of operations performed in parts A and B of Example 7. Synthetic division requires only four multiplications and four additions to find P(00053), whereas the direct evaluation requires ten multiplications and four additions. [Note that evaluating 4(00053)4 actually requires five multiplications.] The difference becomes even larger as the degree of the polynomial increases. Computer programs that involve numerous polynomial evaluations often use synthetic division because of its efficiency. We will find synthetic division and the remainder theorem to be useful tools later in this chapter. The remainder theorem shows that the division algorithm equation, P(x) 0002 (x 0005 r)Q(x) 0003 R can be written in the form where R is replaced by P(r): P(x) 0002 (x 0005 r)Q(x) 0003 P(r) Therefore, x 0005 r is a factor of P(x) if and only if P(r) 0002 0, that is, if and only if r is a zero of the polynomial P(x). This result is called the factor theorem.
Z THEOREM 5 Factor Theorem If r is a zero of the polynomial P(x), then x 0005 r is a factor of P(x). Conversely, if x 0005 r is a factor of P(x), then r is a zero of P(x).
EXAMPLE
8
Factors of Polynomials Use the factor theorem to show that x 0003 1 is a factor of P(x) 0002 x25 0003 1 but is not a factor of Q(x) 0002 x25 0005 1.
SOLUTION
Because P(00051) 0002 (00051)25 0003 1 0002 00051 0003 1 0002 0 x 0005 (00051) 0002 x 0003 1 is a factor of x25 0003 1. On the other hand, Q(00051) 0002 (00051)25 0005 1 0002 00051 0005 1 0002 00052 and x 0003 1 is not a factor of x25 0005 1.
MATCHED PROBLEM 8
0002
Use the factor theorem to show that x 0005 i is a factor of P(x) 0002 x8 0005 1 but is not a factor of Q(x) 0002 x8 0003 1. 0002 One consequence of the factor theorem is Theorem 6 (a proof is outlined in Problem 88 in Exercises 4-1).
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Z THEOREM 6 Zeros of Polynomials A polynomial of degree n has at most n zeros.
Theorem 6 says that the graph of a polynomial of degree n with real coefficients has at most n real zeros (Property 3 of Theorem 1). The polynomial H(x) 0002 x6 0005 7x4 0003 12x2 0005 x 0005 2 for example, has degree 6 and the maximum number of zeros [see Fig. 5(f ), p. 263]. Of course, polynomials of degree 6 may have fewer than six real zeros. In fact, p(x) 0002 x6 0003 1 has no real zeros. However, it can be shown that the polynomial p(x) 0002 x6 0003 1 has exactly six complex zeros.
Z Mathematical Modeling and Data Analysis In Chapters 2 and 3 we saw that linear and quadratic functions can be useful models for certain sets of data. For some data, however, no linear function and no quadratic function can provide a reasonable model. In that case, we investigate the suitability of polynomial models of degree greater than 2. In Examples 9 and 10 we discuss cubic and quartic models, respectively, for the given data.
EXAMPLE
9
Table 1 Sturgeon
Estimating the Weight of Fish Scientists and fishermen often estimate the weight of a fish from its length. The data in Table 1 give the average weight of North American sturgeon for certain lengths. Because weight is associated with volume, which involves three dimensions, we might expect that weight would be associated with the cube of the length. A cubic model for the data is given by
Length (in.) x
Weight (oz.) y
18
13
22
26
26
46
30
75
34
115
38
166
where y is the weight (in ounces) of a sturgeon that has length x (in inches).
44
282
(A) Use the model to estimate the weight of a sturgeon of length 56 inches.
52
492
60
796
y 0002 0.00526x3 0005 0.117x2 0003 1.43x 0005 5.00
(B) Compare the weight of a sturgeon of length 44 inches as given by Table 1 with the weight given by the model.
Source: www.thefishernet.com
SOLUTIONS
(A) If x 0002 56, then y 0002 0.00526(56)3 0005 0.117(56)2 0003 1.43(56) 0005 5.00 0005 632 ounces (B) If x 0002 44, then y 0002 0.00526(44)3 0005 0.117(44)2 0003 1.43(44) 0005 5.00 0005 279 ounces The weight given by the table, 282 ounces, is 3 ounces greater than the weight given by the model. 0002
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Technology Connections Figure 8 shows the details of constructing the cubic model of Example 9 on a graphing calculator. 1,000
0
70
0
(a) Entering the data
(b) Finding the model
(c) Graphing the data and the model
Z Figure 8
MATCHED PROBLEM 9
Use the cubic model of Example 9. (A) Estimate the weight of a sturgeon of length 65 inches. (B) Compare the weight of a sturgeon of length 30 inches as given by Table 1 with the weight given by the model. 0002
EXAMPLE
10
Table 2
Hydroelectric Power The data in Table 2 gives the annual consumption of hydroelectric power (in quadrillion BTU) in the United States for selected years since 1983. From Table 2 it appears that a polynomial model of the data would have three turning points—near 1989, 1997, and 2001. Because a polynomial with three turning points must have degree at least four, we can model the data with a quartic (fourth-degree) polynomial:
Year
U.S. Consumption of Hydroelectric Power (Quadrillion BTU)
1983
3.90
1985
3.40
1987
3.12
1989
2.99
1991
3.14
1993
3.13
1995
3.48
1997
3.88
1999
3.47
(A) Use the model to predict the consumption of hydroelectric power in 2018.
2001
2.38
2003
2.53
(B) Compare the consumption of hydroelectric power in 2003 (as given by Table 2) to the consumption given by the model.
2005
2.61
y 0004 0.00013x4 0003 0.0067x3 0002 0.107x2 0003 0.59x 0002 4.03 where y is the consumption (in quadrillion BTU) and x is time in years with x 0004 0 representing 1983.
Source: U.S. Department of Energy
SOLUTIONS
(A) If x 0004 35 (which represents the year 2018), then y 0004 0.00013(35)4 0003 0.0067(35)3 0002 0.107(35)2 0003 0.59(35) 0002 4.03 0003 22.3 The model predicts a consumption of 22.3 quadrillion BTU in 2018. However, because the predicted consumption for 2018 is so dramatically greater than earlier consumption levels, it is unlikely to be accurate. This brings up an important point: A model that fits a set of data points well is not automatically a good model for predicting future trends.
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(B) If x 0002 20 (which represents 2003), then y 0002 0.00013(20)4 0005 0.0067(20)3 0003 0.107(20)2 0005 0.59(20) 0003 4.03 0002 2.23 The consumption reported in the table, 2.53 quadrillion BTU, is 0.30 quadrillion BTU greater than the consumption given by the model. 0002
Technology Connections Figure 9 shows the details of constructing the quartic model of Example 10 on a graphing calculator. 5
0
(a)
(b)
30
00052
(c)
Z Figure 9
MATCHED PROBLEM 10
Use the quartic model of Example 10. (A) Estimate the consumption of hydroelectric power in 2000. (B) Compare the consumption of hydroelectric power in 1991 (as given by Table 2) to the consumption given by the model. 0002 ANSWERS TO MATCHED PROBLEMS 1. (A) 00051, 1, 2 (B) The zeros are 00055, 00052, 2, 2i, 00052i, 00051 0003 2i, and 00051 00052i; the x intercepts are 00055, 00052, and 2. 2. (A) Properties 1 and 5 (B) Property 5 (C) Properties 1 and 5 3. (A) P(x) S 00050007 as x S 0007 and P(x) S 0007 as x S 00050007. (B) Q(x) S 0007 as x S 0007 and Q(x) S 0007 as x S 00050007. 4. y 200
00055
5
x
0005200
zeros: 00051, 3; turning point; (2, 000527) 66 5. 9x2 0003 24x 0003 48 0003 x00052
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6. 3x4 0003 11x3 0003 18x 0005 8 0004 (x 0003 4)(3x3 0005 x2 0005 4x 0003 2) 7. P(00032) 0004 00033 for both parts, as it should 8. P(i) 0004 0, so x 0003 i is a factor of x8 0003 1; Q(i) 0004 2, so x 0003 i is not a factor of x8 0005 1 9. (A) 1,038 in. (B) The weight given in the table is 0.38 oz greater than the weight given by the model. 10. (A) 2.86 quadrillion BTU (B) The consumption given in the table is 0.12 quadrillion BTU less than the consumption given by the model.
4-1
Exercises y
y
1. What is a polynomial function? 2. Explain the connection between the zeros of a polynomial and its linear factors. 3. Explain what is wrong with the following setup for dividing x4 0005 5x2 0003 2x 0005 6 by x 0003 2 using synthetic division. 1
5
x
x
00032 6
20 4. Explain what is wrong with the following setup for dividing 3x3 0003 x2 0005 8x 0005 9 by x 0005 4 using synthetic division. 00031 8
3
(c)
(d)
9 In Problems 13–16, list the real zeros and turning points, and state the left and right behavior, of the polynomial function P(x) that has the indicated graph.
40 In Problems 5–8, decide whether the statement is true or false, and explain your answer.
y
13. 5
5. Every quadratic function is a polynomial function. 6. Every polynomial of degree 3 has three x intercepts. 7. If a polynomial has no x intercepts, then it has no zeros.
00035
5
x
8. Every polynomial function is continuous. In Problems 9–12, a is a positive real number. Match each function with one of graphs (a)–(d). 9. f(x) 0004 ax3
10. g(x) 0004 0003ax4
11. h(x) 0004 ax6
12. k(x) 0004 0003ax5
y
00035
y
14. 5
y
00035
x
5
x 00035
(a)
(b)
x
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15.
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275
y
20.
5
3
00055
5
x
00053
x
3
00053
00055
y
16. 5
In Problems 21–24, list all zeros of each polynomial function, and specify those zeros that are x intercepts.
00055
5
21. P(x) 0002 x(x2 0005 9)(x2 0003 4)
x
22. P(x) 0002 (x2 0005 4)(x4 0005 1) 23. P(x) 0002 (x 0003 5)(x2 0003 9)(x2 0003 16) 24. P(x) 0002 (x2 0005 5x 0003 6)(x2 0005 5x 0003 7)
00055
In Problems 17–20, explain why each graph is not the graph of a polynomial function. y
17.
In Problems 25–34, use algebraic long division to find the quotient and the remainder. 25. (3x2 0003 5x 0003 6) (x 0003 1) 26. (2x2 0005 7x 0003 4) (x 0005 2)
2
27. (4m2 0005 1) (m 0005 1) 00052
x
2
28. (y2 0005 9) ( y 0003 3) 29. (6 0005 6x 0003 8x2) (x 0003 1)
00052
30. (11x 0005 2 0003 12x2) (3x 0003 2) 31.
y
18.
x3 0005 1 x00051
32.
a3 0003 27 a00033
33. (3y 0005 y2 0003 2y3 0005 1) ( y 0003 2)
5
34. (3 0003 x3 0005 x) (x 0005 3)
00055
5
x
In Problems 35–40, divide using synthetic division. 35. (x2 0003 3x 0005 7) (x 0005 2) 36. (x2 0003 3x 0005 3) (x 0005 3)
00055
37. (4x2 0003 10x 0005 9) (x 0003 3) 38. (2x2 0003 7x 0005 5) (x 0003 4)
y
19. 3
00053
39.
3
x
2x3 0005 3x 0003 1 x00052
40.
In Problems 41–44, is the given number a zero of the polynomial? Use synthetic division. 41. x2 0003 4x 0005 221; 000517
00053
x3 0003 2x2 0005 3x 0005 4 x00032
42. x2 0005 7x 0005 551; 29 43. 2x3 0003 38x2 0005 x 0003 19; 000519 44. 2x3 0005 397x 0003 70; 14
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In Problems 45–48, determine whether the second polynomial is a factor of the first polynomial without dividing or using synthetic division. 45. x18 0003 1; x 0003 1
46. x18 0003 1; x 0005 1
47. 3x3 0003 7x2 0003 8x 0005 2; x 0005 1
In Problems 73 and 74, divide, using synthetic division. 73. (x3 0003 3x2 0005 x 0003 3) (x 0003 i) 74. (x3 0003 2x2 0005 x 0003 2) (x 0005 i) 75. Let P(x) 0004 x2 0005 2ix 0003 10. Use synthetic division to find: (A) P(2 0003 i)
48. 3x4 0003 2x3 0005 5x 0003 6; x 0003 1
(B) P(5 0003 5i)
Use synthetic division and the remainder theorem in Problems 49–54. 49. Find P(00032), given P(x) 0004 3x2 0003 x 0003 10. 50. Find P(00033), given P(x) 0004 4x2 0005 10x 0003 8. 51. Find P(2), given P(x) 0004 2x3 0003 5x2 0005 7x 0003 7. 52. Find P(5), given P(x) 0004 2x3 0003 12x2 0003 x 0005 30. 4
(C) P(3 0003 i) (D) P(00033 0003 i) 76. Let P(x) 0004 x2 0003 4ix 0003 13. Use synthetic division to find: (A) P(5 0005 6i) (B) P(1 0005 2i) (C) P(3 0005 2i) (D) P(00033 0005 2i)
2
53. Find P(00034), given P(x) 0004 x 0003 10x 0005 25x 0003 2. 54. Find P(00037), given P(x) 0004 x4 0005 5x3 0003 13x2 0003 30. In Problems 55–62, use synthetic division to find the quotient and the remainder. As coefficients get more involved, a calculator should prove helpful. Do not round off. 55. (3x4 0003 x 0003 4) (x 0005 1) 56. (5x4 0003 2x2 0003 3) (x 0003 1) 5
57. (x 0005 1) (x 0005 1) 58. (x4 0003 16) (x 0003 2) 59. (3x4 0005 2x3 0003 4x 0003 1) (x 0005 3) 60. (x4 0003 3x3 0003 5x2 0005 6x 0003 3) (x 0003 4) 61. (2x6 0003 13x5 0005 75x3 0005 2x2 0003 50) (x 0003 5) 62. (4x6 0005 20x5 0003 24x4 0003 3x2 0003 13x 0005 30) (x 0005 6) In Problems 63–68, without graphing, state the left and right behavior, the maximum number of x intercepts, and the maximum number of local extrema. 63. P(x) 0004 x3 0003 5x2 0005 2x 0005 6 64. P(x) 0004 x3 0005 2x2 0003 5x 0003 3 65. P(x) 0004 0003x3 0005 4x2 0005 x 0005 5 66. P(x) 0004 0003x3 0003 3x2 0005 4x 0003 4 67. P(x) 0004 x4 0005 x3 0003 5x2 0003 3x 0005 12 68. P(x) 0004 0003x4 0005 6x2 0003 3x 0003 16 In Problems 69–72, either give an example of a polynomial with real coefficients that satisfies the given conditions or explain why such a polynomial cannot exist. 69. P(x) is a third-degree polynomial with one x intercept. 70. P(x) is a fourth-degree polynomial with no x intercepts. 71. P(x) is a third-degree polynomial with no x intercepts. 72. P(x) is a fourth-degree polynomial with no turning points.
In Problems 77–82, approximate (to two decimal places) the x intercepts and the local extrema. 77. P(x) 0004 40 0005 50x 0003 9x2 0003 x3 78. P(x) 0004 40 0005 70x 0005 18x2 0005 x3 79. P(x) 0004 0.04x3 0003 10x 0005 5 80. P(x) 0004 00030.01x3 0005 2.8x 0003 3 81. P(x) 0004 0.1x4 0005 0.3x3 0003 23x2 0003 23x 0005 90 82. P(x) 0004 0.1x4 0005 0.2x3 0003 19x2 0005 17x 0005 100 83. (A) What is the least number of turning points that a polynomial function of degree 4, with real coefficients, can have? The greatest number? Explain and give examples. (B) What is the least number of x intercepts that a polynomial function of degree 4, with real coefficients, can have? The greatest number? Explain and give examples. 84. (A) What is the least number of turning points that a polynomial function of degree 3, with real coefficients, can have? The greatest number? Explain and give examples. (B) What is the least number of x intercepts that a polynomial function of degree 3, with real coefficients, can have? The greatest number? Explain and give examples. 85. Is every polynomial of even degree an even function? Explain. 86. Is every polynomial of odd degree an odd function? Explain. 87. Prove the remainder theorem (Theorem 4): (A) Write the result of the division algorithm if a polynomial P(x) is divided by x 0003 r. (B) Evaluate both sides of the equation from part (A) when x 0004 r. What can you conclude? 88. In this problem, we will prove that a polynomial of degree n has at most n zeros (Theorem 6). Give a reason for each step. Let P(x) be a polynomial of degree n, and suppose that P has n distinct zeros r1, r2, . . . , rn . We will show that it is impossible for P to have any other zeros. Step 1: We can write P(x) in the form P(x) 0004 (x 0003 r1)Q1(x), where the degree of Q1(x) is n 0003 1.
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Step 2: r2 is a zero of Q1(x). Step 3: We can write Q1(x) in the form Q1(x) 0004 (x 0003 r2)Q2(x), where the degree of Q2(x) is n 0003 2. Step 4: P(x) 0004 (x 0003 r1)(x 0003 r2)Q2(x) Step 5: P(x) 0004 (x 0003 r1)(x 0003 r2). . .(x 0003 rn)Qn(x), where the degree of Qn(x) is 0. Step 6: The only zeros of P are r1, r2, . . . , rn.
(B) Find the volume of the plastic coating to four decimal places if the thickness of the shielding is 0.005 feet. Problems 93–96 require a graphing calculator or a computer that can calculate cubic regression polynomials for a given data set. 93. HEALTH CARE Table 3 shows the total national health care expenditures (in billion dollars) and the per capita expenditures (in dollars) for selected years since 1960.
APPLICATIONS
Table 3 National Health Care Expenditures
89. REVENUE The price–demand equation for 8,000-BTU window air conditioners is given by
Year
p 0004 0.0004x2 0003 x 0002 569
0 0005 x 0005 800
277
Polynomial Functions, Division, and Models
Total Expenditures (Billion $)
Per Capita Expenditures ($)
1960
28
148
where x is the number of air conditioners that can be sold at a price of p dollars each. (A) Find the revenue function. (B) Find the number of air conditioners that must be sold to maximize the revenue, the corresponding price to the nearest dollar, and the maximum revenue to the nearest dollar.
1970
75
356
1980
253
1,100
1990
714
2,814
2000
1,353
4,789
90. PROFIT Refer to Problem 89. The cost of manufacturing 8,000BTU window air conditioners is given by
2007
2,241
7,421
C(x) 0004 10,000 0002 90x where C(x) is the total cost in dollars of producing x air conditioners. (A) Find the profit function. (B) Find the number of air conditioners that must be sold to maximize the profit, the corresponding price to the nearest dollar, and the maximum profit to the nearest dollar. 91. CONSTRUCTION A rectangular container measuring 1 foot by 2 feet by 4 feet is covered with a layer of lead shielding of uniform thickness (see the figure). (A) Find the volume of lead shielding V as a function of the thickness x (in feet) of the shielding. (B) Find the volume of the lead shielding if the thickness of the shielding is 0.05 feet.
4
1 2 Lead shielding
Source: U.S. Census Bureau.
(A) Let x represent the number of years since 1960 and find a cubic regression polynomial for the total national expenditures. (B) Use the polynomial model from part A to estimate the total national expenditures (to the nearest billion) for 2018. 94. HEALTH CARE Refer to Table 3. (A) Let x represent the number of years since 1960 and find a cubic regression polynomial for the per capita expenditures. (B) Use the polynomial model from part A to estimate the per capita expenditures (to the nearest dollar) for 2018. 95. MARRIAGE Table 4 shows the marriage and divorce rates per 1,000 population for selected years since 1950.
Table 4 Marriages and Divorces (per 1,000 Population) Year
Marriages
Divorces
1950
11.1
2.6
1960
8.5
2.2
1970
10.6
3.5
1980
10.6
5.2
1990
9.8
4.7
2000
8.2
4.1
Source: U.S. Census Bureau.
(A) Let x represent the number of years since 1950 and find a cubic regression polynomial for the marriage rate. (B) Use the polynomial model from part A to estimate the marriage rate (to one decimal place) for 2016. 92. MANUFACTURING A rectangular storage container measuring 2 feet by 2 feet by 3 feet is coated with a protective coating of plastic of uniform thickness. (A) Find the volume of plastic V as a function of the thickness x (in feet) of the coating.
96. DIVORCE Refer to Table 4. (A) Let x represent the number of years since 1950 and find a cubic regression polynomial for the divorce rate. (B) Use the polynomial model from part A to estimate the divorce rate (to one decimal place) for 2016.
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4-2
Real Zeros and Polynomial Inequalities Z Upper and Lower Bounds for Real Zeros Z Location Theorem and Bisection Method Z Approximating Real Zeros at Turning Points Z Polynomial Inequalities Z Mathematical Modeling
The real zeros of a polynomial P(x) with real coefficients are just the x intercepts of the graph of P(x). So an obvious strategy for finding the real zeros consists of two steps: 1. 2.
Graph P(x). Approximate each x intercept.
In this section, we develop important tools for carrying out this strategy: the upper and lower bound theorem, which determines an interval [a, b] that is guaranteed to contain all x intercepts of P(x), and the bisection method, which permits approximation of x intercepts to any desired accuracy. We emphasize the approximation of real zeros in this section; the problem of finding zeros exactly, when possible, is considered in Section 4-3.
Z Upper and Lower Bounds for Real Zeros On which interval should you graph a polynomial P(x) in order to see all of its x intercepts? The answer is provided by the upper and lower bound theorem. This theorem explains how to find two numbers: a lower bound, which is less than or equal to all real zeros of the polynomial, and an upper bound, which is greater than or equal to all real zeros of the polynomial. A proof of Theorem 1 is outlined in Problems 67 and 68 of Exercises 4-2.
Z THEOREM 1 Upper and Lower Bound Theorem Let P(x) be a polynomial of degree n 7 0 with real coefficients, an 7 0: 1. Upper bound: A number r 7 0 is an upper bound for the real zeros of P(x) if, when P(x) is divided by x 0005 r by synthetic division, all numbers in the quotient row, including the remainder, are nonnegative. 2. Lower bound: A number r 6 0 is a lower bound for the real zeros of P(x) if, when P(x) is divided by x 0005 r by synthetic division, all numbers in the quotient row, including the remainder, alternate in sign. [Note: In the lower bound test, if 0 appears in one or more places in the quotient row, including the remainder, the sign in front of it can be considered either positive or negative, but not both. For example, the numbers 1, 0, 1 can be considered to alternate in sign, whereas 1, 0, 00051 cannot.]
EXAMPLE
1
Bounding Real Zeros Let P(x) 0002 x4 0005 2x 3 0005 10x 2 0003 40x 0005 90. Find the smallest positive integer and the largest negative integer that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x).
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SOLUTION
400
5
LB
1 2 3 4 5 00031 00032 00033 00034 00035
1 1 1 1 1 1 1 1 1 1 1
00032 00031 0 1 2 3 00033 00034 00035 00036 00037
000310 000311 000310 00037 00032 5 00037 00032 5 14 25
40 29 20 19 32 65 47 44 25 000316 000385
000390 000361 000350 000333 38 This quotient row is nonnegative; 235 ← E 5 is an upper bound (UB). 0003137 0003178 0003165 000326 This quotient row alternates in sign; 335 ← E ⴚ5 is a lower bound (LB).
The graph of P(x) 0004 x4 0003 2x 3 0003 10x 2 0005 40x 0003 90 for 00035 x 5 is shown in Figure 1. Theorem 1 guarantees that all the real zeros of P(x) are between 00035 and 5. We can be certain that the graph does not change direction and cross the x axis somewhere outside the viewing window in Figure 1. 0002
0003200 4 3 Z Figure 1 P(x) 0004 x 0003 2x 0003
10x2 0005 40x 0003 90.
Let P(x) 0004 x4 0003 5x 3 0003 x 2 0005 40x 0003 70. Find the smallest positive integer and the largest negative integer that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x). 0002
MATCHED PROBLEM 1
EXAMPLE
279
We perform synthetic division for r 0004 1, 2, 3, . . . until the quotient row turns nonnegative; then repeat this process for r 0004 00031, 00032, 00033, . . . until the quotient row alternates in sign. We organize these results in the synthetic division table shown below. In a synthetic division table we dispense with writing the product of r with each coefficient in the quotient and simply list the results in the table.
UB
00035
Real Zeros and Polynomial Inequalities
Bounding Real Zeros
2
Let P(x) 0004 x3 0003 30x 2 0005 275x 0003 720. Find the smallest positive integer multiple of 10 and the largest negative integer multiple of 10 that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x). SOLUTION
We construct a synthetic division table to search for bounds for the zeros of P(x). The size of the coefficients in P(x) indicates that we can speed up this search by choosing larger increments between test values.
100
000310
30
UB LB
10 20 30 000310
1 1 1 1 1
000330 000320 000310 0 000340
275 75 75 275 675
0003720 30 780 7,530 00037,470
0003100 3 2 Z Figure 2 P(x) 0004 x 0003 30x 0005
275x 0003 720.
MATCHED PROBLEM 2
Therefore, all real zeros of P(x) 0004 x3 0003 30x2 0005 275x 0003 720 must lie between 000310 and 30, as confirmed by Figure 2. 0002 Let P(x) 0004 x 3 0003 25x 2 0005 170x 0003 170. Find the smallest positive integer multiple of 10 and the largest negative integer multiple of 10 that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x). 0002
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Technology Connections How do you determine the correct viewing window for graphing a function? This is one of the most frequently asked questions about graphing calculators. For polynomial functions, the upper and lower bound theorem gives an answer: let Xmin and Xmax be the lower and upper bounds, respectively, of Theorem 1 (appropriate values
5
00055
for Ymin and Ymax can then be found using TRACE). We can approximate the zeros, all of which appear in the chosen viewing window, using the ZERO command. The upper and lower bound theorem and the ZERO command on a graphing calculator are two important mathematical tools that work very well together.
Z Location Theorem and Bisection Method 5
00055
The graph of every polynomial function is continuous. Because the polynomial function P(x) 0002 x 5 0003 3x 0005 1 is negative when x 0002 0 [P(0) 0002 00051] and positive when x 0002 1 [P(1) 0002 3], the graph of P(x) must cross the x axis at least once between x 0002 0 and x 0002 1 (Fig. 3). This observation is the basis for Theorem 2 and leads to a simple method for approximating zeros.
5 Z Figure 3 P(x) 0002 x 0003 3x 0005 1.
Z THEOREM 2 Location Theorem* Suppose that a function f is continuous on an interval I that contains numbers a and b. If f (a) and f (b) have opposite signs, then the graph of f has at least one x intercept between a and b.
The conclusion of Theorem 2 says that at least one zero of the function is “located” between a and b. There may be more than one zero between a and b: if g(x) 0002 x3 0003 x2 0005 2x 0005 1, then g(00052) and g(2) have opposite signs and there are three zeros between x 0002 00052 and x 0002 2 [Fig. 4(a)]. The converse of Theorem 2 is false: h(x) 0002 x2 has an x intercept at x 0002 0 but does not change sign [Fig. 4(b)]. 5
00055
5
5
00055
5
00055
00055
(a)
(b)
Z Figure 4 Polynomials may or may not change sign at a zero.
ZZZ EXPLORE-DISCUSS 1
When synthetic division is used to divide a polynomial P(x) by x 0005 3 the remainder is 000533. When the same polynomial is divided by x 0005 4 the remainder is 38. Must P(x) have a zero between 3 and 4? Explain.
*The location theorem is a formulation of the important intermediate value theorem of calculus.
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Explore-Discuss 2 will provide an introduction to the repeated systematic application of the location theorem (Theorem 2) called the bisection method. This method forms the basis for the zero approximation routines in many graphing calculators.
ZZZ EXPLORE-DISCUSS 2
Let P(x) 0002 x5 0003 3x 0005 1. Because P(0) is negative and P(1) is positive, the location theorem guarantees that P(x) must have at least one zero in the interval (0, 1). (A) Is P(0.5) positive or negative? Does the location theorem guarantee a zero of P(x) in the interval (0, 0.5) or in (0.5, 1)? (B) Let m be the midpoint of the interval from part A that contains a zero of P(x). Is P(m) positive or negative? What does this tell you about the location of the zero? (C) Explain how this process could be used repeatedly to approximate a zero to any desired accuracy.
The bisection method is a systematic application of the procedure suggested in Explore-Discuss 2: Let P(x) be a polynomial with real coefficients. If P(x) has opposite signs at the endpoints of an interval (a, b), then by the location theorem P(x) has a zero in (a, b). Bisect this interval (that is, find the midpoint m 0002 a 00032 b), check the sign of P(m), and select the interval (a, m) or (m, b) that has opposite signs at the endpoints. We repeat this bisection procedure (producing a set of intervals, each contained in and half the length of the previous interval, and each containing a zero) until the desired accuracy is obtained. If at any point in the process P(m) 0002 0, we stop, because a real zero m has been found. Example 3 illustrates the procedure, and clarifies when the procedure is finished.
EXAMPLE
3
The Bisection Method The polynomial P(x) 0002 x4 0005 2x3 0005 10x2 0003 40x 0005 90 of Example 2 has a zero between 3 and 4. Use the bisection method to approximate it to one-decimal-place accuracy.
SOLUTION
We organize the results of our calculations in Table 1. Because the sign of P(x) changes at the endpoints of the interval (3.5625, 3.625), we conclude that a real zero lies in this interval and is given by r 0002 3.6 to one-decimal place accuracy (each endpoint rounds to 3.6).
Table 1 Bisection Approximation Sign Change Interval (a, b)
Midpoint m
(3, 4)
Sign of P P(a)
P(m)
P(b)
3.5
0005
0005
0003
(3.5, 4)
3.75
0005
0003
0003
(3.5, 3.75)
3.625
0005
0003
0003
(3.5, 3.625)
3.5625
0005
0005
0003
(3.5625, 3.625)
We stop here
0005
0003
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Figure 5 illustrates the nested intervals produced by the bisection method in Table 1. Match each step in Table 1 with an interval in Figure 5. Note how each interval that contains a zero gets smaller and smaller and is contained in the preceding interval that contained the zero. 3.5625
(
3.625
( ()
3
3.5
)
3.75
)
4
x
Z Figure 5 Nested intervals produced by the bisection method in Table 1.
If we had wanted two-decimal-place accuracy, we would have continued the process in Table 1 until the endpoints of a sign change interval rounded to the same two-decimal-place number. 0002 MATCHED PROBLEM 3
The polynomial P(x) 0002 x4 0005 2x3 0005 10x2 0003 40x 0005 90 of Example 1 has a zero between 00055 and 00054. Use the bisection method to approximate it to one-decimal-place accuracy. 0002
Z Approximating Real Zeros at Turning Points The bisection method for approximating zeros fails if a polynomial has a turning point at a zero, because the polynomial does not change sign at such a zero. Most graphing calculators use methods that are more sophisticated than the bisection method. Nevertheless, it is not unusual to get an error message when using the zero command to approximate a zero that is also a turning point. In this case, we can use the maximum or minimum command, as appropriate, to approximate the turning point, and the zero.
EXAMPLE
4
Approximating Zeros at Turning Points Let P(x) 0002 x5 0003 6x4 0003 4x3 0005 24x2 0005 16x 0003 32. Find the smallest positive integer and the largest negative integer that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x). Approximate the zeros to two decimal places, using maximum or minimum commands to approximate any zeros at turning points.
SOLUTION
The pertinent rows of a synthetic division table show that 2 is the upper bound and 00056 is the lower bound:
1 2 00055 00056
1 1 1 1 1
6 7 8 1 0
4 11 20 00051 4
000524 000513 16 000519 000548
000516 000529 16 79 272
32 3 64 0005363 00051600
Examining the graph of P(x) we find three zeros: the zero 00053.24, found using the MAXIMUM command [Fig. 6(a)]; the zero 00052, found using the ZERO command [Fig. 6(b)]; and the zero 1.24, found using the MINIMUM command [Fig. 6(c)].
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40
40
00056
2
40
00056
2
00056
000540
000540
(a)
(b)
(c)
0002
Z Figure 6 Zeros of P(x) 0002 x 0003 6x 0003 4x 0005 24x 0005 16x 0003 32.
MATCHED PROBLEM 4
2
000540
5
4
3
283
2
Let P(x) 0002 x5 0005 6x4 0003 40x2 0005 12x 0005 72. Find the smallest positive integer and the largest negative integer that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x). Approximate the zeros to two decimal places, using maximum or minimum commands to approximate any zeros at turning points. 0002
Z Polynomial Inequalities We can apply the techniques we have introduced for finding real zeros to solve polynomial inequalities. Consider, for example, the inequality x3 0005 2x2 0005 5x 0003 6 7 0 The real zeros of P(x) 0002 x3 0005 2x2 0005 5x 0003 6 are easily found to be 00052, 1, and 3. They partition the x axis into four intervals (00050007, 00052), (00052, 1), (1, 3),
and
(3, 0007)
On any one of these intervals, the graph of P is either above the x axis or below the x axis, because, by the location theorem, a continuous function can change sign only at a zero. One way to decide whether the graph of P is above or below the x axis on a given interval, say (00052, 1), is to choose a “test number” that belongs to the interval, 0, for example, and evaluate P at the test number. Because P(0) 0002 6 0, the graph of P is above the x axis throughout the interval (00052, 1). A second way to decide whether the graph of P is above or below the x axis on (00052, 1) is to simply inspect the graph of P. Each technique has its advantages, and both are illustrated in the solutions to Examples 5 and 6.
EXAMPLE
5
Solving Polynomial Inequalities Solve the inequality x3 0005 2x2 0005 5x 0003 6 0.
SOLUTION
Let P(x) 0002 x3 0005 2x2 0005 5x 0003 6. Then P(1) 0002 13 0005 2(12) 0005 5 0003 6 0002 0 so 1 is a zero of P and x 0005 1 is a factor. Dividing P(x) by x 0005 1 (details omitted) gives the quotient x2 – x 0005 6. Therefore, P(x) 0002 (x 0005 1)(x2 0005 x 0005 6) 0002 (x 0005 1)(x 0003 2)(x 0005 3) The zeros of P are 00052, 1, and 3. They partition the x axis into the four intervals shown in the table on page 284. A test number is chosen from each interval as indicated to determine whether P(x) is positive (above the x axis) or negative (below the x axis) on that interval.
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Interval Test number x P(x) Sign of P
(00030006, 00032)
(00032, 1)
(3, 0006)
00033
0
2
4
000324
6
00034
18
0003
0002
0003
0002
(1, 3)
We conclude that the solution set of the inequality is the intervals where P(x) is positive: (00032, 1) ´ (3, 0006) MATCHED PROBLEM 5
0002
Solve the inequality x3 0002 x2 0003 x 0003 1 0007 0. 0002
EXAMPLE
6
Solving Polynomial Inequalities with a Graphing Calculator Solve 3x2 0002 12x 0003 4 2x3 0003 5x2 0002 7 to three decimal places.
SOLUTION
Subtracting the right-hand side gives the equivalent inequality P(x) 0004 00032x3 0002 8x2 0002 12x 0003 11 0 The zeros of P(x), to three decimal places, are 00031.651, 0.669, and 4.983 [Fig. 7(a)]. By inspecting the graph of P we see that P is above the x axis on the intervals (00030006, 00031.651) and (0.669, 4.983). So the solution set of the inequality is (00030006, 00031.651] ´ [0.669, 4.983] The square brackets indicate that the endpoints of the intervals—the zeros of the polynomial— also satisfy the inequality. An alternative to inspecting the graph of P is to inspect the graph of f (x) 0004
P(x) 0004 P(x) 0004
The function f (x) has the value 1 if P(x) is positive, because then the absolute value of P(x) is equal to P(x). Similarly, f(x) has the value 00031 if P(x) is negative. This technique makes it easy to identify the solution set of the original inequality [Fig. 7(b)] and often eliminates difficulties in choosing appropriate window variables. 100
000310
10
10
0003100
(a) P(x) ⴝ ⴚ2x3 ⴙ 8x2 ⴙ 12x ⴚ 11
Z Figure 7
MATCHED PROBLEM 6
000310
10
000310
(b) f (x) ⴝ
P(x) 0004P(x)0004
0002
Solve to three decimal places 5x3 0003 13x 0007 4x2 0002 10x 0003 5. 0002
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Z Mathematical Modeling EXAMPLE
7
Construction An oil tank is in the shape of a right circular cylinder with a hemisphere at each end (Fig. 8). The cylinder is 55 inches long, and the volume of the tank is 11,000 cubic inches (approximately 20 cubic feet). Let x denote the common radius of the hemispheres and the cylinder.
x
x
55 inches
Z Figure 8
(A) Find a polynomial equation that x must satisfy. (B) Approximate x to one decimal place. SOLUTIONS
(A) If x is the common radius of the hemispheres and the cylinder in inches, then °
Volume Volume Volume of ¢ 0004 ° of two ¢ 0002 ° of ¢ tank hemispheres cylinder 4 3 11,000 0004 0002 55 x2 3 x 33,000 0004 4x3 0002 165x2 0 0004 4x3 0002 165x2 0003 33,000
Multiply by 30005 . Subtract 33,000 from both sides.
The radius we are looking for (x) must be a positive zero of P(x) 0004 4x3 0002 165x2 0003 33,000 (B) Because the coefficients of P(x) are large, we use larger increments in the synthetic division table:
70,000
0
20
UB
165 205 245
0 2,050 4,900
000333,000 000312,500 65,000
Applying the bisection method to the interval [10, 20] (nine midpoints are calculated; details omitted) or graphing y 0004 P(x) for 0 0005 x 0005 20 (Fig. 9), we see that x 0004 12.4 inches (to one decimal place). 0002
000370,000
P(x) 0004 4x 0002 165x2 0003 33,000. Z Figure 9 3
MATCHED PROBLEM 7
10 20
4 4 4
Repeat Example 7 if the volume of the tank is 44,000 cubic inches. 0002 ANSWERS TO MATCHED PROBLEMS 1. 3. 5. 7.
Lower bound: 00033; upper bound: 6 2. Lower bound: 000310; upper bound: 30 x 0004 00034.1 4. Lower bound: 00032; upper bound: 6; 00031.65, 2, 3.65 6. (00030006, 00031.899) 傼 (0.212, 2.488) (00030006, 00031) 傼 (00031, 1) (A) P(x) 0004 4x3 0002 165x2 0003 132,000 0004 0 (B) 22.7 inches
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Exercises
1. Given a polynomial of degree n 0007 0, explain why there must exist an upper bound and a lower bound for its real zeros. 2. State the location theorem in your own words. 3. A polynomial P has degree 6 and leading coefficient 1. If synthetic division by x 0003 5 results in all positive numbers in the quotient row, is 10 an upper bound for the real zeros of P? Explain. 4. A polynomial has degree 12 and leading coefficient 1. If synthetic division by x 0005 5 results in numbers that alternate in sign in the quotient row, is 000310 a lower bound for the real zeros of P? Explain. 5. Explain the basic steps in the bisection method. 6. If you use the bisection method to approximate a real root to three decimal place accuracy, explain how you can tell when the method is finished. In Problems 7–10, approximate the real zeros of each polynomial to three decimal places.
21. P(x) 0004 x4 0003 3x3 0005 4x2 0005 2x 0003 9 22. P(x) 0004 x4 0003 4x3 0005 6x2 0003 4x 0003 7 23. P(x) 0004 x5 0003 3x3 0005 3x2 0005 2x 0003 2 24. P(x) 0004 x5 0003 3x4 0005 3x2 0005 2x 0003 1
In Problems 25–30, (A) use the location theorem to explain why the polynomial function has a zero in the indicated interval; and (B) determine the number of additional intervals required by the bisection method to obtain a one-decimal-place approximation to the zero and state the approximate value of the zero. 25. P(x) 0004 x3 0003 2x2 0003 5x 0005 4; (3, 4) 26. P(x) 0004 x3 0005 x2 0003 4x 0003 1; (1, 2) 27. P(x) 0004 x3 0003 2x2 0003 x 0005 5; (00032, 00031) 28. P(x) 0004 x3 0003 3x2 0003 x 0003 2; (3, 4)
7. P(x) 0004 x2 0005 5x 0003 2
29. P(x) 0004 x4 0003 2x3 0003 7x2 0005 9x 0005 7; (3, 4)
8. P(x) 0004 3x2 0003 7x 0005 1
30. P(x) 0004 x4 0003 x3 0003 9x2 0005 9x 0005 4; (2, 3)
9. P(x) 0004 2x3 0003 5x 0005 2 10. P(x) 0004 x3 0003 4x2 0003 8x 0005 3 In Problems 11–14, use the graph of P(x) to write the solution set for each inequality.
In Problems 31–36, (A) find the smallest positive integer and largest negative integer that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x); and (B) use the bisection method to approximate a real zero of each polynomial to one decimal place. 31. P(x) 0004 x3 0003 2x2 0005 3x 0003 8
20
32. P(x) 0004 x3 0005 3x2 0005 4x 0005 5 00035
5
33. P(x) 0004 2x3 0005 x2 0005 2x 0005 1 34. P(x) 0004 2x3 0003 x2 0005 4x 0003 2 35. P(x) 0004 x4 0005 x2 0003 6
36. P(x) 0004 x4 0003 2x2 0003 3
000320
11. P(x) 0
12. P(x) 0
13. P(x) 0007 0
14. P(x) 0
Problems 37–40, refer to the polynomial P(x) 0004 (x 0003 1)2(x 0003 2)(x 0003 3)4
In Problems 15–18, solve each polynomial inequality to three decimal places (note the connection with Problems 7–10).
37. Can the zero at x 0004 1 be approximated by the bisection method? Explain.
15. x2 0005 5x 0003 2 0007 0
16. 3x2 0003 7x 0005 1 0
17. 2x3 0003 5x 0005 2 0
18. x3 0003 4x2 0003 8x 0005 3 0
38. Can the zero at x 0004 2 be approximated by the bisection method? Explain.
Find the smallest positive integer and largest negative integer that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of each of the polynomials given in Problems 19–24. 19. P(x) 0004 x3 0003 3x 0005 1
20. P(x) 0004 x3 0003 4x2 0005 4
39. Can the zero at x 0004 3 be approximated by the bisection method? Explain. 40. Which of the zeros can be approximated by a maximum approximation routine? By a minimum approximation routine? By the zero approximation routine on your graphing calculator?
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In Problems 41–46, approximate the zeros of each polynomial function to two decimal places, using maximum or minimum commands to approximate any zeros at turning points. 41. P(x) 0004 x4 0003 4x3 0003 10x2 0002 28x 0002 49 42. P(x) 0004 x4 0002 4x3 0003 4x2 0003 16x 0002 16 43. P(x) 0004 x5 0003 6x4 0002 4x3 0002 24x2 0003 16x 0003 32 5
4
3
2
44. P(x) 0004 x 0003 6x 0002 2x 0002 28x 0003 15x 0002 2 45. P(x) 0004 x5 0003 6x4 0002 11x3 0003 4x2 0003 3.75x 0003 0.5 46. P(x) 0004 x5 0002 12x4 0002 47x3 0002 56x2 0003 15.75x 0002 1
Real Zeros and Polynomial Inequalities
287
67. Give a reason for each step in the proof of the upper bound case of Theorem 1 on page 278. Step 1: P(x) can be written in the form P(x) 0004 (x 0003 r)Q(x) 0002 R, where the coefficients of Q(x) and R are positive. Step 2: Suppose s r 0. Then P(s) 0. Step 3: r is an upper bound for the real zeros of P(x). 68. Give a reason for each step in the proof of the lower bound case of Theorem 1 on page 278. Step 1: P(x) can be written in the form P(x) 0004 (x 0003 r)Q(x) 0002 R, where the coefficients of Q(x) and R alternate in sign. Step 2: Suppose s 0007 r 0007 0. If P has even degree, then P(s) 0; if P has odd degree, then P(s) 0007 0. Step 3: r is a lower bound for the real zeros of P(x).
In Problems 47–52, solve each polynomial inequality. 47. x2 9
48. 1 0003 x2 0005 0
49. x3 0005 16x
50. 2x x2 0002 x3
51. x4 0002 4 5x2
69. Let P(x) be a polynomial of degree n 0 such that all of the coefficients of P(x) are nonnegative. Explain why 0 is an upper bound for the real zeros of P(x).
52. 2 0002 x 0002 x2 0002 x3 0007 x4 In Problems 53–58, solve each polynomial inequality to three decimal places. 53. x2 0002 7x 0003 3 0005 x3 0002 x 0002 4
Problems 69 and 70 explore the cases in which 0 is an upper bound or lower bound for the real zeros of a polynomial. These cases are not covered by Theorem 1, the upper and lower bound theorem, as formulated on page 278.
54. x4 0002 1 3x2
55. x4 0007 8x3 0003 17x2 0002 9x 0003 2 56. x3 0002 5x 2x3 0003 4x2 0002 6 57. (x2 0002 2x 0003 2)2 2 58. 5 0002 2x 0007 (x2 0003 4)2 In Problems 59–64, (A) find the smallest positive integer multiple of 10 and largest negative integer multiple of 10 that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of each polynomial; and (B) approximate the real zeros of each polynomial to two decimal places. 59. P(x) 0004 x3 0003 24x2 0003 25x 0002 10 60. P(x) 0004 x3 0003 37x2 0002 70x 0003 20 61. P(x) 0004 x4 0002 12x3 0003 900x2 0002 5,000 62. P(x) 0004 x4 0003 12x3 0003 425x2 0002 7,000
70. Let P(x) be a polynomial of degree n 0 such that an 0 and the coefficients of P(x) alternate in sign (as in Theorem 1, a coefficient 0 can be considered either positive or negative, but not both). Explain why 0 is a lower bound for the real zeros of P(x).
APPLICATIONS Express the solutions to Problems 71–76 as the roots of a polynomial equation of the form P(x) ⫽ 0 and approximate these solutions to one decimal place. 71. GEOMETRY Find all points on the graph of y 0004 x2 that are one unit away from the point (1, 2). [Hint: Use the distance formula from Section 2-2.] 72. GEOMETRY Find all points on the graph of y 0004 x2 that are one unit away from the point (2, 1). 73. MANUFACTURING A box is to be made out of a piece of cardboard that measures 18 by 24 inches. Squares, x inches on a side, will be cut from each corner, and then the ends and sides will be folded up (see the figure). Find the value of x that would result in a box with a volume of 600 cubic inches.
63. P(x) 0004 x4 0003 100x2 0003 1,000x 0003 5,000
24 in.
64. P(x) 0004 x4 0003 5x3 0003 50x2 0003 500x 0002 7,000
66. When synthetic division is used to divide a polynomial Q(x) by x 0002 4 the remainder is 10. When the same polynomial is divided by x 0002 5 the remainder is 8. Could Q(x) have a zero between 00035 and 00034? Explain.
18 in.
65. When synthetic division is used to divide a polynomial P(x) by x 0002 4 the remainder is 10. When the same polynomial is divided by x 0002 5 the remainder is 00038. Must P(x) have a zero between 00035 and 00034? Explain.
x x
74. MANUFACTURING A box with a hinged lid is to be made out of a piece of cardboard that measures 20 by 40 inches. Six squares, x inches on a side, will be cut from each corner and the middle, and then the ends and sides will be folded up to form the box and its lid
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(see the figure). Find the value of x that would result in a box with a volume of 500 cubic inches.
20 in.
40 in. x
x
76. SHIPPING A shipping box is reinforced with steel bands in all three directions (see the figure). A total of 20.5 feet of steel tape is to be used, with 6 inches of waste because of a 2-inch overlap in each direction. If the box has a square base and a volume of 2 cubic feet, find the side length of the base.
75. CONSTRUCTION A propane gas tank is in the shape of a right circular cylinder with a hemisphere at each end (see the figure). If the overall length of the tank is 10 feet and the volume is 20 cubic feet, find the common radius of the hemispheres and the cylinder.
y x x
x x
10 feet
4-3
Complex Zeros and Rational Zeros of Polynomials Z The Fundamental Theorem of Algebra Z Factors of Polynomials with Real Coefficients Z Graphs of Polynomials with Real Coefficients Z Rational Zeros
The graph of the polynomial function P(x) 0002 x2 0003 4 does not cross the x axis, so P(x) has no real zeros. It does, however, have complex zeros, 2i and 00052i; by the factor theorem, x2 0003 4 0002 (x 0005 2i)(x 0003 2i). The fundamental theorem of algebra guarantees that every nonconstant polynomial with real or complex coefficients has a complex zero; as a result, such a polynomial can be factored as a product of linear factors. In Section 4-3, we study the fundamental theorem and its implications, including results on the graphs of polynomials with real coefficients. Finally, we consider a problem that has led to important advances in mathematics and its applications: When can zeros of a polynomial be found exactly?
Z The Fundamental Theorem of Algebra The fundamental theorem of algebra was proved by Karl Friedrich Gauss (1777–1855), one of the greatest mathematicians of all time, in his doctoral thesis. A proof of the theorem is beyond the scope of this book, so we will state and use it without proof.
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Z THEOREM 1 Fundamental Theorem of Algebra Every polynomial of degree n > 0 with complex coefficients has a complex zero.
If P(x) is a polynomial of degree n 0 with complex coefficients, then by Theorem 1 it has a zero r1. So x 0005 r1 is a factor of P(x) by Theorem 5 of Section 4-1, and P(x) 0002 (x 0005 r1)Q(x), deg Q(x) 0002 n 0005 1 Now, if deg Q(x) 0, then, applying the fundamental theorem to Q(x), Q(x) has a root r2 and therefore a factor x 0005 r2. (It is possible that r2 is equal to r1.) By continuing this reasoning we obtain a proof of Theorem 2.
Z THEOREM 2 n Linear Factors Theorem Every polynomial of degree n 0 with complex coefficients can be factored as a product of n linear factors.
Suppose that a polynomial P(x) is factored as a product of n linear factors. Any zero r of P(x) must be a zero of one or more of the factors. The number of linear factors that have zero r is said to be the multiplicity of r. For example, the polynomial P(x) 0002 (x 0005 5)3(x 0003 1)2(x 0005 6i)(x 0003 2 0003 3i)
(1)
has degree 7 and is written as a product of seven linear factors. P(x) has just four zeros, namely 5, 00051, 6i, and 00052 0005 3i. Because the factor x 0005 5 appears to the power 3, we say that the zero 5 has multiplicity 3. Similarly, 00051 has multiplicity 2, 6i has multiplicity 1, and 00052 0005 3i has multiplicity 1. A zero of multiplicity 2 is called a double zero, and a zero of multiplicity 3 is called triple zero. Note that the sum of the multiplicities is always equal to the degree of the polynomial: for P(x) in equation (1), 3 0003 2 0003 1 0003 1 0002 7.
EXAMPLE
1
Multiplicities of Zeros Find the zeros and their multiplicities: (A) P(x) 0002 (x 0003 2)7(x 0005 4)8(x2 0003 1) (B) Q(x) 0002 (x 0003 1)3(x2 0005 1)(x 0003 1 0005 i)
SOLUTIONS
(A) Note that x2 0003 1 0002 0 has the solutions i and 0005i. The zeros of P(x) are 00052 (multiplicity 7), 4 (multiplicity 8), i and 0005i (each multiplicity 1). (B) Note that x2 0005 1 0002 (x 0005 1)(x 0003 1), so x 0003 1 appears four times as a factor of Q(x). The zeros of Q(x) are 00051 (multiplicity 4), 1 (multiplicity 1), and 00051 0003 i (multiplicity 1). 0002
MATCHED PROBLEM 1
Find the zeros and their multiplicities: (A) P(x) 0002 (x 0005 5)3(x 0003 3)2(x2 0003 16) (B) Q(x) 0002 (x2 0005 25)3(x 0003 5)(x 0005 i) 0002
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Z Factors of Polynomials with Real Coefficients If p 0002 qi is a zero of P(x) 0004 ax2 0002 bx 0002 c, where a, b, c, p, and q are real numbers, then P( p 0002 qi) 0004 0 a( p 0002 qi) 0002 b( p 0002 qi) 0002 c 0004 0 a( p 0002 qi)2 0002 b( p 0002 qi) 0002 c 0004 0 a ( p 0002 qi)2 0002 b ( p 0002 qi) 0002 c 0004 0 a( p 0003 qi)2 0002 b( p 0003 qi) 0002 c 0004 0 2
Take the conjugate of both sides. z ⴙ w ⴝ z ⴙ w, zw ⴝ z w z ⴝ z if z is real, p ⴙ qi ⴝ p ⴚ qi
P( p 0003 qi) 0004 0 Therefore, p 0003 qi is also a zero of P(x). This method of proof can be applied to any polynomial P(x) of degree n 0 with real coefficients, justifying Theorem 3.
Z THEOREM 3 Imaginary Zeros of Polynomials with Real Coefficients Imaginary zeros of polynomials with real coefficients, if they exist, occur in conjugate pairs.
If a polynomial P(x) of degree n 0 has real coefficients and a linear factor of the form x 0003 ( p 0002 qi) where q 0, then, by Theorem 3, P(x) also has the linear factor x 0003 ( p 0003 qi). But [x 0003 ( p 0002 qi)][x 0003 ( p 0003 qi)] 0004 x2 0003 2px 0002 p2 0003 q2 which is a quadratic factor of P(x) with real coefficients and imaginary zeros. By this reasoning we can prove Theorem 4.
Z THEOREM 4 Linear and Quadratic Factors Theorem* If P(x) is a polynomial of degree n 0 with real coefficients, then P(x) can be factored as a product of linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros).
EXAMPLE
2
Factors of Polynomials Factor P(x) 0004 x3 0002 x2 0002 4x 0002 4 in two ways: (A) As a product of linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros) (B) As a product of linear factors with complex coefficients
SOLUTIONS
(A) Note that P(00031) 0004 0, so 00031 is a zero of P(x) (or graph P(x) and note that 00031 is an x intercept). Therefore, x 0002 1 is a factor of P(x). Using synthetic division, the quotient is x2 0002 4, which has imaginary roots. Therefore, P(x) 0004 (x 0002 1)(x2 0002 4) *Theorem 4 underlies the technique of decomposing a rational function into partial fractions, which is useful in calculus. See Appendix A, Section A-2.
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An alternative solution is to factor by grouping: x3 0003 x2 0003 4x 0003 4 0002 x2(x 0003 1) 0003 4(x 0003 1) 0002 (x2 0003 4)(x 0003 1) (B) Because x2 0003 4 has roots 2i and 00052i, P(x) 0002 (x 0003 1)(x 0005 2i)(x 0003 2i)
0002
Factor P(x) 0002 x5 0005 x4 0005 x 0003 1 in two ways:
MATCHED PROBLEM 2
(A) As a product of linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros) (B) As a product of linear factors with complex coefficients 0002
Z Graphs of Polynomials with Real Coefficients The factorization described in Theorem 4 gives additional information about the graphs of polynomial functions with real coefficients. For certain polynomials the factorization of Theorem 4 will involve only linear factors; for others, only quadratic factors. Of course if only quadratic factors are present, then the degree of the polynomial P(x) must be even. In other words, a polynomial P(x) of odd degree with real coefficients must have a linear factor with real coefficients. This proves Theorem 5.
Z THEOREM 5 Real Zeros and Polynomials of Odd Degree Every polynomial of odd degree with real coefficients has at least one real zero, and consequently at least one x intercept.
ZZZ EXPLORE-DISCUSS 1 3
00053
3
The graph of the polynomial P(x) 0002 x(x 0005 1)2(x 0003 1)4(x 0005 2)3 is shown in Figure 1. Find the real zeros of P(x) and their multiplicities. How can a real zero of even multiplicity be distinguished from a real zero of odd multiplicity using only the graph?
For polynomials with real coefficients, as suggested by Explore-Discuss 1, you can easily distinguish real zeros of even multiplicity from those of odd multiplicity using only the graph. Theorem 6, which we state without proof, tells how to do that.
00053
Z Figure 1 Graph of P(x) 0002 x(x 0005 1)2(x 0003 1)4(x 0005 2)3.
Z THEOREM 6 Zeros of Even or Odd Multiplicity Let P(x) be a polynomial with real coefficients: 1. If r is a real zero of P(x) of even multiplicity, then P(x) has a turning point at r and does not change sign at r. (The graph just touches the x axis, then changes direction.) 2. If r is a real zero of P(x) of odd multiplicity, then P(x) does not have a turning point at r and changes sign at r. (The graph continues through to the opposite side of the x axis.)
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3
Multiplicities from Graphs Figure 2 shows the graph of a polynomial function of degree 6. Find the real zeros and their multiplicities. 5
00055
5
00055
Z Figure 2 SOLUTION
Figure 3 shows the graph of a polynomial function of degree 7. Find the real zeros and their multiplicities. 0002
MATCHED PROBLEM 3 5
00054
4
000510
Z Figure 3 10
000510
The numbers 00052, 00051, 1, and 2 are real zeros (x intercepts). The graph has turning points at x 0002 00061 but not at x 0002 00062. Therefore, by Theorem 6, the zeros 00051 and 1 have even multiplicity, and 00052 and 2 have odd multiplicity. Because the sum of the multiplicities must equal 6 (the degree), the zeros 00051 and 1 each have multiplicity 2, and the zeros 00052 and 2 each have multiplicity 1. 0002
10
000510 2 00059 Z Figure 4 P(x) 0002 x 0005 (4 0003 10 ).
Z Rational Zeros From a graphical perspective, finding a zero of a polynomial means finding a good approximation to an actual zero. A graphing calculator, for example, might give 2 as a zero of P(x) 0002 x2 0005 (4 0003 1000059) even though P(2) is equal to 00051000059, not 0 (Fig. 4). It is natural, however, to want to find zeros exactly. Although this is impossible in general, we will adopt an algebraic strategy to find exact zeros in a special case, that of rational zeros of polynomials with rational coefficients. We will find a graphing calculator to be helpful in carrying out the algebraic strategy. First note that a polynomial with rational coefficients can always be written as a constant times a polynomial with integer coefficients. For example, 1 3 2 2 7 x 0005 x 0003 x00035 2 3 4 1 0002 (6x3 0005 8x2 0003 21x 0003 60) 12
P(x) 0002
Because the zeros of P(x) are the zeros of 6x3 0005 8x2 0003 21x 0003 60, it is sufficient, for the purpose of finding rational zeros of polynomials with rational coefficients, to study just the polynomials with integer coefficients. We introduce the rational zero theorem by examining the following quadratic polynomial whose zeros can be found easily by factoring: P(x) 0002 6x2 0005 13x 0005 5 0002 (2x 0005 5)(3x 0003 1) 5 00051 1 Zeros of P(x): and 0005 0002 2 3 3
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Notice that the numerators 5 and 00051 of the zeros are both integer factors of 00055, the constant term in P(x). The denominators 2 and 3 of the zeros are both integer factors of 6, the coefficient of the highest-degree term in P(x). These observations are generalized in Theorem 7 (a proof is outlined in Problem 89 of Exercises 4-3). Z THEOREM 7 Rational Zero Theorem If the rational number b0002c, in lowest terms, is a zero of the polynomial P(x) 0002 an xn 0003 an00051 xn00051 0003 . . . 0003 a1x 0003 a0
an 0004 0
with integer coefficients, then b must be an integer factor of a0 and c must be an integer factor of an. P(x) ⴝ anxn ⴙ anⴚ1xnⴚ1 ⴙ . . . ⴙ a1x ⴙ a0 b c c must be a factor of an
b must be a factor of a0
Theorem 7 enables us to construct a finite list of possible rational zeros of P(x). Each number in the list must then be tested to determine whether or not it is actually a zero. As Example 4 illustrates, a graphing calculator can often reduce the effort required to locate rational zeros.
EXAMPLE
Finding Rational Zeros
4
Find all the rational zeros for P(x) 0002 2x3 0003 9x2 0003 7x 0005 6. SOLUTION
If b0002c in lowest terms is a rational zero of P(x), then b must be a factor of 00056 and c must be a factor of 2. Possible values of b are the integer factors of 00056: 00061, 00062, 00063, 00066 Possible values of c are the integer factors of 2: 00061, 00062
(2) (3)
Writing all possible fractions b0002c where b is from (2) and c is from (3), we have Possible rational zeros for P(x): 00061, 00062, 00063, 00066, 000612, 000632
(4)
[Note that all fractions are in lowest terms and duplicates like 00066000e00062 0002 00063 are not repeated.] If P(x) has any rational zeros, they must be in list (4). We can test each number r in this list simply by evaluating P(r). However, exploring the graph of y 0002 P(x) first will usually indicate which numbers in the list are the most likely candidates for zeros. Examining a graph of P(x), we see that there are zeros near 00053, near 00052, and between 0 and 1, so we begin by evaluating P(x) at 00053, 00052, and 12 (Fig. 5). 10
00055
Z Figure 5
10
5
00055
10
5
00055
5
000510
000510
000510
(a)
(b)
(c)
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Therefore, 00053, 00052, and 12 are rational zeros of P(x). Because a third-degree polynomial can have at most three zeros, we have found all the rational zeros. There is no need to test the remaining candidates in list (4). 0002 Find all rational zeros for P(x) 0002 2x3 0003 x2 0005 11x 0005 10.
MATCHED PROBLEM 4
0002 As we saw in the solution of Example 4, rational zeros can be located by simply evaluating the polynomial. However, if we want to find multiple zeros, imaginary zeros, or exact values of irrational zeros, we need to consider reduced polynomials. If r is a zero of a polynomial P(x), then we can write P(x) 0002 (x 0005 r)Q(x) where Q(x) is a polynomial of degree one less than the degree of P(x). The quotient polynomial Q(x) is called a reduced polynomial for P(x). In Example 4, after determining that 00053 is a zero of P(x), we can write 9 7 00056 00056 00059 6 00053 2 3 00052 0 3 2 P(x) 0002 2x 0003 9x 0003 7x 0005 6 0002 (x 0003 3)(2x2 0003 3x 0005 2) 0002 (x 0003 3)Q(x) 2
Because the reduced polynomial Q(x) 0002 2x2 0003 3x 0005 2 is a quadratic, we can find its zeros by factoring or the quadratic formula. We get P(x) 0002 (x 0003 3)(2x2 0003 3x 0005 2) 0002 (x 0003 3)(x 0003 2)(2x 0005 1) and we see that the zeros of P(x) are 00053, 00052, and 12, as before.
EXAMPLE
Finding Rational and Irrational Zeros
5
Find all zeros exactly for P(x) 0002 2x3 0005 7x2 0003 4x 0003 3. SOLUTION
00061, 00063, 000612, 000632
5
00055
First, list the possible rational zeros: Examining the graph of y 0002 P(x) (Fig. 6), we see that there is a zero between 00051 and 0, another between 1 and 2, and a third between 2 and 3. We test the only likely candidates, 000512 and 32: P(000512) 0002 00051
5
So 00055
Z Figure 6
3 2
and
P(32) 0002 0
is a zero, but 000512 is not. Using synthetic division (details omitted), we can write P(x) 0002 (x 0005 32)(2x2 0005 4x 0005 2)
Because the reduced polynomial is quadratic, we can use the quadratic formula to find the exact values of the remaining zeros: 2x2 0005 4x 0005 2 0002 0 x2 0005 2x 0005 1 0002 0 2 0006 14 0005 4(1)(00051) x0002 2 2 0006 212 0002 0002 1 0006 12 2
Divide both sides by 2. Use the quadratic formula.
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So the exact zeros of P(x) are
3 2
Complex Zeros and Rational Zeros of Polynomials
and 1 12.*
295 0002
Find all zeros exactly for P(x) 0004 3x3 0003 10x2 0005 5x 0005 4.
MATCHED PROBLEM 5
0002
EXAMPLE
6
Finding Rational and Imaginary Zeros Find all zeros exactly for P(x) 0004 x4 0003 6x3 0005 14x2 0003 14x 0005 5.
SOLUTION 5
The possible rational zeros are 1 and 5. Examining the graph of P(x) (Fig. 7), we see that 1 is a zero. Because the graph of P(x) does not appear to change sign at 1, this may be a multiple root. Using synthetic division (details omitted), we find that P(x) 0004 (x 0003 1)(x3 0003 5x2 0005 9x 0003 5)
00031
5
The possible rational zeros of the reduced polynomial Q(x) 0004 x3 0003 5x2 0005 9x 0003 5
00031
Z Figure 7
are 1 and 5. Examining the graph of Q(x) (Fig. 8), we see that 1 is a rational zero. After a division, we have a quadratic reduced polynomial: 5
00031
Q(x) 0004 (x 0003 1)Q1(x) 0004 (x 0003 1)(x2 0003 4x 0005 5) 5
We use the quadratic formula to find the zeros of Q1(x): x2 0003 4x 0005 5 0004 0 4 116 0003 4(1)(5) x0004 2 4 100034 00042 i 0004 2
00035
Z Figure 8
So the exact zeros of P(x) are 1 (multiplicity 2), 2 0003 i, and 2 0005 i. Find all zeros exactly for P(x) 0004 x4 0005 4x3 0005 10x2 0005 12x 0005 5.
MATCHED PROBLEM 6
0002
0002
REMARK
We were successful in finding all the zeros of the polynomials in Examples 5 and 6 because we could find sufficient rational zeros to reduce the original polynomial to a quadratic. This is not always possible. For example, the polynomial
50
00035
5
000350 3 Z Figure 9 P(x) 0004 x 0005 6x 0003 2.
P(x) 0004 x3 0005 6x 0003 2 has no rational zeros, but does have an irrational zero at x ⬇ 0.32748 (Fig. 9). The other two zeros are imaginary. The techniques we have developed will not find the exact value of these roots. *By analogy with Theorem 3 (imaginary zeros of polynomials with real coefficients occur in conjugate pairs), it can be shown that if r 0005 s1t is a zero of a polynomial with rational coefficients, where r, s, and t are rational but t is not the square of a rational, then r 0003 s1t is also a zero.
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ANSWERS TO MATCHED PROBLEMS 1. (A) 5 (multiplicity 3), 00033 (multiplicity 2), 4i and 00034i (each multiplicity 1) (B) 00035 (multiplicity 4), 5 (multiplicity 3), i (multiplicity 1) 2. (A) (x 0002 1)(x 0003 1)2(x2 0002 1) (B) (x 0002 1)(x 0003 1)2(x 0002 i)(x 0003 i) 3. 00033 (multiplicity 2), 00032 (multiplicity 1), 00031 (multiplicity 1), 0 (multiplicity 2), 1 (multiplicity 1) 4. 00032, 00031, 52 5. 43, 1 0003 12, 1 0002 12 6. 00031 (multiplicity 2), 00031 0003 2i, 00031 0002 2i
4-3
Exercises
1. Explain in your own words what the fundamental theorem of algebra says.
zeros are integers. Write the polynomial as a product of linear factors. Indicate the degree of the polynomial.
2. Does every polynomial of degree 0 with real coefficients have a real zero? Explain.
19.
P (x)
20.
P(x) 15
15
3. What is meant by the multiplicity of a zero of a polynomial? 4. If P(x) is a polynomial with integer coefficients and leading coefficient 1, explain why every rational zero of P(x) is actually an integer.
00035
Write the zeros of each polynomial in Problems 5–12, and indicate the multiplicity of each. What is the degree of each polynomial? 5. P(x) 0004 (x 0002 8)3(x 0003 6)2 6. P(x) 0004 (x 0003 5)(x 0002 7)2
5
x
00035
P (x)
22.
P(x) 15
15
7. P(x) 0004 3(x 0002 4)3(x 0003 3)2(x 0002 1)
x
000315
000315
21.
5
8. P(x) 0004 5(x 0003 2)3(x 0002 3)2(x 0003 1) 9. P(x) 0004 x3(2x 0002 1)2
00035
5
x
00035
5
x
10. P(x) 0004 6x2(5x 0003 4)(3x 0002 2) 11. P(x) 0004 (x2 0002 4)3(x2 0003 4)5(x 0002 2i)
000315
000315
12. P(x) 0004 (x2 0002 7x 0002 10)2(x2 0002 6x 0002 10)3 In Problems 13–18, find a polynomial P(x) of lowest degree, with leading coefficient 1, that has the indicated set of zeros. Write P(x) as a product of linear factors. Indicate the degree of P(x).
23.
P (x)
24.
P(x) 15
15
13. 3 (multiplicity 2) and 00034 14. 00032 (multiplicity 3) and 1 (multiplicity 2)
00035
5
x
00035
5
15. 00037 (multiplicity 3), 00033 0002 12, 00033 0003 12 16. 13 (multiplicity 2), 5 0002 17, 5 0003 17
000315
000315
17. (2 0003 3i), (2 0002 3i), 00034 (multiplicity 2) 18. i 13 (multiplicity 2), 0003i13 (multiplicity 2), and 4 (multiplicity 3) In Problems 19–24, find a polynomial of lowest degree, with leading coefficient 1, that has the indicated graph. Assume all
In Problems 25–28, factor each polynomial in two ways: (A) as a product of linear factors (with real coefficients) and
x
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quadratic factors (with real coefficients and imaginary zeros); and (B) as a product of linear factors with complex coefficients. 4
2
25. P(x) 0002 x 0003 5x 0003 4 4
297
58. P(x) 0002 x3 0005 8x2 0003 17x 0005 4 59. P(x) 0002 4x4 0005 4x3 0005 9x2 0003 x 0003 2 60. P(x) 0002 2x4 0003 3x3 0005 4x2 0005 3x 0003 2
2
26. P(x) 0002 x 0003 18x 0003 81 27. P(x) 0002 x3 0005 x2 0003 25x 0005 25 5
Complex Zeros and Rational Zeros of Polynomials
In Problems 61–68, find a polynomial P(x) that satisfies all of the given conditions. Write the polynomial using only real coefficients.
4
28. P(x) 0002 x 0003 x 0005 x 0005 1
61. 2 0005 5i is a zero; leading coefficient 1; degree 2 In Problems 29–34, list all possible rational zeros (Theorem 7) of a polynomial with integer coefficients that has the given leading coefficient an and constant term a0.
62. 4 0003 3i is a zero; leading coefficient 1; degree 2
29. an 0002 1, a0 0002 00054
30. an 0002 1, a0 0002 9
64. 1 0005 4i is a zero; P(0) 0002 51; degree 2
31. an 0002 10, a0 0002 1
32. an 0002 6, a0 0002 00051
65. 00055 and 8i are zeros; leading coefficient 1; degree 3
33. an 0002 7, a0 0002 00052
34. an 0002 3, a0 0002 8
66. 7 and 00052i are zeros; leading coefficient 1; degree 3
When searching for zeros of a polynomial, a graphing calculator often can be helpful in eliminating from consideration certain candidates for rational zeros.
63. 6 0003 i is a zero; P(0) 0002 74; degree 2
67. i and 1 0005 i are zeros; P(1) 0002 10; degree 4 68. 0005i and 3 0003 i are zeros; P(1) 0002 20; degree 4
In Problems 35–40, write P(x) as a product of linear factors.
In Problems 69–74, multiply.
35. P(x) 0002 x3 0003 9x2 0003 24x 0003 16; 00051 is a zero
69. [x 0005 (4 0005 5i)][x 0005 (4 0003 5i)]
36. P(x) 0002 x3 0005 4x2 0005 3x 0003 18; 3 is a double zero
70. [x 0005 (2 0005 3i)][x 0005 (2 0003 3i)]
37. P(x) 0002 x4 0003 2x2 0003 1; i is a double zero
71. [x 0005 (3 0003 4i)][x 0005 (3 0005 4i)]
38. P(x) 0002 x4 0005 1; 1 and 00051 are zeros
72. [x 0005 (5 0003 2i)][x 0005 (5 0005 2i)]
39. P(x) 0002 2x3 0005 17x2 0003 90x 0005 41; 12 is a zero
73. [x 0005 (a 0003 bi)][x 0005 (a 0005 bi)]
40. P(x) 0002 3x3 0005 10x2 0003 31x 0003 26; 000523 is a zero
74. (x 0005 bi)(x 0003 bi)
In Problems 41–48, find all roots exactly (rational, irrational, and imaginary) for each polynomial equation.
In Problems 75–80, find all other zeros of P(x), given the indicated zero.
41. 2x3 0005 5x2 0003 1 0002 0
75. P(x) 0002 x3 0005 5x2 0003 4x 0003 10; 3 0005 i is one zero
42. 2x3 0005 10x2 0003 12x 0005 4 0002 0
43. x4 0003 4x3 0005 x2 0005 20x 0005 20 0002 0
76. P(x) 0002 x3 0003 x2 0005 4x 0003 6; 1 0003 i is one zero
44. x4 0005 4x2 0005 4x 0005 1 0002 0
77. P(x) 0002 x3 0005 3x2 0003 25x 0005 75; 00055i is one zero
45. x4 0005 2x3 0005 5x2 0003 8x 0003 4 0002 0
78. P(x) 0002 x3 0003 2x2 0003 16x 0003 32; 4i is one zero
46. x4 0005 2x2 0005 16x 0005 15 0002 0
79. P(x) 0002 x4 0005 4x3 0003 3x2 0003 8x 0005 10; 2 0003 i is one zero
47. x4 0003 10x2 0003 9 0002 0
80. P(x) 0002 x4 0005 2x3 0003 7x2 0005 18x 0005 18; 00053i is one zero
48. x4 0003 29x2 0003 100 0002 0
In Problems 49–54, find all zeros exactly (rational, irrational, and imaginary) for each polynomial.
In Problems 81–86, final all zeros (rational, irrational, and imaginary) exactly.
49. P(x) 0002 x3 0005 19x 0003 30
81. P(x) 0002 3x3 0005 37x2 0003 84x 0005 24
4
51. P(x) 0002 x 0005 4
21 3 10 x
3
0003
53. P(x) 0002 x 0005 5x 0003
50. P(x) 0002 x3 0005 7x2 0003 36
2 5x 15 2 2x
4
52. P(x) 0002 x 0003
7 3 6x
0005
7 2 3x
0005
5 2x
0005 2x 0005 2
54. P(x) 0002 x4 0005 134x2 0005 52x 0005 14
83. P(x) 0002 4x4 0003 4x3 0003 49x2 0003 64x 0005 240 84. P(x) 0002 6x4 0003 35x3 0003 2x2 0005 233x 0005 360
In Problems 55–60, write each polynomial as a product of linear factors. 55. P(x) 0002 6x3 0003 13x2 0005 4
82. P(x) 0002 2x3 0005 9x2 0005 2x 0003 30
56. P(x) 0002 6x3 0005 17x2 0005 4x 0003 3
57. P(x) 0002 x3 0003 2x2 0005 9x 0005 4
85. P(x) 0002 4x4 0005 44x3 0003 145x2 0005 192x 0003 90 86. P(x) 0002 x5 0005 6x4 0003 6x3 0003 28x2 0005 72x 0003 48 87. The solutions to the equation x3 0005 1 0002 0 are all the cube roots of 1. (A) 1 is obviously a cube root of 1; find all others. (B) How many distinct cube roots of 1 are there?
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88. The solutions to the equation x3 0003 8 0004 0 are all the cube roots of 8. (A) 2 is obviously a cube root of 8; find all others. (B) How many distinct cube roots of 8 are there? 89. Give a reason for each step in the proof of the rational zero theorem, assuming that P(x) has degree two. Step 1: a2 (bc)2 0005 a1(bc) 0005 a0 0004 0 Step 2: a2b2 0005 a1bc 0005 a0c2 0004 0 Step 3: a2b2 0005 a1bc 0004 0003a0c2 Step 4: b is a factor of 0003a0c2, so b is a factor of a0. Step 5: Modify steps 3 and 4 to conclude that c is a factor of a2. 90. Explain how the ideas in Problem 89 can be adapted to give a proof of the rational zero theorem for P(x) of degree n.
much should this amount be to create a new storage unit with volume 10 times the old? 94. CONSTRUCTION A rectangular box has dimensions 1 by 1 by 2 feet. If each dimension is increased by the same amount, how much should this amount be to create a new box with volume six times the old? 95. PACKAGING An open box is to be made from a rectangular piece of cardboard that measures 8 by 5 inches, by cutting out squares of the same size from each corner and bending up the sides (see the figure). If the volume of the box is to be 14 cubic inches, how large a square should be cut from each corner? [Hint: Determine the domain of x from physical considerations before starting.] x
91. Given P(x) 0004 x2 0005 2ix 0003 5 with 2 0003 i a zero, show that 2 0005 i is not a zero of P(x). Does this contradict Theorem 3? Explain. 92. If P(x) and Q(x) are two polynomials of degree n, and if P(x) 0004 Q(x) for more than n values of x, then how are P(x) and Q(x) related? [Hint: Consider the polynomial D(x) 0004 P(x) 0003 Q(x).]
x
x
x x
APPLICATIONS Find all rational solutions exactly, and find irrational solutions to one decimal place. 93. STORAGE A rectangular storage unit has dimensions 1 by 2 by 3 feet. If each dimension is increased by the same amount, how
4-4
x
x
x
96. FABRICATION An open metal chemical tank is to be made from a rectangular piece of stainless steel that measures 10 by 8 feet, by cutting out squares of the same size from each corner and bending up the sides (see the figure for Problem 95). If the volume of the tank is to be 48 cubic feet, how large a square should be cut from each corner?
Rational Functions and Inequalities Z Rational Functions and Properties of Their Graphs Z Vertical and Horizontal Asymptotes Z Analyzing the Graph of a Rational Function Z Rational Inequalities
In Section 4-4, we will apply our knowledge of graphs and zeros of polynomial functions to study the graphs of rational functions, that is, functions that are quotients of polynomials. Our goal will be to produce hand sketches that clearly show all of the important features of the graph.
Z Rational Functions and Properties of Their Graphs The number 137 is called a rational number because it is a quotient (or ratio) of integers. The function f (x) 0004
x00051 x 0003x00036 2
is called a rational function because it is a quotient of polynomials.
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Rational Functions and Inequalities
299
Z DEFINITION 1 Rational Function
5
A function f is a rational function if it can be written in the form 00035
5
f (x) 0002
x
(1, 2)
p(x) q(x)
where p(x) and q(x) are polynomials.
00035
(a) f(x) ⴝ
(x ⴚ 1)(x2 ⴚ 3)
When working with rational functions, we will assume that the coefficients of p(x) and q(x) are real numbers, and that the domain of f is the set of all real numbers x such that q(x) ⴝ 0.
xⴚ1 y 5
If a real number c is a zero of both p(x) and q(x), then, by the factor theorem, x 0003 c is a factor of both p(x) and q(x). The graphs of
00035
5
x
f (x) 0002
p(x) (x 0003 c)pr (x) 0002 q(x) (x 0003 c)qr(x)
and
fr(x) 0002
pr(x) qr(x)
are then identical, except possibly for a “hole” at x 0002 c (Fig. 1). Later in this section we will explain how to handle the minor complication caused by common real zeros of p(x) and q(x). But to avoid that complication now,
00035
(b) f(x) ⴝ x2 ⴚ 3
unless stated to the contrary, we will assume that for any rational function f we consider, p(x) and q(x) have no real zero in common.
Z Figure 1
Because a polynomial q(x) of degree n has at most n real zeros, there are at most n real numbers that are not in the domain of f (x) 0002 P(x)0004q(x). Because a fraction equals 0 only if its numerator is 0, the x intercepts of the graph of f are the real zeros of a polynomial p(x) of degree m. So the number of x intercepts of f is at most m.
EXAMPLE
1
Domain and x Intercepts Find the domain and x intercepts for f (x) 0002
SOLUTION
f (x) 0002
2x2 0003 2x 0003 4 . x2 0003 9
p(x) 2(x 0003 2)(x 1) 2x2 0003 2x 0003 4 0002 0002 2 q(x) (x 0003 3)(x 3) x 00039
Because q(x) 0002 0 for x 0002 3 and x 0002 00033, the domain of f is x 0006 00073
(00030005, 00033) 傼 (00033, 3) 傼 (3, 0005)
or
Because p(x) 0002 0 for x 0002 2 and x 0002 00031, the zeros of f, and the x intercepts of f, are 00031 and 2. 0002
MATCHED PROBLEM 1
Find the domain and x intercepts for f (x) 0002
3x2 0003 12 . x 2x 0003 3 2
0002 The graph of the rational function f (x) 0002 is shown in Figure 2 on the next page.
x2 0003 1.44 x3 0003 x
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Z Figure 2 f (x) 0002
x2 0003 1.44 x3 0003 x
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POLYNOMIAL AND RATIONAL FUNCTIONS y
. 5
00035
5
x
00035
The domain of f consists of all real numbers except x 0002 00031, x 0002 0, and x 0002 1 (the zeros of the denominator x3 0003 x). The dotted vertical lines at x 0002 00071 indicate that those values of x are excluded from the domain (a dotted vertical line at x 0002 0 would coincide with the y axis and is omitted). The graph is discontinuous at x 0002 00031, x 0002 0, and x 0002 1, but is continuous elsewhere and has no sharp corners. The zeros of f are the zeros of the numerator x2 0003 1.44, namely x 0002 00031.2 and x 0002 1.2. The graph of f has four turning points. Its left and right behavior is the same as that of the function g (x) 0002 1x (the graph is close to the x axis for very large and very small values of x). The graph of f illustrates the general properties of rational functions that are listed in Theorem 1. We have already justified Property 3; the other properties are established in calculus.
Z THEOREM 1 Properties of Rational Functions Let f (x) 0002 p(x)0004q(x) be a rational function where p(x) and q(x) are polynomials of degrees m and n, respectively. Then the graph of f(x): 1. 2. 3. 4. 5.
Is continuous with the exception of at most n real numbers Has no sharp corners Has at most m real zeros Has at most m n 0003 1 turning points Has the same left and right behavior as the quotient of the leading terms of p(x) and q(x)
Figure 3 shows graphs of several rational functions, illustrating the properties of Theorem 1. y
y
Z Figure 3 Graphs of rational functions.
5
3
00035
5
x
2
00033
3
1 x
(b) g(x) ⴝ
x
00032
2
00032
00033
00035
(a) f(x) ⴝ
y
1 x2 ⴚ 1
(c) h(x) ⴝ
1 x2 ⴙ 1
x
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SECTION 4–4 y
y
Z Figure 3 (continued) 15
5
x
EXAMPLE
2
2
00033
x
3
000315
(d) F (x) ⴝ
y
3
00035
000310
10
(e) G(x) ⴝ
x
00032
00033
x2 ⴙ 3x xⴚ1
301
Rational Functions and Inequalities
ⴚx ⴚ 1
(f) H(x) ⴝ
x ⴚ 4x 3
x2 ⴙ x ⴙ 1 x2 ⴙ 1
Properties of Graphs of Rational Functions Use Theorem 1 to explain why each graph is not the graph of a rational function.
y
(A)
y
(B)
3
y
(C)
3
00033
3
x
3
00033
00033
x
3
00033
00033
SOLUTIONS
3
x
00033
(A) The graph has a sharp corner when x 0002 0, so Property 2 is not satisfied. (B) The graph has an infinite number of turning points, so Property 4 is not satisfied. (C) The graph has an infinite number of zeros (all values of x between 0 and 1, inclusive, are zeros), so Property 3 is not satisfied. 0002 Use Theorem 1 to explain why each graph is not the graph of a rational function.
MATCHED PROBLEM 2
y
(A)
y
(B) 3
3
00033
3
00033
y
(C)
x
00033
3
3
00033
x
00033
3
x
00033
0002
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y
Z Vertical and Horizontal Asymptotes
5
00065
5
x
The graphs of Figure 3 on pages 300–301 exhibit similar behaviors near points of discontinuity that can be described using the concept of vertical asymptote. Consider, for example, the rational function f (x) 0002 1x of Figure 3(a). As x approaches 0 from the right, the points (x, 1x ) on the graph have larger and larger y coordinates—that is, 1x increases without bound—as confirmed by Table 1. We write this symbolically as
00065
1 (a) f(x) ⴝ x Z Figure 3(a) Graphs of rational functions.
1 S0004 x
as
x S 00003
and say that the line x 0002 0 (the y axis) is a vertical asymptote for the graph of f.
Table 1 Behavior of 1兾x as x S 00003 x
1
0.1
0.01
0.001
0.0001
0.000 01
0.000 001
. . .
x approaches 0 from the right (x S 0 ⴙ )
1兾x
1
10
100
1,000
10,000
100,000
1,000,000
. . .
1兾x increases without bound (1 0005x S 0004)
If x approaches 0 from the left, the points (x, 1x ) on the graph have smaller and smaller y coordinates—that is, 1x decreases without bound—as confirmed by Table 2. We write this symbolically as 1 S 00060004 x
as
x S 00006
Table 2 Behavior of 1兾x as x S 00006 x
00061
00060.1
00060.01
00060.001
00060.0001
00060.000 01
00060.000 001
. . .
x approaches 0 from the left (x S 0ⴚ)
1兾x
00061
000610
0006100
00061,000
000610,000
0006100,000
00061,000,000
. . .
1兾x decreases without bound (1 0005x S ⴚ0004)
ZZZ EXPLORE-DISCUSS 1
Construct tables similar to Tables 1 and 2 for g(x) 0002 x12 and discuss the behavior of the graph of g(x) near x 0002 0.
Z DEFINITION 2 Vertical Asymptote The vertical line x 0002 a is a vertical asymptote for the graph of y 0002 f (x) if f(x) S 0004
or f(x) S 00060004
as x S a0003
or as x S a0006
(that is, if f (x) either increases or decreases without bound as x approaches a from the right or from the left).
Z THEOREM 2 Vertical Asymptotes of Rational Functions Let f (x) 0002 p(x)0005q(x) be a rational function. If a is a zero of q(x), then the line x 0002 a is a vertical asymptote of the graph of f.*
*Recall that we are assuming that p(x) and q(x) have no real zero in common. Theorem 2 is not valid without this assumption.
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303
For example, x2 0006 1.44 x2 0006 1.44 0002 3 x(x 0006 1)(x 0003 1) x 0006x
f (x) 0002
has three vertical asymptotes, x 0002 00061, x 0002 0, and x 0002 1 (see Fig. 2 on p. 300). The left and right behavior of some, but not all, rational functions can be described using the concept of horizontal asymptote. Consider f (x) 0002 1x . As values of x get larger and larger— that is, as x increases without bound—the points (x, 1x ) have y coordinates that are positive and approach 0, as confirmed by Table 3. Similarly, as values of x get smaller and smaller— that is, as x decreases without bound—the points (x, 1x ) have y coordinates that are negative and approach 0, as confirmed by Table 4. We write these facts symbolically as 1 S0 x
as
xS0004
and as
x S 00060004
and say that the line y 0002 0 (the x axis) is a horizontal asymptote for the graph of f.
Table 3 Behavior of 1/x as x S 0004 x
1
10
100
1,000
10,000
100,000
1,000,000
. . .
x increases without bound (x S 0004)
1兾x
1
0.1
0.01
0.001
0.0001
0.000 01
0.000 001
. . .
1兾x approaches 0 (1兾x S 0)
Table 4 Behavior of 1/x as x S 00060004 x
00061
000610
0006100
00061,000
000610,000
0006100,000
00061,000,000
. . .
x decreases without bound (x S ⴚ0004)
1兾x
00061
00060.1
00060.01
00060.001
00060.0001
00060.000 01
00060.000 001
. . .
1兾x approaches 0 (1兾x S 0)
ZZZ EXPLORE-DISCUSS 2
Construct tables similar to Tables 3 and 4 for each of the following functions, and discuss the behavior of each as x S 0004 and as x S 00060004: (A) f (x) 0002
3x 2 x 00031
(B) g(x) 0002
3x2 x2 0003 1
(C) h(x) 0002
3x3 x2 0003 1
Z DEFINITION 3 Horizontal Asymptote The horizontal line y 0002 b is a horizontal asymptote for the graph of y 0002 f (x) if f (x) S b
as
x S 00060004
or as
xS0004
(that is, if f (x) approaches b as x increases without bound or as x decreases without bound).
A rational function f (x) 0002 p(x)0005q(x) has the same left and right behavior as the quotient of the leading terms of p(x) and q(x) (Property 5 of Theorem 1). Consequently, a rational function has at most one horizontal asymptote. Moreover, we can determine easily whether a rational function has a horizontal asymptote, and if it does, find its equation. Theorem 3 gives the details.
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Z THEOREM 3 Horizontal Asymptotes of Rational Functions Consider the rational function f (x) 0002
am xm p a1x a0 bn xn p b1x b0
where am 0006 0, bn 0006 0. 1. If m n, the line y 0002 0 (the x axis) is a horizontal asymptote. 2. If m 0002 n, the line y 0002 am 0004bn is a horizontal asymptote. 3. If m n, there is no horizontal asymptote. In 1 and 2, the graph of f approaches the horizontal asymptote both as x S 0005 and as x S 00030005.
EXAMPLE
3
Finding Vertical and Horizontal Asymptotes for a Rational Function Find all vertical and horizontal asymptotes for f (x) 0002
SOLUTION
p(x) 2x2 0003 2x 0003 4 0002 q(x) x2 0003 9
Because q(x) 0002 x2 0003 9 0002 (x 0003 3)(x 3), the graph of f (x) has vertical asymptotes at x 0002 3 and x 0002 00033 (Theorem 1). Because p(x) and q(x) have the same degree, the line y
0002
a2 * 2 0002 00022 b2 1
a2 ⴝ 2, b2 ⴝ 1
is a horizontal asymptote (Theorem 3, part 2). MATCHED PROBLEM 3
0002
Find all vertical and horizontal asymptotes for f (x) 0002
3x2 0003 12 x 2x 0003 3 2
0002
Z Analyzing the Graph of a Rational Function We now use the techniques for locating asymptotes, along with other graphing aids discussed in the text, to graph several rational functions. First, we outline a systematic approach to the problem of graphing rational functions.
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
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Z ANALYZING AND SKETCHING THE GRAPH OF A RATIONAL FUNCTION: f(x) ⴝ p(x)冒q(x) Step 1. Intercepts. Find the real solutions of the equation p(x) 0002 0 and use these solutions to plot any x intercepts of the graph of f. Evaluate f (0), if it exists, and plot the y intercept. Step 2. Vertical Asymptotes. Find the real solutions of the equation q(x) 0002 0 and use these solutions to determine the domain of f, the points of discontinuity, and the vertical asymptotes. Sketch any vertical asymptotes as dashed lines. Step 3. Horizontal Asymptotes. Determine whether there is a horizontal asymptote and, if so, sketch it as a dashed line. Step 4. Complete the Sketch. For each interval in the domain of f, plot additional points and join them with a smooth continuous curve.
EXAMPLE
4
Graphing a Rational Function Graph f (x) 0002
2x . x00063 f (x) 0002
SOLUTION
p(x) 2x 0002 x00063 q(x)
Step 1. Intercepts. Find real zeros of p(x) 0002 2x and find f(0): 2x 0002 0 x00020 f (0) 0002 0
x intercept y intercept
The graph crosses the coordinate axes only at the origin. Plot this intercept, as shown in Figure 4. y Horizontal asymptote
10
000610
x and y intercepts
Vertical asymptote
10
x
000610
Intercepts and asymptotes
Z Figure 4 y
Step 2. Vertical Asymptotes. Find real zeros of q(x) 0002 x 0006 3:
10
000610
10
2x f(x) 0002 x00063 000610
Z Figure 5
x0006300020 x00023 x
The domain of f is (00060004, 3) 傼 (3, 0004), f is discontinuous at x 0002 3, and the graph has a vertical asymptote at x 0002 3. Sketch this asymptote, as shown in Figure 4. Step 3. Horizontal Asymptote. Because p(x) and q(x) have the same degree, the line y 0002 2 is a horizontal asymptote, as shown in Figure 4. Step 4. Complete the Sketch. By plotting a few additional points, we obtain the graph in Figure 5. Notice that the graph is a smooth continuous curve over the interval
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(00030005, 3) and over the interval (3, 0005). As expected, there is a break in the graph at x 0002 3. 0002 MATCHED PROBLEM 4
Proceed as in Example 4 and sketch the graph of f (x) 0002
3x . x2 0002
Technology Connections Refer to Example 4. When f (x) ⴝ 2x0004(x ⴚ 3) is graphed on a graphing calculator [Fig. 6(a)], it appears that the graphing calculator has also drawn the vertical asymptote at x ⴝ 3, but this is not the case. Many graphing calculators, when set in connected mode, calculate points on a graph and connect these points with line segments. The last point plotted to the left of the asymptote and the first plotted to the right of the asymptote will usually have very large y coordinates. If these y coordinates have opposite signs, then the graphing
calculator may connect the two points with a nearly vertical line segment, which gives the appearance of an asymptote. If you wish, you can set the calculator in dot mode to plot the points without the connecting line segments [Fig. 6(b)]. Depending on the scale, a graph may even appear to be continuous at a vertical asymptote [Fig. 6(c)]. It is important to always locate the vertical asymptotes as we did in step 2 before turning to the graphing calculator graph to complete the sketch. 10
10
000310
10
40
000310
10
000340
40
000310
000310
000340
(a) Connected mode
(b) Dot mode
(c) Connected mode
2x Z Figure 6 Graphing calculator graphs of f (x) 0002 x 0003 3 .
In Examples 5 and 6 we will just list the results of each step in the graphing strategy and omit the computational details.
EXAMPLE
5
Graphing a Rational Function Graph f (x) 0002
SOLUTION
x2 0003 6x 9 . x2 x 0003 2 f (x) 0002
(x 0003 3)2 x2 0003 6x 9 0002 (x 2)(x 0003 1) x2 x 0003 2
x intercept: x 0002 3 y intercept: f (0) 0002 000392 0002 00034.5 Domain: (00030005, 00032) 傼 (00032, 1) 傼 (1, 0005) Points of discontinuity: x 0002 00032 and x 0002 1 Vertical asymptotes: x 0002 00032 and x 0002 1 Horizontal asymptote: y 0002 1 Locate the intercepts, draw the asymptotes, and plot additional points in each interval of the domain of f to complete the graph (Fig. 7).
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307
y 10
000310
f (x) 0002
x 2 0003 6x 9 x2 x 0003 2
10
x
0002
Z Figure 7
MATCHED PROBLEM 5
ZZZ
Graph f (x) 0002
x2 . x2 0003 7x 10
The graph of a function cannot cross a vertical asymptote, but the same statement is not true for horizontal asymptotes. The rational function
CAUTION ZZZ
f (x) 0002
2x6 x5 0003 5x3 4x 2 x6 1
has the line y 0002 2 as a horizontal asymptote. The graph of f in Figure 8 clearly shows that the graph of a function can cross a horizontal asymptote. The definition of a horizontal asymptote requires f (x) to approach b as x increases or decreases without bound, but it does not preclude the possibility that f (x) 0002 b for one or more values of x. y 4
f (x) 0002
2x 6 x 5 0003 5x 3 4x 2 x6 1 y 0002 2 is a horizontal asymptote
00035
5
x
Z Figure 8 Multiple intersections of a graph and a horizontal asymptote.
EXAMPLE
6
Graphing a Rational Function Graph f (x) 0002
SOLUTION
x2 0003 3x 0003 4 . x00032
(x 1)(x 0003 4) x2 0003 3x 0003 4 0002 x00032 x00032 x intercepts: x 0002 00031 and x 0002 4 y intercept: f (0) 0002 2 Domain: (00030005, 2) 傼 (2, 0005) Points of discontinuity: x 0002 2 Vertical asymptote: x 0002 2 No horizontal asymptote f (x) 0002
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Although the graph of f does not have a horizontal asymptote, we can still gain some useful information about the behavior of the graph as x S 00060004 and as x S 0004 if we first perform a long division: x00061 x 0006 2冄 x2 0006 3x 0006 4 x2 0006 2x 0006x 0006 4 0006x 0003 2 00066
Quotient
Remainder
This shows that f (x) 0002
x2 0006 3x 0006 4 6 0002x000610006 x00062 x00062
As x S 00060004 or x S 0004, 60005(x 0006 2) S 0 and the graph of f approaches the line y 0002 x 0006 1. This line is called an oblique asymptote for the graph of f. The asymptotes and intercepts are shown in Figure 9, and the graph of f is sketched in Figure 10. y y
10 10
000610
Oblique asymptote y0002x00061
10
x 000610
10
f (x) 0002
000610
x
x 2 0006 3x 0006 4 x00062
000610
Intercepts and asymptotes
Z Figure 9
y0002x00061
Z Figure 10
Generalizing the results of Example 6, we have Theorem 4.
Z THEOREM 4 Oblique Asymptotes and Rational Functions If f (x) 0002 p(x) 0005q(x), where p(x) and q(x) are polynomials and the degree of p(x) is 1 more than the degree of q(x), then f(x) can be expressed in the form f (x) 0002 mx 0003 b 0003
r(x) q(x)
where the degree of r(x) is less than the degree of q(x). The line y 0002 mx 0003 b is an oblique asymptote for the graph of f. That is, [ f (x) 0006 (mx 0003 b)] S 0
as
x S 00060004
or
xS0004 0002
MATCHED PROBLEM 6
Graph, including any oblique asymptotes, f (x) 0002
x2 0003 5 . x00031 0002
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At the beginning of this section we made the assumption that for a rational function f (x) 0002 p(x)0004q(x), the polynomials p(x) and q(x) have no common real zero. Now we abandon that assumption. Suppose that p(x) and q(x) have one or more real zeros in common. Then, by the factor theorem, p(x) and q(x) have one or more linear factors in common, so f(x) can be “reduced.” We proceed to divide out common linear factors in f (x) 0002
p(x) q(x)
fr(x) 0002
pr(x) qr(x)
until we obtain a rational function
in which pr(x) and qr(x) have no common real zero. We analyze and graph fr(x), then insert “holes” as required in the graph of fr to obtain the graph of f. Example 7 illustrates the details.
EXAMPLE
7
Graphing Arbitrary Rational Functions Graph f (x) 0002
SOLUTION
2x5 0003 4x4 0003 6x3 . x5 0003 3x4 0003 3x3 7x2 6x
The real zeros of p(x) 0002 2x5 0003 4x4 0003 6x3 (obtained by graphing or factoring) are 00031, 0, and 3. The real zeros of q(x) 0002 x5 0003 3x4 0003 3x3 7x2 6x are 00031, 0, 2, and 3. The common zeros are 00031, 0, and 3. Factoring and dividing out common linear factors gives f (x) 0002
2x3(x 1)(x 0003 3) x(x 1)2(x 0003 2)(x 0003 3)
and
fr (x) 0002
2x2 (x 1)(x 0003 2)
We analyze fr (x) as usual: x intercept: x 0002 0 y intercept: fr(0) 0002 0 Domain: (00030005, 00031) 傼 (00031, 2) 傼 (2, 0005) Points of discontinuity: x 0002 00031, x 0002 2 Vertical asymptotes: x 0002 00031, x 0002 2 Horizontal asymptote: y 0002 2 The graph of f is identical to the graph of fr except possibly at the common real zeros 00031, 0, and 3. We consider each common zero separately. x 0002 00031: Both f and fr are undefined (no difference in their graphs). x 0002 0: f is undefined but fr(0) 0002 0, so the graph of f has a hole at (0, 0). x 0002 3: f is undefined but fr(3) 0002 4.5, so the graph of f has a hole at (3, 4.5). Therefore, f (x) has the following analysis: x intercepts: none y intercepts: none Domain: (00030005, 00031) 傼 (00031, 0) 傼 (0, 2) 傼 (2, 3) 傼 (3, 0005) Points of discontinuity: x 0002 00031, x 0002 0, x 0002 2, x 0002 3
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Vertical asymptotes: x 0002 00031, x 0002 2 Horizontal asymptote: y 0002 2 Holes: (0, 0), (3, 4.5) Figure 11 shows the graphs of f and fr. y
y
5
5
00035
x
5
00035
2x ⴚ 4x ⴚ 6x 4
3
(b) fr(x) ⴝ
x ⴚ 3x ⴚ 3x ⴙ 7x ⴙ 6x 5
5
4
3
2
2x2 (x ⴙ 1)(x ⴚ 2)
0002
Z Figure 11
MATCHED PROBLEM 7
Graph f (x) 0002
x
00035 5
(a) f (x) ⴝ
00035
x3 0003 x . x4 0003 x2 0002
Z Rational Inequalities A rational function f (x) 0002 p(x)0004q(x) can change sign at a real zero of p(x) (where f has an x intercept) or at a real zero of q(x) (where f is discontinuous), but nowhere else (because f is continuous except where it is not defined). Rational inequalities can therefore be solved in the same way as polynomial inequalities, except that the partition of the x axis is determined by the zeros of p(x) and the zeros of q(x).
EXAMPLE
8
Solving Rational Inequalities Solve
SOLUTION
x3 4x2 6 0. x2 0003 4
Let f (x) 0002
p(x) x3 4x2 0002 2 q(x) x 00034
The zeros of p(x) 0002 x3 4x2 0002 x2(x 4) are 0 and 00034. The zeros of q(x) 0002 x2 0003 4 0002 (x 2)(x 0003 2) are 00032 and 2. These four zeros partition the x axis into the five intervals shown in the table. A test number is chosen from each interval as indicated to determine whether f (x) is positive or negative.
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Interval
Test number x
f (x)
Sign of f
(0003 0005, 00034)
00035
000325/21
0003
(00034, 00032)
00033
9/5
(00032, 0)
00031
(0, 2)
1
(2, 0005)
3
00031
0003
00035/3
0003
311
63/5
We conclude that the solution set of the inequality is (00030005, 00034) 傼 (00032, 0) 傼 (0, 2)
Solve
MATCHED PROBLEM 8
EXAMPLE
9
x2 0003 1 0. x2 0003 9
0002
Solving Rational Inequalities with a Graphing Calculator Solve 1
SOLUTION
9x 0003 9 to three decimal places. x x00033 2
First we convert the inequality to an equivalent inequality in which one side is 0:
10
1000b
000310
10003
10
9x 0003 9 x2 x 0003 3
9x 0003 9 0 x x00033
x 2 ⴚ 8x ⴙ 6
Subtract
9x ⴚ 9 x2 ⴙ x ⴚ 3
from both sides.
Find a common denominator.
2
x2 x 0003 3 9x 0003 9 0003 2 0 x2 x 0003 3 x x00033
000310
(a) f (x) ⴝ
0002
Simplify.
x2 0003 8x 6 0 x2 x 0003 3
x2 ⴙ x ⴚ 3
The zeros of x2 0003 8x 6, to three decimal places, are 0.838 and 7.162. The zeros of x2 x 0003 3 are 00032.303 and 1.303. These four zeros partition the x axis into five intervals:
10
000310
(00030005, 00032.303), (00032.303, 0.838), (0.838, 1.303), (1.303, 7.162), and (7.162, 0005)
10
We graph f (x) 0002
000310
(b) g(x) ⴝ
f (x) 冟 f (x) 冟
Z Figure 12
x2 0003 8x 6 x2 x 0003 3
and
g(x) 0002
f (x) 冟 f (x) 冟
(Fig. 12) and observe that the graph of f is above the x axis on the intervals (00030005, 00032.303), (0.838, 1.303), and (7.162, 0005). So the solution set of the inequality is (00030005, 00032.303) 傼 [0.838, 1.303) 傼 [7.162, 0005) Note that the endpoints that are zeros of f are included in the solution set of the inequality, but not the endpoints at which f is undefined. 0002
MATCHED PROBLEM 9
Solve
x3 4x2 0003 7 0 to three decimal places. x2 0003 5x 0003 1 0002
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ANSWERS TO MATCHED PROBLEMS 1. Domain: (00030005, 00033) ´ (00033, 1) ´ (1, 0005); x intercepts: x 0002 00032, x 0002 2 2. (A) Properties 3 and 4 are not satisfied. (B) Property 1 is not satisfied. (C) Properties 1 and 3 are not satisfied. 3. Vertical asymptotes: x 0002 00033, x 0002 1; horizontal asymptote: y 0002 3 y
4.
y
5. 10
10
000310
10
f (x) 0002
x
000310
3x x2
000310
f (x) 0002
6.
y
x
10
x2 x 2 0003 7x 10
7.
y0002x00031
y 5
f (x) 0002 000310
10
f (x) 0002
x
00035
x3 0003 x x4 0003 x2 5
x
x2 5 x1 00035
8. (00030005, 00033) 傼 [00031, 1] 傼 (3, 0005)
4-4
9. [ 00033.391, 00031.773] 傼 (00030.193, 1.164] 傼 (5.193, 0005)
Exercises
1. Is every polynomial function a rational function? Explain. 2. Is every rational function a polynomial function? Explain.
y
7.
8.
y 10
10
3. Explain in your own words what a vertical asymptote is. 4. Explain in your own words what a horizontal asymptote is. 10
5. Explain in your own words what an oblique asymptote is. 6. Explain why a rational function can’t have both a horizontal asymptote and an oblique asymptote. In Problems 7–10, match each graph with one of the following functions: 2x 0003 4 x2 2x 4 h(x) 0002 x00032
f (x) 0002
2x 4 20003x 4 0003 2x k(x) 0002 x2 g(x) 0002
000310
x
10 000310
x
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9.
y
10.
10
10
000310
10
x
000310
10
Rational Functions and Inequalities
313
27. p(x) 0002
x2 2x 1 x
28. q(x) 0002
x3 0003 1 x1
29. h(x) 0002
3x2 8 2x2 6x
30. k(x) 0002
6x2 0003 5x 1 7x2 0003 28x
x
In Problems 31–34, explain why each graph is not the graph of a rational function. 000310
y
31. 5
2x 0003 4 . Complete each statement: 11. Let f (x) 0002 x2 0003 (A) As x S 00032 , f (x) S ? (B) As x S 00032, f (x) S ? (C) As x S 00030005, f (x) S ? (D) As x S 0005, f (x) S ? 2x 4 . Complete each statement: 20003x 0003 (A) As x S 2 , g(x) S ? (B) As x S 2, g(x) S ? (C) As x S 00030005, g(x) S ? (D) As x S 0005, g(x) S ?
00035
y
32. 5
13. Let h(x) 0002
4 0003 2x 14. Let k(x) 0002 . Complete each statement: x2 0003 (A) As x S 2 , k(x) S ? (B) As x S 2, k(x) S ? (C) As x S 00030005, k(x) S ? (D) As x S 0005, k(x) S ?
00035
3x 0003 9 x
16. g(x) 0002
2x 10 x1
17. h(x) 0002
x6 x2 0003 4
18. k(x) 0002
x2 0003 9 x
19. r(x) 0002
x2 0003 3x 0003 4 x2 1
20. s(x) 0002
x2 4x 0003 5 x2 4
21. F(x) 0002
x4 16 x2 0003 36
22. G(x) 0002
x4 x2 1 x2 0003 25
In Problems 23–30, find all vertical and horizontal asymptotes. 5x 1 23. f (x) 0002 x2 25. s(x) 0002
2x 0003 3 x2 0003 16
7x 0003 2 24. g(x) 0002 x00033 26. t(x) 0002
3x 4 x2 0003 49
5
x
00035
y
33. 3
00033
x
3
00033
In Problems 15–22, find the domain and x intercepts. 15. f (x) 0002
x
00035
12. Let g(x) 0002
2x 4 . Complete each statement: x00032 (A) As x S 20003, h(x) S ? (B) As x S 2, h(x) S ? (C) As x S 00030005, h(x) S ? (D) As x S 0005, h(x) S ?
5
y
34. 5
00035
5
x
00035
In Problems 35–38, explain how the graph of f differs from the graph of g. 35. f (x) 0002
x2 2x ; g(x) 0002 x 2 x
36. f (x) 0002
1 x5 ; g(x) 0002 x00035 x2 0003 25
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37. f (x) 0002
1 x2 ; g(x) 0002 x8 x2 10x 16
67.
9 5 0003 2 1 x x
68.
x4 7 2 x2 1
38. f (x) 0002
x2 0003 x 0003 12 ; g(x) 0002 x 3 x00034
69.
3x 2 7 10 x00035
70.
x 0.5 x2 5x 0003 6
71.
4 7 x x1
72.
x2 1 6 x2 0003 1 x4 1
In Problems 39–52, use the graphing strategy outlined in the text to sketch the graph of each function. 39. f (x) 0002
1 x00034
40. g(x) 0002
1 x3
In Problems 73–78, find all vertical, horizontal, and oblique asymptotes.
41. f (x) 0002
x x1
42. f (x) 0002
3x x00033
73. f (x) 0002
2x2 x00031
74. g(x) 0002
3x2 x2
43. g(x) 0002
1 0003 x2 x2
44. f (x) 0002
x2 1 x2
75. p(x) 0002
x3 x 1
76. q(x) 0002
x5 x 00038
45. f (x) 0002
9 x2 0003 9
46. g(x) 0002
6 x2 0003 x 0003 6
77. r(x) 0002
2x2 0003 3x 5 x
78. s(x) 0002
00033x2 5x 9 x
47. f (x) 0002
x x 00031
48. p(x) 0002
x 1 0003 x2
49. g(x) 0002
2 x2 1
50. f (x) 0002
x x2 1
12x2 (3x 5)2
52. f (x) 0002
51. f (x) 0002
2
7x2 (2x 0003 3)2
In Problems 53–56, give an example of a rational function that satisfies the given conditions. 53. Real zeros: 00032, 00031, 1, 2; vertical asymptotes: none; horizontal asymptote: y 0002 3 54. Real zeros: none; vertical asymptotes: x 0002 4; horizontal asymptote: y 0002 00032 55. Real zeros: none; vertical asymptotes: x 0002 10; oblique asymptote: y 0002 2x 5 56. Real zeros: 1, 2, 3; vertical asymptotes: none; oblique asymptote: y 0002 2 0003 x In Problems 57–64, solve each rational inequality.
2
3
In Problems 79–84, use the graphing strategy outlined in the text to sketch the graph of each function. Write the equations of all vertical, horizontal, and oblique asymptotes. 79. f (x) 0002
x2 1 x
80. g(x) 0002
x2 0003 1 x
81. k(x) 0002
x2 0003 4x 3 2x 0003 4
82. h(x) 0002
x2 x 0003 2 2x 0003 4
83. F(x) 0002
8 0003 x3 4x2
84. G(x) 0002
x4 1 x3
In calculus, it is often necessary to consider rational functions that are not in lowest terms, such as the functions given in Problems 85–88. For each function, state the domain. Write the equations of all vertical and horizontal asymptotes, and sketch the graph. 85. f (x) 0002
x2 0003 4 x00032
86. g(x) 0002
x2 0003 1 x1
87. r(x) 0002
x2 x2 0003 4
88. s(x) 0002
x00031 x2 0003 1
57.
x 0 x00032
58.
2x 0003 1 7 0 x3
APPLICATIONS
59.
x2 0003 16 7 0 5x 0003 2
60.
x00034 0 x2 0003 9
61.
x2 4x 0003 20 4 3x
62.
3x 0003 7 6 2 x2 6x
89. EMPLOYEE TRAINING A company producing electronic components used in television sets has established that on the average, a new employee can assemble N(t) components per day after t days of on-the-job training, as given by
9 5x 6 63. 2 x x 00031
1 1 64. 2 x x 8x 12
In Problems 65–72, solve each rational inequality to three decimal places. 65.
x2 7x 3 7 0 x2
66.
x3 4 0 x x00033 2
N(t) 0002
50t t4
t000b0
Sketch the graph of N, including any vertical or horizontal asymptotes. What does N approach as t S 0005? 90. PHYSIOLOGY In a study on the speed of muscle contraction in frogs under various loads, researchers W. O. Fems and J. Marsh found that the speed of contraction decreases with increasing loads. More precisely, they found that the relationship between
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speed of contraction S (in centimeters per second) and load w (in grams) is given approximately by S(w) 0002
26 0.06w w
w 5
Sketch the graph of S, including any vertical or horizontal asymptotes. What does S approach as w S 0005? 91. RETENTION An experiment on retention is conducted in a psychology class. Each student in the class is given 1 day to memorize the same list of 40 special characters. The lists are turned in at the end of the day, and for each succeeding day for 20 days each student is asked to turn in a list of as many of the symbols as can be recalled. Averages are taken, and it is found that a good approximation of the average number of symbols, N(t), retained after t days is given by N(t) 0002
5t 30 t
t 1
Sketch the graph of N, including any vertical or horizontal asymptotes. What does N approach as t S 0005? 92. LEARNING THEORY In 1917, L. L. Thurstone, a pioneer in quantitative learning theory, proposed the function f (x) 0002
a(x c) (x c) b
to describe the number of successful acts per unit time that a person could accomplish after x practice sessions. Suppose that for a particular person enrolling in a typing class, f (x) 0002
50(x 1) x5
x 0
where f (x) is the number of words per minute the person is able to type after x weeks of lessons. Sketch the graph of f, including any vertical or horizontal asymptotes. What does f approach as x S 0005?
Variation and Modeling
315
93. REPLACEMENT TIME A desktop office copier has an initial price of $2,500. A maintenance/service contract costs $200 for the first year and increases $50 per year thereafter. It can be shown that the total cost of the copier after n years is given by C(n) 0002 2,500 175n 25n2 The average cost per year for n years is C(n) 0002 C(n)0004n. (A) Find the rational function C. (B) When is the average cost per year a minimum? (This is frequently referred to as the replacement time for this piece of equipment.) (C) Sketch the graph of C, including any asymptotes. 94. AVERAGE COST The total cost of producing x units of a certain product is given by C(x) 0002 15 x2 2x 2,000 The average cost per unit for producing x units is C(x) 0002 C(x)0004x. (A) Find the rational function C. (B) At what production level will the average cost per unit be minimal? (C) Sketch the graph of C, including any asymptotes. 95. CONSTRUCTION A rectangular dog pen is to be made to enclose an area of 225 square feet. (A) If x represents the width of the pen, express the total length L of the fencing material required for the pen in terms of x. (B) Considering the physical limitations, what is the domain of the function L? (C) Find the dimensions of the pen that will require the least amount of fencing material. (D) Graph the function L, including any asymptotes. 96. CONSTRUCTION Rework Problem 95 with the added assumption that the pen is to be divided into two sections, as shown in the figure. (Approximate dimensions to three decimal places.)
In Problems 93–96, use the fact from calculus that a function of the form c q(x) 0002 ax b , a 7 0, c 7 0, x 7 0 x
x x x
has its minimum value when x 0002 1c0004a.
4-5
Variation and Modeling Z Direct Variation Z Inverse Variation Z Joint and Combined Variation
If you work more hours at a part-time job, then you will get more pay. If you increase your average speed in a bicycle race, then you will decrease the time required to finish. The relationship between hours and pay in the first instance, and between average speed and finishing time in the second, are expressed by saying “Pay is directly proportional to
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hours worked, but average speed is inversely proportional to finishing time.” Such statements, which describe how one quantity varies with respect to another, have a precise mathematical meaning. The purpose of this section is to explain the terminology of variation and how it is used in engineering and the sciences.
Z Direct Variation The perimeter of a square is a constant multiple of the side length, and the circumference of a circle is a constant multiple of the radius. These are examples of direct variation. Z DEFINITION 1 Direct Variation Let x and y be variables. The statement y is directly proportional to x (or y varies directly as x) means y 0002 kx for some nonzero constant k, called the constant of proportionality (or constant of variation).
y
y 0002 kx, k 0006 0 x
Z Figure 1 Direct variation.
EXAMPLE
1
The perimeter P of a square is directly proportional to the side length x; the constant of proportionality is 4 and the equation of variation is P 0002 4x. Similarly, the circumference C of a circle is directly proportional to the radius r; the constant of proportionality is 2 and the equation of variation is C 0002 2 r. Note that the equation of direct variation y 0002 kx, k 0006 0, gives a linear model with nonzero slope that passes through the origin (Fig. 1).
Direct Variation The force F exerted by a spring is directly proportional to the distance x that it is stretched (Hooke’s law). Find the constant of proportionality and the equation of variation if F 0002 12 pounds when x 0002 13 foot.
SOLUTION
The equation of variation has the form F 0002 kx. To find the constant of proportionality, substitute F 0002 12 and x 0002 13 and solve for k. F 0002 kx k (13)
12 0002 k 0002 36
Let F ⴝ 12 and x ⴝ 13 . Multiply both sides by 3.
Therefore, the constant of proportionality is k 0002 36 and the equation of variation is F 0002 36x MATCHED PROBLEM 1
0002
Find the constant of proportionality and the equation of variation if p is directly proportional to v, and p 0002 200 when v 0002 8. 0002
Z Inverse Variation If variables x and y are inversely proportional, the functional relationship between them is a constant multiple of the rational function y 0002 1兾x.
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Z DEFINITION 2 Inverse Variation Let x and y be variables. The statement y is inversely proportional to x (or y varies inversely as x) means y0002
k x
for some nonzero constant k, called the constant of proportionality (or constant of variation).
The rate r and time t it takes to travel a distance of 100 miles are inversely proportional (recall that distance equals rate times time, d 0002 rt). The equation of variation is
y
t0002 y 0002 k/x, k 0006 0 x
Z Figure 2 Inverse variation.
EXAMPLE
2
100 r
and the constant of proportionality is 100. The equation of inverse variation, y 0002 k兾x, determines a rational function having the y axis as a vertical asymptote and the x axis as a horizontal asymptote (Fig. 2). In most applications, the constant k of proportionality will be positive, and only the portion of the graph in Quadrant I will be relevant. If x is very large, then y is close to 0; if x is close to 0, then y is very large.
Inverse Variation The note played by each pipe in a pipe organ is determined by the frequency of vibration of the air in the pipe. The fundamental frequency f of vibration of air in an organ pipe is inversely proportional to the length L of the pipe. (This is why the low frequency notes come from the long pipes.) (A) Find the constant of proportionality and the equation of variation if the fundamental frequency of an 8-foot pipe is 64 vibrations per second. (B) Find the fundamental frequency of a 1.6-foot pipe.
SOLUTIONS
(A) The equation has the form f 0002 k兾L. To find the constant of proportionality, substitute L 0002 8 and f 0002 64 and solve for k. f0002
k L
Let f ⴝ 64 and L ⴝ 8.
64 0002
k 8
Multiply both sides by 8.
k 0002 512 The constant of proportionality is k 0002 512 and the equation of variation is f0002
512 L
(B) If L 0002 1.6, then f 0002 512 1.6 0002 320 vibrations per second. MATCHED PROBLEM 2
0002
Find the constant of proportionality and the equation of variation if P is inversely proportional to V, and P 0002 56 when V 0002 3.5. 0002
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Z Joint and Combined Variation The area of a rectangle is the product of its length and width. This is an example of joint variation. Z DEFINITION 3 Joint Variation Let x, y, and w be variables. The statement w is jointly proportional to x and y (or w varies jointly as x and y) means w 0002 kxy for some nonzero constant k, called the constant of proportionality (or constant of variation).
The area of a rectangle, for example, is jointly proportional to its length and width with constant of proportionality 1; the equation of variation is A 0002 LW. The concept of joint variation can be extended to apply to more than three variables. For example, the volume of a box is jointly proportional to its length, width, and height: V 0002 LWH. Similarly, the concepts of direct and inverse variation can be extended. For example, the area of a circle is directly proportional to the square of its radius; the constant of proportionality is and the equation of variation is A 0002 r2. The three basic types of variation also can be combined. For example, Newton’s law of gravitation, “The force of attraction F between two objects is jointly proportional to their masses m1 and m2 and inversely proportional to the square of the distance d between them,” has the equation F0002k
EXAMPLE
3
m1m2 d2
Joint Variation The volume V of a right circular cone is jointly proportional to the square of its radius r and its height h. Find the constant of proportionality and the equation of variation if a cone of height 8 inches and radius 3 inches has a volume of 24 cubic inches.
SOLUTION
The equation of variation has the form V 0002 kr 2h. To find the constant of proportionality k, substitute V 0002 24 , r 0002 3, and h 0002 8. V 0002 kr2h 24 0002 k(3)28 24 0002 72k
k0002 3
Let V ⴝ 24 , r ⴝ 3, and h ⴝ 8. Simplify. Divide both sides by 72.
The constant of proportionality is k 0002
and the equation of variation is 3
V0002
MATCHED PROBLEM 3
2 rh 3
0002
The volume V of a box with a square base is jointly proportional to the square of a side x of the base and the height h. Find the constant of proportionality and the equation of variation. 0002
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EXAMPLE
4
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319
Combined Variation The frequency f of a vibrating guitar string is directly proportional to the square root of the tension T and inversely proportional to the length L. What is the effect on the frequency if the length is doubled and the tension is quadrupled?
SOLUTION
The equation of variation has the form f0002k
1T L
Let f1, T1, and L1 denote the initial frequency, tension, and length, respectively. Then L2 0002 2L1 and T2 0002 4T1. Therefore, 1T2 L2
Let L2 ⴝ 2L1, and T2 ⴝ 4T1.
0002k
14T1 2L1
Simplify the radical.
0002k
21T1 2L1
Cancel and use the equation of variation.
f2 0002 k
0002 f1 We conclude that there is no effect on the frequency—the pitch remains the same.
MATCHED PROBLEM 4
0002
Refer to Example 4. What is the effect on the frequency if the tension is increased by a factor of 4 and the length is cut in half ? 0002
Refer to the equation of variation in Example 4. Explain why the frequency f, for fixed T, is a rational function of L, but f is not, for fixed L, a rational function of T.
ZZZ EXPLORE-DISCUSS 1
ANSWERS TO MATCHED PROBLEMS 196 V 4. The frequency is increased by a factor of 4. 1. k 0002 25; p 0002 25v
4-5
2. k 0002 196; P 0002
3. k 0002 1; V 0002 x2h
Exercises
1. Suppose that y is directly proportional to x and that the constant of proportionality is positive. If x increases, what happens to y? Explain.
3. Suppose that y is inversely proportional to x and that the constant of proportionality is positive. If x increases, what happens to y? Explain.
2. Suppose that y is directly proportional to x and that the constant of proportionality is negative. If x increases, what happens to y? Explain.
4. Explain what it means for w to be jointly proportional to x and y. 5. Suppose that y varies directly with x. What is the value of y when x 0002 0? Explain.
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6. Suppose that y varies inversely with x. What is the value of y when x 0002 1? Explain.
31. The maximum safe load L for a horizontal beam varies jointly as its width w and the square of its height h, and inversely as its length x.
In Problems 7–22, translate each statement into an equation using k as the constant of proportionality.
32. The number N of long-distance phone calls between two cities varies jointly as the populations P1 and P2 of the two cities, and inversely as the distance d between the two cities.
7. F is inversely proportional to x. 8. y is directly proportional to the square of x. 9. R is jointly proportional to S and T. 10. u is inversely proportional to v. 11. L is directly proportional to the cube of m. 12. W is jointly proportional to X, Y, and Z. 13. A varies jointly as the square of c and d. 14. q varies inversely as t. 15. P varies directly as x. 16. f varies directly as the square of b. 17. h varies inversely as the square root of s.
33. The f-stop numbers N on a camera, known as focal ratios, are directly proportional to the focal length F of the lens and inversely proportional to the diameter d of the effective lens opening. 34. The time t required for an elevator to lift a weight is jointly proportional to the weight w and the distance d through which it is lifted, and inversely proportional to the power P of the motor. 35. Suppose that f varies directly as x. Show that the ratio x1 兾x2 of two values of x is equal to f1 兾f2, the ratio of the corresponding values of f. 36. Suppose that f varies inversely as x. Show that the ratio x1 兾x2 of two values of x is equal to f2 兾f1, the reciprocal of the ratio of corresponding values of f.
18. C varies jointly as the square of x and cube of y. 19. R varies directly as m and inversely as the square of d.
APPLICATIONS
20. T varies jointly as p and q and inversely as w.
37. PHYSICS The weight w of an object on or above the surface of the Earth varies inversely as the square of the distance d between the object and the center of Earth. If a girl weighs 100 pounds on the surface of Earth, how much would she weigh (to the nearest pound) 400 miles above Earth’s surface? (Assume the radius of Earth is 4,000 miles.)
21. D is jointly proportional to x and the square of y and inversely proportional to z. 22. S is directly proportional to the square root of u and inversely proportional to v. 23. u varies directly as the square root of v. If u 0002 3 when v 0002 4, find u when v 0002 10. 24. y varies directly as the cube of x. If y 0002 48 when x 0002 4, find y when x 0002 8. 25. L is inversely proportional to the square of M. If L 0002 9 when M 0002 9, find L when M 0002 6. 26. I is directly proportional to the cube root of y. If I 0002 5 when y 0002 64, find I when y 0002 8. 27. Q varies jointly as m and the square of n, and inversely as P. If Q 0002 2 when m 0002 3, n 0002 6, and P 0002 12, find Q when m 0002 4, n 0002 18, and P 0002 2. 28. w varies jointly as x, y, and z. If w 0002 36 when x 0002 2, y 0002 8, and z 0002 12, find w when x 0002 1, y 0002 2, and z 0002 4. In Problems 29–34, translate each statement into an equation using k as the constant of variation.
38. PHYSICS A child was struck by a car in a crosswalk. The driver of the car had slammed on his brakes and left skid marks 160 feet long. He told the police he had been driving at 30 miles/hour. The police know that the length of skid marks L (when brakes are applied) varies directly as the square of the speed of the car v, and that at 30 miles/hour (under ideal conditions) skid marks would be 40 feet long. How fast was the driver actually going before he applied his brakes? 39. ELECTRICITY Ohm’s law states that the current I in a wire varies directly as the electromotive forces E and inversely as the resistance R. If I 0002 22 amperes when E 0002 110 volts and R 0002 5 ohms, find I if E 0002 220 volts and R 0002 11 ohms. 40. ANTHROPOLOGY Anthropologists, in their study of race and human genetic groupings, often use an index called the cephalic index. The cephalic index C varies directly as the width w of the head and inversely as the length l of the head (both when viewed from the top). If an Indian in Baja California (Mexico) has measurements of C 0002 75, w 0002 6 inches, and l 0002 8 inches, what is C for an Indian in northern California with w 0002 8.1 inches and l 0002 9 inches?
29. The biologist René Réaumur suggested in 1735 that the length of time t that it takes fruit to ripen is inversely proportional to the sum T of the average daily temperatures during the growing season.
41. PHYSICS If the horsepower P required to drive a speedboat through water is directly proportional to the cube of the speed v of the boat, what change in horsepower is required to double the speed of the boat?
30. The erosive force P of a swiftly flowing stream is directly proportional to the sixth power of the velocity v of the water.
42. ILLUMINATION The intensity of illumination E on a surface is inversely proportional to the square of its distance d from a light source. What is the effect on the total illumination on a book if the distance between the light source and the book is doubled?
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43. MUSIC The frequency of vibration f of a musical string is directly proportional to the square root of the tension T and inversely proportional to the length L of the string. If the tension of the string is increased by a factor of 4 and the length of the string is doubled, what is the effect on the frequency? 44. PHYSICS In an automobile accident the destructive force F of a car is (approximately) jointly proportional to the weight w of the car and the square of the speed v of the car. (This is why accidents at high speed are generally so serious.) What would be the effect on the destructive forces of a car if its weight were doubled and its speed were doubled? 45. SPACE SCIENCE The length of time t a satellite takes to complete a circular orbit of Earth varies directly as the radius r of the orbit and inversely as the orbital velocity v of the satellite. If t 0002 1.42 hours when r 0002 4,050 miles and v 0002 18,000 miles0003hour (Sputnik I), find t to two decimal places for r 0002 4,300 miles and v 0002 18,500 miles0003hour. 46. GENETICS The number N of gene mutations resulting from xray exposure varies directly as the size of the x-ray dose r. What is the effect on N if r is quadrupled? 47. BIOLOGY In biology there is an approximate rule, called the bioclimatic rule for temperate climates, which states that the difference d in time for fruit to ripen (or insects to appear) varies directly as the change in altitude h. If d 0002 4 days when h 0002 500 feet, find d when h 0002 2,500 feet. 48. PHYSICS Over a fixed distance d, speed r varies inversely as time t. Police use this relationship to set up speed traps. If in a given speed trap r 0002 30 miles0003hour when t 0002 6 seconds, what would be the speed of a car if t 0002 4 seconds? 49. PHYSICS The length L of skid marks of a car’s tires (when the brakes are applied) is directly proportional to the square of the speed v of the car. How is the length of skid marks affected by doubling the speed? 50. PHOTOGRAPHY In taking pictures using flashbulbs, the lens opening (f-stop number) N is inversely proportional to the distance d from the object being photographed. What adjustment should you make on the f-stop number if the distance between the camera and the object is doubled? 51. ENGINEERING The total pressure P of the wind on a wall is jointly proportional to the area of the wall A and the square of the velocity of the wind v. If P 0002 120 pounds when A 0002 100 square feet
CHAPTER
4-1
4
321
and v 0002 20 miles/hour, find P if A 0002 200 square feet and v 0002 30 miles/hour. 52. ENGINEERING The thrust T of a given type of propeller is jointly proportional to the fourth power of its diameter d and the square of the number of revolutions per minute n it is turning. What happens to the thrust if the diameter is doubled and the number of revolutions per minute is cut in half? 53. PSYCHOLOGY In early psychological studies on sensory perception (hearing, seeing, feeling, and so on), the question was asked: “Given a certain level of stimulation S, what is the minimum amount of added stimulation 0004S that can be detected?” A German physiologist, E. H. Weber (1795–1878) formulated, after many experiments, the famous law that now bears his name: “The amount of change 0004S that will be just noticed varies directly as the magnitude S of the stimulus.” (A) Write the law as an equation of variation. (B) If a person lifting weights can just notice a difference of 1 ounce at the 50-ounce level, what will be the least difference she will be able to notice at the 500-ounce level? (C) Determine the just noticeable difference in illumination a person is able to perceive at 480 candlepower if he is just able to perceive a difference of 1 candlepower at the 60-candle-power level. 54. PSYCHOLOGY Psychologists in their study of intelligence often use an index called IQ. IQ varies directly as mental age MA and inversely as chronological age CA (up to the age of 15). If a 12-yearold boy with a mental age of 14.4 has an IQ of 120, what will be the IQ of an 11-year-old girl with a mental age of 15.4? 55. GEOMETRY The volume of a sphere varies directly as the cube of its radius r. What happens to the volume if the radius is doubled? 56. GEOMETRY The surface area S of a sphere varies directly as the square of its radius r. What happens to the area if the radius is cut in half? 57. MUSIC The frequency of vibration of air in an open organ pipe is inversely proportional to the length of the pipe. If the air column in an open 32-foot pipe vibrates 16 times per second (low C), then how fast would the air vibrate in a 16-foot pipe? 58. MUSIC The frequency of pitch f of a musical string is directly proportional to the square root of the tension T and inversely proportional to the length l and the diameter d. Write the equation of variation using k as the constant of variation. (It is interesting to note that if pitch depended on only length, then pianos would have to have strings varying from 3 inches to 38 feet.)
Review
Polynomial Functions and Models
A function that can be written in the form P(x) 0002 anxn 0005 an00061xn00061 0005 . . . 0005 a1x 0005 a0, an 0007 0, is a polynomial function of degree n. In this chapter, when not specified otherwise, the coefficients an, an00061, . . . , a1, a0 are complex numbers and the domain of P is the set of complex numbers. A number r is said to be a zero (or root) of a function P(x) if P(r) 0002 0.
The real zeros of P(x) are just the x intercepts of the graph of P(x). A point on a continuous graph that separates an increasing portion from a decreasing portion, or vice versa, is called a turning point. If P(x) is a polynomial of degree n 7 0 with real coefficients, then the graph of P(x): 1. Is continuous for all real numbers 2. Has no sharp corners
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3. Has at most n real zeros 4. Has at most n –1 turning points
n Linear Factors Theorem P(x) can be factored as a product of n linear factors.
5. Increases or decreases without bound as x S 0005 and as x S 00030005
If P(x) is factored as a product of linear factors, the number of linear factors that have zero r is said to be the multiplicity of r.
The left and right behavior of such a polynomial P(x) is determined by its highest degree or leading term: As x S 00070005, both an xn and P(x) approach 00070005, with the sign depending on n and the sign of an. For any polynomial P(x) of degree n, we have the following important results:
Imaginary Zeros Theorem Imaginary zeros of polynomials with real coefficients, if they exist, occur in conjugate pairs.
Division Algorithm P(x) 0002 (x 0003 r)Q(x) R where the quotient Q(x) and remainder R are unique; x – r is the divisor. Remainder Theorem P(r) 0002 R Factor Theorem x – r is a factor of P(x) if and only if R 0002 0. Zeros of Polynomials P(x) has at most n zeros. Synthetic division is an efficient method for dividing polynomials by linear terms of the form x – r.
4-2
Real Zeros and Polynomials of Odd Degree If P(x) has odd degree and real coefficients, then the graph of P has at least one x intercept. Zeros of Even or Odd Multiplicity Let P(x) have real coefficients: 1. If r is a real zero of P(x) of even multiplicity, then P(x) has a turning point at r and does not change sign at r. 2. If r is a real zero of P(x) of odd multiplicity, then P(x) does not have a turning point at r and changes sign at r. Rational Zero Theorem If the rational number b/c, in lowest terms, is a zero of the polynomial P(x) 0002 an xn an00031xn00031 # # # a1x a0
Real Zeros and Polynomial Inequalities
The following theorems are useful in locating and approximating all real zeros of a polynomial P(x) of degree n 7 0 with real coefficients, an 7 0: Upper and Lower Bound Theorem 1. Upper bound: A number r 7 0 is an upper bound for the real zeros of P(x) if, when P(x) is divided by x – r using synthetic division, all numbers in the quotient row, including the remainder, are nonnegative. 2. Lower bound: A number r 6 0 is a lower bound for the real zeros of P(x) if, when P(x) is divided by x – r using synthetic division, all numbers in the quotient row, including the remainder, alternate in sign. Location Theorem Suppose that a function f is continuous on an interval I that contains numbers a and b. If f (a) and f (b) have opposite signs, then the graph of f has at least one x intercept between a and b. The bisection method uses the location theorem repeatedly to approximate real zeros to any desired accuracy. Polynomial inequalities can be solved by finding the zeros and inspecting the graph of an appropriate polynomial with real coefficients.
4-3
Linear and Quadratic Factors Theorem If P(x) has real coefficients, then P(x) can be factored as a product of linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros).
Complex Zeros and Rational Zeros of Polynomials
If P(x) is a polynomial of degree n 7 0 we have the following important theorems: Fundamental Theorem of Algebra P(x) has at least one zero.
an 0006 0
with integer coefficients, then b must be an integer factor of a0 and c must be an integer factor of an. If P(x) 0002 (x 0003 r)Q(x), then Q(x) is called a reduced polynomial for P(x).
4-4
Rational Functions and Inequalities
A function f is a rational function if it can be written in the form f (x) 0002
p(x) q(x)
where p(x) and q(x) are polynomials of degrees m and n, respectively. The graph of a rational function f(x): 1. Is continuous with the exception of at most n real numbers 2. Has no sharp corners 3. Has at most m real zeros 4. Has at most m n – 1 turning points 5. Has the same left and right behavior as the quotient of the leading terms of p(x) and q(x) The vertical line x 0002 a is a vertical asymptote for the graph of y 0002 f (x) if f(x) S 0005 or f(x) S 00030005 as x S a or as x S a0003. The horizontal line y 0002 b is a horizontal asymptote for the graph of y 0002 f(x) if f (x) S b as x S 00030005 or as x S 0005. The line y 0002 mx b is an oblique asymptote if [ f(x) 0003 (mx b)] S 0 as x S 00030005 or as x S 0005.
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Review Exercises
Let f (x) 0002
am xm # # # a1x a0 , am 0006 0, bn 0006 0. bn xn # # # b1x b0
1. If m 6 n, the line y 0002 0 (the x axis) is a horizontal asymptote.
4-5
Variation and Modeling
Let x and y be variables. The statement: 1. y is directly proportional to x (or y varies directly as x) means
2. If m 0002 n, the line y 0002 am兾bn is a horizontal asymptote. 3. If m 7 n, there is no horizontal asymptote.
y 0002 kx for some nonzero constant k;
Analyzing and Sketching the Graph of a Rational Function: f (x) ⴝ p(x)兾q(x) Step 1. Intercepts. Find the real solutions of the equation p(x) 0002 0 and use these solutions to plot any x intercepts of the graph of f. Evaluate f(0), if it exists, and plot the y intercept. Step 2. Vertical Asymptotes. Find the real solutions of the equation q(x) 0002 0 and use these solutions to determine the domain of f, the points of discontinuity, and the vertical asymptotes. Sketch any vertical asymptotes as dashed lines. Step 3. Horizontal Asymptotes. Determine whether there is a horizontal asymptote and, if so, sketch it as a dashed line. Step 4. Complete the Sketch. For each interval in the domain of f, plot additional points and join them with a smooth continuous curve. Rational inequalities can be solved by finding the zeros of p(x) and q(x), for an appropriate rational function f (x) 0002 p(x)0004q(x), and inspecting the graph of f.
2. y is inversely proportional to x (or y varies inversely as x) means y0002
4
k x
for some nonzero constant k; 3. w is jointly proportional to x and y (or w varies jointly as x and y) means w 0002 kxy for some nonzero constant k. In each case the nonzero constant k is called the constant of proportionality (or constant of variation). The three basic types of variation also can be combined. For example, Newton’s law of gravitation, “The force of attraction F between two objects is jointly proportional to their masses m1 and m2 and inversely proportional to the square of the distance d between them” has the equation F0002k
CHAPTER
323
m1m2 d2
Review Exercises
Work through all the problems in this chapter review, and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. 1. List the real zeros and turning points, and state the left and right behavior, of the polynomial function that has the indicated graph. y 5
3. If P(x) 0002 x5 0003 4x4 9x2 0003 8, find P(3) using the remainder theorem and synthetic division. 4. What are the zeros of P(x) 0002 3(x 0003 2)(x 4)(x 1)? 5. If P(x) 0002 x2 0003 2x 2 and P(1 i) 0002 0, find another zero of P(x). 6. Let P(x) be the polynomial whose graph is shown in the following figure. (A) Assuming that P(x) has integer zeros and leading coefficient 1, find the lowest-degree equation that could produce this graph. (B) Describe the left and right behavior of P(x). P (x)
00035
5
5
x
00035
2. Use synthetic division to divide P(x) 0002 2x3 3x2 0003 1 by D(x) 0002 x 2, and write the answer in the form P(x) 0002 D(x)Q(x) R.
00035
5
00035
x
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7. According to the upper and lower bound theorem, which of the following are upper or lower bounds of the zeros of P(x) 0002 x3 0003 4x2 0004 2?
25. Determine all rational zeros of P(x) 0002 2x3 0003 3x2 0003 18x 0003 8. 26. Factor the polynomial in Problem 25 into linear factors. 27. Find all rational zeros of P(x) 0002 x3 0003 3x2 0004 5.
00032, 00031, 3, 4 8. How do you know that P(x) 0002 2x3 0003 3x2 0004 x 0003 5 has at least one real zero between 1 and 2?
28. Find all zeros (rational, irrational, and imaginary) exactly for P(x) 0002 2x4 0003 x3 0004 2x 0003 1.
9. List all possible rational zeros of a polynomial with integer coefficients that has leading coefficient 5 and constant term 000315.
29. Factor the polynomial in Problem 28 into linear factors.
10. Find all rational zeros for P(x) = 5x2 0004 74x 0003 15. 11. Find the domain and x intercepts for: 6x (A) f (x) 0002 x00035 7x 0004 3 (B) g (x) 0002 2 x 0004 2x 0003 8
31. Factor P(x) 0002 x4 0004 5x2 0003 36 in two ways: (A) As a product of linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros) (B) As a product of linear factors with complex coefficients
12. Find the horizontal and vertical asymptotes for the functions in Problem 11. 13. Explain why the graph is not the graph of a polynomial function. y 5
00035
30. If P(x) 0002 (x 0003 1)2(x 0004 1)3(x2 0003 1)(x2 0004 1), what is its degree? Write the zeros of P(x), indicating the multiplicity of each if greater than 1.
5
x
00035
In Problems 14–19, translate each statement into an equation using k as the constant of proportionality. 14. F is directly proportional to the square root of x. 15. G is jointly proportional to x and the square of y. 16. H is inversely proportional to the cube of z. 17. R varies jointly as the square of x and the square of y. 18. S varies inversely as the square of u. 19. T varies directly as v and inversely as w. 20. Let P(x) 0002 x3 0003 3x2 0003 3x 0004 4. (A) Graph P(x) and describe the graph verbally, including the number of x intercepts, the number of turning points, and the left and right behavior. (B) Approximate the largest x intercept to two decimal places. 21. If P(x) 0002 8x4 0003 14x3 0003 13x2 0003 4x 0004 7, find Q(x) and R such that P(x) 0002 (x 0003 14)Q(x) 0004 R. What is P(14)?
32. Let P(x) 0002 x5 0003 10x4 0004 30x3 0003 20x2 0003 15x 0003 2. (A) Approximate the zeros of P(x) to two decimal places and state the multiplicity of each zero. (B) Can any of these zeros be approximated with the bisection method? A maximum command? A minimum command? Explain. 33. Let P(x) 0002 x4 0003 2x3 0003 30x2 0003 25. (A) Find the smallest positive and largest negative integers that, by Theorem 1 in Section 4-2, are upper and lower bounds, respectively, for the real zeros of P(x). (B) If (k, k 0004 1), k an integer, is the interval containing the largest real zero of P(x), determine how many additional intervals are required in the bisection method to approximate this zero to one decimal place. (C) Approximate the real zeros of P(x) of two decimal places. x00031 . 2x 0004 2 (A) Find the domain and the intercepts for f. (B) Find the vertical and horizontal asymptotes for f. (C) Sketch a graph of f. Draw vertical and horizontal asymptotes with dashed lines.
34. Let f (x) 0002
35. Solve each polynomial inequality to three decimal places: (A) x3 0003 5x 0004 4 0005 0 (B) x3 0003 5x 0004 4 0005 2 36. Explain why the graph is not the graph of a rational function. y 5
00035
5
x
22. If P(x) 0002 4x3 0003 8x2 0003 3x 0003 3, find P(000312) using the remainder theorem and synthetic division. 23. Use the quadratic formula and the factor theorem to factor P(x) 0002 x2 0003 2x 0003 1. 24. Is x 0004 1 a factor of P(x) 0002 9x26 0003 11x17 0004 8x11 0003 5x4 0003 7? Explain, without dividing or using synthetic division.
00035
37. B varies inversely as the square root of c. If B 0002 5 when c 0002 4, find B when c 0002 25. 38. D is jointly proportional to x and y. If D 0002 10 when x 0002 3 and y 0002 2, find D when x 0002 9 and y 0002 8.
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39. Use synthetic division to divide P(x) 0002 x3 0003 3x 0003 2 by [x 0006 (1 0003 i)]. Write the answer in the form P(x) 0002 D(x)Q(x) 0003 R. 40. Find a polynomial of lowest degree with leading coefficient 1 that has zeros 000612 (multiplicity 2), 00063, and 1 (multiplicity 3). (Leave the answer in factored form.) What is the degree of the polynomial?
APPLICATIONS In Problems 55–58, express the solutions as the roots of a polynomial equation of the form P(x) 0002 0. Find rational solutions exactly and irrational solutions to one decimal place.
41. Repeat Problem 40 for a polynomial P(x) with zeros 00065, 2 0006 3i, and 2 0003 3i.
55. ARCHITECTURE An entryway is formed by placing a rectangular door inside an arch in the shape of the parabola with graph y 0002 16 0006 x2, x and y in feet (see the figure). If the area of the door is 48 square feet, find the dimensions of the door.
42. Find all zeros (rational, irrational, and imaginary) exactly for P(x) 0002 2x5 0006 5x4 0006 8x3 0003 21x2 0006 4.
y
43. Factor the polynomial in Problem 42 into linear factors.
16
y 0002 16 0006 x 2
44. Let P(x) 0002 x4 0003 16x3 0003 47x2 0006 137x 0003 73. Approximate (to two decimal places) the x intercepts and the local extrema. 45. What is the minimal degree of a polynomial P(x), given that P(00061) 0002 00064, P(0) 0002 2, P(1) 0002 00065, and P(2) 0002 3? Justify your conclusion. 46. If P(x) is a cubic polynomial with integer coefficients and if 1 0003 2i is a zero of P(x), can P(x) have an irrational zero? Explain. 47. The solutions to the equation x3 0006 27 0002 0 are the cube roots of 27. (A) How many cube roots of 27 are there? (B) 3 is obviously a cube root of 27; find all others. 48. Let P(x) 0002 x4 0003 2x3 0006 500x2 0006 4,000. (A) Find the smallest positive integer multiple of 10 and the largest negative integer multiple of 10 that, by Theorem 1 in Section 4-2, are upper and lower bounds, respectively, for the real zeros of P(x). (B) Approximate the real zero of P(x) to two decimal places.
00064
x
56. CONSTRUCTION A grain silo is formed by attaching a hemisphere to the top of a right circular cylinder (see the figure). If the cylinder is 18 feet high and the volume of the silo is 486 cubic feet, find the common radius of the cylinder and the hemisphere.
49. Graph
x
x2 0003 2x 0003 3 f (x) 0002 x00031 Indicate any vertical, horizontal, or oblique asymptotes with dashed lines.
x
18 feet
50. Use a graphing calculator to find any horizontal asymptotes for f (x) 0002
4
2x 2
2x 0003 3x 0003 4
51. Solve each rational inequality: x00062 5 17 (A) (B) 0 7 x 50006x x00033 52. Solve each rational inequality to three decimal places: x2 0006 3 (A) 3 0 x 0006 3x 0003 1 x2 0006 3 5 (B) 3 7 2 x 0006 3x 0003 1 x
57. MANUFACTURING A box is to be made out of a piece of cardboard that measures 15 by 20 inches. Squares, x inches on a side, will be cut from each corner, and then the ends and sides will be folded up (see the figure). Find the value of x that would result in a box with a volume of 300 cubic inches. 20 in.
53. If P(x) 0002 x3 0006 x2 0006 5x 0003 4, determine the number of real zeros of P(x) and explain why P(x) has no rational zeros. 15 in.
54. Give an example of a rational function f(x) that satisfies the following conditions: the real zeros of f are 00063, 0, and 2; the vertical asymptotes of f are the line x 0002 00061 and x 0002 4; and the line y 0002 5 is a horizontal asymptote.
x x
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58. PHYSICS The centripetal force F of a body moving in a circular path at constant speed is inversely proportional to the radius r of the path. What happens to F if r is doubled? 59. PHYSICS The Maxwell–Boltzmann equation says that the average velocity v of a molecule varies directly as the square root of the absolute temperature T and inversely as the square root of its molecular weight w. Write the equation of variation using k as the constant of variation. 60. WORK The amount of work A completed varies jointly as the number of workers W used and the time t they spend. If 10 workers can finish a job in 8 days, how long will it take 4 workers to do the same job? 61. SIMPLE INTEREST The simple interest I earned in a given time is jointly proportional to the principal p and the interest rate r. If $100 at 4% interest earns $8, how much will $150 at 3% interest earn in the same period?
CHAPTER
ZZZ
Number of Refrigerators y
10
270
20
430
25
525
30
630
45
890
48
915
63. CRIME STATISTICS According to data published by the FBI, the crime index in the United States has shown a downward trend since the early 1990s. The crime index is defined as the number of crimes per 100,000 inhabitants.
Problems 62 and 63 require a graphing calculator or a computer that can calculate cubic regression polynomials for a given data set. 62. ADVERTISING A chain of appliance stores uses television ads to promote the sale of refrigerators. Analyzing past records produced the data in the table, where x is the number of ads placed monthly and y is the number of refrigerators sold that month. (A) Find a cubic regression equation for these data using the number of ads as the independent variable. (B) Estimate (to the nearest integer) the number of refrigerators that would be sold if 15 ads are placed monthly. (C) Estimate (to the nearest integer) the number of ads that should be placed to sell 750 refrigerators monthly.
Number of Ads x
Year
Crime index
1987
5,550
1992
5,660
1997
4,930
2002
4,119
2007
3,016
Source: Federal Bureau of Investigation
(A) Find a cubic regression model for the crime index if x 0002 0 represents 1987. (B) Use the cubic regression model to predict the crime index in 2020. (C) Do you expect the model to give accurate predictions after 2020? Explain.
4
GROUP ACTIVITY Interpolating Polynomials
How could you find a polynomial whose graph passes through the points (1, 1) and (2, 3)? You could use the point-slope form of the equation of a line. How could you find a polynomial P(x) whose graph passes through all four of the points (1, 1), (2, 3), (3, 00033), and (4, 1)? Such a polynomial is called an interpolating polynomial for the four points. The key is to write the unknown polynomial P(x) in the form P(x) 0002 a0 a1(x 0003 1) a2(x 0003 1)(x 0003 2) a3(x 0003 1)(x 0003 2)(x 0003 3) To find a0 , substitute 1 for x. Next, to find a1, substitute 2 for x. Then, to find a2, substitute 3 for x. Finally, to find a3, substitute 4 for x.
(A) Find a0, a1, a2, and a3. (B) Expand P(x) and verify that P(x) 0002 3x3 0003 22x2 47x 0003 27. (C) Explain why P(x) is the only polynomial of degree 3 whose graph passes through the four given points. (D) Give an example to show that the interpolating polynomial for a set of n 1 points may have degree less than n. (E) Find the interpolating polynomial for the five points (00032, 00033), (00031, 0), (0, 5), (1, 0), and (2, 00033).
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CHAPTER
Exponential and Logarithmic Functions
5
C
OUTLINE
MOST of the functions we’ve worked with so far have been polynomial
or rational functions, with a few others involving roots. Functions that can be expressed in terms of addition, subtraction, multiplication, division, and roots of variables and constants are called algebraic functions. In Chapter 5, we will study exponential and logarithmic functions. These functions are not algebraic; they belong to the class of transcendental functions. Exponential and logarithmic functions are used to model a surprisingly wide variety of real-world phenomena: growth of populations of people, animals, and bacteria; decay of radioactive substances; epidemics; magnitudes of sounds and earthquakes. These and many other applications will be studied in this chapter.
5-1
Exponential Functions
5-2
Exponential Models
5-3
Logarithmic Functions
5-4
Logarithmic Models
5-5
Exponential and Logarithmic Equations Chapter 5 Review Chapter 5 Group Activity: Comparing Regression Models
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5-1
Exponential Functions Z Defining Exponential Functions Z Graphs of Exponential Functions Z Additional Exponential Properties Z The Exponential Function with Base e Z Compound Interest Z Interest Compounded Continuously
Many of the functions we’ve studied so far have included exponents. But in every case, the exponent was a constant, and the base was often a variable. In this section, we will reverse those roles. In an exponential function, the variable appears in an exponent. As we’ll see, this has a significant effect on the properties and graphs of these functions. A review of the basic properties of exponents in Section R-2, would be very helpful before moving on. y
Z Defining Exponential Functions
10
Let’s start by noting that the functions f and g given by f (x) 0002 2x
y 0002 x2
00035
5
x
(a)
y
g(x) 0002 x2
are not the same function. Whether a variable appears as an exponent with a constant base or as a base with a constant exponent makes a big difference. The function g is a quadratic function, which we have already discussed. The function f is an exponential function. The graphs of f and g are shown in Figure 1. As expected, they are very different. We know how to define the values of 2x for many types of inputs. For positive integers, it’s simply repeated multiplication: 22 0002 2 ⴢ 2 0002 4;
10
and
23 0002 2 ⴢ 2 ⴢ 2 0002 8; 24 0002 2 ⴢ 2 ⴢ 2 ⴢ 2 0002 16
For negative integers, we use properties of negative exponents: y 0002 2x
00035
5
x
200032 0002
1 1 0002 ; 2 4 2
200033 0002
1 1 0002 3 8 2
For rational numbers, a calculator comes in handy: 1
(b)
Z Figure 1
1 200031 0002 ; 2
22 0002 12 ⬇ 1.4;
3
9
4 9 22 0002 223 ⬇ 2.8; 24 0002 2 2 ⬇ 4.8
The only catch is that we don’t know how to define 2x for all real numbers. For example, what does 212 mean? Your calculator can give you a decimal approximation, but where does it come from? That question is not easy to answer at this point. In fact, a precise definition of 212 must wait for more advanced courses. For now, we will simply state that for any positive real number b, the expression bx is defined for all real values of x, and the output is a real number as well. This enables us to draw the continuous graph for f (x) 0002 2x in Figure 1. In Problems 79 and 80 in Exercises 5-1, we will explore a method for defining bx for irrational x values like 12.
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Exponential Functions
329
Z DEFINITION 1 Exponential Function The equation f (x) 0002 bx
b 7 0, b 0005 1
defines an exponential function for each different constant b, called the base. The independent variable x can assume any real value.
The domain of f is the set of all real numbers, and it can be shown that the range of f is the set of all positive real numbers. We require the base b to be positive to avoid imaginary numbers such as (00032)100042. Problems 53 and 54 in Exercises 5-1 explore why b 0002 0 and b 0002 1 are excluded.
Z Graphs of Exponential Functions ZZZ EXPLORE-DISCUSS 1
y 10
y3 0002 5x y2 0002 3x y1 0002 2x
5
00035
5
x
x Z Figure 2 y 0002 b for b 0002 2, 3, 5.
y2 0002
冢 13 冣
x
冢 15 冣
x
5
00035
冢 12 冣
Z THEOREM 1 Properties of Graphs of Exponential Functions
1. 2. 3. 4.
10
y1 0002
The graphs of y 0002 bx for b 0002 2, 3, and 5 are shown in Figure 2. Note that all three have the same basic shape, and pass through the point (0, 1). Also, the x axis is a horizontal asymptote for each graph, but only as x S 00030006. The main difference between the graphs is their steepness. Next, let’s look at the graphs of y 0002 bx for b 0002 12, 13, and 15 (Fig. 3). Again, all three have the same basic shape, pass through (0, 1), and have horizontal asymptote y 0002 0, but we can see that for b 6 1, the asymptote is only as x S 0006. In general, for bases less than 1, the graph is a reflection through the y axis of the graphs for bases greater than 1. The graphs in Figures 2 and 3 suggest that the graphs of exponential functions have the properties listed in Theorem 1, which we state without proof.
Let f (x) 0002 bx be an exponential function, b 7 0, b 0005 1. Then the graph of f (x):
y
y3 0002
Compare the graphs of f (x) 0002 3x and g(x) 0002 2x by plotting both functions on the same coordinate system. Find all points of intersection of the graphs. For which values of x is the graph of f above the graph of g? Below the graph of g? Are the graphs of f and g close together as x S 0006? As x S 00030006? Discuss.
x
5
x
Is continuous for all real numbers Has no sharp corners Passes through the point (0, 1) Lies above the x axis, which is a horizontal asymptote either as x S 0006 or x S 00030006, but not both 5. Increases as x increases if b 7 1; decreases as x increases if 0 6 b 6 1 6. Intersects any horizontal line at most once (that is, f is one-to-one)
1 1 1 x Z Figure 3 y 0002 b for b 0002 2, 3, 5.
These properties indicate that the graphs of exponential functions are distinct from the graphs we have already studied. (Actually, property 4 is enough to ensure that graphs of exponential functions are different from graphs of polynomials and rational functions.) Property 6 is important because it guarantees that exponential functions have inverses. Those inverses, called logarithmic functions, are the subject of Section 5-3.
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Transformations of exponential functions are very useful in modeling real-world phenomena, like population growth and radioactive decay. These are among the applications we’ll study in Section 5-2. It is important to understand how the graphs of those functions are related to the graphs of the exponential functions in this section. In Example 1, we will use the transformations we studied in Section 3-3 to examine this relationship.
EXAMPLE
1
Transformations of Exponential Functions For the function g(x) 0002 14 (2x ), use transformations to explain how the graph of g is related to the graph of f (x) 0002 2x in Figure 1(b). Find the intercepts and asymptotes, and draw the graph of g.
SOLUTION
The graph of g is a vertical shrink of the graph of f by a factor of 14. So like f, g(x) 0007 0 for all real numbers x, and g(x) S 0 as x S 00030006. In other words, there are no x intercepts, and the x axis is a horizontal asymptote. Since g(0) 0002 14 (20) 0002 14, 14 is the y intercept. Plotting the intercept and a few more points, we obtain the graph of g shown in the figure, with a portion magnified to illustrate the behavior better. y y
10
1
5 0.5
⫺3
MATCHED PROBLEM 1
⫺2
⫺1
x
⫺5
5
x
0002
Let g(x) 0002 12 (40003x ). Use transformations to explain how the graph of g is related to the graph of the exponential function f (x) 0002 4x. Find the intercepts and asymptotes, and sketch the graph of g. 0002
Z Additional Exponential Properties Exponential functions whose domains include irrational numbers obey the familiar laws of exponents for rational exponents. We summarize these exponent laws here and add two other useful properties. Z EXPONENTIAL FUNCTION PROPERTIES For a and b positive, a 0005 1, b 0005 1, and x and y real: 1. Exponent laws: a xa y 0002 a xy
(a x) y 0002 a xy
a x ax a b 0002 x b b
ax 0002 a x0003y ay
(ab)x 0002 a xb x 25x 27x
* ⴝ 25xⴚ7x
ⴝ 2ⴚ2x
2. ax 0002 a y if and only if x 0002 y. If 64x ⫽ 62xⴙ4, then 4x ⫽ 2x ⴙ 4, and x ⫽ 2. 3. For x 0005 0, ax 0002 bx if and only if a 0002 b. If a4 ⫽ 34, then a ⫽ 3.
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
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331
Property 2 is another way to express the fact that the exponential function f(x) 0002 ax is oneto-one (see property 6 of Theorem 1). Because all exponential functions of the form f(x) 0002 ax pass through the point (0, 1) (see property 3 of Theorem 1), property 3 indicates that the graphs of exponential functions with different bases do not intersect at any other points.
EXAMPLE
2
Using Exponential Function Properties Solve 4x00033 0002 8 for x.
SOLUTION
Express both sides in terms of the same base, and use property 2 to equate exponents. 4x00033 0002 8 2 x00033
Express 4 and 8 as powers of 2. 3
(2 ) 00022 2x00036 2 0002 23 2x 0003 6 0002 3 2x 0002 9 x0002
Use the property (ax)y ⴝ axy. Use property 2 to set exponents equal. Add 6 to both sides. Divide both sides by 2.
9 2
✓ 4(900042)00033 0002 4300042 0002 ( 14)3 0002 23 0002 8
CHECK
Technology Connections 4xⴚ3 ⴝ 8. Graph y1 ⴝ 4xⴚ3 and y2 ⴝ 8, then use the intersect command to obtain x ⴝ 4.5 (Fig. 4).
As an alternative to the algebraic method of Example 2, you can use a graphing calculator to solve the equation 10
⫺10
10
⫺10
Z Figure 4
0002 MATCHED PROBLEM 2
Solve 27x1 0002 9 for x.
0002
Z The Exponential Function with Base e Surprisingly, among the exponential functions it is not the function g(x) 0002 2x with base 2 or the function h(x) 0002 10x with base 10 that is used most frequently in mathematics. Instead, the most commonly used base is a number that you may not be familiar with.
ZZZ EXPLORE-DISCUSS 2
(A) Calculate the values of [1 (1/x)] x for x 0002 1, 2, 3, 4, and 5. Are the values increasing or decreasing as x gets larger? (B) Graph y 0002 [1 (1/x)] x and discuss the behavior of the graph as x increases without bound.
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1 a1 ⴙ b x 1
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Table 1 x
7:35 PM
2
10
2.593 74 …
100
2.704 81 …
1,000
2.716 92 …
10,000
2.718 14 …
100,000
2.718 27 …
1,000,000
2.718 28 …
By calculating the value of [1 (10004x)] x for larger and larger values of x (Table 1), it looks like [1 (10004x)] x approaches a number close to 2.7183. In a calculus course, we can show that as x increases without bound, the value of [1 (10004x)] x approaches an irrational number that we call e. Just as irrational numbers such as and 12 have unending, nonrepeating decimal representations, e also has an unending, nonrepeating decimal representation. To 12 decimal places, 兹2
e ⴝ 2.718 281 828 459
⫺2
⫺1
0
1
e 2
3
4
Don’t let the symbol “e” intimidate you! It’s just a number. Exactly who discovered e is still being debated. It is named after the great Swiss mathematician Leonhard Euler (1707–1783), who computed e to 23 decimal places using [1 (1 0004x)] x. The constant e turns out to be an ideal base for an exponential function because in calculus and higher mathematics many operations take on their simplest form using this base. This is why you will see e used extensively in expressions and formulas that model realworld phenomena.
Z DEFINITION 2 Exponential Function with Base e y
For x a real number, the equation
20
f (x) 0002 ex defines the exponential function with base e.
10
y ⫽ e ⫺x ⫺5
y ⫽ ex 5
x
Z Figure 5 Exponential functions.
EXAMPLE
3
The exponential function with base e is used so frequently that it is often referred to as the exponential function. The graphs of y 0002 e x and y 0002 e0003x are shown in Figure 5.
Analyzing a Graph Let g(x) 0002 4 0003 e x兾2. Use transformations to explain how the graph of g is related to the graph of f1(x) 0002 e x. Determine whether g is increasing or decreasing, find any asymptotes, and sketch the graph of g.
SOLUTION
The graph of g can be obtained from the graph of f1 by a sequence of three transformations: f1(x) 0002 e x
S Horizontal stretch
f2(x) 0002 e x兾2
S Reflection in x axis
f3(x) 0002 0003e x兾2
S
g(x) 0002 4 0003 e x兾2
Vertical translation
[See Fig. 6(a) for the graphs of f1, f2, and f3, and Fig. 6(b) for the graph of g.] The function g is decreasing for all x. Because e x兾2 S 0 as x S 00030006, it follows that g(x) 0002 4 0003 e x兾2 S 4 as x S 00030006. Therefore, the line y 0002 4 is a horizontal asymptote [indicated by the dashed line in Fig. 6(b)]; there are no vertical asymptotes. [To check that the graph of g (as obtained by graph transformations) is correct, plot a few points.]
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SECTION 5–1 y
f1
5
5
5
⫺5
x
y⫽4
⫺5
5
x
g(x) ⫽ 4 ⫺ e x/2
⫺5
f3 (a)
(b)
0002
Z Figure 6
MATCHED PROBLEM 3
333
y
f2
⫺5
Exponential Functions
Let g(x) 0002 2e x兾2 0003 5. Use transformations to explain how the graph of g is related to the graph of f1(x) 0002 e x. Describe the increasing/decreasing behavior, find any asymptotes, and sketch the graph of g. 0002
Z Compound Interest The fee paid to use someone else’s money is called interest. It is usually computed as a percentage, called the interest rate, of the original amount (or principal) over a given period of time. At the end of the payment period, the interest paid is usually added to the principal amount, so the interest in the next period is earned on both the original amount, as well as the interest previously earned. Interest paid on interest previously earned and reinvested in this manner is called compound interest. Suppose you deposit $1,000 in a bank that pays 8% interest compounded semiannually. How much will be in your account at the end of 2 years? “Compounded semiannually” means that the interest is paid to your account at the end of each 6-month period, and the interest will in turn earn more interest. To calculate the interest rate per period, we take the annual rate r, 8% (or 0.08), and divide by the number m of compounding periods per year, in this case 2. If A1 represents the amount of money in the account after one compounding period (6 months), then Principal ⴙ 4% of principal
A1 0002 $1,000 $1,000 a
0.08 b 2
Factor out $1,000.
0002 $1,000(1 0.04) We will next use A2, A3, and A4 to represent the amounts at the end of the second, third, and fourth periods. (Note that the amount we’re looking for is A4.) A2 is calculated by multiplying the amount at the beginning of the second compounding period (A1) by 1.04. A2 0002 A1(1 0.04) 0002 [$1,000(1 0.04)](1 0.04) 0002 $1,000(1 0.04)2 A3 0002 A2(1 0.04) 0002 [$1,000(1 0.04)2 ](1 0.04) 0002 $1,000(1 0.04)3 A4 0002 A3(1 0.04)
Substitute our expression for A1. Multiply. r 2 P a1 ⴙ b m Substitute our expression for A2. Multiply. r 3 P a1 ⴙ b m Substitute our expression for A3.
0002 [$1,000(1 0.04)3 ](1 0.04) 4
0002 $1,000(1 0.04) 0002 $1,169.86
Multiply. P a1 ⴙ
r 4 b m
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What do you think the savings and loan will owe you at the end of 6 years (12 compounding periods)? If you guessed A 0002 $1,000(1 0.04)12 you have observed a pattern that is generalized in the following compound interest formula:
Z COMPOUND INTEREST If a principal P is invested at an annual rate r compounded m times a year, then the amount A in the account at the end of n compounding periods is given by A 0002 Pa1
r n b m
Note that the annual rate r must be expressed in decimal form, and that n 0002 mt, where t is years.
EXAMPLE
4
Compound Interest If you deposit $5,000 in an account paying 9% compounded daily,* how much will you have in the account in 5 years? Compute the answer to the nearest cent.
SOLUTION
We will use the compound interest formula with P 0002 5,000, r 0002 0.09, (which is 9% written as a decimal), m 0002 365, and n 0002 5(365) 0002 1,825: A 0002 P a1
r n b m
0002 5,000 a1
0.09 1,825 b 365
Let P ⴝ 5,000, r ⴝ 0.09, m ⴝ 365, n ⴝ 5(365), or 1,825
Calculate to nearest cent.
0002 $7,841.13 MATCHED PROBLEM 4
EXAMPLE
5
0002
If $1,000 is invested in an account paying 10% compounded monthly, how much will be in the account at the end of 10 years? Compute the answer to the nearest cent. 0002
Comparing Investments If $1,000 is deposited into an account earning 10% compounded monthly and, at the same time, $2,000 is deposited into an account earning 4% compounded monthly, will the first account ever be worth more than the second? If so, when?
SOLUTION
Let y1 and y2 represent the amounts in the first and second accounts, respectively, then y1 0002 1,000(1 0.10000412)x y2 0002 2,000(1 0.04000412)x
P ⴝ 1,000, r ⴝ 0.10, m ⴝ 12 P ⴝ 2,000, r ⴝ 0.04, m ⴝ 12
where x is the number of compounding periods (months). Examining the graphs of y1 and y2 [Fig. 7(a)], we see that the graphs intersect at x ⬇ 139.438 months. Because compound *In all problems involving interest that is compounded daily, we assume a 365-day year.
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interest is paid at the end of each compounding period, we compare the amount in the accounts after 139 months and after 140 months [Fig. 7(b)]. The first account is worth more than the second for x 140 months, or after 11 years and 8 months. 5,000
0
240
0
(a)
(b)
0002
Z Figure 7
MATCHED PROBLEM 5
If $4,000 is deposited into an account earning 10% compounded quarterly and, at the same time, $5,000 is deposited into an account earning 6% compounded quarterly, when will the first account be worth more than the second? 0002
Z Interest Compounded Continuously If $1,000 is deposited in an account that earns compound interest at an annual rate of 8% for 2 years, how will the amount A change if the number of compounding periods is increased? If m is the number of compounding periods per year, then A 0002 1,000a1
0.08 2m b m
The amount A is computed for several values of m in Table 2. Notice that the largest gain appears in going from annually to semiannually. Then, the gains slow down as m increases. In fact, it appears that A might be approaching something close to $1,173.50 as m gets larger and larger.
Table 2 Effect of Compounding Frequency Compounding Frequency
A ⴝ 100a1 ⴙ
m
0.08 2m b m
Annually
1
$1,166.400
Semiannually
2
1,169.859
Quarterly
4
1,171.659
52
1,173.367
365
1,173.490
8,760
1,173.501
Weekly Daily Hourly
We now return to the general problem to see if we can determine what happens to A 0002 P[1 (r/m)] mt as m increases without bound. A little algebraic manipulation of the
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compound interest formula will lead to an answer and a significant result in the mathematics of finance: r mt b m 1 (m/r)rt 0002 Pa1 b m/r
A 0002 Pa1
Replace
m r 1 with , and mt with ⴢ rt. m m/r r
Replace
m with variable x. r
1 x rt 0002 P c a1 b d x Does the expression within the square brackets look familiar? Recall from the first part of this section that 1 x a1 b S e x
as
xS0006
Because the interest rate r is fixed, x 0002 m/r S 0006 as m S 0006. So (1 1x )x S e, and Pa1
1 x rt r mt b 0002 P c a1 b d S Pert m x
as
mS0006
This is known as the continuous compound interest formula, a very important and widely used formula in business, banking, and economics.
Z CONTINUOUS COMPOUND INTEREST FORMULA If a principal P is invested at an annual rate r compounded continuously, then the amount A in the account at the end of t years is given by A 0002 Pert The annual rate r must be expressed as a decimal.
EXAMPLE
6
Continuous Compound Interest If $1,000 is invested at an annual rate of 8% compounded continuously, what amount, to the nearest cent, will be in the account after 2 years?
SOLUTION
Use the continuous compound interest formula to find A when P 0002 $1,000, r 0002 0.08, and t 0002 2: A 0002 Pert 0002 $1,000e(0.08)(2) 0002 $1,173.51
8% is equivalent to r ⴝ 0.08. Calculate to nearest cent.
Notice that the values calculated in Table 2 get closer to this answer as m gets larger. MATCHED PROBLEM 6
0002
What amount will an account have after 5 years if $1,000 is invested at an annual rate of 12% compounded annually? Quarterly? Continuously? Compute answers to the nearest cent. 0002
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ANSWERS TO MATCHED PROBLEMS 1. The graph of g is the same as the graph of f reflected in the y axis and vertically shrunk by a factor of 21. x intercepts: none y intercept: 12 horizontal asymptote: y ⫽ 0 (x axis) vertical asymptotes: none y 40
30
20 1 10 1
⫺5
2
3
x
5
2. x ⫽ ⫺13 3. The graph of g is the same as the graph of f1 stretched horizontally by a factor of 2, stretched vertically by a factor of 2, and shifted 5 units down; g is increasing. horizontal asymptote: y ⫽ ⫺5 vertical asymptote: none y
g
10
⫺5
x
5
y ⫽ ⫺5 ⫺10
4. $2,707.04 5. After 23 quarters 6. Annually: $1,762.34; quarterly: $1,806.11; continuously: $1,822.12
5-1
Exercises
1. What is an exponential function? 2. What is the significance of the symbol e in the study of exponential functions? 3. For a function f (x) ⫽ bx, explain how you can tell if the graph increases or decreases without looking at the graph.
4. Explain why f (x) ⫽ (1/4)x and g(x) ⫽ 4⫺x are really the same function. Can you use this fact to add to your answer for Problem 3? 5. How do we know that the equation e x ⫽ 0 has no solution? 6. Define the following terms related to compound interest: principal, interest rate, compounding period.
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7. Match each equation with the graph of f, g, m, or n in the figure. (A) y 0002 (0.2)x (B) y 0002 2x (C) y 0002 (13)x (D) y 0002 4x f
g
m
6
n 00032
2
0
8. Match each equation with the graph of f, g, m, or n in the figure. (A) y 0002 e00031.2x (B) y 0002 e0.7x 00030.4x (C) y 0002 e (D) y 0002 e1.3x g
m n
6
f 00034
4
37. (45)6x00041 0002 54
38. (73)20003x 0002 37
39. (1 0003 x)5 0002 (2x 0003 1)5
40. 53 0002 (x 0004 2)3
41. 2xe0003x 0002 0
42. (x 0003 3)e x 0002 0
43. x2e x 0003 5xe x 0002 0
44. 3xe0003x 0004 x2e0003x 0002 0
2
2
45. 9x 0002 33x00031
46. 4x 0002 2 x00043
47. 25x00043 0002 125x
48. 45x00041 0002 162x00031
49. 42x00047 0002 8x00042
50. 1002x00043 0002 1,000x00045
51. Find all real numbers a such that a2 0002 a00032. Explain why this does not violate the second exponential function property in the box on page 330. 52. Find real numbers a and b such that a 0006 b but a4 0002 b4. Explain why this does not violate the third exponential function property in the box on page 330. 53. Evaluate y 0002 1x for x 0002 00033, 00032, 00031, 0, 1, 2, and 3. Why is b 0002 1 excluded when defining the exponential function y 0002 bx? 54. Evaluate y 0002 0x for x 0002 00033, 00032, 00031, 0, 1, 2, and 3. Why is b 0002 0 excluded when defining the exponential function y 0002 bx?
0
In Problems 9–16, use a calculator to compute answers to four significant digits. 9. 513
10. 3000312
In Problems 55–64, use transformations to explain how the graph of g is related to the graph of the given exponential function f. Determine whether g is increasing or decreasing, find any asymptotes, and sketch the graph of g.
11. e2 0004 e00032
12. e 0003 e00031
55. g(x) 0002 0003(12)x; f (x) 0002 (12)x
13. 1e
14. e12
56. g(x) 0002 0003(13)0003x; f (x) 0002 (13)x
0005
15.
00030005
0005
2 00042 2
16.
00030005
3 00033 2
57. g(x) 0002 (14)x2 0004 3; f (x) 0002 (14)x 58. g(x) 0002 5 0003 (23)3x; f (x) 0002 (23)x
In Problems 17–24, simplify. 17. 10
3x00031
40003x
10
18. (4 )
x
19.
60. g(x) 0002 1,000(1.03)x; f (x) 0002 1.03x
x00033
3 310003x
20.
4x 3z 21. a y b 5 23.
59. g(x) 0002 500(1.04)x; f (x) 0002 1.04x
3x 2y
5 5x00034
61. g(x) 0002 1 0004 2ex00033; f (x) 0002 e x 62. g(x) 0002 4ex00041 0003 7; f(x) 0002 e x
22. (2x3y)z
e5x
24.
e2x00041
63. g(x) 0002 3 0003 4e20003x; f (x) 0002 e x 64. g(x) 0002 00032 0003 5e40003x; f (x) 0002 e x
e400033x e200035x
In Problems 25–32, use transformations to explain how the graph of g is related to the graph of f(x) 0002 e x. Determine whether g is increasing or decreasing, find the asymptotes, and sketch the graph of g. 25. g(x) 0002 3e
0003x
x
26. g(x) 0002 2e
1 0003x 3e
27. g(x) 0002
29. g(x) 0002 2 0004 e 31. g(x) 0002 e
28. g(x) 0002 x
x00042
1 x 5e
30. g(x) 0002 00034 0004 e x 32. g(x) 0002 e
x00031
In Problems 33–50, solve for x. 33. 53x 0002 54x00032 x2
2x00043
35. 7 0002 7
34. 10200033x 0002 105x00036 36. 4
5x0003x2
00036
00024
In Problems 65–68, simplify. 65.
00032x3e00032x 0003 3x2e00032x x6
66.
5x4e5x 0003 4x3e5x x8
67. (e x 0004 e0003x )2 0004 (e x 0003 e0003x )2 68. e x(e0003x 0004 1) 0003 e0003x(e x 0004 1) In Problems 69–76, use a graphing calculator to find local extrema, y intercepts, and x intercepts. Investigate the behavior as x S 0007 and as x 00030007 and identify any horizontal asymptotes. Round any approximate values to two decimal places. 69. f(x) 0002 2 0004 e x00032 2
71. s(x) 0002 e0003x
70. g(x) 0002 00033 0004 e10004x 2
72. r(x) 0002 e x
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73. F(x) 0002
200 1 3e0003x
74. G(x) 0002
100 1 e0003x
75. f (x) 0002
2x 20003x 2
76. g(x) 0002
3x 30003x 2
77. Use a graphing calculator to investigate the behavior of f (x) 0002 (1 x)1兾x as x approaches 0. 78. Use a graphing calculator to investigate the behavior of f(x) 0002 (1 x)1兾x as x approaches 0006. 79. The irrational number 12 is approximated by 1.414214 to six decimal places. Each of x 0002 1.4, 1.41, 1.414, 1.4142, 1.41421, and 1.414214 is a rational number, so we know how to define 2x for each. Compute the value of 2x for each of these x values, and use your results to estimate the value of 212. Then compute 212 using your calculator to check your estimate. 80. The irrational number 13 is approximated by 1.732051 to six decimal places. Each of x 0002 1.7, 1.73, 1.732, 1.7321, 1.73205, and 1.732051 is a rational number, so we know how to define 3x for each. Compute the value of 3x for each of these x values, and use your results to estimate the value of 313. Then compute 313 using your calculator to check your estimate. It is common practice in many applications of mathematics to approximate nonpolynomial functions with appropriately selected polynomials. For example, the polynomials in Problems 81–84, called Taylor polynomials, can be used to approximate the exponential function f(x) 0002 e x. To illustrate this approximation graphically, in each problem graph f(x) 0002 e x and the indicated polynomial in the same viewing window, 00034 x 4 and 00035 y 50. 81. P1(x) 0002 1 x 12x2 82. P2(x) 0002 1 x 12x2 16x3 83. P3(x) 0002 1 x 12x2 16x3 241 x4 1 5 x 84. P4(x) 0002 1 x 12x2 16x3 241 x4 120
85. Investigate the behavior of the functions f1(x) 0002 x兾e x, f2(x) 0002 x2兾e x, and f3(x) 0002 x3兾e x as x S 0006 and as x S 00030006, and find any horizontal asymptotes. Generalize to functions of the form fn(x) 0002 x n兾e x, where n is any positive integer. 86. Investigate the behavior of the functions g1(x) 0002 xe x, g2(x) 0002 x2e x, and g3(x) 0002 x3e x as x S 0006 and as x S 00030006, and find any horizontal asymptotes. Generalize to functions of the form gn(x) 0002 x ne x, where n is any positive integer.
APPLICATIONS* 87. FINANCE A couple just had a new child. How much should they invest now at 6.25% compounded daily to have $100,000 for the child’s education 17 years from now? Compute the answer to the nearest dollar. 88. FINANCE A person wants to have $25,000 cash for a new car 5 years from now. How much should be placed in an account now if the account pays 4.75% compounded weekly? Compute the answer to the nearest dollar. *Round monetary amounts to the nearest cent unless specified otherwise. In all problems involving interest that is compounded daily, assume a 365-day year.
Exponential Functions
339
89. MONEY GROWTH If you invest $5,250 in an account paying 6.38% compounded continuously, how much money will be in the account at the end of (A) 6.25 years? (B) 17 years? 90. MONEY GROWTH If you invest $7,500 in an account paying 5.35% compounded continuously, how much money will be in the account at the end of (A) 5.5 years? (B) 12 years? 91. FINANCE If $3,000 is deposited into an account earning 8% compounded daily and, at the same time, $5,000 is deposited into an account earning 5% compounded daily, will the first account ever be worth more than the second? If so, when? 92. FINANCE If $4,000 is deposited into an account earning 9% compounded weekly and, at the same time, $6,000 is deposited into an account earning 7% compounded weekly, will the first account ever be worth more than the second? If so, when? 93. FINANCE Will an investment of $10,000 at 4.9% compounded daily ever be worth more at the end of any quarter than an investment of $10,000 at 5% compounded quarterly? Explain. 94. FINANCE A sum of $5,000 is invested at 7% compounded semiannually. Suppose that a second investment of $5,000 is made at interest rate r compounded daily. Both investments are held for 1 year. For which values of r, to the nearest tenth of a percent, is the second investment better than the first? Discuss. 95. PRESENT VALUE A promissory note will pay $30,000 at maturity 10 years from now. How much should you pay for the note now if the note gains value at a rate of 6% compounded continuously? 96. PRESENT VALUE A promissory note will pay $50,000 at maturity 512 years from now. How much should you pay for the note now if the note gains value at a rate of 5% compounded continuously? 97. MONEY GROWTH The website Bankrate.com publishes a weekly list of the top savings deposit yields. In the category of 3-year certificates of deposit, the following were listed: Flagstar Bank, FSB UmbrellaBank.com Allied First Bank
3.12% (CQ) 3.00% (CD) 2.96% (CM)
where CQ represents compounded quarterly, CD compounded daily, and CM compounded monthly. Find the value of $5,000 invested in each account at the end of 3 years. 98. Refer to Problem 97. In the 1-year certificate of deposit category, the following accounts were listed: GMAC Bank UFBDirect.com
2.91% (CD) 2.86% (CM)
Find the value of $10,000 invested in each account at the end of 1 year. 99. FINANCE Suppose $4,000 is invested at 6% compounded weekly. How much money will be in the account in (A) 12 year? (B) 10 years? 100. FINANCE Suppose $2,500 is invested at 4% compounded quarterly. How much money will be in the account in (A) 34 year? (B) 15 years?
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5-2
Exponential Models Z Mathematical Modeling Z Data Analysis and Regression Z A Comparison of Exponential Growth Phenomena
One of the best reasons for studying exponential functions is the fact that many things that occur naturally in our world can be modeled accurately by these functions. In this section, we will study a wide variety of applications, including growth of populations of people, animals, and bacteria; radioactive decay; spread of epidemics; propagation of rumors; light intensity; atmospheric pressure; and electric circuits. The regression techniques we used in Chapter 1 to construct linear and quadratic models will be extended to construct exponential models.
Z Mathematical Modeling Populations tend to grow exponentially and at different rates. A convenient and easily understood measure of growth rate is the doubling time—that is, the time it takes for a population to double. Over short periods the doubling time growth model is often used to model population growth: A ⴝ A02t0004d where
A 0002 Population at time t A0 0002 Population at time t 0002 0 d 0002 Doubling time
Note that when t 0002 d, A 0002 A02d兾d 0002 A02 and the population is double the original, as it should be. We will use this model to solve a population growth problem in Example 1.
EXAMPLE
1
Population Growth According to a 2008 estimate, the population of Nicaragua was about 5.7 million, and that population is growing due to a high birth rate and relatively low mortality rate. If the population continues to grow at the current rate, it will double in 37 years. If the growth remains steady, what will the population be in (A) 15 years?
(B) 40 years?
Calculate answers to three significant digits. SOLUTIONS
We can use the doubling time growth model, A 0002 A0(2)t兾d with A0 0002 5.7 and d 0002 37: A 0002 5.7(2)t兾37
See Figure 1.
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341
A (millions) 20 16 12 8 4 10
20
30
40
50
t
t兾37 Z Figure 1 A 0002 5.7(2)
(A) Find A when t 0002 15 years: A 0002 5.7(2)15000437 0002 7.55 million
To 3 significant digits
(B) Find A when t 0002 40 years: A 0002 5.7(2)40000437 0002 12.1 million MATCHED PROBLEM 1
To 3 significant digits
0002
Before the great housing bust, Palm Coast, Florida, was the fastest-growing city in America. Its population was about 34,000 in 2000, and it doubled in 6.6 years. If the population had continued growing at that rate, what would it be in (A) 2010?
(B) 2020?
Calculate answers to three significant digits. 0002
ZZZ EXPLORE-DISCUSS 1
The doubling time growth model would not be expected to give accurate results over long periods. According to the doubling time growth model of Example 1, what was the population of Nicaragua 500 years ago when it was settled as a Spanish colony? What will the population of Nicaragua be 200 years from now? Explain why these results are unrealistic. Discuss factors that affect human populations that are not taken into account by the doubling time growth model.
The doubling time model is not the only one used to model populations. An alternative model based on the continuous compound interest formula will be used in Example 2. In this case, the formula is written as A 0002 A0ekt where
A 0002 Population at time t A0 0002 Population at time t 0002 0 k 0002 Relative growth rate
The relative growth rate is written as a percentage in decimal form. For example, if a population is growing so that at any time the population is increasing at 3% of the current population per year, the relative growth rate k would be 0.03.
EXAMPLE
2
Medicine—Bacteria Growth Cholera, an intestinal disease, is caused by a cholera bacterium that multiplies exponentially by cell division as modeled by A 0002 A0e1.386t
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where A is the number of bacteria present after t hours and A0 is the number of bacteria present at t 0002 0. If we start with 1 bacterium, how many bacteria will be present in (A) 5 hours?
(B) 12 hours?
Calculate the answers to three significant digits. SOLUTIONS
(A) Use A0 0002 1 and t 0002 5: A 0002 A0e1.386t 0002 e1.386(5) ⬇ 1,020
Let A0 ⴝ 1 and t ⴝ 5. Calculate to three significant digits.
(B) Use A0 0002 1 and t 0002 12: A 0002 A0e1.386t 0002 e1.386(12) 0002 16,700,000 MATCHED PROBLEM 2
Let A0 ⴝ 1 and t ⴝ 12. Calculate to three significant digits.
0002
Repeat Example 2 if A 0002 A0e0.783t and all other information remains the same.
0002
Exponential functions can also be used to model radioactive decay, which is sometimes referred to as negative growth. Radioactive materials are used extensively in medical diagnosis and therapy, as power sources in satellites, and as power sources in many countries. If we start with an amount A0 of a particular radioactive substance, the amount declines exponentially over time. The rate of decay varies depending on the particular radioactive substance. A convenient and easily understood measure of the rate of decay is the half-life of the material—that is, the time it takes for half of a particular material to decay. We can use the following half-life decay model: A ⴝ A0(12)t0004h ⴝ A020003t0004h where
A 0002 Amount at time t A0 0002 Amount at time t 0002 0 h 0002 Half-life
Note that when the amount of time passed is equal to the half-life (t 0002 h), A 0002 A020003h0004h 0002 A0200031 0002 A0 ⴢ 12 and the amount of radioactive material is half the original amount, as it should be.
EXAMPLE
3
Radioactive Decay The radioactive isotope gallium 67 (67Ga), used in the diagnosis of malignant tumors, has a biological half-life of 46.5 hours. If we start with 100 milligrams of the isotope, how many milligrams will be left after (A) 24 hours?
(B) 1 week?
Calculate answers to three significant digits.
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SOLUTIONS
Exponential Models
343
We can use the half-life decay model: A 0002 A0(12)t0004h 0002 A020003t0004h Using A0 0002 100 and h 0002 46.5, we obtain
A (milligrams)
A 0002 100(20003t兾46.5)
100
See Figure 2.
(A) Find A when t 0002 24 hours: A 0002 100(2000324/46.5) 0002 69.9 milligrams
50
100
200
t
Hours
(B) Find A when t 0002 168 hours (1 week 0002 168 hours): A 0002 100(20003168/46.5) 0002 8.17 milligrams
0003t兾46.5 ). Z Figure 2 A 0002 100(2
MATCHED PROBLEM 3
Calculate to three significant digits.
Be careful about units! Half-life was given in hours.
Calculate to three significant digits.
0002
Radioactive gold 198 (198Au), used in imaging the structure of the liver, has a half-life of 2.67 days. If we start with 50 milligrams of the isotope, how many milligrams will be left after: (A)
1 2
day?
(B) 1 week?
Calculate answers to three significant digits. 0002 In Example 2, we saw that a base e exponential function can be used as an alternative to the doubling time model. Not surprisingly, the same can be said for the half-life model. In this case, the formula will be A 0002 A0e0003kt where
A 0002 the amount of radioactive material at time t A0 0002 the amount at time t 0002 0 k 0002 a positive constant specific to the type of material
Our atmosphere is constantly being bombarded with cosmic rays. These rays produce neutrons, which in turn react with nitrogen to produce radioactive carbon-14. Radioactive carbon-14 enters all living tissues through carbon dioxide, which is first absorbed by plants. As long as a plant or animal is alive, carbon-14 is maintained in the living organism at a constant level. Once the organism dies, however, carbon-14 decays according to the equation A 0002 A0e00030.000124t
Carbon-14 decay equation
where A is the amount of carbon-14 present after t years and A0 is the amount present at time t 0002 0. This can be used to calculate the approximate age of fossils.
EXAMPLE
4
Carbon-14 Dating If 1,000 milligrams of carbon-14 are present in the tissue of a recently deceased animal, how many milligrams will be present in (A) 10,000 years?
(B) 50,000 years?
Calculate answers to three significant digits.
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SOLUTIONS
Substituting A0 0002 1,000 in the decay equation, we have A 0002 1,000e00030.000124t
A
See Figure 3.
(A) Solve for A when t 0002 10,000:
1,000
A 0002 1,000e00030.000124(10,000) 0002 289 milligrams 500
Calculate to three significant digits.
(B) Solve for A when t 0002 50,000: A 0002 1,000e00030.000124(50,000) 0002 2.03 milligrams
t
50,000
Z Figure 3
Calculate to three significant digits.
More will be said about carbon-14 dating in Exercises 5-5, where we will be interested in solving for t after being given information about A and A0. 0002
MATCHED PROBLEM 4
Referring to Example 4, how many milligrams of carbon-14 would have to be present at the beginning to have 10 milligrams present after 20,000 years? Compute the answer to four significant digits. 0002 One of the problems with using exponential functions to model things like population is that the growth is completely unlimited in the long term. But in real life, there is often some reasonable maximum value, like the largest population that space and resources allow. We can use modified versions of exponential functions to model such phenomena more realistically. One such type of function is called a learning curve since it can be used to model the performance improvement of a person learning a new task. Learning curves are functions of the form A 0002 c(1 0003 e0003kt ), where c and k are positive constants.
EXAMPLE
5
A
Learning Curve People assigned to assemble circuit boards for a computer manufacturing company undergo on-the-job training. From past experience, it was found that the learning curve for the average employee is given by
50 40
A 0002 40(1 0003 e00030.12t )
30 20
where A is the number of boards assembled per day after t days of training (Fig. 4).
10 10
20
30
40
50
t
Days 00030.12t ). Z Figure 4 A 0002 40(1 0003 e
SOLUTION
(A) How many boards can an average employee produce after 3 days of training? After 5 days of training? Round answers to the nearest integer. (B) Does A approach a limiting value as t increases without bound? Explain. (A) When t 0002 3, A 0002 40(1 0003 e00030.12(3) ) 0002 12
Rounded to nearest integer
so the average employee can produce 12 boards after 3 days of training. Similarly, when t 0002 5, A 0002 40(1 0003 e00030.12(5) ) 0002 18
Rounded to nearest integer
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(B) Because e00030.12t 0002
1 0.12t
e
Exponential Models
345
approaches 0 as t increases without bound,
A 0002 40(1 0003 e00030.12t ) S 40(1 0003 0) 0002 40 So the limiting value of A is 40 boards per day. (Note the horizontal asymptote with equation A 0002 40 that is indicated by the dashed line in Fig. 4.) 0002 MATCHED PROBLEM 5
A company is trying to expose as many people as possible to a new product through television advertising in a large metropolitan area with 2 million potential viewers. A model for the number of people A, in millions, who are aware of the product after t days of advertising was found to be A 0002 2(1 0003 e00030.037t ) (A) How many viewers are aware of the product after 2 days? After 10 days? Express answers as integers, rounded to three significant digits. (B) Does A approach a limiting value as t increases without bound? Explain. 0002 Another limited-growth model is useful for phenomena such as the spread of an epidemic or the propagation of a rumor. It is called the logistic equation, and is given by A0002
M 1 ce0003kt
where M, c, and k are positive constants. Logistic growth, illustrated in Example 6, also approaches a limiting value as t increases without bound.
EXAMPLE
6
Logistic Growth in an Epidemic A certain community consists of 1,000 people. One individual who has just returned from another community has a particularly contagious strain of influenza. Assume the community has not had influenza shots and all are susceptible. The spread of the disease in the community is predicted to be given by the logistic curve A(t) 0002
1,000 1 999e00030.3t
where A is the number of people who have contracted the flu after t days. (A) How many people have contracted the flu after 10 days? After 20 days? (B) Does A approach a limiting value as t increases without bound? Explain.
SOLUTIONS
(A) When t 0002 10, A0002
1,000 1 999e00030.3(10)
0002 20
Rounded to nearest integer
so 20 people have contracted the flu after 10 days. Similarly, when t 0002 20, A0002
1,000 1 999e00030.3(20)
0002 288
Rounded to nearest integer
so 288 people have contracted the flu after 20 days.
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(B) Because e00030.3t approaches 0 as t increases without bound, A0002
1,000 1,000 S 0002 1,000 00030.3t 1 999(0) 1 999e
So the limiting value is 1,000 individuals (everyone in the community will eventually get the flu). (Note the horizontal asymptote with equation A 0002 1,000 that is indicated by the dashed line in Fig. 5.) A 1,500 1,200 900 600 300 10
20
30
40
t
50
Days
Z Figure 5 A 0002
MATCHED PROBLEM 6
1,000 1 999e00030.3t
.
0002
A group of 400 parents, relatives, and friends are waiting anxiously at Kennedy Airport for a charter flight returning students after a year in Europe. It is stormy and the plane is late. A particular parent thought he heard that the plane’s radio had gone out and related this news to some friends, who in turn passed it on to others. The propagation of this rumor is predicted to be given by A(t) 0002
400 1 399e00030.4t
where A is the number of people who have heard the rumor after t minutes. (A) How many people have heard the rumor after 10 minutes? After 20 minutes? Round answers to the nearest integer. (B) Does A approach a limiting value as t increases without bound? Explain. 0002
Z Data Analysis and Regression Many graphing calculators have options for exponential and logistic regression. We can use exponential regression to fit a function of the form y 0002 abx to a set of data points, and logistic regression to fit a function of the form y0002
c 1 ae0003bx
to a set of data points. The techniques are similar to those introduced in Chapters 2 and 3 for linear and quadratic functions.
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EXAMPLE
7
Exponential Models
347
Infectious Diseases The U.S. Department of Health and Human Services published the data in Table 1. Table 1 Reported Cases of Infectious Diseases Year
Mumps
Rubella
1970
104,953
56,552
1980
8,576
3,904
1990
5,292
1,125
1995
906
128
2000
323
152
2005
314
11
An exponential model for the data on mumps is given by A 0002 81,082(0.844)t where A is the number of reported cases of mumps and t is time in years with t 0002 0 representing 1970. (A) Use the model to predict the number of reported cases of mumps in 2010. (B) Compare the actual number of cases of mumps reported in 1980 to the number given by the model. SOLUTIONS
(A) The year 2010 is represented by t 0002 40. Evaluating A 0002 81,082(0.844)t at t 0002 40 gives a prediction of 92 cases of mumps in 2010. (B) The year 1980 is represented by t 0002 10. Evaluating A 0002 81,082(0.844)t at t 0002 10 gives 14,871 cases in 1980. The actual number of cases reported in 1980 was 8,576, nearly 6,300 less than the number given by the model.
Technology Connections Figure 6 shows the details of constructing the exponential model of Example 7 on a graphing calculator.
110,000
00035
45
000310,000
(a) Entering the data
(b) Finding the model
(c) Graphing the data and the model
Z Figure 6
0002
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MATCHED PROBLEM 7
An exponential model for the data on rubella in Table 1 is given by A 0002 54,988(0.799)t where A is the number of reported cases of rubella and t is time in years with t 0002 0 representing 1970. (A) Use the model to predict the number of reported cases of rubella in 2010. (B) Compare the actual number of cases of rubella reported in 1980 to the number given by the model. 0002
EXAMPLE
8
AIDS Cases and Deaths The U.S. Department of Health and Human Services published the data in Table 2. Table 2 Acquired Immunodeficiency Syndrome (AIDS) Cases and Deaths in the United States Year
Cases Diagnosed to Date
Known Deaths to Date
1985
23,185
12,648
1988
107,755
62,468
1991
261,259
159,294
1994
493,713
296,507
1997
672,970
406,179
2000
774,467
447,648
2005
944,306
529,113
A logistic model for the data on AIDS cases is given by A0002
947,000 1 17.3e00030.313t
where A is the number of AIDS cases diagnosed by year t with t 0002 0 representing 1985. (A) Use the model to predict the number of AIDS cases diagnosed by 2010. (B) Compare the actual number of AIDS cases diagnosed by 2005 to the number given by the model. SOLUTIONS
(A) The year 2010 is represented by t 0002 25. Evaluating A0002
947,000 1 17.3e00030.313t
at t 0002 25 gives a prediction of approximately 940,000 cases of AIDS diagnosed by 2010.
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(B) The year 2005 is represented by t 0002 20. Evaluating A0002
947,000 1 0004 17.3e00030.313t
at t 0002 20 gives 916,690 cases in 2005. The actual number of cases diagnosed by 2005 was 944,306, nearly 28,000 greater than the number given by the model.
Technology Connections Figure 7 shows the details of constructing the logistic model of Example 8 on a graphing calculator. 1,000,000
00035
20
0
(a) Entering the data
(b) Finding the model
(c) Graphing the data and the model
Z Figure 7
0002
MATCHED PROBLEM 8
A logistic model for the data on deaths from AIDS in Table 2 is given by A0002
521,000 1 0004 18.8e00030.349t
where A is the number of known deaths from AIDS by year t with t 0002 0 representing 1985. (A) Use the model to predict the number of known deaths from AIDS by 2010. (B) Compare the actual number of known deaths from AIDS by 2005 to the number given by the model. 0002
Z A Comparison of Exponential Growth Phenomena The equations and graphs given in Table 3 compare several widely used growth models. These are divided basically into two groups: unlimited growth and limited growth. Following each equation and graph is a short, incomplete list of areas in which the models are used. We have only touched on a subject that has been extensively developed and that you are likely to study in greater depth in the future.
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Table 3 Exponential Growth and Decay Description
Equation
Unlimited growth
A 0002 A0ekt k00070
Graph
Short List of Uses
A
Short-term population growth (people, bacteria, etc.); growth of money at continuous compound interest
c 0
Exponential decay
A 0002 A0e0003kt k00070
A
A 0002 c(1 0003 e0003kt ) c, k 0007 0
M 1 ce0003kt c, k, M 7 0
t
A
Learning skills; sales fads; company growth; electric circuits
c
0
Logistic growth
Radioactive decay; light absorption in water, glass, and the like; atmospheric pressure; electric circuits
c
0
Limited growth
t
t
A
A0002
Long-term population growth; epidemics; sales of new products; spread of rumors; company growth
M
0
t
ANSWERS TO MATCHED PROBLEMS 1. (A) 97,200 (B) 278,000 2. (A) 50 bacteria (B) 12,000 bacteria 3. (A) 43.9 milligrams (B) 8.12 milligrams 4. 119.4 milligrams 5. (A) 143,000 viewers; 619,000 viewers (B) A approaches an upper limit of 2 million, the number of potential viewers 6. (A) 48 individuals; 353 individuals (B) A approaches an upper limit of 400, the number of people in the entire group. 7. (A) 7 cases (B) The actual number of cases was 1,927 less than the number given by the model. 8. (A) 519,000 deaths (B) The actual number of known deaths was approximately 17,000 greater than the number given by the model.
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5-2
Exponential Models
351
Exercises
1. Define the terms “doubling time” and “half-life” in your own words. 2. One of the models below represents positive growth, and the other represents negative growth. Classify each, and explain how you decided on your answer. (Assume that k 7 0.) A 0002 A0e0003kt
A 0002 A0ekt
3. Explain the difference between exponential growth and limited growth. 4. Explain why a limited growth model would be more accurate than regular exponential growth in modeling the long-term population of birds on an island in Lake Erie. In Problems 5–8, write an exponential equation describing the given population at any time t. 5. Initial population 200; doubling time 5 months 6. Initial population 5,000; doubling time 3 years 7. Initial population 2,000; continuous growth at 2% per year 8. Initial population 500; continuous growth at 3% per week In Problems 9–12, write an exponential equation describing the amount of radioactive material present at any time t. 9. Initial amount 100 grams; half-life 6 hours 10. Initial amount 5 pounds; half-life 1,300 years 11. Initial amount 4 kilograms; continuous decay at 12.4% per year 12. Initial amount 50 milligrams; continuous decay at 0.03% per year
APPLICATIONS 13. GAMING A person bets on red and black on a roulette wheel using a Martingale strategy. That is, a $2 bet is placed on red, and the bet is doubled each time until a win occurs. The process is then repeated. If black occurs n times in a row, then L 0002 2n dollars is lost on the nth bet. Graph this function for 1 n 10. Although the function is defined only for positive integers, points on this type of graph are usually joined with a smooth curve as a visual aid. 14. BACTERIAL GROWTH If bacteria in a certain culture double every 12 hour, write an equation that gives the number of bacteria A in the culture after t hours, assuming the culture has 100 bacteria at the start. Graph the equation for 0 t 5. 15. POPULATION GROWTH Because of its short life span and frequent breeding, the fruit fly Drosophila is used in some genetic studies. Raymond Pearl of Johns Hopkins University, for example, studied 300 successive generations of descendants of a single pair
of Drosophila flies. In a laboratory situation with ample food supply and space, the doubling time for a particular population is 2.4 days. If we start with 5 male and 5 female flies, how many flies should we expect to have in (A) 1 week? (B) 2 weeks? 16. POPULATION GROWTH It was estimated in 2008 that Kenya had a population of about 38,000,000 people, and a doubling time of 25 years. If growth continues at the same rate, find the population in (A) 2012 (B) 2040 Calculate answers to two significant digits. 17. COMPUTER DESIGN In 1965, Gordon Moore, founder of Intel, predicted that the number of transistors that could be placed on a computer chip would double every 2 years. This has come to be known as Moore’s law. In 1970, 2,200 transistors could be placed on a chip. Use Moore’s law to predict the number of transistors in (A) 1990 (B) 2005 18. HISTORY OF TECHNOLOGY The earliest mechanical clocks appeared around 1350 in Europe, and would gain or lose an average of 30 minutes per day. After that, accuracy roughly doubled every 30 years. Find the predicted accuracy of clocks in (A) 1700 (B) 2000 19. INSECTICIDES The use of the insecticide DDT is no longer allowed in many countries because of its long-term adverse effects. If a farmer uses 25 pounds of active DDT, assuming its half-life is 12 years, how much will still be active after (A) 5 years? (B) 20 years? Compute answers to two significant digits. 20. RADIOACTIVE TRACERS The radioactive isotope technetium99m (99mTc) is used in imaging the brain. The isotope has a halflife of 6 hours. If 12 milligrams are used, how much will be present after (A) 3 hours? (B) 24 hours? Compute answers to three significant digits. 21. POPULATION GROWTH According to the CIA World Factbook, the population of the world was estimated to be about 6.8 billion people in 2008, and the population was growing continuously at a relative growth rate of 1.188%. If this growth rate continues, what would the population be in 2020 to two significant digits? 22. POPULATION GROWTH According to the CIA World Factbook, the population of Mexico was about 100 million in 2008, and was growing continuously at a relative growth rate of 1.142%. If that growth continues, what will the population be in 2015 to three significant digits?
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23. POPULATION GROWTH In 2005 the population of Russia was 143 million and the population of Nigeria was 129 million. If the populations of Russia and Nigeria grow continuously at relative growth rates of 00030.37% and 2.56%, respectively, in what year did Nigeria have a greater population than Russia? Use the Internet to find if the prediction was accurate. 24. POPULATION GROWTH In 2005 the population of Germany was 82 million and the population of Egypt was 78 million. If the populations of Germany and Egypt grow continuously at relative growth rates of 0% and 1.78%, respectively, in what year did Egypt have a greater population than Germany? Use the Internet to find if the prediction was accurate. 25. SPACE SCIENCE Radioactive isotopes, as well as solar cells, are used to supply power to space vehicles. The isotopes gradually lose power because of radioactive decay. On a particular space vehicle the nuclear energy source has a power output of P watts after t days of use as given by P 0002 75e00030.0035t Graph this function for 0 t 100. 26. EARTH SCIENCE The atmospheric pressure P, in pounds per square inch, decreases exponentially with altitude h, in miles above sea level, as given by P 0002 14.7e00030.21h Graph this function for 0 h 10. 27. MARINE BIOLOGY Marine life is dependent upon the microscopic plant life that exists in the photic zone, a zone that goes to a depth where about 1% of the surface light still remains. Light intensity I relative to depth d, in feet, for one of the clearest bodies of water in the world, the Sargasso Sea in the West Indies, can be approximated by
continues at the rate, find the number of people that will be living with HIV in (A) 2014 (B) 2020 30. AIDS EPIDEMIC The World Health Organization estimated that there were 3.25 million deaths from AIDS in 2007, and that the number had been growing continuously at a relative growth rate of 3.0%. If the growth continues at this rate, find the number of expected deaths from AIDS in (A) 2012 (B) 2030 31. NEWTON’S LAW OF COOLING This law states that the rate at which an object cools is proportional to the difference in temperature between the object and its surrounding medium. The temperature T of the object t hours later is given by T 0002 Tm 0004 (T0 0003 Tm)e0003kt where Tm is the temperature of the surrounding medium and T0 is the temperature of the object at t 0002 0. Suppose a bottle of wine at a room temperature of 72°F is placed in the refrigerator to cool before a dinner party. If the temperature in the refrigerator is kept at 40°F and k 0002 0.4, find the temperature of the wine, to the nearest degree, after 3 hours. (In Exercises 5-5 we will find out how to determine k.) 32. NEWTON’S LAW OF COOLING Refer to Problem 31. What is the temperature, to the nearest degree, of the wine after 5 hours in the refrigerator? 33. PHOTOGRAPHY An electronic flash unit for a camera is activated when a capacitor is discharged through a filament of wire. After the flash is triggered, and the capacitor is discharged, the circuit (see the figure) is connected and the battery pack generates a current to recharge the capacitor. The time it takes for the capacitor to recharge is called the recycle time. For a particular flash unit using a 12-volt battery pack, the charge q, in coulombs, on the capacitor t seconds after recharging has started is given by
I 0002 I0e00030.00942d where I0 is the intensity of light at the surface. To the nearest percent, what percentage of the surface light will reach a depth of (A) 50 feet? (B) 100 feet? 28. MARINE BIOLOGY Refer to Problem 27. In some waters with a great deal of sediment, the photic zone may go down only 15 to 20 feet. In some murky harbors, the intensity of light d feet below the surface is given approximately by I 0002 I0e00030.23d What percentage of the surface light will reach a depth of (A) 10 feet? (B) 20 feet? 29. AIDS EPIDEMIC The World Health Organization estimated that there were 33.2 million people worldwide living with the HIV infection in 2007, and that the number had been growing continuously at a relative growth rate of 2.37%. If the growth
q 0002 0.0009(1 0003 e00030.2t ) Find the value that q approaches as t increases without bound and interpret. R I
V
C S
34. MEDICINE An electronic heart pacemaker uses the same type of circuit as the flash unit in Problem 33, but it is designed so that the capacitor discharges 72 times a minute. For a particular pacemaker, the charge on the capacitor t seconds after it starts recharging is given by q 0002 0.000 008(1 0003 e00032t ) Find the value that q approaches as t increases without bound and interpret.
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35. WILDLIFE MANAGEMENT A herd of 20 white-tailed deer is introduced to a coastal island where there had been no deer before. Their population is predicted to increase according to the logistic curve 100 1 ⫹ 4e⫺0.14t
A⫽
Table 5
36. TRAINING A trainee is hired by a computer manufacturing company to learn to test a particular model of a personal computer after it comes off the assembly line. The learning curve for an average trainee is given by 200 4 ⫹ 21e⫺0.1t
where A is the number of computers an average trainee can test per day after t days of training. (A) How many computers can an average trainee be expected to test after 3 days of training? After 6 days? Round answers to the nearest integer. (B) How many days will it take until an average trainee can test 30 computers per day? Round answer to the nearest integer. (C) Does A approach a limiting value as t increases without bound? Explain. Problems 37–40 require a graphing calculator or a computer that can calculate exponential and logistic regression models for a given data set. 37. DEPRECIATION Table 4 gives the market value of a minivan (in dollars) x years after its purchase. Find an exponential regression model of the form y ⫽ abx for this data set. Round to four significant digits. Estimate the purchase price of the van. Estimate the value of the van 10 years after its purchase. Round answers to the nearest dollar.
Table 4 x
Value ($)
1
12,575
2
9,455
3
8,115
4
6,845
5
5,225
6
4,485
Source: Kelley Blue Book
353
38. DEPRECIATION Table 5 gives the market value of an SUV (in dollars) x years after its purchase. Find an exponential regression model of the form y ⫽ abx for this data set. Estimate the purchase price of the SUV. Estimate the value of the SUV 10 years after its purchase. Round answers to the nearest dollar.
where A is the number of deer expected in the herd after t years. (A) How many deer will be present after 2 years? After 6 years? Round answers to the nearest integer. (B) How many years will it take for the herd to grow to 50 deer? Round answer to the nearest integer. (C) Does A approach a limiting value as t increases without bound? Explain.
A⫽
Exponential Models
x
Value ($)
1
23,125
2
19,050
3
15,625
4
11,875
5
9,450
6
7,125
Source: Kelley Blue Book
39. NUCLEAR POWER Table 6 gives data on nuclear power generation by region for the years 1980–2005.
Table 6 Nuclear Power Generation (Billion Kilowatt-Hours) Year
North America
Central and South America
1980
287.0
2.2
1985
440.8
8.4
1990
649.0
9.0
1995
774.4
9.5
2000
830.9
10.9
2005
879.7
16.3
Source: U.S. Energy Information Administration
(A) Let x represent time in years with x ⫽ 0 representing 1980. Find a logistic regression model ( y ⫽ 1 ⫹ cae⫺bx) for the generation of nuclear power in North America. (Round the constants a, b, and c to three significant digits.) (B) Use the logistic regression model to predict the generation of nuclear power in North America in 2010 and 2020. 40. NUCLEAR POWER Refer to Table 6. (A) Let x represent time in years with x ⫽ 0 representing 1980. Find a logistic regression model ( y ⫽ 1 ⫹ cae⫺bx) for the generation of nuclear power in Central and South America. (Round the constants a, b, and c to three significant digits.) (B) Use the logistic regression model to predict the generation of nuclear power in Central and South America in 2010 and 2020.
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5-3
Logarithmic Functions Z Defining Logarithmic Functions Z Converting Between Logarithmic Form and Exponential Form Z Properties of Logarithmic Functions Z Common and Natural Logarithms Z The Change-of-Base Formula
Solving an equation like 3x 0003 9 is easy: We know that 32 0003 9, so x 0003 2 is the solution. But what about an equation like 3x 0003 20? There probably is an exponent x between 2 and 3 for which 3x is 20, but its exact value is not at all clear. Compare this situation to an equation like x2 0003 9. This is easy to solve because we know that 32 and (00023)2 are both 9. But what about x2 0003 20? To solve this equation, we needed to introduce a new function to be the opposite of the squaring function. This, of course, is the function f (x) 0003 1x. In this section, we will do something very similar with exponential functions. In the first section of this chapter, we learned that exponential functions are one-to-one, so we can define their inverses. These are known as the logarithmic functions.
Z Defining Logarithmic Functions The exponential function f (x) 0003 bx for b 7 0, b 0004 1, is a one-to-one function, and therefore has an inverse. Its inverse, denoted f 00021(x) 0003 logb x (read “log to the base b of x”) is called the logarithmic function with base b. Just like exponentials, there are different logarithmic functions for each positive base other than 1. A point (x, y) is on the graph of f 00021 0003 logb x if and only if the point (y, x) is on the graph of f 0003 bx. In other words, y 0003 logb x if and only if x 0003 b y In a specific example, y 0003 log2 x if and only if x 0003 2y, and log2 x is the power to which 2 must be raised to obtain x: 2log2 x 0003 2y 0003 x. We can use this fact to learn some things about the logarithmic functions from our knowledge of exponential functions. For example, the graph of f 00021 (x) 0003 logb x is the graph of f (x) 0003 bx reflected through the line y 0003 x. Also, the domain of f 00021 (x) 0003 logb x is the range of f (x) 0003 bx, and vice versa. In Example 1, we will use information about f (x) 0003 2x to graph its inverse, 00021 f (x) 0003 log2 x.
EXAMPLE
1
Graphing a Logarithmic Function Make a table of values for f (x) 0003 2x and reverse the ordered pairs to obtain a table of values for f 00021(x) 0003 log2 x. Then use both tables to graph f (x) and f 00021(x) on the same set of axes.
SOLUTION
We chose to evaluate f for integer values from 00023 to 3. The tables are shown here, along with the graph (Fig. 1). Note the important comments about domain and range below the graph.
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y
f y 0003 2x
5
f 00021
y 0003 log2 x 5
10
y 0002 2x
x
y 0002 log2 x
00023
1 8
1 8
00023
00022
1 4
1 4
00022
00021
1 2
1 2
00021
0
1
1
0
1
2
2
1
2
4
4
2
3
8
8
3
x
10
00025
f 00021
f y0003x
x
00025
355
Logarithmic Functions
Ordered pairs reversed
DOMAIN of f 0002 (00030004, 0004) 0002 RANGE of f 00031 RANGE of f 0002 (0, 0004) 0002 DOMAIN of f 00031
0002
Z Figure 1 Logarithmic function with base 2.
MATCHED PROBLEM 1
Repeat Example 1 for f (x) 0003 (12)x and f 00021(x) 0003 log100052 x.
0002
Z DEFINITION 1 Logarithmic Function For b 7 0, b 0004 1, the inverse of f (x) 0003 bx, denoted f 00021(x) 0003 logb x, is the logarithmic function with base b. Logarithmic form
y 0003 logb x
Exponential form
is equivalent to
x 0003 by
The log to the base b of x is the exponent to which b must be raised to obtain x. For example, y 0003 log10 x y 0003 loge x
y y 0003 logb x 00006b00061 0
x 0003 10 y x 0003 ey
Remember: A logarithm is an exponent. x
1
DOMAIN 0003 (0, 0007) RANGE 0003 (00020007, 0007) (a)
y y 0003 logb x b1 0
is equivalent to is equivalent to
x
1
DOMAIN 0003 (0, 0007) RANGE 0003 (00020007, 0007) (b)
Z Figure 2 Typical logarithmic graphs.
It is very important to remember that the equations y 0002 logb x and x 0002 b y define the same function, and as such can be used interchangeably. Because the domain of an exponential function includes all real numbers and its range is the set of positive real numbers, the domain of a logarithmic function is the set of all positive real numbers and its range is the set of all real numbers. For example, log10 3 is defined, but log10 0 and log10 (00025) are not defined. In short, the function y 0003 logb x for any b is only defined for positive x values. Typical logarithmic curves are shown in Figure 2. Notice that in each case, the y axis is a vertical asymptote for the graph. The graphs in Example 1 and Figure 2 suggest that logarithmic graphs share some common properties. Several of these properties are listed in Theorem 1. It might be helpful in understanding them to review Theorem 1 in Section 5-1. Each of these properties is a consequence of a corresponding property of exponential graphs.
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Z THEOREM 1 Properties of Graphs of Logarithmic Functions Let f (x) 0003 logb x be a logarithmic function, b 7 0, b 0004 1. Then the graph of f (x): 1. 2. 3. 4. 5. 6.
ZZZ EXPLORE-DISCUSS 1
Is continuous on its domain (0, 0007) Has no sharp corners Passes through the point (1, 0) Lies to the right of the y axis, which is a vertical asymptote Is increasing as x increases if b 7 1; is decreasing as x increases if 0 6 b 6 1 Intersects any horizontal line exactly once, so is one-to-one
For the exponential function f (x) 0003 (23)x, graph f and y 0003 x on the same coordinate system. Then sketch the graph of f 00021. Discuss the domains and ranges of f and its inverse. By what other name is f 00021 known?
Z Converting Between Logarithmic Form and Exponential Form We now look into the matter of converting logarithmic forms to equivalent exponential forms, and vice versa. Throughout the remainder of the chapter, it will be useful to sometimes convert a logarithmic expression into the equivalent exponential form. At other times, it will be useful to do the reverse.
EXAMPLE
2
Logarithmic–Exponential Conversions Change each logarithmic form to an equivalent exponential form. (B) log25 5 0003 12
(A) log2 8 0003 3 SOLUTIONS
(C) log2 (14) 0003 00022
(A) log2 8 0003 3
is equivalent to
8 0003 23.
(B) log25 5 0003 12
is equivalent to
5 0003 25100052.
(C) log2 (14) 0003 00022
is equivalent to
1 4
0003 200022.
Note that in each case, the base of the logarithm matches the base of the corresponding exponent. 0002 MATCHED PROBLEM 2
Change each logarithmic form to an equivalent exponential form. (A) log3 27 0003 3
EXAMPLE
3
(B) log36 6 0003 12
(C) log3 (19) 0003 00022
Logarithmic–Exponential Conversions Change each exponential form to an equivalent logarithmic form. (A) 49 0003 72
(B) 3 0003 19
(C) 15 0003 500021
0002
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SOLUTIONS
(A) 49 0002 72
is equivalent to
log7 49 0002 2.
(B) 3 0002 19
is equivalent to
log9 3 0002 12.
0002 500031
is equivalent to
log5 (15) 0002 00031.
(C)
1 5
Logarithmic Functions
Recall that 19 0002 9100042.
Again, the bases match. MATCHED PROBLEM 3
357
0002
Change each exponential form to an equivalent logarithmic form. (A) 64 0002 43
3 (B) 2 0002 18
(C)
1 16
0002 400032 0002
To gain a little deeper understanding of logarithmic functions and their relationship to the exponential functions, we will consider a few problems where we want to find x, b, or y in y 0002 logb x, given the other two values. All values were chosen so that the problems can be solved without a calculator. In each case, converting to the equivalent exponential form is useful.
EXAMPLE
4
Solutions of the Equation y 0002 logb x Find x, b, or y as indicated. (A) Find y: y 0002 log4 8.
SOLUTIONS
(B) Find x: log3 x 0002 00032.
(C) Find b: logb 81 0002 4.
(A) Write y 0002 log4 8 in equivalent exponential form. 8 0002 4y 23 0002 22y 2y 0002 3 y 0002 32 We conclude that
3 2
Write each number to the same base 2. Recall that bm 0002 bn if and only if m 0002 n.
0002 log4 8.
(B) Write log3 x 0002 00032 in equivalent exponential form. x 0002 300032 1 1 0002 20002 9 3 We conclude that log3 (19) 0002 00032. (C) Write logb 81 0002 4 in equivalent exponential form: 81 0002 b4 34 0002 b4 b00023
Write 81 as a fourth power. b could be 3 or 00033, but the base of a logarithm must be positive.
We conclude that log3 81 0002 4. MATCHED PROBLEM 4
0002
Find x, b, or y as indicated. (A) Find y: y 0002 log9 27.
(B) Find x: log2 x 0002 00033.
(C) Find b: logb 100 0002 2. 0002
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Z Properties of Logarithmic Functions Some of the properties of exponential functions that we studied in Section 5-1 can be used to develop corresponding properties of logarithmic functions. Several of these important properties of logarithmic functions are listed in Theorem 2. We will justify them individually.
Z THEOREM 2 Properties of Logarithmic Functions If b, M, and N are positive real numbers, b 0006 1, and p and x are real numbers, then 1. logb 1 0002 0 2. logb b 0002 1 3. logb bx 0002 x 4. blogb x 0002 x, x 7 0
ZZZ
5. logb M 0002 logb N if and only if 6. logb MN 0002 logb M 0005 logb N M 7. logb 0002 logb M 0003 logb N N 8. logb M p 0002 p logb M
M0002N
1. In properties 3 and 4, it’s essential that the base of the exponential and the base of the logarithm are the same. 2. Properties 6 and 7 are often misinterpreted, so you should examine them carefully.
CAUTION ZZZ
log b M 0003 logb N 0002 log b
logb M 0006 logb M 0003 logb N logb N
logb M logb N
logb (M 0005 N) 0006 logb M 0005 logb N
M ; N
cannot be simplified.
logb M 0004 logb N 0002 logb MN; logb (M 0004 N) cannot be simplified.
Now we will justify properties in Theorem 2. b0 0002 1. 1. logb 1 0002 0 because b1 0002 b. 2. logb b 0002 1 because 3 and 4. These are simply another way to state that f (x) 0002 bx and f 00031(x) 0002 logb x are inverse functions. Property 3 can be written as f 00031( f (x)) 0002 x for all x in the domain of f. Property 4 can be written as f ( f 00031(x)) 0002 x for all x in the domain of f 00031. This matches our characterization of inverse functions in Theorem 5, Section 3-6. Together, these properties say that if you apply an exponential function and a logarithmic function with the same base consecutively (in either order) you end up with the same value you started with. 5. This follows from the fact that logarithmic functions are one-to-one. Properties 6, 7, and 8 are used often in manipulating logarithmic expressions. We will justify them in Problems 111 and 112 in Exercises 5-3, and Problem 69 in the Chapter 5 Review Exercises.
EXAMPLE
5
Using Logarithmic Properties Simplify, using the properties in Theorem 2. (A) loge 1
(B) log10 10
(C) loge e2x00051
(D) log10 0.01
(E) 10log10 7
(F) eloge x
2
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SOLUTIONS
(A) loge 1 0003 0
Property 1
(B) log10 10 0003 1
Property 2
(C) loge e2x 1 0003 2x 1
Property 3
(D) log10 0.01 0003 log10 1000022 0003 00022
Property 3
log10 7
(E) 10 MATCHED PROBLEM 5
Logarithmic Functions
00037
(F) e
Property 4
loge x2
2
0003x
Property 4
359
0002
Simplify, using the properties in Theorem 2. (A) log10 1000025 m n
(D) loge e
(B) log5 25
(C) log10 1 4
log10 4
(F) eloge (x
(E) 10
1)
0002
Z Common and Natural Logarithms To work with logarithms effectively, we will need to be able to calculate (or at least approximate) the logarithms of any positive number to a variety of bases. Historically, tables were used for this purpose, but now calculators are used because they are faster and can find far more values than any table can possibly include. Of all possible bases, there are two that are used most often. Common logarithms are logarithms with base 10. Natural logarithms are logarithms with base e. Most calculators have a function key labeled “log” and a function key labeled “ln.” The former represents the common logarithmic function and the latter the natural logarithmic function. In fact, “log” and “ln” are both used in most math books, and whenever you see either used in this book without a base indicated, they should be interpreted as follows:
Z LOGARITHMIC FUNCTIONS y 0003 log x 0003 log10 x y 0003 ln x 0003 loge x
ZZZ EXPLORE-DISCUSS 2
Common logarithmic function Natural logarithmic function
(A) Sketch the graph of y 0003 10 x, y 0003 log x, and y 0003 x in the same coordinate system and state the domain and range of the common logarithmic function. (B) Sketch the graph of y 0003 ex, y 0003 ln x, and y 0003 x in the same coordinate system and state the domain and range of the natural logarithmic function.
EXAMPLE
6
Calculator Evaluation of Logarithms Use a calculator to evaluate each to six decimal places. (A) log 3,184
SOLUTIONS
(B) ln 0.000 349
(A) log 3,184 0003 3.502 973
(C) log (00023.24)
(B) ln 0.000 349 0003 00027.960 439
(C) log (00023.24) 0003 Error Why is an error indicated in part C? Because 00023.24 is not in the domain of the log function. [Note: Calculators display error messages in various ways. Some calculators use a more advanced definition of logarithmic functions that involves complex numbers. They will
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display an ordered pair, representing a complex number, as the value of log (00023.24), rather than an error message. You should interpret such a display as indicating that the number entered is not in the domain of the logarithmic function as we have defined it.] 0002 MATCHED PROBLEM 6
Use a calculator to evaluate each to six decimal places. (A) log 0.013 529
(B) ln 28.693 28
(C) ln (00020.438) 0002
When working with common and natural logarithms, we will follow the common practice of using the equal sign “0003” where it might be technically correct to use the approximately equal sign “⬇.” No harm is done as long as we keep in mind that in a statement such as log 3.184 0003 0.503, the number on the right is only assumed accurate to three decimal places and is not exact.
Graphs of the functions f (x) 0003 log x and g(x) 0003 ln x are shown in the graphing calculator display of Figure 3. Which graph belongs to which function? It appears from the display that one of the functions might be a constant multiple of the other. Is that true? Find and discuss the evidence for your answer.
ZZZ EXPLORE-DISCUSS 3
2
0
5
00022
Z Figure 3
EXAMPLE
7
Calculator Evaluation of Logarithms Use a calculator to evaluate each expression to three decimal places.
SOLUTIONS
(A)
log 2 log 1.1
(A)
log 2 0003 7.273 log 1.1
(B) log
(B) log
2 0003 0.260 1.1
2 1.1
(C) log 2 0002 log 1.1
Enter as (log 2) 0006 (log 1.1).
Enter as log (2 0006 1.1).
(C) log 2 0002 log 1.1 0003 0.260. Note that log MATCHED PROBLEM 7
log 2 0004 log 2 0002 log 1.1, but log 1.1
2 0003 log 2 0002 log 1.1 (see Theorem 2). 1.1
0002
Use a calculator to evaluate each to three decimal places. (A)
ln 3 ln 1.08
(B) ln
3 1.08
(C) ln 3 0002 ln 1.08 0002
We now turn to the opposite problem: Given the logarithm of a number, find the number. To solve this problem, we make direct use of the logarithmic–exponential relationships, and change logarithmic expressions into exponential form.
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Z LOGARITHMIC–EXPONENTIAL RELATIONSHIPS log x 0003 y ln x 0003 y
EXAMPLE
8
is equivalent to is equivalent to
x 0003 10 y. x 0003 e y.
Solving logb x 0002 y for x Find x to three significant digits, given the indicated logarithms. (A) log x 0003 00029.315
SOLUTIONS
(B) ln x 0003 2.386
(A) log x 0003 00029.315 x 0003 1000029.315 0003 4.84 10000210
Change to exponential form (Definition 1).
Notice that the answer is displayed in scientific notation in the calculator. (B) ln x 0003 2.386 Change to exponential form (Definition 1). x 0003 e2.386 0003 10.9 MATCHED PROBLEM 8
0002
Find x to four significant digits, given the indicated logarithms. (A) ln x 0003 00025.062
(B) log x 0003 12.0821 0002
ZZZ EXPLORE-DISCUSS 4
Example 8 was solved algebraically using logarithmic–exponential relationships. Use the INTERSECT command on a graphing calculator to solve this problem graphically. Discuss the relative merits of the two approaches.
Z The Change-of-Base Formula How would you find the logarithm of a positive number to a base other than 10 or e? For example, how would you find log3 5.2? In Example 9 we evaluate this logarithm using several properties of logarithms. Then we develop a change-of-base formula to find such logarithms more easily.
EXAMPLE
9
Evaluating a Base 3 Logarithm Evaluate log3 5.2 to four decimal places.
SOLUTION
Let y 0003 log3 5.2 and proceed as follows: log3 5.2 0003 y 5.2 0003 3y ln 5.2 0003 ln 3 y ln 5.2 0003 y ln 3 ln 5.2 y0003 ln 3
Change to exponential form. Apply the natural log (or common log) to each side. Use log b M p 0002 p log b M, which brings the exponent y in front of ln 3 as a factor. Solve for y.
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Replace y with log3 5.2 from the first step, and use a calculator to evaluate the right side: log3 5.2 0003
MATCHED PROBLEM 9
ln 5.2 0003 1.5007 ln 3
0002
Evaluate log0.5 0.0372 to four decimal places. 0002 If we repeat the process we used in Example 9 on a generic logarithm, something interesting happens. The goal is to evaluate logb N, where b is any acceptable base, and N is any positive real number. As in Example 9, let y 0003 logb N. log b N 0003 y N 0003 by ln N 0003 ln b y ln N 0003 y ln b ln N y0003 ln b
Write in exponential form. Apply natural log to each side. Use ln b y 0002 y ln b (property 8, Theorem 2). Solve for y.
This provides a formula for evaluating a logarithm to any base by using natural log: logb N 0003
ln N ln b
We could also have used log base 10 rather than natural log, and developed an alternative formula: logb N 0003
log N log b
In fact, the same approach would enable us to rewrite logb N in terms of a logarithm with any base we choose! Z THE CHANGE-OF-BASE FORMULA For any b 7 0, b 0004 1, and any positive real number N, logb N 0003
loga N loga b
where a is any positive number other than 1.
ZZZ EXPLORE-DISCUSS 5
If b is any positive real number different from 1, the change-of-base formula shows that the function y 0003 logb x is a constant multiple of the natural logarithmic function; that is, logb x 0003 k ln x for some k. (A) Graph the functions y 0003 ln x, y 0003 2 ln x, y 0003 0.5 ln x, and y 0003 00023 ln x. (B) Write each function of part A in the form y 0003 logb x by finding the base b to two decimal places. (C) Is every exponential function y 0003 bx a constant multiple of y 0003 ex? Explain.
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363
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ANSWERS TO MATCHED PROBLEMS 1.
f 00031
f 1 x y0002a b 2
x
y 0002 log1/2 x
00033
8
8
00033
00032
4
4
00032
00031
2
2
00031
0
1
1
0
1
1 2
1 2
1
2
1 4
1 4
2
3
1 8
1 8
3
x
2. 3. 4. 5. 6. 7. 8.
5-3
(A) (A) (A) (A) (A) (A) (A)
3. Why are logarithmic functions undefined for zero and negative inputs? 4. Why is logb 1 0002 0 for any base? 5. Explain how to calculate log5 3 on a calculator that only has log buttons for base 10 and base e. 6. Using the word “inverse,” explain why log b b x 0002 x for any x and any acceptable base b. Rewrite Problems 7–12 in equivalent exponential form. 7. log3 81 0002 4 9. log10 0.001 0002 00033 0002 00032
x
y
y0002x
10
5
00035
5
x
10
y 0002 log1/2 x f 00031
00035
Exercises
2. Explain why there are infinitely many different logarithmic functions.
11.
f
冢 12 冣
27 0002 33 (B) 6 0002 36100042 (C) 19 0002 300032 1 (B) log8 2 0002 3 (C) log4 (161 ) 0002 00032 log4 64 0002 3 3 1 (B) x 0002 8 (C) b 0002 10 y00022 (B) 2 (C) 0 (D) m 0005 n (E) 4 (F) x4 0005 1 00035 (B) 3.356 663 (C) Not possible 00031.868 734 14.275 (B) 1.022 (C) 1.022 (B) x 0002 1.208 0007 1012 9. 4.7486 x 0002 0.006 333
1. Describe the relationship between logarithmic functions and exponential functions in your own words.
log 6 361
y0002
8. log5 125 0002 3 10. log10 1,000 0002 3 12.
log2 641
0002 00036
Rewrite Problems 13–18 in equivalent logarithmic form. 13. 8 0002 4300042
14. 9 0002 27200043
15. 12 0002 320003100045
16. 18 0002 200033
17. (23)3 0002 278
18. (52)00032 0002 0.16
In Problems 19–22, make a table of values similar to the one in Example 1, then use it to graph both functions by hand. 19. f (x) 0002 3x
f 00031(x) 0002 log3 x
20. f (x) 0002 (13)x
f 00031(x) 0002 log1/3 x
21. f (x) 0002 (23)x
f 00031(x) 0002 log2/3 x
22. f (x) 0002 10 x
f 00031(x) 0002 log x
In Problems 23–38, simplify each expression using Theorem 2. 23. log16 1
24. log25 1
25. log0.5 0.5
26. log7 7
27. loge e4
28. log10 105
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31. log3 27
30. log10 100 33. log1Ⲑ2 2
34.
log1Ⲑ5 (251 ) 3
loge 10
37. log5 15
36. e
In Problems 83–86, rewrite the expression as a single log. 83. ln x ⫺ ln y
84. log3 x ⫹ log3 y
85. 2 ln x ⫹ 5 ln y ⫺ ln z
86. log a ⫺ 2 log b ⫹ 3 log c
38. log2 18 In Problems 87–90, given that log x ⫽ ⫺2 and log y ⫽ 3, find: In Problems 39–46, evaluate to four decimal places. 39. log 49,236
40. log 691,450
41. ln 54.081
42. ln 19.722
43. log7 13
44. log 9 78
45. log5 120.24
46. log17 304.66
In Problems 47–54, evaluate x to four significant digits.
x 88. log a b y
87. log (xy) 89. log a
1x b y3
90. log (x5y3)
In Problems 91–98, use transformations to explain how the graph of g is related to the graph of the given logarithmic function f. Determine whether g is increasing or decreasing, find its domain and asymptote, and sketch the graph of g.
47. log x ⫽ 5.3027
48. log x ⫽ 1.9168
49. log x ⫽ ⫺3.1773
50. log x ⫽ ⫺2.0411
91. g (x) ⫽ 3 ⫹ log2 x; f (x) ⫽ log2 x
51. ln x ⫽ 3.8655
52. ln x ⫽ 5.0884
92. g (x) ⫽ ⫺4 ⫹ log3 x; f (x) ⫽ log3 x
53. ln x ⫽ ⫺0.3916
54. ln x ⫽ ⫺4.1083
93. g (x) ⫽ log1Ⲑ3 (x ⫺ 2); f (x) ⫽ log1Ⲑ3 x
Find x, y, or b, as indicated in Problems 55–72. 55. log2 x ⫽ 2
56. log3 x ⫽ 3
57. log4 16 ⫽ y
58. log8 64 ⫽ y
59. logb 16 ⫽ 2
60. logb 10⫺3 ⫽ ⫺3
61. logb 1 ⫽ 0
62. logb b ⫽ 1
63. log4 x ⫽ 12
64. log8 x ⫽ 13
65. log1Ⲑ3 9 ⫽ y
66. log49 (17) ⫽ y
67. logb 1,000 ⫽ 32
68. logb 4 ⫽ 23
69. log8 x ⫽ ⫺43
70. log25 x ⫽ ⫺32
71. log16 8 ⫽ y
72. log9 27 ⫽ y
In Problems 73–78, evaluate to three decimal places. 73.
log 2 log 1.15
74.
log 2 log 1.12
75.
ln 3 ln 1.15
76.
ln 4 ln 1.2
77.
ln 150 2 ln 3
78.
log 200 3 log 2
81. log (x4y3)
80. log (xy) 82. log a
95. g (x) ⫽ ⫺1 ⫺ log x; f (x) ⫽ log x 96. g (x) ⫽ 2 ⫺ log x; f (x) ⫽ log x 97. g (x) ⫽ 5 ⫺ 3 ln x; f (x) ⫽ ln x 98. g (x) ⫽ ⫺3 ⫺ 2 ln x; f (x) ⫽ ln x In Problems 99–102, find f ⫺1. 99. f (x) ⫽ log5 x
100. f (x) ⫽ log1Ⲑ3 x
101. f (x) ⫽ 4 log3 (x ⫹ 3)
102. f (x) ⫽ 2 log2 (x ⫺ 5)
103. Let f (x) ⫽ log3 (2 ⫺ x). (A) Find f ⫺1. (B) Graph f ⫺1. (C) Reflect the graph of f ⫺1 in the line y ⫽ x to obtain the graph of f. 104. Let f (x) ⫽ log2 (⫺3 ⫺x). (A) Find f ⫺1. (B) Graph f ⫺1. (C) Reflect the graph of f ⫺1 in the line y ⫽ x to obtain the graph of f. 105. What is wrong with the following “proof ” that 3 is less than 2?
In Problems 79–82, rewrite the expression in terms of log x and log y. x 79. log a b y
94. g (x) ⫽ log1Ⲑ2 (x ⫹ 3); f (x) ⫽ log1Ⲑ2 x
2
x b 1y
1 6 3 1 27 1 27 1 3 (3) log (13)3 3 log 13
6 6 6
Divide both sides by 27.
3 27 1 9 (13)2
6 log (13)2 6 2 log 13
3 6 2
Divide both sides by log 13 .
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3 log 12 log (12)3 (12)3 1 8
7 7 7 7
2 log 12 log (12)2 (12)2 1 4
107. P1(x) 0003 x 0002 12 x2
Multiply both sides by log 12 .
108. P2(x) 0003 x 0002 12 x2 13 x3
109. P3(x) 0003 x 0002 12 x2 13 x3 0002 14 x4 110. P4(x) 0003 x 0002 12 x2 13 x3 0002 14 x4 15 x5 111. Prove that for any positive M, N, and b (b 0004 1), logb (MN) 0003 logb M 0002 logb N. (Hint: Start by writing u 0003 logb M and v 0003 logb N and changing each to exponential form.)
Multiply both sides by 8.
1 7 2
112. Prove that for any positive integer p and any positive b and M (b 0004 1), logb M p 0003 p logb M. [Hint: Write M p as M 0007 M 0007 p M ( p factors).]
The polynomials in Problems 107–110, called Taylor polynomials, can be used to approximate the function g(x) 0003 ln (1 x). To illustrate this approximation graphically, in each problem, graph
5-4
365
g(x) 0003 ln (1 x) and the indicated polynomial in the same viewing window, 00021 x 3 and 00022 y 2.
106. What is wrong with the following “proof ” that 1 is greater than 2? 3 7 2
Logarithmic Models
Logarithmic Models Z Logarithmic Scales Z Data Analysis and Regression
Logarithmic functions occur naturally as the inverses of exponential functions. But that’s not to say that they are not useful in their own right. Some of these uses are probably familiar to you, but you might not have realized that they involved logarithmic functions. In this section, we will study logarithmic scales that are used to compare the intensity of sounds, the severity of earthquakes, and the brightness of distant stars. We will also look at using regression to model data with a logarithmic function, and discuss what sort of data is likely to fit such a model.
Z Logarithmic Scales Table 1 Typical Sound Intensities Sound Intensity (W兾m2)
Sound
1.0 10000212
Threshold of hearing
5.2 10000210
Whisper
00026
3.2 10
Normal conversation
8.5 1000024
Heavy traffic
3.2 1000023
Jackhammer
0
1.0 10
Threshold of pain
8.3 102
Jet plane
The human ear is able to hear sound over a very wide range of intensities. The loudest sound a healthy person can hear without damage to the eardrum has an intensity 1 trillion (1,000,000,000,000) times that of the softest sound a person can hear. If we were to use these intensities as a scale for measuring volume, we would be stuck using numbers from zero all the way to the trillions, which seems cumbersome, if not downright silly. In the last section, we saw that logarithmic functions increase very slowly. We can take advantage of this to create a scale for sound intensity that is much more condensed, and therefore more manageable. The decibel scale for sound intensity is an example of such a scale. The decibel, named after the inventor of the telephone, Alexander Graham Bell (1847–1922), is defined as follows:
SOUND INTENSITY:
D 0002 10 log
I I0
Decibel scale
(1)
where D is the decibel level of the sound, I is the intensity of the sound measured in watts per square meter (W/m2), and I0 is the intensity of the least audible sound that an average healthy young person can hear. The latter is standardized to be I0 0003 10000212 watts per square meter. Table 1 lists some typical sound intensities from familiar sources. In Example 1 and Problems 5 and 6 in Exercises 5-4, we will calculate the decibel levels for these sounds.
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1
Sound Intensity (A) Find the number of decibels from a whisper with sound intensity 5.2 10000210 watts per square meter, then from heavy traffic at 8.5 1000024 watts per square meter. Round your answers to two decimal places. (B) How many times larger is the sound intensity of heavy traffic compared to a whisper?
SOLUTIONS
(A) We can use the decibel formula (1) with I0 0003 10000212. First, we use I 0003 5.2 10000210: I I0 5.2 10000210 0003 10 log 10000212 0003 10 log 520 0003 27.16 decibels
D 0003 10 log
Substitute I 0002 5.2 10000310, I0 0002 10000312.
Simplify the fraction.
Next, for heavy traffic: D 0003 10 log
I I0
Substitute I 0002 8.5 1000034, I0 0002 10000312.
8.5 1000024 10000212 0003 10 log 850,000,000 0003 89.29 decibels 0003 10 log
Simplify the fraction.
(B) Dividing the larger intensity by the smaller, 8.5 1000024 0003 1,634,615.4 5.2 10000210 we see that the sound intensity of heavy traffic is more than 1.6 million times as great as the intensity of a whisper! MATCHED PROBLEM 1
ZZZ EXPLORE-DISCUSS 1
0002
Find the number of decibels from a jackhammer with sound intensity 3.2 1000023 watts per square meter. Compute the answer to two decimal places. 0002 Suppose that you are asked to draw a graph of the data in Table 1, with sound intensities on the x axis, and the corresponding decibel levels on the y axis. (A) What would be the coordinates of the point corresponding to a jackhammer (see Matched Problem 1)? (B) Suppose the axes of this graph are labeled as follows: Each tick mark on the x axis corresponds to the intensity of the least audible sound (10000212 watts per square meter), and each tick mark on the y axis corresponds to 1 decibel. If there is 18 inch between all tick marks, how far away from the x axis is the point you found in part A? From the y axis? (Give the first answer in inches and the second in miles!) Discuss your result.
EARTHQUAKE INTENSITY: The energy released by the largest earthquake recorded, measured in joules, is about 100 billion (100,000,000,000) times the energy released by a small earthquake that is barely felt. In 1935 the California seismologist Charles Richter devised a logarithmic
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Table 2 The Richter Scale Magnitude on Richter Scale M 6 4.5
Destructive Power
Moderate
5.5 6 M 6 6.5
Large
6.5 6 M 6 7.5
Major
7.5 6 M
Great
EXAMPLE
367
scale that bears his name and is still widely used in the United States. The magnitude of an earthquake M on the Richter scale* is given as follows: M0002
Small
4.5 6 M 6 5.5
Logarithmic Models
2 E log 3 E0
Richter scale
(2)
where E is the energy released by the earthquake, measured in joules, and E0 is the energy released by a very small reference earthquake, which has been standardized to be E0 0003 104.40 joules The destructive power of earthquakes relative to magnitudes on the Richter scale is indicated in Table 2.
2
Earthquake Intensity The 1906 San Francisco earthquake released approximately 5.96 1016 joules of energy. Another quake struck the Bay Area just before game 3 of the 1989 World Series, releasing 1.12 1015 joules of energy. (A) Find the magnitude of each earthquake on the Richter scale. Round your answers to two decimal places. (B) How many times more energy did the 1906 earthquake release than the one in 1989?
SOLUTIONS
(A) We can use the magnitude formula (2) with E0 0003 104.40. First, for the 1906 earthquake, E 0003 5.96 1016: 2 E log 3 E0 2 5.96 1016 0003 log 3 104.40 0003 8.25
M0003
Substitute E 0002 5.96 1016, E0 0002 104.40.
Next, for the 1989 earthquake, E 0003 1.12 1015 2 E log 3 E0 2 1.12 1015 0003 log 3 104.40 0003 7.1
M0003
Substitute E 0002 1.12 1015, E0 0002 104.40.
(B) Dividing the larger energy release by the smaller, 5.96 1016 0003 53.2 1.12 1015 we see that the 1906 earthquake released 53.2 times as much energy as the 1989 quake. MATCHED PROBLEM 2
0002
A 1985 earthquake in central Chile released approximately 1.26 1016 joules of energy. What was its magnitude on the Richter scale? Compute the answer to two decimal places. 0002 *Originally, Richter defined the magnitude of an earthquake in terms of logarithms of the maximum seismic wave amplitude, in thousandths of a millimeter, measured on a standard seismograph. Equation (2) gives essentially the same magnitude that Richter obtained for a given earthquake but in terms of logarithms of the energy released by the earthquake.
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3
Earthquake Intensity If the energy release of one earthquake is 1,000 times that of another, how much larger is the Richter scale reading of the larger than the smaller?
SOLUTION
Let M1 0003
E1 2 log 3 E0
and
M2 0003
E2 2 log 3 E0
be the Richter equations for the smaller and larger earthquakes, respectively. Since the larger earthquake released 1,000 times as much energy, we can write E2 0003 1,000E1. M2 0003 0003 0003 0003 0003 0003
E2 2 log 3 E0 1,000E1 2 log 3 E0 E1 2 alog 1,000 log b 3 E0 E1 2 a3 log b 3 E0 E1 2 2 (3) log 3 3 E0 2 M1
Substitute 1,000E1 for E2.
Use log (MN) 0002 log M 0005 log N;
1,000E1 E0
0002 1,000 0007
E1 E0
log 1,000 0002 log 103 0002 3
Distribute. E1 2 log is M1! 3 E0
An earthquake with 1,000 times the energy of another has a Richter scale reading of 2 more than the other. 0002 MATCHED PROBLEM 3
If the energy release of one earthquake is 10,000 times that of another, how much larger is the Richter scale reading of the larger than the smaller? 0002 ROCKET FLIGHT: The theory of rocket flight uses advanced mathematics and physics to show that the velocity v of a rocket at burnout (depletion of fuel supply) is given by
v 0002 c ln
Wt Wb
Rocket equation
(3)
where c is the exhaust velocity of the rocket engine, Wt is the takeoff weight (fuel, structure, and payload), and Wb is the burnout weight (structure and payload). Because of the Earth’s atmospheric resistance, a launch vehicle velocity of at least 9.0 kilometers per second is required to achieve the minimum altitude needed for a stable orbit. Formula (3) indicates that to increase velocity v, either the weight ratio Wt0005Wb must be increased or the exhaust velocity c must be increased. The weight ratio can be increased by the use of solid fuels, and the exhaust velocity can be increased by improving the fuels, solid or liquid.
EXAMPLE
4
Rocket Flight Theory A typical single-stage, solid-fuel rocket may have a weight ratio Wt0005Wb 0003 18.7 and an exhaust velocity c 0003 2.38 kilometers per second. Would this rocket reach a launch velocity of 9.0 kilometers per second?
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SOLUTION
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369
We can use the rocket equation (3) with c 0003 2.38 and Wt0005Wb 0003 18.7: Wt Wb 0003 2.38 ln 18.7 0003 6.97 kilometers per second
v 0003 c ln
The velocity of the launch vehicle is far short of the 9.0 kilometers per second required to achieve orbit. This is why multiple-stage launchers are used—the deadweight from a preceding stage can be jettisoned into the ocean when the next stage takes over. 0002 MATCHED PROBLEM 4
A launch vehicle using liquid fuel, such as a mixture of liquid hydrogen and liquid oxygen, can produce an exhaust velocity of c 0003 4.7 kilometers per second. However, the weight ratio Wt0005Wb must be low—around 5.5 for some vehicles—because of the increased structural weight to accommodate the liquid fuel. How much more or less than the 9.0 kilometers per second required to reach orbit will be achieved by this vehicle? 0002
Z Data Analysis and Regression Based on the logarithmic graphs we studied in the last section, when a quantity increases relatively rapidly at first, but then levels off and increases very slowly, it might be a good candidate to be modeled by a logarithmic function. Most graphing calculators with regression commands can fit functions of the form y 0003 a b ln x to a set of data points using the same techniques we used earlier for other types of regression.
EXAMPLE
5
Table 3 Home Ownership Rates Year
Home Ownership Rate (%)
1940 1950 1960 1970 1980 1990 2000
43.6 55.0 61.9 62.9 64.4 64.2 67.4 SOLUTIONS
Home Ownership Rates The U.S. Census Bureau published the data in Table 3 on home ownership rates. A logarithmic model for the data is given by R 0003 000236.7 23.0 ln t where R is the home ownership rate and t is time in years with t 0003 0 representing 1900. (A) Use the model to predict the home ownership rate in 2015. (B) Compare the actual home ownership rate in 1950 to the rate given by the model.
(A) The year 2015 is represented by t 0003 115. Evaluating R 0003 000236.7 23.0 ln t at t 0003 115 predicts a home ownership rate of 72.4% in 2015. (B) The year 1950 is represented by t 0003 50. Evaluating R 0003 000236.7 23.0 ln t at t 0003 50 gives a home ownership rate of 53.3% in 1950. The actual home ownership rate in 1950 was 55%, approximately 1.7% greater than the number given by the model.
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Technology Connections Figure 1 shows the details of constructing the logarithmic model of Example 5 on a graphing calculator. 100
0
120
0
(a) Entering the data
(b) Finding the model
(c) Graphing the data and the model
Z Figure 1
0002 MATCHED PROBLEM 5
Refer to Example 5. The home ownership rate in 2008 was 67.8%. If this data is added to Table 3, a logarithmic model for the expanded data is given by R ⫽ ⫺30.6 ⫹ 21.5 ln t where R is the home ownership rate and t is time in years with t ⫽ 0 representing 1900. (A) Use the model to predict the home ownership rate in 2015. (B) Compare the actual home ownership rate in 1950 to the rate given by the model. 0002 ANSWERS TO MATCHED PROBLEMS 1. 95.05 decibels 2. 7.80 3. 2.67 4. 1 kilometer per second less 5. (A) 70.5% (B) The actual rate was 1.5% greater than the rate given by the model.
5-4
Exercises
1. Describe the decibel scale in your own words. 2. Describe the Richter scale in your own words. 3. Explain why logarithms are a good choice for describing sound intensity and earthquake magnitude. 4. Think of a real-life quantity that is likely to be modeled well by a logarithmic function, and explain your reasoning.
APPLICATIONS 5. SOUND What is the decibel level of (A) The threshold of hearing, 1.0 ⫻ 10⫺12 watts per square meter?
(B) The threshold of pain, 1.0 watt per square meter? Compute answers to two significant digits. 6. SOUND What is the decibel level of (A) A normal conversation, 3.2 ⫻ 10⫺6 watts per square meter? (B) A jet plane with an afterburner, 8.3 ⫻ 102 watts per square meter? Compute answers to two significant digits. 7. SOUND If the intensity of a sound from one source is 1,000 times that of another, how much more is the decibel level of the louder sound than the quieter one?
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8. SOUND If the intensity of a sound from one source is 10,000 times that of another, how much more is the decibel level of the louder sound than the quieter one? 9. EARTHQUAKES One of the strongest recorded earthquakes to date was in Colombia in 1906, with an energy release of 1.99 1017 joules. What was its magnitude on the Richter scale? Compute the answer to one decimal place. 10. EARTHQUAKES Anchorage, Alaska, had a major earthquake in 1964 that released 7.08 1016 joules of energy. What was its magnitude on the Richter scale? Compute the answer to one decimal place. 11. EARTHQUAKES The 1933 Long Beach, California, earthquake had a Richter scale reading of 6.3, and the 1964 Anchorage, Alaska, earthquake had a Richter scale reading of 8.3. How many times more powerful was the Anchorage earthquake than the Long Beach earthquake? 12. EARTHQUAKES Generally, an earthquake requires a magnitude of over 5.6 on the Richter scale to inflict serious damage. How many times more powerful than this was the great 1906 Colombia earthquake, which registered a magnitude of 8.6 on the Richter scale? 13. EXPLOSIVE ENERGY The atomic bomb dropped on Nagasaki, Japan, on August 9, 1945, released about 1.34 1014 joules of energy. What would be the magnitude of an earthquake that released that much energy? 14. EXPLOSIVE ENERGY The largest and most powerful nuclear weapon ever detonated was tested by the Soviet Union on October 30, 1961, on an island in the Arctic Sea. The blast was so powerful there were reports of windows breaking in Finland, over 700 miles away. The detonation released about 2.1 1017 joules of energy. What would be the magnitude of an earthquake that released that much energy? 15. ASTRONOMY A moderate-size solar flare observed on the sun on July 9, 1996, released enough energy to power the United States for almost 23,000 years at 2001 consumption levels, 2.38 1021 joules. What would be the magnitude of an earthquake that released that much energy? 16. CONSTRUCTION The energy released by a typical construction site explosion is about 7.94 105 joules. What would be the magnitude of an earthquake that released that much energy? 17. SPACE VEHICLES A new solid-fuel rocket has a weight ratio Wt0005Wb 0003 19.8 and an exhaust velocity c 0003 2.57 kilometers per second. What is its velocity at burnout? Compute the answer to two decimal places. 18. SPACE VEHICLES A liquid-fuel rocket has a weight ratio Wt0005Wb 0003 6.2 and an exhaust velocity c 0003 5.2 kilometers per second. What is its velocity at burnout? Compute the answer to two decimal places. 19. CHEMISTRY The hydrogen ion concentration of a substance is related to its acidity and basicity. Because hydrogen ion concentrations vary over a very wide range, logarithms are used to create a compressed pH scale, which is defined as follows:
pH 0003 0002log [ H ]
Logarithmic Models
371
where [H ] is the hydrogen ion concentration, in moles per liter. Pure water has a pH of 7, which means it is neutral. Substances with a pH less than 7 are acidic, and those with a pH greater than 7 are basic. Compute the pH of each substance listed, given the indicated hydrogen ion concentration. Also, indicate whether each substance is acidic or basic. Compute answers to one decimal place. (A) Seawater, 4.63 1000029 (B) Vinegar, 9.32 1000024 20. CHEMISTRY Refer to Problem 19. Compute the pH of each substance below, given the indicated hydrogen ion concentration. Also, indicate whether it is acidic or basic. Compute answers to one decimal place. (A) Milk, 2.83 1000027 (B) Garden mulch, 3.78 1000026 21. ECOLOGY Refer to Problem 19. Many lakes in Canada and the United States will no longer sustain some forms of wildlife because of the increase in acidity of the water from acid rain and snow caused by sulfur dioxide emissions from industry. If the pH of a sample of rainwater is 5.2, what is its hydrogen ion concentration in moles per liter? Compute the answer to two significant digits. 22. ECOLOGY Refer to Problem 19. If normal rainwater has a pH of 5.7, what is its hydrogen ion concentration in moles per liter? Compute the answer to two significant digits. 23. ASTRONOMY The brightness of stars is expressed in terms of magnitudes on a numerical scale that increases as the brightness decreases. The magnitude m is given by the formula m 0003 6 0002 2.5 log
L L0
where L is the light flux of the star and L0 is the light flux of the dimmest stars visible to the naked eye. (A) What is the magnitude of the dimmest stars visible to the naked eye? (B) How many times brighter is a star of magnitude 1 than a star of magnitude 6? 24. ASTRONOMY An optical instrument is required to observe stars beyond the sixth magnitude, the limit of ordinary vision. However, even optical instruments have their limitations. The limiting magnitude L of any optical telescope with lens diameter D, in inches, is given by L 0003 8.8 5.1 log D (A) Find the limiting magnitude for a homemade 6-inch reflecting telescope. (B) Find the diameter of a lens that would have a limiting magnitude of 20.6. Compute answers to three significant digits.
Problems 25 and 26 require a graphing calculator or a computer program that can calculate a logarithmic regression model for a given data set. 25. INTERNET ACCESS Table 4 on page 372 shows the percentage of Americans that had access to the Internet either at home or at work between 2000 and 2006. Let x represent years since 1995.
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Table 4 Internet Access in the United States Year
Percentage with Home Access
Percentage with Work Access
2000
46.9
35.2
2001
58.4
37.5
2002
59.3
40.2
2003
65.1
49.6
2005
66.2
55.1
2006
68.1
55.8
5-5
(A) Find a logarithmic regression model (y 0003 a b ln x) for the percentage with home access. Round a and b to three significant digits. Use your model to estimate the percentage in 2008 and 2015. (B) Examine the model for larger and larger values of x. Does it remain reasonable in the long term? 26. INTERNET ACCESS Refer to Table 4. (A) Find a logarithmic regression model (y 0003 a b ln x) for the percentage with work access. (Keep in mind that x represents years since 1995.) Round a and b to three significant digits. Use your model to estimate the percentage in 2008 and 2015. (B) Examine the model for larger and larger values of x. Does it remain reasonable in the long term?
Exponential and Logarithmic Equations Z Solving Exponential Equations Z Solving Logarithmic Equations
We have seen that many quantities can be modeled by exponential or logarithmic functions. So it’s not surprising that equations involving exponential or logarithmic expressions, like those shown next, are useful in studying those quantities. 23x00022 0003 5
and
log (x 3) log x 0003 1
Equations like these are called exponential and logarithmic equations, respectively. The properties of logarithms that we studied in Section 5-3 will play a key role in solving both types of equations.
Z Solving Exponential Equations The distinguishing feature of exponential equations is that the variable appears in an exponent. Before defining logarithms, we didn’t have a reliable method for removing variables from an exponent: Now we do. We’ll illustrate how these properties are helpful in Examples 1-4.
EXAMPLE
1
Solving an Exponential Equation Find all solutions to 23x00022 0003 5 to four decimal places.
SOLUTION
In order to have any chance of solving for x, we will first need to get x out of the exponent. This is where logs come in very handy. 23x00022 0003 5 log 23x00022 0003 log 5 (3x 0002 2) log 2 0003 log 5 log 5 3x 0002 2 0003 log 2
Take the common or natural log of both sides. Use logb N p 0002 p logb N to get 3x 0003 2 out of the exponent position. Divide both sides by log 2. Add 2 to both sides.
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log 5 log 2 log 5 1 x 0002 a2 0005 b 3 log 2 0002 1.4406
3x 0002 2 0005
MATCHED PROBLEM 1
Exponential and Logarithmic Equations
373
1 Divide both sides by 3, or multiply both sides by . 3
Use a calculator. Solution to four decimal places
0002
Solve 35100032x 0002 7 for x to four decimal places. 0002
EXAMPLE
2
Compound Interest Recall that when an amount of money P (principal) is invested at an annual rate r compounded annually, the amount of money A in the account after n years, assuming no withdrawals, is given by A 0002 P a1 0005
r n b 0002 P(1 0005 r)n m
m 0002 1 for annual compounding.
How many years to the nearest year will it take the money to double if it is invested at 6% compounded annually? SOLUTION
The interest rate is r 0002 0.06, and we want the amount A to be twice the principal, or 2P. So we substitute r 0002 0.06 and A 0002 2P, and solve for n. 2P 0002 P(1.06)n 2 0002 1.06n log 2 0002 log 1.06n log 2 0002 n log 1.06 log 2 n0002 log 1.06 0002 12 years
MATCHED PROBLEM 2
Divide both sides by P to isolate (1.06)n. Take the common or natural log of both sides. Note how log properties are used to get n out of the exponent position. Divide both sides by log 1.06 (which is just a number!). Calculate to the nearest year.
0002
Repeat Example 2, changing the interest rate to 9% compounded annually. 0002
EXAMPLE
3
Atmospheric Pressure The atmospheric pressure P, in pounds per square inch, at x miles above sea level is given approximately by P 0002 14.7e00030.21x At what height will the atmospheric pressure be half the sea-level pressure? Compute the answer to two significant digits.
SOLUTION
Since x is miles above sea level, sea-level pressure is the pressure at x 0002 0, which is 14.7e0, or 14.7. One-half of sea level pressure is 14.7兾2 0002 7.35. Now our problem is to find x so that P 0002 7.35; that is, we solve 7.35 0002 14.7e00030.21x for x: 7.35 0002 14.7e00030.21x 0.5 0002 e00030.21x ln 0.5 0002 ln e00030.21x
Divide both sides by 14.7 to isolate the exponential. Because the base is e, take the natural log of both sides. In ea 0002 a, so ln e00030.21 x 0002 00030.21x
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ln 0.5 0003 00020.21x x0003
Divide both sides by 00030.21.
ln 0.5 00020.21
Calculate to two significant digits.
0003 3.3 miles MATCHED PROBLEM 3 y
y0003
e x e0002x 2
10
0002
Using the formula in Example 3, find the altitude in miles so that the atmospheric pressure will be one-eighth that at sea level. Compute the answer to two significant digits. 0002 The graph of y0002
5
00025
5
e x 0005 e0003x 2
(1)
is a curve called a catenary (Fig. 1). A uniform cable suspended between two fixed points is a physical example of such a curve, which resembles a parabola, but isn’t.
x
Z Figure 1 Catenary.
EXAMPLE
4
Solving an Exponential Equation In equation (1), find x when y 0003 2.5. Compute the answer to four decimal places.
SOLUTION
e x e0002x 2 x e e0002x 2.5 0003 2 x 5 0003 e e0002x 5e x 0003 e2x 1 e2x 0002 5e x 1 0003 0 y0003
Let y 0002 2.5.
Multiply both sides by 2. Multiply both sides by ex. Subtract 5ex from both sides. This is a quadratic in ex.
Let u 0003 e x; then u2 0002 5u 1 0003 0 5 125 0002 4(1)(1) u0003 2 5 121 0003 2 5 121 ex 0003 2 5 121 ln e x 0003 ln 2 5 121 x 0003 ln 2 0003 00021.5668, 1.5668
Use the quadratic formula. Simplify.
Replace u with ex and solve for x.
Take the natural log of both sides (both values on the right are positive). logb bx 0002 x, so ln ex 0002 x.
Exact solutions Rounded to four decimal places.
Note that the method produces exact solutions, an important consideration in certain calculus applications (see Problems 57–60 of Exercises 5-5). 0002 MATCHED PROBLEM 4
Given y 0003 (e x 0002 e0002x)00052, find x for y 0003 1.5. Compute the answer to three decimal places. 0002
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Z Solving Logarithmic Equations We will now illustrate the solution of several types of logarithmic equations.
EXAMPLE
5
Solving a Logarithmic Equation Solve log (x 0005 3) 0005 log x 0002 1, and check.
SOLUTION
First use properties of logarithms to express the left side as a single logarithm, then convert to exponential form and solve for x. log (x 0005 3) 0005 log x 0002 1 log [x(x 0005 3)] 0002 1 x(x 0005 3) 0002 101 x2 0005 3x 0003 10 0002 0 (x 0005 5)(x 0003 2) 0002 0 x 0002 00035, 2
Combine left side using log M 0004 log N 0002 log MN. Change to equivalent exponential form (the base is 10). Write in ax2 0004 bx 0004 c 0002 0 form and solve. Factor. If ab 0002 0, then a 0002 0 or b 0002 0.
x 0002 00035: log (00035 0005 3) 0005 log (00035) is not defined because the domain of the log function is (0, ). x 0002 2: log (2 0005 3) 0005 log 2 0002 log 5 0005 log 2 ✓ 0002 log (5 0005 2) 0002 log 10 00021
CHECK
The only solution to the original equation is x 0002 2. Extraneous solutions are common in log equations, so answers should always be checked in the original equation to see whether any should be discarded. 0002 MATCHED PROBLEM 5
EXAMPLE
6
Solve log (x 0003 15) 0002 2 0003 log x, and check. 0002
Solving a Logarithmic Equation Solve (ln x)2 0002 ln x2.
SOLUTION
There are no logarithmic properties for simplifying (ln x)2. However, we can simplify ln x2, obtaining an equation involving ln x and (ln x)2. (ln x)2 0002 ln x2 (ln x)2 0002 2 ln x (ln x)2 0003 2 ln x 0002 0 (ln x)(ln x 0003 2) 0002 0 ln x 0002 0 or ln x 0003 2 0002 0 0 x0002e ln x 0002 2 00021
ln M p 0002 p ln M, so ln x2 0002 2 ln x. This is a quadratic equation in ln x. Move all nonzero terms to the left. Factor out ln x. If ab 0002 0, then a 0002 0 or b 0002 0. If ln x 0002 a, x 0002 ea.
x 0002 e2
Checking that both x 0002 1 and x 0002 e2 are solutions to the original equation is left to you. 0002 MATCHED PROBLEM 6
Solve log x2 0002 (log x)2. 0002
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Note that
CAUTION ZZZ
(logb x)2 0006 logb x2
(logb x)2 0002 (logb x)(logb x) logb x2 0002 2 logb x
You might find it helpful to keep these straight by writing logb x2 as logb (x2).
EXAMPLE
7
Earthquake Intensity Recall from Section 5-4 that the magnitude of an earthquake on the Richter scale is given by M0002
E 2 log 3 E0
Solve for E in terms of the other symbols. 2 E log 3 E0 E 3M log 0002 E0 2 E 0002 103M00042 E0 E 0002 E0103M00042 M0002
SOLUTION
MATCHED PROBLEM 7
Multiply both sides by 32 and switch sides.
Change to exponential form with base 10.
Multiply both sides by E0.
0002
Solve the rocket equation from Section 5-4 for Wb in terms of the other symbols: v 0002 c ln
Wt Wb 0002
ANSWERS TO MATCHED PROBLEMS 1. x 0002 0.2263 3. 9.9 miles
5-5
2. More than double in 9 years, but not quite double in 8 years 4. x 0002 1.195 5. x 0002 20 6. x 0002 1,100 7. Wb 0002 Wt e0003v0004c
Exercises
1. Which property of logarithms do you think is most useful in solving exponential equations? Explain.
6. Can you use a logarithm with the same base to solve both equations below? Explain. ex 0002 10
2. Which properties of logarithms do you think are most useful in solving equations with more than one logarithm? Explain. 3. If u and v represent expressions with variable x, how can you solve equations of the form logb u 0002 logb v for x? Explain why this works.
and
5x 0002 8
In Problems 7–16, solve to three significant digits. 7. 100003x 0002 0.0347 9. 103x00051 0002 92
8. 10x 0002 14.3 10. 105x00032 0002 348
4. Why is it especially important to check answers when solving logarithmic equations?
11. e x 0002 3.65
12. e0003x 0002 0.0142
5. Explain the difference between (ln x)2 and ln x2.
13. e2x00031 0005 68 0002 207
14. 13 0005 e3x00055 0002 23
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15. 2320002x 0003 0.426
16. 3430002x 0003 0.089
17. log5 x 0003 2
18. log3 y 0003 4
19. log (t 0002 4) 0003 00021
20. ln (2x 3) 0003 0
21. log 5 log x 0003 2
22. log x 0002 log 8 0003 1
23. log x log (x 0002 3) 0003 1 24. log (x 0002 9) log 100x 0003 3 25. log (x 1) 0002 log (x 0002 1) 0003 1 26. log (2x 1) 0003 1 log (x 0002 2)
28. 3 0003 1.06x
29. e00021.4x 5 0003 0
30. e0.32x 0.47 0003 0
31. 123 0003 500e00020.12x
32. 438 0003 200e0.25x
0002x 2
33. e
x2
0003 0.23
34. e 0003 125
In Problems 35–48, solve exactly. 35. log (5 0002 2x) 0003 log (3x 1) 36. log (x 3) 0003 log (6 4x) 37. log x 0002 log 5 0003 log 2 0002 log (x 0002 3) 38. log (6x 5) 0002 log 3 0003 log 2 0002 log x 39. ln x 0003 ln (2x 0002 1) 0002 ln (x 0002 2) 40. ln (x 1) 0003 ln (3x 1) 0002 ln x 41. log (2x 1) 0003 1 0002 log (x 0002 1) 42. 1 0002 log (x 0002 2) 0003 log (3x 1) 43. ln (x 1) 0003 ln (3x 3) 44. 1 ln (x 1) 0003 ln (x 0002 1) 45. (ln x)3 0003 ln x4
46. (log x)3 0003 log x4
47. ln (ln x) 0003 1
48. log (log x) 0003 1
Solve Problems 49–56 for the indicated variable in terms of the remaining symbols. Use the natural log for solving exponential equations. 49. A 0003 Pert for r (finance) r nt 50. A 0003 P a1 b for t (finance) n 51. D 0003 10 log 52. t 0003
I for I (sound) I0
00021 (ln A 0002 ln A0) for A (decay) k
53. M 0003 6 0002 2.5 log
I for I (astronomy) I0
E (1 0002 e0002Rt0005L) for t (circuitry) R
56. S 0003 R
(1 i)n 0002 1 for n (annuity) i
The following combinations of exponential functions define four of six hyperbolic functions, a useful class of functions in calculus and higher mathematics. Solve Problems 57–60 for x in terms of y. The results are used to define inverse hyperbolic functions, another useful class of functions in calculus and higher mathematics. 57. y 0003
e x e0002x 2
58. y 0003
e x 0002 e0002x 2
59. y 0003
e x 0002 e0002x e x e0002x
60. y 0003
e x e0002x e x 0002 e0002x
In Problems 27–34, solve to three significant digits. 27. 2 0003 1.05x
377
54. L 0003 8.8 5.1 log D for D (astronomy) 55. I 0003
In Problems 17–26, solve exactly.
Exponential and Logarithmic Equations
In Problems 61–68, use a graphing calculator to approximate to two decimal places any solutions of the equation in the interval 0 x 1. None of these equations can be solved exactly using any step-by-step algebraic process. 61. 20002x 0002 2x 0003 0
62. 30002x 0002 3x 0003 0
63. e0002x 0002 x 0003 0
64. xe2x 0002 1 0003 0
65. ln x 2x 0003 0
66. ln x x2 0003 0
67. ln x e x 0003 0
68. ln x x 0003 0
APPLICATIONS 69. COMPOUND INTEREST How many years, to the nearest year, will it take a sum of money to double if it is invested at 7% compounded annually? 70. COMPOUND INTEREST How many years, to the nearest year, will it take money to quadruple if it is invested at 6% compounded annually? 71. COMPOUND INTEREST At what annual rate compounded continuously will $1,000 have to be invested to amount to $2,500 in 10 years? Compute the answer to three significant digits. 72. COMPOUND INTEREST How many years will it take $5,000 to amount to $8,000 if it is invested at an annual rate of 9% compounded continuously? Compute the answer to three significant digits. 73. IMMIGRATION According to the U.S. Office of Immigration Statistics, there were 10.5 million illegal immigrants in the United States in May 2005, and that number had grown to 11.3 million by May 2007. (A) Find the relative growth rate if we use the P 0003 P0ert model for population growth. Round to three significant digits. (B) Use your answer from part A to write a function describing the illegal immigrant population in millions in terms of years after May 2005, and use it to predict when the illegal immigrant population should reach 20 million.
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74. POPULATION GROWTH According to U.S. Census Bureau estimates, the population of the United States was 227.2 million on July 1, 1980, and 249.5 million on July 1, 1990. (A) Find the relative growth rate if we use the P 0003 P0ert model for population growth. Round to three significant digits. (B) Use your answer from part A to write a function describing the population of the United States in millions in terms of years after July 1980, and use it to predict when the population should reach 400 million. (C) Use your function from part B to estimate the population of the United States today, then compare your estimate to the one found at www.census.gov/population/www/popclockus.html. 75. WORLD POPULATION A mathematical model for world population growth over short periods is given by P 0003 P0ert where P is the population after t years, P0 is the population at t 0003 0, and the population is assumed to grow continuously at the annual rate r. How many years, to the nearest year, will it take the world population to double if it grows continuously at an annual rate of 1.14%? 76. WORLD POPULATION Refer to Problem 75. Starting with a world population of 6.8 billion people (the estimated population in March 2009) and assuming that the population grows continuously at an annual rate of 1.14%, how many years, to the nearest year, will it be before there is only 1 square yard of land per person? Earth contains approximately 1.7 1014 square yards of land. 77. MEDICAL RESEARCH A medical researcher is testing a radioactive isotope for use in a new imaging process. She finds that an original sample of 5 grams decays to 1 gram in 6 hours. Find the half-life of the sample to three significant digits. [Recall that the half-life model is A 0003 A0(12)t/h, where A0 is the original amount and h is the half-life.] 78. CARBON-14 DATING If 90% of a sample of carbon-14 remains after 866 years, what is the half-life of carbon-14? (See Problem 77 for the half-life model.) As long as a plant or animal remains alive, carbon-14 is maintained in a constant amount in its tissues. Once dead, however, the plant or animal ceases taking in carbon, and carbon-14 diminishes by radioactive decay. The amount remaining can be modeled by the equation A 0003 A0e00020.000124t, where A is the amount after t years, and A0 is the amount at time t 0003 0. Use this model to solve Problems 79–82. 79. CARBON-14 DATING In 2003, Japanese scientists announced the beginning of an effort to bring the long-extinct woolly mammoth back to life using modern cloning techniques. Their efforts were focused on an especially well-preserved specimen discovered frozen in the Siberian ice. Nearby samples of plant material were found to have 28.9% of the amount of carbon-14 in a living sample. What was the approximate age of these samples? 80. CARBON-14 DATING In 2004, archaeologist Al Goodyear discovered a site in South Carolina that contains evidence of the earliest human settlement in North America. Carbon dating of burned plant material indicated 0.2% of the amount of carbon-14 in a live sample. How old was that sample? 81. CARBON-14 DATING Many scholars believe that the earliest nonnative settlers of North America were Vikings who sailed from Iceland.
If a fragment of a wooden tool found and dated in 2004 had 88.3% of the amount of carbon-14 in a living sample, when was this tool made? 82. CARBON-14 DATING In 1998, the Shroud of Turin was examined by researchers, who found that plant fibers in the fabric had 92.1% of the amount of carbon-14 in a living sample. If this is accurate, when was the fabric made? 83. PHOTOGRAPHY An electronic flash unit for a camera is activated when a capacitor is discharged through a filament of wire. After the flash is triggered and the capacitor is discharged, the circuit (see the figure) is connected and the battery pack generates a current to recharge the capacitor. The time it takes for the capacitor to recharge is called the recycle time. For a particular flash unit using a 12-volt battery pack, the charge q, in coulombs, on the capacitor t seconds after recharging has started is given by q 0003 0.0009(1 0002 e00020.2t ) How many seconds will it take the capacitor to reach a charge of 0.0007 coulomb? Compute the answer to three significant digits. R I
V
C S
84. ADVERTISING A company is trying to expose as many people as possible to a new product through television advertising in a large metropolitan area with 2 million possible viewers. A model for the number of people N, in millions, who are aware of the product after t days of advertising was found to be N 0003 2(1 – e00020.037t ) How many days, to the nearest day, will the advertising campaign have to last so that 80% of the possible viewers will be aware of the product? 85. NEWTON’S LAW OF COOLING This law states that the rate at which an object cools is proportional to the difference in temperature between the object and its surrounding medium. The temperature T of the object t hours later is given by T 0003 Tm (T0 0002 Tm)e0002kt where Tm is the temperature of the surrounding medium and T0 is the temperature of the object at t 0003 0. Suppose a bottle of wine at a room temperature of 72°F is placed in a refrigerator at 40°F to cool before a dinner party. After an hour the temperature of the wine is found to be 61.5°F. Find the constant k, to two decimal places, and the time, to one decimal place, it will take the wine to cool from 72 to 50°F. 86. MARINE BIOLOGY Marine life is dependent upon the microscopic plant life that exists in the photic zone, a zone that goes to a depth where about 1% of the surface light still remains. Light intensity is reduced according to the exponential function I 0003 I0e0002kd where I is the intensity d feet below the surface and I0 is the intensity at the surface. The constant k is called the coefficient of extinction. At Crystal Lake in Wisconsin it was found that half the surface light remained at a depth of 14.3 feet. Find k, and find the depth of the photic zone. Compute answers to three significant digits.
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379
Problems 87–90 are based on the Richter scale equation from Section 5-4, M 0003 23 log 10E4.40, where M is the magnitude and E is the amount of energy in joules released by the earthquake. Round all calculations to three significant digits.
(B) The total average daily consumption of energy for the entire United States in 2006 was 2.88 1014 joules. How many days could the energy released by a magnitude 9.0 earthquake power the United States?
87. EARTHQUAKES There were 12 earthquakes recorded worldwide in 2008 with magnitude at least 7.0. (A) How much energy is released by a magnitude 7.0 earthquake? (B) The total average daily consumption of energy for the entire United States in 2006 was 2.88 1014 joules. How many days could the energy released by a magnitude 7.0 earthquake power the United States?
89. EARTHQUAKES There were 12 earthquakes worldwide in 2008 with magnitudes between 7.0 and 7.9. Assume that these earthquakes had an average magnitude of 7.5. How long could the total energy released by these 12 earthquakes power the United States, which had a total energy consumption of 1.05 1017 joules in 2006?
88. EARTHQUAKES On December 26, 2004, a magnitude 9.0 earthquake struck in the Indian Ocean, causing a massive tsunami that resulted in over 230,000 deaths. (A) How much energy was released by this earthquake?
5
CHAPTER
5-1
Review
Exponential Functions x
The equation f(x) 0003 b , b 0, b 0004 1, defines an exponential function with base b. The domain of f is (00020007, 0007) and the range is (0, 0007). The graph of f is a continuous curve that has no sharp corners; passes through (0, 1); lies above the x axis, which is a horizontal asymptote; increases as x increases if b 1; decreases as x increases if b 0006 1; and intersects any horizontal line at most once. The function f is one-to-one and has an inverse. We often use the following exponential function properties: 1. a xa y 0003 a x y a x ax a b 0003 x b b
90. EARTHQUAKES There were 166 earthquakes worldwide in 2008 with magnitudes between 6.0 and 6.9. Assume that these earthquakes had an average magnitude of 6.5. How long could the total energy released by these 166 earthquakes power the United States, which had a total energy consumption of 1.05 1017 joules in 2006?
(a x) y 0003 a xy
(ab)x 0003 a xb x
ax 0003 a x0002y ay
2. a x 0003 a y if and only if x 0003 y. 3. For x 0004 0, a x 0003 b x if and only if a 0003 b.
the time it takes for the population to double. Another model of population growth, A 0003 A0ekt, where A0 is the population at time zero and k is a positive constant called the relative growth rate, uses the exponential function with base e. This model is used for many other types of quantities that exhibit exponential growth as well. 2. Radioactive decay can be modeled by using the half-life decay model A 0003 A0(12)t0005h 0003 A020002t0005h, where A is the amount at time t, A0 is the amount at time t 0003 0, and h is the half-life—the time it takes for half the material to decay. Another model of radioactive decay, A 0003 A0e0002kt , where A0 is the amount at time zero and k is a positive constant, uses the exponential function with base e. This model can be used for other types of quantities that exhibit negative exponential growth as well. 3. Limited growth—the growth of a company or proficiency at learning a skill, for example—can often be modeled by the equation y 0003 A(1 0002 e0002kt ), where A and k are positive constants.
As x approaches 0007, the expression [1 (1兾x)]x approaches the irrational number e ⬇ 2.718 281 828 459. The function f(x) 0003 e x is called the exponential function with base e. The growth of money in an account paying compound interest is described by A 0003 P(1 r兾m)n, where P is the principal, r is the annual rate, m is the number of compounding periods in 1 year, and A is the amount in the account after n compounding periods. If the account pays continuous compound interest, the amount A in the account after t years is given by A 0003 Pert.
Logistic growth is another limited growth model that is useful for modeling phenomena like the spread of an epidemic, or sales of a new product. The logistic model is A 0003 M/(1 ce0002kt ), where c, k, and M are positive constants. A good comparison of these different exponential models can be found in Table 3 at the end of Section 5-2. Exponential regression can be used to fit a function of the form y 0003 ab x to a set of data points. Logistic regression can be used to find a function of the form y 0003 c0005(1 ae0002bx ).
5-2
5-3
Exponential Models
Exponential functions are used to model various types of growth: 1. Population growth can be modeled by using the doubling time growth model A 0003 A02t0005d, where A is the population at time t, A0 is the population at time t 0003 0, and d is the doubling time—
Logarithmic Functions
The logarithmic function with base b is defined to be the inverse of the exponential function with base b and is denoted by y 0003 logb x. So y 0003 logb x if and only if x 0003 b y, b 0, b 0004 1. The domain of a logarithmic function is (0, 0007) and the range is (00020007, 0007). The graph of a logarithmic function is a continuous curve that always passes
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through the point (1, 0) and has the y axis as a vertical asymptote. The following properties of logarithmic functions are useful: 1. logb 1 ⫽ 0 2. logb b ⫽ 1
3. The velocity v of a rocket at burnout is given by the rocket equation v ⫽ c ln (Wt兾Wb), where c is the exhaust velocity, Wt is the takeoff weight, and Wb is the burnout weight.
x
3. logb b ⫽ x 4. blogb x ⫽ x, x ⬎ 0
Logarithmic regression can be used to fit a function of the form y ⫽ a ⫹ b ln x to a set of data points.
5. logb MN ⫽ logb M ⫹ logb N 6. logb
M ⫽ logb M ⫺ logb N N
5-5
p
7. log b M ⫽ p log b M 8. logb M ⫽ logb N if and only if M ⫽ N Logarithms to the base 10 are called common logarithms and are denoted by log x. Logarithms to the base e are called natural logarithms and are denoted by ln x. So log x ⫽ y is equivalent to x ⫽ 10 y, and ln x ⫽ y is equivalent to x ⫽ e y. The change-of-base formula, logb N ⫽ (loga N)兾(loga b), relates logarithms to two different bases and can be used, along with a calculator, to evaluate logarithms to bases other than e or 10.
5-4
2. The magnitude M of an earthquake on the Richter scale is given by M ⫽ 23 log (EⲐE0), where E is the energy released by the earthquake and E0 ⫽ 104.40 joules is a standardized energy level.
Logarithmic Models
The following applications involve logarithmic functions:
Exponential and Logarithmic Equations
Exponential equations are equations in which the variable appears in an exponent. If the exponential expression is isolated, applying a logarithmic function to both sides and using the property logb N p ⫽ p logb N will enable you to remove the variable from the exponent. If the exponential expression is not isolated, we can use previously developed techniques to first solve for the exponential, then solve as above. Logarithmic equations are equations in which the variable appears inside a logarithmic function. In most cases, the key to solving them is to change the equation to the equivalent exponential expression. For equations with multiple log expressions, properties of logarithms can be used to combine the expressions before solving.
1. The decibel is defined by D ⫽ 10 log (I兾I0), where D is the decibel level of the sound, I is the intensity of the sound, and I0 ⫽ 10⫺12 watts per square meter is a standardized sound level.
CHAPTER
5
Review Exercises
Work through all the problems in this chapter review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. 1. Match each equation with the graph of f, g, m, or n in the figure. (A) y ⫽ log2 x (B) y ⫽ 0.5x (C) y ⫽ log0.5 x (D) y ⫽ 2x f
3
g m
⫺4.5
4.5
n ⫺3
3. Write in logarithmic form using base e: x ⫽ e y. Write the expression in Problems 4 and 5 in exponential form. 4. log x ⫽ y
5. ln y ⫽ x
6. (A) Plot at least five points, then draw a hand sketch of the graph of y ⫽ (43)x. (B) Use your result from part A to sketch the graph of y ⫽ log4Ⲑ3 x. In Problems 7 and 8, simplify. 7.
7x⫹2 72⫺x
8. a
ex x b e ⫺x
In Problems 9–11, solve for x exactly. n
2. Write in logarithmic form using base 10: m ⫽ 10 .
9. log2 x ⫽ 3
10. logx 25 ⫽ 2
11. log3 27 ⫽ x
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In Problems 12–15, solve for x to three significant digits. x
In Problems 54–57, use transformations to explain how the graph of g is related to the graph of the given logarithmic function f. Determine whether g is increasing or decreasing, find its domain and any asymptotes, and sketch the graph of g.
x
12. 10 0003 17.5
13. e 0003 143,000
14. ln x 0003 00020.015 73
15. log x 0003 2.013
381
54. g (x) 0003 3 0002 13 2x; f (x) 0003 2x Evaluate the expression in Problems 16–19 to four significant digits using a calculator. 16. ln
17. log (0002e)
18.
20. Write as a single log: 2 log a 0002
58. If the graph of y 0003 e x is reflected in the line y 0003 x, the graph of the function y 0003 ln x is obtained. Discuss the functions that are obtained by reflecting the graph of y 0003 e x in the x axis and the y axis.
1 log b log c 3 a5 1b
21. Write in terms of ln a and ln b: ln
59. (A) Explain why the equation e0002x兾3 0003 4 ln (x 1) has exactly one solution. (B) Find the solution of the equation to three decimal places.
In Problems 22–35, solve for x exactly. x
2x
22. 3 0003 120
23. 10
24. log2 (4x 0002 5) 0003 5
25. ln (x 0002 5) 0003 0
60. Approximate all real zeros of f(x) 0003 4 0002 x2 ln x to three decimal places.
0003 500
61. Find the coordinates of the points of intersection of f(x) 0003 10x00023 and g(x) 0003 8 log x to three decimal places.
26. ln (2x 0002 1) 0003 ln (x 3)
In Problems 62–65, solve for the indicated variable in terms of the remaining symbols.
27. log (x2 0002 3) 0003 2 log (x 0002 1) 2
28. e x 00023 0003 e2x
29. 4x00021 0003 210002x
30. 2x2e0002x 0003 18e0002x
31. log1兾4 16 0003 x
32. log x 9 0003 00022
33. log16 x 0003 32
34. log x e5 0003 5
35. 10log10x 0003 33
62. D 0003 10 log 63. y 0003
38. ln x 0003 00023.218
I for I (sound intensity) I0
2 1 e 0002x 00052 for x (probability) 12
1 I 64. x 0003 0002 ln for I (X-ray intensity) k I0
In Problems 36–45, solve for x to three significant digits. 36. x 0003 2(101.32)
56. g(x) 0003 00022 log4 x; f(x) 0003 log4 x 57. g(x) 0003 1 2 log1兾3 x; f(x) 0003 log1兾3 x
e e0002 19. 2
ln 2
55. g(x) 0003 2e x 0002 4; f(x) 0003 e x
37. x 0003 log5 23 00027
39. x 0003 log (2.156 10 )
65. r 0003 P
i for n (finance) 1 0002 (1 i) 0002n
41. 25 0003 5(2x)
66. Write ln y 0003 00025t ln c in an exponential form free of logarithms; then solve for y in terms of the remaining symbols.
42. 4,000 0003 2,500(e0.12x)
43. 0.01 0003 e00020.05x
67. For f 0003 5(x, y) 冟 y 0003 log2 x6, graph f and f 00021 on the same coordinate system. What are the domains and ranges for f and f 00021?
44. 52x00023 0003 7.08
45.
40. x 0003
ln 4 ln 2.31
ex 0002 e 0002x 00031 2
In Problems 46–51, solve for x exactly. 46. log 3x 2 0002 log 9x 0003 2 47. log x 0002 log 3 0003 log 4 0002 log (x 4) 48. ln (x 3) 0002 ln x 0003 2 ln 2 49. ln (2x 1) 0002 ln (x 0002 1) 0003 ln x 50. (log x)3 0003 log x9
51. ln (log x) 0003 1
In Problems 52 and 53, simplify. 52. (e x 1)(e0002x 0002 1) 0002 e x(e0002x 0002 1) 53. (e x e0002x)(e x 0002 e0002x) 0002 (e x 0002 e0002x)2
68. Explain why 1 cannot be used as a logarithmic base. 69. Prove that logb (MN) 0003 logb M logb N.
APPLICATIONS 70. POPULATION GROWTH Many countries have a population growth rate of 3% (or more) per year. At this rate, how many years will it take a population to double? Use the annual compounding growth model P 0003 P0(1 r)t. Compute the answer to three significant digits. 71. POPULATION GROWTH Repeat Problem 70 using the continuous compounding growth model P 0003 P0e rt. 72. CARBON 14-DATING How many years will it take for carbon-14 to diminish to 1% of the original amount after the death of a plant or animal? Use the formula A 0003 A0e00020.000124t. Compute the answer to three significant digits.
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73. MEDICINE One leukemic cell injected into a healthy mouse will divide into two cells in about 12 day. At the end of the day these two cells will divide into four. This doubling continues until 1 billion cells are formed; then the animal dies with leukemic cells in every part of the body. (A) Write an equation that will give the number N of leukemic cells at the end of t days. (B) When, to the nearest day, will the mouse die? 74. MONEY GROWTH Assume $1 had been invested at an annual rate of 3% compounded continuously in the year A.D. 1. What would be the value of the account in the year 2011? Compute the answer to two significant digits.
Problems 81 and 82 require a graphing calculator or a computer that can calculate exponential, logarithmic, and logistic regression models for a given data set. 81. MEDICARE The annual expenditures for Medicare (in billions of dollars) by the U.S. government for selected years since 1980 are shown in Table 1. Let x represent years since 1980. (A) Find an exponential regression model of the form y 0003 abx for these data. Round to three significant digits. Estimate (to the nearest billion) the total expenditures in 2010 and in 2020. (B) When (to the nearest year) will the total expenditures reach $900 billion?
75. PRESENT VALUE Solving A 0003 Pert for P, we obtain P 0003 Ae0002rt, which is the present value of the amount A due in t years if money is invested at a rate r compounded continuously. (A) Graph P 0003 1,000(e00020.08t ), 0 t 30. (B) What does it appear that P tends to as t tends to infinity? [Conclusion: The longer the time until the amount A is due, the smaller its present value, as we would expect.]
Table 1 Medicare Expenditures
76. EARTHQUAKES The 1971 San Fernando, California, earthquake released 1.99 1014 joules of energy. Compute its magnitude on the Richter scale using the formula M 0003 23 log (E0005E0), where E0 0003 104.40 joules. Compute the answer to one decimal place. 77. EARTHQUAKES Refer to Problem 76. If the 1906 San Francisco earthquake had a magnitude of 8.3 on the Richter scale, how much energy was released? Compute the answer to three significant digits. 78. SOUND If the intensity of a sound from one source is 100,000 times that of another, how much more is the decibel level of the louder sound than the softer one? Use the formula D 0003 10 log (I兾I0). 79. MARINE BIOLOGY The intensity of light entering water is reduced according to the exponential function I 0003 I0e0002kd where I is the intensity d feet below the surface, I0 is the intensity at the surface, and k is the coefficient of extinction. Measurements in the Sargasso Sea in the West Indies have indicated that half the surface light reaches a depth of 73.6 feet. Find k, and find the depth at which 1% of the surface light remains. Compute answers to three significant digits. 80. WILDLIFE MANAGEMENT A lake formed by a newly constructed dam is stocked with 1,000 fish. Their population is expected to increase according to the logistic curve N0003
30 1 29e00021.35t
where N is the number of fish, in thousands, expected after t years. The lake will be open to fishing when the number of fish reaches 20,000. How many years, to the nearest year, will this take?
Year
Billion $
1980
37
1985
72
1990
111
1995
181
2000
225
2005
342
Source: U.S. Bureau of the Census
82. Table 2 lists the number of cell phone subscribers in the United States for selected years from 1994 to 2006. Let x 0003 0 correspond to 1990 and round all coefficients to four significant digits. (A) Find a logarithmic regression model of the form y 0003 a b ln x for the data, then use the model to predict the number of subscribers in 2015. (B) Repeat part A, this time finding a logistic regression model of the form y 0003 c0005(1 ae 0002bx). (C) Which of the models do you think models the data better? Explain. Consider how well it fits the points from the table, as well as how well you think it predicts long-term trends.
Table 2 Cell Phone Subscribers in the U.S. Year
Subscribers in millions
1994
24.13
1997
55.31
2000
109.5
2003
158.8
2006
233.0
Source: CTIA—The Wireless Association
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CHAPTER
ZZZ
5
GROUP ACTIVITY Comparing Regression Models
We have used polynomial, exponential, and logarithmic regression models to fit curves to data sets. How can we determine which equation provides the best fit for a given set of data? There are two principal ways to select models. The first is to use information about the type of data to help make a choice. For example, we expect the weight of a fish to be related to the cube of its length. And we expect most populations to grow exponentially, at least over the short term. The second method for choosing among equations involves developing a measure of how closely an equation fits a given data set. This is best introduced through an example. Consider the data set in Figure 1, where L1 represents the x coordinates and L2 represents the y coordinates. The graph of this data set is shown in Figure 2. Suppose we arbitrarily choose the equation y1 ⫽ 0.6x ⫹ 1 to model these data (Fig. 3). 10
0
10
0
Z Figure 1
Z Figure 2 10
0
10
0
Z Figure 3 y1 ⫽ 0.6x ⫹ 1. To measure how well the graph of y1 fits these data, we examine the difference between the y coordinates in the data set and the corresponding y coordinates on the graph of y1 (L3 in Figs. 4 and 5). 10
0
10
0
Z Figure 4
Z Figure 5 Here ⫹ is L2 and 䊐 is L3.
Each of these differences is called a residual. Note that three of the residuals are positive and one is negative (three of the points lie above the line, one lies below). The most commonly accepted measure of the fit provided by a given model is the sum of the squares of the residuals (SSR). When squared, each residual (whether positive or negative or zero) makes a nonnegative contribution to the SSR. SSR ⫽ (4 ⫺ 2.2)2 ⫹ (5 ⫺ 3.4)2 ⫹ (3 ⫺ 4.6)2 ⫹ (7 ⫺ 5.8)2 ⫽ 9.8 (A) A linear regression model for the data in Figure 1 is given by y2 ⫽ 0.35x ⫹ 3 Compute the SSR for the data and y2, and compare it to the one we computed for y1. It turns out that among all possible linear polynomials, the linear regression model minimizes the sum of the squares of the residuals. For this reason, the linear regression model is often called the least-squares line. A similar statement can be made for polynomials of any fixed degree. That is, the quadratic regression model minimizes the SSR over all quadratic polynomials, the cubic regression model minimizes the SSR over all cubic polynomials, and so on. The same statement cannot be made for exponential or logarithmic regression models. Nevertheless, the SSR can still be used to compare exponential, logarithmic, and polynomial models. (B) Find the exponential and logarithmic regression models for the data in Figure 1, compute their SSRs, and compare with the linear model. (C) National annual advertising expenditures for selected years since 1950 are shown in Table 1 where x is years since 1950 and y is total expenditures in billions of dollars. Which regression model would fit this data best: a quadratic model, a cubic model, or an exponential model? Use the SSRs to support your choice.
Table 1 Annual Advertising Expenditures, 1950–2000 x (years)
0
10
20
30
y (billion $)
5.7
12.0
19.6
53.6
Source: U.S. Bureau of the Census.
40
50
128.6
247.5
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Additional Topics in Analytic Geometry
6
C
OUTLINE
ANALYTIC geometry is the study of geometric objects using algebraic
techniques. René Descartes (1596–1650), the French philosopher and mathematician, is generally recognized as the founder of the subject. We used analytic geometry in Chapter 2 to obtain equations of lines and circles. In Chapter 6, we take a similar approach to the study of parabolas, ellipses, and hyperbolas. Each of these geometric objects is a conic section, that is, the intersection of a plane and a cone. We will derive equations for the conic sections and explore a wealth of applications in architecture, communications, engineering, medicine, optics, and space science.
6-1
Conic Sections; Parabola
6-2
Ellipse
6-3
Hyperbola Chapter 6 Review Chapter 6 Group Activity: Focal Chords
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Conic Sections; Parabola Z Conic Sections Z Definition of a Parabola Z Drawing a Parabola Z Standard Equations and Their Graphs Z Applications
In Section 6-1 we introduce the general concept of a conic section and then discuss the particular conic section called a parabola. In Sections 6-2 and 6-3 we will discuss two other conic sections called ellipses and hyperbolas.
Z Conic Sections In Section 2-3 we found that the graph of a first-degree equation in two variables, Ax 0002 By 0003 C
(1)
where A and B are not both 0, is a straight line, and every straight line in a rectangular coordinate system has an equation of this form. What kind of graph will a second-degree equation in two variables, Ax2 0002 Bxy 0002 Cy2 0002 Dx 0002 Ey 0002 F 0003 0
(2)
where A, B, and C are not all 0, yield for different sets of values of the coefficients? The graphs of equation (2) for various choices of the coefficients are plane curves obtainable by intersecting a cone* with a plane, as shown in Figure 1. These curves are called conic sections. Z Figure 1 Conic sections.
L Circle
Constant V Nappe
Ellipse
Parabola
Hyperbola
If a plane cuts clear through one nappe, then the intersection curve is called a circle if the plane is perpendicular to the axis and an ellipse if the plane is not perpendicular to the axis. If a plane cuts only one nappe, but does not cut clear through, then the intersection curve is called *Starting with a fixed line L and a fixed point V on L, the surface formed by all straight lines through V making a constant angle 0004 with L is called a right circular cone. The fixed line L is called the axis of the cone, and V is its vertex. The two parts of the cone separated by the vertex are called nappes.
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a parabola. Finally, if a plane cuts through both nappes, but not through the vertex, the resulting intersection curve is called a hyperbola. A plane passing through the vertex of the cone produces a degenerate conic—a point, a line, or a pair of lines. Conic sections are very useful and are readily observed in your immediate surroundings: wheels (circle), the path of water from a garden hose (parabola), some serving platters (ellipses), and the shadow on a wall from a light surrounded by a cylindrical or conical lamp shade (hyperbola) are some examples (Fig. 2). We will discuss many applications of conics throughout the remainder of this chapter. Z Figure 2 Examples of conics.
Wheel (circle) (a)
Water from garden hose (parabola) (b)
Lamp light shadow (hyperbola) (d)
Serving platter (ellipse) (c)
Z Definition of a Parabola The following definition of a parabola is a coordinate-free definition. It does not depend on the coordinates of points in any coordinate system. Z DEFINITION 1 Parabola A parabola is the set of all points in a plane equidistant from a fixed point F and a fixed line L (not containing F) in the plane. The fixed point F is called the focus, and the fixed line L is called the directrix. A line through the focus perpendicular to the directrix is called the axis of symmetry, and the point on the axis of symmetry halfway between the directrix and focus is called the vertex.
L
d1
P
d1 0003 d2 Axis of symmetry
d2 V (Vertex)
F (Focus) Parabola Directrix
Z Drawing a Parabola Using Definition 1, we can draw a parabola with fairly simple equipment—a straightedge, a right-angle drawing triangle, a piece of string, a thumbtack, and a pencil. Referring to Figure 3 on the next page, tape the straightedge along the line AB and place the thumbtack above the line AB. Place one leg of the triangle along the straightedge as indicated, then take a piece of string the same length as the other leg, tie one end to the thumbtack, and fasten the other end with tape at C on the triangle. Now press the string to the edge of the triangle, and keeping the string taut, slide the triangle along the straightedge. Because DE will always equal DF, the resulting curve will be part of a parabola with directrix AB lying along the straightedge and focus F at the thumbtack.
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Z Figure 3 Drawing a parabola.
C
D F E A
ZZZ EXPLORE-DISCUSS 1
B
The line through the focus F that is perpendicular to the axis of symmetry of a parabola intersects the parabola in two points G and H. Explain why the distance from G to H is twice the distance from F to the directrix of the parabola.
Z Standard Equations and Their Graphs Using the definition of a parabola and the distance formula d 0003 2(x2 0007 x1)2 0002 ( y2 0007 y1)2
(3)
we can derive simple standard equations for a parabola located in a rectangular coordinate system with its vertex at the origin and its axis of symmetry along a coordinate axis. We start with the axis of symmetry of the parabola along the x axis and the focus at F 0003 (a, 0). We locate the parabola in a coordinate system as in Figure 4 and label key lines and points. This is an important step in finding an equation of a geometric figure in a coordinate system. Note that the parabola opens to the right if a 0005 0 and to the left if a 0006 0. The vertex is at the origin, the directrix is x 0003 0007a, and the coordinates of M are (0007a, y). y
Z Figure 4 Parabola with vertex at the origin and axis of symmetry the x axis. M 0003 (0007a, y)
0007a
y d1
P 0003 (x, y)
d2 Focus F 0003 (a, 0)
Directrix x 0003 0007a
a ⬎ 0, focus on positive x axis (a)
x
d1 P 0003 (x, y) d2 Focus F 0003 (a, 0)
M 0003 (0007a, y)
0007a
x
Directrix x 0003 0007a
a ⬍ 0, focus on negative x axis (b)
The point P 0003 (x, y) is a point on the parabola if and only if d1 0003 d2 d(P, M ) 0003 d(P, F) 2 2(x 0002 a) 0002 ( y 0007 y)2 0003 2(x 0007 a)2 0002 ( y 0007 0)2 (x 0002 a)2 0003 (x 0007 a)2 0002 y2 2 x 0002 2ax 0002 a2 0003 x2 0007 2ax 0002 a2 0002 y2 y2 ⴝ 4ax
Use equation (3). Square both sides. Simplify.
(4)
Equation (4) is the standard equation of a parabola with vertex at the origin, axis of symmetry the x axis, and focus at (a, 0).
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By a similar derivation (see Problem 57 in Exercises 6-1), the standard equation of a parabola with vertex at the origin, axis of symmetry the y axis, and focus at (0, a) is given by equation (5). x2 0003 4ay
(5)
Looking at Figure 5, note that the parabola opens upward if a 0005 0 and downward if a 0006 0. y
Z Figure 5 Parabola with vertex at the origin and axis of symmetry the y axis.
y Directrix y 0003 0007a
N 0003 (x, 0007a) 0007a F 0003 (0, a) Focus
d2
P 0003 (x, y) d1
Directrix y 0003 0007a
0007a
x
d1 d2
P 0003 (x, y)
x
F 0003 (0, a) Focus
N 0003 (x, 0007a)
a ⬎ 0, focus on positive y axis (a)
a ⬍ 0, focus on negative y axis (b)
We summarize these results for easy reference in Theorem 1. Z THEOREM 1 Standard Equations of a Parabola with Vertex at (0, 0) 1. y2 0003 4ax Vertex: (0, 0) Focus: (a, 0) Directrix: x 0003 0007a Symmetric with respect to the x axis Axis of symmetry the x axis 2. x2 0003 4ay Vertex: (0, 0) Focus: (0, a) Directrix: y 0003 0007a Symmetric with respect to the y axis Axis of symmetry the y axis
EXAMPLE
1
y
F 0
y
x
F
x
0
a ⬍ 0 (opens left)
a ⬎ 0 (opens right)
y
y
0
F
x
F 0
a ⬍ 0 (opens down)
x
a ⬎ 0 (opens up)
Graphing a Parabola Locate the focus and directrix and sketch the graph of y2 0003 16x.
SOLUTION
The equation y2 0003 16x has the form y2 0003 4ax with 4a 0003 16, so a 0003 4. Therefore, the focus is (4, 0) and the directrix is the line x 0003 00074. To sketch the graph, we choose some values of x that make the right side of the equation a perfect square and solve for y. x
0
1
4
y
0
4
8
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Note that x must be greater than or equal to 0 for y to be a real number. Then we plot the resulting points. Because a ⬎ 0, the parabola opens to the right (Fig. 6). y 10
Directrix x ⫽ ⫺4
Focus F ⫽ (4, 0)
⫺10
10
x
⫺10
Z Figure 6
Technology Connections To graph y 2 ⴝ 16x on a graphing calculator, we solve the equation for y.
y ⴝ 16x 2
Directrix x ⫽ ⫺4
10
Take square roots.
y ⴝ ⴞ41x
⫺10
This results in two functions, y ⴝ 4 1x and y ⴝ ⴚ4 1x. Entering these functions in a graphing calculator (Fig. 7) and graphing in a standard viewing window produces the graph of the parabola (Fig. 8).
10
⫺10
Z Figure 8
Z Figure 7
Focus F ⫽ (4, 0)
0002 MATCHED PROBLEM 1
ZZZ
A common error in making a quick sketch of y2 ⫽ 4ax or x2 ⫽ 4ay is to sketch the first with the y axis as its axis of symmetry and the second with the x axis as its axis of symmetry. The graph of y2 ⫽ 4ax is symmetric with respect to the x axis, and the graph of x2 ⫽ 4ay is symmetric with respect to the y axis, as a quick symmetry check will reveal.
CAUTION ZZZ
EXAMPLE
Graph y2 ⫽ ⫺8x, and locate the focus and directrix.
2
Finding the Equation of a Parabola (A) Find the equation of a parabola having the origin as its vertex, the y axis as its axis of symmetry, and (⫺10, ⫺5) on its graph. (B) Find the coordinates of its focus and the equation of its directrix.
SOLUTIONS
(A) Because the axis of symmetry of the parabola is the y axis, the parabola has an equation of the form x2 ⫽ 4ay. Because (⫺10, ⫺5) is on the graph, we have x2 ⫽ 4ay (⫺10)2 ⫽ 4a(⫺5) 100 ⫽ ⫺20a a ⫽ ⫺5
Substitute x ⴝ ⴚ10 and y ⴝ ⴚ5. Simplify. Divide both sides by ⴚ20.
0002
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Therefore, the equation of the parabola is x2 0003 4(00075)y x2 0003 000720y (B) Focus: F 0003 (0, a) 0003 (0, 00075) Directrix: y 0003 0007a y00035 MATCHED PROBLEM 2
0002
(A) Find the equation of a parabola having the origin as its vertex, the x axis as its axis of symmetry, and (4, 00078) on its graph. (B) Find the coordinates of its focus and the equation of its directrix. 0002
Z Applications If you are observant, you will find many applications of parabolas in the physical world. Parabolas are key to the design of suspension bridges, arch bridges, microphones, symphony shells, satellite antennas, radio and optical telescopes, radar equipment, solar furnaces, and searchlights. Figure 9(a) illustrates a parabolic reflector used in all reflecting telescopes—from 3- to 6-inch home types to the 200-inch research instrument on Mount Palomar in California. Parallel light rays from distant celestial bodies are reflected to the focus off a parabolic mirror. If the light source is the sun, then the parallel rays are focused at F and we have a solar furnace. Temperatures of over 6,000 C have been achieved by such furnaces. If we locate a light source at F, then the rays in Figure 9(a) reverse, and we have a spotlight or a searchlight. Automobile headlights can use parabolic reflectors with special lenses over the light to diffuse the rays into useful patterns. Figure 9(b) shows a suspension bridge, such as the Golden Gate Bridge in San Francisco. The suspension cable is a parabola. It is interesting to note that a free-hanging cable, such as a telephone line, does not form a parabola. It forms another curve called a catenary. Figure 9(c) shows a concrete arch bridge. If all the loads on the arch are to be compression loads (concrete works very well under compression), then using physics and advanced mathematics, it can be shown that the arch must be parabolic.
Parallel light rays
Parabola
F Parabola
Parabolic reflector
Suspension bridge
Arch bridge
(a)
(b)
(c)
Z Figure 9 Uses of parabolic forms.
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Parabolic Reflector A paraboloid is formed by revolving a parabola about its axis of symmetry. A spotlight in the form of a paraboloid 5 inches deep has its focus 2 inches from the vertex. Find, to one decimal place, the radius R of the opening of the spotlight.
SOLUTION
Step 1. Locate a parabolic cross section containing the axis of symmetry in a rectangular coordinate system, and label all known parts and parts to be found. This is a very important step and can be done in infinitely many ways. We can make things simpler for ourselves by locating the vertex at the origin and choosing a coordinate axis as the axis of symmetry. We choose the y axis as the axis of symmetry of the parabola with the parabola opening upward (Fig. 10). y 5
(R, 5) R F 0003 (0, 2) Spotlight
00075
x
5
Z Figure 10
Step 2. Find the equation of the parabola in the figure. Because the parabola has the y axis as its axis of symmetry and the vertex at the origin, the equation is of the form x2 0003 4ay We are given F 0003 (0, a) 0003 (0, 2); so a 0003 2, and the equation of the parabola is x2 0003 8y Step 3. Use the equation found in step 2 to find the radius R of the opening. Because (R, 5) is on the parabola, we have R2 0003 8(5) R 0003 140 ⬇ 6.3 inches MATCHED PROBLEM 3
0002
Repeat Example 3 with a paraboloid 12 inches deep and a focus 9 inches from the vertex. 0002
ANSWERS TO MATCHED PROBLEMS y
1. Focus: (00072, 0) Directrix: x 0003 2 x
0
00072
y
0
4
5
(00072, 0) 00075
F
Directrix x00032 5
x
00075
2. (A) y2 0003 16x (B) Focus: (4, 0); Directrix: x 0003 00074 3. R 0003 20.8 inches
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Exercises
1. List the seven different types of conic sections. 2. Explain how each of the seven types of conic sections can be obtained as the intersection of a cone and a plane. 3. What is a degenerate conic? 4. Give a coordinate-free definition of a parabola in your own words. 5. What happens to light rays that are parallel to the axis of a parabolic mirror when they hit the mirror? 6. What happens to light rays that are emitted from the focus of a parabolic mirror when they hit the mirror? In Problems 7–10, a parabola has its vertex at the origin and the given directrix. Find the coordinates of the focus. 7. x 0003 8 9. y 0003 000710
8. x 0003 00075 10. y 0003 6
In Problems 11–14, a parabola has its vertex at the origin and the given focus. Find the equation of the directrix. 11. (0, 000715)
12. (0, 9)
13. (25, 0)
14. (000721, 0)
In Problems 15–24, graph each equation, and locate the focus and directrix. 15. y2 0003 4x
16. y2 0003 8x
17. x2 0003 8y
18. x2 0003 4y
19. y2 0003 000712x
20. y2 0003 00074x
21. x2 0003 00074y
22. x2 0003 00078y
23. y2 0003 000720x
24. x2 0003 000724y
38. Focus (00074, 0)
37. Focus (2, 0)
In Problems 39–44, find the equation of the parabola having its vertex at the origin, its axis of symmetry as indicated, and passing through the indicated point. 39. y axis; (4, 2)
40. x axis; (4, 8)
41. x axis; (00073, 6)
42. y axis; (00075, 10)
43. y axis; (00076, 00079)
44. x axis; (00076, 000712)
In Problems 45–48, find the first-quadrant points of intersection for each pair of parabolas to three decimal places. 45. x2 0003 4y y2 0003 4x
46. y2 0003 3x x2 0003 3y
47. y2 0003 6x x2 0003 5y
48. x2 0003 7y y2 0003 2x
49. Consider the parabola with equation x2 0003 4ay. (A) How many lines through (0, 0) intersect the parabola in exactly one point? Find their equations. (B) Find the coordinates of all points of intersection of the parabola with the line through (0, 0) having slope m 0. 50. Find the coordinates of all points of intersection of the parabola with equation x2 0003 4ay and the parabola with equation y2 0003 4bx. 51. The line segment AB through the focus in the figure is called a focal chord of the parabola. Find the coordinates of A and B. y F 0003 (0, a) A
B
In Problems 25–30, find the coordinates to two decimal places of the focus of the parabola. 25. y2 0003 39x
26. x2 0003 58y
27. x2 0003 0007105y
28. y2 0003 000793x
29. y2 0003 000777x
30. x2 0003 0007205y
32. Directrix y 0003 4
33. Focus (0, 00077)
34. Focus (0, 5)
35. Directrix x 0003 6
36. Directrix x 0003 00079
x
0
52. The line segment AB through the focus in the figure is called a focal chord of the parabola. Find the coordinates of A and B. y
y 2 0003 4ax B
In Problems 31–38, find the equation of a parabola with vertex at the origin, axis of symmetry the x or y axis, and 31. Directrix y 0003 00073
x 2 0003 4ay
F 0003 (a, 0) 0
A
x
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In Problems 53–56, use the definition of a parabola and the distance formula to find the equation of a parabola with 53. Directrix y 0003 00074 and focus (2, 2) 54. Directrix y 0003 2 and focus (00073, 6)
(A) Find the equation of the parabola after inserting an xy coordinate system with the vertex at the origin and the y axis (pointing upward) the axis of symmetry of the parabola. (B) How far is the focus from the vertex? 61. SPACE SCIENCE A designer of a 200-foot-diameter parabolic electromagnetic antenna for tracking space probes wants to place the focus 100 feet above the vertex (see the figure).
55. Directrix x 0003 2 and focus (6, 00074) 56. Directrix x 0003 00073 and focus (1, 4) 57. Use the definition of a parabola and the distance formula to derive the equation of a parabola with focus F 0003 (0, a) and directrix y 0003 0007a for a 0. 58. Let F be a fixed point and let L be a fixed line in the plane that contains F. Describe the set of all points in the plane that are equidistant from F and L.
200 ft Focus
APPLICATIONS 59. ENGINEERING The parabolic arch in the concrete bridge in the figure must have a clearance of 50 feet above the water and span a distance of 200 feet. Find the equation of the parabola after inserting a coordinate system with the origin at the vertex of the parabola and the vertical y axis (pointing upward) along the axis of symmetry of the parabola.
100 ft Radio telescope
(A) Find the equation of the parabola using the axis of symmetry of the parabola as the y axis (up positive) and vertex at the origin. (B) Determine the depth of the parabolic reflector. 62. SIGNAL LIGHT A signal light on a ship is a spotlight with parallel reflected light rays (see the figure). Suppose the parabolic reflector is 12 inches in diameter and the light source is located at the focus, which is 1.5 inches from the vertex. 60. ASTRONOMY The cross section of a parabolic reflector with 6-inch diameter is ground so that its vertex is 0.15 inch below the rim (see the figure).
Signal light
Focus 6 inches
0.15 inch
Parabolic reflector
(A) Find the equation of the parabola using the axis of symmetry of the parabola as the x axis (right positive) and vertex at the origin. (B) Determine the depth of the parabolic reflector.
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Ellipse
395
Ellipse Z Definition of an Ellipse Z Drawing an Ellipse Z Standard Equations of Ellipses and Their Graphs Z Applications
We start our discussion of the ellipse with a coordinate-free definition. Using this definition, we show how an ellipse can be drawn and we derive standard equations for ellipses specially located in a rectangular coordinate system.
Z Definition of an Ellipse The following is a coordinate-free definition of an ellipse:
Z DEFINITION 1 Ellipse An ellipse is the set of all points P in a plane d1 0002 d2 0003 Constant such that the sum of the distances from P to B two fixed points in the plane is a constant (the V d1 P constant is required to be greater than the F distance between the two fixed points). Each of d2 the fixed points, F and F, is called a focus, and together they are called foci. Referring to F V the figure, the line segment V V through the B foci is the major axis. The perpendicular bisector B B of the major axis is the minor axis. Each end of the major axis, V and V, is called a vertex. The midpoint of the line segment F F is called the center of the ellipse.
Z Drawing an Ellipse An ellipse is easy to draw. All you need is a piece of string, two thumbtacks, and a pencil or pen (see Figure 1 on the next page). Place the two thumbtacks in a piece of cardboard. These form the foci of the ellipse. Take a piece of string longer than the distance between the two thumbtacks—this represents the constant in the definition—and tie each end to a thumbtack. Finally, catch the tip of a pencil under the string and move it while keeping the string taut. The resulting figure is by definition an ellipse. Ellipses of different shapes result, depending on the placement of thumbtacks and the length of the string joining them.
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Note that d1 0006 d2 always adds up to the length of the string, which does not change.
P d2
d1 Focus
String
Focus
Z Figure 1 Drawing an ellipse.
Z Standard Equations of Ellipses and Their Graphs Using the definition of an ellipse and the distance formula, we can derive standard equations for an ellipse located in a rectangular coordinate system. We start by placing an ellipse in the coordinate system with the foci on the x axis at F0002 0003 (0004c, 0) and F 0003 (c, 0) with c 0005 0 (Fig. 2). By definition 1 the constant sum d1 0006 d2 is required to be greater than 2c (the distance between F and F0002). Therefore, the ellipse intersects the x axis at points V0002 0003 (0004a, 0) and V 0003 (a, 0) with a 0005 c 0005 0, and it intersects the y axis at points B0002 0003 (0004b, 0) and B 0003 (b, 0) with b 0005 0.
y b
P 0003 (x, y)
d1 0004a
d2
F0002 0003 (0004c, 0)
0
F 0003 (c, 0) a
x
0004b d1 0006 d2 0003 Constant 0005 d(F, F0002) c00050
Z Figure 2 Ellipse with foci on x axis.
Study Figure 2: Note first that if P 0003 (a, 0), then d1 0006 d2 0003 2a. (Why?) Therefore, the constant sum d1 0006 d2 is equal to the distance between the vertices. Second, if P 0003 (0, b), then d1 0003 d2 0003 a and a2 0003 b2 0006 c2 by the Pythagorean theorem; in particular, a 0005 b. Referring again to Figure 2, the point P 0003 (x, y) is on the ellipse if and only if d1 0006 d2 0003 2a Using the distance formula for d1 and d2, eliminating radicals, and simplifying (see Problem 49 in Exercises 6-2), we obtain the equation of the ellipse pictured in Figure 2: y2 x2 2 ⴙ 2 ⴝ 1 a b By similar reasoning (see Problem 50 in Exercises 6-2) we obtain the equation of an ellipse centered at the origin with foci on the y axis. Both cases are summarized in Theorem 1.
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397
Ellipse
Z THEOREM 1 Standard Equations of an Ellipse with Center at (0, 0) 1.
y2 x2 a0005b00050 0002 00031 a2 b2 x intercepts: a (vertices) y intercepts: b Foci: F 0003 (0007c, 0), F 0003 (c, 0)
y b
0007a
c2 0003 a2 0007 b2
F 0007c
F c
0
Major axis length 0003 2a Minor axis length 0003 2b y2 x2 2. 2 0002 2 0003 1 a0005b00050 b a x intercepts: b y intercepts: a (vertices) Foci: F 0003 (0, 0007c), F 0003 (0, c)
a
x
0007b y a c F
c2 0003 a2 0007 b2 Major axis length 0003 2a Minor axis length 0003 2b [Note: Both graphs are symmetric with respect to the x axis, y axis, and origin. Also, the major axis is always longer than the minor axis.]
ZZZ EXPLORE-DISCUSS 1
EXAMPLE
1
0
0007b
b
x
0007c F 0007a
The line through a focus F of an ellipse that is perpendicular to the major axis intersects the ellipse in two points G and H. For each of the two standard equations of an ellipse with center (0, 0), find an expression in terms of a and b for the distance from G to H.
Graphing an Ellipse Find the coordinates of the foci, find the lengths of the major and minor axes, and graph the following equation: 9x2 0002 16y2 0003 144
SOLUTION
First, write the equation in standard form by dividing both sides by 144 and determine a and b: 9x2 0002 16y2 0003 144 16y2 9x2 144 0002 0003 144 144 144
Divide both sides by 144. * Simplify.
y2 x2 0002 00031 16 9 a00034
and
b00033
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
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y
x intercepts: 4 y intercepts: 3
3
F 0007c
8:46 PM
0
F c
4
Major axis length: 2(4) 0003 8 Minor axis length: 2(3) 0003 6 Foci: c2 0003 a2 0007 b2 0003 16 0007 9 00037
x
c 0003 17
00073
Substitute a ⴝ 4 and b ⴝ 3.
c must be positive.
So the foci are F¿ 0003 (0007 17, 0) and F 0003 (17, 0). Plot the foci and intercepts and sketch the ellipse (Fig. 3).
Z Figure 3
MATCHED PROBLEM 1
Find the coordinates of the foci, find the lengths of the major and minor axes, and graph the following equation: x2 0002 4y2 0003 4
EXAMPLE
2
0002
0002
Graphing an Ellipse Find the coordinates of the foci, find the lengths of the major and minor axes, and graph the following equation: 2x2 0002 y2 0003 10
SOLUTION
First, write the equation in standard form by dividing both sides by 10 and determine a and b: 2x2 0002 y2 0003 10 y2 2x2 10 0002 0003 10 10 10
Divide both sides by 10.
Simplify.
y2 x2 0002 00031 5 10 a 0003 110 y intercepts: 110 ⬇ 3.16 x intercepts: 15 ⬇ 2.24
and
b 0003 15
Major axis length: 2110 ⬇ 6.32 Minor axis length: 215 ⬇ 4.47
Foci: c2 0003 a2 0007 b2 0003 10 0007 5 00035 c 0003 15
Substitute a ⴝ 110, b ⴝ 15.
c must be positive.
So the foci are F¿ 0003 (0, 0007 15) and F 0003 (0, 15). Plot the foci and intercepts and sketch the ellipse (Fig. 4). y 兹10
c F 0
0007兹5
0007c F 0007兹10
Z Figure 4
兹5
x
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Ellipse
399
Technology Connections To graph the ellipse of Example 2 on a graphing calculator, solve the original equation for y:
2x 2 ⴙ y 2 ⴝ 10
near the x intercepts; they are due to the relatively low resolution of the graphing calculator screen. 4
Subtract 2x2 from both sides.
y 2 ⴝ 10 ⴚ 2x 2 y ⴝ ⴞ210 ⴚ 2x
Take square roots of both sides. 2
00076
This produces two functions, y1 ⴝ 210 ⴚ 2x 2 and y2 ⴝ ⴚ 210 ⴚ 2x 2, which are graphed in Figure 5. Notice that we used a squared viewing window to avoid distorting the shape of the ellipse. Also note the gaps in the graph
6
00074
Z Figure 5
0002 MATCHED PROBLEM 2
Find the coordinates of the foci, find the lengths of the major and minor axes, and graph the following equation: 3x2 0002 y2 0003 18
EXAMPLE
3
0002
Finding the Equation of an Ellipse Find an equation of an ellipse in the form y2 x2 0002 00031 M N
M, N 7 0
if the center is at the origin, the major axis is along the y axis, and (A) Length of major axis 0003 20 Length of minor axis 0003 12 SOLUTIONS
(B) Length of major axis 0003 10 Distance of foci from center 0003 4
(A) Compute x and y intercepts and make a rough sketch of the ellipse, as shown in Figure 6. y2 x2 0002 00031 b2 a2 20 a0003 0003 10 2 y2 x2 0002 00031 36 100 y 10
000710
10
000710
Z Figure 6
x
b0003
12 00036 2
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(B) Make a rough sketch of the ellipse, as shown in Figure 7; locate the foci and y intercepts, then determine the x intercepts using the fact that a2 0003 b 2 0002 c2: y2 x2 0002 00031 b2 a2 a0003
10 00035 2
b2 0003 52 0007 42 0003 25 0007 16 0003 9 b00033
y2 x2 0002 00031 9 25 y 5 4
0
0007b
b
x
00075
0002
Z Figure 7
MATCHED PROBLEM 3
Find an equation of an ellipse in the form y2 x2 0002 00031 M N
M, N 7 0
if the center is at the origin, the major axis is along the x axis, and (A) Length of major axis 0003 50 Length of minor axis 0003 30
(B) Length of minor axis 0003 16 Distance of foci from center 0003 6
0002
Z Applications Ellipses have many applications: orbits of satellites, planets, and comets; shapes of galaxies; gears and cams, some airplane wings, boat keels, and rudders; tabletops; public fountains; and domes in buildings are a few examples (Fig. 8).
Planet Sun
F
F
Planetary motion
Elliptical gears
Elliptical dome
(a)
(b)
(c)
Z Figure 8 Uses of elliptical forms.
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Ellipse
401
Johannes Kepler (1571–1630), a German astronomer, discovered that planets move in elliptical orbits, with the sun at a focus, and not in circular orbits as had been thought before [Fig. 8(a)]. Figure 8(b) shows a pair of elliptical gears with pivot points at foci. Such gears transfer constant rotational speed to variable rotational speed, and vice versa. Figure 8(c) shows an elliptical dome. An interesting property of such a dome is that a sound or light source at one focus will reflect off the dome and pass through the other focus. One of the chambers in the Capitol Building in Washington, D.C., has such a dome, and is referred to as a whispering room because a whispered sound at one focus can be easily heard at the other focus. A fairly recent application in medicine is the use of elliptical reflectors and ultrasound to break up kidney stones. A device called a lithotripter is used to generate intense sound waves that break up the stone from outside the body, eliminating the need for surgery. To be certain that the waves do not damage other parts of the body, the reflecting property of the ellipse is used to design and correctly position the lithotripter.
EXAMPLE
4
Medicinal Lithotripsy A lithotripter is formed by rotating the portion of an ellipse below the minor axis around the major axis (Fig. 9). The lithotripter is 20 centimeters wide and 16 centimeters deep. If the ultrasound source is positioned at one focus of the ellipse and the kidney stone at the other, then all the sound waves will pass through the kidney stone. How far from the kidney stone should the point V on the base of the lithotripter be positioned to focus the sound waves on the kidney stone? Round the answer to one decimal place. Kidney stone
Ultrasound source
Base
20 cm
V
16 cm
Z Figure 9 Lithotripter.
SOLUTION
From Figure 9 we see that a 0003 16 and b 0003 10 for the ellipse used to form the lithotripter. So the distance c from the center to either the kidney stone or the ultrasound source is given by c 0003 2a2 0007 b2 0003 2162 0007 102 0003 2156 ⬇ 12.5 and the distance from the base of the lithotripter to the kidney stone is 16 0002 12.5 0003 28.5 centimeters. 0002
MATCHED PROBLEM 4
Because lithotripsy is an external procedure, the lithotripter described in Example 4 can be used only on stones within 12.5 centimeters of the surface of the body. Suppose a kidney stone is located 14 centimeters from the surface. If the diameter is kept fixed at 20 centimeters, how deep must a lithotripter be to focus on this kidney stone? Round answer to one decimal place. 0002
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ANSWERS TO MATCHED PROBLEMS 1.
y 1
Foci: F 0003 (0007兹3, 0), F 0003 (兹3, 0) Major axis length 0003 4 Minor axis length 0003 2
F
F 0
00072
2
x
00071
y
2. 兹18
F
0007兹6
Foci: F 0003 (0, 0007兹12), F 0003 (0, 兹12) Major axis length 0003 2兹18 ⬇ 8.49 Minor axis length 0003 2兹6 ⬇ 4.90 兹6
x
F 0007兹18
3. (A)
6-2
y2 x2 0002 00031 625 225
(B)
y2 x2 0002 00031 100 64
4. 17.2 centimeters
Exercises
1. Give a coordinate-free definition of an ellipse in your own words. 2. Explain how the major axis of an ellipse differs from the minor axis. 3. Given the major axis of an ellipse and the foci, describe a procedure for drawing the ellipse. 4. Is the graph of an ellipse the graph of a function? Explain. 5. Is a circle an ellipse? Explain. 6. Using the definition of an ellipse, explain why the minor axis is shorter than the major axis. In Problems 7–10, find the distance between the foci of the ellipse.
In Problems 11–14, find the length of the major axis of the ellipse. 11. Distance between foci 0003 14 Minor axis length 0003 48 12. Distance between foci 0003 10 Minor axis length 0003 1 13. Distance between foci 0003 5 Minor axis length 0003 5 14. Distance between foci 0003 3 Minor axis length 0003 313 In Problems 15–20, sketch a graph of each equation, find the coordinates of the foci, and find the lengths of the major and minor axes.
7. Major axis length 0003 10 Minor axis length 0003 8
15.
y2 x2 0002 00031 25 4
16.
8. Major axis length 0003 26 Minor axis length 0003 10
18.
y2 x2 0002 00031 4 9
19. x2 0002 9y2 0003 9
9. Major axis length 0003 2 Minor axis length 0003 1
In Problems 21–24, match each equation with one of graphs (a)–(d).
10. Major axis length 0003 4 Minor axis length 0003 3
21. 9x2 0002 16y2 0003 144
y2 x2 0002 00031 9 4
17.
y2 x2 0002 00031 4 25
20. 4x2 0002 y2 0003 4
22. 16x2 0002 9y2 0003 144
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23. 4x2 0002 y2 0003 16 y
y
5
403
y
33. The graph is
24. x2 0002 4y2 0003 16
Ellipse
10
5
000710 00075
5
x
00075
5
10
x
x 000710
00075
(b)
y
y
5
y
34. The graph is
00075
(a)
10
5 000710
00075
5
x
00075
00075
5
In Problems 25–30, sketch a graph of each equation, find the coordinates of the foci, and find the lengths of the major and minor axes. 25. 25x2 0002 9y2 0003 225
26. 16x2 0002 25y2 0003 400
27. 2x2 0002 y2 0003 12
28. 4x2 0002 3y2 0003 24
29. 4x2 0002 7y2 0003 28
30. 3x2 0002 2y2 0003 24
In Problems 31–42, find an equation of an ellipse in the form M, N 7 0
if the center is at the origin, and
10
000710
10
x
000710
10
000710
38. Major axis on y axis Major axis length 0003 24 Minor axis length 0003 18 39. Major axis on x axis Major axis length 0003 16 Distance of foci from center 0003 6
41. Major axis on y axis Minor axis length 0003 20 Distance of foci from center 0003 170 42. Major axis on x axis Minor axis length 0003 14 Distance of foci from center 0003 1200
44. Consider all ellipses having (0, 1) as the ends of the minor axis. Describe the connection between the elongation of the ellipse and the distance from a focus to the origin.
10
000710
37. Major axis on y axis Major axis length 0003 22 Minor axis length 0003 16
43. Explain why an equation whose graph is an ellipse does not define a function.
y
32. The graph is
36. Major axis on x axis Major axis length 0003 14 Minor axis length 0003 10
40. Major axis on y axis Major axis length 0003 24 Distance of foci from center 0003 10
y
31. The graph is
000710
35. Major axis on x axis Major axis length 0003 10 Minor axis length 0003 6
(d)
y2 x2 0002 00031 M N
x
x
00075
(c)
10
x
45. Find an equation of the set of points in a plane, each of whose distance from (2, 0) is one-half its distance from the line x 0003 8. Identify the geometric figure. 46. Find an equation of the set of points in a plane, each of whose distance from (0, 9) is three-fourths its distance from the line y 0003 16. Identify the geometric figure.
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47. Let F and F⬘ be two points in the plane and let c denote the constant d(F, F⬘). Describe the set of all points P in the plane such that the sum of the distances from P to F and F⬘ is equal to the constant c. 48. Let F and F⬘ be two points in the plane and let c be a constant such that 0 ⬍ c ⬍ d(F, F⬘). Describe the set of all points P in the plane such that the sum of the distances from P to F and F⬘ is equal to the constant c. 49. Study the following derivation of the standard equation of an ellipse with foci (⫾c, 0), x intercepts (⫾a, 0), and y intercepts (0, ⫾b). Explain why each equation follows from the equation that precedes it. [Hint: Recall from Figure 2 on page 396 that a2 ⫽ b2 ⫹ c2.] d1 ⫹ d2 ⫽ 2a 2(x ⫹ c)2 ⫹ y2 ⫽ 2a ⫺ 2(x ⫺ c)2 ⫹ y2 (x ⫹ c)2 ⫹ y2 ⫽ 4a2 ⫺ 4a 2(x ⫺ c)2 ⫹ y2 ⫹ (x ⫺ c)2 ⫹ y2 cx 2(x ⫺ c)2 ⫹ y2 ⫽ a ⫺ a (x ⫺ c)2 ⫹ y2 ⫽ a2 ⫺ 2cx ⫹ a1 ⫺
c2x2 a2
c2 2 b x ⫹ y2 ⫽ a2 ⫺ c2 a2 y2 x2 ⫹ 2⫽1 2 a b
Elliptical bridge
52. DESIGN A 4 ⫻ 8 foot elliptical tabletop is to be cut out of a 4 ⫻ 8 foot rectangular sheet of teak plywood (see the figure). To draw the ellipse on the plywood, how far should the foci be located from each edge and how long a piece of string must be fastened to each focus to produce the ellipse (see Fig. 1 on page 396)? Compute the answer to two decimal places.
String
F⬘
F Elliptical table
50. Study the following derivation of the standard equation of an ellipse with foci (0, ⫾c), y intercepts (0, ⫾a), and x intercepts (⫾b, 0). Explain why each equation follows from the equation that precedes it. [Hint: Recall from Figure 2 on page 396 that a2 ⫽ b2 ⫹ c2.]
53. AERONAUTICAL ENGINEERING Of all possible wing shapes, it has been determined that the one with the least drag along the trailing edge is an ellipse. The leading edge may be a straight line, as shown in the figure. One of the most famous planes with this design was the World War II British Spitfire. The plane in the figure has a wingspan of 48.0 feet. Leading edge
d1 ⫹ d2 ⫽ 2a 2
2x ⫹ ( y ⫹ c)2 ⫽ 2a ⫺ 2x2 ⫹ ( y ⫺ c)2 x2 ⫹ ( y ⫹ c)2 ⫽ 4a2 ⫺ 4a 2x2 ⫹ ( y ⫺ c)2 ⫹ x2 ⫹ ( y ⫺ c)2 cy 2x2 ⫹ ( y ⫺ c)2 ⫽ a ⫺ a x2 ⫹ ( y ⫺ c)2 ⫽ a2 ⫺ 2cy ⫹
c2y2
Trailing edge
Elliptical wings and tail
a2
2
c b y2 ⫽ a2 ⫺ c2 a2 y2 x2 ⫹ ⫽1 b2 a2
x2 ⫹ a1 ⫺
Fuselage
APPLICATIONS 51. ENGINEERING The semielliptical arch in the concrete bridge in the figure must have a clearance of 12 feet above the water and span a distance of 40 feet. Find the equation of the ellipse after inserting a coordinate system with the center of the ellipse at the origin and the major axis on the x axis. The y axis points up, and the x axis points to the right. How much clearance above the water is there 5 feet from the bank?
(A) If the straight-line leading edge is parallel to the major axis of the ellipse and is 1.14 feet in front of it, and if the leading edge is 46.0 feet long (including the width of the fuselage), find the equation of the ellipse. Let the x axis lie along the major axis (positive right), and let the y axis lie along the minor axis (positive forward). (B) How wide is the wing in the center of the fuselage (assuming the wing passes through the fuselage)? Compute quantities to three significant digits. 54. NAVAL ARCHITECTURE Currently, many high-performance racing sailboats use elliptical keels, rudders, and main sails for the reasons stated in Problem 53—less drag along the trailing edge. In the accompanying figure, the ellipse containing the keel has a 12.0foot major axis. The straight-line leading edge is parallel to the ma-
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SECTION 6–3
jor axis of the ellipse and 1.00 foot in front of it. The chord is 1.00 foot shorter than the major axis.
Hyperbola
405
(A) Find the equation of the ellipse. Let the y axis lie along the minor axis of the ellipse, and let the x axis lie along the major axis, both with positive direction upward. (B) What is the width of the keel, measured perpendicular to the major axis, 1 foot up the major axis from the bottom end of the keel? Compute quantities to three significant digits.
Rudder
6-3
Keel
Hyperbola Z Definition of a Hyperbola Z Drawing a Hyperbola Z Standard Equations and Their Graphs Z Applications
As before, we start with a coordinate-free definition of a hyperbola. Using this definition, we show how a hyperbola can be drawn and we derive standard equations for hyperbolas specially located in a rectangular coordinate system.
Z Definition of a Hyperbola The following is a coordinate-free definition of a hyperbola: Z DEFINITION 1 Hyperbola A hyperbola is the set of all points P in a plane 兩d 0007 d 兩 0003 Constant 1 2 such that the absolute value of the difference of the distances from P to two fixed points in the P d2 plane is a positive constant (the constant is d1 required to be less than the distance between the F V two fixed points). Each of the fixed points, F¿ and V F F, is called a focus. The intersection points V¿ and V of the line through the foci and the two branches of the hyperbola are called vertices, and each is called a vertex. The line segment V¿V is called the transverse axis. The midpoint of the transverse axis is the center of the hyperbola.
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Z Drawing a Hyperbola Thumbtacks, a straightedge, string, and a pencil are all that are needed to draw a hyperbola (Fig. 1). Place two thumbtacks in a piece of cardboard—these form the foci of the hyperbola. Rest one corner of the straightedge at the focus F¿ so that it is free to rotate about this point. Cut a piece of string shorter than the length of the straightedge, and fasten one end to the straightedge corner A and the other end to the thumbtack at F. Now push the string with a pencil up against the straightedge at B. Keeping the string taut, rotate the straightedge about F¿, keeping the corner at F¿. The resulting curve will be part of a hyperbola. Other parts of the hyperbola can be drawn by changing the position of the straightedge and string. To see that the resulting curve meets the conditions of the definition, note that the difference of the distances BF¿ and BF is BF¿ 0007 BF 0003 BF¿ 0002 BA 0007 BF 0007 BA 0003 AF¿ 0007 (BF 0002 BA) Straightedge String 0003a b0007a b length length 0003 Constant
Z Figure 1 Drawing a hyperbola.
A B
String
F
F
Z Standard Equations of Hyperbolas and Their Graphs Using the definition of a hyperbola and the distance formula, we can derive standard equations for a hyperbola located in a rectangular coordinate system. We start by placing a hyperbola in the coordinate system with the foci on the x axis at F 0003 (0007c, 0) and F 0003 (c, 0) with
y
Z Figure 2 Hyperbola with foci on the x axis.
P 0003 (x, y) d2 x a F 0003 (c, 0)
d1 F 0003 (0007c, 0) 0007a
c00050 兩d1 0007 d2 兩 0003 Positive constant 0006 d(F, F )
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c 0005 0 (Fig. 2). By definition 1, the constant difference | d1 0007 d2 | is required to be less than 2c (the distance between F and F ). Therefore, the hyperbola intersects the x axis at points V 0003 (0007a, 0) and V 0003 (a, 0) with c 0005 a 0005 0. The hyperbola does not intersect the y axis, because the constant difference | d1 0007 d2 | is required to be positive by definition 1. Study Figure 2: Note that if P 0003 (a, 0), then | d1 0007 d2 | 0003 2a. (Why?) Therefore, the constant |d1 0007 d2 | is equal to the distance between the vertices. It is convenient to let b 0003 2c2 0007 a2, so that c2 0003 a2 0002 b2. (Unlike the situation for ellipses, b may be greater than or equal to a.) Referring again to Figure 2, the point P 0003 (x, y) is on the hyperbola if and only if |d1 0007 d2 | 0003 2a Using the distance formula for d1 and d2, eliminating radicals, and simplifying (see Problem 57 in Exercises 6-3), we obtain the equation of the hyperbola pictured in Figure 2: y2 x2 ⴚ ⴝ1 a2 b2 Although the hyperbola does not intersect the y axis, the points (0, b) and (0, 0007b) are significant; the line segment joining them is called the conjugate axis of the hyperbola. Note that the conjugate axis is perpendicular to the transverse axis, that is, the line segment joining the vertices (a, 0) and (0007a, 0). The rectangle with corners (a, b), (a, 0007b), (0007a, 0007b), and (0007a, b) is called the asymptote rectangle because its extended diagonals are asymptotes for the hyperbola (Fig. 3). In other words, the hyperbola approaches the lines y 0003 ba x as | x| becomes larger (see Problems 53 and 54 in Exercises 6-3). As a result, it is helpful to include the asymptote rectangle and its extended diagonals when sketching the graph of a hyperbola.
Asymptote b y00030007 x a
Asymptote b y0003 x a
y
x2
b
a2 0007a
0
a
0007 x
y2 b2
00031
0007b
Z Figure 3 Asymptotes.
Note that the four corners of the asymptote rectangle (Fig. 3) are equidistant from the origin, at distance 2a2 0002 b2 0003 c. Therefore, A circle, with center at the origin, that passes through all four corners of the asymptote rectangle of a hyperbola also passes through its foci. By similar reasoning (see Problem 58 in Exercises 6-3) we obtain the equation of a hyperbola centered at the origin with foci on the y axis. Both cases are summarized in Theorem 1.
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Z THEOREM 1 Standard Equations of a Hyperbola with Center at (0, 0) 1.
y2 x2 0007 00031 a2 b2 x intercepts: a (vertices) y intercepts: none Foci: F 0003 (0007c, 0), F 0003 (c, 0)
y
b
F
0007c
c2 0003 a2 0002 b2
2
a
F
c
x
0007b
Transverse axis length 0003 2a Conjugate axis length 0003 2b b Asymptotes: y 0003 x a y2 x2 2. 2 0007 2 0003 1 a b x intercepts: none y intercepts: a (vertices) Foci: F 0003 (0, 0007c), F 0003 (0, c) 2
c
0007a
y c
F
a c 0007b
2
c 0003a 0002b
b
x
0007a 0007c
Transverse axis length 0003 2a Conjugate axis length 0003 2b a Asymptotes: y 0003 x b
F
[Note: Both graphs are symmetric with respect to the x axis, y axis, and origin.]
ZZZ EXPLORE-DISCUSS 1
EXAMPLE
1
The line through a focus F of a hyperbola that is perpendicular to the transverse axis intersects the hyperbola in two points G and H. For each of the two standard equations of a hyperbola with center (0, 0), find an expression in terms of a and b for the distance from G to H.
Graphing Hyperbolas Find the coordinates of the foci, find the lengths of the transverse and conjugate axes, find the equations of the asymptotes, and graph the following equation: 9x2 0007 16y2 0003 144
SOLUTION
First, write the equation in standard form by dividing both sides by 144 and determine a and b: 9x2 0007 16y2 0003 144 16y2 9x2 144 0007 0003 144 144 144 y2 x2 0007 00031 16 9 a00034 and
Divide both sides by 144.
Simplify.
b00033
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x intercepts: 4 y intercepts: none
Hyperbola
409
Transverse axis length 0003 2(4) 0003 8 Conjugate axis length 0003 2(3) 0003 6 Foci: c2 0003 a2 0002 b2 0003 16 0002 9 0003 25 c00035
Substitute a 0003 4 and b 0003 3.
So the foci are F 0003 (00075, 0) and F 0003 (5, 0). Plot the foci and x intercepts, sketch the asymptote rectangle and the asymptotes, then sketch the hyperbola (Fig. 4). The equations of the asymptotes are y 0003 34 x (note that the diagonals of the asymptote rectangle have slope 34). y 5
c
0007c 00076
F
c F
6
x
00075
0002
Z Figure 4
MATCHED PROBLEM 1
Find the coordinates of the foci, find the lengths of the transverse and conjugate axes, and graph the following equation: 16x2 0007 25y2 0003 400
EXAMPLE
2
0002
Graphing Hyperbolas Find the coordinates of the foci, find the lengths of the transverse and conjugate axes, find the equations of the asymptotes, and graph the following equation: 16y2 0007 9x2 0003 144
SOLUTION
Write the equation in standard form: 16y2 0007 9x2 0003 144 y2 x2 0007 00031 9 16 a00033 and y intercepts: 3 x intercepts: none
Divide both sides by 144.
b00034
Transverse axis length 0003 2(3) 0003 6 Conjugate axis length 0003 2(4) 0003 8 Foci: c2 0003 a2 0002 b2 0003 9 0002 16 0003 25 c00035
Substitute a ⴝ 3 and b ⴝ 4.
So the foci are F 0003 (0, 00075) and F 0003 (0, 5). Plot the foci and y intercepts, sketch the asymptote rectangle and the asymptotes, then sketch the hyperbola (Fig. 5). The equations of the asymptotes are y 0003 34 x (note that the diagonals of the asymptote rectangle have slope 34).
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ADDITIONAL TOPICS IN ANALYTIC GEOMETRY y 6
c
F c
00076
6
0007c
00076
x
F
0002
Z Figure 5
MATCHED PROBLEM 2
Find the coordinates of the foci, find the lengths of the transverse and conjugate axes, and graph the following equation: 25y2 0007 16x2 0003 400
0002
Two hyperbolas of the form y2 x2 0007 00031 M N
and
y2 x2 0007 00031 N M
M, N 7 0
are called conjugate hyperbolas. In Examples 1 and 2 and in Matched Problems 1 and 2, the hyperbolas are conjugate hyperbolas—they share the same asymptotes.
ZZZ
When making a quick sketch of a hyperbola, it is a common error to have the hyperbola opening up and down when it should open left and right, or vice versa. The mistake can be avoided if you first locate the intercepts accurately.
CAUTION ZZZ
EXAMPLE
3
Graphing Hyperbolas Find the coordinates of the foci, find the lengths of the transverse and conjugate axes, and graph the following equation: 2x2 0007 y2 0003 10 2x2 0007 y2 0003 10 y2 x2 0007 00031 5 10 a 0003 15 and
SOLUTION
x intercepts: 15 y intercepts: none
y 5
c 0007c
c
F
F
00075
00075
Z Figure 6
5
x
Divide both sides by 10.
b 0003 110
Transverse axis length 0003 215 ⬇ 4.47 Conjugate axis length 0003 2110 ⬇ 6.32 Foci: c2 0003 0003 0003 c0003
a2 0002 b2 5 0002 10 15 115
Substitute a ⴝ 15 and b ⴝ 110.
