ISBN·13" 9-8'{)· 3·232903-3 O- 3·232903-.! ISBN-l0 i
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Upper Saddle River, New Jersey 07458 www.prenhall.com
9
801 3 2u3 2 9::: 3:; -------'
Eighth Edition
Algebra
&
Trigonometry
Michael Sullivan Chicago State University
Upper Saddle River, New Jersey 07458
Library of Congress Cataloging-in-Publication Data
Sullivan, Michael Algebra
&
trigonometry / Michael Sullivan . ....,<;th ed.
p. cm. ISBN 0-13-232903-4 (alk. paper)
L. Algebra-Textbooks. 2. Trigonometr)�Textbooks. I. Title. II. Title:
Algebra and trigonometry. QAI54.3.S73 2008 512' .l3---
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Contents Preface to the Instructor list of Applications Photo Credits To the Student R
1
xix xxv xxvii
Review
1
R.1 Real Numbers
2
R.2 Algebra Essentials
17
R.3 Geometry Essentials
30
R.4 Polynomials
39
R.S Factoring Polynomials
49
R.6 Synthetic Division
57
R.7 Rational Expressions
61
R.8 nth Roots; Rational Exponents
72
Chapter Review
80
Chapter Test
84
Equations and Inequalities
85
1.1
Linear E quations
86
1.2
Quadratic Equations
97
1.3
Complex Numbers; Quadratic E quations in the Complex Number System
109
Radical Equations; Equations Quadratic in Form; Factorable Equations
118
1.5
Solving Inequalities
124
1.6
E quations and Inequalities Involving Absolute Val ue
135
1 .7
Problem Solving: Interest , Mixture, Uniform Motion, and Constant Rate Job Applications
139
Chapter Review
148
1.4
2
xiii
Chapter Test
152
Chapter Projects
153
Graphs
155
2.1
The Distance and Midpoint Formulas
156
2.2
Graphs of Equations in Two Variables; Intercepts; Symmetry
163
2.3
Lines
173
2.4
Circles
189
2.5
Variation
195
vii
viii
Contents
3
201
Chapter Test
204
Cumulative Review
204
Chapter Project
205
Functions and Their Graphs
4
222
3.3
P roperties of Functions
231
3.4
Library of Functions; P iecewise-defined Functions
242
3.5
Graphing Techniques: Transformations
252
3.6
Mathematical Models: Building Functions
264
208
Chapter Review
269
Chapter Test
274
Cumulative Review
2 75
Chapter Projects
276
Linear and Quadratic Functions
.
207
Functions The Graph of a Function
3.1 3.2
5
Chapter Review
277
4.1
Linear Functions and Their Properties
278
4.2
Building Linear Functions from Data
287
4.3
Quadratic Functions and Their P roperties
293
4.4
Quadratic Models; Building Quadratic Functions from Data
305
4.5
Inequalities Involving Quadratic Functions
314
Chapter Review
318
Chapter Test
320
Cumulative Review
3 21
Chapter Projects
322
-
-
Polynomial and Rational Functions
323
5 .1
Polynomial Functions and Models
324
5 .2
Properties of Rational Functions
344
5 .3
The Graph of a Rational Function
355
5.4 5.5
Polynomial and Rational Inequalities
369
The Real Zeros of a Polynomial Function
375
5.6
Complex Zeros; Fundamental Theorem of Algebra
389
Chapter Review
394
Chapter Test
398
Cumulative Review
398
Chapter Projects
399
Contents
G
Exponential and Logarithmic Functions 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9
7
8
Composite Functions One-to-One Functions; Inverse Functions E xponential Functions Logarithmic Functions P roperties of Logarithms Logarithmic and Exponential Equations Compound Interest Exponential Growth and Decay Models; Newton's Law; Logistic Growth and Decay Models Building Exponential, Logarithmic, and Logistic Models from Data
487
Chapter Review
494
Chapter Test
500
Cumulative Review
501
Chapter Projects
502
Trigonometric Functions 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8
402 409 423 437 450 459 465 475
503
Angles and Their Measure Right Triangle Trigonometry Computing the Values of Trigonometric Functions of Acute Angles Trigonometric Functions of General Angles Unit Circle Approach; P roperties of the Trigonometric Functions Graphs of the Sine and Cosine Functions Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions Phase Shift; Sinusoidal Curve Fitting
581
Chapter Review
592
504 517 529 540 550 560 574
Chapter Test
598
Cumulative Review
598
Chapter Projects
599
Analytic Trigonometry 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8
401
The Inverse Sine, Cosine, and Tangent Functions The Inverse Trigonometric Functions (Continued) Trigonometric Identities Sum and Difference Formulas Double-angle and Half-angle Formulas P roduct-to-Sum and Sum-to-Product Formulas Trigonometric Equations (I) Trigonometric Equations (II)
601 602 614 620 627 637 646 649 656
Chapter Review
663
Chapter Test
666
Cumulative Review
667
Chapter Projects
667
ix
X
Contents
9
10
11
12
Applications of Trigonometric Functions
669
9.1
Applications Involving Right Triangles
670
9.2
The Law of Sines
675
9.3
The Law of Cosines
686
9.4
Area of a Triangle
691
9.5
Simple Harmonic Motion; Damped Motion; Combining Waves
697
Chapter Review
707
Chapter Test
710
Cumulative Review
711
Chapter Projects
711
Polar Coordinates; Vectors
71 3
10.1 Polar Coordinates
714
10.2 Polar E quations and Graphs
722
10.3 The Complex Plane; DeMoivre's Theorem
738
10.4 Vectors
746
10.5 The Dot Product
757
Chapter Review
765
Chapter Test
768
Cumulative Review
769
Chapter Projects
769
Analytic Geometry
771
1 1 . 1 Conics
772
1 1 .2 The P arabola
773
1 1. 3 The Ellipse
782
1 1 .4 The Hyperbola
792
1 1 .5 Rotation of Axes; General Form of a Conic
804
11.6 Polar E quations of Conics
812
1 1.7 P lane Curves and P arametric E quations
818
Chapter Review
830
Chapter Test
833
Cumulative Review
833
Chapter Projects
834
Systems of Equations and Inequalities
835
12.1 Systems of Linear Equations: Substitution and Elimination
836
12.2 Systems of Linear Equations: Matrices
850
12.3 Systems of Linear Equations: Determinants
865
1 2.4 Matrix Algebra
875
12.5 P artial Fraction Decomposition
891
xi
Contents
13
14
12.6 Systems of Nonlinear Equations 12.7 Systems of Inequalities
898
12.8 Linear P rogramming
915
Chapter Review
921
907
Chapter Test
925
Cumulative Review
926
Chapter Projects
927
Seq'uences;...l'duction; the Binomial Theorem '�:, .
��<��.�
929
13.1 Sequences
930
13.2 Arithmetic Sequences
939
13.3 Geometric Sequences; Geometric Series
945
13.4 Mathematical Induction
957
13.5 The Binomial Theorem
961
Chapter Review
967
Chapter Test
970
Cumulative Review
970
Chapter Projects
971
.
.
.
. Counting and.:Pro,b��ility:
973
1 4.1 Counting
974
14.2 P ermutations and Combinations
979
14.3 P robability
988
Chapter Review
998
Chapter Test
1000
Cumulative Review
1001
Chapter Projects
1001
Al
The Viewing Rectangle Using a Graphing Utility to Graph Equations Using a Graphing Utility to Locate Intercepts and Check for Symmetry
A6
4
Using a Graphing Utility to Solve Equations
AS
5 6
Square Screens U sing a Graphing Utility to Using a Graphing Utility to Linear Equations Using a Graphing Utility to Using a Graphing Utility to
1 2 3
7 8
9
Answers Index
Al A3
Al0
Graph Inequalities Solve Systems of
All A12
Graph a Polar Equation Graph P arametric Equations
AB A14 AN l 11
Three Distinct Series Students have different goals, learning styles, and levels of preparation. Instructors have different teaching philosophies, styles, and techniques. Rather than write one series to fit all, the Sullivans have written three distinct series. All share the same goal-to develop a high level of mathematical understanding and an appreciation for the way mathematics can describe the world around us. The manner of reaching that goal, however, differs from series to series.
Contemporary Series, Eighth Edition The Contemporary Series is the most traditional in approach, yet modern in its treat ment of precalculus mathematics. Graphing utility coverage is optional and can be in cluded or excluded at the discretion of the instructor: College Algebra, College Algebra
Essentials, Algebra & Trigonometry, Trigonometry, Precalculus.
Enhanced with Graphing Utilities Series, Fourth Edition This series provides a more thorough integration of graphing utilities into topics, al lowing students to explore mathematical concepts and foreshadow ideas usually stud ied in later courses. Using technology, the approach to solving certain problems differs from the Contemporary series, while the emphasis on understanding concepts and building strong skills does not: College Algebra, College Algebra Essentials, Algebra &
Trigonometry, Trigonometry, Precalculus, Precalculus Essentials.
Concepts through Functions Series, First Edition This series differs from the others, utilizing a functions approach that serves as the or ganizing principle tying concepts together. Functions are introduced early in various formats. This approach supports the Rule of Four, which states that functions are rep resented symbolically, numerically, graphically, and verbally. E ach chapter introduces a new type of function and then develops all concepts pertaining to that particular function. The solutions of equations and inequalities, instead of being developed as stand-alone topics, are developed in the context of the underlying functions. Graph.. ing utility coverage is optional and can be included or excluded at the discretion of the instructor: College Algebra; Precalculus, with a Unit Circle Approach to Trigonom
etry; Precalculus, with a Right Triangle Approach to Trigonometry.
xii
Preface to the Instructor s a professor of mathematics at an urban public uni versity for 35 years, I understand the varied needs of algebra and trigonometry students. Students range from being under-prepared, with little mathematical back ground and a f ear of mathematics, to being highly pre pared and motivated. For some, this is their final course in mathematics. For others, it is preparation for future mathematics courses. I have written this text with both groups in mind. A tremendous benefit of authoring a successful series is the broad-based feedback I receive from teachers and students who have used previous editions. I am sincerely grateful for their support. Virtually every change to this edi tion is the result of their thoughtful comments and sugges tions. I hope that I have been able to take their ideas, and, building upon a successful foundation of the seventh edi tion, make this series an even better learning and teaching tool for students and teachers.
E xamples and exercises that involve a more in-depth analysis of graphing exponential and logarithmic func tions is provided. Examples and exercises that involve graphing a wider variety of inverse trigonometric functions is provided.
A
•
Organizational Changes in the Eighth Edition •
Features in the Eighth Edition Rather than provide a list of features here, that informa tion can be found on the foldout in the front cover of this book. This places the features in their proper context, as build ing blocks of an overall learning system that has been care fully crafted over the years to help students get the most out of the time they put into studying. Please take the time to review this, and to discuss it with your students at the be ginning of your course. My experience has been that when students utilize these features, they are more successful in the course.
New to the Eighth Edition •
•
•
•
are now part of the end of chapter mate rial . These along with the Cumulative Review found at the end of each chapter provide ample opportunity for students to prepare for tests and continually review earlier material. Exercise Sets at the end of each section have been clas sified according to purpose. Where appropriate, more problems to challenge the better student have been added . Applied problems have been updated and many new problems involving sourced information as well as data sets have been added to bring relevance and timeliness to these exercises. Chapter Tests
New to this edition, this chapter gives more emphasis to Linear Functions than before, with more applications, and allows for a fuller discussion of the Quadratic Function, its prop erties, models involving it, and building such functions from data. A full section on quadratic inequalities is also provided. Chapter 4 Linear and Quadratic Functions
•
•
•
Polynomial division is now part of the dis cussion of polynomials, leaving synthetic division as a stand-alone (optional) section. Chapter 1 Parallel and Perpendicular Lines is now part of the section on Lines. The section on Lines now pre cedes the section on Circles. Chapter 2 Scatter diagrams; linear curve fitting has been removed and moved to the new chapter on Linear and Quadratic Functions Trigonometric Functions now contains applications in volving right triangles near the beginning of the chapter. Chapter R
Using the Eighth Edition Effectively with Your Syllabus To meet the varied needs of diverse syllabi, this book con tains more content than is likely to be covered in an Algebra & Trigonometry course. As the chart illustrates, this book has been organized with flexibility of use in mind. Within a given chapter, certain sections are optional (see the detail following the flow chart) and can be omitted without loss of continuity.
New Content
Section R.3, Geometry E ssentials, now contains a dis cussion of congruent and similar triangles. The analysis of the graph of a polynomial function now includes the behavior of the graph near an x-intercept.
C h a pter R Review
This chapter consists of review material. It may be used as the first part of the course or later as a just-in-time review when xiii
xiv
Preface to the Instructor
the content is required. Specific references to this chapter occur throughout the book to assist in the review process. C h a pter 1
C ha pter 14
Counting and Probability
The sections follow in sequence.
Equations a n d Inequ a l ities
Primarily a review of Intermediate Algebra topics, this ma terial is prerequisite for later topics. The coverage of com plex numbers and quadratic equations with a negative discriminant is optional and may be postponed or skipped entirely without loss of continuity. C h a pter 2
G ra p h s
C h a pter 3
F u n ctio n s and Their G ra p h s
C hapter 4
Linear a n d Quadratic F u n ctio n s
Chapter 5
Polynomia l and Rational F u n ctions
C h a pter 6
Exponential a n d Logarith mic F u n ctions
This chapter lays the foundation for functions. Section 2.5 is optional. Perhaps the most important chapter. Section 3.6 is optional.
Acknowledgments Textbooks are written by authors, but evolve from an idea to final form through the efforts of many people. It was Don Dellen who first suggested this book and series to me. Don is remembered for his extensive contributions to publish ing and mathematics. Thanks are due the following people for their assistance and encouragement in the preparation of this edition: •
Topic selection depends on your syllabus. Sections 4.2 and 4A may be omitted without a loss of continuity. Topic selection depends on your syllabus. Sections 6.1-6.6 follow in sequence. Sections 6.7, 6.8, and 6.9 are optional. C ha pter 7
Trig o nometric Fu nctio n s
C h a pter 8
Analytic Trigonometry
C ha pter 9
Applications of Trigonometric F u n ction s
•
Section 7.8 may be omitted in a brief course. Sections 8.2, 8.6, and 8.8 may be omitted in a brief course.
Sections 9.4 and 9.5 may be omitted in a brief course. Chapter 10
Pol a r Coordinates; Vectors
C h a pter 11
Analytic Geometry
Sections 10.1-10.3 and Secti ons 10A-10.S are independent and may be covered separately. Sections 11.1-11.4 follow in sequence. Sections 1 1 .5, 1 1 .6, and 11.7 are independent of each other, but each requires Sections 11.1-1 1 .4. C h apter 12
Systems of Equations a n d Inequalities
C h apter 13
Sequences; Induction; The Binomial
Sections 12.2-12.7 may be covered in any order, but each requires Section 12.1. Section 12.8 requires Section 12.7.
•
From Prentice Hall: Adam Jaworski for his substantial contributions, ideas, and enthusiasm; Patrice Jones, who remains a huge fan and supporter; Dawn Murrin, for her unmatched talent at getting the details right; Bob Walters for his superb organizational skills in directing production; Sally Y agan for her continued support and genuine interest; and the Prentice Hall Sales team, for their continued confidence and personal support of my books. Accuracy checkers: Kathleen Miranda, for reading the entire manuscript and providing insightful suggestions; Kevin B odden and Randy G allaher, for creating the Solutions Manuals and accuracy checking answers; LaurelTech's editors, Teri Lovelace, Kurt Norlin, George Seki, and J ill McClain-Wardynski, who read the entire manuscript and checked all the answers. Reviewers: Debra Kopcso, Louisiana State University; Lynn Marecek, Santa Ana College; K athleen Miranda, SUNY at Old Westbury; Karla Neal, Louisiana State University; Leticia Oropesa, University of Miami; Laura Pyzdrowski, West Virginia University; Mike Rosenthal, Florida International University; Phoebe Rouse, Louisiana State University; Brenda Santistevan, Salt Lake Commu nity College; Larissa Williamson, University of Florida.
Finally, I offer my grateful thanks to the dedicated users and reviewers of my books, whose collective insights form the backbone of each textbook revision. My list of indebtedness just grows and grows. And, if I've forgotten anyone, please accept my apology. Thank you all.
Theorem
There are three independent parts: Sections 13.1-13.3; Sec tion 13.4; and Section 13.5. James Africh, College of DuPage
Sudeshna Basu, Howard University
Bob Bradshaw, Ohlone College
Steve Agronsky, Cal Poly State University
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Trudy Bratten, Grossmont College
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Joby Milo Anthony, University of CentralFlorida
College
Joanne Brunner, Joliet Junior College
William H. Beyer, University of Akron
Warren Burch, Brevard Community College
Annette Blackwelder, Florida State University
Mary Butler, Lincoln Public Schools
Adel Arshaghi, Center for Educational Merit
Richelle Blair, Lakeland Community College
Jim Butterbach, Joliet Junior College
Carolyn Autray, University of West Georgia Agnes Azzolino, Middlesex County College
Kevin Bodden, Lewis and Clark College Larry Bouldin, Roane State Community
James E. Arnold, University of WisconsinMilwaukee
Wilson P Banks, Illi nois State University
College
Melanie Butler, West Virginia University William 1. Cable, University of WisconsinStevens Point
Lois Calamia, Brookdale Community College
Preface to the Instructor
Jim Campbell, Lincoln Public Schools Roger Carlsen, Moraine Valley Community College Elena Catoiu, Joliet Junior College Mathews Chakkanakuzhi, Palomar College John Collado, South S u burban College Nelson Collins, Joliet Junior College
Lynda Hollingsworth, Northwest Missouri State University Charla Holzbog, Denison High School Lee Hruby, Naperville North High School Miles Hubbard, St. Cloud State University Kim Hughes, California State College-San Bernardino
XV
Maria Montoya, Our Lady of the Lake University Craig Morse, Naperville North High School Samad Mortabit, Metropolitan State University Pat Mower, Washburn University A. Muhundan, Manatee Community College
Jim Cooper, Joliet Junior College
Ron Jamison, Brigham Young University
Jane Murphy, Middlesex Community College
Denise Corbett, East Carolina University
Richard A. Jensen, Manatee Community
Richard Nadel, Florida International
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College Sandra G. Johnson, St. Cloud State University Tuesday Johnson, New Mexico State
University Gabriel Nagy, Kansas State University Bill Naegele, South Suburban College
Theodore C. Coskey, South Seattle Community College Donna Costello, Plano Senior High School Paul Crittenden, University of Nebraska at Lincoln John Davenport, East Texas State University
University
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College Donna Katula, Joliet Junior College
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College Cheryl Doolittle, Iowa State University Karen R. Dougan, University o f Florida Louise Dyson, Clark College Paul D. East, Lexington Community College
CoLlege Keith Kuchar, Manatee Community College Tor Kwembe, Chicago State University Linda 1. Kyle, Tarrant Country Jr. College H.E. Lacey, Texas A & M University
Don Edmondson, University of Texas-Austin
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College
Seth F. Oppenheimer, Mississippi State University
Leticia Oropesa, University of Miami Linda Padilla, Joliet Junior College E. James Peake, Iowa State University Kelly Pearson, Murray State University Philip Pina, Florida Atlantic University Michael Prophet, University of Northern Iowa Laura Pyzdrowski, West Virginia University
Erica Egizio, Joliet Junior College
James Lapp, Fort Lewis College
Neal C. Raber, University of Akron
Jason Eltrevoog, Joliet Junior College
Matt Larson, Lincoln Public Schools
T homas Radin, San Joaquin Delta College
Christopher Ennis, University o f Minnesota
Christopher Lattin, Oakton Community
Ken A. Rager, Metropolitan S tate College
Kathy Eppler, Salt Lake Community College
College Julia Ledet, Lousiana State University
Kenneth D. Reeves, San Antonio College
Adele LeGere, Oakton Community College
Elsi Reinhardt, Truckee Meadows Community College
Scott Fallstrom, Shoreline Community
Kevin Leith, University of Houston
Jane Ringwald, Iowa State University
College Pete Falzone, Pensacola Junior College
JoAnn Lewin, Edison College Jeff Lewis, Johnson County Community
Stephen Rodi, Austin Community College William Rogge, Lincoln Northeast High
Ralph Esparza, Jr., Richland College
Garret 1. Etgen, University o f Houston
w.A. Ferguson, University o f Illinois Urbana/Champaign Iris B. Fetta, Clemson University Mason Flake, student at Edison Community College
College Janice C. Lyon, Tallahassee Community College Jean McArthur, Joliet Junior College
Edward Rozema, University of Tennessee at
Karla McCavit, Albion College M i chael McClendon, University of Central
Community College
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Randy Gallaher, Lewis and Clark College Dawit Getachew, Chicago State University
Oklahoma Technology Marilyn McCollum, North Carolina State University Jill McGowan, Howard University
Wayne Gibson, Rancho Santiago College
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University Joan Goliday, Sante Fe Community College Frederic Gooding, Goucher College Donald Goral, Northern Virginia Community
University Jay A. Malmstrom, Oklahoma City Community College Rebecca Mann, Apollo High School
Sue Graupner, Lincoln Public Schools
Lynn Marecek, Santa Ana College Sherry Martina, Naperville North High
Jennifer L. Grimsley, University of Charleston
School Alec Matheson, Lamar University
Ken Gurganus, University o f North Carolina
Nancy Matthews, University of Oklahoma
James E. Hall, University of Wisconsin
James Maxwell, Oklahoma State University-
College
Madison
University Phoebe Rouse, Lousiana State University
Timothy W. Flood, Pittsburg State University
Tina Garn, University of Arizona
Mike Rosenthal, Florida International
Virginia McCarthy, Iowa State University
Merle Friel, Humboldt State University Richard A. Fritz, Moraine Valley Dewey Furness, Ricke College
School Howard L. Rolf, Baylor University
Stillwater
Chattanooga Dennis C. Runde, Manatee Community College Alan Saleski, Loyola University of Chicago Susan Sandmeyer, Jamestown Community College Brenda Santistevan, Salt Lake Community College Linda Schmidt, GreenviLle Technical College A.K. Shamma, University o f West Florida Martin Sherry, Lower Columbia College Tatrana Shubin, San Jose State University Anita S ikes, Delgado Community College Timothy Sipka, Alma College Lori Smellegar, Manatee Community College Gayle Smith, Loyola Blakefield John Spellman, Southwest Texas State University Karen Spike, University of North Carolina Rajalakshmi Sri ram, Okaloosa-Walton Community College
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Brother Herron, Brother Rice High School
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University
School
Sonya Stephens, Florida A&M Univeristy Matthew TenHuisen, University of North Carolina, Wilmington Christopher Terry, Augusta State University Diane Tesar, South Suburban College Tommy Thompson, Brookhaven College
xvi
Preface to the Instructor
Martha K. T ietze, Shawnee Mission
Jennifer Walsh, Daytona Beach Community
Northwest High School Richard J. Tondra, Iowa State University Suzanne Topp, Salt Lake Community College Marvel Townsend, University of Florida Jim Trudnowski, Carroll College Robert Tuskey, Joliet Junior CoUege Richard G. Vinson, University of South Alabama
College Donna Wandke, Naperville North High School Kathryn Wetzel, Amarillo College Darlene W hitkenack, Northern Illinois University Suzanne Williams, Central Piedmont Community College
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Michael Sullivan Chicago Stale University
STUDENT R ESOU R C ES Student Study-Pack
Everything a student needs to succeed in one place. The Student Study-Pack is available for purchase standalone. Study-Pack contains: •
Student Solutions Manual
Fully worked solutions to odd-numbered exercises.
•
Pearson Tutor Center
Tutors provide one-on-one tutoring for any problem with an answer at the back of the book. Students access the Tutor Center via toll-free phone, fax, or email. Available only to college students in the U.S. and Canada.
•
Algebra Review
Four chapters of Intermediate Algebra review. Perfect for a sloweI:.paced course or for individual review. CD Lecture Series
A comprehensive set of CD-ROMs, tied to the textbook, containing short video clips of an instructor working key book examples. The CD Lecture Series can be packaged with the book, or is available for purchase standalone. Chapter Test Prep Video CD
Packaged at the back of the text, this video CD provides students with step-by-step solutions for each of the exercises in the book's chapter tests. MathXL
@
MathXL® is a powerful online homework, tutorial, and assessment system that accompanies your textbook. Instructors can create, edit, and assign online homework and tests using algorithmically generated exercises correlated at the objective level to the text book. Student work is tracked in an online gradebook. Students can take chapter tests and receive personalized study plans based on their results. The study plan diagnoses weaknesses and links students to tutorial exercises for objectives they need to study. Students can also access video clips from selected exercises. MathXL® is available to qualified adopters. For more information, visit our web site at www.mathxI.com. or contact your Prentice Hall sales representative for a demonstration. MyMathLab
MyMathLab is a text-specific, customizable online course for your textbooks. MyMathLab is powered by CourseCompass™-Pearson Education's online teaching and learning environment- and by MathXL ®-our online homework, tutorial, and assessment system. MyMathLab gives you the tools you need to deliver all or a portion of your course online, whether your students are in a lab setting or working from home. MyMathLab provides a rich and flexible set of course materials, featuring free-response exercises that are algorithmically generated for unlimited practice. Students can use online tools such as video lectures and a multimedia textbook to improve their performance. Instructors can use MyMathLab's homework and test managers to select and assign online exercises correlated to the textbook, and can import TestGen tests for added flexibility. The online gradebook-designed specifically for mathematics-automatically tracks students ' homework and test results and gives the instructor control over how to calculate final grades. MyMathLab is available to qualified adopters. For more information, visit our website at www.mymathlab.com or contact your Prentice Hall sales representa tive for a product demonstration. @ MathXL Tutorials on CD
This interactive tutorial CD-ROM provides algorithmically generated practice exercises that are correlated at the objective level to the exercises in the textbook. Every practice exercise is accompanied by an example and a guided solution designed to involve stu dents in the solution process. Selected exercises may also include a video clip to help students visualize concepts. The software pro vides helpful feedback for incorrect answers and can generate printed summaries of students' progress. InterAct Math Tutorial Website: www.interactmath.com
Get practice and tutorial help online! This interactive tutorial website provides algorithmically generated practice exercises that cor relate directly to the exercises in the textbook. Students can retry an exercise as many times as they like with new values each time for unlimited practice and mastery. Every exercise is accompanied by an interactive guided solution that provides helpful feedback for incorrect answers, and students can also view a worked-out sample problem that steps them through an exercise similar to the one they're working on.
xvii
List of Applications Acoustics
Art
amplifying sound, 498 loudness of sound, 449 loudspeaker,705 tuning fork,705,706 whispering galleries, 789,832
fine decorative pieces, 537 framing a painting, 151
Aerodynamics
modeling aircraft motion, 769 Aeronautics Challenger
disaster,486
Agriculture
crop allocation,925 crop yields and fertilizer,309-10 farm management,920 field enclosure, 906 grazing area for cow,696 minimizing cost, 920 removing stump, 757 watering a field,108 Air travel
bearing of aircraft, 674 frequent flyer miles,683 holding pattern,655 parking at O' Hare International Airport, 250 revising a flight plan, 690 speed and direction of aircraft,759-60, 764,768 Archaeology
age of ancient tools,478-79 age of fossil, 485 age of tree, 485 date of prehistoric man's death,499 Architecture
brick staircase, 945,969 floor design,943-44, 969 football stadium seating, 945 Freedom Tower,539 mosaic design, 945, 969 Norman window,37, 312 parabolic arch, 311 racetrack design,791 special window, 312, 319 stadium construction,945 Taipei, 101, 31 window design,312 window dimensions, 108 Area. See also Geometry
of Bermuda Triangle, 696 under a curve,613-14 of isosceles triangle,645 of sector of circle, 5 1 4 o f segment of circle,709
Astronomy
angle of elevation of Sun,673 distance from Earth to its moon,29 distances of planets from Sun, 939 light-year,29 planetary orbits, 788 Earth,791 Jupiter, 791 Mars,791 Mercury,817 Neptune,834 Pluto, 791, 834 Aviation
modeling aircraft motion,769 orbital launches, 848 Biology
alcohol and driving, 444, 449 bacterial growth, 476-78, 491 E-coli,241, 279 blood types,979 bone length,320 gestation period, life expectancy vs.,292 healing of wounds, 435, 448 maternal age versus Down syndrome, 293 yeast biomass as function of time, 490 Business
advertising, 292,320 automobile production, 409, 864 blending coffee, 146 candy bar size, 108 car rentals, 285 checkout lines,997-98 cigarette exports,492-93 cigarette production,492 clothing store, 999 commissions, 319 cookie orders,925 copying machines, 152 cost of can, 365-66,368 of charter bus, 152 of commodity,409 of manufacturing, 29, 83, 146,374,914 marginal,304,319 minimizing,319,920 of production,240, 409, 890,925 of transporting goods,250-51 cost equation, 187, 200 cost function,286 average,226 demand for candy,200
for jeans,292 for pes, 493 demand equation, 310-11,319,399 discount pricing,96 drive-thru rate at Burger King,431-32 at Citibank,435, 448 at McDonald 's, 436 equipment depreciation,955 expense computation, 147 IBM personal computer price and demand,493 Jiffy Lube 's car arrival rate,435,448 managing a meat market, 920 market penetration of Intel's coprocessor, 486 McDonald 's scalding coffee case,401 mixing candy,146 mixing nuts, 146 orange juice production,864 precision ball bearings,29 price markup,96 of new car,134 product design, 921 production scheduling, 920 product promotion,188 profit, 890-91 cigar company, 264 on figurines, 925 maximizing,918-19, 920-21 profit function,222 quarterly corporate earnings, 84 rate of return on, 473 restaurant management, 848 revenue, 146,304,317 advertising and, 292 airline, 921 of clothing store, 880 daily,304-5 from digital music,263 maximizing,304 monthly,304 theater, 849 revenue equation, 200 revenue function, 203 RV rental,321 salary, 945 gross,221 increases in,955, 969 sales commission on,133-34 of movie theater ticket,836, 841, 848 net, 163 salvage value, 499 straight-line depreciation, 282, 286 supply and demand, 282-83,285-86 tax, 374 theater attendance, 96 toy truck manufacturing, 914
xix
XX
List of Applications
transporting goods, 914 truck rentals, 187, 286 unemployment, 1000 wages of car salesperson, 1 87 hourly, 93, 96 women 's weekly earnings, 108 Ca lculus
area under a curve, 613-14 carrying a ladder around a corner, 662-63 maximizing rain gutter construction, 662 projectile motion, 662 Simpson 's rule, 312 Carpentry. See also Construction
pitch, 189 Chemistry, 95
alpha particles, 803 decomposition reactions, 485 drug concentration, 368 gas laws, 200 mixing acids, 151 pH, 448 purity of gold, 147 radioactive decay, 484, 485, 491-92,499 radioactivity from Chernobyl, 485 reactions, 312 salt solutions, 147, 151 sugar molecules, 147 volume of gas, 133 Combinatorics
airport codes, 980 binary codes, 1000 birthday permutations, 982, 986, 987, 993-94, 998, 1000 blouses and skirts combinations, 978 book arrangements, 986, 999 box stacking, 986 code formation, 986 combination locks, 987 committee formation, 984, 986, 987, 1000 Senate committees, 987 flag arrangement, 985, 1000 letter codes, 980, 1000 letter combinations, 999 license plate possibilities, 986, 999, 1001 light bulb wattage, 1000 lining up people, 981, 986 number formation, 978, 986, 987, 1000 objects selection, 987 seating arrangements, 999 shirts and ties combinations, 978 telephone numbers, 999 two-symbol codewords, 978 word formation, 984-85, 987, 1000 Communications
cell phone service, 250, 276 cell phone usage, 488-89, 494
installing cable TV, 269 long distance, 286 comparing phone companies, 319 phone charges, 285 portable phones, 207 satellite dish, 778-79, 781 spreading of rumors, 435, 449 Touch-Tone phones, 649, 706 Computers and computing
graphics, 756 home computers, 486 IBM PCs, 493 Internet searches, 95 Internet users, 96 laser printers, 147 Construction
of border around a garden, 108 of border around a pool, 108 of box, 105-6, 108, 905 open, 240 of brick staircase, 969 of can, 397 of circular pond, 84 of coffee can, 148 of cylindrical tube, 906, 909 of enclosures around garden, 147 around pond, 147 maximizing area of, 306-7, 311, 3 1 9 o f fencing, 306-7, 3 1 1 , 3 1 9 , 906 minimum cost for, 368 of flashlight, 781 of headlight, 781 of highway, 539, 684, 708 home choices, 999 installing cable TV, 269 material for steel drum, 274 patio dimensions, 108 pitch of roof, 674 of playpen, 266-67 of rain gutter, 3 1 1 , 533-34, 662 of ramp, 683 access ramp, 188 of rectangular field enclosure, 311 of stadium, 3 12, 945 of steel drum, 368-69 of swimming pool, 37, 38, 84 of swing set, 691 of tent, 695 TV dish, 781 vent pipe installation, 791 Crime
male murder victims, 304 violent, 319-20 Cryptography
matrices in, 891 Decorating
Christmas tree, 32
Demographics
death rates, 1000 diversity index, 448 marital status, 979 mosquito colony growth, 484 population. See Population rabbit colony growth, 938 Design
of awning, 684 of box with minimum surface area, 368 of fine decorative pieces, 537 of Little League Field, 516 of water sprinkler, 515 of window, 84 Direction
of aircraft, 759-60, 764, 768 compass heading, 764 for crossing a river, 764 of fireworks display, 803 of lightning strikes, 803 of motorboat, 764 of swimmer, 768 Distance
Bermuda Triangle, 38 bicycle riding, 230 from Chicago to Honolulu, 614 circumference of Earth, 516 between cities, 510-1 1 , 5 1 5 between Earth a n d Mercury, 685 between Earth and Venus, 685 of explosion, 803 height of aircraft, 498, 683, 684 of bouncing ball, 955, 969 of bridge, 683 of building, 597, 674 of cloud, 535 of CN Tower, 538 of Eiffel Tower, 538 of embankment, 674 of Ferris Wheel rider, 654 of Freedom Tower, 539 of Great Pyramid of Cheops, 38, 684 of helicopter, 708 of hot-air balloon, 538 of Lincoln's caricature on Mt. Rushmore, 538 of mountain, 498, 679-80,683 of Mt. Everest, 29 of Sears Tower, 674 of statue on a building, 535-36 of tower, 538 of Washington Monument, 538-39 from home, 230 from Honolulu to Melbourne, Australia, 614 of hot-air balloon to airport, 710 from intersection, 162 between houses, 682-83
List of Applications
from intersection, 268 length of guy wire, 538, 690-91 of lake, 596 of mountain trail, 539 of ski lift, 683 limiting magnitude of telescope, 498-99 pendulum swings, 951-52, 955 to plateau, 538 across a pond, 538 range of airplane, 147 reach of ladder, 538 of rotating beacon, 580-81 at sea, 684, 708 of search and rescue, 1 5 1 t o shore, 538, 597, 684 between skyscrapers, 674 sound to measure, 123-24 of storm, 1 5 1 surveying, 708 to tower, 684 traveled by wheel, 37 between two moving vehicles, 1 62 toward intersection, 268 between two objects, 538, 674 visibility of Gibb's Hill Lighthouse beam, 38, 671-72, 675 visual, 38 of pilot, 84 walking, 230 width of gorge, 538 of Mississippi River, 674 of river, 534, 597 Economics
Consumer Price Index (CPI), 475 demand equations, 399 federal deficit, 474 IBM personal computer price and demand, 493 inflation, 474 IS-LM model in, 849 marginal propensity to consume, 291, 956 marginal propensity to save, 291 multiplier, 956 participation rate, 222 relative income of child, 891 unemployment, 1000 Education
advanced degrees, 304 age distribution of community college, 1001 college costs, 474, 956 college tuition and fees, 108, 890 computing grades, 134 doctoral degrees awarded, 976, 997 faculty composition, 998 funding a college education, 499 grades, 96 learning curve, 436, 449 maximum level achieved, 927
xxi
multiple-choice test, 986 Spring break, 920 student loan, 275 interest on, 890 test design, 999 true/false test, 986
rods and pistons, 691 safe load for a beam, 201 searchlight, 627, 781, 832 whispering galleries, 791, 832
Electricity, 95
Demon Roller Coaster customer rate, 436 DVD ownership, 486 movie theater, 613 theater revenues, 849, 850
alternating current (ac) , 597, 636 alternating current (ac) circuits, 572, 590 alternating current (ac) generators, 572-73 charging a capacitor, 706 cost of, 248 current in RC circuit, 436 current in RL circuit, 436, 449 impedance, 1 17 Kirchhoff's Rules, 849, 864-65 light bulbs, 1001 Ohm 's law, 131 parallel circuits, 1 17 resistance in, 354 rates for, 1 34, 1 87-88 resistance, 69, 71, 200, 201 , 354 voltage alternating, 597, 636 foreign, 29 household, 138 U.S., 29 wiring, 999 Electronics
comparing TVs, 108-9 loudspeakers, 705 microphones, 1 73 sawtooth curve, 645, 706 Energy
nuclear power plant, 803 solar, 173, 763-64 solar heat, 781 thermostat control, 263 Engineering
bridges clearance, 573 Golden Gate, 308-9 parabolic arch, 320, 781, 832 semielliptical arch, 791, 832 suspension, 31 1 , 781 crushing load, 124 drive wheel, 709 electrical, 527 Gateway Arch (St. Louis), 781 grade of road, 1 89 horsepower, 200 lean of Leaning Tower of Pisa, 683-84 maximum weight supportable by pine, 197 moment of inertia, 649 piston engines, 537 product of inertia, 645 road system, 721-22
Entertainment
cable subscribers, 494
Environment
lake pollution control laws, 938 oil leakage, 409 Finance, 95. See also Investment(s)
balancing a checkbook, 29 bills in wallet, 1000 clothes shopping, 926 college costs, 474, 956 computer system purchase, 473 consumption and disposable income, 291 cost of car, 96 of car rental, 251 of electricity, 248 of fast food, 848 of land, 709 minimizing, 319, 368 of natural gas, 250 of pizza, 96 of Sunday newspaper home delivery, 187 of trans-Atlantic travel, 221, 229 of triangular lot, 695 cost equation, 200 cost function, 286 cost minimization, 304 credit cards debt, 938 interest on, 473 payment, 25 1 , 938 demand equation, 310-1 1 depreciation, 435 of car, 465, 502 division of money, 92, 95 electricity rates, 1 87-88 financial planning, 141-42, 848 foreign exchange, 409 gross salary, 221 health care expenditures, 285, 304 inheritance, 152 life cycle hypothesis, 312-13 loans, 146 car, 938 interest on, 141, 151, 275, 890 repayment of, 473 student, 321 , 890 mortgages interest rates on, 85, 153, 474 payments, 1 96, 1 99, 203 second, 474
xxii
List of Applications
phone charges, long distance, 286 price appreciation of homes, 473 prices of fast food, 850 refunds, 849 revenue equation, 200 revenue function, 203 revenue maximization, 304, 306 rich man's promise, 956 salary options, 957 sales commission, 133-34 saving for a car, 473 for a home, 956 savings accounts interest, 473 sinking fund, 956 taxes, 286 e-filing returns, 241 federal income, 250, 422 luxury, 286 used-car purchase, 473 water bills, 134 Food and nutrition
animal, 921 candy, 291 color mix of candy, 1000 cooler contents, 1001 cooling time of Beerstein, 485 cooling time of pizza, 485 fast food, 848, 850 Girl Scout cookies, 997 hospital diet, 850, 864 "light" foods, 134 number of possible meals, 977 pig roasts, 485-86 raisins, 291 Forestry
wood product classification, 483-84 Games
die rolling, 1001 grains of wheat on a chess board, 956 Powerball, 1001 Gardens and gardening. See also Landscaping
border around, 108 enclosure for, 147 Geography
area of Bermuda Triangle, 696 area of lake, 695, 709 inclination of hill, 765 inclination of mountain trail, 671, 708 Geology
earthquakes, 449 Geometry
angle between two lines, 636 balloon volume, 409 circle area of, 146, 695 circumference of, 28, 146
inscribed in square, 268 length of chord of, 691 radius of, 905 collinear points, 874 cone volume, 200, 409 cube length of edge of, 388 surface area of, 29 volume of, 29 cylinder inscribing in cone, 269 inscribing in sphere, 268 volume of, 200, 409 Descartes's method of equal roots, 906 equation of line, 874 ladder angle, 710 polygon, diagonals of, 109 quadrilateral area, 710 rectangle area of, 28, 221, 265 dimensions of, 96, 108, 151, 905, 906 inscribed in semicircle, 268, 645 perimeter of, 28 pleasing proportion for, 151 semicircle inscribed in, 268 semicircle area, 695, 710 sphere surface area of, 28 volume of, 28, 274 square area of, 37, 146 perimeter of, 146 surface area of balloon, 409 of cube, 29 of sphere, 28 triangle area of, 28, 695, 710, 874 circumscribing, 685 equilateral, 28 inscribed in circle, 268 isosceles, 221, 710, 905, 906 lengths of the legs, 151 Pascal's, 938 right, 537-38, 673 sides of, 710
Insurance
claims for catastrophic losses, 312 disability benefits, 285 Investment(s), 92-93, 95, 146, 151
annuity, 952-53, 956 in bonds, 921 Treasuries, 864, 865, 912, 914, 915 zero-coupon, 471, 474 in CDs, 921 compound interest on, 466-67, 468, 469, 473 diversified, 850 doubling of, 471-72, 474 finance charges, 473 in fixed-income securities, 474, 921 401K, 956, 969 growth rate for, 473 IRA, 469-70, 473, 499, 953, 956, 969 return on, 473, 920, 921 in stock analyzing, 322 appreciation, 473 NASDAQ stocks, 986 NYSE stocks, 986 portfolios of, 979 price of, 956 time to reach goal, 473, 474 tripling of, 471-72, 474 Landscaping. See also Gardens and gardening
pond enclosure, 319 rectangular pond border, 319 removing stump, 757 tree planting, 864 watering lawn, 515 Law and law enforcement
motor vehicle thefts, 997 violent crimes, 222 Leisure and recreation
cable TV, 269 community skating rink, 275 Ferris wheel, 194, 654, 685, 705
Government
Marketing
federal deficit, 499 federal income tax, 222, 250, 422 e-filing returns, 241 federal tax withholding, 134
IBM personal computer price and demand, 493
Health. See also Medicine
allergies, 399--400 cigarette use among teens, 188 exercising, 134 expenditures on, 222, 285, 304 ideal body weight, 422 life cycle hypothesis, 312-13 life expectancy, 133 Home improvement. See also Construction
painting a house, 850
Measurement
optical methods of, 627 of rainfall, 764 Mechanics, 95. See also Physics Medicine. See also Health
drug concentration, 240, 368 drug medication, 435, 448 healing of wounds, 435, 448 spreading of disease, 500 Meteorology
weather balloon height and atmospheric pressure, 489-90
List of Applications
Military
M usic
active duty personnel, 343
revenues from, 263
Miscellaneous
Navigation
bending wire, 906 biorhythms, 573 carrying a ladder around a corner, 527, 580, 662-63 citrus ladders, 945 cross-sectional area of beam, 221 , 229 curve fitting, 849, 864, 924 diameter of copper wire, 29 drafting error, 162-63 McDonald's scalding coffee case, 502 motor, 29 pet ownership, 997 reading books, 138 rescue at sea, 680-81, 682 surface area of balloon, 409 surveillance satellites, 674-75 volume of balloon, 409 wire enclosure area, 268
avoiding a tropical storm, 690 bearing, 672, 689 of aircraft, 674 of ship, 674, 709 commercial, 683 compass heading, 764 error in correcting, 688, 708 time lost due to, 683 rescue at sea, 680-81, 682 revising a flight plan, 690
Mixtures. See also Chemistry
blending coffees, 142-43, 146, 152, 914, 925 blending teas, 146 cement, 1 48 mixed nuts, 146, 848, 914, 925 mixing candy, 146 water and antifreeze, 147 Motion, 706. See also Physics
catching a train, 833 on a circle, 515 of Ferris Wheel rider, 654 of golf ball, 229 minute hand of clock, 514, 596 objects approaching intersection,829 of pendulum, 706 revolutions of circular disk, 37 simulating, 823-24 tortoise and the hare race, 905 uniform, 143-44, 146, 829 Motor veh icles
alcohol and driving, 444-45 , 449 approaching intersection, 829 automobile production, 409, 864 average car speed, 148 brake repair with tune-up, 1000 braking load, 764 crankshafts, 684 depreciation of, 465, 502 with Global Positioning System CGPS), 499 loans for, 938 markup of new car, 134 miles per gallon, 313-14 runaway car, 317 spin balancing tires, 5 1 6, 596 stopping distance, 312, 422 tire rotation, 654-55 used-car purchase, 473
Oceanography
tides, 591 Optics
angle of refraction,655-56 bending light, 656 index of refraction,655-56 intensity of light, 200 laser beam, 673 laser projection, 645 lensmaker's equation, 71 light obliterated through glass, 435 mirrors, 803 reflecting telescope, 781 Pediatrics
height vs. head circumference, 292, 422 Pharmacy
vitamin intake, 849, 865 Photography
camera distance, 539 Physics, 95
angle of elevation of Sun, 673 bouncing balls, 969 braking load, 764 damped motion, 700-702 diameter of atom, 29 Doppler effect, 368 effect of elevation on weight, 229 falling objects, 1 99 force, 146 resultant, 753, 756 of wind on a window, 1 98, 200 gravity, 354, 374 on Earth, 221, 422 on Jupiter, 22 1 harmonic motion, 699-700 heat loss through a wall, 198 heat transfer, 662 horsepower, 200 inclination of mountain trail, 671 intensity of light, 151, 200 Kepler's Third Law of Planetary Motion, 203 kinetic energy, 146, 200
xxiii
maximum weight supportable by pine, 197 moment of inertia, 649 motion of object, 700 Newton's law, 199 Ohm's law, 131 pendulum motion, 124, 514, 706, 951-52 period, 264, 422 simple pendulum, 1 99 pressure, 146, 200 product of inertia, 645 projectile motion, 108, 307-8, 311, 536-37, 549, 640, 645 , 655, 662, 663, 822, 828-30, 833 artillery, 317, 619 thrown object, 240, 828 safe load for a beam, 201 simulating motion, 823-24 sound to measure distance, 123-24 speed of sound, 138 static equilibrium, 753-54, 756, 757, 768 stress of materials, 201 stretching a spring, 199 tension, 753-54, 756, 768 thrown object, 151 ball, 313, 317 truck pulls, 757 uniform motion, 143-44, 146, 151, 829, 833 velocity down inclined planes, 79 vertically propelled object, 317 vibrating string, 200 wavelength of visible light, 29 weight, 200, 203 work, 1 46 Play
swinging, 710 wagon pulling, 756, 762-63 Population. See also Demographics
bacterial, 484, 486, 491 decline in, 485 E-coli growth, 241, 279 of endangered species, 486 of fruit fly, 482-83 as function of age, 221 growth in, 485, 486 insect, 354, 484 of trout, 938 of United States, 83, 464, 493, 971 of world, 465, 493, 499, 929 Probability
checkout lines, 997-98 classroom composition, 998 "Deal or No Deal" TV show, 973 exponential, 431-32, 435, 448 household annual income, 997 Monty Hall Game, 1001-2 Poisson, 436 "Price is Right" games, 997 of winning a lottery, 998 Psychometrics
IQ tests, 134
xxiv
List of Applications
linear, 5 12-13 on Earth, 515 of merry-go-rounds, 596 of Moon, 515 of motorboat, 768 revolutions per minute of pulley, 515 of rotation of lighthouse beacons, 596 of swimmer, 768 of truck, 673 of wheel pulling cable cars, 516 wind, 848
Publishing
page design, 274 Pyrotechnics
fireworks display, 803 Rate. See also Speed
of car, 515 catching a bus, 828 catching a train, 828 current of stream, 849 of emptying oil tankers, 148, 151 a pool, 148 a tub, 148 to fill tank, 151 to keep up with the Sun, 516 miles per gallon, 313-14 revolutions per minute of bicycle wheels, 515 of pulleys, 5 1 7 speed average, 148 of current, 146 of cyclist, 148 of motorboat, 146 of moving walkways, 146 of plane, 148 of sound, 138
Sports
Real estate
commission, 133-34 cost of land, 709 cost of triangular lot, 695 Recreation
bungee jumping, 374
Demon Roller Coaster customer
online gambling, 997
rate, 436
Security
security cameras, 673
baseball, 829, 987, 999 diamond, 162 dimensions of home plate, 695 field, 690, 691 Little League, 1 62, 516 stadium, 691 World Series, 987 basketball, 987 free throws, 228, 674 granny shots, 229 biathlon, 148 bungee jumping, 374 calculating pool shots, 539 exacta betting, 1001 football, 147, 791 golf, 229 distance to the green, 689 sand bunkers, 619 hammer throw, 598 hunting, 304 Olympic heroes, 148 Olympic performance, 205 races, 147, 1 52, 903, 905 relay runners, 999 soccer, penalties in 2006 Fifa World Cup, 861-62 swimming, 710, 768 tennis, 146 Tour de France, 667 Statistics. See Probability
Seismology
S u rveying
calibrating instruments, 833
distance between houses, 682, 708
Sequences. See also Combinatorics
Su rveys
ceramic tile floor design, 943--44 Drury Lane Theater, 944 football stadium seating, 945 seats in amphitheater, 945
of appliance purchases, 978 data analysis, 975-76, 978 stock portfolios, 979 of summer session attendance, 978 of TV sets in a house, 997
Speed
of aircraft, 759-60, 764, 768 angldar, 515, 596 of current, 515-1 6, 925 as function of time, 230, 268 of glider, 596 of jet stream, 925
Temperature
of air parcel, 945 body, 29, 138 conversion of, 409, 422 cooling time of Beerstein, 485 cooling time of pizza, 485
Fahrenheit from Celsius conversion, 91-92 measuring, 1 88 after midnight, 343 monthly, 590-91 , 597 of portable heater, 499-500 relationship between scales, 264 sinusoidal function from, 585-86 of skillet, 499 wind chill factor, 500 Time
for Beerstein to cool, 485 for block to slide down inclined plane, 537 Ferris Wheel rider height as function of, 654 to go from an island to a town, 269 hours of daylight, 587-88, 591, 597, 613 for pizza to cool, 485 for rescue at sea, 151 of sunr�� 516, 613 of trip, 527, 537 Transportation
deicing salt, 619 Falls Incline Railway, 674 Travel . See also A i r travel; Navigation
bearing, 709 drivers stopped by the police, 501 driving to school, 200 parking at O'Hare International Airport, 250 Volume
of gasoline in tank, 79 of water in cone, 269 Weapons
artillery, 317, 619 Weather
atmospheric pressure, 435, 448 avoiding a tropical storm, 690 cooling air, 945 hurricanes, 342 lightning and thunder, 1 5 1 lightning strikes, 800, 803 rainfall measurement, 764 relative humidity, 436 weather satellites, 194--95 wind chill, 251 , 500 Work, 762
computing, 762-63, 768 constant rate jobs, 925 pulling a wagon, 762-63 ramp angle, 764-65 wheel barrow push, 756 working together, 145, 147, 151
Photo Credits
CHAPTER R
Page 1, Michael Newman/PhotoEdit Inc.; Page 3 1 , AP Wide World Photos;
Page 38, Whole Sprit Press/Diggins, Julia E., illustrations by Cory don Bell,
" String, Straightedge & Shadow, The Story of Geometry;" Page 38, Courtesy of Red River Pess, Inc.; Page 84, Corbis/B ettmann, ©Charless O ' Rear/CORBIS.
CHAPTER 1
Pages 85 and 153, Getty Images Inc. - Stone Allstock; Page 96, Bill
Aron/PhotoEdit Inc.; Page 1 08, David Young-Wolff/PhotoEdit Inc.; Page 138,
Nancy R. Cohen/Getty Images, Inc . - PhotoDisc; Page 1 5 1 , Kent Wood/Photo Researchers, Inc.
CHAPTER 2
Pages 1 55 and 205, Popperfoto/Alamy Images; Page 1 73, ©Kendra
Luck/CORBlS, All Rights Reserved; Page 187, ©Kathy McLaughlinrrhe Image
Works; Page 1 94, AP Wide World Photos.
CHAPTER 3
Pages 207 and 276, Photollibrary.Com; Page 221 , Courtesy of NASA/Jet
Propulsion Laboratory/Caltech; Page 229, Jamie Squire/Getty Images; Page 264,
Tony Freeman/PhotoEdit Inc.
CHAPTER 4 CHAPTER 5 CHAPTER G
Pages 277 and 322, Stephen Chernin/Getty Images; Page 312, Erin Garvey/Index Stock Imagery, Inc.;
Pages 323 and 399, Damien Lovegrove/Photo Researchers, Inc.; Page 368, Corbis
Digital Stock.
Pages 401 and 502, Kim Karpeles/Alamy Images; Page 456, The Granger Collection; NASA/GFSC/Tom Stack & Associates, Inc.; Page 465, Tony
Freeman/Photo Edit Inc.; Page 470, Jim PickerelllThe Image Works; Page 486, ©Theo AJlofs/Corbis/Bettmann.
CHAPTER 7
Pages 503 and 599, Texas Automated B uoy System; Page 5 1 5 , Doug
Pensinger/Getty Images; Page 538, © Digital Vision/eStock Photography LLC; Page 573, P. Bern dt/Custom Medical Stock Photo, Inc.
CHAPTER 8
Pages 601 and 667, Gemunu Amarasinghe/AP Wide World Photos.
CHAPTER 9
Pages 669 and 7 1 1 , Getty Images/Photodisc; Page 696, Nelson Hancock/Rough Guides Dorling Kindersley; Page 698, Richard Megna/Fundamental
Photographs, NYC.
CHAPTER 1 0
Pages 7 1 3 and 769, Ken Reid/Image B ank/Getty Images; Page 735,
CHAPTER 1 1
Pages 771 and 834, NASA/Photo Researchers, Inc.; Page 789, Scott McDonald/
CHAPTER 1 2
Pages 835 and 927, A P Wide World Photos; Page 888, Corbis/Bettmann
CHAPTER 1 3 CHAPTER 1 4
Corbis/Bettmann; Page 744, Corbis/Bettmann; Page 754, Library of Congress.
Hedrich Blessing.
Pages 929 and 971 , NASA/GFSClTom Stack & Associates, Inc.; Page 953, TIle
Granger Collection; Page 966, Corbis/B ettmann.
Pages 973 and 1 00 1 , Photofest; Page 994, The Granger Collection
xxv
To the Student As you begin, you may feel anxious about the number of theorems, definitions, pro cedures, and equations. You may wonder if you can learn it all in time. Don't worry, your concerns are normal. This textbook was written with you in mind. If you attend class, work hard, and read and study this book, you will build the knowledge and skills you need to be successful. Here's how you can use the book to your benefit.
Read Carefully When you get busy, it's easy to skip reading and go right to the problems. Don't. . . the book has a large number of examples and clear explanations to help you break down the mathematics into easy-to-understand steps. Reading will provide you with a clearer understanding, beyond simple memorization. Read before class (not after) so you can ask questions about anything you didn't understand. You'll be amazed at how much more you'll get out of class if you do this.
Use the Features I use many different methods in the classroom to communicate. Those methods, when incorporated into the book, are called "features". The features serve many purposes, from providing timely review of material you learned before Gust when you need it), to providing organized review sessions to help you prepare for quizzes and tests. Take advantage of the features and you will master the material. To make this easier, I've provided a brief guide to getting the most from this book. Refer to the 'Prepare for Class', 'Practice', and 'Review' pages on the inside front cover of this book. Spend fifteen minutes reviewing the guide and familiarizing your self with the features by flipping to the page numbers provided. Then, as you read, use them. This is the best way to make the most of your textbook. Please do not hesitate to contact me, through Prentice Hall, with any questions, suggestions, or comments that would improve this text. I look forward to hearing from you, and good luck with all of your studies.
Best Wishes!
Michael Sullivan
xxvii
Review
A Look Ahead
Chapter R, a s the title states, conta i n s review materia l . Yo u r instructor may choose to
Outline R.l Real Numbers
cover all or part of it a s a reg u l a r c h a pter at the beg i n n i n g of you r course or later as
R.2 Algebra Essentials
a j ust-in-t i m e review when the content i s req u i red. R eg a r d l ess, when i nformation i n
R.3 Geometry Essentials
this cha pter i s needed, a specific refe re n ce t o t h i s c h a pte r wi l l b e made s o you can review.
R.4 Polynomials R.S Factoring Polynomials R.6 Synthetic Division R.7 Rational Expressions R.B nth Roots; Rational Exponents Chapter Review Chapter Test
1
2
CHAPTER R
Review
R.l Rea l N u mbers PREPARING FOR THIS BOOK
of this book.
Before getting started, read "To the Student" on page xxvii at the front
OBJECTIVES 1 Work with Sets (p. 2)
2 Class ify N u m bers (p. 4)
3 Evaluate N u merical Expressions (p. 8)
4 Work with Properties of Real N u m bers (p. 9)
1
Wor k with Sets
A set is a well-defined collection of distinct objects. The objects of a set are called its elements. By well-defined, we mean that there is a rule that enables us to deter mine whether a given object is an element of the set. If a set has no elements, it is called the empty set, or null set, and is denoted by the symbol 0. For example, the set of digits consists of the collection of numbers 0, 1 , 2, 3, 4, 5, 6, 7, 8, and 9. If we use the symbol D to denote the set of digits, then we can write D
{O, 1, 2, 3, 4, 5, 6, 7, 8, 9 }
=
I n this notation, the braces { } are used to enclose the objects, o r elements, in the set. This method of denoting a set is called the roster method. A second way to denote a set is to use set-builder notation, where the set D of digits is written as D
l'
=
t
T
x is a digitj
���
Read as
E XA M P L E 1
"D is
the set of all x such that x is a d igit."
Using Set-bu i lder Notation and the Roster M ethod
(a) E (b) 0
=
=
{xix is an even digit} {xix i s an odd digit}
=
=
{O, 2, 4, 6, 8 } { 1 , 3 , 5 , 7, 9 } •
Because the elements of a set are distinct, we never repeat elements. For exam ple, we would never write { l , 2, 3, 2 } ; the correct listing is { l , 2, 3 } . Because a set is a collection, the order in which the elements are listed is immaterial. { l , 2, 3 } , { 1, 3 , 2 } , {2, 1, 3 } , and so on, all represent the same set. If every element of a set A is also an element of a set B, then we say that A is a subset of B and write A � B. If two sets A and B have the same elements, then we say that A equals B and write A = B. For example, { l , 2, 3} � { l , 2, 3, 4, 5} and { l , 2, 3} = {2, 3, l } . DEFINITION
E XA M P L E 2
If A and B are sets, the intersection of A with B , denoted A n B, i s the set consisting of elements that belong to both A and B. The union of A with B, denoted A U B, is the set consisting of elements that belong to either A or B, or both.
F i n d i ng the I ntersection and U nion of Sets
Let A
=
{ 1, 3, 5, 8 } , B
(a) A n B
=
{3, 5, 7}, and C
(b) A U B
=
{2, 4, 6, 8 } . Find:
(c) B n (A U C)
SECTION R.1
Solution
Real N u m bers
(a) A n B = { I, 3, 5, 8 } n {3, 5, 7 } = {3, 5 } (b) A U B = { I , 3 , 5 , 8 } U {3, 5 , 7 } = { I , 3 , 5 , 7, 8 } (c) B n ( A U C ) = {3, 5, 7 } n [ { 1 , 3, 5, 8 } U {2, 4, 6, 8} ] = {3, 5, 7 } n { I , 2, 3, 4, 5 , 6, 8 } = {3, 5 } CIl!C = = -
Now Work
3
•
PROBLEM 1 3
Usually, in working with sets, we designate a universal set U, the set consisting of all the elements that we wish to consider. Once a universal set has been desig nated, we can consider elements of the universal set not found in a given set. DEFINITION
E XA M P L E 3
If A is a set, the complement of A, denoted A, is the set consisting of all the elements in the universal set that are not in A . '" F i n d i n g the Com plement of a Set
If the universal set is V = { I , 2, 3, 4, 5, 6, 7, 8, 9} and if A A {2, 4, 6, 8 } .
=
{ I , 3, 5, 7, 9 } , then
=
It follows that A U A _11!!1 = 1::111 ""
Figure 1
Universal set
Now Work
=
U and A n A =
•
0. Do you see why?
PROBLEM 1 7
It is often helpful to draw pictures of sets. Such pictures, called Venn diagrams, represent sets as circles enclosed in a rectangle, which represents the universal set. Such diagrams often help us to visualize various relationships among sets. See Figure l . I f w e know that A � B , w e might use the Venn diagram i n Figure 2(a). I f we know that A and B have no elements in common, that is, if A n B = 0, we might use the Venn diagram in Figure 2(b). The sets A and B in Figure 2(b) are said to be disjoint.
Figure 2
Universal set
Universal set
@ (a)
00
A <;;; 8
(b) A n 8 0 disjoint sets =
subset
Figures 3(a), 3(b), and 3(c) use Venn diagrams to illustrate the definitions of intersection, union, and complement, respectively. Figure 3
Universal set
(a)
A n 8
intersection
Un ive rsal set
(b) A U 8 union
':' Some books use t h e notation A' [or t h e complement of A.
Universal set
(c) A
complement
4
CHAPTER R
Review
2
Classify N u m bers
It is helpful to classify the various kinds of numbers that we deal with as sets. The or natural numbers, are the numbers in the set { I, 2, 3, 4 . . . } . (The three dots, called an ellipsis, indicate that the pattern continues indefinitely.) As their name implies, these numbers are often used to count things. For example, there are 26 letters in our alphabet; there are 100 cents in a dollar. The whole numbers are the numbers in the set { O, 1 , 2, 3, . . . } , that is, the counting numbers together with O.
counting numbers,
The
DEFINITION
integers
are the set of numbers { . . . , -3, -2, -1 , 0, 1, 2, 3, . . . } .
These numbers are useful i n many situations. For example, if your checking account has $ 10 in it and you write a check for $ 15, you can represent the current balance as - $5. Notice that the set of counting numbers is a subset of the set of whole numbers. Each time we expand a number system, such as from the whole numbers to the inte gers, we do so in order to be able to handle new, and usually more complicated, problems. The integers allow us to solve problems requiring both positive and neg ative counting numbers, such as profit/loss, height above/below sea level, tempera ture above/below OaF, and so on. But integers alone are not sufficient for all problems. For example, they do not answer the question "What part of a dollar is 38 cents?" To answer such a question, 38 we enlarge our number system to include rational numbers. For example, 100 answers the question "What part of a dollar is 38 cents?" A
DEFINITION
rational number
a
is a number that can be expressed as a quotient b of two
integers. The integer a is called the numerator, and the integer b, which cannot be 0, is called the denominator. The rational numbers are the numbers in the set { x i x =
� , where a, b are integers and b
'*
O}.
3 5 0 Examples of rational numbers are 4" '2' 4"
2 100 . a ' and 3 ' Smce '1 = a for any 3 integer a, it follows that the set of integers is a subset of the set of rational numbers. Rational numbers may be represented as decimals. For example, the rational 3 5 2 7 . numbers 4" '2 ' - "3 ' and may be represented as deClmals by merely carrying out
66
the indicated division: 3 4
-
=
0 .75
5 2
-
=
2.5
2 3
-0.666 . . .
Notice that the decimal representations of
-0.6
7
66
=
0.1060606 . . .
=
-
0. 106
� and % terminate, or end. The decimal
2 7 . . . but they do exhIbIt a pattern of d a not termmate, ' representatlOns af - "3 an d
66
2 repetition. For - "3 ' the 6 repeats indefinitely, as indicated by the bar over the 6; for 7
66 ' the block
06 repeats indefinitely, as indicated by the bar over the 06. It can be
shown that every rational number may be represented by a decimal that either ter minates or is nonterminating with a repeating block of digits, and vice versa. On the other hand, some decimals do not fit into either of these categories. Such decimals represent irrational numbers. Every irrational number may be represented by a decimal that neither repeats nor terminates. In other words, irrational numbers cannot be written in the form
�, where a, b are integers and b '* O.
SECTION R.1
Real N u m bers
5
Irrational numbers occur naturally. For example, consider the isosceles right tri angle whose legs are each of length 1. See Figure 4 . The length of the hypotenuse is \12, an irrational number. Also, the number that equals the ratio of the circumference C to the diameter d of any circle, denoted by the symbol 7r (the Greek letter pi), is an irrational number. See Figure 5 . Figure 4
DEFINITION
Figure 5
c
w =
d
The set of real numbers is the union of the set of rational numbers with the set of irrational numbers. Figure 6 shows the relationship of various types of numbers. *
Figure 6
Rational numbers
Irrational numbers
Integers Whole numbers
Natu ral or counting numbers
Real numbers
EXAM P L E 4
C l assifying the N u m bers i n a Set
List the numbers in the set { 3, -
� 0.12, V2, ,
7r,
10, 2.151515 .
. .
( where the block 15 repeats) }
that are (a) Natural numbers (d) Irrational numbers Solution
(b) Integers (e) Real numbers
(c) Rational numbers
(a) 10 is the only natural number. (b) 3 and 10 are integers. -
4
(c) -3, 10, 3' 0.12, and 2.151515 . are rational numbers. .
.
(d) \12 and 7r are irrational numbers. (e) All the numbers listed are real numbers. 1l!l1J:i= = >-
Now Work
• PROBLEM 23
'" The set of real numbers is a subset of the set of complex numbers. We discuss complex numbers in
Chapter 1, Section 1 .3 .
6
CHAPTER R
Review
Approxi m ations
Every decimal may be represented by a real number (either rational or irrational), and every real number may be represented by a decimal. In practice, the decimal representation of an irrational number is given as an approximation. For example, using the symbol ;::; (read as " approximately equal to"), we can write
V2 ;::;
1 .4142
7T
;::;
3.1416
In approximating decimals, we either round off or truncate to a given number of decimal places.* The number of places establishes the location of the final digit in the decimal approximation. Truncation:
Drop all the digits that follow the specified final digit in the
decimal. Rounding: Identify the specified final digit in the decimal. If the next digit is 5 or more, add 1 to the final digit; if the next digit is 4 or less, leave the final digit as it is. Then truncate following the final digit.
EXAM P L E 5
Approximating a Decimal to Two P l aces
Approximate 20.98752 to two decimal places by (a) Truncating (b) Rounding Solution
For 20.98752, the final digit is 8, since it is two decimal places from the decimal point. (a) To truncate, we remove all digits following the final digit 8. The truncation of 20.98752 to two decimal places is 20.98. (b) The digit following the final digit 8 is the digit 7. Since 7 is 5 or more, we add 1 to the final digit 8 and truncate. The rounded form of 20.98752 to two deci mal places is 20.99. n
E XA M P L E 6
Approximating a Decimal to Two and Four Places
Number
Rounded to Two Decimal Places
Rounded to Four Decimal Places
Truncated to Two Decimal Places
Truncated to Four Decimal Places
(a) 3.14159 (b) 0.056128 (c) 893.46125
3.14 0.06 893.46
3.1416 0.0561 893.4613
3.14 0.05 893.46
3.1415 0.0561 893.4612
=-""'
Now Work
.. PROBLEM 27
Calcu lators
Calculators are finite machines. As a result, they are incapable of displaying deci mals that contain a large number of digits. For example, some calculators are capa ble of displaying only eight digits. When a number requires more than eight digits, '" Sometimes we say "correct to a given number of decimal places" instead o f "truncate."
SECTION R.l
7
Real N u m bers
the calculator either truncates or rounds. To see how your calculator handles deci mals, divide 2 by 3. How many digits do you see? Is the last digit a 6 or a 7? If it is a 6, your calculator truncates; if it is a 7, your calculator rounds. There are different kinds of calculators. An arithmetic calculator can only add, subtract, multiply, and divide numbers; therefore, this type is not adequate for this course. Scientific calculators have all the capabilities of arithmetic calculators and also contain function keys labeled In, log, sin, cos, tan, xY , inv, and so on. As you pro ceed through this text, you will discover how to use many of the function keys. Graphing calculators have all the capabilities of scientific calculators and contain a screen on which graphs can be displayed. For those who have access to a graphing calculator, we have included comments, examples, and exercises marked with a ii� , indicating that a graphing calculator is required. We have also includ,E. d an appendix that explains some of the capabilities of a graphing calculator. The ·t comments, examples, and exercises may be omitted without loss of continuity, if so desired. O pe rations
In algebra, we use letters such as x, y, a, b, and c to represent numbers. The symbols used in algebra for the operations of addition, subtraction, multiplication, and divi sion are + , - , . , and /. The words used to describe the results of these operations are sum, difference, product, and quotient. Table 1 summarizes these ideas. Table 1
Operation
Symbol
Words
Addition
a
Sum:
Subtraction M u ltiplication
Division
+
b
a - b
a plus b
a ' b, (a) . b, a . (b), (a) . (b),
Difference:
a minus b
Product:
times
a a/b or t;
Quotient: a
ab, (a)b, a(b), (a)(b)
a
b
d i vid ed by b
In algebra, we generally avoid using the multiplication sign X and the division sign -;- so familiar in arithmetic. Notice also that when two expressions are placed next to each other without an operation symbol, as in ab, or in parentheses, as in ( a ) (b), it is understood that the expressions, called factors, are to be multiplied. We also prefer not to use mixed numbers in algebra. When mixed numbers are 3
3
means 2 + 4 ' In algebra, use of a 4 mixed number may be confusing because the absence of an operation symbol used, addition is understood; for example, 2
3
between two terms is generally taken to mean multiplication. The expression 2 is 4 11 therefore written instead as 2.75 or as 4 ' The symbol called an equal sign and read as "equals" or "is," is used to express the idea that the number or expression on the left of the equal sign is equiv alent to the number or expression on the right. =
E XA M P L E 7
,
Writing Statements Using Sym bols
(a) The sum of 2 and 7 equals 9. In symbols, this statement is written as 2 + 7 9. (b) The product of 3 and 5 is 15. In symbols, this statement is written as 3 · 5 15 . =
=
•
C'J
- Now Work
PROBLEM 39
8
CHAPTER R
Review
3
('
r
r
Eva l u ate N u merica l Express io n s
Consider the expression 2 + 3 · 6. It is not clear whether we should add 2 and 3 to get 5, and then multiply by 6 to get 30; or first multiply 3 and 6 to get 18, and then add 2 to get 20. To avoid this ambiguity, we have the following agreement. We agree that whenever the two operations of addition and multiplication separate three numbers, the multiplication operation always will be performed first, followed by the addition operation.
In Words
M u ltiply first, then add.
For 2 + 3 · 6, we have 2 + 3 · 6 = 2 + 18 = 20 EXA M P LE 8
F i n d i ng the Value of an Expressi o n
Evaluate each expression. ( a) 3 + 4 · 5 Sol ution
(c) 2 + 2 · 2
(b) 8 · 2 + 1
(b) 8 · 2 + 1
(a) 3 + 4 · 5 = 3 + 20 = 23 i
t
16 + 1 = 17
M u ltiply fi rst
M u ltiply fi rst
(c) 2 + 2 · 2 = 2 + 4 = 6 .1 "'iii
.,, - Now Work
PROBLEM S 1
To first add 3 and 4 and then multiply the result by 5 , we use parentheses and write (3 + 4) · 5. Whenever parentheses appear in an expression, it means "perform the operations within the parentheses first! " E XAM P L E 9
F i n d i ng the Val ue o f an Expression
(a) ( 5 + 3 ) . 4 = 8 · 4 = 32 (b) (4 + 5 ) · (8 - 2) = 9 · 6 = 54 II
When we divide two expressions, as in 2 + 3 4 + 8 it is understood that the division bar acts like parentheses; that is, 2 + 3 4 + 8
(2 + 3 ) (4 + 8)
The following list gives the rules for the order of operations. Rules for the Order of Operations 1. Begin with the innermost parentheses and work outward. Remember that
in dividing two expressions the numerator and denominator are treated as if they were enclosed in parentheses. 2. Perform multiplications and divisions, working from left to right. 3. Perform additions and subtractions, working from left to right.
SECTION R.1
EXAM P L E 1 0
Real N u mbers
9
F in d i n g the Val ue of an Expression
Evaluate each expression. (a) 8 ' 2 + 3 2 + 5 (c) 2 + 4,7 Solution
(b) 5 ' (3 + 4) + 2 (d) 2 + [4 + 2 · (10 + 6 ) ]
(a) 8 · 2 + 3 = 16 + 3 i
(b) 5 · ( 3
=
19
M u lti ply first
+
4) + 2 = 5 · 7 + 2 = 35 + 2 = 37 i i Pa rentheses first
M u lti ply before adding
7 2 + 5 2 + 5 ( c) 2 + 4·7 2 + 28 30 (d) 2 + [4 + 2 ' ( 10 + 6 ) ] = 2 + [ 4 + 2 ' ( 1 6 ) ] = 2 + [ 4 + 32] = 2 + [36] = 38 • Figure 7
B e careful if you use a calculator. For Example 10(c), you need to use paren theses. See Figure 7.* If you don't, the calculator will compute the expression
( 2+5 ) / ( 2+4*7 ) . 2333333333 Ans � Fr.ac.
2
+
5 "2 + 4 · 7 = 2 + 2.5 + 28 = 32.5
glvmg a wrong answer. am:: = = >-
4
Now Work
PROBLEMS 57 AND 65
Work with Properties of Rea l N u m be rs
We have used the equal sign to mean that one expression is equivalent to another. Four important properties of equality are listed next. In this list, a, b, and c repre sent real numbers. 1.
The reflexive property states that a number always equals itself; that is, = a. The symmetric property states that if a = b then b = a. The transitive property states that if a = b and b = c then a = c. The principle of substitution states that if a = b then we may substitute b for a in any expression containing a . a
2. 3. 4.
-1
Now, let's consider some other properties of real numbers. We begin with an example. E XA M P L E 1 1
C o m mutative P roperties
(a) 3 + 5 5 + 3 3 + 5
= =
=
8 8 5 +3
(b) 2 · 3 = 6 3·2 = 6 2,3 = 3'2
•
This example illustrates the commutative property of real numbers, which states that the order in which addition or multiplication takes place will not affect the final result. '" Notice that we converted the decimal to i ts fraction form. Consult your manual to see how your calculator does this.
10
CHAPTER R
Review
Commutative Properties
a+b=b+a a-b = boa
(la) (lb)
Here, and in the properties listed next and on pages 1 1-13, a, b, and c represent real numbers. EXAM P L E 1 2
Associative P roperties
(a) 2 + (3 + 4) (2 + 3 ) + 4 2 + (3 + 4)
=
=
=
(b) 2 - (3 - 4) 2 - 12 = 24 (2 - 3 ) - 4 = 6 - 4 = 24 2 - (3 - 4) = (2 - 3) - 4
2 +7 9 5 +4 9 (2 + 3) + 4
=
=
=
•
The way we add or multiply three real numbers will not affect the final result. Expressions such as 2 + 3 + 4 and 3 - 4 - 5 present no ambiguity, even though addi tion and multiplication are performed on one pair of numbers at a time. This prop erty is called the associative property_ Associative Properties
a + (b + c) = (a + b) + c a + b + c a - (b - c) = (a - b) - c = a - b - c =
(2a) (2b)
The next property is perhaps the most important. Distributive Property
a - (b + c) (a + b) - c
=
=
a-b + a-c a-c + b-c
(3a) (3b)
The distributive property may be used in two different ways. E XA M P L E 1 3
D i stributive P roperty
(a) 2 - (x + 3 ) = 2 - x + 2 - 3 = 2x + 6 Use to remove parentheses. (b) 3x + 5x = (3 + 5)x = 8x Use to combine two expressions. (c) (x + 2 ) ( x + 3 ) = x(x + 3 ) + 2(x + 3 ) (x2 + 3x) + (2x + 6) 2 x + (3x + 2x) + 6 = x2 + 5x + 6 =
=
Cl!l!1C = �-
Now Work
q
PROBLEM 87
The real numbers 0 and 1 have unique properties. EXAM P LE 1 4
I dentity P roperties
(a) 4 + 0 = 0 + 4 = 4
(b) 3 - 1 = 1 - 3 = 3
The properties of 0 and 1 illustrated in Example 14 are called the properties_
.. identity
SECTION R .1
Real N umbers
1 1
Identity Properties
O+a=a+O=a a'1 = 1'a = a
(4a) (4b)
We call 0 the additive identity and 1 the multiplicative identity. For each real number a, there is a real number -a, called the additive inverse of a, having the following property: Additive I nverse Property
a + ( -a) = -a + a = 0
E XA M P L E 1 S
( Sa)
F i n d i ng an Additive I nverse
(a) (b )
The additive inverse of 6 is -6, because 6 + ( -6 ) = O. The additive inverse of -8 is - ( -8) = 8, because -8 + 8 = O.
•
The additive inverse of a, that is, -a, is often called the negative of a or the opposite of a. The use of such terms can be dangerous, because they suggest that the additive inverse is a negative number, which may not be the case. For example, the additive inverse of - 3, or ( -3), equals 3, a positive number. -
For each nonzero real number a, there is a real number l, called the multiplicata of a, having the following property:
ive inverse
M ultiplicative Inverse Property
1 1 a'- = -'a = 1 a a
if a
'*
0
(5b)
1
The multiplicative inverse - of a nonzero real number a is also referred to as the a reciprocal of a. E XA M P L E 1 6
F i n d i n g a Reci procal
(a)
The reciprocal of 6 is
�, because 6 � = 1 . .
1 1 - 3 ' because 3 · - 3 = 1 . 2 3 2 3 ( c ) The reciprocal of '3 IS. '2' because '3 '2 = 1 .
(b )
.
The reciprocal of -3
.
IS
.
•
With these properties for adding and multiplying real numbers, we can now define the operations of subtraction and division as follows: DEFINITION
The
difference
a - b, also read "a less b" or "a minus b," is defined as a - b = a + ( -b)
(6)
I�
�--------------------------------�
12
CHAPTER R
Review
To subtract b from a, add the opposite of b to a. a If b is a nonzero real number, the quotient - , also read as "a divided by b" b or "the ratio of a to b," is defined as
DEFINITION
I
a 1 - = a'if b oF 0 (7) b b � � ---------------------------------- .� EXAM P L E 1 7
Working with Diffe rences and Quotients
(a) 8 - 5 = 8 + ( -5 ) = 3 (b) 4 - 9 = 4 + ( -9) = -5 5 1 (c) - = 5 ' 8 8 • r
r
r ,....
For any number a, the product of a times 0 is always 0; that is, In Words
M ultiplication by Zero
The result of m u ltiplying by zero is zero.
a'O = 0
(8)
For a nonzero number a, Division Properties
a - = 1 a
(9)
if a oF O
2 NOTE Division by 0 is not defined. One rea son is to avoid the following difficu lty: find
2 o
x
such that
- = �
o·
x
=
2.
But
o·x
equ a l s 0 for all
x,
o
= x
so there is no u n ique n u m ber
x
means to such that
•
Rules of Signs
a( -b) = - (ab) - ( -a) = a
E XA M P L E 1 8
( -a)b = - ( ab) a -a a -b b b
( -a)( -b) = ab -a a -b b
(10)
Applying the Rules of Signs
(a) 2 ( -3 ) = - ( 2 ' 3 ) = -6 -3 3 3 (c) = -2 = ' 2 -2 I I x (e) = = - 2x x ' _2 -2
(b) ( -3 ) ( -5 ) -4 4 (d) =9 -9
=
3 · 5 = 15
,.
SECTION R.1
Real N umbers
13
Cancellation Properties
ae = be implies a = b if e "* 0 ae a if b "* 0, e "* 0 be b
E XA M P L E 1 9
U s i ng the Cancel l ation P roperties
(a) If 2x = 6, then
NOTE We follow the common prac tice of using slash ma rks to i n d icate
r r r
r
(b)
•
cancellations.
In Words
3 ·S i 18 = = 12 2'S 2 i Cancel
2x = 6 2x = 2 · 3 x =3
Factor 6. Cancel the 2's.
the 6's.
•
Zero-Product Property
If a product equ a l s 0, then one o r
If ab
both o f the factors is O.
EXAM P L E 2 0
= 0, then a = 0, or b = 0, or both.
(12)
U s i ng the Zero- Product Property
If 2x = 0, then either 2 = 0 or x = O. Since 2 "* 0, it follows that x = O. Arithmetic of Quotients
---
a e ad be ad + be -+-=-+-= b d bd bd bd a e ae b d bd a b a d ad - = -.- = e b e be d
E XA M P L E 2 1
(11)
if b "* 0, d "* 0
(13 )
if b "* 0, d "* 0
(14)
if b "* 0, e "* 0, d "* 0
(IS)
Addi ng, Su btracting, M u lti plyi ng, and Dividing Quotients
(a)
(b)
2 5 + 3 2
-
-
= i
3·5 2·2 + 3·2 3·2
-
i
_
�)
By equation (6)
i
=
By equation (13)
� �= �+( _
-
= i
2·2 + 3·5 3·2
4 + 15 6
19 6
� + �2
By equation (10)
3 · 3 + 5 ' ( -2) 5·3 By equation (13)
9 + ( - 10) 15
-1 15
-
1 15
-
•
14
CHAPTER R
Review
NOTE S l a nting the cancellation ma rks i n d ifferent d i rectio n s for d ifferent fac
( c)
tors, as shown here, is a good p ractice
� . 15 3 4
to follow, since it wi l l help in checking for
•
errors.
i
8 · 15 3·4
2. = 2 . 2. "2 5 7 9
=
By equation
3 (d)
=
r
2 · .4" '3- ' 5 '3- ' .4' · 1
=
i
(14)
2·5 = 10 1
By equation
(11)
3·9 27 = i s ' 7 35
=
By equatio n
By equation
(15)
(14) ..
NOTE In writing quotients, we s h a l l follow the usual convention and write the quotient in lowest terms. That is, we write it so that a ny common factors of the n u merator and the denominator have been removed using the cancellation properties, equation
15 15 · ..6 4 ' ..6 4 4 · ..6 · x · )(. 3 · ..6 · )(.
90
24 24>f 18x - Now Work
4x 3
(11).
x
As exa m ples,
•
# 0
PROBLEMS 67, 7 1 , AND 81
Sometimes it is easier to add two fractions using least common multiples (LCM). The LCM of two numbers is the smallest number that each has as a com mon multiple. E XA M P L E 2 2
F i nd i n g the Least Common M u lti ple of Two N u mbers
Find the least common multiple of 15 and 12. Sol ution
To find the LCM of 15 and 12, we look at multiples of 1 5 and 12. 15, 30, 45, 60, 75, 90, 105, 120, . . . 12, 24, 36, 48, 60, 72, 84, 96, 108,
120, . . .
The common multiples are in blue. The least common multiple is 60. •
E XA M P L E 2 3
Using the least Common M u ltiple to Add Two F ractio n s
8 5 F·In d : - + 12 15 Solution
We use the LCM of the denominators of the fractions and rewrite each fraction using the LCM as a common denominator. The LCM of the denominators (12 and 15) is 60. Rewrite each fraction using 60 as the denominator. 8 15
- +
5 12
8 .4 +5 .5 15 4 12 5 32 25 = + 60 60 32 + 25 60 57 60 19 20
=
-
= = = -
Now Work
PROB
l
EM 75
-
•
Real Num bers
S ECTION R . l
15
�is;torical Feature
T
he real n u m be r system h a s a history that stretches back at least
turned away from the n u m be r concept, expressing facts about whole
to the a n cient Babyl o n i a n s { 1 800 Bc). l t is rem a rka ble how much
n u m bers in terms of line segments.
the a n cient Babylonian attitudes resemble o u r own. As we stated
I n astronomy, however, Babylo n i a n methods, including the Babylon
in the text, the fu nda mental difficulty with irrational n u m bers is that
ian n u m ber system, conti nued to be used. Simon Stevin ( 1 548-1 620),
they can not be written a s quotients of integers or, equivalently, as re peating or term i n ating deci mals. The Babyl o n i a n s wrote their n u m bers in a system based o n 60 i n the same way that we write ours based on
1 0. They wou l d carry as m a ny p l a ces for 7T as the accuracy of the prob
lem demanded, just as we now use 7T �
1 37
7T �
or
or
7T �
3.141 6
or
7T �
3 . 1 4 1 592 6 53 58979
probably using the Babyl o n i a n system as a model, invented the decimal
system, complete with rules of calcu lation, in 1 585. [Others, for example,
a l-Kashi of Samarkand (d. 1 429), had made some progress in the same
d i rection.] The decimal system so effectively conceals the difficulties that the need for more logical precision began to be felt only i n the early 1 800s. Around 1 880, Georg Cantor ( 1 845- 1 9 1 8) and Rich ard Dedekind
( 1 83 1 - 1 9 1 6) gave precise definitions of real n u mbers. Cantor's defini
3 . 1 41 59
tion, a lthough more a bstract and precise, has its roots in the decimal (and hence Babylo n ia n ) n u merica l system.
depending on how accu rate we need to be.
Sets and set theory were a spi n-off of the research that went into
Things were very different for the Greeks, whose n u m ber system a l lowed only rational n u m bers. W h e n it was d iscovered that
V2 was
not
cla rifyi ng the fou n d ations of the real n u mber system. Set theory has de veloped into a large disci p l i n e of its own, and many mathematicians re
a rational n u m ber, this was regarded as a fu n d a mental flaw i n the num
gard it as the fou ndation u po n which modern mathematics is built.
ber concept. So serious was the matter that the Pythagorean Brother
Cantor's discoveries that infinite sets can also be counted and that there
hood (an early mathematical society) is s a id to have drowned one of its
are diffe rent sizes of infin ite sets a re among the most astounding results
members for revea l i n g this terrible secret. Greek mathematicians then
of modern mathematics.
H i storical Problem s
The Babylonian n u m ber system was based on 60. Thus 2,30 means 2
+
30
60
2. What are the fo llowing Babylonian n u mbers when written as frac tions a n d as deci m a l s ?
2.5, a n d 4,2 5, 1 4 means
=
4
+
25 60
+
14 2 60
=
4 +
(b) 4,52,30
(a) 2,20
1 51 4 =
3 600
(e) 3,8,29,44
4.42 0 5 5 5 5 5 . .
1 . What a re the fol lowing n u m bers i n Babylo n i a n notation? 1
5
(a) 1 -
(b) 2 6
3
R.l Assess You r Understandi ng Concepts a nd Voca b u lary 1.
The numbers in the set {x I x and b
*'
O } , are called
a
b ' where a, b are integers
=
____
2.
The value of the expression 4
3.
The fact that 2x + 3x Property.
=
+
5. True or False
Rational numbers have decimals that either terminate or are nonterminating with a repeating block of digits.
numbers.
5·6
-
3 is
6. True or False
The Zero-Product Property states that the product of any number and zero equals zero.
_ _ _ _ _
(2 + 3)x is a consequence of the
____
4.
"The product of 5 and x
+
3 equals 6" may be written as
7. True or False
The least common multiple of 12 and 18 is 6.
8. True or False
No real number is both rational and irrational.
Skill B uilding In Problems 9-20, use U
=
universal set
each set.
=
{O, 1, 2, 3, 4, 5, 6, 7, 8, 9 } , A
=
{ I , 3, 4, 5, 9 } , B
=
{2,
4,
6, 7, 8 } , and e
=
AUB
10.
Aue
11.
AnB
12.
Ane
13.
(A U B) n e
14.
(A n B) U e
15.
A
16.
e
17.
AnB
18.
Bue
19.
AUB
20.
Bne
9.
{ I , 3, 4,6 }
to find
16
CHAPTER R
Review
In Problems 21-26, list the numbers in each set that are (a) Natural numbers, (b) Integers, (c) Rational numbers, (d) Irrational numbers,
{ { 2I ' 3I ' 4I } { \12, \12
(e) Real numbers.
21.
A
23.
C
25.
E
_
-
=
=
1 , -6 ' 2 , - 1 .333 . . . (the 3 s repeat ) , ?T, 2, 5
}
0, 1,
?T,
+ 1, ?T +
�}
{ �,
22.
B
-
24.
D
=
26.
F
=
=
2.060606 . . . (the block 06 repeats ), 1 .25, 0, 1, Vs
{ - I , - 1.1 , - 1 .2, - 1 .3 }
{
- \I2 ' ?T +
\12,�
+ 10.3
}
In Problems 2 7-38, approximate each number (a) rounded and (b) truncated to three decimal places.
27. 1 8.9526
28. 25 .86134
33. 9.9985
34. 1 .0006
29. 28.65319 3 35. "7
30. 99.05249 5 36. "9
32. 0.05388 81 38. 5
31. 0.06291 521 37. 15
In Problems 39-48, write each statement using symbols.
39. The sum of 3 and 2 equals 5.
40. The product of 5 and 2 equals 10.
41. The sum of x and 2 is the product of 3 and 4.
42. The sum of 3 and y is the sum of 2 and 2.
43. The product of 3 and y is the sum of 1 and 2.
44. The product of 2 and x is the product of 4 and 6.
45. The difference x less 2 equals 6.
46. The difference 2 less y equals 6.
47. The quotient x divided by 2 is 6.
48. The quotient 2 divided by x is 6.
In Problems 49-86, evaluate each expression.
49. 9 - 4 + 2
50. 6 - 4 + 3
53. 4 + 5 - 8 ' 57. 6 - [3 ' 5 + 2 ' (3 - 2 ) ]
51 . -6 + 4 · 3
52. 8 - 4 · 2
54. 8 - 3 - 4
1 55. 4 + 3
1 56. 2 - 2
58. 2 · [8 - 3 (4 + 2 ) ] - 3
59. 2 · (3 - 5 ) + 8 · 2 - 1
60. 1 - (4 ' 3 - 2 + 2 )
"
61. 10 - [6 - 2 · 2 + (8 - 3 ) J · 2 1
62. 2 - 5 · 4 - [6 · (3 - 4) J
63. (5 - 3 ) 2
1 64. (5 + 4) 3
65.
4 + 8 5 - 3
66.
2 - 4 5 - 3
3 10 67 . 5 ' 21
5 3 68. - . 9 10
69.
� 10 . 25 27
70.
21 . 100 25 3
3 2 71. 4 + 5
4 1 72. - + 3 2
5 9 73. (5 + 5
15 8 74. - + 9 2
5 1 75. - + 12 18
2 8 76. - + 15 9
1 7 77. - - 18 30
3 2 78. - - 21 14
3 2 79. - - 15 20
6 3 80. - - 35 14
81. 11
4 1 2 84. - + - ' 3 5 6
3 3 85. 2 · - + 4 8
1 3 7 83. - . - + 2 5 10
5 18
5 82. � 2 35
27
5 1 86. 3 · - - 6 2
1n Problems 87-98, use the Distributive Property to remove the parentheses.
", 87. 6(x + 4)
(
91. 2 � r - 1:. 4� 2
)
95. (x - 2)(x + 1 )
88. 4(2x - 1 )
(
92. 3 �X + 1:. 3 6 96.
(x
)
- 4 ) (x + 1 )
89. x(x - 4) 93.
(x
+ 2)( x + 4 )
97. ( x - 8 ) (x - 2)
90. 4x(x + 3) 94.
(x
+ 5 ) (x + 1 )
98. ( x - 4 ) ( x - 2)
}
SECTION R.2
Algebra Essentials
17
Discussion and Writing 99. 100.
Explain to a friend how the D istributive Property is used to justify the fact that 2x + 3 x = 5x. Explain to a friend (2 + 3 ) ' 4 = 20.
why
2 + 3·4
=
14, whereas
101.
Explain why 2(3 . 4) is not equal to (2 · 3) . (2 ' 4).
102.
Explam why 2
103.
Is subtraction commutative? Support your conclusion with an example.
104. 105. 106. 107. 108.
4+3. 4 3 5 I S not equal to 2 + "5 ' +
.
109.
110. 111.
Is subtraction associative? Support your conclusion with an example. Is division commutative? Support your conclusion with an example. Is division associative? Support your conclusion with an example. If 2
=
x, why does x
If x
=
5, why does
=
x2
+
2? x =
3 0?
112. 113.
Are there any real numbers that are both rational and irra tional? Are there any real numbers that are neither? Explain your reasoning. Explain why the sum of a rational number and an irrational number must be irrational. A rational number is defined as the quotient of two integers. When written a s a decimal, the decimal will either repeat or terminate. By looking at the denominator of the rational number, there is a way to tell in advance whether its decimal representation will repeat or terminate. Make a list of ratio nal numbers and their decimals. See if you can discover the pattern. Confirm your conclusion by consulting books on number theory at the library. Write a brief essay on your findings. The current time is 12 noon CST. What time (CST) will it be 1 2 ,997 hours from now? a . 0 . Both 0 ( a "* 0 ) and 0 are undefmed, but for dIfferent reasons. Write a paragraph or two explaining the different reasons.
R.2 Algebra Essentials OBJECTIVES
G ra p h I n e q u a l ities (p.
1 8)
2 Find Distance on the Real N u m ber Line (p. 1 9) 3 Eva l u ate Algebra ic Expressions (p. 20)
4 Determine the Domain of a Va riable (p. 2 1 ) 5 Use the Laws of Exponents (p. 2 1 )
6 Eva l u ate S q u are Roots (p. 23)
7 Use a Ca l c u l ator to Eva l u ate Exponents (p. 24)
8 Use Scie ntific Notation (p. 24)
The Rea l N u m ber Line
Figure 8
Rea l n u m ber l i n e
I(
2 units
Scale
�
-
3
-2
1
-1
I
-!
1 Unit o I
0
I
!
I I
•
I
1� 2
II
3TI
J
The real numbers can be represented by points on a line called the real number line. There is a one-to-one correspondence between real numbers and points on a line. That is, every real number corresponds to a point on the line, and each point on the line has a unique real number associated with it. Pick a point on the line somewhere in the center, and label it O. This point, called the origin, corresponds to the real number O. See Figure 8. The point 1 unit to the right of 0 corresponds to the number 1. The distance between 0 and 1 deter mines the scale of the number line. For example, the point associated with the number 2 is twice as far from 0 as 1 . Notice that an arrowhead on the right end of the line indicates the direction in which the numbers increase. Points to the left of the origin correspond to the real numbers - 1 , -2, and so on. Figure 8 also shows the points associated with the rational numbers numbers V2 and 7T.
DEFINITION
-
� and � and with the irrational
TIle real number associated with a point P is called the coordinate of P, and the line whose points have been assigned coordinates is called the real number line'--1 1l'I!lilI 1mII =-
Now Work
PROBLEM 1 1
18
CHAPTER R
The real number line consists of three classes of real numbers, as shown in Figure 9.
Figure 9
-3
[
-
Review
!
!
2 -� -1
I -
�
o !
O
I
�
[
1
I
J
�
2
I
I
3
1. The negative real numbers are the coordinates of points to the left of the origin O.
I
-y-------' • '------v--
Negative rea l n u mbers
I
The real number zero is the coordinate of the origin O. 3. The positive real numbers are the coordinates of points to the right of the origin O.
Positive
Zero
2.
rea. n u mbers
Negative and positive numbers have the following multiplication properties: M u ltiplication Properties of Positive and Negative Numbers
The product of two positive numbers is a positive number. The product of two negative numbers is a positive number. 3. The product of a positive number and a negative number is a negative number. 1.
2.
1
Figure 1 0
<9
<9
a
(a) a < b
b
Oil
a b (b) a
=
b
<9
b
9
(c) a > b
a
E XA M P L E 1
G ra p h I nequalities
An important property of the real number line follows from the fact that, given two numbers (points) a and b, either a is to the left of b, or a is at the same location as b, or a is to the right of b. See Figure 10. If a is to the left of b, we say that "a is less than b" and write a < b. If a is to the right of b, we say that "a is greater than b" and write a > b. If a is at the same loca tion as b, then a = b. If a is either less than or equal to b, we write a :::; b. Similarly, a 2': b means that a is either greater than or equal to b. Collectively, the symbols < , > , :::; , and 2': are called inequality symbols. Note that a < b and b > a mean the same thing. It does not matter whether we write 2 < 3 or 3 > 2. Furthermore, if a < b or if b > a, then the difference b - a is positive. Do you see why? U s i ng I nequality Sym bols
(a) 3 < 7 (d) -8 < -4
(b) -8 > - 16 (e) 4 > - 1
(c) - 6 < 0 (f) 8 > 0 •
In Example l (a), we conclude that 3 < 7 either because 3 is to the left of 7 on the real number line or because the difference, 7 - 3 = 4, is a positive real number. Similarly, we conclude in Example l (b) that -8 > -16 either because -8 lies to the right of -16 on the real number line or because the difference, -8 - ( -16 ) = -8 + 16 = 8, is a positive real number. Look again at Example 1. Note that the inequality symbol always points in the direction of the smaller number. Statements of the form a < b or b > a are called strict inequalities, whereas statements of the form a :::; b or b 2': a are called nonstrict inequalities. An inequali1iY is a statement in which two expressions are related by an inequality symbol. The expressions are referred to as the sides of the inequality. B ased on the discussion so far, we conclude that a > 0 is equivalent to a is positive a < 0 is equivalent to a is negative
SECTION R.2
We sometimes read a > 0 by saying that "a is positive." If a or a = 0, and we may read this as "a is nonnegative." '-'''"
;z;...-
Now Work
2:
19
Algebra Essentials
0, then either a > 0
PROBLEMS 1 5 AND 25
We shall find it useful in later work to graph inequalities on the real number line. G rap h i n g I nequalities
EXA M P L E 2
(a) On the real number line, graph all numbers x for which x > 4. (b) On the real number line, graph all numbers x for which x :::; 5. (a) See Figure 11. Notice that we use a left parenthesis to indicate that the number 4 is not part of the graph. (b) See Figure 12. Notice that we use a right bracket to indicate that the number 5 is part of the graph.
Solution Figure 1 1
-2 - 1
0
2
3
� 4
Figure 1 2 If I
-2 - 1
0
I
2
3
I
3 5
6
�. 7
• 6
I
7
I
t;I!l!!: = =--
2
Figure 1 3
_
4
5
4 units
•
• -5 - 4 -3 -2 - 1
1_ 3
• 0
1
u n its 2
.1
E!; 3
I.
4
DEFINITION
Now Work
PROBLEM 3 1
Find D ista n ce on t h e Rea l N u m ber Line
The absolute value of a number a is the distance from 0 to a on the number line. For example, -4 is 4 units from 0, and 3 is 3 units from O. See Figure 13. Thus, the absolute value of -4 is 4, and the absolute value of 3 is 3. A more formal definition of absolute value is given next.
The absolute value of a real number a, denoted by the symbol l a l , is defined by the rules
lal = a
if a
2:
0
and
l a l = -a
if a
<
0
I
--�----------�--------��
�
For example, smce -4 - ( -4) = 4. EXA M P L E 3
<
0, the second rule must be used to get 1 -41 =
C o mputing Absol ute Val ue
(a) 181 = 8
(b) 101 = 0
(c) 1 - 1 5 1 =
-
( 15 ) = 15 -
•
Look again at Figure 13. The distance from -4 to 3 is 7 units. This distance is the difference 3 - ( -4 ) , obtained by subtracting the smaller coordinate from the larger. However, since 13 - ( -4 ) 1 = 171 = 7 and 1 -4 - 31 1 -7 1 = 7 , we can use absolute value to calculate the distance between two points without being con cerned about which is smaller. =
DEFINITION
If P and Q are two points on a real number line with coordinates a and b respectively, the distance between P and Q, denoted by d ( P, Q), is
I
d ( P, Q) = I b - a l ----------------------��
�
Since Ib - a l = l a - b l , it follows that d ( P , Q) = d (Q, P ) .
20
CHAPTER R
Review
F i n d i n g Distance on a N u m be r Line
EXA M P L E 4
Let P, Q, and R be points on a real number line with coordinates -5, 7, and -3, respectively. Find the distance (b) between Q and R
(a) between P and Q See Figure 14.
Sol ution
R
P
Figure 1 4
..
-5
-4
.. -3 -2
-1
2
o
1..-<---- d(P, 0)
l- d(O, R)
=
=
3
1 7 - ( - 5) 1 1
-3
-7 1
5
4 =
=
6
o ..
7
I
1 2 -1
1 0 -1
(a) d ( P, Q) = 1 7 - ( -5 ) 1 1 12 1 = 12 (b) d ( Q, R ) = 1 -3 - 7 1 = 1 - 101 = 10 =
q l =>;> -
3
Now Work
•
PROBLEM 37
Eva l u ate Algebra ic Expressions
Remember, in algebra we use letters such as x, y, a, b, and c to represent numbers. If the letter used is to represent any number from a given set of numbers, it is called a variable. A constant is either a fixed number, such as 5 or V3, or a letter that rep resents a fixed (possibly unspecified) number. Constants and variables are combined using the operations of addition, sub traction, multiplication, and division to form algebraic expressions. Examples of algebraic expressions include
3 1 -t
x+3
7x - 2y
To evaluate an algebraic expression, substitute for each variable its numerical value. E XA M P L E 5
Evalu ating an Algebraic Expression
Evaluate each expression if x = 3 and y = - l . (a) x + 3 y
Sol ution
(b) 5xy
(c)
3y 2 - 2x
(d) 1 -4x + y l
(a) Substitute 3 for x and - 1 for y in the expression x + 3y.
x + 3y x =
(b) If x = 3 and y ( c) If x
(d) If x
=
=
i
3, y
3 + 3 ( - 1 ) = 3 + ( - 3) =
=
0
-1
-1, then 5xy = 5 ( 3 ) ( - 1 ) = -15 3 and y = -1, then =
3( -1) 3y 2 - 2x 2 - 2 ( 3) 3 and y = - 1 , then 1 - 4x + y l
tlJ! �-
=
Now Work
=
-3 2-6
-3 -4
3 4
1 - 4( 3) + ( - 1 ) 1 = 1 - 12 + ( - 1 ) 1 = 1 - 13 1 = 13 ..
PROBLEMS 39 AND 47
SECTION R.2
4
Algebra Essentials
21
Dete r m i n e the Domain of a Va riable
In working with expressions or formulas involving variables, the variables may be allowed to take on values from only a certain set of numbers. For example, in the formula for the area A of a circle of radius r, A = 7Tr2, the variable r is necessarily restricted to the positive real numbers. In the expression 2., the variable x cannot x take on the value 0, since division by 0 is not defined. The set of values that a variable may assume is called the
DEFINITION
domain of the
variable.