So the foci are F¿ 0003 (0007 115, 0) and F 0003 ( 115, 0). Plot the foci and x intercepts, sketch the asymptote rectangle and the asymptotes, then sketch the hyperbola (Fig. 6). 0002
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SECTION 6–3
MATCHED PROBLEM 3
Hyperbola
411
Find the coordinates of the foci, find the lengths of the transverse and conjugate axes, and graph the following equation: y2 0007 3x2 0003 12
EXAMPLE
4
0002
Finding the Equation of a Hyperbola Find an equation of a hyperbola in the form y2 x2 0007 00031 M N
M, N 7 0
if the center is at the origin, and: (A) Length of transverse axis is 12 Length of conjugate axis is 20 SOLUTIONS
(B) Length of transverse axis is 6 Distance of foci from center is 5
(A) Start with y2 x2 0007 00031 a2 b2 and find a and b: a0003
12 00036 2
and
b0003
20 0003 10 2
So the equation is y2 x2 0007 00031 36 100 (B) Start with y2 x2 2 0007 2 0003 1 a b and find a and b: a0003
To find b, sketch the asymptote rectangle (Fig. 7), label known parts, and use the Pythagorean theorem:
y 5
6 00033 2
F 5
0007b
b2 0003 52 0007 32 0003 16 b00034
3 b
x
So the equation is 00075
F
Z Figure 7 Asymptote rectangle.
y2 x2 0007 00031 9 16
0002
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MATCHED PROBLEM 4
Find an equation of a hyperbola in the form y2 x2 0007 00031 M N
M, N 7 0
if the center is at the origin, and: (A) Length of transverse axis is 50 Length of conjugate axis is 30
(B) Length of conjugate axis is 12 Distance of foci from center is 9 0002
ZZZ EXPLORE-DISCUSS 2
(A) Does the line with equation y 0003 x intersect the hyperbola with equation x2 0007 (y2兾4) 0003 1? If so, find the coordinates of all intersection points. (B) Does the line with equation y 0003 3x intersect the hyperbola with equation x2 0007 ( y2兾4) 0003 1? If so, find the coordinates of all intersection points. (C) For which values of m does the line with equation y 0003 mx intersect the hypery2 x2 bola 2 0007 2 0003 1? Find the coordinates of all intersection points. a b
Z Applications You may not be aware of the many important uses of hyperbolic forms. They are encountered in the study of comets; the loran system of navigation for pleasure boats, ships, and aircraft; sundials; capillary action; nuclear reactor cooling towers; optical and radio telescopes; and contemporary architectural structures. The TWA building at Kennedy Airport is a hyperbolic paraboloid, and the St. Louis Science Center Planetarium is a hyperboloid. With such structures, thin concrete shells can span large spaces [Fig. 8(a)]. Some comets from outer space occasionally enter the sun’s gravitational field, follow a hyperbolic path around the sun (with the sun as a focus), and then leave, never to be seen again [Fig. 8(b)]. Example 5 illustrates the use of hyperbolas in navigation.
Z Figure 8 Uses of hyperbolic forms. Comet Sun
EXAMPLE
5
St. Louis Planetarium
Comet around sun
(a)
(b)
Navigation A ship is traveling on a course parallel to and 60 miles from a straight shoreline. Two transmitting stations, S1 and S2, are located 200 miles apart on the shoreline (Fig. 9). By timing radio signals from the stations, the ship’s navigator determines that the ship is between the two stations and 50 miles closer to S2 than to S1. Find the distance from the ship to each station. Round answers to one decimal place.
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SECTION 6–3
d1
60 miles
Hyperbola
413
d2
S2
S1
200 miles
Z Figure 9 d1 0007 d2 0003 50. SOLUTION
If d1 and d2 are the distances from the ship to S1 and S2, respectively, then d1 0007 d2 0003 50 and the ship must be on the hyperbola with foci at S1 and S2 and fixed difference 50, as illustrated in Figure 10. In the derivation of the equation of a hyperbola, we represented the fixed difference as 2a. So for the hyperbola in Figure 10 we have c 0003 100 a 0003 12 (50) 0003 25 b 0003 21002 0007 252 0003 29,375 y 200
S1
(x, 60) S2
0007100
100
x
Z Figure 10
The equation for this hyperbola is y2 x2 0007 00031 625 9,375 Substitute y 0003 60 and solve for x (see Fig. 10): x2 602 0007 00031 625 9,375 3,600 x2 0003 00021 625 9,375 3,600 0002 9,375 x2 0003 625 9,375 0003 865
Add
602 to both sides. 9,375
Multiply both sides by 625.
Simplify.
So x 0003 1865 ⬇ 29.41 (The negative square root is discarded, because the ship is closer to S2 than to S1.) Distance from ship to S1 2
Distance from ship to S2 2
d1 0003 2(29.41 0002 100) 0002 60 0003 120,346.9841 ⬇ 142.6 miles
d2 0003 2(29.41 0007 100)2 0002 602 0003 18,582.9841 ⬇ 92.6 miles
Notice that the difference between these two distances is 50, as it should be. MATCHED PROBLEM 5
Repeat Example 5 if the ship is 80 miles closer to S2 than to S1.
0002
0002
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Example 5 illustrates a simplified form of the loran (LOng RAnge Navigation) system. In practice, three transmitting stations are used to send out signals simultaneously (Fig. 11), instead of the two used in Example 5. A computer onboard a ship will record these signals and use them to determine the differences of the distances that the ship is to S1 and S2, and to S2 and S3. Plotting all points so that these distances remain constant produces two branches, p1 and p2, of a hyperbola with foci S1 and S2, and two branches, q1 and q2, of a hyperbola with foci S2 and S3. It is easy to tell which branches the ship is on by comparing the signals from each station. The intersection of a branch of each hyperbola locates the ship and the computer expresses this in terms of longitude and latitude.
Ship S3 q2
S1
S2
p1
q1
p2
Z Figure 11 Loran navigation.
ANSWERS TO MATCHED PROBLEMS 1.
y
y2 x2 0007 00031 25 16 Foci: F 0003 (0007兹41, 0), F 0003 (兹41, 0) Transverse axis length 0003 10 Conjugate axis length 0003 8
10
c
F 000710 0007c
F c
x
10
000710
2.
y2 x2 0007 00031 16 25 Foci: F 0003 (0, 0007兹41), F 0003 (0, 兹41) Transverse axis length 0003 8 Conjugate axis length 0003 10
y 10
c
F c
000710
x
10
0007c F 000710
y
3.
y2 x2 0007 00031 12 4 Foci: F 0003 (0, 00074), F 0003 (0, 4) Transverse axis length 0003 2兹12 ⬇ 6.93 Conjugate axis length 0003 4
6
c F c 00075
5
x
0007c F 00076
4. (A)
y2 x2 0007 00031 625 225
(B)
y2 x2 0007 00031 45 36
5. d1 0003 159.5 miles, d2 0003 79.5 miles
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SECTION 6–3
6-3
Hyperbola
Exercises Sketch a graph of each equation in Problems 15–26, find the coordinates of the foci, and find the lengths of the transverse and conjugate axes.
1. Give a coordinate-free definition of a hyperbola in your own words. 2. Explain how the transverse axis of a hyperbola differs from the conjugate axis. 3. Given the transverse axis and foci of a hyperbola, describe a procedure for drawing the hyperbola. 4. Is the graph of a hyperbola the graph of a function? Explain.
15.
y2 x2 0007 00031 9 4
16.
y2 x2 0007 00031 9 25
17.
y2 x2 0007 00031 4 9
18.
y2 x2 0007 00031 25 9
19. 4x2 0007 y2 0003 16
5. Is the conjugate axis of a hyperbola always shorter then the transverse axis? Explain.
20. x2 0007 9y2 0003 9
21. 9y2 0007 16x2 0003 144
6. Explain what an asymptote rectangle is, and how it is related to the graph of a hyperbola.
22. 4y2 0007 25x2 0003 100 23. 3x2 0007 2y2 0003 12
In Problems 7–10, find the distance between the foci of the hyperbola. 7. Transverse axis length = 24 Conjugate axis length = 18
24. 3x2 0007 4y2 0003 24 25. 7y2 0007 4x2 0003 28 26. 3y2 0007 2x2 0003 24
8. Transverse axis length = 25 Conjugate axis length = 60
In Problems 27–38, find an equation of a hyperbola in the form
9. Transverse axis length = 1 Conjugate axis length = 3
y2 x2 0007 00031 M N
10. Transverse axis length = 7 Conjugate axis length = 1
y2 x2 0007 00031 N M
or
11. x2 0007 y2 0003 1
12. y2 0007 x2 0003 1
13. y2 0007 x2 0003 4
14. x2 0007 y2 0003 4
27. The graph is y 10
y
y
(5, 4)
5
5
000710
5
x
M, N 7 0
if the center is at the origin, and:
In Problems 11–14, match each equation with one of graphs (a)–(d).
00075
00075
5
x
00075
00075
(b)
y
y
5
10
x
000710
28. The graph is
(a)
y 10
5
(4, 5) 00075
5
00075
x 00075
5
00075
(c)
415
(d)
x
000710
10
000710
x
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29. The graph is 41. y
y2 x2 0007 00031 4 16
43. 9x2 0007 y2 0003 9
10
42.
y2 x2 0007 00031 9 25
44. x2 0007 4y2 0003 4
45. 2y2 0007 3x2 0003 1 (3, 5) 000710
10
x
000710
30. The graph is
46. 5y2 0007 6x2 0003 1 47. (A) How many hyperbolas have center at (0, 0) and a focus at (1, 0)? Find their equations. (B) How many ellipses have center at (0, 0) and a focus at (1, 0)? Find their equations. (C) How many parabolas have center at (0, 0) and focus at (1, 0)? Find their equations. 48. How many hyperbolas have the lines y 0003 2x as asymptotes? Find their equations.
y
49. Find all intersection points of the graph of the hyperbola x2 0007 y2 0003 1 with the graph of each of the following lines: (A) y 0003 0.5x (B) y 0003 2x
10
(5, 3) 000710
10
x
000710
31. Transverse axis on x axis Transverse axis length 0003 14 Conjugate axis length 0003 10 32. Transverse axis on x axis Transverse axis length 0003 8 Conjugate axis length 0003 6 33. Transverse axis on y axis Transverse axis length 0003 24 Conjugate axis length 0003 18 34. Transverse axis on y axis Transverse axis length 0003 16 Conjugate axis length 0003 22
For what values of m will the graph of the hyperbola and the graph of the line y 0003 mx intersect? Find the coordinates of these intersection points. 50. Find all intersection points of the graph of the hyperbola y2 0007 x2 0003 1 with the graph of each of the following lines: (A) y 0003 0.5x (B) y 0003 2x For what values of m will the graph of the hyperbola and the graph of the line y 0003 mx intersect? Find the coordinates of these intersection points. 51. Find all intersection points of the graph of the hyperbola y2 0007 4x2 0003 1 with the graph of each of the following lines: (A) y 0003 x (B) y 0003 3x For what values of m will the graph of the hyperbola and the graph of the line y 0003 mx intersect? Find the coordinates of these intersection points.
35. Transverse axis on x axis Transverse axis length 0003 18 Distance of foci from center 0003 11
52. Find all intersection points of the graph of the hyperbola 4x2 0007 y2 0003 1 with the graph of each of the following lines: (A) y 0003 x (B) y 0003 3x
36. Transverse axis on x axis Transverse axis length 0003 16 Distance of foci from center 0003 10
For what values of m will the graph of the hyperbola and the graph of the line y 0003 mx intersect? Find the coordinates of these intersection points.
37. Conjugate axis on x axis Conjugate axis length 0003 14 Distance of foci from center 0003 1200 38. Conjugate axis on x axis Conjugate axis length 0003 10 Distance of foci from center 0003 170 In Problems 39–46, find the equations of the asymptotes of each hyperbola. 39.
y2 x2 0007 00031 25 4
40.
y2 x2 0007 00031 16 36
53. Consider the hyperbola with equation y2 x2 0007 200031 2 a b 2
(A) Show that y 0003 ba x 21 0007 ax2 . (B) Explain why the hyperbola approaches the lines y 0003 ba x as |x| becomes larger. (C) Does the hyperbola approach its asymptotes from above or below? Explain.
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SECTION 6–3
54. Consider the hyperbola with equation y2 2
a
0007
x2 00031 b2 2
(A) Show that y 0003 ab x 21 0002 bx2 . (B) Explain why the hyperbola approaches the lines y 0003 ab x as |x| becomes larger. (C) Does the hyperbola approach its asymptotes from above or below? Explain. 55. Let F and F be two points in the plane and let c be a constant such that c 0005 d(F, F ). Describe the set of all points P in the plane such that the absolute value of the difference of the distances from P to F and F is equal to the constant c. 56. Let F and F be two points in the plane and let c denote the constant d(F, F ). Describe the set of all points P in the plane such that the absolute value of the difference of the distances from P to F and F is equal to the constant c. 57. Study the following derivation of the standard equation of a hyperbola with foci (c, 0), x intercepts (a, 0), and endpoints of the conjugate axis (0, b). Explain why each equation follows from the equation that precedes it. [Hint: Recall that c2 0003 a2 0002 b2.]
Hyperbola
417
ECCENTRICITY Problems 59 and 60 (and Problems 45 and 46 in Exercises 6-2) are related to a property of conics called eccentricity, which is denoted by a positive real number E. Parabolas, ellipses, and hyperbolas all can be defined in terms of E, a fixed point called a focus, and a fixed line not containing the focus called a directrix as follows: The set of points in a plane each of whose distance from a fixed point is E times its distance from a fixed line is an ellipse if 0 0006 E 0006 1, a parabola if E 0003 1, and a hyperbola if E 0005 1.
59. Find an equation of the set of points in a plane each of whose distance from (3, 0) is three-halves its distance from the line x 0003 43. Identify the geometric figure. 60. Find an equation of the set of points in a plane each of whose distance from (0, 4) is four-thirds its distance from the line y 0003 94. Identify the geometric figure.
APPLICATIONS 61. ARCHITECTURE An architect is interested in designing a thinshelled dome in the shape of a hyperbolic paraboloid, as shown in Figure (a). Find the equation of the hyperbola located in a coordinate system [Fig. (b)] satisfying the indicated conditions. How far is the hyperbola above the vertex 6 feet to the right of the vertex? Compute the answer to two decimal places.
|d1 0007 d2| 0003 2a 2(x 0002 c)2 0002 y2 0003 2a 0002 2(x 0007 c)2 0002 y2
Hyperbola
(x 0002 c)2 0002 y2 0003 4a2 4a 2(x 0007 c)2 0002 y2 0002 (x 0007 c)2 0002 y2 2(x 0007 c)2 0002 y2 0003 a 0007
cx a
(x 0007 c)2 0002 y2 0003 a2 0007 2cx 0002 a1 0007
c2x2 a2
c2 2 b x 0002 y2 0003 a2 0007 c2 a2
Parabola
y2 x2 0007 200031 2 a b
Hyperbolic paraboloid (a)
58. Study the following derivation of the standard equation of a hyperbola with foci (0, c), y intercepts (0, a), and endpoints of the conjugate axis (b, 0). Explain why each equation follows from the equation that precedes it. [Hint: Recall that c2 0003 a2 0002 b2.]
y
(8, 12) 10
|d1 0007 d2| 0003 2a 2
2x 0002 ( y 0002 c)2 0003 2a 0002 2x2 0002 ( y 0007 c)2 2
2
2
2
2
2
x 0002 ( y 0002 c) 0003 4a 4a2x 0002 ( y 0007 c) 0002 x 0002 ( y 0007 c) cy 2x2 0002 ( y 0007 c)2 0003 a 0007 a x2 0002 (y 0007 c)2 0003 a2 0007 2cy 0002 x2 0002 a1 0007
c2 2 b y 0003 a2 0007 c2 a2
2
y
a
2
2
0007
x 00031 b2
c2y2 a2
2
000710
10
x
Hyperbola part of dome (b)
62. NUCLEAR POWER A nuclear reactor cooling tower is a hyperboloid, that is, a hyperbola rotated around its conjugate axis, as shown in Figure (a) on page 418. The equation of the hyperbola in Figure (b) used to generate the hyperboloid is y2 x2 0007 00031 1002 1502
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ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Incoming wave Common focus F Hyperbola
Hyperbola focus
Parabola
Receiving cone
Nuclear reactor cooling tower (a)
F
(a)
y 500
0007500
500
x Radio telescope
0007500
Hyperbola part of dome (b)
If the tower is 500 feet tall, the top is 150 feet above the center of the hyperbola, and the base is 350 feet below the center, what is the radius of the top and the base? What is the radius of the smallest circular cross section in the tower? Compute answers to three significant digits. 63. SPACE SCIENCE In tracking space probes to the outer planets, NASA uses large parabolic reflectors with diameters equal to twothirds the length of a football field. Needless to say, many design problems are created by the weight of these reflectors. One weight problem is solved by using a hyperbolic reflector sharing the parabola’s focus to reflect the incoming electromagnetic waves to the other focus of the hyperbola where receiving equipment is installed (see the figure).
CHAPTER
6-1
6
(b)
For the receiving antenna shown in the figure, the common focus F is located 120 feet above the vertex of the parabola, and focus F (for the hyperbola) is 20 feet above the vertex. The vertex of the reflecting hyperbola is 110 feet above the vertex for the parabola. Introduce a coordinate system by using the axis of the parabola as the y axis (up positive), and let the x axis pass through the center of the hyperbola (right positive). What is the equation of the reflecting hyperbola? Write y in terms of x.
Review
Conic Sections; Parabola
The plane curves obtained by intersecting a right circular cone with a plane are called conic sections. If the plane cuts clear through one nappe, then the intersection curve is called a circle if the plane is perpendicular to the axis and an ellipse if the plane is not perpendicular to the axis. If a plane cuts only one nappe, but does not cut
clear through, then the intersection curve is called a parabola. If a plane cuts through both nappes, but not through the vertex, the resulting intersection curve is called a hyperbola. A plane passing through the vertex of the cone produces a degenerate conic—a point, a line, or a pair of lines. The figure illustrates the four nondegenerate conics.
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419
Review y
y
F
F
x
0
a ⬍ 0 (opens left) Circle
x
0
a ⬎ 0 (opens right)
Ellipse
2. x2 0003 4ay Vertex: (0, 0) Focus: (0, a) Directrix: y 0003 0007a Symmetric with respect to the y axis Axis of symmetry the y axis y
y
0
x
F
F
x
0
Parabola
Hyperbola
The graph of a ⬍ 0 (opens down)
Ax2 0002 Bxy 0002 Cy2 0002 Dx 0002 Ey 0002 F 0003 0 is a conic, a degenerate conic, or the empty set. The following is a coordinate-free definition of a parabola:
6-2
a ⬎ 0 (opens up)
Ellipse
The following is a coordinate-free definition of an ellipse:
Parabola A parabola is the set of all points in a plane equidistant from a fixed point F and a fixed line L (not containing F) in the plane. The fixed point F is called the focus, and the fixed line L is called the directrix. A line through the focus perpendicular to the directrix is called the axis of symmetry, and the point on the axis halfway between the directrix and focus is called the vertex.
L
d1
P
d1 0003 d2 Axis of symmetry
Ellipse An ellipse is the set of all points P in a plane such that the sum of the distances from P to two fixed points in the plane is a constant (the constant is required to be greater than the distance between the two fixed points). Each of the fixed points, F and F, is called a focus, and together they are called foci. Referring to the figure, the line segment V V through the foci is the major axis. The perpendicular bisector B B of the major axis is the minor axis. Each end of the major axis, V and V, is called a vertex. The midpoint of the line segment F F is called the center of the ellipse.
d2 V(Vertex)
d1 0002 d2 0003 Constant
F(Focus)
B V
Parabola
d1
P
F
d2
Directrix F
From the definition of a parabola, we can obtain the following standard equations: Standard Equations of a Parabola with Vertex at (0, 0) 1. y2 0003 4ax Vertex: (0, 0) Focus: (a, 0) Directrix: x 0003 0007a Symmetric with respect to the x axis Axis of symmetry the x axis
V
B
From the definition of an ellipse, we can obtain the following standard equations: Standard Equations of an Ellipse with Center at (0, 0) 2
1.
y x2 a0005b00050 0002 200031 2 a b x intercepts: a (vertices) y intercepts: b
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Foci: F 0003 (0007c, 0), F 0003 (c, 0) Major axis length 0003 2a Minor axis length 0003 2b
c2 0003 a2 0007 b2
兩 d1 0007 d2兩 0003 Constant P
d2
d1 y F
b
0007a
F 0007c
F c
0
V
V
F
x
a
From the definition of a hyperbola, we can obtain the following standard equations:
0007b
Standard Equations of a Hyperbola with Center at (0, 0) 2
2
2.
1.
2
y x 0002 200031 a0005b00050 2 b a x intercepts: b y intercepts: a (vertices) Foci: F 0003 (0, 0007c), F 0003 (0, c) Major axis length 0003 2a Minor axis length 0003 2b
c2 0003 a2 0007 b2
y x2 0007 200031 a2 b x intercepts: a (vertices) y intercepts: none Foci: F 0003 (0007c, 0), F 0003 (c, 0) Transverse axis length 0003 2a Conjugate axis length 0003 2b b Asymptotes: y 0003 x a
y
c2 0003 a2 0002 b2
y
a c F
b
F
0007c
0007b
0
b
c
0007a
a
F
c
x
0007b
x
0007c F 0007a
[Note: Both graphs are symmetric with respect to the x axis, y axis, and origin. Also, the major axis is always longer than the minor axis.]
6-3
Hyperbola
The following is a coordinate-free definition of a hyperbola:
2.
y2
x2 00031 a2 b2 x intercepts: none y intercepts: a (vertices) Foci: F 0003 (0, 0007c), F 0003 (0, c) Transverse axis length 0003 2a Conjugate axis length 0003 2b a Asymptotes: y 0003 x b 0007
y
Hyperbola A hyperbola is the set of all points P in a plane such that the absolute value of the difference of the distances from P to two fixed points in the plane is a positive constant (the constant is required to be less than the distance between the two fixed points). Each of the fixed points, F and F, is called a focus. The intersection points V and V of the line through the foci and the two branches of the hyperbola are called vertices, and each is called a vertex. The line segment V V is called the transverse axis. The midpoint of the transverse axis is the center of the hyperbola. The line segment perpendicular to the transverse axis that goes through the center is called the conjugate axis.
c2 0003 a2 0002 b2
c
F
a c 0007b
b
x
0007a 0007c
F
[Note: Both graphs are symmetric with respect to the x axis, y axis, and origin.]
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6
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Review Exercises
Work through all the problems in this chapter review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. In Problems 1–6, graph each equation and locate foci. Locate the directrix for any parabolas. Find the lengths of major, minor, transverse, and conjugate axes where applicable. 1. 9x2 0002 25y2 0003 225
2. x2 0003 000712y
3. 25y2 0007 9x2 0003 225
4. x2 0007 y2 0003 16
5. y2 0003 8x
6. 2x2 0002 y2 0003 8
7. Find the equation of the parabola having its vertex at the origin, its axis of symmetry the x axis, and (00074, 00072) on its graph. In Problems 8 and 9, find the equation of the ellipse in the form y2 x2 M, N 7 0 0002 00031 M N if the center is at the origin, and: 8. Major axis on x axis Major axis length 0003 12 Minor axis length 0003 10 9. Major axis on y axis Minor axis length 0003 12 Distance between foci 0003 16 In Problems 10 and 11, find the equation of the hyperbola in the form y2 y2 x2 x2 or M, N 0005 0 0007 00031 0007 00031 M N M N if the center is at the origin, and: 10. Transverse axis on y axis Conjugate axis length 0003 6 Distance between foci 0003 8 11. Transverse axis on x axis Transverse axis length 0003 14 Conjugate axis length 0003 16 12. Find the equation of the parabola having directrix y 0003 5 and focus (0, 00075). 13. Find the foci of the ellipse through the point (00076, 0) if the center is at the origin, the major axis is on the x axis, and the major axis has twice the length of the minor axis. 14. Find the y intercepts of a hyperbola if the center is at the origin, the conjugate axis is on the x axis and has length 4, and (0, 00073) is a focus.
15. Find the directrix of a parabola having its vertex at the origin and focus (00074, 0). 16. Find the points of intersection of the parabolas x2 0003 8y and y2 0003 0007x. 17. Find the x intercepts of an ellipse if the center is at the origin, the major axis is on the y axis and has length 14, and (0, 00071) is a focus. 18. Find the foci of the hyperbola through the point (0, 00074) if the center is at the origin, the transverse axis is on the y axis, and the conjugate axis has twice the length of the transverse axis. 19. Use the definition of a parabola and the distance formula to find the equation of a parabola with directrix x 0003 6 and focus at (2, 4). 20. Find an equation of the set of points in a plane each of whose distance from (4, 0) is twice its distance from the line x 0003 1. Identify the geometric figure. 21. Find an equation of the set of points in a plane each of whose distance from (4, 0) is two-thirds its distance from the line x 0003 9. Identify the geometric figure. In Problems 22–24, find the equations of the asymptotes of each hyperbola. 22.
y2 x2 0007 00031 49 25
23.
y2 x2 0007 00031 64 4
24. 4x2 0007 y2 0003 1
APPLICATIONS 25. COMMUNICATIONS A parabolic satellite television antenna has a diameter of 8 feet and is 1 foot deep. How far is the focus from the vertex? 26. ENGINEERING An elliptical gear is to have foci 8 centimeters apart and a major axis 10 centimeters long. Letting the x axis lie along the major axis (right positive) and the y axis lie along the minor axis (up positive), write the equation of the ellipse in the standard form y2 x2 0002 200031 2 a b 27. SPACE SCIENCE A hyperbolic reflector for a radio telescope (such as that illustrated in Problem 63, Exercises 6-3) has the equation y2 402
0007
x2 00031 302
If the reflector has a diameter of 30 feet, how deep is it? Compute the answer to three significant digits.
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6
GROUP ACTIVITY Focal Chords
Many of the applications of the conic sections are based on their reflective or focal properties. One of the interesting algebraic properties of the conic sections concerns their focal chords. If a line through a focus F contains two points G and H of a conic section, then the line segment GH is called a focal chord. Let G 0003 (x1, y1) and H 0003 (x2, y2) be points on the graph of x2 0003 4ay such that GH is a focal chord. Let u denote the length of GF and v the length of FH (Fig. 1).
(A) Use the distance formula to show that u 0003 y1 0002 a. (B) Show that G and H lie on the line y 0007 a 0003 mx, where m 0003 ( y2 0007 y1)兾(x2 0007 x1). (C) Solve y 0007 a 0003 mx for x and substitute in x2 0003 4ay, obtaining a quadratic equation in y. Explain why y1 y2 0003 a2. (D) Show that
1 1 1 0002 0003 . u v a
(E) Show that u 0002 v 0007 4a 0003 y
G
F u
v
H (2a, a) x
Z Figure 1 Focal chord GH of the parabola x2 0003 4ay.
(u 0007 2a)2 . Explain why this u0007a
implies that u 0002 v 000e 4a, with equality if and only if u 0003 v 0003 2a. (F) Which focal chord is the shortest? Is there a longest focal chord? (G) Is
1 1 0002 a constant for focal chords of the ellipse? For u v
focal chords of the hyperbola? Obtain evidence for your answers by considering specific examples.
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7
C
OUTLINE
WE have seen many real-world situations where solving an equation
is valuable. But the world is a very complicated place, and many more situations lead to more than one variable. In that case, solving a system of equations becomes important. In this chapter, we will study a variety of methods for solving systems of equations. We will begin with linear systems with two or three variables using algebraic techniques similar to those we used for solving individual equations. Then we will introduce a variety of matrix methods for solving linear systems. These methods can be applied to very large systems that model very complicated real-world problems.
7-1
Systems of Linear Equations
7-2
Solving Systems of Linear Equations Using Gauss–Jordan Elimination
7-3
Matrix Operations
7-4
Solving Systems of Linear Equations Using Matrix Inverse Methods
7-5
Determinants and Cramer’s Rule Chapter 7 Review Chapter 7 Group Activity: Modeling with Systems of Linear Equations
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7-1
Systems of Linear Equations Z Systems of Equations Z Solving by Graphing Z Solving by Substitution Z Solving Using Elimination by Addition Z Applications
We have seen a wide variety of real-world problems that can be solved by writing and solving an equation. But a lot of problems have extra conditions that makes writing a single equation impractical. In this case, two or more equations might be needed to model the situation. In this section, we’ll examine how to solve two or more equations together, then see how to apply what we learn.
Z Systems of Equations To illustrate the basic concepts, we’ll use a simple example. At one campus coffee shop, muffins cost $2 each, and lattes are $3 each. If a total of seven items are sold for $18, how many of each item were sold? There are two natural variables in the problem: the number of muffins, which we’ll call x, and the number of lattes, which we’ll call y. Then x0003 y0002 7 2x 0003 3y 0002 18
Seven items total Total cost is $18.
This is called a system of linear equations in two variables. The solution to the problem is found by finding all pairs of numbers x and y that make both equations true. In general, we will study solving linear systems of the type ax 0003 by 0002 h cx 0003 dy 0002 k
System of two linear equations in two variables
where x and y are variables, a, b, c, and d are real numbers called the coefficients of x and y, and h and k are real numbers called the constant terms in the equations. A pair of numbers x 0002 x0 and y 0002 y0 is a solution of this system if each equation is satisfied by the pair. The set of all such pairs of numbers is called the solution set for the system. To solve a system is to find its solution set.
Z Solving by Graphing Recall that the graph of a linear equation is the line consisting of all ordered pairs that satisfy the equation. To solve the coffee shop problem by graphing, we will graph both equations in the same coordinate system. The coordinates of any points that the lines have in common must be solutions to the system, because they must satisfy both equations.
EXAMPLE
1
Solving a System by Graphing Solve the coffee shop problem by graphing:
x0003 y0002 7 2x 0003 3y 0002 18
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SOLUTION
Systems of Linear Equations
425
Find the x and y intercepts for each line. x⫹y⫽7
2x ⫹ 3y ⫽ 18
x
y
x
y
0
7
0
6
7
0
9
0
Plot these points, graph the two lines, estimate the intersection point visually (Fig. 1), and check the estimate. y 10
5
(3, 4) 2x ⫹ 3y ⫽ 18 10 5
x
x⫹y⫽7
Z Figure 1
x⫽3 y⫽4 CHECK
x⫹y⫽7 ? 3⫹4⫽7
Muffins* Lattes
2x ⫹ 3y ⫽ 18 ? 2(3) ⫹ 3(4) ⫽ 18
✓
✓
7⫽7 MATCHED PROBLEM 1
Solve by graphing:
18 ⫽ 18
0002
x⫺ y⫽ 3 x ⫹ 2y ⫽ ⫺3
0002
Technology Connections To solve Example 1 with a graphing calculator, first solve each equation for y: xⴙyⴝ7
From Figure 3, we see that the solution is
Subtract x from both sides.
xⴝ3
Muffins
yⴝ4
Lattes
yⴝ7ⴚx
10
2x ⴙ 3y ⴝ 18
Subtract 2x from both sides.
3y ⴝ 18 ⴚ 2x yⴝ 6ⴚ
Divide both sides by 3.
⫺10
10
2 3x
Next, enter these functions in the equation editor of a graphing calculator (Fig. 2) and use the intersect command to find the intersection point (Fig. 3).
⫺10
Z Figure 2
Z Figure 3
*When the solution set for a linear system is a single point, we will follow the common practice of writing the solution as (3, 4) or as x ⫽ 3, y ⫽ 4, rather than the more formal expression 5(3, 4)6.
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It is clear that Example 1 has exactly one solution, because the lines have exactly one point of intersection. In general, lines in a rectangular coordinate system are related to each other in one of three ways, as illustrated in Example 2.
EXAMPLE
2
Determining the Nature of Solutions Match each of the following systems with one of the graphs in Figure 4 and discuss the nature of the solutions: (A) 2x 0004 3y 0002 2 x 0003 2y 0002 8
(B) 4x 0003 6y 0002 12 2x 0003 3y 0002 00046
(C) 2x 0004 3y 0002 00046 0004x 0003 32 y 0002 3
y
y
y
5
5
5
(4, 2)
00045
x
5
00045
x
5
00045
00045
00045
(a)
5
x
00045
(b)
(c)
Z Figure 4 SOLUTIONS
(A) Write each equation in slope–intercept form: 2x 0004 3y 0002 2 00043y 0002 00042x 0003 2 y 0002 23 x 0004 23
x 0003 2y 0002 8 2y 0002 0004x 0003 8 y 0002 000412 x 0003 4
One positive slope, one negative
The graphs of these two lines match graph (b). There is exactly one solution: x 0002 4, y 0002 2. (B) 4x 0003 6y 0002 12 6y 0002 00044x 0003 12 y 0002 000423 x 0003 2
2x 0003 3y 0002 00046 3y 0002 00042x 0004 6 y 0002 000423x 0004 2
Slopes are equal.
The graphs of these parallel lines match graph (c). There is no solution. 0004x 0003 32 y 0002 3
(C) 2x 0004 3y 0002 00046 00043y 0002 00042x 0004 6 y 0002 23 x 0003 2
3 2y
0002x00033 y 0002 23 x 0003 2
Same line!
The graph of these identical lines match graph (a). There are an infinite number of solutions. 0002 MATCHED PROBLEM 2
Solve each of the following systems by graphing: (A) 2x 0003 3y 0002 12 x 0004 3y 0002 00043
(B)
x 0004 3y 0002 00043 00042x 0003 6y 0002 12
(C) 2x 0004 3y 0002 12 0004x 0003 32 y 0002 00046
0002
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Next, we’ll define some terms that can be used to describe the different types of solutions to systems of equations illustrated in Example 2. Z SYSTEMS OF LINEAR EQUATIONS: BASIC TERMS A system of linear equations is consistent if it has one or more solutions and inconsistent if no solutions exist. Furthermore, a consistent system is said to be independent if it has exactly one solution (often referred to as the unique solution) and dependent if it has more than one solution.
Referring to the three systems in Example 2, the system in part A [Fig. 4(b)] is consistent and independent, with the unique solution x 0002 4 and y 0002 2. The system in part B [Fig. 4(c)] is inconsistent, with no solution. And the system in part C [Fig. 4(a)] is consistent and dependent, with an infinite number of solutions: all the points on the two coinciding lines. ZZZ EXPLORE-DISCUSS 1
Can a consistent and dependent linear system have exactly two solutions? Exactly three solutions? Explain.
In general, any two lines in a rectangular coordinate plane either intersect in exactly one point, or are parallel, or coincide (have identical graphs). So, the systems in Example 2 illustrate the only three possible types of solutions for systems of two linear equations in two variables. These ideas are summarized in Theorem 1. Z THEOREM 1 Possible Solutions to a Linear System A system of linear equations must have 1. Exactly one solution or 2. No solution or 3. Infinitely many solutions
Consistent and independent
Inconsistent
Consistent and dependent
Note: While the geometric discussion presented here only applies to systems of equations with two variables, the same three possibilities remain for systems of linear equations with more than two variables.
Z Solving by Substitution The accuracy of solutions found by graphing depends a lot on how accurate the graph is when the graphs are drawn by hand. If the solutions are found using a graphing calculator, you will likely get very accurate solutions, but they probably won’t be exact. Worse still, the solutions can be very difficult to find, depending on the window settings that you choose. Also, for systems with more than two variables, the geometry gets extremely complicated. For all of these reasons, we will next turn our attention to solving systems algebraically. There are a number of different techniques that can be used. One of the simplest is the substitution method.
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We will return to the coffee shop problem from page 424 to illustrate the substitution method.
EXAMPLE
3
Solving a System by Substitution Use substitution to solve the coffee shop problem: x 0003 y 0002 7 2x 0003 3y 0002 18
SOLUTION
Step 1: Solve either equation for one variable. It will be easy to solve the first equation for y in terms of x: x0003y00027 y000270004x
Solve the first equation for y in terms of x. Substitute into the second equation.
Step 2: Substitute 7 0004 x for y in the second equation. 2x 0003 3y 0002 18 2x 0003 3(7 0004 x) 0002 18 2x 0003 21 0004 3x 0002 18 0004x 0002 00043 xⴝ3
y ⴝ 7 ⴚ x, so replace y with 7 ⴚ x. Multiply out parentheses. Collect x terms on the left and constant terms on the right. Multiply both sides by ⴚ1.
Step 3: Replace x with 3 in y 0002 7 0004 x: y000270004x y0002700043 yⴝ4 The solution is 3 muffins and 4 lattes, as we found and checked earlier. MATCHED PROBLEM 3
Solve by substitution and check:
x0004 y0002 3 x 0003 2y 0002 00043
0002
0002
The following box summarizes the steps for solving a system using the substitution method. Z SOLVING SYSTEMS OF TWO LINEAR EQUATIONS IN TWO VARIABLES: THE SUBSTITUTION METHOD 1. Choose one of the two equations and solve it for one of the two variables. (Make a choice that avoids fractions, if possible.) 2. Substitute the result of step 1 into the equation that was not used in step 1 and solve the resulting linear equation in one variable. 3. Substitute the result of step 2 into the expression obtained in step 1 to find the value of the second variable.
ZZZ EXPLORE-DISCUSS 2
Use substitution to solve each of the following systems. Discuss the nature of the solution sets you obtain. x 0003 3y 0002 4 2x 0003 6y 0002 7
x 0003 3y 0002 4 2x 0003 6y 0002 8
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Z Solving Using Elimination by Addition Now we turn to elimination by addition. This is probably the most important method of solution, since it is readily generalized to larger systems. The method involves the replacement of systems of equations with simpler equivalent systems, by performing appropriate operations, until we obtain a system with an obvious solution. Equivalent systems of equations are, as you would expect, systems that have exactly the same solution set. Theorem 2 lists operations that produce equivalent systems. Z THEOREM 2 Elementary Equation Operations Producing Equivalent Systems A system of linear equations is transformed into an equivalent system if: 1. Two equations are interchanged. 2. An equation is multiplied by a nonzero constant. 3. A constant multiple of another equation is added to a given equation.
We’ll return one more time to the coffee shop problem to illustrate why elimination by addition works so well. The system of equations was x0003 y 0002 7 2x 0003 3y 0002 18 Notice that if we use the third operation in Theorem 2, adding 00042 times the first equation to the second one, we get 00042x 0004 2y 0002 000414 2x 0003 3y 0002 18 y00024 This eliminated x, and left behind an equation with only y. We could then easily substitute back in to find x. We will rely mostly on operations 2 and 3 for now, but operation 1 will come in especially handy later in the section. Examples 4 and 5 illustrate the use of elimination by addition on two and three variable systems.
EXAMPLE
4
Solving a System Using Elimination by Addition Solve using elimination by addition:
SOLUTION
3x 0004 2y 0002 8 2x 0003 5y 0002 00041
We will use Theorem 2 to eliminate one of the variables and get an easy equation with one variable. 3x 0004 2y 0002 8 2x 0003 5y 0002 00041 15x 0004 10y 0002 40 4x 0003 10y 0002 00042 19x 0002 38 xⴝ2
If we multiply the top equation by 5, the bottom by 2, and then add, we can eliminate y.
Now solve for x.
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The equation x 0002 2 paired with either of the two original equations produces an equivalent system. So, we can substitute x 0002 2 back into either of the two original equations to solve for y. We choose the second equation. 2(2) 0003 5y 0002 00041 5y 0002 00045 y ⴝ ⴚ1 SOLUTION
x 0002 2, y 0002 00041, or (2, 00041). CHECK
MATCHED PROBLEM 4
3x 0004 2y 0002 8 ? 3(2) 0004 2(00041) 0002 8 ✓ 800028
Solve using elimination by addition:
2x 0003 5y 0002 00041 ? 2(2) 0003 5(00041) 0002 00041 ✓ 00041 0002 00041
0002
6x 0003 3y 0002 3 5x 0003 4y 0002 7
0002
When a system has three equations, we will use elimination to reduce to a system with two equations and two variables, then solve like we did in Example 4. To help you follow a solution, we will number the equations as E1, E2, and so on.
EXAMPLE
5
Solution Using Elimination by Addition x 0003 2y 0003 3z 0002 2 3x 0004 5y 0004 4z 0002 15 00042x 0004 3y 0003 2z 0002 2
SOLUTION
E1 E2 E3
Since the coefficient of x in E1 is 1, our calculations will be simplified if we use E1 to eliminate x from the other equations. First we eliminate x from E2 by multiplying E1 by 00043 and adding the result to E2. Equivalent System
00043x 0004 6y 0004 9z 0002 00046 3x 0004 5y 0004 4z 0002 15 000411y 0004 13z 0002 9
ⴚ3E1 E2 E4
x 0003 2y 0003 3z 0002 2 000411y 0004 13z 0002 9 00042x 0004 3y 0003 2z 0002 2
E1 E4 E3
Now we use E1 to eliminate x (the same variable eliminated above) from E3 by multiplying E1 by 2 and adding the result to E3. Equivalent System
2x 0003 4y 0003 6z 0002 4 00042x 0004 3y 0003 2z 0002 2 y 0003 8z 0002 6
2E1 E3 E5
x 0003 2y 0003 3z 0002 2 000411y 0004 13z 0002 9 y 0003 8z 0002 6
E1 E4 E5
Notice that E4 and E5 form a system of two equations with two variables. Next we use E5 to eliminate y from E4 and replace E4 with the result. 11y 0003 88z 0002 66 000411y 0004 13z 0002 9 75z 0002 75
11E5 E4 E6
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Now we can easily solve for z. 75z 0002 75 zⴝ1
E6
Next substitute z 0002 1 in E4 or E5 and solve for y. y 0003 8z 0002 6 y 0003 8(1) 0002 6 y ⴝ ⴚ2
E5
Finally, substitute y 0002 00042 and z 0002 1 in any of E1, E2, or E3 and solve for x. x 0003 2y 0003 3z 0002 2 x 0003 2(00042) 0003 3(1) 0002 2 xⴝ3
E1
The solution to the original system is (3, 00042, 1) or x 0002 3, y 0002 00042, z 0002 1. CHECK
x 0003 2y 0003 3z 0002 2 ? 3 0003 2(00042) 0003 3(1) 0002 2 ✓ 2 00022
E1
MATCHED PROBLEM 5
To check the solution, we must check each equation in the original system:
3x 0004 5y 0004 4z 0002 15 ? 3(3) 0004 5(00042) 0004 4(1) 0002 15 ✓ 15 0002 15
E2
00042x 0004 3y 0003 2z 0002 2 ? 00042(3) 0004 3(00042) 0003 2(1) 0002 2 ✓ 200022
E3
0002
Solve: 2x 0003 3y 0004 5z 0002 000412 3x 0004 2y 0003 2z 0002 1 4x 0004 5y 0004 4z 0002 000412
0002
Let’s see what happens in the solution process when a system either has no solution or has infinitely many solutions. Consider the solutions to the following system: 2x 0003 6y 0002 00043 x 0003 3y 0002 2 Solution by Substitution
Solution by Elimination
Solve the second equation for x and substitute in the first equation.
Multiply the second equation by 00042 and add to the first equation.
x 0002 2 0004 3y 2(2 0004 3y) 0003 6y 0002 00043 4 0004 6y 0003 6y 0002 00043 4 0002 00043
2x 0003 6y 0002 00043 00042x 0004 6y 0002 00044 0 0002 00047
Both methods of solution lead to a contradiction (a statement that is false). An assumption that the original system has solutions must be false. This tells us that the system has no solution. The graphs of the equations are parallel and the system is inconsistent. Now consider the system x 0004 12 y 0002 4 00042x 0003 y 0002 00048
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Solution by Substitution
Solution by Elimination
Solve the first equation for x and substitute in the second equation.
Multiply the first equation by 2 and add to the second equation.
x 0002 12 y 0003 4 00042 (12 y 0003 4) 0003 y 0002 00048 0004y 0004 8 0003 y 0002 00048 00048 0002 00048
2x 0004 y 0002 8 00042x 0003 y 0002 00048 00002 0
This time both solution methods lead to a statement that is always true. This means that the two original equations are equivalent. That is, their graphs coincide. The system is dependent and has an infinite number of solutions. There are many different ways to represent this infinite solution set. For example, S1 0002 5(x, y) ƒ y 0002 2x 0004 8, x any real number} and S2 0002 5(x, y) ƒ x 0002 12 y 0003 4, y any real number} both represent the solutions to this system. For reasons that will become apparent later, it is customary to introduce a new variable, called a parameter, and express both variables in terms of this new variable. If we let x 0002 s and y 0002 2s 0004 8 in S1, we can express the solution set as 5(s, 2s 0004 8) ƒ s any real number} Some particular solutions to this system are obtained by choosing particular values for the parameter.
EXAMPLE
6
s ⴝ ⴚ1
sⴝ2
sⴝ5
s ⴝ 9.4
(00041, 000410)
(2, 00044)
(5, 2)
(9.4, 10.8)
Using Elimination by Addition Solve: x0003 y0003 z00023 x 0004 y 0004 5z 0002 1 2x 0003 3y 0003 5z 0002 6
SOLUTION
E1 E2 E3
Use E1 to eliminate z from E2 and replace E2 with the result. Equivalent System
5x 0003 5y 0003 5z 0002 15 x 0004 y 0004 5z 0002 1 6x 0003 4y 0002 16
x0003 y0003 z0002 3 6x 0003 4y 0002 16 2x 0003 3y 0003 5z 0002 6
5E1 E2 E4
E1 E4 E3
Use E1 to eliminate z from E3 and replace E3 with the result. Equivalent System
00045x 0004 5y 0004 5z 0002 000415 2x 0003 3y 0003 5z 0002 6 00043x 0004 2y 0002 00049
ⴚ5E1 E3 E5
x0003 y0003z0002 3 6x 0003 4y 0002 16 00043x 0004 2y 0002 00049
E1 E4 E5
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Now treat E4 and E5 as a system of two equations, and eliminate y. 6x 0003 4y 0002
16
00046x 0004 4y 0002 000418 0 0002 00042
E4 2E5 E6
Stop! We have obtained a contradiction. The original system is inconsistent and has no solution. (Note: It’s impossible to check in this case.) 0002 MATCHED PROBLEM 6
Solve: 2x 0003 3y 0004 5z 0002 3 3x 0004 2y 0003 2z 0002 2 x 0004 5y 0003 7z 0002 1
EXAMPLE
7
0002
Using Elimination by Addition Solve: x0003y0003 z00021 2x 0003 y 0004 z 0002 3 3x 0003 y 0004 3z 0002 5
SOLUTION
E1 E2 E3
Use E1 to eliminate y from E2 and replace E2 with the result. Equivalent System
0004x 0004 y 0004 z 0002 00041 2x 0003 y 0004 z 0002 3 x 0004 2z 0002 2
x0003y0003 z00021 x 0004 2z 0002 2 3x 0003 y 0004 3z 0002 5
0004E1 E2 E4
E1 E4 E3
Use E1 to eliminate y from E3 and replace E3 with the result. Equivalent System
0004x 0004 y 0004 z 0002 00041 3x 0003 y 0004 3z 0002 5 2x 0004 4z 0002 4
x0003y0003 z00021 x 0004 2z 0002 2 2x 0004 4z 0002 4
ⴚE1 E3 E5
E1 E4 E5
Use E4 to eliminate z from E5 and replace E5 with the result. Equivalent System
00042x 2x
0003 4z 0002 00044 0004 4z 0002 4 00002 0
ⴚ2E4 E5
x0003y0003 z00021 x 0004 2z 0002 2
E1 E4
E6
Since E6 is true for all x, y, and z, it provides no information about the systems’ solution set and can be discarded. The solutions to the last equivalent system can be described by introducing a parameter. If we let z 0002 s, then, using E4, we can write x 0002 2s 0003 2. Substituting for x and z in E1 and solving for y, we have x0003y0003z00021 2s 0003 2 0003 y 0003 s 0002 1 y 0002 00043s 0004 1
E1
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The solution set is given by
5(2s 0003 2, 00043s 0004 1, s) | s any real number6
The check is left to the reader. MATCHED PROBLEM 7
0002
Solve: 3x 0003 2y 0003 4z 0002 5 2x 0003 y 0003 5z 0002 2 x 0003 6z 0002 00041
ZZZ EXPLORE-DISCUSS 3
0002
Refer to the solution to Example 7. The given representation of the solution set is not the only one. Which of the following is a representation of the solution set? Justify your answer. (A) {(t, 2 0004 1.5t, 0.5t 0004 1) | t any real number} (B) {(2u 0003 4, 00042u 0004 3, u) | u any real number} Let y 0002 v, where v is any real number, express x and z in terms of v, and find another representation of the solution set for Example 7.
Z Applications Examples 8–10 illustrate the advantages of using systems of equations in solving word problems.
EXAMPLE
8
Airspeed An airplane makes the 2,400-mile trip from Washington, D.C. to San Francisco in 7.5 hours and makes the return trip in 6 hours. Assuming that the plane travels at a constant airspeed and that the wind blows at a constant rate from west to east, find the plane’s airspeed and the wind rate.
SOLUTION
San 2,400 Francisco miles Washington, D.C.
Let x represent the airspeed of the plane and let y represent the rate at which the wind is blowing (both in miles per hour). The plane’s speed relative to the ground is determined by combining these two rates; that is, x 0004 y 0002 Ground speed flying east to west (airspeed 0004 wind) x 0003 y 0002 Ground speed flying west to east (airspeed 0003 wind) Applying the familiar formula D 0002 RT to each leg of the trip leads to the following system of equations: 2,400 0002 7.5(x 0004 y) 2,400 0002 6(x 0003 y)
Washington to San Francisco: 7.5 hr, 2,400 mi San Francisco to Washington: 6 hr, 2,400 mi
After simplification, we have x 0004 y 0002 320 x 0003 y 0002 400 Add these two equations to eliminate y: 2x 0002 720 x ⴝ 360 mph
Airspeed
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Systems of Linear Equations
435
Substitute for x in the second equation: x 0003 y 0002 400 360 0003 y 0002 400 y ⴝ 40 mph 2,400 0002 7.5(x 0004 y) ? 2,400 0002 7.5(360 0004 40) ✓ 2,400 0002 2,400
CHECK
MATCHED PROBLEM 8
Wind rate
2,400 0002 6(x 0003 y) ? 2,400 0002 6(360 0003 40) ✓ 2,400 0002 2,400
0002
A boat takes 8 hours to travel 80 miles upstream and 5 hours to return to its starting point. Find the speed of the boat in still water and the speed of the current. 0002 The quantity of a product that people are willing to buy (known as the demand) during some period of time depends on its price. Generally, the higher the price, the less the demand; the lower the price, the greater the demand. Similarly, the quantity of a product that a supplier is willing to sell during some period of time (known as the supply) also depends on the price. Generally, a supplier will be willing to supply more of a product at higher prices and less of a product at lower prices. The simplest supply and demand model is a linear model. If the demand for a product is greater than the supply, the price tends to rise. If the demand is less than the supply, the price tends to fall. So the price tends to stabilize at an equilibrium price; at that price, the supply and demand are equal, and that common quantity is called the equilibrium quantity. Example 9 illustrates the basic concepts of supply and demand.
9
Supply and Demand Using collected data and regression analysis, an analyst arrives at the following price–demand and price–supply equations for the sale of cherries each day in a major urban area. p 0002 00040.2q 0003 5.6 0.1q 0003 1.7
p0002
Demand equation (consumer) Supply equation (supplier)
where q represents the quantity of cherries in thousands of pounds and p represents the price in dollars per pound. For example, we see (Fig. 5) that consumers will purchase 11 thousand pounds (q 0002 11) when the price is p 0002 00040.2(11) 0003 5.6 0002 $3.40 per pound. On the other hand, suppliers will be willing to supply 17 thousand pounds of cherries at $3.40 per pound (solve 3.4 0002 0.1q 0003 1.7 for q). So, at $3.40 per pound the suppliers are willing to supply more cherries than the consumers are willing to purchase. The supply exceeds the demand at that price, and the price will come down. Find the equilibrium quantity and the equilibrium price. p
Price per pound ($)
EXAMPLE
Demand
5
(17, 3.4) 3.4
(11, 3.4) Supply
11
17 20
Thousands of pounds
Z Figure 5
q
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SOLUTION
To find the equilibrium quantity, we solve the linear system p ⫽ ⫺0.2q ⫹ 5.6 p ⫽ 0.1q ⫹ 1.7
Demand equation (consumer) Supply equation (supplier)
using substitution (substituting p ⫽ ⫺0.2q ⫹ 5.6 into the second equation). p ⫽ 0.1q ⫹ 1.7 ⫺0.2q ⫹ 5.6 ⫽ 0.1q ⫹ 1.7 5.6 ⫽ 0.3q ⫹ 1.7 3.9 ⫽ 0.3q q ⴝ 13 thousand pounds
Price per pound ($)
p
Equilibrium point (13, 3)
5
3
Demand
13
20
Thousands of pounds
Z Figure 6
MATCHED PROBLEM 9
Add 0.2q to both sides. Subtract 1.7 from both sides. Divide both sides by 0.3. Equilibrium quantity
Now substitute q ⫽ 13 back into either of the original equations in the system and solve for p (we choose the second equation):
Supply q
Substitute p ⴝ ⴚ0.2q ⴙ 5.6.
p ⫽ 0.1(13) ⫹ 1.7 p ⴝ $3 per pound
Equilibrium price
So if the price of cherries is $3 per pound, then the supplier would supply 13,000 pounds of cherries and the consumer would demand (purchase) 13,000 pounds of cherries. In other words, the market would be in equilibrium (see Fig. 6). 0002 The price–demand and price–supply equations for strawberries in a certain city are p ⫽ ⫺0.2q ⫹ 4 p ⫽ 0.04q ⫹ 1.84
Demand equation Supply equation
where q represents the quantity in thousands of pounds and p represents the price in dollars. Find the equilibrium quantity and the equilibrium price. 0002
EXAMPLE
10
Production Scheduling A garment industry manufactures three shirt styles. Each style shirt requires the services of three departments as listed in the table. The cutting, sewing, and packaging departments have available a maximum of 1,160, l,560, and 480 labor-hours per week, respectively. How many of each style shirt must be produced each week for the plant to operate at full capacity?
SOLUTION
Style A
Style B
Style C
Time Available
Cutting department
0.2 hr
0.4 hr
0.3 hr
1,160 hr
Sewing department
0.3 hr
0.5 hr
0.4 hr
1,560 hr
Packaging department
0.1 hr
0.2 hr
0.1 hr
480 hr
Let x ⫽ Number of style A shirts produced per week y ⫽ Number of style B shirts produced per week z ⫽ Number of style C shirts produced per week Then 0.2x ⫹ 0.4y ⫹ 0.3z ⫽ 1,160 0.3x ⫹ 0.5y ⫹ 0.4z ⫽ 1,560 0.1x ⫹ 0.2y ⫹ 0.1z ⫽ 480
Cutting department Sewing department Packaging department
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Systems of Linear Equations
We can clear the system of decimals by multiplying each side of each equation by 10: 2x ⫹ 4y ⫹ 3z ⫽ 11,600 3x ⫹ 5y ⫹ 4z ⫽ 15,600 x ⫹ 2y ⫹ z ⫽ 4,800
E1 E2 E3
Use E3 to eliminate z from E1 and replace E1 with the result. Equivalent System
2x ⫹ 4y ⫹ 3z ⫽ 11,600 ⫺3x ⫺ 6y ⫺ 3z ⫽ ⫺14,400 ⫺x ⫺ 2y ⫽ ⫺2,800
⫺x ⫺ 2y ⫽ ⫺2,800 3x ⫹ 5y ⫹ 4z ⫽ 15,600 x ⫹ 2y ⫹ z ⫽ 4,800
E1 ⴚ3E3 E4
E4 E2 E3
Use E3 to eliminate z from E2 and replace E2 with the result. Equivalent System
3x ⫹ 5y ⫹ 4z ⫽ 15,600 ⫺4x ⫺ 8y ⫺ 4z ⫽ ⫺19,200 ⫺x ⫺ 3y ⫽ ⫺3,600
⫺x ⫺ 2y ⫺x ⫺ 3y
E2 ⴚ4E3
⫽ ⫺2,800 ⫽ ⫺3,600
x ⫹ 2y ⫹ z ⫽
E5
4,800
E4 E5 E3
Now treat E4 and E5 as a system of two equations; eliminate x. x ⫹ 2y ⫺x ⫺ 3y ⫺y
⫽ 2,800 ⫽ ⫺3,600 ⫽ ⫺800
ⴚE4 E5 E6
From E6 we see that y ⴝ 800 Substitute y ⫽ 800 in E4 or E5 and solve for x. ⫺x ⫺ 2y ⫽ ⫺2,800 ⫺x ⫺ 2(800) ⫽ ⫺2,800 x ⴝ 1,200
E4
Substitute x ⫽ 1,200 and y ⫽ 800 in E1, E2, or E3 and solve for z. x ⫹ 2y ⫹ z ⫽ 4,800 1,200 ⫹ 2(800) ⫹ z ⫽ 4,800 z ⴝ 2,000
E3
Each week, the company should produce 1,200 style A shirts, 800 style B shirts, and 2,000 style C shirts to operate at full capacity. You should check this solution. 0002 MATCHED PROBLEM 10
Repeat Example 10 with the cutting, sewing, and packaging departments having available a maximum of 1,180, 1,560, and 510 labor-hours per week, respectively. 0002
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ANSWERS TO MATCHED PROBLEMS 1.
y 5
x0002y00033
x 0004 2y 0003 00023 00025
5
(1, 00022)
x
x 0003 1, y 0003 00022 x0002y00033 Check: ? 1 0002 (00022) 0003 3 ✓ 300033 x 0004 2y 0003 00023 ? 1 0004 2(00022) 0003 00023 ✓ 00023 0003 00023
00025
2. 3. 4. 5. 6. 7. 8. 9. 10.
7-1
(A) (3, 2) or x 0003 3 and y 0003 2 (B) No solutions (C) Infinite number of solutions x 0003 1, y 0003 00022 x 0003 00021, y 0003 3 (00021, 0, 2) or x 0003 00021, y 0003 0, z 0003 2 Inconsistent system with no solution {(00026s 0002 1, 7s 0004 4, s) | s any real number} Boat: 13 mph; current: 3 mph Equilibrium quantity 0003 9 thousand pounds; Equilibrium price 0003 $2.20 per pound Each week, the company should produce 900 style A shirts, 1,300 style B shirts, and 1,600 style C shirts to operate at full capacity.
Exercises
1. Explain in your own words how to solve a system of two linear equations by graphing.
y 5
2. Explain in your own words how to solve a system of two linear equations by substitution. 3. Explain in your own words how to solve a system of two linear equations using elimination by addition.
5
9. 2x 0002 y 0003 5 3x 0004 2y 0003 00023
00025
5
x
00025
(a)
(b)
y
y
6. Describe how the solution sets differ for systems of linear equations that are consistent, inconsistent, and dependent.
8. x 0004 y 0003 3 x 0002 2y 0003 0
x
00025
5. Can a system of two linear equations have exactly two solutions? Explain.
7. 2x 0002 4y 0003 8 x 0002 2y 0003 0
5
00025
4. Which of the three solving techniques is the best choice for a system of three equations? Why?
Match each system in Problems 7–10 with one of the following graphs, and use the graph to solve the system.
y
5
00025
5
x 00025
5
00025
10. 4x 0002 2y 0003 10 2x 0002 y 0003 5
00025
(c)
(d)
x
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Solve the system of equations in Problems 11–46.
Systems of Linear Equations
439
Problems 49 and 50 refer to the system
11. x 0002 y 0003 7 x0004y00033
12. x 0004 y 0003 2 x0002y00034
13. 3x 0004 2y 0003 12 7x 0002 2y 0003 8
14. 3x 0004 y 0003 2 x 0002 2y 0003 10
15. 3u 0002 5v 0003 15 6u 0002 10v 0003 000430
16. m 0002 2n 0003 4 2m 0002 4n 0003 00048
49. Solve the system for x and y in terms of the constants a, b, c, d, h, and k. Clearly state any assumptions you must make about the constants during the solution process.
3x 0004 y 0003 00042 00049x 0002 3y 0003 6
18. 2x 0004 8y 0003 10 8x 0004 32y 0003 40
50. Discuss the nature of solutions to systems that do not satisfy the assumptions you made in Problem 49.
17.
19. x 0004 y 0003 4 x 0002 3y 0003 12
20. 3x 0004 y 0003 7 2x 0002 3y 0003 1
21. 4x 0002 3y 0003 26 3x 0004 11y 0003 00047
22. 9x 0004 3y 0003 24 11x 0002 2y 0003 1
23. 7m 0002 12n 0003 00041 5m 0004 3n 0003 7
24. 3p 0002 8q 0003 4 15p 0002 10q 0003 000410
25. y 0003 0.08x y 0003 100 0002 0.04x
26. 0.2u 0004 0.5v 0003 0.07 0.8u 0004 0.3v 0003 0.79
27. 25x 0002 32 y 0003 2 7 5 3 x 0004 4 y 0003 00045
28. 5x 0004 2y 0003 8 2x 0002 3y 0003 000410
29. 00042.3y 0002 4.1z 0003 000414.21 30. 5.4x 0002 4.2y 0003 000412.9 10.1y 0004 2.9z 0003 26.15 3.7x 0002 6.4y 0003 00044.5 0003 2 31. 00042x x 0004 3y 0003 2 0004x 0002 2y 0002 3z 0003 00047 33.
2y 0004 z 0003 2 00044y 0002 2z 0003 1 x 0004 2y 0002 3z 0003 0
32.
2y 0002 z 0003 00044 x 0004 3y 0002 2z 0003 9 0004y 0003 3
34. x 0002 y 0004 z 0003 3 x 0004 2z 0003 1 y0002 z00032
0003 2 35. x 0004 3y 2y 0002 z 0003 00041 x0004 y0002z0003 1
00031 36. 00044x 0002 3y 8x 0004 6y 00034 2x 0004 4y 0002 3z 0003 6
z 0003 00045 37. 2x 0002 x0004 3z 0003 00046 4x 0002 2y 0004 z 0003 00049
38.
39. x 0004 y 0002 z 0003 1 2x 0002 y 0002 z 0003 6 7x 0004 y 0002 5z 0003 15
40. 2x 0004 y 0002 3z 0003 7 x 0002 2y 0004 z 0003 00043 3x 0002 y 0002 2z 0003 2
41. 2a 0002 4b 0002 3c 0003 00046 a 0004 3b 0002 2c 0003 000415 0004a 0002 2b 0004 c 0003 9
42. 3u 0004 2v 0002 3w 0003 11 2u 0002 3v 0004 2w 0003 00045 u 0002 4v 0004 w 0003 00045
43. 2x 0004 3y 0002 3z 0003 00045 3x 0002 2y 0004 5z 0003 34 5x 0004 4y 0004 2z 0003 23
44.
45. 0004x 0002 2y 0004 z 0003 00044 2x 0002 5y 0004 4z 0003 000416 x 0002 y 0004 z 0003 00044
46. x 0004 8y 0002 2z 0003 00041 x 0004 3y 0002 z 0003 1 2x 0004 11y 0002 3z 0003 2
x 0004 3y 0002 z 0003 4 0004x 0002 4y 0004 4z 0003 1 2x 0004 y 0002 5z 0003 00043
x 0002 2y 0002 z 0003 2 00042x 0002 3y 0004 2z 0003 00043 x 0004 5y 0002 z 0003 2
In Problems 47 and 48, solve each system for p and q in terms of x and y. Explain how you could check your solution and then perform the check. 47. x 0003 2 0002 p 0004 2q y 0003 3 0004 p 0002 3q
48. x 0003 00041 0002 2p 0004 q y0003 40004 p0002q
ax 0002 by 0003 h cx 0002 dy 0003 k where x and y are variables and a, b, c, d, h, and k are real constants.
APPLICATIONS 51. AIRSPEED It takes a private airplane 8.75 hours to make the 2,100-mile flight from Atlanta to Los Angeles and 5 hours to make the return trip. Assuming that the wind blows at a constant rate from Los Angeles to Atlanta, find the airspeed of the plane and the wind rate. 52. AIRSPEED A plane carries enough fuel for 20 hours of flight at an airspeed of 150 miles per hour. How far can it fly into a 30 mph headwind and still have enough fuel to return to its starting point? (This distance is called the point of no return.) 53. RATE–TIME A crew of eight can row 20 kilometers per hour in still water. The crew rows upstream and then returns to its starting point in 15 minutes. If the river is flowing at 2 km/h, how far upstream did the crew row? 54. RATE–TIME It takes a boat 2 hours to travel 20 miles down a river and 3 hours to return upstream to its starting point. What is the rate of the current in the river? 55. BUSINESS A company that supplies bulk candy to bakeries has one batch of chocolate chips that are 50% dark chocolate and 50% milk chocolate. They have another batch that is 80% dark chocolate and 20% milk chocolate. One of their customers sends in a rush order for 100 lb of a mix that is 68% dark chocolate. How many pounds from each batch should be mixed to meet this order? 56. BUSINESS A jeweler has two bars of gold alloy in stock, one of 12 carats and the other of 18 carats (24-carat gold is pure gold, 1212 carat is 24 pure, 18-carat gold is 18 24 pure, and so on). How many grams of each alloy must be mixed to obtain 10 grams of 14-carat gold?
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57. BREAK-EVEN ANALYSIS It costs a small recording company $17,680 to prepare a compact disc. This is a one-time fixed cost that covers recording, package design, and so on. Variable costs, including such things as manufacturing, marketing, and royalties, are $4.60 per CD. If the CD is sold to music shops for $8 each, how many must be sold for the company to break even?
(B) Find the supply and the demand (to the nearest unit) if baseball caps are priced at $8 each. Discuss the stability of the baseball cap market at this price level. (C) Find the equilibrium price and quantity. (D) Graph the two equations in the same coordinate system and identify the equilibrium point, supply curve, and demand curve.
58. FINANCE Suppose you have $12,000 to invest. If part is invested at 10% and the rest at 15%, how much should be invested at each rate to yield 12% on the total amount invested?
63. SUPPLY AND DEMAND At $0.60 per bushel, the daily supply for wheat is 450 bushels and the daily demand is 645 bushels. When the price is raised to $0.90 per bushel, the daily supply increases to 750 bushels and the daily demand decreases to 495 bushels. Assume that the supply and demand equations are linear. (A) Find the supply equation. (B) Find the demand equation. (C) Find the equilibrium price and quantity.
59. PRODUCTION A supplier for the electronics industry manufactures keyboards and screens for graphing calculators at plants in Mexico and Taiwan. The hourly production rates at each plant are given in the table. How many hours should each plant be operated to fill an order for exactly 4,000 keyboards and exactly 4,000 screens? Plant
Keyboards
Screens
Mexico
40
32
Taiwan
20
32
60. PRODUCTION A company produces Italian sausages and bratwursts at plants in Green Bay and Sheboygan. The hourly production rates at each plant are given in the table. How many hours should each plant be operated to exactly fill an order for 62,250 Italian sausages and 76,500 bratwursts? Plant
Italian Sausage
Bratwurst
Green Bay
800
800
Sheboygan
500
1,000
61. SUPPLY AND DEMAND Suppose the supply and demand equations for printed T-shirts in a resort town for a particular week are p0002
0.007q 0003 3
p 0002 00040.018q 0003 15
Supply equation Demand equation
where p is the price in dollars and q is the quantity. (A) Find the supply and the demand (to the nearest unit) if T-shirts are priced at $4 each. Discuss the stability of the T-shirt market at this price level. (B) Find the supply and the demand (to the nearest unit) if T-shirts are priced at $8 each. Discuss the stability of the T-shirt market at this price level. (C) Find the equilibrium price and quantity. (D) Graph the two equations in the same coordinate system and identify the equilibrium point, supply curve, and demand curve. 62. SUPPLY AND DEMAND Suppose the supply and demand equations for printed baseball caps in a resort town for a particular week are p0002
0.006q 0003 2
p 0002 00040.014q 0003 13
Supply equation
64. SUPPLY AND DEMAND At $1.40 per bushel, the daily supply for soybeans is 1,075 bushels and the daily demand is 580 bushels. When the price falls to $1.20 per bushel, the daily supply decreases to 575 bushels and the daily demand increases to 980 bushels. Assume that the supply and demand equations are linear. (A) Find the supply equation. (B) Find the demand equation. (C) Find the equilibrium price and quantity. 65. EARTH SCIENCE An earthquake emits a primary wave and a secondary wave. Near the surface of the Earth the primary wave travels at about 5 miles per second and the secondary wave at about 3 miles per second. From the time lag between the two waves arriving at a given receiving station, it is possible to estimate the distance to the quake. (The epicenter can be located by obtaining distance bearings at three or more stations.) Suppose a station measured a time difference of 16 seconds between the arrival of the two waves. How long did each wave travel, and how far was the earthquake from the station? 66. EARTH SCIENCE A ship using sound-sensing devices above and below water recorded a surface explosion 6 seconds sooner by its underwater device than its above-water device. Sound travels in air at about 1,100 feet per second and in seawater at about 5,000 feet per second. (A) How long did it take each sound wave to reach the ship? (B) How far was the explosion from the ship? 67. PRODUCTION SCHEDULING A company manufactures three products; lawn mowers, snowblowers, and chain saws. The labor, material, and shipping costs for manufacturing one unit of each product are given in the table. The weekly allocations for labor, materials, and shipping are $35,000, $50,000, and $20,000, respectively. How many of each type of product should be manufactured each week in order to exactly use the weekly allocations? Product
Labor
Materials
Shipping
Lawn mower
$20
$35
$15
Snowblower
$30
$50
$25
Chain saw
$45
$40
$10
Demand equation
where p is the price in dollars and q is the quantity in hundreds. (A) Find the supply and the demand (to the nearest unit) if baseball caps are priced at $4 each. Discuss the stability of the baseball cap market at this price level.
68. PRODUCTION SCHEDULING A company manufactures three products; desk chairs, file cabinets, and printer stands. The labor, material, and shipping costs for manufacturing one unit of each product are given in the table. The weekly allocations for labor, materials, and shipping are $21,100, $31,500, and $11,900, respec-
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tively. How many of each type of product should be manufactured each week in order to exactly use the weekly allocations? Product
Desk Chair
File Cabinet
Printer Stand
Labor
$30
$35
$40
Materials
$45
$60
$55
Shipping
$25
$20
$15
69. PRODUCTION SCHEDULING A company has plants located in Michigan, New York, and Ohio where it manufactures laptop computers, desktop computers, and servers. The number of units of each product that can be produced per day at each plant are given in the table below. The company has orders for 2,150 laptop computers, 2,300 desktop computers, and 2,500 servers. How many days should the company operate each plant in order to exactly fill these orders? Plant
Michigan
New York
Ohio
Laptop
10
70
60
Desktop
20
50
80
Server
40
30
90
70. PRODUCTION SCHEDULING A company has plants located in Maine, Utah, and Oregon where it manufactures stoves, refrigerators, and dishwashers. The number of units of each product that can be produced per day at each plant are given in the table. The company has orders for 1,500 stoves, 2,350 refrigerators, and 2,400 dishwashers.
7-2
441
Solving Systems of Linear Equations Using Gauss–Jordan Elimination
How many days should the company operate each plant in order to exactly fill these orders? Set up a system of equations whose solution would answer this question and solve the system. Plant
Stoves
Refrigerators
Dishwashers
Maine
30
70
60
Utah
20
50
50
Oregon
40
30
40
71. INVESTMENT Due to recent volatility in the stock market, Catalina’s financial advisor suggests that she reallocate $70,000 of her retirement fund to bonds. He recommends a mix of treasury bonds earning 4% annually, municipal bonds earning 3.5% annually, and corporate bonds earning 4.5% annually. For tax reasons, he also recommends that the amount invested in treasury bonds should be equal to the sum of the amount invested in the other categories. If Catalina follows these recommendations, and the goal is to produce $2,900 in annual interest income, how much will she invest in each of the three types of bonds? 72. INVESTMENT When the real estate market begins to rebound, Catalina (see Problem 71) decides to reallocate her investment mix. At this point, her investment has grown to $76,000. She’ll leave some money in treasury and corporate bonds, but will replace municipal bonds with a real estate investment trust that guarantees a 6.5% annual return. If she plans to leave as much in treasury bonds as the sum of the other two investments, how much should she invest in each to reach her new goal of earning an annual interest income of $3,600?
Solving Systems of Linear Equations Using Gauss–Jordan Elimination Z Matrices and Row Operations Z Reduced Matrices Z Solving Systems by Gauss–Jordan Elimination Z Application
In this section, we introduce Gauss–Jordan elimination, a step-by-step procedure for solving systems of linear equations. This procedure works for any system of linear equations and is easily implemented on a computer. In fact, the TI-84 has a built-in procedure for performing Gauss–Jordan elimination.
Z Matrices and Row Operations In solving systems of equations using elimination by addition, the coefficients of the variables and the constant terms played a central role. The process can be made more efficient
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by the introduction of a mathematical form called a matrix. A matrix (plural matrices) is a rectangular array of numbers written within brackets. Two examples are 1 A0002 c 5
00043 0
7 d 00044
00045 0 B0002 ≥ 00042 00043
4 1 12 0
11 6 ¥ 8 00041
(1)
Each number in a matrix is called an element of the matrix. Matrix A has six elements arranged in two rows and three columns. Matrix B has 12 elements arranged in four rows and three columns. If a matrix has m rows and n columns, it is called an m ⴛ n matrix (read “m by n matrix”). The expression m 0005 n is called the size of the matrix, and the numbers m and n are called the dimensions of the matrix. It is important to note that the number of rows is always given first. Referring to equations (1), A is a 2 0005 3 matrix and B is a 4 0005 3 matrix. A matrix with n rows and n columns is called a square matrix of order n. A matrix with only one column is called a column matrix, and a matrix with only one row is called a row matrix. These definitions are illustrated by the following: 3 ⴛ 3
0.5 £ 0.0 0.7
0.2 0.3 0.0
4 ⴛ 1
3 00042 ≥ ¥ 1 0
1.0 0.5 § 0.2
Square matrix of order 3
Column matrix
1 ⴛ 4
[2
1 2
0
000423 ]
Row matrix
The position of an element in a matrix is the row and column containing the element. This is usually denoted using double subscript notation aij, where i is the row and j is the column containing the element aij, as illustrated next: 1 A0002 £ 6 00042
5 0 3
00043 00044 4
4 1§ 7
a11 0002 00041, a12 0002 5, a13 0002 00043, a14 0002 4 a21 0002 00046, a22 0002 0, a23 0002 00044, a24 0002 1 a31 0002 00042, a32 0002 3, a33 0002 00044, a34 0002 7
Note that a12 is read “a sub one two,” not “a sub twelve.” The elements a11 0002 1, a22 0002 0, and a33 0002 4 make up the principal diagonal of A. In general, the principal diagonal of a matrix A consists of the elements aii, i 0002 1, 2, . . . n.
Technology Connections Most graphing calculators are capable of storing and manipulating matrices. Figure 1 shows matrix A displayed in the matrix editing screen of a TI-84 graphing calculator. The size of the matrix is given at the top of the screen, and the position of the currently selected element is given at the bottom. Notice that a comma is used in the notation for the position. This is common practice on graphing calculators but it’s almost never written or typed that way.
Z Figure 1 Matrix notation on a TI-84 graphing calculator.*
*The onscreen display of A was too large to fit on the screen of a TI-84, so we pasted together two screen shots to form Figure 1. When this happens on your graphing calculator, you will have to scroll left and right and/or up and down to see the entire matrix.
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Now we turn our attention to the connection between matrices and systems of equations. Consider the system of equations x 0003 5y 0004 3z 0002 4 6x 0004 4z 0002 1 00042x 0003 3y 0003 4z 0002 7
(2)
If we remove the variables and leave behind the numbers, we can think of the result as a matrix: 1 5 00043 4 £ 6 0 00044 † 1 § 00042 3 4 7 Coefficient matrix
1 £ 6 00042
5 0 3
00043 00044 § 4
Constant matrix
4 £1§ 7
Z Figure 2
This is known as the augmented coefficient matrix for the system. We can also define the coefficient matrix and the constant matrix for the system, as shown in Figure 2. The augmented coefficient matrix contains all of the information about the system needed to solve it. Note that we put in a coefficient of zero for the missing y in the second equation, and that we drew a vertical bar to separate the coefficients from the constants. (Matrices displayed on a graphing calculator won’t have that line.) Since we would like to be able to use matrices to solve large systems with many variables, moving forward we will use x1, x2, x3, and the like, rather than x, y, z, and so on. In this notation, we will rewrite system (2) as x1 0003 5x2 0004 3x3 0002 4 6x1 0004 4x3 0002 1 00042x1 0003 3x2 0003 4x3 0002 7 In Section 7-1, we used Ei to denote the equations in a linear system. Now we use Ri to denote the rows and Ci to denote the columns, respectively, in a matrix, as illustrated below for system (2). C1
C2
1 £ 6 00042
5 0 3
C3
C4
00043 4 00044 † 1 § 4 7
R1 R2
(3)
R3
Our goal will be to learn how to perform the basic steps we used to solve systems using elimination by addition, but on an augmented matrix. This enables us to focus on the numbers without being concerned about algebraic manipulations.
EXAMPLE
1
Writing an Augmented Coefficient Matrix Write the augmented coefficient matrix corresponding to each of the following systems. (A)
SOLUTIONS
MATCHED PROBLEM 1
2x1 0004 4x2 0002 5 00043x1 0003 x2 0002 00046
(A) c
2 00043
00044 5 ` d 1 00046
(B) 00043x1 0003 2x3 0002 00044 7x1 0004 5x2 0003 3x3 0002 0
(B) c
00043 7
0 2 00044 d ` 00045 3 0
(C) 2x1 0004 x2 0002 4 3x1 0004 5x3 0002 6 00042x2 0003 x3 0002 00043
2 00041 0 4 (C) £ 3 0 00045 † 6 § 0 00042 1 00043
0002
Write the augmented coefficient matrix corresponding to each of the following systems. (A) 0004x1 0003 2x2 0002 00043 3x1 0004 5x2 0002 8
(B)
00042x2 0003 2x3 0002 00044 7x1 00045x2 0003 3x3 0002 0
(C) 2x1 0004 x2 0003 x3 0002 4 3x1 0003 4x2 0002 6 x1 0003 5x3 0002 00043
0002
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Recall that two linear systems are said to be equivalent if they have the same solution set. In Theorem 2, Section 7-1, we used the operations listed next to transform linear systems into equivalent systems: (A) Two equations are interchanged. (B) An equation is multiplied by a nonzero constant. (C) A constant multiple of one equation is added to another equation. Paralleling this approach, we now say that two augmented matrices are row-equivalent, denoted by the symbol ⬃ between the two matrices, if they are augmented matrices of equivalent systems of equations. How do we transform augmented matrices into row-equivalent matrices? We use Theorem 1, which gives the matrix analogs of operations (A), (B), and (C).
Z THEOREM 1 Elementary Row Operations Producing Row-Equivalent Matrices An augmented matrix is transformed into a row-equivalent matrix if any of the following row operations is performed: 1. Two rows are interchanged (Ri 4 Rj). 2. A row is multiplied by a nonzero constant (kRi S Ri). 3. A constant multiple of one row is added to another row (kRj 0003 Ri S Ri). [Note: The arrow means “replaces.”]
EXAMPLE
2
Row Operations Perform each of the indicated row operations on the following augmented coefficient matrix. c (A) R1 4 R2
SOLUTIONS
MATCHED PROBLEM 2
(A) c
2 1
(B) 12R2 S R2
4 00048 d ` 00044 3
(B) c
1 2
00044 3 d ` 4 00048
(C) (00042)R 1 0003 R2 S R2 1 1
00044 3 d ` 2 00044
(C) c
1 00044 3 d ` 0 12 000414
0002
Perform each of the indicated row operations on the following augmented coefficient matrix. c (A) R1 4 R2
(B) 13R2 S R2
1 3
00042 3 d ` 00046 00043
(C) (00043)R 1 0003 R2 S R2
0002
Z Reduced Matrices The goal of the elimination process is to transform a system of equations into an equivalent system whose solution is easy to find. Now our goal is to use a sequence of matrix row operations to transform an augmented coefficient matrix into a simpler equivalent matrix that corresponds to a system with an obvious solution. Example 3 illustrates the process of interpreting the solution of a system given its augmented coefficient matrix.
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3
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Interpreting an Augmented Coefficient Matrix Write the system corresponding to each of the following augmented coefficient matrices and find its solution. 1 0 0 00044 (A) £ 0 1 0 † 6§ 0 0 1 0
SOLUTIONS
1 0 (B) £ 0 1 0 0
2 00044 00043 † 6 § 0 1
1 0 2 00044 (C) £ 0 1 00043 † 6 § 0 0 0 0
(A) The corresponding system is x1 0002 00044 x2 0002 00046 x3 0002 00040 and (00044, 6, 0) is the solution. (B) The corresponding system is
1 ⴢ x1 ⴙ 0 ⴢ x2 ⴙ 0 ⴢ x3 ⴝ x1
0003 2x3 0002 00044 x2 0004 3x3 0002 6 0 ⴢ x1 0003 0 ⴢ x2 0003 0 ⴢ x3 0002 1 x1
The third equation, 0 0002 1, is a contradiction, so the system has no solutions. (C) The first two rows of this augmented coefficient matrix correspond to the system x1
0003 2x3 0002 00044 x2 0004 3x3 0002 6
The third row corresponds to the equation 0 ⴝ 0, which is always true and can be discarded.
This is a dependent system with an infinite number of solutions. Introducing a parameter s, we can write x1
0003 2s 0002 00044 x2 0004 3s 0002 6 or x3 0002 s
x1 0002 00042s 0004 4 x2 0002 3s 0003 6 x3 0002 s
So the solution set is {(00042s 0004 4, 3s 0003 6, s) | s any real number} MATCHED PROBLEM 3
Write the system corresponding to each of the following augmented coefficient matrices and find its solution. 1 (A) £ 0 0
ZZZ EXPLORE-DISCUSS 1
0002
0 1 0
0 5 0 † 00047 § 1 0
1 (B) £ 0 0
0 1 0
00043 5 4 † 00047 § 0 0
1 (C) £ 0 0
0 1 0
00043 5 4 † 00047 § 0 1 0002
If an augmented coefficient matrix contains a row where every element on the left of the vertical line is 0 and the single element on the right is a nonzero number, what can you say about the solution of the corresponding system?
Next, we will define a particular matrix form that makes it simple to find solutions of the corresponding system.
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Z DEFINITION 1 Reduced Matrix A matrix is in reduced form* if: 1. Each row consisting entirely of 0’s is below any row having at least one nonzero element. 2. The leftmost nonzero element in each row is 1. 3. The column containing the leftmost 1 of a given row has 0’s above and below the 1. 4. The leftmost 1 in any row is to the right of the leftmost 1 in the preceding row.
For example, each of the following matrices is in reduced form. Before moving on, you should verify that each matrix satisfies all four conditions in Definition 1. 1 c 0
EXAMPLE
4
1 0 2 ` d £0 1 00043 0
0 1 0
0 2 1 0 † 00041 § £ 0 1 3 0
0 3 1 1 † 00041 § £ 0 0 0 0
4 0 0
0 1 0
0 3 1 0 † 2§ £0 1 6 0
0 1 0
4 0 3 † 0§ 0 1
Reduced Forms The matrices shown next are not in reduced form. Indicate which condition in the definition is violated for each matrix. State the row operation(s) required to transform the matrix to reduced form, and find the reduced form. 0 1
1 00042 ` d 0 3
(B) c
1 0
2 0
00042 3 ` d 1 00041
1 (C) £ 0 0
0 00043 0 † 0§ 1 00042
1 (D) £ 0 0
0 2 0
0 00041 0 † 3§ 1 00045
(A) c
SOLUTIONS
(A) Condition 4 is violated: The leftmost 1 in row 2 is not to the right of the leftmost 1 in row 1. Perform the row operation R1 4 R2 to obtain the reduced form: c
0 1
1 00042 1 ` d R1 4 R2 c 0 3 0
0 3 ` d 1 00042
(B) Condition 3 is violated: The column containing the leftmost 1 in row 2 does not have a zero above the 1. Perform the row operation 2R2 0003 R1 S R1 to obtain the reduced form: c
1 0
2 0
00042 3 1 ` d 2R2 0003 R1 S R1 c 1 00041 0
2 0
0 1 ` d 1 00041
(C) Condition 1 is violated: The second row contains all zeros, and it is not below any row having at least one nonzero element. Perform the row operation R2 4 R3 to obtain the reduced form: 1 £0 0
0 00043 1 0 † 0 § R2 4 R3 £ 0 1 00042 0
0 00043 1 † 00042 § 0 0
*The reduced form we have defined here is sometimes called the reduced row echelon form, and most graphing calculators use the abbreviation rref to refer to it. There are other reduced forms that can be used to solve systems of equations, but we will use the term “reduced form” for simplicity.
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(D) Condition 2 is violated: The leftmost nonzero element in row 2 is not a 1. Perform the row operation 12R2 S R2 to obtain the reduced form: 1 £0 0 MATCHED PROBLEM 4
0 2 0
1 0 00041 0 † 3 § 12R2 S R2 £ 0 0 1 00045
0 1 0
0 00041 0 † 32 § 1 00045
0002
The matrices below are not in reduced form. Indicate which condition in the definition is violated for each matrix. State the row operation(s) required to transform the matrix to reduced form and find the reduced form. (A) c
1 0
0 (C) £ 1 0
1 (B) £ 0 0
0 2 ` d 3 00046 1 0 0
5 4 3 1 2 † 00041 § 0 0 0 1 2 0 3 (D) £ 0 0 0 † 0 § 0 0 1 4
0 00043 0 † 0§ 1 2
0002
Z Solving Systems by Gauss–Jordan Elimination We are now ready to outline the Gauss–Jordan elimination method for solving systems of linear equations. The method systematically transforms an augmented matrix into a reduced form. The system corresponding to a reduced augmented coefficient matrix is called a reduced system. As we will see, reduced systems are easy to solve. The Gauss–Jordan elimination method is named after the German mathematician Carl Friedrich Gauss (1777–1855) and the German geodesist Wilhelm Jordan (1842–1899). Gauss, one of the greatest mathematicians of all time, used a method of solving systems of equations that was later generalized by Jordan to solve problems in large-scale surveying.
EXAMPLE
5
Solving a System Using Gauss–Jordan Elimination Solve by Gauss–Jordan elimination: 2x1 0004 2x2 0003 x3 0002 3 3x1 0003 x2 0004 x3 0002 7 x1 0004 3x2 0003 2x3 0002 0
SOLUTION
Write the augmented matrix and follow the steps indicated at the right to produce a reduced form.
Need a 1 here.
冤
2 3 1
00042 1 00043
1 00041 2
Need 0’s here.
1 ⬃ 3 2
00043 1 00042
2 00041 1
冤
ⱍ冥 ⱍ冥 3 7 0 0 7 3
R1 ↔ R3
(ⴚ3)R1 ⴙ R2 → R2 (ⴚ2)R1 ⴙ R3 → R3
Step 1: Choose the leftmost nonzero column and get a 1 at the top.
Step 2: Use multiples of the row containing the 1 from step 1 to get zeros in all remaining places in the column containing this 1.
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ⱍ冥 ⱍ冥 ⱍ冥 ⱍ冥
Need a 1 here.
1 ⬃ 0 0
冤
00043 10 4
2 00047 00043
Need 0’s here.
1 ⬃ 0 0
冤
00043 1 4
2 00040.7 00043
0 0.7 3
1 ⬃ 0 0
冤
0 1 0
00040.1 00040.7 00040.2
2.1 0.7 0.2
1 ⬃ 0 0
冤
0 1 0
00040.1 00040.7 1
2.1 0.7 00041
冤
0 1 0
0 0 1
Need a 1 here.
Need 0’s here.
1 ⬃ 0 0
0 7 3
0.1R2 → R2
3R2 ⴙ R1 → R1
Step 4: Repeat step 2 with the entire matrix.
(ⴚ4)R2 ⴙ R3 → R3
(ⴚ5)R3 → R3
0.1R3 ⴙ R1 → R1 0.7R3 ⴙ R2 → R2
ⱍ冥 2 0 00041
x1 x2
Step 3: Repeat step 1 with the submatrix formed by (mentally) deleting the top (shaded) row.
Step 3: Repeat step 1 with the submatrix formed by (mentally) deleting the top two (shaded) rows.
Step 4: Repeat step 2 with the entire matrix.
The matrix is now in reduced form, and we can proceed to solve the corresponding reduced system.
0002 2 0002 0 x3 0002 00041
The solution to this system is x1 0002 2, x2 0002 0, x3 0002 00041. You should check this solution in the original system.
Z GAUSS–JORDAN ELIMINATION Step 1. Choose the leftmost nonzero column and use appropriate row operations to get a 1 at the top. Step 2. Use multiples of the row containing the 1 from step 1 to get zeros in all remaining places in the column containing this 1. Step 3. Repeat step 1 with the submatrix formed by (mentally) deleting the row used in step 2 and all rows above this row. Step 4. Repeat step 2 with the entire matrix, including the mentally deleted rows. Continue this process until the entire matrix is in reduced form. [Note: If at any point in this process we obtain a row with all zeros to the left of the vertical line and a nonzero number to the right, we can stop, since we will have a contradiction: 0 0002 n, n 0006 0. We can then conclude that the system has no solution.] 0002 MATCHED PROBLEM 5
Solve by Gauss–Jordan elimination:
3x1 0003 x2 0004 2x3 0002 2 x1 0004 2x2 0003 x3 0002 3 2x1 0004 x2 0004 3x3 0002 3
0002
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Technology Connections Figure 3 illustrates the solution of Example 5 on a TI-84 graphing calculator using the built-in rref (reduced rowechelon form) routine for finding reduced forms. Notice that in row 2 and column 4 of the reduced form the graphing calculator has displayed the very small number -3.5E-13 instead of the exact value 0. This is a common occurrence caused by rounding error on a graphing calculator and causes no problems. Just replace any very small numbers displayed in scientific notation with 0.
EXAMPLE
6
Z Figure 3 Using rref on a TI-84 graphing calculator.
Solving a System Using Gauss–Jordan Elimination Solve by Gauss–Jordan elimination:
SOLUTION
2x1 0004 4x2 0003 x3 0002 00044 4x1 0004 8x2 0003 7x3 0002 2 00042x1 0003 4x2 0004 3x3 0002 5
2 £ 4 00042
00044 1 00044 00048 7 † 2§ 4 00043 5
1 ⬃£ 4 00042
00042 0.5 00042 00048 7 † 2§ 4 00043 5
1 ⬃£0 0
00042 0 0
0.5 00042 5 † 10 § 00042 1
1 ⬃£0 0
00042 0 0
0.5 00042 1 † 2§ 00042 1
1 ⬃£0 0
00042 0 00043 0 1 † 2§ 0 0 5
0.5R1 S R1 (To get 1 in upper left corner)
(Next, get zeros below that 1.) (ⴚ4)R1 ⴙ R2 S R2 2R1 ⴙ R3 S R3 0.2R2 S R2 Note that column 3 is the leftmost nonzero column in this submatrix.
(ⴚ0.5)R2 ⴙ R1 S R1
2R2 ⴙ R3 S R3 We stop the Gauss–Jordan elimination, even though the matrix is not in reduced form, since the last row produces a contradiction
The system is inconsistent and has no solution. MATCHED PROBLEM 6
Solve by Gauss–Jordan elimination:
2x1 0004 4x2 0004 x3 0002 00048 4x1 0004 8x2 0003 3x3 0002 4 00042x1 0003 4x2 0003 x3 0002 11
0002
0002
Note that if we were to use rref on a graphing calculator for Example 6, it would continue reducing further. But the final reduced form would still show a contradiction.
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7
Solving a System Using Gauss–Jordan Elimination Solve by Gauss–Jordan elimination:
SOLUTION
3x1 0003 6x2 0004 9x3 0002 15 2x1 0003 4x2 0004 6x3 0002 10 00042x1 0004 3x2 0003 4x3 0002 00046
3 £ 2 00042
6 00049 15 4 00046 † 10 § 00043 4 00046
1 ⬃£ 2 00042
2 00043 5 4 00046 † 10 § 00043 4 00046
1 ⬃£0 0
2 0 1
00043 5 0 † 0§ 00042 4
1 ⬃£0 0
2 1 0
00043 5 00042 † 4 § 0 0
1 0 1 00043 ⬃ £ 0 1 00042 † 4 § 0 0 0 0 x1 0003 x3 0002 00043 x2 0004 2x3 0002 4
1 3 R1 S R1
(ⴚ2)R1 ⴙ R2 S R2 2R1 ⴙ R3 S R3 R2 4 R3 Note that we must interchange rows 2 and 3 to obtain a nonzero entry at the top of the second column of this submatrix. (ⴚ2)R2 ⴙ R1 S R1
This matrix is now in reduced form. Write the corresponding reduced system and solve.
We discard the equation corresponding to the third (all 0) row in the reduced form, since it is satisfied by all values of x1, x2, and x3.
Note that the leftmost variable in each equation appears in one and only one equation. We solve for the leftmost variables x1 and x2 in terms of the remaining variable x3: x1 0002 0004x3 0004 3 x2 0002 2x3 0003 4 This dependent system has an infinite number of solutions. We will use a parameter to represent all the solutions. If we let x3 0002 t, then for any real number t, x1 0002 0004t 0004 3 x2 0002 2t 0003 4 x3 0002 t is a solution. You should check that (0004t 0004 3, 2t 0003 4, t) is a solution of the original system for any real number t. Some particular solutions are
MATCHED PROBLEM 7
t ⴝ 0
t ⴝ ⴚ2
t ⴝ 3.5
(00043, 4, 0)
(00041, 0, 00042)
(00046.5, 11, 3.5)
Solve by Gauss–Jordan elimination:
2x1 0004 2x2 0004 4x3 0002 00042 3x1 0004 3x2 0004 6x3 0002 00043 00042x1 0003 3x2 0003 x3 0002 7
In general, If the number of leftmost 1’s in a reduced augmented coefficient matrix is less than the number of variables in the system and there are no contradictions, then the system is dependent and has infinitely many solutions.
0002
0002
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There are many different ways to use the reduced augmented coefficient matrix to describe the infinite number of solutions of a dependent system. We will always proceed as follows: Solve each equation in a reduced system for its leftmost variable and then introduce a different parameter for each remaining variable. Example 8 illustrates a dependent system where two parameters are required to describe the solution.
EXAMPLE
8
Solving a System Using Gauss–Jordan Elimination Solve by Gauss–Jordan elimination:
SOLUTION
x1 0003 2x2 0003 4x3 0003 x4 0004 x5 0002 1 2x1 0003 4x2 0003 8x3 0003 3x4 0004 4x5 0002 2 x1 0003 3x2 0003 7x3 0003 3x5 0002 00042
00041 1 00044 † 2 § 3 00042
1 £2 1
2 4 3
4 8 7
1 3 0
1 ⬃£0 0
2 0 1
4 0 3
1 1 00041
00041 1 00042 † 0 § 4 00043
1 ⬃£0 0
2 1 0
4 3 0
1 00041 1
00041 1 4 † 00043 § 00042 0
1 ⬃£0 0
0 1 0
00042 3 0
3 00049 7 00041 4 † 00043 § 1 00042 0
1 ⬃£0 0
0 1 0
00042 3 0
0 0 1
00043 7 2 † 00043 § 00042 0
x1 0004 2x3 x2 0003 3x3
(ⴚ2)R1 ⴙ R2 S R2 (ⴚ1)R1 ⴙ R3 S R3
R2 4 R3
(ⴚ2)R2 ⴙ R1 S R1
(ⴚ3)R3 ⴙ R1 S R1 R3 ⴙ R2 S R2
Matrix is in reduced form.
0004 3x5 0002 7 0003 2x5 0002 00043 x4 0004 2x5 0002 0
Solve for the leftmost variables x1, x2, and x4 in terms of the remaining variables x3 and x5: x1 0002 2x3 0003 3x5 0003 7 x2 0002 00043x3 0004 2x5 0004 3 x4 0002 2x5 If we let x3 0002 s and x5 0002 t, then for any real numbers s and t, x1 x2 x3 x4 x5
0002 2s 0003 3t 0003 7 0002 00043s 0004 2t 0004 3 0002s 0002 2t 0002t
is a solution. The check is left for you to perform. MATCHED PROBLEM 8
Solve by Gauss–Jordan elimination:
0004 2x5 0002 3 x1 0004 x2 0003 2x3 00042x1 0003 2x2 0004 4x3 0004 x4 0003 x5 0002 00045 3x1 0004 3x2 0003 7x3 0003 x4 0004 4x5 0002 6
0002
0002
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Z Application Dependent systems probably seem very abstract to you—a solution like the one in Example 8 doesn’t seem like it would apply to any real–world situations. But in Example 9, we will solve a problem where a dependent system leads to real solutions.
EXAMPLE
9
Purchasing A chemical manufacturer plans to purchase a fleet of 24 railroad tank cars with a combined carrying capacity of 250,000 gallons. Tank cars with three different carrying capacities are available: 6,000 gallons, 8,000 gallons, and 18,000 gallons. How many of each type of tank car should be purchased?
SOLUTION
Let x1 0002 Number of 6,000-gallon tank cars x2 0002 Number of 8,000-gallon tank cars x3 0002 Number of 18,000-gallon tank cars Then x1 0003 x2 0003 x3 0002 24 6,000x1 0003 8,000x2 0003 18,000x3 0002 250,000
Total number of tank cars Total carrying capacity
Now we can form the augmented matrix of the system and solve by using Gauss–Jordan elimination: 1 6,000
1 8,000
⬃ c
1 6
1 8
1 24 ` d 18 250
(ⴚ6)R1 ⴙ R2 S R2
⬃ c
1 0
1 2
1 24 ` d 12 106
1 2 R2
⬃ c
1 0
1 1
1 24 ` d 6 53
(ⴚ1)R2 ⴙ R1 S R1
⬃ c
1 0
0 1
00045 000429 ` d 6 53
c
x1
1 24 ` d 18,000 250,000
1 1,000 R2
S R2 (simplify R2)
S R2
Matrix is in reduced form.
0004 5x3 0002 000429 x2 0003 6x3 0002 53
or or
x1 0002 5x3 0004 29 x2 0002 00046x3 0003 53
Let x3 0002 t. Then for t any real number, x1 0002 5t 0004 29 x2 0002 00046t 0003 53 x3 0002 t is a solution—or is it? Since the variables in this system represent the number of tank cars purchased, the values of x1, x2, and x3 must be nonnegative integers. The third equation requires that t must be a nonnegative integer. The first equation requires that 5t 0004 29 0007 0, so t must be at least 6. The middle equation requires that 00046t 0003 53 0007 0, so t can be no larger than 8.
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So, 6, 7, and 8 are the only possible values for t. There are three different possible combinations that meet the company’s specifications of 24 tank cars with a total carrying capacity of 250,000 gallons, as shown in Table 1: Table 1
t
6,000-Gallon Tank Cars x1
8,000-Gallon Tank Cars x2
18,000-Gallon Tank Cars x3
6
1
17
6
7
6
11
7
8
11
5
8
The final choice would probably be influenced by other factors. For example, the company might want to minimize the cost of the 24 tank cars. 0002 MATCHED PROBLEM 9
A commuter airline plans to purchase a fleet of 30 airplanes with a combined carrying capacity of 960 passengers. The three available types of planes carry 18, 24, and 42 passengers, respectively. How many of each type of plane should be purchased? 0002 ANSWERS TO MATCHED PROBLEMS 1. (A) c
00041 2 00043 ` d 3 00045 8
(B) c
0 7
00042 00045
2 (C) £ 3 1
2 00044 ` d 3 0
00041 1 4 4 0 † 6§ 0 5 00043
00046 00043 1 00042 3 1 00042 3 ` d ` d ` d (B) c (C) c 00042 3 1 00042 00041 0 0 000412 3. (A) x1 0002 5, x2 0002 00047, x3 0002 0 or (5, 00047, 0) (B) x1 0002 3s 0003 5, x2 0002 00044s 0004 7, x3 0002 s, s any real number; or {(3s 0003 5, 00044s 0004 7, s) | s any real number} (C) No solution 4. (A) Condition 2 is violated: The 3 in row 2 and column 2 should be a 1. Perform the operation 1 3 R2 S R2 to obtain: 2. (A) c
3 1
c
1 0
0 2 ` d 1 00042
(B) Condition 3 is violated: The 5 in row 1 and column 2 should be a 0. Perform the operation (00045)R2 0003 R1 S R1 to obtain: 1 £0 0
00046 8 2 † 00041 § 0 0
0 1 0
(C) Condition 4 is violated: The leftmost 1 in the second row is not to the right of the leftmost 1 in the first row. Perform the operation R1 4 R2 to obtain: 1 £0 0
0 1 0
0 0 0 † 00043 § 1 2
(D) Condition 1 is violated: The all-zero second row should be at the bottom. Perform the operation R2 4 R3 to obtain: 1 £0 0
2 0 0
0 3 1 † 4§ 0 0
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x1 0002 1, x2 0002 00041, x3 0002 0 or (1, 00041, 0) No solution x1 0002 5t 0003 4, x2 0002 3t 0003 5, x3 0002 t, t any real number; or {(5t 0003 4, 3t 0003 5, t) | t any real number} x1 0002 s 0003 7, x2 0002 s, x3 0002 t 0004 2, x4 0002 00043t 0004 1, x5 0002 t, s and t any real numbers; or {(s 0003 7, s, t 0004 2, 00043t 0004 1, t) | s and t any real numbers} 18-Passenger 24-Passenger 42-Passenger 9. Planes Planes Planes t x1 x2 x3
5. 6. 7. 8.
7-2
14
2
14
14
15
5
10
15
16
8
6
16
17
11
2
17
Exercises
1. What is the size of a matrix?
In Problems 19–26, write the linear system corresponding to each reduced augmented matrix and solve.
2. What is a row matrix? What is its size? 3. What is a column matrix? What is its size? 4. What is a square matrix? 5. What does aij mean? 6. What is the principal diagonal of a matrix? 7. What is an augmented coefficient matrix? 8. What operations can you perform on an augmented coefficient matrix to produce a row-equivalent matrix? 9. What is a reduced matrix and how is it used to solve a system of linear equations? 10. Describe the Gauss–Jordan elimination process in your own words. In Problems 11–18, indicate whether each matrix is in reduced form. 11. c
1 0
0 13. £ 0 0
0 00041 ` d 2 6 1 0 0
00042 0 0 † 1§ 0 0
0 15. £ 0 1
0 1 0
1 2 0 † 00045 § 0 4
0 17. c 0
1 0
6 0
0 00048 ` d 1 1
12. c
1 0
1 14. £ 0 0
0 1 0
0 00042 0 † 3§ 1 0
1 0 20. ≥ 0 0
0 1 0 0
1 21. £ 0 0
0 1 0
00042 3 1 † 00045 § 0 0
1 22. £ 0 0
00042 0 00043 0 1 † 5§ 0 0 0
1 23. £ 0 0
0 0 1 † 0§ 0 1
1 24. £ 0 0
0 5 1 † 00043 § 0 0
1 0
00042 0 0 1
25. c
00043 00045 ` d 3 2
1 16. £ 0 0
00042 4 1 0 1 † 00043 § 0 0 0
0 18. c 0
0 0
1 0 ` d 0 0
1 0
0 1
00042 00041
3 4 ` d 2 00041
Perform each of the row operations indicated in Problems 27–38 on the following matrix:
0 5 ` d 1 00043 00041 4 0 0 0 † 0§ 0 0 1
26. c
0 0 1 0
0 00042 0 0 ¥ ∞ 1 0 3 1
1 19. £ 0 0
c
1 4
00043 2 ` d 00046 00048
27. R1 4 R2
28. 12R2 S R2
29. 00044R1 S R1
30. 00042R1 S R1
31. 2R2 S R2
32. 00041R2 S R2
(000412)R2
33. (00044)R1 0003 R2 S R2
34.
0003 R1 S R1
35. (00042)R1 0003 R2 S R2
36. (00043)R1 0003 R2 S R2
37. (00041)R1 0003 R2 S R2
38. 1R1 0003 R2 S R2
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SECTION 7–2
Solving Systems of Linear Equations Using Gauss–Jordan Elimination
Use row operations to change each matrix in Problems 39–44 to reduced form. 39. c
1 0
2 00041 ` d 1 3
1 41. £ 0 0
0 1 0
1 43. £ 0 0
2 3 00041
40. c
00043 1 2 † 0§ 3 00046 00042 00041 00046 † 1 § 2 000413
1 0
3 1 ` d 2 00044
1 42. £ 0 0
0 1 0
0 44. £ 2 0
00042 00042 00041
4 0 00043 † 00041 § 00042 2 8 1 6 † 00044 § 1 4 2
63.
x1 0004 4x2 0002 00042 00042x1 0003 x2 0002 00043
47. x1 0003 2x2 0002 4 2x1 0003 4x2 0002 00048 49.
3x1 0004 6x2 0002 00049 00042x1 0003 4x2 0002 6
46.
x1 0004 3x2 0002 00045 00043x1 0004 x2 0002 5
48.
2x1 0004 3x2 0002 00042 00044x1 0003 6x2 0002 7
50.
2x1 0004 4x2 0002 00042 00043x1 0003 6x2 0002 3
65. x1 0003 2x2 0004 4x3 0004 x4 0002 7 2x1 0003 5x2 0004 9x3 0004 4x4 0002 16 x1 0003 5x2 0004 7x3 0004 7x4 0002 13 66. 2x1 0003 4x2 0003 5x3 0003 4x4 0002 8 x1 0003 2x2 0003 2x3 0003 x4 0002 3 67.
x1 0004 00042x1 0003 3x1 0004 4x1 0004
68.
x1 0003 x2 0003 4x3 0003 x4 0002 1.3 0002 1.1 0004x1 0003 x2 0004 x3 2x1 0003 x3 0003 3x4 0002 00044.4 2x1 0003 5x2 0003 11x3 0003 3x4 0002 5.6
69.
x1 0004 2x2 0003 00042x1 0003 4x2 0003 3x1 0004 6x2 0003 0004x1 0003 2x2 0003
70.
x1 0004 3x2 0003 x3 0003 x4 0003 2x5 0002 2 0004x1 0003 5x2 0003 2x3 0003 2x4 0004 2x5 0002 0 2x1 0004 6x2 0003 2x3 0003 2x4 0003 4x5 0002 4 0004x1 0003 3x2 0004 x3 0004 x5 0002 00043
51. 2x1 0003 4x2 0004 10x3 0002 00042 3x1 0003 9x2 0004 21x3 0002 0 x1 0003 5x2 0004 12x3 0002 1 52. 3x1 0003 5x2 0004 x3 0002 00047 x1 0003 x2 0003 x3 0002 00041 2x1 0003 11x3 0002 7 53. 3x1 0003 8x2 0004 x3 0002 000418 8 2x1 0003 x2 0003 5x3 0002 2x1 0003 4x2 0003 2x3 0002 00044
x2 0003 4x2 0004 x2 0003 3x2 0003
3x3 0004 2x4 0002 1 3x3 0003 x4 0002 0.5 10x3 0004 4x4 0002 2.9 8x3 0004 2x4 0002 0.6
x3 0003 2x3 0003 x3 0003 3x3 0003
x4 0003 2x5 0002 2 2x4 0004 2x5 0002 0 x4 0003 5x5 0002 4 x4 0003 x5 0002 3
71. Consider a consistent system of three linear equations in three variables. Discuss the nature of the solution set for the system if the reduced form of the augmented coefficient matrix has (A) One leftmost 1 (B) Two leftmost 1’s (C) Three leftmost 1’s (D) Four leftmost 1’s
54. 2x1 0003 7x2 0003 15x3 0002 000412 4x1 0003 7x2 0003 13x3 0002 000410 3x1 0003 6x2 0003 12x3 0002 00049 55. 2x1 0004 x2 0004 3x3 0002 8 x1 0004 2x2 00027 56. 2x1 0003 4x2 0004 6x3 0002 10 3x1 0003 3x2 0004 3x3 0002 6 57. 2x1 0004 x2 0002 0 3x1 0003 2x2 0002 7 x1 0004 x2 0002 00041
7 2x1 0004 5x2 0004 3x3 0002 6 00044x1 0003 10x2 0003 2x3 0002 6x1 0004 15x2 0004 x3 0002 000419
64. 5x1 0004 3x2 0003 2x3 0002 13 2x1 0004 x2 0004 3x3 0002 1 4x1 0004 2x2 0003 4x3 0002 12
Solve Problems 45–70 using Gauss–Jordan elimination. 45.
455
58. 2x1 0004 x2 0002 0 3x1 0003 2x2 0002 7 x1 0004 x2 0002 00042
59. 3x1 0004 4x2 0004 x3 0002 1 2x1 0004 3x2 0003 x3 0002 1 x1 0004 2x2 0003 3x3 0002 2 60. 00042x1 0003 x2 0003 3x3 0002 00047 x1 0004 4x2 0003 2x3 0002 0 x1 0004 3x2 0003 x3 0002 1 61.
2x1 0004 2x2 0004 4x3 0002 00042 00043x1 0003 3x2 0003 6x3 0002 3
62.
3 4x1 0004 x2 0003 2x3 0002 00044x1 0003 x2 0004 3x3 0002 000410 8x1 0004 2x2 0003 9x3 0002 00041
72. Consider a system of three linear equations in three variables. Give examples of two reduced forms that are not row equivalent if the system is (A) Consistent and dependent (B) Inconsistent
APPLICATIONS 73. BUYING Suppose that you have a $129 credit on your account at Amazon.com, and you want to spend it all on sale CDs at $10 each, sale DVDs at $12 each, and sale books at $7 each. If you buy 13 items total, how many will you buy of each? 74. PETTY CRIME Shady Grady finds a parking meter with a broken lock and scoops out the change inside. The meter accepts nickels, dimes, and quarters, and there were 32 coins inside with a total value of $6.80. How many of each type of coin did Grady get? 75. CHEMISTRY A chemist has two solutions of sulfuric acid: a 20% solution and an 80% solution. How much of each should be used to obtain 100 liters of a 62% solution?
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76. CHEMISTRY A chemist has two solutions: one containing 40% alcohol and another containing 70% alcohol. How much of each should be used to obtain 80 liters of a 49% solution? 77. GEOMETRY Find a, b, and c so that the graph of the parabola with equation y ⫽ a ⫹ bx ⫹ cx2 passes through the points (⫺2, 3), (⫺1, 2), and (1, 6). 78. GEOMETRY Find a, b, and c so that the graph of the parabola with equation y ⫽ a ⫹ bx ⫹ cx2 passes through the points (1, 3), (2, 2), and (3, 5). 79. PRODUCTION SCHEDULING A small manufacturing plant makes three types of inflatable boats: one-person, two-person, and four-person models. Each boat requires the services of three departments, as listed in the table. The cutting, assembly, and packaging departments have available a maximum of 380, 330, and 120 laborhours per week, respectively. How many boats of each type must be produced each week for the plant to operate at full capacity? One-Person Boat
Two-Person Boat
Four-Person Boat
Cutting department
0.5 h
1.0 h
1.5 h
Assembly department
0.6 h
0.9 h
1.2 h
Packaging department
0.2 h
0.3 h
0.5 h
80. PRODUCTION SCHEDULING Repeat Problem 79 assuming the cutting, assembly, and packaging departments have available a maximum of 350, 330, and 115 labor-hours per week, respectively. 81. PRODUCTION SCHEDULING Rework Problem 79 assuming the packaging department is no longer used. 82. PRODUCTION SCHEDULING Rework Problem 80 assuming the packaging department is no longer used. 83. PRODUCTION SCHEDULING Rework Problem 79 assuming the four-person boat is no longer produced. 84. PRODUCTION SCHEDULING Rework Problem 80 assuming the four-person boat is no longer produced. 85. NUTRITION A dietitian in a hospital is to arrange a special diet using three basic foods. The diet is to include exactly 340 units of calcium, 180 units of iron, and 220 units of vitamin A. The number of units per ounce of each special ingredient for each of the foods is indicated in the table. How many ounces of each food must be used to meet the diet requirements?
88. NUTRITION Solve Problem 86 with the assumption that food C is no longer available. 89. NUTRITION Solve Problem 85 assuming the vitamin A requirement is deleted. 90. NUTRITION Solve Problem 86 assuming the vitamin A requirement is deleted. 91. SOCIOLOGY Two sociologists have grant money to study school busing in a particular city. They wish to conduct an opinion survey using 600 telephone contacts and 400 house contacts. Survey company A has personnel to do 30 telephone and 10 house contacts per hour; survey company B can handle 20 telephone and 20 house contacts per hour. How many hours should be scheduled for each firm to produce exactly the number of contacts needed? 92. SOCIOLOGY Repeat Problem 91 if 650 telephone contacts and 350 house contacts are needed. 93. DELIVERY CHARGES United Express, a nationwide package delivery service, charges a base price for overnight delivery of packages weighing 1 pound or less and a surcharge for each additional pound (or fraction thereof). A customer is billed $27.75 for shipping a 5-pound package and $64.50 for shipping a 20-pound package. Find the base price and the surcharge for each additional pound. 94. DELIVERY CHARGES Refer to Problem 93. Federated Shipping, a competing overnight delivery service, informs the customer in Problem 93 that it would ship the 5-pound package for $29.95 and the 20-pound package for $59.20. (A) If Federated Shipping computes its cost in the same manner as United Express, find the base price and the surcharge for Federated Shipping. (B) Devise a simple rule that the customer can use to choose the cheaper of the two services for each package shipped. Justify your answer. 95. RESOURCE ALLOCATION A coffee manufacturer uses Colombian and Brazilian coffee beans to produce two blends, robust and mild. A pound of the robust blend requires 12 ounces of Colombian beans and 4 ounces of Brazilian beans. A pound of the mild blend requires 6 ounces of Colombian beans and 10 ounces of Brazilian beans. Coffee is shipped in 132-pound burlap bags. The company has 50 bags of Colombian beans and 40 bags of Brazilian beans on hand. How many pounds of each blend should it produce in order to use all the available beans?
Units per Ounce Food A
Food B
Food C
Calcium
30
10
20
Iron
10
10
20
Vitamin A
10
30
20
86. NUTRITION Repeat Problem 85 if the diet is to include exactly 400 units of calcium, 160 units of iron, and 240 units of vitamin A. 87. NUTRITION Solve Problem 85 with the assumption that food C is no longer available.
96. RESOURCE ALLOCATION Refer to Problem 95. (A) If the company decides to discontinue production of the robust blend and only produce the mild blend, how many pounds of the mild blend can it produce and how many beans of each type will it use? Are there any beans that are not used? (B) Repeat part A if the company decides to discontinue production of the mild blend and only produce the robust blend.
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7-3
Matrix Operations
457
Matrix Operations Z Adding and Subtracting Matrices Z Multiplying a Matrix by a Number Z Finding the Product of Two Matrices
In Section 7-2, we introduced basic matrix terminology and solved systems of equations by performing row operations on augmented coefficient matrices. Matrices have many other useful applications and possess an interesting mathematical structure in their own right. As we will see, matrix addition and multiplication are similar to real number addition and multiplication in many respects, but there are some important differences.
Z Adding and Subtracting Matrices Before we can discuss arithmetic operations for matrices, we have to define equality for matrices. Two matrices are equal if they have the same size and their corresponding elements are equal. For example, 200023
c
200023
a b c u v w d 0002 c d d e f x y z
if and only if
a0002u b0002v c0002w d0002x e0002y f0002z
The sum of two matrices of the same size is a matrix with elements that are the sums of the corresponding elements of the two given matrices. Addition is not defined for matrices of different sizes.
EXAMPLE
1
Matrix Addition Add:
SOLUTIONS
2 (A) c 1
00033 2
0 3 d 0004 c 00035 00033
2 (A) c 1
00033 2
0 3 1 2 d 0004 c d 00035 00033 2 5
1 2
2 d 5
2 (B) c 3
2 3
1 2
0 4 d 0004 £ 00033 00033 00031
(2 0004 3) (00033 0004 1) 0002 c (1 0003 3) (2 0004 2) 0002 c
(B) c
1 2
2 5§ 4
(0 0004 2) d (00035 0004 5)
*
5 00032 2 d 00032 4 0
0 2 4 d 0004 £ 00033 5 § 00033 00031 4
Because the first matrix is 2 0005 3 and the second is 3 0005 2, this sum is not defined. *Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
0002
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MATCHED PROBLEM 1
Add: 3 2 00032 3 (A) £ 00031 00031 § 0004 £ 1 00031 § 0 3 2 00032
(B) [1 00032 7] 0004 [ 00032 4
3
00031] 0002
Technology Connections Graphing calculators can be used to solve problems involving matrix operations. Figure 1 illustrates the solutions to Example 1A and 1B on a graphing calculator.
(a) Example 1A
(b) Example 1B
Z Figure 1 Matrix addition on a graphing calculator.
Because we add two matrices by adding their corresponding elements (which are real numbers), it follows from the properties of real numbers that matrices of the same size are commutative and associative relative to addition. That is, if A, B, and C are matrices of the same size, then A0004B0002B0004A (A 0004 B) 0004 C 0002 A 0004 (B 0004 C) 0 £0 0
0 0 0
0 0 0
M0002 c
a b d c d
then
0 d 0
[0 0
Associative
A matrix with elements that are all 0’s is called a zero matrix. Examples of zero matrices are shown in Figure 2. [Note: “0” is often used to denote the zero matrix of any size.] The negative of a matrix M, denoted by 0003M, is a matrix with elements that are the negatives of the elements in M. So if
0 0§ 0
0 0 ≥ ¥ 0 0 0 c 0
Commutative
0003M 0002 c 0]
0003a 0003c
0003b d 0003d
Based on our definition of addition, M 0004 (0003M) 0002 0 (a zero matrix). If A and B are matrices of the same size, then we define subtraction as follows.
Z Figure 2 Zero matrices.
A 0003 B 0002 A 0004 (0003B) To subtract matrix B from matrix A, we subtract corresponding elements.
EXAMPLE
2
Matrix Subtraction Subtract: c
3 5
00032 00032 d 0003 c 0 3
2 d 4
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SECTION 7–3
c
SOLUTION
MATCHED PROBLEM 2
EXAMPLE
3
00032 00032 d 0003 c 0 3
3 5
00033
Subtract: [2
Matrix Operations
2 3 0003 (00032) 00032 0003 2 5 d 0002 c d 0002 c 4 500033 000034 2
5] 0003 [3
00032
459
00034 d 00034
0002
1] 0002
Matrix Equations Find a, b, c, and d so that c
c
3 d 4
3 d 4
Subtract the matrices on the left side.
a00032 b 0003 (00031) 4 d 0002 c c 0003 (00035) d 0003 6 00032
3 d 4
Simplify.
c
a00032 b00041 4 3 d 0002 c d c00045 d00036 00032 4
a0003200024 a00036 MATCHED PROBLEM 3
00031 4 d 0002 c 6 00032
a b 2 00031 4 d 0003 c d 0002 c c d 00035 6 00032
c
SOLUTION
a b 2 d 0003 c c d 00035
b0004100023 b00032
Set corresponding elements equal to each other.
c 0004 5 0002 00032 c 0003 00047
d000360002 4 d 0003 10
0002
Find a, b, c, and d so that c
a b 00034 2 00032 5 d 0003 c d 0002 c d c d 1 00033 8 2 0002
Z Multiplying a Matrix by a Number The product of a number k and a matrix M, denoted by kM, is a matrix formed by multiplying each element of M by k.
EXAMPLE
4
Multiplying a Matrix by a Number 3 Multiply: 00032 £ 00032 0
SOLUTION
MATCHED PROBLEM 4
3 00032 £ 00032 0
00031 1 00031
00031 1 00031
0 3§ 00032
0 00036 3§ 0002 £ 4 00032 0
2 00032 2
0 00036 § 4
0002
1.3 Multiply: 10 £ 0.2 § 3.5 0002
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ZZZ EXPLORE-DISCUSS 1
Multiplication of two numbers can be interpreted as repeated addition if one of the numbers is a positive integer. That is, 2a 0002 a 0004 a
3a 0002 a 0004 a 0004 a
4a 0002 a 0004 a 0004 a 0004 a
and so on. How does this apply to multiplication of a matrix by a number?
Matrix operations have many applications, particularly in business.
EXAMPLE
5
Sales and Commissions Ms. Fong and Mr. Petris are salespeople for a new car agency that sells only two models. August was the last month for this year’s models, and next year’s models were introduced in September. Gross dollar sales for each month are given in the following matrices: AUGUST SALES Compact
c
Fong Petris
SEPTEMBER SALES
Luxury
$36,000 $72,000
Compact
$72,000 d 0002A $0
c
Luxury
$144,000 $180,000
$288,000 d 0002B $216,000
For example, Ms. Fong had $36,000 in compact sales in August and Mr. Petris had $216,000 in luxury car sales in September. (A) What were the combined dollar sales in August and September for each salesperson and each model? (B) What was the increase in dollar sales from August to September? (C) If both salespeople receive a 3% commission on gross dollar sales, compute the commission for each salesperson for each model sold in September. SOLUTIONS
We use matrix addition for part A, matrix subtraction for part B, and multiplication of a matrix by a number for part C. Compact
Luxury
(A) A 0004 B 0002 c
$180,000 $252,000
$360,000 d $216,000
(B) B 0003 A 0002 c
$108,000 $108,000
$216,000 d $216,000
Compact
Fong
Sum of sales for August and September
Petris September sales 0004 August sales
Fong Petris Luxury
(0.03)($144,000) (0.03)($288,000) d (0.03)($180,000) (0.03)($216,000) Fong $4,320 $8,640 0002 c d Petris $5,400 $6,480
(C) 0.03B 0002 c
MATCHED PROBLEM 5
3% of September sales
0002
Repeat Example 5 with A0002 c
$72,000 $36,000
$72,000 d $72,000
and
B0002 c
$180,000 $144,000
$216,000 d $216,000
0002
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Example 5 involved an agency with only two salespeople and two models. A more realistic problem might involve 20 salespeople and 15 models. Problems of this size are often solved using spreadsheets on a computer. Figure 3 illustrates a spreadsheet solution to Example 5. A 1
B
2
Luxury
September Commissions
September Sales
August Sales
G
Compact
Luxury
Compact
Luxury
F
E
D
C
Compact
3
Fong
$36,000
$72,000
$144,000
$288,000
$4,320
$8,640
4
Petris
$72,000
$0
$180,000
$216,000
$5,400
$6,480
5
Sales Increases
Combined Sales
6
Fong
$180,000
$360,000
$108,000
$216,000
7
Petris
$252,000
$216,000
$108,000
$216,000
Z Figure 3
Z Finding the Product of Two Matrices Next we will define a way to multiply two matrices. It will probably seem strange to you at first; eventually you will see examples of why it is useful in many problems. In particular, matrix multiplication will help us to develop an alternative method for solving linear systems that have the same number of variables and equations. We start by defining the product of two special matrices, a row matrix and a column matrix. Z DEFINITION 1 Product of a Row Matrix and a Column Matrix The product of a 1 0005 n row matrix and an n 0005 1 column matrix is a 1 0005 1 matrix given by n00021
b1 b2 [ a1 a2 . . . an ] ≥ ¥ 0002 [ a1b1 0004 a2b2 0004 . . . 0004 anbn ] o bn 10002n
Note that the number of elements in the row matrix and in the column matrix must be the same for the product to be defined.
EXAMPLE
6
Product of a Row Matrix and a Column Matrix 00035 Multiply: [2 00033 0] £ 2 § 00032
SOLUTION
00035 [2 00033 0] £ 2 § 0002 [(2)(00035) 0004 (00033)(2) 0004 (0)(00032)] 00032 0002 [000310 0003 6 0004 0] 0002 [ 000316]
MATCHED PROBLEM 6 Multiply: [00031 0
3
2 3 2] ≥ ¥ 4 00031
0002
0002
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The answer to Example 6 is a 1 ⫻ 1 matrix, which we represented with [⫺16]. From now on, if the result of a calculation is a 1 ⫻ 1 matrix, we’ll usually omit the brackets and write the answer as a real number.
EXAMPLE
7
Production Scheduling A factory produces a slalom water ski that requires 4 labor-hours in the fabricating department and 1 labor-hour in the finishing department. Fabricating personnel receive $10 per hour, and finishing personnel receive $8 per hour. Find the total labor cost per ski.
SOLUTION
Total labor cost per ski is given by the product [4 1] c
MATCHED PROBLEM 7
10 d ⫽ [(4)(10) ⫹ (1)(8)] ⫽ [40 ⫹ 8] ⫽ [48] or $48 per ski 8
0002
If the factory in Example 7 also produces a trick water ski that requires 6 labor-hours in the fabricating department and 1.5 labor-hours in the finishing department. Find the totallabor cost per ski by multiplying an appropriate row matrix and column matrix. 0002 We will now use the product of a 1 ⫻ n row matrix and an n ⫻ 1 column matrix to extend the definition of matrix product to more general matrices.
Z DEFINITION 2 Matrix Product If A is an m ⫻ p matrix and B is a p ⫻ n matrix, then the matrix product of A and B, denoted AB, is an m ⫻ n matrix whose element in the ith row and jth column is the real number obtained from the product of the ith row of A and the jth column of B. If the number of columns in A does not equal the number of rows in B, then the matrix product AB is not defined.
Must be the same (b ⫽ c) a⫻b c⫻d
Size of product a⫻d
A • B ⴝ AB
It is important to check sizes before starting the multiplication process. If A is an a ⫻ b matrix and B is a c ⫻ d matrix, then if b ⫽ c, the product AB will exist and will be an a ⫻ d matrix (see Fig. 4). If b ⫽ c, then the product AB does not exist. The definition is not as complicated as it looks. An example should help clarify the process. For
Z Figure 4
2 A⫽ c ⫺2
3 1
⫺1 d 2
and
1 B⫽ £ 2 ⫺1
3 0§ 2
A is 2 ⫻ 3, B is 3 ⫻ 2, and so AB is 2 ⫻ 2. To find the first row of AB, we take the product of the first row of A with every column of B and write each result as a real number, not a 1 ⫻ 1 matrix. The second row of AB is computed in the same manner. The four products of row and column matrices used to produce the four elements in AB are shown in the dashed box below. These products are usually calculated mentally, or with the aid of a calculator, and need not be written out. The shaded portions highlight the steps involved in computing the element in the first row and second column of AB.
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300052
200053
冤
2 00032
00031 2
冥
3 1
冤
3 0 2
1 2 00031
0002
冤 冥 冤 冥
1 3 00031兴 2 00031 1 1 2兴 2 00031
关2
冥
0002
关 00032
(2)(1) 0004 (3)(2) 0004 (00031)(00031) 冤 (00032)(1) 0004 (1)(2) 0004 (2)(00031)
463
Matrix Operations
关2
3
关 00032
1
00031兴
2兴
冤 冥 冤 冥 3 0 2 3 0 2
(2)(3) 0004 (3)(0) 0004 (00031)(2) (00032)(3) 0004 (1)(0) 0004 (2)(2)
冥
200052
0002
EXAMPLE
8
冤000329
冥
4 00032
Matrix Multiplication Given 2 A0002 £ 1 00031
1 0§ 2
B0002 c
00031 1
1 2
0 2
1 d 0
C0002 c
2 00031
6 d 00033
D0002 c
1 3
2 d 6
Find each product that is defined: (A) AB
(B) BA
(C) CD
300022
SOLUTIONS
2 (A) AB 0002 £ 1 00031
200024
1 1 0§ c 2 2
00031 1
(2)(1) 0004 (1)(2) 0002 £ (1)(1) 0004 (0)(2) (00031)(1) 0004 (2)(2) 4 0002 £1 3
300024
00031 2 00031 0 3 4 00031 1
0 2
1 d 0
(2)(00031) 0004 (1)(1) (1)(00031) 0004 (0)(1) (00031)(00031) 0004 (2)(1)
(2)(0) 0004 (1)(2) (1)(0) 0004 (0)(2) (00031)(0) 0004 (2)(2)
(2)(1) 0004 (1)(0) (1)(1) 0004 (0)(0) § (00031)(1) 0004 (2)(0)
2 1§ 00031 300022
200024
1 (B) BA 0002 c 2
(D) DC
2 1 d£ 1 0 00031
0 2
1 0§ 2
Product is not defined. 200022
2 (C) CD 0002 c 00031
200022
6 1 dc 00033 3
2 d 6
0002c
(2)(1) 0004 (6)(3) (2)(2) 0004 (6)(6) d (00031)(1) 0004 (00033)(3) (00031)(2) 0004 (00033)(6) 200022
20 40 0002c d 000310 000320
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1 (D) DC 0002 c 3
MATCHED PROBLEM 8
200022
2 2 dc 6 00031
200022
6 d 00033
(1)(2) 0004 (2)(00031) (1)(6) 0004 (2)(00033) 0002 c d (3)(2) 0004 (6)(00031) (3)(6) 0004 (6)(00033)
0 0002 c 0
0 d 0
0002
Find each product, if it is defined: (A) c (C) c
00031 1
1 00031
00031 1 00031 1 00032 00031 0 3 00032 d £ 2 3§ (B) £ 2 3 § c d 0 1 2 2 0 1 0 1 0 2 00032 4 00032 4 1 2 dc d (D) c dc d 00032 1 00032 1 00032 00031 00032 0 2
3 2
0002
In the arithmetic of real numbers, it doesn’t matter in which order we multiply; for example, 5 0005 7 0002 7 0005 5. In matrix multiplication, however, it does make a difference. That is, AB does not always equal BA, even if both multiplications are defined and both products are the same size (see Examples 8C and 8D). In other words, Matrix multiplication is not commutative. Also, AB may be zero with neither A nor B equal to zero (see Example 8D). That is, The zero property does not hold for matrix multiplication. (See Section R-1 for a discussion of the zero property for real numbers.) Just as we used the familiar algebraic notation AB to represent the product of matrices A and B, we use the notation A2 for AA (the product of A with itself ), A3 for AAA, and so on.
ZZZ EXPLORE-DISCUSS 2
In addition to the commutative and zero properties, there are other significant differences between real number multiplication and matrix multiplication. (A) In real number multiplication, the only real number whose square is 0 is the real number 0 (02 0002 0). Find at least one 2 0005 2 matrix A with all elements nonzero such that A2 0002 0, where 0 is the 2 0005 2 zero matrix. (B) In real number multiplication, the only nonzero real number that is equal to its square is the real number 1 (12 0002 1). Find at least one 2 0005 2 matrix A with all elements nonzero such that A2 0002 A.
We’ll return to our study of the properties of matrix multiplication in Section 7-4. We will conclude this section with an application of matrix multiplication.
EXAMPLE
9
Labor Costs If we combine the time requirements for making slalom and trick water skis discussed in Example 7 and Matched Problem 7, we get Labor-hours per ski Assembly Finishing department department Trick ski Slalom ski
c
6h 4h
1.5 h d 0002L 1h
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465
Now suppose that the company has two manufacturing plants, X and Y, in different parts of the country and that the hourly rates for each department are given in the following matrix: Hourly Wages Plant Plant X Y Assembly department Finishing department
c
$10 $ 8
$12 d 0002H $10
Find the matrix products HL and LH, and decide if either matrix has a meaningful interpretation in terms of ski production. SOLUTION
Since H and L are both 2 0005 2 matrices, we can find the product of H and L in either order and the result will be a 2 0005 2 matrix: HL 0002 c
10 8
12 6 dc 10 4
1.5 108 d 0002 c 1 88
LH 0002 c
6 4
1.5 10 dc 1 8
12 72 d 0002 c 10 48
27 d 22 87 d 58
How can we interpret the elements in these products? Let’s begin with the product HL. The element 108 in the first row and first column of HL is the product of the first row matrix of H and the first column matrix of L: Plant Plant X Y
6 [10 12] c d 4
Trick Slalom
0002 10(6) 0004 12(4) 0002 60 0004 48 0002 108
Notice that $60 is the labor cost for assembling a trick ski at Plant X and $48 is the labor cost for assembling a slalom ski at Plant Y. Although both numbers represent labor costs, it makes no sense to add them together. They do not pertain to the same type of ski or to the same plant. So, even though the product HL happens to be defined mathematically, it has no useful interpretation in this problem. Now let’s consider the product LH. The element 72 in the first row and first column of LH is given by the following product: Assembly Finishing
[6
1.5] c
10 d 8
Assembly Finishing
0002 6(10) 0004 1.5(8) 0002 60 0004 12 0002 72
where $60 is the labor cost for assembling a trick ski at Plant X and $12 is the labor cost for finishing a trick ski at Plant X. The sum is the total labor cost for producing a trick ski at Plant X. The other elements in LH also represent total labor costs, as indicated by the row and column labels shown below: Labor costs per ski Plant Plant X Y
LH 0002 c MATCHED PROBLEM 9
$72 $48
$87 d $58
Trick ski
0002
Slalom ski
Refer to Example 9. The company wants to know how many hours to schedule in each department in order to produce 1,000 trick skis and 2,000 slalom skis. These production requirements can be represented by either of the following matrices: Trick skis
P 0002 [1,000
Slalom skis
2,000]
Q0002 c
1,000 d 2,000
Trick skis Slalom skis
Using the labor-hour matrix L from Example 9, find PL or LQ, whichever has a meaningful interpretation for this problem, and label the rows and columns accordingly. 0002
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Example 9 and Matched Problem 9 illustrate an important point about matrix multiplication. Even if you are using a graphing calculator to perform the calculations in a matrix product, you will still need to know the definition of matrix multiplication so that you can interpret the results correctly.
CAUTION ZZZ
ANSWERS TO MATCHED PROBLEMS 1 5 1. (A) £ 0 ⫺2 § (B) Not defined 2. [⫺1 ⫺1 4 ] 3. a ⫽ ⫺6, b ⫽ 7, c ⫽ 9, d ⫽ ⫺1 2 1 13 $252,000 $288,000 $108,000 $144,000 $5,400 $6,480 4. £ 2 § 5. (A) c (B) c (C) c d d d $180,000 $288,000 $108,000 $144,000 $4,320 $6,480 35 10 6. [8] 7. [6 1.5] c d ⫽ [72] or $72 8 2 2 ⫺1 2 0 0 ⫺6 ⫺12 8. (A) Not defined (B) £ 1 6 (C) c (D) c 12 ⫺4 § d d 0 0 3 6 ⫺1 0 3 ⫺2 9. Assembly Finishing Labor hours PL ⫽ [14,000 3,500]
7-3
Exercises
1. What conditions must matrics A and B satisfy so that A ⫹ B exists? 2. What conditions must matrices A and B satisfy so that AB exists? 3. What conditions must matrices A and B satisfy so that BA exists? 4. What conditions must matrices A and B satisfy so that both AB and BA exist? 5. What is the negative of a matrix? 6. How do you subtract two matrices?
0 ⫺1 3§ ⫹ £ 0 1 4
4 13. £ ⫺2 8 14. c
6 4
⫺2 3 3 d ⫹ c ⫺8 ⫺7 6
4 15. £ ⫺2 8
0 ⫺1 3§ ⫹ c 2 1
9. If A is a 1 ⫻ n matrix and B is an n ⫻ 1 matrix, how do you find the product BA? What is the size of BA? 10. Describe the operation of matrix multiplication in your own words. Perform the indicated operations in Problems 11–24, if possible. 5 11. c 3
⫺2 ⫺3 d ⫹ c 0 1
7 d ⫺6
0 12. c 2
8 9 d ⫹ c ⫺1 7
⫺4 d 5
9 ⫺2
0 5
3 3 d ⫹ £ 9 ⫺7 ⫺1
16. c
6 4
⫺2 ⫺8
17. c
5 4
⫺1 0 2 d ⫺ c 6 3 3
4 5
2 0 1 § ⫺ £ ⫺7 0 ⫺1
5 2§ 0
7. How do you multiply a matrix by a number? 8. If A is a 1 ⫻ n matrix and B is an n ⫻ 1 matrix, how do you find the product AB? What is the size of AB?
2 5§ ⫺6
6 18. £ ⫺4 3 4 19. £ 10 ⫺13 1
20. c 25 4
⫺7 4 11 § ⫺ c ⫺7 ⫺9
9 ⫺34 2 3d ⫺ c ⫺2 ⫺74
1 4 1d 2
⫺1 d 4
4 d ⫺6 6 ⫺2 § 4 ⫺6 d ⫺5
10 11
⫺13 d ⫺9
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00032.8 0
21. c
2.4 00031.6
22. c
10 d 0003 [20 20
24. 5 c
00037 4
23. 4 c
10]
3 00035
00032.2 00033.2
3.9 7 d 0003 c 4.2 00033.2
0 6
c
7 d 5
c
27. c
00035 d [4 00033
29. [3
28. c
1 00034 ] £ 2 § 00033
00032
1 31. £ 2 § [3 00032 00034 ] 00033 00036 3 1 dc d 33. c 2 00035 3 35. c
5 4
37. c
8 00035
1 2 dc 6 3
0 d 6
3x 00031
5 2y d 0002 c 4x 00036
2 0
00031 4
00031 C0004 £ 4 00032
3 d 00032
B0004 c
36. c
00032 3
38. c
7 0
0 2 00033 1 § 3 5
00033 2
3 D 0004 £0 1
00033 7 d 0004 c 0003y 00037
00032
2 2] £ 00031 § 1
00031 d 5
7 4 dc 00031 0 0 9 dc 3 00034
3 0 w d 0002 c 00037 000311 y
00032 d 00031
c
x 00031 4y d 0003 c 00032 y 5
2 d 2
x 9 d 0004 c z 4
1 d 6
1 d 5
4 12 d 0004 c 3x 00034
00035 d 000314
In Problems 63 and 64, let a, b, and c be any nonzero real numbers, and let A0004 c
b 1 d and I 0004 c 0003a 0
a c
0 d 1
63. If A2 0004 0, how are a, b, and c related? Use this relationship to provide several examples of 2 0005 2 matrices with no zero entries whose square is the zero matrix. 64. If A2 0004 I, how are a, b, and c related? Use this relationship to provide several examples of 2 0005 2 matrices with no zero entries whose square is the matrix I. Problems 65 and 66 refer to the matrices A0004 c
Problems 39–56 refer to the following matrices. A0004 c
00032 d 00034
62. Find x and y so that
00031]
2 32. £ 00031 § [1 00032 2] 1 3 7 4 d dc 34. c 00031 00039 00031
0 d 8
00033 2 dc 3 0
3 d [2 00034
30. [1
c
3 4] c d 00038
26. [ 00032
00032 ]
00033 1 d 0004 c 1 3
61. Find w, x, y, and z so that
Find the products in Problems 25–38. 25. [5
a b 2 d 0002 c c d 0
60. Find x and y so that
9 d 2
4 3] c d 7
467
59. Find a, b, c, and d so that
00032.2 d 1 00034 9
3 00032
Matrix Operations
a b 1 d and B 0004 c c d 1
1 d 1
65. If AB 0004 0, how are a, b, c, and d related? Use this relationship to provide several examples of 2 0005 2 matrices A with no zero entries that satisfy AB 0004 0.
00032 00031 § 2
Perform the indicated operations, if possible. 39. CA
40. AC
41. BA
42. AB
43. C 2
44. B2
45. C 0002 DA
46. B 0002 AD
47. 0.2CD
48. 0.1DB
49. 2DB 0002 5CD
50. 3BA 0002 4AC
51. (00031)AC 0002 3DB
52. (00032)BA 0002 6CD
53. CDA
54. ACD
55. DBA
56. BAD
66. If BA 0004 0, how are a, b, c, and d related? Use this relationship to provide several examples of 2 0005 2 matrices A with no zero entries that satisfy BA 0004 0. 67. Find x and y so that c
1 00032
3 x dc 00032 3
1 y d 0004 c 2 y
7 d 00036
68. Find x and y so that c
x 1
00031 2 dc 0 4
1 y d 0004 c 1 2
y d 1
69. Find a, b, c, and d so that
In Problems 57 and 58, use a graphing calculator to calculate B, B 2, B 3, . . . and AB, AB 2, AB 3, . . . . Describe any patterns you observe in each sequence of matrices. 57. A 0004 [ 0.3
0.4 0.7 ] and B 0004 c 0.2
0.6 d 0.8
58. A 0004 [ 0.4
0.6 ] and B 0004 c
0.9 0.3
0.1 d 0.7
c
1 1
3 a b 6 dc d 0004 c 4 c d 7
00035 d 00037
70. Find a, b, c, and d so that c
1 2
1 00032 a b dc d 0004 c 3 00033 c d
0 d 2
71. A square matrix is a diagonal matrix if all elements not on the principal diagonal are zero. So a 2 0005 2 diagonal matrix has the form a 0 d A0004 c 0 d
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where a and d are any real numbers. Discuss the validity of each of the following statements. If the statement is always true, explain why. If not, give examples. (A) If A and B are 2 0005 2 diagonal matrices, then A 0004 B is a 2 0005 2 diagonal matrix. (B) If A and B are 2 0005 2 diagonal matrices, then A 0004 B 0002 B 0004 A. (C) If A and B are 2 0005 2 diagonal matrices, then AB is a 2 0005 2 diagonal matrix. (D) If A and B are 2 0005 2 diagonal matrices, then AB 0002 BA. 72. A square matrix is an upper triangular matrix if all elements below the principal diagonal are zero. So a 2 0005 2 upper triangular matrix has the form A0002 c
a b d 0 d
where a, b, and d are any real numbers. Discuss the validity of each of the following statements. If the statement is always true, explain why. If not, give examples. (A) If A and B are 2 0005 2 upper triangular matrices, then A 0004 B is a 2 0005 2 upper triangular matrix. (B) If A and B are 2 0005 2 upper triangular matrices, then A 0004 B 0002 B 0004 A. (C) If A and B are 2 0005 2 upper triangular matrices, then AB is a 2 0005 2 upper triangular matrix. (D) If A and B are 2 0005 2 upper triangular matrices, then AB 0002 BA. 73. A company with two different plants makes satellite radios and GPS units. The production costs for each item are given in the following matrices: Plant X Radio GPS Materials Labor
c
$30 $60
Plant Y Radio GPS
$25 d 0002A $80
c
$36 $54
$27 d 0002B $74
Find the matrix 12 (A 0004 B), and explain what information it provides. 74. Suppose that the company in Problem 73 experiences an increase in the cost of both labor and materials at plant X. Find the matrix 12 (1.2A 0004 B). If it provides the average cost of production for the two plants, by how much were the costs at plant X increased? 75. MARKUP An import car dealer sells three models of a car. Current dealer invoice price (cost) and the retail price for the basic models and the indicated options are given in the following two matrices (where “Air” means air conditioning): Basic Car Model A Model B Model C
$10,400 £ $12,500 $16,400 Basic Car
Model A Model B Model C
$13,900 £ $15,000 $18,300
Dealer Invoice Price CD Air changer
$682 $721 $827
$215 $295 $443
Retail Price CD Air changer
$783 $838 $967
$263 $395 $573
Cruise Control
$182 $182 § 0002 M $192 Cruise Control
$215 $236 § 0002 N $248
We define the markup matrix to be N 0003 M (markup is the difference between the retail price and the dealer invoice price). Suppose the value of the dollar has had a sharp decline and the dealer invoice price is to have an across-the-board 15% increase next year. To stay competitive with domestic cars, the dealer increases the retail prices only 10%. Calculate a markup matrix for next year’s models and the indicated options. (Compute results to the nearest dollar.) 76. MARKUP Referring to Problem 75, what is the markup matrix resulting from a 20% increase in dealer invoice prices and an increase in retail prices of 15%? (Compute results to the nearest dollar.) 77. LABOR COSTS A company with manufacturing plants located in different parts of the country has labor-hour and wage requirements for the manufacturing of three types of inflatable boats as given in the following two matrices: Labor-Hours per Boat Cutting Assembly Packaging Department Department Department
0.6 h M 0002 £ 1.0 h 1.5 h
0.6 h 0.9 h 1.2 h
0.2 h 0.3 h § 0.4 h
One-person boat Two-person boat Four-person boat
Hourly Wages Plant I Plant II
$8 N 0002 £ $10 $5
$9 $12 § $6
Cutting department Assembly department Packaging department
(A) Find the labor costs for a one-person boat manufactured at plant I. (B) Find the labor costs for a four-person boat manufactured at plant II. (C) Discuss possible interpretations of the elements in the matrix products MN and NM. (D) If either of the products MN or NM has a meaningful interpretation, find the product and label its rows and columns. 78. INVENTORY VALUE A personal computer retail company sells five different computer models through three stores located in a large metropolitan area. The inventory of each model on hand in each store is summarized in matrix M. Wholesale (W ) and retail (R) values of each model computer are summarized in matrix N. A
4 M0002 £ 2 10
B
Model C D
2 3 4
3 5 3
W
$700 $1,400 N 0002 E$1,800 $2,700 $3,500
7 0 4
E
1 6§ 3
Store 1 Store 2 Store 3
R
$840 $1,800 $2,400U $3,300 $4,900
A B C D E
(A) What is the retail value of the inventory at store 2? (B) What is the wholesale value of the inventory at store 3? (C) Discuss possible interpretations of the elements in the matrix products MN and NM.
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(D) If either of the products MN or NM has a meaningful interpretation, find the product and label its rows and columns. (E) Discuss methods of matrix multiplication that can be used to find the total inventory of each model on hand at all three stores. State the matrices that can be used, and perform the necessary operations. (F) Discuss methods of matrix multiplication that can be used to find the total inventory of all five models at each store. State the matrices that can be used, and perform the necessary operations. 79. AIRFREIGHT A nationwide airfreight service has connecting flights between five cities, as illustrated in the figure. To represent this schedule in matrix form, we construct a 5 ⫻ 5 incidence matrix A, where the rows represent the origins of each flight and the columns represent the destinations. We place a 1 in the ith row and jth column of this matrix if there is a connecting flight from the ith city to the jth city and a 0 otherwise. We also place 0s on the principal diagonal, because a connecting flight with the same origin and destination does not make sense. Atlanta 1
Baltimore 2
1
3 Chicago
Origin
1 2 3 4 5
4 Denver
0 0 E1 0 0
Destination 2 3 4
5
1 0 0 0 0
0 0 1U ⫽ A 0 0
0 1 0 1 0
1 0 0 0 1
5 El Paso
Now that the schedule has been represented in the mathematical form of a matrix, we can perform operations on this matrix to obtain information about the schedule. (A) Find A2. What does the 1 in row 2 and column 1 of A2 indicate about the schedule? What does the 2 in row 1 and column 3 indicate about the schedule? In general, how would you interpret each element off the principal diagonal of A2? [Hint: Examine the diagram for possible connections between the ith city and the jth city.] (B) Find A3. What does the 1 in row 4 and column 2 of A3 indicate about the schedule? What does the 2 in row 1 and column 5 indicate about the schedule? In general, how would you interpret each element off the principal diagonal of A3? (C) Compute A, A ⫹ A2, A ⫹ A2 ⫹ A3, . . . , until you obtain a matrix with no zero elements (except possibly on the principal diagonal), and interpret. 80. AIRFREIGHT Refer to Problem 79. Find the incidence matrix A for the flight schedule illustrated in the figure. Compute A, A ⫹ A2, A ⫹ A2 ⫹ A3, . . . , until you obtain a matrix with no zero elements (except possibly on the principal diagonal), and interpret. Louisville 1
Milwaukee 2
3 Newark
4 Phoenix
5 Oakland
Matrix Operations
469
81. POLITICS In a local election, a group hired a public relations firm to promote its candidate in three ways: telephone, house calls, and direct mail. The cost per contact is given in matrix M: Cost per Contact
$0.80 M ⫽ £ $1.50 § $0.40
Telephone House Call Mail
The number of contacts of each type made in two adjacent cities is given in matrix N: Telephone
House Call
Mail
1,000 2,000
500 800
5,000 d 8,000
N⫽ c
Berkeley Oakland
(A) Find the total amount spent in Berkeley. (B) Find the total amount spent in Oakland. (C) Discuss possible interpretations of the elements in the matrix products MN and NM. (D) If either of the products MN or NM has a meaningful interpretation, find the product and label its rows and columns. (E) Discuss methods of matrix multiplication that can be used to find the total number of telephone calls, house calls, and letters. State the matrices that can be used, and perform the necessary operations. (F) Discuss methods of matrix multiplication that can be used to find the total number of contacts in Berkeley and in Oakland. State the matrices that can be used, and perform the necessary operations. 82. NUTRITION A nutritionist for a cereal company blends two cereals in different mixes. The amounts of protein, carbohydrate, and fat (in grams per ounce) in each cereal are given by matrix M. The amounts of each cereal used in the three mixes are given by matrix N. Cereal A
4 g Ⲑoz M ⫽ £ 20 g Ⲑoz 3 g Ⲑoz N⫽ c
Cereal B
2 g Ⲑoz 16 g Ⲑoz § 1 g Ⲑoz
Mix X
Mix Y
Mix Z
15 oz 5 oz
10 oz 10 oz
5 oz d 15 oz
Protein Carbohydrate Fat
Cereal A Cereal B
(A) Find the amount of protein in mix X. (B) Find the amount of fat in mix Z. (C) Discuss possible interpretations of the elements in the matrix products MN and NM. (D) If either of the products MN or NM has a meaningful interpretation, find the product and label its rows and columns. 83. TOURNAMENT SEEDING To rank players for an upcoming tennis tournament, a club decides to have each player play one set with every other player. The results are given in the table.
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Defeated
Player
Defeated
1. Aaron
Charles, Dan, Elvis
1. Anne
Diane
2. Bart
Aaron, Dan, Elvis
2. Bridget
Anne, Carol, Diane
3. Charles
Bart, Dan
3. Carol
Anne
4. Dan
Frank
5. Elvis
Charles, Dan, Frank
4. Diane
Carol, Erlene
6. Frank
Aaron, Bart, Charles
5. Erlene
Anne, Bridget, Carol
(A) Express the outcomes as an incidence matrix A by placing a 1 in the ith row and jth column of A if player i defeated player j, and a 0 otherwise (see Problem 79). (B) Compute the matrix B 0002 A 0004 A2 (C) Discuss matrix multiplication methods that can be used to find the sum of each of the rows in B. State the matrices that can be used and perform the necessary operations. (D) Rank the players from strongest to weakest. Explain the reasoning behind your ranking.
(A) Express the outcomes as an incidence matrix A by placing a 1 in the ith row and jth column of A if player i defeated player j, and a 0 otherwise (see Problem 79). (B) Compute the matrix B 0002 A 0004 A2 (C) Discuss matrix multiplication methods that can be used to find the sum of each of the rows in B. State the matrices that can be used and perform the necessary operations. (D) Rank the players from strongest to weakest. Explain the reasoning behind your ranking.
84. PLAYER RANKING Each member of a chess team plays one match with every other player. The results are given in the table.
7-4
Solving Systems of Linear Equations Using Matrix Inverse Methods Z The Identity Matrix for Multiplication Z Finding the Inverse of a Square Matrix Z Matrix Equations Z Matrix Equations and Systems of Linear Equations Z Application: Cryptography
Now that we know a bit about matrix multiplication, we will see how it can be used to solve certain systems of equations.
Z The Identity Matrix for Multiplication We know that for any real number a, 1 0005 a 0002 a 0005 1 0002 a. The number 1 is called the identity for real number multiplication. Is there a matrix analog? That is, if M is an arbitrary matrix, is there a matrix I with the property that IM 0002 MI 0002 M? It turns out that, in general, the answer is no. But the set of square matrices of order n (matrices with n rows and n columns) does have an identity.
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ZZZ EXPLORE-DISCUSS 1
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471
(A) Pick any 2 ⫻ 2 matrix you like, and multiply it by the following matrix in both possible orders. c
1 0
0 d 1
(B) Repeat (A) for any 3 ⫻ 3 matrix you like, but multiply by the matrix 1 £0 0
0 1 0
0 0§ 1
What can you conclude?
Z DEFINITION 1 Identity Matrix The identity matrix for multiplication for the set of all square matrices of order n is the square matrix of order n, denoted by I, with 1’s along the principal diagonal (from upper left corner to lower right corner) and 0’s elsewhere.
In Explore-Discuss 1, we saw that 1 c 0
0 d 1
and
1 £0 0
0 1 0
0 0§ 1
are the identity matrices for square matrices of order 2 and 3, respectively. We will show in Exercises 7-4 that if M is any square matrix of order n and I is the identity matrix of order n, then IM ⫽ MI ⫽ M Note: If M is an m ⫻ n matrix that is not square (m ⫽ n), then it is still possible to multiply M on the left and on the right by an identity matrix, but not with the same-size identity matrix. To avoid the complications involved with associating two different identity matrices with each nonsquare matrix, we will restrict our attention in this section to square matrices.
Z Finding the Inverse of a Square Matrix In the set of real numbers, we know that for each real number a, except 0, there exists a real number a⫺1 such that a⫺1a ⫽ 1 The number a⫺1 is called the inverse of the number a relative to multiplication, or the multiplicative inverse of a. For example, 2⫺1 is the multiplicative inverse of 2, since 2⫺1(2) ⫽ 1. We will use this idea to define the inverse of a square matrix.
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Z DEFINITION 2 Inverse of a Square Matrix If A is a square matrix of order n and if there exists a matrix A00031 (read “A inverse”) such that A00031A 0002 AA00031 0002 I then A00031 is called the multiplicative inverse of A or, more simply, the inverse of A. If no such matrix exists, then A is said to be a singular matrix.
Let A 0002 c
ZZZ EXPLORE-DISCUSS 2
4 2
2 d 2
1⁄
1⁄4 B 0002 c 1⁄ 2
1⁄
2 2
d
C0002 c
1⁄
2
00031⁄
2
00031⁄ 2 d 1
(A) How are the entries in A and B related? (B) Find AB. Is B the inverse of A? (C) Find AC. Is C the inverse of A?
The multiplicative inverse of a nonzero real number a also can be written as 1/ a, but this notation is never used for matrix inverses. Let’s use Definition 2 to find A00031, if it exists, for A0002 c
2 1
3 d 2
We are looking for a matrix A00031 0002 c
a b
c d d
such that AA00031 0002 A00031A 0002 I We can write A
2 c 1
A00041
3 a dc 2 b
I
c 1 d 0002 c d 0
0 d 1
and try to find a, b, c, and d so that the product of A and A00031 is the identity matrix I. Multiplying A and A00031 on the left side, we get c
(2a 0004 3b) (2c 0004 3d ) 1 d 0002 c (a 0004 2b) (c 0004 2d) 0
0 d 1
which is true only if 2a 0004 3b 0002 1 a 0004 2b 0002 0 c
2 1
3 1 ` d 2 0
c
1 2
2 0 ` d 3 1
2c 0004 3d 0002 0 c 0004 2d 0002 1 R1 4 R2
00042R1 0006 R2 S R2
Use Gauss–Jordan elimination to solve each system.
c
2 1
3 0 ` d 2 1
R1 4 R2
c
1 2
2 ` 3
00042R1 0006 R2 S R2
1 d 0
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c
1 2 0 ` d 0 00031 1
c
1 0
2 0 d ` 1 00031
c
1 0
0 2 d ` 1 00031
Solving Systems of Linear Equations Using Matrix Inverse Methods
(00041)R2 S R2
(00041)R2 0006 R1 S R1
a 0002 2, b 0002 00031 A00031 0002 c
c
1 0
2 1 d ` 00031 00032
c
1 0
2 1 ` d 1 2
c
1 0
0 00033 d ` 1 2
473
(00041)R2 S R2
(00041)R2 0006 R1 S R1
c 0002 00033, d 0002 2
a b 2 00033 d 0002 c d c d 00031 2
CHECK
A00041
A
c
2 1
3 2 dc 2 00031
A00041
I
00033 1 d 0002 c 2 0
0 2 d 0002 c 1 00031
A
00033 2 dc 2 1
3 d 2
Unlike nonzero real numbers, inverses do not always exist for nonzero square matrices. For example, if B0002 c
2 4
1 d 2
then, proceeding as before, we are led to the systems 2a 0004 b 0002 1 4a 0004 2b 0002 0 c
2 4
1 1 ` d 2 0
c
2 0
1 0 d ` 0 00032
2c 0004 d 0002 0 4c 0004 2 d 0002 1 (00042)R1 0006 R2
c
2 4
1 0 ` d 2 1
c
2 0
1 0 ` d 0 1
Use Gauss–Jordan elimination to solve each system. (00042)R1 0006 R2
The last row of each augmented coefficient matrix contains a contradiction. So each system is inconsistent and has no solution. We conclude that B00031 does not exist and B is a singular matrix.
Technology Connections Most graphing calculators can find matrix inverses and can identify singular matrices. Figure 1 shows the calculation of A00041 for the matrix A discussed earlier. Figure 2 shows the error message that results when the inverse operation is applied to the singular matrix B discussed earlier.
Z Figure 1
Z Figure 2
Note that the inverse operation is performed by pressing the x00041 key. Entering [A]^(00041) results in an error message (Fig. 3).
Z Figure 3
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Being able to find inverses, when they exist, leads to direct and simple solutions to many practical problems. The algebraic method outlined for finding the inverse, if it exists, gets very involved for matrices of order larger than 2. Now that we know what we are looking for, we can use augmented matrices, as in Section 7-2, to make the process more efficient. Details are illustrated in Example 1.
EXAMPLE
1
Finding an Inverse Find the inverse, if it exists, of 1 A ⫽ £0 2
SOLUTION
⫺1 2 3
1 ⫺1 § 0
We start as before and write Aⴚ1
A
1 £0 2
⫺1 2 3
1 a ⫺1 § £ b 0 c
I
d g 1 e h§ ⫽ £0 f i 0
0 1 0
0 0§ 1
Equating corresponding terms, we see that this is true only if a⫺ b⫹c⫽1 2b ⫺ c ⫽ 0 2a ⫹ 3b ⫽0
d⫺ e⫹f⫽0 2e ⫺ f ⫽ 1 2d ⫹ 3e ⫽0
g⫺ h⫹i⫽0 2h ⫺ i ⫽ 0 2g ⫹ 3h ⫽1
Now we write augmented matrices for each of the three systems: First
Second
Third
1 ⫺1 1 1 £0 2 ⫺1 † 0 § 2 3 0 0
1 ⫺1 1 0 £0 2 ⫺1 † 1 § 2 3 0 0
1 ⫺1 1 0 £0 2 ⫺1 † 0 § 2 3 0 1
If you look carefully at the side-by-side solutions on pages 472 and 473, you will see that the exact same row operations were performed on each augmented matrix. The same would happen here; all three preceding augmented matrices have the same coefficient matrix. To save time, we’ll combine all three into one, as shown next. 1 ⫺1 1 1 0 0 £0 2 ⫺1 † 0 1 0 § ⫽ [A | I ] 2 3 0 0 0 1
(1)
We now try to perform row operations on matrix (1) until we obtain a row-equivalent matrix that looks like matrix (2): I
1 £0 0
0 1 0
B
0 a d g 0 † b e h § ⫽ [I | B] 1 c f i
(2)
If this can be done, then the new matrix to the right of the vertical bar is A⫺1! Now let’s try to transform matrix (1) into a form like that of matrix (2). We follow the same
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sequence of steps as in the solution of linear systems by Gauss–Jordan elimination (see Section 7-2): A
I
1 00031 1 1 0 0 £0 2 00031 † 0 1 0 § 2 3 0 0 0 1 1 00031 1 1 0 0 ⬃ £0 2 00031 † 0 1 0 § 0 5 00032 00032 0 1 1 ⬃ £0 0
00031 1 5
1 1 0 0 000312 † 0 12 0 § 00032 00032 0 1
1 1 0 1 2 ⬃ £ 0 1 000312 † 0 1 0 0 00032 2
1 2 1 2 000352
1 1 1 0 2 1 ⬃ £ 0 1 00032 † 0 0 0 1 00034
1 2 1 2
0 0§ 1
0 0§ 00035 2
(00042)R1 0006 R3 S R3
1 2 R2
S R2
R2 0006 R1 S R1
(00045)R2 0006 R3 S R3
2R3 S R3
(000412 )R3 0006 R1 S R1 1 2 R3
0006 R2 S R2
1 0 0 3 3 00031 ⬃ £ 0 1 0 † 00032 00032 1 § 0002 [I 0 B] 0 0 1 00034 00035 2 We suspect that matrix B is actually A00031, but we should check. CHECK
Because the definition of matrix inverse requires that A00031A 0002 I
and
AA00031 0002 I
(3)
it appears that we must compute both A00031A and AA00031 to check our work. However, it can be shown that if one of the equations in (3) is satisfied, then the other is also satisfied. So, for checking purposes it’s enough to compute either A00031A or AA00031—we don’t need to do both. 3 A00031A 0002 £ 00032 00034
MATCHED PROBLEM 1
3 Let A 0002 £ 00031 1
3 00031 1 00032 1§ £0 00035 2 2
00031 2 3
1 1 00031 § 0002 £ 0 0 0
0 1 0
0 0§ 0002 I 1
0002
00031 1 1 0§ 0 1
(A) Form the augmented matrix [A | I ]. (B) Use row operations to transform [A | I ] into [ I | B ]. (C) Verify by multiplication that B 0002 A00031.
0002
The procedure used in Example 1 can be used to find the inverse of any square matrix if the inverse exists, and will also indicate when the inverse does not exist. These ideas are summarized in Theorem 1.
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Z THEOREM 1 Inverse of a Square Matrix A If [A | I ] is transformed by row operations into [I | B ], then the resulting matrix B is A00031. If, however, we obtain all 0s in one or more rows to the left of the vertical line, then A00031 does not exist.
EXAMPLE
2
Finding a Matrix Inverse Find A00031, given A 0002 c
4 00036
00031 d 2 c
SOLUTION
4 00036
00031 1 ` 2 0 1 4
1 4
0 d 1
⬃ c
1 ` 00036 2 0
0 d 1
⬃ c
1 0
000314
⬃ c
1 0
000314
⬃ c
1 0
0 1 12 ` d 1 3 2
1 2
`
1 4 3 2
0 d 1
`
1 4
0 d 2
1 3
A00031 0002 c
1 4 R1
S R1
6R1 0006 R2 S R2
2R2 S R2 1 4 R2
0006 R1 S R1
1 12 d 3 2
You should check our work by showing that A00031A 0002 I. MATCHED PROBLEM 2
Find A00031, given A 0002 c
0002
00036 d 00032
2 1
0002
EXAMPLE
3
Finding an Inverse Find B00031, if it exists, given B 0002 c
SOLUTION
c
10 00035
10 00035
00032 1 ` 1 0
00032 d 1 0 1 000315 101 0 d ⬃ c ` d 1 00035 1 0 1 1 000315 101 0 ⬃ c ` d 0 0 12 1
We have all 0s in the second row to the left of the vertical line. Therefore, B00031 does not exist. 0002 MATCHED PROBLEM 3
Find B00031, if it exists, given B 0002 c
6 00032
00033 d 1 0002
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Z Matrix Equations Before we discuss the solution of matrix equations, you might find it helpful to briefly review the basic properties of real numbers discussed in Section R-1.
ZZZ EXPLORE-DISCUSS 3
Let a, b, and c be real numbers, with a 0006 0. Solve each equation for x. (A) ax 0002 b
(B) ax 0004 b 0002 c
Solving simple matrix equations follows very much the same procedures used in solving real number equations. We have, however, less freedom with matrix equations, because matrix multiplication is not commutative. In solving matrix equations, we will be guided by the properties of matrices summarized in Theorem 2. (Some of these properties were introduced previously.)
Z THEOREM 2 Basic Properties of Matrices Assuming all products and sums are defined for the indicated matrices A, B, C, I, and 0, then Addition Properties Associative: Commutative: Additive Identity: Additive Inverse: Multiplication Properties Associative Property: Multiplicative Identity: Multiplicative Inverse:
(A 0004 B) 0004 C 0002 A 0004 (B 0004 C ) A0004B0002B0004A A00040000200004A0002A A 0004 (0003A) 0002 (0003A) 0004 A 0002 0 A(BC ) 0002 (AB)C AI 0002 IA 0002 A If A is a square matrix and A00031 exists, then AA00031 0002 A00031A 0002 I.
Combined Properties Left Distributive: Right Distributive:
A(B 0004 C ) 0002 AB 0004 AC (B 0004 C )A 0002 BA 0004 CA
Equality Addition: Left Multiplication: Right Multiplication:
If A 0002 B, then A 0004 C 0002 B 0004 C. If A 0002 B, then CA 0002 CB. If A 0002 B, then AC 0002 BC.
The process of solving certain types of simple matrix equations is best illustrated by an example.
EXAMPLE
4
Solving a Matrix Equation Given an n 0005 n matrix A and n 0005 1 column matrices B and X, solve AX 0002 B for X. Assume all necessary inverses exist.
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SOLUTION
We are interested in finding a column matrix X that satisfies the matrix equation AX 0002 B. To solve this equation, we multiply both sides, on the left, by A00031, assuming it exists, to isolate X on the left side. AX 0002 B A (AX ) 0002 A00031B (A00031A)X 0002 A00031B IX 0002 A00031B X 0002 A00031B 00031
ZZZ
Use the left multiplication property. Associative property A00041A 0003 I IX 0003 X
0002
1. Do not mix the left multiplication property and the right multiplication property. If AX 0002 B, then
CAUTION ZZZ
A00031(AX ) 0006 BA00031 2. Matrix division is not defined. If a, b, and x are real numbers, then the solution of ax 0002 b can be written either as x 0002 a00031b or as x 0002 ba. But if A, B, and X are matrices, the solution of AX 0002 B must be written as X 0002 A00031B. The expression B A is not defined for matrices.
MATCHED PROBLEM 4
Given an n 0005 n matrix A and n 0005 1 column matrices B, C, and X, solve AX 0004 C 0002 B for X. Assume all necessary inverses exist. 0002
Z Matrix Equations and Systems of Linear Equations We will now show how independent systems of linear equations with the same number of variables as equations can be solved by first converting the system into a matrix equation of the form AX 0002 B and using X 0002 A00031B, as obtained in Example 4.
EXAMPLE
5
Using Inverses to Solve Systems of Equations Use matrix inverse methods to solve the system x1 0003 x2 0004 x3 0002 1 2x2 0003 x3 0002 1 00021 2x1 0004 3x2
SOLUTION
(4)
First, we will convert the system of equations (4) into a matrix equation: A
1 £0 2
00031 2 3
X
B
1 x1 1 00031 § £ x2 § 0002 £ 1 § 0 x3 1
(5)
You should check that the matrix equation (5) is equivalent to the original system of equations (4) by performing the multiplication on the left side, and then equating corresponding elements. If we can find the column matrix X, it will provide a solution to the system. In Example 4, we found that if AX 0002 B and A00031 exists, then X 0002 A00031B. So our job is to find A00031
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and multiply it by the constant matrix B on the left. In Example 1, we found that the inverse of matrix A is 3 3 A00031 0002 £ 00032 00032 00034 00035
00031 1§ 2
So the equation X 0002 A00031B is A00041
X
3 x1 £ x2 § 0002 £ 00032 x3 00034
B
3 00031 1 5 00032 1 § £ 1 § 0002 £ 00033 § 00035 2 1 00037
and we can conclude that x1 0002 5, x2 0002 00033, and x3 0002 00037. Check this result in system (4). 0002
MATCHED PROBLEM 5
Use matrix inverse methods to solve the system: 3x1 0003 x2 0004 x3 0002 1 0003x1 0004 x2 00023 0004 x3 0002 2 x1 [Note: The inverse of the coefficient matrix was found in Matched Problem 1.] 0002
Z USING INVERSE METHODS TO SOLVE SYSTEMS OF EQUATIONS If the number of equations in a system equals the number of variables and the coefficient matrix has an inverse, then the system will always have a unique solution that can be found by using the inverse of the coefficient matrix to solve the corresponding matrix equation. Matrix equation
Solution
AX 0002 B
X 0002 A00031B
At first, matrix inverse methods don’t seem any better than Gauss–Jordan elimination— both require applying row operations to an augmented matrix. The advantage of the inverse method becomes apparent when solving a number of systems with a common coefficient matrix, as in Example 6.
EXAMPLE
6
Using Inverses to Solve Systems of Equations Use matrix inverse methods to solve each of the following systems: (A)
SOLUTIONS
x1 0003 x2 0004 x3 0002 3 2x2 0003 x3 0002 1 2x1 0004 3x2 00024
(B)
x1 0003 x2 0004 x3 0002 00035 2x2 0003 x3 0002 2 0002 00033 2x1 0004 3x2
Notice that both systems have the same coefficient matrix A as system (4) in Example 5. Only the constant terms have been changed. So we can use A00031 to solve these systems just as we did in Example 5.
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(A)
X
x1 3 £ x2 § 0002 £ 00032 x3 00034
A00041
B
3 00031 3 8 00032 1 § £ 1 § 0002 £ 00034 § 00035 2 4 00039
The solution is, x1 0002 8, x2 0002 00034, and x3 0002 00039 (B)
X
3 x1 £ x2 § 0002 £ 00032 x3 00034
A00041
B
3 00031 00035 00036 00032 1§ £ 2§ 0002 £ 3§ 00035 2 00033 4
The solution is, x1 0002 00036, x2 0002 3, and x3 0002 4
MATCHED PROBLEM 6
0002
Use matrix inverse methods to solve each of the following systems (see Matched Problem 5): (A) 3x1 0003 x2 0004 x3 0002 3 0003x1 0004 x2 0002 00033 x1 0004 x3 0002 2
(B) 3x1 0003 x2 0004 x3 0002 00035 0003x1 0004 x2 0002 1 0004 x3 0002 00034 x1 0002
As Examples 5 and 6 illustrate, inverse methods are very convenient for hand calculations because once the inverse is found, it can be used to solve any new system formed by changing only the constant terms. Since most graphing calculators can compute the inverse of a matrix, this method also adapts readily to graphing calculator solutions. However, if your graphing calculator also has a built-in procedure for finding the reduced form of an augmented coefficient matrix, then it is just as convenient to use Gauss–Jordan elimination. Furthermore, Gauss–Jordan elimination can be used in all cases and, as noted previously, matrix inverse methods cannot always be used. The application in Example 7 illustrates the usefulness of matrix inverses.
EXAMPLE
7
Investment Allocation An investment adviser currently has two types of investments available for clients: an investment A that pays 4% per year and an investment B of higher risk that pays 8% per year. Clients may divide their investments between the two to achieve any total return desired between 4 and 8%. However, the higher the desired return, the higher the risk. How should each client listed in the table invest to achieve the indicated return? Client
SOLUTION
1
2
3
k
Total investment
$20,000
$50,000
$10,000
k1
Annual return desired
$1,200
$3,750
$500
k2
(6%)
(7.5%)
(5%)
We will first solve the problem for an arbitrary client k using inverses, and then apply the result to the three specific clients. Let x1 0002 Amount invested in A x2 0002 Amount invested in B
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Then x1 0004 x2 0002 k1 0.04x1 0004 0.08x2 0002 k2
Total invested Total annual return (4% of X1 0004 8% of X2)
Write as a matrix equation: A
c
1 0.04
X
B
1 x1 k1 dc d0002c d 0.08 x2 k2
We now find A00031 by starting with [A | I ] and proceeding as discussed earlier. 1 1 1 0 ` d 0.04 0.08 0 1 1 1 1 0 c ` d 4 8 0 100 1 1 1 0 c ` d 0 4 00034 100 1 1 1 0 c ` d 0 1 00031 25 1 0 2 000325 c ` d 0 1 00031 25 c ⬃ ⬃ ⬃ ⬃
100 R2 : R2 (To eliminate decimals)
00034R1 0004 R2 : R2
1/4
R2 : R2
(00031)R2 0004 R1 : R1
So A has an inverse, and A00031 0002 c A00041
CHECK
c
2 00031
000325 d 25
2 00031 A
I
000325 1 1 1 dc d 0002 c 25 0.04 0.08 0
0 d 1
Also, A00041
X
c
x1 2 d0002c x2 00031
B
000325 k1 dc d 25 k2
To solve each client’s investment problem, we replace k1 and k2 with appropriate values from the table and multiply by A00031. Client 1
x1 2 c d0002c x2 00031
000325 20,000 10,000 dc d0002c d 25 1,200 10,000
To draw $1,200 interest, invest $10,000 at 4% and $10,000 at 8%. Client 2
c
x1 2 d0002c x2 00031
000325 50,000 6,250 dc d0002c d 25 3,750 43,750
To draw $3,750 interest, invest $6,250 at 4% and $43,750 at 8%. Client 3
c
2 x1 d0002c x2 00031
000325 10,000 7,500 dc d0002c d 25 500 2,500
To draw $500 interest, invest $7,500 at 4% and $2,500 at 8%.
0002
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MATCHED PROBLEM 7
Repeat Example 7 with investment A paying 5% and investment B paying 9%. 0002
Z Application: Cryptography Matrix inverses can be used to provide a simple and effective procedure for encoding and decoding messages. To begin, we assign the numbers 1 to 26 to the letters in the alphabet, as shown. We also assign the number 27 to a blank to provide for space between words. (A more sophisticated code could include both uppercase and lowercase letters and punctuation symbols.) A 1
B 2
C 3
D 4
E 5
F 6
G 7
H 8
I 9
J 10
K 11
L 12
M 13
O 15
P 16
Q 17
R 18
S 19
T 20
U 21
V 22
W 23
X 24
Y 25
Z 26
Blank 27
N 14
The message SPRING BREAK corresponds to the sequence 19 16
18
9
14
7
27
2
18
5
1
11
Any matrix whose elements are positive integers and whose inverse exists can be used as an encoding matrix. For example, to use the 2 ⫻ 2 matrix A⫽c
4 5
3 d 4
to encode the preceding message, first we divide the numbers in the sequence into groups of 2 and use these groups as the columns of a matrix with 2 rows. (We would have added an extra blank in the last entry if the last column had an empty space.) Then we multiply this matrix on the left by A: c
4 5
3 19 dc 4 16
18 9
14 7
27 2
18 5
1 124 d⫽c 11 159
99 126
77 98
114 143
87 110
37
49
37 d 49
The coded message is 124
159
99
126
77
98
114
143
87 110
This message can be decoded simply by putting it back into matrix form and multiplying on the left by the decoding matrix A⫺1. Since A⫺1 is easily determined if A is known, the encoding matrix A is the only key needed to decode messages encoded in this manner. Although simple in concept, codes of this type can be very difficult to crack.
EXAMPLE
8
Cryptography The message 31 54 69 37 64 82 23 50 66 51 69 75 23 30 36 65 84 84 was encoded with the matrix A shown next. Use a graphing calculator to decode this message. 0 A ⫽ £1 2
SOLUTION
2 2 1
1 1§ 1
We begin by entering the 3 ⫻ 3 encoding matrix A (Fig. 4). Then we enter the coded message in the columns of a matrix C with three rows (Fig. 4). If B is the matrix containing the uncoded message, then B and C are related by C ⫽ AB. To find B, we multiply both sides of the equation C ⫽ AB by A⫺1 (Fig. 5).
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Z Figure 5
Z Figure 4
Writing the numbers in the columns of this matrix in sequence and using the correspondence between numbers and letters noted earlier produces the decoded message: 23 W
8 H
15 O
27
9 I
19 S
27
11 K
1 A
18 R
12 L
27
7 G
1 A
21 U
19 S
19 S
27
The answer to this question can be found somewhere in this chapter. MATCHED PROBLEM 8
0002
The message 46 84 85 55 101 100 59 95 132 25 42 53 52 91 90 43 71 83 19 37 25 was encoded with the matrix A shown here. Decode this message. 1 A 0002 £2 2
1 1 3
1 2§ 1
ANSWERS TO MATCHED PROBLEMS 3 1. (A) £ 00031 1
1 1 0 † 0 1 0
0 1 0
0 0§ 1
1 (C) £ 1 00031
1 00031 3 00031 2 00031 § £ 00031 1 00031 2 1 0
00031 3 d 000312 1
3. Does not exit
2. c 4.
00031 1 0
1 (B) £ 0 0
0 1 0
0 1 1 00031 0 † 1 2 00031 § 1 00031 00031 2
1 1 0§ 0002 £0 1 0
0 1 0
0 0§ 1
AX 0004 C 0002 B (AX 0004 C ) 0003 C 0002 B 0003 C AX 0004 (C 0003 C ) 0002 B 0003 C AX 0004 0 0002 B 0003 C AX 0002 B 0003 C A00031(AX ) 0002 A00031(B 0003 C ) (A00031A) X 0002 A00031(B 0003 C ) IX 0002 A00031(B 0003 C ) X 0002 A00031(B 0003 C )
5. x1 0002 2, x2 0002 5, x3 0002 0 6. (A) x1 0002 00032, x2 0002 00035, x3 0002 4 (B) x1 0002 0, x2 0002 1, x3 0002 00034 2.25 000325 7. A00031 0002 c d ; Client 1: $15,000 in A and $5,000 in B; Client 2: $18,750 in A and 00031.25 25 $31,250 in B; Client 3: $10,000 in A 8. WHO IS WILHELM JORDAN
0002
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Exercises
1. What is an identity matrix? 2. What is the (multiplicative) inverse of a real number? Does every real number have an inverse? 3. What is the (multiplicative) inverse of a matrix? Does every matrix have an inverse? 4. What is a singular matrix? 5. Describe the process for finding the inverse of a matrix by hand. 6. Explain how inverse matrices can be used to solve systems of linear equations by hand. 7. Explain how inverse matrices can be used to solve systems of linear equations on a graphing calculator. 8. How would you solve a linear system that has more variables than equations? 9. How would you solve a linear system that has fewer variables than equations? 10. How would you solve a linear system if the number of variables and the number of equations are equal?
1 0
1 13. £ 0 0 ⫺2 14. £ 2 5
0 2 dc 1 4
⫺3 d 5
12. c
0 ⫺2 0§ £ 2 1 5
0 1 0 1 4 1
1 4 1
3 1 ⫺2 § £ 0 0 0
2 4
⫺3 1 dc 5 0
0 d 1
3 ⫺2
⫺4 3 d; c 3 2
0 0§ 1
17. c
2 ⫺1
1 2 d; c ⫺1 ⫺1
19. c
⫺5 ⫺8
2 3 d; c 8 3
4 d 3 1 d ⫺1
⫺2 d ⫺5
⫺1 1 ⫺1 § ; £ ⫺3 0 0
0 1 0
⫺1 ⫺2 § 1
Write the matrix equations in Problems 25–28 as systems of linear equations without matrices. 25. c
2 1
⫺1 x1 3 dc d ⫽ c d 3 x2 ⫺2
⫺2 27. £ 1 0
0 2 1
1 28. £ ⫺3 2
⫺2 1 0
26. c
⫺3 ⫺1
1 x1 ⫺2 dc d ⫽ c d 2 x2 5
1 x1 3 1 § £ x2 § ⫽ £ ⫺4 § ⫺1 x3 2 3 0 x1 ⫺1 § £ x2 § ⫽ £ ⫺2 § 5 4 x3
30.
x1 ⫺ 2x2 ⫹ x3 ⫽ ⫺1 ⫺x1 ⫹ x2 ⫽ 2 2x1 ⫹ 3x2 ⫹ x3 ⫽ ⫺3
x1 ⫺ 2x2 ⫽ 7 ⫺3x1 ⫹ x2 ⫽ ⫺3
⫹ 3x3 ⫽ 5 32. 2x1 x1 ⫺ 2x2 ⫹ x3 ⫽ ⫺4 ⫺x1 ⫹ 3x2 ⫽ 2
In Problems 33–40, find x1 and x2. 33. c 35 c
x1 3 d ⫽ c x2 1
x1 ⫺2 d ⫽ c 2 x2
x1 ⫺2 ⫺2 ⫺2 dc d 34. c d ⫽ c 4 1 x2 ⫺1 3 3 dc d ⫺1 2
36. c
x1 3 d ⫽ c 0 x2
1 3 dc d 2 ⫺2 ⫺1 ⫺2 dc d 2 1
16. c
⫺2 ⫺4
1 ⫺1 d; c 2 2
⫺1 d ⫺2
37. c
1 1
⫺1 x1 5 dc d ⫽ c d ⫺2 x2 7
38. c
1 1
3 x1 9 dc d ⫽ c d 4 x2 6
18. c
5 ⫺2
⫺7 3 d; c 2 3
7 d 5
39. c
1 2
1 x1 15 dc d ⫽ c d ⫺3 x2 10
40. c
1 3
1 x1 10 dc d ⫽ c d ⫺2 x2 20
20. c
3 4 7 4 d; c d ⫺5 ⫺3 ⫺5 ⫺7
1 21. £ 0 ⫺1
2 0 1 1 0§; £0 ⫺1 1 1
⫺2 0 1 0§ ⫺1 0
1 22. £ ⫺3 0
0 1 0
0 1 0
1 1 ⫺2 § ; £ 3 1 0
0 1 0
31.
In Problems 15–24, examine the product of the two matrices to determine if each is the inverse of the other. 15. c
1 24. £ 3 0
1 3 3 ⫺1 ⫺1 § ; £ ⫺2 ⫺2 1§ 0 ⫺4 ⫺5 2
29. 4x1 ⫺ 3x2 ⫽ 2 x1 ⫹ 2x2 ⫽ 1
3 ⫺2 § 0 0 1 0
⫺1 2 3
Write each system in Problems 29–32 as a matrix equation of the form AX ⫽ B.
Perform the indicated operations in Problems 11–14. 11. c
1 23. £ 0 2
⫺1 ⫺1 § 1
In Problems 41–60, given A, find A⫺1, if it exists. Check each inverse by showing A⫺1A ⫽ I. 41. c
1 0
9 d 1
42. c
0 ⫺1
⫺1 d 3
43. c
⫺1 2
44. c
3 ⫺4 d ⫺2 3
45. c
⫺5 7 d 2 ⫺3
46. c
11 3
⫺2 d 5 4 d 1
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47. c
3 2
9 d 6
48. c
2 00033
00034 d 6
49. c
2 3
3 d 5
50. c
00035 4
4 d 00033
1 00031 0 1 00031 § 51. £ 00031 0 00031 1
2 52. £ 0 1
1 53. £ 3 1
1 54. £ 00032 3
00031 1 3 2§ 00033 2 2 00031 1 00031 § 00031 1
2 5 1
5 9§ 00032
00031 0 1 1§ 0 1
2 55. £ 0 00031
2 00031 4 00031 § 00032 1
4 56. £ 1 00033
2 57. £ 1 00031
1 1 1 0§ 00031 0
1 58. £ 2 0
00031 0 00031 1 § 1 1
10 4§ 15
1 60. £ 0 1
00035 1 00034
1 59. £ 0 1
5 1 6
Solving Systems of Linear Equations Using Matrix Inverse Methods
000310 6§ 00033
Write each system in Problems 61–68 as a matrix equation and solve using inverses. [ Note: the inverse of each coefficient matrix was found earlier in this exercise in the indicated problem. ] 61. 0003x1 0003 2x2 0002 k1 2x1 0004 5x2 0002 k2 (A) k1 0002 2, k2 0002 5 (B) k1 0002 00034, k2 0002 1 (C) k1 0002 00033, k2 0002 00032 (see Problem 43.) 62.
3x1 0003 4x2 0002 k1 00032x1 0004 3x2 0002 k2 (A) k1 0002 3, k2 0002 00031 (B) k1 0002 6, k2 0002 5 (C) k1 0002 0, k2 0002 00034 (see Problem 44.)
63. 00035x1 0004 7x2 0002 k1 2x1 0003 3x2 0002 k2 (A) k1 0002 00035, k2 0002 1 (B) k1 0002 8, k2 0002 00034 (C) k1 0002 6, k2 0002 0 (see Problem 45.) 64. 11x1 0004 4x2 0002 k1 3x1 0004 x2 0002 k2 (A) k1 0002 00032, k2 0002 00033 (B) k1 0002 00031, k2 0002 9 (C) k1 0002 4, k2 0002 5 (see Problem 46.) 65.
x1 0003 x2 0002 k1 0003x1 0004 x2 0003 x3 0002 k2 0003 x2 0004 x3 0002 k3 (A) k1 0002 1, k2 0002 1, k3 0002 2 (B) k1 0002 00031, k2 0002 0, k3 0002 00034 (C) k1 0002 3, k2 0002 00032, k3 0002 0 (see Problem 51.)
485
66. 2x1 0003 x2 0002 k1 x2 0004 x3 0002 k2 0004 x3 0002 k3 x1 (A) k1 0002 00032, k2 0002 4, k3 0002 00031 (B) k1 0002 2, k2 0002 00033, k3 0002 1 (C) k1 0002 00031, k2 0002 2, k3 0002 00035 (see Problem 52.) 67. x1 0004 2x2 0004 5x3 0002 k1 3x1 0004 5x2 0004 9x3 0002 k2 x1 0004 x2 0003 2x3 0002 k3 (A) k1 0002 0, k2 0002 1, k3 0002 4 (B) k1 0002 5, k2 0002 00031, k3 0002 0 (C) k1 0002 00036, k2 0002 0, k3 0002 2 (see Problem 53.) 68.
x1 0003 x2 0004 x3 0002 k1 00032x1 0004 3x2 0004 2x3 0002 k2 3x1 0003 3x2 0004 2x3 0002 k3 (A) k1 0002 3, k2 0002 00031, k3 0002 0 (B) k1 0002 0, k2 0002 4, k3 0002 5 (C) k1 0002 00032, k2 0002 0, k3 0002 1 (see Problem 54.)
For n 0005 n matrices A and B and n 0005 1 matrices C, D, and X, solve each matrix equation in Problems 69–74 for X. Assume all necessary inverses exist. 69. AX 0002 BX 0004 C
70. AX 0004 BX 0002 C 0004 D
71. X 0002 AX 0004 C
72. X 0004 C 0002 AX 0003 BX
73. AX 0004 C 0002 3X
74. AX 0004 C 0002 BX 0003 7X 0004 D
75. Discuss the existence of A00031 for 2 0005 2 diagonal matrices of the form A0002 c
a 0
0 d d
76. Discuss the existence of A00031 for 2 0005 2 upper triangular matrices of the form A0002 c
a 0
b d d
77. Find A00031 and A2 for each of the following matrices. (A) A 0002 c
3 00034
2 d 00033
(B) A 0002 c
00032 3
00031 d 2
78. Based on your observations in Problem 77, if A 0002 A00031 for a square matrix A, what is A2? Give a mathematical argument to support your conclusion. 79. Find (A00031)00031 for each of the following matrices. (A) A 0002 c
4 1
2 d 3
(B) A 0002 c
5 00031
5 d 3
80. Based on your observations in Problem 79, if A00031 exists for a square matrix A, what is (A00031)00031? Give a mathematical argument to support your conclusion. 81. Find (AB)00031, A00031B00031, and B00031A00031 for each of the following pairs of matrices. (A) A 0002 c
3 2
4 d 3
and
B0002 c
3 2
7 d 5
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B0002 c
and
6 2
sold (assuming that all seats can be sold) to bring in each of the returns indicated in the table? Use decimals in computing the inverse.
2 d 1
Concert
82. Based on your observations in Problems 81, which of the following is a true statement? Give a mathematical argument to support your conclusion. (A) (AB)00031 0002 A00031B00031
Return required
APPLICATIONS Problems 83–86 refer to the encoding matrix A 0002 c
3 1
5 d 2
84. CRYPTOGRAPHY Encode the message KOBE BRYANT with the matrix A. 85. CRYPTOGRAPHY The following message was encoded with the matrix A. Decode the message.
41
57 20 150 192 73
159
59
103
39
62
61
22
47
18
0 1 1 0 1
1 1 1 1 1
0 0 1 0 2
1 3 1U 2 1
89. CRYPTOGRAPHY The following message was encoded with the matrix B. Decode the message. 32 25 55 19 41 51 64 103 39 100 62 109 114 62 92 58 115 105 73 113 39 110 85 65 111 90. CRYPTOGRAPHY The following message was encoded with the matrix B. Decode the message. 88 77
29 46
$240,000
$250,000
$270,000
Labor Cost
Material Cost
A
$30
$20
B
$40
$30
If a total of $3,000 a week is allowed for labor and material, how many of each model should be produced each week to exactly use each of the allocations of the $3,000 indicated in the following table? Use decimals in computing the inverse.
82 51 61 86 108 61 72 65
45
1
2
3
Labor
$1,800
$1,750
$1,720
Material
$1,200
$1,250
$1,280
93. CIRCUIT ANALYSIS A direct current electric circuit consisting of conductors (wires), resistors, and batteries is diagrammed in the figure. V1
88. CRYPTOGRAPHY Encode the message PITTSBURGH STEELERS with the matrix B.
45 37
10,000
V2
0004 0003
87. CRYPTOGRAPHY Encode the message NEW ENGLAND PATRIOTS with the matrix B.
44 74
10,000
Weekly Allocation
Problems 87–90 require the use of a graphing calculator. To use the 5 0005 5 encoding matrix B given below, form a matrix with 5 rows and as many columns as necessary to accommodate each message. 1 0 B 0002 E2 0 1
10,000
Guitar Model
86. CRYPTOGRAPHY The following message was encoded with the matrix A. Decode the message. 49 18 103 105 41
3
92. PRODUCTION SCHEDULING Labor and material costs for manufacturing two guitar models are given in the following table:
83. CRYPTOGRAPHY Encode the message LEBRON JAMES with the matrix A.
55 22
2
Tickets sold
(B) (AB)00031 0002 B00031A00031
31 12 150 160 61 61
1
84
35
63
Solve Problems 91–97 using systems of equations and matrix inverses. 91. RESOURCE ALLOCATION A concert hall has 10,000 seats. If tickets are $20 and $30, how many of each type of ticket should be
0003 0004
1 ohm
1 ohm
I1
I2
2 ohms I3
If I1, I2, and I3 are the currents (in amperes) in the three branches of the circuit and V1 and V2 are the voltages (in volts) of the two batteries, then Kirchhoff’s* laws can be used to show that the currents satisfy the following system of equations: I1 0003 I2 0004 I3 0002 0 I1 0004 I2 0002 V1 I2 0004 2I3 0002 V2 Solve this system for: (A) V1 0002 10 volts, V2 0002 10 volts (B) V1 0002 10 volts, V2 0002 15 volts (C) V1 0002 15 volts, V2 0002 10 volts *Gustav Kirchhoff (1824–1887), a German physicist, was among the first to apply theoretical mathematics to physics. He is best known for his development of certain properties of electric circuits, which are now known as Kirchhoff’s laws.
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94. CIRCUIT ANALYSIS Repeat Problem 93 for the electric circuit shown in the figure.
96. GEOMETRY Repeat Problem 95 if the graph passes through the points (00031, k1), (0, k2), and (1, k3).
I1 0003 I2 0004 I3 0002 0
97. DIETS A biologist has available two commercial food mixes with the following percentages of protein and fat:
I1 0004 2I2 0002 V1 2I2 0004 2I3 0002 V2 V1
V2
0004 0003
0003 0004
1 ohm
2 ohms
I1
I2
Protein (%)
Fat (%)
A
20
2
B
10
6
How many ounces of each mix should be used to prepare each of the diets listed in the following table?
2 ohms
Diet
I3
95. GEOMETRY The graph of f(x) 0002 ax2 0004 bx 0004 c passes through the points (1, k1), (2, k2), and (3, k3). Determine a, b, and c for: (A) k1 0002 00032, k2 0002 1, k3 0002 6 (B) k1 0002 4, k2 0002 3, k3 0002 00032 (C) k1 0002 8, k2 0002 00035, k3 0002 4
7-5
Mix
Protein
1
2
3
20 oz
10 oz
10 oz
6 oz
4 oz
6 oz
Fat
Determinants and Cramer’s Rule Z Defining First- and Second-Order Determinants Z Evaluating Third-Order Determinants Z Using Cramer’s Rule to Solve Systems of Equations
In this section, we’ll study one more method for solving systems of linear equations using matrices. Like the inverse method, it works only for systems with the same number of equations and variables. The biggest advantage is that it’s purely computational—it requires very little symbol manipulation. The method is based on determinants.
Z Defining First- and Second-Order Determinants For any square matrix A, the determinant of A is a real number denoted by det (A) or 冟 A 冟*. If A is a square matrix of order n, then det (A) is called a determinant of order n. If A 0002 [a11 ] is a square matrix of order 1, then det (A) 0002 a11 is a first-order determinant. Now we proceed to define determinants of higher order. a11 a12 d , the second-order determinant Given a second-order square matrix A 0002 c a21 a22 of A is det (A) 0002 `
a11 a21
a12 ` 0002 a11a22 0003 a21a12 a22
(1)
*The absolute value notation will now have two interpretations: the absolute value of a real number or the determinant of a square matrix. These concepts are not the same. You must always interpret 冟 A 冟 in terms of the context in which it is used.
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Formula (1) is easily remembered if you notice that the expression on the right is the product of the elements on the principal diagonal, from upper left to lower right, minus the product of the elements on the secondary diagonal, from lower left to upper right.
EXAMPLE
1
Evaluating a Second-Order Determinant Find `
⫺1 ⫺3
⫺1 2 ` ⫺3 ⫺4 ⫽ (⫺1)(⫺4) ⫺ (⫺3)(2) ⫽ 4 ⫺ (⫺6) ⫽ 10
det (A) ⫽ `
SOLUTIONS
MATCHED PROBLEM 1
ZZZ
CAUTION ZZZ
2 `. ⫺4
Find `
3 4
⫺5 `. ⫺2
0002
0002
⫺5 3 ⫺5 d is a matrix, but ` ` represents a real number, ⫺2 4 ⫺2 3 ⫺5 ` as a determinant, and refer to the determinant of A. We will often refer to ` 4 ⫺2 the process of finding the real number it represents as “evaluating the determinant.” Remember that A ⫽ c
3 4
Technology Connections Most graphing calculators have a command to calculate determinants. On the TI-84, it is on the MATRIX-MATH menu. In Figure 1, the determinant from Example 1 is calculated.
Z Figure 1
Z Evaluating Third-Order Determinants a11 Given the matrix A ⫽ C a21 a31 a11 det (A) ⫽ † a21 a31
a12 a22 a32
a12 a22 a32
a13 a23 S, the third-order determinant of A is a33
a13 a23 † ⫽ a11a22a33 ⫺ a11a32a23 ⫹ a21a32a13 ⫺ a21a12a33 a33 ⫹ a31a12a23 ⫺ a31a22a13
(2)
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489
Don’t panic! You don’t need to memorize formula (2). After we introduce the ideas of minor and cofactor below, we will state a theorem that can be used to obtain the same result with much less trouble. The minor of an element in a third-order determinant is a second-order determinant obtained by deleting the row and column that contains the element. For example, in the determinant in formula (2), Deletions are usually done mentally.
a11 a31
a12 ` a32
a11 † a21 a31
a12 a22 a32
a13 a23 † 0003 a11a32 0002 a31a12 a33
a11 0003 ` a21
a13 ` a23
a11 † a21 a31
a12 a22 a32
a13 a23 † 0003 a11a23 0002 a21a13 a33
Minor of a23 0003 `
Minor of a32
Write the minors of the other seven elements in the determinant in formula (2).
ZZZ EXPLORE-DISCUSS 1
A quantity closely associated with the minor of an element is the cofactor of an element aij (from the ith row and jth column), which is defined as the product of the minor of aij and (00021)i0004j.
Z DEFINITION 1 Cofactor Cofactor of aij 0003 (00021)i0004j (Minor of aij)
So a cofactor is just a minor with either a positive or negative sign. The sign is determined by raising 00021 to a power that is the sum of the numbers indicating the row and column in which the element appears. Note that (00021)i0004j is 1 if i 0004 j is even and 00021 if i 0004 j is odd. So if we are given the determinant a11 † a21 a31
a12 a22 a32
a13 a23 † a33
then
EXAMPLE
2
Cofactor of a23 0003 (00021)200043 `
a11 a31
a12 a11 ` 00030002` a32 a31
Cofactor of a11 0003 (00021)100041 `
a22 a32
a22 a23 ` 0003 ` a33 a32
a12 ` 0003 0002(a11a32 0002 a31a12) a32 a23 ` 0003 a22a33 0002 a32a23 a33
Finding Cofactors Find the cofactors of 00022 and 5 in the determinant 00022 † 1 00021
0 00026 2
3 5† 0
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00026 5 00026 5 ` 0003 ` ` 2 0 2 0 0003 (00026)(0) 0002 (2)(5) 0003 000210 00022 0 00022 0 Cofactor of 5 0003 (00021)200043 ` ` 00030002` ` 00021 2 00021 2 0003 0002[(00022)(2) 0002 (00021)(0)] 0003 4
Cofactor of 00022 0003 (00021)100041 `
SOLUTION
MATCHED PROBLEM 2
ⴚ2 is a11
5 is a23
0002
Find the cofactors of 2 and 3 in the determinant in Example 2. 0002 [Note: The sign in front of the minor, (00021)i0004j, can be determined rather mechanically by using a checkerboard pattern of 0004 and 0002 signs over the determinant, starting with 0004 in the upper left-hand corner: 0004 0002 0004
0002 0004 0002
0004 0002 0004
Use either the checkerboard or the exponent method—whichever is easier for you—to determine the sign in front of the minor.] Theorem 1 will give us a step-by-step procedure for finding third-order determinants without having to memorize formula (2).
Z THEOREM 1 Computing a Third-Order Determinant The value of a determinant of order 3 is the sum of three products obtained by multiplying each element in any row or any column by its cofactor. This is called expanding along a row or column.
Proving Theorem 1 requires six different calculations: expanding an arbitrary thirdorder determinant along each of the rows and columns, and showing that the result matches formula (2). You will be asked to complete a couple of those cases in the exercises.
EXAMPLE
3
Evaluating a Third-Order Determinant 2 00022 Evaluate † 00023 1 1 00023
SOLUTION
0 2† 00021
We can choose any row or column to expand along. We will choose the first row because of the zero: we won’t need to find that cofactor because it will be multiplied by zero. 2 00022 † 00023 1 1 00023
0 Cofactor Cofactor Cofactor 2 † 0003 a11 a of a b 0004 a12 a of a b 0004 a13 a of a b 11 12 13 00021 0003 2 c (00021)100041 `
1 00023
2 00023 ` d 0004 (00022) c (00021)100042 ` 00021 1
2 `d 0004 0 00021
0003 (2)(1)[(1)(00021) 0002 (00023)(2)] 0004 (00022)(00021)[(00023)(00021) 0002 (1)(2)] 0003 (2)(5) 0004 (2)(1) 0003 12
0002
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MATCHED PROBLEM 3
2 Evaluate † 00022 00021
Determinants and Cramer’s Rule
491
1 00021 00023 0† 2 1 0002
It’s important to note that the determinant will work out the same regardless of which row or column you choose to expand along. So if possible, you should choose a row or column with one or more zeros to minimize the number of computations.
Z Using Cramer’s Rule to Solve Systems of Equations Now we will see how determinants can be used to solve systems of equations. We’ll start by investigating two equations in two variables, and then extend our results to three equations in three variables. Instead of thinking of each system of linear equations in two variables as a different problem, let’s see what happens when we attempt to solve the general system a11x 0004 a12y 0003 k1 a21x 0004 a22y 0003 k2
(3A) (3B)
once and for all, in terms of the unspecified real constants a11, a12, a21, a22, k1, and k2. We proceed by multiplying equations (3A) and (3B) by suitable constants so that when the resulting equations are added, left side to left side and right side to right side, one of the variables drops out. Suppose we choose to eliminate y. What constant should we use to make the coefficients of y the same except for the signs? Multiply equation (3A) by a22 and (3B) by 0002a12; then add: a22(3A): ⴚa12(3B):
a11a22x 0004 a12a22 y 0003
k1a22
0002a21a12x 0002 a12a22 y 0003 0002k2a12 a11a22x 0002 a21a12x 0004 0y 0003 k1a22 0002 k2a12 (a11a22 0002 a21a12)x 0003 k1a22 0002 k2a12 x0003
k1a22 0002 k2a12 a11a22 0002 a21a12
y is eliminated. Factor out x. Solve for x. a11a22 ⴚ a21a12 ⴝ 0
At this point, the numerator and denominator might remind you of second-order determinants. In fact, the value of x can be written as
x0003
`
k1 a12 ` k2 a22 a11 a21
`
a12 ` a22
Similarly, starting with system (3A) and (3B) and eliminating x (this is left as an exercise), we obtain
y0003
` `
a11 a21
k1 ` k2
a11 a21
a12 ` a22
These results are summarized in Theorem 2, Cramer’s rule, which is named after the Swiss mathematician Gabriel Cramer (1704–1752).
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Z THEOREM 2 Cramer’s Rule for Two Equations and Two Variables Given the system
a11x 0004 a12 y 0003 k1 a21x 0004 a22 y 0003 k2
with
D0003 `
a11 a21
a12 ` 00050 a22
then
x0003
`
k1 k2
a12 ` a22 D
and
a11 k1 ` a21 k2 y0003 D `
The determinant D is called the coefficient determinant. If D 0005 0, then the system has exactly one solution, which is given by Cramer’s rule. If, on the other hand, D 0003 0, then it can be shown that the system is either inconsistent and has no solutions or is dependent and has an infinite number of solutions. In that case, we would need to use other methods to determine the exact nature of the solutions.
EXAMPLE
4
Solving a Two-Variable System with Cramer’s Rule Solve using Cramer’s rule:
SOLUTIONS
3x 0002 5y 0003 2 00024x 0004 3y 0003 00021
First find the determinant of the coefficient matrix: D0003 `
3 00024
00025 ` 0003 9 0002 20 0003 000211 3
Now replace the x column with the constants and find the determinant, then divide by 000211. 2 00025 ` 00021 3 600025 1 x0003 0003 00030002 000211 000211 11 `
Now repeat, this time replacing the y column with the constants. 3 2 ` 00023 0002 (00028) 00024 00021 5 y0003 0003 00030002 000211 000211 11 `
The solution to the system is x 0003 0002
MATCHED PROBLEM 4
Solve using Cramer’s rule:
1 5 ,y00030002 . 11 11
3x 0004 2y 0003 00024 00024x 0004 3y 0003 000210
0002
0002
Cramer’s rule can be generalized completely for any size linear system that has the same number of variables as equations. However, it cannot be used to solve systems where the number of variables is not equal to the number of equations. In Theorem 3 we state without proof Cramer’s rule for three equations in three variables.
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Z THEOREM 3 Cramer’s Rule for Three Equations in Three Variables Given the system a11x 0004 a12 y 0004 a13z 0003 k1 a21x 0004 a22 y 0004 a23z 0003 k2 a31x 0004 a32 y 0004 a33z 0003 k3
a11 D 0003 † a21 a31
with
a12 a22 a32
a13 a23 † 0005 0 a33
then
x0003
k1 † k2 k3
a12 a22 a32 D
a13 a23 † a33
y0003
k1 a13 k2 a23 † k3 a33 D
a11 † a21 a31
z0003
a11 † a21 a31
a12 a22 a32 D
k1 k2 † k3
You can easily remember these determinant formulas for x, y, and z if you observe the following: 1. 2. 3.
EXAMPLE
5
Determinant D is formed from the coefficients of x, y, and z, keeping the same relative position in the determinant as found in the system of equations. Determinant D appears in the denominators for x, y, and z. The numerator for x can be obtained from D by replacing the coefficients of x (a11, a21, and a31) with the constants k1, k2, and k3, respectively. Similar statements can be made for the numerators for y and z.
Solving a Three-Variable System with Cramer’s Rule Solve using Cramer’s rule:
SOLUTION
x0004 y 0003 2 3y 0002 z 0003 00024 x 0004z0003 3 1 D0003 †0 1
x0003
y0003
z0003
2 † 00024 3 1 †0 1 1 †0 1
1 3 0
0 00021 † 0003 2 1 1 3 0 2
2 00024 3 2
0 00021 † 1 0 00021 † 1
0003
7 2
00030002
3 2
1 2 3 00024 † 0 3 1 00030002 2 2
0002
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Solve using Cramer’s rule:
MATCHED PROBLEM 5
3x 0002z00035 x0002y0004z00030 x0004y 00031
0002
Cofactor expansion can be used to find determinants of orders higher than 3, so Cramer’s rule can be used for systems with more than three variables. For large systems, however, the Gauss-Jordan method, which involves fewer arithmetic operations than Cramer’s Rule, is a more practical choice.
ANSWERS TO MATCHED PROBLEMS 1. 14 2. Cofactor of 2 0003 13; cofactor of 3 0003 00024 4. x 0003 178 , y 0003 000246 5. x 0003 65, y 0003 000215, z 0003 000275 17
7-5
3. 3
Exercises
1. Explain the difference between c
a11 a21
a12 a11 d and ` a22 a21
a12 `. a22
2. Explain the difference between a matrix and a minor. 3. Explain the difference between a minor and a cofactor. 4. How do you evaluate a third-order determinant? 5. If A is the 2 0006 2 coefficient matrix for a linear system and det (A) 0003 0, what can you conclude about the solution set for the system?
17. 2x 0004 y 0003 1 5x 0004 3y 0003 2
18. x 0004 3y 0003 1 2x 0004 8y 0003 0
19. 2x 0002 y 0003 00023 0002x 0004 3y 0003 3
20. 00023x 0004 2y 0003 1 2x 0002 3y 0003 00023
21. 4x 0002 3y 0003 4 3x 0004 2y 0003 00022
22. 5x 0004 2y 0003 00021 2x 0002 3y 0003 2
Problems 23–30 pertain to the following determinant: 5 †3 0
6. Can you use Cramer’s rule to solve a linear system with a 3 0006 2 coefficient matrix? Explain. 7. Can you use Cramer’s rule to solve a linear system with a 4 0006 4 coefficient matrix? Explain. 8. List all the possible solution methods for linear systems that we have discussed in this chapter. Which is your favorite and why? Evaluate each second-order determinant in Problems 9–14. 5 9. ` 2
4 ` 3
11. `
3 00027 ` 00025 6
13. `
4.3 00025.1
00021.2 ` 3.7
8 10. ` 4
00023 ` 1
12. `
9 4
00022 ` 0
14. `
00020.7 1.9
00023 6† 8
Write the minor of each element given in Problems 23–26. Leave the answer in determinant form. 23. a11
24. a33
25. a23
26. a12
Write the cofactor of each element given in Problems 27–30, and evaluate each. 27. a11
28. a33
30. a12
29. a23
Evaluate the determinant in Problems 31–40 using cofactors. 00022.3 ` 00024.8
Solve the system in Problems 15–22 using Cramer’s rule. 15. x 0004 2y 0003 1 x 0004 3y 0003 00021
00021 4 00022
16. x 0004 2y 0003 3 x 0004 3y 0003 5
1 31. † 00022 5 0 33. † 3 0
0 4 00022
0 3† 1
1 5 00027 6† 00022 00023
2 32. † 0 0
00023 00023 6
5 1† 2
4 34. † 9 1
00022 5 2
0 4† 0
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00021 35. † 00022 4
2 00023 0 00026 † 00023 2
1 37. † 1 2
4 1 1
1 39. † 2 3
4 3 1 6† 00022 9
0 2 00021 36. † 00026 3 1† 7 00029 00022
1 00022 † 00021
3 2 1 38. † 00021 5 1 † 2 3 1 4 40. † 00021 5
00026 3 4 1† 00026 3
In Problems 59 and 60, use Cramer’s rule to solve for z only. 59.
3x 0002 4y 0004 5z 0003 18 60. 13x 0004 11y 0004 10z 0003 2 00029x 0004 8y 0004 7z 0003 000213 10x 0004 8y 0004 7z 0003 1 5x 0002 7y 0004 10z 0003 33 8x 0004 5y 0004 4z 0003 4
If A is a 3 0005 3 matrix, det A can be evaluated by the following diagonal expansion. Form a 3 0005 5 matrix by augmenting A on the right with its first two columns, and compute the diagonal products p1, p2, . . . , p6 indicated by the arrows: a11 a12 C a21 a22 a31 a32
Solve Problems 41–44 to two significant digits using Cramer’s rule. 41. 0.9925x 0002 0.9659y 0003 0 0.1219x 0004 0.2588y 0003 2,500
p4
42. 0.9877x 0002 0.9744y 0003 0 0.1564x 0004 0.2250y 0003 1,900
495
Determinants and Cramer’s Rule
p5
a13 a11 a23 a21 a33 a31
p6
a12 a22 S a32 p1
p2
Diagonal expansion formula
p3
The determinant of A is given by [compare with formula (2)] det A 0003 p1 0004 p2 0004 p3 0002 p4 0002 p5 0002 p6
43. 0.9954x 0002 0.9942y 0003 0 0.0958x 0004 0.1080y 0003 155
0003 a11a22a33 0004 a12a23a31 0004 a13a21a32 0002 a13a22a31 0002 a11a23a32 0002 a12a21a33
44. 0.9973x 0002 0.9957y 0003 0 0.0732x 0004 0.0924y 0003 112
[Caution: The diagonal expansion procedure works only for 3 0005 3 matrices. Do not apply it to matrices of any other size.]
Solve Problems 45–52 using Cramer’s rule:
Use the diagonal expansion formula to evaluate the determinants in Problems 61 and 62.
45.
x0004 y 0003 0 2y 0004 z 0003 00025 0002x 0004 z 0003 00023
47. x 0004 y 00031 2y 0004 z 0003 0 0002y 0004 z 0003 1
46.
48.
x0004 y 0003 00024 2y 0004 z 0003 0 0002x 0004z0003 5 x 0004 3y 0003 00023 2y 0004 z 0003 3 0002x 0004 3z 0003 7
49.
3y 0004 z 0003 00021 x 0004 2z 0003 3 x 0002 3y 0003 00022
50. x 0002z0003 3 2x 0002 y 0003 00023 x0004y0004z0003 1
51.
2y 0002 z 0003 00023 x0002 y0002 z0003 2 x 0002 y 0004 2z 0003 4
52. 2x 0004 y 0003 2 x 0002 y 0004 z 0003 00021 x0004y0004z0003 2
Discuss the number of solutions for the systems in Problems 53 and 54 where a and b are real numbers. Use Cramer’s rule where appropriate and Gauss–Jordan elimination otherwise. 53. ax 0004 3y 0003 b 2x 0004 4y 0003 5
54. 2x 0004 ay 0003 b 3x 0004 4y 0003 7
2 61. † 5 00024
6 00021 3 00027 † 00022 1
A square matrix is called an upper triangular matrix if all elements below the principal diagonal are zero. In Problems 63–66, determine whether the statement is true or false. If true, explain why. If false, give a counterexample. 63. If the determinant of an upper triangular matrix is 0, then the elements on the principal diagonal are all 0. 64. If A and B are upper det (A 0004 B) 0003 det A 0004 det B.
2x 0002 3y 0004 z 0003 00023 00024x 0004 3y 0004 2z 0003 000211 x0002 y0002 z0003 3
56.
x 0004 4y 0002 3z 0003 25 3x 0004 y 0002 z 0003 2 00024x 0004 y 0004 2z 0003 1
triangular
matrices,
then
65. The determinant of an upper triangular matrix is the product of the elements on the principal diagonal. 66. If A and B are upper det (AB) 0003 (det A)(det B).
triangular
matrices,
then
67. Show that the expansion of the determinant a11 a12 † a21 a22 a31 a32
In Problems 55 and 56, use Cramer’s rule to solve for x only. 55.
1 00025 2 00026 † 00021 7
4 62. † 1 00023
a13 a23 † a33
by the first column is the same as its expansion by the third row, and that both match formula (2).
In Problems 57 and 58, use Cramer’s rule to solve for y only.
68. Repeat Problem 67, using the second row and the third column.
57. 12x 0002 14y 0004 11z 0003 5 58. 2x 0002 y 0004 4z 0003 15 15x 0004 7y 0002 9z 0003 000213 0002x 0004 y 0004 2z 0003 5 5x 0002 3y 0004 2z 0003 0 3x 0004 4y 0002 2z 0003 4
69. If A0003 c
2 1
3 d 00022
and
show that det (AB) 0003 (det A)(det B).
B0003 c
00021 2
3 d 1
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70. If A0003 c
a c
b d d
and
B0003 c
w y
x d z
show that det (AB) 0003 (det A)(det B). It is clear that x 0003 0, y 0003 0, z 0003 0 is a solution to each of the systems given in Problem 71. Use Cramer’s rule to determine whether this solution is unique. [Hint: If D 0005 0, what can you conclude? If D 0003 0, what can you conclude?] 71. (a). x 0002 4y 0004 9z 0003 0 4x 0002 y 0004 6z 0003 0 x 0002 y 0004 3z 0003 0
(b).
3x 0002 y 0004 3z 0003 0 5x 0004 5y 0002 9z 0003 0 00022x 0004 y 0002 3z 0003 0
(A) To analyze the effect of price changes on the daily revenue, an economist wants to express the daily revenue R in terms of p and q only. Use system (1) to eliminate x and y in the equation for R, expressing the daily revenue in terms of p and q. (B) To analyze the effect of changes in demand on the daily revenue, the economist now wants to express the daily revenue in terms of x and y only. Use Cramer’s rule to solve system (1) for p and q in terms of x and y and then express the daily revenue R in terms of x and y. 74. REVENUE ANALYSIS A company manufactures ten-speed and three-speed bicycles. The weekly demand equations are p 0003 230 0002 10x 0004 5y q 0003 130 0004 4x 0002 4y
72. Prove Theorem 2 for y.
(2)
where $p is the price of a ten-speed bicycle, $q is the price of a three-speed bicycle, x is the weekly demand for ten-speed bicycles, and y is the weekly demand for three-speed bicycles. The weekly revenue R is given by
APPLICATIONS 73. REVENUE ANALYSIS A supermarket sells two brands of coffee: brand A at $p per pound and brand B at $q per pound. The daily demand equations for brands A and B are, respectively, x 0003 200 0002 6p 0004 4q y 0003 300 0004 2p 0002 3q
(1)
(both in pounds). The daily revenue R is given by
R 0003 xp 0004 yq (A) Use system (2) to express the daily revenue in terms of x and y only. (B) Use Cramer’s rule to solve system (2) for x and y in terms of p and q, and then express the daily revenue R in terms of p and q only.
R 0003 xp 0004 yq
CHAPTER
7-1
7
Review
Systems of Linear Equations
A system of two linear equations in two variables is a system of the form ax 0004 by 0003 h cx 0004 dy 0003 k
(1)
where x and y are variables; a, b, c, and d are real numbers called the coefficients of x and y, and h and k are real numbers called the constant terms in the equations. The ordered pair of numbers (x0, y0) is a solution to system (1) if each equation is satisfied by the pair. The set of all such ordered pairs of numbers is called the solution set for the system. To solve a system is to find its solution set. In general, a system of linear equations has exactly one solution, no solution, or infinitely many solutions. A system of linear equations is consistent if it has one or more solutions and inconsistent if no solutions exist. A consistent system is said to be independent if it has exactly one solution and dependent if it has more than one solution. To solve a system by substitution, solve either equation for either variable, substitute in the other equation, solve the resulting linear equation in one variable, and then substitute this value into the expression obtained in the first step to find the other variable.
Two systems of equations are equivalent if both have the same solution set. To solve a system of equations using elimination by addition, use Theorem 2 to find a simpler equivalent system whose solution is obvious. As stated in Theorem 2, a system of linear equations is transformed into an equivalent system if: 1. Two equations are interchanged. 2. An equation is multiplied by a nonzero constant. 3. A constant multiple of another equation is added to a given equation. The solution set S of a dependent system is often expressed in terms of a parameter. Any element in S is called a particular solution. Any equation that can be written in the form ax 0004 by 0004 cz 0003 k where a, b, c, and k are constants (not all a, b, and c zero) is called a linear equation in three variables. The method of elimination by addition can be used for systems of linear equations in three variables.
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7-2
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497
3. The column containing the leftmost 1 of a given row has 0’s above and below the 1.
The method of solution using elimination by addition can be transformed into a more efficient method for larger-scale systems by the introduction of an augmented matrix. A matrix is a rectangular array of numbers written within brackets. Each number in a matrix is called an element of the matrix. If a matrix has m rows and n columns, it is called an m ⴛ n matrix (read “m by n matrix”). The expression m 0006 n is called the size of the matrix, and the numbers m and n are called the dimensions of the matrix. A matrix with n rows and n columns is called a square matrix of order n. A matrix with only one column is called a column matrix, and a matrix with only one row is called a row matrix. The position of an element in a matrix is the row and column containing the element. This is usually denoted using double subscript notation aij, where i is the row and j is the column containing the element aij. The principal diagonal of a matrix A consists of the elements aii, i 0003 1, 2, . . . , n. Rather than using x, y, and z to denote variables, we will use subscript notation x1, x2, and x3. Related to the system
4. The leftmost 1 in any row is to the right of the leftmost 1 in the preceding row.
x1 0004 5x2 0002 3x3 0003 4
If at any point in the preceding process we obtain a row with all 0’s to the left of the vertical line and a nonzero number n to the right, we can stop, since we have a contradiction: 0 0003 n, n 0005 0. We can then conclude that the system has no solution. If this does not happen and we obtain an augmented matrix in reduced form without any contradictions, the solution can be found by converting back to equation form.
0002 4x3 0003 1
6x1
00022x1 0004 3x2 0004 4x3 0003 7 are the following matrices: Coefficient matrix
1 £ 6 00022
5 0 3
00023 00024 § 4
Constant matrix
4 £1§ 7
Augmented coefficient matrix
1 £ 6 00022
5 0 3
00023 4 00024 † 1 § 4 7
Two augmented matrices are row-equivalent, denoted by the symbol ⬃ between the two matrices, if they are augmented matrices of equivalent systems of equations. An augmented matrix is transformed into a row-equivalent matrix if any of the following row operations is performed: 1. Two rows are interchanged. 2. A row is multiplied by a nonzero constant. 3. A constant multiple of another row is added to a given row. These correspond to the operations on equations from Theorem 2 in Section 7–1. The following symbols are used to describe these row operations: 1. Ri 4 Rj means “interchange row i with row j.” 2. kRi S Ri means “multiply row i by the constant k.” 3. kRj 0004 Ri S Ri means “multiply row j by the constant k and add to row i.” As before, our objective is to start with the augmented matrix of a linear system and transform it using row operations into a simple form where the solution can be found easily. The simple form, called the reduced form, is achieved if: 1. Each row consisting entirely of 0’s is below any row having at least one nonzero element. 2. The leftmost nonzero element in each row is 1.
A reduced system is a system of linear equations that corresponds to a reduced augmented matrix. When a reduced system has more variables than equations and contains no contradictions, the system is dependent and has infinitely many solutions. The Gauss–Jordan elimination procedure for solving a system of linear equations is given in step-by-step form as follows: Step 1. Choose the leftmost nonzero column, and use appropriate row operations to get a 1 at the top. Step 2. Use multiples of the row containing the 1 from step 1 to get zeros in all remaining places in the column containing this 1. Step 3. Repeat step 1 with the submatrix formed by (mentally) deleting the row used in step 2 and all rows above this row. Step 4. Repeat step 2 with the entire matrix, including the mentally deleted rows. Continue this process until the entire matrix is in reduced form.
7-3
Matrix Operations
Two matrices are equal if they are the same size and their corresponding elements are equal. The sum of two matrices of the same size is a matrix with elements that are the sums of the corresponding elements of the two given matrices. Matrix addition is commutative and associative. A matrix with all zero elements is called the zero matrix. The negative of a matrix M, denoted 0002M, is a matrix with elements that are the negatives of the elements in M. If A and B are matrices of the same size, then we define subtraction as follows: A 0002 B 0003 A 0004 (0002B). The product of a number k and a matrix M, denoted by kM, is a matrix formed by multiplying each element of M by k. The product of a 1 0006 n row matrix and an n 0006 1 column matrix is a 1 0006 1 matrix given by nⴛ1 1ⴛn
[a1
a2
...
b1 1ⴛ1 b2 an ] ≥ ¥ 0003 [a1b1 0004 a2b2 0004 # # # 0004 anbn ] o bn
If A is an m 0006 p matrix and B is a p 0006 n matrix, then the matrix product of A and B, denoted AB, is an m 0006 n matrix whose element in the ith row and jth column is the real number obtained from the product of the ith row of A and the jth column of B. If the number of columns in A does not equal the number of rows in B, then the matrix product AB is not defined. Matrix multiplication is not commutative, and the zero property does not hold for matrix multiplication. That is, for matrices A and B, the matrix product AB can be zero without either A or B being the zero matrix.
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Solving Systems of Linear Equations Using Matrix Inverse Methods
The identity matrix for multiplication for the set of all square matrices of order n is the square matrix of order n, denoted by I, with 1’s along the principal diagonal (from upper left corner to lower right corner) and 0’s elsewhere. If M is a square matrix of order n and I is the identity matrix of order n, then IM 0003 MI 0003 M If M is a square matrix of order n and if there exists a matrix M 00021 (read “M inverse”) such that M 00021M 0003 MM 00021 0003 I then M 00021 is called the multiplicative inverse of M or, more simply, the inverse of M. If the augmented matrix [ M | I] is transformed by row operations into [I | B], then the resulting matrix B is M 00021. If, however, we obtain all 0’s in one or more rows to the left of the vertical line, then M 00021 does not exist and M is called a singular matrix. A system of linear equations with the same number of variables as equations such as a11x1 0004 a12 x2 0004 a13x3 0003 k1
A determinant of order n is a determinant with n rows and n columns. The value of a second-order determinant is the real number given by `
a11 a11x 0004 a12 y 0004 a13 z 0003 k1 a21x 0004 a22 y 0004 a23 z 0003 k2 with D 0003 † a21 a31x 0004 a32 y 0004 a33 z 0003 k3 a31
a11 £ a21 a31
a12 a22 a32
x0003 X
B
a13 x1 k1 a23 § £ x2 § 0003 £ k2 § a33 x3 k3
If the inverse of A exists, then the matrix equation has a unique solution given by X 0003 A00021B After multiplying B by A00021 on the left, it is easy to read the solution to the original system of equations.
7-5
Determinants and Cramer’s Rule
Associated with each square matrix A is a real number called the determinant of the matrix. The determinant of A is denoted by det A, or simply by writing the array of elements in A using vertical lines in place of square brackets. For example, det c
CHAPTER
a11 a21
7
a12 a11 d 0003 ` a22 a21
a12 a22 a32
a13 a23 † 0005 0 a33
then
a31x1 0004 a32x2 0004 a33x3 0003 k3
A
a12 ` 0003 a11a22 0002 a21a12 a22
The value of a third-order determinant is the sum of three products obtained by multiplying each element of any one row (or each element of any one column) by its cofactor. The cofactor of an element aij (from the ith row and jth column) is the product of the minor of aij and (00021)i0004j. The minor of an element aij is the determinant remaining after deleting the ith row and jth column. Systems of equations having the same number of variables as equations can also be solved using determinants and Cramer’s rule. Cramer’s rule for three equations and three variables is as follows: Given the system
a21x1 0004 a22x2 0004 a23x3 0003 k2 can be written as the matrix equation
a11 a21
k1 † k2 k3
a12 a13 a11 a22 a23 † † a21 a32 a33 a31 y0003 D
k1 k2 k3 D
a13 a23 † a33
z0003
a11 † a21 a31
k1 k2 † k3
Cramer’s rule can be generalized completely for any size linear system that has the same number of variables as equations. The formulas are easily remembered if you observe the following: 1. Determinant D is formed from the coefficients of x, y, and z, keeping the same relative position in the determinant as found in the system of equations. 2. Determinant D appears in the denominators for x, y, and z. 3. The numerator for x can be obtained from D by replacing the coefficients of x (a11, a21, and a31) with the constants k1, k2, and k3, respectively. Similar statements can be made for the numerators for y and z. Cramer’s rule is rarely used to solve systems of order higher than 3 by hand, because more efficient methods are available. Cramer’s rule, however, is a valuable tool in more advanced theoretical and applied mathematics.
a12 ` a22
Review Exercises
Work through all the problems in this chapter review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text.
a12 a22 a32 D
Solve the system in Problems 1–5 using substitution or elimination by addition. 1. 2x 0004 y 0003 7 3x 0002 2y 0003 0
2.
3x 0002 6y 0003 5 00022x 0004 4y 0003 1
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3.
4x 0002 3y 0003 00028 00022x 0004 32 y 0003 4
4.
x 0002 3y 0004 z 0003 4 0002x 0004 4y 0002 4z 0003 1 2x 0002 y 0004 5z 0003 00023
5. 2x 0004 y 0002 z 0003 5 x 0002 2y 0002 2z 0003 4 3x 0004 4y 0004 3z 0003 3
26. Solve the system using Cramer’s rule: 3x 0002 2y 0003
x 0004 3y 0003 00021 x1 0002 x2 0003 4 2x1 0004 x2 0003 2 Then write the linear system represented by each augmented matrix in your solution, and solve each of these systems graphically. Discuss the relationship between the solutions of these systems.
Perform each of the row operations indicated in Problems 7–9 on the following augmented matrix: c
28. Use an intersection routine on a graphing calculator to approximate the solution of the following system to two decimal places:
00024 5 ` d 00026 12
1 3
x 0004 3y 0003 9 00022x 0004 7y 0003 10
8. 13R2 S R2
7. R1 4 R2
Solve the system in Problems 29–34 using Gauss–Jordan elimination.
9. (00023)R1 0004 R2 S R2 In Problems 10–12, write the linear system corresponding to each reduced augmented matrix and solve. 10. c
1 0
4 0 d ` 1 00027
12. c
1 0
00021 4 ` d 0 0
11. c
1 0
00021 4 ` d 0 1
In Problems 13–21, perform the operations that are defined, given the following matrices: 00022 00021 d B0003 c 3 00024
5 d C 0003 [00021 6
3 4] D 0003 c d 00022
13. AB
14. CD
15. CB
16. AD
17. A 0004 B
18. C 0004 D
19. A 0004 C
20. 2A 0002 5B
21. CA 0004 C
22. Find the inverse of A0003 c
4 7 d 00021 00022
29. 3x1 0004 2x2 0003 3 x1 0004 3x2 0003 8
30. x1 0004 x2 0003 1 x1 0002 x3 0003 00022 x2 0004 2x3 0003 4
31. x1 0004 2x2 0004 3x3 0003 1 2x1 0004 3x2 0004 4x3 0003 3 x1 0004 2x2 0004 x3 0003 3
32. x1 0004 2x2 0002 x3 0003 2 2x1 0004 3x2 0004 x3 0003 00023 3x1 0004 5x2 0003 00021
33. x1 0002 2x2 0003 1 2x1 0002 x2 0003 0 x1 0002 3x2 0003 00022
34. x1 0004 2x2 0002 x3 0003 2 3x1 0002 x2 0004 2x3 0003 00023
In Problems 35–40, perform the operations that are defined, given the following matrices: 1 A0003 £ 4 00023
2 5§ 00021
D0003 c
7 0
00025 d 00022
9 00026
E0003 c
36. DA
37. BC
38. CB
39. DE
40. ED
1 A 0003 £ 00022 4
3x1 0004 2x2 0003 k1 4x1 0004 3x2 0003 k2
00021]
00023 d 2
0 4 1 0§ 00021 4
Show that AA00021 0003 I.
as a matrix equation, and solve using matrix inverse methods for: (A) k1 0003 3, k2 0003 5 (B) k1 0003 7, k2 0003 10 (C) k1 0003 4, k2 0003 2 Evaluate the determinants in Problems 24 and 25. 2 25. † 0 1
C 0003 [2 4
35. AD
23. Write the system
00023 ` 00021
0 8
6 B 0003 £ 0§ 00024
41. Find the inverse of
Show that A00021A 0003 I.
2 24. ` 00025
8
27. Use Gauss–Jordan elimination to solve the system
6. Solve the system by graphing. 3x 0002 2y 0003 8 x 0004 3y 0003 00021
4 A0003 c 0
499
3 5 00024
00024 0† 00022
42. Write the system x1 0004 2x2 0004 3x3 0003 k1 2x1 0004 3x2 0004 4x3 0003 k2 x1 0004 2x2 0004 x3 0003 k3 as a matrix equation, and solve using matrix inverse methods for: (A) k1 0003 1, k2 0003 3, k3 0003 3 (B) k1 0003 0, k2 0003 0, k3 0003 00022 (C) k1 0003 00023, k2 0003 00024, k3 0003 1
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Evaluate the determinants in Problems 43 and 44. 43. `
000214 1 2
2 44. † 00023 1
3 2 2` 3
00021 5 00022
APPLICATIONS 1 2† 4
45. Solve for y only using Cramer’s rule:
55. DIET A laboratory assistant needs a food mix that contains, among other things, 27 grams of protein, 5.4 grams of fat, and 19 grams of moisture. He has available mixes A, B, and C with the compositions listed in the table. How many grams of each mix should be used to get the desired diet mix? Set up a system of equations and solve using Gauss–Jordan elimination.
x 0002 2y 0004 z 0003 00026 y0002z0003
4
2x 0004 2y 0004 z 0003
2
46. Solve using Gauss–Jordan elimination: x1 0004
x2 0004
x3 0003 7,000
0.04x1 0004 0.05x2 0004 0.06x3 0003
360
0.04x1 0004 0.05x2 0002 0.06x3 0003
120
47. Show that u ` w
v u 0004 kv ` 0003 ` w 0004 kx x
v ` x
48. Discuss the number of solutions for the system corresponding to the reduced form shown here if (A) m 0005 0 (B) m 0003 0 and n 0005 0 (C) m 0003 0 and n 0003 0 1 £0 0
00023 4 2 † 5§ m n
0 1 0
50. If A is a nonzero square matrix of order n satisfying A2 0003 0, can A00021 exist? Explain. 51. For n 0006 n matrices A and C and n 0006 1 column matrices B and X, solve for X assuming all necessary inverses exist: AX 0002 B 0003 CX
Moisture (%)
A
30
3
10
B
20
5
20
C
10
4
10
56. RESOURCE ALLOCATION A Colorado mining company operates mines at Big Bend and Saw Pit. The Big Bend mine produces ore that is 5% nickel and 7% copper. The Saw Pit mine produces ore that is 3% nickel and 4% copper. How many tons of ore should be produced at each mine to obtain the amounts of nickel and copper listed in the table? Set up a matrix equation and solve using matrix inverses. Copper
(A) 3.6 tons
5 tons
(B) 3 tons
4.1 tons
(C) 3.2 tons
4.4 tons
57. LABOR COSTS A company with manufacturing plants in North and South Carolina has labor-hour and wage requirements for the manufacturing of computer desks and printer stands as given in matrices L and H:
1.7 h 0.9 h
2.4 h 1.8 h
0.8 h d 0.6 h
Desk Stand
Hourly wages North South Carolina Carolina plant plant
Show that A A 0003 I. 53. Clear the decimals in the system 0.04x1 0004 0.05x2 0004 0.06x3 0003
360
0.04x1 0004 0.05x2 0002 0.06x3 0003
120
x2 0004
Fat (%)
L0003 c
6 00026 § 1
00021
x1 0004
Protein (%)
Labor-hour requirements Fabricating Assembly Packaging department department department
52. Find the inverse of 5 5 1
Mix
Nickel
49. Discuss the number of solutions for a system of n equations in n variables if the coefficient matrix: (A) Has an inverse. (B) Does not have an inverse.
4 A 0003 £4 1
54. BUSINESS A container holds 120 packages. Some of the packages weigh 12 pound each, and the rest weigh 13 pound each. If the total contents of the container weigh 48 pounds, how many are there of each type of package?
x3 0003 7,000
by multiplying the first two equations by 100. Then write the resulting system as a matrix equation and solve using the inverse found in Problem 52.
$11.50 H 0003 £ $9.50 $5.00
$10.00 $8.50 § $4.50
Fabricating department Assembly department Packaging department
(A) Find the labor cost for producing one printer stand at the South Carolina plant. (B) Discuss possible interpretations of the elements in the matrix products HL and LH. (C) If either of the products HL or LH has a meaningful interpretation, find the product and label its rows and columns.
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58. LABOR COSTS The monthly production of computer desks and printer stands for the company in Problem 57 for the months of January and February are given in matrices J and F: January production North South Carolina Carolina plant plant
1,500 J0002 c 850
Desks
1,810 d 740
Desks Stands
(A) Find the average monthly production for the months of January and February.
ZZZ
21 21 27 30 28 31 29 34 50 46 35 62 19 21 39 52 52 79
Stands
February production North South Carolina Carolina plant plant
CHAPTER
(B) Find the increase in production from January to February. 1 (C) Find J c d and interpret. 1 59. CRYPTOGRAPHY The following message was encoded with the matrix B shown below. Decode the message:
1,650 d 700
1,700 F0002 c 930
501
1 B 0002 £1 1
1 0 1
0 1§ 1
60. PUZZLE A piggy bank contains 30 coins worth $1.90. (A) If the bank contains only nickels and dimes, how many coins of each type does it contain? (B) If the bank contains nickels, dimes, and quarters, how many coins of each type does it contain?
7
GROUP ACTIVITY Modeling with Systems of Linear Equations
In this group activity, we will consider two real-world problems that can be solved using systems of linear equations: heat conduction and traffic flow. Both problems involve using a grid and a basic assumption to construct the model (the system of equations). Gauss–Jordan elimination is then used to solve the model. In the heat conduction problem, the solution of the model is easily interpreted in terms of the original problem. The system in the second problem is dependent, and the solution requires a more careful interpretation.
I HEAT CONDUCTION A metal grid consists of four thin metal bars. The end of each bar of the grid is kept at a constant temperature, as shown in Figure 1. We assume that the temperature at each intersection point in the grid is the average of the temperatures at the four 00004
400004 400004
x1
x2
x3
x4
200004
Z Figure 1
400004
200004
400004 00004
adjacent points in the grid (adjacent points are either other intersection points or ends of bars). So the temperature x1 at the intersection point in the upper left-hand corner of the grid must satisfy Left
x1 0002 14(40
Above
0003
0
Right
0003
x2 0003
Below
x3)
Find equations for the temperature at the other three intersection points, and solve the resulting system to find the temperature at each intersection point in the grid.
II TRAFFIC FLOW The rush-hour traffic flow for a network of four one-way streets in a city is shown in Figure 2 on page 506. The numbers next to each street indicate the number of vehicles per hour that enter and leave the network on that street. The variables x1, x2, x3, and x4 represent the flow of traffic between the four intersections in the network. For a smooth flow of traffic, we assume that the number of vehicles entering each intersection should always equal the number leaving. For example, since 1,500 vehicles enter the intersection of 5th Street and Washington Avenue each hour and x1 0003 x4 vehicles leave this intersection, we see that x1 0003 x4 0002 1,500. (A) Find the equations determined by the traffic flow at each of the other three intersections. (B) Find the solution to the system in part A.
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800
700 600 x1
x4
x2
900 400 5th St.
600 Washington Ave.
x3 500 6th St.
500
Lincoln Ave.
Z Figure 2 (C) What is the maximum number of vehicles that can travel from Washington Avenue to Lincoln Avenue on 5th Street? What is the minimum number?
(D) If traffic lights are adjusted so that 1,000 vehicles per hour travel from Washington Avenue to Lincoln Avenue on 5th Street, determine the flow around the rest of the network.
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7-6
Systems of Nonlinear Equations
1
Systems of Nonlinear Equations Z Solving by Substitution Z Solving by Other Methods
If a system of equations contains any equations that are not linear, then the system is called a nonlinear system. In this section we will investigate nonlinear systems involving seconddegree terms, like these: x2 0004 y2 0003 5 3x 0004 y 0003 1
x2 0002 2y2 0003 2 xy 0003 2
x2 0004 3xy 0004 y2 0003 20 xy 0002 y2 0003 0
It can be shown that such systems have at most four solutions, some of which may be imaginary. Since we are interested in finding both real and imaginary solutions to the systems we consider, we will assume that the replacement set for each variable is the set of complex numbers, rather than the set of real numbers.
Z Solving by Substitution The substitution method we used to solve linear systems of two equations in two variables is also an effective method for solving nonlinear systems. This process is best illustrated by examples.
EXAMPLE
1
Solving a Nonlinear System by Substitution Solve the system:
SOLUTION
x2 0004 y2 0003 5 3x 0004 y 0003 1
As usual, can start with either equation. But since the y term in the second equation is a first-degree term with coefficient 1, our calculations will be simplified if we start by solving for y in terms of x in the second equation. Next we substitute for y in the first equation to obtain an equation that involves only x: Add 00023x to both sides. Substitute for y in the second equation.
f
3x 0004 y 0003 1 y 0003 1 0002 3x T
x2 0004 y2 0003 5 2 x 0004 (1 0002 3x)2 0003 5 10x2 0002 6x 0002 4 0003 0 5x2 0002 3x 0002 2 0003 0 (x 0002 1)(5x 0004 2) 0003 0 x 0003 1,
Multiply parentheses and collect like terms on the left side. Divide both sides by 2. Factor. Use the zero property.
000225
If we substitute these values back into the equation y 0003 1 0002 3x, we obtain two solutions to the system: x00031 y 0003 1 0002 3(1) 0003 00022
x 0003 000225 y 0003 1 0002 3(000225) 0003 115
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y
A check, which you should provide, verifies that (1, 00022) and (000225, 115) are both solutions to the system. These solutions are illustrated in Figure 1. However, if we substitute the values of x back into the equation x2 0004 y2 0003 5, we obtain
5
x2 0004 y2 0003 5 00025
5
x00031 2 1 0004 y2 0003 5 y2 0003 4 y 0003 00052
x
00025
x 0003 000225 (000225)2 0004 y2 0003 5 y2 0003 121 25 y 0003 0005115
It appears that we have found two additional solutions, (1, 2) and (000225, 0002115). But neither of these solutions satisfies the equation 3x 0004 y 0003 1, which you should verify. So, neither is a solution of the original system. We have produced two extraneous roots, apparent solutions that do not actually satisfy both equations in the system. This is a common occurrence when solving nonlinear systems.
Z Figure 1
It is always very important to check the solutions of any nonlinear system to ensure that extraneous roots have not been introduced.
MATCHED PROBLEM 1
ZZZ EXPLORE-DISCUSS 1
Solve the system:
x2 0004 y2 0003 10 2x 0004 y 0003 1 0002
In Example 1, we saw that the line 3x 0004 y 0003 1 intersects the circle x2 0004 y2 0003 5 in two points. (A) Consider the system x2 0004 y2 0003 5 3x 0004 y 0003 10 Graph both equations in the same coordinate system. Are there any real solutions to this system? Are there any complex solutions? Find any real or complex solutions. (B) Consider the family of lines given by 3x 0004 y 0003 b
b any real number
What do all these lines have in common? Illustrate graphically the lines in this family that intersect the circle x2 0004 y2 0003 5 in exactly one point. How many such lines are there? What are the corresponding value(s) of b? What are the intersection points? How are these lines related to the circle?
EXAMPLE
2
Solving a Nonlinear System by Substitution Solve:
x2 0002 2y2 0003 2 xy 0003 2
0002
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SECTION 7–6
SOLUTION
Systems of Nonlinear Equations
3
Solve the second equation for y, substitute in the first equation, and proceed as before. xy 0003 2
Divide both sides by x.
2 y0003 x
Substitute for y in the first equation.
2 2 x2 0002 2 a b 0003 2 x
Simplify the left side.
8 00032 x2 x4 0002 2x2 0002 8 0003 0 u2 0002 2u 0002 8 0003 0 (u 0002 4)(u 0004 2) 0003 0 u 0003 4, 00022 x2 0002
Multiply both sides by x2 and simplify. Substitute u 0003 x2 (see Section 1-6). Factor. Use the zero property.
Replacing u with x2, we get x2 0003 4 x 0003 00052
y 5
xy 0003 00022
or
x2 0003 00022 x 0003 0005 100022 0003 0005i12
2 0003 1. 2 2 For x 0003 00022, y 0003 0003 00021. 00022 For x 0003 2, y 0003
5
x
x 2 0002 2y 2 0003 2
Z Figure 2
MATCHED PROBLEM 2
ZZZ EXPLORE-DISCUSS 2
2 0003 0002i 12. i12 2 For x 0003 0002i12, y 0003 0003 i12. 0002i 12 For x 0003 i 12, y 0003
There are four solutions to this system: (2, 1), (00022, 00021), (i 12, 0002i 12), and (0002i12, i12). Notice that two of the solutions involve imaginary numbers. These imaginary solutions cannot be illustrated graphically (see Fig. 2); however, they do satisfy both equations in the system (verify this). 0002 Solve:
3x2 0002 y2 0003 6 xy 0003 3
0002
(A) Refer to the system in Example 2. Could a graphing calculator be used to find the real solutions of this system? The imaginary solutions? (B) In general, explain why graphic approximation techniques can be used to approximate the real solutions of a system, but not the complex solutions.
EXAMPLE
3
Design An engineer is hired to design a rectangular computer screen with a 19-inch diagonal and a 175-square-inch area. Find the dimensions of the screen to the nearest tenth of an inch.
SOLUTION
Sketch a rectangle letting x be the width and y the height (Fig. 3). We obtain the following system using the Pythagorean theorem and the formula for the area of a rectangle: x2 0004 y2 0003 192 xy 0003 175
Pythagorean theorem width 0004 height 0003 Area
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This system is solved using the procedures outlined in Example 2. However, in this case, we are only interested in real solutions. We start by solving the second equation for y in terms of x and substituting the result into the first equation.
y
y0003 x
175 x
1752 0003 192 x2 x4 0004 30,625 0003 361x2 4 x 0002 361x2 0004 30,625 0003 0 x2 0004
Z Figure 3
Multiply both sides by x2 and simplify. Subtract 361x2 from each side. Quadratic in x2
Solve the last equation for x2 using the quadratic formula, then solve for x: x0003
361 0005 23612 0002 4(1)(30,625) B 2
⬇ 15.0 inches or 11.7 inches Substitute each choice of x into y 0003 175兾x to find the corresponding y values: For x 0003 15.0 inches, 175 y0003 0003 11.7 inches 15
For x 0003 11.7 inches, 175 0003 15.0 inches y0003 11.7
In either case, the dimensions are 15.0 by 11.7 inches.
0002
An engineer is hired to design a rectangular television screen with a 21-inch diagonal and a 209-square-inch area. Find the dimensions of the screen to the nearest tenth of an inch. 0002
MATCHED PROBLEM 3
Technology Connections the upper half of the circle and another for the lower half. [Note: since x and y must be nonnegative real numbers, we ignore the intersection points in the third quadrant—see Fig 4(a).]
In Example 3, we’re only concerned with real solutions, so graphic techniques can also be used to approximate the solutions (see Fig. 4). As we saw in Section 2-2, graphing a circle on a graphing calculator requires two functions, one for 40
000260
16
60
000240
(a) y1 0003 2361 0002 x2 y2 0003 0002 2361 0002 x2 175 y3 0003 x
8
16
18
10
(b) Intersection point: (11.7, 15.0)
2 2 2 Z Figure 4 Graphic solution of x 0004 y 0003 19 , xy 0003 175.
8
18
10
(c) Intersection point: (15.0, 11.7)
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5
Z Solving by Other Methods We’ll conclude the section by looking at some other techniques for solving nonlinear systems of equations.
EXAMPLE
4
Solving a Nonlinear System by Elimination Solve:
SOLUTION
x2 0002 y2 0003 5 x2 0004 2y2 0003 17
Because both x and y have the same power in each equation, elimination by addition is a good choice here. (We could also use substitution.) Multiply the second equation by 00021 and add: x2 0002 y2 0003 5 2 2 0002x 0002 2y 0003 000217 00023y2 0003 000212 y2 0003 4 y 0003 00052 Now substitute y 0003 2 and y 0003 00022 back into either original equation to find x. For y 0003 2, 2
For y 0003 00022, x2 0002 (00022)2 0003 5 x2 0003 9
2
x 0002 (2) 0003 5 x2 0003 9 x 0003 00053
x 0003 00053
There are four solutions to the system: (3, 00022), (3, 2), (00023, 00022), and (00023, 2). The check of the solutions is left to you. 0002 MATCHED PROBLEM 4
Solve: 2x2 0002 3y2 0003 5 3x2 0004 4y2 0003 16 0002
EXAMPLE
5
Solving a Nonlinear System Using Factoring and Substitution Solve:
SOLUTION
x2 0004 3xy 0004 y2 0003 20 xy 0002 y2 0003 0
We begin by solving the second equation for y: xy 0002 y2 0003 0 y(x 0002 y) 0003 0 y00030 or
Factor the left side. Use the zero property.
y0003x
Now, the original system is equivalent to the two systems: y0003 0 x 0004 3xy 0004 y2 0003 20
or
2
These systems can be solved by substitution.
y0003 x x 0004 3xy 0004 y2 0003 20 2
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y0003 0 x 0004 3xy 0004 y2 0003 20 x2 0004 3x(0) 0004 (0)2 0003 20
FIRST SYSTEM
2
Substitute y 0003 0 in the second equation, and solve for x.
x2 0003 20 x 0003 0005 120 0003 0005215 (2 15, 0) (00022 15, 0) y0003 x x2 0004 3xy 0004 y2 0003 20 x2 0004 3xx 0004 x2 0003 20 5x2 0003 20 x2 0003 4 x 0003 00052 (2, 2) (00022, 00022)
Take the square root of each side.
Solutions to the first system
SECOND SYSTEM
Substitute y 0003 x in the second equation and solve for x.
Divide both sides by 5. Take the square root of each side. Substitute these values back into y 0003 x to find y. Solutions to the second system.
Combining the solutions for the first system with the solutions for the second system, the solutions for the original system are (2 15, 0), (0002215, 0), (2, 2), and (00022, 00022). The check of the solutions is left to you. 0002 MATCHED PROBLEM 5
Solve:
x2 0002 xy 0004 y2 0003 9 2x2 0002 xy 0003 0
0002
Example 5 is somewhat specialized. However, it suggests a procedure that is effective for some problems.
EXAMPLE
6
Graphic Approximations of Real Solutions Use a graphing calculator to approximate real solutions to two decimal places: x2 0002 4xy 0004 y2 0003 12 2x2 0004 2xy 0004 y2 0003 6
SOLUTION
Before we can enter these equations in our calculator, we must solve for y: x2 0002 4xy 0004 y2 0003 12 y2 0002 4xy 0004 (x2 0002 12) 0003 0
a 0003 1, b 0003 00024x, c 0003 x2 0002 12
2x2 0004 2xy 0004 y2 0003 6 y2 0004 2xy 0004 (2x2 0002 6) 0003 0
a 0003 1, b 0003 2x, c 0003 2x2 0002 6
Applying the quadratic formula to each equation, we have y0003 0003
4x 0005 216x2 0002 4(x2 0002 12) 2 4x 0005 212x2 0004 48 2
0003 2x 0005 23x2 0004 12
y0003 0003
00022x 0005 24x2 0002 4(2x2 0002 6) 2 00022x 0005 224 0002 4x2 2
0003 0002x 0005 26 0002 x2
Since each equation has two solutions, we must enter four functions in the graphing calculator, as shown in Figure 5(a). Examining the graph in Figure 5(b), we see that there are four intersection points. Using the INTERSECT command repeatedly (details omitted), we find that the solutions to two decimal places are (00022.10, 0.83), (00020.37, 2.79), (0.37, 00022.79), and (2.10, 00020.83).
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7
Systems of Nonlinear Equations
5
00027.6
7.6
00025
(a)
(b)
Z Figure 5
MATCHED PROBLEM 6
0002
Use a graphing calculator to approximate real solutions to two decimal places: x2 0004 8xy 0004 y2 0003 70 2x2 0002 2xy 0004 y2 0003 20 0002
ANSWERS TO MATCHED PROBLEMS 1. (00021, 3), (95, 0002135) 2. (13, 13), (000213, 0002 13) (i, 00023i), (0002i, 3i) 3. 17.1 by 12.2 in. 4. (2, 1), (2, 00021), (00022, 1), (00022, 00021) 5. (0, 3), (0, 00023), ( 13, 213), (000213, 0002213) 6. (00023.89, 00021.68), (00020.96, 00025.32), (0.96, 5.32), (3.89, 1.68)
7-6
Exercises
*Additional answers can be found in the Instructor Answer Appendix.
A
1. How can you tell when a system of equations is nonlinear? 2. What is an extraneous solution? Why are they important in the study of solving nonlinear systems? 3. Would you choose substitution or elimination to solve the following nonlinear system? ax 0004 by 0003 c a, b, c, d, e, and f 0006 0 dx2 0004 ey2 0003 f Justify your answer by describing the steps you would take to solve this system? 4. Repeat Problem 3 for the following nonlinear system. ax2 0004 by2 0003 c a, b, c, d, e, and f 0006 0 dx2 0004 ey2 0003 f
Solve each system in Problems 5–16. 2
2
2
2
5. x 0004 y 0003 169 x 0003 000212
6. x 0004 y 0003 25 y 0003 00024
7. 8x2 0002 y2 0003 16 y 0003 2x
8. y2 0003 2x x 0003 y 0002 12
(000212, 5), (000212, 00025)
(2, 4), (00022, 00024)
(3, 00024), (00023, 00024)
(12 , 1)
9. 3x2 0002 2y2 0003 25 x0004y00030 (5, 00025), (00025, 5)
10. x2 0004 4y2 0003 32 x 0004 2y 0003 0 (00024, 2), (4, 00022)
y2 0003 x x 0002 2y 0003 2
12. x2 0003 2y 3x 0003 y 0004 2
(3 0004 15, 7 0004 315), (3 0002 15, 7 0002 315)
13. 2x2 0004 y2 0003 24 x2 0002 y2 0003 000212
14. x2 0002 y2 0003 3 x2 0004 y2 0003 5
(2, 1), (2, 00021), (00022, 00021), (00022, 1)
11.
15.
x2 0004 y2 0003 10 16x2 0004 y2 0003 25
(1, 3), (1, 00023), (00021, 3), (00021, 00023)
16. x2 0002 2y2 0003 1 x2 0004 4y2 0003 25 (3, 2), (3, 00022), (00023, 2), (00023, 00022)
B Solve each system in Problems 17–28. 17. xy 0002 4 0003 0 x0002y00032
18. xy 0002 6 0003 0 x0002y00034
19. x2 0004 2y2 0003 6 xy 0003 2
20. 2x2 0004 y2 0003 18 xy 0003 4
21. 2x2 0004 3y2 0003 00024 4x2 0004 2y2 0003 8
22. 2x2 0002 3y2 0003 10 x2 0004 4y2 0003 000217
23. x2 0002 y2 0003 2 y2 0003 x
24. x2 0004 y2 0003 20 x2 0003 y
25. x2 0004 y2 0003 9 x2 0003 9 0002 2y
26. x2 0004 y2 0003 16 y2 0003 4 0002 x
(2, 4), (00022, 4), (15i, 00025), (000215i, 00025)
(4, 0), (00023, 17), (00023, 000217)
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27. x2 0002 y2 0003 3 xy 0003 2
28. y2 0003 5x2 0004 1 xy 0003 2
(i, 00022i ), (0002i, 2i ), 00022 15 (2 15 5 , 15), ( 5 , 000215)
(2, 1), (00022, 00021), (i, 00022i ), (0002i, 2i )
An important type of calculus problem is to find the area between the graphs of two functions. To solve some of these problems it is necessary to find the coordinates of the points of intersections of the two graphs. In Problems 29–36, find the coordinates of the points of intersections of the two given equations. 29. y 0003 5 0002 x2, y 0003 2 0002 2x
30. y 0003 5x 0002 x2, y 0003 x 0004 3
31. y 0003 x2 0002 x, y 0003 2x
32. y 0003 x2 0004 2x, y 0003 3x
(00021, 4), (3, 00024)
(1, 4), (3, 6)
(0, 0), (3, 6)
(1, 3), (0, 0) (1, 4), (4, 1)
2
33. y 0003 x 0002 6x 0004 9, y 0003 5 0002 x 34. y 0003 x2 0004 2x 0004 3, y 0003 2x 0004 4 35. y 0003 8 0004 4x 0002 x2, y 0003 x2 0002 2x
C
Solve each system in Problems 39–46.
2
2
2
(1
0002 15), 12 (3 0004 15)
0004 15 00021 0004 15 , ), (1 00022 15 , 00021 00022 15 ) 2 2
57. DESIGN An engineer is designing a portable DVD player. According to the design specifications, the set must have a rectangular screen with a 7.5-inch diagonal and an area of 27 square inches. Find the dimensions of the screen. 6 by 4.5 in. 58. DESIGN An artist is designing a logo for a business in the shape of a circle with an inscribed rectangle. The diameter of the circle is 6.5 inches, and the area of the rectangle is 15 square inches. Find the dimensions of the rectangle. 2x 0003 6 in. by 2y 0003 2.5 in.
6.5 inches
59. CONSTRUCTION A rectangular swimming pool with a deck 5 feet wide is enclosed by a fence as shown in the figure. The surface area of the pool is 572 square feet, and the total area enclosed by the fence (including the pool and the deck) is 1,152 square feet. Find the dimensions of the pool. 22 by 26 ft
Fence
(3, 3 ), (2 , 2) 2
42. x 0004 xy 0002 y 0003 00025 y0002x00033
43. 2x2 0002 xy 0004 y2 0003 8 x2 0002 y2 0003 0
44. x2 0004 2xy 0004 y2 0003 36 x2 0002 xy 0003 0
45. x2 0004 xy 0002 3y2 0003 3 x2 0004 4xy 0004 3y2 0003 0
46. x2 0002 2xy 0004 2y2 0003 16 x2 0002 y2 0003 0
(00023, 1), (3, 00021), (0002i, i ), (i, 0002i )
1 2 (3
54. NUMBERS Find two numbers that differ by 1 and have product 1. (Let x be the larger number and y the smaller number.)
40. 2x 0004 3y 0004 xy 0003 16 xy 0002 55 00035 0
41. x 0002 2xy 0004 y 0003 1 x 0002 2y 0003 2 (0, 00021), (00024, 00023)
53. NUMBERS Find two numbers with sum 3 and product 1.
2 meters by 16 meters (4, 000210), (00023, 11)
38. Consider the circle with equation x2 0004 y2 0003 25 and the family of lines given by 3x 0004 4y 0003 b, where b is any real number. (A) Illustrate graphically the lines in this family that intersect the circle in exactly one point, and describe the relationship between the circle and these lines. (B) Find the values of b corresponding to the lines in part A, and find the intersection points of the lines and the circle. (C) How is the line with equation 4x 0002 3y 0003 0 related to this family of lines? How could this line be used to find the intersection points and the values of b in part B?
(00025, 00025 ), (00022 , 00022)
APPLICATIONS
5 in. and 12 in.
37. Consider the circle with equation x2 0004 y2 0003 5 and the family of lines given by 2x 0002 y 0003 b, where b is any real number. (A) Illustrate graphically the lines in this family that intersect the circle in exactly one point, and describe the relationship between the circle and these lines. (B) Find the values of b corresponding to the lines in part A, and find the intersection points of the lines and the circle. (C) How is the line with equation x 0004 2y 0003 0 related to this family of lines? How could this line be used to find the intersection points in part B?
39. 2x 0004 5y 0004 7xy 0003 8 xy 30002 3 0003 0 3
(00022.09, 00020.66), (0.18, 00023.64), (0.58, 2.83), (1.73, 1.28)
56. GEOMETRY Find the dimensions of a rectangle with an area of 32 square meters if its perimeter is 36 meters long.
(00021, 3), (4, 8)
36. y 0003 x 0002 4x 0002 10, y 0003 14 0002 2x 0002 x2
(00022.96, 00023.47), (00020.89, 00023.76), (1.39, 4.05), (2.46, 4.18)
52. 2x2 0004 2xy 0004 y2 0003 12 4x2 0002 4xy 0004 y2 0004 x 0004 2y 0003 9
55. GEOMETRY A right triangle with an area of 30 square inches has a hypotenuse that is 13 inches long. Find the lengths of the two legs.
(1, 6), (00021, 2)
2
51. 2x2 0002 2xy 0004 y2 0003 9 4x2 0002 4xy 0004 y2 0004 x 0003 3
5 ft 5 ft
(4, 7), (00021, 2)
Pool
(0, 6), (0, 00026), (3, 3), (00023, 00023)
00024 15 00024 15 4 15 (4, 4), (00024, 00024), (4 15 5 , 5 ), ( 5 , 5 )
In Problems 47–52, use a graphing calculator to approximate the real solutions of each system to two decimal places. 47. 0002x2 0004 2xy 0004 y2 0003 1 3x2 0002 4xy 0004 y2 0003 2
48. 0002x2 0004 4xy 0004 y2 0003 2 8x2 0002 2xy 0004 y2 0003 9
49. 3x2 0002 4xy 0002 y2 0003 2 2x2 0004 2xy 0004 y2 0003 9
50. 5x2 0004 4xy 0004 y2 0003 4 4x2 0002 2xy 0004 y2 0003 16
5 ft
5 ft
60. CONSTRUCTION An open-topped rectangular box is formed by cutting a 6-inch square from each corner of a rectangular piece of cardboard and bending up the ends and sides. The area of the cardboard before the corners are removed is 768 square inches, and the volume of the box is 1,440 cubic inches. Find the dimensions of the original piece of cardboard. 24 in. by 32 in.
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6 in.
6 in.
6 in.
6 in. 6 in.
62. TRANSPORTATION Bus A leaves Milwaukee at noon and travels west on Interstate 94. Bus B leaves Milwaukee 30 minutes later, travels the same route, and overtakes bus A at a point 210 miles west of Milwaukee. If the average speed of bus B is 10 miles per hour greater than the average speed of bus A, at what time did bus B overtake bus A? 3:30 P.M.
6 in.
61. TRANSPORTATION Two boats leave Bournemouth, England, at the same time and follow the same route on the 75-mile trip across
7-7
9
the English Channel to Cherbourg, France. The average speed of boat A is 5 miles per hour greater than the average speed of boat B. Consequently, boat A arrives at Cherbourg 30 minutes before boat B. Find the average speed of each boat. Boat A: 30 mph; boat B: 25 mph
6 in.
6 in.
Systems of Linear Inequalities in Two Variables
Systems of Linear Inequalities in Two Variables Z Graphing Linear Inequalities in Two Variables Z Solving Systems of Linear Inequalities Graphically Z Application
Earlier in the book, we saw that in many applications, it’s more natural to consider inequalities than equations. The same is true for systems, so here and in Sections 7-8 we will study systems of linear inequalities. A graph is often the most convenient way to represent the solutions of a system of inequalities in two variables. In this section, we discuss techniques for graphing both a single linear inequality in two variables and a system of linear inequalities in two variables.
Z Graphing Linear Inequalities in Two Variables We know how to graph first-degree equations such as y 0003 2x 0002 3
and
2x 0002 3y 0003 5
but how do we graph first-degree inequalities, like the ones below? y 0007 2x 0002 3
and
2x 0002 3y 5
Actually, graphing these inequalities is almost as easy as graphing the equations. But before we begin, we should discuss some important subsets of a plane in a rectangular coordinate system. A line divides a plane into two halves called half-planes. A vertical line divides a plane into left and right half-planes [Fig. 1(a)]; a nonvertical line divides a plane into upper and lower half-planes [Fig. 1(b)]. y
Z Figure 1 Half-planes.
Left half-plane
y
Right half-plane
Upper half-plane x
x Lower half-plane
(a)
(b)
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Consider the following linear equation and related linear inequalities:
ZZZ EXPLORE-DISCUSS 1
(1) 2x 0002 3y 0003 12
(2) 2x 0002 3y 12
(3) 2x 0002 3y 12
(A) Graph the line with equation (1). (B) Find the point on this line with x coordinate 3 and draw a vertical line through this point. Discuss the relationship between the y coordinates of the points on this line and statements (1), (2), and (3). (C) Repeat part B for x 0003 00023. For x 0003 9. (D) Based on your observations in parts B and C, write a verbal description of all the points in the plane that satisfy equation (1), those that satisfy inequality (2), and those that satisfy inequality (3).
Now let’s investigate the half-planes determined by the linear equation y 0003 2x 0002 3. We start by graphing y 0003 2x 0002 3 (Fig. 2). For any given value of x, there is exactly one value for y such that (x, y) lies on the line. For the same x, if the point (x, y) is below the line, then y 2x 0002 3. So the lower half-plane corresponds to the solution of the inequality y 2x 0002 3. Similarly, the upper half-plane corresponds to the solution of the inequality y 2x 0002 3, as shown in Figure 2. y
Z Figure 2
y 0003 2x 0002 3 (4, y) y 2(4) 0002 3 0003 5; point in upper half-plane (4, y) y 0003 2(4) 0002 3 0003 5; point on line
5
(4, y) y 2(4) 0002 3 0003 5; point in lower half-plane 00025
5
10
x
00025
We can form four different inequalities from y 0003 2x 0002 3 by replacing the 0003 sign with , , 0007, and . They are y 2x 0002 3
y 2x 0002 3
y 0007 2x 0002 3
y 2x 0002 3
The graph of each is a half-plane. The line y 0003 2x 0002 3, called the boundary line for the half-plane, is included for and 0007 and excluded for and . In Figure 3, the half-planes are indicated with small arrows on the graph of y 0003 2x 0002 3 and then graphed as shaded regions. Included boundary lines are shown as solid lines, and excluded boundary lines are shown as dashed lines. Z Figure 3
y
0
y 2x 0002 3 (a)
y
x
0
y 2x 0002 3 (b)
y
x
0
y 0007 2x 0002 3 (c)
y
x
0
y 2x 0002 3 (d)
x
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11
Z THEOREM 1 Graphs of Linear Inequalities in Two Variables The graph of a linear inequality Ax 0004 By C
or
Ax 0004 By C
with B 0006 0, is either the upper half-plane or the lower half-plane (but not both) determined by the line Ax 0004 By 0003 C. If B 0003 0, then the graph of Ax C
or
Ax C
is either the left half-plane or the right half-plane (but not both) determined by the line Ax 0003 C.
As a consequence of Theorem 1, we can build a simple and fast mechanical procedure for graphing linear inequalities.
Z PROCEDURE FOR GRAPHING LINEAR INEQUALITIES IN TWO VARIABLES Step 1. Graph Ax 0004 By 0003 C as a dashed line if equality is not included in the original statement or as a solid line if equality is included. Step 2. Choose any point not on the line and substitute the coordinates into the inequality. [It’s usually easiest to use the origin (0,0) if you’re sure that it is not on the line.] Step 3. If the inequality is true when substituting in the test point, the graph of the original inequality is the half-plane containing that point. If the inequality is false, the graph of the original inequality is the half-plane not containing that point.
EXAMPLE
1
Graphing a Linear Inequality Graph: 3x 0002 4y 0007 12
SOLUTION y 5
3x 0002 4y 0003 12
Step 1. Graph 3x 0002 4y 0003 12 as a solid line, since equality is included in the original statement (Fig. 4). Step 2. Pick a convenient test point above or below the line. The point (0, 0) will be easy to test. Substituting (0, 0) into the inequality 3x 0002 4y 0007 12
5
00025
Z Figure 4
x
3(0) 0002 4(0) 0003 0 0007 12 produces a true statement; therefore, (0, 0) is in the solution set. Step 3. The line 3x 0002 4y 0003 12 and the half-plane containing the origin form the graph of 3x 0002 4y 0007 12 (Fig. 5).
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Test point x
5
00025
0002
Z Figure 5
MATCHED PROBLEM 1
EXAMPLE
2
Graph: 2x 0004 3y 6
0002
Graphing a Linear Inequality Graph: (A) y 00023
SOLUTIONS
(B) 2x 0007 5
(A) The solution to y 00023 is the set of all points with y coordinates greater than 00023. Its graph is shown in Figure 6. (B) The inequality 2x 0007 5 is equivalent to x 0007 5/2. This is the set of all points with x coordinates less than or equal to 5/2, as shown in Figure 7. y
y
5
5
Test point 00025
5
Test point
x
00025
00025
Graph: (A) y 0007 2
x
00025
Z Figure 6
MATCHED PROBLEM 2
5
0002
Z Figure 7
(B) 3x 00028
0002
Z Solving Systems of Linear Inequalities Graphically We will now consider systems of linear inequalities such as x0004y 6 2x 0002 y 0
and
2x 0004 y 0007 22 x 0004 y 0007 13 2x 0004 5y 0007 50 x 0 y 0
We will solve such systems graphically—that is, to find the graph of all ordered pairs of real numbers (x, y) that simultaneously satisfy all the inequalities in the system. The graph is called the solution region for the system. To find the solution region, we graph each
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13
Systems of Linear Inequalities in Two Variables
inequality in the system and then find the intersection of all the graphs. To simplify the discussion that follows: We will consider only systems of linear inequalities where equality is included in each statement in the system.
EXAMPLE
3
Solving a System of Linear Inequalities Graphically Solve the following system of linear inequalities graphically: x0004y 6 2x 0002 y 0
SOLUTION
First, graph the line x 0004 y 0003 6 and shade the region that satisfies the inequality x 0004 y 6. This region is shaded in blue in Figure 8(a) on the next page. Next, graph the line 2x 0002 y 0003 0 and shade the region that satisfies the inequality 2x 0002 y 0. This region is shaded in red in Figure 8(a). The solution region for the system of inequalities is the intersection of these two regions. This is the region shaded in both red and blue in Figure 8(a), which is redrawn in Figure 8(b) with only the solution region shaded for clarity. The coordinates of any point in the shaded region of Figure 8(b) specify a solution to the system. For example, the points (2, 4), (6, 3), and (7.43, 8.56) are three of infinitely many solutions, as can be easily checked. The intersection point (2, 4) can be obtained by solving the system of equations x 0004 y 0003 6 and 2x 0002 y 0003 0. y
2x 0002 y 0003 0
y
10
2x 0002 y 0003 0
10
Solution region 5
5
00025
5
10
(2, 4)
x
00025
x0004y00036
00025
5
00025
(a)
10
x
x0004y00036 (b)
0002
Z Figure 8
MATCHED PROBLEM 3
ZZZ EXPLORE-DISCUSS 2
Solve the following system of linear inequalities graphically:
3x 0004 y 0007 21 x 0002 2y 0007 0
0002
Refer to Example 3. Graph each boundary line and shade the regions obtained by reversing each inequality. That is, shade the region of the plane that corresponds to the inequality x 0004 y 6 and then shade the region that corresponds to the inequality 2x 0002 y 0. What portion of the plane is left unshaded? Compare this method with the one used in the solution to Example 3.
The points of intersection of the lines that form the boundary of a solution region play a fundamental role in the solution of linear programming problems, which are discussed in Section 7-8.
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Z DEFINITION 1 Corner Point A corner point of a solution region is a point in the solution region that is the intersection of two boundary lines.
The point (2, 4) is the only corner point of the solution region in Example 3; see Figure 8(b).
EXAMPLE
4
Solving a System of Linear Inequalities Graphically Solve the following system of linear inequalities graphically, and find the corner points. 2x 0004 y 0007 22 x 0004 y 0007 13 2x 0004 5y 0007 50 x 0 y 0
SOLUTION
The inequalities x 0 and y 0, called nonnegative restrictions, occur frequently in applications involving systems of inequalities since x and y often represent quantities that can’t be negative—number of units produced, number of hours worked, and the like. Because both x and y are nonnegative, the solution region lies in the first quadrant, and we can restrict our attention to that portion of the plane. First, we graph the lines 2x 0004 y 0003 22 x 0004 y 0003 13 2x 0004 5y 0003 50
Find the x and y intercepts of each line; then sketch the line through these points, as shown in Figure 9.
Next, choosing (0, 0) as a test point, we see that the graph of each of the first three inequalities in the system consists of its corresponding line and the half-plane lying below the line, as indicated by the arrows in Figure 9. So the solution region of the system consists of the points in the first quadrant that simultaneously lie on or below all three of these lines—see Figure 9. y
2x 0004 y 0003 22
x 0004 y 0003 13
20
2x 0004 5y 0003 50 (5, 8) (9, 4)
(0, 10)
00025
(0, 0)
Z Figure 9
(11, 0) 00025
20
x
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15
The corner points (0, 0), (0, 10), and (11, 0) can be found by just looking at the graph. The other two corner points are determined as follows: Solve the system
Solve the system
2x 0004 5y 0003 50 x 0004 y 0003 13 to obtain (5, 8).
2x 0004 y 0003 22 x 0004 y 0003 13 to obtain (9, 4).
Note that the lines 2x 0004 5y 0003 50 and 2x 0004 y 0003 22 also intersect, but the intersection point is not part of the solution region, and is not a corner point. 0002 MATCHED PROBLEM 4
Solve the following system of linear inequalities graphically, and find the corner points: 5x 0004 y 20 x 0004 y 12 x 0004 3y 18 x 0 y 0
0002
If we compare the solution regions of Examples 3 and 4, we see that there is a fundamental difference between these two regions. We could draw a circle that encloses the entire solution region in Example 4. However, it is impossible to include all the points in the solution region in Example 3 in any circle, no matter how large we draw it. This leads to the following definition.
Z DEFINITION 2 Bounded and Unbounded Solution Regions A solution region of a system of linear inequalities is bounded if it can be enclosed within a circle. If it cannot be enclosed within a circle, then it is unbounded.
So the solution region for Example 4 is bounded and the solution region for Example 3 is unbounded. This definition will be important in Section 7-8.
Z Application EXAMPLE
5
Production Scheduling A manufacturer of surfboards makes a standard model and a competition model. Each standard board requires 6 labor-hours for fabricating and 1 labor-hour for finishing. Each competition board requires 8 labor-hours for fabricating and 3 labor-hours for finishing. The maximum labor-hours available per week in the fabricating and finishing departments are 120 and 30, respectively. What combinations of boards can be produced each week so as not to exceed the number of labor-hours available in each department per week?
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To organize all of the information provided, a table is helpful.
Standard Model (labor-hours per board)
Competition Model (labor-hours per board)
Maximum Labor-Hours Available per Week
Fabricating
6
8
120
Finishing
1
3
30
Let x ⫽ Number of standard boards produced per week y ⫽ Number of competition boards produced per week These variables are restricted as follows: Fabricating department restriction: a
Weekly fabricating time Weekly fabricating time Maximum labor-hours b0002a bⱕa b available per week for x standard boards for y competition boards
6x
⫹
8y
ⱕ
120
Finishing department restriction: a
Weekly finishing time Weekly finishing time Maximum labor-hours b0002a bⱕa b for x standard boards for y competition boards available per week
1x
⫹
3y
ⱕ
30
Since it is not possible to manufacture a negative number of boards, x and y also must satisfy the nonnegative restrictions xⱖ0 yⱖ0
© Corbis RF
The production levels x and y must satisfy the following system of linear inequalities: 6x ⫹ 8y ⱕ 120 x ⫹ 3y ⱕ 30 xⱖ0 yⱖ0
Fabricating department restriction Finishing department restriction Nonnegative restriction Nonnegative restriction
Graphing this system of linear inequalities, we obtain the set of feasible solutions, or the feasible region, as shown in Figure 10. For problems of this type and for the linear programming problems we consider in the next section, solution regions are often referred to as feasible regions. Any point within the shaded area, including the boundary lines, represents a possible production schedule. Any point outside the shaded area represents an impossible schedule. For example, it would be possible to produce 12 standard boards and 5 competition boards per week, but it would not be possible to produce 12 standard boards and 7 competition boards per week (see the figure).
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17
y
20
Remark In Example 5, how do we interpret a production schedule of 10.5 standard boards and 4.3 competition boards? It is not possible to manufacture a fraction of a board. But it is possible to average 10.5 standard and 4.3 competition boards per week. In general, we will assume that all points in the feasible region represent acceptable solutions, even though noninteger solutions might require special interpretation.
MATCHED PROBLEM 5
Fabricating capacity line 6x 0004 8y 0003 120 (12, 7) (12, 6) (12, 5) Feasible region 00025
10
x
20
Finishing capacity line x 0004 3y 0003 30
00025
0002
Z Figure 10
Repeat Example 5 using 5 hours for fabricating a standard board and a maximum of 27 labor-hours for the finishing department. 0002 ANSWERS TO MATCHED PROBLEMS y
1. 5
00025
5
x
00025
y
2. (A)
y
(B)
5
5
00025
5
x
00025
00025
5
00025
y
3.
20
Solution region 10
x 0002 2y 0003 0
(6, 3) 000210
10 00025
x
3x 0004 y 0003 21
x
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4.
(0, 20) Solution region (2, 10)
10
(9, 3) (18, 0) 00025
x
10 00025
5x 0004 y 0003 20
x 0004 3y 0003 18
x 0004 y 0003 12
y
5. 20
, 15 冣 冢 144 7 7 00025
10
x
20
x 0004 3y 0003 27
00025
5x 0004 8y 0003 120
7-7
Exercises
*Additional answers can be found in the Instructor Answer Appendix.
A
1. If m 0006 0, describe geometrically the graphs of (A) y 0003 mx 0004 b (B) y mx 0004 b (C) y mx 0004 b
In Problems 15–18, match the solution region of each system of linear inequalities with one of the four regions shown in the figure.
2. If m 0006 0, describe geometrically the graphs of (A) y 0007 mx 0004 b (B) y mx 0004 b 3. What are nonnegativity restrictions? How do they affect the graph of a system of linear inequalities?
y II x 0004 2y 0003 8
4. In an applied problem, if x is the number of tables produced and y is the number of chairs produced, how would you interpret the solution x 0003 12.6 and y 0003 23.2?
6. 3x 0004 4y 12
7. 3x 0004 2y 18
8. 3y 0002 2x 24
9. y 0007
2 3x
00045
10. y 13x 0002 2
11. y 8
12. x 00025
13. 00023 0007 y 2
14. 00021 x 0007 3
I
5
(2, 3)
III
In Problems 5–14, graph the solution region of each inequality and write a verbal description of the solution region. 5. 2x 0002 3y 6
3x 0002 2y 0003 0
00025
5
IV 00025
15. x 0004 2y 0007 8 3x 0002 2y 0
16. x 0004 2y 8 3x 0002 2y 0007 0
17. x 0004 2y 8 3x 0002 2y 0
18. x 0004 2y 0007 8 3x 0002 2y 0007 0
Region IV
Region I
Region II
Region III
x
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SECTION 7–7
In Problems 19–24, solve each system of linear inequalities graphically. 19. x 5 y00076
20. x 0007 4 y 2
21. 3x 0004 y 6 x00074
22. 3x 0004 4y 0007 12 y 00023
23. x 0002 2y 0007 12 2x 0004 y 4
24. 2x 0004 5y 0007 20 x 0002 5y 0007 00025
B
In Problems 25–28, match the solution region of each system of linear inequalities with one of the four regions shown in the figure. Identify the corner points of each solution region. y
(0, 16)
10
whether each solution region is bounded or unbounded. Find the coordinates of each corner point. 41. x 0004 y 0007 11 5x 0004 y 15 x 0004 2y 12
42. 4x 0004 y 0007 32 x 0004 3y 0007 30 5x 0004 4y 51
43. 3x 0004 2y 24 3x 0004 y 0007 15 x 4
44. 3x 0004 4y 0007 48 x 0004 2y 24 y00079
The feasible region is empty.
45.
x 0004 y 0007 10 3x 0004 5y 15 3x 0002 2y 0007 15 00025x 0004 2y 0007 6
The feasible region is empty.
46. 3x 0002 y 1 0002x 0004 5y 9 x0004 y00079 y00075 48.
8x 0004 4y 0007 41 000215x 0004 5y 0007 19 2x 0004 6y 37
APPLICATIONS II (6, 4) III
(18, 0)
IV (8, 0)
00025
19
C In Problems 41–48, solve the systems graphically, and indicate
47. 16x 0004 13y 0007 119 12x 0004 16y 101 00024x 0004 3y 0007 11
I
(0, 6) (0, 0)
Systems of Linear Inequalities in Two Variables
10
x
x 0004 3y 0003 18
00025
2x 0004 y 0003 16
25. x 0004 3y 0007 18 2x 0004 y 16 x 0 y 0
26. x 0004 3y 0007 18 2x 0004 y 0007 16 x 0 y 0
Region III; corner points: (0, 0), (0, 6), (6, 4), (8, 0)
27. x 0004 3y 18 2x 0004 y 16 x 0 y 0
28. x 0004 3y 18 2x 0004 y 0007 16 x 0 y 0
Region II; corner points: (6, 4), (0, 6), (0, 16)
Region I, corner points: (0, 16), (6, 4), (18, 0)
In Problems 29–40, solve the systems graphically, and indicate whether each solution region is bounded or unbounded. Find the coordinates of each corner point. 29. 2x 0004 3y 0007 6 x 0 y 0
30. 4x 0004 3y 0007 12 x 0 y 0
31. 4x 0004 5y 20 x 0 y 0
32. 5x 0004 6y 30 x 0 y 0
33. 2x 0004 y 0007 8 x 0004 3y 0007 12 x 0 y 0
34. x 0004 2y 0007 10 3x 0004 y 0007 15 x 0 y 0
35. 4x 0004 3y 24 2x 0004 3y 18 x 0 y 0
36. x 0004 2y 8 2x 0004 y 10 x 0 y 0
37. 2x 0004 y 0007 12 x0004 y00077 x 0004 2y 0007 10 x 0 y 0
38. 3x 0004 y 0007 21 x0004 y00079 x 0004 3y 0007 21 x 0 y 0
39. x 0004 2y 16 x 0004 y 12 2x 0004 y 14 x 0 y 0
40. 3x 0004 y 30 x 0004 y 16 x 0004 3y 24 x 0 y 0
49. MANUFACTURING—RESOURCE ALLOCATION A manufacturing company makes two types of water skis: a trick ski and a slalom ski. The trick ski requires 6 labor-hours for fabricating and 1 labor-hour for finishing. The slalom ski requires 4 labor-hours for fabricating and 1 labor-hour for finishing. The maximum labor-hours available per day for fabricating and finishing are 108 and 24, respectively. If x is the number of trick skis and y is the number of slalom skis produced per day, write a system of inequalities that indicates appropriate restraints on x and y. Find the set of feasible solutions graphically for the number of each type of ski that can be produced. 50. MANUFACTURING—RESOURCE ALLOCATION A furniture manufacturing company manufactures dining room tables and chairs. A table requires 8 labor-hours for assembling and 2 labor-hours for finishing. A chair requires 2 labor-hours for assembling and 1 labor-hour for finishing. The maximum labor-hours available per day for assembly and finishing are 400 and 120, respectively. If x is the number of tables and y is the number of chairs produced per day, write a system of inequalities that indicates appropriate restraints on x and y. Find the set of feasible solutions graphically for the number of tables and chairs that can be produced. 51. MANUFACTURING—RESOURCE ALLOCATION Refer to Problem 49. The company makes a profit of $50 on each trick ski and a profit of $60 on each slalom ski. (A) If the company makes 10 trick and 10 slalom skis per day, the daily profit will be $1,100. Are there other feasible production schedules that will result in a daily profit of $1,100? How are these schedules related to the graph of the line 50x 0004 60y 0003 1,100? (B) Find a feasible production schedule that will produce a daily profit greater than $1,100 and repeat part A for this schedule. (C) Discuss methods for using lines like those in parts A and B to find the largest possible daily profit. 52. MANUFACTURING—RESOURCE ALLOCATION Refer to Problem 50. The company makes a profit of $50 on each table and a profit of $15 on each chair. (A) If the company makes 20 tables and 20 chairs per day, the daily profit will be $1,300. Are there other feasible production schedules that will result in a daily profit of $1,300? How are these schedules related to the graph of the line 50x 0004 15y 0003 1,300?
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(B) Find a feasible production schedule that will produce a daily profit greater than $1,300 and repeat part A for this schedule. (C) Discuss methods for using lines like those in parts A and B to find the largest possible daily profit. 53. NUTRITION—PLANTS A farmer can buy two types of plant food, mix A and mix B. Each cubic yard of mix A contains 20 pounds of phosphoric acid, 30 pounds of nitrogen, and 5 pounds of potash. Each cubic yard of mix B contains 10 pounds of phosphoric acid, 30 pounds of nitrogen, and 10 pounds of potash. The minimum requirements are 460 pounds of phosphoric acid, 960 pounds of nitrogen, and 220 pounds of potash. If x is the number of cubic yards of mix A used and y is the number of cubic yards of mix B used, write a system of inequalities that indicates appropriate restraints on x and y. Find the set of feasible solutions graphically for the amount of mix A and mix B that can be used. 54. NUTRITION A dietitian in a hospital is to arrange a special diet using two foods. Each ounce of food M contains 30 units of calcium, 10 units of iron, and 10 units of vitamin A. Each ounce of food N contains 10 units of calcium, 10 units of iron, and 30 units of vitamin A. The minimum requirements in the diet are 360 units of calcium, 160 units of iron, and 240 units of vitamin A. If x is the number of ounces of food M used and y is the number of ounces of food N used, write a system of linear inequalities that reflects the conditions indicated. Find the set of feasible solutions graphically for the amount of each kind of food that can be used.
7-8
55. SOCIOLOGY A city council voted to conduct a study on innercity community problems. A nearby university was contacted to provide sociologists and research assistants. Each sociologist will spend 10 hours per week collecting data in the field and 30 hours per week analyzing data in the research center. Each research assistant will spend 30 hours per week in the field and 10 hours per week in the research center. The minimum weekly labor-hour requirements are 280 hours in the field and 360 hours in the research center. If x is the number of sociologists hired for the study and y is the number of research assistants hired for the study, write a system of linear inequalities that indicates appropriate restrictions on x and y. Find the set of feasible solutions graphically. 56. PSYCHOLOGY In an experiment on conditioning, a psychologist uses two types of Skinner (conditioning) boxes with mice and rats. Each mouse spends 10 minutes per day in box A and 20 minutes per day in box B. Each rat spends 20 minutes per day in box A and 10 minutes per day in box B. The total maximum time available per day is 800 minutes for box A and 640 minutes for box B. We are interested in the various numbers of mice and rats that can be used in the experiment under the conditions stated. If x is the number of mice used and y is the number of rats used, write a system of linear inequalities that indicates appropriate restrictions on x and y. Find the set of feasible solutions graphically.
Linear Programming Z A Linear Programming Problem Z Linear Programming—A General Description Z Application
Several problems in Section 10-7 are related to the general type of problems called linear programming problems. Linear programming is a mathematical process that was developed to help management in decision making, and it has become one of the most widely used and best-known tools of management science and industrial engineering. We will use an intuitive graphical approach based on the techniques discussed in Section 10-7 to illustrate this process for problems involving two variables.
Z A Linear Programming Problem We will begin our discussion with an example that leads to a general procedure for solving linear programming problems in two variables.
EXAMPLE
1
Production Scheduling A manufacturer of fiberglass camper tops for pickup trucks makes a compact model and a regular model. Each compact top requires 5 hours from the fabricating department and 2 hours from the finishing department. Each regular top requires 4 hours from the fabricating department and 3 hours from the finishing department. The maximum labor-hours available per week in the fabricating department and the finishing department are 200 and 108,
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respectively. If the company makes a profit of $40 on each compact top and $50 on each regular top, how many tops of each type should be manufactured each week to maximize the total weekly profit, assuming all tops can be sold? What is the maximum profit? SOLUTION
This is an example of a linear programming problem. To organize the given information, we summarize the manufacturing requirements, objectives, and restrictions in the table: Compact Model (Labor-Hours per Top)
Regular Model (Labor-Hours per Top)
Maximum Labor-Hours Available per Week
Fabricating
5
4
200
Finishing
2
3
108
$40
$50
Profit per top
Now we will formulate a mathematical model for the problem and then to solve it using graphical methods. OBJECTIVE FUNCTION The objective of management is to decide how many of each camper top model should be produced each week in order to maximize profit. Let
x 0003 Number of compact tops produced per week f y 0003 Number of regular tops produced per week
Decision variables
The following function gives the total profit P for x compact tops and y regular tops manufactured each week: $40 for each compact, and $50 for each regular. P 0003 40x 0004 50y
Objective function
Mathematically, management needs to decide on values for the decision variables (x and y) that achieve its objective, which is maximizing the objective function (profit) P 0003 40x 0004 50y. You might think that the profit can be made as large as we like by manufacturing more and more tops—or can it? Any manufacturing company, no matter how large or small, has manufacturing limits imposed by available resources, plant capacity, demand, and so forth. These limits are referred to as problem constraints.
CONSTRAINTS
Fabricating department constraint: Weekly fabricating Weekly fabricating Maximum labor-hours ° time for x ¢ 0005 ° time for y ¢0002a b available per week compact tops regular tops
5x
0004
4y
0007
200
Finishing department constraint: Weekly finishing Weekly finishing Maximum labor-hours ° time for x ¢ 0005 ° time for y ¢0002a b available per week compact tops regular tops
2x
+
3y
0007
108
Nonnegative constraints: It is not possible to manufacture a negative number of tops, so we have the nonnegative constraints x 0 y 0 which we usually write in the form x, y 0
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MATHEMATICAL MODEL
We now have a mathematical model for the problem under con-
sideration: Maximize P 0003 40x 0004 50y Subject to 5x 0004 4y 0007 200 2x 0004 3y 0007 108
冎
x, y 0
Objective function Problem constraints Nonnegative constraints
Solving the system of linear inequality constraints graphically, as in Section 10-7, we obtain the feasible region for production schedules, as shown in Figure 1.
GRAPHIC SOLUTION
y
Fabricating capacity line 5x 0004 4y 0003 200
All lines are restricted to the first quadrant because of the nonnegative constraints x, y 0003 0.
(0, 36)
(24, 20)
20
Feasible region
(0, 0)
20
Finishing capacity line 2x 0004 3y 0003 108
(40, 0)
x
Z Figure 1
Any production schedule (x, y) in the feasible region is possible—but which is the best? That is, which produces the largest profit? We could try some schedules in the region to see what the profit is; for x 0003 24 and y 0003 10, the weekly profit is P 0003 40(24) 0004 50(10) 0003 $1,460 For a different point in the feasible region, (15, 20), the profit is P 0003 40(15) 0004 50(20) 0003 $1,600 But how do we know when we’ve found the largest profit? Such a schedule, if it exists, is called an optimal solution to the problem because it produces the maximum value of the objective function and is in the feasible region. It is not practical to use point-by-point checking to find the optimal solution. Even if we consider only points with integer coordinates, there are over 800 such points in the feasible region for this problem. Instead, we use the theory that has been developed to solve linear programming problems. Using advanced techniques, it can be shown that
Corner Point (x, y)
Objective Function P 0003 40x 0005 50y
(0, 0)
0
(0, 36)
1,800
(24, 20)
1,960
(40, 0)
1,600
Maximum value of P
If the feasible region is bounded, then one or more of the corner points of the feasible region is an optimal solution to the problem. The maximum value of the objective function is unique; however, there can be more than one feasible production schedule that will produce this unique value. We will have more to say about this later in this section. Since the feasible region for this problem is bounded, at least one of the corner points, (0, 0), (0, 36), (24, 20), or (40, 0), is an optimal solution. To find which one, we evaluate P 0003 40x 0004 50y at each corner point and choose the corner point that produces the largest value of P. It is convenient to organize these calculations in a table, as shown in the margin.
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23
Examining the values in the table, we see that the maximum value of P at a corner point is P 0003 1,960 at x 0003 24 and y 0003 20. Since the maximum value of P over the entire feasible region must always occur at a corner point, we conclude that the maximum profit is $1,960 when 24 compact tops and 20 regular tops are produced each week. 0002 MATCHED PROBLEM 1
We will now convert the surfboard problem discussed in Section 7-7 into a linear programming problem. A manufacturer of surfboards makes a standard model and a competition model. Each standard board requires 6 labor-hours for fabricating and 1 labor-hour for finishing. Each competition board requires 8 labor-hours for fabricating and 3 labor-hours for finishing. The maximum labor-hours available per week in the fabricating and finishing departments are 120 and 30, respectively. If the company makes a profit of $40 on each standard board and $75 on each competition board, how many boards of each type should be manufactured each week to maximize the total weekly profit? (A) Identify the decision variables. (B) Write the objective function P. (C) Write the problem constraints and the nonnegative constraints. (D) Graph the feasible region, identify the corner points, and evaluate P at each corner point. (E) How many boards of each type should be manufactured each week to maximize the profit? What is the maximum profit? 0002
ZZZ EXPLORE-DISCUSS 1
Refer to Example 1. If we assign the profit function P in P 0003 40x 0004 50y a particular value and plot the resulting equation in the coordinate system shown in Figure 1, we obtain a constant-profit line (isoprofit line). Every point in the feasible region on this line represents a production schedule that will produce the same profit. Figure 2 shows the constant-profit lines for P 0003 $1,000 and P 0003 $1,500. y (0, 36)
P 20
P
0003
0003
(24, 20)
$1
,5
$1
00
,0
(0, 0)
00 20
(40, 0)
x
Z Figure 2
(A) How are all the constant-profit lines related? (B) Place a straightedge along the constant-profit line for P 0003 $1,000 and slide it as far as possible in the direction of increasing profit without changing its slope and without leaving the feasible region. Explain how this process can be used to identify the optimal solution to a linear programming problem. (C) If P is changed to P 0003 25x 0004 75y, graph the constant-profit lines for P 0003 $1,000 and P 0003 $1,500, and use a straightedge to identify the optimal solution. Check your answer by evaluating P at each corner point. (D) Repeat part C for P 0003 75x 0004 25y.
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Z Linear Programming—A General Description The linear programming problems considered in Example 1 and Matched Problem 1 were maximization problems, where we wanted to maximize profits. The same technique can be used to solve minimization problems, where, for example, we may want to minimize costs. Before considering additional examples, we will state a few general definitions. A linear programming problem is one that is concerned with finding the optimal value (maximum or minimum value) of a linear objective function of the form z 0003 ax 0004 by where the decision variables x and y are subject to problem constraints in the form of linear inequalities and the nonnegative constraints x, y 0. The set of points satisfying both the problem constraints and the nonnegative constraints is called the feasible region for the problem. Any point in the feasible region that produces the optimal value of the objective function over the feasible region is called an optimal solution. Theorem 1 is fundamental to the solving of linear programming problems. Z THEOREM 1 Fundamental Theorem of Linear Programming Let S be the feasible region for a linear programming problem, and let z 0003 ax 0004 by be the objective function. If S is bounded, then z has both a maximum and a minimum value on S and each of these occurs at a corner point of S. If S is unbounded, then a maximum or minimum value of z on S may not exist. However, if either does exist, then it must occur at a corner point of S.
We will not consider any problems with unbounded feasible regions in this brief introduction. If a feasible region is bounded, then Theorem 1 provides the basis for the following simple procedure for solving the associated linear programming problem: Z SOLUTION OF LINEAR PROGRAMMING PROBLEMS Step 1. Form a mathematical model for the problem: (A) Introduce decision variables and write a linear objective function. (B) Write problem constraints in the form of linear inequalities. (C) Write nonnegative constraints. Step 2. Graph the feasible region and find the corner points. Step 3. Evaluate the objective function at each corner point to determine the optimal solution.
Before considering additional applications, we use this procedure to solve a linear programming problem where the model has already been determined.
EXAMPLE
2
Solving a Linear Programming Problem Minimize and maximize z 0003 5x 0004 15y Subject to x 0004 3y 0007 60 x 0004 y 10 x0002 y0007 0 x, y 0
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25
This problem is a combination of two linear programming problems—a minimization problem and a maximization problem. Since the feasible region is the same for both problems, we can solve these problems together. To begin, we graph the feasible region S, as shown in Figure 3, and find the coordinates of each corner point. y (0, 20)
x 0004 3y 0003 60 (15, 15)
S (0, 10)
x0002y00030
x 0004 y 0003 10 5
(5, 5) x
5
Z Figure 3
Next, we evaluate the objective function at each corner point, with the results given in the table: Corner Point (x, y)
Objective Function z 0003 5x 0005 15y
(0, 10)
150
(0, 20)
300
(15, 15)
300
Maximum value ⎫ Multiple ⎬ optimal Maximum value ⎭ solutions
(5, 5)
100
Minimum value
Examining the values in the table, we see that the minimum value of z on the feasible region S is 100 at (5, 5), So (5, 5) is the optimal solution to the minimization problem. The maximum value of z on the feasible region S is 300, which occurs at (0, 20) and at (15, 15); the maximization problem has multiple optimal solutions. In general: If two corner points are both optimal solutions of the same type (both produce the same maximum value or both produce the same minimum value) to a linear programming problem, then any point on the line segment joining the two corner points is also an optimal solution of that type. It can be shown that this is the only time that an optimal value occurs at more than one point. 0002 MATCHED PROBLEM 2
Minimize and maximize z 0003 10x 0004 5y Subject to 2x 0004 y 40 3x 0004 y 0007 150 2x 0002 y 0 x, y 0
0002
Z Application Now we will look at another application where we first find the mathematical model and then find its solution.
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3 Pounds per Cubic Yard
Mix A Mix B Nitrogen
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10
5
Potash
8
24
Phosphoric acid
9
6
SOLUTION
Agriculture A farmer can use two types of plant food, mix A and mix B. The amounts (in pounds) of nitrogen, phosphoric acid, and potash in a cubic yard of each mix are given in the table. Tests performed on the soil in a large field indicate that the field needs at least 840 pounds of potash and at least 350 pounds of nitrogen. The tests also indicate that no more than 630 pounds of phosphoric acid should be added to the field. A cubic yard of mix A costs $7, and a cubic yard of mix B costs $9. How many cubic yards of each mix should the farmer add to the field in order to supply the necessary nutrients at minimal cost? Let x 0003 Number of cubic yards of mix A added to the field y 0003 Number of cubic yards of mix B added to the field
冎
Decision variables
We form the linear objective function C 0003 7x 0004 9y which gives the cost of adding x cubic yards of mix A and y cubic yards of mix B to the field. Using the data in the table and proceeding as in Example 1, we formulate the mathematical model for the problem: Minimize Subject to
C 0003 7x 0004 9y 10x 0004 5y 350 8x 0004 24y 840 9x 0004 6y 0007 630 x, y 0
Objective function Nitrogen constraint Potash constraint Phosphoric acid constraint Nonnegative constraints
Solving the system of constraint inequalities graphically, we obtain the feasible region S shown in Figure 4, and then we find the coordinates of each corner point.
y (0, 105)
9x 0004 6y 0003 630 (0, 70) 60
S
10x 0004 5y 0003 350
Corner Point (x, y)
Objective Function C 0003 7x 0005 9y
(0, 105)
945
(0, 70)
630
(21, 28)
399
(60, 15)
555
(21, 28)
(60, 15)
8x 0004 24y 0003 840 60
x
Z Figure 4 Minimum value of C
Next, we evaluate the objective function at each corner point, as shown in the table in the margin.
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The optimal value is C 0003 399 at the corner point (21, 28). The farmer should add 21 cubic yards of mix A and 28 cubic yards of mix B at a cost of $399. This will result in adding the following nutrients to the field: Nitrogen: Potash: Phosphoric acid:
10(21) 0004 5(28) 0003 350 pounds 8(21) 0004 24(28) 0003 840 pounds 9(21) 0004 6(28) 0003 357 pounds
All the nutritional requirements are satisfied. MATCHED PROBLEM 3
0002
Repeat Example 3 if the tests indicate that the field needs at least 400 pounds of nitrogen with all other conditions remaining the same. 0002 The American mathematician George B. Dantzig (1914–2005) formulated the first linear programming problem in 1947 and introduced a solution technique, called the simplex method, that does not rely on graphing and is readily adaptable to computer solutions. Today, it is quite common to use a computer to solve applied linear programming problems involving thousands of variables and thousands of inequalities. ANSWERS TO MATCHED PROBLEMS 1. (A) x 0003 Number of standard boards manufactured each week y 0003 Number of competition boards manufactured each week (B) P 0003 40x 0004 75y (C) 6x 0004 8y 0007 120 Fabricating constraint x 0004 3y 0007 30 Finishing constraint x, y 0 Nonnegative constraints (D)
Corner Point (x, y)
y (0, 10)
(0, 0)
(12, 6) 5
(0, 0)
Feasible region (20, 0)
5
x
Objective Function P 0003 40x 0005 75y 0
(0, 10)
750
(12, 6)
930
(20, 0)
800
(E) 12 standard boards and 6 competition boards for a maximum profit of $930 2. Max z 0003 600 at (30, 60); min z 0003 200 at (10, 20) and (20, 0) (multiple optimal solutions) 3. 27 cubic yards of mix A, 26 cubic yards of mix B; min C 0003 $423
7-8
Exercises
*Additional answers can be found in the Instructor Answer Appendix.
A
In Problems 1–7, explain how each term is used in the description of a linear programming program. 1. Objective function 2. Decision variables 3. Problem constraints
4. Nonnegativity constraints 5. Feasible region 6. Optimal value and optimal solution 7. Corner point
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8. Can a linear programming problem have more than one optimal value? Explain. In Problems 9–12, find the maximum value of each objective function over the feasible region S shown in the figure. y
value. Check your answer by evaluating the objective function at each corner point. 21. z 0003 x 0004 2y
22. z 0003 2x 0004 y
23. z 0003 5x 0004 4y
24. z 0003 2x 0004 8y
Minimum value of z on T is 32 at both (0, 8) and (4, 3).
25. Maximize Subject to (7, 9)
26. Maximize Subject to
S
5
(10, 0) (0, 0)
5
10. z 0003 4x 0004 y
11. z 0003 3x 0004 7y
27. Minimize Subject to
x
zmax 0003 40 at (10, 0)
12. z 0003 9x 0004 3y
zmax 0003 90 at (7, 9) and (10, 0) (multiple optimal solutions)
Problems 13–16 refer to the feasible region S shown and the constant-profit lines discussed in Explore-Discuss 1. For each objective function, draw the line that passes through the feasible point (5, 5) and use the straightedge method from Explore-Discuss 1 to find the maximum value. Check your answer by evaluating the objective function at each corner point. 13. z 0003 x 0004 2y
14. z 0003 3x 0004 y
15. z 0003 7x 0004 2y
16. z 0003 2x 0004 8y
Maximum value of z on S is 70 at (10, 0).
Maximum value of z on S is 30 at both (7, 9) and (10, 0) Maximum value of z on S is 96 at (0, 12)
28. Minimize Subject to
29. Maximize Subject to
30. Maximize Subject to
In Problems 17–20, find the minimum value of each objective function over the feasible region T shown in the figure. 31. Minimize Subject to
y
(0, 12)
32. Minimize Subject to
(0, 8) 5
Minimum value of z on T is 24 at (12, 0).
B In Problems 25–38, solve the linear programming problems. (0, 12)
9. z 0003 x 0004 y
Minimum value of z on T is 8 at (0, 8).
T (4, 3)
(12, 0) 5
17. z 0003 7x 0004 4y
18. z 0003 7x 0004 9y
19. z 0003 3x 0004 8y
20. z 0003 5x 0004 4y
x
zmin 0003 55 at (4, 3)
zmin 0003 32 at (0, 8) and (4, 3) (multiple optimal solutions)
Problems 21–24 refer to the feasible region T shown for problems 17–20. For each objective function, draw the constant-value line that passes through the feasible point (5, 5) and use the straightedge method from Explore-Discuss 1 to find the minimum
z 0003 3x 0004 2y x 0004 2y 0007 10 3x 0004 y 0007 15 x, y 0
Maximum value of z on S is 18 at (4, 3).
z 0003 4x 0004 5y 2x 0004 y 0007 12 x 0004 3y 0007 21 x, y 0
zmax 0003 42 at (3, 6)
z 0003 3x 0004 4y 2x 0004 y 8 x 0004 2y 0007 10 x, y 0
Minimum value of z on S is 12 at (4, 0).
z 0003 2x 0004 y 4x 0004 3y 24 4x 0004 y 0007 16 x, y 0
zmin 0003 8 at (0, 8)
z 0003 3x 0004 4y x 0004 2y 0007 24 x 0004 y 0007 14 2x 0004 y 0007 24 x, y 0
Maximum value of z on S is 52 at (4, 10).
z 0003 5x 0004 3y 3x 0004 y 0007 24 x 0004 y 0007 10 x 0004 3y 0007 24 x, y 0
zmax 0003 44 at (7, 3)
z 0003 5x 0004 6y x 0004 4y 20 4x 0004 y 20 x 0004 y 0007 20 x, y 0
Minimum value of z on S is 44 at (4, 4).
z 0003 x 0004 2y 2x 0004 3y 30 3x 0004 2y 30 x 0004 y 0007 15 x, y 0
zmin 0003 15 at (15, 0)
33. Minimize and maximize z 0003 25x 0004 50y Subject to x 0004 2y 0007 120 The minimum value of z on S is x 0004 y 60 1,500 at (60, 0). The maximum value x 0002 2y 0 of z on S is 3,000 at (60, 30) and x, y 0 (120, 0) (multiple optimal solutions). 34. Minimize and maximize z 0003 15x 0004 30y Subject to x 0004 2y 100 2x 0002 y 0007 0 zmax 0003 6000 at (0, 200); 2x 0004 y 0007 200 zmin 0003 1500 at (0, 50) and (20, 40) x, y 0 (multiple optimal solutions)
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35. Minimize and maximize z 0003 25x 0004 15y Subject to 4x 0004 5y 100 3x 0004 4y 0007 240 x 0007 60 The minimum value of z on S is y 0007 45 300 at (0, 20). The maximum x, y 0 value of z on S is 1,725 at (60, 15). 36. Minimize and maximize z 0003 25x 0004 30y Subject to 2x 0004 3y 120 3x 0004 2y 0007 360 x 0007 80 y 0007 120 zmin 0003 1200 at (0, 40); x, y 0 zmax 0003 4600 at (40, 120) P 0003 525x1 0004 478x2 275x1 0004 322x2 0007 3,381 350x1 0004 340x2 0007 3,762 425x1 0004 306x2 0007 4,114 x1, x2 0
Max P 0003 5,507 at x1 0003 6.62 and x2 0003 4.25
38. Maximize P 0003 300x1 0004 460x2 Subject to 245x1 0004 452x2 0007 4,181 290x1 0004 379x2 0007 3,888 390x1 0004 299x2 0007 4,407 x1, x2 0
Max P 0003 4,484 at x1 0003 4.52 and x2 0003 6.8
37. Maximize Subject to
C
39. The corner points for the feasible region determined by the problem constraints 2x 0004 y 0007 10 x 0004 3y 0007 15 x, y 0 are O 0003 (0, 0), A 0003 (5, 0), B 0003 (3, 4), and C 0003 (0, 5). If z 0003 ax 0004 by and a, b 0, determine conditions on a and b that ensure that the maximum value of z occurs (A) Only at A (B) Only at B (C) Only at C (D) At both A and B (E) At both B and C
40. The corner points for the feasible region determined by the problem constraints x0004 y 4 x 0004 2y 6 2x 0004 3y 0007 12 x, y 0 are A 0003 (6, 0), B 0003 (2, 2), and C 0003 (0, 4). If z 0003 ax 0004 by and a, b 0, determine conditions on a and b that ensure that the minimum value of z occurs (A) Only at A 2a b (B) Only at B 12 b 6 a 6 b (C) Only at C a b (D) At both A and B b 0003 2a (E) At both B and C b 0003 a
APPLICATIONS 41. RESOURCE ALLOCATION A manufacturing company makes two types of water skis, a trick ski and a slalom ski. The relevant manufacturing data are given in the table. (A) If the profit on a trick ski is $40 and the profit on a slalom ski is $30, how many of each type of ski should be manufactured each day to realize a maximum profit? What is the maximum profit? (B) Discuss the effect on the production schedule and the maximum profit if the profit on a slalom ski decreases to $25 and all other data remains the same.
29
Linear Programming
(C) Discuss the effect on the production schedule and the maximum profit if the profit on a slalom ski increases to $45 and all other data remain the same. Trick Ski (labor-hours per ski)
Slalom Ski (labor-hours per ski)
Maximum Labor-Hours Available per Day
Fabricating department
6
4
108
Finishing department
1
1
24
42. PSYCHOLOGY In an experiment on conditioning, a psychologist uses two types of Skinner boxes with mice and rats. The amount of time (in minutes) each mouse and each rat spends in each box per day is given in the table. What is the maximum total number of mice and rats that can be used in this experiment? How many mice and how many rats produce this maximum?
Mice (minutes)
Rats (minutes)
Max. Time Available per Day (minutes)
Skinner box A
10
20
800
Skinner box B
20
10
640 48; 16 mice, 32 rats
43. PURCHASING A trucking firm wants to purchase a maximum of 15 new trucks that will provide at least 36 tons of additional shipping capacity. A model A truck holds 2 tons and costs $15,000. A model B truck holds 3 tons and costs $24,000. How many trucks of each model should the company purchase to provide the additional shipping capacity at minimal cost? What is the minimal cost?
9 model A trucks and 6 model B trucks to realize the minimum cost of $279,000
44. TRANSPORTATION The officers of a student group are planning to rent buses and vans for a class trip. Each bus can transport 40 students, requires 3 chaperones, and costs $1,200 to rent. Each van can transport 8 students, requires 1 chaperone, and costs $100 to rent. The officers want to be able to accommodate at least 400 students with no more than 36 chaperones. How many vehicles of each type should they rent to minimize the transportation costs? What are the minimal transportation costs? 7 buses, 15 vans; $9,900 45. RESOURCE ALLOCATION A furniture company manufactures dining room tables and chairs. Each table requires 8 hours from the assembly department and 2 hours from the finishing department and contributes a profit of $90. Each chair requires 2 hours from the assembly department and 1 hour from the finishing department and contributes a profit of $25. The maximum labor-hours available each day in the assembly and finishing departments are 400 and 120, respectively. (A) How many tables and how many chairs should be manufactured each day to maximize the daily profit? What is the maximum daily profit? (B) Discuss the effect on the production schedule and the maximum profit if the marketing department of the company decides that the number of chairs produced should be at least four times the number of tables produced. 46. RESOURCE ALLOCATION An electronics firm manufactures two types of personal computers, a desktop model and a laptop. The
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production of a desktop computer requires a capital expenditure of $400 and 40 hours of labor. The production of a laptop computer requires a capital expenditure of $250 and 30 hours of labor. The firm has $20,000 capital and 2,160 labor-hours available for production of desktop and laptop computers. (A) What is the maximum number of computers the company is capable of producing? (B) If each desktop computer contributes a profit of $320 and each laptop contributes a profit of $220, how much profit will the company make by producing the maximum number of computers determined in part A? Is this the maximum profit? If not, what is the maximum profit? 47. POLLUTION CONTROL Because of new federal regulations on pollution, a chemical plant introduced a new process to supplement or replace an older process used in the production of a particular chemical. The older process emitted 20 grams of sulfur dioxide and 40 grams of particulate matter into the atmosphere for each gallon of chemical produced. The new process emits 5 grams of sulfur dioxide and 20 grams of particulate matter for each gallon produced. The company makes a profit of 60¢ per gallon and 20¢ per gallon on the old and new processes, respectively. (A) If the regulations allow the plant to emit no more than 16,000 grams of sulfur dioxide and 30,000 grams of particulate matter daily, how many gallons of the chemical should be produced by each process to maximize daily profit? What is the maximum daily profit? (B) Discuss the effect on the production schedule and the maximum profit if the regulations restrict emissions of sulfur dioxide to 11,500 grams daily and all other data remains unchanged. (C) Discuss the effect on the production schedule and the maximum profit if the regulations restrict emissions of sulfur dioxide to 7,200 grams daily and all other data remains unchanged. 48. SOCIOLOGY A city council voted to conduct a study on innercity community problems. A nearby university was contacted to provide a maximum of 40 sociologists and research assistants. Allocation of time and cost per week are given in the table. (A) How many sociologists and research assistants should be hired to meet the weekly labor-hour requirements and minimize the weekly cost? What is the weekly cost? (B) Discuss the effect on the solution in part A if the council decides that they should not hire more sociologists than research assistants and all other data remain unchanged.
Sociologist (labor-hours)
Research Assistant (labor-hours)
Minimum Labor-Hours Needed per Week
Fieldwork
10
30
280
Research center
30
10
360
$500
$300
Cost per week
49. PLANT NUTRITION A fruit grower can use two types of fertilizer in her orange grove, brand A and brand B. The amounts (in pounds) of nitrogen, phosphoric acid, potash, and chloride in a bag of each mix are given in the table. Tests indicate that the grove needs at least 480 pounds of phosphoric acid, at least 540 pounds of potash, and at most 620 pounds of chloride. If the grower always uses a combination of bags of brand A and brand B that will satisfy the constraints for phosphoric acid, potash, and chloride, discuss the effect that this will have on the amount of nitrogen added to the field. Pounds per Bag Brand A
Brand B
Nitrogen
6
7
Phosphoric acid
2
4
Potash
6
3
Chloride
3
4
50. DIET A dietitian in a hospital is to arrange a special diet composed of two foods, M and N. Each ounce of food M contains 16 units of calcium, 5 units of iron, 6 units of cholesterol, and 8 units of vitamin A. Each ounce of food N contains 4 units of calcium, 25 units of iron, 4 units of cholesterol, and 4 units of vitamin A. The diet requires at least 320 units of calcium, at least 575 units of iron, and at most 300 units of cholesterol. If the dietitian always selects a combination of foods M and N that will satisfy the constraints for calcium, iron, and cholesterol, discuss the effects that this will have on the amount of vitamin A in the diet. The amount of vitamin A will range from a minimum of 200 units when 15 ounces of food M and 20 ounces of food N are used to a maximum of 380 units when 40 ounces of food M and 15 ounces of food N are used.
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8
C
OUTLINE
THE lists
8-1
Sequences and Series
and
8-2
Mathematical Induction
3, 6, 3, 1, 4, 2, 1, 4, . . . are examples of sequences. In the first sequence, a pattern is noticeable: You probably recognize it as the sequence of perfect squares. Its terms are increasing, and as we will see, the differences between terms form a clear pattern. You probably don’t recognize the second sequence because the terms don’t suggest an obvious pattern. In fact, we obtained the second sequence by recording the results of repeatedly tossing a single die. Sequences, and the related concept of series, are useful tools in almost all areas of mathematics. In this chapter, they will play roles in the development of several topics: a method of proof called mathematical induction, techniques for counting, and probability.
8-3
Arithmetic and Geometric Sequences
8-4
Multiplication Principle, Permutations, and Combinations
8-5
Sample Spaces and Probability
8-6
The Binomial Formula
1, 4, 9, 16, 25, 36, 49, 64, . . .
Chapter 8 Review Chapter 8 Group Activity: Sequences Specified by Recursion Formulas
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Sequences and Series Z Defining Sequences Z Defining Series
In this section, we introduce special notation and formulas for representing and generating sequences and sums of sequences.
Z Defining Sequences Consider the following list of numbers: 1, 3, 5, 7, 9, . . . . This is an example of a sequence, which can be defind informally as a list of numbers in a specific order. This particular sequence is the sequence of positive odd integers. Now consider the function f given by f (n) 0002 2n 0003 1
(1)
where the domain of f is {1, 2, 3, . . .} (that is, the set of natural numbers N). Note that f (1) 0002 2(1) 0003 1 0002 1 f (2) 0002 2(2) 0003 1 0002 3 f (3) 0002 2(3) 0003 1 0002 5 The outputs of the function f form the same list of odd positive integers that we started with above. This provides an alternative (and more precise) definition of sequence: A sequence is a function whose domain is a set of successive integers. While the function f above is a perfectly good way to describe a sequence, a special notation for describing sequences with formulas has evolved over the years. Our first order of business should be to become familiar with this notation. To start, the range value f (n) is usually symbolized more compactly with a symbol such as an. So in place of equation (1) we write an 0002 2n 0003 1 The domain is understood to be the set of natural numbers N unless stated to the contrary or the context indicates otherwise. The elements in the range are called terms of the sequence: a1 is the first term, a2 the second term, and an the nth term, or the general term: a1 0002 2(1) 0003 1 0002 1 a2 0002 2(2) 0003 1 0002 3 a3 0002 2(3) 0003 1 0002 5 o o
First term Second term Third term
The ordered list of elements 1, 3, 5, . . . , 2n 0003 1, . . . in which the terms of a sequence are written in their natural order with respect to the domain values, is often informally referred to as a sequence. A sequence is also represented in the abbreviated form {an}, where a symbol for the nth term is placed between braces. For example, we can refer to the sequence 1, 3, 5, . . . , 2n 0003 1, . . . as the sequence {2n 0003 1}.
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505
If the domain of a function is a finite set of successive integers, then the sequence is called a finite sequence. If the domain is an infinite set of successive integers, then the sequence is called an infinite sequence. The preceding sequence {2n 0003 1} is an example of an infinite sequence.
Technology Connections Some graphing calculators have a special sequence mode that can be useful when studying sequences. Figure 1(a) shows the sequence {2n ⴚ 1} entered in the sequence editor.
Figure 1(b) shows the graph of the sequence. Figure 1(c) displays the sequence in a table.
20
0
10
0
(a)
(b)
(c)
Z Figure 1
Some sequences are specified by a recursion formula—that is, a formula that defines each term in terms of one or more preceding terms. The sequence we have chosen to illustrate a recursion formula is a very famous sequence in the history of mathematics called the Fibonacci sequence. It is named after the most celebrated mathematician of the thirteenth century, Leonardo Fibonacci from Italy (1180?–1250?).
EXAMPLE
1
Fibonacci Sequence List the first seven terms of the sequence specified by a1 0002 1 a2 0002 1 an 0002 an00032 0004 an00031
SOLUTION
a1 a2
00021 00021
a3 a4 a5 a6 a7
0002 0002 0002 0002
0004 0004 0004 0004 0002 a5 0004 a1 a2 a3 a4
a2 a3 a4 a5 a6
0002 0002 0002 0002 0002
1 1 2 3 5
0004 0004 0004 0004 0004
n00053
1 2 3 5 8
*
0002 0002 0002 0002
2 3 5 8
0002 13
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
0002
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MATCHED PROBLEM 1
List the first seven terms of the sequence specified by a1 0002 1 a2 0002 1 an 0002 an00032 0003 an00031
ZZZ EXPLORE-DISCUSS 1
n00053
0002
A multiple-choice test question asked for the next term in the sequence: 1, 3, 9, . . . and gave the following choices: (A) 16
(B) 19
(C) 27
Which is the correct answer? Compare the first four terms of the following sequences: (A) an 0002 3n00031
(B) bn 0002 1 0004 2(n 0003 1)2
(C) cn 0002 8n 0004
12 0003 19 n
Now which of the choices appears to be correct?
Now we consider the reverse problem. That is, can a sequence be defined just by listing the first three or four terms of the sequence? And can we then use these initial terms to find a formula for the nth term? In general, without other information, the answer to the first question is no. As Explore-Discuss 1 illustrates, many different sequences may start off with the same terms. Simply listing the first three terms, or any other finite number of terms, does not specify a particular sequence. In fact, it can be shown that given any list of m numbers, there are an infinite number of sequences whose first m terms agree with these given numbers. What about the second question? That is, given a few terms, can we find the general formula for at least one sequence whose first few terms agree with the given terms? The answer to this question is a qualified yes. If we can observe a simple pattern in the given terms, then we may be able to construct a general term that will produce the pattern. Example 2 illustrates this approach.
EXAMPLE
2
Finding the General Term of a Sequence Find the general term of a sequence whose first four terms are (A) 5, 6, 7, 8, . . .
SOLUTIONS
(B) 2, 00034, 8, 000316, . . .
(A) Because these terms are consecutive integers, one solution is an 0002 n, n 0005 5. If we want the domain of the sequence to be all natural numbers, then another solution is bn 0002 n 0004 4. (B) Each of these terms can be written as the product of a power of 2 and a power of 00031: 2 0002 (00031)021 00034 0002 (00031)122 8 0002 (00031)223 000316 0002 (00031)324 If we choose the domain to be all natural numbers, then a solution is an 0002 (00031)n000312n
0002
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MATCHED PROBLEM 2
Sequences and Series
507
Find the general term of a sequence whose first four terms are (A) 2, 4, 6, 8, . . .
(B) 1, 000312, 14, 000318, . . .
0002
In general, there is usually more than one way of representing the nth term of a given sequence. This was seen in the solution of Example 2, part A. However, unless stated to the contrary, we assume the domain of the sequence is the set of natural numbers N. ZZZ EXPLORE-DISCUSS 2
15 1 0004 15 n a b is closely related to the 5 2 Fibonacci sequence. Compute the first 20 terms of both sequences and discuss the relationship. [The first seven values of bn are shown in Fig. 2(b)]. The sequence with general term bn 0002
(a)
(b)
Z Figure 2
Z Defining Series If a1, a2, a3, . . . , an, . . . is a sequence, then the expression a1 0004 a2 0004 a3 0004 . . . 0004 an 0004 . . . is called a series. If the sequence is finite, the corresponding series is a finite series. If the sequence is infinite, the corresponding series is an infinite series. For example, 1, 2, 4, 8, 16 1 0004 2 0004 4 0004 8 0004 16
Finite sequence Finite series
We will restrict our discussion to finite series in this section. Series are often represented in a compact form called summation notation using the symbol a, which is a stylized version of the Greek letter sigma. Consider the following examples: 4
a ak 0002 a1 0004 a2 0004 a3 0004 a4
k00021 7
a bk 0002 b3 0004 b4 0004 b5 0004 b6 0004 b7
k00023 n
. . . 0004 cn a ck 0002 c0 0004 c1 0004 c2 0004
k00020
Domain is the set of integers k satisfying 0 ⱕ k ⱕ n.
The terms on the right are obtained from the expression on the left by successively replacing the summing index k with integers, starting with the first number indicated below a and ending with the number that appears above a. For example, if we are given the sequence 1 1 1 ... 1 , , , , n 2 4 8 2
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the corresponding series is n 1 1 1 1 ... 1 0004 n a 2k 0002 2 0004 4 0004 8 0004 2 k00021
EXAMPLE
3
Writing the Terms of a Series 5
k00031 k k00021
Write without summation notation: a SOLUTION
MATCHED PROBLEM 3
5
k00031 100031 200031 300031 400031 500031 0002 0004 0004 0004 0004 k 1 2 3 4 5 k00021 1 2 3 4 000200004 0004 0004 0004 2 3 4 5 a
0002
5 (00031)k Write without summation notation: a k 0002 0 2k 0004 1
0002 If the terms of a series are alternately positive and negative, it is called an alternating series. Example 4 deals with the representation of such a series.
EXAMPLE
4
Writing a Series in Summation Notation Write the following series using summation notation: 10003
1 1 1 1 1 0004 0003 0004 0003 2 3 4 5 6
(A) Start the summing index at k 0002 1. (B) Start the summing index at k 0002 0. SOLUTIONS
(A) (00031)k00031 provides the alternation of sign, and 10002k provides the other part of each term. So we can write (00031)k00031 a k k00021 6
as can be easily checked. (B) (00031)k provides the alternation of sign, and 10002(k 0004 1) provides the other part of each term. We write the series as (00031)k a k00020 k 0004 1 5
as can be checked. MATCHED PROBLEM 4
0002
Write the following series using summation notation: 10003 (A) Start with k 0002 1.
4 8 16 2 0004 0003 0004 3 9 27 81
(B) Start with k 0002 0.
0002
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509
ANSWERS TO MATCHED PROBLEMS 1. 1, 1, 0, 1, 00031, 2, 00033 3. 1 0003 13 0004 15 0003 17 0004 19 0003 111
8-1
1 n00031 (B) an 0002 (00031)n00031a b 2 5 k00031 4 2 2 k 4. (A) a (00031)k00031a b (B) a (00031)k a b 3 3
2. (A) an 0002 2n
k00021
k00020
Exercises
1. Explain the difference between a sequence and a series.
27. an 0002 (000312)n00031
2. What is a recursion formula?
29. a1 0002 7; an 0002 an00031 0003 4, n 0005 2
3. Explain how the Fibonacci sequence can be defined by means of a recursion formula.
30. a1 0002 3; an 0002 an00031 0004 5, n 0005 2
4. Explain summation notation. 5. Explain why the following statement is not true: The general term of the sequence 1, 3, 7, . . . is 2n 0003 1. 6. Explain why at least one term must be provided when defining a sequence recursively. Write the first four terms for each sequence in Problems 7–12. 7. an 0002 n 0003 2 9. an 0002
n00031 n00041
11. an 0002 (00032)n00041
8. an 0002 n 0004 3
31. a1 0002 4; an 0002 14an00031, n 0005 2 32. a1 0002 2; an 0002 2an00031, n 0005 2 In Problems 33–36, write the first seven terms of each sequence. 33. a1 0002 1, a2 0002 2, an 0002 an00032 0004 2an00031, n 0005 3 34. a1 0002 1, a2 0002 00031, an 0002 an00032 0003 an00031, n 0005 3 35. a1 0002 00031, a2 0002 2, an 0002 2an00032 0004 an00031, n 0005 3 36. a1 0002 2, a2 0002 1, an 0002 0003an00032 0004 an00031, n 0005 3
n
1 10. an 0002 a1 0004 b n 12. an 0002
In Problems 37–48, find a general term an for the given sequence a1, a2, a3, a4, . . .
(00031)n00041 n2
13. Write the eighth term in the sequence in Problem 7. 14. Write the tenth term in the sequence in Problem 8. 15. Write the one-hundredth term in the sequence in Problem 9. 16. Write the two-hundredth term in the sequence in Problem 10.
37. 00032, 00031, 0, 1, . . .
38. 10, 11, 12, 13, . . .
39. 5, 7, 9, 11, . . .
40. 1, 00031, 00033, 00035, . . .
41. 00031, 1, 00031, 1, . . .
42. 1, 000312, 13, 000314, . . .
43. 2, 32, 43, 54, . . .
44. 13, 24, 35, 46, . . .
45. 00033, 9, 000327, 81, . . .
46. 5, 25, 125, 625, . . .
2
47. x, In Problems 17–22, write each series in expanded form without summation notation. 5
4
17. a k
18. a k 2
3 1 19. a k k 0002 1 10
5 1 k 20. a a b k00021 3
k00021
21. a (00031)k k00021
6
22. a (00031)k00041k k00021
Write the first five terms of each sequence in Problems 23–32. 23. an 0002 (00031)n00041n2 25. an 0002
1 1 a1 0003 n b 3 10
24. an 0002 (00031)n00041a
3
4
x x x , , ,... 2 3 4
48. x, 0003x3, x5, 0003x7, . . .
In Problems 49–54: (A) Find the first four terms of the sequence. (B) Find a general term bn for a different sequence that has the same first three terms as the given sequence.
k00021
4
28. an 0002 (000332)n00031
1 b 2n
26. an 0002 n[1 0003 (00031)n]
49. an 0002 n2 0003 n 0004 2
50. an 0002 9n2 0003 21n 0004 14
51. an 0002 6n2 0003 11n 0004 6
52. an 0002 25n2 0003 60n 0004 36
53. an 0002 2n2 0003 8n 0004 7
54. an 0002 00034n2 0004 15n 0003 12
In Problems 55–58, use a graphing calculator to graph the first 20 terms of each sequence. 55. an 0002 10002n
56. an 0002 2 0004 0006n n
57. an 0002 (00030.9)
58. a1 0002 00031, an 0002 23 an00031 0004 12
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In Problems 59–64, write each series in expanded form without summation notation. (00032)k00041 k k00021 4
5
60. a (00031)k00041(2k 0003 1)2
59. a
k00021
3
5
1 61. a xk00041 k00021 k
62. a xk00031
(00031)k00041 k x 63. a k k00021
(00031)kx2k00041 64. a 2k 0004 1 k00020
In calculus, it can be shown that
k00021
5
4
In Problems 65–72, write each series using summation notation with the summing index k starting at k 0002 1. 65. 12 0004 22 0004 32 0004 42 66. 2 0004 3 0004 4 0004 5 0004 6 67.
77. Approximate e0.2 using the first five terms of the series. Compare this approximation with your calculator evaluation of e0.2.
n
1 1 1 0004 0003 2 3 4
k00021 n
n k00021
n
n
80. Show that a (ak 0004 bk) 0002 a ak 0004 a bk
1 1 1 0004 2 0004...0004 2 22 3 n
70. 2 0004
3 4 n00041 0004 0004...0004 n 2 3
k00021
k00021
k00021
APPLICATIONS 81. PHYSICS Suppose that a rubber ball is dropped from a height of 20 feet. If it bounces 10 times, with each bounce going half as high as the one before, the heights of these bounces can be described by the sequence an 0002 10(12)n00031 (1 0007 n 0007 10). (A) How high is the fifth bounce? The tenth?
71. 1 0003 4 0004 9 0003 . . . 0004 (00031)n00041n2 (00031)n00041 1 1 1 0003 0004 0003...0004 2 4 8 2n
10
(B) Find the value of the series a an. What does this number n00021 represent?
The sequence an 0002
where the larger n is, the better the approximation. Problems 77 and 78 refer to this series. Note that n!, read “n factorial,” is defined by 0! 0002 1 and n! 0002 1 ⴢ 2 ⴢ 3 ⴢ . . . ⴢ n for n 苸 N.
79. Show that a cak 0002 c a ak
69. 1 0004
72.
x x2 x3 . . . xn xk ex 0002 a 000310004 0004 0004 0004 0004 1! 2! 3! n! k 0002 0 k!
78. Approximate e00030.5 using the first five terms of the series. Compare this approximation with your calculator evaluation of e00030.5.
1 1 1 1 1 0004 20004 30004 40004 5 2 2 2 2 2
68. 1 0003
76. Define sequences {un} and {vn} by u1 0002 1, v1 0002 0, un 0002 un00031 0004 vn00031 and vn 0002 un00031 for n 0005 2. Find the first 10 terms of each sequence, and explain their relationship to the Fibonacci sequence.
a2n00031 0004 M 2an00031
n 0005 2, M a positive real number
can be used to find 1M to any decimal-place accuracy desired. To start the sequence, choose a1 arbitrarily from the positive real numbers. Problems 73 and 74 are related to this sequence. 73. (A) Find the first four terms of the sequence a1 0002 3
an 0002
a2n00031 0004 2 2an00031
n00052
(B) Compare the terms with 12 from a calculator. (C) Repeat parts A and B letting a1 be any other positive number, say 1. 74. (A) Find the first four terms of the sequence a1 0002 2
an 0002
a2n00031 0004 5 2an00031
n00052
(B) Find 15 with a calculator, and compare with the results of part A. (C) Repeat parts A and B letting a1 be any other positive number, say 3. 75. Let {an} denote the Fibonacci sequence and let {bn} denote the sequence defined by b1 0002 1, b2 0002 3, and bn 0002 bn00031 0004 bn00032 for n 0005 3. Compute 10 terms of the sequence {cn}, where cn 0002 bn0002an. Describe the terms of {cn} for large values of n.
82. PHYSICS A bungee jumper dives off a bridge that is 300 feet above the ground. He bounces back 100 feet on the first bounce, then continues to bounce nine more times before coming to rest, with each bounce 1/3 as high as the previous. The heights of these bounces can be described by the sequence an 0002 100(13)n00031 (1 0007 n 0007 10). (A) How high is the fifth bounce? The tenth? 10
(B) Find the value of the series a an. What does this number n00021 represent? 83. SALARY INCREMENT Suppose that you are offered a job with a starting annual salary of $40,000 and annual increases of 4% of the current salary. (A) Write out the first six terms of a sequence an whose terms describe your salary in the first 6 years on this job. (B) Write the general term of the sequence in part A. 6
(C) Find the value of the series a an. What does this number n00021 represent? 84. SALARY INCREMENT A marketing firm is advertising entrylevel positions with a starting annual salary of $24,000 and annual increments of 3% of the current salary. (A) Write out the first six terms of a sequence an whose terms describe the salary for this position in the first 6 years on this job. (B) Write the general term of the sequence in part A. 6
(C) Find the value of the series a an. What does this number n00021 represent?
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Mathematical Induction Z Using Counterexamples Z Using Mathematical Induction Z Additional Examples of Mathematical Induction Z Three Famous Problems
Many of the most important facts and formulas in this book have been stated as theorems. But a theorem is not a theorem until it has been proved, and proving theorems is one of the most challenging tasks in mathematics. There is a big difference between being pretty sure that a statement is true, and proving that statement. Let’s look at an example. Suppose that we are interested in the sum of the first n consecutive odd integers, where n is a positive integer. We can begin by writing the sums for the first few values of n to see if we can observe a pattern: 1⫽ 1 1⫹3⫽ 4 1⫹3⫹5⫽ 9 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫽ 16 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫽ 25
nⴝ1 nⴝ2 nⴝ3 nⴝ4 nⴝ5
Is there any pattern to the sums 1, 4, 9, 16, and 25? You most likely noticed that each is a perfect square and, in fact, each is the square of the number of terms in the sum. So the following conjecture* seems reasonable: CONJECTURE P :
For each positive integer n, 1 ⫹ 3 ⫹ 5 ⫹ . . . ⫹ (2n ⫺ 1) ⫽ n2
(Recall that the general term 2n ⫺ 1 was used to list the odd positive integers in the last section.) At this point, you may be pretty sure that our conjecture is true. You might even look at the previous five calculations and think that we have proved our conjecture. But in actuality, all we have proved is that the conjecture is true for n ⫽ 1, 2, 3, 4, and 5. We are trying to prove that it is true for every positive integer, not just those five! With that in mind, continuing to check the conjecture for specific n’s like 6, 7, 8, . . . is pointless: You can keep trying for the rest of your life, but you will never be able to check every positive integer. Instead, in this section, we will use a much more powerful tool called mathematical induction to prove conjectures. Before we learn about this method of proof, we first consider how to prove that a conjecture is false.
Z Using Counterexamples
Table 1 n
n2 ⴚ n ⴙ 41
1
Prime?
Consider the following conjecture:
41
Yes
2
43
Yes
3
47
Yes
4
53
Yes
For each positive integer n, the number n2 ⫺ n ⫹ 41 is a prime number. Since the conjecture states that this fact is true for every positive integer n, if we can find even one positive integer n for which it is false, then the conjecture will be proved false. A single case or example for which a conjecture fails is called a counterexample. We checked the conjecture for a few particular cases in Table 1. From the table, it certainly appears
5
61
Yes
CONJECTURE Q:
*A conjecture is a statement that is believed to be true, but has not been proved.
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that conjecture Q has a good chance of being true. You may want to check a few more cases. If you persist, you will find that conjecture Q is true for n up to 40. Most students would guess that the statement is always true long before getting to n 0002 41. But then something interesting happens at n 0002 41: 412 0003 41 0004 41 0002 412 which is not prime. Because n 0002 41 provides a counterexample, conjecture Q is false. Here we see the danger of generalizing without proof from a few special cases, even if that “few” is 40 cases! This example was discovered by Euler (1701–1783), the same mathematician that introduced the number e as the base of the natural exponential function.
EXAMPLE
1
Finding a Counterexample Prove that the following conjecture is false by finding a counterexample: For every positive integer n 0005 2, at least half of the positive integers less than or equal to n are prime.
SOLUTION
We will check the conjecture for positive integer values of n starting at 2.
n
Primes less than or equal to n
Fraction of positive integers less than or equal to n that are prime
True or false
2
2
1/2
True
3
2, 3
2/3
True
4
2, 3
2/4
True
5
2, 3, 5
3/5
True
6
2, 3, 5
3/6
True
7
2, 3, 5, 7
4/7
True
8
2, 3, 5, 7
4/8
True
9
2, 3, 5, 7
4/9
False
Since n 0002 9 provides a counterexample, the conjecture is false. MATCHED PROBLEM 1
0002
Prove that the following conjecture is false by finding a counterexample: For every positive integer n, the last digit of n3 is less than 9. 0002
Z Using Mathematical Induction To begin our study of proving conjectures, we will state the principle of mathematical induction, which forms the basis for all of our work in this section. Z THEOREM 1 Principle of Mathematical Induction Let Pn be a statement associated with each positive integer n, and suppose the following conditions are satisfied: 1. P1 is true. 2. For any positive integer k, if Pk is true, then Pk00041 is also true. Then the statement Pn is true for all positive integers n.
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Theorem 1 must be read very carefully. At first glance, it seems to say that if we assume a statement is true, then it is true. But that is not the case at all. If the two conditions in Theorem 1 are satisfied, then we can reason as follows: P1 P2 P3 P4
is true.
Condition 1
is true, because P1 is true. is true, because P2 is true. is true, because P3 is true. . . .
Condition 2 Condition 2 Condition 2
. . .
Because this chain of implications never ends, we will eventually reach Pn for any positive integer n. This is not the same as checking each case separately: The truth of any case follows from knowing that the previous one is true once we have established condition 2. To help visualize this process, picture a row of dominoes that goes on forever (Fig. 1) and interpret the conditions in Theorem 1 as follows: Condition 1 says that the first domino can be pushed over. Condition 2 says that if the kth domino falls, then so does the (k 0004 1)st domino. Together, these two conditions imply that all the dominoes must fall.
Condition 1: The first domino can be pushed over. (a)
Condition 2: If the kth domino falls, then so does the (k ⴙ 1)st. (b)
Conclusion: All the dominoes will fall. (c)
Z Figure 1 Interpreting mathematical induction.
In Example 2 we illustrate proof by mathematical induction by returning to our conjecture P from the beginning of the section.
EXAMPLE
2
Proving a Conjecture Using Induction Prove that for all positive integers n, 1 0004 3 0004 5 0004 . . . 0004 (2n 0003 1) 0002 n2
SOLUTION
State Pn: Pn: 1 0004 3 0004 5 0004 . . . 0004 (2n 0003 1) 0002 n2 CONDITION 1 Show that P1 is true.
P1: 1 0002 12 CONDITION 2 Show that if Pk is true, then Pk00041 must be true.
It’s a good idea to always write out both Pk and Pk00041 at the beginning of this step to see what we can use, and what we need to prove.
Pk: 1 0004 3 0004 5 0004 . . . 0004 (2k 0003 1) 0002 k2 We assume this is a true statement. Pk00041: 1 0004 3 0004 5 0004 . . . 0004 (2k 0003 1) 0004 [2(k 0004 1) 0003 1] 0002 (k 0004 1)2
We need to show that this is also true.
Note that Pk00041 can be simplified a bit: Pk00041: 1 0004 3 0004 5 0004 . . . 0004 (2k 0003 1) 0004 (2k 0004 1) 0002 (k 0004 1)2
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We will perform algebraic operations on the equation Pk (which we know is true) with a goal of obtaining Pk⫹1. Note that the left side of Pk⫹1 is the left side of Pk plus the addition term 2k ⫹ 1. 1 ⫹ 3 ⫹ 5 ⫹ . . . ⫹ (2k ⫺ 1) ⫽ k2 Add 2k ⴙ 1 to both sides. 1 ⫹ 3 ⫹ 5 ⫹ . . . ⫹ (2k ⫺ 1) ⫹ (2k ⴙ 1) ⫽ k2 ⫹ 2k ⴙ 1 Factor the right side. 1 ⫹ 3 ⫹ 5 ⫹ . . . ⫹ (2k ⫺ 1) ⫹ (2k ⫹ 1) ⫽ (k ⫹ 1)2 This is Pkⴙ1! Pk⫹1 was obtained by adding the same number to both sides of Pk, so if Pk is true, then Pk⫹1 must be as well. CONCLUSION
Both conditions of Theorem 1 are satisfied, so Pn is true for all positive integers n. MATCHED PROBLEM 2
0002
Prove that for all positive integers n 1⫹2⫹3⫹...⫹n⫽
n(n ⫹ 1) 2 0002
Z Additional Examples of Mathematical Induction Now we will consider some additional examples of proof by induction. The first is another summation formula. Mathematical induction is the primary tool for proving that formulas of this type are true.
EXAMPLE
3
Proving a Summation Formula Prove that for all positive integers n 1 1 1 2n ⫺ 1 1 ⫹ ⫹ ⫹...⫹ n ⫽ 2 4 8 2 2n PROOF State Pn:
Pn :
1 1 1 1 2n ⫺ 1 ⫹ ⫹ ⫹...⫹ n ⫽ 2 4 8 2 2n
PART 1 Show that P1 is true.
P1:
1 21 ⫺ 1 ⫽ 2 21 ⫽
1 2
So P1 is true. PART 2 Show that if Pk is true, then Pk⫹1 is true. Again, it is a good idea to always write
out both Pk and Pk⫹1 at the beginning of any induction proof to see what is assumed and what must be proved: 1 1 1 1 2k ⫺ 1 ⫹ ⫹ ⫹... ⫹ k ⫽ 2 4 8 2 2k k⫹1 1 1 1 1 1 2 ⫺1 ⫹ ⫹ ⫹ . . . ⫹ k ⫹ k⫹1 ⫽ k⫹1 2 4 8 2 2 2 Pk :
Pk⫹1:
We assume Pk is true.
We must show that Pkⴙ1 follows from Pk.
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We start with the true statement Pk, add 1 Ⲑ2k⫹1 to both sides, and simplify the right side: 1 1 1 1 2k ⫺ 1 ⫹ ⫹ ⫹...⫹ k ⫽ 2 4 8 2 2k 1 1 1 1 1 2k ⫺ 1 1 ⫹ ⫹ ⫹ . . . ⫹ k ⴙ kⴙ1 ⫽ ⴙ kⴙ1 k 2 4 8 2 2 2 2 ⫽ ⫽ ⫽
2k ⫺ 1
ⴢ
k
2
2 1 ⫹ k⫹1 2 2
2k⫹1 ⫺ 2 ⫹ 1 2k⫹1
Add
1 2k⫹1
to both sides.
Find common denominator for right-hand side.
Write as single fraction.
Simplify.
2k⫹1 ⫺ 1 2k⫹1
So 1 1 1 1 1 2k⫹1 ⫺ 1 ⫹ ⫹ ⫹ . . . ⫹ k ⫹ k⫹1 ⫽ 2 4 8 2 2 2k⫹1
Pkⴙ1
and we have shown that if Pk is true, then Pk⫹1 is true. CONCLUSION
Both conditions in Theorem 1 are satisfied. Therefore, Pn is true for all positive integers n. 0002 MATCHED PROBLEM 3
Prove that for all positive integers n 2 2 2 3n ⫺ 1 2 ⫹ ⫹ ⫹p⫹ n⫽ 3 9 27 3 3n 0002 Example 4 provides a proof of a law of exponents that previously we had to assume was true. First we redefine an for n a positive integer, using a recursion formula: n Z DEFINITION 1 Recursive Definition of a
For n a positive integer a1 ⫽ a an⫹1 ⫽ ana
EXAMPLE
4
n 7 1
Proving a Law of Exponents Prove that (xy)n ⫽ xnyn for all positive integers n. PROOF State Pn:
Pn: (xy)n ⫽ x ny n PART 1 Show that P1 is true.
(xy)1 ⫽ xy ⫽ x1y1 So P1 is true.
Definition 1 Definition 1
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PART 2
Show that if Pk is true, then Pk00041 is true. Pk : (xy)k 0002 xkyk Pk00041: (xy)k00041 0002 xk00041yk00041
Assume Pk is true. Show that Pkⴙ1 follows from Pk.
Here we start with the left side of Pk00041 and use Pk to find the right side of Pk00041: (xy)k00041 0002 (xy)k(xy)1 0002 x ky kxy 0002 (x kx)(y ky) 0002 x k00041y k00041
Use Pk: (xy) k ⴝ x ky k Use properties of real numbers. Use Definition 1 twice.
So (xy)k00041 0002 x k00041y k00041, and we have shown that if Pk is true, then Pk00041 is true. CONCLUSION
Both conditions in Theorem 1 are satisfied. Therefore, Pn is true for all positive integers n. 0002 MATCHED PROBLEM 4
Prove that (x/y)n 0002 x n/y n for all positive integers n.
0002
Example 5 deals with factors of integers. Before we start, recall that an integer p is divisible by an integer q if p 0002 qr for some integer r.
EXAMPLE
5
Proving a Divisibility Property Prove that 42n 0003 1 is divisible by 5 for all positive integers n. PROOF Use the definition of divisibility to state Pn as follows:
Pn: 42n 0003 1 0002 5r
for some integer r
PART 1 Show that P1 is true.
P1: 42 0003 1 0002 15 0002 5 ⴢ 3 So P1 is true. PART 2 Show that if Pk is true, then Pk00041 is true.
Pk: 42k 0003 1 0002 5r Pk00041: 42(k00041) 0003 1 0002 5s
for some integer r for some integer s
Assume Pk is true. Show that Pkⴙ1 must follow.
As before, we start with the true statement Pk: 42k 0003 1 0002 5r 42(42k 0003 1) 0002 42(5r) 42k00042 0003 16 0002 80r 42(k00041) 0003 1 0002 80r 0004 15 0002 5(16r 0004 3) So
42(k00041) 0003 1 0002 5s
Multiply both sides by 42. Simplify. Add 15 to both sides. Factor out 5.
Pkⴙ1
where s 0002 16r 0004 3 is an integer, and we have shown that if Pk is true, then Pk00041 is true. CONCLUSION
Both conditions in Theorem 1 are satisfied. Therefore, Pn is true for all positive integers n. 0002
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MATCHED PROBLEM 5
Mathematical Induction
517
Prove that 8n 0003 1 is divisible by 7 for all positive integers n.
0002
In some cases, a conjecture may be true only for n 0005 m, where m is a positive integer, rather than for all n 0005 0. For example, see Problems 53 and 54 in Exercises 8-2. The principle of mathematical induction can be extended to cover cases like this as follows:
Z THEOREM 2 Extended Principle of Mathematical Induction Let m be a positive integer, let Pn be a statement associated with each integer n 0005 m, and suppose the following conditions are satisfied: 1. Pm is true. 2. For any integer k 0005 m, if Pk is true, then Pk00041 is also true. Then the statement Pn is true for all integers n 0005 m.
Z Three Famous Problems The problem of determining whether a certain statement about the positive integers is true may be extremely difficult. Proofs may require remarkable insight and ingenuity and the development of techniques far more advanced than mathematical induction. Consider, for example, the famous problems of proving the following statements: 1. 2. 3.
Lagrange’s Four Square Theorem, 1772: Each positive integer can be expressed as the sum of four or fewer squares of positive integers. Fermat’s Last Theorem, 1637: For n 7 2, xn 0004 yn 0002 z n does not have solutions in the natural numbers. Goldbach’s Conjecture, 1742: Every positive even integer greater than 2 is the sum of two prime numbers.
The first statement was considered by the early Greeks and finally proved in 1772 by Lagrange. Fermat’s last theorem, defying the best mathematical minds for over 350 years, finally succumbed to a 200-page proof by Professor Andrew Wiles of Princeton University in 1993. To this date no one has been able to prove or disprove Goldbach’s conjecture.
ZZZ EXPLORE-DISCUSS 1
(A) Explain the difference between a theorem and a conjecture. (B) Why is “Fermat’s last theorem” a misnomer? Suggest more accurate names for the result.