F i nd i n g the Domain of a Variable
EXAM P L E 6
The domain of the variable x in the expression 5 x - 2 is {xix
=1=
2 } , since, if x
=
2, the denominator becomes 0, which is not defined .
•
C i rc u mference of a C ircle
E XA M P L E 7
In the formula for the circumference C of a circle of radius
r,
C = 2 7Tr the domain of the variable r, representing the radius of the circle, is the set of pos itive real numbers. The domain of the variable C, representing the circumference of the circle, is also the set of positive real numbers. •
In describing the domain of a variable, we may use either set notation or words, whichever is more convenient. � = = -
5
Now Work
PROBLEM 57
Use the Laws of Exponents
Integer exponents provide a shorthand device for representing repeated multiplica tions of a real number. For example, 34
•
•
= 3·3.3 3 .
=
81
Additionally, many formulas have exponents. For example, The formula for the horsepower rating H of an engine is
where D is the diameter of a cylinder and N is the number of cylinders. A formula for the resistance R of blood flowing in a blood vessel is
where L is the length of the blood vessel, constant.
r
is the radius, and C is a positive
22
CHAPTER R
Review
DEFINITION
If a is a real number and n is a positive integer, then the symbol a " represents the product of n factors of a. That is,
a" = a · a · ... · a
(1)
'----v-----' n factors
Here it is understood that a1 = a. Then a2 = a · a, a3 = a · a . a, and so on. In the expression all, a is called the base and n is called the exponent, or power. We read all as "a raised to the power n" or as "a to the nth power." We usually read a2 as "a squared" and a3 as "a cubed." In working with exponents, the operation of raising to a power is performed before any other operation. As examples, 4 . 32 = 4 · 9 = 36
-24 = -16
22 + 32 = 4 + 9 = 13 5 . 32 + 2 · 4 = 5 . 9 + 2 · 4 = 45 + 8 = 53
Parentheses are used to indicate operations to be performed first. For example,
( _2 ) 4 DEFINITION
=
( - 2 ) ( - 2 ) ( -2 ) ( -2) = 16
If a *- 0, we define
I�
aD = 1 if a *- O
� ---------�
DEFINITION
If a *- O and if n is a positive integer, then we define
I
1 a-n = -n if a *- O a
�
�------------�
Whenever you encounter a negative exponent, think "reciprocal."
E XA M P L E 8
Eval uating Expressions Contai n i ng Negative Exponents
(a) T3 - .l - � 23 - 8
= (i!rr: ::; =�
Now Work
(b) x-4 =
.l x4
1 -2 (c) ( 5 ) -
_
(�y 1
;
1 _ - 25 5 ..
PROBLEMS 75 AND 95
The following properties, called the Laws of Exponents, can be proved using the preceding definitions. In the list, a and b are real numbers, and m and n are integers. THEOREM
Laws of Exponents
(am ) " = all1" am 1 = al11-n = n - In if a *- O all a
(ab )" = a"b" 1 ( a ) 1 = a" if b *- 0 b bl1
SECTION
EXAMPLE 9
23
R.2 Algebra Essentials
X35 x X X-3·2 x-6 x6
Using the Laws of Exponents
( a)
x-3 x5
(b)
(x-3/
•
fiI!l!];;==_ -
EXAMPLE 1 0
=
=
-
+
2
=
=
=F 0
=
Now Work
l:.- x =F 0
• PROBLEM 77
Using the Laws of Exponents
Write each expression so that all exponents are positive. (b)
(3Xy-�1 t
2
x
=F 0,
Y =F 0
Solution
• I-
6
Now Work
PROBLEM 87
Eva l u ate Squ a re Roots
A real number is squared when it is raised to the power 2. The inverse of squaring is finding a square root. For example, since 62 = 36 and (-6)2 = 36, the numbers 6 and 6 are square roots of 36. The symbol .y, called a radical sign, is used to denote the principal, or non negative, square root. For example, V36 = 6. -
DEFINITION
If a is a nonnegative real number, the nonnegative number b, such that b2 is the principal square root of a, is denoted by b = Va.
=
a
-.J
The following comments are noteworthy: 1. Negative numbers do not have square roots (in the real number system), be
cause the square of any real number is nonnegative. For example, v'=4 is not a real number, because there is no real number whose square is -4. 2. The principal square root of 0 is 0, since 02 = O. That is, YO = O. 3. The principal square root of a positive number is positive.
4. If c EXAMPLE 11
2::
0, then (vc?
= c.
For example, (V2? =
2 and
(V3?
=
3.
Evaluating Square Roots
(a)
V64 =
8
(b)
0= 1 'f16 "4
( c)
(V1.4y = 1.4
•
24
CHAPTER R
Review
()
Examples l1(a) and (b) are examples of square roots of perfect squares, since 1 1 2 64 82 and 16 "4 . Consider the expression W. Since a2 2: 0, the principal square root of a2 is defined whether a > 0 or a < O. However, since the principal square root is non negative, we need an absolute value to ensure the nonnegative result. That is,
=
=
W= lal
a any real number
(2)
Using Equation ( 2)
EXAMPLE 1 2
V(2.3)2 12.31 2.3 (b) V( -2.3)2 = 1-2.31 2.3 (c) # = Ixl (a)
=
=
=
� Now Work
7
a
P R O B L E M 83
Use a Calcu lator to Eva l u ate Exponents
G or an [2] key, which is used for computa
Your calculator has either a caret key, tions involving exponents.
II
EXAM PLE 1 3
I
Exponents on a Graphing Calculator
(2.3)5
Evaluate:
Figure 15 shows the result using a T I-84 graphing calculator.
Solution Figure 15
�=>
64.36343
8
-
Now Work
a
PROBLEM 1 13
Use Scientific Notation
Measurements of physical quantities can range from very small to very large. For exa mple, the mass of a proton is approximately 0.00000000000000000000000000167 kilo gram and the mass of Earth is about 5,980,000,000,000,000,000,000,000 kilograms. These numbers obviously are tedious to write down and difficult to read, so we use exponents to rewrite each. DEFINITION
W hen a number has been written as the product of a number x, where 1 ::; x < 10, times a power of 10, it is said to be written in scientific notation,...]
In scientific notation, Mass of a proton Mass of Earth
= 1.67 X 10-27 kilogram
=
5.98 X 1024 kilograms
SECTION R.2 Algebra Essentials
25
Conve rting a Deci mal to Scientific Notation
To change a positive number into scientific notation: 1. Count the number N of places that the decimal point must be moved to arrive at a number x, where 1 � x < 10. 2. If the original number is greater than or equal to 1, the scientific notation
is x X ION. If the original number is between 0 and 1, the scientific no tation is x X lO-N.
EXAMPLE 14
Using Scientific Notation
Write each number in scientific notation. (a) 9582 Solution
(b) 1.245
(c) 0.285
(d) 0.000561
(a) The decimal point in 9582 follows the 2. We count left from the decimal point 9 5 8 2 t t t 3
2
1
=
9.582 X 103
stopping after three moves, because 9.582 is a number between 1 and 10. Since 9582 is greater than 1, we write 9582
(b) The decimal point in 1.245 is between the 1 and 2. Since the number is already between 1 and 10, the scientific notation for it is 1.245 X 10° = 1.245. (c) The decimal point in 0.285 is between the 0 and the 2. We count o
2 8 5
4-J'
.
stopping after one move, because 2.85 is a number between 1 and 10. Since 0.285 is between 0 and 1, we write 0.285 = 2.85 X 10-1 (d) The decimal point in 0.000561 is moved as follows: 0 . 0 0 0 5 6 1 I) t t t t 1
As a result,
t;m::;== -
EXAM PLE 1 5
2
4
0.000561 = 5.61 X 10-4
Now Work
•
PROBLEM 1 1 9
Changing from Scientific Notation to Decimals
W rite each number as a decimal. (b) 3.26 X 10-5 (a) 2.1 X 104 Solution
3
(a) 2.1 X 104
=
(b) 3.26 X 10-5
2
=
(c) 1 X 10-2 = 0
� == >'--
1
I)
0
t
t
0
Now Work
2
0
1:
5
1
t t
0 2
0 4
(I
t 1:
0 3
0 3
t 1:
0 4
0 2
t 1:
(c) 1 X 10-2
X 104 = 21,000 3
(I
2
6 X 10-5
=
0.0000326
o X 10-2 = 0.01
PROBLEM 1 2 7
•
26
CHAPTER R
Review Using Scientific Notation
EXAMPLE 16
(a) The diameter of the smallest living cell is only about 0.00001 centimeter (cm).* Express this number in scientific notation. (b) The surface area of Earth is about 1.97 X 108 square miles.t Express the sur face area as a whole number. (a) 0.00001 cm = 1 X 10-5 cm because the decimal point is moved five places and the number is less than 1. (b) 1.97 X 108 square miles = 197,000,000 square miles.
Solution
•
Now Work
== "L.'l!l:
COMMENT
P R O B L E M 1 53
On a calculator, a n u m ber such as
3.615 X 1012 is usually displayed as 13.615E12·1
•
'" Powers of Ten, Philip and Phylis Morrison.
t1998 Informalion Please Almanac.
f.-li�toriC81 Feature al-jabr. This
number i s added t o one side o f a n equation, then i t must also b e added
word is a part of the title of a ninth century work, "Hisab al-jabr
to the other side in order to "restore" the equality. The title of the work,
T
he word
algebra
I
is derived from the Arabic word
w'al-muqabalah," written by Mohammed ibn Musa al-Khowarizmi.
The word
freely translated, is "The Science of Reduction and Cancellation." Of
al-jabr means "a restoration," a reference to the fact that, if a
course, today, algebra has come to mean a great deal more.
R.2 Assess Your Understanding Concepts a n d Vocabulary
1.
A(n)
7.
is a letter used in algebra to represent any
number from a given set of numbers.
2.
On the real number line, the real number zero is the coordinate of the
8.
3. An inequality of the form
a
>
b
is called a(n)
9.
____
inequality. and 4 is called the
5. 6.
2\
True or False
True or False
The absolute value of a real number is always
the number
2
is called the
1234.5678
When a number is expressed in scientific no
and a power of 10.
____
10. =
Trite or False
tation, it is expressed as the product of a number x, 0 :s x < 1,
____
In scientific notation,
The distance between two distinct points on
greater than zero.
___ _
4. In the expression
Trite or False
the real n umber line is always greater than zero.
___ _
True or False
To multiply two expressions having the same
base, retain the base and multiply the exponents.
The product of two negative real numbers is
always greater than zero.
Skill B u i lding
11.
On the real number line, label the points with coordinates 0, 1, -1,
12.
Repeat Problem
In Problems 13.
18.
�
13-22,
?0
v2? 1.41
11
for the coordinates
0,
-2,2, -1.5,
replace the question mark by 14.
5
?6
19.
�?
0.5
<, >,
or
�,�, � and
=,
whichever is correct.
.15. - 1 ? -2 20.
�, -2 . 5 , �, and 0.25. 2 4
�
? 0.33
16. -3?. - 2 5
2 21. "3?.0. 67
17.
7T? 3.14
22.
�
? 0.25
SECTION R.2 Algebra Essentials
In Problems 23-28, write each statement as an inequality. 24. z is negative 23. x is positive 26.
Y is greater than-5
27
. 25. x i s less than 2
27. x is less than or equal to 1
In Problems 29-32, graph the numbers x on the real number line. 29. x � -2 30. x < 4
28. x is greater than or equal to 2
-1
31. x>
32. x:=:;7
In Problems 33-38, use the given real number line to compute each distance. A
33.
d(C, D )
B
C
-4 -3 -2 -1
0
34. d(C, A)
D
35. d(D,E)
In Problems 39-46, evaluate each expression if x=-2 and y= 3. 39. x+2y 40. 3x+y 43.
2x x-y
-
44.
x+y x-y
-
E 2
3
4
hl y
.37. d(A, E)
41. 5xy +2 45.
53. 1 4x- 5yl
6
36. d(C, E)
3x + 2y 2+Y
54. 1 3x+ 2yl
38. d(D, B)
42. -2x+xy 46.
---
In Problems 47-56, find the value of each expression if x= 3 and y= -2. 47. Ix+yl 48. Ix - yl 49. Ixl+ Iyl 52.
5
2x- 3
--
Y
50. Ixl- Iyl
Ixl 51. x
55. 11 4xl- 15yll
56. 31xl+2 1yl
In Problems 57-64, determine which of the valuers)(a)through (d),if any, must be excluded from the domain of the variable in each expression: (a)x= 3 (c)x=O (b)x=1 (d)x= -1 x x x2- 1 x2+1 59. -2- 60. -1 57. - 58. -x 9 x-+ 9 x x 62.
x3 x -I
2-
63.
x2+ 5x- 10 x3-x
-----
64.
In Problems 65-68, determine the domain of the variable x in each expression. x 4 67. - 65. - 66. � x+ 4 x-5 x+4
68.
-9x2-X+ 1 ----:;, -- x�+x
X -? x-6
--
%
In Problems 69-72, use the formulaC = (F - 3 2) for converting degreesFahrenheit into degreesCelsius to find theCelsius measure of eachFahrenheit temperature. 72. F = -4° 70. F =212° 69. F = 3 2° 71. F = 77° In Problems 73-84, simplify each expression. 75. 4-2 74. -42 73. (-4? 81. v2s
76. -4-2 82. V36
83.
�
84.
�
In Problems 85-94, simplify each expression. Express the answer so that all exponents are positive. Whenever an exponent is 0 or negative, we assume that the base is not O. x2l 89. --4 86. (-4x2rl 85. (8x3)2 xy 93.
(3X-1 )-2 4y-1
94.
(5[2)-3 6y-2
28
CHAPTER R Review
In Problems 95-106, find the value of each expression if x 96. -3[1 Y ,. 95. 2xy-l
99. (xy) 2
100. (x
+
=
2 and y
=
-l.
97. 101.
y?
x2 + l
0
103.
yx2 + l
107.
Find the value of the expression
2x3 - 3x2 + 5 x - 4 if x 2. What is the value if x
lOS.
Find the value of the expression
4x3 + 3x2 - + 2 if x
109.
What is the value o f
104.
# + Vl
106. yt
105. xl' =
X
=
1. What is the value if
--?
(666)4 (222)4
110.
x
=
=
I?
2?
What is the value of
(0.1 )3(20)3?
In Problems 111-118, use a calculator to evaluate each expression. Round your answer to three decimal places. 114. (2.2f5 111. (8.2) 6 113. (6.1) -3 112. (3.7)5
115. "
(-2.8) 6
116.
- (2.8) 6
In Problems 119-126, write each number in scientific notation. 119. 454.2 120. 32.14
123.
32,155
124.
21,210
In Problems 127-134, write each number as a decimal. 127. 6.15 x 104 12S. 9.7 x 103
131.
1.1 X
108
132.
4.112 x 102
117.
(-8.11f4
11S.
-(8.1 1f4
121.
0.013
122.
0.00421
125.
0.000423
126.
0.0514
129.
1.214 x 10-3
130.
9.88 X
133.
8.1 x 10-2
134.
6.453 x 10-1
10-4
Appl ications a n d Extensions
1n Problems 135-144, express each statement as an equation involving the indicated variables. 139. Area of an Equilateral Triangle The area A of an equilat135. Area of a Rectangle The area A of a rectangle is the product of its length I and its width w. V3 eral triangle is - times the square of the length x of one
� �w
136.
Perimeter of a Rectangle
137.
Circllmference of a Circle
13S.
7T
Area of a Triangle
product of its base
P of a rectangle is w.
The perimeter
twice the sum o f its length I and its width
is the product o f
4
side.
T h e circumference
and its diameter
d.
g
x
C of a circle
140.
141.
Perimeter of an Equilateral Triangle
The perimeter
eq uilateral triangle is
3 times the length
Volume of a Sphere
The volume V o f a sphere is
times the cube of the radius
x
P of an
of one side.
r.
� times 7T 3
The area A of a triangle is one-half the
b and its height h.
b
142.
Surface Area of a Sphere
4 times
7T
The surface area
times the square of the radius
r.
S of a sphere is
SECTION R.2 Algebra Essentials
143.
The volume V of a
Volume of a Cube
0-0'
cube is the cube of the length x of a side.
of this stated radius are acceptable. If x is the radius of a ball bearing, a formula describing this situation is Ix - 31
x
144.
The surface area
Surface Area of a Cube
S of a
cube is
6 times the square of the length x of a side.
145.
Manufacturing Cost
where the variable
146.
The weekly production cost
ufacturing x watches is given by the formula
C
=
152.
+
=
2.999 acceptable?
(b) Is a ball bearing of radius x
=
2.89 acceptable?
Body Temperature
�
1.5
(a) What is the cost of producing 1000 watches?
(a) Show that a temperature of 97°F is unhealthy.
(b) What is the cost of producing 2000 watches?
(b) Show that a temperature of 100°F is not unhealthy.
At the beginning of the month,
Balancing a Checkbook
and the other for $32. He was also assessed a monthly ser vice charge of $5. What was his balance at the end of the month?
In Problems 147 and 148, write an inequality using an absolute value to describe each statement. 147. x is at least 6 units from 4.
153.
x is more than 5 units from 2. U.S. Voltage
154. 155.
Ix - 1101
s
156.
158.
Diameter of an Atom
159.
Diameter of Copper Wire
The smallest commercial copper
Smallest Motor
The smallest motor ever made is less than
Astronomy
One light-year is defined by astronomers to be
the distance that a beam of light will travel in 1 year (365 days). If the speed of light is 186,000 miles per second, how many miles are in a light-year? Express your answer in
x
to differ from normal by at most 8 volts. A formula that describes this is
scientific notation.
160.
Astronomy
8
Earth? Express your answer in seconds, using scientific notation.
(b) Show that a voltage of 209 volts is not acceptable. The F ireBall Company
161.
Does
162.
Does
manufactures ball bearings for precision equipment. One of its products is a ball bearing with a stated radius of 3 cen timeters (cm). Only ball bearings with a radius within 0.01 cm
How long does it take a beam of light to reach
Earth from the Sun when the Sun is 93,000,000 miles from
(a) Show that a voltage of 214 volts is acceptable.
Making PI'ecision Ball Bearings
The diameter of an atom is about
1 X 10-10 meter.* Express this diameter as a decimal.
notation.
5
voltage is 220 volts. It is acceptable for the actual voltage
s
The wavelength of visible
using scientific notation.
In other countries, normal household
- 2201
Wavelength of V isible Light
light is about 5 X 10-7 meter.* Express this wavelength as a
wire is about 0.0005 inch in diameter.T Express this diameter
(b) Show that a voltage of 104 volts is not acceptable.
Ix
The height of Mt. Everest is 8872 me
0.05 centimeter wide.t Express this width using scientific
(a) Show that a voltage of 108 volts is acceptable.
Foreign Voltage
Height of Mt. Everest
decimal.
In the United States, normal household volt
this is
The distance from Earth
ters: Express this height in scientific notation.
157.
fer from normal by at most 5 volts. A formula that describes
Distance from Earth to Its Moon
to the Moon is about 4 X 108 meters.* Express this distance
as a whole number.
age is 110 volts. It is acceptable for the actual voltage x to dif
151.
Normal human body temperature is
Ix - 98.61
the next month, he deposited $80, wrote a check for $120,
150.
(a) Is a ball bearing of radius x
C is in dollars.
made another deposit of $25, wrote two checks: one for $60
149.
0.01
1.5°F is considered unhealthy.A formula that describes this is
2x,
Mike had a balance of $210 in his checking account. During
148.
s
98.6°F. A temperature x that differs from normal by at least
C of man
4000
29
� .J
�
equal 0.333? If not, which is larger? By how much?
equal 0.666? If not, which is larger? By how much?
Discussion and Writing
163. 164.
Is there a positive real number "closest" to O? Number game
165.
I'm thinking of a number! It lies between 1
to"
and 10; its square is rational and lies between 1 and 10. The number is larger than
7T.
Correct to two decimal places (that
is, truncated to two decimal places) name the number. Now think of your own number, describe it, and challenge a fellow student to name it.
Morrison. 1998 Information Please Almanac.
':' Powers of Ten, Philip and Phylis t
Write a brief paragraph that illustrates the similarities and differences between "less than"
166.
(s).
«) and "less than or equal
Give a reason why the statement 5 < 8 is true.
30
CHAPTER R
Review
R.3 Geometry Essentia ls OBJECTIVES
1 Use the Pythagorean T h eorem and Its Converse (p. 30) 2
Know Geometry Formulas (p. 31)
3
Understand Congruent Triangles and Similar Triangles (p. 32)
In this section we review some topics studied in geometry that we shall need for our study of algebra.
1
Figure 16
b Leg a
Leg
PYTHAGOREAN THEOREM
Use the Pythagorean Theorem a n d Its Converse
The Pythagorean Theorem is a statement about right triangles. A right triangle is one that contains a right angle, that is, an angle of 90°. The side of the triangle opposite the 90° angle is called the hypotenuse; the remaining two sides are called legs. In Figure 16 we have used c to represent the length of the hypotenuse and a and b to represent the lengths of the legs. Notice the use of the symboli to show the 90° angle. We now state the Pythagorean Theorem. In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. That is, in the right triangle shown in Figure 16, (1)
I�
�--------------------------------�.
A proof of the Pythagorean Theorem is given at the end of this section. EXAMPLE 1
Finding the Hypotenuse of a Right Triangle
In a right triangle, one leg has length 4 and the other has length 3. What is the length of the hypotenuse? Solution
Since the triangle is a right triangle, we use the Pythagorean Theorem with a = 4 and b = 3 to find the length c of the hypotenuse. From equation (1), we have
c2 = a2 + b2 c2 = 42 + 32 = 16 c = \I25 = 5 � ==> .-
Now Work
+ 9
= 25 •
P R O BLE M 1 3
The converse of the Pythagorean Theorem is also true. CONVERSE OF THE PYTHAGOREAN THEOREM
In a triangle, if the square of the length of one side equals the sum of the squares of the lengths of the other two sides, the triangle is a right triangle. The 90° angle is opposite the longest side.
�
A proof is given at the end of this section. EXAMPLE 2
Verifying That a Triangle Is a Right Tri angle
Show that a triangle whose sides are of lengths 5,12, and 13 is a right triangle. Iden tify the hypotenuse. Solution
We square the lengths of the sides.
52
=
25,
122 = 144,
SECTION R.3 Geometry Essentials Figure 17
31
Notice that the sum of the first two squares (25 and 144) equals the third square (169). Hence, the triangle is a right triangle. The longest side, 13, is the hypotenuse. See Figure 17.
5
"i!=,-.
Now Work
•
PROBLEM 2 1
12
EXAMPLE 3
Applying the Pythagorean Theorem
Excluding antenna, the tallest inhabited building in the world is Taipei 101 in Taipei, Taiwan. If the indoor observation deck is 1437 feet above ground level, how far can a person standing on the observation deck see ( with the aid of a telescope)? Use 3960 miles for the radius of Earth. See Figure 18.
Figure 18
Source: Council o n Tall Buildings and Urban Habitat, 2006.
Solution From the center of Earth, draw two radii: one through Taipei 101 and the other to the farthest point a person can see from the observation deck. See Figure 19. Apply the Pythagorean Theorem to the right triangle.
Since 1 mile
=
5280 feet, then 1437 feet =
d2 + (3960)2 d2
=
=
( 3960 ( 3960
+ +
) )
1437 2 5280 1437 2 5280
��!� miles. So we have (3960? :::::: 2155.57
d :::::: 46.43 A person can see about 46 miles from the observation tower. Figure 19
d
1437 It
• 1l'I!>:=="'"
2
Now Work
P R O B L E M 53
Know Geometry Form u las
Certain formulas from geometry are useful in solving algebra problems. We list some of these next. For a rectangle of length l and width w, Area = lw For a triangle with base
b
b
Perimeter
and altitude Area =
h,
1
2
-bh
=
2l + 2w
32
CHAPTER R Review
For a circle of radius r (diameter d Area
=
71-,2
=
2r),
Circumference
2'1Tr
=
=
'1Td
For a closed rectangular box of length l, width w, and height h, Volume
G
=
lwh
Surface area
=
2lh
+
2wh
+
2lw
For a sphere of radius r, Surface area
=
4'1Tr2
For a right circular cylinder of height h and radius r, Surface area
h C.'!Jn:==-
Now Work
=
2'1Tr2 + 2'1Trh
P R O BLE M 2 9
Using Geometry Formulas
EXAMPLE 4
A Christmas tree ornament is in the shape of a semicircle on top of a triangle. How many square centimeters (cm) of copper is required to make the ornament if the height of the triangle is 6 cm and the base is 4 cm? See Figure 20. The amount of copper required equals the shaded area. This area is the sum of the area of the triangle and the semicircle. The triangle has height h = 6 and base b = 4. The semicircle has diameter d = 4, so its radius is r = 2.
Solution Figure 20
Area
=
Area of triangle + Area of semicircle
= -bh =
1 1 ? 1 1 2 + -'1Tr = -(4)(6) + -'1T · 2 2 2 2 2 2 12 + 2'1T 18.28 cm
b
=
4; h
=
6;
r =
2
�
About 18.28 cm2 of copper is required. ...'!I!l:===-- -
3 ('
r r
In Word s
Two triangles are congruent if r they have the same size and r shape.
DEFINITION
Now Work
•
PROBLEM 47
U n derstand Con g ruent Triang les a n d S i m ila r Tria ng les
Throughout the text we will make reference to triangles. We begin with a discussion of congruent triangles. According to dictionary. com, the word congruent means coinciding exactly when superimposed. For example, two angles are congruent if they have the same measure and two line segments are congruent if they have the same length. Two triangles are congruent if each of the corresponding angles is the same measure and each of the corresponding sides is the same length.
.J
In Figure 21, corresponding angles are equal and the lengths of the corre sponding sides are equal: a = d, b = e, and c = f. We conclude that these triangles are congruent.
SECTION R.3 Geometry Essentials
33
Figure 21
Congruent triangles
c
It is not necessary to verify that all three angles and all three sides are the same measure to determine whether two triangles are congruent. Determining Con g ruent Tria ngles
Two triangles are congruent if two of the an gles are equal and the lengths of the corresponding sides between the two angles are equal. For example, in Figure 22 ( a ) , the two triangles are congruent because two angles and the included side are equal. 2. Side-Side-Side Case Two triangles are congruent if the lengths of the corresponding sides of the triangles are equal. For example, in Figure 22 (b) , the two triangles are congruent because the three corresponding sides are all equal. 3. Side-Angle-Side Case Two triangles are congruent if the lengths of two corresponding sides are equal and the angles between the two sides are the same. For example, in Figure 22 ( c) , the two triangles are congruent because two sides and the included angle are equal. 1. Angle-Side-Angle Case
Figure 22
(b)
(e)
We contrast congruent triangles with similar triangles. DEFINITION
Two triangles are similar if the corresponding angles are equal and the lengths of the corresponding sides are proportional.
..J
r
r
r r
r
In
Word s
Two triangles a re similar i f they have the same shape. but (possi bly) different sizes.
For example, the triangles in Figure 23 are similar because the corresponding angles are equal. In addition, the lengths of the corresponding sides are proportional because each side in the triangle on the right is twice as long as each corresponding side in the triangle on the left. That is, the ratio of the corresponding sides is a constant:
d
-
a
=
e
-
b
=
f
-
c
=
2.
34
CHAPTER R
Review Figure 23
It is not necessary to verify that all three angles are equal and all three sides are proportional to determine whether two triangles are congruent. Determ ining Similar Triang les 1. Angle-Angle Case
Two triangles are similar if two of the correspond ing angles are equal. For example, in Figure 24(a), the two triangles are similar because two angles are equal. 2. Side-Side-Side Case Two triangles are similar if the lengths of all three sides of each triangle are proportional. For example, in Figure 24(b), the two triangles are similar because 10 30
6
5 15
18
1 3
3. Side-Angle-Side Case Two triangles are similar if two corresponding sides are proportional and the angles between the two sides are equal. For example, in Figure 24(c), the two triangles are similar because 4 12 2 . = "3 and the angles between the sIdes are equal. = "6 18 Figure 24
//
80
K. EXAMPLE 5
(a)
(b)
(c)
Using Similar Triangles
Given that the triangles in Figure 25 are similar, find the missing length angles A, B, and C. Figure 25
60� 90°
x
and the
3S
SECTION R.3 Geometry Essentials Solution
Because the triangles are similar, corresponding angles are equal. So A
and C 30°. Also, the corresponding sides are proportional. That is, � 5 . th·IS equatlOn for x. =
:)X _
3 5 3 • -
5
=
90°, B
=
60°,
i. We solve x
6 x 5x'
=
3x = 30 x
=
6 x
-
Simplify
10
=
Multiply both sides by 5x.
Divide both sides by 3.
The missing length is 10 units. L'l!l: ==_ -
Now Work
•
PROBLEM 41
We begin with a square, each side of length
Proof of the Pythagorean Theorem
a
b. In this square, we can form four right triangles, each having legs equal in length to a and b. See Figure 26. All these triangles are congruent (two sides and +
their included angle are equal). As a result, the hypotenuse of each is the same, say
c, and the pink shading in Figure 26 indicates a square with an area equal to c2. Figure 26
Area
�ab
=
b
a
a
rr--------�----"
b
b Area
=
�ab
a
b
The area of the original square with sides a + b equals the sum of the areas of the 1 four triangles (each of area 'lab) plus the area of the square with side c. That is,
(a + b)?
a2 + 2ab a2
+ +
b2 b2
= = =
1 1 1 1 ab + ab + ab + ab + c2 2 2 2 2 -
-
-
2ab + c2
c2
The proof is complete.
•
Figure 27
b a
(a)
We begin with two trian gles: one a right triangle with legs a and b and the other a triangle with sides a, b, and c for which c2 = a2 + b2. See Figure 27. By the Pythagorean Theorem, the length x of the third side of the first triangle is Proof of the Converse of the Pythagorean Theorem
x2
b
-
=
a2
+
x
=
c
b2
The two triangles have the same sides and are therefore congruent. This means corre sponding angles are equal, so the angle opposite side c of the second triangle equals 90°. • The proof is complete.
36
CHAPTER R Review
R.3 Assess Your Understanding Concepts and Voca bulary 1. A(n)
2.For a triangle with base area A is
9. True or False
The triangles shown are similar.
10. True or False
The triangles shown are similar.
triangle is one that contains an angle of
90 degrees. The longest side is called the
____
b and altitude h, a formula for the
____
3.The formula for the circumference
4.Two triangles are
C of a circle of radius r is
if corresponding angles
are equal and the lengths of the corresponding sides are proportional. 5. True or False
In a right triangle, the square of the length of
the longest side equals the sum of the squares of the lengths of the other two sides. 6. True or False
The triangle with sides of length 6, 8, and 10
is a right triangle. 4
r is 37Tr2.
7. True or False
The volume of a sphere of radius
8. True or False
The triangles shown are congruent.
Ski l l Building
In Problems 11-16, the lengths of the legs of a right triangle are given. Find the hypotenuse. 12. a 6, b 8 11. a 5, b 12 3 14. a 4, b 15. a 7, b 24 =
=
=
=
=
=
=
=
13. a 16. a
=
=
10,
b
14,
b
=
24
=
48
In Problems 17-24, the lengths of the sides of a triangle are given. Determine which are right triangles. For those that are, identify the hypotenuse. 17.3, 4, 5
18.6, 8, 10
19.4, 5, 6
20.2, 2, 3
21. 7, 24, 25
22.10, 24,26
23.6, 4, 3
24.5, 4, 7
25.Find the area A of a rectangle with length 4 inches and width 2 inches. 26.Find the area A of a rectangle with length 9 centimeters and width 4 centimeters. 27.Find the area A of a triangle with height 4 inches and base 2 inches. 28.Find the area A of a triangle with height 9 centimeters and base 4 centimeters. 29. Find the area A and circumference
C of a circle of radius 5 meters.
30.Find the area A and circumference
C of a circle of radius 2 feet.
31.Find the volume V and surface area
S of a rectangular box with length 8 feet, width 4 feet, and height 7 feet.
32.Find the volume V and surface area
S of a rectangular box with length 9 inches, width 4 inches, and height 8 inches.
33.Find the volume V and surface area
S of a sphere of radius 4 centimeters.
34.Find the volume V and surface area
S of a sphere of radius 3 feet.
35.Find the volume V and surface area
S of a right circular cylinder with radius 9 inches and height 8 inches.
36.Find the volume V and surface area
S of a right circular cylinder with radius 8 inches and height 9 inches.
SECTION R.3 Geometry Essentials
37
In Problems 37-40, find the area of the shaded region. 37. 38. 2 2
'0
'0
60 t? 90° 30°
10
16
In Problems 41-44, each pair of triangles is simila!: Find the missing length x and the missing angles A, B, and C. 41. 42. 44. 43. 4
v7 6y A
Application s a n d Extensions
45.
How many feet does a wheel with a diameter o f 1 6 inches travel after four revolutions?
46.
49.
In the figure shown,
ABCD
needed to enclose the window?
is a square, with each side of
length 6 feet. The width of the border (shaded portion) be
tween the outer square
6'
EFGH and ABCD is 2 feet. Find the
area of the border.
E
F A
B
0
H
48.
��
0
Refer to the figure. Square
C
-+-
G
ABCD has an area of 100 square
CGF?
o
c
11--+---11
4'
50.
feet; square BEFG has an area of 1 6 square feet. What is the
area of the triangle
A Norman window consists of a rectangle sur
dow shown in the illustration. How much wood frame is
How many revolutions will a circular disk with a diameter of 4 feet have completed after it has rolled 20 feet?
47.
Architecture
mounted by a semicircle. Find the area of the Norman win
Construction
A circular swimming pool, 20 feet in diame
ter, is enclosed by a wooden deck that is 3 feet wide. What is the area of the deck? How much fence is required to enclose the deck?
38
51.
CHAPTER R
Review
HowT.'111 Is the Great Pyramid'?
The ancient Greek philoso
52.
The Bermuda Triangle
Karen is doing research on the
Bermuda Triangle which she defines roughly by Hamilton,
pher Thales of Miletus is reported on one occasion to have visited Egypt and calculated the height of the Great Pyramid
Bermuda; San Juan, Puerto Rico; and Fort Lauderdale,
of Cheops by means of shadow reckoning. Thales knew that
Florida. On her atlas Karen measures the straight-line dis tances from Hamilton to Fort Lauderdale, Fort Lauderdale
each side of the base of the pyramid was 252 paces and that
to San Juan, and San Juan to Hamilton to be approximately
his own height was 2 paces. He measured the length of the pyramid's shadow to be 114 paces and determined the length
57 millimeters (mm), 58 mm, and 53.5 mm respectively. If
of his shadow to be 3 paces. See the illustration. Using simi
the actual distance from Fort Lauderdale to San Juan is
lar triangles, determine the height of the Great Pyramid in
1046 miles, approximate the actual distances from San Juan
terms of the number of paces.
to Hamilton and from Hamilton to Fort Lauderdale. Source:
Source: www.anselm.edulhomepageidbanachl thales. htm.This site
Source: www.en. wikipedia. orglwikilBermuda_Triangle.
references another source: Selections, from Julia E. Diggins, String,
Straightedge, and Shadow, Viking Press, New York,
Illustrations by Corydon Bell.
www. worldatlas.com
1965,
In Problems 53-55, use the facts that the radius of Earth is 3960 miles and 1 mile 5280 feet. How far can a person see from the bridge, which is 150 feet 53. How Far Can You See? The conning tower of the U.S.S. above sea level? Silvers ides, a World War II submarine now permanently sta tioned in Muskegon, Michigan, is approximately 20 feet above 56. Suppose that 117 and n are positive integers with 117 > n. If sea level. How far can you see from the conning tower? a 1172 - n2, b 2mn, and c 1172 + /12, show that a, b, 54. How Far Can You See'? A person who is 6 feet tall is stand and c are the lengths of the sides of a right triangle. (This for =
=
ing on the beach in Fort Lauderdale, Florida, and looks out
integers, such as 3,4, 5; 5,12, 13; and so on. Such triplets of
zon. How far is the ship from shore? How Far Can You See?
=
mula can be used to find the sides of a right triangle that are
onto the Atlantic Ocean. Suddenly, a ship appears on the hori
55.
=
integers are called P ythagorean triples.)
The deck of a destroyer is 100 feet
above sea level. How far can a person see from the deck?
Discussion and Writing
57.
You have 1000 feet o f flexible pool siding and wish to con struct a swimming pool. Experiment with rectangular-shaped
58.
The Gibb's Hill Lighthouse, Southampton, Bermuda, in op
eration since 1846, stands 117 feet high on a hill 245 feet high,
pools with perimeters of 1000 feet. How do their areas vary?
so its beam of light is 362 feet above sea level. A brochure
What is the shape of the rectangle with the largest area? Now
states that the light itself can be seen on the horizon about
compute the area enclosed by a circular pool with a perime
26 miles distant. Verify the correctness of this information.
ter (circumference) of 1000 feet. What would be your choice
The brochure further states that ships 40 miles away can see
of shape for the pool? If rectangular, what is your preference
the light and planes flying at 10,000 feet can see it 120 miles
for dimensions? Justify your choice. If your only considera
away. Verify the accuracy of these statements. What assump
tion is to have a pool that encloses the most area, what shape
tion did the brochure make about the height of the ship?
should you use?
SECTION R04 Polynomials
39
R.4 Polynomia ls OBJECTIVES
1 Recognize Monomials (po 39) 2
Recognize Polynomials (po 40)
3
Add and Subtract Polynomials (p.4l)
4
Multiply Polynomials (p. 42)
5
Know Formulas for Special Products (p.43)
6
Divide Polynomials Using Long Division (po 44)
7
Work with Polynomials in Two Variables (po 47)
We have described algebra as a generalization of arithmetic in which letters are used to represent real numbers. From now on, we shall use the letters at the end of the alphabet, such as x, y, and z, to represent variables and the letters at the begin ning of the alphabet, such as a, b, and c, to represent constants. In the expressions 3x + 5 and ax + b, it is understood that x is a variable and that a and b are con stants, even though the constants a and b are unspecified. As you will find out, the context usually makes the intended meaning clear.
1
Recog n ize Monom ials
DEFINITION
A monomial in one variable is the product of a constant and a variable raised to a nonnegative integer power. A monomial is of the form
NOTE
The nonnegative i ntegers a re the integers 0, 1, 2, 3, ... . •
k ax
where a is a constant, x is a variable, and k ;:::: 0 is an integer. The constant a is called the coefficient of the monomial. If a *' 0, then k is called the degree of the monomial.
EXAMPLE 1
Examples of Monomials
(a) 6x2
Coefficient
Monomial
(b) - \I2x3
(c)
'" .)
(d) - 5 x (e)
x4
Degree
6
2
'" .)
o
Since 3 = 301
-5
1
Since -5x
-\12
1
3
4
=
=
3;;, x
"* 0
-5x1 •
Now let's look at some expressions that are not monomials.
EXAMPLE 2
Examples of Nonmonomial Expressions
(a) 3X1/2 is not a monomial, since the exponent of the variable x is 2" and 2" is not a nonnegative integer. (b) 4x-3 is not a monomial, since the exponent of the variable x is -3 and -3 is not a nonnegative integer. �
-
Now Work
1
1
•
PROBLEM 7
40
CHAPTER R
Review
2
Recognize Polynom ials
Two monomials with the same variable raised to the same power are called like terms. For example, 2X4 and -5x4 are like terms. In contrast, the monomials 2x3 and 2x5 are not like terms. We can add or subtract like terms using the Distributive Property. For example,
The sum or difference of two monomials having different degrees is called a binomial. The sum or difference of three monomials with three different degrees is called a trinomial. For example,
x2 - 2 is a binomial. x3 - 3x + 5 is a trinomial. 2x2 + 5x2 + 2 = 7x2 + 2 is a binomial. DEFINITION
r r
r
A polynomial in one variable is an algebraic expression of the form ( 1)
In Words
where an , al1 - 1 , . . . , al , ao are constants,* called the coefficients of the poly nomial, n ;::: ° is an integer, and x is a variable. If an =f. 0, it is called the leading coefficient, and n is called the degree of the polynomial.
A
polynomial is a sum of monomials.
.J
The monomials that make up a polynomial are called its terms. If all the coeffi cients are 0, the polynomial is called the zero polynomial, which has no degree. Polynomials are usually written in standard form, beginning with the nonzero term of highest degree and continuing with terms in descending order according to degree. If a power of x is missing, it is because its coefficient is zero. EXAMPLE 3
Examples of Polynomi als
Polynomial
Coefficients
Degree
8x3 + 4x2 - 6x + 2 3x2 - 5 = 3x2 + 0 · x + ( -5 ) 8 - 2x + x2 = 1 · x2 + ( -2)x + 8 5x + v2 = 5x1 + v2 3 = 3 · 1 = 3 · xo
-8, 4, -6, 2 3, 0, -5 1, -2, 8 5, v2 3
3 2 2 1
-
°
°
°
No degree
•
Although we have been using x to represent the variable, letters such as y or z are also commonly used.
3x4 - x2 + 2 is a polynomial (in x) of degree 4. 9y3 - 21 + y - 3 is a polynomial (in y) of degree 3.
Z5 +
7T
is a polynomial (in z) of degree 5.
Algebraic expressions such as 1
x ':' The notation
a"
is read as
"
a
sub
"
n.
x +5
and
The number
n
is called a subscript and should not be confused
with an exponent. We use subscripts to distinguish one constant from another when a large or undeter mined number of constants is required.
SECTION R.4 Polynomials
are not polynomials. The first i s not a polynomial because 1:.
x
=
41
X-I has an exponent
that is not a nonnegative integer. Although the second expression is the quotient of two polynomials, the polynomial in the denominator has degree greater than 0, so the expression cannot be a polynomial.
- Now Work
'1'l
3
PROBLEM 1 7
Add a n d S u btract Polynomials
Polynomials are added and subtracted by combining like terms.
EXAMPLE 4
Adding Polynomials
Find the sum of the polynomials:
8x3 - 2x2 + 6x - 2 and 3x4 - 2x3 + x2 + X
Solution
We shall find the sum in two ways. Horizontal Addition: The idea here is to group the like terms and then combine
them.
(8x3 - 2x2 + 6 x - 2) + (3X4 - 2x3 + x2 + x ) = 3x4 + (8x� - 2 x3 ) + ( - 2x- + x-) + (6x + x ) - 2 3x4 + 6x3 - x2 + 7 x - 2 ?
0
?
=
Vertical Addition: The idea here is to vertically line up the like terms in each poly nomial and then add the coefficients.
8x3 - 2X2 + 6x - 2 + 3x4 - 2x3 + x2 + X 3x4 + 6x3 - x2 + 7 x - 2
•
We can subtract two polynomials horizontally or vertically as well.
EXAMPLE 5
Subtracting Polynomials
Find the difference: Solution
(3X4 - 4x3 + 6x2 - 1 ) - (2X4 - 8x2 - 6x + 5 )
Horizontal Subtraction:
(3x4 - 4x3 + 6x2 - 1 ) - ( 2X4 - 8x2 - 6x + 5 ) =
3x4 - 4x3 + 6x2 - 1 + ,( _ 2X4 + 8x + 6x - 5)
�
,
Be s u re to change the sig n of each
term in the second polynomial.
=
i
(3X4 - 2X4 ) + ( -4x3 ) + (6x2 + 8x2 ) + 6x + ( - 1 - 5 )
Group like terms. =
X4 - 4x3 + 14x2 + 6x - 6
42
CHAPTER R
Review
COMMENT Vertical subtraction will be _ used when we divide polynomials,
Vertical Subtraction: We line up like terms, change the sign of each coefficient of the second polynomial, and add.
3x4 - 4x3 + 6X2 - 1 - 8X2 - 6x + 5 ] [2X4
=
+
=
3x4 - 4x3 + 6x2 - 1 + 8x2 + 6x - 5 -2x4 x4 - 4x3 + 14x2 + 6x - 6
•
The choice of which of these methods to use for adding and subtracting poly nomials is left to you. To save space, we shall most often use the horizontal format. 1.1'111::
4
\1"'-
Now Work
PROBLEM 29
M ulti ply Polynom ials
Two monomials may be multiplied using the Laws of Exponents and the Commutative and Associative Properties. For example,
Products of polynomials are found by repeated use of the Distributive Property and the Laws of Exponents. Again, you have a choice of horizontal or vertical format.
EXAMPLE 6
Multiplying Polynom i als
Find the product: Solution
(2x + 5 ) (x2 - X + 2 )
Horizontal Multiplication:
( 2x
+
5 ) (x2 - X + 2 ) = 2x (x2 - X
i
+
2)
+
5 ( x2 - X
2)
+
Distributive Property
i
=
(2x · x2 - 2x · x + 2x · 2 )
+
(5 · x2 - 5 . x + 5 . 2)
Distributive Property =
i
(2x3 - 2x2
+
4x) + (5x2 - 5x + 10)
Law of Exponents =
i
2x3
+
3x2 - X + 10
Combine like terms.
Vertical Multiplication: The idea here is very much like multiplying a two-digit number by a three-digit number.
x2 - x + 2 2x + 5 This lin e is 2x(J2 x + 2) , 2x3 - 2X2 + 4x 5x2 - 5x + 10 This line i s 5(J2 x + 2) . (+) 2x3 + 3x2 - X + 10 S u m o f t h e above two lin es. -
-
'I!�
Now Work
PROBLEM 4S
•
SECTION R.4 Polynomials
5
43
Know Form u l a s for Special Prod u cts
Certain products, which we call special products, occur frequently in algebra. We can calculate them easily using the FOIL (First, Outer, Inner, Last) method of multiplying two binomials.
c l i
� l I
Outer
F i rst
(ax + b)(ex + d) = ax(ex + d) + b(ex + d)
Il L
l n ne r
EXAMPLE 7
l
Last
J
Fi rst
Outer
�
�
Inner
,------"---;
Last
,------"---;
= ax . ex + ax . d + b . ex + b . d = aex2 + adx + bex + bd = aex2 + (ad + be)x + bd
Using FO i l
(a) ( x - 3 ) (x + 3 ) = x2 + 3x - 3x - 9 = x2 - 9
F
(b) (c) (d) (e)
( x + 2)2 = (x + 2 ) (x + ( x - 3 )2 = (x - 3 ) (x (x + 3 ) (x + 1 ) = x2 + (2x + 1 ) (3x + 4) = 6x2
0
I
L
2) x2 + 2x + 2x + 4 = x2 + 4x + 4 3) x2 - 3x - 3x + 9 = x2 - 6x + 9 + 3x + 3 = x2 + 4x + 3 + 8x + 3x + 4 6x2 + l lx + 4 =
=
X
==c- 't;'!I!l:
Now Work
=
•
PROBLEMS 47 AND 55
Some products have been given special names because of their form. The fol lowing special products are based on Examples 7(a), (b), and (c). Difference of Two Squares
(2)
(x - a) (x + a) = x2 - a2
Squares of Binomials, or Perfect Squares
(x + a f (x - a) 2
EXAMPLE 8
=
=
x2 + 2ax + a2 x2 - 2ax + a2
(3a) (3b)
Using Special Product Form ulas
(a) (x - 5 ) (x + 5 ) = x2 - 52 = x2 - 25 (b) (x + 7f = x2 + 2 ' 7 ' x + 72 = x2 + 14x + 49 (c) ( 2x + 1 ) 2 = (2x) 2 + 2 · 1 · 2x + 12 = 4x2 + 4x + 1 (d) (3x - 4) 2 = (3x i - 2 · 4 · 3x + 42 = 9x2 - 24x + 16 � _ &::i"" -
Now Work
Difference of two squares Square of a binomial Notice that we used 2x in place of x in formula (3a). Replace x by 3 x in formula (3b),
PROBLEMS 65, 67, AND 69
Let's look at some more examples that lead to general formulas.
•
44
CHAPTER R
Review Cubing a Binomial
EXAMPLE 9
(a ) ( x + 2 ? = (x
+
2)(x
+
2) 2
(b) (x - 1 )3 = (x - l ) (x - 1 ) 2
(x + 2)(x2 + 4x + 4) Formula (3a) = (x3 + 4x2 + 4x) + (2x2 + 8x + 8) = x3 + 6x2 + 1 2x + 8 =
= =
=
(x - 1 ) (x2 - 2x + 1 ) Formula (3b) (x3 - 2x2 + x) - (x2 - 2x + 1 ) x3 - 3x2 + 3x - 1
111
Cu bes of Binomials, or Perfect Cubes
f.'!f = fii: =- -
(x + a)3
=
(x - a)3
=
Now Work
x3 + 3ax2 + 3a2 x + a3 x3 - 3ax2 + 3a2 x - a3
(4a ) (4b)
PROBLEM 85
Forming the Difference of Two C ubes
EXAMPLE 1 0
(x - 1 ) (x2 +
X
+
1)
=
=
=
x (x2 + + 1 ) - 1 (x2 + + 1 ) x3 + x2 + - x2 - x - I x3 - 1 X
X
X
Forming the Sum of Two C ubes
EXAMPLE 1 1
(x + 2) (x2 - 2x
+
4) = x (x2 - 2x + 4) + 2( x2 - 2x + 4) = x3 - 2x2 + 4x + 2x2 - 4x + 8 = x3 + 8
•
Examples 10 and 1 1 lead to two more special products.
Difference of Two Cubes
(5)
S u m of Two Cubes
(6)
6
Divide Polynomials Using long Division
The procedure for dividing two polynomials is similar to the procedure for dividing two integers.
SECTION R.4 Polynomials EXAMPLE 1 2
4S
Dividing Two I ntegers
Divide 842 by 15. Solu tion
Divisor
->
56 15)842 75 92 90
<-
Quotient
<- Dividend <- 5 ' 15 (su btract) <- 6 ' 15 (su btract)
2
<-
So, - = 56 + -
Remainder
2 15 '
842 15
•
In the long division process detailed in Example 12, the number 15 is called the 842 is called the dividend, the number 56 is called the quotient, and the number 2 is called the remainder. To check the answer obtained in a division problem, multiply the quotient by the divisor and add the remainder. The answer should be the dividend. divisor, the number
( Quotient ) ( Divisor ) + Remainder
=
Dividend
For example, we can check the results obtained in Example 12 as follows:
(56) (15) + 2
=
840 + 2
=
842
To divide two polynomials, we first must write each polynomial in standard form. The process then follows a pattern similar to that of Example 12. The next example illustrates the procedure.
EXAMPLE 1 3
Dividing Two Polynomials
Find the quotient and the remainder when
3x3 + 4x2 Solution
+ X
+ 7 is divided by x2 + 1
Each polynomial is in standard form. The dividend is 3x3 + 4x2 + divisor is x2 + 1 .
X
+ 7, and the
STEP 1: Divide the leading term o f the dividend,
3x3 , by the leading term of the divisor, x2 . Enter the result, 3x, over the term 3x3 , as follows: 3x x2 + I hx3 + 4x2 +
STEP 2: Multiply
X
+7
3x by x2 + 1 and enter the result below the dividend. 3x <-
f
3x ' (>f + 1)
=
3;(5 + 3x
Notice that we a lig n the 3x term under the x to make the next step easier.
STEP 3: Subtract and bring down the remaining terms.
3x
x- + 1 ) 3x� + 4x- + x + 7 ?
0
?