ANSWERS TO MATCHED PROBLEMS 1. The last digit of 93 0002 729 is greater than 8. 2. Sketch of proof. n(n 0004 1) Pn: 1 0004 2 0004 3 0004 . . . 0004 n 0002 2 1(1 0004 1) Condition 1. 1 0002 . P1 is true. 2
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Condition 2. Show that if Pk is true, then Pk⫹1 is true. k(k ⫹ 1) 1⫹2⫹3⫹...⫹k⫽ Pk 2 k(k ⫹ 1) 1 ⫹ 2 ⫹ 3 ⫹ . . . ⫹ k ⴙ (k ⴙ 1) ⫽ ⴙ (k ⴙ 1) 2 (k ⫹ 1)(k ⫹ 2) ⫽ Pkⴙ1 2 Conclusion: Pn is true for all positive integers n. 2 2 2 2 3n ⫺ 1 ⫹ ###⫹ n⫽ 3. Sketch of proof. Pn: ⫹ ⫹ 3 9 27 3 3n 2 31 ⫺ 1 . P1 is true. Part 1. ⫽ 3 3 Part 2. Show that if Pk is true, then Pk⫹1 is true. 2 2 2 2 3k ⫺ 1 Pk ⫹ ⫹ ⫹p⫹ k⫽ 3 9 27 3 3k 2 2 2 2 2 3k ⫺ 1 2 ⫹ k⫹1 ⫹ ⫹ ⫹ p ⫹ k ⫹ k⫹1 ⫽ 3 9 27 3 3 3k 3 3k⫹1 ⫺ 1 Pk ⴙ 1 ⫽ 3k Conclusion: Pn is true for all positive integers n. x n xn 4. Sketch of proof. Pn: a b ⫽ n y y x 1 x x1 Part 1. a b ⫽ ⫽ 1 . P1 is true. y y y Part 2. Show that if Pk is true, then Pk⫹1 is true. xk⫹1 x k⫹1 x k x xk x xkx a b ⫽ a b a b ⫽ k a b ⫽ k ⫽ k⫹1 y y y y y yy y Conclusion: Pn is true for all positive integers n. 5. Sketch of proof. Pn: 8n ⫺ 1 ⫽ 7r for some integer r Part 1. 81 ⫺ 1 ⫽ 7 ⫽ 7 ⴢ 1. P1 is true. Part 2. Show that if Pk is true, then Pk⫹1 is true. Pk 8k ⫺ 1 ⫽ 7r 8(8k ⫺ 1) ⫽ 8(7r) 8k⫹1 ⫺ 1 ⫽ 56r ⫹ 7 ⫽ 7(8r ⫹ 1) ⫽ 7s Pkⴙ1 Conclusion: Pn is true for all positive integers n.
8-2
Exercises
1. What is a counterexample? 2. Explain how falling dominoes can be compared to the principle of mathematical induction. 3. In Theorem 1 (principle of mathematical induction), what do Pk and Pk⫹1 represent? 4. The number n2 ⫺ n ⫹ 41 is prime for n ⫽ 1, 2, . . . , 40. Does this prove that n2 ⫺ n ⫹ 41 is prime for every natural number n? Explain.
In Problems 5–8, find the first positive integer n that causes the statement to fail. 5. (3 ⫹ 5)n ⫽ 3n ⫹ 5n 2
7. n ⫽ 3n ⫺ 2
6. n ⬍ 10 8. n3 ⫹ 11n ⫽ 6n2 ⫹ 6
Verify each statement Pn in Problems 9–14 for n ⫽ 1, 2, and 3. 9. Pn: 2 ⫹ 6 ⫹ 10 ⫹ ⭈ ⭈ ⭈ ⫹ (4n ⫺ 2) ⫽ 2n2 10. Pn: 4 ⫹ 8 ⫹ 12 ⫹ ⭈ ⭈ ⭈ ⫹ 4n ⫽ 2n(n ⫹ 1)
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11. Pn: a5an 0002 a50004n
12. Pn: (a5)n 0002 a5n
Mathematical Induction
519
43. x2n 0003 1 is divisible by x 0003 1; x 1
13. Pn: 9n 0003 1 is divisible by 4
44. x2n 0003 1 is divisible by x 0004 1; x 00031
14. Pn: 4n 0003 1 is divisible by 3
45. 13 0004 23 0004 33 0004 0004 n3 0002 (1 0004 2 0004 3 0004 0004 n)2 [Hint: See Matched Problem 2 following Example 2.]
Write Pk and Pk00041 for Pn as indicated in Problems 15–20. 15. Pn in Problem 9
16. Pn in Problem 10
17. Pn in Problem 11
18. Pn in Problem 12
19. Pn in Problem 13
20. Pn in Problem 14
In Problems 21–26, use mathematical induction to prove that each Pn holds for all positive integers n. 21. Pn in Problem 9
22. Pn in Problem 10
23. Pn in Problem 11
24. Pn in Problem 12
25. Pn in Problem 13
26. Pn in Problem 14
In Problems 27–30, prove the statement is false by finding a counterexample. 27. If n 2, then any polynomial of degree n has at least one real zero.
46.
1 1 1 0004 0004 0004 ### 1ⴢ2ⴢ3 2ⴢ3ⴢ4 3ⴢ4ⴢ5 0004
n(n 0004 3) 1 0002 n(n 0004 1)(n 0004 2) 4(n 0004 1)(n 0004 2)
In Problems 47–50, suggest a formula for each expression, and prove your conjecture using mathematical induction, n 僆 N. 47. 2 0004 4 0004 6 0004 0004 2n 48.
1 1 1 1 0004 0004 0004 ###0004 1ⴢ2 2ⴢ3 3ⴢ4 n(n 0004 1)
49. The number of lines determined by n points in a plane, no three of which are collinear 50. The number of diagonals in a polygon with n sides Prove Problems 51–54 true for all integers n as specified. 51. If a 1, then an 1; n 僆 N
28. Any positive integer n 7 can be written as the sum of three or fewer squares of positive integers.
52. If 0 6 a 6 1, then 0 6 an 6 1; n 僆 N
29. If n is a positive integer, then there is at least one prime number p such that n p n 0004 6.
54. 2n n2; n 0005 5
30. If a, b, c, and d are positive integers such that a2 0004 b2 0002 c2 0004 d 2, then a 0002 c or a 0002 d. In Problems 31–46, use mathematical induction to prove each proposition for all positive integers n, unless restricted otherwise. 31. 2 0004 22 0004 23 0004 0004 2n 0002 2n00041 0003 2 32.
33. 12 0004 32 0004 52 0004 # # # 0004 (2n 0003 1)2 0002 13 (4n3 0003 n) 2
34. 1 0004 8 0004 16 0004 0004 8(n 0003 1) 0002 (2n 0003 1) ; n 1 n(n 0004 1)(2n 0004 1) 6
n(n 0004 1)(n 0004 2) 36. 1 ⴢ 2 0004 2 ⴢ 3 0004 3 ⴢ 4 0004 # # # 0004 n(n 0004 1) 0002 3 37.
an 0002 an00033; n 7 3 a3
38.
a5 1 0002 n00035 ; n 7 5 an a
39. aman 0002 am0004n; m, n 僆 N [Hint: Choose m as an arbitrary element of N, and then use induction on n.] n m
mn
40. (a ) 0002 a ; m, n 僆 N n
41. x 0003 1 is divisible by x 0003 1, x 1 [Hint: Divisible means that xn 0003 1 0002 (x 0003 1)Q(x) for some polynomial Q(x).] 42. xn 0003 yn is divisible by x 0003 y; x y
In Problems 55–58, determine whether the statement is true or false. If true, prove using mathematical induction. If false, find a counterexample. 55. If n is a positive integer, then 1 0003 2 0004 3 0003 0004 (2n 0003 1) 0002 n (that is, the alternating sum of the first 2n 0003 1 positive integers is equal to n).
1 1 n 1 1 1 0004 0004 0004...0004 n 000210003a b 2 4 8 2 2
35. 12 0004 22 0004 32 0004 . . . 0004 n2 0002
53. n2 2n; n 0005 3
56. If n is a positive integer, then (00031)n00041n(n 0004 1) 12 0003 22 0004 32 0003 # # # 0004 (00031)n00041n2 0002 2 57. If n is a positive integer, then 3n00041 0004 4n00041 0004 0004 (n 0004 3)n00041 0002 (n 0004 4)n00041 58. If n is a positive integer, then n2 0004 21n 0004 1 is a prime number. If {an} and {bn} are two sequences, we write {an} 0002 {bn} if and only if an 0002 bn for all n 僆 N. In Problems 59–62, use mathematical induction to show that {an} 0002 {bn}. 59. a1 0002 1, an 0002 an00031 0004 2; bn 0002 2n 0003 1 60. a1 0002 2, an 0002 an00031 0004 2; bn 0002 2n 61. a1 0002 2, an 0002 22an00031; bn 0002 22n00031 62. a1 0002 2, an 0002 3an00031; bn 0002 2 ⴢ 3n00031
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Arithmetic and Geometric Sequences Z Arithmetic and Geometric Sequences Z Developing nth-Term Formulas Z Developing Sum Formulas for Finite Arithmetic Series Z Developing Sum Formulas for Finite Geometric Series Z Developing a Sum Formula for Infinite Geometric Series
For most sequences, it is difficult to add up an arbitrary number of terms of the sequence without adding the terms one at a time. In this section, we will study two special types of sequences, arithmetic sequences and geometric sequences. One of the things that make them special is that we can develop formulas for the sum of the corresponding series.
Z Arithmetic and Geometric Sequences Consider the sequence defined by the general term an 0002 5 0004 2(n 0003 1), n 0005 1. The first five terms are 5, 7, 9, 11, and 13. It’s not hard to see that after starting at 5, every term is obtained by adding 2 to the previous term. This is an example of an arithmetic sequence.
Z DEFINITION 1 Arithmetic Sequence A sequence a1, a2, a3, . . . , an, . . . is called an arithmetic sequence, or arithmetic progression, if there exists a constant d, called the common difference, such that an 0003 an00031 0002 d That is, an 0002 an00031 0004 d
for every n 1
In short, a sequence is arithmetic when every term is obtained by adding some fixed number to the previous term. This fixed number is called the common difference, and is usually represented by the letter d. Now consider the sequence with general term an 0002 5(2)n00031. The first five terms are 5, 10, 20, 40, and 80. It also starts at 5, but this time every term is obtained by multiplying the previous term by 2. This is an example of a geometric sequence.
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521
Z DEFINITION 2 Geometric Sequence A sequence a1, a2, a3, . . . , an, . . . is called a geometric sequence, or geometric progression, if there exists a nonzero constant r, called the common ratio, such that an 0002r an00031 That is, an 0002 ran00031
for every n 1
In short, a sequence is geometric when every term is obtained by multiplying the previous term by some fixed number. This fixed number is called the common ratio, and is usually represented by the letter r.
ZZZ EXPLORE-DISCUSS 1
(A) Graph the arithmetic sequence 5, 7, 9, . . . . Describe the graphs of all arithmetic sequences with common difference 2. (B) Graph the geometric sequence 5, 10, 20, . . . . Describe the graphs of all geometric sequences with common ratio 2.
EXAMPLE
1
Recognizing Arithmetic and Geometric Sequences Which of the following can be the first four terms of an arithmetic sequence? Of a geometric sequence?
SOLUTIONS
(A) 1, 2, 3, 5, . . .
(B) 00031, 3, 00039, 27, . . .
(C) 3, 3, 3, 3, . . .
(D) 10, 8.5, 7, 5.5, . . .
(A) Because 2 0003 1 5 0003 3, there is no common difference, so the sequence is not an arithmetic sequence. Because 21 32, there is no common ratio, so the sequence is not geometric either. (B) The sequence is geometric with common ratio 00033, but it is not arithmetic. (C) The sequence is arithmetic with common difference 0 and it is also geometric with common ratio 1. (D) The sequence is arithmetic with common difference 00031.5, but it is not geometric.
MATCHED PROBLEM 1
0002
Which of the following can be the first four terms of an arithmetic sequence? Of a geometric sequence? (A) 8, 2, 0.5, 0.125, . . .
(B) 00037, 00032, 3, 8, . . .
(C) 1, 5, 25, 100, . . . 0002
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Z Developing nth-Term Formulas If {an} is an arithmetic sequence with common difference d, then a2 0002 a1 0004 d a3 0002 a2 0004 d 0002 a1 0004 2d a4 0002 a3 0004 d 0002 a1 0004 3d This suggests Theorem 1, which can be proved by mathematical induction (see Problem 67 in Exercises 8-3). Z THEOREM 1 The nth Term of an Arithmetic Sequence an 0002 a1 0004 (n 0003 1)d
for every n 1
Similarly, if {an} is a geometric sequence with common ratio r, then a2 0002 a1r a3 0002 a2r 0002 a1r 2 a4 0002 a3r 0002 a1r 3 This suggests Theorem 2, which can also be proved by mathematical induction (see Problem 71 in Exercises 8-3). Z THEOREM 2 The nth Term of a Geometric Sequence an 0002 a1r n00031
EXAMPLE
2
for every n 1
Finding Terms in Arithmetic and Geometric Sequences (A) If the first and tenth terms of an arithmetic sequence are 3 and 30, respectively, find the fiftieth term of the sequence. (B) If the first and tenth terms of a geometric sequence are 1 and 4, find the seventeenth term to three decimal places.
SOLUTIONS
(A) First use Theorem 1 with a1 0002 3 and a10 0002 30 to find d: an 0002 a1 0004 (n 0003 1)d a10 0002 a1 0004 (10 0003 1)d 30 0002 3 0004 9d d00023
Substitute n ⴝ 10. Substitute a10 ⴝ 30 and a1 ⴝ 3. Solve for d.
Now find a50: a50 0002 a1 0004 (50 0003 1)3 0002 3 0004 49 ⴢ 3 0002 150
Substitute a1 ⴝ 3. Simplify.
(B) First let n 0002 10, a1 0002 1, a10 0002 4 and use Theorem 2 to find r. an 0002 a1rn00031 4 0002 1r1000031 r 0002 4100029
Substitute n ⴝ 10, a10 ⴝ 4, and a1 ⴝ 1. Solve for r.
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Now use Theorem 2 again, this time with n 0002 17. a17 0002 a1r16 0002 1(4100029)16 0002 41600029 0003 11.758 MATCHED PROBLEM 2
0002
(A) If the first and fifteenth terms of an arithmetic sequence are 00035 and 23, respectively, find the seventy-third term of the sequence. 1 1 1 (B) Find the eighth term of the geometric sequence , 0003 , , . . . . 64 32 16 0002
Z Developing Sum Formulas for Finite Arithmetic Series If a1, a2, a3, . . . , an is a finite arithmetic sequence, then the corresponding series a1 0004 a2 0004 a3 0004 . . . 0004 an is called an arithmetic series. We will derive two simple and very useful formulas for the sum of an arithmetic series. Let d be the common difference of the arithmetic sequence a1, a2, a3, . . . , an and let Sn denote the sum of the series a1 0004 a2 0004 a3 0004 . . . 0004 an. Then Sn 0002 a1 0004 (a1 0004 d) 0004 . . . 0004 [a1 0004 (n 0003 2)d ] 0004 [a1 0004 (n 0003 1)d] Reversing the order of the sum, we obtain Sn 0002 [a1 0004 (n 0003 1)d ] 0004 [a1 0004 (n 0003 2)d ] 0004 . . . 0004 (a1 0004 d ) 0004 a1 Adding the left sides of these two equations and corresponding elements of the right sides, we see that 2Sn 0002 [2a1 0004 (n 0003 1)d ] 0004 [2a1 0004 (n 0003 1)d] 0004 . . . 0004 [2a1 0004 (n 0003 1)d ] 0002 n[2a1 0004 (n 0003 1)d ] This can be restated as in Theorem 3.
Z THEOREM 3 Sum of an Arithmetic Series—First Form n Sn 0002 [2a1 0004 (n 0003 1)d] 2
By replacing a1 0004 (n 0003 1)d with an, we obtain a second useful formula for the sum.
Z THEOREM 4 Sum of an Arithmetic Series—Second Form n Sn 0002 (a1 0004 an) 2
The proof of the first sum formula by mathematical induction is left as an exercise (see Problem 68 in Exercises 8-3).
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3
Finding the Sum of an Arithmetic Series Find the sum of the first 26 terms of an arithmetic series if the first term is 00037 and d 0002 3.
SOLUTION
Let n 0002 26, a1 0002 00037, d 0002 3, and use Theorem 3. n [2a1 0004 (n 0003 1)d] 2 0002 262 [2(00037) 0004 (26 0003 1)3] 0002 793
Sn 0002 S26
MATCHED PROBLEM 3
EXAMPLE
4
Substitute n ⴝ 26, a1 ⴝ ⴚ7, and d ⴝ 3. Simplify.
0002
Find the sum of the first 52 terms of an arithmetic series if the first term is 23 and d 0002 00032. 0002
Finding the Sum of an Arithmetic Series Find the sum of all the odd numbers between 51 and 99, inclusive.
SOLUTION
First, use a1 0002 51, an 0002 99, and Theorem 1 to find n: an 0002 a1 0004 (n 0003 1)d 99 0002 51 0004 (n 0003 1)2 n 0002 25
Substitute an ⴝ 99, a1 ⴝ 51, and d ⴝ 2. Solve for n.
Now use Theorem 4 to find S25: n Sn 0002 (a1 0004 an) 2 S25 0002 252 (51 0004 99)
Substitute n ⴝ 25, a1 ⴝ 51, and an ⴝ 99.
0002 1,875 MATCHED PROBLEM 4
EXAMPLE
5
0002
Find the sum of all the even numbers between 000322 and 52, inclusive.
0002
Prize Money A 16-team bowling league has $8,000 to be awarded as prize money. If the last-place team is awarded $275 in prize money and the award increases by the same amount for each successive finishing place, how much will the first-place team receive?
SOLUTION
If a1 is the award for the first-place team, a2 is the award for the second-place team, and so on, then the prize money awards form an arithmetic sequence with n 0002 16, a16 0002 275, and S16 0002 8,000. Use Theorem 4 to find a1. n Sn 0002 (a1 0004 an) 2 16 8,000 0002 2 (a1 0004 275) a1 0002 725
Substitute n ⴝ 16, S16 ⴝ 8,000, a16 ⴝ 275. Solve for a1.
The first-place team receives $725. MATCHED PROBLEM 5
0002
Refer to Example 5. How much prize money is awarded to the second-place team? 0002
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Z Developing Sum Formulas for Finite Geometric Series If a1, a2, a3, . . . , an is a finite geometric sequence, then the corresponding series a1 0004 a2 0004 a3 0004 . . . 0004 an is called a geometric series. As with arithmetic series, we can derive two simple and very useful formulas for the sum of a geometric series. Let r be the common ratio of the geometric sequence a1, a2, a3, . . . , an and let Sn denote the sum of the series a1 0004 a2 0004 a3 0004 . . . 0004 an. Then Sn 0002 a1 0004 a1r 0004 a1r 2 0004 a1r 3 0004 . . . 0004 a1r n00032 0004 a1r n00031 Multiply both sides of this equation by r to obtain rSn 0002 a1r 0004 a1r 2 0004 a1r 3 0004 . . . 0004 a1r n00031 0004 a1r n Now subtract the left side of the second equation from the left side of the first, and the right side of the second equation from the right side of the first to obtain Sn 0003 rSn 0002 a1 0003 a1r n Sn(1 0003 r) 0002 a1 0003 a1r n
Factor out Sn
Solving for Sn, we obtain the following formula for the sum of a geometric series: Z THEOREM 5 Sum of a Geometric Series—First Form Sn 0002
a1 0003 a1r n 10003r
r 1
Because an 0002 a1r n00031, or ran 0002 a1r n, the sum formula also can be written in the following form: Z THEOREM 6 Sum of a Geometric Series—Second Form Sn 0002
a1 0003 ran 10003r
r 1
The proof of the first sum formula (Theorem 5) by mathematical induction is left as an exercise (see Problem 72, Exercises 8-3). If r 0002 1, then Sn 0002 a1 0004 a1(1) 0004 a1(12) 0004 . . . 0004 a1(1n00031) 0002 na1
EXAMPLE
6
Finding the Sum of a Geometric Series Find the sum of the first 20 terms of a geometric series if the first term is 1 and r 0002 2.
SOLUTION
Let n 0002 20, a1 0002 1, r 0002 2, and use Theorem 5. Sn 0002 0002
a1 0003 a1r n 10003r 1 0003 1 ⴢ 220 0002 1,048,575 100032
Substitute n ⴝ 20, a1 ⴝ 1, and r ⴝ 2.
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Technology Connections To calculate the sum of a series with a graphing calculator, first generate the sequence using the sequence command, then find its sum using the sum command. Figure 1 shows the solution to Example 6.
Z Figure 1
0002 MATCHED PROBLEM 6
Find the sum, to two decimal places, of the first 14 terms of a geometric series if the first term is 641 and r 0002 00032. 0002
Z Developing a Sum Formula for Infinite Geometric Series Consider a geometric series with a1 0002 5 and r 0002 12. What happens to the sum Sn as n increases? To answer this question, we first write the sum formula in the more convenient form Sn 0002
a1 0003 a1r n a1 a1r n 0002 0003 10003r 10003r 10003r
(1)
For a1 0002 5 and r 0002 12, 1 n Sn 0002 10 0003 10 a b 2 Let’s look at some of the Sn s: 1 S2 0002 10 0003 10 a b 0002 7.5 4 1 S3 0002 10 0003 10 a b 0002 8.75 8 1 S4 0002 10 0003 10 a b 0002 9.375 16 o S20 0002 10 0003 10 a
1 b 0002 9.999990 p 1,048,576
It appears that (12)n becomes smaller and smaller as n increases and that the sum gets closer and closer to 10. In general, it is possible to show that, if 0004 r 0004 6 1, then r n will get closer and closer to 0 as n increases. Symbolically, r n S 0 as n S . So the term a1r n 10003r in equation (1) will tend to 0 as n increases, and Sn will tend to a1 10003r
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In other words, if 0004 r 0004 6 1, then Sn can be made as close to a1 10003r as we wish by taking n sufficiently large. So we can define the sum of an infinite geometric series by the following formula:
Z DEFINITION 3 Sum of an Infinite Geometric Series S 0002
a1 10003r
0004r0004 6 1
If 0004 r 0004 0005 1, an infinite geometric series has no sum.
EXAMPLE
7
Expressing a Repeating Decimal as a Fraction Represent the repeating decimal 0.454 545 . . . 0002 0.45 as the quotient of two integers. Recall that a repeating decimal names a rational number and that any rational number can be represented as the quotient of two integers.
SOLUTION
0.45 0002 0.45 0004 0.0045 0004 0.000 045 0004 . . . The right side of the equation is an infinite geometric series with a1 0002 0.45 and r 0002 0.01. The sum is S 0002 This shows that, 0.45 and dividing 5 by 11.
MATCHED PROBLEM 7
EXAMPLE
8
5 11
a1 0.45 0.45 5 0002 0002 0002 10003r 1 0003 0.01 0.99 11
name the same rational number. You can check the result by 0002
Repeat Example 7 for 0.818 181 . . . 0002 0.81.
0002
Economy Stimulation A state government uses proceeds from a lottery to provide a tax rebate for property owners. Suppose an individual receives a $500 rebate and spends 80% of this, and each of the recipients of the money spent by this individual also spends 80% of what he or she receives, and this process continues without end. According to the multiplier doctrine in economics, the effect of the original $500 tax rebate on the economy is multiplied many times. What is the total amount spent if the process continues as indicated?
SOLUTION
The individual receives $500 and spends 0.8(500) 0002 $400. The recipients of this $400 spend 0.8(400) 0002 $320, the recipients of this $320 spend 0.8(320) 0002 $256, and so on. The total spending generated by the $500 rebate is 400 0004 320 0004 256 0004 . . . 0002 400 0004 0.8(400) 0004 (0.8)2(400) 0004 . . .
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which we recognize as an infinite geometric series with a1 0002 400 and r 0002 0.8. The total amount spent is S 0002
MATCHED PROBLEM 8
a1 400 400 0002 0002 0002 $2,000 10003r 1 0003 0.8 0.2
0002
Repeat Example 8 if the tax rebate is $1,000 and the percentage spent by all recipients is 90%. 0002
(A) Find an infinite geometric series with a1 0002 10 whose sum is 1,000.
ZZZ EXPLORE-DISCUSS 2
(B) Find an infinite geometric series with a1 0002 10 whose sum is 6. (C) Suppose that an infinite geometric series with a1 0002 10 has a sum. Explain why that sum must be greater than 5.
ANSWERS TO MATCHED PROBLEMS 1. (A) The sequence is geometric with r 0002 14, but not arithmetic. (B) The sequence is arithmetic with d 0002 5, but not geometric. (C) The sequence is neither arithmetic nor geometric. 2. (A) 139 (B) 00032 3. 00031,456 4. 570 5. $695 6. 000385.33 7. 119 8. $9,000
8-3
Exercises
1. What is an arithmetic sequence?
Let a1, a2, a3, . . . , an , . . . be an arithmetic sequence. In Problems 9–16, find the indicated quantities.
2. What is a geometric sequence? 3. Explain the terms “common difference” and “common ratio.” 4. Explain how a repeating decimal can be viewed as a geometric series.
9. a1 0002 00035, d 0002 4; a2 0002 ?, a3 0002 ?, a4 0002 ? 10. a1 0002 000318, d 0002 3; a2 0002 ?, a3 0002 ?, a4 0002 ? 11. a1 0002 00033, d 0002 5; a15 0002 ?, S11 0002 ?
5. Which infinite arithmetic series have a sum?
12. a1 0002 3, d 0002 4; a22 0002 ?, S21 0002 ?
6. Which infinite geometric series have a sum?
13. a1 0002 1, a2 0002 5; S21 0002 ? 14. a1 0002 5, a2 0002 11; S11 0002 ?
In Problems 7 and 8, determine whether the following can be the first three terms of an arithmetic or geometric sequence, and, if so, find the common difference or common ratio and the next two terms of the sequence. 7. (A) 000311, 000316, 000321, . . . (C) 1, 4, 9, . . .
(B) 2, 00034, 8, . . . (D)
1 1 1 2 , 6 , 18 ,
...
15. a1 0002 7, a2 0002 5; a15 0002 ? 16. a1 0002 00033, d 0002 00034; a10 0002 ? Let a1, a2, a3 , . . . , an , . . . be a geometric sequence. In Problems 17–22, find each of the indicated quantities.
8. (A) 5, 20, 100, . . .
(B) 00035, 00035, 00035, . . .
17. a1 0002 00036, r 0002 000312; a2 0002 ?, a3 0002 ?, a4 0002 ?
(C) 7, 6.5, 6, . . .
(D) 512, 256, 128, . . .
18. a1 0002 12, r 0002 23; a2 0002 ?, a3 0002 ?, a4 0002 ?
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19. a1 0002 81, r 0002 13; a10 0002 ? 20. a1 0002 64, r 0002 12; a13 0002 ? 21. a1 0002 3, a7 0002 2,187, r 0002 3; S7 0002 ? 22. a1 0002 1, a7 0002 729, r 0002 00033; S7 0002 ?
Arithmetic and Geometric Sequences
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50. Show that the sum of the first n even natural numbers is n 0004 n2, using appropriate formulas from Section 8-3. In Problems 51–60, find the sum of each infinite geometric series that has a sum. 51. 2 0004 12 0004 18 0004 . . .
Let a1, a2, a3, . . . , an , . . . be an arithmetic sequence. In Problems 23–30, find the indicated quantities. 23. a1 0002 3, a20 0002 117; d 0002 ?, a101 0002 ? 24. a1 0002 7, a8 0002 28; d 0002 ?, a25 0002 ? 25. a1 0002 000312, a40 0002 22; S40 0002 ? 26. a1 0002 24, a24 0002 000328; S24 0002 ? 27. a1 0002 13, a2 0002 12; a11 0002 ?, S11 0002 ? 28. a1 0002 16, a2 0002 14; a19 0002 ?, S19 0002 ? 29. a3 0002 13, a10 0002 55; a1 0002 ? 30. a9 0002 000312, a13 0002 3; a1 0002 ? Let a1, a2, a3, . . . , an , . . . be a geometric sequence. Find each of the indicated quantities in Problems 31–42. 31. a1 0002 8, a2 0002 2; r 0002 ? 32. a1 0002 00036, a2 0002 2; r 0002 ? 33. a1 0002 120, a4 0002 000315; r 0002 ? 34. a1 0002 12, a6 0002 8; r 0002 ? 35. a1 0002 9, r 0002 23; S10 0002 ? 36. a1 0002 3, r 0002 5; S9 0002 ? 37. a1 0002 1, a8 0002 2,187; S8 0002 ? 38. a1 0002 12, a12 0002 1,024; S12 0002 ? 39. a3 0002 72, a6 0002 0003243; a1 0002 ? 40. a4 0002 8, a5 0002 6; a1 0002 ? 41. a1 0002 1, a4 0002 00031; a100 0002 ? 42. a1 0002 00031, a8 0002 1; a99 0002 ? 51
43. S51 0002 a (3k 0004 3) 0002 ? k00021 40
44. S40 0002 a (2k 0003 3) 0002 ?
52. 6 0004 2 0004 23 0004 . . . 53. 3 0003 1 0004 13 0003 . . . 54. 1 0004 43 0004 169 0004 . . . 55. 1 0004 0.1 0004 0.01 0004 . . . 56. 10 0003 2 0004 0.4 0003 . . . 57. 00031 0004 12 0003 14 0004 . . . 58. 00036 0004 4 0003 83 0004 . . . 59. 1 0003 1 0004 10003 . . . 60. 0003100 0003 80 0003 64 0003 . . . In Problems 61–66, represent each repeating decimal as the quotient of two integers. 61. 0.7 0002 0.7777 . . .
62. 0.5 0002 0.5555 . . .
63. 0.54 0002 0.545 454 . . .
64. 0.27 0002 0.272 727 . . .
65. 3.216 0002 3.216 216 216 . . . 66. 5.63 0002 5.636 363 . . . 67. Prove, using mathematical induction, that if {an} is an arithmetic sequence, then an 0002 a1 0004 (n 0003 1)d
68. Prove, using mathematical induction, that if {an} is an arithmetic sequence, then n Sn 0002 [2a1 0004 (n 0003 1)d] 2 69. If in a given sequence, a1 0002 00032 and an 0002 00033an00031, n 7 1, find an in terms of n. n
70. For the sequence in Problem 69, find Sn 0002 a ak in terms k00021 of n. 71. Prove, using mathematical induction, that if {an} is a geometric sequence, then
k00021 7
an 0002 a1r n00031 k00031
45. S7 0002 a (00033) k00021
0002?
k00021
47. Find the sum of all the even integers between 21 and 135.
n僆N
72. Prove, using mathematical induction, that if {an} is a geometric sequence, then
7
46. S7 0002 a 3k 0002 ?
for every n 7 1
Sn 0002
a1 0003 a1r n 10003r
n 僆 N, r 1
48. Find the sum of all the odd integers between 100 and 500.
73. Is there an arithmetic sequence that is also geometric? Explain.
49. Show that the sum of the first n odd natural numbers is n2, using appropriate formulas from Section 8-3.
74. Is there an infinite geometric sequence with a1 0002 1 that has sum equal to 12? Explain.
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APPLICATIONS 75. BUSINESS In investigating different job opportunities, you find that firm A will start you at $25,000 per year and guarantee you a raise of $1,200 each year whereas firm B will start you at $28,000 per year but will guarantee you a raise of only $800 each year. Over a period of 15 years, how much would you receive from each firm?
84. PHYSICS The first swing of a bob on a pendulum is 10 inches. If on each subsequent swing it travels 0.9 as far as on the preceding swing, how far will the bob travel before coming to rest?
76. BUSINESS In Problem 75, what would be your annual salary at each firm for the tenth year? 77. ECONOMICS The government, through a subsidy program, distributes $1,000,000. If we assume that each individual or agency spends 0.8 of what is received, and 0.8 of this is spent, and so on, how much total increase in spending results from this government action? 78. ECONOMICS Because of reduced taxes, an individual has an extra $600 in spendable income. If we assume that the individual spends 70% of this on consumer goods, that the producers of these goods in turn spend 70% of what they receive on consumer goods, and that this process continues indefinitely, what is the total amount spent on consumer goods? 79. BUSINESS If $P is invested at r% compounded annually, the amount A present after n years forms a geometric sequence with a common ratio 1 0004 r. Write a formula for the amount present after n years. How long will it take a sum of money P to double if invested at 6% interest compounded annually? 80. POPULATION GROWTH If a population of A0 people grows at the constant rate of r% per year, the population after t years forms a geometric sequence with a common ratio 1 0004 r. Write a formula for the total population after t years. If the world’s population is increasing at the rate of 2% per year, how long will it take to double? 81. FINANCE Eleven years ago an investment earned $7,000 for the year. Last year the investment earned $14,000. If the earnings from the investment have increased the same amount each year, what is the yearly increase and how much income has accrued from the investment over the past 11 years? 82. AIR TEMPERATURE As dry air moves upward, it expands. In so doing, it cools at the rate of about 5°F for each 1,000-foot rise. This is known as the adiabatic process. (A) Temperatures at altitudes that are multiples of 1,000 feet form what kind of a sequence? (B) If the ground temperature is 80°F, write a formula for the temperature Tn in terms of n, if n is in thousands of feet. 83. ENGINEERING A rotating flywheel coming to rest rotates 300 revolutions the first minute (see the figure). If in each subsequent minute it rotates two-thirds as many times as in the preceding minute, how many revolutions will the wheel make before coming to rest?
85. FOOD CHAIN A plant is eaten by an insect, an insect by a trout, a trout by a salmon, a salmon by a bear, and the bear is eaten by you. If only 20% of the energy is transformed from one stage to the next, how many calories must be supplied by plant food to provide you with 2,000 calories from the bear meat? 86. GENEALOGY If there are 30 years in a generation, how many direct ancestors did each of us have 600 years ago? By direct ancestors we mean parents, grandparents, great-grandparents, and so on. 87. PHYSICS An object falling from rest in a vacuum near the surface of the Earth falls 16 feet during the first second, 48 feet during the second second, 80 feet during the third second, and so on. (A) How far will the object fall during the eleventh second? (B) How far will the object fall in 11 seconds? (C) How far will the object fall in t seconds? 88. PHYSICS In Problem 87, how far will the object fall during: (A) The twentieth second? (B) The t th second? 89. BACTERIA GROWTH A single cholera bacterium divides every 1 2 hour to produce two complete cholera bacteria. If we start with a colony of A0 bacteria, how many bacteria will we have in t hours, assuming adequate food supply? 90. CELL DIVISION One leukemic cell injected into a healthy mouse will divide into two cells in about 12 day. At the end of the day these two cells will divide again, with the doubling process continuing each 12 day until there are 1 billion cells, at which time the mouse dies. On which day after the experiment is started does this happen? 91. ASTRONOMY Ever since the time of the Greek astronomer Hipparchus, second century B.C., the brightness of stars has been measured in terms of magnitude. The brightest stars, excluding the sun, are classed as magnitude 1, and the dimmest visible to the eye are classed as magnitude 6. In 1856, the English astronomer N. R. Pogson showed that first-magnitude stars are 100 times brighter than sixth-magnitude stars. If the ratio of brightness between consecutive magnitudes is constant, find this ratio. [Hint: If bn is the brightness of an nth-magnitude star, find r for the geometric sequence b1, b2, b3, . . . , given b1 0002 100b6.] 92. PUZZLE If a sheet of very thin paper 0.001-inch thick is torn in half, and each half is again torn in half, and this process is repeated for a total of 32 times, how high will the stack of paper be if the pieces are placed one on top of the other? Give the answer to the nearest mile.
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531
93. PUZZLE If you place 1¢ on the first square of a chessboard, 2¢ on the second square, 4¢ on the third, and so on, continuing to double the amount until all 64 squares are covered, how much money will be on the sixty-fourth square? How much money will there be on the whole board?
95. ATMOSPHERIC PRESSURE If atmospheric pressure decreases roughly by a factor of 10 for each 10-mile increase in altitude up to 60 miles, and if the pressure is 15 pounds per square inch at sea level, what will the pressure be 40 miles up?
94. MUSIC The notes on a piano, as measured in cycles per second, form a geometric sequence. (A) If A is 400 cycles per second and A¿, 12 notes higher, is 800 cycles per second, find the constant ratio r. (B) Find the cycles per second for C, three notes higher than A.
97. GEOMETRY If the midpoints of the sides of an equilateral triangle are joined by straight lines, the new figure will be an equilateral triangle with a perimeter equal to half the original. If we start with an equilateral triangle with perimeter 1 and form a sequence of “nested” equilateral triangles proceeding as described, what will be the total perimeter of all the triangles that can be formed in this way?
96. ZENO’S PARADOX Visualize a hypothetical 440-yard oval racetrack that has tapes stretched across the track at the halfway point and at each point that marks the halfway point of each remaining distance thereafter. A runner running around the track has to break the first tape before the second, the second before the third, and so on. From this point of view it appears that he will never finish the race. This famous paradox is attributed to the Greek philosopher Zeno (495– 435 B.C.). If we assume the runner runs at 440 yards per minute, the times between tape breakings form an infinite geometric sequence. What is the sum of this sequence?
98. PHOTOGRAPHY The shutter speeds and f-stops on a camera are given as follows: 1 1 1 , 250 , 500 Shutter speeds: 1, 12, 14, 18, 151 , 301 , 601 , 125 f-stops: 1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22
These are very close to being geometric sequences. Estimate their common ratios. 99. GEOMETRY We know that the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, . . . sides form an arithmetic sequence. Find the sum of the interior angles for a 21-sided polygon.
8-4
Multiplication Principle, Permutations, and Combinations Z Counting with the Multiplication Principle Z Using Factorial Notation Z Counting Permutations Z Counting Combinations
Section 8-4 introduces some new mathematical tools that are usually referred to as counting techniques. In general, a counting technique is a mathematical method of determining the number of objects in a set without actually enumerating the objects in the set as 1, 2, 3, . . . . For example, we can count the number of squares in a checkerboard
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(Fig. 1) by counting 1, 2, 3, . . . , 64. This is enumeration. Or we can note that there are eight rows with eight squares in each row. So the total number of squares must be 8 0002 8 0003 64. This is a very simple counting technique. Now consider the problem of assigning telephone numbers. How many different sevendigit telephone numbers can be formed? As we will soon see, the answer is 107 0003 10,000,000, a number that is much too large to obtain by enumeration. This shows that counting techniques are essential tools if the number of elements in a set is very large. The techniques developed in this section will be applied to a brief introduction to probability theory in Section 8-5, and to a famous algebraic formula in Section 8-6.
Z Figure 1
Z Counting with the Multiplication Principle We start with an example.
EXAMPLE
1
Combined Outcomes Suppose we flip a coin and then throw a single die (Fig. 2). What are the possible combined outcomes?
SOLUTION Heads
One way to solve this problem is to use a tree diagram:
Tails Coin Die Outcomes Outcomes
Coin outcomes
Combined Outcomes
H
1 2 3 4 5 6
(H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6)
T
1 2 3 4 5 6
(T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6)
Start Die outcomes
Z Figure 2 Coin and die outcomes.
There are 12 possible combined outcomes—two ways in which the coin can come up followed by six ways in which the die can come up. 0002 MATCHED PROBLEM 1
Use a tree diagram to determine the number of possible outcomes of throwing a single die followed by flipping a coin. 0002
Now suppose you are asked, “From the 26 letters in the alphabet, how many ways can 3 letters appear in a row on a license plate if no letter is repeated?” To try to count the possibilities using a tree diagram would be extremely tedious, to say the least. The following multiplication principle, also called the fundamental counting principle, enables us to solve this problem easily. In addition, it forms the basis for several other counting techniques developed later in this section.
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Z MULTIPLICATION PRINCIPLE 1. If two operations O1 and O2 are performed in order with N1 possible outcomes for the first operation and N2 possible outcomes for the second operation, then there are N1 ⴢ N2 possible combined outcomes of the first operation followed by the second. 2. In general, if n operations O1, O2, . . . , On, are performed in order, with possible number of outcomes N1, N2, . . . , Nn, respectively, then there are N1 ⴢ N2 ⴢ . . . ⴢ Nn possible combined outcomes of the operations performed in the given order.
In Example 1, we see that there are two possible outcomes from the first operation of flipping a coin and six possible outcomes from the second operation of throwing a die. So by the multiplication principle, there are 2 ⴢ 6 0003 12 possible combined outcomes of flipping a coin followed by throwing a die. (Now try using the multiplication principle to solve Matched Problem 1.) To answer the license plate question, we reason as follows: There are 26 ways the first letter can be chosen. After a first letter is chosen, 25 letters remain, so there are 25 ways a second letter can be chosen. And after 2 letters are chosen, there are 24 ways a third letter can be chosen. Using the multiplication principle, there are 26 ⴢ 25 ⴢ 24 0003 15,600 possible ways 3 letters can be chosen from the alphabet without allowing any letter to repeat. By not allowing any letter to repeat, earlier selections affect the choice of subsequent selections. If we allow letters to repeat, then earlier selections do not affect the choice in subsequent selections, and there are 26 possible choices for each of the 3 letters. So, if we allow letters to repeat, there are 26 ⴢ 26 ⴢ 26 0003 263 0003 17,576 possible ways the 3 letters can be chosen from the alphabet.
EXAMPLE
2
Computer-Generated Tests Many universities and colleges are now using computer-assisted testing procedures. Suppose a screening test is to consist of five questions, and a computer stores five equivalent questions for the first test question, eight equivalent questions for the second, six for the third, five for the fourth, and ten for the fifth. How many different five-question tests can the computer select? Two tests are considered different if they differ in one or more questions.
SOLUTION
O1: O2: O3: O4: O5:
Select Select Select Select Select
the the the the the
first question second question third question fourth question fifth question
N1: N2: N3: N4: N5:
five ways eight ways six ways five ways ten ways
The computer can generate 5 ⴢ 8 ⴢ 6 ⴢ 5 ⴢ 10 0003 12,000 different tests MATCHED PROBLEM 2
0002
Each question on a multiple-choice test has five choices. If there are five such questions on a test, how many different response sheets are possible if only one choice is marked for each question? 0002
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3
Counting Code Words How many three-letter code words are possible using the first eight letters of the alphabet if: (A) No letter can be repeated?
(B) Letters can be repeated?
(C) Adjacent letters cannot be alike? SOLUTIONS
(A) No letter can be repeated. O1: Select first letter O2: Select second letter O3: Select third letter
N1: eight ways N2: seven ways N3: six ways
Because one letter has been used Because two letters have been used
There are 8 ⴢ 7 ⴢ 6 0003 336 possible code words (B) Letters can be repeated. O1: Select first letter O2: Select second letter O3: Select third letter
N1: eight ways N2: eight ways N3: eight ways
Repeats are allowed. Repeats are allowed.
There are 8 ⴢ 8 ⴢ 8 0003 83 0003 512 possible code words (C) Adjacent letters cannot be alike. O1: Select first letter O2: Select second letter O3: Select third letter
N1: eight ways N2: seven ways N3: seven ways
There are
Cannot be the same as the first Cannot be the same as the second, but can be the same as the first
8 ⴢ 7 ⴢ 7 0003 392 possible code words MATCHED PROBLEM 3
ZZZ EXPLORE-DISCUSS 1
0002
How many four-letter code words are possible using the first ten letters of the alphabet under the three conditions stated in Example 3? 0002
The postal service of a developing country is choosing a five-character postal code consisting of letters (of the English alphabet) and digits. At least a half a million postal codes must be accommodated. Which format would you recommend to make the codes easy to remember?
The multiplication principle can be used to develop two additional counting techniques that are extremely useful in more complicated counting problems. Both of these methods use factorial notation, which we introduce next.
Z Using Factorial Notation For n a natural number, n factorial—denoted by n!—is the product of the first n natural numbers. Zero factorial is defined to be 1.
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Z DEFINITION 1 n Factorial For n a natural number n! 0003 n(n 0004 1) ⴢ . . . ⴢ 2 ⴢ 1 1! 0003 1 0! 0003 1
It is also useful to note that Z THEOREM 1 Recursion Formula for n Factorial n! 0003 n ⴢ (n 0004 1)!
EXAMPLE
4
Evaluating Factorials Evaluate each expression: (A) 4!
SOLUTIONS
(B) 5!
(C)
(A) 4! 0003 4 ⴢ 3 ⴢ 2 ⴢ 1 0003 24
Find (A) 6!
(B)
6! 5!
(D)
8! 5!
(E)
9! 6!3!
(B) 5! 0003 5 ⴢ 4 ⴢ 3 ⴢ 2 ⴢ 1 0003 120
8! 8 ⴢ 7 ⴢ 6 ⴢ 5! (D) 0003 0003 336 5! 5! MATCHED PROBLEM 4
7! 6!
(C)
3
(C)
7! 7 ⴢ 6! 0003 00037 6! 6!
4
9! 9 ⴢ 8 ⴢ 7 ⴢ 6! (E) 0003 0003 84 6!3! 6! 3 ⴢ 2 ⴢ 1 9! 6!
(D)
0002
10! 7!3! 0002
ZZZ
CAUTION ZZZ
When reducing fractions involving factorials, don’t confuse the single integer n with the symbol n!, which represents the product of n consecutive integers. 6! 0005 2! 3!
ZZZ EXPLORE-DISCUSS 2
6! 6 ⴢ 5 ⴢ 4 ⴢ 3! ⴝ ⴝ 6 ⴢ 5 ⴢ 4 ⴝ 120 3! 3!
A student used a calculator* to solve Matched Problem 4, as shown in Figure 3. Check these answers. If any are incorrect, explain why and find a correct calculator solution.
Z Figure 3 *The factorial symbol ! and related symbols can be found under the MATH-PROB menus on a TI-84 or TI-86.
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It is interesting and useful to note that n! grows very rapidly. Compare the following: 5! 0003 120
10! 0003 3,628,800
15! 0003 1,307,674,368,000
If n! is too large for a calculator to store and display, an error message is displayed. Find the value of n such that your calculator will evaluate n!, but not (n 0006 1)!.
Z Counting Permutations Suppose four pictures are to be arranged from left to right on one wall of an art gallery. How many arrangements are possible? Using the multiplication principle, there are four ways of selecting the first picture. After the first picture is selected, there are three ways of selecting the second picture. After the first two pictures are selected, there are two ways of selecting the third picture. And after the first three pictures are selected, there is only one way to select the fourth. So, the number of arrangements possible for the four pictures is 4 ⴢ 3 ⴢ 2 ⴢ 1 0003 4!
or
24
In general, we refer to a particular arrangement, or ordering, of n objects without repetition as a permutation of the n objects. How many permutations of n objects are there? From the preceding reasoning, there are n ways in which the first object can be chosen, there are n 0004 1 ways in which the second object can be chosen, and so on. Applying the multiplication principle, we have Theorem 2.
Z THEOREM 2 Permutations of n Objects The number of permutations of n objects, denoted by Pn,n, is given by Pn,n 0003 n ⴢ (n 0004 1) ⴢ . . . ⴢ 1 0003 n!
Now suppose the director of the art gallery decides to use only two of the four available pictures on the wall, arranged from left to right. How many arrangements of two pictures can be formed from the four? There are four ways the first picture can be selected. After selecting the first picture, there are three ways the second picture can be selected. So, the number of arrangements of two pictures from four pictures, denoted by P4,2, is given by P4,2 0003 4 ⴢ 3 0003 12 Or, in terms of factorials, multiplying 4 ⴢ 3 by 1 in the form 2!00022!, we have P4,2 0003 4 ⴢ 3 0003
4! 4 ⴢ 3 ⴢ 2! 0003 2! 2!
This last form gives P4,2 in terms of factorials, which is useful in some cases. A permutation of a set of n objects taken r at a time is an arrangement of the r objects in a specific order. So, reasoning in the same way as in the preceding example, we find that the number of permutations of n objects taken r at a time, 0 0007 r 0007 n, denoted by Pn,r, is given by Pn,r 0003 n(n 0004 1)(n 0004 2) ⴢ . . . ⴢ (n 0004 r 0006 1) Multiplying the right side of this equation by 1 in the form (n 0004 r)!0002(n 0004 r)!, we obtain a factorial form for Pn,r: Pn,r 0003 n(n 0004 1)(n 0004 2) ⴢ . . . ⴢ (n 0004 r 0006 1)
(n 0004 r)! (n 0004 r)!
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But n(n 0004 1)(n 0004 2) ⴢ . . . ⴢ (n 0004 r 0006 1)(n 0004 r)! 0003 n! We have developed Theorem 3.
Z THEOREM 3 Permutation of n Objects Taken r at a Time The number of permutations of n objects taken r at a time is given by Pn,r 0003 n(n 0004 1)(n 0004 2) ⴢ . . . ⴢ (n 0004 r 0006 1)
y r factors
or Pn,r 0003
n! (n 0004 r)!
00007r0007n
Note that if r 0003 n, then the number of permutations of n objects taken n at a time is Pn,n 0003
n! n! 0003 0003 n! (n 0004 n)! 0!
Recall, 0! ⴝ 1.
which agrees with Theorem 2, as it should. The permutation symbol Pn,r also can be denoted by P nr, n Pr, or P(n, r). Many calculators use n Pr to denote the function that evaluates the permutation symbol.
EXAMPLE
5
Selecting Officers From a committee of eight people, in how many ways can we choose a chair and a vicechair, assuming one person cannot hold more than one position?
SOLUTION
We are actually asking for the number of permutations of eight objects taken two at a time— that is, P8,2: P8,2 0003
MATCHED PROBLEM 5
EXAMPLE
6
8! 8! 8 ⴢ 7 ⴢ 6! 0003 0003 0003 56 (8 0004 2)! 6! 6!
0002
From a committee of ten people, in how many ways can we choose a chair, vice-chair, and secretary, assuming one person cannot hold more than one position? 0002
Evaluating Pn,r Find the number of permutations of 25 objects taken (A) Two at a time (B) Four at a time (C) Eight at a time
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Figure 4 shows the solution on a calculator.
0002
Z Figure 4
MATCHED PROBLEM 6
Find the number of permutations of 30 objects taken (A) Two at a time
(B) Four at a time
(C) Six at a time 0002
Z Counting Combinations Now suppose that an art museum owns eight paintings by a given artist and another art museum hopes to borrow three of these paintings for a special show. How many ways can three paintings be selected for shipment out of the eight available? Here, the order of the items selected doesn’t matter. What we are actually interested in is how many subsets of three objects can be formed from a set of eight objects. We call such a subset a combination of eight objects taken three at a time. The total number of combinations is denoted by the symbol C8,3
or
8 a b 3
To find the number of combinations of eight objects taken three at a time, C8,3, we make use of the formula for Pn,r and the multiplication principle. We know that the number of permutations of eight objects taken three at a time is given by P8,3, and we have a formula for computing this quantity. Now suppose we think of P8,3 in terms of two operations: O1: Select a subset of three objects (paintings) N1: C8,3 ways O2: Arrange the subset in a given order N2: 3! ways The combined operation, O1 followed by O2, produces a permutation of eight objects taken three at a time. So, P8,3 0003 C8,3 ⴢ 3! To find C8,3, we replace P8,3 in the preceding equation with 8!0002(8 0004 3)! and solve for C8,3: 8! 0003 C8,3 ⴢ 3! (8 0004 3)! 8! 8 ⴢ 7 ⴢ 6 ⴢ 5! 0003 0003 56 C8,3 0003 3!(8 0004 3)! 3 ⴢ 2 ⴢ 1 ⴢ 5! The museum can make 56 different selections of three paintings from the eight available. A combination of a set of n objects taken r at a time is an r-element subset of the n objects. Reasoning in the same way as in the example, the number of combinations of n
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objects taken r at a time, 0 0007 r 0007 n, denoted by Cn,r, can be obtained by solving for Cn,r in the relationship Pn,r 0003 Cn,r ⴢ r! Cn,r 0003 0003
Pn,r r! n! r!(n 0004 r)!
Pn,r ⴝ
n! (n ⴚ r)!
Z THEOREM 4 Combination of n Objects Taken r at a Time The number of combinations of n objects taken r at a time is given by Pn,r n n! 0003 Cn,r 0003 a b 0003 r r! r!(n 0004 r)!
00007r0007n
n The combination symbols Cn,r and a b also can be denoted by C nr, nCr, or C(n, r). r
EXAMPLE
7
Selecting Subcommittees From a committee of eight people, in how many ways can we choose a subcommittee of two people?
SOLUTION
Notice how this example differs from Example 5, where we wanted to know how many ways a chair and a vice-chair can be chosen from a committee of eight people. In Example 5, ordering matters. In choosing a subcommittee of two people, the ordering does not matter. So, we are actually asking for the number of combinations of eight objects taken two at a time. The number is given by 8 8! 8 ⴢ 7 ⴢ 6! 0003 0003 28 C8,2 0003 a b 0003 2 2!(8 0004 2)! 2 ⴢ 1 ⴢ 6!
MATCHED PROBLEM 7
EXAMPLE
8
0002
How many subcommittees of three people can be chosen from a committee of eight people? 0002
Evaluating Cn,r Find the number of combinations of 25 objects taken (A) Two at a time
SOLUTION
(B) Four at a time
(C) Eight at a time
Figure 5 shows the solution on a calculator. Compare these results with Example 6.
Z Figure 5
0002
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MATCHED PROBLEM 8
Find the number of combinations of 30 objects taken (A) Two at a time
(B) Four at a time
(C) Six at a time 0002
Remember: In a permutation, order counts. In a combination, order does not count. To determine whether a permutation or combination is needed, decide whether rearranging the collection or listing makes a difference. If so, use permutations. If not, use combinations.
ZZZ EXPLORE-DISCUSS 3
2
3
3
(C) The newly elected president names his cabinet members.
9 10 J Q K A 6 78
(D) The president selects a delegation of three cabinet members to attend the funeral of a head of state.
9 10 J Q K A 67 8 45
(E) An orchestra conductor chooses three pieces of music for a symphony program.
4 5
9 10 J Q K A 6 78
A
2
3
4 5
(B) A baseball manager names his starting lineup.
A
2
3
(A) A student checks out three books from the library.
9 10 J Q K A 67 8 45
A
2
Each of the following is a selection without repetition. Would you consider the selection to be a combination? A permutation? Discuss your reasoning.
A
Z Figure 6 A standard deck of cards.
EXAMPLE
9
A standard deck of 52 cards (Fig. 6) has four 13-card suits: diamonds, hearts, clubs, and spades. Each 13-card suit contains cards numbered from 2 to 10, a jack, a queen, a king, and an ace. The jack, queen, and king are called face cards. Depending on the game, the ace may be counted as the lowest and/or the highest card in the suit. Example 9, as well as other examples and exercises in Chapter 8, refer to this standard deck.
Counting Card Hands Out of a standard 52-card deck, how many 5-card hands will have three aces and two kings?
SOLUTION
O1: N1: O2: N2:
Choose three aces out of four possible C4,3 Choose two kings out of four possible C4,2
Order is not important.
Order is not important.
Using the multiplication principle, we have Number of hands 0003 C4,3 ⴢ C4,2 0003 4 ⴢ 6 0003 24 MATCHED PROBLEM 9
EXAMPLE
10
0002
From a standard 52-card deck, how many 5-card hands will have three hearts and two spades? 0002
Counting Serial Numbers Serial numbers for a product are to be made using two letters followed by three numbers. If the letters are to be taken from the first eight letters of the alphabet with no repeats and the numbers from the 10 digits 0 through 9 with no repeats, how many serial numbers are possible?
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O1: N1: O2: N2:
SOLUTION
Multiplication Principle, Permutations, and Combinations
Choose two letters out of eight available P8,2 Choose three numbers out of ten available P10,3
541
Order is important.
Order is important.
Using the multiplication principle, we have Number of serial numbers 0003 P8,2 ⴢ P10,3 0003 40,320 MATCHED PROBLEM 10
0002
Repeat Example 10 under the same conditions, except the serial numbers are now to have three letters followed by two digits with no repeats. 0002 ANSWERS TO MATCHED PROBLEMS 1. H T H T H T H T H T H T There are 12 outcomes. 1
2
3
4
5
2. 55, or 3,125
6
Start
3. (A) 10 ⴢ 9 ⴢ 8 ⴢ 7 0003 5,040 (B) 10 ⴢ 10 ⴢ 10 ⴢ 10 0003 10,000 (C) 10 ⴢ 9 ⴢ 9 ⴢ 9 0003 7,290 4. (A) 720 (B) 6 (C) 504 (D) 120 10! 5. P10,3 0003 6. (A) 870 (B) 657,720 (C) 427,518,000 0003 720 (10 0004 3)! 8! 7. C8,3 0003 8. (A) 435 (B) 27,405 (C) 593,775 0003 56 3!(8 0004 3)! 9. C13,3 ⴢ C13,2 0003 22,308 10. P8,3 ⴢ P10,2 0003 30,240
8-4
Exercises
1. What is a permutation?
17. The figure shows calculator solutions to Problems 11, 13, and 15. Check these answers. If any are incorrect, explain why and find a correct calculator solution.
2. What is a combination? 3. Explain how n! can be defined by means of a recursion formula. 4. State the multiplication principle for counting in your own words. 5. Explain how permutations and combinations differ with respect to order. 6. Explain how permutations and combinations are alike with respect to repetition. Evaluate the expression in Problems 7–16: 7. 9! 10. 12! 13.
5! 2!3!
16.
8! 3!(8 0004 3)!
8. 10!
9. 11!
11.
11! 8!
12.
14! 12!
14.
6! 4!2!
15.
7! 4!(7 0004 4)!
18. The figure shows calculator solutions to Problems 12, 14, and 16. Check these answers. If any are incorrect, explain why and find a correct calculator solution.
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In Problems 19–26, evaluate. 19. P13,4
20. C20,10
21. P13,9
22. C20,4
23. C15,8
24. P11,3
25. C15,12
26. P11,8
numbers are possible, assuming no digit is repeated? Assuming digits can be repeated? 40. A small combination lock on a suitcase has three wheels, each labeled with digits from 0 to 9. How many opening combinations of three numbers are possible, assuming no digit is repeated? Assuming digits can be repeated?
In Problems 27 and 28, would you consider the selection to be a combination or a permutation? Explain your reasoning.
41. From a standard 52-card deck, how many 5-card hands will have all hearts?
27. (A) The recently elected chief executive officer (CEO) of a company named three new vice-presidents, of marketing, research, and manufacturing. (B) The CEO selected three of her vice-presidents to attend the dedication ceremony of a new plant.
42. From a standard 52-card deck, how many 5-card hands will have all face cards? All face cards, but no kings? Consider only jacks, queens, and kings to be face cards.
28. (A) An individual rented four DVDs from a rental store to watch over a weekend. (B) The same individual did some holiday shopping by buying four DVDs, one for his father, one for his mother, one for his younger sister, and one for his older brother. 29. A particular new car model is available with five choices of color, three choices of transmission, four types of interior, and two types of engine. How many different variations of this model car are possible? 30. A deli serves sandwiches with the following options: three kinds of bread, five kinds of meat, and lettuce or sprouts. How many different sandwiches are possible, assuming one item is used out of each category? 31. In a horse race, how many different finishes among the first three places are possible for a 10-horse race? Exclude ties. 32. In a long-distance foot race, how many different finishes among the first five places are possible for a 50-person race? Exclude ties. 33. How many ways can a subcommittee of three people be selected from a committee of seven people? How many ways can a president, vice-president, and secretary be chosen from a committee of seven people? 34. Suppose nine cards are numbered with the nine digits from 1 to 9. A three-card hand is dealt, one card at a time. How many hands are possible where: (A) Order is taken into consideration? (B) Order is not taken into consideration? 35. There are 10 teams in a league. If each team is to play every other team exactly once, how many games must be scheduled? 36. Given seven points, no three of which are on a straight line, how many lines can be drawn joining two points at a time? 37. How many four-letter code words are possible from the first six letters of the alphabet, with no letter repeated? Allowing letters to repeat? 38. How many five-letter code words are possible from the first seven letters of the alphabet, with no letter repeated? Allowing letters to repeat? 39. A combination lock has five wheels, each labeled with the 10 digits from 0 to 9. How many opening combinations of five
43. How many different license plates are possible if each contains three letters followed by three digits? How many of these license plates contain no repeated letters and no repeated digits? 44. How may five-digit zip codes are possible? How many of these codes contain no repeated digits? 45. From a standard 52-card deck, how many 7-card hands have exactly five spades and two hearts? 46. From a standard 52-card deck, how many 5-card hands will have two clubs and three hearts? 47. A catering service offers eight appetizers, ten main courses, and seven desserts. A banquet chairperson is to select three appetizers, four main courses, and two desserts for a banquet. How many ways can this be done? 48. Three research departments have 12, 15, and 18 members, respectively. If each department is to select a delegate and an alternate to represent the department at a conference, how many ways can this be done? 49. (A) Use a graphing calculator to display the sequences P10,0, P10,1, . . . , P10,10 and 0!, 1!, . . . , 10! in table form, and show that P10,r r! for r 0003 0, 1, . . . , 10. (B) Find all values of r such that P10,r 0003 r! (C) Explain why Pn,r r! whenever 0 0007 r 0007 n. P10,0 P10,1 P10,10 , ,..., 50. (A) How are the sequences and C10,0, 0! 1! 10! C10,1, . . . , C10,10 related? (B) Use a graphing calculator to graph each sequence and confirm the relationship of part A. 51. A sporting goods store has 12 pairs of ski gloves of 12 different brands thrown loosely in a bin. The gloves are all the same size. In how many ways can a left-hand glove and a right-hand glove be selected that do not match relative to brand? 52. A sporting goods store has six pairs of running shoes of six different styles thrown loosely in a basket. The shoes are all the same size. In how many ways can a left shoe and a right shoe be selected that do not match? 53. Eight distinct points are selected on the circumference of a circle. (A) How many chords can be drawn by joining the points in all possible ways? (B) How many triangles can be drawn using these eight points as vertices? (C) How many quadrilaterals can be drawn using these eight points as vertices?
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54. Five distinct points are selected on the circumference of a circle. (A) How many chords can be drawn by joining the points in all possible ways? (B) How many triangles can be drawn using these five points as vertices? 55. How many ways can two people be seated in a row of five chairs? Three people? Four people? Five people? 56. Each of two countries sends five delegates to a negotiating conference. A rectangular table is used with five chairs on each long side. If each country is assigned a long side of the table, how many seating arrangements are possible? [Hint: Operation 1 is assigning a long side of the table to each country.] 57. A basketball team has five distinct positions. Out of eight players, how many starting teams are possible if (A) The distinct positions are taken into consideration? (B) The distinct positions are not taken into consideration? (C) The distinct positions are not taken into consideration, but either Mike or Ken, but not both, must start?
8-5
58. How many committees of four people are possible from a group of nine people if (A) There are no restrictions? (B) Both Juan and Mary must be on the committee? (C) Either Juan or Mary, but not both, must be on the committee? 59. A 5-card hand is dealt from a standard 52-card deck. Which is more likely: the hand contains exactly one king or the hand contains no hearts? 60. A 10-card hand is dealt from a standard 52-card deck. Which is more likely: all cards in the hand are red or the hand contains all four aces?
Sample Spaces and Probability Z Sample Spaces and Events Z Finding the Probability of an Event Z Making Equally Likely Assumptions Z Finding or Approximating Empirical Probability
This section provides an introduction to probability. It’s going to need to be a relatively brief one, because probability is a topic to which entire books and courses are devoted. Probability involves many subtle notions, and care must be taken at the beginning to understand the fundamental concepts on which the subject is based. Our development of probability, because of space limitations, must be somewhat informal. More formal and precise treatments can be found in books on probability.
Z Sample Spaces and Events Our first step in constructing a mathematical model for probability studies is to describe the type of experiments on which probability studies are based. Some types of experiments do not yield the same results, no matter how carefully they are repeated under the same conditions. These experiments are called random experiments. Some standard examples of random experiments are flipping coins, rolling dice, observing the frequency of defective items from an assembly line, or observing the frequency of deaths in a certain age group. Probability theory is a branch of mathematics that has been developed to deal with outcomes of random experiments. In the work that follows, the word experiment will be used to mean a random experiment.
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The outcomes of experiments are typically described in terms of sample spaces and events. Our second step in constructing a mathematical model for probability studies is to define these two terms. Consider the experiment, “A single six-sided die is rolled.” What outcomes might we observe? We might be interested in the number of dots facing up, or whether the number of dots facing up is an even number, or whether the number of dots facing up is divisible by 3, and so on. The list of possible outcomes appears endless. In general, there is no unique method of analyzing all possible outcomes of an experiment. Therefore, before conducting an experiment, it is important to decide just what outcomes are of interest. In the die experiment, suppose we limit our interest to the number of dots facing up when the die comes to rest. Having decided what to observe, we make a list of outcomes of the experiment, called simple events, such that in each trial of the experiment, one and only one of the results on the list will occur. The set of simple events for the experiment is called a sample space for the experiment. The sample space S we have chosen for the die-rolling experiment is S 0003 {1, 2, 3, 4, 5, 6} Now consider the outcome, “The number of dots facing up is an even number.” This outcome is not a simple event, because it will occur whenever 2, 4, or 6 dots appear, that is, whenever an element in the subset E 0003 {2, 4, 6} occurs. Subset E is called a compound event. In general, we have the following definition:
Z DEFINITION 1 Event Given a sample space S for an experiment, we define an event E to be any subset of S. If an event E has only one element in it, it is called a simple event. If event E has more than one element, it is called a compound event. We say that an event E occurs if any of the simple events in E occurs.
EXAMPLE
1
Choosing a Sample Space A nickel and a dime are tossed. How will we identify a sample space for this experiment?
SOLUTIONS
There are a number of possibilities, depending on our interest. We will consider three. (A) If we are interested in whether each coin falls heads (H) or tails (T), then, using a tree diagram, we can easily determine an appropriate sample space for the experiment: Nickel Outcomes H Start T
Dime Outcomes H T H T
The sample space is S1 0003 {HH, HT, TH, TT} and there are four simple events in the sample space.
Combined Outcomes HH HT TH TT
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(B) If we are interested only in the number of heads that appear on a single toss of the two coins, then we can let S2 0003 {0, 1, 2} and there are three simple events in the sample space. (C) If we are interested in whether the coins match (M) or don’t match (D), then we can let S3 0003 {M, D} and there are only two simple events in the sample space.
MATCHED PROBLEM 1
0002
An experiment consists of recording the boy–girl composition of families with two children. (A) What is an appropriate sample space if we are interested in the gender of each child in the order of their births? Draw a tree diagram. (B) What is an appropriate sample space if we are interested only in the number of girls in a family? (C) What is an appropriate sample space if we are interested only in whether the genders are alike (A) or different (D)? (D) What is an appropriate sample space for all three interests expressed above? 0002
In Example 1, sample space S1 contains more information than either S2 or S3. If we know which outcome has occurred in S1, then we know which outcome has occurred in S2 and S3. However, the reverse is not true. In this sense, we say that S1 is a more fundamental sample space than either S2 or S3. Important Remark: There is no one correct sample space for a given experiment. When specifying a sample space for an experiment, we include as much detail as necessary to answer all questions of interest regarding the outcomes of the experiment. If in doubt, include more elements in the sample space rather than fewer. Now let’s return to the two-coin problem in Example 1 and the sample space S1 0003 {HH, HT, TH, TT} Suppose we are interested in the outcome, “Exactly 1 head is up.” Looking at S1, we find that it occurs if either of the two simple events HT or TH occurs.* So, to say that the event, “Exactly 1 head is up” occurs is the same as saying the experiment has an outcome in the set E 0003 {HT, TH} This is a subset of the sample space S1. The event E is a compound event.
*Technically, we should write {HT} and {TH}, because there is a logical distinction between an element of a set and a subset consisting of only that element. But we will just keep this in mind and drop the braces for simple events to simplify the notation.
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2
Rolling Two Dice Consider an experiment of rolling two dice. A convenient sample space that will enable us to answer many questions about events of interest is shown in Figure 1. Let S be the set of all ordered pairs listed in the figure. Note that the simple event (3, 2) is to be distinguished from the simple event (2, 3). The former indicates a 3 turned up on the first die and a 2 on the second, whereas the latter indicates a 2 turned up on the first die and a 3 on the second. What is the event that corresponds to each of the following outcomes? (A) A sum of 7 turns up.
(B) A sum of 11 turns up.
(C) A sum less than 4 turns up.
(D) A sum of 12 turns up.
FIRST DIE
SECOND DIE
(1, 1)
(1, 2)
(1, 3)
(1, 4)
(1, 5)
(1, 6)
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(2, 5)
(2, 6)
(3, 1)
(3, 2)
(3, 3)
(3, 4)
(3, 5)
(3, 6)
(4, 1)
(4, 2)
(4, 3)
(4, 4)
(4, 5)
(4, 6)
(5, 1)
(5, 2)
(5, 3)
(5, 4)
(5, 5)
(5, 6)
(6, 1)
(6, 2)
(6, 3)
(6, 4)
(6, 5)
(6, 6)
Z Figure 1 A sample space for rolling two dice.
SOLUTIONS
(A) By “A sum of 7 turns up,” we mean that the sum of all dots on both turned-up faces is 7. This outcome corresponds to the event {(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)} (B) “A sum of 11 turns up” corresponds to the event {(6, 5), (5, 6)} (C) “A sum less than 4 turns up” corresponds to the event {(1, 1), (2, 1), (1, 2)} (D) “A sum of 12 turns up” corresponds to the event {(6, 6)}
MATCHED PROBLEM 2
0002
Refer to the sample space in Example 2 (Fig. 1). What is the event that corresponds to each of the following outcomes? (A) A sum of 5 turns up. (B) A sum that is a prime number greater than 7 turns up. 0002 Informally, to facilitate discussion, we often use the terms event and outcome of an experiment interchangeably. So, in Example 2 we might say “the event ‘A sum of 11 turns up’ ” in place of “the outcome ‘A sum of 11 turns up,’ ” or even write E 0003 A sum of 11 turns up 0003 {(6, 5), (5, 6)}
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Technically speaking, an event is the mathematical counterpart of an outcome of an experiment.
Z Finding the Probability of an Event The next step in developing our mathematical model for probability studies is the introduction of a probability function. This is a function that assigns to an arbitrary event associated with a sample space a real number between 0 and 1, inclusive. We start by discussing ways in which probabilities are assigned to simple events in S.
Z DEFINITION 2 Probabilities for Simple Events Given a sample space S 0003 {e1, e2, . . . , en} with n simple events, to each simple event ei we assign a real number, denoted by P(ei), that is called the probability of the event ei. These numbers may be assigned in an arbitrary manner as long as the following two conditions are satisfied: 1. 0 0007 P(ei) 0007 1 2. P(e1) 0006 P(e2) 0006 . . . 0006 P(en) 0003 1
The sum of the probabilities of all simple events in the sample space is 1.
Any probability assignment that meets conditions 1 and 2 is said to be an acceptable probability assignment.
Our mathematical theory does not explain how acceptable probabilities are assigned to simple events. These assignments are generally based on the expected or actual percentage of times a simple event occurs when an experiment is repeated a large number of times. Assignments based on this principle are called reasonable. Let an experiment be the flipping of a single coin, and let us choose a sample space S to be S 0003 {H, T} If a coin appears to be fair, we are inclined to assign probabilities to the simple events in S as follows: P(H) 0003
1 2
and
P(T) 0003 12
These assignments are based on reasoning that, because there are two ways a coin can land, in the long run a head will turn up half the time and a tail will turn up half the time. These probability assignments are acceptable, because both of the conditions for acceptable probability assignments in Definition 2 are satisfied: 1. 0 0007 P(H) 0007 1, 0 0007 P(T) 0007 1 2. P(H) 0006 P(T) 0003 12 0006 12 0003 1 But there are other acceptable assignments. Maybe after flipping a coin 1,000 times we find that the head turns up 376 times and the tail turns up 624 times. With this result, we might suspect that the coin is not fair and assign the simple events in the sample space S the probabilities P(H) 0003 .376
and
P(T) 0003 .624
This is also an acceptable assignment. But the probability assignment P(H) 0003 1
and
P(T) 0003 0
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though acceptable, is not reasonable, unless the coin has two heads. The assignment P(H) 0003 .6
P(T) 0003 .8
and
is not acceptable, because .6 0006 .8 0003 1.4, which violates condition 2 in Definition 2. In probability studies, the 0 to the left of the decimal is usually omitted; we write .8 and not 0.8. It is important to keep in mind that out of the infinitely many possible acceptable probability assignments to simple events in a sample space, we are generally inclined to choose one assignment over another based on reasoning or experimental results. Given an acceptable probability assignment for simple events in a sample space S, how do we define the probability of an arbitrary event E associated with S?
Z DEFINITION 3 Probability of an Event E Given an acceptable probability assignment for the simple events in a sample space S, we define the probability of an arbitrary event E, denoted by P(E ), as follows: 1. If E is the empty set, then P(E) 0003 0. 2. If E is a simple event, then P(E) has already been assigned. 3. If E is a compound event, then P(E) is the sum of the probabilities of all the simple events in E. 4. If E is the sample space S, then P(E ) 0003 P(S) 0003 1. This is a special case of 3.
EXAMPLE
3
Finding Probabilities of Events Let’s return to Example 1, the tossing of a nickel and dime, and the sample space S 0003 {HH, HT, TH, TT} Because there are four simple outcomes and the coins are assumed to be fair, it appears that each outcome should occur in the long run 25% of the time. Let’s assign the same probability of 14 to each simple event in S: Simple event, ei P(ei)
HH
HT
TH
TT
1 4
1 4
1 4
1 4
This is an acceptable assignment according to Definition 2 and a reasonable assignment for ideal coins that are perfectly balanced or coins close to ideal. (A) What is the probability of getting exactly one head? (B) What is the probability of getting at least one head? (C) What is the probability of getting a head or a tail? (D) What is the probability of getting three heads? SOLUTIONS
(A) E1 0003 Getting one head 0003 {HT, TH} Because E1 is a compound event, we use item 3 in Definition 3 and find P(E1) by adding the probabilities of the simple events in E1. P(E1) 0003 P(HT) 0006 P(TH) 0003 14 0006 14 0003 12
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(B) E2 0003 Getting at least 1 head 0003 {HH, HT, TH} P(E2) 0003 P(HH) 0006 P(HT) 0006 P(TH) 0003 14 0006 14 0006 14 0003 34 (C) E3 0003 {HH, HT, TH, TT} 0003 S P(E3) 0003 P(S) 0003 1
1 4
(D) E3 0003 Getting three heads 0003 P( ) 0003 0
ⴙ 14 ⴙ 14 ⴙ 14 ⴝ 1
Empty set
Z STEPS FOR FINDING PROBABILITIES OF EVENTS Step 1. Set up an appropriate sample space S for the experiment. Step 2. Assign acceptable probabilities to the simple events in S. Step 3. To obtain the probability of an arbitrary event E, add the probabilities of the simple events in E.
The function P defined in steps 2 and 3 is called a probability function. The domain of this function is all possible events in the sample space S, and the range is a set of real numbers between 0 and 1, inclusive. 0002
MATCHED PROBLEM 3
Return to Matched Problem 1, recording the boy–girl composition of families with two children and the sample space S 0003 {BB, BG, GB, GG} Statistics from the U.S. Census Bureau indicate that an acceptable and reasonable probability for this sample space is Simple event, ei
BB
BG
GB
GG
P(ei)
.26
.25
.25
.24
Find the probabilities of the following events: (A) E1 0003 Having at least one girl in the family (B) E2 0003 Having at most one girl in the family (C) E3 0003 Having two children of the same sex in the family
0002
Z Making Equally Likely Assumptions In tossing a nickel and dime (Example 3), we assigned the same probability, 14, to each simple event in the sample space S 0003 {HH, HT, TH, TT}. By assigning the same probability to each simple event in S, we are actually making the assumption that each simple event is as likely to occur as any other. We refer to this as an equally likely assumption. In general, we have Definition 4.
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Z DEFINITION 4 Probability of a Simple Event Under an Equally Likely Assumption If, in a sample space S 0003 {e1, e2, . . . , en} with n elements, we assume each simple event ei is as likely to occur as any other, then we assign the probability 10002n to each. That is, P(ei) 0003
1 n
Under an equally likely assumption, we can develop a very useful formula for finding probabilities of arbitrary events associated with a sample space S. Consider the following example. If a single die is rolled and we assume each face is as likely to come up as any other, then for the sample space S 0003 {1, 2, 3, 4, 5, 6} we assign a probability of the probability of
1 6
to each simple event, because there are six simple events. Then
E 0003 Rolling a prime number 0003 {2, 3, 5} is P(E) 0003 P(2) 0006 P(3) 0006 P(5) 0003 16 0006 16 0006 16 0003 36 0003 12 So, under the assumption that each simple event is as likely to occur as any other, the computation of the probability of the occurrence of any event E in a sample space S is the number of elements in E divided by the number of elements in S.
Z THEOREM 1 Probability of an Arbitrary Event Under an Equally Likely Assumption If we assume each simple event in sample space S is as likely to occur as any other, then the probability of an arbitrary event E in S is given by P(E) 0003
EXAMPLE
4
n(E ) Number of elements in E 0003 Number of elements in S n(S )
Finding Probabilities of Events If in rolling two dice we assume each simple event in the sample space shown in Figure 1 on page 546 is as likely as any other, find the probabilities of the following events: (A) E1 0003 A sum of 7 turns up
(B) E2 0003 A sum of 11 turns up
(C) E3 0003 A sum less than 4 turns up
(D) E4 0003 A sum of 12 turns up
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SOLUTIONS
MATCHED PROBLEM 4
Sample Spaces and Probability
551
Referring to Figure 1, we see that: (A) P(E1) 0003
n(E1) 6 1 0003 0003 n(S) 36 6
(B) P(E2) 0003
n(E2) 2 1 0003 0003 n(S) 36 18
(C) P(E3) 0003
n(E3) 3 1 0003 0003 n(S) 36 12
(D) P(E4) 0003
n(E4) 1 0003 n(S) 36
0002
Under the conditions in Example 4, find the probabilities of the following events: (A) E5 0003 A sum of 5 turns up (B) E6 0003 A sum that is a prime number greater than 7 turns up
ZZZ EXPLORE-DISCUSS 1
0002
A box contains four red balls and seven green balls. A ball is drawn at random and then, without replacing the first ball, a second ball is drawn. Discuss whether or not the equally likely assumption would be appropriate for the sample space S 0003 {RR, RG, GR, GG}.
We now turn to some examples that make use of the counting techniques developed in Section 8-4.
EXAMPLE
5
Drawing Cards In drawing 5 cards from a 52-card deck without replacement, what is the probability of getting five spades?
SOLUTION
Let the sample space S be the set of all 5-card hands from a 52-card deck. Because the order in a hand does not matter, n(S ) 0003 C52,5. The event we seek is E 0003 Set of all 5-card hands from 13 spades Again, the order does not matter and n(E ) 0003 C13,5. Assuming that each 5-card hand is as likely as any other, P(E ) 0003
MATCHED PROBLEM 5
EXAMPLE
6
C13,5 n(E ) 13! 5!8! 13! 5!47! 0003 0003 0003 ⴢ 0003 .0005 n(S ) C52,5 52! 5!47! 5!8! 52!
0002
In drawing 7 cards from a 52-card deck without replacement, what is the probability of getting seven hearts? 0002
Selecting Committees The board of regents of a university is made up of 12 men and 16 women. If a committee of six is chosen at random, what is the probability that it will contain three men and three women?
SOLUTION
Let S 0003 Set of all 6-person committees out of 28 people: n(S ) 0003 C28,6
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Let E 0003 Set of all 6-person committees with 3 men and 3 women. To find n(E ), we use the multiplication principle and the following two operations: O1: Select 3 men out of the 12 available O2: Select 3 women out of the 16 available
N1: C12,3 N2: C16,3
So n(E) 0003 C12,3 ⴢ C16,3 and P(E) 0003
MATCHED PROBLEM 6
C12,3 ⴢ C16,3 n(E) 0003 0003 .327 n(S) C28,6
0002
What is the probability that the committee in Example 6 will have four men and two women? 0002
Z Finding or Approximating Empirical Probability In the earlier examples in this section we made a reasonable assumption about an experiment and used deductive reasoning to assign probabilities. For example, it is reasonable to assume that an ordinary coin will come up heads about as often as it will come up tails. Probabilities determined in this manner are called theoretical probabilities. No experiments are ever conducted. But what if the theoretical probabilities are not obvious? Then we assign probabilities to simple events based on the results of actual experiments. Probabilities determined from the results of actually performing an experiment are called empirical probabilities. As an experiment is repeated over and over, the percentage of times an event occurs may get closer and closer to a single fixed number. If so, this single fixed number is generally called the actual probability of the event. ZZZ EXPLORE-DISCUSS 2
Like a coin, a thumbtack tossed into the air will land in one of two positions, point up or point down [Fig. 2(a)]. Unlike a coin, we would not expect both events to occur with the same frequency. Indeed, the frequencies of landing point up and point down may well vary from one thumbtack to another [Fig. 2(b)]. Find two thumbtacks of different sizes and guess which one is likely to land point up more frequently. Then toss each tack 100 times and record the number of times each lands point up. Did the experiment confirm your initial guess?
(a) Point up or point down
(b) Two different tacks
Z Figure 2
Suppose when tossing one of the thumbtacks in Explore-Discuss 2, we observe that the tack lands point up 43 times and point down 57 times. Based on this experiment, it seems reasonable to say that for this particular thumbtack P(Point up) 0003
43 0003 .43 100
P(Point down) 0003
57 0003 .57 100
Probability assignments based on the results of repeated trials of an experiment are called approximate empirical probabilities.
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In general, if we conduct an experiment n times and an event E occurs with frequency f (E ), then the ratio f (E )0002n is called the relative frequency of the occurrence of event E in n trials. We define the empirical probability of E, denoted by P(E ), by the number, if it exists, that the relative frequency f (E )0002n approaches as n gets larger and larger. Of course, for any particular n, the relative frequency f (E )0002n is generally only approximately equal to P(E ). However, as n increases, we expect the approximation to improve. Z DEFINITION 5 Empirical Probability If f(E) is the frequency of event E in n trials, then P(E ) 0003
Frequency of occurrence of E f (E ) 0003 n Total number of trials
If we can also deduce theoretical probabilities for an experiment, then we expect the approximate empirical probabilities to approach the theoretical probabilities. If this does not happen, then we should begin to suspect the manner in which the theoretical probabilities were computed. If P(E ) is the theoretical probability of an event E and the experiment is performed n times, then the expected frequency of the occurrence of E is n ⴢ P(E ).
EXAMPLE
7
Finding Approximate Empirical and Theoretical Probabilities Two coins are tossed 500 times with the following frequencies of outcomes: Two heads: 121 One head: 262 Zero heads: 117 (A) Compute the approximate empirical probability for each outcome. (B) Compute the theoretical probability for each outcome. (C) Compute the expected frequency for each outcome.
SOLUTIONS
(A) P(two heads) 0003
121 0003 .242 500
262 0003 .524 500 117 0003 .234 P(zero heads) 0003 500 P(one head) 0003
(B) A sample space of equally likely simple events is S 0003 {HH, HT, TH, TT}. Let E1 0003 two heads 0003 5HH6 E2 0003 one head 0003 5HT, TH6 E3 0003 zero heads 0003 5TT6
Then n(E1) 1 0003 0003 .25 n(S) 4 n(E2) 2 P(E2) 0003 0003 0003 .50 n(S) 4 n(E3) 1 0003 0003 .25 P(E3) 0003 n(S) 4 P(E1) 0003
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(C) The expected frequencies are E1: 500(.25) 0003 125 E2: 500(.5) 0003 250 E3: 500(.25) 0003 125 The actual frequencies obtained from performing the experiment are reasonably close to the expected frequencies. Increasing the number of trials of the experiment would most likely produce even better approximations. 0002 MATCHED PROBLEM 7
One die is rolled 500 times with the following frequencies of outcomes: Outcome
1
2
3
4
5
6
Frequency
89
83
77
91
72
88
(A) Compute the approximate empirical probability for each outcome. (B) Compute the theoretical probability for each outcome. (C) Compute the expected frequency for each outcome. 0002
Technology Connections The data in Example 7 were not generated by tossing two coins 500 times. Instead, the experiment was simulated by a random number generator on a graphing calculator. The command randint (0, 1, 500) produces a random sequence of 500 terms; each term is 0 or 1 with equal liklihood. Thinking of 1 as heads and 0 as tails, such a sequence represents 500 tosses of a single coin. Adding two such sequences together produces a sequence of 500 terms in which each term represents the number of heads in a toss of two coins
[see Fig. 3(a)]. We determine the frequency of each outcome (0, 1, or 2 heads) in 500 tosses of two coins as follows: first, we construct a histogram [Figs. 3(b) and 3(c)], then we use the TRACE command to read off the frequencies [Figs. 3(d), 3(e), and 3(f)]. Compare with the data of Example 7. If you perform the same simulation on your graphing calculator, you are not likely to get exactly the same results. But the approximate empirical probabilities you obtain will be close to the theoretical probabilities.
(a) Generating the random numbers
(b) Setting up the histogram
(c) Selecting the window variables
(d) 0 heads: 117
(e) 1 head: 262
(f) 2 heads: 121
Z Figure 3 Simulating 500 tosses of two coins.
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Empirical Probabilities for an Insurance Company An insurance company selected 1,000 drivers at random in a particular city to determine a relationship between age and accidents. The data obtained are listed in Table 1. Compute the approximate empirical probabilities of the following events for a driver chosen at random in the city: (A) E1: being under 20 years old and having exactly three accidents in 1 year (B) E2: being 30–39 years old and having one or more accidents in 1 year (C) E3: having no accidents in 1 year (D) E4: being under 20 years old or* having exactly three accidents in 1 year Table 1 Accidents in 1 Year
SOLUTIONS
Age
0
1
2
3
Over 3
Under 20
50
62
53
35
20
20–29
64
93
67
40
36
30–39
82
68
32
14
4
40–49
38
32
20
7
3
Over 49
43
50
35
28
24
(A) P(E1) 0003
35 0003 .035 1,000
(B) P(E2) 0003
68 0006 32 0006 14 0006 4 0003 .118 1,000
(C) P(E3) 0003
50 0006 64 0006 82 0006 38 0006 43 0003 .277 1,000
(D) P(E4) 0003
50 0006 62 0006 53 0006 35 0006 20 0006 40 0006 14 0006 7 0006 28 0003 .309 1,000
Notice that in this type of problem, which is typical of many realistic problems, approximate empirical probabilities are the only type we can compute. 0002 MATCHED PROBLEM 8
Referring to Table 1 in Example 8, compute the approximate empirical probabilities of the following events for a driver chosen at random in the city: (A) E1: being under 20 years old with no accidents in 1 year (B) E2: being 20–29 years old and having fewer than two accidents in 1 year (C) E3: not being over 49 years old 0002 Approximate empirical probabilities are often used to test theoretical probabilities. Equally likely assumptions may not be justified in reality. In addition to this use, there are many situations in which it is either very difficult or impossible to compute the theoretical *Interpret “or” in its inclusive sense, as customary in mathematics (a driver who is both under 20 and has three accidents must be counted once in the frequency of E4).
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probabilities for given events. For example, insurance companies use past experience to establish approximate empirical probabilities to predict future accident rates; baseball teams use batting averages, which are approximate empirical probabilities based on past experience, to predict the future performance of a player; and pollsters use approximate empirical probabilities to predict outcomes of elections. ANSWERS TO MATCHED PROBLEMS 1. (A) S1 0003 {BB, BG, GB, GG};
Sex of First Child B G
Sex of Second Child B G B G
Combined Outcomes BB BG GB GG
(C) S3 0003 {A, D} (D) The sample space in part A. (B) S2 0003 {0, 1, 2} 2. (A) {(4, 1), (3, 2), (2, 3), (1, 4)} (B) {(6, 5), (5, 6)} 3. (A) .74 (B) .76 (C) .5 4. (A) P(E5) 0003 19 (B) P(E6) 0003 181 5. C13,70002C52,7 0003 .000013 6. (C12,4 ⴢ C16,2)0002C28,6 0003 .158 7. (A) P(E1) 0003 .178, P(E2) 0003 .166, P(E3) 0003 .154, P(E4) 0003 .182, P(E5) 0003 .144, P(E6) 0003 .176 (B) 16 0003 .167 for each (C) 83.3 for each 8. (A) P(E1) 0003 .05 (B) P(E2) 0003 .157 (C) P(E3) 0003 .82
8-5
Exercises
1. What is a sample space? 2. Explain in your own words how the probability of an event is defined in terms of probabilities of simple events. 3. Explain the difference between a theoretical probability and an empirical probability.
14. A single card is drawn from a standard 52-card deck. What is the probability of getting a numbered card (that is, a two through ten)? 15. A fair coin is tossed three times. What is the probability of getting exactly two tails?
4. What is an equally likely assumption?
16. A fair coin is tossed three times. What is the probability of getting three tails?
5. A single fair die is rolled. What is the probability of getting a one or a six?
17. How would you interpret P(E) 0003 1?
6. A single fair die is rolled. What is the probability of getting a number greater than three? 7. A single card is drawn from a standard 52-card deck. What is the probability of getting a red card? 8. A single card is drawn from a standard 52-card deck. What is the probability of getting a club? 9. A fair coin is tossed twice. What is the probability of getting two heads? 10. A fair coin is tossed twice. What is the probability of getting at least one head? 11. Two fair dice are rolled. What is the probability of getting doubles? 12. Two fair dice are rolled. What is the probability of getting double sixes? 13. A single card is drawn from a standard 52-card deck. What is the probability of getting a king or a queen?
18. How would you interpret P(E) 0003 0? 19. A spinner can land on four different colors: red (R), green (G), yellow (Y), and blue (B). If we do not assume each color is as likely to turn up as any other, which of the following probability assignments have to be rejected, and why? (A) P(R) 0003 .15, P(G) 0003 0004.35, P(Y ) 0003 .50, P(B) 0003 .70 (B) P(R) 0003 .32, P(G) 0003 .28, P(Y ) 0003 .24, P(B) 0003 .30 (C) P(R) 0003 .26, P(G) 0003 .14, P(Y ) 0003 .30, P(B) 0003 .30 20. Under the probability assignments in Problem 19, part C, what is the probability that the spinner will not land on blue? 21. Under the probability assignments in Problem 19, part C, what is the probability that the spinner will land on red or yellow? 22. Under the probability assignments in Problem 19, part C, what is the probability that the spinner will not land on red or yellow? 23. A ski jumper has jumped over 300 feet in 25 out of 250 jumps. What is the approximate empirical probability of the next jump being over 300 feet?
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24. In a certain city there are 4,000 youths between 16 and 20 years old who drive cars. If 560 of them were involved in accidents last year, what is the approximate empirical probability of a youth in this age group being involved in an accident this year?
In Problems 39–50, an experiment consists of rolling two fair dice. Let a and b denote the numbers of dots on the two sides facing up. Use the sample space shown in Figure 1 on page 546 to find the probability of each event.
25. Out of 420 times at bat, a baseball player gets 189 hits. What is the approximate empirical probability that the player will get a hit next time at bat?
39. The sum of a and b is 3.
26. In a medical experiment, a new drug is found to help 2,400 out of 3,000 people. If a doctor prescribes the drug for a particular patient, what is the approximate empirical probability that the patient will be helped?
40. The sum of a and b is 5. 41. The sum of a and b is greater than 9. 42. The sum of a and b is less than 6. 43. The product of a and b is 12.
27. A small combination lock on a suitcase has three wheels, each labeled with the 10 digits from 0 to 9. If an opening combination is a particular sequence of three digits with no repeats, what is the probability of a person guessing the right combination?
44. The product of a and b is 6.
28. A combination lock has five wheels, each labeled with the 10 digits from 0 to 9. If an opening combination is a particular sequence of five digits with no repeats, what is the probability of a person guessing the right combination?
48. a 0005 b
Problems 29–34 involve an experiment consisting of dealing 5 cards from a standard 52-card deck. In Problems 29–32, what is the probability of being dealt:
51. Five thousand people work in a large auto plant. An individual is selected at random and his or her birthday (month and day, not year) is recorded. Set up an appropriate sample space for this experiment and assign acceptable probabilities to the simple events.
29. Five black cards 30. Five hearts 31. Five face cards if an ace is considered to be a face card. 32. Five nonface cards if an ace is considered to be a one and not a face card. 33. If we are interested in the number of aces in a 5-card hand, would S 0003 {0, 1, 2, 3, 4} be an acceptable sample space? Would it be an equally-likely sample space? Explain. 34. If we are interested in the number of black cards in a 5-card hand, would S 0003 {0, 1, 2, 3, 4, 5} be an acceptable sample space? Would it be an equally-likely sample space? Explain. 35. If four-digit numbers less than 5,000 are randomly formed from the digits 1, 3, 5, 7, and 9, what is the probability of forming a number divisible by 5? Digits may be repeated; for example, 1,355 is acceptable. 36. If code words of four letters are generated at random using the letters A, B, C, D, E, and F, what is the probability of forming a word without a vowel in it? Letters may be repeated. 37. Suppose five thank-you notes are written and five envelopes are addressed. Accidentally, the notes are randomly inserted into the envelopes and mailed without checking the addresses. What is the probability that all five notes will be inserted into the correct envelopes? 38. Suppose six people check their coats in a checkroom. If all claim checks are lost and the six coats are randomly returned, what is the probability that all six people will get their own coats back?
45. The product of a and b is less than 5. 46. The product of a and b is greater than 15. 47. a 0003 b
49. At least one of a or b is a 6. 50. Exactly one of a or b is a 6.
52. In a hotly contested three-way race for governor of Minnesota, the leading candidates are running neck-and-neck while the third candidate is receiving half the support of either of the others. Registered voters are chosen at random and are asked for which of the three they are most likely to vote. Set up an appropriate sample space for the random survey experiment and assign acceptable probabilities to the simple events. 53. A pair of dice is rolled 500 times with the following frequencies: Sum Frequency
2 3 4 5 6 7 8 9 10 11 12 11 35 44 50 71 89 72 52 36 26 14
(A) Compute the approximate empirical probability for each outcome. (B) Compute the theoretical probability for each outcome, assuming fair dice. (C) Compute the expected frequency of each outcome. (D) Describe how a random number generator could be used to simulate this experiment. If your graphing calculator has a random number generator, use it to simulate 500 tosses of a pair of dice and compare your results with part C. 54. Three coins are flipped 500 times with the following frequencies of outcomes: Three heads: 58 One head: 190
Two heads: 198 Zero heads: 54
(A) Compute the approximate empirical probability for each outcome. (B) Compute the theoretical probability for each outcome, assuming fair coins. (C) Compute the expected frequency of each outcome.
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(D) Describe how a random number generator could be used to simulate this experiment. If your graphing calculator has a random number generator, use it to simulate 500 tosses of three coins and compare your results with part C. 55. (A) Is it possible to get 29 heads in 30 flips of a fair coin? Explain. (B) If you flip a coin 50 times and get 42 heads, would you suspect that the coin was unfair? Why or why not? If you suspect an unfair coin, what empirical probabilities would you assign to the simple events of the sample space? 56. (A) Is it possible to get nine double sixes in 12 rolls of a pair of fair dice? Explain. (B) If you roll a pair of dice 40 times and get 14 double sixes, would you suspect that the dice were unfair? Why or why not? If you suspect unfair dice, what empirical probability would you assign to the event of rolling a double six? An experiment consists of tossing three fair coins, but one of the three coins has a head on both sides. Compute the probabilities of obtaining the indicated results in Problems 57–62. 57. One head
58. Two heads
59. Three heads
60. Zero heads
61. More than one head
62. More than one tail
An experiment consists of rolling two fair dice and adding the dots on the two sides facing up. Each die has one dot on two opposite faces, two dots on two opposite faces, and three dots on two opposite faces. Compute the probabilities of obtaining the indicated sums in Problems 63–70. 63. 2
64. 3
65. 4
68. 7
69. An odd sum
66. 5
67. 6
70. An even sum
An experiment consists of dealing 5 cards from a standard 52-card deck. In Problems 71–78, what is the probability of being dealt the following cards? 71. Five cards, jacks through aces 72. Five cards, 2 through 10 73. Four aces 74. Four of a kind
8-6
75. Straight flush, ace high; that is, 10, jack, queen, king, ace in one suit 76. Straight flush, starting with 2; that is, 2, 3, 4, 5, 6 in one suit 77. Two aces and three queens 78. Two kings and three aces
APPLICATIONS 79. MARKET ANALYSIS A company selected 1,000 households at random and surveyed them to determine a relationship between income level and the number of television sets in a home. The information gathered is listed in the table: Televisions per Household Yearly Income ($)
0
1
2
3
Above 3
Less than 12,000
0
40
51
11
0
12,000–19,999
0
70
80
15
1
20,000–39,999
2
112
130
80
12
40,000–59,999
10
90
80
60
21
60,000 or more
30
32
28
25
20
Compute the approximate empirical probabilities: (A) Of a household earning $12,000–$19,999 per year and owning exactly three television sets (B) Of a household earning $20,000–$39,999 per year and owning more than one television set (C) Of a household earning $60,000 or more per year or owning more than three television sets (D) Of a household not owning zero television sets 80. MARKET ANALYSIS Use the sample results in Problem 79 to compute the approximate empirical probabilities: (A) Of a household earning $40,000–$59,999 per year and owning zero television sets (B) Of a household earning $12,000–$39,999 per year and owning more than two television sets (C) Of a household earning less than $20,000 per year or owning exactly two television sets (D) Of a household not owning more than three television sets
The Binomial Formula Z Using Pascal’s Triangle Z The Binomial Formula Z Proving the Binomial Formula
In a surprising number of areas in math, it turns out to be useful to expand expressions of the form (a 0006 b)n, where n is a natural number. This is known as a binomial expansion. Expanding a binomial is pretty straightforward for small values of n, but gets hard very
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quickly as n increases. The good news is that it turns out that the coefficients in such an expansion are related to counting techniques that we have already learned about.
Z Using Pascal’s Triangle Let’s begin by expanding (a 0006 b)n for the first few values of n. We include n 0003 0, which is not a natural number, for reasons of completeness that will become apparent later. (a 0006 b)0 0003 1 (a 0006 b)1 0003 a 0006 b (a 0006 b)2 0003 a2 0006 2ab 0006 b2 (a 0006 b)3 0003 a3 0006 3a2b 0006 3ab2 0006 b3
ZZZ EXPLORE-DISCUSS 1
(1)
Based on the expansions in equations (1), how many terms would you expect (a 0006 b)n to have? What is the first term? What is the last term?
1 1 1 1
1 2
3
1 3
1
Now let’s examine just the coefficients of the expansions in equations (1) arranged in a form that is usually referred to as Pascal’s triangle (Fig. 1). It is convenient to refer to the top row of Pascal’s triangle (containing a single 1) as row 0. Then row 1 is “1 1,” row 2 is “1 2 1,” and row 3 is “1 3 3 1.” For n a natural number, the first two entries of row n are 1 and n.
Z Figure 1 Pascal’s triangle.
ZZZ EXPLORE-DISCUSS 2
Refer to Figure 1. (A) How is the middle element of row 2 related to the elements in the row above? (B) How are the two inner elements of row 3 related to the elements in the row above? (C) Based on your observations in parts A and B, conjecture the entries of row 4 and row 5. Check your conjecture by expanding (a 0006 b)4 and (a 0006 b)5.
Z Figure 2
Many students find Pascal’s triangle a useful tool for determining the coefficients in the expansion of (a 0006 b)n, especially for small values of n. Figure 2 contains output from a program called PASCAL.* You should recognize the output in the table—it is the first six lines of Pascal’s triangle. The major drawback of using this triangle, whether it is generated by hand or on a graphing calculator, is that to find the elements in a given row, you must write out all the preceding rows. It would be useful to find a formula that gives the coefficients for a binomial expansion directly. Fortunately, such a formula exists— the combination formula Cn,r introduced in Section 8-4.
Z The Binomial Formula When working with binomial expansions, it is customary to use the notation ( nr ) for Cn,r. Recall the combination formula from Section 8-4.
*Programs for TI-84 and TI-86 graphing calculators can be found at the website for this book.
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COMBINATION FORMULA For nonnegative integers r and n, 0 0007 r 0007 n,
n n! a b 0003 Cn,r 0003 r r!(n 0004 r)! Theorem 1 establishes that the coefficients in a binomial expansion can always be expressed in terms of the combination formula. This is a very important theoretical result and a very practical tool. As we will see, using this theorem in conjunction with a graphing calculator provides a very efficient method for expanding binomials. Z THEOREM 1 Binomial Formula For n a positive integer n n (a 0006 b)n 0003 a a b an0004k bk k00030 k
We defer the proof of Theorem 1 until the end of this section. Because the values of the combination formula are the coefficients in a binomial expansion, it is natural to call them binomial coefficients.
EXAMPLE
1
Using the Binomial Formula Use the binomial formula to expand (x 0006 y)6. 6 6 (x 0006 y)6 0003 a a b x60004ky k k00030 k
6 6 6 6 6 6 6 0003 a b x6y0 0006 a b x5y 0006 a b x4y2 0006 a b x3y3 0006 a b x2y4 0006 a b xy5 0006 a b x0y6 0 1 2 3 4 5 6 0003 x6 0006 6x5y 0006 15x4y2 0006 20x3y3 0006 15x2y4 0006 6xy5 0006 y6 Note that the coefficients (1, 6, 15, 20, 15, 6, 1) are the entries of row 6 of Pascal’s triangle. 0002 MATCHED PROBLEM 1
EXAMPLE
2
Use the binomial formula to expand (x 0006 1)5.
0002
Using the Binomial Formula Use the binomial formula to expand (3p 0004 2q)4.
SOLUTION
(3p 0004 2q)4 0003 [(3p) 0006 (00042q)] 4 4 4 0003 a a b (3p)40004k(00042q)k k00030 k 4 4 0003 a a b 340004k(00042)kp40004kqk k00030 k
0003 1(3)4(00042)0p4q0 0006 4(3)3(00042)p3q 0006 6(3)2(00042)2p2q2 0006 4(3)(00042)3pq3 0006 1(3)0(00042)4p0q4 0003 81p4 0004 216p3q 0006 216p2q2 0004 96pq3 0006 16q4 Note that the coefficients (81, 0004216, 216, 000496, 16) are formed by multiplying the entries in row 4 of Pascal’s triangle (1, 4, 6, 4, 1) by the appropriate powers of 3 and 00042.
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Technology Connections The table feature on a graphing calculator provides an efficient alternative to calculating the coefficients of Example 2 one by one (Fig. 3).
40004x x Z Figure 3 y1 0003 C4,x3 (00042) .
0002 MATCHED PROBLEM 2
Use the binomial formula to expand (2m 0004 5n)3.
0002
(A) Compute each term and also the sum of the alternating series
ZZZ EXPLORE-DISCUSS 3
6 6 6 6 a b0004a b0006a b0004...0006a b 0 1 2 6 (B) What result about an alternating series can be deduced by letting a 0003 1 and b 0003 00041 in the binomial formula?
EXAMPLE
3
Using the Binomial Formula Find the term containing x9 in the expansion of (x 0006 3)14.
SOLUTION
In the expansion 14 14 (x 0006 3)14 0003 a a b x140004k3k k k00030
the exponent of x is 9 when k 0003 5. So the term containing x9 is a MATCHED PROBLEM 3
EXAMPLE
4
14 9 5 b x 3 0003 (2,002)(243)x9 0003 486,486x9 5
Find the term containing y8 in the expansion of (2 0006 y)14.
0002
0002
Using the Binomial Formula If the terms in the expansion of (x 0004 2)20 are arranged in decreasing powers of x, find the fourth term and the sixteenth term.
SOLUTION
In the expansion of (a 0006 b)n, the exponent of b in the rth term is r 0004 1 and the exponent of a is n 0004 (r 0004 1). Therefore Fourth term:
Sixteenth term:
20 a b x17(00042)3 3
a
20 ⴢ 19 ⴢ 18 17 x (00048) 3ⴢ2ⴢ1 0003 00049,120x17 0003
20 5 b x (00042)15 15 20 ⴢ 19 ⴢ 18 ⴢ 17 ⴢ 16 5 x (000432,768) 5ⴢ4ⴢ3ⴢ2ⴢ1 0003 0004508,035,072 x5
0003
0002
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If the terms in the expansion of (u 0004 1)18 are arranged in decreasing powers of u, find the fifth term and the twelfth term. 0002
Z Proving the Binomial Formula We now prove that the binomial formula holds for all natural numbers n using mathematical induction. PROOF
State the conjecture. n n Pn: (a 0006 b)n 0003 a a b an0004jb j j00030 j
PART 1
Show that P1 is true. 1
1 10004j j 1 1 1 a a j b a b 0003 a 0 b a 0006 a 1 b b 0003 a 0006 b 0003 (a 0006 b)
j00030
P1 is true. PART 2
Show that if Pk is true, then Pk00061 is true. k k Pk: (a 0006 b)k 0003 a a b ak0004jb j j00030 j k 00061
Pk00061: (a 0006 b)k 00061 0003 a a j00030
k 0006 1 k 000610004j j ba b j
Assume Pk is true.
Show Pk
ⴙ 1
is true.
We begin by multiplying both sides of Pk by (a 0006 b): k k (a 0006 b)k(a 0006 b) 0003 c a a b ak 0004jb j d (a 0006 b) j00030 j
The left side of this equation is the left side of Pk00061. Now we multiply out the right side of the equation and try to obtain the right side of Pk 00061: k k k k (a 0006 b)k 00061 0003 c a b ak 0006 a b ak 00041b 0006 a b ak 00042b2 0006 . . . 0006 a b bk d (a 0006 b) 0 1 2 k k k k k 0003 c a b ak 00061 0006 a b akb 0006 a b ak 00041b2 0006 . . . 0006 a b abk d 1 2 k 0 k k k k 00041 2 . . . k k 0006 c a ba b 0006 a ba b 0006 0006a b abk 0006 a b bk 00061 d 0 1 k00041 k k k k k k 0003 a b ak 00061 0006 c a b 0006 a b d akb 0006 c a b 0006 a b d ak00041b2 0006 . . . 0 0 1 1 2 k k k 0006 ca b 0006 a b d abk 0006 a b bk 00061 k00041 k k
Use the distributive property.
Combine like terms.
We now use the following facts (the proofs are left as exercises; see Problems 63–65, Exercises 8-6). a
k k k00061 b0006a b0003a b r00041 r r
k k00061 a b0003a b 0 0
k k00061 a b0003a b k k00061
to rewrite the right side as a
k 0006 1 k 00061 k00061 k k 0006 1 k 00041 2 . . . ba 0006a ba b 0006 a ba b 0006 0 1 2 0006a
k00061 k 0006 1 k 00061 k 0006 1 k 000610004j j k00061 bb 0003 a a ba b b abk 0006 a k00061 j k j0003 0
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Because the right side of the last equation is the right side of Pk00061, we have shown that Pk 00061 follows from Pk. CONCLUSION
Pn is true. That is, the binomial formula holds for all positive integers n. ANSWERS TO MATCHED PROBLEMS 1. x5 0006 5x4 0006 10x3 0006 10x2 0006 5x 0006 1 2. 8m3 0004 60m2n 0006 150mn2 0004 125n3 8 14 7 3. 192,192y 4. 3,060u ; 000431,824u
8-6
Exercises
1. What is a binomial?
45. (2m 0006 n)12; eleventh term
46. (x 0006 2y)20; third term
2. What is a binomial coefficient?
47. [(w00022) 0004 2]12; seventh term
48. (x 0004 3)10; fourth term
49. (3x 0004 2y)8; sixth term
50. (2p 0004 3q)7; fourth term
3. Explain how the entries in Pascal’s triangle are generated. 5
4. How can Pascal’s triangle be used to expand (a 0006 b) ? In Problems 51–54, use the binomial formula to expand and simplify the difference quotient
In Problems 5–12, use Pascal’s triangle to evaluate each expression. 8 5. a b 3
8 6. a b 4
9 7. a b 6
9 8. a b 7
9. C7,5
10. C7,3
11. C9,0
12. C10,10
In Problems 13–20, evaluate each expression. 13. a
13 b 3
14. a
13 b 9
18. C52,4
17. C52,3
15. a
12 b 4
19. C12,6
16. a
12 b 11
20. C12,11
f (x 0006 h) 0004 f (x) h for the indicated function f. Discuss the behavior of the simplified form as h approaches 0. 51. f (x) 0003 x3
52. f(x) 0003 x4
53. f(x) 0003 x5
54. f(x) 0003 x6
In Problems 55–58, use a graphing calculator to graph each sequence and to display it in table form. 55. Find the number of terms of the sequence
Expand Problems 21–32 using the binomial formula. 21. (m 0006 n)3
22. (x 0006 2)3
23. (2x 0004 3y)3
24. (3u 0006 2v)3
25. (x 0004 2)4
26. (x 0004 y)4
27. (m 0006 3n)4
28. (3p 0004 q)4
29. (2x 0004 y)5
5
30. (2x 0004 1)
6
32. (2x 0004 y)
In Problems 33–42, find the term of the binomial expansion containing the given power of x. 33. (x 0006 1)7; x4
34. (x 0006 1)8; x5
35. (2x 0004 1)11; x6
36. (3x 0006 1)12; x7
37. (2x 0006 3)18; x14
38. (3x 0004 2)17; x5
39. (x2 0004 1)6; x8
40. (x2 0004 1)9; x7
41. (x2 0006 1)9; x11
56. Find the number of terms of the sequence a
40 40 40 40 b, a b, a b, . . . , a b 1 2 40 0
that are greater than one-half of the largest term. 57. (A) Find the largest term of the sequence a0, a1, a2, . . . , a10 to three decimal places, where ak 0003 a
10 b (0.6)100004k(0.4)k k
(B) According to the binomial formula, what is the sum of the series a0 0006 a1 0006 a2 0006 . . . 0006 a10?
42. (x2 0006 1)10; x14 In Problems 43–50, find the indicated term in each expansion if the terms of the expansion are arranged in decreasing powers of the first term in the binomial. 43. (u 0006 v)15; seventh term
20 20 20 20 b, a b, a b, . . . , a b 0 1 2 20
that are greater than one-half of the largest term.
6
31. (m 0006 2n)
a
44. (a 0006 b)12; fifth term
58. (A) Find the largest term of the sequence a0, a1, a2, . . . , a10 to three decimal places, where ak 0003 a
10 b (0.3)100004k(0.7)k k
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(B) According to the binomial formula, what is the sum of the series a0 0006 a1 0006 a2 0006 . . . 0006 a10?
n 66. Show that: a b is given by the recursion formula r
59. Evaluate (1.01)10 to four decimal places, using the binomial formula. [Hint: Let 1.01 0003 1 0006 0.01.]
n n n0004r00061 a b a b0003 r r00041 r
60. Evaluate (0.99)6 to four decimal places, using the binomial formula. n n 61. Show that: a b 0003 a b r n0004r
n where a b 0003 1. 0 67. Write 2n 0003 (1 0006 1)n and expand, using the binomial formula to obtain
n n 62. Show that: a b 0003 a b 0 n 63. Show that: a
n n n n 2n 0003 a b 0006 a b 0006 a b 0006 . . . 0006 a b 0 1 2 n
k k k00061 b0006a b0003a b r00041 r r
68. Write 0 0003 (1 0004 1)n and expand, using the binomial formula, to obtain n n n n 0 0003 a b 0004 a b 0006 a b 0004 . . . 0006 (00041)na b 0 1 2 n
k k00061 64. Show that: a b 0003 a b 0 0 k k00061 65. Show that: a b 0003 a b k k00061
CHAPTER
8-1
8
Review
Sequences and Series
A sequence is a function with the domain a set of successive integers. The symbol an, called the nth term, or general term, represents the range value associated with the domain value n. Unless specified otherwise, the domain is understood to be the set of natural numbers. A finite sequence has a finite domain, and an infinite sequence has an infinite domain. A recursion formula defines each term of a sequence in terms of one or more of the preceding terms. For example, the Fibonacci sequence is defined by an 0003 an00041 0006 an00042 for n 3, where a1 0003 a2 0003 1. If a1, a2, . . . , an, . . . is a sequence, then the expression a1 0006 a2 0006 . . . 0006 an 0006 . . . is called a series. A finite sequence produces a finite series, and an infinite sequence produces an infinite series. Series can be represented using the summation notation:
To use mathematical induction to prove statements involving laws of exponents, it is convenient to state a recursive definition of an: a1 0003 a
and
an00061 0003 ana
for any integer n 1
To deal with conjectures that may be true only for n m, where m is a positive integer, we use the extended principle of mathematical induction: Let m be a positive integer, let Pn be a statement associated with each integer n m, and suppose the following conditions are satisfied: 1. Pm is true. 2. For any integer k m, if Pk is true, then Pk 00061 is also true. Then the statement Pn is true for all integers n m.
n
. . . 0006 an a ak 0003 am 0006 am00061 0006
k0003m
where k is called the summing index. If the terms in the series are alternately positive and negative, the series is called an alternating series.
8-2
Mathematical Induction
A wide variety of statements can be proven using the principle of mathematical induction: Let Pn be a statement associated with each positive integer n and suppose the following conditions are satisfied: 1. P1 is true. 2. For any positive integer k, if Pk is true, then Pk 00061 is also true. Then the statement Pn is true for all positive integers n.
8-3
Arithmetic and Geometric Sequences
A sequence is called an arithmetic sequence, or arithmetic progression, if there exists a constant d, called the common difference, such that or an 0004 an00041 0003 d an 0003 an00041 0006 d for every n 7 1 The following formulas are useful when working with arithmetic sequences and their corresponding series: an 0003 a1 0006 (n 0004 1)d n Sn 0003 [2a1 0006 (n 0004 1)d] 2 n Sn 0003 (a1 0006 an) 2
nth-Term Formula Sum Formula—First Form Sum Formula—Second Form
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A sequence is called a geometric sequence, or a geometric progression, if there exists a nonzero constant r, called the common ratio, such that an 0003r an00041
an 0003 ran00041
or
for every n 7 1
The following formulas are useful when working with geometric sequences and their corresponding series: an 0003 a1r n00041
nth-Term Formula
Sn 0003
a1 0004 a1r 10004r
Sn 0003
a1 0004 ran 10004r
a1 S 0003 10004r
8-4
n
r00051 r00051
Sum Formula—First Form Sum Formula—Second Form
0004r0004 6 1
Sum of an Infinite Geometric Series
Multiplication Principle, Permutations, and Combinations
A counting technique is a mathematical method of determining the number of objects in a set without actually enumerating them. Given a sequence of operations, tree diagrams are often used to list all the possible combined outcomes. To count the number of combined outcomes without listing them, we use the multiplication principle (also called the fundamental counting principle): 1. If operations O1 and O2 are performed in order with N1 possible outcomes for the first operation and N2 possible outcomes for the second operation, then there are N1 ⴢ N2 possible outcomes of the first operation followed by the second. 2. In general, if n operations O1, O2, . . . , On are performed in order, with possible number of outcomes N1, N2, . . . , Nn, respectively, then there are N1 ⴢ N2 ⴢ . . . ⴢ Nn possible combined outcomes of the operations performed in the given order. The symbol n! is read n factorial and 0! is defined to be 1. A particular arrangement or ordering of n objects without repetition is called a permutation. The number of permutations of n objects is given by Pn,n 0003 n ⴢ (n 0004 1) ⴢ . . . ⴢ 1 0003 n! A permutation of a set of n objects taken r at a time is an arrangement of the r objects in a specific order. The number of permutations of n objects taken r at a time is given by Pn,r 0003
n! (n 0004 r)!
00007r0007n
A combination of a set of n objects taken r at a time is an r-element subset of the n objects. The number of combinations of n objects taken r at a time is given by Pn,r
n n! Cn,r 0003 a b 0003 0003 r r! r!(n 0004 r)!
00007r0007n
In a permutation, order is important. In a combination, order is not important.
8-5
565
Sample Spaces and Probability
The outcomes of an experiment are called simple events if one and only one of these results will occur in each trial of the experiment. The set of all simple events is called the sample space. Any subset of the sample space is called an event. An event is a simple event if it has only one element in it and a compound event if it has more than one element in it. We say that an event E occurs if any of the simple events in E occurs. A sample space S1 is more fundamental than a second sample space S2 if knowledge of which event occurs in S1 tells us which event in S2 occurs, but not conversely. Given a sample space S 0003 {e1, e2, . . . , en} with n simple events, to each simple event ei we assign a real number denoted by P(ei), that is called the probability of the event ei and satisfies: 1. 0 0007 P(ei) 0007 1 2. P(e1) 0006 P(e2) 0006 . . . 0006 P(en) 0003 1 Any probability assignment that meets conditions 1 and 2 is said to be an acceptable probability assignment. Given an acceptable probability assignment for the simple events in a sample space S, the probability of an arbitrary event E is defined as follows: 1. If E is the empty set, then P(E ) 0003 0. 2. If E is a simple event, then P(E ) has already been assigned. 3. If E is a compound event, then P(E ) is the sum of the probabilities of all the simple events in E. 4. If E is the sample space S, then P(E ) 0003 P(S ) 0003 1. If each of the simple events in a sample space S 0003 {e1, e2, . . . , en} with n simple events is equally likely to occur, then we assign the probability 10002n to each. If E is an arbitrary event in S, then P(E ) 0003
n(E ) Number of elements in E 0003 Number of elements in S n(S )
If we conduct an experiment n times and event E occurs with frequency f (E ), then the ratio f(E )0002n is called the relative frequency of the occurrence of event E in n trials. As n increases, f (E )0002n usually approaches a number that is called the empirical probability P(E ). So f(E )0002n is used as an approximate empirical probability for P(E ). If P(E ) is the theoretical probability of an event E and the experiment is performed n times, then the expected frequency of the occurrence of E is n ⴢ P(E ).
8-6
Binomial Formula
Pascal’s triangle is a triangular array of coefficients for the expansion of the binomial (a 0006 b)n, where n is a positive integer. Notation for the combination formula is n n! a b 0003 Cn,r 0003 r r!(n 0004 r)! For n a positive integer, the binomial formula is n n (a 0006 b)n 0003 a a b an0004kbk k00030 k
n The numbers a b, 0 0007 k 0007 n, are called binomial coefficients. k
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Review Exercises
Work through all the problems in this chapter review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. 1. Determine whether each of the following can be the first three terms of a geometric sequence, an arithmetic sequence, or neither. (A) 16, 00048, 4, . . . (B) 5, 7, 9, . . . (C) 00048, 00045, 00042, . . . (D) 2, 3, 5, . . . (E) 00041, 2, 00044, . . . In Problems 2–5: (A) Write the first four terms of each sequence. (B) Find a10. (C) Find S10. 2. an 0003 2n 0006 3
3. an 0003 32(12)n
4. a1 0003 00048; an 0003 an00041 0006 3, n 2 5. a1 0003 00041, an 0003 (00042)an00041, n 2 6. Find S in Problem 3.
9.
7! 2!(7 0004 2)!
19. Pn: 49n 0004 1 is divisible by 6 In Problems 20–22, write Pk and Pk 0006 1. 20. For Pn in Problem 17
21. For Pn in Problem 18
22. For Pn in Problem 19 23. Either prove the statement is true or prove it is false by finding a counterexample: If n is a positive integer, then the 1 1 1 sum of the series 1 0006 0006 0006 . . . 0006 is less than 4. n 2 3 Write Problems 24 and 25 without summation notation, and find the sum. 10
24. S10 0003 a (2k 0004 8) k00031
7 16 25. S7 0003 a k k00031 2
26. S 0003 27 0004 18 0006 12 0006 . . . 0003 ?
Evaluate the expression in Problems 7–10. 7. 6!
18. Pn: 2 0006 4 0006 8 0006 . . . 0006 2n 0003 2n00061 0004 2
27. Write
22! 8. 19! 10. C6,2 and P6,2
11. A single die is rolled and a coin is flipped. How many combined outcomes are possible? Solve (A) By using a tree diagram (B) By using the multiplication principle 12. How many seating arrangements are possible with six people and six chairs in a row? Solve by using the multiplication principle. 13. Solve Problem 12 using permutations or combinations, whichever is applicable. 14. In a single deal of 5 cards from a standard 52-card deck, what is the probability of being dealt five clubs? 15. Betty and Bill are members of a 15-person ski club. If the president and treasurer are selected by lottery, what is the probability that Betty will be president and Bill will be treasurer? A person cannot hold more than one office. 16. A drug has side effects for 50 out of 1,000 people in a test. What is the approximate empirical probability that a person using the drug will have side effects? Verify the statement Pn in Problems 17–19 for n 0003 1, 2, and 3. 17. Pn: 5 0006 7 0006 9 0006 . . . 0006 (2n 0006 3) 0003 n2 0006 4n
Sn 0003
(00041)n00061 1 1 1 0004 0006 0006...0006 3 9 27 3n
using summation notation, and find S . 28. Someone tells you that the following approximate empirical probabilities apply to the sample space {e1, e2, e3, e4}: P(e1) 0003 .1, P(e2) 0003 0004.2, P(e3) 0003 .6, P(e4) 0003 2. There are three reasons why P cannot be a probability function. Name them. 29. Six distinct points are selected on the circumference of a circle. How many triangles can be formed using these points as vertices? 30. In an arithmetic sequence, a1 0003 13 and a7 0003 31. Find the common difference d and the fifth term a5. 31. How many three-letter code words are possible using the first eight letters of the alphabet if no letter can be repeated? If letters can be repeated? If adjacent letters cannot be alike? 32. Two coins are flipped 1,000 times with the following frequencies: Two heads:
210
One head:
480
Zero heads:
310
(A) Compute the empirical probability for each outcome. (B) Compute the theoretical probability for each outcome. (C) Using the theoretical probabilities computed in part B, compute the expected frequency of each outcome, assuming fair coins.
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33. From a standard deck of 52 cards, what is the probability of obtaining a 5-card hand: (A) Of all diamonds? (B) Of three diamonds and two spades?
51. How many ways can two people be seated in a row of four chairs?
Write answers in terms of Cn,r or Pn,r, as appropriate. Do not evaluate.
53. If three people are selected from a group of seven men and three women, what is the probability that at least one woman is selected?
34. A group of 10 people includes one married couple. If four people are selected at random, what is the probability that the married couple is selected?
52. Expand (x 0006 i)6, where i is the imaginary unit, using the binomial formula.
54. Three fair coins are tossed 1,000 times with the following frequencies of outcomes:
35. A spinning device has three numbers, 1, 2, 3, each as likely to turn up as the other. If the device is spun twice, what is the probability that: (A) The same number turns up both times? (B) The sum of the numbers turning up is 5?
Number of heads Frequency
38.
20! 18!(20 0004 18)!
39. a
16 b 12
40. a
11 b 11
5
41. Expand (x 0004 y) using the binomial formula. 42. Find the term containing x6 in the expansion of (x 0006 2)9. 43. If the terms in the expansion of (2x 0004 y)12 are arranged in descending powers of x, find the tenth term. Establish each statement in Problems 44–46 for all natural numbers using mathematical induction. 44. Pn in Problem 17
45. Pn in Problem 18
46. Pn in Problem 19 In Problems 47 and 48, find the smallest positive integer n such that an bn by graphing the sequences {an} and {bn} with a graphing calculator. Check your answer by using a graphing calculator to display both sequences in table form. 47. an 0003 C50,n, bn 0003 3n 48. a1 0003 100, an 0003 0.99an00041 0006 5, bn 0003 9 0006 7n 49. How many different families with five children are possible, excluding multiple births, where the sex of each child in the order of their birth is taken into consideration? How many families are possible if the order pattern is not taken into account? 50. A free-falling body travels g/2 feet in the first second, 3g00022 feet during the next second, 5g00022 feet the next, and so on. Find the distance fallen during the twenty-fifth second and the total distance fallen from the start to the end of the twenty-fifth second.
2
3
120
360
350
170
Prove that each statement in Problems 55–59 holds for all positive integers using mathematical induction. n
n
2
55. a k3 0003 a a kb k00031 2n
56. x 57.
Evaluate Problems 38–40.
1
(A) What is the approximate empirical probability of obtaining two heads? (B) What is the theoretical probability of obtaining two heads? (C) What is the expected frequency of obtaining two heads?
36. Use the formula for the sum of an infinite geometric series to write 0.727 272 . . . 0003 0.72 as the quotient of two integers. 37. Solve the following problems using Pn,r or Cn,r, as appropriate: (A) How many three-digit opening combinations are possible on a combination lock with six digits if the digits cannot be repeated? (B) Suppose five tennis players have made the finals. If each of the five players is to play every other player exactly once, how many games must be scheduled?
0
k00031
2n
0004 y is divisible by x 0004 y, x 0005 y
an 0003 an0004m; n 7 m; n, m positive integers am
58. {an} 0003 {bn}, where an 0003 an00041 0006 2, a1 0003 00043, bn 0003 00045 0006 2n 59. (1!)1 0006 (2!)2 0006 (3!)3 0006 . . . 0006 (n!)n 0003 (n 0006 1)! 0004 1 (From U.S.S.R. Mathematical Olympiads, 1955–1956, Grade 10.)
APPLICATIONS 60. LOAN REPAYMENT You borrow $7,200 and agree to pay 1% of the unpaid balance each month for interest. If you decide to pay an additional $300 each month to reduce the unpaid balance, how much interest will you pay over the 24 months it will take to repay this loan? 61. ECONOMICS Due to reduced taxes, an individual has an extra $2,400 in spendable income. If we assume that the individual spends 75% of this on consumer goods, and the producers of those consumer goods in turn spend 75% on consumer goods, and that this process continues indefinitely, what is the total amount (to the nearest dollar) spent on consumer goods? 62. COMPOUND INTEREST If $500 is invested at 6% compounded annually, the amount A present after n years forms a geometric sequence with common ratio 1 0006 0.06 0003 1.06. Use a geometric sequence formula to find the amount A in the account (to the nearest cent) after 10 years; after 20 years. 63. TRANSPORTATION A distribution center A wishes to distribute its products to five different retail stores, B, C, D, E, and F, in a city. How many different route plans can be constructed so that a single truck can start from A, deliver to each store exactly once, and then return to the center? 64. MARKET ANALYSIS A DVD distributor selected 1,000 persons at random and surveyed them to determine a relationship between age of purchaser and annual DVD purchases. The results are given in the table on page 568.
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DVDs Purchased Annually Age
0
1
2
Above 2
Under 12
60
70
30
10
170
12–18
30
100
100
60
290
19–25
70
110
120
30
330
Over 25
100
50
40
20
210
Totals
260
330
290
120
1,000
CHAPTER
ZZZ
Totals
Find the empirical probability that a person selected at random (A) Is over 25 and buys exactly two DVDs annually. (B) Is 12–18 years old and buys more than one DVD annually. (C) Is 12–18 years old or buys more than one DVD annually. 65. QUALITY CONTROL Twelve precision parts, including two that are substandard, are sent to an assembly plant. The plant manager selects four at random and will return the whole shipment if one or more of the samples are found to be substandard. What is the probability that the shipment will be returned?
8
GROUP ACTIVITY Sequences Specified by Recursion Formulas
The recursion formula* an 0003 5an00041 0004 6an00042, together with the initial values a1 0003 4, a2 0003 14, specifies the sequence {an} whose first several terms are 4, 14, 46, 146, 454, 1,394, . . . . The sequence {an} is neither arithmetic nor geometric. Nevertheless, because it satisfies a simple recursion formula, it is possible to obtain an nth-term formula for {an} that is analogous to the nthterm formulas for arithmetic and geometric sequences. Such an nth-term formula is valuable because it allows us to estimate a term of a sequence without computing all the preceding terms. If the geometric sequence {rn} satisfies the preceding recursion formula, then rn 0003 5rn00041 0004 6rn00042. Dividing both sides by rn00042 leads to the quadratic equation r2 0004 5r 0006 6 0003 0, whose solutions are r 0003 2 and r 0003 3. Now it is easy to check that the geometric sequences {2n} 0003 2, 4, 8, 16, . . . and {3n} 0003 3, 9, 27, 81, . . . satisfy the recursion formula. Therefore, any sequence of the form {u2n 0006 v3n}, where u and v are constants, will satisfy the same recursion formula. We now find u and v so that the first two terms of {u2n 0006 v3n} are a1 0003 4, a2 0003 14. Letting n 0003 1 and n 0003 2 we see that u and v must satisfy the following linear system:
Solving the system gives u 0003 00041, v 0003 2. Therefore, an nth-term formula for the original sequence is an 0003 (00041)2n 0006 (2)3n. Note that the nth-term formula was obtained by solving a quadratic equation and a system of two linear equations in two variables. (A) Compute (00041)2n 0006 (2)3n for n 0003 1, 2, . . . , 6, and compare with the terms of {an}. (B) Estimate the one-hundredth term of {an}. (C) Show that any sequence of the form {u2n 0006 v3n}, where u and v are constants, satisfies the recursion formula an 0003 5an00041 0004 6an00042. (D) Find an nth-term formula for the sequence {bn} that is specified by b1 0003 5, b2 0003 55, bn 0003 3bn00041 0006 4bn00042. (E) Find an nth-term formula for the Fibonacci sequence. (F) Find an nth-term formula for the sequence {cn} that is specified by c1 0003 00043, c2 0003 15, c3 0003 99, cn 0003 6cn00041 0004 3cn00042 0004 10cn00043. (Because the recursion formula involves the three terms that precede cn, our method will involve the solution of a cubic equation and a system of three linear equations in three variables.)
2u 0006 3v 0003 4 4u 0006 9v 0003 14 *The program RECUR, found at the website for this book, evaluates the terms in any sequence defined by this type of recursion formula.
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11 A
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APPENDIX A
CHAPTERS
1–3
Cumulative Review Exercises
Work through all the problems in this cumulative review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. 1. Solve for x:
Problems 16–18 refer to the function f given by the graph: f (x) 5
7x 3 0003 2x x 0002 10 0002 0007 00032 5 2 3
00025
5
x
In Problems 2–4, solve and graph the inequality. 2. 2(3 0002 y) 0003 4 0004 5 0002 y
16. Find the domain and range of f. Express answers in interval notation.
4. x2 0003 3x 0006 10 5. Perform the indicated operations and write the answer in standard form: (A) (2 0002 3i) 0002 (00025 0003 7i) (B) (1 0003 4i)(3 0002 5i) 50003i (C) 2 0003 3i In Problems 6–9, solve the equation. 2
6. 3x 0007 000212x 2
8. x 0002 6x 0003 2 0007 0
00025
3. 冟 x 0002 2冟 0005 7
2
17. Is f an even function, an odd function, or neither? Explain. 18. Use the graph of f to sketch a graph of the following: (A) y 0007 0002f(x 0003 1) (B) y 0007 2f (x) 0002 2 In Problems 19–21, solve the equation. 19.
7. 4x 0002 20 0007 0 9. x 0002 112 0002 x 0007 0
10. Given the points A 0007 (3, 2) and B 0007 (5, 6), find: (A) Distance between A and B. (B) Slope of the line through A and B. (C) Slope of a line perpendicular to the line through A and B. 11. Find the equation of the circle with radius 12 and center: (A) (0, 0) (B) (00023, 1) 12. Graph 2x 0002 3y 0007 6 and indicate its slope and intercepts. 13. Indicate whether each set defines a function. Find the domain and range of each function. (A) {(1, 1), (2, 1), (3, 1)} (B) {(1, 1), (1, 2), (1, 3)} (C) {(00022, 2), (00021, 00021), (0, 0), (1, 00021), (2, 2)} 14. For f (x) 0007 x2 0002 2x 0003 5 and g(x) 0007 3x 0002 2, find: (A) f (00022) 0003 g(3) (B) ( f 0003 g)(x) (C) ( f ° g)(x) f (a 0003 h) 0002 f (a) (D) h 15. How are the graphs of the following related to the graph of y 0007 冟 x 冟? (A) y 0007 2冟 x 冟 (B) y 0007 冟 x 0002 2 冟 (C) y 0007 冟 x 冟 0002 2
x00033 5x 0003 2 5 0003 0007 2x 0003 2 3x 0003 3 6
20.
3 1 6 0007 0002 x x00031 x00021
21. 2x 0003 1 0007 312x 0002 1 In Problems 22–24, solve and graph the inequality. 22. 冟 4x 0002 9 冟 7 3 24.
23. 2(3m 0002 4)2 0004 2
x00031 7 x00022 2
25. For what real values of x does the following expression represent a real number? 1x 0002 2 x00024 26. Perform the indicated operations and write the final answers in standard form: (A) (2 0002 3i)2 0002 (4 0002 5i)(2 0002 3i) 0002 (2 0003 10i) 1 (B) 35 0003 45i 0003 3 4 (C) i35 5 0003 5i 27. Convert to a 0003 bi form, perform the indicated operations, and write the final answers in standard form: (A) (5 0003 2100029) 0002 (2 0002 31000216) 12 0002 1000264 2 0003 7 1000225 (B) (C) 3 0002 100021 100024 In Problems 28–31, solve the equation. 28. 1 0003
14 6 0007 2 y y
29. 4x2/3 0002 4x1/3 0002 3 0007 0
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A-3
y
30. u4 0003 u2 0002 12 0007 0 5
31. 18t 0002 2 0002 2 1t 0007 1
Use a calculator to solve the equation or inequality in Problems 32 and 33. Compute answers to two decimal places.
5
x
32. 00023.45 6 1.86 0002 0.33x 0004 7.92 33. 2.35x2 0003 10.44x 0002 16.47 0007 0
00025
34. Solve for y in terms of x: 2y 0003 1 x00022 0007 x00031 y00022 35. Find each of the following for the function f given by the graph shown in the figure. (A) The domain of f (B) The range of f (C) f (00023) 0003 f(00022) 0003 f (2) (D) The intervals over which f is increasing (E) The x coordinates of any points of discontinuity f (x)
44. Let f (x) 0007 1x 0003 4 (A) Find f 00021(x). (B) Find the domain and range of f and f 00021. (C) Graph f, f 00021, and y 0007 x on the same coordinate system. Check by graphing f, f 00021, and y 0007 x in a squared window on a graphing calculator. 45. Find the center and radius of the circle given by the equation x2 0002 6x 0003 y2 0003 2y 0007 0. Graph the circle and show the center and the radius. 46. Discuss symmetry with respect to the x axis, y axis, and the origin for the equation xy 0003 冟 xy 冟 0007 5
5
00025
5
x
47. Write an equation for the graph in the figure in the form y 0007 a(x 0002 h)2 0003 k, where a is either 00021 or 00031 and h and k are integers. y 5
00025
36. Write equations of the lines (A) Parallel to (B) Perpendicular to the line 3x 0003 2y 0007 12 and passing through the point (00026, 1). Write the final answers in the slope–intercept form y 0007 mx 0003 b. 37. Find the domain of g(x) 0007 1x 0003 4. 38. Graph f(x) 0007 x2 0002 2x 0002 8. Show the axis of symmetry and vertex, and find the range, intercepts, and maximum or minimum value of f(x).
00025
5
00025
48. Solve for y in terms of x: x0003y 00071 x0003y y0002 x0002y
39. Given f (x) 0007 1兾(x 0002 2) and g(x) 0007 (x 0003 3)兾x, find f ⴰ g. What is the domain of f ⴰ g?
49. Find all roots: 3x2 0007 212x 0002 1.
40. Find f 00021(x) for f(x) 0007 2x 0003 5.
50. Consider the quadratic equation
41. Graph, finding the domain, range, and any points of discontinuity: f (x) 0007 e 42. Graph: (A) y 0007 2 1x 0003 1 (B) y 0007 0002 1x 0003 1
x00021 x2 0003 1
if x 6 0 if x 0006 0
43. The graph in the figure is the result of applying a sequence of transformations to the graph of y 0007 冟 x 冟. Describe the transformations verbally and write an equation for the graph in the figure.
x
x2 0003 bx 0003 1 0007 0 where b is a real number. Discuss the relationship between the values of b and the three types of roots listed in Table 1 in Section 1-5. 51. Find all solutions: 13 0002 2x 0002 1x 0003 7 0007 1x 0003 4. 52. Write in standard form:
a 0003 bi , a, b 0. a 0002 bi
53. Given f (x) 0007 x2 and g(x) 0007 24 0002 x2, find: (A) Domain of g (B) f兾g and its domain (C) f ⴰ g and its domain
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54. Let f(x) ⫽ x2 ⫺ 2x ⫺ 3, x ⱖ 1. (A) Find f ⫺1(x). (B) Find the domain and range of f ⫺1. (C) Graph f, f ⫺1, and y ⫽ x on the same coordinate system. Check by graphing f, f ⫺1, and y ⫽ x in a squared window on a graphing calculator.
63. COMPUTER SCIENCE Let f (x) ⫽ x ⫺ 2冀 xⲐ2冁 . This function can be used to determine if an integer is odd or even. (A) Find f(1), f(2), f(3), and f(4). (B) Find f (n) for any integer n. [Hint: Consider two cases, n ⫽ 2k and n ⫽ 2k ⫹ 1, k an integer.]
55. NUMBERS Find a number such that the number exceeds its reciprocal by 32.
64. DEPRECIATION Office equipment was purchased for $20,000 and is assumed to depreciate linearly to a scrap value of $4,000 after 8 years. (A) Find a linear function v ⫽ d(t) that relates value v in dollars to time t in years. (B) Find t ⫽ d ⫺1(v).
56. RATE–TIME A boat travels upstream for 35 miles and then returns to its starting point. If the round-trip took 4.8 hours and the boat’s speed in still water is 15 miles per hour, what is the speed of the current?
65. PROFIT AND LOSS ANALYSIS At a price of $p per unit, the marketing department at a company estimates that the weekly cost C and the weekly revenue R, in thousands of dollars, will be given by the equations
57. CHEMISTRY How many gallons of distilled water must be mixed with 24 gallons of a 90% sulfuric acid solution to obtain a 60% solution? 58. BREAK-EVEN ANALYSIS The publisher’s fixed costs for the production of a new study guide are $41,800. Variable costs are $4.90 per book. If the book is sold to bookstores for $9.65, how many must be sold for the publisher to break even? 59. FINANCE An investor instructs a broker to buy a certain stock whenever the price per share p of the stock is within $10 of $200. Express this instruction as an absolute value inequality. 60. PRICE AND DEMAND The weekly demand for mouthwash in a chain of drugstores is 1,160 bottles at a price of $3.79 each. If the price is lowered to $3.59, the weekly demand increases to 1,340 bottles. Assuming that the relationship between the weekly demand x and the price per bottle p is linear, express x as a function of p. How many bottles would the store sell each week if the price were lowered to $3.29? 61. BUSINESS—PRICING A telephone company begins a new pricing plan that charges customers for local calls as follows: The first 60 calls each month are 6 cents each, the next 90 are 5 cents each, the next 150 are 4 cents each, and any additional calls are 3 cents each. If C is the cost, in dollars, of placing x calls per month, write a piecewise definition of C as a function of x and graph. 62. CONSTRUCTION A homeowner has 80 feet of chain-link fencing to be used to construct a dog pen adjacent to a house (see the figure). (A) Express the area A(x) enclosed by the pen as a function of the width x. (B) From physical considerations, what is the domain of the function A? (C) Graph A and determine the dimensions of the pen that will make the area maximum. x
x
C ⫽ 88 ⫺ 12p
Cost equation
R ⫽ 15p ⫺ 2p2
Revenue equation
Find the prices for which the company has: (A) A profit (B) A loss 66. SHIPPING A ship leaves port A, sails east to port B, and then north to port C, a total distance of 115 miles. The next day the ship sails directly from port C back to port A, a distance of 85 miles. Find the distance between ports A and B and between ports B and C. 67. PHYSICS The distance s above the ground (in feet) of an object dropped from a hot-air balloon t seconds after it is released is given by s ⫽ a ⫹ bt2 where a and b are constants. Suppose the object is 2,100 feet above the ground 5 seconds after its release and 900 feet above the ground 10 seconds after its release. (A) Find the constants a and b. (B) How high is the balloon? (C) How long does the object fall? 68. PRICE AND DEMAND The demand for barley q (in thousands of bushels) and the corresponding price p (in cents) at a midwestern grain exchange are shown in the figure. p 350
Price (in cents)
APPLICATIONS
340 330 320 310 10
20
30
40
50
q
Barley (thousands of bushels)
(A) What is the demand (to the nearest thousand bushels) when the price is 325 cents per bushel? (B) Does the demand increase or decrease if the price is increased to 340 cents per bushel? By how much? (C) Does the demand increase or decrease if the price is decreased to 315 cents per bushel? By how much? (D) Write a brief description of the relationship between price and demand illustrated by this graph.
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(E) Use the graph to estimate the price (to the nearest cent) when the demand is 20, 25, 30, 35, and 40 thousand bushels. Use these data to find a quadratic regression model for the price of barley using the demand as the independent variable.
Table 1 Skid Marks Speed (mph)
Length of Skid Marks (feet)
69. STOPPING DISTANCE Table 1 contains data related to the length of the skid marks left by an automobile when making an emergency stop. A model for the skid mark length L (in feet) is
20
24
30
48
L 0007 f (s) 0007 0.05s2 0002 0.2s 0003 6.5, s 0006 20
40
77
where s is speed in miles per hour. (A) Graph L 0007 f(s) and the data for skid mark length on the same axes. (B) Find s 0007 f 00021(L) and find its domain and range. (C) An insurance investigator finds skid marks 220 feet long at the scene of an accident involving this automobile. How fast (to the nearest mile per hour) was the automobile traveling when it made these skid marks?
50
115
60
187
70
246
80
312
4–5
CHAPTERS
A-5
Cumulative Review Exercises
Work through all the problems in this cumulative review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text.
3. For P(x) 0007 3x3 0003 5x2 0002 18x 0002 3 and D(x) 0007 x 0003 3, use synthetic division to divide P(x) by D(x), and write the answer in the form P(x) 0007 D(x)Q(x) 0003 R.
1. Let P(x) be the polynomial whose graph is shown in the figure. (A) Assuming that P(x) has integer zeros and leading coefficient 1, find the lowest-degree equation that could produce this graph. (B) Describe the left and right behavior of P(x).
5. Let P(x) 0007 4x3 0002 5x2 0002 3x 0002 1. How do you know that P(x) has at least one real zero between 1 and 2?
P (x)
7. Solve for x. (A) y 0007 10 x (A) (2e x )3
00025
5
2. Match each equation with the graph of f, g, m, or n in the figure. (A) y 0007 (34)x (B) y 0007 (43)x 3 x 4 x (C) y 0007 (4) 0003 (3) (D) y 0007 (43)x 0002 (34)x 3
g
00024.5
(B)
e3x e00022x
10. Solve for x to three significant digits. (A) 10 x 0007 2.35 (B) ex 0007 87,500 (C) log x 0007 00021.25 (D) ln x 0007 2.75 In Problems 11 and 12, translate each statement into an equation using k as the constant of proportionality. 11. E varies directly as p and inversely as the cube of x.
4.5
00023
(B) y 0007 ln x
9. Solve for x exactly. Do not use a calculator or a table. (A) log3 x 0007 2 (B) log3 81 0007 x (C) logx 4 0007 00022
x
00025
f
6. Let P(x) 0007 x3 0003 x2 0002 10x 0003 8. Find all rational zeros for P(x).
8. Simplify.
5
mn
4. Let P(x) 0007 2(x 0003 2)(x 0002 3)(x 0002 5). What are the zeros of P(x)?
12. F is jointly proportional to q1 and q2 and inversely proportional to the square of r.
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13. Explain why the graph in the figure is not the graph of a polynomial function. y
26. Final all zeros (rational, irrational, and imaginary) exactly for P(x) 0007 x4 0003 5x3 0003 x2 0002 15x 0002 12, and factor P(x) into linear factors.
5
00025
25. Find all zeros (rational, irrational, and imaginary) exactly for P(x) 0007 4x3 0002 20x2 0003 29x 0002 15.
5
x
In Problems 27–36, solve for x exactly. Do not use a calculator or a table. 27. 2x 0007 4x00034
28. 2x2e0002x 0003 xe0002x 0007 e0002x
29. eln x 0007 2.5
30. logx 104 0007 4
2
00025
14. Explain why the graph in the figure is not the graph of a rational function. 15. The function f subtracts the square root of the domain element from three times the natural log of the domain element. Write an algebraic definition of f. 16. Write a verbal description of the function f(x) 0007 100e0.5x 0002 50. 2x 0003 8 . x00032 (A) Find the domain and the intercepts for f. (B) Find the vertical and horizontal asymptotes for f. (C) Sketch the graph of f. Draw vertical and horizontal asymptotes with dashed lines.
17. Let f (x) 0007
18. Find all zeros of P(x) 0007 (x3 0003 4x)(x 0003 4), and specify those zeros that are x intercepts.
31. log9 x 0007 000232 32. ln (x 0003 4) 0002 ln (x 0002 4) 0007 2 ln 3 33. ln (2x2 0003 2) 0007 2 ln (2x 0002 4) 34. log x 0003 log (x 0003 15) 0007 2 35. log (ln x) 0007 00021
36. 4 (ln x)2 0007 ln x2
In Problems 37–41, solve for x to three significant digits. 37. x 0007 log3 41 x
39. 4(2 ) 0007 20 41.
38. ln x 0007 1.45 40. 10e00020.5x 0007 1.6
1 e x 0002 e0002x 0007 e x 0003 e0002x 2
19. Solve (x3 0003 4x)(x 0003 4) 0004 0.
42. G is directly proportional to the square of x. If G 0007 10 when x 0007 5, find G when x 0007 7.
20. If P(x) 0007 2x3 0002 5x2 0003 3x 0003 2, find P(12) using the remainder theorem and synthetic division.
43. H varies inversely as the cube of r. If H 0007 162 when r 0007 2, find H when r 0007 3.
21. Which of the following is a factor of P(x)? P(x) 0007 x25 0002 x20 0003 x15 0003 x10 0002 x5 0003 1 (A) x 0002 1
(B) x 0003 1 4
2
In Problems 44–50, find the domain, range, and the equations of any horizontal or vertical asymptotes. 44. f(x) 0007 3 0003 2x
22. Let P(x) 0007 x 0002 8x 0003 3. (A) Graph P(x) and describe the graph verbally, including the number of x intercepts, the number of turning points, and the left and right behavior. (B) Approximate the largest x intercept to two decimal places.
45. f(x) 0007 2 0002 log3 (x 0002 1)
23. Let P(x) 0007 x5 0002 8x4 0003 17x3 0003 2x2 0002 20x 0002 8. (A) Approximate the zeros of P(x) to two decimal places and state the multiplicity of each zero. (B) Can any of these zeros be approximated with the bisection method? The MAXIMUM or MINIMUM commands? Explain.
48. f (x) 0007
4
3
2
24. Let P(x) 0007 x 0003 2x 0002 20x 0002 30. (A) Find the smallest positive and largest negative integers that, by Theorem 1 in Section 4-2, are upper and lower bounds, respectively, for the real zeros of P(x). (B) If (k, k 0003 1), k an integer, is the interval containing the largest real zero of P(x), determine how many additional intervals are required in the bisection method to approximate this zero to one decimal place. (C) Approximate the real zeros of P(x) to two decimal places.
46. f(x) 0007 5 0002 4x3 47. f(x) 0007 3 0003 2x4 5 x00033
49. f(x) 0007 20e0002x 0002 15 50. f (x) 0007 8 0003 ln (x 0003 2) 51. If the graph of y 0007 ln x is reflected in the line y 0007 x, the graph of the function y 0007 e x is obtained. Discuss the functions that are obtained by reflecting the graph of y 0007 ln x in the x axis and in the y axis. 52. (A) Explain why the equation e0002x 0007 ln x has exactly one solution. (B) Approximate the solution of the equation to two decimal places.
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In Problems 53 and 54, factor each polynomial in two ways: (A) As a product of linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros). (B) As a product of linear factors with complex coefficients.
A-7
67. Solve (to three decimal places) 4x 6 3 x 00021 2
53. P(x) 0007 x4 0003 9x2 0003 18 54. P(x) 0007 x4 0002 23x2 0002 50
APPLICATIONS
55. Graph f and indicate any horizontal, vertical, or oblique asymptotes with dashed lines:
68. SHIPPING A mailing service provides customers with rectangular shipping containers. The length plus the girth of one of these containers is 10 feet (see the figure). If the end of the container is square and the volume is 8 cubic feet, find the side length of the end. Find solutions exactly; round irrational solutions to two decimal places.
f (x) 0007
x2 0003 4x 0003 8 x00032
56. Let P(x) 0007 x4 0002 28x3 0003 262x2 0002 922x 0003 1.083. Approximate (to two decimal places) the x intercepts and the local extrema. 57. Find a polynomial of lowest degree with leading coefficient 1 that has zeros 00021 (multiplicity 2), 0 (multiplicity 3), 3 0003 5i, and 3 0002 5i. Leave the answer in factored form. What is the degree of the polynomial? 58. If P(x) is a fourth-degree polynomial with integer coefficients and if i is a zero of P(x), can P(x) have any irrational zeros? Explain. 4
3
2
59. Let P(x) 0007 x 0003 9x 0002 500x 0003 20,000. (A) Find the smallest positive integer multiple of 10 and the largest negative integer multiple of 10 that, by Theorem 1 in Section 4-2, are upper and lower bounds, respectively, for the real zeros of P(x). (B) Approximate the real zeros of P(x) to two decimal places. 60. Find all zeros (rational, irrational, and imaginary) exactly for P(x) 0007 x5 0002 4x4 0003 3x3 0003 10x2 0002 10x 0002 12 and factor P(x) into linear factors. 61. Find rational roots exactly and irrational roots to two decimal places for P(x) 0007 x5 0003 4x4 0003 x3 0002 11x2 0002 8x 0003 4 62. Give an example of a rational function f(x) that satisfies the following conditions: the real zeros of f are 5 and 8; x 0007 1 is the only vertical asymptote; and the line y 0007 3 is a horizontal asymptote. 63. Use natural logarithms to solve for n. A0007P
(1 0003 i)n 0002 1 i
64. Solve ln y 0007 5x 0003 ln A for y. Express the answer in a form that is free of logarithms. 65. Solve for x. y0007 66. Solve
x3 0002 x 0006 0. x3 0002 8
ex 0002 2e 0002x 2
gth
Len
x
Girth
x
y
69. GEOMETRY The diagonal of a rectangle is 2 feet longer than one of the sides, and the area of the rectangle is 6 square feet. Find the dimensions of the rectangle to two decimal places. 70. POPULATION GROWTH If the Democratic Republic of the Congo has a population of about 60 million people and a doubling time of 23 years, find the population in (A) 5 years (B) 30 years Compute answers to three significant digits. 71. COMPOUND INTEREST How long will it take money invested in an account earning 7% compounded annually to double? Use the annual compounding growth model P 0007 P0(1 0003 r)t, and compute the answer to three significant digits. 72. COMPOUND INTEREST Repeat Problem 71 using the continuous compound interest model P 0007 P0ert. 73. EARTHQUAKES If the 1906 and 1989 San Francisco earthquakes registered 8.3 and 7.1, respectively, on the Richter scale, how many times more powerful was the 1906 earthquake than the 1989 earthquake? Use the formula M 0007 23 log (E000bE0), where E 0 0007 104.40 joules, and compute the answer to one decimal place. 74. SOUND If the decibel level at a rock concert is 88, find the intensity of the sound at the concert. Use the formula D 0007 10 log (I兾I0), where I0 0007 10000212 watts per square meter, and compute the answer to two significant digits. 75. ASTRONOMY The square of the time t required for a planet to make one orbit around the sun varies directly as the cube of its mean (average) distance d from the sun. Write the equation of variation, using k as the constant of variation.
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76. PHYSICS Atoms and molecules that make up the air constantly fly about like microscopic missiles. The velocity v of a particular particle at a fixed temperature varies inversely as the square root of its molecular weight w. If an oxygen molecule in air at room temperature has an average velocity of 0.3 mile/second, what will be the average velocity of a hydrogen molecule, given that the hydrogen molecule is one-sixteenth as heavy as the oxygen molecule? Problems 77 and 78 require a graphing calculator or a computer that can calculate linear, quadratic, cubic, and exponential regression models for a given data set. 77. Table 1 shows the life expectancy (in years) at birth for residents of the United States from 1970 to 1995. Let x represent years since 1970. Use the indicated regression model to estimate the life expectancy (to the nearest tenth of a year) for a U.S. resident born in 2010. (A) Linear regression (B) Quadratic regression (C) Cubic regression (D) Exponential regression
CHAPTERS
6–8
Table 1 Year
Life Expectancy
1970
70.8
1975
72.6
1980
73.7
1985
74.7
1990
75.4
1995
75.9
2000
77.0
2005
77.7
Source: U.S. Census Bureau
78. Refer to Problem 77. The Census Bureau projected the life expectancy for a U.S. resident born in 2010 to be 77.9 years. Which of the models in Problem 77 is closest to the Census Bureau projection?
Cumulative Review Exercises
Work through all the problems in this cumulative review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. Note that Problems 4, 15, 16, 40, 41, 48, 49, and 88 are from sections that appear online. 1. Solve using substitution or elimination by addition: 3x 0002 5y 0007 11 2x 0003 3y 0007 1 2. Solve by graphing: 2x 0002 y 0007 00024 3x 0003 y 0007 00021 3. Solve by substitution or elimination by addition: 00026x 0003 3y 0007 2 2x 0002 y 0007 1 4. Solve by graphing: 3x 0003 5y 0004 15 x, y 0006 0 5. Determine whether each of the following can be the first three terms of an arithmetic sequence, a geometric sequence, or neither. (A) 20, 15, 10, . . . (B) 5, 25, 125, . . . (C) 5, 25, 50, . . . (D) 27, 00029, 3, . . . (E) 00029, 00026, 00023, . . .
In Problems 6–8: (A) Write the first four terms of each sequence. (B) Find a8. (C) Find S8. 6. an 0007 2 ⴢ 5n
7. an 0007 3n 0002 1
8. a1 0007 100; an 0007 an00021 0002 6, n 0006 2 9. Evaluate each of the following: 32! 9! (A) 8! (B) (C) 30! 3!(9 0002 3)! 10. Evaluate each of the following: 7 (A) a b (B) C7,2 (C) P7,2 2 In Problems 11–13, graph each equation and locate foci. Locate the directrix for any parabolas. Find the lengths of major, minor, transverse, and conjugate axes where applicable. 11. 25x2 0002 36y2 0007 900
12. 25x2 0003 36y2 0007 900
13. 25x2 0002 36y 0007 0 14. Find each determinant: 00023 5 5 (A) ` (B) ` ` 2 00022 00025 15. Solve x2 0003 y2 0007 2 2x 0002 y 0007 1
3 ` 00023
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16. Find the maximum and minimum value of z 0007 2x 0003 3y over the feasible region S:
(0, 10) (6, 7) S
x1 0003 3x2 0007 10 (5, 0)
2x1 0002 x2 0007 00021
x
5
17. Perform the operations that are defined, given the following matrices: M0007 c
2 1
1 d 00023
P 0007 [1 2] (B) P 0003 Q (E) PN
N0007 c
1 00021
2 d 3
00021 d 2 (C) PQ (F) QM
19. How many ways can four distinct books be arranged on a shelf? Solve (A) By using the multiplication principle (B) By using permutations or combinations, whichever is applicable 20. In a single deal of 3 cards from a standard 52-card deck, what is the probability of being dealt three diamonds? 21. Each of the 10 digits 0 through 9 is printed on 1 of 10 different cards. Four of these cards are drawn in succession without replacement. What is the probability of drawing the digits 4, 5, 6, and 7 by drawing 4 on the first draw, 5 on the second draw, 6 on the third draw, and 7 on the fourth draw? What is the probability of drawing the digits 4, 5, 6, and 7 in any order? 22. A thumbtack lands point down in 38 out of 100 tosses. What is the approximate empirical probability of the tack landing point up? 23. Write the linear system corresponding to each augmented matrix and solve: 1 00022 3 1 0 3 (A) c (B) c ` d ` d 0 0 0 0 1 00024 1 0
00022 3 ` d 0 1
24. Given the system:
Then write the linear system represented by each augmented matrix in your solution, and solve each of these systems graphically. Discuss the relationship between the solutions of these systems. 27. Solve graphically to two decimal places: 00022x 0003 3y 0007 7
Q0007 c
18. A coin is flipped three times. How many combined outcomes are possible? Solve (A) By using a tree diagram (B) By using the multiplication principle
(C) c
2x1 0002 5x2 0007 k2
26. Use Gauss–Jordan elimination to solve the system
(0, 4)
(A) M 0002 2N (D) MN
25. Given the system: x1 0002 3x2 0007 k1 (A) Write the system as a matrix equation of the form AX 0007 B. (B) Find the inverse of the coefficient matrix A. (C) Use A00021 to find the solution for k1 0007 00022 and k2 0007 1. (D) Use A00021 to find the solution for k1 0007 1 and k2 0007 00022.
y
5
A-9
x1 0003 x2 0007 3 0002x1 0003 x2 0007 5
(A) Write the augmented matrix for the system. (B) Transform the augmented matrix into reduced form. (C) Write the solution to the system.
3x 0003 4y 0007 18 Verify the statement Pn in Problems 28 and 29 for n 0007 1, 2, and 3. 28. Pn: 1 0003 5 0003 9 0003 . . . 0003 (4n 0002 3) 0007 n(2n 0002 1) 29. Pn: n2 0003 n 0003 2 is divisible by 2 In Problems 30 and 31, write Pk and Pk 00031. 30. For Pn in Problem 28
31. For Pn in Problem 29
32. Find the equation of the parabola having its vertex at the origin, its axis the y axis, and (2, 00028) on its graph. 33. Find an equation of an ellipse in the form y2 x2 0003 00071 M N
M, N 7 0
if the center is at the origin, the major axis is the x axis, the major axis length is 10, and the distance of the foci from the center is 3. 34. Find an equation of a hyperbola in the form y2 x2 0002 00071 M N
M, N 7 0
if the center is at the origin, the transverse axis length is 16, and the distance of the foci from the center is 189. Solve Problems 35–37 using Gauss–Jordan elimination. 35. x1 0003 2x2 0002 x3 0007
3
36. x1 0003 x2 0002 x3 0007 2
x2 0003 x3 0007 00022
4x2 0003 6x3 0007 00021
2x1 0003 3x2 0003 x3 0007
0
37. x1 0002 2x2 0003 x3 0007
1
6x2 0003 9x3 0007 0
3x1 0002 2x2 0002 x3 0007 00025 1 38. Given M 0007 [1 2 00021] and N 0007 £ 00021 § . Find: 2 (A) MN (B) NM
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39. Given L0007 c
1 M 0007 £ 00021 1
00021 0 d 2 1
2 1
Find, if defined: (A) LM 0002 2N
2 0§ 1
centers are selected? Express the answer in terms of Cn,r or Pn,r, as appropriate, and evaluate. N0007 c
2 00021
1 d 0
(B) ML 0003 N
In Problems 40 and 41, solve the system. 40. x2 0002 3xy 0003 3y2 0007 1
41. x2 0002 3xy 0003 y2 0007 00021 x2 0002 xy 0007 0
xy 0007 1
In Problems 42 and 43, find the determinant. 1 42. 3 2 3
0 5 0
00024 43. 3 3 2
4 00021 3 00026
5 00022 4
00026 00021 3 6
44. Find all real solutions to two decimal places x2 0003 2xy 0002 y2 0007 1 9x2 0003 4xy 0003 y2 0007 15
53. A single die is rolled 1,000 times with the frequencies of outcomes shown in the table. (A) What is the approximate empirical probability that the number of dots showing is divisible by 3? (B) What is the theoretical probability that the number of dots showing is divisible by 3? Number of dots facing up Frequency
1
2
3
4
5
6
160
155
195
180
140
170
54. Let an 0007 100(0.9)n and bn 0007 10 0003 0.03n. Find the least positive integer n such that an 0005 bn by graphing the sequences {an} and {bn} with a graphing calculator. Check your answer by using a graphing calculator to display both sequences in table form. 55. Evaluate each of the following: (A) P25,5
(C) a
(B) C(25, 5)
5
45. Write a kk without summation notation and find the sum. k00071
2 22 23 24 25 26 0002 0003 0002 0003 0002 using sum2! 3! 4! 5! 6! 7! mation notation with the summation index k starting at k 0007 1.
46. Write the series
47. Find S for the geometric series 108 0002 36 0003 12 0002 4 0003 . . .. 48. Graph the solution region and indicate whether the solution region is bounded or unbounded. Find the coordinates of each corner point. 3x 0003 2y 0006 12 x 0003 2y 0006 8 x, y 0006 0 49. Solve the linear programming problem: Maximize
z 0007 4x 0003 9y
Subject to
x 0003 2y 0004 14
56. Expand (a 0003 12b)6 using the binomial formula. 57. Find the fifth and the eighth terms in the expansion of (3x 0002 y)10. Prove each statement in Problems 58 and 59 for all positive integers using mathematical induction. 58. Pn in Problem 28
61. Use the formula for the sum of an infinite geometric series to write 2.45 0007 2.454 545 . . . as the quotient of two integers. 30 b (0.1)300002k(0.9)k for k 0007 0, 1, . . ., 30. Use a k graphing calculator to find the largest term of the sequence {ak} and the number of terms that are greater than 0.01.
62. Let ak 0007 a
63. Use Cramer’s rule to solve the system for x only: 00022x
2x1 0003 5x2 0003 2x3 0007 k3 (A) Write the system as a matrix equation of the form AX 0007 B. (B) Find the inverse of the coefficient matrix A. (C) Use A00021 to solve the system when k1 0007 00021, k2 0007 2, and k3 0007 1. (D) Use A00021 to solve the system when k1 0007 2, k2 0007 0, and k3 0007 00021. 51. How many four-letter code words are possible using the first six letters of the alphabet if no letter can be repeated? If letters can be repeated? If adjacent letters cannot be alike? 52. A basketball team with 12 members has two centers. If 5 players are selected at random, what is the probability that both
0003 3z 0007 000213
x 0002 6y 0003 5z 0007 000216
x, y 0006 0 2x1 0003 6x2 0003 3x3 0007 k2
59. Pn in Problem 29
60. Find the sum of all the odd integers between 50 and 500.
2x 0003 y 0004 16 50. Given the system: x1 0003 4x2 0003 2x3 0007 k1
25 b 20
0002x 0003 2y 0007 00021 64. Use Cramer’s rule to solve the system in Problem 63 for y. 65. Use Cramer’s rule to solve the system in Problem 63 for z. 66. How many nine-digit zip codes are possible? How many of these have no repeated digits? 67. Use mathematical induction to prove that the following statement holds for all positive integers: Pn:
1 1 1 0003 0003 0003... 1ⴢ3 3ⴢ5 5ⴢ7 0003
1 n 0007 (2n 0002 1)(2n 0003 1) 2n 0003 1
68. Three-digit numbers are randomly formed from the digits 1, 2, 3, 4, and 5. What is the probability of forming an even number if digits cannot be repeated? If digits can be repeated?
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69. Discuss the number of solutions for the system corresponding to the reduced form shown below if (A) m 0007 0 and n 0007 0 (B) m 0007 0 and n 0 (C) m 0 1 £0 0
0 1 0
00025 2 3 † 6§ m n
A-11
84. Use mathematical induction to show that {an} 0007 {bn}, where a1 0007 3, an 0007 2an00021 0002 1 for n 1, and bn 0007 2n 0003 1, n 0006 1. 85. Find an equation of the set of points in the plane each of whose distance from (1, 4) is three times its distance from the x axis. Write the equation in the form Ax2 0003 Cy2 0003 Dx 0003 Ey 0003 F 0007 0, and identify the curve.
70. If a square matrix A satisfies the equation A2 0007 A, find A. Assume that A00021 exists.
86. A box of 12 lightbulbs contains 4 defective bulbs. If three bulbs are selected at random, what is the probability of selecting at least one defective bulb?
71. Which of the following augmented matrices are in reduced form?
APPLICATIONS
1 L 0007 £0 0
0 1 0
0 2 0 † 0§ 1 00021
0 N 0007 £1 0
0 0 0 † 2§ 1 00023
1 M 0007 £0 0
0 1 0
3 3 00022 † 2 § 0 0
1 0
0 1
2 00022 ` d 1 3
P0007 c
2 0
Recall that a square matrix is called upper triangular if all elements below the principal diagonal are zero, and it is called diagonal if all elements not on the principal diagonal are zero. A square matrix is called lower triangular if all elements above the principal diagonal are zero. In Problems 72–77, determine whether the statement is true or false. If true, explain why. If false, give a counterexample. 72. The sum of two upper triangular matrices is upper triangular.
87. ECONOMICS The government, through a subsidy program, distributes $2,000,000. If we assume that each individual or agency spends 75% of what it receives, and 75% of this is spent, and so on, how much total increase in spending results from this government action? 88. GEOMETRY Find the dimensions of a rectangle with perimeter 24 meters and area 32 square meters. 89. ENGINEERING An automobile headlight contains a parabolic reflector with a diameter of 8 inches. If the light source is located at the focus, which is 1 inch from the vertex, how deep is the reflector? 90. ARCHITECTURE A sound whispered at one focus of a whispering chamber can be easily heard at the other focus. Suppose that a cross section of this chamber is a semielliptical arch that is 80 feet wide and 24 feet high (see the figure). How far is each focus from the center of the arch? How high is the arch above each focus?
73. The product of two lower triangular matrices is lower triangular. 74. The sum of an upper triangular matrix and a lower triangular matrix is a diagonal matrix.
24 feet
75. The product of an upper triangular matrix and a lower triangular matrix is a diagonal matrix. 76. A matrix that is both upper triangular and lower triangular is a diagonal matrix. 77. If a diagonal matrix has no zero elements on the principal diagonal, then it has an inverse. 78. Use the binomial formula to expand (x 0002 2i)6, where i is the imaginary unit. 79. Use the definition of a parabola and the distance formula to find the equation of a parabola with directrix y 0007 3 and focus (6, 1). 80. An ellipse has vertices (4, 0) and foci (2, 0). Find the y intercepts. 81. A hyperbola has vertices (2, 3) and foci (2, 5). Find the length of the conjugate axis. 82. Seven distinct points are selected on the circumference of a circle. How many triangles can be formed using these seven points as vertices? 83. Use mathematical induction to prove that 2n 0005 n! for all integers n 3.
80 feet
91. FINANCE An investor has $12,000 to invest. If part is invested at 8% and the rest in a higher-risk investment at 14%, how much should be invested at each rate to produce the same yield as if all had been invested at 10%? 92. DIET In an experiment involving mice, a zoologist needs a food mix that contains, among other things, 23 grams of protein, 6.2 grams of fat, and 16 grams of moisture. She has on hand mixes of the following compositions: Mix A contains 20% protein, 2% fat, and 15% moisture, mix B contains 10% protein, 6% fat, and 10% moisture; and mix C contains 15% protein, 5% fat, and 5% moisture. How many grams of each mix should be used to get the desired diet mix? 93. PURCHASING A soft-drink distributor has budgeted $300,000 for the purchase of 12 new delivery trucks. If a model A truck costs $18,000, a model B truck costs $22,000, and a model C truck costs $30,000, how many trucks of each model should the distributor purchase to use exactly all the budgeted funds?
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94. MANUFACTURING A manufacturer makes two types of day packs, a standard model and a deluxe model. Each standard model requires 0.5 labor-hour from the fabricating department and 0.3 labor-hour from the sewing department. Each deluxe model requires 0.5 labor-hour from the fabricating department and 0.6 labor-hour from the sewing department. The maximum number of labor-hours available per week in the fabricating department and the sewing department are 300 and 240, respectively. (A) If the profit on a standard day pack is $8 and the profit on a deluxe day pack is $12, how many of each type of pack should be manufactured each day to realize a maximum profit? What is the maximum profit? (B) Discuss the effect on the production schedule and the maximum profit if the profit on a standard day pack decreases by $3 and the profit on a deluxe day pack increases by $3. (C) Discuss the effect on the production schedule and the maximum profit if the profit on a standard day pack increases by $3 and the profit on a deluxe day pack decreases by $3. 95. AVERAGING TESTS A teacher has given four tests to a class of five students and stored the results in the following matrix:
Ann Bob Carol Dan Eric
1 78 91 G 95 75 83
Tests 2 3 84 81 65 84 90 92 82 87 88 81
4 86 92 91 W 0007 M 91 76
Discuss methods of matrix multiplication that the teacher can use to obtain the indicated information in parts A–C. In each case, state the matrices to be used and then perform the necessary multiplications. (A) The average on all four tests for each student, assuming that all four tests are given equal weight (B) The average on all four tests for each student, assuming that the first three tests are given equal weight and the fourth is given twice this weight (C) The class average on each of the four tests 96. POLITICAL SCIENCE A random survey of 1,000 residents in a state produced the following results: Party Affiliation Age
Democrat
Republican
Independent
Under 30
130
80
40
250
30–39
120
90
20
230
40–49
70
80
20
170
50–59
50
60
10
120
Over 59
90
110
30
230
460
420
120
1,000
Totals
Totals
Find the empirical probability that a person selected at random: (A) Is under 30 and a Democrat (B) Is under 40 and a Republican (C) Is over 59 or is an Independent
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APPENDIX
Special Topics
11 B OUTLINE B-1
Scientific Notation and Significant Digits
B-2
Partial Fractions
B-3
Parametric Equations
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SPECIAL TOPICS
Scientific Notation and Significant Digits Z Significant Digits Z Rounding Convention
Z Significant Digits Most calculations involving problems in the real world deal with numbers that are only approximate. It therefore seems reasonable to assume that a final answer should not be any more accurate than the least accurate number used in the calculation. This is an important point, because calculators tend to give the impression that greater accuracy is achieved than is warranted. Suppose we want to compute the length of the diagonal of a rectangular field from measurements of its sides of 237.8 meters and 61.3 meters. Using the Pythagorean theorem and a calculator, we find d 0002 2237.82 0003 61.32
d
0002 245.573 878 . . .
61.3 meters
237.8 meters
The calculator answer suggests an accuracy that is not justified. What accuracy is justified? To answer this question, we introduce the idea of significant digits. Whenever we write a measurement such as 61.3 meters, we assume that the measurement is accurate to the last digit written. So the measurement 61.3 meters indicates that the measurement was made to the nearest tenth of a meter. That is, the actual width is between 61.25 meters and 61.35 meters. In general, the digits in a number that indicate the accuracy of the number are called significant digits. If all the digits in a number are nonzero, then they are all significant. So the measurement 61.3 meters has three significant digits, and the measurement 237.8 meters has four significant digits. What are the significant digits in the number 7,800? The accuracy of this number is not clear. It could represent a measurement with any of the following accuracies: Between 7,750 and 7,850 Between 7,795 and 7,805 Between 7,799.5 and 7,800.5
Correct to the hundreds place Correct to the tens place Correct to the units place
To give a precise definition of significant digits that resolves this ambiguity, we use scientific notation.
Z DEFINITION 1 Significant Digits If a number x is written in scientific notation as x 0002 a 0004 10n
1 0005 a 0006 10, n an integer
then the number of significant digits in x is the number of digits in a.
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SECTION B–1
Scientific Notation and Significant Digits
A-15
Using this definition, 7.8 0004 103 7.80 0004 103 7.800 0004 103
has two significant digits has three significant digits has four significant digits
All three of these measurements have the same decimal representation (7,800), but each represents a different accuracy. Definition 1 tells us how to write a number so that the number of significant digits is clear, but it does not tell us how to interpret the accuracy of a number that is not written in scientific notation. We will use the following convention for numbers that are written as decimal fractions:
Z SIGNIFICANT DIGITS IN DECIMAL FRACTIONS The number of significant digits in a number with no decimal point is found by counting the digits from left to right, starting with the first digit and ending with the last nonzero digit. The number of significant digits in a number containing a decimal point is found by counting the digits from left to right, starting with the first nonzero digit and ending with the last digit.
Applying this rule to the number 7,800, we conclude that this number has two significant digits. If we want to indicate that it has three or four significant digits, we must use scientific notation.
EXAMPLE
1
Significant Digits in Decimal Fractions Underline the significant digits in the following numbers:
SOLUTIONS
MATCHED PROBLEM 1
(A) 70,007
(B) 82,000
(C) 5.600
(D) 0.0008
(E) 0.000 830
(A) 70,007
(B) 82,000
(C) 5.600
(D) 0.0008
(E) 0.000 830
0002
Underline the significant digits in the following numbers: (A) 5,009
(B) 12,300
(C) 23.4000
(D) 0.00050
(E) 0.0012 0002
Z Rounding Convention In calculations involving multiplication, division, powers, and roots, we adopt the following convention:
Z ROUNDING CALCULATED VALUES The result of a calculation is rounded to the same number of significant digits as the number used in the calculation that has the least number of significant digits.
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SPECIAL TOPICS
So, in computing the length of the diagonal of the rectangular field shown earlier, we write the answer rounded to three significant digits because the width has three significant digits and the length has four significant digits: d 0002 246 meters
Three significant digits
One Final Note: In rounding a number that is exactly halfway between a larger and a smaller number, we use the convention of making the final result even.
EXAMPLE
2
Rounding Numbers Round each number to three significant digits. (A) 43.0690
SOLUTIONS
(B) 48.05
(D) 8.017 632 0004 1000073
(C) 48.15
(A) 43.1 (B) 48.0 ⎫ ⎬ (C) 48.2 ⎭
Use the convention of making the digit before the 5 even if it is odd, or leaving it alone if it is even.
(D) 8.02 0004 1000073 MATCHED PROBLEM 2
0002
Round each number to three significant digits. (A) 3.1495
(B) 0.004 135
(C) 32,450
(D) 4.314 764 09 0004 1012
0002
ANSWERS TO MATCHED PROBLEMS 1. (A) 5,009 2. (A) 3.15
B-1
(B) 12,300 (B) 0.004 14
(C) 23.4000 (C) 32,400
(D) 0.00050 (E) 0.0012 (D) 4.31 0004 1012
Exercises
In Problems 1–12, underline the significant digits in each number. 1. 123,005
2. 3,400,002
3. 20,040
4. 300,600
5. 6.0
6. 7.00
7. 80.000
8. 900.0000
9. 0.012
10. 0.0015
11. 0.000 960
12. 0.000 700
In Problems 13–22, round each number to three significant digits. 13. 3.0780
14. 4.0240
15. 924,300
16. 643,820
17. 23.65
18. 23.75
19. 2.816 743 0004 103 20. 56.114 0004 104 21. 6.782 045 0004 1000074 22. 5.248 102 0004 1000073 In Problems 23 and 24, find the diagonal of the rectangle with the indicated side measurements. Round answers to the number of significant digits appropriate for the given measurements. 23. 25 feet by 20 feet 24. 2,900 yards by 1,570 yards
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SECTION B–2
B-2
Partial Fractions
A-17
Partial Fractions Z Basic Theorems Z Partial Fraction Decomposition
You have now had considerable experience combining two or more rational expressions into a single rational expression. For example, problems such as 2(x 0007 4) 0003 3(x 0003 5) 2 3 5x 0003 7 0003 0002 0002 x00035 x00074 (x 0003 5)(x 0007 4) (x 0003 5)(x 0007 4) should seem routine. Frequently in more advanced courses, particularly in calculus, it is useful to be able to reverse this process—that is, to be able to express a rational expression as the sum of two or more simpler rational expressions called partial fractions. As is often the case with reverse processes, the process of decomposing a rational expression into partial fractions is more difficult than combining rational expressions. Basic to the process is the factoring of polynomials, so many of the topics discussed in Chapter 4 can be put to effective use. Partial fraction decomposition is usually accomplished by solving a related system of linear equations. If you are familiar with basic techniques for solving linear systems discussed earlier in this book, such as Gauss–Jordan elimination, inverse matrix solutions, or Cramer’s rule, you may use these as you see fit. However, all of the linear systems encountered in this section can also be solved by some special techniques developed here. Mathematically equivalent to the techniques mentioned, these special techniques are generally easier to use in partial fraction decomposition problems. We confine our attention to rational expressions of the form P(x)兾D(x), where P(x) and D(x) are polynomials with real coefficients. In addition, we assume that the degree of P(x) is less than the degree of D(x). If the degree of P(x) is greater than or equal to that of D(x), we have only to divide P(x) by D(x) to obtain R(x) P(x) 0002 Q(x) 0003 D(x) D(x) where the degree of R(x) is less than that of D(x). For example, 00076x 0003 2 x4 0007 3x3 0003 2x2 0007 5x 0003 1 0002 x2 0007 x 0007 1 0003 2 2 x 0007 2x 0003 1 x 0007 2x 0003 1 If the degree of P(x) is less than that of D(x), then P(x)兾D(x) is called a proper fraction.
Z Basic Theorems Our task now is to establish a systematic way to decompose a proper fraction into the sum of two or more partial fractions. Theorems 1, 2, and 3 take care of the problem completely. Z THEOREM 1 Equal Polynomials Two polynomials are equal to each other if and only if the coefficients of terms of like degree are equal.
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SPECIAL TOPICS
For example, if Equate the constant terms.
⎫ ⎪ ⎬ ⎪ ⎭
(A 0003 2B)x 0003 B 0002 5x 0007 3 Equate the coefficients of x.
then B 0002 00073 A 0003 2B 0002 5 A 0003 2(ⴚ3) 0002 5 A 0002 11 ZZZ EXPLORE-DISCUSS 1
Substitute B ⴝ ⴚ3 into the second equation to solve for A.
If x 0003 5 0002 A(x 0003 1) 0003 B(x 0007 3)
(1)
is a polynomial identity (that is, both sides represent the same polynomial), then equating coefficients produces the system 10002A0003B 5 0002 A 0007 3B
Equating coefficients of x Equating constant terms
(A) Solve this system graphically. (B) For an alternate method of solution, substitute x 0002 3 in equation (1) to find A and then substitute x 0002 00071 in equation (1) to find B. Explain why this method is valid.
The Linear and Quadratic Factors Theorem from Chapter 4 (page 290) underlies the technique of decomposing a rational function into partial fractions. We restate the theorem here. Z THEOREM 2 Linear and Quadratic Factors Theorem For a polynomial of degree n 0 with real coefficients, there always exists a factorization involving only linear and/or quadratic factors with real coefficients in which the quadratic factors have imaginary zeros.
The quadratic formula can be used to determine easily whether a given quadratic factor ax2 0003 bx 0003 c, with real coefficients, has imaginary zeros. If b2 0007 4ac 0006 0, then ax2 0003 bx 0003 c has imaginary zeros. Otherwise its zeros are real. Therefore, ax2 0003 bx 0003 c has imaginary zeros if and only if it cannot be factored as a product of linear factors with real coefficients.
Z Partial Fraction Decomposition We are now ready to state Theorem 3, which forms the basis for partial fraction decomposition.
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SECTION B–2
Partial Fractions
A-19
Z THEOREM 3 Partial Fraction Decomposition Any proper fraction P(x)兾D(x) reduced to lowest terms can be decomposed into the sum of partial fractions as follows: 1. If D(x) has a nonrepeating linear factor of the form ax 0003 b, then the partial fraction decomposition of P(x)兾D(x) contains a term of the form A ax 0003 b
A a constant
2. If D(x) has a k-repeating linear factor of the form (ax 0003 b)k, then the partial fraction decomposition of P(x)兾D(x) contains k terms of the form Ak A1 A2 0003 0003...0003 2 ax 0003 b (ax 0003 b) (ax 0003 b)k
A1, A2, . . . , Ak constants
3. If D(x) has a nonrepeating quadratic factor of the form ax2 0003 bx 0003 c that has imaginary zeros, then the partial fraction decomposition of P(x)兾D(x) contains a term of the form Ax 0003 B ax 0003 bx 0003 c 2
A, B constants
4. If D(x) has a k-repeating quadratic factor of the form (ax2 0003 bx 0003 c)k, where ax2 0003 bx 0003 c has imaginary zeros, then the partial fraction decomposition of P(x)兾D(x) contains k terms of the form Ak x 0003 Bk A1x 0003 B1 A2x 0003 B2 0003...0003 0003 ax2 0003 bx 0003 c (ax2 0003 bx 0003 c)2 (ax2 0003 bx 0003 c)k A1, . . . , Ak, B1, . . . , Bk constants
Let’s see how the theorem is used to obtain partial fraction decompositions in several examples.
EXAMPLE
1
Nonrepeating Linear Factors Decompose into partial fractions:
SOLUTION
5x 0003 7 . x 0003 2x 0007 3 2
We first try to factor the denominator. If it can’t be factored in the real numbers, then we can’t go any further. In this example, the denominator factors, so we apply part 1 from Theorem 3: 5x 0003 7 A B 0002 0003 (x 0007 1)(x 0003 3) x00071 x00033
(2)
To find the constants A and B, we combine the fractions on the right side of equation (2) to obtain A(x 0003 3) 0003 B(x 0007 1) 5x 0003 7 0002 (x 0007 1)(x 0003 3) (x 0007 1)(x 0003 3) Because these fractions have the same denominator, their numerators must be equal. So 5x 0003 7 0002 A(x 0003 3) 0003 B(x 0007 1)
(3)
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SPECIAL TOPICS
We could multiply the right side and find A and B by using Theorem 1, but in this case it is easier to take advantage of the fact that equation (3) is an identity—that is, it must hold for all values of x. In particular, we note that if we let x 0002 1, then the second term of the right side drops out and we can solve for A: 5 ⴢ 1 0003 7 0002 A(1 0003 3) 0003 B(1 0007 1) 12 0002 4A A00023 Similarly, if we let x 0002 00073, the first term drops out and we find 00078 0002 00074B B00022 Now we have the decomposition: 3 2 5x 0003 7 0002 0003 x 0007 1 x 0003 3 x 0003 2x 0007 3 2
as can easily be checked by adding the two fractions on the right.
MATCHED PROBLEM 1
Decompose into partial fractions:
7x 0003 6 . x 0003x00076
0002
(Fig. 1). Discuss how the TRACE command on the graphing calculator can be used to check that the graphing calculator is displaying two identical graphs. 10
000710
10
000710
Z Figure 1
EXAMPLE
2
Repeating Linear Factors Decompose into partial fractions:
SOLUTION
0002
2
Technology Connections A graphing calculator can also be used to check a partial fraction decomposition. To check Example 1, we graph the left and right sides of equation (4) in a graphing calculator
(4)
6x2 0007 14x 0007 27 . (x 0003 2)(x 0007 3)2
Using parts 1 and 2 from Theorem 3, we write 6x2 0007 14x 0007 27 A B C 0002 0003 0003 x00032 x00073 (x 0003 2)(x 0007 3)2 (x 0007 3)2 A(x 0007 3)2 0003 B(x 0003 2)(x 0007 3) 0003 C(x 0003 2) 0002 (x 0003 2)(x 0007 3)2
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SECTION B–2
Partial Fractions
A-21
So for all x, 6x2 0007 14x 0007 27 0002 A(x 0007 3)2 0003 B(x 0003 2)(x 0007 3) 0003 C(x 0003 2) If x 0002 3, then
If x 0002 00072, then
000715 0002 5C C 0002 00073
25 0002 25A A00021
There are no other values of x that will cause terms on the right to drop out. Because any value of x can be substituted to produce an equation relating A, B, and C, we let x 0002 0 and obtain 000727 0002 9A 0007 6B 0003 2C 000727 0002 9 0007 6B 0007 6 B00025
Substitute A ⴝ 1 and C ⴝ ⴚ3.
Therefore, 1 5 3 6x2 0007 14x 0007 27 0002 0003 0007 2 x 0003 2 x 0007 3 (x 0003 2)(x 0007 3) (x 0007 3)2
MATCHED PROBLEM 2
EXAMPLE
3
Decompose into partial fractions:
0002
Nonrepeating Linear and Quadratic Factors Decompose into partial fractions:
SOLUTION
x2 0003 11x 0003 15 . (x 0007 1)(x 0003 2)2
0002
5x2 0007 8x 0003 5 . (x 0007 2)(x2 0007 x 0003 1)
First, we see that the quadratic in the denominator can’t be factored further in the real numbers. Then, we use parts 1 and 3 from Theorem 3 to write A 5x2 0007 8x 0003 5 Bx 0003 C 0002 0003 2 2 x00072 (x 0007 2)(x 0007 x 0003 1) x 0007x00031 0002
A(x2 0007 x 0003 1) 0003 (Bx 0003 C)(x 0007 2) (x 0007 2)(x2 0007 x 0003 1)
So for all x, 5x2 0007 8x 0003 5 0002 A(x2 0007 x 0003 1) 0003 (Bx 0003 C)(x 0007 2) If x 0002 2, then 9 0002 3A A00023 If x 0002 0, then, using A 0002 3, we have 5 0002 3 0007 2C C 0002 00071 If x 0002 1, then, using A 0002 3 and C 0002 00071, we have 2 0002 3 0003 (B 0007 1)(00071) B00022
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SPECIAL TOPICS
Therefore, 3 2x 0007 1 5x2 0007 8x 0003 5 0002 0003 2 x00072 (x 0007 2)(x2 0007 x 0003 1) x 0007x00031
MATCHED PROBLEM 3
EXAMPLE
4
Decompose into partial fractions:
7x2 0007 11x 0003 6 . (x 0007 1)(2x2 0007 3x 0003 2)
0002
Repeating Quadratic Factors Decompose into partial fractions:
SOLUTION
0002
x3 0007 4x2 0003 9x 0007 5 . (x2 0007 2x 0003 3)2
Because x2 0007 2x 0003 3 can’t be factored further in the real numbers, we proceed to use part 4 from Theorem 3 to write Cx 0003 D x3 0007 4x2 0003 9x 0007 5 Ax 0003 B 0003 2 0002 2 (x2 0007 2x 0003 3)2 x 0007 2x 0003 3 (x 0007 2x 0003 3)2 0002
(Ax 0003 B)(x2 0007 2x 0003 3) 0003 Cx 0003 D (x2 0007 2x 0003 3)2
So for all x, x3 0007 4x2 0003 9x 0007 5 0002 (Ax 0003 B)(x2 0007 2x 0003 3) 0003 Cx 0003 D Because the substitution of carefully chosen values of x doesn’t lead to the immediate determination of A, B, C, or D, we multiply and rearrange the right side to obtain x3 0007 4x2 0003 9x 0007 5 0002 Ax3 0003 (B 0007 2A)x2 0003 (3A 0007 2B 0003 C)x 0003 (3B 0003 D) Now we use Theorem 1 to equate coefficients of terms of like degree: A00021 B 0007 2A 0002 00074 3A 0007 2B 0003 C 0002 9 3B 0003 D 0002 00075
1x 3
ⴚ4x 2
ⴙ9x
ⴚ5
Ax 3 ⴙ (B ⴚ 2A)x 2 ⴙ (3A ⴚ 2B ⴙ C )x ⴙ (3B ⴙ D)
From these equations we easily find that A 0002 1, B 0002 00072, C 0002 2, and D 0002 1. Now we can write 2x 0003 1 x3 0007 4x2 0003 9x 0007 5 x00072 0003 2 0002 2 2 2 (x 0007 2x 0003 3) x 0007 2x 0003 3 (x 0007 2x 0003 3)2
MATCHED PROBLEM 4
Decompose into partial fractions:
3x3 0007 6x2 0003 7x 0007 2 . (x2 0007 2x 0003 2)2
ANSWERS TO MATCHED PROBLEMS 3 4 3 2 1 2. 0003 0007 0003 x00072 x00033 x00071 x00032 (x 0003 2)2 x00072 3x 2 3x 0007 2 0003 2 3. 4. 2 0003 2 x00071 2x 0007 3x 0003 2 x 0007 2x 0003 2 (x 0007 2x 0003 2)2 1.
0002
0002
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SECTION B–3
B-2
2. 3. 4.
6. 7. 8. 9.
10.
Cx 0003 D 3x3 0007 3x2 0003 10x 0007 4 Ax 0003 B 0003 2 0002 2 (x2 0007 x 0003 3)2 x 0007x00033 (x 0007 x 0003 3)2
In Problems 11–30, decompose into partial fractions. 11.
0007x 0003 22 x 0007 2x 0007 8
12.
0007x 0007 21 x 0003 2x 0007 15
13.
3x 0007 13 6x 0007 x 0007 12
14.
11x 0007 11 6x2 0003 7x 0007 3
15.
x2 0007 12x 0003 18 x3 0007 6x2 0003 9x
16.
5x2 0007 36x 0003 48 x(x 0007 4)2
17.
5x2 0003 3x 0003 6 x3 0003 2x2 0003 3x
18.
2x3 0003 7x 0003 5 6x2 0007 15x 0003 16 19. 4 3 2 x 0007 3x 0003 4x x 0003 4x2 0003 4
20.
00075x2 0003 7x 0007 18 x4 0003 6x2 0003 9
22.
x3 0003 x2 0007 13x 0003 11 x2 0003 2x 0007 15
23.
B A C 3x2 0003 7x 0003 1 0002 0003 0003 2 x x 0003 1 x(x 0003 1) (x 0003 1)2
7x 0007 14 A B 0002 0003 (x 0007 4)(x 0003 3) x00074 x00033 9x 0003 21 A B 0002 0003 (x 0003 5)(x 0007 3) x00035 x00073 17x 0007 1 A B 0002 0003 (2x 0007 3)(3x 0007 1) 2x 0007 3 3x 0007 1
2
2
2
21.
x3 0007 7x2 0003 17x 0007 17 x2 0007 5x 0003 6
4x2 0003 5x 0007 9 x3 0007 6x 0007 9
24.
4x2 0007 8x 0003 1 x3 0007 x 0003 6
25.
26.
x2 0007 6x 0003 11 A B C 0002 0003 0003 x00031 x00072 (x 0003 1)(x 0007 2)2 (x 0007 2)2
x2 0003 16x 0003 18 x 0003 2x2 0007 15x 0007 36
5x2 0007 18x 0003 1 x 0007 x2 0007 8x 0003 12
27.
3x2 0003 x Bx 0003 C A 0003 2 0002 x00072 (x 0007 2)(x2 0003 3) x 00033
0007x2 0003 x 0007 7 x 4 0007 5x3 0003 9x2 0007 8x 0003 4
28.
A 5x2 0007 9x 0003 19 Bx 0003 C 0002 0003 2 2 x00074 (x 0007 4)(x 0003 5) x 00035
00072x3 0003 12x2 0007 20x 0007 10 x 0007 7x3 0003 17x2 0007 21x 0003 18
29.
Cx 0003 D Ax 0003 B 2x2 0003 4x 0007 1 0003 2 0002 2 (x2 0003 x 0003 1)2 x 0003x00031 (x 0003 x 0003 1)2
4x5 0003 12x4 0007 x3 0003 7x2 0007 4x 0003 2 4x 4 0003 4x3 0007 5x2 0003 5x 0007 2
30.
6x5 0007 13x 4 0003 x3 0007 8x2 0003 2x 6x 4 0007 7x3 0003 x2 0003 x 0007 1
x 0007 11 A B 0002 0003 (3x 0003 2)(2x 0007 1) 3x 0003 2 2x 0007 1
In Problems 5–10, find A, B, C, and D, so that the right side is equal to the left. 5.
A-23
Exercises
In Problems 1–4, find A and B so that the right side is equal to the left. After cross-multiplying to produce a polynomial equation, solve each problem two ways (see Explore-Discuss 1). First, equate the coefficients of both sides to determine a linear system for A and B and solve this system. Second, solve for A and B by evaluating both sides for selected values of x. 1.
Parametric Equations
B-3
3
3
4
Parametric Equations Z Parametric Equations and Plane Curves Z Projectile Motion
Z Parametric Equations and Plane Curves Consider the two equations x0002t00031 y 0002 t 2 0007 2t
0007 6 t 6
(1)
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SPECIAL TOPICS
Each value of t determines a value of x, a value of y, and therefore, an ordered pair (x, y). To graph the set of ordered pairs (x, y) determined by letting t assume all real values, we construct Table 1 listing selected values of t and the corresponding values of x and y. Then we plot the ordered pairs (x, y) and connect them with a continuous curve, as shown in Figure 1. The variable t is called a parameter and does not appear on the graph. Equations (1) are called parametric equations because both x and y are expressed in terms of the parameter t. The graph of the ordered pairs (x, y) is called a plane curve.
10
5
x
Table 1 t
0
1
2
3
4
ⴚ1
ⴚ2
x
1
2
3
4
5
0
00071
y
0
00071
0
3
8
3
8
Z Figure 1 Graph of x 0002 t 0003 1, y 0002 t2 0007 2t, 0007 0006 t 0006 .
Technology Connections Parametric equations can also be graphed on a graphing calculator. Figure 2(a) shows the Parametric mode selected on a Texas Instruments TI-84 calculator. Figure 2(b) shows the equation editor with the parametric equations in (1) en-
tered as x1T and y1T. In Figure 2(c), notice that there are three new window variables, Tmin, Tmax, and Tstep, that must be entered by the user.
10
00073
7
00072
(a)
(b)
(c)
(d)
Z Figure 2 Graphing parametric equations on a graphing calculator.
In some cases, it is possible to eliminate the parameter by solving one of the equations for t and substituting into the other. In the example just considered, solving the first equation for t in terms of x, we have t0002x00071 Then, substituting the result into the second equation, we obtain y 0002 (x 0007 1)2 0007 2(x 0007 1) 0002 x2 0007 4x 0003 3 We recognize this as the equation of a parabola, as we would guess from Figure 1. In other cases, it may not be easy or possible to eliminate the parameter to obtain an equation in just x and y. For example, for x 0002 t 0003 log t y 0002 t 0007 et
t 7 0
you will not find it possible to solve either equation for t in terms of functions we have considered.
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SECTION B–3
Parametric Equations
A-25
Is there more than one parametric representation for a plane curve? The answer is yes. In fact, there is an unlimited number of parametric representations for the same plane curve. The following are two additional representations of the parabola in Figure 1. x0002t00033 y 0002 t 2 0003 2t
0007 6 t 6
(2)
x0002t y 0002 t 2 0007 4t 0003 3
0007 6 t 6
(3)
The concepts introduced in the preceding discussion are summarized in Definition 1.
Z DEFINITION 1 Parametric Equations and Plane Curves A plane curve is the set of points (x, y) determined by the parametric equations x 0002 f (t) y 0002 g (t) where the parameter t varies over an interval I and the functions f and g are both defined on the interval I.
Why are we interested in parametric representations of plane curves? It turns out that this approach is more general than using equations with two variables as we have been doing. In addition, the approach generalizes to curves in three- and higher-dimensional spaces. Other important reasons for using parametric representations of plane curves will be brought out in the discussion and examples that follow.
EXAMPLE
1
Eliminating the Parameter Eliminate the parameter and identify the plane curve given parametrically by x 0002 1t y 0002 19 0007 t
SOLUTION
00005t00059
(4)
To eliminate the parameter t, we solve each equation (4) for t: x 0002 1t x2 0002 t
y 5
y 0002 19 0007 t y2 0002 9 0007 t t 0002 9 0007 y2
Equating the last two equations, we have 00075
5
x
00075
Z Figure 3
MATCHED PROBLEM 1
x2 0002 9 0007 y2 x 0003 y2 0002 9 2
A circle of radius 3 centered at (0, 0)
As the parameter t increases from 0 to 9, x will increase from 0 to 3 and y will decrease from 3 to 0. So the graph of the parametric equations in (4) is the quarter of the circle of radius 3 centered at the origin that lies in the first quadrant (Fig. 3). 0002 Eliminate the parameter and identify the plane curve given parametrically by x 0002 14 0007 t, y 0002 0007 1t, 0 0005 t 0005 4. 0002
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SPECIAL TOPICS
Z Projectile Motion Newton’s laws and advanced mathematics can be used to determine the path of a projectile. If v0 is the vertical speed of the projectile, h0 is the horizontal speed, and a0 is the initial altitude of the projectile (Fig. 4), then, neglecting air resistance, the path of the projectile is given by x 0002 h0t y 0002 a0 0003 v0t 0007 4.9t 2
00005t0005b
(5)
y v0 ␣ a0
v0
h0 x
Z Figure 4 Projectile motion.
The parameter t represents time in seconds, and x and y are distances measured in meters. Solving the first equation in equations (5) for t in terms of x, substituting into the second equation, and simplifying produces the following equation: y 0002 a0 0003
v0 4.9 x 0007 2 x2 h0 h0
(6)
You should verify this by supplying the omitted details. We recognize equation (6) as a parabola. This equation in x and y describes the path the projectile follows but tells us little else about its flight. On the other hand, the parametric equations (5) not only determine the path of the projectile but also tell us where it is at any time t. Furthermore, using concepts from physics and calculus, the parametric equations can be used to determine the velocity and acceleration of the projectile at any time t. This illustrates another advantage of using parametric representations of plane curves.
EXAMPLE
2
Projectile Motion An automobile drives off a 50-meter cliff traveling at 25 meters per second (Fig. 5). When (to the nearest tenth of a second) will the automobile strike the ground? How far (to the nearest meter) from the base of the cliff is the point of impact?
50 m
Z Figure 5
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SECTION B–3
SOLUTION
Parametric Equations
A-27
At the instant the automobile leaves the cliff, the vertical speed is 0, the horizontal speed is 25 meters per second, and the altitude is 50 meters. Substituting these values in equations (5), the parametric equations for the path of the automobile are x 0002 25t y 0002 50 0007 4.9t2 The automobile strikes the ground when y 0002 0. Using the parametric equation for y, we have y 0002 50 0007 4.9t2 0002 0 00074.9t2 0002 000750 t0002
000750 ⬇ 3.2 seconds B 00074.9
The distance from the base of the cliff is the same as the value of x. Substituting t 0002 3.2 in the first parametric equation, the distance from the base of the cliff at the point of impact is x 0002 25(3.2) 0002 80 meters. 0002 MATCHED PROBLEM 2
A gardener is holding a hose in a horizontal position 1.5 meters above the ground. Water is leaving the hose at a speed of 5 meters per second. What is the distance (to the nearest tenth of a meter) from the gardener’s feet to the point where the water hits the ground? 0002
ANSWERS TO MATCHED PROBLEMS 1. The quarter of the circle of radius 2 centered at the origin that lies in the fourth quadrant. 2. 2.8 meters
B-3
Exercises
1. If x 0002 t 2 and y 0002 t 2 0007 2, then y 0002 x 0007 2. Discuss the differences between the graph of the parametric equations and the graph of the line y 0002 x 0007 2. 2. If x 0002 t 2 and y 0002 t 4 0007 2, then y 0002 x2 0007 2. Discuss the differences between the graph of the parametric equations and the graph of the parabola y 0002 x2 0007 2. In Problems 3–12, the interval for the parameter is the whole real line. For each pair of parametric equations, eliminate the parameter t and find an equation for the curve in terms of x and y. Identify and graph the curve.
In Problems 13–20, obtain an equation in x and y by eliminating the parameter. Identify the curve. 13. x 0002 t 0007 2, y 0002 4 0007 2t 14. x 0002 t 0007 1, y 0002 2t 0003 2 15. x 0002 t 0007 1, y 0002 1t, t 0 16. x 0002 1t, y 0002 t 0003 1, t 0 17. x 0002 1t, y 0002 2 116 0007 t, 0 0005 t 0005 16 18. x 0002 000731t, y 0002 125 0007 t, 0 0005 t 0005 25
3. x 0002 0007t, y 0002 2t 0007 2
4. x 0002 t, y 0002 t 0003 1
19. x 0002 0007 1t 0003 1, y 0002 0007 1t 0007 1, t 1
5. x 0002 0007t 2, y 0002 2t 2 0007 2
6. x 0002 t 2, y 0002 t 2 0003 1
20. x 0002 12 0007 t, y 0002 0007 14 0007 t, t 0005 2
7. x 0002 3t, y 0002 00072t
8. x 0002 2t, y 0002 t
21. If A ≠ 0, C 0002 0, and E ≠ 0, find parametric equations for Ax2 0003 Cy2 0003 Dx 0003 Ey 0003 F 0002 0. Identify the curve.
9. x 0002 14t 2, y 0002 t
10. x 0002 2t, y 0002 t 2
11. x 0002 14t 4, y 0002 t2
12. x 0002 2t 2, y 0002 t 4
22. If A 0002 0, C ≠ 0, and D ≠ 0, find parametric equations for Ax2 0003 Cy2 0003 Dx 0003 Ey 0003 F 0002 0. Identify the curve.
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SPECIAL TOPICS
In Problems 23–28, the interval for the parameter is the entire real line. Obtain an equation in x and y by eliminating the parameter and identify the curve. 2
2
2
2
23. x 0002 2t 0003 1, y 0002 2t 0003 9 24. x 0002 2t 0003 4, y 0002 2t 0003 1 25. x 0002 26. x 0002
2 2t 2 0003 1 3t 2
2t 0003 1
,y0002 ,y0002
2t 2t 2 0003 1 3
30. Consider the following two pairs of parametric equations: 1. x1 0002 t, y1 0002 log t, t 0 2. x2 0002 log t, y2 0002 t, t 0 (A) Graph both pairs of parametric equations in a squared viewing window and discuss the relationship between the graphs. (B) Eliminate the parameter and express each equation as a function of x. How are these functions related?
2
2t 0003 1
8 4t ,y0002 2 27. x 0002 2 t 00034 t 00034 28. x 0002
(B) Eliminate the parameter and express each equation as a function of x. How are these functions related?
4t 4t 2 ,y0002 2 t 00031 t 00031 2
29. Consider the following two pairs of parametric equations: 1. x1 0002 t, y1 0002 et, 0007 0006 t 0006 2. x2 0002 et, y2 0002 t, 0007 0006 t 0006 (A) Graph both pairs of parametric equations in a squared viewing window and discuss the relationship between the graphs.
APPLICATIONS 31. PROJECTILE MOTION An airplane flying at an altitude of 1,000 meters is dropping medical supplies to hurricane victims on an island. The path of the plane is horizontal, the speed is 125 meters per second, and the supplies are dropped at the instant the plane crosses the shoreline. How far inland (to the nearest meter) will the supplies land? 32. PROJECTILE MOTION One stone is dropped vertically from the top of a tower 40 meters high. A second stone is thrown horizontally from the top of the tower with a speed of 30 meters per second. How far apart (to the nearest tenth of a meter) are the stones when they land?
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APPENDIX
Geometric Formulas
11 C
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APPENDIX C
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GEOMETRIC FORMULAS
Z Similar Triangles (A) Two triangles are similar if two angles of one triangle have the same measure as two angles of the other. (B) If two triangles are similar, their corresponding sides are proportional: a b c ⫽ ⫽ a¿ b¿ c¿
a
b
b⬘
a⬘
c c⬘
Z Pythagorean Theorem c2 ⫽ a2 ⫹ b2
c
a b
Z Rectangle A ⫽ ab P ⫽ 2a ⫹ 2b
Area
b
Perimeter
a
Z Parallelogram h ⫽ Height A ⫽ ah ⫽ ab sin P ⫽ 2a ⫹ 2b
Area Perimeter
h
b
a
Z Triangle h ⫽ Height A ⫽ 12 hc P⫽a⫹b⫹c s ⫽ 12 (a ⫹ b ⫹ c) A ⫽ 1s(s ⫺ a)(s ⫺ b)(s ⫺ c)
b
Area
h
Perimeter
c
Semiperimeter
b
Area—Heron’s formula
c
Z Trapezoid Base a is parallel to base b. h ⫽ Height A ⫽ 12 (a ⫹ b)h
a
Area
h b
a
a
h
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APPENDIX C
A-31
GEOMETRIC FORMULAS
Z Circle R ⫽ Radius D ⫽ Diameter D ⫽ 2R A ⫽ R2 ⫽ 14 D2 C ⫽ 2R ⫽ D C ⫽ D ⬇ 3.141 59
D R
Area Circumference For all circles
Z Rectangular Solid V ⫽ abc T ⫽ 2ab ⫹ 2ac ⫹ 2bc
Volume Total surface area
c a
b
Z Right Circular Cylinder R ⫽ Radius of base h ⫽ Height V ⫽ R2h S ⫽ 2Rh T ⫽ 2R(R ⫹ h)
h
Volume Lateral surface area
R
Total surface area
Z Right Circular Cone R ⫽ Radius of base h ⫽ Height s ⫽ Slant height V ⫽ 13 R2h S ⫽ Rs ⫽ R2R2 ⫹ h2
Volume
R
Lateral surface area 2
2
T ⫽ R(R ⫹ s) ⫽ R(R ⫹ 2R ⫹ h )
Total surface area
Z Sphere R ⫽ Radius D ⫽ Diameter D ⫽ 2R V ⫽ 43 R3 ⫽ 16 D3 S ⫽ 4R2 ⫽ D2
s
h
D Volume Surface area
R
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STUDENT ANSWER APPENDIX CHAPTER R Exercises R-1 35. 37.
(A) 51, 11446 (B) 5⫺3, 0, 1, 11446 (C) 5⫺3, ⫺23 , 0, 1, 95 , 11446 (D) 5 136 (A) 0.888 888 p ; repeating; repeated digit: 8 (B) 0.272 727 p ; repeating; repeated digits: 27 (C) 2.236 067 977 p ; nonrepeating and nonterminating
1.375; terminating
(D)
Exercises R-2 61.
n8
3 3 69. 6 1 5 ⫺ 1 25
67. ⫺2 13
m12
3 51 2
71.
12 Ⲑ2 or 12 12
83.
Exercises R-4 59.
⫺x(x ⫹ y) y
CHAPTER 1 Exercises 1-2 5.
⫺8 ⱕ x ⱕ 7
9.
x ⱖ ⫺6
[
⫺10 ⫺5
[
⫺10 ⫺5
0
15.
(⫺⬁, ⫺2]
33.
N 6 ⫺8; (⫺⬁, ⫺8)
39.
B ⱖ ⫺4; [⫺4, ⬁)
45. (2, 4)
(
2
⫺10 ⫺5
(
4
5
5
10
0
5
(
[
x
77.
(A) and (C)
11. x
(⫺2, 6]
29.
[
[
⫺42
(
30
x
)
[
(
(
q
6
(
(B) and (D)
[
)[ (
⫺1 3
7
0
( (
[
⫺5
x
x
7
51.
(1, 5) [
(
1
(
x
(
x
5
x
4.5
[
x ⱖ 8; [8, ⬁)
69.
x
10
m
3
63. ⫺8 ⱕ x 6 ⫺3; [⫺8, ⫺3)
x
(
5
y
x ⱖ 4.5; [4.5, ⬁)
x
12
10
⫺10 ⫺5
2
43. (⫺5, 7]
57. (
(
x 6 10; (⫺⬁, 10)
a ⬎ 0 and b ⬎ 0, or a ⬍ 0 and b ⬍ 0
t
3
⫺14
)
(⫺7, 8)
37. m 7 3; (3, ⬁) [
⫺2
13.
31. y ⱖ 2
t
2
61. 6 6 x 6 12
67.
x x
5
x
10
49. (⫺⬁, ⫺1) 傼 [3, 7)
x
x
5
10
(
41. ⫺2 6 t ⱕ 3; (⫺2, 3]
20
5
(
(
0
[
0
x 6 5; (⫺⬁, 5)
35. t 7 2; (2, ⬁)
B
⫺20
⫺10 ⫺5
⫺10 ⫺5
55. q 6 ⫺14; (⫺⬁, ⫺14)
59. ⫺20 ⱕ x ⱕ 20; [⫺20, 20]
[
7. ⫺6 ⱕ x 6 6
47. (⫺⬁, ⬁)
6
65. ⫺42 ⱕ x 6 30
x
N
⫺4
x
x
10
10
⫺8
[
53. (⫺⬁, 6]
[
[
0
[
⫺8
⫺3
x
8
a ⬎ 0 and b ⬍ 0, or a ⬍ 0 and b ⬎ 0
Exercises 1-3 31.
y is 3 units from 5; y ⫽ 2, 8
2
8
y
33.
y is less than 3 units from 5; 2 ⬍ y ⬍ 8; (2, 8) (
35. y is more than 3 units from 5; y ⬍ 2 or y ⬎ 8; (⫺⬁, 2) ´ (8, ⬁)
2
[
( [
u is no more than 3 units from ⫺8; ⫺11 ⱕ u ⱕ ⫺5; [⫺11, ⫺5]
41.
u is at least 3 units from ⫺8; u ⱕ ⫺11 or u ⱖ ⫺5; (⫺⬁, ⫺11] ´ [⫺5, ⬁)
51. u ⱕ ⫺11 or u ⱖ ⫺6; (⫺⬁, ⫺11] 傼 [⫺6, ⬁)
⫺5
The distance from x to 3 is between zero and 0.1; (2.9, 3) ´ (3, 3.1);
[
[
⫺11 ⫺5
67.
The distance from x to a is between 0 and 1Ⲑ10; aa ⫺
( 2.9
1 1 , ab ´ aa, a ⫹ b 10 10
(
y
8 ⫺11
⫺5
u
u u
53. ⫺35 6 C 6 ⫺59 ; (⫺35, ⫺59 )
65.
2
37. u is 3 units from ⫺8; u ⫽ ⫺11, ⫺5
8
39.
⫺11
y
(
( 3
⫺2 ⬍ x ⬍ 2; (⫺2, 2)
55.
57. ⫺13 ⱕ t ⱕ 1; [⫺13 , 1]
x
3.1
( a⫺
(
x
1 1 a a ⫹ 10 10
SA-1
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Student Answer Appendix
Exercises 1-4 9. 15.
3 2 (A) 4000b
(A)
5 i 6 (B) 0
0002
3 5 0002 0002 i 2 6 (C) 4000b
(B)
11.
(C)
(A) 6.5
(C) 6.5 0002 2.1i
(B) 2.1i
13.
(A) 0
000bi
(B)
(C)
0002000bi
Exercises 1-5 19. z 4 12 5 25. Two real roots: x 1 12 27. No real roots: x 1 i 12 29. No real roots: t (3 i17) 2 31. Two real roots: t (3 17) 2 33. x 2 15 35. r (00025 13) 2 37. u (00022 i 111) 2 43. y (3 15) 2 45. x (3 113) 2
Exercises 1-6 27.
2u2 0002 4u 0, u x00023
51.
y
1 i 12
3 3
53.
29. t
10 1 4u 0002 7u2 0, u 2 9 x 5 113 x (four roots) B 6
12 , 12 2
63.
35. m 13, i 15
31.
Not of quadratic type
y 000264, 278
39.
Chapter 1 Review Exercises (
5. 000214 0007 y 0007 00024; (000214, 00024) (1-3) 19.
]
x 0004 000219; [000219, 0005) (1-2)
(
000214
y
x
000219
x 272 or 12 114
9.
00024
x (1 143) 3
27.
(1-5)
I (E 2E 2 0002 4PR) (2R)
37.
(1-5)
13. m 000212 ( 13 2)i (1-5) (1-5)
CHAPTER 2 Exercises 2-1 y
15.
y
17.
5
5
(00024, 2)
(4, 4) (5, 0)
00025
5
(00022, 1) x
00025
5
(00021, 00023)
(3, 00022) 00025
x
(0, 00022) (4, 00025)
00025
Points: A (2, 4), B (3, 00021), C (00024, 0), D (00025, 2) Reflections: A (00022, 4), B (00023, 00021), C (4, 0), D (5, 2) 21. Points: A (00023, 00023), B (0, 4), C (00023, 2), D (5, 00021) Reflections: A (3, 3), B (0, 00024), C (3, 00022), D (00025, 1) 23. No symmetry with respect 25. Symmetric with respect 27. to x axis, y axis, or origin to the origin 19.
y
y
5
5
x
35.
5
x
5
00025
5
00025
x
x
y
5
5
00025
(D)
y
5
x
00025
00025
00025
(C)
y
5
x
00025
(B)
y
5
00025
00025
(A)
y
5
00025
00025
Symmetric with respect to the x axis, y axis, and origin
y
5
00025
29.
Symmetric with respect to the x axis
5
00025 5
00025
x
00025
5
00025
x
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Student Answer Appendix 47.
49.
Symmetric with respect to the x axis y
x
5
5
x
57.
y
59.
00025
5
00025
63.
61.
y
A
x
10
A
x
C
B 000210
B
B
C
B
A
x
10
A C
000210
000210
00025
y 23 0002 2x
65.
y 00021 2x 2 0002 4
10
67.
10
69.
Symmetric with respect to the y axis
Symmetric with respect to the origin
y 000210
10
000210
10
000210
y
5
000210
00025
5
x
73.
No symmetry with respect to the x axis, y axis, or origin
75.
Symmetric with respect to the x axis, y axis, and origin
y
00025
Symmetric with respect to the x axis, y axis, and origin
y
10
y
5
x
5
00025
00025
71.
x
10
C 000210
x
10
000210
10
y
5
000210
y
5
00025
x
5
00025
Symmetric with respect to the y axis
5
10
00025
00025
Symmetric with respect to the y axis
y
5
00025
00025
Symmetric with respect to the origin
y
5
00025
53.
Symmetric with respect to the x axis, y axis, and origin
y
5
55.
51.
Symmetric with respect to the y axis
81.
R 30
5 20
000210
10
x
00025
5
x
00025
00025
83. 87.
(A) 3,000 cases (A) y
5
10 00025
(B) Demand decreases by 400 cases
(C) Demand increases by 600 cases
1
1
2
x
Exercises 2-2 21.
x2 y2 4
23.
(x 0002 1)2 y2 1
y
00025
(x 2)2 ( y 0002 1)2 9 y
5
5
00025
25.
y
5
x
00025
5
5
00025
x
x
00025
5
00025
x
0
5
10
p
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Student Answer Appendix
33.
The set of all points that are two units from the point (0, 2). x2 (y 0002 2)2 4 35. The set of all points that are four units from the point (1, 1). (x 0002 1)2 ( y 0002 1)2 16 43. Center: (0, 00022); radius: 3 45. Center: (00024, 2); radius: 17 y 5
5
x
y 10
10
x
000210
10
x
000210
10
000210
000210
00025
53. y 23 0002 x2
Center: (00024, 3); radius: 117
Center: (3, 2); radius: 7
y
3
00025
49.
Center: (00023, 0); radius: 5
5
00025
51.
47.
y
55.
y 00021 22 0002 (x 3)2
3.1
3.1
y 00024.7
4.7
00024.7
4.7
5 00023.1 00025
00023.1
x
73. (A) A (0, 0), B (0, 13.5), C (0, 27), D (60, 27), E (78, 27), F (78, 13.5), G (78, 0) 77. (A) (x 12)2 ( y 5)2 262; center: (000212, 00025); radius: 26 (B) 13.5 miles y (000212, 00025)
Town B (36, 15)
25
Town A 000250
(B) 62 feet, 79 feet
000225
x
25
000225
Exercises 2-3 19.
Slope 000235
21.
Slope 000234
y
Slope 00022
y
10
x
00025
000210
5
29.
x
y 10
00025
5
x
31.
Slope 0
y
y
5
5
10
000210
10
x
00025
5
x
00025
00025
41. 69.
47. y 000225 x 2 67. y 32 x 232 (slope AB )(slope BC ) (000234 )(43 ) 00021
5
00025
slope AB 000234 slope DC
000210
10
000210
00025
Slope not defined
y
Slope 45
y
00025
Slope 2
25.
5
5
000210
27.
23.
x
x
x
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Student Answer Appendix 75.
3x 4y 25
77.
x 0002 y 10
y
79. 232 5x 0002 12y y
y
10
10
10
x
000210
000210
x
10
000210
SA-5
10
20
x
000210
000210
000220
81.
(A)
x
0
5,000
10,000
15,000
20,000
25,000
30,000
B
212
203
194
185
176
167
158
(B) The boiling point drops 9°F for each 5,000-ft increase in altitude. 87.
(A) F 95 C 32
(B) 68°F, 30°C
Exercises 2-4 C 2,147 75x The rate of change of cost with respect to production is $75. Increasing production by 1 unit increases cost by $75 7. The rate of change of height with respect to DBH is 4.06 feet per inch. Increasing DBH by 1 inch increases height by 4.06 feet. 73 feet 19 inches 9. Robinson: The rate of change of weight with respect to height is 3.7 pounds per inch. Miller: The rate of change of weight with respect to height is 3 pounds per inch. (B) Robinson: 130.2 pounds; Miller: 135 pounds (C) Robinson: 5 9000e; Miller: 5 8000e 11. s 0.75t 717; speed increases 0.75 mph for each 1°F change in temperature. 15. (A) V 142,000 0002 7,500t (B) The tractor’s value is decreasing at the rate of $7,500 per year. (C) $97,000 17. (A) R 1.4C 0002 7 (B) The slope is 1.4; this is the rate of change of retail price with respect to cost. (C) $137 23. (A) y 5.
(A) (B) (C) (A) (B) (C) (D) (A)
1.0
0.5
0.5
1.0
x
(B) 0.97 million (C) 1.3 million
Chapter 2 Review Exercises 1.
(2-1)
3. (A) Symmetric with respect to the origin
y
(B) No symmetry with respect to the x axis, y axis, or origin
y
5
y
5
A 00025
C
5
B
5
x 00025
5
x
00025
5
00025 00025
00025
x
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Student Answer Appendix
(C) Symmetric with respect to the y axis
(D) Symmetric with respect to the x axis
y 5
5
00025
5
x
00025
5
00025
9.
(A)
(2-1)
y
x
00025
(2-2)
y 5 C
B 00025
5
x
A 00025
(B) d(A, C) 2 110, d(B, C) 110, d(A, B) 150, perimeter 16.56 (C) d(A, C)2 d(B, C)2 d(A, B)2; right triangle (D) Midpoint of side AC (0, 1), of side BC (2.5, 3.5), and of side AB (1.5, 0.5) 11. Slope 000232 (2-3) 15. Symmetric with respect to the y axis (2-1) y
y 5
5 00025 00025
5
5
x
x 00025
17.
Symmetric with respect to the x axis, y axis, and origin
27. y 0002x 7 (1-5, 2-2) y
(2-1)
y 5
6
(4, 3) 00025
5
x 6
x
00025
29.
(A)
(2-1)
(B)
y 5
00025
35.
(A) (B) (C) 37. (A)
x
00025
y
5
5
00025
(D)
y
5
5
00025
(C)
y
x
00025
5
5
00025
x
00025
5
00025
The rate of change of body surface area with respect to weight is 0.3433. Body surface area increases by 34.33 cm2. 6,470.5 cm2 (2-4) H 0.7(220 0002 A) (B) H 140 beats per minute (C) A 40 years old (2-4)
CHAPTER 3 Exercises 3-1 39. Not a function; for example, when x 0, y 2 41. A function with domain all real numbers 43. Not a function; for example, when x 0, y 7 45. A function with domain all real numbers 59. [00024, 1) 傼 (1, 0005); 00024 0003 x 6 1 or x 7 1 67. Function f multiplies the square of the domain element by 2 then adds 5 to the result. 69. Function z divides the sum of four times the domain element and 5 by the square root of the domain element. 1 1 79. (A) 00028x 3 0002 4h (B) 00024x 0002 4a 3 81. (A) (B) 1x h 2 1x 2 1x 2 1a 2 00024 00024 83. (A) (B) 91. The cost is a flat $17 per month, plus $2.40 for each hour of airtime. x(x h) ax
x
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Student Answer Appendix 93.
(A) s(0) 0, s(1) 16, s(2) 64, s(3) 144 (C) Let q(h) [s(2 h) 0002 s(2)]0002h h q(h)
SA-7
(B) 64 16h
00021
00020.1
00020.01
00020.001
0.001
0.01
0.1
1
48
62.4
63.84
63.984
64.016
64.16
65.6
80
(D) q(h), the average velocity from 2 to 2 h seconds, approaches 64 feet per second 97.
F 8x (2500002x) 0002 12; x
4
5
6
7
F
82.5
78
77.7
79.7
Exercises 3-2 9. (A) [00024, 4) (B) [00023, 3) (C) 0 (D) 0 (E) [00024, 4) (F) None (G) None (H) None 11. (A) (00020005, 0005) (B) [00024, 0005] (C) 00023, 1 (D) 00023 (E) [00021, 0005) (F) (00020005, 00021] (G) None (H) None 13. (A) (00020005, 2) 傼 (2, 0005) (B) (00020005, 00021) 傼 [ 1, 0005) (C) None (D) 1 (E) None (F) (00020005, 00022], (2, 0005) (G) [00022, 2) (H) x 2 21. One possible answer: 23. One possible answer: 25. One possible answer: 27. Slope 2, x intercept 00022, f(x) f(x) f(x) y intercept 4 5
5
00025
x
5
5
00025
00025
5
x
f(x)
00025
5
00025
x 5
00025 00025
29.
Slope 000212 , x intercept 0002103 , y intercept 000253
31.
5
x
Slope 00022.3, x intercept 3.1, y intercept 7.1
f(x)
y
5
10
x
00025
000210
00025
10
x
000210
37. Domain: 5x ƒ x 000f 000226; x intercept: 4; y intercept: 00023 39. Domain: 5x ƒ x 000f 54 6; x intercept: 23 ; y intercept: 25 41. Domain: 5x ƒ x 000f 26; x intercept: 0; y intercept: 0 43. Domain: 5x ƒ x 000f 00023, 36; x intercept: 4; y intercept: 169 45. Domain: 5x ƒ x 000f 00025, 56; no x intercept; y intercept: 0002257 47. (A) f (00021) 0, f (0) 1, f (1) 0 49. (A) f (00022) 00022, f (00021) is not defined, f (2) 4 (B)
(B)
y
y
2.0
5
(2, 4)
(0, 1) 00025
(00021, 0) 00021
0.2
00025
1
(A) f (00022) 0, f (00021) is not defined, f (0) 00022
x
(00023, 00022)
(1, 0) x
(C) Domain: [00023, 00021) 傼 (00021, 2]; range: {00022, 4}; discontinuous at x 00021
(C) Domain: [00021, 1]; range: [0, 1]; continuous on its domain 51.
5
(B)
(C) Domain: (00020005, 00021) 傼 (00021, 0005); range: R; discontinuous at x 00021
y 5
(00022, 0) 00025
5
(0, 00022) 00025
x
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Student Answer Appendix
(A)
f(00023) 0, f (00022) 00022, f (0) 00022, f (3) 00022, f (4) 4
(B)
55.
f (00023) 000232 , f (00022) 1, f (0) 1, f (3) 1, f (4) 52
(B)
y 5
y 5
(4, 4) (00022, 1)
(00023, 0) 00025
5
(00022, 00022) 00025
x
(3, 00022) (0, 00022)
(00023,
(3, 1) 5
3 00022 )
63.
0003
(C) Domain: R; range: R; continuous on its domain 3 if x 6 00022 f (x) 00022 x 0002 2 if 00022 6 x 6 1 00021 if x 7 1
y 10 ) 3
(00021,
5
(1, 52 )
00025
5
(3,
x
3 00022 )
00025
(C) Domain: (00020005, 0) 傼 (0, 2) 傼 (2, 0005); range: (00020005, 4); discontinuous at x 0 and x 2 10002x if x 6 0 0002x 2 65. f (x) 67. f (x) 1 x if x 0004 0 x00022
0003
0003
y 5
5
x
00025
00025
69.
Domain: R; range: [ 1, 0005); continuous on its domain (A) One possible answer:
x
71.
Domain: R; range: [ 0, 0005); continuous on its domain (A) One possible answer:
f (x)
f(x)
5
5
10
x
00025
00025
5
x
00025
(B) The graph must cross the x axis exactly once. Graphs of f and g
(B) The graph must cross the x axis at least twice. There is no upper limit on the number of times it can cross the x axis. Graph of m Graph of n
10
000210
10
10
000210
000210
75.
5
00025
5
73.
if x 6 2 if x 0004 2
y
5
00025
10
10
000210
10
10
000210
000210
Graphs of f and g
000210
x
00025
(A) f (00021) 103 , f (0) is not defined, f (1) 52 , f (2) is not defined, f (3) 000232 (B)
(4, 52 ) (0, 1)
00025
(C) Domain: R; range: [00022, 0005); continuous on its domain 57.
(A)
10
000210
Graph of m
Graph of n
10
10
000210
10
000210
000210
10
000210
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Student Answer Appendix 77.
Graphs of f and g
Graph of m
Graph of n
10
10
10
000210
10
000210
10
000210
81. R(x)
0003
10
0003
000210
000210
200 if 0 0003 x 0003 3,000 83. E(x) 80 0.04x if 3,000 6 x 6 8,000 if x 0004 8,000 180 0.04x Discontinuous at x 8,000 E(5,750) $310, E(9,200) $548
if 0 0003 x 0003 100 if x 7 100
32 16 0.16x
000210
E(x)
500
5,000
85.
89.
10,000
x
x
4
00024
6
00026
24
25
247
0002243
0002245
0002246
f(x)
0
0
10
000210
20
30
250
0002240
0002240
0002250
15 18 21 (A) C(x) f 24 27 30
0 1 2 3 4 5
6 6 6 6 6 6
x x x x x x
0003 0003 0003 0003 0003 0003
1 2 3 4 5 6
C(x) $30
$15
0
1
2
3
4
5
(B) No, since f (x) 000f C(x) at x 1, 2, 3, 4, 5, or 6 0.03x if 0 0003 x 0003 10,000 91. T(x) 0.05x 0002 200 if x 7 10,000 93.
6
x
0003
0003
0.0535x T(x) 0.0705x 0002 338.25 0.0785x 0002 860.41 T(10,000) $535 T(30,000) $1,776.75 T(100,000) $6,989.60
y 1,000
24
; f rounds numbers to the tens place.
0 0003 x 0003 19,890 19,890 6 x 0003 65,330 x 7 65,330
x
Exercises 3-3 5. 11.
Domain: [ 0, 0005); Range: (00020005, 0 ] Domain: [ 00022, 2 ] ; range: [0, 4]
13.
Domain: [ 00022, 2 ] ; range: [1, 3]
y
00022
15.
Domain: [0, 4]; range: [ 00022, 2]
y
4
4
2
2
2
x
00022
y 2 2
2
x
00022
4
x
SA-9
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Student Answer Appendix
Domain: [ 00024, 0 ] ; range: [ 00021, 1 ]
19.
Domain: [ 00022, 2 ] ; range: [ 00022, 2 ]
y 2
00024
y 2
x
00022
25.
Domain: [ 00022, 2 ] ; range: [ 00022, 2 ]
x 00022
49.
y
00022
5
00025
5
x
55. 10
5
5
x
00025
x
5
x
00025
5
00025
x
00025
3 (A) f is a horizontal shrink of y 1x by a factor of 1 8. g is a vertical stretch of y 1x by a factor of 2. 3 3 3 3 (C) f (x) 18x 18 ⴢ 1x 2 1x (A) The graphs are different; order is significant. (B) i. f (x) 0002(x2 0002 5) ii. f (x) 0002x2 0002 5 93. g(x) f(x) 3
5
5
5
00025
x
y
00025
95.
y
5
5
00025
x
5
5
61.
y
5
57.
00025
5
81. 91.
00025
00025
y
10
10
x
00025
y
5
79.
y
5
5
x
2
51.
y
10
00025
x
2
00022
59.
y 5
2
2
00025
x
2
45.
y
2
53.
00022
00022
y
47.
x
2
Domain: [ 00021, 1 ] ; range: [ 00021, 1 ]
00022
Domain: [ 00022, 2 ] ; range: [ 00022, 2]
2
00022
23.
21.
y
x
00025
5
x
00025
Conclusion: any function can be written as the sum of two other functions, one even and the other odd.
(B) The graphs are identical.
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Student Answer Appendix 97.
99. Each graph is a vertical translation of the graph of y 0.004(x 0002 10)3.
f(x)
35
Total production costs
$150,000 10
25
100,000 10 50,000
500
0
101.
1,000
x
Units produced
Each graph is a portion of the graph of a horizontal translation followed by a vertical shrink (except for C 8) of the graph of y t2. Larger values of C correspond to a smaller opening. V 70
0
2
4
6
8
t
Exercises 3-4 7.
Vertex: (00023, 00024); axis: x 00023
9.
3 3 Vertex: a ,00025b; axis: x 2 2
y
000210
11.
Vertex: (000210, 20); axis: x 000210 y
y
10
40
10
10
x
000210
0004
000210
3 2
(000210, 20)
x
10
(00023, 00024)
13. 17. 25.
, 000250005
000220
5 000210
000210
The graph is shifted 2 units right and 1 unit up. 15. The graph is reflected in the x axis, then shifted 1 unit left. The graph is shifted 2 units right and 3 units down. f (x) (x 0002 2)2 1; 27. h(x) 0002(x 1)2 0002 2; 29. m(x) 2(x 0002 3)2 4; vertex: (2, 1); axis: x 2 vertex: (00021, 00022); axis: x 00021 vertex: (3, 4); axis: x 3 y
y
y
10
10
10 (3, 4)
(2, 1) 10
000210
x
000210(00021, 00022)
10
x
000210
000210
000210
31.
SA-11
1 f(x) (x 3)2 0002 8; 2 vertex: (00023, 00028); axis: x 00023
33.
y
10
30 (6,18)
000210
10
35.
Vertex: (00024, 00028); The graph is symmetric about the axis, x 00024. It decreases until reaching a minimum at (00024, 00028), then increases. The range is [00028, 0005). y
x 000210
(00023,00028)
x
000210
f (x) 2(x 0002 6)2 18; vertex: (6, 18); axis: x 6
y
10
00025
10
10
x
000210 000210 (00024, 00028)
10
000210
x
x
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Student Answer Appendix
Vertex: (000572 , 654 ); The graph is symmetric about the axis, x 0002 000572 . It increases until reaching a maximum at (000572 , 654 ), then decreases. The range is (00050003, 654 ].
39.
Vertex: (94 , 194 ); The graph is symmetric about the axis, x 0002 94 . It decreases until reaching a minimum at (94 , 194 ), then increases. The range is [ 194 , 0003).
y
y
20
(00057/2, 65/4)
14
000510
x
10
(9/4, 19/4) 00055
5
000520
41.
00056
(52 , 149 2 );
5 2.
Vertex: The graph is symmetric about the axis, x 0002 It increases until reaching a maximum at (52 , 149 2 ), then decreases. The range is (00050003, 149 2 ].
43.
Vertex: (000532 , 000594 ); axis of symmetry: x 0002 0; domain: range: [ 000594 , 0003); min f (x) 0002 f (000532 ) 0002 000594 ; decreasing on (00050003, 000532 ); increasing on (000532 , 0003)
(00050003, 0003);
y
y 100
x
(5/2, 149/2)
00056
5
x
6
00055
5
x
(0005 32 , 0005 94 )
0005100
00055 3
45.
Vertex: (2, 00059); axis of symmetry: x 0002 2; domain: (00050003, 0003); range: [00059, 0003); min f (x) 0002 f (2) 0002 00059; decreasing on (00050003, 2); increasing on (2, 0003)
x 0002 00052
y 10
000510
10
000510
x
(2, 00059) x00022
81. The minimum product is 0005225 for the numbers 15 and 000515. There is no maximum product. 83. 26 employees; $322,800 85. (A) 2003 (B) The domain values should be whole numbers. y 97. (B) (C) 56 mph 200
80
105.
x
(A) R(x) 0002 3.5x 0005 0.00007x2; domain: [0, 50,000]; C(x) 0002 24,500 0004 0.35x, domain: [ 0, 0003) (B) x 0002 10,000 and x 0002 35,000 y 50,000
y 0002 C(x) y 0002 R(x)
50,000
x
(C) The company makes a profit for those sales levels for which the graph of the revenue function is above the graph of the cost function, that is, if the sales are between 10,000 and 35,000 gallons. The company suffers a loss for those sales levels for which the graph of the revenue function is below the graph of the cost function, that is, if the sales are between 0 and 10,000 gallons or between 35,000 and 50,000 gallons. (D) The maximum profit is $10,937.50 when 22,500 gallons are sold at a price of $1.92 per gallon.
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Student Answer Appendix
Exercises 3-5 7.
x
00023
00022
00021
0
1
2
3
3
3
1
00021
00023
00023
00023
( f g)(x)
9.
x ( fg)(x)
y
00023
00022
00021
0
1
2
3
2
0
00022
00022
0
2
2
y
5
5
00025
5
x
00025
00025
5
x
00025
27. ( f ° g)(00027) 3; ( f ° g)(0) 9; ( f ° g)(4) 000210 29. ( f g)(x) 5x 1; ( f 0002 g)(x) 3x 0002 1; ( fg)(x) 4x 2 4x; f 4x a b(x) ; domain f g, f 0002 g, fg (00020005, 0005); domain of f0002g (00020005, 00021) 傼 (00021, 0005) g x 1 2 31. ( f g)(x) 3x 1; ( f 0002 g)(x) x2 0002 1; ( fg)(x) 2x4 2x2; f 2x2 a b(x) 2 ; domain of each function: (00020005, 0005) g x 1 2 33. ( f g)(x) x 3x 4; ( f 0002 g)(x) 0002x2 3x 6; ( fg)(x) 3x3 5x2 0002 3x 0002 5; f 3x 5 ; domain f g, f 0002 g, fg: (00020005, 0005); domain of f0002g: (00020005, 00021) 傼 (00021, 1) 傼 (1, 0005) a b(x) 2 g x 00021 35. ( f g)(x) 12 0002 x 1x 3; ( f 0002 g)(x) 22 0002 x 0002 2x 3; ( fg)(x) 26 0002 x 0002 x2; f f 20002x a b(x) . The domain of the functions f g, f 0002 g, and fg is [00023, 2]. The domain of is (00023, 2]. g Ax 3 g 37. ( f g)(x) 21x 0002 2; ( f 0002 g)(x) 6; ( fg)(x) x 0002 2 1x 0002 8; f f 1x 2 . The domain of f g, f 0002 g, and fg is [0, 0005). Domain of [ 0, 16) 傼 (16, 0005). a b(x) g g 1x 0002 4 39. ( f g)(x) 2x2 x 0002 6 27 6x 0002 x2; ( f 0002 g)(x) 2x2 x 0002 6 0002 27 6x 0002 x2; ( fg)(x) 20002x4 5x3 19x2 0002 29x 0002 42; f f x2 x 0002 6 a b(x) . The domain of the functions f g, f 0002 g, and fg is [2, 7]. The domain of is [2, 7). g B 7 6x 0002 x2 g f 2 1 x2 1 41. ( f g)(x) 2x; ( f 0002 g)(x) ; ( fg)(x) x2 0002 2 ; a b(x) 2 The domain of f g, f 0002 g, and fg is (00020005, 0) 傼 (0, 0005). x g x x 00021 f is (00020005, 00021) 傼 (00021, 0) 傼 (0, 1) 傼 (1, 0005). g 2 ( f ° g)(x) (x 0002 x 1)3; domain: (00020005, 0005); (g ° f )(x) x6 0002 x3 1; domain: (00020005, 0005) ( f ° g)(x) 0006 2x 4 0006; domain: (00020005, 0005); (g ° f )(x) 20006 x 1 0006 3; domain: (00020005, 0005) ( f ° g)(x) (2x3 4)1 3; domain: (00020005, 0005); (g ° f )(x) 2x 4; domain: (00020005, 0005) ( f ° g)(x) 1x 0002 4; domain: [4, 0005); (g ° f )(x) 1x 0002 4; domain: [0, 0005) 1 1 ( f ° g)(x) 2; domain: (00020005, 0) 傼 (0, 0005); (g ° f )(x) ; domain: (00020005, 00022) 傼 (00022, 0005) x x 2 2 ( f ° g)(x) 24 0002 x ; domain of f ° g is [ 00022, 2 ] ; (g ° f )(x) 4 0002 x; domain of g ° f is (00020005, 4 ] . 6x 0002 10 x 5 ( f ° g)(x) ; domain of f ° g is (00020005, 0) 傼 (0, 2) 傼 (2, 0005); (g ° f )(x) ; domain of g ° f is (00020005, 0) 傼 (0, 5) 傼 (5, 0005). x 50002x ( f ° g)(x) x; domain: (00020005, 2) 傼 (2, 0005); (g ° f )(x) x; domain: (00020005, 0) 傼 (0, 0005) ( f ° g)(x) 216 0002 x2; domain of f ° g is [ 00024, 4 ] ; (g ° f )(x) 234 0002 x2; domain of g ° f is [00025, 5 ] . ( f ° g)(x) (g ° f )(x) x; the graphs of f and g are 67. ( f ° g)(x) (g ° f )(x) x; the graphs of f and g are symmetric with respect to the line y x. symmetric with respect to the line y x. The domain of
43. 45. 47. 49. 51. 53. 55. 57. 59. 65.
4
00026
4
6
00024
00026
6
00024
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Student Answer Appendix
( f ° g)(x) x, (g ° f )(x) x; the graphs of f and g are symmetric with respect to the line y x. y
( f ° g)(x) x, (g ° f )(x) x; the graphs of f and g are symmetric with respect to the line y x.
71.
y y g(x)
y f (x) y x
5
10
y g(x)
00025
5
yx
x
000210
85.
87.
y f(x) x
000210
00025
73. 77.
10
g(x) 2x 0002 7; f (x) x 4; h(x) ( f ° g)(x) f (x) x 7; g(x) 3x 0002 5; h(x) (g ° f )(x)
75. 79.
g(x) 4 2x; f (x) x1 2; h(x) ( f ° g)(x) f (x) x00021 2; g(x) 4x 3; h(x) (g ° f )(x)
1 2 ( f g)(x) 2x; (f 0002 g)(x) ; (fg)(x) x2 0002 2 ; x x f x2 1 (x) 2 The domain of f g, f 0002 g, and fg is g x 00021 f (00020005, 0) 傼 (0, 0005). The domain of is g (00020005, 00021) 傼 (00021, 0) 傼 (0, 1) 傼 (1, 0005). f 00022x ( f g)(x) 2; ( f 0002 g)(x) ; ( fg)(x) 0; a b(x) 0 |x| g The domain of f g, f 0002 g, and fg is (00020005, 0) 傼 (0, 0005). f Domain of is (0, 0005). g
Exercises 3-6 7. 9. 11. 41.
The original set and the reversed set are both one-to-one functions. The original set is a function. The reversed set is not a function. Neither set is a function. Domain of f [00024, 4] 43. Domain of f [00025, 3] range of f [1, 5] range of f [00023, 5] domain of f 00021 [1, 5] domain of f 00021 [00023, 5] range of f 00021 [00024, 4] range of f 00021 [00025, 3] y
45.
y
5
5
f 00021(x) 13 x domain of f (00020005, 0005) range of f (00020005, 0005) domain of f 00021 (00020005, 0005) range of f 00021 (00020005, 0005) y
yx
yx
10
y f (x) 00025
5
y f 00021(x)
x
00025
5
y f 00021(x)
yx
y f00021(x)
x
y f(x)
00025
000210
10
x
00025 000210
y f(x)
47.
f 00021(x) (x 3)00024 domain of f (00020005, 0005) range of f (00020005, 0005) domain of f 00021 (00020005, 0005) range of f 00021 (00020005, 0005)
49.
f 00021(x) 5x 0002 2 domain of f (00020005, 0005) range of f (00020005, 0005) domain of f 00021 (00020005, 0005) range of f 00021 (00020005, 0005)
y 10
y f(x) 000210
10
f 00021(x) (x 0002 3)2, x 0004 3 domain of f [0, 0005) range of f [3, 0005) domain of f 00021 [3, 0005) range of f 00021 [0, 0005)
y y f00021(x)
yx
10
y f00021(x) 000210
51.
x
y 10
yx
y f(x)
y f(x) 000210
10
000210
x
y f 00021(x) yx
000210
10
000210
x
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Student Answer Appendix 53.
f 00021(x) 16 0002 4x 2, x 0004 0 domain of f (00020005, 16] range of f [0, 0005) domain of f 00021 [0, 0005) range of f 00021 (00020005, 16]
55. f 00021(x) (3 0002 x)2 1, x 0003 3 domain of f [1, 0005) range of f (00020005, 3] domain of f 00021 (00020005, 3] range of f 00021 [1, 0005)
y
y yx
20
5
y
y f(x)
10
yx
20
yf
x
00025
5
x
000210
(x)
000220
00025
59. f (x) 000214 0002 x domain of f (00020005, 0] range of f (00020005, 4] domain of f 00021 (00020005, 4] range of f 00021 (00020005, 0]
61.
000210
00021
f (x) 1x 16 0002 4 domain of f [00024, 0005) range of f [000216, 0005) domain of f 00021 [000216, 0005) range of f 00021 [00024, 0005)
y
y yx
20
10
x
000210
63. f (x) 2 0002 1x domain of f (00020005, 2] range of f [0, 0005) domain of f 00021 [0, 0005) range of f 00021 (00020005, 2] y
y f(x)
y f(x)
20
000210
000220
f 00021(x) 1 1x 0002 2 domain of f [1, 0005) range of f [2, 0005) domain of f 00021 [2, 0005) range of f 00021 [1, 0005) y
y f (x)
69.
f 00021(x) 29 0002 x 2 domain of f [0, 3] range of f [00023, 0] domain of f 00021 [00023, 0] range of f 00021 [0, 3]
y yx
10
y yx
10
y f(x)
x
5
10
x
00025
5
y f 00021(x)
73.
y yx
yx 2
y f (x) 00025
5
x
y f 00021(x) 00025
00025
f 00021(x) 0002 22x 0002 x2 domain of f [00021, 0] range of f [0, 1] domain of f 00021 [0, 1] range of f 00021 [00021, 0]
y 5
x
y f(x)
000210
71. f 00021(x) 0002 29 0002 x2 domain of f [00023, 0] range of f [0, 3] domain of f 00021 [0, 3] range of f 00021 [00023, 0]
yx
y f 00021(x)
000210
000210
x
10
y f 00021(x) 000210
67. f 00021(x) 00021x 3 0002 1 domain of f (00020005, 00021] range of f [00023, 0005) domain of f 00021 [00023, 0005) range of f 00021 (00020005, 00021]
y f 00021(x) 000210
yx
10
y f 00021(x) x
000220
y f 00021(x)
00021
yx
y f (x)
10
x
10
00021
00021
10
yx
y f 00021(x)
y f(x)
000220
65.
f 00021(x) 1x 0002 5 domain of f [0, 0005) range of f [5, 0005) domain of f 00021 [5, 0005) range of f 00021 [0, 0005)
y f 00021(x)
y f (x)
000210
57.
y f(x) 00022
x
2 00022
yf
00021
(x)
2 2 x x 4x 5 77. f 00021(x) 79. f 00021(x) 81. f 00021(x) 83. f 00021(x) (4 0002 x)5 0002 2 30002x x 20002x 3x 0002 2 85. The x intercept of f is the y intercept of f 00021 and the y intercept of f is the x intercept of f 00021. 89. One possible answer: domain x 0003 2, f00021(x) 2 0002 1x 91. One possible answer: domain 0 0003 x 0003 2, f00021(x) 2 0002 24 0002 x2 15,000 0002 5; domain: [200, 1,000]; range: [10, 70] 95. (A) [200, 1,000] (B) d 00021(q) q 75.
f 00021(x)
(A) r m(w) 1.25w 3; domain: [0, 0005); range: [3, 0005) (B) w m00021(r) 0.8r 0002 2.4; domain: [3, 0005); range: [0, 0005) 50 (L 0002 20); domain: [20, 0005); range: [10, 0005) 99. s f 00021(L) 10 A3
97.
SA-15
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Student Answer Appendix
Chapter 3 Review Exercises 1. 3. 23.
(A) Function (B) Function (C) Not a function (3-1) If there is at least one team that has won more than one Super Bowl, then the correspondence is not a function because one input (team) will correspond with more than one output (year). There are several teams that have won at least two Super Bowls, so it is not a function. (3-1) 00023
00022
00021
0
1 2
000212
00026
00022
0
0 26
x ( fg)(x)
3
y
(3-5) 5
00025
5
x
000210
31.
(3-3)
y
33.
5
5
x
(3-3)
y 5
00025
5
00025
49. 51. 53. 57. 61. 67.
35.
5
00025
39.
(3-3)
y
x
00025
00025
5
x
00025
2
(B) (g0002f )(x) (x 3)0002(x 2 0002 4); (A) ( f0002g)(x) (x 0002 4)0002(x 3); domain of f0002g (00020005, 00023) 傼 (00023, 0005) domain of g f (00020005, 00022) 傼 (00022, 2) 傼 (2, 0005) (C) ( f ⴰ g)(x) x2 6x 5; domain of f ⴰ g (00020005, 0005) (D) (g ⴰ f )(x) x2 0002 1; domain of g ⴰ f (00020005, 0005) (3-5) The function f multiplies the square of the domain element by 3, adds 4 times the domain element, and then subtracts 6. (3-1) This equation does not define a function. For example, the ordered pairs (2, 2) and (2, 00022) both satisfy the equation. (3-1) Domain: [0, 0005); y intercept: 2; no x intercepts (3-1, 3-2) 55. Domain: (00020005, 3); y intercept: 0; x intercept: 0 (3-1, 3-2) Domain: [0, 16) 傼 (16, 0005); y intercept: 14 ; no x intercepts (3-1, 3-2) (A) (f ⴰ g)(x) 1|x| 0002 8, (g ⴰ f )(x) | 1x 0002 8| (B) Domain of f ⴰ g (00020005, 0005), domain of g ⴰ f [ 0, 0005) (3-5) g(x) 5 0002 3|x 0002 2| (3-3) y 5
x
5
3 The graph of y 2x is vertically stretched by a factor of 2, reflected through the x axis, shifted 1 unit left and 1 unit down. 3 Equation: y 000222x 1 0002 1 (3-3) 2 y 73. t(x) 0.25x x 0002 3 (3-3) 75. (3-3) 77. (3-3) y
69.
y
10
5
10
000210
10
00025
x
5
x
00025
10
000210
y
79.
(3-3)
81.
(3-3)
y
5
10
00025
x
x
00025
00025
x
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Student Answer Appendix 83. x 0007 00022 or x 0006 6; (00020005, 00022) 傼 (6, 0005) (3-4) 85. (A) x2 11 0002 x, domain (00020005, 1] (B) x2 11 0002 x, domain (00020005, 1) (C) 1 0002 x, domain (00020005, 1] (D) 21 0002 x2, domain [00021, 1] (3-5) 87. (A) f 00021(x) x2 1 (B) Domain of f [1, 0005) Range of f 00021 Range of f [0, 0005) Domain of f 00021 (3-6) (C) y 5
f 00021 f
00025
x
5
yx 00025
89.
(A)
(B)
y
(3-3)
y
5
5
00025
x
5
00025
00025
5
x
00025
if 0 0003 x 0003 2,000 if 2,000 6 x 0003 5,000 if x 7 5,000 (D) $24.99 (3-6) 93. $168 (C) c f 00021(r ) 0.625r; domain: [16, 0005); range: [10, 0005) 4,500 00021 95. (A) [1, 3] (B) q g (p) (C) R(p) 4,500 0002 500p 0002 500; domain: [1, 3]; range: [1,000, 4,000] p (D) R(q) 9q0002(1 0.002q) (3-6) 97. (A) A(x) 60x 0002 32 x2 (B) 0 0007 x 0007 40 (C) x 20, y 15 (3-4) 91.
120 (A) E(x) • 0.1x 0002 80 0.1x 170 (A) f f (c) 1.6c (B)
0003
if 0 0003 x 0003 3,000 if 3,000 6 x 0003 5,000 if 5,000 6 x 0003 17,000 if 17,000 0003 x
0.02x 0.03x 0002 30 99. T(x) 0.05x 0002 130 0.0575x 0002 257.5 x T(x)
$2,000
$4,000
$40
$10,000
$90.00
$370
$30,000 $1,467.50 (3-2)
CHAPTER 4 Exercises 4-1 37. 63. 65. 67. 77. 79. 81. 93.
11 4x2 10x 0002 9 3 2x3 0002 3x 1 4x 0002 2 0002 2x2 4x 5 39. x 3 x 3 x00022 x00022 P(x) S 0005 as x S 0005 and P(x) S 00020005 as x S 00020005; three intercepts and two local extrema P(x) S 00020005 as x S 0005 and P(x) S 0005 as x S 00020005; three intercepts and two local extrema P(x) S 0005 as x S 0005 and as x S 00020005; four intercepts and three local extrema x intercepts: 000212.69, 00020.72, 4.41; local maximum: P(2.07) 0007 96.07; local minimum: P(00028.07) 0007 0002424.07 x intercepts: 000216.06, 0.50, 15.56; local maximum: P(00029.13) 0007 65.86; local minimum: P(9.13) 0007 000255.86 x intercepts: 000216.15, 00022.53, 1.56, 14.12; local minimum: P(000211.68) 0007 00021,395.99; local maximum: P(00020.50) 0007 95.72; local minimum: P(9.92) 0007 00021,140.27 (A) 95. (A)
(B) $4,062 billion
(B)
00023.6 (implausible estimate)
Exercises 4-2 35. (A) Upper bound: 2; lower bound: 00022
(B) 1.4 (or 00021.4)
Exercises 4-3 9. 0 (multiplicity 3), 000212 (multiplicity 2); degree of P(x) is 5 11. 2i (multiplicity 3); 00022i (multiplicity 4); 00022 (multiplicity 5); 2 (multiplicity 5); degree of P(x) is 17 15. P(x) (x 7)3 [ x 0002 (00023 12)][x 0002 (00023 0002 12)]; degree 5
SA-17
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Student Answer Appendix
17. P(x) [x 0002 (2 0002 3i)][x 0002 (2 0002 3i)](x 4)2; degree 4 1 13 87. (A) 000212 13 (B) 3 2 i, 00022 0002 2 i 91. No, because P(x) is not a polynomial with real coefficients (the coefficient of x is the imaginary number 2i ).
Exercises 4-4 15. Domain: all real numbers except 0; x intercept: 3 17. Domain: all real numbers except 2; x intercept: 00026 19. Domain: all real numbers; x intercepts: 00024, 1 21. Domain: all real numbers except 6; x intercepts: none 23. Vertical asymptote: x 00022; horizontal asymptote: y 5 25. Vertical asymptotes: x 00024, x 4; horizontal asymptote: y 0 3 27. Vertical asymptote: x 0; horizontal asymptote: none 29. Vertical asymptotes: x 00023, x 0; horizontal asymptote: y 2 37. The graph of f is the same as the graph of g except that f has a hole at (00022, 16 ). 39.
41.
y 10
x
00025
5
000210
47.
49.
y
5
5
x
51.
y
77. 79.
5
x
10
73.
Vertical asymptote: x 1; oblique asymptote: y 2x 2
83.
y
5
5
x
00025
00025
5
x
00025
5
00025
87.
Vertical asymptote: x 0 Oblique asymptote: y 000214 x
Domain: x 000f 2, 00022 or (00020005, 00022) 傼 (00022, 2) 傼 (2, 0005); r(x)
y
y
5
00025
5
5
x
5
00025
91.
N
Vertical asymptote: x 2 Horizontal asymptote: y 0
As t S 0005, N S 5 N
50
50
25
25
25
x
00025
As t S 0005, N S 50
50
t
0
x
00025
Vertical asymptote: x 2 Oblique asymptote: y 12 x 0002 1
Domain: x 000f 2, or (00020005, 2) 傼 (2, 0005); f (x) x 2
0
x
000210
(2x 5)(x 0002 10) 100 55. f (x) x 0002 10 x4 1 Vertical asymptote: x 0; oblique asymptote: y 2x 0002 3 81. y y
Vertical asymptote: x 0 Oblique asymptote: y x
89.
00025
000210
3(x2 0002 1)(x2 0002 4)
00025
85.
5
10
00025
5
00025
y
00025
53. f (x)
x
00025
5
00025
5
00025
x
00025
5
y
45. 5
5
10
000210
y
43.
y
25
50
t
1 x00022
x
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Student Answer Appendix
93.
(A)
C(n) 25n 175
2,500 n
95.
(A)
L(x) 2x
450 x
(B) (0, 0005) (C) 15 ft by 15 ft (D) L
(B) 10 yr (C) C 2,000
200 1,000 100
0
25
50
n 0
25
50
x
Chapter 4 Review Exercise 1. 9. 11. 13. 29. 33. 47.
Zeros: 00021, 3; turning points: (00021, 0), (1, 2), (3, 0); P(x) S 0005 as x S 0005 and P(x) S 0005 as x S 00020005 (4-1) 1 3
1, 3, 5, 15, , (4-3) 5 5 3 (A) Domain: all real numbers except 5; x intercept: 0 (B) Domain: all real numbers except 00024 and 2; x intercept: 0002 7 The graph does not increase or decrease without bound as x S 0005 and as x S 00020005 (4-1) 1 i 13 1 0002 i 13 (x 1)(2x 0002 1) ax 0002 b ax 0002 b (4-3) 2 2 (A) Upper bound: 7; lower bound: 00025 (B) Four intervals (C) 00024.67, 6.62 (4-2) 3 3i 13 (A) 3 (B) 0002 (4-3) 49. (4-4) y 2 2
(4-4)
5
00025
5
x
00025
53. 3: None of the candidates for rational zeros ( 1, 2, and 4) are actually zeros. 63. (A)
(4-3)
59. v k
1T 1w
(4-5)
(B) 1,915
CHAPTER 5 Exercises 5-1 25.
The graph of g is the same as the graph of f stretched vertically by a factor of 3; g is increasing; horizontal asymptote: y 0
27.
The graph of g is the same as the graph of f reflected through the y axis and shrunk vertically by a factor of 13 ; g is decreasing; horizontal asymptote: y 0
y
y
10
10
00025
5
x
00025
000210
29.
5
000210
The graph of g is the same as the graph of f shifted upward 2 units; g is increasing; horizontal asymptote: y 2
31.
The graph of g is the same as the graph of f shifted 2 units to the left; g is increasing; horizontal asymptote: y 0
y
y
10
00025
10
5
000210
x
x
00025
5
000210
x
SA-19
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Student Answer Appendix
In every case, y 1. The function y 1x is simply the constant function y 1.
55.
The graph of g is the same as the graph of f reflected through the x axis; g is increasing; horizontal asymptote: y 0 y 10
00025
5
x
000210
57.
59. The graph of g is the same as the graph of f stretched vertically by a factor of 500; g is increasing; horizontal asymptote: y 0
The graph of g is the same as the graph of f stretched horizontally by a factor of 2 and shifted upward 3 units; g is decreasing; horizontal asymptote: y 3 y
y
10
2500
00025
5
x
000210
61.
000250
63.
The graph of g is the same as the graph of f shifted 3 units to the right, stretched vertically by a factor of 2, and shifted upward 1 unit; g is increasing; horizontal asymptote: y 1
50
The graph of g is the same as the graph of f shifted 2 units to the right, reflected through the origin, stretched vertically by a factor of 4, and shifted upward 3 units; g is increasing; horizontal asymptote: y 3
y
y
10
10
00025
5
x
00025
000210
65. 71. 73. 75. 79.
x
5
x
000210
e00022x(00022x 0002 3)
69. No local extrema; no x intercept; y intercept: 2.14; horizontal asymptote: y 2 x4 Local maximum: s(0) 1; no x intercepts: y intercept: 1: horizontal asymptote: x axis No local extrema; no x intercept; y intercept: 50; horizontal asymptotes: x axis and y 200 Local minimum: f(0) 1; no x intercepts; no horizontal asymptotes 21.4 2.6390; 21.41 2.6574; 21.414 2.6648; 21.4142 2.6651; 21.41421 2.6651; 21.414214 2.6651; 212 0007 2.6651
81.
83.
50
00024
4
50
00024
00025
4
00025
As x S 0005, fn (x) S 0; the line y 0 is a horizontal asymptote. As x S 00020005, f1(x) S 00020005 and f3(x) S 00020005, while f2(x) S 0005. As x S 00020005, fn(x) S 0005 if n is even and fn(x) S 00020005 if n is odd. 97. Flagstar: $5,488.61; UmbrellaBank.com: $5,470.85; Allied First Bank: $5,463.71
85.
Exercises 5-2 L
13. 1,000
500
5
10
n
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Student Answer Appendix 25.
SA-21
P 100
50
0
50
t
100
33. q approaches 0.000 9 coulombs, the upper limit for the charge on the capacitor. 35. (C) A approaches 100 deer, the upper limit for the number of deer the island can support. 37. y 14,910(0.8163)x; estimated purchase price: $14,910; estimated value after 10 years: $1,959 906 39. (A) y (B) 2010: 893.3 billion 2020: 903.6 billion 1 2.27e00020.169x
Exercises 5-3 19.
x 00023 00022 00021 0 1 2 3
21.
x
y 3x
x
y log3x
1 27 1 9 1 3
1 27 1 9 1 3
1 3 9 27
1 3 9 27
00023 00022 00021 0 1 2 3
y ( 23 ) x
00023 00022 00021 0 1 2 3
x
27 8 9 4 3 2
27 8 9 4 3 2
1
1
2 3 4 9 8 27
2 3 4 9 8 27
y 30
y 3x
y log3 x 00025 00025
30
y
y log2 3x 00023 00022 00021 0 1 2 3 79.
x
5
2 x
y 000430005 5
00025
00025
log x 0002 log y
61.
b is any positive real number except 1.
91.
The graph of g is the same as the graph of f shifted upward 3 units; g is increasing. Domain: (0, 0005); vertical asymptote: x 0
x
y log2/3 x
81.
x2y5 x 83. ln a b 85. ln a b y z 93. The graph of g is the same as the graph of f shifted 2 units to the right; g is decreasing. Domain: (2, 0005); vertical asymptote: x 2 4 log x 3 log y
y
y
10
10
000210
10
x
000210
000210
95.
10
000210
The graph of g is the same as the graph of f reflected through the x axis and shifted downward 1 unit; g is decreasing. Domain: (0, 0005); vertical asymptote: x 0
97.
The graph of g is the same as the graph of f reflected through the x axis, stretched vertically by a factor of 3, and shifted upward 5 units; g is decreasing. Domain: (0, 0005); vertical asymptote: x 0
y
y
10
000210
10
10
000210
x
x
000210
10
000210
x
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Student Answer Appendix
(A) f 00021(x) 2 0002 3x (B)
107.
109.
2
2
(C)
y
y
10
00021
3
00021
3
10
f 00021 000210
x
10
000210
00022 10
f
000210
00022
x
000210
Exercises 5-4 25.
(A)
y 11.9 24.1 ln x; 2008: 73.7%; 2015: 84.1%
(B) No; the predicted percentage goes over 100 sometime around 2034.
Exercises 5-5 87.
(A)
7.94 0010 1014 joules
Chapter 5 Review Exercises 55.
57.
The graph of g is the same as the graph of f stretched vertically by a factor of 2 and shifted downward 4 units; g is increasing. Domain: all real numbers Horizontal asymptote: y 00024 (5-1) y
The graph of g is the same as the graph of f stretched vertically by a factor of 2 and shifted upward 1 unit; g is decreasing. Domain: (0, 0005); Vertical asymptote: x 0 (5-3) y
10
000210
10
10
x
000210
10
000210
67.
000210
Domain f (0, 0005) Range f 00021 Range f (00020005, 0005) Domain f 00021
75.
(A) P
(5-3)
y 10
x
1,000
f 00021: y 2x 500
5
f : y log2 x
5
00025
10
0
x
25
50
t
00025
81.
(A)
y 43.3(1.09)x; 2010: $574 billion; 2020: $1,360 billion
CHAPTER 6 Exercises 6-1 y
15. Directrix x 00021 00025
y
17.
5
5
F (1, 0) 5
x
5
00025
00025
Directrix y 00022
y
19.
F (0, 2)
y
21.
5
x
F (00023, 0) 00025
5
Directrix x3 5
x
F (0, 00021)
Directrix y1 5
00025
00025
x
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Student Answer Appendix y
23.
y
45.
10
y
47.
5
10
Directrix x5
F (00025, 0)
10
000210
(4, 4)
x
(5.313, 5.646) x
5
00025
57.
2
2(x 0002 x) ( y a) ( y a)2 y2 2ay a2 x2
2
x
(0, 0)
00025
2
10
000210
(0, 0) 000210
SA-23
000210 2
2(x 0002 0) (y 0002 a) x2 (y 0002 a)2 x2 y2 0002 2ay a2 4ay
Exercises 6-2 15 .
Foci: F ¿ (0002 121, 0), F ( 121, 0); major axis length 10; minor axis length 4 y
17. Foci: F¿ (0, 0002 121), F (0, 121); major axis length 10; minor axis length 4 y
5
5
F
y
F
5
F
F x
5
00025
5
00025
00025
25.
19. Foci: F¿ (0002 18, 0), F ( 18, 0); major axis length 6; minor axis length 2
y
y
12
F
5
00025
29. Foci: F¿ (0002 13, 0), F ( 13, 0); major axis length 217 0007 5.29; minor axis length 4 y
5
5
F
00026
x
x
5
00025
27. Foci: F¿ (0, 0002 16), F (0, 16); minor axis length 2 112 0007 6.93; major axis length 216 0007 4.90
5
F
00025
F
00025
Foci: F¿ (0, 00024), F (0, 4); major axis length 10; minor axis length 6
x
5
00025
7
00027
6
F
x
F
00025
x
5
F F
000212 00025
00025 2
35.
y x2 1 25 9
2
37.
y x2 1 64 121
00025
2
51.
y x2 1; 7.94 feet approximately 400 144
53.
(A)
y2 x2 1 576 15.9
(B) 5.13 feet
Exercises 6-3 15.
Foci: F¿ (0002 113, 0), F ( 113, 0); transverse axis length 6; conjugate axis length 4
17.
Foci: F¿ (0, 0002 113), F (0, 113); transverse axis length 4; conjugate axis length 6
y c
F 0002c
F c 5
F
x
5
00025
0002c
y
10
23.
F 0002c
x
F
00025
Foci: F¿ (0, 00025), F (0, 5); transverse axis length 8; conjugate axis length 6
c
c
00025
F c 10
0002c F
x
Foci: F¿ (0002110, 0), F ( 110, 0); transverse axis length 4; conjugate axis length 216 0007 4.90 y
F 00025 0002c
F c
x
00025
25.
Foci: F¿ (0, 0002 111), F (0, 111); transverse axis length 4; conjugate axis length 217 0007 5.29 y
5
6 5
000210
000210
y
5
5
c
c
Foci: F¿ (0002 120, 0), F ( 120, 0); transverse axis length 4; conjugate axis length 8
y
5
21.
19.
c
00025 00026
F c 5
00027
c F c
x
0002c F 00025
7
x
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Student Answer Appendix
45. y
13 x 12 x2
y2
(A) Infinitely many;
0002
1 (0 0007 a 0007 1)
x2
y2
(C) One; y2 4x
1 (a 0006 1) a2 1 0002 a2 a2 a2 0002 1 49. (A) (2 13, 1 13), (00022 13, 00021 13) (B) No intersection points The graphs intersect at x 1 ( 21 0002 m 2) and y m ( 21 0002 m2) for 00021 0007 m 0007 1. 51. (A) No intersection points (B) (1 15, 3 15), (00021 15, 00023 15) The graphs intersect at x 1 ( 2m 2 0002 4) and y m ( 2m2 0002 4) for m 0007 00022 or m 0006 2. 2 y2 x 59. 0002 1; hyperbola 4 5
47.
(B) Infinitely many;
Chapter 6 Review Exercises 1.
Foci: F (00024, 0), F (4, 0); major axis length 10; minor axis length 6 (6-2)
3.
y
Foci: F¿ (0, 0002134), F (0, 134); transverse axis length 6; conjugate axis length 10 (6-3) 10
5
00025
x
10
000210
F (2, 0) x 5
00025
F F
(6-1)
5
y
5
F
y
5.
Directrix x 00022 00025
x
F 00025
21.
000210
y2 x2 1; ellipse (6-2) 36 20
CHAPTER 7 Exercises 7-1 {(3s 2, s, 00022s 0002 1) | s any real number} dh 0002 bk ak 0002 ch 49. x ,y , ad 0002 bc 000f 0 ad 0002 bc ad 0002 bc
35.
39.
{(00022s 5, s, 3s 0002 4) | s any real number}
45.
5(13 s 0002 43 , 23 s 0002 83 , s) ƒ s any real number6
61. (A) Supply: 143 T-shirts; demand: 611 T-shirts (B) Supply: 714 T-shirts; demand: 389 T-shirts (C) Equilibrium price: $6.36; equilibrium quantity: 480 T-shirts p (D)
Price ($)
20
Equilibrium point (480, 6.36) Supply curve
10
0
400
Demand curve q
800
Quantity
71.
$35,000 treasury bonds; $7,500 municipal bonds; $27,500 corporate bonds
Exercises 7-2 21.
x1 2t 3, x2 0002t 0002 5, x3 t, t any real number
27.
c
4 1
00026 00028 ` d 00023 2
29.
c
00024 4
12 00028 ` d 00026 00028
31.
c
1 8
00023 2 ` d 000212 000216
33.
c
1 0
00023 2 ` d 6 000216
35.
c
1 2
00023 2 ` d 0 000212
1 0 2 000253 1 0 0 00025 41. £ 0 1 0 † 4 § 43. £ 0 1 00022 † 13 § 49. Infinitely many solutions; for any real number s, x2 s, x1 2s 0002 3 0 0 0 0 0 1 00022 0 73. Either 11 CDs, 1 DVD and 1 book; 6 CDs, 4 DVDs, and 3 books; or 1 CD, 7 DVDs, and 5 books 81. One-person boats: t 0002 80; two-person boats: 00022t 420; four-person boats: t, 80 0003 t 0003 210, t an integer 83. No solution; no production schedule will use all the labor-hours in all departments.
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Student Answer Appendix
SA-25
Exercises 7-3 2 4
5 d 00026
11.
c
47.
00020.2 £ 2.6 00020.2
73.
31.
1.2 00020.6 § 2.2
3 £ 6 00029 49.
00022 £ 26 00022
53.
000211 d 00024
7 18
25 000225 45
00023 £ 000218 4
43.
000215 45 § 000225
55.
6 12 6 000226 £ 00024 2
8 10 § 24 000215 000218 43
45.
5 £4 0
000211 00027 10
15 3§ 4
000225 4§ 000219
77.
(A) $11.80 (B) $30.30 (C) MN gives the labor costs per boat at each plant. (D) Plant I Plant II $11.80 $13.80 One-person boat MN £ $18.50 $21.60 § Two-person boat $26.00 $30.30 Four-person boat
0 0 2 0 0 1 0 0 0 1 (A) A2 G 0 1 0 2 0 W 1 0 0 0 1 0 0 1 0 0 There is one way to travel from Baltimore to Atlanta with one intermediate connection; there are two ways to travel from Atlanta to Chicago with one intermediate connection. In general, the elements in A2 indicate the number of different ways to travel from the ith city to the jth city with one intermediate connection.
(B)
(C)
(A) (D)
(E)
(F)
83.
00026 4
Markup AM/FM Cruise Air radio control $77 $42 $27 $93 $95 $50 § $113 $121 $52
Basic car Model A $3,330 Model B £ $2,125 Model C $1,270
81.
c
41.
$33 $26 1 (A B) c d 2 $57 $77 This is the average cost of materials and labor for each product at the two plants.
75.
79.
00022 00024 00024 00028 § 6 12 000231 16 £ 61 000225 § 00023 77
(A)
2 0 0 0 2 0 1 0 2 0 A3 G 0 0 3 0 0 W 0 1 0 2 0 1 0 0 0 1 There is one way to travel from Denver to Baltimore with two intermediate connections; there are two ways to travel from Atlanta to El Paso with two intermediate connections. In general, the elements in A3 indicate the number of different ways to travel from the ith city to the jth city with two intermediate connections. 2 3 2 5 2 1 1 4 2 1 A A2 A3 A4 G 4 1 3 2 4 W 1 1 4 2 1 1 1 1 3 1 It is possible to travel from any origin to any destination with at most three intermediate connections. $3,550 (B) $6,000 (C) NM gives the total cost per town. Cost/town $3,550 Berkeley NM c d $6,000 Oakland Telephone House call call Letter 1,300 13,000 ] [1 1] N [ 3,000 Total contacts 1 6,500 Berkeley N £1§ c d 10,800 Oakland 1 0 1 0 G 0 0 1
0 0 1 0 0 1
1 0 0 0 1 1
1 1 1 0 1 0
1 1 0 0 0 0
0 0 0 W 1 1 0
(B)
0 1 1 G 1 1 2
1 0 1 1 2 2
2 2 0 1 2 2
3 3 2 0 2 3
1 2 1 0 0 2
2 2 1 W 1 2 0
(C)
9 1 10 1 6 1 BC G W where C G W 4 1 9 1 11 1
00023 x1 2 dc d c d 2 x2 1
31.
1 £ 00021 2
(D) Frank, Bart, Aaron and Elvis (tie), Charles, Dan
Exercises 7-4 11.
c
2 4
00023 d 5
25.
2x1 0002 x2 3 x1 3x2 00022
29.
c
4 1
00021 00022 1 x1 1 0 § £ x2 § £ 2 § 00023 3 1 x3
41.
c
1 0
00029 d 1
bar19499_saa_SA1-SA30.qxd
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81.
(A)
5:16 PM
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Student Answer Appendix
00025 00022 d 2 1
43.
11/20/09
c
45.
(AB)00021 c
00023 00022
00027 d 00025
51.
0 00021 00021 £ 00021 00021 00021 § 00021 00021 0
29 000212
000241 23 000233 d , A00021B00021 c d, 17 000216 23
29 000212
000241 d 17
B00021A00021 c
000219 £ 15 00022
53.
9 00027 00027 6§ 1 00021
1 0 1 £ 12 21 1 § 2 1 4
55.
59.
00029 000215 10 £ 4 5 00024 § 00021 00021 1
83. 61 22 96 38 115 43 131 47 68 27 110 43 85. BEYONCE KNOWLES 87. 42 43 88 33 101 40 61 62 40 49 40 103 72 56 69 52 81 99 53 101 89. RAIDERS OF THE LOST ARK 91. Concert 1: 6,000 $20 tickets, 4,000 $30 tickets Concert 2: 5,000 $20 tickets, 5,000 $30 tickets Concert 3: 3,000 $20 tickets, 7,000 $30 tickets
Exercises 7-5 19.
x 000265 , y 35
x 172 , y 000220 17
21.
4 00022
23. `
6 ` 8
25. `
00021 ` 00022
5 0
27.
(00021)1 1 `
4 00022
6 ` 44 8
29.
(00021)2 3 `
5 0
00021 ` 10 00022
x 43 , y 000213 , z 23 49. x 00029, y 000273 , z 6 51. x 32 , y 000276 , z 23 3 15 If a 2 and b 4 , there are an infinite number of solutions. If a 32 and b 000f 154 , there are no solutions. If a 000f 83 , there is one solution. (A) Since D 0, the system either has no solution or infinitely many. Since x 0, y 0, z 0 is a solution, the second case must hold. (B) Since D 000f 0, by Cramer’s rule, x 0, y 0, z 0 is the only solution. 73. (A) R 200p 300q 0002 6p2 6pq 0002 3q2 (B) p 00020.3x 0002 0.4y 180, q 00020.2x 0002 0.6y 220, R 180x 220y 0002 0.3x2 0002 0.6xy 0002 0.6y2
47. 53. 71.
Chapter 7 Review Exercises 3.
Infinitely many solutions [t, (4t 8)00023], for any real number t (7-1)
3 00026 12 4 8 3 3 ` d (7-2) d (7-3) d (7-3) 13. c 17. c 1 00024 5 000212 18 00024 9 23. (A) x1 00021, x2 3 (B) x1 1, x2 2 (C) x1 8, x2 000210 (7-4) 27. x1 2, x2 00022; each pair of lines has the same intersection point. (7-1, 7-2) 7.
c
x2
x2
x2
5
5
5
00025
x1
5
00025
5
(2, 00022)
x2
x1
5
00025
5
x1
00025
5
(2, 00022)
(2, 00022)
(2, 00022)
x1
00025
x1 0002 x2 4 3x2 00026
x1 0002 x2 4 2x1 x2 2 35. 57.
16 00029 40 000230 § 00028 17
7 £ 28 000221
(7-3)
37.
12 £ 0 00028
24 0 000216
x1 0002 x2 4 x2 00022 00026 0§ 4
x1 2 x2 00022
(7-3)
(A) $27 (C)
(B) Elements in LH give the total cost of manufacturing each product at each plant. North South Carolina Carolina $46.35 $41.00 Desks LH c d (7-3) Stands $30.45 $27.00
CHAPTER 8 Exercises 8-1 15.
99 101
19.
1 10
1 1 100 1,000
25.
0.3, 0.33, 0.333, 0.3333, 0.33333
55.
57.
1
0
20
00020.3
59.
4 1
8 2
0002
16 3
0002
32 4
73.
(A) 3, 1.83, 1.46, 1.415
(B) Calculator 12 1.4142135. . .
1.5
0
20
00021.5
(C)
a1 1; 1, 1.5, 1.417, 1.414
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Student Answer Appendix 81. 83.
(A) 0.625 ft; 0.02 ft (B) 19.98 (A) 40,000, 41,600, 43,264, 44,998.56, 46,794.34, 48,666.12
(B)
40,000(1.04)n00021
(C)
SA-27
265,319.02
Exercises 8-2 11. P1: a5a1 a5 1; P2: a5a2 a5(a1a) (a5a)a a6a a7 a5 2; P3: a5a3 a5(a2a) a5(a1a)a [(a5a)a]a a8 a5 3 13. P1: 91 0002 1 8 is divisible by 4; P2: 92 0002 1 80 is divisible by 4; P3: 93 0002 1 728 is divisible by 4 15. Pk: 2 6 10 . . . (4k 0002 2) 2k2; Pk 1: 2 6 10 . . . (4k 0002 2) (4k 2) 2(k 1)2 n(n 0002 1) 17. Pk: a5ak a5 k; Pk 1: a5ak 1 a5 k 1 49. 1 2 3 . . . (n 0002 1) ,n00042 2
Exercises 8-3 7. (A) Arithmetic with d 00025; 000226, 000231
(B) Geometric with r 00022; 000216, 32
(C) Neither
1 (D) Geometric with r 13 , 541 , 162
Exercises 8-4 39. 43. 49. 55. 57. 59.
No repeats: 10 ⴢ 9 ⴢ 8 ⴢ 7 ⴢ 6 30,240; with repeats: 10 ⴢ 10 ⴢ 10 ⴢ 10 ⴢ 10 100,000 26 ⴢ 26 ⴢ 26 ⴢ 10 ⴢ 10 ⴢ 10 17,576,000 possible license plates; no repeats: 26 ⴢ 25 ⴢ 24 ⴢ 10 ⴢ 9 ⴢ 8 11,232,000 (B) r 0, 10 (C) Each is the product of r consecutive integers, the largest of which is n for Pn,r and r for r! Two people: 5 ⴢ 4 20; three people: 5 ⴢ 4 ⴢ 3 60; four people: 5 ⴢ 4 ⴢ 3 ⴢ 2 120; five people: 5 ⴢ 4 ⴢ 3 ⴢ 2 ⴢ 1 120 (A) P8,5 6,720 (B) C8,5 56 (C) C2,1 ⴢ C6,4 30 There are C4,1 ⴢ C48,4 778,320 hands that contain exactly one king, and C39,5 575,757 hands containing no hearts, so the former is more likely.
Exercises 8-5 (A) No probability can be negative (B) P(R) P(G) P(Y ) P(B) 000f 1 (C) Is an acceptable probability assignment. C16,5 0007 .0017 31. C52,5 53. (A) P(2) 0007 .022, P(3) 0007 .07, P(4) 0007 .088, P(5) 0007 .1, P(6) 0007 .142, P(7) 0007 .178, P(8) 0007 .144, P(9) 0007 .104, P(10) 0007 .072, P(11) 0007 .052, P(12) 0007 .028 (B) P(2) 361 , P(3) 362 , P(4) 363 , P(5) 364 , P(6) 365 , P(7) 366 , P(8) 365 , P(9) 364 , P(10) 363 , P(11) 362 , P(12) 361
19.
(C) Sum
Expected frequency
2 3 4 5 6 7
13.9 27.8 41.7 55.6 69.4 83.3
Sum
Expected frequency
8 9 10 11 12
69.4 55.6 41.7 27.8 13.9
Exercises 8-6 21. m3 3m2n 3mn2 n3 23. 8x3 0002 36x2y 54xy2 0002 27y3 25. x4 0002 8x3 24x2 0002 32x 16 4 3 2 2 3 4 5 4 27. m 12m n 54m n 108mn 81n 29. 32x 0002 80x y 80x3y2 0002 40x2y3 10xy4 0002 y5 31. m6 12m5n 60m4n2 160m3n3 240m2n4 192mn5 64n6 51. 3x 2 3xh h 2; approaches 3x 2 4 3 2 2 3 4 4 53. 5x 10x h 10x h 5xh h ; approaches 5x
Chapter 8 Review Exercises 1. (A) Geometric (B) Arithmetic (C) Arithmetic (D) Neither (E) Geometric (8-1, 8-3) 3. (A) 16, 8, 4, 2 (B) a10 321 (C) S10 3131 (8-1, 8-3) 9. 21 (8-4) 32 11. (A) 12 combined outcomes: (B) 6 ⴢ 2 12 (8-5) (1, H) H 1
T
(1, T)
2
H T
(2, H) (2, T)
3
H T
(3, H) (3, T)
4
H T
(4, H) (4, T)
5
H T
(5, H) (5, T)
6
H T
(6, H) (6, T)
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17. P1: 5 12 4 ⴢ 1 5; P2: 5 7 22 4 ⴢ 2; P3: 5 7 9 32 4 ⴢ 3 (8-2) 21. Pk: 2 4 8 . . . 2k 2k 1 0002 2; Pk 1: 2 4 8 . . . 2k 2k 1 2k 2 0002 2 n (00021)k 1 C13,5 C13,3 ⴢ C13,2 1 ; S0005 27. Sn a (8-3) 33. (A) (B) (8-5) k 4 C C52,5 3 52,5 k 1
(8-2)
APPENDIX A Cumulative Review Exercise: Chapters 1–3 3.
00025 0007 x 0007 9 (00025, 9) (
(
00025
(1-3)
x
9
13. (A) Function; domain: {1, 2, 3}; range: {1} (B) Not a function 2 23. 3 0003 m 0003 2 (1-3) 25. x 0004 2, x 000f 4 (1-2) [ 2, 4) ´ (4, 0005) [23 , 2] [
[
]
m
2
2 3
2
4
(C) Function; domain: {00022, 00021, 0, 1, 2}; range: {00021, 0, 2} (3-1)
x
35. (A) All real numbers (B) {00022} 傼 [1, 0005) (C) 1 (D) [00023, 00022] and [2, 0005) x ; Domain: x 000f 0, 3 (3-5) 39. ( f ⴰ g)(x) 30002x 41. Domain: (00020005, 0005) (3-2) 45. Center: (3, 00021); radius: 110 (2-2) Range: (00020005, 00021) ´ [1, 0005) f(x) Discontinuous at x 0
(E) 00022, 2 (3-1, 3-2)
5
y 5
5
00025
5
x
(3, 00021)
x 00025
00025
53.
(A) Domain g: [ 00022, 2 ]
f f x2 ; Domain a b: (00022, 2) (B) a b(x) g g 24 0002 x2
(C) (f ⴰ g)(x) 4 0002 x2; Domain (f ⴰ g): [00022, 2]
(3-5)
if 0 0003 x 0003 60 if 60 6 x 0003 150 if 150 6 x 0003 300 if 300 6 x (3-2)
0.06x 0.05x 0.6 61. C(x) μ 0.04x 2.1 0.03x 5.1 C (x)
20
10
60 150
63.
(A)
300
500
x
1 if n is an odd integer (3-2) if n is an even integer 0 (B) Loss: $0 0003 p 0007 5.5 or p 0006 $8 or [$0, $5.5) 傼 ($8, 0005) (3-4) (B) s f 00021 (L) 2 120L 0002 126; domain: [22.5, 0005); range: [20, 0005)
f(1) f(3) 1, f(2) f(4) 0 (B) f (n) e
65. (A) Profit: $5.5 0007 p 0007 $8 or ($5.5, $8) 69. (A) (3-4) L 400
80
s
(C) 67 mph
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SA-29
Cumulative Review for Chapters 4 and 5 17.
(A) Domain: x 000f 00022; x intercept: x 00024; y intercept: y 4 (C) (4-4) y
(B) Vertical asymptote: x 00022; horizontal asymptote: y 2
10
000210
10
x
000210
(A) 00020.56 (double zero); 2 (simple zero); 3.56 (double zero) (B) 00020.56 can be approximated with a maximum routine; 2 can be approximated with the bisection: 3.56 can be approximated with a minimum routine (4-2) 51. A reflection through the x axis transforms the graph of y In x into the graph of y 0002ln x. A reflection through the y axis transforms the graph of y ln x into the graph of y ln (0002x). (5-3) y 55. Vertical asymptote: x 00022; (4–4) oblique asymptote: y x 2 10
23.
000210
10
x
000210
Cumulative Review for Chapters 6–8 5. (A) Arithmetic (B) Geometric (C) Neither (D) Geometric (E) Arithmetic 7. (A) 2, 5, 8, 11 (B) a8 23 (C) S8 100 (8-3) 11. Foci: F (0002 161, 0), F ( 161, 0); 13. (6-1) transverse axis length 12; y conjugate axis length 10 (6-3) 4
y 10
2
F
F
10
000210
(8-3)
x
00022
F 00040,
9 25
0005
x Directrix 9 y 0002 25
000210
17. 23. 25. 31. 39. 69. 93.
95.
0 00023 1 7 d d (B) Not defined (C) [3] (D) c (E) [00021, 8] (F) Not defined (7-4) 3 00029 4 00027 (B) x1 2t 3, x2 t, t any real number. (C) No solution (7-3) (A) x1 3, x2 00024 1 00023 x1 k1 00025 3 (A) c (B) A00021 c (C) x1 13, x2 5 dc d c d d (D) x1 000211, x2 00024 (7-5) k2 2 00025 x2 00022 1 2 2 Pk: k k 2 2r for some integer r; Pk 1: (k 1) (k 1) 2 2s for some integer s (8-2) 00021 2 d (A) c (B) Not defined (7-4) 41. (0, i ), (0, 0002i), (1, 1), (00021, 00021) (7-6) 2 3 (A) Infinite number of solutions (B) No solution (C) Unique solution (7-3) 1 model A truck, 6 model B trucks, and 5 model C trucks; or 3 model A trucks, 3 model B trucks, and 6 model C trucks; or 5 model A trucks and 7 model C trucks. (7-3) 82.25 Ann 83 Ann 0.25 0.2 83 Bob 84.8 Bob 0.25 0.2 (A) M ≥ (B) M ≥ ¥ G 92 W Carol ¥ G 91.8 W Carol 0.25 0.2 83.75 Dan 85.2 Dan 0.25 0.4 82 Eric 80.8 Eric (C) Class averages Test 1 Test 2 Test 3 Test 4 [0.2 0.2 0.2 0.2 0.2]M [84.4 81.8 85 87.2] (7-4) (A)
c
APPENDIX B Exercises B-2 13.
3 1 0002 3x 4 2x 0002 3
15.
2 1 3 0002 0002 x x00023 (x 0002 3)2
17.
2 3x 0002 1 2 x x 2x 3
19.
2x x2 2
3x 5 (x 2 2)2
23.
2x 5 2 2 x00023 x 3x 3
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2 1 3 0002 x00024 x 3 (x 3)2
29. x 2 0002
1 x00021 2 2 x 2 2x 0002 1 2x 0002 x 1
Exercises B-3 3.
y 00022x 0002 2; straight line
5.
y 00022x 0002 2, x 0003 0; a ray (part of a straight line) y
y
00025
5
x
5
x
y2 4x; parabola
11.
00025
5
00025
00025
y
y2 4x, y 0004 0; parabola (upper half) y
5
5
00025
5
00025
x
00025
5
x
00025 2
29.
y
5
00025
00025
At Dt F , 00020005 6 t 6 0005; parabola 0002E (A) The graphs are symmetric about the line y x. (B) 1. y ex 2. x ex or y ln x Function 2 is the inverse of function 1.
21. x t, y
y 000223 x; straight line
5
5
9.
7.
23. y2 0002 x2 8, x 0004 1, y 0004 3; part of a hyperbola
x
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SUBJECT INDEX Abscissa, 110 Absolute value definitions for, 65 distance and, 66 method to find, 65–66 to solve radical inequalities, 71–72 Absolute value equations geometric interpretation of, 67–68 method to solve, 66–70, 99–100 verbal statements as, 68–69 Absolute value functions, 188 Absolute value inequalities geometric interpretation of, 67–68 method to solve, 66–70 Absolute value problems method to solve, 69–71 solved geometrically, 67–68 with two cases, 71 Acceptable probability assignment, 547 Actual probability, 552 Addition associative property of, 4, 6 commutative property of, 4, 6 of complex numbers, 76–77 elimination by, 429–434 explanation of, 3–4 of matrices, 457–458 of polynomials, 23 of rational expressions, 34–36 of real numbers, 3–7 Addition properties of equality, 45 of matrices, 477 of real numbers, 6 Additive identity, 4, 77 Additive inverse, 4, 6 Adiabatic process, 530 Algebra, 1 Algebraic equations. See also Equations algebraic expressions vs., 49 explanation of, 44 Algebraic expressions algebraic equations vs., 49 containing radicals, 17 explanation of, 21 factor of, 25 Algorithm, division, 267 Analytic geometry basic problems studied in, 122 fundamental theorem of, 110 Approximation by rational numbers, 5 of real zeros, 282–283 Arithmetic sequences explanation of, 520, 564–565 method to find terms in, 522–523 method to recognize, 521 nth term of, 522 Arithmetic series, 523–524 Associative property of addition, 4, 6 of multiplication, 4 Asymptote rectangle, 407
Asymptotes on graphing calculator, 306 horizontal, 303–304 oblique, 308 vertical, 302–304 Augmented matrices explanation of, 443 Gauss-Jordan elimination and, 447 interpretation of, 445 method to write, 443–444 reduced, 450, 451 Axis of cone, 386n conjugate, 407–409 of ellipse, 395 of hyperbola, 405, 407 of parabola, 1111 of symmetry, 205, 387 transverse, 405 Base of exponent, 11 of exponential functions, 329, 331–333 Bell, Alexander Graham, 365 Binomial coefficients, 22–23, 560 Binomial expansion, 558–559 Binomial formula explanation of, 559–560, 565 proof of, 562–563 use of, 560–562 Binomials, 22. See also Polynomials Bisection method, 281–282 Briggsian logarithms. See Common logarithms Calculators. See Graphing calculators Carbon-14 decay equation, 343–344 Cardano, Girolamo, 108 Cardano’s formula, 108 Cartesian coordinate system, 110, 157–158 Catenary curve, 374, 391 Center of circle, 127, 129 of ellipse, 395 of hyperbola, 405, 408 Change-of-base formula, 361–362 Circles equations of, 126–128 explanation of, 127, 386 formulas for, 599 graphs of, 126–128 Closure property, 6 Coefficient determinant, 492 Coefficient matrix, 443 Coefficients binomial, 22–23, 560 in linear systems, 424 of polynomial functions, 260–261 real, 290–291 Cofactor of element explanation of, 489 method to find, 489–490 Column matrices, 442, 461–462
I-1
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SUBJECT INDEX
Combinations, 538–541, 565 Combined properties, of matrices, 477 Combined variation, 319 Common difference, 520 Common factors, 26, 33 Common logarithms, 359, 360 Common ratio, 521 Commutative property, 4, 6 Completing the square, 86–87 Complex numbers addition of, 76–77 division of, 78–79 explanation of, 74–76, 105 historical background of, 74 multiplication of, 77–78 operations with, 76–79 radicals and, 80–81 set of, 75 solving equations involving, 81–82 subtraction of, 76–77 types of, 75 zero of, 77 Composite numbers, 25 Composition of functions, 226–230, 252 inverse functions and, 240 Compound events, 544 Compound fractions, 36, 37 Compound interest applications of, 334, 373 continuous, 335–336 explanation of, 333–334 Conditional equations, 45 Conic sections. See also Circles; Ellipses; Hyperbolas; Parabolas explanation of, 386–387 review of, 418–420 Conjecture, 511–512, 517 Conjugate, of a ⫹ bi, 75 Conjugate axis, 407–409 Conjugate hyperbolas, 410 Consistent systems, 427 Constant in term of polynomial, 22 of variation, 316–318 Constant functions, 178, 179 Constant matrix, 443 Constant terms, 424 Continuous compound interest, 335–336 Continuous compound interest formula, 336 Continuous graphs, 181 Contraction. See Shrinking Coordinate, 3, 110 Coordinate axis, 110. See also x axis; y axis Correspondence, 162, 167 Counterexamples, 511–512 Counting techniques combinations and, 538–541 explanation of, 531–532 factorial notation and, 534–536 multiplication principle and, 532–534 permutations and, 536–538 Cramer, Gabriel, 491 Cramer’s rule explanation of, 491–492, 498 to solve three-variable system, 493–494 to solve two-variable system, 492–494 for three equations in three variables, 493 Cube functions, 189
Cube root functions, 189 Cube roots, 14 Cubic equations, 108 Cubic models, 272 Curve fitting, 151 Curves catenary, 374, 391 explanation of, 151 plane, 592, 593 Data analysis examples of, 271–273 regression and, 346–349, 369 Decibels, 365, 366 Decimal expansions, 5 Decoding matrix, 482 Decreasing functions, 178, 238 Degenerate conic, 387 Degree, of polynomials, 22, 260 Demand, 93, 435 Denominator explanation of, 9 least common, 35 rationalizing the, 18–19 Dependent variables, 164 Descartes, René, 11 Determinants coefficient, 492 explanation of, 487, 498 first-order, 487–488 second-order, 487, 488 to solve systems of equations, 491–494 third-order, 488–491 Diagonal expansion, 495 Difference function, 224–225 Difference of cubes formula, 28, 29 Difference of square formula, 28, 29 Difference quotient, 170 Dimensions, of matrix, 442 Directrix, of parabola, 387 Direct variation, 316 Discriminant, 90–91 Distance absolute value and, 66 in plane, 123–129, 158 between two points, 123–124 Distance formula explanation of, 124 use of, 124, 388, 396, 406–407 Divisibility property, 516 Division of complex numbers, 78 long, 5, 266–267 polynomial, 266–269 of rational expressions, 33–34 of real numbers, 7 synthetic, 268–269 Division algorithm, 267 Division properties, of equality, 45 Divisor, 267 Domain of exponential functions, 355 of functions, 163, 164, 166–167, 169–170, 176–177, 204, 225, 229, 230 implied, 166 of rational functions, 299–300 of variables, 44–45
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SUBJECT INDEX
Double inequalities, 61, 67 Double subscript notation, 442 Double zero, 289 Doubling time, 340 Doubling time growth model, 340, 341 Eccentricity, 417 Element of matrix, 442 of set, 2 Elimination, by addition, 429–434 Ellipses applications of, 400–401 equations of, 396–400 explanation of, 386, 395, 418–420 graphs of, 396–400 method to draw, 395–396 Empirical probability application of, 555–556 approximate, 552–554 explanation of, 552, 553 method to find, 553–554 Empty set, 2 Endpoints, 57 Equality explanation of, 76 properties of, 45, 477 symbols for, 57 Equallly likely assumptions, 549–550 Equal polynomials, 585–586 Equations. See also Linear equations; Systems of linear equations; specific types of equations of circles, 126–128 conditional, 45 cubic, 108 defining functions by, 164–167 of ellipse, 396–400 equivalent, 45 explanation of, 44, 56 exponential, 372–374, 380 graphs of, 111–112, 118 of hyperbolas, 406–408 as identities, 45 involving complex numbers, 81–82 involving radicals, 97–99 of lines, 132, 133, 137–140, 158 logarithmic, 375–376, 380 matrix, 477–480 of parabola, 209 parametric, 591–595 price-demand, 93 properties of, 69 quadratic, 84–93, 105 of quadratic type, 101–102, 105 solution set of, 44, 111 solutions of, 44, 111 squaring operation on, 98 in two variables, 111 Equilibrium price, 435 Equilibrium quantity, 435 Equivalent equations, 45 Equivalent inequalities, 59–60 Equivalent systems of equations, 429 Euler, Leonhard, 74 Even functions, 196–197 Events compound, 544
explanation of, 546–547 probability of, 547–551 simple, 544, 547 Expected frequency, 553 Experiments, 543–544 Exponential decay, 350 Exponential equations explanation of, 372, 380 method to solve, 372–374 Exponential functions with base e, 331–333 compound interest and, 333–336 domain of, 355 explanation of, 328–329, 379 graphs of, 329–333 inverse of, 329 (See also Logarithmic functions) properties of, 330–331 transformations of, 330 Exponential growth/decay, 349, 350 Exponential models application of, 379 data analysis and regression and, 346–349 exponential growth phenomena and, 349–350 on graphing calculator, 347 mathematical, 340–346 Exponents explanation of, 11, 39 integer, 11–13 rational, 16 Extended principle of mathematical induction, 517 Extraneous solutions, 98, 105 Extrapolation, 153 Face cards, 540 Factorials, 534–536 Factoring explanation of, 25 by grouping, 26–27 of polynomials, 25–29 to solve quadratic equations, 84–86 Factoring formulas, 28, 29 Factors of algebraic expression, 25 common, 26 explanation of, 25 of polynomials with real coefficients, 290–291 Factor theorem, 270 Fermat’s last theorem, 517 Fibonacci, Leonardo, 505 Fibonacci sequences, 505 Finite sequences arithmetic, 523–524 explanation of, 505 Finite series arithmetic, 523–524 explanation of, 507 geometric, 526–527 Finite sets, arithmetic, 523–524 First-degree equations. See Linear equations First-degree functions. See Linear functions First-order determinants, 487–488 Focal chords, 393, 422 Focus of ellipse, 395–398 of hyperbola, 405 of parabola, 387 Fractional expressions, 32
I-3
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SUBJECT INDEX
Fractions compound, 36–37 explanation of, 9 fundamental property of, 32 partial, 585–590 proper, 585 properties of, 9 raised to higher terms, 32 reduced to lowest terms, 32 significant digits in decimal, 583 simple, 36 solving inequalities involving, 61 Frequency, 553 Functions. See also specific types of functions absolute value, 188 applications of, 170–171 composition of, 226–230, 240, 252 constant, 178, 179 cube, 189 cube root, 189 decreasing, 178, 238 defined by equations, 164–166 definition of, 163, 235, 250 difference, 224–225 difference quotient and, 170 domains of, 166–167, 169–170, 176–177, 204, 225, 229, 230 even, 196–197 exponential, 328–336, 379 graphs of, 175–184, 188–199, 250–251 greatest integer, 182, 183 identity, 179, 188 increasing, 178, 238 inverse, 235–246, 252 linear, 178–180 logarithmic, 354–362, 379–380 notation for, 167–168 odd, 196–197 one-to-one, 235–238, 358 operations on, 223–226 overview of, 162 piecewise-defined, 180–181 polynomial, 260–266 product, 224–225 quadratic, 203–211, 251–252 quotient, 224–225 range of, 163, 166, 177 rational, 298–310 set form of definition of, 163 square, 188, 203, 204 square root, 189 sum, 224–225 transformations of, 188–197, 251 vertical line test for, 166 Fundamental counting principle. See Multiplication principle Fundamental property of fractions, 32 Fundamental sample space, 545 Fundamental theorem of algebra, 288–289 Fundamental theorem of analytic geometry, 110 Fundamental theorem of arithmetic, 25 Gauss, Carl Friedrich, 288, 447 Gauss-Jordan elimination explanation of, 441, 447 on graphing calculators, 449 to solve linear systems, 447–451, 497 use of, 475
General form, of quadratic function, 204 Geometric sequences explanation of, 521, 564–565 method to find terms in, 522–523 method to recognize, 521 nth term of, 522 Geometric series sum formulas for finite, 525 sum formulas for infinite, 526–527 Goldbach’s conjecture, 517 Graphing calculator features INTERSECT, 361 MATRIX-MATH, 488 maximum and minimum, 209 random number generator, 554 rref on, 449 table, 561 TRACE, 134, 280, 588 viewing window, 127 ZERO command, 280 ZSquare, 127 Graphing calculators asymptotes on, 306 circles on, 127 cubic models on, 272 domain of functions on, 225 ellipses on, 399 exponential functions on, 328, 331 exponential models on, 347 graphs of equations on, 112, 118, 143 greatest integer functions on, 183 interest rate on, 335 inverse functions on, 246 linear systems on, 425 logarithms on, 359–360, 370 logistic models on, 349 matrices on, 442, 458, 473, 488 parabolas on, 390 parametric equations on, 592 partial fraction decomposition on, 588 polynomial inequalities on, 284 quadratic regression on, 215, 216 quartic model on, 273 rational inequalities on, 311 reduced echelon form on, 449 regression on, 153 scientific notation on, 13–14 sequences on, 505, 507 sum of series on, 526 Graphs/graphing of circles, 126–128 continuous, 181 of ellipses, 396–400 of equation in two variables, 111 explanation of, 111 of exponential functions, 329–333 of functions, 175–184, 188–199, 250–251 horizontal and vertical shifts in, 189–191 of hyperbolas, 406–412 of inequalities, 58, 59 of intervals, 58, 59 of inverse functions, 244–246 line, 57 of linear functions, 179–180 of lines, 132–133 of logarithmic functions, 354–356, 359–361 multiplicities from, 292
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SUBJECT INDEX
of parabolas, 111, 389–390 point-by-point plotting on, 111 of polynomial functions, 260–266, 280 of polynomials, 266, 291–292 of quadratic functions, 204–209 of rational functions, 299–301, 304–310 reflections of, 114, 191–193 stretching and shrinking in, 193–196 symmetry as aid in, 113–117 of systems of linear equations, 424–425 Greatest integer, 182 Greatest integer functions, 182, 183
Half-life, 342 Half-life decay model, 342 Horizontal asymptotes, of rational functions, 303–304 Horizontal axis, 110. See also x axis Horizontal lines, 139, 140 Horizontal line test, 237 Horizontal shifts, 189–191, 195 Horizontal shrinks, 194, 195 Horizontal stretches, 194, 195 Hyperbolas applications of, 412–414 conjugate, 410 equations of, 406–408 explanation of, 387, 405, 420 graphs of, 406–412 method to draw, 406 Hyperbolic paraboloids, 412 Hyperboloids, 412
Identities, 45 Identity functions, 179 Identity matrix, for multiplication, 470–471 Identity property, 6 Imaginary numbers, 75 Imaginary unit, 74–75 Imaginary zeros, of polynomials, 290, 295 Implied domain, 166 Inconsistent systems, in two variables, 427 Increasing functions, 178, 238 Independent systems, 427 Independent variables, 164, 165 Index, 15 Induction. See Mathematical induction Inequalities absolute value, 66–70 applications for, 61–62 double, 61, 67 equivalent, 59 explanation of, 57 graphs of, 58, 59 linear, 56–62, 105 polynomial, 283–284, 322 properties of, 60, 69 quadratic, 211–214, 252 radical, 71–72 rational, 310–311, 322–323 solution set for, 59–60 symbols for, 57 Infinite sequences explanation of, 505 geometric, 526–527
I-5
Infinite series explanation of, 507 geometric, 526–527 Infinity, symbol for, 57 Integer exponents explanation of, 11–12 properties of, 12–13 Integers explanation of, 2, 3 greatest, 182, 183 set of, 2 Intercepts. See also x intercepts; y intercepts of functions, 176–177 to graph lines, 133 of rational functions, 305 Interest compound, 333–336, 373 explanation of, 333 Interest rate, 333 Interpolation, 153 Intersections, 59 Intervals explanation of, 57 graphs of, 58, 59 notation for, 57–58, 177 Inverse additive, 4, 6 of functions, 238–242, 245–246 method to find, 476 multiplicative, 4, 6, 11, 471–472 to solve linear systems, 478–480, 498 of square matrix, 471–473, 476 Inverse functions explanation of, 235, 252 graphs of, 244–246 method for finding inverse and, 238–242 modeling with, 242–243 one-to-one, 235–238 properties of, 239 Inverse variation, 316–317 Irrational numbers explanation of, 2, 5 historical background of, 74 Joint variation, 318 Lagranges’ four square theorem, 517 Leading term, 264 Learning curves, 344–345 Least common denominator (LCD), 35 Least-squares line, 383 Like terms, of polynomials, 23 Limited growth, 350 Linear and quadratic factors theorem, 290, 586 Linear equations. See also Equations; Systems of linear equations explanation of, 104 with more than one variable, 46–47 in one variable, 45–46 Linear factors theorem, 289 Linear functions. See also Functions explanation of, 178–179 graphs of, 179–180 Linear inequalities. See also Inequalities applications for, 61–62 explanation of, 56, 57, 105 graphs of, 59–62
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SUBJECT INDEX
Linear models, construction of, 149–151 Linear regression examples of, 152–154 explanation of, 151 Linear systems. See Systems of linear equations Line graph, 57 Lines equations of, 132, 133, 137–140, 158 graphs of, 132–133 horizontal, 139, 140 parallel, 141–142 perpendicular, 141–142 regression, 153 slope-intercept form of, 137–138 slope of, 134–136 vertical, 139, 140, 166 Line segment length of, 66 midpoint of, 124–126 Location theorem, 280–281 Logarithmic equations explanation of, 372, 380 method to solve, 375–376 Logarithmic-exponential conversions, 356–357 Logarithmic-exponential relationships, 360–361 Logarithmic functions change-of-base formula and, 361–362 conversions of, 356–357 explanation of, 329, 354, 379–380 graphs of, 354–356 properties of, 358–359, 380 Logarithmic models applications of, 380 data analysis and regression, 369 logarithmic scales, 365–369 Logarithmic scales, 365–369 Logarithms common, 359, 360 on graphing calculator, 359–361, 370 natural, 359, 360 Logistic growth, 350 Logistic models, 349 Long division explanation of, 5 polynomial, 266–267 Lower triangular matrix, 571 Lowest terms, 32–33 Magnitude, 367 Mathematical induction examples of, 513–517 explanation of, 512–513, 564 extended principle of, 517 principle of, 512 Mathematical models applications of, 230–231, 242–243 explanation of, 147–148 exponential, 340–350 polynomial, 271–273, 285 quadratic, 210–211, 214–215 Matrices addition of, 457–458 applications of, 460–462, 464–465 augmented, 443–445, 447 column, 461–462 decoding, 482 explanation of, 442–443, 497 Gauss-Jordan elimination and, 447–451, 475
on graphing calculators, 442, 458, 473, 488 identity, 470–471 inverse methods to solve linear systems, 498 inverse of square, 471–473 lower triangular, 571 multiplication of, 459–466 negative of, 458 principal diagonal of, 442 properties of, 477 reduced, 444–447 row, 442, 461–462 row-equivalent, 444, 474 singular, 472 size of, 442 square, 442 subtraction of, 458–459 upper triangular, 495, 571 zero, 458 Matrix equations explanation of, 477 method to solve, 477–478 systems of linear equations and, 478–480 Midpoint, of line segment, 124–126 Midpoint formula explanation of, 124 use of, 125–126 Minor of element, in third-order determinant, 489 Mixture problems, 52–53 Models. See Mathematical models Monomials, 22. See also Polynomials Multiplication associative property of, 4, 6 commutative property of, 4, 6 of complex numbers, 76–78 identity matrix for, 470–471 of matrices, 459–466 of polynomials, 24 of rational expressions, 33–34 of real numbers, 3–7 Multiplication principle application of, 533–534 explanation of, 532–533, 565 Multiplication properties of equality, 45 of matrices, 477 of real numbers, 6 Multiplicative identity, 4, 78 Multiplicative inverse, 4, 6, 11, 471–472 Multiplicities from graphs, 292 of zero, 289, 291, 292 Multiplier doctrine, 527 Napierian logarithms. See Natural logarithms Nappes, of cone, 386n Natural logarithms, 359, 360 Natural numbers, 2, 79 Negative growth, 342 Negative real numbers explanation of, 3 principal square root of, 80 properties of, 7, 8 n factorial, 534–535 Nonrepeating linear factors, 587–588 Nonrigid transformations, 193 Notation/symbols absolute value, 65 composition of function, 226, 228
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SUBJECT INDEX
double subscript, 442 empty set, 2 equality and inequality, 57 exponent, 11 factorial, 534–536 function, 167–169, 226, 228 infinity, 57 interval, 57–58, 177 parallel, 141 perpendicular, 141 radical, 15 real number, 2 scientific, 13–14 summation, 507, 508 nth root explanation of, 14–15 principal, 15–16 nth-term formulas, 522–523 Null set, 2 Number line, real, 3 Numbers. See also Integers complex, 74–82, 105 composite, 25 imaginary, 75 irrational, 2, 5, 74 natural, 2, 79 pure imaginary, 75 rational, 2, 3–7, 298 real, 2–9, 75 Numerator, 9 Numerical coefficient, 22. See also Coefficients Oblique asymptotes, 308 Odd functions, 196–197 One-to-one functions explanation of, 235–236, 258 identification of, 236–238 Ordered pairs explanation of, 110n, 111 functions as sets of, 163–164 Ordering, 536 Ordinate, 110 Origin explanation of, 3, 110 reflection through, 114, 192, 193 symmetry and, 114, 115 Parabolas. See also Quadratic functions applications of, 391–392 coordinate-free definition of, 387 equation of, 209, 388–391 explanation of, 111, 204, 387, 418–419 focal chord of, 393 graphs of, 111, 389–390 method to draw, 387–388 vertex of, 205–208 Paraboloids explanation of, 391, 392 hyperbolic, 412 Parallel lines, 141–142 Parallelograms, 598 Parameter, 432, 592–593 Parametric equations application of, 594–595 explanation of, 591–593 on graphing calculator, 592
Partial fraction decomposition, 586–590 Partial fractions, 585 Particular solutions, 432 PASCAL, 559 Pascal’s triangle, 559 Path of projectile, 594 Perfect square formula, 28, 29 Permutations, 536–538, 565 Piecewise-defined functions, 180–181 Plane, distance in, 123–129, 158 Plane curve, 592, 593 Point, coordinate of, 3 Point-by-point plotting, 111 Point-slope form, 138–140 Polynomial functions explanation of, 260, 321–322 graphs of, 260–266, 280 left and right behavior of, 265 Polynomial inequalities explanation of, 283, 322 on graphing calculators, 284 method to solve, 283–284 Polynomials addition of, 23 bisection method and, 281–282 degree of, 22 division of, 266–269 equal, 585–586 evaluation of, 269–270 explanation of, 21–23, 40 factoring, 25–29 factors of, 270, 290–291 factor theorem and, 270 fundamental theorem of algebra and, 288–289 graphs of, 266, 291–292 location theorem and, 280–281 multiplication of, 24 in one variable, 22 prime, 25, 26 rational zeros of, 292–293, 322 with real coefficients, 290–291 real zeros of, 278–279 reduced, 294 remainder theorem and, 269–279 second-degree, 27–28 subtraction of, 24 Taylor, 365 in two variables, 22 zeros of, 266, 271 Positive real numbers, 3, 81 Predictions, 153 Price-demand equation, 93 Prime numbers, 25 Prime polynomials, 25, 26 Principal, 333, 334 Principal diagonal, 442, 488 Principal nth root, 15–16 Principle square root, of negative real number, 80 Probability actual, 552 empirical, 552–556 of events, 547–551 explanation of, 543 theoretical, 552 Probability function, 547 Problem solving. See Word problems Product function, 224–225 Proper fractions, 585
I-7
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SUBJECT INDEX
Pure imaginary numbers, 75 Pythagoreans, 74 Pythagorean theorem, 92, 411, 598 Quadrants, 110 Quadratic equations applications for, 91–93 completing the square to solve, 87–89 explanation of, 84, 105 factoring to solve, 84–86 methods to solve, 100–102 quadratic formula to solve, 89–91 square root property to solve, 86–87 Quadratic factors, 589–590 Quadratic formula explanation of, 90 to solve quadratic equations, 89–90 use of, 294–295, 586 Quadratic functions explanation of, 204, 251–252 general form of, 204 graphs of, 204–209 modeling with, 210–211 properties of, 206 Quadratic inequalities explanation of, 211–212, 252 method to solve, 212–214 Quadratic regression, 214–215 Quadratic solving techniques applications using, 102 direct solution, 100 example of, 101 substitution method as, 101 Quantity-rate-time formula, 50 Quartic models, 273 Quotient function, 224–225 Quotients difference, 170 explanation of, 267 of functions, 226 Radical inequalities, 71–72 Radicals equations involving, 97–99 explanation of, 15, 39–40 properties of, 17, 81 in simplified form, 17–19 use of, 16–17 Radius, of circle, 127, 129 Random experiments, 543–546 Range, of functions, 163, 166, 177 Rate of change, 148–149 Rational exponents explanation of, 15–16 use of, 16–17 Rational expressions addition and subtraction of, 34–36 compound fraction, 36–37 explanation of, 32, 40 multiplication and division of, 33–34 reduced to lowest terms, 32–33 Rational functions domain and x intercepts of, 299 explanation of, 298–299, 322–323 graphs of, 299–301, 304–310 oblique asymptotes of, 308
properties of, 300–301 vertical and horizontal asymptotes of, 302–304 Rational inequalities explanation of, 310, 322–323 on graphing calculators, 311 method to solve, 310–311 Rationalizing factor, 18 Rationalizing the denominator, 18–19 Rational numbers addition and multiplication of, 3–7 explanation of, 2, 298 Rational zeros explanation of, 292–293, 322 method for finding, 293–295 Rational zero theorem, 293 Real number line, 3 Real numbers addition of, 3–7 division of, 7 explanation of, 2, 39, 75 multiplication of, 3–7 negative, 3, 7, 8, 80 positive, 3 properties of, 6 roots of, 14–15 set of, 2–3, 6, 8, 164 subtraction of, 7 Real root, 84 Real zeros approximation of, 282–283 explanation of, 278, 322 upper and lower bound for, 278–279 Reciprocals, 78–79 Rectangles, 407, 598 Rectangular coordinate system. See Cartesian coordinate system Rectangular solids, 599 Recursion formulas explanation of, 505 use of, 515–516 Recursive formula n factorial, 535 Reduced augmented coefficient, 450, 451 Reduced matrices, 444–447 Reduced polynomials, 294 Reduced system, 447 Reflections explanation of, 114 of graphs of functions, 191–193, 195 Regression on graphing calculators, 153 linear, 151–154 logarithmic, 369 quadratic, 214–215 Regression analysis, 151 Regression line, 153 Regression models, 383 Relative frequency, 553 Relative growth rate, 341 Remainder, 267 Remainder theorem, 269–270 Replacement set, 44. See also Domain Residuals, 383 Revenue, 93 Richter scale, 367 Right circular cones, 386n, 599 Right circular cylinders, 599 Rigid transformations, 193
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SUBJECT INDEX
Rise, 134 Rocket equation, 368 Roots. See also Square roots cube, 14 of equation, 176 of functions, 261–262 nth, 14–16 real, 84 of real numbers, 14–15 Rounding convention, 583–584 Row-equivalent matrices, 444 Row matrices, 442, 461–462 Row operations, 443 Run, 134 Sample spaces example of, 546 explanation of, 543–544 fundamental, 545 method to choose, 544–545 Scatter plots, 152 Scientific notation, 13–14 Second-degree polynomial functions. See Quadratic functions Second-degree polynomials, factoring, 27–28 Second diagonal, 488 Second-order determinants, 487, 488 Sequences arithmetic, 520, 522–523 explanation of, 504–505, 564 Fibonacci, 505–506 finite, 505 general term of, 506–507 geometric, 521–523 on graphing calculators, 505, 507 infinite, 505 terms of, 504 Series explanation of, 507, 564 finite, 507 infinite, 507, 526–527 sum formulas for finite arithmetic, 523–524 sum formulas for geometric, 525–527 in summation notation, 508 terms of, 508 Sets of complex numbers, 75 empty or null, 2 equal, 3 of integers, 2 intersection of, 59 of real numbers, 2–3, 6, 8, 164 replacement, 44 union of, 59 Shrinking, in graphs, 193–196 Significant digits, 582–583 Simple events, 544, 547, 550 Simple fractions, 36 Singular matrix, 472 Slope explanation of, 134 geometric interpretation of, 135 method to find, 134–136 of parallel lines, 141–142 of perpendicular lines, 141–142 as rate of change, 148–149 Slope-intercept form, 137–138, 140
Solutions of equations, 44, 111 extraneous, 98, 105 of linear systems, 424, 426–427 particular, 432 unique, 427 Solution set of equations, 44, 111 of inequalities, 59–60 of linear systems, 424 of quadratic inequalities, 211 Speed, 148. See also Rate of change Spheres, 599 Square functions, 188, 203, 204 Square matrices explanation of, 442 inverse of, 471–473, 476 of order n, 470–471 Square root functions, 189 Square root property, 86–87 Square roots, 14, 80 Squaring operation on equations, 98 Standard deck, 540 Standard form of complex numbers, 80 of equation of circle, 128 of equation of line, 133, 140 of linear equations, 45 of quadratic equations, 84, 100 quadratic inequalities in, 211 Stretching, in graphs, 193–196 Subset, 2 Substitution to solve equations of quadratic type, 101 to solve linear systems, 427–428, 431, 432 Substitution property, of equality, 45 Subtraction of complex numbers, 76–77 of matrices, 458–459 of polynomials, 24 of rational expressions, 34–36 of real numbers, 7 Subtraction properties, of equality, 45 Sum formulas for finite arithmetic series, 523–524 for finite geometric series, 525 for infinite geometric series, 526–527 Sum function, 224–225 Summation formula, proof of, 514–515 Summation notation, 507, 508 Summing index, 507 Sum of cubes formula, 28, 29 Sum of the squares of the residuals (SSR), 383 Supply, 435 Symbols. See Notation/symbols Symmetry as aid in graphing, 113–117 axis of, 205 in even and odd functions, 197 tests for, 115–116 Symmetry property, 244–245 Synthetic division, 268–269 Synthetic division table, 279 Systems of linear equations applications of, 434–437, 452–454 basic terms of, 427 Cramer’s rule to solve, 491–494
I-9
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SUBJECT INDEX
Systems of linear equations—Cont. elimination by addition to solve, 429–434 equivalent, 429 explanation of, 496 Gauss-Jordan elimination to solve, 441, 447–453, 497 graphs of, 424–425 matrices and row operations and, 441–447, 498 matrix equations and, 478–480 modeling with, 501–502 substitution method to solve, 427–428, 431, 432 in two variables, 424 Taylor polynomials, 365 Technology Connections. See Graphing calculators Theorems, 511 Theoretical probability explanation of, 552 method to find, 553–554 Third-order determinants, 488–491 Transformations combining graph, 196 even and odd functions and, 196–197 explanation of, 189, 251 of exponential functions, 330 nonrigid, 193 reflections and, 191–193, 195 rigid, 193 stretching and shrinking and, 193–195 vertical and horizontal shifts and, 189–191, 195 Transverse axis, of hyperbola, 405 Trapezoids, 598 Tree diagrams, 532, 544 Triangles formulas for, 598 Pascal’s, 559 similar, 598 Trinomials, 22. See also Polynomials Triple zero, 289 Turning points approximating real zeros at, 282–283 explanation of, 262 Union, of sets, 59 Unique solution, 427 Unlimited growth, 350 Upper and lower bound theorem, 278, 279 Upper triangular matrix, 495, 571 Variables dependent, 164 domains of, 44–45 independent, 164, 165 Variation combined, 319 direct, 316 explanation of, 323 inverse, 316–317 joint, 318 Velocity, 148, 368. See also Rate of change
Vertex form, of quadratic functions, 204 Vertical asymptotes, 302–304 Vertical axis, 110. See also y axis Vertical lines, graphs of, 139, 140, 166 Vertical line test, 166 Vertical shifts, 189–191, 195 Vertical shrinks, 194, 195 Vertical stretches, 194, 195 Vertices of cone, 386n of ellipse, 395 of hyperbola, 405 of parabola, 205, 206, 208, 387
Wiles, Andrew, 517 Word problems method to set up, 48, 91 mixture, 52–53 rate, 50–52 strategies to solve, 47, 104 using diagrams in solution of, 48–49
x axis reflection through, 114, 192, 193 symmetry and, 114, 115 x coordinate, 110, 176 x intercepts explanation of, 133 of functions, 176–177 of polynomial functions, 261–262 of rational functions, 299–300
y axis reflection through, 114, 192, 193 symmetry and, 114, 115 y coordinate, 110, 176 y intercepts explanation of, 133 of functions, 176–177 on graphing calculator, 134
Zero factorial, 534–535 Zero matrix, 458 Zero product property, 84–85 Zero property, of real numbers, 8 Zeros complex, 77, 322 double, 289 of functions, 176, 261–262 imaginary, 290, 295 irrational, 294–295 multiplicities of, 289, 291, 292 of polynomials, 266, 271, 278–279 rational, 292–295, 322 real, 278–279, 282–283 triple, 289
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Sets a苸A aⰻA ⭋ 5x 0 p(x)6 A ( B A´B A傽B
Inequalities and Intervals a is an element of set A a is not an element of set A Empty or null set Set of all x such that p(x) is true A is a subset of B 5x 0 x 苸 A or x 苸 B6, union 5x 0 x 苸 A and x 苸 B6, intersection
a ⬍ b a is less than b a ⱕ b a is less than or equal to b a ⬎ b a is greater than b a ⱖ b a is greater than or equal to b (a, b) Open interval; 5x 0 a 6 x 6 b6 (a, b] Half-open interval; 5x 0 a 6 x ⱕ b6 [a, b) Half-open interval; 5x 0 a ⱕ x 6 b6 [a, b] Closed interval; 5x 0 a ⱕ x ⱕ b6
Number Systems Absolute Value N Natural numbers Z Integers Q Rational numbers R Real numbers C Complex numbers N傺Z傺Q傺R傺C
x if x ⱖ 0 ⫺x if x 6 0 0 x 0 2 ⫽ x2 2x2 ⫽ 0 x 0 0 x 0 6 p if and only if ⫺p 6 x 6 p; p 7 0 0 x 0 7 p if and only if x 6 ⫺p or x 7 p; p 7 0 0x0 ⫽ e
Real Number Properties Quadratic Formula For all real numbers a, b, and c: a ⫹ b ⫽ b ⫹ a;
ab ⫽ ba
a ⫹ (b ⫹ c) ⫽ (a ⫹ b) ⫹ c;
a(bc) ⫽ (ab)c
a(b ⫹ c) ⫽ ab ⫹ ac a ⫹ 0 ⫽ a; 1 ⴢ a ⫽ a a ⫹ (⫺a) ⫽ 0; a(1Ⲑa) ⫽ 1, a ⫽ 0 ab ⫽ 0 if and only if a ⫽ 0 or b ⫽ 0
Exponents and Radicals an ⫽ a ⴢ a . . . a (n factors of a), n 苸 N a0 ⫽ 1, a ⫽ 0 1 a⫺n ⫽ n , a ⫽ 0, n 苸 R a n bm/n ⫽ 2bm (nth root of bm)
Special Products (a ⫺ b)(a ⫹ b) ⫽ a2 ⫺ b2 (a ⫹ b)2 ⫽ a2 ⫹ 2ab ⫹ b2 (a ⫺ b)2 ⫽ a2 ⫺ 2ab ⫹ b2 (a ⫺ b)(a2 ⫹ ab ⫹ b2) ⫽ a3 ⫺ b3 (a ⫹ b)(a2 ⫺ ab ⫹ b2) ⫽ a3 ⫹ b3
Commutative properties Associative properties Distributive property Identities Inverses Zero property
If ax2 ⫹ bx ⫹ c ⫽ 0, a ⫽ 0, then: x⫽
⫺b ⫾ 2b2 ⫺ 4ac 2a
Rectangular Coordinates (x1, y1) d ⫽ 2(x2 ⫺ x1)2 ⫹ ( y2 ⫺ y1)2 x1 ⫹ x2 y1 ⫹ y2 , b 2 2 y2 ⫺ y1 m⫽ , x1 ⫽ x2 x2 ⫺ x1 a
Coordinates of point P1 Distance between P1(x1, y1) and P2(x2, y2) Midpoint of line joining P1 and P2 Slope of line through P1 and P2
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Function Notation
Arithmetic Sequence Value of f at x Composite function Value of inverse of f at x
f (x) ( f ° g)(x) ⫽ f [g(x)] f ⫺1(x)
Linear Equations and Functions y ⫽ mx ⫹ b ( y ⫺ y1) ⫽ m(x ⫺ x1) f (x) ⫽ mx ⫹ b y⫽b x⫽a
Slope–intercept form Point–slope form Linear function Horizontal line Vertical line
Polynomial and Rational Forms f (x) ⫽ ax2 ⫹ bx ⫹ c f (x) ⫽ an x n ⫹ an⫺1x n⫺1 ⫹ . . . ⫹ a1x ⫹ a0, an ⫽ 0, n a nonnegative integer p(x) f (x) ⫽ , p and q polynomial q(x) functions, q(x) ⫽ 0
Quadratic function Polynomial function Rational function
a1, a2, . . . , an, . . . an ⫺ an⫺1 ⫽ d an ⫽ a1 ⫹ (n ⫺ 1)d
Common difference nth-term formula n Sn ⫽ a1 ⫹ . . . ⫹ an ⫽ [2a1 ⫹ (n ⫺ 1)d ] 2 n Sn ⫽ (a1 ⫹ an) 2
Sum of n terms
Geometric Sequence a1, a2, . . . , an, . . . an ⫽r Common ratio an⫺1 n⫺1 an ⫽ a1r nth-term formula a1 ⫺ a1r n , r⫽1 Sn ⫽ a1 ⫹ . . . ⫹ an ⫽ 1⫺r a1 ⫺ ran , r⫽1 Sn ⫽ 1⫺r a1 , 0r0 6 1 S⬁ ⫽ a1 ⫹ a2 ⫹ . . . ⫽ 1⫺r
Sum of n terms
Sum of infinitely many terms
Factorial and Binomial Formulas Exponential and Logarithmic Functions f (x) ⫽ b x, b 7 0, b ⫽ 1 Exponential function f (x) ⫽ logb x, b 7 0, b ⫽ 1 Logarithmic function y ⫽ logb x if and only if x ⫽ b y, b 7 0, b ⫽ 1
n factorial n! ⫽ n(n ⫺ 1) . . . 2 ⴢ 1, n 苸 N 0! ⫽ 1 n n! , 0ⱕrⱕn a b⫽ r!(n ⫺ r)! r n n (a ⫹ b)n ⫽ a a ba n⫺k b k, n ⱖ 1 Binomial formula k⫽0 k
Matrices and Determinants a d
b e
c d f
a †d g
b e h
c f† i
c
Matrix
Determinant
Permutations and Combinations For 0 ⱕ r ⱕ n: n! Pn,r ⫽ (n ⫺ r)! n n! Cn,r ⫽ a b ⫽ r r!(n ⫺ r)!
Permutation Combination
(Continued on back endpaper)
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Hyperbola
Circle (x ⫺ h)2 ⫹ ( y ⫺ k)2 ⫽ r 2 x2 ⫹ y2 ⫽ r 2
y2 x2 ⫺ ⫽1 a2 b2 2 y x2 ⫺ 2⫽1 2 a b
Center at (h, k); radius r Center at (0, 0); radius r
Foci: F¿(⫺c, 0), F(c, 0); c2 ⫽ a2 ⫹ b2 Foci: F¿(0, ⫺c), F(0, c); c2 ⫽ a2 ⫹ b2
Parabola y2 ⫽ 4ax,
a ⬎ 0, opens right; a ⬍ 0, opens left Focus: (a, 0); Directrix: x ⫽ ⫺a a ⬎ 0, opens up; a ⬍ 0, opens down Focus: (0, a); Directrix: y ⫽ ⫺a
x2 ⫽ 4ay,
Translation Formulas x ⫽ x¿ ⫹ h, y ⫽ y¿ ⫹ k ; New origin (h, k)
x¿ ⫽ x ⫺ h, y¿ ⫽ y ⫺ k
Ellipse y2 x2 ⫹ ⫽ 1, a2 b2 2 y x2 ⫹ 2 ⫽ 1, 2 b a
a 7 b 7 0 Foci: F¿(⫺c, 0), F(c, 0); c2 ⫽ a2 ⫺ b2 a 7 b 7 0 Foci: F¿(0, ⫺c), F(0, c); c2 ⫽ a2 ⫺ b2
Metric Units Standard Units of Metric Measure
Important Prefixes
Meter (m): length (approximately 3.28 ft)
kilo (⫻ 1,000)
deci (⫻ 101 )
Liter (L):
volume (approximately 1.06 ft)
hecto (⫻ 100)
1 centi (⫻ 100 )
Gram (g):
weight (approximately 0.035 oz)
deka (⫻ 10)
1 milli (⫻ 1,000 )
Abbreviations Length m km hm dam
meter kilometer hectometer dekameter
Volume
dm decimeter cm centimeter mm millimeter
L kL hL daL
liter kiloliter hectoliter dekaliter
Weight
dL deciliter cL centiliter mL milliliter
g kg hg dag
gram kilogram hectogram dekagram
dg decigram cg centigram mg milligram
English-Metric Conversion Length
Volume (U.S.)
Weight
Length
Volume (U.S.)
1 in ⫽ 2.540 cm 1 ft ⫽ 30.48 cm 1 yd ⫽ 0.9144 m 1 mi ⫽ 1.609 km
1 pt ⫽ 0.4732 L 1 qt ⫽ 0.9464 L 1 gal ⫽ 3.785 L
1 oz ⫽ 28.35 g 1 lb ⫽ 453.6 kg 1 lb ⫽ 0.4536 kg
1 cm ⫽ 0.3937 in 1 cm ⫽ 0.03281 ft 1 m ⫽ 1.0936 yd 1 km ⫽ 0.6215 mi
1 L ⫽ 2.1133 pt 1 L ⫽ 1.0567 qt 1 L ⫽ 0.2642 gal
Weight 1 g ⫽ 0.0353 oz 1 g ⫽ 0.002205 lb 1 kg ⫽ 2.205 lb
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College Algebra
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NINTH EDITION
College Algebra Raymond A. Barnett Merritt College
Michael R. Ziegler Marquette University
Karl E. Byleen Marquette University
Dave Sobecki Miami University Hamilton
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COLLEGE ALGEBRA, NINTH EDITION Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © 2011 by The McGraw-Hill Companies, Inc. All rights reserved. Previous editions © 2008, 2001, and 1999. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 DOW/DOW 1 0 9 8 7 6 5 4 3 2 1 0 ISBN 978–0–07–351949–4 MHID 0–07–351949–9 ISBN 978–0–07–729713–8 (Annotated Instructor’s Edition) MHID 0–07–729713–X Editorial Director: Stewart K. Mattson Sponsoring Editor: John R. Osgood Director of Development: Kristine Tibbetts Developmental Editor: Christina A. Lane Marketing Manager: Kevin M. Ernzen Lead Project Manager: Sheila M. Frank Senior Production Supervisor: Kara Kudronowicz Senior Media Project Manager: Sandra M. Schnee Designer: Tara McDermott Cover/Interior Designer: Ellen Pettergell (USE) Cover Image: © Bob Pool/Getty Images Senior Photo Research Coordinator: Lori Hancock Supplement Producer: Mary Jane Lampe Compositor: Aptara®, Inc. Typeface: 10/12 Times Roman Printer: R. R. Donnelley All credits appearing on page or at the end of the book are considered to be an extension of the copyright page. Chapter R Opener: © Corbis RF; p. 31: © The McGraw-Hill Companies, Inc./John Thoeming photographer. Chapter 1 Opener: © Corbis RF; p. 56: © Vol. 71/Getty RF; p. 92: © Getty RF. Chapter 2 Opener: © Vol. 88/Getty RF; p. 142: © Big Stock Photo; p. 147: © Corbis RF; p. 151: © Vol. 112/Getty RF. Chapter 3 Opener: © Getty RF; p. 170: © Getty RF; p. 187: © Vol. 88/Getty RF; p. 220: © Corbis RF; p. 250: © The McGraw-Hill Companies, Inc./Andrew Resek photographer. Chapter 4 Opener: © Corbis RF; p. 271: © Corbis RF; p. 272: © Vol. 4/Getty RF. Chapter 5 Opener: © Getty RF; p. 333: © Vol. 68/Getty RF; p. 345: © Corbis RF. Chapter 6 Opener: © Brand X/SuperStock RF; p. 401: © California Institute of Technology; Chapter 7 Opener: © Corbis RF; p. 435: Courtesy of Bill Tapenning, USDA; p. 439: © Vol. 5/Getty RF; p. 456: © Vol. 48/Getty RF; p. 460: © Getty RF. Chapter 8 Opener: © Vol.6/Getty RF; p. 531: © ThinkStock/PictureQuest RF; p. 543: © Corbis RF. Library of Congress Cataloging-in-Publication Data College algebra / Raymond A. Barnett ... [et al.]. — 9th ed. p. cm. Rev. ed. of: College algebra. 8th ed. / Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen. Includes index. ISBN 978-0-07-351949-4 — ISBN 0-07-351949-9 (hard copy : alk. paper) 1. Algebra–Textbooks. I. Barnett, Raymond A. QA154.3.B365 2011 512.9–dc22 2009019471
www.mhhe.com
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The Barnett, Ziegler, Byleen, and Sobecki Precalculus Series College Algebra, Ninth Edition This book is the same as Precalculus without the three chapters on trigonometry. ISBN 0-07-351949-9, ISBN 978-0-07-351-949-4
Precalculus, Seventh Edition This book is the same as College Algebra with three chapters of trigonometry added. The trigonometry functions are introduced by a unit circle approach. ISBN 0-07-351951-0, ISBN 978-0-07-351-951-7
College Algebra with Trigonometry, Ninth Edition This book differs from Precalculus in that College Algebra with Trigonometry uses right triangle trigonometric to introduce the trigonometric functions. ISBN 0-07-735010-3, ISBN 978-0-07-735010-9
College Algebra: Graphs and Models, Third Edition This book is the same as Precalculus: Graphs and Models without the three chapters on trigonometry. This text assumes the use of a graphing calculator. ISBN 0-07-305195-0, ISBN 978-0-07-305195-6
Precalculus: Graphs and Models, Third Edition This book is the same as College Algebra: Graphs and Models with three additional chapters on trigonometry. The trigonometric functions are introduced by a unit circle approach. This text assumes the use of a graphing calculator. ISBN 0-07-305196-9, ISBN 978-0-07-305-196-3
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About the Authors
Raymond A. Barnett, a native of and educated in California, received his B.A. in mathematical statistics from the University of California at Berkeley and his M.A. in mathematics from the University of Southern California. He has been a member of the Merritt College Mathematics Department and was chairman of the department for four years. Associated with four different publishers, Raymond Barnett has authored or co-authored 18 textbooks in mathematics, most of which are still in use. In addition to international English editions, a number of the books have been translated into Spanish. Co-authors include Michael Ziegler, Marquette University; Thomas Kearns, Northern Kentucky University; Charles Burke, City College of San Francisco; John Fujii, Merritt College; Karl Byleen, Marquette University; and Dave Sobecki, Miami University Hamilton. Michael R. Ziegler received his B.S. from Shippensburg State College and his M.S. and Ph.D. from the University of Delaware. After completing postdoctoral work at the University of Kentucky, he was appointed to the faculty of Marquette University where he held the rank of Professor in the Department of Mathematics, Statistics, and Computer Science. Dr. Ziegler published more than a dozen research articles in complex analysis and co-authored more than a dozen undergraduate mathematics textbooks with Raymond Barnett and Karl Byleen before passing away unexpectedly in 2008. Karl E. Byleen received his B.S., M.A., and Ph.D. degrees in mathematics from the University of Nebraska. He is currently an Associate Professor in the Department of Mathematics, Statistics, and Computer Science of Marquette University. He has published a dozen research articles on the algebraic theory of semigroups and co-authored more than a dozen undergraduate mathematics textbooks with Raymond Barnett and Michael Ziegler. Dave Sobecki earned a B.A. in math education from Bowling Green State University, then went on to earn an M.A. and a Ph.D. in mathematics from Bowling Green. He is an associate professor in the Department of Mathematics at Miami University in Hamilton, Ohio. He has written or co-authored five journal articles, eleven books and five interactive CD-ROMs. Dave lives in Fairfield, Ohio with his wife (Cat) and dogs (Macleod and Tessa). His passions include Ohio State football, Cleveland Indians baseball, heavy metal music, travel, and home improvement projects.
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Dedicated to the memory of Michael R. Ziegler, trusted author, colleague, and friend.
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Brief Contents
Preface xiv Features xvii Application Index xxviii
R CHAPTER 1 CHAPTER 2 CHAPTER 3 CHAPTER 4 CHAPTER 5 CHAPTER 6 CHAPTER 7 CHAPTER 8
CHAPTER
Basic Algebraic Operations 1 Equations and Inequalities 43 Graphs 109 Functions 161 Polynomial and Rational Functions 259 Exponential and Logarithmic Functions 327 Additional Topics in Analytic Geometry 385 Systems of Equations and Matrices 423 Sequences, Induction, and Probability 503 Appendix A Cumulative Review Exercises A-1 Appendix B Special Topics A-13 Appendix C Geometric Formulas A-29 Student Answers SA-1 Subject Index I-1
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SECTION 11–1
Systems of Linear Equations in Two Variables
xii
Contents Preface xiv Features xvii Applications Index xxviii
CHAPTER R-1 R-2 R-3 R-4
1-1 1-2 1-3 1-4 1-5 1-6
xii
3
Functions 161 Functions 162 Graphing Functions 175 Transformations of Functions 188 Quadratic Functions 203 Operations on Functions; Composition 223
Inverse Functions 235 Chapter 3 Review 250 Chapter 3 Review Exercises 252 Chapter 3 Group Activity: Mathematical Modeling: Choosing a Cell Phone Plan 257
CHAPTER 4-1 4-2 4-3 4-4 4-5
5-1 5-2 5-3 5-4 5-5
5
Exponential and Logarithmic Functions 327 Exponential Functions 328 Exponential Models 340 Logarithmic Functions 354 Logarithmic Models 365 Exponential and Logarithmic Equations 372 Chapter 5 Review 379 Chapter 5 Review Exercises 380 Chapter 5 Group Activity: Comparing Regression Models 383
CHAPTER 6-1 6-2 6-3
4
Polynomial and Rational Functions 259 Polynomial Functions, Division, and Models 260 Real Zeros and Polynomial Inequalities 278 Complex Zeros and Rational Zeros of Polynomials 288 Rational Functions and Inequalities 298 Variation and Modeling 315 Chapter 4 Review 321 Chapter 4 Review Exercises 323 Chapter 4 Group Activity: Interpolating Polynomials 326
CHAPTER
2
Graphs 109 Cartesian Coordinate Systems 110 Distance in the Plane 122 Equation of a Line 132 Linear Equations and Models 147 Chapter 2 Review 157 Chapter 2 Review Exercises 158 Chapter 2 Group Activity: Average Speed 160
CHAPTER 3-1 3-2 3-3 3-4 3-5
1
Equations and Inequalities 43 Linear Equations and Applications 44 Linear Inequalities 56 Absolute Value in Equations and Inequalities 65 Complex Numbers 74 Quadratic Equations and Applications 84 Additional Equation-Solving Techniques 97 Chapter 1 Review 104 Chapter 1 Review Exercises 106 Chapter 1 Group Activity: Solving a Cubic Equation 108
CHAPTER 2-1 2-2 2-3 2-4
R
Basic Algebraic Operations 1 Algebra and Real Numbers 2 Exponents and Radicals 11 Polynomials: Basic Operations and Factoring 21 Rational Expressions: Basic Operations 32 Chapter R Review 39 Chapter R Review Exercises 40
CHAPTER
3-6
6
Additional Topics in Analytic Geometry 385 Conic Sections; Parabola 386 Ellipse 395 Hyperbola 405 Chapter 6 Review 418 Chapter 6 Review Exercises 421 Chapter 6 Group Activity: Focal Chords 422
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SECTION 11–1
CHAPTER 7-1 7-2 7-3 7-4 7-5
7
Systems of Equations and Matrices 423 Systems of Linear Equations 424 Solving Systems of Linear Equations 441 Matrix Operations 457 Solving Systems of Linear Equations Using Matrix Inverse Methods 470 Determinants and Cramer’s Rule 487
Additional Topics Available Online: (Visit www.mhhe.com/barnett) 7-6 Systems of Nonlinear Equations 7-7 Systems of Linear Inequalities in Two Variables 7-8 Linear Programming Chapter 7 Review 496 Chapter 7 Review Exercises 498 Chapter 7 Group Activity: Modeling with Systems of Linear Equations 501
Systems of Linear Equations in Two Variables
CHAPTER 8-1 8-2 8-3 8-4 8-5 8-6
xiii
8
Sequences, Induction, and Probability 503 Sequences and Series 504 Mathematical Induction 511 Arithmetic and Geometric Sequences 520 Multiplication Principle, Permutations, and Combinations 531 Sample Spaces and Probability 543 The Binomial Formula 558 Chapter 8 Review 564 Chapter 8 Review Exercises 566 Chapter 8 Group Activity: Sequences Specified by Recursion Formulas 568
Appendix A Cumulative Review Exercises A-1 Appendix B Special Topics A-13 Appendix C Geometric Formulas A-29 Student Answers SA-1 Subject Index I-1
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Preface Enhancing a Tradition of Success The ninth edition of College Algebra represents a substantial step forward in student accessibility. Every aspect of the revision of this classic text focuses on making the text more accessible to students while retaining the precise presentation of the mathematics for which the Barnett name is renowned. Extensive work has been done to enhance the clarity of the exposition, improving the overall presentation of the content. This in turn has decreased the length of the text. Specifically, we concentrated on the areas of writing, exercises, worked examples, design, and technology. Based on numerous reviews, advice from expert consultants, and direct correspondence with the many users of previous editions, this edition is more relevant and accessible than ever before. Writing Without sacrificing breadth or depth or coverage, we have rewritten explanations to make them clearer and more direct. As in previous editions, the text emphasizes computational skills, essential ideas, and problem solving rather than theory. Exercises Over twenty percent of the exercises in the ninth edition are new. These exercises encompass both a variety of skill levels as well as increased content coverage, ensuring a gradual increase in difficulty level throughout. In addition, brand new writing exercises have been included at the beginning of each exercise set in order to encourage a more thorough understanding of key concepts for students. Examples Color annotations accompany many examples, encouraging the learning process for students by explaining the solution steps in words. Each example is then followed by a similar matched problem for the student to solve. Answers to the matched problems are located at the end of each section for easy reference. This active involvement in learning while reading helps students develop a more thorough understanding of concepts and processes. Technology Instructors who use technology to teach college algebra, whether it be exploring mathematics with a graphing calculator or assigning homework and quizzes online, will find the ninth edition to be much improved. Refined “Technology Connections” boxes included at appropriate points in the text illustrate how problems previously introduced in an algebraic context may be solved using a graphing calculator. Exercise sets include calculator-based exercises marked with a calculator icon. Note, however, that the use of graphing technology is completely optional with this text. We understand that at many colleges a single text must serve the purposes of teachers with widely divergent views on the proper use of graphing and scientific calculators in college algebra, and this text remains flexible regarding the degree of calculator integration. Additionally, McGraw-Hill’s MathZone offers a complete online homework system for mathematics and statistics. Instructors can assign textbook-specific content as well as customize the level of feedback students receive, including the ability to have students show their work for any given exercise. Assignable content for the ninth edition of College Algebra includes an array of videos and other multimedia along with algorithmic exercises, providing study tools for students with many different learning styles.
A Central Theme In the Barnett series, the function concept serves as a unifying theme. A brief look at the table of contents reveals this emphasis. A major objective of this book is the development of a library of elementary functions, including their important properties and uses. Employing this library as a basic working tool, students will be able to proceed through this book with greater confidence and understanding. xiv
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Preface
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Reflecting trends in the way college algebra is taught, the ninth edition emphasizes functions modeled in the real world more strongly than previous editions. In some cases, data are provided and the student is asked to produce an approximate corresponding function using regression on a graphing calculator. However, as with previous editions, the use of a graphing calculator remains completely optional and any such examples or exercises can be easily omitted without loss of continuity.
Key Features The revised full-color design gives the book a more contemporary feel and will appeal to students who are accustomed to high production values in books, magazines, and nonprint media. The rich color palette, streamlined calculator explorations, and use of color to signify important steps in problem material work in conjunction to create a more visually appealing experience for students. An emphasis on mathematical modeling is evident in section titles such as “Linear Equations and Models” and “Exponential Models.” These titles reflect a focus on the relationship between functions and real-world phenomena, especially in examples and exercises. Modeling problems vary from those where only the function model is given (e.g., when the model is a physical law such as F ⫽ ma), through problems where a table of data and the function are provided, to cases where the student is asked to approximate a function from data using the regression function of a calculator or computer. Matched problems following worked examples encourage students to practice problem solving immediately after reading through a solution. Answers to the matched problems are located at the end of each section for easy reference. Interspersed throughout each section, Explore-Discuss boxes foster conceptual understanding by asking students to think about a relationship or process before a result is stated. Verbalization of mathematical concepts, results, and processes is strongly encouraged in these explanations and activities. Many Explore-Discuss boxes are appropriate for group work. Refined Technology Connections boxes employ graphing calculators to show graphical and numerical alternatives to pencil-and-paper symbolic methods for problem solving—but the algebraic methods are not omitted. Screen shots are from the TI-84 Plus calculator, but the Technology Connections will interest users of any automated graphing utility. Think boxes (color dashed boxes) are used to enclose steps that, with some experience, many students will be able to perform mentally. Balanced exercise sets give instructors maximum flexibility in assigning homework. A wide variety of easy, moderate, and difficult level exercises presented in a range of problem types help to ensure a gradual increase in difficulty level throughout each exercise set. The division of exercise sets into A (routine, easy mechanics), B (more difficult mechanics), and C (difficult mechanics and some theory) is explicitly presented only in the Annotated Instructor’s Edition. This is due to our attempt to avoid fueling students’ anxiety about challenging exercises. This book gives the student substantial experience in modeling and solving applied problems. Over 500 application exercises help convince even the most skeptical student that mathematics is relevant to life outside the classroom. An Applications Index is included following the Guided Tour to help locate particular applications. Most exercise sets include calculator-based exercises that are clearly marked with a calculator icon. These exercises may use real or realistic data, making them computationally heavy, or they may employ the calculator to explore mathematics in a way that would be impractical with paper and pencil. As many students will use this book to prepare for a calculus course, examples and exercises that are especially pertinent to calculus are marked with an icon. A Group Activity is located at the end of each chapter and involves many of the concepts discussed in that chapter. These activities require students to discuss and write about mathematical concepts in a complex, real-world context.
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Preface
Changes to this Edition A more modernized, casual, and student-friendly writing style has been infused throughout the chapters without radically changing the tone of the text overall. This directly works toward a goal of increasing motivation for students to actively engage with their textbooks, resulting in higher degrees of retention. A significant revision to the exercise sets in the new edition has produced a variety of important changes for both students and instructors. As a result, over twenty percent of the exercises are new. These exercises encompass both a variety of skill levels as well as increased content coverage, ensuring a gradual increase in difficulty level throughout. In addition, brand new writing exercises have been included at the beginning of each exercise set in order to encourage a more thorough understanding of key concepts for students. Specific changes include: • The addition of hundreds of new writing exercises to the beginning of each exercise set. These exercises encourage students to think about the key concepts of the sections before attempting the computational and application exercises, ensuring a more thorough understanding of the material. • An update to the data in many application exercises to reflect more current statistics in topics that are both familiar and highly relevant to today’s students. • A significant increase the amount of moderate skill level problems throughout the text in response to the growing need expressed by instructors. The number of colored annotations that guide students through worked examples has been increased throughout the text to add clarity and guidance for students who are learning critical concepts. New instructional videos on graphing calculator operations posted on MathZone help students master the most essential calculator skills used in the college algebra course. The videos are closed-captioned for the hearing impaired, subtitled in Spanish, and meet the Americans with Disabilities Act Standards for Accessible Design. Though these are an entirely optional ancillary, instructors may use them as resources in a learning center, for online courses, and to provide extra help to students who require extra practice. Chapter R, “Basic Algebraic Operations,” has been extensively rewritten based upon feedback from reviewers to provide a streamlined review of basic algebra in four sections rather than six. Exponents and radicals are now covered in a single section (R-2), and the section covering operations on polynomials (R-3) now includes factoring. Chapter 7, “Systems of Equations and Matrices,” has been reorganized to focus on systems of linear equations, rather than on systems of inequalities or nonlinear systems. A section on determinants and Cramer’s rule (10-5) has been added. Three additional sections on systems of nonlinear equations, systems of linear inequalities, and linear programming are also available online.
Design: A Refined Look with Your Students in Mind The McGraw-Hill Mathematics Team has gathered a great deal of information about how to create a student-friendly textbook in recent years by going directly to the source—your students. As a result, two significant changes have been made to the design of the ninth edition based upon this feedback. First, example headings have been pulled directly out into the margins, making them easy for students to find. Additionally, we have modified the design of one of our existing features—the caution box—to create a more powerful tool for your students. Described by students as one of the most useful features in a math text, these boxes now demand attention with bold red headings pulled out into the margin, alerting students to avoid making a common mistake. These fundamental changes have been made entirely with the success of your students in mind and we are confident that they will improve your students’ overall reaction to and enjoyment of the course.
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Features Examples and Matched Problems Integrated throughout the text, completely worked examples and practice problems are used to introduce concepts and demonstrate problem-solving techniques—algebraic, graphical, and numerical. Each example is followed by a similar Matched Problem for the student to work through while reading the material. Answers to EXAMPLE the matched problems are located at the end of each section for easy reference. This active involvement in the learning process helps students develop a more thorough understanding of algebraic concepts and processes.
2
Using the Distance Formula Find the distance between the points (00033, 5) and (00032, 00038).*
SOLUTION
Let (x1, y1) 0002 (ⴚ3, 5) and (x2, y2) 0002 (ⴚ2, ⴚ8). Then, d 0002 2[(ⴚ2) 0003 (ⴚ3)] 2 0004 [(ⴚ8) 0003 5 ] 2 0002 2(00032 0004 3)2 0004 (00038 0003 5)2 0002 212 0004 (000313)2 0002 21 0004 169 0002 2170 Notice that if we choose (x1, y1) 0002 (00032, 00038) and (x2, y2) 0002 (00033, 5), then d 0002 2 [(00033) 0003 (00032)] 2 0004 [5 0003 (00038) ] 2 0002 21 0004 169 0002 2170 so it doesn’t matter which point we designate as P1 or P2.
MATCHED PROBLEM 2
Find the distance between the points (6, 00033) and (00037, 00035).
Z Midpoint of a Line Segment The midpoint of a line segment is the point that is equidistant from each of the endp A formula for finding the midpoint is given in Theorem 2. The proof is discussed i exercises.
Exploration and Discussion Would you like to incorporate more discovery learning in your course? Interspersed at appropriate places in every section, Explore-Discuss boxes encourage students to think critically about mathematics and to explore key concepts in more detail. Verbalization of mathematical concepts, results, and processes is ZZZ EXPLORE-DISCUSS 1 encouraged in these Explore-Discuss boxes, as well as in some matched problems, and in problems marked with color numerals in almost every exercise set. Explore-Discuss material can be used in class or in an out-of-class activity.
To graph the equation y ⫽ ⫺x3 ⫹ 2x, we use point-by-point plotting to obtain the graph in Figure 5. (A) Do you think this is the correct graph of the equation? If so, why? If not, why? (B) Add points on the graph for x ⫽ ⫺2, ⫺0.5, 0.5, and 2. (C) Now, what do you think the graph looks like? Sketch your version of the graph, adding more points as necessary. (D) Write a short statement explaining any conclusions you might draw from parts A, B, and C.
y 5
x
y
⫺1 ⫺1 0 0 1 1
⫺5
5
x
⫺5
Z Figure 5
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Applications One of the primary objectives of this book is to give the student substantial experience in modeling and solving real-world problems. Over 500 application exercises help convince even the most skeptical student that mathematics is relevant to everyday life. An Applications 15. 2 2 0002 0.426 16. 3 3 Index is included following the features to help In Problems 17–26, solve exactly. locate particular applications. 3 0005x
4 0005x
4.23
17. log5 x 0002 2
0002 0.089
55. I 0002
18. log3 y 0002 4
x 0002 25
19. log (t 0005 4) 0002 00051
t 0002 41 10
21. log 5 0003 log x 0002 2
20
y 0002 81
20. ln (2x 0003 3) 0002 0
x 0002 00051
22. log x 0005 log 8 0002 1
23. log x 0003 log (x 0005 3) 0002 1
54. L 0002 8.8 0003 5.1 log D for D (astronomy)
6.20
80
5
24. log (x 0005 9) 0003 log 100x 0002 3
10
25. log (x 0003 1) 0005 log (x 0005 1) 0002 1
11 9
26. log (2x 0003 1) 0002 1 0003 log (x 0005 2)
21 8
(1 0003 i)n 0005 1 56. S 0002 R for n (annuity) i
27. 2 0002 1.05x
28. 3 0002 1.06x
14.2
29. e00051.4x 0003 5 0002 0 No solution
31. 123 0002 500e00050.12x 0005x 2
33. e
0002 0.23
00071.21
11.7
32. 438 0002 200e0.25x x2
34. e 0002 125
00072.20
B In Problems 35–48, solve exactly. 35. log (5 0005 2x) 0002 log (3x 0003 1) 36. log (x 0003 3) 0002 log (6 0003 4x)
59. y 0002 No solution
e x 0005 e0005x e x 0003 e0005x
3.14
5
38. log (6x 0003 5) 0005 log 3 0002 log 2 0005 log x
2 0003 13
40. ln (x 0003 1) 0002 ln (3x 0003 1) 0005 ln x
1 0003 12 1 0003 189 4
42. 1 0005 log (x 0005 2) 0002 log (3x 0003 1)
3
n0002
ln(1 0003 i )
60. y 0002
x 0002 12 ln
e x 0005 e0005x 2
e x 0003 e0005x e x 0005 e0005x
1 y00031 x 0002 ln 2 y00051
10005y
In Problems 61–68, use a graphing calculator to approximate to two decimal places any solutions of the equation in the interval 0 0004 x 0004 1. None of these equations can be solved exactly using any step-by-step algebraic process. 0.38
62. 30005x 0005 3x 0002 0 64. xe2x 0005 1 0002 0
0.57
x 0002 0.25 x 0002 0.43
65. ln x 0003 2x 0002 0
0.43
66. ln x 0003 x2 0002 0
67. ln x 0003 e x 0002 0
0.27
68. ln x 0003 x 0002 0
x 0002 0.65 x 0002 0.57
2 3
39. ln x 0002 ln (2x 0005 1) 0005 ln (x 0005 2) 41. log (2x 0003 1) 0002 1 0005 log (x 0005 1)
RI L ln al 0005 b R E
ln(Si R 0003 1)
x 0002 ln [y 0003 2y 2 0003 1]
10003y
63. e0005x 0005 x 0002 0
00051
No solution
44. 1 0003 ln (x 0003 1) 0002 ln (x 0005 1)
58. y 0002
61. 20005x 0005 2x 0002 0 4 5
37. log x 0005 log 5 0002 log 2 0005 log (x 0005 3)
43. ln (x 0003 1) 0002 ln (3x 0003 3)
e x 0003 e0005x 2
x 0002 ln (y 0007 2y 2 0005 1)
18.9
30. e0.32x 0003 0.47 0002 0
D 0002 10(L00058.8)00065.1
t00020005
The following combinations of exponential functions define four of six hyperbolic functions, a useful class of functions in calculus and higher mathematics. Solve Problems 57–60 for x in terms of y. The results are used to define inverse hyperbolic functions, another useful class of functions in calculus and higher mathematics. 57. y 0002
In Problems 27–34, solve to three significant digits.
E (1 0005 e0005Rt0006L) for t (circuitry) R
No solution
APPLICATIONS 69. COMPOUND INTEREST How many years, to the nearest year, will it take a sum of money to double if it is invested at 7% compounded annually? 10 years 70. COMPOUND INTEREST How many years, to the nearest year, will it take money to quadruple if it is invested at 6% compounded annually? 24 years
Technology Connections Technology Connections Technology Connections boxes integrated at appropriate points in the text illustrate how conFigure 1 shows the details of constructing the logarithmic model of Example 5 on a graphing calculator. cepts previously introduced in an algebraic context may be approached using a graphing calculator. Students always learn the algebraic methods first so that they develop a solid grasp of these methods and do not become calculatordependent. The exercise sets contain calculatorZ Figure 1 based exercises that are clearly marked with a calculator icon. The use of technology is 62. g(x) 0002 4e 0003 7; f (x) 0002 e completely optional with this text. All technology 63. g(x) 0002 3 0003 4e ; f (x) 0002 e features and exercises may be omitted without sacrificing 64. g(x) 0002 00032 0003 5e ; f (x) 0002 e content coverage. 100
0
120
0
(a) Entering the data
(b) Finding the model
(c) Graphing the data and the model
x00041
0002
x
20003x
40003x
x
x
In Problems 65–68, simplify. 65.
00032x3e00032x 0003 3x2e00032x x6
66.
5x4e5x 0003 4x3e5x x8
67. (e x 0004 e0003x )2 0004 (e x 0003 e0003x )2
2e2x 0004 2e00032x
68. e x(e0003x 0004 1) 0003 e0003x(e x 0004 1)
ex 0003 e0003x
In Problems 69–76, use a graphing calculator to find local extrema, y intercepts, and x intercepts. Investigate the behavior as x S 0007 and as x 00030007 and identify any horizontal asymptotes. Round any approximate values to two decimal places. 69. f (x) 0002 2 0004 e x00032
70. g(x) 0002 00033 0004 e10004x
71. s(x) 0002 e0003x
72. r(x) 0002 e x
2
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Group Activities A Group Activity is located at the end of each chapter and involves many of the concepts discussed in that chapter. These activities strongly encourage the verbalization of mathematical concepts, results, and processes. All of these special activities are highlighted to emphasize their importance.
CHAPTER
ZZZ
5
GROUP ACTIVITY Comparing Regression Models
We have used polynomial, exponential, and logarithmic regression models to fit curves to data sets. How can we determine which equation provides the best fit for a given set of data? There are two principal ways to select models. The first is to use information about the type of data to help make a choice. For example, we expect the weight of a fish to be related to the cube of its length. And we expect most populations to grow exponentially, at least over the short term. The second method for choosing among equations involves developing a measure of how closely an equation fits a given data set. This is best introduced through an example. Consider the data set in Figure 1, where L1 represents the x coordinates and L2 represents the y coordinates. The graph of this data set is shown in Figure 2. Suppose we arbitrarily choose the equation y1 ⫽ 0.6x ⫹ 1 to model these data (Fig. 3).
Each of these differences is called a residual. Note that three of the residuals are positive and one is negative (three of the points lie above the line, one lies below). The most commonly accepted measure of the fit provided by a given model is the sum of the squares of the residuals (SSR). When squared, each residual (whether positive or negative or zero) makes a nonnegative contribution to the SSR. SSR ⫽ (4 ⫺ 2.2)2 ⫹ (5 ⫺ 3.4)2 ⫹ (3 ⫺ 4.6)2 ⫹ (7 ⫺ 5.8)2 ⫽ 9.8 (A) A linear regression model for the data in Figure 1 is given by y2 ⫽ 0.35x ⫹ 3 Compute the SSR for the data and y2, and compare it to the one we computed for y1.
10
0
10
0
Z Figure 1
Z Figure 2 10
0
It turns out that among all possible linear polynomials, the linear regression model minimizes the sum of the squares of the residuals. For this reason, the linear regression model is often called the least-squares line. A similar statement can be made for polynomials of any fixed degree. That is, the quadratic regression model minimizes the SSR over all quadratic polynomials, the cubic regression model minimizes the SSR over all cubic polynomials, and so on. The same statement cannot be made for exponential or logarithmic regression models. Nevertheless, the SSR can still be used to compare exponential, logarithmic, and polynomial models. (B) Find the exponential and logarithmic regression models for the data in Figure 1, compute their SSRs, and compare with the linear model. (C) National annual advertising expenditures for selected years since 1950 are shown in Table 1 where x is years since 1950 and y is total expenditures in billions of dollars. Which regression model would fit this data best: a quadratic model, a cubic model, or an exponential model? Use the SSRs to sup-
10
0
Z Figure 3 y1 ⫽ 0.6x ⫹ 1.
Foundation for Calculus As many students will use this book to prepare for a calculus course, examples and exercises that are especially pertinent to calculus are marked with an icon. EXAMPLE
6
Evaluating and Simplifying a Difference Quotient For f(x) ⫽ x2 ⫹ 4x ⫹ 5, find and simplify: (A) f(x ⫹ h)
SOLUTIONS
(B) f(x ⫹ h) ⫺ f(x)
(C)
f(x ⫹ h) ⫺ f(x) ,h⫽0 h
(A) To find f(x ⫹ h), we replace x with x ⫹ h everywhere it appears in the equation that defines f and simplify: f(x ⴙ h) ⫽ (x ⴙ h)2 ⫹ 4(x ⴙ h) ⫹ 5 ⫽ x2 ⫹ 2xh ⫹ h2 ⫹ 4x ⫹ 4h ⫹ 5 (B) Using the result of part A, we get f(x ⴙ h) ⫺ f(x) ⫽ x2 ⴙ 2xh ⴙ h2 ⴙ 4x ⴙ 4h ⴙ 5 ⫺ (x2 ⴙ 4x ⴙ 5) ⫽ x2 ⫹ 2xh ⫹ h2 ⫹ 4x ⫹ 4h ⫹ 5 ⫺ x2 ⫺ 4x ⫺ 5 ⫽ 2xh ⫹ h2 ⫹ 4h (C)
f(x ⫹ h) ⫺ f(x) 2xh ⫹ h2 ⫹ 4h ⫽ h h ⫽ 2x ⫹ h ⫹ 4
⫽
h(2x ⫹ h ⫹ 4) h
Divide numerator and denominator by h ⴝ 0.
0002
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Student Aids
The domain of f is all x values except ⫺52, or (⫺⬁, ⫺52 ) 傼 (⫺52, ⬁). The value of a fraction is 0 if and only if the numerator is zero:
Annotation of examples and explanations, in small colored type, is found throughout the text to help students through critical stages. Think Boxes are dashed boxes used to enclose steps that students may be encouraged to perform mentally.
Screen Boxes are used to highlight important definitions, theorems, results, and step-by-step processes.
4 ⫺ 3x ⫽ 0
Subtract 4 from both sides.
⫺3x ⫽ ⫺4 x⫽
Divide both sides by ⴚ3.
4 3
The x intercept of f is 43. The y intercept is f(0) ⫽
4 ⫺ 3(0) 2(0) ⫹ 5
4 ⫽ . 5
0002
Z COMPOUND INTEREST If a principal P is invested at an annual rate r compounded m times a year, then the amount A in the account at the end of n compounding periods is given by A 0002 P a1 0003
r n b m
Note that the annual rate r must be expressed in decimal form, and that n 0002 mt, where t is years.
Z DEFINITION 1 Increasing, Decreasing, and Constant Functions Let I be an interval in the domain of function f. Then, 1. f is increasing on I and the graph of f is rising on I if f(x1) 6 f(x2) whenever x1 6 x2 in I. 2. f is decreasing on I and the graph of f is falling on I if f(x1) 7 f(x2) whenever x1 6 x2 in I. 3. f is constant on I and the graph of f is horizontal on I if f(x1) 0002 f (x2) whenever x1 6 x2 in I.
Z THEOREM 1 Tests for Symmetry
Caution Boxes appear throughout the text to indicate where student errors often occur.
Symmetry with respect to the:
An equivalent equation results if:
y axis
x is replaced with ⫺x
x axis
y is replaced with ⫺y
Origin
x and y are replaced with ⫺x and ⫺y
ZZZ CAUTION ZZZ
A very common error occurs about now—students tend to confuse algebraic expressions involving fractions with algebraic equations involving fractions. Consider these two problems: (A) Solve:
x x ⫹ ⫽ 10 2 3
(B) Add:
x x ⫹ ⫹ 10 2 3
The problems look very much alike but are actually very different. To solve the equation in (A) we multiply both sides by 6 (the LCD) to clear the fractions. This works so well for equations that students want to do the same thing for problems like (B). The only catch is that (B) is not an equation, and the multiplication property of equality does not apply. If we multiply (B) by 6, we simply obtain an expression 6 times as large as the original! Compare these correct solutions: x x ⫹ ⫽ 10 2 3
(A) 6ⴢ
x x ⫹ 6 ⴢ ⫽ 6 ⴢ 10 2 3 3x ⫹ 2x ⫽ 60 5x ⫽ 60 x ⫽ 12
xx
(B)
x x ⫹ ⫹ 10 2 3 ⫽
3ⴢx 2ⴢx 6 ⴢ 10 ⫹ ⫹ 3ⴢ2 2ⴢ3 6ⴢ1
3x 2x 60 ⫹ ⫹ 6 6 6 5x ⫹ 60 ⫽ 6
⫽
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Chapter Review sections are provided at the end of each chapter and include a thorough review of all the important terms and symbols. This recap is followed by a comprehensive set of review exercises.
j
CHAPTER
5-1
5
Review
Exponential Functions
The equation f (x) 0003 bx, b 0004 0, b 0005 1, defines an exponential function with base b. The domain of f is (00060007, 0007) and the range is (0, 0007). The graph of f is a continuous curve that has no sharp corners; passes through (0, 1); lies above the x axis, which is a horizontal asymptote; increases as x increases if b 0004 1; decreases as x increases if b 1; and intersects any horizontal line at most once. The function f is one-to-one and has an inverse. We often use the following exponential function properties: 1. a xa y 0003 a x y
(a x) y 0003 a xy
a x ax a b 0003 x b b
(ab)x 0003 a xb x
ax 0003 a x0006y ay
2. a x 0003 a y if and only if x 0003 y. 3. For x 0005 0, a x 0003 b x if and only if a 0003 b. As x approaches 0007, the expression [1 (1兾x)]x approaches the irrational number e ⬇ 2.718 281 828 459. The function f (x) 0003 e x is called the exponential function with base e. The growth of money in an account paying compound interest is described by A 0003 P(1 r兾m)n, where P is the principal, r is the annual rate, m is the number of compounding periods in 1 year, and A is the amount in the account after n compounding periods. If the account pays continuous compound interest, the amount A in the account after t years is given by A 0003 Pert.
5-2
1. Population growth can be modeled by using the doubling time growth model A 0003 A02t d, where A is the population at time t, A0 is the population at time t 0003 0, and d is the doubling time—
CHAPTERS
1–3
3. Limited growth—the growth of a company or proficiency at learning a skill, for example—can often be modeled by the equation y 0003 A(1 0006 e0006kt ), where A and k are positive constants. Logistic growth is another limited growth model that is useful for modeling phenomena like the spread of an epidemic, or sales of a new product. The logistic model is A 0003 M/(1 ce0006kt ), where c, k, and M are positive constants. A good comparison of these different exponential models can be found in Table 3 at the end of Section 5-2. Exponential regression can be used to fit a function of the form y 0003 ab x to a set of data points. Logistic regression can be used to find a function of the form y 0003 c (1 ae0006bx ).
Logarithmic Functions
The logarithmic function with base b is defined to be the inverse of the exponential function with base b and is denoted by y 0003 logb x. So y 0003 logb x if and only if x 0003 b y, b 0004 0, b 0005 1. The domain of a logarithmic function is (0, 0007) and the range is (00060007, 0007). The graph of a logarithmic function is a continuous curve that always passes
Cumulative Review Exercises
*Additional answers can be found in the Instructor Answer Appendix.
Work through all the problems in this cumulative review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. 1. Solve for x:
2. Radioactive decay can be modeled by using the half-life decay model A 0003 A0(12)t h 0003 A020006t h, where A is the amount at time t, A0 is the amount at time t 0003 0, and h is the half-life—the time it takes for half the material to decay. Another model of radioactive decay, A 0003 A0e0006kt , where A0 is the amount at time zero and k is a positive constant, uses the exponential function with base e. This model can be used for other types of quantities that exhibit negative exponential growth as well.
5-3
Exponential Models
Exponential functions are used to model various types of growth:
Cumulative Review Exercise sets are provided in Appendix A for additional reinforcement of key concepts.
the time it takes for the population to double. Another model of population growth, A 0003 A0ekt, where A0 is the population at time zero and k is a positive constant called the relative growth rate, uses the exponential function with base e. This model is used for many other types of quantities that exhibit exponential growth as well.
7x 3 0003 2x x 0002 10 0002 0007 00032 5 2 3
x 0007 52
Problems 16–18 refer to the function f given by the graph: f(x) 5
(1-1)
00025
5
x
In Problems 2–4, solve and graph the inequality. 2. 2(3 0002 y) 0003 4 0004 5 0002 y
00025
3. 冟x 0002 2冟 0005 7
16. Find the domain and range of f. Express answers in interval notation. Domain: [00022, 3]; range: [00021, 2] (3-2)
4. x2 0003 3x 0006 10 5. Perform the indicated operations and write the answer in standard form: (A) (2 0002 3i) 0002 (00025 0003 7i) (B) (1 0003 4i)(3 0002 5i) 50003i (C) (A) 7 0002 10i (B) 23 0003 7i (C) 1 0002 i (1-4) 2 0003 3i In Problems 6–9, solve the equation. 7. 4x2 0002 20 0007 0
8. x2 0002 6x 0003 2 0007 0
9. x 0002 112 0002 x 0007 0
x 0007 000215, 15 (1-5)
x 0007 3 (1-6)
x 0007 3 17 (1-5)
10. Given the points A 0007 (3, 2) and B 0007 (5, 6), find: (A) Distance between A and B. (B) Slope of the line through A and B. (C) Slope of a line perpendicular to the line through A and B. (A) 215
(B) 2
(C)
000212
(2-2, 2-3)
11. Find the equation of the circle with radius 12 and center: (A) (0, 0) (B) (00023, 1) (A) x2 0003 y2 0007 2
Neither (3-3)
18. Use the graph of f to sketch a graph of the following: (A) y 0007 0002f(x 0003 1) (B) y 0007 2f (x) 0002 2 In Problems 19–21, solve the equation. 19.
6. 3x2 0007 000212x
x 0007 00024, 0 (1-5)
17. Is f an even function, an odd function, or neither? Explain.
(B) (x 0003 3)2 0003 (y 0002 1)2 0007 2 (2-2)
12. Graph 2x 0002 3y 0007 6 and indicate its slope and intercepts. 13. Indicate whether each set defines a function. Find the domain and range of each function. (A) {(1, 1), (2, 1), (3, 1)}
x00033 5x 0003 2 5 0003 0007 2x 0003 2 3x 0003 3 6
No solution
20.
21. 2x 0003 1 0007 312x 0002 1 x 0007 1, 52
3 6 1 0007 0002 x x00031 x00021
x 0007 12 , 3 (1-1)
(1-1)
(1-6)
In Problems 22–24, solve and graph the inequality. 22. 冟4x 0002 9冟 7 3
23. 2(3m 0002 4)2 0004 2
x00031 24. 7 x00022 2 25. For what real values of x does the following expression represent a real number? 1x 0002 2 x00024 26 P f
th i di t d
ti
d
it th fi l
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Experience Student Success! ALEKS is a unique online math tool that uses adaptive questioning and artificial intelligence to correctly place, prepare, and remediate students . . . all in one product! Institutional case studies have shown that ALEKS has improved pass rates by over 20% versus traditional online homework and by over 30% compared to using a text alone. By offering each student an individualized learning path, ALEKS directs students to work on the math topics that they are ready to learn. Also, to help students keep pace in their course, instructors can correlate ALEKS to their textbook or syllabus in seconds. To learn more about how ALEKS can be used to boost student performance, please visit www.aleks.com/highered/math or contact your McGraw-Hill representative.
ALEKS Pie
Easy Graphing Utility!
Each student is given their own individualized learning path.
Students can answer graphing problems with ease!
Course Calendar Instructors can schedule assignments and reminders for students.
xxii
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New ALEKS Instructor Module Enhanced Functionality and Streamlined Interface Help to Save Instructor Time The new ALEKS Instructor Module features enhanced functionality and streamlined interface based on research with ALEKS instructors and homework management instructors. Paired with powerful assignment driven features, textbook integration, and extensive content flexibility, the new ALEKS Instructor Module simplifies administrative tasks and makes ALEKS more powerful than ever.
New Gradebook! Instructors can seamlessly track student scores on automatically graded assignments. They can also easily adjust the weighting and grading scale of each assignment.
Gradebook view for all students
Gradebook view for an individual student
Track Student Progress Through Detailed Reporting Instructors can track student progress through automated reports and robust reporting features.
Automatically Graded Assignments Instructors can easily assign homework, quizzes, tests, and assessments to all or select students. Deadline extensions can also be created for select students.
Learn more about ALEKS by visiting www.aleks.com/highered/math or contact your McGraw-Hill representative. Select topics for each assignment xxiii
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Preface
Supplements ALEKS (Assessment and Learning in Knowledge Spaces) is a dynamic online learning system for mathematics education, available over the Web 24/7. ALEKS assesses students, accurately determines their knowledge, and then guides them to the material that they are most ready to learn. With a variety of reports, Textbook Integration Plus, quizzes, and homework assignment capabilities, ALEKS offers flexibility and ease of use for instructors. • ALEKS uses artificial intelligence to determine exactly what each student knows and is ready to learn. ALEKS remediates student gaps and provides highly efficient learning and improved learning outcomes • ALEKS is a comprehensive curriculum that aligns with syllabi or specified textbooks. Used in conjunction with McGraw-Hill texts, students also receive links to text-specific videos, multimedia tutorials, and textbook pages. • ALEKS offers a dynamic classroom management system that enables instructors to monitor and direct student progress towards mastery of course objectives. ALEKS Prep/Remediation: • Helps instructors meet the challenge of remediating under prepared or improperly placed students. • Assesses students on their pre-requisite knowledge needed for the course they are entering (i.e. Calculus students are tested on Precalculus knowledge) and prescribes unique and efficient learning paths specific to their strengths and weaknesses. • Students can address pre-requisite knowledge gaps outside of class freeing the instructor to use class time pursuing course outcomes.
McGraw-Hill’s MathZone is a complete online homework system for mathematics and statistics. Instructors can assign textbook-specific content from over 40 McGraw-Hill titles as well as customize the level of feedback students receive, including the ability to have students show their work for any given exercise. Assignable content includes an array of videos and other multimedia along with algorithmic exercises, providing study tools for students with many different learning styles. MathZone also helps ensure consistent assignment delivery across several sections through a course administration function and makes sharing courses with other instructors easy. In addition, instructors can also take advantage of a virtual whiteboard by setting up a Live Classroom for online office hours or a review session with students. For more information, visit the book’s website (www.mhhe.com/barnett) or contact your local McGraw-Hill sales representative (www.mhhe.com/rep).
Tegrity Campus is a service that makes class time available all the time by automatically capturing every lecture in a searchable format for students to review when they study and complete assignments. With a simple one-click start and stop process, you capture all computer screens and corresponding audio. Students replay any part of any class with easy-touse browser-based viewing on a PC or Mac.
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Preface
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Educators know that the more students can see, hear, and experience class resources, the better they learn. With Tegrity Campus, students quickly recall key moments by using Tegrity Campus’s unique search feature. This search helps students efficiently find what they need, when they need it across an entire semester of class recordings. Help turn all your students’ study time into learning moments immediately supported by your lecture. To learn more about Tegrity watch a 2 minute Flash demo at http://tegritycampus.mhhe.com. Instructor Solutions Manual Prepared by Fred Safier of City College of San Francisco, this supplement provides detailed solutions to exercises in the text. The methods used to solve the problems in the manual are the same as those used to solve the examples in the textbook.
Student Solutions Manual Prepared by Fred Safier of City College of San Francisco, the Student’s Solutions Manual provides complete worked-out solutions to odd-numbered exercises from the text. The procedures followed in the solutions in the manual match exactly those shown in worked examples in the text. Lecture and Exercise Videos The video series is based on exercises from the textbook. J. D. Herdlick of St. Louis Community College-Meramec introduces essential definitions, theorems, formulas, and problem-solving procedures. Professor Herdlick then works through selected problems from the textbook, following the solution methodologies employed by the authors. The video series is available on DVD or online as part of MathZone. The DVDs are closed-captioned for the hearing impaired, subtitled in Spanish, and meet the Americans with Disabilities Act Standards for Accessible Design. NetTutor Available through MathZone, NetTutor is a revolutionary system that enables students to interact with a live tutor over the web. NetTutor’s web-based, graphical chat capabilities enable students and tutors to use mathematical notation and even to draw graphs as they work through a problem together. Students can also submit questions and receive answers, browse previously answered questions, and view previous sessions. Tutors are familiar with the textbook’s objectives and problem-solving styles. Computerized Test Bank (CTB) Online Available through the book’s website, this computerized test bank, utilizing Brownstone Diploma® algorithm-based testing software, enables users to create customized exams quickly. This user-friendly program enables instructors to search for questions by topic, format, or difficulty level; to edit existing questions or to add new ones; and to scramble questions and answer keys for multiple versions of the same test. Hundreds of text-specific open-ended and multiple-choice questions are included in the question bank. Sample chapter tests and a sample final exam in Microsoft Word® and PDF formats are also provided.
Acknowledgments In addition to the authors, many others are involved in the successful publication of a book. We wish to thank personally the following people who reviewed the text and offered invaluable advice for improvements: Marwan Abu-Sawwa, Florida Community College at Jacksonville Gerardo Aladro, Florida International University Eugene Allevato, Woodbury University Joy Becker, University of Wisconsin–Stout Susan Bradley, Angelina College Ellen Brook, Cuyahoga Community College, Eastern Campus Kelly Brooks, Pierce College Denise Brown, Collin County Community College Cheryl Davids, Central Carolina Technical College
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Preface
Timothy Delworth, Purdue University Marcial Echenique, Broward Community College Gay Ellis, Missouri State University Jackie English, Northern Oklahoma College Enid Mike Everett, Santa Ana College Nicki Feldman, Pulaski Technical College James Fightmaster, Virginia Western Community College Perry Gillespie, University of North Carolina at Fayetteville Vanetta Grier-Felix, Seminole Community College David Gurney, Southeastern Louisiana University Celeste Hernandez, Richland College Fredrick Hoffman, Florida Atlantic University Syed Hossain, University of Nebraska at Kearney Glenn Jablonski, Triton College Sarah Jackson, Pratt Community College Charles Johnson, South Georgia College Larry Johnson, Metropolitan State College of Denver Cheryl Kane, University of Nebraska Lincoln Raja Khoury, Collin County Community College Betty Larson, South Dakota State University Owen Mertens, Southwest Missouri State University Dana Nimic, Southeast Community College, Lincoln Campus Lyn Noble, Florida Community College at Jacksonville Luke Papademas, DeVry University, DeVry Chicago Campus David Phillips, Georgia State University Margaret Rosen, Mercer County Community College Patty Schovanec, Texas Tech University Eleanor Storey, Front Range Community College, Westminster Campus Linda Sundbye, Metropolitan State College of Denver Cynthia Woodburn, Pittsburg State University Martha Zimmerman, University of Louisville Bob Martin, Tarrant County College Susan Walker, Montana Tech of the University of Montana Lynn Cleaveland, University of Arkansas Michael Wodzak, Viterbo University Ryan Kasha, Valencia Community College Frank Juric, Brevard Community College Jerry Mayfield, North Lake College Andrew Shiers, Dakota State University Richard Avery, Dakota State University Mike Everett, Santa Ana College Greg Boyd, Murray State College Sarah Cook, Washburn University Nga Wai Liu, Bowling Green State University Donald Bennett, Murray State University Sharon Suess, Asheville-Buncombe Technical Community College Dale Rohm, University of Wisconsin at Stevens Point George Anastassiou, The University of Memphis Bill White, University of South Carolina Upstate Linda Sundbye, Metropolitan State College of Denver Khaled Hussein, University of Wisconsin Diane Cook, Okaloosa Walton College Celeste Hernandez, Richland College Thomas Riedel, University of Louisville Thomas English, College of the Mainland Hayward Allan Edwards, West Virginia University at Parkersburg
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Preface
xxvii
Debra Lehman, State Fair Community College Nancy Ressler, Oakton Community College Marwan Zabdawi, Gordon College Ianna West, Nicholls State University Tzu-Yi Alan Yang, Columbus State Community College Patricia Jones, Methodist University Kay Geving, Belmont University Linda Horner, Columbia State Community College Martha Zimmerman, University of Louisville Faye Childress, Central Piedmont Community College Bradley Thiessen, Saint Ambrose University Pamela Lasher, Edinboro University of Pennsylvania We also wish to thank Carrie Green for providing a careful and thorough check of all the mathematical calculations in the book (a tedious but extremely important job). Fred Safier for developing the supplemental manuals that are so important to the success of a text. Mitchel Levy for scrutinizing our exercises in the manuscript and making recommendations that helped us to build balanced exercise sets. Tony Palermino for providing excellent guidance in making the writing more direct and accessible to students. Pat Steele for carefully editing and correcting the manuscript. Christina Lane for editorial guidance throughout the revision process. Sheila Frank for guiding the book smoothly through all publication details. All the people at McGraw-Hill who contributed their efforts to the production of this book. Producing this new edition with the help of all these extremely competent people has been a most satisfying experience.
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APPLICATION INDEX Advertising, 326, 378 Aeronautical engineering, 404 AIDS epidemics, 348–349, 352 Airfreight, 469 Air search, 95–96 Airspeed, 434–435, 439 Air temperature, 146, 530 Alcohol consumption, 95, 221 Anthropology, 320 Approximation, 74 Architecture, 96, 132, 160, 174–175, 325, 417, 572 Astronomy, 371, 394, 530, 579 Atmospheric pressure, 373–374, 531 Automobile rental, 187 Average cost, 315 Bacterial growth, 341–342, 351, 530 Biology, 321 Body surface area, 148–149, 160 Body weight, 155, 249 Boiling point of water, 146 Braking distance, 20 Break-even analysis, 107, 213–214, 222, 440, 576 Business, 55, 64, 74, 121, 439, 500, 530, 576 Business markup policy, 149 Buying, 455 Carbon-14 dating, 343–344, 378, 381 Card hands, 540, 551 Cell division, 530 Cell phone cost, 174 Cell phone subscribers, 382 Chemistry, 55, 61–62, 74, 107, 371, 455–456, 576 Circuit analysis, 486–487 City planning, 147 Code words, 534 Coin problem, 31 Coin toss, 532–533, 548–549, 553, 557–558 Committee selection, 551–552 Communications, 421 Compound interest, 333–335, 373, 377, 567, 579 Computer design, 351 Computer-generated tests, 533 Computer science, 183, 187, 257, 576 Construction, 31, 42, 95, 97, 104, 132, 170–171, 175, 220, 256–257, 277, 285, 288, 298, 315, 325, 371, 576 Continuous compound interest, 335–336 Cost analysis, 107, 142, 146, 155, 160 Cost functions, 174, 202 Cost of high speed internet, 174 Counting card hands, 540 Counting code words, 534 Counting serial numbers, 540–541 Court design, 102 xxviii
Crime statistics, 326 Cryptography, 482–483, 486, 501 Data analysis, 160, 346 Delivery charges, 187, 456 Demographics, 147 Depreciation, 155–156, 159, 353, 576 Design, 104, 107, 404 Diamond prices, 152–153 Dice roles, 546, 554, 557 Diet, 500, 572 Distance-rate-time problems, 50–51, 92 Divorce, 277 Earthquakes, 366–368, 371, 376, 379, 382, 579 Earth science, 55, 64, 352, 440 Ecology, 371 Economics, 20, 42, 55, 64, 530, 567, 572 Economy stimulation, 527–528 Electric circuit, 42 Electricity, 320 Employee training, 314 Energy, 64 Engineering, 96, 132, 220–221, 321, 394, 404, 421, 530, 572 Epidemics, 345–346 Evaporation, 203, 234 Explosive energy, 371 Fabrication, 298 Falling object, 220, 256 Finance, 339, 440, 530, 572, 576 Flight conditions, 156 Flight navigation, 156 Fluid flow, 203, 234 Food chain, 530 Forestry, 155, 160 Gaming, 351 Gas mileage, 220 Genealogy, 530 Genetics, 321 Geometry, 31, 55, 103, 287, 321, 456, 487, 531, 572, 579 Health care, 277 Heat conduction, 501 History of technology, 351 Home ownership rates, 369 Hydroelectric power, 272–273 Illumination, 320 Immigration, 377 Income, 256
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APPLICATION INDEX
Income tax, 181–182 Infectious diseases, 347 Insecticides, 351 Insurance company data, 555–556 Internet access, 371–372 Inventory value, 468–469 Investment, 441 Investment allocation, 480–481 Labor costs, 464–465, 468, 500–501 Learning curve, 344–345 Learning theory, 315 Licensed drivers, 156 Life science, 55 Linear depreciation, 159 Loan repayment, 567 Logistic growth, 345–346 Manufacturing, 103, 174, 277, 287–288, 325, 572 Marine biology, 352, 378, 382 Market analysis, 558, 567–568 Market research, 234, 256 Markup policy, 156, 250, 256, 468 Marriage, 277 Maximizing revenue, 222, 277 Maximum area, 210–211 Medical research, 378 Medicare, 382 Medicinal lithotripsy, 401 Medicine, 352, 382 Mixing antifreeze, 150 Mixture problems, 52–53, 150 Money growth, 339, 382 Movie industry revenue, 220 Music, 56, 321, 531 Naval architecture, 404–405 Navigation, 95, 412–414 Newton’s law of cooling, 352, 378 Nuclear power, 353, 417–418 Numbers, 107, 576 Nutrition, 456, 469 Oceanography, 146–147 Officer selection, 537 Olympic games, 157 Ozone levels, 113 Packaging, 31, 298 Parachutes, 156 Pendulum, 21 Petty crime, 455 Photography, 321, 352, 378, 531 Physics, 122, 146, 320–321, 326, 510, 530, 576, 580 Physiology, 314–315 Player ranking, 470 Political science, 572 Politics, 107, 469
xxix
Pollution, 234 Population growth, 340–341, 351–352, 378, 381, 530, 579 Present value, 339, 382 Price and demand, 93, 95, 121, 249–250, 256, 576–577 Price and supply, 121, 250 Prize money, 524 Production costs, 202, 468 Production rates, 440 Production scheduling, 436–437, 440–441, 456, 462, 486 Profit analysis, 213–214, 220, 221–222, 230–231 Profit and loss analysis, 213–214, 576 Projectile flight, 220 Projectile motion, 211, 594–595, 596 Psychology, 56, 64, 321 Purchasing, 452–453, 572 Puzzle, 501, 530–531 Quality control, 568 Quantity-rate-time problems, 51–52 Radioactive decay, 342–343 Radioactive tracers, 351 Rate of descent, 156 Rate problems, 174 Rate-time, 576 Rate-time problems, 55–56, 107, 439 Regression, 346 Relativistic mass, 21 Replacement time, 315 Resource allocation, 456, 486, 500 Retention, 315 Revenue, 242–243, 277 Revenue analysis, 496 Rocket flight, 368–369 Safety research, 203 Salary increment, 510 Sales and commissions, 187, 460–461 Serial numbers, 540–541 Service charges, 187 Shipping, 288, 576, 579 Signal light, 394 Significant digits, 74 Simple interest, 326 Smoking statistics, 155 Sociology, 456 Sound, 365–366, 371, 382, 579 Space science, 321, 352, 394, 418, 421 Space vehicles, 371 Speed of sound, 155 Sports, 131–132 Sports medicine, 107, 160 State income tax, 187, 257 Statistics, 74 Stopping distance, 214–215, 221, 250, 256, 576 Storage, 298 Subcommittee selection, 539 Supply and demand, 157, 435–436, 440
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APPLICATION INDEX
Telephone charges, 187 Telephone expenditures, 153–154 Temperature, 122 Test averaging, 572 Thumbtack toss, 552 Timber harvesting, 202–203 Tournament seeding, 469–470 Traffic flow, 501–502 Training, 353 Transportation, 96, 567
Underwater pressure, 151 Weather, 175 Weather balloon, 234 Weight of fish, 271 Well depth, 103 Wildlife management, 353, 382 Work, 326 Zeno’s paradox, 531
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College Algebra
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CHAPTER
Basic Algebraic Operations
R
C
OUTLINE
ALGEBRA is “generalized arithmetic.” In arithmetic we add, subtract,
multiply, and divide specific numbers. In algebra we use all that we know about arithmetic, but, in addition, we work with symbols that represent one or more numbers. In this chapter we review some important basic algebraic operations usually studied in earlier courses.
R-1
Algebra and Real Numbers
R-2
Exponents and Radicals
R-3
Polynomials: Basic Operations and Factoring
R-4
Rational Expressions: Basic Operations Chapter R Review
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BASIC ALGEBRAIC OPERATIONS
Algebra and Real Numbers Z The Set of Real Numbers Z The Real Number Line Z Addition and Multiplication of Real Numbers Z Further Operations and Properties
3 The numbers 14, 00033, 0, 73, 12, and 1 6 are examples of real numbers. Because the symbols we use in algebra often stand for real numbers, we will discuss important properties of the real number system.
Z The Set of Real Numbers Informally, a real number is any number that has a decimal representation. So the real numbers are the numbers you have used for most of your life. The set of real numbers, denoted by R, is the collection of all real numbers. The notation 12 僆 R (read “ 12 is an element of R”) expresses the fact that 12 is a real number. The set Z 0002 {. . . , 00032, 00031, 0, 1, 2, . . .} of the natural numbers, along with their negatives and zero, is called the set of integers. We write Z ( R (read “Z is a subset of R”) to express the fact that every element of Z is an element of R; that is, that every integer is a real number. Table 1 describes the set of real numbers and some of its important subsets. Study Table 1 and note in particular that N ( Z ( Q ( R. No real number is both rational and irrational, so the intersection (overlap) of the sets Q and I is the empty set (or null set), denoted by 0004. The empty set contains no elements,
Table 1 The Set of Real Numbers Symbol
Name
Description
Examples
N
Natural numbers
Counting numbers (also called positive integers)
1, 2, 3, . . .
Z
Integers
Natural numbers, their negatives, and 0 (also called whole numbers)
. . . , 00032, 00031, 0, 1, 2, . . .
Q
Rational numbers
Numbers that can be represented as a兾b, where a and b are integers and b 0006 0; decimal representations are repeating or terminating
00034, 0, 1, 25, 000335, 23, 3.67, 00030.333,* 5.272727
I
Irrational numbers
Numbers that can be represented as nonrepeating and nonterminating decimal numbers
3 12, 0005, 1 7, 1.414213 . . . ,† 2.71828182 . . .†
R
Real numbers
Rational numbers and irrational numbers
*The overbar indicates that the number (or block of numbers) repeats indefinitely. †Note that the ellipsis does not indicate that a number (or block of numbers) repeats indefinitely.
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SECTION R–1
3
Algebra and Real Numbers
so it is true that every element of the empty set is an element of any given set. In other words, the empty set is a subset of every set. Two sets are equal if they have exactly the same elements. The order in which the elements of a set are listed does not matter. For example, {1, 2, 3, 4} 0002 {3, 1, 4, 2}
Z The Real Number Line A one-to-one correspondence exists between the set of real numbers and the set of points on a line. That is, each real number corresponds to exactly one point, and each point to exactly one real number. A line with a real number associated with each point, and vice versa, as in Figure 1, is called a real number line, or simply a real line. Each number associated with a point is called the coordinate of the point. The point with coordinate 0 is called the origin. The arrow on the right end of the line indicates a positive direction. The coordinates of all points to the right of the origin are called positive real numbers, and those to the left of the origin are called negative real numbers. The real number 0 is neither positive nor negative.
4 0003 3 Origin
0003兹27 000310
00035
0
7.64
5
10
Z Figure 1 A real number line.
Z Addition and Multiplication of Real Numbers How do you add or multiply two real numbers that have nonrepeating and nonterminating decimal expansions? The answer to this difficult question relies on a solid understanding of the arithmetic of rational numbers. The rational numbers are numbers that can be written in the form a兾b, where a and b are integers and b 0006 0 (see Table 1 on page 2). The numbers 7兾5 and 00032兾3 are rational, and any integer a is rational because it can be written in the form a兾1. Two rational numbers a兾b and c兾d are equal if ad 0002 bc; for example, 35兾10 0002 7兾2. Recall how the sum and product of rational numbers are defined:
Z DEFINITION 1 Addition and Multiplication of Rationals For rational numbers a兾b and c兾d, where a, b, c, and d are integers and b 0006 0, d 0006 0: Addition: Multiplication:
a 0007 b a ⴢ b
c ad 0007 bc 0002 d bd c ac 0002 d bd
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BASIC ALGEBRAIC OPERATIONS
Addition and multiplication of rational numbers are commutative; changing the order in which two numbers are added or multiplied does not change the result. 3 0007 2 3 ⴢ 2
5 5 3 0002 0007 7 7 2 5 5 3 0002 ⴢ 7 7 2
Addition is commutative.
Multiplication is commutative.
Addition and multiplication of rational numbers is also associative; changing the grouping of three numbers that are added or multiplied does not change the result: 3 5 0007a 0007 2 7 3 5 ⴢa ⴢ 2 7
9 3 5 9 b0002a 0007 b0007 4 2 7 4 9 3 5 9 b0002a ⴢ bⴢ 4 2 7 4
Addition is associative.
Multiplication is associative.
Furthermore, the operations of addition and multiplication are related in that multiplication distributes over addition: 3 5 9 ⴢa 0007 b0002 2 7 4 9 3 5 a 0007 bⴢ 0002 7 4 2
3 5 3 9 ⴢ 0007 ⴢ 2 7 2 4 5 3 9 3 ⴢ 0007 ⴢ 7 2 4 2
Left distributive law
Right distributive law
The rational number 0 is an additive identity; adding 0 to a number does not change it. The rational number 1 is a multiplicative identity; multiplying a number by 1 does not change it. Every rational number r has an additive inverse, denoted 0003r; the additive inverse of 4兾5 is 00034兾5, and the additive inverse of 00033兾2 is 3兾2. The sum of a number and its additive inverse is 0. Every nonzero rational number r has a multiplicative inverse, denoted r00031; the multiplicative inverse of 4兾5 is 5兾4, and the multiplicative inverse of 00033兾2 is 00032兾3. The product of a number and its multiplicative inverse is 1. The rational number 0 has no multiplicative inverse.
EXAMPLE
1
Arithmetic of Rational Numbers Perform the indicated operations. (A)
1 6 0007 3 5
(C) (0003179)00031 SOLUTIONS
(B)
8 5 ⴢ 3 4
(D) (00036 0007 92)00031
(A)
1 6 5 0007 18 23 0007 0002 0002 3 5 15 15
(B)
8 5 40 10 ⴢ 0002 0002 3 4 12 3
40 10 ⴝ 12 3
because
40 ⴢ 3 ⴝ 12 ⴢ 10
(C) (0003179)00031 0002 0003917 (D) (00036 0007 92)00031 0002 a
00036 9 00031 000312 0007 9 00031 00033 00031 2 0007 b 0002a b 0002a b 00020003 1 2 2 2 3
0002
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MATCHED PROBLEM 1*
Algebra and Real Numbers
5
Perform the indicated operations. (A) 0003(52 0007 73) (C)
21 15 ⴢ 20 14
00031 (B) 0003(817)
(D) 5 ⴢ (12 0007 13) 0002
Rational numbers have decimal expansions that are repeating or terminating. For example, using long division, 2 0002 0.666 3 22 0002 3.142857 7 13 0002 1.625 8
The number 6 repeats indefinitely.
The block 142857 repeats indefinitely.
Terminating expansion
Conversely, any decimal expansion that is repeating or terminating represents a rational number (see Problems 49 and 50 in Exercise R-1). The number 12 is irrational because it cannot be written in the form a兾b, where a and b are integers, b 0006 0 (for an explanation, see Problem 89 in Section R-3). Similarly, 13 is irrational. But 14, which is equal to 2, is a rational number. In fact, if n is a positive integer, then 1n is irrational unless n belongs to the sequence of perfect squares 1, 4, 9, 16, 25, . . . (see Problem 90 in Section R-3). We now return to our original question: how do you add or multiply two real numbers that have nonrepeating and nonterminating decimal expansions? Although we will not give a detailed answer to this question, the key idea is that every real number can be approximated to any desired precision by rational numbers. For example, the irrational number 12 ⬇ 1.414 213 562 . . . is approximated by the rational numbers 14 10 141 100 1,414 1,000 14,142 10,000 141,421 100,000
0002 1.4 0002 1.41 0002 1.414 0002 1.4142 0002 1.41421 .. .
Using the idea of approximation by rational numbers, we can extend the definitions of rational number operations to include real number operations. The following box summarizes the basic properties of real number operations. *Answers to matched problems in a given section are found near the end of the section, before the exercise set.
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Z BASIC PROPERTIES OF THE SET OF REAL NUMBERS Let R be the set of real numbers, and let x, y, and z be arbitrary elements of R. Addition Properties Closure:
x 0007 y is a unique element in R.
Associative:
(x 0007 y) 0007 z 0002 x 0007 (y 0007 z)
Commutative:
x0007y0002y0007x
Identity:
0 is the additive identity; that is, 0 0007 x 0002 x 0007 0 0002 x for all x in R, and 0 is the only element in R with this property.
Inverse:
For each x in R, 0003x is its unique additive inverse; that is, x 0007 (0003x) 0002 (0003x) 0007 x 0002 0, and 0003x is the only element in R relative to x with this property.
Multiplication Properties Closure:
xy is a unique element in R.
Associative:
(xy)z 0002 x( yz)
Commutative:
xy 0002 yx
Identity:
1 is the multiplicative identity; that is, for all x in R, (1)x 0002 x(1) 0002 x, and 1 is the only element in R with this property.
Inverse:
For each x in R, x 0006 0, x00031 is its unique multiplicative inverse; that is, xx00031 0002 x00031x 0002 1, and x00031 is the only element in R relative to x with this property.
Combined Property Distributive:
EXAMPLE
2
x(y 0007 z) 0002 xy 0007 xz
(x 0007 y)z 0002 xz 0007 yz
Using Real Number Properties Which real number property justifies the indicated statement?
SOLUTIONS
(A) (B) (C) (D) (E)
(7x)y 0002 7(xy) a(b 0007 c) 0002 (b 0007 c)a (2x 0007 3y) 0007 5y 0002 2x 0007 (3y 0007 5y) (x 0007 y)(a 0007 b) 0002 (x 0007 y)a 0007 (x 0007 y)b If a 0007 b 0002 0, then b 0002 0003a.
(A) (B) (C) (D) (E)
Associative (ⴢ) Commutative (ⴢ) Associative (0007) Distributive Inverse (0007)
0002
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MATCHED PROBLEM 2
Algebra and Real Numbers
7
Which real number property justifies the indicated statement? (A) 4 0007 (2 0007 x) 0002 (4 0007 2) 0007 x (C) 3x 0007 7x 0002 (3 0007 7)x (E) If ab 0002 1, then b 0002 1a.
(B) (a 0007 b) 0007 c 0002 c 0007 (a 0007 b) (D) (2x 0007 3y) 0007 0 0002 2x 0007 3y 0002
Z Further Operations and Properties Subtraction of real numbers can be defined in terms of addition and the additive inverse. If a and b are real numbers, then a 0003 b is defined to be a 0007 (0003b). Similarly, division can be defined in terms of multiplication and the multiplicative inverse. If a and b are real numbers and b 0006 0, then a b (also denoted a兾b) is defined to be a ⴢ b00031.
Z DEFINITION 2 Subtraction and Division of Real Numbers For all real numbers a and b: Subtraction:
a 0003 b 0002 a 0007 (0003b)
Division:
a b 0002 a ⴢ b00031
5 ⴚ 3 ⴝ 5 ⴙ (ⴚ3) ⴝ 2
b00060
3 ⴜ 2 ⴝ 3 ⴢ 2ⴚ1 ⴝ 3 ⴢ
1 ⴝ 1.5 2
It is important to remember that Division by 0 is never allowed.
ZZZ EXPLORE-DISCUSS 1
(A) Give an example that shows that subtraction of real numbers is not commutative. (B) Give an example that shows that division of real numbers is not commutative.
The basic properties of the set of real numbers, together with the definitions of subtraction and division, lead to the following properties of negatives and zero.
Z THEOREM 1 Properties of Negatives For all real numbers a and b: 0003(0003a) 0002 a (0003a)b 0002 0003(ab) 0002 a(0003b) 0002 0003ab (0003a)(0003b) 0002 ab (00031)a 0002 0003a a a 0003a 5. 00020003 0002 b00060 b b 0003b 0003a 0003a a a 6. b00060 00020003 00020003 0002 0003b b 0003b b 1. 2. 3. 4.
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Z THEOREM 2 Zero Properties For all real numbers a and b: 1. a ⴢ 0 0002 0 ⴢ a 0002 0 2. ab 0002 0 if and only if*
a 0002 0 or b 0002 0 or both
Note that if b 0006 0, then 0b 0002 0 ⴢ b00031 0002 0 by Theorem 2. In particular, 03 0002 0; but the expressions 30 and 00 are undefined.
EXAMPLE
3
Using Negative and Zero Properties Which real number property or definition justifies each statement? (A) 3 0003 (00032) 0002 3 0007 [0003(00032)] 0002 5 (B) 0003(00032) 0002 2 00033 3 0002 (C) 0003 2 2 5 5 00020003 (D) 00032 2 (E) If (x 0003 3)(x 0007 5) 0002 0, then either x 0003 3 0002 0 or x 0007 5 0002 0.
SOLUTIONS
MATCHED PROBLEM 3
(A) (B) (C) (D) (E)
Subtraction (Definition 1 and Theorem 1, part 1) Negatives (Theorem 1, part 1) Negatives (Theorem 1, part 6) Negatives (Theorem 1, part 5) Zero (Theorem 2, part 2)
0002
Which real number property or definition justifies each statement? (A)
3 1 0002 3a b 5 5
(B) (00035)(2) 0002 0003(5 ⴢ 2)
(D)
7 00037 00020003 9 9
(E) If x 0007 5 0002 0, then (x 0003 3)(x 0007 5) 0002 0.
(C) (00031)3 0002 00033
0002
ZZZ EXPLORE-DISCUSS 2
A set of numbers is closed under an operation if performing the operation on numbers of the set always produces another number in the set. For example, the set of odd integers is closed under multiplication, but is not closed under addition. (A) Give an example that shows that the set of irrational numbers is not closed under addition. (B) Explain why the set of irrational numbers is closed under taking multiplicative inverses.
*Given statements P and Q, “P if and only if Q” stands for both “if P, then Q” and “if Q, then P.”
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Algebra and Real Numbers
9
If a and b are real numbers, b 0006 0, the quotient a b, when written in the form a兾b, is called a fraction. The number a is the numerator, and b is the denominator. It can be shown that fractions satisfy the following properties. (Note that some of these properties, under the restriction that numerators and denominators are integers, were used earlier to define arithmetic operations on the rationals.)
Z THEOREM 3 Fraction Properties For all real numbers a, b, c, d, and k (division by 0 excluded): 1.
a c 0002 b d 4 6 ⴝ 6 9
2.
since
4ⴢ9ⴝ6ⴢ6
ka a 0002 kb b
3.
7ⴢ3 3 ⴝ 7ⴢ5 5
5.
ad 0002 bc
if and only if
c a0007c a 0007 0002 b b b 3 4 3ⴙ4 7 ⴙ ⴝ ⴝ 6 6 6 6
a c ac ⴢ 0002 b d bd
4.
3 7 3ⴢ7 21 ⴢ ⴝ ⴝ 5 8 5ⴢ8 40
6.
c a0003c a 0003 0002 b b b 7 2 7ⴚ2 5 ⴚ ⴝ ⴝ 8 8 8 8
a c a d 0002 ⴢ b d b c 2 5 2 7 14 ⴜ ⴝ ⴢ ⴝ 3 7 3 5 15
7.
c ad 0007 bc a 0007 0002 b d bd 2 1 2ⴢ5ⴙ3ⴢ1 13 ⴙ ⴝ ⴝ 3 5 3ⴢ5 15
ANSWERS TO MATCHED PROBLEMS 1. (A) 0003296 (B) 000317 8 (C) 9 8 (D) 256 2. (A) Associative (0007) (B) Commutative (0007) (C) Distributive (D) Identity (0007) (E) Inverse (ⴢ) 3. (A) Division (Definition 1) (B) Negatives (Theorem 1, part 2) (C) Negatives (Theorem 1, part 4) (D) Negatives (Theorem 1, part 5) (E) Zero (Theorem 2, part 1)
R-1
Exercises
In Problems 1–16, perform the indicated operations, if defined. If the result is not an integer, express it in the form a/b, where a and b are integers. 1 1 1. 0007 3 5
1 1 2. 0007 2 7
3.
3 4 0003 4 3
4.
8 4 0003 9 5
5.
2 4 ⴢ 3 7
6. a0003
1 3 bⴢ 10 8
7.
11 1 5 3
9. 100 0 3 5 11. a0003 b a0003 b 5 3 13.
17 2 ⴢ 8 7
3 00031 15. a b 0007 200031 8
8.
7 2 9 5
10. 0 0 12.
6 4 a3 0003 b 7 2
2 5 14. a0003 b a0003 b 3 6 16. 0003(400031 0007 3)
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In Problems 17–28, each statement illustrates the use of one of the following properties or definitions. Indicate which one. Commutative (0007) Commutative (ⴢ) Associative (0007) Associative (ⴢ) Distributive Identity (0007) Identity (ⴢ)
Inverse (0007) Inverse (ⴢ) Subtraction Division Negatives (Theorem 1) Zero (Theorem 2)
17. x 0007 ym 0002 x 0007 my 1 21. (00032)(00032 )00021
18. 7(3m) 0002 (7 ⴢ 3)m u u 20. 0003 0002 0003v v 22. 8 0003 12 0002 8 0007 (000312)
23. w 0007 (0003w) 0002 0
1 24. 5 (00036) 0002 5(00036 )
19. 7u 0007 9u 0002 (7 0007 9)u
40. Indicate true (T) or false (F), and for each false statement find real number replacements for a, b, and c that will provide a counterexample. For all real numbers a, b, and c: (A) (a 0007 b) 0007 c 0002 a 0007 (b 0007 c) (B) (a 0003 b) 0003 c 0002 a 0003 (b 0003 c) (C) a(bc) 0002 (ab)c (D) (a b) c 0002 a (b c) In Problems 41–48, indicate true (T) or false (F), and for each false statement give a specific counterexample.
25. 3(xy 0007 z) 0007 0 0002 3(xy 0007 z) 26. ab(c 0007 d ) 0002 abc 0007 abd 0003x x 27. 0002 0003y y
39. Indicate true (T) or false (F), and for each false statement find real number replacements for a and b that will provide a counterexample. For all real numbers a and b: (A) a 0007 b 0002 b 0007 a (B) a 0003 b 0002 b 0003 a (C) ab 0002 ba (D) a b 0002 b a
28. (x 0007 y) ⴢ 0 0002 0
41. The difference of any two natural numbers is a natural number. 42. The quotient of any two nonzero integers is an integer. 43. The sum of any two rational numbers is a rational number.
29. If ab 0002 0, does either a or b have to be 0?
44. The sum of any two irrational numbers is an irrational number.
30. If ab 0002 1, does either a or b have to be 1?
45. The product of any two irrational numbers is an irrational number.
31. Indicate which of the following are true: (A) All natural numbers are integers. (B) All real numbers are irrational. (C) All rational numbers are real numbers. 32. Indicate which of the following are true: (A) All integers are natural numbers. (B) All rational numbers are real numbers. (C) All natural numbers are rational numbers. 33. Give an example of a rational number that is not an integer. 34. Give an example of a real number that is not a rational number. In Problems 35 and 36, list the subset of S consisting of (A) natural numbers, (B) integers, (C) rational numbers, and (D) irrational numbers. 35. S 0002 500033, 000323, 0, 1, 13, 95, 11446
36. S 0002 50003 15, 00031, 000312, 2, 17, 6, 16259, 00056
46. The product of any two rational numbers is a rational number. 47. The multiplicative inverse of any irrational number is an irrational number. 48. The multiplicative inverse of any nonzero rational number is a rational number. 49. If c 0002 0.151515 . . . , then 100c 0002 15.1515 . . . and 100c 0003 c 0002 15.1515 . . . 0003 0.151515 . . . 99c 0002 15 5 c 0002 15 99 0002 33
Proceeding similarly, convert the repeating decimal 0.090909 . . . into a fraction. (All repeating decimals are rational numbers, and all rational numbers have repeating decimal representations.) 50. Repeat Problem 49 for 0.181818. . . .
In Problems 37 and 38, use a calculator* to express each number in decimal form. Classify each decimal number as terminating, repeating, or nonrepeating and nonterminating. Identify the pattern of repeated digits in any repeating decimal numbers. 37. (A) 98
(B) 113
38. (A) 136
(B) 121
(C) 15 (C) 167
(D) 118 29 (D) 111
*Later in the book you will encounter optional exercises that require a graphing calculator. If you have such a calculator, you can certainly use it here. Otherwise, any scientific calculator will be sufficient for the problems in this chapter.
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R-2
Exponents and Radicals
11
Exponents and Radicals Z Integer Exponents Z Scientific Notation Z Roots of Real Numbers Z Rational Exponents and Radicals Z Simplifying Radicals
The French philosopher/mathematician René Descartes (1596–1650) is generally credited with the introduction of the very useful exponent notation “x n.” This notation as well as other improvements in algebra may be found in his Geometry, published in 1637. If n is a natural number, x n denotes the product of n factors, each equal to x. In this section, the meaning of x n will be expanded to allow the exponent n to be any rational number. Each of the following expressions will then represent a unique real number: 500034
75
3.140
612
14000353
Z Integer Exponents If a is a real number, then a6 0002 a ⴢ a ⴢ a ⴢ a ⴢ a ⴢ a
6 factors of a
In the expression a6, 6 is called an exponent and a is called the base. Recall that a00031, for a 0006 0, denotes the multiplicative inverse of a (that is, 1 a). To generalize exponent notation to include negative integer exponents and 0, we define a00036 to be the multiplicative inverse of a6, and we define a0 to be 1. n Z DEFINITION 1 a , n an Integer and a a Real Number
For n a positive integer and a a real number: an 0002 a ⴢ a ⴢ . . . ⴢ a 1 a0003n 0002 n a a0 0002 1
EXAMPLE
1
n factors of a
(a 0006 0) (a 0006 0)
Using the Definition of Integer Exponents Write parts (A) and (B) in decimal form and parts (C) and (D) using positive exponents. Assume all variables represent nonzero real numbers. (A) (u3v2)0
(B) 1000033
(C) x00038
(D)
x00033 y00035
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SOLUTIONS
(A) (u3v2)0 0002 1 (C) x00038 0002
MATCHED PROBLEM 1
1 x8
(B) 1000033 0002
1 1 0002 0.001 3 0002 1,000 10
x00033 y00035
x00033 1 1 y5 ⴢ 00035 0002 3 ⴢ 1 1 y x
(D)
0002
* 0002
y5 x3
0002
Write parts (A) and (B) in decimal form and parts (C) and (D) using positive exponents. Assume all variables represent nonzero real numbers. (B) 1000035
(A) (x2)0
(C)
1
u00037 v00033
(D)
00034
x
0002 To calculate with exponents, it is helpful to remember Definition 1. For example: 23 ⴢ 24 0002 (2 ⴢ 2 ⴢ 2)(2 ⴢ 2 ⴢ 2 ⴢ 2) 0002 2300074 0002 27 (23)4 0002 (2 ⴢ 2 ⴢ 2)4 0002 (2 ⴢ 2 ⴢ 2)(2 ⴢ 2 ⴢ 2)(2 ⴢ 2 ⴢ 2)(2 ⴢ 2 ⴢ 2) 0002 23ⴢ4 0002 212 These are instances of Properties 1 and 2 of Theorem 1. Z THEOREM 1 Properties of Integer Exponents For n and m integers and a and b real numbers: 1. aman 0002 am0007n
a5a ⴚ7
ⴝ a5ⴙ(ⴚ7)
2. (an)m 0002 amn 3. (ab)m 0002 ambm a m am 4. a b 0002 m b b
(a3)ⴚ2
ⴝ a(ⴚ2)3
再
am0003n a 1 5. n 0002 a an0003m m
EXAMPLE
2
ⴝ aⴚ2 ⴝ aⴚ6
(ab)3 ⴝ a3b3 a 4 a4 a b ⴝ 4 b b
b00060
a3 aⴚ2
a00060
a3 aⴚ2
ⴝ a3ⴚ(ⴚ2) ⴝ a5 ⴝ
1 aⴚ2ⴚ3
ⴝ
1 aⴚ5
Using Exponent Properties Simplify using exponent properties, and express answers using positive exponents only.†
SOLUTIONS
6x00032 8x00035
(A) (3a5)(2a00033)
(B)
(C) 00034y3 0003 (00034y)3
(D) (2a00033b2)00032
(A) (3a5)(2a00033)
(B)
6x00032 8x00035
0002
0002 (3 ⴢ 2)(a5a00033)
3x000320003(00035) 4
0002
0002 6a2
3x3 4
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally. †
By “simplify” we mean eliminate common factors from numerators and denominators and reduce to a minimum the number of times a given constant or variable appears in an expression. We ask that answers be expressed using positive exponents only in order to have a definite form for an answer. Later (in this section and elsewhere) we will encounter situations where we will want negative exponents in a final answer.
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SECTION R–2
(C) 00034y3 0003 (00034y)3 0002 00034y3 0003 (00034)3y3
Exponents and Radicals
13
0002 00034y3 0003 (000364)y3
0002 00034y3 0007 64y3 0002 60y3 (D) (2a00033b2)00032 0002 200032a6b00034 0002
MATCHED PROBLEM 2
a6 4b4
0002
Simplify using exponent properties, and express answers using positive exponents only. (A) (5x00033)(3x4)
(B)
9y00037 6y00034
(C) 2x4 0003 (00032x)4
(D) (3x4y00033)00032 0002
Z Scientific Notation Scientific work often involves the use of very large numbers or very small numbers. For example, the average cell contains about 200,000,000,000,000 molecules, and the diameter of an electron is about 0.000 000 000 0004 centimeter. It is generally troublesome to write and work with numbers of this type in standard decimal form. The two numbers written here cannot even be entered into most calculators as they are written. However, each can be expressed as the product of a number between 1 and 10 and an integer power of 10: 200,000,000,000,000 0002 2 1014 0.000 000 000 0004 0002 4 10000313 In fact, any positive number written in decimal form can be expressed in scientific notation, that is, in the form a 10n
EXAMPLE
3
1 a 6 10, n an integer, a in decimal form
Scientific Notation (A) Write each number in scientific notation: 6,430; 5,350,000; 0.08; 0.000 32 (B) Write in standard decimal form: 2.7 102; 9.15 104; 5 1000033; 8.4 1000035
SOLUTIONS
MATCHED PROBLEM 3
(A) 6,430 0002 6.43 103; 5,350,000 0002 5.35 106; 0.08 0002 8 1000032; 0.000 32 0002 3.2 1000034 (B) 270; 91,500; 0.005; 0.000 084
0002
(A) Write each number in scientific notation: 23,000; 345,000,000; 0.0031; 0.000 000 683 (B) Write in standard decimal form: 4 103; 5.3 105; 2.53 1000032; 7.42 1000036
0002
Most calculators express very large and very small numbers in scientific notation. Consult the manual for your calculator to see how numbers in scientific notation are entered in your calculator. Some common methods for displaying scientific notation on a calculator are shown here. Number Represented
Typical Scientific Calculator Display
Typical Graphing Calculator Display
5.427 493 10000317
5.427493 – 17
5.427493E – 17
2.359 779 1012
2.359779 12
2.359779E12
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4
Using Scientific Notation on a Calculator 325,100,000,000 by writing each number in scientific notation and then 0.000 000 000 000 0871 using your calculator. (Refer to the user’s manual accompanying your calculator for the procedure.) Express the answer to three significant digits* in scientific notation. Calculate
SOLUTION
325,100,000,000 3.251 1011 0002 0.000 000 000 000 0871 8.71 10000314 0002 3.732491389E24 24
0002 3.73 10
Z Figure 1
MATCHED PROBLEM 4
Calculator display To three significant digits
Figure 1 shows two solutions to this problem on a graphing calculator. In the first solution we entered the numbers in scientific notation, and in the second we used standard decimal notation. Although the multiple-line screen display on a graphing calculator enables us to enter very long standard decimals, scientific notation is usually more efficient and less prone to errors in data entry. Furthermore, as Figure 1 shows, the calculator uses scientific notation to display the answer, regardless of the manner in which the numbers are entered. 0002 Repeat Example 4 for: 0.000 000 006 932 62,600,000,000 0002
Z Roots of Real Numbers The solutions of the equation x2 0002 64 are called square roots of 64 and the solutions of x3 0002 64 are the cube roots of 64. So there are two real square roots of 64 (00038 and 8) and one real cube root of 64 (4 is a cube root, but 00034 is not). Note that 000364 has no real square root (x2 0002 000364 has no real solution because the square of a real number can’t be negative), but 00034 is a cube root of 000364 because (00034)3 0002 000364. In general:
Z DEFINITION 2 Definition of an nth Root For a natural number n and a and b real numbers: a is an nth root of b if an 0002 b
3 is a fourth root of 81, since 34 ⴝ 81.
The number of real nth roots of a real number b is either 0, 1, or 2, depending on whether b is positive or negative, and whether n is even or odd. Theorem 2 gives the details, which are summarized in Table 1.
*For those not familiar with the meaning of significant digits, see Appendix A for a brief discussion of this concept.
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Table 1 Number of Real nth Roots of b n even
n odd
b 7 0
2
1
b00020
1
1
b 6 0
0
1
Exponents and Radicals
15
Z THEOREM 2 Number of Real nth Roots of a Real Number b Let n be a natural number and let b be a real number: 1. b 7 0: If n is even, then b has two real nth roots, each the negative of the other; if n is odd, then b has one real nth root. 2. b 0002 0: 0 is the only nth root of b 0002 0. 3. b 6 0: If n is even, then b has no real nth root; if n is odd, then b has one real nth root.
Z Rational Exponents and Radicals To denote nth roots, we can use rational exponents or we can use radicals. For example, the square root of a number b can be denoted by b12 or 1b. To avoid ambiguity, both expressions denote the positive square root when there are two real square roots. Furthermore, both expressions are undefined when there is no real square root. In general:
Z DEFINITION 3 Principal nth Root For n a natural number and b a real number, the principal nth root of b, n denoted by b1n or 1b, is: 1. The real nth root of b if there is only one. 2. The positive nth root of b if there are two real nth roots. 3. Undefined if b has no real nth root.
n
In the notation 1b, the symbol 1 is called a radical, n is called the index, and b is the 2 radicand. If n 0002 2, we write 1b in place of 1 b.
EXAMPLE
5
Principal nth Roots Evaluate each expression: (A) 912
SOLUTIONS
MATCHED PROBLEM 5
(B) 1121
(A) 912 0002 3 3 (C) 10003125 0002 00035 (E) 2713 0002 3
3 (C) 1 0003125
(D) (000316)14
(E) 2713
5 (F) 1 32
(B) 1121 0002 11 (D) (000316)14 is undefined (not a real number). 5 (F) 132 0002 2
0002
Evaluate each expression: (A) 813
(B) 100034
4 (C) 110,000
(D) (00031)15
3
(E) 1000327
(F) 018 0002
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How should a symbol such as 723 be defined? If the properties of exponents are to hold for rational exponents, then 723 0002 (713)2; that is, 723 must represent the square of the cube root of 7. This leads to the following general definition: m兾n and bⴚm兾n, Rational Number Exponent Z DEFINITION 4 b
For m and n natural numbers and b any real number (except b cannot be negative when n is even): bmn 0002 (b1n)m 432 ⴝ (412)3 ⴝ 23 ⴝ 8
b0003mn 0002
and 4ⴚ32 ⴝ
1
ⴝ
432
1 8
1 bmn
(ⴚ4)32 is not real
(ⴚ32)35 ⴝ [(ⴚ32)15 ] 3 ⴝ (ⴚ2)3 ⴝ ⴚ8
We have now discussed bmn for all rational numbers mn and real numbers b. It can be shown, though we will not do so, that all five properties of exponents listed in Theorem 1 continue to hold for rational exponents as long as we avoid even roots of negative numbers. With the latter restriction in effect, the following useful relationship is an immediate consequence of the exponent properties: Z THEOREM 3 Rational Exponent/Radical Property For m and n natural numbers and b any real number (except b cannot be negative when n is even): (b1n)m 0002 (bm)1n
ZZZ EXPLORE-DISCUSS 1
and
n
n
(1b)m 0002 2bm
Find the contradiction in the following chain of equations: 00031 0002 (00031)22 0002 3(00031)2 4 12 0002 112 0002 1
(1)
Where did we try to use Theorem 3? Why was this not correct?
EXAMPLE
6
Using Rational Exponents and Radicals Simplify and express answers using positive exponents only. All letters represent positive real numbers. (A) 823
SOLUTIONS
4 (B) 2312
3 (C) (3 1 x)(2 1x)
(A) 823 0002 (813)2 0002 22 0002 4 or 4 12 12 1/4 3 (B) 23 0002 (3 ) 0002 3 0002 27
(D) a
4x13 12 b x12
823 0002 (82)13 0002 6413 0002 4
3 (C) (3 1x)(2 1x) 0002 (3x13)(2x12) 0002 6x13000712 0002 6x56
(D) a
2 2 4x13 12 412x16 b 0002 0002 14000316 0002 112 12 14 x x x x
0002
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MATCHED PROBLEM 6
Exponents and Radicals
17
Simplify and express answers using positive exponents only. All letters represent positive real numbers. (A) (⫺8)5Ⲑ3
5 (B) 2 324
(D) a
4 3 3 (C) (5 2 y )(2 1 y)
8x1Ⲑ2 1Ⲑ3 b x2Ⲑ3
0002
Z Simplifying Radicals The exponent properties considered earlier lead to the following properties of radicals.
Z THEOREM 4 Properties of Radicals For n a natural number greater than 1, and x and y positive real numbers: n
1. 2xn ⫽ x n n n 2. 2xy ⫽ 2x2y 3.
n
x 2x ⫽ n By 2y n
3 3 2 x ⫽x 5 5 5 2 xy ⫽ 2 x2 y 4 x 1 x ⫽ 4 Ay 1y 4
An algebraic expression that contains radicals is said to be in simplified form if all four of the conditions listed in the following definition are satisfied.
Z DEFINITION 5 Simplified (Radical) Form 1. No radicand (the expression within the radical sign) contains a factor to a power greater than or equal to the index of the radical. For example, 2x5 violates this condition.
2. No power of the radicand and the index of the radical have a common factor other than 1. 6 4 For example, 2 x violates this condition.
3. No radical appears in a denominator. For example, y/ 1x violates this condition.
4. No fraction appears within a radical. For example, 235 violates this condition.
EXAMPLE
7
Finding Simplified Form Write in simplified radical form. (A) 212x5y2
6 (B) 216x4y2
(C)
6 12x
(D)
8x4 B y 3
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SOLUTIONS
(A) Condition 1 is violated. First we convert to rational exponent form. 212x5y2 0002 (12x5y2)12 0002 1212x52y 0002 (4 ⴢ 3)12x2x12y 0002 213 x2 1x y 0002 2x2y13x
Use (ab)m ⴝ ambm and (an)m ⴝ amn. 12 ⴝ 4 ⴢ 3, x52 0002 x2x12 Write in radical form. Use commutative property and radical property 2.
(B) Condition 2 is violated. First we convert to rational exponent form. 6 2 16x4y2 0002 (16x4y2)16 0002 1616x23y13 0002 223x23y13 3 0002 2 4x2y
Use (ab)m ⴝ ambm and (an)m ⴝ amn. 16 ⴝ 24 Write in radical form.
(C) Condition 3 is violated. We multiply numerator and denominator by 12x; the effect is to multiply the expression by 1, so its value is unchanged, but the denominator is left free of radicals. 6 6 12x 612x 312x 0002 ⴢ 0002 0002 x 2x 12x 12x 12x (D) Condition 4 is violated. First we convert to rational exponent form. 8x4 813x43 0002 B y y13
y23
3
Multiply by
y23
ⴝ 1.
0002
2x43y23 y
x 43 ⴝ xx 13
0002
2xx13y23 y
Write in radical form.
0002
2x 2xy2 y
3
MATCHED PROBLEM 7
0002
Write in simplified radical form. (A) 218x4y3
9 (B) 2 8x6y3
(C)
30 1 16x 4
(D)
5x3 B y 0002
Eliminating a radical from a denominator [as in Example 7(C)] is called rationalizing the denominator. To rationalize the denominator, we multiply the numerator and denominator by a suitable factor that will leave the denominator free of radicals. This factor is called a rationalizing factor. If the denominator is of the form 1a 0007 1b, then 1a 0003 1b is a rationalizing factor because (1a 0007 1b)(1a 0003 1b) 0002 a 0003 b Similarly, if the denominator is of the form 1a 0003 1b, then 1a 0007 1b is a rationalizing factor.
EXAMPLE
8
Rationalizing Denominators Rationalize the denominator and write the answer in simplified radical form. (A)
8 16 0007 15
(B)
1x 0007 1y 1x 0003 1y
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SOLUTIONS
Exponents and Radicals
19
(A) Multiply numerator and denominator by the rationalizing factor 16 ⫺ 15. 8 8 16 ⫺ 15 ⫽ ⴢ 16 ⫹ 15 16 ⫹ 15 16 ⫺ 15 ⫽
8(16 ⫺ 15) 6⫺5
(1a ⴙ 1b) ( 1a ⴚ 1b) ⴝ a ⴚ b
Simplify.
⫽ 8( 16 ⫺ 15) (B) Multiply numerator and denominator by the rationalizing factor 1x ⫹ 1y. 1x ⫹ 1y 1x ⫹ 1y 1x ⫹ 1y ⫽ ⴢ 1x ⫺ 1y 1x ⫺ 1y 1x ⫹ 1y
MATCHED PROBLEM 8
⫽
x ⫹ 1x1y ⫹ 1y1x ⫹ y x⫺y
⫽
x ⫹ 21xy ⫹ y x⫺y
Expand numerator and denominator.
Combine like terms.
0002
Rationalize the denominator and write the answer in simplified radical form. (A)
6 1 ⫺ 13
(B)
21x ⫺ 31y 1x ⫹ 1y
0002
ANSWERS TO MATCHED PROBLEMS 1. (A) 1 (B) 0.000 01 (C) x4 (D) v3Ⲑu7 3 4 2. (A) 15x (B) 3Ⲑ(2y ) (C) ⫺14x (D) y6 Ⲑ(9x8) 4 8 ⫺3 3. (A) 2.3 ⫻ 10 ; 3.45 ⫻ 10 ; 3.1 ⫻ 10 ; 6.83 ⫻ 10⫺7 (B) 4,000; 530,000; 0.0253; 0.000 007 42 4. 1.11 ⫻ 10⫺19 5. (A) 2 (B) Not real (C) 10 (D) ⫺1 (E) ⫺3 6. (A) ⫺32 (B) 16 (C) 10y13Ⲑ12 (D) 2 Ⲑx1Ⲑ18 4 3 x 15xy 152 x 3 7. (A) 3x2y 12y (B) 2 (C) (D) 2x2y x y 2x ⫺ 51xy ⫹ 3y 8. (A) ⫺3 ⫺ 313 (B) x⫺y
R-2
(F) 0
Exercises
All variables represent positive real numbers and are restricted to prevent division by 0. In Problems 1–14, evaluate each expression. If the answer is not an integer, write it in fraction form. 1 8 1. 37 2. 56 3. a b 2 3 3 4. a b 5. 6⫺3 6. 2⫺6 5
7. (⫺5)4
8. (⫺4)5
10. (⫺7)⫺2
11. ⫺7⫺2
1 0 13. a b 3
14. a
1 ⫺1 b 10
9. (⫺3)⫺1 12. ⫺100
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In Problems 15–20, write the numbers in scientific notation. 15. 58,620,000
75.
16. 4,390
1 215
76.
2
17. 0.027
18. 0.11
19. 0.000 000 064
20. 0.000 0325
78.
In Problems 21–26, write each number in standard decimal form. 21. 4 1000033
22. 5 1000036
23. 2.99 105
24. 7.75 1011
25. 3.1 1000037
26. 8.167 1000034
12y
16y
5 6 7 11 81. x 2 3xy
84.
31y 21y 0003 3
79.
1
77.
3 1 7
4 16 0003 2
3 82. 2a2 8a8b13
85.
215 0007 312 515 0007 212
3 3
254
80.
12 16 0007 2
83.
12m15 120m
86.
312 0003 213 313 0003 212
2
In Problems 27–32, change to radical form. Do not simplify. 27. 3215
28. 62534
29. 4x000312
30. 32y000325
31. x13 0003 y13
32. (x 0003 y)13
87. What is the result of entering 23 on a calculator? 2
88. Refer to Problem 87. What is the difference between 2(3 ) and 3 2 32 (2 ) ? Which agrees with the value of 2 obtained with a calculator?
APPLICATIONS In Problems 33–38, change to rational exponent form. Do not simplify. 33. 1361
3 34. 2172
35. 4x 2y3
5
4 36. 27x3y2
3 37. 2x2 0007 y2
3 3 38. 2x2 0007 2y2
In Problems 39–50, evaluate each expression that represents a real number. 39. 10012
40. 16912
41. 1121
42. 1361
13
44. 27
3 45. 1000327
3 46. 164
N 0002 10x34y14
4
6
48. 100031
000332
50. 64000343
47. 1000316 49. 9
Estimate how many units of a finished product will be produced using 256 units of labor and 81 units of capital.
In Problems 51–64, simplify and express answers using positive exponents only. 51. x5x00032
52. y6y00038
53. (2y)(3y2)(5y4)
54. (6x3)(4x7)(x00035)
55. (a2b3)5
56. (2c4d00032)00033
13 53
57. u
000315 65
u
58. v
v
60. (49a4b00032)12
61. a
m00032n3 2 b m4n00031
w4 000312 b 9x00032
64. a
8a00034b3 13 b 27a2b00033
63. a
00033 16
59. (x ) 62. a
6mn00032 00033 b 3m00031n2
In Problems 65–86, write in simplified radical form. 65. 0003 1128
66. 00031125
67. 127 0003 5 13 3
3
68. 2 18 0007 118
69. 25 0003 225 0007 2625
3 3 70. 220 0007 240 0003 25
3 3 71. 225 210
72. 16114
4 8
73. 216m y
3
90. ECONOMICS If in the United States in 2007 the gross domestic product (GDP) was about $14,074,000,000,000 and the population was about 301,000,000, estimate to three significant digits the GDP per person. Write your answer in scientific notation and in standard decimal form. 91. ECONOMICS The number of units N of a finished product produced from the use of x units of labor and y units of capital for a particular Third World country is approximated by
23
43. 125
89. ECONOMICS If in the United States in 2007 the national debt was about $8,868,000,000,000 and the population was about 301,000,000, estimate to three significant digits each individual’s share of the national debt. Write your answer in scientific notation and in standard decimal form.
4 74. 216m4n8
92. ECONOMICS The number of units N of a finished product produced by a particular automobile company where x units of labor and y units of capital are used is approximated by N 0002 50x12y12 Estimate how many units will be produced using 256 units of labor and 144 units of capital. 93. BRAKING DISTANCE R. A. Moyer of Iowa State College found, in comprehensive tests carried out on 41 wet pavements, that the braking distance d (in feet) for a particular automobile traveling at v miles per hour was given approximately by d 0002 0.0212v73 Approximate the braking distance to the nearest foot for the car traveling on wet pavement at 70 miles per hour. 94. BRAKING DISTANCE Approximately how many feet would it take the car in Problem 93 to stop on wet pavement if it were traveling at 50 miles per hour? (Compute answer to the nearest foot.)
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and back again is called the period T and is given by
95. PHYSICS—RELATIVISTIC MASS The mass M of an object moving at a velocity v is given by
L Ag
T 0002 20005
M0
M0002 A
10003
v2 c2
where g is the gravitational constant. Show that T can be written in the form
where M0 0002 mass at rest and c 0002 velocity of light. The mass of an object increases with velocity and tends to infinity as the velocity approaches the speed of light. Show that M can be written in the form M0002
21
T0002
200051gL g
M0c2c2 0003 v2 c2 0003 v2
96. PHYSICS—PENDULUM A simple pendulum is formed by hanging a bob of mass M on a string of length L from a fixed support (see the figure). The time it takes the bob to swing from right to left
R-3
Polynomials: Basic Operations and Factoring Z Polynomials Z Addition and Subtraction Z Multiplication Z Factoring
In this section, we review the basic operations on polynomials. Polynomials are expressions such as x4 0003 5x2 0007 1 or 3xy 0003 2x 0007 5y 0007 6 that are built from constants and variables using only addition, subtraction, and multiplication (the power x4 is the product x ⴢ x ⴢ x ⴢ x). Polynomials are used throughout mathematics to describe and approximate mathematical relationships.
Z Polynomials Algebraic expressions are formed by using constants and variables and the algebraic operations of addition, subtraction, multiplication, division, raising to powers, and taking roots. Some examples are 3 3 2 x 00075 x00035 x2 0007 2x 0003 5
5x4 0007 2x2 0003 7 1 10007 1 10007 x
An algebraic expression involving only the operations of addition, subtraction, multiplication, and raising to natural number powers is called a polynomial. (Note that raising to a natural number power is repeated multiplication.) Some examples are 2x 0003 3 x 0003 2y
4x2 0003 3x 0007 7 x3 0003 3x2y 0007 xy2 0007 2y7
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In a polynomial, a variable cannot appear in a denominator, as an exponent, or within a radical. Accordingly, a polynomial in one variable x is constructed by adding or subtracting constants and terms of the form axn, where a is a real number and n is a natural number. A polynomial in two variables x and y is constructed by adding and subtracting constants and terms of the form axmyn, where a is a real number and m and n are natural numbers. Polynomials in three or more variables are defined in a similar manner. Polynomials can be classified according to their degree. If a term in a polynomial has only one variable as a factor, then the degree of that term is the power of the variable. If two or more variables are present in a term as factors, then the degree of the term is the sum of the powers of the variables. The degree of a polynomial is the degree of the nonzero term with the highest degree in the polynomial. Any nonzero constant is defined to be a polynomial of degree 0. The number 0 is also a polynomial but is not assigned a degree.
EXAMPLE
1
Polynomials and Nonpolynomials (A) Which of the following are polynomials? 2x 0002 5 0003
1 x
x2 0003 3x 0002 2
2x3 0003 4x 0002 1
x4 0002 12
(B) Given the polynomial 2x3 0003 x6 0002 7, what is the degree of the first term? The third term? The whole polynomial? (C) Given the polynomial x3y2 0002 2x2y 0002 1, what is the degree of the first term? The second term? The whole polynomial? SOLUTIONS
MATCHED PROBLEM 1
(A) x2 0003 3x 0002 2 and x4 0002 12 are polynomials. (The others are not polynomials since a variable appears in a denominator or within a radical.) (B) The first term has degree 3, the third term has degree 0, and the whole polynomial has degree 6. (C) The first term has degree 5, the second term has degree 3, and the whole polynomial has degree 5. 0002 (A) Which of the following are polynomials? 3x2 0003 2x 0002 1
1x 0003 3
x2 0003 2xy 0002 y2
x00031 x2 0002 2
(B) Given the polynomial 3x5 0003 6x3 0002 5, what is the degree of the first term? The second term? The whole polynomial? (C) Given the polynomial 6x4y2 0003 3xy3, what is the degree of the first term? The second term? The whole polynomial? 0002 In addition to classifying polynomials by degree, we also call a single-term polynomial a monomial, a two-term polynomial a binomial, and a three-term polynomial a trinomial. 5 2 3 2x y 3
x 0002 4.7 x4 0003 12x2 0002 9
Monomial Binomial Trinomial
A constant in a term of a polynomial, including the sign that precedes it, is called the numerical coefficient, or simply, the coefficient, of the term. If a constant doesn’t appear, or
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23
only a 0002 sign appears, the coefficient is understood to be 1. If only a 0003 sign appears, the coefficient is understood to be 00031. So given the polynomial 2x4 0003 4x3 0002 x2 0003 x 0002 5
2x4 ⴙ (ⴚ4)x3 ⴙ 1x2 ⴙ (ⴚ1)x ⴙ 5
the coefficient of the first term is 2, the coefficient of the second term is 00034, the coefficient of the third term is 1, the coefficient of the fourth term is 00031, and the coefficient of the last term is 5. Two terms in a polynomial are called like terms if they have exactly the same variable factors to the same powers. The numerical coefficients may or may not be the same. Since constant terms involve no variables, all constant terms are like terms. If a polynomial contains two or more like terms, these terms can be combined into a single term by making use of distributive properties. Consider the following example: 5x3y 0003 2xy 0003 x3y 0003 2x3y
0004 5x3y 0003 x3y 0003 2x3y 0003 2xy 0004 (5x3y 0003 x3y 0003 2 x3y) 0003 2xy 0004 (5 0003 1 0003 2) x3y 0003 2xy
Group like terms. Use the distributive property Simplify.
0004 2x3y 0003 2xy It should be clear that free use has been made of the real number properties discussed earlier. The steps done in the dashed box are usually done mentally, and the process is quickly done as follows: Like terms in a polynomial are combined by adding their numerical coefficients.
Z Addition and Subtraction Addition and subtraction of polynomials can be thought of in terms of removing parentheses and combining like terms. Horizontal and vertical arrangements are illustrated in the next two examples. You should be able to work either way, letting the situation dictate the choice.
EXAMPLE
2
Adding Polynomials Add: x4 0003 3x3 0002 x2,
SOLUTION
0003x3 0003 2x2 0002 3x,
and
3x2 0003 4x 0003 5
Add horizontally: (x4 0003 3x3 0002 x2) 0002 (0003x3 0003 2x2 0002 3x) 0002 (3x2 0003 4x 0003 5) 0004 x4 0003 3x3 0002 x2 0003 x3 0003 2x2 0002 3x 0002 3x2 0003 4x 0003 5 0004 x4 0003 4x3 0002 2x2 0003 x 0003 5
Remove parentheses. Combine like terms.
Or vertically, by lining up like terms and adding their coefficients: x4 0003 3x3 0002 x2 0003 x3 0003 2x2 0002 3x 3x2 0003 4x 0003 5 4 3 x 0003 4x 0002 2x2 0003 x 0003 5 MATCHED PROBLEM 2
0002
Add horizontally and vertically: 3x4 0003 2x3 0003 4x2,
x3 0003 2x2 0003 5x,
and
x2 0002 7x 0003 2
0002
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3
Subtracting Polynomials Subtract:
SOLUTION
MATCHED PROBLEM 3
ZZZ
Page 24
4x2 ⫺ 3x ⫹ 5
(x2 ⫺ 8) ⫺ (4x2 ⫺ 3x ⫹ 5) ⫽ x2 ⫺ 8 ⫺ 4x2 ⫹ 3x ⫺ 5 ⫽ ⫺3x2 ⫹ 3x ⫺ 13
from
x2 ⫺ 8 2 ⫺4x ⫹ 3x ⫺ 5 ⫺3x2 ⫹ 3x ⫺ 13
or
2x2 ⫺ 5x ⫹ 4
Subtract:
x2 ⫺ 8
from
d Change signs and add.
5x2 ⫺ 6
0002
0002
When you use a horizontal arrangement to subtract a polynomial with more than one term, you must enclose the polynomial in parentheses. For example, to subtract 2x ⫹ 5 from 4x ⫺ 11, you must write
CAUTION ZZZ
4x ⫺ 11 ⫺ (2x ⫹ 5)
and not
4x ⫺ 11 ⫺ 2x ⫹ 5
Z Multiplication Multiplication of algebraic expressions involves extensive use of distributive properties for real numbers, as well as other real number properties.
EXAMPLE
4
Multiplying Polynomials (2x ⫺ 3)(3x2 ⫺ 2x ⫹ 3)
Multiply: (2x ⫺ 3)(3x2 ⫺ 2x ⫹ 3)
SOLUTION
⫽ 2x(3x2 ⫺ 2x ⫹ 3) ⫺ 3(3x2 ⫺ 2x ⫹ 3) ⫽ 6x3 ⫺ 4x2 ⫹ 6x ⫺ 9x2 ⫹ 6x ⫺ 9 ⫽ 6x3 ⫺ 13x2 ⫹ 12x ⫺ 9
Distribute, multiply out parentheses.
Combine like terms.
Or, using a vertical arrangement, 3x2 ⫺ 2x ⫹ 3 2x ⫺ 3 6x3 ⫺ 4x2 ⫹ 6x ⫺ 9x2 ⫹ 6x ⫺ 9 6x3 ⫺ 13x2 ⫹ 12x ⫺ 9 MATCHED PROBLEM 4
0002
Multiply: (2x ⫺ 3)(2x2 ⫹ 3x ⫺ 2)
0002
To multiply two polynomials, multiply each term of one by each term of the other, and combine like terms.
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25
Z Factoring A factor of a number is one of two or more numbers whose product is the given number. Similarly, a factor of an algebraic expression is one of two or more algebraic expressions whose product is the given algebraic expression. For example, 30 0004 2 ⴢ 3 ⴢ 5 x2 0003 4 0004 (x 0003 2)(x 0002 2)
2, 3, and 5 are each factors of 30. (x ⴚ 2) and (x ⴙ 2) are each factors of x2 ⴚ 4.
The process of writing a number or algebraic expression as the product of other numbers or algebraic expressions is called factoring. We start our discussion of factoring with the positive integers. An integer such as 30 can be represented in a factored form in many ways. The products 6ⴢ5
(12)(10)(6)
15 ⴢ 2
2ⴢ3ⴢ5
all yield 30. A particularly useful way of factoring positive integers greater than 1 is in terms of prime numbers. An integer greater than 1 is prime if its only positive integer factors are itself and 1. So 2, 3, 5, and 7 are prime, but 4, 6, 8, and 9 are not prime. An integer greater than 1 that is not prime is called a composite number. The integer 1 is neither prime nor composite. A composite number is said to be factored completely if it is represented as a product of prime factors. The only factoring of 30 that meets this condition, except for the order of the factors, is 30 0004 2 ⴢ 3 ⴢ 5. This illustrates an important property of integers.
Z THEOREM 1 The Fundamental Theorem of Arithmetic Each integer greater than 1 is either prime or can be expressed uniquely, except for the order of factors, as a product of prime factors.
We can also write polynomials in completely factored form. A polynomial such as 2x2 0003 x 0003 6 can be written in factored form in many ways. The products (2x 0002 3)(x 0003 2)
2(x2 0003 12x 0003 3)
2(x 0002 32)(x 0003 2)
all yield 2x2 0003 x 0003 6. A particularly useful way of factoring polynomials is in terms of prime polynomials.
Z DEFINITION 1 Prime Polynomials A polynomial of degree greater than 0 is said to be prime relative to a given set of numbers if: (1) all of its coefficients are from that set of numbers; and (2) it cannot be written as a product of two polynomials (excluding constant polynomials that are factors of 1) having coefficients from that set of numbers. Relative to the set of integers: x2 ⴚ 2 is prime x2 ⴚ 9 is not prime, since x2 ⴚ 9 ⴝ (x ⴚ 3)(x ⴙ 3)
[Note: The set of numbers most frequently used in factoring polynomials is the set of integers.]
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A nonprime polynomial is said to be factored completely relative to a given set of numbers if it is written as a product of prime polynomials relative to that set of numbers. In Examples 5 and 6 we review some of the standard factoring techniques for polynomials with integer coefficients.
EXAMPLE
5
Factoring Out Common Factors Factor out, relative to the integers, all factors common to all terms: (A) 2x3y ⫺ 8x2y2 ⫺ 6xy3
SOLUTIONS
(A) 2x3y ⫺ 8x2y2 ⫺ 6xy3
(B) 2x(3x ⫺ 2) ⫺ 7(3x ⫺ 2) ⫽ (2xy)x2 ⫺ (2xy)4xy ⫺ (2xy)3y2
Factor out 2xy.
⫽ 2xy(x2 ⫺ 4xy ⫺ 3y2) (B) 2x(3x ⫺ 2) ⫺ 7(3x ⫺ 2)
⫽ 2x(3x ⫺ 2) ⫺ 7(3x ⫺ 2)
Factor out 3x ⴚ 2.
⫽ (2x ⫺ 7)(3x ⫺ 2) MATCHED PROBLEM 5
0002
Factor out, relative to the integers, all factors common to all terms: (A) 3x3y ⫺ 6x2y2 ⫺ 3xy3
(B) 3y(2y ⫹ 5) ⫹ 2(2y ⫹ 5) 0002
The polynomials in Example 6 can be factored by first grouping terms to find a common factor.
EXAMPLE
6
Factoring by Grouping Factor completely, relative to the integers, by grouping: (A) 3x2 ⫺ 6x ⫹ 4x ⫺ 8 (C) 3ac ⫹ bd ⫺ 3ad ⫺ bc
SOLUTIONS
(B) wy ⫹ wz ⫺ 2xy ⫺ 2xz
(A) 3x2 ⫺ 6x ⫹ 4x ⫺ 8 Group the first two and last two terms. ⫽ (3x2 ⫺ 6x) ⫹ (4x ⫺ 8) Remove common factors from each group. ⫽ 3x(x ⫺ 2) ⫹ 4(x ⫺ 2) Factor out the common factor (x ⴚ 2). ⫽ (3x ⫹ 4)(x ⫺ 2) (B) wy ⫹ wz ⫺ 2xy ⫺ 2xz Group the first two and last two terms—be careful of signs. ⫽ (wy ⫹ wz) ⫺ (2xy ⫹ 2xz) Remove common factors from each group. ⫽ w( y ⫹ z) ⫺ 2x(y ⫹ z) Factor out the common factor ( y ⴙ z). ⫽ (w ⫺ 2x)(y ⫹ z) (C) 3ac ⫹ bd ⫺ 3ad ⫺ bc In parts (A) and (B) the polynomials are arranged in such a way that grouping the first two terms and the last two terms leads to common factors. In this problem neither the first two terms nor the last two terms have a common factor. Sometimes rearranging terms will lead to a factoring by grouping. In this case, we interchange
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27
the second and fourth terms to obtain a problem comparable to part (B), which can be factored as follows: 3ac ⫺ bc ⫺ 3ad ⫹ bd ⫽ (3ac ⫺ bc) ⫺ (3ad ⫺ bd) ⫽ c(3a ⫺ b) ⫺ d(3a ⫺ b)
Factor out c, d. Factor out 3a ⴚ b.
⫽ (c ⫺ d)(3a ⫺ b) MATCHED PROBLEM 6
0002
Factor completely, relative to the integers, by grouping: (A) 2x2 ⫹ 6x ⫹ 5x ⫹ 15 (C) 6wy ⫺ xz ⫺ 2xy ⫹ 3wz
(B) 2pr ⫹ ps ⫺ 6qr ⫺ 3qs 0002
Example 7 illustrates an approach to factoring a second-degree polynomial of the form 2x2 ⫺ 5x ⫺ 3
2x2 ⫹ 3xy ⫺ 2y2
or
into the product of two first-degree polynomials with integer coefficients.
EXAMPLE
7
Factoring Second-Degree Polynomials Factor each polynomial, if possible, using integer coefficients: (A) 2x2 ⫹ 3xy ⫺ 2y2
SOLUTIONS
(B) x2 ⫺ 3x ⫹ 4
(A) 2x2 ⫹ 3xy ⫺ 2y2 ⫽ (2x ⫹ y)(x ⫺ y) c
c
?
?
(C) 6x2 ⫹ 5xy ⫺ 4y2 Put in what we know. Signs must be opposite. (We can reverse this choice if we get ⴚ3xy instead of ⴙ3xy for the middle term.)
Now, what are the factors of 2 (the coefficient of y2)? 2 1ⴢ2 2ⴢ1
(2x ⴙ y)(x ⴚ 2y) ⴝ 2x2 ⴚ 3xy ⴚ 2y2 (2x ⴙ 2y)(x ⴚ y) ⴝ 2x2 ⴚ 2y2
The first choice gives us ⫺3xy for the middle term—close, but not there—so we reverse our choice of signs to obtain 2x2 ⫹ 3xy ⫺ 2y2 ⫽ (2x ⫺ y)(x ⫹ 2y) (B) x2 ⫺ 3x ⫹ 4 ⫽ (x ⫺ )(x ⫺ ) 4 2ⴢ2 1ⴢ4 4ⴢ1
Signs must be the same because the third term is positive and must be negative because the middle term is negative.
(x ⴚ 2)(x ⴚ 2) ⴝ x2 ⴚ 4x ⴙ 4 (x ⴚ 1)(x ⴚ 4) ⴝ x2 ⴚ 5x ⴙ 4 (x ⴚ 4)(x ⴚ 1) ⴝ x2 ⴚ 5x ⴙ 4
No choice produces the middle term; so x2 ⫺ 3x ⫹ 4 is not factorable using integer coefficients. (C) 6x2 ⫹ 5xy ⫺ 4y2 ⫽ ( x ⫹ y)( x ⫺ y) c
c
c
c
?
?
?
?
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The signs must be opposite in the factors, because the third term is negative. We can reverse our choice of signs later if necessary. We now write all factors of 6 and of 4: 6 2ⴢ3 3ⴢ2 1ⴢ6 6ⴢ1
4 2ⴢ2 1ⴢ4 4ⴢ1
and try each choice on the left with each on the right—a total of 12 combinations that give us the first and last terms in the polynomial 6x2 0002 5xy 0003 4y2. The question is: Does any combination also give us the middle term, 5xy? After trial and error and, perhaps, some educated guessing among the choices, we find that 3 ⴢ 2 matched with 4 ⴢ 1 gives us the correct middle term. 6x2 0002 5xy 0003 4y2 0004 (3x 0002 4y)(2x 0003 y) If none of the 24 combinations (including reversing our sign choice) had produced the middle term, then we would conclude that the polynomial is not factorable using integer coefficients. 0002 MATCHED PROBLEM 7
Factor each polynomial, if possible, using integer coefficients: (A) x2 0003 8x 0002 12
(B) x2 0002 2x 0002 5
(C) 2x2 0002 7xy 0003 4y2
(D) 4x2 0003 15xy 0003 4y2
0002
The special factoring formulas listed here will enable us to factor certain polynomial forms that occur frequently.
Z SPECIAL FACTORING FORMULAS 1. u2 0002 2uv 0002 v2 0004 (u 0002 v)2 2
2
Perfect Square
2
2. u 0003 2uv 0002 v 0004 (u 0003 v)
Perfect Square
2
2
3
3
2
2
Difference of Cubes
3
3
2
2
Sum of Cubes
3. u 0003 v 0004 (u 0003 v)(u 0002 v)
Difference of Squares
4. u 0003 v 0004 (u 0003 v)(u 0002 uv 0002 v ) 5. u 0002 v 0004 (u 0002 v)(u 0003 uv 0002 v )
The formulas in the box can be established by multiplying the factors on the right.
ZZZ EXPLORE-DISCUSS 1
Explain why there is no formula for factoring a sum of squares u2 0002 v2 into the product of two first-degree polynomials with real coefficients.
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EXAMPLE
8
Polynomials: Basic Operations and Factoring
29
Using Special Factoring Formulas Factor completely relative to the integers: (A) x2 0002 6xy 0002 9y2
SOLUTIONS
(A) x2 0002 6xy 0002 9y2
(B) 9x2 0003 4y2
(C) 8m3 0003 1
0004 x2 0002 2(x)(3y) 0002 (3y)2
(D) x3 0002 y3z3
0004 (x 0002 3y)2
(B) 9x2 0003 4y2
0004 (3x)2 0003 (2y)2
(C) 8m3 0003 1
0004 (2m)3 0003 13 0004 (2m 0003 1) 冤(2m)2 0002 (2m)(1) 0002 12冥
0004 (3x 0003 2y)(3x 0002 2y)
Perfect square
Difference of squares
Difference of cubes Simplify.
0004 (2m 0003 1)(4m2 0002 2m 0002 1) (D) x3 0002 y3z3
0004 x3 0002 ( yz)3
Sum of cubes
0004 (x 0002 yz)(x2 0003 xyz 0002 y2z2) MATCHED PROBLEM 8
0002
Factor completely relative to the integers: (A) 4m2 0003 12mn 0002 9n2
(B) x2 0003 16y2
(C) z3 0003 1
(D) m3 0002 n3
0002
ANSWERS TO MATCHED PROBLEMS (A) 3x2 0003 2x 0002 1, x2 0003 2xy 0002 y2 (B) 5, 3, 5 (C) 6, 4, 6 3. 3x2 0002 5x 0003 10 4. 4x3 0003 13x 0002 6 3x4 0003 x3 0003 5x2 0002 2x 0003 2 (A) 3xy(x2 0003 2xy 0003 y2) (B) (3y 0002 2)(2y 0002 5) (A) (2x 0002 5)(x 0002 3) (B) (p 0003 3q)(2r 0002 s) (C) (3w 0003 x)(2y 0002 z) (A) (x 0003 2)(x 0003 6) (B) Not factorable using integers (C) (2x 0003 y)(x 0002 4y) (D) (4x 0002 y)(x 0003 4y) 8. (A) (2m 0003 3n)2 (B) (x 0003 4y)(x 0002 4y) (C) (z 0003 1)(z2 0002 z 0002 1) 2 2 (D) (m 0002 n)(m 0003 mn 0002 n ) 1. 2. 5. 6. 7.
R-3
Exercises
Problems 1–8 refer to the polynomials (a) x2 0002 1 and (b) x4 0003 2x 0002 1. 1. What is the degree of (a)?
In Problems 9–14, is the algebraic expression a polynomial? If so, give its degree. 9. 4 0003 x2
10. x3 0003 5x6 0002 1
2. What is the degree of (b)?
11. x3 0003 7x 0002 81x
12. x4 0002 3x 0003 15
3. What is the degree of the sum of (a) and (b)?
13. x5 0003 4x2 0002 600032
14. 3x4 0003 2x00031 0003 10
4. What is the degree of the product of (a) and (b)? 5. Multiply (a) and (b). 6. Add (a) and (b). 7. Subtract (b) from (a). 8. Subtract (a) from (b).
In Problems 15–22, perform the indicated operations and simplify. 15. 2(x 0003 1) 0002 3(2x 0003 3) 0003 (4x 0003 5) 16. 2y 0003 3y [4 0003 2( y 0003 1)] 17. (m 0003 n)(m 0002 n)
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18. (5y ⫺ 1)(3 ⫺ 2y)
19. (3x ⫹ 2y)(x ⫺ 3y)
65. 2(x ⫹ h)2 ⫺ 3(x ⫹ h) ⫺ (2x2 ⫺ 3x)
20. (4x ⫺ y)2
21. (a ⫹ b)(a2 ⫺ ab ⫹ b2)
66. ⫺4(x ⫹ h)2 ⫹ 6(x ⫹ h) ⫺ (⫺4x2 ⫹ 6x)
2
2
22. (a ⫺ b)(a ⫹ ab ⫹ b )
67. (x ⫹ h)3 ⫺ 2(x ⫹ h)2 ⫺ (x3 ⫺ 2x2)
In Problems 23–28, factor out, relative to the integers, all factors common to all terms.
68. (x ⫹ h)3 ⫹ 3(x ⫹ h) ⫺ (x3 ⫹ 3x)
23. 6x4 ⫺ 8x3 ⫺ 2x2
24. 3x5 ⫹ 6x3 ⫹ 9x
Problems 69–74 are calculus-related. Factor completely, relative to the integers.
25. x2y ⫹ 2xy2 ⫹ x2y2
26. 8u3v ⫺ 6u2v2 ⫹ 4uv3
69. 2x(x ⫹ 1)4 ⫹ 4x2(x ⫹ 1)3
27. 2w( y ⫺ 2z) ⫺ x( y ⫺ 2z)
70. (x ⫺ 1)3 ⫹ 3x(x ⫺ 1)2
28. 2x(u ⫺ 3v) ⫹ 5y(u ⫺ 3v)
71. 6(3x ⫺ 5)(2x ⫺ 3)2 ⫹ 4(3x ⫺ 5)2(2x ⫺ 3)
In Problems 29–34, factor completely, relative to the integers. 2
2
29. x ⫹ 4x ⫹ x ⫹ 4 2
30. 2y ⫺ 6y ⫹ 5y ⫺ 15 2
31. x ⫺ xy ⫹ 3xy ⫺ 3y
32. 3a2 ⫺ 12ab ⫺ 2ab ⫹ 8b2
33. 8ac ⫹ 3bd ⫺ 6bc ⫺ 4ad
In Problems 35–42, perform the indicated operations and simplify. 35. 2x ⫺ 35x ⫹ 2 3x ⫺ (x ⫹ 5) 4 ⫹ 16
78. 15ac ⫺ 20ad ⫹ 3bc ⫺ 4bd
38. (x2 ⫺ 3xy ⫹ y2)(x2 ⫹ 3xy ⫹ y2)
79. 3x2 ⫺ 2xy ⫺ 4y2
2
39. (3u ⫺ 2v) ⫺ (2u ⫺ 3v)(2u ⫹ 3v)
80. 5u2 ⫹ 4uv ⫺ v2
40. (2a ⫺ b)2 ⫺ (a ⫹ 2b)2 3
42. (3a ⫹ 2b)
In Problems 43–62, factor completely, relative to the integers. If a polynomial is prime relative to the integers, say so. 43. 2x2 ⫹ x ⫺ 3
44. 3y2 ⫺ 8y ⫺ 3
45. x2 ⫹ 5xy ⫺ 14y2
46. x2 ⫹ 4y2
47. 4x2 ⫺ 20x ⫹ 25
48. a2b2 ⫺ c2
2
49. a b ⫹ c 2
2
50. 9x ⫺ 4
51. 4x ⫹ 9
52. 16x2 ⫺ 25
53. 6x2 ⫹ 48x ⫹ 72
54. 3z2 ⫺ 28z ⫹ 48
55. 2x4 ⫺ 24x3 ⫹ 40x2
56. 16x2y ⫺ 8xy ⫹ y
57. 6m2 ⫺ mn ⫺ 12n2
58. 4u3v ⫺ uv3
59. 3m3 ⫺ 6m2 ⫹ 15m
60. 2x3 ⫺ 2x2 ⫹ 8x
61. m3 ⫹ n3
62. 8x3 ⫺ 125
Problems 63–68 are calculus-related. Perform the indicated operations and simplify. 63. 3(x ⫹ h) ⫺ 7 ⫺ (3x ⫺ 7) 64. (x ⫹ h)2 ⫺ x2
75. (a ⫺ b)2 ⫺ 4(c ⫺ d )2 77. 2am ⫺ 3an ⫹ 2bm ⫺ 3bn
37. (2x2 ⫺ 3x ⫹ 1)(x2 ⫹ x ⫺ 2)
2 2
74. 3x4(x ⫺ 7)2 ⫹ 4x3(x ⫺ 7)3
76. (x ⫹ 2)2 ⫹ 9
36. m ⫺ 5m ⫺ 3m ⫺ (m ⫺ 1)4 6
41. (2m ⫺ n)
73. 5x4(9 ⫺ x)4 ⫺ 4x5(9 ⫺ x)3
In Problems 75–86, factor completely, relative to the integers. In polynomials involving more than three terms, try grouping the terms in various combinations as a first step. If a polynomial is prime relative to the integers, say so.
34. 3ux ⫺ 4vy ⫹ 3vx ⫺ 4uy
3
72. 2(x ⫺ 3)(4x ⫹ 7)2 ⫹ 8(x ⫺ 3)2(4x ⫹ 7)
81. x3 ⫺ 3x2 ⫺ 9x ⫹ 27
82. t3 ⫺ 2t 2 ⫹ t ⫺ 2
83. 4(A ⫹ B)2 ⫺ 5(A ⫹ B) ⫺ 5 84. x4 ⫹ 6x2 ⫹ 8
85. m4 ⫺ n4
86. y4 ⫺ 3y2 ⫺ 4 87. Show by example that, in general, (a ⫹ b)2 ⫽ a2 ⫹ b2. Discuss possible conditions on a and b that would make this a valid equation. 88. Show by example that, in general, (a ⫺ b)2 ⫽ a2 ⫺ b2. Discuss possible conditions on a and b that would make this a valid equation. 89. To show that 12 is an irrational number, explain how the assumption that 12 is rational leads to a contradiction of Theorem 1, the fundamental theorem of arithmetic, by the following steps: (A) Suppose that 12 ⫽ aⲐb, where a and b are positive integers, b ⫽ 0. Explain why a2 ⫽ 2b2. (B) Explain why the prime number 2 appears an even number of times (possibly 0 times) as a factor in the prime factorization of a2. (C) Explain why the prime number 2 appears an odd number of times as a factor in the prime factorization of 2b2. (D) Explain why parts (B) and (C) contradict the fundamental theorem of arithmetic.
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90. To show that 1n is an irrational number unless n is a perfect square, explain how the assumption that 1n is rational leads to a contradiction of the fundamental theorem of arithmetic by the following steps: (A) Assume that n is not a perfect square, that is, does not belong to the sequence 1, 4, 9, 16, 25, . . . . Explain why some prime number p appears an odd number of times as a factor in the prime factorization of n. (B) Suppose that 1n 0004 a0006b, where a and b are positive integers, b 0005 0. Explain why a2 0004 nb2. (C) Explain why the prime number p appears an even number of times (possibly 0 times) as a factor in the prime factorization of a2. (D) Explain why the prime number p appears an odd number of times as a factor in the prime factorization of nb2. (E) Explain why parts (C) and (D) contradict the fundamental theorem of arithmetic.
APPLICATIONS 91. GEOMETRY The width of a rectangle is 5 centimeters less than its length. If x represents the length, write an algebraic expression in terms of x that represents the perimeter of the rectangle. Simplify the expression. 92. GEOMETRY The length of a rectangle is 8 meters more than its width. If x represents the width of the rectangle, write an algebraic expression in terms of x that represents its area. Change the expression to a form without parentheses. 93. COIN PROBLEM A parking meter contains nickels, dimes, and quarters. There are 5 fewer dimes than nickels, and 2 more quarters than dimes. If x represents the number of nickels, write an algebraic expression in terms of x that represents the value of all the coins in the meter in cents. Simplify the expression.
Polynomials: Basic Operations and Factoring
31
0.3 centimeters thick, write an algebraic expression in terms of x that represents the volume of the plastic used to construct the container. Simplify the expression. [Recall: The volume V of a sphere of radius r is given by V 0004 430007r3.] 96. PACKAGING A cubical container for shipping computer components is formed by coating a metal mold with polystyrene. If the metal mold is a cube with sides x centimeters long and the polystyrene coating is 2 centimeters thick, write an algebraic expression in terms of x that represents the volume of the polystyrene used to construct the container. Simplify the expression. [Recall: The volume V of a cube with sides of length t is given by V 0004 t3.] 97. CONSTRUCTION A rectangular open-topped box is to be constructed out of 20-inch-square sheets of thin cardboard by cutting x-inch squares out of each corner and bending the sides up as indicated in the figure. Express each of the following quantities as a polynomial in both factored and expanded form. (A) The area of cardboard after the corners have been removed. (B) The volume of the box.
20 inches x
x
x
x
20 inches
x
x x
x
94. COIN PROBLEM A vending machine contains dimes and quarters only. There are 4 more dimes than quarters. If x represents the number of quarters, write an algebraic expression in terms of x that represents the value of all the coins in the vending machine in cents. Simplify the expression. 95. PACKAGING A spherical plastic container for designer wristwatches has an inner radius of x centimeters (see the figure). If the plastic shell is
0.3 cm x cm
Figure for 95
98. CONSTRUCTION A rectangular open-topped box is to be constructed out of 9- by 16-inch sheets of thin cardboard by cutting x-inch squares out of each corner and bending the sides up. Express each of the following quantities as a polynomial in both factored and expanded form. (A) The area of cardboard after the corners have been removed. (B) The volume of the box.
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Rational Expressions: Basic Operations Z Reducing to Lowest Terms Z Multiplication and Division Z Addition and Subtraction Z Compound Fractions
A quotient of two algebraic expressions, division by 0 excluded, is called a fractional expression. If both the numerator and denominator of a fractional expression are polynomials, the fractional expression is called a rational expression. Some examples of rational expressions are the following (recall that a nonzero constant is a polynomial of degree 0): x⫺2 2 2x ⫺ 3x ⫹ 5
1 4 x ⫺1
3 x
x2 ⫹ 3x ⫺ 5 1
In this section, we discuss basic operations on rational expressions, including multiplication, division, addition, and subtraction. Since variables represent real numbers in the rational expressions we are going to consider, the properties of real number fractions summarized in Section R-1 play a central role in much of the work that we will do. Even though not always explicitly stated, we always assume that variables are restricted so that division by 0 is excluded.
Z Reducing to Lowest Terms We start this discussion by restating the fundamental property of fractions (from Theorem 3 in Section R-1):
Z FUNDAMENTAL PROPERTY OF FRACTIONS If a, b, and k are real numbers with b, k ⫽ 0, then ka a ⫽ kb b
2ⴢ3 3 ⴝ 2ⴢ4 4
(x ⴚ 3)2
2 ⴝ (x ⴚ 3)x x x ⴝ 0, x ⴝ 3
Using this property from left to right to eliminate all common factors from the numerator and the denominator of a given fraction is referred to as reducing a fraction to lowest terms. We are actually dividing the numerator and denominator by the same nonzero common factor. Using the property from right to left—that is, multiplying the numerator and the denominator by the same nonzero factor—is referred to as raising a fraction to higher terms. We will use the property in both directions in the material that follows. We say that a rational expression is reduced to lowest terms if the numerator and denominator do not have any factors in common. Unless stated to the contrary, factors will be relative to the integers.
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EXAMPLE
1
Rational Expressions: Basic Operations
33
Reducing Rational Expressions Reduce each rational expression to lowest terms. (A)
SOLUTIONS
(A)
x2 ⫺ 6x ⫹ 9 x2 ⫺ 9
(B)
x3 ⫺ 1 x2 ⫺ 1
(x ⫺ 3)2 x2 ⫺ 6x ⫹ 9 ⫽ (x ⫺ 3)(x ⫹ 3) x2 ⫺ 9 x⫺3 ⫽ x⫹3
Factor numerator and denominator completely. Divide numerator and denominator by (x ⫺ 3); this is a valid operation as long as x ⴝ 3.
1
(x ⫺ 1)(x2 ⫹ x ⫹ 1) x3 ⫺ 1 (B) 2 ⫽ (x ⫺ 1)(x ⫹ 1) x ⫺1
Dividing numerator and denominator by (x ⴚ 1) can be indicated by drawing lines through both (x ⴚ 1)’s and writing the resulting quotients, 1’s.
1
x2 ⫹ x ⫹ 1 ⫽ x⫹1 MATCHED PROBLEM 1
CAUTION ZZZ
0002
Reduce each rational expression to lowest terms. (A)
ZZZ
x ⴝ ⴚ1 and x ⴝ 1
6x2 ⫹ x ⫺ 2 2x2 ⫹ x ⫺ 1
(B)
x4 ⫺ 8x 3x ⫺ 2x2 ⫺ 8x 3
0002
Remember to always factor the numerator and denominator first, then divide out any common factors. Do not indiscriminately eliminate terms that appear in both the numerator and the denominator. For example, 1
2x3 ⫹ y2 2x3 ⫹ y2 ⫽ 2 y y2 1
2x3 ⫹ y2 ⫽ 2x3 ⫹ 1 y2 Since the term y2 is not a factor of the numerator, it cannot be eliminated. In fact, (2x3 ⫹ y2)Ⲑy2 is already reduced to lowest terms.
Z Multiplication and Division Since we are restricting variable replacements to real numbers, multiplication and division of rational expressions follow the rules for multiplying and dividing real number fractions (Theorem 3 in Section R-1).
Z MULTIPLICATION AND DIVISION If a, b, c, and d are real numbers with b, d ⫽ 0, then: 1.
a c ac ⴢ ⫽ b d bd
2.
a c a d ⫼ ⫽ ⴢ b d b c
2 x 2x ⴢ ⴝ 3 xⴚ1 3(x ⴚ 1)
c⫽0
2 x 2 xⴚ1 ⴜ ⴝ ⴢ 3 xⴚ1 3 x
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2
Multiplying and Dividing Rational Expressions Perform the indicated operations and reduce to lowest terms. (A)
10x3y x2 ⫺ 9 ⴢ 2 3xy ⫹ 9y 4x ⫺ 12x
(C)
x3 ⫹ y3 2x3 ⫺ 2x2y ⫹ 2xy2 ⫼ 2 3 3 x y ⫺ xy x ⫹ 2xy ⫹ y2 5x2
SOLUTIONS
4 ⫺ 2x ⫼ (x ⫺ 2) 4
(B)
1ⴢ1
10x3y 10x3y (x ⫺ 3)(x ⫹ 3) x2 ⫺ 9 (A) ⴢ 2 ⫽ ⴢ 3xy ⫹ 9y 4x ⫺ 12x 3y(x ⫹ 3) 4x(x ⫺ 3) 3ⴢ1
2ⴢ1
Factor numerators and denominators; then divide any numerator and any denominator with a like common factor.
2
⫽
5x 6
1
2(2 ⫺ x) 4 ⫺ 2x 1 ⫼ (x ⫺ 2) ⫽ ⴢ (B) 4 4 x⫺2
x ⴚ 2 is the same as
xⴚ2 . 1
2
⫺1
⫺(x ⫺ 2) 2⫺x ⫽ ⫽ 2(x ⫺ 2) 2(x ⫺ 2)
b ⴚ a ⴝ ⴚ(a ⴚ b), a useful change in some problems.
1
⫽⫺ (C)
1 2
2x3 ⫺ 2x2y ⫹ 2xy2 x3 ⫹ y3 ⫼ x3y ⫺ xy3 x2 ⫹ 2xy ⫹ y2 2
1
a c a d ⴜ ⴝ ⴢ b d b c
1
2x(x2 ⫺ xy ⫹ y2) (x ⫹ y)2 ⫽ ⴢ xy(x ⫹ y)(x ⫺ y) (x ⫹ y)(x2 ⫺ xy ⫹ y2) y
1
1
Divide out common factors.
1
2 ⫽ y(x ⫺ y)
MATCHED PROBLEM 2
0002
Perform the indicated operations and reduce to lowest terms. (A)
12x2y3 y2 ⫹ 6y ⫹ 9 ⴢ 2xy2 ⫹ 6xy 3y3 ⫹ 9y2
(C)
m3n ⫺ m2n2 ⫹ mn3 m3 ⫹ n3 ⫼ 2m2 ⫹ mn ⫺ n2 2m3n2 ⫺ m2n3
(B) (4 ⫺ x) ⫼
x2 ⫺ 16 5
0002
Z Addition and Subtraction Again, because we are restricting variable replacements to real numbers, addition and subtraction of rational expressions follow the rules for adding and subtracting real number fractions (Theorem 3 in Section R-1).
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Rational Expressions: Basic Operations
35
Z ADDITION AND SUBTRACTION For a, b, and c real numbers with b ⫽ 0: 1.
a c a⫹c ⫹ ⫽ b b b
2.
a c a⫺c ⫺ ⫽ b b b
x 2 xⴙ2 ⴙ ⴝ xⴚ3 xⴚ3 xⴚ3 x 2xy
2
ⴚ
xⴚ4 2xy
2
ⴝ
x ⴚ (x ⴚ 4) 2xy 2
So we add rational expressions with the same denominators by adding or subtracting their numerators and placing the result over the common denominator. If the denominators are not the same, we raise the fractions to higher terms, using the fundamental property of fractions to obtain common denominators, and then proceed as described. Even though any common denominator will do, our work will be simplified if the least common denominator (LCD) is used. Often, the LCD is obvious, but if it is not, the steps in the box describe how to find it.
Z THE LEAST COMMON DENOMINATOR (LCD) The LCD of two or more rational expressions is found as follows: 1. Factor each denominator completely. 2. Identify each different prime factor from all the denominators. 3. Form a product using each different factor to the highest power that occurs in any one denominator. This product is the LCD.
EXAMPLE
3
Adding and Subtracting Rational Expressions Combine into a single fraction and reduce to lowest terms. (A)
SOLUTIONS
3 5 11 ⫹ ⫺ 10 6 45
(B)
4 5x ⫺ 2⫹1 9x 6y
(C)
x⫹3 x⫹2 5 ⫺ 2 ⫺ 3⫺x x ⫺ 6x ⫹ 9 x ⫺9 2
(A) To find the LCD, factor each denominator completely:
冎
10 ⫽ 2 ⴢ 5 6 ⫽ 2 ⴢ 3 LCD ⫽ 2 ⴢ 32 ⴢ 5 ⫽ 90 45 ⫽ 32 ⴢ 5 Now use the fundamental property of fractions to make each denominator 90: 3 5 11 9ⴢ3 15 ⴢ 5 2 ⴢ 11 ⫹ ⫺ ⫽ ⫹ ⫺ 10 6 45 9 ⴢ 10 15 ⴢ 6 2 ⴢ 45 ⫽
27 75 22 ⫹ ⫺ 90 90 90
⫽
27 ⫹ 75 ⫺ 22 80 8 ⫽ ⫽ 90 90 9
Multiply.
Combine into a single fraction.
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(B)
冎
9x ⫽ 32x LCD ⫽ 2 ⴢ 32xy2 ⫽ 18xy2 6y2 ⫽ 2 ⴢ 3y2 2y2 ⴢ 4 18xy2 4 5x 3x ⴢ 5x ⫺ 2⫹1⫽ 2 ⫹ ⫺ 9x 6y 2y ⴢ 9x 3x ⴢ 6y2 18xy2 ⫽
(C)
Multiply, combine.
8y2 ⫺ 15x2 ⫹ 18xy2 18xy2
x⫹3 x⫹2 5 x⫹3 x⫹2 5 ⫺ 2 ⫺ ⫽ ⫺ ⫹ 3⫺x (x ⫺ 3)(x ⫹ 3) x⫺3 x2 ⫺ 6x ⫹ 9 x ⫺9 (x ⫺ 3)2 Note: ⫺
5 5 5 ⫽⫺ ⫽ 3⫺x ⫺(x ⫺ 3) x⫺3
We have again used the fact that a ⴚ b ⴝ ⴚ(b ⴚ a).
The LCD ⫽ (x ⫺ 3)2(x ⫹ 3). (x ⫹ 3)2 (x ⫺ 3)(x ⫹ 2) 5(x ⫺ 3)(x ⫹ 3) ⫺ ⫹ 2 2 (x ⫺ 3) (x ⫹ 3) (x ⫺ 3) (x ⫹ 3) (x ⫺ 3)2(x ⫹ 3)
MATCHED PROBLEM 3
Expand numerators.
⫽
(x2 ⫹ 6x ⫹ 9) ⫺ (x2 ⫺ x ⫺ 6) ⫹ 5(x2 ⫺ 9) (x ⫺ 3)2(x ⫹ 3)
Be careful of sign errors here.
⫽
x2 ⫹ 6x ⫹ 9 ⫺ x2 ⫹ x ⫹ 6 ⫹ 5x2 ⫺ 45 (x ⫺ 3)2(x ⫹ 3)
Combine like terms.
⫽
5x2 ⫹ 7x ⫺ 30 (x ⫺ 3)2(x ⫹ 3)
0002
Combine into a single fraction and reduce to lowest terms. (A)
5 1 6 ⫺ ⫹ 28 10 35
(C)
y⫹2 y⫺3 2 ⫺ 2 ⫺ 2⫺y y2 ⫺ 4 y ⫺ 4y ⫹ 4
(B)
1 2x ⫹ 1 3 ⫹ 2 ⫺ 3 12x 4x 3x
0002 ZZZ EXPLORE-DISCUSS 1
What is the result of entering 16 ⫼ 4 ⫼ 2 on a calculator? What is the difference between 16 ⫼ (4 ⫼ 2) and (16 ⫼ 4) ⫼ 2? How could you use fraction bars to distinguish between these two cases when 16 writing 4 ? 2
Z Compound Fractions A fractional expression with fractions in its numerator, denominator, or both is called a compound fraction. It is often necessary to represent a compound fraction as a simple fraction—that is (in all cases we will consider), as the quotient of two polynomials. The process does not involve any new concepts. It is a matter of applying old concepts and processes in the right sequence. We will illustrate two approaches to the problem, each with its own merits, depending on the particular problem under consideration.
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SECTION R–4
EXAMPLE
4
Rational Expressions: Basic Operations
37
Simplifying Compound Fractions Express as a simple fraction reduced to lowest terms: 2 ⫺1 x 4 ⫺1 x2
SOLUTION
Method 1. Multiply the numerator and denominator by the LCD of all fractions in the numerator and denominator—in this case, x2. (We are multiplying by 1 ⫽ x2Ⲑx2.) 2 x2a ⫺ 1b x x2a
4 ⫺ 1b x2
2 x2 ⫺ x2 x
⫽
x2
4 ⫺ x2 x2
1
x(2 ⫺ x) 2x ⫺ x2 ⫽ 2 ⫽ (2 ⫹ x)(2 ⫺ x) 4⫺x 1
⫽
x 2⫹x
Method 2. Write the numerator and denominator as single fractions. Then treat as a quotient. 2 ⫺1 x
2⫺x x
1
x
2⫺x 4 ⫺ x2 2⫺x x2 ⫽ ⫽ ⫼ ⫽ ⴢ x x 4 4 ⫺ x2 (2 ⫺ x)(2 ⫹ x) x2 1 1 2 ⫺ 1 2 x x ⫽
MATCHED PROBLEM 4
x 2⫹x
0002
Express as a simple fraction reduced to lowest terms. Use the two methods described in Example 4. 1⫹
1 x
x⫺
1 x 0002
ANSWERS TO MATCHED PROBLEMS 3x ⫹ 2 x2 ⫹ 2x ⫹ 4 ⫺5 (B) 2. (A) 2x (B) (C) mn x⫹1 3x ⫹ 4 x⫹4 2 2 2y ⫺ 9y ⫺ 6 1 3x ⫺ 5x ⫺ 4 1 3. (A) (B) (C) 4. 3 2 4 x⫺1 12x ( y ⫺ 2) ( y ⫹ 2)
1. (A)
R-4
Exercises
In Problems 1–10, reduce each rational expression to lowest terms. 1.
17 85
2.
91 26
3.
360 288
4.
63 105
5.
x⫹1 x ⫹ 3x ⫹ 2 2
6.
x2 ⫺ 2x ⫺ 24 x⫺6
7.
x2 ⫺ 9 x ⫹ 3x ⫺ 18 2
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3x2y3
10.
x4y
2a2b4c6 6a5b3c
In Problems 11–36, perform the indicated operations and reduce answers to lowest terms. Represent any compound fractions as simple fractions reduced to lowest terms. 7 19 12. ⫹ 10 25
5 11 11. ⫹ 6 15
41. 42.
⫺2x(x ⫹ 4)3 ⫺ 3(3 ⫺ x2)(x ⫹ 4)2 (x ⫹ 4)6 3x2(x ⫹ 1)3 ⫺ 3(x3 ⫹ 4)(x ⫹ 1)2 (x ⫹ 1)6
In Problems 43–54, perform the indicated operations and reduce answers to lowest terms. Represent any compound fractions as simple fractions reduced to lowest terms. y
2 1 ⫹ 2 y2 ⫺ 5y ⫹ 4 y ⫹y⫺2
13.
1 1 ⫺ 8 9
14.
9 8 ⫺ 8 9
43.
15.
1 1 ⫺ n m
16.
m n ⫺ n m
44.
x⫺8 x x⫹4 ⫹ ⫹ x ⫺ 6 x ⫺3 x ⫺ 9x ⫹ 18
17.
5 3 ⫼ 12 4
18.
10 5 ⫼ 3 2
45.
16 ⫺ m2 m⫺1 ⴢ m ⫹ 3m ⫺ 4 m ⫺ 4
y2 ⫺ 2y ⫺ 8
⫺
2
2
19. a
25 5 4 ⫼ bⴢ 8 16 15
20.
25 5 4 ⫼a ⴢ b 8 16 15
46.
x⫹1 x2 ⫺ 2x ⫹ 1 ⴢ x(1 ⫺ x) x2 ⫺ 1
21. a
b2 b a ⫼ 2b ⴢ 2a 3b a
22.
b2 b a ⫼a 2ⴢ b 2a a 3b
47.
y⫹9 x⫹7 ⫹ ax ⫺ bx by ⫺ ay
23.
x2 ⫺ 1 x⫹1 ⫼ 2 x⫹2 x ⫺4
24.
x⫺3 x2 ⫺ 9 ⫼ 2 x⫺1 x ⫺1
48.
c⫺2 c c⫹2 ⫺ ⫹ 5c ⫺ 5 3c ⫺ 3 1⫺c
25.
1 1 1 ⫹ ⫹ c a b
26.
1 1 1 ⫹ ⫹ ac bc ab
49.
x2 ⫺ 13x ⫹ 36 x2 ⫺ 16 ⫼ 2x ⫹ 10x ⫹ 8 x3 ⫹ 1
2a ⫺ b 2a ⫹ 3b ⫺ 2 27. 2 2 a ⫺b a ⫹ 2ab ⫹ b2 28.
x⫺2 x⫹2 ⫺ x2 ⫺ 1 (x ⫺ 1)2
29. m ⫹ 2 ⫺
m⫺2 m⫺1
31.
3 2 ⫺ x⫺2 2⫺x
33.
4y 3 2 ⫹ ⫺ 2 y⫹2 y⫺2 y ⫺4
34.
4x 3 2 ⫹ ⫺ 2 2 x ⫺ y x ⫹ y x ⫺y
x2 ⫺1 y2 35. x ⫹1 y
38. 39. 40.
50. a
x3 ⫺ y3
51. a
1 x 4 ⫺ b⫼ x⫹4 x⫹4 x2 ⫺ 16 3 1 x⫹4 ⫺ b⫼ x⫺2 x⫹1 x⫺2
y3
ⴢ
y x2 ⫹ xy ⫹ y2 b⫼ x⫺y y2
30.
x⫹1 ⫹x x⫺1
52. a
32.
1 2 ⫺ a⫺3 3⫺a
2 15 ⫺ 2 x x 53. 4 5 1⫹ ⫺ 2 x x 1⫹
y x ⫺2⫹ y x 54. y x ⫺ y x
Problems 55–58 are calculus-related. Perform the indicated operations and reduce answers to lowest terms. Represent any compound fractions as simple fractions reduced to lowest terms. 4 ⫺x x 36. 2 ⫺1 x
Problems 37–42 are calculus-related. Reduce each fraction to lowest terms. 37.
2
1 1 ⫺ x x⫹h 55. h
1 1 ⫺ 2 (x ⫹ h)2 x 56. h
(x ⫹ h)2 x2 ⫺ x⫹h⫹2 x⫹2 57. h
2x ⫹ 2h ⫹ 3 2x ⫹ 3 ⫺ x x⫹h 58. h
6x3(x2 ⫹ 2)2 ⫺ 2x(x2 ⫹ 2)3 x4 4x4(x2 ⫹ 3) ⫺ 3x2(x2 ⫹ 3)2 x6 2x(1 ⫺ 3x)3 ⫹ 9x2(1 ⫺ 3x)2 (1 ⫺ 3x)6 2x(2x ⫹ 3)4 ⫺ 8x2(2x ⫹ 3)3 (2x ⫹ 3)8
In Problems 59–62, perform the indicated operations and reduce answers to lowest terms. Represent any compound fractions as simple fractions reduced to lowest terms. y2 y⫺x 59. x2 1⫹ 2 y ⫺ x2 y⫺
s2 ⫺s s⫺t 60. 2 t ⫹t s⫺t
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Review
1
61. 2 ⫺
1
62. 1 ⫺
2 1⫺ a⫹2
Discuss possible conditions of a and b that would make this a valid equation.
1
1⫺
1⫺
1 x
64. Show by example that, in general, a2 ⫹ b2 ⫽a⫹b a⫹b
63. Show by example that, in general, a⫹b ⫽a⫹1 b
CHAPTER
R-1
R
Review 6.
A real number is any number that has a decimal representation. There is a one-to-one correspondence between the set of real numbers and the set of points on a line. Important subsets of the real numbers include the natural numbers, integers, and rational numbers. A rational number can be written in the form aⲐb, where a and b are integers and b ⫽ 0. A real number can be approximated to any desired precision by rational numbers. Consequently, arithmetic operations on rational numbers can be extended to operations on real numbers. These operations satisfy basic real number properties, including associative properties: x ⫹ ( y ⫹ z) ⫽ (x ⫹ y) ⫹ z and x( yz) ⫽ (xy)z; commutative properties: x ⫹ y ⫽ y ⫹ x and xy ⫽ yx; identities: 0 ⫹ x ⫽ x ⫹ 0 ⫽ x and (1)x ⫽ x(1) ⫽ x; inverses: ⫺x is the additive inverse of x and, if x ⫽ 0, x⫺1 is the multiplicative inverse of x; and distributive property: x( y ⫹ z) ⫽ xy ⫹ xz. Subtraction is defined by a ⫺ b ⫽ a ⫹ (⫺b) and division by aⲐb ⫽ ab⫺1. Division by 0 is never allowed. Additional properties include properties of negatives:
2. (⫺a)b ⫽ ⫺(ab) ⫽ a(⫺b) ⫽ ⫺ab 3. (⫺a)(⫺b) ⫽ ab 4. (⫺1)a ⫽ ⫺a
6.
⫺a a a ⫺a ⫽⫺ ⫽⫺ ⫽ ⫺b b ⫺b b
b⫽0 b⫽0
zero properties: 1. a ⴢ 0 ⫽ 0 2. ab ⫽ 0
if and only if
a⫽0
or
b⫽0
or both.
and fraction properties (division by 0 excluded): 1.
a c ⫽ b d
2.
ka a ⫽ kb b
3.
a c ac ⴢ ⫽ b d bd
4.
a c a d ⫼ ⫽ ⴢ b d b c
5.
a c a⫹c ⫹ ⫽ b b b
a c a⫺c ⫺ ⫽ b b b
R-2
7.
a c ad ⫹ bc ⫹ ⫽ b d bd
Exponents and Radicals
The notation an, in which the exponent n is an integer, is defined as follows. For n a positive integer and a a real number: an ⫽ a ⴢ a ⴢ . . . ⴢ a (n factors of a) a⫺n ⫽
1 an
a0 ⫽ 1
(a ⫽ 0) (a ⫽ 0)
Properties of integer exponents (division by 0 excluded): 1. aman ⫽ am⫹n
2. (an)m ⫽ amn
3. (ab)m ⫽ ambm
a m am 4. a b ⫽ m b b
5.
1. ⫺(⫺a) ⫽ a
⫺a a a ⫽⫺ ⫽ b b ⫺b
(assume a ⫽ ⫺b)
Discuss possible conditions of a and b that would make this a valid equation.
(assume b ⫽ 0)
Algebra and Real Numbers
5.
39
1 am ⫽ am⫺n ⫽ n⫺m an a
Any positive number written in decimal form can be expressed in scientific notation, that is, in the form a ⫻ 10n 1 ⱕ a 6 10, n an integer, a in decimal form. For n a natural number, a and b real numbers: a is an nth root of b if an ⫽ b. The number of real nth roots of a real number b is either 0, 1, or 2, depending on whether b is positive or negative, and whether n is even or odd. The principal nth root of b, denoted by n b1/n or 1b, is the real nth root of b if there is only one, and the posn itive nth root of b if there are two real nth roots. In the notation 1b, the symbol 1 is called a radical, n is called the index, and b is the 2 radicand. If n ⫽ 2 we write 1b in place of 1 b. We extend exponent notation so that exponents can be rational numbers, not just integers, as follows. For m and n natural numbers and b any real number (except b can't be negative when n is even), bmⲐn ⫽ (b1Ⲑn)m and b⫺mⲐn ⫽
if and only if ad ⫽ bc
1 bmⲐn
Rational exponent/radical property: (b1Ⲑn)m ⫽ (bm)1Ⲑn
and
n
n
(1b)m ⫽ 2bm
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Properties of radicals (x 7 0, y 7 0): n
1. 2xn ⫽ x
n
n
n
2. 2xy ⫽ 2x2y
A radical is in simplified form if:
3.
n
x 1x ⫽ n y A 1y n
1. No radicand contains a factor to a power greater than or equal to the index of the radical. 2. No power of the radicand and the index of the radical have a common factor other than 1. 3. No radical appears in a denominator.
number otherwise. Each composite number can be factored uniquely into a product of prime numbers. A polynomial is prime relative to a given set of numbers (usually the set of integers) if (1) all its coefficients are from that set of numbers, and (2) it cannot be written as a product of two polynomials of positive degree having coefficients from that set of numbers. A nonprime polynomial is factored completely relative to a given set of numbers if it is written as a product of prime polynomials relative to that set of numbers. Common factors can be factored out by applying the distributive properties. Grouping can be used to identify common factors. Second-degree polynomials can be factored by trial and error. The following special factoring formulas are useful:
4. No fraction appears within a radical.
1. u2 ⫹ 2uv ⫹ v2 ⫽ (u ⫹ v)2
Perfect Square
Eliminating a radical from a denominator is called rationalizing the denominator. To rationalize the denominator, we multiply the numerator and denominator by a suitable factor that will leave the denominator free of radicals. This factor is called a rationalizing factor. For example, if the denominator is of the form 1a ⫹ 1b, then 1a ⫺ 1b is a rationalizing factor.
2. u2 ⫺ 2uv ⫹ v2 ⫽ (u ⫺ v)2
Perfect Square
R-3
Polynomials: Basic Operations and Factoring
An algebraic expression is formed by using constants and variables and the operations of addition, subtraction, multiplication, division, raising to powers, and taking roots. A polynomial is an algebraic expression formed by adding and subtracting constants and terms of the form axn (one variable), axnym (two variables), and so on. The degree of a term is the sum of the powers of all variables in the term, and the degree of a polynomial is the degree of the nonzero term with highest degree in the polynomial. Polynomials with one, two, or three terms are called monomials, binomials, and trinomials, respectively. Like terms have exactly the same variable factors to the same powers and can be combined by adding their coefficients. Polynomials can be added, subtracted, and multiplied by repeatedly applying the distributive property and combining like terms. A number or algebraic expression is factored if it is expressed as a product of other numbers or algebraic expressions, which are called factors. An integer greater than 1 is a prime number if its only positive integer factors are itself and 1, and a composite
CHAPTER
R
⫺1 ⫺1
3. 7 9 5.
5 1 ⫼ a ⫺ 3⫺1 b 7 3
2
3
3
2
2
Difference of Cubes
3
3
2
2
Sum of Cubes
3. u ⫺ v ⫽ (u ⫺ v)(u ⫹ v)
Difference of Squares
4. u ⫺ v ⫽ (u ⫺ v)(u ⫹ uv ⫹ v ) 5. u ⫹ v ⫽ (u ⫹ v)(u ⫺ uv ⫹ v )
There is no factoring formula relative to the real numbers for u2 ⫹ v2.
R-4
Rational Expressions: Basic Operations
A fractional expression is the ratio of two algebraic expressions, and a rational expression is the ratio of two polynomials. The rules for adding, subtracting, multiplying, and dividing real number fractions (see Section R-1 in this review) all extend to fractional expressions with the understanding that variables are always restricted to exclude division by zero. Fractions can be reduced to lowest terms or raised to higher terms by using the fundamental property of fractions: ka a ⫽ kb b
with b, k ⫽ 0
A rational expression is reduced to lowest terms if the numerator and denominator do not have any factors in common relative to the integers. The least common denominator (LCD) is useful for adding and subtracting fractions with different denominators and for reducing compound fractions to simple fractions.
Review Exercises
In Problems 1–6, perform the indicated operations, if defined. If the result is not an integer, express it in the form a/b, where a and b are integers. 5 3 1. ⫹ 6 4
2
4 2 2. ⫺ 3 9 6 10 4. a⫺ b a⫺ b 3 5 6.
11 3 ⫼ a⫺ b 12 4
Problems 7–12 refer to the polynomials (a) x4 ⫹ 3x2 ⫹ 1 and (b) 4 ⫺ x4. 7. What is the degree of (a)? 8. What is the degree of (b)? 9. What is the degree of the sum of (a) and (b)? 10. What is the degree of the product of (a) and (b)? 11. Multiply (a) and (b). 12. Add (a) and (b).
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In Problems 13–18, evaluate each expression that results in a rational number. 13. 2891Ⲑ2
14. 2161Ⲑ3
15. 8⫺2Ⲑ3
16. (⫺64)5Ⲑ3
9 ⫺1Ⲑ2 17. a b 16
1Ⲑ2
18. (121
⫹ 25
)
20. (3m ⫺ 5n)(3m ⫹ 5n)
21. (2x ⫹ y)(3x ⫺ 4y)
22. (2a ⫺ 3b)2
In Problems 23–25, write each polynomial in a completely factored form relative to the integers. If the polynomial is prime relative to the integers, say so. 24. t 2 ⫺ 4t ⫺ 6
25. 6n3 ⫺ 9n2 ⫺ 15n In Problems 26–29, perform the indicated operations and reduce to lowest terms. Represent all compound fractions as simple fractions reduced to lowest terms. 26.
28.
2 4 1 ⫺ 3⫺ 2 2 5b 3a 6a b y⫺2 2
y ⫺ 4y ⫹ 4
27.
⫼
1 3x ⫹ 6x 3x2 ⫺ 12x
1 u 29. 1 1⫺ 2 u u⫺
2
y ⫹ 2y 2
y ⫹ 4y ⫹ 4
Simplify Problems 30–35, and write answers using positive exponents only. All variables represent positive real numbers. 8 6
9u v 3u4v8
30. 6(xy3)5
31.
32. (2 ⫻ 105)(3 ⫻ 10⫺3)
33. (x⫺3y2)⫺2
5Ⲑ3 2Ⲑ3
34. u
35. (9a b ) 2Ⲑ5
36. Change to radical form: 3x
3 37. Change to rational exponent form: ⫺3 2 (xy)2
40.
6ab 13a
39. 22x2y5 218x3y2 41.
15 3 ⫺ 15
a a ⫽⫺ ⫺(b ⫺ c) b⫺c
48. 3xy ⫹ 0 ⫽ 3xy
49. Indicate true (T) or false (F): (A) An integer is a rational number and a real number. (B) An irrational number has a repeating decimal representation. 50. Give an example of an integer that is not a natural number. 51. Given the algebraic expressions: (a) 2x2 ⫺ 3x ⫹ 5 (b) x2 ⫺ 1x ⫺ 3 ⫺3 ⫺2 ⫺1 (c) x ⫹ x ⫺ 3x (d) x2 ⫺ 3xy ⫺ y2 (A) Identify all second-degree polynomials. (B) Identify all third-degree polynomials. In Problems 52–55, perform the indicated operations and simplify. 52. (2x ⫺ y)(2x ⫹ y) ⫺ (2x ⫺ y)2 53. (m2 ⫹ 2mn ⫺ n2)(m2 ⫺ 2mn ⫺ n2) 54. 5(x ⫹ h)2 ⫺ 7(x ⫹ h) ⫺ (5x2 ⫺ 7x)
55. ⫺2x5(x2 ⫹ 2)(x ⫺ 3) ⫺ x [x ⫺ x(3 ⫺ x)] 6 In Problems 56–61, write in a completely factored form relative to the integers. 56. (4x ⫺ y)2 ⫺ 9x2
57. 2x2 ⫹ 4xy ⫺ 5y2
58. 6x3y ⫹ 12x2y2 ⫺ 15xy3
59. (y ⫺ b)2 ⫺ y ⫹ b
60. y3 ⫹ 2y2 ⫺ 4y ⫺ 8
61. 2x(x ⫺ 4)3 ⫹ 3x2(x ⫺ 4)2
62.
Simplify Problems 38–42, and express answers in simplified form. All variables represent positive real numbers. 3 5 4 38. 3x 2 xy
47.
In Problems 62–65, perform the indicated operations and reduce to lowest terms. Represent all compound fractions as simple fractions reduced to lowest terms.
4 ⫺2 1Ⲑ2
u
44. 3y ⫹ (2x ⫹ 5) ⫽ (2x ⫹ 5) ⫹ 3y 46. 3 ⴢ (5x) ⫽ (3 ⴢ 5)x
1Ⲑ2 ⫺3Ⲑ4
19. 5x ⫺ 3x[4 ⫺ 3(x ⫺ 2) ]
23. 9x2 ⫺ 12x ⫹ 4
43. (⫺3) ⫺ (⫺2) ⫽ (⫺3) ⫹ [⫺(⫺2)] 45. (2x ⫹ 3)(3x ⫹ 5) ⫽ (2x ⫹ 3)3x ⫹ (2x ⫹ 3)5
In Problems 19–22, perform the indicated operations and simplify. 2
41
63.
64. 8 42. 2y6
In Problems 43–48, each statement illustrates the use of one of the following real number properties or definitions. Indicate which one. Commutative (⫹) Identity (⫹) Commutative (ⴢ) Identity (ⴢ) Division Associative (⫹) Inverse (⫹) Associative (ⴢ) Inverse (ⴢ) Zero Distributive Subtraction Negatives
3x2(x ⫹ 2)2 ⫺ 2x(x ⫹ 2)3 x4 m⫹3 2 m⫺1 ⫹ 2 ⫹ 2⫺m m ⫺ 4m ⫹ 4 m ⫺4 2
y x2
⫼a
1⫺
x3y ⫺ x2y x2 ⫹ 3x ⫼ b 2x2 ⫹ 5x ⫺ 3 2x2 ⫺ 3x ⫹ 1 1
1⫹ 65. 1⫺
x y
1 1⫺
x y
66. Convert to scientific notation and simplify: 0.000 000 000 52 (1,300)(0.000 002)
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BASIC ALGEBRAIC OPERATIONS
In Problems 67–75, perform the indicated operations and express answers in simplified form. All radicands represent positive real numbers. 5
6 7 11
2x2
67. ⫺2x 23 x y
68.
9 70. 2 8x6y12
3 71. 21 4x4
69.
3 2 4x
5
3y2
B 8x2
(C) What is the effect on production of doubling the units of labor and capital at any production level? 79. ELECTRIC CIRCUIT If three electric resistors with resistances R1, R2, and R3 are connected in parallel, then the total resistance R for the circuit shown in the figure is given by R⫽
72. (2 1x ⫺ 5 1y)( 1x ⫹ 1y) 73.
75.
3 1x 2 1x ⫺ 1y
74.
2 1u ⫺ 3 1v 2 1u ⫹ 3 1v
1 1 1 1 ⫹ ⫹ R1 R2 R3
Represent this compound fraction as a simple fraction.
y2
R1
2y2 ⫹ 4 ⫺ 2
R2 R3
APPLICATIONS 76. CONSTRUCTION A circular fountain in a park includes a concrete wall that is 3 ft high and 2 ft thick (see the figure). If the inner radius of the wall is x feet, write an algebraic expression in terms of x that represents the volume of the concrete used to construct the wall. Simplify the expression.
2 feet
x feet
80. CONSTRUCTION A box with a hinged lid is to be made out of a piece of cardboard that measures 16 by 30 inches. Six squares, x inches on a side, will be cut from each corner and the middle, and then the ends and sides will be folded up to form the box and its lid (see the figure). Express each of the following quantities as a polynomial in both factored and expanded form. (A) The area of cardboard after the corners have been removed. (B) The volume of the box.
3 feet
30 in. x
77. ECONOMICS If in the United States in 2007 the total personal income was about $11,580,000,000,000 and the population was about 301,000,000, estimate to three significant digits the average personal income. Write your answer in scientific notation and in standard decimal form. 78. ECONOMICS The number of units N produced by a petroleum company from the use of x units of capital and y units of labor is approximated by N ⫽ 20x1Ⲑ2y1Ⲑ2 (A) Estimate the number of units produced by using 1,600 units of capital and 900 units of labor. (B) What is the effect on production if the number of units of capital and labor are doubled to 3,200 units and 1,800 units, respectively?
16 in.
x
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CHAPTER
Equations and Inequalities
1
C
OUTLINE
SOLVING equations and inequalities is one of the most important
skills in algebra because it can be applied to solving a boundless supply of real-world problems. In this chapter, we will begin with a look at techniques for solving linear equations and inequalities. After a study of complex numbers, we’ll return to equations, learning how to solve a variety of nonlinear equations. For each type of equation and inequality we solve, we will look at some real-world problems that can be solved using those solution techniques. This doesn’t close the book on solving equations, though—we will learn how to solve new types of equations in many of the remaining chapters.
1-1
Linear Equations and Applications
1-2
Linear Inequalities
1-3
Absolute Value in Equations and Inequalities
1-4
Complex Numbers
1-5
Quadratic Equations and Applications
1-6
Additional Equation-Solving Techniques Chapter 1 Review Chapter 1 Group Activity: Solving a Cubic Equation
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Linear Equations and Applications Z Understanding Basic Terms Z Solving Linear Equations Z Solving Number and Geometric Problems Z Solving Rate–Time Problems Z Solving Mixture Problems
We begin this section with a quick look at what an equation is and what it means to solve one. After solving some linear equations, we move on to the main topic: using linear equations to solve word problems.
Z Understanding Basic Terms An algebraic equation is a mathematical statement that two algebraic expressions are equal. Some examples of equations with variable x are 3x 0002 2 0004 7 2x2 0002 3x 0003 5 0004 0
1 x 0004 10003x x00022 1x 0003 4 0004 x 0002 1
The replacement set, or domain, for a variable is defined to be the set of numbers that are permitted to replace the variable.
Z ASSUMPTION On Domains of Variables Unless stated to the contrary, we assume that the domain for a variable in an algebraic expression or equation is the set of those real numbers for which the algebraic expressions involving the variable are real numbers.
For example, the domain for the variable x in the expression 2x 0002 4 is R, the set of all real numbers, since 2x 0002 4 represents a real number for all replacements of x by real numbers. The domain of x in the equation 1 2 0004 x x00023 is the set of all real numbers except 0 and 3. These values are excluded because the expression on the left is not defined for x 0004 0 and the expression on the right is not defined for x 0004 3. Both expressions represent real numbers for all other replacements of x by real numbers. The solution set for an equation is defined to be the set of all elements in the domain of the variable that make the equation true. Each element of the solution set is called a solution, or root, of the equation. To solve an equation is to find the solution set for the equation.
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45
An equation is called an identity if the equation is true for all elements from the domain of the variable. An equation is called a conditional equation if it is true for certain domain values and false for others. For example, 2x 0002 4 0004 2(x 0002 2)
and
5 5 0004 x(x 0002 3) x2 0002 3x
are identities, since both equations are true for all elements from the respective domains of their variables. On the other hand, the equations 3x 0002 2 0004 5
and
1 2 0004 x x00021
are conditional equations, since, for example, neither equation is true for the domain value 2. Knowing what we mean by the solution set of an equation is one thing; finding it is another. We introduce the idea of equivalent equations to help us find solutions. We will call two equations equivalent if they both have the same solution set. To solve an equation, we perform operations on the equation to produce simpler equivalent equations. We stop when we find an equation whose solution is obvious. Then we check this obvious solution in the original equation. Any of the properties of equality given in Theorem 1 can be used to produce equivalent equations.
Z THEOREM 1 Properties of Equality For a, b, and c any real numbers: 1. If a 0004 b, then a 0003 c 0004 b 0003 c. 2. If a 0004 b, then a 0002 c 0004 b 0002 c. 3. If a 0004 b and c 0005 0, then ca 0004 cb. b a 4. If a 0004 b and c 0005 0, then 0004 . c c
Addition Property Subtraction Property Multiplication Property
5. If a 0004 b, then either may replace the other in any statement without changing the truth or falsity of the statement.
Substitution Property
Division Property
Z Solving Linear Equations We now turn our attention to methods of solving first-degree, or linear, equations in one variable.
Z DEFINITION 1 Linear Equation in One Variable Any equation that can be written in the form ax ⴙ b ⴝ 0
aⴝ0
Standard Form
where a and b are real constants and x is a variable, is called a linear, or firstdegree, equation in one variable. 5x ⴚ 1 ⴝ 2(x ⴙ 3) is a linear equation because after simplifying, it can be written in the standard form 3x ⴚ 7 ⴝ 0.
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1
Solving a Linear Equation Solve 5x ⫺ 9 ⫽ 3x ⫹ 7 and check.
SOLUTION
We will use the properties of equality to transform the given equation into an equivalent equation whose solution is obvious. 5x ⫺ 9 ⫽ 3x ⫹ 7 5x ⫺ 9 ⫹ 9 ⫽ 3x ⫹ 7 ⫹ 9 5x ⫽ 3x ⫹ 16 5x ⫺ 3x ⫽ 3x ⫹ 16 ⫺ 3x 2x ⫽ 16 2x 16 ⫽ 2 2 x⫽8
Add 9 to both sides. Combine like terms. Subtract 3x from both sides. Combine like terms. Divide both sides by 2. Simplify.
The solution set for this last equation is obvious: Solution set: {8} And since the equation x ⫽ 8 is equivalent to all the preceding equations in our solution, {8} is also the solution set for all these equations, including the original equation. [Note: If an equation has only one element in its solution set, we generally use the last equation (in this case, x ⫽ 8) rather than set notation to represent the solution.] CHECK
MATCHED PROBLEM 1
A ⫽ lw
5x ⫺ 9 ⫽ 3x ⫹ 7 ? 5(8) ⫺ 9 ⫽ 3(8) ⫹ 7 ? 40 ⫺ 9 ⫽ 24 ⫹ 7 ✓ 31 ⫽ 31
Simplify each side.
A true statement
Solve and check: 7x ⫺ 10 ⫽ 4x ⫹ 5
0002
0002
We often encounter equations involving more than one variable. For example, if l and w are the length and width of a rectangle, respectively, the area of the rectangle is given by A ⫽ lw (see Fig. 1). Depending on the situation, we may want to solve this equation for l or w. To solve for w, we simply consider A and l to be constants and w to be a variable. Then the equation A ⫽ lw becomes a linear equation in w that can be solved easily by dividing both sides by l:
w
l
Z Figure 1 Area of a rectangle.
w⫽
EXAMPLE
Substitute x ⴝ 8.
2
A l
l⫽0
Solving an Equation with More Than One Variable Solve for P in terms of the other variables: A ⫽ P ⫹ Prt
SOLUTION
A ⫽ P ⫹ Prt A ⫽ P(1 ⫹ rt)
Factor to isolate P. Divide both sides by 1 ⴙ rt.
A ⫽P 1 ⫹ rt P⫽
A 1 ⫹ rt
Restriction: 1 ⴙ rt ⴝ 0
0002
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MATCHED PROBLEM 2
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47
Solve for F in terms of C: C 0004 59(F 0002 32) 0002 A great many practical problems can be solved using algebraic techniques—so many, in fact, that there is no one method of attack that will work for all. However, we can put together a strategy that will help you organize your approach.
Z STRATEGY FOR SOLVING WORD PROBLEMS 1. Read the problem slowly and carefully, more than once if necessary. Write down information as you read the problem the first time to help you get started. Identify what it is that you are asked to find. 2. Use a variable to represent an unknown quantity in the problem, usually what you are asked to find. Then try to represent any other unknown quantities in terms of that variable. It’s pretty much impossible to solve a word problem without this step. 3. If it helps to visualize a situation, draw a diagram and label known and unknown parts. 4. Write an equation relating the quantities in the problem. Often, you can accomplish this by finding a formula that connects those quantities. Try to write the equation in words first, then translate to symbols. 5. Solve the equation, then answer the question in a sentence by rephrasing the question. Make sure that you’re answering all of the questions asked. 6. Check to see if your answers make sense in the original problem, not just the equation you wrote.
ZZZ EXPLORE-DISCUSS 1
Translate each of the following sentences involving two numbers into an equation. (A) The first number is 10 more than the second number. (B) The first number is 15 less than the second number. (C) The first number is half the second number. (D) The first number is three times the second number. (E) Ten times the first number is 15 more than the second number.
The remaining examples in this section contain solutions to a variety of word problems illustrating both the process of setting up word problems and the techniques used to solve the resulting equations. As you read an example, try covering up the solution and working the problem yourself. If you need a hint, uncover just part of the solution and try to work out the rest. After you successfully solve an example problem, try the matched problem. If you work through the remainder of the section in this way, you will already have experience with a wide variety of word problems.
Z Solving Number and Geometric Problems Example 3 introduces the process of setting up and solving word problems in a simple mathematical context. Examples 4–8 are more realistic.
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3
Setting Up and Solving a Word Problem Find four consecutive even integers so that the sum of the first three is 8 more than the fourth.
SOLUTION
Let x ⫽ the first even integer; then x
x⫹2
x⫹4
and
x⫹6
represent four consecutive even integers starting with the even integer x. (Remember, even integers are separated by 2.) The phrase “the sum of the first three is 8 more than the fourth” translates into an equation: Sum of the first three ⫽ Fourth ⫹ 8 x ⫹ (x ⫹ 2) ⫹ (x ⫹ 4) ⫽ (x ⫹ 6) ⫹ 8 3x ⫹ 6 ⫽ x ⫹ 14 2x ⫽ 8 x⫽4
Combine like terms. Subtract 6 and x from both sides. Divide both sides by 2.
The first even integer is 4, so the four consecutive integers are 4, 6, 8, and 10. CHECK
MATCHED PROBLEM 3
ZZZ EXPLORE-DISCUSS 2
4 ⫹ 6 ⫹ 8 ⫽ 18 ⫽ 10 ⫹ 8
Sum of first three is 8 more than the fourth.
0002
Find three consecutive odd integers so that 3 times their sum is 5 more than 8 times the middle one. 0002
According to Part 3 of Theorem 1, multiplying both sides of an equation by a nonzero number always produces an equivalent equation. By what number would you choose to multiply both sides of the following equation to eliminate all the fractions? x⫹1 x 1 ⫺ ⫽ 3 4 2 If you did not choose 12, the LCD of all the fractions in this equation, you could still solve the resulting equation, but with more effort. (For a discussion of LCDs and how to find them, see Section R-4.)
EXAMPLE
4
Using a Diagram in the Solution of a Word Problem A landscape designer plans a series of small triangular gardens outside a new office building. Her plans call for one side to be one-third of the perimeter, and another side to be onefifth of the perimeter. The space allotted for each will allow the third side to be 7 meters. Find the perimeter of the triangle.
SOLUTION p 5
p 3 7 meters
Z Figure 2
Draw a triangle, and label one side 7 meters. Let p ⫽ the perimeter: then the remaining sides are one-third p, or p Ⲑ3, and one-fifth p, or pⲐ 5 (see Fig. 2). Perimeter ⫽ Sum of the side lengths p⫽
p p ⫹ ⫹7 3 5
Multiply both sides by 15, the LCD. Make sure to multiply every term by 15!
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p p 15 ⴢ p 0004 15 ⴢ a 0003 0003 7b 3 5 p p 15p 0004 15 ⴢ 0003 15 ⴢ 0003 15 ⴢ 7 3 5 15p 0004 5p 0003 3p 0003 105 15p 0004 8p 0003 105 7p 0004 105 p 0004 15
Linear Equations and Applications
49
*
Combine like terms. Subtract 8p from both sides. Divide both sides by 7.
The perimeter is 15 meters. p 15 0004 00045 3 3 p 15 0003 0004 00043 5 5 0003 7 15 meters
CHECK
MATCHED PROBLEM 4
ZZZ
CAUTION ZZZ
Side 1
Side 2 Side 3 Perimeter
0002
If one side of a triangle is one-fourth the perimeter, the second side is 7 centimeters, and the third side is two-fifths the perimeter, what is the perimeter? 0002
A very common error occurs about now—students tend to confuse algebraic expressions involving fractions with algebraic equations involving fractions. Consider these two problems: (A) Solve:
x x 0003 0004 10 2 3
(B) Add:
x x 0003 0003 10 2 3
The problems look very much alike but are actually very different. To solve the equation in (A) we multiply both sides by 6 (the LCD) to clear the fractions. This works so well for equations that students want to do the same thing for problems like (B). The only catch is that (B) is not an equation, and the multiplication property of equality does not apply. If we multiply (B) by 6, we simply obtain an expression 6 times as large as the original! Compare these correct solutions: x x 0003 0004 10 2 3
(A) 6ⴢ
x x 0003 6 ⴢ 0004 6 ⴢ 10 2 3 3x 0003 2x 0004 60 5x 0004 60 x 0004 12
(B)
x x 0003 0003 10 2 3 0004
3ⴢx 2ⴢx 6 ⴢ 10 0003 0003 3ⴢ2 2ⴢ3 6ⴢ1
0004
3x 2x 60 0003 0003 6 6 6
0004
5x 0003 60 6
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
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There are many problems in which a rate plays a key role. For example, if you’re losing weight at the rate of 2 lb per week, you can use that rate to find a total weight loss for some period of time. Rate problems can often be solved using the following basic formula: Z QUANTITY-RATE-TIME FORMULA The change in a quantity is the rate at which it changes times the time passed: Quantity 0004 Rate 0007 Time, or Q 0004 RT. If the quantity is distance, then D 0004 RT. The formulas can be solved for R or T to get a related formula to find the rate or the time. [Note: R is an average or uniform rate.]
ZZZ EXPLORE-DISCUSS 3
EXAMPLE
5
(A) If you drive at an average rate of 65 miles per hour, how far do you go in 3 hours? (B) If you make $750 for 2 weeks of part-time work, what is your weekly rate of pay? (C) If you eat at the rate of 1,900 calories per day, how long will it take you to eat 7,600 calories?
A Distance–Rate–Time Problem The distance along a shipping route between San Francisco and Honolulu is 2,100 nautical miles. If one ship leaves San Francisco at the same time another leaves Honolulu, and if the former travels at 15 knots* and the latter at 20 knots, how long will it take the two ships to rendezvous? How far will they be from Honolulu and San Francisco at that time?
SOLUTION
Let T 0004 number of hours until both ships meet. Draw a diagram and label known and unknown parts. Both ships will have traveled the same amount of time when they meet.
San Francisco 2,100
H
miles
D1 0004 20T
D2 0004 15T
20 knots
15 knots Meeting
D 0004 RT D1 0004 20 knots ⴢ Time D2 0004 15 knots ⴢ Time
SF
Distance ship 1 Distance ship 2 from Honolulu from San Francisco ± ≤ 0003 ± ≤ travels to travels to meeting point meeting point D1 0003 D2 20T 0003 15T 35T T
Honolulu
Total distance 0004 ° from Honolulu ¢ to San Francisco 0004 0004 0004 0004
2,100 2,100 2,100 60
Therefore, it takes 60 hours, or 2.5 days, for the ships to meet. Distance from Honolulu 0004 20 ⴢ 60 0004 1,200 nautical miles Distance from San Francisco 0004 15 ⴢ 60 0004 900 nautical miles CHECK
1,200 0003 900 0004 2,100 nautical miles
0002
*15 knots means 15 nautical miles per hour. There are 6,076.1 feet in 1 nautical mile, and 5,280 feet in 1 statute mile.
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51
Linear Equations and Applications
An old piece of equipment can print, stuff, and label 38 mailing pieces per minute. A newer model can handle 82 per minute. How long will it take for both pieces of equipment to prepare a mailing of 6,000 pieces? [Hint: Use Quantity 0004 Rate 0007 Time for each machine.] 0002 Some equations involving variables in a denominator can be transformed into linear equations. We can proceed in essentially the same way as in Example 5; however, we need to exclude any value of the variable that will make a denominator 0. With these values excluded, we can multiply through by the LCD even though it contains a variable, and, according to Theorem 1, the new equation will be equivalent to the old.
EXAMPLE
6
A Distance–Rate–Time Problem An excursion boat takes 1.5 times as long to go 360 miles up a river as to return. If the boat cruises at 15 miles per hour in still water, what is the rate of the current?
SOLUTION
360 miles
Let x 0004 Rate of current (in miles per hour) 15 0002 x 0004 Rate of boat upstream 15 0003 x 0004 Rate of boat downstream Time upstream 0004 (1.5)(Time downstream) Distance upstream Distance downstream 0004 (1.5) Rate upstream Rate downstream 360 360 0004 (1.5) 15 0002 x 15 0003 x 360 540 0004 15 0002 x 15 0003 x 360(15 0003 x) 0004 540(15 0002 x) 5,400 0003 360x 0004 8,100 0002 540x 5,400 0003 900x 0004 8,100 900x 0004 2,700 x00043
What we were asked to find.
Faster downstream.
Because D 0004 RT, T ⴝ
x cannot be 15 or 000215 Multiply both sides by the LCD, (15 ⴚ x)(15 ⴙ x). Multiply out parentheses. Add 540x to both sides. Subtract 5,400 from both sides. Divide both sides by 900.
The rate of the current is 3 miles per hour. The check is left to the reader. MATCHED PROBLEM 6
EXAMPLE
7
D R
0002
A jetliner takes 1.2 times as long to fly from Paris to New York (3,600 miles) as to return. If the jet cruises at 550 miles per hour in still air, what is the average rate of the wind blowing in the direction of Paris from New York? 0002
A Quantity–Rate–Time Problem An advertising firm has an old computer that can prepare a whole mailing in 6 hours. With the help of a newer model the job is complete in 2 hours. How long would it take the newer model to do the job alone?
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SOLUTION
Let x 0004 time (in hours) for the newer model to do the whole job alone. a
Part of job completed b 0004 (Rate)(Time) in a given length of time 1 job per hour 6 1 Rate of new model 0004 job per hour x Rate of old model 0004
Part of job completed Part of job completed ° by old model ¢ 0003 ° by new model ¢ 0004 1 whole job in 2 hours in 2 hours Rate of Time of Rate of Time of Recall: Q ⴝ RT a ba b0003a ba b00041 old model old model new model new model 1 (2) 6 1 3
0003
1 (2) x
00041
0003
2 x
00041
1 3x a b 3
0003
2 3x a b x
0004 3x
x
0003
6
0004 3x
6
0004 2x
3
0004x
x cannot be zero.
Multiply both sides by 3x, the LCD.
Subtract x from both sides. Divide both sides by 2.
Therefore, the new computer could do the job alone in 3 hours. CHECK
MATCHED PROBLEM 7
Part of job completed by old model in 2 hours 0004 2(16) 0004 13 0003 Part of job completed by new model in 2 hours 0004 2(13) 0004 23 Part of job completed by both models in 2 hours 0004 1
0002
Two pumps are used to fill a water storage tank at a resort. One pump can fill the tank by itself in 9 hours, and the other can fill it in 6 hours. How long will it take both pumps operating together to fill the tank? 0002
Z Solving Mixture Problems A variety of applications can be classified as mixture problems. Even though the problems come from different areas, their mathematical treatment is essentially the same.
EXAMPLE
8
A Mixture Problem How many liters of a mixture containing 80% alcohol should be added to 5 liters of a 20% solution to yield a 30% solution?
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SOLUTION
Linear Equations and Applications
53
Let x 0004 amount of 80% solution used. BEFORE MIXING 80% solution
AFTER MIXING 30% solution
20% solution
0003
0004 5 liters
x liters
(x 0003 5) liters
Amount of Amount of Amount of ° alcohol in ¢ 0003 ° alcohol in ¢ 0004 ° alcohol in ¢ first solution second solution mixture 0.8x 0003 0.2(5) 0004 0.3(x 0003 5) 0.8x 0003 1 0004 0.3x 0003 1.5 0.5x 0004 0.5 x00041 Add 1 liter of the 80% solution. CHECK First solution Second solution Mixture
Liters of solution
Liters of alcohol 0.8(1) 0004 0.8 0003 0.2(5) 0004 1 1.8
1 00035 6
Percent alcohol 80 or 0.8兾1 20 or 1兾5 1.8兾6 0004 0.3, or 30%
0002 MATCHED PROBLEM 8
A chemical storeroom has a 90% acid solution and a 40% acid solution. How many centiliters of the 90% solution should be added to 50 centiliters of the 40% solution to yield a 50% solution? 0002 ANSWERS TO MATCHED PROBLEMS 1. x 0004 5 2. F 0004 95C 0003 32 3. 3, 5, 7 4. 20 centimeters 5. 50 minutes 6. 50 miles per hour 7. 3.6 hours 8. 12.5 centiliters
1-1
Exercises
1. What does it mean to solve an equation? 2. Describe the difference between an equation and an expression. 3. How can you tell if an equation is linear?
4. In one or two sentences, describe what parts 1– 4 in Theorem 1 say about working with equations. 5. How can you check your solution to an equation? 6. How do you check your solution to a word problem?
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7. Explain why the following does not make sense: Solve the equation P 0002 2l 0003 2w.
41.
1 1 1 0003 for f (simple lens formula) 0002 f d1 d2
8. Explain why the following does not make sense: Solve y y 0004 0003 1. 4 5
42.
1 1 1 0003 0002 for R1 (electric circuit) R R1 R2
44. A 0002 2ab 0003 2ac 0003 2bc for c
In Problems 9–34, solve each equation. 9. 10 x 0004 7 0002 4x 0004 25
10. 11 0003 3y 0002 5y 0004 5
11. 3(x 0003 2) 0002 5(x 0004 6)
12. 3(y 0004 4) 0003 2y 0002 18
13. 5 0003 4(t 0004 2) 0002 2(t 0003 7) 0003 1 14. 4 0004 3(t 0003 2) 0003 t 0002 5(t 0004 1) 0004 7t 15. 5 0004 17.
3a 0004 4 7 0004 2a 0002 5 2
16. 5 0004
x00033 x00044 3 0004 0002 4 2 8
18.
43. A 0002 2ab 0003 2ac 0003 2bc for a (surface area of a rectangular solid)
2x 0004 1 x00032 0002 4 3
x 3x 0004 1 6x 0003 5 0003 0002 5 2 4
45. y 0002
21. 0.35(s 0003 0.34) 0003 0.15s 0002 0.2s 0004 1.66 22. 0.35(u 0003 0.34) 0004 0.15u 0002 0.2u 0004 1.66 2 2 5 23. 0003 0002 4 0004 y 2 3y 25. 27.
z 1 0002 00032 z00041 z00041
30003w 1 4 0002 0003 24. 6w 2w 3 26.
y y 0004 10 2y 0004 2 0003 0002 00043 3 5 4
29. 1 0004
x00043 2x 0004 3 0002 x00042 x00042
6 5 000310002 31. y00034 2y 0003 8
t 2 0002 00032 t00041 t00041 28.
30.
z00034 z z00038 0003 0002 00035 7 6 3
2x 0004 3 3x 0004 1 000220004 x00031 x00031
4y 12 000350002 32. y00043 y00043
33.
3 3 3a 0004 1 0004 2 0002 a a2 0003 4a 0003 4 a 0003 2a
34.
1 1 10 0002 0004 2 b00045 b 0003 5 b 0004 5b 0003 25
46. x 0002
47.
x 2x 0004 3 000340002 x00043 x00043 x 0003 4x 0004 12 0002 2x 0004 3 x00023
1 x 00023 49. 1 10003 x
x0004
x0004
x000310004 51. 10004
2 x
1 x
50.
2.34 5.67x 0003 5.67 0002 x x00034
x2 0003 4x 0004 3 x2 0003 1 0002 x00041 x00041 x2 0003 1 0002 x2 0003 4x 0004 3 x00021
00021
0002x00032
3 y x 0002a b 10004y 10004x
a
53. Solve for x in terms of y: y 0002 10003
b x0003c
54. Let m and n be real numbers with m larger than n. Then there exists a positive real number p such that m 0002 n 0003 p. Find the fallacy in the following argument: m0002n0003p (m 0004 n)m 0002 (m 0004 n)(n 0003 p) m2 0004 mn 0002 mn 0003 mp 0004 n2 0004 np 2
m 0004 mn 0004 mp 0002 mn 0004 n2 0004 np m(m 0004 n 0004 p) 0002 n(m 0004 n 0004 p) m0002n
36. 1.73y 0003 0.279(y 0004 3) 0002 2.66y 38.
1 x
2 x000310004 x
52. Solve for y in terms of x:
35. 3.142x 0004 0.4835(x 0004 4) 0002 6.795
2.32x 3.76 0004 0002 2.32 x x00042
48.
In Problems 49–51, solve the equation.
In Problems 35–38, use a calculator to solve each equation to three significant digits.*
37.
3y 0003 2 for y y00043
In Problems 47 and 48, imagine that the indicated “solutions” were given to you by a student whom you were tutoring in this class. Is the solution right or wrong? If the solution is wrong, explain what is wrong and show a correct solution.
19. 0.1(t 0003 0.5) 0003 0.2t 0002 0.3(t 0004 0.4) 20. 0.1(w 0003 0.5) 0003 0.2w 0002 0.2(w 0004 0.4)
2x 0004 3 for x 3x 0003 5
APPLICATIONS These problems are grouped according to subject area.
In Problems 39–46, solve for the indicated variable in terms of the other variables. 39. an 0002 a1 0003 (n 0004 1)d for d (arithmetic progressions) 40. F 0002 95C 0003 32 for C (temperature scale) *Appendix A contains a brief discussion of significant digits.
Numbers 55. Find a number so that 10 less than two-thirds the number is one-fourth the number. 56. Find a number so that 6 more than one-half the number is twothirds the number.
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57. Find four consecutive even integers so that the sum of the first three is 2 more than twice the fourth. 58. Find three consecutive even integers so that the first plus twice the second is twice the third. Geometry 59. Find the perimeter of a triangle if one side is 16 feet, another side is two-sevenths the perimeter, and the third side is one-third the perimeter. 60. Find the perimeter of a triangle if one side is 11 centimeters, another side is two-fifths the perimeter, and the third side is onehalf the perimeter. 61. A new game show requires a playing field with a perimeter of 54 yards and length 3 yards less than twice the width. What are the dimensions? 62. A celebrity couple wants to have a rectangular pool put in the backyard of their vacation home. They want it to be 24 meters long, and they insist that it have at least as much area as the neighbor’s pool, which is a square 12 meters on a side. Find the dimensions of the smallest pool that meets these criteria. Business and Economics 63. The sale price of an MP3 player after a 30% discount was $140. What was the original price? 64. A sporting goods store marks up each item it sells 60% above wholesale price. What is the wholesale price on a snowboard that sells for $144? 65. One employee of a computer store is paid a base salary of $2,150 a month plus an 8% commission on all sales over $7,000 during the month. How much must the employee sell in 1 month to earn a total of $3,170 for the month? 66. A second employee of the computer store in Problem 65 is paid a base salary of $1,175 a month plus a 5% commission on all sales during the month. (A) How much must this employee sell in 1 month to earn a total of $3,170 for the month? (B) Determine the sales level where both employees receive the same monthly income. If employees can select either of these payment methods, how would you advise an employee to make this selection? Earth Science 67. In 1970, Russian geologists began drilling a very deep borehole in the Kola Peninsula. Their goal was to reach a depth of 15 kilometers, but high temperatures in the borehole forced them to stop in 1994 after reaching a depth of 12 kilometers. They found that below 3 kilometers the temperature T increased 2.5°C for each additional 100 meters of depth. (A) If the temperature at 3 kilometers is 30°C and x is the depth of the hole in kilometers, write an equation using x that will give the temperature T in the hole at any depth beyond 3 kilometers. (B) What would the temperature be at 12 kilometers? (C) At what depth (in kilometers) would they reach a temperature of 200°C?
Linear Equations and Applications
55
68. An earthquake emits a primary wave and a secondary wave. Near the surface of the Earth the primary wave travels at about 5 miles per second, and the secondary wave travels at about 3 miles per second. From the time lag between the two waves arriving at a given seismic station, it is possible to estimate the distance to the quake. Suppose a station measures a time difference of 12 seconds between the arrival of the two waves. How far is the earthquake from the station? (The epicenter can be located by obtaining distance bearings at three or more stations.) Life Science 69. The kangaroo rat is an endangered species native to California. In order to keep track of their population size in a state nature preserve, a conservation biologist trapped, tagged, and released 80 individuals from the population. After waiting 2 weeks for the animals to mix back in with the general population, she again caught 80 individuals and found that 22 of them were tagged. Assuming that the ratio of tagged animals to total animals in the second sample is the same as the ratio of all tagged animals to the total population in the preserve, estimate the total number of kangaroo rats in the preserve. 70. Repeat Problem 69 with a first (marked) sample of 70 and a second sample of 30 with only 11 marked animals. Chemistry 71. How many gallons of distilled water must be mixed with 50 gallons of 30% alcohol solution to obtain a 25% solution? 72. How many gallons of hydrochloric acid must be added to 12 gallons of a 30% solution to obtain a 40% solution? 73. A chemist mixes distilled water with a 90% solution of sulfuric acid to produce a 50% solution. If 5 liters of distilled water are used, how much 50% solution is produced? 74. A fuel oil distributor has 120,000 gallons of fuel with 0.9% sulfur content, which exceeds pollution control standards of 0.8% sulfur content. How many gallons of fuel oil with a 0.3% sulfur content must be added to the 120,000 gallons to obtain fuel oil that will comply with the pollution control standards? Rate–Time 75. An old computer can do the weekly payroll in 5 hours. A newer computer can do the same payroll in 3 hours. The old computer starts on the payroll, and after 1 hour the newer computer is brought on-line to work with the older computer until the job is finished. How long will it take both computers working together to finish the job? (Assume the computers operate independently.) 76. One pump can fill a gasoline storage tank in 8 hours. With a second pump working simultaneously, the tank can be filled in 3 hours. How long would it take the second pump to fill the tank operating alone? 77. The cruising speed of an airplane is 150 miles per hour (relative to the ground). You plan to hire the plane for a 3-hour sightseeing trip. You instruct the pilot to fly north as far as she can and still return to the airport at the end of the allotted time. (A) How far north should the pilot fly if the wind is blowing from the north at 30 miles per hour? (B) How far north should the pilot fly if there is no wind?
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78. Suppose you are at a river resort and rent a motor boat for 5 hours starting at 7 A.M. You are told that the boat will travel at 8 miles per hour upstream and 12 miles per hour returning. You decide that you would like to go as far up the river as you can and still be back at noon. At what time should you turn back, and how far from the resort will you be at that time?
82. A minor chord is composed of notes whose frequencies are in the ratio 10:12:15. If the first note of a minor chord is A, with a frequency of 220 hertz, what are the frequencies of the other two notes?
79. A two-woman rowing team can row 1,200 meters with the current in a river in the same amount of time it takes them to row 1,000 meters against that same current. In each case, their average rowing speed without the effect of the current is 3 meters per second. Find the speed of the current.
83. In an experiment on motivation, Professor Brown trained a group of rats to run down a narrow passage in a cage to receive food in a goal box. He then put a harness on each rat and connected it to an overhead wire attached to a scale. In this way, he could place the rat different distances from the food and measure the pull (in grams) of the rat toward the food. He found that the relationship between motivation (pull) and position was given approximately by the equation
80. The winners of the men’s 1,000-meter double sculls event in the 2008 Olympics rowed at an average of 11.3 miles per hour. If this team were to row this speed for a half mile with a current in 80% of the time they were able to row that same distance against the current, what would be the speed of the current? Music 81. A major chord in music is composed of notes whose frequencies are in the ratio 4:5:6. If the first note of a chord has a frequency of 264 hertz (middle C on the piano), find the frequencies of the other two notes. [Hint: Set up two proportions using 4:5 and 4:6.]
Psychology
p 0004 000215d 0003 70
30 d 170
where pull p is measured in grams and distance d in centimeters. When the pull registered was 40 grams, how far was the rat from the goal box? 84. Professor Brown performed the same kind of experiment as described in Problem 83, except that he replaced the food in the goal box with a mild electric shock. With the same kind of apparatus, he was able to measure the avoidance strength relative to the distance from the object to be avoided. He found that the avoidance strength a (measured in grams) was related to the distance d that the rat was from the shock (measured in centimeters) approximately by the equation a 0004 000243d 0003 230
30 d 170
If the same rat were trained as described in this problem and in Problem 83, at what distance (to one decimal place) from the goal box would the approach and avoidance strengths be the same? (What do you think the rat would do at this point?)
1-2
Linear Inequalities Z Understanding Inequality and Interval Notation Z Solving Linear Inequalities Z Applying Linear Inequalities
An equation is a statement that two expressions are equal. Sometimes it is useful to find when one expression is more or less than another, so in this section we turn our attention to linear inequalities in one variable, like 3x 0003 5 7 x 0002 10
and
00024 6 3 0002 2x 6 7
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57
Z Understanding Inequality and Interval Notation The preceding mathematical statements use the inequality, or order, relations, more commonly known as “greater than” and “less than.” Just as we use the symbol “0004” to replace the words “is equal to,” we use the inequality symbols and to replace “is less than” and “is greater than,” respectively. You probably have a natural understanding of how to compare numbers using these symbols, but to be precise about using inequality symbols, we should have a clear definition of what they mean.
Z DEFINITION 1 a < b and b > a For two real numbers a and b, we say that a is less than b, and write a b, if there is a positive real number p so that a 0003 p 0004 b. The statement b a, read b is greater than a, means exactly the same as a b.
This definition basically says that if you add a positive number to any number, the sum is larger than the original number. When we write a b we mean a 6 b or a 0004 b and say a is less than or equal to b. When we write a b we mean a 7 b or a 0004 b and say a is greater than or equal to b. The inequality symbols 6 and 7 have a very clear geometric interpretation on the real number line. If a 6 b, then a is to the left of b; if c 7 d, then c is to the right of d (Fig. 1). This is called a line graph.
a
d
b
c
Z Figure 1 a b, c d.
If we want to state that some number x is between a and b, we could use two inequalities: x a and x b. Instead, we will write one double inequality, a x b. For example, the inequality 00022 x 5 indicates that x is between 00022 and 5, and could be equal to 5, but not 00022. The set of all real numbers that satisfy this inequality is called an interval, and is commonly represented by (00022, 5]. In general, (a, b] 0004 5x ƒ a 6 x b6*
The number a is called the left endpoint of the interval, and the symbol “(” indicates that a is not included in the interval. The number b is called the right endpoint of the interval, and the symbol “]” indicates that b is included in the interval. An interval is closed if it contains its endpoint(s) and open if it does not contain any endpoint. Other types of intervals of real numbers are shown in Table 1. Note that the symbol “ ,” read “infinity,” used in Table 1 is not a numeral. When we write [b, ), we are simply referring to the interval starting at b and continuing indefinitely to the right. We would never write [b, ] or b x , because cannot be used as an endpoint of an interval. The interval (0002 , ) represents the set of real numbers R, since its graph is the entire real number line. *In general, 5x ƒ P(x)6 represents the set of all x such that statement P(x) is true. To express this set verbally, just read the vertical bar as “such that.”
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Table 1 Interval Notation Interval notation
Inequality notation
[a, b]
aⱕxⱕb
[a, b)
aⱕx⬍b
Line graph
(a, b]
a⬍xⱕb
(a, b)
a⬍x⬍b
[b, ⬁)
xⱖb
(b, ⬁)
x⬎b
(⫺⬁, a]
xⱕa
(⫺⬁, a)
x⬍a
Type
[
]
x
Closed
)
x
Half-open
]
x
Half-open
)
x
Open
[
x
Closed*
(
x
Open
]
x
Closed*
)
x
Open
a
b
[
a
b
(
a
b
(
a
b b b a a
*These intervals are closed because they contain all of their endpoints; they have only one endpoint.
ZZZ
It is important to note that
CAUTION ZZZ
5 7 x ⱖ ⫺3
is equivalent to [⫺3, 5) and not to (5, ⫺3]
In interval notation, the smaller number is always written to the left. It may be useful to rewrite the inequality as ⫺3 ⱕ x 6 5 before rewriting it in interval notation. The symbol (5, ⫺3] is meaningless.
EXAMPLE
1
Graphing Intervals and Inequalities Write each of the following in inequality notation and graph on a real number line: (A) [⫺2, 3)
SOLUTIONS
(C) [ ⫺2, ⬁)
(B) (⫺4, 2)
(A) ⫺2 ⱕ x 6 3 (B) ⫺4 6 x 6 2 (C) x ⱖ ⫺2 (D) x 6 3
[
⫺5
⫺2
(
⫺5 ⫺4 ⫺5
[
⫺5
MATCHED PROBLEM 1
3
x
5
)
2
x
5
0
x
5 0
) 3
5
0002
x
Write each of the following in interval notation and graph on a real number line: (A) ⫺3 6 x ⱕ 3
ZZZ EXPLORE-DISCUSS 1
)
0 0
⫺2
(D) (⫺⬁, 3)
(B) 2 ⱖ x ⱖ ⫺1
(C) x 7 1
(D) x ⱕ 2
Example 1C shows the graph of the inequality x ⱖ ⫺2. What is the graph of x 6 ⫺2? What is the corresponding interval? Describe the relationship between these sets.
0002
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59
Since intervals are sets of real numbers, the set operations of union and intersection are often useful when working with intervals. The union of sets A and B, denoted by A 傼 B, is the set formed by combining all the elements of A and all the elements of B. The intersection of sets A and B, denoted by A 傽 B, is the set of elements of A that are also in B. Symbolically: Z DEFINITION 2 Union and Intersection A 傼 B 0004 5x ƒ x is in A or x is in B6
Union:
{1, 2, 3} ´ {2, 3, 4, 5} ⴝ {1, 2, 3, 4, 5}
Intersection: A 傽 B 0004 5x ƒ x is in A and x is in B6 {1, 2, 3} 傽 {2, 3, 4, 5} ⴝ {2, 3}
EXAMPLE
2
Graphing the Union and Intersection of Intervals If A 0004 (00022, 5] and B 0004 (1, ), graph the sets A 傼 B and A 傽 B and write them in interval notation. )
00022
1
00022
1
5
1
5
)
[
)
00022
ZZZ EXPLORE-DISCUSS 2
5
)
00022
MATCHED PROBLEM 2
[
SOLUTION
1
5
x
A 0004 (00022, 5]
x
B 0004 (1, )
x
A 傼 B 0004 (00022, )
x
A 傽 B 0004 (1, 5]
0002
If C 0004 [ 00024, 3) and D 0004 (0002 , 00021] , graph the sets C 傼 D and C 傽 D and write them in interval notation. 0002 Replace ? with 6 or 7 in each of the following. (A) 00021 ? 3
and
2(00021) ? 2(3)
(B) 00021 ? 3
and
00022(00021) ? 00022(3)
(C) 12 ? 00028
and
12 00028 ? 4 4
(D) 12 ? 00028
and
12 00028 ? 00024 00024
Based on your results, describe verbally the effect of multiplying or dividing both sides of an inequality by a number.
Z Solving Linear Inequalities We now turn to the problem of solving linear inequalities in one variable, such as 2(2x 0003 3) 6 6(x 0002 2) 0003 10
and
00023 6 2x 0003 3 9
The solution set for an inequality is the set of all values of the variable that make the inequality a true statement. Each element of the solution set is called a solution of the inequality. To solve an inequality is to find its solution set. Two inequalities are equivalent
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if they have the same solution set. Just as with equations, we perform operations on inequalities that produce simpler equivalent inequalities, and continue the process until an inequality is reached whose solution is obvious. The properties of inequalities given in Theorem 1 can be used to produce equivalent inequalities. Z THEOREM 1 Inequality Properties An equivalent inequality will result and the sense (or direction) will remain the same if each side of the original inequality • Has the same real number added to or subtracted from it • Is multiplied or divided by the same positive number An equivalent inequality will result and the sense (or direction) will reverse if each side of the original inequality • Is multiplied or divided by the same negative number Note: Multiplication by 0 and division by 0 are not permitted.
Theorem 1 tells us that we can perform essentially the same operations on inequalities that we perform on equations, with the exception that the sense (or direction) of the inequality reverses if we multiply or divide both sides by a negative number: Otherwise the sense of the inequality does not change. Now let’s see how the inequality properties are used to solve linear inequalities. Examples 3, 4, and 5 will illustrate the process.
EXAMPLE
3
Solving a Linear Inequality Solve and graph: 2(2x 0003 3) 0002 10 6 6(x 0002 2)
SOLUTION
2(2x 0003 3) 0002 10 6 6(x 0002 2) 4x 0003 6 0002 10 6 6x 0002 12
Multiply out parentheses. Combine like terms.
4x 0002 4 6 6x 0002 12
Add 4 to both sides.
4x 0002 4 0003 4 6 6x 0002 12 0003 4 4x 6 6x 0002 8
Subtract 6x from both sides.
4x 0002 6x 6 6x 0002 8 0002 6x 00022x 6 00028
Divide both sides by ⴚ2. Note that direction reverses because ⴚ2 is negative.
00022x 00028 7 00022 00022 x 7 4 2
3
( 4
5
6
7
8
(4, )
or 9
x
Graph of solution set
MATCHED PROBLEM 3
Solution set
0002
Solve and graph: 3(x 0002 1) 5(x 0003 2) 0002 5 0002
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EXAMPLE
4
61
Solving a Linear Inequality Involving Fractions Solve and graph:
4x 2x 0002 3 00036 20003 4 3
2x 0002 3 4x 00036 20003 4 3
SOLUTION
12 ⴢ
Multiply both sides by 12, the LCD.
2x 0002 3 4x 0003 12 ⴢ 6 12 ⴢ 2 0003 12 ⴢ 4 3
Direction doesn’t change: we multiplied by a positive number.
3(2x 0002 3) 0003 72 24 0003 4(4x) 6x 0002 9 0003 72 24 0003 16x 6x 0003 63 24 0003 16x 6x 000239 0003 16x 000210x 000239 x 3.9 ]
3.9
MATCHED PROBLEM 4
Linear Inequalities
Solve and graph:
or
Multiply out parentheses. Combine like terms. Subtract 63 from both sides. Subtract 16x from both sides.
(0002 , 3.9]
Order reverses when both sides are divided by ⴚ10, a negative number.
x
0002
Graph of solution set
3x 4x 0002 3 00038 6 60003 3 2 0002
EXAMPLE
5
Solving a Double Inequality Solve and graph: 00023 4 0002 7x 6 18
SOLUTION
We proceed as before, except we try to isolate x in the middle with a coefficient of 1, being sure to perform operations on all three parts of the inequality. 00023 4 0002 7x 6 18
Subtract 4 from each member.
00023 0002 4 4 0002 7x 0002 4 6 18 0002 4 00027 00027x 6 14
Divide each member by ⴚ7 and reverse each inequality.
00027x 14 00027 7 00027 00027 00027 1 x 7 00022 (
00022
MATCHED PROBLEM 5
00022 6 x 1
or ]
1
x
or
Graph of solution set
(00022, 1] 0002
Solve and graph: 00023 6 7 0002 2x 7 0002
Z Applying Linear Inequalities to Chemistry EXAMPLE
6
Chemistry In a chemistry experiment, a solution of hydrochloric acid is to be kept between 30°C and 35°C—that is, 30 C 35. What is the range in temperature in degrees Fahrenheit if the Celsius/Fahrenheit conversion formula is C 0004 59 (F 0002 32)?
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30 C 35 5 30 (F 0002 32) 35 9
SOLUTION
Replace C with
5 (F ⴚ 32). 9
9 Multiply each member by 5 to clear fractions.
9 9 5 9 ⴢ 30 ⴢ (F 0002 32) ⴢ 35 5 5 9 5 54 F 0002 32 63
Add 32 to each member.
54 0003 32 F 0002 32 0003 32 63 0003 32 86 F 95 The range of the temperature is from 86°F to 95°F, inclusive. MATCHED PROBLEM 6
0002
A film developer is to be kept between 68°F and 77°F—that is, 68 F 77. What is the range in temperature in degrees Celsius if the Celsius/Fahrenheit conversion formula is F 0004 95C 0003 32? 0002
ANSWERS TO MATCHED PROBLEMS
(B) [ 00021, 2]
[
] 2
( ]
) 3
00021
3
3. x 00024 or (0002 , 00024] 5
2
x x x
5
0
00021
5 5
0 1
00025
4. x 7 6 or (6, )
3
00021 0
[
[
00024
0
00025
(D) (0002 , 2]
00024
00023
00025
(C) (1, )
2.
)
00025
[
1. (A) (00023, 3]
5
x
x
C 傼 D 0004 (0002 , 3)
x
C 傽 D 0004 [ 00024, 00021] ]
00027
00024
( 6
0 12
5. 5 7 x 0 or 0 x 6 5 or [0, 5)
00021
[ 0
x
x
) 5
6
x
6. 20 C 25: the range in temperature is from 20°C to 25°C
1-2
Exercises
1. Explain in your own words what it means to solve an inequality.
3. What is the main difference between the procedures for solving linear equations and linear inequalities?
2. Explain why the “interval” [5, 00023) is meaningless.
4. Describe how to graph the solution set of an inequality.
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45. [ ⫺1, 4) 傽 (2, 6]
46. [⫺1, 4) 傼 (2, 6]
47. (⫺⬁, 1) 傼 (⫺2, ⬁)
48. (⫺⬁, 1) 傽 (2, ⬁)
49. (⫺⬁, ⫺1) 傼 [3, 7)
50. (1, 6] ´ [9, ⬁)
51. [2, 3] 傼 (1, 5)
52. [2, 3] 傽 (1, 5)
In Problems 11–16, rewrite in interval notation and graph on a real number line.
53. (⫺⬁, 4) 傼 (⫺1, 6]
54. (⫺3, 2) 傼 [0, ⬁)
11. ⫺2 6 x ⱕ 6
12. ⫺5 ⱕ x ⱕ 5
13. ⫺7 6 x 6 8
In Problems 55–70, solve and graph.
14. ⫺4 ⱕ x 6 5
15. x ⱕ ⫺2
16. x 7 3
55.
q q⫺4 ⫺3 7 ⫹1 7 3
57.
2x 1 3 2x ⫺ (x ⫺ 3) ⱕ ⫺ (x ⫹ 2) 5 2 3 10
In Problems 5–10, rewrite in inequality notation and graph on a real number line. 5. [ ⫺8, 7 ]
6. (⫺4, 8)
7. [ ⫺6, 6)
8. (⫺3, 3 ]
9. [⫺6, ⬁)
10. (⫺⬁, 7)
In Problems 17–20, write in interval and inequality notation. 17. 18. 19. 20.
⫺10 ⫺10
[
[
⫺5
⫺10
⫺10
⫺5
0 0
5 5
]
⫺5
⫺5
)
0
0
(
] 5
5
x
10 10
x x
10
10
x
In Problems 21–28, replace each ? with ⬎ or ⬍ to make the resulting statement true. and
12 ⫹ 5 ? 6 ⫹ 5
22. ⫺4 ? ⫺2
and
⫺4 ⫺ 7 ? ⫺2 ⫺ 7
23. ⫺6 ? ⫺8
and
⫺6 ⫺ 3 ? ⫺8 ⫺ 3
21.
12 ? 6
56.
2 62. 24 ⱕ (x ⫺ 5) 6 36 3 63. 16 6 7 ⫺ 3x ⱕ 31 64. 19 ⱕ 7 ⫺ 6x 6 49 1 65. ⫺8 ⱕ ⫺ (2 ⫺ x) ⫹ 3 6 10 4
4?9
and
4⫹2?9⫹2
25.
2 ? ⫺1
and
⫺2(2) ? ⫺2(⫺1)
66. 0 6
and
4(⫺3) ? 4(2)
67. 0.1(x ⫺ 7) 6 0.8 ⫺ 0.05x
27.
2?6
28. ⫺10 ? ⫺15
and
2 6 ? 2 2
and
⫺10 ⫺15 ? 5 5
In Problems 29–42, solve and graph. 29. 7x ⫺ 8 6 4x ⫹ 7
30. 5x ⫺ 21 ⱖ 3x ⫹ 5
31. 12 ⫺ y ⱖ 2(9 ⫺ 2y)
32. 4(y ⫹ 1) ⫺ 7 6 ⫺9 ⫺ 2y
N 33. 7 4 ⫺2
Z 34. ⱕ3 ⫺10
35. ⫺5t 6 ⫺10
36. ⫺20m ⱖ 100
37. 3 ⫺ m 6 4(m ⫺ 3)
38. 6(5 ⫺ 2k) ⱖ 6 ⫺ 8k
39. ⫺2 ⫺
B 1⫹B ⱕ 4 3
41. ⫺4 6 5t ⫹ 6 ⱕ 21
40.
t t⫺2 ⫹2 7 5 3
42. ⫺2 ⱕ 4r ⫺ 14 6 2
In Problems 43–54, graph the indicated set and write as a single interval, if possible. 43. (⫺5, 5) 傼 [ 4, 7 ]
44. (⫺5, 5) 傽 [4, 7]
p⫺2 p p ⫺ ⱕ ⫺4 3 2 4
x 1 x 2 58. (x ⫹ 7) ⫺ 7 (3 ⫺ x) ⫹ 3 4 2 6 9 4 59. ⫺4 ⱕ x ⫹ 32 ⱕ 68 60. 2 ⱕ z ⫹ 6 6 18 5 5 5 61. ⫺20 6 (4 ⫺ x) 6 ⫺5 2
24.
26. ⫺3 ? 2
63
1 (4 ⫺ x) ⫺ 10 ⱕ 16 3
68. 0.4(x ⫹ 5) 7 0.3x ⫹ 17 69. 0.3x ⫺ 2.04 ⱖ 0.04(x ⫹ 1) 70. 0.02x ⫺ 5.32 ⱕ 0.5(x ⫺ 2) Problems 71–76 are calculus-related. For what real number(s) x does each expression represent a real number? 71. 11 ⫺ x
72. 1x ⫹ 5
73. 13x ⫹ 5
74. 17 ⫺ 2x
75.
1
76.
4
22x ⫹ 3
1 4
25 ⫺ 6x
77. What can be said about the signs of the numbers a and b in each case? (A) ab 7 0 (B) ab 6 0 a a (C) 7 0 (D) 6 0 b b 78. What can be said about the signs of the numbers a, b, and c in each case? ab (A) abc 7 0 (B) 6 0 c (C)
a 7 0 bc
(D)
a2 6 0 bc
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79. Replace each question mark with 6 or 7, as appropriate: (A) If a 0002 b 0004 1, then a ? b. (B) If u 0002 v 0004 00022, then u ? v. 80. For what p and q is p 0003 q 6 p 0002 q? 81. If both a and b are negative numbers and b0006a is greater than 1, then is a 0002 b positive or negative? 82. If both a and b are positive numbers and b0006a is greater than 1, then is a 0002 b positive or negative? 83. Indicate true (T) or false (F): (A) If p 7 q and m 7 0, then mp 6 mq. (B) If p 6 q and m 6 0, then mp 7 mq. (C) If p 7 0 and q 6 0, then p 0003 q 7 q. 84. Assume that m 7 n 7 0; then mn 7 n2 mn 0002 m2 7 n2 0002 m2 m(n 0002 m) 7 (n 0003 m)(n 0002 m) m 7 n0003m 0 7 n But it was assumed that n 7 0. Find the error. Prove each inequality property in Problems 85–88, given a, b, and c are arbitrary real numbers. 85. If a 6 b, then a 0003 c 6 b 0003 c. 86. If a 6 b, then a 0002 c 6 b 0002 c. 87. (A) If a 6 b and c is positive, then ca 6 cb. (B) If a 6 b and c is negative, then ca 7 cb. b a 6 . c c a b (B) If a 6 b and c is negative, then 7 . c c
88. (A) If a 6 b and c is positive, then
APPLICATIONS Write all your answers using inequality notation. 89. EARTH SCIENCE In 1970, Russian geologists began drilling a very deep borehole in the Kola Peninsula. Their goal was to reach a depth of 15 kilometers, but high temperatures in the borehole forced them to stop in 1994 after reaching a depth of 12 kilometers. They found that the approximate temperature x kilometers below the surface of the Earth is given by T 0004 30 0003 25(x 0002 3)
3 x 12
where T is temperature in degrees Celsius. At what depth is the temperature between 150°C and 250°C, inclusive? 90. EARTH SCIENCE As dry air moves upward it expands, and in so doing it cools at a rate of about 5.5°F for each 1,000-foot rise up to about 40,000 feet. If the ground temperature is 70°F, then the temperature T at height h is given approximately by T 0004 70 0002 0.0055h.
For what range in altitude will the temperature be between 26°F and 000240°F, inclusive? 91. BUSINESS AND ECONOMICS An electronics firm is planning to market a new graphing calculator. The fixed costs are $650,000 and the variable costs are $47 per calculator. The wholesale price of the calculator will be $63. For the company to make a profit, it is clear that revenues must be greater than costs. (A) How many calculators must be sold for the company to make a profit? (B) How many calculators must be sold for the company to break even? (C) Discuss the relationship between the results in parts A and B. 92. BUSINESS AND ECONOMICS A video game manufacturer is planning to market a handheld version of its game machine. The fixed costs are $550,000 and the variable costs are $120 per machine. The wholesale price of the machine will be $140. (A) How many game machines must be sold for the company to make a profit? (B) How many game machines must be sold for the company to break even? (C) Discuss the relationship between the results in parts A and B. 93. BUSINESS AND ECONOMICS The electronics firm in Problem 91 finds that rising prices for parts increases the variable costs to $50.50 per calculator. (A) Discuss possible strategies the company might use to deal with this increase in costs. (B) If the company continues to sell the calculators for $63, how many must they sell now to make a profit? (C) If the company wants to start making a profit at the same production level as before the cost increase, how much should they increase the wholesale price? 94. BUSINESS AND ECONOMICS The video game manufacturer in Problem 92 finds that unexpected programming problems increases the fixed costs to $660,000. (A) Discuss possible strategies the company might use to deal with this increase in costs. (B) If the company continues to sell the game machines for $140, how many must they sell now to make a profit? (C) If the company wants to start making a profit at the same production level as before the cost increase, how much should they increase the wholesale price? 95. ENERGY If the power demands in a 110-volt electric circuit in a home vary between 220 and 2,750 watts, what is the range of current flowing through the circuit? (W 0004 EI, where W 0004 Power in watts, E 0004 Pressure in volts, and I 0004 Current in amperes.) 96. PSYCHOLOGY A person’s IQ is given by the formula IQ 0004
MA 100 CA
where MA is mental age and CA is chronological age. If 80 IQ 140 for a group of 12-year-old children, find the range of their mental ages.
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65
Absolute Value in Equations and Inequalities Z Relating Absolute Value and Distance Z Solving Absolute Value Equations and Inequalities Z Using Absolute Value to Solve Radical Inequalities
We can express the distance between two points on a number line using the concept of absolute value. As a result, absolute values often appear in equations and inequalities that are associated with distance. In this section, we define absolute value and we show how to solve equations and inequalities that involve absolute value.
Z Relating Absolute Value and Distance We start with a geometric definition of absolute value. If a is the coordinate of a point on a real number line, then the distance from the origin to a is represented by |a| and is referred to as the absolute value of a. So |5| 0002 5, since the point with coordinate 5 is five units from the origin, and 冟 00036 冟 0002 6, since the point with coordinate 00036 is six units from the origin (Fig. 1).
兩00036兩 0002 6 00036
兩5兩 0002 5 0
5
x
Z Figure 1 Absolute value.
We can use symbols to write a formal definition of absolute value:
Z DEFINITION 1 Absolute Value 冟x冟 0002
再
0003x x
if x 6 0 if x 0004 0
For example, 冟00033冟
0002 0003(00033)
00023
For example, 冟4冟 0002 4
[Note: 0003x is positive if x is negative.]
Both the geometric and algebraic definitions of absolute value are useful, as will be seen in the material that follows. Remember: The absolute value of a number is never negative.
EXAMPLE
1
Finding Absolute Value Write without the absolute value sign: (A) 冟 0005 0003 3 冟
(B) 冟 3 0003 0005 冟
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(A) 冟 0005 0003 3 冟 0002 0005 0003 3
Because 0005 ⬇ 3.14, 0005 0003 3 is positive.
(B) 冟 3 0003 0005 冟 0002 0003(3 0003 0005) 0002 0005 0003 3
Because 3 0003 0005 is negative.
0002
Write without the absolute value sign: (A) 冟 8 冟
3 (B) 冟 29 0003 2 冟
(C) 冟 000312 冟
3 (D) 冟 2 0003 29 冟
0002
Notice that the solution in both parts of Example 1 was the same. This suggests Theorem 1, which will be proved in Problem 81.
Z THEOREM 1 For All Real Numbers a and b, 冟b 0003 a冟 0002 冟a 0003 b冟
To find the distance between two numbers, we subtract, larger minus smaller. But if we don’t know which is larger, we can use absolute value; Theorem 1 tells us that the order is immaterial.
Z DEFINITION 2 Distance Between Points A and B Let A and B be two points on a real number line with coordinates a and b, respectively. The distance between A and B is given by d(A, B) 0002 冟 b 0003 a 冟 This distance is also called the length of the line segment joining A and B.
It will come in very handy to observe that an expression like 冟 b 0003 a 冟 can always be interpreted as the distance between two numbers a and b, and that the order of the subtraction doesn’t matter.
Z Solving Absolute Value Equations and Inequalities The connection between algebra and geometry is an important tool when working with equations and inequalities involving absolute value. For example, the algebraic statement 冟x 0003 1冟 0002 2 can be interpreted geometrically as stating that the distance from x to 1 is 2.
ZZZ EXPLORE-DISCUSS 1
Write geometric interpretations of the following algebraic statements: (A) 冟 x 0003 1 冟 6 2
(B) 0 6 冟 x 0003 1 冟 6 2
(C) 冟 x 0003 1 冟 7 2
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EXAMPLE
2
Absolute Value in Equations and Inequalities
67
Solving Absolute Value Problems Geometrically Interpret geometrically, solve, and graph. For inequalities, write solutions in both inequality and interval notation.
SOLUTIONS
(A) 冟 x 0003 3 冟 0002 5
(B) 冟 x 0003 3 冟 6 5
(C) 0 6 冟 x 0003 3 冟 6 5
(D) 冟 x 0003 3 冟 7 5
(A) The expression |x 0003 3| represents the distance between x and 3, so the solutions to |x 0003 3| 0002 5 are all numbers that are exactly 5 units away from 3 on a number line. x 0002 3 5 0002 00032 or 8 The solution set is {00032, 8}. 5
This is not interval notation.
5
00032
3
x
8
(B) Solutions to |x 0003 3| 0006 5 are all numbers whose distance from 3 is less than 5. These are the numbers between 00032 and 8: 00032 6 x 6 8 The solution set is (00032, 8). (
00032
This is interval notation.
)
3
x
8
(C) Expressions like 0 0006 |x 0003 3| 0006 5 are important in calculus. The solutions are all numbers whose distance from 3 is less than 5, and is not zero. This excludes 3 itself from the solution set: 00032 6 x 6 8
x3
or
(00032, 3) 0004 (3, 8)
Hole
(
00032
)
3
x
8
(D) The solutions to 冟 x 0003 3 冟 7 5 are all numbers whose distance from 3 is greater than 5; that is, x 6 00032 )
00032
ZZZ
CAUTION ZZZ
3
or
x 7 8
(00030007, 00032) 0004 (8, 0007)
(
8
The pair of inequalities 00032 0006 x and x 0006 8 can be written as a double inequality: 00032 6 x 6 8 or in interval notation (00032, 8) But the pair x 0006 00032 or x 8 from Example 2(D) cannot be written as a double inequality, or as a single interval: no number is both less than 00032 and greater than 8.
0002
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MATCHED PROBLEM 2
Interpret geometrically, solve, and graph. For inequalities, write solutions in both inequality and interval notation. [Hint: |x ⫹ 2| ⫽ |x ⫺ (⫺2)|.] (A) 冟 x ⫹ 2 冟 ⫽ 6
(B) 冟 x ⫹ 2 冟 6 6
(C) 0 6 冟 x ⫹ 2 冟 6 6
(D) 冟 x ⫹ 2 冟 7 6
0002
The preceding results are summarized in Table 1. Table 1 Geometric Interpretation of Absolute Value Equations and Inequalities Form (d ⬎ 0)
冟x ⫺ c冟 ⫽ d
Geometric interpretation
Solution
Graph
c⫺d
冟x ⫺ c冟 6 d
Distance between x and c is less than d.
(c ⫺ d, c ⫹ d )
0 6 冟x ⫺ c冟 6 d
Distance between x and c is less than d, but x ⫽ c.
(c ⫺ d, c) 傼 (c, c ⫹ d )
Distance between x and c is greater than d.
(⫺⬁, c ⫺ d ) 傼 (c ⫹ d, ⬁)
冟x ⫺ c冟 7 d
EXAMPLE
3
d
5c ⫺ d, c ⫹ d6
Distance between x and c is equal to d.
(
c⫺d
(
c⫺d
)
c⫺d
d c
c⫹d
c
c⫹d
c
c⫹d
c
c⫹d
x
)
x
)
x
(
x
Interpreting Verbal Statements Algebraically Express each verbal statement as an absolute value equation or inequality. (A) x is 4 units from 2. (B) y is less than 3 units from ⫺5. (C) t is no more than 5 units from 7. (D) w is no less than 2 units from ⫺1.
SOLUTIONS
MATCHED PROBLEM 3
⫽
|x ⫺ 2| ⫽ 4
The distance from x to 2 is 4.
(B)
d( y, ⫺5) ⫽
| y ⫹ 5| 6 3
The distance from y to 00025 is 0003 3.
(C)
d(t, 7)
⫽
|t ⫺ 7| ⱕ 5
The distance from t to 7 is 0004 5.
(D) d(w, ⫺1) ⫽
|w ⫹ 1| ⱖ 2
The distance from w to 00021 is 0004 2.
(A) d(x, 2)
0002
Express each verbal statement as an absolute value equation or inequality. (A) x is 6 units from 5. (B) y is less than 7 units from ⫺6. (C) w is no less than 3 units from ⫺2. (D) t is no more than 4 units from 3. 0002
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ZZZ EXPLORE-DISCUSS 2
Absolute Value in Equations and Inequalities
69
Describe the set of numbers that satisfies each of the following: (A) 2 7 x 7 1
(B) 2 7 x 6 1
(C) 2 6 x 7 1
(D) 2 6 x 6 1
Explain why we never write double inequalities with inequality symbols pointing in different directions.
The results of Example 2 can be generalized as Theorem 2. [Note: |x| 0002 |x 0003 0|.]
Z THEOREM 2 Properties of Equations and Inequalities Involving 冟x冟 For p 0:
p has to be positive!
1. 冟 x 冟 0002 p is equivalent to
x0002p
or
The distance from x to zero is p.
x 0002 0003p.
0003p
2. 冟 x 冟 6 p is equivalent to 0003p 6 x 6 p.
(
0003p
The distance from x to zero is less than p.
3. 冟 x 冟 7 p is equivalent to
x 6 0003p
or
x 7 p.
)
0003p
The distance from x to zero is greater than p.
0
p
0
p
0
p
x
)
x
(
x
If we replace x in Theorem 2 with ax b, we obtain the more general Theorem 3.
Z THEOREM 3 Properties of Equations and Inequalities Involving |ax b| For p 0:
EXAMPLE
4
p has to be positive!
1. 冟 ax b 冟 0002 p
is equivalent to
ax b 0002 p
2. 冟 ax b 冟 6 p
is equivalent to
0003p 6 ax b 6 p.
3. 冟 ax b 冟 7 p
is equivalent to
ax b 6 0003p
ax b 0002 0003p.*
or or
ax b 7 p.
Solving Absolute Value Problems Solve each equation or inequality. For inequalities, write solutions in both inequality and interval notation. (A) 冟 3x 5 冟 0002 4
(B) 冟 x 冟 6 5
(C) 冟 2x 0003 1 冟 6 3
(D) 冟 7 0003 3x 冟 2
*This can be more concisely written as ax b 0002 ; p.
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SOLUTIONS
(A) 冟 3x 5 冟 0002 4 Use Theorem 3, part 1 3x 5 0002 4 3x 0002 00035 4 00035 4 x0002 3 x 0002 00033, 000313
(B) 冟 x 冟 6 5 Use Theorem 2, part 2 00035 6 x 6 5 or (00035, 5)
or 500033, 000313 6 (C) 冟 2x 0003 1 冟 6 3 Use Theorem 3, part 2 00033 6 2x 0003 1 6 3 00032 6 2x 6 4 00031 6 x 6 2 or (00031, 2)
MATCHED PROBLEM 4
5
0002
Solve each equation or inequality. For inequalities, write solutions in both inequality and interval notation. (A) 冟 2x 0003 1 冟 0002 8
EXAMPLE
(D) 冟 7 0003 3x 冟 2 Use Theorem 3, part 2 00032 7 0003 3x 2 00039 00033x 00035 3 0004 x 0004 53 5 3 x 3 or [ 53, 3]
(B) 冟 x 冟 7
(C) 冟 3x 3 冟 9
(D) 冟 5 0003 2x 冟 6 9
0002
Solving Absolute Value Inequalities Solve, and write solutions in both inequality and interval notation. (A) 冟 x 冟 7 3
SOLUTIONS
(B) 冟 2x 0003 1 冟 0004 3
(A) 冟 x 冟 7 3 or x 7 3 x 6 00033 (00030007, 00033) 0004 (3, 0007) (B) 冟 2x 0003 1 冟 0004 3 2x 0003 1 00033 2x 00032 x 00031 (00030007, 00031] 0004 (C) 冟 7 0003 3x 冟 7 2 7 0003 3x 6 00032 00033x 6 00039
Use Theorem 2, part 3. Solution in inequality notation Solution in interval notation Use Theorem 3, part 3.
or 2x 0003 1 0004 3 or 2x 0004 4 or x00042 [2, 0007)
Add 1 to both sides. Divide both sides by 2. Solution in inequality notation Solution in interval notation Use Theorem 3, part 3.
or or
7 0003 3x 7 2 00033x 7 00035
x 7 3 or (00030007, 53) 0004 (3, 0007)
MATCHED PROBLEM 5
(C) 冟 7 0003 3x 冟 7 2
x 6
5 3
Subtract 7 from both sides. Divide both sides by 00033 and reverse the direction of the inequality. Solution in inequality notation Solution in interval notation
0002
Solve, and write solutions in both inequality and interval notation. (A) 冟 x 冟 0004 5
(B) 冟 4x 0003 3 冟 7 5
(C) 冟 6 0003 5x 冟 7 16
0002
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EXAMPLE
6
Absolute Value in Equations and Inequalities
71
An Absolute Value Problem with Two Cases Solve: 冟 x 4 冟 0002 3x 0003 8
SOLUTION
We can’t use Theorem 3 directly, because we don’t know that 3x 0003 8 is positive. However, we can use the definition of absolute value and two cases: x 4 0004 0 and x 4 0006 0. Case 1. x 4 0004 0 (in which case, x 0004 00034) For this case, the only acceptable values of x are greater than or equal to 00034. 冟 x 4 冟 0002 3x 0003 8 x 4 0002 3x 0003 8 00032x 0002 000312 x00026
If x 0005 4 is positive, |x 0005 4| 0002 x 0005 4. Subtract 3x and 4 from both sides. Divide both sides by 00032. A solution, because 6 is among the acceptable values of x (6 0004 00024).
Case 2. x 4 0006 0 (in which case, x 0006 00034) In this case, the only acceptable values of x are less than 00034. 冟 x 4 冟 0002 3x 0003 8 0003(x 4) 0002 3x 0003 8 0003x 0003 4 0002 3x 0003 8 00034x 0002 00034 x00021
If x 0005 4 is negative, |x 0005 4| 0002 0003(x 0005 4). Distribute 00021. Subtract 3x and add 4 to both sides. Divide both sides by 00034. Not a solution, since 1 is not among the acceptable values of x (1 0005 00024).
Combining both cases, we see that the only solution is x 0002 6. As a final check, we substitute x 0002 6 and x 0002 1 in the original equation. 冟 x 4 冟 0002 3x 0003 8 ? 冟 6 4 冟 0002 3(6) 0003 8 ✓ 10 000210 MATCHED PROBLEM 6
冟 x 4 冟 0002 3x 0003 8 ? 冟 1 4 冟 0002 3(1) 0003 8 5 00035
Solve: 冟 3x 0003 4 冟 0002 x 5
0002
0002
Z Using Absolute Value to Solve Radical Inequalities In Section R-2, we found that if x is positive or zero, 2x2 0002 x. But what if x is negative? Let’s look at an example: 2(00032)2 0002 14 0002 2 We see that for negative x, 2x2 0002 0003x. So for any real number, 2x2 0002
0003x x
冦
if x 6 0 if x 0004 0
But this is exactly how we defined 冟 x 冟 at the beginning of this section (see Definition 1). So for any real number x, 2x2 0002 冟 x 冟
(1)
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7
Solving a Radical Inequality Solve the inequality. Write your answer in both inequality and interval notation. 2(x 0003 2)2 5 2(x 0003 2)2 5 冟x 0003 2冟 5
SOLUTION
Use equation (1): 2(x 0003 2)2 0002 冟x 0003 2冟 Use Theorem 3, part 2.
00035 x 0003 2 5 00033 x 7 or [00033, 7] MATCHED PROBLEM 7
Add 2 to each member. Solution in inequality notation
0002
Solution in interval notation
Solve the inequality. Write your answers in both inequality and interval notation. 2(x 2)2 6 3 0002
ANSWERS TO MATCHED PROBLEMS 3 3 1. (A) 8 (B) 2 (C) 12 (D) 2 900032 900032 2. (A) x is a number whose distance from 00032 is 6. x 0002 00038, 4 or 500038, 46
00038
00032
x
4
(B) x is a number whose distance from 00032 is less than 6. 00038 6 x 6 4 or (00038, 4) ( ) 00038
00032
4
x
(C) x is a number whose distance from 00032 is less than 6, but x cannot equal 00032. 00038 6 x 6 4, x 00032, or (00038, 00032) 0004 (00032, 4) ( ) (D) x is a number whose distance from 00032 is greater than 6. x 6 00038 or x 7 4, or (00030007, 00038) 0004 (4, 0007) ) 00038
00038
00032
00032
( 4
4
x
x
3. (A) 冟 x 0003 5 冟 0002 6 (B) 冟 y 6 冟 6 7 (C) 冟 w 2 冟 0004 3 (D) 冟 t 0003 3 冟 4 4. (A) x 0002 000372, 92 or 5000372, 92 6 (B) 00037 x 7 or [ 00037, 7] (C) 00034 x 2 or [ 00034, 2] (D) 00032 6 x 6 7 or (00032, 7) 5. (A) x 00035 or x 0004 5, or (00030007, 00035] 0004 [5, 0007) (B) x 6 000312 or x 7 2, or (00030007, 000312) 0004 (2, 0007) 22 22 (C) x 6 00032 or x 7 5 , or (00030007, 00032) 0004 ( 5 , 0007) 6. x 0002 000314, 92 or 5000314, 92 6 7. 00035 6 x 6 1 or (00035, 1)
1-3
Exercises
1. Describe how to find the absolute value of a number, then explain how your description matches Definition 1.
4. Repeat Problem 3 for the inequalities |x 0003 5| 0006 10 and |x 0003 5| 10.
2. Explain what the expression |x 0003 5| represents geometrically, and why.
5. Explain why it is incorrect to say that 2x2 0002 x.
3. Describe the equation |x 0003 5| 0002 10 in terms of your answer to Problem 2, then explain how that helps you to solve it.
6. Why can’t the following be a legitimate solution to an inequality? x 0006 1 and x 5.
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SECTION 1–3
In Problems 7–14, simplify, and write without absolute value signs. Do not replace radicals with decimal approximations. 8. 冟 ⫺34 冟
7. 冟 15 冟 9. 冟 (⫺6) ⫺ (⫺2) 冟
10. 冟 (⫺2) ⫺ (⫺6) 冟
11. 冟 5 ⫺ 15 冟
12. 冟 17 ⫺ 2 冟
13. 冟 15 ⫺ 5 冟
14. 冟 2 ⫺ 17 冟
Absolute Value in Equations and Inequalities
55. 2x2 6 2
56. 2m2 7 3
57. 2(1 ⫺ 3t)2 ⱕ 2
58. 2(3 ⫺ 2x)2 6 5
59. 2(2t ⫺ 3)2 7 3
60. 2(3m ⫹ 5)2 ⱖ 4
73
In Problems 61–64, solve and write answers in inequality notation. Round decimals to three significant digits. 61. 冟 2.25 ⫺ 1.02x 冟 ⱕ 1.64
In Problems 15–20, use the number line shown to find the indicated distances. A
B
⫺10
⫺5
O
C
0
5
D 10
15. d(B, O)
16. d(A, B)
17. d(O, B)
18. d(B, A)
19. d(B, C)
20. d(D, C)
62. 冟 0.962 ⫺ 0.292x 冟 ⱕ 2.52 63. 冟 21.7 ⫺ 11.3x 冟 ⫽ 15.2 x
Write each of the statements in Problems 21–30 as an absolute value equation or inequality. 21. x is 4 units from 3. 22. y is 3 units from 1.
64. 冟 195 ⫺ 55.5x 冟 ⫽ 315 Problems 65⫺68 involve expressions that are important in the study of limits in calculus. First, provide a verbal translation of the inequality. Then solve and graph, writing your solution in interval notation. 65. 0 6 冟 x ⫺ 3 冟 6 0.1 1 67. 0 6 冟 x ⫺ a 冟 6 10
66. 0 6 冟 x ⫹ 5 冟 6 0.5 68. 0 6 冟 x ⫺ 8 冟 6 d
23. m is 5 units from ⫺2.
In Problems 69–76, for what values of x does each hold?
24. n is 7 units from ⫺5.
69. 冟 x ⫺ 2 冟 ⫽ 2x ⫺ 7
70. 冟 x ⫹ 4 冟 ⫽ 3x ⫹ 8
25. x is less than 5 units from 3.
71. 冟 3x ⫹ 5 冟 ⫽ 2x ⫹ 6
72. 冟 7 ⫺ 2x 冟 ⫽ 5 ⫺ x
26. z is less than 8 units from ⫺2.
73. 冟 x 冟 ⫹ 冟 x ⫹ 3 冟 ⫽ 3
74. 冟 x 冟 ⫺ 冟 x ⫺ 5 冟 ⫽ 5
27. p is more than 6 units from ⫺2.
75. 冟 3 ⫺ x 冟 ⫽ 2 (4 ⫹ x)
28. c is no greater than 7 units from ⫺3.
76. 冟 5 ⫺ 2x 冟 ⫽ 4(x ⫺ 5)
29. q is no less than 2 units from 1.
77. What are the possible values of
x ? 冟x冟
78. What are the possible values of
冟x ⫺ 1冟 ? x⫺1
30. d is no more than 4 units from 5. In Problems 31–42, solve, interpret geometrically, and graph. When applicable, write answers using both inequality notation and interval notation.
79. Explain why 冟 ax ⫹ b 冟 6 ⫺3 has no solution for any values of a and b.
31. 冟 y ⫺ 5 冟 ⫽ 3
32. 冟 t ⫺ 3 冟 ⫽ 4
33. 冟 y ⫺ 5 冟 6 3
34. 冟 t ⫺ 3 冟 6 4
35. 冟 y ⫺ 5 冟 7 3
36. 冟 t ⫺ 3 冟 7 4
80. Explain why 冟 ax ⫹ b 冟 7 ⫺3 has solution all real numbers for any values of a and b.
37. 冟 u ⫹ 8 冟 ⫽ 3
38. 冟 x ⫹ 1 冟 ⫽ 5
39. 冟 u ⫹ 8 冟 ⱕ 3
81. Prove that 冟 b ⫺ a 冟 ⫽ 冟 a ⫺ b 冟 for all real numbers a and b. [Hint: Apply Definition 1 and use cases.]
40. 冟 x ⫹ 1 冟 ⱕ 5
41. 冟 u ⫹ 8 冟 ⱖ 3
42. 冟 x ⫹ 1 冟 ⱖ 5
82. Prove that 冟 x 冟2 ⫽ x2 for all real numbers x.
In Problems 43–60, solve the equation or inequality. Write solutions to inequalities using both inequality and interval notation. 43. 冟 2x ⫺ 11 冟 ⱕ 13
44. 冟 5x ⫹ 20 冟 ⱖ 5
45. 冟 100 ⫺ 40t 冟 7 60
46. 冟 150 ⫺ 20y 冟 6 10
47. 冟 4x ⫺ 7 冟 ⫽ 13
48. 冟 ⫺8x ⫹ 3 冟 ⱕ 91
49. 冟 12w
50. 冟 13z ⫹ 56 冟 ⫽ 1
⫺
3 4冟
6 2
51. 冟 0.2u ⫹ 1.7 冟 ⱖ 0.5 53. 冟 95 C
⫹ 32 冟 6 31
52. 冟 0.5v ⫺ 2.5 冟 7 1.6 54.
冟 59 (F
⫺ 32) 冟 6 40
83. Prove that the average of two numbers is between the two numbers; that is, if m 6 n, then m 6
m⫹n 6 n 2
84. Prove that for m 6 n, d am,
m⫹n m⫹n b ⫽ da , nb 2 2
85. Prove that 冟 ⫺m 冟 ⫽ 冟 m 冟. 86. Prove that 冟 m 冟 ⫽ 冟 n 冟 if and only if m ⫽ n or m ⫽ ⫺n.
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87. Prove that for n 0, `
冟m冟 m `0002 冟n冟 n
88. Prove that 冟 mn 冟 0002 冟 m 冟冟 n 冟.
94. CHEMISTRY In order to manufacture a polymer for soft drink containers, a chemical reaction must take place within 10°C of 200°C. Write this temperature restriction as an absolute value inequality, then solve to find the acceptable temperatures.
89. Prove that 0003冟 m 冟 m 冟 m 冟. 90. Prove the triangle inequality: 冟m n冟 冟m冟 冟n冟 Hint: Use Problem 89 to show that 0003冟 m 冟 0003 冟 n 冟 m n 冟 m 冟 冟 n 冟
APPLICATIONS 91. STATISTICS Inequalities of the form `
x0003m ` 6 n s
occur frequently in statistics. If m 0002 45.4, s 0002 3.2, and n 0002 1, solve for x. 92. STATISTICS Repeat Problem 91 for m 0002 28.6, s 0002 6.5, and n 0002 2.
1-4
93. BUSINESS The daily production P in an automobile assembly plant is always within 20 units of 500 units. Write the daily production as an absolute value inequality, then solve to find the range of daily productions possible.
;
95. APPROXIMATION The area A of a region is approximately equal to 12.436. The error in this approximation is less than 0.001. Describe the possible values of this area both with an absolute value inequality and with interval notation. 96. APPROXIMATION The volume V of a solid is approximately equal to 6.94. The error in this approximation is less than 0.02. Describe the possible values of this volume both with an absolute value inequality and with interval notation. 97. SIGNIFICANT DIGITS If N 0002 2.37 represents a measurement, then we assume an accuracy of 2.37 0.005. Express the accuracy assumption using an absolute value inequality. 98. SIGNIFICANT DIGITS If N 0002 3.65 000e 1000033 is a number from a measurement, then we assume an accuracy of 3.65 000e 1000033 5 000e 1000036. Express the accuracy assumption using an absolute value inequality.
Complex Numbers Z Understanding Complex Number Terminology Z Performing Operations with Complex Numbers Z Relating Complex Numbers and Radicals Z Solving Equations Involving Complex Numbers
The idea of inventing new numbers might seem odd to you, but think about this example: A group of mathematicians known as the Pythagoreans proved over 2,000 years ago that the equation x2 0002 2 has no solutions that are rational numbers. You may be thinking that the solutions are 12 and 000312, but at the time, those numbers had not been defined, so the Pythagoreans invented a new kind of number—irrational numbers, like 12 and 000312. Now consider the similar equation x2 0002 00031. To be a solution, a number has to result in 00031 when squared. But we know that the square of any real number cannot be negative, so again a new type of number is invented—a number whose square is 00031. The concept of square roots of negative numbers had been kicked around for a few centuries, but in 1748, the Swiss mathematician Leonhard Euler (pronounced “Oiler”) used the letter i to represent a square root of 00031. From this simple beginning, it is possible to build a new system of numbers called the complex number system.
Z Understanding Complex Number Terminology The number i, whose square is 00031, is called the imaginary unit. Complex numbers are defined in terms of the imaginary unit.
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Complex Numbers
75
Z DEFINITION 1 Complex Number A complex number is a number of the form a bi, where a and b are real numbers, and i is the imaginary unit (a square root of 00031). A complex number written this way is said to be in standard form. The real number a is called the real part, and bi is called the imaginary part.
Some examples of complex numbers are 3 0003 2i 0 3i
1 2
2 0003 13i 0 0i
5i 5 0i
The notation 3 0003 2i is shorthand for 3 (00032)i. Particular kinds of complex numbers are given special names as follows:
Z DEFINITION 2 Special Terms a 0005 bi b0 0 0005 bi 0002 bi b0 a 0005 0i 0002 a 0 0002 0 0005 0i a 0003 bi
EXAMPLE
1
Imaginary Number Pure Imaginary Number Real Number Zero Conjugate of a 0005 bi
Complex Numbers Identify the real part, the imaginary part, and the conjugate of each of the following numbers: (A) 3 0003 2i
SOLUTIONS
(B) 2 5i
(C) 7i
(D) 6
(A) Real part: 3; imaginary part: 00032i; conjugate: 3 2i (B) Real part: 2; imaginary part: 5i; conjugate: 2 0003 5i (C) Real part: 0; imaginary part: 7i; conjugate: 00037i (D) Real part: 6; imaginary part: 0; conjugate: 6
MATCHED PROBLEM 1
0002
Identify the real part, the imaginary part, and the conjugate of each of the following numbers: (A) 6 7i
(B) 00033 0003 8i
(C) 00034i
(D) 00039 0002
We will identify a complex number of the form a 0i with the real number a, a complex number of the form 0 bi, b 0, with the pure imaginary number bi, and the complex number 0 0i with the real number 0. So a real number is also a complex number, just as a rational number is also a real number. Any complex number that is not a real number is called an imaginary number. If we combine the set of all real numbers with the set of all imaginary numbers, we obtain C, the set of complex numbers. The relationship of the complex number system to the other number systems we have studied is shown in Figure 1.
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Z Figure 1 Natural numbers (N ) Zero Negative integers
N0006Z0006Q0006R0006C Integers (Z ) Noninteger rational numbers
Rational numbers (Q) Irrational numbers (I )
Real numbers (R ) Imaginary numbers
Complex numbers (C)
Z Performing Operations with Complex Numbers To work with complex numbers, we will need to know how to add, subtract, multiply, and divide them. We start by defining equality, addition, and multiplication.
Z DEFINITION 3 Equality and Basic Operations 1. Equality: a bi 0002 c di if and only if a 0002 c and b 0002 d 2. Addition: (a bi) (c di) 0002 (a c) (b d )i 3. Multiplication: (a bi)(c di) 0002 (ac 0003 bd) (ad bc)i
In Section R-1 we listed the basic properties of the real number system. Using Definition 3, it can be shown that the complex number system possesses the same properties. That is, 1. 2. 3. 4. 5.
Addition and multiplication of complex numbers are commutative and associative operations. There is an additive identity and a multiplicative identity for complex numbers. Every complex number has an additive inverse or negative. Every nonzero complex number has a multiplicative inverse or reciprocal. Multiplication distributes over addition.
This is actually really good news: it tells us that we don’t have to memorize the formulas for adding and multiplying complex numbers in Definition 3. Instead: We can treat complex numbers of the form a 0005 bi exactly as we treat algebraic expressions of the form a 0005 bx. We just need to remember that in this case, i stands for the imaginary unit; it is not a variable that represents a real number. The first two arithmetic operations we consider are addition and subtraction.
EXAMPLE
2
Addition and Subtraction of Complex Numbers Carry out each operation and express the answer in standard form: (A) (2 0003 3i) (6 2i) (C) (7 0003 3i) 0003 (6 2i)
(B) (00035 4i) (0 0i) (D) (00032 7i) (2 0003 7i)
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SOLUTIONS
Complex Numbers
77
(A) We could apply the definition of addition directly, but it is easier to use complex number properties. (2 0003 3i) (6 2i) 0002 2 0003 3i 6 2i
Use the commutative property.
0002 (2 6) (00033 2)i
Combine like terms.
000280003i (B) (00035 4i) (0 0i) 0002 00035 4i 0 0i 0002 00035 4i (C) (7 0003 3i) 0003 (6 2i) 0002 7 0003 3i 0003 6 0003 2i 0002 1 0003 5i
Make sure you distribute the negative sign!
(D) (00032 7i) (2 0003 7i) 0002 00032 7i 2 0003 7i 0002 0
MATCHED PROBLEM 2
0002
Carry out each operation and express the answer in standard form: (A) (3 2i) (6 0003 4i)
(B) (0 0i) (7 0003 5i)
(C) (3 0003 5i) 0003 (1 0003 3i)
(D) (00034 9i) (4 0003 9i)
0002
Example 2, part B, illustrates the following general property: For any complex number a bi, (a bi) (0 0i) 0002 a bi and
(0 0i) (a bi) 0002 a bi
That is, 0 0i is the additive identity or zero for the complex numbers. This is why we identify 0 0i with the real number zero in Definition 2. Example 2, part D, illustrates a different result: In general, the additive inverse or negative of a bi is 0003a 0003 bi because (a bi) (0003a 0003 bi) 0002 0 and (0003a 0003 bi) (a bi) 0002 0 Now we turn our attention to multiplication. Just like addition and subtraction, multiplication of complex numbers can be carried out by treating a bi in the same way we treat the algebraic expression a bx. The key difference is that we replace i2 with 00031 each time it occurs.
EXAMPLE
3
Multiplying Complex Numbers Carry out each operation and express the answer in standard form:
SOLUTIONS
(A) (2 0003 3i)(6 2i)
(B) 1(3 0003 5i)
(C) i(1 i)
(D) (3 4i)(3 0003 4i)
(A) (2 0003 3i)(6 2i)
(B) 1(3 0003 5i)
0002 12 4i 0003 18i 0003 6i2 0002 12 0003 14i 0003 6(00031) 0002 18 0003 14i
0002 1 0007 3 0003 1 0007 5i
Replace i 2 with 00031. 00036(00031) 0002 6; combine like terms.
0002 3 0003 5i
(C) i(1 i) 0002 i i2 0002 i 0003 1 0002 00031 i (D) (3 4i)(3 0003 4i) 0002 9 0003 12i 12i 0003 16i2 0002 9 16 0002 25
Answer in standard form. 000316i 2 0002 000316(00031) 0002 16
0002
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MATCHED PROBLEM 3
Carry out each operation and express the answer in standard form: (A) (5 2i)(4 0003 3i)
(B) 3(00032 6i)
(C) i(2 0003 3i)
(D) (2 3i)(2 0003 3i) 0002
For any complex number a bi, 1(a bi) 0002 a bi and (a bi)1 0002 a bi (see Example 3, part B). This indicates that 1 is the multiplicative identity for complex numbers, just as it is for real numbers. Earlier we stated that every nonzero complex number has a multiplicative inverse or reciprocal. We will denote this as a fraction, just as we do with real numbers: 1 a bi
a bi
is the reciprocal of
a bi 0
The following important property of the conjugate of a complex number is used to express reciprocals and quotients in standard form. (See Example 3, part D)
Z THEOREM 1 Product of a Complex Number and Its Conjugate (a bi)(a 0003 bi) 0002 a2 b2
A real number
To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator.
EXAMPLE
4
Reciprocals and Quotients Write each expression in standard form: (A)
SOLUTIONS
1 2 3i
(B)
7 0003 3i 1000bi
(A) Multiply numerator and denominator by the conjugate of the denominator: 1 1 2 0003 3i 0002 0007 2 3i 2 3i 2 0003 3i 0002
0002
2 3 2 0003 3i 0002 0003 i 13 13 13
2 0003 3i 2 0003 3i 0002 4000b9 4 0003 9i2 Answer in standard form.
This answer can be checked by multiplication: CHECK
(2 3i) a
2 3 4 6 6 9 2 0003 ib 0002 0003 i i0003 i 13 13 13 13 13 13 4 9 0002 00021 ✓ 13 13
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(B)
7 ⫺ 3i 7 ⫺ 3i 1 0003 i ⫽ 0002 1⫹i 1⫹i 10003i ⫽
CHECK
MATCHED PROBLEM 4
⫽
7 ⫺ 7i ⫺ 3i ⫹ 3i2 1 ⫺ i2
4 ⫺ 10i ⫽ 2 ⫺ 5i 2
Complex Numbers
79
3i 2 0004 00033
Answer in standard form.
(1 ⫹ i)(2 ⫺ 5i) ⫽ 2 ⫺ 5i ⫹ 2i ⫺ 5i2 ⫽ 7 ⫺ 3i
0002
Carry out each operation and express the answer in standard form: (A)
1 4 ⫹ 2i
(B)
6 ⫹ 7i 2⫺i 0002
EXAMPLE
5
Combined Operations Carry out the indicated operations and write each answer in standard form: (A) (3 ⫺ 2i)2 ⫺ 6(3 ⫺ 2i) ⫹ 13
SOLUTIONS
2 ⫺ 3i 2i
(B)
(A) (3 ⫺ 2i)2 ⫺ 6(3 ⫺ 2i) ⫹ 13 ⫽ 9 ⫺ 12i ⫹ 4i2 ⫺ 18 ⫹ 12i ⫹ 13 ⫽ 9 ⫺ 12i ⫺ 4 ⫺ 18 ⫹ 12i ⫹ 13 ⫽0 (B) If a complex number is divided by a pure imaginary number, we can make the denominator real by multiplying numerator and denominator by i. (We could also multiply by the conjugate of 2i, which is ⫺2i.) 2i ⫺ 3i2 2i ⫹ 3 3 2 ⫺ 3i i 0002 ⫽ ⫽ ⫽⫺ ⫺i 2 2i i ⫺2 2 2i
MATCHED PROBLEM 5
0002
Carry out the indicated operations and write each answer in standard form: (A) (3 ⫹ 2i)2 ⫺ 6(3 ⫹ 2i) ⫹ 13
4⫺i 3i
(B)
0002
ZZZ EXPLORE-DISCUSS 1
Natural number powers of i take on particularly simple forms: i i 2 ⫽ ⫺1 i 3 ⫽ i 2 0002 i ⫽ (⫺1)i ⫽ ⫺i i 4 ⫽ i 2 0002 i 2 ⫽ (⫺1)(⫺1) ⫽ 1
i5 i6 i7 i8
⫽ ⫽ ⫽ ⫽
i4 i4 i4 i4
0002 0002 0002 0002
i ⫽ (1)i ⫽ i i 2 ⫽ 1(⫺1) ⫽ ⫺1 i 3 ⫽ 1(⫺i) ⫽ ⫺i i4 ⫽ 1 0002 1 ⫽ 1
In general, what are the possible values for i n, n a natural number? Explain how you could easily evaluate i n for any natural number n. Then evaluate each of the following: (A) i 17
(B) i 24
(C) i 38
(D) i 47
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Z Relating Complex Numbers and Radicals Recall that we say that a is a square root of b if a2 0002 b. If x is a positive real number, then x has two square roots, the principal square root, denoted by 1x, and its negative, 00031x (Section R-2). If x is a negative real number, then x still has two square roots, but now these square roots are imaginary numbers.
Z DEFINITION 4 Principal Square Root of a Negative Real Number The principal square root of a negative real number, denoted by 10003a, where a is positive, is defined by 10003a 0002 i 1a
For example 100033 0002 i 13; 100039 0002 i 19 0002 3i
The other square root of 0003a, a 7 0, is 0003 10003a 0002 0003i1a.
Note in Definition 4 that we wrote i1a and i 13 in place of the standard forms 1ai and 13i. We follow this convention to avoid confusion over whether the i should or should not be under the radical. (Notice that 13i and 13i look a lot alike, but are not the same number.)
EXAMPLE
6
Complex Numbers and Radicals Write in standard form: (A) 100034 (C)
SOLUTIONS
(B) 4 100035
00033 0003 100035 2
(D)
1 1 0003 100039
(A) 100034 0002 i 14 0002 2i (B) 4 100035 0002 4 i 15 (C)
00033 0003 100035 00033 0003 i 15 3 15 0002 00020003 0003 i 2 2 2 2
(D)
1 0007 (1 0005 3i ) 1 1 0002 0002 1 0003 3i (1 0003 3i) 0007 (1 0005 3i ) 1 0003 100039 0002
MATCHED PROBLEM 6
1 3i 1 3 1 3i 0002 0002 i 10 10 10 1 0003 9i2
Answer in standard form.
Standard form
0002
Write in standard form: (A) 1000316 (C)
00035 0003 100032 2
(B) 5 100037 (D)
1 3 0003 100034 0002
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ZZZ EXPLORE-DISCUSS 2
81
Complex Numbers
From Theorem 4 in Section R-2, we know that if a and b are positive real numbers, then 1a 1b 0002 1ab
(1)
So we can evaluate expressions like 19 14 two ways: 19 14 0002 1(9)(4) 0002 136 0002 6
and
19 14 0002 (3)(2) 0002 6
Evaluate each of the following two ways. Is equation (1) a valid property to use in all cases? (A) 19 100034
ZZZ
(B) 100039 14
(C) 100039 100034
Note that in Example 6, part D, we wrote 1 0003 100039 0002 1 0003 3i before proceeding with the simplification. This is a necessary step because some of the properties of radicals that are true for real numbers turn out not to be true for complex numbers. In particular, for positive real numbers a and b,
CAUTION ZZZ
1a 1b 0002 1ab
but
10003a 10003b 1(0003a)(0003b)
(See Explore-Discuss 2.) To avoid having to worry about this, always convert expressions of the form 10003a to the equivalent form in terms of i before performing any operations.
Z Solving Equations Involving Complex Numbers EXAMPLE
7
Equations Involving Complex Numbers (A) Solve for real numbers x and y: (3x 2) (2y 0003 4)i 0002 00034 6i (B) Solve for complex number z: (3 2i)z 0003 3 6i 0002 8 0003 4i
SOLUTIONS
(A) This equation is really a statement that two complex numbers are equal: (3x 2) (2y 0003 4)i, and 00034 6i. In order for these numbers to be equal, the real parts must be the same, and the imaginary parts must be the same as well. 3x 2 0002 00034 3x 0002 00036 x 0002 00032
and
2y 0003 4 0002 6 2y 0002 10 y00025
(B) Solve for z, then write the answer in standard form. (3 2i)z 0003 3 6i 0002 8 0003 4i (3 2i)z 0002 11 0003 10i 11 0003 10i 3 2i (11 0003 10i)(3 0003 2i) 0002 (3 2i)(3 0003 2i)
z0002
0002
13 0003 52i 13
Add 3 and subtract 6i from both sides. Divide both sides by 3 0005 2i. Multiply numerator and denominator by the conjugate of the denominator.
Simplify.
Write in standard form.
0002 1 0003 4i A check is left to the reader.
0002
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(A) Solve for real numbers x and y: (2y 0003 7) (3x 4)i 0002 1 i (B) Solve for complex number z: (1 3i)z 4 0003 5i 0002 3 2i 0002 The truth is that the numbers we studied in this section weren’t received very well when they were invented, as you can guess from the names they were given: complex and imaginary. These names are not exactly ringing endorsements. Still, complex numbers eventually came into widespread use in areas like electrical engineering, physics, chemistry, statistics, and aeronautical engineering. Our first application of complex numbers will be in solving second-degree equations in Section 1-5.
ANSWERS TO MATCHED PROBLEMS 1. (A) Real part: 6; imaginary part: 7i; conjugate: 6 0003 7i (B) Real part: 00033; imaginary part: 00038i; conjugate: 00033 8i (C) Real part: 0; imaginary part: 00034i; conjugate: 4i (D) Real part: 00039; imaginary part: 0; conjugate: 00039 2. (A) 9 0003 2i (B) 7 0003 5i (C) 2 0003 2i (D) 0 3. (A) 26 0003 7i (B) 00036 18i (C) 3 2i (D) 13 4. (A) 15 0003 101 i (B) 1 4i 5. (A) 0 (B) 000313 0003 43 i 6. (A) 4i (B) 5 i17 (C) 000352 0003 (12 2)i (D) 133 132 i 7. (A) x 0002 00031, y 0002 4 (B) z 0002 2 i
1-4
Exercises 4 3 i 5 5
1. Do negative real numbers have square roots? Explain.
10. 4.2 0003 9.7i
11. 6.5 2.1i
12.
2. Arrange the following sets of numbers so that each one contains the one that comes before it in the list: rational numbers, complex numbers, integers, real numbers, natural numbers.
13. i0005
14. 60005
15. 40005
16. 000320005i
17. 00035 i12
18. 4 0003 i 17
3. Is it possible to square an imaginary number and get a real number? Explain. 4. What is the conjugate of a complex number? How do we use conjugates?
In Problems 19–44, perform the indicated operations and write each answer in standard form. 19. (3 5i) (2 4i)
20. (4 i) (5 3i)
5. Which statement is false, and which is true? Justify your response. (A) Every real number is a complex number. (B) Every complex number is a real number.
21. (8 0003 3i) (00035 6i)
22. (00031 2i) (4 0003 7i)
23. (9 5i) 0003 (6 2i)
24. (3 7i) 0003 (2 5i)
25. (3 0003 4i) 0003 (00035 6i)
26. (00034 0003 2i) 0003 (1 i)
6. Is it possible to add a real number and an imaginary number? If so, what kind of number is the result?
27. 2 (3i 5)
28. (2i 7) 0003 4i
29. (2i)(4i)
31. 00032i(4 0003 6i)
For each number in Problems 7–18, find the (A) real part, (B) imaginary part, and (C) conjugate. 7. 2 0003 9i
8. 00036i 4
3 5 9. 0003 i 2 6
30. (3i)(5i)
32. (00034i)(2 0003 3i)
33. (1 2i)(3 0003 4i)
34. (2 0003 i)(00035 6i)
35. (3 0003 i)(4 i)
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36. (5 2i)(4 0003 3i)
37. (2 9i)(2 0003 9i)
38. (3 8i)(3 0003 8i)
39.
40.
i 3000bi
43.
7000bi 2000bi
41.
4 3i 1 2i
1 2 4i
42.
3 0003 5i 20003i
44.
00035 10i 3 4i
Complex Numbers
77.
(1 x) ( y 0003 2)i 000220003i 1000bi
78.
(2 x) ( y 3)i 0002 00033 i 10003i
83
In Problems 79–82, solve for z and write your answer in standard form. 79. (10 0003 2i)z (5 i) 0002 2i
In Problems 45–52, evaluate and express results in standard form.
80. (3 0003 2i)z (4i 6) 0002 8i
45. 12 18
46. 13 112
81. (4 2i)z (7 0003 2i) 0002 (4 0003 i)z (3 5i)
47. 12 100038
48. 100033 112
82. (00032 3i) (4 5i)z 0002 (1 i) 0003 (00034 2i)z
49. 100032 18
50. 13 1000312
83. Show that 2 0003 i and 00032 i are square roots of 3 0003 4i.
51. 100032 100038
52. 100033 1000312
84. Show that 00033 2i and 3 0003 2i are square roots of 5 0003 12i.
In Problems 53–62, convert imaginary numbers to standard form, perform the indicated operations, and express answers in standard form.
85. Explain what is wrong with the following “proof ” that 00031 0002 1: 00031 0002 i2 0002 100031 100031 0002 1(00031)(00031) 0002 11 0002 1 86. Explain what is wrong with the following “proof ” that 1 i 0002 i. What is the correct value of 1 i?
53. (2 0003 100034) (5 0003 100039) 54. (3 0003 100034) (00038 1000325)
1 1 11 1 0002 0002 0002 0002 100031 0002 i i 100031 100031 A 00031
55. (9 0003 100039) 0003 (12 0003 1000325) 56. (00032 0003 1000336) 0003 (4 1000349)
87. Show that i 4k 0002 1, k a natural number
57. (3 0003 100034)(00032 1000349)
88. Show that i 4k000b1 0002 i, k a natural number
58. (2 0003 100031)(5 100039) 59.
5 0003 100034 7
60.
6 0003 1000364 2
Supply the reasons in the proofs for the theorems stated in Problems 89 and 90.
61.
1 2 0003 100039
62.
1 3 0003 1000316
89. Theorem: The complex numbers are commutative under addition. Proof: Let a bi and c di be two arbitrary complex numbers; then:
In Problems 63–68, write the complex number in standard form. 63. 0003
5 i
64.
65. (2i)2 0003 5(2i) 6
1 10i
66. (i13)4 2(i13)2 15
67. (5 2i)2 0003 4(5 2i) 0003 1 68. (7 0003 3i)2 8(7 0003 3i) 0003 30 69. Evaluate x2 0003 2x 2 for x 0002 1 0003 i. 70. Evaluate x2 0003 2x 2 for x 0002 1 i. In Problems 71–74, for what real values of x does each expression represent an imaginary number? 71. 13 0003 x
72. 15 x
73. 12 0003 3x
74. 13 2x
In Problems 75–78, solve for x and y. 75. (2x 0003 1) (3y 2)i 0002 5 0003 4i 76. 3x ( y 0003 2)i 0002 (5 0003 2x) (3y 0003 8)i
Statement 1. (a bi) (c di) 0002 (a c) (b d )i 2. 0002 (c a) (d b)i 3. 0002 (c di) (a bi) Reason 1. 2. 3. 90. Theorem: The complex numbers are commutative under multiplication. Proof: Let a bi and c di be two arbitrary complex numbers; then: Statement 1. (a bi) 0007 (c di) 0002 (ac 0003 bd ) (ad bc)i 0002 (ca 0003 db) (da cb)i 2. 0002 (c di)(a bi) 3. Reason 1. 2. 3.
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Letters z and w are often used as complex variables, where z 0002 x yi, w 0002 u vi, and x, y, u, v are real numbers. The conjugates of z and w, denoted by z and w, respectively, are given by z 0002 x 0003 yi and w 0002 u 0003 vi. In Problems 91–98, express each property of conjugates verbally and then prove the property. 91. zz is a real number.
1-5
93. z 0002 z if and only if z is real.
94. z 0002 z
95. z w 0002 z w
96. z 0003 w 0002 z 0003 w
97. zw 0002 z 0007 w
98. z w 0002 z w
92. z z is a real number.
Quadratic Equations and Applications Z Using Factoring to Solve Quadratic Equations Z Using the Square Root Property to Solve Quadratic Equations Z Using Completing the Square to Solve Quadratic Equations Z Using the Quadratic Formula to Solve Quadratic Equations Z Solving Applications Involving Quadratic Equations
The next class of equations we consider are the second-degree polynomial equations in one variable, called quadratic equations.
Z DEFINITION 1 Quadratic Equation A quadratic equation in one variable is any equation that can be written in the form ax2 0005 bx 0005 c 0002 0
a0
Standard Form
where x is a variable and a, b, and c are constants.
Now that we have discussed the complex number system, we can use complex numbers when solving equations. Recall that a solution of an equation is also called a root of the equation. A real number solution of an equation is called a real root, and an imaginary number solution is called an imaginary root. In this section, we develop methods for finding all real and imaginary roots of a quadratic equation.
Z Using Factoring to Solve Quadratic Equations There is one single reason why factoring is so important in solving equations. It’s called the zero product property.
ZZZ EXPLORE-DISCUSS 1
(A) Write down a pair of numbers whose product is zero. Is one of them zero? Can you think of two nonzero numbers whose product is zero? (B) Choose any number other than zero and call it a. Write down a pair of numbers whose product is a. Is one of them a? Can you think of a pair, neither of which is a, whose product is a?
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Z ZERO PRODUCT PROPERTY If m and n are complex numbers, then m0002n00020
m 0002 0 or n 0002 0 (or both)
if and only if
It is very helpful to think about what this says in words: If the product of two factors is zero, then at least one of those factors has to be zero. It’s also helpful to observe that zero is the only number for which this is true.
EXAMPLE
1
Solving Quadratic Equations by Factoring Solve by factoring: (B) 6x2 0003 19x 0003 7 0002 0 (D) 2x2 0002 3x
(A) (x 0003 5)(x 0004 3) 0002 0 (C) x2 0003 6x 0004 5 0002 00034 SOLUTIONS
(A) The product of two factors is zero, so by the zero product property, one of the two must be zero. This enables us to write two easier equations to solve. (x 0003 5)(x 0004 3) 0002 0 x0003500020 x00025 (B)
6x2 0003 19x 0003 7 0002 0 (2x 0003 7)(3x 0004 1) 0002 0 2x 0003 7 0002 0 x0002
(C)
Solution set: {00033, 5}.
Factor the left side. Use the zero product property.
or
3x 0004 1 0002 0
7 2
x2 0003 6x 0004 5 0002 00034 x2 0003 6x 0004 9 0002 0 (x 0003 3)(x 0003 3) 0002 0 x0003300020 x00023
x0004300020 x 0002 00033
or
x 0002 000313
Solution set: 5000313 , 72 6.
Add 4 to both sides. Factor left side. Use the zero product property.
Solution set: {3}.
The equation has one root, 3. But because it came from two factors, we call 3 a double root or a root of multiplicity 2. (D)
2x2 0002 3x 2x 0003 3x 0002 0 x(2x 0003 3) 0002 0 x00020 2
MATCHED PROBLEM 1
Subtract 3x from both sides. Factor the left side. Use the zero product property.
or
2x 0003 3 0002 0 x 0002 32
Solution set: 50, 32 6
0002
Solve by factoring: (A) (2x 0004 4)(x 0003 7) 0002 0 (C) 4x2 0004 12x 0004 9 0002 0
(B) 3x2 0004 7x 0003 20 0002 0 (D) 4x2 0002 5x 0002
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1. One side of an equation must be 0 before the zero product property can be applied. So
CAUTION ZZZ
x2 0003 6x 0004 5 0002 00034 (x 0003 1)(x 0003 5) 0002 00034 does not mean that x 0003 1 0002 00034 or x 0003 5 0002 00034. See Example 1, part C, for the correct solution of this equation. 2. The equations 2x2 0002 3x
2x 0002 3
and
are not equivalent. The first has solution set 50, 32 6, but the second has solution set 5 32 6. The root x 0002 0 is lost when each member of the first equation is divided by the variable x. See Example 1, part D, for the correct solution of this equation.
Never divide both sides of an equation by an expression containing the variable for which you are solving. You may be dividing by 0, which of course is not allowed.
Z Using the Square Root Property to Solve Quadratic Equations We now turn our attention to quadratic equations that do not have the first-degree term— that is, equations of the special form ax2 0004 c 0005 0
a00060
The method of solution of this special form makes direct use of the square root property:
Z SQUARE ROOT PROPERTY If A2 0002 C, then A 0002 00051C.
The use of the square root property is illustrated in Example 2.
EXAMPLE
2
Using the Square Root Property Solve using the square root property: (A) 9x2 0003 7 0002 0
SOLUTIONS
(B) 3x2 0004 27 0002 0
(A) 9x2 0003 7 0002 0 9x2 0002 7 7 x2 0002 9 x00020005
(C) (x 0004 12)2 0002 54
Add 7 to both sides. Divide both sides by 9. Apply the square root property; don’t forget the 0007!
7 17 00020005 B9 3
Solution set: e
17 000317 , f 3 3
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(B) 3x2 0004 27 0002 0 x2 0002 00039 x 0002 0005 100039 000200053i (C) (x
MATCHED PROBLEM 2
0004 12)2 x 0004 12
5 4
0002 0002 0005 254 1 15 x00020003 0005 2 2 00031 0005 15 0002 2
Quadratic Equations and Applications
87
Solve for x2. Apply the square root property. Solution set: 500033i, 3i6
Apply the square root property. Subtract
1 2
from both sides, and simplify 254 .
Combine fractions with common denominators.
0002
Solve using the square root property: (A) 9x2 0003 5 0002 0
(B) 2x2 0004 8 0002 0
(C) (x 0004 13)2 0002 29 0002
Note: It is common practice to represent solutions of quadratic equations informally by the last equation (Example 2, part C) rather than by writing a solution set using set notation (Example 2, parts A and B). From now on, we will follow this practice unless we need to make a special point.
Z Using Completing the Square to Solve Quadratic Equations The methods of square root property and factoring are generally fast when they apply; however, there are equations, such as x2 0004 6x 0003 2 0002 0, that cannot be solved directly by these methods. A more general procedure must be developed to take care of this type of equation. One is called the method of completing the square.* This method is based on the process of transforming the standard quadratic equation ax2 0004 bx 0004 c 0002 0 into the form (x 0004 A)2 0002 B where A and B are constants. Equations of this form can easily be solved by using the square root property. But how do we transform the first equation into the second? We will need to find a way to make the left side factor as a perfect square.
ZZZ EXPLORE-DISCUSS 2
Replace ? in each of the following with a number that makes the equation valid. (A) (x 0004 1)2 0002 x2 0004 2x 0004 ?
(B) (x 0004 2)2 0002 x2 0004 4x 0004 ?
(C) (x 0004 3)2 0002 x2 0004 6x 0004 ?
(D) (x 0004 4)2 0002 x2 0004 8x 0004 ?
Replace ? in each of the following with a number that makes the expression a perfect square of the form (x 0004 h)2. (E) x2 0004 10x 0004 ?
(F) x2 0004 12x 0004 ?
(G) x2 0004 bx 0004 ?
Given the quadratic expression x2 0004 bx *We will find many other uses for this important method.
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what number should be added to this expression to make it a perfect square? To find out, consider the square of the following expression:
{
{
(x 0004 m)2 0002 x2 0004 2mx 0004 m2
m2 is the square of one-half the coefficient of x.
We see that the third term on the right side of the equation is the square of one-half the coefficient of x in the second term on the right; that is, m2 is the square of 12(2m). This observation leads to the following rule:
Z COMPLETING THE SQUARE To complete the square of a quadratic expression of the form x2 0004 bx, add the square of one-half the coefficient of x; that is, add (b兾2)2, or b2兾4. The resulting expression factors as a perfect square,
EXAMPLE
3
x2 0004 bx
For example, x2 0004 5x
b 2 b 2 x2 0004 bx 0004 a b 0002 ax 0004 b 2 2
5 2 5 2 x2 0004 5x 0004 a b 0005 ax 0004 b 2 2
Completing the Square Complete the square for each of the following: (A) x2 0003 3x
SOLUTIONS
00033 2 9 b ; that is, and factor. 2 4
(A) x2 0003 3x x2 0003 3x 0004 94 0002 (x 0003 32)2
Add a
(B) x2 0003 bx
Add a
x2 0003 bx 0004
MATCHED PROBLEM 3
(B) x2 0003 bx
b 2 b2 0002 ax 0003 b 4 2
0003b 2 b2 b ; that is, and factor. 2 4
0002
Complete the square for each of the following: (A) x2 0003 5x
(B) x2 0004 mx
0002
You should note that the rule for completing the square applies only if the coefficient of the second-degree term is 1. This causes little trouble, however, as you will see. To solve equations by completing the square, we will add b2兾4 to both sides after moving the constant term to the right side.
EXAMPLE
4
Solution by Completing the Square Solve by completing the square: (A) x2 0004 6x 0003 2 0002 0
(B) 2x2 0003 4x 0004 3 0002 0
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SOLUTIONS
(A) x2 0004 6x 0003 2 0002 0 x2 0004 6x 0002 2
Complete the square on the left side and add (2b )2 0005 (62 )2 0005 9 to both sides. Factor the left side; add on the right. Use the square root property. Don’t forget the ⫾! Add 00033 to both sides.
(B) 2x2 0003 4x 0004 3 0002 0 x2 0003 2x 0004 32 0002 0
Make the leading coefficient 1 by dividing both sides by 2. Subtract 000332 from both sides. 2 Complete the square on the left side and add (2b )2 0005 (00032 2 ) 0005 1 to both sides.
x2 0003 2x 0002 000332 x2 0003 2x 0004 1 0002 000332 0004 1 (x 0003 1)2 0002 000312 x 0003 1 0002 0005 2000312 x 0002 1 0005 i212
MATCHED PROBLEM 4
89
Add 2 to both sides to obtain the form x2 0004 bx on the left side.
x2 0004 6x 0004 9 0002 2 0004 9 (x 0004 3)2 0002 11 x 0004 3 0002 0005 111 x 0002 00033 0005 111
000210005
Quadratic Equations and Applications
Factor the left side; add on the right. Use the square root property. Add 1 to both sides and simplify 2000312 .
12 i 2
Answer in a 0004 bi form.
0002
Solve by completing the square: (A) x2 0004 8x 0003 3 0002 0
(B) 3x2 0003 12x 0004 13 0002 0
0002
Z Using the Quadratic Formula to Solve Quadratic Equations If we solve a generic quadratic equation using the method of completing the square, the result will be a formula for solving any quadratic equation. ax2 0004 bx 0004 c 0002 0 b c x2 0004 x 0004 0002 0 a a
a00060
b c x2 0004 x 0002 0003 a a x2 0004
b b2 b2 c x0004 20002 20003 a a 4a 4a
b 2 b2 0003 4ac b 0002 2a 4a2 b b2 0003 4ac x0004 00020005 2a B 4a2 b 2b2 0003 4ac x00020003 0005 2a 2a 0003b 0005 2b2 0003 4ac x0002 2a
ax 0004
Make the leading coefficient 1 by dividing by a. Subtract
c from both sides. a
Complete the square on the left side and add b2 b 2 a b 0005 to both sides. 2a 4a2 Factor the left side and combine terms on the right side, getting a common denominator. Use the square root property. b2 0003 4ac b to both sides and simplify B 4a2 2a (see Problem 75 in Exercises 1-5). Add 0003
Combine terms on the right side.
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The result is known as the quadratic formula: Z THEOREM 1 Quadratic Formula If ax2 0004 bx 0004 c 0002 0, a 0006 0, then 0003b 0007 2b2 0003 4ac 2a
x0005
The quadratic formula should be memorized and used to solve quadratic equations when other methods fail, or are more difficult to apply.
EXAMPLE
5
Using the Quadratic Formula Solve 2x 0004 32 0002 x2 using the quadratic formula. Leave the answer in simplest radical form. 2x 0004 32 0002 x2 4x 0004 3 0002 2x2
SOLUTION
Multiply both sides by 2. Write in standard form.
Identify a, b, and c and use the quadratic 2x2 0003 4x 0003 3 0002 0 formula: a 0005 2, b 0005 00034, c 0005 00033 2 0003b 0005 2b 0003 4ac x0002 2a 0003(00034) 0005 2(00034)2 0003 4(2)(00033) 0002 2(2) 4 0005 2110 2 0005 110 4 0005 140 0002 0002 0002 4 4 2
ZZZ
1. 000342 0006 (00034)2
CAUTION ZZZ
2. 2 0004 3.
MATCHED PROBLEM 5
110 2 0004 110 0006 2 2
4 0005 2110 0006 00052110 4
0002
000342 0005 000316 and (00034)2 0005 16 20004
110 4 0004 110 0005 2 2
2(2 0007 110) 4 0007 2 110 2 0007 110 0005 0005 4 4 2
Solve x2 0003 52 0002 00033x by use of the quadratic formula. Leave the answer in simplest radical form. 0002 The expression under the square root in the quadratic formula, b2 0003 4ac, is called the discriminant. It gives us useful information about the corresponding roots, as shown in Table 1. Table 1 Discriminant and Roots Discriminant b2 0003 4ac
Roots of ax2 0004 bx 0004 c 0005 0 a, b, and c real numbers, a 0006 0
Positive
Two distinct real roots
0
One real root (a double root)
Negative
Two imaginary roots, one the conjugate of the other
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EXAMPLE
6
Quadratic Equations and Applications
91
Using the Discriminant Find the number of real roots of each quadratic equation. (A) 2x2 0002 4x 0003 1 0004 0
SOLUTIONS
(B) 2x2 0002 4x 0003 2 0004 0
(C) 2x2 0002 4x 0003 3 0004 0
(A) b2 0002 4ac 0004 (00024)2 0002 4(2)(1) 0004 8 7 0; two real roots (B) b2 0002 4ac 0004 (00024)2 0002 4(2)(2) 0004 0; one real (double) root (C) b2 0002 4ac 0004 (00024)2 0002 4(2)(3) 0004 00028 6 0; no real roots (two imaginary roots)
MATCHED PROBLEM 6
0002
Find the number of real roots of each quadratic equation. (A) 3x2 0002 6x 0003 5 0004 0
(B) 3x2 0002 6x 0003 1 0004 0
(C) 3x2 0002 6x 0003 3 0004 0
0002
Z Solving Applications Involving Quadratic Equations Now that we’re good at solving quadratic equations, we can use them to solve many applied problems. It would be a good idea to review the problem-solving strategy on page 47 before beginning.
EXAMPLE
7
Setting Up and Solving a Word Problem The sum of a number and its reciprocal is
SOLUTION
Find all such numbers.
Let x 0004 the number we’re asked to find; then its reciprocal is 1x . x0003
13 1 0004 x 6
Multiply both sides by the LCD, 6x. [Note: x cannot be zero.]
1 13 (6x)x 0003 (6x) 0004 (6x) x 6 6x2 0003 6 0004 13x 6x 0002 13x 0003 6 0004 0 (2x 0002 3)(3x 0002 2) 0004 0 2x 0002 3 0004 0 x 0004 32
Make sure to multiply every term by 6x.
Subtract 13x from both sides.
2
These are two such numbers: CHECK
MATCHED PROBLEM 7
13 6.
3 2
Factor the left side. Use the zero product property.
3x 0002 2 0004 0 x 0004 23
or
3 2
Solve each equation for x.
and 23.
0003 23 0004 136
2 3
0003 32 0004 136
0002
The sum of two numbers is 23 and their product is 132. Find the two numbers. [Hint: If one number is x, then the other number is 23 0002 x.] 0002
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8
A Distance–Rate–Time Problem A casino boat takes 1.6 hours longer to go 36 miles up a river than to return. If the rate of the current is 4 miles per hour, what is the speed of the boat in still water?
SOLUTION
Let x 0002 Speed of boat in still water x 0004 4 0002 Speed downstream x 0003 4 0002 Speed upstream
a
Time Time b0003a b 0002 1.6 upstream downstream 36 x00034
0003
36(x 0004 4) 0003
36 x00044
Use Time 0005
Multiply both sides by (x 0003 4)(x 0004 4), the LCD.
0002 1.6
36(x 0003 4)
Distance . Rate
0002 1.6(x 0003 4)(x 0004 4)
Multiply out parentheses. Combine like terms and isolate 1.6x 2 on one side of the equation.
2
36x 0004 144 0003 36x 0004 144 0002 1.6x 0003 25.6 1.6x2 0002 313.6
Divide both sides by 1.6.
2
x 0002 196
Use the square root property.
x 0002 0005 1196 0002 000514 The speed in still water is 14 miles per hour. (The negative answer is thrown out, because it doesn’t make sense in the problem to have a negative speed.) CHECK
Time upstream 0002 0003 Time downstream 0002
D 36 0002 0002 3.6 R 14 0003 4 36 D 0002 00022 R 14 0004 4 1.6
MATCHED PROBLEM 8
Difference of times
0002
Two boats travel at right angles to each other after leaving a dock at the same time. One hour later they are 25 miles apart. If one boat travels 5 miles per hour faster than the other, what is the rate of each? [Hint: Use the Pythagorean theorem,* remembering that distance equals rate times time.] 0002 In Example 9, we introduce some concepts from economics that will be used throughout this book. The quantity of a product that people are willing to buy during some period
*Pythagorean theorem: In a right triangle, the square of the length of the longest side is equal to the sum of the squares of the lengths of the two shorter sides.
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93
of time is called the demand for that product. The price p of a product and the demand q for that product are often related by a price–demand equation of the following form: q 0002 a 0003 bp
q is the number of items that can be sold at $p per item.
The constants a and b in a price–demand equation are usually determined by using historical data and statistical analysis. The amount of money received from the sale of q items at $p per item is called the revenue and is given by R 0002 (Number of items sold) 0007 (Price per item) 0002 qp 0002 (a 0003 bp)p
EXAMPLE
9
Using the price-demand equation
Price and Demand The daily price–demand equation for whole milk in a chain of supermarkets is q 0002 5,600 0003 800p where p is the price per gallon and q is the number of gallons sold per day. Find the price(s) that will produce a revenue of $9,500. Round answer(s) to two decimal places.
SOLUTION
The revenue equation is R 0002 qp 0002 (5,600 0003 800p)p 0002 5,600p 0003 800p2 To get a revenue of $9,500, we substitute 9,500 for R: 5,600p 0003 800p2 0002 9,500 00039,500 0004 5,600p 0003 800p2 0002 0 p2 0003 7p 0004 11.875 0002 0 7 0005 11.5 2 0002 2.89, 4.11
Subtract 9,500 from both sides. Divide both sides by 0003800. Use the quadratic formula with a 0005 1, b 0005 00037, and c 0005 11.875.
p0002
Selling whole milk for either $2.89 per gallon or $4.11 per gallon will produce a revenue of $9,500. 0002 MATCHED PROBLEM 9
If the price–demand equation for milk is q 0002 4,800 0003 600p, find the price that will produce revenues of (A) $9,300
(B) $10,500
ANSWERS TO MATCHED PROBLEMS 1. (A) x 0002 00032, 7 (B) x 0002 00034, 53 (C) x 0002 000332 (a double root) (D) x 0002 0, 54 2. (A) x 0002 0005 15 2 (B) x 0002 00052i (C) x 0002 (00031 0005 12)3 3. (A) x2 0003 5x 0004 254 0002 (x 0003 52)2 (B) x2 0004 mx 0004 (m24) 0002 [x 0004 (m2)] 2 4. (A) x 0002 000340005 119 (B) x 0002 (6 0005 i13)3 or 20005 (133)i 5. x 0002 (000330005 119)2 6. (A) No real roots (two imaginary roots) (B) Two real roots (C) One real (double) root 7. 11 and 12 8. 15 and 20 miles per hour 9. (A) $3.29 or $4.71 (B) Not possible
0002
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Exercises
Leave all answers involving radicals in simplified radical form unless otherwise stated.
In Problems 41–56, solve by any method. 41. 12x2 ⫹ 7x ⫽ 10
42. 9x2 ⫹ 9x ⫽ 4
1. How can you tell when an equation is quadratic?
43. (2y ⫺ 3)2 ⫽ 5
44. (3m ⫹ 2)2 ⫽ ⫺4
2. What do a, b, and c in the quadratic formula stand for?
45. x2 ⫽ 3x ⫹ 1
46. x2 ⫹ 2x ⫽ 2
3. Explain what the zero product property says in your own words.
47. 7n2 ⫽ ⫺4n
48. 8u2 ⫹ 3u ⫽ 0
4. Explain what the square root property says in your own words. 5. If you could only use one of factoring, completing the square, and quadratic formula on an important test featuring a variety of quadratic equations, which would you choose, and why?
49. 1 ⫹
9. ⫺8 ⫽ 22t ⫺ 6t2 11. 3w2 ⫹ 13w ⫽ 10
10. 25z2 ⫽ ⫺10z 12. 36x2 ⫽ ⫺12x ⫺ 1
In Problems 13–24, solve by using the square root property.
2 3 ⫽ 2⫹1 u u
53.
4 1 2 ⫽ ⫺ x⫺2 x⫺3 x⫹1
24 24 ⫹1⫽ 10 ⫹ m 10 ⫺ m
52.
1.2 1.2 ⫽1 ⫹ y y⫺1
54.
2 4 3 ⫺ ⫽ x⫺1 x⫹3 x⫺2
55.
x⫺1 x⫹2 x2 ⫽1⫺ ⫺ 2 x⫹3 3⫺x x ⫺9
56.
x⫹3 2x ⫺ 3 11 ⫹ ⫽ 2⫺x x⫹2 x2 ⫺ 4
In Problems 7–12, solve by factoring. 8. 3y2 ⫽ y ⫹ 10
50.
51.
6. Does every quadratic equation have two solutions? Explain.
7. 2x2 ⫽ 8x
8 4 ⫽ x x2
13. m2 ⫺ 25 ⫽ 0
14. n2 ⫹ 16 ⫽ 0
15. c2 ⫹ 9 ⫽ 0
16. d 2 ⫺ 36 ⫽ 0
In Problems 57–60, solve for the indicated variable in terms of the other variables. Use positive square roots only.
17. 4y2 ⫹ 9 ⫽ 0
18. 9x2 ⫺ 25 ⫽ 0
57. s ⫽ 12gt 2
2
2
19. 25z ⫺ 32 ⫽ 0 2
21. (2k ⫺ 5) ⫽ 16 2
23. (n ⫺ 3) ⫽ ⫺4
20. 16w ⫹ 27 ⫽ 0 2
22. (t ⫺ 2) ⫽ ⫺3
26. y2 ⫺ 4y ⫹ 7 ⫽ 0
27. x2 ⫺ 2x ⫹ 3 ⫽ 0
28. y2 ⫺ 4y ⫹ 1 ⫽ 0
2
30. 9s ⫹ 2 ⫽ 12s
2
32. 9s2 ⫹ 7 ⫽ 12s
31. 2t ⫹ 1 ⫽ 6t
2
In Problems 33–40, solve by completing the square. 2
33. x ⫺ 4x ⫺ 1 ⫽ 0
2
34. y ⫹ 4y ⫺ 3 ⫽ 0
2
36. 2s ⫺ 6s ⫹ 7 ⫽ 0
2
38. 4v2 ⫹ 16v ⫹ 23 ⫽ 0
35. 2r ⫹ 10r ⫹ 11 ⫽ 0 37. 4u ⫹ 8u ⫹ 15 ⫽ 0 2
39. 3w ⫹ 4w ⫹ 3 ⫽ 0
2
2
40. 3z ⫺ 8z ⫹ 1 ⫽ 0
60. A ⫽ P(1 ⫹ r)2 for r
61. Consider the quadratic equation
24. (5m ⫺ 6) ⫽ 7
25. x2 ⫺ 2x ⫺ 1 ⫽ 0 29. 2t ⫹ 8 ⫽ 6t
59. P ⫽ EI ⫺ RI 2 for I
2
In Problems 25–32, use the discriminant to determine the number of real roots of each equation and then solve each equation using the quadratic formula.
58. a2 ⫹ b2 ⫽ c2 for a
for t
x2 ⫹ 4x ⫹ c ⫽ 0 where c is a real number. Discuss the relationship between the values of c and the three types of roots listed in Table 1. 62. Consider the quadratic equation x2 ⫺ 2x ⫹ c ⫽ 0 where c is a real number. Discuss the relationship between the values of c and the three types of roots listed in Table 1. Solve the equation in Problems 63–66 and leave answers in simplified radical form (i is the imaginary unit). 63. x2 ⫹ 3ix ⫺ 2 ⫽ 0
64. x2 ⫺ 7ix ⫺ 10 ⫽ 0
65. x2 ⫹ 2ix ⫽ 3
66. x2 ⫽ 2ix ⫺ 3
In Problems 67 and 68, find all solutions. 67. x3 ⫺ 1 ⫽ 0
68. x4 ⫺ 1 ⫽ 0
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69. Prove that when the discriminant of a quadratic equation with real coefficients is negative, the equation has two imaginary solutions. 70. Prove that when the discriminant of a quadratic equation with real coefficients is zero, the equation has one real solution.
Quadratic Equations and Applications
83. CONSTRUCTION A gardener has a 30 foot by 20 foot rectangular plot of ground. She wants to build a brick walkway of uniform width on the border of the plot (see the figure). If the gardener wants to have 400 square feet of ground left for planting, how wide (to two decimal places) should she build the walkway? x
71. Can a quadratic equation with rational coefficients have one rational root and one irrational root? Explain. 72. Can a quadratic equation with real coefficients have one real root and one imaginary root? Explain. 73. Show that if r1 and r2 are the two roots of ax2 0004 bx 0004 c 0002 0, then r1r2 0002 ca. 74. For r1 and r2 in Problem 73, show that r1 0004 r2 0002 0003ba. 75. In one stage of the derivation of the quadratic formula, we replaced the expression 00052(b2 0003 4ac)4a2 00052b2 0003 4ac2a What justifies using 2a in place of 冟 2a 冟? 76. Find the error in the following “proof ” that two arbitrary numbers are equal to each other: Let a and b be arbitrary numbers such that a 0006 b. Then 2
2
20 feet 30 feet
84. CONSTRUCTION Refer to Problem 83. The gardener buys enough bricks to build 160 square feet of walkway. Is this sufficient to build the walkway determined in Problem 83? If not, how wide (to two decimal places) can she build the walkway with these bricks? 85. CONSTRUCTION A 1,200 square foot rectangular garden is enclosed with 150 feet of fencing. Find the dimensions of the garden to the nearest tenth of a foot.
with
2
95
2
2
(a 0003 b) 0002 a 0003 2ab 0004 b 0002 b 0003 2ab 0004 a
86. CONSTRUCTION The intramural fields at a small college will cover a total area of 140,000 square feet, and the administration has budgeted for 1,600 feet of fence to enclose the rectangular field. Find the dimensions of the field. 87. PRICE AND DEMAND The daily price–demand equation for hamburgers at a fast-food restaurant is
(a 0003 b)2 0002 (b 0003 a)2
q 0002 1,600 0003 200p
a0003b0002b0003a
where q is the number of hamburgers sold daily and p is the price of one hamburger (in dollars). Find the demand and the revenue when the price of a hamburger is $3.
2a 0002 2b a0002b 77. Find two numbers such that their sum is 21 and their product is 104. 78. Find all numbers with the property that when the number is added to itself the sum is the same as when the number is multiplied by itself. 79. Find two consecutive positive even integers whose product is 168. 80. The sum of a number and its reciprocal is
10 3.
Find the number.
APPLICATIONS 81. ALCOHOL CONSUMPTION The beer consumption by Americans for the years 1960–2005 can be modeled by the equation y 0002 00030.0665x2 0004 3.58x 0004 122, where x is the number of years after 1960, and y is the number of ounces of beer consumed per person in that year. Find the per person consumption in 1960, then find in what year the model predicts that consumption will return to that level. 82. ALCOHOL CONSUMPTION The wine consumption by Americans for the years 1985–2005 can be modeled by the equation y 0002 0.0951x2 0003 2.06x 0004 49.0, where x is the number of years after 1985, and y is the number of ounces of wine consumed per person in that year. In what year does the model predict that consumption will reach the 1960 level of beer consumption (see Problem 81)?
88. PRICE AND DEMAND The weekly price–demand equation for medium pepperoni pizzas at a fast-food restaurant is q 0002 8,000 0003 400p where q is the number of pizzas sold weekly and p is the price of one medium pepperoni pizza (in dollars). Find the demand and the revenue when the price is $8. 89. PRICE AND DEMAND Refer to Problem 87. Find the price p that will produce each of the following revenues. Round answers to two decimal places. (A) $2,800 (B) $3,200 (C) $3,400 90. PRICE AND DEMAND Refer to Problem 88. Find the price p that will produce each of the following revenues. Round answers to two decimal places. (A) $38,000 (B) $40,000 (C) $42,000 91. NAVIGATION Two planes travel at right angles to each other after leaving the same airport at the same time. One hour later they are 260 miles apart. If one travels 140 miles per hour faster than the other, what is the rate of each? 92. NAVIGATION A speedboat takes 1 hour longer to go 24 miles up a river than to return. If the boat cruises at 10 miles per hour in still water, what is the rate of the current? 93. AIR SEARCH A search plane takes off from an airport at 6:00 A.M. and travels due north at 200 miles per hour. A second plane leaves that airport at the same time and travels due east at 170 miles
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per hour. The planes carry radios with a maximum range of 500 miles. When (to the nearest minute) will these planes no longer be able to communicate with each other? 94. AIR SEARCH If the second plane in Problem 93 leaves at 6:30 A.M. instead of 6 A.M., when (to the nearest minute) will the planes lose communication with each other? 95. ENGINEERING One pipe can fill a tank in 5 hours less than another. Together they can fill the tank in 5 hours. How long would it take each alone to fill the tank? Compute the answer to two decimal places.
(B) A potential buyer for the building needs to have a floor area of 25,000 square feet. Can the builder accommodate them? 100. ARCHITECTURE An architect is designing a small A-frame cottage for a resort area. A cross section of the cottage is an isosceles triangle with an area of 98 square feet. The front wall of the cottage must accommodate a sliding door that is 6 feet wide and 8 feet high (see the figure). Find the width and height of the cross section of the cottage. [Recall: The area of a triangle with base b and altitude h is bh兾2.]
96. ENGINEERING Two gears rotate so that one completes 1 more revolution per minute than the other. If it takes the smaller gear 1 second less than the larger gear to complete 15 revolution, how many revolutions does each gear make in 1 minute? 97. PHYSICS—ENGINEERING For a car traveling at a speed of v miles per hour, under the best possible conditions the shortest distance d necessary to stop it (including reaction time) is given by the formula d 0002 0.044v2 0004 1.1v, where d is measured in feet. Estimate the speed of a car that requires 165 feet to stop in an emergency. 98. PHYSICS—ENGINEERING If a projectile is shot vertically into the air (from the ground) with an initial velocity of 176 feet per second, its distance y (in feet) above the ground t seconds after it is shot is given by y 0002 176t 0003 16t 2 (neglecting air resistance). (A) Find the times when y is 0, and interpret the results physically. (B) Find the times when the projectile is 16 feet off the ground. Compute answers to two decimal places.
REBEKAH DRIVE
200 feet
99. ARCHITECTURE A developer wants to erect a rectangular building on a triangular-shaped piece of property that is 200 feet wide and 400 feet long (see the figure).
Property A
8 feet
6 feet
101. TRANSPORTATION A delivery truck leaves a warehouse and travels north to factory A. From factory A the truck travels east to factory B and then returns directly to the warehouse (see the figure). The driver recorded the truck’s odometer reading at the warehouse at both the beginning and the end of the trip and also at factory B, but forgot to record it at factory A (see the table). The driver does recall that it was farther from the warehouse to factory A than it was from factory A to factory B. Since delivery charges are based on distance from the warehouse, the driver needs to know how far factory A is from the warehouse. Find this distance.
Property Line
l Proposed Building
Factory A
Factory B
w
FIRST STREET 400 feet
(A) Building codes require that industrial buildings on lots that size have a floor area of at least 15,000 square feet. Find the dimensions that will yield the smallest building that meets code. [Hint: Use Euclid’s theorem* to find a relationship between the length and width of the building.]
*Euclid’s theorem: If two triangles are similar, their corresponding sides are proportional: c
a b
a
c b
b c a 0002 0002 a¿ b¿ c¿
Warehouse
Odometer readings Warehouse
5 2 8 4 6
Factory A
5 2 ? ? ?
Factory B
5 2 9 3 7
Warehouse
5 3 0 0 2
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102. CONSTRUCTION A 14-mile track for racing stock cars consists of two semicircles connected by parallel straightaways (see the figure). In order to provide sufficient room for pit crews, emergency vehicles, and spectator parking, the track must enclose an area of 100,000 square feet. Find the length of the straightaways and the diameter of the semicircles to the nearest foot. [Recall: The area A and circumference C of a circle of diameter d are given by A 0002 d 24 and c 0002 d. ]
1-6
97
100,000 square feet
Additional Equation-Solving Techniques Z Solving Equations Involving Radicals Z Revisiting Equations Involving Absolute Value Z Solving Equations of Quadratic Type
In this section, we’ll study equations that are not quadratic but can be transformed into quadratic equations. We can then solve the quadratic equation, and with a little bit of interpretation, use the solutions to solve the original equation.
Z Solving Equations Involving Radicals In solving an equation involving a radical, like x 0002 1x 0004 2 it seems reasonable that we can remove the radical by squaring each side and then proceed to solve the resulting quadratic equation. Let’s give it a try: Square both sides. x 0002 1x 0004 2 2 2 Recall that ( 1a)2 0005 a if a 0. x 0002 ( 1x 0004 2) 2 Subtract x 0004 2 from both sides. x 0002x00042 2 Factor the left side. x 0003x0003200020 Use the zero product property. (x 0003 2)(x 0004 1) 0002 0 or x0003200020 x0004100020 or x00022 x 0002 00031
Now we check these results in the original equation. Check: x 0002 2
Check: x 0002 00031
x 0002 1x 0004 2 ? 2 0002 12 0004 2 ? 2 0002 14 ✓ 200022
x0002 ? 00031 0002 ? 00031 0002 00031 0006
1x 0004 2 100031 0004 2 11 1
That’s interesting: 2 is a solution, but 00031 is not. These results are a special case of Theorem 1.
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Z THEOREM 1 Squaring Operation on Equations If both sides of an equation are squared, then the solution set of the original equation is a subset of the solution set of the new equation. Equation x00053 x2 0005 9
Solution Set {3} {00033, 3}
This theorem provides us with a method of solving some equations involving radicals. It is important to remember that any new equation obtained by raising both sides of an equation to the same power may have solutions, called extraneous solutions, that are not solutions of the original equation. Fortunately though, any solution of the original equation must be among those of the new equation. When raising both sides of an equation to a power, checking solutions is not just a good idea—it is essential to identify any extraneous solutions.
Squaring both sides of the equations x 0002 1x and x 0002 0003 1x produces the new equation x2 0002 x. Find the solutions to the new equation and then check for extraneous solutions in each of the original equations.
ZZZ EXPLORE-DISCUSS 1
EXAMPLE
1
Solving Equations Involving Radicals Solve: (A) x 0004 1x 0003 4 0002 4
SOLUTIONS
(A)
(B) 12x 0004 3 0003 1x 0003 2 0002 2
x 0004 1x 0003 4 0002 4 1x 0003 4 0002 4 0003 x
Isolate radical on one side. Square both sides.
(1x 0003 4)2 0002 (4 0003 x)2
See the upcoming caution on squaring the right side.
x 0003 4 0002 16 0003 8x 0004 x2 x 0003 9x 0004 20 0002 0 (x 0003 5)(x 0003 4) 0002 0 2
CHECK
Write in standard form. Factor left side. Use the zero product property.
x0003500020
or
x0003400020
x00025
or
x00024
x00025 x 0004 1x 0003 4 0002 4 ? 5 0004 15 0003 4 0002 4 600064
x00024 x 0004 1x 0003 4 0002 4 ? 4 0004 14 0003 4 0002 4 ✓ 400024
This shows that 4 is a solution to the original equation and 5 is extraneous. The only solution is x 0002 4.
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99
(B) To solve an equation that contains more than one radical, isolate one radical at a time and square both sides to eliminate the isolated radical. Repeat this process until all the radicals are eliminated. 12x 0004 3 0003 1x 0003 2 0002 2 12x 0004 3 0002 1x 0003 2 0004 2
Isolate one of the radicals. Square both sides. See the upcoming caution on squaring the right side.
(12x 0004 3)2 0002 ( 1x 0003 2 0004 2)2 2x 0004 3 0002 x 0003 2 0004 41x 0003 2 0004 4 x 0004 1 0002 41x 0003 2
Isolate the remaining radical. Square both sides.
(x 0004 1)2 0002 (4 1x 0003 2)2 x2 0004 2x 0004 1 0002 16(x 0003 2) x2 0003 14x 0004 33 0002 0 (x 0003 3)(x 0003 11) 0002 0
CHECK
x0003300020
or
x00023
or
x00023 12x 0004 3 0003 1x 0003 2 0002 2 ? 12(3) 0004 3 0003 13 0003 2 0002 2 ✓ 200022
Write in standard form. Factor left side. Use the zero property.
x 0003 11 0002 0 x 0002 11 x 0002 11 12x 0004 3 0003 1x 0003 2 0002 2 ? 12(11) 0004 3 0003 111 0003 2 0002 2 ✓ 2 00022
Both solutions check, so there are two solutions: x 0002 3, 11. MATCHED PROBLEM 1
Solve: (A) x 0003 5 0002 1x 0003 3
ZZZ
0002
(B) 12x 0004 5 0004 1x 0004 2 0002 5 0002
1. When squaring both sides, it is very important to isolate the radical first. 2. Be sure to square binomials like (4 0003 x) by first writing as (4 0003 x)(4 0003 x) and then multiplying. Remember: (4 0003 x)2 0006 42 0003 x2.
CAUTION ZZZ
Z Revisiting Equations Involving Absolute Value Squaring both sides of an equation can be a useful operation even if the equation does not involve any radicals. Because |x|2 0002 x2 for any x, squaring can be helpful in some absolute value equations.
EXAMPLE
2
Absolute Value Equations Revisited Solve the following equation by squaring both sides: |x 0004 4| 0002 3x 0003 8
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|x 0004 4| 0002 3x 0003 8
SOLUTION
Square both sides.
|x 0004 4|2 0002 (3x 0003 8)2
Use |x 0004 4|2 0005 (x 0004 4)2 and expand each side.
x2 0004 8x 0004 16 0002 9x2 0003 48x 0004 64 8x2 0003 56x 0004 48 0002 0 x2 0003 7x 0004 6 0002 0 (x 0003 1)(x 0003 6) 0002 0
Divide both sides by 8. Factor the left side. Use the zero product property.
x0003100020
or
x0003600020
x00021
or
x00026
x00021 |x 0004 4| 0002 3x 0003 8 ? |1 0004 4| 0002 3(1) 0003 8 ? |5| 0002 00035 5 0006 00035
CHECK
Write in standard form.
x00026 冟 x 0004 4 冟 0002 3x 0003 8 ? |6 0004 4| 0002 3(6) 0003 8 ? 冟 10 冟 0002 10 ✓ 10 0002 10
The only solution is x 0002 6. Compare this solution with the solution of Example 6, Section 1-3. Squaring both sides eliminates the need to consider two separate cases. 0002 MATCHED PROBLEM 2
Solve the following equation by squaring both sides: 冟 3x 0003 4 冟 0002 x 0004 4 0002
Z Solving Equations of Quadratic Type Quadratic equations in standard form have two terms with the variable; one has power 2, the other power 1. When equations have two variable terms where the larger power is twice the smaller, we can use quadratic solving techniques.
EXAMPLE
3
Solving an Equation of Quadratic Type Solve x23 0003 x13 0003 6 0002 0.
SOLUTIONS
Method I. Direct solution: Note that the larger power (23) is twice the smaller. Using the properties of exponents from basic algebra, we can write x23 as (x13)2 and solve by factoring. (x13)2 0003 x13 0003 6 0002 0 (x13 0003 3)(x13 0004 2) 0002 0 or x13 0002 3 x13 0002 00032 13 3
(x
3
) 00023
x 0002 27 The solution is x 0002 27, 00038
13 3
(x
Factor left side. Use the zero product property. Cube both sides. 3
) 0002 (00032) x 0002 00038
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101
Method II. Using substitution: Replace x100053 (the smaller power) with a new variable u. Then the larger power x200053 is u2. This gives us a quadratic equation with variable u. u2 0004 u 0004 6 0003 0 (u 0004 3)(u 0002 2) 0003 0 u 0003 3, 00042
Factor. Use the zero product property.
This is not the solution! We still need to find the values of x that correspond to u 0003 3 and u 0003 00042. Replacing u with x100053, we obtain x100053 0003 3 x 0003 27
or
x100053 0003 00042 x 0003 00048
Cube both sides.
The solution is x 0003 27, 00048. MATCHED PROBLEM 3
0002
Solve algebraically using both Method I and Method II: x100052 0004 5x100054 0002 6 0003 0.
0002
In general, if an equation that is not quadratic can be transformed to the form au2 0002 bu 0002 c 0003 0 where u is an expression in some other variable, then the equation is called an equation of quadratic type. Equations of quadratic type often can be solved using quadratic methods.
ZZZ EXPLORE-DISCUSS 2
Which of the following can be transformed into a quadratic equation by making a substitution of the form u 0003 xn? What is the resulting quadratic equation? (A) 3x00044 0002 2x00042 0002 7 0003 0
(B) 7x5 0004 3x2 0002 3 0003 0
(C) 2x5 0002 4x2 1x 0004 6 0003 0
(D) 8x00042 1x 0004 5x00041 1x 0004 2 0003 0
In general, if a, b, c, m, and n are nonzero real numbers, when can an equation of the form axm 0002 bxn 0002 c 0003 0 be transformed into an equation of quadratic type?
EXAMPLE
4
Solving an Equation of Quadratic Type Solve: 3x4 0004 5x2 0002 1 0003 0
SOLUTION
If we let u 0003 x2, then u2 0003 x4, and the equation becomes 3u2 0004 5u 0002 1 0003 0 5 0006 113 u0003 6 x2 0003
5 0006 113 6
x00030006
MATCHED PROBLEM 4
5 0006 113 B 6
Use the quadratic formula with a ⴝ 3, b ⴝ ⴚ5, c ⴝ 1. Substitute x2 back in for u.
Use the square root property to solve for x.
There are four solutions.
0002
Solve: 2x4 0002 3x2 0004 4 0003 0 0002
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Many applied problems result in equations that can be solved using the techniques in this section.
EXAMPLE
5
An Application: Court Design A hardcourt version of the game broomball becomes popular on college campuses because it enables people to hit each other with a stick. The court is a rectangle with diagonal 30 feet and area 400 square feet. Find the dimensions to one decimal place.
SOLUTION
t
30
fee
Draw a rectangle and label the dimensions as shown in Figure 1. The area is given by A 0002 xy. Also, x2 0004 y2 0002 302 (Pythagorean theorem), and we can solve for y to get y 0002 2900 0003 x2. Now substitute in for y in our area equation, then set area equal to 400 and solve.
y
x 2900 0003 x2 0002 400 x2(900 0003 x2) 0002 160,000 2
4
900x 0003 x 0002 160,000
x
2 2
2
(x ) 0003 900x 0004 160,000 0002 0
Z Figure 1
Square both sides. Multiply out parentheses. Write in standard quadratic form.
Use quadratic formula with a 0005 1, b 0005 0003900, and c 0005 160,000.
900 0005 2(0003900)2 0003 4(1)(160,000) x2 0002 2 0002
900 0005 1170,000 2
Simplify inside the square root.
Use a calculator.
x2 ⬇ 656.2, 243.8 x 0002 1656.2 ⬇ 25.6 or 1243 ⬇ 15.6
Use square root property; discard negative solutions.
If x 0002 25.6, then y 0002 2900 0003 25.62 ⬇ 15.6. If x 0002 15.6, then y 0002 2900 0003 15.62 ⬇ 25.6. In either case, the dimensions are 25.6 feet by 15.6 feet. CHECK Area: 25.6 0007 15.6 0002 399.36 ⬇ 400
Diagonal: 225.62 0004 15.62 ⬇ 30 MATCHED PROBLEM 5
0002
If the area of a right triangle is 24 square inches and the hypotenuse is 12 inches, find the lengths of the legs of the triangle correct to one decimal place. 0002 ANSWERS TO MATCHED PROBLEMS 1. (A) x 0002 7 (B) x 0002 2 2. x 0002 0, 4 3. x 0002 16, 81 0005200033 0005 141 4. x 0002 5. 11.2 inches by 4.3 inches 2
1-6
Exercises
1. What is meant by the term “extraneous solution”? 2. When is it necessary to check for extraneous solutions? 3. How can squaring both sides help in solving absolute value equations?
4. How can you recognize when an equation is of quadratic type? In Problems 5–12, determine the validity of each statement. If a statement is false, explain why. 5. If x2 0002 5, then x 0002 0005 15.
6. 125 0002 00055
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7. (x 0004 5)2 0002 x2 0004 25 9. (1x 0003 1 0004 1)2 0002 x 11. If x3 0002 2, then x 0002 8.
8. (2x 0003 1)2 0002 4x2 0003 1 10. (1x 0003 1)2 0004 1 0002 x 12. If x1兾3 0002 8, then x 0002 2.
Additional Equation-Solving Techniques
59. t 0003 111t 0004 18 0002 0
103
60. x 0002 15 0003 21x
In Problems 61–68, solve the equation. 61. 17 0003 2x 0003 1x 0004 2 0002 1x 0004 5 62. 11 0004 3x 0003 12x 0003 1 0002 1x 0004 2
In Problems 13–26, solve the equation. 13. 1x 0004 2 0002 4
14. 1x 0003 4 0002 2
63. 3 0004 x00034 0002 5x00032
64. 2 0004 4x00034 0002 7x00032
15. 13y 0003 5 0004 10 0002 0
16. 14 0003 x 0004 5 0002 0
65. 21x 0004 5 0002 0.01x 0004 2.04
17. 13y 0003 2 0002 y 0003 2
18. 14y 0004 1 0002 5 0003 y
66. 3 1x 0003 1 0002 0.05x 0004 2.9
19. 15w 0004 6 0003 w 0002 2
20. 12w 0003 3 0004 w 0002 1
67. 2x000325 0003 5x000315 0004 1 0002 0
21. 冟 2x 0004 1 冟 0002 x 0004 2
22. 冟 2x 0004 2 冟 0002 5 0003 x
68. x000325 0003 3x000315 0004 1 0002 0
23. 冟 x 0003 5 冟 0002 7 0003 2x
24. 冟 x 0004 7 冟 0002 1 0003 2x
69. Explain why the following “solution” is incorrect:
25. 冟 3x 0003 4 冟 0002 2x 0003 5
26. 冟 3x 0003 1 冟 0002 x 0003 1
1x 0004 3 0004 5 0002 12 x 0004 3 0004 25 0002 144
In Problems 27–32, transform each equation of quadratic type into a quadratic equation in u and state the substitution used in the transformation. If the equation is not an equation of quadratic type, say so. 4 3 6 27. 2x00036 0003 4x00033 0002 0 28. 0003 0004 2 0002 0 x 7 x 29. 3x3 0003 4x 0004 9 0002 0
30. 7x00031 0004 3x00031/2 0004 2 0002 0
10 4 7 0004 20003 400020 9 x x
32. 3x3/2 0003 5x1/2 0004 12 0002 0
31.
In Problems 33–56, solve the equation. 33. 13t 0003 2 0002 1 0003 2 1t
34. 15t 0004 4 0003 21t 0002 1
35. m4 0004 2m2 0003 15 0002 0
36. m4 0004 4m2 0003 12 0002 0
37. 3x 0002 2x2 0003 2
38. x 0002 25x2 0004 9
23
39. 2y
13
0004 5y
0003 12 0002 0
2
2
2
2
2
2
23
40. 3y
13
0004 2y
0003800020
41. (m 0003 2m) 0004 2(m 0003 2m) 0002 15 42. (m 0004 2m) 0003 6(m 0004 2m) 0002 16 43. 12t 0004 3 0004 2 0002 1t 0003 2
70. Explain why the following “solution” is incorrect. 2x2 0003 16 0002 2x 0004 3 x 0003 4 0002 2x 0004 3 00037 0002 x
APPLICATIONS 71. PHYSICS—WELL DEPTH When a stone is dropped into a deep well, the number of seconds until the sound of a splash is heard is x 1x given by the formula t 0002 0004 , where x is the depth of the 4 1,100 well in feet. For one particular well, the splash is heard 14 seconds after the stone is released. How deep (to the nearest foot) is the well? 72. PHYSICS—WELL DEPTH Refer to Problem 71. For a different well, the sound of the splash is heard 2 seconds after the stone is released. How deep (to the nearest foot) is the well? 73. GEOMETRY The diagonal of a rectangle is 10 inches and the area is 45 square inches. Find the dimensions of the rectangle, correct to one decimal place. 74. GEOMETRY The hypotenuse of a right triangle is 12 inches and the area is 24 square inches. Find the dimensions of the triangle, correct to one decimal place.
44. 12x 0003 1 0003 1x 0003 5 0002 3 45. 1w 0004 3 0004 12 0003 w 0002 3
75. MANUFACTURING A lumber mill cuts rectangular beams from circular logs (see the figure). If the diameter of the log is 16 inches and the cross-sectional area of the beam is 120 square inches, find the dimensions of the cross section of the beam correct to one decimal place.
46. 1w 0004 7 0002 2 0004 13 0003 w 47. 18 0003 z 0002 1 0004 1z 0004 5 48. 13z 0004 1 0004 2 0002 1z 0003 1 49. 24x2 0004 12x 0004 1 0003 6x 0002 9 50. 6x 0003 24x2 0003 20x 0004 17 0002 15 51. y00032 0003 2y00031 0004 3 0002 0
52. y00032 0003 3y00031 0004 4 0002 0
53. 2t00034 0003 5t00032 0004 2 0002 0
54. 15t00034 0003 23t00032 0004 4 0002 0
55. 3z00031 0003 3z00031/2 0004 1 0002 0
56. 2z00031 0003 3z00031/2 0004 2 0002 0
Solve Problems 57–60 two ways: by squaring and by substitution. 57. m 0003 7 1m 0004 12 0002 0
x 0002 116
58. y 0003 6 0004 1y 0002 0
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76. DESIGN A food-processing company packages an assortment of their products in circular metal tins 12 inches in diameter. Four identically sized rectangular boxes are used to divide the tin into eight compartments (see the figure). If the cross-sectional area of each box is 15 square inches, find the dimensions of the boxes correct to one decimal place.
78. DESIGN A paper drinking cup in the shape of a right circular cone is constructed from 125 square centimeters of paper (see the figure). If the height of the cone is 10 centimeters, find the radius correct to two decimal places. r
h
77. CONSTRUCTION A water trough is constructed by bending a 4- by 6-foot rectangular sheet of metal down the middle and attaching triangular ends (see the figure). If the volume of the trough is 9 cubic feet, find the width correct to two decimal places.
Lateral surface area: S ⫽ r 兹r 2 ⫹ h 2
6 feet
2 feet
CHAPTER
1-1
1
Review
Linear Equations and Applications
Solving an equation is the process of finding all values of the variable that make the equation a true statement. An equation that is true for some values of the variable is called a conditional equation. An equation that is true for all permissible values of the variable is called an identity. An equation that is false for all permissible values of the variable is called a contradiction, and has no solution. An equation that can be written in the standard form ax ⫹ b ⫽ 0, a ⫽ 0, is a linear or first-degree equation. Linear
equations are solved by performing algebraic steps that result in equivalent equations until the result is an equation whose solution is obvious. When an equation has fractions, begin by multiplying both sides by the least common denominator of all the fractions. The formula Quantity ⫽ Rate ⫻ Time is useful in modeling problems that involve a rate of change, like speed.
Z STRATEGY FOR SOLVING WORD PROBLEMS 1. Read the problem slowly and carefully, more than once if
4. Write an equation relating the quantities in the problem.
necessary. Write down information as you read the problem the first time to help you get started. Identify what it is that you are asked to find.
Often, you can accomplish this by finding a formula that connects those quantities. Try to write the equation in words first, then translate to symbols.
2. Use a variable to represent an unknown quantity in the
5. Solve the equation, then answer the question in a sentence
problem, usually what you are asked to find. Then try to represent any other unknown quantities in terms of that variable. It’s pretty much impossible to solve a word problem without this step.
by rephrasing the question. Make sure that you’re answering all of the questions asked.
3. If it helps to visualize a situation, draw a diagram and label known and unknown parts.
6. Check to see if your answers make sense in the original problem, not just the equation you wrote.
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Linear Inequalities
The inequality symbols , , 000e, are used to express inequality relations. Line graphs, interval notation, and the set operations of union and intersection are used to describe inequality relations. A solution of a linear inequality in one variable is a value of the variable that makes the inequality a true statement. Two inequalities are equivalent if they have the same solution set. Linear inequalities can be solved using the same basic procedure as linear equations, with one important difference: the direction of an inequality reverses if we multiply or divide both sides by a negative number.
1-3
Absolute Value in Equations and Inequalities
The absolute value of a number x is the distance on a real number line from the origin to the point with coordinate x and is given by 冟x冟 0002
0003x x
冦
if x 6 0 if x 0
The distance between points A and B with coordinates a and b, respectively, is d(A, B) 0002 冟 b 0003 a 冟, which has the following geometric interpretations: 冟 x 0003 c 冟 0002 d Distance between x and c is equal to d.
Because complex numbers obey the same commutative, associative, and distributive properties as real numbers, most operations with complex numbers are performed by using these properties in the same way that algebraic operations are performed on the expression a 0004 bx. Keep in mind that i2 0002 00031. The property of conjugates, (a 0004 bi)(a 0003 bi) 0002 a2 0004 b2 can be used to find reciprocals and quotients. To divide by a complex number, we multiply the numerator and denominator by the conjugate of the denominator. This enables us to write the result in a 0004 bi form. If a 7 0, then the principal square root of the negative real number 0003a is 10003a 0002 i1a. To solve equations involving complex numbers, set the real and imaginary parts equal to each other and solve.
1-5
冟 x 0003 c 冟 7 d Distance between x and c is greater than d.
ax2 0004 bx 0004 c 0002 0
1. Factoring and using the zero product property: m0002n00020
m 0002 0 or n 0002 0 (or both)
If A2 0002 C, then A 0002 00051C 3. Completing the square: b 2 b 2 x2 0004 bx 0004 a b 0002 ax 0004 b 2 2 4. Using the quadratic formula: x0002
3. 冟 x 冟 7 p is equivalent to x 6 0003p or x 7 p.
1-4
if and only if
2. Using the square root property:
2. 冟 x 冟 6 p is equivalent to 0003p 6 x 6 p. These relationships also hold if x is replaced with ax 0004 b. For x any real number, 2x2 0002 冟 x 冟.
a00060
where x is a variable and a, b, and c are constants. Methods of solution include:
Equations and inequalities involving absolute values are solved using the following relationships for p 0: 1. 冟 x 冟 0002 p is equivalent to x 0002 p or x 0002 0003p.
Quadratic Equations and Applications
A quadratic equation is an equation that can be written in the standard form
冟 x 0003 c 冟 6 d Distance between x and c is less than d. 0 6 冟 x 0003 c 冟 6 d Distance between x and c is less than d, but x 0006 c.
105
0003b 0005 2b2 0003 4ac 2a
If the discriminant b2 0003 4ac is positive, the equation has two distinct real roots; if the discriminant is 0, the equation has one real double root; and if the discriminant is negative, the equation has two imaginary roots, each the conjugate of the other.
Complex Numbers
A complex number in standard form is a number in the form a 0004 bi where a and b are real numbers and i denotes a square root of 00031. The number i is known as the imaginary unit. For a complex number a 0004 bi, a is the real part and bi is the imaginary part. If b 0006 0 then a 0004 bi is also called an imaginary number. If a 0002 0 then 0 0004 bi 0002 bi is also called a pure imaginary number. If b 0002 0 then a 0004 0i 0002 a is a real number. The complex zero is 0 0004 0i 0002 0. The conjugate of a 0004 bi is a 0003 bi. Equality, addition, and multiplication are defined as follows:
1. a 0004 bi 0002 c 0004 di if and only if a 0002 c and b 0002 d 2. (a 0004 bi) 0004 (c 0004 di) 0002 (a 0004 c) 0004 (b 0004 d)i 3. (a 0004 bi)(c 0004 di) 0002 (ac 0003 bd) 0004 (ad 0004 bc)i
1-6
Additional Equation-Solving Techniques
A square root radical can be eliminated from an equation by isolating the radical on one side of the equation and squaring both sides of the equation. The new equation formed by squaring both sides may have extraneous solutions. Consequently, every solution of the new equation must be checked in the original equation to eliminate extraneous solutions. If an equation contains more than one radical, then the process of isolating a radical and squaring both sides can be repeated until all radicals are eliminated. If a substitution transforms an equation into the form au2 0004 bu 0004 c 0002 0, where u is an expression in some other variable, then the equation is an equation of quadratic type that can be solved by quadratic methods.
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Review Exercises
Work through all the problems in this chapter review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text.
1. 8x 0004 10 0002 4x 0003 30 2. 4 0003 3(x 0004 2) 0002 5x 0003 7(4 0003 x) y 0004 10 y00041 1 1 0003 0002 0003 15 5 6 10
Solve the equation in Problems 25–30. 25. ay 0004
Solve and graph the inequality in Problems 4–6. 4. 3(2 0003 x) 0003 2 000e 2x 0003 1
5. 冟 y 0004 9 冟 6 5
26. 1 0004
6. 冟 3 0003 2x 冟 000e 5 7. Find the real part, the imaginary part, and the conjugate: (A) 9 0003 4i (B) 5i (C) 000310
27.
8. Perform the indicated operations and write the answer in standard form. (A) (4 0004 7i) 0004 (00032 0003 3i) (B) (00033 0004 5i) 0003 (4 0003 8i) (C) (1 0003 2i)(3 0004 4i) 21 0004 9i (D) 5 0003 2i
12. 2x2 0002 7x 0003 3
13. m2 0004 m 0004 1 0002 0
14. y2 0002 32 ( y 0004 1)
2 3 0002 u u2
2 x 0003 0002 3 28. 2x23 0003 5x13 0003 12 0002 0 x00033 x 0003x00036 2
29. m4 0004 5m2 0003 36 0002 0
30. 1y 0003 2 0003 15y 0004 1 0002 00033
Solve the equation or inequality in Problems 31–35, and round answers to three significant digits if necessary. 31. 2.15x 0003 3.73(x 0003 0.930) 0002 6.11x 32. 00031.52 000e 0.770 0003 2.04x 000e 5.33
4 1 8 34. 2 0003 t2 000e 3 5 2
10. 5x2 0004 20 0002 0
11. 2x2 0002 4x
11 2 b 0002 20 3
33. 冟 9.71 0003 3.62x 冟 7 5.48
Solve the equation in Problems 9–15. 9. 2x2 0003 7 0002 0
23. Perform the indicated operations and write the final answers in standard form: (A) (3 0004 i)2 0003 2(3 0004 i) 0004 3 (B) i 27 24. Convert to a 0004 bi forms, perform the indicated operations, and write the final answers in standard form: (A) (2 0003 100034) 0003 (3 0003 100039) 4 0004 1000325 2 0003 100031 (B) (C) 3 0004 100034 100034
In Problems 1–3, solve the equation.
3.
22. If points A, B, and C have coordinates on a number line of 5, 20, and 00038 respectively, find (A) d(A, B) (B) d(A, C) (C) d(B, C)
35. 6.09x2 0004 4.57x 0003 8.86 0002 0
15. 15x 0003 6 0003 x 0002 0 16. For what values of x does the expression 115 0004 6x represent a real number?
Solve the equation in Problems 36–38 for the indicated variable in terms of the other variables. 36. P 0002 M 0003 Mdt for M (mathematics of finance) 37. P 0002 EI 0003 RI 2 for I (electrical engineering)
Solve the equation in Problems 17 and 18. 7 10 0003 4x 17. 0002 2 20003x x 0004 3x 0003 10
38. x 0002
u00033 1 10003u 18. 0002 0003 2u 0003 2 6 3u 0003 3
x00043 20003x 000e50003 8 3
21. 2(1 0003 2m)2 000e 3
for y
39. Find the error in the following “solution” and then find the correct solution. 3 4 0002 2 x2 0003 4x 0004 3 x 0003 3x 0004 2
Solve and graph the inequality in Problems 19–21. 19.
4y 0004 5 2y 0004 1
20. 冟 3x 0003 8 冟 7 2 [
00031
[
2
m
4x2 0003 12x 0004 8 0002 3x2 0003 12x 0004 9 x2 0002 1 or x00021 x 0002 00031
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40. Consider the quadratic equation x2 ⫺ 8x ⫹ c ⫽ 0 , where c is a real number. Describe the number and type of solutions for c ⫽ ⫺16, 16, and 32. Use your result to make a general statement about the number and type of solutions for certain values of c, then use an inequality to prove your statement. 41. For what values of a and b is the inequality a ⫹ b 6 b ⫺ a true? 42. If a and b are negative numbers and a 7 b, then is aⲐb greater than 1 or less than 1? 1
43. Solve for x in terms of y: y ⫽ 1⫺
1 1⫺x
44. Solve and graph: 0 6 冟 x ⫺ 6 冟 6 d Solve the equation in Problems 45–47. 45. 2x2 ⫽ 13x ⫺ 12 46. 4 ⫽ 8x⫺2 ⫺ x⫺4 47. 2ix2 ⫹ 3ix ⫺ 5i ⫽ 0 48. Evaluate: (a ⫹ bi) a
a b ⫺ 2 ib, a, b ⫽ 0 a2 ⫹ b2 a ⫹ b2
(C) If the crew wants to increase their still-water speed to 18 km/h, how fast must they make the round-trip? Express answer in minutes rounded to one decimal place. 54. COST ANALYSIS Cost equations for manufacturing companies are often quadratic in nature. If the cost equation for manufacturing inexpensive calculators is C ⫽ x2 ⫺ 10x ⫹ 31, where C is the cost of manufacturing x units per week (both in thousands), find: (A) The output for a $15 thousand weekly cost (B) The output for a $6 thousand weekly cost 55. BREAK-EVEN ANALYSIS The manufacturing company in Problem 54 sells its calculators to wholesalers for $3 each. So its revenue equation is R ⫽ 3x, where R is revenue and x is the number of units sold per week (both in thousands). Find the break-even point(s) for the company—that is, the output at which revenue equals cost. 56. POLITICS Before the 2008 presidential election, one news outlet estimated that the percentage of voters casting a vote for Barack Obama would be within 1.2% of 54%. Express this range as an absolute value inequality, then solve the inequality. 57. DESIGN The pages of a textbook have uniform margins of 2 centimeters on all four sides (see the figure). If the area of the entire page is 480 square centimeters and the area of the printed portion is 320 square centimeters, find the dimensions of the page.
APPLICATIONS 49. NUMBERS Find a number such that subtracting its reciprocal 16 from the number gives 15 . 50. SPORTS MEDICINE The following quotation was found in a sports medicine handout: “The idea is to raise and sustain your heart rate to 70% of its maximum safe rate for your age. One way to determine this is to subtract your age from 220 and multiply by 0.7.” (A) If H is the maximum safe sustained heart rate (in beats per minute) for a person of age A (in years), write a formula relating H and A. (B) What is the maximum safe sustained heart rate for a 20-yearold? (C) If the maximum safe sustained heart rate for a person is 126 beats per minute, how old is the person? 51. CHEMISTRY A chemical storeroom has an 80% alcohol solution and a 30% alcohol solution. How many milliliters of each should be used to obtain 50 milliliters of a 60% solution? 52. RATE–TIME A student group flies to Cancun for spring break, a distance of 1,200 miles. The plane used for both trips has an average cruising speed of 300 miles per hour in still air. The trip down is with the prevailing winds and takes 112 hours less than the trip back, against the same strength wind. What is the wind speed? 53. RATE–TIME A crew of four practices by rowing up a river for a fixed distance and then returning to their starting point. The river has a current of 3 km/h. (A) Currently the crew can row 15 km/h in still water. If it takes them 25 minutes to make the round-trip, how far upstream did they row? (B) After some additional practice the crew cuts the round-trip time to 23 minutes. What is their still-water speed now? Round answers to one decimal place.
107
2
2
2
2
2
2 2
2
Figure for 57.
58. DESIGN A landscape designer uses 8-foot timbers to form a pattern of isosceles triangles along the wall of a building (see the figure). If the area of each triangle is 24 square feet, find the base correct to two decimal places.
8 feet
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1
GROUP ACTIVITY Solving a Cubic Equation
If a, b, and c are real numbers with a 0006 0, then the quadratic equation ax2 0004 bx 0004 c 0002 0 can be solved by a variety of methods, including the quadratic formula. How can we solve the cubic equation 3
2
ax 0004 bx 0004 cx 0004 d 0002 0,
a00060
(1)
and is there a formula for the roots of this equation? The first published solution of equation (1) is generally attributed to the Italian mathematician Girolamo Cardano (1501–1576) in 1545. His work led to a complicated formula for the roots of equation (1) that involves topics that are discussed later in this text. For now, we will use Cardano’s method to find a real solution in special cases of equation (1). Note that because a is nonzero, we can always multiply both sides of (1) by 1 a to make the coefficient of x3 equal to 1.
Let x3 0004 bx2 0004 cx 0004 d 0002 0 Example problem: x3 0003 6x2 0004 6x 0003 5 0005 0. Steps will be in red.
Step 1. Substitute x 0002 y 0003 b兾3 to obtain the reduced cubic y3 0004 my 0002 n. 00036 x0005y0003 or x 0005 y 0004 2. The equation becomes 3 (y 0004 2)3 0003 6(y 0004 2) 0004 6(y 0004 2) 0003 5 0005 0,
3
Step 2. Define u and v by m 0002 3uv and n 0002 u 0003 v . Use v 0002 write
m 3 b 3u
Step 3. Using either of the solutions found in step 2, x0002y0003
b b 0002u0003v0003 3 3
is a solution to x3 0004 bx2 0004 cx 0004 d 0002 0 For u 0005 2, v 0005 00031, x 0005 2 0003 (00031) 0003
00036 0005 5 3
(Solution)
x3 0003 6x2 0003 3x 0003 8 0002 0
m 3u
to
(2)
Use a calculator to find a decimal approximation of your solution and check your answer by substituting this approximate value in equation (2). (C) Use Cardano’s method to solve x3 0003 6x2 0004 9x 0003 6 0002 0
which simplifies to y3 0003 6y 0005 9: m 0005 00036, n 0005 9.
n 0002 u3 0003 a
00036 2 2 3 8 or v 0005 0003 ; 9 0005 u3 0003 a0003 b 0005 u3 0004 Multiply both sides by 3u u u u3 3 6 3 u to obtain u 0003 9u 0004 8 0005 0; solve by factoring to get u 0005 2 (in which case v 0005 00031) or u 0005 1 (in which case v 0005 00032). v0005
(A) The key to Cardano’s method is to recognize that if u and v are defined as in step 2, then y 0002 u 0003 v is a solution of the reduced cubic. Verify this by substituting y 0002 u 0003 v, m 0002 3uv, and n 0002 u3 0003 y3 in y3 0004 my 0002 n and show that the result is an identity. (B) Use Cardano’s method to solve
CARDANO’S METHOD FOR SOLVING A CUBIC EQUATION
3
Multiply both sides by u3 to obtain an equation quadratic in u3. Solve for u3 by factoring or by using the quadratic formula. Then solve for u, and find the associated value of v.
(3)
Use a calculator to find a decimal approximation of your solution and check your answer by substituting this approximate value in equation (3). (D) In step 2 of Cardano’s method, show that u3 is real if n 2 0003m 3 a b a b. 2 3
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Graphs C EQUATIONS and inequalities are algebraic objects. A graph, on the other hand, is a geometric object such as a line, circle, or parabola. The idea of visualizing an equation or inequality by means of a graph was crucial to the development of analytic geometry, a subject that combines algebra and geometry. In this chapter, we study the fundamentals of analytic geometry: The Cartesian coordinate system, named after the French mathematician and philosopher René Descartes (1596–1650); the calculation of distances in the plane; and equations of lines and circles. We conclude the chapter by applying linear models to solve real-world problems.
2 OUTLINE 2-1
Cartesian Coordinate Systems
2-2
Distance in the Plane
2-3
Equations of a Line
2-4
Linear Equations and Models Chapter 2 Review Chapter 2 Group Activity: Average Speed
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2-1
Cartesian Coordinate Systems Z Reviewing Cartesian Coordinate Systems Z Graphing: Point by Point Z Using Symmetry as an Aid in Graphing
In Chapter 1, we discussed algebraic methods for solving equations. In this section we show how to find a geometric representation ( graph) of an equation. Examining the graph of an equation often results in additional insight into the nature of the equation’s solutions.
Z Reviewing Cartesian Coordinate Systems y 10
II
I
000210
x
10
III
IV
000210
Z Figure 1 Cartesian coordinate system.
y 10
R 0003 (5, 10)
Q 0003 (000210, 5) a 000210
Origin b (0, 0)
10
P 0003 (a, b)
000210
Z Figure 2 Coordinates in a plane.
x
Just as a real number line is formed by establishing a one-to-one correspondence between the points on a line and the elements in the set of real numbers, we can form a real plane by establishing a one-to-one correspondence between the points in a plane and elements in the set of all ordered pairs of real numbers. This can be done by means of a Cartesian coordinate system. To form a Cartesian or rectangular coordinate system, we select two real number lines, one horizontal and one vertical, and let them cross through their origins, as indicated in Figure 1. Up and to the right are the usual choices for the positive directions. These two number lines are called the horizontal axis and the vertical axis, or, together, the coordinate axes. The horizontal axis is usually referred to as the x axis and the vertical axis as the y axis, and each is labeled accordingly. Other labels may be used in certain situations. The coordinate axes divide the plane into four parts called quadrants, which are numbered counterclockwise from I to IV (see Fig. 1). Given an arbitrary point P in the plane, pass horizontal and vertical lines through the point (Fig. 2). The vertical line will intersect the horizontal axis at a point with coordinate a, and the horizontal line will intersect the vertical axis at a point with coordinate b. These two numbers written as the ordered pair* (a, b) form the coordinates of the point P. The first coordinate a is called the abscissa of P; the second coordinate b is called the ordinate of P. The abscissa of Q in Figure 2 is 000210, and the ordinate of Q is 5. The coordinates of a point can also be referenced in terms of the axis labels. The x coordinate of R in Figure 2 is 5, and the y coordinate of R is 10. The point with coordinates (0, 0) is called the origin. The procedure we have just described assigns to each point P in the plane a unique pair of real numbers (a, b). Conversely, if we are given an ordered pair of real numbers (a, b), then, reversing this procedure, we can determine a unique point P in the plane. There is a one-to-one correspondence between the points in a plane and the elements in the set of all ordered pairs of real numbers. This correspondence is often referred to as the fundamental theorem of analytic geometry. Because of this correspondence, we regularly speak of the point (a, b) when we are referring to the point with coordinates (a, b). We also write P 0003 (a, b) to identify the coordinates of the point P. In Figure 2, referring to Q as the point (000210, 5) and writing R 0003 (5, 10) are both acceptable statements.
*An ordered pair of real numbers is a pair of numbers in which the order is specified. We now use (a, b) as the coordinates of a point in a plane. In Chapter 1, we used (a, b) to represent an interval on a real number line. These concepts are not the same. You must always interpret the symbol (a, b) in terms of the context in which it is used.
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111
Z Graphing: Point by Point Given any set of ordered pairs of real numbers S, the graph of S is the set of points in the plane corresponding to the ordered pairs in S. The fundamental theorem of analytic geometry enables us to look at an algebraic object (a set of ordered pairs) geometrically and to look at a geometric object (a graph) algebraically. We begin by considering an equation in two variables: y 0003 x2 0002 4
(1)
A solution to equation (1) is an ordered pair of real numbers (a, b) such that b 0003 a2 0002 4. The solution set of equation (1) is the set of all its solutions. To find a solution to equation (1) we simply replace one of the variables with a number and solve for the other variable. For example, if x 0003 2, then y 0003 22 0002 4 0003 0, and the ordered pair (2, 0) is a solution. Similarly, if y 0003 5, then 5 0003 x2 0002 4, x2 0003 9, x 0003 00043, and the ordered pairs (3, 5) and (00023, 5) are solutions. Sometimes replacing one variable with a number and solving for the other variable will introduce imaginary numbers. For example, if y 0003 00025 in equation (1), then 00025 0003 x2 0002 4 x2 0003 00021 x 0003 0004 100021 0003 0004i So (0002i, 00025) and (i, 00025) are solutions to y 0003 x2 0002 4. However, the coordinates of a point in a rectangular coordinate system must be real numbers. For that reason, when graphing an equation, we consider only those values of the variables that produce real solutions to the equation. The graph of an equation in two variables is the graph of its solution set. In equation (1), we find that its solution set will have infinitely many elements and its graph will extend off any paper we might choose, no matter how large. To sketch the graph of an equation, we include enough points from its solution set so that the total graph is apparent. This process is called point-by-point plotting.
EXAMPLE
1
Graphing an Equation Using Point-by-Point Plotting Sketch a graph of y 0003 x2 0002 4.
SOLUTION y
y 0003 x2 0002 4
We make a table of solutions—ordered pairs of real numbers that satisfy the given equation.
15
(00024, 12)
(4, 12) 10
(00023, 5)
5
(00022, 0)
(3, 5)
(2, 0)
00025
5
(00021, 00023) 00025
Z Figure 3
(1, 00023) (0, 00024)
x
x
00024
00023
00022
00021
0
1
2
3
4
y
12
5
0
00023
00024
00023
0
5
12
After plotting these solutions, if there are any portions of the graph that are unclear, we plot additional points until the shape of the graph is apparent. Then we join all these plotted points with a smooth curve, as shown in Figure 3. Arrowheads are used to indicate that the graph continues beyond the portion shown here with no significant changes in shape. The resulting figure is called a parabola. Notice that if we fold the paper along the y axis, the right side will match the left side. We say that the graph is symmetric with respect to the y axis and call the y axis the axis of the parabola. More will be said about parabolas later in the text. 0002
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MATCHED PROBLEM 1
Sketch a graph of y2 0003 x. 0002 This book contains a number of activities that use a graphing calculator or computer with appropriate software. All of these activities are clearly marked and easily omitted if no such device is available.
Technology Connections To graph the equation in Example 1 on a graphing calculator, we first enter the equation in the calculator’s equation editor* [Fig. 4(a)]. Using Figure 3 as a guide, we next enter values for the window variables [Fig. 4(b)], and then we graph the equation [Fig. 4(c)]. The values of the window variables, shown in red in Figure 4(c), are not displayed on the calculator screen. We add them to give you additional insight into the graph.
Compare the graphs in Figure 3 and Figure 4(c). They are similar in shape, but they are not identical. The discrepancy is due to the difference in the axes scales. In Figure 3, one unit on the x axis is equal to one unit on the y axes. In Figure 4(c), one unit on the x axis is equal to about three units on the y axis. We will have more to say about axes scales later in this section. 15
00025
5
00025
Enter the equation. (a)
Enter the window variables. (b)
Graph the equation. (c)
Z Figure 4
*See the Technology Index for a list of graphing calculator terms used in this book.
ZZZ EXPLORE-DISCUSS 1
To graph the equation y 0003 0002x3 0005 2x, we use point-by-point plotting to obtain the graph in Figure 5. (A) Do you think this is the correct graph of the equation? If so, why? If not, why? (B) Add points on the graph for x 0003 00022, 00020.5, 0.5, and 2. (C) Now, what do you think the graph looks like? Sketch your version of the graph, adding more points as necessary. (D) Write a short statement explaining any conclusions you might draw from parts A, B, and C.
y 5
x
y
00021 00021 0 0 1 1
00025
5
00025
Z Figure 5
x
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Graphs illustrate the relationship between two quantities, one represented by x coordinates and the other by y coordinates. If no equation for the graph is available, we can find specific examples of this relationship by estimating coordinates of points on the graph. Example 2 illustrates this process.
EXAMPLE
2
Ozone Levels The ozone level during a 12-hour period in a suburb of Milwaukee, Wisconsin, on a particular summer day is given in Figure 6, where L is ozone in parts per billion and t is time in hours. Use this graph to estimate the following ozone levels to the nearest integer and times to the nearest quarter hour. (A) The ozone level at 6 P.M. (B) The highest ozone level and the time when it occurs. (C) The time(s) when the ozone level is 90 ppb. L 120
Parts per billion (ppb)
100
80
60
40
20
0 Noon 1
2
3
4
5
6
7
8
9
10
11
12
t
Z Figure 6 Ozone level. SOLUTIONS
MATCHED PROBLEM 2
(A) The L coordinate of the point on the graph with t coordinate 6 is approximately 97 ppb. (B) The highest ozone level is approximately 109 ppb at 3 P.M. (C) The ozone level is 90 ppb at about 12:30 P.M. and again at 10 P.M. 0002 Use Figure 6 to estimate the following ozone levels to the nearest integer and times to the nearest quarter hour. (A) The ozone level at 7 P.M. (B) The time(s) when the ozone level is 100 ppb. 0002
Z Using Symmetry as an Aid in Graphing We noticed that the graph of y 0003 x2 0002 4 in Example 1 is symmetric with respect to the y axis; that is, the two parts of the graph coincide if the paper is folded along the y axis. Similarly, we say that a graph is symmetric with respect to the x axis if the parts above and
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below the x axis coincide when the paper is folded along the x axis. To make the intuitive idea of folding a graph along a line more concrete, we introduce two related concepts— reflection and symmetry.
Z DEFINITION 1 Reflection 1. The reflection through the y axis of the point (a, b) is the point (0002a, b). 2. The reflection through the x axis of the point (a, b) is the point (a, 0002b). 3. The reflection through the origin of the point (a, b) is the point (0002a, 0002b). 4. To reflect a graph just reflect each point on the graph.
EXAMPLE
3
Reflections In a Cartesian coordinate system, plot the point P 0003 (4, 00022) along with its reflection through (A) the y axis, (B) the x axis, (C) and the origin. y
SOLUTION 5
C 0003 (00024, 2)
B 0003 (4, 2)
00025
5
A 0003 (00024, 00022)
x
P 0003 (4, 00022) 00025
MATCHED PROBLEM 3
0002
In a Cartesian coordinate system, plot the point P 0003 (00023, 5) along with its reflection through (A) the x axis, (B) the y axis, and (C) the origin. 0002
Z DEFINITION 2 Symmetry A graph is symmetric with respect to 1. The x axis if (a, 0002b) is on the graph whenever (a, b) is on the graph— reflecting the graph through the x axis does not change the graph. 2. The y axis if (0002a, b) is on the graph whenever (a, b) is on the graph— reflecting the graph through the y axis does not change the graph. 3. The origin if (0002a, 0002b) is on the graph whenever (a, b) is on the graph—reflecting the graph through the origin does not change the graph.
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Figure 7 illustrates these three types of symmetry. y (0002a, b)
y
y (a, b)
Symmetry with respect to x axis (b)
Symmetry with respect to y axis (a)
(a, b)
x
x (a, 0002b)
(0002a, b)
(a, b)
(a, b) x
y
x (0002a, 0002b)
(0002a, 0002b)
(a, 0002b)
Symmetry with respect to y axis, x axis, and origin (d)
Symmetry with respect to origin (c)
Z Figure 7 Symmetry.
ZZZ EXPLORE-DISCUSS 2
If a graph possesses two of the three types of symmetry in Definition 1, must it also possess the third? Explain.
Given an equation, if we can determine the symmetry properties of its graph ahead of time, we can save a lot of time and energy in sketching the graph. For example, we know that the graph of y 0003 x2 0002 4 in Example 1 is symmetric with respect to the y axis, so we can carefully sketch only the right side of the graph; then reflect the result through the y axis to obtain the whole sketch—the point-by-point plotting is cut in half! The tests for symmetry are given in Theorem 1. These tests are easily applied and are very helpful aids to graphing. Recall, two equations are equivalent if they have the same solution set.
Z THEOREM 1 Tests for Symmetry
EXAMPLE
4
Symmetry with respect to the:
An equivalent equation results if:
y axis
x is replaced with 0002x
x axis
y is replaced with 0002y
Origin
x and y are replaced with 0002x and 0002y
Using Symmetry as an Aid to Graphing Test the equation y 0003 x3 for symmetry and sketch its graph.
SOLUTION
Test y Axis Replace x with 0002x:
Test x Axis Replace y with 0002y:
Test Origin Replace x with 0002x and y with 0002y:
y 0003 (0002x)3 y 0003 0002x3
0002y 0003 x3 y 0003 0002x3
0002y 0003 (0002x)3 0002y 0003 0002x3 y 0003 x3
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The only test that produces an equivalent equation is replacing x with 0002x and y with 0002y. So the only symmetry property for the graph of y 0003 x3 is symmetry with respect to the origin. Note that positive values of x produce positive values for y and negative values of x produce negative values for y. So the graph is in the first and third quadrants. First we make a careful sketch in the first quadrant [Fig. 8(a)]. It is easier to perform a reflection through the origin if you first reflect through one axis [Fig. 8(b)] and then through the other axis [Fig. 8(c)].
x
0
1
2
y
0
1
8
y
y
y 0003 x3
10
10
(2, 8) (1, 1)
00025
5
y
x
10
(2, 8) (1, 1)
00025
(1, 00021)
5
(1, 1)
x
00025
(00021, 00021)
(b)
EXAMPLE
5
x
(c)
0002
Z Figure 8
MATCHED PROBLEM 4
5
000210
000210
(a)
(2, 8)
(00022, 00028)
(2, 00028) 000210
y 0003 x3
Test the equation y 0003 x for symmetry and sketch its graph.
0002
Using Symmetry as an Aid to Graphing Test the equation y 0003 冟x冟 for symmetry and sketch its graph.
SOLUTION
Test y Axis Replace x with 0002x:
Test x Axis Replace y with 0002y:
Test Origin Replace x with 0002x and y with 0002y:
y 0003 冟 0002x 冟 y 0003 冟x冟
0002y 0003 冟 x 冟 y 0003 0002冟 x 冟
0002y 0003 冟 0002x 冟 0002y 0003 冟 x 冟 y 0003 0002冟 x 冟
The only symmetry property for the graph of y 0003 冟x冟 is symmetry with respect to the y axis. Since 冟x冟 is never negative, this graph is in the first and second quadrants. We make a careful sketch in the first quadrant; then reflect this graph through the y axis to obtain the complete sketch shown in Figure 9.
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117
y 5
y 0003 兩x兩 00025
x
0
2
4
y
0
2
4
5
x
00025
0002
Z Figure 9
MATCHED PROBLEM 5
EXAMPLE
6
Test the equation y 0003 0002冟 x 冟 for symmetry and sketch its graph.
0002
Using Symmetry as an Aid to Graphing Test the equation y2 0002 x2 0003 4 for symmetry and sketch its graph.
SOLUTION
Since (0002x)2 0003 x2 and (0002y)2 0003 y2, the equation y2 0002 x2 0003 4 will be unchanged if x is replaced with 0002x or if y is replaced with 0002y. So the graph is symmetric with respect to the y axis, the x axis, and the origin. We need to make a careful sketch in only the first quadrant, reflect this graph through the y axis, and then reflect everything through the x axis. To find quadrant I solutions, we solve the equation for either y in terms of x or x in terms of y. We choose to solve for y. y2 0002 x2 0003 4 y2 0003 x2 0005 4 y 0003 0004 2x2 0005 4 To obtain the quadrant I portion of the graph, we sketch y 0003 2x2 0005 4 for x 0003 0, 1, 2, . . . . The final graph is shown in Figure 10.
x
0
1
2
3
4
y
2
15 ⬇ 2.2
18 ⬇ 2.8
113 ⬇ 3.6
120 ⬇ 4.5
y 5
y 2 0002 x2 0003 4
(3, √13)
(4, √20)
(2, √8) (0, 2) (1, √5) 00025
5
x
00025
Z Figure 10
MATCHED PROBLEM 6
0002
Test the equation 2y2 0002 x2 0003 2 for symmetry and sketch its graph. 0002
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Technology Connections 5
To graph y2 0002 x2 0003 4 on a graphing calculator, we enter both 2x2 0005 4 and 00022x2 0005 4 in the equation editor [Fig. 11(a)] and graph. 00025
5
00025
(a)
(b)
Z Figure 11
ANSWERS TO MATCHED PROBLEMS y
1.
2. (A) 96 ppb 3.
5
(1, 1) (0, 0)
5
00025
(1, 00021)
(4, 00022)
00025
P 0003 (00023, 5)
(9, 3)
(4, 2)
10
(B) 1:45 P.M. and 5 P.M. y 5
x 00025
5
00025
4. Symmetric with respect to the origin
y
5
5
x
00025
00025
5
00025
6. Symmetric with respect to the x axis, the y axis, and the origin y 5
00025
5
00025
C 0003 (3, 00025)
5. Symmetric with respect to the y axis
y
5
x
(9, 00023) A 0003 (00023, 00025)
00025
B 0003 (3, 5)
x
x
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2-1
Cartesian Coordinate Systems
119
Exercises
1. Describe the one-to-one correspondence between points in the plane and ordered pairs of real numbers.
20. Reflect A, B, C, and D through the x axis. y
2. Explain how to graph an equation in two variables using pointby-point plotting.
5
A
3. Explain how to sketch the reflection of a graph through the y axis. 00025
4. How can you tell whether the graph of an equation is symmetric with respect to the origin?
5
5. 5(x, y) ƒ x 0003 06
7. 5(x, y) ƒ x 6 0, y 6 06 9. 5(x, y) ƒ x 7 0, y 6 06
11. 5(x, y) ƒ x 7 0, y 0006 06 13. 5(x, y) ƒ xy 6 06
6. 5(x, y) ƒ x 7 0, y 7 06
x
D
B
In Problems 5–14, give a verbal description of the indicated subset of the plane in terms of quadrants and axes.
C
00025
21. Reflect A, B, C, and D through the origin. y
8. 5(x, y) ƒ y 0003 06
5
10. 5(x, y) ƒ y 6 0, x 0006 06
B
12. 5(x, y) ƒ x 6 0, y 7 06
C
14. 5(x, y) ƒ xy 7 06
00025
D
[Hint: In Problems 13 and 14, consider two cases.]