+ 3x 4x2 - 2x + 7
<-
Subtract (change the signs and add).
<-
Bring down the 4x2 and the 7.
46
CHAPTER R Review STEP 4: Repeat Steps
1-3 using 4x2 - 2x + 7 as the dividend.
3x + 4 x2 + 1 hx3 + 4x2 + X 3x3 + 3x 4x2 - 2x 4x2 -2x
+7 +7 +4 +3
J
<-
D;,;d, 4x' by x' to get 4. M ultiply x2 + 1 by 4; subtract.
Since x2 does not divide -2x evenly (that is, the result is not a monomial), the process ends. The quotient is 3x + 4, and the remainder is -2x + 3. Check: ( Quotient)( Divisor)
+
Remainder
= (3x + 4)(x2 + 1 ) + ( -2x + 3)
= 3x3 + 3x + 4x2 + 4 + ( -2x + 3) =
3x3 + 4x2 + X + 7 = Dividend
Then
The next example combines the steps involved in long division.
EXAMPLE 1 4
Dividing Two Polynomials
Find the quotient and the remainder when
x4 - 3x3 + 2x - 5 is divided by x2 - X +
Solution
1
In setting up this division problem, it is necessary to leave. a space for the missing
x2 term in the dividend. Divisor
-->
Subtract
-->
Su btract
-->
Subtract
-->
x2 - 2x - 3 x2 - X + 1 )x4 - 3x3 X4 - x3 + x2 -2x3 - x2 -2x3 + 2x2 -3x2 -3x2
+ 2x - 5 + + +
2x 2x 4x 3x x
<-
Quotient
<-
Dividend
<-
Remainder
-5 -5 -3 -2
Check: ( Quotient)(Divisor)
+ Remainder = (x2 - 2x - 3) (x2 - X + 1 ) + x - 2 = X4 - x3 + x2 - 2x3 + 2x2 - 2x - 3x2 + 3x - 3 + x - 2 = X4 - 3x 3 + 2x - 5 Dividend =
As a result,
X4 - 3x3 + 2x - 5 �-x+1
------- = x2 - 2x - 3 +
x-2 �-x+l
----
•
SECTION R.4 Polynomials
47
The process of dividing two polynomials leads to the following result: THEOREM
Let Q be a polynomial of positive degree and let P be a polynomial whose degree is greater than or equal to the degree of Q. The remainder after dividing P by Q is either the zero polynomial or a polynomial whose degree is less than the degree of the divisor Q.
-.J
�=--,," - Now Work
7
P RO BL EM 9 3
Wo rk with Po lyno m i a l s i n Two Va riab l es
A monomial in two variables x and y has the form ax"y''', where a is a constant, x and y are variables, and n and m are nonnegative integers. The degree of a mono mial is the sum of the powers of the variables. For example,
are monomials, each of which has degree 4. A polynomial in two variables x and y is the sum of one or more monomials in two variables. The degree of a polynomial in two variables is the highest degree of all the monomials with nonzero coefficients.
Exam ples of Polynom i als in Two Variables
EXAMPLE 1 5
3x2
+
2x3y + 5
Two varia bles, degree is 4.
7TX3
-
i
Two varia bles, degree is 3.
X4 +
4x3y - xy3 + l
Two varia bles, degree is 4.
•
Multiplying polynomials in two variables is handled in the same way as polyno mials in one variable.
Using a Special Product Formula
EXAMPLE 16
To multiply (2x - y)2, use the Squares of Binomials formula (3b) with 2x instead of x and y instead of a.
(2x - y)2
= =
� = = .....
Now Work
(2X)2 - 2 · y ' 2x + i 4x2 - 4xy + i
•
PROBLEM 79
\
R.4 Assess Your Understanding Concepts and Voca bulary
1.
The polynomial
3x4 - 2x3 + 13x2
The leading coefficient is
(x2 - 4 ) (x2 + 4 ) 3. (x - 2) (x2 + 2x + 4) 2.
=
.
__
.
- 5 is of degree
.
__
4. 5.
__ .
True or False
4[2 is a monomial of degree - 2
.
The degree of the product of two nonzero
polynomials equals the sum of their degrees.
__
=
True or False
6.
True or False
(x + a)(x2 + ax + a)
=
x3 + a3
48
CHAPTER R
Review
Skill Building
In Problems 7-16, tell whether the expression is a monomial. If it is, name the variable(s) and the coefficient and give the degree of the monomial. If it is not a monomial ,state why not. 11. -2xi 10. -2x-3 9. � 8. -4x2 7. 2x3 x 13.
8x y
14.
2X2
16.
3 --
Y
3x2 + 4
In Problems 1 7-26, tell whether the expression is a polynomial. 1f it is, give its degree. If it is not, state why not. 21. 3x2-� 20. - 7j 19. 5 18. 1 - 4x 1 7. 3x2- 5 X 22.
3 +2 x
23.
-
2y - v2 3
24.
1 0z2+z
25.
3x3+ 2x-1 �r-+x+1
x2+ 5 x -I
3
26. ----=?:-----
In Problems 27-46, add, subtract, or multiply, as indicated. Express your answer as a single polynomial in standard form. 28. ( x3 + 3x2 + 2)+( x2- 4x+ 4) 27. ( x2+ 4x+ 5)+( 3x-3) '- 29.
( x2 - 3x - 4)-( x3-3x2 + X+ 5)
( x3- 2x2+ 5x+ 10)-( 2x2- 4x+3)
30.
( x2-3x + 1 )+ 2( 3x2+ X- 4)
34. -2(
33.
36. 8(
( x2-x+ 2)+( 2x2-3x + 5)-( x2+ 1 )
38.
39.
9( i-3y + 4)-6( 1 -i)
40. 8(
x( x2+x- 4)
44.
5x\3x - 4)
45.
4x3-3x2 - 1 )-6( 4x3+8x- 2)
( x2+ 1 )-( 4x2+ 5)+( x2+X- 2)
37.
41.
x2+x+ 1)+(-5x2-X+ 2)
1 -i)+ 4( 1 +y+i+i)
( x+ 1)( x2+ 2x - 4)
46.
( 2x-3)( x2+X+ 1)
In Problems 47-64, multiply the polynomials using the FOIL method. Express your answer as a single polynomial in standard form. 49. ( 2x+ 5)( x+ 2) 48. ( x + 3)( x+ 5) 47. ( x+ 2)( x+ 4) ( 3x+ 1)( 2x+ 1 )
51.
( x- 4)( x+ 2)
52.
( x+ 4)( x - 2)
53.
( x-3)( x- 2)
54.
( x- 5)( x - 1)
55.
( 2x+3)( x- 2)
56.
( 2x - 4)( 3x+ 1)
57.
( - 2x+3)( x- 4)
58.
( -3x - l)( x+ 1)
59.
(-x- 2)(-2x- 4)
60.
(-2x-3)( 3-x)
61.
( x - 2y)( x+y)
62.
( 2x+3y)( x - y)
63.
(-2x-3y)( 3x+ 2y)
64.
( x-3y)(-2x +y)
50.
In Problems 65-88, multiply the polynomials using the special product formulas. Express your answer as a single polynomial in standard form. 68. ( 3x+ 2)( 3x- 2) 67. ( 2x+3)( 2x - 3) 66. ( x - l)( x+ 1 ) 65. ( x - 7)( x +7) 69.
( x + 4)2
73.
( 3x
77.
(x+y)(x - y)
+
4)( 3x- 4)
70.
( x+ 5f
71.
( x - 4)2
72.
( x- 5)2
74.
( 5x - 3)( 5x+3)
75.
( 2x - 3f
76.
( 3x- 4f
78.
( x+3y)( x - 3y)
79.
( 3x
80.
( 3x+ 4y)( 3x - 4y)
+
y)( 3x - y)
SECTION R.S Factoring Polynomials +
81.
(x
y)2
82.
(x - y?
85.
(x - 2 )3
86.
(x
+
1 )3
+
83.
(x - 2y)2
84.
(2x
87.
(2x
+
88.
(3x - 2)3
1 )3
49
3y)2
In Problems 89-104, find the quotient and the remainda Check your work by verifying that (Quotient)(Divisor) 2
89.
4x3 - 3x2
+
X
+
1
divided by
x
91.
4x3 - 3x2
+
X
+
1
divided by
x2
' 93.
5x4 - 3x2
+
X
+ 1
95.
4x5 - 3x2
+
X
+
97.
2X4 - 3x3
+
X
+
99.
-4x3
101. 1 103.
+
- x2
x3 - a3
x2 - 4 +
4 X
1
1
+
Remainder Dividend 90. 3x 3 - x2 + X - 2 =
divided by
x
+
2
+
2
92.
3x3 - x2
+
X
-
2
divided by
x2
2
94.
5x4 - x2
+
X
-
2
divided by
x2
divided by
2x3 - 1
96.
3x5 - x2
+
X
-
2
divided by
3x3 - 1
divided by
2x2
98.
3x4 - x3
+
X-2
divided by
3x2
divided by
divided by
divided by
divided by
+
x2
x2
+
+
X
+
1
x- 1
100.
+
102.
X
+ 1
x-a
104.
- 3x4 - 2x - 1
1
-
x2
+
x4
divided by
divided by
X
+
1
- 1
x2 - X
+
1
x-a
divided by
x5 - a5
x
+
Discussion a n d Writing 105. Explain why the degree o f the product o f two nonzero poly
108. Do you prefer adding two polynomials using the horizontal
nomials equals the sum of their degrees.
method or the vertical method? Write a brief position paper defending your choice.
106. Explain why the degree of the sum of two polynomials of dif
109. Do you prefer to memorize the rule for the square of a bi nomial
ferent degrees equals the larger of their degrees.
(x
+
a)2 or to use F OIL to obtain the product? Write
a brief position paper defending your choice. 107. Give a careful statement about the degree of the sum of two polynomials of the same degree.
R.S Factoring Polynomials OBJECTIVES
1 Factor the Difference of Two Sq uares and the Sum and the Difference
of Two Cubes (p. 50) 2
Factor Perfect Squares (p. 5 1 )
3
Factor a Second-Degree Polynomial: x2
4
Factor by Grouping (p. 53)
5
+
Bx
+
Factor a Second-Degree Polynomial: Ax2 + Bx
Consider the following product: ( 2x + 3 ) (x
-
4)
=
2x2
-
C
+
5x
(p. 52)
C, A *- 1
-
(p. 54)
12
The two polynomials on the left side are called factors of the polynomial on the right side. Expressing a given polynomial as a product of other polynomials, that is, finding the factors of a polynomial, is called factoring. We shall restrict our discussion here to factoring polynomials in one variable into products of polynomials in one variable, where all coefficients are integers. We call this factoring over the integers.
SO
CHAPTER R
COMMENr
Review
Over the real numbers, 3x + 4 factors into 3(x + It is the non integer that causes 3x + 4 to be _ prime over the integers.
�
Any polynomial can be written as the product of 1 times itself or as - 1 times its additive inverse. If a polynomial cannot be written as the product of two other poly nomials (excluding 1 and - 1 ) , then the polynomial is said to be prime. When a polynomial has been written as a product consisting only of prime factors, it is said to be factored completely. Examples of prime polynomials (over the integers) are
�).
2, 3, 5 , x, x + 1 , x - I , 3x + 4 x2 + 4 The first factor to look for in a factoring problem is a common monomial fac tor present in each term of the polynomial. If one is present, use the Distributive Property to factor it out.
I dentifying Common Monomial Factors
EXAMPLE 1
Polynomial
2x + 4 3x - 6 2x2 - 4x + 8 8x - 12 x2 + X x3 - 3x2 6x2 + 9x
Common Monomial Factor
2 3 2 4 X
x2 3x
Remaining Factor
Factored Form
x+2 x-2 x2 - 2x + 4 2x - 3 X + 1 X - 3 2x + 3
2x + 4 = 2(x + 2) 3x - 6 = 3(x - 2) 2x2 - 4x + 8 = 2 (x2 - 2x + 4) 8x - 12 = 4(2x - 3 ) x2 + X = X (x + 1 ) x3 - 3x2 = x2( X - 3 ) 6x2 + 9x = 3x(2x + 3 )
III
Notice that, once all common monomial factors have been removed from a polynomial, the remaining factor is either a prime polynomial of degree 1 or a poly nomial of degree 2 or higher. (Do you see why?) "" j� -
1
Now Work
PROBLEM 5
Factor the Difference of Two Squ a res a n d the S u m a n d Difference o f Two Cu bes
When you factor a polynomial, first check for common monomial factors. Then see whether you can use one of the special formulas discussed in the previous section.
x2 - a2 = (x - a) (x + a) x2 + 2ax + a2 = (x + a) 2
Difference of Two Squares Perfect Squares
x2 - 2ax + a2 = (x - a)2 x3 + a3 = (x + a) (x2 - ax + a2 ) x3 - a3 = (x - a) (x2 + ax + a2 )
Sum of Two Cubes Difference of Two Cubes
EXAM P L E 2
F actoring the Difference of Two Sq uares
Factor completely: Sol ution
x2
-
4
We notice that x2 - 4 is the difference of two squares, x2 and 22.
x2 - 4
=
(x
-
2) (x + 2 )
III
SECTION R.S Factoring Polynomials EXAMPLE 3
51
Factoring the D ifference of Two C u bes
Factor completely: Solution
x3 - 1
Because x3 - 1 is the difference of two cubes, x3 and 13, we find that
x3 - 1 EXAMPLE 4
=
(x - 1 ) (x2 + X + 1 )
•
Factoring the Sum o f Two Cubes
Factor completely: Solution
x3 + 8
Because x3 + 8 is the sum of two cubes, x3 and 23, we have
x3 + 8 EXAMPLE 5
=
(x + 2) (x2 - 2x + 4)
•
Factoring the D i fference of Two Sq u ares
Factor completely: Solution
X4 - 16
Because X4 - 16 is the difference of two squares, X4
x4 - 16
=
=
(x2 ) 2 and 16
=
42, we have
(x2 - 4) (x2 + 4)
But x2 - 4 i s also the difference o f two squares. Thus,
x4 - 16 = (x2 - 4) (x2 + 4)
= = = ;> .-
2
Now Work
=
(x - 2 ) (x + 2 ) (x2 + 4)
•
P R O B L E M S 1 5 A N D 33
Factor Perfect Squa res
When the first term and third term of a trinomial are both positive and are perfect squares, such as x2, 9x2, 1, and 4, check to see whether the trinomial is a perfect square. EXAMPLE 6
Factoring Perfect Sq uares
Factor completely: Solution
x2 + 6x + 9
The first term, x2, and the third term, 9 = 32, are perfect squares. Because the middle term 6x is twice the product of x and 3, we have a perfect square.
x2 + 6x + 9 EXAMPLE 7
(x + 3)2
9x2 - 6x + 1
The first term, 9x2 (3x f, and the third term, 1 = 12, are perfect squares. Because the middle term, -6x, is -2 times the product of 3x and 1, we have a perfect square. =
9x2 - 6x + 1 = (3x - 1 )2
EXAMPLE 8
•
Factoring Perfect Squ ares
Factor completely: Solution
•
Factoring Perfect Squ ares
Factor completely: Solution
=
25x2 + 30x + 9
The first term, 25x2 (5x f, and the third term, 9 = 32, are perfect squares. Because the middle term, 30x, is twice the product of 5x and 3, we have a perfect square. =
25x2 + 30x + 9 = (5x + 3 f
'q'=> -
Now Work
P R O B L E M S 2 5 A N D 93
•
52
CHAPTER R
Review
If a trinomial is not a perfect square, it may be possible to factor it using the technique discussed next. 3
Factor a Second-deg ree Polynom i a l : x2 + Bx + C
The idea behind factoring a second-degree polynomial like x2 + Bx + C is to see whether it can be made equal to the product of two, possibly equal, first-degree polynomials. For example, we know that
(x + 3 ) ( x
+
4)
=
x2 + 7x + 12
The factors of x2 + 7 x + 12 are x + 3 and x + 4. Notice the following:
x2 + 7x + 12
L
ab
In general, if x2 + Bx + C C and a + b = B.
=
(x + 3)(x + 4)
L- 12 is t h e product o f 3 and 4 7 is the s u m of 3 a n d 4
=
(x + a)(x + b )
x2 + (a + b)x + ab, then
=
=
To factor a second-degree polynomial x2 + Bx + C, find integers whose prod uct is C and whose sum is B. That is, if there are numbers a, b, where ab = C and a + b B, then =
x2 + Bx + C = (x + a ) (x + b)
EXAMPLE 9
Factoring Trinomials
Factor completely: Solution
x2 + 7x + 1 0
First, determine all integers whose product is 10 and then compute their sums. Integers whose product is 1 0
1, 10
-1, -10
2, 5
-
Sum
11
-11
7
-7
2, - 5
The integers 2 and 5 have a product of 10 and add up to 7, the coefficient of the middle term. As a result, we have
x2 + 7x + 1 0 = (x + 2 ) (x + 5 )
EXAMPLE 1 0
Factoring Trinomials
Factor completely: Solution
•
x2
-
6x + 8
First, determine all integers whose product is 8 and then compute each sum. Integers whose product is 8
1,8
- 1 , -8
2, 4
- 2, -4
Sum
9
-9
6
-6
Since -6 is the coefficient o f the middle term,
x2 - 6x + 8
=
(x
-
2) ( x
-
4)
•
SECTION R.S
Factoring Polynomials
S3
Factoring Trinomials
EXAMPLE 1 1
Factor completely:
x2 - x - 12
First, determine all integers whose product is - 1 2 and then compute each sum.
Solution
Integers whose product is
-
12
Sum
1. - 1 2
-1. 12
2. - 6
- 2. 6
3. - 4
- 3.
-11
11
-4
4
-1
1
4
Since - 1 is the coefficient of the middle term,
x2 - x - 1 2
=
(x + 3 ) (x - 4)
•
Factoring Trinomials
EXAMPLE 1 2
Factor completely:
x2 + 4x - 12
The integers -2 and 6 have a product of -12 and have the sum 4. Thus,
Solution
x2 + 4x - 12
=
(x - 2 ) (x + 6)
•
To avoid errors in factoring, always check your answer by mUltiplying it out to see if the result equals the original expression. When none of the possibilities works, the polynomial is prime. I dentifying Prime Polynomials
EXAMPLE 1 3
Show that x2 + 9 is prime. First, list the integers whose product is 9 and then compute their sums.
Solution
Integers whose product is 9
1,9
-1. -9
Sum
10
-10
3, 3
6
- 3. - 3 -6
Since the coefficient o f the middle term in x2 + 9 = x2 + Ox + 9 i s 0 and none of the sums equals 0, we conclude that x2 + 9 is prime. •
Example 13 demonstrates a more general result: THEOREM
Any polynomial of the form x2 + a2, a real, is prime. W!l: = = >-
4
Now Work
.J
PROBLEMS 39 AND 77
Factor by Grouping
Sometimes a common factor does not occur in every term of the polynomial, but in each of several groups of terms that together make up the polynomial. When this happens, the common factor can be factored out of each group by means of the Distributive Property. This technique is called factoring by grouping. EXAMPLE 1 4
Factoring by Grouping
Factor completely by grouping: Solution
(x2 + 2)x + (x2 + 2) 3 .
Notice the common factor x2 + 2 . By applying the Distributive Property, we have
(x2 + 2) x + (x2 + 2) . 3
=
(x2 + 2) (x + 3 )
Since x2 + 2 and x + 3 are prime, the factorization is complete.
•
54
CHAPTER R
Review (f. The next example shows a factoring problem that occurs in calculus. Factoring by Grouping
EXAMPLE 1 5
Factor completely by grouping:
3(x - 1 )2(x + 2)4 + 4(x - 1 )3(x + 2)3
Here, (x - 1 f(x + 2)3 is a common factor of 3(x - 1 )2(x + 2)4 and of 4(x - 1?(x + 2)3. As a result,
Solution
3(x - 1 f(x + 2)4 + 4(x - 1 )3(x + 2?
(x - 1 ?(x + 2)3 [ 3(x + 2) + 4(x - 1 ) J (x - 1 )2(X + 2?[3x + 6 + 4x - 4 J = (x - 1 f(x + 2?(7x + 2) •
=
=
Factoring by Grouping
EXAMPLE 16
Factor completely by grouping:
x3 - 4x2 + 2x - 8
To see if factoring by grouping will work, group the first two terms and the last two terms. Then look for a common factor in each group. In this example, we can factor x2 from x3 - 4x2 and 2 from 2x - 8. The remaining factor in each case is the same, x - 4. This means that factoring by grouping will work, as follows:
Solution
x3 - 4x2 + 2x - 8 = (x3 - 4x2) + (2x - 8 ) x2( X - 4) + 2( x - 4) = (x - 4 ) (x2 + 2) =
Since x2 + 2 and x - 4 are prime, the factorization i s complete. "I! �-
5
Now Work
PROBLEMS 5 1
•
AND 1 21
Factor a Second-deg ree Polynom i a l : Ax2 + Bx + C, A =1= 1
To factor a second-degree polynomial Ax2 + Bx + C, when A C have no common factors, follow these steps:
1=
1 and A, B, and
Steps for Factori ng Ax2 + Bx + C, when A -=1= 1 and A, B, and C Have No Com mon Factors STEP 1: Find the value of AC. STEP 2: Find integers whose product is AC that add up to
and b so that ab = AC and a + b = B. STEP 3: Write Ax2 + Bx + C = Ax2 + ax + bx + C . STEP 4 : Factor this last expression b y grouping. EXAMPLE 1 7
Factoring Trinom ials
Factor completely: Solution
B. That is, find a
2x2 + 5x + 3
Comparing 2x2 + 5x + 3 to Ax2 + Bx + C, we find that A = 2, B = 5, and C
STEP 1: The value of AC is 2 · 3 = 6. STEP 2: Determine the integers whose product is AC
=
=
3.
6 and compute their sums.
Integers whose product is 6
1,6
- 1 , -6
2, 3
- 2, - 3
Sum
7
-7
5
-5
SECTION R.S STEP 3: The integers whose product is 6 that add up to B
J
2X2 + 5x + 3
STEP 4: Factor by grouping.
=
J
=
SS
Factoring Polynomials
5 are 2 and 3.
l'
t
w
2X2 + 2x + 3x + 3
r
2x2 + 2x + 3x + 3 = (2x2 + 2x) + (3x + 3 ) = 2x(x + 1 ) + 3 ( x + 1 )
= ( x + 1 ) (2x + 3 )
As a result,
EXAMPLE 1 8
•
Factoring Trinomials
Factor completely: Sol u tion
2X2 + 5x + 3 = (x + 1 ) ( 2x + 3 )
2x2 - x - 6
Comparing 2x2 - x - 6 to Ax2 + Bx + C, we find that A = 2, B C = -6.
STEP 1: The value of AC is 2 . ( -6) = - 12. STEP 2: Determine the integers whose product is AC
=
=
- 1 , and
-12 and compute their sums.
Integers whose
product is - 1 2
1, -12
-1, 12
2, - 6
Sum
-1 1
11
-4
- 2, 6
3, - 4
- 3, 4
4
-1
1
STEP 3: The integers whose product is - 12 that add up to B
J
2X2 - X - 6
STEP 4: Factor by grouping.
J
=
=
- 1 are -4 and 3. w
l'
t
2X2 - 4x + 3x - 6
r
2x2 - 4x + 3x - 6 = (2X2 - 4x) + (3x - 6 ) = 2x( x - 2 ) + 3 ( x - 2) =
(x - 2 ) (2x + 3 )
As a result, 2X2 - X - 6 " "� -
Now Work
=
(x - 2) (2x + 3 )
•
PROBLEM 57
SUM MARY Type of Polynomial
Method
Example
Any polynomial
Look for common monomial factors. ( Always do this first!)
6x2 + 9 x
Binomials of degree 2 or higher
Check for a special product: Difference of two squares, x2 - a2 Difference of two cubes, x3 - a3 Sum of two cubes, x3 + a3
Trinomials of degree 2
Check for a perfect square, (x
Three or more terms
±
Factoring x2 + Bx + C (p. 52) Factoring Ax2 + Bx + C (p. 54) Grouping
af
=
3x(2x + 3 )
x2 - 1 6 = ( x - 4 ) ( x + 4 ) x 3 - 6 4 = ( x - 4 ) (x2 + 4 x + 16) x3 + 27 = (x + 3 ) (x2 - 3x + 9) x2 + 8x + 16 = ( x + 4)2 x2 - lOx + 25 = ( x - 5 ) 2 x 2 - X - 2 = ( x - 2) (x + 1 ) 6x2 + x - I = (2x + 1 ) (3x - 1 ) 2x3 - 3x2 + 4x - 6
=
(2x - 3 ) (x2 + 2)
S6
CHAPTER R
Review
R.S Assess You r Understa nding Concepts and Voca b u l a ry
)
\>�
If factored completely,
3x3 - 12x =
� True or False
__.
)1', True or False
If a polynomial cannot be written as the product of two other polynomials (excluding
1 and - 1 ) , then the polynomial is said
x2 + 4 is prime. 3x3 - 2x2 - 6x + 4 = (3x - 2)(x2 + 2).
The polynomial
to be
Skill B u i lding
In Problems 5-14, factor each polynomial by removing the common monomial factor. 8. 3x + 6 6. 7x - 14 ax2 + a
�
�
12.
ax - a
3x2 - 3x
In Problems 15-22, factor the difference of two squares. t'( x2 - 1 16. x2 - 4
K x2 - 16
20.
x2 - 25
In Problems 23-32, factor the perfect squares.
�x2 + 2x + 1 28.
24.
� x2 + 4x + 4
x2 - 4x + 4
� 4x2 + 4x + 1
x2 + lOx + 25
30.
18.
9x2 - 1
22.
36x2 - 9
�X2 - lax + 25
M. 16x2 + 8x + 1
9x2 + 6x + 1
'I'
32.
\
25x2 + lax + 1
In Problems 33-38, factor the sum or difference of two cubes.
. � x3 - 27
34.
� x3 + 27
x3 + 125
In Problems 39-50, factor each polynomial.
� x2 + 5x + 6 '2m. x2 + 7x + 10 V '-
40.
x2 + 6x + 8
44.
x2 + llx + 10
48.
x2 - 2x - 8
In Problems 51-56, factor by grouping.
� 2x2 + 4x + 3x + 6 54.
52.
58. 62. 66.
2x2 + 3x +
1
3x2 + lax + 8 3x2 - 14x + 8
38.
�. x2 + 7x + 6
,5\ x2 - lax + 16 )� x2 + 7x - 8
64 - 27x3
42.
x2 + 9x + 8
46.
x2 - 17 x + 16
50.
x2 + 2x - 8
� 2X2 - 4x + x - 2
3x2 - 3x + 2x - 2
In Problems 57-68, factor each polynomial.
��Ux2 + 14x + 8
27 - 8x3
)'t 6x2 + 9x + 4x + 6
3x2 + 6x - x - 2
�3X2 + 4x + 1 �. 3x2 + 2x - 8
36.
56.
)� 2z2 + 5z + 3 � 3x2 - 2x - 8 /?Z..,3x2 + lax - 8
9x2 - 6x + 3x - 2 60.
pz,2 + 5z + 1
64.
3x2 - lax + 8
68.
3x2 - lax - 8
I
In Problems 69-11 6, factor completely each polynomial. If the polynomial cannot be factored, say it is prime.
�ti.. x2 - 36 x2 + llx + 10 � 4x2 - 8x + 32 •. 15 + 2x - x2 �/ l + ll i + 30i .� 6x2 + 8x + 2 . � x6 - 2x3 + 1 16x2 + 24x + 9
� •
�
70.
x2 - 9
74.
x2 + 5x + 4
78.
3x2 - 12x + 15
82.
14 + 6x - x2
86.
3i - 18i - 48y
94.
8x2 + 6x - 2 x6 + 2x3 + 1
98.
9x2 - 24x + 16
90.
IJ(. 2 - 8x2 '7J x2 - lax + 21 ....
?{ x2 + 4x + 16 �x2 - 12x - 36 );t. 4x2 + 12x + 9 X X4 - 81
)� x7 - x5 5 + 16x - 16x2
�
72.
3 - 27x2
76.
x2 - 6x + 8
80.
x2 + 12x + 36
84.
x3 + 8x2 - 20x
88.
9x2 - 12x + 4
92.
X4 - 1
96.
x8 - xS
100.
5 + llx - 16x2
SECTION R.6 Synthetic Division
�. 41 - 16y 15 � x(x 3 ) - 6(x + 3 ) �'9. (3x - 2)3 - 27 +
+
112.
7(x2 - 6x
+
+
9)
+
102.
91
106.
5 (3x - 7) 110.
5(x -
+
x(3x - 7 )
(5x +
�. x3
3)
�
9y - 4
+
W.
1 - 8x2 - 9x4 (x
2f - 5(x + 2)
+
�. 3 (x2
1 )3 - 1
2x2 -
X
-2
114.
116.
x3
X4 +
+
X
104.
4 - 14x2 - 8x4
108.
( x - 1 )2 - 2 ( x - 1 )
l Ox
+
25)
x3 - 3x2 -
X
+
+
- 4(x
+
57
5)
3
+1
Applications a n d Extensions
(fi In Problems 1 1 7-126, expressions that occur in calculus are given. Faclor completely each expression. 117. 2(3x + 4)2 + (2x + 3 ) ' 2(3x + 4) ' 3 118 . 5 (2x + 1 )2 + (5x - 6 ) · 2 (2x 119. . 1 21.
2x(2x 2(x
+
+
5) +
(4x - 3 )2
125.
2 (3x - 5) ' 3 (2x x2
'
+
+
(x
2 ( 4x -
123.
127. Show that
x
1) ' 2
x2 . 2
3 ) (x - 2)3 +
+
+
1 )3
+ 3 )2 . 3 (x - 2)2 3) · 4
+
(3x
- 5)2 . 3 ( 2x
+
1 )2 · 2
4 i s prime.
+
122 .
4(x
124.
3x2(3x
126.
3(4x
5)3(x _
+
+
4)2
+
1 )2
+
x3 . 2(3x
5f · 4 ( 5x +
128. Show that
x2
+
x
+
1 )2
5 t · 2 (x -
+
(x
+
+
1)
4) · 3
(4x
+
5)3 · 2(5x +
1) ' 5
1 i s prime.
Discussion and Writing 129. M a k e u p a polynomial t h a t factors i n t o a perfect square.
1 30. Explain to a fel l o w student what you look for first when presented with a factoring problem. What do you do next?
OBJECTIVE
1
1
Divide Polynomials Using Synthetic Division (p. 57)
Divide Pol ynomials Using Synth etic Division
To find the quotient as well as the remainder when a polynomial of degree 1 or higher is divided by x - c, a shortened version of long division, called synthetic division, makes the task simpler. To see how synthetic division works, we will use long division to divide the polynomial 2x3 - x2 + 3 by x - 3. 2X2 + 5x + 15 + 3 x - 3 hx3 - x2 2 2x3 - 6x 5x2 5x2 - 15x 15x + 3 15x - 45 48
v"CHECK:
<"-
<"-
Q u otien t
Remainder
(Divisor ) ' ( Quotient) + Remainder 2 = (x - 3 ) (2x + 5x + 1 5 ) + 48 2 = 2x3 + 5x + 15x - 6x2 - 15x - 45 2 = 2x3 - x + 3
+
48
58
CHAPTER R
Review
The process of synthetic division arises from rewriting the long division in a more compact form, using simpler notation. For example, in the long division above, the terms in blue are not really necessary because they are identical to the terms directly above them. With these terms removed, we have x
-
2X2 + 5x + 1 5 + 3 3 hx3 - x2 6 x2 5x2 -
- 15x 15x
- 45 48 Most of the x's that appear in this process can also be removed, provided that we are careful about positioning each coefficient. In this regard, we will need to use 0 as the coefficient of x in the dividend, because that power of x is missing. Now we have x
-
2X2 + 5x + 15 1 0 3)2
3
-
- 6
5 - 15 15 - 45 48 We can make this display more compact by moving the lines up until the numbers in blue align horizontally. x
-
2X2 + 5x + 15 3h - 1 0 - 6 5
o
-
3
1 5 - 45 15 48
Row 1
Row 2 Row 3 Row 4
B ecause the leading coefficient of the divisor is always 1 , we know that the leading coefficient of the dividend will also be the leading coefficient of the quotient. So we place the leading coefficient of the quotient, 2, in the circled position. Now, the first three numbers in row 4 are precisely the coefficients of the quotient, and the last number in row 4 is the remainder. Thus, row 1 is not really needed, so we can compress the process to three rows, where the bottom row contains both the coef ficients of the quotient and the remainder. o
- 1
- 6
-
5
2
3
1 5 - 45 15 48
Row 1 Row 2 (su btract) Row 3
Recall that the entries in row 3 are obtained by subtracting the entries in row 2 from those in row 1 . Rather than subtracting the entries in row 2, we can change the sign of each entry and add. With this modification, our display will look like this: x
-
3h
- 1
6
0
15 15
3 45 48
-------
2
5
Row 1 Row 2 (add) Row 3
Notice that the entries in row 2 are three times the prior entries in row 3. Our last modification to the display replaces the x 3 by 3. The entries in row 3 give the quotient and the remainder, as shown next. -
SECTION R.6
3) 2
2
-1 6 5
3 45 48
0 15 15
�
,-"------, 2x2 + 5x + 15 Quotient
48
Row
Row
Synthetic Division
S9
1
2 (add)
Row 3
"" /
Remai nder
Let's go through an example step by step. Using Synthetic Division to Find the Quotient and Remainder
EXA M P LE 1
Use synthetic division to find the quotient and remainder when x3 - 4x2 - 5
is divided by
x - 3
STEP 1: Write the dividend in descending powers of x. Then copy the coefficients,
Solution
remembering to insert a 0 for any missing powers of x. 1
-4
-5
0
Row 1
STEP 2: Insert the usual division symbol. In synthetic division, the divisor is of the
form x - c, and c is the number placed to the left of the division symbol. Here, since the divisor is x - 3, we insert 3 to the left of the division symbol. -4 0
3h
-5
Row 1
STEP 3: Bring the 1 down two rows, and enter it in row 3.
3h
1
-4 0
-5
Row 1 Row 2
Row 3 1 STEP 4: Multiply the latest entry in row 3 by 3, and place the result in row 2, one column over to the right.
311 -4 3
0
-5
Row 1 Row
2
Row 3
STEP 5: Add the entry in row 2 to the entry above it in row 1 , and enter the sum in
row 3.
3 11 -4 3 lY -l
0
-5
Row 1
Row 2
Row 3
STEP 6: Repeat Steps 4 and 5 until no more entries are available in row l .
0 -5 311 -4 3 -3 -9 l Y-IY-3 Y- 14
Row 1
Row
2
Row 3
STEP 7: The final entry in row 3, the - 14, is the remainder; the other entries in row 3,
the 1 , - 1 , and - 3, are the coefficients (in descending order) of a polynomial whose degree is 1 less than that of the dividend. This is the quotient. Thus,
"
Quotient = x2 -
X
- 3
Check: (Divisor) ( Quotient) + Remainder
Remainder = - 1 4
2 = (x - 3) (x -
X
= x3 - 4x2 - 5
=
=
- 3 ) + ( - 14 ) (x3 - x2 - 3x - 3x2 + 3x + 9) + ( - 14) Dividend
Let's do an example in which all seven steps are combined.
•
60
CHAPTER R
Review Using Synthetic Division to Verify a Factor
EXAMPLE 2
Use synthetic division to show that x + 3 is a factor of
2xs
+
5x4 - 2x3
2X2 - 2x
+
3
+
The divisor is x + 3 = x - (-3), so we place -3 to the left of the division symbol. Then the row 3 entries will be multiplied by -3, entered in row 2, and added to row 1.
Solution
-3 )2
5 -2 2 -2 3 3 -3 3 -3 -----2 -1 1 -1 1 °
Row 1 Row 2
-6
Row 3
Because the remainder is 0, we have (Divisor) ( Quotient) + Remainder
= (x + 3) (2x4 - x3 + x2 - X + 1 )
2x5
=
+
5x4 - 2x3
+
2x2 - 2x
+
3
As we see, x + 3 is a factor of 2xs + 5x4 - 2x3 + 2X2 - 2x + 3. As Example 2 illustrates, the remainder after division gives information about whether the divisor is, or is not, a factor. We shall have more to say about this in Chapter 5. �;.
-
Now Work
PROBLEMS 7 AND 1 7
R.6 Assess Your U nderstanding Concepts and Voca bulary
" To check division, the expression that is being divided, the dividend, should equal the p{6duct of the
2.
To divide
2x3 - 5x +
" True or False •
4. True or False
1 by
x + 3
using synthetic division, the first step is to write
____
In using synthetic division, the divisor is always a polynomial of degree
-2 )5
3 -10 -7
5
2 14 16
1 means
-32 -31
5x3 + 3x2 + 2x + l x + 2
)
=
5x- - 7x +
16
____
and the
).
1 , whose leading coefficient is 1 . +
-31 -. x + 2
Skill Building
In Problems 5-16, use synthetic division to find the quotient and remainder when: \:f,\ x3 - x2 + 2x + 4 is divided by x - 2 6. x3 + 2x2 - 3x + 1
. � 3x3 + 2x2 - X + 3
�
x5 - 4x3 + x
is divided by
)$� 0.lx3 + 0.2x � x5 1
x + 3
is divided by
4x6 - 3x4 + x2 + 5
is divided by
is divided by
is divided by
x - 3
x + 1 .1
x - 1
8. -4x3 + 2x2 - X + 10. X4 + x2 + 2
x - 1
14. 0.1x2 - 0.2 16. x 5 + 1
is divided by 1
x - 2
is divided b y
is divided by
is divided b y
)§.�
�
3x4 - 6x3 - 5x + 10; 3x6 + 82x3 + 27;
x - 2 x - 2
x +3
18.
-4x3 + 5x2 + 8;
x + 1
x + 2.1
x + 1
In Problems 1 7-26, use synthetic division to determine whether x - c is a factor of the given polynomial.
,� 4x3 - 3x2 - 8x + 4;
x + 1
is divided by
is divided by
12. x5 + 5x3 - 10
____
x + 3
20. 4X4 - 1 5x2 - 4; x - 2 22. 2x6 - 1 8x4 + x2 - 9; x + 3
x + 2
plus the
SECTION R.7
23. 4x6 - 64x4 + x2
'lfi: 2x .
,
4
-
- x3 + 2x - 1 ;
15;
x + 4
Rational Expressions
61
24. x 6 - 1 6x4 + x2 - 1 6; x + 4 1 3
1 x -2
26. 3x4 + x3 - 3x + 1 ; x + -
Appl ications a n d Extensions
X
Find the s u m of a,
b, C, a n d d i f x 3 - 2X2 + 3 x + x + 2
5
-------
=
?
ax- + bx +
C
+
d
--
x + 2
Discussion a n d Writin g
28.
When dividing a polynomial by
x -
c,
do you prefer to use long division or synthetic division? Does the value of C make a differ
ence to you in choosing? Give reasons.
R.7 Rational Expressions OBJECTIVES
1
'
' .. "
.. ... ..
.
'
1 Reduce a Rational Expression to Lowest Terms (p. 61) 2
Multiply and Divide Rational Expressions (p. 62)
3
Add and Subtract Rational Expressions (p. 63)
4
Use the Least Common Multiple Method (p. 65)
5
Simplify Complex Rational Expressions (p. 67)
Red u ce a Rational Expression to Lowest Terms
If we form the quotient of two polynomials, the result is called a rational expression. Some examples of rational expressions are
xl (x y )2 Expressions ( a) , ( b ) , and (c) are rational expressions in one variable, x, whereas ( d ) is a rational expression in two variables, x and y . ( a)
x3 + 1 -x-
3x2 + X 2 x2 + 5
(b)
( c)
-
x2
x
-
1
(d)
_
Rational expressions are described in the same manner as rational numbers. In expression ( a ) , the polynomial x3 + 1 is called the numerator, and x is called the denominator. When the numerator and denominator of a rational expression con tain no common factors ( except 1 and - 1), we say that the rational expression is reduced to lowest terms, or simplified. The polynomial in the denominator of a rational expression cannot be equal to a x3 + 1 because division by a is not defined. For example, for the expression --- x cannot x take on the value O. The domain of the variable x is {xix =1= a } . A rational expression is reduced to lowest terms by factoring completely the numerator and the denominator and canceling any common factors by using the Cancellation Property: ,
ae be EXAMPLE 1
if b =1= 0,
C
=1= °
( 1)
Reducing a Rational Expression to Lowest Terms
Reduce to lowest terms: Sol ution
a b
x2 x2
+ +
4x 3x
+
+
4
2
We begin by factoring the numerator and the denominator.
x2 + 4x x2 + 3x
+ +
4 2
=
=
(x (x
+ +
2)(x 2)(x
+
2)
+ 1)
62
CHAPTER R
Review
WA RNING
Since a common factor, x + 2, appears, the original expression is not in lowest terms. To reduce it to lowest terms, we use the Cancellation Property:
Apply the Cancel lation Property only to rational expressions written in factored form. Be sure to • cancel only common factors!
x2 + 4x + 4 x2 + 3x + 2
+X---+-2J' (x + 2) �(x + 1 )
x + 2 x + 1
x
=F-
-2, - 1 ..
Reducing Rational Expressions t o Lowest Terms
EXAMPLE 2
Reduce each rational expression to lowest terms.
Solution
( a)
x3 - 8 x3 2x2
(b)
( a)
x3 - 8 x3 - 2x2
¥--2} ( x2 + 2x + 4)
(b)
8 - 2x x2 - x - 12
CJ1JR ;
2
_
mOl>-
x2¥--2}
8 - 2x x2 - x - 12 x2 + 2x + 4 x2
2(4 - x)
2 ( - 1 )�
(x - 4 ) ( x + 3)
�(x + 3 )
Now Work
x =F- 0, 2 -2 x+3
x
=F-
-3, 4 •
PROBLEM 5
M u ltiply a n d Divide Ration al Expressions
The rules for multiplying and dividing rational expressions are the same as the rules for multiplying and dividing rational numbers. If b and d ' b =F- 0, d =F- 0, are two rational expressions, then
a
a e b d a b e d
_
==
a d b e
_ e _
ae bd
=
ad be
_
e
if b =F- 0, d =F- °
if b =F- 0, e =F- 0, d =F- °
(2)
(3)
In using equations (2) and (3) with rational expressions, be sure first to factor each polynomial completely so that common factors can be canceled. Leave your answer in factored form. EXAMPLE 3
M ultiplying and Dividing Rational Expressions Perform the indicated operation and simplify the result. Leave your answer in fac tored form.
( a)
x2 - 2x + 1 4x2 + 4 3 x + X x2 + X - 2
x + 3 x2 - 4 (b) x2 - X 12 3 x - 8 -
SECTION R.7
( a)
Solution
Rational Expressions
63
(x - 1 )2 4(x2 + 1 ) x(x2 + 1 ) ( x + 2 ) ( x - 1 ) ( x - 1 )� (4)�
x2 - 2x + 1
x �-rr ( x + 2)f.r--1+ 4(x - 1 ) x(x + 2)
x *" -2, 0, 1
x+3 x+3 x3 - 8 x2 - 4 (b ) . x2 - X - 12 x2 - 4 x2 - X - 1 2 --
-,:--
(x - 2) (x2 + 2x + 4) x+3 (x - 2)(x + 2) (x - 4) (x + 3) ��--zn x2 + 2x + 4) ��(x + 2)(x - 4)� x2 + 2x + 4 x *" -3, -2, 2, 4 (x + 2)(x - 4) ""=>:> -
3 ('
r
In
Word s
�
(' To add (or subtract) two rational (' expressions with the sa m e r denominator, keep t h e common denominator and add (or sub (' tract) the numerators.
Now Work
PROBLEMS 1 7 AND 25
Add a n d Su btract Rational Expressions
The rules for adding and subtracting rational expressions are the same as the rules for adding and subtracting rational numbers. So, if the denominators of two rational expressions to be added ( or subtracted ) are equal, we add ( or subtract) the numer ators and keep the common denominator. If
a
b
and
c
b
. . are two ratIOnal expressIOns, th en
a
+ b
c
a c --
-+-= b b
EXAMPLE 4
•
a
b
c
b
a - c
b
if b *" 0
(4)
Adding and Subtracting Rational Expressions with Equal Denominators
Perform the indicated operation and simplify the result. Leave your answer in fac tored form.
3x + 2 x 5 x *" - (b) _ x-3 2 x-3 (2x2 - 4) + (x + 3 ) x+3 2x2 - 4 + ( a) 2x + 5 2x + 5 2x + 5 2X2 + X - 1 (2x - 1 ) (x + 1 ) 2x + 5 2x + 5 x - 3x - 2 x 3x + 2 x - (3x + 2) - --(b ) x-3 x 3 x 3 x-3 -2x - 2 -2(x + 1 ) x-3 x-3 (a)
Sol ution
2X2 - 4 x+3 + 2x + 5 2x + 5
_
_
x *" 3
-
•
64
CHAPTER R
Review
EXA M P L E 5
Adding Rational Expressions Whose Denominators Are Additive Inverses of Each Other Perform the indicated operation and simplify the result. Leave your answer in fac tored form.
5 2x + x-3 3-x
--
Solution
-
x
*
3
Notice that the denominators of the two rational expressions are different. However, the denominator of the second expression is the additive inverse of the denominator of the first. That is,
3 - x = -x + 3 =
-
1 (x - 3 ) = - (x - 3 ) '
Then
5 5 2x 2x -5 2x = + + = + x-3 3-x x - 3 -(x - 3 ) x - 3 x - 3
--
--
--
--
i
a
3 - x = - (x - 3)
Ii''''
, ;;; .;p..
Now Work
i
--
-a
b
-b
2x + ( -5) x-3
2x - 5 x-3
•
P R O B L E M S 3 7 A N D 43
If the denominators of two rational expressions to be added or subtracted are not equal, we can use the general formulas for adding and subtracting quotients.
a c a · d b · c ad + bc - + - = -- + -- = b d b·d b·d bd a c a · d b · c ad - bc -- - -b d b·d b·d bd
EXAM P L E 6
if b
0, d
*
0
( Sa)
if b #- 0, d
*
0
(5b)
*
Adding and Subtracting Rational Expressions with Unequal Denominators Perform the indicated operation and simplify the result. Leave your answer in fac tored form. (a ) (b )
Solution
(a )
x-3 x + x * -4, 2 x+4 x-2 x2 1 -� x * -2, 0, 2 x2 4 x-3 x x - 3 . x - 2 x + 4 .x + + x+4 x-2 x+4 x-2 x+4 x-2
--
--
_
--
--
=
I
--
(5a )
--
--
(x - 3 ) (x - 2 ) + (x + 4) (x) (x + 4) (x - 2) x2 - 5x + 6 + x2 + 4x (x + 4)(x - 2)
2x2 - X + 6 (x + 4) (x - 2)
SECTION R.7
(b )
x2 x2 - 4
--
-
1
X
-
r
=
x2 X x2 - 4 X
-- . -
-
x2 - 4 1 x2 - 4 X
-- . -
=
Rational Expressions
65
x2 ( x) - (x2 - 4) ( 1 ) (x2 - 4)(x)
------'-'------'----'---'-�
(5b)
(x - 2)(x + 2) (x) li'l=i;> -
4
Now Work
•
PROBLEM 47
Use the Least Common M u ltiple M ethod
If the denominators of two rational expressions to be added ( or subtracted) have common factors, we usually do not use the general rules given by equations (Sa) and ( 5b ) . Just as with fractions, we apply the least common multiple (LCM) method. The LCM method uses the polynomial of least degree that has each denominator poly nomial as a factor. The LCM Method for Adding or S u btracting Rational Expressions
The Least Common Multiple ( LCM) Method requires four steps:
STEP 1: Factor completely the polynomial in the denominator of each rational
expression.
STEP 2: The LCM of the denominator is the product of each of these factors
raised to a power equal to the greatest number of times that the factor occurs in the polynomials. STEP 3: Write each rational expression using the LCM as the common denominator. STEP 4: Add or subtract the rational expressions using equation (4).
Let's work an example that only requires Steps 1 and 2. EXAMPLE 7
Finding the Least Common Multiple Find the least common multiple of the following pair of polynomials:
x(x - 1 )2(x + 1 ) and 4(x - l ) (x S o l utio n
+
1 )3
STEP 1: The polynomials are already factored completely as
x(x - 1 )2(x + 1 ) and 4(x - l ) (x + 1 )3
STEP 2: Start by writing the factors of the left-hand polynomial. ( Or you could start
with the one on the right. )
Now look at the right-hand polynomial. Its first factor, 4, does not appear in our list, so we insert it.
4x(x - 1 )2(x + 1 ) The next factor, x - I, is already in our list, so no change is necessary. The final factor is (x + 1 )3. Since our list has x + 1 to the first power only, we replace x + 1 in the list by (x + 1 )3. The LCM is
4x(x - 1 )2(x + 1 )3
•
66
CHAPTER R
Review
Notice that the LCM is, in fact, the polynomial of least degree that contains
x(x - 1 )2(x + 1 ) and 4(x - 1 ) (x + 1 )3 as factors. ';!J! �
E XA M P L E 8
Now Work
PROBLEM 53
Using the Least Common M u ltiple to Add Rational Expressions Perform the indicated operation and simplify the result. Leave your answer in fac tored form.
2x - 3 x + x2 + 3x + 2 x2 - 1
-----
Solution
x
---
-2, -1, 1
*
STEP 1: Factor completely the polynomials in the denominators.
x2 + 3x + 2 x2 - 1
= =
(x + 2) (x + 1 ) ( x - 1 )( x + 1 )
STEP 2: The LCM is ( x + 2)(x + 1 ) (x - 1 ) . Do you see why? STEP 3: Write each rational expression using the LCM as the denominator.
x x2 + 3x + 2
2x - 3 x2 - 1
x (x + 2)(x + 1 )
r
x x- I . (x + 2) (x + 1 ) x - I
------
--
=
- 1 ) -x(x "------'-(x + 2 ) ( x + 1 ) (x - 1 )
-
M ultiply numerator and denominator by x - 1 to get the LCM in the denominator.
(2x - 3 ) (x + 2) 2x - 3 2x - 3 x+2 (x - 1 ) (x + 1 ) i (x - 1 ) (x + 1 ) x + 2 (x - 1 ) (x + 1 ) (x + 2) I M ultiply n u merator and ------ . --
--''--'-'-
denominator by x + 2 to get the LCM in the denominator.
STEP 4: Now we can add by using equation
2x - 3 x + x2 + 3x + 2 x2 - 1
----=-
(4). x( x - I ) (2x - 3 ) ( x + 2) + (x + 2)(x + l ) (x - l ) (x + 2) (x + l ) (x - l )
------
=
(x2 - x) + (2x2 + x - 6) (x + 2)(x + 1 ) (x - 1 ) (x + 2)(x + 1 ) (x - 1 )
E XA M P L E 9
3(x2 - 2) ( x + 2)(x + 1 ) (x - 1 )
iii!
Using the Least Common M ultiple to Subtract Rational Expressions Perform the indicated operations and simplify the result. Leave your answer in fac tored form.
3 x2 + Solution
X
x+4 x2 + 2x + 1
x
*
-1 , 0
STEP 1: Factor completely the polynomials in the denominators.
x2 + X x2 + 2x + 1 STEP 2: The LCM is
x(x + I f
(x + 1 ) (x + I ?