5
x
A 00025
In Problems 15–18, plot the given points in a rectangular coordinate system. 15. (5, 0), (3, 00022), (00024, 2), (4, 4)
22. Reflect A, B, C, and D through the x axis and then through the y axis. y
16. (0, 4), (00023, 2), (5, 00021), (00022, 00024)
5
17. (0, 00022), (00021, 00023), (4, 00025), (00022, 1)
C
18. (00022, 0), (3, 2), (1, 00024), (00023, 5)
A D
00025
In Problems 19–22, find the coordinates of points A, B, C, and D and the coordinates of the indicated reflections.
00025
y
Test each equation in Problems 23–30 for symmetry with respect to the x axis, y axis, and the origin. Sketch the graph of the equation.
5
00025
A
C
B
00025
x
B
19. Reflect A, B, C, and D through the y axis.
D
5
5
x
23. y 0003 2x 0002 4
24. y 0003 12x 0005 1
25. y 0003 12x
26. y 0003 2x
27. 冟 y 冟 0003 x
28. 冟 y 冟 0003 0002x
29. 冟 x 冟 0003 冟 y 冟
30. y 0003 0002x
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In Problems 31–34, use the graph to estimate to the nearest integer the missing coordinates of the indicated points. (Be sure you find all possible answers.) 31. (A) (8, ?) (D) (?, 6)
(B) (00025, ?) (E) (?, 00025)
(C) (0, ?) (F) (?, 0) y
The figures in Problems 35 and 36 show a portion of a graph. Extend the given graph to one that exhibits the indicated type of symmetry.
10
x
10
(C) (0, ?) (F) (?, 0)
x
5
00025
y
36. (A) x axis only (B) y axis only (C) origin only (D) x axis, y axis, and origin
000210
(B) (00025, ?) (E) (?, 00024)
5
00025
000210
32. (A) (3, ?) (D) (?, 3)
y
35. (A) x axis only (B) y axis only (C) origin only (D) x axis, y axis, and origin
5
00025
y
5
x
10 00025
000210
10
x
Test each equation in Problems 37–46 for symmetry with respect to the x axis, the y axis, and the origin. Do not sketch the graph. 37. 2x 0005 7y 0003 0 38. x2 0005 6y 0005 y2 0003 25
000210
33. (A) (1, ?) (D) (?, 00026)
(B) (00028, ?) (E) (?, 4)
39. x2 0002 4xy2 0003 3
(C) (0, ?) (F) (?, 0)
40. 3x 0002 5y 0003 2 41. x4 0002 5x2y 0005 y4 0003 1
y
42. x4 0002 y4 0003 16
10
43. x3 0002 y3 0003 8 000210
10
x
44. x2 0005 2xy 0005 3y2 0003 12 45. x4 0002 4x2y2 0005 y4 0003 81 46. x3 0002 4y2 0003 1
000210
34. (A) (6, ?) (D) (?, 00022)
(B) (00026, ?) (E) (?, 1)
Test each equation in Problems 47–58 for symmetry with respect to the x axis, the y axis, and the origin. Sketch the graph of the equation.
(C) (0, ?) (F) (?, 0)
y 10
000210
10
000210
x
47. y2 0003 x 0005 2
48. y2 0003 x 0002 2
49. y 0003 x2 0005 1
50. y 0005 2 0003 x2
51. 4y2 0002 x2 0003 1
52. 4x2 0002 y2 0003 1
53. y3 0003 x
54. y 0003 x4
55. y 0003 0.6x2 0002 4.5
56. x 0003 0.8y2 0002 3.5
57. y 0003 x2兾3
58. y2兾3 0003 x
59. (A) Graph the triangle with vertices A 0003 (1, 1), B 0003 (7, 2), and C 0003 (4, 6). (B) Now graph the triangle with vertices A0007 0003 (1, 00021), B0007 0003 (7, 00022), and C0007 0003 (4, 00026) in the same coordinate system. (C) How are these two triangles related? How would you describe the effect of changing the sign of the y coordinate of all the points on a graph?
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61. (A) Graph the triangle with vertices A 0002 (1, 1), B 0002 (7, 2), and C 0002 (4, 6). (B) Now graph the triangle with vertices A0003 0002 (00041, 00041), B0003 0002 (00047, 00042), and C0003 0002 (00044, 00046) in the same coordinate system. (C) How are these two triangles related? How would you describe the effect of changing the signs of the x and y coordinates of all the points on a graph? 62. (A) Graph the triangle with vertices A 0002 (1, 2), B 0002 (1, 4), and C 0002 (3, 4). (B) Now graph y 0002 x and the triangle obtained by reversing the coordinates for each vertex of the original triangle: A0003 0002 (2, 1), B0003 0002 (4, 1), B0003 0002 (4, 3). (C) How are these two triangles related? How would you describe the effect of reversing the coordinates of each point on a graph? In Problems 63–66, solve for y, producing two equations, and then graph both of these equations in the same viewing window. 63. 2x 0005 y2 0002 3
64. x3 0005 y2 0002 8
65. x 2 0004 ( y 0005 1)2 0002 4
66. ( y 0004 2)2 0004 x 2 0002 9
Test each equation in Problems 67–76 for symmetry with respect to the x axis, the y axis, and the origin. Sketch the graph of the equation. 68. 冟 y 冟 0002 x3
67. y3 0002 冟 x 冟
69. xy 0002 1 2
70. xy 0002 00041
71. y 0002 6x 0004 x
72. y 0002 x2 0004 6x
73. y2 0002 冟 x 冟 0005 1
74. y2 0002 4冟 x 冟 0005 1
75. 冟 xy 冟 0005 2冟 y 冟 0002 6
76. 冟 xy 冟 0005 冟 y 冟 0002 4
77. If a graph is symmetric with respect to the x axis and to the origin, must it be symmetric with respect to the y axis? Explain. 78. If a graph is symmetric with respect to the y axis and to the origin, must it be symmetric with respect to the x axis? Explain.
R 0002 np 0002 (10 0004 p)p
82. BUSINESS Repeat Problem 81 for the demand equation n000280004p
81. BUSINESS After extensive surveys, the marketing research department of a producer of popular compact discs developed the demand equation n 0002 10 0004 p
5 0006 p 0006 10
40006p00068
83. PRICE AND DEMAND The quantity of a product that consumers are willing to buy during some period of time depends on its price. The price p and corresponding weekly demand q for a particular brand of diet soda in a city are shown in the figure. Use this graph to estimate the following demands to the nearest 100 cases. (A) What is the demand when the price is $6.00 per case? (B) Does the demand increase or decrease if the price is increased from $6.00 to $6.30 per case? By how much? (C) Does the demand increase or decrease if the price is decreased from $6.00 to $5.70? By how much? (D) Write a brief description of the relationship between price and demand illustrated by this graph. p $7
$6
$5 2,000
3,000
4,000
q
Number of cases
84. PRICE AND SUPPLY The quantity of a product that suppliers are willing to sell during some period of time depends on its price. The price p and corresponding weekly supply q for a particular brand of diet soda in a city are shown in the figure. Use this graph to estimate the following supplies to the nearest 100 cases. (A) What is the supply when the price is $5.60 per case? (B) Does the supply increase or decrease if the price is increased from $5.60 to $5.80 per case? By how much? (C) Does the supply increase or decrease if the price is decreased from $5.60 to $5.40 per case? By how much? (D) Write a brief description of the relationship between price and supply illustrated by this graph. p $7
Price per case
APPLICATIONS
5 0006 p 0006 10
Graph the revenue equation for the indicated values of p.
79. If a graph is symmetric with respect to the origin, must it be symmetric with respect to the x axis? Explain. 80. If a graph is symmetric with respect to the origin, must it be symmetric with respect to the y axis? Explain.
121
where n is the number of units (in thousands) retailers are willing to buy per day at $p per disc. The company’s daily revenue R (in thousands of dollars) is given by
Price per case
60. (A) Graph the triangle with vertices A 0002 (1, 1), B 0002 (7, 2), and C 0002 (4, 6). (B) Now graph the triangle with vertices A0003 0002 (00041, 1), B0003 0002 (00047, 2), and C0003 0002 (00044, 6) in the same coordinate system. (C) How are these two triangles related? How would you describe the effect of changing the sign of the x coordinate of all the points on a graph?
Cartesian Coordinate Systems
$6
$5 2,000
3,000
4,000
Number of cases
q
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85. TEMPERATURE The temperature during a spring day in the Midwest is given in the figure. Use this graph to estimate the following temperatures to the nearest degree and times to the nearest hour. (A) The temperature at 9:00 A.M. (B) The highest temperature and the time when it occurs. (C) The time(s) when the temperature is 49°F. 70
x
(A) Graph v for 0 x 2. (B) Describe the relationship between this graph and the physical behavior of the ball as it swings back and forth.
60
88. PHYSICS The speed (in meters per second) of a ball oscillating at the end of a spring is given by 50
v 0003 4 225 0002 x2
40 Midnight
where x is the vertical displacement (in centimeters) of the ball from its position at rest (positive displacement measured downward—see the figure). 6 AM
Noon
6 PM
Midnight
86. TEMPERATURE Use the graph in Problem 85 to estimate the following temperatures to the nearest degree and times to the nearest half hour. (A) The temperature at 7:00 P.M. (B) The lowest temperature and the time when it occurs. (C) The time(s) when the temperature is 52°F.
x000b0
87. PHYSICS The speed (in meters per second) of a ball swinging at the end of a pendulum is given by v 0003 0.5 12 0002 x where x is the vertical displacement (in centimeters) of the ball from its position at rest (see the figure).
2-2
x 0
(A) Graph v for 00025 x 5. (B) Describe the relationship between this graph and the physical behavior of the ball as it oscillates up and down.
Distance in the Plane Z Distance Between Two Points Z Midpoint of a Line Segment Z Circles
Two basic problems studied in analytic geometry are 1. 2.
Given an equation, find its graph. Given a figure (line, circle, parabola, ellipse, etc.) in a coordinate system, find its equation.
The first problem was discussed in Section 2-1. In this section, we introduce some tools that are useful when studying the second problem.
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123
Z Distance Between Two Points Given two points P1 and P2 in a rectangular coordinate system, we denote the distance between P1 and P2 by d(P1, P2). We begin with an example.
EXAMPLE
1
Distance Between Two Points Find the distance between the points P1 ⫽ (1, 2) and P2 ⫽ (4, 6).
SOLUTION
Connecting the points P1, P2, and P3 ⫽ (4, 2) with straight line segments forms a right triangle (Fig. 1). y
P1 ⫽ (1, 2)
P
兩6 ⫺ 2兩 ⫽ 4
d(
P
1,
5
2)
P2 ⫽ (4, 6)
P3 ⫽ (4, 2) 兩4 ⫺ 1兩 ⫽ 3 5
10
x
Z Figure 1
From the figure, we see that the lengths of the legs of the triangle are d(P1, P3) ⫽ 冟 4 ⫺ 1 冟 ⫽ 3 and d(P3, P2) ⫽ 冟 6 ⫺ 2 冟 ⫽ 4 The length of the hypotenuse is d(P1, P2), the distance we are seeking. Applying the Pythagorean theorem (see Appendix B), we get [d(P1, P2)] 2 ⫽ ⫽ ⫽ ⫽
[ d(P1, P3)] 2 ⫹ [ d(P3, P2)] 2 32 ⫹ 42 9 ⫹ 16 25
Therefore, d(P1, P2) ⫽ 125 ⫽ 5
MATCHED PROBLEM 1
0002
Find the distance between the points P1 ⫽ (1, 2) and P2 ⫽ (13, 7).
0002
The ideas used in Example 1 can be applied to any two distinct points in the plane. If P1 ⫽ (x1, y1) and P2 ⫽ (x2, y2 ) are two points in a rectangular coordinate system (Fig. 2), then [d(P1, P2)] 2 ⫽ 冟 x2 ⫺ x1 冟2 ⫹ 冟 y2 ⫺ y1 冟2 ⫽ (x2 ⫺ x1)2 ⫹ ( y2 ⫺ y1)2 Taking square roots gives the distance formula.
Because 冟N冟2 ⴝ N2
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Z Figure 2 Distance between two points.
P1 0003 (x1, y1)
,
1
兩x2 0002 x1兩 x1
P2 0003 (x2, y2) y2 兩y2 0002 y1兩
d(P
P 2)
x
y1 (x2, y1) x2
Z THEOREM 1 Distance Formula The distance between P1 0003 (x1, y1) and P2 0003 (x2, y2) is d(P1, P2) 0003 2(x2 0002 x1)2 0005 ( y2 0002 y1)2
EXAMPLE
2
Using the Distance Formula Find the distance between the points (00023, 5) and (00022, 00028).*
SOLUTION
Let (x1, y1) 0003 (ⴚ3, 5) and (x2, y2) 0003 (ⴚ2, ⴚ8). Then, d 0003 2 [(ⴚ2) 0002 (ⴚ3)] 2 0005 [(ⴚ8) 0002 5] 2 0003 2(00022 0005 3)2 0005 (00028 0002 5)2 0003 212 0005 (000213)2 0003 21 0005 169 0003 2170 Notice that if we choose (x1, y1) 0003 (00022, 00028) and (x2, y2) 0003 (00023, 5), then d 0003 2[(00023) 0002 (00022)] 2 0005 [5 0002 (00028)] 2 0003 21 0005 169 0003 2170 so it doesn’t matter which point we designate as P1 or P2.
MATCHED PROBLEM 2
Find the distance between the points (6, 00023) and (00027, 00025).
0002
0002
Z Midpoint of a Line Segment The midpoint of a line segment is the point that is equidistant from each of the endpoints. A formula for finding the midpoint is given in Theorem 2. The proof is discussed in the exercises. Z THEOREM 2 Midpoint Formula The midpoint of the line segment joining P1 0003 (x1, y1) and P2 0003 (x2, y2) is M0003a
x1 0005 x2 y1 0005 y2 , b 2 2
The point M is the unique point satisfying 1 d(P1, M ) 0003 d(M, P2) 0003 d(P1, P2) 2
*We often speak of the point (a, b) when we are referring to the point with coordinates (a, b). This shorthand, though not technically accurate, causes little trouble, and we will continue the practice.
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Note that the coordinates of the midpoint are simply the averages of the respective coordinates of the two given points.
EXAMPLE
3
Using the Midpoint Formula Find the midpoint M of the line segment joining A 0003 (00023, 2) and B 0003 (4, 00025). Plot A, B, and M and verify that d(A, M ) 0003 d(M, B) 0003 12d(A, B).
SOLUTION
We use the midpoint formula with (x1, y1) 0003 (00023, 2) and (x2, y2) 0003 (4, 00025) to obtain the coordinates of the midpoint M. M0003a
x1 0005 x2 y1 0005 y2 , b 2 2
00023 0005 4 2 0005 (00025) , b 2 2 1 00023 0003a , b 2 2 0003a
Substitute x1 ⴝ ⴚ3, y1 ⴝ 2, x2 ⴝ 4, and y2 ⴝ ⴚ5.
Simplify.
0003 (0.5, 00021.5) We plot the three points (Fig. 3) and compute the distances d(A, M ), d(M, B), and d(A, B):
y 5
d(A, M ) 0003 2(00023 0002 0.5)2 0005 [2 0002 (00021.5)] 2 0003 212.25 0005 12.25 0003 224.5
A 0003 (00023, 2) 00025
5
x
M 0003 冢2 , 0002 2 冣 1
00025
3
d(A, B) 0003 2(00023 0002 4)2 0005 [2 0002 (00025)] 2 0003 249 0005 49 0003 298 1 1 98 d(A, B) 0003 198 0003 0003 124.5 0003 d(A, M ) 0003 d(M, B) 2 2 B4
B 0003 (4, 00025)
Z Figure 3
This verifies that M is the midpoint of the line segment joining A and B.
MATCHED PROBLEM 3
EXAMPLE
d(M, B) 0003 2(0.5 0002 4)2 0005 [ 00021.5 0002 (00025)] 2 0003 212.25 0005 12.25 0003 224.5
4
0002
Find the midpoint M of the line segment joining A 0003 (4, 1) and B 0003 (00023, 00025). Plot A, B, and M and verify that d(A, M ) 0003 d(M, B) 0003 12 d(A, B). 0002
Using the Midpoint Formula If M 0003 (1, 1) is the midpoint of the line segment joining A 0003 (00023, 00021) and B 0003 (x, y), find the coordinates of B.
SOLUTION
From the midpoint formula, we have M 0003 (1, 1) 0003 a
00023 0005 x 00021 0005 y , b 2 2
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We equate the corresponding coordinates and solve the resulting equations for x and y: 00021 0005 y 2 2 0003 00021 0005 y
00023 0005 x 2 2 0003 00023 0005 x 10003
10003
2 0005 3 0003 00023 0005 x 0005 3
*
2 0005 1 0003 00021 0005 y 0005 1
50003x
30003y
Therefore, B 0003 (5, 3). MATCHED PROBLEM 4
0002
If M 0003 (1, 00021) is the midpoint of the line segment joining A 0003 (00021, 00025) and B 0003 (x, y), find the coordinates of B. 0002
Z Circles The distance formula would be helpful if its only use were to find actual distances between points, such as in Example 2. However, its more important use is in finding equations of figures in a rectangular coordinate system. We start with an example.
EXAMPLE
5
Equations and Graphs of Circles Write an equation for the set of all points that are 5 units from the origin. Graph your equation.
SOLUTION
The distance between a point (x, y) and the origin is d 0003 2(x 0002 0)2 0005 ( y 0002 0)2 0003 2x2 0005 y2 So, an equation for the set of points that are 5 units from the origin is 2x2 0005 y2 0003 5 We square both sides of this equation to obtain an equation that does not contain any radicals. x2 0005 y2 0003 25 Because (0002x)2 0003 x2 and (0002y)2 0003 y2, the graph will be symmetric with respect to the x axis, y axis, and origin. We make up a table of solutions, sketch the curve in the first quadrant, and use symmetry properties to produce a familiar geometric object—a circle (Fig. 4). x
y
0
5
3
4
4
3
5
0
y (00023, 4) (00024, 3) (00025, 0)
(00024, 00023) (00023, 00024)
(0, 5)
(3, 4) (4, 3) (5, 0) x
(4, 00023) (3, 00024) (0, 00025)
Z Figure 4
MATCHED PROBLEM 5
0002
Write an equation for the set of all points that are three units from the origin. Graph your equation. 0002 *Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
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Technology Connections Refer to Example 5. To graph this circle on a graphing calculator, first we solve x 2 ⫹ y 2 ⫽ 25 for y:
x2 ⴙ y2 ⴝ 25 y2 ⴝ 25 ⴚ x2 y ⴝ ⴞ225 ⴚ x2 Next we enter y ⴝ 225 ⴚ x2 and y ⴝ ⴚ 225 ⴚ x2 in the equation editor of a graphing calculator [Fig. 5(a)], enter appropriate window variables [Fig. 5(b)], and graph [Fig. 5(c)].
The graph in Figure 5(c) doesn’t look like a circle. (A circle is as wide as it is tall.) This distortion is caused by the difference between axes scales. One unit on the x axis appears to be longer than one unit on the y axis. Most graphing calculators have an option called ZSquare under the zoom menu [Fig. 6(a)] that automatically adjusts the x axis scale [Fig. 6(b)] to produce a squared viewing window. The graph of a circle in a squared viewing window is not distorted [Fig. 6(c)]. 5
00025
5
00025
(a)
(b)
(c)
Z Figure 5 5
00027.6
7.6
00025
(a)
(b)
(c)
Z Figure 6
In Example 5, we began with a verbal description of a set of points, produced an algebraic equation that these points must satisfy, constructed a numerical table listing some of these points, and then drew a graphical representation of this set of points. The interplay between verbal, algebraic, numerical, and graphical concepts is one of the central themes of this book. Now we generalize the ideas introduced in Example 5. Z DEFINITION 1 Circle y
A circle is the set of all points in a plane equidistant from a fixed point. The fixed distance is called the radius, and the fixed point is called the center. r C 0003 (h, k)
Z Figure 7 Circle.
P 0003 (x, y)
x
Let’s find the equation of a circle with radius r (r 0) and center C at (h, k) in a rectangular coordinate system (Fig. 7). The circle consists of all points P 0003 (x, y) satisfying d(P, C ) 0003 r; that is, all points satisfying 2(x 0002 h)2 0005 ( y 0002 k)2 0003 r
r 7 0
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or, equivalently, (x ⫺ h)2 ⫹ ( y ⫺ k)2 ⫽ r 2 r 7 0
Z THEOREM 3 Standard Form of the Equation of a Circle The standard form of a circle with radius r and center at (h, k) is: (x ⫺ h)2 ⫹ ( y ⫺ k)2 ⫽ r2 r 7 0
EXAMPLE
6
Equations and Graphs of Circles Find the equation of a circle with radius 4 and center at C ⫽ (⫺3, 6). Graph the equation. C ⫽ (h, k) ⫽ (⫺3, 6) and r ⫽ 4 (x ⫺ h)2 ⫹ ( y ⫺ k)2 ⫽ r2 Substitute h ⴝ ⴚ3, k ⴝ 6 [x ⫺ (⫺3)] 2 ⫹ ( y ⫺ 6)2 ⫽ 42 Simplify 2 2 (x ⫹ 3) ⫹ ( y ⫺ 6) ⫽ 16
SOLUTION
To graph the equation, plot the center and a few points on the circle (the easiest points to plot are those located 4 units from the center in either the horizontal or vertical direction), then draw a circle of radius 4 (Fig. 8). y (⫺3, 10)
10
C ⫽ (⫺3, 6)
(⫺7, 6)
r⫽4
5
(1, 6)
(⫺3, 2) ⫺5
x
(x ⫹ 3)2 ⫹ (y ⫺ 6)2 ⫽ 16
Z Figure 8
MATCHED PROBLEM 6
ZZZ EXPLORE-DISCUSS 1
0002
Find the equation of a circle with radius 3 and center at C ⫽ (3, ⫺2). Graph the equation. 0002
Explain how to find the equation of the circle with diameter AB, if A ⫽ (3, 8) and B ⫽ (11, 12).
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EXAMPLE
7
Distance in the Plane
129
Finding the Center and Radius of a Circle Find the center and radius of the circle with equation x2 0005 y2 0005 6x 0002 4y 0003 23.
SOLUTION
We transform the equation into the form (x 0002 h)2 0005 (y 0002 k)2 0003 r2 by completing the square relative to x and relative to y (see Section 1-5). From this standard form we can determine the center and radius. x2 0005 y2 0005 6x 0002 4y 0003 23
Group together the terms involving x and those involving y.
(x2 0005 6x ) 0005 ( y2 0002 4y ) 0003 23 2 2 (x 0005 6x 0005 9) 0005 ( y 0002 4y 0005 4) 0003 23 0005 9 0005 4 (x 0005 3)2 0005 ( y 0002 2)2 0003 36 [ x 0002 (00023)] 2 0005 ( y 0002 2)2 0003 62 (h, k) 0003 (00023, 2) Center: Radius: r 0003 136 0003 6 MATCHED PROBLEM 7
Complete the squares. Factor each trinomial. Write ⴙ3 as ⴚ(ⴚ3) to identify h.
0002
Find the center and radius of the circle with equation x2 0005 y2 0002 8x 0005 10y 0003 000225.
ANSWERS TO MATCHED PROBLEMS 1. 13 2. 1173 3. M 0003 (12, 00022) 0003 (0.5, 00022); d(A, B) 0003 185; d(A, M ) 0003 121.25 0003 d(M, B) 0003 12 d(A, B) y 5
A 0003 (4, 1) 00025
x
5
00025
B 0003 (00023, 00025)
4. B 0003 (3, 3) 5. x2 + y2 0003 9
6. (x 0002 3)2 0005 ( y 0005 2)2 0003 9 y
y 5
(0, 3) (00023, 0)
(3, 1) (3, 0)
00025
5
(0, 00023)
x
(0, 00022) 00025
00025
7. (x 0002 4)2 0005 ( y 0005 5)2 0003 16; radius: 4, center: (4, 00025)
C (3, 00022) (3, 00025)
5
x (6, 00022)
0002
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Exercises
1. State the Pythagorean theorem.
29. Find x such that (x, 7) is 10 units from (00024, 1).
2. Explain how to calculate the distance between two points in the plane if you know their coordinates.
30. Find x such that (x, 2) is 4 units from (3, 00023). 31. Find y such that (2, y) is 3 units from (00021, 4).
3. Explain how to calculate the midpoint of a line segment if you know the coordinates of the endpoints.
32. Find y such that (3, y) is 13 units from (00029, 2).
4. Explain how to find the standard form of the equation of the circle with center (1, 5) and radius 12.
In Problems 33–36, write a verbal description of the graph and then write an equation that would produce the graph.
In Problems 5–12, find the distance between each pair of points and the midpoint of the line segment joining the points. Leave distance in radical form, if applicable. 5. (1, 0), (4, 4)
6. (0, 1), (3, 5)
7. (0, 00022), (5, 10)
8. (3, 0), (00022, 00023)
9. (00026, 00024), (3, 4)
10. (00025, 4), (6, 00021)
11. (00026, 00023), (00022, 00021)
y
33. 5
00025
13. C 0003 (0, 0), r 0003 7
14. C 0003 (0, 0), r 0003 5
15. C 0003 (2, 3), r 0003 6
16. C 0003 (5, 6), r 0003 2
17. C 0003 (00024, 1), r 0003 17
18. C 0003 (00025, 6), r 0003 111
19. C 0003 (00023, 00024), r 0003 12
20. C 0003 (4, 00021), r 0003 15
x
00025
12. (00025, 00022), (00021, 2)
In Problems 13–20, write the equation of a circle with the indicated center and radius.
5
y
34. 5
00025
5
x
00025
In Problems 21–26, write an equation for the given set of points. Graph your equation.
y
35. 5
21. The set of all points that are two units from the origin. 22. The set of all points that are four units from the origin. 23. The set of all points that are one unit from (1, 0).
00025
5
x
24. The set of all points that are one unit from (0, 00021). 25. The set of all points that are three units from (00022, 1).
00025
26. The set of all points that are two units from (3, 00022). 27. Let M be the midpoint of A and B, where
y
36.
A 0003 (a1, a2), B 0003 (1, 3), and M 0003 (00022, 6).
5
(A) Use the fact that 00022 is the average of a1 and 1 to find a1. (B) Use the fact that 6 is the average of a2 and 3 to find a2. (C) Find d(A, M ) and d(M, B). 28. Let M be the midpoint of A and B, where
00025
5
A 0003 (00023, 5), B 0003 (b1, b2), and M 0003 (4, 00022). (A) Use the fact that 4 is the average of 00023 and b1 to find b1. (B) Use the fact that 00022 is the average of 5 and b2 to find b2. (C) Find d(A, M ) and d(M, B).
00025
x
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In Problems 37–42, M is the midpoint of A and B. Find the indicated point. Verify that d(A, M) 0003 d(M, B) 0003 12d(A, B). 37. A 0003 (00024.3, 5.2), B 0003 (9.6, 00021.7), M 0003 ? 38. A 0003 (2.8, 00023.5), B 0003 (00024.1, 7.6), M 0003 ?
131
Distance in the Plane
62. A parallelogram ABCD is shown in the figure. (A) Find the midpoint of the line segment joining A and C. (B) Find the midpoint of the line segment joining B and D. (C) What can you conclude about the diagonals of the parallelogram?
39. A 0003 (25, 10), M 0003 (00025, 00022), B 0003 ?
y
40. M 0003 (2.5, 3.5), B 0003 (12, 10), A 0003 ?
B 0003 (a, b)
41. M 0003 (00028, 00026), B 0003 (2, 4), A 0003 ? 42. A 0003 (00024, 00022), M 0003 (00021.5, 00024.5), B 0003 ?
A 0003 (0, 0)
C 0003 (a 0005 c, b)
x D 0003 (c, 0)
In Problems 43–52, find the center and radius of the circle with the given equation. Graph the equation. 43. x2 0005 ( y 0005 2)2 0003 9 44. (x 0002 5)2 0005 y2 0003 16 45. (x 0005 4)2 0005 (y 0002 2)2 0003 7 46. (x 0002 5)2 0005 (y 0005 7)2 0003 15 47. x2 0005 6x 0005 y2 0003 16 48. x2 0005 y2 0002 8y 0003 9 49. x2 0005 y2 0002 6x 0002 4y 0003 36 50. x2 0005 y2 0002 2x 0002 10y 0003 55 51. 3x2 0005 3y2 0005 24x 0002 18y 0005 24 0003 0 2
2
52. 2x 0005 2y 0005 8x 0005 20y 0005 30 0003 0
In Problems 63–68, find the standard form of the equation of the circle that has a diameter with the given endpoints. 63. (00024, 3), (6, 3) 64. (5, 00021), (5, 7) 65. (4, 0), (0, 10) 66. (00026, 0), (0, 00028) 67. (11, 00022), (3, 00024) 68. (00028, 9), (12, 15) In Problems 69–72, find the standard form of the equation of the circle with the given center that passes through the given point. 69. Center: (0, 5); point on circle: (2, 00024)
In Problems 53–56, solve for y, producing two equations, and then graph both of these equations in the same viewing window. 53. x2 0005 y2 0003 3 54. x2 0005 y2 0003 5
70. Center: (00023, 0); point on circle: (6, 1) 71. Center: (00022, 9); point on circle: (8, 00027) 72. Center: (7, 000212); point on circle: (13, 8)
55. (x 0005 3)2 0005 (y 0005 1)2 0003 2
APPLICATIONS
56. (x 0002 2)2 0005 (y 0002 1)2 0003 3
73. SPORTS A singles court for lawn tennis is a rectangle 27 feet wide and 78 feet long (see the figure). Points B and F are the midpoints of the end lines of the court.
In Problems 57 and 58, show that the given points are the vertices of a right triangle (see the Pythagorean theorem in Appendix B). Find the length of the line segment from the midpoint of the hypotenuse to the opposite vertex. 57. (00023, 2), (1, 00022), (8, 5)
18 feet B
C
A
18 feet
58. (00021, 3), (3, 5), (5, 1)
D 78 feet
Find the perimeter (to two decimal places) of the triangle with the vertices indicated in Problems 59 and 60. 59. (00023, 1), (1, 00022), (4, 3) 60. (00022, 4), (3, 1), (00023, 00022) x1 0005 x2 y1 0005 y2 , b, 2 2 show that d(P1, M ) 0003 d(M, P2) 0003 12d(P1, P2). (This is one step in the proof of Theorem 2.)
61. If P1 0003 (x1, y1), P2 0003 (x2, y2) and M 0003 a
27 feet
E F
G
(A) Sketch a graph of the court with A at the origin of your coordinate system, C on the positive y axis, and G on the positive x axis. Find the coordinates of points A through G. (B) Find d(B, D) and d(F, C ) to the nearest foot.
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74. SPORTS Refer to Problem 73. Find d(A, D) and d(C, G) to the nearest foot. 75. ARCHITECTURE An arched doorway is formed by placing a circular arc on top of a rectangle (see the figure). If the doorway is 4 feet wide and the height of the arc above its ends is 1 foot, what is the radius of the circle containing the arc? [Hint: Note that (2, r 0002 1) must satisfy x2 0005 y2 0003 r 2.] y
77. CONSTRUCTION Town B is located 36 miles east and 15 miles north of town A (see the figure). A local telephone company wants to position a relay tower so that the distance from the tower to town B is twice the distance from the tower to town A. (A) Show that the tower must lie on a circle, find the center and radius of this circle, and graph. (B) If the company decides to position the tower on this circle at a point directly east of town A, how far from town A should they place the tower? Compute answer to one decimal place.
(2, r 0002 1)
y
r
25
x
Town B
Tower
(36, 15)
(x, y) 4 feet
Town A
Arched doorway
76. ENGINEERING The cross section of a rivet has a top that is an arc of a circle (see the figure). If the ends of the arc are 12 millimeters apart and the top is 4 millimeters above the ends, what is the radius of the circle containing the arc?
25
x
78. CONSTRUCTION Repeat Problem 77 if the distance from the tower to town A is twice the distance from the tower to town B.
Rivet
2-3
Equations of a Line Z Graphing Lines Z Finding the Slope of a Line Z Determining Special Forms of the Equation of a Line Z Finding Slopes of Parallel or Perpendicular Lines
In this section, we consider one of the most basic geometric figures—a line. When we use the term line in this book, we mean straight line. We will learn how to recognize and graph a line and how to use information concerning a line to find its equation.
Z Graphing Lines With your past experience in graphing equations in two variables, you probably remember that first-degree equations in two variables, such as y 0003 00023x 0005 5
3x 0002 4y 0003 9
y 0003 000223 x
have graphs that are lines. This fact is stated in Theorem 1.
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Z THEOREM 1 The Equation of a Line If A, B, and C are constants, with A and B not both 0, and x and y are variables, then the graph of the equation Ax ⴙ By ⴝ C
Standard Form
(1)
is a line. Any line in a rectangular coordinate system has an equation of this form.
Also, the graph of any equation of the form y ⴝ mx ⴙ b
(2)
where m and b are constants, is a line. Equation (2), which we will discuss in detail later, is simply a special case of equation (1) for B 0002 0. This can be seen by solving equation (1) for y in terms of x: C A y00030004 x0005 B B
B00020
To graph either equation (1) or (2), we plot any two points from the solution set and use a straightedge to draw a line through these two points. The points where the line crosses the axes are convenient to use and easy to find. The y intercept* is the y coordinate of the point where the graph crosses the y axis, and the x intercept is the x coordinate of the point where the graph crosses the x axis. To find the y intercept, let x = 0 and solve for y; to find the x intercept, let y = 0 and solve for x. It is often advisable to find a third point as a checkpoint. All three points must lie on the same line or a mistake has been made.
EXAMPLE
1
Using Intercepts to Graph a Line Graph the equation 3x 0004 4y 0003 12.
SOLUTION
Find intercepts, a third checkpoint (optional), and draw a line through the two (three) points (Fig. 1). y 5
(8, 3) (4, 0)
00045
5
y intercept is 00043
MATCHED PROBLEM 1
x
0
4
8
y
00043
0
3
Graph the equation 4x 0005 3y 0003 12.
Checkpoint 10
x
x intercept is 4 (0, 00043)
00045
Z Figure 1
0002
0002
*If the x intercept is a and the y intercept is b, then the graph of the line passes through the points (a, 0) and (0, b). It is common practice to refer to both the numbers a and b and the points (a, 0) and (0, b) as the x and y intercepts of the line.
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Technology Connections To solve Example 1 on a graphing calculator, we first solve the equation for y:
3x ⴚ 4y ⴝ 12 ⴚ4y ⴝ ⴚ3x ⴙ 12 y ⴝ 0.75x ⴚ 3 To find the y intercept of this line, we graph the preceding equation, press TRACE, and then enter 0 for x [Fig. 2(a)]. The displayed y value is the y intercept.
The x intercept can be found by using the zero option on the CALC menu. After selecting the zero option, you will be asked to provide three x values: a left bound (a number less than the zero), a right bound (a number greater than the zero), and a guess (a number between the left and right bounds). You can enter the three values from the keypad, but most find it easier to use the cursor. The zero or x intercept is displayed at the bottom of the screen [Fig. 2(b)].
5
00045
5
10
00045
10
00045
00045
(a) y intercept
(b) x intercept
Z Figure 2
Z Finding the Slope of a Line If we take two different points P1 0003 (x1, y1) and P2 0003 (x2, y2) on a line, then the ratio of the change in y to the change in x as we move from point P1 to point P2 is called the slope of the line. Roughly speaking, slope is a measure of the “steepness” of a line. Sometimes the change in x is called the run and the change in y is called the rise. Z DEFINITION 1 Slope of a Line If a line passes through two distinct points P1 0003 (x1, y1) and P2 0003 (x2, y2), then its slope m is given by the formula m0003 0003
y2 0004 y1 x2 0004 x1
y
x1 0002 x2
P2 0003 (x2, y2)
Vertical change (rise) Horizontal change (run)
y2 0004 y1 Rise x
P1 0003 (x1, y1) x2 0004 x1 Run
(x2, y1)
For a horizontal line, y doesn’t change as x changes, so its slope is 0. On the other hand, for a vertical line, x doesn’t change as y changes, so its slope is not defined: y2 0004 y1 y2 0004 y1 0003 x2 0004 x1 0
For a vertical line, slope is not defined.
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In general, the slope of a line may be positive, negative, 0, or not defined. Each of these cases is interpreted geometrically as shown in Table 1. Table 1 Geometric Interpretation of Slope Line
Slope
Example y
Rising as x moves from left to right y values are increasing
Positive
x
y
Falling as x moves from left to right y values are decreasing
Negative
x
y
Horizontal y values are constant
0
x
y
Vertical x values are constant
Not defined
x
In using the formula to find the slope of the line through two points, it doesn’t matter which point is labeled P1 or P2, because changing the labeling will change the sign in both the numerator and denominator of the slope formula: y2 0004 y1 y1 0004 y2 0003 x2 0004 x1 x1 0004 x2
b0006 a0006
For example, the slope of the line through the points (3, 2) and (7, 5) is
b
500042 3 00043 200045 0003 0003 0003 700043 4 00044 300047
a m0003
b0006 b 0003 a0006 a
In addition, it is important to note that the definition of slope doesn’t depend on the two points chosen on the line as long as they are distinct. This follows from the fact that the ratios of corresponding sides of similar triangles are equal (Fig. 3).
Z Figure 3
EXAMPLE
2
Finding Slopes For each line in Figure 4, find the run, the rise, and the slope. (All the horizontal and vertical line segments have integer lengths.) y
y
5
5
00045
5
00045
00045
5
00045
(a)
Z Figure 4
x
(b)
x
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SOLUTION
MATCHED PROBLEM 2
In Figure 4(a), the run is 3, the rise is 6 and the slope is 63 0003 2. In Figure 4(b), the run is 6, 2 the rise is 00044 and the slope is 00044 0002 6 0003 00043 . For each line in Figure 5, find the run, the rise, and the slope. (All the horizontal and vertical line segments have integer lengths.) y
y
5
5
00045
5
x
00045
5
00045
x
00045
(a)
(b)
Z Figure 5
0002
EXAMPLE
3
Finding Slopes Sketch a line through each pair of points and find the slope of each line.
SOLUTIONS
(A) (00043, 00044), (3, 2)
(B) (00042, 3), (1, 00043)
(C) (00044, 2), (3, 2)
(D) (2, 4), (2, 00043) y
(A)
y
(B)
5
5
(00042, 3)
(3, 2) 00045
x
5
00045
5
(1, 00043)
(00043, 00044) 00045
m0003
x
00045
2 0004 (00044) 6 0003 00031 3 0004 (00043) 6
m0003
y
(C)
y
(D)
5
(00044, 2)
00043 0004 3 00046 0003 0003 00042 1 0004 (00042) 3
5
(2, 4)
(3, 2)
00045
5
x
00045
5
x
(2, 00043) 00045
m0003
200042 0 0003 00030 3 0004 (00044) 7
00045
00043 0004 4 00047 0003 ; 200042 0 slope is not defined m0003
0002
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MATCHED PROBLEM 3
137
Equations of a Line
Find the slope of the line through each pair of points. Do not graph. (A) (00043, 00043), (2, 00043)
(B) (00042, 00041), (1, 2)
(C) (0, 4), (2, 00044)
(D) (00043, 2), (00043, 00041)
0002
Z Determining Special Forms of the Equation of a Line We start by investigating why y 0003 mx 0005 b is called the slope–intercept form for a line.
ZZZ EXPLORE-DISCUSS 1
(A) Graph y 0003 x 0005 b for b 0003 00045, 00043, 0, 3, and 5 simultaneously in the same coordinate system. Verbally describe the geometric significance of b. (B) Graph y 0003 mx 0004 1 for m 0003 00042, 00041, 0, 1, and 2 simultaneously in the same coordinate system. Verbally describe the geometric significance of m.
As you see from the preceding exploration, constants m and b in y = mx 0005 b have special geometric significance. If we let x = 0, then y = b and the graph of y = mx 0005 b crosses the y axis at (0, b). So the constant b is the y intercept. For example, the y intercept of the graph of y = 2x – 7 is 00047. We have already seen that the point (0, b) is on the graph of y = mx 0005 b. If we let x = 1, then it follows that the point (1, m 0005 b) is also on the graph (Fig. 6). Because the graph of y = mx 0005 b is a line, we can use these two points to compute the slope:
f (x)
(0, b)
(1, m 0005 b) x
Slope 0003
Z Figure 6
y2 0004 y1 (m 0005 b) 0004 b 0003 0003m x2 0004 x1 100040
(x1, y1) ⴝ (0, b) (x2, y2) ⴝ (1, m ⴙ b)
So m is the slope of the line with equation y = mx 0005 b.
Z THEOREM 2 Slope–Intercept Form An equation of the line with slope m and y intercept b is y 0003 mx 0005 b
y y 0003 mx 0005 b
which is called the slope–intercept form.
EXAMPLE
4
m0003
Rise
y intercept b
Run
x
Using the Slope–Intercept Form (A) Write the slope–intercept form of a line with slope (B) Find the slope and y intercept, and graph y 0003
3 4x
2 3
and y intercept 00045.
0004 1.
Rise Run
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SOLUTIONS
(A) Substitute m 0003 23 and b 0003 00045 in y = mx 0005 b to obtain y 0003 23x 0004 5. (B) The y intercept of y 0003 34 x 0004 1 is 00041 and the slope is 34. If we start at the point (0, 00041) and move four units to the right (run), then the y coordinate of a point on the line must move up three units (rise) to the point (4, 2). Drawing a line through these two points produces the graph shown in Figure 7. y 5
3 00045
5
x
4 00045
0002
Z Figure 7
MATCHED PROBLEM 4
Write the slope–intercept form of the line with slope equation.
5 4
and y intercept 00042. Graph the 0002
y (x, y) x (x1, y 1)
Z Figure 8
In Example 4 we found the equation of a line with a given slope and y intercept. It is also possible to find the equation of a line passing through a given point with a given slope or to find the equation of a line containing two given points. Suppose a line has slope m and passes through the point (x1, y1). If (x, y) is any other point on the line (Fig. 8), then y 0004 y1 0003m x 0004 x1
(x, y 1)
that is, y 0004 y1 0003 m(x 0004 x1)
(3)
Because the point (x1, y1) also satisfies equation (3), we can conclude that equation (3) is an equation of a line with slope m that passes through (x1, y1).
Z THEOREM 3 Point–Slope Form An equation of the line with slope m that passes through (x1, y1) is y 0004 y1 0003 m(x 0004 x1) which is called the point–slope form.
If we are given the coordinates of two points on a line, we can use the given coordinates to find the slope and then use the point–slope form with either of the given points to find the equation of the line.
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EXAMPLE
5
Equations of a Line
139
Point–Slope Form (A) Find an equation for the line that has slope 23 and passes through the point (00042, 1). Write the final answer in the form Ax 0005 By 0003 C. (B) Find an equation for the line that passes through the two points (4, 00041) and (00048, 5). Write the final answer in the form y 0003 mx 0005 b.
SOLUTIONS
(A) If m 0003 23 and (x1, y1) 0003 (00042, 1), then y 0004 y1 0003 m(x 0004 x1) y000410003
Substitute y1 ⴝ 1, x1 ⴝ ⴚ2, and m ⴝ 23 .
2 [x 0004 (00042)] 3
Multiply both sides by 3.
3( y 0004 1) 0003 2(x 0005 2) 3y 0004 3 0003 2x 0005 4 or 00042x 0005 3y 0003 7
Distribute. Write in standard form.
2x 0004 3y 0003 00047
(B) First use the slope formula to find the slope of the line: m0003
y2 0004 y1 5 0004 (00041) 6 1 0003 0003 00030004 x2 0004 x1 00048 0004 4 000412 2
Substitute x1 ⴝ 4, y1 ⴝ ⴚ1, x2 ⴝ ⴚ8, and y2 ⴝ 5 in the slope formula.
Now we choose (x1, y1) 0003 (4, 00041) and proceed as in part A: y 0004 y1 0003 m(x 0004 x1) 1 y 0004 (00041) 0003 0004 (x 0004 4) 2 1 y0005100030004 x00052 2 1 y00030004 x00051 2
1 Substitute x1 ⴝ 4, y1 ⴝ ⴚ1, and m ⴝ ⴚ . 2 y ⴚ (ⴚ1) ⴝ y ⴙ 1; Distribute on right side.
Subtract 1 from both sides.
You may want to verify that choosing (x1, y1) = (00048, 5), the other given point, produces the same equation. 0002 MATCHED PROBLEM 5
(A) Find an equation for the line that has slope 000425 and passes through the point (3, 00042). Write the final answer in the form Ax 0005 By 0003 C. (B) Find an equation for the line that passes through the two points (00043, 1) and (7, 00043). Write the final answer in the form y 0003 mx 0005 b. 0002 The simplest equations of lines are those for horizontal and vertical lines. Consider the following two equations: x 0005 0y 0003 a 0x 0005 y 0003 b
or or
x0003a y0003b
(4) (5)
In equation (4), y can be any number as long as x 0003 a. So the graph of x 0003 a is a vertical line crossing the x axis at (a, 0). In equation (5), x can be any number as long as y 0003 b.
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So the graph of y 0003 b is a horizontal line crossing the y axis at (0, b). We summarize these results as follows:
Z THEOREM 4 Vertical and Horizontal Lines Equation x0003a (short for x 0005 0y 0003 a) y0003b
Graph Vertical line through (a, 0) (Slope is undefined.) Horizontal line through (0, b) (Slope is 0.)
(short for 0x 0005 y 0003 b) y
x0003a
y0003b
b a
EXAMPLE
6
x
Graphing Horizontal and Vertical Lines Graph the line x 0003 00042 and the line y 0003 3. y
SOLUTION 5
y00033
00045
5
x
x 0003 00042 00045
MATCHED PROBLEM 6
0002
Graph the line x 0003 4 and the line y 0003 00042. 0002 The various forms of the equation of a line that we have discussed are summarized in Table 2 for convenient reference. Table 2 Equations of a Line Standard form
Ax 0005 By 0003 C
A and B not both 0
Slope–intercept form
y 0003 mx 0005 b
Slope: m; y intercept: b
Point–slope form
y 0004 y1 0003 m(x 0004 x1)
Slope: m; Point: (x1, y1)
Horizontal line
y0003b
Slope: 0
Vertical line
x0003a
Slope: Undefined
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141
Z Finding Slopes of Parallel or Perpendicular Lines From geometry, we know that two vertical lines are parallel to each other and that a horizontal line and a vertical line are perpendicular to each other. How can we tell when two nonvertical lines are parallel or perpendicular to each other? Theorem 5, which we state without proof, provides a convenient test.
Z THEOREM 5 Parallel and Perpendicular Lines Given two nonvertical lines L1 and L2 with slopes m1 and m2, respectively, then L1 储 L2 L1 ⬜ L2
if and only if if and only if
m1 0003 m2 m1m2 0003 00041
The symbols 储 and ⬜ mean, respectively, “is parallel to” and “is perpendicular to.” In the case of perpendicularity, the condition m1m2 = 00041 also can be written as m2 0003 0004
1 m1
or
m1 0003 0004
1 m2
Therefore, Two nonvertical lines are perpendicular if and only if their slopes are the negative reciprocals of each other.
EXAMPLE
7
Parallel and Perpendicular Lines Given the line L: 3x 0004 2y = 5 and the point P 0003 (00043, 5), find an equation of a line through P that is (A) Parallel to L
(B) Perpendicular to L
Write the final answers in the slope–intercept form y = mx 0005 b. SOLUTIONS
First, find the slope of L by writing 3x 0004 2y = 5 in the equivalent slope–intercept form y = mx 0005 b: 3x 0004 2y 0003 5 00042y 0003 00043x 0005 5 y 0003 32 x 0004 52 So the slope of L is 32. The slope of a line parallel to L is the same, 32, and the slope of a line perpendicular to L is 000423. We now can find the equations of the two lines in parts A and B using the point–slope form. (A) Parallel (m 0003 32): y 0004 y1 0003 m(x 0004 x1) y 0004 5 0003 32 (x 0005 3) y 0004 5 0003 32 x 0005 92 y 0003 32 x 0005 192
(B) Perpendicular (m 0003 000423): y 0004 y1 0003 m(x 0004 x1) y 0004 5 0003 000423 (x 0005 3) y 0004 5 0003 000423x 0004 2 y 0003 000423x 0005 3
Substitute for x1, y1, and m. Distribute. Add 5 to both sides.
0002
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MATCHED PROBLEM 7
Given the line L: 4x 0005 2y = 3 and the point P = (2, 00043), find an equation of a line through P that is (A) Parallel to L
(B) Perpendicular to L
Write the final answers in the slope–intercept form y = mx 0005 b.
EXAMPLE
8
0002
Cost Analysis A hot dog vendor pays $25 per day to rent a pushcart and $1.25 for the ingredients in one hot dog. (A) Find the cost of selling x hot dogs in 1 day. (B) What is the cost of selling 200 hot dogs in 1 day? (C) If the daily cost is $355, how many hot dogs were sold that day?
SOLUTIONS
(A) The rental charge of $25 is the vendor’s fixed cost—a cost that is accrued every day and does not depend on the number of hot dogs sold. The cost of the ingredients for x hot dogs is $1.25x. This is the vendor’s variable cost—a cost that depends on the number of hot dogs sold. The total cost for selling x hot dogs is C(x) 0003 1.25x 0005 25
Total Cost ⴝ Variable Cost ⴙ Fixed Cost
(B) The cost of selling 200 hot dogs in 1 day is C(200) 0003 1.25(200) 0005 25 0003 $275 (C) The number of hot dogs that can be sold for $355 is the solution of the equation 1.25x 0005 25 0003 355 1.25x 0003 330 330 x0003 1.25 0003 264 hot dogs MATCHED PROBLEM 8
Subtract 25 from each side. Divide both sides by 1.25. Simplify.
0002
It costs a pretzel vendor $20 per day to rent a cart and $0.75 for each pretzel. (A) Find the cost of selling x pretzels in 1 day. (B) What is the cost of selling 150 pretzels in 1 day? (C) If the daily cost is $275, how many pretzels were sold that day? 0002
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143
Technology Connections A graphing calculator can be used to solve equations like 1.25x ⴙ 25 ⴝ 355 (see Example 8). First enter both sides of the equation in the equation editor [Fig. 9(a)] and choose window variables [Fig. 9(b)] so that the graphs of both equations appear on the screen. There is no “right” choice for the window variables. Any choice that displays the intersection point will do. (Here is how we chose our window variables: We chose Ymax ⴝ 600 to place the graph of the horizontal
line below the top of the window. We chose Ymin ⴝ ⴚ200 to place the graph of the x axis above the text displayed at the bottom of the screen. Since x cannot be negative, we chose Xmin ⴝ 0. We used trial and error to determine a reasonable choice for Xmax.) Now choose intersect on the CALC menu, and respond to the prompts from the calculator. The coordinates of the intersection point of the two graphs are shown at the bottom of the screen [Fig. 9(c)]. 600
0
(a)
400
0004200
(b)
(c)
Z Figure 9
ANSWERS TO MATCHED PROBLEMS 2. (A) Run 0003 5, rise 0003 4, slope 0003 45 (B) Run 0003 3, rise 0003 00046, slope 0003 00046 3 0003 00042 3. (A) m 0003 0 (B) m 0003 1 (C) m 0003 00044 (D) m is not defined
y
1.
00045
5
x
00045
4. y 0003 54 x 0004 2
5. (A) 2x 0005 5y 0003 00044 y
y
6.
5
5
5 5
00045
x
4 00045
7. (A) y 0003 00042x 0005 1 (B) y 0003 12 x 0004 4 8. (A) C(x) 0003 0.75x 0005 20 (B) $132.50
(B) y 0003 000425 x 0004 15
x00034
00045
5
y 0003 00042 00045
(C) 340 pretzels
x
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Exercises
1. Explain how to find the x and y intercepts of a line if its equation is written in standard form.
y
10. 6
2. Given the graph of a line, explain how to determine whether the slope is negative. 3. Explain why y ⫽ mx ⫹ b is called the slope–intercept form. ⫺6
4. Explain why y ⫺ y1 ⫽ m(x ⫺ x1) is called the point–slope form. 5. Given the equations of two lines in standard form, explain how to determine whether the lines are parallel. 6. Given the equations of two lines in standard form, explain how to determine whether the lines are perpendicular.
6
x
⫺6
y
11. 5
In Problems 7–12, use the graph of each line to find the rise, run, and slope. Write the equation of each line in the standard form Ax ⫹ By ⫽ C, A ⱖ 0. (All the horizontal and vertical line segments have integer lengths.)
⫺5
5
x
y
7.
⫺5
5
y
12. ⫺5
5
5
x
⫺5
⫺5
5
x
y
8. 5
⫺5
⫺5
5
x
In Problems 13–18, use the graph of each line to find the x intercept, y intercept, and slope, if they exist. Write the equation of each line, using the slope–intercept form whenever possible. y
13.
⫺5
5
y
9. 5
⫺5
⫺5
5
⫺5
x
5
⫺5
x
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SECTION 2–3 y
14.
25. 4x 0004 5y 0003 000424
5
27.
00045
5
x
145
26. 6x 0004 7y 0003 000449 28.
y x 0004 00031 6 5
29. x 0003 00043
30. y 0003 00042
31. y 0003 3.5
32. x 0003 2.5
In Problems 33–38, find an equation of the line with the indicated slope and y intercept, and write it in the form Ax 0005 By 0003 C, A 0007 0, where A, B, and C are integers.
00045
y
15.
y x 0004 00031 8 4
Equations of a Line
33. Slope 0003 00043; y intercept 0003 7
5
34. Slope 0003 4; y intercept 0003 000410 35. Slope 0003 72; y intercept 0003 000413
00045
5
x
36. Slope 0003 000454; y intercept 0003 115 37. Slope 0003 0; y intercept 0003 23 38. Slope 0003 0; y intercept 0003 0
00045
In Problems 39–44, find the equation of the line passing through the given point with the given slope. Write the final answer in the slope–intercept form y 0003 mx 0005 b.
y
16. 5
39. (0, 3); m 0003 00042 00045
5
x
43. (00042, 00043); m 0003 000412
y
45. (0, 4); m 0003 00043
5
47. (00045, 4); m 0003 00045
5
x
00045
y
18. 5
00045
40. (4, 0); m 0003 3 42. (2, 00043); m 0003 000445 44. (2, 1); m 0003
4 3
In Problem 45–58, write the equation of the line that contains the indicated point(s), and/or has the given slope or intercepts; use either the slope–intercept form y 0003 mx 0005 b, or the form x 0003 c.
00045
17.
41. (00045, 4); m 0003
3 2
5
x
000425
46. (2, 0); m 0003 2 48. (00044, 00042); m 0003 12
49. (1, 6); (5, 00042)
50. (00043, 4); (6, 1)
51. (00044, 8); (2, 0)
52. (2, 00041); (10, 5)
53. (00043, 4); (5, 4)
54. (0, 00042); (4, 00042)
55. (4, 6); (4, 00043)
56. (00043, 1); (00043, 00044)
57. x intercept 00044; y intercept 3
58. x intercept 00044; y intercept 00045
In Problems 59–66, write an equation of the line that contains the indicated point and meets the indicated condition(s). Write the final answer in the standard form Ax 0005 By 0003 C, A 0007 0. 59. (00043, 4); parallel to y 0003 3x 0004 5 60. (00044, 0); parallel to y 0003 00042x 0005 1
00045
61. (2, 00043); perpendicular to y 0003 000413 x
Graph each equation in Problems 19–32, and indicate the slope, if it exists.
62. (00042, 00044); perpendicular to y 0003 23 x 0004 5 63. (5, 0); parallel to 3x 0004 2y 0003 4
19. y 0003 000435 x 0005 4
20. y 0003 000432 x 0005 6
64. (3, 5); parallel to 3x 0005 4y 0003 8
21. y 0003 000434 x
22. y 0003 23 x 0004 3
65. (0, 00044); perpendicular to x 0005 3y 0003 9
23. 4x 0005 2y 0003 0
24. 6x 0004 2y 0003 0
66. (00042, 4); perpendicular to 4x 0005 5y 0003 0
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Problems 67–72 refer to the quadrilateral with vertices A 0003 (0, 2), B 0003 (4, 00041), C 0003 (1, 00045), and D 0003 (00043, 00042).
(A) Complete Table 4.
Table 4
67. Show that AB 储 DC.
68. Show that DA 储 CB.
69. Show that AB ⬜ BC.
70. Show that AD ⬜ DC.
x
71. Find an equation of the perpendicular bisector* of AD.
A
72. Find an equation of the perpendicular bisector of AB. 73. Prove that if a line L has x intercept (a, 0) and y intercept (0, b), then the equation of L can be written in the intercept form y x 0005 00031 a b
a, b 0002 0
74. Prove that if a line L passes through P1 0003 (x1, y1) and P2 0003 (x2, y2), then the equation of L can be written in the twopoint form ( y 0004 y1)(x2 0004 x1) 0003 ( y2 0004 y1)(x 0004 x1)
75. x2 0005 y2 0003 25, (3, 4)
76. x2 0005 y2 0003 100, (00048, 6)
77. x2 0005 y2 0003 50, (5, 00045)
78. x2 0005 y2 0003 80, (00044, 00048)
79. (x 0004 3)2 0005 ( y 0005 4)2 0003 169, (8, 000416) 80. (x 0005 5)2 0005 ( y 0004 9)2 0003 289, (000413, 00046)
APPLICATIONS 81. BOILING POINT OF WATER At sea level, water boils when it reaches a temperature of 212°F. At higher altitudes, the atmospheric pressure is lower and so is the temperature at which water boils. The boiling point B in degrees Fahrenheit at an altitude of x feet is given approximately by B 0003 212 0004 0.0018x (A) Complete Table 3.
Table 3 0
5,000
10,000
15,000
20,000
1
2
3
4
5
(B) Based on the information in the table, write a brief verbal description of the relationship between altitude and air temperature. 83. COST ANALYSIS A doughnut shop has a fixed cost of $124 per day and a variable cost of $0.12 per doughnut. Find the total daily cost of producing x doughnuts. How many doughnuts can be produced for a total daily cost of $250? 84. COST ANALYSIS A small company manufactures picnic tables. The weekly fixed cost is $1,200 and the variable cost is $45 per table. Find the total weekly cost of producing x picnic tables. How many picnic tables can be produced for a total weekly cost of $4,800? 85. PHYSICS Hooke’s law states that the relationship between the stretch s of a spring and the weight w causing the stretch is linear (a principle upon which all spring scales are constructed). For a particular spring, a 5-pound weight causes a stretch of 2 inches, while with no weight the stretch of the spring is 0. (A) Find a linear equation that expresses s in terms of w. (B) What is the stretch for a weight of 20 pounds? (C) What weight will cause a stretch of 3.6 inches?
Problems 75–80 are calculus related. Recall that a line tangent to a circle at a point is perpendicular to the radius drawn to that point (see the figure). Find the equation of the line tangent to the circle at the indicated point. Write the final answer in the standard form Ax 0005 By 0003 C, A 0007 0. Graph the circle and the tangent line on the same coordinate system.
x
0
25,000
30,000
B (B) Based on the information in the table, write a brief verbal description of the relationship between altitude and the boiling point of water. 82. AIR TEMPERATURE As dry air moves upward, it expands and cools. The air temperature A in degrees Celsius at an altitude of x kilometers is given approximately by A 0003 25 0004 9x *The perpendicular bisector of a line segment is a line perpendicular to the segment and passing through its midpoint.
86. PHYSICS The distance d between a fixed spring and the floor is a linear function of the weight w attached to the bottom of the spring. The bottom of the spring is 18 inches from the floor when the weight is 3 pounds and 10 inches from the floor when the weight is 5 pounds. (A) Find a linear equation that expresses d in terms of w. (B) Find the distance from the bottom of the spring to the floor if no weight is attached. (C) Find the smallest weight that will make the bottom of the spring touch the floor. (Ignore the height of the weight.) 87. PHYSICS The two most widespread temperature scales are Fahrenheit* (F) and Celsius† (C). It is known that water freezes at 32°F or 0°C and boils at 212°F or 100°C. (A) Find a linear equation that expresses F in terms of C. (B) If a European family sets its house thermostat at 20°C, what is the setting in degrees Fahrenheit? If the outside temperature in Milwaukee is 86°F, what is the temperature in degrees Celsius? 88. PHYSICS Two other temperature scales, used primarily by scientists, are Kelvin‡ (K) and Rankine** (R). Water freezes at 273 K or 492°R and boils at 373 K or 672°R. Find a linear equation that expresses R in terms of K. 89. OCEANOGRAPHY After about 9 hours of a steady wind, the height of waves in the ocean is approximately linearly related to *Invented in 1724 by Daniel Gabriel Fahrenheit (1686–1736), a German physicist. † Invented in 1742 by Anders Celsius (1701–1744), a Swedish astronomer. ‡ Invented in 1848 by Lord William Thompson Kelvin (1824–1907), a Scottish mathematician and physicist. Note that the degree symbol “ ° ” is not used with degrees Kelvin. **Invented in 1859 by John Maquorn Rankine (1820–1872), a Scottish engineer and physicist.
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Express all calculated quantities to three significant digits. 90. OCEANOGRAPHY Refer to Problem 89. A steady 25-knot wind produces a wave 7 feet high after 9 hours and 11 feet high after 25 hours. (A) Write a linear equation that expresses height h in terms of time t. (B) How long will the wind have been blowing for the waves to be 20 feet high? 91. DEMOGRAPHICS Life expectancy in the United States has increased from about 49.2 years in 1900 to about 77.3 years in 2000. The growth in life expectancy is approximately linear with respect to time. (A) If L represents life expectancy and t represents the number of years since 1900, write a linear equation that expresses L in terms of t. (B) What is the predicted life expectancy in the year 2020? Express all calculated quantities to three significant digits.
147
92. DEMOGRAPHICS The average number of persons per household in the United States has been shrinking steadily for as long as statistics have been kept and is approximately linear with respect to time. In 1900, there were about 4.76 persons per household and in 2000, about 2.59. (A) If N represents the average number of persons per household and t represents the number of years since 1900, write a linear equation that expresses N in terms of t. (B) What is the predicted household size in the year 2025? Express all calculated quantities to three significant digits. 93. CITY PLANNING The design of a new subdivision calls for three parallel streets connecting First Street with Main Street (see the figure). Find the distance d1 (to the nearest foot) from Avenue A to Avenue B. First Street Avenue A
Distance in feet
the duration of time the wind has been blowing. During a storm with 50-knot winds, the wave height after 9 hours was found to be 23 feet, and after 24 hours it was 40 feet. (A) If t is time after the 50-knot wind started to blow and h is the wave height in feet, write a linear equation that expresses height h in terms of time t. (B) How long will the wind have been blowing for the waves to be 50 feet high?
Linear Equations and Models
5,000
Avenue B
Avenue C
d2
d1
0
5,000
Main Street
Distance in feet
94. CITY PLANNING Refer to Problem 93. Find the distance d2 (to the nearest foot) from Avenue B to Avenue C.
2-4
Linear Equations and Models Z Slope as a Rate of Change Z Linear Models Z Linear Regression
Mathematical modeling is the process of using mathematics to solve real-world problems. This process can be broken down into three steps (Fig. 1): Step 1. Construct the mathematical model, a mathematics problem that, when solved, will provide information about the real-world problem.
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Step 2. Solve the mathematical model. Step 3. Interpret the solution to the mathematical model in terms of the original real-world problem.
Real-world problem
3. I
t uc str
nt er p
on
re t
C 1.
Mathematical solution
2. Solve
Mathematical model
Z Figure 1
In more complex problems, this cycle may have to be repeated several times to obtain the required information about the real-world problem. In this section, we discuss one of the simplest mathematical models, a linear equation. With the aid of a graphing calculator, we also learn how to analyze a linear model based on real-world data.
Z Slope as a Rate of Change If x and y are related by the equation y 0003 mx 0005 b, where m and b are constants with m 0002 0, then x and y are linearly related. If (x1, y1) and (x2, y2) are two distinct points on this line, then the slope of the line is m0003
Change in y y2 0004 y1 0003 x2 0004 x1 Change in x
(1)
In applications, ratio (1) is called the rate of change of y with respect to x. Since the slope of a line is unique, the rate of change of two linearly related variables is constant. Here are some examples of familiar rates of change: miles per hour, revolutions per minute, price per pound, passengers per plane, etc. If y is distance and x is time, then the rate of change is also referred to as speed or velocity. If the relationship between x and y is not linear, ratio (1) is called the average rate of change of y with respect to x.
EXAMPLE
1
Estimating Body Surface Area Appropriate doses of medicine for both animals and humans are often based on body surface area (BSA). Since weight is much easier to determine than BSA, veterinarians use the weight of an animal to estimate BSA. The following linear equation expresses BSA for canines in terms of weight*: a 0003 16.21w 0005 375.6 where a is BSA in square inches and w is weight in pounds. (A) Interpret the slope of the BSA equation. (B) What is the effect of a 1-pound increase in weight?
SOLUTIONS
(A) The rate of change BSA with respect to weight is 16.21 square inches per pound. (B) Since slope is the ratio of rise to run, increasing w by 1 pound (run) increases a by 16.21 square inches (rise). 0002 *Based on data from Veterinary Oncology Consultants, PTY LTD.
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MATCHED PROBLEM 1
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149
The following linear equation expresses BSA for felines in terms of weight: a ⫽ 28.55w ⫹ 118.7 where a is BSA in square inches and w is weight in pounds. (A) Interpret the slope of the BSA equation. (B) What is the effect of a 1-pound increase in weight? 0002
Z Linear Models We can use our experience with lines in Section 2-3 to construct linear models for applications involving linearly related quantities. This process is best illustrated through examples.
EXAMPLE
2
Business Markup Policy A sporting goods store sells a fishing rod that cost $60 for $82 and a pair of cross-country ski boots that cost $80 for $106. (A) If the markup policy of the store for items that cost more than $30 is assumed to be linear, find a linear model that express the retail price P in terms of the wholesale cost C. (B) What is the effect on the price of a $1 increase in cost for any item costing over $30? (C) Use the model to find the retail price for a pair of running shoes that cost $40.
SOLUTIONS
(A) If price P is linearly related to cost C, then we are looking for the equation of a line whose graph passes through (C1, P1) ⫽ (60, 82) and (C2, P2) ⫽ (80, 106). We find the slope, and then use the point–slope form to find the equation. m⫽
P2 ⫺ P1 106 ⫺ 82 24 ⫽ ⫽ ⫽ 1.2 C2 ⫺ C1 80 ⫺ 60 20
Substitute P1 ⴝ 82, C1 ⴝ 60, and m ⴝ 1.2 into the point–slope formula.
P ⫺ P1 ⫽ m(C ⫺ C1) P ⫺ 82 ⫽ 1.2(C ⫺ 60) P ⫺ 82 ⫽ 1.2C ⫺ 72 P ⫽ 1.2C ⫹ 10
Substitute C1 ⴝ 60, P1 ⴝ 82, C2 ⴝ 80, and P2 ⴝ 106 into the slope formula.
Distribute Add 82 to both sides.
C 7 30
Linear model
(B) If the cost is increased by $1, then the price will increase by 1.2(1) ⫽ $1.20. (C) P ⫽ 1.2(40) ⫹ 10 ⫽ $58. MATCHED PROBLEM 2
ZZZ EXPLORE-DISCUSS 1
0002
The sporting goods store in Example 2 is celebrating its twentieth anniversary with a 20% off sale. The sale price of a mountain bike is $380. What was the presale price of the bike? How much did the bike cost the store? 0002 The wholesale supplier for the sporting goods store in Example 2 offers the store a 15% discount on all items. The store decides to pass on the savings from this discount to the consumer. Which of the following markup policies is better for the consumer? 1. Apply the store’s markup policy to the discounted cost. 2. Apply the store’s markup policy to the original cost and then reduce this price by 15%. Support your choice with examples.
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3
Mixing Antifreeze Ethylene glycol and propylene glycol are liquids used in antifreeze and deicing solutions. Ethylene glycol is listed as a hazardous chemical by the Environmental Protection Agency, while propylene glycol is generally regarded as safe. Table 1 lists solution concentration percentages and the corresponding freezing points for each chemical. Table 1 Concentration
Ethylene Glycol
Propylene Glycol
20%
15°F
17°F
50%
000436°F
000428°F
(A) Assume that the concentration and the freezing point for ethylene glycol are linearly related. Construct a linear model for the freezing point. (B) Interpret the slope in part (A). (C) What percentage (to one decimal place) of ethylene glycol will result in a freezing point of 000410°F? SOLUTIONS
(A) We begin by defining appropriate variables: Let p 0003 percentage of ethylene glycol in the antifreeze solution f 0003 freezing point of the antifreeze solution From Table 1, we see that (20, 15) and (50, ⴚ36) are two points on the line relating p and f. The slope of this line is m0003
f2 0004 f1 15 0004 (ⴚ36) 51 0003 0003 0003 00041.7 p2 0004 p1 20 0004 50 000430
and its equation is f 0004 15 0003 00041.7( p 0004 20) f 0003 00041.7p 0005 49
Linear model
(B) The rate of change of the freezing point with respect to the percentage of ethylene glycol in the antifreeze solution is 00041.7 degrees per percentage of ethylene glycol. Increasing the amount of ethylene glycol by 1% will lower the freezing point by 1.7°F. (C) We must find p when f is 000410°. f 0003 00041.7p 0005 49 000410 0003 00041.7p 0005 49 1.7p 0003 59 59 p0003 0003 34.7% 1.7
MATCHED PROBLEM 3
Add 10 ⴙ 1.7p to both sides. Divide both sides by 1.7.
0002
Refer to Table 1. (A) Assume that the concentration and the freezing point for propylene glycol are linearly related. Construct a linear model for the freezing point. (B) Interpret the slope in part (A). (C) What percentage (to one decimal place) of propylene glycol will result in a freezing point of 000415°F? 0002
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EXAMPLE
4
Linear Equations and Models
151
Underwater Pressure The pressure at sea level is 14.7 pounds per square inch. As you descend into the ocean, the pressure increases linearly at a rate of about 0.445 pounds per square foot. (A) Find the pressure p at a depth of d feet. (B) If a diver’s equipment is rated to be safe up to a pressure of 40 pounds per square foot, how deep (to the nearest foot) is it safe to use this equipment?
SOLUTIONS
(A) Let p ⫽ md ⫹ b. At the surface, d ⫽ 0 and p ⫽ 14.7, so b ⫽ 14.7. The slope m is the given rate of change, m ⫽ 0.445. So the pressure at a depth of d feet is p ⫽ 0.445d ⫹ 14.7 (B) The safe depth is the solution of the equation 0.445d ⫹ 14.7 ⫽ 40 0.445d ⫽ 25.3 25.3 d⫽ 0.445 ⬇ 57 feet
MATCHED PROBLEM 4
Subtract 14.7 from each side. Divide both sides by 0.445. Simplify.
0002
The rate of change of pressure in fresh water is 0.432 pounds per square foot. Repeat Example 4 for a body of fresh water. 0002
Technology Connections 80
Figure 2 shows the solution of Example 4(B) on a graphing calculator. 0
100
⫺20
Z Figure 2 y1 ⫽ 0.445x ⫹ 14.7, y2 ⫽ 40
Z Linear Regression In real-world applications we often encounter numerical data in the form of a table. The very powerful mathematical tool, regression analysis, can be used to analyze numerical data. In general, regression analysis is a process for finding an equation that provides a useful model for a set of data points. Graphs of equations are often called curves and regression analysis is also referred to as curve fitting. In Example 5, we use a linear model obtained by using linear regression on a graphing calculator.
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5
Table 2 Round-Shaped Diamond Prices Weight (Carats)
Price
0.5
$1,340
0.6
$1,760
0.7
$2,540
0.8
$3,350
0.9
$4,130
1.0
$4,920
Source: www.tradeshop.com
SOLUTIONS
Diamond Prices Prices for round-shaped diamonds taken from an online trader are given in Table 2. (A) A linear model for the data in Table 2 is given by p ⫽ 7,380c ⫺ 2,530
(2)
where p is the price of a diamond weighing c carats. (We will discuss the source of models like this later in this section.) Plot the points in Table 2 on a Cartesian coordinate system, producing a scatter plot, and graph the model on the same axes. (B) Interpret the slope of the model in equation (2). (C) Use the model to estimate the cost of a 0.85-carat diamond and the cost of a 1.2-carat diamond. Round answers to the nearest dollar. (D) Use the model to estimate the weight of a diamond that sells for $3,000. Round the answer to two significant digits. (A) A scatter plot is simply a plot of the points in Table 2 [Fig. 3(a)]. To add the graph of the model to the scatter plot, we find any two points that satisfy equation (2) [we choose (0.4, 422) and (1.4, 7,802)]. Plotting these points and drawing a line through them gives us Figure 3(b). p
p
$8,000
$8,000
Price
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$4,000
0.5
1
1.5
$4,000
c
0.5
Carats
1
1.5
c
Carats (b) Linear model
(a) Scatter plot
Z Figure 3
(B) The rate of change of the price of a diamond with respect to its weight is 7,380. Increasing the weight by 1 carat will increase the price by about $7,380. (C) The graph of the model [Fig. 3(b)] does not pass through any of the points in the scatter plot, but it comes close to all of them. [Verify this by evaluating equation (2) at c ⫽ 0.5, 0.6, . . . , 1.] So we can use equation (2) to approximate points not in Table 2. c ⫽ 0.85 p ⫽ 7,380(0.85) ⫺ 2,530 ⫽ $3,743
c ⫽ 1.2 p ⫽ 7,380(1.2) ⫺ 2,530 ⫽ $6,326
A 0.85-carat diamond will cost about $3,743 and a 1.2-carat diamond will cost about $6,326. (D) To find the weight of a $3,000 diamond, we solve the following equation for c: 7,380c ⫺ 2,530 ⫽ 3,000 7,380c ⫽ 3,000 ⫹ 2,530 ⫽ 5,530 5,530 ⫽ 0.75 c⫽ 7,380
To two significant digits
A $3,000 diamond will weigh about 0.75 carats.
0002
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MATCHED PROBLEM 5 Table 3 Emerald-Shaped Diamond Prices Weight (Carats)
Price
0.5
$1,350
0.6
$1,740
0.7
$2,610
0.8
$3,320
0.9
$4,150
1.0
$4,850
Source: www.tradeshop.com
Linear Equations and Models
153
Prices for emerald-shaped diamonds taken from an online trader are given in Table 3. Repeat Example 5 for this data with the linear model p ⫽ 7,270c ⫺ 2,450 where p is the price of an emerald-shaped diamond weighing c carats. 0002 The model we used in Example 5 was obtained by using a technique called linear regression and the model is called the regression line. This technique produces a line that is the best fit for a given data set. We will not discuss the theory behind this technique, nor the meaning of “best fit.” Although you can find a linear regression line by hand, we prefer to leave the calculations to a graphing calculator or a computer. Don’t be concerned if you don’t have either of these electronic devices. We will supply the regression model in the applications we discuss, as we did in Example 5.
Technology Connections If you want to use a graphing calculator to construct regression lines, you should consult your user’s manual.* The process varies from one calculator to another. Figure 4
shows three of the screens related to the construction of the model in Example 5 on a Texas Instruments TI-84 Plus. 8,000
0
1.5
⫺1,000
(a) Entering the data.
(b) Finding the model.
(c) Graphing the data and the model.
Z Figure 4 *User’s manuals for the most popular graphing calculators are readily available on the Internet.
In Example 5, we used the regression line to approximate points that were not given in Table 2, but would fit between points in the table. This process is called interpolation. In the next example we use a regression model to approximate points outside the given data set. This process is called extrapolation and the approximations are often referred to as predictions.
EXAMPLE
6
Telephone Expenditures Table 4 gives information about expenditures for residential and cellular phone service. The linear regression model for residential service is r ⫽ 722 ⫺ 33.1t where r is the average annual expenditure (in dollars per consumer unit) on residential service and t is time in years with t ⫽ 0 corresponding to 2000. (A) Interpret the slope of the regression line as a rate of change. (B) Use the regression line to predict expenditures for residential service in 2018.
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Table 4 Average Annual Telephone Expenditures (dollars per consumer unit) 2001
2003
2005
2007
Residential
686
620
570
482
Cellular
210
316
455
608
Source: Bureau of Labor Statistics
SOLUTIONS
(A) The slope m 0003 000433.1 is the rate of change of expenditures with respect to time. Because the slope is negative, the expenditures for residential service are decreasing at a rate of $33.10 per year. (B) If t 0003 18, then r 0003 722 0004 33.1(18) 0003 $126 So the model predicts that expenditures for residential phone service will be approximately $126 in 2018. Repeat Example 6 using the following linear regression model for cellular service: c 0003 66.7t 0005 131 where c is the average annual expenditure (in dollars per consumer unit) on cellular service and t is time in years with t = 0 corresponding to 2000. 0002
ANSWERS TO MATCHED PROBLEMS 1. (A) The rate of change of BSA with respect to weight is 28.55 square inches per pound. (B) Increasing w by 1 pound increases a by 28.55 square inches. 2. Presale price is $475. Cost is $387.50 3. (A) f 0003 00041.5p 0005 47 (B) The rate of change of the freezing point with respect to the percentage of propylene glycol in the antifreeze solution is 00041.5. Increasing the percentage of propylene glycol by 1% will lower the freezing point by 1.5°F. (C) 41.3% 4. (A) p 0003 0.432d 0005 14.7 (B) 59 ft p 5. (A) $8,000
Price
MATCHED PROBLEM 6
0002
$4,000
0.5
1
1.5
c
Carats
(B) The rate of change of the price of a diamond with respect to the size is 7,270. Increasing the size by 1 carat will increase the price by about $7,270. (C) $3,730; $6,274 (D) 0.75 carats 6. (A) The expenditures for cellular service are increasing at a rate of $66.70 per year. (B) $1,332.
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2-4
Linear Equations and Models
155
Exercises
1. Explain the steps that are involved in the process of mathematical modeling.
9. Dr. J. D. Robinson and Dr. D. R. Miller published the following models for estimating the weight of a woman:
2. If two variables x and y are linearly related, explain how to calculate the rate of change.
Robinson: w 0003 108 0005 3.7h
3. If two variables x and y are not linearly related, explain how to calculate the average rate of change from x 0003 x1 to x 0003 x2.
where w is weight (in pounds) and h is height over 5 feet (in inches). (A) Interpret the slope of each model. (B) If a woman is 500066 tall, what does each model predict her weight to be? (C) If a woman weighs 140 pounds, what does each model predict her height to be?
4. Explain the difference between interpolation and extrapolation in the context of regression analysis.
APPLICATIONS 5. COST ANALYSIS A plant can manufacture 80 golf clubs per day for a total daily cost of $8,147 and 100 golf clubs per day for a total daily cost of $9,647. (A) Assuming that daily cost and production are linearly related, find the total daily cost of producing x golf clubs. (B) Interpret the slope of this cost equation. (C) What is the effect of a 1 unit increase in production? 6. COST ANALYSIS A plant can manufacture 50 tennis rackets per day for a total daily cost of $4,174 and 60 tennis rackets per day for a total daily cost of $4,634. (A) Assuming that daily cost and production are linearly related, find the total daily cost of producing x tennis rackets. (B) Interpret the slope of this cost equation. (C) What is the effect of a 1 unit increase in production? 7. FORESTRY Forest rangers estimate the height of a tree by measuring the tree’s diameter at breast height (DBH) and then using a model constructed for a particular species.* A model for white spruce trees is h 0003 4.06d 0005 24.1 where d is the DBH in inches and h is the tree height in feet. (A) Interpret the slope of this model. (B) What is the effect of a 1-inch increase in DBH? (C) How tall is a white spruce with a DBH of 12 inches? Round answer to the nearest foot. (D) What is the DBH of a white spruce that is 100 feet tall? Round answer to the nearest inch. 8. FORESTRY A model for black spruce trees is h 0003 2.27d 0005 33.1 where d is the DBH in inches and h is the tree height in feet. (A) Interpret the slope of this model. (B) What is the effect of a 1-inch increase in DBH? (C) How tall is a black spruce with a DBH of 12 inches? Round answer to the nearest foot. (D) What is the DBH of a black spruce that is 100 feet tall? Round answer to the nearest inch. *Models in Problems 7 and 8 are based on data found at http://flash.lakeheadu.ca/~fluckai/htdbh04.xls
Miller: w 0003 117 0005 3.0h
10. Dr. J. D. Robinson and Dr. D. R. Miller also published the following models for estimating the weight of a man: Robinson: w 0003 115 0005 4.2h Miller: w 0003 124 0005 3.1h where w is weight (in pounds) and h is height over 5 feet (in inches). (A) Interpret the slope of each model. (B) If a man is 5000610 tall, what does each model predict his weight to be? (C) If a man weighs 160 pounds, what does each model predict his height to be? 11. SPEED OF SOUND The speed of sound through the air near sea level is linearly related to the temperature of the air. If sound travels at 741 mph at 32°F and at 771 mph at 72°F, construct a linear model relating the speed of sound (s) and the air temperature (t). Interpret the slope of this model. 12. SPEED OF SOUND The speed of sound through the air near sea level is linearly related to the temperature of the air. If sound travels at 337 mps (meters per second) at 10°C and at 343 mps at 20°C, construct a linear model relating the speed of sound (s) and the air temperature (t). Interpret the slope of this model. 13. SMOKING STATISTICS The percentage of male cigarette smokers in the United States declined from 25.7% in 2000 to 23.9% in 2006. Find a linear model relating the percentage m of male smokers to years t since 2000. Use the model to predict the first year for which the percentage of male smokers will be less than or equal to 18%. 14. SMOKING STATISTICS The percentage of female cigarette smokers in the United States declined from 21.0% in 2000 to 18.0% in 2006. Find a linear model relating the percentage f of female smokers to years t since 2000. Use the model to predict the first year for which the percentage of female smokers will be less than or equal to 10%. 15. BUSINESS—DEPRECIATION A farmer buys a new tractor for $142,000 and assumes that it will have a trade-in value of $67,000 after 10 years. The farmer uses a constant rate of depreciation (commonly called straight-line depreciation—one of several methods permitted by the IRS) to determine the annual value of the tractor. (A) Find a linear model for the depreciated value V of the tractor t years after it was purchased.
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(B) Interpret the slope of this model. (C) What is the depreciated value of the tractor after 6 years? 16. BUSINESS—DEPRECIATION A charter fishing company buys a new boat for $154,900 and assumes that it will have a trade-in value of $46,100 after 16 years. (A) Use straight-line depreciation (See Problem 15) to find a linear model for the depreciated value V of the boat t years after it was purchased. (B) Interpret the slope of this model. (C) In which year will the depreciated value of the boat fall below $100,000?
23. LICENSED DRIVERS Table 5 contains the state population and the number of licensed drivers in the state (both in millions) for the states with population under 1 million. The regression model for this data is y 0003 0.72x 0005 0.03 where x is the state population and y is the number of licensed drivers in the state.
Table 5 Licensed Drivers in 2006 State
Population
Licensed Drivers
17. BUSINESS—MARKUP POLICY A drugstore sells a drug costing $85 for $112 and a drug costing $175 for $238. (A) If the markup policy of the drugstore is assumed to be linear, write an equation that expresses retail price R in terms of cost C (wholesale price). (B) What is the slope of the graph of the equation found in part A? Interpret verbally. (C) What does a store pay (to the nearest dollar) for a drug that retails for $185?
Alaska
0.67
0.49
Delaware
0.85
0.62
Montana
0.94
0.72
North Dakota
0.64
0.47
South Dakota
0.78
0.58
Vermont
0.62
0.53
18. BUSINESS—MARKUP POLICY A clothing store sells a shirt costing $20 for $33 and a jacket costing $60 for $93. (A) If the markup policy of the store for items costing over $10 is assumed to be linear, write an equation that expresses retail price R in terms of cost C (wholesale price). (B) What is the slope of the equation found in part A? Interpret verbally. (C) What does a store pay for a suit that retails for $240?
Wyoming
0.52
0.39
19. FLIGHT CONDITIONS In stable air, the air temperature drops about 5 F for each 1,000-foot rise in altitude. (A) If the temperature at sea level is 70°F and a commercial pilot reports a temperature of 000420 F at 18,000 feet, write a linear equation that expresses temperature T in terms of altitude A (in thousands of feet). (B) How high is the aircraft if the temperature is 0 F? 20. FLIGHT NAVIGATION An airspeed indicator on some aircraft is affected by the changes in atmospheric pressure at different altitudes. A pilot can estimate the true airspeed by observing the indicated airspeed and adding to it about 2% for every 1,000 feet of altitude. (A) If a pilot maintains a constant reading of 200 miles per hour on the airspeed indicator as the aircraft climbs from sea level to an altitude of 10,000 feet, write a linear equation that expresses true airspeed T (miles per hour) in terms of altitude A (thousands of feet). (B) What would be the true airspeed of the aircraft at 6,500 feet? 21. RATE OF DESCENT—PARACHUTES At low altitudes, the altitude of a parachutist and time in the air are linearly related. A jump at 2,880 ft using the U.S. Army’s T-10 parachute system lasts 120 seconds. (A) Find a linear model relating altitude a (in feet) and time in the air t (in seconds). (B) The rate of descent is the speed at which the jumper falls. What is the rate of descent for a T-10 system? 22. RATE OF DESCENT—PARACHUTES The U.S. Army is considering a new parachute, the ATPS system. A jump at 2,880 ft using the ATPS system lasts 180 seconds. (A) Find a linear model relating altitude a (in feet) and time in the air t (in seconds). (B) What is the rate of descent for an ATPS system parachute?
Source: Bureau of Transportation Statistics
(A) Plot the data in Table 5 and the model on the same axes. (B) If the population of New Hampshire in 2006 was about 1.3 million, use the model to estimate the number of licensed drivers in New Hampshire. (C) If the population of Nebraska in 2006 was about 1.8 million, use the model to estimate the number of licensed drivers in Nebraska. 24. LICENSED DRIVERS Table 6 contains the state population and the number of licensed drivers in the state (both in millions) for several states with population over 10 million. The regression model for this data is y 0003 0.60x 0005 1.15 where x is the state population and y is the number of licensed drivers in the state.
Table 6 Licensed Drivers in 2006 State
Population
Licensed Drivers
California
36
23
Florida
18
14
Illinois
13
8
Michigan
10
7
New York
19
11
Ohio
11
8
Pennsylvania
12
9
Texas
24
15
Source: Bureau of Transportation Statistics
(A) Plot the data in Table 6 and the model on the same axes. (B) If the population of Georgia in 2006 was about 9.4 million, use the model to estimate the number of licensed drivers in Georgia. (C) If the population of New Jersey in 2006 was about 8.7 million, use the model to estimate the number of licensed drivers in New Jersey.
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Problems 25–28 require a graphing calculator or a computer that can calculate the linear regression line for a given data set. 25. OLYMPIC GAMES Find a linear regression model for the men’s 100-meter freestyle data given in Table 7, where x is years since 1968 and y is winning time (in seconds). Do the same for the women’s 100-meter freestyle data. (Round regression coefficients to four significant digits.) Do these models indicate that the women will eventually catch up with the men?
Table 7 Winning Times in Olympic Swimming Events 100-Meter Freestyle
200-Meter Backstroke
Men
Women
Men
Women
1968
52.20
60.0
2:09.60
2:24.80
1976
49.99
55.65
1:59.19
2:13.43
1984
49.80
55.92
2:00.23
2:12.38
1992
49.02
54.65
1:58.47
2:07.06
2000
48.30
53.83
1:56.76
2:08.16
2008
47.21
53.12
1:53.94
2:05.24
26. OLYMPIC GAMES Find a linear regression model for the men’s 200-meter backstroke data given in Table 7 where x is years since 1968 and y is winning time (in seconds). Do the same for the women’s 200-meter backstroke data. (Round regression coefficients to five significant digits.) Do these models indicate that the women will eventually catch up with the men? 27. SUPPLY AND DEMAND Table 8 contains price–supply data and price–demand data for corn. Find a linear regression model for the price–supply data where x is supply (in billions of bushels) and y is price (in dollars). Do the same for the price–demand data. (Round regression coefficients to three significant digits.) Find the price at which supply and demand are equal. (In economics, this price is referred to as the equilibrium price.)
2-1
2
Price ($/bu.)
Supply (Billion bu.)
Price ($/bu.)
Demand (Billion bu.)
2.15
6.29
2.07
9.78
2.29
7.27
2.15
9.35
2.36
7.53
2.22
8.47
2.48
7.93
2.34
8.12
2.47
8.12
2.39
7.76
2.55
8.24
2.47
6.98
Source: www.usda.gov/nass/pubs/histdata.htm
28. SUPPLY AND DEMAND Table 9 contains price–supply data and price–demand data for soybeans. Find a linear regression model for the price–supply data where x is supply (in billions of bushels) and y is price (in dollars). Do the same for the price–demand data. (Round regression coefficients to three significant digits.) Find the equilibrium price for soybeans.
Table 9 Supply and Demand for U.S. Soybeans
Source: www.infoplease.com
CHAPTER
Table 8 Supply and Demand for U.S. Corn
Price ($/bu.)
Supply (Billion bu.)
Price ($/bu.)
Demand (Billion bu.)
5.15
1.55
4.93
2.60
5.79
1.86
5.48
2.40
5.88
1.94
5.71
2.18
6.07
2.08
6.07
2.05
6.15
2.15
6.40
1.95
6.25
2.27
6.66
1.85
Source: www.usda.gov/nass/pubs/histdata.htm
Review
Cartesian Coordinate System
A Cartesian or rectangular coordinate system is formed by the intersection of a horizontal real number line and a vertical real number line at their origins. These lines are called the coordinate axes. The horizontal axis is often referred to as the x axis and the vertical axis as the y axis. These axes divide the plane into four quadrants. Each point in the plane corresponds to its coordinates— an ordered pair (a, b) determined by passing horizontal and vertical lines through the point. The abscissa or x coordinate a is the coordinate of the intersection of the vertical line with the horizontal axis, and the ordinate or y coordinate b is the coordinate of the intersection of the horizontal line with the vertical axis. The point (0, 0) is
called the origin. A solution of an equation in two variables is an ordered pair of real numbers that makes the equation a true statement. The solution set of an equation is the set of all its solutions. The graph of an equation in two variables is the graph of its solution set formed using point-by-point plotting or with the aid of a graphing calculator. The reflection of the point (a, b) through the y axis is the point (0004a, b), through the x axis is the point (a, 0004b), and through the origin is the point (0004a, 0004b). The reflection of a graph is the reflection of each point on the graph. If reflecting a graph through the y axis, x axis, or origin does not change its shape, the graph is said to be symmetric with respect to the y axis, x axis, or origin, respectively. To test an equation for symmetry, determine if the equation is unchanged when y is replaced with 0004y (x axis symmetry), x is replaced
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with 0004x ( y axis symmetry), or both x and y are replaced with 0004x and 0004y (origin symmetry).
The slope is not defined for a vertical line where x1 0003 x2. Two lines with slopes m1 and m2 are parallel if and only if m1 0003 m2 and perpendicular if and only if m1m2 0003 00041.
2-2
Equations of a Line Standard form Ax 0005 By 0003 C
Distance in the Plane
The distance between the two points P1 0003 (x1, y1) and P2 0003 (x2, y2) is d(P1, P2) 0003 2(x2 0004 x1)2 0005 ( y2 0004 y1)2 and the midpoint of the line segment joining P1 0003 (x1, y1) and P2 0003 (x2, y2) is M0003a
x1 0005 x2 y1 0005 y2 , b 2 2
The standard form for the equation of a circle with radius r and center at (h, k) is (x 0004 h)2 0005 ( y 0004 k)2 0003 r2,
2-3
r 7 0
Equations of a Line
The standard form for the equation of a line is Ax 0005 By 0003 C, where A, B, and C are constants, A and B not both 0. The y intercept is the y coordinate of the point where the graph crosses the y axis, and the x intercept is the x coordinate of the point where the graph crosses the x axis. The slope of the line through the points (x1, y1) and (x2, y2) is m0003
CHAPTER
y2 0004 y1 x2 0004 x1
2
if x1 0002 x2
Slope–intercept form y 0003 mx 0005 b
A and B not both 0 Slope: m; y intercept: b
Point–slope form
y 0004 y1 0003 m(x 0004 x1) Slope: m; Point: (x1, y1)
Horizontal line
y0003b
Slope: 0
Vertical line
x0003a
Slope: Undefined
2-4
Linear Equations and Models
A mathematical model is a mathematics problem that, when solved, will provide information about a real-world problem. If y 0003 mx 0005 b, then the variables x and y are linearly related and the rate of change of y with respect to x is the constant m. If x and y are not linearly related, the ratio ( y2 0004 y1)兾(x2 0004 x1) is called the average rate of change of y with respect to x. Regression analysis produces an equation whose graph is a curve that fits (approximates) a set of data points. A scatter plot is the graph of the points in a data set. Linear regression produces a regression line that is the best fit for a given data set. Graphing calculators or other electronic devices are frequently used to find regression lines.
Review Exercises
Work through all the problems in this chapter review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text.
y 5
00045
5
x
1. Plot A 0003 (00044, 1), B 0003 (2, 00043), and C 0003 (00041, 00042) in a rectangular coordinate system. 2. Refer to Problem 1. Plot the reflection of A through the x axis, the reflection of B through the y axis, and the reflection of C through the origin. 3. Test each equation for symmetry with respect to the x axis, y axis, and origin and sketch its graph. (A) y 0003 2x (B) y 0003 2x 0004 1 (C) y 0003 2|x| (D) | y| 0003 2x 4. Use the following graph to estimate to the nearest integer the missing coordinates of the indicated points. (Be sure you find all possible answers.) (A) (0, ?) (B) (?, 0) (C) (?, 4)
00045
5. Given the points A 0003 (00042, 3) and B 0003 (4, 0), find: (A) Distance between A and B (B) Slope of the line through A and B (C) Slope of a line perpendicular to the line through A and B 6. Write the equation of a circle with radius 17 and center: (A) (0, 0) (B) (3, 00042) 7. Find the center and radius of the circle given by (x 0005 3)2 0005 ( y 0004 2)2 0003 5
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8. Let M be the midpoint of A and B, where A 0003 (a1, a2), B 0003 (2, 00045), and M 0003 (00044, 3). (A) Use the fact that 00044 is the average of a1 and 2 to find a1. (B) Use the fact that 3 is the average of a2 and 00045 to find a2. (C) Find d(A, M ) and d(M, B). 9. (A) Graph the triangle with vertices A 0003 (00041, 00042), B 0003 (4, 3), and C 0003 (1, 4). (B) Find the perimeter to two decimal places. (C) Use the Pythagorean theorem to determine if the triangle is a right triangle. (D) Find the midpoint of each side of the triangle. 10. Use the graph of the linear function in the figure to find the rise, run, and slope. Write the equation of the line in the form Ax 0005 By 0003 C, where A, B, and C are integers with A 0. (The horizontal and vertical line segments have integer lengths.) y
159
21. Write the slope–intercept form of the equation of the line that passes through the point (00042, 1) and is (A) parallel to the line 6x 0005 3y 0003 5 (B) perpendicular to the line 6x 0005 3y 0003 5 22. Find the equation of a circle that passes through the point (00041, 4) with center at (3, 0). 23. Find the center and radius of the circle given by x2 0005 y2 0005 4x 0004 6y 0003 3 24. Find the equation of the set of points equidistant from (3, 3) and (6, 0). What is the name of the geometric figure formed by this set? 25. Are the graphs of mx 0004 y 0003 b and x 0005 my 0003 b parallel, perpendicular, or neither? Justify your answer. 26. Use completing the square to find the center and radius of the circle with equation: x2 0004 4x 0005 y2 0004 2y 0004 3 0003 0
5
00045
5
x
27. Refer to Problem 26. Find the equation of the line tangent to the circle at the point (4, 3). Graph the circle and the line on the same coordinate system. 28. Find the equation of a circle with center (4, 00043) whose graph passes through the point (1, 2).
00045
11. Graph 3x 0005 2y 0003 9 and indicate its slope. 12. Write an equation of a line with x intercept 6 and y intercept 4. Write the final answer in the standard form Ax 0005 By 0003 C, where A, B, and C are integers.
29. Extend the following graph to one that exhibits the indicated symmetry: (A) x axis only (B) y axis only (C) origin only (D) x axis, y axis, and origin y 5
13. Write the slope–intercept form of the equation of the line with slope 000423 and y intercept 2. 14. Write the equations of the vertical and horizontal lines passing through the point (00043, 4). What is the slope of each?
00045
5
00045
Test each equation in Problems 15–18 for symmetry with respect to the x axis, y axis, and the origin. Sketch the graph of the equation. 15. y 0003 x2 0004 2
16. y2 0003 x 0004 2
17. 9y2 0005 4x2 0003 36
18. 9y2 0004 4x2 0003 36
x
Problems 30 and 31 refer to a triangle with base b and height h (see the figure). Write a mathematical expression in terms of b and h for each of the verbal statements in Problems 30 and 31.
19. Write a verbal description of the graph shown in the figure and then write an equation that would produce the graph.
h
y
b
30. The base is five times the height.
5
31. The height is one-fourth of the base. 00045
5
x
00045
20. (A) Find an equation of the line through P 0003 (00044, 3) and Q 0003 (0, 00043). Write the final answer in the standard form Ax 0005 By 0003 C, where A, B, and C are integers with A 0. (B) Find d(P, Q).
APPLICATIONS 32. LINEAR DEPRECIATION A computer system was purchased by a small company for $12,000 and is assumed to have a depreciated value of $2,000 after 8 years. If the value is depreciated linearly from $12,000 to $2,000: (A) Find the linear equation that relates value V (in dollars) to time t (in years). (B) What would be the depreciated value of the system after 5 years?
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33. COST ANALYSIS A video production company is planning to produce an instructional CD. The producer estimates that it will cost $24,900 to produce the CD and $5 per unit to copy and distribute the CD. The budget for this project is $62,000. How many CDs can be produced without exceeding the budget? 34. FORESTRY Forest rangers estimate the height of a tree by measuring the tree’s diameter at breast height (DBH) and then using a model constructed for a particular species. A model for sugar maples is h 0003 2.9d 0005 30.2 where d is the DBH in inches and h is the tree height in feet. (A) Interpret the slope of this model. (B) What is the effect of a 1-inch increase in DBH? (C) How tall is a sugar maple with a DBH of 3 inches? Round answer to the nearest foot. (D) What is the DBH of a sugar maple that is 45 feet tall? Round answer to the nearest inch. 35. ESTIMATING BODY SURFACE AREA An important criterion for determining drug dosage for children is the patient’s body surface area (BSA). John D. Current published the following useful model for estimating BSA*: BSA 0003 1,321 0005 0.3433 Wt where BSA is given in square centimeters and Wt in grams. (A) Interpret the slope of this model. (B) What is the effect of a 100-gram increase in weight? (C) What is the BSA for a child that weighs 15 kilograms?
*“Body Surface Area in Infants and Children,” The Internet Journal of Anesthesiology, 1998, Volume 2, Number 2.
CHAPTER
ZZZ
36. ARCHITECTURE A circular arc forms the top of an entryway with 6-foot vertical sides 8 feet apart. If the top of the arc is 2 feet above the ends, what is the radius of the arc? 37. SPORTS MEDICINE The following quotation was found in a sports medicine handout: “The idea is to raise and sustain your heart rate to 70% of its maximum safe rate for your age. One way to determine this is to subtract your age from 220 and multiply by 0.7.” (A) If H is the maximum safe sustained heart rate (in beats per minute) for a person of age A (in years), write a formula relating H and A. (B) What is the maximum safe sustained heart rate for a 20-year-old? (C) If the maximum safe sustained heart rate for a person is 126 beats per minute, how old is the person? 38. DATA ANALYSIS Winning times in the men’s Olympic 400-meter freestyle event in minutes for selected years are given in Table 1. A mathematical model for these data is y 0003 00040.021x 0005 5.57 where x is years since 1900. (A) Compare the model and the data graphically and numerically. (B) Estimate (to three decimal places) the winning time in 2024.
Table 1 Year
Time
1912
5.41
1932
4.81
1952
4.51
1972
4.00
1992
3.75
2
GROUP ACTIVITY Average Speed
If you score 40 on the first exam and 80 on the second, then your average score for the two exams is (40 0005 80) 2 0003 60. The number 60 is the arithmetic average of 40 and 80. On the other hand, if you drive 100 miles at a speed of 40 mph, and then drive an additional 100 miles at 80 mph, your average speed for the entire trip is not 60 mph. Average speed is defined to be the constant speed at which you could drive the same distance in the same length of time. So to calculate average speed, total distance (200 miles) must be divided by total time: The time t1 it takes to drive 100 miles at 40 mph is t1 0003 (100 miles) (40 mph) 0003 2.5 hours. Similarly, the time t2 it takes to drive 100 miles at 80 mph is t2 0003 (100 miles) (80 mph) 0003 1.25 hours. Therefore, your average speed is 200 200 200 miles 0003 0003 0003 53.3 mph t1 0005 t2 2.5 0005 1.25 3.75
(A) You bicycle 15 miles at 21 mph, then 20 miles at 18 mph, and finally 30 miles at 12 mph. Find the average speed. (B) You bicycle for 2 hours at 18 mph, then 2 more hours at 12 mph. Find the average speed. (C) You run a 10-mile race by running at a pace of 8 minutes per mile for 1 hour, and after that at a pace of 9 minutes per mile. Define average pace, find it (to the nearest second) for the 10-mile race, and discuss the connection between average pace (in minutes per mile) and average speed (in miles per hour).
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CHAPTER
Functions
3
C
OUTLINE
THE function concept is one of the most important ideas in mathe-
matics. To study math beyond the elementary level, you absolutely need to have a solid understanding of functions and their graphs. In this chapter, you’ll learn the fundamentals of what functions are all about, and how to apply them. As you work through this and subsequent chapters, this will pay off as you study specific types of functions in depth. Everything you learn in this chapter will increase your chance of success in this course, and in almost any other course you may take that involves mathematics.
3-1
Functions
3-2
Graphing Functions
3-3
Transformations of Functions
3-4
Quadratic Functions
3-5
Operations on Functions; Composition
3-6
Inverse Functions Chapter 3 Review Chapter 3 Group Activity: Mathematical Modeling: Choosing a Cell Phone Plan
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FUNCTIONS
Functions Z Definition of Function Z Defining Functions by Equations Z Using Function Notation Z Application
The idea of correspondence plays a really important role in understanding the concept of functions, which is easily one of the most important ideas in this book. The good news is that you have already had years of experience with correspondences in everyday life. For example, For For For For For
every every every every every
person, there is a corresponding age. item in a store, there is a corresponding price. football season, there is a corresponding Super Bowl champion. circle, there is a corresponding area. number, there is a corresponding cube.
One of the most basic and important ways that math can be applied to other areas of study is the establishment of correspondence among various types of phenomena. In many cases, once a correspondence is known, it can be used to make important decisions and predictions. An engineer can use a formula to predict the weight capacity of a stadium grandstand. A political operative decides how many resources to allocate to a race given current polling results. A computer scientist can use formulas to compare the efficiency of algorithms for sorting data stored on a computer. An economist would like to be able to predict interest rates, given the rate of change of the money supply. And the list goes on and on.
Z Definition of a Function What do all of the preceding examples have in common? Each describes the matching of elements from one set with elements from a second set. Consider the correspondences in Tables 1 and 2. Table 1 Top Four Weekly Average Primetime Network Viewers for the 2007–2008 Season
Table 2 Top Four Best Selling Automobiles in the United States for 2008
Network
Manufacturer
Model
Viewers (Millions)
Fox
10.9
Toyota
Camry
CBS
10.1
Honda
Accord
ABC
8.9
Toyota
Corolla
NBC
7.8
Honda
Civic
Source: tvbythenumbers.com
Source: www.2-speed.com
Table 1 specifies a function, but Table 2 does not. Why not? The definition of function will explain.
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SECTION 3–1
Functions
163
Z DEFINITION 1 Definition of Function A function is a correspondence between two sets of elements such that to each element in the first set there corresponds one and only one element in the second set. The first set is called the domain and the set of all corresponding elements in the second set is called the range.
Table 1 specifies a function with domain {Fox, CBS, ABC, NBC} and range {10.9, 10.1, 8.9, 7.8} because every network in the first set corresponds with exactly one number in the second set. Table 2 does not specify a function, because each manufacturer in the first set corresponds to two different models in the second set. Functions can also be specified by using ordered pairs of elements, where the first component represents an element from the domain, and the second component represents the corresponding element from the range. The function in Table 1 can be written as F 0002 {(Fox, 10.9), (CBS, 10.1), (ABC, 8.9), (NBC, 7.8)} Notice that no two ordered pairs have the same first component and different second component. On the other hand, if we list the set H of ordered pairs determined by Table 2, we get H 0002 {(Toyota, Camry), (Honda, Accord), (Toyota, Corolla), (Honda, Civic)} In this case, there are ordered pairs with the same first component but different second components. This means that H does not specify a function. This ordered pair approach leads to a second (but equivalent) way to define a function.
Z DEFINITION 2 Set Form of the Definition of Function A function is a set of ordered pairs with the property that no two ordered pairs have the same first component and different second components. The set of all first components in a function is called the domain of the function, and the set of all second components is called the range.
EXAMPLE
1
Functions Specified as Sets of Ordered Pairs Determine whether each set specifies a function. If it does, then state the domain and range. (A) S 0002 5(1, 4), (2, 3), (3, 2), (4, 3), (5, 4)6 (B) T 0002 5(1, 4), (2, 3), (3, 2), (2, 4), (1, 5)6
SOLUTIONS
(A) Because all the ordered pairs in S have distinct first components, this set specifies a function. The domain and range are Domain 0002 51, 2, 3, 4, 56 Range 0002 52, 3, 46
Set of first components Set of second components written with no repeats
(B) Because there are ordered pairs in T with the same first component [for example, (1, 4) and (1, 5)], this set does not specify a function.
0002
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FUNCTIONS
Determine whether each set defines a function. If it does, then state the domain and range. (A) S 0002 5(00042, 1), (00041, 2), (0, 0), (00041, 1), (00042, 2)6 (B) T 0002 5(00042, 1), (00041, 2), (0, 0), (1, 2), (2, 1)6
0002
Z Defining Functions by Equations So far, we have described a particular function in various ways: (1) by a verbal description, (2) by a table, and (3) by a set of ordered pairs. We will see that if the domain and range are sets of numbers, we can also define a function by an equation, or by a graph. If the domain of a function is a large or infinite set, it may be impractical or impossible to actually list all of the ordered pairs that belong to the function, or to display the function in a table. Such a function can often be defined by a verbal description of the “rule of correspondence” that clearly specifies the element of the range that corresponds to each element of the domain. One example is “to each real number corresponds its square.” When the domain and range are sets of numbers, the algebraic and graphical analogs of the verbal description are the equation and graph, respectively. We will find it valuable to be able to view a particular function from multiple perspectives—algebraic (in terms of an equation), graphical (in terms of a graph), and numeric (in terms of a table or ordered pairs). Both versions of our definition of function are very general. The objects in the domain and range can be pretty much anything, and there is no restriction on the number of elements in each. In this text, we are primarily interested, however, in functions with real number domains and ranges. Unless otherwise indicated, the domain and range of a function will be sets of real numbers. For such a function we often use an equation with two variables to specify both the rule of correspondence and the set of ordered pairs. Consider the equation y 0002 x2 0003 2x
x any real number
(1)
This equation assigns to each domain value x exactly one range value y. For example, If x 0002 4, If x 0002 000413,
then then
y 0002 (4)2 0003 2(4) 0002 24 y 0002 (000413)2 0003 2(000413) 0002 000459
We can view equation (1) as a function with rule of correspondence y 0002 x2 0003 2x
any x corresponds to x 2 ⴙ 2x
The variable x is called an independent variable, indicating that values can be assigned “independently” to x from the domain. The variable y is called a dependent variable, indicating that the value of y “depends” on the value assigned to x and on the given equation. In general, any variable used as a placeholder for domain values is called an independent variable; any variable used as a placeholder for range values is called a dependent variable. We often refer to a value of the independent variable as the input of the function, and the corresponding value of the dependent variable as the associated output. In this regard, a function can be thought of as a process that accepts an input from the domain and outputs an appropriate range element. We next address the question of which equations can be used to define functions.
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SECTION 3–1
Functions
165
Z FUNCTIONS DEFINED BY EQUATIONS In an equation with two variables, if to each value of the independent variable there corresponds exactly one value of the dependent variable, then the equation defines a function. If there is any value of the independent variable to which there corresponds more than one value of the dependent variable, then the equation does not define a function.
Since an equation is just one way to represent a function, we will say “an equation defines a function” rather than “an equation is a function.”
EXAMPLE
2
Determining if an Equation Defines a Function Determine if each equation defines a function with independent variable x. (A) y ⫽ x2 ⫺ 4
SOLUTIONS
(B) x2 ⫹ y2 ⫽ 16
(A) For any real number x, the square of x is a unique real number. When you subtract 4, the result is again unique. So for any input x, there is exactly one output y, and the equation defines a function. (B) In this case, it will be helpful to solve the equation for the dependent variable. x2 ⫹ y2 ⫽ 16 y2 ⫽ 16 ⫺ x2 y ⫽ ⫾ 216 ⫺ x2
Subtract x2 from both sides. Take the square root of both sides.
For any x that provides an output (when 16 ⫺ x2 ⱖ 0), there are two choices for y, one positive and one negative. The equation has more than one output for some inputs, so does not define a function. 0002 MATCHED PROBLEM 2
Determine if each equation defines a function with independent variable x. (A) y2 ⫹ x4 ⫽ 4
(B) y3 ⫺ x3 ⫽ 3
0002
It is very easy to determine whether an equation defines a function if you have the graph of the equation. The two equations we considered in Example 2 are graphed next in Figure 1. y
Z Figure 1 Graphs of equations and the vertical line test.
y
5
5
y ⫽ x2 ⫺ 4
(2, 2兹3) x 2 ⫹ y 2 ⫽ 16
⫺5
5
x
⫺5
5
(1, ⫺3) ⫺5
(2, ⫺2兹3) ⫺5
(a)
x
(b)
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In Figure 1(a), any vertical line will intersect the graph of y 0002 x2 0004 4 exactly once. This shows that every value of the independent variable x corresponds to exactly one value of the dependent variable y, and confirms our conclusion that y 0002 x2 0004 4 defines a function. But in Figure 1(b), there are many vertical lines that intersect the graph of x2 0003 y2 0002 16 in two points. This shows that there are values of the independent variable x that correspond to two different values of the dependent variable y, which confirms our conclusion that x2 0003 y2 0002 16 does not define a function. These observations lead to Theorem 1.
Z THEOREM 1 Vertical Line Test for a Function An equation defines a function if each vertical line in a rectangular coordinate system passes through at most one point on the graph of the equation. If any vertical line passes through two or more points on the graph of an equation, then the equation does not define a function.
ZZZ EXPLORE-DISCUSS 1
The definition of a function specifies that to each element in the domain there corresponds one and only one element in the range. (A) Give an example of a function such that to each element of the range there correspond exactly two elements of the domain. (B) Give an example of a function such that to each element of the range there corresponds exactly one element of the domain.
Sometimes when a function is defined by an equation, a domain is specified, as in f (x) 0002 2x2 0003 5, x 7 0 The “x 7 0” tells us that the domain is all positive real numbers. More often, a function is defined by an equation with no domain specified. Unless a domain is specified, we will use the following convention regarding domains and ranges for functions defined by equations.
Z AGREEMENT ON DOMAINS AND RANGES If a function is defined by an equation and the domain is not stated explicitly, then we assume that the implied domain is the set of all real number replacements of the independent variable that produce real values for the dependent variable. The range is the set of all values of the dependent variable corresponding to the domain values.
EXAMPLE
3
Finding the Domain of a Function Find the domain of the function defined by the equation y 0002 1x 0004 3, assuming x is the independent variable.
SOLUTION
For y to be real, x 0004 3 must be greater than or equal to 0. That is, x0004300050
The domain is 5x ƒ x 0005 36, or [3, 0007).
or
x00053 0002
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MATCHED PROBLEM 3
Functions
167
Find the domain of the function defined by the equation y 0002 1x 0003 5, assuming x is the independent variable. 0002
Z Using Function Notation We will use letters to name functions and to provide a very important and convenient notation for defining functions. For example, if f is the name of the function defined by the equation y 0002 2x 0003 1, we could use the formal representations f : y 0002 2x 0003 1
Rule of correspondence
or f :5(x, y) | y 0002 2x 0003 16
Set of ordered pairs
But instead, we will simply write f (x) 0002 2x 0003 1
Function notation
The symbol f (x) is read “f of x,” “f at x,” or “the value of f at x” and represents the number in the range of the function f (the output) that is paired with the domain value x (the input).
ZZZ
CAUTION ZZZ
The symbol “f (x)” should never be read as “f times x.” The notation does not represent a product. It tells us that the function named f has independent variable x. f (x) is the value of the function f at x. 2(x) 0002 2x is algebraic multiplication.
Using function notation, f (3) is the output for the function f associated with the input 3. We find this range value by replacing x with 3 wherever x occurs in the function definition f(x) 0002 2x 0003 1
f x
f (x)
and evaluating the right side, f (3) 0002 2 ⴢ 3 0003 1 0002 6 0003 1 0002 7
DOMAIN
RANGE
The function f “maps” the domain value x into the range value f (x).
Z Figure 2 Function notation.
The statement f(3) 0002 7 indicates in a concise way that the function f assigns the range value 7 to the domain value 3 or, equivalently, that the ordered pair (3, 7) belongs to f. The symbol f : x S f(x), read “f maps x into f (x),” is also used to denote the relationship between the domain value x and the range value f (x) (Fig. 2). Letters other than f and x can be used to represent functions and independent variables. For example, g(t) 0002 t 2 0004 3t 0003 7 defines g as a function of the independent variable t. To find g(00042), we replace t by 00042 wherever t occurs in the equation g(t) 0002 t 2 0004 3t 0003 7 and evaluate the right side: g(ⴚ2) 0002 (ⴚ2)2 0004 3(ⴚ2) 0003 7 000240003600037 0002 17 The function g assigns the range value 17 (output) to the domain value 00042 (input); the ordered pair (00042, 17) belongs to g.
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It is important to understand and remember the definition of the symbol f(x): Z DEFINITION 3 The Symbol f(x) The symbol f(x), read “f of x,” represents the real number in the range of the function f corresponding to the domain value x. The symbol f (x) is also called the value of the function f at x. The ordered pair (x, f (x)) belongs to the function f. If x is a real number that is not in the domain of f, then f is undefined at x and f (x) does not exist.
EXAMPLE
4
Evaluating Functions (A) Find f(6), f(a), and f(6 0003 a) for f (x) 0002
15 . x00043
(B) Find g(7), g(h), and g(7 0003 h) for g(x) 0002 16 0003 3x 0004 x2. (C) Find k(9), 4k(a), and k(4a) for k(x) 0002
SOLUTIONS
(A)
f (6)
0002
15 600043
* 0002
15 00025 3
Substitute 6 for x.
15 a00043 15 15 f (6 ⴙ a) 0002 0002 (6 ⴙ a) 0004 3 30003a f (a) 0002
(B)
0002 16 0003 3(7) 0004 (7)2
g(7)
2 . 1x 0004 2
Substitute a for x.
Substitute (6 ⫹ a) for x and simplify.
0002 16 0003 21 0004 49 0002 000412
g(h) 0002 16 0003 3h 0004 h2 g(7 ⴙ h) 0002 16 0003 3(7 ⴙ h) 0004 (7 ⴙ h)2 0002 16 0003 21 0003 3h 0004 (49 0003 14h 0003 h2) 0002 37 0003 3h 0004 49 0004 14h 0004 h2 0002 000412 0004 11h 0004 h2 (C) k (9)
0002
2 19 0004 2
0002
2 00022 300042
2 8 0002 1a 0004 2 1a 0004 2 2 k(4a) 0002 14a 0004 2 2 0002 21a 0004 2 1 0002 1a 0004 1
Multiply out the first set of parentheses and square (7 ⫹ h). Combine like terms and distribute the negative through the parentheses. Combine like terms.
19 ⴝ 3, not ⴞ3.
4k(a) 0002 4
14a ⴝ 141a ⴝ 2 1a.
Divide numerator and denominator by 2.
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
0002
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MATCHED PROBLEM 4
Functions
169
4 . 20004x (B) Find G(3), G(h), and G(3 0003 h) for G(x) 0002 x2 0003 5x 0004 2. 6 (C) Find K(4), K(9x), and 9K(x) for K(x) 0002 . 3 0004 1x (A) Find F(4), F(4 0003 h), and F(4) 0003 F(h) for F(x) 0002
0002
EXAMPLE
5
Finding Domains of Functions Find the domain of each of the following functions. Express the answer in both set notation and inequality notation.* (A) f (x) 0002
SOLUTIONS
15 x00043
(B) g(x) 0002 16 0003 3x 0004 x2
(C) k(x) 0002
2 1x 0004 2
(A) The rational expression 15兾(x 0004 3) represents a real number for all replacements of x by real numbers except x 0002 3, since division by 0 is not defined. So f(3) does not exist, and the domain of f is 5x ƒ x 36
or
R
or
(00040007, 3) 傼 (3, 0007)
(B) Since 16 0003 3x 0004 x2 represents a real number for all replacements of x by real numbers, the domain of g is (00040007, 0007)
(C) Since 1x is not a real number for negative real numbers x, x must be a nonnegative real number. Because division by 0 is not defined, we must exclude any values of x that make the denominator 0. Set the denominator equal to zero and solve: 2 0004 1x 0002 0 2 0002 1x 40002x
Add 1x to both sides. Square both sides.
The domain of f is all nonnegative real numbers except 4. This can be written as 5x ƒ x 0005 0, x 46
MATCHED PROBLEM 5
[0, 4) 傼 (4, 0007)
0002
Find the domain of each of the following functions. Express the answer in both set notation and inequality notation. (A) F(x) 0002
ZZZ EXPLORE-DISCUSS 2
or
4 20004x
(B) G(x) 0002 x2 0003 5x 0004 2
(C) K(x) 0002
6 3 0004 1x
Let x and h be real numbers. (A) If f(x) 0002 4x 0003 3, which of the following is true: (1) f (x 0003 h) 0002 4x 0003 3 0003 h (2) f(x 0003 h) 0002 4x 0003 4h 0003 3 (3) f (x 0003 h) 0002 4x 0003 4h 0003 6 (B) If g(x) 0002 x2, which of the following is true: (1) g(x 0003 h) 0002 x2 0003 h (2) g(x 0003 h) 0002 x2 0003 h2 (3) g(x 0003 h) 0002 x2 0003 2hx 0003 h2 (C) If M(x) 0002 x2 0003 4x 0003 3, describe the operations that must be performed to evaluate M(x 0003 h). *A review of Table 1 in Section 1-2 might prove to be helpful at this point.
0002
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In addition to evaluating functions at specific numbers, it is useful to be able to evaluate functions at expressions that involve one or more variables. For example, the difference quotient f (x 0003 h) 0004 f (x) h
x and x 0003 h in the domain of f, h 0
is very important in calculus courses.
EXAMPLE
6
Evaluating and Simplifying a Difference Quotient For f(x) 0002 x2 0003 4x 0003 5, find and simplify: (A) f(x 0003 h)
SOLUTIONS
(B) f(x 0003 h) 0004 f(x)
(C)
f (x 0003 h) 0004 f (x) ,h0 h
(A) To find f(x 0003 h), we replace x with x 0003 h everywhere it appears in the equation that defines f and simplify: f (x ⴙ h) 0002 (x ⴙ h)2 0003 4(x ⴙ h) 0003 5 0002 x2 0003 2xh 0003 h2 0003 4x 0003 4h 0003 5 (B) Using the result of part A, we get f (x ⴙ h) 0004 f (x) 0002 x2 ⴙ 2xh ⴙ h2 ⴙ 4x ⴙ 4h ⴙ 5 0004 (x2 ⴙ 4x ⴙ 5) 0002 x2 0003 2xh 0003 h2 0003 4x 0003 4h 0003 5 0004 x2 0004 4x 0004 5 0002 2xh 0003 h2 0003 4h (C)
f (x 0003 h) 0004 f (x) 2xh 0003 h2 0003 4h 0002 h h
0002
h(2x 0003 h 0003 4) h
Divide numerator and denominator by h ⴝ 0.
0002 2x 0003 h 0003 4 MATCHED PROBLEM 6
ZZZ
Repeat Example 6 for f (x) 0002 x2 0003 3x 0003 7.
1. Remember, f(x 0003 h) is not a multiplication! 2. In general, f(x 0003 h) is not equal to f(x) 0003 f(h), nor is it equal to f(x) 0003 h.
CAUTION ZZZ
Z Application EXAMPLE
7
Construction A rectangular feeding pen for cattle is to be made with 100 meters of fencing. (A) If x represents the width of the pen, express its area A in terms of x. (B) What is the domain of the function A (determined by the physical restrictions)?
0002
0002
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SOLUTIONS
171
Functions
(A) Draw a figure and label the sides.
x ( Width)
Perimeter ⴝ 100 meters of fencing. Half the perimeter ⴝ 50. If x ⴝ Width, then 50 ⴚ x ⴝ Length.
50 0004 x (Length)
A 0002 (Width)(Length) 0002 x(50 0004 x) (B) To have a pen, x must be positive, but x must also be less than 50 (or the length will not exist). So the domain is 5x ƒ 0 6 x 6 506
(0, 50) MATCHED PROBLEM 7
Inequality notation
0002
Interval notation
Rework Example 7 with the added assumption that a large barn is to be used as one of the sides that run the length of the pen. 0002 ANSWERS TO MATCHED PROBLEMS 1. (A) S does not define a function. (B) T defines a function with domain {00042, 00041, 0, 1, 2} and range {0, 1, 2}. 2. (A) Does not define a function (B) Defines a function 3. 5x ƒ x 0005 000456 or [ 00045, 0007) 4 2h 4. (A) F(4) 0002 00042, F(4 0003 h) 0002 0004 , F(4) 0003 F(h) 0002 20003h 20004h (B) G(3) 0002 22, G(h) 0002 h2 0003 5h 0004 2, G(3 0003 h) 0002 22 0003 11h 0003 h2 2 54 (C) K(4) 0002 6, K(9x) 0002 , 9K(x) 0002 1 0004 1x 3 0004 1x 5. (A) 5x ƒ x 26 or (00040007, 2) ´ (2, 0007) (B) R or (00040007, 0007) (C) 5x ƒ x 0005 0, x 96 or [0, 9) 傼 (9, 0007) 6. (A) x2 0003 2xh 0003 h2 0003 3x 0003 3h 0003 7 2 (B) 2xh 0003 h 0003 3h (C) 2x 0003 h 0003 3 7. (A) A 0002 x(100 0004 2x) (B) Domain: 5x ƒ 0 6 x 6 506 or (0, 50)
3-1
Exercises
1. Is every correspondence between two sets a function? Why or why not? 2. Describe four different ways that we represented functions in this section. 3. Explain what the domain and range of a function are. Don’t just think about functions defined by equations.
6. Describe how to determine if an equation defines a function by looking at the graph of the equation. Indicate whether each table in Problems 7–12 defines a function. 7. Domain
Range
8. Domain
Range
4. What do the terms “input” and “output” refer to when working with functions?
00041
1
2
1
0
2
4
3
5. If 2(x 0003 h) 0002 2x 0003 2h, why doesn’t f (x 0003 h) 0002 f (x) 0003 f (h), where f is a function?
1
3
6
5
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9. Domain
Range
1
3
00041
0
3
5
00042
5
7
00043
8
5
10. Domain
Range
y
21. 10
000410
10
x
9
11. Domain
Range
12. Domain
000410
Range
English
A
Auburn
Tigers
Math
B
Memphis
Tigers
Sociology
A
Georgia
Bulldogs
Chemistry
B
Fresno State
Bulldogs
y
22. 10
000410
Indicate whether each set in Problems 13–18 defines a function. Find the domain and range of each function. 13. {(2, 4), (3, 6), (4, 8), (5, 10)}
10
x
000410
y
23.
14. {(00041, 4), (0, 3), (1, 2), (2, 1)}
10
15. {(10, 000410), (5, 00045), (0, 0), (5, 5), (10, 10)} 16. {(0, 1), (1, 1), (2, 1), (3, 2), (4, 2), (5, 2)} 000410
17. {(Ohio, Obama), (Alabama, McCain), (West Virginia, McCain), (California, Obama)} 18. {(Democrat, Obama), (Republican, Bush), (Democrat, Clinton), (Republican, Reagan)}
10
000410
y
24. Indicate whether each graph in Problems 19–24 is the graph of a function. 19.
x
10
y 10
000410
000410
10
x
y 10
000410
10
000410
x
000410
In Problems 25 and 26, which of the indicated correspondences define functions? Explain.
000410
20.
10
x
25. Let F be the set of all faculty teaching Math 125 at Enormous State University, and let S be the set of all students taking that course. (A) Students from set S correspond to their Math 125 instructors. (B) Faculty from set F correspond to the students in their Math 125 class. 26. Let A be the set of floor advisors in Hoffmann Hall, a dorm at Enormous State. Assume that each floor has one floor advisor. Let R be the set of residents of that dorm. (A) Floor advisors from set A correspond to the residents on their floor. (B) Students from set R correspond to their floor advisor.
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27. Let f (x) 0002 3x 0004 5. Find (A) f(3) (B) f (h) (C) f(3) 0003 f(h) (D) f (3 0003 h)
173
In Problems 47–62, find the domain of the indicated function. Express answers in both interval notation and inequality notation.
28. Let g(y) 0002 7 0004 2y. Find (A) g(4) (B) g(h) (C) g(4) 0003 g(h) (D) g(4 0003 h) 29. Let F(w) 0002 0004w2 0003 2w. Find (A) F(4) (B) F(00044) (C) F(4 0003 a) (D) F(2 0004 a)
47. f(x) 0002 4 0004 9x 0003 3x2
48. g(t) 0002 1 0003 7t 0004 2t2
49. L(u) 0002 23u2 0003 4
50. M(w) 0002
2 40004z
51. h(z) 0002
30. Let G(t) 0002 5t 0004 t2. Find (A) G(8) (B) G(00048) (C) G(00041 0003 h) (D) G(6 0004 t) 2
31. Let f(t) 0002 2 0004 3t . Find (A) f(00042) (B) f(0004t) (C) 0004f(t) (D) 0004f(0004t) 32. Let k(z) 0002 40 0003 20z2. Find (A) k(00042) (B) k(0004z) (C) 0004k(z) (D) 0004k(0004z)
52. k(z) 0002
w00045 23 0003 2w2
z z00043
53. g(t) 0002 1t 0004 4
54. h(t) 0002 16 0004 t
55. k(w) 0002 17 0003 3w
56. j(w) 0002 19 0003 4w
57. H(u) 0002
u u2 0003 4
58. G(u) 0002
u u2 0004 4
59. M(x) 0002
1x 0003 4 x00041
60. N(x) 0002
1x 0004 3 x00032
61. s(t) 0002
33. Let F(u) 0002 u2 0004 u 0004 1. Find (A) F(10) (B) F(u2) (C) F(5u) (D) 5F(u)
1 3 0004 1t
62. r(t) 0002
1 1t 0004 4
The verbal statement “function f multiplies the square of the domain element by 3 and then subtracts 7 from the result” and the algebraic statement “f(x) 0002 3x2 0004 7” define the same function. In Problems 63–66, translate each verbal definition of a function into an algebraic definition.
34. Let G(u) 0002 4 0004 3u 0004 u2. Find (A) G(00048) (B) G(u2) (C) G(00042u) (D) 00042G(u) Problems 35–36 refer to the following graph of a function f.
63. Function g subtracts 5 from twice the cube of the domain element. 64. Function f multiplies the square of the domain element by 10 then adds 1,000 to the result.
f (x) y 0002 f (x)
65. Function F multiplies the square root of the domain element by 8, then subtracts the product of 4 and the sum of the domain element and two.
10
000410
Functions
10
x
000410
66. Function G divides the sum of the domain element and 7 by the cube root of the domain element. In Problems 67–70, translate each algebraic definition of the function into a verbal definition.
35. (A) Find f (00042) to the nearest integer. (B) Find all values of x, to the nearest integer, so that f (x) 0002 00044.
67. f(x) 0002 2x2 0003 5
36. (A) Find f(4) to the nearest integer. (B) Find all values of x, to the nearest integer, so that f (x) 0002 0.
69. z(x) 0002
Determine which of the equations in Problems 37–46 define a function with independent variable x. For those that do, find the domain. For those that do not, find a value of x to which there corresponds more than one value of y. 37. y 0004 x2 0002 1
38. y2 0004 x 0002 1
39. 2x3 0003 y2 0002 4
40. 3x2 0003 y3 0002 8
41. x3 0004 y 0002 2
42. x3 0003 冟 y 冟 0002 6
43. 2x 0003 冟 y 冟 0002 7
44. y 0004 2冟 x 冟 0002 3
45. 3y 0003 2|x| 0002 12
46. x| y| 0002 x 0003 1
68. g(x) 0002 00042x 0003 7
4x 0003 5 1x
70. M(t) 0002 5t 0004 21t
71. If F(s) 0002 3s 0003 15, find:
F(2 0003 h) 0004 F(2) h
72. If K(r) 0002 7 0004 4r, find:
K(1 0003 h) 0004 K(1) h
73. If g(x) 0002 2 0004 x2, find:
g(3 0003 h) 0004 g(3) h
74. If P(m) 0002 2m2 0003 3, find:
P(2 0003 h) 0004 P(2) h
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In Problems 75–84, find and simplify: (A)
f (x 0004 h) 0003 f (x) h
(B)
75. f (x) 0002 4x 0003 7
f (x) 0003 f (a) x0003a 76. f (x) 0002 00035x 0004 2
2
77. f(x) 0002 2x 0003 4
78. f(x) 0002 5 0003 3x2
79. f (x) 0002 00034x2 0004 3x 0003 2
80. f (x) 0002 3x2 0003 5x 0003 9
81. f (x) 0002 1x 0004 2
82. f (x) 0002 1x 0003 1
83. f (x) 0002
4 x
84. f (x) 0002
s(2 0004 h) 0003 s(2) . h (C) Evaluate the expression in part (B) for h 0002 00051, 00050.1, 00050.01, 00050.001. (D) What happens in part (C) as h gets closer and closer to 0? Interpret physically. (B) Find and simplify
3 x00042
85. The area of a rectangle is 64 square inches. Express the perimeter P as a function of the width w and state the domain. 86. The perimeter of a rectangle is 50 inches. Express the area A as a function of the width w and state the domain. 87. The altitude of a right triangle is 5 meters. Express the hypotenuse h as a function of the base b and state the domain. 88. The altitude of a right triangle is 4 meters. Express the base b as a function of the hypotenuse h and state the domain.
94. PHYSICS—RATE An automobile starts from rest and travels along a straight and level road. The distance in feet traveled by the automobile is given by s(t) 0002 10t2, where t is time in seconds. (A) Find: s(8), s(9), s(10), and s(11). s(11 0004 h) 0003 s(11) (B) Find and simplify . h (C) Evaluate the expression in part (B) for h 0002 00051, 00050.1, 00050.01, 00050.001. (D) What happens in part (C) as h gets closer and closer to 0? Interpret physically. 95. MANUFACTURING A candy box is to be made out of a piece of cardboard that measures 8 by 12 inches. Squares, x inches on a side, will be cut from each corner, and then the ends and sides will be folded down (see the figure). Find a formula for the volume of the box V in terms of x. What is the domain of the function V that makes sense in this problem?
APPLICATIONS Most of the applications in this section are calculus-related. That is, similar problems will appear in a calculus course, but additional analysis of the functions will be performed. 89. COST FUNCTION The fixed costs per day for a doughnut shop are $300, and the variable costs are $1.75 per dozen doughnuts produced. If x dozen doughnuts are produced daily, express the daily cost C(x) as a function of x. 90. COST FUNCTION A manufacturer of MP3 players has fixed daily costs of 15,700 Chinese yuan, and it costs 178 yuan to produce one MP3 player. If the manufacturer produces x players daily, express the daily cost C in yuan as a function of x. 91. CELL PHONE COST Since Don usually borrows his roommate’s cell phone for long-distance calls, he chooses an inexpensive plan for his own phone with a monthly access charge, and a variable charge for each hour of calls used. The function C(h) 0002 17 0004 2.40h is used to calculate Don’s monthly bill, where C is the cost in dollars and h is hours of airtime used. Translate this equation into a verbal statement that you could use to explain Don’s monthly charge. 92. COST OF HIGH SPEED INTERNET A college offers highspeed Internet in dorm rooms. The monthly access fee in dollars is calculated using the function
x CITR
US D
x
ELIG
HTS
x
x
CITRUS DELIGHTS
CITRUS DELIGHTS
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x
x
x x
96. CONSTRUCTION A rancher has 20 miles of fencing to fence a rectangular piece of grazing land along a straight river. If no fence is required along the river and the sides perpendicular to the river are x miles long, find a formula for the area A of the rectangle in terms of x. What is the domain of the function A that makes sense in this problem? 97. CONSTRUCTION The manager of an animal clinic wants to construct a kennel with four identical pens, as indicated in the figure. State law requires that each pen have a gate 3 feet wide and an area of 50 square feet. If x is the width of one pen, express the total amount of fencing F (excluding the gates) required for the construction of the kennel as a function of x. Complete the following table (round values of F to one decimal place): 4
x
5
6
7
F x
A(m) 0002 15 0004 0.02m where m is the number of minutes spent online. Translate this equation into a verbal statement that can be used to explain the monthly charges to an incoming freshman. 93. PHYSICS—RATE The distance in feet that an object falls (ignoring air resistance) is given by s(t) 0002 16t2, where t is time in seconds. (A) Find: s(0), s(1), s(2), and s(3).
3 feet
98. ARCHITECTURE An architect wants to design a window with an area of 24 square feet in the shape of a rectangle with a
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semicircle on top, as indicated in the figure. If x is the width of the window, express the perimeter P of the window as a function of x. Complete the following table (round each value of P to one decimal place): 5
6
Island Lake 8 miles
4
x
175
Graphing Functions
7
Pipe
Freshwater source
Land
P
x
20 0004 x 20 miles
Figure for 99
100. WEATHER An observation balloon is released at a point 10 miles from the station that receives its signal and rises vertically, as indicated in the figure. Express the distance d between the balloon and the receiving station as a function of the altitude h of the balloon.
x
99. CONSTRUCTION A freshwater pipeline is to be run from a source on the edge of a lake to a small resort community on an island 8 miles offshore, as indicated in the figure. It costs $10,000 per mile to lay the pipe on land and $15,000 per mile to lay the pipe in the lake. Express the total cost C of constructing the pipeline as a function of x. From practical considerations, what is the domain of the function C ?
d
h
10 miles Figure for 100
3-2
Graphing Functions Z Basic Concepts Z Linear Functions Z Piecewise-Defined Functions
One of the ways we represented functions in Section 3-1 was with sets of ordered pairs. If these ordered pairs reminded you of points on a graph, you already understand the most important idea in this section—that graphs are a natural fit for functions because a graph matches up a pair of numbers in exactly the same way a function matches up a pair of objects. y or f (x) y intercept
(x, y) or (x, f (x))
Z Basic Concepts
f
y or f (x) x x intercept
Z Figure 1 Graph of a function.
When we graph a function whose domain and range are both sets of numbers, we are drawing a visual representation of the pairs of numbers matched up by that function. We will associate domain values with the horizontal axis, and range values with the vertical axis. The graph of a function f (x) is the set of all points whose first coordinate is an element of the domain of f, and whose second coordinate is the associated element of the range. We can use the symbol y or f (x) to represent the dependent variable. See Figure 1. Since it is
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typical to use the variables x and y for the independent and dependent variables, respectively, we usually refer to the first coordinate of a point as the x coordinate, and the second coordinate as the y coordinate. The x coordinate of a point where the graph of a function intersects the x axis is called an x intercept or zero of the function. An x intercept is also a real solution or root of the equation f (x) 0002 0. The y coordinate of a point where the graph of a function crosses the y axis is called the y intercept of the function. The y intercept is given by f (0), provided 0 is in the domain of f. Note that a function can have more than one x intercept but can never have more than one y intercept—a consequence of the vertical line test from Section 3-1.
EXAMPLE
1
Finding the Domain and Intercepts of a Function Find the domain, x intercept, and y intercept of f (x) 0002
SOLUTION
4 0004 3x . 2x 0003 5
The rational expression (4 0004 3x)兾(2x 0003 5) is defined for every x except those that make the denominator zero: 2x 0003 5 0002 0 2x 0002 00045 x 0002 000452
Subtract 5 from both sides. Divide both sides by 2.
The domain of f is all x values except 000452, or (00040007, 000452 ) 傼 (000452, 0007). The value of a fraction is 0 if and only if the numerator is zero: 4 0004 3x 0002 0 00043x 0002 00044 x 0002 43
Subtract 4 from both sides. Divide both sides by ⴚ3.
The x intercept of f is 43. The y intercept is f (0) 0002
MATCHED PROBLEM 1
4 0004 3(0) 2(0) 0003 5
4 0002 . 5
Find the domain, x intercept, and y intercept of f (x) 0002
0002 4x 0003 5 . 3x 0004 2 0002
The domain of a function is the set of all the x coordinates of points on the graph of the function and the range is the set of all the y coordinates. It is very useful to view the domain and range as subsets of the coordinate axes as in Figure 2 on the next page. Note the effective use of interval notation in describing the domain and range of the functions in this figure. In Figure 2(a) a solid dot is used to indicate that a point is on the graph of the function and in Figure 2(b) an open dot is used to indicate that a point is not on the graph of the function. An open or solid dot at the end of a graph indicates that the graph terminates there, whereas an arrowhead indicates that the graph continues indefinitely beyond the portion shown with no significant changes of direction [see Fig. 2(b) and note that the arrowhead indicates that the domain extends infinitely far to the right, and the range extends infinitely far downward].
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]
(
d
a
[
]
x
b
[
177
f (x)
f (x) d
Graphing Functions
(
x
a
c Domain f 0002 (a, 0007) Range f 0002 (00040007, d )
Domain f 0002 [a, b] Range f 0002 [c, d ] (a)
(b)
Z Figure 2 Domain and range.
EXAMPLE
2
Finding the Domain and Range from a Graph (A) Find the domain and range of the function f whose graph is shown in Figure 3. (B) Find f(1), f (3), and f(5). y or f (x) 4
1 00043
3
5
x
y 0002 f (x) 00044 00045
Z Figure 3 SOLUTIONS y or f (x)
3
Domain: 00043 6 x 6 0007
y 0002 f (x) 1
5
00044
00044
Z Figure 4
MATCHED PROBLEM 2
(A) The dot at the left end of the graph indicates that the graph terminates at that point, while the arrowhead on the right end indicates that the graph continues infinitely far to the right. So the x coordinates on the graph go from 00043 to 0007. The open dot at (00043, 4) indicates that 00043 is not in the domain of f.
x
or
(00043, 0007)
The least y coordinate on the graph is 00045, and there is no greatest y coordinate. (The arrowhead tells us that the graph continues infinitely far upward.) The closed dot at (3, 00045) indicates that 00045 is in the range of f. Range: 00045 y 6 0007
or
[00045, 0007)
(B) The point on the graph with x coordinate 1 is (1, 00044), so f(1) 0002 00044. Likewise, (3, 00045) and (5, 00044) are on the graph, so f (3) 0002 00045 and f (5) 0002 00044.
0002
(A) Find the domain and range of the function f given by the graph in Figure 4. (B) Find f(–4), f (0), and f(2). 0002
ZZZ
CAUTION ZZZ
When using interval notation to describe domain and range, make sure that you always write the least number first! You should find the domain by working left to right along the x axis, and find the range by working bottom to top along the y axis.
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Z Identifying Increasing and Decreasing Functions We will now take a look at increasing and decreasing properties of functions. Informally, a function is increasing over an interval if its graph rises as the x coordinate increases (moves from left to right) over that interval. A function is decreasing over an interval if its graph falls as the x coordinate increases over that interval. A function is constant on an interval if its graph is horizontal (i.e., the height doesn’t change) over that interval (Fig. 5). g(x)
f (x)
5
5
f (x) 0002 0004x 3
g(x) 0002 2x 0003 2 00045
5
x
00045
00045
5
x
00045
(a) Increasing on (ⴚⴥ, ⴥ)
(b) Decreasing on (ⴚⴥ, ⴥ)
h(x)
p (x)
5
5
h(x) 0002 2 00045
5
x
00045
(c) Constant on (ⴚⴥ, ⴥ)
p (x) 0002 x 2 0004 1 x
00045
5
00045
(d) Decreasing on (ⴚⴥ, 0 ] Increasing on [0, ⴥ)
Z Figure 5 Increasing, decreasing, and constant functions.
More formally, we define increasing, decreasing, and constant functions as follows:
Z DEFINITION 1 Increasing, Decreasing, and Constant Functions Let I be an interval in the domain of function f. Then, 1. f is increasing on I and the graph of f is rising on I if f(x1) 6 f(x2) whenever x1 6 x2 in I. 2. f is decreasing on I and the graph of f is falling on I if f(x1) 7 f(x2) whenever x1 6 x2 in I. 3. f is constant on I and the graph of f is horizontal on I if f(x1) 0002 f(x2) whenever x1 6 x2 in I.
Z Linear Functions In Section 2-3, we studied the slope–intercept form of the equation of a line: y 0002 mx 0003 b, where m is the slope, and b is the y intercept. We can carry over what we learned to the study of linear functions.
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Z DEFINITION 2 Linear Function A function of the form f (x) 0002 mx 0003 b is called a linear function. If m 0002 0, the result is f(x) 0002 b, which is called a constant function. If m 0002 1 and b 0002 0, then the result is f(x) 0002 x, which is called the identity function. The domain of any linear function is all real numbers. If m 0, then the range is also all real numbers. If m 0002 0, the function is constant and the range is {b}.
Z GRAPH PROPERTIES OF f(x) ⴝ mx ⴙ b The graph of a linear function is a line with slope m and y intercept b. f (x)
b
f (x)
f(x)
b
b
x
mⴝ0 Constant on (ⴚ0007, 0007) Domain: (ⴚ0007, 0007) Range: {b}
m⬍0 Decreasing on (ⴚ0007, 0007) Domain: (ⴚ0007, 0007) Range: (ⴚ0007, 0007)
ZZZ EXPLORE-DISCUSS 1
EXAMPLE
3
x
x
m⬎0 Increasing on (ⴚ0007, 0007) Domain: (ⴚ0007, 0007) Range: (ⴚ0007, 0007)
(A) Is it possible for a linear function to have two x intercepts? No x intercepts? If either of your answers is yes, give an example. (B) Is it possible for a linear function to have two y intercepts? No y intercept? If either of your answers is yes, give an example.
Graphing a Linear Function Find the slope and intercepts, and then sketch the graph of the linear function defined by f (x) 0002 000423 x 0003 4
SOLUTION f(x)
The y intercept is f(0) 0002 4, and the slope is 000423. To find the x intercept, we solve the equation f(x) 0002 0 for x: 000423 x
5
5
x
f (x) 0002 0 0003400020 000423 x 0002 00044 x0002
00044 0002 (000432)(00044) 0002 6 000423
The graph of f is shown in Figure 6. Z Figure 6
Substitute ⴚ23 x ⴙ 4 for f(x). Subtract 4 from both sides Divide both sides by ⴚ23 . x intercept
0002
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MATCHED PROBLEM 3
Find the slope and intercepts, and then sketch the graph of the linear function defined by f (x) 0002 32 x 0004 6 0002
Z Piecewise-Defined Functions The absolute value function can be defined using the definition of absolute value from Section 1-3: f (x) 0002 冟 x 冟 0002
再
0004x x
if x 6 0 if x 0005 0
Notice that this function is defined by different expressions for different parts of its domain. Functions whose definitions involve more than one expression are called piecewise-defined functions. Example 4 will show you how to work with a piecewise-defined function.
EXAMPLE
4
Analyzing a Piecewise-Defined Function The function f is defined by
再
4x 0003 11 f (x) 0002 3 000412 x 0003 72
if x 6 00042 if 00042 x 1 if x 7 1
(A) Find f (00043), f (00042), f (1), and f(3). (B) Graph f. (C) Find the domain, range, and intervals where f is increasing, decreasing, or constant. SOLUTIONS
(A) Since 00043 is an x value less than 00042, we use the formula 4x 0003 11 to calculate f(00043). f (ⴚ3) 0002 4(ⴚ3) 0003 11 0002 000412 0003 11 0002 00041 Since both 00042 and 1 are in the interval 00042 x 1, the output is 3 for both. f (ⴚ2) 0002 3
f(1) 0002 3
and
Since 3 is an x value greater than 1, we use the formula 000412 x 0003 72 to calculate f (3). f (3) 0002 000412 (3) 0003 72 0002 000432 0003 72 0002 42 0002 2 (B) To graph f, we graph each expression in the definition of f over the appropriate interval. That is, we graph y 0002 4x 0003 11 y00023 y 0002 000412 x 0003 72
for x 6 00042 for 00042 x 1 for x 7 1
y y00023
5
(1, 3)
(00042, 3)
1
y 0002 00042x 0003
00045
5
y 0002 4x 0003 11 00045
x
7 2
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181
We used a solid dot at the point (00042, 3) to indicate that y 0002 4x 0003 11 and y 0002 3 agree at x 0002 00042. The solid dot at the point (1, 3) indicates that y 0002 3 and y 0002 000412 x 0003 72 agree at x 0002 1. (C) The domain of a piecewise-defined function is the union of the intervals used in its definition: Domain of f: (00040007, 00042) ´ [00042, 1] ´ (1, 0007) 0002 (00040007, 0007) The graph of f shows that the range of f is (00040007, 3]. The function f is increasing on (00040007, 00042), constant on [00042, 1], and decreasing on (1, 0007). 0002 MATCHED PROBLEM 4
The function f is defined by
再
000413 x 0004 73 f (x) 0002 00042 5x 0004 17
if x 00041 if 00041 6 x 6 3 if x 0005 3
(A) Find f(00044), f(00041), f(3), and f(4). (B) Graph f. (C) Find the domain, range, and intervals where f is increasing, decreasing, or constant. 0002 Notice that the graph of f in Example 4 contains no breaks. Informally, a graph (or portion of a graph) is said to be continuous if it contains no breaks or gaps. (A formal presentation of continuity can be found in calculus texts.) Piecewise-defined functions occur naturally in many applications, especially ones involving money. A very useful example is income tax.
EXAMPLE
5
Income Tax Table 1 contains a recent tax rate chart for a single filer in the state of Oregon. If T(x) is the tax on an income of $x, write a piecewise definition for T. Find the tax on each of the following incomes: $2,000, $5,000, and $9,000. Table 1 2009 Tax Rate Chart for Persons Filing Single, or Married Filing Separately If the taxable income is:
The tax is:
Not over $3,050
5% of taxable income
Over $3,050 but not over $7,600
$153 plus 7% of the excess over $3,050
Over $7,600
$471 plus 9% of the excess over $7,600
Source: Oregon Department of Revenue
SOLUTION
Since taxes are computed differently on [0, 3,050], (3,050, 7,600] and (7,600, 0007), we must find an expression for the tax on incomes in each of these intervals. [0, 3,050]: Tax is 0.05x. (3,050, 7,600]: Tax is $153 0003 0.07(x 0004 3,050) 0002 0.07x – 61* (7,600, 0007): Tax is $471 0003 0.09(x 0004 7,600) 0002 0.09x 0004 213
*In the Oregon tax rate chart, dollar amounts ending with 0.50 were rounded up to the next dollar. We will do the same.
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Combining the three intervals with the preceding linear expressions, we can write
再
if 0 x 3,050 if 3,050 6 x 7,600 if x 7 7,600
0.05x T(x) 0002 0.07x 0004 61 0.09x 0004 213 Using the piecewise definition of T, we have
T(2,000) 0002 0.05(2,000) 0002 $100 T(5,000) 0002 0.07(5,000) 0004 61 0002 $289 T(9,000) 0002 0.09(9,000) 0004 213 0002 $597 MATCHED PROBLEM 5
0002
Table 2 contains a recent tax rate chart for persons filing a joint return in the state of Oregon. If T(x) is the tax on an income of $x, write a piecewise definition for T. Find the tax on each of the following incomes: $4,000, $10,000, and $18,000. Table 2 2009 Tax Rate Chart for Persons Filing Jointly If the taxable income is:
The tax is:
Not over $6,100
5% of taxable income
Over $6,100 but not over $15,200
$305 plus 7% of the excess over $6,100
Over $15,200
$942 plus 9% of the excess over $15,200
0002 We will conclude the section with a look at a particular piecewise function that is especially useful in computer science. It is called the greatest integer function. The greatest integer for a real number x, denoted by 冀x冁, is the integer n such that n x n 0003 1; that is, 冀x冁 is the largest integer less than or equal to x. For example, 冀3.45 冁 0002 3 冀7 冁 0002 7
5
f(x) 0002 冚x 军 00045
5
00045
Z Figure 7 Greatest integer function.
x
冀5.99冁 0002 5 冀0冁 0002 0
冀00042.13 冁 0002 00043 冀00048 冁 0002 00048 冀00043.79 冁 0002 00044
Not ⴚ2
The greatest integer function f is defined by the equation f(x) 0002 冀x冁. A piecewise definition of f for 00042 x 3 is shown below, and a sketch of the graph of f for 00045 x 5 is shown in Figure 7. Since the domain of f is all real numbers, the piecewise definition continues indefinitely in both directions, as does the stairstep pattern in the figure. So the range of f is the set of all integers.
f (x) 0002 冀x冁 0002
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭
f (x)
o 00042 00041 0 1 2 o
if 00042 if 00041 if 0 if 1 if 2
x x x x x
6 00041 6 0 6 1 6 2 6 3
Notice in Figure 7 that at each integer value of x there is a break in the graph, and between integer values of x there is no break. In other words, the greatest integer function is discontinuous at each integer n and continuous on each interval of the form [n, n 0003 1).
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Graphing Functions
183
Technology Connections graph and Figure 7. If your graphing calculator supports both a connected mode and a dot mode for graphing functions (consult your manual), which mode is preferable for this graph?
Most graphing calculators denote the greatest integer function as int (x), although not all define it the same way we have here. Graph y ⴝ int (x) for ⴚ5 ⱕ x ⱕ 5 and ⴚ5 ⱕ y ⱕ 5 and discuss any differences between your
EXAMPLE
6
Computer Science Let f (x) 0002
冀10x 0003 0.5冁 10
Find: (A) f(6)
(B) f (1.8)
(C) f (3.24)
(D) f(4.582)
(E) f(00042.68)
What operation does this function perform? SOLUTIONS
Table 3 x
f(x)
6
6
1.8
1.8
3.24
3.2
4.582
4.6
00042.68
00042.7
MATCHED PROBLEM 6
(A) f (6) 0002
冀60.5冁 60 0002 00026 10 10
(C) f (3.24) 0002
(B) f (1.8) 0002
冀32.9冁 32 0002 0002 3.2 10 10
(E) f (00042.68) 0002
冀18.5冁 18 0002 0002 1.8 10 10
(D) f (4.582) 0002
冀46.32冁 46 0002 0002 4.6 10 10
冀000426.3冁 000427 0002 0002 00042.7 10 10
Comparing the values of x and f (x) in Table 3 in the margin, we conclude that this function rounds decimal fractions to the nearest tenth. The greatest integer function is used in programming (spreadsheets, for example) to round numbers to a specified accuracy. 0002 Let f(x) 0002 冀x 0003 0.5冁. Find: (A) f(6)
(B) f (1.8)
(C) f(3.24)
(D) f(00044.3)
(E) f(00042.69)
What operation does this function perform? 0002 ANSWERS TO MATCHED PROBLEMS 1. Domain: (00040007, 23) 傼 (23, 0007); x intercept: 000454; y intercept: f (0) 0002 000452 2. (A) Domain: (00044, 5); range: (00044, 3] (B) f (00044) 0002 1, f (0) 0002 3, f (2) 0002 2 3. y intercept: f(0) 0002 00046 y x intercept: 4 3 Slope: 2 5
00045
x
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4. (A) f (00044) 0002 00041; f (00041) 0002 00042; f (3) 0002 00042; f (4) 0002 3 (B) (C) Domain: (00040007, 0007); y range: [00042, 0007); 5 increasing: [3, 0007); y 0002 5x 0003 17 decreasing: (00040007, 00041]; constant: (00041, 3) 1
y 0002 00043x 0004
7 3
00045
y 0002 00042
(00041, 00042)
再
x
5
(3, 00042)
00045
0.05x x 6,100 5. T(x) 0002 0.07x 0004 122 6,100 6 x 15,200 0.09x 0004 426 x 7 15,200 T(4,000) 0002 $200; T(10,000) 0002 $578; T(18,000) 0002 $1,194 T(4,000) 0002 $200; T(10,000) 0002 $594; T(18,000) 0002 $1,248 6. (A) 6 (B) 2 (C) 3 (D) 00044 (E) 00043; f rounds decimal fractions to the nearest integer.
3-2
Exercises
1. Describe in your own words what the graph of a function is.
13. Repeat Problem 9 for the function p.
2. Explain how to find the domain and range of a function from its graph.
14. Repeat Problem 9 for the function q.
3. How many y intercepts can a function have? What about x intercepts? Explain. 4. True or false: On any interval in its domain, every function is either increasing or decreasing. Explain. 5. Explain in your own words what it means to say that a function is increasing on an interval.
f (x)
g(x)
5
5
00045
6. Explain in your own words what it means to say that a function is decreasing on an interval.
5
x
00045
00045
00045
h (x)
7. What does it mean for a function to be defined piecewise?
x
5
k (x)
5
5
8. Explain how the output of the greatest integer function is calculated for any real number input. Problems 9–20 refer to functions f, g, h, k, p, and q given by the following graphs.
00045
9. For the function f, find: (A) Domain (B) Range (C) x intercepts (D) y intercept (E) Intervals over which f is increasing (F) Intervals over which f is decreasing (G) Intervals over which f is constant (H) Any points of discontinuity 10. Repeat Problem 9 for the function g.
5
x
00045
00045
5
00045
p (x)
q(x)
5
5
00045
5
x
00045
5
11. Repeat Problem 9 for the function h. 12. Repeat Problem 9 for the function k.
00045
x
00045
x
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15. Find f (00044), f (0), and f (4).
Graphing Functions
185
17. Find h(00043), h(0), and h(2).
In Problems 47–58, (A) find the indicated values of f; (B) graph f and label the points from part A, if they exist; and (C) find the domain, range, and the values of x in the domain of f at which f is discontinuous.
18. Find k (0), k(2), and k(4).
47. f(00041), f(0), f(1)
16. Find g (00045), g(0), and g(5).
f (x) 0002 e
19. Find p(00042), p(2), and p (5). 20. Find q(00044), q(00043), and q (1).
x00031 0004x 0003 1
if 00041 x 6 0 if 0 x 1
48. f(00042), f(1), f(2) Problems 21–26 describe the graph of a continuous function f over the interval [00045, 5]. Sketch the graph of a function that is consistent with the given information. 21. The function f is increasing on [00045, 00042], constant on [00042, 2], and decreasing on [2, 5]. 22. The function f is decreasing on [00045, 00042], constant on [00042, 2], and increasing on [2, 5]. 23. The function f is decreasing on [00045, 00042], constant on [00042, 2], and decreasing on [2, 5]. 24. The function f is increasing on [00045, 00042], constant on [00042, 2], and increasing on [2, 5]. 25. The function f is decreasing on [00045, 00042], increasing on [00042, 2], and decreasing on [2, 5]. 26. The function f is increasing on [00045, 00042], decreasing on [00042, 2], and increasing on [2, 5]. In Problems 27–32, find the slope and intercepts, and then sketch the graph. 27. f(x) 0002 2x 0003 4
28. f (x) 0002 3x 0004 3
29. f (x) 0002 000412 x 0004 53
30. f (x) 0002 000434 x 0003 65
31. f (x) 0002 00042.3x 0003 7.1
32. f (x) 0002 5.2x 0004 3.4
In Problems 33–36, find a linear function f satisfying the given conditions. 33. f (00042) 0002 2 and f(0) 0002 10
f (x) 0002 e
if 00042 x 6 1 if 1 x 2
x 0004x 0003 2
49. f (00043), f(00041), f(2) f (x) 0002 e
00042 4
if 00043 x 6 00041 if 00041 6 x 2
50. f (00042), f(2), f(5) f (x) 0002 e
if 00042 x 6 2 if 2 6 x 5
1 00043
51. f(00042), f(00041), f(0) f (x) 0002 e
x00032 x00042
if x 6 00041 if x 7 00041
52. f(0), f(2), f(4) f (x) 0002 e
00041 0004 x 50004x
if x 6 2 if x 7 2
53. f(00043), f(00042), f(0), f(3), f(4)
再 再 再 再 再 再
00042x 0004 6 f (x) 0002 00042 6x 0004 20
if x 6 00042 if 00042 x 6 3 if x 0005 3
54. f(00042), f(00041), f(0), f(2), f(3) 2 3x
0003 113
f (x) 0002 3 000432 x 0003 6
if x 00041 if 00041 6 x 2 if x 7 2
55. f(00043), f(00042), f(0), f(3), f(4)
34. f(4) 0002 00047 and f(0) 0002 5
5 2x
00036 f (x) 0002 1 3 7 2x 0004 2
35. f (00042) 0002 7 and f (4) 0002 00042 36. f (00043) 0002 00042 and f(5) 0002 4
if x 6 00042 if 00042 x 3 if x 7 3
56. f(00043), f(00042), f(0), f(1), f(2) In Problems 37–46, find the domain, x intercept, and y intercept. 3x 0004 12 37. f (x) 0002 2x 0003 4 39. f (x) 0002
3x 0004 2 4x 0004 5
4x 41. f (x) 0002 (x 0004 2)2 43. f (x) 0002
x2 0004 16 x2 0004 9
2x 0003 9 38. f (x) 0002 x00043 40. f (x) 0002
2x 42. f (x) 0002 (x 0003 1)2 44. f (x) 0002
2
45. f (x) 0002
x 00037 x2 0004 25
2x 0003 7 5x 0003 8
x2 0004 4 x2 0003 10 2
46. f (x) 0002
x 0003 11 x2 0003 5
3 f (x) 0002 000413 x 0003 73 00043x 0003 5
if x 00042 if 00042 6 x 6 1 if x 0005 1
57. f(00041), f(0), f(1), f(2), f(3) f (x) 0002
2 3x 000412 x 000412 x
00034 00033
if x 6 0 if 0 6 x 6 2 if x 7 2
58. f(00043), f(00042), f(0), f(2), f(3) 000432 x 0004 2 3 1 f (x) 0002 4x 0004 2 3 5 4x 0004 2
if x 6 00042 if 00042 6 x 6 2 if x 7 2
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FUNCTIONS
In Problems 59–64, use the graph of f to find a piecewise definition for f.
f (x)
64.
5
f (x)
59.
(2, 4)
5
(2, 0) 00045
(00042, 3) 00045
x
5
(0, 00041)
In Problems 65–68, find a piecewise definition of f that does not involve the absolute value function. (Hint: Use the definition of absolute value on page 180 to consider cases.) Sketch the graph of f, and find the domain, range, and the values of x at which f is discontinuous.
00045
f (x) 5
(0, 2)
(2, 2)
00045
x
5
65. f (x) 0002 1 0003 冟 x 冟
66. f (x) 0002 2 0004 冟 x 冟
67. f (x) 0002 冟 x 0004 2 冟
68. f (x) 0002 冟 x 0003 1 冟
69. The function f is continuous and increasing on the interval [1, 9] with f (1) 0002 00045 and f(9) 0002 4. (A) Sketch a graph of f that is consistent with the given information. (B) How many times does your graph cross the x axis? Could the graph cross more times? Fewer times? Support your conclusions with additional sketches and/or verbal arguments.
(0, 00042) 00045
61.
x
(00044, 00043) (00043, 00043) 00045
(2, 00043)
(00042, 00042)
5
(00043, 00041)
(0, 1)
60.
(4, 4)
f (x) 5
(00044, 3) (00041, 3)
70. Repeat Problem 69 if the function is not continuous. 00045
x
5
71. The function f is continuous on the interval [00045, 5] with f(00045) 0002 00044, f (1) 0002 3, and f (5) 0002 00042. (A) Sketch a graph of f that is consistent with the given information. (B) How many times does your graph cross the x axis? Could the graph cross more times? Fewer times? Support your conclusions with additional sketches and/or verbal arguments.
(1, 00043) 00045
f (x)
62.
5
72. Repeat Problem 71 if f is continuous on [00048, 8] with f (00048) 0002 00046, f(00044) 0002 3, f(3) 0002 00042, and f (8) 0002 5. (4, 1)
00045
x
5
In Problems 73–78, first graph functions f and g in the same viewing window, then graph m(x) and n(x) in their own viewing windows:
(00044, 00042) (00042, 00042) 00045
m(x) 0002 0.5[ f (x) 0003 g(x) 0003 冟 f (x) 0004 g(x) 冟 ] n(x) 0002 0.5[ f (x) 0003 g(x) 0004 冟 f (x) 0004 g(x) 冟 ]
f (x)
63.
Problems 73–80 require the use of a graphing calculator.
5
73. f(x) 0002 00042x, g(x) 0002 0.5x
(00044, 3) (00042, 3)
74. f(x) 0002 3x 0003 1, g(x) 0002 00040.5x 0004 4
(00042, 2) 00045
5
x
(1, 00041) (4, 00041) 00045
(1, 00044)
75. f(x) 0002 5 0004 0.2x2, g(x) 0002 0.3x2 0004 4 76. f (x) 0002 0.15x2 0004 5, g(x) 0002 5 0004 1.5冟 x 冟 77. f (x) 0002 0.2x2 0004 0.4x 0004 5, g(x) 0002 0.3x 0004 3 78. f (x) 0002 8 0003 1.5x 0004 0.4x2, g(x) 0002 00040.2x 0003 5 79. How would you characterize the relationship between f, g, and m in Problems 73–78? [Hint: Use the trace feature on the calculator and the up/down arrows to examine all 3 graphs at several points.]
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SECTION 3–2
80. How would you characterize the relationship between f, g, and n in Problems 73–78? [Hint: Use the trace feature on the calculator and the up/down arrows to examine all 3 graphs at several points.]
Graphing Functions
187
(B) Can the function f defined by f (x) 0002 15 0003 3冀x冁 be used to compute the delivery charges for all x, 0 x 6? Justify your answer.
APPLICATIONS Table 4 contains daily automobile rental rates from a New Jersey firm.
Table 4 Vehicle Type
Daily Charge
Included Miles
Mileage Charge*
Compact
$32.00
100/Day
$0.16/mile
Midsize
$41.00
200/Day
$0.18/mile
*Mileage charge does not apply to included miles.
81. AUTOMOBILE RENTAL Use the data in Table 4 to construct a piecewise-defined model for the daily rental charge for a compact automobile that is driven x miles. 82. AUTOMOBILE RENTAL Use the data in Table 4 to construct a piecewise-defined model for the daily rental charge for a midsize automobile that is driven x miles. 83. SALES COMMISSIONS A high-volume website pays salespeople to solicit advertisements for placement on their site. The sales staff each gets $200 per week in salary, and a commission of 4% on all sales over $3,000 for the week. In addition, if the weekly sales are $8,000 or more, the salesperson gets a $100 bonus. Find a piecewise definition for the weekly earnings E (in dollars) in terms of the weekly sales x (in dollars). Graph this function and find the values of x at which the function is discontinuous. Find the weekly earnings for sales of $5,750 and of $9,200. 84. SERVICE CHARGES On weekends and holidays, an emergency plumbing repair service charges $2.00 per minute for the first 30 minutes of a service call and $1.00 per minute for each additional minute. Express the total service charge S (in dollars) as a piecewise-defined function of the duration of a service call x (in minutes). Graph this function and find the values of x at which the function is discontinuous. Find the charge for a 25-minute service call and for a 45-minute service call. 85. COMPUTER SCIENCE Let f (x) 0002 10 冀 0.5 0003 x 10冁 . Evaluate f at 4, 00044, 6, 00046, 24, 25, 247, 0004243, 0004245, and 0004246. What operation does this function perform? 86. COMPUTER SCIENCE Let f (x) 0002 100 冀 0.5 0003 x 100冁 . Evaluate f at 40, 000440, 60, 000460, 740, 750, 7,551, 0004601, 0004649, and 0004651. What operation does this function perform? 87. COMPUTER SCIENCE Use the greatest integer function to define a function f that rounds real numbers to the nearest hundredth. 88. COMPUTER SCIENCE Use the greatest integer function to define a function f that rounds real numbers to the nearest thousandth. 89. DELIVERY CHARGES A nationwide package delivery service charges $15 for overnight delivery of packages weighing 1 pound or less. Each additional pound (or fraction thereof ) costs an additional $3. Let C be the charge for overnight delivery of a package weighing x pounds. (A) Write a piecewise definition of C for 0 x 6, and sketch the graph of C.
90. TELEPHONE CHARGES Calls to 900 numbers are charged to the caller. A 900 number hot line for gambling advice on college football games charges $4 for the first minute of the call and $2 for each additional minute (or fraction thereof). Let C be the charge for a call lasting x minutes. (A) Write a piecewise definition of C for 0 x 6, and sketch the graph of C. (B) Can the function f defined by f(x) 0002 4 0003 2冀x冁 be used to compute the charges for all x, 0 x 6? Justify your answer. 91. STATE INCOME TAX The Connecticut state income taxes for an individual filing a single return are 3% for the first $10,000 of taxable income and 5% on the taxable income in excess of $10,000. Find a piecewise-defined function for the taxes owed by a single filer with an income of x dollars and graph this function. 92. STATE INCOME TAX The Connecticut state income taxes for an individual filing a head of household return are 3% for the first $16,000 of taxable income and 5% on the taxable income in excess of $16,000. Find a piecewise-defined function for the taxes owed by a head of household filer with an income of x dollars and graph this function. Table 5 contains income tax rates for Minnesota in a recent year. Table 5
Status
Taxable Income Over
But Not Over
Tax Is
Of the Amount Over
Single
$0
$19,890
5.35%
$0
19,890
65,330
$1,064 0003 7.05%
19,890
65,330
...
4,268 0003 7.85%
65,330
0
29,070
5.35%
0
29,070
115,510
1,555 0003 7.05%
29,070
115,510
...
7,649 0003 7.85%
115,510
Married
93. STATE INCOME TAX Use the schedule in Table 5 to construct a piecewise-defined model for the taxes due for a single taxpayer with a taxable income of x dollars. Find the tax on the following incomes: $10,000, $30,000, $100,000. 94. STATE INCOME TAX Use the schedule in Table 5 to construct a piecewise-defined model for the taxes due for a married taxpayer with a taxable income of x dollars. Find the tax on the following incomes: $20,000, $60,000, $200,000.
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Transformations of Functions Z A Library of Elementary Graphs Z Shifting Graphs Horizontally and Vertically Z Reflecting Graphs Z Stretching and Shrinking Graphs Z Even and Odd Functions
We have seen that the graph of a function can provide valuable insight into the information provided by that function. But there is a seemingly endless variety of functions out there, and it seems like an insurmountable task to learn about so many different graphs. In this section, we will see that relationships between the formulas for certain functions lead to relationships between their graphs as well. For example, the functions g(x) 0002 x2 0003 2
h(x) 0002 (x 0003 2)2
k(x) 0002 2x2
can be expressed in terms of the function f(x) 0002 x2 as follows: g(x) 0002 f (x) 0003 2
h(x) 0002 f (x 0003 2)
k(x) 0002 2f(x)
We will see that the graphs of functions g, h, and k are closely related to the graph of function f. Once we understand these relationships, knowing the graph of a very simple function like f (x) 0002 x2 will enable us to learn about the graphs of many related functions.
Z A Library of Elementary Graphs As you progress through this book, you will encounter a number of basic functions that you will want to add to your library of elementary functions. Figure 1 shows six basic functions that you will encounter frequently. You should know the definition, domain, and range of each of these functions, and be able to draw their graphs.
f (x)
g(x)
h(x)
5
5
5
00045
5
x
00045
5
x
00045
00045
(a) Identity function f(x) 0002 x Domain: R Range: R
(b) Absolute value function g(x) 0002 |x| Domain: R Range: [0, 0003)
(c) Square function h(x) 0002 x2 Domain: R Range: [0, 0003)
5
x
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SECTION 3–3 m (x)
n (x)
5
p(x)
5
00045
5
x
5
x
5
00045
00045
5
00045
(d) Cube function m(x) 0002 x3 Domain: R Range: R
189
Transformations of Functions
x
00045
(f) Cube root function 3 p(x) 0002 1x Domain: R Range: R
(e) Square root function n(x) 0002 1x Domain: [0, 0003) Range: [0, 0003)
Z Figure 1 Some basic functions and their graphs. [Note: Letters used to designate these functions may vary from context to context; R represents the set of all real numbers.]
Z Shifting Graphs Vertically and Horizontally If a new function is formed by performing an operation on a given function, then the graph of the new function is called a transformation of the graph of the original function. For example, if we add a constant k to f (x), then the graph of y 0002 f (x) is transformed into the graph of y 0002 f (x) 0003 k.
ZZZ EXPLORE-DISCUSS 1
The following activities refer to the graph of f shown in Figure 2 and the corresponding points on the graph shown in Table 1. (A) Use the points in Table 1 to construct a similar table and then sketch a graph for each of the following functions: y 0002 f (x) 0003 2, y 0002 f(x) 0004 3. Describe the relationship between the graph of y 0002 f (x) and the graph of y 0002 f(x) 0003 k for k any real number. (B) Use the points in Table 1 to construct a similar table and then sketch a graph for each of the following functions: y 0002 f (x 0003 2), y 0002 f(x 0004 3). [Hint: Choose values of x so that x 0003 2 or x 0004 3 is in Table 1.] Describe the relationship between the graph of y 0002 f (x) and the graph of y 0002 f (x 0003 h) for h any real number. y
Table 1
5
x
B
00045
E
C
A
5
x
y 0002 f (x) D 00045
Z Figure 2
f(x)
A
00044
0
B
00042
3
C
0
0
D
2
00043
E
4
0
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1
Vertical and Horizontal Shifts (A) How are the graphs of y 0002 x2 0003 2 and y 0002 x2 0004 3 related to the graph of y 0002 x2? Confirm your answer by graphing all three functions in the same coordinate system. (B) How are the graphs of y 0002 (x 0003 2)2 and y 0002 (x 0004 3)2 related to the graph of y 0002 x2? Confirm your answer by graphing all three functions in the same coordinate system.
SOLUTIONS
(A) Note that the output of y 0002 x2 0003 2 is always exactly two more than the output of y 0002 x2. Consequently, the graph of y 0002 x2 0003 2 is the same as the graph of y 0002 x2 shifted upward two units, and the graph of y 0002 x2 0004 3 is the same as the graph of y 0002 x2 shifted downward three units. Figure 3 confirms these conclusions. (It appears that the graph of y 0002 f(x) 0003 k is the graph of y 0002 f (x) shifted up if k is positive and down if k is negative.) y 5
y 0002 x2 0003 2 y 0002 x2 x
00045
5
y 0002 x2 0004 3 00045
Z Figure 3 Vertical shifts.
(B) Note that the output of y 0002 (x 0003 2)2 is zero for x 0002 00042, while the output of y 0002 x2 is zero for x 0002 0. This suggests that the graph of y 0002 (x 0003 2)2 is the same as the graph of y 0002 x2 shifted to the left two units, and the graph of y 0002 (x 0004 3)2 is the same as the graph of y 0002 x2 shifted to the right three units. Figure 4 confirms these conclusions. It appears that the graph of y 0002 f(x 0003 h) is the graph of y 0002 f(x) shifted right if h is negative and left if h is positive. y
y 0002 x2
y 0002 (x 0004 3)2
5
y 0002 (x 0003 2)2 00045
Z Figure 4 Horizontal shifts.
MATCHED PROBLEM 1
5
x
0002
(A) How are the graphs of y 0002 1x 0003 3 and y 0002 1x 0004 1 related to the graph of y 0002 1x? Confirm your answer by graphing all three functions in the same coordinate system. (B) How are the graphs of y 0002 1x 0003 3 and y 0002 1x 0004 1 related to the graph of y 0002 1x? Confirm your answer by graphing all three functions in the same coordinate system. 0002 To summarize our experiences in Explore-Discuss 1 and Example 1: We can graph y 0002 f(x) 0003 k by vertically shifting the graph of y 0002 f(x) upward k units if k is positive
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SECTION 3–3
Transformations of Functions
191
and downward 冟 k 冟 units if k is negative. We can graph y 0002 f(x 0003 h) by horizontally shifting the graph of y 0002 f (x) left h units if h is positive and right 冟 h 冟 units if h is negative.
EXAMPLE
2
Vertical and Horizontal Shifts The graphs in Figure 5 are either horizontal or vertical shifts of the graph of f(x) 0002 |x|. Write appropriate equations for functions H, G, M, and N in terms of f. y
G
y
f
5
f
H
00045
5
M
5
N
x 00045
x
5
00045
Z Figure 5 Vertical and horizontal shifts. SOLUTION
The graphs of functions H and G are 3 units lower and 1 unit higher, respectively, than the graph of f, so H and G are vertical shifts given by H(x) 0002 冟 x 冟 0004 3
G(x) 0002 冟 x 冟 0003 1
The graphs of functions M and N are 2 units to the left and 3 units to the right, respectively, of the graph of f, so M and N are horizontal shifts given by M(x) 0002 冟 x 0003 2 冟 MATCHED PROBLEM 2
N(x) 0002 冟 x 0004 3 冟
0002
The graphs in Figure 6 are either horizontal or vertical shifts of the graph of f (x) 0002 x3. Write appropriate equations for functions H, G, M, and N in terms of f. G y
y MfN
f H 5
5
00045
5
x
00045
5
x
Z Figure 6 Vertical and horizontal shifts.
0002
Z Reflecting Graphs In Section 2-1, we discussed reflections of graphs and developed symmetry properties that we used as an aid in graphing equations. Now we will consider reflection as an operation that transforms the graph of a function.
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ZZZ EXPLORE-DISCUSS 2
The following activities refer to the graph of f shown in Figure 7 and the corresponding points on the graph shown in Table 2. (A) Construct a similar table for y 0002 0004f (x) and then sketch the graph of y 0002 0004f(x). Describe the relationship between the graph of y 0002 f(x) and the graph of y 0002 0004f(x) in terms of reflections. (B) Construct a similar table for y 0002 f(0004x) and then sketch the graph of y 0002 f(0004x). [Hint: Choose x values so that 0004x is in Table 2.] Describe the relationship between the graph of y 0002 f (x) and the graph of y 0002 f (0004x) in terms of reflections. (C) Construct a similar table for y 0002 0004f(0004x) and then sketch the graph of y 0002 0004f(0004x). [Hint: Choose x values so that 0004x is in Table 2.] Describe the relationship between the graph of y 0002 f(x) and the graph of y 0002 0004f(0004x) in terms of reflections. y
Table 2
5
A
E
x
y 0002 f (x) 00045
D
B
00045
5
x
C
Z Figure 7
EXAMPLE
3
f(x)
A
00042
5
B
00041
0
C
1
00044
D
3
0
E
4
5
Reflecting the Graph of a Function Let f(x) 0002 (x 0004 1)2. (A) How are the graphs of y 0002 f (x) and y 0002 0004f(x) related? Confirm your answer by graphing both functions in the same coordinate system. (B) How are the graphs of y 0002 f (x) and y 0002 f (0004x) related? Confirm your answer by graphing both functions in the same coordinate system. (C) How are the graphs of y 0002 f (x) and y 0002 0004f(0004x) related? Confirm your answer by graphing both functions in the same coordinate system.
SOLUTIONS
Refer to Definition 1 in Section 2-1. (A) The graph of y 0002 0004f (x) can be obtained from the graph of y 0002 f (x) by changing the sign of each y coordinate. This has the effect of moving every point to the opposite side of the x axis. So the graph of y 0002 0004f (x) is the reflection through the x axis of the graph of y 0002 f (x) [Fig. 8(a)]. (B) The graph of y 0002 f (0004x) can be obtained from the graph of y 0002 f (x) by changing the sign of each x coordinate. This has the effect of moving every point to the opposite side of the y axis. So the graph of y 0002 f (0004x) is the reflection through the y axis of the graph of y 0002 f (x) [Fig. 8(b)]. (C) The graph of y 0002 0004f (0004x) can be obtained from the graph of y 0002 f (x) by changing the sign of each x and y coordinate. So the graph of y 0002 0004f (0004x) is the reflection through the origin of the graph of y 0002 f (x) [Fig. 8(c)].
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SECTION 3–3 y 5
00045
5
x
5
5
(00043, 4)
y 0002 f(x)
y 0002 0004f(x)
y
y (3, 4)
(3, 4) y 0002 f(x)
y 0002 f(0004x) 00045
5
y 0002 f(x) x
00045
(00043, 00044) 00045
(a) y 0002 f(x) and y 0002 0004f(x); reflection through the x axis
(3, 4)
5
x
y 0002 0004f(0004x)
(3, 00044)
00045
193
Transformations of Functions
(b) y 0002 f(x) and y 0002 f(0004x); reflection through the y axis
00045
(c) y 0002 f(x) and y 0002 0004f(0004x); reflection through the origin
0002
Z Figure 8
MATCHED PROBLEM 3
Repeat Example 3 for f (x) 0002 |x 0003 2|.
0002
Z Stretching and Shrinking Graphs Horizontal shifts, vertical shifts, and reflections are called rigid transformations because they do not change the shape of a graph, only its location. Now we consider some nonrigid transformations that change the shape by stretching or shrinking a graph.
ZZZ EXPLORE-DISCUSS 3 y
(A) Use the points in Table 3 to construct a similar table and sketch a graph for each of the following functions: y 0002 2f(x) and y 0002 12 f (x). If A 0005 1, does multiplying f by A stretch or shrink the graph of y 0002 f(x) in the vertical direction? What happens if 0 0006 A 0006 1?
8
A
E B
D C
00043
The following activities refer to the graph of f shown in Figure 9 and the corresponding points on the graph shown in Table 3.
x
7
(B) Use the points in Table 3 to complete the following tables and then sketch a graph of y 0002 f (2x) and of y 0002 f (12x):
00042
x
2x
f (2x)
x
Z Figure 9
Table 3 x
f(x)
A
00042
5
B
0
3
C
2
1
D
4
3
E
6
5
00041
00044
0
0
1
4
2
8
3
12
1 2x
f (12x)
If A 0005 1, is the graph of y 0002 f(Ax) a horizontal stretch or a horizontal shrink of the graph of y 0002 f(x)? What if 0 0006 A 0006 1?
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In general, the graph of y 0002 Af(x) can be obtained from the graph of y 0002 f(x) by multiplying the y coordinate of each point on the graph f by A. This vertically stretches the graph of y 0002 f(x) if A 0005 1 and vertically shrinks the graph if 0 0006 A 0006 1. The graph of y 0002 f(Ax) can be obtained from the graph of y 0002 f(x) by multiplying the x coordinate of each point on the graph f by 1兾A. This horizontally stretches the graph of y 0002 f(x) if 0 0006 A 0006 1 and horizontally shrinks the graph if A 0005 1. Another common name for a stretch is an expansion and another common name for a shrink is a contraction.
EXAMPLE
4
Stretching or Shrinking a Graph Let f(x) 0002 1 0003 x2. (A) How are the graphs of y 0002 2f(x) and y 0002 12 f (x) related to the graph of y 0002 f(x)? Confirm your answer by graphing all three functions in the same coordinate system. (B) How are the graphs of y 0002 f(2x) and y 0002 f (12 x) related to the graph of y 0002 f(x)? Confirm your answer by graphing all three functions in the same coordinate system.
SOLUTIONS
(A) The graph of y 0002 2f(x) 0002 2 0003 2x2 can be obtained from the graph of f by multiplying each y value by 2. This stretches the graph of f vertically (away from the x axis) by a factor of 2. The graph of y 0002 12 f (x) 0002 12 0003 12 x2 can be obtained from the graph of f by multiplying each y value by 12. This shrinks the graph of f vertically (toward the x axis) by a factor of 12 [Fig. 10(a)]. (B) The graph of y 0002 f(2x) 0002 1 0003 4x2 can be obtained from the graph of f by multiplying each x value by 12. This shrinks the graph of f horizontally (toward the y axis) by a factor of 12. The graph of y 0002 f (12 x) 0002 1 0003 14 x2 can be obtained from the graph of f by multiplying each x value by 2. This stretches the graph of f horizontally (away from the y axis) by a factor of 2 [Fig. 10(b)].
y 7
y 0002 2 0003 2x 2 y 0002 1 0003 x2 1 1 y 0002 2 0003 2 x2
y 7
(1, 4)
5
x
00043
(a) Vertical stretching and shrinking
y 0002 1 0003 x2 1 y 0002 1 0003 4 x2 (4, 5) (2, 5) (1, 5) x
(1, 2) (1, 1) 00045
y 0002 1 0003 4x 2
00045
5
00043
(b) Horizontal stretching and shrinking
Z Figure 10
MATCHED PROBLEM 4
0002
Let f(x) 0002 4 0004 x2. (A) How are the graphs of y 0002 2f(x) and y 0002 12 f (x) related to the graph of y 0002 f(x)? Confirm your answer by graphing all three functions in the same coordinate system. (B) How are the graphs of y 0002 f(2x) and y 0002 f (12 x) related to the graph of y 0002 f(x)? Confirm your answer by graphing all three functions in the same coordinate system. 0002 Plotting points with the same x coordinate will help you recognize vertical stretches and shrinks [Fig. 10(a)]. And plotting points with the same y coordinate will help you recognize horizontal stretches and shrinks [Fig. 10(b)].
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Note that for some functions, a horizontal stretch or shrink can also be interpreted as a vertical stretch or shrink. For example, if y 0002 f(x) 0002 x2, then y 0002 4f (x) 0002 4x2 0002 (2x)2 0002 f (2x) So the graph of y 0002 4x2 is both a vertical stretch and a horizontal shrink of the graph of y 0002 x2. The transformations we’ve studied are summarized next for easy reference.
Z GRAPH TRANSFORMATIONS (SUMMARY) Z Figure 11 Graph transformations.
Vertical Shift [Fig. 11(a)]:
y 5
g
00045
5
再
y 0002 f (x) 0003 k
Shift graph of y 0002 f (x) up k units Shift graph of y 0002 f (x) down 冟 k 冟 units
k 7 0 k 6 0
f
Horizontal Shift [Fig. 11(b)]:
x h
y 0002 f (x 0003 h)
再
h 7 0 h 6 0
Shift graph of y 0002 f (x) left h units Shift graph of y 0002 f (x) right 冟 h 冟 units
Vertical Stretch and Shrink [Fig. 11(c)]:
00045
(a) Vertical translation g(x) 0002 f(x) 0005 2 h(x) 0002 f(x) 0004 3
g y
f
再 再
y 0002 Af (x)
A 7 1
Vertically stretch the graph of y 0002 f (x) by multiplying each y value by A
0 6 A 6 1
Vertically shrink the graph of y 0002 f (x) by multiplying each y value by A
Horizontal Stretch and Shrink [Fig. 11(d)]:
h
A 7 1
5
y 0002 f (Ax) 00045
0 6 A 6 1
x
5
Horizontally shrink the graph of y 0002 f (x) by multiplying each x value by A1 Horizontally stretch the graph of y 0002 f (x) by multiplying each x value by A1
Reflection [Fig. 11(e)]: y 0002 0004f (x) y 0002 f (0004x) y 0002 0004f(0004x)
00045
(b) Horizontal translation g(x) 0002 f(x 0005 3) h(x) 0002 f(x 0004 2)
y
y g f
g
Reflect the graph of y 0002 f(x) through the x axis Reflect the graph of y 0002 f(x) through the y axis Reflect the graph of y 0002 f(x) through the origin
y
g h
f
5
5
f
5
h 00045
5
00045
5
x
00045
5
x 00045 00045
(c) Vertical expansion and contraction g(x) 0002 2f(x) h(x) 0002 12 f (x)
(d) Horizontal expansion and contraction g(x) 0002 f(2x) h(x) 0002 f(12 x)
k
h (e) Reflection g(x) 0002 f(0004x) h(x) 0002 0004f(x) k(x) 0002 0004f(0004x)
x
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EXAMPLE
5
Combining Graph Transformations The graph of y 0002 g(x) in Figure 12 is a transformation of the graph of y 0002 x2. Find an equation for the function g. y 5
y 0002 g(x) 00045
5
x
Z Figure 12 SOLUTION
To transform the graph of y 0002 x2 [Fig. 13(a)] into the graph of y 0002 g(x), we first reflect the graph of y 0002 x2 through the x axis [Fig. 13(b)], then shift it to the right two units [Fig. 13(c)]. An equation for the function g is g(x) 0002 0004(x 0004 2)2 y
y 5
5
5
x
00045
5
x
00045
y 0002 0004(x 0004 2)2 x
(b) y 0002 0004x2
(c) y 0002 0004(x 0004 2)2
Z Figure 13
MATCHED PROBLEM 5 y 5
5
00045
Z Figure 14
0002
The graph of y 0002 h(x) in Figure 14 is a transformation of the graph of y 0002 x3. Find an equation for the function h. 0002
Z Even and Odd Functions
y 0002 h(x)
00045
5
00045
00045
(a) y 0002 x2
5
y 0002 0004x 2
y 0002 x2 00045
y
x
Certain transformations leave the graphs of some functions unchanged. For example, reflecting the graph of y 0002 x2 through the y axis does not change the graph. Functions with this property are called even functions. Similarly, reflecting the graph of y 0002 x3 through the origin does not change the graph. Functions with this property are called odd functions. More formally, we have the following definitions. Z EVEN AND ODD FUNCTIONS If f(x) 0002 f (0004x) for all x in the domain of f, then f is an even function. If f(0004x) 0002 0004f(x) for all x in the domain of f, then f is an odd function.
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The graph of an even function is symmetric with respect to the y axis and the graph of an odd function is symmetric with respect to the origin (Fig. 15). f (x)
f (x) f
f
f (x)
f (⫺x) ⫽ f (x) ⫺x
x
f (x)
⫺x
x
f (⫺x) ⫽ ⫺f (x)
x
x
Even function (symmetric with respect to y axis)
Odd function (symmetric with respect to origin)
Z Figure 15 Even and odd functions.
EXAMPLE
6
Testing for Even and Odd Functions Determine whether the functions f, g, and h are even, odd, or neither. (A) f(x) ⫽ x4 ⫹ 1
SOLUTIONS
(B) g(x) ⫽ x3 ⫹ 1
(C) h(x) ⫽ x5 ⫹ x
It will be useful to note the following: if n is an even integer, then (⫺x)n ⫽ (⫺1)n xn ⫽ xn because (⫺1)n ⫽ 1 if n is even. But if n is an odd integer, (⫺x)n ⫽ (⫺1)n xn ⫽ ⫺xn because (⫺1)n ⫽ ⫺1 when n is odd. (A)
f (x) ⫽ x4 ⫹ 1 f (⫺x) ⫽ (⫺x)4 ⫹ 1 ⫽ x4 ⫹ 1 ⫽ f(x)
(⫺x)4 ⫽ x4 because 4 is even.
This shows that f is even. (B)
g(x) ⫽ x3 ⫹ 1 g(⫺x) ⫽ (⫺x)3 ⫹ 1 ⫽ ⫺x3 ⫹ 1 ⫺g(x) ⫽ ⫺(x3 ⫹ 1) ⫽ ⫺x3 ⫺ 1
(⫺x)3 ⫽ ⫺x3 because 3 is odd.
Distribute the negative.
The function g(⫺x) is neither g(x) nor ⫺g(x), so g is neither even nor odd. (C)
h(x) ⫽ x5 ⫹ x h(⫺x) ⫽ (⫺x)5 ⫹ (⫺x) ⫽ ⫺x5 ⫺ x ⫺h(x) ⫽ ⫺(x5 ⫹ x) ⫽ ⫺x5 ⫺ x
(⫺x)5 ⫽ ⫺x5 because 5 is odd.
Distribute the negative.
Since h(⫺x) ⫽ ⫺h(x), h is odd. MATCHED PROBLEM 6
0002
Determine whether the functions F, G, and H are even, odd, or neither: (A) F(x) ⫽ x3 ⫺ 2x
(B) G(x) ⫽ x2 ⫹ 1
(C) H(x) ⫽ 2x ⫹ 4
0002
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ANSWERS TO MATCHED PROBLEMS 1. (A) The graph of y 0002 1x 0003 3 is the same as the graph of y 0002 1x shifted upward 3 units, and the graph of y 0002 1x 0004 1 is the same as the graph of y 0002 1x shifted downward 1 unit. The figure confirms these conclusions. (B) The graph of y 0002 1x 0003 3 is the same as the graph of y 0002 1x shifted to the left 3 units, and the graph of y 0002 1x 0004 1 is the same as the graph of y 0002 1x shifted to the right 1 unit. The figure confirms these conclusions. y
y y 0002 兹x 0003 3
5
00045
y 0002 兹x y 0002 兹x 0004 1 x
5
5
y 0002 兹x 0003 3 y 0002 兹x y 0002 兹x 0004 1
00045
00045
5
x
00045
2. G(x) 0002 (x 0003 3)3, H(x) 0002 (x 0004 1)3, M(x) 0002 x3 0003 3, N(x) 0002 x3 0004 4 (B) The graph of y 0002 f(0004x) is the 3. (A) The graph of y 0002 0004f(x) is the reflection through the y axis of the reflection through the x axis of graph of y 0002 f(x). the graph of y 0002 f(x). y
y 5
5
(00042, 4)
(2, 4) y 0002 f(x) 00045
5
y 0002 0004f(x)
x
(2, 4) y 0002 f(0004x)
y 0002 f(x)
5
00045
x
(2, 00044) 00045
00045
(C) The graph of y 0002 0004f (0004x) is the reflection through the origin of the graph of y 0002 f(x). y 5
y 0002 f(x) (2, 4)
5
00045
(00042, 00044) 00045
x
y 0002 0004f(0004x)
4. (A) The graph of y 0002 2f (x) is a vertical stretch of the graph of y 0002 f (x) by a factor of 2. The graph of y 0002 12 f (x) is a vertical shrink of the graph of y 0002 f (x) by a factor of 12.
(B) The graph of y 0002 f(2x) is a horizontal shrink of the graph of y 0002 f(x) by a factor of 12. The graph of y 0002 f (12 x) is a horizontal stretch of the graph of y 0002 f(x) by a factor of 2.
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199
Transformations of Functions
y
10
10
(1, 6)
y 0002 8 0003 2x 2 y 0002 4 0003 x2 1 y 0002 2 0003 2 x2
(1, 3) (1, 32 )
00035
5
(1, 0) (2, 0) (4, 0)
1
y 0002 4 0003 4 x2 x
00035
5
x
y 0002 4 0003 x2 000310
000310
y 0002 4 0003 4x 2
5. The graph of function h is a reflection through the x axis and a horizontal translation of three units to the left of the graph of y 0002 x3. An equation for h is h(x) 0002 0003(x 0004 3)3. 6. (A) Odd (B) Even (C) Neither
3-3
Exercises
1. Explain why the graph of y 0002 f (x) 0004 k is the same as the graph of y 0002 f (x) moved upward k units when k is positive. 2. Explain why the graph of y 0002 Af (x) is a vertical stretch of the graph of y 0002 f (x) when A 0005 1, and a vertical shrink when A 0006 1.
18. h(x) 0002 f(x 0003 1)
g (x)
19. h(x) 0002 0003f(x)
5
20. h(x) 0002 0003g(x)
3. Explain why the graph of y 0002 0003f (x) is a reflection of the graph of y 0002 f (x) about the x axis, and why the graph of y 0002 f (0003x) is a reflection about the y axis.
21. h(x) 0002 2g(x)
4. Is every function either even or odd? Explain your answer.
23. h(x) 0002 g(2x)
In Problems 5–10, without looking back in the text, indicate the domain and range of each of the following functions. (Making rough sketches on scratch paper may help.) 3
5. h(x) 0002 0003 1x
00035
5
x
00035
1 24. h(x) 0002 f a xb 2 25. h(x) 0002 f (0003x) 26. h(x) 0002 0003g(0003x)
6. m(x) 0002 0003 1 x
7. g(x) 0002 00032x2
1 22. h(x) 0002 f (x) 2
8. f (x) 0002 00030.5|x| 3
9. F(x) 0002 00030.5x
Indicate whether each function in Problems 27–36 is even, odd, or neither.
10. G(x) 0002 4x3
Problems 11–26 refer to the functions f and g given by the graphs below. The domain of each function is [00032, 2], the range of f is [00032, 2], and the range of g is [00031, 1]. Use the graph of f or g, as required, to graph the function h and state the domain and range of h. 11. h(x) 0002 f (x) 0004 2
f (x)
12. h(x) 0002 g(x) 0003 1
5
13. h(x) 0002 g(x) 0004 2 14. h(x) 0002 f(x) 0003 1 15. h(x) 0002 f(x 0003 2)
00035
5
16. h(x) 0002 g(x 0003 1) 17. h(x) 0002 g(x 0004 2)
00035
x
27. g(x) 0002 x3 0004 x
28. f(x) 0002 x5 0003 x
29. m(x) 0002 x4 0004 3x2
30. h(x) 0002 x4 0003 x2
31. F(x) 0002 x5 0004 1
32. f (x) 0002 x5 0003 3
33. G(x) 0002 x4 0004 2
34. P(x) 0002 x4 0003 4
35. q(x) 0002 x2 0004 x 0003 3
36. n(x) 0002 2x 0003 3
In Problems 37–44, the graph of the function g is formed by applying the indicated sequence of transformations to the given function f. Find an equation for the function g. Check your work by graphing f and g in a standard viewing window. 3 37. The graph of f (x) 0002 1x is shifted four units to the left and five units down.
38. The graph of f (x) 0002 x3 is shifted five units to the right and four units up.
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39. The graph of f (x) 0002 1x is shifted six units up, reflected in the x axis, and vertically shrunk by a factor of 0.5.
y
65. 5
40. The graph of f (x) 0002 1x is shifted two units down, reflected in the x axis, and vertically stretched by a factor of 4. 41. The graph of f (x) 0002 x2 is reflected in the x axis, vertically stretched by a factor of 2, shifted four units to the left, and shifted two units down. 42. The graph of f (x) 0002 冟 x 冟 is reflected in the x axis, vertically shrunk by a factor of 0.5, shifted three units to the right, and shifted four units up.
00045
y
66. 5
3 44. The graph of f (x) 0002 1x is horizontally shrunk by a factor of 2, shifted three units up, and reflected in the y axis.
45. f (x) 0002 4x
1 46. g(x) 0002 1x 3
47. h(x) 0002 |x 0003 2|
48. k(x) 0002 |x 0004 4|
49. m(x) 0002 0004|4x 0004 8|
50. n(x) 0002 0004|9 0003 3x|
51. p(x) 0002 3 0004 1x
52. q(x) 0002 00042 0003 1x 0003 3
53. r(x) 0002 3 1x 0004 1 0003 2
54. s(x) 0002 1x 0004 1 0003 2
2
2
55. h(x) 0002 x 0003 3
56. h(x) 0002 4 0004 x2
57. k(x) 0002 2x3 0003 1
58. h(x) 0002 3x3 0004 1
59. n(x) 0002 (x 0003 2)2
60. m(x) 0002 (x 0004 4)2
61. q(x) 0002 4 0004
1 (x 0004 2)2 2
x
00045
43. The graph of f (x) 0002 1x is horizontally stretched by a factor of 0.5, reflected in the y axis, and shifted two units to the left.
Use graph transformations to sketch the graph of each function in Problems 45–62.
5
00045
5
x
00045
y
67. 5
00045
5
x
00045
y
68. 5
2 62. p(x) 0002 5 0004 (x 0003 3)2 3
00045
5
x
Each graph in Problems 63–78 is a transformation of one of the six basic functions in Figure 1. Find an equation for the given graph. 00045
y
63.
y
69.
5
5
00045
5
x 00045
5
x
00045 00045
y
64.
y
70.
5
5
00045
5
x 00045
5
00045 00045
x
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SECTION 3–3 y
71.
5
5
x
00045
y
5
00045
5
x
00045
00045
5
00045
5
x
00045
5
00045
5
x
00045
y 5
5
00045
00045
5
x
00045
y
00045
x
y
78.
5
75.
5
00045
y
74.
x
y
77.
5
73.
5
00045
00045
72.
201
y
76.
5
00045
Transformations of Functions
x
3 3 79. Consider the graphs of f (x) 0002 18x and g (x) 0002 2 1x . 3 (A) Describe each as a stretch or shrink of y 0002 1x. (B) Graph both functions in the same viewing window on a graphing calculator. What do you notice? (C) Rewrite the formula for f algebraically to show that f and g are in fact the same function. (This shows that for some functions, a horizontal stretch or shrink can also be interpreted as a vertical stretch or shrink.)
80. Consider the graphs of f (x) 0002 (3x)3 and g(x) 0002 27x3. (A) Describe each as a stretch or shrink of y 0002 x3. (B) Graph both functions in the same viewing window on a graphing calculator. What do you notice? (C) Rewrite the formula for f algebraically to show that f and g are in fact the same function. (This shows that for some functions, a horizontal stretch or shrink can also be interpreted as a vertical stretch or shrink.) 81. (A) Starting with the graph of y 0002 x2, apply the following transformations. (i) Shift downward 5 units, then reflect in the x axis. (ii) Reflect in the x axis, then shift downward 5 units. What do your results indicate about the significance of order when combining transformations? (B) Write a formula for the function corresponding to each of the above transformations. Discuss the results of part A in terms of order of operations.
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82. (A) Starting with the graph of y 0002 冟 x 冟, apply the following transformations. (i) Stretch vertically by a factor of 2, then shift upward 4 units. (ii) Shift upward 4 units, then stretch vertically by a factor of 2. What do your results indicate about the significance of order when combining transformations? (B) Write a formula for the function corresponding to each of the above transformations. Discuss the results of part A in terms of order of operations.
95. Let f be any function with the property that 0004x is in the domain of f whenever x is in the domain of f, and let E and O be the functions defined by
83. Based on the graphs of the six elementary functions in Figure 1, which are odd, which are even, and which are neither? Use the definitions of odd and even functions to prove your answers.
96. Let f be any function with the property that –x is in the domain of f whenever x is in the domain of f, and let g(x) 0002 xf(x). (A) If f is even, is g even, odd, or neither? (B) If f is odd, is g even, odd, or neither?
84. Based on the results of Example 6, why do you think the terms “even” and “odd” are used to describe functions with particular symmetry properties? Changing the order in a sequence of transformations may change the final result. Investigate each pair of transformations in Problems 85–90 to determine if reversing their order can produce a different result. Support your conclusions with specific examples and/or mathematical arguments.
E(x) 0002 12 [ f (x) 0003 f (0004x)] and O(x) 0002 12 [ f (x) 0004 f (0004x)] (A) Show that E is always even. (B) Show that O is always odd. (C) Show that f (x) 0002 E(x) 0003 O(x). What is your conclusion?
APPLICATIONS 97. PRODUCTION COSTS Total production costs for a product can be broken down into fixed costs, which do not depend on the number of units produced, and variable costs, which do depend on the number of units produced. So, the total cost of producing x units of the product can be expressed in the form C(x) 0002 K 0003 f(x)
85. Vertical shift, horizontal shift
where K is a constant that represents the fixed costs and f (x) is a function that represents the variable costs. Use the graph of the variable-cost function f(x) shown in the figure to graph the total cost function if the fixed costs are $30,000.
86. Vertical shift, reflection in y axis 87. Vertical shift, reflection in x axis 88. Vertical shift, expansion
f (x)
89. Horizontal shift, reflection in x axis
150,000
Problems 91–94 refer to two functions f and g with domain [00045, 5] and partial graphs as shown here. f (x)
g (x)
5
5
00045
5
00045
x
00045
5
x
00045
91. Complete the graph of f over the interval [00045, 0], given that f is an even function. 92. Complete the graph of f over the interval [00045, 0], given that f is an odd function. 93. Complete the graph of g over the interval [00045, 0], given that g is an odd function. 94. Complete the graph of g over the interval [00045, 0], given that g is an even function.
Variable production costs
90. Horizontal shift, contraction
100,000
50,000
500
1,000
x
Units produced
98. COST FUNCTIONS Refer to the variable-cost function f (x) in Problem 97. Suppose construction of a new production facility results in a 25% decrease in the variable cost at all levels of output. If F is the new variable-cost function, use the graph of f to graph y 0002 F(x), then graph the total cost function for fixed costs of $30,000. 99. TIMBER HARVESTING To determine when a forest should be harvested, forest managers often use formulas to estimate the number of board feet a tree will produce. A board foot equals 1 square foot of wood, 1 inch thick. Suppose that the number of board feet y yielded by a tree can be estimated by y 0002 f (x) 0002 C 0003 0.004(x 0004 10)3
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where x is the diameter of the tree in inches measured at a height of 4 feet above the ground and C is a constant that depends on the species being harvested. Graph y 0002 f(x) for C 0002 10, 15, and 20 simultaneously in the viewing window with Xmin 0002 10, Xmax 0002 25, Ymin 0002 10, and Ymax 0002 35. Write a brief verbal description of this collection of functions.
Quadratic Functions
203
a brief verbal description of this collection of functions. Based on the graphs, do larger values of C correspond to a larger or smaller opening? 4 feet
100. SAFETY RESEARCH If a person driving a vehicle slams on the brakes and skids to a stop, the speed v in miles per hour at the time the brakes are applied is given approximately by
4 feet
4 feet
v 0002 f (x) 0002 C 1x where x is the length of the skid marks and C is a constant that depends on the road conditions and the weight of the vehicle. The table lists values of C for a midsize automobile and various road conditions. Graph v 0002 f (x) for the values of C in the table simultaneously in the viewing window with Xmin 0002 0, Xmax 0002 100, Ymin 0002 0, and Ymax 0002 60. Write a brief verbal description of this collection of functions. Road Condition
C
Wet (concrete)
3.5
Wet (asphalt)
4
Dry (concrete)
5
Dry (asphalt)
5.5
101. FLUID FLOW A cubic tank is 4 feet on a side and is initially full of water. Water flows out an opening in the bottom of the tank at a rate proportional to the square root of the depth (see the figure). Using advanced concepts from mathematics and physics, it can be shown that the volume of the water in the tank t minutes after the water begins to flow is given by 64 V(t) 0002 2 (C 0004 t)2 C
Figure for 101
102. EVAPORATION A water trough with triangular ends is 9 feet long, 4 feet wide, and 2 feet deep (see the figure). Initially, the trough is full of water, but due to evaporation, the volume of the water in the trough decreases at a rate proportional to the square root of the volume. Using advanced concepts from mathematics and physics, it can be shown that the volume after t hours is given by V(t) 0002
0 t 6 |C|
where C is a constant. Sketch by hand the graphs of y 0002 V(t) for C 0002 00044, 00045, and 00046. Write a brief verbal description of this collection of functions. Based on the graphs, do values of C with a larger absolute value correspond to faster or slower evaporation? 4 feet 9 feet
0tC
where C is a constant that depends on the size of the opening. Sketch by hand the graphs of y 0002 V(t) for C 0002 1, 2, 4, and 8. Write
3-4
1 (t 0003 6C)2 C2
2 feet
Quadratic Functions Z Graphing Quadratic Functions Z Modeling with Quadratic Functions Z Solving Quadratic Inequalities Z Modeling with Quadratic Regression
The graph of the squaring function h(x) 0002 x2 is shown in Figure 1 on page 204. Notice that h is an even function; that is, the graph of h is symmetric with respect to the y axis. Also, the lowest point on the graph is (0, 0). Let’s explore the effect of applying a sequence of basic transformations to the graph of h.
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Indicate how the graph of each function is related to the graph of h(x) 0002 x2. Discuss the symmetry of the graphs and find the highest or lowest point, whichever exists, on each graph.
ZZZ EXPLORE-DISCUSS 1
(A) f (x) 0002 (x 0004 3)2 0004 7 0002 x2 0004 6x 0003 2 (B) g(x) 0002 0.5(x 0003 2)2 0003 3 0002 0.5x2 0003 2x 0003 5 (C) m(x) 0002 0004(x 0004 4)2 0003 8 0002 0004x2 0003 8x 0004 8 (D) n(x) 0002 00043(x 0003 1)2 0004 1 0002 00043x2 0004 6x 0004 4
00045
h(x)
Z Graphing Quadratic Functions
5
Graphing the functions in Explore-Discuss 1 produces figures similar in shape to the graph of the squaring function in Figure 1. These figures are called parabolas. The functions that produced these parabolas are examples of the important class of quadratic functions, which we will now define. 5
x
Z DEFINITION 1 Quadratic Functions Z Figure 1 Squaring function
If a, b, and c are real numbers with a 0, then the function
h(x) 0002 x2.
f (x) 0002 ax2 0003 bx 0003 c is called a quadratic function and its graph is called a parabola. This is known as the general form of a quadratic function.
Because the expression ax2 0003 bx 0003 c represents a real number no matter what number we substitute for x, the domain of a quadratic function is the set of all real numbers. We will discuss methods for determining the range of a quadratic function later in this section. Typical graphs of quadratic functions are illustrated in Figure 2. y
Z Figure 2 Graphs of quadratic functions.
10
00045
f(x) 0002 x2 0004 4 (a)
10
10
5
000410
y
y
x
00045
5
x
000410
g(x) 0002 3x 2 0004 12x 0005 14 (b)
00045
5
x
000410
h(x) 0002 3 0004 2x 0004 x2 (c)
We will begin our detailed study of quadratic functions by examining some in a special form, which we will call the vertex form:* f (x) 0002 a(x 0004 h)2 0003 k *In Problem 75 of Exercises 3-4, you will be asked to show that any function of this form fits the definition of quadratic function in Definition 1.
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We’ll see where the name comes from in a bit. For now, refer to Explore-Discuss 1. Any function of this form is a transformation of the basic squaring function g(x) ⫽ x2, so we can use transformations to analyze the graph.
EXAMPLE
1
The Graph of a Quadratic Function Use transformations of g(x) ⫽ x2 to graph the function f (x) ⫽ 2(x ⫺ 3)2 ⫹ 4. Use your graph to determine the graphical significance of the constants 2, 3, and 4 in this function.
SOLUTION
Multiplying by 2 vertically stretches the graph by a factor of 2. Subtracting 3 inside the square moves the graph 3 units to the right. Adding 4 outside the square moves the graph 4 units up. The graph of f is shown in Figure 3, along with the graph of g(x) ⫽ x2.
y ⫽ x2
y 10
y ⫽ 2(x ⫺ 3)2 ⫹ 4
5
(3, 4)
⫺5
5
x
Z Figure 3
The lowest point on the graph of f is (3, 4), so h ⫽ 3 and k ⫽ 4 determine the key point where the graph changes direction. The constant a ⫽ 2 affects the width of the parabola. 0002 MATCHED PROBLEM 1
Use transformations of g(x) ⫽ x2 to graph the function f (x) ⫽ ⫺12(x ⫺ 2)2 ⫹ 5. Use your graph to determine the significance of the constants ⫺12, 2, and 5 in this function. 0002 Every parabola has a point where the graph reaches a maximum or minimum and changes direction. We will call that point the vertex of the parabola. Finding the vertex is key to many of the things we’ll do with parabolas. Example 1 and Explore-Discuss 1 demonstrate that if a quadratic function is in the form f (x) ⴝ a(x ⴚ h)2 ⴙ k, then the vertex is the point (h, k). Next, notice that the graph of h(x) ⫽ x2 is symmetric about the y axis. As a result, the transformation f (x) ⫽ 2(x ⫺ 3)2 ⫹ 4 is symmetric about the vertical line x ⫽ 3 (which runs through the vertex). We will call this vertical line of symmetry the axis, or axis of symmetry of a parabola. If the page containing the graph of f is folded along the line x ⫽ 3, the two halves of the graph will match exactly. Finally, Explore-Discuss 1 illustrates the significance of the constant a in f (x) ⫽ a(x ⫺ h)2 ⫹ k. If a is positive, the graph has a minimum and opens upward. But if a is negative, the graph will be a vertical reflection of h(x) ⫽ x2 and will have a maximum and open downward. The size of a determines the width of the parabola: if 冟 a 冟 7 1, the graph is narrower than h(x) ⫽ x2, and if 冟 a 冟 6 1, it is wider. These properties of a quadratic function in vertex form are summarized next.
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Z PROPERTIES OF A QUADRATIC FUNCTION AND ITS GRAPH Given a quadratic function in vertex form f (x) 0002 a(x 0004 h)2 0003 k
a 0
we summarize general properties as follows: 1. The graph of f is a parabola: f (x)
f (x)
Axis x0002h
Axis x0002h Vertex (h, k)
k
Max f(x)
Vertex (h, k) k
Min f (x) h
x
a00050 Opens upward
h
x
a00060 Opens downward
2. Vertex: (h, k) (parabola rises on one side of the vertex and falls on the other). 3. Axis (of symmetry): x 0002 h (parallel to y axis). 4. f (h) 0002 k is the minimum if a 7 0 and the maximum if a 6 0. 5. Domain: all real numbers; range: (00040007, k] if a 6 0 or [ k, 0007) if a 7 0. 6. The graph of f is the graph of g(x) 0002 ax2 translated horizontally h units and vertically k units.
Now that we can recognize the key properties of quadratic functions in vertex form, the obvious question is “What if a quadratic function is not in vertex form?” More often than not, the quadratic functions we will encounter will be in the form f(x) 0002 ax2 0003 bx 0003 c. The method of completing the square, which we studied in Section 1-5, can be used to find the vertex form in this case.
EXAMPLE
2
Finding the Vertex Form of a Parabola Find the vertex form of f(x) 0002 2x2 0004 8x 0003 4 by completing the square, then write the vertex and the axis.
SOLUTION
We will begin by separating the first two terms with parentheses; then we will complete the square to factor part of f as a perfect square. f (x) 0002 2x2 0004 8x 0003 4 0002 (2x2 0004 8x) 0003 4 0002 2(x2 0004 4x) 0003 4 0002 2(x2 0004 4x 0003 ?) 0003 4 0002 2(x2 0004 4x 0003 4) 0003 4 0004 8 0002 2(x 0004 2)2 0004 4
Group first two terms. Factor out 2. (b a)2 0002 (00042)2 0002 4 Add 4 inside parentheses; because of the 2 in front, we really added 8, so subtract 8 as well. Factor inside parentheses; simplify 4 0004 8.
The vertex form is f (x) 0002 2(x 0004 2)2 0004 4; the vertex is (2, 00044) and the axis is x 0002 2.
0002
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MATCHED PROBLEM 2
EXAMPLE
3
Quadratic Functions
207
Find the vertex form of g(x) 0002 3x2 0004 18x 0003 2 by completing the square, then write the vertex and axis. 0002
Graphing a Quadratic Function Let f(x) 0002 00040.5x2 0004 x 0003 2. (A) Use completing the square to find the vertex form of f. State the vertex and the axis of symmetry. (B) Graph f and find the maximum or minimum of f (x), the domain, the range, and the intervals where f is increasing or decreasing.
SOLUTIONS
(A) Complete the square: f (x) 0002 00040.5x2 0004 x 0003 2 0002 (00040.5x2 0004 x) 0003 2
Group first two terms Factor out 00040.5 Add 1 inside the parentheses to complete the square and 0.5 outside the parentheses.
0002 00040.5(x2 0003 2x 0003 ?) 0003 2 0002 00040.5(x2 0003 2x 0003 1) 0003 2 0003 0.5 0002 00040.5(x 0003 1)2 0003 2.5
Factor the trinomial and combine like terms.
From this last form we see that h 0002 00041 and k 0002 2.5, so the vertex is (00041, 2.5) and the axis of symmetry is x 0002 00041. (B) To graph f, locate the axis and vertex; then plot several points on either side of the axis Axis x 0002 00041
y 5
00045
Vertex (00041, 2.5)
5
00045
x
x
f(x)
00044
00042
00042
2
00041
2.5
0
2
2
00042
The domain of f is (00040007, 0007). From the graph we see that the maximum value is f(00041) 0002 2.5 and that f is increasing on (00040007, 00041] and decreasing on [00041, 0007). Also, y 0002 f(x) can be any number less than or equal to 2.5; the range of f is y 2.5 or (00040007, 2.5]. 0002 MATCHED PROBLEM 3
Let f (x) 0002 0004x2 0003 4x 0003 2. (A) Use completing the square to find the vertex form of f. State the vertex and the axis of symmetry. (B) Graph f and find the maximum or minimum of f (x), the domain, the range, and the intervals where f is increasing or decreasing. 0002
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We can develop a simple formula for finding the vertex of a parabola if we apply completing the square to f (x) ⫽ ax2 ⫹ bx ⫹ c. f (x) ⫽ ax2 ⫹ bx ⫹ c ⫽ a ax2 ⫹
Factor a out of the first two terms. Add a
b ⫹ ?b ⫹ c a
b 2 b inside the parentheses and 2a
subtract a a
b2 b b2 ⫹ 2b ⫹ c ⫺ a 4a 4a 2 2 b b ⫽ a ax ⫹ b ⫹ c ⫺ 2a 4a
⫽ a ax2 ⫹
b2 b 2 outside the parentheses. b ⴝ 4a 2a
Factor the trinomial.
This is in vertex form, and the x coordinate of the vertex is ⫺bⲐ2a. Z FINDING THE VERTEX OF A PARABOLA When a quadratic function is written in the form f (x) ⫽ ax2 ⫹ bx ⫹ c, the first coordinate of the vertex can be found using the formula x⫽⫺
b 2a
The second coordinate can then be found by evaluating f at the first coordinate.
EXAMPLE
4
Graphing a Quadratic Function Let f (x) ⫽ x2 ⫺ 6x ⫹ 4. (A) Use the vertex formula to find the vertex and the axis of symmetry of f. (B) Graph f and find the maximum or minimum of f (x), the domain, the range, and the intervals where f is increasing or decreasing.
SOLUTIONS
(A) Using a ⫽ 1 and b ⫽ ⫺6 in the vertex formula, x⫽⫺
b ⫺6 ⫽⫺ ⫽ 3; f (3) ⫽ 32 ⫺ 6(3) ⫹ 4 ⫽ ⫺5 2a 2
The vertex is (3, ⫺5) and the axis of symmetry is x ⫽ 3. (B) Locate the axis of symmetry, the vertex, and several points on either side of the axis of symmetry, and graph f. y
x
f (x)
0
4
2
⫺4
3
⫺5
4
⫺4
6
4
x⫽3 9
⫺2
8
⫺5
Vertex (3, ⫺5)
x
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The minimum of f(x) is 00045, the domain is (00040007, 0007), the range is [00045, 0007), f is decreasing on (00040007, 3] and increasing on [3, 0007).
0002
Let f (x) 0002 14 x2 0003 12x 0004 5.
MATCHED PROBLEM 4
(A) Use the vertex formula to find the vertex and the axis of symmetry of f. (B) Graph f and find the maximum or minimum of f (x), the domain, the range, and the intervals where f is increasing or decreasing. 0002
EXAMPLE
5
Finding the Equation of a Parabola Find the equation of the parabola with vertex (3, 00042) and x intercept 4. Since the vertex is (3, 00042), the vertex form for the equation is
SOLUTION
f (x) 0002 a(x 0004 3)2 0004 2
h 0002 3, k 0002 00042 in a(x 0004 h)2 0005 k
Since 4 is an x intercept, f (4) 0002 0. Substituting x 0002 4 and f (x) 0002 0 into the vertex formula, we have f (4) 0002 a(4 0004 3)2 0004 2 0002 0 a00022
Add 2 to both sides.
The equation of this parabola is f (x) 0002 2(x 0004 3)2 0004 2 0002 2x2 0004 12x 0003 16 Find the equation of the parabola with vertex (4, 00042) and y intercept 2.
MATCHED PROBLEM 5
0002
0002
We have presented two methods for locating the vertex of a parabola: completing the square and evaluating the vertex formula. You may prefer to use the completing the square process or to remember the formula. Unless directed otherwise, we will leave this choice to you. If you have a graphing calculator, there is a third approach.
Technology Connections The maximum and minimum options on the CALC menu of a graphing calculator can be used to find the vertex of a parabola. After selecting the appropriate option (maximum or minimum), you will be asked to provide three x values: a
left bound, a right bound, and a guess. The maximum or minimum is displayed at the bottom of the screen. Figure 4(a) locates the vertex of the parabola in Example 1 and Figure 4(b) locates the vertex of the parabola in Example 4. 10
5
00045
5
8
000410
00045
(a) f(x) 0002 00040.5x 0004 x 0005 2 2
Z Figure 4
00042
(b) f(x) 0002 x 2 0004 6x 0005 4
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Z Modeling with Quadratic Functions We will now look at some applications that can be modeled using quadratic functions.
EXAMPLE
6
Maximum Area A dairy farm has a barn that is 150 feet long and 75 feet wide. The owner has 240 feet of fencing and plans to use all of it in the construction of two identical adjacent outdoor pens, with part of the long side of the barn as one side of the pens, and a common fence between the two (Fig. 5). The owner wants the pens to be as large as possible.
150 feet
x x 75 feet
y
x
Z Figure 5
(A) Construct a mathematical model for the combined area of both pens in the form of a function A(x) (see Fig. 5) and state the domain of A. (B) Find the value of x that produces the maximum combined area. (C) Find the dimensions and the area of each pen. SOLUTIONS
(A) The combined area of the two pens is A 0002 xy Adding up the lengths of all four segments of fence, we find that building the pens will require 3x 0003 y feet of fencing. We have 240 feet of fence to use, so 3x 0003 y 0002 240 y 0002 240 0004 3x Because the distances x and y must be nonnegative, x and y must satisfy x 0 and y 0002 240 0004 3x 0. It follows that 0 x 80. Substituting for y in the combined area equation, we have the following model for this problem: A(x) 0002 x(240 0004 3x) 0002 240x 0004 3x2
0 x 80
(B) The function A(x) 0002 240x 0004 3x2 is a parabola that opens downward, so the maximum value of area will occur at the vertex. b 240 00020004 0002 40; 2a 2(00043) A(40) 0002 240(40) 0004 3(40)2 0002 4,800 x00020004
A value of x 0002 40 gives a maximum area of 4,800 square feet.
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(C) When x 0002 40, y 0002 240 0004 3(40) 0002 120. Each pen is x by y 2, or 40 feet by 60 feet. The area of each pen is 40 feet 60 feet 0002 2,400 square feet. 0002 MATCHED PROBLEM 6
Repeat Example 6 with the owner constructing three identical adjacent pens instead of two. 0002 The great sixteenth-century astronomer and physicist Galileo was the first to discover that the distance an object falls is proportional to the square of the time it has been falling. This makes quadratic functions a natural fit for modeling falling objects. Neglecting air resistance, the quadratic function h(t) 0002 h0 0004 16t2 represents the height of an object t seconds after it is dropped from an initial height of h0 feet. The constant 000416 is related to the force of gravity and is dependent on the units used. That is, 000416 only works for distances measured in feet and time measured in seconds. If the object is thrown either upward or downward, the quadratic model will also have a term involving t. (See Problems 93 and 94 in Exercises 3-4.)
EXAMPLE
7
Projectile Motion As a publicity stunt, a late-night talk show host drops a pumpkin from a rooftop that is 200 feet high. When will the pumpkin hit the ground? Round your answer to two decimal places.
SOLUTION
Because the initial height is 200 feet, the quadratic model for the height of the pumpkin is h(t) 0002 200 0004 16t2 Because h(t) 0002 0 when the pumpkin hits the ground, we must solve this equation for t. h(t) 0002 200 0004 16t2 0002 0 Add 16t 2 to both sides. 2 16t 0002 200 Divide both sides by 16. 200 t2 0002 0002 12.5 Take the square root of both sides. 16 t 0002 112.5 Only the positive solution is relevant. ⬇ 3.54 seconds
MATCHED PROBLEM 7
0002
A watermelon is dropped from a rooftop that is 300 feet high. When will the melon hit the ground? Round your answer to two decimal places. 0002
Z Solving Quadratic Inequalities Given a quadratic function f (x) 0002 ax2 0003 bx 0003 c, a 0, the zeros of f are the solutions of the quadratic equation ax2 0003 bx 0003 c 0002 0
(1)
(see Section 1-5). If the equal sign in equation (1) is replaced with 0006, 0005, , or , the result is a quadratic inequality in standard form. Just as was the case with linear inequalities (see Section 1-2), the solution set for a quadratic inequality is the subset of the real number line that makes the inequality a true statement. We can identify this subset by examining the graph of a quadratic function. We begin with a specific example and then generalize the results. The graph of f (x) 0002 x2 0004 2x 0004 3 0002 (x 0004 3)(x 0003 1)
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is shown in Figure 6. Information obtained from the graph is listed in Table 1. y
Table 1
5
f(x) > 0 f(x) > 0 (00040007, 00041) (00041, 3) (3, 0007) x )( )(
00045
5
00045
f(x) < 0
Z Figure 6
x
f (x)
00040007 0006 x 0006 00041
Positive
x 0002 00041
Zero
00041 0006 x 0006 3
Negative
x00023
Zero
30006x00060007
Positive
y 0002 f(x) 0002 x 0004 2x 0004 3 0002 (x 0004 3)(x 0003 1) 2
Because we now know where the output of f is positive, negative, and zero, we can use the graph or the table to solve a number of related inequalities involving f. For example, f (x) 7 0 on (00040007, 00041) 0006 (3, 0007)
and
f (x) 0 on [ 00041, 3]
The key steps in the preceding process are summarized in the box. Z SOLVING A QUADRATIC INEQUALITY 1. Write the inequality in standard form (a form where one side of the inequality defines a quadratic function f and the other side is 0). 2. Find the zeros of f. 3. Graph f and plot its zeros. 4. Use the graph to identify the intervals on the x axis that satisfy the original inequality.
EXAMPLE
8
Solving a Quadratic Inequality Solve: x2 0004 4x 14
SOLUTION
Step 1. Write in standard form. x2 0004 4x 14 x2 0004 4x 0004 14 0 f (x) 0002 x2 0004 4x 0004 14 0
y f(x) 0002 x2 0004 4x 0004 14
2
Step 2. Solve: f(x) 0002 x 0004 4x 0004 14 0002 0
10
Subtract 14 from both sides. Write using function notation. Standard form
Use the quadratic formula with a 0002 1, b 0002 00044, and c 0002 000414.
2
x0002 2 0004 3√2 000410
2 0003 3√2 x 10
(0, 000414) 000420
Z Figure 7
(4, 000414) (2, 000418)
0004b 2b 0004 4ac 2a
0004(00044) 2(00044)2 0004 4(1)(000414) 2(1) 4 172 4 612 0002 0002 2 2 0002 2 312
0002
Divide both terms in numerator by 2.
The zeros of f are 2 0004 312 ⬇ 00042.24 and 2 0003 312 ⬇ 6.24. Step 3. Plot these zeros, along with a few other points, and graph f (Figure 7).
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Step 4. We need to identify intervals where f (x) 0. From the graph we see that f(x) 0 for x 2 0004 312 and for x 2 0003 312. Returning to the original inequality, the solution to x2 0004 4x 14 MATCHED PROBLEM 8
EXAMPLE
9
(00040007, 2 0004 312] 0006 [2 0003 312, 0007)
is
Solve: x2 0003 6x 6
0002 0002
Break-Even, Profit, and Loss
Table 2 Price–Demand Data Table 2 contains price–demand data for a paint manufacturer. A linear regression model for this data is Weekly Sales (in gallons)
Price per Gallon
1,400
$43.00
2,550
$37.25
3,475
$32.60
4,856
$25.72
5,625
$21.88
6,900
$15.50
SOLUTIONS
p 0002 50 0004 0.005x
Price–demand equation
where x is the weekly sales (in gallons) and $p is the price per gallon. The manufacturer has weekly fixed costs of $58,500 and variable costs of $3.50 per gallon produced. (A) Find the weekly revenue function R and weekly cost function C as functions of the sales x. What is the domain of each function? (B) Graph R and C on the same coordinate axes and find the level of sales for which the company will break even. (C) Describe verbally and graphically the sales levels that result in a profit and those that result in a loss. (A) If x gallons of paint are sold weekly at a price of $p per gallon, then the weekly revenue is R 0002 xp 0002 x(50 0004 0.005x) 0002 50x 0004 0.005x2 Since the sales x and the price p cannot be negative, x must satisfy x000b0
and
p 0002 50 0004 0.005x 0
Subtract 50 from both sides.
0004 0.005x 000450 000450 x 0002 10,000 00040.005
Divide both sides by 00040.005 and reverse the inequality. Simplify.
The revenue function and its domain are R(x) 0002 50x 0004 0.005x2
0 x 10,000
The cost of producing x gallons of paint weekly is C(x) 0002 58,500 0003 3.5x
x000b0
Fixed costs 0005 $3.50 times number of gallons
(B) The graph of C is a line and the graph of R is a parabola opening downward. Using the vertex formula, b 50 00020004 0002 5,000 2a 2(00040.005) R(5,000) 0002 50(5,000) 0004 0.005(5,000)2 0002 125,000 x00020004
The vertex is (5,000, 125,000).
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After plotting a few points (Table 3), we sketch the graphs of R and C (Fig. 8). y
Table 3 x
R(x)
C(x)
150,000
0
0
58,500
5,000
125,000
76,000
10,000
0
93,500
y ⫽ R(x) ⫽ 50x ⫺ 0.005x2
100,000
y ⫽ C(x) ⫽ 58,500 ⫹ 3.5x 50,000
Break-even points (
Loss
Profit
)( 1,500
Loss )( ) 7,800 10,000
x
Z Figure 8 Profit when R ⬎ C; loss when R ⬍ C
The company breaks even if cost equals revenue: C(x) ⫽ R(x) 58,500 ⫹ 3.5x ⫽ 50x ⫺ 0.005x2 0.005x2 ⫺ 46.5x ⫹ 58,000 ⫽ 0
x⫽
46.5 ⫾ 246.52 ⫺ 4(0.005)(58,500) 2(0.005)
⫽
Use the quadratic formula with a ⴝ 0.005, b ⴝ ⴚ46.5, and c ⴝ 58,000.
46.5 ⫾ 1992.25 46.5 ⫾ 31.5 ⫽ 0.01 0.01
⫽ 1,500 or 7,800 Now we find the corresponding points on the graph: C(1,500) ⫽ R(1,500) ⫽ $63,750 C(7,800) ⫽ R(7,800) ⫽ $85,800 The graphs of C and R intersect at the points (1,500, 63,750) and (7,800, 85,800) (see Figure 8). These intersection points are called the break-even points. (C) If the company produces and sells between 1,500 and 7,800 gallons of paint weekly, then R ⬎ C and the company will make a profit. These sales levels are shown in blue in Figure 8. If it produces and sells between 0 and 1,500 gallons or between 7,800 and 10,000 gallons of paint, then R ⬍ C and the company will lose money. These sales levels are shown in red in Figure 8. 0002 MATCHED PROBLEM 9
Refer to Example 9. (A) Find the profit function P and state its domain. (B) Find the sales levels for which P(x) ⬎ 0 and those for which P(x) ⬍ 0. (C) Find the maximum profit and the sales level at which it occurs. 0002
Z Modeling with Quadratic Regression We obtained the linear model for the price–demand data in Example 9 by applying linear regression to the data in Table 2. Regression is not limited to just linear functions. In Example 10 we will use a quadratic model obtained by applying quadratic regression to a data set.
EXAMPLE
10
Stopping Distance Automobile accident investigators often use the length of skid marks to approximate the speed of vehicles involved in an accident. The skid mark length depends on a number of factors, including the make and weight of the vehicle, the road surface, and the road
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Table 4 Length of Skid Marks (in feet) Speed (mph)
Wet Asphalt
Dry Concrete
20
22
16
30
49
33
40
84
61
50
137
94
60
197
133
Quadratic Functions
215
conditions at the time of the accident. Investigators conduct tests to determine skid mark length for various vehicles under varying conditions. Some of the test results for a particular vehicle are listed in Table 4. Using the quadratic regression feature on a graphing calculator, (see the Technology Connections following this example) we find a model for the skid mark length on wet asphalt: L(x) 0002 0.06x2 0003 0.42x 0004 6.6 where x is speed in miles per hour. (A) Graph y 0002 L(x) and the data for skid mark length on wet asphalt on the same axes. (B) How fast (to the nearest mile) was the vehicle traveling if it left skid marks 100 feet long? y L(x) 0002 0.06x2 0003 0.42x 0004 6.6
(A) Skid mark length (feet)
SOLUTIONS
300
(60, 197) (50, 137) (40, 84) (30, 49) (20, 22)
50 10
80
x
Speed (mph)
(B) To approximate the speed from the skid mark length, we solve L(x) 0002 100 0.06x 0003 0.42x 0004 6.6 0002 100 0.06x2 0003 0.42x 0003 93.4 0002 0 2
x0002 0002
Use the quadratic formula.
0003(00030.42) 0005 2(00030.42)2 0003 4(0.06)(000393.4) 2(0.06) 0.42 0005 122.5924 0.12
x ⬇ 43 mph
MATCHED PROBLEM 10
Subtract 100 from both sides.
The negative root was discarded.
0002
A model for the skid mark length on dry concrete in Table 4 is M(x) 0002 0.035x2 0004 0.15x 0003 1.6 where x is speed in miles per hour. (A) Graph y 0002 L(x) and the data for skid mark length on dry concrete on the same axes. (B) How fast (to the nearest mile) was the vehicle traveling if it left skid marks 100 feet long? 0002
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Technology Connections we expand our library of functions, we will see that regression can be used to construct models involving these new functions.
Figure 9 shows three of the screens related to the construction of the quadratic model in Example 10 on a Texas Instruments TI-84 Plus. The use of regression to construct mathematical models is not limited to just linear and quadratic models. As
240
0
(a) Enter the data.
(b) Use the QuadReg option on a calculator.
80
0
(c) Graph the data and the model.
Z Figure 9
ANSWERS TO MATCHED PROBLEMS 1.
y
y 0002 x2 10
(2, 5) 000410
10
000410
x
1
y 0002 0004 2 (x 0004 2)2 0003 5
The 000412 makes the graph open downward and vertically shrinks it by a factor of 12, the 2 moves it 2 units right, and the 5 moves it 5 units up. 2. g(x) 0002 3(x 0004 3)2 0004 25; vertex: (3, 000425); axis: x 0002 3 3. (A) Vertex form: f (x) 0002 0004(x 0004 2)2 0003 6; vertex: (2, 6); axis of symmetry: x 0002 2. y (B) 7
00043
x00022
Vertex (2, 6)
7
x
00043
Max f (x) 0002 f (2) 0002 6; domain: (00040007, 0007); range: (00040007, 6]; increasing on (00040007, 2]; decreasing on [2, 0007)
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217
Quadratic Functions
4. (A) Vertex: (00041, 0004214); axis of symmetry: x 0002 00041 y (B) 10
000410
x
10
000410
Vertex (00041, 000421/4)
x 0002 00041
Skid mark length (feet)
Min f (x) 0002 f (00041) 0002 0004214; domain: (00040007, 0007); range: [ 0004214, 0007); decreasing on (00040007, 00041]; increasing on [00041, 0007) 5. y 0002 14(x 0004 4)2 0004 2 0002 0.25x2 0004 2x 0003 2 6. (A) A(x) 0002 (240 0004 4x)x, 0 x 60 (B) The maximum combined area of 3,600 ft.2 occurs at x 0002 30 feet. (C) Each pen is 30 feet by 40 feet with area 1,200 ft.2 7. 4.33 seconds 8. [00043 0004 115, 00043 0003 115] 9. (A) P(x) 0002 46.5x 0004 0.005x2 0004 58,500, 0 x 10,000 (B) Profit is positive for sales between 1,500 and 7,800 gallons per week and negative for sales less than 1,500 or for sales between 7,800 and 10,000. (C) The maximum profit is $49,612.50 at a sales level of 4,650 gallons. y M(x) 0002 0.035x2 0003 0.15x 0004 1.6 10. (A) 200
(60, 133) (50, 94) (40, 61)
50
(30, 33) (20, 16) 10
80
x
Speed (mph)
(B) 52 mph
3-4
Exercises
1. Describe the graph of any quadratic function. 2. How can you tell from a quadratic function whether its graph opens up or down?
In Problems 7–12, find the vertex and axis of the parabola, then draw the graph. 7. f (x) 0002 (x 0003 3)2 0004 4
3. True or False: Every quadratic function has a maximum. Explain.
3 2 9. f (x) 0002 0004ax 0004 b 0004 5 2
4. Using transformations, explain why the vertex of f (x) 0002 a(x 0004 h)2 0003 k is (h, k).
11. f (x) 0002 2(x 0003 10)2 0003 20
5. What information does the constant a provide about the graph of a function of the form f (x) 0002 ax2 0003 bx 0003 c? 6. Explain how to find the maximum or minimum value of a quadratic function.
8. f (x) 0002 (x 0003 2)2 0004 2 10. f (x) 0002 0004ax 0004
11 2 b 00033 2
1 12. f (x) 0002 0004 (x 0003 8)2 0003 12 2
In Problems 13–18, write a brief verbal description of the relationship between the graph of the indicated function and the graph of y 0002 x2. 13. f (x) 0002 (x 0004 2)2 0003 1
14. g(x) 0002 0004(x 0003 1)2 0004 2
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FUNCTIONS
15. h(x) 0002 0004(x 0003 1)2
16. k(x) 0002 (x 0004 2)2
17. m(x) 0002 (x 0004 2)2 0004 3
18. n(x) 0002 0004(x 0003 1)2 0003 4
In Problems 19–24, match each graph with one of the functions in Problems 13–18.
y
24. 5
00045
5
x
y
19. 5
00045
00045
5
x
00045
y
20. 5
00045
5
25. f (x) 0002 x2 0004 4x 0003 5
26. g(x) 0002 x2 0004 6x 0003 1
27. h(x) 0002 0004x2 0004 2x 0004 3
28. k(x) 0002 0004x2 0004 10x 0003 3
29. m(x) 0002 2x2 0004 12x 0003 22 1 7 31. f (x) 0002 x2 0003 3x 0004 2 2 33. f (x) 0002 2x2 0004 24x 0003 90
30. n(x) 0002 3x2 0003 6x 0004 2 3 11 32. g(x) 0002 0004 x2 0003 9x 0003 2 2 34. g(x) 0002 3x2 0003 24x 0003 30
x
In Problems 35–46, use the formula x 0002 0004b 2a to find the vertex. Then write a description of the graph using all of the following words: axis, increases, decreases, range, and maximum or minimum. Finally, draw the graph.
00045
35. f (x) 0002 x2 0003 8x 0003 8
y
21.
In Problems 25–34, complete the square and find the vertex form of each quadratic function, then write the vertex and the axis and draw the graph.
36. f (x) 0002 x2 0003 10x 0003 10
5
37. f (x) 0002 0004x2 0004 7x 0003 4 00045
5
x
38. f (x) 0002 0004x2 0004 11x 0003 1 39. f (x) 0002 4x2 0004 18x 0003 25 40. f (x) 0002 5x2 0003 30x 0004 17 41. f (x) 0002 000410x2 0003 50x 0003 12
00045
42. f (x) 0002 00048x2 0004 24x 0003 16
y
22.
43. f(x) 0002 x2 0003 3x
5
44. f (x) 0002 4x 0004 x2 00045
5
x
y 5
00045
5
00045
46. f(x) 0002 0.4x2 0003 4x 0003 4 In Problems 47–60, solve and write the answer using interval notation.
00045
23.
45. f(x) 0002 0.5x2 0004 2x 0004 7
x
47. x2 0006 10 0004 3x
48. x2 0003 x 0006 12
49. x2 0003 21 0005 10x
50. x2 0003 7x 0003 10 0005 0
51. x2 8x
52. x2 0003 6x 0
53. x2 0003 5x 0
54. x2 4
55. x2 0003 1 0006 2x
56. x2 0003 25 0006 10x
57. x2 0006 3x 0004 3
58. x2 0003 3 0005 2x
59. x2 0004 1 4x
60. 2x 0003 2 0005 x2
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SECTION 3–4
In Problems 61–68, find the standard form of the equation for the quadratic function whose graph is shown.
Quadratic Functions
219
y
66. 9
y
61. 5
(00041, 4)
(0, 5)
(3, 4)
00045
(5, 0) x
(00041, 0)
x
5
00045
(1, 00044)
5
00041
y
67.
00045 5
62.
y
(0, 2.5)
5
(5, 0)
(00041, 0) 00043
00045
(00043, 00041)
x
5
(00041, 00041)
00045
y
68.
(00042, 00044)
x
7
00045 5
y
63.
(0, 2.5)
5
(00041, 4)
(00045, 0)
(00041, 0)
00048
00045
x
5
00045
00045
In Problems 69–74, find the equation of a quadratic function whose graph satisfies the given conditions.
y 5
x
(1, 2)
(00043, 2)
64.
2
69. Vertex: (4, 8); x intercept: 6 70. Vertex: (00042, 000412); x intercept: 00044
(3, 3) (6, 0)
(0, 0) 00042
8
71. Vertex: (00044, 12); y intercept: 4 x
72. Vertex: (5, 8); y intercept: 00042 73. Vertex: (00045, 000425); additional point on graph: (00042, 20) 74. Vertex: (6, 000440); additional point on graph: (3, 50)
00045
75. For f (x) 0002 a(x 0004 h)2 0003 k, expand the parentheses and simplify to write in the form f (x) 0002 ax2 0003 bx 0003 c. This proves that any function in vertex form is a quadratic function as defined in Definition 1.
y
65. 5
(3, 0)
(00041, 0) 00045
5
(0, 00043) 00045
x
76. Find a general formula for the constant term c when expanding f (x) 0002 a(x 0004 h)2 0003 k into the form f (x) 0002 ax2 0003 bx 0003 c. 77. Let g(x) 0002 x2 0003 kx 0003 1. Graph g for several different values of k and discuss the relationship between these graphs. 78. Confirm your conclusions in Problem 77 by finding the vertex form for g.
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79. Let f(x) 0002 (x 0004 1)2 0003 k. Discuss the relationship between the values of k and the number of x intercepts for the graph of f. Generalize your comments to any function of the form f (x) 0002 a(x 0004 h)2 0003 k, a 7 0 80. Let f (x) 0002 0004(x 0004 2)2 0003 k. Discuss the relationship between the values of k and the number of x intercepts for the graph of f. Generalize your comments to any function of the form f (x) 0002 a(x 0004 h)2 0003 k, a 6 0
Horse barn 50 feet x Corral y
81. Find the minimum product of two numbers whose difference is 30. Is there a maximum product? Explain. 82. Find the maximum product of two numbers whose sum is 60. Is there a minimum product? Explain.
APPLICATIONS 83. PROFIT ANALYSIS A consultant hired by a small manufacturing company informs the company owner that their annual profit can be modeled by the function P(x) 0002 00041.2x2 0003 62.5x 0004 491, where x represents the number of employees and P is profit in thousands of dollars. How many employees should the company have to maximize annual profit? What is the maximum annual profit they can expect in that case? 84. PROFIT ANALYSIS The annual profits (in thousands of dollars) from 2000 to 2009 for the company in Problem 83 can be modeled by the function P(t) 0002 6.8t2 0004 80.5t 0003 427.3, 0 t 9, where t is years after 2000. How much profit did the company make in their worst year? 85. MOVIE INDUSTRY REVENUE The annual U.S. box office revenue in billions of dollars for a span of years beginning in 2002 can be modeled by the function B(x) 0002 00040.19x2 0003 1.2x 0003 7.6, 0 x 7, where x is years after 2002. (A) In what year was box office revenue at its highest in that time span? (B) Explain why you should not use the exact vertex in answering part A in this problem. 86. GAS MILEAGE The speed at which a car is driven can have a big effect on gas mileage. Based on EPA statistics for compact cars, the function m(x) 0002 00040.025x2 0003 2.45x 0004 30, 30 x 65, models the average miles per gallon for compact cars in terms of the speed driven x (in miles per hour). (A) At what speed should the owner of a compact car drive to maximize miles per gallon? (B) If one compact car has a 14-gallon gas tank, how much farther could you drive it on one tank of gas driving at the speed you found in part A than if you drove at 65 miles per hour? 87. CONSTRUCTION A horse breeder plans to construct a corral next to a horse barn that is 50 feet long, using all of the barn as one side of the corral (see the figure). He has 250 feet of fencing available and wants to use all of it. (A) Express the area A(x) of the corral as a function of x and indicate its domain. (B) Find the value of x that produces the maximum area. (C) What are the dimensions of the corral with the maximum area?
88. CONSTRUCTION Repeat Problem 87 if the horse breeder has only 140 feet of fencing available for the corral. Does the maximum value of the area function still occur at the vertex? Explain. Problems 89–92 use the falling object function described on page 211. 89. FALLING OBJECT A sandbag is dropped off a high-altitude balloon at an altitude of 10,000 ft. When will the sandbag hit the ground? 90. FALLING OBJECT A prankster drops a water balloon off the top of a 144-ft.-high building. When will the balloon hit the ground? 91. FALLING OBJECT A cliff diver hits the water 2.5 seconds after diving off the cliff. How high is the cliff? 92. FALLING OBJECT A forest ranger drops a coffee cup off a fire watchtower. If the cup hits the ground 1.5 seconds later, how high is the tower? 93. PROJECTILE FLIGHT An arrow shot vertically into the air reaches a maximum height of 484 feet after 5.5 seconds of flight. Let the quadratic function d(t) represent the distance above ground (in feet) t seconds after the arrow is released. (If air resistance is neglected, a quadratic model provides a good approximation for the flight of a projectile.) (A) Find d(t) and state its domain. (B) At what times (to two decimal places) will the arrow be 250 feet above the ground?
94. PROJECTILE FLIGHT Repeat Problem 93 if the arrow reaches a maximum height of 324 feet after 4.5 seconds of flight. 95. ENGINEERING The arch of a bridge is in the shape of a parabola 14 feet high at the center and 20 feet wide at the base (see the figure).
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SECTION 3–4
h(x)
221
(B) Graph y ⫽ L(x) and the data for skid mark length on the same axes. (C) How fast (to the nearest mile per hour) was the car traveling if it left skid marks 150 feet long?
14 ft
x 20 ft
(A) Express the height of the arch h(x) in terms of x and state its domain. (B) Can a truck that is 8 feet wide and 12 feet high pass through the arch? (C) What is the tallest 8-ft.-wide truck that can pass through the arch? (D) What (to two decimal places) is the widest 12-ft.-high truck that can pass through the arch? 96. ENGINEERING The roadbed of one section of a suspension bridge is hanging from a large cable suspended between two towers that are 200 feet apart (see the figure). The cable forms a parabola that is 60 feet above the roadbed at the towers and 10 feet above the roadbed at the lowest point. 200 feet d(x)
Quadratic Functions
98. STOPPING DISTANCE (A) Use the quadratic regression feature on a graphing calculator to find a quadratic model M(x) for the skid mark length for Car B, where x is speed in miles per hour. (Round to two significant digits.) (B) Graph y ⫽ M(x) and the data for skid mark length on the same axes. (C) How fast (to the nearest mile) was the car traveling if it left skid marks 100 feet long? 99. ALCOHOL CONSUMPTION Table 6 contains data related to the per capita ethanol consumption in the United States from 1960 to 2000 (Source: NIAAA). A quadratic regression model for the per capita beer consumption is B(x) ⫽ ⫺0.0006x2 ⫹ 0.03x ⫹ 1 (A) If beer consumption continues to follow the trend exhibited in Table 6, when (to the nearest year) would the consumption return to the 1960 level? (B) What does this model predict for beer consumption in the year 2005? Use the Internet or a library to compare the predicted results with the actual results.
60 feet
Table 6 Per Capita Alcohol Consumption (in gallons) x feet
(A) Express the vertical distance d(x) (in feet) from the roadbed to the suspension cable in terms of x and state the domain of d. (B) The roadbed is supported by seven equally spaced vertical cables (see the figure). Find the combined total length of these supporting cables. 97. STOPPING DISTANCE Table 5 contains data related to the length of the skid marks left by two different cars when making emergency stops.
Year
Beer
Wine
1960
0.99
0.22
1970
1.14
0.27
1980
1.38
0.34
1990
1.34
0.33
2000
1.22
0.31
100. ALCOHOL CONSUMPTION Refer to Table 6. A quadratic regression model for the per capita wine consumption is W(x) ⫽ ⫺0.00016x2 ⫹ 0.009x ⫹ 0.2
Table 5 Length of Skid Marks (in feet)
Speed (mph)
Car A
Car B
20
26
38
30
45
62
40
73
102
50
118
158
60
171
230
(A) If wine consumption continues to follow the trend exhibited in Table 6, when (to the nearest year) would the consumption return to the 1960 level? (B) What does this model predict for wine consumption in the year 2005? Use the Internet or a library to compare the predicted results with the actual results. 101. PROFIT ANALYSIS A screen printer produces custom silkscreen apparel. The cost C(x) of printing x custom T-shirts and the revenue R(x) from the sale of x T-shirts (both in dollars) are given by C(x) ⫽ 245 ⫹ 1.6x R(x) ⫽ 10x ⫺ 0.04x2
(A) Use the quadratic regression feature on a graphing calculator to find a quadratic model L(x) for the skid mark length for Car A, where x is speed in miles per hour. (Round to two significant digits.)
Find the break-even points and determine the sales levels x (to the nearest integer) that will result in the printer showing a profit.
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102. PROFIT ANALYSIS Refer to Problem 101. Determine the sales levels x (to the nearest integer) that will result in the printer showing a profit of at least $60. 103. MAXIMIZING REVENUE A company that manufactures beer mugs has collected the price–demand data in Table 7. A linear regression model for this data is p 0002 d(x) 0002 9.3 0004 0.15x where x is the number of mugs (in thousands) that the company can sell at a price of $p. Find the price that maximizes the company’s revenue from the sale of beer mugs.
where x is the number of gallons of orange juice that can be sold at a price of $p. (A) Find the revenue and cost functions as functions of the sales x. What is the domain of each function? (B) Graph R and C on the same coordinate axes and find the sales levels for which the company will break even. (C) Describe verbally and graphically the sales levels that result in a profit and those that result in a loss. (D) Find the sales and the price that will produce the maximum profit. Find the maximum profit.
Table 9 Orange Juice
Table 7 Demand
Price
45,800
$2.43
40,500
$3.23
37,900
$3.67
34,700
$4.10
30,400
$4.74
28,900
$4.97
25,400
$5.49
Demand
Price
21,800
$1.97
24,300
$1.80
26,700
$1.63
28,900
$1.48
29,700
$1.42
33,700
$1.14
34,800
$1.06
104. MAXIMIZING REVENUE A company that manufactures inexpensive flash drives has collected the price–demand data in Table 8. A linear regression model for this data is
106. BREAK-EVEN ANALYSIS Table 10 contains weekly price– demand data for grapefruit juice for a fruit-juice producer. The producer has weekly fixed cost of $4,500 and variable cost of $0.15 per gallon of grapefruit juice produced. A linear regression model for the data in Table 10 is
p 0002 d(x) 0002 12.3 0004 0.15x
p 0002 d(x) 0002 3 0004 0.0003x
where x is the number of drives (in thousands) that the company can sell at a price of $p. Find the price that maximizes the company’s revenue from the sale of flash drives.
where x is the number of gallons of grapefruit juice that can be sold at a price of $p. (A) Find the revenue and cost functions as functions of the sales x. What is the domain of each function? (B) Graph R and C on the same coordinate axes and find the sales levels for which the company will break even. (C) Describe verbally and graphically the sales levels that result in a profit and those that result in a loss. (D) Find the sales and the price that will produce the maximum profit. Find the maximum profit.
Table 8 Demand
Price
47,800
$5.13
45,600
$5.46
42,700
$5.90
39,600
$6.36
34,700
$7.10
31,600
$7.56
27,800
$8.13
105. BREAK-EVEN ANALYSIS Table 9 contains weekly price– demand data for orange juice for a fruit-juice producer. The producer has weekly fixed cost of $24,500 and variable cost of $0.35 per gallon of orange juice produced. A linear regression model for the data in Table 9 is p 0002 d(x) 0002 3.5 0004 0.00007x
Table 10 Grapefruit Juice Demand
Price
2,130
$2.36
2,480
$2.26
2,610
$2.22
2,890
$2.13
3,170
$2.05
3,640
$1.91
4,350
$1.70
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SECTION 3–5
3-5
Operations on Functions; Composition
223
Operations on Functions; Composition Z Performing Operations on Functions Z Composition Z Mathematical Modeling
Perhaps the most basic thing you’ve done in math classes is operations on numbers: things like addition, subtraction, multiplication, and division. In this section, we will explore the concept of operations on functions. In many cases, combining functions will enable us to model more complex and useful situations. If two functions f and g are both defined at some real number x, then f (x) and g(x) are both real numbers, so it makes sense to perform the four basic arithmetic operations with f(x) and g(x). Furthermore, if g(x) is a number in the domain of f, then it is also possible to evaluate f at g(x). We will see that operations on the outputs of the functions can be used to define operations on the functions themselves.
Z Performing Operations on Functions The functions f and g given by f (x) 0002 2x 0003 3 and g(x) 0002 x2 0004 4 are both defined for all real numbers. Note that f(3) 0002 9 and g(3) 0002 5, so it would seem reasonable to assign the value 9 0003 5, or 14, to a new function ( f 0003 g)(x). Based on this idea, for any real x we can perform the operation f(x) 0003 g(x) 0002 (2x 0003 3) 0003 (x2 0004 4) 0002 x2 0003 2x 0004 1 Similarly, we can define other operations on functions: f (x) 0004 g(x) 0002 (2x 0003 3) 0004 (x2 0004 4) 0002 0004x2 0003 2x 0003 7 f (x)g(x) 0002 (2x 0003 3)(x2 0004 4) 0002 2x3 0003 3x2 0004 8x 0004 12 For x 0005 00062 (to avoid zero in the denominator) we can also form the quotient f (x) 2x 0003 3 0002 2 g(x) x 00044
x 0005 00062
Notice that the result of each operation is a new function. So, we have ( f 0003 g)(x) 0002 f(x) 0003 g(x) 0002 x2 0003 2x 0004 1 ( f 0004 g)(x) 0002 f(x) 0004 g(x) 0002 0004x2 0003 2x 0003 7 ( fg)(x) 0002 f(x)g(x) 0002 2x3 0003 3x2 0004 8x 0004 12 f (x) f 2x 0003 3 0002 2 a b(x) 0002 g g(x) x 00044
x 0005 00062
Sum Difference Product
Quotient
The sum, difference, and product functions are defined for all values of x, as were the original functions f and g, but the domain of the quotient function must be restricted to exclude those values where g(x) 0002 0.
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Z DEFINITION 1 Operations on Functions The sum, difference, product, and quotient of the functions f and g are the functions defined by ( f ⫹ g)(x) ⫽ f (x) ⫹ g(x) ( f ⫺ g)(x) ⫽ f(x) ⫺ g(x) ( fg)(x) ⫽ f(x)g(x) f(x) f a b(x) ⫽ g g(x)
Sum function Difference function Product function
g(x) ⫽ 0
Quotient function
The domain of each function consists of all elements in the domains of both f and g, with the exception that the values of x where g(x) ⫽ 0 must be excluded from the domain of the quotient function.
ZZZ EXPLORE-DISCUSS 1
The following activities refer to the graphs of f and g shown in Figure 1 and the corresponding points on the graph shown in Table 1. Table 1
y
x
10
y ⫽ f (x)
y ⫽ g(x)
10
x
Z Figure 1
f(x)
g(x)
0
8
0
2
7
2
4
6
3
6
5
3
8
4
2
10
3
0
For each of the following functions, construct a table of values, sketch a graph, and state the domain and range. (A) ( f ⫹ g)(x)
EXAMPLE
1
(B) ( f ⫺ g)(x)
(C) ( fg)(x)
f (D) a b(x) g
Finding the Sum, Difference, Product, and Quotient Functions Let f (x) ⫽ 14 ⫺ x and g(x) ⫽ 13 ⫹ x. Find the functions f ⫹ g, f ⫺ g, fg, and f Ⲑg, and find their domains.
SOLUTION
( f ⫹ g)(x) ⫽ f (x) ⫹ g(x) ⫽ ( f ⫺ g)(x) ⫽ f (x) ⫺ g(x) ⫽ ( fg)(x) ⫽ f (x)g(x) ⫽ ⫽ ⫽
14 ⫺ x ⫹ 13 ⫹ x 14 ⫺ x ⫺ 13 ⫹ x 14 ⫺ x 13 ⫹ x 1(4 ⫺ x)(3 ⫹ x) 212 ⫹ x ⫺ x2
f (x) f 14 ⫺ x 4⫺x ⫽ a b(x) ⫽ ⫽ g g(x) 13 ⫹ x A 3 ⫹ x
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SECTION 3–5 Domain of f 00043
4
x
Domain of f: x 4 or (0004, 4] [Fig. 2(a)] Domain of g: x 00043 or [00043, ) [Fig. 2(b)]
(a)
Domain of g
[
4
0
x
The intersection of these domains is shown in Figure 2(c): (0004, 4] 傽 [00043, ) 0002 [00043, 4]
(b)
Domain of f 0003 g, f 0004 g, and fg
[
00043
225
The domains of f and g are [
0
00043
Operations on Functions; Composition
[ 4
0
x
This is the domain of the functions f 0003 g, f 0004 g, and fg. Since g(00043) 0002 0, x 0002 00043 must be excluded from the domain of the quotient function, and
(c)
Domain of
Z Figure 2
MATCHED PROBLEM 1
f : (00043, 4] g
0002
Let f (x) 0002 1x and g(x) 0002 110 0004 x. Find the functions f 0003 g, f 0004 g, fg, and f兾g, and find their domains. 0002
Technology Connections A graphing calculator can be used to check the domains in the solution of Example 1. To check the domain of f ⴙ g, we enter y1 ⴝ 14 ⴚ x, y2 ⴝ 13 ⴙ x, and y3 ⴝ y1 ⴙ y2 in the equation editor of a graphing calculator and graph y3 (Fig. 3).
5
00045
5
00045
5
Z Figure 5 00045
5
Figures 6 and 7 indicate that y3 is not defined for x 0002 4. This confirms that the domain of y3 ⴝ f ⴙ g is [ⴚ3, 4]. 5
00045
Z Figure 3 00045
Next we press TRACE and enter ⴚ3 (Fig. 4). Pressing the left cursor indicates that y3 is not defined for x 0003 ⴚ3 (Fig. 5).
5
00045
Z Figure 6 5
5
00045
5
00045
00045
00045
Z Figure 4
5
Z Figure 7
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2
Finding the Quotient of Two Functions Let f (x) 0002
SOLUTION
f x00044 x . Find the function and find its domain. and g(x) 0002 g x00041 x00033
Because division by 0 must be excluded, the domain of f is all x except x 0002 1 and the domain of g is all x except x 0002 00043. Now we find f兾g. x f (x) f x00041 a b(x) 0002 0002 g g (x) x00044 x00033 x x00033 0002 ⴢ x00041 x00044 x(x 0003 3) 0002 (x 0004 1)(x 0004 4)
(1)
The fraction in equation (1) indicates that 1 and 4 must be excluded from the domain of f兾g to avoid division by 0. But equation (1) does not indicate that 00043 must be excluded also. Although the fraction in equation (1) is defined at x 0002 00043, 00043 was excluded from the domain of g, so it must be excluded from the domain of f兾g also. The domain of f兾g is all real numbers x except 00043, 1, and 4. 0002 MATCHED PROBLEM 2
Let f (x) 0002
f 1 x00045 . Find the function and find its domain. and g (x) 0002 x g x00032
0002
Z Composition Consider the functions f and g given by f (x) 0002 1x and
g(x) 0002 4 0004 2x
Note that g(0) 0002 4 0004 2(0) 0002 4 and f(4) 0002 14 0002 2. So if we apply these two functions consecutively, we get f (g(0)) 0002 f (4) 0002 2 In a diagram, this would look like
x00020
g(x)
4
f (x)
2
When two functions are applied consecutively, we call the result the composition of functions. We will use the symbol f 0002 g to represent the composition of f and g, which we formally define now.
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227
Z DEFINITION 2 Composition of Functions The composition of a function f with another function g is denoted by f 0002 g (read “f composed with g”) and is defined by ( f 0002 g)(x) 0002 f(g(x))
EXAMPLE
3
Computing Composition From a Table Functions f and g are defined by Table 2. Find ( f 0002 g)(2), ( f 0002 g)(5), and ( f 0002 g)(00043). Table 2
SOLUTION
x
f(x)
g(x)
00045
00048
11
00043
00046
2
0
00041
00046
2
5
00043
5
12
0
We will use the formula provided by Definition 2. (f 0002 g)(2) 0002 f (g(2)) 0002 f(00043) 0002 00046 ( f 0002 g)(5) 0002 f (g(5)) 0002 f (0) 0002 00041 ( f 0002 g)(00043) 0002 f(g(00043)) 0002 f(2) 0002 5
MATCHED PROBLEM 3
0002
Functions h and k are defined by Table 3. Find (h 0002 k)(10), (h 0002 k)(00048), and (h 0002 k)(0). Table 3 x
h(x)
k(x)
00048
12
0
00044
18
22
0
40
00044
10
52
00048
20
70
000430
0002
ZZZ
CAUTION ZZZ
When computing f 0002 g, it’s important to keep in mind that the first function that appears in the notation ( f, in this case) is actually the second function that is applied. For this reason, some people read f 0002 g as “f following g.”
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ZZZ EXPLORE-DISCUSS 2
Refer to the functions f and g on page 226, and let h(x) ⫽ (f ⴰ g)(x). Complete Table 4 and graph h. Table 4 x 0
g(x)
h(x) ⴝ f(g(x))
g(0) ⫽ 4 h(0) ⫽ f (g(0)) ⫽ f (4) ⫽ 2
1 2 3 4
The domain of f is {x ƒ x ⱖ 0} and the domain of g is the set of all real numbers. What is the domain of h?
So far, we have looked at composition on a point-by-point basis. Using algebra, we can find a formula for the composition of two functions.
EXAMPLE
4
Finding the Composition of Two Functions Find ( f ⴰ g)(x) for f(x) ⫽ x2 ⫺ x and g(x) ⫽ 3 ⫹ 2x.
SOLUTION
We again use the formula in Definition 2. (f ⴰ g)(x) ⫽ f(g(x)) ⫽ f(3 ⫹ 2x) ⫽ (3 ⫹ 2x)2 ⫺ (3 ⫹ 2x) ⫽ 9 ⫹ 12x ⫹ 4x2 ⫺ 3 ⫺ 2x ⫽ 4x2 ⫹ 10x ⫹ 6
MATCHED PROBLEM 4
0002
Find (h ⴰ k)(x) for h(x) ⫽ 11 ⫹ x2 and k(x) ⫽ 4x ⫺ 1. 0002
ZZZ EXPLORE-DISCUSS 3
(A) For f (x) ⫽ x ⫺ 10 and g(x) ⫽ 3 ⫹ 7x, find ( f ⴰ g)(x) and (g ⴰ f )(x). Based on this result, what do you think is the relationship between f ⴰ g and g ⴰ f in general? x⫺1 . Does this change your thoughts 2 on the relationship between f ⴰ g and g ⴰ f ? (B) Repeat for f (x) ⫽ 2x ⫹ 1 and g(x) ⫽
Explore-Discuss 3 tells us that order is important in composition. Sometimes f ⴰ g and g ⴰ f are equal, but more often they are not. Finding the domain of a composition of functions can sometimes be a bit tricky. Based on the definition ( f ⴰ g)(x) ⫽ f (g(x)), we can see that for an x value to be in the domain of f ⴰ g, two things must occur. First, x must be in the domain of g so that g(x) is defined. Second, g(x) must be in the domain of f, so that f (g(x)) is defined.
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EXAMPLE
5
Operations on Functions; Composition
229
Finding the Composition of Two Functions Find ( f ⴰ g)(x) and (g ⴰ f )(x) and their domains for f (x) ⫽ x10 and g(x) ⫽ 3x4 ⫺ 1.
SOLUTION
( f ⴰ g)(x) ⫽ f (g(x)) ⫽ f(3x4 ⫺ 1) ⫽ (3x4 ⫺ 1)10 (g ⴰ f )(x) ⫽ g( f (x)) ⫽ g(x10) ⫽ 3(x10)4 ⫺ 1 ⫽ 3x40 ⫺ 1 Note that the functions f and g are both defined for all real numbers. If x is any real number, then x is in the domain of g, so g(x) is a real number. This then tells us that g(x) is in the domain of f, which means that f (g(x)) is a real number. In other words, every real number is in the domain of f ⴰ g. Using similar reasoning, we can conclude that the domain of g ⴰ f is also the set of all real numbers. 0002
MATCHED PROBLEM 5
3 Find ( f ⴰ g)(x) and (g ⴰ f )(x) and their domains for f (x) ⫽ 1 x and g(x) ⫽ 7x ⫹ 5.
0002 The line of reasoning used in Example 5 can be used to deduce the following fact: If two functions are both defined for all real numbers, then so is their composition. If either function in a composition is not defined for some real numbers, then, as Example 6 illustrates, the domain of the composition may not be what you first think it should be.
EXAMPLE
6
Finding the Composition of Two Functions Find ( f ⴰ g)(x) for f(x) ⫽ 24 ⫺ x2 and g(x) ⫽ 13 ⫺ x, then find the domain of f ⴰ g.
SOLUTION
We begin by stating the domains of f and g, which is a good idea in any composition problem: Domain f : ⫺2 ⱕ x ⱕ 2 Domain g: x ⱕ 3 or
or [⫺2, 2] (⫺⬁, 3]
Next we find the composition: ( f ⴰ g)(x) ⫽ f(g(x)) ⫽ f (13 ⫺ x) ⫽ 24 ⫺ (13 ⫺ x)2 ⫽ 24 ⫺ (3 ⫺ x) ⫽ 11 ⫹ x
Substitute 13 ⴚ x for g(x). Square: (1t)2 ⴝ t as long as t 0002 0. Subtract.
Although 11 ⫹ x is defined for all x ⱖ ⫺1, we must restrict the domain of f ⴰ g to those values that also are in the domain of g. Domain f ⴰ g: x ⱖ ⫺1 and x ⱕ 3
MATCHED PROBLEM 6
or
[⫺1, 3]
Find f ⴰ g for f (x) ⫽ 29 ⫺ x2 and g(x) ⫽ 1x ⫺ 1, then find the domain of f ⴰ g.
0002
0002
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FUNCTIONS
The domain of f 0002 g cannot always be determined simply by examining the final form of ( f 0002 g)(x). Any numbers that are excluded from the domain of g must also be excluded from the domain of f 0002 g.
CAUTION ZZZ
In calculus, it is not only important to be able to find the composition of two functions, but also to recognize when a given function is the composition of simpler functions.
EXAMPLE
7
Recognizing Composition Forms Express h as a composition of two simpler functions for h(x) 0002 21 0003 3x4
SOLUTION
MATCHED PROBLEM 7
If we were to evaluate this function for some x value, say, x 0002 1, we would do so in two stages. First, we would find the value of 1 0003 3(1)4, which is 4. Then we would apply the square root to get 2. This shows that h can be thought of as two consecutive functions: First, g(x) 0002 1 0003 3x4, then f(x) 0002 1x. So h(x) 0002 f (g(x)), and we have written h as f 0002 g. 0002 Express h as the composition of two simpler functions for h(x) 0002 (4x3 0004 7)4.
0002
The answers to Example 7 and Matched Problem 7 are not unique. For example, if f(x) 0002 11 0003 3x and g(x) 0002 x4, then f (g(x)) 0002 21 0003 3g(x) 0002 21 0003 3x4 0002 h(x)
Z Mathematical Modeling The operations discussed in this section can be applied in many different situations. Example 8 shows how they are used to construct a model in economics.
EXAMPLE
8
Modeling Profit The research department for an electronics firm estimates that the weekly demand for a certain brand of headphones is given by x 0002 f( p) 0002 20,000 0004 1,000p
0 p 20
Demand function
This function describes the number x of pairs of headphones retailers are likely to buy per week at p dollars per pair. The research department also has determined that the total cost (in dollars) of producing x pairs per week is given by C(x) 0002 25,000 0003 3x
Cost function
and the total weekly revenue (in dollars) obtained from the sale of these headphones is given by R(x) 0002 20x 0004 0.001x2
Revenue function
Express the firm’s weekly profit as a function of the price p and find the price that produces the largest profit. What is the largest possible profit?
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SOLUTION
Operations on Functions; Composition
231
The basic economic principle we are using is that profit is revenue minus cost. So the profit function P is the difference of the revenue function R and the cost function C. P(x) ⫽ (R ⫺ C)(x) ⫽ R(x) ⫺ C(x) ⫽ (20x ⫺ 0.001x2) ⫺ (25,000 ⫹ 3x) ⫽ 17x ⫺ 0.001x2 ⫺ 25,000 This is a function of the demand x. We were asked to find the profit P as a function of the price p; we can accomplish this using composition, because x ⫽ f( p). (P ⴰ f )( p) ⫽ P( f ( p)) ⫽ P(20,000 ⫺ 1,000p) ⫽ 17(20,000 ⫺ 1,000p) ⫺ 0.001(20,000 ⫺ 1,000p)2 ⫺ 25,000 ⫽ 340,000 ⫺ 17,000p ⫺ 400,000 ⫹ 40,000p ⫺ 1,000p2 ⫺ 25,000 ⫽ ⫺85,000 ⫹ 23,000p ⫺ 1,000p2 Technically, P ⴰ f and P are different functions, because the first has independent variable p and the second has independent variable x. However, because both functions represent the same quantity (the profit), it is customary to use the same symbol to name each function. So P( p) ⫽ ⫺85,000 ⫹ 23,000p ⫺ 1,000p2 expresses the weekly profit P as a function of price p. Now we can use the vertex formula to find the maximum. p⫽⫺
23,000 b ⫽⫺ ⫽ 11.5 2a ⫺2,000
P(11.5) ⫽ ⫺85,000 ⫹ 23,000(11.5) ⫺ 1,000(11.5)2 ⫽ 47,250 Since a ⬍ 0, the parabola opens downward, and the maximum value of P occurs at the vertex. So the largest profit is $47,250 and it will occur when the price of the headphones is $11.50. 0002
MATCHED PROBLEM 8
Repeat Example 8 for the functions x ⫽ f( p) ⫽ 10,000 ⫺ 1,000p 0 ⱕ p ⱕ 10 C(x) ⫽ 10,000 ⫹ 2x R(x) ⫽ 10x ⫺ 0.001x2
ANSWERS TO MATCHED PROBLEMS 1. ( f ⫹ g)(x) ⫽ 1x ⫹ 110 ⫺ x, ( f ⫺ g)(x) ⫽ 1x ⫺ 110 ⫺ x, ( fg)(x) ⫽ 210x ⫺ x2, ( fⲐg)(x) ⫽ 1xⲐ(10 ⫺ x); the functions f ⫹ g, f ⫺ g, and fg have domain:[0, 10] , the domain of fⲐg is [0, 10) f x 2. a b(x) ⫽ ; domain: all real numbers x except ⫺2, 0, and 5 g (x ⫹ 2)(x ⫺ 5) 3. (h ⴰ k)(10) ⫽ 12; (h ⴰ k)(⫺8) ⫽ 40; (h ⴰ k)(0) ⫽ 18 4. (h ⴰ k)(x) ⫽ 16x2 ⫺ 8x ⫹ 12 3 5. ( f ⴰ g)(x) ⫽ 1 7x ⫹ 5, domain: (⫺⬁, ⬁) 3 (g ⴰ f )(x) ⫽ 71 x ⫹ 5, domain: (⫺⬁, ⬁) 6. ( f ⴰ g)(x) ⫽ 110 ⫺ x; domain: x ⱖ 1 and x ⱕ 10 or [1, 10] 7. h(x) ⫽ ( f ⴰ g)(x) where f (x) ⫽ x4 and g(x) ⫽ 4x3 ⫺ 7 8. P( p) ⫽ ⫺30,000 ⫹ 12,000p ⫺ 1,000p2. The largest profit is $6,000 and occurs when the price is $6.
0002
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Exercises 27. Functions f and g are defined by Table 5. Find ( f ° g)(00027), ( f ° g)(0), and ( f ° g)(4).
1. Explain how to find the sum of two functions. 2. Explain how to find the product of two functions. 3. Describe in your own words what the composition of two functions means. Don’t focus on how to find composition, but rather on what it really means. 4. Is the domain of f0005g always the same as the intersection of the domains of f and g? Explain.
28. Functions h and k are defined by Table 6. Find ( h ° k)(000215), ( h ° k)(000210), and ( h ° k)(15).
Table 5 x
Table 6 f (x)
g(x)
x
h (x)
k(x)
5. When composing two functions, why can’t you always find the domain by simply looking at the simplified form of the composition?
00027
5
4
000220
0002100
30
6. Describe a real-world situation where the composition of two functions would have significance.
00022
9
10
000215
0002200
5
0
0
00022
000210
0002300
15
4
3
6
5
0002150
8
6
000210
00023
15
000290
000210
Problems 7–18 refer to functions f and g whose graphs are shown below. f(x)
g (x)
5
5
00025
5
x
00025
In Problems 29–42, for the indicated functions f and g, find the functions f 0003 g, f 0002 g, fg, and f兾g, and find their domains.
5
x
29. f (x) 0004 4x;
g(x) 0004 x 0003 1
30. f (x) 0004 3x;
g(x) 0004 x 0002 2
2
31. f (x) 0004 2x ; 00025
00025
In Problems 7–10 use the graphs of f and g to construct a table of values and sketch the graph of the indicated function. 7. ( f 0003 g)(x) 9. ( fg)(x)
8. (g 0002 f )(x) 10. ( f 0002 g)(x)
32. f(x) 0004 3x;
g(x) 0004 x2 0003 1 g(x) 0004 x2 0003 4
33. f(x) 0004 3x 0003 5;
g(x) 0004 x2 0002 1
34. f(x) 0004 2x 0002 7;
g(x) 0004 9 0002 x2
35. f (x) 0004 12 0002 x; g(x) 0004 1x 0003 3 36. f (x) 0004 1x 0003 4; g(x) 0004 13 0002 x 37. f (x) 0004 1x 0003 2; g(x) 0004 1x 0002 4
In Problems 11–18, use the graphs of f and g to find each of the following: 11. ( f ° g)(00021)
12. ( f ° g)(2)
13. ( g ° f )(00022)
14. ( g ° f )(3)
15. f (g(1))
16. f(g(0))
17. g( f (2))
18. g( f (00023))
38. f (x) 0004 1 0002 1x; g(x) 0004 2 0002 1x 39. f (x) 0004 2x2 0003 x 0002 6; g(x) 0004 27 0003 6x 0002 x2 40. f (x) 0004 28 0003 2x 0002 x2; g(x) 0004 2x2 0002 7x 0003 10 1 1 41. f (x) 0004 x 0003 ; g(x) 0004 x 0002 x x 42. f (x) 0004 x 0002 1; g(x) 0004 x 0002
In Problems 19–26, find the indicated function value, if it exists, given f(x) 0004 2 0002 x and g(x) 0004 13 0002 x. 19. ( f 0003 g)(00023)
20. (g 0002 f )(00025)
23. ( f ° g)(00022)
f 22. a b(3) g 24. ( f ° g)(1)
25. (g ° f )(1)
26. (g ° g)(00027)
21. ( fg)(00021)
6 x00021
In Problems 43–60, for the indicated functions f and g, find the functions f ° g, and g ° f, and find their domains. 43. f(x) 0004 x3;
g(x) 0004 x2 0002 x 0003 1
44. f(x) 0004 x2;
g(x) 0004 x3 0003 2x 0003 4
45. f(x) 0004 |x 0003 1|;
g(x) 0004 2x 0003 3
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46. f (x) 0002 |x 0004 4|;
y
g(x) 0002 3x 0003 2
100073
47. f (x) 0002 x
3
; g(x) 0002 2x 0003 4
200073
48. f (x) 0002 x
y
5
5
3
; g(x) 0002 8 0004 x
49. f (x) 0002 1x; g(x) 0002 x 0004 4
00045
50. f (x) 0002 1x; g(x) 0002 2x 0003 5 51. f (x) 0002 x 0003 2; g(x) 0002
1 x
52. f (x) 0002 x 0004 3; g(x) 0002
1 x2
5
x
00045
00045
x
(d)
In Problems 65–72, find f ° g and g ° f. Graph f, g, f ° g, and g ° f in the same coordinate system and describe any apparent symmetry between these graphs.
54. f (x) 0002 1x 0004 1; g(x) 0002 x2 55. f (x) 0002
x00035 ; x
56. f (x) 0002
x 2x 0004 4 ; g(x) 0002 x x00041
66. f (x) 0002 3x 0003 2; g(x) 0002 13 x 0004 23
57. f (x) 0002
2x 0003 1 1 ; g(x) 0002 x x00042
68. f (x) 0002 00042x 0003 3; g(x) 0002 000412 x 0003 32
2 2 0004 3x ; g(x) 0002 x x00033
69. f (x) 0002
g(x) 0002
5
00045
(c)
53. f (x) 0002 14 0004 x; g(x) 0002 x2
58. f (x) 0002
233
Operations on Functions; Composition
x x00042
65. f (x) 0002 12 x 0003 1; g(x) 0002 2x 0004 2 67. f (x) 0002 000423 x 0004 53; g(x) 0002 000432 x 0004 52
3 g(x) 0002 22 x
3 70. f (x) 0002 3 2x; g(x) 0002
59. f (x) 0002 225 0004 x2; g(x) 0002 29 0003 x2 60. f (x) 0002 2x2 0004 9; g(x) 0002 2x2 0003 25
x3 27
3 71. f (x) 0002 2x 0004 2; g(x) 0002 x3 0003 2
Use the graphs of functions f and g shown below to match each function in Problems 61–64 with one of graphs (a)–( d). y 0002 f (x) y 5
x3 ; 8
3 72. f (x) 0002 x3 0004 3; g(x) 0002 2x 0003 3
In Problems 73–80, express h as a composition of two simpler functions f and g. 73. h(x) 0002 (2x 0004 7)4
y 0002 g (x)
74. h(x) 0002 (3 0004 5x)7 00045
5
75. h(x) 0002 14 0003 2x
x
76. h(x) 0002 13x 0004 11 77. h(x) 0002 3x7 0004 5
00045
78. h(x) 0002 5x6 0003 3
61. ( f 0003 g)(x)
62. ( f 0004 g)(x)
63. ( g 0004 f )(x)
64. ( fg)(x)
y
79. h(x) 0002
80. h(x) 0002 0004 y
5
5
00045
x
82. Are the functions f ° g and g ° f identical? Justify your answer.
00045
5
x
83. Is there a function g that satisfies f ° g 0002 g ° f 0002 f for all functions f ? If so, what is it? 84. Is there a function g that satisfies fg 0002 gf 0002 f for all functions f ? If so, what is it?
00045
(a)
2 00031 1x
81. Are the functions fg and gf identical? Justify your answer.
5
00045
4 00033 1x
(b)
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In Problems 85–88, for the indicated functions f and g, find the functions f 0003 g, f 0004 g, fg, and f0007g, and find their domains. 1 1 85. f (x) 0002 x 0003 ; g(x) 0002 x 0004 x x 86. f (x) 0002 x 0004 1; g(x) 0002 x 0004 87. f (x) 0002 1 0004
6 x00041
x x ; g(x) 0002 1 0003 冟x冟 冟x冟
92. WEATHER BALLOON A weather balloon is rising vertically. An observer is standing on the ground 100 meters from the point where the weather balloon was released. (A) Express the distance d between the balloon and the observer as a function of the balloon’s distance h above the ground. (B) If the balloon’s distance above ground after t seconds is given by h 0002 5t, express the distance d between the balloon and the observer as a function of t. 93. FLUID FLOW A conical paper cup with diameter 4 inches and height 4 inches is initially full of water. A small hole is made in the bottom of the cup and the water begins to flow out of the cup. Let h and r be the height and radius, respectively, of the water in the cup t minutes after the water begins to flow.
88. f (x) 0002 x 0003 冟 x 冟; g(x) 0002 x 0004 冟 x 冟
APPLICATIONS
4 inches
89. MARKET RESEARCH The demand x and the price p (in dollars) for new release CDs for a large online retailer are related by x 0002 f ( p) 0002 4,000 0004 200p
0 p 20
The revenue (in dollars) from the sale of x units is given by R(x) 0002 20x 0004
r
1 2 x 200
4 inches h
and the cost (in dollars) of producing x units is given by C(x) 0002 2x 0003 8,000 Express the profit as a function of the price p and find the price that produces the largest profit. 90. MARKET RESEARCH The demand x and the price p (in dollars) for portable iPod speakers at a national electronics store are related by x 0002 f(p) 0002 5,000 0004 100p
0 p 50
The revenue (in dollars) from the sale of x units and the cost (in dollars) of producing x units are given, respectively, by R(x) 0002 50x 0004
1 2 x 100
C(x) 0002 20x 0003 40,000
and
Express the profit as a function of the price p and find the price that produces the largest profit.
1
V 0002 3 r 2h
(A) Express r as a function of h. (B) Express the volume V as a function of h. (C) If the height of the water after t minutes is given by h(t) 0002 4 0004 0.51t express V as a function of t. 94. EVAPORATION A water trough with triangular ends is 6 feet long, 4 feet wide, and 2 feet deep. Initially, the trough is full of water, but due to evaporation, the volume of the water is decreasing. Let h and w be the height and width, respectively, of the water in the tank t hours after it began to evaporate.
91. POLLUTION An oil tanker aground on a reef is leaking oil that forms a circular oil slick about 0.1 foot thick (see the figure). The radius of the slick (in feet) t minutes after the leak first occurred is given by r(t) 0002 0.4t100073 Express the volume of the oil slick as a function of t.
r
4 feet 6 feet
2 feet
w h
V 0002 3wh
(A) Express w as a function of h. (B) Express V as a function of h. (C) If the height of the water after t hours is given by h(t) 0002 2 0004 0.21t express V as a function of t.
A 0002 r 2 V 0002 0.1A
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3-6
Inverse Functions
235
Inverse Functions Z One-to-One Functions Z Finding the Inverse of a Function Z Mathematical Modeling Z Graphing Inverse Functions
We have seen that many important mathematical relationships can be expressed in terms of functions. For example, C 0002 d
The circumference of a circle is a function of the diameter d.
V 0002 s3 d 0002 1,000 0004 100p 9 F 0002 C 0003 32 5
The volume of a cube is a function of length s of the edges. The demand for a product is a function of the price p. Temperature measured in °F is a function of temperature in °C.
In many cases, we are interested in reversing the correspondence determined by a function. For our examples, C 3 s0002 1 V
d0002
p 0002 10 0004
The diameter of a circle is a function of the circumference C. The length of the edge of a cube is a function of the volume V.
1 d 100
5 C 0002 (F 0004 32) 9
The price of a product is a function of the demand d.
Temperature measured in °C is a function of temperature in °F.
As these examples illustrate, reversing the correspondence between two quantities often produces a new function. This new function is called the inverse of the original function. Later in this text we will see that many important functions are actually defined as the inverses of other functions. In this section, we develop techniques for determining whether the inverse of a function exists, some general properties of inverse functions, and methods for finding the rule of correspondence that defines the inverse function. A review of function basics in Section 3-1 would be very helpful at this point.
Z One-to-One Functions Recall the set form of the definition of function: A function is a set of ordered pairs with the property that no two ordered pairs have the same first component and different second components. However, it is possible that two ordered pairs in a function could have the same second component and different first components. If this does not happen, then we call the function a one-to-one function. In other words, a function is one-to-one if there are no duplicates among the second components.
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Z DEFINITION 1 One-to-One Function A function is one-to-one if no two ordered pairs in the function have the same second component and different first components.
ZZZ EXPLORE-DISCUSS 1
Given the following sets of ordered pairs: f 0002 5(0, 1), (0, 2), (1, 1), (1, 2)6 g 0002 5(0, 1), (1, 1), (2, 2), (3, 2)6 h 0002 5(0, 1), (1, 2), (2, 3), (3, 0)6 (A) Which of these sets represent functions? (B) Which of the functions are one-to-one functions? (C) For each set that is a function, form a new set by reversing each ordered pair in the set. Which of these new sets represent functions? (D) What do these results tell you about the result of reversing the ordered pairs for functions that are one-to-one, and for functions that are not one-to-one?
Explore-Discuss 1 illustrates an important idea that we will examine later: Only oneto-one functions have inverses.
EXAMPLE
1
Determining Whether a Function Is One-to-One Determine whether f is a one-to-one function for (A) f (x) 0002 x2
SOLUTIONS
(B) f (x) 0002 2x 0004 1
(A) To show that a function is not one-to-one, all we have to do is find two different ordered pairs in the function with the same second component and different first components. Because f (2) 0002 22 0002 4
and
f (00042) 0002 (00042)2 0002 4
the ordered pairs (2, 4) and (00042, 4) both belong to f, and f is not one-to-one. (Note that there’s nothing special about 2 and 00042 here: Any real number and its negative can be used in the same way.) (B) To show that a function is one-to-one, we have to show that no two ordered pairs have the same second component and different first components. To do this, we’ll show that if any two ordered pairs (a, f (a)) and (b, f(b)) in f have the same second components, then the first components must also be the same. That is, we show that f (a) 0002 f (b) implies a 0002 b. We proceed as follows: f (a) 0002 f (b) 2a 0004 1 0002 2b 0004 1 2a 0002 2b a0002b
Assume second components are equal. Evaluate f(a) and f(b).
Simplify. Conclusion: f is one-to-one.
By Definition 1, f is a one-to-one function.
0002
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MATCHED PROBLEM 1
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237
Determine whether f is a one-to-one function for (A) f (x) 0002 4 0004 x2
(B) f (x) 0002 4 0004 2x
0002
The methods used in the solution of Example 1 can be stated as a theorem.
Z THEOREM 1 One-to-One Functions 1. If f (a) 0002 f (b) for at least one pair of domain values a and b, a 0005 b, then f is not one-to-one. 2. If the assumption f (a) 0002 f (b) always implies that the domain values a and b are equal, then f is one-to-one.
Applying Theorem 1 is not always easy—try testing f (x) 0002 x 3 0003 2x 0003 3, for example. (Good luck!) However, the graph of a function can help us develop a simple procedure for determining if a function is one-to-one. If any horizontal line intersects the graph in more than one point [as shown in Fig. 1(a)], then there is a second component (height) that corresponds to two different first components (x values). This shows that the function is not one-to-one. On the other hand, if every horizontal line intersects the graph in just one point or not at all [as shown in Fig. 1(b)], the function is one-to-one. These observations form the basis of the horizontal line test. y
y
y 0002 f (x) (a, f (a))
(b, f (b))
(a, f (a))
y 0002 f (x) a
b
f(a) ⴝ f(b) for a 0005 b f is not one-to-one (a)
x
a
x
Only one point has second component f (a); f is one-to-one (b)
Z Figure 1 Intersections of graphs and horizontal lines.
Z THEOREM 2 Horizontal Line Test A function is one-to-one if and only if every horizontal line intersects the graph of the function in at most one point.
The graphs of the functions considered in Example 1 are shown in Figure 2 on page 238. Applying the horizontal line test to each graph confirms the results we obtained in Example 1. A function that is increasing throughout its domain or decreasing throughout its domain will always pass the horizontal line test [Figs. 3(a) and 3(b)]. This gives us the following theorem.
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FUNCTIONS y
y 5
5
(00042, 4)
(2, 4)
00045
00045
5
5
x
x 00045
f(x) ⴝ 2x 0006 1 passes the horizontal line test; f is one-to-one (b)
f(x) ⴝ x2 does not pass the horizontal line test; f is not one-to-one (a)
Z Figure 2 Applying the horizontal line test.
Z THEOREM 3 Increasing and Decreasing Functions If a function f is increasing throughout its domain or decreasing throughout its domain, then f is a one-to-one function.
y
y
y
x
x
An increasing function is always one-to-one (a)
A decreasing function is always one-to-one (b)
x
A one-to-one function is not always increasing or decreasing (c)
Z Figure 3 Increasing, decreasing, and one-to-one functions.
Figure 3(c) shows that a function can still be one-to-one even if it is neither increasing nor decreasing. The function illustrated is increasing on [ 0004, 0] and decreasing on (0, ).
Z Finding the Inverse of a Function Now we will demonstrate how we can form a new function by reversing the correspondence determined by a given function. Let g be the function defined as follows: g 0002 5(00043, 9), (0, 0), (3, 9)6
g is not one-to-one.
Notice that g is not one-to-one because the domain elements 00043 and 3 both correspond to the range element 9. We can reverse the correspondence determined by function g simply by reversing the components in each ordered pair in g, producing the following set: G 0002 5(9, 00043), (0, 0), (9, 3)6
G is not a function.
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But the result is not a function because the domain element 9 corresponds to two different range elements, ⫺3 and 3. On the other hand, if we reverse the ordered pairs in the function f ⫽ 5(1, 2), (2, 4), (3, 9)6
f is one-to-one; all second components are distinct.
we obtain F ⫽ 5(2, 1), (4, 2), (9, 3)6
F is a function.
This time f is a one-to-one function, and the set F turns out to be a function also. This new function F, formed by reversing all the ordered pairs in f, is called the inverse of f and is usually denoted by f ⫺1 (this is read as “inverse f ” or “the inverse of f ”): f ⫺1 ⫽ 5(2, 1), (4, 2), (9, 3)6
The inverse of f
Notice that f ⫺1 is also a one-to-one function and that the following relationships hold: Domain of f ⫺1 ⫽ 52, 4, 96 ⫽ Range of f Range of f ⫺1 ⫽ 51, 2, 36 ⫽ Domain of f We conclude that reversing all the ordered pairs in a one-to-one function forms a new one-to-one function and reverses the domain and range in the process. We are now ready to present a formal definition of the inverse of a function.
Z DEFINITION 2 Inverse of a Function If f is a one-to-one function, then the inverse of f, denoted f ⫺1, is the function formed by reversing all the ordered pairs in f. That is, f ⫺1 ⫽ 5( y, x) | (x, y) is in f } If f is not one-to-one, then f does not have an inverse and f ⫺1 does not exist.
ZZZ
CAUTION ZZZ
Be careful not to confuse inverse notation and reciprocal notation. For numbers, a 1 superscript of ⫺1 means reciprocal: 2⫺1 ⫽ . For functions, a superscript of ⫺1 2 1 means inverse: f ⫺1(x) is the inverse of f (x), which is not the same as . f (x)
The following properties of inverse functions follow directly from the definition.
Z THEOREM 4 Properties of Inverse Functions For a given function f, if f ⫺1 exists, then 1. f ⫺1 is a one-to-one function. 2. The domain of f ⫺1 is the range of f. 3. The range of f ⫺1 is the domain of f.
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ZZZ EXPLORE-DISCUSS 2
(A) For the function f 0002 5(3, 5), (7, 11), (11, 17)6, find f 00041.
(B) What do you think would be the result of composing f with f 00041? Justify your answer using Definition 2. (C) Check your conjecture from part B by finding both f 0002 f 00041 and f 00041 0002 f. Were you correct?
Explore-Discuss 2 brings up an important point: If you apply a function to any number in its domain, then apply the inverse of that function to the result, you’ll get right back where you started. This leads to the following theorem. Z THEOREM 5 Inverse Functions and Composition If f 00041 exists, then 1. f( f 00041(x)) 0002 x for all x in the domain of f 00041. 2. f 00041( f (x)) 0002 x for all x in the domain of f. If f and g are one-to-one functions satisfying f(g(x)) 0002 x for all x in the domain of g and g( f (x)) 0002 x for all x in the domain of f then f and g are inverses of one another.
We can use Theorem 5 to see if two functions defined by equations are inverses.
EXAMPLE
2
Deciding If Two Functions Are Inverses Use Theorem 5 to decide if these two functions are inverses. f (x) 0002 3x 0004 7
SOLUTION
g(x) 0002
x00037 3
The domain of both functions is all real numbers. For any x, f (g(x)) 0002 f a 0002 3a
x00037 b 3
Substitute into f(x).
x00037 b00047 3
Multiply.
0002x0003700047 0002x g( f(x)) 0002 g(3x 0004 7) 3x 0004 7 0003 7 0002 3 0002
3x 3
Add.
Substitute into g(x). Add.
Simplify.
0002x By Theorem 5, f and g are inverses.
0002
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MATCHED PROBLEM 2
Inverse Functions
241
Use Theorem 5 to decide if these two functions are inverses. 2 f(x) 0002 (11 0004 x) 5
5 g(x) 0002 0004 x 0003 11 2 0002
There is one obvious question that remains: when a function is defined by an equation, how can we find the inverse? Given a function y 0002 f(x), the first coordinates of points on the graph are represented by x, and the second coordinates are represented by y. Finding the inverse by reversing the order of the coordinates would then correspond to switching the variables x and y. This leads us to the following procedure, which can be applied whenever it is possible to solve y 0002 f (x) for x in terms of y. Z FINDING THE INVERSE OF A FUNCTION f Step 1. Find the domain of f and verify that f is one-to-one. If f is not one-to-one, then stop, because f 00041 does not exist. Step 2. If the function is written with function notation, like f (x), replace the function symbol with the letter y. Then interchange x and y. Step 3. Solve the resulting equation for y. The result is f 00041(x). Step 4. Find the domain of f 00041. Remember, the domain of f 00041 must be the same as the range of f. You can check your work using Theorem 5.
EXAMPLE
3
Finding the Inverse of a Function Find f 00041 for f(x) 0002 1x 0004 1.
SOLUTION
y 5
y 0002 1x 0004 1 x 0002 1y 0004 1
y 0002 f (x) 5
00045
f(x) 0002 兹x 0004 1, x 1
Z Figure 4
Step 1. Find the domain of f and verify that f is one-to-one. Since 1x 0004 1 is defined only for x 0004 1 0, the domain of f is [1, ). The graph of f in Figure 4 shows that f is one-to-one, so f 00041 exists. Step 2. Replace f (x) with y, then interchange x and y.
x
Interchange x and y.
Step 3. Solve the equation for y. x 0002 1y 0004 1 x2 0002 y 0004 1 2 x 000310002y
Square both sides. Add 1 to each side.
The inverse is f 00041(x) 0002 x2 0003 1. Step 4. Find the domain of f 00041. The equation we found for f 00041 is defined for all x, but the domain should be the range of f. From Figure 4, we see that the range of f is [0, ) so that is the domain of f 00041. Therefore, f 00041(x) 0002 x2 0003 1
x 0
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Find the composition of f with the alleged inverse (in both orders!). For x in [1, ), the domain of f, we have
CHECK
f 00041( f (x)) 0002 f 00041(1x 0004 1) 0002 ( 1x 0004 1)2 0003 1 0002x0004100031 ✓ 0002x
Substitute 1x 0004 1 into f ⴚ1. Square 1x 0004 1. Add.
For x in [0, ), the domain of f 00041, we have f( f 00041(x)) 0002 f (x2 0003 1) 0002 2(x2 0003 1) 0004 1 0002 2x2 0002 冟x冟
Substitute x2 ⴙ 1 into f. Add. 2x2 ⴝ 円x円 for any real number x. 円x円 ⴝ x for x 0.
✓
0002x MATCHED PROBLEM 3
0002
Find f 00041 for f (x) 0002 1x 0003 2.
0002
The technique of finding an inverse by interchanging x and y leads to the following property of inverses that comes in very handy later in the course.
Z THEOREM 6 A Property of Inverses If f 00041 exists, then x 0002 f 00041( y) if and only if y 0002 f (x).
Z Mathematical Modeling Example 4 shows how an inverse function is used in constructing a revenue model. It is based on Example 8 in Section 3-5.
EXAMPLE
4
Modeling Revenue The research department for an electronics firm estimates that the weekly demand for a certain brand of headphones is given by x 0002 f ( p) 0002 20,000 0004 1,000p
Demand function
where x is the number of pairs retailers are likely to buy per week at p dollars per pair. Express the revenue as a function of the demand x and state its domain. SOLUTION
If x pairs of headphones are sold at p dollars each, the total revenue is Revenue 0002 (Number of pairs)(price of each pair) R 0002 xp To express the revenue as a function of the demand x, we need to express the price in terms of x. That is, we must find the inverse of the demand function.
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Step 1. Find the domain of f and verify that f is one-to-one. Price and demand are never negative, so p 0002 0 and x 0005 20,000 0004 1,000p 0005 1,000(20 0004 p) 0002 0 20 0004 p 0002 0 20 0002 p
Factor. Divide both sides by 1,000. Add p to both sides.
p 0003 20
or
Since p must satisfy both p 0002 0 and p 0003 20, the domain of f is [0, 20]. The graph of f (Fig. 5) shows that f is one-to-one. x 20,000
x 0005 20,000 0004 1,000p
0
p
20
Z Figure 5
Step 2. Since x and p have specific meaning in the context of this problem, interchanging them does not apply here. Step 3. Solve the equation x 0005 20,000 0004 1,000p for p. x 0005 20,000 0004 1,000p x 0004 20,000 0005 00041,000p 00040.001x 0006 20 0005 p
Subtract 20,000 from both sides. Divide both sides by ⴚ1,000.
The inverse of the demand function is p 0005 f 00041(x) 0005 20 0004 0.001x Step 4. From Figure 5, we see that the range of f is [0, 20,000], so this must also be the domain of f 00041. p 0005 f 00041(x) 0005 20 0004 0.001x We should check that f ( f the reader.
00041
(x)) 0005 x and f
0 0003 x 0003 20,000 00041
( f ( p)) 0005 p, but we will leave that to
The revenue R is given by R 0005 xp R(x) 0005 x(20 0004 0.001x) 0005 20x 0004 0.001x2 and the domain of R is [0, 20,000]. MATCHED PROBLEM 4
0002
Repeat Example 3 for the demand function x 0005 f ( p) 0005 10,000 0004 1,000p
0 0003 p 0003 10 0002
The demand function in Example 4 was defined with independent variable p and dependent variable x. When we found the inverse function, we did not rewrite it with independent variable p. Because p represents price and x represents number of players, to interchange these variables would be confusing. In most applications, the variables have specific meaning and should not be interchanged as part of the inverse process.
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Z Graphing Inverse Functions The following activities refer to the graph of f in Figure 6 and Tables 1 and 2.
ZZZ EXPLORE-DISCUSS 3
y 0002 f (x)
Table 1
Table 2
x
5
f(x)
f ⴚ1(x)
x
00044 00045
5
00042
x
0 2
00045
Z Figure 6
(A) Complete the second column in Table 1. (B) Reverse the ordered pairs in Table 1 and list the results in Table 2. (C) Add the points in Table 2 to Figure 6 (or a copy of the figure) and sketch the graph of f 00041. (D) Discuss any symmetry you observe between the graphs of f and f 00041.
Explore-Discuss 3 is based on an important relationship between the graph of any function and its inverse. In a rectangular coordinate system, the points (a, b) and (b, a) are symmetric with respect to the line y 0002 x [Fig. 7(a)]. Theorem 6 is an immediate consequence of this observation.
y
Z Figure 7 Symmetry with respect to the line y 0002 x.
5
y
y0002x (1, 4)
y 0002 f (x)
y
y0002x
5
y 0002 f 00041(x)
y 0002 f 00041(x)
y0002x
10
(00043, 2) (4, 1) x
00045
5
00045
5
x
y 0002 f(x)
(00045, 00042) (2, 00043) 00045
(00042, 00045)
(a, b) and (b, a) are symmetric with respect to the line y ⴝ x (a)
00045
10
f(x) ⴝ 2x ⴚ 1 f ⴚ1(x) ⴝ 12 x ⴙ 12
f (x) ⴝ 1x ⴚ 1 f ⴚ1(x) ⴝ x 2 ⴙ 1, x 0004 0
(b)
(c)
00041 Z THEOREM 7 Symmetry Property for the Graphs of f and f
The graphs of y 0002 f (x) and y 0002 f 00041(x) are symmetric with respect to the line y 0002 x.
x
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Inverse Functions
Knowledge of this symmetry property allows us to graph f 00041 if the graph of f is known, and vice versa. Figures 7(b) and 7(c) illustrate this property for the two inverse functions we found earlier. If a function is not one-to-one, we can usually restrict the domain of the function to produce a new function that is one-to-one. Then we can find an inverse for the restricted function. Suppose we start with f (x) 0002 x2 0004 4. Because f is not one-to-one, f 00041 does not exist [Fig. 8(a)]. But there are many ways the domain of f can be restricted to obtain a oneto-one function. Figures 8(b) and 8(c) illustrate two such restrictions. In essence, we are “forcing” the function to be one-to-one by throwing out a portion of the graph that would make it fail the horizontal line test. y
Z Figure 8 Restricting the domain of a function.
y
y 0002 f (x)
5
y
y 0002 h(x)
y0002x
5
y0002x
5
y 0002 g00041(x) 00045
5
x
00045
00045
5
00045
f(x) ⴝ x2 ⴚ 4 f ⴚ1 does not exist (a)
x
00045
5
y 0002 g(x)
00045
x
y 0002 h 00041(x)
h(x) ⴝ x2 ⴚ 4, x 0007 0 hⴚ1(x) ⴝ ⴚ1x ⴙ 4, x 0004 ⴚ4 (c)
g(x) ⴝ x 2 ⴚ 4, x 0004 0 g ⴚ1(x) ⴝ 1x ⴙ 4, x 0004 ⴚ4 (b)
Recall from Theorem 3 that increasing and decreasing functions are always one-to-one. This provides the basis for a convenient method of restricting the domain of a function: If the domain of a function f is restricted to an interval on the x axis over which f is increasing (or decreasing), then the new function determined by this restriction is one-to-one and has an inverse. We used this method to form the functions g and h in Figure 8.
EXAMPLE
5
Finding the Inverse of a Function Find the inverse of f(x) 0002 4x 0004 x2, x 2. Graph f, f 00041, and the line y 0002 x in the same coordinate system.
SOLUTION
Step 1. Find the domain of f and verify that f is one-to-one. We are given that the domain of f is (0004, 2]. The graph of y 0002 4x 0004 x2 is a parabola opening downward with vertex (2, 4) (Fig. 9). The graph of f is the left side of this parabola (Fig. 10). From the graph of f, we see that f is increasing and one-to-one on (0004, 2]. y 5
00045
y 0002 4x 0004 x2
5
00045
Z Figure 9
y 5
x
00045
y 0002 f(x)
5
00045
Z Figure 10
x
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Step 2. Replace f (x) with y, then interchange x and y. y 0002 4x 0004 x2 x 0002 4y 0004 y2 Step 3. Solve the equation for y. x 0002 4y 0004 y2 y2 0004 4y 0002 0004x y2 0004 4x 0003 4 0002 0004x 0003 4 (y 0004 2)2 0002 4 0004 x y 0004 2 0002 0006 14 0004 x y 0002 2 000614 0004 x
Rewrite so that the coefficient of y 2 is ⴙ1. Add 4 to both sides to complete the square. Factor the left side. Take the square root of both sides. Add 2 to both sides.
Now we have two possible solutions. The domain of f was (–, 2], and this should be the range of f 00041. In other words, the output of the inverse is never greater than 2. But y 0002 2 0003 14 0004 x would always be greater than or equal to 2, so we must instead choose y 0002 2 0004 14 0004 x. y 5
y0002x y 0002 f (x)
5
y0002
f 00041(x)
f 00041(x) 0002 2 0004 14 0004 x Step 4. The domain of f 00041 is the range of f. We can see from Figure 10 that this is (0004, 4]. Notice that the equation we found for f 00041(x) is defined for these values. Our final answer is f 00041(x) 0002 2 0004 14 0004 x
x
x 4
The check is again left for the reader. The graphs of f, f 00041, and y 0002 x are shown in Figure 11. To aid in graphing f 00041, we plotted several points on the graph of f and then reflected these points in the line y 0002 x. 0002
Z Figure 11
MATCHED PROBLEM 5
Find the inverse of f(x) 0002 4x 0004 x2, x 2. Graph f, f 00041, and y 0002 x in the same coordinate system. 0002
Technology Connections To reproduce Figure 11 on a graphing calculator, first enter
y1 ⴝ (4x ⴚ x2)0007(x 0007 2) in the equation editor (Fig. 12) and graph (Fig. 13). (For graphs involving both f and f 00061 it is best to use a squared viewing window.) The Boolean expression (x 0007 2) is
assigned the value 1 if the inequality is true and 0 if it is false. The calculator recognizes that division by 0 is an undefined operation and no graph is drawn for x 0002 2. Now enter
y2 ⴝ 2 ⴚ 14 ⴚ x
5
7.6
00047.6
00045
Z Figure 12
Z Figure 13
y3 ⴝ x
in the equation editor and graph (Fig. 14).
5
00047.6
and
7.6
00045
Z Figure 14
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ANSWERS TO MATCHED PROBLEMS 1. (A) Not one-to-one (B) One-to-one 2. They are inverses. 3. f ⫺1(x) ⫽ x2 ⫺ 2, x ⱖ 0 4. R(x) ⫽ 10x ⫺ 0.001x2
5. f ⫺1(x) ⫽ 2 ⫹ 14 ⫺ x, x ⱕ 4 y ⫽ f ⫺1(x)
y
y⫽x
5
⫺5
5
⫺5
3-6
x
y ⫽ f (x)
Exercises
1. When a function is defined by ordered pairs, how can you tell if it is one-to-one?
In Problems 13–30, determine if the function is one-to-one. 13. Domain
2. When you have the graph of a function, how can you tell if it is one-to-one?
Range
14. Domain
⫺2
⫺4
⫺2
3. Why does a function fail to have an inverse if it is not one-toone? Give an example using ordered pairs to illustrate your answer.
⫺1
⫺2
⫺1
0
0
0
4. True or False: Any function whose graph changes direction is not one-to-one. Explain.
1
1
1
2
5
2
5. What is the result of composing a function with its inverse? Why does this make sense? 6. What is the relationship between the graphs of two functions that are inverses?
15. Domain
For each set of ordered pairs in Problems 7–12, determine if the set is a function, a one-to-one function, or neither. Reverse all the ordered pairs in each set and determine if this new set is a function, a one-to-one function, or neither.
Range
17.
9
Range 5
2
2
3
3
1
4
4
2
5
5
4
7
f (x)
10. {(5, 4), (4, 3), (3, 2), (2, 1)} 11. 5(1, 2), (1, 4), (⫺3, 2), (⫺3, 4)6 12. 5(0, 5), (⫺4, 5), (⫺4, 2), (0, 2)6
7
1
8. {(⫺1, 0), (0, 1), (1, ⫺1), (2, 1)6 9. {(5, 4), (4, 3), (3, 3), (2, 4)}
⫺3
1
3
7. {(1, 2), (2, 1), (3, 4), (4, 3)}
16. Domain
Range
x
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FUNCTIONS
g (x)
r (x)
23.
x
19.
x
s(x)
24.
h(x)
x
x
20.
k(x)
x
21.
m(x)
x
25. F(x) 0002 12x 0003 2
26. G(x) 0002 000413x 0003 1
27. H(x) 0002 4 0004 x2
28. K(x) 0002 14 0004 x
29. M(x) 0002 1x 0003 1
30. N(x) 0002 x2 0004 1
In Problems 31–40, determine if g is the inverse of f. 31. f (x) 0002 3x 0003 5;
g(x) 0002 13x 0004 53
32. f (x) 0002 2x 0004 4;
g(x) 0002 12x 0004 2
33. f (x) 0002 2 0004 (x 0003 1)3;
3 g(x) 0002 2 30004x00041
34. f (x) 0002 (x 0004 3)3 0003 4;
3 g(x) 0002 2 x0004400033
35. f (x) 0002
2x 0004 3 ; x00034
g(x) 0002
3 0003 4x 20004x
36. f (x) 0002
x00031 ; 2x 0004 3
g(x) 0002
3x 0003 1 2x 0003 1
37. f (x) 0002 4 0003 x2, x 0; 38. f (x) 0002 1x 0003 2;
g(x) 0002 1x 0004 4
g(x) 0002 x2 0004 2, x 0
39. f (x) 0002 1 0004 x2, x 0; 40. f (x) 0002 0004 1x 0004 2; 22.
g(x) 0002 0004 11 0004 x
g(x) 0002 x2 0003 2, x 0
n(x)
In Problems 41–44, find the domain and range of f, sketch the graph of f 00041, and find the domain and range of f 00041. y
41. x
y0002x
5
y 0002 f (x) 00045
5
00045
x
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SECTION 3–6 y
42.
Inverse Functions
249
67. f (x) ⫽ x2 ⫹ 2x ⫺ 2, x ⱕ ⫺1
y⫽x
68. f (x) ⫽ x2 ⫹ 8x ⫹ 7, x ⱖ ⫺4
5
69. f (x) ⫽ ⫺ 29 ⫺ x2, 0 ⱕ x ⱕ 3 ⫺5
5
70. f (x) ⫽ 29 ⫺ x2, 0 ⱕ x ⱕ 3
x
71. f (x) ⫽ 29 ⫺ x2, ⫺3 ⱕ x ⱕ 0
y ⫽ f (x)
72. f (x) ⫽ ⫺ 29 ⫺ x2, ⫺3 ⱕ x ⱕ 0
⫺5
73. f (x) ⫽ 1 ⫺ 21 ⫺ x2, ⫺1 ⱕ x ⱕ 0 y
43.
74. f (x) ⫽ 1 ⫹ 21 ⫺ x2, ⫺1 ⱕ x ⱕ 0
y⫽x
5
The functions in Problems 75–84 are one-to-one. Find f ⫺1. 75. f (x) ⫽ 3 ⫺
⫺5
y ⫽ f (x)
5
x
y⫽x
78. f (x) ⫽
2x ⫹ 5 3x ⫺ 4
82. f (x) ⫽
5 83. f (x) ⫽ 4 ⫺ 2x ⫹ 2
5
4 x
3 x⫹4 4x 80. f (x) ⫽ 2⫺x
2 x⫺1 2x 79. f (x) ⫽ x⫹1 81. f (x) ⫽
y
76. f (x) ⫽ 5 ⫹
77. f (x) ⫽
⫺5
44.
2 x
5 ⫺ 3x 7 ⫺ 4x
3 84. f (x) ⫽ 2x ⫹ 3 ⫺ 2
85. How are the x and y intercepts of a function and its inverse related? ⫺5
5
x
86. Does a constant function have an inverse? Explain.
y ⫽ f (x)
87. Are the functions f (x) ⫽ x2 and g(x) ⫽ 1x inverses? Why or why not?
⫺5
In Problems 45–74, graph f and verify that f is a one-to-one function. Find f ⫺1and add the graph of f ⫺1 and the line y ⫽ x to the graph of f. State the domain and range of f and the domain and range of f ⫺1. 45. f (x) ⫽ 3x 47. f (x) ⫽ 4x ⫺ 3
1 46. f (x) ⫽ x 2 1 5 48. f (x) ⫽ ⫺ x ⫹ 3 3
49. f (x) ⫽ 0.2x ⫹ 0.4
50. f (x) ⫽ 0.25x ⫹ 2.25
51. f (x) ⫽ 1x ⫹ 3
52. f (x) ⫽ 2 ⫺ 1x
53. f (x) ⫽
1 116 ⫺ x 2
54. f (x) ⫽
1 136 ⫺ x 3
55. f (x) ⫽ 3 ⫺ 1x ⫺ 1
56. f (x) ⫽ 2 ⫹ 15 ⫺ x
57. f (x) ⫽ x2 ⫹ 5, x ⱖ 0
58. f (x) ⫽ x2 ⫹ 5, x ⱕ 0
59. f (x) ⫽ 4 ⫺ x2, x ⱕ 0
60. f (x) ⫽ 4 ⫺ x2, x ⱖ 0
61. f (x) ⫽ x2 ⫹ 8x, x ⱖ ⫺4 62. f (x) ⫽ x2 ⫹ 8x, x ⱕ ⫺4 63. f (x) ⫽ (2 ⫺ x)2, x ⱕ 2 64. f (x) ⫽ (2 ⫺ x)2, x ⱖ 2 65. f (x) ⫽ (x ⫺ 1)2 ⫹ 2, x ⱖ 1 66. f (x) ⫽ 3 ⫺ (x ⫺ 2)2, x ⱕ 2
3 x inverses? Why or 88. Are the functions f (x) ⫽ x3 and g(x) ⫽ 1 why not?
In Problems 89–92, the given function is not one-to-one. Find a way to restrict the domain so that the function is one-to-one, then find the inverse of the function with that domain. 89. f (x) ⫽ (2 ⫺ x)2
90. f (x) ⫽ (1 ⫹ x)2
91. f (x) ⫽ 24x ⫺ x2
92. f (x) ⫽ 26x ⫺ x2
APPLICATIONS 93. BODY WEIGHT Two formulas for estimating body weight as a function of height that are commonly used are Women: p ⫽ W(h) ⫽ 100 ⫹ 5h Men: p ⫽ M(h) ⫽ 110 ⫹ 5h where p is weight in pounds and h is height over 5 feet (in inches). Find h ⫽ W ⫺1(p) and state its domain. 94. BODY WEIGHT Refer to Problem 93. Find h ⫽ M ⫺1( p) and state its domain. 95. PRICE AND DEMAND The number q of CD players consumers are willing to buy per week from a retail chain at a price of $p is given approximately by (see the figure) q ⫽ d(p) ⫽
3,000 0.2p ⫹ 1
10 ⱕ p ⱕ 70
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(A) Find the range of d. (B) Find p ⫽ d⫺1(q), and find its domain and range. q 1,000
q ⫽ d(p) q ⫽ s(p) 70
p
Figure for 95–96
96. PRICE AND SUPPLY The number q of CD players a retail chain is willing to supply at a price of $p is given approximately by (see the figure) q ⫽ s( p) ⫽
900p p ⫹ 20
10 ⱕ p ⱕ 70
(A) Find the range of s. (B) Find p ⫽ s⫺1(q), and find its domain and range.
97. BUSINESS—MARKUP POLICY A bookstore sells a book with a wholesale price of $6 for $10.50 and one with a wholesale price of $10 for $15.50. (A) If the markup policy for the store is assumed to be linear, find a function r ⫽ m(w) that expresses the retail price r as a function of the wholesale price w and find its domain and range. (B) Find w ⫽ m⫺1(r) and find its domain and range. 98. BUSINESS—MARKUP POLICY Repeat Problem 97 if the second book has a wholesale price of $11 and sells for $18.50. Problems 99 and 100 are related to Problems 97 and 98 in Exercises 3-4. 99. STOPPING DISTANCE A model for the length L (in feet) of the skid marks left by a particular automobile when making an emergency stop is L ⫽ f (s) ⫽ 0.06s2 ⫺ 1.2s ⫹ 26, s ⱖ 10 where s is speed in miles per hour. Find s ⫽ f ⫺1(L) and find its domain and range. 100. STOPPING DISTANCE A model for the length L (in feet) of the skid marks left by a second automobile when making an emergency stop is L ⫽ f (s) ⫽ 0.08s2 ⫺ 1.6s ⫹ 38, s ⱖ 10 where s is speed in miles per hour. Find s ⫽ f ⫺1(L) and find its domain and range.
CHAPTER
3-1
3
Review
Functions
A function is a correspondence between two sets of elements such that to each element in the first set there corresponds one and only one element in the second set. The first set is called the domain and the set of all corresponding elements in the second set is called the range. Equivalently, a function is a set of ordered pairs with the property that no two ordered pairs have the same first component and different second components. The domain is the set of all first components, and the range is the set of all second components. An equation in two variables defines a function if to each value of the independent variable, the placeholder for domain values, there corresponds exactly one value of the dependent variable, the placeholder for range values. The vertical line test states that a vertical line will intersect the graph of a function in at most one point. Unless otherwise specified, the implied domain of a function defined by an equation is assumed to be the set of all real number replacements for the independent variable that produce real values for the dependent variable. The symbol f(x) represents the real number in the range of the function f corresponding to the domain value x. Equivalently, the ordered pair (x, f (x)) belongs to the function f.
3-2
Graphing Functions
The graph of a function f is the set of all points (x, f(x)), where x is in the domain of f and f(x) is the associated output. This is also the same as the graph of the equation y ⫽ f (x). The first coordinate of a point where the graph of a function intersects the x axis is called an x intercept or real zero of the function. The x intercept is also a real solution or root of the equation f (x) ⫽ 0. The second coordinate of a point where the graph of a function crosses the y axis is called the y intercept of the function. The y intercept is given by f(0), provided 0 is in the domain of f. A solid dot on a graph of a function indicates a point that belongs to the graph and an open dot indicates a point that does not belong to the graph. Dots are also used to indicate that a graph terminates at a point, and arrows are used to indicate that the graph continues indefinitely with no significant changes in direction. Let I be an interval in the domain of a function f. Then, 1. f is increasing on I and the graph of f is rising on I if f (x1) 6 f (x2) whenever x1 6 x2 in I. 2. f is decreasing on I and the graph of f is falling on I if f (x1) 7 f (x2) whenever x1 6 x2 in I.
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3. f is constant on I and the graph of f is horizontal on I if f (x1) ⫽ f (x2) whenever x1 6 x2 in I. A function of the form f(x) ⫽ mx ⫹ b, where m and b are constants, is a linear function. If m ⫽ 0, then f(x) ⫽ b is a constant function, and if m ⫽ 1 and b ⫽ 0, then f(x) ⫽ x is the identity function. A piecewise-defined function is a function whose definition involves more than one formula. The absolute value function is a piecewise-defined function. The graph of a function is continuous if it has no holes or breaks and discontinuous at any point where it has a hole or break. Intuitively, the graph of a continuous function can be sketched without lifting a pen from the paper. The greatest integer for a real number x, denoted by 冀x冁 , is the largest integer less than or equal to x; that is, 冀x 冁 ⫽ n, where n is an integer, n ⱕ x 6 n ⫹ 1. The greatest integer function f is defined by the equation f (x) ⫽ 冀x 冁.
3-3
Transformations of Functions
The first six basic functions in a library of elementary functions are defined by f (x) ⫽ x (identity function), g(x) ⫽ 冟 x 冟 (absolute value function), h(x) ⫽ x2 (square function), m(x) ⫽ x3 (cube function), 3 n(x) ⫽ 1x (square root function), and p(x) ⫽ 2 x (cube root function) (see Figure 1, Section 3-3). Performing an operation on a function produces a transformation of the graph of the function. The basic transformations are the following:
251
A function f is called an even function if f (x) ⫽ f (⫺x) for all x in the domain of f and an odd function if f (⫺x) ⫽ ⫺f (x) for all x in the domain of f. The graph of an even function is said to be symmetric with respect to the y axis and the graph of an odd function is said to be symmetric with respect to the origin.
3-4
Quadratic Functions
If a, b, and c are real numbers with a ⫽ 0, then the function f (x) ⫽ ax2 ⫹ bx ⫹ c is a quadratic function and its graph is a parabola. Completing the square of the quadratic expression x2 ⫹ bx produces a perfect square: b 2 b 2 x2 ⫹ bx ⫹ a b ⫽ ax ⫹ b 2 2 Completing the square for f (x) ⫽ ax2 ⫹ bx ⫹ c produces the vertex form f (x) ⫽ a(x ⫺ h)2 ⫹ k and gives the following properties:
1. The graph of f is a parabola: f (x)
Axis of symmetry x⫽h
Vertex (h, k)
Vertical Translation: k 7 0 Shift graph of y ⫽ f(x) up k units y ⫽ f(x) ⫹ k e k 6 0 Shift graph of y ⫽ f(x) down 冟 k 冟 units
e
Min f(x) h
h 7 0 Shift graph of y ⫽ f (x) left h units h 6 0 Shift graph of y ⫽ f (x) right 冟 h 冟 units
f (x)
Reflection: y ⫽ ⫺f (x) y ⫽ f (⫺x) y ⫽ ⫺f (⫺x)
x
a⬎0 Opens upward
Horizontal Translation: y ⫽ f(x ⫹ h)
k
Reflect the graph of y ⫽ f (x) through the x axis Reflect the graph of y ⫽ f (x) through the y axis Reflect the graph of y ⫽ f (x) through the origin
Axis of symmetry x⫽h Vertex (h, k)
k
Max f(x)
Vertical Stretch and Shrink: A 7 1 y ⫽ Af (x) f
0 6 A 6 1
Vertically stretch the graph of y ⫽ f (x) by multiplying each y value by A Vertically shrink the graph of y ⫽ f (x) by multiplying each y value by A
Horizontal Stretch and Shrink: A 7 1
y ⫽ f (Ax) h
0 6 A 6 1
Horizontally shrink the graph of y ⫽ f (x) by multiplying 1 each x value by A Horizontally stretch the graph of y ⫽ f (x) by multiplying 1 each x value by A
h
x
a⬍0 Opens downward
2. Vertex: (h, k) (Parabola increases on one side of the vertex and decreases on the other.)
3. Axis (of symmetry): x ⫽ h (parallel to y axis) 4. f (h) ⫽ k is the minimum if a 7 0 and the maximum if a 6 0. 5. Domain: All real numbers
Range: (⫺⬁, k] if a 6 0 or [k, ⬁) if a 7 0
6. The graph of f is the graph of g(x) ⫽ ax2 translated horizontally h units and vertically k units.
The first coordinate of the vertex of a parabola in standard form can be located using the formula x ⫽ ⫺b/2a. This can then be substituted into the function to find the second coordinate. The vertex
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form of a parabola can be used to find the equation when the vertex and one other point on the graph are known. Replacing the equal sign in a quadratic equation with 6, 7 ,
, or produces a quadratic inequality. The set of all values of the variable that make the inequality a true statement is the solution set.
3-5
Combining Functions; Composition
The sum, difference, product, and quotient of the functions f and g are defined by ( f 0003 g)(x) 0002 f (x) 0003 g (x)
( f 0004 g)(x) 0002 f (x) 0004 g (x)
( fg)(x) 0002 f (x)g (x)
f (x) f a b (x) 0002 g g (x)
3
Inverse Functions
A function is one-to-one if no two ordered pairs in the function have the same second component and different first components. According to the horizontal line test, a horizontal line will intersect the graph of a one-to-one function in at most one point. A function that is increasing (or decreasing) throughout its domain is one-to-one. The inverse of the one-to-one function f is the function f 00041 formed by reversing all the ordered pairs in f. If f is a one-to-one function, then: 1. f 00041 is one-to-one. 2. Domain of f 00041 0002 Range of f.
g(x) 0005 0
The domain of each function is the intersection of the domains of f and g, with the exception that values of x where g(x) 0002 0 must be excluded from the domain of f兾g. The composition of functions f and g is defined by ( f ° g)(x) 0002 f ( g (x)). The domain of f ° g is the set of all real numbers x in the domain of g such that g(x) is in the domain of f. The domain of f ° g is always a subset of the domain of g.
CHAPTER
3-6
3. Range of f 00041 0002 Domain of f. 4. x 0002 f 00041( y) if and only if y 0002 f (x). 5. f 00041 ( f (x)) 0002 x for all x in the domain of f. 6. f ( f 00041(x)) 0002 x for all x in the domain of f 00041. 7. To find f 00041, solve the equation y 0002 f (x) for x. Interchanging x and y at this point is an option. 8. The graphs of y 0002 f (x) and y 0002 f 00041 (x) are symmetric with respect to the line y 0002 x.
Review Exercises
Work through all the problems in this review and check answers in the back of the book. Answers to most review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. 1. Indicate whether each table defines a function. (A) Domain
Range
(B) Domain
1
4
7
3
6
8
5
8
9
Range
0
years during which a Super Bowl was played. If each team corresponds to the year or years in which they won the Super Bowl, does this correspondence define a function? Explain your answer. 4. Indicate whether each graph specifies a function: (A)
(C) Domain
Range
5
1
10
2
y
x
20
2. Indicate whether each set defines a function. Indicate whether any of the functions are one-to-one. Find the domain and range of each function. Find the inverse of any one-to-one functions. Find the domain and range of any inverse functions. (A) {(1, 1), (2, 4), (3, 9)} (B) {(1, 1), (1, 00041), (2, 2), (2, 00042)} (C) {(Albany, New York), (Utica, New York), (Akron, Ohio), (Dayton, Ohio)} (D) {(Albany, New York),(Akron, Ohio), (Tucson, Arizona), (Atlanta, Georgia), (Muncie, Indiana)} 3. Let T be the set of teams in the National Football League that have won at least one Super Bowl, and let Y be the set of
(B)
y
x
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Review Exercises y
(C)
17. Find f(00044), f(0), f(3), and f(5). 18. Find all values of x for which f (x) 0002 00042. 19. Find the domain and range of f. 20. Find the intervals over which f is increasing and decreasing.
x
21. Find any points of discontinuity. Problems 22–29 refer to the graphs of f and g shown here. y
(D)
x
f (x)
g(x)
5
5
00045
5
x
00045
5
00045
5. Which of the following equations define functions? (A) y 0002 x (B) y2 0002 x 3 (C) y 0002 x (D) 冟 y 冟 0002 x Problems 6–15 refer to the functions f, g, k, and m given by: g(x) 0002 4 0004 x2
f (x) 0002 3x 0003 5
k(x) 0002 5 m(x) 0002 2冟 x 冟 0004 1
8.
f (2 0003 h) 0004 f (2) h
m(00042) 0003 1 g (2) 0003 4
9.
g (a 0003 h) 0004 g (a) h
11. ( f 0004 g)(x)
12. ( fg)(x)
f 13. a b (x) g
14. ( f ° g)(x)
15. (g ° f )(x)
In Problems 24–27, use the graphs of f and g to find: 24. ( f ° g)(00041)
25. (g ° f )(00042)
26. f [g(1)]
27. g[ f(00043)]
Problems 31–36 refer to the graph of the function f used in Problems 17–21. Sketch the graph of each of the following.
16. For f (x) 0002 x 0004 2x, find (B) f (00044)
23. Construct a table of values of ( fg)(x) for x 0002 00043, 00042, 00041, 0, 1, 2, and 3, and sketch the graph of fg.
30. Indicate whether each function is even, odd, or neither: (A) f (x) 0002 x5 0003 6x (B) g(t) 0002 t 4 0003 3t 2 (C) h(z) 0002 z5 0003 4z2
2
(A) f(1)
22. Construct a table of values of ( f 0004 g)(x) for x 0002 00043, 00042, 00041, 0, 1, 2, and 3, and sketch the graph of f 0004 g.
29. Is g a one-to-one function?
7.
10. ( f 0003 g)(x)
00045
28. Is f a one-to-one function?
Find the indicated quantities or expressions. 6. f (2) 0003 g (00042) 0003 k (0)
(C) f (2) ⴢ f (00041)
(D)
f (0) f (3)
Problems 17–21 refer to the function f given by the following graph.
31. f (x) 0003 1
32. f (x 0003 1)
33. 0004f (x)
34. 0.5f (x)
35. f (2x)
36. 0004f (0004x)
37. Match each equation with a graph of one of the functions f, g, m, or n in the figure. Each graph is a graph of one of the equations. (B) y 0002 0004(x 0003 2)2 0003 4 (A) y 0002 (x 0004 2)2 0004 4 2 (C) y 0002 0004(x 0004 2) 0003 4 (D) y 0002 (x 0003 2)2 0004 4 y
f
g
5
f(x) 5
00045
5
00045
x
00045
5
m
n
x
x
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38. Referring to the graph of function f in the figure for Problem 37 and using known properties of quadratic functions, find each of the following to the nearest integer: (A) Intercepts (B) Vertex (C) Maximum or minimum (D) Range (E) Interval of increase (F) Interval of decrease 39. Let f (x) ⫽ x 2 ⫺ 4 and g(x) ⫽ x ⫹ 3. Find each of the following functions and find their domains. (A) f兾g (B) g兾f (C) f ° g (D) g ° f 40. For each function, find the maximum or minimum value without graphing. Then write the coordinates of the vertex. (A) f (x) ⫽ ⫺2(x ⫹ 4) 2 ⫺ 10 (B) f (x) ⫽ x 2 ⫺ 6x ⫹ 11 41. Complete the square to write the quadratic function in vertex form: q (x) ⫽ 2x 2 ⫺ 14x ⫹ 3
In Problems 52–57, find the domain, y intercept (if it exists), and any x intercepts. 52. m(x) ⫽ x2 ⫺ 4x ⫹ 5
53. r(x) ⫽ 2 ⫹ 31x
2
54. p(x) ⫽
1⫺x x3
55. f (x) ⫽
x 13 ⫺ x
56. g(x) ⫽
2x ⫹ 3 x2 ⫺ 4
57. h(x) ⫽
1 4 ⫺ 1x
58. Let f (x) ⫽ 0.5x2 ⫺ 4x ⫹ 5. (A) Sketch the graph of f and label the axis and the vertex. (B) Where is f increasing? Decreasing? What is the range? (Express answers in interval notation.) 59. Find the equations of the linear function g and the quadratic function f whose graphs are shown in the figure. This line is called the tangent line to the graph of f at the point (⫺1, 0).
42. How are the graphs of the following related to the graph of y ⫽ x2? (A) y ⫽ ⫺x2 (B) y ⫽ x2 ⫺ 3 (C) y ⫽ (x ⫹ 3)2
y 5
y ⫽ g (x)
y ⫽ f (x) ⫺5
5
x
Problems 43–49 refer to the function q given by the following graph. ⫺5
q(x)
60. Let
5
f (x) ⫽ e ⫺5
5
x
⫺5
43. Find y to the nearest integer: (A) y ⫽ q(0) (B) y ⫽ q(1) (C) y ⫽ q(2) (D) y ⫽ q(⫺2) 44. Find x to the nearest integer: (A) q(x) ⫽ 0 (B) q(x) ⫽ 1 (C) q(x) ⫽ ⫺3 (D) q(x) ⫽ 3 45. Find the domain and range of q. 46. Find the intervals over which q is increasing, decreasing, and constant. 47. Identify any points of discontinuity. 48. The function f multiplies the cube of the domain element by 4 and then subtracts the square root of the domain element. Write an algebraic definition of f. 49. Write a verbal description of the function f(x) ⫽ 3x2 ⫹ 4x ⫺ 6. In Problems 50 and 51, determine if the indicated equation defines a function. Justify your answer. 50. x ⫹ 2y ⫽ 10
51. x ⫹ 2y2 ⫽ 10
⫺x ⫺ 5 0.2x2
for ⫺4 ⱕ x 6 0 for 0 ⱕ x ⱕ 5
(A) Find f (⫺4), f(⫺2), f(0), f(2), and f(5). (B) Sketch the graph of y ⫽ f (x). (C) Find the domain and range. (D) Find any points of discontinuity. (E) Find the intervals over which f is increasing, decreasing, and constant. 61. Given f (x) ⫽ 1x ⫺ 8 and g(x) ⫽ 冟 x 冟: (A) Find f ° g and g ° f. (B) Find the domains of f ° g and g ° f. 62. Which of the following functions are one-to-one? (A) f(x) ⫽ x3 (B) g(x) ⫽ (x ⫺ 2)2 (C) h(x) ⫽ 2x ⫺ 3 (D) F(x) ⫽ (x ⫹ 3)2, x ⱖ ⫺3 63. Is u(x) ⫽ 4x ⫺ 8 the inverse of v(x) ⫽ 0.25x ⫹ 2? 64. The function f(x) ⫽ 2(x ⫺ 3)2 is not one-to-one. (A) Graph f using transformations of y ⫽ x2. (B) Restrict the domain of f to make it a one-to-one function. (C) Find the inverse of the one-to-one function. 65. Given f (x) ⫽ 3x ⫺ 7: (A) Find f ⫺1(x). (B) Find f ⫺1(5). (C) Find f ⫺1 [f (x)]. (D) Is f increasing, decreasing, or constant on (⫺⬁, ⬁)?
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66. The following graph is the result of applying a sequence of transformations to the graph of y 0002 x2. Describe the transformations verbally and write an equation for the given graph.
255
y
(A) 5
Check by graphing your equation on a graphing calculator. y
00045
5
x
5
00045 00045
5
x y
(B) 5
00045
67. The graph of f (x) 0002 冟 x 冟 is vertically stretched by a factor of 3, reflected through the x axis, and shifted 2 units to the right and 5 units up to form the graph of the function g. Find an equation for the function g and graph g. 68. Write an equation for the following graph in the form y 0002 a(x 0004 h)2 0003 k, where a is either 00041 or 00031 and h and k are integers. Check by graphing your equation on a graphing calculator. y
00045
5
x
00045
72. The graph of f (x) 0002 冟 x 冟 is stretched vertically by a factor of 3, reflected through the x axis, shifted four units to the right and eight units up to form the graph of the function g. Find an equation for the function g and graph g. 73. The graph of m(x) 0002 x2 is stretched horizontally by a factor of 2, shifted two units to the left and four units down to form the graph of the function t. Find an equation for the function t and graph t.
5
00045
5
x
00045
69. The following graph is the result of applying a sequence of 3 transformations to the graph of y 0002 1 x. Describe the transformations verbally, and write an equation for the given graph. y
Use graph transformations to sketch the graph of each equation in Problems 74–81: 74. y 0002 冟 x 0003 1 冟
3 75. y 0002 1 0003 1 10004x
76. y 0002 冟 x 冟 0004 2
77. y 0002 9 0004 3 1x
78. y 0002 12 冟 x 冟
3 79. y 0002 1 4 0004 0.5x
80. y 0002 2 0004 3(x 0004 1)3
81. y 0002 0004冟 x 0003 1 冟 0004 1
Solve Problems 82 and 83. Express answers in interval notation.
5
82. x2 0003 x 6 20
83. x2 7 4x 0003 12
84. Find the domain of f (x) 0002 225 0004 x2. 00045
5
x
00045
85. Given f (x) 0002 x 2 and g(x) 0002 11 0004 x, find each function and its domain. (A) fg (B) f兾g (C) f ° g (D) g ° f 86. For the one-to-one function f given by
Check by graphing your equation on a graphing calculator. 70. How is the graph of f(x) 0002 0004(x 0004 2)2 0004 1 related to the graph of g(x) 0002 x2? 71. Each of the following graphs is the result of applying one or more transformations to the graph of one of the six basic functions in Figure 1, Section 3-3. Find an equation for the graph. Check by graphing the equation on a graphing calculator.
f (x) 0002 (A) Find f 00041(x). (B) Find f 00041(3). (C) Find f 00041 [ f (x)].
x00032 x00043
87. Given f (x) 0002 1x 0004 1: (A) Find f 00041(x). (B) Find the domain and range of f and f 00041. (C) Graph f, f 00041, and y 0002 x on the same coordinate system.
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Check by graphing f, f 00041, and y 0002 x in a squared window on a graphing calculator. 88. Given f(x) 0002 x2 0004 1, x 0: (A) Find the domain and range of f and f 00041. (B) Find f 00041(x). (C) Find f 00041(3). (D) Find f 00041[ f (4)]. (E) Find f 00041[ f (x)].
94. STOPPING DISTANCE Table 1 contains data related to the length of the skid marks left by an automobile when making an emergency stop. A model for the skid mark length L (in feet) of the auto is L 0002 f (s) 0002 0.06s2 0004 2.4s 0003 50, s 20 where s is speed in miles per hour.
Table 1
Check by graphing f, f 00041, and y 0002 x in a squared window on a graphing calculator. 89. A partial graph of the function f is shown in the figure. Complete the graph of f over the interval [0, 5] given that: (A) f is symmetric with respect to the y axis. (B) f is symmetric with respect to the origin. y 5
00045
5
x
00045
90. The function f is decreasing on [00045, 5] with f(00045) 0002 4 and f(5) 0002 00043. (A) If f is continuous on [00045, 5], how many times can the graph of f cross the x axis? Support your conclusion with examples and/or verbal arguments. (B) Repeat part A if the function does not have to be continuous.
APPLICATIONS 91. INCOME Megan works 20 hours per week at an electronics store to help pay for tuition and rent. She gets a base salary of $6 per hour, a commission of 10% on all sales over $2,000 for the week, and a bonus of $250 if her weekly sales are over $5,000. (A) Write a function that describes Megan’s weekly earnings, where x represents her weekly sales. (B) Find Megan’s weekly earnings if her sales are $2,000, $4,000, and $6,000. (C) If Megan needs to average at least $400 per week to cover her tuition and rent, how much does she need to sell on average each week? 92. On the set of a movie, a stuntman will be jumping from a helicopter that is hovering at a height of 120 feet, and landing in a moving truck full of chicken feathers. How many seconds after he jumps does the truck need to be in position? 93. BUSINESS—MARKUP POLICY A sporting goods store sells tennis shorts that cost $30 for $48 and sunglasses that cost $20 for $32. (A) If the markup policy of the store for items that cost over $10 is assumed to be linear and is reflected in the pricing of these two items, find a function r 0002 f(c) that expresses retail price r as a function of cost c. (B) What should be the retail price of a pair of skis that cost $105? (C) Find c 0002 f 00041(r) and find its domain and range. (D) What is the cost of a box of golf balls that retail for $39.99?
Speed (mph)
Length of Skid Marks (feet)
20
26
30
32
40
49
50
80
60
122
70
176
80
242
(A) Graph L 0002 f(s) and the data for skid mark length on the same axes. (B) Find s 0002 f 00041(L) and find its domain and range. (C) How fast (to the nearest mile) was the auto traveling if it left skid marks 200 feet long? 95. PRICE AND DEMAND The price $p per hot dog at which q hot dogs can be sold during a baseball game is given approximately by 9 p 0002 g(q) 0002 1,000 q 4,000 1 0003 0.002q (A) Find the range of g. (B) Find q 0002 g00041( p) and find its domain and range. (C) Express the revenue as a function of p. (D) Express the revenue as a function of q. 96. MARKET RESEARCH A market research firm is hired to study demand for a new blanket that looks an awful lot like a bathrobe worn backwards. They determine that if x units are produced each week and sold at a price of $p per unit, then the weekly demand, revenue, and cost equations are, respectively x 0002 500 0004 10p R(x) 0002 50x 0004 0.1x2 C(x) 0002 10x 0003 1,500 Express the weekly profit as a function of the price p and find the price that produces the largest profit. 97. CONSTRUCTION A farmer has 120 feet of fencing to be used in the construction of two identical rectangular pens sharing a common side (see the figure).
x
y y
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Group Activity
(A) Express the total area A(x) enclosed by both pens as a function of the width x. (B) From physical considerations, what is the domain of the function A? (C) Find the dimensions of the pens that will make the total enclosed area maximum. 98. COMPUTER SCIENCE In computer programming, it is often necessary to check numbers for certain properties (even, odd, perfect square, etc.). The greatest integer function provides a convenient method for determining some of these properties. Consider the function f (x) ⫽ x ⫺ ( 冀 1x冁)2 (A) Evaluate f for x ⫽ 1, 2, . . . , 16. (B) Find f (n2), where n is a positive integer. (C) What property of x does this function determine?
CHAPTER
ZZZ GROUP
99. Use the schedule in Table 2 to construct a piecewise-defined model for the taxes due for a single taxpayer in Virginia with a taxable income of x dollars. Find the tax on the following incomes: $2,000, $4,000, $10,000, $30,000.
Table 2 Virginia Tax Rate Schedule
Status Single
Taxable Income Over $
But Not Over
Tax Is
Of the Amount Over
0
$ 3,000
2%
$ 3,000
$ 5,000
$ 60 ⫹ 3%
$
$ 3,000
$ 5,000
$17,000
$120 ⫹ 5%
$ 5,000
$17,000
—
$720 ⫹ 5.75%
$17,000
3 ACTIVITY Mathematical Modeling: Choosing a Cell Phone Plan
The number of companies offering cellular telephone service has grown rapidly in recent years. The plans they offer vary greatly and it can be difficult to select the plan that is best for you. Here are five typical plans: Plan 1: A flat fee of $50 per month for unlimited calls. Plan 2: A $30 per month fee for a total of 30 hours of calls and an additional charge of $0.01 per minute for all minutes over 30 hours. Plan 3: A $5 per month fee and a charge of $0.04 per minute for all calls. Plan 4: A $2 per month fee and a charge of $0.045 per minute for all calls; the fee is waived if the charge for calls is $20 or more. Plan 5: A charge of $0.05 per minute for all calls; there are no additional fees.
(A) Construct a mathematical model for each plan that gives the total monthly cost in terms of the total number of minutes of calls placed in a month. (B) Compare plans 1 and 2. Determine how many minutes per month would make plan 1 cheaper and how many would make plan 2 cheaper. (C) Repeat part (B) for plans 1 and 3; plans 1 and 4; plans 1 and 5. (D) Repeat part (B) for plans 2 and 3; plans 2 and 4; plans 2 and 5. (E) Repeat part (B) for plans 3 and 4; plans 3 and 5. (F) Repeat part (B) for plans 4 and 5. (G) Is there one plan that is always better than all the others? Based on your personal calling history, which plan would you choose and why?
0
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4
C IN Chapters 2 and 3, we used lines and parabolas to model a variety of situations. But the graph of a line doesn't change direction, and the graph of a parabola has just one turning point. So to model more complicated phenomena, we will study the more general class of polynomial functions in Chapter 4. A polynomial function can have many turning points. We will investigate the graphs and zeros of polynomials and apply that knowledge to study functions that can be written as quotients of polynomials, that is, the rational functions. Finally, we will use the language of variation to describe a wide range of mathematical models used in engineering and the physical, social, and health sciences.
OUTLINE 4-1
Polynomial Functions, Division, and Models
4-2
Real Zeros and Polynomial Inequalities
4-3
Complex Zeros and Rational Zeros of Polynomials
4-4
Rational Functions and Inequalities
4-5
Variation and Modeling Chapter 4 Review Chapter 4 Group Activity: Interpolating Polynomials
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Polynomial Functions, Division, and Models Z Graphs of Polynomial Functions Z Polynomial Division Z Remainder and Factor Theorems Z Mathematical Modeling and Data Analysis
In this section, we will study polynomial functions, a class that includes the linear and quadratic functions of Chapter 3. Graphs of polynomials exhibit much greater variety than just lines and parabolas. We will examine the properties of the graphs of polynomial functions, and we will use tools from algebra (division and factorization) to understand those properties. We also will show how polynomials are used to model data for which linear and quadratic functions are unsuitable.
Z Graphs of Polynomial Functions In Chapter 3 we introduced linear and quadratic functions and their graphs (Fig. 1): f (x) 0002 ax 0003 b, a00040 f (x) 0002 ax2 0003 bx 0003 c, a00040
Linear function Quadratic function
10
10
000510
10
000510
10
000510
000510
Z Figure 1 Graphs of linear and quadratic functions.
A function such as g(x) 0002 7x4 0005 5x3 0003 (2 0003 9i)x2 0003 3x 0005 1.95 which is the sum of a finite number of terms, each of the form axk, where a is a number and k is a nonnegative integer, is called a polynomial function. The polynomial function g(x) is said to have degree 4 because x4 is the highest power of x that appears among the terms of g(x). Therefore, linear and quadratic functions are polynomial functions of degrees 1 and 2, respectively. The two functions h(x) 0002 x00051 and k(x) 0002 x100022, however, are not polynomial functions (the exponents 00051 and 12 are not nonnegative integers). Z DEFINITION 1 Polynomial Function If n is a nonnegative integer, a function that can be written in the form P(x) 0002 an xn 0003 an00051xn00051 0003 . . . 0003 a1x 0003 a0,
an 0004 0
is called a polynomial function of degree n. The numbers an, an00051, . . ., a1, a0 are called the coefficients of P(x).
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We will assume that the coefficients of a polynomial function are complex numbers, or real numbers, or rational numbers, or integers, depending on our interest. Similarly, the domain of a polynomial function can be the set of complex numbers, the set of real numbers, or an appropriate subset of either, depending on the situation. According to Definition 1, a nonzero constant function like f (x) 0002 5 has degree 0 (it can be written as f (x) 0002 5x0). The constant function with value 0 is considered to be a polynomial but is not assigned a degree.
Z DEFINITION 2 Zeros or Roots A number r is said to be a zero or root of a function P(x) if P(r) 0002 0.
The zeros of P(x) are the solutions of the equation P(x) 0002 0. So if the coefficients of a polynomial P(x) are real numbers, then the real zeros of P(x) are just the x intercepts of the graph of P(x). For example, the real zeros of the polynomial P(x) 0002 x2 0005 4 are 2 and 00052, the x intercepts of the graph of P(x) [Fig. 2(a)]. However, a polynomial may have zeros that are not x intercepts. Q(x) 0002 x2 0003 4, for example, has zeros 2i and 00052i, but its graph has no x intercepts [Fig. 2(b)]. 10
10
000510
10
000510
10
000510
000510
(a)
(b)
Z Figure 2 Real zeros are x intercepts.
EXAMPLE
1
Zeros and x Intercepts (A) Figure 3 shows the graph of a polynomial function of degree 5. List its real zeros. 200
00055
5
0005200
Z Figure 3
(B) List all zeros of the polynomial function P(x) 0002 (x 0005 4)(x 0003 7)3(x2 0003 9)(x2 0005 2x 0003 2) Which zeros of P(x) are x intercepts?
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(A) The real zeros are the x intercepts: 00054, 00052, 0, and 3. (B) Note first that P(x) is a polynomial because it can be written in the form of Definition 1 (it is not necessary to actually multiply out P(x) to find that form). The zeros of P(x) are the solutions to the equation P(x) 0002 0. Because a product equals 0 if and only if one of the factors equals 0, we can find the zeros by solving each of the following equations (the last was solved using the quadratic formula): x0005400020 x00024
(x 0003 7)3 0002 0 x 0002 00057
x2 0003 9 0002 0 x 0002 00063i
x2 0005 2x 0003 2 0002 0 x000210006i
Therefore, the zeros of P(x), are 4, 00057, 3i, 00053i, 1 0003 i, and 1 0005 i. Only two of the six zeros are real numbers and therefore x intercepts: 4 and 00057. 0002 MATCHED PROBLEM 1
(A) Figure 4 shows the graph of a polynomial function of degree 4. List its real zeros. 5
00055
5
00055
Z Figure 4
(B) List all zeros of the polynomial function P(x) 0002 (x 0003 5)(x2 0005 4)(x2 0003 4)(x2 0003 2x 0003 5) Which zeros of P(x) are x intercepts? 0002
A point on a continuous graph that separates an increasing portion from a decreasing portion, or vice versa, is called a turning point. The vertex of a parabola, for example, is a turning point. Linear functions with real coefficients have exactly one real zero and no turning points; quadratic functions with real coefficients have at most two real zeros and exactly one turning point.