= X =
SECTION
STEP
67
3: Write each rational expression using the LCM as the denominator.
3 x2 + x+4 x2 + 2x + 1
3 x(x + 1 )
X
STEP
R.7 Rational Expressions
x+4 (x + 1 )2
3(x + 1 ) 3 x+ 1 x(x + 1 ) x + 1 x(x + 1 )2 x(x + 4) + 4 x (x + 1 )2 X x(x + 1 )2 .-
--:-
X
4: Subtract, using equation
(4). 3 x+4 x2 + x x2 + 2x + l
---
3(x + l ) x(x + 4) x(x + l )2 x(x + l )2 3(x + 1 ) - x(x + 4) x(x + 1 )2 3x + 3 - x2 - 4x x(x + 1)2 -x2 - + 3 x(x + 1 )2 X
,-== � ,....
5
Now Work
•
PROBLEM 63
S i m p l ify Co mplex Rational Expressions
When sums and/or differences of rational expressions appear as the numerator and/or denominator of a quotient, the quotient is called a complex rational expres sion.'" For example,
1 1 + x and 1 I- x
x-3 -1 x+2
--
are complex rational expressions. To simplify a complex rational expression means to write it as a rational expression reduced to lowest terms. This can be accom plished in either of two ways. S i m p l ifyi ng a Complex Rational Expression METHOD
1: Treat the numerator and denominator of the complex rational ex
pression separately, performing whatever operations are indicated and simplifying the results. Follow this by simplifying the result ing rational expression. METHOD 2: Find the LCM of the denominators of all rational expressions that appear in the complex rational expression. Multiply the numerator and denominator of the complex rational expression by the LCM and simplify the result. We will use both methods in the next example. By carefully studying each method, you can discover situations in which one method may be easier to use than the other. E XA M P L E 1 0
Simplifying a Complex Rational Expression
1 3 -+2 x Simplify: x+3 4
x
=F
-3, 0
'" Some texts use the term complex fraction.
68
CHAPTER R
Review
Solution
Method 1: First, we perform the indicated operation in the numerator, and then we
divide.
-1 + -3 2 x x+3 -4-
1
l'x + 2'3 2·x x+3 4
x+6 2x x+3 -4-
--
=
--
Ru le for addi ng quotients
1
=
1
x+6 4 2x x + 3
-- . --
Rule for d ividing quotients
Z ' 2 ' (x + 6) Z ' x ' (x + 3)
(x + 6) , 4 2 ' X ' (x + 3 )
2(x + 6) x(x + 3)
Rule for multiplying quotients
Method 2 : The rational expressions that appear in the complex rational expression
are
1 3 x+3 2' x' 4 The LCM of their denominators is 4x. We multiply the numerator and de nominator of the complex rational expression by 4x and then simplify.
1 3 -+2 x x+3 4
(� + �) (X +4 3 ) 4x ' --
4x '
1
1
1 3 4x ' - + 4x ' x 2 4x · (x + 3 ) 4
M ultiply the Use the Distributive Property n umerator and in the n u merator. denominator by 4x. '2- •
1 3 2x . - + 4a" ' a" '2.4"x · (x + 3 ) .4"
1
2x + 12 x(x + 3)
Simplify
EXAM P L E 1 1
Factor
•
Simplifying a Complex Rational Expression Simplify:
Solution
2(x + 6)
1 x(x + 3 )
x2 -+2 x-4 2x - 2 - 1 --x
x *- 0, 2, 4
We will use Method 1 .
x2 -+2 x-4
2x - 2 --- - 1 x
= {I!l!l;; =-
Now Work
2(x - 4) x2 + 2x - 8 x2 -- + --'-'-x-4 x-4 x-4 2x - 2 x 2x - 2 - x x x x (x + 4) (x - 2) (x + 4)� x-4 x-2 x-4 x (x + 4) ' x x-4
PROBLEM 73
x
•
Rational Expressions
SECTION R.7
69
Appl ication
EXAM P L E 1 2
Solving an Application in Electricity An electrical circuit contains two resistors connected in parallel, as shown in Fig ure 28. If the resistance of each is Rl and R2 ohms, respectively, their combined resistance R is given by the formula
Figure 2 8
R1
1 R = ---1 1 - +
Rl
R2
Express R as a rational expression; that is, simplify the right-hand side of this for mula. Evaluate the rational expression if R] = 6 ohms and R2 = 10 ohms. We will use Method 2. If we consider 1 as the fraction expressions in the complex rational expression are
Solution
T' then the rational
The LCM of the denominators is R 1 R2 . We multiply the numerator and denomi nator of the complex rational expression by R l R2 and simplify. 1 1 -
Rl
+
1
R2
Thus,
If
Rl = 6
and
R2 = 10,
then 6 · 10 R = --10 + 6
60
15
16
4
ohms
•
R.7 Assess Your Understanding Concepts a n d Vocabulary
� When the numerator and denominator of a rational expres sion contain no common factors (except tional expression is in
2.
1
and
- 1 ) , the
\
�
LCM is an abbreviation for __ __ __
The rational expression
to lowest terms.
ra-
__ ___ .
True or False
4. Trite or False
4x3 (x + 1 ) .
The LCM of
2x3 - 4x i s reduced x-2
2x3 + 6x2
and
6x4
Ski l l Building
In Problems 5-16, reduce each rational expression to lowest terms. 6. 4x2 + 8x 12x + 24 12x- - 6x x 2 + 4x - 5 , . x2 - 2x + 1
10.
x 2 + 4x + 4 X2 - 4
1 4.
-=x2 +x -2
x - x2
\)f
x2 - 2x r" 3 x - 6
)
•
l - 25 2 2y - 8y - 10 x2 + 5x - 14 2-x
8.
15x2 + 24x 3x-
--::-?-
31 - y - 2 3y- + 5 y + 2 2x2 + 5x - 3 1 6. 1 - 2x 12.
-": ?:---'----
+
4x3
is
70
CHAPTER R
Review
In Problems 1 7-34, pelform the indicated operation and simplify the result. Leave your answer in factored form. x 'WI. 3x + 6 12 x3 + 1 4x2 x3 - 64 3 x2 20. 18. �. � . x2 - 4 2x 2x ' 6x + 1 0 x2 + X 4x - 2 x2 16
\m. 4x - 8
12 K � ' 1 2 - 6x
24 •
\:2 + x - 6
.
-
22.
-----
25.
x2 + 4x - 5 x2 + 2x - 1 5 8x
27.
28.
x2 + 7x + 1 2 31.
6x - 27 2 5x 4x - 18
0 ---
23.
6x
x2 - 25
x2 - 1 lOx x + 1
x2 - 4 3x - 9 2x + 4
26.
�--=--'
x 2 - 7x + 1 2 x2 + X - 12
32.
x2 - x - 1 2
x2 - 3x - 1 0 x2 x2 + 2x - 35 x2
+
+
4x - 21 9x + 14
1 2x 5x + 20
--= -
4x2
x2 - 16
x - 2 4x
29.
x2 - 4x + 4 12x
4 - x 4 + x 4x
---
30.
3 + x 3 - x X2- 9
x2 - 1 6
x2 + 7x + 6
.:..-'.
•
_
•
9x3 9x2 + 3x - 2
2x2 - x - 2 8
x2 + x - 6 x2 + 5x - 6
- -33. -,----
3x2 - x - 2 4x2 + 1 6x + 7
- -34. --::--
x2 + 5x + 6
3x2 + 1 1 x + 6
8x2 - lOx - 3
1 2x2 + 5x - 2 9x2 - 6x + 1
[n Problems 35-52, perform [he indicated operation and simplify the result. Leave your answer in factored form. 35.
-
x 5 +2 2
3 x
39.
41.
-3x + 5 2x - 1
2x - 4 2x - 1
- --
50
42.
--- -6 x - I
x I - x
44. -- -
. 47.
6 x
36. - - -
.�-
-- --- -x + 1 x - 3
2x - 3 x - 3
+
5x - 4 3x + 4
-
4 x - I
40.
x + 1 3x + 4
43.
2 x + 2
46.
3x 2x + -x - 4 x + 3
49.
45. -- - --
x 2x - 3 + x + l x - I
2x - 3
37.
48.
2x + 1 x + l
51.
---
1
x
+x x2 - 4
52.
-- --- -x2 4 2x - 3 2x - 3 2x - 5 3x + 2
+
x + 4 3x + 2
4 x -- + -x - 2 2 - x
-- --- --- -2 x + 5
-
5 x - 5
x - 3 x + 2
-
x + 4 x - 2
x - I x3
+
x
x2 + 1
[n Problems 53-60, find the LCM of the given polynomials. 53.
x2 - 4,
56.
3x2 - 27,
x2 -
X
-
2
54.
x2 - x - 12,
x2 - 8x + 1 6
2x2 - X - 15
58.
x - 3,
x2 + 3x,
x3 - 9x
In Problems 61-72, pelform the indicated operations and simplify the result. Leave your answer in factored form. 61.
64.
X
x
x2 - 7x + 6
x2 - 2x - 24
�
x - I
-
x - 4 x2 - 2x + l
x + 4
2x + 3
r - x - 2
x2 + 2x - 8
, --67. -
62.
65.
68.
-x x - 3
x + 1
- -=,-
r + 5x - 24
3
(x - 1 )2 ( x + 1 ) 2x - 3 x2 + 8x + 7 1
71 . -
-
(-- ) 1
h x+h
1 x
- -
+
2
(x - 1 ) (x + 1 )2
---f x -2
(x + 1
66.
69.
72
2 ( x + 2 )2 ( x - 1 )
.!. x
.!.
. h
_
[
_
6 ( x + 2 ) ( x - 1 )2
2 3 _ + x2 + x x3 - X2
_
1 (x +
hi
-
�J x2
SECTION R.7
Rational Expressions
71
In Problems 73-84, perform the indicated operations and simplify the result. Leave your answer in factored form. x _ o_ x + 1 1 __ 1+� 4 + -.!... 2_ X + 1 2 x X x 76 . 74. -. 73. -75. x- 1 1 1 x - 1 2 - -1 -3 - -2 3+ x x x + 1 x x+4 x-3 x-2 x+ 1 x+ 1
-- - --
77.
81. 1 -
x - 2 -x x+ 1 x-2 x+3
79.
x-2 x-1 + - x +2 x + 1 ------=x 2x - 3 x+ 1 x
83.
--=------'-
--
--
78.
1
1 1 - x
--
82. 1 -
1
----
1 -
1 1 -x
--
80.
2(x - 1 ) - 1 + 3 3(x - 1 t ) + 2
2x + 5 -x x x-3 (x + 1 ) 2 x-? -X - 3 x+3 ---
4(x + 2 ) - 1 - 3 3 (x + 2)- - 1
84. --=-------'-- ) -
!fi In Problems 85-92, expressions that occur in calculus are given. Reduce each expression to lowest terms. �. (2x + 3) · 3 - (3 - 5 ) · 2 x · 2x - (x2 + 1 ) · 1 (4x + 1 ) · 5 - (5x - 2) · 4 � � ? (3x - 5 )(5x - 2) 2 (x2 + 1 t �
�
88.
x . 2x - ( Xl - 4) . 1
91.
(x2 + 1 ) · 3 - (3x + 4) · 2x
?
(3x + 1 ) · 2x - x2 • 3 � (3x + 1 )2
89.
(x2 - 4t
--
90.
-
92.
?
(x2 + 1 t
(2x - 5 ) · 3x2 - x3 . 2 (2x - 5)-?
( Xl + 9) · 2 - (2x - 5 ) · 2x ? (x2 + 9 t
Appl ication s a nd Extensions
A
The Lensmaker's Equation
index of refraction
1 f
-
where
R)
and
Rl
The focal length
n is =
[
1 1 ( n - 1 ) -- + -RI R 2
f o f a lens with
sistors connected in parallel. If the resistance of each is
]
R1 , Rz , and R3 ohms, respectively, their combined resistance R is given by the formula
and back surfaces of the lens. Express
f
expression.
expression
n
=
1 .5, R)
=
0.1
1 1 1 1 - = -- + -- + - R RJ R2 R3
are the radii of curvature of the front
Evaluate
the
meter, and
rational
Rl
=
A n e lectrical circuit contains three re
94. Electrical Circuits
as a rational
R as a rational expression. Evaluate R 5 ohms, R2 4 ohms, and R3 10 ohms.
Express
for
RJ
0.2 meter.
=
=
for
=
Discussion a n d Writing
�
The following expressions are called
continued fractions:
1 1 +- 1 + x'
1
--
1' 1 + x
1 +
1
------
1 +
1
--
1
1 + x
1 +
-----=---
1 +
1
1
------
1 +
1
--
1 1 + x
Each simplifies to an expression of the form
ax + b bx + c Trace the successive values of
a, b, and c as you "continue" the fraction. Can you discover the patterns that these values follow? Go
to the library and research Fibonacci numbers. Write a report on your findings.
96. Explain to a fellow student when you would use the LCM
97. Which of the two methods given in the text for simplifying
method to add two rational expressions. Give two examples
complex rational expressions do you prefer? Write a brief
of adding two rational expressions, one in which you use the
paragraph stating the reasons for your choice.
LCM and the other in which you do not.
72
CHAPTER R
Review
PREPARING FOR THIS SECTION •
Before getting started, review the following:
Exponents, Square Roots (Section R.2, pp. 21-24)
'NOW Work
the 'Are You Prepared?' problems on page 77.
OBJECTIVES
1
1 Work with nth Roots (p. 72) 2
Simplify Radicals (p. 73)
3
Rationalize Denominators (p. 74)
4
Simplify Expressions with Rational Exponents (p. 75)
Work with nth Roots
DEFINITION
The principal nth root of a real number Va , is defined as follows: where
a
;:::
0 and b
Va
;:::
=b
a, n
means
;:::
2 an integer, symbolized by
a = bIt
0 if n is even and a, b are any real numbers if n is odd.�
Notice that if a is negative and n is even then Va is not defined. When it is defined, the principal nth root of a number is unique. The symbol Va for the principal nth root of a is called a radical; the integer n is called the index, and a is called the radicand. If the index of a radical is 2, we call Va the square root of a and omit the index 2 by simply writing va . If the index is 3, we call -0a the cube root of a.
EXA M P LE 1
Simplifying Principal nth Roots (a) (c)
Vs = V23 = 2
V(=4)3 = - 4 ( d ) � = 1-21 = 2 (b)
� = 1(�r = �
V=64 =
•
These are examples of perfect roots, since each simplifies to a rational number. Notice the absolute value in Example I( d). If n is even, the principal nth root must be nonnegative.
In general, if n
;:::
2 is an integer and a is a real number, we have
yi;t = a yi;t = l al """
$- -
Now Work
if n
;:::
3 is odd
(la)
if n
;:::
2 is even
(lb )
PROBLEM 7
Radicals provide a way of representing many irrational real numbers. For example, there is no rational number whose square is 2. Using radicals, we can say that v2 is the positive number whose square is 2.
SECTION R.S
i3 I
EXAMPLE 2
5 "'.f 16
73
Using a Calculator to Approximate Roots
-YI6.
Use a calculator to approximate Figure 29
nth Roots; Rational Exponents
Solution
Figure
29
shows the result using a TI-S4 plus graphing calculator.
Now Work
1.741101127 2
•
1 01
PRO B L E M
Simplify Radicals Let n 2:::
2
and
m
2:::
2
denote positive integers, and let
a and b represent real num
bers. Assuming that all radicals are defined, we have the following properties: Properties of Radicals
(2a) (2b) (2c)
When used in reference to radicals, the direction to "simplify" will mean to remove from the radicals any perfect roots that occur as factors. Let's look at some examples of how the preceding rules are applied to simplify radicals.
EXAMPLE 3
Simplify ing Radicals (a)
v'32
VJ:6-2
=
i
=
i
Vi6. V2
=
4V2
(2a) Factor out 16, a perfect square. (b)
\Yi6 = \Y8-2 i
Factor out 8, a perfect cube.
(c)
=
i
VB· V2 \)'23. V2 = 2V2 =
(2a)
V-16x4 = �-S·2'x3.x = \y(-8x3)(2x) i i Group perfect Factor perfect cubes inside radical. cubes.
=
\y(-2x?·2x \y(-2x?' � -2x� =
=
i
(2a)
(d)
�l�t �24;44X 4 C;y·x 1( 2; y .\Yx 1 2;1 \Yx =
r. '!l!DII5I IIIC>< IIO .-
Now Work
=
=
=
11
PRO B L EMS
AND
•
1 7
Two or more radicals can be combined, provided that they have the same index and the same radicand. Such radicals are called
EXAMPLE 4
like radicals.
Combining Like Radicals (a)
-SV12
+
v3
-8V4-3 v3 = -8' V4 v3 v3 +
=
=
-16v3
+
+
v3
=
-15v3
74
CHAPTER R
Review
( b)
\)8x4 + -0CX +4�=V2Yx + V-1·x +4�
=v(2x)3 0X + v=1 . 0X +4\.Y33. 0X =2x0X- 1 0X + 120X · = (2x + 1l)0X .
== &1'J1l: -
3
Now Work
•
PRO B L EM 3 3
Rationalize Denominators When radicals occur in quotients, it is customary to rewrite the quotient so that the new denominator contains no radicals. This process is referred to as rationalizing the
denominator. The idea is to multiply by an appropriate expression so that the new denomi nator contains no radicals. For example:
If a Denominator Contains the Factor
Multiply by
To Obtain a Denominator Free of Radicals
v3 v3 + 1 V2-3 vs-v3 �
v3 v3 -1 V2 +3 vs+v3 V2
(v3? =3 (v3)2- 12=3-1=2 (V2?-32=2-9= -7 (VS)2-(v3)2=5-3=2 �·V2=�=2
In rationalizing the denominator of a quotient, be sure to multiply both the numerator and the denominator by the expression .
EXAMPLE 5
Rationalizing Denominators Rationalize the denominator of each expression:
Solution
5 ( b) _ _ 4V2
1
( a)
-
( a)
The denominator contains the factor
( b)
v3
denominator by
V3 to
(c) V3,
V2 v3-3V2
so we multiply the numerator and
obtain
1 1 = . v3 = v3 =-v3 3 v3 v3 v3 (v3? The denominator contains the factor denominator by
V2 to obtain
V2,
so we multiply the numerator and
5 = 5 V2= 5V2 5V2 5V2 8 4V2 4V2· V2 4(V2? 4·2 ( c) The denominator contains the factor V3-3 V2, so we multiply the numera tor and denominator by V3 + 3V2 to obtain V2 v3 +3V2 V2( v3 +3V2) v3-3V2· v3 +3V2 (v3? -(3V2)2 V2 v3 +3(V2y V6 +6 6 + V6 3-18 -15 15 �-
Now Work
PRO B LEM
47
•
SECTION R.8
4
nth Roots; Rational Exponents
75
Simplify Expressions with Rational Exponents Radicals are used to define rational exponents.
DEFINITION
If
a is a real number and
n 2::
2
is an integer, then
(3) provided that
Va exists.
Note that if n is even and
EXAMPLE 6
a < 0 then Va and al/n do not exist.
Writing Expressions Containing Fractional Exponents as Radicals (a) (c)
41/2 = \i4 (_27)1/3
=
DEFINITION
If
2
=
�
(b)
-3
=
(d)
81/2 Vs = 2\12 161/3 = V'16 = 2V2 =
•
a is a real number and m and n are integers containing no common factors,
with
n
2:: 2, then
(4) provided that
Va exists.
We have two comments about equation
1. The exponent
m
n
(4):
must be in lowest terms and
2. In simplifying the rational expression
n
must be positive.
am/n, either � or (Va)111 may be used,
the choice depending on which is easier to simplify. Generally, taking the root first, as in
EXAMPLE 7
(Va),,', is easier.
Using Equation (4) (a) (c)
=
(\i4y 23 (32r2/5 = ( 032 ) -2
43/2 =
,'i�'
' 1� -
Now Work
=
8
=
1 r2 = 4"
(v=st (_2)4 16 (V2s/ = 53 = 125 (d) 256/4 = 253/2 (b)
(_8)4/3 =
=
=
=
•
PROB L E M 55
It can be shown that the Laws of Exponents hold for rational exponents. The next example illustrates using the Law of Exponents to simplify.
EXAMPLE 8
Simplifying Expressions Containing Rational Exponents Simplify each expression. Express your answer so that only positive exponents occur. Assume that the variables are positive. (c)
(
)
9X2//3 1/2 X1/3 Y
76
CHAPTER R
Review
Solution
( a)
(x2/3y)(x-2y/12 = (x2/3y)[(x-2/12)11/2] =x2/3yx-lyl/2 = (X2/3 .[1)(y . )11/2) x-1/3112 112 xV3 / /13 3 (/13)3 2xl/3 -3 = l /13 2x /3 (2xl/3)3 23 (xI/3)3 =
(b)
( ) ( )
=
/ 8x
=
• �I"!
==> -
Now Work
PRO B L E M 7 1
The next two examples illustrate some algebra that you will need to know for certain calculus problems.
EXAMPLE 9
Writing an Expression as a Single Quotient Write the following expression as a single quotient in which only positive exponents appear.
(x-+ 1)1/2 +X' -1 (x-+ 1r l/2 . 2x 2 x2 1 2 + 1r1/7-'Zx=(x7 +1) 1/-+ 7 (x-+ 1) 1/7-+x'-(x Z (x2 + 1)1/-7 (x2 + 1) l/2 (x2 + 1) 1/2 +x2 (x2 + 1)l/2 (x2 + 1) +x2 (x2 + 1//2 2x2 + 1 (x2 + 1)1/2 7
Solution
7
&(IJ !_ we·
EXAMPLE 10
- Now Work
•
PRO B L E M 77
Factoring an Expression Containing Rational Exponents
4 x1/3 (2x + 1) +2x4/3 .) We begin by writing 2x4/3 as a fraction with 3 as denominator. /3 4 x1/.. ' (2x 6x4/3 = 4xl/\2x + 1) +6x4/3 /3 = 4xl (2x + 1) +-+ 1) + 2x4 3 3 3 3 Factor:
Solution
7
::;-
0
----'-----'-----
i
Add the two fractions
2 and
1l'1!I::=� - Now Work
)13
i
2x1/3[2(2x + 1) +3x] 3
are common factors
PRO B L E M 89
i
2x1/\7x +2) 3
Simplify
•
SECTION R.S
nth Roots; Rational Exponents
77
f-H�torica l Feature
T
he radical sign,
V, was first used in print by Christoff Rudolff in
1525. It is thought to be the manuscript form of the letter =
the Latin word radix
r
(for
root),although this is not quite conclusively
confirmed. It took a long time for
V to become the standard symbol
for a square root and much longer to standardize
-0', {/, �,
3
and
-0'8. The notation
V V16 was popular for
The bar on top of the present radical symbol, as follows,
Va2 + 2ab + b2
is the last survivor of the vinculum, a bar placed atop an expression to
indicate what we would now indicate with parentheses. For example,
V8 3
ab + c = a(b + c)
R.S Assess Your Understanding
Answers are given at the end of these exercises. If you get a wrong answel; read the pages in red. ; -32 ; v(=4)2 (pp. 2 1-24) Vi6 = (pp. 21-24)
'Are You Prepared?'
( )2 )( -3
=
__
=
\
__
=
__
__
Concepts and Vocabulary
�
In the symbol -Va, the integer n is called the
\We call .;ya the
__ __
.
__
of a.
4. True or False
�
6 . True o r False
�
=
-2 =
-3
Skill Building
In Problems 7-42, simplify each expression. Assume thai all variables are positive when they appear.
'K Vii :R: Vs
ts� � f
K�
NVs�)2
�3v2 + 4v2 P\ (\1'3 + 3)(\1'3 '[t (Vx ) - 1 2
1
)
8.
fu
jf, �
10.
\V=l
12.
\154
}� �
14.
�
16.
�
20.
y'9x5
24.
(\13 ViOt
32.
(Vs - 2) (Vs
36.
(Vx + Vs)2
)( v£3 - 3v5(h 'r\�-3X�+5�
18.
( V3.?VUx X. (3V6)(2v2) Jf: -V18 + 2Vs +
3)
22.
26.
2vTI - 3V27 34. V24 - V'sl 9
X·�-� 40. 3xv9Y + 4 v25.Y 8xy - V25x2/
(5Vs)( -3\1'3)
30.
�. 5\12 - 2\154
42.
\& y'2Ox3
+
38.
\Y32x +
�
V8x3i
In Problems 43-54, rationalize the denominator of each expression. Assume that all variables are positive when they appear. -\1'3 -\1'3 2 1 46.
� v2
44.
48.
\1'3
v2
r.:.
v7 + 2
,
ffl.
By the 1700s, the index had settled where we now put it.
and so on.
The indexes of the root were placed in every conceivable position, with
V8, V@8,
all being variants for
�
Vs
Vs
50.
\1'3t::. 2 v3 + 3 ,
- 1
CHAPTER R
78
Review
52.
"
-2
\Y9
In ProbLems 55-66, simpLify each expression. 2--6. 43/2 � 82/3 62.
Vx+h. - \IX Vx+h.+ \IX
�. � -27) 1/3
58.
+� 5 4. Vx+h � - ,vx ,vx+h � -h
163/4
60.
253/2
( )
66.
(-8 ) -2/3
27 2/3 . 8
64
16-3/2
27
In Problems 67-74, simplify each expression. Express your answer so that only positive exponents occur. Assume that the variables are positive. � x3/4XI/3x-I/2
Applications and Extensions
In Problems 75-88, expressions thaI occur in calculus are given. Write each expression as a single quotient in which only positive exponents and/or radicals appear. \. X 1 +x 7 + 2(1+x) /2 x> -1 7J£. 6 +x1- /2 x>o f' (1+x)l/2x 1/2 I
I
?
•
. (x+ 1)1/3+x. '!3'(x+ 1t2/3
78
�.�.
1 +�. 1 2� 5V4x+3
� - x ' -= -1= 2� "T' (, 1+x
¥-
(x+4)1/2 - 2x(x +4tl/2 x+4
x"* 2,x"* - 8'1
x>5 ,V� x"+ 1
x>-1
82.
2x - x ·----=== 2� r+1
-� � -
---
-3 < x
x>-4
x < -1
x"* -1
or
<
3
x> 1
x"*-1,x"*1
x>O
In Problems 89-98, expressions that occur in calculus are given. Factor each expression. Express your answer so that only positive exponents occur. � (x + 1)3/2+x' (x+1) /2 X -1 90. (x2+4)4/3+x' (x2 + 4//3 '2x
�
1
2:
�
SECTION R.8
92.6x1/2 (2x + 3)
x?:O
+
nth Roots; Rational Exponents
79
x3/2. 8
94.2x(3x + 4)413 + x2. 4(3x + 4) 1/3 .. •
?A
4(3x + 5)1/3(2x + 3)3/2 + 3(3x + 5)4/3(2x + 3)1/2 "
3 2
r> -
-
96. 6(6x + 1)1/3(4x - 3)3/2 + 6(6x + 1)4/3 (4x - 3)1/2 " - 3 4
r> -
9S. 8xl/3 - 4x-2/3
x>O
x '1-0
In Problems 99-106, use a calculator to approximate each radical. Round your answer to two decimal places. 99. V2 100. V7 V4 102. v=s l.j
1. 2 + \13 3-
104.
Vs
Vs
lrtz � 3Vs - V2
2 V2 + 4 -
5.
-=---
--
\13
106.
2\13V2 - V4
Applications and Extensions
'f/:( Calculating the Amount of Gasoline in a Tank
1
r
A Shell station stores its gasoline in underground tanks that are right circular cylinders lying on their sides. See the illustration. The volume V of gasoline in the tank (in gallons) is given by the formula V
=
40h2
)�: - 0.608
where h is the height of the gasoline (in inches) as measured on a depth stick. (a) If h = 12 inches, how many gallons of gasoline are in the tank? (b) If h = 1 inch, how many gallons of gasoline are in the tank? lOS. Inclined Planes
height
h feet is
The final velocity v =
V64h +
v
of an object in feet per second (ft/sec) after it slides down a frictionless inclined plane of
V
B
where Va is the initial velocity (in ft/sec) of the object. (a) What is the final velocity v of an object that slides down a frictionless inclined plane of height 4 feet? Assume that the initial velocity is O. (b) What is the final velocity v of an object that slides down a frictionless inclined plane of height 16 feet? Assume that the initial velocity is O. (c) What is the finaL-velocity v of an object that slides down a frictionless inclined plane of height 2 feet with an initial velocity of 4 ftlsec?
Problems
109-112
require the following information. The period T, in seconds, of a pendulum of length l, in feet, may be approximated using the formula
Period of a Pendulum
T
=
27r
-ff
In Problems 109-112, express your answer both as a square roo/ and as a decimal. T of a pendulum whose length is 64 feet. 110. Find the period T of a pendulum whose length is 16 feet. W.. Find the period T of a pendulum whose length is 8 inches. 112.Find the period T of a pendulum whose length is 4 inches.
�. Find the period
Discussion and Writing
�
GiVe an example to show that
W is not equal to a. Use it to explain why W = lal ·
'Are you Prepared?' Answers 1. 9 ; - 9
�'<:::""----'
2.4;4
h
80
CHAPTER R
Review
CHAPTER REVIEW Things to Know
Well-defined collection of distinct objects, called elements
Set (pp. 2-3)
Empty set or null set
o
Set that has no elements
Equality
A=B
A and B have the same elements.
Subset
ACB
Each element of A is also an element of B.
Intersection
An B
Set consisting of elements that belong to both A and B
Union
AUB
Set consisting of elements that belong to either A or B, or both
U
Set consisting of all the elements that we wish to consider
Universal set
Set consisting of elements of the universal set that are not in A
A
Complement Classify numbers (pp. 4-5)
Integers
1,2,3 ... 0,1,2,3 . .. ... , -2, -1,0,1,2,
Rational numbers
Quotients of two integers (denominator not equal to
Irrational numbers
Nonrepeating nonterminating decimals
Real numbers
Rational or irrational numbers
Counting numbers Whole numbers
. . .
0); terminating or repeating decimals
Work with properties of real numbers (pp. 9-14)
Commutative properties
a + b =b + a, a . b=b·a
Associative properties
a + (b + c)= (a + b) + c, a·(b·c)=(a·b)·c
Distributive property
a'(b + c)=a·b + a'c
Identity properties
a+O=a
Inverse properties
a + (-a)= 0 a' -= 1 where a =1= 0 a
a'1=a
1
If ab = 0,then a = 0 or b= 0 or both.
Zero-Product Property
lal =a if a
Absolute value (p. 19)
2:
0
lal=-a if a
Exponents (p. 22)
an=a' a' .... a n
n
<
1 _ a n=an
a positive integer
0 a
=1=
0,
n
a positive integer
factors
Laws of exponents (p. 22)
(al1l)1'I =amn
(a·b)n=a"·bn
Pythagorean Theorem (p. 30)
In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. Converse of the Pythagorean Theorem (p. 30)
In a triangle, if the square of the length of one side equals the sum of the squares of the lengths of the other two sides, then the tri angle is a right triangle. Polynomial (p. 40)
Algebraic expression of the form a x" + a,,_jXn-1 + ... + ajX + ao, n
n
a nonnegative integer
Special products/factoring formulas (pp. 43-44)
(x - a)(x + a)=x2
-
a2
(x + a) 2=x 2 + 2ax + a2 (x + a)(x + b) = x 2 +
(x - a) 2=x 2
(a + b)x
-
2ax
+ a2
+ ab
(ax + b)(cx + d)=acx 2 + (ad + bc)x + bd (x + a)3=x3 + 3ax2 + 3a 2x + a3 (x - a)3=x3 3ax2 + 3a2x a3 (x + a)(x2 - ax + a2) =x3 + a 3 (x - a)(x2 + ax + a2) =x3 - a3 -
-
Chapter Review
81
Radicals; Rational exponents (pp. 72 and 75)
%
alll/"
=
b means a
=
.,;yam
Objectives Section
=
=
b" where
(\Ya)1II
In,
n
a
:2':
:2':
0 and b
:2':
0 if n
:2':
2 is even and
min in lowest terms
2
a, b are any real numbers if n
2
3
4
R.2 2
3
4 5
6
3 is odd.
--------� You should be able to ...
R.l
:2':
Review Exercises
Work with sets (p. 2 )
1-8 9,10 11-14,109
Classify numbers (p. 4 )
Evaluate numerical expressions (p. 8 )
Work with properties of real numbers (p. 9 )
17,18
Graph inequalities (p. 18 ) Find distance on the real number line (p. 19 ) Evaluate algebraic expressions (p. 20) Determine the domain of a variable (p. 21 ) Use the Laws of Exponents (p. 21 ) Evaluate square roots (p. 23 )
19,20 21,22 23-27,108 29,30 83,84 15,27
Use a calculator to evaluate exponents (p. 24 ) Use scientific notation (p. 24 )
33 36,105
3
Use the Pythagorean Theorem and its converse (p. 30 ) Know geometry formulas (p. 31 ) Understand congruent triangles and similar triangles (p. 32 )
106,107,112 110,111 31,32
2
Recognize monomials (p. 39 ) Recognize polynomials (p. 40)
34 35
7
8
R.3 2
RA 3
4
Add and subtract polynomials (p. 41 ) Multiply polynomials (p. 42 )
37,38 39-44
6
Know formulas for special products (p. 43 ) Divide polynomials using long division (p. 44 ) Work with polynomials in two variables (p. 47 )
41,42 45-50 39-40
2
Factor the difference of two squares and the sum and difference of two cubes (p. 50) Factor perfect squares (p. 51 ) Factor a second-degree polynomial: x2 + Bx + e (p. 52 )
57,58,61,62 65,66 51,52,64 59,60 53-56,63
5
7
R.5 3
4 5
R.6 R.7 2
3
4 5
R.8 2
3
4
Factor by grouping (p. 53 ) Factor a second-degree polynomial: Ax2 + Bx
+
e, A
'"
1 (p. 54 )
Divide polynomials using synthetic division (p. 57 )
45,46,49,50
Reduce a rational expression to lowest terms (p. 61 ) Multiply and divide rational expressions (p. 62 ) Add and subtract rational expressions (p. 63 ) Use the least common multiple method (p. 65 ) Simplify complex rational expressions (p. 67 )
67,68 69,70,75 71-74 71-74 76
Work with nth roots (p. 72 ) Simplify radicals (p. 73 ) Rationalize denominators (p. 74 ) Simplify expressions with rational exponents (p. 75 )
28,87-90 77-82,87-90,101,102 91-96 85,86,97-100,103,104
Review Exercises
In Problems 1-8, use each set.
U
=
universal set
=
{1, 2, 3, 4, 5, 6, 7, 8, 9 } , A
=
{ I, 3, 5, 7 } , B
=
{ 3, 5, 6, 7, 8 } , and e
=
{ 2, 3,7, 8, 9 } to find
2. BUe
. Ane
4. AnB
6. B
"Bne
8. AUB
82
CHAPTER R
Review
In Problems 9 and 10, list the numbers in the set that are (a) Natural numbers, (b) Integers, (c) Rational numbers, (d) Irrational numbers, (e) Real numbers.
1/" A
=
{ -10, 0.65,1.343434 . . . , V7, �}
10. B
In Problems 11-16, evaluate each expression. 5 + 1 12. \C -6+4'(8 - 3) 18 12
=
{ 0,-5,� ,0.59,1.333 . . . ,2V2,� }
5
16. (-3)5 11 27 In Problems 17 and 18, use the Distributive Property to remove the parentheses. 18. (x - 2)(3x + 1) lR: 4(x - 3) In Problems 1 9 and 20, graph the numbers x on the real number line. �. x > 3 20. x 5 In Problems 21 and 22, let P, Q, and R be points on a real number line with coordinates -2,3, and 9, respectively. 22. Find the distance between Q and R. � Find the distance between P and Q. '-
14.�
I
:s;
In Problems 23-28, evaluate each expression if x
�. x+ ""'- �
=
-5 and y
=
28. �
26. 2x2y-2
24. 12x - 3yl
Y
7.
In Problems 29 and 30, determine the domain of the variable. I
x+1 x+5 In Problems 31 and 32, the triangles are similal: Find the missing length x and the missing angles A, B, and C. 3
R-x - 6
30.
-
32.
Y/' c
�. Use a calculator to evaluate (1.5)4. j�.
34. Identify the coefficient and degree of the monomial -5x3.
Identify the coefficients and degree of the polynomial
\ 3x5
+
4X4 - 2x3 + 5x - 12.
In Problems 37-44, perform the indicated operation. "tf-. (2x4 - 8x3 + 5x - 1) +(6x3 + x2 + 4)
1t.. (2x + y) (3x - 5y)
36. Write 3.275
lOs as an integer.
38. (x3 + 8x2 - 3x + 4) - (4x3 - 7x2 - 2x + 3)
4j( (4x + 1) (4x - 1)
40. (2x - 5y) (3x +2y)
¥ (x + l)(x + 2) (x - 3)
42. (5x + 2)2
X
44. (x + l)(x + 3)(x - 5)
In Problems 45-50, find the quotient and remainder. Check your work by verifying that (Quotient) (Divisor) + Remainder Dividend 46. 2x3 - 3x2 + X + 1 4ft,- 3x3 - x2 + X + 4 divided by x - 3 =
�. -3x4 + x2 + 2 )t x5 + 1
divided by
divided by
x+1
x2 + 1
divided by
x -2
48. -4x3 + x2 - 2 divided by x2 - 1 50. x5 - 1 divided by x -I
In Problems 51-66, factor each polynomial completely (over the integers). If the polynomial cannot be factored, say it is prime. x2 + 5x - 14 52. x2 - 9x + 14 6x2 - 5x - 6 54. 6x2 + x - 2 §. 56. 2x3 + 18x2 + 28x �. 58. 27x3 - 8 "- 3x2 - 15x - 42 1'<. 8x3 + 1 "I v� 2x3 + 3x2 - 2x - 3 60. 2x3 + 3x2 + 2x + 3 � 25r - 4 62. 16x2 - 1 \wi. 64. x2 - x + 1 x2 + 8x + 16 C6i, 66. 4x2 + 12x + 9 r� 9x2 + 1 f\
� • I
�
?
Chapter Review
�
83
In Problems 67-76, perform the indicated operation and simplify. Leave your answer in factored form. 2x2+ llx+14 x2 - 5x - 14 9x2 - 1 3x - 9 68. 2-4 4- 2 x2 - 9 9x2+6x+1 x2 - 25 x2+ x x 2x 70. 72. -- - -3 - 4r - 5x 0 --1 - x2 x+1 x+2 �7.;{ 3x+4 2x - 3 x2 x2 +- --74. ----,--of\: x2 - 4 x2+4x+4 ? -2 2r+5x - 3 2x2 - 5x+ X
•
X
�
•
X
• ::-----
?
_
x+4 3 76.-1-+ 2 4 -x
x2 - 5x+6 x+ 1 x-2
In Problems 77-86, simplify each expression. All exponents must be positive. Assume that x
R V32 82.
78.
V75
11- v=16
80.
>
0 and y
>
1164
0, when they appem: �( 5Vs - 2 V32
4V12+ 5V27
In Problems 8 7-90, simplify each expression. x2 �t1 0, Y > 0 871 . - - x .." 25/
HI;
2:
88.
90.
YhvSx; ,v;:;-;;lOy
x
2:
0, Y > 0
� x > 0,Y > 0 , 4 ;::;-Q V 3x/
In Problems 91-96, rationalize the denominator of each expression. -2 92. �� V3 v· Vs -4 94. ��1+Vs � 1+ ,v3 1 - Vs
. -/ 2 J� 1 - v'2
96.
4V3+ 2 2V3+ 1
In Problems 9 7-102, write each expression as a single quotient in which only positive exponents and/or radicals appear. � (2+x2)112+x . (2+x2)-1I2·2x 98. (x2 + 4)2/ + x. (x2+4 r1/3·2x
�
3
(x2
+
�
�
4) 1 /2·2 x - x2. (x2+4r /2 . 2x
100.
1
x2+ 4 4+x2 - 2xVX 2VX (4+ ? ) ?
-102.
In Problems 103 and 104, factor each expression. t'I 6. 3(x2+4)4/3+ x·4(x2+4) 1/3·2x "
�. U.S. Population
According to the US. Census Bureau, the "- US. population in 2000 was 281,421,906. Write the US. population in scientific notation. 106. Find the hypotenuse of a right triangle whose legs are of lengths 5 and 8. 1 The lengths of the sides of a triangle are 12,1 6, and 20. Is this f a right triangle?
�
104.
r -
2x(3x+5)5/3+x2. 4(3x+5)2/3 The weekly production cost C of man
108. Manufacturing Cost
x calculators is given by the formula x2 C 3000+6x - 1000 What is the cost of producing 1000 calculators? What is the cost of producing 3000 calculators?
ufacturing
=
(a) (b)
CHAPTER R
84
Review
� Quarterly Corporate Earnings
In the first quarter of its fis cal year, a company posted earnings of $1.20 per share. Dur ing the second and third quarters, it posted losses of $0.75 per share and $0.30 per share, respectively. In the fourth quarter, it earned a modest $0.20 per share. What were the annual earnings per share of this company? 110. Design A window consists of a rectangle surmounted by a trian gle. Find the area of the window shown in the illustration. How much wood frame is needed to enclose the window?
On a recent flight to San Francisco, the pilot announced that we were 139 miles from the city, flying at an altitude of 35,000 feet. The pilot claimed that he could see the Golden Gate Bridge and beyond. Was he telling the truth? How far could he see?
112. How far can a pilot see'!
6 f! I}---I-t----I
�.
5f!
113. Use the material in this chapter to create a problem that uses
each of the following words: (a) Simplify (b) Factor
A statue with a circular base of radius 3 feet is enclosed by a circular pond as shown in the illustration. What is the surface area of the pond? How much fence is required to enclose the pond? Construction
(c) Reduce
CHAPTER TEST
� List the numbers in the set that are (a) Natural numbers,
(b) Integers, (c) Rational numbers, (d) Irrational numbers, (e) Real numbers.
{
0,1.2,
� Evaluate each expression if x (a) 3x-1i
� }
\12, 7, , 1T =
y
-3 and
y
=
4.
�
� A
erform the indicated operation. (a) (-2x3 + 4 + 10) (b) (2x - 3)(-5x + 2)
x2 - 6x
� Factor each polynomial. (a) x2 - 6x + 8 (b) 4x2 - 25 (c) 6x2 - 19x - 7
'\
Simplify each expression.
- (6x3 - 7x2 + 8x
V27
(b)
-0'=8
(c)
(x;2)�3 (x-y) 32
.. Rationalize the denominator of
(c) Vx2+7 (d) 5x3 - 3x2 + 2x - 6 \:i."The triangles below are similar. Find the missing length x and t the missing angles A, E, and C.
4 0°
divided by x - 2.
(a)
(b) 12x - 3 l
2
� Find the quotient and remainder if x3 - 3x2
C
-
1)
(d)
\13.
j\
b
=
x
>
-3.
The Zero-Product Property states that if 0 then either or , or both.
Fill in the blanks
a·
x
8x - 10 is
(16x4/3y-2/3)3/2
5 + 3 � Graph the numbers on the real number line:
�
+
__
__
Construction A rectangular swimming pool, 20 feet long and 10 feet wide, is enclosed by a wooden deck that is 3 feet wide. What is the area of the deck? How much fence is re quired to enclose the deck?
Equations and Inequalities
Interest Rates Fall on 15- and 30-Year Loans From the Associated Press May 12, 2006 Rates on 3D-year fixed-rate mortgages averaged 6.58% this week, down from 6.59% last week, mortgage company Freddie Mac said. Rates on 15-year, fixed-rate mortgages, a popular choice for refi nancing a home mortgage, fell to 6.17%, down from 6.22% last week. One-year adjustable-rate mortgages (ARM) fell to 5.62%, down from 5.67%. Rates on 5-year hybrid adjustable-rate mortgages averaged 6.22%, up slightly from 6.21% last week. The mortgage rates do not include add-on fees known as points. The 1-year ARM carried a nationwide average fee of 0.7 point, and the three other mortgage categories had an average fee of 0.5 point. Source: Adapted from Martin Cn.llsingel; "Rates on 30-year Mongages Edge Down Slightly." Associated Press Financial Wire, May 11, 2005. ©2006 The
Associated Press.
-See the Chapter Project-
A Look Ahead
Outline
Algebra. If your instructor decides to exclude complex numbers from the course,
1 .2
Chapter 1, Equations and Inequalities, reviews many topics covered in Intermediate
don't be alarmed. The book has been designed so that the topic of complex numbers can be included or excluded without any confusion later on.
1.1 linear Equations
Quadratic Equations
1.3 Complex Numbers; Quadratic Equations in the Complex Number System 1.4 Radical Equations; Equations Quadratic
1 .5
in Form; Factorable Equations Solving Inequalities
1.6 Equations and Inequalities Involving Absolute Value 1.7 Problem Solving: Interest, Mixture, Uniform Motion, and Constant Rate Job Applications Chapter Review Chapter Test Chapter Projects
85
86
CHAPTER 1
Equations and Inequalities
1.1 Linear Equations Before getting started, review the following:
PREPARING FOR THIS SECTION •
•
Properties of Real Numbers (Section R.l, pp. 9-14) Now Work
Domain of a Variable (Section R.2, p. 21)
the 'Are You Prepared?, problems on page 94. 1 Solve a Linear Equation (p.
OBJECTIVES
88)
2 Solve Equations That Lead to Linear Equations (p.90)
3 Solve Applied Problems Involving Linear Equations (p. 92)
An
equation in one variable is a statement in which two expressions, at least one sides of the equa
containing the variable, are equal. The expressions are called the
tion. Since an equation is a statement, it may be true or false, depending on the value of the variable. Unless otherwise restricted, the admissible values of the variable are those in the domain of the variable. These admissible values of the variable, if any, that result in a true statement are called solutions, or roots, of the equation. To solve an equation means to find all the solutions of the equation. For example, the following are all equations in one variable, x:
x +5 = 9
x2 + 5x = 2x - 2
x2-4 =0 x + 1
,V� x2 + 9=5
---
x + 5= 9, is true when x =4 and false for any other x + 5 = 9. We also say that 4 x + 5= 9, because, when we substitute4 for x, a true state
The first of these statements, choice of
x. That
is,4 is a solution of the equation
satisfies the equation ment results.
Sometimes an equation will have more than one solution. For example, the equation
x2-4 =0 x + 1
--
has
x = -2
and
x= 2
as solutions.
Usually, we will write the solution of an equation in set notation. This set is called the
solution set of the equation. For example, the solution set of the equation
x2
Ois
-
9=
{-3,3}.
Some equations have no real solution. For example,
x2 + 9=5
solution, because there is no real number whose square when added to
has no real
9equals5.
An equation that is satisfied for every value of the variable for which both sides are defined is called an
identity. For example, the equation
3x + 5= x + 3 + 2x + 2 is an identity, because this statement is true for any real number
x.
One method for solving an equation is to replace the original equation by a succession of equivalent equations until an equation with an obvious solution is obtained. For example, all the following equations are equivalent.
2x + 3 = 13 2x = 10 x=5
We conclude that the solution set of the original equation is
{5}.
How do we obtain equivalent equations? In general, there are five ways.
SECTION 1.1
Linear Equations
87
Procedures That Result in Equivalent Equations
1. Interchange the two sides of the equation: Replace
3
by
x
=
x
=
3
2. Simplify the sides of the equation by combining like terms, eliminating parentheses, and so on:
(x + 2) + 6 x + 8
Replace by
= =
2x + (x + 1) 3x + 1
3. Add or subtract the same expression on both sides of the equation:
5 (3x - 5) + 5
Replace
3x
by
-
=
=
4 4 +5
4. Multiply or divide both sides of the equation by the same nonzero expression:
6 x -I
3x x -I
Replace
x*- 1
3x 6 ·(x - 1) '(x -1) x -I x -I equation is 0 and the other side can be
by
S. If one side of the
--
=
--
factored, then
we may use the Zero-Product Property" and set each factor equal to WARNING Squaring both sides of an
Replace
equation does not necessarily lead to
by
_
an equivalent equation.
x(x - 3) x
=
0
or
x
-
3
=
0
=
0
0:
Whenever it is possible to solve an equation in your head, do so. For example,
8 is x
The solution of
2x
The solution of
3x - 15
� == ....
Now Work
=
=
=
4.
0 is x
=
5.
PRO B L EM 9
Often, though, some rearrangement is necessary.
EXAMPLE 1
Solving an Equation Solve the equation:
Solution
3x - 5
=
4
We replace the original equation by a succession of equivalent equations.
(3x
-
3x - 5 5) + 5 3x 3x 3 x
The last equation, lent, so
3
x
=
3,
=
4
=
4 + 5
=
9
=
9 3 3
Add 5 to both sides. Simplify. Divide both sides by 3. Simplify.
has the single solution
3. All these equations 3x - 5 4.
is the only solution of the original equation,
'" The Zero-Product Property says that if ab
=
0 then
a =
0 or b
=
=
0 or both equal O.
are equiva
88
CHAPTER 1
Equations and Inequalities
"'CheCk:
It is a good practice to check the solution by substituting
3 for x
in the
original equation.
3x -5 = 4 3(3) 51, 4 -
9-51, 4 4= 4
The solution checks. The solution set is OJD'
§ >-
Now Work
PRO B L E M
{3}.
•
23
Steps for Solving Equations STEP 1: List any restrictions on the domain of the variable. STEP 2: Simplify the equation by replacing the original equation by a succes sion of equivalent equations following the procedures listed earlier.
STEP 3: If the result of Step 2 is a product of factors equal to 0, use the Zero Product Property and set each factor equal to
STEP 4: Check your solution ( s) .
1
° (procedure5).
Solve a Linear Equation Linear equations are equations such as 3x + 12
=
°
-2x + 5= °
1
-x 2
A general definition is given next.
DEFINITION
A
linear equation in one variable is equivalent to an equation of the form
ax + b = ° where
a
and
b
are real numbers and
a
i= 0.
Sometimes, a linear equation is called a side is a polynomial in
x of degree l.
first-degree equation, because the left
It is relatively easy to solve a linear equation. The idea is to
ax + b
=
°
ax = -b x= The linear equation formula
EXAMPLE 2
b x = --. a
-b a
-
ax + b = 0, a
a of- 0
Subtract b from both sides. Divide both sides by a, a of- O.
i= 0, has the single solution given by the
Solving a Linear Equation Solve the equation:
1
-(x + 5) - 4 2
isolate the variable:
=
1
-(2x - 1) 3
SECTION 1.1
Solution
Linear Equations
89
To clear the equation of fractions, we multiply both sides by 6, the least common multiple of the denominators of the fractions
1 1 - (x + 5 ) - 4 = - (2 x - 1 ) 2 3 -
6
[� (x
] [� (2x - 1 ) ]
+ 5) - 4 =
3(x + 5 ) - 6·4
=
3x + 15 - 24 = 3x - 9 = 3x - 9 + 9 = 3x = 3x - 4x = -x = x= Check:
� and �.
6
M ultiply both sides by 6, the LCM o f 2 a nd 3.
2(2x - 1)
Use the Distributive Property on the left a nd the Associative Property on the right.
4x 4x 4x 4x + 4x + 7 -7
2 2 2+9 7 7 - 4x
Use the Distributive Property. Combine like terms. Add 9 to each side. Sim p lify. Subtract 4x from each side. Simplify M u ltiply both sides by -1.
1 1 1 - (x + 5 ) - 4 = - ( -7 + 5 ) - 4 = - ( -2) - 4 = -1 - 4 = -5 2 2 2 1 1 1 1 - (2x - 1 ) -[2( -7) - 1] = - ( - 14 - 1 ) - ( - 1 5 ) = -5 3 3 3 3 =
=
Since the two expressions are equal, the solution x = -7 checks and the solution set is { -7 } . � ==....
EXAM P L E 3
Now Work
•
PR O B L E M 3 3
Solving a Linear Equation Using a Calculator
.
2.78x +
Solve the equatIOn:
2 = 54.06 17.931
Round the answer to two decimal places. Solution
To avoid rounding errors, we solve for x before using the calculator.
2.78x +
2 = 54 . 06 17.931
2 17 . 931 2 54.06 17.93 1 x= 2.78
2.78x
=
54.06 -
Subtract
2
--
17.931
from each side.
--
------
Divide each side by 2.78.
Now use your calculator. The solution, rounded to two decimal places, is 19.41. Check: We store the unrounded solution
2 to evaluate 2.78x + 17.931 ' (2.78)(19.405921 34)
1J!l!l:-== =:�
Now Work
PR O B L E M
65
1 9.40592134 in memory and proceed
�
+ 17. 31 = 54.06
•
90
CHAPTER 1
Equations a n d Inequalities
2
Solve Equations That Lead to Linear Eq uations
The next three examples illustrate equations that lead to linear equations upon simplification. Solving Equations
EXAM P L E 4
( 2y + l ) (y - 1 ) = (y + 5) (2y - 5) (2y + l)(y - 1 ) = (y + 5 ) ( 2 y - 5 ) 21 - y - 1 = 21 + 5 y - 25 Multiply and combine like terms. Subtract 21 from each side. -y - 1 = 5y - 25 Add 1 to each side. -y= 5y - 24 Su btract 5y from each side. -6y= -24 Divide both sides by -6 y= 4 Check: (2y + l ) (y - 1 ) = [2 (4) + 1 ] (4 - 1) = (8 + 1) ( 3) = (9) (3) = 27 (y + 5) (2y - 5 ) = (4 + 5 )[2(4) - 5] = ( 9 ) ( 8 - 5 ) = (9)(3) = 27
Solve the equation: Solution
Since the two expressions are equal, the solution The solution set is {4}. EXAM P L E 5
y= 4 checks. •
Solving Equations
.
Solve the equatIOn:
3 1 + 7 -= -x - 2 x - 1 (x - l ) (x - 2 )
First, we note that the domain of the variable is {xix *" 1, x *" 2}. We clear the equation of fractions by multiplying both sides by the least common multiple of the denominators of the three fractions, (x - 1 ) (x - 2) .
Solution
3 1 7 -- = -- + ----x - 2 x - I (x - l) ( x- 2 ) ( x-
�= ( x - 1 ) ( x - 2) [ x � 1 + ( x - ;( x - 2) ]
Multiply both sides by (x - 1)(x - 2). Ca ncel on the left.
1 + �7 � ---3x - 3 = �(x - 2 ) .x---r
Use the Distributive Property on each side; cancel on the right.
1 )�
1
3x - 3 = ( x - 2 ) + 7 3x - 3 = x + 5 2x = 8
Combine l ike terms. Add 3 to each side. S ubtract x from each side.
x= 4
Divide by 2.
3 = 3 = -3 -x - 2 -4-2 2 1 + 7 1 7 1 +-= 7 2 +-=-= 7 9 -3 -= -- + = x - I (x - 1 ) (x - 2 ) 4 - 1 (4 - 1) (4 - 2) 3 3·2 6 6 6 2 Since the two expressions are equal, the solution x= 4 checks. The solution set is {4}. Check:
• �== -
Now Work
PROB L E M
59
The next example is of an equation that has no solution.
SECTION 1.1
EXA M P L E 6
91
A n Equation with No Solution . S oI ve the equatIOn:
Solution
Linear Equations
3x + 2 - 3 -x- 1 x -1 =
First, we note that the domain of the variable is {x I x "* 1}. Since the two quotients in the equation have the same denominator, x - 1, we can simplify by multiplying both sides by x - 1. The resulting equation is equivalent to the original equation, since we are multiplying by x - 1, which is not 0. (Remember, x -:/= 1).
3 3x -x - 1 + 2= - x-1 3 3X + 2 '(x - 1 (--'� ) ) x =
- 1
Multiply both sides by 1; cancel on the right.
.x----r
x
Use the Distributive Property on the left side; cancel on the left..
3x '� + 2'(x - 1 ) = 3 -.x----r 3x + 2x - 2 = 3 5x 2 = 3 5x = 5 x=1
Simplify.
-
Combine like terms. Add
2 to
each side.
Divide both sides by 5.
The solution appears to be 1. But recall that able. The equation has no solution.
x
=
1 is not in the domain of the vari •
'."
EXA M P L E 7
";..,..- Now Work
PROB
LEM 49
Converting to Fahrenheit from Celsius In the United States we measure temperature in both degrees Fahrenheit (OF) and degrees Celsius (0C), which are related by the formula C
=
t (F - 32 ) . What are
the Fahrenheit temperatures corresponding to Celsius temperatures of 0°, and 30°C? Solution
We could solve four equations for F by replacing C each time by
10°, 20°,
0, 10, 20, and 30. Instead, it is much easier and faster first to solve the equation C = (F - 32) for F and then substitute in the values of C. 5 -(F - 32) 9
C
=
9C
=
9C
=
5(F - 32 ) 5F - 160
=
9C
=
9C
=
9 -C
5F - 160 5F F
Multiply both sides by 9. Use the Distributive Property. Intercha nge sides.
+ 160
5 + 32
Add 160 to each side. Divide both sides by 5.
t
92
CHAPTER 1
Equations and Ineq ualities
We can now do the required arithmetic. O°C: lOoC: 20°C: 30°C: 3
9 + 32 = 32°F -(0) 5 F = -59 (10) + 32 = 50°F
F
=
-9 (20) + 5 9 F = -(3 5 0) +
F
=
=
68°F
32 =
86°F
32
•
Solve Applied Problems I nvolving Linear Equations
Although each situation has its unique features, we can provide an outline of the steps to follow to solve applied problems. Steps for Solving Applied Problems STEP
1: Read the problem carefully, perhaps two or three times. Pay particu
STEP
2:
STEP
3:
STEP
4:
STEP
5:
lar attention to the question being asked in order to identify what you are looking for. If you can, determine realistic possibilities for the answer. Assign a letter (variable) to represent what you are looking for, and, if necessary, express any remaining unknown quantities in terms of this variable. Make a list of all the known facts, and translate them into mathe matical expressions. These may take the form of an equation (or, later, an inequality) involving the variable. If possible, draw an appropri ately labeled diagram to assist you. Sometimes a table or chart helps. Solve the equation for the variable, and then answer the question, usually using a complete sentence. Check the answer with the facts in the problem. If it agrees, congrat ulations! If it does not agree, try again.
Let's look at two examples. EXAM P LE 8
Investments
A total of $18,000 is invested, some in stocks and some in bonds. If the amount in vested in bonds is half that invested in stocks, how much is invested in each category? Solution
STEP
1: We are being asked to find the amount of two investments. These amounts
must total STEP
STEP
$18,000. (Do you see why?)
2: If we let x equal the amounts invested in stocks, then the rest of the money, 18,000 - x, is the amount invested in bonds. Do you see why? Look at Step 3. 3 : We set up a table: Amount in Stocks
Amount in Bonds
x
18,000
Reason
-x
Total invested is $18,000
We also know that Total amount invested in bonds
18,000
- x
is
one-half that in stocks
1(
-
2
x
)
SECTION STEP
X = -x21 18,000 = X + "21 x 3 18,000 = "2x (�) 18,000 = (�)(%X) 12,000 = X
Solution
93
Add
x
to both sides.
Simplify.
2
Multiply both sides by 3' Simplify.
So, $12,000 is invested in stocks and $18,000 - $12,000 = $6000 is invested in bonds. The total invested is $12,000 + $6000 $18,000, and the amount in bonds, $6000, is half that in stocks, $12,000.
=
� ==-
EXA M P L E 9
Linear Equations
18,000 -
4:
STEP 5:
1.1
•
Now Work
PROB
LEM 8 3
Determining an Hourly Wage
Shannon grossed $435 one week by working 52 hours. Her employer pays time-and a-half for all hours worked in excess of 40 hours. With this information, can you de termine Shannon's regular hourly wage? STEP 1: We are looking for an hourly wage. Our answer will be in dollars per hour. STEP 2: Let x represent the regular hourly wage; x is measured in dollars per hour. STEP 3: We set up a table: Hours Worked
STEP
4:
STEP 5:
Salary
Hourly Wage
Regular
40
x
40x
Overtime
12
l.5x
12(1.5x)
=
=
18x
The sum of regular salary plus overtime salary will equal $435. From the table, 40x + 18x 435. 40x + 18x 435 58x = 435 x 7.50 Shannon's regular hourly wage is $7.50 per hour. Forty hours yields a salary of 40(7.50 ) = $300, and 12 hours of overtime yields a salary of 12(1.5) (7.50) = $135, for a total of $435.
�== O>-
Now Work
P ROB
= =
LEM 87
•
Steps for Solving a Linear Equation To solve a linear equation, follow these steps: STEP 1: List any restrictions on the variable. STEP 2: If necessary, clear the equation of fractions by multiplying both sides by the least common multiple CLCM) of the denominators of all the fractions. STEP 3: Remove all parentheses and simplify. STEP 4: Collect all terms containing the variable on one side and all remaining terms on the other side. STEP 5: Simplify and solve. STEP 6: Check your solution(s) . SUMMARY
94
C H A PTER 1
�
Equations and Inequal ities
OIVing equations is among the oldest of mathematical activities,
The solution of this problem u s i ng only words is the earliest form of
and efforts to systematize this activity determined much of the
algebra. Such problems were solved exactly this way i n Babylonia i n BC
W e know almost nothing o f mathematical work before this
shape of modern mathematics.
1800
Consider the followi n g problem and its solution u s i ng only words:
date, although most authorities bel ieve the sophistication of the earli
Solve the problem of how many apples Jim has, given that
est known texts i n d i cates that a long period of previo u s d evelopment must have occurred. The method of writing out equations in words
"Bob's five apples and Jim's apples together make twelve apples" by
persisted for thousands of years, and although it now seems extremely
thinking,
cumbersome, it was used very effectively by many generations of math
"Jim's apples are all twelve apples less Bob's five apples" and then
ematicians. The Arabs d eveloped a good deal of the theory of cubic equation s while writing out all the equation s in words. About
concluding,
AD
1500,
the tendency to abbreviate words in the written equations began to
"Jim has seven apples."
lead in the direction of modern notation; for example, the Latin word et
The mental steps translated into algebra are
5
+x
=
(mean i n g and) developed i nto the plus sign, +. Although the occasional use of letters to represent variables dates back to
12
x = 12 = 7
-
d i d not become common u ntil about
5
AD
AD
1200, the practice
1600. Development thereafter
was rapid, and by 1635 algebraic notation d id not differ essentially from what we use now.
1.1 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
� The
.�The
fact that 2(x + 3 ) = 2x + 6 is because of the Property. (p. 10) � ��... The fact that 3x = 0 implies that x = 0 is a result of the V Property. (p. 13) ____
x domain of the variable in the expression _ _ is x - 4
. (p.21)
____
Concepts and Vocabulary
4.
\/..
True or False Multiplying both sides of an equation by any number results in an equivalent equation.
True or False
. 3
IS
An equation that is satisfied for every value of the variable for "\ which both sides are defined is called a(n) . 6. An equation of the form ax + b equation or a(n) _____
=
8.
0 is called a(n) equation.
The solution of the equation 3x - 8
.
S
True or False
Some equations have no solution.
Skill Building In
}( �. � :
....�7 ... x
Problems 9-16, mentally solve each equation. = 21 10. 6x = - 24
�2X -
In
3
Problems
\t 3x + 4 20. 5Y + 6
=
14. 3x + 4
0
17-64,
=
x
=
-18
solve each equation.
- Y
" ¥- 3 + 2n = 4n + 7 26. 3(2 - x)
=
2x
-
=
3x + 15
0
x
18. 2x + 9 121. 6 - x \1"-
=
24. 6 - 2m
1
X
=
1 30. "3x
=
2 2 - "3x
2 16. - x 3
2
2t - 6
=
22. 3 - 2x
=
'\
3m + 1
8x - (3x + 2)
12. 6x + 18
0
\1'9.
5x
2x + 9 =
=
=
=
3x - 10
9 2
=-
3 - t
� 2(3 + 2x) 28. 7 - (2x
=
2 - x
-
=
1)
3(x - 4) =
10
0
=
0
SECTION 1.1
32.
1
-2:1 x= 6 36.
1 1=-4 2 --p 3 3 �. x ; 1+x ; 2= 2
0 .9t 1+t =
2x- -+ + 1 16=3x 3
)i �+= � 3
(x + 2)(x- 3) (x+3)2
-x3 13 61 3) (2x + l)(x-4) �\ x(2x- = 42•
44.
95
34.-
lit' 0.9t =0.4+O.lt ��
38.
Linear Equations
=
� Z ( Z2+1)=3+
Z3
�•
2x -6 x+3 x+3
50. -=---2
-4 -3 -x+4-=x+6 4 7 -3 �. -x-2=-x+5+---(x+5)(x-2) -5 +-4 =--3 62. 5z-11 2z - 3 5-z
=
-
2x 4 3 x+2 x2-4--x2-4--3x=2 54. -x-I S � 6t+7 =2t3t+ -4 �\ 4t-1 1 1-= 60. � +2x+3 x-I (2x+3)(x-1) ,
.
56.
- -
40.
--
�,,-' x2-1- x x
x+3 -_ -3 x2 + 2
- X
4 -5 5 2y y
-
=-
� (x+7)(x ,M.
-
1)
=
(x+1)2
=(2x-l)(x-2) x(l+2x) x +3= -2 . �. -x-2 x-2 x 4 3 52. --+-x2-9 x+3=r-9 46.
-? -
5
�. 2x _ 3= x ! 5
Sw+5 4w- 3 -7 5w+7
58. --lOw
64.
X
x+1 x+4 +2x x2+
-3 x2+3x+2
-- - --2 X X
In Problems 65-68, use a calculator to solve each equation. Round the solution to two decimal places.
JJ. 'V\.
66. 6.2x - -S19.1 3-= .72 0.195 21.2=-x 14 68. lS.6 3x-2.6 2. 32 -20
21.3 =19.2 3 3.2x+65.S71
IS
2.11 +2.4 �. 14.72-21.5Sx= -x Applications and Extensions
In Problems 69-74, solve each equation. The letters a, b, and c are constants.
� x = a - b
72.
�
0
a '*
c,
a -+ -xb c, x =
70.
1 - ax =b,
a '*
Find �he number equatI On
a '*
0 , b '* 0
,
a '* -b
74. -=-- ,
0
c '*
14 x-+= ' .'\ a xb- c'
0
a
for which
x= 4
is a solution of the
76. Find
b + c b- c x+a x-a c '* 0, a '* 0 the number b for which = x 2 is a solution of the
equation
x+2b= x-4+2bx
x+2a =16+ax-6a
Problems 77-82 list some formulas that occur in applications. Solve each formula for the indicated variable.
\--l], .
EI ect' 'tY rICI
1= 1 +1
-
R
RJ
R2
= A P(l+rt) for r mv2 lro � Mechanics = F for R R 78.
80.
Finance
Chemistry
PV
!:t{ Mathematics
=
nRT a __
for T
= 1-r for r 82. Mechanics = -gt+ for t � Finance A total of $20,000 is to be invested, some in bonds K
S
v
$ 3000, how
invested in bonds is to exceed that in CDs by much will be invested in each type of investment?
for R
Vo
and some in certificates of deposit ( CDs ) . If the amount
84.
$10,000
Finance A total of is to be divided between Sean and George, with George to receive less than Sean. How much will each receive?
��. Internet Searches /
$ 3000
2005,
In November Google and Yahoo! search engines were used to conduct a total of billion online searches. Google was used to conduct billion more searches than Yahoo!. How many searches were conducted on each search engine?
0.5 3
Source: comScore Networks
3.57
96
CHAPTER 1
Equations and I nequalities
86. Sharing the Cost of a Pizza
certain four-door model has a discounted price of $8000, what was its list price? How much can be saved by purchasing last year's model? 93. Business: Marking up the Price of Books A college book store marks up the price that it pays the publisher for a book by 35 %. If the selling price of a book is $92.00, how much did the bookstore pay for this book?
Judy and Tom agree to share the cost of an $18 pizza based on how much each ate. If Tom 2 ate:3 the amount that Judy ate, how much should each pay? [Hint: Some pizza may be left.] Tom's portion
Judy's portion
87.
88.
89.
90.
91.
92.
Computing Hourly Wages Sandra, who is paid time-and-a half for hours worked in excess of 40 hours, had gross weekly wages of $442 for 48 hours worked. What is her regular hourly rate? Computing Hourly Wages Leigh is paid time-and-a-half for hours worked in excess of 40 hours and double-time for hours worked on Sunday. If Leigh had gross weekly wages of $342 for working 50 hours, 4 of which were on Sunday, what is her regular hourly rate? Computing Grades Going into the final exam, which will count as two tests, Brooke has test scores of 80, 83, 71, 61, and 95. What score does Brooke need on the final in order to have an average score of 80? Computing Grades Going into the final exam, which will count as two-thirds of the final grade, Mike has test scores of 86, 80, 84, and 90. What score does Mike need on the final in order to earn a B, which requires an average score of 80? What does he need to earn an A, which requires an average of 90? Business: Discount Pricing A builder of tract homes re duced the price of a model by 1 5 % . If the new price is $425,000, what was its original price? How much can be saved by purchasing the model? Business: D iscOlmt Pricing A car dealer, at a year-end clear ance, reduces the list price of last year's models by 15% . If a
94. Personal Finance: Cost of a Car
95.
96.
97. 98. 99.
Source: comScore Networks
Discussion and Writing
100. One step in the following list contains an error. Identify it
and explain what is wrong. x= 2 3x - 2x= 2 3x = 2x+2 x2 + 3x= x2 + 2x + 2 x2 + 3x - 1 0 x2 + 2x - 8 (x - 2 ) (x + 5 ) = (x - 2) ( x+4 ) x + 5= x + 4 1= 0 =
'Are You Prepared?' Answers
1. Distributive
2. Zero-Product
101. The equation
5 8+x + 3= x+3 x+3
(1) (2) (3) (4)
--
-
has no solution, yet when we go through the process of solv ing it we obtain x -3. Write a brief paragraph to explain what causes this to happen. =
(5)
(6)
102. Make up an equation that has no solution and give it to a fel
(7)
low student to solve. Ask the fellow student to write a cri tique of your equation.
(8)
3. {xix
The suggested list price of a new car is $ 18,000. The dealer's cost is 85% of list. How much will you pay if the dealer is willing to accept $100 over cost for the car? Business: Theater Attendance The manager of the Coral Theater wants to know whether the majority of its patrons are adults or children. One day in July, 5200 tickets were sold and the receipts totaled $29,961 . The adult admission is $7.50, and the children's admission is $4.50. How many adult patrons were there? Business: Discount Pricing A wool suit, discounted by 30% for a clearance sale, has a price tag of $399. What was the suit's original price? Geometry The perimeter of a rectangle is 60 feet. Find its length and width if the length is 8 feet longer than the width. Geometry The perimeter of a rectangle is 42 meters. Find its length and width if the length is twice the width. Internet Users In March 2006, 152 million people in the United States were Internet users, which accounted for 21.9% of the world's online audience. How many people worldwide were Internet users in March 2006?
-t=
4}
SECTION 1.2
Quadratic Equations
97
1.2 Quadratic Equations
Before getting started, review the following:
PREPARING FOR THIS SECTION •
•
•
Factoring Polynomials (Section R.5, pp. 49-55) Zero-Product Property (Section R.I, p. 13) Now Work
Square Roots (Section R.2, pp. 23-24)
the 'Are You Prepared?' problems on page 106.
OBJECTIVES
1
Solve a Quadratic Equation by Factoring (p. 97)
2
Know How to Complete the Square (p. 99)
3
Solve a Quadratic Equation by Completing the Square (p. 1 00)
4
Solve a Quadratic Equation Using the Quadratic Formu la (p. 1 02)
5
Solve Applied Problems Involving Quadratic Equations (p. 1 05)
Quadratic equations are equations such as 2x2 + x + 8 = 0 3x2 - 5x + 6 = 0 x2 - 9 = 0 A DEFINITION
general definition is given next. A quadratic equation
is an equation equivalent to one of the form
+ bx + c = 0 where a, b, and c are real numbers and a "* O. ax2
A
quadratic equation written in the form ax2
standard form.
(1)
+
bx
+
c=0
is said to be in
Sometimes, a quadratic equation is called a second-degree equation, because the left side is a polynomial of degree 2. We shall discuss three ways of solving qua dratic equations: by factoring, by completing the square, and by using the quadratic formula.
1
EXAM P L E 1
Solution
S olve a Quadratic Equation by Factoring
When a quadratic equation is written in standard form ax2 + bx + c = 0, it may be possible to factor the expression on the left side into the product of two first-degree polynomials. Then, by using the Zero-Product Property and setting each factor equal to 0, we can solve the resulting linear equations and obtain the solutions of the quadratic equation. Let's look at an example. Solving a Quadratic Equation by Factoring
Solve the equation: (b) 2x2 = x + 3 (a) x2 + 6x = 0 (a) The equation is in the standard form specified in equation ( 1 ) . The left side may be factored as x2 + 6x = 0 x(x + 6) = 0
Factor.
98
CHAPTER 1
Equations and Inequalities
Using the Zero-Product Property, we set each factor equal to 0 and then solve the resulting first-degree equations. x= 0 or x + 6= 0 Zero-Product Property X= 0 or x= -6 Solve. The solution set is {O, -6}. (b) We put the equation 2x2 = x + 3 in standard form by adding -x - 3 to both sides. 2X2= X + 3 Add 3 to both sides. 2X2 - X - 3= 0 The left side may now be factored as (2x - 3)(x + 1 ) = 0 Factor. so that 2x - 3= 0 or x + 1= 0 Zero-Product Property x = -1 Solve. x= -32 -
.
x
-
. { 3}
The solutIOn set IS -1'"2 .
•
When the left side factors into two linear equations with the same solution, the quadratic equation is said to have a repeated solution. We also call this solu tion a root of multiplicity 2, or a double root. EXAM P LE 2 Solution
Solving a Quadratic Equation by Factoring
Solve the equation: 9x2 - 6x + 1 = 0 This equation is already in standard form, and the left side can be factored. 9x2 - 6x + 1 = 0 (3x - 1 ) (3x - 1 ) = 0 so 1 x= -3 or x= -31 This equation has only the repeated solution
�. The solution set is {�}.
• I! =:;> -
Now Work P R O B L E M S 1 1
AND
21
The Square Root Method
Suppose that we wish to solve the quadratic equation x2= p (2) where p 2: 0 is a nonnegative number. We proceed as in the earlier examples. Put in sta nda rd form. x2 - p= 0 (x - vp ) (x + vp ) = 0 Factor (over the rea l numbers) . x= vp or x = -vp Solve. We have the following result: If x2= p and p
2:
0, then x= vp or x
= - vp.
(3 )
SECTION 1.2
Quadratic Equations
99
When statement (3) is used, it is called the Square Root Method. In statement (3), note that if p > 0 the equation x2 = p has two solutions, x = vp and x = -vp. We usually abbreviate these solutions as x = ± vp, read as "x equals plus or minus the square root ofp." For example, the two solutions of the equation x2 = 4
are
x = ± V4
and, since V4 = 2, we have
x = ±2
The solution set is { -2, 2}. EXA M P L E 3
Use the Square Root Method.
Solving a Quadratic Equation Using the Square Root M ethod
Solve each equation. (a) x2 = 5 (b) (x - 2) 2 = 1 6 (a) We use the Square Root Method to get
Solution
x2 = 5 x = ± Vs x = Vs or x =
- Vs
Use the Square Root Method.
The solution set is { - v'5 , v'5}. (b) We use the Square Root Method to get (x - 2) 2 x-2 X - 2 x-2 x
The solution set
= 16 = ± Vi6 = ±4 = 4 or x - 2 = -4 x = -2 = 6 or is { -2, 6}.
Use the Square Root Method.
•
31 2
Know How to Complete the Square
We now introduce the method of completing the square. The idea behind this method is to adjust the left side of a quadratic equation, ax2 + bx + c = 0, so that it becomes a perfect square, that is, the square of a first-degree polynomial. For example, x2 + 6x + 9 and x2 - 4x + 4 are perfect squares because x2 + 6x + 9 = (x + 3 ) 2 and x2 - 4x + 4 = (x - 2 ) 2 How do we adjust the left side? We do it by adding the appropriate number to the left side to create a perfect square. For example, to make x2 + 6x a perfect square, we add 9. Let's look at several examples of completing the square when the coefficient of x2 is 1 : Start
x2 + 8x x2 + 2x x2 - 6x x2 + x
Add
16 1 9 1 4
Result
x2 + 8x + 1 6 = (x + 4) 2 x2 + 2x + 1 = (x + 1 ) 2 x2 - 6x + 9 = (x - 3 ) 2 1 1 x2 + x + 4 = x + 2
(
Y
100
CHAPTER 1
Equations and Inequalities
Do you see the pattern? Provided that the coefficient of square by adding the square of of the coefficient of
�
x.
x2 is 1, we complete the
Procedure for Completing a Square
m)2 x2 mx (m)2 2 = (x 2
Result
Add
Start
+
+
+
Completing the Square
EXAM P L E 4
Determine the number that must be added to each expression to complete the square. Then factor the expression. Add
Start
? x2
+
G·8r G Y G·(-20)Y C2·(-5)Y
By
=
+12x
·12
02 - 200 p2 - 5p
16
=
36
Factored Form
y2+8y+16
(y+ 4)2
x2
=
=
Result
+
12x + 36
02 - 200+100
100
25 p2 -5p+4
25
4"
(x+ 6)2 (0 - 10)2
( %y p-
•
Notice that the factored form of a perfect square is either 4
Y
y 4
Area
Area
x2 + bx (%y (x + %y =
+
Figure 1
=
=
y2
'!I't
Area =
4y
3
Solution
x2
-
bx +
(%y = ( x %y -
Now Work P R O B l E M 3 5
Are you wondering why we call making an expression a perfect square "com pleting the square"? Look at the square in Figure 1. Its area is (y + 4 f The yellow area is l and each orange area is 4y (for a total area of 8y) . The sum of these areas is l + 8y. To complete the square we need to add the area of the green region: 4·4 = 16.As a result, l + 8y + 16 = (y + 4l
4y
EXAM P L E 5
, >-
or
Solve a Quadratic Eq uation by Completing the Square
Solving a Quadratic Equation by Completing the Square
5x
Solve by completing the square: x2 + + 4 = 0 We always begin this procedure by rearranging the equation so that the constant is on the right side.
x2 5x x2 + 5x +
x2
+ 4 =
0
= -4
Since the coefficient of is 1, we can complete the square on the left side by adding 1 = 4. Of course, in an equation, whatever we add to the left side also must be added to the right side. So we add to both sides.
( 5)2 25 "2.
2:
SECTION 1.2
25 -4 x2 + 5x + -= 4
25 4
Add
+ -
9
25 -
4
Quadratic Equations
10 1
to both sides.
Factor.
4
f2. x + �2 =± \/"4 5 3 x + -= 2 ±-2 5 3 x= - -±2 2
Use the Squa re Root Method.
5 3 -4 x = - - + -3 = -l or x = - -5 - -= 2 2 2 2 The solution set is { -4, I } -
•
.
I' !'l!'i:==""- THE SOLUTION OF THE EQUATION
CAN BE OBTAINED BY
IN EXAMPLE
FACTORING. REWORK
5
ALSO
EXAMPLE
5
USING THIS TECHNIQUE
The next example illustrates an equation that cannot be solved by factoring. EXAM P L E 6 Solution
Solving a Quadratic Equation by Completing the Square
Solve by completing the square: 2x2 - 8x - 5 = 0 First, we rewrite the equation. 2X2 - 8x - 5 = 0 2x2 - 8x = 5 Next, we divide both sides by 2 so that the coefficient of x2 is to complete the square at the next step.) x2 - 4x = -52 Finally, we complete the square by adding 4 to both sides. ?
1.
(This enables us
5 4 = -2 + 4 13 (x - 2? = 2
x- - 4x
+
x - 2= ±
.J¥
Use the Square Root Method.
v26 x - 2 = ±-2
x= 2± v26 2
(13= 'V 2
v13 Vz
v13 Vz Vz Vz .
=
=
Vz6 2
--
NOTE If w e wanted a n approximation,
say rounded to two decimal places, of these solutions, we would use a calcula • tor to get { -0.55,4.55}.
{
v26 The solution set is 2 - 2- , 2 I.l'l. = _-
Now Work
PROBL EM
+
v26 2-
41
} •
102
CHAPTER 1
Equations and Inequal ities
4
Solve a Quadratic Eq uation Using the Quadratic Form u l a
We can use the method of completing the square to obtain a general formula for solving any quadratic equation ax2 + bx + c NOTE There i s no loss i n generality to assume that a > 0, since if a < ° we can multiply by -1 to obtain an equiva lent equation with a positive leading coefficient. •
=0
a*" O
As in Examples 5 and 6, we rearrange the terms as ax2 + bx = -c
Since a
>
0, we can
a > 0
divide both sides by a to get b x2 + - x a
=
c a
--
Now the coefficient of x2 is 1. To complete the square on the left side, add the square of of the coefficient of x; that is, add
�
to both sides. Then if
b2 - 4ac 4a2
Provided that b2 - 4ac
)
2::
0, we now
if
c
4ac
( 4)
can use the Square Root Method to get
b2 - 4ac b x + - =± 2a 4a2 b ± Yb2 - 4ac x + - = ----2a 2a yb2 - 4ac b x = --± 2a 2a -b ± yb2 - 4ac 2a
-----
The square root of a quotient equals the quotient of th e square roots. Also,
�
=
b
2a since a > O.
Add - - to both sides. 2a Combine the quotients on the right.
What if b2 - 4ac is negative? Then equation (4) states that the left expression (a real number squared) equals the right expression (a negative number). Since this occurrence is impossible for real numbers, we conclude that if b2 - 4ac < 0 the quadratic equation has no real solution. (We discuss quadratic equations for which the quantity b2 - 4ac < 0 in detail in the next section.) We now state the quadratic formula. THEOREM
Consider the quadratic equation ax2 + bx + c = 0
a*" O
If b2 - 4ac < 0, this equation has no real solution. If b2 - 4ac 2:: 0, the real solution(s) of this equation is (are) given by the
quadratic formula. Quadratic Formula
x=
-b±
yb2 - 4ac 2a
( 5)
I
�--------------------------------��
SECTION 1.2
Quadratic Equations
103
The quantity b2 - 4ac is called the discriminant of the quadratic equation, because its value tells us whether the equation has real solutions. In fact, it also tells us how many solutions to expect. Discriminant of a Quadratic Equation
For a quadratic equation ax2 + bx + c
= 0:
b2 - 4ac > 0, there are two unequal real solutions. 2. If b2 - 4ac = 0, there is a repeated solution, a root of multiplicity 2. 3. If b2 - 4ac < 0, there is no real solution. 1. If
When asked to find the real solutions, if any, of a quadratic equation, always evaluate the discriminant first to see how many real solutions there are. EXA M P L E 7
Solving a Quadratic Equation Using the Q u adratic Formula Use the quadratic formula to find the real solutions, if any, of the equation
+ 1 =
3x2 - 5x
Solution
0
The equation is in standard form, so we compare it to ax2 + bx + c = b, and c.
+ 1 =0 + bx + c = 0
0
to find a,
3x2 - 5x ax
With a =
3,
b=
-5,
2
and c =
1,
> 0,
3, b
4(3)(1)
=
-( -5) ±
{
.
The solutIOn set is
=
-5, C
1
=
25 - 12
4a
c.
= 13
there are two real solutions, which can be found using the qua
x=
E XA M P L E 8
=
we evaluate the discriminant b2 -
b2 - 4ac = ( -5? Since b2 - 4ac dratic formula.
a
5
-
Vi3
6
5
+
'
Vi3}
6
2(3)
Vi3
5
± Vi3 6
.
•
Solving a Quadratic Equation Using the Quadratic Formula Use the quadratic formula to find the real solutions, if any, of the equation 25 -x2 - 30x 2
Solution
+ 18 =
0
The equation is given in standard form. However, to simplify the arithmetic, we clear the fractions. 25
x2 -
2 -
30x + 18 = 0
+ 36 = 0 + bx + c = 0
2 25x - 60x ax2
With a =
25,
b=
-60,
and c =
36,
b2 - 4ac = (-60)2
Clear fractions; multiply by
2.
Compare to standard form.
we evaluate the discriminant.
- 4(25)(36)
= 3600
- 3600
=
0
104
C HAPTER 1
Equations and I nequalities
The equation has a repeated solution, which we find by using the quadratic formula.
x= The solution set is
EXAM P LE 9
-b ±
yb2 - 4ac
60± Yo 50
2a
{�}.
60 50
6 5 •
Solving a Quadratic Equation Using the Quadratic Form u la Use the quadratic formula to find the real solutions, if any, of the equation
3x2 + 2 = 4x Solution
The equation, as given, is not in standard form.
3x2 + 2 = 4x 2 3x - 4x + 2 = 0 ax2 + bx + c = 0
Put in standard form. Compa re to standard form.
With a = 3, b = -4, and c = 2, we find
b2 - 4ac = ( -4f - 4(3) (2) = 16 - 24 = -8 Since b2 - 4ac < 0, the equation has no real solution. 1JI\OiZ . _ >-
Now Work
PROBLEMS
51
AND
•
61
Sometimes a given equation can be transformed into a quadratic equation so that it can be solved using the quadratic formula. EXAM P L E 1 0
Solving a Quadratic Equation Using the Quadratic Formula Find the real solutions, if any, of the equation:
Solution
2 3 9 + -- - = Ox7"oO ' x x2
In its present form, the equation
3 2 9 + - -- = 0 x x2 is not a quadratic equation. However, it can be transformed into one by multiply ing each side by x2 . The result is
9x2 + 3x - 2 = 0 Although we multiplied each side by x2 , we know that x2
* 0 ( do you see why? ) , so this quadratic equation is equivalent to the original equation. Using a = 9, b = 3, and c = -2, the discriminant is
b2 - 4ac = 32 - 4(9 ) ( -2) = 9 + 72 = 81 Since b2 - 4ac
0, the new equation has two real solutions.
>
x= x=
-b±
Yb2 - 4ac 2a
-3 + 9 18
The solution set is
6 18
{ -�, �}.
-3 ± \I8i -3 ± 9 2(9) 18 1 -3 - 9 -12 or x = 3 18 18
2 3
•
SECTION 1.2
SUMMARY
Quadratic Equations
105
Procedure for Solving a Quadratic Equation
To solve a quadratic equation, first put it in standard form:
ax2
+
bx +
c =
0
Then:
STEP 1: Identify a, b, and c. STEP 2: Evaluate the discriminant, b2 - 4ac. STEP 3: (a) If the discriminant is negative, the equation has no real solution. (b) If the discriminant is zero, the equation has one real solution, a repeated root. (c) If the discriminant is positive, the equation has two distinct real solutions. If you can easily spot factors, use the factoring method to solve the equation. Otherwise, use the quadratic formula or the method of completing the square.
5
So lve Applied Problems I nvolving Q u a d ratic Eq uations
Many applied problems require the solution of a quadratic equation. Let's look at one that you will probably see again in a slightly different form if you study calculus.
Constructing a Box
EXAM P L E 1 1
From each corner of a square piece of sheet metal, remove a square of side 9 cen timeters. Turn up the edges to form an open box. If the box is to hold 144 cubic centimeters (cm 3), what should be the dimensions of the piece of sheet metal? Solution
We use Figure 2 as a guide. We have labeled by x the length of a side of the square piece of sheet metal. The box will be of height 9 centimeters, and its square base will measure x - 18 on each side. The volume V ( Length X Width X Height) of the box is therefore v =
Figure
2
(x - 18)(x - 18) · 9
=
9(x - 18?
I��---- xem ---+-. I i I
1
I I
1 I I
gem:
gem
_______
1
I I I
gem
9 em
x- 18 Volume 9(x - 18)(x- 18) =
Since the volume of the box is to be 144 cm3 , we have
9(x - 18) 2 (x - 18) 2 x - 18 x
= = =
=
144 16 ±4 18 ± 4
x = 22
or
V = 144
Divide each side by 9. Use the Square Root Method. x =
14
We discard the solution x = 14 (do you see why?) and conclude that the sheet metal should be 22 centimeters by 22 centimeters.
106
CHAPTER 1
Equations and Inequalities
Check: If we begin with a piece of sheet metal
22 centimeters by 22 centime ters, cut out a 9 centimeter square from each corner, and fold up the edges, we get a box whose dimensions are 9 by 4 by 4, with volume 9 X 4 X 4 = 144 cm3, as required.
•
C/q'=>-
Now Work
PRO
BLEM 1 05
{-ti�torical Feature
P
roblems using quadratic equation s are found in the oldest
Thomas Harriot (1560-1621) introduced the method of factoring to ob
known mathematical literature. Babylonia n s and Egyptia n s were
tain solutions, a n d Franc;:ois Viete (1540-1603) introduced a method that
solving such problems before 1800 Be. Euclid solved quadratic
is essentia lly completing the square.
equations geometrica l ly in his Data (300 BC). a n d the Hindus a n d Arabs
Until modern times it was usual to neglect the negative roots (if there
gave rules for solving a ny quadratic equation with real roots. Because
were a ny). a n d equation s involving square roots of negative quantities
negative n umbers were not freely used before
were regarded as unsolvable until the 1500s.
AD
1500, there were sev
eral differen t types of quadratic equations, each with its own rule.
Historical Problems 1. One of al-Khwdiizmi solutions
2
+
=
85 by draw
is a difference of two squares. If you factor this difference of two
ing the square shown .The area of the four white rectangles and the 2 yellow square is x + 12x. We then set this expression equal to 85
moreover, the quadratic expression is factored, which is sometimes
to get the equation
useful.
x2
We solve x
12x
+ 12x= 85. If we ad d the four blue squares,
squares, you will easily be able to get the quadratic formula, and,
we wil l have a larger square of known area. Complete the solution. We solve x2
2. Viete's method Then
(u + u2
+
(2z
d
+
Now select z so that 2z
+
12x - 85= 0 by letting x
+ 12( u + z) - 85
12) u +
(Z2 +
+ 12= 0 a n d
u
+
,
z.
3:J�
___
0
finish the solution.
( Vb2 - 4ac )2
Look a t equation
2a
I
x
_________
:�� : 3
I I I I I I I IX Area = I I I I I I I I I I X ___ �___________ J___ _
12z - 85)= 0
3. Another method to get the quadratic formula on page 102. Rewrite the right side as
=
=
(4)
3
a n d then
subtract it from each side. The right side is now 0 a n d the left side
i3 I
Area
x2 x
=
3x 31 i
3
1.2 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red.
�.
Factor: x2-5x - 6 (pp. 49-55)
The solution set of the equation (x -3)(3x . (p. 13)
� True or False
)'\. Factor:2x2-x -3 (pp. 49-55)
______
�
=
+
5)
=
0 is
Ixl. (pp. 23-24)
Concepts and Vocabulary
'?'\ To complete the square of the expression x2 the number _____
would
+
5x, you
6. The quantity b2 - 4ac is called the of a quadratic equation. If it is , the equation has no real solution.
V. True or False Quadratic equations always have two real solutions.
8. True or False If the discriminant of a quadratic equation is positive, then the equation has two solutions that are nega tives of one another.
Skill Building
In Problems
9-28, solve
'i1lx2-9x �.\ Z2 Z -
=
+
6
0 =
0
each equation by factoring.
10. x2 + 4x = 0 14. v2 + 7v
+ 6
=
0
¥(x2-25= O �. 2x2 - 5x -3
12. x2- 9 =
0
=
16. 3x2 + 5x
0 +2
=0
SECTION 1.2
'¥.z
� 4x2 + 9 'i�. f'\
It x(x - 8) + 12 = 0
18. 2/ - 50= 0
3t 2 - 48 = 0
22. 2 5x2 + 1 6 = 40x
= 12 x
. 6 ( p 2 - 1)
12 26. x + - = 7
6x - 5 = .§. x
27.
x
4(x -2) x -3
=
+
�
=
)( (2y
32. ( x + 2)2 = 1
24. 2(2u2 - 4u ) + 3 = 0
-3 3 - = --x x(x - 3)
):(.
.
5 3 = 4 + -x + 4 x -2
--
28.
(x - 1)2 = 4
34. (3 z - 2)2
+ 3)2 = 9
107
20. x(x + 4) = 12
5p
In Proble ms 29-34, so lve each eq uatio n by the Sq uare Root Method. x2 25 30. x2 = 3 6
Quadratic Equations
=
4
I n Prob le ms 35-40, what number sho uld be added to co mp lete the sq uare ofeach e xpre ssio n? x2 - 8x 36. x2 - 4x
)i. x2 + � X
�
1 3
38. x2 - -x I n Proble ms
40
41-46,
1\. x2 + 4x = 2 1
2
2
- -7
5�
42. x2 - 6x= 13 46. 2 X2 -3 x - 1 = 0
3
3
7 �
solve each eq uatio n by co mpleti ng the sq uare.
2 1 44. v-? + -"v -- = O .>
.
I n Prob le ms 47-70, fi nd the rea l so lutio ns, ifa ny , ofeach eq uatio n. Use the q uadratic formula .
'
"
W. x2 - 4x - 1 = 0
48. x2 + 4x + 2 = 0
x2 - 4x + 2 = 0
51. 2 x2 - 5x + 3 = 0
52. 2X2 + 5x + 3
55. 4x2 = 1 - 2x
56. 2X2
59. 9t 2 - 6t +
1
60. 4u2 - 6u + 9
5 63. :;-x2 - x .)
1 3
= 0
3
=
1
64. 5x2 - x
= -
68. 4 +
=
0
-2 x
=
=
54. 4t 2 + t + 1= 0
57. 4x2 = 9 x
58. 5x = 4x2
3 1 , 61. -x2 --x 4
1
5
4
1 -
2
-
=
2
62. 3" x-? - x -3= 0
0
65. 2 x( x + 2) = 3
�
-x1 - ?1 = 0 .r
53. 41 -y + 2 = 0 ""
0
50. x2 + 6x + 1 = 0
.
-
66. 3 x( x + 2) =
3x 1 +-=4 x -2 x
--
1
1 2x +-=4 x -3 x
--
70.
I n Prob le ms 71-78, fi nd the rea l so lutio ns, i fa ny , of each eq uatio n. Use the q uadratic formula a nd a ca lc ulato l: Expre ss a ny so lutio ns ro unded to two deci ma l place s.
71. x2 - 4. 1 x + 2 .2 = 0
72. x2 + 3.9 x + 1.8 = 0
73. x2 + V3 x -3= 0
74. x2 + V2 x -2 = 0
75. 7TX2 - x - 7T = 0
76. 7TX2 +
78. 7Tx2 - 1 5 V2x + 2 0
77. 3 x2 + 87TX + v29 = 0 I n P rob le ms 79. x2 - 5
79-92, fi nd the
=
TTX
-2 = 0
0
rea l so lutio ns, i fa ny , o feach eq uatio n. Use a ny method.
80. x2 - 6 = 0
81. 16x2 - 8x + 1 = 0
82. 9 x2 - 12 x + 4 = 0
83. 10x2 - 19 x - 15 = 0
84. 6x2 + 7x - 2 0 = 0
85. 2 + z = 6z2
86. 2 = y
88. 1:. x2 = V2 x + 1 2 x 2 91. -- +
89. x2 + x = 4
x -2
=
0
--
x + l
=
+
1
;-;:. x = 2 87. r? + .V2
61
90. x2 +
7x + 1 x2 - x -2
-:-
92.
X =
1
3x 1 4 - 7x + -- = ----=--- x +2 x-I x2 + x - 2
--
I n Prob le ms 93-98, use the di scri mi na nt to determi ne whether each q uadratic eq ua cio n ha s two uneq ua l rea l so lutio ns, a re peated rea l sol utio n, or 11. 0 rea l so lutio n, wi tho ut solvi ng the eq uatio n.
93. 2 x2 - 6x + 7 = 0
94. x2
96. 2 5x2 - 2 0x
97. 3 x2 + 5x - 8 = 0
+
4=0
+
4x
+
7 =0
95. 9 x2 -3 0x + 2 5 = 0 98. 2 x2 - 3 x - 7 = 0
108
CHAPTER 1
Equations and Inequalities
Applications and Extensions
99.
College Tuition and Fees The average annual published undergraduate tuition-and-fee charges C, in dollars, for pub lic four-year institutions from academic years 2000- 2001 through 2005- 2006 can be estimated by the equation C 20.2x 2 +3 14.5x +3 467.6, where x is the number of years afer the 2000- 2001 academic year. Assuming the model will remain valid beyond 2005- 2006, in what academic year will average annual tuition-and-fee charges be $8000?
109.
=
Source: College Board, Trendsin College P ri cing 100.
2005
The median weekly earnings E, in dollars, for full-time women wage and salary workers ages 16 years and older from 2000 through 2004 can be estimated by the equation E 0. 14x 2 + 7.8x + 5 40, where x is the nurnber of years after 2000. Assuming the model will remain valid beyond 2004, in what year will the median weekly earn ings be $63 2?
Women's Weekly Earnings
(b) When will it strike the ground? (c) Will the object reach a height of 100 meters? Reducing the Size of a Candy Bar A jumbo chocolate bar with a rectangular shape measures 12 centimeters in length, 7 centimeters in width, and 3 centimeters in thickness. Due to escalating costs of cocoa, management decides to reduce the volume of the bar by 10%. To accomplish this reduction, management decides that the new bar should have the same 3 centimeter thickness, but the length and width of each should be reduced an equal number of centimeters. What should be the dimensions of the new candy bar?
=
Source: U.S. Department of Labor, Highlight s of Women 's Ea rning s in
1. 102.
2004,
September 2005
The area of the opening of a rec tangular window is to be 143 square feet. If the length is to be 2 feet more than the width, what are the dimensions? Dimensions of a Window
110.
1 04.
Watering a Field
A circular pool mea-
V\ sures 10 feet across. One cubic yard of concrete is to be used to create a circular border of uniform width around the pool. If the border is to have a depth of3 inches, how wide will the border be? (1 cubic yard 27 cubic feet) See the illustration.
The area of a rectangular window is to be 3 06 square centimeters. If the length exceeds the width by 1centimeter, what are the dimensions? Geometry Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters.
Rework Problem 109 if
ill. Constructing a Border around a Pool
Dimensions of a Window
103.
Reducing the Size of a Candy Bar
the reduction is to be 20% .
=
An adjustable water sprinkler that sprays water in a circular pattern is placed at the center of a square field whose area is 125 0 square feet (see the figure). What is the shortest radius setting that can be used if the field is to be completely enclosed within the circle?
112.
Constructing a Border arollnd a Pool
113.
Constructing a Border around a Garden A landscaper, who just completed a rectangular flower garden measuring 6 feet by 10 feet, orders 1 cubic yard of premixed cement, all of which is to be used to create a border of uniform width around the garden. If the border is to have a depth of3 inches, how wide will the border be? ( 1 cubic yard 27 cubic feet)
if the depth of the border is 4 inches.
Rework Problem 111
=
l�. Constructing a Box An open box is to be constructed from 1J\... a square piece of sheet metal by removmg a square of Side
1 foot from each corner and turning up the edges. If the box is to hold 4 cubic feet, what should be the dimensions of the sheet metal?
106.
Rework Problem 105 if the piece of sheet metal is a rectangle whose length is twice its width.
Constructing a Box
�. Physics A ball is thrown vertically upward from the top of � a building 9 6 feet tall with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s 9 6 + 80t - 16r2. (a) After how many seconds does the ball strike the ground? (b) After how many seconds will the ball pass the top of the building on its way down? =
108.
Physics An object is propelled vertically upward with an initial velocity of 20 meters per second. The distance s (ill me ters) of the object from the ground after t seconds is s -4.9t 2 + 20t . (a) When will the object be 15 meters above the ground? =
114.
�.
Dimensions of a Patio A contractor orders 8 cubic yards of premixed cement, all of which is to be used to pour a patio that will be 4 inches thick. If the length of the patio is speci fied to be twice the width, what will be the patio dimensions? (1 cubic yard 27 cubic feet) Comparing TVs The screen size of a television is deter mined by the length of the diagonal of the rectangular screen. =
trad itional
4 :3
LCD 1 6:9
SECTION 1.3
Complex Numbers; Quadratic Equations in the Complex Number System
Traditional TVs come in a 4 : 3 format, meaning the ratio of the length to the width of the rectangular is 4 to 3. What is the area of a 37-inch traditional TV screen? What is the area of a 37-inch LCD TV whose screen is in a 16 : 9 format? Which screen is larger? 2 [Hint: If x is the length of a 4 : 3 format screen, then x is 4 the width.]
116.
�.
c a
121. Find k such that the equation kx2 + x + k
� -
=
0 has a
=
0 has a
repeated real solution.
122. Find k such that the equation x 2 - kx + 4 repeated real solution.
123. Show
that the real solutions of the equation ax 2 + bx + c 0 are the negatives of the real solutions of the equationax 2 -bx + c O. Assume thatb 2 - 4ac ;:0: O. =
=
124. Show that the real
solutions of the equation ax 2 + bx + c 0 are the reciprocals of the real solutions of the equation cx 2 + bx + a O. Assume thatb2- 4ac ;:0: O.
starting with 1, must be added to get a sum of 666? If a polygon ofn sides has n (n
-�.
. lS - .
Comparing TVs Refer to Problem 1 15 . Find the screen area of a traditional 50-inch TV and compare it with a 50-inch Plasma TV whose screen is in a 16 : 9 format. Which screen is larger?
Geometry
Show that the sum o f the roots of a quadratic equation is
120. Show that the product of the roots of a quadratic equation
. The sum of the consecutive integers 1, 2 , 3, . . . , n is given by 1 the formula "2n (n + 1). How many consecutive integers,
118.
109
=
3) diagonals,
=
how many sides will a polygon with 65 diagonals have? Is there a polygon with SO diagonals? Discussion and Writing
128. Create three quadratic equations: one having two distinct so
. Which of the following pairs of equations are equivalent? Explain. (a) x 2 = 9; x 3 (b) x v9 ; x 3 (c) (x - 1)(x -2) (x - 1 f; x - 2 x - 1 126. Describe three ways that you might solve a quadratic equa tion. State your preferred method; explain why you chose it.
lutions, one having no real solution, and one having exactly one real solution.
=
=
129. The word quadra tic seems to imply four (quad), yet a qua
=
=
dratic equation is an equation that involves a polynomial of degree 2 . I nvestigate the origin of the term quadra tic as it is used in the expressionquadra tic equa tion . Write a brief essay on your findings.