ZZZ EXPLORE-DISCUSS 1
Examine Figures 2(a), 2(b), 3, and 4, which show the graphs of polynomial functions of degree 2, 2, 5, and 4, respectively. In each figure, all real zeros and all turning points of the function appear in the given viewing window. (A) Is the number of real zeros ever less than the degree? Equal to the degree? Greater than the degree? How is the number of real zeros of a polynomial related to its degree? (B) Is the number of turning points ever less than the degree? Equal to the degree? Greater than the degree? How is the number of turning points of a polynomial related to its degree?
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Explore-Discuss 1 suggests that graphs of polynomial functions with real coefficients have the properties listed in Theorem 1, which we accept now without proof. Property 3 is proved later in this section. The other properties are established in calculus.
Z THEOREM 1 Properties of Graphs of Polynomial Functions Let P(x) be a polynomial of degree n 0 with real coefficients. Then the graph of P(x): 1. 2. 3. 4. 5.
Is continuous for all real numbers Has no sharp corners Has at most n real zeros Has at most n 0005 1 turning points Increases or decreases without bound as x → 0007 and as x → 00050007*
Figure 5 shows graphs of representative polynomial functions of degrees 1 through 6, illustrating the five properties of Theorem 1.
y
y 5
5
00055
5
x
5
00055
x
5
00055
00055
00055
(c) h(x) ⴝ x5 ⴚ 6x3 ⴙ 8x ⴙ 1
y
y
y
5
5
5
x
00055
(d) F(x) ⴝ x2 ⴚ x ⴙ 1
5
5
x
00055
00055
x
5
00055
(b) g(x) ⴝ x3 ⴙ 5x
(a) f(x) ⴝ x ⴚ 2
00055
y
(e) G(x) ⴝ 2x4 ⴚ 7x2 ⴙ x ⴙ 3
00055
5
x
00055
(f) H(x) ⴝ x6 ⴚ 7x4 ⴙ 12 x 2 ⴚ x ⴚ 2
Z Figure 5 Graphs of polynomial functions.
*Remember that 0007 and 00050007 are not real numbers. The statement the graph of P(x) increases without bound as x → 00050007 means that for any horizontal line y 0002 b there is some interval (00050007, a] 0002 {x0003 x a} on which the graph of P(x) is above the horizontal line.
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2
Properties of Graphs of Polynomials Explain why each graph is not the graph of a polynomial function by listing the properties of Theorem 1 that it fails to satisfy. (A)
(B)
y
5
5
00055
5
x
MATCHED PROBLEM 2
y 5
00055
5
x
00055
00055
00055
SOLUTIONS
(C)
y
5
x
00055
(A) The graph has a sharp corner when x 0002 0. Property 2 fails. (B) There are no points on the graph with x coordinate less then or equal to 0, so properties 1 and 5 fail. (C) There are an infinite number of zeros and an infinite number of turning points, so properties 3 and 4 fail. Furthermore, the graph is bounded by the horizontal lines y 0002 00061, so property 5 fails.
0002
Explain why each graph is not the graph of a polynomial function by listing the properties of Theorem 1 that it fails to satisfy. (A)
(B)
y 5
(C)
y 5
00055
5
x
5
00055
00055
y
5
x
00055
00055
5
x
00055
0002 The shape of the graph of a polynomial function with real coefficients is similar to the shape of the graph of the leading term, that is, the term of highest degree. Figure 6 compares the graph of the polynomial h(x) 0002 x5 0005 6x3 0003 8x 0003 1 from Figure 5 with the graph of its leading term p(x) 0002 x5. The graphs are dissimilar near the origin, but as we zoom out, the shapes of the two graphs become quite similar. The leading term in the polynomial dominates all other terms combined. Because the graph of p(x) increases without bound as x → 0007, the same is true of the graph of h(x). And because the graph of p(x) decreases without bound as x → 00050007, the same is true of the graph of h(x). y ph
5 Z Figure 6 p(x) 0002 x ,
h(x) 0002 x 0005 6x 0003 8x 0003 1. 5
3
y
5
ph
500
ZOOM OUT 00055
5
00055
x
00055
5
0005500
x
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The left and right behavior of a polynomial function with real coefficients is determined by the left and right behavior of its leading term (see Fig. 6). Property 5 of Theorem 1 can therefore be refined. The various possibilities are summarized in Theorem 2.
Z THEOREM 2 Left and Right Behavior of Polynomial Functions Let P(x) 0004 an xn 0005 an00031xn00031 0005 . . . 0005 a1x 0005 a0 be a polynomial function with real coefficients, an 0006 0, n 0007 0. 1. an > 0, n even: The graph of P(x) increases without bound as x S 0002 and increases without bound as x S 00030002 (like the graphs of x2, x4, x6, etc.). 2. an > 0, n odd: The graph of P(x) increases without bound as x S 0002 and decreases without bound as x S 00030002 (like the graphs of x, x3, x5, etc.). 3. an < 0, n even: The graph of P(x) decreases without bound as x S 0002 and decreases without bound as x S 00030002 (like the graphs of 0003x2, 0003x4, 0003x6, etc.). 4. an < 0, n odd: The graph of P(x) decreases without bound as x S 0002 and increases without bound as x S 00030002 (like the graphs of 0003x, 0003x3, 0003x5, etc.). y
y
y
Case 1
x
x
x
Case 2
y
Case 3
x
Case 4
It is convenient to write P(x) → 0002 as an abbreviation for the phrase the graph of P(x) increases without bound. Using this notation, the left and right behavior in Case 4 of Theorem 2, for example, is P(x) → 00030002 as x → 0002 and P(x) → 0002 as x → 00030002.
EXAMPLE
3
Left and Right Behavior of Polynomials Determine the left and right behavior of each polynomial. (A) The degree of P(x) 0004 3 0003 x2 0005 4x3 0003 x4 0003 2x6 (B) The degree of Q(x) 0004 4x5 0005 8x3 0005 5x 0003 1
SOLUTIONS
(A) The degree P(x) is 6 (even) and the coefficient a6 is 00032 (negative), so the left and right behavior is the same as that of 0003x6 (Case 3 of Theorem 2): P(x) → 00030002 as x → 0002 and P(x) → 00030002 as x → 00030002. (B) The degree Q(x) is 5 (odd) and the coefficient a5 is 4 (positive), so the left and right behavior is the same as that of x5 (Case 2 of Theorem 2): P(x) → 0002 as x → 0002 and P(x) → 00030002 as x → 00030002. 0002
MATCHED PROBLEM 3
Determine the left and right behavior of each polynomial. (A) P(x) 0004 4x9 0003 3x11 0005 5 (B) Q(x) 0004 1 0003 2x50 0005 x100
0002
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EXAMPLE
4
Graphing a Polynomial Graph the polynomial P(x) 0002 x3 0005 12x 0005 16, 00055 x 5. List the real zeros and turning points.
SOLUTION
First we construct a table of values by calculating P(x) for each integer x, 00055 x 5. For example, P(00055) 0002 (00055)3 0005 12(00055) 0005 16 0002 000581
y 100
00055
5
x
0005100 3 Z Figure 7 P(x) 0002 x 0005 12x 0005 16.
MATCHED PROBLEM 4
ZZZ
x
P(x)
x
P(x)
00055
000581
1
000527
00054
000532
2
000532
00053
00057
3
000525
00052
0
4
0
00051
00055
5
49
0
000516
Then we plot the points in the table and join them with a smooth curve (Fig. 7). The zeros are 00052 and 4. The turning points are (00052, 0) and (2, 000532). Note that P(x) has the maximum number of turning points for a polynomial of degree 3, but one fewer than the maximum number of real zeros. 0002 Graph P(x) 0002 x4 0005 6x2 0005 8x 0005 3, 00054 x 4. List the real zeros and turning points. 0002 Finding the real zeros and turning points of a polynomial is usually more difficult than suggested by Example 4. In Example 4, how did we know that the real zeros were between 00055 and 5 rather than between, say, 95 and 105? Could there be another real zero just to the left or right of 00052? How do we know that (00052, 0) and (2, 000532), rather than nearby points having noninteger coordinates, are the turning points? To answer such questions we must view polynomials from an algebraic perspective. Polynomials can be factored. So next we will study the division and factorization of polynomials.
CAUTION ZZZ
Z Polynomial Division We can find quotients of polynomials by a long-division process similar to the one used in arithmetic. Example 5 will illustrate the process.
EXAMPLE
5
Polynomial Long Division Divide P(x) 0002 3x3 0005 5 0003 2x4 0005 x by 2 0003 x.
SOLUTION
First, rewrite the dividend P(x) in descending powers of x, inserting 0 as the coefficient for any missing terms of degree less than 4: P(x) 0002 2x4 0003 3x3 0003 0x2 0005 x 0005 5
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Similarly, rewrite the divisor 2 0002 x in the form x 0002 2. Then divide the first term x of the divisor into the first term 2x4 of the dividend. Multiply the result, 2x3, by the divisor, obtaining 2x4 0002 4x3. Line up like terms, subtract as in arithmetic, and bring down 0x2. Repeat the process until the degree of the remainder is less than the degree of the divisor. 2x3 0003 x2 0002 2x 0003 5 x 0002 2 0002 2x4 0002 3x3 0002 0x2 0003 x 0003 5 2x4 0002 4x3 0003x3 0002 0x2 0003x3 0003 2x2 2x2 0003 x
Divisor
Quotient Dividend Subtract
Subtract
2x2 0002 4x 00035x 0003 5 00035x 0003 10 5
Subtract
Subtract Remainder
Therefore, 5 2x4 0002 3x3 0003 x 0003 5 0004 2x3 0003 x2 0002 2x 0003 5 0002 x00022 x00022 CHECK
You can always check division using multiplication: 5 d x00022 0004 (x 0002 2)(2x3 0003 x2 0002 2x 0003 5) 0002 5 0004 2x4 0002 3x3 0003 x 0003 5
(x 0002 2) c 2x3 0003 x2 0002 2x 0003 5 0002
MATCHED PROBLEM 5
Multiply and collect like terms
0002
Divide 6x2 0003 30 0002 9x3 by x 0003 2. 0002 The procedure illustrated in Example 5 is called the division algorithm. The concluding equation of Example 5 (before the check) may be multiplied by the divisor x 0002 2 to give the following form: Dividend 4
ⴝ
3
Divisor ⴢ Quotient 3
ⴙ Remainder 2
2x 0002 3x 0003 x 0003 5 0004 (x 0002 2)(2x 0003 x 0002 2x 0003 5) 0002 5 This last equation is an identity: it is true for all replacements of x by real or complex numbers including x 0004 00032. Theorem 3, which we state without proof, gives the general result of applying the division algorithm when the divisor has the form x 0003 r.
Z THEOREM 3 Division Algorithm For each polynomial P(x) of degree greater than 0 and each number r, there exists a unique polynomial Q(x) of degree 1 less than P(x) and a unique number R such that P(x) 0004 (x 0003 r)Q(x) 0002 R The polynomial Q(x) is called the quotient, x 0003 r is the divisor, and R is the remainder. Note that R may be 0.
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There is a shortcut called synthetic division for the long division of Example 5. First write the coefficients of the dividend and the negative of the constant term of the divisor in the format shown below at the left. Bring down the 2 as indicated next on the right, multiply by 00052, and record the product 00054. Add 3 and 00054, bringing down their sum 00051. Repeat the process until the coefficients of the quotient and the remainder are obtained. Dividend coefficients
Dividend coefficients
2
3
0
00051
00055
00052 Negative of constant term of divisor
2
3
0
00051
00055
00052 2
00054 00051
2 2
00054 00055
10 5
Quotient coefficients
Remainder
Compare the preceding synthetic division to the long division shown below, in which the essential numerals appear in color, to convince yourself that synthetic division produces the correct quotient and remainder. (In synthetic division we use the negative of the constant term of the divisor so we can add rather than subtract.)
Divisor
2x3 ⴚ 1x2 ⴙ 2x ⴚ 5 x ⴙ 20004 2x4 0003 3x3 0003 0x2 ⴚ 1x ⴚ 5 2x4 0003 4x3 ⴚ1x3 0003 0x2 00051x3 ⴚ 2x2 2x2 0005 1x 2x2 ⴙ 4x ⴚ5x 0005 5 00055x ⴚ 10 5
Quotient Dividend
Remainder
Z KEY STEPS IN THE SYNTHETIC DIVISION PROCESS To divide the polynomial P(x) by x 0005 r: Step 1. Arrange the coefficients of P(x) in order of descending powers of x. Write 0 as the coefficient for each missing power. Step 2. After writing the divisor in the form x 0005 r, use r to generate the second and third rows of numbers as follows. Bring down the first coefficient of the dividend and multiply it by r; then add the product to the second coefficient of the dividend. Multiply this sum by r, and add the product to the third coefficient of the dividend. Repeat the process until a product is added to the constant term of P(x). Step 3. The last number to the right in the third row of numbers is the remainder. The other numbers in the third row are the coefficients of the quotient, which is of degree 1 less than P(x).
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6
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269
Synthetic Division Use synthetic division to divide P(x) 0004 4x5 0003 30x3 0003 50x 0003 2 by x 0002 3. Find the quotient and remainder. Write the conclusion in the form P(x) 0004 (x 0003 r)Q(x) 0002 R of Theorem 3.
SOLUTION
Because x 0002 3 0004 x 0003 (00033), we have r 0004 00033, and 4 00033 4
4
0 000312 000312
000330 36 6
0 000318 000318
000350 54 4
00032 000312 000314
3
The quotient is 4x 0003 12x 0002 6x2 0003 18x 0002 4 with a remainder of 000314. So 4x5 0003 30x3 0003 50x 0003 2 0004 (x 0002 3)(4x4 0003 12x3 0002 6x2 0003 18x 0002 4) 0003 14 MATCHED PROBLEM 6
0002
Repeat Example 6 with P(x) 0004 3x4 0003 11x3 0003 18x 0002 8 and divisor x 0003 4. 0002
Z Remainder and Factor Theorems ZZZ EXPLORE-DISCUSS 2
Let P(x) 0004 x3 0003 3x2 0003 2x 0002 8. (A) Evaluate P(x) for (i) x 0004 00032 (ii) x 0004 1
(iii) x 0004 3
(B) Use synthetic division to find the remainder when P(x) is divided by (i) x 0002 2 (ii) x 0003 1 (iii) x 0003 3 What conclusion does a comparison of the results in parts A and B suggest? Explore-Discuss 2 suggests that when a polynomial P(x) is divided by x 0003 r, the remainder is equal to P(r), the value of the polynomial P(x) at x 0004 r. In Problem 87 of Exercises 4-1, you are asked to complete a proof of this fact, which is called the remainder theorem. Z THEOREM 4 Remainder Theorem If R is the remainder after dividing the polynomial P(x) by x 0003 r, then P(r) 0004 R
EXAMPLE
7
Two Methods for Evaluating Polynomials If P(x) 0004 4x4 0002 10x3 0002 19x 0002 5, find P(00033) by (A) Using the remainder theorem and synthetic division (B) Evaluating P(00033) directly
SOLUTIONS
(A) Use synthetic division to divide P(x) by x 0003 (00033). 4 00033
4
10 000312 00032
0 6 6
19 000318 1
5 00033 2 0004 R 0004 P(00033)
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(B) P(00053) 0002 4(00053)4 0003 10(00053)3 0003 19(00053) 0003 5 00022
MATCHED PROBLEM 7
0002
Repeat Example 7 for P(x) 0002 3x4 0005 16x2 0005 3x 0005 7 and x 0002 00052. 0002 You might think the remainder theorem is not a very effective tool for evaluating polynomials. But let’s consider the number of operations performed in parts A and B of Example 7. Synthetic division requires only four multiplications and four additions to find P(00053), whereas the direct evaluation requires ten multiplications and four additions. [Note that evaluating 4(00053)4 actually requires five multiplications.] The difference becomes even larger as the degree of the polynomial increases. Computer programs that involve numerous polynomial evaluations often use synthetic division because of its efficiency. We will find synthetic division and the remainder theorem to be useful tools later in this chapter. The remainder theorem shows that the division algorithm equation, P(x) 0002 (x 0005 r)Q(x) 0003 R can be written in the form where R is replaced by P(r): P(x) 0002 (x 0005 r)Q(x) 0003 P(r) Therefore, x 0005 r is a factor of P(x) if and only if P(r) 0002 0, that is, if and only if r is a zero of the polynomial P(x). This result is called the factor theorem.
Z THEOREM 5 Factor Theorem If r is a zero of the polynomial P(x), then x 0005 r is a factor of P(x). Conversely, if x 0005 r is a factor of P(x), then r is a zero of P(x).
EXAMPLE
8
Factors of Polynomials Use the factor theorem to show that x 0003 1 is a factor of P(x) 0002 x25 0003 1 but is not a factor of Q(x) 0002 x25 0005 1.
SOLUTION
Because P(00051) 0002 (00051)25 0003 1 0002 00051 0003 1 0002 0 x 0005 (00051) 0002 x 0003 1 is a factor of x25 0003 1. On the other hand, Q(00051) 0002 (00051)25 0005 1 0002 00051 0005 1 0002 00052 and x 0003 1 is not a factor of x25 0005 1.
MATCHED PROBLEM 8
0002
Use the factor theorem to show that x 0005 i is a factor of P(x) 0002 x8 0005 1 but is not a factor of Q(x) 0002 x8 0003 1. 0002 One consequence of the factor theorem is Theorem 6 (a proof is outlined in Problem 88 in Exercises 4-1).
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Z THEOREM 6 Zeros of Polynomials A polynomial of degree n has at most n zeros.
Theorem 6 says that the graph of a polynomial of degree n with real coefficients has at most n real zeros (Property 3 of Theorem 1). The polynomial H(x) 0002 x6 0005 7x4 0003 12x2 0005 x 0005 2 for example, has degree 6 and the maximum number of zeros [see Fig. 5(f ), p. 263]. Of course, polynomials of degree 6 may have fewer than six real zeros. In fact, p(x) 0002 x6 0003 1 has no real zeros. However, it can be shown that the polynomial p(x) 0002 x6 0003 1 has exactly six complex zeros.
Z Mathematical Modeling and Data Analysis In Chapters 2 and 3 we saw that linear and quadratic functions can be useful models for certain sets of data. For some data, however, no linear function and no quadratic function can provide a reasonable model. In that case, we investigate the suitability of polynomial models of degree greater than 2. In Examples 9 and 10 we discuss cubic and quartic models, respectively, for the given data.
EXAMPLE
9
Table 1 Sturgeon
Estimating the Weight of Fish Scientists and fishermen often estimate the weight of a fish from its length. The data in Table 1 give the average weight of North American sturgeon for certain lengths. Because weight is associated with volume, which involves three dimensions, we might expect that weight would be associated with the cube of the length. A cubic model for the data is given by
Length (in.) x
Weight (oz.) y
18
13
22
26
26
46
30
75
34
115
38
166
where y is the weight (in ounces) of a sturgeon that has length x (in inches).
44
282
(A) Use the model to estimate the weight of a sturgeon of length 56 inches.
52
492
60
796
y 0002 0.00526x3 0005 0.117x2 0003 1.43x 0005 5.00
(B) Compare the weight of a sturgeon of length 44 inches as given by Table 1 with the weight given by the model.
Source: www.thefishernet.com
SOLUTIONS
(A) If x 0002 56, then y 0002 0.00526(56)3 0005 0.117(56)2 0003 1.43(56) 0005 5.00 0005 632 ounces (B) If x 0002 44, then y 0002 0.00526(44)3 0005 0.117(44)2 0003 1.43(44) 0005 5.00 0005 279 ounces The weight given by the table, 282 ounces, is 3 ounces greater than the weight given by the model. 0002
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Technology Connections Figure 8 shows the details of constructing the cubic model of Example 9 on a graphing calculator. 1,000
0
70
0
(a) Entering the data
(b) Finding the model
(c) Graphing the data and the model
Z Figure 8
MATCHED PROBLEM 9
Use the cubic model of Example 9. (A) Estimate the weight of a sturgeon of length 65 inches. (B) Compare the weight of a sturgeon of length 30 inches as given by Table 1 with the weight given by the model. 0002
EXAMPLE
10
Table 2
Hydroelectric Power The data in Table 2 gives the annual consumption of hydroelectric power (in quadrillion BTU) in the United States for selected years since 1983. From Table 2 it appears that a polynomial model of the data would have three turning points—near 1989, 1997, and 2001. Because a polynomial with three turning points must have degree at least four, we can model the data with a quartic (fourth-degree) polynomial:
Year
U.S. Consumption of Hydroelectric Power (Quadrillion BTU)
1983
3.90
1985
3.40
1987
3.12
1989
2.99
1991
3.14
1993
3.13
1995
3.48
1997
3.88
1999
3.47
(A) Use the model to predict the consumption of hydroelectric power in 2018.
2001
2.38
2003
2.53
(B) Compare the consumption of hydroelectric power in 2003 (as given by Table 2) to the consumption given by the model.
2005
2.61
y 0004 0.00013x4 0003 0.0067x3 0002 0.107x2 0003 0.59x 0002 4.03 where y is the consumption (in quadrillion BTU) and x is time in years with x 0004 0 representing 1983.
Source: U.S. Department of Energy
SOLUTIONS
(A) If x 0004 35 (which represents the year 2018), then y 0004 0.00013(35)4 0003 0.0067(35)3 0002 0.107(35)2 0003 0.59(35) 0002 4.03 0003 22.3 The model predicts a consumption of 22.3 quadrillion BTU in 2018. However, because the predicted consumption for 2018 is so dramatically greater than earlier consumption levels, it is unlikely to be accurate. This brings up an important point: A model that fits a set of data points well is not automatically a good model for predicting future trends.
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(B) If x 0002 20 (which represents 2003), then y 0002 0.00013(20)4 0005 0.0067(20)3 0003 0.107(20)2 0005 0.59(20) 0003 4.03 0002 2.23 The consumption reported in the table, 2.53 quadrillion BTU, is 0.30 quadrillion BTU greater than the consumption given by the model. 0002
Technology Connections Figure 9 shows the details of constructing the quartic model of Example 10 on a graphing calculator. 5
0
(a)
(b)
30
00052
(c)
Z Figure 9
MATCHED PROBLEM 10
Use the quartic model of Example 10. (A) Estimate the consumption of hydroelectric power in 2000. (B) Compare the consumption of hydroelectric power in 1991 (as given by Table 2) to the consumption given by the model. 0002 ANSWERS TO MATCHED PROBLEMS 1. (A) 00051, 1, 2 (B) The zeros are 00055, 00052, 2, 2i, 00052i, 00051 0003 2i, and 00051 00052i; the x intercepts are 00055, 00052, and 2. 2. (A) Properties 1 and 5 (B) Property 5 (C) Properties 1 and 5 3. (A) P(x) S 00050007 as x S 0007 and P(x) S 0007 as x S 00050007. (B) Q(x) S 0007 as x S 0007 and Q(x) S 0007 as x S 00050007. 4. y 200
00055
5
x
0005200
zeros: 00051, 3; turning point; (2, 000527) 66 5. 9x2 0003 24x 0003 48 0003 x00052
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6. 3x4 0003 11x3 0003 18x 0005 8 0004 (x 0003 4)(3x3 0005 x2 0005 4x 0003 2) 7. P(00032) 0004 00033 for both parts, as it should 8. P(i) 0004 0, so x 0003 i is a factor of x8 0003 1; Q(i) 0004 2, so x 0003 i is not a factor of x8 0005 1 9. (A) 1,038 in. (B) The weight given in the table is 0.38 oz greater than the weight given by the model. 10. (A) 2.86 quadrillion BTU (B) The consumption given in the table is 0.12 quadrillion BTU less than the consumption given by the model.
4-1
Exercises y
y
1. What is a polynomial function? 2. Explain the connection between the zeros of a polynomial and its linear factors. 3. Explain what is wrong with the following setup for dividing x4 0005 5x2 0003 2x 0005 6 by x 0003 2 using synthetic division. 1
5
x
x
00032 6
20 4. Explain what is wrong with the following setup for dividing 3x3 0003 x2 0005 8x 0005 9 by x 0005 4 using synthetic division. 00031 8
3
(c)
(d)
9 In Problems 13–16, list the real zeros and turning points, and state the left and right behavior, of the polynomial function P(x) that has the indicated graph.
40 In Problems 5–8, decide whether the statement is true or false, and explain your answer.
y
13. 5
5. Every quadratic function is a polynomial function. 6. Every polynomial of degree 3 has three x intercepts. 7. If a polynomial has no x intercepts, then it has no zeros.
00035
5
x
8. Every polynomial function is continuous. In Problems 9–12, a is a positive real number. Match each function with one of graphs (a)–(d). 9. f(x) 0004 ax3
10. g(x) 0004 0003ax4
11. h(x) 0004 ax6
12. k(x) 0004 0003ax5
y
00035
y
14. 5
y
00035
x
5
x 00035
(a)
(b)
x
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15.
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275
y
20.
5
3
00055
5
x
00053
x
3
00053
00055
y
16. 5
In Problems 21–24, list all zeros of each polynomial function, and specify those zeros that are x intercepts.
00055
5
21. P(x) 0002 x(x2 0005 9)(x2 0003 4)
x
22. P(x) 0002 (x2 0005 4)(x4 0005 1) 23. P(x) 0002 (x 0003 5)(x2 0003 9)(x2 0003 16) 24. P(x) 0002 (x2 0005 5x 0003 6)(x2 0005 5x 0003 7)
00055
In Problems 17–20, explain why each graph is not the graph of a polynomial function. y
17.
In Problems 25–34, use algebraic long division to find the quotient and the remainder. 25. (3x2 0003 5x 0003 6) (x 0003 1) 26. (2x2 0005 7x 0003 4) (x 0005 2)
2
27. (4m2 0005 1) (m 0005 1) 00052
x
2
28. (y2 0005 9) ( y 0003 3) 29. (6 0005 6x 0003 8x2) (x 0003 1)
00052
30. (11x 0005 2 0003 12x2) (3x 0003 2) 31.
y
18.
x3 0005 1 x00051
32.
a3 0003 27 a00033
33. (3y 0005 y2 0003 2y3 0005 1) ( y 0003 2)
5
34. (3 0003 x3 0005 x) (x 0005 3)
00055
5
x
In Problems 35–40, divide using synthetic division. 35. (x2 0003 3x 0005 7) (x 0005 2) 36. (x2 0003 3x 0005 3) (x 0005 3)
00055
37. (4x2 0003 10x 0005 9) (x 0003 3) 38. (2x2 0003 7x 0005 5) (x 0003 4)
y
19. 3
00053
39.
3
x
2x3 0005 3x 0003 1 x00052
40.
In Problems 41–44, is the given number a zero of the polynomial? Use synthetic division. 41. x2 0003 4x 0005 221; 000517
00053
x3 0003 2x2 0005 3x 0005 4 x00032
42. x2 0005 7x 0005 551; 29 43. 2x3 0003 38x2 0005 x 0003 19; 000519 44. 2x3 0005 397x 0003 70; 14
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In Problems 45–48, determine whether the second polynomial is a factor of the first polynomial without dividing or using synthetic division. 45. x18 0003 1; x 0003 1
46. x18 0003 1; x 0005 1
47. 3x3 0003 7x2 0003 8x 0005 2; x 0005 1
In Problems 73 and 74, divide, using synthetic division. 73. (x3 0003 3x2 0005 x 0003 3) (x 0003 i) 74. (x3 0003 2x2 0005 x 0003 2) (x 0005 i) 75. Let P(x) 0004 x2 0005 2ix 0003 10. Use synthetic division to find: (A) P(2 0003 i)
48. 3x4 0003 2x3 0005 5x 0003 6; x 0003 1
(B) P(5 0003 5i)
Use synthetic division and the remainder theorem in Problems 49–54. 49. Find P(00032), given P(x) 0004 3x2 0003 x 0003 10. 50. Find P(00033), given P(x) 0004 4x2 0005 10x 0003 8. 51. Find P(2), given P(x) 0004 2x3 0003 5x2 0005 7x 0003 7. 52. Find P(5), given P(x) 0004 2x3 0003 12x2 0003 x 0005 30. 4
(C) P(3 0003 i) (D) P(00033 0003 i) 76. Let P(x) 0004 x2 0003 4ix 0003 13. Use synthetic division to find: (A) P(5 0005 6i) (B) P(1 0005 2i) (C) P(3 0005 2i) (D) P(00033 0005 2i)
2
53. Find P(00034), given P(x) 0004 x 0003 10x 0005 25x 0003 2. 54. Find P(00037), given P(x) 0004 x4 0005 5x3 0003 13x2 0003 30. In Problems 55–62, use synthetic division to find the quotient and the remainder. As coefficients get more involved, a calculator should prove helpful. Do not round off. 55. (3x4 0003 x 0003 4) (x 0005 1) 56. (5x4 0003 2x2 0003 3) (x 0003 1) 5
57. (x 0005 1) (x 0005 1) 58. (x4 0003 16) (x 0003 2) 59. (3x4 0005 2x3 0003 4x 0003 1) (x 0005 3) 60. (x4 0003 3x3 0003 5x2 0005 6x 0003 3) (x 0003 4) 61. (2x6 0003 13x5 0005 75x3 0005 2x2 0003 50) (x 0003 5) 62. (4x6 0005 20x5 0003 24x4 0003 3x2 0003 13x 0005 30) (x 0005 6) In Problems 63–68, without graphing, state the left and right behavior, the maximum number of x intercepts, and the maximum number of local extrema. 63. P(x) 0004 x3 0003 5x2 0005 2x 0005 6 64. P(x) 0004 x3 0005 2x2 0003 5x 0003 3 65. P(x) 0004 0003x3 0005 4x2 0005 x 0005 5 66. P(x) 0004 0003x3 0003 3x2 0005 4x 0003 4 67. P(x) 0004 x4 0005 x3 0003 5x2 0003 3x 0005 12 68. P(x) 0004 0003x4 0005 6x2 0003 3x 0003 16 In Problems 69–72, either give an example of a polynomial with real coefficients that satisfies the given conditions or explain why such a polynomial cannot exist. 69. P(x) is a third-degree polynomial with one x intercept. 70. P(x) is a fourth-degree polynomial with no x intercepts. 71. P(x) is a third-degree polynomial with no x intercepts. 72. P(x) is a fourth-degree polynomial with no turning points.
In Problems 77–82, approximate (to two decimal places) the x intercepts and the local extrema. 77. P(x) 0004 40 0005 50x 0003 9x2 0003 x3 78. P(x) 0004 40 0005 70x 0005 18x2 0005 x3 79. P(x) 0004 0.04x3 0003 10x 0005 5 80. P(x) 0004 00030.01x3 0005 2.8x 0003 3 81. P(x) 0004 0.1x4 0005 0.3x3 0003 23x2 0003 23x 0005 90 82. P(x) 0004 0.1x4 0005 0.2x3 0003 19x2 0005 17x 0005 100 83. (A) What is the least number of turning points that a polynomial function of degree 4, with real coefficients, can have? The greatest number? Explain and give examples. (B) What is the least number of x intercepts that a polynomial function of degree 4, with real coefficients, can have? The greatest number? Explain and give examples. 84. (A) What is the least number of turning points that a polynomial function of degree 3, with real coefficients, can have? The greatest number? Explain and give examples. (B) What is the least number of x intercepts that a polynomial function of degree 3, with real coefficients, can have? The greatest number? Explain and give examples. 85. Is every polynomial of even degree an even function? Explain. 86. Is every polynomial of odd degree an odd function? Explain. 87. Prove the remainder theorem (Theorem 4): (A) Write the result of the division algorithm if a polynomial P(x) is divided by x 0003 r. (B) Evaluate both sides of the equation from part (A) when x 0004 r. What can you conclude? 88. In this problem, we will prove that a polynomial of degree n has at most n zeros (Theorem 6). Give a reason for each step. Let P(x) be a polynomial of degree n, and suppose that P has n distinct zeros r1, r2, . . . , rn . We will show that it is impossible for P to have any other zeros. Step 1: We can write P(x) in the form P(x) 0004 (x 0003 r1)Q1(x), where the degree of Q1(x) is n 0003 1.
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Step 2: r2 is a zero of Q1(x). Step 3: We can write Q1(x) in the form Q1(x) 0004 (x 0003 r2)Q2(x), where the degree of Q2(x) is n 0003 2. Step 4: P(x) 0004 (x 0003 r1)(x 0003 r2)Q2(x) Step 5: P(x) 0004 (x 0003 r1)(x 0003 r2). . .(x 0003 rn)Qn(x), where the degree of Qn(x) is 0. Step 6: The only zeros of P are r1, r2, . . . , rn.
(B) Find the volume of the plastic coating to four decimal places if the thickness of the shielding is 0.005 feet. Problems 93–96 require a graphing calculator or a computer that can calculate cubic regression polynomials for a given data set. 93. HEALTH CARE Table 3 shows the total national health care expenditures (in billion dollars) and the per capita expenditures (in dollars) for selected years since 1960.
APPLICATIONS
Table 3 National Health Care Expenditures
89. REVENUE The price–demand equation for 8,000-BTU window air conditioners is given by
Year
p 0004 0.0004x2 0003 x 0002 569
0 0005 x 0005 800
277
Polynomial Functions, Division, and Models
Total Expenditures (Billion $)
Per Capita Expenditures ($)
1960
28
148
where x is the number of air conditioners that can be sold at a price of p dollars each. (A) Find the revenue function. (B) Find the number of air conditioners that must be sold to maximize the revenue, the corresponding price to the nearest dollar, and the maximum revenue to the nearest dollar.
1970
75
356
1980
253
1,100
1990
714
2,814
2000
1,353
4,789
90. PROFIT Refer to Problem 89. The cost of manufacturing 8,000BTU window air conditioners is given by
2007
2,241
7,421
C(x) 0004 10,000 0002 90x where C(x) is the total cost in dollars of producing x air conditioners. (A) Find the profit function. (B) Find the number of air conditioners that must be sold to maximize the profit, the corresponding price to the nearest dollar, and the maximum profit to the nearest dollar. 91. CONSTRUCTION A rectangular container measuring 1 foot by 2 feet by 4 feet is covered with a layer of lead shielding of uniform thickness (see the figure). (A) Find the volume of lead shielding V as a function of the thickness x (in feet) of the shielding. (B) Find the volume of the lead shielding if the thickness of the shielding is 0.05 feet.
4
1 2 Lead shielding
Source: U.S. Census Bureau.
(A) Let x represent the number of years since 1960 and find a cubic regression polynomial for the total national expenditures. (B) Use the polynomial model from part A to estimate the total national expenditures (to the nearest billion) for 2018. 94. HEALTH CARE Refer to Table 3. (A) Let x represent the number of years since 1960 and find a cubic regression polynomial for the per capita expenditures. (B) Use the polynomial model from part A to estimate the per capita expenditures (to the nearest dollar) for 2018. 95. MARRIAGE Table 4 shows the marriage and divorce rates per 1,000 population for selected years since 1950.
Table 4 Marriages and Divorces (per 1,000 Population) Year
Marriages
Divorces
1950
11.1
2.6
1960
8.5
2.2
1970
10.6
3.5
1980
10.6
5.2
1990
9.8
4.7
2000
8.2
4.1
Source: U.S. Census Bureau.
(A) Let x represent the number of years since 1950 and find a cubic regression polynomial for the marriage rate. (B) Use the polynomial model from part A to estimate the marriage rate (to one decimal place) for 2016. 92. MANUFACTURING A rectangular storage container measuring 2 feet by 2 feet by 3 feet is coated with a protective coating of plastic of uniform thickness. (A) Find the volume of plastic V as a function of the thickness x (in feet) of the coating.
96. DIVORCE Refer to Table 4. (A) Let x represent the number of years since 1950 and find a cubic regression polynomial for the divorce rate. (B) Use the polynomial model from part A to estimate the divorce rate (to one decimal place) for 2016.
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4-2
Real Zeros and Polynomial Inequalities Z Upper and Lower Bounds for Real Zeros Z Location Theorem and Bisection Method Z Approximating Real Zeros at Turning Points Z Polynomial Inequalities Z Mathematical Modeling
The real zeros of a polynomial P(x) with real coefficients are just the x intercepts of the graph of P(x). So an obvious strategy for finding the real zeros consists of two steps: 1. 2.
Graph P(x). Approximate each x intercept.
In this section, we develop important tools for carrying out this strategy: the upper and lower bound theorem, which determines an interval [a, b] that is guaranteed to contain all x intercepts of P(x), and the bisection method, which permits approximation of x intercepts to any desired accuracy. We emphasize the approximation of real zeros in this section; the problem of finding zeros exactly, when possible, is considered in Section 4-3.
Z Upper and Lower Bounds for Real Zeros On which interval should you graph a polynomial P(x) in order to see all of its x intercepts? The answer is provided by the upper and lower bound theorem. This theorem explains how to find two numbers: a lower bound, which is less than or equal to all real zeros of the polynomial, and an upper bound, which is greater than or equal to all real zeros of the polynomial. A proof of Theorem 1 is outlined in Problems 67 and 68 of Exercises 4-2.
Z THEOREM 1 Upper and Lower Bound Theorem Let P(x) be a polynomial of degree n 7 0 with real coefficients, an 7 0: 1. Upper bound: A number r 7 0 is an upper bound for the real zeros of P(x) if, when P(x) is divided by x 0005 r by synthetic division, all numbers in the quotient row, including the remainder, are nonnegative. 2. Lower bound: A number r 6 0 is a lower bound for the real zeros of P(x) if, when P(x) is divided by x 0005 r by synthetic division, all numbers in the quotient row, including the remainder, alternate in sign. [Note: In the lower bound test, if 0 appears in one or more places in the quotient row, including the remainder, the sign in front of it can be considered either positive or negative, but not both. For example, the numbers 1, 0, 1 can be considered to alternate in sign, whereas 1, 0, 00051 cannot.]
EXAMPLE
1
Bounding Real Zeros Let P(x) 0002 x4 0005 2x 3 0005 10x 2 0003 40x 0005 90. Find the smallest positive integer and the largest negative integer that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x).
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SOLUTION
400
5
LB
1 2 3 4 5 00031 00032 00033 00034 00035
1 1 1 1 1 1 1 1 1 1 1
00032 00031 0 1 2 3 00033 00034 00035 00036 00037
000310 000311 000310 00037 00032 5 00037 00032 5 14 25
40 29 20 19 32 65 47 44 25 000316 000385
000390 000361 000350 000333 38 This quotient row is nonnegative; 235 ← E 5 is an upper bound (UB). 0003137 0003178 0003165 000326 This quotient row alternates in sign; 335 ← E ⴚ5 is a lower bound (LB).
The graph of P(x) 0004 x4 0003 2x 3 0003 10x 2 0005 40x 0003 90 for 00035 x 5 is shown in Figure 1. Theorem 1 guarantees that all the real zeros of P(x) are between 00035 and 5. We can be certain that the graph does not change direction and cross the x axis somewhere outside the viewing window in Figure 1. 0002
0003200 4 3 Z Figure 1 P(x) 0004 x 0003 2x 0003
10x2 0005 40x 0003 90.
Let P(x) 0004 x4 0003 5x 3 0003 x 2 0005 40x 0003 70. Find the smallest positive integer and the largest negative integer that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x). 0002
MATCHED PROBLEM 1
EXAMPLE
279
We perform synthetic division for r 0004 1, 2, 3, . . . until the quotient row turns nonnegative; then repeat this process for r 0004 00031, 00032, 00033, . . . until the quotient row alternates in sign. We organize these results in the synthetic division table shown below. In a synthetic division table we dispense with writing the product of r with each coefficient in the quotient and simply list the results in the table.
UB
00035
Real Zeros and Polynomial Inequalities
Bounding Real Zeros
2
Let P(x) 0004 x3 0003 30x 2 0005 275x 0003 720. Find the smallest positive integer multiple of 10 and the largest negative integer multiple of 10 that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x). SOLUTION
We construct a synthetic division table to search for bounds for the zeros of P(x). The size of the coefficients in P(x) indicates that we can speed up this search by choosing larger increments between test values.
100
000310
30
UB LB
10 20 30 000310
1 1 1 1 1
000330 000320 000310 0 000340
275 75 75 275 675
0003720 30 780 7,530 00037,470
0003100 3 2 Z Figure 2 P(x) 0004 x 0003 30x 0005
275x 0003 720.
MATCHED PROBLEM 2
Therefore, all real zeros of P(x) 0004 x3 0003 30x2 0005 275x 0003 720 must lie between 000310 and 30, as confirmed by Figure 2. 0002 Let P(x) 0004 x 3 0003 25x 2 0005 170x 0003 170. Find the smallest positive integer multiple of 10 and the largest negative integer multiple of 10 that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x). 0002
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Technology Connections How do you determine the correct viewing window for graphing a function? This is one of the most frequently asked questions about graphing calculators. For polynomial functions, the upper and lower bound theorem gives an answer: let Xmin and Xmax be the lower and upper bounds, respectively, of Theorem 1 (appropriate values
5
00055
for Ymin and Ymax can then be found using TRACE). We can approximate the zeros, all of which appear in the chosen viewing window, using the ZERO command. The upper and lower bound theorem and the ZERO command on a graphing calculator are two important mathematical tools that work very well together.
Z Location Theorem and Bisection Method 5
00055
The graph of every polynomial function is continuous. Because the polynomial function P(x) 0002 x 5 0003 3x 0005 1 is negative when x 0002 0 [P(0) 0002 00051] and positive when x 0002 1 [P(1) 0002 3], the graph of P(x) must cross the x axis at least once between x 0002 0 and x 0002 1 (Fig. 3). This observation is the basis for Theorem 2 and leads to a simple method for approximating zeros.
5 Z Figure 3 P(x) 0002 x 0003 3x 0005 1.
Z THEOREM 2 Location Theorem* Suppose that a function f is continuous on an interval I that contains numbers a and b. If f (a) and f (b) have opposite signs, then the graph of f has at least one x intercept between a and b.
The conclusion of Theorem 2 says that at least one zero of the function is “located” between a and b. There may be more than one zero between a and b: if g(x) 0002 x3 0003 x2 0005 2x 0005 1, then g(00052) and g(2) have opposite signs and there are three zeros between x 0002 00052 and x 0002 2 [Fig. 4(a)]. The converse of Theorem 2 is false: h(x) 0002 x2 has an x intercept at x 0002 0 but does not change sign [Fig. 4(b)]. 5
00055
5
5
00055
5
00055
00055
(a)
(b)
Z Figure 4 Polynomials may or may not change sign at a zero.
ZZZ EXPLORE-DISCUSS 1
When synthetic division is used to divide a polynomial P(x) by x 0005 3 the remainder is 000533. When the same polynomial is divided by x 0005 4 the remainder is 38. Must P(x) have a zero between 3 and 4? Explain.
*The location theorem is a formulation of the important intermediate value theorem of calculus.
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Explore-Discuss 2 will provide an introduction to the repeated systematic application of the location theorem (Theorem 2) called the bisection method. This method forms the basis for the zero approximation routines in many graphing calculators.
ZZZ EXPLORE-DISCUSS 2
Let P(x) 0002 x5 0003 3x 0005 1. Because P(0) is negative and P(1) is positive, the location theorem guarantees that P(x) must have at least one zero in the interval (0, 1). (A) Is P(0.5) positive or negative? Does the location theorem guarantee a zero of P(x) in the interval (0, 0.5) or in (0.5, 1)? (B) Let m be the midpoint of the interval from part A that contains a zero of P(x). Is P(m) positive or negative? What does this tell you about the location of the zero? (C) Explain how this process could be used repeatedly to approximate a zero to any desired accuracy.
The bisection method is a systematic application of the procedure suggested in Explore-Discuss 2: Let P(x) be a polynomial with real coefficients. If P(x) has opposite signs at the endpoints of an interval (a, b), then by the location theorem P(x) has a zero in (a, b). Bisect this interval (that is, find the midpoint m 0002 a 00032 b), check the sign of P(m), and select the interval (a, m) or (m, b) that has opposite signs at the endpoints. We repeat this bisection procedure (producing a set of intervals, each contained in and half the length of the previous interval, and each containing a zero) until the desired accuracy is obtained. If at any point in the process P(m) 0002 0, we stop, because a real zero m has been found. Example 3 illustrates the procedure, and clarifies when the procedure is finished.
EXAMPLE
3
The Bisection Method The polynomial P(x) 0002 x4 0005 2x3 0005 10x2 0003 40x 0005 90 of Example 2 has a zero between 3 and 4. Use the bisection method to approximate it to one-decimal-place accuracy.
SOLUTION
We organize the results of our calculations in Table 1. Because the sign of P(x) changes at the endpoints of the interval (3.5625, 3.625), we conclude that a real zero lies in this interval and is given by r 0002 3.6 to one-decimal place accuracy (each endpoint rounds to 3.6).
Table 1 Bisection Approximation Sign Change Interval (a, b)
Midpoint m
(3, 4)
Sign of P P(a)
P(m)
P(b)
3.5
0005
0005
0003
(3.5, 4)
3.75
0005
0003
0003
(3.5, 3.75)
3.625
0005
0003
0003
(3.5, 3.625)
3.5625
0005
0005
0003
(3.5625, 3.625)
We stop here
0005
0003
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Figure 5 illustrates the nested intervals produced by the bisection method in Table 1. Match each step in Table 1 with an interval in Figure 5. Note how each interval that contains a zero gets smaller and smaller and is contained in the preceding interval that contained the zero. 3.5625
(
3.625
( ()
3
3.5
)
3.75
)
4
x
Z Figure 5 Nested intervals produced by the bisection method in Table 1.
If we had wanted two-decimal-place accuracy, we would have continued the process in Table 1 until the endpoints of a sign change interval rounded to the same two-decimal-place number. 0002 MATCHED PROBLEM 3
The polynomial P(x) 0002 x4 0005 2x3 0005 10x2 0003 40x 0005 90 of Example 1 has a zero between 00055 and 00054. Use the bisection method to approximate it to one-decimal-place accuracy. 0002
Z Approximating Real Zeros at Turning Points The bisection method for approximating zeros fails if a polynomial has a turning point at a zero, because the polynomial does not change sign at such a zero. Most graphing calculators use methods that are more sophisticated than the bisection method. Nevertheless, it is not unusual to get an error message when using the zero command to approximate a zero that is also a turning point. In this case, we can use the maximum or minimum command, as appropriate, to approximate the turning point, and the zero.
EXAMPLE
4
Approximating Zeros at Turning Points Let P(x) 0002 x5 0003 6x4 0003 4x3 0005 24x2 0005 16x 0003 32. Find the smallest positive integer and the largest negative integer that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x). Approximate the zeros to two decimal places, using maximum or minimum commands to approximate any zeros at turning points.
SOLUTION
The pertinent rows of a synthetic division table show that 2 is the upper bound and 00056 is the lower bound:
1 2 00055 00056
1 1 1 1 1
6 7 8 1 0
4 11 20 00051 4
000524 000513 16 000519 000548
000516 000529 16 79 272
32 3 64 0005363 00051600
Examining the graph of P(x) we find three zeros: the zero 00053.24, found using the MAXIMUM command [Fig. 6(a)]; the zero 00052, found using the ZERO command [Fig. 6(b)]; and the zero 1.24, found using the MINIMUM command [Fig. 6(c)].
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40
40
00056
2
40
00056
2
00056
000540
000540
(a)
(b)
(c)
0002
Z Figure 6 Zeros of P(x) 0002 x 0003 6x 0003 4x 0005 24x 0005 16x 0003 32.
MATCHED PROBLEM 4
2
000540
5
4
3
283
2
Let P(x) 0002 x5 0005 6x4 0003 40x2 0005 12x 0005 72. Find the smallest positive integer and the largest negative integer that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x). Approximate the zeros to two decimal places, using maximum or minimum commands to approximate any zeros at turning points. 0002
Z Polynomial Inequalities We can apply the techniques we have introduced for finding real zeros to solve polynomial inequalities. Consider, for example, the inequality x3 0005 2x2 0005 5x 0003 6 7 0 The real zeros of P(x) 0002 x3 0005 2x2 0005 5x 0003 6 are easily found to be 00052, 1, and 3. They partition the x axis into four intervals (00050007, 00052), (00052, 1), (1, 3),
and
(3, 0007)
On any one of these intervals, the graph of P is either above the x axis or below the x axis, because, by the location theorem, a continuous function can change sign only at a zero. One way to decide whether the graph of P is above or below the x axis on a given interval, say (00052, 1), is to choose a “test number” that belongs to the interval, 0, for example, and evaluate P at the test number. Because P(0) 0002 6 0, the graph of P is above the x axis throughout the interval (00052, 1). A second way to decide whether the graph of P is above or below the x axis on (00052, 1) is to simply inspect the graph of P. Each technique has its advantages, and both are illustrated in the solutions to Examples 5 and 6.
EXAMPLE
5
Solving Polynomial Inequalities Solve the inequality x3 0005 2x2 0005 5x 0003 6 0.
SOLUTION
Let P(x) 0002 x3 0005 2x2 0005 5x 0003 6. Then P(1) 0002 13 0005 2(12) 0005 5 0003 6 0002 0 so 1 is a zero of P and x 0005 1 is a factor. Dividing P(x) by x 0005 1 (details omitted) gives the quotient x2 – x 0005 6. Therefore, P(x) 0002 (x 0005 1)(x2 0005 x 0005 6) 0002 (x 0005 1)(x 0003 2)(x 0005 3) The zeros of P are 00052, 1, and 3. They partition the x axis into the four intervals shown in the table on page 284. A test number is chosen from each interval as indicated to determine whether P(x) is positive (above the x axis) or negative (below the x axis) on that interval.
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Interval Test number x P(x) Sign of P
(00030006, 00032)
(00032, 1)
(3, 0006)
00033
0
2
4
000324
6
00034
18
0003
0002
0003
0002
(1, 3)
We conclude that the solution set of the inequality is the intervals where P(x) is positive: (00032, 1) ´ (3, 0006) MATCHED PROBLEM 5
0002
Solve the inequality x3 0002 x2 0003 x 0003 1 0007 0. 0002
EXAMPLE
6
Solving Polynomial Inequalities with a Graphing Calculator Solve 3x2 0002 12x 0003 4 2x3 0003 5x2 0002 7 to three decimal places.
SOLUTION
Subtracting the right-hand side gives the equivalent inequality P(x) 0004 00032x3 0002 8x2 0002 12x 0003 11 0 The zeros of P(x), to three decimal places, are 00031.651, 0.669, and 4.983 [Fig. 7(a)]. By inspecting the graph of P we see that P is above the x axis on the intervals (00030006, 00031.651) and (0.669, 4.983). So the solution set of the inequality is (00030006, 00031.651] ´ [0.669, 4.983] The square brackets indicate that the endpoints of the intervals—the zeros of the polynomial— also satisfy the inequality. An alternative to inspecting the graph of P is to inspect the graph of f (x) 0004
P(x) 0004 P(x) 0004
The function f (x) has the value 1 if P(x) is positive, because then the absolute value of P(x) is equal to P(x). Similarly, f(x) has the value 00031 if P(x) is negative. This technique makes it easy to identify the solution set of the original inequality [Fig. 7(b)] and often eliminates difficulties in choosing appropriate window variables. 100
000310
10
10
0003100
(a) P(x) ⴝ ⴚ2x3 ⴙ 8x2 ⴙ 12x ⴚ 11
Z Figure 7
MATCHED PROBLEM 6
000310
10
000310
(b) f (x) ⴝ
P(x) 0004P(x)0004
0002
Solve to three decimal places 5x3 0003 13x 0007 4x2 0002 10x 0003 5. 0002
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Z Mathematical Modeling EXAMPLE
7
Construction An oil tank is in the shape of a right circular cylinder with a hemisphere at each end (Fig. 8). The cylinder is 55 inches long, and the volume of the tank is 11,000 cubic inches (approximately 20 cubic feet). Let x denote the common radius of the hemispheres and the cylinder.
x
x
55 inches
Z Figure 8
(A) Find a polynomial equation that x must satisfy. (B) Approximate x to one decimal place. SOLUTIONS
(A) If x is the common radius of the hemispheres and the cylinder in inches, then °
Volume Volume Volume of ¢ 0004 ° of two ¢ 0002 ° of ¢ tank hemispheres cylinder 4 3 11,000 0004 0002 55 x2 3 x 33,000 0004 4x3 0002 165x2 0 0004 4x3 0002 165x2 0003 33,000
Multiply by 30005 . Subtract 33,000 from both sides.
The radius we are looking for (x) must be a positive zero of P(x) 0004 4x3 0002 165x2 0003 33,000 (B) Because the coefficients of P(x) are large, we use larger increments in the synthetic division table:
70,000
0
20
UB
165 205 245
0 2,050 4,900
000333,000 000312,500 65,000
Applying the bisection method to the interval [10, 20] (nine midpoints are calculated; details omitted) or graphing y 0004 P(x) for 0 0005 x 0005 20 (Fig. 9), we see that x 0004 12.4 inches (to one decimal place). 0002
000370,000
P(x) 0004 4x 0002 165x2 0003 33,000. Z Figure 9 3
MATCHED PROBLEM 7
10 20
4 4 4
Repeat Example 7 if the volume of the tank is 44,000 cubic inches. 0002 ANSWERS TO MATCHED PROBLEMS 1. 3. 5. 7.
Lower bound: 00033; upper bound: 6 2. Lower bound: 000310; upper bound: 30 x 0004 00034.1 4. Lower bound: 00032; upper bound: 6; 00031.65, 2, 3.65 6. (00030006, 00031.899) 傼 (0.212, 2.488) (00030006, 00031) 傼 (00031, 1) (A) P(x) 0004 4x3 0002 165x2 0003 132,000 0004 0 (B) 22.7 inches
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Exercises
1. Given a polynomial of degree n 0007 0, explain why there must exist an upper bound and a lower bound for its real zeros. 2. State the location theorem in your own words. 3. A polynomial P has degree 6 and leading coefficient 1. If synthetic division by x 0003 5 results in all positive numbers in the quotient row, is 10 an upper bound for the real zeros of P? Explain. 4. A polynomial has degree 12 and leading coefficient 1. If synthetic division by x 0005 5 results in numbers that alternate in sign in the quotient row, is 000310 a lower bound for the real zeros of P? Explain. 5. Explain the basic steps in the bisection method. 6. If you use the bisection method to approximate a real root to three decimal place accuracy, explain how you can tell when the method is finished. In Problems 7–10, approximate the real zeros of each polynomial to three decimal places.
21. P(x) 0004 x4 0003 3x3 0005 4x2 0005 2x 0003 9 22. P(x) 0004 x4 0003 4x3 0005 6x2 0003 4x 0003 7 23. P(x) 0004 x5 0003 3x3 0005 3x2 0005 2x 0003 2 24. P(x) 0004 x5 0003 3x4 0005 3x2 0005 2x 0003 1
In Problems 25–30, (A) use the location theorem to explain why the polynomial function has a zero in the indicated interval; and (B) determine the number of additional intervals required by the bisection method to obtain a one-decimal-place approximation to the zero and state the approximate value of the zero. 25. P(x) 0004 x3 0003 2x2 0003 5x 0005 4; (3, 4) 26. P(x) 0004 x3 0005 x2 0003 4x 0003 1; (1, 2) 27. P(x) 0004 x3 0003 2x2 0003 x 0005 5; (00032, 00031) 28. P(x) 0004 x3 0003 3x2 0003 x 0003 2; (3, 4)
7. P(x) 0004 x2 0005 5x 0003 2
29. P(x) 0004 x4 0003 2x3 0003 7x2 0005 9x 0005 7; (3, 4)
8. P(x) 0004 3x2 0003 7x 0005 1
30. P(x) 0004 x4 0003 x3 0003 9x2 0005 9x 0005 4; (2, 3)
9. P(x) 0004 2x3 0003 5x 0005 2 10. P(x) 0004 x3 0003 4x2 0003 8x 0005 3 In Problems 11–14, use the graph of P(x) to write the solution set for each inequality.
In Problems 31–36, (A) find the smallest positive integer and largest negative integer that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of P(x); and (B) use the bisection method to approximate a real zero of each polynomial to one decimal place. 31. P(x) 0004 x3 0003 2x2 0005 3x 0003 8
20
32. P(x) 0004 x3 0005 3x2 0005 4x 0005 5 00035
5
33. P(x) 0004 2x3 0005 x2 0005 2x 0005 1 34. P(x) 0004 2x3 0003 x2 0005 4x 0003 2 35. P(x) 0004 x4 0005 x2 0003 6
36. P(x) 0004 x4 0003 2x2 0003 3
000320
11. P(x) 0
12. P(x) 0
13. P(x) 0007 0
14. P(x) 0
Problems 37–40, refer to the polynomial P(x) 0004 (x 0003 1)2(x 0003 2)(x 0003 3)4
In Problems 15–18, solve each polynomial inequality to three decimal places (note the connection with Problems 7–10).
37. Can the zero at x 0004 1 be approximated by the bisection method? Explain.
15. x2 0005 5x 0003 2 0007 0
16. 3x2 0003 7x 0005 1 0
17. 2x3 0003 5x 0005 2 0
18. x3 0003 4x2 0003 8x 0005 3 0
38. Can the zero at x 0004 2 be approximated by the bisection method? Explain.
Find the smallest positive integer and largest negative integer that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of each of the polynomials given in Problems 19–24. 19. P(x) 0004 x3 0003 3x 0005 1
20. P(x) 0004 x3 0003 4x2 0005 4
39. Can the zero at x 0004 3 be approximated by the bisection method? Explain. 40. Which of the zeros can be approximated by a maximum approximation routine? By a minimum approximation routine? By the zero approximation routine on your graphing calculator?
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In Problems 41–46, approximate the zeros of each polynomial function to two decimal places, using maximum or minimum commands to approximate any zeros at turning points. 41. P(x) 0004 x4 0003 4x3 0003 10x2 0002 28x 0002 49 42. P(x) 0004 x4 0002 4x3 0003 4x2 0003 16x 0002 16 43. P(x) 0004 x5 0003 6x4 0002 4x3 0002 24x2 0003 16x 0003 32 5
4
3
2
44. P(x) 0004 x 0003 6x 0002 2x 0002 28x 0003 15x 0002 2 45. P(x) 0004 x5 0003 6x4 0002 11x3 0003 4x2 0003 3.75x 0003 0.5 46. P(x) 0004 x5 0002 12x4 0002 47x3 0002 56x2 0003 15.75x 0002 1
Real Zeros and Polynomial Inequalities
287
67. Give a reason for each step in the proof of the upper bound case of Theorem 1 on page 278. Step 1: P(x) can be written in the form P(x) 0004 (x 0003 r)Q(x) 0002 R, where the coefficients of Q(x) and R are positive. Step 2: Suppose s r 0. Then P(s) 0. Step 3: r is an upper bound for the real zeros of P(x). 68. Give a reason for each step in the proof of the lower bound case of Theorem 1 on page 278. Step 1: P(x) can be written in the form P(x) 0004 (x 0003 r)Q(x) 0002 R, where the coefficients of Q(x) and R alternate in sign. Step 2: Suppose s 0007 r 0007 0. If P has even degree, then P(s) 0; if P has odd degree, then P(s) 0007 0. Step 3: r is a lower bound for the real zeros of P(x).
In Problems 47–52, solve each polynomial inequality. 47. x2 9
48. 1 0003 x2 0005 0
49. x3 0005 16x
50. 2x x2 0002 x3
51. x4 0002 4 5x2
69. Let P(x) be a polynomial of degree n 0 such that all of the coefficients of P(x) are nonnegative. Explain why 0 is an upper bound for the real zeros of P(x).
52. 2 0002 x 0002 x2 0002 x3 0007 x4 In Problems 53–58, solve each polynomial inequality to three decimal places. 53. x2 0002 7x 0003 3 0005 x3 0002 x 0002 4
Problems 69 and 70 explore the cases in which 0 is an upper bound or lower bound for the real zeros of a polynomial. These cases are not covered by Theorem 1, the upper and lower bound theorem, as formulated on page 278.
54. x4 0002 1 3x2
55. x4 0007 8x3 0003 17x2 0002 9x 0003 2 56. x3 0002 5x 2x3 0003 4x2 0002 6 57. (x2 0002 2x 0003 2)2 2 58. 5 0002 2x 0007 (x2 0003 4)2 In Problems 59–64, (A) find the smallest positive integer multiple of 10 and largest negative integer multiple of 10 that, by Theorem 1, are upper and lower bounds, respectively, for the real zeros of each polynomial; and (B) approximate the real zeros of each polynomial to two decimal places. 59. P(x) 0004 x3 0003 24x2 0003 25x 0002 10 60. P(x) 0004 x3 0003 37x2 0002 70x 0003 20 61. P(x) 0004 x4 0002 12x3 0003 900x2 0002 5,000 62. P(x) 0004 x4 0003 12x3 0003 425x2 0002 7,000
70. Let P(x) be a polynomial of degree n 0 such that an 0 and the coefficients of P(x) alternate in sign (as in Theorem 1, a coefficient 0 can be considered either positive or negative, but not both). Explain why 0 is a lower bound for the real zeros of P(x).
APPLICATIONS Express the solutions to Problems 71–76 as the roots of a polynomial equation of the form P(x) ⫽ 0 and approximate these solutions to one decimal place. 71. GEOMETRY Find all points on the graph of y 0004 x2 that are one unit away from the point (1, 2). [Hint: Use the distance formula from Section 2-2.] 72. GEOMETRY Find all points on the graph of y 0004 x2 that are one unit away from the point (2, 1). 73. MANUFACTURING A box is to be made out of a piece of cardboard that measures 18 by 24 inches. Squares, x inches on a side, will be cut from each corner, and then the ends and sides will be folded up (see the figure). Find the value of x that would result in a box with a volume of 600 cubic inches.
63. P(x) 0004 x4 0003 100x2 0003 1,000x 0003 5,000
24 in.
64. P(x) 0004 x4 0003 5x3 0003 50x2 0003 500x 0002 7,000
66. When synthetic division is used to divide a polynomial Q(x) by x 0002 4 the remainder is 10. When the same polynomial is divided by x 0002 5 the remainder is 8. Could Q(x) have a zero between 00035 and 00034? Explain.
18 in.
65. When synthetic division is used to divide a polynomial P(x) by x 0002 4 the remainder is 10. When the same polynomial is divided by x 0002 5 the remainder is 00038. Must P(x) have a zero between 00035 and 00034? Explain.
x x
74. MANUFACTURING A box with a hinged lid is to be made out of a piece of cardboard that measures 20 by 40 inches. Six squares, x inches on a side, will be cut from each corner and the middle, and then the ends and sides will be folded up to form the box and its lid
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(see the figure). Find the value of x that would result in a box with a volume of 500 cubic inches.
20 in.
40 in. x
x
76. SHIPPING A shipping box is reinforced with steel bands in all three directions (see the figure). A total of 20.5 feet of steel tape is to be used, with 6 inches of waste because of a 2-inch overlap in each direction. If the box has a square base and a volume of 2 cubic feet, find the side length of the base.
75. CONSTRUCTION A propane gas tank is in the shape of a right circular cylinder with a hemisphere at each end (see the figure). If the overall length of the tank is 10 feet and the volume is 20 cubic feet, find the common radius of the hemispheres and the cylinder.
y x x
x x
10 feet
4-3
Complex Zeros and Rational Zeros of Polynomials Z The Fundamental Theorem of Algebra Z Factors of Polynomials with Real Coefficients Z Graphs of Polynomials with Real Coefficients Z Rational Zeros
The graph of the polynomial function P(x) 0002 x2 0003 4 does not cross the x axis, so P(x) has no real zeros. It does, however, have complex zeros, 2i and 00052i; by the factor theorem, x2 0003 4 0002 (x 0005 2i)(x 0003 2i). The fundamental theorem of algebra guarantees that every nonconstant polynomial with real or complex coefficients has a complex zero; as a result, such a polynomial can be factored as a product of linear factors. In Section 4-3, we study the fundamental theorem and its implications, including results on the graphs of polynomials with real coefficients. Finally, we consider a problem that has led to important advances in mathematics and its applications: When can zeros of a polynomial be found exactly?
Z The Fundamental Theorem of Algebra The fundamental theorem of algebra was proved by Karl Friedrich Gauss (1777–1855), one of the greatest mathematicians of all time, in his doctoral thesis. A proof of the theorem is beyond the scope of this book, so we will state and use it without proof.
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Z THEOREM 1 Fundamental Theorem of Algebra Every polynomial of degree n > 0 with complex coefficients has a complex zero.
If P(x) is a polynomial of degree n 0 with complex coefficients, then by Theorem 1 it has a zero r1. So x 0005 r1 is a factor of P(x) by Theorem 5 of Section 4-1, and P(x) 0002 (x 0005 r1)Q(x), deg Q(x) 0002 n 0005 1 Now, if deg Q(x) 0, then, applying the fundamental theorem to Q(x), Q(x) has a root r2 and therefore a factor x 0005 r2. (It is possible that r2 is equal to r1.) By continuing this reasoning we obtain a proof of Theorem 2.
Z THEOREM 2 n Linear Factors Theorem Every polynomial of degree n 0 with complex coefficients can be factored as a product of n linear factors.
Suppose that a polynomial P(x) is factored as a product of n linear factors. Any zero r of P(x) must be a zero of one or more of the factors. The number of linear factors that have zero r is said to be the multiplicity of r. For example, the polynomial P(x) 0002 (x 0005 5)3(x 0003 1)2(x 0005 6i)(x 0003 2 0003 3i)
(1)
has degree 7 and is written as a product of seven linear factors. P(x) has just four zeros, namely 5, 00051, 6i, and 00052 0005 3i. Because the factor x 0005 5 appears to the power 3, we say that the zero 5 has multiplicity 3. Similarly, 00051 has multiplicity 2, 6i has multiplicity 1, and 00052 0005 3i has multiplicity 1. A zero of multiplicity 2 is called a double zero, and a zero of multiplicity 3 is called triple zero. Note that the sum of the multiplicities is always equal to the degree of the polynomial: for P(x) in equation (1), 3 0003 2 0003 1 0003 1 0002 7.
EXAMPLE
1
Multiplicities of Zeros Find the zeros and their multiplicities: (A) P(x) 0002 (x 0003 2)7(x 0005 4)8(x2 0003 1) (B) Q(x) 0002 (x 0003 1)3(x2 0005 1)(x 0003 1 0005 i)
SOLUTIONS
(A) Note that x2 0003 1 0002 0 has the solutions i and 0005i. The zeros of P(x) are 00052 (multiplicity 7), 4 (multiplicity 8), i and 0005i (each multiplicity 1). (B) Note that x2 0005 1 0002 (x 0005 1)(x 0003 1), so x 0003 1 appears four times as a factor of Q(x). The zeros of Q(x) are 00051 (multiplicity 4), 1 (multiplicity 1), and 00051 0003 i (multiplicity 1). 0002
MATCHED PROBLEM 1
Find the zeros and their multiplicities: (A) P(x) 0002 (x 0005 5)3(x 0003 3)2(x2 0003 16) (B) Q(x) 0002 (x2 0005 25)3(x 0003 5)(x 0005 i) 0002
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Z Factors of Polynomials with Real Coefficients If p 0002 qi is a zero of P(x) 0004 ax2 0002 bx 0002 c, where a, b, c, p, and q are real numbers, then P( p 0002 qi) 0004 0 a( p 0002 qi) 0002 b( p 0002 qi) 0002 c 0004 0 a( p 0002 qi)2 0002 b( p 0002 qi) 0002 c 0004 0 a ( p 0002 qi)2 0002 b ( p 0002 qi) 0002 c 0004 0 a( p 0003 qi)2 0002 b( p 0003 qi) 0002 c 0004 0 2
Take the conjugate of both sides. z ⴙ w ⴝ z ⴙ w, zw ⴝ z w z ⴝ z if z is real, p ⴙ qi ⴝ p ⴚ qi
P( p 0003 qi) 0004 0 Therefore, p 0003 qi is also a zero of P(x). This method of proof can be applied to any polynomial P(x) of degree n 0 with real coefficients, justifying Theorem 3.
Z THEOREM 3 Imaginary Zeros of Polynomials with Real Coefficients Imaginary zeros of polynomials with real coefficients, if they exist, occur in conjugate pairs.
If a polynomial P(x) of degree n 0 has real coefficients and a linear factor of the form x 0003 ( p 0002 qi) where q 0, then, by Theorem 3, P(x) also has the linear factor x 0003 ( p 0003 qi). But [x 0003 ( p 0002 qi)][x 0003 ( p 0003 qi)] 0004 x2 0003 2px 0002 p2 0003 q2 which is a quadratic factor of P(x) with real coefficients and imaginary zeros. By this reasoning we can prove Theorem 4.
Z THEOREM 4 Linear and Quadratic Factors Theorem* If P(x) is a polynomial of degree n 0 with real coefficients, then P(x) can be factored as a product of linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros).
EXAMPLE
2
Factors of Polynomials Factor P(x) 0004 x3 0002 x2 0002 4x 0002 4 in two ways: (A) As a product of linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros) (B) As a product of linear factors with complex coefficients
SOLUTIONS
(A) Note that P(00031) 0004 0, so 00031 is a zero of P(x) (or graph P(x) and note that 00031 is an x intercept). Therefore, x 0002 1 is a factor of P(x). Using synthetic division, the quotient is x2 0002 4, which has imaginary roots. Therefore, P(x) 0004 (x 0002 1)(x2 0002 4) *Theorem 4 underlies the technique of decomposing a rational function into partial fractions, which is useful in calculus. See Appendix A, Section A-2.
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An alternative solution is to factor by grouping: x3 0003 x2 0003 4x 0003 4 0002 x2(x 0003 1) 0003 4(x 0003 1) 0002 (x2 0003 4)(x 0003 1) (B) Because x2 0003 4 has roots 2i and 00052i, P(x) 0002 (x 0003 1)(x 0005 2i)(x 0003 2i)
0002
Factor P(x) 0002 x5 0005 x4 0005 x 0003 1 in two ways:
MATCHED PROBLEM 2
(A) As a product of linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros) (B) As a product of linear factors with complex coefficients 0002
Z Graphs of Polynomials with Real Coefficients The factorization described in Theorem 4 gives additional information about the graphs of polynomial functions with real coefficients. For certain polynomials the factorization of Theorem 4 will involve only linear factors; for others, only quadratic factors. Of course if only quadratic factors are present, then the degree of the polynomial P(x) must be even. In other words, a polynomial P(x) of odd degree with real coefficients must have a linear factor with real coefficients. This proves Theorem 5.
Z THEOREM 5 Real Zeros and Polynomials of Odd Degree Every polynomial of odd degree with real coefficients has at least one real zero, and consequently at least one x intercept.
ZZZ EXPLORE-DISCUSS 1 3
00053
3
The graph of the polynomial P(x) 0002 x(x 0005 1)2(x 0003 1)4(x 0005 2)3 is shown in Figure 1. Find the real zeros of P(x) and their multiplicities. How can a real zero of even multiplicity be distinguished from a real zero of odd multiplicity using only the graph?
For polynomials with real coefficients, as suggested by Explore-Discuss 1, you can easily distinguish real zeros of even multiplicity from those of odd multiplicity using only the graph. Theorem 6, which we state without proof, tells how to do that.
00053
Z Figure 1 Graph of P(x) 0002 x(x 0005 1)2(x 0003 1)4(x 0005 2)3.
Z THEOREM 6 Zeros of Even or Odd Multiplicity Let P(x) be a polynomial with real coefficients: 1. If r is a real zero of P(x) of even multiplicity, then P(x) has a turning point at r and does not change sign at r. (The graph just touches the x axis, then changes direction.) 2. If r is a real zero of P(x) of odd multiplicity, then P(x) does not have a turning point at r and changes sign at r. (The graph continues through to the opposite side of the x axis.)
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3
Multiplicities from Graphs Figure 2 shows the graph of a polynomial function of degree 6. Find the real zeros and their multiplicities. 5
00055
5
00055
Z Figure 2 SOLUTION
Figure 3 shows the graph of a polynomial function of degree 7. Find the real zeros and their multiplicities. 0002
MATCHED PROBLEM 3 5
00054
4
000510
Z Figure 3 10
000510
The numbers 00052, 00051, 1, and 2 are real zeros (x intercepts). The graph has turning points at x 0002 00061 but not at x 0002 00062. Therefore, by Theorem 6, the zeros 00051 and 1 have even multiplicity, and 00052 and 2 have odd multiplicity. Because the sum of the multiplicities must equal 6 (the degree), the zeros 00051 and 1 each have multiplicity 2, and the zeros 00052 and 2 each have multiplicity 1. 0002
10
000510 2 00059 Z Figure 4 P(x) 0002 x 0005 (4 0003 10 ).
Z Rational Zeros From a graphical perspective, finding a zero of a polynomial means finding a good approximation to an actual zero. A graphing calculator, for example, might give 2 as a zero of P(x) 0002 x2 0005 (4 0003 1000059) even though P(2) is equal to 00051000059, not 0 (Fig. 4). It is natural, however, to want to find zeros exactly. Although this is impossible in general, we will adopt an algebraic strategy to find exact zeros in a special case, that of rational zeros of polynomials with rational coefficients. We will find a graphing calculator to be helpful in carrying out the algebraic strategy. First note that a polynomial with rational coefficients can always be written as a constant times a polynomial with integer coefficients. For example, 1 3 2 2 7 x 0005 x 0003 x00035 2 3 4 1 0002 (6x3 0005 8x2 0003 21x 0003 60) 12
P(x) 0002
Because the zeros of P(x) are the zeros of 6x3 0005 8x2 0003 21x 0003 60, it is sufficient, for the purpose of finding rational zeros of polynomials with rational coefficients, to study just the polynomials with integer coefficients. We introduce the rational zero theorem by examining the following quadratic polynomial whose zeros can be found easily by factoring: P(x) 0002 6x2 0005 13x 0005 5 0002 (2x 0005 5)(3x 0003 1) 5 00051 1 Zeros of P(x): and 0005 0002 2 3 3
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Notice that the numerators 5 and 00051 of the zeros are both integer factors of 00055, the constant term in P(x). The denominators 2 and 3 of the zeros are both integer factors of 6, the coefficient of the highest-degree term in P(x). These observations are generalized in Theorem 7 (a proof is outlined in Problem 89 of Exercises 4-3). Z THEOREM 7 Rational Zero Theorem If the rational number b0002c, in lowest terms, is a zero of the polynomial P(x) 0002 an xn 0003 an00051 xn00051 0003 . . . 0003 a1x 0003 a0
an 0004 0
with integer coefficients, then b must be an integer factor of a0 and c must be an integer factor of an. P(x) ⴝ anxn ⴙ anⴚ1xnⴚ1 ⴙ . . . ⴙ a1x ⴙ a0 b c c must be a factor of an
b must be a factor of a0
Theorem 7 enables us to construct a finite list of possible rational zeros of P(x). Each number in the list must then be tested to determine whether or not it is actually a zero. As Example 4 illustrates, a graphing calculator can often reduce the effort required to locate rational zeros.
EXAMPLE
Finding Rational Zeros
4
Find all the rational zeros for P(x) 0002 2x3 0003 9x2 0003 7x 0005 6. SOLUTION
If b0002c in lowest terms is a rational zero of P(x), then b must be a factor of 00056 and c must be a factor of 2. Possible values of b are the integer factors of 00056: 00061, 00062, 00063, 00066 Possible values of c are the integer factors of 2: 00061, 00062
(2) (3)
Writing all possible fractions b0002c where b is from (2) and c is from (3), we have Possible rational zeros for P(x): 00061, 00062, 00063, 00066, 000612, 000632
(4)
[Note that all fractions are in lowest terms and duplicates like 00066000e00062 0002 00063 are not repeated.] If P(x) has any rational zeros, they must be in list (4). We can test each number r in this list simply by evaluating P(r). However, exploring the graph of y 0002 P(x) first will usually indicate which numbers in the list are the most likely candidates for zeros. Examining a graph of P(x), we see that there are zeros near 00053, near 00052, and between 0 and 1, so we begin by evaluating P(x) at 00053, 00052, and 12 (Fig. 5). 10
00055
Z Figure 5
10
5
00055
10
5
00055
5
000510
000510
000510
(a)
(b)
(c)
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Therefore, 00053, 00052, and 12 are rational zeros of P(x). Because a third-degree polynomial can have at most three zeros, we have found all the rational zeros. There is no need to test the remaining candidates in list (4). 0002 Find all rational zeros for P(x) 0002 2x3 0003 x2 0005 11x 0005 10.
MATCHED PROBLEM 4
0002 As we saw in the solution of Example 4, rational zeros can be located by simply evaluating the polynomial. However, if we want to find multiple zeros, imaginary zeros, or exact values of irrational zeros, we need to consider reduced polynomials. If r is a zero of a polynomial P(x), then we can write P(x) 0002 (x 0005 r)Q(x) where Q(x) is a polynomial of degree one less than the degree of P(x). The quotient polynomial Q(x) is called a reduced polynomial for P(x). In Example 4, after determining that 00053 is a zero of P(x), we can write 9 7 00056 00056 00059 6 00053 2 3 00052 0 3 2 P(x) 0002 2x 0003 9x 0003 7x 0005 6 0002 (x 0003 3)(2x2 0003 3x 0005 2) 0002 (x 0003 3)Q(x) 2
Because the reduced polynomial Q(x) 0002 2x2 0003 3x 0005 2 is a quadratic, we can find its zeros by factoring or the quadratic formula. We get P(x) 0002 (x 0003 3)(2x2 0003 3x 0005 2) 0002 (x 0003 3)(x 0003 2)(2x 0005 1) and we see that the zeros of P(x) are 00053, 00052, and 12, as before.
EXAMPLE
Finding Rational and Irrational Zeros
5
Find all zeros exactly for P(x) 0002 2x3 0005 7x2 0003 4x 0003 3. SOLUTION
00061, 00063, 000612, 000632
5
00055
First, list the possible rational zeros: Examining the graph of y 0002 P(x) (Fig. 6), we see that there is a zero between 00051 and 0, another between 1 and 2, and a third between 2 and 3. We test the only likely candidates, 000512 and 32: P(000512) 0002 00051
5
So 00055
Z Figure 6
3 2
and
P(32) 0002 0
is a zero, but 000512 is not. Using synthetic division (details omitted), we can write P(x) 0002 (x 0005 32)(2x2 0005 4x 0005 2)
Because the reduced polynomial is quadratic, we can use the quadratic formula to find the exact values of the remaining zeros: 2x2 0005 4x 0005 2 0002 0 x2 0005 2x 0005 1 0002 0 2 0006 14 0005 4(1)(00051) x0002 2 2 0006 212 0002 0002 1 0006 12 2
Divide both sides by 2. Use the quadratic formula.
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So the exact zeros of P(x) are
3 2
Complex Zeros and Rational Zeros of Polynomials
and 1 12.*
295 0002
Find all zeros exactly for P(x) 0004 3x3 0003 10x2 0005 5x 0005 4.
MATCHED PROBLEM 5
0002
EXAMPLE
6
Finding Rational and Imaginary Zeros Find all zeros exactly for P(x) 0004 x4 0003 6x3 0005 14x2 0003 14x 0005 5.
SOLUTION 5
The possible rational zeros are 1 and 5. Examining the graph of P(x) (Fig. 7), we see that 1 is a zero. Because the graph of P(x) does not appear to change sign at 1, this may be a multiple root. Using synthetic division (details omitted), we find that P(x) 0004 (x 0003 1)(x3 0003 5x2 0005 9x 0003 5)
00031
5
The possible rational zeros of the reduced polynomial Q(x) 0004 x3 0003 5x2 0005 9x 0003 5
00031
Z Figure 7
are 1 and 5. Examining the graph of Q(x) (Fig. 8), we see that 1 is a rational zero. After a division, we have a quadratic reduced polynomial: 5
00031
Q(x) 0004 (x 0003 1)Q1(x) 0004 (x 0003 1)(x2 0003 4x 0005 5) 5
We use the quadratic formula to find the zeros of Q1(x): x2 0003 4x 0005 5 0004 0 4 116 0003 4(1)(5) x0004 2 4 100034 00042 i 0004 2
00035
Z Figure 8
So the exact zeros of P(x) are 1 (multiplicity 2), 2 0003 i, and 2 0005 i. Find all zeros exactly for P(x) 0004 x4 0005 4x3 0005 10x2 0005 12x 0005 5.
MATCHED PROBLEM 6
0002
0002
REMARK
We were successful in finding all the zeros of the polynomials in Examples 5 and 6 because we could find sufficient rational zeros to reduce the original polynomial to a quadratic. This is not always possible. For example, the polynomial
50
00035
5
000350 3 Z Figure 9 P(x) 0004 x 0005 6x 0003 2.
P(x) 0004 x3 0005 6x 0003 2 has no rational zeros, but does have an irrational zero at x ⬇ 0.32748 (Fig. 9). The other two zeros are imaginary. The techniques we have developed will not find the exact value of these roots. *By analogy with Theorem 3 (imaginary zeros of polynomials with real coefficients occur in conjugate pairs), it can be shown that if r 0005 s1t is a zero of a polynomial with rational coefficients, where r, s, and t are rational but t is not the square of a rational, then r 0003 s1t is also a zero.
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ANSWERS TO MATCHED PROBLEMS 1. (A) 5 (multiplicity 3), 00033 (multiplicity 2), 4i and 00034i (each multiplicity 1) (B) 00035 (multiplicity 4), 5 (multiplicity 3), i (multiplicity 1) 2. (A) (x 0002 1)(x 0003 1)2(x2 0002 1) (B) (x 0002 1)(x 0003 1)2(x 0002 i)(x 0003 i) 3. 00033 (multiplicity 2), 00032 (multiplicity 1), 00031 (multiplicity 1), 0 (multiplicity 2), 1 (multiplicity 1) 4. 00032, 00031, 52 5. 43, 1 0003 12, 1 0002 12 6. 00031 (multiplicity 2), 00031 0003 2i, 00031 0002 2i
4-3
Exercises
1. Explain in your own words what the fundamental theorem of algebra says.
zeros are integers. Write the polynomial as a product of linear factors. Indicate the degree of the polynomial.
2. Does every polynomial of degree 0 with real coefficients have a real zero? Explain.
19.
P (x)
20.
P(x) 15
15
3. What is meant by the multiplicity of a zero of a polynomial? 4. If P(x) is a polynomial with integer coefficients and leading coefficient 1, explain why every rational zero of P(x) is actually an integer.
00035
Write the zeros of each polynomial in Problems 5–12, and indicate the multiplicity of each. What is the degree of each polynomial? 5. P(x) 0004 (x 0002 8)3(x 0003 6)2 6. P(x) 0004 (x 0003 5)(x 0002 7)2
5
x
00035
P (x)
22.
P(x) 15
15
7. P(x) 0004 3(x 0002 4)3(x 0003 3)2(x 0002 1)
x
000315
000315
21.
5
8. P(x) 0004 5(x 0003 2)3(x 0002 3)2(x 0003 1) 9. P(x) 0004 x3(2x 0002 1)2
00035
5
x
00035
5
x
10. P(x) 0004 6x2(5x 0003 4)(3x 0002 2) 11. P(x) 0004 (x2 0002 4)3(x2 0003 4)5(x 0002 2i)
000315
000315
12. P(x) 0004 (x2 0002 7x 0002 10)2(x2 0002 6x 0002 10)3 In Problems 13–18, find a polynomial P(x) of lowest degree, with leading coefficient 1, that has the indicated set of zeros. Write P(x) as a product of linear factors. Indicate the degree of P(x).
23.
P (x)
24.
P(x) 15
15
13. 3 (multiplicity 2) and 00034 14. 00032 (multiplicity 3) and 1 (multiplicity 2)
00035
5
x
00035
5
15. 00037 (multiplicity 3), 00033 0002 12, 00033 0003 12 16. 13 (multiplicity 2), 5 0002 17, 5 0003 17
000315
000315
17. (2 0003 3i), (2 0002 3i), 00034 (multiplicity 2) 18. i 13 (multiplicity 2), 0003i13 (multiplicity 2), and 4 (multiplicity 3) In Problems 19–24, find a polynomial of lowest degree, with leading coefficient 1, that has the indicated graph. Assume all
In Problems 25–28, factor each polynomial in two ways: (A) as a product of linear factors (with real coefficients) and
x
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quadratic factors (with real coefficients and imaginary zeros); and (B) as a product of linear factors with complex coefficients. 4
2
25. P(x) 0002 x 0003 5x 0003 4 4
297
58. P(x) 0002 x3 0005 8x2 0003 17x 0005 4 59. P(x) 0002 4x4 0005 4x3 0005 9x2 0003 x 0003 2 60. P(x) 0002 2x4 0003 3x3 0005 4x2 0005 3x 0003 2
2
26. P(x) 0002 x 0003 18x 0003 81 27. P(x) 0002 x3 0005 x2 0003 25x 0005 25 5
Complex Zeros and Rational Zeros of Polynomials
In Problems 61–68, find a polynomial P(x) that satisfies all of the given conditions. Write the polynomial using only real coefficients.
4
28. P(x) 0002 x 0003 x 0005 x 0005 1
61. 2 0005 5i is a zero; leading coefficient 1; degree 2 In Problems 29–34, list all possible rational zeros (Theorem 7) of a polynomial with integer coefficients that has the given leading coefficient an and constant term a0.
62. 4 0003 3i is a zero; leading coefficient 1; degree 2
29. an 0002 1, a0 0002 00054
30. an 0002 1, a0 0002 9
64. 1 0005 4i is a zero; P(0) 0002 51; degree 2
31. an 0002 10, a0 0002 1
32. an 0002 6, a0 0002 00051
65. 00055 and 8i are zeros; leading coefficient 1; degree 3
33. an 0002 7, a0 0002 00052
34. an 0002 3, a0 0002 8
66. 7 and 00052i are zeros; leading coefficient 1; degree 3
When searching for zeros of a polynomial, a graphing calculator often can be helpful in eliminating from consideration certain candidates for rational zeros.
63. 6 0003 i is a zero; P(0) 0002 74; degree 2
67. i and 1 0005 i are zeros; P(1) 0002 10; degree 4 68. 0005i and 3 0003 i are zeros; P(1) 0002 20; degree 4
In Problems 35–40, write P(x) as a product of linear factors.
In Problems 69–74, multiply.
35. P(x) 0002 x3 0003 9x2 0003 24x 0003 16; 00051 is a zero
69. [x 0005 (4 0005 5i)][x 0005 (4 0003 5i)]
36. P(x) 0002 x3 0005 4x2 0005 3x 0003 18; 3 is a double zero
70. [x 0005 (2 0005 3i)][x 0005 (2 0003 3i)]
37. P(x) 0002 x4 0003 2x2 0003 1; i is a double zero
71. [x 0005 (3 0003 4i)][x 0005 (3 0005 4i)]
38. P(x) 0002 x4 0005 1; 1 and 00051 are zeros
72. [x 0005 (5 0003 2i)][x 0005 (5 0005 2i)]
39. P(x) 0002 2x3 0005 17x2 0003 90x 0005 41; 12 is a zero
73. [x 0005 (a 0003 bi)][x 0005 (a 0005 bi)]
40. P(x) 0002 3x3 0005 10x2 0003 31x 0003 26; 000523 is a zero
74. (x 0005 bi)(x 0003 bi)
In Problems 41–48, find all roots exactly (rational, irrational, and imaginary) for each polynomial equation.
In Problems 75–80, find all other zeros of P(x), given the indicated zero.
41. 2x3 0005 5x2 0003 1 0002 0
75. P(x) 0002 x3 0005 5x2 0003 4x 0003 10; 3 0005 i is one zero
42. 2x3 0005 10x2 0003 12x 0005 4 0002 0
43. x4 0003 4x3 0005 x2 0005 20x 0005 20 0002 0
76. P(x) 0002 x3 0003 x2 0005 4x 0003 6; 1 0003 i is one zero
44. x4 0005 4x2 0005 4x 0005 1 0002 0
77. P(x) 0002 x3 0005 3x2 0003 25x 0005 75; 00055i is one zero
45. x4 0005 2x3 0005 5x2 0003 8x 0003 4 0002 0
78. P(x) 0002 x3 0003 2x2 0003 16x 0003 32; 4i is one zero
46. x4 0005 2x2 0005 16x 0005 15 0002 0
79. P(x) 0002 x4 0005 4x3 0003 3x2 0003 8x 0005 10; 2 0003 i is one zero
47. x4 0003 10x2 0003 9 0002 0
80. P(x) 0002 x4 0005 2x3 0003 7x2 0005 18x 0005 18; 00053i is one zero
48. x4 0003 29x2 0003 100 0002 0
In Problems 49–54, find all zeros exactly (rational, irrational, and imaginary) for each polynomial.
In Problems 81–86, final all zeros (rational, irrational, and imaginary) exactly.
49. P(x) 0002 x3 0005 19x 0003 30
81. P(x) 0002 3x3 0005 37x2 0003 84x 0005 24
4
51. P(x) 0002 x 0005 4
21 3 10 x
3
0003
53. P(x) 0002 x 0005 5x 0003
50. P(x) 0002 x3 0005 7x2 0003 36
2 5x 15 2 2x
4
52. P(x) 0002 x 0003
7 3 6x
0005
7 2 3x
0005
5 2x
0005 2x 0005 2
54. P(x) 0002 x4 0005 134x2 0005 52x 0005 14
83. P(x) 0002 4x4 0003 4x3 0003 49x2 0003 64x 0005 240 84. P(x) 0002 6x4 0003 35x3 0003 2x2 0005 233x 0005 360
In Problems 55–60, write each polynomial as a product of linear factors. 55. P(x) 0002 6x3 0003 13x2 0005 4
82. P(x) 0002 2x3 0005 9x2 0005 2x 0003 30
56. P(x) 0002 6x3 0005 17x2 0005 4x 0003 3
57. P(x) 0002 x3 0003 2x2 0005 9x 0005 4
85. P(x) 0002 4x4 0005 44x3 0003 145x2 0005 192x 0003 90 86. P(x) 0002 x5 0005 6x4 0003 6x3 0003 28x2 0005 72x 0003 48 87. The solutions to the equation x3 0005 1 0002 0 are all the cube roots of 1. (A) 1 is obviously a cube root of 1; find all others. (B) How many distinct cube roots of 1 are there?
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88. The solutions to the equation x3 0003 8 0004 0 are all the cube roots of 8. (A) 2 is obviously a cube root of 8; find all others. (B) How many distinct cube roots of 8 are there? 89. Give a reason for each step in the proof of the rational zero theorem, assuming that P(x) has degree two. Step 1: a2 (bc)2 0005 a1(bc) 0005 a0 0004 0 Step 2: a2b2 0005 a1bc 0005 a0c2 0004 0 Step 3: a2b2 0005 a1bc 0004 0003a0c2 Step 4: b is a factor of 0003a0c2, so b is a factor of a0. Step 5: Modify steps 3 and 4 to conclude that c is a factor of a2. 90. Explain how the ideas in Problem 89 can be adapted to give a proof of the rational zero theorem for P(x) of degree n.
much should this amount be to create a new storage unit with volume 10 times the old? 94. CONSTRUCTION A rectangular box has dimensions 1 by 1 by 2 feet. If each dimension is increased by the same amount, how much should this amount be to create a new box with volume six times the old? 95. PACKAGING An open box is to be made from a rectangular piece of cardboard that measures 8 by 5 inches, by cutting out squares of the same size from each corner and bending up the sides (see the figure). If the volume of the box is to be 14 cubic inches, how large a square should be cut from each corner? [Hint: Determine the domain of x from physical considerations before starting.] x
91. Given P(x) 0004 x2 0005 2ix 0003 5 with 2 0003 i a zero, show that 2 0005 i is not a zero of P(x). Does this contradict Theorem 3? Explain. 92. If P(x) and Q(x) are two polynomials of degree n, and if P(x) 0004 Q(x) for more than n values of x, then how are P(x) and Q(x) related? [Hint: Consider the polynomial D(x) 0004 P(x) 0003 Q(x).]
x
x
x x
APPLICATIONS Find all rational solutions exactly, and find irrational solutions to one decimal place. 93. STORAGE A rectangular storage unit has dimensions 1 by 2 by 3 feet. If each dimension is increased by the same amount, how
4-4
x
x
x
96. FABRICATION An open metal chemical tank is to be made from a rectangular piece of stainless steel that measures 10 by 8 feet, by cutting out squares of the same size from each corner and bending up the sides (see the figure for Problem 95). If the volume of the tank is to be 48 cubic feet, how large a square should be cut from each corner?
Rational Functions and Inequalities Z Rational Functions and Properties of Their Graphs Z Vertical and Horizontal Asymptotes Z Analyzing the Graph of a Rational Function Z Rational Inequalities
In Section 4-4, we will apply our knowledge of graphs and zeros of polynomial functions to study the graphs of rational functions, that is, functions that are quotients of polynomials. Our goal will be to produce hand sketches that clearly show all of the important features of the graph.
Z Rational Functions and Properties of Their Graphs The number 137 is called a rational number because it is a quotient (or ratio) of integers. The function f (x) 0004
x00051 x 0003x00036 2
is called a rational function because it is a quotient of polynomials.
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Z DEFINITION 1 Rational Function
5
A function f is a rational function if it can be written in the form 00035
5
f (x) 0002
x
(1, 2)
p(x) q(x)
where p(x) and q(x) are polynomials.
00035
(a) f(x) ⴝ
(x ⴚ 1)(x2 ⴚ 3)
When working with rational functions, we will assume that the coefficients of p(x) and q(x) are real numbers, and that the domain of f is the set of all real numbers x such that q(x) ⴝ 0.
xⴚ1 y 5
If a real number c is a zero of both p(x) and q(x), then, by the factor theorem, x 0003 c is a factor of both p(x) and q(x). The graphs of
00035
5
x
f (x) 0002
p(x) (x 0003 c)pr (x) 0002 q(x) (x 0003 c)qr(x)
and
fr(x) 0002
pr(x) qr(x)
are then identical, except possibly for a “hole” at x 0002 c (Fig. 1). Later in this section we will explain how to handle the minor complication caused by common real zeros of p(x) and q(x). But to avoid that complication now,
00035
(b) f(x) ⴝ x2 ⴚ 3
unless stated to the contrary, we will assume that for any rational function f we consider, p(x) and q(x) have no real zero in common.
Z Figure 1
Because a polynomial q(x) of degree n has at most n real zeros, there are at most n real numbers that are not in the domain of f (x) 0002 P(x)0004q(x). Because a fraction equals 0 only if its numerator is 0, the x intercepts of the graph of f are the real zeros of a polynomial p(x) of degree m. So the number of x intercepts of f is at most m.
EXAMPLE
1
Domain and x Intercepts Find the domain and x intercepts for f (x) 0002
SOLUTION
f (x) 0002
2x2 0003 2x 0003 4 . x2 0003 9
p(x) 2(x 0003 2)(x 1) 2x2 0003 2x 0003 4 0002 0002 2 q(x) (x 0003 3)(x 3) x 00039
Because q(x) 0002 0 for x 0002 3 and x 0002 00033, the domain of f is x 0006 00073
(00030005, 00033) 傼 (00033, 3) 傼 (3, 0005)
or
Because p(x) 0002 0 for x 0002 2 and x 0002 00031, the zeros of f, and the x intercepts of f, are 00031 and 2. 0002
MATCHED PROBLEM 1
Find the domain and x intercepts for f (x) 0002
3x2 0003 12 . x 2x 0003 3 2
0002 The graph of the rational function f (x) 0002 is shown in Figure 2 on the next page.
x2 0003 1.44 x3 0003 x
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Z Figure 2 f (x) 0002
x2 0003 1.44 x3 0003 x
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POLYNOMIAL AND RATIONAL FUNCTIONS y
. 5
00035
5
x
00035
The domain of f consists of all real numbers except x 0002 00031, x 0002 0, and x 0002 1 (the zeros of the denominator x3 0003 x). The dotted vertical lines at x 0002 00071 indicate that those values of x are excluded from the domain (a dotted vertical line at x 0002 0 would coincide with the y axis and is omitted). The graph is discontinuous at x 0002 00031, x 0002 0, and x 0002 1, but is continuous elsewhere and has no sharp corners. The zeros of f are the zeros of the numerator x2 0003 1.44, namely x 0002 00031.2 and x 0002 1.2. The graph of f has four turning points. Its left and right behavior is the same as that of the function g (x) 0002 1x (the graph is close to the x axis for very large and very small values of x). The graph of f illustrates the general properties of rational functions that are listed in Theorem 1. We have already justified Property 3; the other properties are established in calculus.
Z THEOREM 1 Properties of Rational Functions Let f (x) 0002 p(x)0004q(x) be a rational function where p(x) and q(x) are polynomials of degrees m and n, respectively. Then the graph of f(x): 1. 2. 3. 4. 5.
Is continuous with the exception of at most n real numbers Has no sharp corners Has at most m real zeros Has at most m n 0003 1 turning points Has the same left and right behavior as the quotient of the leading terms of p(x) and q(x)
Figure 3 shows graphs of several rational functions, illustrating the properties of Theorem 1. y
y
Z Figure 3 Graphs of rational functions.
5
3
00035
5
x
2
00033
3
1 x
(b) g(x) ⴝ
x
00032
2
00032
00033
00035
(a) f(x) ⴝ
y
1 x2 ⴚ 1
(c) h(x) ⴝ
1 x2 ⴙ 1
x
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y
Z Figure 3 (continued) 15
5
x
EXAMPLE
2
2
00033
x
3
000315
(d) F (x) ⴝ
y
3
00035
000310
10
(e) G(x) ⴝ
x
00032
00033
x2 ⴙ 3x xⴚ1
301
Rational Functions and Inequalities
ⴚx ⴚ 1
(f) H(x) ⴝ
x ⴚ 4x 3
x2 ⴙ x ⴙ 1 x2 ⴙ 1
Properties of Graphs of Rational Functions Use Theorem 1 to explain why each graph is not the graph of a rational function.
y
(A)
y
(B)
3
y
(C)
3
00033
3
x
3
00033
00033
x
3
00033
00033
SOLUTIONS
3
x
00033
(A) The graph has a sharp corner when x 0002 0, so Property 2 is not satisfied. (B) The graph has an infinite number of turning points, so Property 4 is not satisfied. (C) The graph has an infinite number of zeros (all values of x between 0 and 1, inclusive, are zeros), so Property 3 is not satisfied. 0002 Use Theorem 1 to explain why each graph is not the graph of a rational function.
MATCHED PROBLEM 2
y
(A)
y
(B) 3
3
00033
3
00033
y
(C)
x
00033
3
3
00033
x
00033
3
x
00033
0002
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y
Z Vertical and Horizontal Asymptotes
5
00065
5
x
The graphs of Figure 3 on pages 300–301 exhibit similar behaviors near points of discontinuity that can be described using the concept of vertical asymptote. Consider, for example, the rational function f (x) 0002 1x of Figure 3(a). As x approaches 0 from the right, the points (x, 1x ) on the graph have larger and larger y coordinates—that is, 1x increases without bound—as confirmed by Table 1. We write this symbolically as
00065
1 (a) f(x) ⴝ x Z Figure 3(a) Graphs of rational functions.
1 S0004 x
as
x S 00003
and say that the line x 0002 0 (the y axis) is a vertical asymptote for the graph of f.
Table 1 Behavior of 1兾x as x S 00003 x
1
0.1
0.01
0.001
0.0001
0.000 01
0.000 001
. . .
x approaches 0 from the right (x S 0 ⴙ )
1兾x
1
10
100
1,000
10,000
100,000
1,000,000
. . .
1兾x increases without bound (1 0005x S 0004)
If x approaches 0 from the left, the points (x, 1x ) on the graph have smaller and smaller y coordinates—that is, 1x decreases without bound—as confirmed by Table 2. We write this symbolically as 1 S 00060004 x
as
x S 00006
Table 2 Behavior of 1兾x as x S 00006 x
00061
00060.1
00060.01
00060.001
00060.0001
00060.000 01
00060.000 001
. . .
x approaches 0 from the left (x S 0ⴚ)
1兾x
00061
000610
0006100
00061,000
000610,000
0006100,000
00061,000,000
. . .
1兾x decreases without bound (1 0005x S ⴚ0004)
ZZZ EXPLORE-DISCUSS 1
Construct tables similar to Tables 1 and 2 for g(x) 0002 x12 and discuss the behavior of the graph of g(x) near x 0002 0.
Z DEFINITION 2 Vertical Asymptote The vertical line x 0002 a is a vertical asymptote for the graph of y 0002 f (x) if f(x) S 0004
or f(x) S 00060004
as x S a0003
or as x S a0006
(that is, if f (x) either increases or decreases without bound as x approaches a from the right or from the left).
Z THEOREM 2 Vertical Asymptotes of Rational Functions Let f (x) 0002 p(x)0005q(x) be a rational function. If a is a zero of q(x), then the line x 0002 a is a vertical asymptote of the graph of f.*
*Recall that we are assuming that p(x) and q(x) have no real zero in common. Theorem 2 is not valid without this assumption.
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303
For example, x2 0006 1.44 x2 0006 1.44 0002 3 x(x 0006 1)(x 0003 1) x 0006x
f (x) 0002
has three vertical asymptotes, x 0002 00061, x 0002 0, and x 0002 1 (see Fig. 2 on p. 300). The left and right behavior of some, but not all, rational functions can be described using the concept of horizontal asymptote. Consider f (x) 0002 1x . As values of x get larger and larger— that is, as x increases without bound—the points (x, 1x ) have y coordinates that are positive and approach 0, as confirmed by Table 3. Similarly, as values of x get smaller and smaller— that is, as x decreases without bound—the points (x, 1x ) have y coordinates that are negative and approach 0, as confirmed by Table 4. We write these facts symbolically as 1 S0 x
as
xS0004
and as
x S 00060004
and say that the line y 0002 0 (the x axis) is a horizontal asymptote for the graph of f.
Table 3 Behavior of 1/x as x S 0004 x
1
10
100
1,000
10,000
100,000
1,000,000
. . .
x increases without bound (x S 0004)
1兾x
1
0.1
0.01
0.001
0.0001
0.000 01
0.000 001
. . .
1兾x approaches 0 (1兾x S 0)
Table 4 Behavior of 1/x as x S 00060004 x
00061
000610
0006100
00061,000
000610,000
0006100,000
00061,000,000
. . .
x decreases without bound (x S ⴚ0004)
1兾x
00061
00060.1
00060.01
00060.001
00060.0001
00060.000 01
00060.000 001
. . .
1兾x approaches 0 (1兾x S 0)
ZZZ EXPLORE-DISCUSS 2
Construct tables similar to Tables 3 and 4 for each of the following functions, and discuss the behavior of each as x S 0004 and as x S 00060004: (A) f (x) 0002
3x 2 x 00031
(B) g(x) 0002
3x2 x2 0003 1
(C) h(x) 0002
3x3 x2 0003 1
Z DEFINITION 3 Horizontal Asymptote The horizontal line y 0002 b is a horizontal asymptote for the graph of y 0002 f (x) if f (x) S b
as
x S 00060004
or as
xS0004
(that is, if f (x) approaches b as x increases without bound or as x decreases without bound).
A rational function f (x) 0002 p(x)0005q(x) has the same left and right behavior as the quotient of the leading terms of p(x) and q(x) (Property 5 of Theorem 1). Consequently, a rational function has at most one horizontal asymptote. Moreover, we can determine easily whether a rational function has a horizontal asymptote, and if it does, find its equation. Theorem 3 gives the details.
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Z THEOREM 3 Horizontal Asymptotes of Rational Functions Consider the rational function f (x) 0002
am xm p a1x a0 bn xn p b1x b0
where am 0006 0, bn 0006 0. 1. If m n, the line y 0002 0 (the x axis) is a horizontal asymptote. 2. If m 0002 n, the line y 0002 am 0004bn is a horizontal asymptote. 3. If m n, there is no horizontal asymptote. In 1 and 2, the graph of f approaches the horizontal asymptote both as x S 0005 and as x S 00030005.
EXAMPLE
3
Finding Vertical and Horizontal Asymptotes for a Rational Function Find all vertical and horizontal asymptotes for f (x) 0002
SOLUTION
p(x) 2x2 0003 2x 0003 4 0002 q(x) x2 0003 9
Because q(x) 0002 x2 0003 9 0002 (x 0003 3)(x 3), the graph of f (x) has vertical asymptotes at x 0002 3 and x 0002 00033 (Theorem 1). Because p(x) and q(x) have the same degree, the line y
0002
a2 * 2 0002 00022 b2 1
a2 ⴝ 2, b2 ⴝ 1
is a horizontal asymptote (Theorem 3, part 2). MATCHED PROBLEM 3
0002
Find all vertical and horizontal asymptotes for f (x) 0002
3x2 0003 12 x 2x 0003 3 2
0002
Z Analyzing the Graph of a Rational Function We now use the techniques for locating asymptotes, along with other graphing aids discussed in the text, to graph several rational functions. First, we outline a systematic approach to the problem of graphing rational functions.
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
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Z ANALYZING AND SKETCHING THE GRAPH OF A RATIONAL FUNCTION: f(x) ⴝ p(x)冒q(x) Step 1. Intercepts. Find the real solutions of the equation p(x) 0002 0 and use these solutions to plot any x intercepts of the graph of f. Evaluate f (0), if it exists, and plot the y intercept. Step 2. Vertical Asymptotes. Find the real solutions of the equation q(x) 0002 0 and use these solutions to determine the domain of f, the points of discontinuity, and the vertical asymptotes. Sketch any vertical asymptotes as dashed lines. Step 3. Horizontal Asymptotes. Determine whether there is a horizontal asymptote and, if so, sketch it as a dashed line. Step 4. Complete the Sketch. For each interval in the domain of f, plot additional points and join them with a smooth continuous curve.
EXAMPLE
4
Graphing a Rational Function Graph f (x) 0002
2x . x00063 f (x) 0002
SOLUTION
p(x) 2x 0002 x00063 q(x)
Step 1. Intercepts. Find real zeros of p(x) 0002 2x and find f(0): 2x 0002 0 x00020 f (0) 0002 0
x intercept y intercept
The graph crosses the coordinate axes only at the origin. Plot this intercept, as shown in Figure 4. y Horizontal asymptote
10
000610
x and y intercepts
Vertical asymptote
10
x
000610
Intercepts and asymptotes
Z Figure 4 y
Step 2. Vertical Asymptotes. Find real zeros of q(x) 0002 x 0006 3:
10
000610
10
2x f(x) 0002 x00063 000610
Z Figure 5
x0006300020 x00023 x
The domain of f is (00060004, 3) 傼 (3, 0004), f is discontinuous at x 0002 3, and the graph has a vertical asymptote at x 0002 3. Sketch this asymptote, as shown in Figure 4. Step 3. Horizontal Asymptote. Because p(x) and q(x) have the same degree, the line y 0002 2 is a horizontal asymptote, as shown in Figure 4. Step 4. Complete the Sketch. By plotting a few additional points, we obtain the graph in Figure 5. Notice that the graph is a smooth continuous curve over the interval
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(00030005, 3) and over the interval (3, 0005). As expected, there is a break in the graph at x 0002 3. 0002 MATCHED PROBLEM 4
Proceed as in Example 4 and sketch the graph of f (x) 0002
3x . x2 0002
Technology Connections Refer to Example 4. When f (x) ⴝ 2x0004(x ⴚ 3) is graphed on a graphing calculator [Fig. 6(a)], it appears that the graphing calculator has also drawn the vertical asymptote at x ⴝ 3, but this is not the case. Many graphing calculators, when set in connected mode, calculate points on a graph and connect these points with line segments. The last point plotted to the left of the asymptote and the first plotted to the right of the asymptote will usually have very large y coordinates. If these y coordinates have opposite signs, then the graphing
calculator may connect the two points with a nearly vertical line segment, which gives the appearance of an asymptote. If you wish, you can set the calculator in dot mode to plot the points without the connecting line segments [Fig. 6(b)]. Depending on the scale, a graph may even appear to be continuous at a vertical asymptote [Fig. 6(c)]. It is important to always locate the vertical asymptotes as we did in step 2 before turning to the graphing calculator graph to complete the sketch. 10
10
000310
10
40
000310
10
000340
40
000310
000310
000340
(a) Connected mode
(b) Dot mode
(c) Connected mode
2x Z Figure 6 Graphing calculator graphs of f (x) 0002 x 0003 3 .
In Examples 5 and 6 we will just list the results of each step in the graphing strategy and omit the computational details.
EXAMPLE
5
Graphing a Rational Function Graph f (x) 0002
SOLUTION
x2 0003 6x 9 . x2 x 0003 2 f (x) 0002
(x 0003 3)2 x2 0003 6x 9 0002 (x 2)(x 0003 1) x2 x 0003 2
x intercept: x 0002 3 y intercept: f (0) 0002 000392 0002 00034.5 Domain: (00030005, 00032) 傼 (00032, 1) 傼 (1, 0005) Points of discontinuity: x 0002 00032 and x 0002 1 Vertical asymptotes: x 0002 00032 and x 0002 1 Horizontal asymptote: y 0002 1 Locate the intercepts, draw the asymptotes, and plot additional points in each interval of the domain of f to complete the graph (Fig. 7).
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y 10
000310
f (x) 0002
x 2 0003 6x 9 x2 x 0003 2
10
x
0002
Z Figure 7
MATCHED PROBLEM 5
ZZZ
Graph f (x) 0002
x2 . x2 0003 7x 10
The graph of a function cannot cross a vertical asymptote, but the same statement is not true for horizontal asymptotes. The rational function
CAUTION ZZZ
f (x) 0002
2x6 x5 0003 5x3 4x 2 x6 1
has the line y 0002 2 as a horizontal asymptote. The graph of f in Figure 8 clearly shows that the graph of a function can cross a horizontal asymptote. The definition of a horizontal asymptote requires f (x) to approach b as x increases or decreases without bound, but it does not preclude the possibility that f (x) 0002 b for one or more values of x. y 4
f (x) 0002
2x 6 x 5 0003 5x 3 4x 2 x6 1 y 0002 2 is a horizontal asymptote
00035
5
x
Z Figure 8 Multiple intersections of a graph and a horizontal asymptote.
EXAMPLE
6
Graphing a Rational Function Graph f (x) 0002
SOLUTION
x2 0003 3x 0003 4 . x00032
(x 1)(x 0003 4) x2 0003 3x 0003 4 0002 x00032 x00032 x intercepts: x 0002 00031 and x 0002 4 y intercept: f (0) 0002 2 Domain: (00030005, 2) 傼 (2, 0005) Points of discontinuity: x 0002 2 Vertical asymptote: x 0002 2 No horizontal asymptote f (x) 0002
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Although the graph of f does not have a horizontal asymptote, we can still gain some useful information about the behavior of the graph as x S 00060004 and as x S 0004 if we first perform a long division: x00061 x 0006 2冄 x2 0006 3x 0006 4 x2 0006 2x 0006x 0006 4 0006x 0003 2 00066
Quotient
Remainder
This shows that f (x) 0002
x2 0006 3x 0006 4 6 0002x000610006 x00062 x00062
As x S 00060004 or x S 0004, 60005(x 0006 2) S 0 and the graph of f approaches the line y 0002 x 0006 1. This line is called an oblique asymptote for the graph of f. The asymptotes and intercepts are shown in Figure 9, and the graph of f is sketched in Figure 10. y y
10 10
000610
Oblique asymptote y0002x00061
10
x 000610
10
f (x) 0002
000610
x
x 2 0006 3x 0006 4 x00062
000610
Intercepts and asymptotes
Z Figure 9
y0002x00061
Z Figure 10
Generalizing the results of Example 6, we have Theorem 4.
Z THEOREM 4 Oblique Asymptotes and Rational Functions If f (x) 0002 p(x) 0005q(x), where p(x) and q(x) are polynomials and the degree of p(x) is 1 more than the degree of q(x), then f(x) can be expressed in the form f (x) 0002 mx 0003 b 0003
r(x) q(x)
where the degree of r(x) is less than the degree of q(x). The line y 0002 mx 0003 b is an oblique asymptote for the graph of f. That is, [ f (x) 0006 (mx 0003 b)] S 0
as
x S 00060004
or
xS0004 0002
MATCHED PROBLEM 6
Graph, including any oblique asymptotes, f (x) 0002
x2 0003 5 . x00031 0002
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At the beginning of this section we made the assumption that for a rational function f (x) 0002 p(x)0004q(x), the polynomials p(x) and q(x) have no common real zero. Now we abandon that assumption. Suppose that p(x) and q(x) have one or more real zeros in common. Then, by the factor theorem, p(x) and q(x) have one or more linear factors in common, so f(x) can be “reduced.” We proceed to divide out common linear factors in f (x) 0002
p(x) q(x)
fr(x) 0002
pr(x) qr(x)
until we obtain a rational function
in which pr(x) and qr(x) have no common real zero. We analyze and graph fr(x), then insert “holes” as required in the graph of fr to obtain the graph of f. Example 7 illustrates the details.
EXAMPLE
7
Graphing Arbitrary Rational Functions Graph f (x) 0002
SOLUTION
2x5 0003 4x4 0003 6x3 . x5 0003 3x4 0003 3x3 7x2 6x
The real zeros of p(x) 0002 2x5 0003 4x4 0003 6x3 (obtained by graphing or factoring) are 00031, 0, and 3. The real zeros of q(x) 0002 x5 0003 3x4 0003 3x3 7x2 6x are 00031, 0, 2, and 3. The common zeros are 00031, 0, and 3. Factoring and dividing out common linear factors gives f (x) 0002
2x3(x 1)(x 0003 3) x(x 1)2(x 0003 2)(x 0003 3)
and
fr (x) 0002
2x2 (x 1)(x 0003 2)
We analyze fr (x) as usual: x intercept: x 0002 0 y intercept: fr(0) 0002 0 Domain: (00030005, 00031) 傼 (00031, 2) 傼 (2, 0005) Points of discontinuity: x 0002 00031, x 0002 2 Vertical asymptotes: x 0002 00031, x 0002 2 Horizontal asymptote: y 0002 2 The graph of f is identical to the graph of fr except possibly at the common real zeros 00031, 0, and 3. We consider each common zero separately. x 0002 00031: Both f and fr are undefined (no difference in their graphs). x 0002 0: f is undefined but fr(0) 0002 0, so the graph of f has a hole at (0, 0). x 0002 3: f is undefined but fr(3) 0002 4.5, so the graph of f has a hole at (3, 4.5). Therefore, f (x) has the following analysis: x intercepts: none y intercepts: none Domain: (00030005, 00031) 傼 (00031, 0) 傼 (0, 2) 傼 (2, 3) 傼 (3, 0005) Points of discontinuity: x 0002 00031, x 0002 0, x 0002 2, x 0002 3
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Vertical asymptotes: x 0002 00031, x 0002 2 Horizontal asymptote: y 0002 2 Holes: (0, 0), (3, 4.5) Figure 11 shows the graphs of f and fr. y
y
5
5
00035
x
5
00035
2x ⴚ 4x ⴚ 6x 4
3
(b) fr(x) ⴝ
x ⴚ 3x ⴚ 3x ⴙ 7x ⴙ 6x 5
5
4
3
2
2x2 (x ⴙ 1)(x ⴚ 2)
0002
Z Figure 11
MATCHED PROBLEM 7
Graph f (x) 0002
x
00035 5
(a) f (x) ⴝ
00035
x3 0003 x . x4 0003 x2 0002
Z Rational Inequalities A rational function f (x) 0002 p(x)0004q(x) can change sign at a real zero of p(x) (where f has an x intercept) or at a real zero of q(x) (where f is discontinuous), but nowhere else (because f is continuous except where it is not defined). Rational inequalities can therefore be solved in the same way as polynomial inequalities, except that the partition of the x axis is determined by the zeros of p(x) and the zeros of q(x).
EXAMPLE
8
Solving Rational Inequalities Solve
SOLUTION
x3 4x2 6 0. x2 0003 4
Let f (x) 0002
p(x) x3 4x2 0002 2 q(x) x 00034
The zeros of p(x) 0002 x3 4x2 0002 x2(x 4) are 0 and 00034. The zeros of q(x) 0002 x2 0003 4 0002 (x 2)(x 0003 2) are 00032 and 2. These four zeros partition the x axis into the five intervals shown in the table. A test number is chosen from each interval as indicated to determine whether f (x) is positive or negative.
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Interval
Test number x
f (x)
Sign of f
(0003 0005, 00034)
00035
000325/21
0003
(00034, 00032)
00033
9/5
(00032, 0)
00031
(0, 2)
1
(2, 0005)
3
00031
0003
00035/3
0003
311
63/5
We conclude that the solution set of the inequality is (00030005, 00034) 傼 (00032, 0) 傼 (0, 2)
Solve
MATCHED PROBLEM 8
EXAMPLE
9
x2 0003 1 0. x2 0003 9
0002
Solving Rational Inequalities with a Graphing Calculator Solve 1
SOLUTION
9x 0003 9 to three decimal places. x x00033 2
First we convert the inequality to an equivalent inequality in which one side is 0:
10
1000b
000310
10003
10
9x 0003 9 x2 x 0003 3
9x 0003 9 0 x x00033
x 2 ⴚ 8x ⴙ 6
Subtract
9x ⴚ 9 x2 ⴙ x ⴚ 3
from both sides.
Find a common denominator.
2
x2 x 0003 3 9x 0003 9 0003 2 0 x2 x 0003 3 x x00033
000310
(a) f (x) ⴝ
0002
Simplify.
x2 0003 8x 6 0 x2 x 0003 3
x2 ⴙ x ⴚ 3
The zeros of x2 0003 8x 6, to three decimal places, are 0.838 and 7.162. The zeros of x2 x 0003 3 are 00032.303 and 1.303. These four zeros partition the x axis into five intervals:
10
000310
(00030005, 00032.303), (00032.303, 0.838), (0.838, 1.303), (1.303, 7.162), and (7.162, 0005)
10
We graph f (x) 0002
000310
(b) g(x) ⴝ
f (x) 冟 f (x) 冟
Z Figure 12
x2 0003 8x 6 x2 x 0003 3
and
g(x) 0002
f (x) 冟 f (x) 冟
(Fig. 12) and observe that the graph of f is above the x axis on the intervals (00030005, 00032.303), (0.838, 1.303), and (7.162, 0005). So the solution set of the inequality is (00030005, 00032.303) 傼 [0.838, 1.303) 傼 [7.162, 0005) Note that the endpoints that are zeros of f are included in the solution set of the inequality, but not the endpoints at which f is undefined. 0002
MATCHED PROBLEM 9
Solve
x3 4x2 0003 7 0 to three decimal places. x2 0003 5x 0003 1 0002
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ANSWERS TO MATCHED PROBLEMS 1. Domain: (00030005, 00033) ´ (00033, 1) ´ (1, 0005); x intercepts: x 0002 00032, x 0002 2 2. (A) Properties 3 and 4 are not satisfied. (B) Property 1 is not satisfied. (C) Properties 1 and 3 are not satisfied. 3. Vertical asymptotes: x 0002 00033, x 0002 1; horizontal asymptote: y 0002 3 y
4.
y
5. 10
10
000310
10
f (x) 0002
x
000310
3x x2
000310
f (x) 0002
6.
y
x
10
x2 x 2 0003 7x 10
7.
y0002x00031
y 5
f (x) 0002 000310
10
f (x) 0002
x
00035
x3 0003 x x4 0003 x2 5
x
x2 5 x1 00035
8. (00030005, 00033) 傼 [00031, 1] 傼 (3, 0005)
4-4
9. [ 00033.391, 00031.773] 傼 (00030.193, 1.164] 傼 (5.193, 0005)
Exercises
1. Is every polynomial function a rational function? Explain. 2. Is every rational function a polynomial function? Explain.
y
7.
8.
y 10
10
3. Explain in your own words what a vertical asymptote is. 4. Explain in your own words what a horizontal asymptote is. 10
5. Explain in your own words what an oblique asymptote is. 6. Explain why a rational function can’t have both a horizontal asymptote and an oblique asymptote. In Problems 7–10, match each graph with one of the following functions: 2x 0003 4 x2 2x 4 h(x) 0002 x00032
f (x) 0002
2x 4 20003x 4 0003 2x k(x) 0002 x2 g(x) 0002
000310
x
10 000310
x
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9.
y
10.
10
10
000310
10
x
000310
10
Rational Functions and Inequalities
313
27. p(x) 0002
x2 2x 1 x
28. q(x) 0002
x3 0003 1 x1
29. h(x) 0002
3x2 8 2x2 6x
30. k(x) 0002
6x2 0003 5x 1 7x2 0003 28x
x
In Problems 31–34, explain why each graph is not the graph of a rational function. 000310
y
31. 5
2x 0003 4 . Complete each statement: 11. Let f (x) 0002 x2 0003 (A) As x S 00032 , f (x) S ? (B) As x S 00032, f (x) S ? (C) As x S 00030005, f (x) S ? (D) As x S 0005, f (x) S ? 2x 4 . Complete each statement: 20003x 0003 (A) As x S 2 , g(x) S ? (B) As x S 2, g(x) S ? (C) As x S 00030005, g(x) S ? (D) As x S 0005, g(x) S ?
00035
y
32. 5
13. Let h(x) 0002
4 0003 2x 14. Let k(x) 0002 . Complete each statement: x2 0003 (A) As x S 2 , k(x) S ? (B) As x S 2, k(x) S ? (C) As x S 00030005, k(x) S ? (D) As x S 0005, k(x) S ?
00035
3x 0003 9 x
16. g(x) 0002
2x 10 x1
17. h(x) 0002
x6 x2 0003 4
18. k(x) 0002
x2 0003 9 x
19. r(x) 0002
x2 0003 3x 0003 4 x2 1
20. s(x) 0002
x2 4x 0003 5 x2 4
21. F(x) 0002
x4 16 x2 0003 36
22. G(x) 0002
x4 x2 1 x2 0003 25
In Problems 23–30, find all vertical and horizontal asymptotes. 5x 1 23. f (x) 0002 x2 25. s(x) 0002
2x 0003 3 x2 0003 16
7x 0003 2 24. g(x) 0002 x00033 26. t(x) 0002
3x 4 x2 0003 49
5
x
00035
y
33. 3
00033
x
3
00033
In Problems 15–22, find the domain and x intercepts. 15. f (x) 0002
x
00035
12. Let g(x) 0002
2x 4 . Complete each statement: x00032 (A) As x S 20003, h(x) S ? (B) As x S 2, h(x) S ? (C) As x S 00030005, h(x) S ? (D) As x S 0005, h(x) S ?
5
y
34. 5
00035
5
x
00035
In Problems 35–38, explain how the graph of f differs from the graph of g. 35. f (x) 0002
x2 2x ; g(x) 0002 x 2 x
36. f (x) 0002
1 x5 ; g(x) 0002 x00035 x2 0003 25
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37. f (x) 0002
1 x2 ; g(x) 0002 x8 x2 10x 16
67.
9 5 0003 2 1 x x
68.
x4 7 2 x2 1
38. f (x) 0002
x2 0003 x 0003 12 ; g(x) 0002 x 3 x00034
69.
3x 2 7 10 x00035
70.
x 0.5 x2 5x 0003 6
71.
4 7 x x1
72.
x2 1 6 x2 0003 1 x4 1
In Problems 39–52, use the graphing strategy outlined in the text to sketch the graph of each function. 39. f (x) 0002
1 x00034
40. g(x) 0002
1 x3
In Problems 73–78, find all vertical, horizontal, and oblique asymptotes.
41. f (x) 0002
x x1
42. f (x) 0002
3x x00033
73. f (x) 0002
2x2 x00031
74. g(x) 0002
3x2 x2
43. g(x) 0002
1 0003 x2 x2
44. f (x) 0002
x2 1 x2
75. p(x) 0002
x3 x 1
76. q(x) 0002
x5 x 00038
45. f (x) 0002
9 x2 0003 9
46. g(x) 0002
6 x2 0003 x 0003 6
77. r(x) 0002
2x2 0003 3x 5 x
78. s(x) 0002
00033x2 5x 9 x
47. f (x) 0002
x x 00031
48. p(x) 0002
x 1 0003 x2
49. g(x) 0002
2 x2 1
50. f (x) 0002
x x2 1
12x2 (3x 5)2
52. f (x) 0002
51. f (x) 0002
2
7x2 (2x 0003 3)2
In Problems 53–56, give an example of a rational function that satisfies the given conditions. 53. Real zeros: 00032, 00031, 1, 2; vertical asymptotes: none; horizontal asymptote: y 0002 3 54. Real zeros: none; vertical asymptotes: x 0002 4; horizontal asymptote: y 0002 00032 55. Real zeros: none; vertical asymptotes: x 0002 10; oblique asymptote: y 0002 2x 5 56. Real zeros: 1, 2, 3; vertical asymptotes: none; oblique asymptote: y 0002 2 0003 x In Problems 57–64, solve each rational inequality.
2
3
In Problems 79–84, use the graphing strategy outlined in the text to sketch the graph of each function. Write the equations of all vertical, horizontal, and oblique asymptotes. 79. f (x) 0002
x2 1 x
80. g(x) 0002
x2 0003 1 x
81. k(x) 0002
x2 0003 4x 3 2x 0003 4
82. h(x) 0002
x2 x 0003 2 2x 0003 4
83. F(x) 0002
8 0003 x3 4x2
84. G(x) 0002
x4 1 x3
In calculus, it is often necessary to consider rational functions that are not in lowest terms, such as the functions given in Problems 85–88. For each function, state the domain. Write the equations of all vertical and horizontal asymptotes, and sketch the graph. 85. f (x) 0002
x2 0003 4 x00032
86. g(x) 0002
x2 0003 1 x1
87. r(x) 0002
x2 x2 0003 4
88. s(x) 0002
x00031 x2 0003 1
57.
x 0 x00032
58.
2x 0003 1 7 0 x3
APPLICATIONS
59.
x2 0003 16 7 0 5x 0003 2
60.
x00034 0 x2 0003 9
61.
x2 4x 0003 20 4 3x
62.
3x 0003 7 6 2 x2 6x
89. EMPLOYEE TRAINING A company producing electronic components used in television sets has established that on the average, a new employee can assemble N(t) components per day after t days of on-the-job training, as given by
9 5x 6 63. 2 x x 00031
1 1 64. 2 x x 8x 12
In Problems 65–72, solve each rational inequality to three decimal places. 65.
x2 7x 3 7 0 x2
66.
x3 4 0 x x00033 2
N(t) 0002
50t t4
t000b0
Sketch the graph of N, including any vertical or horizontal asymptotes. What does N approach as t S 0005? 90. PHYSIOLOGY In a study on the speed of muscle contraction in frogs under various loads, researchers W. O. Fems and J. Marsh found that the speed of contraction decreases with increasing loads. More precisely, they found that the relationship between
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speed of contraction S (in centimeters per second) and load w (in grams) is given approximately by S(w) 0002
26 0.06w w
w 5
Sketch the graph of S, including any vertical or horizontal asymptotes. What does S approach as w S 0005? 91. RETENTION An experiment on retention is conducted in a psychology class. Each student in the class is given 1 day to memorize the same list of 40 special characters. The lists are turned in at the end of the day, and for each succeeding day for 20 days each student is asked to turn in a list of as many of the symbols as can be recalled. Averages are taken, and it is found that a good approximation of the average number of symbols, N(t), retained after t days is given by N(t) 0002
5t 30 t
t 1
Sketch the graph of N, including any vertical or horizontal asymptotes. What does N approach as t S 0005? 92. LEARNING THEORY In 1917, L. L. Thurstone, a pioneer in quantitative learning theory, proposed the function f (x) 0002
a(x c) (x c) b
to describe the number of successful acts per unit time that a person could accomplish after x practice sessions. Suppose that for a particular person enrolling in a typing class, f (x) 0002
50(x 1) x5
x 0
where f (x) is the number of words per minute the person is able to type after x weeks of lessons. Sketch the graph of f, including any vertical or horizontal asymptotes. What does f approach as x S 0005?
Variation and Modeling
315
93. REPLACEMENT TIME A desktop office copier has an initial price of $2,500. A maintenance/service contract costs $200 for the first year and increases $50 per year thereafter. It can be shown that the total cost of the copier after n years is given by C(n) 0002 2,500 175n 25n2 The average cost per year for n years is C(n) 0002 C(n)0004n. (A) Find the rational function C. (B) When is the average cost per year a minimum? (This is frequently referred to as the replacement time for this piece of equipment.) (C) Sketch the graph of C, including any asymptotes. 94. AVERAGE COST The total cost of producing x units of a certain product is given by C(x) 0002 15 x2 2x 2,000 The average cost per unit for producing x units is C(x) 0002 C(x)0004x. (A) Find the rational function C. (B) At what production level will the average cost per unit be minimal? (C) Sketch the graph of C, including any asymptotes. 95. CONSTRUCTION A rectangular dog pen is to be made to enclose an area of 225 square feet. (A) If x represents the width of the pen, express the total length L of the fencing material required for the pen in terms of x. (B) Considering the physical limitations, what is the domain of the function L? (C) Find the dimensions of the pen that will require the least amount of fencing material. (D) Graph the function L, including any asymptotes. 96. CONSTRUCTION Rework Problem 95 with the added assumption that the pen is to be divided into two sections, as shown in the figure. (Approximate dimensions to three decimal places.)
In Problems 93–96, use the fact from calculus that a function of the form c q(x) 0002 ax b , a 7 0, c 7 0, x 7 0 x
x x x
has its minimum value when x 0002 1c0004a.
4-5
Variation and Modeling Z Direct Variation Z Inverse Variation Z Joint and Combined Variation
If you work more hours at a part-time job, then you will get more pay. If you increase your average speed in a bicycle race, then you will decrease the time required to finish. The relationship between hours and pay in the first instance, and between average speed and finishing time in the second, are expressed by saying “Pay is directly proportional to
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hours worked, but average speed is inversely proportional to finishing time.” Such statements, which describe how one quantity varies with respect to another, have a precise mathematical meaning. The purpose of this section is to explain the terminology of variation and how it is used in engineering and the sciences.
Z Direct Variation The perimeter of a square is a constant multiple of the side length, and the circumference of a circle is a constant multiple of the radius. These are examples of direct variation. Z DEFINITION 1 Direct Variation Let x and y be variables. The statement y is directly proportional to x (or y varies directly as x) means y 0002 kx for some nonzero constant k, called the constant of proportionality (or constant of variation).
y
y 0002 kx, k 0006 0 x
Z Figure 1 Direct variation.
EXAMPLE
1
The perimeter P of a square is directly proportional to the side length x; the constant of proportionality is 4 and the equation of variation is P 0002 4x. Similarly, the circumference C of a circle is directly proportional to the radius r; the constant of proportionality is 2 and the equation of variation is C 0002 2 r. Note that the equation of direct variation y 0002 kx, k 0006 0, gives a linear model with nonzero slope that passes through the origin (Fig. 1).
Direct Variation The force F exerted by a spring is directly proportional to the distance x that it is stretched (Hooke’s law). Find the constant of proportionality and the equation of variation if F 0002 12 pounds when x 0002 13 foot.
SOLUTION
The equation of variation has the form F 0002 kx. To find the constant of proportionality, substitute F 0002 12 and x 0002 13 and solve for k. F 0002 kx k (13)
12 0002 k 0002 36
Let F ⴝ 12 and x ⴝ 13 . Multiply both sides by 3.
Therefore, the constant of proportionality is k 0002 36 and the equation of variation is F 0002 36x MATCHED PROBLEM 1
0002
Find the constant of proportionality and the equation of variation if p is directly proportional to v, and p 0002 200 when v 0002 8. 0002
Z Inverse Variation If variables x and y are inversely proportional, the functional relationship between them is a constant multiple of the rational function y 0002 1兾x.
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317
Z DEFINITION 2 Inverse Variation Let x and y be variables. The statement y is inversely proportional to x (or y varies inversely as x) means y0002
k x
for some nonzero constant k, called the constant of proportionality (or constant of variation).
The rate r and time t it takes to travel a distance of 100 miles are inversely proportional (recall that distance equals rate times time, d 0002 rt). The equation of variation is
y
t0002 y 0002 k/x, k 0006 0 x
Z Figure 2 Inverse variation.
EXAMPLE
2
100 r
and the constant of proportionality is 100. The equation of inverse variation, y 0002 k兾x, determines a rational function having the y axis as a vertical asymptote and the x axis as a horizontal asymptote (Fig. 2). In most applications, the constant k of proportionality will be positive, and only the portion of the graph in Quadrant I will be relevant. If x is very large, then y is close to 0; if x is close to 0, then y is very large.
Inverse Variation The note played by each pipe in a pipe organ is determined by the frequency of vibration of the air in the pipe. The fundamental frequency f of vibration of air in an organ pipe is inversely proportional to the length L of the pipe. (This is why the low frequency notes come from the long pipes.) (A) Find the constant of proportionality and the equation of variation if the fundamental frequency of an 8-foot pipe is 64 vibrations per second. (B) Find the fundamental frequency of a 1.6-foot pipe.
SOLUTIONS
(A) The equation has the form f 0002 k兾L. To find the constant of proportionality, substitute L 0002 8 and f 0002 64 and solve for k. f0002
k L
Let f ⴝ 64 and L ⴝ 8.
64 0002
k 8
Multiply both sides by 8.
k 0002 512 The constant of proportionality is k 0002 512 and the equation of variation is f0002
512 L
(B) If L 0002 1.6, then f 0002 512 1.6 0002 320 vibrations per second. MATCHED PROBLEM 2
0002
Find the constant of proportionality and the equation of variation if P is inversely proportional to V, and P 0002 56 when V 0002 3.5. 0002
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Z Joint and Combined Variation The area of a rectangle is the product of its length and width. This is an example of joint variation. Z DEFINITION 3 Joint Variation Let x, y, and w be variables. The statement w is jointly proportional to x and y (or w varies jointly as x and y) means w 0002 kxy for some nonzero constant k, called the constant of proportionality (or constant of variation).
The area of a rectangle, for example, is jointly proportional to its length and width with constant of proportionality 1; the equation of variation is A 0002 LW. The concept of joint variation can be extended to apply to more than three variables. For example, the volume of a box is jointly proportional to its length, width, and height: V 0002 LWH. Similarly, the concepts of direct and inverse variation can be extended. For example, the area of a circle is directly proportional to the square of its radius; the constant of proportionality is and the equation of variation is A 0002 r2. The three basic types of variation also can be combined. For example, Newton’s law of gravitation, “The force of attraction F between two objects is jointly proportional to their masses m1 and m2 and inversely proportional to the square of the distance d between them,” has the equation F0002k
EXAMPLE
3
m1m2 d2
Joint Variation The volume V of a right circular cone is jointly proportional to the square of its radius r and its height h. Find the constant of proportionality and the equation of variation if a cone of height 8 inches and radius 3 inches has a volume of 24 cubic inches.
SOLUTION
The equation of variation has the form V 0002 kr 2h. To find the constant of proportionality k, substitute V 0002 24 , r 0002 3, and h 0002 8. V 0002 kr2h 24 0002 k(3)28 24 0002 72k
k0002 3
Let V ⴝ 24 , r ⴝ 3, and h ⴝ 8. Simplify. Divide both sides by 72.
The constant of proportionality is k 0002
and the equation of variation is 3
V0002
MATCHED PROBLEM 3
2 rh 3
0002
The volume V of a box with a square base is jointly proportional to the square of a side x of the base and the height h. Find the constant of proportionality and the equation of variation. 0002
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319
Combined Variation The frequency f of a vibrating guitar string is directly proportional to the square root of the tension T and inversely proportional to the length L. What is the effect on the frequency if the length is doubled and the tension is quadrupled?
SOLUTION
The equation of variation has the form f0002k
1T L
Let f1, T1, and L1 denote the initial frequency, tension, and length, respectively. Then L2 0002 2L1 and T2 0002 4T1. Therefore, 1T2 L2
Let L2 ⴝ 2L1, and T2 ⴝ 4T1.
0002k
14T1 2L1
Simplify the radical.
0002k
21T1 2L1
Cancel and use the equation of variation.
f2 0002 k
0002 f1 We conclude that there is no effect on the frequency—the pitch remains the same.
MATCHED PROBLEM 4
0002
Refer to Example 4. What is the effect on the frequency if the tension is increased by a factor of 4 and the length is cut in half ? 0002
Refer to the equation of variation in Example 4. Explain why the frequency f, for fixed T, is a rational function of L, but f is not, for fixed L, a rational function of T.
ZZZ EXPLORE-DISCUSS 1
ANSWERS TO MATCHED PROBLEMS 196 V 4. The frequency is increased by a factor of 4. 1. k 0002 25; p 0002 25v
4-5
2. k 0002 196; P 0002
3. k 0002 1; V 0002 x2h
Exercises
1. Suppose that y is directly proportional to x and that the constant of proportionality is positive. If x increases, what happens to y? Explain.
3. Suppose that y is inversely proportional to x and that the constant of proportionality is positive. If x increases, what happens to y? Explain.
2. Suppose that y is directly proportional to x and that the constant of proportionality is negative. If x increases, what happens to y? Explain.
4. Explain what it means for w to be jointly proportional to x and y. 5. Suppose that y varies directly with x. What is the value of y when x 0002 0? Explain.
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6. Suppose that y varies inversely with x. What is the value of y when x 0002 1? Explain.
31. The maximum safe load L for a horizontal beam varies jointly as its width w and the square of its height h, and inversely as its length x.
In Problems 7–22, translate each statement into an equation using k as the constant of proportionality.
32. The number N of long-distance phone calls between two cities varies jointly as the populations P1 and P2 of the two cities, and inversely as the distance d between the two cities.
7. F is inversely proportional to x. 8. y is directly proportional to the square of x. 9. R is jointly proportional to S and T. 10. u is inversely proportional to v. 11. L is directly proportional to the cube of m. 12. W is jointly proportional to X, Y, and Z. 13. A varies jointly as the square of c and d. 14. q varies inversely as t. 15. P varies directly as x. 16. f varies directly as the square of b. 17. h varies inversely as the square root of s.
33. The f-stop numbers N on a camera, known as focal ratios, are directly proportional to the focal length F of the lens and inversely proportional to the diameter d of the effective lens opening. 34. The time t required for an elevator to lift a weight is jointly proportional to the weight w and the distance d through which it is lifted, and inversely proportional to the power P of the motor. 35. Suppose that f varies directly as x. Show that the ratio x1 兾x2 of two values of x is equal to f1 兾f2, the ratio of the corresponding values of f. 36. Suppose that f varies inversely as x. Show that the ratio x1 兾x2 of two values of x is equal to f2 兾f1, the reciprocal of the ratio of corresponding values of f.
18. C varies jointly as the square of x and cube of y. 19. R varies directly as m and inversely as the square of d.
APPLICATIONS
20. T varies jointly as p and q and inversely as w.
37. PHYSICS The weight w of an object on or above the surface of the Earth varies inversely as the square of the distance d between the object and the center of Earth. If a girl weighs 100 pounds on the surface of Earth, how much would she weigh (to the nearest pound) 400 miles above Earth’s surface? (Assume the radius of Earth is 4,000 miles.)
21. D is jointly proportional to x and the square of y and inversely proportional to z. 22. S is directly proportional to the square root of u and inversely proportional to v. 23. u varies directly as the square root of v. If u 0002 3 when v 0002 4, find u when v 0002 10. 24. y varies directly as the cube of x. If y 0002 48 when x 0002 4, find y when x 0002 8. 25. L is inversely proportional to the square of M. If L 0002 9 when M 0002 9, find L when M 0002 6. 26. I is directly proportional to the cube root of y. If I 0002 5 when y 0002 64, find I when y 0002 8. 27. Q varies jointly as m and the square of n, and inversely as P. If Q 0002 2 when m 0002 3, n 0002 6, and P 0002 12, find Q when m 0002 4, n 0002 18, and P 0002 2. 28. w varies jointly as x, y, and z. If w 0002 36 when x 0002 2, y 0002 8, and z 0002 12, find w when x 0002 1, y 0002 2, and z 0002 4. In Problems 29–34, translate each statement into an equation using k as the constant of variation.
38. PHYSICS A child was struck by a car in a crosswalk. The driver of the car had slammed on his brakes and left skid marks 160 feet long. He told the police he had been driving at 30 miles/hour. The police know that the length of skid marks L (when brakes are applied) varies directly as the square of the speed of the car v, and that at 30 miles/hour (under ideal conditions) skid marks would be 40 feet long. How fast was the driver actually going before he applied his brakes? 39. ELECTRICITY Ohm’s law states that the current I in a wire varies directly as the electromotive forces E and inversely as the resistance R. If I 0002 22 amperes when E 0002 110 volts and R 0002 5 ohms, find I if E 0002 220 volts and R 0002 11 ohms. 40. ANTHROPOLOGY Anthropologists, in their study of race and human genetic groupings, often use an index called the cephalic index. The cephalic index C varies directly as the width w of the head and inversely as the length l of the head (both when viewed from the top). If an Indian in Baja California (Mexico) has measurements of C 0002 75, w 0002 6 inches, and l 0002 8 inches, what is C for an Indian in northern California with w 0002 8.1 inches and l 0002 9 inches?
29. The biologist René Réaumur suggested in 1735 that the length of time t that it takes fruit to ripen is inversely proportional to the sum T of the average daily temperatures during the growing season.
41. PHYSICS If the horsepower P required to drive a speedboat through water is directly proportional to the cube of the speed v of the boat, what change in horsepower is required to double the speed of the boat?
30. The erosive force P of a swiftly flowing stream is directly proportional to the sixth power of the velocity v of the water.
42. ILLUMINATION The intensity of illumination E on a surface is inversely proportional to the square of its distance d from a light source. What is the effect on the total illumination on a book if the distance between the light source and the book is doubled?
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43. MUSIC The frequency of vibration f of a musical string is directly proportional to the square root of the tension T and inversely proportional to the length L of the string. If the tension of the string is increased by a factor of 4 and the length of the string is doubled, what is the effect on the frequency? 44. PHYSICS In an automobile accident the destructive force F of a car is (approximately) jointly proportional to the weight w of the car and the square of the speed v of the car. (This is why accidents at high speed are generally so serious.) What would be the effect on the destructive forces of a car if its weight were doubled and its speed were doubled? 45. SPACE SCIENCE The length of time t a satellite takes to complete a circular orbit of Earth varies directly as the radius r of the orbit and inversely as the orbital velocity v of the satellite. If t 0002 1.42 hours when r 0002 4,050 miles and v 0002 18,000 miles0003hour (Sputnik I), find t to two decimal places for r 0002 4,300 miles and v 0002 18,500 miles0003hour. 46. GENETICS The number N of gene mutations resulting from xray exposure varies directly as the size of the x-ray dose r. What is the effect on N if r is quadrupled? 47. BIOLOGY In biology there is an approximate rule, called the bioclimatic rule for temperate climates, which states that the difference d in time for fruit to ripen (or insects to appear) varies directly as the change in altitude h. If d 0002 4 days when h 0002 500 feet, find d when h 0002 2,500 feet. 48. PHYSICS Over a fixed distance d, speed r varies inversely as time t. Police use this relationship to set up speed traps. If in a given speed trap r 0002 30 miles0003hour when t 0002 6 seconds, what would be the speed of a car if t 0002 4 seconds? 49. PHYSICS The length L of skid marks of a car’s tires (when the brakes are applied) is directly proportional to the square of the speed v of the car. How is the length of skid marks affected by doubling the speed? 50. PHOTOGRAPHY In taking pictures using flashbulbs, the lens opening (f-stop number) N is inversely proportional to the distance d from the object being photographed. What adjustment should you make on the f-stop number if the distance between the camera and the object is doubled? 51. ENGINEERING The total pressure P of the wind on a wall is jointly proportional to the area of the wall A and the square of the velocity of the wind v. If P 0002 120 pounds when A 0002 100 square feet
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and v 0002 20 miles/hour, find P if A 0002 200 square feet and v 0002 30 miles/hour. 52. ENGINEERING The thrust T of a given type of propeller is jointly proportional to the fourth power of its diameter d and the square of the number of revolutions per minute n it is turning. What happens to the thrust if the diameter is doubled and the number of revolutions per minute is cut in half? 53. PSYCHOLOGY In early psychological studies on sensory perception (hearing, seeing, feeling, and so on), the question was asked: “Given a certain level of stimulation S, what is the minimum amount of added stimulation 0004S that can be detected?” A German physiologist, E. H. Weber (1795–1878) formulated, after many experiments, the famous law that now bears his name: “The amount of change 0004S that will be just noticed varies directly as the magnitude S of the stimulus.” (A) Write the law as an equation of variation. (B) If a person lifting weights can just notice a difference of 1 ounce at the 50-ounce level, what will be the least difference she will be able to notice at the 500-ounce level? (C) Determine the just noticeable difference in illumination a person is able to perceive at 480 candlepower if he is just able to perceive a difference of 1 candlepower at the 60-candle-power level. 54. PSYCHOLOGY Psychologists in their study of intelligence often use an index called IQ. IQ varies directly as mental age MA and inversely as chronological age CA (up to the age of 15). If a 12-yearold boy with a mental age of 14.4 has an IQ of 120, what will be the IQ of an 11-year-old girl with a mental age of 15.4? 55. GEOMETRY The volume of a sphere varies directly as the cube of its radius r. What happens to the volume if the radius is doubled? 56. GEOMETRY The surface area S of a sphere varies directly as the square of its radius r. What happens to the area if the radius is cut in half? 57. MUSIC The frequency of vibration of air in an open organ pipe is inversely proportional to the length of the pipe. If the air column in an open 32-foot pipe vibrates 16 times per second (low C), then how fast would the air vibrate in a 16-foot pipe? 58. MUSIC The frequency of pitch f of a musical string is directly proportional to the square root of the tension T and inversely proportional to the length l and the diameter d. Write the equation of variation using k as the constant of variation. (It is interesting to note that if pitch depended on only length, then pianos would have to have strings varying from 3 inches to 38 feet.)
Review
Polynomial Functions and Models
A function that can be written in the form P(x) 0002 anxn 0005 an00061xn00061 0005 . . . 0005 a1x 0005 a0, an 0007 0, is a polynomial function of degree n. In this chapter, when not specified otherwise, the coefficients an, an00061, . . . , a1, a0 are complex numbers and the domain of P is the set of complex numbers. A number r is said to be a zero (or root) of a function P(x) if P(r) 0002 0.
The real zeros of P(x) are just the x intercepts of the graph of P(x). A point on a continuous graph that separates an increasing portion from a decreasing portion, or vice versa, is called a turning point. If P(x) is a polynomial of degree n 7 0 with real coefficients, then the graph of P(x): 1. Is continuous for all real numbers 2. Has no sharp corners
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3. Has at most n real zeros 4. Has at most n –1 turning points
n Linear Factors Theorem P(x) can be factored as a product of n linear factors.
5. Increases or decreases without bound as x S 0005 and as x S 00030005
If P(x) is factored as a product of linear factors, the number of linear factors that have zero r is said to be the multiplicity of r.
The left and right behavior of such a polynomial P(x) is determined by its highest degree or leading term: As x S 00070005, both an xn and P(x) approach 00070005, with the sign depending on n and the sign of an. For any polynomial P(x) of degree n, we have the following important results:
Imaginary Zeros Theorem Imaginary zeros of polynomials with real coefficients, if they exist, occur in conjugate pairs.
Division Algorithm P(x) 0002 (x 0003 r)Q(x) R where the quotient Q(x) and remainder R are unique; x – r is the divisor. Remainder Theorem P(r) 0002 R Factor Theorem x – r is a factor of P(x) if and only if R 0002 0. Zeros of Polynomials P(x) has at most n zeros. Synthetic division is an efficient method for dividing polynomials by linear terms of the form x – r.
4-2
Real Zeros and Polynomials of Odd Degree If P(x) has odd degree and real coefficients, then the graph of P has at least one x intercept. Zeros of Even or Odd Multiplicity Let P(x) have real coefficients: 1. If r is a real zero of P(x) of even multiplicity, then P(x) has a turning point at r and does not change sign at r. 2. If r is a real zero of P(x) of odd multiplicity, then P(x) does not have a turning point at r and changes sign at r. Rational Zero Theorem If the rational number b/c, in lowest terms, is a zero of the polynomial P(x) 0002 an xn an00031xn00031 # # # a1x a0
Real Zeros and Polynomial Inequalities
The following theorems are useful in locating and approximating all real zeros of a polynomial P(x) of degree n 7 0 with real coefficients, an 7 0: Upper and Lower Bound Theorem 1. Upper bound: A number r 7 0 is an upper bound for the real zeros of P(x) if, when P(x) is divided by x – r using synthetic division, all numbers in the quotient row, including the remainder, are nonnegative. 2. Lower bound: A number r 6 0 is a lower bound for the real zeros of P(x) if, when P(x) is divided by x – r using synthetic division, all numbers in the quotient row, including the remainder, alternate in sign. Location Theorem Suppose that a function f is continuous on an interval I that contains numbers a and b. If f (a) and f (b) have opposite signs, then the graph of f has at least one x intercept between a and b. The bisection method uses the location theorem repeatedly to approximate real zeros to any desired accuracy. Polynomial inequalities can be solved by finding the zeros and inspecting the graph of an appropriate polynomial with real coefficients.
4-3
Linear and Quadratic Factors Theorem If P(x) has real coefficients, then P(x) can be factored as a product of linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros).
Complex Zeros and Rational Zeros of Polynomials
If P(x) is a polynomial of degree n 7 0 we have the following important theorems: Fundamental Theorem of Algebra P(x) has at least one zero.
an 0006 0
with integer coefficients, then b must be an integer factor of a0 and c must be an integer factor of an. If P(x) 0002 (x 0003 r)Q(x), then Q(x) is called a reduced polynomial for P(x).
4-4
Rational Functions and Inequalities
A function f is a rational function if it can be written in the form f (x) 0002
p(x) q(x)
where p(x) and q(x) are polynomials of degrees m and n, respectively. The graph of a rational function f(x): 1. Is continuous with the exception of at most n real numbers 2. Has no sharp corners 3. Has at most m real zeros 4. Has at most m n – 1 turning points 5. Has the same left and right behavior as the quotient of the leading terms of p(x) and q(x) The vertical line x 0002 a is a vertical asymptote for the graph of y 0002 f (x) if f(x) S 0005 or f(x) S 00030005 as x S a or as x S a0003. The horizontal line y 0002 b is a horizontal asymptote for the graph of y 0002 f(x) if f (x) S b as x S 00030005 or as x S 0005. The line y 0002 mx b is an oblique asymptote if [ f(x) 0003 (mx b)] S 0 as x S 00030005 or as x S 0005.
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Review Exercises
Let f (x) 0002
am xm # # # a1x a0 , am 0006 0, bn 0006 0. bn xn # # # b1x b0
1. If m 6 n, the line y 0002 0 (the x axis) is a horizontal asymptote.
4-5
Variation and Modeling
Let x and y be variables. The statement: 1. y is directly proportional to x (or y varies directly as x) means
2. If m 0002 n, the line y 0002 am兾bn is a horizontal asymptote. 3. If m 7 n, there is no horizontal asymptote.
y 0002 kx for some nonzero constant k;
Analyzing and Sketching the Graph of a Rational Function: f (x) ⴝ p(x)兾q(x) Step 1. Intercepts. Find the real solutions of the equation p(x) 0002 0 and use these solutions to plot any x intercepts of the graph of f. Evaluate f(0), if it exists, and plot the y intercept. Step 2. Vertical Asymptotes. Find the real solutions of the equation q(x) 0002 0 and use these solutions to determine the domain of f, the points of discontinuity, and the vertical asymptotes. Sketch any vertical asymptotes as dashed lines. Step 3. Horizontal Asymptotes. Determine whether there is a horizontal asymptote and, if so, sketch it as a dashed line. Step 4. Complete the Sketch. For each interval in the domain of f, plot additional points and join them with a smooth continuous curve. Rational inequalities can be solved by finding the zeros of p(x) and q(x), for an appropriate rational function f (x) 0002 p(x)0004q(x), and inspecting the graph of f.
2. y is inversely proportional to x (or y varies inversely as x) means y0002
4
k x
for some nonzero constant k; 3. w is jointly proportional to x and y (or w varies jointly as x and y) means w 0002 kxy for some nonzero constant k. In each case the nonzero constant k is called the constant of proportionality (or constant of variation). The three basic types of variation also can be combined. For example, Newton’s law of gravitation, “The force of attraction F between two objects is jointly proportional to their masses m1 and m2 and inversely proportional to the square of the distance d between them” has the equation F0002k
CHAPTER
323
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Review Exercises
Work through all the problems in this chapter review, and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. 1. List the real zeros and turning points, and state the left and right behavior, of the polynomial function that has the indicated graph. y 5
3. If P(x) 0002 x5 0003 4x4 9x2 0003 8, find P(3) using the remainder theorem and synthetic division. 4. What are the zeros of P(x) 0002 3(x 0003 2)(x 4)(x 1)? 5. If P(x) 0002 x2 0003 2x 2 and P(1 i) 0002 0, find another zero of P(x). 6. Let P(x) be the polynomial whose graph is shown in the following figure. (A) Assuming that P(x) has integer zeros and leading coefficient 1, find the lowest-degree equation that could produce this graph. (B) Describe the left and right behavior of P(x). P (x)
00035
5
5
x
00035
2. Use synthetic division to divide P(x) 0002 2x3 3x2 0003 1 by D(x) 0002 x 2, and write the answer in the form P(x) 0002 D(x)Q(x) R.
00035
5
00035
x
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7. According to the upper and lower bound theorem, which of the following are upper or lower bounds of the zeros of P(x) 0002 x3 0003 4x2 0004 2?
25. Determine all rational zeros of P(x) 0002 2x3 0003 3x2 0003 18x 0003 8. 26. Factor the polynomial in Problem 25 into linear factors. 27. Find all rational zeros of P(x) 0002 x3 0003 3x2 0004 5.
00032, 00031, 3, 4 8. How do you know that P(x) 0002 2x3 0003 3x2 0004 x 0003 5 has at least one real zero between 1 and 2?
28. Find all zeros (rational, irrational, and imaginary) exactly for P(x) 0002 2x4 0003 x3 0004 2x 0003 1.
9. List all possible rational zeros of a polynomial with integer coefficients that has leading coefficient 5 and constant term 000315.
29. Factor the polynomial in Problem 28 into linear factors.
10. Find all rational zeros for P(x) = 5x2 0004 74x 0003 15. 11. Find the domain and x intercepts for: 6x (A) f (x) 0002 x00035 7x 0004 3 (B) g (x) 0002 2 x 0004 2x 0003 8
31. Factor P(x) 0002 x4 0004 5x2 0003 36 in two ways: (A) As a product of linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros) (B) As a product of linear factors with complex coefficients
12. Find the horizontal and vertical asymptotes for the functions in Problem 11. 13. Explain why the graph is not the graph of a polynomial function. y 5
00035
30. If P(x) 0002 (x 0003 1)2(x 0004 1)3(x2 0003 1)(x2 0004 1), what is its degree? Write the zeros of P(x), indicating the multiplicity of each if greater than 1.
5
x
00035
In Problems 14–19, translate each statement into an equation using k as the constant of proportionality. 14. F is directly proportional to the square root of x. 15. G is jointly proportional to x and the square of y. 16. H is inversely proportional to the cube of z. 17. R varies jointly as the square of x and the square of y. 18. S varies inversely as the square of u. 19. T varies directly as v and inversely as w. 20. Let P(x) 0002 x3 0003 3x2 0003 3x 0004 4. (A) Graph P(x) and describe the graph verbally, including the number of x intercepts, the number of turning points, and the left and right behavior. (B) Approximate the largest x intercept to two decimal places. 21. If P(x) 0002 8x4 0003 14x3 0003 13x2 0003 4x 0004 7, find Q(x) and R such that P(x) 0002 (x 0003 14)Q(x) 0004 R. What is P(14)?
32. Let P(x) 0002 x5 0003 10x4 0004 30x3 0003 20x2 0003 15x 0003 2. (A) Approximate the zeros of P(x) to two decimal places and state the multiplicity of each zero. (B) Can any of these zeros be approximated with the bisection method? A maximum command? A minimum command? Explain. 33. Let P(x) 0002 x4 0003 2x3 0003 30x2 0003 25. (A) Find the smallest positive and largest negative integers that, by Theorem 1 in Section 4-2, are upper and lower bounds, respectively, for the real zeros of P(x). (B) If (k, k 0004 1), k an integer, is the interval containing the largest real zero of P(x), determine how many additional intervals are required in the bisection method to approximate this zero to one decimal place. (C) Approximate the real zeros of P(x) of two decimal places. x00031 . 2x 0004 2 (A) Find the domain and the intercepts for f. (B) Find the vertical and horizontal asymptotes for f. (C) Sketch a graph of f. Draw vertical and horizontal asymptotes with dashed lines.
34. Let f (x) 0002
35. Solve each polynomial inequality to three decimal places: (A) x3 0003 5x 0004 4 0005 0 (B) x3 0003 5x 0004 4 0005 2 36. Explain why the graph is not the graph of a rational function. y 5
00035
5
x
22. If P(x) 0002 4x3 0003 8x2 0003 3x 0003 3, find P(000312) using the remainder theorem and synthetic division. 23. Use the quadratic formula and the factor theorem to factor P(x) 0002 x2 0003 2x 0003 1. 24. Is x 0004 1 a factor of P(x) 0002 9x26 0003 11x17 0004 8x11 0003 5x4 0003 7? Explain, without dividing or using synthetic division.
00035
37. B varies inversely as the square root of c. If B 0002 5 when c 0002 4, find B when c 0002 25. 38. D is jointly proportional to x and y. If D 0002 10 when x 0002 3 and y 0002 2, find D when x 0002 9 and y 0002 8.
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39. Use synthetic division to divide P(x) 0002 x3 0003 3x 0003 2 by [x 0006 (1 0003 i)]. Write the answer in the form P(x) 0002 D(x)Q(x) 0003 R. 40. Find a polynomial of lowest degree with leading coefficient 1 that has zeros 000612 (multiplicity 2), 00063, and 1 (multiplicity 3). (Leave the answer in factored form.) What is the degree of the polynomial?
APPLICATIONS In Problems 55–58, express the solutions as the roots of a polynomial equation of the form P(x) 0002 0. Find rational solutions exactly and irrational solutions to one decimal place.
41. Repeat Problem 40 for a polynomial P(x) with zeros 00065, 2 0006 3i, and 2 0003 3i.
55. ARCHITECTURE An entryway is formed by placing a rectangular door inside an arch in the shape of the parabola with graph y 0002 16 0006 x2, x and y in feet (see the figure). If the area of the door is 48 square feet, find the dimensions of the door.
42. Find all zeros (rational, irrational, and imaginary) exactly for P(x) 0002 2x5 0006 5x4 0006 8x3 0003 21x2 0006 4.
y
43. Factor the polynomial in Problem 42 into linear factors.
16
y 0002 16 0006 x 2
44. Let P(x) 0002 x4 0003 16x3 0003 47x2 0006 137x 0003 73. Approximate (to two decimal places) the x intercepts and the local extrema. 45. What is the minimal degree of a polynomial P(x), given that P(00061) 0002 00064, P(0) 0002 2, P(1) 0002 00065, and P(2) 0002 3? Justify your conclusion. 46. If P(x) is a cubic polynomial with integer coefficients and if 1 0003 2i is a zero of P(x), can P(x) have an irrational zero? Explain. 47. The solutions to the equation x3 0006 27 0002 0 are the cube roots of 27. (A) How many cube roots of 27 are there? (B) 3 is obviously a cube root of 27; find all others. 48. Let P(x) 0002 x4 0003 2x3 0006 500x2 0006 4,000. (A) Find the smallest positive integer multiple of 10 and the largest negative integer multiple of 10 that, by Theorem 1 in Section 4-2, are upper and lower bounds, respectively, for the real zeros of P(x). (B) Approximate the real zero of P(x) to two decimal places.
00064
x
56. CONSTRUCTION A grain silo is formed by attaching a hemisphere to the top of a right circular cylinder (see the figure). If the cylinder is 18 feet high and the volume of the silo is 486 cubic feet, find the common radius of the cylinder and the hemisphere.
49. Graph
x
x2 0003 2x 0003 3 f (x) 0002 x00031 Indicate any vertical, horizontal, or oblique asymptotes with dashed lines.
x
18 feet
50. Use a graphing calculator to find any horizontal asymptotes for f (x) 0002
4
2x 2
2x 0003 3x 0003 4
51. Solve each rational inequality: x00062 5 17 (A) (B) 0 7 x 50006x x00033 52. Solve each rational inequality to three decimal places: x2 0006 3 (A) 3 0 x 0006 3x 0003 1 x2 0006 3 5 (B) 3 7 2 x 0006 3x 0003 1 x
57. MANUFACTURING A box is to be made out of a piece of cardboard that measures 15 by 20 inches. Squares, x inches on a side, will be cut from each corner, and then the ends and sides will be folded up (see the figure). Find the value of x that would result in a box with a volume of 300 cubic inches. 20 in.
53. If P(x) 0002 x3 0006 x2 0006 5x 0003 4, determine the number of real zeros of P(x) and explain why P(x) has no rational zeros. 15 in.
54. Give an example of a rational function f(x) that satisfies the following conditions: the real zeros of f are 00063, 0, and 2; the vertical asymptotes of f are the line x 0002 00061 and x 0002 4; and the line y 0002 5 is a horizontal asymptote.
x x
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58. PHYSICS The centripetal force F of a body moving in a circular path at constant speed is inversely proportional to the radius r of the path. What happens to F if r is doubled? 59. PHYSICS The Maxwell–Boltzmann equation says that the average velocity v of a molecule varies directly as the square root of the absolute temperature T and inversely as the square root of its molecular weight w. Write the equation of variation using k as the constant of variation. 60. WORK The amount of work A completed varies jointly as the number of workers W used and the time t they spend. If 10 workers can finish a job in 8 days, how long will it take 4 workers to do the same job? 61. SIMPLE INTEREST The simple interest I earned in a given time is jointly proportional to the principal p and the interest rate r. If $100 at 4% interest earns $8, how much will $150 at 3% interest earn in the same period?
CHAPTER
ZZZ
Number of Refrigerators y
10
270
20
430
25
525
30
630
45
890
48
915
63. CRIME STATISTICS According to data published by the FBI, the crime index in the United States has shown a downward trend since the early 1990s. The crime index is defined as the number of crimes per 100,000 inhabitants.
Problems 62 and 63 require a graphing calculator or a computer that can calculate cubic regression polynomials for a given data set. 62. ADVERTISING A chain of appliance stores uses television ads to promote the sale of refrigerators. Analyzing past records produced the data in the table, where x is the number of ads placed monthly and y is the number of refrigerators sold that month. (A) Find a cubic regression equation for these data using the number of ads as the independent variable. (B) Estimate (to the nearest integer) the number of refrigerators that would be sold if 15 ads are placed monthly. (C) Estimate (to the nearest integer) the number of ads that should be placed to sell 750 refrigerators monthly.
Number of Ads x
Year
Crime index
1987
5,550
1992
5,660
1997
4,930
2002
4,119
2007
3,016
Source: Federal Bureau of Investigation
(A) Find a cubic regression model for the crime index if x 0002 0 represents 1987. (B) Use the cubic regression model to predict the crime index in 2020. (C) Do you expect the model to give accurate predictions after 2020? Explain.
4
GROUP ACTIVITY Interpolating Polynomials
How could you find a polynomial whose graph passes through the points (1, 1) and (2, 3)? You could use the point-slope form of the equation of a line. How could you find a polynomial P(x) whose graph passes through all four of the points (1, 1), (2, 3), (3, 00033), and (4, 1)? Such a polynomial is called an interpolating polynomial for the four points. The key is to write the unknown polynomial P(x) in the form P(x) 0002 a0 a1(x 0003 1) a2(x 0003 1)(x 0003 2) a3(x 0003 1)(x 0003 2)(x 0003 3) To find a0 , substitute 1 for x. Next, to find a1, substitute 2 for x. Then, to find a2, substitute 3 for x. Finally, to find a3, substitute 4 for x.
(A) Find a0, a1, a2, and a3. (B) Expand P(x) and verify that P(x) 0002 3x3 0003 22x2 47x 0003 27. (C) Explain why P(x) is the only polynomial of degree 3 whose graph passes through the four given points. (D) Give an example to show that the interpolating polynomial for a set of n 1 points may have degree less than n. (E) Find the interpolating polynomial for the five points (00032, 00033), (00031, 0), (0, 5), (1, 0), and (2, 00033).
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5
C
OUTLINE
MOST of the functions we’ve worked with so far have been polynomial
or rational functions, with a few others involving roots. Functions that can be expressed in terms of addition, subtraction, multiplication, division, and roots of variables and constants are called algebraic functions. In Chapter 5, we will study exponential and logarithmic functions. These functions are not algebraic; they belong to the class of transcendental functions. Exponential and logarithmic functions are used to model a surprisingly wide variety of real-world phenomena: growth of populations of people, animals, and bacteria; decay of radioactive substances; epidemics; magnitudes of sounds and earthquakes. These and many other applications will be studied in this chapter.
5-1
Exponential Functions
5-2
Exponential Models
5-3
Logarithmic Functions
5-4
Logarithmic Models
5-5
Exponential and Logarithmic Equations Chapter 5 Review Chapter 5 Group Activity: Comparing Regression Models
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5-1
Exponential Functions Z Defining Exponential Functions Z Graphs of Exponential Functions Z Additional Exponential Properties Z The Exponential Function with Base e Z Compound Interest Z Interest Compounded Continuously
Many of the functions we’ve studied so far have included exponents. But in every case, the exponent was a constant, and the base was often a variable. In this section, we will reverse those roles. In an exponential function, the variable appears in an exponent. As we’ll see, this has a significant effect on the properties and graphs of these functions. A review of the basic properties of exponents in Section R-2, would be very helpful before moving on. y
Z Defining Exponential Functions
10
Let’s start by noting that the functions f and g given by f (x) 0002 2x
y 0002 x2
00035
5
x
(a)
y
g(x) 0002 x2
are not the same function. Whether a variable appears as an exponent with a constant base or as a base with a constant exponent makes a big difference. The function g is a quadratic function, which we have already discussed. The function f is an exponential function. The graphs of f and g are shown in Figure 1. As expected, they are very different. We know how to define the values of 2x for many types of inputs. For positive integers, it’s simply repeated multiplication: 22 0002 2 ⴢ 2 0002 4;
10
and
23 0002 2 ⴢ 2 ⴢ 2 0002 8; 24 0002 2 ⴢ 2 ⴢ 2 ⴢ 2 0002 16
For negative integers, we use properties of negative exponents: y 0002 2x
00035
5
x
200032 0002
1 1 0002 ; 2 4 2
200033 0002
1 1 0002 3 8 2
For rational numbers, a calculator comes in handy: 1
(b)
Z Figure 1
1 200031 0002 ; 2
22 0002 12 ⬇ 1.4;
3
9
4 9 22 0002 223 ⬇ 2.8; 24 0002 2 2 ⬇ 4.8
The only catch is that we don’t know how to define 2x for all real numbers. For example, what does 212 mean? Your calculator can give you a decimal approximation, but where does it come from? That question is not easy to answer at this point. In fact, a precise definition of 212 must wait for more advanced courses. For now, we will simply state that for any positive real number b, the expression bx is defined for all real values of x, and the output is a real number as well. This enables us to draw the continuous graph for f (x) 0002 2x in Figure 1. In Problems 79 and 80 in Exercises 5-1, we will explore a method for defining bx for irrational x values like 12.
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329
Z DEFINITION 1 Exponential Function The equation f (x) 0002 bx
b 7 0, b 0005 1
defines an exponential function for each different constant b, called the base. The independent variable x can assume any real value.
The domain of f is the set of all real numbers, and it can be shown that the range of f is the set of all positive real numbers. We require the base b to be positive to avoid imaginary numbers such as (00032)100042. Problems 53 and 54 in Exercises 5-1 explore why b 0002 0 and b 0002 1 are excluded.
Z Graphs of Exponential Functions ZZZ EXPLORE-DISCUSS 1
y 10
y3 0002 5x y2 0002 3x y1 0002 2x
5
00035
5
x
x Z Figure 2 y 0002 b for b 0002 2, 3, 5.
y2 0002
冢 13 冣
x
冢 15 冣
x
5
00035
冢 12 冣
Z THEOREM 1 Properties of Graphs of Exponential Functions
1. 2. 3. 4.
10
y1 0002
The graphs of y 0002 bx for b 0002 2, 3, and 5 are shown in Figure 2. Note that all three have the same basic shape, and pass through the point (0, 1). Also, the x axis is a horizontal asymptote for each graph, but only as x S 00030006. The main difference between the graphs is their steepness. Next, let’s look at the graphs of y 0002 bx for b 0002 12, 13, and 15 (Fig. 3). Again, all three have the same basic shape, pass through (0, 1), and have horizontal asymptote y 0002 0, but we can see that for b 6 1, the asymptote is only as x S 0006. In general, for bases less than 1, the graph is a reflection through the y axis of the graphs for bases greater than 1. The graphs in Figures 2 and 3 suggest that the graphs of exponential functions have the properties listed in Theorem 1, which we state without proof.
Let f (x) 0002 bx be an exponential function, b 7 0, b 0005 1. Then the graph of f (x):
y
y3 0002
Compare the graphs of f (x) 0002 3x and g(x) 0002 2x by plotting both functions on the same coordinate system. Find all points of intersection of the graphs. For which values of x is the graph of f above the graph of g? Below the graph of g? Are the graphs of f and g close together as x S 0006? As x S 00030006? Discuss.
x
5
x
Is continuous for all real numbers Has no sharp corners Passes through the point (0, 1) Lies above the x axis, which is a horizontal asymptote either as x S 0006 or x S 00030006, but not both 5. Increases as x increases if b 7 1; decreases as x increases if 0 6 b 6 1 6. Intersects any horizontal line at most once (that is, f is one-to-one)
1 1 1 x Z Figure 3 y 0002 b for b 0002 2, 3, 5.
These properties indicate that the graphs of exponential functions are distinct from the graphs we have already studied. (Actually, property 4 is enough to ensure that graphs of exponential functions are different from graphs of polynomials and rational functions.) Property 6 is important because it guarantees that exponential functions have inverses. Those inverses, called logarithmic functions, are the subject of Section 5-3.
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Transformations of exponential functions are very useful in modeling real-world phenomena, like population growth and radioactive decay. These are among the applications we’ll study in Section 5-2. It is important to understand how the graphs of those functions are related to the graphs of the exponential functions in this section. In Example 1, we will use the transformations we studied in Section 3-3 to examine this relationship.
EXAMPLE
1
Transformations of Exponential Functions For the function g(x) 0002 14 (2x ), use transformations to explain how the graph of g is related to the graph of f (x) 0002 2x in Figure 1(b). Find the intercepts and asymptotes, and draw the graph of g.
SOLUTION
The graph of g is a vertical shrink of the graph of f by a factor of 14. So like f, g(x) 0007 0 for all real numbers x, and g(x) S 0 as x S 00030006. In other words, there are no x intercepts, and the x axis is a horizontal asymptote. Since g(0) 0002 14 (20) 0002 14, 14 is the y intercept. Plotting the intercept and a few more points, we obtain the graph of g shown in the figure, with a portion magnified to illustrate the behavior better. y y
10
1
5 0.5
⫺3
MATCHED PROBLEM 1
⫺2
⫺1
x
⫺5
5
x
0002
Let g(x) 0002 12 (40003x ). Use transformations to explain how the graph of g is related to the graph of the exponential function f (x) 0002 4x. Find the intercepts and asymptotes, and sketch the graph of g. 0002
Z Additional Exponential Properties Exponential functions whose domains include irrational numbers obey the familiar laws of exponents for rational exponents. We summarize these exponent laws here and add two other useful properties. Z EXPONENTIAL FUNCTION PROPERTIES For a and b positive, a 0005 1, b 0005 1, and x and y real: 1. Exponent laws: a xa y 0002 a xy
(a x) y 0002 a xy
a x ax a b 0002 x b b
ax 0002 a x0003y ay
(ab)x 0002 a xb x 25x 27x
* ⴝ 25xⴚ7x
ⴝ 2ⴚ2x
2. ax 0002 a y if and only if x 0002 y. If 64x ⫽ 62xⴙ4, then 4x ⫽ 2x ⴙ 4, and x ⫽ 2. 3. For x 0005 0, ax 0002 bx if and only if a 0002 b. If a4 ⫽ 34, then a ⫽ 3.
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
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331
Property 2 is another way to express the fact that the exponential function f(x) 0002 ax is oneto-one (see property 6 of Theorem 1). Because all exponential functions of the form f(x) 0002 ax pass through the point (0, 1) (see property 3 of Theorem 1), property 3 indicates that the graphs of exponential functions with different bases do not intersect at any other points.
EXAMPLE
2
Using Exponential Function Properties Solve 4x00033 0002 8 for x.
SOLUTION
Express both sides in terms of the same base, and use property 2 to equate exponents. 4x00033 0002 8 2 x00033
Express 4 and 8 as powers of 2. 3
(2 ) 00022 2x00036 2 0002 23 2x 0003 6 0002 3 2x 0002 9 x0002
Use the property (ax)y ⴝ axy. Use property 2 to set exponents equal. Add 6 to both sides. Divide both sides by 2.
9 2
✓ 4(900042)00033 0002 4300042 0002 ( 14)3 0002 23 0002 8
CHECK
Technology Connections 4xⴚ3 ⴝ 8. Graph y1 ⴝ 4xⴚ3 and y2 ⴝ 8, then use the intersect command to obtain x ⴝ 4.5 (Fig. 4).
As an alternative to the algebraic method of Example 2, you can use a graphing calculator to solve the equation 10
⫺10
10
⫺10
Z Figure 4
0002 MATCHED PROBLEM 2
Solve 27x1 0002 9 for x.
0002
Z The Exponential Function with Base e Surprisingly, among the exponential functions it is not the function g(x) 0002 2x with base 2 or the function h(x) 0002 10x with base 10 that is used most frequently in mathematics. Instead, the most commonly used base is a number that you may not be familiar with.
ZZZ EXPLORE-DISCUSS 2
(A) Calculate the values of [1 (1/x)] x for x 0002 1, 2, 3, 4, and 5. Are the values increasing or decreasing as x gets larger? (B) Graph y 0002 [1 (1/x)] x and discuss the behavior of the graph as x increases without bound.
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Table 1 x
7:35 PM
2
10
2.593 74 …
100
2.704 81 …
1,000
2.716 92 …
10,000
2.718 14 …
100,000
2.718 27 …
1,000,000
2.718 28 …
By calculating the value of [1 (10004x)] x for larger and larger values of x (Table 1), it looks like [1 (10004x)] x approaches a number close to 2.7183. In a calculus course, we can show that as x increases without bound, the value of [1 (10004x)] x approaches an irrational number that we call e. Just as irrational numbers such as and 12 have unending, nonrepeating decimal representations, e also has an unending, nonrepeating decimal representation. To 12 decimal places, 兹2
e ⴝ 2.718 281 828 459
⫺2
⫺1
0
1
e 2
3
4
Don’t let the symbol “e” intimidate you! It’s just a number. Exactly who discovered e is still being debated. It is named after the great Swiss mathematician Leonhard Euler (1707–1783), who computed e to 23 decimal places using [1 (1 0004x)] x. The constant e turns out to be an ideal base for an exponential function because in calculus and higher mathematics many operations take on their simplest form using this base. This is why you will see e used extensively in expressions and formulas that model realworld phenomena.
Z DEFINITION 2 Exponential Function with Base e y
For x a real number, the equation
20
f (x) 0002 ex defines the exponential function with base e.
10
y ⫽ e ⫺x ⫺5
y ⫽ ex 5
x
Z Figure 5 Exponential functions.
EXAMPLE
3
The exponential function with base e is used so frequently that it is often referred to as the exponential function. The graphs of y 0002 e x and y 0002 e0003x are shown in Figure 5.
Analyzing a Graph Let g(x) 0002 4 0003 e x兾2. Use transformations to explain how the graph of g is related to the graph of f1(x) 0002 e x. Determine whether g is increasing or decreasing, find any asymptotes, and sketch the graph of g.
SOLUTION
The graph of g can be obtained from the graph of f1 by a sequence of three transformations: f1(x) 0002 e x
S Horizontal stretch
f2(x) 0002 e x兾2
S Reflection in x axis
f3(x) 0002 0003e x兾2
S
g(x) 0002 4 0003 e x兾2
Vertical translation
[See Fig. 6(a) for the graphs of f1, f2, and f3, and Fig. 6(b) for the graph of g.] The function g is decreasing for all x. Because e x兾2 S 0 as x S 00030006, it follows that g(x) 0002 4 0003 e x兾2 S 4 as x S 00030006. Therefore, the line y 0002 4 is a horizontal asymptote [indicated by the dashed line in Fig. 6(b)]; there are no vertical asymptotes. [To check that the graph of g (as obtained by graph transformations) is correct, plot a few points.]
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SECTION 5–1 y
f1
5
5
5
⫺5
x
y⫽4
⫺5
5
x
g(x) ⫽ 4 ⫺ e x/2
⫺5
f3 (a)
(b)
0002
Z Figure 6
MATCHED PROBLEM 3
333
y
f2
⫺5
Exponential Functions
Let g(x) 0002 2e x兾2 0003 5. Use transformations to explain how the graph of g is related to the graph of f1(x) 0002 e x. Describe the increasing/decreasing behavior, find any asymptotes, and sketch the graph of g. 0002
Z Compound Interest The fee paid to use someone else’s money is called interest. It is usually computed as a percentage, called the interest rate, of the original amount (or principal) over a given period of time. At the end of the payment period, the interest paid is usually added to the principal amount, so the interest in the next period is earned on both the original amount, as well as the interest previously earned. Interest paid on interest previously earned and reinvested in this manner is called compound interest. Suppose you deposit $1,000 in a bank that pays 8% interest compounded semiannually. How much will be in your account at the end of 2 years? “Compounded semiannually” means that the interest is paid to your account at the end of each 6-month period, and the interest will in turn earn more interest. To calculate the interest rate per period, we take the annual rate r, 8% (or 0.08), and divide by the number m of compounding periods per year, in this case 2. If A1 represents the amount of money in the account after one compounding period (6 months), then Principal ⴙ 4% of principal
A1 0002 $1,000 $1,000 a
0.08 b 2
Factor out $1,000.
0002 $1,000(1 0.04) We will next use A2, A3, and A4 to represent the amounts at the end of the second, third, and fourth periods. (Note that the amount we’re looking for is A4.) A2 is calculated by multiplying the amount at the beginning of the second compounding period (A1) by 1.04. A2 0002 A1(1 0.04) 0002 [$1,000(1 0.04)](1 0.04) 0002 $1,000(1 0.04)2 A3 0002 A2(1 0.04) 0002 [$1,000(1 0.04)2 ](1 0.04) 0002 $1,000(1 0.04)3 A4 0002 A3(1 0.04)
Substitute our expression for A1. Multiply. r 2 P a1 ⴙ b m Substitute our expression for A2. Multiply. r 3 P a1 ⴙ b m Substitute our expression for A3.
0002 [$1,000(1 0.04)3 ](1 0.04) 4
0002 $1,000(1 0.04) 0002 $1,169.86
Multiply. P a1 ⴙ
r 4 b m
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What do you think the savings and loan will owe you at the end of 6 years (12 compounding periods)? If you guessed A 0002 $1,000(1 0.04)12 you have observed a pattern that is generalized in the following compound interest formula:
Z COMPOUND INTEREST If a principal P is invested at an annual rate r compounded m times a year, then the amount A in the account at the end of n compounding periods is given by A 0002 Pa1
r n b m
Note that the annual rate r must be expressed in decimal form, and that n 0002 mt, where t is years.
EXAMPLE
4
Compound Interest If you deposit $5,000 in an account paying 9% compounded daily,* how much will you have in the account in 5 years? Compute the answer to the nearest cent.
SOLUTION
We will use the compound interest formula with P 0002 5,000, r 0002 0.09, (which is 9% written as a decimal), m 0002 365, and n 0002 5(365) 0002 1,825: A 0002 P a1
r n b m
0002 5,000 a1
0.09 1,825 b 365
Let P ⴝ 5,000, r ⴝ 0.09, m ⴝ 365, n ⴝ 5(365), or 1,825
Calculate to nearest cent.
0002 $7,841.13 MATCHED PROBLEM 4
EXAMPLE
5
0002
If $1,000 is invested in an account paying 10% compounded monthly, how much will be in the account at the end of 10 years? Compute the answer to the nearest cent. 0002
Comparing Investments If $1,000 is deposited into an account earning 10% compounded monthly and, at the same time, $2,000 is deposited into an account earning 4% compounded monthly, will the first account ever be worth more than the second? If so, when?
SOLUTION
Let y1 and y2 represent the amounts in the first and second accounts, respectively, then y1 0002 1,000(1 0.10000412)x y2 0002 2,000(1 0.04000412)x
P ⴝ 1,000, r ⴝ 0.10, m ⴝ 12 P ⴝ 2,000, r ⴝ 0.04, m ⴝ 12
where x is the number of compounding periods (months). Examining the graphs of y1 and y2 [Fig. 7(a)], we see that the graphs intersect at x ⬇ 139.438 months. Because compound *In all problems involving interest that is compounded daily, we assume a 365-day year.
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335
interest is paid at the end of each compounding period, we compare the amount in the accounts after 139 months and after 140 months [Fig. 7(b)]. The first account is worth more than the second for x 140 months, or after 11 years and 8 months. 5,000
0
240
0
(a)
(b)
0002
Z Figure 7
MATCHED PROBLEM 5
If $4,000 is deposited into an account earning 10% compounded quarterly and, at the same time, $5,000 is deposited into an account earning 6% compounded quarterly, when will the first account be worth more than the second? 0002
Z Interest Compounded Continuously If $1,000 is deposited in an account that earns compound interest at an annual rate of 8% for 2 years, how will the amount A change if the number of compounding periods is increased? If m is the number of compounding periods per year, then A 0002 1,000a1
0.08 2m b m
The amount A is computed for several values of m in Table 2. Notice that the largest gain appears in going from annually to semiannually. Then, the gains slow down as m increases. In fact, it appears that A might be approaching something close to $1,173.50 as m gets larger and larger.
Table 2 Effect of Compounding Frequency Compounding Frequency
A ⴝ 100a1 ⴙ
m
0.08 2m b m
Annually
1
$1,166.400
Semiannually
2
1,169.859
Quarterly
4
1,171.659
52
1,173.367
365
1,173.490
8,760
1,173.501
Weekly Daily Hourly
We now return to the general problem to see if we can determine what happens to A 0002 P[1 (r/m)] mt as m increases without bound. A little algebraic manipulation of the
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compound interest formula will lead to an answer and a significant result in the mathematics of finance: r mt b m 1 (m/r)rt 0002 Pa1 b m/r
A 0002 Pa1
Replace
m r 1 with , and mt with ⴢ rt. m m/r r
Replace
m with variable x. r
1 x rt 0002 P c a1 b d x Does the expression within the square brackets look familiar? Recall from the first part of this section that 1 x a1 b S e x
as
xS0006
Because the interest rate r is fixed, x 0002 m/r S 0006 as m S 0006. So (1 1x )x S e, and Pa1
1 x rt r mt b 0002 P c a1 b d S Pert m x
as
mS0006
This is known as the continuous compound interest formula, a very important and widely used formula in business, banking, and economics.
Z CONTINUOUS COMPOUND INTEREST FORMULA If a principal P is invested at an annual rate r compounded continuously, then the amount A in the account at the end of t years is given by A 0002 Pert The annual rate r must be expressed as a decimal.
EXAMPLE
6
Continuous Compound Interest If $1,000 is invested at an annual rate of 8% compounded continuously, what amount, to the nearest cent, will be in the account after 2 years?
SOLUTION
Use the continuous compound interest formula to find A when P 0002 $1,000, r 0002 0.08, and t 0002 2: A 0002 Pert 0002 $1,000e(0.08)(2) 0002 $1,173.51
8% is equivalent to r ⴝ 0.08. Calculate to nearest cent.
Notice that the values calculated in Table 2 get closer to this answer as m gets larger. MATCHED PROBLEM 6
0002
What amount will an account have after 5 years if $1,000 is invested at an annual rate of 12% compounded annually? Quarterly? Continuously? Compute answers to the nearest cent. 0002
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ANSWERS TO MATCHED PROBLEMS 1. The graph of g is the same as the graph of f reflected in the y axis and vertically shrunk by a factor of 21. x intercepts: none y intercept: 12 horizontal asymptote: y ⫽ 0 (x axis) vertical asymptotes: none y 40
30
20 1 10 1
⫺5
2
3
x
5
2. x ⫽ ⫺13 3. The graph of g is the same as the graph of f1 stretched horizontally by a factor of 2, stretched vertically by a factor of 2, and shifted 5 units down; g is increasing. horizontal asymptote: y ⫽ ⫺5 vertical asymptote: none y
g
10
⫺5
x
5
y ⫽ ⫺5 ⫺10
4. $2,707.04 5. After 23 quarters 6. Annually: $1,762.34; quarterly: $1,806.11; continuously: $1,822.12
5-1
Exercises
1. What is an exponential function? 2. What is the significance of the symbol e in the study of exponential functions? 3. For a function f (x) ⫽ bx, explain how you can tell if the graph increases or decreases without looking at the graph.
4. Explain why f (x) ⫽ (1/4)x and g(x) ⫽ 4⫺x are really the same function. Can you use this fact to add to your answer for Problem 3? 5. How do we know that the equation e x ⫽ 0 has no solution? 6. Define the following terms related to compound interest: principal, interest rate, compounding period.
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7. Match each equation with the graph of f, g, m, or n in the figure. (A) y 0002 (0.2)x (B) y 0002 2x (C) y 0002 (13)x (D) y 0002 4x f
g
m
6
n 00032
2
0
8. Match each equation with the graph of f, g, m, or n in the figure. (A) y 0002 e00031.2x (B) y 0002 e0.7x 00030.4x (C) y 0002 e (D) y 0002 e1.3x g
m n
6
f 00034
4
37. (45)6x00041 0002 54
38. (73)20003x 0002 37
39. (1 0003 x)5 0002 (2x 0003 1)5
40. 53 0002 (x 0004 2)3
41. 2xe0003x 0002 0
42. (x 0003 3)e x 0002 0
43. x2e x 0003 5xe x 0002 0
44. 3xe0003x 0004 x2e0003x 0002 0
2
2
45. 9x 0002 33x00031
46. 4x 0002 2 x00043
47. 25x00043 0002 125x
48. 45x00041 0002 162x00031
49. 42x00047 0002 8x00042
50. 1002x00043 0002 1,000x00045
51. Find all real numbers a such that a2 0002 a00032. Explain why this does not violate the second exponential function property in the box on page 330. 52. Find real numbers a and b such that a 0006 b but a4 0002 b4. Explain why this does not violate the third exponential function property in the box on page 330. 53. Evaluate y 0002 1x for x 0002 00033, 00032, 00031, 0, 1, 2, and 3. Why is b 0002 1 excluded when defining the exponential function y 0002 bx? 54. Evaluate y 0002 0x for x 0002 00033, 00032, 00031, 0, 1, 2, and 3. Why is b 0002 0 excluded when defining the exponential function y 0002 bx?
0
In Problems 9–16, use a calculator to compute answers to four significant digits. 9. 513
10. 3000312
In Problems 55–64, use transformations to explain how the graph of g is related to the graph of the given exponential function f. Determine whether g is increasing or decreasing, find any asymptotes, and sketch the graph of g.
11. e2 0004 e00032
12. e 0003 e00031
55. g(x) 0002 0003(12)x; f (x) 0002 (12)x
13. 1e
14. e12
56. g(x) 0002 0003(13)0003x; f (x) 0002 (13)x
0005
15.
00030005
0005
2 00042 2
16.
00030005
3 00033 2
57. g(x) 0002 (14)x2 0004 3; f (x) 0002 (14)x 58. g(x) 0002 5 0003 (23)3x; f (x) 0002 (23)x
In Problems 17–24, simplify. 17. 10
3x00031
40003x
10
18. (4 )
x
19.
60. g(x) 0002 1,000(1.03)x; f (x) 0002 1.03x
x00033
3 310003x
20.
4x 3z 21. a y b 5 23.
59. g(x) 0002 500(1.04)x; f (x) 0002 1.04x
3x 2y
5 5x00034
61. g(x) 0002 1 0004 2ex00033; f (x) 0002 e x 62. g(x) 0002 4ex00041 0003 7; f(x) 0002 e x
22. (2x3y)z
e5x
24.
e2x00041
63. g(x) 0002 3 0003 4e20003x; f (x) 0002 e x 64. g(x) 0002 00032 0003 5e40003x; f (x) 0002 e x
e400033x e200035x
In Problems 25–32, use transformations to explain how the graph of g is related to the graph of f(x) 0002 e x. Determine whether g is increasing or decreasing, find the asymptotes, and sketch the graph of g. 25. g(x) 0002 3e
0003x
x
26. g(x) 0002 2e
1 0003x 3e
27. g(x) 0002
29. g(x) 0002 2 0004 e 31. g(x) 0002 e
28. g(x) 0002 x
x00042
1 x 5e
30. g(x) 0002 00034 0004 e x 32. g(x) 0002 e
x00031
In Problems 33–50, solve for x. 33. 53x 0002 54x00032 x2
2x00043
35. 7 0002 7
34. 10200033x 0002 105x00036 36. 4
5x0003x2
00036
00024
In Problems 65–68, simplify. 65.
00032x3e00032x 0003 3x2e00032x x6
66.
5x4e5x 0003 4x3e5x x8
67. (e x 0004 e0003x )2 0004 (e x 0003 e0003x )2 68. e x(e0003x 0004 1) 0003 e0003x(e x 0004 1) In Problems 69–76, use a graphing calculator to find local extrema, y intercepts, and x intercepts. Investigate the behavior as x S 0007 and as x 00030007 and identify any horizontal asymptotes. Round any approximate values to two decimal places. 69. f(x) 0002 2 0004 e x00032 2
71. s(x) 0002 e0003x
70. g(x) 0002 00033 0004 e10004x 2
72. r(x) 0002 e x
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73. F(x) 0002
200 1 3e0003x
74. G(x) 0002
100 1 e0003x
75. f (x) 0002
2x 20003x 2
76. g(x) 0002
3x 30003x 2
77. Use a graphing calculator to investigate the behavior of f (x) 0002 (1 x)1兾x as x approaches 0. 78. Use a graphing calculator to investigate the behavior of f(x) 0002 (1 x)1兾x as x approaches 0006. 79. The irrational number 12 is approximated by 1.414214 to six decimal places. Each of x 0002 1.4, 1.41, 1.414, 1.4142, 1.41421, and 1.414214 is a rational number, so we know how to define 2x for each. Compute the value of 2x for each of these x values, and use your results to estimate the value of 212. Then compute 212 using your calculator to check your estimate. 80. The irrational number 13 is approximated by 1.732051 to six decimal places. Each of x 0002 1.7, 1.73, 1.732, 1.7321, 1.73205, and 1.732051 is a rational number, so we know how to define 3x for each. Compute the value of 3x for each of these x values, and use your results to estimate the value of 313. Then compute 313 using your calculator to check your estimate. It is common practice in many applications of mathematics to approximate nonpolynomial functions with appropriately selected polynomials. For example, the polynomials in Problems 81–84, called Taylor polynomials, can be used to approximate the exponential function f(x) 0002 e x. To illustrate this approximation graphically, in each problem graph f(x) 0002 e x and the indicated polynomial in the same viewing window, 00034 x 4 and 00035 y 50. 81. P1(x) 0002 1 x 12x2 82. P2(x) 0002 1 x 12x2 16x3 83. P3(x) 0002 1 x 12x2 16x3 241 x4 1 5 x 84. P4(x) 0002 1 x 12x2 16x3 241 x4 120
85. Investigate the behavior of the functions f1(x) 0002 x兾e x, f2(x) 0002 x2兾e x, and f3(x) 0002 x3兾e x as x S 0006 and as x S 00030006, and find any horizontal asymptotes. Generalize to functions of the form fn(x) 0002 x n兾e x, where n is any positive integer. 86. Investigate the behavior of the functions g1(x) 0002 xe x, g2(x) 0002 x2e x, and g3(x) 0002 x3e x as x S 0006 and as x S 00030006, and find any horizontal asymptotes. Generalize to functions of the form gn(x) 0002 x ne x, where n is any positive integer.
APPLICATIONS* 87. FINANCE A couple just had a new child. How much should they invest now at 6.25% compounded daily to have $100,000 for the child’s education 17 years from now? Compute the answer to the nearest dollar. 88. FINANCE A person wants to have $25,000 cash for a new car 5 years from now. How much should be placed in an account now if the account pays 4.75% compounded weekly? Compute the answer to the nearest dollar. *Round monetary amounts to the nearest cent unless specified otherwise. In all problems involving interest that is compounded daily, assume a 365-day year.
Exponential Functions
339
89. MONEY GROWTH If you invest $5,250 in an account paying 6.38% compounded continuously, how much money will be in the account at the end of (A) 6.25 years? (B) 17 years? 90. MONEY GROWTH If you invest $7,500 in an account paying 5.35% compounded continuously, how much money will be in the account at the end of (A) 5.5 years? (B) 12 years? 91. FINANCE If $3,000 is deposited into an account earning 8% compounded daily and, at the same time, $5,000 is deposited into an account earning 5% compounded daily, will the first account ever be worth more than the second? If so, when? 92. FINANCE If $4,000 is deposited into an account earning 9% compounded weekly and, at the same time, $6,000 is deposited into an account earning 7% compounded weekly, will the first account ever be worth more than the second? If so, when? 93. FINANCE Will an investment of $10,000 at 4.9% compounded daily ever be worth more at the end of any quarter than an investment of $10,000 at 5% compounded quarterly? Explain. 94. FINANCE A sum of $5,000 is invested at 7% compounded semiannually. Suppose that a second investment of $5,000 is made at interest rate r compounded daily. Both investments are held for 1 year. For which values of r, to the nearest tenth of a percent, is the second investment better than the first? Discuss. 95. PRESENT VALUE A promissory note will pay $30,000 at maturity 10 years from now. How much should you pay for the note now if the note gains value at a rate of 6% compounded continuously? 96. PRESENT VALUE A promissory note will pay $50,000 at maturity 512 years from now. How much should you pay for the note now if the note gains value at a rate of 5% compounded continuously? 97. MONEY GROWTH The website Bankrate.com publishes a weekly list of the top savings deposit yields. In the category of 3-year certificates of deposit, the following were listed: Flagstar Bank, FSB UmbrellaBank.com Allied First Bank
3.12% (CQ) 3.00% (CD) 2.96% (CM)
where CQ represents compounded quarterly, CD compounded daily, and CM compounded monthly. Find the value of $5,000 invested in each account at the end of 3 years. 98. Refer to Problem 97. In the 1-year certificate of deposit category, the following accounts were listed: GMAC Bank UFBDirect.com
2.91% (CD) 2.86% (CM)
Find the value of $10,000 invested in each account at the end of 1 year. 99. FINANCE Suppose $4,000 is invested at 6% compounded weekly. How much money will be in the account in (A) 12 year? (B) 10 years? 100. FINANCE Suppose $2,500 is invested at 4% compounded quarterly. How much money will be in the account in (A) 34 year? (B) 15 years?
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5-2
Exponential Models Z Mathematical Modeling Z Data Analysis and Regression Z A Comparison of Exponential Growth Phenomena
One of the best reasons for studying exponential functions is the fact that many things that occur naturally in our world can be modeled accurately by these functions. In this section, we will study a wide variety of applications, including growth of populations of people, animals, and bacteria; radioactive decay; spread of epidemics; propagation of rumors; light intensity; atmospheric pressure; and electric circuits. The regression techniques we used in Chapter 1 to construct linear and quadratic models will be extended to construct exponential models.
Z Mathematical Modeling Populations tend to grow exponentially and at different rates. A convenient and easily understood measure of growth rate is the doubling time—that is, the time it takes for a population to double. Over short periods the doubling time growth model is often used to model population growth: A ⴝ A02t0004d where
A 0002 Population at time t A0 0002 Population at time t 0002 0 d 0002 Doubling time
Note that when t 0002 d, A 0002 A02d兾d 0002 A02 and the population is double the original, as it should be. We will use this model to solve a population growth problem in Example 1.
EXAMPLE
1
Population Growth According to a 2008 estimate, the population of Nicaragua was about 5.7 million, and that population is growing due to a high birth rate and relatively low mortality rate. If the population continues to grow at the current rate, it will double in 37 years. If the growth remains steady, what will the population be in (A) 15 years?
(B) 40 years?
Calculate answers to three significant digits. SOLUTIONS
We can use the doubling time growth model, A 0002 A0(2)t兾d with A0 0002 5.7 and d 0002 37: A 0002 5.7(2)t兾37
See Figure 1.
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A (millions) 20 16 12 8 4 10
20
30
40
50
t
t兾37 Z Figure 1 A 0002 5.7(2)
(A) Find A when t 0002 15 years: A 0002 5.7(2)15000437 0002 7.55 million
To 3 significant digits
(B) Find A when t 0002 40 years: A 0002 5.7(2)40000437 0002 12.1 million MATCHED PROBLEM 1
To 3 significant digits
0002
Before the great housing bust, Palm Coast, Florida, was the fastest-growing city in America. Its population was about 34,000 in 2000, and it doubled in 6.6 years. If the population had continued growing at that rate, what would it be in (A) 2010?
(B) 2020?
Calculate answers to three significant digits. 0002
ZZZ EXPLORE-DISCUSS 1
The doubling time growth model would not be expected to give accurate results over long periods. According to the doubling time growth model of Example 1, what was the population of Nicaragua 500 years ago when it was settled as a Spanish colony? What will the population of Nicaragua be 200 years from now? Explain why these results are unrealistic. Discuss factors that affect human populations that are not taken into account by the doubling time growth model.
The doubling time model is not the only one used to model populations. An alternative model based on the continuous compound interest formula will be used in Example 2. In this case, the formula is written as A 0002 A0ekt where
A 0002 Population at time t A0 0002 Population at time t 0002 0 k 0002 Relative growth rate
The relative growth rate is written as a percentage in decimal form. For example, if a population is growing so that at any time the population is increasing at 3% of the current population per year, the relative growth rate k would be 0.03.
EXAMPLE
2
Medicine—Bacteria Growth Cholera, an intestinal disease, is caused by a cholera bacterium that multiplies exponentially by cell division as modeled by A 0002 A0e1.386t
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where A is the number of bacteria present after t hours and A0 is the number of bacteria present at t 0002 0. If we start with 1 bacterium, how many bacteria will be present in (A) 5 hours?
(B) 12 hours?
Calculate the answers to three significant digits. SOLUTIONS
(A) Use A0 0002 1 and t 0002 5: A 0002 A0e1.386t 0002 e1.386(5) ⬇ 1,020
Let A0 ⴝ 1 and t ⴝ 5. Calculate to three significant digits.
(B) Use A0 0002 1 and t 0002 12: A 0002 A0e1.386t 0002 e1.386(12) 0002 16,700,000 MATCHED PROBLEM 2
Let A0 ⴝ 1 and t ⴝ 12. Calculate to three significant digits.
0002
Repeat Example 2 if A 0002 A0e0.783t and all other information remains the same.
0002
Exponential functions can also be used to model radioactive decay, which is sometimes referred to as negative growth. Radioactive materials are used extensively in medical diagnosis and therapy, as power sources in satellites, and as power sources in many countries. If we start with an amount A0 of a particular radioactive substance, the amount declines exponentially over time. The rate of decay varies depending on the particular radioactive substance. A convenient and easily understood measure of the rate of decay is the half-life of the material—that is, the time it takes for half of a particular material to decay. We can use the following half-life decay model: A ⴝ A0(12)t0004h ⴝ A020003t0004h where
A 0002 Amount at time t A0 0002 Amount at time t 0002 0 h 0002 Half-life
Note that when the amount of time passed is equal to the half-life (t 0002 h), A 0002 A020003h0004h 0002 A0200031 0002 A0 ⴢ 12 and the amount of radioactive material is half the original amount, as it should be.
EXAMPLE
3
Radioactive Decay The radioactive isotope gallium 67 (67Ga), used in the diagnosis of malignant tumors, has a biological half-life of 46.5 hours. If we start with 100 milligrams of the isotope, how many milligrams will be left after (A) 24 hours?
(B) 1 week?
Calculate answers to three significant digits.
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SOLUTIONS
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343
We can use the half-life decay model: A 0002 A0(12)t0004h 0002 A020003t0004h Using A0 0002 100 and h 0002 46.5, we obtain
A (milligrams)
A 0002 100(20003t兾46.5)
100
See Figure 2.
(A) Find A when t 0002 24 hours: A 0002 100(2000324/46.5) 0002 69.9 milligrams
50
100
200
t
Hours
(B) Find A when t 0002 168 hours (1 week 0002 168 hours): A 0002 100(20003168/46.5) 0002 8.17 milligrams
0003t兾46.5 ). Z Figure 2 A 0002 100(2
MATCHED PROBLEM 3
Calculate to three significant digits.
Be careful about units! Half-life was given in hours.
Calculate to three significant digits.
0002
Radioactive gold 198 (198Au), used in imaging the structure of the liver, has a half-life of 2.67 days. If we start with 50 milligrams of the isotope, how many milligrams will be left after: (A)
1 2
day?
(B) 1 week?
Calculate answers to three significant digits. 0002 In Example 2, we saw that a base e exponential function can be used as an alternative to the doubling time model. Not surprisingly, the same can be said for the half-life model. In this case, the formula will be A 0002 A0e0003kt where
A 0002 the amount of radioactive material at time t A0 0002 the amount at time t 0002 0 k 0002 a positive constant specific to the type of material
Our atmosphere is constantly being bombarded with cosmic rays. These rays produce neutrons, which in turn react with nitrogen to produce radioactive carbon-14. Radioactive carbon-14 enters all living tissues through carbon dioxide, which is first absorbed by plants. As long as a plant or animal is alive, carbon-14 is maintained in the living organism at a constant level. Once the organism dies, however, carbon-14 decays according to the equation A 0002 A0e00030.000124t
Carbon-14 decay equation
where A is the amount of carbon-14 present after t years and A0 is the amount present at time t 0002 0. This can be used to calculate the approximate age of fossils.
EXAMPLE
4
Carbon-14 Dating If 1,000 milligrams of carbon-14 are present in the tissue of a recently deceased animal, how many milligrams will be present in (A) 10,000 years?
(B) 50,000 years?
Calculate answers to three significant digits.
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SOLUTIONS
Substituting A0 0002 1,000 in the decay equation, we have A 0002 1,000e00030.000124t
A
See Figure 3.
(A) Solve for A when t 0002 10,000:
1,000
A 0002 1,000e00030.000124(10,000) 0002 289 milligrams 500
Calculate to three significant digits.
(B) Solve for A when t 0002 50,000: A 0002 1,000e00030.000124(50,000) 0002 2.03 milligrams
t
50,000
Z Figure 3
Calculate to three significant digits.
More will be said about carbon-14 dating in Exercises 5-5, where we will be interested in solving for t after being given information about A and A0. 0002
MATCHED PROBLEM 4
Referring to Example 4, how many milligrams of carbon-14 would have to be present at the beginning to have 10 milligrams present after 20,000 years? Compute the answer to four significant digits. 0002 One of the problems with using exponential functions to model things like population is that the growth is completely unlimited in the long term. But in real life, there is often some reasonable maximum value, like the largest population that space and resources allow. We can use modified versions of exponential functions to model such phenomena more realistically. One such type of function is called a learning curve since it can be used to model the performance improvement of a person learning a new task. Learning curves are functions of the form A 0002 c(1 0003 e0003kt ), where c and k are positive constants.
EXAMPLE
5
A
Learning Curve People assigned to assemble circuit boards for a computer manufacturing company undergo on-the-job training. From past experience, it was found that the learning curve for the average employee is given by
50 40
A 0002 40(1 0003 e00030.12t )
30 20
where A is the number of boards assembled per day after t days of training (Fig. 4).
10 10
20
30
40
50
t
Days 00030.12t ). Z Figure 4 A 0002 40(1 0003 e
SOLUTION
(A) How many boards can an average employee produce after 3 days of training? After 5 days of training? Round answers to the nearest integer. (B) Does A approach a limiting value as t increases without bound? Explain. (A) When t 0002 3, A 0002 40(1 0003 e00030.12(3) ) 0002 12
Rounded to nearest integer
so the average employee can produce 12 boards after 3 days of training. Similarly, when t 0002 5, A 0002 40(1 0003 e00030.12(5) ) 0002 18
Rounded to nearest integer
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(B) Because e00030.12t 0002
1 0.12t
e
Exponential Models
345
approaches 0 as t increases without bound,
A 0002 40(1 0003 e00030.12t ) S 40(1 0003 0) 0002 40 So the limiting value of A is 40 boards per day. (Note the horizontal asymptote with equation A 0002 40 that is indicated by the dashed line in Fig. 4.) 0002 MATCHED PROBLEM 5
A company is trying to expose as many people as possible to a new product through television advertising in a large metropolitan area with 2 million potential viewers. A model for the number of people A, in millions, who are aware of the product after t days of advertising was found to be A 0002 2(1 0003 e00030.037t ) (A) How many viewers are aware of the product after 2 days? After 10 days? Express answers as integers, rounded to three significant digits. (B) Does A approach a limiting value as t increases without bound? Explain. 0002 Another limited-growth model is useful for phenomena such as the spread of an epidemic or the propagation of a rumor. It is called the logistic equation, and is given by A0002
M 1 ce0003kt
where M, c, and k are positive constants. Logistic growth, illustrated in Example 6, also approaches a limiting value as t increases without bound.
EXAMPLE
6
Logistic Growth in an Epidemic A certain community consists of 1,000 people. One individual who has just returned from another community has a particularly contagious strain of influenza. Assume the community has not had influenza shots and all are susceptible. The spread of the disease in the community is predicted to be given by the logistic curve A(t) 0002
1,000 1 999e00030.3t
where A is the number of people who have contracted the flu after t days. (A) How many people have contracted the flu after 10 days? After 20 days? (B) Does A approach a limiting value as t increases without bound? Explain.
SOLUTIONS
(A) When t 0002 10, A0002
1,000 1 999e00030.3(10)
0002 20
Rounded to nearest integer
so 20 people have contracted the flu after 10 days. Similarly, when t 0002 20, A0002
1,000 1 999e00030.3(20)
0002 288
Rounded to nearest integer
so 288 people have contracted the flu after 20 days.
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(B) Because e00030.3t approaches 0 as t increases without bound, A0002
1,000 1,000 S 0002 1,000 00030.3t 1 999(0) 1 999e
So the limiting value is 1,000 individuals (everyone in the community will eventually get the flu). (Note the horizontal asymptote with equation A 0002 1,000 that is indicated by the dashed line in Fig. 5.) A 1,500 1,200 900 600 300 10
20
30
40
t
50
Days
Z Figure 5 A 0002
MATCHED PROBLEM 6
1,000 1 999e00030.3t
.
0002
A group of 400 parents, relatives, and friends are waiting anxiously at Kennedy Airport for a charter flight returning students after a year in Europe. It is stormy and the plane is late. A particular parent thought he heard that the plane’s radio had gone out and related this news to some friends, who in turn passed it on to others. The propagation of this rumor is predicted to be given by A(t) 0002
400 1 399e00030.4t
where A is the number of people who have heard the rumor after t minutes. (A) How many people have heard the rumor after 10 minutes? After 20 minutes? Round answers to the nearest integer. (B) Does A approach a limiting value as t increases without bound? Explain. 0002
Z Data Analysis and Regression Many graphing calculators have options for exponential and logistic regression. We can use exponential regression to fit a function of the form y 0002 abx to a set of data points, and logistic regression to fit a function of the form y0002
c 1 ae0003bx
to a set of data points. The techniques are similar to those introduced in Chapters 2 and 3 for linear and quadratic functions.
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EXAMPLE
7
Exponential Models
347
Infectious Diseases The U.S. Department of Health and Human Services published the data in Table 1. Table 1 Reported Cases of Infectious Diseases Year
Mumps
Rubella
1970
104,953
56,552
1980
8,576
3,904
1990
5,292
1,125
1995
906
128
2000
323
152
2005
314
11
An exponential model for the data on mumps is given by A 0002 81,082(0.844)t where A is the number of reported cases of mumps and t is time in years with t 0002 0 representing 1970. (A) Use the model to predict the number of reported cases of mumps in 2010. (B) Compare the actual number of cases of mumps reported in 1980 to the number given by the model. SOLUTIONS
(A) The year 2010 is represented by t 0002 40. Evaluating A 0002 81,082(0.844)t at t 0002 40 gives a prediction of 92 cases of mumps in 2010. (B) The year 1980 is represented by t 0002 10. Evaluating A 0002 81,082(0.844)t at t 0002 10 gives 14,871 cases in 1980. The actual number of cases reported in 1980 was 8,576, nearly 6,300 less than the number given by the model.
Technology Connections Figure 6 shows the details of constructing the exponential model of Example 7 on a graphing calculator.
110,000
00035
45
000310,000
(a) Entering the data
(b) Finding the model
(c) Graphing the data and the model
Z Figure 6
0002
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MATCHED PROBLEM 7
An exponential model for the data on rubella in Table 1 is given by A 0002 54,988(0.799)t where A is the number of reported cases of rubella and t is time in years with t 0002 0 representing 1970. (A) Use the model to predict the number of reported cases of rubella in 2010. (B) Compare the actual number of cases of rubella reported in 1980 to the number given by the model. 0002
EXAMPLE
8
AIDS Cases and Deaths The U.S. Department of Health and Human Services published the data in Table 2. Table 2 Acquired Immunodeficiency Syndrome (AIDS) Cases and Deaths in the United States Year
Cases Diagnosed to Date
Known Deaths to Date
1985
23,185
12,648
1988
107,755
62,468
1991
261,259
159,294
1994
493,713
296,507
1997
672,970
406,179
2000
774,467
447,648
2005
944,306
529,113
A logistic model for the data on AIDS cases is given by A0002
947,000 1 17.3e00030.313t
where A is the number of AIDS cases diagnosed by year t with t 0002 0 representing 1985. (A) Use the model to predict the number of AIDS cases diagnosed by 2010. (B) Compare the actual number of AIDS cases diagnosed by 2005 to the number given by the model. SOLUTIONS
(A) The year 2010 is represented by t 0002 25. Evaluating A0002
947,000 1 17.3e00030.313t
at t 0002 25 gives a prediction of approximately 940,000 cases of AIDS diagnosed by 2010.
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(B) The year 2005 is represented by t 0002 20. Evaluating A0002
947,000 1 0004 17.3e00030.313t
at t 0002 20 gives 916,690 cases in 2005. The actual number of cases diagnosed by 2005 was 944,306, nearly 28,000 greater than the number given by the model.
Technology Connections Figure 7 shows the details of constructing the logistic model of Example 8 on a graphing calculator. 1,000,000
00035
20
0
(a) Entering the data
(b) Finding the model
(c) Graphing the data and the model
Z Figure 7
0002
MATCHED PROBLEM 8
A logistic model for the data on deaths from AIDS in Table 2 is given by A0002
521,000 1 0004 18.8e00030.349t
where A is the number of known deaths from AIDS by year t with t 0002 0 representing 1985. (A) Use the model to predict the number of known deaths from AIDS by 2010. (B) Compare the actual number of known deaths from AIDS by 2005 to the number given by the model. 0002
Z A Comparison of Exponential Growth Phenomena The equations and graphs given in Table 3 compare several widely used growth models. These are divided basically into two groups: unlimited growth and limited growth. Following each equation and graph is a short, incomplete list of areas in which the models are used. We have only touched on a subject that has been extensively developed and that you are likely to study in greater depth in the future.
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Table 3 Exponential Growth and Decay Description
Equation
Unlimited growth
A 0002 A0ekt k00070
Graph
Short List of Uses
A
Short-term population growth (people, bacteria, etc.); growth of money at continuous compound interest
c 0
Exponential decay
A 0002 A0e0003kt k00070
A
A 0002 c(1 0003 e0003kt ) c, k 0007 0
M 1 ce0003kt c, k, M 7 0
t
A
Learning skills; sales fads; company growth; electric circuits
c
0
Logistic growth
Radioactive decay; light absorption in water, glass, and the like; atmospheric pressure; electric circuits
c
0
Limited growth
t
t
A
A0002
Long-term population growth; epidemics; sales of new products; spread of rumors; company growth
M
0
t
ANSWERS TO MATCHED PROBLEMS 1. (A) 97,200 (B) 278,000 2. (A) 50 bacteria (B) 12,000 bacteria 3. (A) 43.9 milligrams (B) 8.12 milligrams 4. 119.4 milligrams 5. (A) 143,000 viewers; 619,000 viewers (B) A approaches an upper limit of 2 million, the number of potential viewers 6. (A) 48 individuals; 353 individuals (B) A approaches an upper limit of 400, the number of people in the entire group. 7. (A) 7 cases (B) The actual number of cases was 1,927 less than the number given by the model. 8. (A) 519,000 deaths (B) The actual number of known deaths was approximately 17,000 greater than the number given by the model.
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5-2
Exponential Models
351
Exercises
1. Define the terms “doubling time” and “half-life” in your own words. 2. One of the models below represents positive growth, and the other represents negative growth. Classify each, and explain how you decided on your answer. (Assume that k 7 0.) A 0002 A0e0003kt
A 0002 A0ekt
3. Explain the difference between exponential growth and limited growth. 4. Explain why a limited growth model would be more accurate than regular exponential growth in modeling the long-term population of birds on an island in Lake Erie. In Problems 5–8, write an exponential equation describing the given population at any time t. 5. Initial population 200; doubling time 5 months 6. Initial population 5,000; doubling time 3 years 7. Initial population 2,000; continuous growth at 2% per year 8. Initial population 500; continuous growth at 3% per week In Problems 9–12, write an exponential equation describing the amount of radioactive material present at any time t. 9. Initial amount 100 grams; half-life 6 hours 10. Initial amount 5 pounds; half-life 1,300 years 11. Initial amount 4 kilograms; continuous decay at 12.4% per year 12. Initial amount 50 milligrams; continuous decay at 0.03% per year
APPLICATIONS 13. GAMING A person bets on red and black on a roulette wheel using a Martingale strategy. That is, a $2 bet is placed on red, and the bet is doubled each time until a win occurs. The process is then repeated. If black occurs n times in a row, then L 0002 2n dollars is lost on the nth bet. Graph this function for 1 n 10. Although the function is defined only for positive integers, points on this type of graph are usually joined with a smooth curve as a visual aid. 14. BACTERIAL GROWTH If bacteria in a certain culture double every 12 hour, write an equation that gives the number of bacteria A in the culture after t hours, assuming the culture has 100 bacteria at the start. Graph the equation for 0 t 5. 15. POPULATION GROWTH Because of its short life span and frequent breeding, the fruit fly Drosophila is used in some genetic studies. Raymond Pearl of Johns Hopkins University, for example, studied 300 successive generations of descendants of a single pair
of Drosophila flies. In a laboratory situation with ample food supply and space, the doubling time for a particular population is 2.4 days. If we start with 5 male and 5 female flies, how many flies should we expect to have in (A) 1 week? (B) 2 weeks? 16. POPULATION GROWTH It was estimated in 2008 that Kenya had a population of about 38,000,000 people, and a doubling time of 25 years. If growth continues at the same rate, find the population in (A) 2012 (B) 2040 Calculate answers to two significant digits. 17. COMPUTER DESIGN In 1965, Gordon Moore, founder of Intel, predicted that the number of transistors that could be placed on a computer chip would double every 2 years. This has come to be known as Moore’s law. In 1970, 2,200 transistors could be placed on a chip. Use Moore’s law to predict the number of transistors in (A) 1990 (B) 2005 18. HISTORY OF TECHNOLOGY The earliest mechanical clocks appeared around 1350 in Europe, and would gain or lose an average of 30 minutes per day. After that, accuracy roughly doubled every 30 years. Find the predicted accuracy of clocks in (A) 1700 (B) 2000 19. INSECTICIDES The use of the insecticide DDT is no longer allowed in many countries because of its long-term adverse effects. If a farmer uses 25 pounds of active DDT, assuming its half-life is 12 years, how much will still be active after (A) 5 years? (B) 20 years? Compute answers to two significant digits. 20. RADIOACTIVE TRACERS The radioactive isotope technetium99m (99mTc) is used in imaging the brain. The isotope has a halflife of 6 hours. If 12 milligrams are used, how much will be present after (A) 3 hours? (B) 24 hours? Compute answers to three significant digits. 21. POPULATION GROWTH According to the CIA World Factbook, the population of the world was estimated to be about 6.8 billion people in 2008, and the population was growing continuously at a relative growth rate of 1.188%. If this growth rate continues, what would the population be in 2020 to two significant digits? 22. POPULATION GROWTH According to the CIA World Factbook, the population of Mexico was about 100 million in 2008, and was growing continuously at a relative growth rate of 1.142%. If that growth continues, what will the population be in 2015 to three significant digits?
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23. POPULATION GROWTH In 2005 the population of Russia was 143 million and the population of Nigeria was 129 million. If the populations of Russia and Nigeria grow continuously at relative growth rates of 00030.37% and 2.56%, respectively, in what year did Nigeria have a greater population than Russia? Use the Internet to find if the prediction was accurate. 24. POPULATION GROWTH In 2005 the population of Germany was 82 million and the population of Egypt was 78 million. If the populations of Germany and Egypt grow continuously at relative growth rates of 0% and 1.78%, respectively, in what year did Egypt have a greater population than Germany? Use the Internet to find if the prediction was accurate. 25. SPACE SCIENCE Radioactive isotopes, as well as solar cells, are used to supply power to space vehicles. The isotopes gradually lose power because of radioactive decay. On a particular space vehicle the nuclear energy source has a power output of P watts after t days of use as given by P 0002 75e00030.0035t Graph this function for 0 t 100. 26. EARTH SCIENCE The atmospheric pressure P, in pounds per square inch, decreases exponentially with altitude h, in miles above sea level, as given by P 0002 14.7e00030.21h Graph this function for 0 h 10. 27. MARINE BIOLOGY Marine life is dependent upon the microscopic plant life that exists in the photic zone, a zone that goes to a depth where about 1% of the surface light still remains. Light intensity I relative to depth d, in feet, for one of the clearest bodies of water in the world, the Sargasso Sea in the West Indies, can be approximated by
continues at the rate, find the number of people that will be living with HIV in (A) 2014 (B) 2020 30. AIDS EPIDEMIC The World Health Organization estimated that there were 3.25 million deaths from AIDS in 2007, and that the number had been growing continuously at a relative growth rate of 3.0%. If the growth continues at this rate, find the number of expected deaths from AIDS in (A) 2012 (B) 2030 31. NEWTON’S LAW OF COOLING This law states that the rate at which an object cools is proportional to the difference in temperature between the object and its surrounding medium. The temperature T of the object t hours later is given by T 0002 Tm 0004 (T0 0003 Tm)e0003kt where Tm is the temperature of the surrounding medium and T0 is the temperature of the object at t 0002 0. Suppose a bottle of wine at a room temperature of 72°F is placed in the refrigerator to cool before a dinner party. If the temperature in the refrigerator is kept at 40°F and k 0002 0.4, find the temperature of the wine, to the nearest degree, after 3 hours. (In Exercises 5-5 we will find out how to determine k.) 32. NEWTON’S LAW OF COOLING Refer to Problem 31. What is the temperature, to the nearest degree, of the wine after 5 hours in the refrigerator? 33. PHOTOGRAPHY An electronic flash unit for a camera is activated when a capacitor is discharged through a filament of wire. After the flash is triggered, and the capacitor is discharged, the circuit (see the figure) is connected and the battery pack generates a current to recharge the capacitor. The time it takes for the capacitor to recharge is called the recycle time. For a particular flash unit using a 12-volt battery pack, the charge q, in coulombs, on the capacitor t seconds after recharging has started is given by
I 0002 I0e00030.00942d where I0 is the intensity of light at the surface. To the nearest percent, what percentage of the surface light will reach a depth of (A) 50 feet? (B) 100 feet? 28. MARINE BIOLOGY Refer to Problem 27. In some waters with a great deal of sediment, the photic zone may go down only 15 to 20 feet. In some murky harbors, the intensity of light d feet below the surface is given approximately by I 0002 I0e00030.23d What percentage of the surface light will reach a depth of (A) 10 feet? (B) 20 feet? 29. AIDS EPIDEMIC The World Health Organization estimated that there were 33.2 million people worldwide living with the HIV infection in 2007, and that the number had been growing continuously at a relative growth rate of 2.37%. If the growth
q 0002 0.0009(1 0003 e00030.2t ) Find the value that q approaches as t increases without bound and interpret. R I
V
C S
34. MEDICINE An electronic heart pacemaker uses the same type of circuit as the flash unit in Problem 33, but it is designed so that the capacitor discharges 72 times a minute. For a particular pacemaker, the charge on the capacitor t seconds after it starts recharging is given by q 0002 0.000 008(1 0003 e00032t ) Find the value that q approaches as t increases without bound and interpret.
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35. WILDLIFE MANAGEMENT A herd of 20 white-tailed deer is introduced to a coastal island where there had been no deer before. Their population is predicted to increase according to the logistic curve 100 1 ⫹ 4e⫺0.14t
A⫽
Table 5
36. TRAINING A trainee is hired by a computer manufacturing company to learn to test a particular model of a personal computer after it comes off the assembly line. The learning curve for an average trainee is given by 200 4 ⫹ 21e⫺0.1t
where A is the number of computers an average trainee can test per day after t days of training. (A) How many computers can an average trainee be expected to test after 3 days of training? After 6 days? Round answers to the nearest integer. (B) How many days will it take until an average trainee can test 30 computers per day? Round answer to the nearest integer. (C) Does A approach a limiting value as t increases without bound? Explain. Problems 37–40 require a graphing calculator or a computer that can calculate exponential and logistic regression models for a given data set. 37. DEPRECIATION Table 4 gives the market value of a minivan (in dollars) x years after its purchase. Find an exponential regression model of the form y ⫽ abx for this data set. Round to four significant digits. Estimate the purchase price of the van. Estimate the value of the van 10 years after its purchase. Round answers to the nearest dollar.
Table 4 x
Value ($)
1
12,575
2
9,455
3
8,115
4
6,845
5
5,225
6
4,485
Source: Kelley Blue Book
353
38. DEPRECIATION Table 5 gives the market value of an SUV (in dollars) x years after its purchase. Find an exponential regression model of the form y ⫽ abx for this data set. Estimate the purchase price of the SUV. Estimate the value of the SUV 10 years after its purchase. Round answers to the nearest dollar.
where A is the number of deer expected in the herd after t years. (A) How many deer will be present after 2 years? After 6 years? Round answers to the nearest integer. (B) How many years will it take for the herd to grow to 50 deer? Round answer to the nearest integer. (C) Does A approach a limiting value as t increases without bound? Explain.
A⫽
Exponential Models
x
Value ($)
1
23,125
2
19,050
3
15,625
4
11,875
5
9,450
6
7,125
Source: Kelley Blue Book
39. NUCLEAR POWER Table 6 gives data on nuclear power generation by region for the years 1980–2005.
Table 6 Nuclear Power Generation (Billion Kilowatt-Hours) Year
North America
Central and South America
1980
287.0
2.2
1985
440.8
8.4
1990
649.0
9.0
1995
774.4
9.5
2000
830.9
10.9
2005
879.7
16.3
Source: U.S. Energy Information Administration
(A) Let x represent time in years with x ⫽ 0 representing 1980. Find a logistic regression model ( y ⫽ 1 ⫹ cae⫺bx) for the generation of nuclear power in North America. (Round the constants a, b, and c to three significant digits.) (B) Use the logistic regression model to predict the generation of nuclear power in North America in 2010 and 2020. 40. NUCLEAR POWER Refer to Table 6. (A) Let x represent time in years with x ⫽ 0 representing 1980. Find a logistic regression model ( y ⫽ 1 ⫹ cae⫺bx) for the generation of nuclear power in Central and South America. (Round the constants a, b, and c to three significant digits.) (B) Use the logistic regression model to predict the generation of nuclear power in Central and South America in 2010 and 2020.
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5-3
Logarithmic Functions Z Defining Logarithmic Functions Z Converting Between Logarithmic Form and Exponential Form Z Properties of Logarithmic Functions Z Common and Natural Logarithms Z The Change-of-Base Formula
Solving an equation like 3x 0003 9 is easy: We know that 32 0003 9, so x 0003 2 is the solution. But what about an equation like 3x 0003 20? There probably is an exponent x between 2 and 3 for which 3x is 20, but its exact value is not at all clear. Compare this situation to an equation like x2 0003 9. This is easy to solve because we know that 32 and (00023)2 are both 9. But what about x2 0003 20? To solve this equation, we needed to introduce a new function to be the opposite of the squaring function. This, of course, is the function f (x) 0003 1x. In this section, we will do something very similar with exponential functions. In the first section of this chapter, we learned that exponential functions are one-to-one, so we can define their inverses. These are known as the logarithmic functions.
Z Defining Logarithmic Functions The exponential function f (x) 0003 bx for b 7 0, b 0004 1, is a one-to-one function, and therefore has an inverse. Its inverse, denoted f 00021(x) 0003 logb x (read “log to the base b of x”) is called the logarithmic function with base b. Just like exponentials, there are different logarithmic functions for each positive base other than 1. A point (x, y) is on the graph of f 00021 0003 logb x if and only if the point (y, x) is on the graph of f 0003 bx. In other words, y 0003 logb x if and only if x 0003 b y In a specific example, y 0003 log2 x if and only if x 0003 2y, and log2 x is the power to which 2 must be raised to obtain x: 2log2 x 0003 2y 0003 x. We can use this fact to learn some things about the logarithmic functions from our knowledge of exponential functions. For example, the graph of f 00021 (x) 0003 logb x is the graph of f (x) 0003 bx reflected through the line y 0003 x. Also, the domain of f 00021 (x) 0003 logb x is the range of f (x) 0003 bx, and vice versa. In Example 1, we will use information about f (x) 0003 2x to graph its inverse, 00021 f (x) 0003 log2 x.
EXAMPLE
1
Graphing a Logarithmic Function Make a table of values for f (x) 0003 2x and reverse the ordered pairs to obtain a table of values for f 00021(x) 0003 log2 x. Then use both tables to graph f (x) and f 00021(x) on the same set of axes.
SOLUTION
We chose to evaluate f for integer values from 00023 to 3. The tables are shown here, along with the graph (Fig. 1). Note the important comments about domain and range below the graph.
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y
f y 0003 2x
5
f 00021
y 0003 log2 x 5
10
y 0002 2x
x
y 0002 log2 x
00023
1 8
1 8
00023
00022
1 4
1 4
00022
00021
1 2
1 2
00021
0
1
1
0
1
2
2
1
2
4
4
2
3
8
8
3
x
10
00025
f 00021
f y0003x
x
00025
355
Logarithmic Functions
Ordered pairs reversed
DOMAIN of f 0002 (00030004, 0004) 0002 RANGE of f 00031 RANGE of f 0002 (0, 0004) 0002 DOMAIN of f 00031
0002
Z Figure 1 Logarithmic function with base 2.
MATCHED PROBLEM 1
Repeat Example 1 for f (x) 0003 (12)x and f 00021(x) 0003 log100052 x.
0002
Z DEFINITION 1 Logarithmic Function For b 7 0, b 0004 1, the inverse of f (x) 0003 bx, denoted f 00021(x) 0003 logb x, is the logarithmic function with base b. Logarithmic form
y 0003 logb x
Exponential form
is equivalent to
x 0003 by
The log to the base b of x is the exponent to which b must be raised to obtain x. For example, y 0003 log10 x y 0003 loge x
y y 0003 logb x 00006b00061 0
x 0003 10 y x 0003 ey
Remember: A logarithm is an exponent. x
1
DOMAIN 0003 (0, 0007) RANGE 0003 (00020007, 0007) (a)
y y 0003 logb x b1 0
is equivalent to is equivalent to
x
1
DOMAIN 0003 (0, 0007) RANGE 0003 (00020007, 0007) (b)
Z Figure 2 Typical logarithmic graphs.
It is very important to remember that the equations y 0002 logb x and x 0002 b y define the same function, and as such can be used interchangeably. Because the domain of an exponential function includes all real numbers and its range is the set of positive real numbers, the domain of a logarithmic function is the set of all positive real numbers and its range is the set of all real numbers. For example, log10 3 is defined, but log10 0 and log10 (00025) are not defined. In short, the function y 0003 logb x for any b is only defined for positive x values. Typical logarithmic curves are shown in Figure 2. Notice that in each case, the y axis is a vertical asymptote for the graph. The graphs in Example 1 and Figure 2 suggest that logarithmic graphs share some common properties. Several of these properties are listed in Theorem 1. It might be helpful in understanding them to review Theorem 1 in Section 5-1. Each of these properties is a consequence of a corresponding property of exponential graphs.
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Z THEOREM 1 Properties of Graphs of Logarithmic Functions Let f (x) 0003 logb x be a logarithmic function, b 7 0, b 0004 1. Then the graph of f (x): 1. 2. 3. 4. 5. 6.
ZZZ EXPLORE-DISCUSS 1
Is continuous on its domain (0, 0007) Has no sharp corners Passes through the point (1, 0) Lies to the right of the y axis, which is a vertical asymptote Is increasing as x increases if b 7 1; is decreasing as x increases if 0 6 b 6 1 Intersects any horizontal line exactly once, so is one-to-one
For the exponential function f (x) 0003 (23)x, graph f and y 0003 x on the same coordinate system. Then sketch the graph of f 00021. Discuss the domains and ranges of f and its inverse. By what other name is f 00021 known?
Z Converting Between Logarithmic Form and Exponential Form We now look into the matter of converting logarithmic forms to equivalent exponential forms, and vice versa. Throughout the remainder of the chapter, it will be useful to sometimes convert a logarithmic expression into the equivalent exponential form. At other times, it will be useful to do the reverse.
EXAMPLE
2
Logarithmic–Exponential Conversions Change each logarithmic form to an equivalent exponential form. (B) log25 5 0003 12
(A) log2 8 0003 3 SOLUTIONS
(C) log2 (14) 0003 00022
(A) log2 8 0003 3
is equivalent to
8 0003 23.
(B) log25 5 0003 12
is equivalent to
5 0003 25100052.
(C) log2 (14) 0003 00022
is equivalent to
1 4
0003 200022.
Note that in each case, the base of the logarithm matches the base of the corresponding exponent. 0002 MATCHED PROBLEM 2
Change each logarithmic form to an equivalent exponential form. (A) log3 27 0003 3
EXAMPLE
3
(B) log36 6 0003 12
(C) log3 (19) 0003 00022
Logarithmic–Exponential Conversions Change each exponential form to an equivalent logarithmic form. (A) 49 0003 72
(B) 3 0003 19
(C) 15 0003 500021
0002
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SOLUTIONS
(A) 49 0002 72
is equivalent to
log7 49 0002 2.
(B) 3 0002 19
is equivalent to
log9 3 0002 12.
0002 500031
is equivalent to
log5 (15) 0002 00031.
(C)
1 5
Logarithmic Functions
Recall that 19 0002 9100042.
Again, the bases match. MATCHED PROBLEM 3
357
0002
Change each exponential form to an equivalent logarithmic form. (A) 64 0002 43
3 (B) 2 0002 18
(C)
1 16
0002 400032 0002
To gain a little deeper understanding of logarithmic functions and their relationship to the exponential functions, we will consider a few problems where we want to find x, b, or y in y 0002 logb x, given the other two values. All values were chosen so that the problems can be solved without a calculator. In each case, converting to the equivalent exponential form is useful.
EXAMPLE
4
Solutions of the Equation y 0002 logb x Find x, b, or y as indicated. (A) Find y: y 0002 log4 8.
SOLUTIONS
(B) Find x: log3 x 0002 00032.
(C) Find b: logb 81 0002 4.
(A) Write y 0002 log4 8 in equivalent exponential form. 8 0002 4y 23 0002 22y 2y 0002 3 y 0002 32 We conclude that
3 2
Write each number to the same base 2. Recall that bm 0002 bn if and only if m 0002 n.
0002 log4 8.
(B) Write log3 x 0002 00032 in equivalent exponential form. x 0002 300032 1 1 0002 20002 9 3 We conclude that log3 (19) 0002 00032. (C) Write logb 81 0002 4 in equivalent exponential form: 81 0002 b4 34 0002 b4 b00023
Write 81 as a fourth power. b could be 3 or 00033, but the base of a logarithm must be positive.
We conclude that log3 81 0002 4. MATCHED PROBLEM 4
0002
Find x, b, or y as indicated. (A) Find y: y 0002 log9 27.
(B) Find x: log2 x 0002 00033.
(C) Find b: logb 100 0002 2. 0002
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Z Properties of Logarithmic Functions Some of the properties of exponential functions that we studied in Section 5-1 can be used to develop corresponding properties of logarithmic functions. Several of these important properties of logarithmic functions are listed in Theorem 2. We will justify them individually.
Z THEOREM 2 Properties of Logarithmic Functions If b, M, and N are positive real numbers, b 0006 1, and p and x are real numbers, then 1. logb 1 0002 0 2. logb b 0002 1 3. logb bx 0002 x 4. blogb x 0002 x, x 7 0
ZZZ
5. logb M 0002 logb N if and only if 6. logb MN 0002 logb M 0005 logb N M 7. logb 0002 logb M 0003 logb N N 8. logb M p 0002 p logb M
M0002N
1. In properties 3 and 4, it’s essential that the base of the exponential and the base of the logarithm are the same. 2. Properties 6 and 7 are often misinterpreted, so you should examine them carefully.
CAUTION ZZZ
log b M 0003 logb N 0002 log b
logb M 0006 logb M 0003 logb N logb N
logb M logb N
logb (M 0005 N) 0006 logb M 0005 logb N
M ; N
cannot be simplified.
logb M 0004 logb N 0002 logb MN; logb (M 0004 N) cannot be simplified.
Now we will justify properties in Theorem 2. b0 0002 1. 1. logb 1 0002 0 because b1 0002 b. 2. logb b 0002 1 because 3 and 4. These are simply another way to state that f (x) 0002 bx and f 00031(x) 0002 logb x are inverse functions. Property 3 can be written as f 00031( f (x)) 0002 x for all x in the domain of f. Property 4 can be written as f ( f 00031(x)) 0002 x for all x in the domain of f 00031. This matches our characterization of inverse functions in Theorem 5, Section 3-6. Together, these properties say that if you apply an exponential function and a logarithmic function with the same base consecutively (in either order) you end up with the same value you started with. 5. This follows from the fact that logarithmic functions are one-to-one. Properties 6, 7, and 8 are used often in manipulating logarithmic expressions. We will justify them in Problems 111 and 112 in Exercises 5-3, and Problem 69 in the Chapter 5 Review Exercises.
EXAMPLE
5
Using Logarithmic Properties Simplify, using the properties in Theorem 2. (A) loge 1
(B) log10 10
(C) loge e2x00051
(D) log10 0.01
(E) 10log10 7
(F) eloge x
2
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SOLUTIONS
(A) loge 1 0003 0
Property 1
(B) log10 10 0003 1
Property 2
(C) loge e2x 1 0003 2x 1
Property 3
(D) log10 0.01 0003 log10 1000022 0003 00022
Property 3
log10 7
(E) 10 MATCHED PROBLEM 5
Logarithmic Functions
00037
(F) e
Property 4
loge x2
2
0003x
Property 4
359
0002
Simplify, using the properties in Theorem 2. (A) log10 1000025 m n
(D) loge e
(B) log5 25
(C) log10 1 4
log10 4
(F) eloge (x
(E) 10
1)
0002
Z Common and Natural Logarithms To work with logarithms effectively, we will need to be able to calculate (or at least approximate) the logarithms of any positive number to a variety of bases. Historically, tables were used for this purpose, but now calculators are used because they are faster and can find far more values than any table can possibly include. Of all possible bases, there are two that are used most often. Common logarithms are logarithms with base 10. Natural logarithms are logarithms with base e. Most calculators have a function key labeled “log” and a function key labeled “ln.” The former represents the common logarithmic function and the latter the natural logarithmic function. In fact, “log” and “ln” are both used in most math books, and whenever you see either used in this book without a base indicated, they should be interpreted as follows:
Z LOGARITHMIC FUNCTIONS y 0003 log x 0003 log10 x y 0003 ln x 0003 loge x
ZZZ EXPLORE-DISCUSS 2
Common logarithmic function Natural logarithmic function
(A) Sketch the graph of y 0003 10 x, y 0003 log x, and y 0003 x in the same coordinate system and state the domain and range of the common logarithmic function. (B) Sketch the graph of y 0003 ex, y 0003 ln x, and y 0003 x in the same coordinate system and state the domain and range of the natural logarithmic function.
EXAMPLE
6
Calculator Evaluation of Logarithms Use a calculator to evaluate each to six decimal places. (A) log 3,184
SOLUTIONS
(B) ln 0.000 349
(A) log 3,184 0003 3.502 973
(C) log (00023.24)
(B) ln 0.000 349 0003 00027.960 439
(C) log (00023.24) 0003 Error Why is an error indicated in part C? Because 00023.24 is not in the domain of the log function. [Note: Calculators display error messages in various ways. Some calculators use a more advanced definition of logarithmic functions that involves complex numbers. They will
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display an ordered pair, representing a complex number, as the value of log (00023.24), rather than an error message. You should interpret such a display as indicating that the number entered is not in the domain of the logarithmic function as we have defined it.] 0002 MATCHED PROBLEM 6
Use a calculator to evaluate each to six decimal places. (A) log 0.013 529
(B) ln 28.693 28
(C) ln (00020.438) 0002
When working with common and natural logarithms, we will follow the common practice of using the equal sign “0003” where it might be technically correct to use the approximately equal sign “⬇.” No harm is done as long as we keep in mind that in a statement such as log 3.184 0003 0.503, the number on the right is only assumed accurate to three decimal places and is not exact.
Graphs of the functions f (x) 0003 log x and g(x) 0003 ln x are shown in the graphing calculator display of Figure 3. Which graph belongs to which function? It appears from the display that one of the functions might be a constant multiple of the other. Is that true? Find and discuss the evidence for your answer.
ZZZ EXPLORE-DISCUSS 3
2
0
5
00022
Z Figure 3
EXAMPLE
7
Calculator Evaluation of Logarithms Use a calculator to evaluate each expression to three decimal places.
SOLUTIONS
(A)
log 2 log 1.1
(A)
log 2 0003 7.273 log 1.1
(B) log
(B) log
2 0003 0.260 1.1
2 1.1
(C) log 2 0002 log 1.1
Enter as (log 2) 0006 (log 1.1).
Enter as log (2 0006 1.1).
(C) log 2 0002 log 1.1 0003 0.260. Note that log MATCHED PROBLEM 7
log 2 0004 log 2 0002 log 1.1, but log 1.1
2 0003 log 2 0002 log 1.1 (see Theorem 2). 1.1
0002
Use a calculator to evaluate each to three decimal places. (A)
ln 3 ln 1.08
(B) ln
3 1.08
(C) ln 3 0002 ln 1.08 0002
We now turn to the opposite problem: Given the logarithm of a number, find the number. To solve this problem, we make direct use of the logarithmic–exponential relationships, and change logarithmic expressions into exponential form.
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Z LOGARITHMIC–EXPONENTIAL RELATIONSHIPS log x 0003 y ln x 0003 y
EXAMPLE
8
is equivalent to is equivalent to
x 0003 10 y. x 0003 e y.
Solving logb x 0002 y for x Find x to three significant digits, given the indicated logarithms. (A) log x 0003 00029.315
SOLUTIONS
(B) ln x 0003 2.386
(A) log x 0003 00029.315 x 0003 1000029.315 0003 4.84 10000210
Change to exponential form (Definition 1).
Notice that the answer is displayed in scientific notation in the calculator. (B) ln x 0003 2.386 Change to exponential form (Definition 1). x 0003 e2.386 0003 10.9 MATCHED PROBLEM 8
0002
Find x to four significant digits, given the indicated logarithms. (A) ln x 0003 00025.062
(B) log x 0003 12.0821 0002
ZZZ EXPLORE-DISCUSS 4
Example 8 was solved algebraically using logarithmic–exponential relationships. Use the INTERSECT command on a graphing calculator to solve this problem graphically. Discuss the relative merits of the two approaches.
Z The Change-of-Base Formula How would you find the logarithm of a positive number to a base other than 10 or e? For example, how would you find log3 5.2? In Example 9 we evaluate this logarithm using several properties of logarithms. Then we develop a change-of-base formula to find such logarithms more easily.
EXAMPLE
9
Evaluating a Base 3 Logarithm Evaluate log3 5.2 to four decimal places.
SOLUTION
Let y 0003 log3 5.2 and proceed as follows: log3 5.2 0003 y 5.2 0003 3y ln 5.2 0003 ln 3 y ln 5.2 0003 y ln 3 ln 5.2 y0003 ln 3
Change to exponential form. Apply the natural log (or common log) to each side. Use log b M p 0002 p log b M, which brings the exponent y in front of ln 3 as a factor. Solve for y.
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Replace y with log3 5.2 from the first step, and use a calculator to evaluate the right side: log3 5.2 0003
MATCHED PROBLEM 9
ln 5.2 0003 1.5007 ln 3
0002
Evaluate log0.5 0.0372 to four decimal places. 0002 If we repeat the process we used in Example 9 on a generic logarithm, something interesting happens. The goal is to evaluate logb N, where b is any acceptable base, and N is any positive real number. As in Example 9, let y 0003 logb N. log b N 0003 y N 0003 by ln N 0003 ln b y ln N 0003 y ln b ln N y0003 ln b
Write in exponential form. Apply natural log to each side. Use ln b y 0002 y ln b (property 8, Theorem 2). Solve for y.
This provides a formula for evaluating a logarithm to any base by using natural log: logb N 0003
ln N ln b
We could also have used log base 10 rather than natural log, and developed an alternative formula: logb N 0003
log N log b
In fact, the same approach would enable us to rewrite logb N in terms of a logarithm with any base we choose! Z THE CHANGE-OF-BASE FORMULA For any b 7 0, b 0004 1, and any positive real number N, logb N 0003
loga N loga b
where a is any positive number other than 1.
ZZZ EXPLORE-DISCUSS 5
If b is any positive real number different from 1, the change-of-base formula shows that the function y 0003 logb x is a constant multiple of the natural logarithmic function; that is, logb x 0003 k ln x for some k. (A) Graph the functions y 0003 ln x, y 0003 2 ln x, y 0003 0.5 ln x, and y 0003 00023 ln x. (B) Write each function of part A in the form y 0003 logb x by finding the base b to two decimal places. (C) Is every exponential function y 0003 bx a constant multiple of y 0003 ex? Explain.
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ANSWERS TO MATCHED PROBLEMS 1.
f 00031
f 1 x y0002a b 2
x
y 0002 log1/2 x
00033
8
8
00033
00032
4
4
00032
00031
2
2
00031
0
1
1
0
1
1 2
1 2
1
2
1 4
1 4
2
3
1 8
1 8
3
x
2. 3. 4. 5. 6. 7. 8.
5-3
(A) (A) (A) (A) (A) (A) (A)
3. Why are logarithmic functions undefined for zero and negative inputs? 4. Why is logb 1 0002 0 for any base? 5. Explain how to calculate log5 3 on a calculator that only has log buttons for base 10 and base e. 6. Using the word “inverse,” explain why log b b x 0002 x for any x and any acceptable base b. Rewrite Problems 7–12 in equivalent exponential form. 7. log3 81 0002 4 9. log10 0.001 0002 00033 0002 00032
x
y
y0002x
10
5
00035
5
x
10
y 0002 log1/2 x f 00031
00035
Exercises
2. Explain why there are infinitely many different logarithmic functions.
11.
f
冢 12 冣
27 0002 33 (B) 6 0002 36100042 (C) 19 0002 300032 1 (B) log8 2 0002 3 (C) log4 (161 ) 0002 00032 log4 64 0002 3 3 1 (B) x 0002 8 (C) b 0002 10 y00022 (B) 2 (C) 0 (D) m 0005 n (E) 4 (F) x4 0005 1 00035 (B) 3.356 663 (C) Not possible 00031.868 734 14.275 (B) 1.022 (C) 1.022 (B) x 0002 1.208 0007 1012 9. 4.7486 x 0002 0.006 333
1. Describe the relationship between logarithmic functions and exponential functions in your own words.
log 6 361
y0002
8. log5 125 0002 3 10. log10 1,000 0002 3 12.
log2 641
0002 00036
Rewrite Problems 13–18 in equivalent logarithmic form. 13. 8 0002 4300042
14. 9 0002 27200043
15. 12 0002 320003100045
16. 18 0002 200033
17. (23)3 0002 278
18. (52)00032 0002 0.16
In Problems 19–22, make a table of values similar to the one in Example 1, then use it to graph both functions by hand. 19. f (x) 0002 3x
f 00031(x) 0002 log3 x
20. f (x) 0002 (13)x
f 00031(x) 0002 log1/3 x
21. f (x) 0002 (23)x
f 00031(x) 0002 log2/3 x
22. f (x) 0002 10 x
f 00031(x) 0002 log x
In Problems 23–38, simplify each expression using Theorem 2. 23. log16 1
24. log25 1
25. log0.5 0.5
26. log7 7
27. loge e4
28. log10 105
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31. log3 27
30. log10 100 33. log1Ⲑ2 2
34.
log1Ⲑ5 (251 ) 3
loge 10
37. log5 15
36. e
In Problems 83–86, rewrite the expression as a single log. 83. ln x ⫺ ln y
84. log3 x ⫹ log3 y
85. 2 ln x ⫹ 5 ln y ⫺ ln z
86. log a ⫺ 2 log b ⫹ 3 log c
38. log2 18 In Problems 87–90, given that log x ⫽ ⫺2 and log y ⫽ 3, find: In Problems 39–46, evaluate to four decimal places. 39. log 49,236
40. log 691,450
41. ln 54.081
42. ln 19.722
43. log7 13
44. log 9 78
45. log5 120.24
46. log17 304.66
In Problems 47–54, evaluate x to four significant digits.
x 88. log a b y
87. log (xy) 89. log a
1x b y3
90. log (x5y3)
In Problems 91–98, use transformations to explain how the graph of g is related to the graph of the given logarithmic function f. Determine whether g is increasing or decreasing, find its domain and asymptote, and sketch the graph of g.
47. log x ⫽ 5.3027
48. log x ⫽ 1.9168
49. log x ⫽ ⫺3.1773
50. log x ⫽ ⫺2.0411
91. g (x) ⫽ 3 ⫹ log2 x; f (x) ⫽ log2 x
51. ln x ⫽ 3.8655
52. ln x ⫽ 5.0884
92. g (x) ⫽ ⫺4 ⫹ log3 x; f (x) ⫽ log3 x
53. ln x ⫽ ⫺0.3916
54. ln x ⫽ ⫺4.1083
93. g (x) ⫽ log1Ⲑ3 (x ⫺ 2); f (x) ⫽ log1Ⲑ3 x
Find x, y, or b, as indicated in Problems 55–72. 55. log2 x ⫽ 2
56. log3 x ⫽ 3
57. log4 16 ⫽ y
58. log8 64 ⫽ y
59. logb 16 ⫽ 2
60. logb 10⫺3 ⫽ ⫺3
61. logb 1 ⫽ 0
62. logb b ⫽ 1
63. log4 x ⫽ 12
64. log8 x ⫽ 13
65. log1Ⲑ3 9 ⫽ y
66. log49 (17) ⫽ y
67. logb 1,000 ⫽ 32
68. logb 4 ⫽ 23
69. log8 x ⫽ ⫺43
70. log25 x ⫽ ⫺32
71. log16 8 ⫽ y
72. log9 27 ⫽ y
In Problems 73–78, evaluate to three decimal places. 73.
log 2 log 1.15
74.
log 2 log 1.12
75.
ln 3 ln 1.15
76.
ln 4 ln 1.2
77.
ln 150 2 ln 3
78.
log 200 3 log 2
81. log (x4y3)
80. log (xy) 82. log a
95. g (x) ⫽ ⫺1 ⫺ log x; f (x) ⫽ log x 96. g (x) ⫽ 2 ⫺ log x; f (x) ⫽ log x 97. g (x) ⫽ 5 ⫺ 3 ln x; f (x) ⫽ ln x 98. g (x) ⫽ ⫺3 ⫺ 2 ln x; f (x) ⫽ ln x In Problems 99–102, find f ⫺1. 99. f (x) ⫽ log5 x
100. f (x) ⫽ log1Ⲑ3 x
101. f (x) ⫽ 4 log3 (x ⫹ 3)
102. f (x) ⫽ 2 log2 (x ⫺ 5)
103. Let f (x) ⫽ log3 (2 ⫺ x). (A) Find f ⫺1. (B) Graph f ⫺1. (C) Reflect the graph of f ⫺1 in the line y ⫽ x to obtain the graph of f. 104. Let f (x) ⫽ log2 (⫺3 ⫺x). (A) Find f ⫺1. (B) Graph f ⫺1. (C) Reflect the graph of f ⫺1 in the line y ⫽ x to obtain the graph of f. 105. What is wrong with the following “proof ” that 3 is less than 2?
In Problems 79–82, rewrite the expression in terms of log x and log y. x 79. log a b y
94. g (x) ⫽ log1Ⲑ2 (x ⫹ 3); f (x) ⫽ log1Ⲑ2 x
2
x b 1y
1 6 3 1 27 1 27 1 3 (3) log (13)3 3 log 13
6 6 6
Divide both sides by 27.
3 27 1 9 (13)2
6 log (13)2 6 2 log 13
3 6 2
Divide both sides by log 13 .
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3 log 12 log (12)3 (12)3 1 8
7 7 7 7
2 log 12 log (12)2 (12)2 1 4
107. P1(x) 0003 x 0002 12 x2
Multiply both sides by log 12 .
108. P2(x) 0003 x 0002 12 x2 13 x3
109. P3(x) 0003 x 0002 12 x2 13 x3 0002 14 x4 110. P4(x) 0003 x 0002 12 x2 13 x3 0002 14 x4 15 x5 111. Prove that for any positive M, N, and b (b 0004 1), logb (MN) 0003 logb M 0002 logb N. (Hint: Start by writing u 0003 logb M and v 0003 logb N and changing each to exponential form.)
Multiply both sides by 8.
1 7 2
112. Prove that for any positive integer p and any positive b and M (b 0004 1), logb M p 0003 p logb M. [Hint: Write M p as M 0007 M 0007 p M ( p factors).]
The polynomials in Problems 107–110, called Taylor polynomials, can be used to approximate the function g(x) 0003 ln (1 x). To illustrate this approximation graphically, in each problem, graph
5-4
365
g(x) 0003 ln (1 x) and the indicated polynomial in the same viewing window, 00021 x 3 and 00022 y 2.
106. What is wrong with the following “proof ” that 1 is greater than 2? 3 7 2
Logarithmic Models
Logarithmic Models Z Logarithmic Scales Z Data Analysis and Regression
Logarithmic functions occur naturally as the inverses of exponential functions. But that’s not to say that they are not useful in their own right. Some of these uses are probably familiar to you, but you might not have realized that they involved logarithmic functions. In this section, we will study logarithmic scales that are used to compare the intensity of sounds, the severity of earthquakes, and the brightness of distant stars. We will also look at using regression to model data with a logarithmic function, and discuss what sort of data is likely to fit such a model.
Z Logarithmic Scales Table 1 Typical Sound Intensities Sound Intensity (W兾m2)
Sound
1.0 10000212
Threshold of hearing
5.2 10000210
Whisper
00026
3.2 10
Normal conversation
8.5 1000024
Heavy traffic
3.2 1000023
Jackhammer
0
1.0 10
Threshold of pain
8.3 102
Jet plane
The human ear is able to hear sound over a very wide range of intensities. The loudest sound a healthy person can hear without damage to the eardrum has an intensity 1 trillion (1,000,000,000,000) times that of the softest sound a person can hear. If we were to use these intensities as a scale for measuring volume, we would be stuck using numbers from zero all the way to the trillions, which seems cumbersome, if not downright silly. In the last section, we saw that logarithmic functions increase very slowly. We can take advantage of this to create a scale for sound intensity that is much more condensed, and therefore more manageable. The decibel scale for sound intensity is an example of such a scale. The decibel, named after the inventor of the telephone, Alexander Graham Bell (1847–1922), is defined as follows:
SOUND INTENSITY:
D 0002 10 log
I I0
Decibel scale
(1)
where D is the decibel level of the sound, I is the intensity of the sound measured in watts per square meter (W/m2), and I0 is the intensity of the least audible sound that an average healthy young person can hear. The latter is standardized to be I0 0003 10000212 watts per square meter. Table 1 lists some typical sound intensities from familiar sources. In Example 1 and Problems 5 and 6 in Exercises 5-4, we will calculate the decibel levels for these sounds.
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1
Sound Intensity (A) Find the number of decibels from a whisper with sound intensity 5.2 10000210 watts per square meter, then from heavy traffic at 8.5 1000024 watts per square meter. Round your answers to two decimal places. (B) How many times larger is the sound intensity of heavy traffic compared to a whisper?
SOLUTIONS
(A) We can use the decibel formula (1) with I0 0003 10000212. First, we use I 0003 5.2 10000210: I I0 5.2 10000210 0003 10 log 10000212 0003 10 log 520 0003 27.16 decibels
D 0003 10 log
Substitute I 0002 5.2 10000310, I0 0002 10000312.
Simplify the fraction.
Next, for heavy traffic: D 0003 10 log
I I0
Substitute I 0002 8.5 1000034, I0 0002 10000312.
8.5 1000024 10000212 0003 10 log 850,000,000 0003 89.29 decibels 0003 10 log
Simplify the fraction.
(B) Dividing the larger intensity by the smaller, 8.5 1000024 0003 1,634,615.4 5.2 10000210 we see that the sound intensity of heavy traffic is more than 1.6 million times as great as the intensity of a whisper! MATCHED PROBLEM 1
ZZZ EXPLORE-DISCUSS 1
0002
Find the number of decibels from a jackhammer with sound intensity 3.2 1000023 watts per square meter. Compute the answer to two decimal places. 0002 Suppose that you are asked to draw a graph of the data in Table 1, with sound intensities on the x axis, and the corresponding decibel levels on the y axis. (A) What would be the coordinates of the point corresponding to a jackhammer (see Matched Problem 1)? (B) Suppose the axes of this graph are labeled as follows: Each tick mark on the x axis corresponds to the intensity of the least audible sound (10000212 watts per square meter), and each tick mark on the y axis corresponds to 1 decibel. If there is 18 inch between all tick marks, how far away from the x axis is the point you found in part A? From the y axis? (Give the first answer in inches and the second in miles!) Discuss your result.
EARTHQUAKE INTENSITY: The energy released by the largest earthquake recorded, measured in joules, is about 100 billion (100,000,000,000) times the energy released by a small earthquake that is barely felt. In 1935 the California seismologist Charles Richter devised a logarithmic
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Table 2 The Richter Scale Magnitude on Richter Scale M 6 4.5
Destructive Power
Moderate
5.5 6 M 6 6.5
Large
6.5 6 M 6 7.5
Major
7.5 6 M
Great
EXAMPLE
367
scale that bears his name and is still widely used in the United States. The magnitude of an earthquake M on the Richter scale* is given as follows: M0002
Small
4.5 6 M 6 5.5
Logarithmic Models
2 E log 3 E0
Richter scale
(2)
where E is the energy released by the earthquake, measured in joules, and E0 is the energy released by a very small reference earthquake, which has been standardized to be E0 0003 104.40 joules The destructive power of earthquakes relative to magnitudes on the Richter scale is indicated in Table 2.
2
Earthquake Intensity The 1906 San Francisco earthquake released approximately 5.96 1016 joules of energy. Another quake struck the Bay Area just before game 3 of the 1989 World Series, releasing 1.12 1015 joules of energy. (A) Find the magnitude of each earthquake on the Richter scale. Round your answers to two decimal places. (B) How many times more energy did the 1906 earthquake release than the one in 1989?
SOLUTIONS
(A) We can use the magnitude formula (2) with E0 0003 104.40. First, for the 1906 earthquake, E 0003 5.96 1016: 2 E log 3 E0 2 5.96 1016 0003 log 3 104.40 0003 8.25
M0003
Substitute E 0002 5.96 1016, E0 0002 104.40.
Next, for the 1989 earthquake, E 0003 1.12 1015 2 E log 3 E0 2 1.12 1015 0003 log 3 104.40 0003 7.1
M0003
Substitute E 0002 1.12 1015, E0 0002 104.40.
(B) Dividing the larger energy release by the smaller, 5.96 1016 0003 53.2 1.12 1015 we see that the 1906 earthquake released 53.2 times as much energy as the 1989 quake. MATCHED PROBLEM 2
0002
A 1985 earthquake in central Chile released approximately 1.26 1016 joules of energy. What was its magnitude on the Richter scale? Compute the answer to two decimal places. 0002 *Originally, Richter defined the magnitude of an earthquake in terms of logarithms of the maximum seismic wave amplitude, in thousandths of a millimeter, measured on a standard seismograph. Equation (2) gives essentially the same magnitude that Richter obtained for a given earthquake but in terms of logarithms of the energy released by the earthquake.
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3
Earthquake Intensity If the energy release of one earthquake is 1,000 times that of another, how much larger is the Richter scale reading of the larger than the smaller?
SOLUTION
Let M1 0003
E1 2 log 3 E0
and
M2 0003
E2 2 log 3 E0
be the Richter equations for the smaller and larger earthquakes, respectively. Since the larger earthquake released 1,000 times as much energy, we can write E2 0003 1,000E1. M2 0003 0003 0003 0003 0003 0003
E2 2 log 3 E0 1,000E1 2 log 3 E0 E1 2 alog 1,000 log b 3 E0 E1 2 a3 log b 3 E0 E1 2 2 (3) log 3 3 E0 2 M1
Substitute 1,000E1 for E2.
Use log (MN) 0002 log M 0005 log N;
1,000E1 E0
0002 1,000 0007
E1 E0
log 1,000 0002 log 103 0002 3
Distribute. E1 2 log is M1! 3 E0
An earthquake with 1,000 times the energy of another has a Richter scale reading of 2 more than the other. 0002 MATCHED PROBLEM 3
If the energy release of one earthquake is 10,000 times that of another, how much larger is the Richter scale reading of the larger than the smaller? 0002 ROCKET FLIGHT: The theory of rocket flight uses advanced mathematics and physics to show that the velocity v of a rocket at burnout (depletion of fuel supply) is given by
v 0002 c ln
Wt Wb
Rocket equation
(3)
where c is the exhaust velocity of the rocket engine, Wt is the takeoff weight (fuel, structure, and payload), and Wb is the burnout weight (structure and payload). Because of the Earth’s atmospheric resistance, a launch vehicle velocity of at least 9.0 kilometers per second is required to achieve the minimum altitude needed for a stable orbit. Formula (3) indicates that to increase velocity v, either the weight ratio Wt0005Wb must be increased or the exhaust velocity c must be increased. The weight ratio can be increased by the use of solid fuels, and the exhaust velocity can be increased by improving the fuels, solid or liquid.
EXAMPLE
4
Rocket Flight Theory A typical single-stage, solid-fuel rocket may have a weight ratio Wt0005Wb 0003 18.7 and an exhaust velocity c 0003 2.38 kilometers per second. Would this rocket reach a launch velocity of 9.0 kilometers per second?
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SOLUTION
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369
We can use the rocket equation (3) with c 0003 2.38 and Wt0005Wb 0003 18.7: Wt Wb 0003 2.38 ln 18.7 0003 6.97 kilometers per second
v 0003 c ln
The velocity of the launch vehicle is far short of the 9.0 kilometers per second required to achieve orbit. This is why multiple-stage launchers are used—the deadweight from a preceding stage can be jettisoned into the ocean when the next stage takes over. 0002 MATCHED PROBLEM 4
A launch vehicle using liquid fuel, such as a mixture of liquid hydrogen and liquid oxygen, can produce an exhaust velocity of c 0003 4.7 kilometers per second. However, the weight ratio Wt0005Wb must be low—around 5.5 for some vehicles—because of the increased structural weight to accommodate the liquid fuel. How much more or less than the 9.0 kilometers per second required to reach orbit will be achieved by this vehicle? 0002
Z Data Analysis and Regression Based on the logarithmic graphs we studied in the last section, when a quantity increases relatively rapidly at first, but then levels off and increases very slowly, it might be a good candidate to be modeled by a logarithmic function. Most graphing calculators with regression commands can fit functions of the form y 0003 a b ln x to a set of data points using the same techniques we used earlier for other types of regression.
EXAMPLE
5
Table 3 Home Ownership Rates Year
Home Ownership Rate (%)
1940 1950 1960 1970 1980 1990 2000
43.6 55.0 61.9 62.9 64.4 64.2 67.4 SOLUTIONS
Home Ownership Rates The U.S. Census Bureau published the data in Table 3 on home ownership rates. A logarithmic model for the data is given by R 0003 000236.7 23.0 ln t where R is the home ownership rate and t is time in years with t 0003 0 representing 1900. (A) Use the model to predict the home ownership rate in 2015. (B) Compare the actual home ownership rate in 1950 to the rate given by the model.
(A) The year 2015 is represented by t 0003 115. Evaluating R 0003 000236.7 23.0 ln t at t 0003 115 predicts a home ownership rate of 72.4% in 2015. (B) The year 1950 is represented by t 0003 50. Evaluating R 0003 000236.7 23.0 ln t at t 0003 50 gives a home ownership rate of 53.3% in 1950. The actual home ownership rate in 1950 was 55%, approximately 1.7% greater than the number given by the model.
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Technology Connections Figure 1 shows the details of constructing the logarithmic model of Example 5 on a graphing calculator. 100
0
120
0
(a) Entering the data
(b) Finding the model
(c) Graphing the data and the model
Z Figure 1
0002 MATCHED PROBLEM 5
Refer to Example 5. The home ownership rate in 2008 was 67.8%. If this data is added to Table 3, a logarithmic model for the expanded data is given by R ⫽ ⫺30.6 ⫹ 21.5 ln t where R is the home ownership rate and t is time in years with t ⫽ 0 representing 1900. (A) Use the model to predict the home ownership rate in 2015. (B) Compare the actual home ownership rate in 1950 to the rate given by the model. 0002 ANSWERS TO MATCHED PROBLEMS 1. 95.05 decibels 2. 7.80 3. 2.67 4. 1 kilometer per second less 5. (A) 70.5% (B) The actual rate was 1.5% greater than the rate given by the model.
5-4
Exercises
1. Describe the decibel scale in your own words. 2. Describe the Richter scale in your own words. 3. Explain why logarithms are a good choice for describing sound intensity and earthquake magnitude. 4. Think of a real-life quantity that is likely to be modeled well by a logarithmic function, and explain your reasoning.
APPLICATIONS 5. SOUND What is the decibel level of (A) The threshold of hearing, 1.0 ⫻ 10⫺12 watts per square meter?
(B) The threshold of pain, 1.0 watt per square meter? Compute answers to two significant digits. 6. SOUND What is the decibel level of (A) A normal conversation, 3.2 ⫻ 10⫺6 watts per square meter? (B) A jet plane with an afterburner, 8.3 ⫻ 102 watts per square meter? Compute answers to two significant digits. 7. SOUND If the intensity of a sound from one source is 1,000 times that of another, how much more is the decibel level of the louder sound than the quieter one?
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8. SOUND If the intensity of a sound from one source is 10,000 times that of another, how much more is the decibel level of the louder sound than the quieter one? 9. EARTHQUAKES One of the strongest recorded earthquakes to date was in Colombia in 1906, with an energy release of 1.99 1017 joules. What was its magnitude on the Richter scale? Compute the answer to one decimal place. 10. EARTHQUAKES Anchorage, Alaska, had a major earthquake in 1964 that released 7.08 1016 joules of energy. What was its magnitude on the Richter scale? Compute the answer to one decimal place. 11. EARTHQUAKES The 1933 Long Beach, California, earthquake had a Richter scale reading of 6.3, and the 1964 Anchorage, Alaska, earthquake had a Richter scale reading of 8.3. How many times more powerful was the Anchorage earthquake than the Long Beach earthquake? 12. EARTHQUAKES Generally, an earthquake requires a magnitude of over 5.6 on the Richter scale to inflict serious damage. How many times more powerful than this was the great 1906 Colombia earthquake, which registered a magnitude of 8.6 on the Richter scale? 13. EXPLOSIVE ENERGY The atomic bomb dropped on Nagasaki, Japan, on August 9, 1945, released about 1.34 1014 joules of energy. What would be the magnitude of an earthquake that released that much energy? 14. EXPLOSIVE ENERGY The largest and most powerful nuclear weapon ever detonated was tested by the Soviet Union on October 30, 1961, on an island in the Arctic Sea. The blast was so powerful there were reports of windows breaking in Finland, over 700 miles away. The detonation released about 2.1 1017 joules of energy. What would be the magnitude of an earthquake that released that much energy? 15. ASTRONOMY A moderate-size solar flare observed on the sun on July 9, 1996, released enough energy to power the United States for almost 23,000 years at 2001 consumption levels, 2.38 1021 joules. What would be the magnitude of an earthquake that released that much energy? 16. CONSTRUCTION The energy released by a typical construction site explosion is about 7.94 105 joules. What would be the magnitude of an earthquake that released that much energy? 17. SPACE VEHICLES A new solid-fuel rocket has a weight ratio Wt0005Wb 0003 19.8 and an exhaust velocity c 0003 2.57 kilometers per second. What is its velocity at burnout? Compute the answer to two decimal places. 18. SPACE VEHICLES A liquid-fuel rocket has a weight ratio Wt0005Wb 0003 6.2 and an exhaust velocity c 0003 5.2 kilometers per second. What is its velocity at burnout? Compute the answer to two decimal places. 19. CHEMISTRY The hydrogen ion concentration of a substance is related to its acidity and basicity. Because hydrogen ion concentrations vary over a very wide range, logarithms are used to create a compressed pH scale, which is defined as follows:
pH 0003 0002log [ H ]
Logarithmic Models
371
where [H ] is the hydrogen ion concentration, in moles per liter. Pure water has a pH of 7, which means it is neutral. Substances with a pH less than 7 are acidic, and those with a pH greater than 7 are basic. Compute the pH of each substance listed, given the indicated hydrogen ion concentration. Also, indicate whether each substance is acidic or basic. Compute answers to one decimal place. (A) Seawater, 4.63 1000029 (B) Vinegar, 9.32 1000024 20. CHEMISTRY Refer to Problem 19. Compute the pH of each substance below, given the indicated hydrogen ion concentration. Also, indicate whether it is acidic or basic. Compute answers to one decimal place. (A) Milk, 2.83 1000027 (B) Garden mulch, 3.78 1000026 21. ECOLOGY Refer to Problem 19. Many lakes in Canada and the United States will no longer sustain some forms of wildlife because of the increase in acidity of the water from acid rain and snow caused by sulfur dioxide emissions from industry. If the pH of a sample of rainwater is 5.2, what is its hydrogen ion concentration in moles per liter? Compute the answer to two significant digits. 22. ECOLOGY Refer to Problem 19. If normal rainwater has a pH of 5.7, what is its hydrogen ion concentration in moles per liter? Compute the answer to two significant digits. 23. ASTRONOMY The brightness of stars is expressed in terms of magnitudes on a numerical scale that increases as the brightness decreases. The magnitude m is given by the formula m 0003 6 0002 2.5 log
L L0
where L is the light flux of the star and L0 is the light flux of the dimmest stars visible to the naked eye. (A) What is the magnitude of the dimmest stars visible to the naked eye? (B) How many times brighter is a star of magnitude 1 than a star of magnitude 6? 24. ASTRONOMY An optical instrument is required to observe stars beyond the sixth magnitude, the limit of ordinary vision. However, even optical instruments have their limitations. The limiting magnitude L of any optical telescope with lens diameter D, in inches, is given by L 0003 8.8 5.1 log D (A) Find the limiting magnitude for a homemade 6-inch reflecting telescope. (B) Find the diameter of a lens that would have a limiting magnitude of 20.6. Compute answers to three significant digits.
Problems 25 and 26 require a graphing calculator or a computer program that can calculate a logarithmic regression model for a given data set. 25. INTERNET ACCESS Table 4 on page 372 shows the percentage of Americans that had access to the Internet either at home or at work between 2000 and 2006. Let x represent years since 1995.
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Table 4 Internet Access in the United States Year
Percentage with Home Access
Percentage with Work Access
2000
46.9
35.2
2001
58.4
37.5
2002
59.3
40.2
2003
65.1
49.6
2005
66.2
55.1
2006
68.1
55.8
5-5
(A) Find a logarithmic regression model (y 0003 a b ln x) for the percentage with home access. Round a and b to three significant digits. Use your model to estimate the percentage in 2008 and 2015. (B) Examine the model for larger and larger values of x. Does it remain reasonable in the long term? 26. INTERNET ACCESS Refer to Table 4. (A) Find a logarithmic regression model (y 0003 a b ln x) for the percentage with work access. (Keep in mind that x represents years since 1995.) Round a and b to three significant digits. Use your model to estimate the percentage in 2008 and 2015. (B) Examine the model for larger and larger values of x. Does it remain reasonable in the long term?
Exponential and Logarithmic Equations Z Solving Exponential Equations Z Solving Logarithmic Equations
We have seen that many quantities can be modeled by exponential or logarithmic functions. So it’s not surprising that equations involving exponential or logarithmic expressions, like those shown next, are useful in studying those quantities. 23x00022 0003 5
and
log (x 3) log x 0003 1
Equations like these are called exponential and logarithmic equations, respectively. The properties of logarithms that we studied in Section 5-3 will play a key role in solving both types of equations.
Z Solving Exponential Equations The distinguishing feature of exponential equations is that the variable appears in an exponent. Before defining logarithms, we didn’t have a reliable method for removing variables from an exponent: Now we do. We’ll illustrate how these properties are helpful in Examples 1-4.
EXAMPLE
1
Solving an Exponential Equation Find all solutions to 23x00022 0003 5 to four decimal places.
SOLUTION
In order to have any chance of solving for x, we will first need to get x out of the exponent. This is where logs come in very handy. 23x00022 0003 5 log 23x00022 0003 log 5 (3x 0002 2) log 2 0003 log 5 log 5 3x 0002 2 0003 log 2
Take the common or natural log of both sides. Use logb N p 0002 p logb N to get 3x 0003 2 out of the exponent position. Divide both sides by log 2. Add 2 to both sides.
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log 5 log 2 log 5 1 x 0002 a2 0005 b 3 log 2 0002 1.4406
3x 0002 2 0005
MATCHED PROBLEM 1
Exponential and Logarithmic Equations
373
1 Divide both sides by 3, or multiply both sides by . 3
Use a calculator. Solution to four decimal places
0002
Solve 35100032x 0002 7 for x to four decimal places. 0002
EXAMPLE
2
Compound Interest Recall that when an amount of money P (principal) is invested at an annual rate r compounded annually, the amount of money A in the account after n years, assuming no withdrawals, is given by A 0002 P a1 0005
r n b 0002 P(1 0005 r)n m
m 0002 1 for annual compounding.
How many years to the nearest year will it take the money to double if it is invested at 6% compounded annually? SOLUTION
The interest rate is r 0002 0.06, and we want the amount A to be twice the principal, or 2P. So we substitute r 0002 0.06 and A 0002 2P, and solve for n. 2P 0002 P(1.06)n 2 0002 1.06n log 2 0002 log 1.06n log 2 0002 n log 1.06 log 2 n0002 log 1.06 0002 12 years
MATCHED PROBLEM 2
Divide both sides by P to isolate (1.06)n. Take the common or natural log of both sides. Note how log properties are used to get n out of the exponent position. Divide both sides by log 1.06 (which is just a number!). Calculate to the nearest year.
0002
Repeat Example 2, changing the interest rate to 9% compounded annually. 0002
EXAMPLE
3
Atmospheric Pressure The atmospheric pressure P, in pounds per square inch, at x miles above sea level is given approximately by P 0002 14.7e00030.21x At what height will the atmospheric pressure be half the sea-level pressure? Compute the answer to two significant digits.
SOLUTION
Since x is miles above sea level, sea-level pressure is the pressure at x 0002 0, which is 14.7e0, or 14.7. One-half of sea level pressure is 14.7兾2 0002 7.35. Now our problem is to find x so that P 0002 7.35; that is, we solve 7.35 0002 14.7e00030.21x for x: 7.35 0002 14.7e00030.21x 0.5 0002 e00030.21x ln 0.5 0002 ln e00030.21x
Divide both sides by 14.7 to isolate the exponential. Because the base is e, take the natural log of both sides. In ea 0002 a, so ln e00030.21 x 0002 00030.21x
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ln 0.5 0003 00020.21x x0003
Divide both sides by 00030.21.
ln 0.5 00020.21
Calculate to two significant digits.
0003 3.3 miles MATCHED PROBLEM 3 y
y0003
e x e0002x 2
10
0002
Using the formula in Example 3, find the altitude in miles so that the atmospheric pressure will be one-eighth that at sea level. Compute the answer to two significant digits. 0002 The graph of y0002
5
00025
5
e x 0005 e0003x 2
(1)
is a curve called a catenary (Fig. 1). A uniform cable suspended between two fixed points is a physical example of such a curve, which resembles a parabola, but isn’t.
x
Z Figure 1 Catenary.
EXAMPLE
4
Solving an Exponential Equation In equation (1), find x when y 0003 2.5. Compute the answer to four decimal places.
SOLUTION
e x e0002x 2 x e e0002x 2.5 0003 2 x 5 0003 e e0002x 5e x 0003 e2x 1 e2x 0002 5e x 1 0003 0 y0003
Let y 0002 2.5.
Multiply both sides by 2. Multiply both sides by ex. Subtract 5ex from both sides. This is a quadratic in ex.
Let u 0003 e x; then u2 0002 5u 1 0003 0 5 125 0002 4(1)(1) u0003 2 5 121 0003 2 5 121 ex 0003 2 5 121 ln e x 0003 ln 2 5 121 x 0003 ln 2 0003 00021.5668, 1.5668
Use the quadratic formula. Simplify.
Replace u with ex and solve for x.
Take the natural log of both sides (both values on the right are positive). logb bx 0002 x, so ln ex 0002 x.
Exact solutions Rounded to four decimal places.
Note that the method produces exact solutions, an important consideration in certain calculus applications (see Problems 57–60 of Exercises 5-5). 0002 MATCHED PROBLEM 4
Given y 0003 (e x 0002 e0002x)00052, find x for y 0003 1.5. Compute the answer to three decimal places. 0002
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Z Solving Logarithmic Equations We will now illustrate the solution of several types of logarithmic equations.
EXAMPLE
5
Solving a Logarithmic Equation Solve log (x 0005 3) 0005 log x 0002 1, and check.
SOLUTION
First use properties of logarithms to express the left side as a single logarithm, then convert to exponential form and solve for x. log (x 0005 3) 0005 log x 0002 1 log [x(x 0005 3)] 0002 1 x(x 0005 3) 0002 101 x2 0005 3x 0003 10 0002 0 (x 0005 5)(x 0003 2) 0002 0 x 0002 00035, 2
Combine left side using log M 0004 log N 0002 log MN. Change to equivalent exponential form (the base is 10). Write in ax2 0004 bx 0004 c 0002 0 form and solve. Factor. If ab 0002 0, then a 0002 0 or b 0002 0.
x 0002 00035: log (00035 0005 3) 0005 log (00035) is not defined because the domain of the log function is (0, ). x 0002 2: log (2 0005 3) 0005 log 2 0002 log 5 0005 log 2 ✓ 0002 log (5 0005 2) 0002 log 10 00021
CHECK
The only solution to the original equation is x 0002 2. Extraneous solutions are common in log equations, so answers should always be checked in the original equation to see whether any should be discarded. 0002 MATCHED PROBLEM 5
EXAMPLE
6
Solve log (x 0003 15) 0002 2 0003 log x, and check. 0002
Solving a Logarithmic Equation Solve (ln x)2 0002 ln x2.
SOLUTION
There are no logarithmic properties for simplifying (ln x)2. However, we can simplify ln x2, obtaining an equation involving ln x and (ln x)2. (ln x)2 0002 ln x2 (ln x)2 0002 2 ln x (ln x)2 0003 2 ln x 0002 0 (ln x)(ln x 0003 2) 0002 0 ln x 0002 0 or ln x 0003 2 0002 0 0 x0002e ln x 0002 2 00021
ln M p 0002 p ln M, so ln x2 0002 2 ln x. This is a quadratic equation in ln x. Move all nonzero terms to the left. Factor out ln x. If ab 0002 0, then a 0002 0 or b 0002 0. If ln x 0002 a, x 0002 ea.
x 0002 e2
Checking that both x 0002 1 and x 0002 e2 are solutions to the original equation is left to you. 0002 MATCHED PROBLEM 6
Solve log x2 0002 (log x)2. 0002
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Note that
CAUTION ZZZ
(logb x)2 0006 logb x2
(logb x)2 0002 (logb x)(logb x) logb x2 0002 2 logb x
You might find it helpful to keep these straight by writing logb x2 as logb (x2).
EXAMPLE
7
Earthquake Intensity Recall from Section 5-4 that the magnitude of an earthquake on the Richter scale is given by M0002
E 2 log 3 E0
Solve for E in terms of the other symbols. 2 E log 3 E0 E 3M log 0002 E0 2 E 0002 103M00042 E0 E 0002 E0103M00042 M0002
SOLUTION
MATCHED PROBLEM 7
Multiply both sides by 32 and switch sides.
Change to exponential form with base 10.
Multiply both sides by E0.
0002
Solve the rocket equation from Section 5-4 for Wb in terms of the other symbols: v 0002 c ln
Wt Wb 0002
ANSWERS TO MATCHED PROBLEMS 1. x 0002 0.2263 3. 9.9 miles
5-5
2. More than double in 9 years, but not quite double in 8 years 4. x 0002 1.195 5. x 0002 20 6. x 0002 1,100 7. Wb 0002 Wt e0003v0004c
Exercises
1. Which property of logarithms do you think is most useful in solving exponential equations? Explain.
6. Can you use a logarithm with the same base to solve both equations below? Explain. ex 0002 10
2. Which properties of logarithms do you think are most useful in solving equations with more than one logarithm? Explain. 3. If u and v represent expressions with variable x, how can you solve equations of the form logb u 0002 logb v for x? Explain why this works.
and
5x 0002 8
In Problems 7–16, solve to three significant digits. 7. 100003x 0002 0.0347 9. 103x00051 0002 92
8. 10x 0002 14.3 10. 105x00032 0002 348
4. Why is it especially important to check answers when solving logarithmic equations?
11. e x 0002 3.65
12. e0003x 0002 0.0142
5. Explain the difference between (ln x)2 and ln x2.
13. e2x00031 0005 68 0002 207
14. 13 0005 e3x00055 0002 23
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15. 2320002x 0003 0.426
16. 3430002x 0003 0.089
17. log5 x 0003 2
18. log3 y 0003 4
19. log (t 0002 4) 0003 00021
20. ln (2x 3) 0003 0
21. log 5 log x 0003 2
22. log x 0002 log 8 0003 1
23. log x log (x 0002 3) 0003 1 24. log (x 0002 9) log 100x 0003 3 25. log (x 1) 0002 log (x 0002 1) 0003 1 26. log (2x 1) 0003 1 log (x 0002 2)
28. 3 0003 1.06x
29. e00021.4x 5 0003 0
30. e0.32x 0.47 0003 0
31. 123 0003 500e00020.12x
32. 438 0003 200e0.25x
0002x 2
33. e
x2
0003 0.23
34. e 0003 125
In Problems 35–48, solve exactly. 35. log (5 0002 2x) 0003 log (3x 1) 36. log (x 3) 0003 log (6 4x) 37. log x 0002 log 5 0003 log 2 0002 log (x 0002 3) 38. log (6x 5) 0002 log 3 0003 log 2 0002 log x 39. ln x 0003 ln (2x 0002 1) 0002 ln (x 0002 2) 40. ln (x 1) 0003 ln (3x 1) 0002 ln x 41. log (2x 1) 0003 1 0002 log (x 0002 1) 42. 1 0002 log (x 0002 2) 0003 log (3x 1) 43. ln (x 1) 0003 ln (3x 3) 44. 1 ln (x 1) 0003 ln (x 0002 1) 45. (ln x)3 0003 ln x4
46. (log x)3 0003 log x4
47. ln (ln x) 0003 1
48. log (log x) 0003 1
Solve Problems 49–56 for the indicated variable in terms of the remaining symbols. Use the natural log for solving exponential equations. 49. A 0003 Pert for r (finance) r nt 50. A 0003 P a1 b for t (finance) n 51. D 0003 10 log 52. t 0003
I for I (sound) I0
00021 (ln A 0002 ln A0) for A (decay) k
53. M 0003 6 0002 2.5 log
I for I (astronomy) I0
E (1 0002 e0002Rt0005L) for t (circuitry) R
56. S 0003 R
(1 i)n 0002 1 for n (annuity) i
The following combinations of exponential functions define four of six hyperbolic functions, a useful class of functions in calculus and higher mathematics. Solve Problems 57–60 for x in terms of y. The results are used to define inverse hyperbolic functions, another useful class of functions in calculus and higher mathematics. 57. y 0003
e x e0002x 2
58. y 0003
e x 0002 e0002x 2
59. y 0003
e x 0002 e0002x e x e0002x
60. y 0003
e x e0002x e x 0002 e0002x
In Problems 27–34, solve to three significant digits. 27. 2 0003 1.05x
377
54. L 0003 8.8 5.1 log D for D (astronomy) 55. I 0003
In Problems 17–26, solve exactly.
Exponential and Logarithmic Equations
In Problems 61–68, use a graphing calculator to approximate to two decimal places any solutions of the equation in the interval 0 x 1. None of these equations can be solved exactly using any step-by-step algebraic process. 61. 20002x 0002 2x 0003 0
62. 30002x 0002 3x 0003 0
63. e0002x 0002 x 0003 0
64. xe2x 0002 1 0003 0
65. ln x 2x 0003 0
66. ln x x2 0003 0
67. ln x e x 0003 0
68. ln x x 0003 0
APPLICATIONS 69. COMPOUND INTEREST How many years, to the nearest year, will it take a sum of money to double if it is invested at 7% compounded annually? 70. COMPOUND INTEREST How many years, to the nearest year, will it take money to quadruple if it is invested at 6% compounded annually? 71. COMPOUND INTEREST At what annual rate compounded continuously will $1,000 have to be invested to amount to $2,500 in 10 years? Compute the answer to three significant digits. 72. COMPOUND INTEREST How many years will it take $5,000 to amount to $8,000 if it is invested at an annual rate of 9% compounded continuously? Compute the answer to three significant digits. 73. IMMIGRATION According to the U.S. Office of Immigration Statistics, there were 10.5 million illegal immigrants in the United States in May 2005, and that number had grown to 11.3 million by May 2007. (A) Find the relative growth rate if we use the P 0003 P0ert model for population growth. Round to three significant digits. (B) Use your answer from part A to write a function describing the illegal immigrant population in millions in terms of years after May 2005, and use it to predict when the illegal immigrant population should reach 20 million.
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74. POPULATION GROWTH According to U.S. Census Bureau estimates, the population of the United States was 227.2 million on July 1, 1980, and 249.5 million on July 1, 1990. (A) Find the relative growth rate if we use the P 0003 P0ert model for population growth. Round to three significant digits. (B) Use your answer from part A to write a function describing the population of the United States in millions in terms of years after July 1980, and use it to predict when the population should reach 400 million. (C) Use your function from part B to estimate the population of the United States today, then compare your estimate to the one found at www.census.gov/population/www/popclockus.html. 75. WORLD POPULATION A mathematical model for world population growth over short periods is given by P 0003 P0ert where P is the population after t years, P0 is the population at t 0003 0, and the population is assumed to grow continuously at the annual rate r. How many years, to the nearest year, will it take the world population to double if it grows continuously at an annual rate of 1.14%? 76. WORLD POPULATION Refer to Problem 75. Starting with a world population of 6.8 billion people (the estimated population in March 2009) and assuming that the population grows continuously at an annual rate of 1.14%, how many years, to the nearest year, will it be before there is only 1 square yard of land per person? Earth contains approximately 1.7 1014 square yards of land. 77. MEDICAL RESEARCH A medical researcher is testing a radioactive isotope for use in a new imaging process. She finds that an original sample of 5 grams decays to 1 gram in 6 hours. Find the half-life of the sample to three significant digits. [Recall that the half-life model is A 0003 A0(12)t/h, where A0 is the original amount and h is the half-life.] 78. CARBON-14 DATING If 90% of a sample of carbon-14 remains after 866 years, what is the half-life of carbon-14? (See Problem 77 for the half-life model.) As long as a plant or animal remains alive, carbon-14 is maintained in a constant amount in its tissues. Once dead, however, the plant or animal ceases taking in carbon, and carbon-14 diminishes by radioactive decay. The amount remaining can be modeled by the equation A 0003 A0e00020.000124t, where A is the amount after t years, and A0 is the amount at time t 0003 0. Use this model to solve Problems 79–82. 79. CARBON-14 DATING In 2003, Japanese scientists announced the beginning of an effort to bring the long-extinct woolly mammoth back to life using modern cloning techniques. Their efforts were focused on an especially well-preserved specimen discovered frozen in the Siberian ice. Nearby samples of plant material were found to have 28.9% of the amount of carbon-14 in a living sample. What was the approximate age of these samples? 80. CARBON-14 DATING In 2004, archaeologist Al Goodyear discovered a site in South Carolina that contains evidence of the earliest human settlement in North America. Carbon dating of burned plant material indicated 0.2% of the amount of carbon-14 in a live sample. How old was that sample? 81. CARBON-14 DATING Many scholars believe that the earliest nonnative settlers of North America were Vikings who sailed from Iceland.
If a fragment of a wooden tool found and dated in 2004 had 88.3% of the amount of carbon-14 in a living sample, when was this tool made? 82. CARBON-14 DATING In 1998, the Shroud of Turin was examined by researchers, who found that plant fibers in the fabric had 92.1% of the amount of carbon-14 in a living sample. If this is accurate, when was the fabric made? 83. PHOTOGRAPHY An electronic flash unit for a camera is activated when a capacitor is discharged through a filament of wire. After the flash is triggered and the capacitor is discharged, the circuit (see the figure) is connected and the battery pack generates a current to recharge the capacitor. The time it takes for the capacitor to recharge is called the recycle time. For a particular flash unit using a 12-volt battery pack, the charge q, in coulombs, on the capacitor t seconds after recharging has started is given by q 0003 0.0009(1 0002 e00020.2t ) How many seconds will it take the capacitor to reach a charge of 0.0007 coulomb? Compute the answer to three significant digits. R I
V
C S
84. ADVERTISING A company is trying to expose as many people as possible to a new product through television advertising in a large metropolitan area with 2 million possible viewers. A model for the number of people N, in millions, who are aware of the product after t days of advertising was found to be N 0003 2(1 – e00020.037t ) How many days, to the nearest day, will the advertising campaign have to last so that 80% of the possible viewers will be aware of the product? 85. NEWTON’S LAW OF COOLING This law states that the rate at which an object cools is proportional to the difference in temperature between the object and its surrounding medium. The temperature T of the object t hours later is given by T 0003 Tm (T0 0002 Tm)e0002kt where Tm is the temperature of the surrounding medium and T0 is the temperature of the object at t 0003 0. Suppose a bottle of wine at a room temperature of 72°F is placed in a refrigerator at 40°F to cool before a dinner party. After an hour the temperature of the wine is found to be 61.5°F. Find the constant k, to two decimal places, and the time, to one decimal place, it will take the wine to cool from 72 to 50°F. 86. MARINE BIOLOGY Marine life is dependent upon the microscopic plant life that exists in the photic zone, a zone that goes to a depth where about 1% of the surface light still remains. Light intensity is reduced according to the exponential function I 0003 I0e0002kd where I is the intensity d feet below the surface and I0 is the intensity at the surface. The constant k is called the coefficient of extinction. At Crystal Lake in Wisconsin it was found that half the surface light remained at a depth of 14.3 feet. Find k, and find the depth of the photic zone. Compute answers to three significant digits.
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Problems 87–90 are based on the Richter scale equation from Section 5-4, M 0003 23 log 10E4.40, where M is the magnitude and E is the amount of energy in joules released by the earthquake. Round all calculations to three significant digits.
(B) The total average daily consumption of energy for the entire United States in 2006 was 2.88 1014 joules. How many days could the energy released by a magnitude 9.0 earthquake power the United States?
87. EARTHQUAKES There were 12 earthquakes recorded worldwide in 2008 with magnitude at least 7.0. (A) How much energy is released by a magnitude 7.0 earthquake? (B) The total average daily consumption of energy for the entire United States in 2006 was 2.88 1014 joules. How many days could the energy released by a magnitude 7.0 earthquake power the United States?
89. EARTHQUAKES There were 12 earthquakes worldwide in 2008 with magnitudes between 7.0 and 7.9. Assume that these earthquakes had an average magnitude of 7.5. How long could the total energy released by these 12 earthquakes power the United States, which had a total energy consumption of 1.05 1017 joules in 2006?
88. EARTHQUAKES On December 26, 2004, a magnitude 9.0 earthquake struck in the Indian Ocean, causing a massive tsunami that resulted in over 230,000 deaths. (A) How much energy was released by this earthquake?
5
CHAPTER
5-1
Review
Exponential Functions x
The equation f(x) 0003 b , b 0, b 0004 1, defines an exponential function with base b. The domain of f is (00020007, 0007) and the range is (0, 0007). The graph of f is a continuous curve that has no sharp corners; passes through (0, 1); lies above the x axis, which is a horizontal asymptote; increases as x increases if b 1; decreases as x increases if b 0006 1; and intersects any horizontal line at most once. The function f is one-to-one and has an inverse. We often use the following exponential function properties: 1. a xa y 0003 a x y a x ax a b 0003 x b b
90. EARTHQUAKES There were 166 earthquakes worldwide in 2008 with magnitudes between 6.0 and 6.9. Assume that these earthquakes had an average magnitude of 6.5. How long could the total energy released by these 166 earthquakes power the United States, which had a total energy consumption of 1.05 1017 joules in 2006?
(a x) y 0003 a xy
(ab)x 0003 a xb x
ax 0003 a x0002y ay
2. a x 0003 a y if and only if x 0003 y. 3. For x 0004 0, a x 0003 b x if and only if a 0003 b.
the time it takes for the population to double. Another model of population growth, A 0003 A0ekt, where A0 is the population at time zero and k is a positive constant called the relative growth rate, uses the exponential function with base e. This model is used for many other types of quantities that exhibit exponential growth as well. 2. Radioactive decay can be modeled by using the half-life decay model A 0003 A0(12)t0005h 0003 A020002t0005h, where A is the amount at time t, A0 is the amount at time t 0003 0, and h is the half-life—the time it takes for half the material to decay. Another model of radioactive decay, A 0003 A0e0002kt , where A0 is the amount at time zero and k is a positive constant, uses the exponential function with base e. This model can be used for other types of quantities that exhibit negative exponential growth as well. 3. Limited growth—the growth of a company or proficiency at learning a skill, for example—can often be modeled by the equation y 0003 A(1 0002 e0002kt ), where A and k are positive constants.
As x approaches 0007, the expression [1 (1兾x)]x approaches the irrational number e ⬇ 2.718 281 828 459. The function f(x) 0003 e x is called the exponential function with base e. The growth of money in an account paying compound interest is described by A 0003 P(1 r兾m)n, where P is the principal, r is the annual rate, m is the number of compounding periods in 1 year, and A is the amount in the account after n compounding periods. If the account pays continuous compound interest, the amount A in the account after t years is given by A 0003 Pert.
Logistic growth is another limited growth model that is useful for modeling phenomena like the spread of an epidemic, or sales of a new product. The logistic model is A 0003 M/(1 ce0002kt ), where c, k, and M are positive constants. A good comparison of these different exponential models can be found in Table 3 at the end of Section 5-2. Exponential regression can be used to fit a function of the form y 0003 ab x to a set of data points. Logistic regression can be used to find a function of the form y 0003 c0005(1 ae0002bx ).
5-2
5-3
Exponential Models
Exponential functions are used to model various types of growth: 1. Population growth can be modeled by using the doubling time growth model A 0003 A02t0005d, where A is the population at time t, A0 is the population at time t 0003 0, and d is the doubling time—
Logarithmic Functions
The logarithmic function with base b is defined to be the inverse of the exponential function with base b and is denoted by y 0003 logb x. So y 0003 logb x if and only if x 0003 b y, b 0, b 0004 1. The domain of a logarithmic function is (0, 0007) and the range is (00020007, 0007). The graph of a logarithmic function is a continuous curve that always passes
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through the point (1, 0) and has the y axis as a vertical asymptote. The following properties of logarithmic functions are useful: 1. logb 1 ⫽ 0 2. logb b ⫽ 1
3. The velocity v of a rocket at burnout is given by the rocket equation v ⫽ c ln (Wt兾Wb), where c is the exhaust velocity, Wt is the takeoff weight, and Wb is the burnout weight.
x
3. logb b ⫽ x 4. blogb x ⫽ x, x ⬎ 0
Logarithmic regression can be used to fit a function of the form y ⫽ a ⫹ b ln x to a set of data points.
5. logb MN ⫽ logb M ⫹ logb N 6. logb
M ⫽ logb M ⫺ logb N N
5-5
p
7. log b M ⫽ p log b M 8. logb M ⫽ logb N if and only if M ⫽ N Logarithms to the base 10 are called common logarithms and are denoted by log x. Logarithms to the base e are called natural logarithms and are denoted by ln x. So log x ⫽ y is equivalent to x ⫽ 10 y, and ln x ⫽ y is equivalent to x ⫽ e y. The change-of-base formula, logb N ⫽ (loga N)兾(loga b), relates logarithms to two different bases and can be used, along with a calculator, to evaluate logarithms to bases other than e or 10.
5-4
2. The magnitude M of an earthquake on the Richter scale is given by M ⫽ 23 log (EⲐE0), where E is the energy released by the earthquake and E0 ⫽ 104.40 joules is a standardized energy level.
Logarithmic Models
The following applications involve logarithmic functions:
Exponential and Logarithmic Equations
Exponential equations are equations in which the variable appears in an exponent. If the exponential expression is isolated, applying a logarithmic function to both sides and using the property logb N p ⫽ p logb N will enable you to remove the variable from the exponent. If the exponential expression is not isolated, we can use previously developed techniques to first solve for the exponential, then solve as above. Logarithmic equations are equations in which the variable appears inside a logarithmic function. In most cases, the key to solving them is to change the equation to the equivalent exponential expression. For equations with multiple log expressions, properties of logarithms can be used to combine the expressions before solving.
1. The decibel is defined by D ⫽ 10 log (I兾I0), where D is the decibel level of the sound, I is the intensity of the sound, and I0 ⫽ 10⫺12 watts per square meter is a standardized sound level.
CHAPTER
5
Review Exercises
Work through all the problems in this chapter review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. 1. Match each equation with the graph of f, g, m, or n in the figure. (A) y ⫽ log2 x (B) y ⫽ 0.5x (C) y ⫽ log0.5 x (D) y ⫽ 2x f
3
g m
⫺4.5
4.5
n ⫺3
3. Write in logarithmic form using base e: x ⫽ e y. Write the expression in Problems 4 and 5 in exponential form. 4. log x ⫽ y
5. ln y ⫽ x
6. (A) Plot at least five points, then draw a hand sketch of the graph of y ⫽ (43)x. (B) Use your result from part A to sketch the graph of y ⫽ log4Ⲑ3 x. In Problems 7 and 8, simplify. 7.
7x⫹2 72⫺x
8. a
ex x b e ⫺x
In Problems 9–11, solve for x exactly. n
2. Write in logarithmic form using base 10: m ⫽ 10 .
9. log2 x ⫽ 3
10. logx 25 ⫽ 2
11. log3 27 ⫽ x
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In Problems 12–15, solve for x to three significant digits. x
In Problems 54–57, use transformations to explain how the graph of g is related to the graph of the given logarithmic function f. Determine whether g is increasing or decreasing, find its domain and any asymptotes, and sketch the graph of g.
x
12. 10 0003 17.5
13. e 0003 143,000
14. ln x 0003 00020.015 73
15. log x 0003 2.013
381
54. g (x) 0003 3 0002 13 2x; f (x) 0003 2x Evaluate the expression in Problems 16–19 to four significant digits using a calculator. 16. ln
17. log (0002e)
18.
20. Write as a single log: 2 log a 0002
58. If the graph of y 0003 e x is reflected in the line y 0003 x, the graph of the function y 0003 ln x is obtained. Discuss the functions that are obtained by reflecting the graph of y 0003 e x in the x axis and the y axis.
1 log b log c 3 a5 1b
21. Write in terms of ln a and ln b: ln
59. (A) Explain why the equation e0002x兾3 0003 4 ln (x 1) has exactly one solution. (B) Find the solution of the equation to three decimal places.
In Problems 22–35, solve for x exactly. x
2x
22. 3 0003 120
23. 10
24. log2 (4x 0002 5) 0003 5
25. ln (x 0002 5) 0003 0
60. Approximate all real zeros of f(x) 0003 4 0002 x2 ln x to three decimal places.
0003 500
61. Find the coordinates of the points of intersection of f(x) 0003 10x00023 and g(x) 0003 8 log x to three decimal places.
26. ln (2x 0002 1) 0003 ln (x 3)
In Problems 62–65, solve for the indicated variable in terms of the remaining symbols.
27. log (x2 0002 3) 0003 2 log (x 0002 1) 2
28. e x 00023 0003 e2x
29. 4x00021 0003 210002x
30. 2x2e0002x 0003 18e0002x
31. log1兾4 16 0003 x
32. log x 9 0003 00022
33. log16 x 0003 32
34. log x e5 0003 5
35. 10log10x 0003 33
62. D 0003 10 log 63. y 0003
38. ln x 0003 00023.218
I for I (sound intensity) I0
2 1 e 0002x 00052 for x (probability) 12
1 I 64. x 0003 0002 ln for I (X-ray intensity) k I0
In Problems 36–45, solve for x to three significant digits. 36. x 0003 2(101.32)
56. g(x) 0003 00022 log4 x; f(x) 0003 log4 x 57. g(x) 0003 1 2 log1兾3 x; f(x) 0003 log1兾3 x
e e0002 19. 2
ln 2
55. g(x) 0003 2e x 0002 4; f(x) 0003 e x
37. x 0003 log5 23 00027
39. x 0003 log (2.156 10 )
65. r 0003 P
i for n (finance) 1 0002 (1 i) 0002n
41. 25 0003 5(2x)
66. Write ln y 0003 00025t ln c in an exponential form free of logarithms; then solve for y in terms of the remaining symbols.
42. 4,000 0003 2,500(e0.12x)
43. 0.01 0003 e00020.05x
67. For f 0003 5(x, y) 冟 y 0003 log2 x6, graph f and f 00021 on the same coordinate system. What are the domains and ranges for f and f 00021?
44. 52x00023 0003 7.08
45.
40. x 0003
ln 4 ln 2.31
ex 0002 e 0002x 00031 2
In Problems 46–51, solve for x exactly. 46. log 3x 2 0002 log 9x 0003 2 47. log x 0002 log 3 0003 log 4 0002 log (x 4) 48. ln (x 3) 0002 ln x 0003 2 ln 2 49. ln (2x 1) 0002 ln (x 0002 1) 0003 ln x 50. (log x)3 0003 log x9
51. ln (log x) 0003 1
In Problems 52 and 53, simplify. 52. (e x 1)(e0002x 0002 1) 0002 e x(e0002x 0002 1) 53. (e x e0002x)(e x 0002 e0002x) 0002 (e x 0002 e0002x)2
68. Explain why 1 cannot be used as a logarithmic base. 69. Prove that logb (MN) 0003 logb M logb N.
APPLICATIONS 70. POPULATION GROWTH Many countries have a population growth rate of 3% (or more) per year. At this rate, how many years will it take a population to double? Use the annual compounding growth model P 0003 P0(1 r)t. Compute the answer to three significant digits. 71. POPULATION GROWTH Repeat Problem 70 using the continuous compounding growth model P 0003 P0e rt. 72. CARBON 14-DATING How many years will it take for carbon-14 to diminish to 1% of the original amount after the death of a plant or animal? Use the formula A 0003 A0e00020.000124t. Compute the answer to three significant digits.
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73. MEDICINE One leukemic cell injected into a healthy mouse will divide into two cells in about 12 day. At the end of the day these two cells will divide into four. This doubling continues until 1 billion cells are formed; then the animal dies with leukemic cells in every part of the body. (A) Write an equation that will give the number N of leukemic cells at the end of t days. (B) When, to the nearest day, will the mouse die? 74. MONEY GROWTH Assume $1 had been invested at an annual rate of 3% compounded continuously in the year A.D. 1. What would be the value of the account in the year 2011? Compute the answer to two significant digits.
Problems 81 and 82 require a graphing calculator or a computer that can calculate exponential, logarithmic, and logistic regression models for a given data set. 81. MEDICARE The annual expenditures for Medicare (in billions of dollars) by the U.S. government for selected years since 1980 are shown in Table 1. Let x represent years since 1980. (A) Find an exponential regression model of the form y 0003 abx for these data. Round to three significant digits. Estimate (to the nearest billion) the total expenditures in 2010 and in 2020. (B) When (to the nearest year) will the total expenditures reach $900 billion?
75. PRESENT VALUE Solving A 0003 Pert for P, we obtain P 0003 Ae0002rt, which is the present value of the amount A due in t years if money is invested at a rate r compounded continuously. (A) Graph P 0003 1,000(e00020.08t ), 0 t 30. (B) What does it appear that P tends to as t tends to infinity? [Conclusion: The longer the time until the amount A is due, the smaller its present value, as we would expect.]
Table 1 Medicare Expenditures
76. EARTHQUAKES The 1971 San Fernando, California, earthquake released 1.99 1014 joules of energy. Compute its magnitude on the Richter scale using the formula M 0003 23 log (E0005E0), where E0 0003 104.40 joules. Compute the answer to one decimal place. 77. EARTHQUAKES Refer to Problem 76. If the 1906 San Francisco earthquake had a magnitude of 8.3 on the Richter scale, how much energy was released? Compute the answer to three significant digits. 78. SOUND If the intensity of a sound from one source is 100,000 times that of another, how much more is the decibel level of the louder sound than the softer one? Use the formula D 0003 10 log (I兾I0). 79. MARINE BIOLOGY The intensity of light entering water is reduced according to the exponential function I 0003 I0e0002kd where I is the intensity d feet below the surface, I0 is the intensity at the surface, and k is the coefficient of extinction. Measurements in the Sargasso Sea in the West Indies have indicated that half the surface light reaches a depth of 73.6 feet. Find k, and find the depth at which 1% of the surface light remains. Compute answers to three significant digits. 80. WILDLIFE MANAGEMENT A lake formed by a newly constructed dam is stocked with 1,000 fish. Their population is expected to increase according to the logistic curve N0003
30 1 29e00021.35t
where N is the number of fish, in thousands, expected after t years. The lake will be open to fishing when the number of fish reaches 20,000. How many years, to the nearest year, will this take?
Year
Billion $
1980
37
1985
72
1990
111
1995
181
2000
225
2005
342
Source: U.S. Bureau of the Census
82. Table 2 lists the number of cell phone subscribers in the United States for selected years from 1994 to 2006. Let x 0003 0 correspond to 1990 and round all coefficients to four significant digits. (A) Find a logarithmic regression model of the form y 0003 a b ln x for the data, then use the model to predict the number of subscribers in 2015. (B) Repeat part A, this time finding a logistic regression model of the form y 0003 c0005(1 ae 0002bx). (C) Which of the models do you think models the data better? Explain. Consider how well it fits the points from the table, as well as how well you think it predicts long-term trends.
Table 2 Cell Phone Subscribers in the U.S. Year
Subscribers in millions
1994
24.13
1997
55.31
2000
109.5
2003
158.8
2006
233.0
Source: CTIA—The Wireless Association
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Group Activity
CHAPTER
ZZZ
5
GROUP ACTIVITY Comparing Regression Models
We have used polynomial, exponential, and logarithmic regression models to fit curves to data sets. How can we determine which equation provides the best fit for a given set of data? There are two principal ways to select models. The first is to use information about the type of data to help make a choice. For example, we expect the weight of a fish to be related to the cube of its length. And we expect most populations to grow exponentially, at least over the short term. The second method for choosing among equations involves developing a measure of how closely an equation fits a given data set. This is best introduced through an example. Consider the data set in Figure 1, where L1 represents the x coordinates and L2 represents the y coordinates. The graph of this data set is shown in Figure 2. Suppose we arbitrarily choose the equation y1 ⫽ 0.6x ⫹ 1 to model these data (Fig. 3). 10
0
10
0
Z Figure 1
Z Figure 2 10
0
10
0
Z Figure 3 y1 ⫽ 0.6x ⫹ 1. To measure how well the graph of y1 fits these data, we examine the difference between the y coordinates in the data set and the corresponding y coordinates on the graph of y1 (L3 in Figs. 4 and 5). 10
0
10
0
Z Figure 4
Z Figure 5 Here ⫹ is L2 and 䊐 is L3.
Each of these differences is called a residual. Note that three of the residuals are positive and one is negative (three of the points lie above the line, one lies below). The most commonly accepted measure of the fit provided by a given model is the sum of the squares of the residuals (SSR). When squared, each residual (whether positive or negative or zero) makes a nonnegative contribution to the SSR. SSR ⫽ (4 ⫺ 2.2)2 ⫹ (5 ⫺ 3.4)2 ⫹ (3 ⫺ 4.6)2 ⫹ (7 ⫺ 5.8)2 ⫽ 9.8 (A) A linear regression model for the data in Figure 1 is given by y2 ⫽ 0.35x ⫹ 3 Compute the SSR for the data and y2, and compare it to the one we computed for y1. It turns out that among all possible linear polynomials, the linear regression model minimizes the sum of the squares of the residuals. For this reason, the linear regression model is often called the least-squares line. A similar statement can be made for polynomials of any fixed degree. That is, the quadratic regression model minimizes the SSR over all quadratic polynomials, the cubic regression model minimizes the SSR over all cubic polynomials, and so on. The same statement cannot be made for exponential or logarithmic regression models. Nevertheless, the SSR can still be used to compare exponential, logarithmic, and polynomial models. (B) Find the exponential and logarithmic regression models for the data in Figure 1, compute their SSRs, and compare with the linear model. (C) National annual advertising expenditures for selected years since 1950 are shown in Table 1 where x is years since 1950 and y is total expenditures in billions of dollars. Which regression model would fit this data best: a quadratic model, a cubic model, or an exponential model? Use the SSRs to support your choice.
Table 1 Annual Advertising Expenditures, 1950–2000 x (years)
0
10
20
30
y (billion $)
5.7
12.0
19.6
53.6
Source: U.S. Bureau of the Census.
40
50
128.6
247.5
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CHAPTER
Additional Topics in Analytic Geometry
6
C
OUTLINE
ANALYTIC geometry is the study of geometric objects using algebraic
techniques. René Descartes (1596–1650), the French philosopher and mathematician, is generally recognized as the founder of the subject. We used analytic geometry in Chapter 2 to obtain equations of lines and circles. In Chapter 6, we take a similar approach to the study of parabolas, ellipses, and hyperbolas. Each of these geometric objects is a conic section, that is, the intersection of a plane and a cone. We will derive equations for the conic sections and explore a wealth of applications in architecture, communications, engineering, medicine, optics, and space science.
6-1
Conic Sections; Parabola
6-2
Ellipse
6-3
Hyperbola Chapter 6 Review Chapter 6 Group Activity: Focal Chords
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ADDITIONAL TOPICS IN ANALYTIC GEOMETRY
Conic Sections; Parabola Z Conic Sections Z Definition of a Parabola Z Drawing a Parabola Z Standard Equations and Their Graphs Z Applications
In Section 6-1 we introduce the general concept of a conic section and then discuss the particular conic section called a parabola. In Sections 6-2 and 6-3 we will discuss two other conic sections called ellipses and hyperbolas.
Z Conic Sections In Section 2-3 we found that the graph of a first-degree equation in two variables, Ax 0002 By 0003 C
(1)
where A and B are not both 0, is a straight line, and every straight line in a rectangular coordinate system has an equation of this form. What kind of graph will a second-degree equation in two variables, Ax2 0002 Bxy 0002 Cy2 0002 Dx 0002 Ey 0002 F 0003 0
(2)
where A, B, and C are not all 0, yield for different sets of values of the coefficients? The graphs of equation (2) for various choices of the coefficients are plane curves obtainable by intersecting a cone* with a plane, as shown in Figure 1. These curves are called conic sections. Z Figure 1 Conic sections.
L Circle
Constant V Nappe
Ellipse
Parabola
Hyperbola
If a plane cuts clear through one nappe, then the intersection curve is called a circle if the plane is perpendicular to the axis and an ellipse if the plane is not perpendicular to the axis. If a plane cuts only one nappe, but does not cut clear through, then the intersection curve is called *Starting with a fixed line L and a fixed point V on L, the surface formed by all straight lines through V making a constant angle 0004 with L is called a right circular cone. The fixed line L is called the axis of the cone, and V is its vertex. The two parts of the cone separated by the vertex are called nappes.
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Conic Sections; Parabola
387
a parabola. Finally, if a plane cuts through both nappes, but not through the vertex, the resulting intersection curve is called a hyperbola. A plane passing through the vertex of the cone produces a degenerate conic—a point, a line, or a pair of lines. Conic sections are very useful and are readily observed in your immediate surroundings: wheels (circle), the path of water from a garden hose (parabola), some serving platters (ellipses), and the shadow on a wall from a light surrounded by a cylindrical or conical lamp shade (hyperbola) are some examples (Fig. 2). We will discuss many applications of conics throughout the remainder of this chapter. Z Figure 2 Examples of conics.
Wheel (circle) (a)
Water from garden hose (parabola) (b)
Lamp light shadow (hyperbola) (d)
Serving platter (ellipse) (c)
Z Definition of a Parabola The following definition of a parabola is a coordinate-free definition. It does not depend on the coordinates of points in any coordinate system. Z DEFINITION 1 Parabola A parabola is the set of all points in a plane equidistant from a fixed point F and a fixed line L (not containing F) in the plane. The fixed point F is called the focus, and the fixed line L is called the directrix. A line through the focus perpendicular to the directrix is called the axis of symmetry, and the point on the axis of symmetry halfway between the directrix and focus is called the vertex.
L
d1
P
d1 0003 d2 Axis of symmetry
d2 V (Vertex)
F (Focus) Parabola Directrix
Z Drawing a Parabola Using Definition 1, we can draw a parabola with fairly simple equipment—a straightedge, a right-angle drawing triangle, a piece of string, a thumbtack, and a pencil. Referring to Figure 3 on the next page, tape the straightedge along the line AB and place the thumbtack above the line AB. Place one leg of the triangle along the straightedge as indicated, then take a piece of string the same length as the other leg, tie one end to the thumbtack, and fasten the other end with tape at C on the triangle. Now press the string to the edge of the triangle, and keeping the string taut, slide the triangle along the straightedge. Because DE will always equal DF, the resulting curve will be part of a parabola with directrix AB lying along the straightedge and focus F at the thumbtack.
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Z Figure 3 Drawing a parabola.
C
D F E A
ZZZ EXPLORE-DISCUSS 1
B
The line through the focus F that is perpendicular to the axis of symmetry of a parabola intersects the parabola in two points G and H. Explain why the distance from G to H is twice the distance from F to the directrix of the parabola.
Z Standard Equations and Their Graphs Using the definition of a parabola and the distance formula d 0003 2(x2 0007 x1)2 0002 ( y2 0007 y1)2
(3)
we can derive simple standard equations for a parabola located in a rectangular coordinate system with its vertex at the origin and its axis of symmetry along a coordinate axis. We start with the axis of symmetry of the parabola along the x axis and the focus at F 0003 (a, 0). We locate the parabola in a coordinate system as in Figure 4 and label key lines and points. This is an important step in finding an equation of a geometric figure in a coordinate system. Note that the parabola opens to the right if a 0005 0 and to the left if a 0006 0. The vertex is at the origin, the directrix is x 0003 0007a, and the coordinates of M are (0007a, y). y
Z Figure 4 Parabola with vertex at the origin and axis of symmetry the x axis. M 0003 (0007a, y)
0007a
y d1
P 0003 (x, y)
d2 Focus F 0003 (a, 0)
Directrix x 0003 0007a
a ⬎ 0, focus on positive x axis (a)
x
d1 P 0003 (x, y) d2 Focus F 0003 (a, 0)
M 0003 (0007a, y)
0007a
x
Directrix x 0003 0007a
a ⬍ 0, focus on negative x axis (b)
The point P 0003 (x, y) is a point on the parabola if and only if d1 0003 d2 d(P, M ) 0003 d(P, F) 2 2(x 0002 a) 0002 ( y 0007 y)2 0003 2(x 0007 a)2 0002 ( y 0007 0)2 (x 0002 a)2 0003 (x 0007 a)2 0002 y2 2 x 0002 2ax 0002 a2 0003 x2 0007 2ax 0002 a2 0002 y2 y2 ⴝ 4ax
Use equation (3). Square both sides. Simplify.
(4)
Equation (4) is the standard equation of a parabola with vertex at the origin, axis of symmetry the x axis, and focus at (a, 0).
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By a similar derivation (see Problem 57 in Exercises 6-1), the standard equation of a parabola with vertex at the origin, axis of symmetry the y axis, and focus at (0, a) is given by equation (5). x2 0003 4ay
(5)
Looking at Figure 5, note that the parabola opens upward if a 0005 0 and downward if a 0006 0. y
Z Figure 5 Parabola with vertex at the origin and axis of symmetry the y axis.
y Directrix y 0003 0007a
N 0003 (x, 0007a) 0007a F 0003 (0, a) Focus
d2
P 0003 (x, y) d1
Directrix y 0003 0007a
0007a
x
d1 d2
P 0003 (x, y)
x
F 0003 (0, a) Focus
N 0003 (x, 0007a)
a ⬎ 0, focus on positive y axis (a)
a ⬍ 0, focus on negative y axis (b)
We summarize these results for easy reference in Theorem 1. Z THEOREM 1 Standard Equations of a Parabola with Vertex at (0, 0) 1. y2 0003 4ax Vertex: (0, 0) Focus: (a, 0) Directrix: x 0003 0007a Symmetric with respect to the x axis Axis of symmetry the x axis 2. x2 0003 4ay Vertex: (0, 0) Focus: (0, a) Directrix: y 0003 0007a Symmetric with respect to the y axis Axis of symmetry the y axis
EXAMPLE
1
y
F 0
y
x
F
x
0
a ⬍ 0 (opens left)
a ⬎ 0 (opens right)
y
y
0
F
x
F 0
a ⬍ 0 (opens down)
x
a ⬎ 0 (opens up)
Graphing a Parabola Locate the focus and directrix and sketch the graph of y2 0003 16x.
SOLUTION
The equation y2 0003 16x has the form y2 0003 4ax with 4a 0003 16, so a 0003 4. Therefore, the focus is (4, 0) and the directrix is the line x 0003 00074. To sketch the graph, we choose some values of x that make the right side of the equation a perfect square and solve for y. x
0
1
4
y
0
4
8
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Note that x must be greater than or equal to 0 for y to be a real number. Then we plot the resulting points. Because a ⬎ 0, the parabola opens to the right (Fig. 6). y 10
Directrix x ⫽ ⫺4
Focus F ⫽ (4, 0)
⫺10
10
x
⫺10
Z Figure 6
Technology Connections To graph y 2 ⴝ 16x on a graphing calculator, we solve the equation for y.
y ⴝ 16x 2
Directrix x ⫽ ⫺4
10
Take square roots.
y ⴝ ⴞ41x
⫺10
This results in two functions, y ⴝ 4 1x and y ⴝ ⴚ4 1x. Entering these functions in a graphing calculator (Fig. 7) and graphing in a standard viewing window produces the graph of the parabola (Fig. 8).
10
⫺10
Z Figure 8
Z Figure 7
Focus F ⫽ (4, 0)
0002 MATCHED PROBLEM 1
ZZZ
A common error in making a quick sketch of y2 ⫽ 4ax or x2 ⫽ 4ay is to sketch the first with the y axis as its axis of symmetry and the second with the x axis as its axis of symmetry. The graph of y2 ⫽ 4ax is symmetric with respect to the x axis, and the graph of x2 ⫽ 4ay is symmetric with respect to the y axis, as a quick symmetry check will reveal.
CAUTION ZZZ
EXAMPLE
Graph y2 ⫽ ⫺8x, and locate the focus and directrix.
2
Finding the Equation of a Parabola (A) Find the equation of a parabola having the origin as its vertex, the y axis as its axis of symmetry, and (⫺10, ⫺5) on its graph. (B) Find the coordinates of its focus and the equation of its directrix.
SOLUTIONS
(A) Because the axis of symmetry of the parabola is the y axis, the parabola has an equation of the form x2 ⫽ 4ay. Because (⫺10, ⫺5) is on the graph, we have x2 ⫽ 4ay (⫺10)2 ⫽ 4a(⫺5) 100 ⫽ ⫺20a a ⫽ ⫺5
Substitute x ⴝ ⴚ10 and y ⴝ ⴚ5. Simplify. Divide both sides by ⴚ20.
0002
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Therefore, the equation of the parabola is x2 0003 4(00075)y x2 0003 000720y (B) Focus: F 0003 (0, a) 0003 (0, 00075) Directrix: y 0003 0007a y00035 MATCHED PROBLEM 2
0002
(A) Find the equation of a parabola having the origin as its vertex, the x axis as its axis of symmetry, and (4, 00078) on its graph. (B) Find the coordinates of its focus and the equation of its directrix. 0002
Z Applications If you are observant, you will find many applications of parabolas in the physical world. Parabolas are key to the design of suspension bridges, arch bridges, microphones, symphony shells, satellite antennas, radio and optical telescopes, radar equipment, solar furnaces, and searchlights. Figure 9(a) illustrates a parabolic reflector used in all reflecting telescopes—from 3- to 6-inch home types to the 200-inch research instrument on Mount Palomar in California. Parallel light rays from distant celestial bodies are reflected to the focus off a parabolic mirror. If the light source is the sun, then the parallel rays are focused at F and we have a solar furnace. Temperatures of over 6,000 C have been achieved by such furnaces. If we locate a light source at F, then the rays in Figure 9(a) reverse, and we have a spotlight or a searchlight. Automobile headlights can use parabolic reflectors with special lenses over the light to diffuse the rays into useful patterns. Figure 9(b) shows a suspension bridge, such as the Golden Gate Bridge in San Francisco. The suspension cable is a parabola. It is interesting to note that a free-hanging cable, such as a telephone line, does not form a parabola. It forms another curve called a catenary. Figure 9(c) shows a concrete arch bridge. If all the loads on the arch are to be compression loads (concrete works very well under compression), then using physics and advanced mathematics, it can be shown that the arch must be parabolic.
Parallel light rays
Parabola
F Parabola
Parabolic reflector
Suspension bridge
Arch bridge
(a)
(b)
(c)
Z Figure 9 Uses of parabolic forms.
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3
Parabolic Reflector A paraboloid is formed by revolving a parabola about its axis of symmetry. A spotlight in the form of a paraboloid 5 inches deep has its focus 2 inches from the vertex. Find, to one decimal place, the radius R of the opening of the spotlight.
SOLUTION
Step 1. Locate a parabolic cross section containing the axis of symmetry in a rectangular coordinate system, and label all known parts and parts to be found. This is a very important step and can be done in infinitely many ways. We can make things simpler for ourselves by locating the vertex at the origin and choosing a coordinate axis as the axis of symmetry. We choose the y axis as the axis of symmetry of the parabola with the parabola opening upward (Fig. 10). y 5
(R, 5) R F 0003 (0, 2) Spotlight
00075
x
5
Z Figure 10
Step 2. Find the equation of the parabola in the figure. Because the parabola has the y axis as its axis of symmetry and the vertex at the origin, the equation is of the form x2 0003 4ay We are given F 0003 (0, a) 0003 (0, 2); so a 0003 2, and the equation of the parabola is x2 0003 8y Step 3. Use the equation found in step 2 to find the radius R of the opening. Because (R, 5) is on the parabola, we have R2 0003 8(5) R 0003 140 ⬇ 6.3 inches MATCHED PROBLEM 3
0002
Repeat Example 3 with a paraboloid 12 inches deep and a focus 9 inches from the vertex. 0002
ANSWERS TO MATCHED PROBLEMS y
1. Focus: (00072, 0) Directrix: x 0003 2 x
0
00072
y
0
4
5
(00072, 0) 00075
F
Directrix x00032 5
x
00075
2. (A) y2 0003 16x (B) Focus: (4, 0); Directrix: x 0003 00074 3. R 0003 20.8 inches
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6-1
Conic Sections; Parabola
393
Exercises
1. List the seven different types of conic sections. 2. Explain how each of the seven types of conic sections can be obtained as the intersection of a cone and a plane. 3. What is a degenerate conic? 4. Give a coordinate-free definition of a parabola in your own words. 5. What happens to light rays that are parallel to the axis of a parabolic mirror when they hit the mirror? 6. What happens to light rays that are emitted from the focus of a parabolic mirror when they hit the mirror? In Problems 7–10, a parabola has its vertex at the origin and the given directrix. Find the coordinates of the focus. 7. x 0003 8 9. y 0003 000710
8. x 0003 00075 10. y 0003 6
In Problems 11–14, a parabola has its vertex at the origin and the given focus. Find the equation of the directrix. 11. (0, 000715)
12. (0, 9)
13. (25, 0)
14. (000721, 0)
In Problems 15–24, graph each equation, and locate the focus and directrix. 15. y2 0003 4x
16. y2 0003 8x
17. x2 0003 8y
18. x2 0003 4y
19. y2 0003 000712x
20. y2 0003 00074x
21. x2 0003 00074y
22. x2 0003 00078y
23. y2 0003 000720x
24. x2 0003 000724y
38. Focus (00074, 0)
37. Focus (2, 0)
In Problems 39–44, find the equation of the parabola having its vertex at the origin, its axis of symmetry as indicated, and passing through the indicated point. 39. y axis; (4, 2)
40. x axis; (4, 8)
41. x axis; (00073, 6)
42. y axis; (00075, 10)
43. y axis; (00076, 00079)
44. x axis; (00076, 000712)
In Problems 45–48, find the first-quadrant points of intersection for each pair of parabolas to three decimal places. 45. x2 0003 4y y2 0003 4x
46. y2 0003 3x x2 0003 3y
47. y2 0003 6x x2 0003 5y
48. x2 0003 7y y2 0003 2x
49. Consider the parabola with equation x2 0003 4ay. (A) How many lines through (0, 0) intersect the parabola in exactly one point? Find their equations. (B) Find the coordinates of all points of intersection of the parabola with the line through (0, 0) having slope m 0. 50. Find the coordinates of all points of intersection of the parabola with equation x2 0003 4ay and the parabola with equation y2 0003 4bx. 51. The line segment AB through the focus in the figure is called a focal chord of the parabola. Find the coordinates of A and B. y F 0003 (0, a) A
B
In Problems 25–30, find the coordinates to two decimal places of the focus of the parabola. 25. y2 0003 39x
26. x2 0003 58y
27. x2 0003 0007105y
28. y2 0003 000793x
29. y2 0003 000777x
30. x2 0003 0007205y
32. Directrix y 0003 4
33. Focus (0, 00077)
34. Focus (0, 5)
35. Directrix x 0003 6
36. Directrix x 0003 00079
x
0
52. The line segment AB through the focus in the figure is called a focal chord of the parabola. Find the coordinates of A and B. y
y 2 0003 4ax B
In Problems 31–38, find the equation of a parabola with vertex at the origin, axis of symmetry the x or y axis, and 31. Directrix y 0003 00073
x 2 0003 4ay
F 0003 (a, 0) 0
A
x
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In Problems 53–56, use the definition of a parabola and the distance formula to find the equation of a parabola with 53. Directrix y 0003 00074 and focus (2, 2) 54. Directrix y 0003 2 and focus (00073, 6)
(A) Find the equation of the parabola after inserting an xy coordinate system with the vertex at the origin and the y axis (pointing upward) the axis of symmetry of the parabola. (B) How far is the focus from the vertex? 61. SPACE SCIENCE A designer of a 200-foot-diameter parabolic electromagnetic antenna for tracking space probes wants to place the focus 100 feet above the vertex (see the figure).
55. Directrix x 0003 2 and focus (6, 00074) 56. Directrix x 0003 00073 and focus (1, 4) 57. Use the definition of a parabola and the distance formula to derive the equation of a parabola with focus F 0003 (0, a) and directrix y 0003 0007a for a 0. 58. Let F be a fixed point and let L be a fixed line in the plane that contains F. Describe the set of all points in the plane that are equidistant from F and L.
200 ft Focus
APPLICATIONS 59. ENGINEERING The parabolic arch in the concrete bridge in the figure must have a clearance of 50 feet above the water and span a distance of 200 feet. Find the equation of the parabola after inserting a coordinate system with the origin at the vertex of the parabola and the vertical y axis (pointing upward) along the axis of symmetry of the parabola.
100 ft Radio telescope
(A) Find the equation of the parabola using the axis of symmetry of the parabola as the y axis (up positive) and vertex at the origin. (B) Determine the depth of the parabolic reflector. 62. SIGNAL LIGHT A signal light on a ship is a spotlight with parallel reflected light rays (see the figure). Suppose the parabolic reflector is 12 inches in diameter and the light source is located at the focus, which is 1.5 inches from the vertex. 60. ASTRONOMY The cross section of a parabolic reflector with 6-inch diameter is ground so that its vertex is 0.15 inch below the rim (see the figure).
Signal light
Focus 6 inches
0.15 inch
Parabolic reflector
(A) Find the equation of the parabola using the axis of symmetry of the parabola as the x axis (right positive) and vertex at the origin. (B) Determine the depth of the parabolic reflector.
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Ellipse
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Ellipse Z Definition of an Ellipse Z Drawing an Ellipse Z Standard Equations of Ellipses and Their Graphs Z Applications
We start our discussion of the ellipse with a coordinate-free definition. Using this definition, we show how an ellipse can be drawn and we derive standard equations for ellipses specially located in a rectangular coordinate system.
Z Definition of an Ellipse The following is a coordinate-free definition of an ellipse:
Z DEFINITION 1 Ellipse An ellipse is the set of all points P in a plane d1 0002 d2 0003 Constant such that the sum of the distances from P to B two fixed points in the plane is a constant (the V d1 P constant is required to be greater than the F distance between the two fixed points). Each of d2 the fixed points, F and F, is called a focus, and together they are called foci. Referring to F V the figure, the line segment V V through the B foci is the major axis. The perpendicular bisector B B of the major axis is the minor axis. Each end of the major axis, V and V, is called a vertex. The midpoint of the line segment F F is called the center of the ellipse.
Z Drawing an Ellipse An ellipse is easy to draw. All you need is a piece of string, two thumbtacks, and a pencil or pen (see Figure 1 on the next page). Place the two thumbtacks in a piece of cardboard. These form the foci of the ellipse. Take a piece of string longer than the distance between the two thumbtacks—this represents the constant in the definition—and tie each end to a thumbtack. Finally, catch the tip of a pencil under the string and move it while keeping the string taut. The resulting figure is by definition an ellipse. Ellipses of different shapes result, depending on the placement of thumbtacks and the length of the string joining them.
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Note that d1 0006 d2 always adds up to the length of the string, which does not change.
P d2
d1 Focus
String
Focus
Z Figure 1 Drawing an ellipse.
Z Standard Equations of Ellipses and Their Graphs Using the definition of an ellipse and the distance formula, we can derive standard equations for an ellipse located in a rectangular coordinate system. We start by placing an ellipse in the coordinate system with the foci on the x axis at F0002 0003 (0004c, 0) and F 0003 (c, 0) with c 0005 0 (Fig. 2). By definition 1 the constant sum d1 0006 d2 is required to be greater than 2c (the distance between F and F0002). Therefore, the ellipse intersects the x axis at points V0002 0003 (0004a, 0) and V 0003 (a, 0) with a 0005 c 0005 0, and it intersects the y axis at points B0002 0003 (0004b, 0) and B 0003 (b, 0) with b 0005 0.
y b
P 0003 (x, y)
d1 0004a
d2
F0002 0003 (0004c, 0)
0
F 0003 (c, 0) a
x
0004b d1 0006 d2 0003 Constant 0005 d(F, F0002) c00050
Z Figure 2 Ellipse with foci on x axis.
Study Figure 2: Note first that if P 0003 (a, 0), then d1 0006 d2 0003 2a. (Why?) Therefore, the constant sum d1 0006 d2 is equal to the distance between the vertices. Second, if P 0003 (0, b), then d1 0003 d2 0003 a and a2 0003 b2 0006 c2 by the Pythagorean theorem; in particular, a 0005 b. Referring again to Figure 2, the point P 0003 (x, y) is on the ellipse if and only if d1 0006 d2 0003 2a Using the distance formula for d1 and d2, eliminating radicals, and simplifying (see Problem 49 in Exercises 6-2), we obtain the equation of the ellipse pictured in Figure 2: y2 x2 2 ⴙ 2 ⴝ 1 a b By similar reasoning (see Problem 50 in Exercises 6-2) we obtain the equation of an ellipse centered at the origin with foci on the y axis. Both cases are summarized in Theorem 1.
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Ellipse
Z THEOREM 1 Standard Equations of an Ellipse with Center at (0, 0) 1.
y2 x2 a0005b00050 0002 00031 a2 b2 x intercepts: a (vertices) y intercepts: b Foci: F 0003 (0007c, 0), F 0003 (c, 0)
y b
0007a
c2 0003 a2 0007 b2
F 0007c
F c
0
Major axis length 0003 2a Minor axis length 0003 2b y2 x2 2. 2 0002 2 0003 1 a0005b00050 b a x intercepts: b y intercepts: a (vertices) Foci: F 0003 (0, 0007c), F 0003 (0, c)
a
x
0007b y a c F
c2 0003 a2 0007 b2 Major axis length 0003 2a Minor axis length 0003 2b [Note: Both graphs are symmetric with respect to the x axis, y axis, and origin. Also, the major axis is always longer than the minor axis.]
ZZZ EXPLORE-DISCUSS 1
EXAMPLE
1
0
0007b
b
x
0007c F 0007a
The line through a focus F of an ellipse that is perpendicular to the major axis intersects the ellipse in two points G and H. For each of the two standard equations of an ellipse with center (0, 0), find an expression in terms of a and b for the distance from G to H.
Graphing an Ellipse Find the coordinates of the foci, find the lengths of the major and minor axes, and graph the following equation: 9x2 0002 16y2 0003 144
SOLUTION
First, write the equation in standard form by dividing both sides by 144 and determine a and b: 9x2 0002 16y2 0003 144 16y2 9x2 144 0002 0003 144 144 144
Divide both sides by 144. * Simplify.
y2 x2 0002 00031 16 9 a00034
and
b00033
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
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y
x intercepts: 4 y intercepts: 3
3
F 0007c
8:46 PM
0
F c
4
Major axis length: 2(4) 0003 8 Minor axis length: 2(3) 0003 6 Foci: c2 0003 a2 0007 b2 0003 16 0007 9 00037
x
c 0003 17
00073
Substitute a ⴝ 4 and b ⴝ 3.
c must be positive.
So the foci are F¿ 0003 (0007 17, 0) and F 0003 (17, 0). Plot the foci and intercepts and sketch the ellipse (Fig. 3).
Z Figure 3
MATCHED PROBLEM 1
Find the coordinates of the foci, find the lengths of the major and minor axes, and graph the following equation: x2 0002 4y2 0003 4
EXAMPLE
2
0002
0002
Graphing an Ellipse Find the coordinates of the foci, find the lengths of the major and minor axes, and graph the following equation: 2x2 0002 y2 0003 10
SOLUTION
First, write the equation in standard form by dividing both sides by 10 and determine a and b: 2x2 0002 y2 0003 10 y2 2x2 10 0002 0003 10 10 10
Divide both sides by 10.
Simplify.
y2 x2 0002 00031 5 10 a 0003 110 y intercepts: 110 ⬇ 3.16 x intercepts: 15 ⬇ 2.24
and
b 0003 15
Major axis length: 2110 ⬇ 6.32 Minor axis length: 215 ⬇ 4.47
Foci: c2 0003 a2 0007 b2 0003 10 0007 5 00035 c 0003 15
Substitute a ⴝ 110, b ⴝ 15.
c must be positive.
So the foci are F¿ 0003 (0, 0007 15) and F 0003 (0, 15). Plot the foci and intercepts and sketch the ellipse (Fig. 4). y 兹10
c F 0
0007兹5
0007c F 0007兹10
Z Figure 4
兹5
x
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Ellipse
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Technology Connections To graph the ellipse of Example 2 on a graphing calculator, solve the original equation for y:
2x 2 ⴙ y 2 ⴝ 10
near the x intercepts; they are due to the relatively low resolution of the graphing calculator screen. 4
Subtract 2x2 from both sides.
y 2 ⴝ 10 ⴚ 2x 2 y ⴝ ⴞ210 ⴚ 2x
Take square roots of both sides. 2
00076
This produces two functions, y1 ⴝ 210 ⴚ 2x 2 and y2 ⴝ ⴚ 210 ⴚ 2x 2, which are graphed in Figure 5. Notice that we used a squared viewing window to avoid distorting the shape of the ellipse. Also note the gaps in the graph
6
00074
Z Figure 5
0002 MATCHED PROBLEM 2
Find the coordinates of the foci, find the lengths of the major and minor axes, and graph the following equation: 3x2 0002 y2 0003 18
EXAMPLE
3
0002
Finding the Equation of an Ellipse Find an equation of an ellipse in the form y2 x2 0002 00031 M N
M, N 7 0
if the center is at the origin, the major axis is along the y axis, and (A) Length of major axis 0003 20 Length of minor axis 0003 12 SOLUTIONS
(B) Length of major axis 0003 10 Distance of foci from center 0003 4
(A) Compute x and y intercepts and make a rough sketch of the ellipse, as shown in Figure 6. y2 x2 0002 00031 b2 a2 20 a0003 0003 10 2 y2 x2 0002 00031 36 100 y 10
000710
10
000710
Z Figure 6
x
b0003
12 00036 2
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(B) Make a rough sketch of the ellipse, as shown in Figure 7; locate the foci and y intercepts, then determine the x intercepts using the fact that a2 0003 b 2 0002 c2: y2 x2 0002 00031 b2 a2 a0003
10 00035 2
b2 0003 52 0007 42 0003 25 0007 16 0003 9 b00033
y2 x2 0002 00031 9 25 y 5 4
0
0007b
b
x
00075
0002
Z Figure 7
MATCHED PROBLEM 3
Find an equation of an ellipse in the form y2 x2 0002 00031 M N
M, N 7 0
if the center is at the origin, the major axis is along the x axis, and (A) Length of major axis 0003 50 Length of minor axis 0003 30
(B) Length of minor axis 0003 16 Distance of foci from center 0003 6
0002
Z Applications Ellipses have many applications: orbits of satellites, planets, and comets; shapes of galaxies; gears and cams, some airplane wings, boat keels, and rudders; tabletops; public fountains; and domes in buildings are a few examples (Fig. 8).
Planet Sun
F
F
Planetary motion
Elliptical gears
Elliptical dome
(a)
(b)
(c)
Z Figure 8 Uses of elliptical forms.
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Ellipse
401
Johannes Kepler (1571–1630), a German astronomer, discovered that planets move in elliptical orbits, with the sun at a focus, and not in circular orbits as had been thought before [Fig. 8(a)]. Figure 8(b) shows a pair of elliptical gears with pivot points at foci. Such gears transfer constant rotational speed to variable rotational speed, and vice versa. Figure 8(c) shows an elliptical dome. An interesting property of such a dome is that a sound or light source at one focus will reflect off the dome and pass through the other focus. One of the chambers in the Capitol Building in Washington, D.C., has such a dome, and is referred to as a whispering room because a whispered sound at one focus can be easily heard at the other focus. A fairly recent application in medicine is the use of elliptical reflectors and ultrasound to break up kidney stones. A device called a lithotripter is used to generate intense sound waves that break up the stone from outside the body, eliminating the need for surgery. To be certain that the waves do not damage other parts of the body, the reflecting property of the ellipse is used to design and correctly position the lithotripter.
EXAMPLE
4
Medicinal Lithotripsy A lithotripter is formed by rotating the portion of an ellipse below the minor axis around the major axis (Fig. 9). The lithotripter is 20 centimeters wide and 16 centimeters deep. If the ultrasound source is positioned at one focus of the ellipse and the kidney stone at the other, then all the sound waves will pass through the kidney stone. How far from the kidney stone should the point V on the base of the lithotripter be positioned to focus the sound waves on the kidney stone? Round the answer to one decimal place. Kidney stone
Ultrasound source
Base
20 cm
V
16 cm
Z Figure 9 Lithotripter.
SOLUTION
From Figure 9 we see that a 0003 16 and b 0003 10 for the ellipse used to form the lithotripter. So the distance c from the center to either the kidney stone or the ultrasound source is given by c 0003 2a2 0007 b2 0003 2162 0007 102 0003 2156 ⬇ 12.5 and the distance from the base of the lithotripter to the kidney stone is 16 0002 12.5 0003 28.5 centimeters. 0002
MATCHED PROBLEM 4
Because lithotripsy is an external procedure, the lithotripter described in Example 4 can be used only on stones within 12.5 centimeters of the surface of the body. Suppose a kidney stone is located 14 centimeters from the surface. If the diameter is kept fixed at 20 centimeters, how deep must a lithotripter be to focus on this kidney stone? Round answer to one decimal place. 0002
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ANSWERS TO MATCHED PROBLEMS 1.
y 1
Foci: F 0003 (0007兹3, 0), F 0003 (兹3, 0) Major axis length 0003 4 Minor axis length 0003 2
F
F 0
00072
2
x
00071
y
2. 兹18
F
0007兹6
Foci: F 0003 (0, 0007兹12), F 0003 (0, 兹12) Major axis length 0003 2兹18 ⬇ 8.49 Minor axis length 0003 2兹6 ⬇ 4.90 兹6
x
F 0007兹18
3. (A)
6-2
y2 x2 0002 00031 625 225
(B)
y2 x2 0002 00031 100 64
4. 17.2 centimeters
Exercises
1. Give a coordinate-free definition of an ellipse in your own words. 2. Explain how the major axis of an ellipse differs from the minor axis. 3. Given the major axis of an ellipse and the foci, describe a procedure for drawing the ellipse. 4. Is the graph of an ellipse the graph of a function? Explain. 5. Is a circle an ellipse? Explain. 6. Using the definition of an ellipse, explain why the minor axis is shorter than the major axis. In Problems 7–10, find the distance between the foci of the ellipse.
In Problems 11–14, find the length of the major axis of the ellipse. 11. Distance between foci 0003 14 Minor axis length 0003 48 12. Distance between foci 0003 10 Minor axis length 0003 1 13. Distance between foci 0003 5 Minor axis length 0003 5 14. Distance between foci 0003 3 Minor axis length 0003 313 In Problems 15–20, sketch a graph of each equation, find the coordinates of the foci, and find the lengths of the major and minor axes.
7. Major axis length 0003 10 Minor axis length 0003 8
15.
y2 x2 0002 00031 25 4
16.
8. Major axis length 0003 26 Minor axis length 0003 10
18.
y2 x2 0002 00031 4 9
19. x2 0002 9y2 0003 9
9. Major axis length 0003 2 Minor axis length 0003 1
In Problems 21–24, match each equation with one of graphs (a)–(d).
10. Major axis length 0003 4 Minor axis length 0003 3
21. 9x2 0002 16y2 0003 144
y2 x2 0002 00031 9 4
17.
y2 x2 0002 00031 4 25
20. 4x2 0002 y2 0003 4
22. 16x2 0002 9y2 0003 144
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23. 4x2 0002 y2 0003 16 y
y
5
403
y
33. The graph is
24. x2 0002 4y2 0003 16
Ellipse
10
5
000710 00075
5
x
00075
5
10
x
x 000710
00075
(b)
y
y
5
y
34. The graph is
00075
(a)
10
5 000710
00075
5
x
00075
00075
5
In Problems 25–30, sketch a graph of each equation, find the coordinates of the foci, and find the lengths of the major and minor axes. 25. 25x2 0002 9y2 0003 225
26. 16x2 0002 25y2 0003 400
27. 2x2 0002 y2 0003 12
28. 4x2 0002 3y2 0003 24
29. 4x2 0002 7y2 0003 28
30. 3x2 0002 2y2 0003 24
In Problems 31–42, find an equation of an ellipse in the form M, N 7 0
if the center is at the origin, and
10
000710
10
x
000710
10
000710
38. Major axis on y axis Major axis length 0003 24 Minor axis length 0003 18 39. Major axis on x axis Major axis length 0003 16 Distance of foci from center 0003 6
41. Major axis on y axis Minor axis length 0003 20 Distance of foci from center 0003 170 42. Major axis on x axis Minor axis length 0003 14 Distance of foci from center 0003 1200
44. Consider all ellipses having (0, 1) as the ends of the minor axis. Describe the connection between the elongation of the ellipse and the distance from a focus to the origin.