=
127. Explain the benefits of evaluating the discriminant of a qua dratic equation before attempting to solve it. 'Are You Prepared?' Answers
1. (x - 6)(x + 1)
2. (2x - 3)(x + 1 )
3.
{-� 3} 3'
4. True
1 .3 Com plex Num bers; Quadratic Equations in the Com plex Number System* PREPARING FOR T H I S SECTION •
Before getting started, review the following: •
Classification of Numbers (Section R.1, pp. 4-5 ) Now Work the 'Are You
Prepared?' problems
OBJECTIVES
on page
Rationalizing Denominators (Section R.S, p. 74)
1 1 6.
1
Add, Su btract, M u ltiply, and Divide Complex N u mbers (p. 1 1 0)
2
Solve Quadratic Equations in the Complex N u mber System (p. 1 1 4)
Complex N u m bers
One property of a real number is that its square is nonnegative. For example, there is no real number x for which To remedy this situation, we introduce a new number called the imaginary unit. '" This section may be omitted without any loss of continuity.
1 10
CHAPTER 1
Equations and I nequa lities
The imaginary unit, which we denote by i, is the number whose square is - 1 . That is,
DEFINITION
�I�
2 = -1 i_ ___
L-____________________
__ __ __ __ __ __ __ __ __ __
This should not surprise you. If our universe were to consist only of integers, there would be no number x for which 2x = 1 . This unfortunate circumstance was
1
2
remedied by introducing numbers such as "2 and "3 ' the rational numbers. If our universe were to consist only of rational numbers, there would be no x whose square equals 2. That is, there would be no number x for which x2 = 2. To remedy this, we introduced numbers such as V2 and .,ys , the irrational numbers. The real numbers, you will recall, consist of the rational numbers and the irrational numbers. Now, if our universe were to consist only of real numbers, then there would be no number x whose square is - 1 . To remedy this, we introduce a number i, whose square is -1. I n the progression outlined, each time w e encountered a situation that was unsuitable, we introduced a new number system to remedy this situation. And each new number system contained the earlier number system as a subset. The number sys tem that results from introducing the number i is called the complex number system. Complex numbers are numbers of the form a + hi, where a and b are real numbers. The real number a is called the real part of the number a + bi; the real number b is called the imaginary part of a + bi; and i is the imaginary
DEFINITION
unit, so P
=
-1.
�
For example, the complex number -S + 6i has the real part -S and the imagi nary part 6. When a complex number is written in the form a + bi, where a and b are real numbers, we say it is in standard form. However, if the imaginary part of a complex number is negative, such as in the complex number 3 + ( -2)i, we agree to write it instead in the form 3 2i. Also, the complex number a + Oi is usually written merely as a. This serves to remind us that the real numbers are a subset of the complex numbers. The complex number 0 + bi is usually written as bi. Sometimes the complex number bi is called a pure imaginary number. -
1
Add, Su btra ct, Multiply, a n d Divide Complex N u m bers
Equality, addition, subtraction, and multiplication of complex numbers are defined so as to preserve the familiar rules of algebra for real numbers. Two complex num bers are equal if and only if their real parts are equal and their imaginary parts are equal. That is, Equality of Complex Numbers
a + bi
= c +
di
if and only if a
=
c and b
=
d
(1)
Two complex numbers are added by forming the complex number whose real part is the sum of the real parts and whose imaginary part is the sum of the imagi nary parts. That is, Sum of Complex Numbers
(a + bi) + (c + di)
(a + c) + (b + d)i
(2)
SECTION 1.3
111
Complex Numbers; Quadratic Equations i n the Complex Number System
To subtract two complex numbers, we use this rule: Difference o f Complex Numbers
(a
EXAM P L E 1
+
bi) - (e + di)
=
(a - e)
+
(b - d)i
(3)
Adding and Subtracting Complex N u mbers
(a) (3 + 5i) + ( -2 + 3i) = [3 + ( -2 ) ] + (5 + 3)i = 1 + 8i (b) (6 + 4i) - (3 + 6i) = (6 - 3 ) + (4 - 6) i = 3 + (- 2 ) i = 3 - 2i = � = -
Now Work
•
1 3
PROBLE M
Products of complex numbers are calculated as illustrated in Example 2. EXAM P L E 2
M ultiplying Complex N u m bers (5
+
3i) ' (2 + 7i)
5 · (2 + 7i)
=
i
3i(2 + 7i) = 10 + 35i + 6i + 21P
+
i
Distributive Property
Distributive Property
= 10 + 41i + 21( - 1 )
i i2
=
-1
= -11 + 41i
•
Based on the procedure of Example 2, we define the product of two complex numbers as follows: Product of Complex Numbers
(a
+
bi) · (e + di) = (ae - bd) + (ad + be)i
(4)
Do not bother to memorize formula ( 4). Instead, whenever it is necessary to multiply two complex numbers, follow the usual rules for multiplying two binomi als, as in Example 2, remembering that i2 - 1 . For example, =
(2i) (2i)
=
4z2
=
-4
(2 + i)(1 - i) = 2 - 2i + i - P Q!I!l: = = ""
Now Work
PROBLE M
=
3-i
19
Algebraic properties for addition and multiplication, such as the commutative, associative, and distributive properties, hold for complex numbers. The property that every nonzero complex number has a multiplicative inverse, or reciprocal, requires a closer look. DEFINITION
If z = a + bi is a complex number, then its defined as
conjugate,
= a + bi = a - bi _ __ _ _
z__ �___________________
For example, 2 + 3i
=
__ __
__
__
denoted by
Z,
�I�
__ __ __ __ __ __ __ __ __
2 - 3i and -6 - 2i = -6 + 2i.
is
1 12
CHAPTER 1
Equations and Inequa lities
EXA M P L E 3
M u ltiplying a Complex Number by Its Conjugate Find the product of the complex number z
Solution
Since z
zz
=
=
3 + 4i and its conjugate z.
=
3 - 4i, we have
(3 + 4i) (3 - 4i)
=
9 - 12i + 12i - 16P
=
9 + 16
= 25
•
The result obtained in Example 3 has an important generalization. THEOREM
The product of a complex number and its conjugate is a nonnegative real number. That is, if z = a + bi, then
(5)
I
�J�
�__________________________________
If z
Proof
=
zz
a + bi, then
=
(a + bi) (a - bi)
=
a2 - (bi?
=
a2 - b2i2
=
a2
+
b2
•
To express the reciprocal of a nonzero complex number z in standard form,
1
multiply the numerator and denominator of - by z. That is, if z z nonzero complex number, then
1 a + bi
l
z
I z
z
z
z zz
i
EXA M P L E 4
(5).
a a2 + b2
-
b . l a2 + b2
Writing the Reciprocal of a Complex N u mber i n Standard Form Write
Solution
a + bi is a
a - bi a2 + b2
Use
=
=
3
� 4i in standard form a + bi; that is, find the reciprocal of 3 + 4i.
The idea is to multiply the numerator and denominator by the conjugate of 3 + 4i, that is, by the complex number 3 - 4i. The result is
1 3 + 4i
1 3 - 4i 3 + 4i 3 - 4i
-- = -- . -- =
3 - 4i 9 + 16
3
4 .
= - - -l 25 25
•
To express the quotient of two complex numbers in standard form, we multi ply the numerator and denominator of the quotient by the conjugate of the denominator.
EXA M P L E 5
Writing the Quotient of a Complex N u m ber in Standard Form Write each of the following in standard form. (a)
1 + 4i 5 - 12 i
(b)
3i 4 - 3i
2 -
SECTION 1 . 3
Solution
Complex Num bers; Quadratic Equations in the Complex Number System
1 13
1 + 4i 5 + 12i 5 + 12i + 20i + 48i2 5 - 12i 5 + 12i 25 + 144 43 32 -43 + 32i + i 169 169 1 69 2 - 3i 2 - 3i 4 + 3i 8 + 6i - 12i - 9P = . = 4 - 3i 4 - 3i 4 + 3i 16 + 9 17 - 6i = 17 - 6 . l 25 25 25
4i (a) 51 -+ 12i
=
(b)
-
�==- Now Work
E XA M P L E 6
27
Writing Other Expressions i n Standard Form
If z = 2 - 3i and form. (a) Wz
-
Solution
PROBLEM
•
W = 5 + 2i,
write each of the following expressions in standard
(b) z +
W
(c) z + Z
(2 - 3i) (5 - 2i) 10 - 4i - lSi + 6P z·w w · w (5 + 2i) (5 - 2i) 25 + 4 4 19 . 4 - 19i = - - -1 29 29 29 z + w = (2 - 3i) + (5 + 2i) = 7 - i = 7 + i z + Z = (2 - 3i) + (2 + 3i) = 4
(a) wz
-
(b) (c)
•
The conjugate of a complex number has certain general properties that we shall find useful later. For a real number a = a + Oi, the conjugate is a = a + Oi = a - Oi = a. That is, THEOREM
The conjugate of a real number is the real number itself. Other properties of the conjugate that are direct consequences of the definition are given next. In each statement, z and w represent complex numbers.
THEOREM
The conjugate of the conjugate of a complex number is the complex number itself. (z) = z
(6)
I
(7)
I
The conjugate of the sum of two complex numbers equals the sum of their conjugates. z+w=z+w
The conjugate of the product of two complex numbers equals the product of their conjugates. z·w = z·w
(8)
I�
�----------------------------------�
We leave the proofs of equations (6), (7), and (8) as exercises.
1 14
CHAPTER 1
Equations and I nequalities
Powers of ;
The powers of i follow a pattern that is useful to know. is i6
i1 = i P = -1 P = P · i = - I · i = -i
= =
i4 • i = 1 · i = i i4 • P = - 1
i7 = i4 • i3 = - i i8 = i4 • i4 = 1
i4 = p . p = ( - 1 ) ( - 1 ) = 1
And so on. The powers of i repeat with every fourth power. EXA M P L E 7
Evaluating Powers of ; (a) p7 = p4 . p = (i4) 6 . i3 = 16 · i3 = -i (b) i 1 0 1 = i100 • i 1 = ( i4) 25 . i = 125 . i = i
•
Writing the Power of a Complex N u m ber in Standard Form
EXA M P L E 8
Write ( 2 + i)3 in standard form. We use the special product formula for (x + a )3.
Solution
(x + a ) 3 = x3 + 3ax2 + 3a2 x + a3 Using this special product formula, (2 + i)3 = 23 + 3 · i . 22
+
3·P.2
= 8 + 12i + 6( - 1 )
+
+
P
(-i)
= 2 + l 1i.
•
c;rn:==- Now Work P R O B L E M 4 1
2
Solve Quadratic Equations in the Com p lex N um be r System
Quadratic equations with a negative discriminant have no real number solution. However, if we extend our number system to allow complex numbers, quadratic equations will always have a solution. Since the solution to a quadratic equation involves the square root of the discriminant, we begin with a · discussion of square roots of negative numbers. DEFINITION
WARNING In writing
v=N
=
If N is a positive real number, we define the principal square root of denoted by V- N , as
Villi be
sure to place i outside the V symbol.
EXA M P L E 9
_
where i is the imaginary unit and F = - 1 .
- N,
.J
Evaluating the Square Root of a Negative N umber (a) v=I = VIi = i (c) v=s = Y8 i = 2 Y2 i
(b) v=4 = v4 i = 2i •
SECTION 1.3
EXA M P L E 1 0
Solution
1 15
Complex Numbers; Quadratic Equations i n the Complex Number System
Solving Equations
Solve each equation in the complex number system. (a) x2= 4 (b) x2= -9 (a) x2 = 4 x = ± V4 = ± 2 The equation has two solutions, -2 and 2. The solution set is ( - 2, 2). (b) x2 = -9 x = ± \.1-9 = ± V9i = ±3i The equation has two solutions, -3i and 3i. The solution set is {-3i, 3i}.
.... .
-
:z:: """ .- Now Work ''!l!l;:cs:=
PROBLEM S
49
A ND
53
WARNING When working with square roots of negative numbers, do not set the square root of
a product equal to the product of the square roots (which can be done with positive n umbers). To see why, look at this calculation: We know that V1c5O = 10. However, it is also true that 100 so =
10
=
( -25) ( -4),
ViOO
=
V( -25)( -4)
'*
i
V=25 v=4
=
(V25i)(V4i)
=
(5i)(2i)
=
10?
=
-10 •
Here is the error.
Because we have defined the square root of a negative number, we can now restate the quadratic formula without restriction. THEOREM
Quadratic Formula
In the complex number system, the solutions of the quadratic equation ax2 + bx + c = 0, where a, b, and c are real numbers and a =f:. 0, are given by the formula x=
-b ± Vb2 - 4ac 2a
(9)
I
�
� . � -------------------
EXA M P L E 1 1 Solution
Solving Quadratic Equations i n the Complex N u m ber System
Solve the equation x2 - 4x + 8= 0 in the complex number system. Here a = 1, b = -4, c = 8, and b2 - 4ac = 16 - 4( 1 ) (8 ) = -16. Using equa tion (9), we find that - ( -4) ± v=i6 4 ± V16 i 4 ± 4i = 2 ± 2i = -= x= 2 2( 1 ) 2 The equation has two solutions 2 - 2i and 2 + 2i. The solution set is {2 - 2i, 2 + 2i} . Check :
2 + 2i:
(2 + 2i ) 2
2 - 2i:
(2
-
4( 2 + 2i ) + 8
- 2i ) 2 - 4 ( 2
1l'l!l'O=="".- Now Work
PROBLEM
- 2i)
+
4 + '8i. + 4F - .8' - '8i. + .8' =4- 4 = 0 8 = 4 - '8i. + 4F - .8' + '8i. + .8' = 4 - 4= 0 =
•
59
The discriminant b2 - 4ac of a quadratic equation still serves as a way to deter mine the character of the solutions.
1 16
CHAPTER 1
Equations and I nequalities
Character of the Solutions of a Quadratic Equation
In the complex number system, consider a quadratic equation ax2 + bx + c = ° with real coefficients. 1. If b2 - 4ac > 0, the equation has two unequal real solutions. b2 - 4ac = 0, the equation has a repeated real solution, a double root. 3. If b2 - 4ac < 0, the equation has two complex solutions that are not real. The solutions are conjugates of each other.
2. If
The third conclusion in the display is a consequence of the fact that if
b2 - 4ac = - N < ° then, by the quadratic formula, the solutions are x=
-b +
-b
Yb2 - 4ac
-b
2a
and
x=
Yb2 - 4ac
-------
-b -
2a
+
v=N
-
v=N
2a
2a
-b + VN i -b VN = - + -- i 2a 2a 2a
-----
-b
-
VN i = -b VN . l
2a
2a
--
2a
which are conjugates of each other. EXA M P L E 1 2
Determi ning the Character of the Solution of a Quadratic Equation Without solving, determine the character of the solution of each equation. (b) 2x2 + 4x + 1
(a) 3x2 + 4x + 5 = ° Solution
=
°
(c) 9x2 - 6x + 1 = °
(a) Here a = 3, b = 4, and c = 5, so b2 - 4ac = 1 6 - 4 ( 3 ) ( 5 ) = -44. The solutions are two complex numbers that are not real and are conjugates of each other. (b) Here a = 2, b = 4, and c = 1, so b2 - 4ac = 16 - 8 = 8. The solutions are two unequal real numbers. (c) Here a = 9, b = -6, and c = 1 , so b2 - 4ac = 3 6 - 4(9 ) ( 1 ) = 0. The solution is a repeated real number, that is, a double root. •
w = = = "",·-
Now Work
P ROBLEM
73
1.3 Assess Your Understanding 'Are You Prepared?' An swe rs a re g iven at t he end of t he se exe rcise s. If you get a w rong an swe r, read t he page s l isted in re d.
� Name the integers and the rational numbers in
}
Y True or False
the set
6 -3 , 0, v2 ,-, 1T . (pp. 4-5) 5
{
/..
'"
Rational numbers and irrational numbers are in the set of real numbers. (pp. 4 -5 ) 3 ( p 74 ) Rationalize the denominator of 2 + V3 3' .
Concepts and Vocabulary
6. True or False The conjugate of2 + 5 i is -2 - Si. True or False All real numbers are complex numbers. 8. True or False If2 -3 i is a solution of a quadratic equation with real coefficients, then -2 + 3 i is also a solution.
4. In the complex number 5
+ 2 i, the number 5 is called the part; the number 2 is called the ____ part; the number i is called the ____ ____
X The equation x 2
=-4
X
has the solution set ____
Skill Building
In P roblem s 9-46, w rite ea ch ex p re ssion in t he standa rd fo rm a + b i . .t. (2 -3 i )
}\
+ (6 + 8i)
(2 - 5 i) - (8 + 6i)
10. (4 + 5 i ) + (-8 + 2 i) 14. ( - 8 + 4i) - (2 -2 i)
� (-3 + 2 i ) - (4 -4 i ) � 3 (2 - 6i)
12. (3 -4 i) - (-3 -4 i) 16.
-4(2
+
8i)
SECTION 1.3
� 2i(2 - 3i) �. ( -6 + i ) ( -6 - i) Y. 2 7 i 1\ '
_ 2 -_i 26. _
, i23 i6 - 5
34. 38. 42. 46.
"1f- ( 1 + i)3 A i6 + i4 + P + 1 � V-4 50. v=64
47-52,
-
13 5 - 12i 2 + 3i 28. 1 - !
24. ___
' ''' l + i
(� � y
" i)
i
1 17
20. ( 5 + 3i) ( 2 - i )
O
\.� �
-2!
30.
(3 - 4i ) ( 2 + i )
}( 3 � 4i
22. ( -3 + i ) ( 3 + i)
i
In Problems
J(
18. 3i( -3 + 4i)
�G � y " 'j( +
Complex Numbers; Quadratic Equations in the Complex Number System
32. ( 1 - i f
�. ( 1 + i) 2
il4
36. i-23 40. 4i3 - 2P + 1 44. 2i4( 1 + i2)
4 + i3 (3i)4 + 1
P + i5 + P + i
perform the indicated operations and express your answer in the form a + bi.
� V-2s -
48. vC9
�, Y(3 + 4i) (4i - 3)
52. yr(4-) (3i-+ 3i4) --
In Problems 53-72, solve each equation in the complex number system. . ,2 + 4 = 0 54. x2 - 4 = 0 x2 - 16 0 x2 - 6x + 13 = 0 58. x2 + 4x + 8 = 0 x2 - 6x + 10 = 0 8x2 - 4x + 1 = 0 62. l Ox2 + 6x + 1 = 0 . 5x2 + 1 = 2x x2 + X + 1 = 0 66. x2 - x + 1 = 0 . x3 - 8 = 0 I . X4 = 1 6 70. X4 1 X4 + 1 3x2 + 36 = 0
�
�. �
�
•
=
X
·
=
56. 60. 64. 68. 72.
x2 + 25 = 0 x2 - 2x + 5 = 0 1 3x2 + 1 = 6x
x3 + 27
=
0
X4 + 3x2 - 4
=
0
In Problems 73-78, without solving, determine the character of the solutions of each equation in the complex number system. 3x2 - 3x + 4 0 74. 2x2 - 4x + 1 = 0 2x2 + 3x = 4 76. x2 + 6 = 2x 9x2 - 12x + 4 = 0 78. 4x2 + 12x + 9 = 0 2 + 3i is a solution of a quadratic equation with real coefficients. Find the other solution. 80. 4 - i is a solution of a quadratic equation with real coefficients. Find the other solution.
.�
�
In Problems 81-84, z z +Z
'Yk
�
�
=
=
3 - 4i and w 8 + 3i. Write each expression in the standard form a + bi. 82. w - IV zz 84. z - w
�
=
Applications and Extensions
5.
Electrical Circuits
The impedance Z, in ohms, of a circuit element is defined as the ratio of the phasor voltage V, in volts, across
the element to the phasor current I, in amperes, through the elements. That is, Z
=
f.
If the voltage across a circuit element is 18 + i
volts and the current through the element is 3 - 4i amperes, determine the impedance.
86
·
)t
Parallel C ircuits
1 . . . ' . . . I n an ac CirCUit WitI1 two para I I el pathways, the total I mpedance Z, 111 ohms, satls fles the formula Z
=
1 1 + -, ZI Z2
where ZI is the impedance of the first pathway and Z2 is the impedance of the second pathway. Determine the total impedance if the impedances of the two pathways are ZI = 2 + i ohms and Z2 = 4 - 3i ohms. Use z
=
a + bi to show that z +
z
88. Use z
=
=
z.
( Use z
90. Use z
=
=
a + bi to show that z a + bi and w a + bi and w
=
=
=
2a and
z
-z
=
c + di to show that z + w c + di to show that z · w
=
=
2bi.
z+
W.
z · w.
Discussion and Writing
91. Explain to a friend how you would add two complex numbers and how you would multiply two complex numbers. Explain any dif ferences in the two explanations.
92. Write a brief paragraph that compares the method used to rationalize the denominator of a radical expression and the method used to write the quotient of two complex numbers in standard form. 'Are You Prepared?' Answers
{
1. Integers: { -3, O } ; rational numbers: -3, 0,
�}
2. True
3. 3 ( 2 -
\13)
118
CHAPTER 1
Equations and Inequalities
1 .4 Radical Equations; Equations Quadratic in Form; Factorable Equations PREPARING FOR THIS SECTION •
•
Before getting started, review the following: •
Square Roots (Section R.2, pp. 23-24) Factoring Polynomials (Section R.5, pp. 49-55)
,NOW Work
nth Roots; Rational Exponents (Section R.8, pp. 72-76)
the 'Are You Prepared?' problems on page 122.
OBJECTIVES
1
1
Solve Radical Equations (p. 1 1 8)
2
Solve Equations Quadratic in Form (p. 1 1 9)
3
Solve Equations by Factoring (p. 1 2 1 )
Solve Rad ical Equations
When the variable in an equation occurs in a square root, cube root, and so on, that is, when it occurs in a radical, the equation is called a radical equation. Sometimes a suitable operation will change a radical equation to one that is linear or quadratic. A commonly used procedure is to isolate the most complicated radical on one side of the equation and then eliminate it by raising each side to a power equal to the index of the radical. Care must be taken, however, because apparent solutions that are not, in fact, solutions of the original equation may result. These are called extraneous solutions. Therefore, we need to check all answers when working with radical equations. EXA M P L E 1
Solving a Radical Equation Find the real solutions of the equation:
Solution
V2x - 4 - 2
=
0
The equation contains a radical whose index is 3. We isolate it on the left side.
V2x - 4 - 2 V2x - 4
= =
0 2
Now raise each side to the third power (the index of the radical is 3) and solve.
( V2x - 4) 3 = 23 2x - 4 = 8 2x = 12 X = 6
Raise each side to the power 3. Sim plify. Add 4 to both sides. Divide both sides by 2.
Check:
\12 ( 6) - 4 - 2 = V12 - 4 - 2 The solution set is {6}. Li!l!
EXA M P L E 2
>-
Now Work
PRO
VB - 2
=
B LE M 7
Solving a Radical Equation Find the real solutions of the equation:
Solution
=
Vx"="l = x - 7
We square both sides since the index of a square root is 2.
Vx"="l = x - 7 (\/�-=-ly = ( x - 7? X - 1 = x2 - 14x x2 - 15x + 50 = 0
Square both sides. +
49
Remove pa rentheses. Put in standa rd form.
2-2
=
o. •
SECTION 1 . 4
Radical Equations; Equations Quadratic i n Form; Factorable Equations
(x - 10) (x - 5 ) = 0 x = 10 or x = 5
1 19
Factor. Apply the Zero-Product Property and solve.
Check:
VlO-=---i = V9 = 3 and x - 7 = 10 - 7 = 3 = 5: = v'S=J:" = v4 = 2 and x - 7 = 5 - 7 = -2 The solution x = 5 is extraneous; the only solution of the equation is x = 10. The solution set is /10j. •
x x
= 1 0:
� �
&:."l!l:==- Now Work
=
PRO B LEM
19
Sometimes we need to raise each side to a power more than once in order to solve a radical equation. Solving a Radical Equation
EXA M P L E 3
Find the real solutions of the equation: Solution
V2x
+
3
-
Vx+2 2 =
First, we choose to isolate the more complicated radical expression (in this case, v'2x + 3) on the left side.
Now square both sides (the index of the radical on the left is
( V2x + 3) 2 2x + 3 2x + 3 2x + 3
= = = =
(Vx+2 + 2)2 ( Vx+2) 2 + 4 Vx+2 + 4 x + 2 + 4Vx+2 + 4 x + 6 + 4 Vx+2
2).
Square both sides. Remove parentheses. Simplify. Combine like terms.
Because the equation still contains a radical, we isolate the remaining radical on the right side and again square both sides.
x - 3 = 4 Vx+2 ( x - 3 ? = (4 Vx+2)2 x2 - 6x + 9 = 16x + 32 x2 - 22x - 23 = 0 ( x - 23 ) (x + 1 ) = 0 X = 23 or x = -1
Isolate the radica l on the right side. Square both sides. Remove pa rentheses. Put i n standard form. Factor.
The original equation appears to have the solution set { -1, 23}. However, we have not yet checked. Check:
x x
= =
23:
- 1:
V2x + 3 V2x + 3 -
Vx+2 V2(23) + 3 - V23 + 2 Vx+2 V2( - 1 ) + 3 - V + 2
v49 - v'2s = 7 - 5 = 2 -1 = Vi - Vi = 1 - 1 = 0 = The equation has only one solution, 23; the solution -1 is extraneous. The solution set is /23j. • =
===- Now Work 2
=
P RO B L EM
29
Solve Eq uations Q u a d ratic in Form
The equation X4 + x2 - 12 = 0 is not quadratic in x, but it is quadratic in x2. That is, if we let = x , we get + - 12 = 0, a quadratic equation. This equation can be solved for and, in turn, by using = x , we can find the solutions x of the orig inal equation.
u
u
2
u2 u
u
2
120
CHAPTER 1
Equations and I nequalities
In general, if an appropriate substitution u transforms an equation into one of the form
au2 + bu +
0
c =
a *- O
then the original equation is called an equation of the quadratic type or an equation quadratic in form.
The difficulty of solving such an equation lies in the determination that the equation is, in fact, quadratic in form. After you are told an equation is quadratic in form, it is easy enough to see it, but some practice is needed to enable you to rec ognize such equations on your own. E XA M P LE 4
Solution
Solving Equations That Are Quadratic in Form Find the real solutions of the equation:
(x + 2)2 + 1 1 (x + 2) - 12 = 0
For this equation, let u = x + 2. Then u2
=
(x + 2? + 1 1 ( x + 2) - 12
=
(x + 2 ) 2 , and the original equation,
0
becomes
u2 + 1 1u - 12 = 0 (u + 12)(u - 1 ) = 0 u = - 12 or u = 1
Let
u
= x
+ 2.
Then
u2
= (x
+ 2) 2 .
Factor. Solve.
But we want to solve for x. Because u = x + 2, we have
x + 2 = -12 o r x + 2 = 1 x = -1 x = - 14
Check:
x
-14: ( - 14 + 2)2 + 1 1 ( - 14 + 2) - 12 = ( - 12? + 1 1 ( -12) - 12 = 144 - 132 - 12 = 0 x = - 1 : ( - 1 + 2 ) 2 + 1 1 ( - 1 + 2) - 12 = 1 + 1 1 - 12 = 0 =
The original equation has the solution set { -14, - I } .
EXA M P L E 5
Solving Equations That Are Quadratic in Form Find the real solutions of the equation:
Solution
•
(x2 - 1 ) 2 + (x2 - 1 ) - 12
=
0
For the equation (x2 - 1 ) 2 + (x2 - 1 ) - 12 = 0, we let u = x2 - 1 so that u2 = (x2 - 1 ) 2 . Then the original equation,
(x2 - 1 ) 2 + (x2 - 1 ) - 12 = 0 becomes
u2 + u - 12 = 0 (u + 4)(u - 3 ) = 0 u = -4 or u = 3
Let
u
=
�
-
1. Then u2
=
(�
-
1{
Factor. Solve.
But remember that we want to solve for x. Because u = x2 - 1, we have
x2 - 1 = -4 or x2 - 1 = 3 x2 = -3 x2 = 4 The first of these has no real solution; the second has the solution set { -2 2} ,
Check:
x = -2: x = 2:
.
(4 1 )2 + (4 - 1) - 12 = 9 + 3 - 12 = 0 (4 - I ? + (4 - 1 ) - 12 = 9 + 3 - 12 = 0 -
The original equation has the solution set { -2, 2}.
•
SECTION 1.4
EXAM P L E 6
Radical Equations; Equations Quadratic i n Form; Factora ble Equations
12 1
Solving Equations That Are Quadratic i n Form x + 2 Vx - 3 = 0 For the equation x + 2 Vx - 3 = 0, let u = Vx. Then u2 equation,
Find the real solutions of the equation: Solution
x + 2Vx - 3
=
=
x, and the original
0
becomes u2 + 2u - 3 = 0 ( u + 3) (u
1) = 0 u = -3 or u = 1 -
Let u = \IX. Then u2 = x. Factor. Solve.
Since u = Vx , we have Vx = -3 or Vx = 1. The first of these, Vx = - 3, has no real solution, since the square root of a real number is never negative. The second, Vx = 1, has the solution x = 1 . Check: 1
+ 2 v1
-
3 = 1 + 2 - 3 = 0
The original equation has the solution set ( I J . ,,-===,...
ANO T H ER M ET H OD
F OR SOLV I NG EX A M PLE
•
6
WOULD
B E TO T REAT IT AS A R AD ICAL EQUATION. SOLV E WAY
IT T H IS
FOR PRACT I CE .
The idea should now be clear. If an equation contains an expression and that same expression squared, make a substitution for the expression. You may get a quadratic equation. 'e'
3
.�
Now Work
PRO B LE M
S
1
So lve Equations by Factoring
We have already solved certain quadratic equations using factoring. Let's look at examples of other kinds of equations that can be solved by factoring. EXAM P L E 7 Solution
Solvi n g Equations by Factoring Solve the equation: X4 = 4x2 We begin by collecting all terms on one side. This results in 0 on one side and an expression to be factored on the other. X4 x2 ( x2 - 4 ) = 0 or x2 - 4 = 0 x2 = 4 -
x2 = 0 x = 0
or
X4 = 4x2 4x2 = 0
x = -2
or
Factor. Apply the Zero-Prod uct Property.
x = 2
The solution set is { -2, 0, 2}. Check: x = -2:
x = 0: x = 2:
( _2 )4 = 1 6 and 4 ( -2 ? = 16 S o -2 is a solution. 04 0 and 4 · 02 = 0 So 0 is a solution . 2 24 = 16 and 4 · 2 = 1 6 S o 2 i s a solution. =
•
122
CHAPTER 1
Equations and Inequa lities
EXA M P L E 8
Solving Equations by Factoring Solve the equation:
Solution
x3 - x2 - 4x + 4
=
0
Do you recall the method of factoring by grouping? (If not, review pp. 53-54.) We group the terms of x3 - x2 - 4x + 4 = 0 as follows: (x3 - x2) - (4x - 4)
0
=
Factor out x2 from the first grouping and 4 from the second. x2(x - 1 ) - 4( x - 1 )
=
This reveals the common factor (x - 1 ) , so we have (x2 - 4) ( x - 1 ) = 0 (x - 2)(x + 2)(x - 1) = 0 x - 2 = 0 or x + 2 = 0 or x
=
x = -2
2
0
Factor again.
x - I x
= =
0
Set each factor equa l to
1
0.
Solve.
The solution set is { -2, 1, 2 } . Check:
x
=
x
=
x
=
1'"
•
-2:
1:
2: >-
( -2)3 - ( -2f 4( -2) + 4 = -8 - 4 + 8 + 4 1 3 - 12 - 4( 1 ) + 4 = 1 - 1 - 4 + 4 = 0 -
23 - 22 - 4(2) + 4 Now Work
=
PRO B L E M
8 - 4 - 8 + 4
=
=
0
0 -2 is a sol ution. 1 is a sol ution . 2 is a solution .
•
79
1 .4 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
'%
=
True or False The principal square root of any nonnegative real number is always nonnegative. (pp. 23-24)
v=s �� Factor
6x3 -
(pp. 72-76)
2X2 ( pp. 49-55)
Concepts and Vocabulary
6. True or False Radical equations sometimes have extraneous
4. When an apparent solution does not satisfy the original equa-
�
solutions.
tion, it is called a(n ) solution. If u is an expression involving x, the equation au2+bu+ = c 0, a oF 0, is called a(n) equation ____
Skill Building
In Problems
'/(
7-40,
find the real solutions of each equation.
=1 V2t=1
10. v5t+3 =-2
=2 *- � =Vs 16. \Yx2+16 ( VIS - 2x =x 22. = x 2V- x - 1 ¥- 3+� =X
8. v3t+4 =2
)&� VI
- 2x -
= 3 ° 3 -1 14. �2x -= it(= x 8vX 20. � =x
)I: y
x2
- X
x
+2 =x 26. 2+V12 - 2x -4
=
� v3t+4=-6
12. � - 1
�. �
x2 +
=
°
2x=-1 18. = x 3vX 1( X = 2� 24. Y3 x + = x2 X - 2 ,*. � - Vx+l = 1 -
SECTION 1.4
)(.
28. � + Vx+2 = 1
¥ V3 - 2 VX = VX
Radical Equations; Equations Quadratic in Form; Factora ble Equations
�- �=2
30. V3x - 5
�
32. VlO + 3 VX = VX
34. (3x - 5 ) 1/2 = 2
35. (5x - 2 ) 1/3 = 2
37. ( x2 + 9) 1/2 = 5
38. ( x2 - 1 6 ) 1 /2
=
4
2 X - 5x2 - 12
=
0
47. (x + 2 ) 2 + 7(x + 2 ) + 12 = 0
0
=
)(
53. x - 4xVX = 0
60. XI/2 - 3xl/4 + 2
65.
= x
63. x2 + 3x +
1 I ? = __ + 2 x + 1 ( x + 1 )-
66.
68. 2x-2 - 3x-1 - 4 = 0 71.
0
(-V )2 + -2v v+1 v+ 1
=
1
= 0
49. ( 3x + 4f - 6(3x + 4) + 9 = 0 52. 3 ( 1 - y)2 + 5 ( 1 - y) + 2 = 0
3
=
55. x + VX
59. 4X I /2 - 9X I/4 + 4 = 0
\y4 - 5x2
=
54. x + 8VX = 0 57. t l/2 - 2tl/4 + 1
-
46. x6 - 7x3 - 8 = 0
2(s + I f - 5 ( s + 1 )
56. x + VX = 6
62.
40. X3/4 - 9xl/4 = 0 43. 3x4 - 2x2
48. (2x + 5 ) 2 - (2x + 5 ) - 6
50. (2 - x)2 + (2 - x ) - 20 = 0
2
(3x + 1 ) 1/2 = 4
39. x3/2 - 3xl/2 = 0
9
45. x6 + 7x3 - 8
=
36. (2x + 1 ) 1/3 = - 1
In Problems 41-72, find the real solutions of each equation. 41. x4 - 5x2 + 4 = 0 42. X4 - 1 0x2 + 25 = 0 44.
- Vx+7
123
58. Z I /2 - 4 Z I/4
0 0
=
61.
Vx2 + 3x = 6
--
1 1 + x - I ( x - 1 )2
=
72.
\Y5x2
+
4 = 0
- 6 = x
64. x2 - 3x -
12
69. 2x2/3 - 5xl/3 - 3 = 0
= 8
20
=
Vx2 - 3x = 2
67. 3x-2 - 7x-1 - 6 = 0 70. 3x4/3 + 5x2/3 - 2 = 0
(-) (-) + y 2 Y = 6 y - 1 y - 1
In Problems 73-88, find the real solutions of each equation by factoring. 73. x3 - 9x = 0 74. X4 - x2 = 0 75. 4x3
3x2
=
�.
7
77. x3 + x2 - 20x = 0
78. x3 + 6x2 - 7x = 0
80. x3 + 4x2 - X - 4 = 0
81. x3 - 3x2 - 4x + 12 = 0
83. 2x3 + 4 = x2 + 8x
84. 3x3 + 4x2 = 27x + 36
85. 5x3 + 45x = 2x2 + 18
87. x (x2 - 3x) I /3 + 2(x2 - 3x)4/3 = 0
88. 3x(x2 + 2X) I /2 - 2(x2 + 2x)3/2 = 0
86. 3x3 + 1 2x = 5x2
+
20
In Problems 89-94, find the real solutions of each equation. Use a calculator
89. x - 4xl/2 + 2 = 0
95. If k =
-- and x + 3 x - 3
90. x2/3
+
4x l/3 + 2
=
0
82. x3 - 3x2 - X + 3 = 0
LO
express any solutions rounded to two decimal places.
93. 7T ( 1 + t)2 = 7T + 1 + t . k2 - k = 12, fmd x.
x3 + x2 - x - 1 = 0
96. If k =
91. X4 + v3 x2 - 3 = 0
-- and x+ 3 x - 4
94. 7T ( 1 + r)2
=
2 + 7T ( 1 + r)
k2 - 3k = 2 8, find x.
Ap plicatiens
97.
Physics: Using Sound to Measure Distance The distance to the sur face of the water in a well can sometimes be found by dropping an ob ject into the well and measuring the time elapsed until a sound is heard. If I I is the time (measured in seconds) that it takes for the object to strike the water, then t1 will obey the equation s = 16ft , where s is the
distance (measured in feet). It follows that
II
Vs
= 4 . Suppose that t
2
is the time that it takes for the sound of the impact to reach your ears. Because sound waves are known to travel at a speed of approximately 1 1 00 feet per second, the time t to travel the distance s will be
--
s t = See the I. 11 ustratlOn. . 2 1 100 .
2
Sound waves:
Falling object:
t1 = {S "4
t2 =
II�O
124
CHAPTER 1
Equations and Inequal ities
Now tl + tz is the total time that elapses from the moment that the object is dropped to the moment that a sound is heard. We have the equation . Total time elapse d
=
Vs
4+
s 1 00 1
Find the distance to the water's surface if the total time elapsed from dropping a rock to hearing it hit water is 4 seconds.
98.
Crushing Load
load L, in tons, using the model
99.
H, in feet, of a square wooden pillar to its crushing H2 1L If a square wooden pillar is 4 inches thick and feet high, what is its crushing load?
A civil engineer relates the thickness T, in inches, and height T
=
1
25
0
The period of a pendulum is the time it takes the pendulum to make one full swing back and forth. The
Foucault's Pendulum
32: where I is the length, in feet, of the pendulum. In 1 85 1 , lean-Bernard-Leon period T, in seconds, is given by the formula T 271"'YIT =
Foucault demonstrated the axial rotation of Earth using a large pendulum that he hung in the Pantheon in Paris. The period of Foucault's pendulum was approximately 1 6.5 seconds. What was its length?
Discussion and Writing
100. Make up a radical equation that has no solution. 101 . Make up a radical equation that has an extraneous solution. 1 02. Discuss the step in the solving process for radical equations that leads to the possibility of extraneous solutions. Why is there no such
possibility for linear and quadratic equations?
'Are You Prepared?' Answers
2. -2
1. True
PREPARING FOR THIS SECTION •
Before getting started, review the following:
Algebra Essentials (Section R.2, pp. 1 7 1 9)
'\.Now work
-
the 'Are You Prepared?' problems on page 1 32.
OBJECTIVES
1
Use Interval Notation (p. 1 25)
2
Use Properties of Inequalities (p. 1 26)
3
Solve Inequalities (p. 1 28)
4 Solve Combined Inequalities (p. 1 29)
Suppose that a and b are two real numbers and a < b. We shall use the notation a < x < b to mean that x is a number between a and b . The expression a < x < b is equivalent to the two inequalities a < x and x < b. Similarly, the expression a :::; x :::; b is equivalent to the two inequalities a :::; x and x :::; b. The remaining two possibilities, a :::; x < b and a < x :::; b, are defined similarly. Although it is acceptable to write 3 2': X 2': 2, it is preferable to reverse the inequality symbols and write instead 2 :::; x :::; 3 so that, as you read from left to right, the values go from smaller to larger. A statement such as 2 :::; x :::; 1 is false because there is no number x for which 2 :::; x and x :::; 1 . Finally, we never mix inequality symbols, as in 2 :::; x 2': 3.
SECTION 1.5
1
Solving I n equalities
125
Use I nterva l Notation
Let a and b represent two real numbers with a < b. DEFINITION
A closed interval, denoted by [a, b], consists of all real numbers x for which a ::;
x
::;
b.
An open interval, denoted by ( a, b ) , consists of all real numbers x for which
a
< x < b.
The half-open, or half-closed, intervals are (a, b], consisting of all real num bers x for which a < x ::; b, and [a, b) , consisting of all real numbers x for which a ::; x < b.
..J
In each of these definitions, a is called the left endpoint and b the right endpoint of the interval. The symbol 00 (read as "infinity") is not a real number, but a notational device used to indicate unboundedness in the positive direction. The symbol - 00 (read as "negative infinity") also is not a real number, but a notational device used to indi cate unboundedness in the negative direction. Using the symbols 00 and - 00 , we can define five other kinds of intervals:
[ a, 00 ) ( a, 00 ) ( - 00 , a 1 ( - 00 , a) (- 00 , 00 )
Consists of all real numbers x for which x 2: Consists of all real numbers x for which x > Consists of aU real numbers x for which x ::; Consists of all real numbers x for which x < Consists of all real numbers x
a a
a
a
Note that 00 and -00 are never included as endpoints, since neither is a real number. Table 1 summarizes interval notation, corresponding inequality notation, and their graphs. Ta ble 1
I nterval
The open interval ( a, The closed interval
b)
[a, bl
The h a lf-open interval The h a lf-open interval
a < x< b
[a, b)
a :!S x < b
( a, b]
a < x :!S b a
X 2:
( a, co)
x> a
The interval
( - co, a]
x :!S a
The interval
( -co, co)
The interval
The interval ( -co,
a)
Graph
a :!S x :!S b
[a, co)
The interval
EXAM P L E 1
Inequality
x< a
( a
) b
E a
) b
E a
� b � b
( a
[ a
( a
•• :.
.
� a
) a
All rea l numbers
Writing I nequalities Using I nterval N otation Write each inequality using interval notation. (a) 1
Solution
::;
x
::;
3
(b) -4 < x < 0
(c) x > 5
(d) x
::;
1
(a) 1 ::; x ::; 3 describes all numbers x between 1 and 3, inclusive. In interval notation, we write [1, 3 ] ' ( b ) I n interval notation, -4 < x < 0 i s written ( -4, 0).
126
CHAPTER 1
Equations and Inequal ities
5 consists of all numbers x greater than 5. In interval notation, we write (5, 00 ) . (d) In interval notation, x � 1 is written ( - 00, 1 )' X >
(c)
•
EXAM P LE 2
Writing I ntervals Using I nequality N otation Write each interval as an inequality involving x. (a) [ 1 , 4)
Solution
(a) (b) (c) (d)
(0
)
(d) ( - 00 , -3J
(c) [2, 3 ]
[ 1 , 4) consists o f all numbers x for which 1 :::; x < 4. (2, 00 ) consists of all numbers x for which x > 2. [2, 3 J consists of all numbers x for which 2 � x � 3. ( - 00 , -3J consists of all numbers x for which x � -3.
1.£==> -
2
(b) (2,
Now Work
P ROBLEM S
11 , 23,
•
31
A ND
Use Properties of I n eq u a l ities
The product of two positive real numbers is positive, the product of two negative real numbers is positive, and the product of 0 and 0 is O. For any real number a, the value 2 2 of a is 0 or positive; that is, a is nonnegative. This is called the nonnegative property. f
f f
In
Word s
The s quare of a real number is
Nonnegative Property
For any real numbers
a,
never negative.
(1) If we add the same number to both sides of an inequality, we obtain an equiva lent inequality. For example, since 3 < 5, then 3 + 4 < 5 + 4 or 7 < 9. This is called the addition property of inequalities. Addition Property of Inequalities
For real numbers
a,
b, and
c,
If a < b, then a + If a
>
b, then a +
c
c.
(2a)
b+
c.
(2b)
c >
The addition property states that the sense, or direction, of an inequality remains unchanged if the same number is added to each side. Figure 3 illustrates the addition property (2a). In Figure 3 ( a), we see that a lies to the left of b. If c is posi tive, then a + c and b + c each lie c units to the right of a and b, respectively. Consequently, a + c must lie to the left of b + c; that is, a + c < b + c. Figure 3(b) illustrates the situation if c is negative. Figure 3 -c u n its
c u n its
� units
,---A----., e . •
•
a
(a)
b
a+c
b+c
If a < b and c > 0,
then a + c < b + c.
1I!l� - D R A W
�
- c units
� • • •
a+c
b+c
(b)
•
b
If a < b an d c < O, then a + c < b + c.
A N ILLU S T R A T I ON S IMIL A R T O
T H A T ILLU S T R A T E S
a
F IGURE
T HE A D D I T ION P RO P E R T Y
3 (2b)
SECTION 1.5
EXA M P L E 3
Solving Inequalities
127
Addition P roperty of I nequalities
(a) If x < -5, then x + 5 < -5 + 5 or x + 5 < 0. (b) If x > 2, then x + ( -2 ) > 2 + ( -2) or x - 2 > 0. II!IT::=� .-
Now Work
PROB L E M
•
39
We will use two examples to arrive at our next property. E XA M P L E 4
Solution
EXA M P L E 5
Solution
r
r
In Word s
r
reverses the inequality.
r
Multiplying by a negative number
M u ltiplying an I nequality by a Positive N u m ber
Express as an inequality the result of multiplying each side of the inequality 3 < 7 by 2. We begin with 3<7 Multiplying each side by 2 yields the numbers 6 and 14, so we have 6 < 14 • M u ltip lying an I nequality by a Negative N u m ber
Express as an inequality the result of multiplying each side of the inequality 9 > 2 by -4. We begin with 9>2 MUltiplying each side by -4 yields the numbers -36 and -8, so we have - 36 < -8 • Note that the effect of multiplying both sides of 9 > 2 by the negative number -4 is that the direction of the inequality symbol is reversed. Examples 4 and 5 illustrate the following general multiplication properties for
inequalities:
Multiplication Properties for Inequalities
For real numbers a, b, and e, If a If a If a If a
< b and if e > 0, then ae < be. < b and if e < 0, then ae > be.
(3a)
> b and if e > 0, then ae > be. > b and if e < 0, then ae < be.
(3b)
The multiplication properties state that the sense, or direction, of an inequality remains the same if each side is multiplied by a positive real number, whereas the direction is reversed if each side is multiplied by a negative real number. EXA M P L E 6
M ultiplication Property of I nequ alities
1 (a) If 2x < 6, then "21 (2x) < "2 (6) or x < 3. (b) If x > 12, then -3 x < -3 ( 12 ) or x < -36. -3 -3
-
(-)
128
CHAPTER 1
Equations and Inequa lities
-4x -8 > or x > 2. -4 -4 ( d ) If -x > 8, then ( - l ) ( -x) < ( - 1 ) (8) or x < -8.
( c ) If -4x < -8, then
c;1l'!:= : ==-- Now Work
PROB L E M
•
45
The reciprocal property states that the reciprocal of a positive real number is positive and that the reciprocal of a negative real number is negative. Reciprocal Property for Inequalities
1 a
If a > 0, then - > 0
1 a
If a < 0, then - < 0
3
1 a 1 If - < 0, then a < 0 a If - > 0, then a > 0
(4a) (4b)
Solve I nequalities
An inequality in one variable is a statement involving two expressions, at least one containing the variable, separated by one of the inequality symbols < , :5 , > , or ;::: . To solve an inequality means to find all values of the variable for which the state ment is true. These values are called solutions of the inequality. For example, the following are all inequalities involving one variable x:
x+5<8
2x - 3 ;::: 4
x+1 >0 x-2
x2 - 1 :5 3
--
As with equations, one method for solving an inequality is to replace it by a series of equivalent inequalities until an inequality with an obvious solution, such as x < 3, is obtained. We obtain equivalent inequalities by applying some of the same properties as those used to find equivalent equations. The addition property and the multiplication properties form the bases for the following procedures. Procedures That Leave the I nequality Symbol Unchanged
1. Simplify both sides of the inequality by combining like terms and eliminating parentheses: Replace by
(x + 2) + 6 > 2x + 5(x + 1 ) x + 8 > 7x + 5
2. Add or subtract the same expression on both sides of the inequality:
Replace by
3x - 5 < 4 (3x - 5 ) + 5 < 4
+
5
3. Multiply or divide both sides of the inequality by the same positive
expression: Replace
4x > 1 6 by
4x 16 >4 4
-
Procedures That Reverse the Sense or Direction of the Inequality Symbol
1. Interchange the two sides of the inequality: Replace
3 < x by x > 3
2. Multiply or divide both sides of the inequality by the same
expression: Replace
6 -2x -2x > 6 by -- < -2 -2
negative
SECTION 1.S
S o l v i n g Inequalities
129
As the examples that follow illustrate, we solve inequalities using many of the same steps that we would use to solve equations. In writing the solution of an inequality, we may use either set notation or interval notation, whichever is more convenient. Solving an I nequality
E XA M P L E 7
Solve the inequality: 3 - 2x < 5 Graph the solution set. Solution
3 - 2x < 5 3 - 2x - 3 < 5 - 3 -2x < 2 2 -2x -- > -2 -2
Simplify. Divide both sides by - 2. (The sense of the inequal ity symbol is reversed.)
x > -1 Figure 4
-3
-2
(
-1
Simplify
The solution set is { x i x > - I } or, using interval notation, all numbers in the interval ( - 1 , 00 ) . See Figure 4 for the graph.
[ )I ,
2
a
Subtract 3 from both sides.
•
Solving an I neq uality
EXAM P L E 8
Solve the inequality: 4x + 7 2: 2x - 3 Graph the solution set. 4x + 7 2: 2x - 3
Solution
4x + 7 - 7 2: 2x - 3 - 7
Subtract 7 from both sides.
4x 2: 2x - 10
Sim pl ify.
4x - 2x 2: 2x - 10 - 2x
Su btract 2x from both sides.
2x 2: -10
Si m plify.
-10 2x 2 2
- > -X
Figure 5
-6
[
-5
-4
-3
-2
!I
-5
The solution set is { x i x 2: - 5 } or, using interval notation, all numbers in the in terval [ - 5 , 00 ) . See Figure 5 for the graph.
-1
•
�= ::::EC-- '-
4
EXA M P L E 9
2:
Divide both sides by 2. (The direction of the inequality sym bol is unchanged.) Simplify.
Now Work
PROBLEM
53
So lve Com b i ned I n eq u a l ities
Solving Combined I nequalities Solve the inequality: -5 < 3x - 2 < 1 Graph the solution set.
Solution
Recall that the inequality -5 < 3x - 2 < 1 is equivalent to the two inequalities -5 < 3x - 2
and
3x - 2 < 1
130
CHAPTER 1
Equations and I nequa lities
We will solve each of these inequalities separately. -5 < 3x - 2 -5 + 2 < 3x - 2 + 2 -3 < 3x 3x -3 -<3 3 -1 < x
3x - 2 < 1 3x - 2 + 2 < 1 + 2 3x < 3
Add 2 to both sides. Simplify. Divide both sides by 3. Simpl ify.