10
000710
37. Major axis on y axis Major axis length 0003 22 Minor axis length 0003 16
43. Explain why an equation whose graph is an ellipse does not define a function.
y
32. The graph is
36. Major axis on x axis Major axis length 0003 14 Minor axis length 0003 10
40. Major axis on y axis Major axis length 0003 24 Distance of foci from center 0003 10
y
31. The graph is
000710
35. Major axis on x axis Major axis length 0003 10 Minor axis length 0003 6
(d)
y2 x2 0002 00031 M N
x
x
00075
(c)
10
x
45. Find an equation of the set of points in a plane, each of whose distance from (2, 0) is one-half its distance from the line x 0003 8. Identify the geometric figure. 46. Find an equation of the set of points in a plane, each of whose distance from (0, 9) is three-fourths its distance from the line y 0003 16. Identify the geometric figure.
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47. Let F and F⬘ be two points in the plane and let c denote the constant d(F, F⬘). Describe the set of all points P in the plane such that the sum of the distances from P to F and F⬘ is equal to the constant c. 48. Let F and F⬘ be two points in the plane and let c be a constant such that 0 ⬍ c ⬍ d(F, F⬘). Describe the set of all points P in the plane such that the sum of the distances from P to F and F⬘ is equal to the constant c. 49. Study the following derivation of the standard equation of an ellipse with foci (⫾c, 0), x intercepts (⫾a, 0), and y intercepts (0, ⫾b). Explain why each equation follows from the equation that precedes it. [Hint: Recall from Figure 2 on page 396 that a2 ⫽ b2 ⫹ c2.] d1 ⫹ d2 ⫽ 2a 2(x ⫹ c)2 ⫹ y2 ⫽ 2a ⫺ 2(x ⫺ c)2 ⫹ y2 (x ⫹ c)2 ⫹ y2 ⫽ 4a2 ⫺ 4a 2(x ⫺ c)2 ⫹ y2 ⫹ (x ⫺ c)2 ⫹ y2 cx 2(x ⫺ c)2 ⫹ y2 ⫽ a ⫺ a (x ⫺ c)2 ⫹ y2 ⫽ a2 ⫺ 2cx ⫹ a1 ⫺
c2x2 a2
c2 2 b x ⫹ y2 ⫽ a2 ⫺ c2 a2 y2 x2 ⫹ 2⫽1 2 a b
Elliptical bridge
52. DESIGN A 4 ⫻ 8 foot elliptical tabletop is to be cut out of a 4 ⫻ 8 foot rectangular sheet of teak plywood (see the figure). To draw the ellipse on the plywood, how far should the foci be located from each edge and how long a piece of string must be fastened to each focus to produce the ellipse (see Fig. 1 on page 396)? Compute the answer to two decimal places.
String
F⬘
F Elliptical table
50. Study the following derivation of the standard equation of an ellipse with foci (0, ⫾c), y intercepts (0, ⫾a), and x intercepts (⫾b, 0). Explain why each equation follows from the equation that precedes it. [Hint: Recall from Figure 2 on page 396 that a2 ⫽ b2 ⫹ c2.]
53. AERONAUTICAL ENGINEERING Of all possible wing shapes, it has been determined that the one with the least drag along the trailing edge is an ellipse. The leading edge may be a straight line, as shown in the figure. One of the most famous planes with this design was the World War II British Spitfire. The plane in the figure has a wingspan of 48.0 feet. Leading edge
d1 ⫹ d2 ⫽ 2a 2
2x ⫹ ( y ⫹ c)2 ⫽ 2a ⫺ 2x2 ⫹ ( y ⫺ c)2 x2 ⫹ ( y ⫹ c)2 ⫽ 4a2 ⫺ 4a 2x2 ⫹ ( y ⫺ c)2 ⫹ x2 ⫹ ( y ⫺ c)2 cy 2x2 ⫹ ( y ⫺ c)2 ⫽ a ⫺ a x2 ⫹ ( y ⫺ c)2 ⫽ a2 ⫺ 2cy ⫹
c2y2
Trailing edge
Elliptical wings and tail
a2
2
c b y2 ⫽ a2 ⫺ c2 a2 y2 x2 ⫹ ⫽1 b2 a2
x2 ⫹ a1 ⫺
Fuselage
APPLICATIONS 51. ENGINEERING The semielliptical arch in the concrete bridge in the figure must have a clearance of 12 feet above the water and span a distance of 40 feet. Find the equation of the ellipse after inserting a coordinate system with the center of the ellipse at the origin and the major axis on the x axis. The y axis points up, and the x axis points to the right. How much clearance above the water is there 5 feet from the bank?
(A) If the straight-line leading edge is parallel to the major axis of the ellipse and is 1.14 feet in front of it, and if the leading edge is 46.0 feet long (including the width of the fuselage), find the equation of the ellipse. Let the x axis lie along the major axis (positive right), and let the y axis lie along the minor axis (positive forward). (B) How wide is the wing in the center of the fuselage (assuming the wing passes through the fuselage)? Compute quantities to three significant digits. 54. NAVAL ARCHITECTURE Currently, many high-performance racing sailboats use elliptical keels, rudders, and main sails for the reasons stated in Problem 53—less drag along the trailing edge. In the accompanying figure, the ellipse containing the keel has a 12.0foot major axis. The straight-line leading edge is parallel to the ma-
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SECTION 6–3
jor axis of the ellipse and 1.00 foot in front of it. The chord is 1.00 foot shorter than the major axis.
Hyperbola
405
(A) Find the equation of the ellipse. Let the y axis lie along the minor axis of the ellipse, and let the x axis lie along the major axis, both with positive direction upward. (B) What is the width of the keel, measured perpendicular to the major axis, 1 foot up the major axis from the bottom end of the keel? Compute quantities to three significant digits.
Rudder
6-3
Keel
Hyperbola Z Definition of a Hyperbola Z Drawing a Hyperbola Z Standard Equations and Their Graphs Z Applications
As before, we start with a coordinate-free definition of a hyperbola. Using this definition, we show how a hyperbola can be drawn and we derive standard equations for hyperbolas specially located in a rectangular coordinate system.
Z Definition of a Hyperbola The following is a coordinate-free definition of a hyperbola: Z DEFINITION 1 Hyperbola A hyperbola is the set of all points P in a plane 兩d 0007 d 兩 0003 Constant 1 2 such that the absolute value of the difference of the distances from P to two fixed points in the P d2 plane is a positive constant (the constant is d1 required to be less than the distance between the F V two fixed points). Each of the fixed points, F¿ and V F F, is called a focus. The intersection points V¿ and V of the line through the foci and the two branches of the hyperbola are called vertices, and each is called a vertex. The line segment V¿V is called the transverse axis. The midpoint of the transverse axis is the center of the hyperbola.
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Z Drawing a Hyperbola Thumbtacks, a straightedge, string, and a pencil are all that are needed to draw a hyperbola (Fig. 1). Place two thumbtacks in a piece of cardboard—these form the foci of the hyperbola. Rest one corner of the straightedge at the focus F¿ so that it is free to rotate about this point. Cut a piece of string shorter than the length of the straightedge, and fasten one end to the straightedge corner A and the other end to the thumbtack at F. Now push the string with a pencil up against the straightedge at B. Keeping the string taut, rotate the straightedge about F¿, keeping the corner at F¿. The resulting curve will be part of a hyperbola. Other parts of the hyperbola can be drawn by changing the position of the straightedge and string. To see that the resulting curve meets the conditions of the definition, note that the difference of the distances BF¿ and BF is BF¿ 0007 BF 0003 BF¿ 0002 BA 0007 BF 0007 BA 0003 AF¿ 0007 (BF 0002 BA) Straightedge String 0003a b0007a b length length 0003 Constant
Z Figure 1 Drawing a hyperbola.
A B
String
F
F
Z Standard Equations of Hyperbolas and Their Graphs Using the definition of a hyperbola and the distance formula, we can derive standard equations for a hyperbola located in a rectangular coordinate system. We start by placing a hyperbola in the coordinate system with the foci on the x axis at F 0003 (0007c, 0) and F 0003 (c, 0) with
y
Z Figure 2 Hyperbola with foci on the x axis.
P 0003 (x, y) d2 x a F 0003 (c, 0)
d1 F 0003 (0007c, 0) 0007a
c00050 兩d1 0007 d2 兩 0003 Positive constant 0006 d(F, F )
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Hyperbola
407
c 0005 0 (Fig. 2). By definition 1, the constant difference | d1 0007 d2 | is required to be less than 2c (the distance between F and F ). Therefore, the hyperbola intersects the x axis at points V 0003 (0007a, 0) and V 0003 (a, 0) with c 0005 a 0005 0. The hyperbola does not intersect the y axis, because the constant difference | d1 0007 d2 | is required to be positive by definition 1. Study Figure 2: Note that if P 0003 (a, 0), then | d1 0007 d2 | 0003 2a. (Why?) Therefore, the constant |d1 0007 d2 | is equal to the distance between the vertices. It is convenient to let b 0003 2c2 0007 a2, so that c2 0003 a2 0002 b2. (Unlike the situation for ellipses, b may be greater than or equal to a.) Referring again to Figure 2, the point P 0003 (x, y) is on the hyperbola if and only if |d1 0007 d2 | 0003 2a Using the distance formula for d1 and d2, eliminating radicals, and simplifying (see Problem 57 in Exercises 6-3), we obtain the equation of the hyperbola pictured in Figure 2: y2 x2 ⴚ ⴝ1 a2 b2 Although the hyperbola does not intersect the y axis, the points (0, b) and (0, 0007b) are significant; the line segment joining them is called the conjugate axis of the hyperbola. Note that the conjugate axis is perpendicular to the transverse axis, that is, the line segment joining the vertices (a, 0) and (0007a, 0). The rectangle with corners (a, b), (a, 0007b), (0007a, 0007b), and (0007a, b) is called the asymptote rectangle because its extended diagonals are asymptotes for the hyperbola (Fig. 3). In other words, the hyperbola approaches the lines y 0003 ba x as | x| becomes larger (see Problems 53 and 54 in Exercises 6-3). As a result, it is helpful to include the asymptote rectangle and its extended diagonals when sketching the graph of a hyperbola.
Asymptote b y00030007 x a
Asymptote b y0003 x a
y
x2
b
a2 0007a
0
a
0007 x
y2 b2
00031
0007b
Z Figure 3 Asymptotes.
Note that the four corners of the asymptote rectangle (Fig. 3) are equidistant from the origin, at distance 2a2 0002 b2 0003 c. Therefore, A circle, with center at the origin, that passes through all four corners of the asymptote rectangle of a hyperbola also passes through its foci. By similar reasoning (see Problem 58 in Exercises 6-3) we obtain the equation of a hyperbola centered at the origin with foci on the y axis. Both cases are summarized in Theorem 1.
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Z THEOREM 1 Standard Equations of a Hyperbola with Center at (0, 0) 1.
y2 x2 0007 00031 a2 b2 x intercepts: a (vertices) y intercepts: none Foci: F 0003 (0007c, 0), F 0003 (c, 0)
y
b
F
0007c
c2 0003 a2 0002 b2
2
a
F
c
x
0007b
Transverse axis length 0003 2a Conjugate axis length 0003 2b b Asymptotes: y 0003 x a y2 x2 2. 2 0007 2 0003 1 a b x intercepts: none y intercepts: a (vertices) Foci: F 0003 (0, 0007c), F 0003 (0, c) 2
c
0007a
y c
F
a c 0007b
2
c 0003a 0002b
b
x
0007a 0007c
Transverse axis length 0003 2a Conjugate axis length 0003 2b a Asymptotes: y 0003 x b
F
[Note: Both graphs are symmetric with respect to the x axis, y axis, and origin.]
ZZZ EXPLORE-DISCUSS 1
EXAMPLE
1
The line through a focus F of a hyperbola that is perpendicular to the transverse axis intersects the hyperbola in two points G and H. For each of the two standard equations of a hyperbola with center (0, 0), find an expression in terms of a and b for the distance from G to H.
Graphing Hyperbolas Find the coordinates of the foci, find the lengths of the transverse and conjugate axes, find the equations of the asymptotes, and graph the following equation: 9x2 0007 16y2 0003 144
SOLUTION
First, write the equation in standard form by dividing both sides by 144 and determine a and b: 9x2 0007 16y2 0003 144 16y2 9x2 144 0007 0003 144 144 144 y2 x2 0007 00031 16 9 a00034 and
Divide both sides by 144.
Simplify.
b00033
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x intercepts: 4 y intercepts: none
Hyperbola
409
Transverse axis length 0003 2(4) 0003 8 Conjugate axis length 0003 2(3) 0003 6 Foci: c2 0003 a2 0002 b2 0003 16 0002 9 0003 25 c00035
Substitute a 0003 4 and b 0003 3.
So the foci are F 0003 (00075, 0) and F 0003 (5, 0). Plot the foci and x intercepts, sketch the asymptote rectangle and the asymptotes, then sketch the hyperbola (Fig. 4). The equations of the asymptotes are y 0003 34 x (note that the diagonals of the asymptote rectangle have slope 34). y 5
c
0007c 00076
F
c F
6
x
00075
0002
Z Figure 4
MATCHED PROBLEM 1
Find the coordinates of the foci, find the lengths of the transverse and conjugate axes, and graph the following equation: 16x2 0007 25y2 0003 400
EXAMPLE
2
0002
Graphing Hyperbolas Find the coordinates of the foci, find the lengths of the transverse and conjugate axes, find the equations of the asymptotes, and graph the following equation: 16y2 0007 9x2 0003 144
SOLUTION
Write the equation in standard form: 16y2 0007 9x2 0003 144 y2 x2 0007 00031 9 16 a00033 and y intercepts: 3 x intercepts: none
Divide both sides by 144.
b00034
Transverse axis length 0003 2(3) 0003 6 Conjugate axis length 0003 2(4) 0003 8 Foci: c2 0003 a2 0002 b2 0003 9 0002 16 0003 25 c00035
Substitute a ⴝ 3 and b ⴝ 4.
So the foci are F 0003 (0, 00075) and F 0003 (0, 5). Plot the foci and y intercepts, sketch the asymptote rectangle and the asymptotes, then sketch the hyperbola (Fig. 5). The equations of the asymptotes are y 0003 34 x (note that the diagonals of the asymptote rectangle have slope 34).
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c
F c
00076
6
0007c
00076
x
F
0002
Z Figure 5
MATCHED PROBLEM 2
Find the coordinates of the foci, find the lengths of the transverse and conjugate axes, and graph the following equation: 25y2 0007 16x2 0003 400
0002
Two hyperbolas of the form y2 x2 0007 00031 M N
and
y2 x2 0007 00031 N M
M, N 7 0
are called conjugate hyperbolas. In Examples 1 and 2 and in Matched Problems 1 and 2, the hyperbolas are conjugate hyperbolas—they share the same asymptotes.
ZZZ
When making a quick sketch of a hyperbola, it is a common error to have the hyperbola opening up and down when it should open left and right, or vice versa. The mistake can be avoided if you first locate the intercepts accurately.
CAUTION ZZZ
EXAMPLE
3
Graphing Hyperbolas Find the coordinates of the foci, find the lengths of the transverse and conjugate axes, and graph the following equation: 2x2 0007 y2 0003 10 2x2 0007 y2 0003 10 y2 x2 0007 00031 5 10 a 0003 15 and
SOLUTION
x intercepts: 15 y intercepts: none
y 5
c 0007c
c
F
F
00075
00075
Z Figure 6
5
x
Divide both sides by 10.
b 0003 110
Transverse axis length 0003 215 ⬇ 4.47 Conjugate axis length 0003 2110 ⬇ 6.32 Foci: c2 0003 0003 0003 c0003
a2 0002 b2 5 0002 10 15 115
Substitute a ⴝ 15 and b ⴝ 110.
So the foci are F¿ 0003 (0007 115, 0) and F 0003 ( 115, 0). Plot the foci and x intercepts, sketch the asymptote rectangle and the asymptotes, then sketch the hyperbola (Fig. 6). 0002
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MATCHED PROBLEM 3
Hyperbola
411
Find the coordinates of the foci, find the lengths of the transverse and conjugate axes, and graph the following equation: y2 0007 3x2 0003 12
EXAMPLE
4
0002
Finding the Equation of a Hyperbola Find an equation of a hyperbola in the form y2 x2 0007 00031 M N
M, N 7 0
if the center is at the origin, and: (A) Length of transverse axis is 12 Length of conjugate axis is 20 SOLUTIONS
(B) Length of transverse axis is 6 Distance of foci from center is 5
(A) Start with y2 x2 0007 00031 a2 b2 and find a and b: a0003
12 00036 2
and
b0003
20 0003 10 2
So the equation is y2 x2 0007 00031 36 100 (B) Start with y2 x2 2 0007 2 0003 1 a b and find a and b: a0003
To find b, sketch the asymptote rectangle (Fig. 7), label known parts, and use the Pythagorean theorem:
y 5
6 00033 2
F 5
0007b
b2 0003 52 0007 32 0003 16 b00034
3 b
x
So the equation is 00075
F
Z Figure 7 Asymptote rectangle.
y2 x2 0007 00031 9 16
0002
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MATCHED PROBLEM 4
Find an equation of a hyperbola in the form y2 x2 0007 00031 M N
M, N 7 0
if the center is at the origin, and: (A) Length of transverse axis is 50 Length of conjugate axis is 30
(B) Length of conjugate axis is 12 Distance of foci from center is 9 0002
ZZZ EXPLORE-DISCUSS 2
(A) Does the line with equation y 0003 x intersect the hyperbola with equation x2 0007 (y2兾4) 0003 1? If so, find the coordinates of all intersection points. (B) Does the line with equation y 0003 3x intersect the hyperbola with equation x2 0007 ( y2兾4) 0003 1? If so, find the coordinates of all intersection points. (C) For which values of m does the line with equation y 0003 mx intersect the hypery2 x2 bola 2 0007 2 0003 1? Find the coordinates of all intersection points. a b
Z Applications You may not be aware of the many important uses of hyperbolic forms. They are encountered in the study of comets; the loran system of navigation for pleasure boats, ships, and aircraft; sundials; capillary action; nuclear reactor cooling towers; optical and radio telescopes; and contemporary architectural structures. The TWA building at Kennedy Airport is a hyperbolic paraboloid, and the St. Louis Science Center Planetarium is a hyperboloid. With such structures, thin concrete shells can span large spaces [Fig. 8(a)]. Some comets from outer space occasionally enter the sun’s gravitational field, follow a hyperbolic path around the sun (with the sun as a focus), and then leave, never to be seen again [Fig. 8(b)]. Example 5 illustrates the use of hyperbolas in navigation.
Z Figure 8 Uses of hyperbolic forms. Comet Sun
EXAMPLE
5
St. Louis Planetarium
Comet around sun
(a)
(b)
Navigation A ship is traveling on a course parallel to and 60 miles from a straight shoreline. Two transmitting stations, S1 and S2, are located 200 miles apart on the shoreline (Fig. 9). By timing radio signals from the stations, the ship’s navigator determines that the ship is between the two stations and 50 miles closer to S2 than to S1. Find the distance from the ship to each station. Round answers to one decimal place.
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SECTION 6–3
d1
60 miles
Hyperbola
413
d2
S2
S1
200 miles
Z Figure 9 d1 0007 d2 0003 50. SOLUTION
If d1 and d2 are the distances from the ship to S1 and S2, respectively, then d1 0007 d2 0003 50 and the ship must be on the hyperbola with foci at S1 and S2 and fixed difference 50, as illustrated in Figure 10. In the derivation of the equation of a hyperbola, we represented the fixed difference as 2a. So for the hyperbola in Figure 10 we have c 0003 100 a 0003 12 (50) 0003 25 b 0003 21002 0007 252 0003 29,375 y 200
S1
(x, 60) S2
0007100
100
x
Z Figure 10
The equation for this hyperbola is y2 x2 0007 00031 625 9,375 Substitute y 0003 60 and solve for x (see Fig. 10): x2 602 0007 00031 625 9,375 3,600 x2 0003 00021 625 9,375 3,600 0002 9,375 x2 0003 625 9,375 0003 865
Add
602 to both sides. 9,375
Multiply both sides by 625.
Simplify.
So x 0003 1865 ⬇ 29.41 (The negative square root is discarded, because the ship is closer to S2 than to S1.) Distance from ship to S1 2
Distance from ship to S2 2
d1 0003 2(29.41 0002 100) 0002 60 0003 120,346.9841 ⬇ 142.6 miles
d2 0003 2(29.41 0007 100)2 0002 602 0003 18,582.9841 ⬇ 92.6 miles
Notice that the difference between these two distances is 50, as it should be. MATCHED PROBLEM 5
Repeat Example 5 if the ship is 80 miles closer to S2 than to S1.
0002
0002
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Example 5 illustrates a simplified form of the loran (LOng RAnge Navigation) system. In practice, three transmitting stations are used to send out signals simultaneously (Fig. 11), instead of the two used in Example 5. A computer onboard a ship will record these signals and use them to determine the differences of the distances that the ship is to S1 and S2, and to S2 and S3. Plotting all points so that these distances remain constant produces two branches, p1 and p2, of a hyperbola with foci S1 and S2, and two branches, q1 and q2, of a hyperbola with foci S2 and S3. It is easy to tell which branches the ship is on by comparing the signals from each station. The intersection of a branch of each hyperbola locates the ship and the computer expresses this in terms of longitude and latitude.
Ship S3 q2
S1
S2
p1
q1
p2
Z Figure 11 Loran navigation.
ANSWERS TO MATCHED PROBLEMS 1.
y
y2 x2 0007 00031 25 16 Foci: F 0003 (0007兹41, 0), F 0003 (兹41, 0) Transverse axis length 0003 10 Conjugate axis length 0003 8
10
c
F 000710 0007c
F c
x
10
000710
2.
y2 x2 0007 00031 16 25 Foci: F 0003 (0, 0007兹41), F 0003 (0, 兹41) Transverse axis length 0003 8 Conjugate axis length 0003 10
y 10
c
F c
000710
x
10
0007c F 000710
y
3.
y2 x2 0007 00031 12 4 Foci: F 0003 (0, 00074), F 0003 (0, 4) Transverse axis length 0003 2兹12 ⬇ 6.93 Conjugate axis length 0003 4
6
c F c 00075
5
x
0007c F 00076
4. (A)
y2 x2 0007 00031 625 225
(B)
y2 x2 0007 00031 45 36
5. d1 0003 159.5 miles, d2 0003 79.5 miles
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SECTION 6–3
6-3
Hyperbola
Exercises Sketch a graph of each equation in Problems 15–26, find the coordinates of the foci, and find the lengths of the transverse and conjugate axes.
1. Give a coordinate-free definition of a hyperbola in your own words. 2. Explain how the transverse axis of a hyperbola differs from the conjugate axis. 3. Given the transverse axis and foci of a hyperbola, describe a procedure for drawing the hyperbola. 4. Is the graph of a hyperbola the graph of a function? Explain.
15.
y2 x2 0007 00031 9 4
16.
y2 x2 0007 00031 9 25
17.
y2 x2 0007 00031 4 9
18.
y2 x2 0007 00031 25 9
19. 4x2 0007 y2 0003 16
5. Is the conjugate axis of a hyperbola always shorter then the transverse axis? Explain.
20. x2 0007 9y2 0003 9
21. 9y2 0007 16x2 0003 144
6. Explain what an asymptote rectangle is, and how it is related to the graph of a hyperbola.
22. 4y2 0007 25x2 0003 100 23. 3x2 0007 2y2 0003 12
In Problems 7–10, find the distance between the foci of the hyperbola. 7. Transverse axis length = 24 Conjugate axis length = 18
24. 3x2 0007 4y2 0003 24 25. 7y2 0007 4x2 0003 28 26. 3y2 0007 2x2 0003 24
8. Transverse axis length = 25 Conjugate axis length = 60
In Problems 27–38, find an equation of a hyperbola in the form
9. Transverse axis length = 1 Conjugate axis length = 3
y2 x2 0007 00031 M N
10. Transverse axis length = 7 Conjugate axis length = 1
y2 x2 0007 00031 N M
or
11. x2 0007 y2 0003 1
12. y2 0007 x2 0003 1
13. y2 0007 x2 0003 4
14. x2 0007 y2 0003 4
27. The graph is y 10
y
y
(5, 4)
5
5
000710
5
x
M, N 7 0
if the center is at the origin, and:
In Problems 11–14, match each equation with one of graphs (a)–(d).
00075
00075
5
x
00075
00075
(b)
y
y
5
10
x
000710
28. The graph is
(a)
y 10
5
(4, 5) 00075
5
00075
x 00075
5
00075
(c)
415
(d)
x
000710
10
000710
x
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29. The graph is 41. y
y2 x2 0007 00031 4 16
43. 9x2 0007 y2 0003 9
10
42.
y2 x2 0007 00031 9 25
44. x2 0007 4y2 0003 4
45. 2y2 0007 3x2 0003 1 (3, 5) 000710
10
x
000710
30. The graph is
46. 5y2 0007 6x2 0003 1 47. (A) How many hyperbolas have center at (0, 0) and a focus at (1, 0)? Find their equations. (B) How many ellipses have center at (0, 0) and a focus at (1, 0)? Find their equations. (C) How many parabolas have center at (0, 0) and focus at (1, 0)? Find their equations. 48. How many hyperbolas have the lines y 0003 2x as asymptotes? Find their equations.
y
49. Find all intersection points of the graph of the hyperbola x2 0007 y2 0003 1 with the graph of each of the following lines: (A) y 0003 0.5x (B) y 0003 2x
10
(5, 3) 000710
10
x
000710
31. Transverse axis on x axis Transverse axis length 0003 14 Conjugate axis length 0003 10 32. Transverse axis on x axis Transverse axis length 0003 8 Conjugate axis length 0003 6 33. Transverse axis on y axis Transverse axis length 0003 24 Conjugate axis length 0003 18 34. Transverse axis on y axis Transverse axis length 0003 16 Conjugate axis length 0003 22
For what values of m will the graph of the hyperbola and the graph of the line y 0003 mx intersect? Find the coordinates of these intersection points. 50. Find all intersection points of the graph of the hyperbola y2 0007 x2 0003 1 with the graph of each of the following lines: (A) y 0003 0.5x (B) y 0003 2x For what values of m will the graph of the hyperbola and the graph of the line y 0003 mx intersect? Find the coordinates of these intersection points. 51. Find all intersection points of the graph of the hyperbola y2 0007 4x2 0003 1 with the graph of each of the following lines: (A) y 0003 x (B) y 0003 3x For what values of m will the graph of the hyperbola and the graph of the line y 0003 mx intersect? Find the coordinates of these intersection points.
35. Transverse axis on x axis Transverse axis length 0003 18 Distance of foci from center 0003 11
52. Find all intersection points of the graph of the hyperbola 4x2 0007 y2 0003 1 with the graph of each of the following lines: (A) y 0003 x (B) y 0003 3x
36. Transverse axis on x axis Transverse axis length 0003 16 Distance of foci from center 0003 10
For what values of m will the graph of the hyperbola and the graph of the line y 0003 mx intersect? Find the coordinates of these intersection points.
37. Conjugate axis on x axis Conjugate axis length 0003 14 Distance of foci from center 0003 1200 38. Conjugate axis on x axis Conjugate axis length 0003 10 Distance of foci from center 0003 170 In Problems 39–46, find the equations of the asymptotes of each hyperbola. 39.
y2 x2 0007 00031 25 4
40.
y2 x2 0007 00031 16 36
53. Consider the hyperbola with equation y2 x2 0007 200031 2 a b 2
(A) Show that y 0003 ba x 21 0007 ax2 . (B) Explain why the hyperbola approaches the lines y 0003 ba x as |x| becomes larger. (C) Does the hyperbola approach its asymptotes from above or below? Explain.
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SECTION 6–3
54. Consider the hyperbola with equation y2 2
a
0007
x2 00031 b2 2
(A) Show that y 0003 ab x 21 0002 bx2 . (B) Explain why the hyperbola approaches the lines y 0003 ab x as |x| becomes larger. (C) Does the hyperbola approach its asymptotes from above or below? Explain. 55. Let F and F be two points in the plane and let c be a constant such that c 0005 d(F, F ). Describe the set of all points P in the plane such that the absolute value of the difference of the distances from P to F and F is equal to the constant c. 56. Let F and F be two points in the plane and let c denote the constant d(F, F ). Describe the set of all points P in the plane such that the absolute value of the difference of the distances from P to F and F is equal to the constant c. 57. Study the following derivation of the standard equation of a hyperbola with foci (c, 0), x intercepts (a, 0), and endpoints of the conjugate axis (0, b). Explain why each equation follows from the equation that precedes it. [Hint: Recall that c2 0003 a2 0002 b2.]
Hyperbola
417
ECCENTRICITY Problems 59 and 60 (and Problems 45 and 46 in Exercises 6-2) are related to a property of conics called eccentricity, which is denoted by a positive real number E. Parabolas, ellipses, and hyperbolas all can be defined in terms of E, a fixed point called a focus, and a fixed line not containing the focus called a directrix as follows: The set of points in a plane each of whose distance from a fixed point is E times its distance from a fixed line is an ellipse if 0 0006 E 0006 1, a parabola if E 0003 1, and a hyperbola if E 0005 1.
59. Find an equation of the set of points in a plane each of whose distance from (3, 0) is three-halves its distance from the line x 0003 43. Identify the geometric figure. 60. Find an equation of the set of points in a plane each of whose distance from (0, 4) is four-thirds its distance from the line y 0003 94. Identify the geometric figure.
APPLICATIONS 61. ARCHITECTURE An architect is interested in designing a thinshelled dome in the shape of a hyperbolic paraboloid, as shown in Figure (a). Find the equation of the hyperbola located in a coordinate system [Fig. (b)] satisfying the indicated conditions. How far is the hyperbola above the vertex 6 feet to the right of the vertex? Compute the answer to two decimal places.
|d1 0007 d2| 0003 2a 2(x 0002 c)2 0002 y2 0003 2a 0002 2(x 0007 c)2 0002 y2
Hyperbola
(x 0002 c)2 0002 y2 0003 4a2 4a 2(x 0007 c)2 0002 y2 0002 (x 0007 c)2 0002 y2 2(x 0007 c)2 0002 y2 0003 a 0007
cx a
(x 0007 c)2 0002 y2 0003 a2 0007 2cx 0002 a1 0007
c2x2 a2
c2 2 b x 0002 y2 0003 a2 0007 c2 a2
Parabola
y2 x2 0007 200031 2 a b
Hyperbolic paraboloid (a)
58. Study the following derivation of the standard equation of a hyperbola with foci (0, c), y intercepts (0, a), and endpoints of the conjugate axis (b, 0). Explain why each equation follows from the equation that precedes it. [Hint: Recall that c2 0003 a2 0002 b2.]
y
(8, 12) 10
|d1 0007 d2| 0003 2a 2
2x 0002 ( y 0002 c)2 0003 2a 0002 2x2 0002 ( y 0007 c)2 2
2
2
2
2
2
x 0002 ( y 0002 c) 0003 4a 4a2x 0002 ( y 0007 c) 0002 x 0002 ( y 0007 c) cy 2x2 0002 ( y 0007 c)2 0003 a 0007 a x2 0002 (y 0007 c)2 0003 a2 0007 2cy 0002 x2 0002 a1 0007
c2 2 b y 0003 a2 0007 c2 a2
2
y
a
2
2
0007
x 00031 b2
c2y2 a2
2
000710
10
x
Hyperbola part of dome (b)
62. NUCLEAR POWER A nuclear reactor cooling tower is a hyperboloid, that is, a hyperbola rotated around its conjugate axis, as shown in Figure (a) on page 418. The equation of the hyperbola in Figure (b) used to generate the hyperboloid is y2 x2 0007 00031 1002 1502
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ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Incoming wave Common focus F Hyperbola
Hyperbola focus
Parabola
Receiving cone
Nuclear reactor cooling tower (a)
F
(a)
y 500
0007500
500
x Radio telescope
0007500
Hyperbola part of dome (b)
If the tower is 500 feet tall, the top is 150 feet above the center of the hyperbola, and the base is 350 feet below the center, what is the radius of the top and the base? What is the radius of the smallest circular cross section in the tower? Compute answers to three significant digits. 63. SPACE SCIENCE In tracking space probes to the outer planets, NASA uses large parabolic reflectors with diameters equal to twothirds the length of a football field. Needless to say, many design problems are created by the weight of these reflectors. One weight problem is solved by using a hyperbolic reflector sharing the parabola’s focus to reflect the incoming electromagnetic waves to the other focus of the hyperbola where receiving equipment is installed (see the figure).
CHAPTER
6-1
6
(b)
For the receiving antenna shown in the figure, the common focus F is located 120 feet above the vertex of the parabola, and focus F (for the hyperbola) is 20 feet above the vertex. The vertex of the reflecting hyperbola is 110 feet above the vertex for the parabola. Introduce a coordinate system by using the axis of the parabola as the y axis (up positive), and let the x axis pass through the center of the hyperbola (right positive). What is the equation of the reflecting hyperbola? Write y in terms of x.
Review
Conic Sections; Parabola
The plane curves obtained by intersecting a right circular cone with a plane are called conic sections. If the plane cuts clear through one nappe, then the intersection curve is called a circle if the plane is perpendicular to the axis and an ellipse if the plane is not perpendicular to the axis. If a plane cuts only one nappe, but does not cut
clear through, then the intersection curve is called a parabola. If a plane cuts through both nappes, but not through the vertex, the resulting intersection curve is called a hyperbola. A plane passing through the vertex of the cone produces a degenerate conic—a point, a line, or a pair of lines. The figure illustrates the four nondegenerate conics.
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419
Review y
y
F
F
x
0
a ⬍ 0 (opens left) Circle
x
0
a ⬎ 0 (opens right)
Ellipse
2. x2 0003 4ay Vertex: (0, 0) Focus: (0, a) Directrix: y 0003 0007a Symmetric with respect to the y axis Axis of symmetry the y axis y
y
0
x
F
F
x
0
Parabola
Hyperbola
The graph of a ⬍ 0 (opens down)
Ax2 0002 Bxy 0002 Cy2 0002 Dx 0002 Ey 0002 F 0003 0 is a conic, a degenerate conic, or the empty set. The following is a coordinate-free definition of a parabola:
6-2
a ⬎ 0 (opens up)
Ellipse
The following is a coordinate-free definition of an ellipse:
Parabola A parabola is the set of all points in a plane equidistant from a fixed point F and a fixed line L (not containing F) in the plane. The fixed point F is called the focus, and the fixed line L is called the directrix. A line through the focus perpendicular to the directrix is called the axis of symmetry, and the point on the axis halfway between the directrix and focus is called the vertex.
L
d1
P
d1 0003 d2 Axis of symmetry
Ellipse An ellipse is the set of all points P in a plane such that the sum of the distances from P to two fixed points in the plane is a constant (the constant is required to be greater than the distance between the two fixed points). Each of the fixed points, F and F, is called a focus, and together they are called foci. Referring to the figure, the line segment V V through the foci is the major axis. The perpendicular bisector B B of the major axis is the minor axis. Each end of the major axis, V and V, is called a vertex. The midpoint of the line segment F F is called the center of the ellipse.
d2 V(Vertex)
d1 0002 d2 0003 Constant
F(Focus)
B V
Parabola
d1
P
F
d2
Directrix F
From the definition of a parabola, we can obtain the following standard equations: Standard Equations of a Parabola with Vertex at (0, 0) 1. y2 0003 4ax Vertex: (0, 0) Focus: (a, 0) Directrix: x 0003 0007a Symmetric with respect to the x axis Axis of symmetry the x axis
V
B
From the definition of an ellipse, we can obtain the following standard equations: Standard Equations of an Ellipse with Center at (0, 0) 2
1.
y x2 a0005b00050 0002 200031 2 a b x intercepts: a (vertices) y intercepts: b
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Foci: F 0003 (0007c, 0), F 0003 (c, 0) Major axis length 0003 2a Minor axis length 0003 2b
c2 0003 a2 0007 b2
兩 d1 0007 d2兩 0003 Constant P
d2
d1 y F
b
0007a
F 0007c
F c
0
V
V
F
x
a
From the definition of a hyperbola, we can obtain the following standard equations:
0007b
Standard Equations of a Hyperbola with Center at (0, 0) 2
2
2.
1.
2
y x 0002 200031 a0005b00050 2 b a x intercepts: b y intercepts: a (vertices) Foci: F 0003 (0, 0007c), F 0003 (0, c) Major axis length 0003 2a Minor axis length 0003 2b
c2 0003 a2 0007 b2
y x2 0007 200031 a2 b x intercepts: a (vertices) y intercepts: none Foci: F 0003 (0007c, 0), F 0003 (c, 0) Transverse axis length 0003 2a Conjugate axis length 0003 2b b Asymptotes: y 0003 x a
y
c2 0003 a2 0002 b2
y
a c F
b
F
0007c
0007b
0
b
c
0007a
a
F
c
x
0007b
x
0007c F 0007a
[Note: Both graphs are symmetric with respect to the x axis, y axis, and origin. Also, the major axis is always longer than the minor axis.]
6-3
Hyperbola
The following is a coordinate-free definition of a hyperbola:
2.
y2
x2 00031 a2 b2 x intercepts: none y intercepts: a (vertices) Foci: F 0003 (0, 0007c), F 0003 (0, c) Transverse axis length 0003 2a Conjugate axis length 0003 2b a Asymptotes: y 0003 x b 0007
y
Hyperbola A hyperbola is the set of all points P in a plane such that the absolute value of the difference of the distances from P to two fixed points in the plane is a positive constant (the constant is required to be less than the distance between the two fixed points). Each of the fixed points, F and F, is called a focus. The intersection points V and V of the line through the foci and the two branches of the hyperbola are called vertices, and each is called a vertex. The line segment V V is called the transverse axis. The midpoint of the transverse axis is the center of the hyperbola. The line segment perpendicular to the transverse axis that goes through the center is called the conjugate axis.
c2 0003 a2 0002 b2
c
F
a c 0007b
b
x
0007a 0007c
F
[Note: Both graphs are symmetric with respect to the x axis, y axis, and origin.]
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Review Exercises
CHAPTER
6
421
Review Exercises
Work through all the problems in this chapter review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. In Problems 1–6, graph each equation and locate foci. Locate the directrix for any parabolas. Find the lengths of major, minor, transverse, and conjugate axes where applicable. 1. 9x2 0002 25y2 0003 225
2. x2 0003 000712y
3. 25y2 0007 9x2 0003 225
4. x2 0007 y2 0003 16
5. y2 0003 8x
6. 2x2 0002 y2 0003 8
7. Find the equation of the parabola having its vertex at the origin, its axis of symmetry the x axis, and (00074, 00072) on its graph. In Problems 8 and 9, find the equation of the ellipse in the form y2 x2 M, N 7 0 0002 00031 M N if the center is at the origin, and: 8. Major axis on x axis Major axis length 0003 12 Minor axis length 0003 10 9. Major axis on y axis Minor axis length 0003 12 Distance between foci 0003 16 In Problems 10 and 11, find the equation of the hyperbola in the form y2 y2 x2 x2 or M, N 0005 0 0007 00031 0007 00031 M N M N if the center is at the origin, and: 10. Transverse axis on y axis Conjugate axis length 0003 6 Distance between foci 0003 8 11. Transverse axis on x axis Transverse axis length 0003 14 Conjugate axis length 0003 16 12. Find the equation of the parabola having directrix y 0003 5 and focus (0, 00075). 13. Find the foci of the ellipse through the point (00076, 0) if the center is at the origin, the major axis is on the x axis, and the major axis has twice the length of the minor axis. 14. Find the y intercepts of a hyperbola if the center is at the origin, the conjugate axis is on the x axis and has length 4, and (0, 00073) is a focus.
15. Find the directrix of a parabola having its vertex at the origin and focus (00074, 0). 16. Find the points of intersection of the parabolas x2 0003 8y and y2 0003 0007x. 17. Find the x intercepts of an ellipse if the center is at the origin, the major axis is on the y axis and has length 14, and (0, 00071) is a focus. 18. Find the foci of the hyperbola through the point (0, 00074) if the center is at the origin, the transverse axis is on the y axis, and the conjugate axis has twice the length of the transverse axis. 19. Use the definition of a parabola and the distance formula to find the equation of a parabola with directrix x 0003 6 and focus at (2, 4). 20. Find an equation of the set of points in a plane each of whose distance from (4, 0) is twice its distance from the line x 0003 1. Identify the geometric figure. 21. Find an equation of the set of points in a plane each of whose distance from (4, 0) is two-thirds its distance from the line x 0003 9. Identify the geometric figure. In Problems 22–24, find the equations of the asymptotes of each hyperbola. 22.
y2 x2 0007 00031 49 25
23.
y2 x2 0007 00031 64 4
24. 4x2 0007 y2 0003 1
APPLICATIONS 25. COMMUNICATIONS A parabolic satellite television antenna has a diameter of 8 feet and is 1 foot deep. How far is the focus from the vertex? 26. ENGINEERING An elliptical gear is to have foci 8 centimeters apart and a major axis 10 centimeters long. Letting the x axis lie along the major axis (right positive) and the y axis lie along the minor axis (up positive), write the equation of the ellipse in the standard form y2 x2 0002 200031 2 a b 27. SPACE SCIENCE A hyperbolic reflector for a radio telescope (such as that illustrated in Problem 63, Exercises 6-3) has the equation y2 402
0007
x2 00031 302
If the reflector has a diameter of 30 feet, how deep is it? Compute the answer to three significant digits.
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CHAPTER
ZZZ
Page 422
ADDITIONAL TOPICS IN ANALYTIC GEOMETRY
6
GROUP ACTIVITY Focal Chords
Many of the applications of the conic sections are based on their reflective or focal properties. One of the interesting algebraic properties of the conic sections concerns their focal chords. If a line through a focus F contains two points G and H of a conic section, then the line segment GH is called a focal chord. Let G 0003 (x1, y1) and H 0003 (x2, y2) be points on the graph of x2 0003 4ay such that GH is a focal chord. Let u denote the length of GF and v the length of FH (Fig. 1).
(A) Use the distance formula to show that u 0003 y1 0002 a. (B) Show that G and H lie on the line y 0007 a 0003 mx, where m 0003 ( y2 0007 y1)兾(x2 0007 x1). (C) Solve y 0007 a 0003 mx for x and substitute in x2 0003 4ay, obtaining a quadratic equation in y. Explain why y1 y2 0003 a2. (D) Show that
1 1 1 0002 0003 . u v a
(E) Show that u 0002 v 0007 4a 0003 y
G
F u
v
H (2a, a) x
Z Figure 1 Focal chord GH of the parabola x2 0003 4ay.
(u 0007 2a)2 . Explain why this u0007a
implies that u 0002 v 000e 4a, with equality if and only if u 0003 v 0003 2a. (F) Which focal chord is the shortest? Is there a longest focal chord? (G) Is
1 1 0002 a constant for focal chords of the ellipse? For u v
focal chords of the hyperbola? Obtain evidence for your answers by considering specific examples.
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7
C
OUTLINE
WE have seen many real-world situations where solving an equation
is valuable. But the world is a very complicated place, and many more situations lead to more than one variable. In that case, solving a system of equations becomes important. In this chapter, we will study a variety of methods for solving systems of equations. We will begin with linear systems with two or three variables using algebraic techniques similar to those we used for solving individual equations. Then we will introduce a variety of matrix methods for solving linear systems. These methods can be applied to very large systems that model very complicated real-world problems.
7-1
Systems of Linear Equations
7-2
Solving Systems of Linear Equations Using Gauss–Jordan Elimination
7-3
Matrix Operations
7-4
Solving Systems of Linear Equations Using Matrix Inverse Methods
7-5
Determinants and Cramer’s Rule Chapter 7 Review Chapter 7 Group Activity: Modeling with Systems of Linear Equations
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7-1
Systems of Linear Equations Z Systems of Equations Z Solving by Graphing Z Solving by Substitution Z Solving Using Elimination by Addition Z Applications
We have seen a wide variety of real-world problems that can be solved by writing and solving an equation. But a lot of problems have extra conditions that makes writing a single equation impractical. In this case, two or more equations might be needed to model the situation. In this section, we’ll examine how to solve two or more equations together, then see how to apply what we learn.
Z Systems of Equations To illustrate the basic concepts, we’ll use a simple example. At one campus coffee shop, muffins cost $2 each, and lattes are $3 each. If a total of seven items are sold for $18, how many of each item were sold? There are two natural variables in the problem: the number of muffins, which we’ll call x, and the number of lattes, which we’ll call y. Then x0003 y0002 7 2x 0003 3y 0002 18
Seven items total Total cost is $18.
This is called a system of linear equations in two variables. The solution to the problem is found by finding all pairs of numbers x and y that make both equations true. In general, we will study solving linear systems of the type ax 0003 by 0002 h cx 0003 dy 0002 k
System of two linear equations in two variables
where x and y are variables, a, b, c, and d are real numbers called the coefficients of x and y, and h and k are real numbers called the constant terms in the equations. A pair of numbers x 0002 x0 and y 0002 y0 is a solution of this system if each equation is satisfied by the pair. The set of all such pairs of numbers is called the solution set for the system. To solve a system is to find its solution set.
Z Solving by Graphing Recall that the graph of a linear equation is the line consisting of all ordered pairs that satisfy the equation. To solve the coffee shop problem by graphing, we will graph both equations in the same coordinate system. The coordinates of any points that the lines have in common must be solutions to the system, because they must satisfy both equations.
EXAMPLE
1
Solving a System by Graphing Solve the coffee shop problem by graphing:
x0003 y0002 7 2x 0003 3y 0002 18
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Systems of Linear Equations
425
Find the x and y intercepts for each line. x⫹y⫽7
2x ⫹ 3y ⫽ 18
x
y
x
y
0
7
0
6
7
0
9
0
Plot these points, graph the two lines, estimate the intersection point visually (Fig. 1), and check the estimate. y 10
5
(3, 4) 2x ⫹ 3y ⫽ 18 10 5
x
x⫹y⫽7
Z Figure 1
x⫽3 y⫽4 CHECK
x⫹y⫽7 ? 3⫹4⫽7
Muffins* Lattes
2x ⫹ 3y ⫽ 18 ? 2(3) ⫹ 3(4) ⫽ 18
✓
✓
7⫽7 MATCHED PROBLEM 1
Solve by graphing:
18 ⫽ 18
0002
x⫺ y⫽ 3 x ⫹ 2y ⫽ ⫺3
0002
Technology Connections To solve Example 1 with a graphing calculator, first solve each equation for y: xⴙyⴝ7
From Figure 3, we see that the solution is
Subtract x from both sides.
xⴝ3
Muffins
yⴝ4
Lattes
yⴝ7ⴚx
10
2x ⴙ 3y ⴝ 18
Subtract 2x from both sides.
3y ⴝ 18 ⴚ 2x yⴝ 6ⴚ
Divide both sides by 3.
⫺10
10
2 3x
Next, enter these functions in the equation editor of a graphing calculator (Fig. 2) and use the intersect command to find the intersection point (Fig. 3).
⫺10
Z Figure 2
Z Figure 3
*When the solution set for a linear system is a single point, we will follow the common practice of writing the solution as (3, 4) or as x ⫽ 3, y ⫽ 4, rather than the more formal expression 5(3, 4)6.
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It is clear that Example 1 has exactly one solution, because the lines have exactly one point of intersection. In general, lines in a rectangular coordinate system are related to each other in one of three ways, as illustrated in Example 2.
EXAMPLE
2
Determining the Nature of Solutions Match each of the following systems with one of the graphs in Figure 4 and discuss the nature of the solutions: (A) 2x 0004 3y 0002 2 x 0003 2y 0002 8
(B) 4x 0003 6y 0002 12 2x 0003 3y 0002 00046
(C) 2x 0004 3y 0002 00046 0004x 0003 32 y 0002 3
y
y
y
5
5
5
(4, 2)
00045
x
5
00045
x
5
00045
00045
00045
(a)
5
x
00045
(b)
(c)
Z Figure 4 SOLUTIONS
(A) Write each equation in slope–intercept form: 2x 0004 3y 0002 2 00043y 0002 00042x 0003 2 y 0002 23 x 0004 23
x 0003 2y 0002 8 2y 0002 0004x 0003 8 y 0002 000412 x 0003 4
One positive slope, one negative
The graphs of these two lines match graph (b). There is exactly one solution: x 0002 4, y 0002 2. (B) 4x 0003 6y 0002 12 6y 0002 00044x 0003 12 y 0002 000423 x 0003 2
2x 0003 3y 0002 00046 3y 0002 00042x 0004 6 y 0002 000423x 0004 2
Slopes are equal.
The graphs of these parallel lines match graph (c). There is no solution. 0004x 0003 32 y 0002 3
(C) 2x 0004 3y 0002 00046 00043y 0002 00042x 0004 6 y 0002 23 x 0003 2
3 2y
0002x00033 y 0002 23 x 0003 2
Same line!
The graph of these identical lines match graph (a). There are an infinite number of solutions. 0002 MATCHED PROBLEM 2
Solve each of the following systems by graphing: (A) 2x 0003 3y 0002 12 x 0004 3y 0002 00043
(B)
x 0004 3y 0002 00043 00042x 0003 6y 0002 12
(C) 2x 0004 3y 0002 12 0004x 0003 32 y 0002 00046
0002
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Next, we’ll define some terms that can be used to describe the different types of solutions to systems of equations illustrated in Example 2. Z SYSTEMS OF LINEAR EQUATIONS: BASIC TERMS A system of linear equations is consistent if it has one or more solutions and inconsistent if no solutions exist. Furthermore, a consistent system is said to be independent if it has exactly one solution (often referred to as the unique solution) and dependent if it has more than one solution.
Referring to the three systems in Example 2, the system in part A [Fig. 4(b)] is consistent and independent, with the unique solution x 0002 4 and y 0002 2. The system in part B [Fig. 4(c)] is inconsistent, with no solution. And the system in part C [Fig. 4(a)] is consistent and dependent, with an infinite number of solutions: all the points on the two coinciding lines. ZZZ EXPLORE-DISCUSS 1
Can a consistent and dependent linear system have exactly two solutions? Exactly three solutions? Explain.
In general, any two lines in a rectangular coordinate plane either intersect in exactly one point, or are parallel, or coincide (have identical graphs). So, the systems in Example 2 illustrate the only three possible types of solutions for systems of two linear equations in two variables. These ideas are summarized in Theorem 1. Z THEOREM 1 Possible Solutions to a Linear System A system of linear equations must have 1. Exactly one solution or 2. No solution or 3. Infinitely many solutions
Consistent and independent
Inconsistent
Consistent and dependent
Note: While the geometric discussion presented here only applies to systems of equations with two variables, the same three possibilities remain for systems of linear equations with more than two variables.
Z Solving by Substitution The accuracy of solutions found by graphing depends a lot on how accurate the graph is when the graphs are drawn by hand. If the solutions are found using a graphing calculator, you will likely get very accurate solutions, but they probably won’t be exact. Worse still, the solutions can be very difficult to find, depending on the window settings that you choose. Also, for systems with more than two variables, the geometry gets extremely complicated. For all of these reasons, we will next turn our attention to solving systems algebraically. There are a number of different techniques that can be used. One of the simplest is the substitution method.
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We will return to the coffee shop problem from page 424 to illustrate the substitution method.
EXAMPLE
3
Solving a System by Substitution Use substitution to solve the coffee shop problem: x 0003 y 0002 7 2x 0003 3y 0002 18
SOLUTION
Step 1: Solve either equation for one variable. It will be easy to solve the first equation for y in terms of x: x0003y00027 y000270004x
Solve the first equation for y in terms of x. Substitute into the second equation.
Step 2: Substitute 7 0004 x for y in the second equation. 2x 0003 3y 0002 18 2x 0003 3(7 0004 x) 0002 18 2x 0003 21 0004 3x 0002 18 0004x 0002 00043 xⴝ3
y ⴝ 7 ⴚ x, so replace y with 7 ⴚ x. Multiply out parentheses. Collect x terms on the left and constant terms on the right. Multiply both sides by ⴚ1.
Step 3: Replace x with 3 in y 0002 7 0004 x: y000270004x y0002700043 yⴝ4 The solution is 3 muffins and 4 lattes, as we found and checked earlier. MATCHED PROBLEM 3
Solve by substitution and check:
x0004 y0002 3 x 0003 2y 0002 00043
0002
0002
The following box summarizes the steps for solving a system using the substitution method. Z SOLVING SYSTEMS OF TWO LINEAR EQUATIONS IN TWO VARIABLES: THE SUBSTITUTION METHOD 1. Choose one of the two equations and solve it for one of the two variables. (Make a choice that avoids fractions, if possible.) 2. Substitute the result of step 1 into the equation that was not used in step 1 and solve the resulting linear equation in one variable. 3. Substitute the result of step 2 into the expression obtained in step 1 to find the value of the second variable.
ZZZ EXPLORE-DISCUSS 2
Use substitution to solve each of the following systems. Discuss the nature of the solution sets you obtain. x 0003 3y 0002 4 2x 0003 6y 0002 7
x 0003 3y 0002 4 2x 0003 6y 0002 8
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Z Solving Using Elimination by Addition Now we turn to elimination by addition. This is probably the most important method of solution, since it is readily generalized to larger systems. The method involves the replacement of systems of equations with simpler equivalent systems, by performing appropriate operations, until we obtain a system with an obvious solution. Equivalent systems of equations are, as you would expect, systems that have exactly the same solution set. Theorem 2 lists operations that produce equivalent systems. Z THEOREM 2 Elementary Equation Operations Producing Equivalent Systems A system of linear equations is transformed into an equivalent system if: 1. Two equations are interchanged. 2. An equation is multiplied by a nonzero constant. 3. A constant multiple of another equation is added to a given equation.
We’ll return one more time to the coffee shop problem to illustrate why elimination by addition works so well. The system of equations was x0003 y 0002 7 2x 0003 3y 0002 18 Notice that if we use the third operation in Theorem 2, adding 00042 times the first equation to the second one, we get 00042x 0004 2y 0002 000414 2x 0003 3y 0002 18 y00024 This eliminated x, and left behind an equation with only y. We could then easily substitute back in to find x. We will rely mostly on operations 2 and 3 for now, but operation 1 will come in especially handy later in the section. Examples 4 and 5 illustrate the use of elimination by addition on two and three variable systems.
EXAMPLE
4
Solving a System Using Elimination by Addition Solve using elimination by addition:
SOLUTION
3x 0004 2y 0002 8 2x 0003 5y 0002 00041
We will use Theorem 2 to eliminate one of the variables and get an easy equation with one variable. 3x 0004 2y 0002 8 2x 0003 5y 0002 00041 15x 0004 10y 0002 40 4x 0003 10y 0002 00042 19x 0002 38 xⴝ2
If we multiply the top equation by 5, the bottom by 2, and then add, we can eliminate y.
Now solve for x.
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The equation x 0002 2 paired with either of the two original equations produces an equivalent system. So, we can substitute x 0002 2 back into either of the two original equations to solve for y. We choose the second equation. 2(2) 0003 5y 0002 00041 5y 0002 00045 y ⴝ ⴚ1 SOLUTION
x 0002 2, y 0002 00041, or (2, 00041). CHECK
MATCHED PROBLEM 4
3x 0004 2y 0002 8 ? 3(2) 0004 2(00041) 0002 8 ✓ 800028
Solve using elimination by addition:
2x 0003 5y 0002 00041 ? 2(2) 0003 5(00041) 0002 00041 ✓ 00041 0002 00041
0002
6x 0003 3y 0002 3 5x 0003 4y 0002 7
0002
When a system has three equations, we will use elimination to reduce to a system with two equations and two variables, then solve like we did in Example 4. To help you follow a solution, we will number the equations as E1, E2, and so on.
EXAMPLE
5
Solution Using Elimination by Addition x 0003 2y 0003 3z 0002 2 3x 0004 5y 0004 4z 0002 15 00042x 0004 3y 0003 2z 0002 2
SOLUTION
E1 E2 E3
Since the coefficient of x in E1 is 1, our calculations will be simplified if we use E1 to eliminate x from the other equations. First we eliminate x from E2 by multiplying E1 by 00043 and adding the result to E2. Equivalent System
00043x 0004 6y 0004 9z 0002 00046 3x 0004 5y 0004 4z 0002 15 000411y 0004 13z 0002 9
ⴚ3E1 E2 E4
x 0003 2y 0003 3z 0002 2 000411y 0004 13z 0002 9 00042x 0004 3y 0003 2z 0002 2
E1 E4 E3
Now we use E1 to eliminate x (the same variable eliminated above) from E3 by multiplying E1 by 2 and adding the result to E3. Equivalent System
2x 0003 4y 0003 6z 0002 4 00042x 0004 3y 0003 2z 0002 2 y 0003 8z 0002 6
2E1 E3 E5
x 0003 2y 0003 3z 0002 2 000411y 0004 13z 0002 9 y 0003 8z 0002 6
E1 E4 E5
Notice that E4 and E5 form a system of two equations with two variables. Next we use E5 to eliminate y from E4 and replace E4 with the result. 11y 0003 88z 0002 66 000411y 0004 13z 0002 9 75z 0002 75
11E5 E4 E6
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Now we can easily solve for z. 75z 0002 75 zⴝ1
E6
Next substitute z 0002 1 in E4 or E5 and solve for y. y 0003 8z 0002 6 y 0003 8(1) 0002 6 y ⴝ ⴚ2
E5
Finally, substitute y 0002 00042 and z 0002 1 in any of E1, E2, or E3 and solve for x. x 0003 2y 0003 3z 0002 2 x 0003 2(00042) 0003 3(1) 0002 2 xⴝ3
E1
The solution to the original system is (3, 00042, 1) or x 0002 3, y 0002 00042, z 0002 1. CHECK
x 0003 2y 0003 3z 0002 2 ? 3 0003 2(00042) 0003 3(1) 0002 2 ✓ 2 00022
E1
MATCHED PROBLEM 5
To check the solution, we must check each equation in the original system:
3x 0004 5y 0004 4z 0002 15 ? 3(3) 0004 5(00042) 0004 4(1) 0002 15 ✓ 15 0002 15
E2
00042x 0004 3y 0003 2z 0002 2 ? 00042(3) 0004 3(00042) 0003 2(1) 0002 2 ✓ 200022
E3
0002
Solve: 2x 0003 3y 0004 5z 0002 000412 3x 0004 2y 0003 2z 0002 1 4x 0004 5y 0004 4z 0002 000412
0002
Let’s see what happens in the solution process when a system either has no solution or has infinitely many solutions. Consider the solutions to the following system: 2x 0003 6y 0002 00043 x 0003 3y 0002 2 Solution by Substitution
Solution by Elimination
Solve the second equation for x and substitute in the first equation.
Multiply the second equation by 00042 and add to the first equation.
x 0002 2 0004 3y 2(2 0004 3y) 0003 6y 0002 00043 4 0004 6y 0003 6y 0002 00043 4 0002 00043
2x 0003 6y 0002 00043 00042x 0004 6y 0002 00044 0 0002 00047
Both methods of solution lead to a contradiction (a statement that is false). An assumption that the original system has solutions must be false. This tells us that the system has no solution. The graphs of the equations are parallel and the system is inconsistent. Now consider the system x 0004 12 y 0002 4 00042x 0003 y 0002 00048
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Solution by Substitution
Solution by Elimination
Solve the first equation for x and substitute in the second equation.
Multiply the first equation by 2 and add to the second equation.
x 0002 12 y 0003 4 00042 (12 y 0003 4) 0003 y 0002 00048 0004y 0004 8 0003 y 0002 00048 00048 0002 00048
2x 0004 y 0002 8 00042x 0003 y 0002 00048 00002 0
This time both solution methods lead to a statement that is always true. This means that the two original equations are equivalent. That is, their graphs coincide. The system is dependent and has an infinite number of solutions. There are many different ways to represent this infinite solution set. For example, S1 0002 5(x, y) ƒ y 0002 2x 0004 8, x any real number} and S2 0002 5(x, y) ƒ x 0002 12 y 0003 4, y any real number} both represent the solutions to this system. For reasons that will become apparent later, it is customary to introduce a new variable, called a parameter, and express both variables in terms of this new variable. If we let x 0002 s and y 0002 2s 0004 8 in S1, we can express the solution set as 5(s, 2s 0004 8) ƒ s any real number} Some particular solutions to this system are obtained by choosing particular values for the parameter.
EXAMPLE
6
s ⴝ ⴚ1
sⴝ2
sⴝ5
s ⴝ 9.4
(00041, 000410)
(2, 00044)
(5, 2)
(9.4, 10.8)
Using Elimination by Addition Solve: x0003 y0003 z00023 x 0004 y 0004 5z 0002 1 2x 0003 3y 0003 5z 0002 6
SOLUTION
E1 E2 E3
Use E1 to eliminate z from E2 and replace E2 with the result. Equivalent System
5x 0003 5y 0003 5z 0002 15 x 0004 y 0004 5z 0002 1 6x 0003 4y 0002 16
x0003 y0003 z0002 3 6x 0003 4y 0002 16 2x 0003 3y 0003 5z 0002 6
5E1 E2 E4
E1 E4 E3
Use E1 to eliminate z from E3 and replace E3 with the result. Equivalent System
00045x 0004 5y 0004 5z 0002 000415 2x 0003 3y 0003 5z 0002 6 00043x 0004 2y 0002 00049
ⴚ5E1 E3 E5
x0003 y0003z0002 3 6x 0003 4y 0002 16 00043x 0004 2y 0002 00049
E1 E4 E5
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Now treat E4 and E5 as a system of two equations, and eliminate y. 6x 0003 4y 0002
16
00046x 0004 4y 0002 000418 0 0002 00042
E4 2E5 E6
Stop! We have obtained a contradiction. The original system is inconsistent and has no solution. (Note: It’s impossible to check in this case.) 0002 MATCHED PROBLEM 6
Solve: 2x 0003 3y 0004 5z 0002 3 3x 0004 2y 0003 2z 0002 2 x 0004 5y 0003 7z 0002 1
EXAMPLE
7
0002
Using Elimination by Addition Solve: x0003y0003 z00021 2x 0003 y 0004 z 0002 3 3x 0003 y 0004 3z 0002 5
SOLUTION
E1 E2 E3
Use E1 to eliminate y from E2 and replace E2 with the result. Equivalent System
0004x 0004 y 0004 z 0002 00041 2x 0003 y 0004 z 0002 3 x 0004 2z 0002 2
x0003y0003 z00021 x 0004 2z 0002 2 3x 0003 y 0004 3z 0002 5
0004E1 E2 E4
E1 E4 E3
Use E1 to eliminate y from E3 and replace E3 with the result. Equivalent System
0004x 0004 y 0004 z 0002 00041 3x 0003 y 0004 3z 0002 5 2x 0004 4z 0002 4
x0003y0003 z00021 x 0004 2z 0002 2 2x 0004 4z 0002 4
ⴚE1 E3 E5
E1 E4 E5
Use E4 to eliminate z from E5 and replace E5 with the result. Equivalent System
00042x 2x
0003 4z 0002 00044 0004 4z 0002 4 00002 0
ⴚ2E4 E5
x0003y0003 z00021 x 0004 2z 0002 2
E1 E4
E6
Since E6 is true for all x, y, and z, it provides no information about the systems’ solution set and can be discarded. The solutions to the last equivalent system can be described by introducing a parameter. If we let z 0002 s, then, using E4, we can write x 0002 2s 0003 2. Substituting for x and z in E1 and solving for y, we have x0003y0003z00021 2s 0003 2 0003 y 0003 s 0002 1 y 0002 00043s 0004 1
E1
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The solution set is given by
5(2s 0003 2, 00043s 0004 1, s) | s any real number6
The check is left to the reader. MATCHED PROBLEM 7
0002
Solve: 3x 0003 2y 0003 4z 0002 5 2x 0003 y 0003 5z 0002 2 x 0003 6z 0002 00041
ZZZ EXPLORE-DISCUSS 3
0002
Refer to the solution to Example 7. The given representation of the solution set is not the only one. Which of the following is a representation of the solution set? Justify your answer. (A) {(t, 2 0004 1.5t, 0.5t 0004 1) | t any real number} (B) {(2u 0003 4, 00042u 0004 3, u) | u any real number} Let y 0002 v, where v is any real number, express x and z in terms of v, and find another representation of the solution set for Example 7.
Z Applications Examples 8–10 illustrate the advantages of using systems of equations in solving word problems.
EXAMPLE
8
Airspeed An airplane makes the 2,400-mile trip from Washington, D.C. to San Francisco in 7.5 hours and makes the return trip in 6 hours. Assuming that the plane travels at a constant airspeed and that the wind blows at a constant rate from west to east, find the plane’s airspeed and the wind rate.
SOLUTION
San 2,400 Francisco miles Washington, D.C.
Let x represent the airspeed of the plane and let y represent the rate at which the wind is blowing (both in miles per hour). The plane’s speed relative to the ground is determined by combining these two rates; that is, x 0004 y 0002 Ground speed flying east to west (airspeed 0004 wind) x 0003 y 0002 Ground speed flying west to east (airspeed 0003 wind) Applying the familiar formula D 0002 RT to each leg of the trip leads to the following system of equations: 2,400 0002 7.5(x 0004 y) 2,400 0002 6(x 0003 y)
Washington to San Francisco: 7.5 hr, 2,400 mi San Francisco to Washington: 6 hr, 2,400 mi
After simplification, we have x 0004 y 0002 320 x 0003 y 0002 400 Add these two equations to eliminate y: 2x 0002 720 x ⴝ 360 mph
Airspeed
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Substitute for x in the second equation: x 0003 y 0002 400 360 0003 y 0002 400 y ⴝ 40 mph 2,400 0002 7.5(x 0004 y) ? 2,400 0002 7.5(360 0004 40) ✓ 2,400 0002 2,400
CHECK
MATCHED PROBLEM 8
Wind rate
2,400 0002 6(x 0003 y) ? 2,400 0002 6(360 0003 40) ✓ 2,400 0002 2,400
0002
A boat takes 8 hours to travel 80 miles upstream and 5 hours to return to its starting point. Find the speed of the boat in still water and the speed of the current. 0002 The quantity of a product that people are willing to buy (known as the demand) during some period of time depends on its price. Generally, the higher the price, the less the demand; the lower the price, the greater the demand. Similarly, the quantity of a product that a supplier is willing to sell during some period of time (known as the supply) also depends on the price. Generally, a supplier will be willing to supply more of a product at higher prices and less of a product at lower prices. The simplest supply and demand model is a linear model. If the demand for a product is greater than the supply, the price tends to rise. If the demand is less than the supply, the price tends to fall. So the price tends to stabilize at an equilibrium price; at that price, the supply and demand are equal, and that common quantity is called the equilibrium quantity. Example 9 illustrates the basic concepts of supply and demand.
9
Supply and Demand Using collected data and regression analysis, an analyst arrives at the following price–demand and price–supply equations for the sale of cherries each day in a major urban area. p 0002 00040.2q 0003 5.6 0.1q 0003 1.7
p0002
Demand equation (consumer) Supply equation (supplier)
where q represents the quantity of cherries in thousands of pounds and p represents the price in dollars per pound. For example, we see (Fig. 5) that consumers will purchase 11 thousand pounds (q 0002 11) when the price is p 0002 00040.2(11) 0003 5.6 0002 $3.40 per pound. On the other hand, suppliers will be willing to supply 17 thousand pounds of cherries at $3.40 per pound (solve 3.4 0002 0.1q 0003 1.7 for q). So, at $3.40 per pound the suppliers are willing to supply more cherries than the consumers are willing to purchase. The supply exceeds the demand at that price, and the price will come down. Find the equilibrium quantity and the equilibrium price. p
Price per pound ($)
EXAMPLE
Demand
5
(17, 3.4) 3.4
(11, 3.4) Supply
11
17 20
Thousands of pounds
Z Figure 5
q
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SOLUTION
To find the equilibrium quantity, we solve the linear system p ⫽ ⫺0.2q ⫹ 5.6 p ⫽ 0.1q ⫹ 1.7
Demand equation (consumer) Supply equation (supplier)
using substitution (substituting p ⫽ ⫺0.2q ⫹ 5.6 into the second equation). p ⫽ 0.1q ⫹ 1.7 ⫺0.2q ⫹ 5.6 ⫽ 0.1q ⫹ 1.7 5.6 ⫽ 0.3q ⫹ 1.7 3.9 ⫽ 0.3q q ⴝ 13 thousand pounds
Price per pound ($)
p
Equilibrium point (13, 3)
5
3
Demand
13
20
Thousands of pounds
Z Figure 6
MATCHED PROBLEM 9
Add 0.2q to both sides. Subtract 1.7 from both sides. Divide both sides by 0.3. Equilibrium quantity
Now substitute q ⫽ 13 back into either of the original equations in the system and solve for p (we choose the second equation):
Supply q
Substitute p ⴝ ⴚ0.2q ⴙ 5.6.
p ⫽ 0.1(13) ⫹ 1.7 p ⴝ $3 per pound
Equilibrium price
So if the price of cherries is $3 per pound, then the supplier would supply 13,000 pounds of cherries and the consumer would demand (purchase) 13,000 pounds of cherries. In other words, the market would be in equilibrium (see Fig. 6). 0002 The price–demand and price–supply equations for strawberries in a certain city are p ⫽ ⫺0.2q ⫹ 4 p ⫽ 0.04q ⫹ 1.84
Demand equation Supply equation
where q represents the quantity in thousands of pounds and p represents the price in dollars. Find the equilibrium quantity and the equilibrium price. 0002
EXAMPLE
10
Production Scheduling A garment industry manufactures three shirt styles. Each style shirt requires the services of three departments as listed in the table. The cutting, sewing, and packaging departments have available a maximum of 1,160, l,560, and 480 labor-hours per week, respectively. How many of each style shirt must be produced each week for the plant to operate at full capacity?
SOLUTION
Style A
Style B
Style C
Time Available
Cutting department
0.2 hr
0.4 hr
0.3 hr
1,160 hr
Sewing department
0.3 hr
0.5 hr
0.4 hr
1,560 hr
Packaging department
0.1 hr
0.2 hr
0.1 hr
480 hr
Let x ⫽ Number of style A shirts produced per week y ⫽ Number of style B shirts produced per week z ⫽ Number of style C shirts produced per week Then 0.2x ⫹ 0.4y ⫹ 0.3z ⫽ 1,160 0.3x ⫹ 0.5y ⫹ 0.4z ⫽ 1,560 0.1x ⫹ 0.2y ⫹ 0.1z ⫽ 480
Cutting department Sewing department Packaging department
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Systems of Linear Equations
We can clear the system of decimals by multiplying each side of each equation by 10: 2x ⫹ 4y ⫹ 3z ⫽ 11,600 3x ⫹ 5y ⫹ 4z ⫽ 15,600 x ⫹ 2y ⫹ z ⫽ 4,800
E1 E2 E3
Use E3 to eliminate z from E1 and replace E1 with the result. Equivalent System
2x ⫹ 4y ⫹ 3z ⫽ 11,600 ⫺3x ⫺ 6y ⫺ 3z ⫽ ⫺14,400 ⫺x ⫺ 2y ⫽ ⫺2,800
⫺x ⫺ 2y ⫽ ⫺2,800 3x ⫹ 5y ⫹ 4z ⫽ 15,600 x ⫹ 2y ⫹ z ⫽ 4,800
E1 ⴚ3E3 E4
E4 E2 E3
Use E3 to eliminate z from E2 and replace E2 with the result. Equivalent System
3x ⫹ 5y ⫹ 4z ⫽ 15,600 ⫺4x ⫺ 8y ⫺ 4z ⫽ ⫺19,200 ⫺x ⫺ 3y ⫽ ⫺3,600
⫺x ⫺ 2y ⫺x ⫺ 3y
E2 ⴚ4E3
⫽ ⫺2,800 ⫽ ⫺3,600
x ⫹ 2y ⫹ z ⫽
E5
4,800
E4 E5 E3
Now treat E4 and E5 as a system of two equations; eliminate x. x ⫹ 2y ⫺x ⫺ 3y ⫺y
⫽ 2,800 ⫽ ⫺3,600 ⫽ ⫺800
ⴚE4 E5 E6
From E6 we see that y ⴝ 800 Substitute y ⫽ 800 in E4 or E5 and solve for x. ⫺x ⫺ 2y ⫽ ⫺2,800 ⫺x ⫺ 2(800) ⫽ ⫺2,800 x ⴝ 1,200
E4
Substitute x ⫽ 1,200 and y ⫽ 800 in E1, E2, or E3 and solve for z. x ⫹ 2y ⫹ z ⫽ 4,800 1,200 ⫹ 2(800) ⫹ z ⫽ 4,800 z ⴝ 2,000
E3
Each week, the company should produce 1,200 style A shirts, 800 style B shirts, and 2,000 style C shirts to operate at full capacity. You should check this solution. 0002 MATCHED PROBLEM 10
Repeat Example 10 with the cutting, sewing, and packaging departments having available a maximum of 1,180, 1,560, and 510 labor-hours per week, respectively. 0002
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ANSWERS TO MATCHED PROBLEMS 1.
y 5
x0002y00033
x 0004 2y 0003 00023 00025
5
(1, 00022)
x
x 0003 1, y 0003 00022 x0002y00033 Check: ? 1 0002 (00022) 0003 3 ✓ 300033 x 0004 2y 0003 00023 ? 1 0004 2(00022) 0003 00023 ✓ 00023 0003 00023
00025
2. 3. 4. 5. 6. 7. 8. 9. 10.
7-1
(A) (3, 2) or x 0003 3 and y 0003 2 (B) No solutions (C) Infinite number of solutions x 0003 1, y 0003 00022 x 0003 00021, y 0003 3 (00021, 0, 2) or x 0003 00021, y 0003 0, z 0003 2 Inconsistent system with no solution {(00026s 0002 1, 7s 0004 4, s) | s any real number} Boat: 13 mph; current: 3 mph Equilibrium quantity 0003 9 thousand pounds; Equilibrium price 0003 $2.20 per pound Each week, the company should produce 900 style A shirts, 1,300 style B shirts, and 1,600 style C shirts to operate at full capacity.
Exercises
1. Explain in your own words how to solve a system of two linear equations by graphing.
y 5
2. Explain in your own words how to solve a system of two linear equations by substitution. 3. Explain in your own words how to solve a system of two linear equations using elimination by addition.
5
9. 2x 0002 y 0003 5 3x 0004 2y 0003 00023
00025
5
x
00025
(a)
(b)
y
y
6. Describe how the solution sets differ for systems of linear equations that are consistent, inconsistent, and dependent.
8. x 0004 y 0003 3 x 0002 2y 0003 0
x
00025
5. Can a system of two linear equations have exactly two solutions? Explain.
7. 2x 0002 4y 0003 8 x 0002 2y 0003 0
5
00025
4. Which of the three solving techniques is the best choice for a system of three equations? Why?
Match each system in Problems 7–10 with one of the following graphs, and use the graph to solve the system.
y
5
00025
5
x 00025
5
00025
10. 4x 0002 2y 0003 10 2x 0002 y 0003 5
00025
(c)
(d)
x
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Solve the system of equations in Problems 11–46.
Systems of Linear Equations
439
Problems 49 and 50 refer to the system
11. x 0002 y 0003 7 x0004y00033
12. x 0004 y 0003 2 x0002y00034
13. 3x 0004 2y 0003 12 7x 0002 2y 0003 8
14. 3x 0004 y 0003 2 x 0002 2y 0003 10
15. 3u 0002 5v 0003 15 6u 0002 10v 0003 000430
16. m 0002 2n 0003 4 2m 0002 4n 0003 00048
49. Solve the system for x and y in terms of the constants a, b, c, d, h, and k. Clearly state any assumptions you must make about the constants during the solution process.
3x 0004 y 0003 00042 00049x 0002 3y 0003 6
18. 2x 0004 8y 0003 10 8x 0004 32y 0003 40
50. Discuss the nature of solutions to systems that do not satisfy the assumptions you made in Problem 49.
17.
19. x 0004 y 0003 4 x 0002 3y 0003 12
20. 3x 0004 y 0003 7 2x 0002 3y 0003 1
21. 4x 0002 3y 0003 26 3x 0004 11y 0003 00047
22. 9x 0004 3y 0003 24 11x 0002 2y 0003 1
23. 7m 0002 12n 0003 00041 5m 0004 3n 0003 7
24. 3p 0002 8q 0003 4 15p 0002 10q 0003 000410
25. y 0003 0.08x y 0003 100 0002 0.04x
26. 0.2u 0004 0.5v 0003 0.07 0.8u 0004 0.3v 0003 0.79
27. 25x 0002 32 y 0003 2 7 5 3 x 0004 4 y 0003 00045
28. 5x 0004 2y 0003 8 2x 0002 3y 0003 000410
29. 00042.3y 0002 4.1z 0003 000414.21 30. 5.4x 0002 4.2y 0003 000412.9 10.1y 0004 2.9z 0003 26.15 3.7x 0002 6.4y 0003 00044.5 0003 2 31. 00042x x 0004 3y 0003 2 0004x 0002 2y 0002 3z 0003 00047 33.
2y 0004 z 0003 2 00044y 0002 2z 0003 1 x 0004 2y 0002 3z 0003 0
32.
2y 0002 z 0003 00044 x 0004 3y 0002 2z 0003 9 0004y 0003 3
34. x 0002 y 0004 z 0003 3 x 0004 2z 0003 1 y0002 z00032
0003 2 35. x 0004 3y 2y 0002 z 0003 00041 x0004 y0002z0003 1
00031 36. 00044x 0002 3y 8x 0004 6y 00034 2x 0004 4y 0002 3z 0003 6
z 0003 00045 37. 2x 0002 x0004 3z 0003 00046 4x 0002 2y 0004 z 0003 00049
38.
39. x 0004 y 0002 z 0003 1 2x 0002 y 0002 z 0003 6 7x 0004 y 0002 5z 0003 15
40. 2x 0004 y 0002 3z 0003 7 x 0002 2y 0004 z 0003 00043 3x 0002 y 0002 2z 0003 2
41. 2a 0002 4b 0002 3c 0003 00046 a 0004 3b 0002 2c 0003 000415 0004a 0002 2b 0004 c 0003 9
42. 3u 0004 2v 0002 3w 0003 11 2u 0002 3v 0004 2w 0003 00045 u 0002 4v 0004 w 0003 00045
43. 2x 0004 3y 0002 3z 0003 00045 3x 0002 2y 0004 5z 0003 34 5x 0004 4y 0004 2z 0003 23
44.
45. 0004x 0002 2y 0004 z 0003 00044 2x 0002 5y 0004 4z 0003 000416 x 0002 y 0004 z 0003 00044
46. x 0004 8y 0002 2z 0003 00041 x 0004 3y 0002 z 0003 1 2x 0004 11y 0002 3z 0003 2
x 0004 3y 0002 z 0003 4 0004x 0002 4y 0004 4z 0003 1 2x 0004 y 0002 5z 0003 00043
x 0002 2y 0002 z 0003 2 00042x 0002 3y 0004 2z 0003 00043 x 0004 5y 0002 z 0003 2
In Problems 47 and 48, solve each system for p and q in terms of x and y. Explain how you could check your solution and then perform the check. 47. x 0003 2 0002 p 0004 2q y 0003 3 0004 p 0002 3q
48. x 0003 00041 0002 2p 0004 q y0003 40004 p0002q
ax 0002 by 0003 h cx 0002 dy 0003 k where x and y are variables and a, b, c, d, h, and k are real constants.
APPLICATIONS 51. AIRSPEED It takes a private airplane 8.75 hours to make the 2,100-mile flight from Atlanta to Los Angeles and 5 hours to make the return trip. Assuming that the wind blows at a constant rate from Los Angeles to Atlanta, find the airspeed of the plane and the wind rate. 52. AIRSPEED A plane carries enough fuel for 20 hours of flight at an airspeed of 150 miles per hour. How far can it fly into a 30 mph headwind and still have enough fuel to return to its starting point? (This distance is called the point of no return.) 53. RATE–TIME A crew of eight can row 20 kilometers per hour in still water. The crew rows upstream and then returns to its starting point in 15 minutes. If the river is flowing at 2 km/h, how far upstream did the crew row? 54. RATE–TIME It takes a boat 2 hours to travel 20 miles down a river and 3 hours to return upstream to its starting point. What is the rate of the current in the river? 55. BUSINESS A company that supplies bulk candy to bakeries has one batch of chocolate chips that are 50% dark chocolate and 50% milk chocolate. They have another batch that is 80% dark chocolate and 20% milk chocolate. One of their customers sends in a rush order for 100 lb of a mix that is 68% dark chocolate. How many pounds from each batch should be mixed to meet this order? 56. BUSINESS A jeweler has two bars of gold alloy in stock, one of 12 carats and the other of 18 carats (24-carat gold is pure gold, 1212 carat is 24 pure, 18-carat gold is 18 24 pure, and so on). How many grams of each alloy must be mixed to obtain 10 grams of 14-carat gold?
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57. BREAK-EVEN ANALYSIS It costs a small recording company $17,680 to prepare a compact disc. This is a one-time fixed cost that covers recording, package design, and so on. Variable costs, including such things as manufacturing, marketing, and royalties, are $4.60 per CD. If the CD is sold to music shops for $8 each, how many must be sold for the company to break even?
(B) Find the supply and the demand (to the nearest unit) if baseball caps are priced at $8 each. Discuss the stability of the baseball cap market at this price level. (C) Find the equilibrium price and quantity. (D) Graph the two equations in the same coordinate system and identify the equilibrium point, supply curve, and demand curve.
58. FINANCE Suppose you have $12,000 to invest. If part is invested at 10% and the rest at 15%, how much should be invested at each rate to yield 12% on the total amount invested?
63. SUPPLY AND DEMAND At $0.60 per bushel, the daily supply for wheat is 450 bushels and the daily demand is 645 bushels. When the price is raised to $0.90 per bushel, the daily supply increases to 750 bushels and the daily demand decreases to 495 bushels. Assume that the supply and demand equations are linear. (A) Find the supply equation. (B) Find the demand equation. (C) Find the equilibrium price and quantity.
59. PRODUCTION A supplier for the electronics industry manufactures keyboards and screens for graphing calculators at plants in Mexico and Taiwan. The hourly production rates at each plant are given in the table. How many hours should each plant be operated to fill an order for exactly 4,000 keyboards and exactly 4,000 screens? Plant
Keyboards
Screens
Mexico
40
32
Taiwan
20
32
60. PRODUCTION A company produces Italian sausages and bratwursts at plants in Green Bay and Sheboygan. The hourly production rates at each plant are given in the table. How many hours should each plant be operated to exactly fill an order for 62,250 Italian sausages and 76,500 bratwursts? Plant
Italian Sausage
Bratwurst
Green Bay
800
800
Sheboygan
500
1,000
61. SUPPLY AND DEMAND Suppose the supply and demand equations for printed T-shirts in a resort town for a particular week are p0002
0.007q 0003 3
p 0002 00040.018q 0003 15
Supply equation Demand equation
where p is the price in dollars and q is the quantity. (A) Find the supply and the demand (to the nearest unit) if T-shirts are priced at $4 each. Discuss the stability of the T-shirt market at this price level. (B) Find the supply and the demand (to the nearest unit) if T-shirts are priced at $8 each. Discuss the stability of the T-shirt market at this price level. (C) Find the equilibrium price and quantity. (D) Graph the two equations in the same coordinate system and identify the equilibrium point, supply curve, and demand curve. 62. SUPPLY AND DEMAND Suppose the supply and demand equations for printed baseball caps in a resort town for a particular week are p0002
0.006q 0003 2
p 0002 00040.014q 0003 13
Supply equation
64. SUPPLY AND DEMAND At $1.40 per bushel, the daily supply for soybeans is 1,075 bushels and the daily demand is 580 bushels. When the price falls to $1.20 per bushel, the daily supply decreases to 575 bushels and the daily demand increases to 980 bushels. Assume that the supply and demand equations are linear. (A) Find the supply equation. (B) Find the demand equation. (C) Find the equilibrium price and quantity. 65. EARTH SCIENCE An earthquake emits a primary wave and a secondary wave. Near the surface of the Earth the primary wave travels at about 5 miles per second and the secondary wave at about 3 miles per second. From the time lag between the two waves arriving at a given receiving station, it is possible to estimate the distance to the quake. (The epicenter can be located by obtaining distance bearings at three or more stations.) Suppose a station measured a time difference of 16 seconds between the arrival of the two waves. How long did each wave travel, and how far was the earthquake from the station? 66. EARTH SCIENCE A ship using sound-sensing devices above and below water recorded a surface explosion 6 seconds sooner by its underwater device than its above-water device. Sound travels in air at about 1,100 feet per second and in seawater at about 5,000 feet per second. (A) How long did it take each sound wave to reach the ship? (B) How far was the explosion from the ship? 67. PRODUCTION SCHEDULING A company manufactures three products; lawn mowers, snowblowers, and chain saws. The labor, material, and shipping costs for manufacturing one unit of each product are given in the table. The weekly allocations for labor, materials, and shipping are $35,000, $50,000, and $20,000, respectively. How many of each type of product should be manufactured each week in order to exactly use the weekly allocations? Product
Labor
Materials
Shipping
Lawn mower
$20
$35
$15
Snowblower
$30
$50
$25
Chain saw
$45
$40
$10
Demand equation
where p is the price in dollars and q is the quantity in hundreds. (A) Find the supply and the demand (to the nearest unit) if baseball caps are priced at $4 each. Discuss the stability of the baseball cap market at this price level.
68. PRODUCTION SCHEDULING A company manufactures three products; desk chairs, file cabinets, and printer stands. The labor, material, and shipping costs for manufacturing one unit of each product are given in the table. The weekly allocations for labor, materials, and shipping are $21,100, $31,500, and $11,900, respec-
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tively. How many of each type of product should be manufactured each week in order to exactly use the weekly allocations? Product
Desk Chair
File Cabinet
Printer Stand
Labor
$30
$35
$40
Materials
$45
$60
$55
Shipping
$25
$20
$15
69. PRODUCTION SCHEDULING A company has plants located in Michigan, New York, and Ohio where it manufactures laptop computers, desktop computers, and servers. The number of units of each product that can be produced per day at each plant are given in the table below. The company has orders for 2,150 laptop computers, 2,300 desktop computers, and 2,500 servers. How many days should the company operate each plant in order to exactly fill these orders? Plant
Michigan
New York
Ohio
Laptop
10
70
60
Desktop
20
50
80
Server
40
30
90
70. PRODUCTION SCHEDULING A company has plants located in Maine, Utah, and Oregon where it manufactures stoves, refrigerators, and dishwashers. The number of units of each product that can be produced per day at each plant are given in the table. The company has orders for 1,500 stoves, 2,350 refrigerators, and 2,400 dishwashers.
7-2
441
Solving Systems of Linear Equations Using Gauss–Jordan Elimination
How many days should the company operate each plant in order to exactly fill these orders? Set up a system of equations whose solution would answer this question and solve the system. Plant
Stoves
Refrigerators
Dishwashers
Maine
30
70
60
Utah
20
50
50
Oregon
40
30
40
71. INVESTMENT Due to recent volatility in the stock market, Catalina’s financial advisor suggests that she reallocate $70,000 of her retirement fund to bonds. He recommends a mix of treasury bonds earning 4% annually, municipal bonds earning 3.5% annually, and corporate bonds earning 4.5% annually. For tax reasons, he also recommends that the amount invested in treasury bonds should be equal to the sum of the amount invested in the other categories. If Catalina follows these recommendations, and the goal is to produce $2,900 in annual interest income, how much will she invest in each of the three types of bonds? 72. INVESTMENT When the real estate market begins to rebound, Catalina (see Problem 71) decides to reallocate her investment mix. At this point, her investment has grown to $76,000. She’ll leave some money in treasury and corporate bonds, but will replace municipal bonds with a real estate investment trust that guarantees a 6.5% annual return. If she plans to leave as much in treasury bonds as the sum of the other two investments, how much should she invest in each to reach her new goal of earning an annual interest income of $3,600?
Solving Systems of Linear Equations Using Gauss–Jordan Elimination Z Matrices and Row Operations Z Reduced Matrices Z Solving Systems by Gauss–Jordan Elimination Z Application
In this section, we introduce Gauss–Jordan elimination, a step-by-step procedure for solving systems of linear equations. This procedure works for any system of linear equations and is easily implemented on a computer. In fact, the TI-84 has a built-in procedure for performing Gauss–Jordan elimination.
Z Matrices and Row Operations In solving systems of equations using elimination by addition, the coefficients of the variables and the constant terms played a central role. The process can be made more efficient
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by the introduction of a mathematical form called a matrix. A matrix (plural matrices) is a rectangular array of numbers written within brackets. Two examples are 1 A0002 c 5
00043 0
7 d 00044
00045 0 B0002 ≥ 00042 00043
4 1 12 0
11 6 ¥ 8 00041
(1)
Each number in a matrix is called an element of the matrix. Matrix A has six elements arranged in two rows and three columns. Matrix B has 12 elements arranged in four rows and three columns. If a matrix has m rows and n columns, it is called an m ⴛ n matrix (read “m by n matrix”). The expression m 0005 n is called the size of the matrix, and the numbers m and n are called the dimensions of the matrix. It is important to note that the number of rows is always given first. Referring to equations (1), A is a 2 0005 3 matrix and B is a 4 0005 3 matrix. A matrix with n rows and n columns is called a square matrix of order n. A matrix with only one column is called a column matrix, and a matrix with only one row is called a row matrix. These definitions are illustrated by the following: 3 ⴛ 3
0.5 £ 0.0 0.7
0.2 0.3 0.0
4 ⴛ 1
3 00042 ≥ ¥ 1 0
1.0 0.5 § 0.2
Square matrix of order 3
Column matrix
1 ⴛ 4
[2
1 2
0
000423 ]
Row matrix
The position of an element in a matrix is the row and column containing the element. This is usually denoted using double subscript notation aij, where i is the row and j is the column containing the element aij, as illustrated next: 1 A0002 £ 6 00042
5 0 3
00043 00044 4
4 1§ 7
a11 0002 00041, a12 0002 5, a13 0002 00043, a14 0002 4 a21 0002 00046, a22 0002 0, a23 0002 00044, a24 0002 1 a31 0002 00042, a32 0002 3, a33 0002 00044, a34 0002 7
Note that a12 is read “a sub one two,” not “a sub twelve.” The elements a11 0002 1, a22 0002 0, and a33 0002 4 make up the principal diagonal of A. In general, the principal diagonal of a matrix A consists of the elements aii, i 0002 1, 2, . . . n.
Technology Connections Most graphing calculators are capable of storing and manipulating matrices. Figure 1 shows matrix A displayed in the matrix editing screen of a TI-84 graphing calculator. The size of the matrix is given at the top of the screen, and the position of the currently selected element is given at the bottom. Notice that a comma is used in the notation for the position. This is common practice on graphing calculators but it’s almost never written or typed that way.
Z Figure 1 Matrix notation on a TI-84 graphing calculator.*
*The onscreen display of A was too large to fit on the screen of a TI-84, so we pasted together two screen shots to form Figure 1. When this happens on your graphing calculator, you will have to scroll left and right and/or up and down to see the entire matrix.
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Solving Systems of Linear Equations Using Gauss–Jordan Elimination
443
Now we turn our attention to the connection between matrices and systems of equations. Consider the system of equations x 0003 5y 0004 3z 0002 4 6x 0004 4z 0002 1 00042x 0003 3y 0003 4z 0002 7
(2)
If we remove the variables and leave behind the numbers, we can think of the result as a matrix: 1 5 00043 4 £ 6 0 00044 † 1 § 00042 3 4 7 Coefficient matrix
1 £ 6 00042
5 0 3
00043 00044 § 4
Constant matrix
4 £1§ 7
Z Figure 2
This is known as the augmented coefficient matrix for the system. We can also define the coefficient matrix and the constant matrix for the system, as shown in Figure 2. The augmented coefficient matrix contains all of the information about the system needed to solve it. Note that we put in a coefficient of zero for the missing y in the second equation, and that we drew a vertical bar to separate the coefficients from the constants. (Matrices displayed on a graphing calculator won’t have that line.) Since we would like to be able to use matrices to solve large systems with many variables, moving forward we will use x1, x2, x3, and the like, rather than x, y, z, and so on. In this notation, we will rewrite system (2) as x1 0003 5x2 0004 3x3 0002 4 6x1 0004 4x3 0002 1 00042x1 0003 3x2 0003 4x3 0002 7 In Section 7-1, we used Ei to denote the equations in a linear system. Now we use Ri to denote the rows and Ci to denote the columns, respectively, in a matrix, as illustrated below for system (2). C1
C2
1 £ 6 00042
5 0 3
C3
C4
00043 4 00044 † 1 § 4 7
R1 R2
(3)
R3
Our goal will be to learn how to perform the basic steps we used to solve systems using elimination by addition, but on an augmented matrix. This enables us to focus on the numbers without being concerned about algebraic manipulations.
EXAMPLE
1
Writing an Augmented Coefficient Matrix Write the augmented coefficient matrix corresponding to each of the following systems. (A)
SOLUTIONS
MATCHED PROBLEM 1
2x1 0004 4x2 0002 5 00043x1 0003 x2 0002 00046
(A) c
2 00043
00044 5 ` d 1 00046
(B) 00043x1 0003 2x3 0002 00044 7x1 0004 5x2 0003 3x3 0002 0
(B) c
00043 7
0 2 00044 d ` 00045 3 0
(C) 2x1 0004 x2 0002 4 3x1 0004 5x3 0002 6 00042x2 0003 x3 0002 00043
2 00041 0 4 (C) £ 3 0 00045 † 6 § 0 00042 1 00043
0002
Write the augmented coefficient matrix corresponding to each of the following systems. (A) 0004x1 0003 2x2 0002 00043 3x1 0004 5x2 0002 8
(B)
00042x2 0003 2x3 0002 00044 7x1 00045x2 0003 3x3 0002 0
(C) 2x1 0004 x2 0003 x3 0002 4 3x1 0003 4x2 0002 6 x1 0003 5x3 0002 00043
0002
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Recall that two linear systems are said to be equivalent if they have the same solution set. In Theorem 2, Section 7-1, we used the operations listed next to transform linear systems into equivalent systems: (A) Two equations are interchanged. (B) An equation is multiplied by a nonzero constant. (C) A constant multiple of one equation is added to another equation. Paralleling this approach, we now say that two augmented matrices are row-equivalent, denoted by the symbol ⬃ between the two matrices, if they are augmented matrices of equivalent systems of equations. How do we transform augmented matrices into row-equivalent matrices? We use Theorem 1, which gives the matrix analogs of operations (A), (B), and (C).
Z THEOREM 1 Elementary Row Operations Producing Row-Equivalent Matrices An augmented matrix is transformed into a row-equivalent matrix if any of the following row operations is performed: 1. Two rows are interchanged (Ri 4 Rj). 2. A row is multiplied by a nonzero constant (kRi S Ri). 3. A constant multiple of one row is added to another row (kRj 0003 Ri S Ri). [Note: The arrow means “replaces.”]
EXAMPLE
2
Row Operations Perform each of the indicated row operations on the following augmented coefficient matrix. c (A) R1 4 R2
SOLUTIONS
MATCHED PROBLEM 2
(A) c
2 1
(B) 12R2 S R2
4 00048 d ` 00044 3
(B) c
1 2
00044 3 d ` 4 00048
(C) (00042)R 1 0003 R2 S R2 1 1
00044 3 d ` 2 00044
(C) c
1 00044 3 d ` 0 12 000414
0002
Perform each of the indicated row operations on the following augmented coefficient matrix. c (A) R1 4 R2
(B) 13R2 S R2
1 3
00042 3 d ` 00046 00043
(C) (00043)R 1 0003 R2 S R2
0002
Z Reduced Matrices The goal of the elimination process is to transform a system of equations into an equivalent system whose solution is easy to find. Now our goal is to use a sequence of matrix row operations to transform an augmented coefficient matrix into a simpler equivalent matrix that corresponds to a system with an obvious solution. Example 3 illustrates the process of interpreting the solution of a system given its augmented coefficient matrix.
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EXAMPLE
3
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Interpreting an Augmented Coefficient Matrix Write the system corresponding to each of the following augmented coefficient matrices and find its solution. 1 0 0 00044 (A) £ 0 1 0 † 6§ 0 0 1 0
SOLUTIONS
1 0 (B) £ 0 1 0 0
2 00044 00043 † 6 § 0 1
1 0 2 00044 (C) £ 0 1 00043 † 6 § 0 0 0 0
(A) The corresponding system is x1 0002 00044 x2 0002 00046 x3 0002 00040 and (00044, 6, 0) is the solution. (B) The corresponding system is
1 ⴢ x1 ⴙ 0 ⴢ x2 ⴙ 0 ⴢ x3 ⴝ x1
0003 2x3 0002 00044 x2 0004 3x3 0002 6 0 ⴢ x1 0003 0 ⴢ x2 0003 0 ⴢ x3 0002 1 x1
The third equation, 0 0002 1, is a contradiction, so the system has no solutions. (C) The first two rows of this augmented coefficient matrix correspond to the system x1
0003 2x3 0002 00044 x2 0004 3x3 0002 6
The third row corresponds to the equation 0 ⴝ 0, which is always true and can be discarded.
This is a dependent system with an infinite number of solutions. Introducing a parameter s, we can write x1
0003 2s 0002 00044 x2 0004 3s 0002 6 or x3 0002 s
x1 0002 00042s 0004 4 x2 0002 3s 0003 6 x3 0002 s
So the solution set is {(00042s 0004 4, 3s 0003 6, s) | s any real number} MATCHED PROBLEM 3
Write the system corresponding to each of the following augmented coefficient matrices and find its solution. 1 (A) £ 0 0
ZZZ EXPLORE-DISCUSS 1
0002
0 1 0
0 5 0 † 00047 § 1 0
1 (B) £ 0 0
0 1 0
00043 5 4 † 00047 § 0 0
1 (C) £ 0 0
0 1 0
00043 5 4 † 00047 § 0 1 0002
If an augmented coefficient matrix contains a row where every element on the left of the vertical line is 0 and the single element on the right is a nonzero number, what can you say about the solution of the corresponding system?
Next, we will define a particular matrix form that makes it simple to find solutions of the corresponding system.
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Z DEFINITION 1 Reduced Matrix A matrix is in reduced form* if: 1. Each row consisting entirely of 0’s is below any row having at least one nonzero element. 2. The leftmost nonzero element in each row is 1. 3. The column containing the leftmost 1 of a given row has 0’s above and below the 1. 4. The leftmost 1 in any row is to the right of the leftmost 1 in the preceding row.
For example, each of the following matrices is in reduced form. Before moving on, you should verify that each matrix satisfies all four conditions in Definition 1. 1 c 0
EXAMPLE
4
1 0 2 ` d £0 1 00043 0
0 1 0
0 2 1 0 † 00041 § £ 0 1 3 0
0 3 1 1 † 00041 § £ 0 0 0 0
4 0 0
0 1 0
0 3 1 0 † 2§ £0 1 6 0
0 1 0
4 0 3 † 0§ 0 1
Reduced Forms The matrices shown next are not in reduced form. Indicate which condition in the definition is violated for each matrix. State the row operation(s) required to transform the matrix to reduced form, and find the reduced form. 0 1
1 00042 ` d 0 3
(B) c
1 0
2 0
00042 3 ` d 1 00041
1 (C) £ 0 0
0 00043 0 † 0§ 1 00042
1 (D) £ 0 0
0 2 0
0 00041 0 † 3§ 1 00045
(A) c
SOLUTIONS
(A) Condition 4 is violated: The leftmost 1 in row 2 is not to the right of the leftmost 1 in row 1. Perform the row operation R1 4 R2 to obtain the reduced form: c
0 1
1 00042 1 ` d R1 4 R2 c 0 3 0
0 3 ` d 1 00042
(B) Condition 3 is violated: The column containing the leftmost 1 in row 2 does not have a zero above the 1. Perform the row operation 2R2 0003 R1 S R1 to obtain the reduced form: c
1 0
2 0
00042 3 1 ` d 2R2 0003 R1 S R1 c 1 00041 0
2 0
0 1 ` d 1 00041
(C) Condition 1 is violated: The second row contains all zeros, and it is not below any row having at least one nonzero element. Perform the row operation R2 4 R3 to obtain the reduced form: 1 £0 0
0 00043 1 0 † 0 § R2 4 R3 £ 0 1 00042 0
0 00043 1 † 00042 § 0 0
*The reduced form we have defined here is sometimes called the reduced row echelon form, and most graphing calculators use the abbreviation rref to refer to it. There are other reduced forms that can be used to solve systems of equations, but we will use the term “reduced form” for simplicity.
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(D) Condition 2 is violated: The leftmost nonzero element in row 2 is not a 1. Perform the row operation 12R2 S R2 to obtain the reduced form: 1 £0 0 MATCHED PROBLEM 4
0 2 0
1 0 00041 0 † 3 § 12R2 S R2 £ 0 0 1 00045
0 1 0
0 00041 0 † 32 § 1 00045
0002
The matrices below are not in reduced form. Indicate which condition in the definition is violated for each matrix. State the row operation(s) required to transform the matrix to reduced form and find the reduced form. (A) c
1 0
0 (C) £ 1 0
1 (B) £ 0 0
0 2 ` d 3 00046 1 0 0
5 4 3 1 2 † 00041 § 0 0 0 1 2 0 3 (D) £ 0 0 0 † 0 § 0 0 1 4
0 00043 0 † 0§ 1 2
0002
Z Solving Systems by Gauss–Jordan Elimination We are now ready to outline the Gauss–Jordan elimination method for solving systems of linear equations. The method systematically transforms an augmented matrix into a reduced form. The system corresponding to a reduced augmented coefficient matrix is called a reduced system. As we will see, reduced systems are easy to solve. The Gauss–Jordan elimination method is named after the German mathematician Carl Friedrich Gauss (1777–1855) and the German geodesist Wilhelm Jordan (1842–1899). Gauss, one of the greatest mathematicians of all time, used a method of solving systems of equations that was later generalized by Jordan to solve problems in large-scale surveying.
EXAMPLE
5
Solving a System Using Gauss–Jordan Elimination Solve by Gauss–Jordan elimination: 2x1 0004 2x2 0003 x3 0002 3 3x1 0003 x2 0004 x3 0002 7 x1 0004 3x2 0003 2x3 0002 0
SOLUTION
Write the augmented matrix and follow the steps indicated at the right to produce a reduced form.
Need a 1 here.
冤
2 3 1
00042 1 00043
1 00041 2
Need 0’s here.
1 ⬃ 3 2
00043 1 00042
2 00041 1
冤
ⱍ冥 ⱍ冥 3 7 0 0 7 3
R1 ↔ R3
(ⴚ3)R1 ⴙ R2 → R2 (ⴚ2)R1 ⴙ R3 → R3
Step 1: Choose the leftmost nonzero column and get a 1 at the top.
Step 2: Use multiples of the row containing the 1 from step 1 to get zeros in all remaining places in the column containing this 1.
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ⱍ冥 ⱍ冥 ⱍ冥 ⱍ冥
Need a 1 here.
1 ⬃ 0 0
冤
00043 10 4
2 00047 00043
Need 0’s here.
1 ⬃ 0 0
冤
00043 1 4
2 00040.7 00043
0 0.7 3
1 ⬃ 0 0
冤
0 1 0
00040.1 00040.7 00040.2
2.1 0.7 0.2
1 ⬃ 0 0
冤
0 1 0
00040.1 00040.7 1
2.1 0.7 00041
冤
0 1 0
0 0 1
Need a 1 here.
Need 0’s here.
1 ⬃ 0 0
0 7 3
0.1R2 → R2
3R2 ⴙ R1 → R1
Step 4: Repeat step 2 with the entire matrix.
(ⴚ4)R2 ⴙ R3 → R3
(ⴚ5)R3 → R3
0.1R3 ⴙ R1 → R1 0.7R3 ⴙ R2 → R2
ⱍ冥 2 0 00041
x1 x2
Step 3: Repeat step 1 with the submatrix formed by (mentally) deleting the top (shaded) row.
Step 3: Repeat step 1 with the submatrix formed by (mentally) deleting the top two (shaded) rows.
Step 4: Repeat step 2 with the entire matrix.
The matrix is now in reduced form, and we can proceed to solve the corresponding reduced system.
0002 2 0002 0 x3 0002 00041
The solution to this system is x1 0002 2, x2 0002 0, x3 0002 00041. You should check this solution in the original system.
Z GAUSS–JORDAN ELIMINATION Step 1. Choose the leftmost nonzero column and use appropriate row operations to get a 1 at the top. Step 2. Use multiples of the row containing the 1 from step 1 to get zeros in all remaining places in the column containing this 1. Step 3. Repeat step 1 with the submatrix formed by (mentally) deleting the row used in step 2 and all rows above this row. Step 4. Repeat step 2 with the entire matrix, including the mentally deleted rows. Continue this process until the entire matrix is in reduced form. [Note: If at any point in this process we obtain a row with all zeros to the left of the vertical line and a nonzero number to the right, we can stop, since we will have a contradiction: 0 0002 n, n 0006 0. We can then conclude that the system has no solution.] 0002 MATCHED PROBLEM 5
Solve by Gauss–Jordan elimination:
3x1 0003 x2 0004 2x3 0002 2 x1 0004 2x2 0003 x3 0002 3 2x1 0004 x2 0004 3x3 0002 3
0002
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Technology Connections Figure 3 illustrates the solution of Example 5 on a TI-84 graphing calculator using the built-in rref (reduced rowechelon form) routine for finding reduced forms. Notice that in row 2 and column 4 of the reduced form the graphing calculator has displayed the very small number -3.5E-13 instead of the exact value 0. This is a common occurrence caused by rounding error on a graphing calculator and causes no problems. Just replace any very small numbers displayed in scientific notation with 0.
EXAMPLE
6
Z Figure 3 Using rref on a TI-84 graphing calculator.
Solving a System Using Gauss–Jordan Elimination Solve by Gauss–Jordan elimination:
SOLUTION
2x1 0004 4x2 0003 x3 0002 00044 4x1 0004 8x2 0003 7x3 0002 2 00042x1 0003 4x2 0004 3x3 0002 5
2 £ 4 00042
00044 1 00044 00048 7 † 2§ 4 00043 5
1 ⬃£ 4 00042
00042 0.5 00042 00048 7 † 2§ 4 00043 5
1 ⬃£0 0
00042 0 0
0.5 00042 5 † 10 § 00042 1
1 ⬃£0 0
00042 0 0
0.5 00042 1 † 2§ 00042 1
1 ⬃£0 0
00042 0 00043 0 1 † 2§ 0 0 5
0.5R1 S R1 (To get 1 in upper left corner)
(Next, get zeros below that 1.) (ⴚ4)R1 ⴙ R2 S R2 2R1 ⴙ R3 S R3 0.2R2 S R2 Note that column 3 is the leftmost nonzero column in this submatrix.
(ⴚ0.5)R2 ⴙ R1 S R1
2R2 ⴙ R3 S R3 We stop the Gauss–Jordan elimination, even though the matrix is not in reduced form, since the last row produces a contradiction
The system is inconsistent and has no solution. MATCHED PROBLEM 6
Solve by Gauss–Jordan elimination:
2x1 0004 4x2 0004 x3 0002 00048 4x1 0004 8x2 0003 3x3 0002 4 00042x1 0003 4x2 0003 x3 0002 11
0002
0002
Note that if we were to use rref on a graphing calculator for Example 6, it would continue reducing further. But the final reduced form would still show a contradiction.
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7
Solving a System Using Gauss–Jordan Elimination Solve by Gauss–Jordan elimination:
SOLUTION
3x1 0003 6x2 0004 9x3 0002 15 2x1 0003 4x2 0004 6x3 0002 10 00042x1 0004 3x2 0003 4x3 0002 00046
3 £ 2 00042
6 00049 15 4 00046 † 10 § 00043 4 00046
1 ⬃£ 2 00042
2 00043 5 4 00046 † 10 § 00043 4 00046
1 ⬃£0 0
2 0 1
00043 5 0 † 0§ 00042 4
1 ⬃£0 0
2 1 0
00043 5 00042 † 4 § 0 0
1 0 1 00043 ⬃ £ 0 1 00042 † 4 § 0 0 0 0 x1 0003 x3 0002 00043 x2 0004 2x3 0002 4
1 3 R1 S R1
(ⴚ2)R1 ⴙ R2 S R2 2R1 ⴙ R3 S R3 R2 4 R3 Note that we must interchange rows 2 and 3 to obtain a nonzero entry at the top of the second column of this submatrix. (ⴚ2)R2 ⴙ R1 S R1
This matrix is now in reduced form. Write the corresponding reduced system and solve.
We discard the equation corresponding to the third (all 0) row in the reduced form, since it is satisfied by all values of x1, x2, and x3.
Note that the leftmost variable in each equation appears in one and only one equation. We solve for the leftmost variables x1 and x2 in terms of the remaining variable x3: x1 0002 0004x3 0004 3 x2 0002 2x3 0003 4 This dependent system has an infinite number of solutions. We will use a parameter to represent all the solutions. If we let x3 0002 t, then for any real number t, x1 0002 0004t 0004 3 x2 0002 2t 0003 4 x3 0002 t is a solution. You should check that (0004t 0004 3, 2t 0003 4, t) is a solution of the original system for any real number t. Some particular solutions are
MATCHED PROBLEM 7
t ⴝ 0
t ⴝ ⴚ2
t ⴝ 3.5
(00043, 4, 0)
(00041, 0, 00042)
(00046.5, 11, 3.5)
Solve by Gauss–Jordan elimination:
2x1 0004 2x2 0004 4x3 0002 00042 3x1 0004 3x2 0004 6x3 0002 00043 00042x1 0003 3x2 0003 x3 0002 7
In general, If the number of leftmost 1’s in a reduced augmented coefficient matrix is less than the number of variables in the system and there are no contradictions, then the system is dependent and has infinitely many solutions.
0002
0002
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There are many different ways to use the reduced augmented coefficient matrix to describe the infinite number of solutions of a dependent system. We will always proceed as follows: Solve each equation in a reduced system for its leftmost variable and then introduce a different parameter for each remaining variable. Example 8 illustrates a dependent system where two parameters are required to describe the solution.
EXAMPLE
8
Solving a System Using Gauss–Jordan Elimination Solve by Gauss–Jordan elimination:
SOLUTION
x1 0003 2x2 0003 4x3 0003 x4 0004 x5 0002 1 2x1 0003 4x2 0003 8x3 0003 3x4 0004 4x5 0002 2 x1 0003 3x2 0003 7x3 0003 3x5 0002 00042
00041 1 00044 † 2 § 3 00042
1 £2 1
2 4 3
4 8 7
1 3 0
1 ⬃£0 0
2 0 1
4 0 3
1 1 00041
00041 1 00042 † 0 § 4 00043
1 ⬃£0 0
2 1 0
4 3 0
1 00041 1
00041 1 4 † 00043 § 00042 0
1 ⬃£0 0
0 1 0
00042 3 0
3 00049 7 00041 4 † 00043 § 1 00042 0
1 ⬃£0 0
0 1 0
00042 3 0
0 0 1
00043 7 2 † 00043 § 00042 0
x1 0004 2x3 x2 0003 3x3
(ⴚ2)R1 ⴙ R2 S R2 (ⴚ1)R1 ⴙ R3 S R3
R2 4 R3
(ⴚ2)R2 ⴙ R1 S R1
(ⴚ3)R3 ⴙ R1 S R1 R3 ⴙ R2 S R2
Matrix is in reduced form.
0004 3x5 0002 7 0003 2x5 0002 00043 x4 0004 2x5 0002 0
Solve for the leftmost variables x1, x2, and x4 in terms of the remaining variables x3 and x5: x1 0002 2x3 0003 3x5 0003 7 x2 0002 00043x3 0004 2x5 0004 3 x4 0002 2x5 If we let x3 0002 s and x5 0002 t, then for any real numbers s and t, x1 x2 x3 x4 x5
0002 2s 0003 3t 0003 7 0002 00043s 0004 2t 0004 3 0002s 0002 2t 0002t
is a solution. The check is left for you to perform. MATCHED PROBLEM 8
Solve by Gauss–Jordan elimination:
0004 2x5 0002 3 x1 0004 x2 0003 2x3 00042x1 0003 2x2 0004 4x3 0004 x4 0003 x5 0002 00045 3x1 0004 3x2 0003 7x3 0003 x4 0004 4x5 0002 6
0002
0002
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Z Application Dependent systems probably seem very abstract to you—a solution like the one in Example 8 doesn’t seem like it would apply to any real–world situations. But in Example 9, we will solve a problem where a dependent system leads to real solutions.
EXAMPLE
9
Purchasing A chemical manufacturer plans to purchase a fleet of 24 railroad tank cars with a combined carrying capacity of 250,000 gallons. Tank cars with three different carrying capacities are available: 6,000 gallons, 8,000 gallons, and 18,000 gallons. How many of each type of tank car should be purchased?
SOLUTION
Let x1 0002 Number of 6,000-gallon tank cars x2 0002 Number of 8,000-gallon tank cars x3 0002 Number of 18,000-gallon tank cars Then x1 0003 x2 0003 x3 0002 24 6,000x1 0003 8,000x2 0003 18,000x3 0002 250,000
Total number of tank cars Total carrying capacity
Now we can form the augmented matrix of the system and solve by using Gauss–Jordan elimination: 1 6,000
1 8,000
⬃ c
1 6
1 8
1 24 ` d 18 250
(ⴚ6)R1 ⴙ R2 S R2
⬃ c
1 0
1 2
1 24 ` d 12 106
1 2 R2
⬃ c
1 0
1 1
1 24 ` d 6 53
(ⴚ1)R2 ⴙ R1 S R1
⬃ c
1 0
0 1
00045 000429 ` d 6 53
c
x1
1 24 ` d 18,000 250,000
1 1,000 R2
S R2 (simplify R2)
S R2
Matrix is in reduced form.
0004 5x3 0002 000429 x2 0003 6x3 0002 53
or or
x1 0002 5x3 0004 29 x2 0002 00046x3 0003 53
Let x3 0002 t. Then for t any real number, x1 0002 5t 0004 29 x2 0002 00046t 0003 53 x3 0002 t is a solution—or is it? Since the variables in this system represent the number of tank cars purchased, the values of x1, x2, and x3 must be nonnegative integers. The third equation requires that t must be a nonnegative integer. The first equation requires that 5t 0004 29 0007 0, so t must be at least 6. The middle equation requires that 00046t 0003 53 0007 0, so t can be no larger than 8.
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So, 6, 7, and 8 are the only possible values for t. There are three different possible combinations that meet the company’s specifications of 24 tank cars with a total carrying capacity of 250,000 gallons, as shown in Table 1: Table 1
t
6,000-Gallon Tank Cars x1
8,000-Gallon Tank Cars x2
18,000-Gallon Tank Cars x3
6
1
17
6
7
6
11
7
8
11
5
8
The final choice would probably be influenced by other factors. For example, the company might want to minimize the cost of the 24 tank cars. 0002 MATCHED PROBLEM 9
A commuter airline plans to purchase a fleet of 30 airplanes with a combined carrying capacity of 960 passengers. The three available types of planes carry 18, 24, and 42 passengers, respectively. How many of each type of plane should be purchased? 0002 ANSWERS TO MATCHED PROBLEMS 1. (A) c
00041 2 00043 ` d 3 00045 8
(B) c
0 7
00042 00045
2 (C) £ 3 1
2 00044 ` d 3 0
00041 1 4 4 0 † 6§ 0 5 00043
00046 00043 1 00042 3 1 00042 3 ` d ` d ` d (B) c (C) c 00042 3 1 00042 00041 0 0 000412 3. (A) x1 0002 5, x2 0002 00047, x3 0002 0 or (5, 00047, 0) (B) x1 0002 3s 0003 5, x2 0002 00044s 0004 7, x3 0002 s, s any real number; or {(3s 0003 5, 00044s 0004 7, s) | s any real number} (C) No solution 4. (A) Condition 2 is violated: The 3 in row 2 and column 2 should be a 1. Perform the operation 1 3 R2 S R2 to obtain: 2. (A) c
3 1
c
1 0
0 2 ` d 1 00042
(B) Condition 3 is violated: The 5 in row 1 and column 2 should be a 0. Perform the operation (00045)R2 0003 R1 S R1 to obtain: 1 £0 0
00046 8 2 † 00041 § 0 0
0 1 0
(C) Condition 4 is violated: The leftmost 1 in the second row is not to the right of the leftmost 1 in the first row. Perform the operation R1 4 R2 to obtain: 1 £0 0
0 1 0
0 0 0 † 00043 § 1 2
(D) Condition 1 is violated: The all-zero second row should be at the bottom. Perform the operation R2 4 R3 to obtain: 1 £0 0
2 0 0
0 3 1 † 4§ 0 0
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x1 0002 1, x2 0002 00041, x3 0002 0 or (1, 00041, 0) No solution x1 0002 5t 0003 4, x2 0002 3t 0003 5, x3 0002 t, t any real number; or {(5t 0003 4, 3t 0003 5, t) | t any real number} x1 0002 s 0003 7, x2 0002 s, x3 0002 t 0004 2, x4 0002 00043t 0004 1, x5 0002 t, s and t any real numbers; or {(s 0003 7, s, t 0004 2, 00043t 0004 1, t) | s and t any real numbers} 18-Passenger 24-Passenger 42-Passenger 9. Planes Planes Planes t x1 x2 x3
5. 6. 7. 8.
7-2
14
2
14
14
15
5
10
15
16
8
6
16
17
11
2
17
Exercises
1. What is the size of a matrix?
In Problems 19–26, write the linear system corresponding to each reduced augmented matrix and solve.
2. What is a row matrix? What is its size? 3. What is a column matrix? What is its size? 4. What is a square matrix? 5. What does aij mean? 6. What is the principal diagonal of a matrix? 7. What is an augmented coefficient matrix? 8. What operations can you perform on an augmented coefficient matrix to produce a row-equivalent matrix? 9. What is a reduced matrix and how is it used to solve a system of linear equations? 10. Describe the Gauss–Jordan elimination process in your own words. In Problems 11–18, indicate whether each matrix is in reduced form. 11. c
1 0
0 13. £ 0 0
0 00041 ` d 2 6 1 0 0
00042 0 0 † 1§ 0 0
0 15. £ 0 1
0 1 0
1 2 0 † 00045 § 0 4
0 17. c 0
1 0
6 0
0 00048 ` d 1 1
12. c
1 0
1 14. £ 0 0
0 1 0
0 00042 0 † 3§ 1 0
1 0 20. ≥ 0 0
0 1 0 0
1 21. £ 0 0
0 1 0
00042 3 1 † 00045 § 0 0
1 22. £ 0 0
00042 0 00043 0 1 † 5§ 0 0 0
1 23. £ 0 0
0 0 1 † 0§ 0 1
1 24. £ 0 0
0 5 1 † 00043 § 0 0
1 0
00042 0 0 1
25. c
00043 00045 ` d 3 2
1 16. £ 0 0
00042 4 1 0 1 † 00043 § 0 0 0
0 18. c 0
0 0
1 0 ` d 0 0
1 0
0 1
00042 00041
3 4 ` d 2 00041
Perform each of the row operations indicated in Problems 27–38 on the following matrix:
0 5 ` d 1 00043 00041 4 0 0 0 † 0§ 0 0 1
26. c
0 0 1 0
0 00042 0 0 ¥ ∞ 1 0 3 1
1 19. £ 0 0
c
1 4
00043 2 ` d 00046 00048
27. R1 4 R2
28. 12R2 S R2
29. 00044R1 S R1
30. 00042R1 S R1
31. 2R2 S R2
32. 00041R2 S R2
(000412)R2
33. (00044)R1 0003 R2 S R2
34.
0003 R1 S R1
35. (00042)R1 0003 R2 S R2
36. (00043)R1 0003 R2 S R2
37. (00041)R1 0003 R2 S R2
38. 1R1 0003 R2 S R2
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Solving Systems of Linear Equations Using Gauss–Jordan Elimination
Use row operations to change each matrix in Problems 39–44 to reduced form. 39. c
1 0
2 00041 ` d 1 3
1 41. £ 0 0
0 1 0
1 43. £ 0 0
2 3 00041
40. c
00043 1 2 † 0§ 3 00046 00042 00041 00046 † 1 § 2 000413
1 0
3 1 ` d 2 00044
1 42. £ 0 0
0 1 0
0 44. £ 2 0
00042 00042 00041
4 0 00043 † 00041 § 00042 2 8 1 6 † 00044 § 1 4 2
63.
x1 0004 4x2 0002 00042 00042x1 0003 x2 0002 00043
47. x1 0003 2x2 0002 4 2x1 0003 4x2 0002 00048 49.
3x1 0004 6x2 0002 00049 00042x1 0003 4x2 0002 6
46.
x1 0004 3x2 0002 00045 00043x1 0004 x2 0002 5
48.
2x1 0004 3x2 0002 00042 00044x1 0003 6x2 0002 7
50.
2x1 0004 4x2 0002 00042 00043x1 0003 6x2 0002 3
65. x1 0003 2x2 0004 4x3 0004 x4 0002 7 2x1 0003 5x2 0004 9x3 0004 4x4 0002 16 x1 0003 5x2 0004 7x3 0004 7x4 0002 13 66. 2x1 0003 4x2 0003 5x3 0003 4x4 0002 8 x1 0003 2x2 0003 2x3 0003 x4 0002 3 67.
x1 0004 00042x1 0003 3x1 0004 4x1 0004
68.
x1 0003 x2 0003 4x3 0003 x4 0002 1.3 0002 1.1 0004x1 0003 x2 0004 x3 2x1 0003 x3 0003 3x4 0002 00044.4 2x1 0003 5x2 0003 11x3 0003 3x4 0002 5.6
69.
x1 0004 2x2 0003 00042x1 0003 4x2 0003 3x1 0004 6x2 0003 0004x1 0003 2x2 0003
70.
x1 0004 3x2 0003 x3 0003 x4 0003 2x5 0002 2 0004x1 0003 5x2 0003 2x3 0003 2x4 0004 2x5 0002 0 2x1 0004 6x2 0003 2x3 0003 2x4 0003 4x5 0002 4 0004x1 0003 3x2 0004 x3 0004 x5 0002 00043
51. 2x1 0003 4x2 0004 10x3 0002 00042 3x1 0003 9x2 0004 21x3 0002 0 x1 0003 5x2 0004 12x3 0002 1 52. 3x1 0003 5x2 0004 x3 0002 00047 x1 0003 x2 0003 x3 0002 00041 2x1 0003 11x3 0002 7 53. 3x1 0003 8x2 0004 x3 0002 000418 8 2x1 0003 x2 0003 5x3 0002 2x1 0003 4x2 0003 2x3 0002 00044
x2 0003 4x2 0004 x2 0003 3x2 0003
3x3 0004 2x4 0002 1 3x3 0003 x4 0002 0.5 10x3 0004 4x4 0002 2.9 8x3 0004 2x4 0002 0.6
x3 0003 2x3 0003 x3 0003 3x3 0003
x4 0003 2x5 0002 2 2x4 0004 2x5 0002 0 x4 0003 5x5 0002 4 x4 0003 x5 0002 3
71. Consider a consistent system of three linear equations in three variables. Discuss the nature of the solution set for the system if the reduced form of the augmented coefficient matrix has (A) One leftmost 1 (B) Two leftmost 1’s (C) Three leftmost 1’s (D) Four leftmost 1’s
54. 2x1 0003 7x2 0003 15x3 0002 000412 4x1 0003 7x2 0003 13x3 0002 000410 3x1 0003 6x2 0003 12x3 0002 00049 55. 2x1 0004 x2 0004 3x3 0002 8 x1 0004 2x2 00027 56. 2x1 0003 4x2 0004 6x3 0002 10 3x1 0003 3x2 0004 3x3 0002 6 57. 2x1 0004 x2 0002 0 3x1 0003 2x2 0002 7 x1 0004 x2 0002 00041
7 2x1 0004 5x2 0004 3x3 0002 6 00044x1 0003 10x2 0003 2x3 0002 6x1 0004 15x2 0004 x3 0002 000419
64. 5x1 0004 3x2 0003 2x3 0002 13 2x1 0004 x2 0004 3x3 0002 1 4x1 0004 2x2 0003 4x3 0002 12
Solve Problems 45–70 using Gauss–Jordan elimination. 45.
455
58. 2x1 0004 x2 0002 0 3x1 0003 2x2 0002 7 x1 0004 x2 0002 00042
59. 3x1 0004 4x2 0004 x3 0002 1 2x1 0004 3x2 0003 x3 0002 1 x1 0004 2x2 0003 3x3 0002 2 60. 00042x1 0003 x2 0003 3x3 0002 00047 x1 0004 4x2 0003 2x3 0002 0 x1 0004 3x2 0003 x3 0002 1 61.
2x1 0004 2x2 0004 4x3 0002 00042 00043x1 0003 3x2 0003 6x3 0002 3
62.
3 4x1 0004 x2 0003 2x3 0002 00044x1 0003 x2 0004 3x3 0002 000410 8x1 0004 2x2 0003 9x3 0002 00041
72. Consider a system of three linear equations in three variables. Give examples of two reduced forms that are not row equivalent if the system is (A) Consistent and dependent (B) Inconsistent
APPLICATIONS 73. BUYING Suppose that you have a $129 credit on your account at Amazon.com, and you want to spend it all on sale CDs at $10 each, sale DVDs at $12 each, and sale books at $7 each. If you buy 13 items total, how many will you buy of each? 74. PETTY CRIME Shady Grady finds a parking meter with a broken lock and scoops out the change inside. The meter accepts nickels, dimes, and quarters, and there were 32 coins inside with a total value of $6.80. How many of each type of coin did Grady get? 75. CHEMISTRY A chemist has two solutions of sulfuric acid: a 20% solution and an 80% solution. How much of each should be used to obtain 100 liters of a 62% solution?
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76. CHEMISTRY A chemist has two solutions: one containing 40% alcohol and another containing 70% alcohol. How much of each should be used to obtain 80 liters of a 49% solution? 77. GEOMETRY Find a, b, and c so that the graph of the parabola with equation y ⫽ a ⫹ bx ⫹ cx2 passes through the points (⫺2, 3), (⫺1, 2), and (1, 6). 78. GEOMETRY Find a, b, and c so that the graph of the parabola with equation y ⫽ a ⫹ bx ⫹ cx2 passes through the points (1, 3), (2, 2), and (3, 5). 79. PRODUCTION SCHEDULING A small manufacturing plant makes three types of inflatable boats: one-person, two-person, and four-person models. Each boat requires the services of three departments, as listed in the table. The cutting, assembly, and packaging departments have available a maximum of 380, 330, and 120 laborhours per week, respectively. How many boats of each type must be produced each week for the plant to operate at full capacity? One-Person Boat
Two-Person Boat
Four-Person Boat
Cutting department
0.5 h
1.0 h
1.5 h
Assembly department
0.6 h
0.9 h
1.2 h
Packaging department
0.2 h
0.3 h
0.5 h
80. PRODUCTION SCHEDULING Repeat Problem 79 assuming the cutting, assembly, and packaging departments have available a maximum of 350, 330, and 115 labor-hours per week, respectively. 81. PRODUCTION SCHEDULING Rework Problem 79 assuming the packaging department is no longer used. 82. PRODUCTION SCHEDULING Rework Problem 80 assuming the packaging department is no longer used. 83. PRODUCTION SCHEDULING Rework Problem 79 assuming the four-person boat is no longer produced. 84. PRODUCTION SCHEDULING Rework Problem 80 assuming the four-person boat is no longer produced. 85. NUTRITION A dietitian in a hospital is to arrange a special diet using three basic foods. The diet is to include exactly 340 units of calcium, 180 units of iron, and 220 units of vitamin A. The number of units per ounce of each special ingredient for each of the foods is indicated in the table. How many ounces of each food must be used to meet the diet requirements?
88. NUTRITION Solve Problem 86 with the assumption that food C is no longer available. 89. NUTRITION Solve Problem 85 assuming the vitamin A requirement is deleted. 90. NUTRITION Solve Problem 86 assuming the vitamin A requirement is deleted. 91. SOCIOLOGY Two sociologists have grant money to study school busing in a particular city. They wish to conduct an opinion survey using 600 telephone contacts and 400 house contacts. Survey company A has personnel to do 30 telephone and 10 house contacts per hour; survey company B can handle 20 telephone and 20 house contacts per hour. How many hours should be scheduled for each firm to produce exactly the number of contacts needed? 92. SOCIOLOGY Repeat Problem 91 if 650 telephone contacts and 350 house contacts are needed. 93. DELIVERY CHARGES United Express, a nationwide package delivery service, charges a base price for overnight delivery of packages weighing 1 pound or less and a surcharge for each additional pound (or fraction thereof). A customer is billed $27.75 for shipping a 5-pound package and $64.50 for shipping a 20-pound package. Find the base price and the surcharge for each additional pound. 94. DELIVERY CHARGES Refer to Problem 93. Federated Shipping, a competing overnight delivery service, informs the customer in Problem 93 that it would ship the 5-pound package for $29.95 and the 20-pound package for $59.20. (A) If Federated Shipping computes its cost in the same manner as United Express, find the base price and the surcharge for Federated Shipping. (B) Devise a simple rule that the customer can use to choose the cheaper of the two services for each package shipped. Justify your answer. 95. RESOURCE ALLOCATION A coffee manufacturer uses Colombian and Brazilian coffee beans to produce two blends, robust and mild. A pound of the robust blend requires 12 ounces of Colombian beans and 4 ounces of Brazilian beans. A pound of the mild blend requires 6 ounces of Colombian beans and 10 ounces of Brazilian beans. Coffee is shipped in 132-pound burlap bags. The company has 50 bags of Colombian beans and 40 bags of Brazilian beans on hand. How many pounds of each blend should it produce in order to use all the available beans?
Units per Ounce Food A
Food B
Food C
Calcium
30
10
20
Iron
10
10
20
Vitamin A
10
30
20
86. NUTRITION Repeat Problem 85 if the diet is to include exactly 400 units of calcium, 160 units of iron, and 240 units of vitamin A. 87. NUTRITION Solve Problem 85 with the assumption that food C is no longer available.
96. RESOURCE ALLOCATION Refer to Problem 95. (A) If the company decides to discontinue production of the robust blend and only produce the mild blend, how many pounds of the mild blend can it produce and how many beans of each type will it use? Are there any beans that are not used? (B) Repeat part A if the company decides to discontinue production of the mild blend and only produce the robust blend.
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7-3
Matrix Operations
457
Matrix Operations Z Adding and Subtracting Matrices Z Multiplying a Matrix by a Number Z Finding the Product of Two Matrices
In Section 7-2, we introduced basic matrix terminology and solved systems of equations by performing row operations on augmented coefficient matrices. Matrices have many other useful applications and possess an interesting mathematical structure in their own right. As we will see, matrix addition and multiplication are similar to real number addition and multiplication in many respects, but there are some important differences.
Z Adding and Subtracting Matrices Before we can discuss arithmetic operations for matrices, we have to define equality for matrices. Two matrices are equal if they have the same size and their corresponding elements are equal. For example, 200023
c
200023
a b c u v w d 0002 c d d e f x y z
if and only if
a0002u b0002v c0002w d0002x e0002y f0002z
The sum of two matrices of the same size is a matrix with elements that are the sums of the corresponding elements of the two given matrices. Addition is not defined for matrices of different sizes.
EXAMPLE
1
Matrix Addition Add:
SOLUTIONS
2 (A) c 1
00033 2
0 3 d 0004 c 00035 00033
2 (A) c 1
00033 2
0 3 1 2 d 0004 c d 00035 00033 2 5
1 2
2 d 5
2 (B) c 3
2 3
1 2
0 4 d 0004 £ 00033 00033 00031
(2 0004 3) (00033 0004 1) 0002 c (1 0003 3) (2 0004 2) 0002 c
(B) c
1 2
2 5§ 4
(0 0004 2) d (00035 0004 5)
*
5 00032 2 d 00032 4 0
0 2 4 d 0004 £ 00033 5 § 00033 00031 4
Because the first matrix is 2 0005 3 and the second is 3 0005 2, this sum is not defined. *Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
0002
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MATCHED PROBLEM 1
Add: 3 2 00032 3 (A) £ 00031 00031 § 0004 £ 1 00031 § 0 3 2 00032
(B) [1 00032 7] 0004 [ 00032 4
3
00031] 0002
Technology Connections Graphing calculators can be used to solve problems involving matrix operations. Figure 1 illustrates the solutions to Example 1A and 1B on a graphing calculator.
(a) Example 1A
(b) Example 1B
Z Figure 1 Matrix addition on a graphing calculator.
Because we add two matrices by adding their corresponding elements (which are real numbers), it follows from the properties of real numbers that matrices of the same size are commutative and associative relative to addition. That is, if A, B, and C are matrices of the same size, then A0004B0002B0004A (A 0004 B) 0004 C 0002 A 0004 (B 0004 C) 0 £0 0
0 0 0
0 0 0
M0002 c
a b d c d
then
0 d 0
[0 0
Associative
A matrix with elements that are all 0’s is called a zero matrix. Examples of zero matrices are shown in Figure 2. [Note: “0” is often used to denote the zero matrix of any size.] The negative of a matrix M, denoted by 0003M, is a matrix with elements that are the negatives of the elements in M. So if
0 0§ 0
0 0 ≥ ¥ 0 0 0 c 0
Commutative
0003M 0002 c 0]
0003a 0003c
0003b d 0003d
Based on our definition of addition, M 0004 (0003M) 0002 0 (a zero matrix). If A and B are matrices of the same size, then we define subtraction as follows.
Z Figure 2 Zero matrices.
A 0003 B 0002 A 0004 (0003B) To subtract matrix B from matrix A, we subtract corresponding elements.
EXAMPLE
2
Matrix Subtraction Subtract: c
3 5
00032 00032 d 0003 c 0 3
2 d 4
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c
SOLUTION
MATCHED PROBLEM 2
EXAMPLE
3
00032 00032 d 0003 c 0 3
3 5
00033
Subtract: [2
Matrix Operations
2 3 0003 (00032) 00032 0003 2 5 d 0002 c d 0002 c 4 500033 000034 2
5] 0003 [3
00032
459
00034 d 00034
0002
1] 0002
Matrix Equations Find a, b, c, and d so that c
c
3 d 4
3 d 4
Subtract the matrices on the left side.
a00032 b 0003 (00031) 4 d 0002 c c 0003 (00035) d 0003 6 00032
3 d 4
Simplify.
c
a00032 b00041 4 3 d 0002 c d c00045 d00036 00032 4
a0003200024 a00036 MATCHED PROBLEM 3
00031 4 d 0002 c 6 00032
a b 2 00031 4 d 0003 c d 0002 c c d 00035 6 00032
c
SOLUTION
a b 2 d 0003 c c d 00035
b0004100023 b00032
Set corresponding elements equal to each other.
c 0004 5 0002 00032 c 0003 00047
d000360002 4 d 0003 10
0002
Find a, b, c, and d so that c
a b 00034 2 00032 5 d 0003 c d 0002 c d c d 1 00033 8 2 0002
Z Multiplying a Matrix by a Number The product of a number k and a matrix M, denoted by kM, is a matrix formed by multiplying each element of M by k.
EXAMPLE
4
Multiplying a Matrix by a Number 3 Multiply: 00032 £ 00032 0
SOLUTION
MATCHED PROBLEM 4
3 00032 £ 00032 0
00031 1 00031
00031 1 00031
0 3§ 00032
0 00036 3§ 0002 £ 4 00032 0
2 00032 2
0 00036 § 4
0002
1.3 Multiply: 10 £ 0.2 § 3.5 0002
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ZZZ EXPLORE-DISCUSS 1
Multiplication of two numbers can be interpreted as repeated addition if one of the numbers is a positive integer. That is, 2a 0002 a 0004 a
3a 0002 a 0004 a 0004 a
4a 0002 a 0004 a 0004 a 0004 a
and so on. How does this apply to multiplication of a matrix by a number?
Matrix operations have many applications, particularly in business.
EXAMPLE
5
Sales and Commissions Ms. Fong and Mr. Petris are salespeople for a new car agency that sells only two models. August was the last month for this year’s models, and next year’s models were introduced in September. Gross dollar sales for each month are given in the following matrices: AUGUST SALES Compact
c
Fong Petris
SEPTEMBER SALES
Luxury
$36,000 $72,000
Compact
$72,000 d 0002A $0
c
Luxury
$144,000 $180,000
$288,000 d 0002B $216,000
For example, Ms. Fong had $36,000 in compact sales in August and Mr. Petris had $216,000 in luxury car sales in September. (A) What were the combined dollar sales in August and September for each salesperson and each model? (B) What was the increase in dollar sales from August to September? (C) If both salespeople receive a 3% commission on gross dollar sales, compute the commission for each salesperson for each model sold in September. SOLUTIONS
We use matrix addition for part A, matrix subtraction for part B, and multiplication of a matrix by a number for part C. Compact
Luxury
(A) A 0004 B 0002 c
$180,000 $252,000
$360,000 d $216,000
(B) B 0003 A 0002 c
$108,000 $108,000
$216,000 d $216,000
Compact
Fong
Sum of sales for August and September
Petris September sales 0004 August sales
Fong Petris Luxury
(0.03)($144,000) (0.03)($288,000) d (0.03)($180,000) (0.03)($216,000) Fong $4,320 $8,640 0002 c d Petris $5,400 $6,480
(C) 0.03B 0002 c
MATCHED PROBLEM 5
3% of September sales
0002
Repeat Example 5 with A0002 c
$72,000 $36,000
$72,000 d $72,000
and
B0002 c
$180,000 $144,000
$216,000 d $216,000
0002
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Matrix Operations
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Example 5 involved an agency with only two salespeople and two models. A more realistic problem might involve 20 salespeople and 15 models. Problems of this size are often solved using spreadsheets on a computer. Figure 3 illustrates a spreadsheet solution to Example 5. A 1
B
2
Luxury
September Commissions
September Sales
August Sales
G
Compact
Luxury
Compact
Luxury
F
E
D
C
Compact
3
Fong
$36,000
$72,000
$144,000
$288,000
$4,320
$8,640
4
Petris
$72,000
$0
$180,000
$216,000
$5,400
$6,480
5
Sales Increases
Combined Sales
6
Fong
$180,000
$360,000
$108,000
$216,000
7
Petris
$252,000
$216,000
$108,000
$216,000
Z Figure 3
Z Finding the Product of Two Matrices Next we will define a way to multiply two matrices. It will probably seem strange to you at first; eventually you will see examples of why it is useful in many problems. In particular, matrix multiplication will help us to develop an alternative method for solving linear systems that have the same number of variables and equations. We start by defining the product of two special matrices, a row matrix and a column matrix. Z DEFINITION 1 Product of a Row Matrix and a Column Matrix The product of a 1 0005 n row matrix and an n 0005 1 column matrix is a 1 0005 1 matrix given by n00021
b1 b2 [ a1 a2 . . . an ] ≥ ¥ 0002 [ a1b1 0004 a2b2 0004 . . . 0004 anbn ] o bn 10002n
Note that the number of elements in the row matrix and in the column matrix must be the same for the product to be defined.
EXAMPLE
6
Product of a Row Matrix and a Column Matrix 00035 Multiply: [2 00033 0] £ 2 § 00032
SOLUTION
00035 [2 00033 0] £ 2 § 0002 [(2)(00035) 0004 (00033)(2) 0004 (0)(00032)] 00032 0002 [000310 0003 6 0004 0] 0002 [ 000316]
MATCHED PROBLEM 6 Multiply: [00031 0
3
2 3 2] ≥ ¥ 4 00031
0002
0002
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The answer to Example 6 is a 1 ⫻ 1 matrix, which we represented with [⫺16]. From now on, if the result of a calculation is a 1 ⫻ 1 matrix, we’ll usually omit the brackets and write the answer as a real number.
EXAMPLE
7
Production Scheduling A factory produces a slalom water ski that requires 4 labor-hours in the fabricating department and 1 labor-hour in the finishing department. Fabricating personnel receive $10 per hour, and finishing personnel receive $8 per hour. Find the total labor cost per ski.
SOLUTION
Total labor cost per ski is given by the product [4 1] c
MATCHED PROBLEM 7
10 d ⫽ [(4)(10) ⫹ (1)(8)] ⫽ [40 ⫹ 8] ⫽ [48] or $48 per ski 8
0002
If the factory in Example 7 also produces a trick water ski that requires 6 labor-hours in the fabricating department and 1.5 labor-hours in the finishing department. Find the totallabor cost per ski by multiplying an appropriate row matrix and column matrix. 0002 We will now use the product of a 1 ⫻ n row matrix and an n ⫻ 1 column matrix to extend the definition of matrix product to more general matrices.
Z DEFINITION 2 Matrix Product If A is an m ⫻ p matrix and B is a p ⫻ n matrix, then the matrix product of A and B, denoted AB, is an m ⫻ n matrix whose element in the ith row and jth column is the real number obtained from the product of the ith row of A and the jth column of B. If the number of columns in A does not equal the number of rows in B, then the matrix product AB is not defined.
Must be the same (b ⫽ c) a⫻b c⫻d
Size of product a⫻d
A • B ⴝ AB
It is important to check sizes before starting the multiplication process. If A is an a ⫻ b matrix and B is a c ⫻ d matrix, then if b ⫽ c, the product AB will exist and will be an a ⫻ d matrix (see Fig. 4). If b ⫽ c, then the product AB does not exist. The definition is not as complicated as it looks. An example should help clarify the process. For
Z Figure 4
2 A⫽ c ⫺2
3 1
⫺1 d 2
and
1 B⫽ £ 2 ⫺1
3 0§ 2
A is 2 ⫻ 3, B is 3 ⫻ 2, and so AB is 2 ⫻ 2. To find the first row of AB, we take the product of the first row of A with every column of B and write each result as a real number, not a 1 ⫻ 1 matrix. The second row of AB is computed in the same manner. The four products of row and column matrices used to produce the four elements in AB are shown in the dashed box below. These products are usually calculated mentally, or with the aid of a calculator, and need not be written out. The shaded portions highlight the steps involved in computing the element in the first row and second column of AB.
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300052
200053
冤
2 00032
00031 2
冥
3 1
冤
3 0 2
1 2 00031
0002
冤 冥 冤 冥
1 3 00031兴 2 00031 1 1 2兴 2 00031
关2
冥
0002
关 00032
(2)(1) 0004 (3)(2) 0004 (00031)(00031) 冤 (00032)(1) 0004 (1)(2) 0004 (2)(00031)
463
Matrix Operations
关2
3
关 00032
1
00031兴
2兴
冤 冥 冤 冥 3 0 2 3 0 2
(2)(3) 0004 (3)(0) 0004 (00031)(2) (00032)(3) 0004 (1)(0) 0004 (2)(2)
冥
200052
0002
EXAMPLE
8
冤000329
冥
4 00032
Matrix Multiplication Given 2 A0002 £ 1 00031
1 0§ 2
B0002 c
00031 1
1 2
0 2
1 d 0
C0002 c
2 00031
6 d 00033
D0002 c
1 3
2 d 6
Find each product that is defined: (A) AB
(B) BA
(C) CD
300022
SOLUTIONS
2 (A) AB 0002 £ 1 00031
200024
1 1 0§ c 2 2
00031 1
(2)(1) 0004 (1)(2) 0002 £ (1)(1) 0004 (0)(2) (00031)(1) 0004 (2)(2) 4 0002 £1 3
300024
00031 2 00031 0 3 4 00031 1
0 2
1 d 0
(2)(00031) 0004 (1)(1) (1)(00031) 0004 (0)(1) (00031)(00031) 0004 (2)(1)
(2)(0) 0004 (1)(2) (1)(0) 0004 (0)(2) (00031)(0) 0004 (2)(2)
(2)(1) 0004 (1)(0) (1)(1) 0004 (0)(0) § (00031)(1) 0004 (2)(0)
2 1§ 00031 300022
200024
1 (B) BA 0002 c 2
(D) DC
2 1 d£ 1 0 00031
0 2
1 0§ 2
Product is not defined. 200022
2 (C) CD 0002 c 00031
200022
6 1 dc 00033 3
2 d 6
0002c
(2)(1) 0004 (6)(3) (2)(2) 0004 (6)(6) d (00031)(1) 0004 (00033)(3) (00031)(2) 0004 (00033)(6) 200022
20 40 0002c d 000310 000320
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1 (D) DC 0002 c 3
MATCHED PROBLEM 8
200022
2 2 dc 6 00031
200022
6 d 00033
(1)(2) 0004 (2)(00031) (1)(6) 0004 (2)(00033) 0002 c d (3)(2) 0004 (6)(00031) (3)(6) 0004 (6)(00033)
0 0002 c 0
0 d 0
0002
Find each product, if it is defined: (A) c (C) c
00031 1
1 00031
00031 1 00031 1 00032 00031 0 3 00032 d £ 2 3§ (B) £ 2 3 § c d 0 1 2 2 0 1 0 1 0 2 00032 4 00032 4 1 2 dc d (D) c dc d 00032 1 00032 1 00032 00031 00032 0 2
3 2
0002
In the arithmetic of real numbers, it doesn’t matter in which order we multiply; for example, 5 0005 7 0002 7 0005 5. In matrix multiplication, however, it does make a difference. That is, AB does not always equal BA, even if both multiplications are defined and both products are the same size (see Examples 8C and 8D). In other words, Matrix multiplication is not commutative. Also, AB may be zero with neither A nor B equal to zero (see Example 8D). That is, The zero property does not hold for matrix multiplication. (See Section R-1 for a discussion of the zero property for real numbers.) Just as we used the familiar algebraic notation AB to represent the product of matrices A and B, we use the notation A2 for AA (the product of A with itself ), A3 for AAA, and so on.
ZZZ EXPLORE-DISCUSS 2
In addition to the commutative and zero properties, there are other significant differences between real number multiplication and matrix multiplication. (A) In real number multiplication, the only real number whose square is 0 is the real number 0 (02 0002 0). Find at least one 2 0005 2 matrix A with all elements nonzero such that A2 0002 0, where 0 is the 2 0005 2 zero matrix. (B) In real number multiplication, the only nonzero real number that is equal to its square is the real number 1 (12 0002 1). Find at least one 2 0005 2 matrix A with all elements nonzero such that A2 0002 A.
We’ll return to our study of the properties of matrix multiplication in Section 7-4. We will conclude this section with an application of matrix multiplication.
EXAMPLE
9
Labor Costs If we combine the time requirements for making slalom and trick water skis discussed in Example 7 and Matched Problem 7, we get Labor-hours per ski Assembly Finishing department department Trick ski Slalom ski
c
6h 4h
1.5 h d 0002L 1h
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Now suppose that the company has two manufacturing plants, X and Y, in different parts of the country and that the hourly rates for each department are given in the following matrix: Hourly Wages Plant Plant X Y Assembly department Finishing department
c
$10 $ 8
$12 d 0002H $10
Find the matrix products HL and LH, and decide if either matrix has a meaningful interpretation in terms of ski production. SOLUTION
Since H and L are both 2 0005 2 matrices, we can find the product of H and L in either order and the result will be a 2 0005 2 matrix: HL 0002 c
10 8
12 6 dc 10 4
1.5 108 d 0002 c 1 88
LH 0002 c
6 4
1.5 10 dc 1 8
12 72 d 0002 c 10 48
27 d 22 87 d 58
How can we interpret the elements in these products? Let’s begin with the product HL. The element 108 in the first row and first column of HL is the product of the first row matrix of H and the first column matrix of L: Plant Plant X Y
6 [10 12] c d 4
Trick Slalom
0002 10(6) 0004 12(4) 0002 60 0004 48 0002 108
Notice that $60 is the labor cost for assembling a trick ski at Plant X and $48 is the labor cost for assembling a slalom ski at Plant Y. Although both numbers represent labor costs, it makes no sense to add them together. They do not pertain to the same type of ski or to the same plant. So, even though the product HL happens to be defined mathematically, it has no useful interpretation in this problem. Now let’s consider the product LH. The element 72 in the first row and first column of LH is given by the following product: Assembly Finishing
[6
1.5] c
10 d 8
Assembly Finishing
0002 6(10) 0004 1.5(8) 0002 60 0004 12 0002 72
where $60 is the labor cost for assembling a trick ski at Plant X and $12 is the labor cost for finishing a trick ski at Plant X. The sum is the total labor cost for producing a trick ski at Plant X. The other elements in LH also represent total labor costs, as indicated by the row and column labels shown below: Labor costs per ski Plant Plant X Y
LH 0002 c MATCHED PROBLEM 9
$72 $48
$87 d $58
Trick ski
0002
Slalom ski
Refer to Example 9. The company wants to know how many hours to schedule in each department in order to produce 1,000 trick skis and 2,000 slalom skis. These production requirements can be represented by either of the following matrices: Trick skis
P 0002 [1,000
Slalom skis
2,000]
Q0002 c
1,000 d 2,000
Trick skis Slalom skis
Using the labor-hour matrix L from Example 9, find PL or LQ, whichever has a meaningful interpretation for this problem, and label the rows and columns accordingly. 0002
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ZZZ
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Example 9 and Matched Problem 9 illustrate an important point about matrix multiplication. Even if you are using a graphing calculator to perform the calculations in a matrix product, you will still need to know the definition of matrix multiplication so that you can interpret the results correctly.
CAUTION ZZZ
ANSWERS TO MATCHED PROBLEMS 1 5 1. (A) £ 0 ⫺2 § (B) Not defined 2. [⫺1 ⫺1 4 ] 3. a ⫽ ⫺6, b ⫽ 7, c ⫽ 9, d ⫽ ⫺1 2 1 13 $252,000 $288,000 $108,000 $144,000 $5,400 $6,480 4. £ 2 § 5. (A) c (B) c (C) c d d d $180,000 $288,000 $108,000 $144,000 $4,320 $6,480 35 10 6. [8] 7. [6 1.5] c d ⫽ [72] or $72 8 2 2 ⫺1 2 0 0 ⫺6 ⫺12 8. (A) Not defined (B) £ 1 6 (C) c (D) c 12 ⫺4 § d d 0 0 3 6 ⫺1 0 3 ⫺2 9. Assembly Finishing Labor hours PL ⫽ [14,000 3,500]
7-3
Exercises
1. What conditions must matrics A and B satisfy so that A ⫹ B exists? 2. What conditions must matrices A and B satisfy so that AB exists? 3. What conditions must matrices A and B satisfy so that BA exists? 4. What conditions must matrices A and B satisfy so that both AB and BA exist? 5. What is the negative of a matrix? 6. How do you subtract two matrices?
0 ⫺1 3§ ⫹ £ 0 1 4
4 13. £ ⫺2 8 14. c
6 4
⫺2 3 3 d ⫹ c ⫺8 ⫺7 6
4 15. £ ⫺2 8
0 ⫺1 3§ ⫹ c 2 1
9. If A is a 1 ⫻ n matrix and B is an n ⫻ 1 matrix, how do you find the product BA? What is the size of BA? 10. Describe the operation of matrix multiplication in your own words. Perform the indicated operations in Problems 11–24, if possible. 5 11. c 3
⫺2 ⫺3 d ⫹ c 0 1
7 d ⫺6
0 12. c 2
8 9 d ⫹ c ⫺1 7
⫺4 d 5
9 ⫺2
0 5
3 3 d ⫹ £ 9 ⫺7 ⫺1
16. c
6 4
⫺2 ⫺8
17. c
5 4
⫺1 0 2 d ⫺ c 6 3 3
4 5
2 0 1 § ⫺ £ ⫺7 0 ⫺1
5 2§ 0
7. How do you multiply a matrix by a number? 8. If A is a 1 ⫻ n matrix and B is an n ⫻ 1 matrix, how do you find the product AB? What is the size of AB?
2 5§ ⫺6
6 18. £ ⫺4 3 4 19. £ 10 ⫺13 1
20. c 25 4
⫺7 4 11 § ⫺ c ⫺7 ⫺9
9 ⫺34 2 3d ⫺ c ⫺2 ⫺74
1 4 1d 2
⫺1 d 4
4 d ⫺6 6 ⫺2 § 4 ⫺6 d ⫺5
10 11
⫺13 d ⫺9
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00032.8 0
21. c
2.4 00031.6
22. c
10 d 0003 [20 20
24. 5 c
00037 4
23. 4 c
10]
3 00035
00032.2 00033.2
3.9 7 d 0003 c 4.2 00033.2
0 6
c
7 d 5
c
27. c
00035 d [4 00033
29. [3
28. c
1 00034 ] £ 2 § 00033
00032
1 31. £ 2 § [3 00032 00034 ] 00033 00036 3 1 dc d 33. c 2 00035 3 35. c
5 4
37. c
8 00035
1 2 dc 6 3
0 d 6
3x 00031
5 2y d 0002 c 4x 00036
2 0
00031 4
00031 C0004 £ 4 00032
3 d 00032
B0004 c
36. c
00032 3
38. c
7 0
0 2 00033 1 § 3 5
00033 2
3 D 0004 £0 1
00033 7 d 0004 c 0003y 00037
00032
2 2] £ 00031 § 1
00031 d 5
7 4 dc 00031 0 0 9 dc 3 00034
3 0 w d 0002 c 00037 000311 y
00032 d 00031
c
x 00031 4y d 0003 c 00032 y 5
2 d 2
x 9 d 0004 c z 4
1 d 6
1 d 5
4 12 d 0004 c 3x 00034
00035 d 000314
In Problems 63 and 64, let a, b, and c be any nonzero real numbers, and let A0004 c
b 1 d and I 0004 c 0003a 0
a c
0 d 1
63. If A2 0004 0, how are a, b, and c related? Use this relationship to provide several examples of 2 0005 2 matrices with no zero entries whose square is the zero matrix. 64. If A2 0004 I, how are a, b, and c related? Use this relationship to provide several examples of 2 0005 2 matrices with no zero entries whose square is the matrix I. Problems 65 and 66 refer to the matrices A0004 c
Problems 39–56 refer to the following matrices. A0004 c
00032 d 00034
62. Find x and y so that
00031]
2 32. £ 00031 § [1 00032 2] 1 3 7 4 d dc 34. c 00031 00039 00031
0 d 8
00033 2 dc 3 0
3 d [2 00034
30. [1
c
3 4] c d 00038
26. [ 00032
00032 ]
00033 1 d 0004 c 1 3
61. Find w, x, y, and z so that
Find the products in Problems 25–38. 25. [5
a b 2 d 0002 c c d 0
60. Find x and y so that
9 d 2
4 3] c d 7
467
59. Find a, b, c, and d so that
00032.2 d 1 00034 9
3 00032
Matrix Operations
a b 1 d and B 0004 c c d 1
1 d 1
65. If AB 0004 0, how are a, b, c, and d related? Use this relationship to provide several examples of 2 0005 2 matrices A with no zero entries that satisfy AB 0004 0.
00032 00031 § 2
Perform the indicated operations, if possible. 39. CA
40. AC
41. BA
42. AB
43. C 2
44. B2
45. C 0002 DA
46. B 0002 AD
47. 0.2CD
48. 0.1DB
49. 2DB 0002 5CD
50. 3BA 0002 4AC
51. (00031)AC 0002 3DB
52. (00032)BA 0002 6CD
53. CDA
54. ACD
55. DBA
56. BAD
66. If BA 0004 0, how are a, b, c, and d related? Use this relationship to provide several examples of 2 0005 2 matrices A with no zero entries that satisfy BA 0004 0. 67. Find x and y so that c
1 00032
3 x dc 00032 3
1 y d 0004 c 2 y
7 d 00036
68. Find x and y so that c
x 1
00031 2 dc 0 4
1 y d 0004 c 1 2
y d 1
69. Find a, b, c, and d so that
In Problems 57 and 58, use a graphing calculator to calculate B, B 2, B 3, . . . and AB, AB 2, AB 3, . . . . Describe any patterns you observe in each sequence of matrices. 57. A 0004 [ 0.3
0.4 0.7 ] and B 0004 c 0.2
0.6 d 0.8
58. A 0004 [ 0.4
0.6 ] and B 0004 c
0.9 0.3
0.1 d 0.7
c
1 1
3 a b 6 dc d 0004 c 4 c d 7
00035 d 00037
70. Find a, b, c, and d so that c
1 2
1 00032 a b dc d 0004 c 3 00033 c d
0 d 2
71. A square matrix is a diagonal matrix if all elements not on the principal diagonal are zero. So a 2 0005 2 diagonal matrix has the form a 0 d A0004 c 0 d
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where a and d are any real numbers. Discuss the validity of each of the following statements. If the statement is always true, explain why. If not, give examples. (A) If A and B are 2 0005 2 diagonal matrices, then A 0004 B is a 2 0005 2 diagonal matrix. (B) If A and B are 2 0005 2 diagonal matrices, then A 0004 B 0002 B 0004 A. (C) If A and B are 2 0005 2 diagonal matrices, then AB is a 2 0005 2 diagonal matrix. (D) If A and B are 2 0005 2 diagonal matrices, then AB 0002 BA. 72. A square matrix is an upper triangular matrix if all elements below the principal diagonal are zero. So a 2 0005 2 upper triangular matrix has the form A0002 c
a b d 0 d
where a, b, and d are any real numbers. Discuss the validity of each of the following statements. If the statement is always true, explain why. If not, give examples. (A) If A and B are 2 0005 2 upper triangular matrices, then A 0004 B is a 2 0005 2 upper triangular matrix. (B) If A and B are 2 0005 2 upper triangular matrices, then A 0004 B 0002 B 0004 A. (C) If A and B are 2 0005 2 upper triangular matrices, then AB is a 2 0005 2 upper triangular matrix. (D) If A and B are 2 0005 2 upper triangular matrices, then AB 0002 BA. 73. A company with two different plants makes satellite radios and GPS units. The production costs for each item are given in the following matrices: Plant X Radio GPS Materials Labor
c
$30 $60
Plant Y Radio GPS
$25 d 0002A $80
c
$36 $54
$27 d 0002B $74
Find the matrix 12 (A 0004 B), and explain what information it provides. 74. Suppose that the company in Problem 73 experiences an increase in the cost of both labor and materials at plant X. Find the matrix 12 (1.2A 0004 B). If it provides the average cost of production for the two plants, by how much were the costs at plant X increased? 75. MARKUP An import car dealer sells three models of a car. Current dealer invoice price (cost) and the retail price for the basic models and the indicated options are given in the following two matrices (where “Air” means air conditioning): Basic Car Model A Model B Model C
$10,400 £ $12,500 $16,400 Basic Car
Model A Model B Model C
$13,900 £ $15,000 $18,300
Dealer Invoice Price CD Air changer
$682 $721 $827
$215 $295 $443
Retail Price CD Air changer
$783 $838 $967
$263 $395 $573
Cruise Control
$182 $182 § 0002 M $192 Cruise Control
$215 $236 § 0002 N $248
We define the markup matrix to be N 0003 M (markup is the difference between the retail price and the dealer invoice price). Suppose the value of the dollar has had a sharp decline and the dealer invoice price is to have an across-the-board 15% increase next year. To stay competitive with domestic cars, the dealer increases the retail prices only 10%. Calculate a markup matrix for next year’s models and the indicated options. (Compute results to the nearest dollar.) 76. MARKUP Referring to Problem 75, what is the markup matrix resulting from a 20% increase in dealer invoice prices and an increase in retail prices of 15%? (Compute results to the nearest dollar.) 77. LABOR COSTS A company with manufacturing plants located in different parts of the country has labor-hour and wage requirements for the manufacturing of three types of inflatable boats as given in the following two matrices: Labor-Hours per Boat Cutting Assembly Packaging Department Department Department
0.6 h M 0002 £ 1.0 h 1.5 h
0.6 h 0.9 h 1.2 h
0.2 h 0.3 h § 0.4 h
One-person boat Two-person boat Four-person boat
Hourly Wages Plant I Plant II
$8 N 0002 £ $10 $5
$9 $12 § $6
Cutting department Assembly department Packaging department
(A) Find the labor costs for a one-person boat manufactured at plant I. (B) Find the labor costs for a four-person boat manufactured at plant II. (C) Discuss possible interpretations of the elements in the matrix products MN and NM. (D) If either of the products MN or NM has a meaningful interpretation, find the product and label its rows and columns. 78. INVENTORY VALUE A personal computer retail company sells five different computer models through three stores located in a large metropolitan area. The inventory of each model on hand in each store is summarized in matrix M. Wholesale (W ) and retail (R) values of each model computer are summarized in matrix N. A
4 M0002 £ 2 10
B
Model C D
2 3 4
3 5 3
W
$700 $1,400 N 0002 E$1,800 $2,700 $3,500
7 0 4
E
1 6§ 3
Store 1 Store 2 Store 3
R
$840 $1,800 $2,400U $3,300 $4,900
A B C D E
(A) What is the retail value of the inventory at store 2? (B) What is the wholesale value of the inventory at store 3? (C) Discuss possible interpretations of the elements in the matrix products MN and NM.
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(D) If either of the products MN or NM has a meaningful interpretation, find the product and label its rows and columns. (E) Discuss methods of matrix multiplication that can be used to find the total inventory of each model on hand at all three stores. State the matrices that can be used, and perform the necessary operations. (F) Discuss methods of matrix multiplication that can be used to find the total inventory of all five models at each store. State the matrices that can be used, and perform the necessary operations. 79. AIRFREIGHT A nationwide airfreight service has connecting flights between five cities, as illustrated in the figure. To represent this schedule in matrix form, we construct a 5 ⫻ 5 incidence matrix A, where the rows represent the origins of each flight and the columns represent the destinations. We place a 1 in the ith row and jth column of this matrix if there is a connecting flight from the ith city to the jth city and a 0 otherwise. We also place 0s on the principal diagonal, because a connecting flight with the same origin and destination does not make sense. Atlanta 1
Baltimore 2
1
3 Chicago
Origin
1 2 3 4 5
4 Denver
0 0 E1 0 0
Destination 2 3 4
5
1 0 0 0 0
0 0 1U ⫽ A 0 0
0 1 0 1 0
1 0 0 0 1
5 El Paso
Now that the schedule has been represented in the mathematical form of a matrix, we can perform operations on this matrix to obtain information about the schedule. (A) Find A2. What does the 1 in row 2 and column 1 of A2 indicate about the schedule? What does the 2 in row 1 and column 3 indicate about the schedule? In general, how would you interpret each element off the principal diagonal of A2? [Hint: Examine the diagram for possible connections between the ith city and the jth city.] (B) Find A3. What does the 1 in row 4 and column 2 of A3 indicate about the schedule? What does the 2 in row 1 and column 5 indicate about the schedule? In general, how would you interpret each element off the principal diagonal of A3? (C) Compute A, A ⫹ A2, A ⫹ A2 ⫹ A3, . . . , until you obtain a matrix with no zero elements (except possibly on the principal diagonal), and interpret. 80. AIRFREIGHT Refer to Problem 79. Find the incidence matrix A for the flight schedule illustrated in the figure. Compute A, A ⫹ A2, A ⫹ A2 ⫹ A3, . . . , until you obtain a matrix with no zero elements (except possibly on the principal diagonal), and interpret. Louisville 1
Milwaukee 2
3 Newark
4 Phoenix
5 Oakland
Matrix Operations
469
81. POLITICS In a local election, a group hired a public relations firm to promote its candidate in three ways: telephone, house calls, and direct mail. The cost per contact is given in matrix M: Cost per Contact
$0.80 M ⫽ £ $1.50 § $0.40
Telephone House Call Mail
The number of contacts of each type made in two adjacent cities is given in matrix N: Telephone
House Call
1,000 2,000
500 800
5,000 d 8,000
N⫽ c
Berkeley Oakland
(A) Find the total amount spent in Berkeley. (B) Find the total amount spent in Oakland. (C) Discuss possible interpretations of the elements in the matrix products MN and NM. (D) If either of the products MN or NM has a meaningful interpretation, find the product and label its rows and columns. (E) Discuss methods of matrix multiplication that can be used to find the total number of telephone calls, house calls, and letters. State the matrices that can be used, and perform the necessary operations. (F) Discuss methods of matrix multiplication that can be used to find the total number of contacts in Berkeley and in Oakland. State the matrices that can be used, and perform the necessary operations. 82. NUTRITION A nutritionist for a cereal company blends two cereals in different mixes. The amounts of protein, carbohydrate, and fat (in grams per ounce) in each cereal are given by matrix M. The amounts of each cereal used in the three mixes are given by matrix N. Cereal A
4 g Ⲑoz M ⫽ £ 20 g Ⲑoz 3 g Ⲑoz N⫽ c
Cereal B
2 g Ⲑoz 16 g Ⲑoz § 1 g Ⲑoz
Mix X
Mix Y
Mix Z
15 oz 5 oz
10 oz 10 oz
5 oz d 15 oz
Protein Carbohydrate Fat
Cereal A Cereal B
(A) Find the amount of protein in mix X. (B) Find the amount of fat in mix Z. (C) Discuss possible interpretations of the elements in the matrix products MN and NM. (D) If either of the products MN or NM has a meaningful interpretation, find the product and label its rows and columns. 83. TOURNAMENT SEEDING To rank players for an upcoming tennis tournament, a club decides to have each player play one set with every other player. The results are given in the table.
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Defeated
Player
Defeated
1. Aaron
Charles, Dan, Elvis
1. Anne
Diane
2. Bart
Aaron, Dan, Elvis
2. Bridget
Anne, Carol, Diane
3. Charles
Bart, Dan
3. Carol
Anne
4. Dan
Frank
5. Elvis
Charles, Dan, Frank
4. Diane
Carol, Erlene
6. Frank
Aaron, Bart, Charles
5. Erlene
Anne, Bridget, Carol
(A) Express the outcomes as an incidence matrix A by placing a 1 in the ith row and jth column of A if player i defeated player j, and a 0 otherwise (see Problem 79). (B) Compute the matrix B 0002 A 0004 A2 (C) Discuss matrix multiplication methods that can be used to find the sum of each of the rows in B. State the matrices that can be used and perform the necessary operations. (D) Rank the players from strongest to weakest. Explain the reasoning behind your ranking.
(A) Express the outcomes as an incidence matrix A by placing a 1 in the ith row and jth column of A if player i defeated player j, and a 0 otherwise (see Problem 79). (B) Compute the matrix B 0002 A 0004 A2 (C) Discuss matrix multiplication methods that can be used to find the sum of each of the rows in B. State the matrices that can be used and perform the necessary operations. (D) Rank the players from strongest to weakest. Explain the reasoning behind your ranking.
84. PLAYER RANKING Each member of a chess team plays one match with every other player. The results are given in the table.
7-4
Solving Systems of Linear Equations Using Matrix Inverse Methods Z The Identity Matrix for Multiplication Z Finding the Inverse of a Square Matrix Z Matrix Equations Z Matrix Equations and Systems of Linear Equations Z Application: Cryptography
Now that we know a bit about matrix multiplication, we will see how it can be used to solve certain systems of equations.
Z The Identity Matrix for Multiplication We know that for any real number a, 1 0005 a 0002 a 0005 1 0002 a. The number 1 is called the identity for real number multiplication. Is there a matrix analog? That is, if M is an arbitrary matrix, is there a matrix I with the property that IM 0002 MI 0002 M? It turns out that, in general, the answer is no. But the set of square matrices of order n (matrices with n rows and n columns) does have an identity.
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ZZZ EXPLORE-DISCUSS 1
Solving Systems of Linear Equations Using Matrix Inverse Methods
471
(A) Pick any 2 ⫻ 2 matrix you like, and multiply it by the following matrix in both possible orders. c
1 0
0 d 1
(B) Repeat (A) for any 3 ⫻ 3 matrix you like, but multiply by the matrix 1 £0 0
0 1 0
0 0§ 1
What can you conclude?
Z DEFINITION 1 Identity Matrix The identity matrix for multiplication for the set of all square matrices of order n is the square matrix of order n, denoted by I, with 1’s along the principal diagonal (from upper left corner to lower right corner) and 0’s elsewhere.
In Explore-Discuss 1, we saw that 1 c 0
0 d 1
and
1 £0 0
0 1 0
0 0§ 1
are the identity matrices for square matrices of order 2 and 3, respectively. We will show in Exercises 7-4 that if M is any square matrix of order n and I is the identity matrix of order n, then IM ⫽ MI ⫽ M Note: If M is an m ⫻ n matrix that is not square (m ⫽ n), then it is still possible to multiply M on the left and on the right by an identity matrix, but not with the same-size identity matrix. To avoid the complications involved with associating two different identity matrices with each nonsquare matrix, we will restrict our attention in this section to square matrices.
Z Finding the Inverse of a Square Matrix In the set of real numbers, we know that for each real number a, except 0, there exists a real number a⫺1 such that a⫺1a ⫽ 1 The number a⫺1 is called the inverse of the number a relative to multiplication, or the multiplicative inverse of a. For example, 2⫺1 is the multiplicative inverse of 2, since 2⫺1(2) ⫽ 1. We will use this idea to define the inverse of a square matrix.
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Z DEFINITION 2 Inverse of a Square Matrix If A is a square matrix of order n and if there exists a matrix A00031 (read “A inverse”) such that A00031A 0002 AA00031 0002 I then A00031 is called the multiplicative inverse of A or, more simply, the inverse of A. If no such matrix exists, then A is said to be a singular matrix.
Let A 0002 c
ZZZ EXPLORE-DISCUSS 2
4 2
2 d 2
1⁄
1⁄4 B 0002 c 1⁄ 2
1⁄
2 2
d
C0002 c
1⁄
2
00031⁄
2
00031⁄ 2 d 1
(A) How are the entries in A and B related? (B) Find AB. Is B the inverse of A? (C) Find AC. Is C the inverse of A?
The multiplicative inverse of a nonzero real number a also can be written as 1/ a, but this notation is never used for matrix inverses. Let’s use Definition 2 to find A00031, if it exists, for A0002 c
2 1
3 d 2
We are looking for a matrix A00031 0002 c
a b
c d d
such that AA00031 0002 A00031A 0002 I We can write A
2 c 1
A00041
3 a dc 2 b
I
c 1 d 0002 c d 0
0 d 1
and try to find a, b, c, and d so that the product of A and A00031 is the identity matrix I. Multiplying A and A00031 on the left side, we get c
(2a 0004 3b) (2c 0004 3d ) 1 d 0002 c (a 0004 2b) (c 0004 2d) 0
0 d 1
which is true only if 2a 0004 3b 0002 1 a 0004 2b 0002 0 c
2 1
3 1 ` d 2 0
c
1 2
2 0 ` d 3 1
2c 0004 3d 0002 0 c 0004 2d 0002 1 R1 4 R2
00042R1 0006 R2 S R2
Use Gauss–Jordan elimination to solve each system.
c
2 1
3 0 ` d 2 1
R1 4 R2
c
1 2
2 ` 3
00042R1 0006 R2 S R2
1 d 0
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SECTION 7–4
c
1 2 0 ` d 0 00031 1
c
1 0
2 0 d ` 1 00031
c
1 0
0 2 d ` 1 00031
Solving Systems of Linear Equations Using Matrix Inverse Methods
(00041)R2 S R2
(00041)R2 0006 R1 S R1
a 0002 2, b 0002 00031 A00031 0002 c
c
1 0
2 1 d ` 00031 00032
c
1 0
2 1 ` d 1 2
c
1 0
0 00033 d ` 1 2
473
(00041)R2 S R2
(00041)R2 0006 R1 S R1
c 0002 00033, d 0002 2
a b 2 00033 d 0002 c d c d 00031 2
CHECK
A00041
A
c
2 1
3 2 dc 2 00031
A00041
I
00033 1 d 0002 c 2 0
0 2 d 0002 c 1 00031
A
00033 2 dc 2 1
3 d 2
Unlike nonzero real numbers, inverses do not always exist for nonzero square matrices. For example, if B0002 c
2 4
1 d 2
then, proceeding as before, we are led to the systems 2a 0004 b 0002 1 4a 0004 2b 0002 0 c
2 4
1 1 ` d 2 0
c
2 0
1 0 d ` 0 00032
2c 0004 d 0002 0 4c 0004 2 d 0002 1 (00042)R1 0006 R2
c
2 4
1 0 ` d 2 1
c
2 0
1 0 ` d 0 1
Use Gauss–Jordan elimination to solve each system. (00042)R1 0006 R2
The last row of each augmented coefficient matrix contains a contradiction. So each system is inconsistent and has no solution. We conclude that B00031 does not exist and B is a singular matrix.
Technology Connections Most graphing calculators can find matrix inverses and can identify singular matrices. Figure 1 shows the calculation of A00041 for the matrix A discussed earlier. Figure 2 shows the error message that results when the inverse operation is applied to the singular matrix B discussed earlier.
Z Figure 1
Z Figure 2
Note that the inverse operation is performed by pressing the x00041 key. Entering [A]^(00041) results in an error message (Fig. 3).
Z Figure 3
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Being able to find inverses, when they exist, leads to direct and simple solutions to many practical problems. The algebraic method outlined for finding the inverse, if it exists, gets very involved for matrices of order larger than 2. Now that we know what we are looking for, we can use augmented matrices, as in Section 7-2, to make the process more efficient. Details are illustrated in Example 1.
EXAMPLE
1
Finding an Inverse Find the inverse, if it exists, of 1 A ⫽ £0 2
SOLUTION
⫺1 2 3
1 ⫺1 § 0
We start as before and write Aⴚ1
A
1 £0 2
⫺1 2 3
1 a ⫺1 § £ b 0 c
I
d g 1 e h§ ⫽ £0 f i 0
0 1 0
0 0§ 1
Equating corresponding terms, we see that this is true only if a⫺ b⫹c⫽1 2b ⫺ c ⫽ 0 2a ⫹ 3b ⫽0
d⫺ e⫹f⫽0 2e ⫺ f ⫽ 1 2d ⫹ 3e ⫽0
g⫺ h⫹i⫽0 2h ⫺ i ⫽ 0 2g ⫹ 3h ⫽1
Now we write augmented matrices for each of the three systems: First
Second
Third
1 ⫺1 1 1 £0 2 ⫺1 † 0 § 2 3 0 0
1 ⫺1 1 0 £0 2 ⫺1 † 1 § 2 3 0 0
1 ⫺1 1 0 £0 2 ⫺1 † 0 § 2 3 0 1
If you look carefully at the side-by-side solutions on pages 472 and 473, you will see that the exact same row operations were performed on each augmented matrix. The same would happen here; all three preceding augmented matrices have the same coefficient matrix. To save time, we’ll combine all three into one, as shown next. 1 ⫺1 1 1 0 0 £0 2 ⫺1 † 0 1 0 § ⫽ [A | I ] 2 3 0 0 0 1
(1)
We now try to perform row operations on matrix (1) until we obtain a row-equivalent matrix that looks like matrix (2): I
1 £0 0
0 1 0
B
0 a d g 0 † b e h § ⫽ [I | B] 1 c f i
(2)
If this can be done, then the new matrix to the right of the vertical bar is A⫺1! Now let’s try to transform matrix (1) into a form like that of matrix (2). We follow the same
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sequence of steps as in the solution of linear systems by Gauss–Jordan elimination (see Section 7-2): A
I
1 00031 1 1 0 0 £0 2 00031 † 0 1 0 § 2 3 0 0 0 1 1 00031 1 1 0 0 ⬃ £0 2 00031 † 0 1 0 § 0 5 00032 00032 0 1 1 ⬃ £0 0
00031 1 5
1 1 0 0 000312 † 0 12 0 § 00032 00032 0 1
1 1 0 1 2 ⬃ £ 0 1 000312 † 0 1 0 0 00032 2
1 2 1 2 000352
1 1 1 0 2 1 ⬃ £ 0 1 00032 † 0 0 0 1 00034
1 2 1 2
0 0§ 1
0 0§ 00035 2
(00042)R1 0006 R3 S R3
1 2 R2
S R2
R2 0006 R1 S R1
(00045)R2 0006 R3 S R3
2R3 S R3
(000412 )R3 0006 R1 S R1 1 2 R3
0006 R2 S R2
1 0 0 3 3 00031 ⬃ £ 0 1 0 † 00032 00032 1 § 0002 [I 0 B] 0 0 1 00034 00035 2 We suspect that matrix B is actually A00031, but we should check. CHECK
Because the definition of matrix inverse requires that A00031A 0002 I
and
AA00031 0002 I
(3)
it appears that we must compute both A00031A and AA00031 to check our work. However, it can be shown that if one of the equations in (3) is satisfied, then the other is also satisfied. So, for checking purposes it’s enough to compute either A00031A or AA00031—we don’t need to do both. 3 A00031A 0002 £ 00032 00034
MATCHED PROBLEM 1
3 Let A 0002 £ 00031 1
3 00031 1 00032 1§ £0 00035 2 2
00031 2 3
1 1 00031 § 0002 £ 0 0 0
0 1 0
0 0§ 0002 I 1
0002
00031 1 1 0§ 0 1
(A) Form the augmented matrix [A | I ]. (B) Use row operations to transform [A | I ] into [ I | B ]. (C) Verify by multiplication that B 0002 A00031.
0002
The procedure used in Example 1 can be used to find the inverse of any square matrix if the inverse exists, and will also indicate when the inverse does not exist. These ideas are summarized in Theorem 1.
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Z THEOREM 1 Inverse of a Square Matrix A If [A | I ] is transformed by row operations into [I | B ], then the resulting matrix B is A00031. If, however, we obtain all 0s in one or more rows to the left of the vertical line, then A00031 does not exist.
EXAMPLE
2
Finding a Matrix Inverse Find A00031, given A 0002 c
4 00036
00031 d 2 c
SOLUTION
4 00036
00031 1 ` 2 0 1 4
1 4
0 d 1
⬃ c
1 ` 00036 2 0
0 d 1
⬃ c
1 0
000314
⬃ c
1 0
000314
⬃ c
1 0
0 1 12 ` d 1 3 2
1 2
`
1 4 3 2
0 d 1
`
1 4
0 d 2
1 3
A00031 0002 c
1 4 R1
S R1
6R1 0006 R2 S R2
2R2 S R2 1 4 R2
0006 R1 S R1
1 12 d 3 2
You should check our work by showing that A00031A 0002 I. MATCHED PROBLEM 2
Find A00031, given A 0002 c
0002
00036 d 00032
2 1
0002
EXAMPLE
3
Finding an Inverse Find B00031, if it exists, given B 0002 c
SOLUTION
c
10 00035
10 00035
00032 1 ` 1 0
00032 d 1 0 1 000315 101 0 d ⬃ c ` d 1 00035 1 0 1 1 000315 101 0 ⬃ c ` d 0 0 12 1
We have all 0s in the second row to the left of the vertical line. Therefore, B00031 does not exist. 0002 MATCHED PROBLEM 3
Find B00031, if it exists, given B 0002 c
6 00032
00033 d 1 0002
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Z Matrix Equations Before we discuss the solution of matrix equations, you might find it helpful to briefly review the basic properties of real numbers discussed in Section R-1.
ZZZ EXPLORE-DISCUSS 3
Let a, b, and c be real numbers, with a 0006 0. Solve each equation for x. (A) ax 0002 b
(B) ax 0004 b 0002 c
Solving simple matrix equations follows very much the same procedures used in solving real number equations. We have, however, less freedom with matrix equations, because matrix multiplication is not commutative. In solving matrix equations, we will be guided by the properties of matrices summarized in Theorem 2. (Some of these properties were introduced previously.)
Z THEOREM 2 Basic Properties of Matrices Assuming all products and sums are defined for the indicated matrices A, B, C, I, and 0, then Addition Properties Associative: Commutative: Additive Identity: Additive Inverse: Multiplication Properties Associative Property: Multiplicative Identity: Multiplicative Inverse:
(A 0004 B) 0004 C 0002 A 0004 (B 0004 C ) A0004B0002B0004A A00040000200004A0002A A 0004 (0003A) 0002 (0003A) 0004 A 0002 0 A(BC ) 0002 (AB)C AI 0002 IA 0002 A If A is a square matrix and A00031 exists, then AA00031 0002 A00031A 0002 I.
Combined Properties Left Distributive: Right Distributive:
A(B 0004 C ) 0002 AB 0004 AC (B 0004 C )A 0002 BA 0004 CA
Equality Addition: Left Multiplication: Right Multiplication:
If A 0002 B, then A 0004 C 0002 B 0004 C. If A 0002 B, then CA 0002 CB. If A 0002 B, then AC 0002 BC.
The process of solving certain types of simple matrix equations is best illustrated by an example.
EXAMPLE
4
Solving a Matrix Equation Given an n 0005 n matrix A and n 0005 1 column matrices B and X, solve AX 0002 B for X. Assume all necessary inverses exist.
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SOLUTION
We are interested in finding a column matrix X that satisfies the matrix equation AX 0002 B. To solve this equation, we multiply both sides, on the left, by A00031, assuming it exists, to isolate X on the left side. AX 0002 B A (AX ) 0002 A00031B (A00031A)X 0002 A00031B IX 0002 A00031B X 0002 A00031B 00031
ZZZ
Use the left multiplication property. Associative property A00041A 0003 I IX 0003 X
0002
1. Do not mix the left multiplication property and the right multiplication property. If AX 0002 B, then
CAUTION ZZZ
A00031(AX ) 0006 BA00031 2. Matrix division is not defined. If a, b, and x are real numbers, then the solution of ax 0002 b can be written either as x 0002 a00031b or as x 0002 ba. But if A, B, and X are matrices, the solution of AX 0002 B must be written as X 0002 A00031B. The expression B A is not defined for matrices.
MATCHED PROBLEM 4
Given an n 0005 n matrix A and n 0005 1 column matrices B, C, and X, solve AX 0004 C 0002 B for X. Assume all necessary inverses exist. 0002
Z Matrix Equations and Systems of Linear Equations We will now show how independent systems of linear equations with the same number of variables as equations can be solved by first converting the system into a matrix equation of the form AX 0002 B and using X 0002 A00031B, as obtained in Example 4.
EXAMPLE
5
Using Inverses to Solve Systems of Equations Use matrix inverse methods to solve the system x1 0003 x2 0004 x3 0002 1 2x2 0003 x3 0002 1 00021 2x1 0004 3x2
SOLUTION
(4)
First, we will convert the system of equations (4) into a matrix equation: A
1 £0 2
00031 2 3
X
B
1 x1 1 00031 § £ x2 § 0002 £ 1 § 0 x3 1
(5)
You should check that the matrix equation (5) is equivalent to the original system of equations (4) by performing the multiplication on the left side, and then equating corresponding elements. If we can find the column matrix X, it will provide a solution to the system. In Example 4, we found that if AX 0002 B and A00031 exists, then X 0002 A00031B. So our job is to find A00031
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and multiply it by the constant matrix B on the left. In Example 1, we found that the inverse of matrix A is 3 3 A00031 0002 £ 00032 00032 00034 00035
00031 1§ 2
So the equation X 0002 A00031B is A00041
X
3 x1 £ x2 § 0002 £ 00032 x3 00034
B
3 00031 1 5 00032 1 § £ 1 § 0002 £ 00033 § 00035 2 1 00037
and we can conclude that x1 0002 5, x2 0002 00033, and x3 0002 00037. Check this result in system (4). 0002
MATCHED PROBLEM 5
Use matrix inverse methods to solve the system: 3x1 0003 x2 0004 x3 0002 1 0003x1 0004 x2 00023 0004 x3 0002 2 x1 [Note: The inverse of the coefficient matrix was found in Matched Problem 1.] 0002
Z USING INVERSE METHODS TO SOLVE SYSTEMS OF EQUATIONS If the number of equations in a system equals the number of variables and the coefficient matrix has an inverse, then the system will always have a unique solution that can be found by using the inverse of the coefficient matrix to solve the corresponding matrix equation. Matrix equation
Solution
AX 0002 B
X 0002 A00031B
At first, matrix inverse methods don’t seem any better than Gauss–Jordan elimination— both require applying row operations to an augmented matrix. The advantage of the inverse method becomes apparent when solving a number of systems with a common coefficient matrix, as in Example 6.
EXAMPLE
6
Using Inverses to Solve Systems of Equations Use matrix inverse methods to solve each of the following systems: (A)
SOLUTIONS
x1 0003 x2 0004 x3 0002 3 2x2 0003 x3 0002 1 2x1 0004 3x2 00024
(B)
x1 0003 x2 0004 x3 0002 00035 2x2 0003 x3 0002 2 0002 00033 2x1 0004 3x2
Notice that both systems have the same coefficient matrix A as system (4) in Example 5. Only the constant terms have been changed. So we can use A00031 to solve these systems just as we did in Example 5.
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(A)
X
x1 3 £ x2 § 0002 £ 00032 x3 00034
A00041
B
3 00031 3 8 00032 1 § £ 1 § 0002 £ 00034 § 00035 2 4 00039
The solution is, x1 0002 8, x2 0002 00034, and x3 0002 00039 (B)
X
3 x1 £ x2 § 0002 £ 00032 x3 00034
A00041
B
3 00031 00035 00036 00032 1§ £ 2§ 0002 £ 3§ 00035 2 00033 4
The solution is, x1 0002 00036, x2 0002 3, and x3 0002 4
MATCHED PROBLEM 6
0002
Use matrix inverse methods to solve each of the following systems (see Matched Problem 5): (A) 3x1 0003 x2 0004 x3 0002 3 0003x1 0004 x2 0002 00033 x1 0004 x3 0002 2
(B) 3x1 0003 x2 0004 x3 0002 00035 0003x1 0004 x2 0002 1 0004 x3 0002 00034 x1 0002
As Examples 5 and 6 illustrate, inverse methods are very convenient for hand calculations because once the inverse is found, it can be used to solve any new system formed by changing only the constant terms. Since most graphing calculators can compute the inverse of a matrix, this method also adapts readily to graphing calculator solutions. However, if your graphing calculator also has a built-in procedure for finding the reduced form of an augmented coefficient matrix, then it is just as convenient to use Gauss–Jordan elimination. Furthermore, Gauss–Jordan elimination can be used in all cases and, as noted previously, matrix inverse methods cannot always be used. The application in Example 7 illustrates the usefulness of matrix inverses.
EXAMPLE
7
Investment Allocation An investment adviser currently has two types of investments available for clients: an investment A that pays 4% per year and an investment B of higher risk that pays 8% per year. Clients may divide their investments between the two to achieve any total return desired between 4 and 8%. However, the higher the desired return, the higher the risk. How should each client listed in the table invest to achieve the indicated return? Client
SOLUTION
1
2
3
k
Total investment
$20,000
$50,000
$10,000
k1
Annual return desired
$1,200
$3,750
$500
k2
(6%)
(7.5%)
(5%)
We will first solve the problem for an arbitrary client k using inverses, and then apply the result to the three specific clients. Let x1 0002 Amount invested in A x2 0002 Amount invested in B
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Then x1 0004 x2 0002 k1 0.04x1 0004 0.08x2 0002 k2
Total invested Total annual return (4% of X1 0004 8% of X2)
Write as a matrix equation: A
c
1 0.04
X
B
1 x1 k1 dc d0002c d 0.08 x2 k2
We now find A00031 by starting with [A | I ] and proceeding as discussed earlier. 1 1 1 0 ` d 0.04 0.08 0 1 1 1 1 0 c ` d 4 8 0 100 1 1 1 0 c ` d 0 4 00034 100 1 1 1 0 c ` d 0 1 00031 25 1 0 2 000325 c ` d 0 1 00031 25 c ⬃ ⬃ ⬃ ⬃
100 R2 : R2 (To eliminate decimals)
00034R1 0004 R2 : R2
1/4
R2 : R2
(00031)R2 0004 R1 : R1
So A has an inverse, and A00031 0002 c A00041
CHECK
c
2 00031
000325 d 25
2 00031 A
I
000325 1 1 1 dc d 0002 c 25 0.04 0.08 0
0 d 1
Also, A00041
X
c
x1 2 d0002c x2 00031
B
000325 k1 dc d 25 k2
To solve each client’s investment problem, we replace k1 and k2 with appropriate values from the table and multiply by A00031. Client 1
x1 2 c d0002c x2 00031
000325 20,000 10,000 dc d0002c d 25 1,200 10,000
To draw $1,200 interest, invest $10,000 at 4% and $10,000 at 8%. Client 2
c
x1 2 d0002c x2 00031
000325 50,000 6,250 dc d0002c d 25 3,750 43,750
To draw $3,750 interest, invest $6,250 at 4% and $43,750 at 8%. Client 3
c
2 x1 d0002c x2 00031
000325 10,000 7,500 dc d0002c d 25 500 2,500
To draw $500 interest, invest $7,500 at 4% and $2,500 at 8%.
0002
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MATCHED PROBLEM 7
Repeat Example 7 with investment A paying 5% and investment B paying 9%. 0002
Z Application: Cryptography Matrix inverses can be used to provide a simple and effective procedure for encoding and decoding messages. To begin, we assign the numbers 1 to 26 to the letters in the alphabet, as shown. We also assign the number 27 to a blank to provide for space between words. (A more sophisticated code could include both uppercase and lowercase letters and punctuation symbols.) A 1
B 2
C 3
D 4
E 5
F 6
G 7
H 8
I 9
J 10
K 11
L 12
M 13
O 15
P 16
Q 17
R 18
S 19
T 20
U 21
V 22
W 23
X 24
Y 25
Z 26
Blank 27
N 14
The message SPRING BREAK corresponds to the sequence 19 16
18
9
14
7
27
2
18
5
1
11
Any matrix whose elements are positive integers and whose inverse exists can be used as an encoding matrix. For example, to use the 2 ⫻ 2 matrix A⫽c
4 5
3 d 4
to encode the preceding message, first we divide the numbers in the sequence into groups of 2 and use these groups as the columns of a matrix with 2 rows. (We would have added an extra blank in the last entry if the last column had an empty space.) Then we multiply this matrix on the left by A: c
4 5
3 19 dc 4 16
18 9
14 7
27 2
18 5
1 124 d⫽c 11 159
99 126
77 98
114 143
87 110
37
49
37 d 49
The coded message is 124
159
99
126
77
98
114
143
87 110
This message can be decoded simply by putting it back into matrix form and multiplying on the left by the decoding matrix A⫺1. Since A⫺1 is easily determined if A is known, the encoding matrix A is the only key needed to decode messages encoded in this manner. Although simple in concept, codes of this type can be very difficult to crack.
EXAMPLE
8
Cryptography The message 31 54 69 37 64 82 23 50 66 51 69 75 23 30 36 65 84 84 was encoded with the matrix A shown next. Use a graphing calculator to decode this message. 0 A ⫽ £1 2
SOLUTION
2 2 1
1 1§ 1
We begin by entering the 3 ⫻ 3 encoding matrix A (Fig. 4). Then we enter the coded message in the columns of a matrix C with three rows (Fig. 4). If B is the matrix containing the uncoded message, then B and C are related by C ⫽ AB. To find B, we multiply both sides of the equation C ⫽ AB by A⫺1 (Fig. 5).
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Z Figure 5
Z Figure 4
Writing the numbers in the columns of this matrix in sequence and using the correspondence between numbers and letters noted earlier produces the decoded message: 23 W
8 H
15 O
27
9 I
19 S
27
11 K
1 A
18 R
12 L
27
7 G
1 A
21 U
19 S
19 S
27
The answer to this question can be found somewhere in this chapter. MATCHED PROBLEM 8
0002
The message 46 84 85 55 101 100 59 95 132 25 42 53 52 91 90 43 71 83 19 37 25 was encoded with the matrix A shown here. Decode this message. 1 A 0002 £2 2
1 1 3
1 2§ 1
ANSWERS TO MATCHED PROBLEMS 3 1. (A) £ 00031 1
1 1 0 † 0 1 0
0 1 0
0 0§ 1
1 (C) £ 1 00031
1 00031 3 00031 2 00031 § £ 00031 1 00031 2 1 0
00031 3 d 000312 1
3. Does not exit
2. c 4.
00031 1 0
1 (B) £ 0 0
0 1 0
0 1 1 00031 0 † 1 2 00031 § 1 00031 00031 2
1 1 0§ 0002 £0 1 0
0 1 0
0 0§ 1
AX 0004 C 0002 B (AX 0004 C ) 0003 C 0002 B 0003 C AX 0004 (C 0003 C ) 0002 B 0003 C AX 0004 0 0002 B 0003 C AX 0002 B 0003 C A00031(AX ) 0002 A00031(B 0003 C ) (A00031A) X 0002 A00031(B 0003 C ) IX 0002 A00031(B 0003 C ) X 0002 A00031(B 0003 C )
5. x1 0002 2, x2 0002 5, x3 0002 0 6. (A) x1 0002 00032, x2 0002 00035, x3 0002 4 (B) x1 0002 0, x2 0002 1, x3 0002 00034 2.25 000325 7. A00031 0002 c d ; Client 1: $15,000 in A and $5,000 in B; Client 2: $18,750 in A and 00031.25 25 $31,250 in B; Client 3: $10,000 in A 8. WHO IS WILHELM JORDAN
0002
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Exercises
1. What is an identity matrix? 2. What is the (multiplicative) inverse of a real number? Does every real number have an inverse? 3. What is the (multiplicative) inverse of a matrix? Does every matrix have an inverse? 4. What is a singular matrix? 5. Describe the process for finding the inverse of a matrix by hand. 6. Explain how inverse matrices can be used to solve systems of linear equations by hand. 7. Explain how inverse matrices can be used to solve systems of linear equations on a graphing calculator. 8. How would you solve a linear system that has more variables than equations? 9. How would you solve a linear system that has fewer variables than equations? 10. How would you solve a linear system if the number of variables and the number of equations are equal?
1 0
1 13. £ 0 0 ⫺2 14. £ 2 5
0 2 dc 1 4
⫺3 d 5
12. c
0 ⫺2 0§ £ 2 1 5
0 1 0 1 4 1
1 4 1
3 1 ⫺2 § £ 0 0 0
2 4
⫺3 1 dc 5 0
0 d 1
3 ⫺2
⫺4 3 d; c 3 2
0 0§ 1
17. c
2 ⫺1
1 2 d; c ⫺1 ⫺1
19. c
⫺5 ⫺8
2 3 d; c 8 3
4 d 3 1 d ⫺1
⫺2 d ⫺5
⫺1 1 ⫺1 § ; £ ⫺3 0 0
0 1 0
⫺1 ⫺2 § 1
Write the matrix equations in Problems 25–28 as systems of linear equations without matrices. 25. c
2 1
⫺1 x1 3 dc d ⫽ c d 3 x2 ⫺2
⫺2 27. £ 1 0
0 2 1
1 28. £ ⫺3 2
⫺2 1 0
26. c
⫺3 ⫺1
1 x1 ⫺2 dc d ⫽ c d 2 x2 5
1 x1 3 1 § £ x2 § ⫽ £ ⫺4 § ⫺1 x3 2 3 0 x1 ⫺1 § £ x2 § ⫽ £ ⫺2 § 5 4 x3
30.
x1 ⫺ 2x2 ⫹ x3 ⫽ ⫺1 ⫺x1 ⫹ x2 ⫽ 2 2x1 ⫹ 3x2 ⫹ x3 ⫽ ⫺3
x1 ⫺ 2x2 ⫽ 7 ⫺3x1 ⫹ x2 ⫽ ⫺3
⫹ 3x3 ⫽ 5 32. 2x1 x1 ⫺ 2x2 ⫹ x3 ⫽ ⫺4 ⫺x1 ⫹ 3x2 ⫽ 2
In Problems 33–40, find x1 and x2. 33. c 35 c
x1 3 d ⫽ c x2 1
x1 ⫺2 d ⫽ c 2 x2
x1 ⫺2 ⫺2 ⫺2 dc d 34. c d ⫽ c 4 1 x2 ⫺1 3 3 dc d ⫺1 2
36. c
x1 3 d ⫽ c 0 x2
1 3 dc d 2 ⫺2 ⫺1 ⫺2 dc d 2 1
16. c
⫺2 ⫺4
1 ⫺1 d; c 2 2
⫺1 d ⫺2
37. c
1 1
⫺1 x1 5 dc d ⫽ c d ⫺2 x2 7
38. c
1 1
3 x1 9 dc d ⫽ c d 4 x2 6
18. c
5 ⫺2
⫺7 3 d; c 2 3
7 d 5
39. c
1 2
1 x1 15 dc d ⫽ c d ⫺3 x2 10
40. c
1 3
1 x1 10 dc d ⫽ c d ⫺2 x2 20
20. c
3 4 7 4 d; c d ⫺5 ⫺3 ⫺5 ⫺7
1 21. £ 0 ⫺1
2 0 1 1 0§; £0 ⫺1 1 1
⫺2 0 1 0§ ⫺1 0
1 22. £ ⫺3 0
0 1 0
0 1 0
1 1 ⫺2 § ; £ 3 1 0
0 1 0
31.
In Problems 15–24, examine the product of the two matrices to determine if each is the inverse of the other. 15. c
1 24. £ 3 0
1 3 3 ⫺1 ⫺1 § ; £ ⫺2 ⫺2 1§ 0 ⫺4 ⫺5 2
29. 4x1 ⫺ 3x2 ⫽ 2 x1 ⫹ 2x2 ⫽ 1
3 ⫺2 § 0 0 1 0
⫺1 2 3
Write each system in Problems 29–32 as a matrix equation of the form AX ⫽ B.
Perform the indicated operations in Problems 11–14. 11. c
1 23. £ 0 2
⫺1 ⫺1 § 1
In Problems 41–60, given A, find A⫺1, if it exists. Check each inverse by showing A⫺1A ⫽ I. 41. c
1 0
9 d 1
42. c
0 ⫺1
⫺1 d 3
43. c
⫺1 2
44. c
3 ⫺4 d ⫺2 3
45. c
⫺5 7 d 2 ⫺3
46. c
11 3
⫺2 d 5 4 d 1
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47. c
3 2
9 d 6
48. c
2 00033
00034 d 6
49. c
2 3
3 d 5
50. c
00035 4
4 d 00033
1 00031 0 1 00031 § 51. £ 00031 0 00031 1
2 52. £ 0 1
1 53. £ 3 1
1 54. £ 00032 3
00031 1 3 2§ 00033 2 2 00031 1 00031 § 00031 1
2 5 1
5 9§ 00032
00031 0 1 1§ 0 1
2 55. £ 0 00031
2 00031 4 00031 § 00032 1
4 56. £ 1 00033
2 57. £ 1 00031
1 1 1 0§ 00031 0
1 58. £ 2 0
00031 0 00031 1 § 1 1
10 4§ 15
1 60. £ 0 1
00035 1 00034
1 59. £ 0 1
5 1 6
Solving Systems of Linear Equations Using Matrix Inverse Methods
000310 6§ 00033
Write each system in Problems 61–68 as a matrix equation and solve using inverses. [ Note: the inverse of each coefficient matrix was found earlier in this exercise in the indicated problem. ] 61. 0003x1 0003 2x2 0002 k1 2x1 0004 5x2 0002 k2 (A) k1 0002 2, k2 0002 5 (B) k1 0002 00034, k2 0002 1 (C) k1 0002 00033, k2 0002 00032 (see Problem 43.) 62.
3x1 0003 4x2 0002 k1 00032x1 0004 3x2 0002 k2 (A) k1 0002 3, k2 0002 00031 (B) k1 0002 6, k2 0002 5 (C) k1 0002 0, k2 0002 00034 (see Problem 44.)
63. 00035x1 0004 7x2 0002 k1 2x1 0003 3x2 0002 k2 (A) k1 0002 00035, k2 0002 1 (B) k1 0002 8, k2 0002 00034 (C) k1 0002 6, k2 0002 0 (see Problem 45.) 64. 11x1 0004 4x2 0002 k1 3x1 0004 x2 0002 k2 (A) k1 0002 00032, k2 0002 00033 (B) k1 0002 00031, k2 0002 9 (C) k1 0002 4, k2 0002 5 (see Problem 46.) 65.
x1 0003 x2 0002 k1 0003x1 0004 x2 0003 x3 0002 k2 0003 x2 0004 x3 0002 k3 (A) k1 0002 1, k2 0002 1, k3 0002 2 (B) k1 0002 00031, k2 0002 0, k3 0002 00034 (C) k1 0002 3, k2 0002 00032, k3 0002 0 (see Problem 51.)
485
66. 2x1 0003 x2 0002 k1 x2 0004 x3 0002 k2 0004 x3 0002 k3 x1 (A) k1 0002 00032, k2 0002 4, k3 0002 00031 (B) k1 0002 2, k2 0002 00033, k3 0002 1 (C) k1 0002 00031, k2 0002 2, k3 0002 00035 (see Problem 52.) 67. x1 0004 2x2 0004 5x3 0002 k1 3x1 0004 5x2 0004 9x3 0002 k2 x1 0004 x2 0003 2x3 0002 k3 (A) k1 0002 0, k2 0002 1, k3 0002 4 (B) k1 0002 5, k2 0002 00031, k3 0002 0 (C) k1 0002 00036, k2 0002 0, k3 0002 2 (see Problem 53.) 68.
x1 0003 x2 0004 x3 0002 k1 00032x1 0004 3x2 0004 2x3 0002 k2 3x1 0003 3x2 0004 2x3 0002 k3 (A) k1 0002 3, k2 0002 00031, k3 0002 0 (B) k1 0002 0, k2 0002 4, k3 0002 5 (C) k1 0002 00032, k2 0002 0, k3 0002 1 (see Problem 54.)
For n 0005 n matrices A and B and n 0005 1 matrices C, D, and X, solve each matrix equation in Problems 69–74 for X. Assume all necessary inverses exist. 69. AX 0002 BX 0004 C
70. AX 0004 BX 0002 C 0004 D
71. X 0002 AX 0004 C
72. X 0004 C 0002 AX 0003 BX
73. AX 0004 C 0002 3X
74. AX 0004 C 0002 BX 0003 7X 0004 D
75. Discuss the existence of A00031 for 2 0005 2 diagonal matrices of the form A0002 c
a 0
0 d d
76. Discuss the existence of A00031 for 2 0005 2 upper triangular matrices of the form A0002 c
a 0
b d d
77. Find A00031 and A2 for each of the following matrices. (A) A 0002 c
3 00034
2 d 00033
(B) A 0002 c
00032 3
00031 d 2
78. Based on your observations in Problem 77, if A 0002 A00031 for a square matrix A, what is A2? Give a mathematical argument to support your conclusion. 79. Find (A00031)00031 for each of the following matrices. (A) A 0002 c
4 1
2 d 3
(B) A 0002 c
5 00031
5 d 3
80. Based on your observations in Problem 79, if A00031 exists for a square matrix A, what is (A00031)00031? Give a mathematical argument to support your conclusion. 81. Find (AB)00031, A00031B00031, and B00031A00031 for each of the following pairs of matrices. (A) A 0002 c
3 2
4 d 3
and
B0002 c
3 2
7 d 5
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B0002 c
and
6 2
sold (assuming that all seats can be sold) to bring in each of the returns indicated in the table? Use decimals in computing the inverse.
2 d 1
Concert
82. Based on your observations in Problems 81, which of the following is a true statement? Give a mathematical argument to support your conclusion. (A) (AB)00031 0002 A00031B00031
Return required
APPLICATIONS Problems 83–86 refer to the encoding matrix A 0002 c
3 1
5 d 2
84. CRYPTOGRAPHY Encode the message KOBE BRYANT with the matrix A. 85. CRYPTOGRAPHY The following message was encoded with the matrix A. Decode the message.
41
57 20 150 192 73
159
59
103
39
62
61
22
47
18
0 1 1 0 1
1 1 1 1 1
0 0 1 0 2
1 3 1U 2 1
89. CRYPTOGRAPHY The following message was encoded with the matrix B. Decode the message. 32 25 55 19 41 51 64 103 39 100 62 109 114 62 92 58 115 105 73 113 39 110 85 65 111 90. CRYPTOGRAPHY The following message was encoded with the matrix B. Decode the message. 88 77
29 46
$240,000
$250,000
$270,000
Labor Cost
Material Cost
A
$30
$20
B
$40
$30
If a total of $3,000 a week is allowed for labor and material, how many of each model should be produced each week to exactly use each of the allocations of the $3,000 indicated in the following table? Use decimals in computing the inverse.
82 51 61 86 108 61 72 65
45
1
2
3
Labor
$1,800
$1,750
$1,720
Material
$1,200
$1,250
$1,280
93. CIRCUIT ANALYSIS A direct current electric circuit consisting of conductors (wires), resistors, and batteries is diagrammed in the figure. V1
88. CRYPTOGRAPHY Encode the message PITTSBURGH STEELERS with the matrix B.
45 37
10,000
V2
0004 0003
87. CRYPTOGRAPHY Encode the message NEW ENGLAND PATRIOTS with the matrix B.
44 74
10,000
Weekly Allocation
Problems 87–90 require the use of a graphing calculator. To use the 5 0005 5 encoding matrix B given below, form a matrix with 5 rows and as many columns as necessary to accommodate each message. 1 0 B 0002 E2 0 1
10,000
Guitar Model
86. CRYPTOGRAPHY The following message was encoded with the matrix A. Decode the message. 49 18 103 105 41
3
92. PRODUCTION SCHEDULING Labor and material costs for manufacturing two guitar models are given in the following table:
83. CRYPTOGRAPHY Encode the message LEBRON JAMES with the matrix A.
55 22
2
Tickets sold
(B) (AB)00031 0002 B00031A00031
31 12 150 160 61 61
1
84
35
63
Solve Problems 91–97 using systems of equations and matrix inverses. 91. RESOURCE ALLOCATION A concert hall has 10,000 seats. If tickets are $20 and $30, how many of each type of ticket should be
0003 0004
1 ohm
1 ohm
I1
I2
2 ohms I3
If I1, I2, and I3 are the currents (in amperes) in the three branches of the circuit and V1 and V2 are the voltages (in volts) of the two batteries, then Kirchhoff’s* laws can be used to show that the currents satisfy the following system of equations: I1 0003 I2 0004 I3 0002 0 I1 0004 I2 0002 V1 I2 0004 2I3 0002 V2 Solve this system for: (A) V1 0002 10 volts, V2 0002 10 volts (B) V1 0002 10 volts, V2 0002 15 volts (C) V1 0002 15 volts, V2 0002 10 volts *Gustav Kirchhoff (1824–1887), a German physicist, was among the first to apply theoretical mathematics to physics. He is best known for his development of certain properties of electric circuits, which are now known as Kirchhoff’s laws.
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94. CIRCUIT ANALYSIS Repeat Problem 93 for the electric circuit shown in the figure.
96. GEOMETRY Repeat Problem 95 if the graph passes through the points (00031, k1), (0, k2), and (1, k3).
I1 0003 I2 0004 I3 0002 0
97. DIETS A biologist has available two commercial food mixes with the following percentages of protein and fat:
I1 0004 2I2 0002 V1 2I2 0004 2I3 0002 V2 V1
V2
0004 0003
0003 0004
1 ohm
2 ohms
I1
I2
Protein (%)
Fat (%)
A
20
2
B
10
6
How many ounces of each mix should be used to prepare each of the diets listed in the following table?
2 ohms
Diet
I3
95. GEOMETRY The graph of f(x) 0002 ax2 0004 bx 0004 c passes through the points (1, k1), (2, k2), and (3, k3). Determine a, b, and c for: (A) k1 0002 00032, k2 0002 1, k3 0002 6 (B) k1 0002 4, k2 0002 3, k3 0002 00032 (C) k1 0002 8, k2 0002 00035, k3 0002 4
7-5
Mix
Protein
1
2
3
20 oz
10 oz
10 oz
6 oz
4 oz
6 oz
Fat
Determinants and Cramer’s Rule Z Defining First- and Second-Order Determinants Z Evaluating Third-Order Determinants Z Using Cramer’s Rule to Solve Systems of Equations
In this section, we’ll study one more method for solving systems of linear equations using matrices. Like the inverse method, it works only for systems with the same number of equations and variables. The biggest advantage is that it’s purely computational—it requires very little symbol manipulation. The method is based on determinants.
Z Defining First- and Second-Order Determinants For any square matrix A, the determinant of A is a real number denoted by det (A) or 冟 A 冟*. If A is a square matrix of order n, then det (A) is called a determinant of order n. If A 0002 [a11 ] is a square matrix of order 1, then det (A) 0002 a11 is a first-order determinant. Now we proceed to define determinants of higher order. a11 a12 d , the second-order determinant Given a second-order square matrix A 0002 c a21 a22 of A is det (A) 0002 `
a11 a21
a12 ` 0002 a11a22 0003 a21a12 a22
(1)
*The absolute value notation will now have two interpretations: the absolute value of a real number or the determinant of a square matrix. These concepts are not the same. You must always interpret 冟 A 冟 in terms of the context in which it is used.
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Formula (1) is easily remembered if you notice that the expression on the right is the product of the elements on the principal diagonal, from upper left to lower right, minus the product of the elements on the secondary diagonal, from lower left to upper right.
EXAMPLE
1
Evaluating a Second-Order Determinant Find `
⫺1 ⫺3
⫺1 2 ` ⫺3 ⫺4 ⫽ (⫺1)(⫺4) ⫺ (⫺3)(2) ⫽ 4 ⫺ (⫺6) ⫽ 10
det (A) ⫽ `
SOLUTIONS
MATCHED PROBLEM 1
ZZZ
CAUTION ZZZ
2 `. ⫺4
Find `
3 4
⫺5 `. ⫺2
0002
0002
⫺5 3 ⫺5 d is a matrix, but ` ` represents a real number, ⫺2 4 ⫺2 3 ⫺5 ` as a determinant, and refer to the determinant of A. We will often refer to ` 4 ⫺2 the process of finding the real number it represents as “evaluating the determinant.” Remember that A ⫽ c
3 4
Technology Connections Most graphing calculators have a command to calculate determinants. On the TI-84, it is on the MATRIX-MATH menu. In Figure 1, the determinant from Example 1 is calculated.
Z Figure 1
Z Evaluating Third-Order Determinants a11 Given the matrix A ⫽ C a21 a31 a11 det (A) ⫽ † a21 a31
a12 a22 a32
a12 a22 a32
a13 a23 S, the third-order determinant of A is a33
a13 a23 † ⫽ a11a22a33 ⫺ a11a32a23 ⫹ a21a32a13 ⫺ a21a12a33 a33 ⫹ a31a12a23 ⫺ a31a22a13
(2)
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Don’t panic! You don’t need to memorize formula (2). After we introduce the ideas of minor and cofactor below, we will state a theorem that can be used to obtain the same result with much less trouble. The minor of an element in a third-order determinant is a second-order determinant obtained by deleting the row and column that contains the element. For example, in the determinant in formula (2), Deletions are usually done mentally.
a11 a31
a12 ` a32
a11 † a21 a31
a12 a22 a32
a13 a23 † 0003 a11a32 0002 a31a12 a33
a11 0003 ` a21
a13 ` a23
a11 † a21 a31
a12 a22 a32
a13 a23 † 0003 a11a23 0002 a21a13 a33
Minor of a23 0003 `
Minor of a32
Write the minors of the other seven elements in the determinant in formula (2).
ZZZ EXPLORE-DISCUSS 1
A quantity closely associated with the minor of an element is the cofactor of an element aij (from the ith row and jth column), which is defined as the product of the minor of aij and (00021)i0004j.
Z DEFINITION 1 Cofactor Cofactor of aij 0003 (00021)i0004j (Minor of aij)
So a cofactor is just a minor with either a positive or negative sign. The sign is determined by raising 00021 to a power that is the sum of the numbers indicating the row and column in which the element appears. Note that (00021)i0004j is 1 if i 0004 j is even and 00021 if i 0004 j is odd. So if we are given the determinant a11 † a21 a31
a12 a22 a32
a13 a23 † a33
then
EXAMPLE
2
Cofactor of a23 0003 (00021)200043 `
a11 a31
a12 a11 ` 00030002` a32 a31
Cofactor of a11 0003 (00021)100041 `
a22 a32
a22 a23 ` 0003 ` a33 a32
a12 ` 0003 0002(a11a32 0002 a31a12) a32 a23 ` 0003 a22a33 0002 a32a23 a33
Finding Cofactors Find the cofactors of 00022 and 5 in the determinant 00022 † 1 00021
0 00026 2
3 5† 0
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00026 5 00026 5 ` 0003 ` ` 2 0 2 0 0003 (00026)(0) 0002 (2)(5) 0003 000210 00022 0 00022 0 Cofactor of 5 0003 (00021)200043 ` ` 00030002` ` 00021 2 00021 2 0003 0002[(00022)(2) 0002 (00021)(0)] 0003 4
Cofactor of 00022 0003 (00021)100041 `
SOLUTION
MATCHED PROBLEM 2
ⴚ2 is a11
5 is a23
0002
Find the cofactors of 2 and 3 in the determinant in Example 2. 0002 [Note: The sign in front of the minor, (00021)i0004j, can be determined rather mechanically by using a checkerboard pattern of 0004 and 0002 signs over the determinant, starting with 0004 in the upper left-hand corner: 0004 0002 0004
0002 0004 0002
0004 0002 0004
Use either the checkerboard or the exponent method—whichever is easier for you—to determine the sign in front of the minor.] Theorem 1 will give us a step-by-step procedure for finding third-order determinants without having to memorize formula (2).
Z THEOREM 1 Computing a Third-Order Determinant The value of a determinant of order 3 is the sum of three products obtained by multiplying each element in any row or any column by its cofactor. This is called expanding along a row or column.
Proving Theorem 1 requires six different calculations: expanding an arbitrary thirdorder determinant along each of the rows and columns, and showing that the result matches formula (2). You will be asked to complete a couple of those cases in the exercises.
EXAMPLE
3
Evaluating a Third-Order Determinant 2 00022 Evaluate † 00023 1 1 00023
SOLUTION
0 2† 00021
We can choose any row or column to expand along. We will choose the first row because of the zero: we won’t need to find that cofactor because it will be multiplied by zero. 2 00022 † 00023 1 1 00023
0 Cofactor Cofactor Cofactor 2 † 0003 a11 a of a b 0004 a12 a of a b 0004 a13 a of a b 11 12 13 00021 0003 2 c (00021)100041 `
1 00023
2 00023 ` d 0004 (00022) c (00021)100042 ` 00021 1
2 `d 0004 0 00021
0003 (2)(1)[(1)(00021) 0002 (00023)(2)] 0004 (00022)(00021)[(00023)(00021) 0002 (1)(2)] 0003 (2)(5) 0004 (2)(1) 0003 12
0002
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2 Evaluate † 00022 00021
Determinants and Cramer’s Rule
491
1 00021 00023 0† 2 1 0002
It’s important to note that the determinant will work out the same regardless of which row or column you choose to expand along. So if possible, you should choose a row or column with one or more zeros to minimize the number of computations.
Z Using Cramer’s Rule to Solve Systems of Equations Now we will see how determinants can be used to solve systems of equations. We’ll start by investigating two equations in two variables, and then extend our results to three equations in three variables. Instead of thinking of each system of linear equations in two variables as a different problem, let’s see what happens when we attempt to solve the general system a11x 0004 a12y 0003 k1 a21x 0004 a22y 0003 k2
(3A) (3B)
once and for all, in terms of the unspecified real constants a11, a12, a21, a22, k1, and k2. We proceed by multiplying equations (3A) and (3B) by suitable constants so that when the resulting equations are added, left side to left side and right side to right side, one of the variables drops out. Suppose we choose to eliminate y. What constant should we use to make the coefficients of y the same except for the signs? Multiply equation (3A) by a22 and (3B) by 0002a12; then add: a22(3A): ⴚa12(3B):
a11a22x 0004 a12a22 y 0003
k1a22
0002a21a12x 0002 a12a22 y 0003 0002k2a12 a11a22x 0002 a21a12x 0004 0y 0003 k1a22 0002 k2a12 (a11a22 0002 a21a12)x 0003 k1a22 0002 k2a12 x0003
k1a22 0002 k2a12 a11a22 0002 a21a12
y is eliminated. Factor out x. Solve for x. a11a22 ⴚ a21a12 ⴝ 0
At this point, the numerator and denominator might remind you of second-order determinants. In fact, the value of x can be written as
x0003
`
k1 a12 ` k2 a22 a11 a21
`
a12 ` a22
Similarly, starting with system (3A) and (3B) and eliminating x (this is left as an exercise), we obtain
y0003
` `
a11 a21
k1 ` k2
a11 a21
a12 ` a22
These results are summarized in Theorem 2, Cramer’s rule, which is named after the Swiss mathematician Gabriel Cramer (1704–1752).
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Z THEOREM 2 Cramer’s Rule for Two Equations and Two Variables Given the system
a11x 0004 a12 y 0003 k1 a21x 0004 a22 y 0003 k2
with
D0003 `
a11 a21
a12 ` 00050 a22
then
x0003
`
k1 k2
a12 ` a22 D
and
a11 k1 ` a21 k2 y0003 D `
The determinant D is called the coefficient determinant. If D 0005 0, then the system has exactly one solution, which is given by Cramer’s rule. If, on the other hand, D 0003 0, then it can be shown that the system is either inconsistent and has no solutions or is dependent and has an infinite number of solutions. In that case, we would need to use other methods to determine the exact nature of the solutions.
EXAMPLE
4
Solving a Two-Variable System with Cramer’s Rule Solve using Cramer’s rule:
SOLUTIONS
3x 0002 5y 0003 2 00024x 0004 3y 0003 00021
First find the determinant of the coefficient matrix: D0003 `
3 00024
00025 ` 0003 9 0002 20 0003 000211 3
Now replace the x column with the constants and find the determinant, then divide by 000211. 2 00025 ` 00021 3 600025 1 x0003 0003 00030002 000211 000211 11 `
Now repeat, this time replacing the y column with the constants. 3 2 ` 00023 0002 (00028) 00024 00021 5 y0003 0003 00030002 000211 000211 11 `
The solution to the system is x 0003 0002
MATCHED PROBLEM 4
Solve using Cramer’s rule:
1 5 ,y00030002 . 11 11
3x 0004 2y 0003 00024 00024x 0004 3y 0003 000210
0002
0002
Cramer’s rule can be generalized completely for any size linear system that has the same number of variables as equations. However, it cannot be used to solve systems where the number of variables is not equal to the number of equations. In Theorem 3 we state without proof Cramer’s rule for three equations in three variables.
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Z THEOREM 3 Cramer’s Rule for Three Equations in Three Variables Given the system a11x 0004 a12 y 0004 a13z 0003 k1 a21x 0004 a22 y 0004 a23z 0003 k2 a31x 0004 a32 y 0004 a33z 0003 k3
a11 D 0003 † a21 a31
with
a12 a22 a32
a13 a23 † 0005 0 a33
then
x0003
k1 † k2 k3
a12 a22 a32 D
a13 a23 † a33
y0003
k1 a13 k2 a23 † k3 a33 D
a11 † a21 a31
z0003
a11 † a21 a31
a12 a22 a32 D
k1 k2 † k3
You can easily remember these determinant formulas for x, y, and z if you observe the following: 1. 2. 3.
EXAMPLE
5
Determinant D is formed from the coefficients of x, y, and z, keeping the same relative position in the determinant as found in the system of equations. Determinant D appears in the denominators for x, y, and z. The numerator for x can be obtained from D by replacing the coefficients of x (a11, a21, and a31) with the constants k1, k2, and k3, respectively. Similar statements can be made for the numerators for y and z.
Solving a Three-Variable System with Cramer’s Rule Solve using Cramer’s rule:
SOLUTION
x0004 y 0003 2 3y 0002 z 0003 00024 x 0004z0003 3 1 D0003 †0 1
x0003
y0003
z0003
2 † 00024 3 1 †0 1 1 †0 1
1 3 0
0 00021 † 0003 2 1 1 3 0 2
2 00024 3 2
0 00021 † 1 0 00021 † 1
0003
7 2
00030002
3 2
1 2 3 00024 † 0 3 1 00030002 2 2
0002
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Solve using Cramer’s rule:
MATCHED PROBLEM 5
3x 0002z00035 x0002y0004z00030 x0004y 00031
0002
Cofactor expansion can be used to find determinants of orders higher than 3, so Cramer’s rule can be used for systems with more than three variables. For large systems, however, the Gauss-Jordan method, which involves fewer arithmetic operations than Cramer’s Rule, is a more practical choice.
ANSWERS TO MATCHED PROBLEMS 1. 14 2. Cofactor of 2 0003 13; cofactor of 3 0003 00024 4. x 0003 178 , y 0003 000246 5. x 0003 65, y 0003 000215, z 0003 000275 17
7-5
3. 3
Exercises
1. Explain the difference between c
a11 a21
a12 a11 d and ` a22 a21
a12 `. a22
2. Explain the difference between a matrix and a minor. 3. Explain the difference between a minor and a cofactor. 4. How do you evaluate a third-order determinant? 5. If A is the 2 0006 2 coefficient matrix for a linear system and det (A) 0003 0, what can you conclude about the solution set for the system?
17. 2x 0004 y 0003 1 5x 0004 3y 0003 2
18. x 0004 3y 0003 1 2x 0004 8y 0003 0
19. 2x 0002 y 0003 00023 0002x 0004 3y 0003 3
20. 00023x 0004 2y 0003 1 2x 0002 3y 0003 00023
21. 4x 0002 3y 0003 4 3x 0004 2y 0003 00022
22. 5x 0004 2y 0003 00021 2x 0002 3y 0003 2
Problems 23–30 pertain to the following determinant: 5 †3 0
6. Can you use Cramer’s rule to solve a linear system with a 3 0006 2 coefficient matrix? Explain. 7. Can you use Cramer’s rule to solve a linear system with a 4 0006 4 coefficient matrix? Explain. 8. List all the possible solution methods for linear systems that we have discussed in this chapter. Which is your favorite and why? Evaluate each second-order determinant in Problems 9–14. 5 9. ` 2
4 ` 3
11. `
3 00027 ` 00025 6
13. `
4.3 00025.1
00021.2 ` 3.7
8 10. ` 4
00023 ` 1
12. `
9 4
00022 ` 0
14. `
00020.7 1.9
00023 6† 8
Write the minor of each element given in Problems 23–26. Leave the answer in determinant form. 23. a11
24. a33
25. a23
26. a12
Write the cofactor of each element given in Problems 27–30, and evaluate each. 27. a11
28. a33
30. a12
29. a23
Evaluate the determinant in Problems 31–40 using cofactors. 00022.3 ` 00024.8
Solve the system in Problems 15–22 using Cramer’s rule. 15. x 0004 2y 0003 1 x 0004 3y 0003 00021
00021 4 00022
16. x 0004 2y 0003 3 x 0004 3y 0003 5
1 31. † 00022 5 0 33. † 3 0
0 4 00022
0 3† 1
1 5 00027 6† 00022 00023
2 32. † 0 0
00023 00023 6
5 1† 2
4 34. † 9 1
00022 5 2
0 4† 0
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00021 35. † 00022 4
2 00023 0 00026 † 00023 2
1 37. † 1 2
4 1 1
1 39. † 2 3
4 3 1 6† 00022 9
0 2 00021 36. † 00026 3 1† 7 00029 00022
1 00022 † 00021
3 2 1 38. † 00021 5 1 † 2 3 1 4 40. † 00021 5
00026 3 4 1† 00026 3
In Problems 59 and 60, use Cramer’s rule to solve for z only. 59.
3x 0002 4y 0004 5z 0003 18 60. 13x 0004 11y 0004 10z 0003 2 00029x 0004 8y 0004 7z 0003 000213 10x 0004 8y 0004 7z 0003 1 5x 0002 7y 0004 10z 0003 33 8x 0004 5y 0004 4z 0003 4
If A is a 3 0005 3 matrix, det A can be evaluated by the following diagonal expansion. Form a 3 0005 5 matrix by augmenting A on the right with its first two columns, and compute the diagonal products p1, p2, . . . , p6 indicated by the arrows: a11 a12 C a21 a22 a31 a32
Solve Problems 41–44 to two significant digits using Cramer’s rule. 41. 0.9925x 0002 0.9659y 0003 0 0.1219x 0004 0.2588y 0003 2,500
p4
42. 0.9877x 0002 0.9744y 0003 0 0.1564x 0004 0.2250y 0003 1,900
495
Determinants and Cramer’s Rule
p5
a13 a11 a23 a21 a33 a31
p6
a12 a22 S a32 p1
p2
Diagonal expansion formula
p3
The determinant of A is given by [compare with formula (2)] det A 0003 p1 0004 p2 0004 p3 0002 p4 0002 p5 0002 p6
43. 0.9954x 0002 0.9942y 0003 0 0.0958x 0004 0.1080y 0003 155
0003 a11a22a33 0004 a12a23a31 0004 a13a21a32 0002 a13a22a31 0002 a11a23a32 0002 a12a21a33
44. 0.9973x 0002 0.9957y 0003 0 0.0732x 0004 0.0924y 0003 112
[Caution: The diagonal expansion procedure works only for 3 0005 3 matrices. Do not apply it to matrices of any other size.]
Solve Problems 45–52 using Cramer’s rule:
Use the diagonal expansion formula to evaluate the determinants in Problems 61 and 62.
45.
x0004 y 0003 0 2y 0004 z 0003 00025 0002x 0004 z 0003 00023
47. x 0004 y 00031 2y 0004 z 0003 0 0002y 0004 z 0003 1
46.
48.
x0004 y 0003 00024 2y 0004 z 0003 0 0002x 0004z0003 5 x 0004 3y 0003 00023 2y 0004 z 0003 3 0002x 0004 3z 0003 7
49.
3y 0004 z 0003 00021 x 0004 2z 0003 3 x 0002 3y 0003 00022
50. x 0002z0003 3 2x 0002 y 0003 00023 x0004y0004z0003 1
51.
2y 0002 z 0003 00023 x0002 y0002 z0003 2 x 0002 y 0004 2z 0003 4
52. 2x 0004 y 0003 2 x 0002 y 0004 z 0003 00021 x0004y0004z0003 2
Discuss the number of solutions for the systems in Problems 53 and 54 where a and b are real numbers. Use Cramer’s rule where appropriate and Gauss–Jordan elimination otherwise. 53. ax 0004 3y 0003 b 2x 0004 4y 0003 5
54. 2x 0004 ay 0003 b 3x 0004 4y 0003 7
2 61. † 5 00024
6 00021 3 00027 † 00022 1
A square matrix is called an upper triangular matrix if all elements below the principal diagonal are zero. In Problems 63–66, determine whether the statement is true or false. If true, explain why. If false, give a counterexample. 63. If the determinant of an upper triangular matrix is 0, then the elements on the principal diagonal are all 0. 64. If A and B are upper det (A 0004 B) 0003 det A 0004 det B.
2x 0002 3y 0004 z 0003 00023 00024x 0004 3y 0004 2z 0003 000211 x0002 y0002 z0003 3
56.
x 0004 4y 0002 3z 0003 25 3x 0004 y 0002 z 0003 2 00024x 0004 y 0004 2z 0003 1
triangular
matrices,
then
65. The determinant of an upper triangular matrix is the product of the elements on the principal diagonal. 66. If A and B are upper det (AB) 0003 (det A)(det B).
triangular
matrices,
then
67. Show that the expansion of the determinant a11 a12 † a21 a22 a31 a32
In Problems 55 and 56, use Cramer’s rule to solve for x only. 55.
1 00025 2 00026 † 00021 7
4 62. † 1 00023
a13 a23 † a33
by the first column is the same as its expansion by the third row, and that both match formula (2).
In Problems 57 and 58, use Cramer’s rule to solve for y only.
68. Repeat Problem 67, using the second row and the third column.
57. 12x 0002 14y 0004 11z 0003 5 58. 2x 0002 y 0004 4z 0003 15 15x 0004 7y 0002 9z 0003 000213 0002x 0004 y 0004 2z 0003 5 5x 0002 3y 0004 2z 0003 0 3x 0004 4y 0002 2z 0003 4
69. If A0003 c
2 1
3 d 00022
and
show that det (AB) 0003 (det A)(det B).
B0003 c
00021 2
3 d 1
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70. If A0003 c
a c
b d d
and
B0003 c
w y
x d z
show that det (AB) 0003 (det A)(det B). It is clear that x 0003 0, y 0003 0, z 0003 0 is a solution to each of the systems given in Problem 71. Use Cramer’s rule to determine whether this solution is unique. [Hint: If D 0005 0, what can you conclude? If D 0003 0, what can you conclude?] 71. (a). x 0002 4y 0004 9z 0003 0 4x 0002 y 0004 6z 0003 0 x 0002 y 0004 3z 0003 0
(b).
3x 0002 y 0004 3z 0003 0 5x 0004 5y 0002 9z 0003 0 00022x 0004 y 0002 3z 0003 0
(A) To analyze the effect of price changes on the daily revenue, an economist wants to express the daily revenue R in terms of p and q only. Use system (1) to eliminate x and y in the equation for R, expressing the daily revenue in terms of p and q. (B) To analyze the effect of changes in demand on the daily revenue, the economist now wants to express the daily revenue in terms of x and y only. Use Cramer’s rule to solve system (1) for p and q in terms of x and y and then express the daily revenue R in terms of x and y. 74. REVENUE ANALYSIS A company manufactures ten-speed and three-speed bicycles. The weekly demand equations are p 0003 230 0002 10x 0004 5y q 0003 130 0004 4x 0002 4y
72. Prove Theorem 2 for y.
(2)
where $p is the price of a ten-speed bicycle, $q is the price of a three-speed bicycle, x is the weekly demand for ten-speed bicycles, and y is the weekly demand for three-speed bicycles. The weekly revenue R is given by
APPLICATIONS 73. REVENUE ANALYSIS A supermarket sells two brands of coffee: brand A at $p per pound and brand B at $q per pound. The daily demand equations for brands A and B are, respectively, x 0003 200 0002 6p 0004 4q y 0003 300 0004 2p 0002 3q
(1)
(both in pounds). The daily revenue R is given by
R 0003 xp 0004 yq (A) Use system (2) to express the daily revenue in terms of x and y only. (B) Use Cramer’s rule to solve system (2) for x and y in terms of p and q, and then express the daily revenue R in terms of p and q only.
R 0003 xp 0004 yq
CHAPTER
7-1
7
Review
Systems of Linear Equations
A system of two linear equations in two variables is a system of the form ax 0004 by 0003 h cx 0004 dy 0003 k
(1)
where x and y are variables; a, b, c, and d are real numbers called the coefficients of x and y, and h and k are real numbers called the constant terms in the equations. The ordered pair of numbers (x0, y0) is a solution to system (1) if each equation is satisfied by the pair. The set of all such ordered pairs of numbers is called the solution set for the system. To solve a system is to find its solution set. In general, a system of linear equations has exactly one solution, no solution, or infinitely many solutions. A system of linear equations is consistent if it has one or more solutions and inconsistent if no solutions exist. A consistent system is said to be independent if it has exactly one solution and dependent if it has more than one solution. To solve a system by substitution, solve either equation for either variable, substitute in the other equation, solve the resulting linear equation in one variable, and then substitute this value into the expression obtained in the first step to find the other variable.
Two systems of equations are equivalent if both have the same solution set. To solve a system of equations using elimination by addition, use Theorem 2 to find a simpler equivalent system whose solution is obvious. As stated in Theorem 2, a system of linear equations is transformed into an equivalent system if: 1. Two equations are interchanged. 2. An equation is multiplied by a nonzero constant. 3. A constant multiple of another equation is added to a given equation. The solution set S of a dependent system is often expressed in terms of a parameter. Any element in S is called a particular solution. Any equation that can be written in the form ax 0004 by 0004 cz 0003 k where a, b, c, and k are constants (not all a, b, and c zero) is called a linear equation in three variables. The method of elimination by addition can be used for systems of linear equations in three variables.
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Solving Systems of Linear Equations Using Gauss–Jordan Elimination
497
3. The column containing the leftmost 1 of a given row has 0’s above and below the 1.
The method of solution using elimination by addition can be transformed into a more efficient method for larger-scale systems by the introduction of an augmented matrix. A matrix is a rectangular array of numbers written within brackets. Each number in a matrix is called an element of the matrix. If a matrix has m rows and n columns, it is called an m ⴛ n matrix (read “m by n matrix”). The expression m 0006 n is called the size of the matrix, and the numbers m and n are called the dimensions of the matrix. A matrix with n rows and n columns is called a square matrix of order n. A matrix with only one column is called a column matrix, and a matrix with only one row is called a row matrix. The position of an element in a matrix is the row and column containing the element. This is usually denoted using double subscript notation aij, where i is the row and j is the column containing the element aij. The principal diagonal of a matrix A consists of the elements aii, i 0003 1, 2, . . . , n. Rather than using x, y, and z to denote variables, we will use subscript notation x1, x2, and x3. Related to the system
4. The leftmost 1 in any row is to the right of the leftmost 1 in the preceding row.
x1 0004 5x2 0002 3x3 0003 4
If at any point in the preceding process we obtain a row with all 0’s to the left of the vertical line and a nonzero number n to the right, we can stop, since we have a contradiction: 0 0003 n, n 0005 0. We can then conclude that the system has no solution. If this does not happen and we obtain an augmented matrix in reduced form without any contradictions, the solution can be found by converting back to equation form.
0002 4x3 0003 1
6x1
00022x1 0004 3x2 0004 4x3 0003 7 are the following matrices: Coefficient matrix
1 £ 6 00022
5 0 3
00023 00024 § 4
Constant matrix
4 £1§ 7
Augmented coefficient matrix
1 £ 6 00022
5 0 3
00023 4 00024 † 1 § 4 7
Two augmented matrices are row-equivalent, denoted by the symbol ⬃ between the two matrices, if they are augmented matrices of equivalent systems of equations. An augmented matrix is transformed into a row-equivalent matrix if any of the following row operations is performed: 1. Two rows are interchanged. 2. A row is multiplied by a nonzero constant. 3. A constant multiple of another row is added to a given row. These correspond to the operations on equations from Theorem 2 in Section 7–1. The following symbols are used to describe these row operations: 1. Ri 4 Rj means “interchange row i with row j.” 2. kRi S Ri means “multiply row i by the constant k.” 3. kRj 0004 Ri S Ri means “multiply row j by the constant k and add to row i.” As before, our objective is to start with the augmented matrix of a linear system and transform it using row operations into a simple form where the solution can be found easily. The simple form, called the reduced form, is achieved if: 1. Each row consisting entirely of 0’s is below any row having at least one nonzero element. 2. The leftmost nonzero element in each row is 1.
A reduced system is a system of linear equations that corresponds to a reduced augmented matrix. When a reduced system has more variables than equations and contains no contradictions, the system is dependent and has infinitely many solutions. The Gauss–Jordan elimination procedure for solving a system of linear equations is given in step-by-step form as follows: Step 1. Choose the leftmost nonzero column, and use appropriate row operations to get a 1 at the top. Step 2. Use multiples of the row containing the 1 from step 1 to get zeros in all remaining places in the column containing this 1. Step 3. Repeat step 1 with the submatrix formed by (mentally) deleting the row used in step 2 and all rows above this row. Step 4. Repeat step 2 with the entire matrix, including the mentally deleted rows. Continue this process until the entire matrix is in reduced form.
7-3
Matrix Operations
Two matrices are equal if they are the same size and their corresponding elements are equal. The sum of two matrices of the same size is a matrix with elements that are the sums of the corresponding elements of the two given matrices. Matrix addition is commutative and associative. A matrix with all zero elements is called the zero matrix. The negative of a matrix M, denoted 0002M, is a matrix with elements that are the negatives of the elements in M. If A and B are matrices of the same size, then we define subtraction as follows: A 0002 B 0003 A 0004 (0002B). The product of a number k and a matrix M, denoted by kM, is a matrix formed by multiplying each element of M by k. The product of a 1 0006 n row matrix and an n 0006 1 column matrix is a 1 0006 1 matrix given by nⴛ1 1ⴛn
[a1
a2
...
b1 1ⴛ1 b2 an ] ≥ ¥ 0003 [a1b1 0004 a2b2 0004 # # # 0004 anbn ] o bn
If A is an m 0006 p matrix and B is a p 0006 n matrix, then the matrix product of A and B, denoted AB, is an m 0006 n matrix whose element in the ith row and jth column is the real number obtained from the product of the ith row of A and the jth column of B. If the number of columns in A does not equal the number of rows in B, then the matrix product AB is not defined. Matrix multiplication is not commutative, and the zero property does not hold for matrix multiplication. That is, for matrices A and B, the matrix product AB can be zero without either A or B being the zero matrix.
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Solving Systems of Linear Equations Using Matrix Inverse Methods
The identity matrix for multiplication for the set of all square matrices of order n is the square matrix of order n, denoted by I, with 1’s along the principal diagonal (from upper left corner to lower right corner) and 0’s elsewhere. If M is a square matrix of order n and I is the identity matrix of order n, then IM 0003 MI 0003 M If M is a square matrix of order n and if there exists a matrix M 00021 (read “M inverse”) such that M 00021M 0003 MM 00021 0003 I then M 00021 is called the multiplicative inverse of M or, more simply, the inverse of M. If the augmented matrix [ M | I] is transformed by row operations into [I | B], then the resulting matrix B is M 00021. If, however, we obtain all 0’s in one or more rows to the left of the vertical line, then M 00021 does not exist and M is called a singular matrix. A system of linear equations with the same number of variables as equations such as a11x1 0004 a12 x2 0004 a13x3 0003 k1
A determinant of order n is a determinant with n rows and n columns. The value of a second-order determinant is the real number given by `
a11 a11x 0004 a12 y 0004 a13 z 0003 k1 a21x 0004 a22 y 0004 a23 z 0003 k2 with D 0003 † a21 a31x 0004 a32 y 0004 a33 z 0003 k3 a31
a11 £ a21 a31
a12 a22 a32
x0003 X
B
a13 x1 k1 a23 § £ x2 § 0003 £ k2 § a33 x3 k3
If the inverse of A exists, then the matrix equation has a unique solution given by X 0003 A00021B After multiplying B by A00021 on the left, it is easy to read the solution to the original system of equations.
7-5
Determinants and Cramer’s Rule
Associated with each square matrix A is a real number called the determinant of the matrix. The determinant of A is denoted by det A, or simply by writing the array of elements in A using vertical lines in place of square brackets. For example, det c
CHAPTER
a11 a21
7
a12 a11 d 0003 ` a22 a21
a12 a22 a32
a13 a23 † 0005 0 a33
then
a31x1 0004 a32x2 0004 a33x3 0003 k3
A
a12 ` 0003 a11a22 0002 a21a12 a22
The value of a third-order determinant is the sum of three products obtained by multiplying each element of any one row (or each element of any one column) by its cofactor. The cofactor of an element aij (from the ith row and jth column) is the product of the minor of aij and (00021)i0004j. The minor of an element aij is the determinant remaining after deleting the ith row and jth column. Systems of equations having the same number of variables as equations can also be solved using determinants and Cramer’s rule. Cramer’s rule for three equations and three variables is as follows: Given the system
a21x1 0004 a22x2 0004 a23x3 0003 k2 can be written as the matrix equation
a11 a21
k1 † k2 k3
a12 a13 a11 a22 a23 † † a21 a32 a33 a31 y0003 D
k1 k2 k3 D
a13 a23 † a33
z0003
a11 † a21 a31
k1 k2 † k3
Cramer’s rule can be generalized completely for any size linear system that has the same number of variables as equations. The formulas are easily remembered if you observe the following: 1. Determinant D is formed from the coefficients of x, y, and z, keeping the same relative position in the determinant as found in the system of equations. 2. Determinant D appears in the denominators for x, y, and z. 3. The numerator for x can be obtained from D by replacing the coefficients of x (a11, a21, and a31) with the constants k1, k2, and k3, respectively. Similar statements can be made for the numerators for y and z. Cramer’s rule is rarely used to solve systems of order higher than 3 by hand, because more efficient methods are available. Cramer’s rule, however, is a valuable tool in more advanced theoretical and applied mathematics.
a12 ` a22
Review Exercises
Work through all the problems in this chapter review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text.
a12 a22 a32 D
Solve the system in Problems 1–5 using substitution or elimination by addition. 1. 2x 0004 y 0003 7 3x 0002 2y 0003 0
2.
3x 0002 6y 0003 5 00022x 0004 4y 0003 1
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3.
4x 0002 3y 0003 00028 00022x 0004 32 y 0003 4
4.
x 0002 3y 0004 z 0003 4 0002x 0004 4y 0002 4z 0003 1 2x 0002 y 0004 5z 0003 00023
5. 2x 0004 y 0002 z 0003 5 x 0002 2y 0002 2z 0003 4 3x 0004 4y 0004 3z 0003 3
26. Solve the system using Cramer’s rule: 3x 0002 2y 0003
x 0004 3y 0003 00021 x1 0002 x2 0003 4 2x1 0004 x2 0003 2 Then write the linear system represented by each augmented matrix in your solution, and solve each of these systems graphically. Discuss the relationship between the solutions of these systems.
Perform each of the row operations indicated in Problems 7–9 on the following augmented matrix: c
28. Use an intersection routine on a graphing calculator to approximate the solution of the following system to two decimal places:
00024 5 ` d 00026 12
1 3
x 0004 3y 0003 9 00022x 0004 7y 0003 10
8. 13R2 S R2
7. R1 4 R2
Solve the system in Problems 29–34 using Gauss–Jordan elimination.
9. (00023)R1 0004 R2 S R2 In Problems 10–12, write the linear system corresponding to each reduced augmented matrix and solve. 10. c
1 0
4 0 d ` 1 00027
12. c
1 0
00021 4 ` d 0 0
11. c
1 0
00021 4 ` d 0 1
In Problems 13–21, perform the operations that are defined, given the following matrices: 00022 00021 d B0003 c 3 00024
5 d C 0003 [00021 6
3 4] D 0003 c d 00022
13. AB
14. CD
15. CB
16. AD
17. A 0004 B
18. C 0004 D
19. A 0004 C
20. 2A 0002 5B
21. CA 0004 C
22. Find the inverse of A0003 c
4 7 d 00021 00022
29. 3x1 0004 2x2 0003 3 x1 0004 3x2 0003 8
30. x1 0004 x2 0003 1 x1 0002 x3 0003 00022 x2 0004 2x3 0003 4
31. x1 0004 2x2 0004 3x3 0003 1 2x1 0004 3x2 0004 4x3 0003 3 x1 0004 2x2 0004 x3 0003 3
32. x1 0004 2x2 0002 x3 0003 2 2x1 0004 3x2 0004 x3 0003 00023 3x1 0004 5x2 0003 00021
33. x1 0002 2x2 0003 1 2x1 0002 x2 0003 0 x1 0002 3x2 0003 00022
34. x1 0004 2x2 0002 x3 0003 2 3x1 0002 x2 0004 2x3 0003 00023
In Problems 35–40, perform the operations that are defined, given the following matrices: 1 A0003 £ 4 00023
2 5§ 00021
D0003 c
7 0
00025 d 00022
9 00026
E0003 c
36. DA
37. BC
38. CB
39. DE
40. ED
1 A 0003 £ 00022 4
3x1 0004 2x2 0003 k1 4x1 0004 3x2 0003 k2
00021]
00023 d 2
0 4 1 0§ 00021 4
Show that AA00021 0003 I.
as a matrix equation, and solve using matrix inverse methods for: (A) k1 0003 3, k2 0003 5 (B) k1 0003 7, k2 0003 10 (C) k1 0003 4, k2 0003 2 Evaluate the determinants in Problems 24 and 25. 2 25. † 0 1
C 0003 [2 4
35. AD
23. Write the system
00023 ` 00021
0 8
6 B 0003 £ 0§ 00024
41. Find the inverse of
Show that A00021A 0003 I.
2 24. ` 00025
8
27. Use Gauss–Jordan elimination to solve the system
6. Solve the system by graphing. 3x 0002 2y 0003 8 x 0004 3y 0003 00021
4 A0003 c 0
499
3 5 00024
00024 0† 00022
42. Write the system x1 0004 2x2 0004 3x3 0003 k1 2x1 0004 3x2 0004 4x3 0003 k2 x1 0004 2x2 0004 x3 0003 k3 as a matrix equation, and solve using matrix inverse methods for: (A) k1 0003 1, k2 0003 3, k3 0003 3 (B) k1 0003 0, k2 0003 0, k3 0003 00022 (C) k1 0003 00023, k2 0003 00024, k3 0003 1
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Evaluate the determinants in Problems 43 and 44. 43. `
000214 1 2
2 44. † 00023 1
3 2 2` 3
00021 5 00022
APPLICATIONS 1 2† 4
45. Solve for y only using Cramer’s rule:
55. DIET A laboratory assistant needs a food mix that contains, among other things, 27 grams of protein, 5.4 grams of fat, and 19 grams of moisture. He has available mixes A, B, and C with the compositions listed in the table. How many grams of each mix should be used to get the desired diet mix? Set up a system of equations and solve using Gauss–Jordan elimination.
x 0002 2y 0004 z 0003 00026 y0002z0003
4
2x 0004 2y 0004 z 0003
2
46. Solve using Gauss–Jordan elimination: x1 0004
x2 0004
x3 0003 7,000
0.04x1 0004 0.05x2 0004 0.06x3 0003
360
0.04x1 0004 0.05x2 0002 0.06x3 0003
120
47. Show that u ` w
v u 0004 kv ` 0003 ` w 0004 kx x
v ` x
48. Discuss the number of solutions for the system corresponding to the reduced form shown here if (A) m 0005 0 (B) m 0003 0 and n 0005 0 (C) m 0003 0 and n 0003 0 1 £0 0
00023 4 2 † 5§ m n
0 1 0
50. If A is a nonzero square matrix of order n satisfying A2 0003 0, can A00021 exist? Explain. 51. For n 0006 n matrices A and C and n 0006 1 column matrices B and X, solve for X assuming all necessary inverses exist: AX 0002 B 0003 CX
Moisture (%)
A
30
3
10
B
20
5
20
C
10
4
10
56. RESOURCE ALLOCATION A Colorado mining company operates mines at Big Bend and Saw Pit. The Big Bend mine produces ore that is 5% nickel and 7% copper. The Saw Pit mine produces ore that is 3% nickel and 4% copper. How many tons of ore should be produced at each mine to obtain the amounts of nickel and copper listed in the table? Set up a matrix equation and solve using matrix inverses. Copper
(A) 3.6 tons
5 tons
(B) 3 tons
4.1 tons
(C) 3.2 tons
4.4 tons
57. LABOR COSTS A company with manufacturing plants in North and South Carolina has labor-hour and wage requirements for the manufacturing of computer desks and printer stands as given in matrices L and H:
1.7 h 0.9 h
2.4 h 1.8 h
0.8 h d 0.6 h
Desk Stand
Hourly wages North South Carolina Carolina plant plant
Show that A A 0003 I. 53. Clear the decimals in the system 0.04x1 0004 0.05x2 0004 0.06x3 0003
360
0.04x1 0004 0.05x2 0002 0.06x3 0003
120
x2 0004
Fat (%)
L0003 c
6 00026 § 1
00021
x1 0004
Protein (%)
Labor-hour requirements Fabricating Assembly Packaging department department department
52. Find the inverse of 5 5 1
Mix
Nickel
49. Discuss the number of solutions for a system of n equations in n variables if the coefficient matrix: (A) Has an inverse. (B) Does not have an inverse.
4 A 0003 £4 1
54. BUSINESS A container holds 120 packages. Some of the packages weigh 12 pound each, and the rest weigh 13 pound each. If the total contents of the container weigh 48 pounds, how many are there of each type of package?
x3 0003 7,000
by multiplying the first two equations by 100. Then write the resulting system as a matrix equation and solve using the inverse found in Problem 52.
$11.50 H 0003 £ $9.50 $5.00
$10.00 $8.50 § $4.50
Fabricating department Assembly department Packaging department
(A) Find the labor cost for producing one printer stand at the South Carolina plant. (B) Discuss possible interpretations of the elements in the matrix products HL and LH. (C) If either of the products HL or LH has a meaningful interpretation, find the product and label its rows and columns.
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58. LABOR COSTS The monthly production of computer desks and printer stands for the company in Problem 57 for the months of January and February are given in matrices J and F: January production North South Carolina Carolina plant plant
1,500 J0002 c 850
Desks
1,810 d 740
Desks Stands
(A) Find the average monthly production for the months of January and February.
ZZZ
21 21 27 30 28 31 29 34 50 46 35 62 19 21 39 52 52 79
Stands
February production North South Carolina Carolina plant plant
CHAPTER
(B) Find the increase in production from January to February. 1 (C) Find J c d and interpret. 1 59. CRYPTOGRAPHY The following message was encoded with the matrix B shown below. Decode the message:
1,650 d 700
1,700 F0002 c 930
501
1 B 0002 £1 1
1 0 1
0 1§ 1
60. PUZZLE A piggy bank contains 30 coins worth $1.90. (A) If the bank contains only nickels and dimes, how many coins of each type does it contain? (B) If the bank contains nickels, dimes, and quarters, how many coins of each type does it contain?
7
GROUP ACTIVITY Modeling with Systems of Linear Equations
In this group activity, we will consider two real-world problems that can be solved using systems of linear equations: heat conduction and traffic flow. Both problems involve using a grid and a basic assumption to construct the model (the system of equations). Gauss–Jordan elimination is then used to solve the model. In the heat conduction problem, the solution of the model is easily interpreted in terms of the original problem. The system in the second problem is dependent, and the solution requires a more careful interpretation.
I HEAT CONDUCTION A metal grid consists of four thin metal bars. The end of each bar of the grid is kept at a constant temperature, as shown in Figure 1. We assume that the temperature at each intersection point in the grid is the average of the temperatures at the four 00004
400004 400004
x1
x2
x3
x4
200004
Z Figure 1
400004
200004
400004 00004
adjacent points in the grid (adjacent points are either other intersection points or ends of bars). So the temperature x1 at the intersection point in the upper left-hand corner of the grid must satisfy Left
x1 0002 14(40
Above
0003
0
Right
0003
x2 0003
Below
x3)
Find equations for the temperature at the other three intersection points, and solve the resulting system to find the temperature at each intersection point in the grid.
II TRAFFIC FLOW The rush-hour traffic flow for a network of four one-way streets in a city is shown in Figure 2 on page 506. The numbers next to each street indicate the number of vehicles per hour that enter and leave the network on that street. The variables x1, x2, x3, and x4 represent the flow of traffic between the four intersections in the network. For a smooth flow of traffic, we assume that the number of vehicles entering each intersection should always equal the number leaving. For example, since 1,500 vehicles enter the intersection of 5th Street and Washington Avenue each hour and x1 0003 x4 vehicles leave this intersection, we see that x1 0003 x4 0002 1,500. (A) Find the equations determined by the traffic flow at each of the other three intersections. (B) Find the solution to the system in part A.
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800
700 600 x1
x4
x2
900 400 5th St.
600 Washington Ave.
x3 500 6th St.
500
Lincoln Ave.
Z Figure 2 (C) What is the maximum number of vehicles that can travel from Washington Avenue to Lincoln Avenue on 5th Street? What is the minimum number?
(D) If traffic lights are adjusted so that 1,000 vehicles per hour travel from Washington Avenue to Lincoln Avenue on 5th Street, determine the flow around the rest of the network.
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7-6
Systems of Nonlinear Equations
1
Systems of Nonlinear Equations Z Solving by Substitution Z Solving by Other Methods
If a system of equations contains any equations that are not linear, then the system is called a nonlinear system. In this section we will investigate nonlinear systems involving seconddegree terms, like these: x2 0004 y2 0003 5 3x 0004 y 0003 1
x2 0002 2y2 0003 2 xy 0003 2
x2 0004 3xy 0004 y2 0003 20 xy 0002 y2 0003 0
It can be shown that such systems have at most four solutions, some of which may be imaginary. Since we are interested in finding both real and imaginary solutions to the systems we consider, we will assume that the replacement set for each variable is the set of complex numbers, rather than the set of real numbers.
Z Solving by Substitution The substitution method we used to solve linear systems of two equations in two variables is also an effective method for solving nonlinear systems. This process is best illustrated by examples.
EXAMPLE
1
Solving a Nonlinear System by Substitution Solve the system:
SOLUTION
x2 0004 y2 0003 5 3x 0004 y 0003 1
As usual, can start with either equation. But since the y term in the second equation is a first-degree term with coefficient 1, our calculations will be simplified if we start by solving for y in terms of x in the second equation. Next we substitute for y in the first equation to obtain an equation that involves only x: Add 00023x to both sides. Substitute for y in the second equation.
f
3x 0004 y 0003 1 y 0003 1 0002 3x T
x2 0004 y2 0003 5 2 x 0004 (1 0002 3x)2 0003 5 10x2 0002 6x 0002 4 0003 0 5x2 0002 3x 0002 2 0003 0 (x 0002 1)(5x 0004 2) 0003 0 x 0003 1,
Multiply parentheses and collect like terms on the left side. Divide both sides by 2. Factor. Use the zero property.
000225
If we substitute these values back into the equation y 0003 1 0002 3x, we obtain two solutions to the system: x00031 y 0003 1 0002 3(1) 0003 00022
x 0003 000225 y 0003 1 0002 3(000225) 0003 115
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y
A check, which you should provide, verifies that (1, 00022) and (000225, 115) are both solutions to the system. These solutions are illustrated in Figure 1. However, if we substitute the values of x back into the equation x2 0004 y2 0003 5, we obtain
5
x2 0004 y2 0003 5 00025
5
x00031 2 1 0004 y2 0003 5 y2 0003 4 y 0003 00052
x
00025
x 0003 000225 (000225)2 0004 y2 0003 5 y2 0003 121 25 y 0003 0005115
It appears that we have found two additional solutions, (1, 2) and (000225, 0002115). But neither of these solutions satisfies the equation 3x 0004 y 0003 1, which you should verify. So, neither is a solution of the original system. We have produced two extraneous roots, apparent solutions that do not actually satisfy both equations in the system. This is a common occurrence when solving nonlinear systems.
Z Figure 1
It is always very important to check the solutions of any nonlinear system to ensure that extraneous roots have not been introduced.
MATCHED PROBLEM 1
ZZZ EXPLORE-DISCUSS 1
Solve the system:
x2 0004 y2 0003 10 2x 0004 y 0003 1 0002
In Example 1, we saw that the line 3x 0004 y 0003 1 intersects the circle x2 0004 y2 0003 5 in two points. (A) Consider the system x2 0004 y2 0003 5 3x 0004 y 0003 10 Graph both equations in the same coordinate system. Are there any real solutions to this system? Are there any complex solutions? Find any real or complex solutions. (B) Consider the family of lines given by 3x 0004 y 0003 b
b any real number
What do all these lines have in common? Illustrate graphically the lines in this family that intersect the circle x2 0004 y2 0003 5 in exactly one point. How many such lines are there? What are the corresponding value(s) of b? What are the intersection points? How are these lines related to the circle?
EXAMPLE
2
Solving a Nonlinear System by Substitution Solve:
x2 0002 2y2 0003 2 xy 0003 2
0002
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SECTION 7–6
SOLUTION
Systems of Nonlinear Equations
3
Solve the second equation for y, substitute in the first equation, and proceed as before. xy 0003 2
Divide both sides by x.
2 y0003 x
Substitute for y in the first equation.
2 2 x2 0002 2 a b 0003 2 x
Simplify the left side.
8 00032 x2 x4 0002 2x2 0002 8 0003 0 u2 0002 2u 0002 8 0003 0 (u 0002 4)(u 0004 2) 0003 0 u 0003 4, 00022 x2 0002
Multiply both sides by x2 and simplify. Substitute u 0003 x2 (see Section 1-6). Factor. Use the zero property.
Replacing u with x2, we get x2 0003 4 x 0003 00052
y 5
xy 0003 00022
or
x2 0003 00022 x 0003 0005 100022 0003 0005i12
2 0003 1. 2 2 For x 0003 00022, y 0003 0003 00021. 00022 For x 0003 2, y 0003
5
x
x 2 0002 2y 2 0003 2
Z Figure 2
MATCHED PROBLEM 2
ZZZ EXPLORE-DISCUSS 2
2 0003 0002i 12. i12 2 For x 0003 0002i12, y 0003 0003 i12. 0002i 12 For x 0003 i 12, y 0003
There are four solutions to this system: (2, 1), (00022, 00021), (i 12, 0002i 12), and (0002i12, i12). Notice that two of the solutions involve imaginary numbers. These imaginary solutions cannot be illustrated graphically (see Fig. 2); however, they do satisfy both equations in the system (verify this). 0002 Solve:
3x2 0002 y2 0003 6 xy 0003 3
0002
(A) Refer to the system in Example 2. Could a graphing calculator be used to find the real solutions of this system? The imaginary solutions? (B) In general, explain why graphic approximation techniques can be used to approximate the real solutions of a system, but not the complex solutions.
EXAMPLE
3
Design An engineer is hired to design a rectangular computer screen with a 19-inch diagonal and a 175-square-inch area. Find the dimensions of the screen to the nearest tenth of an inch.
SOLUTION
Sketch a rectangle letting x be the width and y the height (Fig. 3). We obtain the following system using the Pythagorean theorem and the formula for the area of a rectangle: x2 0004 y2 0003 192 xy 0003 175
Pythagorean theorem width 0004 height 0003 Area
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This system is solved using the procedures outlined in Example 2. However, in this case, we are only interested in real solutions. We start by solving the second equation for y in terms of x and substituting the result into the first equation.
y
y0003 x
175 x
1752 0003 192 x2 x4 0004 30,625 0003 361x2 4 x 0002 361x2 0004 30,625 0003 0 x2 0004
Z Figure 3
Multiply both sides by x2 and simplify. Subtract 361x2 from each side. Quadratic in x2
Solve the last equation for x2 using the quadratic formula, then solve for x: x0003
361 0005 23612 0002 4(1)(30,625) B 2
⬇ 15.0 inches or 11.7 inches Substitute each choice of x into y 0003 175兾x to find the corresponding y values: For x 0003 15.0 inches, 175 y0003 0003 11.7 inches 15
For x 0003 11.7 inches, 175 0003 15.0 inches y0003 11.7
In either case, the dimensions are 15.0 by 11.7 inches.
0002
An engineer is hired to design a rectangular television screen with a 21-inch diagonal and a 209-square-inch area. Find the dimensions of the screen to the nearest tenth of an inch. 0002
MATCHED PROBLEM 3
Technology Connections the upper half of the circle and another for the lower half. [Note: since x and y must be nonnegative real numbers, we ignore the intersection points in the third quadrant—see Fig 4(a).]
In Example 3, we’re only concerned with real solutions, so graphic techniques can also be used to approximate the solutions (see Fig. 4). As we saw in Section 2-2, graphing a circle on a graphing calculator requires two functions, one for 40
000260
16
60
000240
(a) y1 0003 2361 0002 x2 y2 0003 0002 2361 0002 x2 175 y3 0003 x
8
16
18
10
(b) Intersection point: (11.7, 15.0)
2 2 2 Z Figure 4 Graphic solution of x 0004 y 0003 19 , xy 0003 175.
8
18
10
(c) Intersection point: (15.0, 11.7)
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Systems of Nonlinear Equations
5
Z Solving by Other Methods We’ll conclude the section by looking at some other techniques for solving nonlinear systems of equations.
EXAMPLE
4
Solving a Nonlinear System by Elimination Solve:
SOLUTION
x2 0002 y2 0003 5 x2 0004 2y2 0003 17
Because both x and y have the same power in each equation, elimination by addition is a good choice here. (We could also use substitution.) Multiply the second equation by 00021 and add: x2 0002 y2 0003 5 2 2 0002x 0002 2y 0003 000217 00023y2 0003 000212 y2 0003 4 y 0003 00052 Now substitute y 0003 2 and y 0003 00022 back into either original equation to find x. For y 0003 2, 2
For y 0003 00022, x2 0002 (00022)2 0003 5 x2 0003 9
2
x 0002 (2) 0003 5 x2 0003 9 x 0003 00053
x 0003 00053
There are four solutions to the system: (3, 00022), (3, 2), (00023, 00022), and (00023, 2). The check of the solutions is left to you. 0002 MATCHED PROBLEM 4
Solve: 2x2 0002 3y2 0003 5 3x2 0004 4y2 0003 16 0002
EXAMPLE
5
Solving a Nonlinear System Using Factoring and Substitution Solve:
SOLUTION
x2 0004 3xy 0004 y2 0003 20 xy 0002 y2 0003 0
We begin by solving the second equation for y: xy 0002 y2 0003 0 y(x 0002 y) 0003 0 y00030 or
Factor the left side. Use the zero property.
y0003x
Now, the original system is equivalent to the two systems: y0003 0 x 0004 3xy 0004 y2 0003 20
or
2
These systems can be solved by substitution.
y0003 x x 0004 3xy 0004 y2 0003 20 2
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y0003 0 x 0004 3xy 0004 y2 0003 20 x2 0004 3x(0) 0004 (0)2 0003 20
FIRST SYSTEM
2
Substitute y 0003 0 in the second equation, and solve for x.
x2 0003 20 x 0003 0005 120 0003 0005215 (2 15, 0) (00022 15, 0) y0003 x x2 0004 3xy 0004 y2 0003 20 x2 0004 3xx 0004 x2 0003 20 5x2 0003 20 x2 0003 4 x 0003 00052 (2, 2) (00022, 00022)
Take the square root of each side.
Solutions to the first system
SECOND SYSTEM
Substitute y 0003 x in the second equation and solve for x.
Divide both sides by 5. Take the square root of each side. Substitute these values back into y 0003 x to find y. Solutions to the second system.
Combining the solutions for the first system with the solutions for the second system, the solutions for the original system are (2 15, 0), (0002215, 0), (2, 2), and (00022, 00022). The check of the solutions is left to you. 0002 MATCHED PROBLEM 5
Solve:
x2 0002 xy 0004 y2 0003 9 2x2 0002 xy 0003 0
0002
Example 5 is somewhat specialized. However, it suggests a procedure that is effective for some problems.
EXAMPLE
6
Graphic Approximations of Real Solutions Use a graphing calculator to approximate real solutions to two decimal places: x2 0002 4xy 0004 y2 0003 12 2x2 0004 2xy 0004 y2 0003 6
SOLUTION
Before we can enter these equations in our calculator, we must solve for y: x2 0002 4xy 0004 y2 0003 12 y2 0002 4xy 0004 (x2 0002 12) 0003 0
a 0003 1, b 0003 00024x, c 0003 x2 0002 12
2x2 0004 2xy 0004 y2 0003 6 y2 0004 2xy 0004 (2x2 0002 6) 0003 0
a 0003 1, b 0003 2x, c 0003 2x2 0002 6
Applying the quadratic formula to each equation, we have y0003 0003
4x 0005 216x2 0002 4(x2 0002 12) 2 4x 0005 212x2 0004 48 2
0003 2x 0005 23x2 0004 12
y0003 0003
00022x 0005 24x2 0002 4(2x2 0002 6) 2 00022x 0005 224 0002 4x2 2
0003 0002x 0005 26 0002 x2
Since each equation has two solutions, we must enter four functions in the graphing calculator, as shown in Figure 5(a). Examining the graph in Figure 5(b), we see that there are four intersection points. Using the INTERSECT command repeatedly (details omitted), we find that the solutions to two decimal places are (00022.10, 0.83), (00020.37, 2.79), (0.37, 00022.79), and (2.10, 00020.83).
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7
Systems of Nonlinear Equations
5
00027.6
7.6
00025
(a)
(b)
Z Figure 5
MATCHED PROBLEM 6
0002
Use a graphing calculator to approximate real solutions to two decimal places: x2 0004 8xy 0004 y2 0003 70 2x2 0002 2xy 0004 y2 0003 20 0002
ANSWERS TO MATCHED PROBLEMS 1. (00021, 3), (95, 0002135) 2. (13, 13), (000213, 0002 13) (i, 00023i), (0002i, 3i) 3. 17.1 by 12.2 in. 4. (2, 1), (2, 00021), (00022, 1), (00022, 00021) 5. (0, 3), (0, 00023), ( 13, 213), (000213, 0002213) 6. (00023.89, 00021.68), (00020.96, 00025.32), (0.96, 5.32), (3.89, 1.68)
7-6
Exercises
*Additional answers can be found in the Instructor Answer Appendix.
A
1. How can you tell when a system of equations is nonlinear? 2. What is an extraneous solution? Why are they important in the study of solving nonlinear systems? 3. Would you choose substitution or elimination to solve the following nonlinear system? ax 0004 by 0003 c a, b, c, d, e, and f 0006 0 dx2 0004 ey2 0003 f Justify your answer by describing the steps you would take to solve this system? 4. Repeat Problem 3 for the following nonlinear system. ax2 0004 by2 0003 c a, b, c, d, e, and f 0006 0 dx2 0004 ey2 0003 f
Solve each system in Problems 5–16. 2
2
2
2
5. x 0004 y 0003 169 x 0003 000212
6. x 0004 y 0003 25 y 0003 00024
7. 8x2 0002 y2 0003 16 y 0003 2x
8. y2 0003 2x x 0003 y 0002 12
(000212, 5), (000212, 00025)
(2, 4), (00022, 00024)
(3, 00024), (00023, 00024)
(12 , 1)
9. 3x2 0002 2y2 0003 25 x0004y00030 (5, 00025), (00025, 5)
10. x2 0004 4y2 0003 32 x 0004 2y 0003 0 (00024, 2), (4, 00022)
y2 0003 x x 0002 2y 0003 2
12. x2 0003 2y 3x 0003 y 0004 2
(3 0004 15, 7 0004 315), (3 0002 15, 7 0002 315)
13. 2x2 0004 y2 0003 24 x2 0002 y2 0003 000212
14. x2 0002 y2 0003 3 x2 0004 y2 0003 5
(2, 1), (2, 00021), (00022, 00021), (00022, 1)
11.
15.
x2 0004 y2 0003 10 16x2 0004 y2 0003 25
(1, 3), (1, 00023), (00021, 3), (00021, 00023)
16. x2 0002 2y2 0003 1 x2 0004 4y2 0003 25 (3, 2), (3, 00022), (00023, 2), (00023, 00022)
B Solve each system in Problems 17–28. 17. xy 0002 4 0003 0 x0002y00032
18. xy 0002 6 0003 0 x0002y00034
19. x2 0004 2y2 0003 6 xy 0003 2
20. 2x2 0004 y2 0003 18 xy 0003 4
21. 2x2 0004 3y2 0003 00024 4x2 0004 2y2 0003 8
22. 2x2 0002 3y2 0003 10 x2 0004 4y2 0003 000217
23. x2 0002 y2 0003 2 y2 0003 x
24. x2 0004 y2 0003 20 x2 0003 y
25. x2 0004 y2 0003 9 x2 0003 9 0002 2y
26. x2 0004 y2 0003 16 y2 0003 4 0002 x
(2, 4), (00022, 4), (15i, 00025), (000215i, 00025)
(4, 0), (00023, 17), (00023, 000217)
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27. x2 0002 y2 0003 3 xy 0003 2
28. y2 0003 5x2 0004 1 xy 0003 2
(i, 00022i ), (0002i, 2i ), 00022 15 (2 15 5 , 15), ( 5 , 000215)
(2, 1), (00022, 00021), (i, 00022i ), (0002i, 2i )
An important type of calculus problem is to find the area between the graphs of two functions. To solve some of these problems it is necessary to find the coordinates of the points of intersections of the two graphs. In Problems 29–36, find the coordinates of the points of intersections of the two given equations. 29. y 0003 5 0002 x2, y 0003 2 0002 2x
30. y 0003 5x 0002 x2, y 0003 x 0004 3
31. y 0003 x2 0002 x, y 0003 2x
32. y 0003 x2 0004 2x, y 0003 3x
(00021, 4), (3, 00024)
(1, 4), (3, 6)
(0, 0), (3, 6)
(1, 3), (0, 0) (1, 4), (4, 1)
2
33. y 0003 x 0002 6x 0004 9, y 0003 5 0002 x 34. y 0003 x2 0004 2x 0004 3, y 0003 2x 0004 4 35. y 0003 8 0004 4x 0002 x2, y 0003 x2 0002 2x
C
Solve each system in Problems 39–46.
2
2
2
(1
0002 15), 12 (3 0004 15)
0004 15 00021 0004 15 , ), (1 00022 15 , 00021 00022 15 ) 2 2
57. DESIGN An engineer is designing a portable DVD player. According to the design specifications, the set must have a rectangular screen with a 7.5-inch diagonal and an area of 27 square inches. Find the dimensions of the screen. 6 by 4.5 in. 58. DESIGN An artist is designing a logo for a business in the shape of a circle with an inscribed rectangle. The diameter of the circle is 6.5 inches, and the area of the rectangle is 15 square inches. Find the dimensions of the rectangle. 2x 0003 6 in. by 2y 0003 2.5 in.
6.5 inches
59. CONSTRUCTION A rectangular swimming pool with a deck 5 feet wide is enclosed by a fence as shown in the figure. The surface area of the pool is 572 square feet, and the total area enclosed by the fence (including the pool and the deck) is 1,152 square feet. Find the dimensions of the pool. 22 by 26 ft
Fence
(3, 3 ), (2 , 2) 2
42. x 0004 xy 0002 y 0003 00025 y0002x00033
43. 2x2 0002 xy 0004 y2 0003 8 x2 0002 y2 0003 0
44. x2 0004 2xy 0004 y2 0003 36 x2 0002 xy 0003 0
45. x2 0004 xy 0002 3y2 0003 3 x2 0004 4xy 0004 3y2 0003 0
46. x2 0002 2xy 0004 2y2 0003 16 x2 0002 y2 0003 0
(00023, 1), (3, 00021), (0002i, i ), (i, 0002i )
1 2 (3
54. NUMBERS Find two numbers that differ by 1 and have product 1. (Let x be the larger number and y the smaller number.)
40. 2x 0004 3y 0004 xy 0003 16 xy 0002 55 00035 0
41. x 0002 2xy 0004 y 0003 1 x 0002 2y 0003 2 (0, 00021), (00024, 00023)
53. NUMBERS Find two numbers with sum 3 and product 1.
2 meters by 16 meters (4, 000210), (00023, 11)
38. Consider the circle with equation x2 0004 y2 0003 25 and the family of lines given by 3x 0004 4y 0003 b, where b is any real number. (A) Illustrate graphically the lines in this family that intersect the circle in exactly one point, and describe the relationship between the circle and these lines. (B) Find the values of b corresponding to the lines in part A, and find the intersection points of the lines and the circle. (C) How is the line with equation 4x 0002 3y 0003 0 related to this family of lines? How could this line be used to find the intersection points and the values of b in part B?
(00025, 00025 ), (00022 , 00022)
APPLICATIONS
5 in. and 12 in.
37. Consider the circle with equation x2 0004 y2 0003 5 and the family of lines given by 2x 0002 y 0003 b, where b is any real number. (A) Illustrate graphically the lines in this family that intersect the circle in exactly one point, and describe the relationship between the circle and these lines. (B) Find the values of b corresponding to the lines in part A, and find the intersection points of the lines and the circle. (C) How is the line with equation x 0004 2y 0003 0 related to this family of lines? How could this line be used to find the intersection points in part B?
39. 2x 0004 5y 0004 7xy 0003 8 xy 30002 3 0003 0 3
(00022.09, 00020.66), (0.18, 00023.64), (0.58, 2.83), (1.73, 1.28)
56. GEOMETRY Find the dimensions of a rectangle with an area of 32 square meters if its perimeter is 36 meters long.
(00021, 3), (4, 8)
36. y 0003 x 0002 4x 0002 10, y 0003 14 0002 2x 0002 x2
(00022.96, 00023.47), (00020.89, 00023.76), (1.39, 4.05), (2.46, 4.18)
52. 2x2 0004 2xy 0004 y2 0003 12 4x2 0002 4xy 0004 y2 0004 x 0004 2y 0003 9
55. GEOMETRY A right triangle with an area of 30 square inches has a hypotenuse that is 13 inches long. Find the lengths of the two legs.
(1, 6), (00021, 2)
2
51. 2x2 0002 2xy 0004 y2 0003 9 4x2 0002 4xy 0004 y2 0004 x 0003 3
5 ft 5 ft
(4, 7), (00021, 2)
Pool
(0, 6), (0, 00026), (3, 3), (00023, 00023)
00024 15 00024 15 4 15 (4, 4), (00024, 00024), (4 15 5 , 5 ), ( 5 , 5 )
In Problems 47–52, use a graphing calculator to approximate the real solutions of each system to two decimal places. 47. 0002x2 0004 2xy 0004 y2 0003 1 3x2 0002 4xy 0004 y2 0003 2
48. 0002x2 0004 4xy 0004 y2 0003 2 8x2 0002 2xy 0004 y2 0003 9
49. 3x2 0002 4xy 0002 y2 0003 2 2x2 0004 2xy 0004 y2 0003 9
50. 5x2 0004 4xy 0004 y2 0003 4 4x2 0002 2xy 0004 y2 0003 16
5 ft
5 ft
60. CONSTRUCTION An open-topped rectangular box is formed by cutting a 6-inch square from each corner of a rectangular piece of cardboard and bending up the ends and sides. The area of the cardboard before the corners are removed is 768 square inches, and the volume of the box is 1,440 cubic inches. Find the dimensions of the original piece of cardboard. 24 in. by 32 in.
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6 in.
6 in.
6 in.
6 in. 6 in.
62. TRANSPORTATION Bus A leaves Milwaukee at noon and travels west on Interstate 94. Bus B leaves Milwaukee 30 minutes later, travels the same route, and overtakes bus A at a point 210 miles west of Milwaukee. If the average speed of bus B is 10 miles per hour greater than the average speed of bus A, at what time did bus B overtake bus A? 3:30 P.M.
6 in.
61. TRANSPORTATION Two boats leave Bournemouth, England, at the same time and follow the same route on the 75-mile trip across
7-7
9
the English Channel to Cherbourg, France. The average speed of boat A is 5 miles per hour greater than the average speed of boat B. Consequently, boat A arrives at Cherbourg 30 minutes before boat B. Find the average speed of each boat. Boat A: 30 mph; boat B: 25 mph
6 in.
6 in.
Systems of Linear Inequalities in Two Variables
Systems of Linear Inequalities in Two Variables Z Graphing Linear Inequalities in Two Variables Z Solving Systems of Linear Inequalities Graphically Z Application
Earlier in the book, we saw that in many applications, it’s more natural to consider inequalities than equations. The same is true for systems, so here and in Sections 7-8 we will study systems of linear inequalities. A graph is often the most convenient way to represent the solutions of a system of inequalities in two variables. In this section, we discuss techniques for graphing both a single linear inequality in two variables and a system of linear inequalities in two variables.
Z Graphing Linear Inequalities in Two Variables We know how to graph first-degree equations such as y 0003 2x 0002 3
and
2x 0002 3y 0003 5
but how do we graph first-degree inequalities, like the ones below? y 0007 2x 0002 3
and
2x 0002 3y 5
Actually, graphing these inequalities is almost as easy as graphing the equations. But before we begin, we should discuss some important subsets of a plane in a rectangular coordinate system. A line divides a plane into two halves called half-planes. A vertical line divides a plane into left and right half-planes [Fig. 1(a)]; a nonvertical line divides a plane into upper and lower half-planes [Fig. 1(b)]. y
Z Figure 1 Half-planes.
Left half-plane
y
Right half-plane
Upper half-plane x
x Lower half-plane
(a)
(b)
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Consider the following linear equation and related linear inequalities:
ZZZ EXPLORE-DISCUSS 1
(1) 2x 0002 3y 0003 12
(2) 2x 0002 3y 12
(3) 2x 0002 3y 12
(A) Graph the line with equation (1). (B) Find the point on this line with x coordinate 3 and draw a vertical line through this point. Discuss the relationship between the y coordinates of the points on this line and statements (1), (2), and (3). (C) Repeat part B for x 0003 00023. For x 0003 9. (D) Based on your observations in parts B and C, write a verbal description of all the points in the plane that satisfy equation (1), those that satisfy inequality (2), and those that satisfy inequality (3).
Now let’s investigate the half-planes determined by the linear equation y 0003 2x 0002 3. We start by graphing y 0003 2x 0002 3 (Fig. 2). For any given value of x, there is exactly one value for y such that (x, y) lies on the line. For the same x, if the point (x, y) is below the line, then y 2x 0002 3. So the lower half-plane corresponds to the solution of the inequality y 2x 0002 3. Similarly, the upper half-plane corresponds to the solution of the inequality y 2x 0002 3, as shown in Figure 2. y
Z Figure 2
y 0003 2x 0002 3 (4, y) y 2(4) 0002 3 0003 5; point in upper half-plane (4, y) y 0003 2(4) 0002 3 0003 5; point on line
5
(4, y) y 2(4) 0002 3 0003 5; point in lower half-plane 00025
5
10
x
00025
We can form four different inequalities from y 0003 2x 0002 3 by replacing the 0003 sign with , , 0007, and . They are y 2x 0002 3
y 2x 0002 3
y 0007 2x 0002 3
y 2x 0002 3
The graph of each is a half-plane. The line y 0003 2x 0002 3, called the boundary line for the half-plane, is included for and 0007 and excluded for and . In Figure 3, the half-planes are indicated with small arrows on the graph of y 0003 2x 0002 3 and then graphed as shaded regions. Included boundary lines are shown as solid lines, and excluded boundary lines are shown as dashed lines. Z Figure 3
y
0
y 2x 0002 3 (a)
y
x
0
y 2x 0002 3 (b)
y
x
0
y 0007 2x 0002 3 (c)
y
x
0
y 2x 0002 3 (d)
x
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11
Z THEOREM 1 Graphs of Linear Inequalities in Two Variables The graph of a linear inequality Ax 0004 By C
or
Ax 0004 By C
with B 0006 0, is either the upper half-plane or the lower half-plane (but not both) determined by the line Ax 0004 By 0003 C. If B 0003 0, then the graph of Ax C
or
Ax C
is either the left half-plane or the right half-plane (but not both) determined by the line Ax 0003 C.
As a consequence of Theorem 1, we can build a simple and fast mechanical procedure for graphing linear inequalities.
Z PROCEDURE FOR GRAPHING LINEAR INEQUALITIES IN TWO VARIABLES Step 1. Graph Ax 0004 By 0003 C as a dashed line if equality is not included in the original statement or as a solid line if equality is included. Step 2. Choose any point not on the line and substitute the coordinates into the inequality. [It’s usually easiest to use the origin (0,0) if you’re sure that it is not on the line.] Step 3. If the inequality is true when substituting in the test point, the graph of the original inequality is the half-plane containing that point. If the inequality is false, the graph of the original inequality is the half-plane not containing that point.
EXAMPLE
1
Graphing a Linear Inequality Graph: 3x 0002 4y 0007 12
SOLUTION y 5
3x 0002 4y 0003 12
Step 1. Graph 3x 0002 4y 0003 12 as a solid line, since equality is included in the original statement (Fig. 4). Step 2. Pick a convenient test point above or below the line. The point (0, 0) will be easy to test. Substituting (0, 0) into the inequality 3x 0002 4y 0007 12
5
00025
Z Figure 4
x
3(0) 0002 4(0) 0003 0 0007 12 produces a true statement; therefore, (0, 0) is in the solution set. Step 3. The line 3x 0002 4y 0003 12 and the half-plane containing the origin form the graph of 3x 0002 4y 0007 12 (Fig. 5).
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Test point x
5
00025
0002
Z Figure 5
MATCHED PROBLEM 1
EXAMPLE
2
Graph: 2x 0004 3y 6
0002
Graphing a Linear Inequality Graph: (A) y 00023
SOLUTIONS
(B) 2x 0007 5
(A) The solution to y 00023 is the set of all points with y coordinates greater than 00023. Its graph is shown in Figure 6. (B) The inequality 2x 0007 5 is equivalent to x 0007 5/2. This is the set of all points with x coordinates less than or equal to 5/2, as shown in Figure 7. y
y
5
5
Test point 00025
5
Test point
x
00025
00025
Graph: (A) y 0007 2
x
00025
Z Figure 6
MATCHED PROBLEM 2
5
0002
Z Figure 7
(B) 3x 00028
0002
Z Solving Systems of Linear Inequalities Graphically We will now consider systems of linear inequalities such as x0004y 6 2x 0002 y 0
and
2x 0004 y 0007 22 x 0004 y 0007 13 2x 0004 5y 0007 50 x 0 y 0
We will solve such systems graphically—that is, to find the graph of all ordered pairs of real numbers (x, y) that simultaneously satisfy all the inequalities in the system. The graph is called the solution region for the system. To find the solution region, we graph each
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13
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inequality in the system and then find the intersection of all the graphs. To simplify the discussion that follows: We will consider only systems of linear inequalities where equality is included in each statement in the system.
EXAMPLE
3
Solving a System of Linear Inequalities Graphically Solve the following system of linear inequalities graphically: x0004y 6 2x 0002 y 0
SOLUTION
First, graph the line x 0004 y 0003 6 and shade the region that satisfies the inequality x 0004 y 6. This region is shaded in blue in Figure 8(a) on the next page. Next, graph the line 2x 0002 y 0003 0 and shade the region that satisfies the inequality 2x 0002 y 0. This region is shaded in red in Figure 8(a). The solution region for the system of inequalities is the intersection of these two regions. This is the region shaded in both red and blue in Figure 8(a), which is redrawn in Figure 8(b) with only the solution region shaded for clarity. The coordinates of any point in the shaded region of Figure 8(b) specify a solution to the system. For example, the points (2, 4), (6, 3), and (7.43, 8.56) are three of infinitely many solutions, as can be easily checked. The intersection point (2, 4) can be obtained by solving the system of equations x 0004 y 0003 6 and 2x 0002 y 0003 0. y
2x 0002 y 0003 0
y
10
2x 0002 y 0003 0
10
Solution region 5
5
00025
5
10
(2, 4)
x
00025
x0004y00036
00025
5
00025
(a)
10
x
x0004y00036 (b)
0002
Z Figure 8
MATCHED PROBLEM 3
ZZZ EXPLORE-DISCUSS 2
Solve the following system of linear inequalities graphically:
3x 0004 y 0007 21 x 0002 2y 0007 0
0002
Refer to Example 3. Graph each boundary line and shade the regions obtained by reversing each inequality. That is, shade the region of the plane that corresponds to the inequality x 0004 y 6 and then shade the region that corresponds to the inequality 2x 0002 y 0. What portion of the plane is left unshaded? Compare this method with the one used in the solution to Example 3.
The points of intersection of the lines that form the boundary of a solution region play a fundamental role in the solution of linear programming problems, which are discussed in Section 7-8.
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Z DEFINITION 1 Corner Point A corner point of a solution region is a point in the solution region that is the intersection of two boundary lines.
The point (2, 4) is the only corner point of the solution region in Example 3; see Figure 8(b).
EXAMPLE
4
Solving a System of Linear Inequalities Graphically Solve the following system of linear inequalities graphically, and find the corner points. 2x 0004 y 0007 22 x 0004 y 0007 13 2x 0004 5y 0007 50 x 0 y 0
SOLUTION
The inequalities x 0 and y 0, called nonnegative restrictions, occur frequently in applications involving systems of inequalities since x and y often represent quantities that can’t be negative—number of units produced, number of hours worked, and the like. Because both x and y are nonnegative, the solution region lies in the first quadrant, and we can restrict our attention to that portion of the plane. First, we graph the lines 2x 0004 y 0003 22 x 0004 y 0003 13 2x 0004 5y 0003 50
Find the x and y intercepts of each line; then sketch the line through these points, as shown in Figure 9.
Next, choosing (0, 0) as a test point, we see that the graph of each of the first three inequalities in the system consists of its corresponding line and the half-plane lying below the line, as indicated by the arrows in Figure 9. So the solution region of the system consists of the points in the first quadrant that simultaneously lie on or below all three of these lines—see Figure 9. y
2x 0004 y 0003 22
x 0004 y 0003 13
20
2x 0004 5y 0003 50 (5, 8) (9, 4)
(0, 10)
00025
(0, 0)
Z Figure 9
(11, 0) 00025
20
x
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15
The corner points (0, 0), (0, 10), and (11, 0) can be found by just looking at the graph. The other two corner points are determined as follows: Solve the system
Solve the system
2x 0004 5y 0003 50 x 0004 y 0003 13 to obtain (5, 8).
2x 0004 y 0003 22 x 0004 y 0003 13 to obtain (9, 4).
Note that the lines 2x 0004 5y 0003 50 and 2x 0004 y 0003 22 also intersect, but the intersection point is not part of the solution region, and is not a corner point. 0002 MATCHED PROBLEM 4
Solve the following system of linear inequalities graphically, and find the corner points: 5x 0004 y 20 x 0004 y 12 x 0004 3y 18 x 0 y 0
0002
If we compare the solution regions of Examples 3 and 4, we see that there is a fundamental difference between these two regions. We could draw a circle that encloses the entire solution region in Example 4. However, it is impossible to include all the points in the solution region in Example 3 in any circle, no matter how large we draw it. This leads to the following definition.
Z DEFINITION 2 Bounded and Unbounded Solution Regions A solution region of a system of linear inequalities is bounded if it can be enclosed within a circle. If it cannot be enclosed within a circle, then it is unbounded.
So the solution region for Example 4 is bounded and the solution region for Example 3 is unbounded. This definition will be important in Section 7-8.
Z Application EXAMPLE
5
Production Scheduling A manufacturer of surfboards makes a standard model and a competition model. Each standard board requires 6 labor-hours for fabricating and 1 labor-hour for finishing. Each competition board requires 8 labor-hours for fabricating and 3 labor-hours for finishing. The maximum labor-hours available per week in the fabricating and finishing departments are 120 and 30, respectively. What combinations of boards can be produced each week so as not to exceed the number of labor-hours available in each department per week?
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To organize all of the information provided, a table is helpful.
Standard Model (labor-hours per board)
Competition Model (labor-hours per board)
Maximum Labor-Hours Available per Week
Fabricating
6
8
120
Finishing
1
3
30
Let x ⫽ Number of standard boards produced per week y ⫽ Number of competition boards produced per week These variables are restricted as follows: Fabricating department restriction: a
Weekly fabricating time Weekly fabricating time Maximum labor-hours b0002a bⱕa b available per week for x standard boards for y competition boards
6x
⫹
8y
ⱕ
120
Finishing department restriction: a
Weekly finishing time Weekly finishing time Maximum labor-hours b0002a bⱕa b for x standard boards for y competition boards available per week
1x
⫹
3y
ⱕ
30
Since it is not possible to manufacture a negative number of boards, x and y also must satisfy the nonnegative restrictions xⱖ0 yⱖ0
© Corbis RF
The production levels x and y must satisfy the following system of linear inequalities: 6x ⫹ 8y ⱕ 120 x ⫹ 3y ⱕ 30 xⱖ0 yⱖ0
Fabricating department restriction Finishing department restriction Nonnegative restriction Nonnegative restriction
Graphing this system of linear inequalities, we obtain the set of feasible solutions, or the feasible region, as shown in Figure 10. For problems of this type and for the linear programming problems we consider in the next section, solution regions are often referred to as feasible regions. Any point within the shaded area, including the boundary lines, represents a possible production schedule. Any point outside the shaded area represents an impossible schedule. For example, it would be possible to produce 12 standard boards and 5 competition boards per week, but it would not be possible to produce 12 standard boards and 7 competition boards per week (see the figure).
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17
y
20
Remark In Example 5, how do we interpret a production schedule of 10.5 standard boards and 4.3 competition boards? It is not possible to manufacture a fraction of a board. But it is possible to average 10.5 standard and 4.3 competition boards per week. In general, we will assume that all points in the feasible region represent acceptable solutions, even though noninteger solutions might require special interpretation.
MATCHED PROBLEM 5
Fabricating capacity line 6x 0004 8y 0003 120 (12, 7) (12, 6) (12, 5) Feasible region 00025
10
x
20
Finishing capacity line x 0004 3y 0003 30
00025
0002
Z Figure 10
Repeat Example 5 using 5 hours for fabricating a standard board and a maximum of 27 labor-hours for the finishing department. 0002 ANSWERS TO MATCHED PROBLEMS y
1. 5
00025
5
x
00025
y
2. (A)
y
(B)
5
5
00025
5
x
00025
00025
5
00025
y
3.
20
Solution region 10
x 0002 2y 0003 0
(6, 3) 000210
10 00025
x
3x 0004 y 0003 21
x
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4.
(0, 20) Solution region (2, 10)
10
(9, 3) (18, 0) 00025
x
10 00025
5x 0004 y 0003 20
x 0004 3y 0003 18
x 0004 y 0003 12
y
5. 20
, 15 冣 冢 144 7 7 00025
10
x
20
x 0004 3y 0003 27
00025
5x 0004 8y 0003 120
7-7
Exercises
*Additional answers can be found in the Instructor Answer Appendix.
A
1. If m 0006 0, describe geometrically the graphs of (A) y 0003 mx 0004 b (B) y mx 0004 b (C) y mx 0004 b
In Problems 15–18, match the solution region of each system of linear inequalities with one of the four regions shown in the figure.
2. If m 0006 0, describe geometrically the graphs of (A) y 0007 mx 0004 b (B) y mx 0004 b 3. What are nonnegativity restrictions? How do they affect the graph of a system of linear inequalities?
y II x 0004 2y 0003 8
4. In an applied problem, if x is the number of tables produced and y is the number of chairs produced, how would you interpret the solution x 0003 12.6 and y 0003 23.2?
6. 3x 0004 4y 12
7. 3x 0004 2y 18
8. 3y 0002 2x 24
9. y 0007
2 3x
00045
10. y 13x 0002 2
11. y 8
12. x 00025
13. 00023 0007 y 2
14. 00021 x 0007 3
I
5
(2, 3)
III
In Problems 5–14, graph the solution region of each inequality and write a verbal description of the solution region. 5. 2x 0002 3y 6
3x 0002 2y 0003 0
00025
5
IV 00025
15. x 0004 2y 0007 8 3x 0002 2y 0
16. x 0004 2y 8 3x 0002 2y 0007 0
17. x 0004 2y 8 3x 0002 2y 0
18. x 0004 2y 0007 8 3x 0002 2y 0007 0
Region IV
Region I
Region II
Region III
x
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In Problems 19–24, solve each system of linear inequalities graphically. 19. x 5 y00076
20. x 0007 4 y 2
21. 3x 0004 y 6 x00074
22. 3x 0004 4y 0007 12 y 00023
23. x 0002 2y 0007 12 2x 0004 y 4
24. 2x 0004 5y 0007 20 x 0002 5y 0007 00025
B
In Problems 25–28, match the solution region of each system of linear inequalities with one of the four regions shown in the figure. Identify the corner points of each solution region. y
(0, 16)
10
whether each solution region is bounded or unbounded. Find the coordinates of each corner point. 41. x 0004 y 0007 11 5x 0004 y 15 x 0004 2y 12
42. 4x 0004 y 0007 32 x 0004 3y 0007 30 5x 0004 4y 51
43. 3x 0004 2y 24 3x 0004 y 0007 15 x 4
44. 3x 0004 4y 0007 48 x 0004 2y 24 y00079
The feasible region is empty.
45.
x 0004 y 0007 10 3x 0004 5y 15 3x 0002 2y 0007 15 00025x 0004 2y 0007 6
The feasible region is empty.
46. 3x 0002 y 1 0002x 0004 5y 9 x0004 y00079 y00075 48.
8x 0004 4y 0007 41 000215x 0004 5y 0007 19 2x 0004 6y 37
APPLICATIONS II (6, 4) III
(18, 0)
IV (8, 0)
00025
19
C In Problems 41–48, solve the systems graphically, and indicate
47. 16x 0004 13y 0007 119 12x 0004 16y 101 00024x 0004 3y 0007 11
I
(0, 6) (0, 0)
Systems of Linear Inequalities in Two Variables
10
x
x 0004 3y 0003 18
00025
2x 0004 y 0003 16
25. x 0004 3y 0007 18 2x 0004 y 16 x 0 y 0
26. x 0004 3y 0007 18 2x 0004 y 0007 16 x 0 y 0
Region III; corner points: (0, 0), (0, 6), (6, 4), (8, 0)
27. x 0004 3y 18 2x 0004 y 16 x 0 y 0
28. x 0004 3y 18 2x 0004 y 0007 16 x 0 y 0
Region II; corner points: (6, 4), (0, 6), (0, 16)
Region I, corner points: (0, 16), (6, 4), (18, 0)
In Problems 29–40, solve the systems graphically, and indicate whether each solution region is bounded or unbounded. Find the coordinates of each corner point. 29. 2x 0004 3y 0007 6 x 0 y 0
30. 4x 0004 3y 0007 12 x 0 y 0
31. 4x 0004 5y 20 x 0 y 0
32. 5x 0004 6y 30 x 0 y 0
33. 2x 0004 y 0007 8 x 0004 3y 0007 12 x 0 y 0
34. x 0004 2y 0007 10 3x 0004 y 0007 15 x 0 y 0
35. 4x 0004 3y 24 2x 0004 3y 18 x 0 y 0
36. x 0004 2y 8 2x 0004 y 10 x 0 y 0
37. 2x 0004 y 0007 12 x0004 y00077 x 0004 2y 0007 10 x 0 y 0
38. 3x 0004 y 0007 21 x0004 y00079 x 0004 3y 0007 21 x 0 y 0
39. x 0004 2y 16 x 0004 y 12 2x 0004 y 14 x 0 y 0
40. 3x 0004 y 30 x 0004 y 16 x 0004 3y 24 x 0 y 0
49. MANUFACTURING—RESOURCE ALLOCATION A manufacturing company makes two types of water skis: a trick ski and a slalom ski. The trick ski requires 6 labor-hours for fabricating and 1 labor-hour for finishing. The slalom ski requires 4 labor-hours for fabricating and 1 labor-hour for finishing. The maximum labor-hours available per day for fabricating and finishing are 108 and 24, respectively. If x is the number of trick skis and y is the number of slalom skis produced per day, write a system of inequalities that indicates appropriate restraints on x and y. Find the set of feasible solutions graphically for the number of each type of ski that can be produced. 50. MANUFACTURING—RESOURCE ALLOCATION A furniture manufacturing company manufactures dining room tables and chairs. A table requires 8 labor-hours for assembling and 2 labor-hours for finishing. A chair requires 2 labor-hours for assembling and 1 labor-hour for finishing. The maximum labor-hours available per day for assembly and finishing are 400 and 120, respectively. If x is the number of tables and y is the number of chairs produced per day, write a system of inequalities that indicates appropriate restraints on x and y. Find the set of feasible solutions graphically for the number of tables and chairs that can be produced. 51. MANUFACTURING—RESOURCE ALLOCATION Refer to Problem 49. The company makes a profit of $50 on each trick ski and a profit of $60 on each slalom ski. (A) If the company makes 10 trick and 10 slalom skis per day, the daily profit will be $1,100. Are there other feasible production schedules that will result in a daily profit of $1,100? How are these schedules related to the graph of the line 50x 0004 60y 0003 1,100? (B) Find a feasible production schedule that will produce a daily profit greater than $1,100 and repeat part A for this schedule. (C) Discuss methods for using lines like those in parts A and B to find the largest possible daily profit. 52. MANUFACTURING—RESOURCE ALLOCATION Refer to Problem 50. The company makes a profit of $50 on each table and a profit of $15 on each chair. (A) If the company makes 20 tables and 20 chairs per day, the daily profit will be $1,300. Are there other feasible production schedules that will result in a daily profit of $1,300? How are these schedules related to the graph of the line 50x 0004 15y 0003 1,300?
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(B) Find a feasible production schedule that will produce a daily profit greater than $1,300 and repeat part A for this schedule. (C) Discuss methods for using lines like those in parts A and B to find the largest possible daily profit. 53. NUTRITION—PLANTS A farmer can buy two types of plant food, mix A and mix B. Each cubic yard of mix A contains 20 pounds of phosphoric acid, 30 pounds of nitrogen, and 5 pounds of potash. Each cubic yard of mix B contains 10 pounds of phosphoric acid, 30 pounds of nitrogen, and 10 pounds of potash. The minimum requirements are 460 pounds of phosphoric acid, 960 pounds of nitrogen, and 220 pounds of potash. If x is the number of cubic yards of mix A used and y is the number of cubic yards of mix B used, write a system of inequalities that indicates appropriate restraints on x and y. Find the set of feasible solutions graphically for the amount of mix A and mix B that can be used. 54. NUTRITION A dietitian in a hospital is to arrange a special diet using two foods. Each ounce of food M contains 30 units of calcium, 10 units of iron, and 10 units of vitamin A. Each ounce of food N contains 10 units of calcium, 10 units of iron, and 30 units of vitamin A. The minimum requirements in the diet are 360 units of calcium, 160 units of iron, and 240 units of vitamin A. If x is the number of ounces of food M used and y is the number of ounces of food N used, write a system of linear inequalities that reflects the conditions indicated. Find the set of feasible solutions graphically for the amount of each kind of food that can be used.
7-8
55. SOCIOLOGY A city council voted to conduct a study on innercity community problems. A nearby university was contacted to provide sociologists and research assistants. Each sociologist will spend 10 hours per week collecting data in the field and 30 hours per week analyzing data in the research center. Each research assistant will spend 30 hours per week in the field and 10 hours per week in the research center. The minimum weekly labor-hour requirements are 280 hours in the field and 360 hours in the research center. If x is the number of sociologists hired for the study and y is the number of research assistants hired for the study, write a system of linear inequalities that indicates appropriate restrictions on x and y. Find the set of feasible solutions graphically. 56. PSYCHOLOGY In an experiment on conditioning, a psychologist uses two types of Skinner (conditioning) boxes with mice and rats. Each mouse spends 10 minutes per day in box A and 20 minutes per day in box B. Each rat spends 20 minutes per day in box A and 10 minutes per day in box B. The total maximum time available per day is 800 minutes for box A and 640 minutes for box B. We are interested in the various numbers of mice and rats that can be used in the experiment under the conditions stated. If x is the number of mice used and y is the number of rats used, write a system of linear inequalities that indicates appropriate restrictions on x and y. Find the set of feasible solutions graphically.
Linear Programming Z A Linear Programming Problem Z Linear Programming—A General Description Z Application
Several problems in Section 10-7 are related to the general type of problems called linear programming problems. Linear programming is a mathematical process that was developed to help management in decision making, and it has become one of the most widely used and best-known tools of management science and industrial engineering. We will use an intuitive graphical approach based on the techniques discussed in Section 10-7 to illustrate this process for problems involving two variables.
Z A Linear Programming Problem We will begin our discussion with an example that leads to a general procedure for solving linear programming problems in two variables.
EXAMPLE
1
Production Scheduling A manufacturer of fiberglass camper tops for pickup trucks makes a compact model and a regular model. Each compact top requires 5 hours from the fabricating department and 2 hours from the finishing department. Each regular top requires 4 hours from the fabricating department and 3 hours from the finishing department. The maximum labor-hours available per week in the fabricating department and the finishing department are 200 and 108,
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respectively. If the company makes a profit of $40 on each compact top and $50 on each regular top, how many tops of each type should be manufactured each week to maximize the total weekly profit, assuming all tops can be sold? What is the maximum profit? SOLUTION
This is an example of a linear programming problem. To organize the given information, we summarize the manufacturing requirements, objectives, and restrictions in the table: Compact Model (Labor-Hours per Top)
Regular Model (Labor-Hours per Top)
Maximum Labor-Hours Available per Week
Fabricating
5
4
200
Finishing
2
3
108
$40
$50
Profit per top
Now we will formulate a mathematical model for the problem and then to solve it using graphical methods. OBJECTIVE FUNCTION The objective of management is to decide how many of each camper top model should be produced each week in order to maximize profit. Let
x 0003 Number of compact tops produced per week f y 0003 Number of regular tops produced per week
Decision variables
The following function gives the total profit P for x compact tops and y regular tops manufactured each week: $40 for each compact, and $50 for each regular. P 0003 40x 0004 50y
Objective function
Mathematically, management needs to decide on values for the decision variables (x and y) that achieve its objective, which is maximizing the objective function (profit) P 0003 40x 0004 50y. You might think that the profit can be made as large as we like by manufacturing more and more tops—or can it? Any manufacturing company, no matter how large or small, has manufacturing limits imposed by available resources, plant capacity, demand, and so forth. These limits are referred to as problem constraints.
CONSTRAINTS
Fabricating department constraint: Weekly fabricating Weekly fabricating Maximum labor-hours ° time for x ¢ 0005 ° time for y ¢0002a b available per week compact tops regular tops
5x
0004
4y
0007
200
Finishing department constraint: Weekly finishing Weekly finishing Maximum labor-hours ° time for x ¢ 0005 ° time for y ¢0002a b available per week compact tops regular tops
2x
+
3y
0007
108
Nonnegative constraints: It is not possible to manufacture a negative number of tops, so we have the nonnegative constraints x 0 y 0 which we usually write in the form x, y 0
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MATHEMATICAL MODEL
We now have a mathematical model for the problem under con-
sideration: Maximize P 0003 40x 0004 50y Subject to 5x 0004 4y 0007 200 2x 0004 3y 0007 108
冎
x, y 0
Objective function Problem constraints Nonnegative constraints
Solving the system of linear inequality constraints graphically, as in Section 10-7, we obtain the feasible region for production schedules, as shown in Figure 1.
GRAPHIC SOLUTION
y
Fabricating capacity line 5x 0004 4y 0003 200
All lines are restricted to the first quadrant because of the nonnegative constraints x, y 0003 0.
(0, 36)
(24, 20)
20
Feasible region
(0, 0)
20
Finishing capacity line 2x 0004 3y 0003 108
(40, 0)
x
Z Figure 1
Any production schedule (x, y) in the feasible region is possible—but which is the best? That is, which produces the largest profit? We could try some schedules in the region to see what the profit is; for x 0003 24 and y 0003 10, the weekly profit is P 0003 40(24) 0004 50(10) 0003 $1,460 For a different point in the feasible region, (15, 20), the profit is P 0003 40(15) 0004 50(20) 0003 $1,600 But how do we know when we’ve found the largest profit? Such a schedule, if it exists, is called an optimal solution to the problem because it produces the maximum value of the objective function and is in the feasible region. It is not practical to use point-by-point checking to find the optimal solution. Even if we consider only points with integer coordinates, there are over 800 such points in the feasible region for this problem. Instead, we use the theory that has been developed to solve linear programming problems. Using advanced techniques, it can be shown that
Corner Point (x, y)
Objective Function P 0003 40x 0005 50y
(0, 0)
0
(0, 36)
1,800
(24, 20)
1,960
(40, 0)
1,600
Maximum value of P
If the feasible region is bounded, then one or more of the corner points of the feasible region is an optimal solution to the problem. The maximum value of the objective function is unique; however, there can be more than one feasible production schedule that will produce this unique value. We will have more to say about this later in this section. Since the feasible region for this problem is bounded, at least one of the corner points, (0, 0), (0, 36), (24, 20), or (40, 0), is an optimal solution. To find which one, we evaluate P 0003 40x 0004 50y at each corner point and choose the corner point that produces the largest value of P. It is convenient to organize these calculations in a table, as shown in the margin.
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23
Examining the values in the table, we see that the maximum value of P at a corner point is P 0003 1,960 at x 0003 24 and y 0003 20. Since the maximum value of P over the entire feasible region must always occur at a corner point, we conclude that the maximum profit is $1,960 when 24 compact tops and 20 regular tops are produced each week. 0002 MATCHED PROBLEM 1
We will now convert the surfboard problem discussed in Section 7-7 into a linear programming problem. A manufacturer of surfboards makes a standard model and a competition model. Each standard board requires 6 labor-hours for fabricating and 1 labor-hour for finishing. Each competition board requires 8 labor-hours for fabricating and 3 labor-hours for finishing. The maximum labor-hours available per week in the fabricating and finishing departments are 120 and 30, respectively. If the company makes a profit of $40 on each standard board and $75 on each competition board, how many boards of each type should be manufactured each week to maximize the total weekly profit? (A) Identify the decision variables. (B) Write the objective function P. (C) Write the problem constraints and the nonnegative constraints. (D) Graph the feasible region, identify the corner points, and evaluate P at each corner point. (E) How many boards of each type should be manufactured each week to maximize the profit? What is the maximum profit? 0002
ZZZ EXPLORE-DISCUSS 1
Refer to Example 1. If we assign the profit function P in P 0003 40x 0004 50y a particular value and plot the resulting equation in the coordinate system shown in Figure 1, we obtain a constant-profit line (isoprofit line). Every point in the feasible region on this line represents a production schedule that will produce the same profit. Figure 2 shows the constant-profit lines for P 0003 $1,000 and P 0003 $1,500. y (0, 36)
P 20
P
0003
0003
(24, 20)
$1
,5
$1
00
,0
(0, 0)
00 20
(40, 0)
x
Z Figure 2
(A) How are all the constant-profit lines related? (B) Place a straightedge along the constant-profit line for P 0003 $1,000 and slide it as far as possible in the direction of increasing profit without changing its slope and without leaving the feasible region. Explain how this process can be used to identify the optimal solution to a linear programming problem. (C) If P is changed to P 0003 25x 0004 75y, graph the constant-profit lines for P 0003 $1,000 and P 0003 $1,500, and use a straightedge to identify the optimal solution. Check your answer by evaluating P at each corner point. (D) Repeat part C for P 0003 75x 0004 25y.
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Z Linear Programming—A General Description The linear programming problems considered in Example 1 and Matched Problem 1 were maximization problems, where we wanted to maximize profits. The same technique can be used to solve minimization problems, where, for example, we may want to minimize costs. Before considering additional examples, we will state a few general definitions. A linear programming problem is one that is concerned with finding the optimal value (maximum or minimum value) of a linear objective function of the form z 0003 ax 0004 by where the decision variables x and y are subject to problem constraints in the form of linear inequalities and the nonnegative constraints x, y 0. The set of points satisfying both the problem constraints and the nonnegative constraints is called the feasible region for the problem. Any point in the feasible region that produces the optimal value of the objective function over the feasible region is called an optimal solution. Theorem 1 is fundamental to the solving of linear programming problems. Z THEOREM 1 Fundamental Theorem of Linear Programming Let S be the feasible region for a linear programming problem, and let z 0003 ax 0004 by be the objective function. If S is bounded, then z has both a maximum and a minimum value on S and each of these occurs at a corner point of S. If S is unbounded, then a maximum or minimum value of z on S may not exist. However, if either does exist, then it must occur at a corner point of S.
We will not consider any problems with unbounded feasible regions in this brief introduction. If a feasible region is bounded, then Theorem 1 provides the basis for the following simple procedure for solving the associated linear programming problem: Z SOLUTION OF LINEAR PROGRAMMING PROBLEMS Step 1. Form a mathematical model for the problem: (A) Introduce decision variables and write a linear objective function. (B) Write problem constraints in the form of linear inequalities. (C) Write nonnegative constraints. Step 2. Graph the feasible region and find the corner points. Step 3. Evaluate the objective function at each corner point to determine the optimal solution.
Before considering additional applications, we use this procedure to solve a linear programming problem where the model has already been determined.
EXAMPLE
2
Solving a Linear Programming Problem Minimize and maximize z 0003 5x 0004 15y Subject to x 0004 3y 0007 60 x 0004 y 10 x0002 y0007 0 x, y 0
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25
This problem is a combination of two linear programming problems—a minimization problem and a maximization problem. Since the feasible region is the same for both problems, we can solve these problems together. To begin, we graph the feasible region S, as shown in Figure 3, and find the coordinates of each corner point. y (0, 20)
x 0004 3y 0003 60 (15, 15)
S (0, 10)
x0002y00030
x 0004 y 0003 10 5
(5, 5) x
5
Z Figure 3
Next, we evaluate the objective function at each corner point, with the results given in the table: Corner Point (x, y)
Objective Function z 0003 5x 0005 15y
(0, 10)
150
(0, 20)
300
(15, 15)
300
Maximum value ⎫ Multiple ⎬ optimal Maximum value ⎭ solutions
(5, 5)
100
Minimum value
Examining the values in the table, we see that the minimum value of z on the feasible region S is 100 at (5, 5), So (5, 5) is the optimal solution to the minimization problem. The maximum value of z on the feasible region S is 300, which occurs at (0, 20) and at (15, 15); the maximization problem has multiple optimal solutions. In general: If two corner points are both optimal solutions of the same type (both produce the same maximum value or both produce the same minimum value) to a linear programming problem, then any point on the line segment joining the two corner points is also an optimal solution of that type. It can be shown that this is the only time that an optimal value occurs at more than one point. 0002 MATCHED PROBLEM 2
Minimize and maximize z 0003 10x 0004 5y Subject to 2x 0004 y 40 3x 0004 y 0007 150 2x 0002 y 0 x, y 0
0002
Z Application Now we will look at another application where we first find the mathematical model and then find its solution.
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3 Pounds per Cubic Yard
Mix A Mix B Nitrogen
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10
5
Potash
8
24
Phosphoric acid
9
6
SOLUTION
Agriculture A farmer can use two types of plant food, mix A and mix B. The amounts (in pounds) of nitrogen, phosphoric acid, and potash in a cubic yard of each mix are given in the table. Tests performed on the soil in a large field indicate that the field needs at least 840 pounds of potash and at least 350 pounds of nitrogen. The tests also indicate that no more than 630 pounds of phosphoric acid should be added to the field. A cubic yard of mix A costs $7, and a cubic yard of mix B costs $9. How many cubic yards of each mix should the farmer add to the field in order to supply the necessary nutrients at minimal cost? Let x 0003 Number of cubic yards of mix A added to the field y 0003 Number of cubic yards of mix B added to the field
冎
Decision variables
We form the linear objective function C 0003 7x 0004 9y which gives the cost of adding x cubic yards of mix A and y cubic yards of mix B to the field. Using the data in the table and proceeding as in Example 1, we formulate the mathematical model for the problem: Minimize Subject to
C 0003 7x 0004 9y 10x 0004 5y 350 8x 0004 24y 840 9x 0004 6y 0007 630 x, y 0
Objective function Nitrogen constraint Potash constraint Phosphoric acid constraint Nonnegative constraints
Solving the system of constraint inequalities graphically, we obtain the feasible region S shown in Figure 4, and then we find the coordinates of each corner point.
y (0, 105)
9x 0004 6y 0003 630 (0, 70) 60
S
10x 0004 5y 0003 350
Corner Point (x, y)
Objective Function C 0003 7x 0005 9y
(0, 105)
945
(0, 70)
630
(21, 28)
399
(60, 15)
555
(21, 28)
(60, 15)
8x 0004 24y 0003 840 60
x
Z Figure 4 Minimum value of C
Next, we evaluate the objective function at each corner point, as shown in the table in the margin.
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The optimal value is C 0003 399 at the corner point (21, 28). The farmer should add 21 cubic yards of mix A and 28 cubic yards of mix B at a cost of $399. This will result in adding the following nutrients to the field: Nitrogen: Potash: Phosphoric acid:
10(21) 0004 5(28) 0003 350 pounds 8(21) 0004 24(28) 0003 840 pounds 9(21) 0004 6(28) 0003 357 pounds
All the nutritional requirements are satisfied. MATCHED PROBLEM 3
0002
Repeat Example 3 if the tests indicate that the field needs at least 400 pounds of nitrogen with all other conditions remaining the same. 0002 The American mathematician George B. Dantzig (1914–2005) formulated the first linear programming problem in 1947 and introduced a solution technique, called the simplex method, that does not rely on graphing and is readily adaptable to computer solutions. Today, it is quite common to use a computer to solve applied linear programming problems involving thousands of variables and thousands of inequalities. ANSWERS TO MATCHED PROBLEMS 1. (A) x 0003 Number of standard boards manufactured each week y 0003 Number of competition boards manufactured each week (B) P 0003 40x 0004 75y (C) 6x 0004 8y 0007 120 Fabricating constraint x 0004 3y 0007 30 Finishing constraint x, y 0 Nonnegative constraints (D)
Corner Point (x, y)
y (0, 10)
(0, 0)
(12, 6) 5
(0, 0)
Feasible region (20, 0)
5
x
Objective Function P 0003 40x 0005 75y 0
(0, 10)
750
(12, 6)
930
(20, 0)
800
(E) 12 standard boards and 6 competition boards for a maximum profit of $930 2. Max z 0003 600 at (30, 60); min z 0003 200 at (10, 20) and (20, 0) (multiple optimal solutions) 3. 27 cubic yards of mix A, 26 cubic yards of mix B; min C 0003 $423
7-8
Exercises
*Additional answers can be found in the Instructor Answer Appendix.
A
In Problems 1–7, explain how each term is used in the description of a linear programming program. 1. Objective function 2. Decision variables 3. Problem constraints
4. Nonnegativity constraints 5. Feasible region 6. Optimal value and optimal solution 7. Corner point
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8. Can a linear programming problem have more than one optimal value? Explain. In Problems 9–12, find the maximum value of each objective function over the feasible region S shown in the figure. y
value. Check your answer by evaluating the objective function at each corner point. 21. z 0003 x 0004 2y
22. z 0003 2x 0004 y
23. z 0003 5x 0004 4y
24. z 0003 2x 0004 8y
Minimum value of z on T is 32 at both (0, 8) and (4, 3).
25. Maximize Subject to (7, 9)
26. Maximize Subject to
S
5
(10, 0) (0, 0)
5
10. z 0003 4x 0004 y
11. z 0003 3x 0004 7y
27. Minimize Subject to
x
zmax 0003 40 at (10, 0)
12. z 0003 9x 0004 3y
zmax 0003 90 at (7, 9) and (10, 0) (multiple optimal solutions)
Problems 13–16 refer to the feasible region S shown and the constant-profit lines discussed in Explore-Discuss 1. For each objective function, draw the line that passes through the feasible point (5, 5) and use the straightedge method from Explore-Discuss 1 to find the maximum value. Check your answer by evaluating the objective function at each corner point. 13. z 0003 x 0004 2y
14. z 0003 3x 0004 y
15. z 0003 7x 0004 2y
16. z 0003 2x 0004 8y
Maximum value of z on S is 70 at (10, 0).
Maximum value of z on S is 30 at both (7, 9) and (10, 0) Maximum value of z on S is 96 at (0, 12)
28. Minimize Subject to
29. Maximize Subject to
30. Maximize Subject to
In Problems 17–20, find the minimum value of each objective function over the feasible region T shown in the figure. 31. Minimize Subject to
y
(0, 12)
32. Minimize Subject to
(0, 8) 5
Minimum value of z on T is 24 at (12, 0).
B In Problems 25–38, solve the linear programming problems. (0, 12)
9. z 0003 x 0004 y
Minimum value of z on T is 8 at (0, 8).
T (4, 3)
(12, 0) 5
17. z 0003 7x 0004 4y
18. z 0003 7x 0004 9y
19. z 0003 3x 0004 8y
20. z 0003 5x 0004 4y
x
zmin 0003 55 at (4, 3)
zmin 0003 32 at (0, 8) and (4, 3) (multiple optimal solutions)
Problems 21–24 refer to the feasible region T shown for problems 17–20. For each objective function, draw the constant-value line that passes through the feasible point (5, 5) and use the straightedge method from Explore-Discuss 1 to find the minimum
z 0003 3x 0004 2y x 0004 2y 0007 10 3x 0004 y 0007 15 x, y 0
Maximum value of z on S is 18 at (4, 3).
z 0003 4x 0004 5y 2x 0004 y 0007 12 x 0004 3y 0007 21 x, y 0
zmax 0003 42 at (3, 6)
z 0003 3x 0004 4y 2x 0004 y 8 x 0004 2y 0007 10 x, y 0
Minimum value of z on S is 12 at (4, 0).
z 0003 2x 0004 y 4x 0004 3y 24 4x 0004 y 0007 16 x, y 0
zmin 0003 8 at (0, 8)
z 0003 3x 0004 4y x 0004 2y 0007 24 x 0004 y 0007 14 2x 0004 y 0007 24 x, y 0
Maximum value of z on S is 52 at (4, 10).
z 0003 5x 0004 3y 3x 0004 y 0007 24 x 0004 y 0007 10 x 0004 3y 0007 24 x, y 0
zmax 0003 44 at (7, 3)
z 0003 5x 0004 6y x 0004 4y 20 4x 0004 y 20 x 0004 y 0007 20 x, y 0
Minimum value of z on S is 44 at (4, 4).
z 0003 x 0004 2y 2x 0004 3y 30 3x 0004 2y 30 x 0004 y 0007 15 x, y 0
zmin 0003 15 at (15, 0)
33. Minimize and maximize z 0003 25x 0004 50y Subject to x 0004 2y 0007 120 The minimum value of z on S is x 0004 y 60 1,500 at (60, 0). The maximum value x 0002 2y 0 of z on S is 3,000 at (60, 30) and x, y 0 (120, 0) (multiple optimal solutions). 34. Minimize and maximize z 0003 15x 0004 30y Subject to x 0004 2y 100 2x 0002 y 0007 0 zmax 0003 6000 at (0, 200); 2x 0004 y 0007 200 zmin 0003 1500 at (0, 50) and (20, 40) x, y 0 (multiple optimal solutions)
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35. Minimize and maximize z 0003 25x 0004 15y Subject to 4x 0004 5y 100 3x 0004 4y 0007 240 x 0007 60 The minimum value of z on S is y 0007 45 300 at (0, 20). The maximum x, y 0 value of z on S is 1,725 at (60, 15). 36. Minimize and maximize z 0003 25x 0004 30y Subject to 2x 0004 3y 120 3x 0004 2y 0007 360 x 0007 80 y 0007 120 zmin 0003 1200 at (0, 40); x, y 0 zmax 0003 4600 at (40, 120) P 0003 525x1 0004 478x2 275x1 0004 322x2 0007 3,381 350x1 0004 340x2 0007 3,762 425x1 0004 306x2 0007 4,114 x1, x2 0
Max P 0003 5,507 at x1 0003 6.62 and x2 0003 4.25
38. Maximize P 0003 300x1 0004 460x2 Subject to 245x1 0004 452x2 0007 4,181 290x1 0004 379x2 0007 3,888 390x1 0004 299x2 0007 4,407 x1, x2 0
Max P 0003 4,484 at x1 0003 4.52 and x2 0003 6.8
37. Maximize Subject to
C
39. The corner points for the feasible region determined by the problem constraints 2x 0004 y 0007 10 x 0004 3y 0007 15 x, y 0 are O 0003 (0, 0), A 0003 (5, 0), B 0003 (3, 4), and C 0003 (0, 5). If z 0003 ax 0004 by and a, b 0, determine conditions on a and b that ensure that the maximum value of z occurs (A) Only at A (B) Only at B (C) Only at C (D) At both A and B (E) At both B and C
40. The corner points for the feasible region determined by the problem constraints x0004 y 4 x 0004 2y 6 2x 0004 3y 0007 12 x, y 0 are A 0003 (6, 0), B 0003 (2, 2), and C 0003 (0, 4). If z 0003 ax 0004 by and a, b 0, determine conditions on a and b that ensure that the minimum value of z occurs (A) Only at A 2a b (B) Only at B 12 b 6 a 6 b (C) Only at C a b (D) At both A and B b 0003 2a (E) At both B and C b 0003 a
APPLICATIONS 41. RESOURCE ALLOCATION A manufacturing company makes two types of water skis, a trick ski and a slalom ski. The relevant manufacturing data are given in the table. (A) If the profit on a trick ski is $40 and the profit on a slalom ski is $30, how many of each type of ski should be manufactured each day to realize a maximum profit? What is the maximum profit? (B) Discuss the effect on the production schedule and the maximum profit if the profit on a slalom ski decreases to $25 and all other data remains the same.
29
Linear Programming
(C) Discuss the effect on the production schedule and the maximum profit if the profit on a slalom ski increases to $45 and all other data remain the same. Trick Ski (labor-hours per ski)
Slalom Ski (labor-hours per ski)
Maximum Labor-Hours Available per Day
Fabricating department
6
4
108
Finishing department
1
1
24
42. PSYCHOLOGY In an experiment on conditioning, a psychologist uses two types of Skinner boxes with mice and rats. The amount of time (in minutes) each mouse and each rat spends in each box per day is given in the table. What is the maximum total number of mice and rats that can be used in this experiment? How many mice and how many rats produce this maximum?
Mice (minutes)
Rats (minutes)
Max. Time Available per Day (minutes)
Skinner box A
10
20
800
Skinner box B
20
10
640 48; 16 mice, 32 rats
43. PURCHASING A trucking firm wants to purchase a maximum of 15 new trucks that will provide at least 36 tons of additional shipping capacity. A model A truck holds 2 tons and costs $15,000. A model B truck holds 3 tons and costs $24,000. How many trucks of each model should the company purchase to provide the additional shipping capacity at minimal cost? What is the minimal cost?
9 model A trucks and 6 model B trucks to realize the minimum cost of $279,000
44. TRANSPORTATION The officers of a student group are planning to rent buses and vans for a class trip. Each bus can transport 40 students, requires 3 chaperones, and costs $1,200 to rent. Each van can transport 8 students, requires 1 chaperone, and costs $100 to rent. The officers want to be able to accommodate at least 400 students with no more than 36 chaperones. How many vehicles of each type should they rent to minimize the transportation costs? What are the minimal transportation costs? 7 buses, 15 vans; $9,900 45. RESOURCE ALLOCATION A furniture company manufactures dining room tables and chairs. Each table requires 8 hours from the assembly department and 2 hours from the finishing department and contributes a profit of $90. Each chair requires 2 hours from the assembly department and 1 hour from the finishing department and contributes a profit of $25. The maximum labor-hours available each day in the assembly and finishing departments are 400 and 120, respectively. (A) How many tables and how many chairs should be manufactured each day to maximize the daily profit? What is the maximum daily profit? (B) Discuss the effect on the production schedule and the maximum profit if the marketing department of the company decides that the number of chairs produced should be at least four times the number of tables produced. 46. RESOURCE ALLOCATION An electronics firm manufactures two types of personal computers, a desktop model and a laptop. The
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production of a desktop computer requires a capital expenditure of $400 and 40 hours of labor. The production of a laptop computer requires a capital expenditure of $250 and 30 hours of labor. The firm has $20,000 capital and 2,160 labor-hours available for production of desktop and laptop computers. (A) What is the maximum number of computers the company is capable of producing? (B) If each desktop computer contributes a profit of $320 and each laptop contributes a profit of $220, how much profit will the company make by producing the maximum number of computers determined in part A? Is this the maximum profit? If not, what is the maximum profit? 47. POLLUTION CONTROL Because of new federal regulations on pollution, a chemical plant introduced a new process to supplement or replace an older process used in the production of a particular chemical. The older process emitted 20 grams of sulfur dioxide and 40 grams of particulate matter into the atmosphere for each gallon of chemical produced. The new process emits 5 grams of sulfur dioxide and 20 grams of particulate matter for each gallon produced. The company makes a profit of 60¢ per gallon and 20¢ per gallon on the old and new processes, respectively. (A) If the regulations allow the plant to emit no more than 16,000 grams of sulfur dioxide and 30,000 grams of particulate matter daily, how many gallons of the chemical should be produced by each process to maximize daily profit? What is the maximum daily profit? (B) Discuss the effect on the production schedule and the maximum profit if the regulations restrict emissions of sulfur dioxide to 11,500 grams daily and all other data remains unchanged. (C) Discuss the effect on the production schedule and the maximum profit if the regulations restrict emissions of sulfur dioxide to 7,200 grams daily and all other data remains unchanged. 48. SOCIOLOGY A city council voted to conduct a study on innercity community problems. A nearby university was contacted to provide a maximum of 40 sociologists and research assistants. Allocation of time and cost per week are given in the table. (A) How many sociologists and research assistants should be hired to meet the weekly labor-hour requirements and minimize the weekly cost? What is the weekly cost? (B) Discuss the effect on the solution in part A if the council decides that they should not hire more sociologists than research assistants and all other data remain unchanged.
Sociologist (labor-hours)
Research Assistant (labor-hours)
Minimum Labor-Hours Needed per Week
Fieldwork
10
30
280
Research center
30
10
360
$500
$300
Cost per week
49. PLANT NUTRITION A fruit grower can use two types of fertilizer in her orange grove, brand A and brand B. The amounts (in pounds) of nitrogen, phosphoric acid, potash, and chloride in a bag of each mix are given in the table. Tests indicate that the grove needs at least 480 pounds of phosphoric acid, at least 540 pounds of potash, and at most 620 pounds of chloride. If the grower always uses a combination of bags of brand A and brand B that will satisfy the constraints for phosphoric acid, potash, and chloride, discuss the effect that this will have on the amount of nitrogen added to the field. Pounds per Bag Brand A
Brand B
Nitrogen
6
7
Phosphoric acid
2
4
Potash
6
3
Chloride
3
4
50. DIET A dietitian in a hospital is to arrange a special diet composed of two foods, M and N. Each ounce of food M contains 16 units of calcium, 5 units of iron, 6 units of cholesterol, and 8 units of vitamin A. Each ounce of food N contains 4 units of calcium, 25 units of iron, 4 units of cholesterol, and 4 units of vitamin A. The diet requires at least 320 units of calcium, at least 575 units of iron, and at most 300 units of cholesterol. If the dietitian always selects a combination of foods M and N that will satisfy the constraints for calcium, iron, and cholesterol, discuss the effects that this will have on the amount of vitamin A in the diet. The amount of vitamin A will range from a minimum of 200 units when 15 ounces of food M and 20 ounces of food N are used to a maximum of 380 units when 40 ounces of food M and 15 ounces of food N are used.
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8
C
OUTLINE
THE lists
8-1
Sequences and Series
and
8-2
Mathematical Induction
3, 6, 3, 1, 4, 2, 1, 4, . . . are examples of sequences. In the first sequence, a pattern is noticeable: You probably recognize it as the sequence of perfect squares. Its terms are increasing, and as we will see, the differences between terms form a clear pattern. You probably don’t recognize the second sequence because the terms don’t suggest an obvious pattern. In fact, we obtained the second sequence by recording the results of repeatedly tossing a single die. Sequences, and the related concept of series, are useful tools in almost all areas of mathematics. In this chapter, they will play roles in the development of several topics: a method of proof called mathematical induction, techniques for counting, and probability.
8-3
Arithmetic and Geometric Sequences
8-4
Multiplication Principle, Permutations, and Combinations
8-5
Sample Spaces and Probability
8-6
The Binomial Formula
1, 4, 9, 16, 25, 36, 49, 64, . . .
Chapter 8 Review Chapter 8 Group Activity: Sequences Specified by Recursion Formulas
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Sequences and Series Z Defining Sequences Z Defining Series
In this section, we introduce special notation and formulas for representing and generating sequences and sums of sequences.
Z Defining Sequences Consider the following list of numbers: 1, 3, 5, 7, 9, . . . . This is an example of a sequence, which can be defind informally as a list of numbers in a specific order. This particular sequence is the sequence of positive odd integers. Now consider the function f given by f (n) 0002 2n 0003 1
(1)
where the domain of f is {1, 2, 3, . . .} (that is, the set of natural numbers N). Note that f (1) 0002 2(1) 0003 1 0002 1 f (2) 0002 2(2) 0003 1 0002 3 f (3) 0002 2(3) 0003 1 0002 5 The outputs of the function f form the same list of odd positive integers that we started with above. This provides an alternative (and more precise) definition of sequence: A sequence is a function whose domain is a set of successive integers. While the function f above is a perfectly good way to describe a sequence, a special notation for describing sequences with formulas has evolved over the years. Our first order of business should be to become familiar with this notation. To start, the range value f (n) is usually symbolized more compactly with a symbol such as an. So in place of equation (1) we write an 0002 2n 0003 1 The domain is understood to be the set of natural numbers N unless stated to the contrary or the context indicates otherwise. The elements in the range are called terms of the sequence: a1 is the first term, a2 the second term, and an the nth term, or the general term: a1 0002 2(1) 0003 1 0002 1 a2 0002 2(2) 0003 1 0002 3 a3 0002 2(3) 0003 1 0002 5 o o
First term Second term Third term
The ordered list of elements 1, 3, 5, . . . , 2n 0003 1, . . . in which the terms of a sequence are written in their natural order with respect to the domain values, is often informally referred to as a sequence. A sequence is also represented in the abbreviated form {an}, where a symbol for the nth term is placed between braces. For example, we can refer to the sequence 1, 3, 5, . . . , 2n 0003 1, . . . as the sequence {2n 0003 1}.
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505
If the domain of a function is a finite set of successive integers, then the sequence is called a finite sequence. If the domain is an infinite set of successive integers, then the sequence is called an infinite sequence. The preceding sequence {2n 0003 1} is an example of an infinite sequence.
Technology Connections Some graphing calculators have a special sequence mode that can be useful when studying sequences. Figure 1(a) shows the sequence {2n ⴚ 1} entered in the sequence editor.
Figure 1(b) shows the graph of the sequence. Figure 1(c) displays the sequence in a table.
20
0
10
0
(a)
(b)
(c)
Z Figure 1
Some sequences are specified by a recursion formula—that is, a formula that defines each term in terms of one or more preceding terms. The sequence we have chosen to illustrate a recursion formula is a very famous sequence in the history of mathematics called the Fibonacci sequence. It is named after the most celebrated mathematician of the thirteenth century, Leonardo Fibonacci from Italy (1180?–1250?).
EXAMPLE
1
Fibonacci Sequence List the first seven terms of the sequence specified by a1 0002 1 a2 0002 1 an 0002 an00032 0004 an00031
SOLUTION
a1 a2
00021 00021
a3 a4 a5 a6 a7
0002 0002 0002 0002
0004 0004 0004 0004 0002 a5 0004 a1 a2 a3 a4
a2 a3 a4 a5 a6
0002 0002 0002 0002 0002
1 1 2 3 5
0004 0004 0004 0004 0004
n00053
1 2 3 5 8
*
0002 0002 0002 0002
2 3 5 8
0002 13
*Throughout the book, dashed boxes—called think boxes—are used to represent steps that may be performed mentally.
0002
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MATCHED PROBLEM 1
List the first seven terms of the sequence specified by a1 0002 1 a2 0002 1 an 0002 an00032 0003 an00031
ZZZ EXPLORE-DISCUSS 1
n00053
0002
A multiple-choice test question asked for the next term in the sequence: 1, 3, 9, . . . and gave the following choices: (A) 16
(B) 19
(C) 27
Which is the correct answer? Compare the first four terms of the following sequences: (A) an 0002 3n00031
(B) bn 0002 1 0004 2(n 0003 1)2
(C) cn 0002 8n 0004
12 0003 19 n
Now which of the choices appears to be correct?
Now we consider the reverse problem. That is, can a sequence be defined just by listing the first three or four terms of the sequence? And can we then use these initial terms to find a formula for the nth term? In general, without other information, the answer to the first question is no. As Explore-Discuss 1 illustrates, many different sequences may start off with the same terms. Simply listing the first three terms, or any other finite number of terms, does not specify a particular sequence. In fact, it can be shown that given any list of m numbers, there are an infinite number of sequences whose first m terms agree with these given numbers. What about the second question? That is, given a few terms, can we find the general formula for at least one sequence whose first few terms agree with the given terms? The answer to this question is a qualified yes. If we can observe a simple pattern in the given terms, then we may be able to construct a general term that will produce the pattern. Example 2 illustrates this approach.
EXAMPLE
2
Finding the General Term of a Sequence Find the general term of a sequence whose first four terms are (A) 5, 6, 7, 8, . . .
SOLUTIONS
(B) 2, 00034, 8, 000316, . . .
(A) Because these terms are consecutive integers, one solution is an 0002 n, n 0005 5. If we want the domain of the sequence to be all natural numbers, then another solution is bn 0002 n 0004 4. (B) Each of these terms can be written as the product of a power of 2 and a power of 00031: 2 0002 (00031)021 00034 0002 (00031)122 8 0002 (00031)223 000316 0002 (00031)324 If we choose the domain to be all natural numbers, then a solution is an 0002 (00031)n000312n
0002
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MATCHED PROBLEM 2
Sequences and Series
507
Find the general term of a sequence whose first four terms are (A) 2, 4, 6, 8, . . .
(B) 1, 000312, 14, 000318, . . .
0002
In general, there is usually more than one way of representing the nth term of a given sequence. This was seen in the solution of Example 2, part A. However, unless stated to the contrary, we assume the domain of the sequence is the set of natural numbers N. ZZZ EXPLORE-DISCUSS 2
15 1 0004 15 n a b is closely related to the 5 2 Fibonacci sequence. Compute the first 20 terms of both sequences and discuss the relationship. [The first seven values of bn are shown in Fig. 2(b)]. The sequence with general term bn 0002
(a)
(b)
Z Figure 2
Z Defining Series If a1, a2, a3, . . . , an, . . . is a sequence, then the expression a1 0004 a2 0004 a3 0004 . . . 0004 an 0004 . . . is called a series. If the sequence is finite, the corresponding series is a finite series. If the sequence is infinite, the corresponding series is an infinite series. For example, 1, 2, 4, 8, 16 1 0004 2 0004 4 0004 8 0004 16
Finite sequence Finite series
We will restrict our discussion to finite series in this section. Series are often represented in a compact form called summation notation using the symbol a, which is a stylized version of the Greek letter sigma. Consider the following examples: 4
a ak 0002 a1 0004 a2 0004 a3 0004 a4
k00021 7
a bk 0002 b3 0004 b4 0004 b5 0004 b6 0004 b7
k00023 n
. . . 0004 cn a ck 0002 c0 0004 c1 0004 c2 0004
k00020
Domain is the set of integers k satisfying 0 ⱕ k ⱕ n.
The terms on the right are obtained from the expression on the left by successively replacing the summing index k with integers, starting with the first number indicated below a and ending with the number that appears above a. For example, if we are given the sequence 1 1 1 ... 1 , , , , n 2 4 8 2
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the corresponding series is n 1 1 1 1 ... 1 0004 n a 2k 0002 2 0004 4 0004 8 0004 2 k00021
EXAMPLE
3
Writing the Terms of a Series 5
k00031 k k00021
Write without summation notation: a SOLUTION
MATCHED PROBLEM 3
5
k00031 100031 200031 300031 400031 500031 0002 0004 0004 0004 0004 k 1 2 3 4 5 k00021 1 2 3 4 000200004 0004 0004 0004 2 3 4 5 a
0002
5 (00031)k Write without summation notation: a k 0002 0 2k 0004 1
0002 If the terms of a series are alternately positive and negative, it is called an alternating series. Example 4 deals with the representation of such a series.
EXAMPLE
4
Writing a Series in Summation Notation Write the following series using summation notation: 10003
1 1 1 1 1 0004 0003 0004 0003 2 3 4 5 6
(A) Start the summing index at k 0002 1. (B) Start the summing index at k 0002 0. SOLUTIONS
(A) (00031)k00031 provides the alternation of sign, and 10002k provides the other part of each term. So we can write (00031)k00031 a k k00021 6
as can be easily checked. (B) (00031)k provides the alternation of sign, and 10002(k 0004 1) provides the other part of each term. We write the series as (00031)k a k00020 k 0004 1 5
as can be checked. MATCHED PROBLEM 4
0002
Write the following series using summation notation: 10003 (A) Start with k 0002 1.
4 8 16 2 0004 0003 0004 3 9 27 81
(B) Start with k 0002 0.
0002
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Sequences and Series
509
ANSWERS TO MATCHED PROBLEMS 1. 1, 1, 0, 1, 00031, 2, 00033 3. 1 0003 13 0004 15 0003 17 0004 19 0003 111
8-1
1 n00031 (B) an 0002 (00031)n00031a b 2 5 k00031 4 2 2 k 4. (A) a (00031)k00031a b (B) a (00031)k a b 3 3
2. (A) an 0002 2n
k00021
k00020
Exercises
1. Explain the difference between a sequence and a series.
27. an 0002 (000312)n00031
2. What is a recursion formula?
29. a1 0002 7; an 0002 an00031 0003 4, n 0005 2
3. Explain how the Fibonacci sequence can be defined by means of a recursion formula.
30. a1 0002 3; an 0002 an00031 0004 5, n 0005 2
4. Explain summation notation. 5. Explain why the following statement is not true: The general term of the sequence 1, 3, 7, . . . is 2n 0003 1. 6. Explain why at least one term must be provided when defining a sequence recursively. Write the first four terms for each sequence in Problems 7–12. 7. an 0002 n 0003 2 9. an 0002
n00031 n00041
11. an 0002 (00032)n00041
8. an 0002 n 0004 3
31. a1 0002 4; an 0002 14an00031, n 0005 2 32. a1 0002 2; an 0002 2an00031, n 0005 2 In Problems 33–36, write the first seven terms of each sequence. 33. a1 0002 1, a2 0002 2, an 0002 an00032 0004 2an00031, n 0005 3 34. a1 0002 1, a2 0002 00031, an 0002 an00032 0003 an00031, n 0005 3 35. a1 0002 00031, a2 0002 2, an 0002 2an00032 0004 an00031, n 0005 3 36. a1 0002 2, a2 0002 1, an 0002 0003an00032 0004 an00031, n 0005 3
n
1 10. an 0002 a1 0004 b n 12. an 0002
In Problems 37–48, find a general term an for the given sequence a1, a2, a3, a4, . . .
(00031)n00041 n2
13. Write the eighth term in the sequence in Problem 7. 14. Write the tenth term in the sequence in Problem 8. 15. Write the one-hundredth term in the sequence in Problem 9. 16. Write the two-hundredth term in the sequence in Problem 10.
37. 00032, 00031, 0, 1, . . .
38. 10, 11, 12, 13, . . .
39. 5, 7, 9, 11, . . .
40. 1, 00031, 00033, 00035, . . .
41. 00031, 1, 00031, 1, . . .
42. 1, 000312, 13, 000314, . . .
43. 2, 32, 43, 54, . . .
44. 13, 24, 35, 46, . . .
45. 00033, 9, 000327, 81, . . .
46. 5, 25, 125, 625, . . .
2
47. x, In Problems 17–22, write each series in expanded form without summation notation. 5
4
17. a k
18. a k 2
3 1 19. a k k 0002 1 10
5 1 k 20. a a b k00021 3
k00021
21. a (00031)k k00021
6
22. a (00031)k00041k k00021
Write the first five terms of each sequence in Problems 23–32. 23. an 0002 (00031)n00041n2 25. an 0002
1 1 a1 0003 n b 3 10
24. an 0002 (00031)n00041a
3
4
x x x , , ,... 2 3 4
48. x, 0003x3, x5, 0003x7, . . .
In Problems 49–54: (A) Find the first four terms of the sequence. (B) Find a general term bn for a different sequence that has the same first three terms as the given sequence.
k00021
4
28. an 0002 (000332)n00031
1 b 2n
26. an 0002 n[1 0003 (00031)n]
49. an 0002 n2 0003 n 0004 2
50. an 0002 9n2 0003 21n 0004 14
51. an 0002 6n2 0003 11n 0004 6
52. an 0002 25n2 0003 60n 0004 36
53. an 0002 2n2 0003 8n 0004 7
54. an 0002 00034n2 0004 15n 0003 12
In Problems 55–58, use a graphing calculator to graph the first 20 terms of each sequence. 55. an 0002 10002n
56. an 0002 2 0004 0006n n
57. an 0002 (00030.9)
58. a1 0002 00031, an 0002 23 an00031 0004 12
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In Problems 59–64, write each series in expanded form without summation notation. (00032)k00041 k k00021 4
5
60. a (00031)k00041(2k 0003 1)2
59. a
k00021
3
5
1 61. a xk00041 k00021 k
62. a xk00031
(00031)k00041 k x 63. a k k00021
(00031)kx2k00041 64. a 2k 0004 1 k00020
In calculus, it can be shown that
k00021
5
4
In Problems 65–72, write each series using summation notation with the summing index k starting at k 0002 1. 65. 12 0004 22 0004 32 0004 42 66. 2 0004 3 0004 4 0004 5 0004 6 67.
77. Approximate e0.2 using the first five terms of the series. Compare this approximation with your calculator evaluation of e0.2.
n
1 1 1 0004 0003 2 3 4
k00021 n
n k00021
n
n
80. Show that a (ak 0004 bk) 0002 a ak 0004 a bk
1 1 1 0004 2 0004...0004 2 22 3 n
70. 2 0004
3 4 n00041 0004 0004...0004 n 2 3
k00021
k00021
k00021
APPLICATIONS 81. PHYSICS Suppose that a rubber ball is dropped from a height of 20 feet. If it bounces 10 times, with each bounce going half as high as the one before, the heights of these bounces can be described by the sequence an 0002 10(12)n00031 (1 0007 n 0007 10). (A) How high is the fifth bounce? The tenth?
71. 1 0003 4 0004 9 0003 . . . 0004 (00031)n00041n2 (00031)n00041 1 1 1 0003 0004 0003...0004 2 4 8 2n
10
(B) Find the value of the series a an. What does this number n00021 represent?
The sequence an 0002
where the larger n is, the better the approximation. Problems 77 and 78 refer to this series. Note that n!, read “n factorial,” is defined by 0! 0002 1 and n! 0002 1 ⴢ 2 ⴢ 3 ⴢ . . . ⴢ n for n 苸 N.
79. Show that a cak 0002 c a ak
69. 1 0004
72.
x x2 x3 . . . xn xk ex 0002 a 000310004 0004 0004 0004 0004 1! 2! 3! n! k 0002 0 k!
78. Approximate e00030.5 using the first five terms of the series. Compare this approximation with your calculator evaluation of e00030.5.
1 1 1 1 1 0004 20004 30004 40004 5 2 2 2 2 2
68. 1 0003
76. Define sequences {un} and {vn} by u1 0002 1, v1 0002 0, un 0002 un00031 0004 vn00031 and vn 0002 un00031 for n 0005 2. Find the first 10 terms of each sequence, and explain their relationship to the Fibonacci sequence.
a2n00031 0004 M 2an00031
n 0005 2, M a positive real number
can be used to find 1M to any decimal-place accuracy desired. To start the sequence, choose a1 arbitrarily from the positive real numbers. Problems 73 and 74 are related to this sequence. 73. (A) Find the first four terms of the sequence a1 0002 3
an 0002
a2n00031 0004 2 2an00031
n00052
(B) Compare the terms with 12 from a calculator. (C) Repeat parts A and B letting a1 be any other positive number, say 1. 74. (A) Find the first four terms of the sequence a1 0002 2
an 0002
a2n00031 0004 5 2an00031
n00052
(B) Find 15 with a calculator, and compare with the results of part A. (C) Repeat parts A and B letting a1 be any other positive number, say 3. 75. Let {an} denote the Fibonacci sequence and let {bn} denote the sequence defined by b1 0002 1, b2 0002 3, and bn 0002 bn00031 0004 bn00032 for n 0005 3. Compute 10 terms of the sequence {cn}, where cn 0002 bn0002an. Describe the terms of {cn} for large values of n.
82. PHYSICS A bungee jumper dives off a bridge that is 300 feet above the ground. He bounces back 100 feet on the first bounce, then continues to bounce nine more times before coming to rest, with each bounce 1/3 as high as the previous. The heights of these bounces can be described by the sequence an 0002 100(13)n00031 (1 0007 n 0007 10). (A) How high is the fifth bounce? The tenth? 10
(B) Find the value of the series a an. What does this number n00021 represent? 83. SALARY INCREMENT Suppose that you are offered a job with a starting annual salary of $40,000 and annual increases of 4% of the current salary. (A) Write out the first six terms of a sequence an whose terms describe your salary in the first 6 years on this job. (B) Write the general term of the sequence in part A. 6
(C) Find the value of the series a an. What does this number n00021 represent? 84. SALARY INCREMENT A marketing firm is advertising entrylevel positions with a starting annual salary of $24,000 and annual increments of 3% of the current salary. (A) Write out the first six terms of a sequence an whose terms describe the salary for this position in the first 6 years on this job. (B) Write the general term of the sequence in part A. 6
(C) Find the value of the series a an. What does this number n00021 represent?
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8-2
Mathematical Induction
511
Mathematical Induction Z Using Counterexamples Z Using Mathematical Induction Z Additional Examples of Mathematical Induction Z Three Famous Problems
Many of the most important facts and formulas in this book have been stated as theorems. But a theorem is not a theorem until it has been proved, and proving theorems is one of the most challenging tasks in mathematics. There is a big difference between being pretty sure that a statement is true, and proving that statement. Let’s look at an example. Suppose that we are interested in the sum of the first n consecutive odd integers, where n is a positive integer. We can begin by writing the sums for the first few values of n to see if we can observe a pattern: 1⫽ 1 1⫹3⫽ 4 1⫹3⫹5⫽ 9 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫽ 16 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫽ 25
nⴝ1 nⴝ2 nⴝ3 nⴝ4 nⴝ5
Is there any pattern to the sums 1, 4, 9, 16, and 25? You most likely noticed that each is a perfect square and, in fact, each is the square of the number of terms in the sum. So the following conjecture* seems reasonable: CONJECTURE P :
For each positive integer n, 1 ⫹ 3 ⫹ 5 ⫹ . . . ⫹ (2n ⫺ 1) ⫽ n2
(Recall that the general term 2n ⫺ 1 was used to list the odd positive integers in the last section.) At this point, you may be pretty sure that our conjecture is true. You might even look at the previous five calculations and think that we have proved our conjecture. But in actuality, all we have proved is that the conjecture is true for n ⫽ 1, 2, 3, 4, and 5. We are trying to prove that it is true for every positive integer, not just those five! With that in mind, continuing to check the conjecture for specific n’s like 6, 7, 8, . . . is pointless: You can keep trying for the rest of your life, but you will never be able to check every positive integer. Instead, in this section, we will use a much more powerful tool called mathematical induction to prove conjectures. Before we learn about this method of proof, we first consider how to prove that a conjecture is false.
Z Using Counterexamples
Table 1 n
n2 ⴚ n ⴙ 41
1
Prime?
Consider the following conjecture:
41
Yes
2
43
Yes
3
47
Yes
4
53
Yes
For each positive integer n, the number n2 ⫺ n ⫹ 41 is a prime number. Since the conjecture states that this fact is true for every positive integer n, if we can find even one positive integer n for which it is false, then the conjecture will be proved false. A single case or example for which a conjecture fails is called a counterexample. We checked the conjecture for a few particular cases in Table 1. From the table, it certainly appears
5
61
Yes
CONJECTURE Q:
*A conjecture is a statement that is believed to be true, but has not been proved.
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that conjecture Q has a good chance of being true. You may want to check a few more cases. If you persist, you will find that conjecture Q is true for n up to 40. Most students would guess that the statement is always true long before getting to n 0002 41. But then something interesting happens at n 0002 41: 412 0003 41 0004 41 0002 412 which is not prime. Because n 0002 41 provides a counterexample, conjecture Q is false. Here we see the danger of generalizing without proof from a few special cases, even if that “few” is 40 cases! This example was discovered by Euler (1701–1783), the same mathematician that introduced the number e as the base of the natural exponential function.
EXAMPLE
1
Finding a Counterexample Prove that the following conjecture is false by finding a counterexample: For every positive integer n 0005 2, at least half of the positive integers less than or equal to n are prime.
SOLUTION
We will check the conjecture for positive integer values of n starting at 2.
n
Primes less than or equal to n
Fraction of positive integers less than or equal to n that are prime
True or false
2
2
1/2
True
3
2, 3
2/3
True
4
2, 3
2/4
True
5
2, 3, 5
3/5
True
6
2, 3, 5
3/6
True
7
2, 3, 5, 7
4/7
True
8
2, 3, 5, 7
4/8
True
9
2, 3, 5, 7
4/9
False
Since n 0002 9 provides a counterexample, the conjecture is false. MATCHED PROBLEM 1
0002
Prove that the following conjecture is false by finding a counterexample: For every positive integer n, the last digit of n3 is less than 9. 0002
Z Using Mathematical Induction To begin our study of proving conjectures, we will state the principle of mathematical induction, which forms the basis for all of our work in this section. Z THEOREM 1 Principle of Mathematical Induction Let Pn be a statement associated with each positive integer n, and suppose the following conditions are satisfied: 1. P1 is true. 2. For any positive integer k, if Pk is true, then Pk00041 is also true. Then the statement Pn is true for all positive integers n.
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513
Theorem 1 must be read very carefully. At first glance, it seems to say that if we assume a statement is true, then it is true. But that is not the case at all. If the two conditions in Theorem 1 are satisfied, then we can reason as follows: P1 P2 P3 P4
is true.
Condition 1
is true, because P1 is true. is true, because P2 is true. is true, because P3 is true. . . .
Condition 2 Condition 2 Condition 2
. . .
Because this chain of implications never ends, we will eventually reach Pn for any positive integer n. This is not the same as checking each case separately: The truth of any case follows from knowing that the previous one is true once we have established condition 2. To help visualize this process, picture a row of dominoes that goes on forever (Fig. 1) and interpret the conditions in Theorem 1 as follows: Condition 1 says that the first domino can be pushed over. Condition 2 says that if the kth domino falls, then so does the (k 0004 1)st domino. Together, these two conditions imply that all the dominoes must fall.
Condition 1: The first domino can be pushed over. (a)
Condition 2: If the kth domino falls, then so does the (k ⴙ 1)st. (b)
Conclusion: All the dominoes will fall. (c)
Z Figure 1 Interpreting mathematical induction.
In Example 2 we illustrate proof by mathematical induction by returning to our conjecture P from the beginning of the section.
EXAMPLE
2
Proving a Conjecture Using Induction Prove that for all positive integers n, 1 0004 3 0004 5 0004 . . . 0004 (2n 0003 1) 0002 n2
SOLUTION
State Pn: Pn: 1 0004 3 0004 5 0004 . . . 0004 (2n 0003 1) 0002 n2 CONDITION 1 Show that P1 is true.
P1: 1 0002 12 CONDITION 2 Show that if Pk is true, then Pk00041 must be true.
It’s a good idea to always write out both Pk and Pk00041 at the beginning of this step to see what we can use, and what we need to prove.
Pk: 1 0004 3 0004 5 0004 . . . 0004 (2k 0003 1) 0002 k2 We assume this is a true statement. Pk00041: 1 0004 3 0004 5 0004 . . . 0004 (2k 0003 1) 0004 [2(k 0004 1) 0003 1] 0002 (k 0004 1)2
We need to show that this is also true.
Note that Pk00041 can be simplified a bit: Pk00041: 1 0004 3 0004 5 0004 . . . 0004 (2k 0003 1) 0004 (2k 0004 1) 0002 (k 0004 1)2
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We will perform algebraic operations on the equation Pk (which we know is true) with a goal of obtaining Pk⫹1. Note that the left side of Pk⫹1 is the left side of Pk plus the addition term 2k ⫹ 1. 1 ⫹ 3 ⫹ 5 ⫹ . . . ⫹ (2k ⫺ 1) ⫽ k2 Add 2k ⴙ 1 to both sides. 1 ⫹ 3 ⫹ 5 ⫹ . . . ⫹ (2k ⫺ 1) ⫹ (2k ⴙ 1) ⫽ k2 ⫹ 2k ⴙ 1 Factor the right side. 1 ⫹ 3 ⫹ 5 ⫹ . . . ⫹ (2k ⫺ 1) ⫹ (2k ⫹ 1) ⫽ (k ⫹ 1)2 This is Pkⴙ1! Pk⫹1 was obtained by adding the same number to both sides of Pk, so if Pk is true, then Pk⫹1 must be as well. CONCLUSION
Both conditions of Theorem 1 are satisfied, so Pn is true for all positive integers n. MATCHED PROBLEM 2
0002
Prove that for all positive integers n 1⫹2⫹3⫹...⫹n⫽
n(n ⫹ 1) 2 0002
Z Additional Examples of Mathematical Induction Now we will consider some additional examples of proof by induction. The first is another summation formula. Mathematical induction is the primary tool for proving that formulas of this type are true.
EXAMPLE
3
Proving a Summation Formula Prove that for all positive integers n 1 1 1 2n ⫺ 1 1 ⫹ ⫹ ⫹...⫹ n ⫽ 2 4 8 2 2n PROOF State Pn:
Pn :
1 1 1 1 2n ⫺ 1 ⫹ ⫹ ⫹...⫹ n ⫽ 2 4 8 2 2n
PART 1 Show that P1 is true.
P1:
1 21 ⫺ 1 ⫽ 2 21 ⫽
1 2
So P1 is true. PART 2 Show that if Pk is true, then Pk⫹1 is true. Again, it is a good idea to always write
out both Pk and Pk⫹1 at the beginning of any induction proof to see what is assumed and what must be proved: 1 1 1 1 2k ⫺ 1 ⫹ ⫹ ⫹... ⫹ k ⫽ 2 4 8 2 2k k⫹1 1 1 1 1 1 2 ⫺1 ⫹ ⫹ ⫹ . . . ⫹ k ⫹ k⫹1 ⫽ k⫹1 2 4 8 2 2 2 Pk :
Pk⫹1:
We assume Pk is true.
We must show that Pkⴙ1 follows from Pk.
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We start with the true statement Pk, add 1 Ⲑ2k⫹1 to both sides, and simplify the right side: 1 1 1 1 2k ⫺ 1 ⫹ ⫹ ⫹...⫹ k ⫽ 2 4 8 2 2k 1 1 1 1 1 2k ⫺ 1 1 ⫹ ⫹ ⫹ . . . ⫹ k ⴙ kⴙ1 ⫽ ⴙ kⴙ1 k 2 4 8 2 2 2 2 ⫽ ⫽ ⫽
2k ⫺ 1
ⴢ
k
2
2 1 ⫹ k⫹1 2 2
2k⫹1 ⫺ 2 ⫹ 1 2k⫹1
Add
1 2k⫹1
to both sides.
Find common denominator for right-hand side.
Write as single fraction.
Simplify.
2k⫹1 ⫺ 1 2k⫹1
So 1 1 1 1 1 2k⫹1 ⫺ 1 ⫹ ⫹ ⫹ . . . ⫹ k ⫹ k⫹1 ⫽ 2 4 8 2 2 2k⫹1
Pkⴙ1
and we have shown that if Pk is true, then Pk⫹1 is true. CONCLUSION
Both conditions in Theorem 1 are satisfied. Therefore, Pn is true for all positive integers n. 0002 MATCHED PROBLEM 3
Prove that for all positive integers n 2 2 2 3n ⫺ 1 2 ⫹ ⫹ ⫹p⫹ n⫽ 3 9 27 3 3n 0002 Example 4 provides a proof of a law of exponents that previously we had to assume was true. First we redefine an for n a positive integer, using a recursion formula: n Z DEFINITION 1 Recursive Definition of a
For n a positive integer a1 ⫽ a an⫹1 ⫽ ana
EXAMPLE
4
n 7 1
Proving a Law of Exponents Prove that (xy)n ⫽ xnyn for all positive integers n. PROOF State Pn:
Pn: (xy)n ⫽ x ny n PART 1 Show that P1 is true.
(xy)1 ⫽ xy ⫽ x1y1 So P1 is true.
Definition 1 Definition 1
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PART 2
Show that if Pk is true, then Pk00041 is true. Pk : (xy)k 0002 xkyk Pk00041: (xy)k00041 0002 xk00041yk00041
Assume Pk is true. Show that Pkⴙ1 follows from Pk.
Here we start with the left side of Pk00041 and use Pk to find the right side of Pk00041: (xy)k00041 0002 (xy)k(xy)1 0002 x ky kxy 0002 (x kx)(y ky) 0002 x k00041y k00041
Use Pk: (xy) k ⴝ x ky k Use properties of real numbers. Use Definition 1 twice.
So (xy)k00041 0002 x k00041y k00041, and we have shown that if Pk is true, then Pk00041 is true. CONCLUSION
Both conditions in Theorem 1 are satisfied. Therefore, Pn is true for all positive integers n. 0002 MATCHED PROBLEM 4
Prove that (x/y)n 0002 x n/y n for all positive integers n.
0002
Example 5 deals with factors of integers. Before we start, recall that an integer p is divisible by an integer q if p 0002 qr for some integer r.
EXAMPLE
5
Proving a Divisibility Property Prove that 42n 0003 1 is divisible by 5 for all positive integers n. PROOF Use the definition of divisibility to state Pn as follows:
Pn: 42n 0003 1 0002 5r
for some integer r
PART 1 Show that P1 is true.
P1: 42 0003 1 0002 15 0002 5 ⴢ 3 So P1 is true. PART 2 Show that if Pk is true, then Pk00041 is true.
Pk: 42k 0003 1 0002 5r Pk00041: 42(k00041) 0003 1 0002 5s
for some integer r for some integer s
Assume Pk is true. Show that Pkⴙ1 must follow.
As before, we start with the true statement Pk: 42k 0003 1 0002 5r 42(42k 0003 1) 0002 42(5r) 42k00042 0003 16 0002 80r 42(k00041) 0003 1 0002 80r 0004 15 0002 5(16r 0004 3) So
42(k00041) 0003 1 0002 5s
Multiply both sides by 42. Simplify. Add 15 to both sides. Factor out 5.
Pkⴙ1
where s 0002 16r 0004 3 is an integer, and we have shown that if Pk is true, then Pk00041 is true. CONCLUSION
Both conditions in Theorem 1 are satisfied. Therefore, Pn is true for all positive integers n. 0002
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MATCHED PROBLEM 5
Mathematical Induction
517
Prove that 8n 0003 1 is divisible by 7 for all positive integers n.
0002
In some cases, a conjecture may be true only for n 0005 m, where m is a positive integer, rather than for all n 0005 0. For example, see Problems 53 and 54 in Exercises 8-2. The principle of mathematical induction can be extended to cover cases like this as follows:
Z THEOREM 2 Extended Principle of Mathematical Induction Let m be a positive integer, let Pn be a statement associated with each integer n 0005 m, and suppose the following conditions are satisfied: 1. Pm is true. 2. For any integer k 0005 m, if Pk is true, then Pk00041 is also true. Then the statement Pn is true for all integers n 0005 m.
Z Three Famous Problems The problem of determining whether a certain statement about the positive integers is true may be extremely difficult. Proofs may require remarkable insight and ingenuity and the development of techniques far more advanced than mathematical induction. Consider, for example, the famous problems of proving the following statements: 1. 2. 3.
Lagrange’s Four Square Theorem, 1772: Each positive integer can be expressed as the sum of four or fewer squares of positive integers. Fermat’s Last Theorem, 1637: For n 7 2, xn 0004 yn 0002 z n does not have solutions in the natural numbers. Goldbach’s Conjecture, 1742: Every positive even integer greater than 2 is the sum of two prime numbers.
The first statement was considered by the early Greeks and finally proved in 1772 by Lagrange. Fermat’s last theorem, defying the best mathematical minds for over 350 years, finally succumbed to a 200-page proof by Professor Andrew Wiles of Princeton University in 1993. To this date no one has been able to prove or disprove Goldbach’s conjecture.
ZZZ EXPLORE-DISCUSS 1
(A) Explain the difference between a theorem and a conjecture. (B) Why is “Fermat’s last theorem” a misnomer? Suggest more accurate names for the result.
ANSWERS TO MATCHED PROBLEMS 1. The last digit of 93 0002 729 is greater than 8. 2. Sketch of proof. n(n 0004 1) Pn: 1 0004 2 0004 3 0004 . . . 0004 n 0002 2 1(1 0004 1) Condition 1. 1 0002 . P1 is true. 2
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Condition 2. Show that if Pk is true, then Pk⫹1 is true. k(k ⫹ 1) 1⫹2⫹3⫹...⫹k⫽ Pk 2 k(k ⫹ 1) 1 ⫹ 2 ⫹ 3 ⫹ . . . ⫹ k ⴙ (k ⴙ 1) ⫽ ⴙ (k ⴙ 1) 2 (k ⫹ 1)(k ⫹ 2) ⫽ Pkⴙ1 2 Conclusion: Pn is true for all positive integers n. 2 2 2 2 3n ⫺ 1 ⫹ ###⫹ n⫽ 3. Sketch of proof. Pn: ⫹ ⫹ 3 9 27 3 3n 2 31 ⫺ 1 . P1 is true. Part 1. ⫽ 3 3 Part 2. Show that if Pk is true, then Pk⫹1 is true. 2 2 2 2 3k ⫺ 1 Pk ⫹ ⫹ ⫹p⫹ k⫽ 3 9 27 3 3k 2 2 2 2 2 3k ⫺ 1 2 ⫹ k⫹1 ⫹ ⫹ ⫹ p ⫹ k ⫹ k⫹1 ⫽ 3 9 27 3 3 3k 3 3k⫹1 ⫺ 1 Pk ⴙ 1 ⫽ 3k Conclusion: Pn is true for all positive integers n. x n xn 4. Sketch of proof. Pn: a b ⫽ n y y x 1 x x1 Part 1. a b ⫽ ⫽ 1 . P1 is true. y y y Part 2. Show that if Pk is true, then Pk⫹1 is true. xk⫹1 x k⫹1 x k x xk x xkx a b ⫽ a b a b ⫽ k a b ⫽ k ⫽ k⫹1 y y y y y yy y Conclusion: Pn is true for all positive integers n. 5. Sketch of proof. Pn: 8n ⫺ 1 ⫽ 7r for some integer r Part 1. 81 ⫺ 1 ⫽ 7 ⫽ 7 ⴢ 1. P1 is true. Part 2. Show that if Pk is true, then Pk⫹1 is true. Pk 8k ⫺ 1 ⫽ 7r 8(8k ⫺ 1) ⫽ 8(7r) 8k⫹1 ⫺ 1 ⫽ 56r ⫹ 7 ⫽ 7(8r ⫹ 1) ⫽ 7s Pkⴙ1 Conclusion: Pn is true for all positive integers n.
8-2
Exercises
1. What is a counterexample? 2. Explain how falling dominoes can be compared to the principle of mathematical induction. 3. In Theorem 1 (principle of mathematical induction), what do Pk and Pk⫹1 represent? 4. The number n2 ⫺ n ⫹ 41 is prime for n ⫽ 1, 2, . . . , 40. Does this prove that n2 ⫺ n ⫹ 41 is prime for every natural number n? Explain.
In Problems 5–8, find the first positive integer n that causes the statement to fail. 5. (3 ⫹ 5)n ⫽ 3n ⫹ 5n 2
7. n ⫽ 3n ⫺ 2
6. n ⬍ 10 8. n3 ⫹ 11n ⫽ 6n2 ⫹ 6
Verify each statement Pn in Problems 9–14 for n ⫽ 1, 2, and 3. 9. Pn: 2 ⫹ 6 ⫹ 10 ⫹ ⭈ ⭈ ⭈ ⫹ (4n ⫺ 2) ⫽ 2n2 10. Pn: 4 ⫹ 8 ⫹ 12 ⫹ ⭈ ⭈ ⭈ ⫹ 4n ⫽ 2n(n ⫹ 1)
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11. Pn: a5an 0002 a50004n
12. Pn: (a5)n 0002 a5n
Mathematical Induction
519
43. x2n 0003 1 is divisible by x 0003 1; x 1
13. Pn: 9n 0003 1 is divisible by 4
44. x2n 0003 1 is divisible by x 0004 1; x 00031
14. Pn: 4n 0003 1 is divisible by 3
45. 13 0004 23 0004 33 0004 0004 n3 0002 (1 0004 2 0004 3 0004 0004 n)2 [Hint: See Matched Problem 2 following Example 2.]
Write Pk and Pk00041 for Pn as indicated in Problems 15–20. 15. Pn in Problem 9
16. Pn in Problem 10
17. Pn in Problem 11
18. Pn in Problem 12
19. Pn in Problem 13
20. Pn in Problem 14
In Problems 21–26, use mathematical induction to prove that each Pn holds for all positive integers n. 21. Pn in Problem 9
22. Pn in Problem 10
23. Pn in Problem 11
24. Pn in Problem 12
25. Pn in Problem 13
26. Pn in Problem 14
In Problems 27–30, prove the statement is false by finding a counterexample. 27. If n 2, then any polynomial of degree n has at least one real zero.
46.
1 1 1 0004 0004 0004 ### 1ⴢ2ⴢ3 2ⴢ3ⴢ4 3ⴢ4ⴢ5 0004
n(n 0004 3) 1 0002 n(n 0004 1)(n 0004 2) 4(n 0004 1)(n 0004 2)
In Problems 47–50, suggest a formula for each expression, and prove your conjecture using mathematical induction, n 僆 N. 47. 2 0004 4 0004 6 0004 0004 2n 48.
1 1 1 1 0004 0004 0004 ###0004 1ⴢ2 2ⴢ3 3ⴢ4 n(n 0004 1)
49. The number of lines determined by n points in a plane, no three of which are collinear 50. The number of diagonals in a polygon with n sides Prove Problems 51–54 true for all integers n as specified. 51. If a 1, then an 1; n 僆 N
28. Any positive integer n 7 can be written as the sum of three or fewer squares of positive integers.
52. If 0 6 a 6 1, then 0 6 an 6 1; n 僆 N
29. If n is a positive integer, then there is at least one prime number p such that n p n 0004 6.
54. 2n n2; n 0005 5
30. If a, b, c, and d are positive integers such that a2 0004 b2 0002 c2 0004 d 2, then a 0002 c or a 0002 d. In Problems 31–46, use mathematical induction to prove each proposition for all positive integers n, unless restricted otherwise. 31. 2 0004 22 0004 23 0004 0004 2n 0002 2n00041 0003 2 32.
33. 12 0004 32 0004 52 0004 # # # 0004 (2n 0003 1)2 0002 13 (4n3 0003 n) 2
34. 1 0004 8 0004 16 0004 0004 8(n 0003 1) 0002 (2n 0003 1) ; n 1 n(n 0004 1)(2n 0004 1) 6
n(n 0004 1)(n 0004 2) 36. 1 ⴢ 2 0004 2 ⴢ 3 0004 3 ⴢ 4 0004 # # # 0004 n(n 0004 1) 0002 3 37.
an 0002 an00033; n 7 3 a3
38.
a5 1 0002 n00035 ; n 7 5 an a
39. aman 0002 am0004n; m, n 僆 N [Hint: Choose m as an arbitrary element of N, and then use induction on n.] n m
mn
40. (a ) 0002 a ; m, n 僆 N n
41. x 0003 1 is divisible by x 0003 1, x 1 [Hint: Divisible means that xn 0003 1 0002 (x 0003 1)Q(x) for some polynomial Q(x).] 42. xn 0003 yn is divisible by x 0003 y; x y
In Problems 55–58, determine whether the statement is true or false. If true, prove using mathematical induction. If false, find a counterexample. 55. If n is a positive integer, then 1 0003 2 0004 3 0003 0004 (2n 0003 1) 0002 n (that is, the alternating sum of the first 2n 0003 1 positive integers is equal to n).
1 1 n 1 1 1 0004 0004 0004...0004 n 000210003a b 2 4 8 2 2
35. 12 0004 22 0004 32 0004 . . . 0004 n2 0002
53. n2 2n; n 0005 3
56. If n is a positive integer, then (00031)n00041n(n 0004 1) 12 0003 22 0004 32 0003 # # # 0004 (00031)n00041n2 0002 2 57. If n is a positive integer, then 3n00041 0004 4n00041 0004 0004 (n 0004 3)n00041 0002 (n 0004 4)n00041 58. If n is a positive integer, then n2 0004 21n 0004 1 is a prime number. If {an} and {bn} are two sequences, we write {an} 0002 {bn} if and only if an 0002 bn for all n 僆 N. In Problems 59–62, use mathematical induction to show that {an} 0002 {bn}. 59. a1 0002 1, an 0002 an00031 0004 2; bn 0002 2n 0003 1 60. a1 0002 2, an 0002 an00031 0004 2; bn 0002 2n 61. a1 0002 2, an 0002 22an00031; bn 0002 22n00031 62. a1 0002 2, an 0002 3an00031; bn 0002 2 ⴢ 3n00031
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Arithmetic and Geometric Sequences Z Arithmetic and Geometric Sequences Z Developing nth-Term Formulas Z Developing Sum Formulas for Finite Arithmetic Series Z Developing Sum Formulas for Finite Geometric Series Z Developing a Sum Formula for Infinite Geometric Series
For most sequences, it is difficult to add up an arbitrary number of terms of the sequence without adding the terms one at a time. In this section, we will study two special types of sequences, arithmetic sequences and geometric sequences. One of the things that make them special is that we can develop formulas for the sum of the corresponding series.
Z Arithmetic and Geometric Sequences Consider the sequence defined by the general term an 0002 5 0004 2(n 0003 1), n 0005 1. The first five terms are 5, 7, 9, 11, and 13. It’s not hard to see that after starting at 5, every term is obtained by adding 2 to the previous term. This is an example of an arithmetic sequence.
Z DEFINITION 1 Arithmetic Sequence A sequence a1, a2, a3, . . . , an, . . . is called an arithmetic sequence, or arithmetic progression, if there exists a constant d, called the common difference, such that an 0003 an00031 0002 d That is, an 0002 an00031 0004 d
for every n 1
In short, a sequence is arithmetic when every term is obtained by adding some fixed number to the previous term. This fixed number is called the common difference, and is usually represented by the letter d. Now consider the sequence with general term an 0002 5(2)n00031. The first five terms are 5, 10, 20, 40, and 80. It also starts at 5, but this time every term is obtained by multiplying the previous term by 2. This is an example of a geometric sequence.
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521
Z DEFINITION 2 Geometric Sequence A sequence a1, a2, a3, . . . , an, . . . is called a geometric sequence, or geometric progression, if there exists a nonzero constant r, called the common ratio, such that an 0002r an00031 That is, an 0002 ran00031
for every n 1
In short, a sequence is geometric when every term is obtained by multiplying the previous term by some fixed number. This fixed number is called the common ratio, and is usually represented by the letter r.
ZZZ EXPLORE-DISCUSS 1
(A) Graph the arithmetic sequence 5, 7, 9, . . . . Describe the graphs of all arithmetic sequences with common difference 2. (B) Graph the geometric sequence 5, 10, 20, . . . . Describe the graphs of all geometric sequences with common ratio 2.
EXAMPLE
1
Recognizing Arithmetic and Geometric Sequences Which of the following can be the first four terms of an arithmetic sequence? Of a geometric sequence?
SOLUTIONS
(A) 1, 2, 3, 5, . . .
(B) 00031, 3, 00039, 27, . . .
(C) 3, 3, 3, 3, . . .
(D) 10, 8.5, 7, 5.5, . . .
(A) Because 2 0003 1 5 0003 3, there is no common difference, so the sequence is not an arithmetic sequence. Because 21 32, there is no common ratio, so the sequence is not geometric either. (B) The sequence is geometric with common ratio 00033, but it is not arithmetic. (C) The sequence is arithmetic with common difference 0 and it is also geometric with common ratio 1. (D) The sequence is arithmetic with common difference 00031.5, but it is not geometric.
MATCHED PROBLEM 1
0002
Which of the following can be the first four terms of an arithmetic sequence? Of a geometric sequence? (A) 8, 2, 0.5, 0.125, . . .
(B) 00037, 00032, 3, 8, . . .
(C) 1, 5, 25, 100, . . . 0002
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Z Developing nth-Term Formulas If {an} is an arithmetic sequence with common difference d, then a2 0002 a1 0004 d a3 0002 a2 0004 d 0002 a1 0004 2d a4 0002 a3 0004 d 0002 a1 0004 3d This suggests Theorem 1, which can be proved by mathematical induction (see Problem 67 in Exercises 8-3). Z THEOREM 1 The nth Term of an Arithmetic Sequence an 0002 a1 0004 (n 0003 1)d
for every n 1
Similarly, if {an} is a geometric sequence with common ratio r, then a2 0002 a1r a3 0002 a2r 0002 a1r 2 a4 0002 a3r 0002 a1r 3 This suggests Theorem 2, which can also be proved by mathematical induction (see Problem 71 in Exercises 8-3). Z THEOREM 2 The nth Term of a Geometric Sequence an 0002 a1r n00031
EXAMPLE
2
for every n 1
Finding Terms in Arithmetic and Geometric Sequences (A) If the first and tenth terms of an arithmetic sequence are 3 and 30, respectively, find the fiftieth term of the sequence. (B) If the first and tenth terms of a geometric sequence are 1 and 4, find the seventeenth term to three decimal places.
SOLUTIONS
(A) First use Theorem 1 with a1 0002 3 and a10 0002 30 to find d: an 0002 a1 0004 (n 0003 1)d a10 0002 a1 0004 (10 0003 1)d 30 0002 3 0004 9d d00023
Substitute n ⴝ 10. Substitute a10 ⴝ 30 and a1 ⴝ 3. Solve for d.
Now find a50: a50 0002 a1 0004 (50 0003 1)3 0002 3 0004 49 ⴢ 3 0002 150
Substitute a1 ⴝ 3. Simplify.
(B) First let n 0002 10, a1 0002 1, a10 0002 4 and use Theorem 2 to find r. an 0002 a1rn00031 4 0002 1r1000031 r 0002 4100029
Substitute n ⴝ 10, a10 ⴝ 4, and a1 ⴝ 1. Solve for r.
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523
Now use Theorem 2 again, this time with n 0002 17. a17 0002 a1r16 0002 1(4100029)16 0002 41600029 0003 11.758 MATCHED PROBLEM 2
0002
(A) If the first and fifteenth terms of an arithmetic sequence are 00035 and 23, respectively, find the seventy-third term of the sequence. 1 1 1 (B) Find the eighth term of the geometric sequence , 0003 , , . . . . 64 32 16 0002
Z Developing Sum Formulas for Finite Arithmetic Series If a1, a2, a3, . . . , an is a finite arithmetic sequence, then the corresponding series a1 0004 a2 0004 a3 0004 . . . 0004 an is called an arithmetic series. We will derive two simple and very useful formulas for the sum of an arithmetic series. Let d be the common difference of the arithmetic sequence a1, a2, a3, . . . , an and let Sn denote the sum of the series a1 0004 a2 0004 a3 0004 . . . 0004 an. Then Sn 0002 a1 0004 (a1 0004 d) 0004 . . . 0004 [a1 0004 (n 0003 2)d ] 0004 [a1 0004 (n 0003 1)d] Reversing the order of the sum, we obtain Sn 0002 [a1 0004 (n 0003 1)d ] 0004 [a1 0004 (n 0003 2)d ] 0004 . . . 0004 (a1 0004 d ) 0004 a1 Adding the left sides of these two equations and corresponding elements of the right sides, we see that 2Sn 0002 [2a1 0004 (n 0003 1)d ] 0004 [2a1 0004 (n 0003 1)d] 0004 . . . 0004 [2a1 0004 (n 0003 1)d ] 0002 n[2a1 0004 (n 0003 1)d ] This can be restated as in Theorem 3.
Z THEOREM 3 Sum of an Arithmetic Series—First Form n Sn 0002 [2a1 0004 (n 0003 1)d] 2
By replacing a1 0004 (n 0003 1)d with an, we obtain a second useful formula for the sum.
Z THEOREM 4 Sum of an Arithmetic Series—Second Form n Sn 0002 (a1 0004 an) 2
The proof of the first sum formula by mathematical induction is left as an exercise (see Problem 68 in Exercises 8-3).
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3
Finding the Sum of an Arithmetic Series Find the sum of the first 26 terms of an arithmetic series if the first term is 00037 and d 0002 3.
SOLUTION
Let n 0002 26, a1 0002 00037, d 0002 3, and use Theorem 3. n [2a1 0004 (n 0003 1)d] 2 0002 262 [2(00037) 0004 (26 0003 1)3] 0002 793
Sn 0002 S26
MATCHED PROBLEM 3
EXAMPLE
4
Substitute n ⴝ 26, a1 ⴝ ⴚ7, and d ⴝ 3. Simplify.
0002
Find the sum of the first 52 terms of an arithmetic series if the first term is 23 and d 0002 00032. 0002
Finding the Sum of an Arithmetic Series Find the sum of all the odd numbers between 51 and 99, inclusive.
SOLUTION
First, use a1 0002 51, an 0002 99, and Theorem 1 to find n: an 0002 a1 0004 (n 0003 1)d 99 0002 51 0004 (n 0003 1)2 n 0002 25
Substitute an ⴝ 99, a1 ⴝ 51, and d ⴝ 2. Solve for n.
Now use Theorem 4 to find S25: n Sn 0002 (a1 0004 an) 2 S25 0002 252 (51 0004 99)
Substitute n ⴝ 25, a1 ⴝ 51, and an ⴝ 99.
0002 1,875 MATCHED PROBLEM 4
EXAMPLE
5
0002
Find the sum of all the even numbers between 000322 and 52, inclusive.
0002
Prize Money A 16-team bowling league has $8,000 to be awarded as prize money. If the last-place team is awarded $275 in prize money and the award increases by the same amount for each successive finishing place, how much will the first-place team receive?
SOLUTION
If a1 is the award for the first-place team, a2 is the award for the second-place team, and so on, then the prize money awards form an arithmetic sequence with n 0002 16, a16 0002 275, and S16 0002 8,000. Use Theorem 4 to find a1. n Sn 0002 (a1 0004 an) 2 16 8,000 0002 2 (a1 0004 275) a1 0002 725
Substitute n ⴝ 16, S16 ⴝ 8,000, a16 ⴝ 275. Solve for a1.
The first-place team receives $725. MATCHED PROBLEM 5
0002
Refer to Example 5. How much prize money is awarded to the second-place team? 0002
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Z Developing Sum Formulas for Finite Geometric Series If a1, a2, a3, . . . , an is a finite geometric sequence, then the corresponding series a1 0004 a2 0004 a3 0004 . . . 0004 an is called a geometric series. As with arithmetic series, we can derive two simple and very useful formulas for the sum of a geometric series. Let r be the common ratio of the geometric sequence a1, a2, a3, . . . , an and let Sn denote the sum of the series a1 0004 a2 0004 a3 0004 . . . 0004 an. Then Sn 0002 a1 0004 a1r 0004 a1r 2 0004 a1r 3 0004 . . . 0004 a1r n00032 0004 a1r n00031 Multiply both sides of this equation by r to obtain rSn 0002 a1r 0004 a1r 2 0004 a1r 3 0004 . . . 0004 a1r n00031 0004 a1r n Now subtract the left side of the second equation from the left side of the first, and the right side of the second equation from the right side of the first to obtain Sn 0003 rSn 0002 a1 0003 a1r n Sn(1 0003 r) 0002 a1 0003 a1r n
Factor out Sn
Solving for Sn, we obtain the following formula for the sum of a geometric series: Z THEOREM 5 Sum of a Geometric Series—First Form Sn 0002
a1 0003 a1r n 10003r
r 1
Because an 0002 a1r n00031, or ran 0002 a1r n, the sum formula also can be written in the following form: Z THEOREM 6 Sum of a Geometric Series—Second Form Sn 0002
a1 0003 ran 10003r
r 1
The proof of the first sum formula (Theorem 5) by mathematical induction is left as an exercise (see Problem 72, Exercises 8-3). If r 0002 1, then Sn 0002 a1 0004 a1(1) 0004 a1(12) 0004 . . . 0004 a1(1n00031) 0002 na1
EXAMPLE
6
Finding the Sum of a Geometric Series Find the sum of the first 20 terms of a geometric series if the first term is 1 and r 0002 2.
SOLUTION
Let n 0002 20, a1 0002 1, r 0002 2, and use Theorem 5. Sn 0002 0002
a1 0003 a1r n 10003r 1 0003 1 ⴢ 220 0002 1,048,575 100032
Substitute n ⴝ 20, a1 ⴝ 1, and r ⴝ 2.
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Technology Connections To calculate the sum of a series with a graphing calculator, first generate the sequence using the sequence command, then find its sum using the sum command. Figure 1 shows the solution to Example 6.
Z Figure 1
0002 MATCHED PROBLEM 6
Find the sum, to two decimal places, of the first 14 terms of a geometric series if the first term is 641 and r 0002 00032. 0002
Z Developing a Sum Formula for Infinite Geometric Series Consider a geometric series with a1 0002 5 and r 0002 12. What happens to the sum Sn as n increases? To answer this question, we first write the sum formula in the more convenient form Sn 0002
a1 0003 a1r n a1 a1r n 0002 0003 10003r 10003r 10003r
(1)
For a1 0002 5 and r 0002 12, 1 n Sn 0002 10 0003 10 a b 2 Let’s look at some of the Sn s: 1 S2 0002 10 0003 10 a b 0002 7.5 4 1 S3 0002 10 0003 10 a b 0002 8.75 8 1 S4 0002 10 0003 10 a b 0002 9.375 16 o S20 0002 10 0003 10 a
1 b 0002 9.999990 p 1,048,576
It appears that (12)n becomes smaller and smaller as n increases and that the sum gets closer and closer to 10. In general, it is possible to show that, if 0004 r 0004 6 1, then r n will get closer and closer to 0 as n increases. Symbolically, r n S 0 as n S . So the term a1r n 10003r in equation (1) will tend to 0 as n increases, and Sn will tend to a1 10003r
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In other words, if 0004 r 0004 6 1, then Sn can be made as close to a1 10003r as we wish by taking n sufficiently large. So we can define the sum of an infinite geometric series by the following formula:
Z DEFINITION 3 Sum of an Infinite Geometric Series S 0002
a1 10003r
0004r0004 6 1
If 0004 r 0004 0005 1, an infinite geometric series has no sum.
EXAMPLE
7
Expressing a Repeating Decimal as a Fraction Represent the repeating decimal 0.454 545 . . . 0002 0.45 as the quotient of two integers. Recall that a repeating decimal names a rational number and that any rational number can be represented as the quotient of two integers.
SOLUTION
0.45 0002 0.45 0004 0.0045 0004 0.000 045 0004 . . . The right side of the equation is an infinite geometric series with a1 0002 0.45 and r 0002 0.01. The sum is S 0002 This shows that, 0.45 and dividing 5 by 11.
MATCHED PROBLEM 7
EXAMPLE
8
5 11
a1 0.45 0.45 5 0002 0002 0002 10003r 1 0003 0.01 0.99 11
name the same rational number. You can check the result by 0002
Repeat Example 7 for 0.818 181 . . . 0002 0.81.
0002
Economy Stimulation A state government uses proceeds from a lottery to provide a tax rebate for property owners. Suppose an individual receives a $500 rebate and spends 80% of this, and each of the recipients of the money spent by this individual also spends 80% of what he or she receives, and this process continues without end. According to the multiplier doctrine in economics, the effect of the original $500 tax rebate on the economy is multiplied many times. What is the total amount spent if the process continues as indicated?
SOLUTION
The individual receives $500 and spends 0.8(500) 0002 $400. The recipients of this $400 spend 0.8(400) 0002 $320, the recipients of this $320 spend 0.8(320) 0002 $256, and so on. The total spending generated by the $500 rebate is 400 0004 320 0004 256 0004 . . . 0002 400 0004 0.8(400) 0004 (0.8)2(400) 0004 . . .
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which we recognize as an infinite geometric series with a1 0002 400 and r 0002 0.8. The total amount spent is S 0002
MATCHED PROBLEM 8
a1 400 400 0002 0002 0002 $2,000 10003r 1 0003 0.8 0.2
0002
Repeat Example 8 if the tax rebate is $1,000 and the percentage spent by all recipients is 90%. 0002
(A) Find an infinite geometric series with a1 0002 10 whose sum is 1,000.
ZZZ EXPLORE-DISCUSS 2
(B) Find an infinite geometric series with a1 0002 10 whose sum is 6. (C) Suppose that an infinite geometric series with a1 0002 10 has a sum. Explain why that sum must be greater than 5.
ANSWERS TO MATCHED PROBLEMS 1. (A) The sequence is geometric with r 0002 14, but not arithmetic. (B) The sequence is arithmetic with d 0002 5, but not geometric. (C) The sequence is neither arithmetic nor geometric. 2. (A) 139 (B) 00032 3. 00031,456 4. 570 5. $695 6. 000385.33 7. 119 8. $9,000
8-3
Exercises
1. What is an arithmetic sequence?
Let a1, a2, a3, . . . , an , . . . be an arithmetic sequence. In Problems 9–16, find the indicated quantities.
2. What is a geometric sequence? 3. Explain the terms “common difference” and “common ratio.” 4. Explain how a repeating decimal can be viewed as a geometric series.
9. a1 0002 00035, d 0002 4; a2 0002 ?, a3 0002 ?, a4 0002 ? 10. a1 0002 000318, d 0002 3; a2 0002 ?, a3 0002 ?, a4 0002 ? 11. a1 0002 00033, d 0002 5; a15 0002 ?, S11 0002 ?
5. Which infinite arithmetic series have a sum?
12. a1 0002 3, d 0002 4; a22 0002 ?, S21 0002 ?
6. Which infinite geometric series have a sum?
13. a1 0002 1, a2 0002 5; S21 0002 ? 14. a1 0002 5, a2 0002 11; S11 0002 ?
In Problems 7 and 8, determine whether the following can be the first three terms of an arithmetic or geometric sequence, and, if so, find the common difference or common ratio and the next two terms of the sequence. 7. (A) 000311, 000316, 000321, . . . (C) 1, 4, 9, . . .
(B) 2, 00034, 8, . . . (D)
1 1 1 2 , 6 , 18 ,
...
15. a1 0002 7, a2 0002 5; a15 0002 ? 16. a1 0002 00033, d 0002 00034; a10 0002 ? Let a1, a2, a3 , . . . , an , . . . be a geometric sequence. In Problems 17–22, find each of the indicated quantities.
8. (A) 5, 20, 100, . . .
(B) 00035, 00035, 00035, . . .
17. a1 0002 00036, r 0002 000312; a2 0002 ?, a3 0002 ?, a4 0002 ?
(C) 7, 6.5, 6, . . .
(D) 512, 256, 128, . . .
18. a1 0002 12, r 0002 23; a2 0002 ?, a3 0002 ?, a4 0002 ?
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19. a1 0002 81, r 0002 13; a10 0002 ? 20. a1 0002 64, r 0002 12; a13 0002 ? 21. a1 0002 3, a7 0002 2,187, r 0002 3; S7 0002 ? 22. a1 0002 1, a7 0002 729, r 0002 00033; S7 0002 ?
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50. Show that the sum of the first n even natural numbers is n 0004 n2, using appropriate formulas from Section 8-3. In Problems 51–60, find the sum of each infinite geometric series that has a sum. 51. 2 0004 12 0004 18 0004 . . .
Let a1, a2, a3, . . . , an , . . . be an arithmetic sequence. In Problems 23–30, find the indicated quantities. 23. a1 0002 3, a20 0002 117; d 0002 ?, a101 0002 ? 24. a1 0002 7, a8 0002 28; d 0002 ?, a25 0002 ? 25. a1 0002 000312, a40 0002 22; S40 0002 ? 26. a1 0002 24, a24 0002 000328; S24 0002 ? 27. a1 0002 13, a2 0002 12; a11 0002 ?, S11 0002 ? 28. a1 0002 16, a2 0002 14; a19 0002 ?, S19 0002 ? 29. a3 0002 13, a10 0002 55; a1 0002 ? 30. a9 0002 000312, a13 0002 3; a1 0002 ? Let a1, a2, a3, . . . , an , . . . be a geometric sequence. Find each of the indicated quantities in Problems 31–42. 31. a1 0002 8, a2 0002 2; r 0002 ? 32. a1 0002 00036, a2 0002 2; r 0002 ? 33. a1 0002 120, a4 0002 000315; r 0002 ? 34. a1 0002 12, a6 0002 8; r 0002 ? 35. a1 0002 9, r 0002 23; S10 0002 ? 36. a1 0002 3, r 0002 5; S9 0002 ? 37. a1 0002 1, a8 0002 2,187; S8 0002 ? 38. a1 0002 12, a12 0002 1,024; S12 0002 ? 39. a3 0002 72, a6 0002 0003243; a1 0002 ? 40. a4 0002 8, a5 0002 6; a1 0002 ? 41. a1 0002 1, a4 0002 00031; a100 0002 ? 42. a1 0002 00031, a8 0002 1; a99 0002 ? 51
43. S51 0002 a (3k 0004 3) 0002 ? k00021 40
44. S40 0002 a (2k 0003 3) 0002 ?
52. 6 0004 2 0004 23 0004 . . . 53. 3 0003 1 0004 13 0003 . . . 54. 1 0004 43 0004 169 0004 . . . 55. 1 0004 0.1 0004 0.01 0004 . . . 56. 10 0003 2 0004 0.4 0003 . . . 57. 00031 0004 12 0003 14 0004 . . . 58. 00036 0004 4 0003 83 0004 . . . 59. 1 0003 1 0004 10003 . . . 60. 0003100 0003 80 0003 64 0003 . . . In Problems 61–66, represent each repeating decimal as the quotient of two integers. 61. 0.7 0002 0.7777 . . .
62. 0.5 0002 0.5555 . . .
63. 0.54 0002 0.545 454 . . .
64. 0.27 0002 0.272 727 . . .
65. 3.216 0002 3.216 216 216 . . . 66. 5.63 0002 5.636 363 . . . 67. Prove, using mathematical induction, that if {an} is an arithmetic sequence, then an 0002 a1 0004 (n 0003 1)d
68. Prove, using mathematical induction, that if {an} is an arithmetic sequence, then n Sn 0002 [2a1 0004 (n 0003 1)d] 2 69. If in a given sequence, a1 0002 00032 and an 0002 00033an00031, n 7 1, find an in terms of n. n
70. For the sequence in Problem 69, find Sn 0002 a ak in terms k00021 of n. 71. Prove, using mathematical induction, that if {an} is a geometric sequence, then
k00021 7
an 0002 a1r n00031 k00031
45. S7 0002 a (00033) k00021
0002?
k00021
47. Find the sum of all the even integers between 21 and 135.
n僆N
72. Prove, using mathematical induction, that if {an} is a geometric sequence, then
7
46. S7 0002 a 3k 0002 ?
for every n 7 1
Sn 0002
a1 0003 a1r n 10003r
n 僆 N, r 1
48. Find the sum of all the odd integers between 100 and 500.
73. Is there an arithmetic sequence that is also geometric? Explain.
49. Show that the sum of the first n odd natural numbers is n2, using appropriate formulas from Section 8-3.
74. Is there an infinite geometric sequence with a1 0002 1 that has sum equal to 12? Explain.
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APPLICATIONS 75. BUSINESS In investigating different job opportunities, you find that firm A will start you at $25,000 per year and guarantee you a raise of $1,200 each year whereas firm B will start you at $28,000 per year but will guarantee you a raise of only $800 each year. Over a period of 15 years, how much would you receive from each firm?
84. PHYSICS The first swing of a bob on a pendulum is 10 inches. If on each subsequent swing it travels 0.9 as far as on the preceding swing, how far will the bob travel before coming to rest?
76. BUSINESS In Problem 75, what would be your annual salary at each firm for the tenth year? 77. ECONOMICS The government, through a subsidy program, distributes $1,000,000. If we assume that each individual or agency spends 0.8 of what is received, and 0.8 of this is spent, and so on, how much total increase in spending results from this government action? 78. ECONOMICS Because of reduced taxes, an individual has an extra $600 in spendable income. If we assume that the individual spends 70% of this on consumer goods, that the producers of these goods in turn spend 70% of what they receive on consumer goods, and that this process continues indefinitely, what is the total amount spent on consumer goods? 79. BUSINESS If $P is invested at r% compounded annually, the amount A present after n years forms a geometric sequence with a common ratio 1 0004 r. Write a formula for the amount present after n years. How long will it take a sum of money P to double if invested at 6% interest compounded annually? 80. POPULATION GROWTH If a population of A0 people grows at the constant rate of r% per year, the population after t years forms a geometric sequence with a common ratio 1 0004 r. Write a formula for the total population after t years. If the world’s population is increasing at the rate of 2% per year, how long will it take to double? 81. FINANCE Eleven years ago an investment earned $7,000 for the year. Last year the investment earned $14,000. If the earnings from the investment have increased the same amount each year, what is the yearly increase and how much income has accrued from the investment over the past 11 years? 82. AIR TEMPERATURE As dry air moves upward, it expands. In so doing, it cools at the rate of about 5°F for each 1,000-foot rise. This is known as the adiabatic process. (A) Temperatures at altitudes that are multiples of 1,000 feet form what kind of a sequence? (B) If the ground temperature is 80°F, write a formula for the temperature Tn in terms of n, if n is in thousands of feet. 83. ENGINEERING A rotating flywheel coming to rest rotates 300 revolutions the first minute (see the figure). If in each subsequent minute it rotates two-thirds as many times as in the preceding minute, how many revolutions will the wheel make before coming to rest?
85. FOOD CHAIN A plant is eaten by an insect, an insect by a trout, a trout by a salmon, a salmon by a bear, and the bear is eaten by you. If only 20% of the energy is transformed from one stage to the next, how many calories must be supplied by plant food to provide you with 2,000 calories from the bear meat? 86. GENEALOGY If there are 30 years in a generation, how many direct ancestors did each of us have 600 years ago? By direct ancestors we mean parents, grandparents, great-grandparents, and so on. 87. PHYSICS An object falling from rest in a vacuum near the surface of the Earth falls 16 feet during the first second, 48 feet during the second second, 80 feet during the third second, and so on. (A) How far will the object fall during the eleventh second? (B) How far will the object fall in 11 seconds? (C) How far will the object fall in t seconds? 88. PHYSICS In Problem 87, how far will the object fall during: (A) The twentieth second? (B) The t th second? 89. BACTERIA GROWTH A single cholera bacterium divides every 1 2 hour to produce two complete cholera bacteria. If we start with a colony of A0 bacteria, how many bacteria will we have in t hours, assuming adequate food supply? 90. CELL DIVISION One leukemic cell injected into a healthy mouse will divide into two cells in about 12 day. At the end of the day these two cells will divide again, with the doubling process continuing each 12 day until there are 1 billion cells, at which time the mouse dies. On which day after the experiment is started does this happen? 91. ASTRONOMY Ever since the time of the Greek astronomer Hipparchus, second century B.C., the brightness of stars has been measured in terms of magnitude. The brightest stars, excluding the sun, are classed as magnitude 1, and the dimmest visible to the eye are classed as magnitude 6. In 1856, the English astronomer N. R. Pogson showed that first-magnitude stars are 100 times brighter than sixth-magnitude stars. If the ratio of brightness between consecutive magnitudes is constant, find this ratio. [Hint: If bn is the brightness of an nth-magnitude star, find r for the geometric sequence b1, b2, b3, . . . , given b1 0002 100b6.] 92. PUZZLE If a sheet of very thin paper 0.001-inch thick is torn in half, and each half is again torn in half, and this process is repeated for a total of 32 times, how high will the stack of paper be if the pieces are placed one on top of the other? Give the answer to the nearest mile.
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93. PUZZLE If you place 1¢ on the first square of a chessboard, 2¢ on the second square, 4¢ on the third, and so on, continuing to double the amount until all 64 squares are covered, how much money will be on the sixty-fourth square? How much money will there be on the whole board?
95. ATMOSPHERIC PRESSURE If atmospheric pressure decreases roughly by a factor of 10 for each 10-mile increase in altitude up to 60 miles, and if the pressure is 15 pounds per square inch at sea level, what will the pressure be 40 miles up?
94. MUSIC The notes on a piano, as measured in cycles per second, form a geometric sequence. (A) If A is 400 cycles per second and A¿, 12 notes higher, is 800 cycles per second, find the constant ratio r. (B) Find the cycles per second for C, three notes higher than A.
97. GEOMETRY If the midpoints of the sides of an equilateral triangle are joined by straight lines, the new figure will be an equilateral triangle with a perimeter equal to half the original. If we start with an equilateral triangle with perimeter 1 and form a sequence of “nested” equilateral triangles proceeding as described, what will be the total perimeter of all the triangles that can be formed in this way?
96. ZENO’S PARADOX Visualize a hypothetical 440-yard oval racetrack that has tapes stretched across the track at the halfway point and at each point that marks the halfway point of each remaining distance thereafter. A runner running around the track has to break the first tape before the second, the second before the third, and so on. From this point of view it appears that he will never finish the race. This famous paradox is attributed to the Greek philosopher Zeno (495– 435 B.C.). If we assume the runner runs at 440 yards per minute, the times between tape breakings form an infinite geometric sequence. What is the sum of this sequence?
98. PHOTOGRAPHY The shutter speeds and f-stops on a camera are given as follows: 1 1 1 , 250 , 500 Shutter speeds: 1, 12, 14, 18, 151 , 301 , 601 , 125 f-stops: 1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22
These are very close to being geometric sequences. Estimate their common ratios. 99. GEOMETRY We know that the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, . . . sides form an arithmetic sequence. Find the sum of the interior angles for a 21-sided polygon.
8-4
Multiplication Principle, Permutations, and Combinations Z Counting with the Multiplication Principle Z Using Factorial Notation Z Counting Permutations Z Counting Combinations
Section 8-4 introduces some new mathematical tools that are usually referred to as counting techniques. In general, a counting technique is a mathematical method of determining the number of objects in a set without actually enumerating the objects in the set as 1, 2, 3, . . . . For example, we can count the number of squares in a checkerboard
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(Fig. 1) by counting 1, 2, 3, . . . , 64. This is enumeration. Or we can note that there are eight rows with eight squares in each row. So the total number of squares must be 8 0002 8 0003 64. This is a very simple counting technique. Now consider the problem of assigning telephone numbers. How many different sevendigit telephone numbers can be formed? As we will soon see, the answer is 107 0003 10,000,000, a number that is much too large to obtain by enumeration. This shows that counting techniques are essential tools if the number of elements in a set is very large. The techniques developed in this section will be applied to a brief introduction to probability theory in Section 8-5, and to a famous algebraic formula in Section 8-6.
Z Figure 1
Z Counting with the Multiplication Principle We start with an example.
EXAMPLE
1
Combined Outcomes Suppose we flip a coin and then throw a single die (Fig. 2). What are the possible combined outcomes?
SOLUTION Heads
One way to solve this problem is to use a tree diagram:
Tails Coin Die Outcomes Outcomes
Coin outcomes
Combined Outcomes
H
1 2 3 4 5 6
(H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6)
T
1 2 3 4 5 6
(T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6)
Start Die outcomes
Z Figure 2 Coin and die outcomes.
There are 12 possible combined outcomes—two ways in which the coin can come up followed by six ways in which the die can come up. 0002 MATCHED PROBLEM 1
Use a tree diagram to determine the number of possible outcomes of throwing a single die followed by flipping a coin. 0002
Now suppose you are asked, “From the 26 letters in the alphabet, how many ways can 3 letters appear in a row on a license plate if no letter is repeated?” To try to count the possibilities using a tree diagram would be extremely tedious, to say the least. The following multiplication principle, also called the fundamental counting principle, enables us to solve this problem easily. In addition, it forms the basis for several other counting techniques developed later in this section.
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Z MULTIPLICATION PRINCIPLE 1. If two operations O1 and O2 are performed in order with N1 possible outcomes for the first operation and N2 possible outcomes for the second operation, then there are N1 ⴢ N2 possible combined outcomes of the first operation followed by the second. 2. In general, if n operations O1, O2, . . . , On, are performed in order, with possible number of outcomes N1, N2, . . . , Nn, respectively, then there are N1 ⴢ N2 ⴢ . . . ⴢ Nn possible combined outcomes of the operations performed in the given order.
In Example 1, we see that there are two possible outcomes from the first operation of flipping a coin and six possible outcomes from the second operation of throwing a die. So by the multiplication principle, there are 2 ⴢ 6 0003 12 possible combined outcomes of flipping a coin followed by throwing a die. (Now try using the multiplication principle to solve Matched Problem 1.) To answer the license plate question, we reason as follows: There are 26 ways the first letter can be chosen. After a first letter is chosen, 25 letters remain, so there are 25 ways a second letter can be chosen. And after 2 letters are chosen, there are 24 ways a third letter can be chosen. Using the multiplication principle, there are 26 ⴢ 25 ⴢ 24 0003 15,600 possible ways 3 letters can be chosen from the alphabet without allowing any letter to repeat. By not allowing any letter to repeat, earlier selections affect the choice of subsequent selections. If we allow letters to repeat, then earlier selections do not affect the choice in subsequent selections, and there are 26 possible choices for each of the 3 letters. So, if we allow letters to repeat, there are 26 ⴢ 26 ⴢ 26 0003 263 0003 17,576 possible ways the 3 letters can be chosen from the alphabet.
EXAMPLE
2
Computer-Generated Tests Many universities and colleges are now using computer-assisted testing procedures. Suppose a screening test is to consist of five questions, and a computer stores five equivalent questions for the first test question, eight equivalent questions for the second, six for the third, five for the fourth, and ten for the fifth. How many different five-question tests can the computer select? Two tests are considered different if they differ in one or more questions.
SOLUTION
O1: O2: O3: O4: O5:
Select Select Select Select Select
the the the the the
first question second question third question fourth question fifth question
N1: N2: N3: N4: N5:
five ways eight ways six ways five ways ten ways
The computer can generate 5 ⴢ 8 ⴢ 6 ⴢ 5 ⴢ 10 0003 12,000 different tests MATCHED PROBLEM 2
0002
Each question on a multiple-choice test has five choices. If there are five such questions on a test, how many different response sheets are possible if only one choice is marked for each question? 0002
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3
Counting Code Words How many three-letter code words are possible using the first eight letters of the alphabet if: (A) No letter can be repeated?
(B) Letters can be repeated?
(C) Adjacent letters cannot be alike? SOLUTIONS
(A) No letter can be repeated. O1: Select first letter O2: Select second letter O3: Select third letter
N1: eight ways N2: seven ways N3: six ways
Because one letter has been used Because two letters have been used
There are 8 ⴢ 7 ⴢ 6 0003 336 possible code words (B) Letters can be repeated. O1: Select first letter O2: Select second letter O3: Select third letter
N1: eight ways N2: eight ways N3: eight ways
Repeats are allowed. Repeats are allowed.
There are 8 ⴢ 8 ⴢ 8 0003 83 0003 512 possible code words (C) Adjacent letters cannot be alike. O1: Select first letter O2: Select second letter O3: Select third letter
N1: eight ways N2: seven ways N3: seven ways
There are
Cannot be the same as the first Cannot be the same as the second, but can be the same as the first
8 ⴢ 7 ⴢ 7 0003 392 possible code words MATCHED PROBLEM 3
ZZZ EXPLORE-DISCUSS 1
0002
How many four-letter code words are possible using the first ten letters of the alphabet under the three conditions stated in Example 3? 0002
The postal service of a developing country is choosing a five-character postal code consisting of letters (of the English alphabet) and digits. At least a half a million postal codes must be accommodated. Which format would you recommend to make the codes easy to remember?
The multiplication principle can be used to develop two additional counting techniques that are extremely useful in more complicated counting problems. Both of these methods use factorial notation, which we introduce next.
Z Using Factorial Notation For n a natural number, n factorial—denoted by n!—is the product of the first n natural numbers. Zero factorial is defined to be 1.
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Z DEFINITION 1 n Factorial For n a natural number n! 0003 n(n 0004 1) ⴢ . . . ⴢ 2 ⴢ 1 1! 0003 1 0! 0003 1
It is also useful to note that Z THEOREM 1 Recursion Formula for n Factorial n! 0003 n ⴢ (n 0004 1)!
EXAMPLE
4
Evaluating Factorials Evaluate each expression: (A) 4!
SOLUTIONS
(B) 5!
(C)
(A) 4! 0003 4 ⴢ 3 ⴢ 2 ⴢ 1 0003 24
Find (A) 6!
(B)
6! 5!
(D)
8! 5!
(E)
9! 6!3!
(B) 5! 0003 5 ⴢ 4 ⴢ 3 ⴢ 2 ⴢ 1 0003 120
8! 8 ⴢ 7 ⴢ 6 ⴢ 5! (D) 0003 0003 336 5! 5! MATCHED PROBLEM 4
7! 6!
(C)
3
(C)
7! 7 ⴢ 6! 0003 00037 6! 6!
4
9! 9 ⴢ 8 ⴢ 7 ⴢ 6! (E) 0003 0003 84 6!3! 6! 3 ⴢ 2 ⴢ 1 9! 6!
(D)
0002
10! 7!3! 0002
ZZZ
CAUTION ZZZ
When reducing fractions involving factorials, don’t confuse the single integer n with the symbol n!, which represents the product of n consecutive integers. 6! 0005 2! 3!
ZZZ EXPLORE-DISCUSS 2
6! 6 ⴢ 5 ⴢ 4 ⴢ 3! ⴝ ⴝ 6 ⴢ 5 ⴢ 4 ⴝ 120 3! 3!
A student used a calculator* to solve Matched Problem 4, as shown in Figure 3. Check these answers. If any are incorrect, explain why and find a correct calculator solution.
Z Figure 3 *The factorial symbol ! and related symbols can be found under the MATH-PROB menus on a TI-84 or TI-86.
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It is interesting and useful to note that n! grows very rapidly. Compare the following: 5! 0003 120
10! 0003 3,628,800
15! 0003 1,307,674,368,000
If n! is too large for a calculator to store and display, an error message is displayed. Find the value of n such that your calculator will evaluate n!, but not (n 0006 1)!.
Z Counting Permutations Suppose four pictures are to be arranged from left to right on one wall of an art gallery. How many arrangements are possible? Using the multiplication principle, there are four ways of selecting the first picture. After the first picture is selected, there are three ways of selecting the second picture. After the first two pictures are selected, there are two ways of selecting the third picture. And after the first three pictures are selected, there is only one way to select the fourth. So, the number of arrangements possible for the four pictures is 4 ⴢ 3 ⴢ 2 ⴢ 1 0003 4!
or
24
In general, we refer to a particular arrangement, or ordering, of n objects without repetition as a permutation of the n objects. How many permutations of n objects are there? From the preceding reasoning, there are n ways in which the first object can be chosen, there are n 0004 1 ways in which the second object can be chosen, and so on. Applying the multiplication principle, we have Theorem 2.
Z THEOREM 2 Permutations of n Objects The number of permutations of n objects, denoted by Pn,n, is given by Pn,n 0003 n ⴢ (n 0004 1) ⴢ . . . ⴢ 1 0003 n!
Now suppose the director of the art gallery decides to use only two of the four available pictures on the wall, arranged from left to right. How many arrangements of two pictures can be formed from the four? There are four ways the first picture can be selected. After selecting the first picture, there are three ways the second picture can be selected. So, the number of arrangements of two pictures from four pictures, denoted by P4,2, is given by P4,2 0003 4 ⴢ 3 0003 12 Or, in terms of factorials, multiplying 4 ⴢ 3 by 1 in the form 2!00022!, we have P4,2 0003 4 ⴢ 3 0003
4! 4 ⴢ 3 ⴢ 2! 0003 2! 2!
This last form gives P4,2 in terms of factorials, which is useful in some cases. A permutation of a set of n objects taken r at a time is an arrangement of the r objects in a specific order. So, reasoning in the same way as in the preceding example, we find that the number of permutations of n objects taken r at a time, 0 0007 r 0007 n, denoted by Pn,r, is given by Pn,r 0003 n(n 0004 1)(n 0004 2) ⴢ . . . ⴢ (n 0004 r 0006 1) Multiplying the right side of this equation by 1 in the form (n 0004 r)!0002(n 0004 r)!, we obtain a factorial form for Pn,r: Pn,r 0003 n(n 0004 1)(n 0004 2) ⴢ . . . ⴢ (n 0004 r 0006 1)
(n 0004 r)! (n 0004 r)!
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But n(n 0004 1)(n 0004 2) ⴢ . . . ⴢ (n 0004 r 0006 1)(n 0004 r)! 0003 n! We have developed Theorem 3.
Z THEOREM 3 Permutation of n Objects Taken r at a Time The number of permutations of n objects taken r at a time is given by Pn,r 0003 n(n 0004 1)(n 0004 2) ⴢ . . . ⴢ (n 0004 r 0006 1)
y r factors
or Pn,r 0003
n! (n 0004 r)!
00007r0007n
Note that if r 0003 n, then the number of permutations of n objects taken n at a time is Pn,n 0003
n! n! 0003 0003 n! (n 0004 n)! 0!
Recall, 0! ⴝ 1.
which agrees with Theorem 2, as it should. The permutation symbol Pn,r also can be denoted by P nr, n Pr, or P(n, r). Many calculators use n Pr to denote the function that evaluates the permutation symbol.
EXAMPLE
5
Selecting Officers From a committee of eight people, in how many ways can we choose a chair and a vicechair, assuming one person cannot hold more than one position?
SOLUTION
We are actually asking for the number of permutations of eight objects taken two at a time— that is, P8,2: P8,2 0003
MATCHED PROBLEM 5
EXAMPLE
6
8! 8! 8 ⴢ 7 ⴢ 6! 0003 0003 0003 56 (8 0004 2)! 6! 6!
0002
From a committee of ten people, in how many ways can we choose a chair, vice-chair, and secretary, assuming one person cannot hold more than one position? 0002
Evaluating Pn,r Find the number of permutations of 25 objects taken (A) Two at a time (B) Four at a time (C) Eight at a time
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Figure 4 shows the solution on a calculator.
0002
Z Figure 4
MATCHED PROBLEM 6
Find the number of permutations of 30 objects taken (A) Two at a time
(B) Four at a time
(C) Six at a time 0002
Z Counting Combinations Now suppose that an art museum owns eight paintings by a given artist and another art museum hopes to borrow three of these paintings for a special show. How many ways can three paintings be selected for shipment out of the eight available? Here, the order of the items selected doesn’t matter. What we are actually interested in is how many subsets of three objects can be formed from a set of eight objects. We call such a subset a combination of eight objects taken three at a time. The total number of combinations is denoted by the symbol C8,3
or
8 a b 3
To find the number of combinations of eight objects taken three at a time, C8,3, we make use of the formula for Pn,r and the multiplication principle. We know that the number of permutations of eight objects taken three at a time is given by P8,3, and we have a formula for computing this quantity. Now suppose we think of P8,3 in terms of two operations: O1: Select a subset of three objects (paintings) N1: C8,3 ways O2: Arrange the subset in a given order N2: 3! ways The combined operation, O1 followed by O2, produces a permutation of eight objects taken three at a time. So, P8,3 0003 C8,3 ⴢ 3! To find C8,3, we replace P8,3 in the preceding equation with 8!0002(8 0004 3)! and solve for C8,3: 8! 0003 C8,3 ⴢ 3! (8 0004 3)! 8! 8 ⴢ 7 ⴢ 6 ⴢ 5! 0003 0003 56 C8,3 0003 3!(8 0004 3)! 3 ⴢ 2 ⴢ 1 ⴢ 5! The museum can make 56 different selections of three paintings from the eight available. A combination of a set of n objects taken r at a time is an r-element subset of the n objects. Reasoning in the same way as in the example, the number of combinations of n
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objects taken r at a time, 0 0007 r 0007 n, denoted by Cn,r, can be obtained by solving for Cn,r in the relationship Pn,r 0003 Cn,r ⴢ r! Cn,r 0003 0003
Pn,r r! n! r!(n 0004 r)!
Pn,r ⴝ
n! (n ⴚ r)!
Z THEOREM 4 Combination of n Objects Taken r at a Time The number of combinations of n objects taken r at a time is given by Pn,r n n! 0003 Cn,r 0003 a b 0003 r r! r!(n 0004 r)!
00007r0007n
n The combination symbols Cn,r and a b also can be denoted by C nr, nCr, or C(n, r). r
EXAMPLE
7
Selecting Subcommittees From a committee of eight people, in how many ways can we choose a subcommittee of two people?
SOLUTION
Notice how this example differs from Example 5, where we wanted to know how many ways a chair and a vice-chair can be chosen from a committee of eight people. In Example 5, ordering matters. In choosing a subcommittee of two people, the ordering does not matter. So, we are actually asking for the number of combinations of eight objects taken two at a time. The number is given by 8 8! 8 ⴢ 7 ⴢ 6! 0003 0003 28 C8,2 0003 a b 0003 2 2!(8 0004 2)! 2 ⴢ 1 ⴢ 6!
MATCHED PROBLEM 7
EXAMPLE
8
0002
How many subcommittees of three people can be chosen from a committee of eight people? 0002
Evaluating Cn,r Find the number of combinations of 25 objects taken (A) Two at a time
SOLUTION
(B) Four at a time
(C) Eight at a time
Figure 5 shows the solution on a calculator. Compare these results with Example 6.
Z Figure 5
0002
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MATCHED PROBLEM 8
Find the number of combinations of 30 objects taken (A) Two at a time
(B) Four at a time
(C) Six at a time 0002
Remember: In a permutation, order counts. In a combination, order does not count. To determine whether a permutation or combination is needed, decide whether rearranging the collection or listing makes a difference. If so, use permutations. If not, use combinations.
ZZZ EXPLORE-DISCUSS 3
2
3
3
(C) The newly elected president names his cabinet members.
9 10 J Q K A 6 78
(D) The president selects a delegation of three cabinet members to attend the funeral of a head of state.
9 10 J Q K A 67 8 45
(E) An orchestra conductor chooses three pieces of music for a symphony program.
4 5
9 10 J Q K A 6 78
A
2
3
4 5
(B) A baseball manager names his starting lineup.
A
2
3
(A) A student checks out three books from the library.
9 10 J Q K A 67 8 45
A
2
Each of the following is a selection without repetition. Would you consider the selection to be a combination? A permutation? Discuss your reasoning.
A
Z Figure 6 A standard deck of cards.
EXAMPLE
9
A standard deck of 52 cards (Fig. 6) has four 13-card suits: diamonds, hearts, clubs, and spades. Each 13-card suit contains cards numbered from 2 to 10, a jack, a queen, a king, and an ace. The jack, queen, and king are called face cards. Depending on the game, the ace may be counted as the lowest and/or the highest card in the suit. Example 9, as well as other examples and exercises in Chapter 8, refer to this standard deck.
Counting Card Hands Out of a standard 52-card deck, how many 5-card hands will have three aces and two kings?
SOLUTION
O1: N1: O2: N2:
Choose three aces out of four possible C4,3 Choose two kings out of four possible C4,2
Order is not important.
Order is not important.
Using the multiplication principle, we have Number of hands 0003 C4,3 ⴢ C4,2 0003 4 ⴢ 6 0003 24 MATCHED PROBLEM 9
EXAMPLE
10
0002
From a standard 52-card deck, how many 5-card hands will have three hearts and two spades? 0002
Counting Serial Numbers Serial numbers for a product are to be made using two letters followed by three numbers. If the letters are to be taken from the first eight letters of the alphabet with no repeats and the numbers from the 10 digits 0 through 9 with no repeats, how many serial numbers are possible?
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O1: N1: O2: N2:
SOLUTION
Multiplication Principle, Permutations, and Combinations
Choose two letters out of eight available P8,2 Choose three numbers out of ten available P10,3
541
Order is important.
Order is important.
Using the multiplication principle, we have Number of serial numbers 0003 P8,2 ⴢ P10,3 0003 40,320 MATCHED PROBLEM 10
0002
Repeat Example 10 under the same conditions, except the serial numbers are now to have three letters followed by two digits with no repeats. 0002 ANSWERS TO MATCHED PROBLEMS 1. H T H T H T H T H T H T There are 12 outcomes. 1
2
3
4
5
2. 55, or 3,125
6
Start
3. (A) 10 ⴢ 9 ⴢ 8 ⴢ 7 0003 5,040 (B) 10 ⴢ 10 ⴢ 10 ⴢ 10 0003 10,000 (C) 10 ⴢ 9 ⴢ 9 ⴢ 9 0003 7,290 4. (A) 720 (B) 6 (C) 504 (D) 120 10! 5. P10,3 0003 6. (A) 870 (B) 657,720 (C) 427,518,000 0003 720 (10 0004 3)! 8! 7. C8,3 0003 8. (A) 435 (B) 27,405 (C) 593,775 0003 56 3!(8 0004 3)! 9. C13,3 ⴢ C13,2 0003 22,308 10. P8,3 ⴢ P10,2 0003 30,240
8-4
Exercises
1. What is a permutation?
17. The figure shows calculator solutions to Problems 11, 13, and 15. Check these answers. If any are incorrect, explain why and find a correct calculator solution.
2. What is a combination? 3. Explain how n! can be defined by means of a recursion formula. 4. State the multiplication principle for counting in your own words. 5. Explain how permutations and combinations differ with respect to order. 6. Explain how permutations and combinations are alike with respect to repetition. Evaluate the expression in Problems 7–16: 7. 9! 10. 12! 13.
5! 2!3!
16.
8! 3!(8 0004 3)!
8. 10!
9. 11!
11.
11! 8!
12.
14! 12!
14.
6! 4!2!
15.
7! 4!(7 0004 4)!
18. The figure shows calculator solutions to Problems 12, 14, and 16. Check these answers. If any are incorrect, explain why and find a correct calculator solution.
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In Problems 19–26, evaluate. 19. P13,4
20. C20,10
21. P13,9
22. C20,4
23. C15,8
24. P11,3
25. C15,12
26. P11,8
numbers are possible, assuming no digit is repeated? Assuming digits can be repeated? 40. A small combination lock on a suitcase has three wheels, each labeled with digits from 0 to 9. How many opening combinations of three numbers are possible, assuming no digit is repeated? Assuming digits can be repeated?
In Problems 27 and 28, would you consider the selection to be a combination or a permutation? Explain your reasoning.
41. From a standard 52-card deck, how many 5-card hands will have all hearts?
27. (A) The recently elected chief executive officer (CEO) of a company named three new vice-presidents, of marketing, research, and manufacturing. (B) The CEO selected three of her vice-presidents to attend the dedication ceremony of a new plant.
42. From a standard 52-card deck, how many 5-card hands will have all face cards? All face cards, but no kings? Consider only jacks, queens, and kings to be face cards.
28. (A) An individual rented four DVDs from a rental store to watch over a weekend. (B) The same individual did some holiday shopping by buying four DVDs, one for his father, one for his mother, one for his younger sister, and one for his older brother. 29. A particular new car model is available with five choices of color, three choices of transmission, four types of interior, and two types of engine. How many different variations of this model car are possible? 30. A deli serves sandwiches with the following options: three kinds of bread, five kinds of meat, and lettuce or sprouts. How many different sandwiches are possible, assuming one item is used out of each category? 31. In a horse race, how many different finishes among the first three places are possible for a 10-horse race? Exclude ties. 32. In a long-distance foot race, how many different finishes among the first five places are possible for a 50-person race? Exclude ties. 33. How many ways can a subcommittee of three people be selected from a committee of seven people? How many ways can a president, vice-president, and secretary be chosen from a committee of seven people? 34. Suppose nine cards are numbered with the nine digits from 1 to 9. A three-card hand is dealt, one card at a time. How many hands are possible where: (A) Order is taken into consideration? (B) Order is not taken into consideration? 35. There are 10 teams in a league. If each team is to play every other team exactly once, how many games must be scheduled? 36. Given seven points, no three of which are on a straight line, how many lines can be drawn joining two points at a time? 37. How many four-letter code words are possible from the first six letters of the alphabet, with no letter repeated? Allowing letters to repeat? 38. How many five-letter code words are possible from the first seven letters of the alphabet, with no letter repeated? Allowing letters to repeat? 39. A combination lock has five wheels, each labeled with the 10 digits from 0 to 9. How many opening combinations of five
43. How many different license plates are possible if each contains three letters followed by three digits? How many of these license plates contain no repeated letters and no repeated digits? 44. How may five-digit zip codes are possible? How many of these codes contain no repeated digits? 45. From a standard 52-card deck, how many 7-card hands have exactly five spades and two hearts? 46. From a standard 52-card deck, how many 5-card hands will have two clubs and three hearts? 47. A catering service offers eight appetizers, ten main courses, and seven desserts. A banquet chairperson is to select three appetizers, four main courses, and two desserts for a banquet. How many ways can this be done? 48. Three research departments have 12, 15, and 18 members, respectively. If each department is to select a delegate and an alternate to represent the department at a conference, how many ways can this be done? 49. (A) Use a graphing calculator to display the sequences P10,0, P10,1, . . . , P10,10 and 0!, 1!, . . . , 10! in table form, and show that P10,r r! for r 0003 0, 1, . . . , 10. (B) Find all values of r such that P10,r 0003 r! (C) Explain why Pn,r r! whenever 0 0007 r 0007 n. P10,0 P10,1 P10,10 , ,..., 50. (A) How are the sequences and C10,0, 0! 1! 10! C10,1, . . . , C10,10 related? (B) Use a graphing calculator to graph each sequence and confirm the relationship of part A. 51. A sporting goods store has 12 pairs of ski gloves of 12 different brands thrown loosely in a bin. The gloves are all the same size. In how many ways can a left-hand glove and a right-hand glove be selected that do not match relative to brand? 52. A sporting goods store has six pairs of running shoes of six different styles thrown loosely in a basket. The shoes are all the same size. In how many ways can a left shoe and a right shoe be selected that do not match? 53. Eight distinct points are selected on the circumference of a circle. (A) How many chords can be drawn by joining the points in all possible ways? (B) How many triangles can be drawn using these eight points as vertices? (C) How many quadrilaterals can be drawn using these eight points as vertices?
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54. Five distinct points are selected on the circumference of a circle. (A) How many chords can be drawn by joining the points in all possible ways? (B) How many triangles can be drawn using these five points as vertices? 55. How many ways can two people be seated in a row of five chairs? Three people? Four people? Five people? 56. Each of two countries sends five delegates to a negotiating conference. A rectangular table is used with five chairs on each long side. If each country is assigned a long side of the table, how many seating arrangements are possible? [Hint: Operation 1 is assigning a long side of the table to each country.] 57. A basketball team has five distinct positions. Out of eight players, how many starting teams are possible if (A) The distinct positions are taken into consideration? (B) The distinct positions are not taken into consideration? (C) The distinct positions are not taken into consideration, but either Mike or Ken, but not both, must start?
8-5
58. How many committees of four people are possible from a group of nine people if (A) There are no restrictions? (B) Both Juan and Mary must be on the committee? (C) Either Juan or Mary, but not both, must be on the committee? 59. A 5-card hand is dealt from a standard 52-card deck. Which is more likely: the hand contains exactly one king or the hand contains no hearts? 60. A 10-card hand is dealt from a standard 52-card deck. Which is more likely: all cards in the hand are red or the hand contains all four aces?
Sample Spaces and Probability Z Sample Spaces and Events Z Finding the Probability of an Event Z Making Equally Likely Assumptions Z Finding or Approximating Empirical Probability
This section provides an introduction to probability. It’s going to need to be a relatively brief one, because probability is a topic to which entire books and courses are devoted. Probability involves many subtle notions, and care must be taken at the beginning to understand the fundamental concepts on which the subject is based. Our development of probability, because of space limitations, must be somewhat informal. More formal and precise treatments can be found in books on probability.
Z Sample Spaces and Events Our first step in constructing a mathematical model for probability studies is to describe the type of experiments on which probability studies are based. Some types of experiments do not yield the same results, no matter how carefully they are repeated under the same conditions. These experiments are called random experiments. Some standard examples of random experiments are flipping coins, rolling dice, observing the frequency of defective items from an assembly line, or observing the frequency of deaths in a certain age group. Probability theory is a branch of mathematics that has been developed to deal with outcomes of random experiments. In the work that follows, the word experiment will be used to mean a random experiment.
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The outcomes of experiments are typically described in terms of sample spaces and events. Our second step in constructing a mathematical model for probability studies is to define these two terms. Consider the experiment, “A single six-sided die is rolled.” What outcomes might we observe? We might be interested in the number of dots facing up, or whether the number of dots facing up is an even number, or whether the number of dots facing up is divisible by 3, and so on. The list of possible outcomes appears endless. In general, there is no unique method of analyzing all possible outcomes of an experiment. Therefore, before conducting an experiment, it is important to decide just what outcomes are of interest. In the die experiment, suppose we limit our interest to the number of dots facing up when the die comes to rest. Having decided what to observe, we make a list of outcomes of the experiment, called simple events, such that in each trial of the experiment, one and only one of the results on the list will occur. The set of simple events for the experiment is called a sample space for the experiment. The sample space S we have chosen for the die-rolling experiment is S 0003 {1, 2, 3, 4, 5, 6} Now consider the outcome, “The number of dots facing up is an even number.” This outcome is not a simple event, because it will occur whenever 2, 4, or 6 dots appear, that is, whenever an element in the subset E 0003 {2, 4, 6} occurs. Subset E is called a compound event. In general, we have the following definition:
Z DEFINITION 1 Event Given a sample space S for an experiment, we define an event E to be any subset of S. If an event E has only one element in it, it is called a simple event. If event E has more than one element, it is called a compound event. We say that an event E occurs if any of the simple events in E occurs.
EXAMPLE
1
Choosing a Sample Space A nickel and a dime are tossed. How will we identify a sample space for this experiment?
SOLUTIONS
There are a number of possibilities, depending on our interest. We will consider three. (A) If we are interested in whether each coin falls heads (H) or tails (T), then, using a tree diagram, we can easily determine an appropriate sample space for the experiment: Nickel Outcomes H Start T
Dime Outcomes H T H T
The sample space is S1 0003 {HH, HT, TH, TT} and there are four simple events in the sample space.
Combined Outcomes HH HT TH TT
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(B) If we are interested only in the number of heads that appear on a single toss of the two coins, then we can let S2 0003 {0, 1, 2} and there are three simple events in the sample space. (C) If we are interested in whether the coins match (M) or don’t match (D), then we can let S3 0003 {M, D} and there are only two simple events in the sample space.
MATCHED PROBLEM 1
0002
An experiment consists of recording the boy–girl composition of families with two children. (A) What is an appropriate sample space if we are interested in the gender of each child in the order of their births? Draw a tree diagram. (B) What is an appropriate sample space if we are interested only in the number of girls in a family? (C) What is an appropriate sample space if we are interested only in whether the genders are alike (A) or different (D)? (D) What is an appropriate sample space for all three interests expressed above? 0002
In Example 1, sample space S1 contains more information than either S2 or S3. If we know which outcome has occurred in S1, then we know which outcome has occurred in S2 and S3. However, the reverse is not true. In this sense, we say that S1 is a more fundamental sample space than either S2 or S3. Important Remark: There is no one correct sample space for a given experiment. When specifying a sample space for an experiment, we include as much detail as necessary to answer all questions of interest regarding the outcomes of the experiment. If in doubt, include more elements in the sample space rather than fewer. Now let’s return to the two-coin problem in Example 1 and the sample space S1 0003 {HH, HT, TH, TT} Suppose we are interested in the outcome, “Exactly 1 head is up.” Looking at S1, we find that it occurs if either of the two simple events HT or TH occurs.* So, to say that the event, “Exactly 1 head is up” occurs is the same as saying the experiment has an outcome in the set E 0003 {HT, TH} This is a subset of the sample space S1. The event E is a compound event.
*Technically, we should write {HT} and {TH}, because there is a logical distinction between an element of a set and a subset consisting of only that element. But we will just keep this in mind and drop the braces for simple events to simplify the notation.
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2
Rolling Two Dice Consider an experiment of rolling two dice. A convenient sample space that will enable us to answer many questions about events of interest is shown in Figure 1. Let S be the set of all ordered pairs listed in the figure. Note that the simple event (3, 2) is to be distinguished from the simple event (2, 3). The former indicates a 3 turned up on the first die and a 2 on the second, whereas the latter indicates a 2 turned up on the first die and a 3 on the second. What is the event that corresponds to each of the following outcomes? (A) A sum of 7 turns up.
(B) A sum of 11 turns up.
(C) A sum less than 4 turns up.
(D) A sum of 12 turns up.
FIRST DIE
SECOND DIE
(1, 1)
(1, 2)
(1, 3)
(1, 4)
(1, 5)
(1, 6)
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(2, 5)
(2, 6)
(3, 1)
(3, 2)
(3, 3)
(3, 4)
(3, 5)
(3, 6)
(4, 1)
(4, 2)
(4, 3)
(4, 4)
(4, 5)
(4, 6)
(5, 1)
(5, 2)
(5, 3)
(5, 4)
(5, 5)
(5, 6)
(6, 1)
(6, 2)
(6, 3)
(6, 4)
(6, 5)
(6, 6)
Z Figure 1 A sample space for rolling two dice.
SOLUTIONS
(A) By “A sum of 7 turns up,” we mean that the sum of all dots on both turned-up faces is 7. This outcome corresponds to the event {(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)} (B) “A sum of 11 turns up” corresponds to the event {(6, 5), (5, 6)} (C) “A sum less than 4 turns up” corresponds to the event {(1, 1), (2, 1), (1, 2)} (D) “A sum of 12 turns up” corresponds to the event {(6, 6)}
MATCHED PROBLEM 2
0002
Refer to the sample space in Example 2 (Fig. 1). What is the event that corresponds to each of the following outcomes? (A) A sum of 5 turns up. (B) A sum that is a prime number greater than 7 turns up. 0002 Informally, to facilitate discussion, we often use the terms event and outcome of an experiment interchangeably. So, in Example 2 we might say “the event ‘A sum of 11 turns up’ ” in place of “the outcome ‘A sum of 11 turns up,’ ” or even write E 0003 A sum of 11 turns up 0003 {(6, 5), (5, 6)}
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Technically speaking, an event is the mathematical counterpart of an outcome of an experiment.
Z Finding the Probability of an Event The next step in developing our mathematical model for probability studies is the introduction of a probability function. This is a function that assigns to an arbitrary event associated with a sample space a real number between 0 and 1, inclusive. We start by discussing ways in which probabilities are assigned to simple events in S.
Z DEFINITION 2 Probabilities for Simple Events Given a sample space S 0003 {e1, e2, . . . , en} with n simple events, to each simple event ei we assign a real number, denoted by P(ei), that is called the probability of the event ei. These numbers may be assigned in an arbitrary manner as long as the following two conditions are satisfied: 1. 0 0007 P(ei) 0007 1 2. P(e1) 0006 P(e2) 0006 . . . 0006 P(en) 0003 1
The sum of the probabilities of all simple events in the sample space is 1.
Any probability assignment that meets conditions 1 and 2 is said to be an acceptable probability assignment.
Our mathematical theory does not explain how acceptable probabilities are assigned to simple events. These assignments are generally based on the expected or actual percentage of times a simple event occurs when an experiment is repeated a large number of times. Assignments based on this principle are called reasonable. Let an experiment be the flipping of a single coin, and let us choose a sample space S to be S 0003 {H, T} If a coin appears to be fair, we are inclined to assign probabilities to the simple events in S as follows: P(H) 0003
1 2
and
P(T) 0003 12
These assignments are based on reasoning that, because there are two ways a coin can land, in the long run a head will turn up half the time and a tail will turn up half the time. These probability assignments are acceptable, because both of the conditions for acceptable probability assignments in Definition 2 are satisfied: 1. 0 0007 P(H) 0007 1, 0 0007 P(T) 0007 1 2. P(H) 0006 P(T) 0003 12 0006 12 0003 1 But there are other acceptable assignments. Maybe after flipping a coin 1,000 times we find that the head turns up 376 times and the tail turns up 624 times. With this result, we might suspect that the coin is not fair and assign the simple events in the sample space S the probabilities P(H) 0003 .376
and
P(T) 0003 .624
This is also an acceptable assignment. But the probability assignment P(H) 0003 1
and
P(T) 0003 0
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though acceptable, is not reasonable, unless the coin has two heads. The assignment P(H) 0003 .6
P(T) 0003 .8
and
is not acceptable, because .6 0006 .8 0003 1.4, which violates condition 2 in Definition 2. In probability studies, the 0 to the left of the decimal is usually omitted; we write .8 and not 0.8. It is important to keep in mind that out of the infinitely many possible acceptable probability assignments to simple events in a sample space, we are generally inclined to choose one assignment over another based on reasoning or experimental results. Given an acceptable probability assignment for simple events in a sample space S, how do we define the probability of an arbitrary event E associated with S?
Z DEFINITION 3 Probability of an Event E Given an acceptable probability assignment for the simple events in a sample space S, we define the probability of an arbitrary event E, denoted by P(E ), as follows: 1. If E is the empty set, then P(E) 0003 0. 2. If E is a simple event, then P(E) has already been assigned. 3. If E is a compound event, then P(E) is the sum of the probabilities of all the simple events in E. 4. If E is the sample space S, then P(E ) 0003 P(S) 0003 1. This is a special case of 3.
EXAMPLE
3
Finding Probabilities of Events Let’s return to Example 1, the tossing of a nickel and dime, and the sample space S 0003 {HH, HT, TH, TT} Because there are four simple outcomes and the coins are assumed to be fair, it appears that each outcome should occur in the long run 25% of the time. Let’s assign the same probability of 14 to each simple event in S: Simple event, ei P(ei)
HH
HT
TH
TT
1 4
1 4
1 4
1 4
This is an acceptable assignment according to Definition 2 and a reasonable assignment for ideal coins that are perfectly balanced or coins close to ideal. (A) What is the probability of getting exactly one head? (B) What is the probability of getting at least one head? (C) What is the probability of getting a head or a tail? (D) What is the probability of getting three heads? SOLUTIONS
(A) E1 0003 Getting one head 0003 {HT, TH} Because E1 is a compound event, we use item 3 in Definition 3 and find P(E1) by adding the probabilities of the simple events in E1. P(E1) 0003 P(HT) 0006 P(TH) 0003 14 0006 14 0003 12
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(B) E2 0003 Getting at least 1 head 0003 {HH, HT, TH} P(E2) 0003 P(HH) 0006 P(HT) 0006 P(TH) 0003 14 0006 14 0006 14 0003 34 (C) E3 0003 {HH, HT, TH, TT} 0003 S P(E3) 0003 P(S) 0003 1
1 4
(D) E3 0003 Getting three heads 0003 P( ) 0003 0
ⴙ 14 ⴙ 14 ⴙ 14 ⴝ 1
Empty set
Z STEPS FOR FINDING PROBABILITIES OF EVENTS Step 1. Set up an appropriate sample space S for the experiment. Step 2. Assign acceptable probabilities to the simple events in S. Step 3. To obtain the probability of an arbitrary event E, add the probabilities of the simple events in E.
The function P defined in steps 2 and 3 is called a probability function. The domain of this function is all possible events in the sample space S, and the range is a set of real numbers between 0 and 1, inclusive. 0002
MATCHED PROBLEM 3
Return to Matched Problem 1, recording the boy–girl composition of families with two children and the sample space S 0003 {BB, BG, GB, GG} Statistics from the U.S. Census Bureau indicate that an acceptable and reasonable probability for this sample space is Simple event, ei
BB
BG
GB
GG
P(ei)
.26
.25
.25
.24
Find the probabilities of the following events: (A) E1 0003 Having at least one girl in the family (B) E2 0003 Having at most one girl in the family (C) E3 0003 Having two children of the same sex in the family
0002
Z Making Equally Likely Assumptions In tossing a nickel and dime (Example 3), we assigned the same probability, 14, to each simple event in the sample space S 0003 {HH, HT, TH, TT}. By assigning the same probability to each simple event in S, we are actually making the assumption that each simple event is as likely to occur as any other. We refer to this as an equally likely assumption. In general, we have Definition 4.
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Z DEFINITION 4 Probability of a Simple Event Under an Equally Likely Assumption If, in a sample space S 0003 {e1, e2, . . . , en} with n elements, we assume each simple event ei is as likely to occur as any other, then we assign the probability 10002n to each. That is, P(ei) 0003
1 n
Under an equally likely assumption, we can develop a very useful formula for finding probabilities of arbitrary events associated with a sample space S. Consider the following example. If a single die is rolled and we assume each face is as likely to come up as any other, then for the sample space S 0003 {1, 2, 3, 4, 5, 6} we assign a probability of the probability of
1 6
to each simple event, because there are six simple events. Then
E 0003 Rolling a prime number 0003 {2, 3, 5} is P(E) 0003 P(2) 0006 P(3) 0006 P(5) 0003 16 0006 16 0006 16 0003 36 0003 12 So, under the assumption that each simple event is as likely to occur as any other, the computation of the probability of the occurrence of any event E in a sample space S is the number of elements in E divided by the number of elements in S.
Z THEOREM 1 Probability of an Arbitrary Event Under an Equally Likely Assumption If we assume each simple event in sample space S is as likely to occur as any other, then the probability of an arbitrary event E in S is given by P(E) 0003
EXAMPLE
4
n(E ) Number of elements in E 0003 Number of elements in S n(S )
Finding Probabilities of Events If in rolling two dice we assume each simple event in the sample space shown in Figure 1 on page 546 is as likely as any other, find the probabilities of the following events: (A) E1 0003 A sum of 7 turns up
(B) E2 0003 A sum of 11 turns up
(C) E3 0003 A sum less than 4 turns up
(D) E4 0003 A sum of 12 turns up
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MATCHED PROBLEM 4
Sample Spaces and Probability
551
Referring to Figure 1, we see that: (A) P(E1) 0003
n(E1) 6 1 0003 0003 n(S) 36 6
(B) P(E2) 0003
n(E2) 2 1 0003 0003 n(S) 36 18
(C) P(E3) 0003
n(E3) 3 1 0003 0003 n(S) 36 12
(D) P(E4) 0003
n(E4) 1 0003 n(S) 36
0002
Under the conditions in Example 4, find the probabilities of the following events: (A) E5 0003 A sum of 5 turns up (B) E6 0003 A sum that is a prime number greater than 7 turns up
ZZZ EXPLORE-DISCUSS 1
0002
A box contains four red balls and seven green balls. A ball is drawn at random and then, without replacing the first ball, a second ball is drawn. Discuss whether or not the equally likely assumption would be appropriate for the sample space S 0003 {RR, RG, GR, GG}.
We now turn to some examples that make use of the counting techniques developed in Section 8-4.
EXAMPLE
5
Drawing Cards In drawing 5 cards from a 52-card deck without replacement, what is the probability of getting five spades?
SOLUTION
Let the sample space S be the set of all 5-card hands from a 52-card deck. Because the order in a hand does not matter, n(S ) 0003 C52,5. The event we seek is E 0003 Set of all 5-card hands from 13 spades Again, the order does not matter and n(E ) 0003 C13,5. Assuming that each 5-card hand is as likely as any other, P(E ) 0003
MATCHED PROBLEM 5
EXAMPLE
6
C13,5 n(E ) 13! 5!8! 13! 5!47! 0003 0003 0003 ⴢ 0003 .0005 n(S ) C52,5 52! 5!47! 5!8! 52!
0002
In drawing 7 cards from a 52-card deck without replacement, what is the probability of getting seven hearts? 0002
Selecting Committees The board of regents of a university is made up of 12 men and 16 women. If a committee of six is chosen at random, what is the probability that it will contain three men and three women?
SOLUTION
Let S 0003 Set of all 6-person committees out of 28 people: n(S ) 0003 C28,6
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Let E 0003 Set of all 6-person committees with 3 men and 3 women. To find n(E ), we use the multiplication principle and the following two operations: O1: Select 3 men out of the 12 available O2: Select 3 women out of the 16 available
N1: C12,3 N2: C16,3
So n(E) 0003 C12,3 ⴢ C16,3 and P(E) 0003
MATCHED PROBLEM 6
C12,3 ⴢ C16,3 n(E) 0003 0003 .327 n(S) C28,6
0002
What is the probability that the committee in Example 6 will have four men and two women? 0002
Z Finding or Approximating Empirical Probability In the earlier examples in this section we made a reasonable assumption about an experiment and used deductive reasoning to assign probabilities. For example, it is reasonable to assume that an ordinary coin will come up heads about as often as it will come up tails. Probabilities determined in this manner are called theoretical probabilities. No experiments are ever conducted. But what if the theoretical probabilities are not obvious? Then we assign probabilities to simple events based on the results of actual experiments. Probabilities determined from the results of actually performing an experiment are called empirical probabilities. As an experiment is repeated over and over, the percentage of times an event occurs may get closer and closer to a single fixed number. If so, this single fixed number is generally called the actual probability of the event. ZZZ EXPLORE-DISCUSS 2
Like a coin, a thumbtack tossed into the air will land in one of two positions, point up or point down [Fig. 2(a)]. Unlike a coin, we would not expect both events to occur with the same frequency. Indeed, the frequencies of landing point up and point down may well vary from one thumbtack to another [Fig. 2(b)]. Find two thumbtacks of different sizes and guess which one is likely to land point up more frequently. Then toss each tack 100 times and record the number of times each lands point up. Did the experiment confirm your initial guess?
(a) Point up or point down
(b) Two different tacks
Z Figure 2
Suppose when tossing one of the thumbtacks in Explore-Discuss 2, we observe that the tack lands point up 43 times and point down 57 times. Based on this experiment, it seems reasonable to say that for this particular thumbtack P(Point up) 0003
43 0003 .43 100
P(Point down) 0003
57 0003 .57 100
Probability assignments based on the results of repeated trials of an experiment are called approximate empirical probabilities.
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In general, if we conduct an experiment n times and an event E occurs with frequency f (E ), then the ratio f (E )0002n is called the relative frequency of the occurrence of event E in n trials. We define the empirical probability of E, denoted by P(E ), by the number, if it exists, that the relative frequency f (E )0002n approaches as n gets larger and larger. Of course, for any particular n, the relative frequency f (E )0002n is generally only approximately equal to P(E ). However, as n increases, we expect the approximation to improve. Z DEFINITION 5 Empirical Probability If f(E) is the frequency of event E in n trials, then P(E ) 0003
Frequency of occurrence of E f (E ) 0003 n Total number of trials
If we can also deduce theoretical probabilities for an experiment, then we expect the approximate empirical probabilities to approach the theoretical probabilities. If this does not happen, then we should begin to suspect the manner in which the theoretical probabilities were computed. If P(E ) is the theoretical probability of an event E and the experiment is performed n times, then the expected frequency of the occurrence of E is n ⴢ P(E ).
EXAMPLE
7
Finding Approximate Empirical and Theoretical Probabilities Two coins are tossed 500 times with the following frequencies of outcomes: Two heads: 121 One head: 262 Zero heads: 117 (A) Compute the approximate empirical probability for each outcome. (B) Compute the theoretical probability for each outcome. (C) Compute the expected frequency for each outcome.
SOLUTIONS
(A) P(two heads) 0003
121 0003 .242 500
262 0003 .524 500 117 0003 .234 P(zero heads) 0003 500 P(one head) 0003
(B) A sample space of equally likely simple events is S 0003 {HH, HT, TH, TT}. Let E1 0003 two heads 0003 5HH6 E2 0003 one head 0003 5HT, TH6 E3 0003 zero heads 0003 5TT6
Then n(E1) 1 0003 0003 .25 n(S) 4 n(E2) 2 P(E2) 0003 0003 0003 .50 n(S) 4 n(E3) 1 0003 0003 .25 P(E3) 0003 n(S) 4 P(E1) 0003
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(C) The expected frequencies are E1: 500(.25) 0003 125 E2: 500(.5) 0003 250 E3: 500(.25) 0003 125 The actual frequencies obtained from performing the experiment are reasonably close to the expected frequencies. Increasing the number of trials of the experiment would most likely produce even better approximations. 0002 MATCHED PROBLEM 7
One die is rolled 500 times with the following frequencies of outcomes: Outcome
1
2
3
4
5
6
Frequency
89
83
77
91
72
88
(A) Compute the approximate empirical probability for each outcome. (B) Compute the theoretical probability for each outcome. (C) Compute the expected frequency for each outcome. 0002
Technology Connections The data in Example 7 were not generated by tossing two coins 500 times. Instead, the experiment was simulated by a random number generator on a graphing calculator. The command randint (0, 1, 500) produces a random sequence of 500 terms; each term is 0 or 1 with equal liklihood. Thinking of 1 as heads and 0 as tails, such a sequence represents 500 tosses of a single coin. Adding two such sequences together produces a sequence of 500 terms in which each term represents the number of heads in a toss of two coins
[see Fig. 3(a)]. We determine the frequency of each outcome (0, 1, or 2 heads) in 500 tosses of two coins as follows: first, we construct a histogram [Figs. 3(b) and 3(c)], then we use the TRACE command to read off the frequencies [Figs. 3(d), 3(e), and 3(f)]. Compare with the data of Example 7. If you perform the same simulation on your graphing calculator, you are not likely to get exactly the same results. But the approximate empirical probabilities you obtain will be close to the theoretical probabilities.
(a) Generating the random numbers
(b) Setting up the histogram
(c) Selecting the window variables
(d) 0 heads: 117
(e) 1 head: 262
(f) 2 heads: 121
Z Figure 3 Simulating 500 tosses of two coins.
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Empirical Probabilities for an Insurance Company An insurance company selected 1,000 drivers at random in a particular city to determine a relationship between age and accidents. The data obtained are listed in Table 1. Compute the approximate empirical probabilities of the following events for a driver chosen at random in the city: (A) E1: being under 20 years old and having exactly three accidents in 1 year (B) E2: being 30–39 years old and having one or more accidents in 1 year (C) E3: having no accidents in 1 year (D) E4: being under 20 years old or* having exactly three accidents in 1 year Table 1 Accidents in 1 Year
SOLUTIONS
Age
0
1
2
3
Over 3
Under 20
50
62
53
35
20
20–29
64
93
67
40
36
30–39
82
68
32
14
4
40–49
38
32
20
7
3
Over 49
43
50
35
28
24
(A) P(E1) 0003
35 0003 .035 1,000
(B) P(E2) 0003
68 0006 32 0006 14 0006 4 0003 .118 1,000
(C) P(E3) 0003
50 0006 64 0006 82 0006 38 0006 43 0003 .277 1,000
(D) P(E4) 0003
50 0006 62 0006 53 0006 35 0006 20 0006 40 0006 14 0006 7 0006 28 0003 .309 1,000
Notice that in this type of problem, which is typical of many realistic problems, approximate empirical probabilities are the only type we can compute. 0002 MATCHED PROBLEM 8
Referring to Table 1 in Example 8, compute the approximate empirical probabilities of the following events for a driver chosen at random in the city: (A) E1: being under 20 years old with no accidents in 1 year (B) E2: being 20–29 years old and having fewer than two accidents in 1 year (C) E3: not being over 49 years old 0002 Approximate empirical probabilities are often used to test theoretical probabilities. Equally likely assumptions may not be justified in reality. In addition to this use, there are many situations in which it is either very difficult or impossible to compute the theoretical *Interpret “or” in its inclusive sense, as customary in mathematics (a driver who is both under 20 and has three accidents must be counted once in the frequency of E4).
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probabilities for given events. For example, insurance companies use past experience to establish approximate empirical probabilities to predict future accident rates; baseball teams use batting averages, which are approximate empirical probabilities based on past experience, to predict the future performance of a player; and pollsters use approximate empirical probabilities to predict outcomes of elections. ANSWERS TO MATCHED PROBLEMS 1. (A) S1 0003 {BB, BG, GB, GG};
Sex of First Child B G
Sex of Second Child B G B G
Combined Outcomes BB BG GB GG
(C) S3 0003 {A, D} (D) The sample space in part A. (B) S2 0003 {0, 1, 2} 2. (A) {(4, 1), (3, 2), (2, 3), (1, 4)} (B) {(6, 5), (5, 6)} 3. (A) .74 (B) .76 (C) .5 4. (A) P(E5) 0003 19 (B) P(E6) 0003 181 5. C13,70002C52,7 0003 .000013 6. (C12,4 ⴢ C16,2)0002C28,6 0003 .158 7. (A) P(E1) 0003 .178, P(E2) 0003 .166, P(E3) 0003 .154, P(E4) 0003 .182, P(E5) 0003 .144, P(E6) 0003 .176 (B) 16 0003 .167 for each (C) 83.3 for each 8. (A) P(E1) 0003 .05 (B) P(E2) 0003 .157 (C) P(E3) 0003 .82
8-5
Exercises
1. What is a sample space? 2. Explain in your own words how the probability of an event is defined in terms of probabilities of simple events. 3. Explain the difference between a theoretical probability and an empirical probability.
14. A single card is drawn from a standard 52-card deck. What is the probability of getting a numbered card (that is, a two through ten)? 15. A fair coin is tossed three times. What is the probability of getting exactly two tails?
4. What is an equally likely assumption?
16. A fair coin is tossed three times. What is the probability of getting three tails?
5. A single fair die is rolled. What is the probability of getting a one or a six?
17. How would you interpret P(E) 0003 1?
6. A single fair die is rolled. What is the probability of getting a number greater than three? 7. A single card is drawn from a standard 52-card deck. What is the probability of getting a red card? 8. A single card is drawn from a standard 52-card deck. What is the probability of getting a club? 9. A fair coin is tossed twice. What is the probability of getting two heads? 10. A fair coin is tossed twice. What is the probability of getting at least one head? 11. Two fair dice are rolled. What is the probability of getting doubles? 12. Two fair dice are rolled. What is the probability of getting double sixes? 13. A single card is drawn from a standard 52-card deck. What is the probability of getting a king or a queen?
18. How would you interpret P(E) 0003 0? 19. A spinner can land on four different colors: red (R), green (G), yellow (Y), and blue (B). If we do not assume each color is as likely to turn up as any other, which of the following probability assignments have to be rejected, and why? (A) P(R) 0003 .15, P(G) 0003 0004.35, P(Y ) 0003 .50, P(B) 0003 .70 (B) P(R) 0003 .32, P(G) 0003 .28, P(Y ) 0003 .24, P(B) 0003 .30 (C) P(R) 0003 .26, P(G) 0003 .14, P(Y ) 0003 .30, P(B) 0003 .30 20. Under the probability assignments in Problem 19, part C, what is the probability that the spinner will not land on blue? 21. Under the probability assignments in Problem 19, part C, what is the probability that the spinner will land on red or yellow? 22. Under the probability assignments in Problem 19, part C, what is the probability that the spinner will not land on red or yellow? 23. A ski jumper has jumped over 300 feet in 25 out of 250 jumps. What is the approximate empirical probability of the next jump being over 300 feet?
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24. In a certain city there are 4,000 youths between 16 and 20 years old who drive cars. If 560 of them were involved in accidents last year, what is the approximate empirical probability of a youth in this age group being involved in an accident this year?
In Problems 39–50, an experiment consists of rolling two fair dice. Let a and b denote the numbers of dots on the two sides facing up. Use the sample space shown in Figure 1 on page 546 to find the probability of each event.
25. Out of 420 times at bat, a baseball player gets 189 hits. What is the approximate empirical probability that the player will get a hit next time at bat?
39. The sum of a and b is 3.
26. In a medical experiment, a new drug is found to help 2,400 out of 3,000 people. If a doctor prescribes the drug for a particular patient, what is the approximate empirical probability that the patient will be helped?
40. The sum of a and b is 5. 41. The sum of a and b is greater than 9. 42. The sum of a and b is less than 6. 43. The product of a and b is 12.
27. A small combination lock on a suitcase has three wheels, each labeled with the 10 digits from 0 to 9. If an opening combination is a particular sequence of three digits with no repeats, what is the probability of a person guessing the right combination?
44. The product of a and b is 6.
28. A combination lock has five wheels, each labeled with the 10 digits from 0 to 9. If an opening combination is a particular sequence of five digits with no repeats, what is the probability of a person guessing the right combination?
48. a 0005 b
Problems 29–34 involve an experiment consisting of dealing 5 cards from a standard 52-card deck. In Problems 29–32, what is the probability of being dealt:
51. Five thousand people work in a large auto plant. An individual is selected at random and his or her birthday (month and day, not year) is recorded. Set up an appropriate sample space for this experiment and assign acceptable probabilities to the simple events.
29. Five black cards 30. Five hearts 31. Five face cards if an ace is considered to be a face card. 32. Five nonface cards if an ace is considered to be a one and not a face card. 33. If we are interested in the number of aces in a 5-card hand, would S 0003 {0, 1, 2, 3, 4} be an acceptable sample space? Would it be an equally-likely sample space? Explain. 34. If we are interested in the number of black cards in a 5-card hand, would S 0003 {0, 1, 2, 3, 4, 5} be an acceptable sample space? Would it be an equally-likely sample space? Explain. 35. If four-digit numbers less than 5,000 are randomly formed from the digits 1, 3, 5, 7, and 9, what is the probability of forming a number divisible by 5? Digits may be repeated; for example, 1,355 is acceptable. 36. If code words of four letters are generated at random using the letters A, B, C, D, E, and F, what is the probability of forming a word without a vowel in it? Letters may be repeated. 37. Suppose five thank-you notes are written and five envelopes are addressed. Accidentally, the notes are randomly inserted into the envelopes and mailed without checking the addresses. What is the probability that all five notes will be inserted into the correct envelopes? 38. Suppose six people check their coats in a checkroom. If all claim checks are lost and the six coats are randomly returned, what is the probability that all six people will get their own coats back?
45. The product of a and b is less than 5. 46. The product of a and b is greater than 15. 47. a 0003 b
49. At least one of a or b is a 6. 50. Exactly one of a or b is a 6.
52. In a hotly contested three-way race for governor of Minnesota, the leading candidates are running neck-and-neck while the third candidate is receiving half the support of either of the others. Registered voters are chosen at random and are asked for which of the three they are most likely to vote. Set up an appropriate sample space for the random survey experiment and assign acceptable probabilities to the simple events. 53. A pair of dice is rolled 500 times with the following frequencies: Sum Frequency
2 3 4 5 6 7 8 9 10 11 12 11 35 44 50 71 89 72 52 36 26 14
(A) Compute the approximate empirical probability for each outcome. (B) Compute the theoretical probability for each outcome, assuming fair dice. (C) Compute the expected frequency of each outcome. (D) Describe how a random number generator could be used to simulate this experiment. If your graphing calculator has a random number generator, use it to simulate 500 tosses of a pair of dice and compare your results with part C. 54. Three coins are flipped 500 times with the following frequencies of outcomes: Three heads: 58 One head: 190
Two heads: 198 Zero heads: 54
(A) Compute the approximate empirical probability for each outcome. (B) Compute the theoretical probability for each outcome, assuming fair coins. (C) Compute the expected frequency of each outcome.
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(D) Describe how a random number generator could be used to simulate this experiment. If your graphing calculator has a random number generator, use it to simulate 500 tosses of three coins and compare your results with part C. 55. (A) Is it possible to get 29 heads in 30 flips of a fair coin? Explain. (B) If you flip a coin 50 times and get 42 heads, would you suspect that the coin was unfair? Why or why not? If you suspect an unfair coin, what empirical probabilities would you assign to the simple events of the sample space? 56. (A) Is it possible to get nine double sixes in 12 rolls of a pair of fair dice? Explain. (B) If you roll a pair of dice 40 times and get 14 double sixes, would you suspect that the dice were unfair? Why or why not? If you suspect unfair dice, what empirical probability would you assign to the event of rolling a double six? An experiment consists of tossing three fair coins, but one of the three coins has a head on both sides. Compute the probabilities of obtaining the indicated results in Problems 57–62. 57. One head
58. Two heads
59. Three heads
60. Zero heads
61. More than one head
62. More than one tail
An experiment consists of rolling two fair dice and adding the dots on the two sides facing up. Each die has one dot on two opposite faces, two dots on two opposite faces, and three dots on two opposite faces. Compute the probabilities of obtaining the indicated sums in Problems 63–70. 63. 2
64. 3
65. 4
68. 7
69. An odd sum
66. 5
67. 6
70. An even sum
An experiment consists of dealing 5 cards from a standard 52-card deck. In Problems 71–78, what is the probability of being dealt the following cards? 71. Five cards, jacks through aces 72. Five cards, 2 through 10 73. Four aces 74. Four of a kind
8-6
75. Straight flush, ace high; that is, 10, jack, queen, king, ace in one suit 76. Straight flush, starting with 2; that is, 2, 3, 4, 5, 6 in one suit 77. Two aces and three queens 78. Two kings and three aces
APPLICATIONS 79. MARKET ANALYSIS A company selected 1,000 households at random and surveyed them to determine a relationship between income level and the number of television sets in a home. The information gathered is listed in the table: Televisions per Household Yearly Income ($)
0
1
2
3
Above 3
Less than 12,000
0
40
51
11
0
12,000–19,999
0
70
80
15
1
20,000–39,999
2
112
130
80
12
40,000–59,999
10
90
80
60
21
60,000 or more
30
32
28
25
20
Compute the approximate empirical probabilities: (A) Of a household earning $12,000–$19,999 per year and owning exactly three television sets (B) Of a household earning $20,000–$39,999 per year and owning more than one television set (C) Of a household earning $60,000 or more per year or owning more than three television sets (D) Of a household not owning zero television sets 80. MARKET ANALYSIS Use the sample results in Problem 79 to compute the approximate empirical probabilities: (A) Of a household earning $40,000–$59,999 per year and owning zero television sets (B) Of a household earning $12,000–$39,999 per year and owning more than two television sets (C) Of a household earning less than $20,000 per year or owning exactly two television sets (D) Of a household not owning more than three television sets
The Binomial Formula Z Using Pascal’s Triangle Z The Binomial Formula Z Proving the Binomial Formula
In a surprising number of areas in math, it turns out to be useful to expand expressions of the form (a 0006 b)n, where n is a natural number. This is known as a binomial expansion. Expanding a binomial is pretty straightforward for small values of n, but gets hard very
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quickly as n increases. The good news is that it turns out that the coefficients in such an expansion are related to counting techniques that we have already learned about.
Z Using Pascal’s Triangle Let’s begin by expanding (a 0006 b)n for the first few values of n. We include n 0003 0, which is not a natural number, for reasons of completeness that will become apparent later. (a 0006 b)0 0003 1 (a 0006 b)1 0003 a 0006 b (a 0006 b)2 0003 a2 0006 2ab 0006 b2 (a 0006 b)3 0003 a3 0006 3a2b 0006 3ab2 0006 b3
ZZZ EXPLORE-DISCUSS 1
(1)
Based on the expansions in equations (1), how many terms would you expect (a 0006 b)n to have? What is the first term? What is the last term?
1 1 1 1
1 2
3
1 3
1
Now let’s examine just the coefficients of the expansions in equations (1) arranged in a form that is usually referred to as Pascal’s triangle (Fig. 1). It is convenient to refer to the top row of Pascal’s triangle (containing a single 1) as row 0. Then row 1 is “1 1,” row 2 is “1 2 1,” and row 3 is “1 3 3 1.” For n a natural number, the first two entries of row n are 1 and n.
Z Figure 1 Pascal’s triangle.
ZZZ EXPLORE-DISCUSS 2
Refer to Figure 1. (A) How is the middle element of row 2 related to the elements in the row above? (B) How are the two inner elements of row 3 related to the elements in the row above? (C) Based on your observations in parts A and B, conjecture the entries of row 4 and row 5. Check your conjecture by expanding (a 0006 b)4 and (a 0006 b)5.
Z Figure 2
Many students find Pascal’s triangle a useful tool for determining the coefficients in the expansion of (a 0006 b)n, especially for small values of n. Figure 2 contains output from a program called PASCAL.* You should recognize the output in the table—it is the first six lines of Pascal’s triangle. The major drawback of using this triangle, whether it is generated by hand or on a graphing calculator, is that to find the elements in a given row, you must write out all the preceding rows. It would be useful to find a formula that gives the coefficients for a binomial expansion directly. Fortunately, such a formula exists— the combination formula Cn,r introduced in Section 8-4.
Z The Binomial Formula When working with binomial expansions, it is customary to use the notation ( nr ) for Cn,r. Recall the combination formula from Section 8-4.
*Programs for TI-84 and TI-86 graphing calculators can be found at the website for this book.
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COMBINATION FORMULA For nonnegative integers r and n, 0 0007 r 0007 n,
n n! a b 0003 Cn,r 0003 r r!(n 0004 r)! Theorem 1 establishes that the coefficients in a binomial expansion can always be expressed in terms of the combination formula. This is a very important theoretical result and a very practical tool. As we will see, using this theorem in conjunction with a graphing calculator provides a very efficient method for expanding binomials. Z THEOREM 1 Binomial Formula For n a positive integer n n (a 0006 b)n 0003 a a b an0004k bk k00030 k
We defer the proof of Theorem 1 until the end of this section. Because the values of the combination formula are the coefficients in a binomial expansion, it is natural to call them binomial coefficients.
EXAMPLE
1
Using the Binomial Formula Use the binomial formula to expand (x 0006 y)6. 6 6 (x 0006 y)6 0003 a a b x60004ky k k00030 k
6 6 6 6 6 6 6 0003 a b x6y0 0006 a b x5y 0006 a b x4y2 0006 a b x3y3 0006 a b x2y4 0006 a b xy5 0006 a b x0y6 0 1 2 3 4 5 6 0003 x6 0006 6x5y 0006 15x4y2 0006 20x3y3 0006 15x2y4 0006 6xy5 0006 y6 Note that the coefficients (1, 6, 15, 20, 15, 6, 1) are the entries of row 6 of Pascal’s triangle. 0002 MATCHED PROBLEM 1
EXAMPLE
2
Use the binomial formula to expand (x 0006 1)5.
0002
Using the Binomial Formula Use the binomial formula to expand (3p 0004 2q)4.
SOLUTION
(3p 0004 2q)4 0003 [(3p) 0006 (00042q)] 4 4 4 0003 a a b (3p)40004k(00042q)k k00030 k 4 4 0003 a a b 340004k(00042)kp40004kqk k00030 k
0003 1(3)4(00042)0p4q0 0006 4(3)3(00042)p3q 0006 6(3)2(00042)2p2q2 0006 4(3)(00042)3pq3 0006 1(3)0(00042)4p0q4 0003 81p4 0004 216p3q 0006 216p2q2 0004 96pq3 0006 16q4 Note that the coefficients (81, 0004216, 216, 000496, 16) are formed by multiplying the entries in row 4 of Pascal’s triangle (1, 4, 6, 4, 1) by the appropriate powers of 3 and 00042.
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Technology Connections The table feature on a graphing calculator provides an efficient alternative to calculating the coefficients of Example 2 one by one (Fig. 3).
40004x x Z Figure 3 y1 0003 C4,x3 (00042) .
0002 MATCHED PROBLEM 2
Use the binomial formula to expand (2m 0004 5n)3.
0002
(A) Compute each term and also the sum of the alternating series
ZZZ EXPLORE-DISCUSS 3
6 6 6 6 a b0004a b0006a b0004...0006a b 0 1 2 6 (B) What result about an alternating series can be deduced by letting a 0003 1 and b 0003 00041 in the binomial formula?
EXAMPLE
3
Using the Binomial Formula Find the term containing x9 in the expansion of (x 0006 3)14.
SOLUTION
In the expansion 14 14 (x 0006 3)14 0003 a a b x140004k3k k k00030
the exponent of x is 9 when k 0003 5. So the term containing x9 is a MATCHED PROBLEM 3
EXAMPLE
4
14 9 5 b x 3 0003 (2,002)(243)x9 0003 486,486x9 5
Find the term containing y8 in the expansion of (2 0006 y)14.
0002
0002
Using the Binomial Formula If the terms in the expansion of (x 0004 2)20 are arranged in decreasing powers of x, find the fourth term and the sixteenth term.
SOLUTION
In the expansion of (a 0006 b)n, the exponent of b in the rth term is r 0004 1 and the exponent of a is n 0004 (r 0004 1). Therefore Fourth term:
Sixteenth term:
20 a b x17(00042)3 3
a
20 ⴢ 19 ⴢ 18 17 x (00048) 3ⴢ2ⴢ1 0003 00049,120x17 0003
20 5 b x (00042)15 15 20 ⴢ 19 ⴢ 18 ⴢ 17 ⴢ 16 5 x (000432,768) 5ⴢ4ⴢ3ⴢ2ⴢ1 0003 0004508,035,072 x5
0003
0002
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If the terms in the expansion of (u 0004 1)18 are arranged in decreasing powers of u, find the fifth term and the twelfth term. 0002
Z Proving the Binomial Formula We now prove that the binomial formula holds for all natural numbers n using mathematical induction. PROOF
State the conjecture. n n Pn: (a 0006 b)n 0003 a a b an0004jb j j00030 j
PART 1
Show that P1 is true. 1
1 10004j j 1 1 1 a a j b a b 0003 a 0 b a 0006 a 1 b b 0003 a 0006 b 0003 (a 0006 b)
j00030
P1 is true. PART 2
Show that if Pk is true, then Pk00061 is true. k k Pk: (a 0006 b)k 0003 a a b ak0004jb j j00030 j k 00061
Pk00061: (a 0006 b)k 00061 0003 a a j00030
k 0006 1 k 000610004j j ba b j
Assume Pk is true.
Show Pk
ⴙ 1
is true.
We begin by multiplying both sides of Pk by (a 0006 b): k k (a 0006 b)k(a 0006 b) 0003 c a a b ak 0004jb j d (a 0006 b) j00030 j
The left side of this equation is the left side of Pk00061. Now we multiply out the right side of the equation and try to obtain the right side of Pk 00061: k k k k (a 0006 b)k 00061 0003 c a b ak 0006 a b ak 00041b 0006 a b ak 00042b2 0006 . . . 0006 a b bk d (a 0006 b) 0 1 2 k k k k k 0003 c a b ak 00061 0006 a b akb 0006 a b ak 00041b2 0006 . . . 0006 a b abk d 1 2 k 0 k k k k 00041 2 . . . k k 0006 c a ba b 0006 a ba b 0006 0006a b abk 0006 a b bk 00061 d 0 1 k00041 k k k k k k 0003 a b ak 00061 0006 c a b 0006 a b d akb 0006 c a b 0006 a b d ak00041b2 0006 . . . 0 0 1 1 2 k k k 0006 ca b 0006 a b d abk 0006 a b bk 00061 k00041 k k
Use the distributive property.
Combine like terms.
We now use the following facts (the proofs are left as exercises; see Problems 63–65, Exercises 8-6). a
k k k00061 b0006a b0003a b r00041 r r
k k00061 a b0003a b 0 0
k k00061 a b0003a b k k00061
to rewrite the right side as a
k 0006 1 k 00061 k00061 k k 0006 1 k 00041 2 . . . ba 0006a ba b 0006 a ba b 0006 0 1 2 0006a
k00061 k 0006 1 k 00061 k 0006 1 k 000610004j j k00061 bb 0003 a a ba b b abk 0006 a k00061 j k j0003 0
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Because the right side of the last equation is the right side of Pk00061, we have shown that Pk 00061 follows from Pk. CONCLUSION
Pn is true. That is, the binomial formula holds for all positive integers n. ANSWERS TO MATCHED PROBLEMS 1. x5 0006 5x4 0006 10x3 0006 10x2 0006 5x 0006 1 2. 8m3 0004 60m2n 0006 150mn2 0004 125n3 8 14 7 3. 192,192y 4. 3,060u ; 000431,824u
8-6
Exercises
1. What is a binomial?
45. (2m 0006 n)12; eleventh term
46. (x 0006 2y)20; third term
2. What is a binomial coefficient?
47. [(w00022) 0004 2]12; seventh term
48. (x 0004 3)10; fourth term
49. (3x 0004 2y)8; sixth term
50. (2p 0004 3q)7; fourth term
3. Explain how the entries in Pascal’s triangle are generated. 5
4. How can Pascal’s triangle be used to expand (a 0006 b) ? In Problems 51–54, use the binomial formula to expand and simplify the difference quotient
In Problems 5–12, use Pascal’s triangle to evaluate each expression. 8 5. a b 3
8 6. a b 4
9 7. a b 6
9 8. a b 7
9. C7,5
10. C7,3
11. C9,0
12. C10,10
In Problems 13–20, evaluate each expression. 13. a
13 b 3
14. a
13 b 9
18. C52,4
17. C52,3
15. a
12 b 4
19. C12,6
16. a
12 b 11
20. C12,11
f (x 0006 h) 0004 f (x) h for the indicated function f. Discuss the behavior of the simplified form as h approaches 0. 51. f (x) 0003 x3
52. f(x) 0003 x4
53. f(x) 0003 x5
54. f(x) 0003 x6
In Problems 55–58, use a graphing calculator to graph each sequence and to display it in table form. 55. Find the number of terms of the sequence
Expand Problems 21–32 using the binomial formula. 21. (m 0006 n)3
22. (x 0006 2)3
23. (2x 0004 3y)3
24. (3u 0006 2v)3
25. (x 0004 2)4
26. (x 0004 y)4
27. (m 0006 3n)4
28. (3p 0004 q)4
29. (2x 0004 y)5
5
30. (2x 0004 1)
6
32. (2x 0004 y)
In Problems 33–42, find the term of the binomial expansion containing the given power of x. 33. (x 0006 1)7; x4
34. (x 0006 1)8; x5
35. (2x 0004 1)11; x6
36. (3x 0006 1)12; x7
37. (2x 0006 3)18; x14
38. (3x 0004 2)17; x5
39. (x2 0004 1)6; x8
40. (x2 0004 1)9; x7
41. (x2 0006 1)9; x11
56. Find the number of terms of the sequence a
40 40 40 40 b, a b, a b, . . . , a b 1 2 40 0
that are greater than one-half of the largest term. 57. (A) Find the largest term of the sequence a0, a1, a2, . . . , a10 to three decimal places, where ak 0003 a
10 b (0.6)100004k(0.4)k k
(B) According to the binomial formula, what is the sum of the series a0 0006 a1 0006 a2 0006 . . . 0006 a10?
42. (x2 0006 1)10; x14 In Problems 43–50, find the indicated term in each expansion if the terms of the expansion are arranged in decreasing powers of the first term in the binomial. 43. (u 0006 v)15; seventh term
20 20 20 20 b, a b, a b, . . . , a b 0 1 2 20
that are greater than one-half of the largest term.
6
31. (m 0006 2n)
a
44. (a 0006 b)12; fifth term
58. (A) Find the largest term of the sequence a0, a1, a2, . . . , a10 to three decimal places, where ak 0003 a
10 b (0.3)100004k(0.7)k k
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(B) According to the binomial formula, what is the sum of the series a0 0006 a1 0006 a2 0006 . . . 0006 a10?
n 66. Show that: a b is given by the recursion formula r
59. Evaluate (1.01)10 to four decimal places, using the binomial formula. [Hint: Let 1.01 0003 1 0006 0.01.]
n n n0004r00061 a b a b0003 r r00041 r
60. Evaluate (0.99)6 to four decimal places, using the binomial formula. n n 61. Show that: a b 0003 a b r n0004r
n where a b 0003 1. 0 67. Write 2n 0003 (1 0006 1)n and expand, using the binomial formula to obtain
n n 62. Show that: a b 0003 a b 0 n 63. Show that: a
n n n n 2n 0003 a b 0006 a b 0006 a b 0006 . . . 0006 a b 0 1 2 n
k k k00061 b0006a b0003a b r00041 r r
68. Write 0 0003 (1 0004 1)n and expand, using the binomial formula, to obtain n n n n 0 0003 a b 0004 a b 0006 a b 0004 . . . 0006 (00041)na b 0 1 2 n
k k00061 64. Show that: a b 0003 a b 0 0 k k00061 65. Show that: a b 0003 a b k k00061
CHAPTER
8-1
8
Review
Sequences and Series
A sequence is a function with the domain a set of successive integers. The symbol an, called the nth term, or general term, represents the range value associated with the domain value n. Unless specified otherwise, the domain is understood to be the set of natural numbers. A finite sequence has a finite domain, and an infinite sequence has an infinite domain. A recursion formula defines each term of a sequence in terms of one or more of the preceding terms. For example, the Fibonacci sequence is defined by an 0003 an00041 0006 an00042 for n 3, where a1 0003 a2 0003 1. If a1, a2, . . . , an, . . . is a sequence, then the expression a1 0006 a2 0006 . . . 0006 an 0006 . . . is called a series. A finite sequence produces a finite series, and an infinite sequence produces an infinite series. Series can be represented using the summation notation:
To use mathematical induction to prove statements involving laws of exponents, it is convenient to state a recursive definition of an: a1 0003 a
and
an00061 0003 ana
for any integer n 1
To deal with conjectures that may be true only for n m, where m is a positive integer, we use the extended principle of mathematical induction: Let m be a positive integer, let Pn be a statement associated with each integer n m, and suppose the following conditions are satisfied: 1. Pm is true. 2. For any integer k m, if Pk is true, then Pk 00061 is also true. Then the statement Pn is true for all integers n m.
n
. . . 0006 an a ak 0003 am 0006 am00061 0006
k0003m
where k is called the summing index. If the terms in the series are alternately positive and negative, the series is called an alternating series.
8-2
Mathematical Induction
A wide variety of statements can be proven using the principle of mathematical induction: Let Pn be a statement associated with each positive integer n and suppose the following conditions are satisfied: 1. P1 is true. 2. For any positive integer k, if Pk is true, then Pk 00061 is also true. Then the statement Pn is true for all positive integers n.
8-3
Arithmetic and Geometric Sequences
A sequence is called an arithmetic sequence, or arithmetic progression, if there exists a constant d, called the common difference, such that or an 0004 an00041 0003 d an 0003 an00041 0006 d for every n 7 1 The following formulas are useful when working with arithmetic sequences and their corresponding series: an 0003 a1 0006 (n 0004 1)d n Sn 0003 [2a1 0006 (n 0004 1)d] 2 n Sn 0003 (a1 0006 an) 2
nth-Term Formula Sum Formula—First Form Sum Formula—Second Form
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A sequence is called a geometric sequence, or a geometric progression, if there exists a nonzero constant r, called the common ratio, such that an 0003r an00041
an 0003 ran00041
or
for every n 7 1
The following formulas are useful when working with geometric sequences and their corresponding series: an 0003 a1r n00041
nth-Term Formula
Sn 0003
a1 0004 a1r 10004r
Sn 0003
a1 0004 ran 10004r
a1 S 0003 10004r
8-4
n
r00051 r00051
Sum Formula—First Form Sum Formula—Second Form
0004r0004 6 1
Sum of an Infinite Geometric Series
Multiplication Principle, Permutations, and Combinations
A counting technique is a mathematical method of determining the number of objects in a set without actually enumerating them. Given a sequence of operations, tree diagrams are often used to list all the possible combined outcomes. To count the number of combined outcomes without listing them, we use the multiplication principle (also called the fundamental counting principle): 1. If operations O1 and O2 are performed in order with N1 possible outcomes for the first operation and N2 possible outcomes for the second operation, then there are N1 ⴢ N2 possible outcomes of the first operation followed by the second. 2. In general, if n operations O1, O2, . . . , On are performed in order, with possible number of outcomes N1, N2, . . . , Nn, respectively, then there are N1 ⴢ N2 ⴢ . . . ⴢ Nn possible combined outcomes of the operations performed in the given order. The symbol n! is read n factorial and 0! is defined to be 1. A particular arrangement or ordering of n objects without repetition is called a permutation. The number of permutations of n objects is given by Pn,n 0003 n ⴢ (n 0004 1) ⴢ . . . ⴢ 1 0003 n! A permutation of a set of n objects taken r at a time is an arrangement of the r objects in a specific order. The number of permutations of n objects taken r at a time is given by Pn,r 0003
n! (n 0004 r)!
00007r0007n
A combination of a set of n objects taken r at a time is an r-element subset of the n objects. The number of combinations of n objects taken r at a time is given by Pn,r
n n! Cn,r 0003 a b 0003 0003 r r! r!(n 0004 r)!
00007r0007n
In a permutation, order is important. In a combination, order is not important.
8-5
565
Sample Spaces and Probability
The outcomes of an experiment are called simple events if one and only one of these results will occur in each trial of the experiment. The set of all simple events is called the sample space. Any subset of the sample space is called an event. An event is a simple event if it has only one element in it and a compound event if it has more than one element in it. We say that an event E occurs if any of the simple events in E occurs. A sample space S1 is more fundamental than a second sample space S2 if knowledge of which event occurs in S1 tells us which event in S2 occurs, but not conversely. Given a sample space S 0003 {e1, e2, . . . , en} with n simple events, to each simple event ei we assign a real number denoted by P(ei), that is called the probability of the event ei and satisfies: 1. 0 0007 P(ei) 0007 1 2. P(e1) 0006 P(e2) 0006 . . . 0006 P(en) 0003 1 Any probability assignment that meets conditions 1 and 2 is said to be an acceptable probability assignment. Given an acceptable probability assignment for the simple events in a sample space S, the probability of an arbitrary event E is defined as follows: 1. If E is the empty set, then P(E ) 0003 0. 2. If E is a simple event, then P(E ) has already been assigned. 3. If E is a compound event, then P(E ) is the sum of the probabilities of all the simple events in E. 4. If E is the sample space S, then P(E ) 0003 P(S ) 0003 1. If each of the simple events in a sample space S 0003 {e1, e2, . . . , en} with n simple events is equally likely to occur, then we assign the probability 10002n to each. If E is an arbitrary event in S, then P(E ) 0003
n(E ) Number of elements in E 0003 Number of elements in S n(S )
If we conduct an experiment n times and event E occurs with frequency f (E ), then the ratio f(E )0002n is called the relative frequency of the occurrence of event E in n trials. As n increases, f (E )0002n usually approaches a number that is called the empirical probability P(E ). So f(E )0002n is used as an approximate empirical probability for P(E ). If P(E ) is the theoretical probability of an event E and the experiment is performed n times, then the expected frequency of the occurrence of E is n ⴢ P(E ).
8-6
Binomial Formula
Pascal’s triangle is a triangular array of coefficients for the expansion of the binomial (a 0006 b)n, where n is a positive integer. Notation for the combination formula is n n! a b 0003 Cn,r 0003 r r!(n 0004 r)! For n a positive integer, the binomial formula is n n (a 0006 b)n 0003 a a b an0004kbk k00030 k
n The numbers a b, 0 0007 k 0007 n, are called binomial coefficients. k
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Review Exercises
Work through all the problems in this chapter review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. 1. Determine whether each of the following can be the first three terms of a geometric sequence, an arithmetic sequence, or neither. (A) 16, 00048, 4, . . . (B) 5, 7, 9, . . . (C) 00048, 00045, 00042, . . . (D) 2, 3, 5, . . . (E) 00041, 2, 00044, . . . In Problems 2–5: (A) Write the first four terms of each sequence. (B) Find a10. (C) Find S10. 2. an 0003 2n 0006 3
3. an 0003 32(12)n
4. a1 0003 00048; an 0003 an00041 0006 3, n 2 5. a1 0003 00041, an 0003 (00042)an00041, n 2 6. Find S in Problem 3.
9.
7! 2!(7 0004 2)!
19. Pn: 49n 0004 1 is divisible by 6 In Problems 20–22, write Pk and Pk 0006 1. 20. For Pn in Problem 17
21. For Pn in Problem 18
22. For Pn in Problem 19 23. Either prove the statement is true or prove it is false by finding a counterexample: If n is a positive integer, then the 1 1 1 sum of the series 1 0006 0006 0006 . . . 0006 is less than 4. n 2 3 Write Problems 24 and 25 without summation notation, and find the sum. 10
24. S10 0003 a (2k 0004 8) k00031
7 16 25. S7 0003 a k k00031 2
26. S 0003 27 0004 18 0006 12 0006 . . . 0003 ?
Evaluate the expression in Problems 7–10. 7. 6!
18. Pn: 2 0006 4 0006 8 0006 . . . 0006 2n 0003 2n00061 0004 2
27. Write
22! 8. 19! 10. C6,2 and P6,2
11. A single die is rolled and a coin is flipped. How many combined outcomes are possible? Solve (A) By using a tree diagram (B) By using the multiplication principle 12. How many seating arrangements are possible with six people and six chairs in a row? Solve by using the multiplication principle. 13. Solve Problem 12 using permutations or combinations, whichever is applicable. 14. In a single deal of 5 cards from a standard 52-card deck, what is the probability of being dealt five clubs? 15. Betty and Bill are members of a 15-person ski club. If the president and treasurer are selected by lottery, what is the probability that Betty will be president and Bill will be treasurer? A person cannot hold more than one office. 16. A drug has side effects for 50 out of 1,000 people in a test. What is the approximate empirical probability that a person using the drug will have side effects? Verify the statement Pn in Problems 17–19 for n 0003 1, 2, and 3. 17. Pn: 5 0006 7 0006 9 0006 . . . 0006 (2n 0006 3) 0003 n2 0006 4n
Sn 0003
(00041)n00061 1 1 1 0004 0006 0006...0006 3 9 27 3n
using summation notation, and find S . 28. Someone tells you that the following approximate empirical probabilities apply to the sample space {e1, e2, e3, e4}: P(e1) 0003 .1, P(e2) 0003 0004.2, P(e3) 0003 .6, P(e4) 0003 2. There are three reasons why P cannot be a probability function. Name them. 29. Six distinct points are selected on the circumference of a circle. How many triangles can be formed using these points as vertices? 30. In an arithmetic sequence, a1 0003 13 and a7 0003 31. Find the common difference d and the fifth term a5. 31. How many three-letter code words are possible using the first eight letters of the alphabet if no letter can be repeated? If letters can be repeated? If adjacent letters cannot be alike? 32. Two coins are flipped 1,000 times with the following frequencies: Two heads:
210
One head:
480
Zero heads:
310
(A) Compute the empirical probability for each outcome. (B) Compute the theoretical probability for each outcome. (C) Using the theoretical probabilities computed in part B, compute the expected frequency of each outcome, assuming fair coins.
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Review Exercises
33. From a standard deck of 52 cards, what is the probability of obtaining a 5-card hand: (A) Of all diamonds? (B) Of three diamonds and two spades?
51. How many ways can two people be seated in a row of four chairs?
Write answers in terms of Cn,r or Pn,r, as appropriate. Do not evaluate.
53. If three people are selected from a group of seven men and three women, what is the probability that at least one woman is selected?
34. A group of 10 people includes one married couple. If four people are selected at random, what is the probability that the married couple is selected?
52. Expand (x 0006 i)6, where i is the imaginary unit, using the binomial formula.
54. Three fair coins are tossed 1,000 times with the following frequencies of outcomes:
35. A spinning device has three numbers, 1, 2, 3, each as likely to turn up as the other. If the device is spun twice, what is the probability that: (A) The same number turns up both times? (B) The sum of the numbers turning up is 5?
Number of heads Frequency
38.
20! 18!(20 0004 18)!
39. a
16 b 12
40. a
11 b 11
5
41. Expand (x 0004 y) using the binomial formula. 42. Find the term containing x6 in the expansion of (x 0006 2)9. 43. If the terms in the expansion of (2x 0004 y)12 are arranged in descending powers of x, find the tenth term. Establish each statement in Problems 44–46 for all natural numbers using mathematical induction. 44. Pn in Problem 17
45. Pn in Problem 18
46. Pn in Problem 19 In Problems 47 and 48, find the smallest positive integer n such that an bn by graphing the sequences {an} and {bn} with a graphing calculator. Check your answer by using a graphing calculator to display both sequences in table form. 47. an 0003 C50,n, bn 0003 3n 48. a1 0003 100, an 0003 0.99an00041 0006 5, bn 0003 9 0006 7n 49. How many different families with five children are possible, excluding multiple births, where the sex of each child in the order of their birth is taken into consideration? How many families are possible if the order pattern is not taken into account? 50. A free-falling body travels g/2 feet in the first second, 3g00022 feet during the next second, 5g00022 feet the next, and so on. Find the distance fallen during the twenty-fifth second and the total distance fallen from the start to the end of the twenty-fifth second.
2
3
120
360
350
170
Prove that each statement in Problems 55–59 holds for all positive integers using mathematical induction. n
n
2
55. a k3 0003 a a kb k00031 2n
56. x 57.
Evaluate Problems 38–40.
1
(A) What is the approximate empirical probability of obtaining two heads? (B) What is the theoretical probability of obtaining two heads? (C) What is the expected frequency of obtaining two heads?
36. Use the formula for the sum of an infinite geometric series to write 0.727 272 . . . 0003 0.72 as the quotient of two integers. 37. Solve the following problems using Pn,r or Cn,r, as appropriate: (A) How many three-digit opening combinations are possible on a combination lock with six digits if the digits cannot be repeated? (B) Suppose five tennis players have made the finals. If each of the five players is to play every other player exactly once, how many games must be scheduled?
0
k00031
2n
0004 y is divisible by x 0004 y, x 0005 y
an 0003 an0004m; n 7 m; n, m positive integers am
58. {an} 0003 {bn}, where an 0003 an00041 0006 2, a1 0003 00043, bn 0003 00045 0006 2n 59. (1!)1 0006 (2!)2 0006 (3!)3 0006 . . . 0006 (n!)n 0003 (n 0006 1)! 0004 1 (From U.S.S.R. Mathematical Olympiads, 1955–1956, Grade 10.)
APPLICATIONS 60. LOAN REPAYMENT You borrow $7,200 and agree to pay 1% of the unpaid balance each month for interest. If you decide to pay an additional $300 each month to reduce the unpaid balance, how much interest will you pay over the 24 months it will take to repay this loan? 61. ECONOMICS Due to reduced taxes, an individual has an extra $2,400 in spendable income. If we assume that the individual spends 75% of this on consumer goods, and the producers of those consumer goods in turn spend 75% on consumer goods, and that this process continues indefinitely, what is the total amount (to the nearest dollar) spent on consumer goods? 62. COMPOUND INTEREST If $500 is invested at 6% compounded annually, the amount A present after n years forms a geometric sequence with common ratio 1 0006 0.06 0003 1.06. Use a geometric sequence formula to find the amount A in the account (to the nearest cent) after 10 years; after 20 years. 63. TRANSPORTATION A distribution center A wishes to distribute its products to five different retail stores, B, C, D, E, and F, in a city. How many different route plans can be constructed so that a single truck can start from A, deliver to each store exactly once, and then return to the center? 64. MARKET ANALYSIS A DVD distributor selected 1,000 persons at random and surveyed them to determine a relationship between age of purchaser and annual DVD purchases. The results are given in the table on page 568.
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DVDs Purchased Annually Age
0
1
2
Above 2
Under 12
60
70
30
10
170
12–18
30
100
100
60
290
19–25
70
110
120
30
330
Over 25
100
50
40
20
210
Totals
260
330
290
120
1,000
CHAPTER
ZZZ
Totals
Find the empirical probability that a person selected at random (A) Is over 25 and buys exactly two DVDs annually. (B) Is 12–18 years old and buys more than one DVD annually. (C) Is 12–18 years old or buys more than one DVD annually. 65. QUALITY CONTROL Twelve precision parts, including two that are substandard, are sent to an assembly plant. The plant manager selects four at random and will return the whole shipment if one or more of the samples are found to be substandard. What is the probability that the shipment will be returned?
8
GROUP ACTIVITY Sequences Specified by Recursion Formulas
The recursion formula* an 0003 5an00041 0004 6an00042, together with the initial values a1 0003 4, a2 0003 14, specifies the sequence {an} whose first several terms are 4, 14, 46, 146, 454, 1,394, . . . . The sequence {an} is neither arithmetic nor geometric. Nevertheless, because it satisfies a simple recursion formula, it is possible to obtain an nth-term formula for {an} that is analogous to the nthterm formulas for arithmetic and geometric sequences. Such an nth-term formula is valuable because it allows us to estimate a term of a sequence without computing all the preceding terms. If the geometric sequence {rn} satisfies the preceding recursion formula, then rn 0003 5rn00041 0004 6rn00042. Dividing both sides by rn00042 leads to the quadratic equation r2 0004 5r 0006 6 0003 0, whose solutions are r 0003 2 and r 0003 3. Now it is easy to check that the geometric sequences {2n} 0003 2, 4, 8, 16, . . . and {3n} 0003 3, 9, 27, 81, . . . satisfy the recursion formula. Therefore, any sequence of the form {u2n 0006 v3n}, where u and v are constants, will satisfy the same recursion formula. We now find u and v so that the first two terms of {u2n 0006 v3n} are a1 0003 4, a2 0003 14. Letting n 0003 1 and n 0003 2 we see that u and v must satisfy the following linear system:
Solving the system gives u 0003 00041, v 0003 2. Therefore, an nth-term formula for the original sequence is an 0003 (00041)2n 0006 (2)3n. Note that the nth-term formula was obtained by solving a quadratic equation and a system of two linear equations in two variables. (A) Compute (00041)2n 0006 (2)3n for n 0003 1, 2, . . . , 6, and compare with the terms of {an}. (B) Estimate the one-hundredth term of {an}. (C) Show that any sequence of the form {u2n 0006 v3n}, where u and v are constants, satisfies the recursion formula an 0003 5an00041 0004 6an00042. (D) Find an nth-term formula for the sequence {bn} that is specified by b1 0003 5, b2 0003 55, bn 0003 3bn00041 0006 4bn00042. (E) Find an nth-term formula for the Fibonacci sequence. (F) Find an nth-term formula for the sequence {cn} that is specified by c1 0003 00043, c2 0003 15, c3 0003 99, cn 0003 6cn00041 0004 3cn00042 0004 10cn00043. (Because the recursion formula involves the three terms that precede cn, our method will involve the solution of a cubic equation and a system of three linear equations in three variables.)
2u 0006 3v 0003 4 4u 0006 9v 0003 14 *The program RECUR, found at the website for this book, evaluates the terms in any sequence defined by this type of recursion formula.
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11 A
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CHAPTERS
1–3
Cumulative Review Exercises
Work through all the problems in this cumulative review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. 1. Solve for x:
Problems 16–18 refer to the function f given by the graph: f (x) 5
7x 3 0003 2x x 0002 10 0002 0007 00032 5 2 3
00025
5
x
In Problems 2–4, solve and graph the inequality. 2. 2(3 0002 y) 0003 4 0004 5 0002 y
16. Find the domain and range of f. Express answers in interval notation.
4. x2 0003 3x 0006 10 5. Perform the indicated operations and write the answer in standard form: (A) (2 0002 3i) 0002 (00025 0003 7i) (B) (1 0003 4i)(3 0002 5i) 50003i (C) 2 0003 3i In Problems 6–9, solve the equation. 2
6. 3x 0007 000212x 2
8. x 0002 6x 0003 2 0007 0
00025
3. 冟 x 0002 2冟 0005 7
2
17. Is f an even function, an odd function, or neither? Explain. 18. Use the graph of f to sketch a graph of the following: (A) y 0007 0002f(x 0003 1) (B) y 0007 2f (x) 0002 2 In Problems 19–21, solve the equation. 19.
7. 4x 0002 20 0007 0 9. x 0002 112 0002 x 0007 0
10. Given the points A 0007 (3, 2) and B 0007 (5, 6), find: (A) Distance between A and B. (B) Slope of the line through A and B. (C) Slope of a line perpendicular to the line through A and B. 11. Find the equation of the circle with radius 12 and center: (A) (0, 0) (B) (00023, 1) 12. Graph 2x 0002 3y 0007 6 and indicate its slope and intercepts. 13. Indicate whether each set defines a function. Find the domain and range of each function. (A) {(1, 1), (2, 1), (3, 1)} (B) {(1, 1), (1, 2), (1, 3)} (C) {(00022, 2), (00021, 00021), (0, 0), (1, 00021), (2, 2)} 14. For f (x) 0007 x2 0002 2x 0003 5 and g(x) 0007 3x 0002 2, find: (A) f (00022) 0003 g(3) (B) ( f 0003 g)(x) (C) ( f ° g)(x) f (a 0003 h) 0002 f (a) (D) h 15. How are the graphs of the following related to the graph of y 0007 冟 x 冟? (A) y 0007 2冟 x 冟 (B) y 0007 冟 x 0002 2 冟 (C) y 0007 冟 x 冟 0002 2
x00033 5x 0003 2 5 0003 0007 2x 0003 2 3x 0003 3 6
20.
3 1 6 0007 0002 x x00031 x00021
21. 2x 0003 1 0007 312x 0002 1 In Problems 22–24, solve and graph the inequality. 22. 冟 4x 0002 9 冟 7 3 24.
23. 2(3m 0002 4)2 0004 2
x00031 7 x00022 2
25. For what real values of x does the following expression represent a real number? 1x 0002 2 x00024 26. Perform the indicated operations and write the final answers in standard form: (A) (2 0002 3i)2 0002 (4 0002 5i)(2 0002 3i) 0002 (2 0003 10i) 1 (B) 35 0003 45i 0003 3 4 (C) i35 5 0003 5i 27. Convert to a 0003 bi form, perform the indicated operations, and write the final answers in standard form: (A) (5 0003 2100029) 0002 (2 0002 31000216) 12 0002 1000264 2 0003 7 1000225 (B) (C) 3 0002 100021 100024 In Problems 28–31, solve the equation. 28. 1 0003
14 6 0007 2 y y
29. 4x2/3 0002 4x1/3 0002 3 0007 0
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A-3
y
30. u4 0003 u2 0002 12 0007 0 5
31. 18t 0002 2 0002 2 1t 0007 1
Use a calculator to solve the equation or inequality in Problems 32 and 33. Compute answers to two decimal places.
5
x
32. 00023.45 6 1.86 0002 0.33x 0004 7.92 33. 2.35x2 0003 10.44x 0002 16.47 0007 0
00025
34. Solve for y in terms of x: 2y 0003 1 x00022 0007 x00031 y00022 35. Find each of the following for the function f given by the graph shown in the figure. (A) The domain of f (B) The range of f (C) f (00023) 0003 f(00022) 0003 f (2) (D) The intervals over which f is increasing (E) The x coordinates of any points of discontinuity f (x)
44. Let f (x) 0007 1x 0003 4 (A) Find f 00021(x). (B) Find the domain and range of f and f 00021. (C) Graph f, f 00021, and y 0007 x on the same coordinate system. Check by graphing f, f 00021, and y 0007 x in a squared window on a graphing calculator. 45. Find the center and radius of the circle given by the equation x2 0002 6x 0003 y2 0003 2y 0007 0. Graph the circle and show the center and the radius. 46. Discuss symmetry with respect to the x axis, y axis, and the origin for the equation xy 0003 冟 xy 冟 0007 5
5
00025
5
x
47. Write an equation for the graph in the figure in the form y 0007 a(x 0002 h)2 0003 k, where a is either 00021 or 00031 and h and k are integers. y 5
00025
36. Write equations of the lines (A) Parallel to (B) Perpendicular to the line 3x 0003 2y 0007 12 and passing through the point (00026, 1). Write the final answers in the slope–intercept form y 0007 mx 0003 b. 37. Find the domain of g(x) 0007 1x 0003 4. 38. Graph f(x) 0007 x2 0002 2x 0002 8. Show the axis of symmetry and vertex, and find the range, intercepts, and maximum or minimum value of f(x).
00025
5
00025
48. Solve for y in terms of x: x0003y 00071 x0003y y0002 x0002y
39. Given f (x) 0007 1兾(x 0002 2) and g(x) 0007 (x 0003 3)兾x, find f ⴰ g. What is the domain of f ⴰ g?
49. Find all roots: 3x2 0007 212x 0002 1.
40. Find f 00021(x) for f(x) 0007 2x 0003 5.
50. Consider the quadratic equation
41. Graph, finding the domain, range, and any points of discontinuity: f (x) 0007 e 42. Graph: (A) y 0007 2 1x 0003 1 (B) y 0007 0002 1x 0003 1
x00021 x2 0003 1
if x 6 0 if x 0006 0
43. The graph in the figure is the result of applying a sequence of transformations to the graph of y 0007 冟 x 冟. Describe the transformations verbally and write an equation for the graph in the figure.
x
x2 0003 bx 0003 1 0007 0 where b is a real number. Discuss the relationship between the values of b and the three types of roots listed in Table 1 in Section 1-5. 51. Find all solutions: 13 0002 2x 0002 1x 0003 7 0007 1x 0003 4. 52. Write in standard form:
a 0003 bi , a, b 0. a 0002 bi
53. Given f (x) 0007 x2 and g(x) 0007 24 0002 x2, find: (A) Domain of g (B) f兾g and its domain (C) f ⴰ g and its domain
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54. Let f(x) ⫽ x2 ⫺ 2x ⫺ 3, x ⱖ 1. (A) Find f ⫺1(x). (B) Find the domain and range of f ⫺1. (C) Graph f, f ⫺1, and y ⫽ x on the same coordinate system. Check by graphing f, f ⫺1, and y ⫽ x in a squared window on a graphing calculator.
63. COMPUTER SCIENCE Let f (x) ⫽ x ⫺ 2冀 xⲐ2冁 . This function can be used to determine if an integer is odd or even. (A) Find f(1), f(2), f(3), and f(4). (B) Find f (n) for any integer n. [Hint: Consider two cases, n ⫽ 2k and n ⫽ 2k ⫹ 1, k an integer.]
55. NUMBERS Find a number such that the number exceeds its reciprocal by 32.
64. DEPRECIATION Office equipment was purchased for $20,000 and is assumed to depreciate linearly to a scrap value of $4,000 after 8 years. (A) Find a linear function v ⫽ d(t) that relates value v in dollars to time t in years. (B) Find t ⫽ d ⫺1(v).
56. RATE–TIME A boat travels upstream for 35 miles and then returns to its starting point. If the round-trip took 4.8 hours and the boat’s speed in still water is 15 miles per hour, what is the speed of the current?
65. PROFIT AND LOSS ANALYSIS At a price of $p per unit, the marketing department at a company estimates that the weekly cost C and the weekly revenue R, in thousands of dollars, will be given by the equations
57. CHEMISTRY How many gallons of distilled water must be mixed with 24 gallons of a 90% sulfuric acid solution to obtain a 60% solution? 58. BREAK-EVEN ANALYSIS The publisher’s fixed costs for the production of a new study guide are $41,800. Variable costs are $4.90 per book. If the book is sold to bookstores for $9.65, how many must be sold for the publisher to break even? 59. FINANCE An investor instructs a broker to buy a certain stock whenever the price per share p of the stock is within $10 of $200. Express this instruction as an absolute value inequality. 60. PRICE AND DEMAND The weekly demand for mouthwash in a chain of drugstores is 1,160 bottles at a price of $3.79 each. If the price is lowered to $3.59, the weekly demand increases to 1,340 bottles. Assuming that the relationship between the weekly demand x and the price per bottle p is linear, express x as a function of p. How many bottles would the store sell each week if the price were lowered to $3.29? 61. BUSINESS—PRICING A telephone company begins a new pricing plan that charges customers for local calls as follows: The first 60 calls each month are 6 cents each, the next 90 are 5 cents each, the next 150 are 4 cents each, and any additional calls are 3 cents each. If C is the cost, in dollars, of placing x calls per month, write a piecewise definition of C as a function of x and graph. 62. CONSTRUCTION A homeowner has 80 feet of chain-link fencing to be used to construct a dog pen adjacent to a house (see the figure). (A) Express the area A(x) enclosed by the pen as a function of the width x. (B) From physical considerations, what is the domain of the function A? (C) Graph A and determine the dimensions of the pen that will make the area maximum. x
x
C ⫽ 88 ⫺ 12p
Cost equation
R ⫽ 15p ⫺ 2p2
Revenue equation
Find the prices for which the company has: (A) A profit (B) A loss 66. SHIPPING A ship leaves port A, sails east to port B, and then north to port C, a total distance of 115 miles. The next day the ship sails directly from port C back to port A, a distance of 85 miles. Find the distance between ports A and B and between ports B and C. 67. PHYSICS The distance s above the ground (in feet) of an object dropped from a hot-air balloon t seconds after it is released is given by s ⫽ a ⫹ bt2 where a and b are constants. Suppose the object is 2,100 feet above the ground 5 seconds after its release and 900 feet above the ground 10 seconds after its release. (A) Find the constants a and b. (B) How high is the balloon? (C) How long does the object fall? 68. PRICE AND DEMAND The demand for barley q (in thousands of bushels) and the corresponding price p (in cents) at a midwestern grain exchange are shown in the figure. p 350
Price (in cents)
APPLICATIONS
340 330 320 310 10
20
30
40
50
q
Barley (thousands of bushels)
(A) What is the demand (to the nearest thousand bushels) when the price is 325 cents per bushel? (B) Does the demand increase or decrease if the price is increased to 340 cents per bushel? By how much? (C) Does the demand increase or decrease if the price is decreased to 315 cents per bushel? By how much? (D) Write a brief description of the relationship between price and demand illustrated by this graph.
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(E) Use the graph to estimate the price (to the nearest cent) when the demand is 20, 25, 30, 35, and 40 thousand bushels. Use these data to find a quadratic regression model for the price of barley using the demand as the independent variable.
Table 1 Skid Marks Speed (mph)
Length of Skid Marks (feet)
69. STOPPING DISTANCE Table 1 contains data related to the length of the skid marks left by an automobile when making an emergency stop. A model for the skid mark length L (in feet) is
20
24
30
48
L 0007 f (s) 0007 0.05s2 0002 0.2s 0003 6.5, s 0006 20
40
77
where s is speed in miles per hour. (A) Graph L 0007 f(s) and the data for skid mark length on the same axes. (B) Find s 0007 f 00021(L) and find its domain and range. (C) An insurance investigator finds skid marks 220 feet long at the scene of an accident involving this automobile. How fast (to the nearest mile per hour) was the automobile traveling when it made these skid marks?
50
115
60
187
70
246
80
312
4–5
CHAPTERS
A-5
Cumulative Review Exercises
Work through all the problems in this cumulative review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text.
3. For P(x) 0007 3x3 0003 5x2 0002 18x 0002 3 and D(x) 0007 x 0003 3, use synthetic division to divide P(x) by D(x), and write the answer in the form P(x) 0007 D(x)Q(x) 0003 R.
1. Let P(x) be the polynomial whose graph is shown in the figure. (A) Assuming that P(x) has integer zeros and leading coefficient 1, find the lowest-degree equation that could produce this graph. (B) Describe the left and right behavior of P(x).
5. Let P(x) 0007 4x3 0002 5x2 0002 3x 0002 1. How do you know that P(x) has at least one real zero between 1 and 2?
P (x)
7. Solve for x. (A) y 0007 10 x (A) (2e x )3
00025
5
2. Match each equation with the graph of f, g, m, or n in the figure. (A) y 0007 (34)x (B) y 0007 (43)x 3 x 4 x (C) y 0007 (4) 0003 (3) (D) y 0007 (43)x 0002 (34)x 3
g
00024.5
(B)
e3x e00022x
10. Solve for x to three significant digits. (A) 10 x 0007 2.35 (B) ex 0007 87,500 (C) log x 0007 00021.25 (D) ln x 0007 2.75 In Problems 11 and 12, translate each statement into an equation using k as the constant of proportionality. 11. E varies directly as p and inversely as the cube of x.
4.5
00023
(B) y 0007 ln x
9. Solve for x exactly. Do not use a calculator or a table. (A) log3 x 0007 2 (B) log3 81 0007 x (C) logx 4 0007 00022
x
00025
f
6. Let P(x) 0007 x3 0003 x2 0002 10x 0003 8. Find all rational zeros for P(x).
8. Simplify.
5
mn
4. Let P(x) 0007 2(x 0003 2)(x 0002 3)(x 0002 5). What are the zeros of P(x)?
12. F is jointly proportional to q1 and q2 and inversely proportional to the square of r.
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13. Explain why the graph in the figure is not the graph of a polynomial function. y
26. Final all zeros (rational, irrational, and imaginary) exactly for P(x) 0007 x4 0003 5x3 0003 x2 0002 15x 0002 12, and factor P(x) into linear factors.
5
00025
25. Find all zeros (rational, irrational, and imaginary) exactly for P(x) 0007 4x3 0002 20x2 0003 29x 0002 15.
5
x
In Problems 27–36, solve for x exactly. Do not use a calculator or a table. 27. 2x 0007 4x00034
28. 2x2e0002x 0003 xe0002x 0007 e0002x
29. eln x 0007 2.5
30. logx 104 0007 4
2
00025
14. Explain why the graph in the figure is not the graph of a rational function. 15. The function f subtracts the square root of the domain element from three times the natural log of the domain element. Write an algebraic definition of f. 16. Write a verbal description of the function f(x) 0007 100e0.5x 0002 50. 2x 0003 8 . x00032 (A) Find the domain and the intercepts for f. (B) Find the vertical and horizontal asymptotes for f. (C) Sketch the graph of f. Draw vertical and horizontal asymptotes with dashed lines.
17. Let f (x) 0007
18. Find all zeros of P(x) 0007 (x3 0003 4x)(x 0003 4), and specify those zeros that are x intercepts.
31. log9 x 0007 000232 32. ln (x 0003 4) 0002 ln (x 0002 4) 0007 2 ln 3 33. ln (2x2 0003 2) 0007 2 ln (2x 0002 4) 34. log x 0003 log (x 0003 15) 0007 2 35. log (ln x) 0007 00021
36. 4 (ln x)2 0007 ln x2
In Problems 37–41, solve for x to three significant digits. 37. x 0007 log3 41 x
39. 4(2 ) 0007 20 41.
38. ln x 0007 1.45 40. 10e00020.5x 0007 1.6
1 e x 0002 e0002x 0007 e x 0003 e0002x 2
19. Solve (x3 0003 4x)(x 0003 4) 0004 0.
42. G is directly proportional to the square of x. If G 0007 10 when x 0007 5, find G when x 0007 7.
20. If P(x) 0007 2x3 0002 5x2 0003 3x 0003 2, find P(12) using the remainder theorem and synthetic division.
43. H varies inversely as the cube of r. If H 0007 162 when r 0007 2, find H when r 0007 3.
21. Which of the following is a factor of P(x)? P(x) 0007 x25 0002 x20 0003 x15 0003 x10 0002 x5 0003 1 (A) x 0002 1
(B) x 0003 1 4
2
In Problems 44–50, find the domain, range, and the equations of any horizontal or vertical asymptotes. 44. f(x) 0007 3 0003 2x
22. Let P(x) 0007 x 0002 8x 0003 3. (A) Graph P(x) and describe the graph verbally, including the number of x intercepts, the number of turning points, and the left and right behavior. (B) Approximate the largest x intercept to two decimal places.
45. f(x) 0007 2 0002 log3 (x 0002 1)
23. Let P(x) 0007 x5 0002 8x4 0003 17x3 0003 2x2 0002 20x 0002 8. (A) Approximate the zeros of P(x) to two decimal places and state the multiplicity of each zero. (B) Can any of these zeros be approximated with the bisection method? The MAXIMUM or MINIMUM commands? Explain.
48. f (x) 0007
4
3
2
24. Let P(x) 0007 x 0003 2x 0002 20x 0002 30. (A) Find the smallest positive and largest negative integers that, by Theorem 1 in Section 4-2, are upper and lower bounds, respectively, for the real zeros of P(x). (B) If (k, k 0003 1), k an integer, is the interval containing the largest real zero of P(x), determine how many additional intervals are required in the bisection method to approximate this zero to one decimal place. (C) Approximate the real zeros of P(x) to two decimal places.
46. f(x) 0007 5 0002 4x3 47. f(x) 0007 3 0003 2x4 5 x00033
49. f(x) 0007 20e0002x 0002 15 50. f (x) 0007 8 0003 ln (x 0003 2) 51. If the graph of y 0007 ln x is reflected in the line y 0007 x, the graph of the function y 0007 e x is obtained. Discuss the functions that are obtained by reflecting the graph of y 0007 ln x in the x axis and in the y axis. 52. (A) Explain why the equation e0002x 0007 ln x has exactly one solution. (B) Approximate the solution of the equation to two decimal places.
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In Problems 53 and 54, factor each polynomial in two ways: (A) As a product of linear factors (with real coefficients) and quadratic factors (with real coefficients and imaginary zeros). (B) As a product of linear factors with complex coefficients.
A-7
67. Solve (to three decimal places) 4x 6 3 x 00021 2
53. P(x) 0007 x4 0003 9x2 0003 18 54. P(x) 0007 x4 0002 23x2 0002 50
APPLICATIONS
55. Graph f and indicate any horizontal, vertical, or oblique asymptotes with dashed lines:
68. SHIPPING A mailing service provides customers with rectangular shipping containers. The length plus the girth of one of these containers is 10 feet (see the figure). If the end of the container is square and the volume is 8 cubic feet, find the side length of the end. Find solutions exactly; round irrational solutions to two decimal places.
f (x) 0007
x2 0003 4x 0003 8 x00032
56. Let P(x) 0007 x4 0002 28x3 0003 262x2 0002 922x 0003 1.083. Approximate (to two decimal places) the x intercepts and the local extrema. 57. Find a polynomial of lowest degree with leading coefficient 1 that has zeros 00021 (multiplicity 2), 0 (multiplicity 3), 3 0003 5i, and 3 0002 5i. Leave the answer in factored form. What is the degree of the polynomial? 58. If P(x) is a fourth-degree polynomial with integer coefficients and if i is a zero of P(x), can P(x) have any irrational zeros? Explain. 4
3
2
59. Let P(x) 0007 x 0003 9x 0002 500x 0003 20,000. (A) Find the smallest positive integer multiple of 10 and the largest negative integer multiple of 10 that, by Theorem 1 in Section 4-2, are upper and lower bounds, respectively, for the real zeros of P(x). (B) Approximate the real zeros of P(x) to two decimal places. 60. Find all zeros (rational, irrational, and imaginary) exactly for P(x) 0007 x5 0002 4x4 0003 3x3 0003 10x2 0002 10x 0002 12 and factor P(x) into linear factors. 61. Find rational roots exactly and irrational roots to two decimal places for P(x) 0007 x5 0003 4x4 0003 x3 0002 11x2 0002 8x 0003 4 62. Give an example of a rational function f(x) that satisfies the following conditions: the real zeros of f are 5 and 8; x 0007 1 is the only vertical asymptote; and the line y 0007 3 is a horizontal asymptote. 63. Use natural logarithms to solve for n. A0007P
(1 0003 i)n 0002 1 i
64. Solve ln y 0007 5x 0003 ln A for y. Express the answer in a form that is free of logarithms. 65. Solve for x. y0007 66. Solve
x3 0002 x 0006 0. x3 0002 8
ex 0002 2e 0002x 2
gth
Len
x
Girth
x
y
69. GEOMETRY The diagonal of a rectangle is 2 feet longer than one of the sides, and the area of the rectangle is 6 square feet. Find the dimensions of the rectangle to two decimal places. 70. POPULATION GROWTH If the Democratic Republic of the Congo has a population of about 60 million people and a doubling time of 23 years, find the population in (A) 5 years (B) 30 years Compute answers to three significant digits. 71. COMPOUND INTEREST How long will it take money invested in an account earning 7% compounded annually to double? Use the annual compounding growth model P 0007 P0(1 0003 r)t, and compute the answer to three significant digits. 72. COMPOUND INTEREST Repeat Problem 71 using the continuous compound interest model P 0007 P0ert. 73. EARTHQUAKES If the 1906 and 1989 San Francisco earthquakes registered 8.3 and 7.1, respectively, on the Richter scale, how many times more powerful was the 1906 earthquake than the 1989 earthquake? Use the formula M 0007 23 log (E000bE0), where E 0 0007 104.40 joules, and compute the answer to one decimal place. 74. SOUND If the decibel level at a rock concert is 88, find the intensity of the sound at the concert. Use the formula D 0007 10 log (I兾I0), where I0 0007 10000212 watts per square meter, and compute the answer to two significant digits. 75. ASTRONOMY The square of the time t required for a planet to make one orbit around the sun varies directly as the cube of its mean (average) distance d from the sun. Write the equation of variation, using k as the constant of variation.
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76. PHYSICS Atoms and molecules that make up the air constantly fly about like microscopic missiles. The velocity v of a particular particle at a fixed temperature varies inversely as the square root of its molecular weight w. If an oxygen molecule in air at room temperature has an average velocity of 0.3 mile/second, what will be the average velocity of a hydrogen molecule, given that the hydrogen molecule is one-sixteenth as heavy as the oxygen molecule? Problems 77 and 78 require a graphing calculator or a computer that can calculate linear, quadratic, cubic, and exponential regression models for a given data set. 77. Table 1 shows the life expectancy (in years) at birth for residents of the United States from 1970 to 1995. Let x represent years since 1970. Use the indicated regression model to estimate the life expectancy (to the nearest tenth of a year) for a U.S. resident born in 2010. (A) Linear regression (B) Quadratic regression (C) Cubic regression (D) Exponential regression
CHAPTERS
6–8
Table 1 Year
Life Expectancy
1970
70.8
1975
72.6
1980
73.7
1985
74.7
1990
75.4
1995
75.9
2000
77.0
2005
77.7
Source: U.S. Census Bureau
78. Refer to Problem 77. The Census Bureau projected the life expectancy for a U.S. resident born in 2010 to be 77.9 years. Which of the models in Problem 77 is closest to the Census Bureau projection?
Cumulative Review Exercises
Work through all the problems in this cumulative review and check answers in the back of the book. Answers to all review problems are there, and following each answer is a number in italics indicating the section in which that type of problem is discussed. Where weaknesses show up, review appropriate sections in the text. Note that Problems 4, 15, 16, 40, 41, 48, 49, and 88 are from sections that appear online. 1. Solve using substitution or elimination by addition: 3x 0002 5y 0007 11 2x 0003 3y 0007 1 2. Solve by graphing: 2x 0002 y 0007 00024 3x 0003 y 0007 00021 3. Solve by substitution or elimination by addition: 00026x 0003 3y 0007 2 2x 0002 y 0007 1 4. Solve by graphing: 3x 0003 5y 0004 15 x, y 0006 0 5. Determine whether each of the following can be the first three terms of an arithmetic sequence, a geometric sequence, or neither. (A) 20, 15, 10, . . . (B) 5, 25, 125, . . . (C) 5, 25, 50, . . . (D) 27, 00029, 3, . . . (E) 00029, 00026, 00023, . . .
In Problems 6–8: (A) Write the first four terms of each sequence. (B) Find a8. (C) Find S8. 6. an 0007 2 ⴢ 5n
7. an 0007 3n 0002 1
8. a1 0007 100; an 0007 an00021 0002 6, n 0006 2 9. Evaluate each of the following: 32! 9! (A) 8! (B) (C) 30! 3!(9 0002 3)! 10. Evaluate each of the following: 7 (A) a b (B) C7,2 (C) P7,2 2 In Problems 11–13, graph each equation and locate foci. Locate the directrix for any parabolas. Find the lengths of major, minor, transverse, and conjugate axes where applicable. 11. 25x2 0002 36y2 0007 900
12. 25x2 0003 36y2 0007 900
13. 25x2 0002 36y 0007 0 14. Find each determinant: 00023 5 5 (A) ` (B) ` ` 2 00022 00025 15. Solve x2 0003 y2 0007 2 2x 0002 y 0007 1
3 ` 00023
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APPENDIX A
16. Find the maximum and minimum value of z 0007 2x 0003 3y over the feasible region S:
(0, 10) (6, 7) S
x1 0003 3x2 0007 10 (5, 0)
2x1 0002 x2 0007 00021
x
5
17. Perform the operations that are defined, given the following matrices: M0007 c
2 1
1 d 00023
P 0007 [1 2] (B) P 0003 Q (E) PN
N0007 c
1 00021
2 d 3
00021 d 2 (C) PQ (F) QM
19. How many ways can four distinct books be arranged on a shelf? Solve (A) By using the multiplication principle (B) By using permutations or combinations, whichever is applicable 20. In a single deal of 3 cards from a standard 52-card deck, what is the probability of being dealt three diamonds? 21. Each of the 10 digits 0 through 9 is printed on 1 of 10 different cards. Four of these cards are drawn in succession without replacement. What is the probability of drawing the digits 4, 5, 6, and 7 by drawing 4 on the first draw, 5 on the second draw, 6 on the third draw, and 7 on the fourth draw? What is the probability of drawing the digits 4, 5, 6, and 7 in any order? 22. A thumbtack lands point down in 38 out of 100 tosses. What is the approximate empirical probability of the tack landing point up? 23. Write the linear system corresponding to each augmented matrix and solve: 1 00022 3 1 0 3 (A) c (B) c ` d ` d 0 0 0 0 1 00024 1 0
00022 3 ` d 0 1
24. Given the system:
Then write the linear system represented by each augmented matrix in your solution, and solve each of these systems graphically. Discuss the relationship between the solutions of these systems. 27. Solve graphically to two decimal places: 00022x 0003 3y 0007 7
Q0007 c
18. A coin is flipped three times. How many combined outcomes are possible? Solve (A) By using a tree diagram (B) By using the multiplication principle
(C) c
2x1 0002 5x2 0007 k2
26. Use Gauss–Jordan elimination to solve the system
(0, 4)
(A) M 0002 2N (D) MN
25. Given the system: x1 0002 3x2 0007 k1 (A) Write the system as a matrix equation of the form AX 0007 B. (B) Find the inverse of the coefficient matrix A. (C) Use A00021 to find the solution for k1 0007 00022 and k2 0007 1. (D) Use A00021 to find the solution for k1 0007 1 and k2 0007 00022.
y
5
A-9
x1 0003 x2 0007 3 0002x1 0003 x2 0007 5
(A) Write the augmented matrix for the system. (B) Transform the augmented matrix into reduced form. (C) Write the solution to the system.
3x 0003 4y 0007 18 Verify the statement Pn in Problems 28 and 29 for n 0007 1, 2, and 3. 28. Pn: 1 0003 5 0003 9 0003 . . . 0003 (4n 0002 3) 0007 n(2n 0002 1) 29. Pn: n2 0003 n 0003 2 is divisible by 2 In Problems 30 and 31, write Pk and Pk 00031. 30. For Pn in Problem 28
31. For Pn in Problem 29
32. Find the equation of the parabola having its vertex at the origin, its axis the y axis, and (2, 00028) on its graph. 33. Find an equation of an ellipse in the form y2 x2 0003 00071 M N
M, N 7 0
if the center is at the origin, the major axis is the x axis, the major axis length is 10, and the distance of the foci from the center is 3. 34. Find an equation of a hyperbola in the form y2 x2 0002 00071 M N
M, N 7 0
if the center is at the origin, the transverse axis length is 16, and the distance of the foci from the center is 189. Solve Problems 35–37 using Gauss–Jordan elimination. 35. x1 0003 2x2 0002 x3 0007
3
36. x1 0003 x2 0002 x3 0007 2
x2 0003 x3 0007 00022
4x2 0003 6x3 0007 00021
2x1 0003 3x2 0003 x3 0007
0
37. x1 0002 2x2 0003 x3 0007
1
6x2 0003 9x3 0007 0
3x1 0002 2x2 0002 x3 0007 00025 1 38. Given M 0007 [1 2 00021] and N 0007 £ 00021 § . Find: 2 (A) MN (B) NM
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39. Given L0007 c
1 M 0007 £ 00021 1
00021 0 d 2 1
2 1
Find, if defined: (A) LM 0002 2N
2 0§ 1
centers are selected? Express the answer in terms of Cn,r or Pn,r, as appropriate, and evaluate. N0007 c
2 00021
1 d 0
(B) ML 0003 N
In Problems 40 and 41, solve the system. 40. x2 0002 3xy 0003 3y2 0007 1
41. x2 0002 3xy 0003 y2 0007 00021 x2 0002 xy 0007 0
xy 0007 1
In Problems 42 and 43, find the determinant. 1 42. 3 2 3
0 5 0
00024 43. 3 3 2
4 00021 3 00026
5 00022 4
00026 00021 3 6
44. Find all real solutions to two decimal places x2 0003 2xy 0002 y2 0007 1 9x2 0003 4xy 0003 y2 0007 15
53. A single die is rolled 1,000 times with the frequencies of outcomes shown in the table. (A) What is the approximate empirical probability that the number of dots showing is divisible by 3? (B) What is the theoretical probability that the number of dots showing is divisible by 3? Number of dots facing up Frequency
1
2
3
4
5
6
160
155
195
180
140
170
54. Let an 0007 100(0.9)n and bn 0007 10 0003 0.03n. Find the least positive integer n such that an 0005 bn by graphing the sequences {an} and {bn} with a graphing calculator. Check your answer by using a graphing calculator to display both sequences in table form. 55. Evaluate each of the following: (A) P25,5
(C) a
(B) C(25, 5)
5
45. Write a kk without summation notation and find the sum. k00071
2 22 23 24 25 26 0002 0003 0002 0003 0002 using sum2! 3! 4! 5! 6! 7! mation notation with the summation index k starting at k 0007 1.
46. Write the series
47. Find S for the geometric series 108 0002 36 0003 12 0002 4 0003 . . .. 48. Graph the solution region and indicate whether the solution region is bounded or unbounded. Find the coordinates of each corner point. 3x 0003 2y 0006 12 x 0003 2y 0006 8 x, y 0006 0 49. Solve the linear programming problem: Maximize
z 0007 4x 0003 9y
Subject to
x 0003 2y 0004 14
56. Expand (a 0003 12b)6 using the binomial formula. 57. Find the fifth and the eighth terms in the expansion of (3x 0002 y)10. Prove each statement in Problems 58 and 59 for all positive integers using mathematical induction. 58. Pn in Problem 28
61. Use the formula for the sum of an infinite geometric series to write 2.45 0007 2.454 545 . . . as the quotient of two integers. 30 b (0.1)300002k(0.9)k for k 0007 0, 1, . . ., 30. Use a k graphing calculator to find the largest term of the sequence {ak} and the number of terms that are greater than 0.01.
62. Let ak 0007 a
63. Use Cramer’s rule to solve the system for x only: 00022x
2x1 0003 5x2 0003 2x3 0007 k3 (A) Write the system as a matrix equation of the form AX 0007 B. (B) Find the inverse of the coefficient matrix A. (C) Use A00021 to solve the system when k1 0007 00021, k2 0007 2, and k3 0007 1. (D) Use A00021 to solve the system when k1 0007 2, k2 0007 0, and k3 0007 00021. 51. How many four-letter code words are possible using the first six letters of the alphabet if no letter can be repeated? If letters can be repeated? If adjacent letters cannot be alike? 52. A basketball team with 12 members has two centers. If 5 players are selected at random, what is the probability that both
0003 3z 0007 000213
x 0002 6y 0003 5z 0007 000216
x, y 0006 0 2x1 0003 6x2 0003 3x3 0007 k2
59. Pn in Problem 29
60. Find the sum of all the odd integers between 50 and 500.
2x 0003 y 0004 16 50. Given the system: x1 0003 4x2 0003 2x3 0007 k1
25 b 20
0002x 0003 2y 0007 00021 64. Use Cramer’s rule to solve the system in Problem 63 for y. 65. Use Cramer’s rule to solve the system in Problem 63 for z. 66. How many nine-digit zip codes are possible? How many of these have no repeated digits? 67. Use mathematical induction to prove that the following statement holds for all positive integers: Pn:
1 1 1 0003 0003 0003... 1ⴢ3 3ⴢ5 5ⴢ7 0003
1 n 0007 (2n 0002 1)(2n 0003 1) 2n 0003 1
68. Three-digit numbers are randomly formed from the digits 1, 2, 3, 4, and 5. What is the probability of forming an even number if digits cannot be repeated? If digits can be repeated?
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69. Discuss the number of solutions for the system corresponding to the reduced form shown below if (A) m 0007 0 and n 0007 0 (B) m 0007 0 and n 0 (C) m 0 1 £0 0
0 1 0
00025 2 3 † 6§ m n
A-11
84. Use mathematical induction to show that {an} 0007 {bn}, where a1 0007 3, an 0007 2an00021 0002 1 for n 1, and bn 0007 2n 0003 1, n 0006 1. 85. Find an equation of the set of points in the plane each of whose distance from (1, 4) is three times its distance from the x axis. Write the equation in the form Ax2 0003 Cy2 0003 Dx 0003 Ey 0003 F 0007 0, and identify the curve.
70. If a square matrix A satisfies the equation A2 0007 A, find A. Assume that A00021 exists.
86. A box of 12 lightbulbs contains 4 defective bulbs. If three bulbs are selected at random, what is the probability of selecting at least one defective bulb?
71. Which of the following augmented matrices are in reduced form?
APPLICATIONS
1 L 0007 £0 0
0 1 0
0 2 0 † 0§ 1 00021
0 N 0007 £1 0
0 0 0 † 2§ 1 00023
1 M 0007 £0 0
0 1 0
3 3 00022 † 2 § 0 0
1 0
0 1
2 00022 ` d 1 3
P0007 c
2 0
Recall that a square matrix is called upper triangular if all elements below the principal diagonal are zero, and it is called diagonal if all elements not on the principal diagonal are zero. A square matrix is called lower triangular if all elements above the principal diagonal are zero. In Problems 72–77, determine whether the statement is true or false. If true, explain why. If false, give a counterexample. 72. The sum of two upper triangular matrices is upper triangular.
87. ECONOMICS The government, through a subsidy program, distributes $2,000,000. If we assume that each individual or agency spends 75% of what it receives, and 75% of this is spent, and so on, how much total increase in spending results from this government action? 88. GEOMETRY Find the dimensions of a rectangle with perimeter 24 meters and area 32 square meters. 89. ENGINEERING An automobile headlight contains a parabolic reflector with a diameter of 8 inches. If the light source is located at the focus, which is 1 inch from the vertex, how deep is the reflector? 90. ARCHITECTURE A sound whispered at one focus of a whispering chamber can be easily heard at the other focus. Suppose that a cross section of this chamber is a semielliptical arch that is 80 feet wide and 24 feet high (see the figure). How far is each focus from the center of the arch? How high is the arch above each focus?
73. The product of two lower triangular matrices is lower triangular. 74. The sum of an upper triangular matrix and a lower triangular matrix is a diagonal matrix.
24 feet
75. The product of an upper triangular matrix and a lower triangular matrix is a diagonal matrix. 76. A matrix that is both upper triangular and lower triangular is a diagonal matrix. 77. If a diagonal matrix has no zero elements on the principal diagonal, then it has an inverse. 78. Use the binomial formula to expand (x 0002 2i)6, where i is the imaginary unit. 79. Use the definition of a parabola and the distance formula to find the equation of a parabola with directrix y 0007 3 and focus (6, 1). 80. An ellipse has vertices (4, 0) and foci (2, 0). Find the y intercepts. 81. A hyperbola has vertices (2, 3) and foci (2, 5). Find the length of the conjugate axis. 82. Seven distinct points are selected on the circumference of a circle. How many triangles can be formed using these seven points as vertices? 83. Use mathematical induction to prove that 2n 0005 n! for all integers n 3.
80 feet
91. FINANCE An investor has $12,000 to invest. If part is invested at 8% and the rest in a higher-risk investment at 14%, how much should be invested at each rate to produce the same yield as if all had been invested at 10%? 92. DIET In an experiment involving mice, a zoologist needs a food mix that contains, among other things, 23 grams of protein, 6.2 grams of fat, and 16 grams of moisture. She has on hand mixes of the following compositions: Mix A contains 20% protein, 2% fat, and 15% moisture, mix B contains 10% protein, 6% fat, and 10% moisture; and mix C contains 15% protein, 5% fat, and 5% moisture. How many grams of each mix should be used to get the desired diet mix? 93. PURCHASING A soft-drink distributor has budgeted $300,000 for the purchase of 12 new delivery trucks. If a model A truck costs $18,000, a model B truck costs $22,000, and a model C truck costs $30,000, how many trucks of each model should the distributor purchase to use exactly all the budgeted funds?
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APPENDIX A
94. MANUFACTURING A manufacturer makes two types of day packs, a standard model and a deluxe model. Each standard model requires 0.5 labor-hour from the fabricating department and 0.3 labor-hour from the sewing department. Each deluxe model requires 0.5 labor-hour from the fabricating department and 0.6 labor-hour from the sewing department. The maximum number of labor-hours available per week in the fabricating department and the sewing department are 300 and 240, respectively. (A) If the profit on a standard day pack is $8 and the profit on a deluxe day pack is $12, how many of each type of pack should be manufactured each day to realize a maximum profit? What is the maximum profit? (B) Discuss the effect on the production schedule and the maximum profit if the profit on a standard day pack decreases by $3 and the profit on a deluxe day pack increases by $3. (C) Discuss the effect on the production schedule and the maximum profit if the profit on a standard day pack increases by $3 and the profit on a deluxe day pack decreases by $3. 95. AVERAGING TESTS A teacher has given four tests to a class of five students and stored the results in the following matrix:
Ann Bob Carol Dan Eric
1 78 91 G 95 75 83
Tests 2 3 84 81 65 84 90 92 82 87 88 81
4 86 92 91 W 0007 M 91 76
Discuss methods of matrix multiplication that the teacher can use to obtain the indicated information in parts A–C. In each case, state the matrices to be used and then perform the necessary multiplications. (A) The average on all four tests for each student, assuming that all four tests are given equal weight (B) The average on all four tests for each student, assuming that the first three tests are given equal weight and the fourth is given twice this weight (C) The class average on each of the four tests 96. POLITICAL SCIENCE A random survey of 1,000 residents in a state produced the following results: Party Affiliation Age
Democrat
Republican
Independent
Under 30
130
80
40
250
30–39
120
90
20
230
40–49
70
80
20
170
50–59
50
60
10
120
Over 59
90
110
30
230
460
420
120
1,000
Totals
Totals
Find the empirical probability that a person selected at random: (A) Is under 30 and a Democrat (B) Is under 40 and a Republican (C) Is over 59 or is an Independent
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APPENDIX
Special Topics
11 B OUTLINE B-1
Scientific Notation and Significant Digits
B-2
Partial Fractions
B-3
Parametric Equations
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SPECIAL TOPICS
Scientific Notation and Significant Digits Z Significant Digits Z Rounding Convention
Z Significant Digits Most calculations involving problems in the real world deal with numbers that are only approximate. It therefore seems reasonable to assume that a final answer should not be any more accurate than the least accurate number used in the calculation. This is an important point, because calculators tend to give the impression that greater accuracy is achieved than is warranted. Suppose we want to compute the length of the diagonal of a rectangular field from measurements of its sides of 237.8 meters and 61.3 meters. Using the Pythagorean theorem and a calculator, we find d 0002 2237.82 0003 61.32
d
0002 245.573 878 . . .
61.3 meters
237.8 meters
The calculator answer suggests an accuracy that is not justified. What accuracy is justified? To answer this question, we introduce the idea of significant digits. Whenever we write a measurement such as 61.3 meters, we assume that the measurement is accurate to the last digit written. So the measurement 61.3 meters indicates that the measurement was made to the nearest tenth of a meter. That is, the actual width is between 61.25 meters and 61.35 meters. In general, the digits in a number that indicate the accuracy of the number are called significant digits. If all the digits in a number are nonzero, then they are all significant. So the measurement 61.3 meters has three significant digits, and the measurement 237.8 meters has four significant digits. What are the significant digits in the number 7,800? The accuracy of this number is not clear. It could represent a measurement with any of the following accuracies: Between 7,750 and 7,850 Between 7,795 and 7,805 Between 7,799.5 and 7,800.5
Correct to the hundreds place Correct to the tens place Correct to the units place
To give a precise definition of significant digits that resolves this ambiguity, we use scientific notation.
Z DEFINITION 1 Significant Digits If a number x is written in scientific notation as x 0002 a 0004 10n
1 0005 a 0006 10, n an integer
then the number of significant digits in x is the number of digits in a.
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SECTION B–1
Scientific Notation and Significant Digits
A-15
Using this definition, 7.8 0004 103 7.80 0004 103 7.800 0004 103
has two significant digits has three significant digits has four significant digits
All three of these measurements have the same decimal representation (7,800), but each represents a different accuracy. Definition 1 tells us how to write a number so that the number of significant digits is clear, but it does not tell us how to interpret the accuracy of a number that is not written in scientific notation. We will use the following convention for numbers that are written as decimal fractions:
Z SIGNIFICANT DIGITS IN DECIMAL FRACTIONS The number of significant digits in a number with no decimal point is found by counting the digits from left to right, starting with the first digit and ending with the last nonzero digit. The number of significant digits in a number containing a decimal point is found by counting the digits from left to right, starting with the first nonzero digit and ending with the last digit.
Applying this rule to the number 7,800, we conclude that this number has two significant digits. If we want to indicate that it has three or four significant digits, we must use scientific notation.
EXAMPLE
1
Significant Digits in Decimal Fractions Underline the significant digits in the following numbers:
SOLUTIONS
MATCHED PROBLEM 1
(A) 70,007
(B) 82,000
(C) 5.600
(D) 0.0008
(E) 0.000 830
(A) 70,007
(B) 82,000
(C) 5.600
(D) 0.0008
(E) 0.000 830
0002
Underline the significant digits in the following numbers: (A) 5,009
(B) 12,300
(C) 23.4000
(D) 0.00050
(E) 0.0012 0002
Z Rounding Convention In calculations involving multiplication, division, powers, and roots, we adopt the following convention:
Z ROUNDING CALCULATED VALUES The result of a calculation is rounded to the same number of significant digits as the number used in the calculation that has the least number of significant digits.
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SPECIAL TOPICS
So, in computing the length of the diagonal of the rectangular field shown earlier, we write the answer rounded to three significant digits because the width has three significant digits and the length has four significant digits: d 0002 246 meters
Three significant digits
One Final Note: In rounding a number that is exactly halfway between a larger and a smaller number, we use the convention of making the final result even.
EXAMPLE
2
Rounding Numbers Round each number to three significant digits. (A) 43.0690
SOLUTIONS
(B) 48.05
(D) 8.017 632 0004 1000073
(C) 48.15
(A) 43.1 (B) 48.0 ⎫ ⎬ (C) 48.2 ⎭
Use the convention of making the digit before the 5 even if it is odd, or leaving it alone if it is even.
(D) 8.02 0004 1000073 MATCHED PROBLEM 2
0002
Round each number to three significant digits. (A) 3.1495
(B) 0.004 135
(C) 32,450
(D) 4.314 764 09 0004 1012
0002
ANSWERS TO MATCHED PROBLEMS 1. (A) 5,009 2. (A) 3.15
B-1
(B) 12,300 (B) 0.004 14
(C) 23.4000 (C) 32,400
(D) 0.00050 (E) 0.0012 (D) 4.31 0004 1012
Exercises
In Problems 1–12, underline the significant digits in each number. 1. 123,005
2. 3,400,002
3. 20,040
4. 300,600
5. 6.0
6. 7.00
7. 80.000
8. 900.0000
9. 0.012
10. 0.0015
11. 0.000 960
12. 0.000 700
In Problems 13–22, round each number to three significant digits. 13. 3.0780
14. 4.0240
15. 924,300
16. 643,820
17. 23.65
18. 23.75
19. 2.816 743 0004 103 20. 56.114 0004 104 21. 6.782 045 0004 1000074 22. 5.248 102 0004 1000073 In Problems 23 and 24, find the diagonal of the rectangle with the indicated side measurements. Round answers to the number of significant digits appropriate for the given measurements. 23. 25 feet by 20 feet 24. 2,900 yards by 1,570 yards
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SECTION B–2
B-2
Partial Fractions
A-17
Partial Fractions Z Basic Theorems Z Partial Fraction Decomposition
You have now had considerable experience combining two or more rational expressions into a single rational expression. For example, problems such as 2(x 0007 4) 0003 3(x 0003 5) 2 3 5x 0003 7 0003 0002 0002 x00035 x00074 (x 0003 5)(x 0007 4) (x 0003 5)(x 0007 4) should seem routine. Frequently in more advanced courses, particularly in calculus, it is useful to be able to reverse this process—that is, to be able to express a rational expression as the sum of two or more simpler rational expressions called partial fractions. As is often the case with reverse processes, the process of decomposing a rational expression into partial fractions is more difficult than combining rational expressions. Basic to the process is the factoring of polynomials, so many of the topics discussed in Chapter 4 can be put to effective use. Partial fraction decomposition is usually accomplished by solving a related system of linear equations. If you are familiar with basic techniques for solving linear systems discussed earlier in this book, such as Gauss–Jordan elimination, inverse matrix solutions, or Cramer’s rule, you may use these as you see fit. However, all of the linear systems encountered in this section can also be solved by some special techniques developed here. Mathematically equivalent to the techniques mentioned, these special techniques are generally easier to use in partial fraction decomposition problems. We confine our attention to rational expressions of the form P(x)兾D(x), where P(x) and D(x) are polynomials with real coefficients. In addition, we assume that the degree of P(x) is less than the degree of D(x). If the degree of P(x) is greater than or equal to that of D(x), we have only to divide P(x) by D(x) to obtain R(x) P(x) 0002 Q(x) 0003 D(x) D(x) where the degree of R(x) is less than that of D(x). For example, 00076x 0003 2 x4 0007 3x3 0003 2x2 0007 5x 0003 1 0002 x2 0007 x 0007 1 0003 2 2 x 0007 2x 0003 1 x 0007 2x 0003 1 If the degree of P(x) is less than that of D(x), then P(x)兾D(x) is called a proper fraction.
Z Basic Theorems Our task now is to establish a systematic way to decompose a proper fraction into the sum of two or more partial fractions. Theorems 1, 2, and 3 take care of the problem completely. Z THEOREM 1 Equal Polynomials Two polynomials are equal to each other if and only if the coefficients of terms of like degree are equal.
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SPECIAL TOPICS
For example, if Equate the constant terms.
⎫ ⎪ ⎬ ⎪ ⎭
(A 0003 2B)x 0003 B 0002 5x 0007 3 Equate the coefficients of x.
then B 0002 00073 A 0003 2B 0002 5 A 0003 2(ⴚ3) 0002 5 A 0002 11 ZZZ EXPLORE-DISCUSS 1
Substitute B ⴝ ⴚ3 into the second equation to solve for A.
If x 0003 5 0002 A(x 0003 1) 0003 B(x 0007 3)
(1)
is a polynomial identity (that is, both sides represent the same polynomial), then equating coefficients produces the system 10002A0003B 5 0002 A 0007 3B
Equating coefficients of x Equating constant terms
(A) Solve this system graphically. (B) For an alternate method of solution, substitute x 0002 3 in equation (1) to find A and then substitute x 0002 00071 in equation (1) to find B. Explain why this method is valid.
The Linear and Quadratic Factors Theorem from Chapter 4 (page 290) underlies the technique of decomposing a rational function into partial fractions. We restate the theorem here. Z THEOREM 2 Linear and Quadratic Factors Theorem For a polynomial of degree n 0 with real coefficients, there always exists a factorization involving only linear and/or quadratic factors with real coefficients in which the quadratic factors have imaginary zeros.
The quadratic formula can be used to determine easily whether a given quadratic factor ax2 0003 bx 0003 c, with real coefficients, has imaginary zeros. If b2 0007 4ac 0006 0, then ax2 0003 bx 0003 c has imaginary zeros. Otherwise its zeros are real. Therefore, ax2 0003 bx 0003 c has imaginary zeros if and only if it cannot be factored as a product of linear factors with real coefficients.
Z Partial Fraction Decomposition We are now ready to state Theorem 3, which forms the basis for partial fraction decomposition.
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SECTION B–2
Partial Fractions
A-19
Z THEOREM 3 Partial Fraction Decomposition Any proper fraction P(x)兾D(x) reduced to lowest terms can be decomposed into the sum of partial fractions as follows: 1. If D(x) has a nonrepeating linear factor of the form ax 0003 b, then the partial fraction decomposition of P(x)兾D(x) contains a term of the form A ax 0003 b
A a constant
2. If D(x) has a k-repeating linear factor of the form (ax 0003 b)k, then the partial fraction decomposition of P(x)兾D(x) contains k terms of the form Ak A1 A2 0003 0003...0003 2 ax 0003 b (ax 0003 b) (ax 0003 b)k
A1, A2, . . . , Ak constants
3. If D(x) has a nonrepeating quadratic factor of the form ax2 0003 bx 0003 c that has imaginary zeros, then the partial fraction decomposition of P(x)兾D(x) contains a term of the form Ax 0003 B ax 0003 bx 0003 c 2
A, B constants
4. If D(x) has a k-repeating quadratic factor of the form (ax2 0003 bx 0003 c)k, where ax2 0003 bx 0003 c has imaginary zeros, then the partial fraction decomposition of P(x)兾D(x) contains k terms of the form Ak x 0003 Bk A1x 0003 B1 A2x 0003 B2 0003...0003 0003 ax2 0003 bx 0003 c (ax2 0003 bx 0003 c)2 (ax2 0003 bx 0003 c)k A1, . . . , Ak, B1, . . . , Bk constants
Let’s see how the theorem is used to obtain partial fraction decompositions in several examples.
EXAMPLE
1
Nonrepeating Linear Factors Decompose into partial fractions:
SOLUTION
5x 0003 7 . x 0003 2x 0007 3 2
We first try to factor the denominator. If it can’t be factored in the real numbers, then we can’t go any further. In this example, the denominator factors, so we apply part 1 from Theorem 3: 5x 0003 7 A B 0002 0003 (x 0007 1)(x 0003 3) x00071 x00033
(2)
To find the constants A and B, we combine the fractions on the right side of equation (2) to obtain A(x 0003 3) 0003 B(x 0007 1) 5x 0003 7 0002 (x 0007 1)(x 0003 3) (x 0007 1)(x 0003 3) Because these fractions have the same denominator, their numerators must be equal. So 5x 0003 7 0002 A(x 0003 3) 0003 B(x 0007 1)
(3)
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SPECIAL TOPICS
We could multiply the right side and find A and B by using Theorem 1, but in this case it is easier to take advantage of the fact that equation (3) is an identity—that is, it must hold for all values of x. In particular, we note that if we let x 0002 1, then the second term of the right side drops out and we can solve for A: 5 ⴢ 1 0003 7 0002 A(1 0003 3) 0003 B(1 0007 1) 12 0002 4A A00023 Similarly, if we let x 0002 00073, the first term drops out and we find 00078 0002 00074B B00022 Now we have the decomposition: 3 2 5x 0003 7 0002 0003 x 0007 1 x 0003 3 x 0003 2x 0007 3 2
as can easily be checked by adding the two fractions on the right.
MATCHED PROBLEM 1
Decompose into partial fractions:
7x 0003 6 . x 0003x00076
0002
(Fig. 1). Discuss how the TRACE command on the graphing calculator can be used to check that the graphing calculator is displaying two identical graphs. 10
000710
10
000710
Z Figure 1
EXAMPLE
2
Repeating Linear Factors Decompose into partial fractions:
SOLUTION
0002
2
Technology Connections A graphing calculator can also be used to check a partial fraction decomposition. To check Example 1, we graph the left and right sides of equation (4) in a graphing calculator
(4)
6x2 0007 14x 0007 27 . (x 0003 2)(x 0007 3)2
Using parts 1 and 2 from Theorem 3, we write 6x2 0007 14x 0007 27 A B C 0002 0003 0003 x00032 x00073 (x 0003 2)(x 0007 3)2 (x 0007 3)2 A(x 0007 3)2 0003 B(x 0003 2)(x 0007 3) 0003 C(x 0003 2) 0002 (x 0003 2)(x 0007 3)2
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SECTION B–2
Partial Fractions
A-21
So for all x, 6x2 0007 14x 0007 27 0002 A(x 0007 3)2 0003 B(x 0003 2)(x 0007 3) 0003 C(x 0003 2) If x 0002 3, then
If x 0002 00072, then
000715 0002 5C C 0002 00073
25 0002 25A A00021
There are no other values of x that will cause terms on the right to drop out. Because any value of x can be substituted to produce an equation relating A, B, and C, we let x 0002 0 and obtain 000727 0002 9A 0007 6B 0003 2C 000727 0002 9 0007 6B 0007 6 B00025
Substitute A ⴝ 1 and C ⴝ ⴚ3.
Therefore, 1 5 3 6x2 0007 14x 0007 27 0002 0003 0007 2 x 0003 2 x 0007 3 (x 0003 2)(x 0007 3) (x 0007 3)2
MATCHED PROBLEM 2
EXAMPLE
3
Decompose into partial fractions:
0002
Nonrepeating Linear and Quadratic Factors Decompose into partial fractions:
SOLUTION
x2 0003 11x 0003 15 . (x 0007 1)(x 0003 2)2
0002
5x2 0007 8x 0003 5 . (x 0007 2)(x2 0007 x 0003 1)
First, we see that the quadratic in the denominator can’t be factored further in the real numbers. Then, we use parts 1 and 3 from Theorem 3 to write A 5x2 0007 8x 0003 5 Bx 0003 C 0002 0003 2 2 x00072 (x 0007 2)(x 0007 x 0003 1) x 0007x00031 0002
A(x2 0007 x 0003 1) 0003 (Bx 0003 C)(x 0007 2) (x 0007 2)(x2 0007 x 0003 1)
So for all x, 5x2 0007 8x 0003 5 0002 A(x2 0007 x 0003 1) 0003 (Bx 0003 C)(x 0007 2) If x 0002 2, then 9 0002 3A A00023 If x 0002 0, then, using A 0002 3, we have 5 0002 3 0007 2C C 0002 00071 If x 0002 1, then, using A 0002 3 and C 0002 00071, we have 2 0002 3 0003 (B 0007 1)(00071) B00022
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SPECIAL TOPICS
Therefore, 3 2x 0007 1 5x2 0007 8x 0003 5 0002 0003 2 x00072 (x 0007 2)(x2 0007 x 0003 1) x 0007x00031
MATCHED PROBLEM 3
EXAMPLE
4
Decompose into partial fractions:
7x2 0007 11x 0003 6 . (x 0007 1)(2x2 0007 3x 0003 2)
0002
Repeating Quadratic Factors Decompose into partial fractions:
SOLUTION
0002
x3 0007 4x2 0003 9x 0007 5 . (x2 0007 2x 0003 3)2
Because x2 0007 2x 0003 3 can’t be factored further in the real numbers, we proceed to use part 4 from Theorem 3 to write Cx 0003 D x3 0007 4x2 0003 9x 0007 5 Ax 0003 B 0003 2 0002 2 (x2 0007 2x 0003 3)2 x 0007 2x 0003 3 (x 0007 2x 0003 3)2 0002
(Ax 0003 B)(x2 0007 2x 0003 3) 0003 Cx 0003 D (x2 0007 2x 0003 3)2
So for all x, x3 0007 4x2 0003 9x 0007 5 0002 (Ax 0003 B)(x2 0007 2x 0003 3) 0003 Cx 0003 D Because the substitution of carefully chosen values of x doesn’t lead to the immediate determination of A, B, C, or D, we multiply and rearrange the right side to obtain x3 0007 4x2 0003 9x 0007 5 0002 Ax3 0003 (B 0007 2A)x2 0003 (3A 0007 2B 0003 C)x 0003 (3B 0003 D) Now we use Theorem 1 to equate coefficients of terms of like degree: A00021 B 0007 2A 0002 00074 3A 0007 2B 0003 C 0002 9 3B 0003 D 0002 00075
1x 3
ⴚ4x 2
ⴙ9x
ⴚ5
Ax 3 ⴙ (B ⴚ 2A)x 2 ⴙ (3A ⴚ 2B ⴙ C )x ⴙ (3B ⴙ D)
From these equations we easily find that A 0002 1, B 0002 00072, C 0002 2, and D 0002 1. Now we can write 2x 0003 1 x3 0007 4x2 0003 9x 0007 5 x00072 0003 2 0002 2 2 2 (x 0007 2x 0003 3) x 0007 2x 0003 3 (x 0007 2x 0003 3)2
MATCHED PROBLEM 4
Decompose into partial fractions:
3x3 0007 6x2 0003 7x 0007 2 . (x2 0007 2x 0003 2)2
ANSWERS TO MATCHED PROBLEMS 3 4 3 2 1 2. 0003 0007 0003 x00072 x00033 x00071 x00032 (x 0003 2)2 x00072 3x 2 3x 0007 2 0003 2 3. 4. 2 0003 2 x00071 2x 0007 3x 0003 2 x 0007 2x 0003 2 (x 0007 2x 0003 2)2 1.
0002
0002
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SECTION B–3
B-2
2. 3. 4.
6. 7. 8. 9.
10.
Cx 0003 D 3x3 0007 3x2 0003 10x 0007 4 Ax 0003 B 0003 2 0002 2 (x2 0007 x 0003 3)2 x 0007x00033 (x 0007 x 0003 3)2
In Problems 11–30, decompose into partial fractions. 11.
0007x 0003 22 x 0007 2x 0007 8
12.
0007x 0007 21 x 0003 2x 0007 15
13.
3x 0007 13 6x 0007 x 0007 12
14.
11x 0007 11 6x2 0003 7x 0007 3
15.
x2 0007 12x 0003 18 x3 0007 6x2 0003 9x
16.
5x2 0007 36x 0003 48 x(x 0007 4)2
17.
5x2 0003 3x 0003 6 x3 0003 2x2 0003 3x
18.
2x3 0003 7x 0003 5 6x2 0007 15x 0003 16 19. 4 3 2 x 0007 3x 0003 4x x 0003 4x2 0003 4
20.
00075x2 0003 7x 0007 18 x4 0003 6x2 0003 9
22.
x3 0003 x2 0007 13x 0003 11 x2 0003 2x 0007 15
23.
B A C 3x2 0003 7x 0003 1 0002 0003 0003 2 x x 0003 1 x(x 0003 1) (x 0003 1)2
7x 0007 14 A B 0002 0003 (x 0007 4)(x 0003 3) x00074 x00033 9x 0003 21 A B 0002 0003 (x 0003 5)(x 0007 3) x00035 x00073 17x 0007 1 A B 0002 0003 (2x 0007 3)(3x 0007 1) 2x 0007 3 3x 0007 1
2
2
2
21.
x3 0007 7x2 0003 17x 0007 17 x2 0007 5x 0003 6
4x2 0003 5x 0007 9 x3 0007 6x 0007 9
24.
4x2 0007 8x 0003 1 x3 0007 x 0003 6
25.
26.
x2 0007 6x 0003 11 A B C 0002 0003 0003 x00031 x00072 (x 0003 1)(x 0007 2)2 (x 0007 2)2
x2 0003 16x 0003 18 x 0003 2x2 0007 15x 0007 36
5x2 0007 18x 0003 1 x 0007 x2 0007 8x 0003 12
27.
3x2 0003 x Bx 0003 C A 0003 2 0002 x00072 (x 0007 2)(x2 0003 3) x 00033
0007x2 0003 x 0007 7 x 4 0007 5x3 0003 9x2 0007 8x 0003 4
28.
A 5x2 0007 9x 0003 19 Bx 0003 C 0002 0003 2 2 x00074 (x 0007 4)(x 0003 5) x 00035
00072x3 0003 12x2 0007 20x 0007 10 x 0007 7x3 0003 17x2 0007 21x 0003 18
29.
Cx 0003 D Ax 0003 B 2x2 0003 4x 0007 1 0003 2 0002 2 (x2 0003 x 0003 1)2 x 0003x00031 (x 0003 x 0003 1)2
4x5 0003 12x4 0007 x3 0003 7x2 0007 4x 0003 2 4x 4 0003 4x3 0007 5x2 0003 5x 0007 2
30.
6x5 0007 13x 4 0003 x3 0007 8x2 0003 2x 6x 4 0007 7x3 0003 x2 0003 x 0007 1
x 0007 11 A B 0002 0003 (3x 0003 2)(2x 0007 1) 3x 0003 2 2x 0007 1
In Problems 5–10, find A, B, C, and D, so that the right side is equal to the left. 5.
A-23
Exercises
In Problems 1–4, find A and B so that the right side is equal to the left. After cross-multiplying to produce a polynomial equation, solve each problem two ways (see Explore-Discuss 1). First, equate the coefficients of both sides to determine a linear system for A and B and solve this system. Second, solve for A and B by evaluating both sides for selected values of x. 1.
Parametric Equations
B-3
3
3
4
Parametric Equations Z Parametric Equations and Plane Curves Z Projectile Motion
Z Parametric Equations and Plane Curves Consider the two equations x0002t00031 y 0002 t 2 0007 2t
0007 6 t 6
(1)
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SPECIAL TOPICS
Each value of t determines a value of x, a value of y, and therefore, an ordered pair (x, y). To graph the set of ordered pairs (x, y) determined by letting t assume all real values, we construct Table 1 listing selected values of t and the corresponding values of x and y. Then we plot the ordered pairs (x, y) and connect them with a continuous curve, as shown in Figure 1. The variable t is called a parameter and does not appear on the graph. Equations (1) are called parametric equations because both x and y are expressed in terms of the parameter t. The graph of the ordered pairs (x, y) is called a plane curve.
10
5
x
Table 1 t
0
1
2
3
4
ⴚ1
ⴚ2
x
1
2
3
4
5
0
00071
y
0
00071
0
3
8
3
8
Z Figure 1 Graph of x 0002 t 0003 1, y 0002 t2 0007 2t, 0007 0006 t 0006 .
Technology Connections Parametric equations can also be graphed on a graphing calculator. Figure 2(a) shows the Parametric mode selected on a Texas Instruments TI-84 calculator. Figure 2(b) shows the equation editor with the parametric equations in (1) en-
tered as x1T and y1T. In Figure 2(c), notice that there are three new window variables, Tmin, Tmax, and Tstep, that must be entered by the user.
10
00073
7
00072
(a)
(b)
(c)
(d)
Z Figure 2 Graphing parametric equations on a graphing calculator.
In some cases, it is possible to eliminate the parameter by solving one of the equations for t and substituting into the other. In the example just considered, solving the first equation for t in terms of x, we have t0002x00071 Then, substituting the result into the second equation, we obtain y 0002 (x 0007 1)2 0007 2(x 0007 1) 0002 x2 0007 4x 0003 3 We recognize this as the equation of a parabola, as we would guess from Figure 1. In other cases, it may not be easy or possible to eliminate the parameter to obtain an equation in just x and y. For example, for x 0002 t 0003 log t y 0002 t 0007 et
t 7 0
you will not find it possible to solve either equation for t in terms of functions we have considered.
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SECTION B–3
Parametric Equations
A-25
Is there more than one parametric representation for a plane curve? The answer is yes. In fact, there is an unlimited number of parametric representations for the same plane curve. The following are two additional representations of the parabola in Figure 1. x0002t00033 y 0002 t 2 0003 2t
0007 6 t 6
(2)
x0002t y 0002 t 2 0007 4t 0003 3
0007 6 t 6
(3)
The concepts introduced in the preceding discussion are summarized in Definition 1.
Z DEFINITION 1 Parametric Equations and Plane Curves A plane curve is the set of points (x, y) determined by the parametric equations x 0002 f (t) y 0002 g (t) where the parameter t varies over an interval I and the functions f and g are both defined on the interval I.
Why are we interested in parametric representations of plane curves? It turns out that this approach is more general than using equations with two variables as we have been doing. In addition, the approach generalizes to curves in three- and higher-dimensional spaces. Other important reasons for using parametric representations of plane curves will be brought out in the discussion and examples that follow.
EXAMPLE
1
Eliminating the Parameter Eliminate the parameter and identify the plane curve given parametrically by x 0002 1t y 0002 19 0007 t
SOLUTION
00005t00059
(4)
To eliminate the parameter t, we solve each equation (4) for t: x 0002 1t x2 0002 t
y 5
y 0002 19 0007 t y2 0002 9 0007 t t 0002 9 0007 y2
Equating the last two equations, we have 00075
5
x
00075
Z Figure 3
MATCHED PROBLEM 1
x2 0002 9 0007 y2 x 0003 y2 0002 9 2
A circle of radius 3 centered at (0, 0)
As the parameter t increases from 0 to 9, x will increase from 0 to 3 and y will decrease from 3 to 0. So the graph of the parametric equations in (4) is the quarter of the circle of radius 3 centered at the origin that lies in the first quadrant (Fig. 3). 0002 Eliminate the parameter and identify the plane curve given parametrically by x 0002 14 0007 t, y 0002 0007 1t, 0 0005 t 0005 4. 0002
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APPENDIX B
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SPECIAL TOPICS
Z Projectile Motion Newton’s laws and advanced mathematics can be used to determine the path of a projectile. If v0 is the vertical speed of the projectile, h0 is the horizontal speed, and a0 is the initial altitude of the projectile (Fig. 4), then, neglecting air resistance, the path of the projectile is given by x 0002 h0t y 0002 a0 0003 v0t 0007 4.9t 2
00005t0005b
(5)
y v0 ␣ a0
v0
h0 x
Z Figure 4 Projectile motion.
The parameter t represents time in seconds, and x and y are distances measured in meters. Solving the first equation in equations (5) for t in terms of x, substituting into the second equation, and simplifying produces the following equation: y 0002 a0 0003
v0 4.9 x 0007 2 x2 h0 h0
(6)
You should verify this by supplying the omitted details. We recognize equation (6) as a parabola. This equation in x and y describes the path the projectile follows but tells us little else about its flight. On the other hand, the parametric equations (5) not only determine the path of the projectile but also tell us where it is at any time t. Furthermore, using concepts from physics and calculus, the parametric equations can be used to determine the velocity and acceleration of the projectile at any time t. This illustrates another advantage of using parametric representations of plane curves.
EXAMPLE
2
Projectile Motion An automobile drives off a 50-meter cliff traveling at 25 meters per second (Fig. 5). When (to the nearest tenth of a second) will the automobile strike the ground? How far (to the nearest meter) from the base of the cliff is the point of impact?
50 m
Z Figure 5
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SECTION B–3
SOLUTION
Parametric Equations
A-27
At the instant the automobile leaves the cliff, the vertical speed is 0, the horizontal speed is 25 meters per second, and the altitude is 50 meters. Substituting these values in equations (5), the parametric equations for the path of the automobile are x 0002 25t y 0002 50 0007 4.9t2 The automobile strikes the ground when y 0002 0. Using the parametric equation for y, we have y 0002 50 0007 4.9t2 0002 0 00074.9t2 0002 000750 t0002
000750 ⬇ 3.2 seconds B 00074.9
The distance from the base of the cliff is the same as the value of x. Substituting t 0002 3.2 in the first parametric equation, the distance from the base of the cliff at the point of impact is x 0002 25(3.2) 0002 80 meters. 0002 MATCHED PROBLEM 2
A gardener is holding a hose in a horizontal position 1.5 meters above the ground. Water is leaving the hose at a speed of 5 meters per second. What is the distance (to the nearest tenth of a meter) from the gardener’s feet to the point where the water hits the ground? 0002
ANSWERS TO MATCHED PROBLEMS 1. The quarter of the circle of radius 2 centered at the origin that lies in the fourth quadrant. 2. 2.8 meters
B-3
Exercises
1. If x 0002 t 2 and y 0002 t 2 0007 2, then y 0002 x 0007 2. Discuss the differences between the graph of the parametric equations and the graph of the line y 0002 x 0007 2. 2. If x 0002 t 2 and y 0002 t 4 0007 2, then y 0002 x2 0007 2. Discuss the differences between the graph of the parametric equations and the graph of the parabola y 0002 x2 0007 2. In Problems 3–12, the interval for the parameter is the whole real line. For each pair of parametric equations, eliminate the parameter t and find an equation for the curve in terms of x and y. Identify and graph the curve.
In Problems 13–20, obtain an equation in x and y by eliminating the parameter. Identify the curve. 13. x 0002 t 0007 2, y 0002 4 0007 2t 14. x 0002 t 0007 1, y 0002 2t 0003 2 15. x 0002 t 0007 1, y 0002 1t, t 0 16. x 0002 1t, y 0002 t 0003 1, t 0 17. x 0002 1t, y 0002 2 116 0007 t, 0 0005 t 0005 16 18. x 0002 000731t, y 0002 125 0007 t, 0 0005 t 0005 25
3. x 0002 0007t, y 0002 2t 0007 2
4. x 0002 t, y 0002 t 0003 1
19. x 0002 0007 1t 0003 1, y 0002 0007 1t 0007 1, t 1
5. x 0002 0007t 2, y 0002 2t 2 0007 2
6. x 0002 t 2, y 0002 t 2 0003 1
20. x 0002 12 0007 t, y 0002 0007 14 0007 t, t 0005 2
7. x 0002 3t, y 0002 00072t
8. x 0002 2t, y 0002 t
21. If A ≠ 0, C 0002 0, and E ≠ 0, find parametric equations for Ax2 0003 Cy2 0003 Dx 0003 Ey 0003 F 0002 0. Identify the curve.
9. x 0002 14t 2, y 0002 t
10. x 0002 2t, y 0002 t 2
11. x 0002 14t 4, y 0002 t2
12. x 0002 2t 2, y 0002 t 4
22. If A 0002 0, C ≠ 0, and D ≠ 0, find parametric equations for Ax2 0003 Cy2 0003 Dx 0003 Ey 0003 F 0002 0. Identify the curve.
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APPENDIX B
2:44 PM
Page A-28
SPECIAL TOPICS
In Problems 23–28, the interval for the parameter is the entire real line. Obtain an equation in x and y by eliminating the parameter and identify the curve. 2
2
2
2
23. x 0002 2t 0003 1, y 0002 2t 0003 9 24. x 0002 2t 0003 4, y 0002 2t 0003 1 25. x 0002 26. x 0002
2 2t 2 0003 1 3t 2
2t 0003 1
,y0002 ,y0002
2t 2t 2 0003 1 3
30. Consider the following two pairs of parametric equations: 1. x1 0002 t, y1 0002 log t, t 0 2. x2 0002 log t, y2 0002 t, t 0 (A) Graph both pairs of parametric equations in a squared viewing window and discuss the relationship between the graphs. (B) Eliminate the parameter and express each equation as a function of x. How are these functions related?
2
2t 0003 1
8 4t ,y0002 2 27. x 0002 2 t 00034 t 00034 28. x 0002
(B) Eliminate the parameter and express each equation as a function of x. How are these functions related?
4t 4t 2 ,y0002 2 t 00031 t 00031 2
29. Consider the following two pairs of parametric equations: 1. x1 0002 t, y1 0002 et, 0007 0006 t 0006 2. x2 0002 et, y2 0002 t, 0007 0006 t 0006 (A) Graph both pairs of parametric equations in a squared viewing window and discuss the relationship between the graphs.
APPLICATIONS 31. PROJECTILE MOTION An airplane flying at an altitude of 1,000 meters is dropping medical supplies to hurricane victims on an island. The path of the plane is horizontal, the speed is 125 meters per second, and the supplies are dropped at the instant the plane crosses the shoreline. How far inland (to the nearest meter) will the supplies land? 32. PROJECTILE MOTION One stone is dropped vertically from the top of a tower 40 meters high. A second stone is thrown horizontally from the top of the tower with a speed of 30 meters per second. How far apart (to the nearest tenth of a meter) are the stones when they land?
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APPENDIX
Geometric Formulas
11 C
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APPENDIX C
6:01 PM
Page A-30
GEOMETRIC FORMULAS
Z Similar Triangles (A) Two triangles are similar if two angles of one triangle have the same measure as two angles of the other. (B) If two triangles are similar, their corresponding sides are proportional: a b c ⫽ ⫽ a¿ b¿ c¿
a
b
b⬘
a⬘
c c⬘
Z Pythagorean Theorem c2 ⫽ a2 ⫹ b2
c
a b
Z Rectangle A ⫽ ab P ⫽ 2a ⫹ 2b
Area
b
Perimeter
a
Z Parallelogram h ⫽ Height A ⫽ ah ⫽ ab sin P ⫽ 2a ⫹ 2b
Area Perimeter
h
b
a
Z Triangle h ⫽ Height A ⫽ 12 hc P⫽a⫹b⫹c s ⫽ 12 (a ⫹ b ⫹ c) A ⫽ 1s(s ⫺ a)(s ⫺ b)(s ⫺ c)
b
Area
h
Perimeter
c
Semiperimeter
b
Area—Heron’s formula
c
Z Trapezoid Base a is parallel to base b. h ⫽ Height A ⫽ 12 (a ⫹ b)h
a
Area
h b
a
a
h
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APPENDIX C
A-31
GEOMETRIC FORMULAS
Z Circle R ⫽ Radius D ⫽ Diameter D ⫽ 2R A ⫽ R2 ⫽ 14 D2 C ⫽ 2R ⫽ D C ⫽ D ⬇ 3.141 59
D R
Area Circumference For all circles
Z Rectangular Solid V ⫽ abc T ⫽ 2ab ⫹ 2ac ⫹ 2bc
Volume Total surface area
c a
b
Z Right Circular Cylinder R ⫽ Radius of base h ⫽ Height V ⫽ R2h S ⫽ 2Rh T ⫽ 2R(R ⫹ h)
h
Volume Lateral surface area
R
Total surface area
Z Right Circular Cone R ⫽ Radius of base h ⫽ Height s ⫽ Slant height V ⫽ 13 R2h S ⫽ Rs ⫽ R2R2 ⫹ h2
Volume
R
Lateral surface area 2
2
T ⫽ R(R ⫹ s) ⫽ R(R ⫹ 2R ⫹ h )
Total surface area
Z Sphere R ⫽ Radius D ⫽ Diameter D ⫽ 2R V ⫽ 43 R3 ⫽ 16 D3 S ⫽ 4R2 ⫽ D2
s
h
D Volume Surface area
R
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STUDENT ANSWER APPENDIX CHAPTER R Exercises R-1 35. 37.
(A) 51, 11446 (B) 5⫺3, 0, 1, 11446 (C) 5⫺3, ⫺23 , 0, 1, 95 , 11446 (D) 5 136 (A) 0.888 888 p ; repeating; repeated digit: 8 (B) 0.272 727 p ; repeating; repeated digits: 27 (C) 2.236 067 977 p ; nonrepeating and nonterminating
1.375; terminating
(D)
Exercises R-2 61.
n8
3 3 69. 6 1 5 ⫺ 1 25
67. ⫺2 13
m12
3 51 2
71.
12 Ⲑ2 or 12 12
83.
Exercises R-4 59.
⫺x(x ⫹ y) y
CHAPTER 1 Exercises 1-2 5.
⫺8 ⱕ x ⱕ 7
9.
x ⱖ ⫺6
[
⫺10 ⫺5
[
⫺10 ⫺5
0
15.
(⫺⬁, ⫺2]
33.
N 6 ⫺8; (⫺⬁, ⫺8)
39.
B ⱖ ⫺4; [⫺4, ⬁)
45. (2, 4)
(
2
⫺10 ⫺5
(
4
5
5
10
0
5
(
[
x
77.
(A) and (C)
11. x
(⫺2, 6]
29.
[
[
⫺42
(
30
x
)
[
(
(
q
6
(
(B) and (D)
[
)[ (
⫺1 3
7
0
( (
[
⫺5
x
x
7
51.
(1, 5) [
(
1
(
x
(
x
5
x
4.5
[
x ⱖ 8; [8, ⬁)
69.
x
10
m
3
63. ⫺8 ⱕ x 6 ⫺3; [⫺8, ⫺3)
x
(
5
y
x ⱖ 4.5; [4.5, ⬁)
x
12
10
⫺10 ⫺5
2
43. (⫺5, 7]
57. (
(
x 6 10; (⫺⬁, 10)
a ⬎ 0 and b ⬎ 0, or a ⬍ 0 and b ⬍ 0
t
3
⫺14
)
(⫺7, 8)
37. m 7 3; (3, ⬁) [
⫺2
13.
31. y ⱖ 2
t
2
61. 6 6 x 6 12
67.
x x
5
x
10
49. (⫺⬁, ⫺1) 傼 [3, 7)
x
x
5
10
(
41. ⫺2 6 t ⱕ 3; (⫺2, 3]
20
5
(
(
0
[
0
x 6 5; (⫺⬁, 5)
35. t 7 2; (2, ⬁)
B
⫺20
⫺10 ⫺5
⫺10 ⫺5
55. q 6 ⫺14; (⫺⬁, ⫺14)
59. ⫺20 ⱕ x ⱕ 20; [⫺20, 20]
[
7. ⫺6 ⱕ x 6 6
47. (⫺⬁, ⬁)
6
65. ⫺42 ⱕ x 6 30
x
N
⫺4
x
x
10
10
⫺8
[
53. (⫺⬁, 6]
[
[
0
[
⫺8
⫺3
x
8
a ⬎ 0 and b ⬍ 0, or a ⬍ 0 and b ⬎ 0
Exercises 1-3 31.
y is 3 units from 5; y ⫽ 2, 8
2
8
y
33.
y is less than 3 units from 5; 2 ⬍ y ⬍ 8; (2, 8) (
35. y is more than 3 units from 5; y ⬍ 2 or y ⬎ 8; (⫺⬁, 2) ´ (8, ⬁)
2
[
( [
u is no more than 3 units from ⫺8; ⫺11 ⱕ u ⱕ ⫺5; [⫺11, ⫺5]
41.
u is at least 3 units from ⫺8; u ⱕ ⫺11 or u ⱖ ⫺5; (⫺⬁, ⫺11] ´ [⫺5, ⬁)
51. u ⱕ ⫺11 or u ⱖ ⫺6; (⫺⬁, ⫺11] 傼 [⫺6, ⬁)
⫺5
The distance from x to 3 is between zero and 0.1; (2.9, 3) ´ (3, 3.1);
[
[
⫺11 ⫺5
67.
The distance from x to a is between 0 and 1Ⲑ10; aa ⫺
( 2.9
1 1 , ab ´ aa, a ⫹ b 10 10
(
y
8 ⫺11
⫺5
u
u u
53. ⫺35 6 C 6 ⫺59 ; (⫺35, ⫺59 )
65.
2
37. u is 3 units from ⫺8; u ⫽ ⫺11, ⫺5
8
39.
⫺11
y
(
( 3
⫺2 ⬍ x ⬍ 2; (⫺2, 2)
55.
57. ⫺13 ⱕ t ⱕ 1; [⫺13 , 1]
x
3.1
( a⫺
(
x
1 1 a a ⫹ 10 10
SA-1
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Student Answer Appendix
Exercises 1-4 9. 15.
3 2 (A) 4000b
(A)
5 i 6 (B) 0
0002
3 5 0002 0002 i 2 6 (C) 4000b
(B)
11.
(C)
(A) 6.5
(C) 6.5 0002 2.1i
(B) 2.1i
13.
(A) 0
000bi
(B)
(C)
0002000bi
Exercises 1-5 19. z 4 12 5 25. Two real roots: x 1 12 27. No real roots: x 1 i 12 29. No real roots: t (3 i17) 2 31. Two real roots: t (3 17) 2 33. x 2 15 35. r (00025 13) 2 37. u (00022 i 111) 2 43. y (3 15) 2 45. x (3 113) 2
Exercises 1-6 27.
2u2 0002 4u 0, u x00023
51.
y
1 i 12
3 3
53.
29. t
10 1 4u 0002 7u2 0, u 2 9 x 5 113 x (four roots) B 6
12 , 12 2
63.
35. m 13, i 15
31.
Not of quadratic type
y 000264, 278
39.
Chapter 1 Review Exercises (
5. 000214 0007 y 0007 00024; (000214, 00024) (1-3) 19.
]
x 0004 000219; [000219, 0005) (1-2)
(
000214
y
x
000219
x 272 or 12 114
9.
00024
x (1 143) 3
27.
(1-5)
I (E 2E 2 0002 4PR) (2R)
37.
(1-5)
13. m 000212 ( 13 2)i (1-5) (1-5)
CHAPTER 2 Exercises 2-1 y
15.
y
17.
5
5
(00024, 2)
(4, 4) (5, 0)
00025
5
(00022, 1) x
00025
5
(00021, 00023)
(3, 00022) 00025
x
(0, 00022) (4, 00025)
00025
Points: A (2, 4), B (3, 00021), C (00024, 0), D (00025, 2) Reflections: A (00022, 4), B (00023, 00021), C (4, 0), D (5, 2) 21. Points: A (00023, 00023), B (0, 4), C (00023, 2), D (5, 00021) Reflections: A (3, 3), B (0, 00024), C (3, 00022), D (00025, 1) 23. No symmetry with respect 25. Symmetric with respect 27. to x axis, y axis, or origin to the origin 19.
y
y
5
5
x
35.
5
x
5
00025
5
00025
x
x
y
5
5
00025
(D)
y
5
x
00025
00025
00025
(C)
y
5
x
00025
(B)
y
5
00025
00025
(A)
y
5
00025
00025
Symmetric with respect to the x axis, y axis, and origin
y
5
00025
29.
Symmetric with respect to the x axis
5
00025 5
00025
x
00025
5
00025
x
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SA-3
Student Answer Appendix 47.
49.
Symmetric with respect to the x axis y
x
5
5
x
57.
y
59.
00025
5
00025
63.
61.
y
A
x
10
A
x
C
B 000210
B
B
C
B
A
x
10
A C
000210
000210
00025
y 23 0002 2x
65.
y 00021 2x 2 0002 4
10
67.
10
69.
Symmetric with respect to the y axis
Symmetric with respect to the origin
y 000210
10
000210
10
000210
y
5
000210
00025
5
x
73.
No symmetry with respect to the x axis, y axis, or origin
75.
Symmetric with respect to the x axis, y axis, and origin
y
00025
Symmetric with respect to the x axis, y axis, and origin
y
10
y
5
x
5
00025
00025
71.
x
10
C 000210
x
10
000210
10
y
5
000210
y
5
00025
x
5
00025
Symmetric with respect to the y axis
5
10
00025
00025
Symmetric with respect to the y axis
y
5
00025
00025
Symmetric with respect to the origin
y
5
00025
53.
Symmetric with respect to the x axis, y axis, and origin
y
5
55.
51.
Symmetric with respect to the y axis
81.
R 30
5 20
000210
10
x
00025
5
x
00025
00025
83. 87.
(A) 3,000 cases (A) y
5
10 00025
(B) Demand decreases by 400 cases
(C) Demand increases by 600 cases
1
1
2
x
Exercises 2-2 21.
x2 y2 4
23.
(x 0002 1)2 y2 1
y
00025
(x 2)2 ( y 0002 1)2 9 y
5
5
00025
25.
y
5
x
00025
5
5
00025
x
x
00025
5
00025
x
0
5
10
p
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Student Answer Appendix
33.
The set of all points that are two units from the point (0, 2). x2 (y 0002 2)2 4 35. The set of all points that are four units from the point (1, 1). (x 0002 1)2 ( y 0002 1)2 16 43. Center: (0, 00022); radius: 3 45. Center: (00024, 2); radius: 17 y 5
5
x
y 10
10
x
000210
10
x
000210
10
000210
000210
00025
53. y 23 0002 x2
Center: (00024, 3); radius: 117
Center: (3, 2); radius: 7
y
3
00025
49.
Center: (00023, 0); radius: 5
5
00025
51.
47.
y
55.
y 00021 22 0002 (x 3)2
3.1
3.1
y 00024.7
4.7
00024.7
4.7
5 00023.1 00025
00023.1
x
73. (A) A (0, 0), B (0, 13.5), C (0, 27), D (60, 27), E (78, 27), F (78, 13.5), G (78, 0) 77. (A) (x 12)2 ( y 5)2 262; center: (000212, 00025); radius: 26 (B) 13.5 miles y (000212, 00025)
Town B (36, 15)
25
Town A 000250
(B) 62 feet, 79 feet
000225
x
25
000225
Exercises 2-3 19.
Slope 000235
21.
Slope 000234
y
Slope 00022
y
10
x
00025
000210
5
29.
x
y 10
00025
5
x
31.
Slope 0
y
y
5
5
10
000210
10
x
00025
5
x
00025
00025
41. 69.
47. y 000225 x 2 67. y 32 x 232 (slope AB )(slope BC ) (000234 )(43 ) 00021
5
00025
slope AB 000234 slope DC
000210
10
000210
00025
Slope not defined
y
Slope 45
y
00025
Slope 2
25.
5
5
000210
27.
23.
x
x
x
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Student Answer Appendix 75.
3x 4y 25
77.
x 0002 y 10
y
79. 232 5x 0002 12y y
y
10
10
10
x
000210
000210
x
10
000210
SA-5
10
20
x
000210
000210
000220
81.
(A)
x
0
5,000
10,000
15,000
20,000
25,000
30,000
B
212
203
194
185
176
167
158
(B) The boiling point drops 9°F for each 5,000-ft increase in altitude. 87.
(A) F 95 C 32
(B) 68°F, 30°C
Exercises 2-4 C 2,147 75x The rate of change of cost with respect to production is $75. Increasing production by 1 unit increases cost by $75 7. The rate of change of height with respect to DBH is 4.06 feet per inch. Increasing DBH by 1 inch increases height by 4.06 feet. 73 feet 19 inches 9. Robinson: The rate of change of weight with respect to height is 3.7 pounds per inch. Miller: The rate of change of weight with respect to height is 3 pounds per inch. (B) Robinson: 130.2 pounds; Miller: 135 pounds (C) Robinson: 5 9000e; Miller: 5 8000e 11. s 0.75t 717; speed increases 0.75 mph for each 1°F change in temperature. 15. (A) V 142,000 0002 7,500t (B) The tractor’s value is decreasing at the rate of $7,500 per year. (C) $97,000 17. (A) R 1.4C 0002 7 (B) The slope is 1.4; this is the rate of change of retail price with respect to cost. (C) $137 23. (A) y 5.
(A) (B) (C) (A) (B) (C) (D) (A)
1.0
0.5
0.5
1.0
x
(B) 0.97 million (C) 1.3 million
Chapter 2 Review Exercises 1.
(2-1)
3. (A) Symmetric with respect to the origin
y
(B) No symmetry with respect to the x axis, y axis, or origin
y
5
y
5
A 00025
C
5
B
5
x 00025
5
x
00025
5
00025 00025
00025
x
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Student Answer Appendix
(C) Symmetric with respect to the y axis
(D) Symmetric with respect to the x axis
y 5
5
00025
5
x
00025
5
00025
9.
(A)
(2-1)
y
x
00025
(2-2)
y 5 C
B 00025
5
x
A 00025
(B) d(A, C) 2 110, d(B, C) 110, d(A, B) 150, perimeter 16.56 (C) d(A, C)2 d(B, C)2 d(A, B)2; right triangle (D) Midpoint of side AC (0, 1), of side BC (2.5, 3.5), and of side AB (1.5, 0.5) 11. Slope 000232 (2-3) 15. Symmetric with respect to the y axis (2-1) y
y 5
5 00025 00025
5
5
x
x 00025
17.
Symmetric with respect to the x axis, y axis, and origin
27. y 0002x 7 (1-5, 2-2) y
(2-1)
y 5
6
(4, 3) 00025
5
x 6
x
00025
29.
(A)
(2-1)
(B)
y 5
00025
35.
(A) (B) (C) 37. (A)
x
00025
y
5
5
00025
(D)
y
5
5
00025
(C)
y
x
00025
5
5
00025
x
00025
5
00025
The rate of change of body surface area with respect to weight is 0.3433. Body surface area increases by 34.33 cm2. 6,470.5 cm2 (2-4) H 0.7(220 0002 A) (B) H 140 beats per minute (C) A 40 years old (2-4)
CHAPTER 3 Exercises 3-1 39. Not a function; for example, when x 0, y 2 41. A function with domain all real numbers 43. Not a function; for example, when x 0, y 7 45. A function with domain all real numbers 59. [00024, 1) 傼 (1, 0005); 00024 0003 x 6 1 or x 7 1 67. Function f multiplies the square of the domain element by 2 then adds 5 to the result. 69. Function z divides the sum of four times the domain element and 5 by the square root of the domain element. 1 1 79. (A) 00028x 3 0002 4h (B) 00024x 0002 4a 3 81. (A) (B) 1x h 2 1x 2 1x 2 1a 2 00024 00024 83. (A) (B) 91. The cost is a flat $17 per month, plus $2.40 for each hour of airtime. x(x h) ax
x
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Student Answer Appendix 93.
(A) s(0) 0, s(1) 16, s(2) 64, s(3) 144 (C) Let q(h) [s(2 h) 0002 s(2)]0002h h q(h)
SA-7
(B) 64 16h
00021
00020.1
00020.01
00020.001
0.001
0.01
0.1
1
48
62.4
63.84
63.984
64.016
64.16
65.6
80
(D) q(h), the average velocity from 2 to 2 h seconds, approaches 64 feet per second 97.
F 8x (2500002x) 0002 12; x
4
5
6
7
F
82.5
78
77.7
79.7
Exercises 3-2 9. (A) [00024, 4) (B) [00023, 3) (C) 0 (D) 0 (E) [00024, 4) (F) None (G) None (H) None 11. (A) (00020005, 0005) (B) [00024, 0005] (C) 00023, 1 (D) 00023 (E) [00021, 0005) (F) (00020005, 00021] (G) None (H) None 13. (A) (00020005, 2) 傼 (2, 0005) (B) (00020005, 00021) 傼 [ 1, 0005) (C) None (D) 1 (E) None (F) (00020005, 00022], (2, 0005) (G) [00022, 2) (H) x 2 21. One possible answer: 23. One possible answer: 25. One possible answer: 27. Slope 2, x intercept 00022, f(x) f(x) f(x) y intercept 4 5
5
00025
x
5
5
00025
00025
5
x
f(x)
00025
5
00025
x 5
00025 00025
29.
Slope 000212 , x intercept 0002103 , y intercept 000253
31.
5
x
Slope 00022.3, x intercept 3.1, y intercept 7.1
f(x)
y
5
10
x
00025
000210
00025
10
x
000210
37. Domain: 5x ƒ x 000f 000226; x intercept: 4; y intercept: 00023 39. Domain: 5x ƒ x 000f 54 6; x intercept: 23 ; y intercept: 25 41. Domain: 5x ƒ x 000f 26; x intercept: 0; y intercept: 0 43. Domain: 5x ƒ x 000f 00023, 36; x intercept: 4; y intercept: 169 45. Domain: 5x ƒ x 000f 00025, 56; no x intercept; y intercept: 0002257 47. (A) f (00021) 0, f (0) 1, f (1) 0 49. (A) f (00022) 00022, f (00021) is not defined, f (2) 4 (B)
(B)
y
y
2.0
5
(2, 4)
(0, 1) 00025
(00021, 0) 00021
0.2
00025
1
(A) f (00022) 0, f (00021) is not defined, f (0) 00022
x
(00023, 00022)
(1, 0) x
(C) Domain: [00023, 00021) 傼 (00021, 2]; range: {00022, 4}; discontinuous at x 00021
(C) Domain: [00021, 1]; range: [0, 1]; continuous on its domain 51.
5
(B)
(C) Domain: (00020005, 00021) 傼 (00021, 0005); range: R; discontinuous at x 00021
y 5
(00022, 0) 00025
5
(0, 00022) 00025
x
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(A)
f(00023) 0, f (00022) 00022, f (0) 00022, f (3) 00022, f (4) 4
(B)
55.
f (00023) 000232 , f (00022) 1, f (0) 1, f (3) 1, f (4) 52
(B)
y 5
y 5
(4, 4) (00022, 1)
(00023, 0) 00025
5
(00022, 00022) 00025
x
(3, 00022) (0, 00022)
(00023,
(3, 1) 5
3 00022 )
63.
0003
(C) Domain: R; range: R; continuous on its domain 3 if x 6 00022 f (x) 00022 x 0002 2 if 00022 6 x 6 1 00021 if x 7 1
y 10 ) 3
(00021,
5
(1, 52 )
00025
5
(3,
x
3 00022 )
00025
(C) Domain: (00020005, 0) 傼 (0, 2) 傼 (2, 0005); range: (00020005, 4); discontinuous at x 0 and x 2 10002x if x 6 0 0002x 2 65. f (x) 67. f (x) 1 x if x 0004 0 x00022
0003
0003
y 5
5
x
00025
00025
69.
Domain: R; range: [ 1, 0005); continuous on its domain (A) One possible answer:
x
71.
Domain: R; range: [ 0, 0005); continuous on its domain (A) One possible answer:
f (x)
f(x)
5
5
10
x
00025
00025
5
x
00025
(B) The graph must cross the x axis exactly once. Graphs of f and g
(B) The graph must cross the x axis at least twice. There is no upper limit on the number of times it can cross the x axis. Graph of m Graph of n
10
000210
10
10
000210
000210
75.
5
00025
5
73.
if x 6 2 if x 0004 2
y
5
00025
10
10
000210
10
10
000210
000210
Graphs of f and g
000210
x
00025
(A) f (00021) 103 , f (0) is not defined, f (1) 52 , f (2) is not defined, f (3) 000232 (B)
(4, 52 ) (0, 1)
00025
(C) Domain: R; range: [00022, 0005); continuous on its domain 57.
(A)
10
000210
Graph of m
Graph of n
10
10
000210
10
000210
000210
10
000210
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Student Answer Appendix 77.
Graphs of f and g
Graph of m
Graph of n
10
10
10
000210
10
000210
10
000210
81. R(x)
0003
10
0003
000210
000210
200 if 0 0003 x 0003 3,000 83. E(x) 80 0.04x if 3,000 6 x 6 8,000 if x 0004 8,000 180 0.04x Discontinuous at x 8,000 E(5,750) $310, E(9,200) $548
if 0 0003 x 0003 100 if x 7 100
32 16 0.16x
000210
E(x)
500
5,000
85.
89.
10,000
x
x
4
00024
6
00026
24
25
247
0002243
0002245
0002246
f(x)
0
0
10
000210
20
30
250
0002240
0002240
0002250
15 18 21 (A) C(x) f 24 27 30
0 1 2 3 4 5
6 6 6 6 6 6
x x x x x x
0003 0003 0003 0003 0003 0003
1 2 3 4 5 6
C(x) $30
$15
0
1
2
3
4
5
(B) No, since f (x) 000f C(x) at x 1, 2, 3, 4, 5, or 6 0.03x if 0 0003 x 0003 10,000 91. T(x) 0.05x 0002 200 if x 7 10,000 93.
6
x
0003
0003
0.0535x T(x) 0.0705x 0002 338.25 0.0785x 0002 860.41 T(10,000) $535 T(30,000) $1,776.75 T(100,000) $6,989.60
y 1,000
24
; f rounds numbers to the tens place.
0 0003 x 0003 19,890 19,890 6 x 0003 65,330 x 7 65,330
x
Exercises 3-3 5. 11.
Domain: [ 0, 0005); Range: (00020005, 0 ] Domain: [ 00022, 2 ] ; range: [0, 4]
13.
Domain: [ 00022, 2 ] ; range: [1, 3]
y
00022
15.
Domain: [0, 4]; range: [ 00022, 2]
y
4
4
2
2
2
x
00022
y 2 2
2
x
00022
4
x
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Student Answer Appendix
Domain: [ 00024, 0 ] ; range: [ 00021, 1 ]
19.
Domain: [ 00022, 2 ] ; range: [ 00022, 2 ]
y 2
00024
y 2
x
00022
25.
Domain: [ 00022, 2 ] ; range: [ 00022, 2 ]
x 00022
49.
y
00022
5
00025
5
x
55. 10
5
5
x
00025
x
5
x
00025
5
00025
x
00025
3 (A) f is a horizontal shrink of y 1x by a factor of 1 8. g is a vertical stretch of y 1x by a factor of 2. 3 3 3 3 (C) f (x) 18x 18 ⴢ 1x 2 1x (A) The graphs are different; order is significant. (B) i. f (x) 0002(x2 0002 5) ii. f (x) 0002x2 0002 5 93. g(x) f(x) 3
5
5
5
00025
x
y
00025
95.
y
5
5
00025
x
5
5
61.
y
5
57.
00025
5
81. 91.
00025
00025
y
10
10
x
00025
y
5
79.
y
5
5
x
2
51.
y
10
00025
x
2
00022
59.
y 5
2
2
00025
x
2
45.
y
2
53.
00022
00022
y
47.
x
2
Domain: [ 00021, 1 ] ; range: [ 00021, 1 ]
00022
Domain: [ 00022, 2 ] ; range: [ 00022, 2]
2
00022
23.
21.
y
x
00025
5
x
00025
Conclusion: any function can be written as the sum of two other functions, one even and the other odd.
(B) The graphs are identical.
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Student Answer Appendix 97.
99. Each graph is a vertical translation of the graph of y 0.004(x 0002 10)3.
f(x)
35
Total production costs
$150,000 10
25
100,000 10 50,000
500
0
101.
1,000
x
Units produced
Each graph is a portion of the graph of a horizontal translation followed by a vertical shrink (except for C 8) of the graph of y t2. Larger values of C correspond to a smaller opening. V 70
0
2
4
6
8
t
Exercises 3-4 7.
Vertex: (00023, 00024); axis: x 00023
9.
3 3 Vertex: a ,00025b; axis: x 2 2
y
000210
11.
Vertex: (000210, 20); axis: x 000210 y
y
10
40
10
10
x
000210
0004
000210
3 2
(000210, 20)
x
10
(00023, 00024)
13. 17. 25.
, 000250005
000220
5 000210
000210
The graph is shifted 2 units right and 1 unit up. 15. The graph is reflected in the x axis, then shifted 1 unit left. The graph is shifted 2 units right and 3 units down. f (x) (x 0002 2)2 1; 27. h(x) 0002(x 1)2 0002 2; 29. m(x) 2(x 0002 3)2 4; vertex: (2, 1); axis: x 2 vertex: (00021, 00022); axis: x 00021 vertex: (3, 4); axis: x 3 y
y
y
10
10
10 (3, 4)
(2, 1) 10
000210
x
000210(00021, 00022)
10
x
000210
000210
000210
31.
SA-11
1 f(x) (x 3)2 0002 8; 2 vertex: (00023, 00028); axis: x 00023
33.
y
10
30 (6,18)
000210
10
35.
Vertex: (00024, 00028); The graph is symmetric about the axis, x 00024. It decreases until reaching a minimum at (00024, 00028), then increases. The range is [00028, 0005). y
x 000210
(00023,00028)
x
000210
f (x) 2(x 0002 6)2 18; vertex: (6, 18); axis: x 6
y
10
00025
10
10
x
000210 000210 (00024, 00028)
10
000210
x
x
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Vertex: (000572 , 654 ); The graph is symmetric about the axis, x 0002 000572 . It increases until reaching a maximum at (000572 , 654 ), then decreases. The range is (00050003, 654 ].
39.
Vertex: (94 , 194 ); The graph is symmetric about the axis, x 0002 94 . It decreases until reaching a minimum at (94 , 194 ), then increases. The range is [ 194 , 0003).
y
y
20
(00057/2, 65/4)
14
000510
x
10
(9/4, 19/4) 00055
5
000520
41.
00056
(52 , 149 2 );
5 2.
Vertex: The graph is symmetric about the axis, x 0002 It increases until reaching a maximum at (52 , 149 2 ), then decreases. The range is (00050003, 149 2 ].
43.
Vertex: (000532 , 000594 ); axis of symmetry: x 0002 0; domain: range: [ 000594 , 0003); min f (x) 0002 f (000532 ) 0002 000594 ; decreasing on (00050003, 000532 ); increasing on (000532 , 0003)
(00050003, 0003);
y
y 100
x
(5/2, 149/2)
00056
5
x
6
00055
5
x
(0005 32 , 0005 94 )
0005100
00055 3
45.
Vertex: (2, 00059); axis of symmetry: x 0002 2; domain: (00050003, 0003); range: [00059, 0003); min f (x) 0002 f (2) 0002 00059; decreasing on (00050003, 2); increasing on (2, 0003)
x 0002 00052
y 10
000510
10
000510
x
(2, 00059) x00022
81. The minimum product is 0005225 for the numbers 15 and 000515. There is no maximum product. 83. 26 employees; $322,800 85. (A) 2003 (B) The domain values should be whole numbers. y 97. (B) (C) 56 mph 200
80
105.
x
(A) R(x) 0002 3.5x 0005 0.00007x2; domain: [0, 50,000]; C(x) 0002 24,500 0004 0.35x, domain: [ 0, 0003) (B) x 0002 10,000 and x 0002 35,000 y 50,000
y 0002 C(x) y 0002 R(x)
50,000
x
(C) The company makes a profit for those sales levels for which the graph of the revenue function is above the graph of the cost function, that is, if the sales are between 10,000 and 35,000 gallons. The company suffers a loss for those sales levels for which the graph of the revenue function is below the graph of the cost function, that is, if the sales are between 0 and 10,000 gallons or between 35,000 and 50,000 gallons. (D) The maximum profit is $10,937.50 when 22,500 gallons are sold at a price of $1.92 per gallon.
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Student Answer Appendix
Exercises 3-5 7.
x
00023
00022
00021
0
1
2
3
3
3
1
00021
00023
00023
00023
( f g)(x)
9.
x ( fg)(x)
y
00023
00022
00021
0
1
2
3
2
0
00022
00022
0
2
2
y
5
5
00025
5
x
00025
00025
5
x
00025
27. ( f ° g)(00027) 3; ( f ° g)(0) 9; ( f ° g)(4) 000210 29. ( f g)(x) 5x 1; ( f 0002 g)(x) 3x 0002 1; ( fg)(x) 4x 2 4x; f 4x a b(x) ; domain f g, f 0002 g, fg (00020005, 0005); domain of f0002g (00020005, 00021) 傼 (00021, 0005) g x 1 2 31. ( f g)(x) 3x 1; ( f 0002 g)(x) x2 0002 1; ( fg)(x) 2x4 2x2; f 2x2 a b(x) 2 ; domain of each function: (00020005, 0005) g x 1 2 33. ( f g)(x) x 3x 4; ( f 0002 g)(x) 0002x2 3x 6; ( fg)(x) 3x3 5x2 0002 3x 0002 5; f 3x 5 ; domain f g, f 0002 g, fg: (00020005, 0005); domain of f0002g: (00020005, 00021) 傼 (00021, 1) 傼 (1, 0005) a b(x) 2 g x 00021 35. ( f g)(x) 12 0002 x 1x 3; ( f 0002 g)(x) 22 0002 x 0002 2x 3; ( fg)(x) 26 0002 x 0002 x2; f f 20002x a b(x) . The domain of the functions f g, f 0002 g, and fg is [00023, 2]. The domain of is (00023, 2]. g Ax 3 g 37. ( f g)(x) 21x 0002 2; ( f 0002 g)(x) 6; ( fg)(x) x 0002 2 1x 0002 8; f f 1x 2 . The domain of f g, f 0002 g, and fg is [0, 0005). Domain of [ 0, 16) 傼 (16, 0005). a b(x) g g 1x 0002 4 39. ( f g)(x) 2x2 x 0002 6 27 6x 0002 x2; ( f 0002 g)(x) 2x2 x 0002 6 0002 27 6x 0002 x2; ( fg)(x) 20002x4 5x3 19x2 0002 29x 0002 42; f f x2 x 0002 6 a b(x) . The domain of the functions f g, f 0002 g, and fg is [2, 7]. The domain of is [2, 7). g B 7 6x 0002 x2 g f 2 1 x2 1 41. ( f g)(x) 2x; ( f 0002 g)(x) ; ( fg)(x) x2 0002 2 ; a b(x) 2 The domain of f g, f 0002 g, and fg is (00020005, 0) 傼 (0, 0005). x g x x 00021 f is (00020005, 00021) 傼 (00021, 0) 傼 (0, 1) 傼 (1, 0005). g 2 ( f ° g)(x) (x 0002 x 1)3; domain: (00020005, 0005); (g ° f )(x) x6 0002 x3 1; domain: (00020005, 0005) ( f ° g)(x) 0006 2x 4 0006; domain: (00020005, 0005); (g ° f )(x) 20006 x 1 0006 3; domain: (00020005, 0005) ( f ° g)(x) (2x3 4)1 3; domain: (00020005, 0005); (g ° f )(x) 2x 4; domain: (00020005, 0005) ( f ° g)(x) 1x 0002 4; domain: [4, 0005); (g ° f )(x) 1x 0002 4; domain: [0, 0005) 1 1 ( f ° g)(x) 2; domain: (00020005, 0) 傼 (0, 0005); (g ° f )(x) ; domain: (00020005, 00022) 傼 (00022, 0005) x x 2 2 ( f ° g)(x) 24 0002 x ; domain of f ° g is [ 00022, 2 ] ; (g ° f )(x) 4 0002 x; domain of g ° f is (00020005, 4 ] . 6x 0002 10 x 5 ( f ° g)(x) ; domain of f ° g is (00020005, 0) 傼 (0, 2) 傼 (2, 0005); (g ° f )(x) ; domain of g ° f is (00020005, 0) 傼 (0, 5) 傼 (5, 0005). x 50002x ( f ° g)(x) x; domain: (00020005, 2) 傼 (2, 0005); (g ° f )(x) x; domain: (00020005, 0) 傼 (0, 0005) ( f ° g)(x) 216 0002 x2; domain of f ° g is [ 00024, 4 ] ; (g ° f )(x) 234 0002 x2; domain of g ° f is [00025, 5 ] . ( f ° g)(x) (g ° f )(x) x; the graphs of f and g are 67. ( f ° g)(x) (g ° f )(x) x; the graphs of f and g are symmetric with respect to the line y x. symmetric with respect to the line y x. The domain of
43. 45. 47. 49. 51. 53. 55. 57. 59. 65.
4
00026
4
6
00024
00026
6
00024
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( f ° g)(x) x, (g ° f )(x) x; the graphs of f and g are symmetric with respect to the line y x. y
( f ° g)(x) x, (g ° f )(x) x; the graphs of f and g are symmetric with respect to the line y x.
71.
y y g(x)
y f (x) y x
5
10
y g(x)
00025
5
yx
x
000210
85.
87.
y f(x) x
000210
00025
73. 77.
10
g(x) 2x 0002 7; f (x) x 4; h(x) ( f ° g)(x) f (x) x 7; g(x) 3x 0002 5; h(x) (g ° f )(x)
75. 79.
g(x) 4 2x; f (x) x1 2; h(x) ( f ° g)(x) f (x) x00021 2; g(x) 4x 3; h(x) (g ° f )(x)
1 2 ( f g)(x) 2x; (f 0002 g)(x) ; (fg)(x) x2 0002 2 ; x x f x2 1 (x) 2 The domain of f g, f 0002 g, and fg is g x 00021 f (00020005, 0) 傼 (0, 0005). The domain of is g (00020005, 00021) 傼 (00021, 0) 傼 (0, 1) 傼 (1, 0005). f 00022x ( f g)(x) 2; ( f 0002 g)(x) ; ( fg)(x) 0; a b(x) 0 |x| g The domain of f g, f 0002 g, and fg is (00020005, 0) 傼 (0, 0005). f Domain of is (0, 0005). g
Exercises 3-6 7. 9. 11. 41.
The original set and the reversed set are both one-to-one functions. The original set is a function. The reversed set is not a function. Neither set is a function. Domain of f [00024, 4] 43. Domain of f [00025, 3] range of f [1, 5] range of f [00023, 5] domain of f 00021 [1, 5] domain of f 00021 [00023, 5] range of f 00021 [00024, 4] range of f 00021 [00025, 3] y
45.
y
5
5
f 00021(x) 13 x domain of f (00020005, 0005) range of f (00020005, 0005) domain of f 00021 (00020005, 0005) range of f 00021 (00020005, 0005) y
yx
yx
10
y f (x) 00025
5
y f 00021(x)
x
00025
5
y f 00021(x)
yx
y f00021(x)
x
y f(x)
00025
000210
10
x
00025 000210
y f(x)
47.
f 00021(x) (x 3)00024 domain of f (00020005, 0005) range of f (00020005, 0005) domain of f 00021 (00020005, 0005) range of f 00021 (00020005, 0005)
49.
f 00021(x) 5x 0002 2 domain of f (00020005, 0005) range of f (00020005, 0005) domain of f 00021 (00020005, 0005) range of f 00021 (00020005, 0005)
y 10
y f(x) 000210
10
f 00021(x) (x 0002 3)2, x 0004 3 domain of f [0, 0005) range of f [3, 0005) domain of f 00021 [3, 0005) range of f 00021 [0, 0005)
y y f00021(x)
yx
10
y f00021(x) 000210
51.
x
y 10
yx
y f(x)
y f(x) 000210
10
000210
x
y f 00021(x) yx
000210
10
000210
x
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Student Answer Appendix 53.
f 00021(x) 16 0002 4x 2, x 0004 0 domain of f (00020005, 16] range of f [0, 0005) domain of f 00021 [0, 0005) range of f 00021 (00020005, 16]
55. f 00021(x) (3 0002 x)2 1, x 0003 3 domain of f [1, 0005) range of f (00020005, 3] domain of f 00021 (00020005, 3] range of f 00021 [1, 0005)
y
y yx
20
5
y
y f(x)
10
yx
20
yf
x
00025
5
x
000210
(x)
000220
00025
59. f (x) 000214 0002 x domain of f (00020005, 0] range of f (00020005, 4] domain of f 00021 (00020005, 4] range of f 00021 (00020005, 0]
61.
000210
00021
f (x) 1x 16 0002 4 domain of f [00024, 0005) range of f [000216, 0005) domain of f 00021 [000216, 0005) range of f 00021 [00024, 0005)
y
y yx
20
10
x
000210
63. f (x) 2 0002 1x domain of f (00020005, 2] range of f [0, 0005) domain of f 00021 [0, 0005) range of f 00021 (00020005, 2] y
y f(x)
y f(x)
20
000210
000220
f 00021(x) 1 1x 0002 2 domain of f [1, 0005) range of f [2, 0005) domain of f 00021 [2, 0005) range of f 00021 [1, 0005) y
y f (x)
69.
f 00021(x) 29 0002 x 2 domain of f [0, 3] range of f [00023, 0] domain of f 00021 [00023, 0] range of f 00021 [0, 3]
y yx
10
y yx
10
y f(x)
x
5
10
x
00025
5
y f 00021(x)
73.
y yx
yx 2
y f (x) 00025
5
x
y f 00021(x) 00025
00025
f 00021(x) 0002 22x 0002 x2 domain of f [00021, 0] range of f [0, 1] domain of f 00021 [0, 1] range of f 00021 [00021, 0]
y 5
x
y f(x)
000210
71. f 00021(x) 0002 29 0002 x2 domain of f [00023, 0] range of f [0, 3] domain of f 00021 [0, 3] range of f 00021 [00023, 0]
yx
y f 00021(x)
000210
000210
x
10
y f 00021(x) 000210
67. f 00021(x) 00021x 3 0002 1 domain of f (00020005, 00021] range of f [00023, 0005) domain of f 00021 [00023, 0005) range of f 00021 (00020005, 00021]
y f 00021(x) 000210
yx
10
y f 00021(x) x
000220
y f 00021(x)
00021
yx
y f (x)
10
x
10
00021
00021
10
yx
y f 00021(x)
y f(x)
000220
65.
f 00021(x) 1x 0002 5 domain of f [0, 0005) range of f [5, 0005) domain of f 00021 [5, 0005) range of f 00021 [0, 0005)
y f 00021(x)
y f (x)
000210
57.
y f(x) 00022
x
2 00022
yf
00021
(x)
2 2 x x 4x 5 77. f 00021(x) 79. f 00021(x) 81. f 00021(x) 83. f 00021(x) (4 0002 x)5 0002 2 30002x x 20002x 3x 0002 2 85. The x intercept of f is the y intercept of f 00021 and the y intercept of f is the x intercept of f 00021. 89. One possible answer: domain x 0003 2, f00021(x) 2 0002 1x 91. One possible answer: domain 0 0003 x 0003 2, f00021(x) 2 0002 24 0002 x2 15,000 0002 5; domain: [200, 1,000]; range: [10, 70] 95. (A) [200, 1,000] (B) d 00021(q) q 75.
f 00021(x)
(A) r m(w) 1.25w 3; domain: [0, 0005); range: [3, 0005) (B) w m00021(r) 0.8r 0002 2.4; domain: [3, 0005); range: [0, 0005) 50 (L 0002 20); domain: [20, 0005); range: [10, 0005) 99. s f 00021(L) 10 A3
97.
SA-15
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Student Answer Appendix
Chapter 3 Review Exercises 1. 3. 23.
(A) Function (B) Function (C) Not a function (3-1) If there is at least one team that has won more than one Super Bowl, then the correspondence is not a function because one input (team) will correspond with more than one output (year). There are several teams that have won at least two Super Bowls, so it is not a function. (3-1) 00023
00022
00021
0
1 2
000212
00026
00022
0
0 26
x ( fg)(x)
3
y
(3-5) 5
00025
5
x
000210
31.
(3-3)
y
33.
5
5
x
(3-3)
y 5
00025
5
00025
49. 51. 53. 57. 61. 67.
35.
5
00025
39.
(3-3)
y
x
00025
00025
5
x
00025
2
(B) (g0002f )(x) (x 3)0002(x 2 0002 4); (A) ( f0002g)(x) (x 0002 4)0002(x 3); domain of f0002g (00020005, 00023) 傼 (00023, 0005) domain of g f (00020005, 00022) 傼 (00022, 2) 傼 (2, 0005) (C) ( f ⴰ g)(x) x2 6x 5; domain of f ⴰ g (00020005, 0005) (D) (g ⴰ f )(x) x2 0002 1; domain of g ⴰ f (00020005, 0005) (3-5) The function f multiplies the square of the domain element by 3, adds 4 times the domain element, and then subtracts 6. (3-1) This equation does not define a function. For example, the ordered pairs (2, 2) and (2, 00022) both satisfy the equation. (3-1) Domain: [0, 0005); y intercept: 2; no x intercepts (3-1, 3-2) 55. Domain: (00020005, 3); y intercept: 0; x intercept: 0 (3-1, 3-2) Domain: [0, 16) 傼 (16, 0005); y intercept: 14 ; no x intercepts (3-1, 3-2) (A) (f ⴰ g)(x) 1|x| 0002 8, (g ⴰ f )(x) | 1x 0002 8| (B) Domain of f ⴰ g (00020005, 0005), domain of g ⴰ f [ 0, 0005) (3-5) g(x) 5 0002 3|x 0002 2| (3-3) y 5
x
5
3 The graph of y 2x is vertically stretched by a factor of 2, reflected through the x axis, shifted 1 unit left and 1 unit down. 3 Equation: y 000222x 1 0002 1 (3-3) 2 y 73. t(x) 0.25x x 0002 3 (3-3) 75. (3-3) 77. (3-3) y
69.
y
10
5
10
000210
10
00025
x
5
x
00025
10
000210
y
79.
(3-3)
81.
(3-3)
y
5
10
00025
x
x
00025
00025
x
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Student Answer Appendix 83. x 0007 00022 or x 0006 6; (00020005, 00022) 傼 (6, 0005) (3-4) 85. (A) x2 11 0002 x, domain (00020005, 1] (B) x2 11 0002 x, domain (00020005, 1) (C) 1 0002 x, domain (00020005, 1] (D) 21 0002 x2, domain [00021, 1] (3-5) 87. (A) f 00021(x) x2 1 (B) Domain of f [1, 0005) Range of f 00021 Range of f [0, 0005) Domain of f 00021 (3-6) (C) y 5
f 00021 f
00025
x
5
yx 00025
89.
(A)
(B)
y
(3-3)
y
5
5
00025
x
5
00025
00025
5
x
00025
if 0 0003 x 0003 2,000 if 2,000 6 x 0003 5,000 if x 7 5,000 (D) $24.99 (3-6) 93. $168 (C) c f 00021(r ) 0.625r; domain: [16, 0005); range: [10, 0005) 4,500 00021 95. (A) [1, 3] (B) q g (p) (C) R(p) 4,500 0002 500p 0002 500; domain: [1, 3]; range: [1,000, 4,000] p (D) R(q) 9q0002(1 0.002q) (3-6) 97. (A) A(x) 60x 0002 32 x2 (B) 0 0007 x 0007 40 (C) x 20, y 15 (3-4) 91.
120 (A) E(x) • 0.1x 0002 80 0.1x 170 (A) f f (c) 1.6c (B)
0003
if 0 0003 x 0003 3,000 if 3,000 6 x 0003 5,000 if 5,000 6 x 0003 17,000 if 17,000 0003 x
0.02x 0.03x 0002 30 99. T(x) 0.05x 0002 130 0.0575x 0002 257.5 x T(x)
$2,000
$4,000
$40
$10,000
$90.00
$370
$30,000 $1,467.50 (3-2)
CHAPTER 4 Exercises 4-1 37. 63. 65. 67. 77. 79. 81. 93.
11 4x2 10x 0002 9 3 2x3 0002 3x 1 4x 0002 2 0002 2x2 4x 5 39. x 3 x 3 x00022 x00022 P(x) S 0005 as x S 0005 and P(x) S 00020005 as x S 00020005; three intercepts and two local extrema P(x) S 00020005 as x S 0005 and P(x) S 0005 as x S 00020005; three intercepts and two local extrema P(x) S 0005 as x S 0005 and as x S 00020005; four intercepts and three local extrema x intercepts: 000212.69, 00020.72, 4.41; local maximum: P(2.07) 0007 96.07; local minimum: P(00028.07) 0007 0002424.07 x intercepts: 000216.06, 0.50, 15.56; local maximum: P(00029.13) 0007 65.86; local minimum: P(9.13) 0007 000255.86 x intercepts: 000216.15, 00022.53, 1.56, 14.12; local minimum: P(000211.68) 0007 00021,395.99; local maximum: P(00020.50) 0007 95.72; local minimum: P(9.92) 0007 00021,140.27 (A) 95. (A)
(B) $4,062 billion
(B)
00023.6 (implausible estimate)
Exercises 4-2 35. (A) Upper bound: 2; lower bound: 00022
(B) 1.4 (or 00021.4)
Exercises 4-3 9. 0 (multiplicity 3), 000212 (multiplicity 2); degree of P(x) is 5 11. 2i (multiplicity 3); 00022i (multiplicity 4); 00022 (multiplicity 5); 2 (multiplicity 5); degree of P(x) is 17 15. P(x) (x 7)3 [ x 0002 (00023 12)][x 0002 (00023 0002 12)]; degree 5
SA-17
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Student Answer Appendix
17. P(x) [x 0002 (2 0002 3i)][x 0002 (2 0002 3i)](x 4)2; degree 4 1 13 87. (A) 000212 13 (B) 3 2 i, 00022 0002 2 i 91. No, because P(x) is not a polynomial with real coefficients (the coefficient of x is the imaginary number 2i ).
Exercises 4-4 15. Domain: all real numbers except 0; x intercept: 3 17. Domain: all real numbers except 2; x intercept: 00026 19. Domain: all real numbers; x intercepts: 00024, 1 21. Domain: all real numbers except 6; x intercepts: none 23. Vertical asymptote: x 00022; horizontal asymptote: y 5 25. Vertical asymptotes: x 00024, x 4; horizontal asymptote: y 0 3 27. Vertical asymptote: x 0; horizontal asymptote: none 29. Vertical asymptotes: x 00023, x 0; horizontal asymptote: y 2 37. The graph of f is the same as the graph of g except that f has a hole at (00022, 16 ). 39.
41.
y 10
x
00025
5
000210
47.
49.
y
5
5
x
51.
y
77. 79.
5
x
10
73.
Vertical asymptote: x 1; oblique asymptote: y 2x 2
83.
y
5
5
x
00025
00025
5
x
00025
5
00025
87.
Vertical asymptote: x 0 Oblique asymptote: y 000214 x
Domain: x 000f 2, 00022 or (00020005, 00022) 傼 (00022, 2) 傼 (2, 0005); r(x)
y
y
5
00025
5
5
x
5
00025
91.
N
Vertical asymptote: x 2 Horizontal asymptote: y 0
As t S 0005, N S 5 N
50
50
25
25
25
x
00025
As t S 0005, N S 50
50
t
0
x
00025
Vertical asymptote: x 2 Oblique asymptote: y 12 x 0002 1
Domain: x 000f 2, or (00020005, 2) 傼 (2, 0005); f (x) x 2
0
x
000210
(2x 5)(x 0002 10) 100 55. f (x) x 0002 10 x4 1 Vertical asymptote: x 0; oblique asymptote: y 2x 0002 3 81. y y
Vertical asymptote: x 0 Oblique asymptote: y x
89.
00025
000210
3(x2 0002 1)(x2 0002 4)
00025
85.
5
10
00025
5
00025
y
00025
53. f (x)
x
00025
5
00025
5
00025
x
00025
5
y
45. 5
5
10
000210
y
43.
y
25
50
t
1 x00022
x
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Student Answer Appendix
93.
(A)
C(n) 25n 175
2,500 n
95.
(A)
L(x) 2x
450 x
(B) (0, 0005) (C) 15 ft by 15 ft (D) L
(B) 10 yr (C) C 2,000
200 1,000 100
0
25
50
n 0
25
50
x
Chapter 4 Review Exercise 1. 9. 11. 13. 29. 33. 47.
Zeros: 00021, 3; turning points: (00021, 0), (1, 2), (3, 0); P(x) S 0005 as x S 0005 and P(x) S 0005 as x S 00020005 (4-1) 1 3
1, 3, 5, 15, , (4-3) 5 5 3 (A) Domain: all real numbers except 5; x intercept: 0 (B) Domain: all real numbers except 00024 and 2; x intercept: 0002 7 The graph does not increase or decrease without bound as x S 0005 and as x S 00020005 (4-1) 1 i 13 1 0002 i 13 (x 1)(2x 0002 1) ax 0002 b ax 0002 b (4-3) 2 2 (A) Upper bound: 7; lower bound: 00025 (B) Four intervals (C) 00024.67, 6.62 (4-2) 3 3i 13 (A) 3 (B) 0002 (4-3) 49. (4-4) y 2 2
(4-4)
5
00025
5
x
00025
53. 3: None of the candidates for rational zeros ( 1, 2, and 4) are actually zeros. 63. (A)
(4-3)
59. v k
1T 1w
(4-5)
(B) 1,915
CHAPTER 5 Exercises 5-1 25.
The graph of g is the same as the graph of f stretched vertically by a factor of 3; g is increasing; horizontal asymptote: y 0
27.
The graph of g is the same as the graph of f reflected through the y axis and shrunk vertically by a factor of 13 ; g is decreasing; horizontal asymptote: y 0
y
y
10
10
00025
5
x
00025
000210
29.
5
000210
The graph of g is the same as the graph of f shifted upward 2 units; g is increasing; horizontal asymptote: y 2
31.
The graph of g is the same as the graph of f shifted 2 units to the left; g is increasing; horizontal asymptote: y 0
y
y
10
00025
10
5
000210
x
x
00025
5
000210
x
SA-19
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Student Answer Appendix
In every case, y 1. The function y 1x is simply the constant function y 1.
55.
The graph of g is the same as the graph of f reflected through the x axis; g is increasing; horizontal asymptote: y 0 y 10
00025
5
x
000210
57.
59. The graph of g is the same as the graph of f stretched vertically by a factor of 500; g is increasing; horizontal asymptote: y 0
The graph of g is the same as the graph of f stretched horizontally by a factor of 2 and shifted upward 3 units; g is decreasing; horizontal asymptote: y 3 y
y
10
2500
00025
5
x
000210
61.
000250
63.
The graph of g is the same as the graph of f shifted 3 units to the right, stretched vertically by a factor of 2, and shifted upward 1 unit; g is increasing; horizontal asymptote: y 1
50
The graph of g is the same as the graph of f shifted 2 units to the right, reflected through the origin, stretched vertically by a factor of 4, and shifted upward 3 units; g is increasing; horizontal asymptote: y 3
y
y
10
10
00025
5
x
00025
000210
65. 71. 73. 75. 79.
x
5
x
000210
e00022x(00022x 0002 3)
69. No local extrema; no x intercept; y intercept: 2.14; horizontal asymptote: y 2 x4 Local maximum: s(0) 1; no x intercepts: y intercept: 1: horizontal asymptote: x axis No local extrema; no x intercept; y intercept: 50; horizontal asymptotes: x axis and y 200 Local minimum: f(0) 1; no x intercepts; no horizontal asymptotes 21.4 2.6390; 21.41 2.6574; 21.414 2.6648; 21.4142 2.6651; 21.41421 2.6651; 21.414214 2.6651; 212 0007 2.6651
81.
83.
50
00024
4
50
00024
00025
4
00025
As x S 0005, fn (x) S 0; the line y 0 is a horizontal asymptote. As x S 00020005, f1(x) S 00020005 and f3(x) S 00020005, while f2(x) S 0005. As x S 00020005, fn(x) S 0005 if n is even and fn(x) S 00020005 if n is odd. 97. Flagstar: $5,488.61; UmbrellaBank.com: $5,470.85; Allied First Bank: $5,463.71
85.
Exercises 5-2 L
13. 1,000
500
5
10
n
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Student Answer Appendix 25.
SA-21
P 100
50
0
50
t
100
33. q approaches 0.000 9 coulombs, the upper limit for the charge on the capacitor. 35. (C) A approaches 100 deer, the upper limit for the number of deer the island can support. 37. y 14,910(0.8163)x; estimated purchase price: $14,910; estimated value after 10 years: $1,959 906 39. (A) y (B) 2010: 893.3 billion 2020: 903.6 billion 1 2.27e00020.169x
Exercises 5-3 19.
x 00023 00022 00021 0 1 2 3
21.
x
y 3x
x
y log3x
1 27 1 9 1 3
1 27 1 9 1 3
1 3 9 27
1 3 9 27
00023 00022 00021 0 1 2 3
y ( 23 ) x
00023 00022 00021 0 1 2 3
x
27 8 9 4 3 2
27 8 9 4 3 2
1
1
2 3 4 9 8 27
2 3 4 9 8 27
y 30
y 3x
y log3 x 00025 00025
30
y
y log2 3x 00023 00022 00021 0 1 2 3 79.
x
5
2 x
y 000430005 5
00025
00025
log x 0002 log y
61.
b is any positive real number except 1.
91.
The graph of g is the same as the graph of f shifted upward 3 units; g is increasing. Domain: (0, 0005); vertical asymptote: x 0
x
y log2/3 x
81.
x2y5 x 83. ln a b 85. ln a b y z 93. The graph of g is the same as the graph of f shifted 2 units to the right; g is decreasing. Domain: (2, 0005); vertical asymptote: x 2 4 log x 3 log y
y
y
10
10
000210
10
x
000210
000210
95.
10
000210
The graph of g is the same as the graph of f reflected through the x axis and shifted downward 1 unit; g is decreasing. Domain: (0, 0005); vertical asymptote: x 0
97.
The graph of g is the same as the graph of f reflected through the x axis, stretched vertically by a factor of 3, and shifted upward 5 units; g is decreasing. Domain: (0, 0005); vertical asymptote: x 0
y
y
10
000210
10
10
000210
x
x
000210
10
000210
x
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Student Answer Appendix
(A) f 00021(x) 2 0002 3x (B)
107.
109.
2
2
(C)
y
y
10
00021
3
00021
3
10
f 00021 000210
x
10
000210
00022 10
f
000210
00022
x
000210
Exercises 5-4 25.
(A)
y 11.9 24.1 ln x; 2008: 73.7%; 2015: 84.1%
(B) No; the predicted percentage goes over 100 sometime around 2034.
Exercises 5-5 87.
(A)
7.94 0010 1014 joules
Chapter 5 Review Exercises 55.
57.
The graph of g is the same as the graph of f stretched vertically by a factor of 2 and shifted downward 4 units; g is increasing. Domain: all real numbers Horizontal asymptote: y 00024 (5-1) y
The graph of g is the same as the graph of f stretched vertically by a factor of 2 and shifted upward 1 unit; g is decreasing. Domain: (0, 0005); Vertical asymptote: x 0 (5-3) y
10
000210
10
10
x
000210
10
000210
67.
000210
Domain f (0, 0005) Range f 00021 Range f (00020005, 0005) Domain f 00021
75.
(A) P
(5-3)
y 10
x
1,000
f 00021: y 2x 500
5
f : y log2 x
5
00025
10
0
x
25
50
t
00025
81.
(A)
y 43.3(1.09)x; 2010: $574 billion; 2020: $1,360 billion
CHAPTER 6 Exercises 6-1 y
15. Directrix x 00021 00025
y
17.
5
5
F (1, 0) 5
x
5
00025
00025
Directrix y 00022
y
19.
F (0, 2)
y
21.
5
x
F (00023, 0) 00025
5
Directrix x3 5
x
F (0, 00021)
Directrix y1 5
00025
00025
x
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Student Answer Appendix y
23.
y
45.
10
y
47.
5
10
Directrix x5
F (00025, 0)
10
000210
(4, 4)
x
(5.313, 5.646) x
5
00025
57.
2
2(x 0002 x) ( y a) ( y a)2 y2 2ay a2 x2
2
x
(0, 0)
00025
2
10
000210
(0, 0) 000210
SA-23
000210 2
2(x 0002 0) (y 0002 a) x2 (y 0002 a)2 x2 y2 0002 2ay a2 4ay
Exercises 6-2 15 .
Foci: F ¿ (0002 121, 0), F ( 121, 0); major axis length 10; minor axis length 4 y
17. Foci: F¿ (0, 0002 121), F (0, 121); major axis length 10; minor axis length 4 y
5
5
F
y
F
5
F
F x
5
00025
5
00025
00025
25.
19. Foci: F¿ (0002 18, 0), F ( 18, 0); major axis length 6; minor axis length 2
y
y
12
F
5
00025
29. Foci: F¿ (0002 13, 0), F ( 13, 0); major axis length 217 0007 5.29; minor axis length 4 y
5
5
F
00026
x
x
5
00025
27. Foci: F¿ (0, 0002 16), F (0, 16); minor axis length 2 112 0007 6.93; major axis length 216 0007 4.90
5
F
00025
F
00025
Foci: F¿ (0, 00024), F (0, 4); major axis length 10; minor axis length 6
x
5
00025
7
00027
6
F
x
F
00025
x
5
F F
000212 00025
00025 2
35.
y x2 1 25 9
2
37.
y x2 1 64 121
00025
2
51.
y x2 1; 7.94 feet approximately 400 144
53.
(A)
y2 x2 1 576 15.9
(B) 5.13 feet
Exercises 6-3 15.
Foci: F¿ (0002 113, 0), F ( 113, 0); transverse axis length 6; conjugate axis length 4
17.
Foci: F¿ (0, 0002 113), F (0, 113); transverse axis length 4; conjugate axis length 6
y c
F 0002c
F c 5
F
x
5
00025
0002c
y
10
23.
F 0002c
x
F
00025
Foci: F¿ (0, 00025), F (0, 5); transverse axis length 8; conjugate axis length 6
c
c
00025
F c 10
0002c F
x
Foci: F¿ (0002110, 0), F ( 110, 0); transverse axis length 4; conjugate axis length 216 0007 4.90 y
F 00025 0002c
F c
x
00025
25.
Foci: F¿ (0, 0002 111), F (0, 111); transverse axis length 4; conjugate axis length 217 0007 5.29 y
5
6 5
000210
000210
y
5
5
c
c
Foci: F¿ (0002 120, 0), F ( 120, 0); transverse axis length 4; conjugate axis length 8
y
5
21.
19.
c
00025 00026
F c 5
00027
c F c
x
0002c F 00025
7
x
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Student Answer Appendix
45. y
13 x 12 x2
y2
(A) Infinitely many;
0002
1 (0 0007 a 0007 1)
x2
y2
(C) One; y2 4x
1 (a 0006 1) a2 1 0002 a2 a2 a2 0002 1 49. (A) (2 13, 1 13), (00022 13, 00021 13) (B) No intersection points The graphs intersect at x 1 ( 21 0002 m 2) and y m ( 21 0002 m2) for 00021 0007 m 0007 1. 51. (A) No intersection points (B) (1 15, 3 15), (00021 15, 00023 15) The graphs intersect at x 1 ( 2m 2 0002 4) and y m ( 2m2 0002 4) for m 0007 00022 or m 0006 2. 2 y2 x 59. 0002 1; hyperbola 4 5
47.
(B) Infinitely many;
Chapter 6 Review Exercises 1.
Foci: F (00024, 0), F (4, 0); major axis length 10; minor axis length 6 (6-2)
3.
y
Foci: F¿ (0, 0002134), F (0, 134); transverse axis length 6; conjugate axis length 10 (6-3) 10
5
00025
x
10
000210
F (2, 0) x 5
00025
F F
(6-1)
5
y
5
F
y
5.
Directrix x 00022 00025
x
F 00025
21.
000210
y2 x2 1; ellipse (6-2) 36 20
CHAPTER 7 Exercises 7-1 {(3s 2, s, 00022s 0002 1) | s any real number} dh 0002 bk ak 0002 ch 49. x ,y , ad 0002 bc 000f 0 ad 0002 bc ad 0002 bc
35.
39.
{(00022s 5, s, 3s 0002 4) | s any real number}
45.
5(13 s 0002 43 , 23 s 0002 83 , s) ƒ s any real number6
61. (A) Supply: 143 T-shirts; demand: 611 T-shirts (B) Supply: 714 T-shirts; demand: 389 T-shirts (C) Equilibrium price: $6.36; equilibrium quantity: 480 T-shirts p (D)
Price ($)
20
Equilibrium point (480, 6.36) Supply curve
10
0
400
Demand curve q
800
Quantity
71.
$35,000 treasury bonds; $7,500 municipal bonds; $27,500 corporate bonds
Exercises 7-2 21.
x1 2t 3, x2 0002t 0002 5, x3 t, t any real number
27.
c
4 1
00026 00028 ` d 00023 2
29.
c
00024 4
12 00028 ` d 00026 00028
31.
c
1 8
00023 2 ` d 000212 000216
33.
c
1 0
00023 2 ` d 6 000216
35.
c
1 2
00023 2 ` d 0 000212
1 0 2 000253 1 0 0 00025 41. £ 0 1 0 † 4 § 43. £ 0 1 00022 † 13 § 49. Infinitely many solutions; for any real number s, x2 s, x1 2s 0002 3 0 0 0 0 0 1 00022 0 73. Either 11 CDs, 1 DVD and 1 book; 6 CDs, 4 DVDs, and 3 books; or 1 CD, 7 DVDs, and 5 books 81. One-person boats: t 0002 80; two-person boats: 00022t 420; four-person boats: t, 80 0003 t 0003 210, t an integer 83. No solution; no production schedule will use all the labor-hours in all departments.
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Student Answer Appendix
SA-25
Exercises 7-3 2 4
5 d 00026
11.
c
47.
00020.2 £ 2.6 00020.2
73.
31.
1.2 00020.6 § 2.2
3 £ 6 00029 49.
00022 £ 26 00022
53.
000211 d 00024
7 18
25 000225 45
00023 £ 000218 4
43.
000215 45 § 000225
55.
6 12 6 000226 £ 00024 2
8 10 § 24 000215 000218 43
45.
5 £4 0
000211 00027 10
15 3§ 4
000225 4§ 000219
77.
(A) $11.80 (B) $30.30 (C) MN gives the labor costs per boat at each plant. (D) Plant I Plant II $11.80 $13.80 One-person boat MN £ $18.50 $21.60 § Two-person boat $26.00 $30.30 Four-person boat
0 0 2 0 0 1 0 0 0 1 (A) A2 G 0 1 0 2 0 W 1 0 0 0 1 0 0 1 0 0 There is one way to travel from Baltimore to Atlanta with one intermediate connection; there are two ways to travel from Atlanta to Chicago with one intermediate connection. In general, the elements in A2 indicate the number of different ways to travel from the ith city to the jth city with one intermediate connection.
(B)
(C)
(A) (D)
(E)
(F)
83.
00026 4
Markup AM/FM Cruise Air radio control $77 $42 $27 $93 $95 $50 § $113 $121 $52
Basic car Model A $3,330 Model B £ $2,125 Model C $1,270
81.
c
41.
$33 $26 1 (A B) c d 2 $57 $77 This is the average cost of materials and labor for each product at the two plants.
75.
79.
00022 00024 00024 00028 § 6 12 000231 16 £ 61 000225 § 00023 77
(A)
2 0 0 0 2 0 1 0 2 0 A3 G 0 0 3 0 0 W 0 1 0 2 0 1 0 0 0 1 There is one way to travel from Denver to Baltimore with two intermediate connections; there are two ways to travel from Atlanta to El Paso with two intermediate connections. In general, the elements in A3 indicate the number of different ways to travel from the ith city to the jth city with two intermediate connections. 2 3 2 5 2 1 1 4 2 1 A A2 A3 A4 G 4 1 3 2 4 W 1 1 4 2 1 1 1 1 3 1 It is possible to travel from any origin to any destination with at most three intermediate connections. $3,550 (B) $6,000 (C) NM gives the total cost per town. Cost/town $3,550 Berkeley NM c d $6,000 Oakland Telephone House call call Letter 1,300 13,000 ] [1 1] N [ 3,000 Total contacts 1 6,500 Berkeley N £1§ c d 10,800 Oakland 1 0 1 0 G 0 0 1
0 0 1 0 0 1
1 0 0 0 1 1
1 1 1 0 1 0
1 1 0 0 0 0
0 0 0 W 1 1 0
(B)
0 1 1 G 1 1 2
1 0 1 1 2 2
2 2 0 1 2 2
3 3 2 0 2 3
1 2 1 0 0 2
2 2 1 W 1 2 0
(C)
9 1 10 1 6 1 BC G W where C G W 4 1 9 1 11 1
00023 x1 2 dc d c d 2 x2 1
31.
1 £ 00021 2
(D) Frank, Bart, Aaron and Elvis (tie), Charles, Dan
Exercises 7-4 11.
c
2 4
00023 d 5
25.
2x1 0002 x2 3 x1 3x2 00022
29.
c
4 1
00021 00022 1 x1 1 0 § £ x2 § £ 2 § 00023 3 1 x3
41.
c
1 0
00029 d 1
bar19499_saa_SA1-SA30.qxd
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81.
(A)
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Student Answer Appendix
00025 00022 d 2 1
43.
11/20/09
c
45.
(AB)00021 c
00023 00022
00027 d 00025
51.
0 00021 00021 £ 00021 00021 00021 § 00021 00021 0
29 000212
000241 23 000233 d , A00021B00021 c d, 17 000216 23
29 000212
000241 d 17
B00021A00021 c
000219 £ 15 00022
53.
9 00027 00027 6§ 1 00021
1 0 1 £ 12 21 1 § 2 1 4
55.
59.
00029 000215 10 £ 4 5 00024 § 00021 00021 1
83. 61 22 96 38 115 43 131 47 68 27 110 43 85. BEYONCE KNOWLES 87. 42 43 88 33 101 40 61 62 40 49 40 103 72 56 69 52 81 99 53 101 89. RAIDERS OF THE LOST ARK 91. Concert 1: 6,000 $20 tickets, 4,000 $30 tickets Concert 2: 5,000 $20 tickets, 5,000 $30 tickets Concert 3: 3,000 $20 tickets, 7,000 $30 tickets
Exercises 7-5 19.
x 000265 , y 35
x 172 , y 000220 17
21.
4 00022
23. `
6 ` 8
25. `
00021 ` 00022
5 0
27.
(00021)1 1 `
4 00022
6 ` 44 8
29.
(00021)2 3 `
5 0
00021 ` 10 00022
x 43 , y 000213 , z 23 49. x 00029, y 000273 , z 6 51. x 32 , y 000276 , z 23 3 15 If a 2 and b 4 , there are an infinite number of solutions. If a 32 and b 000f 154 , there are no solutions. If a 000f 83 , there is one solution. (A) Since D 0, the system either has no solution or infinitely many. Since x 0, y 0, z 0 is a solution, the second case must hold. (B) Since D 000f 0, by Cramer’s rule, x 0, y 0, z 0 is the only solution. 73. (A) R 200p 300q 0002 6p2 6pq 0002 3q2 (B) p 00020.3x 0002 0.4y 180, q 00020.2x 0002 0.6y 220, R 180x 220y 0002 0.3x2 0002 0.6xy 0002 0.6y2
47. 53. 71.
Chapter 7 Review Exercises 3.
Infinitely many solutions [t, (4t 8)00023], for any real number t (7-1)
3 00026 12 4 8 3 3 ` d (7-2) d (7-3) d (7-3) 13. c 17. c 1 00024 5 000212 18 00024 9 23. (A) x1 00021, x2 3 (B) x1 1, x2 2 (C) x1 8, x2 000210 (7-4) 27. x1 2, x2 00022; each pair of lines has the same intersection point. (7-1, 7-2) 7.
c
x2
x2
x2
5
5
5
00025
x1
5
00025
5
(2, 00022)
x2
x1
5
00025
5
x1
00025
5
(2, 00022)
(2, 00022)
(2, 00022)
x1
00025
x1 0002 x2 4 3x2 00026
x1 0002 x2 4 2x1 x2 2 35. 57.
16 00029 40 000230 § 00028 17
7 £ 28 000221
(7-3)
37.
12 £ 0 00028
24 0 000216
x1 0002 x2 4 x2 00022 00026 0§ 4
x1 2 x2 00022
(7-3)
(A) $27 (C)
(B) Elements in LH give the total cost of manufacturing each product at each plant. North South Carolina Carolina $46.35 $41.00 Desks LH c d (7-3) Stands $30.45 $27.00
CHAPTER 8 Exercises 8-1 15.
99 101
19.
1 10
1 1 100 1,000
25.
0.3, 0.33, 0.333, 0.3333, 0.33333
55.
57.
1
0
20
00020.3
59.
4 1
8 2
0002
16 3
0002
32 4
73.
(A) 3, 1.83, 1.46, 1.415
(B) Calculator 12 1.4142135. . .
1.5
0
20
00021.5
(C)
a1 1; 1, 1.5, 1.417, 1.414
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Student Answer Appendix 81. 83.
(A) 0.625 ft; 0.02 ft (B) 19.98 (A) 40,000, 41,600, 43,264, 44,998.56, 46,794.34, 48,666.12
(B)
40,000(1.04)n00021
(C)
SA-27
265,319.02
Exercises 8-2 11. P1: a5a1 a5 1; P2: a5a2 a5(a1a) (a5a)a a6a a7 a5 2; P3: a5a3 a5(a2a) a5(a1a)a [(a5a)a]a a8 a5 3 13. P1: 91 0002 1 8 is divisible by 4; P2: 92 0002 1 80 is divisible by 4; P3: 93 0002 1 728 is divisible by 4 15. Pk: 2 6 10 . . . (4k 0002 2) 2k2; Pk 1: 2 6 10 . . . (4k 0002 2) (4k 2) 2(k 1)2 n(n 0002 1) 17. Pk: a5ak a5 k; Pk 1: a5ak 1 a5 k 1 49. 1 2 3 . . . (n 0002 1) ,n00042 2
Exercises 8-3 7. (A) Arithmetic with d 00025; 000226, 000231
(B) Geometric with r 00022; 000216, 32
(C) Neither
1 (D) Geometric with r 13 , 541 , 162
Exercises 8-4 39. 43. 49. 55. 57. 59.
No repeats: 10 ⴢ 9 ⴢ 8 ⴢ 7 ⴢ 6 30,240; with repeats: 10 ⴢ 10 ⴢ 10 ⴢ 10 ⴢ 10 100,000 26 ⴢ 26 ⴢ 26 ⴢ 10 ⴢ 10 ⴢ 10 17,576,000 possible license plates; no repeats: 26 ⴢ 25 ⴢ 24 ⴢ 10 ⴢ 9 ⴢ 8 11,232,000 (B) r 0, 10 (C) Each is the product of r consecutive integers, the largest of which is n for Pn,r and r for r! Two people: 5 ⴢ 4 20; three people: 5 ⴢ 4 ⴢ 3 60; four people: 5 ⴢ 4 ⴢ 3 ⴢ 2 120; five people: 5 ⴢ 4 ⴢ 3 ⴢ 2 ⴢ 1 120 (A) P8,5 6,720 (B) C8,5 56 (C) C2,1 ⴢ C6,4 30 There are C4,1 ⴢ C48,4 778,320 hands that contain exactly one king, and C39,5 575,757 hands containing no hearts, so the former is more likely.
Exercises 8-5 (A) No probability can be negative (B) P(R) P(G) P(Y ) P(B) 000f 1 (C) Is an acceptable probability assignment. C16,5 0007 .0017 31. C52,5 53. (A) P(2) 0007 .022, P(3) 0007 .07, P(4) 0007 .088, P(5) 0007 .1, P(6) 0007 .142, P(7) 0007 .178, P(8) 0007 .144, P(9) 0007 .104, P(10) 0007 .072, P(11) 0007 .052, P(12) 0007 .028 (B) P(2) 361 , P(3) 362 , P(4) 363 , P(5) 364 , P(6) 365 , P(7) 366 , P(8) 365 , P(9) 364 , P(10) 363 , P(11) 362 , P(12) 361
19.
(C) Sum
Expected frequency
2 3 4 5 6 7
13.9 27.8 41.7 55.6 69.4 83.3
Sum
Expected frequency
8 9 10 11 12
69.4 55.6 41.7 27.8 13.9
Exercises 8-6 21. m3 3m2n 3mn2 n3 23. 8x3 0002 36x2y 54xy2 0002 27y3 25. x4 0002 8x3 24x2 0002 32x 16 4 3 2 2 3 4 5 4 27. m 12m n 54m n 108mn 81n 29. 32x 0002 80x y 80x3y2 0002 40x2y3 10xy4 0002 y5 31. m6 12m5n 60m4n2 160m3n3 240m2n4 192mn5 64n6 51. 3x 2 3xh h 2; approaches 3x 2 4 3 2 2 3 4 4 53. 5x 10x h 10x h 5xh h ; approaches 5x
Chapter 8 Review Exercises 1. (A) Geometric (B) Arithmetic (C) Arithmetic (D) Neither (E) Geometric (8-1, 8-3) 3. (A) 16, 8, 4, 2 (B) a10 321 (C) S10 3131 (8-1, 8-3) 9. 21 (8-4) 32 11. (A) 12 combined outcomes: (B) 6 ⴢ 2 12 (8-5) (1, H) H 1
T
(1, T)
2
H T
(2, H) (2, T)
3
H T
(3, H) (3, T)
4
H T
(4, H) (4, T)
5
H T
(5, H) (5, T)
6
H T
(6, H) (6, T)
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Student Answer Appendix
17. P1: 5 12 4 ⴢ 1 5; P2: 5 7 22 4 ⴢ 2; P3: 5 7 9 32 4 ⴢ 3 (8-2) 21. Pk: 2 4 8 . . . 2k 2k 1 0002 2; Pk 1: 2 4 8 . . . 2k 2k 1 2k 2 0002 2 n (00021)k 1 C13,5 C13,3 ⴢ C13,2 1 ; S0005 27. Sn a (8-3) 33. (A) (B) (8-5) k 4 C C52,5 3 52,5 k 1
(8-2)
APPENDIX A Cumulative Review Exercise: Chapters 1–3 3.
00025 0007 x 0007 9 (00025, 9) (
(
00025
(1-3)
x
9
13. (A) Function; domain: {1, 2, 3}; range: {1} (B) Not a function 2 23. 3 0003 m 0003 2 (1-3) 25. x 0004 2, x 000f 4 (1-2) [ 2, 4) ´ (4, 0005) [23 , 2] [
[
]
m
2
2 3
2
4
(C) Function; domain: {00022, 00021, 0, 1, 2}; range: {00021, 0, 2} (3-1)
x
35. (A) All real numbers (B) {00022} 傼 [1, 0005) (C) 1 (D) [00023, 00022] and [2, 0005) x ; Domain: x 000f 0, 3 (3-5) 39. ( f ⴰ g)(x) 30002x 41. Domain: (00020005, 0005) (3-2) 45. Center: (3, 00021); radius: 110 (2-2) Range: (00020005, 00021) ´ [1, 0005) f(x) Discontinuous at x 0
(E) 00022, 2 (3-1, 3-2)
5
y 5
5
00025
5
x
(3, 00021)
x 00025
00025
53.
(A) Domain g: [ 00022, 2 ]
f f x2 ; Domain a b: (00022, 2) (B) a b(x) g g 24 0002 x2
(C) (f ⴰ g)(x) 4 0002 x2; Domain (f ⴰ g): [00022, 2]
(3-5)
if 0 0003 x 0003 60 if 60 6 x 0003 150 if 150 6 x 0003 300 if 300 6 x (3-2)
0.06x 0.05x 0.6 61. C(x) μ 0.04x 2.1 0.03x 5.1 C (x)
20
10
60 150
63.
(A)
300
500
x
1 if n is an odd integer (3-2) if n is an even integer 0 (B) Loss: $0 0003 p 0007 5.5 or p 0006 $8 or [$0, $5.5) 傼 ($8, 0005) (3-4) (B) s f 00021 (L) 2 120L 0002 126; domain: [22.5, 0005); range: [20, 0005)
f(1) f(3) 1, f(2) f(4) 0 (B) f (n) e
65. (A) Profit: $5.5 0007 p 0007 $8 or ($5.5, $8) 69. (A) (3-4) L 400
80
s
(C) 67 mph
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Student Answer Appendix
SA-29
Cumulative Review for Chapters 4 and 5 17.
(A) Domain: x 000f 00022; x intercept: x 00024; y intercept: y 4 (C) (4-4) y
(B) Vertical asymptote: x 00022; horizontal asymptote: y 2
10
000210
10
x
000210
(A) 00020.56 (double zero); 2 (simple zero); 3.56 (double zero) (B) 00020.56 can be approximated with a maximum routine; 2 can be approximated with the bisection: 3.56 can be approximated with a minimum routine (4-2) 51. A reflection through the x axis transforms the graph of y In x into the graph of y 0002ln x. A reflection through the y axis transforms the graph of y ln x into the graph of y ln (0002x). (5-3) y 55. Vertical asymptote: x 00022; (4–4) oblique asymptote: y x 2 10
23.
000210
10
x
000210
Cumulative Review for Chapters 6–8 5. (A) Arithmetic (B) Geometric (C) Neither (D) Geometric (E) Arithmetic 7. (A) 2, 5, 8, 11 (B) a8 23 (C) S8 100 (8-3) 11. Foci: F (0002 161, 0), F ( 161, 0); 13. (6-1) transverse axis length 12; y conjugate axis length 10 (6-3) 4
y 10
2
F
F
10
000210
(8-3)
x
00022
F 00040,
9 25
0005
x Directrix 9 y 0002 25
000210
17. 23. 25. 31. 39. 69. 93.
95.
0 00023 1 7 d d (B) Not defined (C) [3] (D) c (E) [00021, 8] (F) Not defined (7-4) 3 00029 4 00027 (B) x1 2t 3, x2 t, t any real number. (C) No solution (7-3) (A) x1 3, x2 00024 1 00023 x1 k1 00025 3 (A) c (B) A00021 c (C) x1 13, x2 5 dc d c d d (D) x1 000211, x2 00024 (7-5) k2 2 00025 x2 00022 1 2 2 Pk: k k 2 2r for some integer r; Pk 1: (k 1) (k 1) 2 2s for some integer s (8-2) 00021 2 d (A) c (B) Not defined (7-4) 41. (0, i ), (0, 0002i), (1, 1), (00021, 00021) (7-6) 2 3 (A) Infinite number of solutions (B) No solution (C) Unique solution (7-3) 1 model A truck, 6 model B trucks, and 5 model C trucks; or 3 model A trucks, 3 model B trucks, and 6 model C trucks; or 5 model A trucks and 7 model C trucks. (7-3) 82.25 Ann 83 Ann 0.25 0.2 83 Bob 84.8 Bob 0.25 0.2 (A) M ≥ (B) M ≥ ¥ G 92 W Carol ¥ G 91.8 W Carol 0.25 0.2 83.75 Dan 85.2 Dan 0.25 0.4 82 Eric 80.8 Eric (C) Class averages Test 1 Test 2 Test 3 Test 4 [0.2 0.2 0.2 0.2 0.2]M [84.4 81.8 85 87.2] (7-4) (A)
c
APPENDIX B Exercises B-2 13.
3 1 0002 3x 4 2x 0002 3
15.
2 1 3 0002 0002 x x00023 (x 0002 3)2
17.
2 3x 0002 1 2 x x 2x 3
19.
2x x2 2
3x 5 (x 2 2)2
23.
2x 5 2 2 x00023 x 3x 3
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Student Answer Appendix
2 1 3 0002 x00024 x 3 (x 3)2
29. x 2 0002
1 x00021 2 2 x 2 2x 0002 1 2x 0002 x 1
Exercises B-3 3.
y 00022x 0002 2; straight line
5.
y 00022x 0002 2, x 0003 0; a ray (part of a straight line) y
y
00025
5
x
5
x
y2 4x; parabola
11.
00025
5
00025
00025
y
y2 4x, y 0004 0; parabola (upper half) y
5
5
00025
5
00025
x
00025
5
x
00025 2
29.
y
5
00025
00025
At Dt F , 00020005 6 t 6 0005; parabola 0002E (A) The graphs are symmetric about the line y x. (B) 1. y ex 2. x ex or y ln x Function 2 is the inverse of function 1.
21. x t, y
y 000223 x; straight line
5
5
9.
7.
23. y2 0002 x2 8, x 0004 1, y 0004 3; part of a hyperbola
x
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SUBJECT INDEX Abscissa, 110 Absolute value definitions for, 65 distance and, 66 method to find, 65–66 to solve radical inequalities, 71–72 Absolute value equations geometric interpretation of, 67–68 method to solve, 66–70, 99–100 verbal statements as, 68–69 Absolute value functions, 188 Absolute value inequalities geometric interpretation of, 67–68 method to solve, 66–70 Absolute value problems method to solve, 69–71 solved geometrically, 67–68 with two cases, 71 Acceptable probability assignment, 547 Actual probability, 552 Addition associative property of, 4, 6 commutative property of, 4, 6 of complex numbers, 76–77 elimination by, 429–434 explanation of, 3–4 of matrices, 457–458 of polynomials, 23 of rational expressions, 34–36 of real numbers, 3–7 Addition properties of equality, 45 of matrices, 477 of real numbers, 6 Additive identity, 4, 77 Additive inverse, 4, 6 Adiabatic process, 530 Algebra, 1 Algebraic equations. See also Equations algebraic expressions vs., 49 explanation of, 44 Algebraic expressions algebraic equations vs., 49 containing radicals, 17 explanation of, 21 factor of, 25 Algorithm, division, 267 Analytic geometry basic problems studied in, 122 fundamental theorem of, 110 Approximation by rational numbers, 5 of real zeros, 282–283 Arithmetic sequences explanation of, 520, 564–565 method to find terms in, 522–523 method to recognize, 521 nth term of, 522 Arithmetic series, 523–524 Associative property of addition, 4, 6 of multiplication, 4 Asymptote rectangle, 407
Asymptotes on graphing calculator, 306 horizontal, 303–304 oblique, 308 vertical, 302–304 Augmented matrices explanation of, 443 Gauss-Jordan elimination and, 447 interpretation of, 445 method to write, 443–444 reduced, 450, 451 Axis of cone, 386n conjugate, 407–409 of ellipse, 395 of hyperbola, 405, 407 of parabola, 1111 of symmetry, 205, 387 transverse, 405 Base of exponent, 11 of exponential functions, 329, 331–333 Bell, Alexander Graham, 365 Binomial coefficients, 22–23, 560 Binomial expansion, 558–559 Binomial formula explanation of, 559–560, 565 proof of, 562–563 use of, 560–562 Binomials, 22. See also Polynomials Bisection method, 281–282 Briggsian logarithms. See Common logarithms Calculators. See Graphing calculators Carbon-14 decay equation, 343–344 Cardano, Girolamo, 108 Cardano’s formula, 108 Cartesian coordinate system, 110, 157–158 Catenary curve, 374, 391 Center of circle, 127, 129 of ellipse, 395 of hyperbola, 405, 408 Change-of-base formula, 361–362 Circles equations of, 126–128 explanation of, 127, 386 formulas for, 599 graphs of, 126–128 Closure property, 6 Coefficient determinant, 492 Coefficient matrix, 443 Coefficients binomial, 22–23, 560 in linear systems, 424 of polynomial functions, 260–261 real, 290–291 Cofactor of element explanation of, 489 method to find, 489–490 Column matrices, 442, 461–462
I-1
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SUBJECT INDEX
Combinations, 538–541, 565 Combined properties, of matrices, 477 Combined variation, 319 Common difference, 520 Common factors, 26, 33 Common logarithms, 359, 360 Common ratio, 521 Commutative property, 4, 6 Completing the square, 86–87 Complex numbers addition of, 76–77 division of, 78–79 explanation of, 74–76, 105 historical background of, 74 multiplication of, 77–78 operations with, 76–79 radicals and, 80–81 set of, 75 solving equations involving, 81–82 subtraction of, 76–77 types of, 75 zero of, 77 Composite numbers, 25 Composition of functions, 226–230, 252 inverse functions and, 240 Compound events, 544 Compound fractions, 36, 37 Compound interest applications of, 334, 373 continuous, 335–336 explanation of, 333–334 Conditional equations, 45 Conic sections. See also Circles; Ellipses; Hyperbolas; Parabolas explanation of, 386–387 review of, 418–420 Conjecture, 511–512, 517 Conjugate, of a ⫹ bi, 75 Conjugate axis, 407–409 Conjugate hyperbolas, 410 Consistent systems, 427 Constant in term of polynomial, 22 of variation, 316–318 Constant functions, 178, 179 Constant matrix, 443 Constant terms, 424 Continuous compound interest, 335–336 Continuous compound interest formula, 336 Continuous graphs, 181 Contraction. See Shrinking Coordinate, 3, 110 Coordinate axis, 110. See also x axis; y axis Correspondence, 162, 167 Counterexamples, 511–512 Counting techniques combinations and, 538–541 explanation of, 531–532 factorial notation and, 534–536 multiplication principle and, 532–534 permutations and, 536–538 Cramer, Gabriel, 491 Cramer’s rule explanation of, 491–492, 498 to solve three-variable system, 493–494 to solve two-variable system, 492–494 for three equations in three variables, 493 Cube functions, 189
Cube root functions, 189 Cube roots, 14 Cubic equations, 108 Cubic models, 272 Curve fitting, 151 Curves catenary, 374, 391 explanation of, 151 plane, 592, 593 Data analysis examples of, 271–273 regression and, 346–349, 369 Decibels, 365, 366 Decimal expansions, 5 Decoding matrix, 482 Decreasing functions, 178, 238 Degenerate conic, 387 Degree, of polynomials, 22, 260 Demand, 93, 435 Denominator explanation of, 9 least common, 35 rationalizing the, 18–19 Dependent variables, 164 Descartes, René, 11 Determinants coefficient, 492 explanation of, 487, 498 first-order, 487–488 second-order, 487, 488 to solve systems of equations, 491–494 third-order, 488–491 Diagonal expansion, 495 Difference function, 224–225 Difference of cubes formula, 28, 29 Difference of square formula, 28, 29 Difference quotient, 170 Dimensions, of matrix, 442 Directrix, of parabola, 387 Direct variation, 316 Discriminant, 90–91 Distance absolute value and, 66 in plane, 123–129, 158 between two points, 123–124 Distance formula explanation of, 124 use of, 124, 388, 396, 406–407 Divisibility property, 516 Division of complex numbers, 78 long, 5, 266–267 polynomial, 266–269 of rational expressions, 33–34 of real numbers, 7 synthetic, 268–269 Division algorithm, 267 Division properties, of equality, 45 Divisor, 267 Domain of exponential functions, 355 of functions, 163, 164, 166–167, 169–170, 176–177, 204, 225, 229, 230 implied, 166 of rational functions, 299–300 of variables, 44–45
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SUBJECT INDEX
Double inequalities, 61, 67 Double subscript notation, 442 Double zero, 289 Doubling time, 340 Doubling time growth model, 340, 341 Eccentricity, 417 Element of matrix, 442 of set, 2 Elimination, by addition, 429–434 Ellipses applications of, 400–401 equations of, 396–400 explanation of, 386, 395, 418–420 graphs of, 396–400 method to draw, 395–396 Empirical probability application of, 555–556 approximate, 552–554 explanation of, 552, 553 method to find, 553–554 Empty set, 2 Endpoints, 57 Equality explanation of, 76 properties of, 45, 477 symbols for, 57 Equallly likely assumptions, 549–550 Equal polynomials, 585–586 Equations. See also Linear equations; Systems of linear equations; specific types of equations of circles, 126–128 conditional, 45 cubic, 108 defining functions by, 164–167 of ellipse, 396–400 equivalent, 45 explanation of, 44, 56 exponential, 372–374, 380 graphs of, 111–112, 118 of hyperbolas, 406–408 as identities, 45 involving complex numbers, 81–82 involving radicals, 97–99 of lines, 132, 133, 137–140, 158 logarithmic, 375–376, 380 matrix, 477–480 of parabola, 209 parametric, 591–595 price-demand, 93 properties of, 69 quadratic, 84–93, 105 of quadratic type, 101–102, 105 solution set of, 44, 111 solutions of, 44, 111 squaring operation on, 98 in two variables, 111 Equilibrium price, 435 Equilibrium quantity, 435 Equivalent equations, 45 Equivalent inequalities, 59–60 Equivalent systems of equations, 429 Euler, Leonhard, 74 Even functions, 196–197 Events compound, 544
explanation of, 546–547 probability of, 547–551 simple, 544, 547 Expected frequency, 553 Experiments, 543–544 Exponential decay, 350 Exponential equations explanation of, 372, 380 method to solve, 372–374 Exponential functions with base e, 331–333 compound interest and, 333–336 domain of, 355 explanation of, 328–329, 379 graphs of, 329–333 inverse of, 329 (See also Logarithmic functions) properties of, 330–331 transformations of, 330 Exponential growth/decay, 349, 350 Exponential models application of, 379 data analysis and regression and, 346–349 exponential growth phenomena and, 349–350 on graphing calculator, 347 mathematical, 340–346 Exponents explanation of, 11, 39 integer, 11–13 rational, 16 Extended principle of mathematical induction, 517 Extraneous solutions, 98, 105 Extrapolation, 153 Face cards, 540 Factorials, 534–536 Factoring explanation of, 25 by grouping, 26–27 of polynomials, 25–29 to solve quadratic equations, 84–86 Factoring formulas, 28, 29 Factors of algebraic expression, 25 common, 26 explanation of, 25 of polynomials with real coefficients, 290–291 Factor theorem, 270 Fermat’s last theorem, 517 Fibonacci, Leonardo, 505 Fibonacci sequences, 505 Finite sequences arithmetic, 523–524 explanation of, 505 Finite series arithmetic, 523–524 explanation of, 507 geometric, 526–527 Finite sets, arithmetic, 523–524 First-degree equations. See Linear equations First-degree functions. See Linear functions First-order determinants, 487–488 Focal chords, 393, 422 Focus of ellipse, 395–398 of hyperbola, 405 of parabola, 387 Fractional expressions, 32
I-3
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SUBJECT INDEX
Fractions compound, 36–37 explanation of, 9 fundamental property of, 32 partial, 585–590 proper, 585 properties of, 9 raised to higher terms, 32 reduced to lowest terms, 32 significant digits in decimal, 583 simple, 36 solving inequalities involving, 61 Frequency, 553 Functions. See also specific types of functions absolute value, 188 applications of, 170–171 composition of, 226–230, 240, 252 constant, 178, 179 cube, 189 cube root, 189 decreasing, 178, 238 defined by equations, 164–166 definition of, 163, 235, 250 difference, 224–225 difference quotient and, 170 domains of, 166–167, 169–170, 176–177, 204, 225, 229, 230 even, 196–197 exponential, 328–336, 379 graphs of, 175–184, 188–199, 250–251 greatest integer, 182, 183 identity, 179, 188 increasing, 178, 238 inverse, 235–246, 252 linear, 178–180 logarithmic, 354–362, 379–380 notation for, 167–168 odd, 196–197 one-to-one, 235–238, 358 operations on, 223–226 overview of, 162 piecewise-defined, 180–181 polynomial, 260–266 product, 224–225 quadratic, 203–211, 251–252 quotient, 224–225 range of, 163, 166, 177 rational, 298–310 set form of definition of, 163 square, 188, 203, 204 square root, 189 sum, 224–225 transformations of, 188–197, 251 vertical line test for, 166 Fundamental counting principle. See Multiplication principle Fundamental property of fractions, 32 Fundamental sample space, 545 Fundamental theorem of algebra, 288–289 Fundamental theorem of analytic geometry, 110 Fundamental theorem of arithmetic, 25 Gauss, Carl Friedrich, 288, 447 Gauss-Jordan elimination explanation of, 441, 447 on graphing calculators, 449 to solve linear systems, 447–451, 497 use of, 475
General form, of quadratic function, 204 Geometric sequences explanation of, 521, 564–565 method to find terms in, 522–523 method to recognize, 521 nth term of, 522 Geometric series sum formulas for finite, 525 sum formulas for infinite, 526–527 Goldbach’s conjecture, 517 Graphing calculator features INTERSECT, 361 MATRIX-MATH, 488 maximum and minimum, 209 random number generator, 554 rref on, 449 table, 561 TRACE, 134, 280, 588 viewing window, 127 ZERO command, 280 ZSquare, 127 Graphing calculators asymptotes on, 306 circles on, 127 cubic models on, 272 domain of functions on, 225 ellipses on, 399 exponential functions on, 328, 331 exponential models on, 347 graphs of equations on, 112, 118, 143 greatest integer functions on, 183 interest rate on, 335 inverse functions on, 246 linear systems on, 425 logarithms on, 359–360, 370 logistic models on, 349 matrices on, 442, 458, 473, 488 parabolas on, 390 parametric equations on, 592 partial fraction decomposition on, 588 polynomial inequalities on, 284 quadratic regression on, 215, 216 quartic model on, 273 rational inequalities on, 311 reduced echelon form on, 449 regression on, 153 scientific notation on, 13–14 sequences on, 505, 507 sum of series on, 526 Graphs/graphing of circles, 126–128 continuous, 181 of ellipses, 396–400 of equation in two variables, 111 explanation of, 111 of exponential functions, 329–333 of functions, 175–184, 188–199, 250–251 horizontal and vertical shifts in, 189–191 of hyperbolas, 406–412 of inequalities, 58, 59 of intervals, 58, 59 of inverse functions, 244–246 line, 57 of linear functions, 179–180 of lines, 132–133 of logarithmic functions, 354–356, 359–361 multiplicities from, 292
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SUBJECT INDEX
of parabolas, 111, 389–390 point-by-point plotting on, 111 of polynomial functions, 260–266, 280 of polynomials, 266, 291–292 of quadratic functions, 204–209 of rational functions, 299–301, 304–310 reflections of, 114, 191–193 stretching and shrinking in, 193–196 symmetry as aid in, 113–117 of systems of linear equations, 424–425 Greatest integer, 182 Greatest integer functions, 182, 183
Half-life, 342 Half-life decay model, 342 Horizontal asymptotes, of rational functions, 303–304 Horizontal axis, 110. See also x axis Horizontal lines, 139, 140 Horizontal line test, 237 Horizontal shifts, 189–191, 195 Horizontal shrinks, 194, 195 Horizontal stretches, 194, 195 Hyperbolas applications of, 412–414 conjugate, 410 equations of, 406–408 explanation of, 387, 405, 420 graphs of, 406–412 method to draw, 406 Hyperbolic paraboloids, 412 Hyperboloids, 412
Identities, 45 Identity functions, 179 Identity matrix, for multiplication, 470–471 Identity property, 6 Imaginary numbers, 75 Imaginary unit, 74–75 Imaginary zeros, of polynomials, 290, 295 Implied domain, 166 Inconsistent systems, in two variables, 427 Increasing functions, 178, 238 Independent systems, 427 Independent variables, 164, 165 Index, 15 Induction. See Mathematical induction Inequalities absolute value, 66–70 applications for, 61–62 double, 61, 67 equivalent, 59 explanation of, 57 graphs of, 58, 59 linear, 56–62, 105 polynomial, 283–284, 322 properties of, 60, 69 quadratic, 211–214, 252 radical, 71–72 rational, 310–311, 322–323 solution set for, 59–60 symbols for, 57 Infinite sequences explanation of, 505 geometric, 526–527
I-5
Infinite series explanation of, 507 geometric, 526–527 Infinity, symbol for, 57 Integer exponents explanation of, 11–12 properties of, 12–13 Integers explanation of, 2, 3 greatest, 182, 183 set of, 2 Intercepts. See also x intercepts; y intercepts of functions, 176–177 to graph lines, 133 of rational functions, 305 Interest compound, 333–336, 373 explanation of, 333 Interest rate, 333 Interpolation, 153 Intersections, 59 Intervals explanation of, 57 graphs of, 58, 59 notation for, 57–58, 177 Inverse additive, 4, 6 of functions, 238–242, 245–246 method to find, 476 multiplicative, 4, 6, 11, 471–472 to solve linear systems, 478–480, 498 of square matrix, 471–473, 476 Inverse functions explanation of, 235, 252 graphs of, 244–246 method for finding inverse and, 238–242 modeling with, 242–243 one-to-one, 235–238 properties of, 239 Inverse variation, 316–317 Irrational numbers explanation of, 2, 5 historical background of, 74 Joint variation, 318 Lagranges’ four square theorem, 517 Leading term, 264 Learning curves, 344–345 Least common denominator (LCD), 35 Least-squares line, 383 Like terms, of polynomials, 23 Limited growth, 350 Linear and quadratic factors theorem, 290, 586 Linear equations. See also Equations; Systems of linear equations explanation of, 104 with more than one variable, 46–47 in one variable, 45–46 Linear factors theorem, 289 Linear functions. See also Functions explanation of, 178–179 graphs of, 179–180 Linear inequalities. See also Inequalities applications for, 61–62 explanation of, 56, 57, 105 graphs of, 59–62
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SUBJECT INDEX
Linear models, construction of, 149–151 Linear regression examples of, 152–154 explanation of, 151 Linear systems. See Systems of linear equations Line graph, 57 Lines equations of, 132, 133, 137–140, 158 graphs of, 132–133 horizontal, 139, 140 parallel, 141–142 perpendicular, 141–142 regression, 153 slope-intercept form of, 137–138 slope of, 134–136 vertical, 139, 140, 166 Line segment length of, 66 midpoint of, 124–126 Location theorem, 280–281 Logarithmic equations explanation of, 372, 380 method to solve, 375–376 Logarithmic-exponential conversions, 356–357 Logarithmic-exponential relationships, 360–361 Logarithmic functions change-of-base formula and, 361–362 conversions of, 356–357 explanation of, 329, 354, 379–380 graphs of, 354–356 properties of, 358–359, 380 Logarithmic models applications of, 380 data analysis and regression, 369 logarithmic scales, 365–369 Logarithmic scales, 365–369 Logarithms common, 359, 360 on graphing calculator, 359–361, 370 natural, 359, 360 Logistic growth, 350 Logistic models, 349 Long division explanation of, 5 polynomial, 266–267 Lower triangular matrix, 571 Lowest terms, 32–33 Magnitude, 367 Mathematical induction examples of, 513–517 explanation of, 512–513, 564 extended principle of, 517 principle of, 512 Mathematical models applications of, 230–231, 242–243 explanation of, 147–148 exponential, 340–350 polynomial, 271–273, 285 quadratic, 210–211, 214–215 Matrices addition of, 457–458 applications of, 460–462, 464–465 augmented, 443–445, 447 column, 461–462 decoding, 482 explanation of, 442–443, 497 Gauss-Jordan elimination and, 447–451, 475
on graphing calculators, 442, 458, 473, 488 identity, 470–471 inverse methods to solve linear systems, 498 inverse of square, 471–473 lower triangular, 571 multiplication of, 459–466 negative of, 458 principal diagonal of, 442 properties of, 477 reduced, 444–447 row, 442, 461–462 row-equivalent, 444, 474 singular, 472 size of, 442 square, 442 subtraction of, 458–459 upper triangular, 495, 571 zero, 458 Matrix equations explanation of, 477 method to solve, 477–478 systems of linear equations and, 478–480 Midpoint, of line segment, 124–126 Midpoint formula explanation of, 124 use of, 125–126 Minor of element, in third-order determinant, 489 Mixture problems, 52–53 Models. See Mathematical models Monomials, 22. See also Polynomials Multiplication associative property of, 4, 6 commutative property of, 4, 6 of complex numbers, 76–78 identity matrix for, 470–471 of matrices, 459–466 of polynomials, 24 of rational expressions, 33–34 of real numbers, 3–7 Multiplication principle application of, 533–534 explanation of, 532–533, 565 Multiplication properties of equality, 45 of matrices, 477 of real numbers, 6 Multiplicative identity, 4, 78 Multiplicative inverse, 4, 6, 11, 471–472 Multiplicities from graphs, 292 of zero, 289, 291, 292 Multiplier doctrine, 527 Napierian logarithms. See Natural logarithms Nappes, of cone, 386n Natural logarithms, 359, 360 Natural numbers, 2, 79 Negative growth, 342 Negative real numbers explanation of, 3 principal square root of, 80 properties of, 7, 8 n factorial, 534–535 Nonrepeating linear factors, 587–588 Nonrigid transformations, 193 Notation/symbols absolute value, 65 composition of function, 226, 228
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SUBJECT INDEX
double subscript, 442 empty set, 2 equality and inequality, 57 exponent, 11 factorial, 534–536 function, 167–169, 226, 228 infinity, 57 interval, 57–58, 177 parallel, 141 perpendicular, 141 radical, 15 real number, 2 scientific, 13–14 summation, 507, 508 nth root explanation of, 14–15 principal, 15–16 nth-term formulas, 522–523 Null set, 2 Number line, real, 3 Numbers. See also Integers complex, 74–82, 105 composite, 25 imaginary, 75 irrational, 2, 5, 74 natural, 2, 79 pure imaginary, 75 rational, 2, 3–7, 298 real, 2–9, 75 Numerator, 9 Numerical coefficient, 22. See also Coefficients Oblique asymptotes, 308 Odd functions, 196–197 One-to-one functions explanation of, 235–236, 258 identification of, 236–238 Ordered pairs explanation of, 110n, 111 functions as sets of, 163–164 Ordering, 536 Ordinate, 110 Origin explanation of, 3, 110 reflection through, 114, 192, 193 symmetry and, 114, 115 Parabolas. See also Quadratic functions applications of, 391–392 coordinate-free definition of, 387 equation of, 209, 388–391 explanation of, 111, 204, 387, 418–419 focal chord of, 393 graphs of, 111, 389–390 method to draw, 387–388 vertex of, 205–208 Paraboloids explanation of, 391, 392 hyperbolic, 412 Parallel lines, 141–142 Parallelograms, 598 Parameter, 432, 592–593 Parametric equations application of, 594–595 explanation of, 591–593 on graphing calculator, 592
Partial fraction decomposition, 586–590 Partial fractions, 585 Particular solutions, 432 PASCAL, 559 Pascal’s triangle, 559 Path of projectile, 594 Perfect square formula, 28, 29 Permutations, 536–538, 565 Piecewise-defined functions, 180–181 Plane, distance in, 123–129, 158 Plane curve, 592, 593 Point, coordinate of, 3 Point-by-point plotting, 111 Point-slope form, 138–140 Polynomial functions explanation of, 260, 321–322 graphs of, 260–266, 280 left and right behavior of, 265 Polynomial inequalities explanation of, 283, 322 on graphing calculators, 284 method to solve, 283–284 Polynomials addition of, 23 bisection method and, 281–282 degree of, 22 division of, 266–269 equal, 585–586 evaluation of, 269–270 explanation of, 21–23, 40 factoring, 25–29 factors of, 270, 290–291 factor theorem and, 270 fundamental theorem of algebra and, 288–289 graphs of, 266, 291–292 location theorem and, 280–281 multiplication of, 24 in one variable, 22 prime, 25, 26 rational zeros of, 292–293, 322 with real coefficients, 290–291 real zeros of, 278–279 reduced, 294 remainder theorem and, 269–279 second-degree, 27–28 subtraction of, 24 Taylor, 365 in two variables, 22 zeros of, 266, 271 Positive real numbers, 3, 81 Predictions, 153 Price-demand equation, 93 Prime numbers, 25 Prime polynomials, 25, 26 Principal, 333, 334 Principal diagonal, 442, 488 Principal nth root, 15–16 Principle square root, of negative real number, 80 Probability actual, 552 empirical, 552–556 of events, 547–551 explanation of, 543 theoretical, 552 Probability function, 547 Problem solving. See Word problems Product function, 224–225 Proper fractions, 585
I-7
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SUBJECT INDEX
Pure imaginary numbers, 75 Pythagoreans, 74 Pythagorean theorem, 92, 411, 598 Quadrants, 110 Quadratic equations applications for, 91–93 completing the square to solve, 87–89 explanation of, 84, 105 factoring to solve, 84–86 methods to solve, 100–102 quadratic formula to solve, 89–91 square root property to solve, 86–87 Quadratic factors, 589–590 Quadratic formula explanation of, 90 to solve quadratic equations, 89–90 use of, 294–295, 586 Quadratic functions explanation of, 204, 251–252 general form of, 204 graphs of, 204–209 modeling with, 210–211 properties of, 206 Quadratic inequalities explanation of, 211–212, 252 method to solve, 212–214 Quadratic regression, 214–215 Quadratic solving techniques applications using, 102 direct solution, 100 example of, 101 substitution method as, 101 Quantity-rate-time formula, 50 Quartic models, 273 Quotient function, 224–225 Quotients difference, 170 explanation of, 267 of functions, 226 Radical inequalities, 71–72 Radicals equations involving, 97–99 explanation of, 15, 39–40 properties of, 17, 81 in simplified form, 17–19 use of, 16–17 Radius, of circle, 127, 129 Random experiments, 543–546 Range, of functions, 163, 166, 177 Rate of change, 148–149 Rational exponents explanation of, 15–16 use of, 16–17 Rational expressions addition and subtraction of, 34–36 compound fraction, 36–37 explanation of, 32, 40 multiplication and division of, 33–34 reduced to lowest terms, 32–33 Rational functions domain and x intercepts of, 299 explanation of, 298–299, 322–323 graphs of, 299–301, 304–310 oblique asymptotes of, 308
properties of, 300–301 vertical and horizontal asymptotes of, 302–304 Rational inequalities explanation of, 310, 322–323 on graphing calculators, 311 method to solve, 310–311 Rationalizing factor, 18 Rationalizing the denominator, 18–19 Rational numbers addition and multiplication of, 3–7 explanation of, 2, 298 Rational zeros explanation of, 292–293, 322 method for finding, 293–295 Rational zero theorem, 293 Real number line, 3 Real numbers addition of, 3–7 division of, 7 explanation of, 2, 39, 75 multiplication of, 3–7 negative, 3, 7, 8, 80 positive, 3 properties of, 6 roots of, 14–15 set of, 2–3, 6, 8, 164 subtraction of, 7 Real root, 84 Real zeros approximation of, 282–283 explanation of, 278, 322 upper and lower bound for, 278–279 Reciprocals, 78–79 Rectangles, 407, 598 Rectangular coordinate system. See Cartesian coordinate system Rectangular solids, 599 Recursion formulas explanation of, 505 use of, 515–516 Recursive formula n factorial, 535 Reduced augmented coefficient, 450, 451 Reduced matrices, 444–447 Reduced polynomials, 294 Reduced system, 447 Reflections explanation of, 114 of graphs of functions, 191–193, 195 Regression on graphing calculators, 153 linear, 151–154 logarithmic, 369 quadratic, 214–215 Regression analysis, 151 Regression line, 153 Regression models, 383 Relative frequency, 553 Relative growth rate, 341 Remainder, 267 Remainder theorem, 269–270 Replacement set, 44. See also Domain Residuals, 383 Revenue, 93 Richter scale, 367 Right circular cones, 386n, 599 Right circular cylinders, 599 Rigid transformations, 193
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SUBJECT INDEX
Rise, 134 Rocket equation, 368 Roots. See also Square roots cube, 14 of equation, 176 of functions, 261–262 nth, 14–16 real, 84 of real numbers, 14–15 Rounding convention, 583–584 Row-equivalent matrices, 444 Row matrices, 442, 461–462 Row operations, 443 Run, 134 Sample spaces example of, 546 explanation of, 543–544 fundamental, 545 method to choose, 544–545 Scatter plots, 152 Scientific notation, 13–14 Second-degree polynomial functions. See Quadratic functions Second-degree polynomials, factoring, 27–28 Second diagonal, 488 Second-order determinants, 487, 488 Sequences arithmetic, 520, 522–523 explanation of, 504–505, 564 Fibonacci, 505–506 finite, 505 general term of, 506–507 geometric, 521–523 on graphing calculators, 505, 507 infinite, 505 terms of, 504 Series explanation of, 507, 564 finite, 507 infinite, 507, 526–527 sum formulas for finite arithmetic, 523–524 sum formulas for geometric, 525–527 in summation notation, 508 terms of, 508 Sets of complex numbers, 75 empty or null, 2 equal, 3 of integers, 2 intersection of, 59 of real numbers, 2–3, 6, 8, 164 replacement, 44 union of, 59 Shrinking, in graphs, 193–196 Significant digits, 582–583 Simple events, 544, 547, 550 Simple fractions, 36 Singular matrix, 472 Slope explanation of, 134 geometric interpretation of, 135 method to find, 134–136 of parallel lines, 141–142 of perpendicular lines, 141–142 as rate of change, 148–149 Slope-intercept form, 137–138, 140
Solutions of equations, 44, 111 extraneous, 98, 105 of linear systems, 424, 426–427 particular, 432 unique, 427 Solution set of equations, 44, 111 of inequalities, 59–60 of linear systems, 424 of quadratic inequalities, 211 Speed, 148. See also Rate of change Spheres, 599 Square functions, 188, 203, 204 Square matrices explanation of, 442 inverse of, 471–473, 476 of order n, 470–471 Square root functions, 189 Square root property, 86–87 Square roots, 14, 80 Squaring operation on equations, 98 Standard deck, 540 Standard form of complex numbers, 80 of equation of circle, 128 of equation of line, 133, 140 of linear equations, 45 of quadratic equations, 84, 100 quadratic inequalities in, 211 Stretching, in graphs, 193–196 Subset, 2 Substitution to solve equations of quadratic type, 101 to solve linear systems, 427–428, 431, 432 Substitution property, of equality, 45 Subtraction of complex numbers, 76–77 of matrices, 458–459 of polynomials, 24 of rational expressions, 34–36 of real numbers, 7 Subtraction properties, of equality, 45 Sum formulas for finite arithmetic series, 523–524 for finite geometric series, 525 for infinite geometric series, 526–527 Sum function, 224–225 Summation formula, proof of, 514–515 Summation notation, 507, 508 Summing index, 507 Sum of cubes formula, 28, 29 Sum of the squares of the residuals (SSR), 383 Supply, 435 Symbols. See Notation/symbols Symmetry as aid in graphing, 113–117 axis of, 205 in even and odd functions, 197 tests for, 115–116 Symmetry property, 244–245 Synthetic division, 268–269 Synthetic division table, 279 Systems of linear equations applications of, 434–437, 452–454 basic terms of, 427 Cramer’s rule to solve, 491–494
I-9
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SUBJECT INDEX
Systems of linear equations—Cont. elimination by addition to solve, 429–434 equivalent, 429 explanation of, 496 Gauss-Jordan elimination to solve, 441, 447–453, 497 graphs of, 424–425 matrices and row operations and, 441–447, 498 matrix equations and, 478–480 modeling with, 501–502 substitution method to solve, 427–428, 431, 432 in two variables, 424 Taylor polynomials, 365 Technology Connections. See Graphing calculators Theorems, 511 Theoretical probability explanation of, 552 method to find, 553–554 Third-order determinants, 488–491 Transformations combining graph, 196 even and odd functions and, 196–197 explanation of, 189, 251 of exponential functions, 330 nonrigid, 193 reflections and, 191–193, 195 rigid, 193 stretching and shrinking and, 193–195 vertical and horizontal shifts and, 189–191, 195 Transverse axis, of hyperbola, 405 Trapezoids, 598 Tree diagrams, 532, 544 Triangles formulas for, 598 Pascal’s, 559 similar, 598 Trinomials, 22. See also Polynomials Triple zero, 289 Turning points approximating real zeros at, 282–283 explanation of, 262 Union, of sets, 59 Unique solution, 427 Unlimited growth, 350 Upper and lower bound theorem, 278, 279 Upper triangular matrix, 495, 571 Variables dependent, 164 domains of, 44–45 independent, 164, 165 Variation combined, 319 direct, 316 explanation of, 323 inverse, 316–317 joint, 318 Velocity, 148, 368. See also Rate of change
Vertex form, of quadratic functions, 204 Vertical asymptotes, 302–304 Vertical axis, 110. See also y axis Vertical lines, graphs of, 139, 140, 166 Vertical line test, 166 Vertical shifts, 189–191, 195 Vertical shrinks, 194, 195 Vertical stretches, 194, 195 Vertices of cone, 386n of ellipse, 395 of hyperbola, 405 of parabola, 205, 206, 208, 387
Wiles, Andrew, 517 Word problems method to set up, 48, 91 mixture, 52–53 rate, 50–52 strategies to solve, 47, 104 using diagrams in solution of, 48–49
x axis reflection through, 114, 192, 193 symmetry and, 114, 115 x coordinate, 110, 176 x intercepts explanation of, 133 of functions, 176–177 of polynomial functions, 261–262 of rational functions, 299–300
y axis reflection through, 114, 192, 193 symmetry and, 114, 115 y coordinate, 110, 176 y intercepts explanation of, 133 of functions, 176–177 on graphing calculator, 134
Zero factorial, 534–535 Zero matrix, 458 Zero product property, 84–85 Zero property, of real numbers, 8 Zeros complex, 77, 322 double, 289 of functions, 176, 261–262 imaginary, 290, 295 irrational, 294–295 multiplicities of, 289, 291, 292 of polynomials, 266, 271, 278–279 rational, 292–295, 322 real, 278–279, 282–283 triple, 289
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Sets a苸A aⰻA ⭋ 5x 0 p(x)6 A ( B A´B A傽B
Inequalities and Intervals a is an element of set A a is not an element of set A Empty or null set Set of all x such that p(x) is true A is a subset of B 5x 0 x 苸 A or x 苸 B6, union 5x 0 x 苸 A and x 苸 B6, intersection
a ⬍ b a is less than b a ⱕ b a is less than or equal to b a ⬎ b a is greater than b a ⱖ b a is greater than or equal to b (a, b) Open interval; 5x 0 a 6 x 6 b6 (a, b] Half-open interval; 5x 0 a 6 x ⱕ b6 [a, b) Half-open interval; 5x 0 a ⱕ x 6 b6 [a, b] Closed interval; 5x 0 a ⱕ x ⱕ b6
Number Systems Absolute Value N Natural numbers Z Integers Q Rational numbers R Real numbers C Complex numbers N傺Z傺Q傺R傺C
x if x ⱖ 0 ⫺x if x 6 0 0 x 0 2 ⫽ x2 2x2 ⫽ 0 x 0 0 x 0 6 p if and only if ⫺p 6 x 6 p; p 7 0 0 x 0 7 p if and only if x 6 ⫺p or x 7 p; p 7 0 0x0 ⫽ e
Real Number Properties Quadratic Formula For all real numbers a, b, and c: a ⫹ b ⫽ b ⫹ a;
ab ⫽ ba
a ⫹ (b ⫹ c) ⫽ (a ⫹ b) ⫹ c;
a(bc) ⫽ (ab)c
a(b ⫹ c) ⫽ ab ⫹ ac a ⫹ 0 ⫽ a; 1 ⴢ a ⫽ a a ⫹ (⫺a) ⫽ 0; a(1Ⲑa) ⫽ 1, a ⫽ 0 ab ⫽ 0 if and only if a ⫽ 0 or b ⫽ 0
Exponents and Radicals an ⫽ a ⴢ a . . . a (n factors of a), n 苸 N a0 ⫽ 1, a ⫽ 0 1 a⫺n ⫽ n , a ⫽ 0, n 苸 R a n bm/n ⫽ 2bm (nth root of bm)
Special Products (a ⫺ b)(a ⫹ b) ⫽ a2 ⫺ b2 (a ⫹ b)2 ⫽ a2 ⫹ 2ab ⫹ b2 (a ⫺ b)2 ⫽ a2 ⫺ 2ab ⫹ b2 (a ⫺ b)(a2 ⫹ ab ⫹ b2) ⫽ a3 ⫺ b3 (a ⫹ b)(a2 ⫺ ab ⫹ b2) ⫽ a3 ⫹ b3
Commutative properties Associative properties Distributive property Identities Inverses Zero property
If ax2 ⫹ bx ⫹ c ⫽ 0, a ⫽ 0, then: x⫽
⫺b ⫾ 2b2 ⫺ 4ac 2a
Rectangular Coordinates (x1, y1) d ⫽ 2(x2 ⫺ x1)2 ⫹ ( y2 ⫺ y1)2 x1 ⫹ x2 y1 ⫹ y2 , b 2 2 y2 ⫺ y1 m⫽ , x1 ⫽ x2 x2 ⫺ x1 a
Coordinates of point P1 Distance between P1(x1, y1) and P2(x2, y2) Midpoint of line joining P1 and P2 Slope of line through P1 and P2
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Function Notation
Arithmetic Sequence Value of f at x Composite function Value of inverse of f at x
f (x) ( f ° g)(x) ⫽ f [g(x)] f ⫺1(x)
Linear Equations and Functions y ⫽ mx ⫹ b ( y ⫺ y1) ⫽ m(x ⫺ x1) f (x) ⫽ mx ⫹ b y⫽b x⫽a
Slope–intercept form Point–slope form Linear function Horizontal line Vertical line
Polynomial and Rational Forms f (x) ⫽ ax2 ⫹ bx ⫹ c f (x) ⫽ an x n ⫹ an⫺1x n⫺1 ⫹ . . . ⫹ a1x ⫹ a0, an ⫽ 0, n a nonnegative integer p(x) f (x) ⫽ , p and q polynomial q(x) functions, q(x) ⫽ 0
Quadratic function Polynomial function Rational function
a1, a2, . . . , an, . . . an ⫺ an⫺1 ⫽ d an ⫽ a1 ⫹ (n ⫺ 1)d
Common difference nth-term formula n Sn ⫽ a1 ⫹ . . . ⫹ an ⫽ [2a1 ⫹ (n ⫺ 1)d ] 2 n Sn ⫽ (a1 ⫹ an) 2
Sum of n terms
Geometric Sequence a1, a2, . . . , an, . . . an ⫽r Common ratio an⫺1 n⫺1 an ⫽ a1r nth-term formula a1 ⫺ a1r n , r⫽1 Sn ⫽ a1 ⫹ . . . ⫹ an ⫽ 1⫺r a1 ⫺ ran , r⫽1 Sn ⫽ 1⫺r a1 , 0r0 6 1 S⬁ ⫽ a1 ⫹ a2 ⫹ . . . ⫽ 1⫺r
Sum of n terms
Sum of infinitely many terms
Factorial and Binomial Formulas Exponential and Logarithmic Functions f (x) ⫽ b x, b 7 0, b ⫽ 1 Exponential function f (x) ⫽ logb x, b 7 0, b ⫽ 1 Logarithmic function y ⫽ logb x if and only if x ⫽ b y, b 7 0, b ⫽ 1
n factorial n! ⫽ n(n ⫺ 1) . . . 2 ⴢ 1, n 苸 N 0! ⫽ 1 n n! , 0ⱕrⱕn a b⫽ r!(n ⫺ r)! r n n (a ⫹ b)n ⫽ a a ba n⫺k b k, n ⱖ 1 Binomial formula k⫽0 k
Matrices and Determinants a d
b e
c d f
a †d g
b e h
c f† i
c
Matrix
Determinant
Permutations and Combinations For 0 ⱕ r ⱕ n: n! Pn,r ⫽ (n ⫺ r)! n n! Cn,r ⫽ a b ⫽ r r!(n ⫺ r)!
Permutation Combination
(Continued on back endpaper)
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Hyperbola
Circle (x ⫺ h)2 ⫹ ( y ⫺ k)2 ⫽ r 2 x2 ⫹ y2 ⫽ r 2
y2 x2 ⫺ ⫽1 a2 b2 2 y x2 ⫺ 2⫽1 2 a b
Center at (h, k); radius r Center at (0, 0); radius r
Foci: F¿(⫺c, 0), F(c, 0); c2 ⫽ a2 ⫹ b2 Foci: F¿(0, ⫺c), F(0, c); c2 ⫽ a2 ⫹ b2
Parabola y2 ⫽ 4ax,
a ⬎ 0, opens right; a ⬍ 0, opens left Focus: (a, 0); Directrix: x ⫽ ⫺a a ⬎ 0, opens up; a ⬍ 0, opens down Focus: (0, a); Directrix: y ⫽ ⫺a
x2 ⫽ 4ay,
Translation Formulas x ⫽ x¿ ⫹ h, y ⫽ y¿ ⫹ k ; New origin (h, k)
x¿ ⫽ x ⫺ h, y¿ ⫽ y ⫺ k
Ellipse y2 x2 ⫹ ⫽ 1, a2 b2 2 y x2 ⫹ 2 ⫽ 1, 2 b a
a 7 b 7 0 Foci: F¿(⫺c, 0), F(c, 0); c2 ⫽ a2 ⫺ b2 a 7 b 7 0 Foci: F¿(0, ⫺c), F(0, c); c2 ⫽ a2 ⫺ b2
Metric Units Standard Units of Metric Measure
Important Prefixes
Meter (m): length (approximately 3.28 ft)
kilo (⫻ 1,000)
deci (⫻ 101 )
Liter (L):
volume (approximately 1.06 ft)
hecto (⫻ 100)
1 centi (⫻ 100 )
Gram (g):
weight (approximately 0.035 oz)
deka (⫻ 10)
1 milli (⫻ 1,000 )
Abbreviations Length m km hm dam
meter kilometer hectometer dekameter
Volume
dm decimeter cm centimeter mm millimeter
L kL hL daL
liter kiloliter hectoliter dekaliter
Weight
dL deciliter cL centiliter mL milliliter
g kg hg dag
gram kilogram hectogram dekagram
dg decigram cg centigram mg milligram
English-Metric Conversion Length
Volume (U.S.)
Weight
Length
Volume (U.S.)
1 in ⫽ 2.540 cm 1 ft ⫽ 30.48 cm 1 yd ⫽ 0.9144 m 1 mi ⫽ 1.609 km
1 pt ⫽ 0.4732 L 1 qt ⫽ 0.9464 L 1 gal ⫽ 3.785 L
1 oz ⫽ 28.35 g 1 lb ⫽ 453.6 kg 1 lb ⫽ 0.4536 kg
1 cm ⫽ 0.3937 in 1 cm ⫽ 0.03281 ft 1 m ⫽ 1.0936 yd 1 km ⫽ 0.6215 mi
1 L ⫽ 2.1133 pt 1 L ⫽ 1.0567 qt 1 L ⫽ 0.2642 gal
Weight 1 g ⫽ 0.0353 oz 1 g ⫽ 0.002205 lb 1 kg ⫽ 2.205 lb