3x 3 - < 3 3 x < 1
The solution set of the original pair of inequalities consists of all x for which -1 < x
Figure 6
-3
-2
(
-1
o
)
and
x < 1
This may be written more compactly as { x 1 - 1 < x < I } . In interval notation, the solution is ( - 1 , 1 ) . See Figure 6 for the graph.
2
•
We observe in the preceding process that the two inequalities we solved required exactly the same steps. A shortcut to solving the original inequality alge braically is to deal with the two inequalities at the same time, as follows: -5 -5 + 2 -3 -3 3 -1
< 3x - 2 < 1 < 3x - 2 + 2 < 1 + 2 < 3x < 3 3x 3 < <3 3 < < 1 x
Add 2 to each part. Sim plify. Divide each part by 3. Sim plify.
We use this shortcut in the next example.
EXAM P L E 1 0
Solving Combined Inequalities Solve the inequality:
-
3 - 5x < - 9 2
1 < -
Graph the solution set. Solution
Figure 7
-4
[
-3
-2
-1
o
-1
�
2( - 1 )
�
-2
�
-2 - 3
2
3 - 5x 2 5X
�
9
�
2(9)
M u ltiply each part by 2 to remove the denominator.
3 - 5x
�
18
�
3 - 5x - 3
�
18 - 3
-5
�
-5x
�
15
Simplify. Su btract 3 from each part to isolate the term containing x. Simplify.
-
-5 -5
2:
2: -
15 -5
Divide each part by -5 (reverse the sense of each i nequal ity symbol) .
1
2:
x
2:
-3
Simplify.
-3
�
x
�
1
Reverse the order so that the num bers get larger as you read from left to right.
2
e� ) -5x --5
The solution set is {x l -3 illustrates the graph. "'" I
}�
�
x
�
I } , that is, all x in the interval [ -3, 1 ]' Figure 7
Now Work P R O B L E M 7 3
•
SECTION 1.5
EXA M P L E 1 1
13 1
Using the Reciprocal P roperty to Solve an I nequality Solve the inequality: Graph the solution set.
Solution
So l ving Inequalities
(4x - 1 r1 > 0
(4x - 1 )-1 = 4x 1- 1 -a1 > 0 a > 0, (4x - l rl > 0 1 4x 1 > 0 4x - 1 > 0 4x > 1 x > -41 {xix > l}, x 4 Since
and since the Reciprocal Property states that when
then
we have
---
Reciproca l Property
Figure
o
8
(
The solution set is illustrates the graph.
1
4
that is, all
"'!!t:==- Now Work
EXAM P L E 1 2
(�, 00).
Figure 8 •
83
C reating Equivalent I nequalities If - 1
Solution
PRO B L E M
in the interval
x
< x < 4, a find
and b so that
a < 2x + 1 < b.
The idea here is to change the middle part of the combined inequality from 2 + 1, using properties of inequalities.
Now we see that pm--�
< x <4 < 2x < 8 < 2x + 1 < 9 a = - 1 b = 9. -1 -2 -1
Now Work
x
M u ltiply each part by 2. Add 1 to eac h part.
and
PRO B L E M
to
•
91
Appl ication
Let's look at an applied problem involving inequalities. EXA M P L E 1 3
Solution
P hysics: Ohm's Law
In electricity, Ohm's law states that E = I R, where E is the voltage (in volts ) , I is the current ( in amperes ) , and R is the resistance (in ohms ) . An air-conditioning unit is rated at a resistance of 10 ohms. If the voltage varies from 110 to 120 volts, in clusive, what corresponding range of current will the air conditioner draw? The voltage lies between 110 and 120, inclusive, so
1 10 1 10 110 110 10 11
� E � 120 � I R � 120 � 1 ( 10) � 120 � �
1 ( 10)
----w-
1
�
120
:tO
Ohm's law, E R
=
=
IR
10
Divide each part by 10,
12 Sim plify, The air conditioner will draw between 1 1 and 12 amperes of current, inclusive. �
•
132
CHAPTER 1
Equations and I n eq ua lities
1.5 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Graph the inequality: x
:2::
2. True or False
-2. (pp. 1 7- 1 9 )
-5 > - 3 (pp. 1 7-1 9)
Concepts and Vocabulary
3. If each side of an inequality is multiplied by a(n)
In Problems 6-9, assume that a < b and c < o. 6. Tl'ue 01' False a + c < b + c
number, then the sense of the inequality sym bol is reversed.
____
, denoted [ a, b], consists of all
4. A(n)
real numbers x for which a ::; x ::; b.
5. The
state that the sense, or direction, of an inequality remains the same if each side is multiplied by a positive number, while the direction is reversed if each side is multiplied by a negative number.
7. Tl'Ue or False 8. Tl'Ue 01' False
a - c < b - c
9. True or False
a b - < -
ac > bc C
C
10. True or False The square of any real number is always nonnegative.
Skill Building In Problems II-16, express the graph shown in blue using interval notation. A lso express each as an inequality involving x.
'- 11.
14.
]
[
-1
0
2
12.
3
� � � I____�__� ] _�__L-� 2 -1 -2 0
(
-2
-1
-1
o
)
0
13.
2
[--�---.� I --� )�� 1 5. --�---r 3
2
16.
-1
0
-1
o
€
[
I
2
3
2
3
••
[n Problems 1 7-22, an inequality is given. Write the inequality obtained by: (a) A dding 3 to each side of the given inequality. (b) Subtracting 5 from each side of the given inequality. (c) Multiplying each side of the given inequality by 3. (d) Multiplying each side of the given inequality by -2.
17. "
18.
3 < 5
2 >
19.
1
4 > -3
20.
- 3 > -5
21.
2x + 1 < 2
22. 1 - 2x > 5
In Problems 23-30, write each inequality using interval notation, and illustrate each inequality using the real number line.
23. 0 27.
::; x ::; 4
24. - 1 28.
x :2:: 4
< x < 5
x ::; 5
25. 4 29.
::; x < 6
x < -4
26.
-2 < x <
30.
x > 1
0
In Problems 31-38, write each interval as an inequality involving x, and illustrate each inequality using the real number line.
' 31.
35.
[2, 5 ]
32.
( 1 , 2)
33. (-3 , - 2 )
34. [0, 1 )
[ 4, 00 )
36.
( - 00 , 2 ]
37.
38.
( - 00 , -3 )
In Problems 39-52, fill in the blank with the correct inequality symbol.
' 39.
40. If x
< -4, then x + 4 ___
O.
42. If x
> 6, then x - 6
-4, then 3 x
- 12.
44. If x
::; 3 , then 2x
If x > 6, then -2x
- 12.
46. If x
> -2, then -4x
-20.
48. If x
::;
41. If x
> -4, then x + 4
43. If x
:2::
' 45.
O.
If x < 5, then x - 5
47. If x
:2:: 5, then -4x
49.
I f 2x >
51.
It
.
1
- "2 x
6, then x ::; 3 , then x
3.
_
50. If 3 x -6.
( -8, 00 )
1
52. If - "4 x
o.
6. 8.
-4, then - 3 x
::; 1 2 , then x > 1 , then x
O.
12.
4. -4.
SECTION 1.5
Solvi ng Inequalities
133
In Problems 53-88, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set.
54. x - 6 < 1
55. 1 - 2x :::; 3
56. 2 - 3x :::; 5
57. 3 x - 7 > 2
58. 2x + 5 > 1
59. 3 x - 1 2: 3 + x
60. 2x - 2
62. -3( 1
63. 4 - 3 ( 1 - x) :::; 3
53. x
+
1< 5
-
x) < 12
1
2:
3 +x
61. - 2 ( x + 3 ) < 8 64. 8 - 4(2 - x) :::; -2x
1
65. "2 (x - 4) > x + 8
66. 3 x
68. ::
69. 0 :::; 2x - 6 :::; 4
70. 4 :::; 2x + 2 :::; 10
72. -3 :::; 3 - 2x :::; 9
73. -3 <
3
2:
2 + :: 6
71. -5 :::; 4 - 3x :::; 2 3x + 2 2
+
67. ::
4 > 3 (x - 2)
2
1
77. (x + 2 ) (x - 3 ) > ( x - l ) ( x + 1)
78. (x - l ) ( x + 1) > (x - 3 ) ( x + 4)
-
5 ) :::; (3x - 1) 2
2 3
2
3
>
2x - 1 < 0 4
--
79. x(4x + 3 ) :::; ( 2x + 1)2 +
1
3 4
82. 3 < -- :::; 3
0
85. 0 < - < x 5
1
< -
x
2
2
2
1
87. 0 < (2x - 4r l < "2
86. 0 < - < x
1
x
--
84. (2x - 1)- 1
83. (4x + 2rl < 0 4
+
1
81. - :::;
4
76. 0 < 1 - 3 x < 1
75. 1 < 1
80. x(9x
1 - ::
1
"2 x < 4
74. 0 < -- < 4
-
2:
3
1 3
88. 0 < (3x + 6r 1 < -
Applications and Extensions
expect to live at least 49 .66 more years and an average 30-year-old female in 2005 could expect to live at least 53.58 more years. (a) To what age can an average 30-year-old male expect to live? Express your answer as an inequality. (b) To what age can an average 30-year-old female expect to live? Express your answer as an inequality. (c) Who can expect to live longer, a male or a female? By how many years?
In Problems 89-98, find a and b.
89. If -1 < x < 1, then a < x + 4 < b. 90. If -3 < x < 2, then a < x - 6 < b. 91. If 2 < x < 3, then a < -4x < b. 1
92. If -4 < x < 0, then a < "2 x < b. 93. If 0 < x < 4, then a < 2x + 3 < b. 94. If -3 < x < 3, then a < 1 - 2x < b.
Source: Actuarial Study No. 120, August 2005
1 95. If -3 < x < 0, then a < -- < b.
x+4
96. If 2 < x < 4, then a <
x
1
_
6
< b.
97. If 6 < 3x < 12 , then a < � < b. 98. If 0 < 2x < 6, then a < .-? < b. 99. What is the domain of the variable in the expression � ? 100. What is the domain of the variable in the expression Vs+2x ? 101. A young adult may be defined as someone older than 21, but less than 30 years of age. Express this statement using inequalities. 102. Middle-aged may be defined as being 40 or more and less than 60. Express this statement using inequalities. 103. Life Expectancy The Social Security Administration determined that an average 30-year-old male in 2005 could
104.
General Chemistry For a certain ideal gas, the volume V (in cubic centimeters) equals 20 times the temperature T (in degrees Celsius). If the temperature varies from 80° to 120° C inclusive, what is the corresponding range of the volume of the gas? 105. Real Estate A real estate agent agrees to sell an apartment complex according to the following commission schedule: $45,000 plus 25 % of the selling price in excess of $900,000. As suming that the complex will sell at some price between
134
CHAPTER 1
Equations and Ineq ualities
$900,000 and $ 1 ,100,000 inclusive, over what range does the agent's commission vary? How does the commission vary as a percent of selling price?
106.
A used car salesperson is paid a com mission of $25 plus 40% of the selling price in excess of owner's cost. The owner claims that used cars typically sell for at least owner's cost plus $200 and at most owner's cost plus $3000. For each sale made, over what range can the sales person expect the commission to vary?
107.
The percentage method of with holding for federal income tax (2006) states that a single per son whose weekly wages, after subtracting withholding allowances, are over $620, but not over $1409, shall have $78.30 plus 25 % of the excess over $620 withheld. Over what range does the amount withheld vary if the weekly wages vary from $700 to $900 inclusive? Source: Employer's Tax Guide. Department of the Treasury, Internal Revenue Service, Publication 2006.
109.
Electricity Rates Commonwealth Edison Company's charge for electricity in May 2006 is 8.275¢ per kilowatt-hour. In ad dition, each monthly bill contains a customer charge of $7.58. If last year's bills ranged from a low of $63 .47 to a high of $214.53, over what range did usage vary (in kilowatt-hours)? Source: Commonwealth Edison Co., Chicago, Illinois, 2006.
112.
114.
"Light" Foods For food products to be labeled "light," the U.S. Food and Drug Administration requires that the altered product must either contain one-third or fewer calories than the regular product or it must contain one-half or less fat than the regular product. If a serving of Miracle Whip® Light con tains 20 calories and 1.5 grams of fat, then what must be true about either the number of calories or the grams of fat in a serving of regular Miracle Whip®? a + b Arithmetic Mean If a < b, show that a < -- < b. The 2 a+b. number -- I S ca II e d t he art'thmefIC mean af a an d b . 2 Refer to Problem 1 1 5 . Show that the arithmetic mean of a and b is equidistant from a and b. Geometric Mean If 0 < a < b, show that a < vaJj < b. The number vaJj is called the geometric mean of a and b. Refer to Problems 1 1 5 and 1 17 . Show that the geometric mean of a and b is less than the arithmetic mean of a and b. Harmonic Mean For 0 < a < b, let h be defined by
Federal Tax Withholding
Exercising
111.
Computing Grades In your Economics 101 class, you have scores of 68, 82, 87, and 89 on the first four of five tests. To get a grade of B, the average of the first five test scores must be greater than or equal to 80 and less than 90. (a) Solve an inequality to find the range of the score that you need on the last test to get a B. (b) What score do you need if the fifth test counts double?
Sales Commission
108.
110.
113.
Sue wants to lose weight. For healthy weight loss, the American College of Sports Medicine (ACSM) recom mends 200 to 300 minutes of exercise per week. For the first six days of the week, Sue exercised 40, 45, 0, 50, 25, and 35 minutes. How long should Sue exercise on the seventh day in order to stay within the ACSM guidelines?
The Village of Oak Lawn charges homeowners $28.84 per quarter-year plus $2.28 per 1000 gallons for water usage in excess of 12,000 gallons. In 2006 one homeowner's quarterly bill ranged from a high of $74.44 to a low of $42.52. Over what range did water usage vary? Source: Village of Oak Lawn, Illinois, April 2006. Water Bills
115. 116. 117. 118. 119.
! h
The markup over dealer's cost of a new car ranges from 1 2 % to 1 8 % . If the sticker price is $18,000, over what range will the dealer's cost vary? Markup of a New Car
IQ Tests A standard intelligence test has an average score of 100. According to statistical theory, of the people who take the test, the 2.5% with the highest scores will have scores of more than 1 .960' above the average, where 0' (sigma, a num ber called the standard deviation) depends on the nature of the test. If 0' 12 for this test and there is (in principle) no upper limit to the score possible on the test, write the inter val of possible test scores of the people in the top 2.5 % . =
120. 121.
=
(
! !+! 2 a
b
)
Show that a < h < b. The number h is called the harmonic mean of a and b. Refer to Problems 1 15, 1 17, and 1 19. Show that the harmonic mean of a and b equals the geometric mean squared divided by the arithmetic mean. Another Reciprocal PrOI)erty Prove that if 0 < a < b, then 1 1 0 < - < -. a b
Discussion and Writing
122. Make up an inequality that has no solution. Make up one that has exactly one solution.
123. The inequality y} + 1 < -5 has no real solution. Explain why. 124. Do you prefer to use inequality notation or interval notation to express the solution to an inequality? Give your reasons. Are there particular circumstances when you prefer one to the other? Cite examples. 'Are You Prepared?' Answers [ !
1.
-4
-2
0
.. •
2. False
125. How would you explain to a fellow student the underly ing reason for the multiplication properties for inequalities (page 1 27), that is, the sense or direction of an inequality remains the same if each side is multiplied by a positive real number, whereas the direction is reversed if each side is mul tiplied by a negative real number.
SECTION 1.6
Equations and Inequa lities Involving Absolute Va lue
135
1.6 Equations and Inequalities Involving Absolute Value PREPARING FOR THIS SECTION •
Before getting started, review the following:
Algebra Essentials (Chapter R, Section R.2, pp. 1 7-26) Now Work
the 'Are You Prepared?' problems on page 1 3 7.
OBJECTIVES
Solve Equations I nvolving Absol ute Va l u e (p. 1 35)
2 Solve I neq ual ities I nvolving Absol ute Va l u e (p. 1 35 )
1
Solve Equations I nvo lving Absol ute Val u e
Recall that, o n the real number line, the absolute value o f a equals the distance from the origin to the point whose coordinate is a. For example, there are two points whose distance from the origin is 5 units, -5 and 5. So the equation Ixl = 5 will have the solution set { -5, 5 } . This leads to the following result: THEOREM
If a is a positive real number and if u is any algebraic expression, then
lui =
a
is equivalent to
u = a or u = - a
(1)
I�
� -----------------�
Solving an E quation I nvolving Absolute Value
EXAM P L E 1
Solve the equations:
( a ) Ix + 41 = 13
(b ) 12x - 31 + 2 = 7
( a ) This follows the form of equation (1), where u = x + 4. There are two
Solution
possibilities.
x + 4 = 13 or x + 4 = - 13 x = 9 or x = - 17 The solution set is { - 17, 9 } . ( b) The equation 12x - 31 + 2 = 7 is not in the form of equation (1). We proceed as follows:
12x - 31 + 2 12x - 31 2x - 3 = 5 or 2x - 3 2x 2x = 8 or x = 4 or x
= = = = =
7 Subtract 2 from each side. 5 -5 Apply (1). -2 -1
The solution set is { - 1, 4 } . "-"J!l:= : =-- Now Work 2
S olve I n eq u a l ities I nvolving Absol ute Val ue
Solve the inequality: Solution
Less than 4 units I�.-- from origin 0 � -5
(
- 4 -3 - 2 - 1
o I
0
1
2
3
PROBLEM 9
Solving an I nequality I nvolving Absolute Value
EXAM P L E 2
Figure 9
•
I
) 4
•
Ixl < 4
We are looking for all points whose coordinate x is a distance less than 4 units from the origin. See Figure 9 for an illustration. B ecause any number x between -4 and 4 satisfies the condition Ixl < 4, the solution set consists of all numbers x for which -4 < x < 4, that is, all x in the interval ( -4, 4 ) .
•
136
CHAPTER 1
Equations and Inequalities
We are led to the following results: THEOREM
Figure 1 0
[
lui
-a
:5 a, a > t
a
If a is a positive number and if u is an algebraic expression, then
lui < a is equivalent to - a < u < a
(2)
lui :5 a is equivalent to -a :5 u :5 a
(3)
In other words, lui < a is equivalent to -a < u and u < a.
a
] .
..J
See Figure 10 for an illustration of statement (3).
a
Solving an I nequality I nvolving Absolute Value
EXAM P L E 3
Solve the inequality: 12x + 41 :5 3 Graph the solution set. This follows the form of statement (3); the expression u 2x + 4 is inside the a bsolute value bars.
12x + 4 1 :5 3
Solution
=
Apply statement (3).
-3 :5 2x + 4 :5 3 -3 - 4 :5 2x + 4 - 4 :5 3 - 4 :5 - 1 2x :5
-7 -7 7
-
2
2x
:5
Figure 1 1
-5
I
EI 7 -2 2
I3 I -1 a
I
2
2
4
•
-1
2
TlIe soIutlOn · set
IS .
Divide each part by 2.
1
X
2
Sim plify.
:5 2
-
- - :5
Subtract 4 from each part.
7
I
Sim plify.
:5 - 2
{ X - 2 :5
X
1
:5
-
2
}
.
.
.
[ 7 1]
, that IS, all x m the mterval - 2' - 2 . See
Figure 1 1 for the graph of the solution set.
•
Solving an I nequality I nvolving Absolute Value
EXA M P L E 4
Solve the inequality: 1 1 - 4xl < 5 Graph the solution set.
11 - 4xl < 5
Solution
This expression follows the form of statement (2); the expression u = 1 4x is inside the a bsolute value bars. Apply statement (2). -
-5 -5 - 1 -6 -6 -4
-
< < < >
3
Figure 1 2
-5 -4
(
-3 -2 -1
a
I
)
I
H 2 2
3
I
4
.
1
1 - 4x - 4x -4x -4x -4
1
<5 <5 <4 4 > -4
-
1
Subtract 1 from each part. Sim pl ify.
Divide each part by 4 which reverses the sense of the i nequa lity symbols. -
-
-> 2
X
> -1
-1 <
X
<2
Sim plify.
3
The solution set is { x l - 1 <
X
Rearrange the ordering.
<
%},
that is, all x in the interval
Figure 12 for the graph of the solution set. "'"
i "-
Now Work
PRO
B L
,
EM 39
•
SECTION 1.6
E XA M P L E 5
Solving a n I nequality I nvolving Absolute Value
Solution F i g u re 1 3
I. ! ) -5 -4 - 3 - 2 - 1
0
2
( 3
,- . 4
Solve the inequality: Ixl > 3 Graph the solution set. We are looking for all points whose coordinate x is a distance greater than 3 units from the origin. Figure 13 illustrates the situation. We conclude that any number x less than -3 or greater than 3 satisfies the condition Ixl > 3. Consequently, the solu tion set consists of all numbers x for which x < -3 or x > 3, that is, all x in ( - 00 , -3 ) U (3, ) • (0
THEOREM
. '"
If a is a positive number and u is an algebraic expression, then
lui > a is equivalent to
Figure 1 4
IuI
-a
< -a or > a lui 2:: a is equivalent to u :::; -a or u 2:: a os a, a > !
0
a
EXA M P L E 6
(4)
(5 )
�
See Figure 14 for an illustration of statement (5). Solve the inequality: 12x - 51 > 3 Graph the solution set. 12x - 5 1 > 3 This fol lows the form of statement (4); the expression u
2x
==
2x
-
5 is inside the a bsol ute val ue bars.
2x - 5 < -3 or 2x - 5 > 3 - 5 + 5 < -3 + 5 or 2x - 5 + 5 > 3 + 5 2x < 2 or 2x > 8 2x 2 < -2 2
or
2x
8
> -2 2
Apply statement (4). Add 5 to each part. Simplify. Divide each part by 2.
Simplify. x>4 <1 or The solution set is {xix < 1 or x > 4 } , that is, all x in ( - 00 , 1 ) U (4, Figure 15 for the graph of the solution set.
x
Figure 1 5 •!
u
Solving an I nequality Involving Absolute Value
Solution
-2 - 1
u
�------�
[ ..
o
137
Equations and I nequalities Involving Absolute Value
)
0
1
2
3
4
5
6
7
(0
).
See •
WA RNING A common error to be avoided is to attempt to write the solution
the com bined inequal ity 1 and x > 4. C!ll!:= : ::::f.i> -
Now Work
" Recall that the symbol
>x>
x < 1 or x > 4 as 4, which is incorrect, since there are no n um bers x for which 1 > x •
PROBLEM 43
U stands for the union o f two sets. Refer t o p. 2 i f necessary.
1.6 Assess Your Understanding 'Are You Prepared?' A nswers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red.
1.
1 -2 1
==
2. True or False Ixl � 0 for any real number x. (p. 19)
(p. 19)
Concepts and Vocabulary
3. The solution set of the equation Ixl 5 is { 4. The solution set of the inequality Ixl < 5 is { xl }. ==
____
}.
5. True or False The equation Ixl == -2 has no solution. has the set of real 6. True or False The inequality Ixl � numbers as solution set.
-2
138
CHAPTER 1
Equations and Inequalities
Skill Building
,,
In Problems 7-34, solve each equation.
11. 1 1 - 4t l
+
S=
13
15. 1 -21x = 4 19.
12. 1 1 - 2z1 16. 1 3 1 x
1 � �I = 2 +
23. 4 - 12xl = 3 27. Ix2 - 2xl = 3 31.
" 9. 12x
8. 13xl = 1 2
7. 12xl = 6
1 32xX -- 23 1 = 2
=
6
+
9
2
I�2 - �I3 1 24. 5 - I� X I = 3 20•
32.
= 12
+ xl
1 23x 41 1 = x
+
--
=
=
2
=
11 1 9
21. l u - 21 = - -
22. 12 - vi = - 1
25. I x2 - 91 = 0
26. I x2 - 1 6 1
29. I x2
+
x - II = 1
30. Ix2
2
+
3x l
33. I x
1
+
3 18. 4 1 x l
9
=
1 2
=
28. Ix2
14. I - x l
13. 1 -2xl = l s i 17. 3" l x l
9
=
10. 13 x - 1 1
31 = 5
+
=
I x2 - 2x l
+
=
0
3x - 2 1 = 2
34. Ix2 - 2x l = I x
2
+
6x l
In Problems 35-62, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set.
35. 12xl < S ' 39. I x - 21
37. 1 3 x l
36. 13xl < 15
+
2 <3
'- 43. 12x - 31 � 2
40. I x
+
44. 13x
41
+
-
+
3 <5
41 � 2
38. 12xl
12
>
:5
41. 13t - 21
42. 12u
4
> +
6 51
:5
7
45. 11 - 4xl - 7 < -2
46. 1 1 - 2xl - 4 < - 1
49. 1 -4xl
so.
:5
47. 1 1 - 2x l > 3
48. 12
51. 1 -2x l
52. I -x - 21 � 1
53. -12x - 1 1 � -3
54. - 1 1 - 2xl � -3
56. 13xl � 0
57. 1 5 x l
58. 1 6x l < -2
>
1 -31
55. 12xl < -1
3xl
60. 3 - I x
+
>
1
1
1 1 < "2
61. 5
+
�
+ Ix
I -51
:5 1
-1 - 11
>
1 "2
62.
I - xl - 141
2
1 2X 2- 3 -31 1 ---
+
> 1
Applications and Extensions
63.
64.
Body Temperature "Normal" human body temperature is 9S.6°F. If a temperature x that differs from normal by at least 1 .5° is considered unhealthy, write the condition for an un healthy temperature x as an inequality involving an absolute value, and solve for x.
In the United States, normal household voltage is 110 volts. However, it is not uncommon for actual voltage to differ from normal voltage by at most 5 volts. Ex press this situation as an inequality involving an absolute value. Use x as the actual voltage and solve for x. 65. Reading Books A Gallup poll conducted May 20-22, 2005, found that Americans read an average of 13.4 books per year. Gallup is 99% confident that the result from this poll is off by fewer than 1.35 books from the actual average x. Express this situation as an inequality involving absolute value, and solve Household Voltage
the inequality for x to determine the interval in which the ac tual average is likely to fall. Note: In statistics, this interval is called a 99% confidence interval.
66.
According to data from the Hill Aerospace Museum (Hill Air Force Base, Utah), the speed of sound varies depending on altitude, barometric pressure, and tem perature. For example, at 20,000 feet, 13.75 inches of mercury, and -12.3°F, the speed of sound is about 707 miles per hour, but the speed can vary from this result by as much as 55 miles per hour as conditions change. (a) Express this situation as an inequality involving an ab solute value. (b) Using x for the speed of sound, solve for x to find an in terval for the speed of sound. 67. Express the fact that x differs from 3 by less than � as an inequality involving an absolute value. Solve for x. 2 68. Express the fact that x differs from -4 by less than 1 as an inequality involving an absolute value. Solve for x. 69. Express the fact that x differs from -3 by more than 2 as an inequality involving an absolute value. Solve for x. 70. Express the fact that x differs from 2 by more than 3 as an in equality involving an absolute value. Solve for x. SI)eed of Sound
SECTION 1.7
Problem Solving: Interest, Mixture, Un iform Motion, Constant Rate Job Applications
In Problems 71-76, find a and b. 71. If Ix 1 1 < 3, then a < x + 4 < b.
80.
-
72.
If Ix + 21 <
73.
If Ix
74.
If Ix - 31 ::; 1, then a ::; 3x
75.
If Ix
76.
77.
78. 79.
If Ix
+
-
5,
then a < x
-
2 < b.
41 ::; 2, then a ::; 2x - 3 ::; b.
21 ::;
7,
then a ::;
x
+
1
1 10
--
::;
81.
b.
::; b.
1 + 11 ::; 3, then a ::; -- ::; b. x+5
82.
and va < Vb, then a < b. [Hint: b - a = (Vb - va)(Vb + ya)] Show that if a>
0,
b>
0,
1 39
Prove that la - bl 2: lal - Ibl. [Hint: Apply the triangle inequality from Problem to lal = I(a - b) + bl·] If a>
79
0,
show that the solution set of the inequality x2 < a consists of all numbers x for which -va < x < va If a> 0, show that the solution set of the inequality x2> a consists of all numbers x for which
x<
-
va or x> va
Show that a ::; lal. Prove the triangle inequality la + bl ::; lal + Ibl. 2 2 [Hint: Expand la + bl = (a + b ) , and use the result of Problem 78.]
In Problems 83-90, use the results found in Problems to solve each inequality. 2 2 84. x < 4 83. x < 1
81
and 82 89.
x2> 4
91.
Solve 13x - 12x
95.
The inequality Ixl> why.
+
111 = 4.
90.
x2> 16
92.
Solve Ix + 13x - 211
=
2.
Discussion and Writing 93.
94.
The equation Ixl = -2 has no solution. Explain why. The inequality /x/ > -0.5 has all real numbers as the solution. Explain why.
0 has as
solution set {xix
=F
O}. Explain
'Are You Prepared?' Answers 1.
2
1.7
2.
True
Problem Solvi ng: Interest, Mixture, Un iform Motion, Con stant Rate Job Applications OBJECTIVES
1 Translate Verba l Descriptions i nto Mathematical Expressions (p. 140) 2
Solve I nterest Problems (p. 141)
3
Solve M ixture Problems (p. 142)
5
Solve Constant Rate Job Problems (p. 145)
4
Solve U n iform Motion Problems (p. 143)
Applied (word) problems do not come in the form "Solve the equation . . . . " Instead, they supply information using words, a verbal description of the real prob lem. So, to solve applied problems, we must be able to translate the verbal descrip tion into the language of mathematics. We do this by using variables to represent unknown quantities and then finding relationships (such as equations) that involve these variables. The process of doing all this is called mathematical modeling. Any solution to the mathematical problem must be checked against the math ematical problem, the verbal description, and the real problem. See Figure 16 on page 140 for an illustration of the modeling process.
140
CHAPTER 1
Equations a nd Inequalities
Figure 1 6 Verbal description
�
\ \ \ \
I I \
\, Check , "
1
"
"
\ Check "
..
....
....
.... _-
"
..
.... ..
.. .. _-
-------- ---- --
--- ---
-
..--
Translate Verbal Descriptions into Mathematical Expressions
Let's look at a few examples that will help you to translate certain words into math ematical symbols. E XA M P L E 1
Translating Verbal Descriptions i nto M athematical Expressions
(a) For uniform motion, the velocity of an object equals the distance traveled di vided by the time required. Translation: If v is the velocity, s the distance, and t the time, then v = �.
t
(b) Let x denote a number. The number 5 times as large as x is 5x. The number 3 less than x is x - 3. The number that exceeds x by 4 is x + 4. The number that, when added to x, gives 5 is 5 'I'�
- x.
•
Now Work PROBLEM 7
Always check the units used to measure the variables of an applied problem. In Example l(a), if v is measured in miles per hour, then the distance s must be expressed in miles and the time t must be expressed in hours. It is a good practice to check units to be sure that they are consistent and make sense. The steps to follow for setting up applied problems, given earlier, are repeated next: Steps for Setting up Applied Problems STEP 1: Read the problem carefully, perhaps two or three times. Pay particu lar attention to the question being asked in order to identify what you are looking for. If you can, determine realistic possibilities for the answer. STEP 2: Assign a letter (variable) to represent what you are looking for, and, if necessary, express any remaining unknown quantities in terms of this variable. STEP 3: Make a list of all the known facts, and translate them into mathe matical expressions. These may take the form of an equation or an inequality involving the variable. If possible, draw an appropriately labeled diagram to assist you. Sometimes a table or chart helps. STEP 4: Solve the equation for the variable, and then answer the question. STEP 5: Check the answer with the facts in the problem. If it agrees, congrat ulations! If it does not agree, try again.
2
SECTION 1.7
Problem Solving: I nterest, Mixture, Uniform Motion, Constant Rate Job Applications
141
Solve Interest Problems Interest is money paid for the use of money. The total amount borrowed (whether by an individual from a bank in the form of a loan or by a bank from an individual in the form of a savings account) is called the principal. The rate of interest, expressed as a percent, is the amount charged for the use of the principal for a given period of time, usually on a yearly (that is, on a per annum) basis.
Simple Interest Formula
If a principal of P dollars is borrowed for a period of t years at a per annum interest rate r, expressed as a decimal, the interest I charged is ( 1)
1 = Prt
Interest charged according to formula (1) is called simple interest. When using for mula (1), be sure to express r as a decimal. EXAM P L E 2
F i n an ce: Computing I nterest on a Loan
Suppose that Juanita borrows $500 for 6 months at the simple interest rate of 9% per annum. What is the interest that Juanita will be charged on the loan? How much does Juanita owe after 6 months? Sol uti o n
The rate of interest is given per annum, so the actual time that the money is borrowed must be expressed in years. The interest charged would be the principal, $500, times the rate of interest (9% = 0.09) times the time in years, Interest charged = I = Prt = ( 500) (0.09)
( �)
�:
= $22.50
After 6 months, Juanita will owe what she borrowed plus the interest: $500
E XA M P L E 3
+
$22.50 = $522.50
•
F i n ancial Planning
Candy has $70,000 to invest and requires an overall rate of return of 9%. She can invest in a safe, government-insured certificate of deposit, but it only pays 8%. To obtain 9%, she agrees to invest some of her money in noninsured corporate bonds paying 12%. How much should be placed in each investment to achieve her goal? S o lution
STEP 1:
STEP 2:
STEP 3:
The question is asking for two dollar amounts: the principal to invest in the corporate bonds and the principal to invest in the certificate of deposit. We let x represent the amount (in dollars) to be invested in the bonds. Then 70,000 x is the amount that will be invested in the certificate. (Do you see why?) We set up a table:
Bonds
-
Principal ($)
Rate
x
1 2%
Time (yr) =
Interest ($)
0.1 2
0.12x
Certificate
70,000 - x
8%
=
0.08
0.08(70,000
Total
70,000
9%
=
0.09
0.09(70,000)
�
�
x) 6300
142
CHAPTER 1
Equations and Inequalities
STEP 4:
STEP
5:
Since the total interest from the investments is equal to 0.09(70,000) = 6300, we must have the equation 0.12x + 0.08(70,000 - x) = 6300 (Note that the units are consistent: the unit is dollars on each side. ) 0.12x + 5600 - 0.08x = 6300 0.04x = 700 x = 17,500 Candy should place $17,500 in the bonds and $70,000 - $ 1 7,500 = $52,500 in the certificate. The interest on the bonds after 1 year is 0.12 ( $ 17,500) = $ 2100; the inter est on the certificate after 1 year is 0.08( $ 52,500) = $4200. The total annual interest is $6300, the required amount.
•
=
3
Now Work P R O B L E M 1 7
Solve Mixture Problems
Oil refineries sometimes produce gasoline that is a blend of two or more types of fuel; bakeries occasionally blend two or more types of flour for their bread. These problems are referred to as mixture problems because they combine two or more quantities to form a mixture. E XA M P L E 4
Ble n d i n g Coffees
The manager of a Starbucks store decides to experiment with a new blend of cof fee. She will mix some B grade Colombian coffee that sells for $5 per pound with some A grade Arabica coffee that sells for $10 per pound to get 100 pounds of the new blend. The selling price of the new blend is to be $7 per pound, and there is to be no difference in revenue from selling the new blend versus selling the other types. How many pounds of the B grade Colombian and A grade Arabica coffees are required? Solution
Let x represent the number of pounds of the B grade Colombian coffee. Then 1 00 - x equals the number of pounds of the A grade Arabica coffee. See Figure 17. $5 per pound
Figure 17
$10 per pound
$7 per pound
A Grade Arabica 100 x pounds
Blend
+
B Grade Colombian x pounds
{
+
100 pounds
-
Since there is to be no difference in revenue between selling the A and B grades separately versus the blend, we have
Price per pound of B grade $5
}{
# poundS B grade x
} { +
price per pound of A grade
+
$ 10
}{
# poundS A grade
} {
-( 100 - x)
=
price per pound of blend $7
}{
# poundS blend 100
}
SECTION 1.7
Problem Solving: Interest, Mixtu re, Uniform Motion, Constant Rate Job Applications
We have the equation
5x + 10( 100 - x) 5x + 1000 - lOx -5x x
= = = =
143
700 700 -300 60
The manager should blend 60 pounds of B grade Colombian coffee with 100 - 60 = 40 pounds of A grade Arabica coffee to get the desired blend. Check: The 60 pounds of B grade coffee would sell for ( $5) ( 60) = $300, and the 40 pounds of A grade coffee would sell for ( $ 10) ( 40) = $400; the total rev enue, $700, equals the revenue obtained from selling the blend, as desired .
•
�=�.4
Now Work PROB L E M 2 1
Solve Uniform Motion Problems
Objects that move at a constant velocity (or speed) are said to be in uniform motion: When the average velocity of an object is known, it can be interpreted as its constant velocity. For example, a bicyclist traveling at an average velocity of 25 miles per hour is in uniform motion. Uniform Motion Formula
If an object moves at an average velocity v, the distance s covered in time t is given by the formula (2)
s = vt
That is, Distance = Velocity· Time. E XA M P L E 5
P hysics: U n iform M otion
Tanya, who is a long-distance runner, runs at an average velocity of 8 miles per hour (mi/hr). Two hours after Tanya leaves your house, you leave in your Honda and follow the same route. If your average velocity is 40 milhr, how long will it be be fore you catch up to Tanya? How far will each of you be from your home? Solutio n
Figure 1 8
Refer to Figure 18. We use t to represent the time (in hours) that it takes the Honda to catch up to Tanya. When this occurs, the total time elapsed for Tanya is t + 2 hours.
�t
2 hr
t= 0
-------1�
Time t
ft
-+------Time t-------l�
t= 0
Set up the following table: Velocity mi/hr
Time hr
Distance mi
Tanya
8
t + 2
8(t + 2)
Honda
40
40t
':' Technically, velocity has both magnitude (speed) and direction. Since all velocities in this section are in the positive direction, we make no distinction between speed and velocity.
144
CHAPTER 1
Equations a n d Ineq ualities
Since the distance traveled is the same, we are led to the following equation:
8(t 2) 40t 8t 40t 32t t
+ = + 16 = = 16 1 = "2 hour
20
It will take the Honda miles.
Check: In
2.5
1. hour to catch up to Tanya. Each of you will have gone
2
Honda travels a distance of
E XA M P L E 6
(2.5)(8) 20 (�}40) 20 =
hours, Tanya travels a distance of
=
1 miles. In "2 hour, the
miles.
•
24
P hysi cs: U niform M otion
A motorboat heads upstream a distance of miles on a river whose current is running at 3 miles per hour (mi/hr). The trip up and back takes 6 hours. Assum ing that the motorboat maintained a constant speed relative to the water, what was its speed? Solution
.- - 24 miles --+.1 1.;... ... -
See Figure 19. We use v to represent the constant speed of the motorboat relative to the water. Then the true speed going upstream is v - 3 mi/hr, and the true speed going downstream is v + 3 mi/hr. Since Distance = Velocity X Time, then Distance . . We set up a table. Time = VelOCIty
Figure 1 9
Velocity mi/hr
Distance mi
-
24
v
Upstream
v
Downstream
3
+ 3
Time hr
Distance Velocity
=
24 v
-
3
24
24
v
+ 3
Since the total time up and back is 6 hours, we have
24
24
-- +-- = 6
v - 3 v+3 3) 24(v + 3) + 24(v ---'-----=6 (v - 3 ) (v+3 ) � =6 v2 - 9 = 6(v2 - 9) 2 6v = v2 9 = (v - 9) (v+ 1 ) = v = 9 or v = -1
48v 48v - 54 0 8v 0 0
Add the quotients on the left.
Simplify. Multipl y both sides by
v2
-
9.
P l a ce in standard for m. Divide by 6. Factor. Ap pl y the Zero-Product Property and solve.
We discard the solution v = - 1 mi/hr, so the speed of the motorboat relative to the water is 9 mi/hr. a:.==:::> -
•
Now Work PRO B L E M 2 7
SECTION 1.7
5
145
Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
Solve Constant Rate Job Problems
This section involves jobs that are performed at a constant rate. Our assumption is 1 that, if a job can be done in t units of time, then - of the job is done in 1 unit of time. t Let's look at an example. E XA M P L E 7
Working Together to Do a Job
At 10 AM Danny is asked by his father to weed the garden. From past experience, Danny knows that this will take him 4 hours, working alone. His older brother, Mike, when it is his turn to do this job, requires 6 hours. Since Mike wants to go golfing with Danny and has a reservation for 1 PM, he agrees to belp Danny. Assuming no gain or loss of efficiency, when will they finish if they work together? Can they make the golf date? Soluti o n
Table 2 Hours to DoJob
Danny Mike
4 6
Part of Job Done in 1 Hou r 4
-1
We set up Table 2. In 1 hour, Danny does
� of the job, and in 1 hour, Mike does �
of the j ob. Let t be the time (in hours) that it takes them to do the job together. In ! 1 hour, then, of the job is completed. We reason as follows: t
(
Part done by Danny in 1 hour
) ( +
part ? one by Mike III 1 hour
) ( =
part ? one tOgether III 1 hour
)
From Table 2, 1 1 1 -+-=4 6 t 3 + 2 1 = 12 12 t 1 S 12 St = 12 12 t =S
6
Together
-
-
-
Working together, the job can be done in
�
2
hours, or 2 hours, 24 minutes. They
should make the golf date, since they will finish at 12 : 24 PM.
•
Now Work P R O B L E M 3 3
1.7 Assess Your Understanding
Concepts and Vocabulary 1.
The process of using variables to represent unknown quan tities and then finding relationships that involve these variables is referred to as The money paid for the use of money is Objects that move at a constant velocity are said to be in
5. True or False v,
s =
____ ___ _
2. 3.
_ _ __
The amount charged for the use of principal for a given period of time is called the rate of interest.
4. True or False
I f an object moves at an average velocity the distance s covered in time t is given by the formula
6.
vt.
Suppose that you want to mix two coffees in order to obtain 100 pounds of the blend. If x represents the number of pounds of coffee A, write an algebraic expression that represents the number of pounds of coffee B.
146
CHAPTER 1
Equations a n d Inequalities
Applications and Extensions In Problems 7-16, translate each sentence into a mathematical equation. Be sure to identify the meaning of all symbols. 60 pounds of almonds with some cashews and sell the mixture 7. Geometry The area of a circle is the product of the number for $7.50 per pound. How many pounds of cashews should be 7T and the square of the radius. mixed with the almonds to ensure no change in the profit? 8. Geometry The circumference of a circle is the product of the number 7T and twice the radius. 24. Business: Mixing Candy A candy store sells boxes of candy containing caramels and cremes. Each box sells for $12.50 9. Geometry The area of a square is the square of the length and holds 30 pieces of candy (all pieces are the same size). If of a side. the caramels cost $0.25 to produce and the cremes cost $0.45 10. Geometry The perimeter of a square is four times the to produce, how many of each should be in a box to make a length of a side. profit of $3? 11. Physics Force equals the product of mass and acceleration. 25. Physics: Uniform Motion A motorboat can maintain a con 12. Physics Pressure is force per unit area. stant speed of 16 miles per hour relative to the water. The 13. Physics Work equals force times distance. boat makes a trip upstream to a certain point in 20 minutes; Kinetic energy is one-half the product of the mass 14. Physics the return trip takes 15 minutes. What is the speed of the cur and the square of the velocity. rent? See the figure. 15. Business The total variable cost of manufacturing x dish washers is $150 per dishwasher times the number of dish washers manufactured. 16. Business The total revenue derived from selling x dish washers is $250 per dishwasher times the number of dish washers sold. 17. Financial Planning Betsy, a recent retiree, requires $6000 per year in extra income. She has $50,000 to invest and can in 26. Physics: Uniform Motion A motorboat heads upstream on vest in B-rated bonds paying 15% per year or in a certificate a river that has a current of 3 miles per hour. The trip up of deposit (CD) paying 7 % per year. How much money stream takes 5 hours, and the return trip takes 2.5 hours. What should be invested in each to realize exactly $6000 in inter is the speed of the motorboat? (Assume that the motorboat est per year? maintains a constant speed relative to the water.) 18. Financial Planning After 2 years, Betsy (see Problem 17) 27. Physics: Uniform Motion A motorboat maintained a con finds that she will now require $7000 per year. Assuming that stant speed of 15 miles per hour relative to the water in going the remaining information is the same, how should the money 10 miles upstream and then returning. The total time for the be reinvested? trip was 1.5 hours. Use this information to find the speed of 19. Banking A bank loaned out $12,000, part of it at the rate of the current. 8% per year and the rest at the rate of 18% per year. If the 28. Physics: Uniform Motion Two cars enter the Florida Turn interest received totaled $1000, how much was loaned at 8%? pike at Commercial Boulevard at 8:00 AM, each heading for 20. Banking Wendy, a loan officer at a bank, has $ 1,000,000 to Wildwood. One car's average speed is 10 miles per hour lend and is required to obtain an average return of 18% per more than the other's. The faster car arrives at Wildwood at year. If she can lend at the rate of 1 9 % or at the rate of 16%, 1 1 1 :00 AM, 2" hour before the other car. What was the average how much can she lend at the 16% rate and still meet her re quirement? speed of each car? How far did each travel? 21. Blending Teas The manager of a store that specializes in 29. Moving Walkways The speed of a moving walkway is typi selling tea decides to experiment with a new blend. She will cally about 2.5 feet per second. Walking on such a moving mix some Earl Grey tea that sells for $5 per pound with some walkway, it takes Karen a total of 40 seconds to travel 50 feet Orange Pekoe tea that sells for $3 per pound to get 100 pounds with the movement of the walkway and then back again of the new blend. The selling price of the new blend is to be against the movement of the walkway. What is Karen's nor $4.50 per pound, and there is to be no difference in revenue mal walking speed? from selling the new blend versus selling the other types. How many pounds of the Earl Grey tea and Orange Pekoe tea are Source: A nswers. com required? 30. Moving Walkways The Gare Montparnasse train station in 22. Business: Blending Coffee A coffee manufacturer wants to Paris has a high-speed version of a moving walkway. If he market a new blend of coffee that sells for $3.90 per pound walks while riding this moving walkway, Jean Claude can by mixing two coffees that sell for $2.75 and $5 per pound, re travel 200 meters in 30 seconds less time than if he stands still spectively. What amounts of each coffee should be blended to on the moving walkway. If Jean Claude walks at a normal obtain the desired mixture? rate of 1.5 meters per second, what is the speed of the Gare Montparnasse walkway? [ Hint: Assume that the total weight of the desired blend is Source: A nswers. com 100 pounds.] 31. Tennis Anyone'! A regulation doubles tennis court has an 23. Business: Mixing Nuts A nut store normally sells cashews area of 2808 square feet. If it is 6 feet longer than twice its for $9.00 per pound and almonds for $3.50 per pound. But at width, determine the dimensions of the court. the end of the month the almonds had not sold well, so, in order to sell 60 pounds of almonds, the manager decided to mix the Source: United States Tennis Association
SECTION 1.7
32.
Problem Solving: I n terest, Mixtu re, Uniform Motion, Constant Rate Job Applications
Laser Printers It takes an H P LaserJet 1300 laser printer 10 minutes longer to complete a 600-page print job by itself than it takes an HP LaserJet 2420 to complete the same job by itself. Together the two printers can complete the job in 12 minutes. How long does it take each printer to complete the print job alone? What is the speed of each printer?
38.
Source: Hewlett-Packard 33.
34.
35.
36.
37.
Working Together on a Job Trent can deliver his newspa pers in 30 minutes. It takes Lois 20 minutes to do the same route. How long would it take them to deliver the newspapers if they work together? Working Together on a Job Patrice, by himself, can paint four rooms in 10 hours. If he hires April to help, they can do the same job together in 6 hours. If he lets April work alone, how long will it take her to paint four rooms? Enclosing a Garden A gardener has 46 feet of fencing to be used to enclose a rectangular garden that has a border 2 feet wide surrounding it. See the figure. (a) If the length of the garden is to be twice its width, what will be the dimensions of the garden? (b) What is the area of the garden? (c) If the length and width of the garden are to be the same, what would be the dimensions of the garden? (d) What would be the area of the square garden?
Construction A pond is enclosed by a wooden deck that is 3 feet wide. The fence surrounding the deck is 100 feet long. (a) If the pond is square, what are its dimensions? (b) If the pond is rectangular and the length of the pond is to be three times its width, what are its dimensions? (c) If the pond is circular, what is its diameter? (d) Which pond has the most area? Football A tight end can run the 100-yard dash in 12 sec onds. A defensive back can do it in 10 seconds. The tight end catches a pass at his own 20-yard line with the defensive back at the 1 5-yard line. (See the figure.) If no other players are nearby, at what yard line will the defensive back catch up to the tight end? [Hint: At time t = 0, the defensive back is 5 yards behind the tight end.]
39.
147
Computing Business Expense Therese, an outside sales person, uses her car for both business and pleasure. Last year, she traveled 30,000 miles, using 900 gallons of gasoline. Her car gets 40 miles per gallon on the highway and 25 in the city. She can deduct all highway travel, but no city travel, on her taxes. How many miles should Therese be allowed as a busi ness expense? Mixing Water and Antifreeze How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60% antifreeze?
40.
Mixing Water and Antifreeze The cooling system of a cer tain foreign-made car has a capacity of 15 liters. If the sys tem is filled with a mixture that is 40% antifreeze,how much of this mixture should be drained and replaced by pure an tifreeze so that the system is filled with a solution that is 60% antifreeze?
41.
Chemistry: Salt Solutions How much water must be evap orated from 32 ounces of a 4% salt solution to make a 6% salt solution?
42.
Chemistry: Salt Solutions How much water must be evap orated from 240 gallons of a 3% salt solution to produce a 5 % salt solution?
43.
Purity of Gold The purity of gold is measured in karats, with pure gold being 24 karats. Other purities of gold are expressed as proportional parts of pure gold. Thus, IS-karat 12 IS gold i s ' o r 75 % pure gold; 12-karat gold i s ' o r 5 0 % pure 24 24 gold; and so on. How much 12-karat gold should be mixed with pure gold to obtain 60 grams of 16-karat gold?
44.
Chemistry: Sugar Molecules A sugar molecule has twice as many atoms of hydrogen as it does oxygen and one more atom of carbon than oxygen. If a sugar molecule has a total of 45 atoms, how many are oxygen? How many are hydrogen?
45.
Running a Race Mike can run the mile in 6 minutes, and Dan can run the mile in 9 minutes. If Mike gives Dan a head start of 1 minute, how far from the start will Mike pass Dan? How long does it take? See the figure.
46.
Range of an Airplane An air rescue plane averages 300 miles per hour in still air. It carries enough fuel for 5 hours of flying time. If, upon takeoff, it encounters a head wind of 30 mi/hr, how far can it fly and return safely? (Assume that the wind remains constant.)
148
CHAPTER 1
Equations and Inequalities
47.Emptying Oil Tankers An oil tanker can be emptied by the main pump in 4 hours. An auxiliary pump can empty the tanker in 9 hours. If the main pump is started at 9 AM, when should the auxiliary pump be started so that the tanker is emptied by noon? 48. Cement Mix A 20-pound bag of Economy brand cement mix contains 25% cement and 75% sand. How much pure cement must be added to produce a cement mix that is 40% cement? 49.Emptying a Tub A bathroom tub will fill in 15 minutes with both faucets open and the stopper in place. With both faucets closed and the stopper removed, the tub will empty in 20 min utes. How long will it take for the tub to fill if both faucets are open and the stopper is removed? 50.Using Two Pumps A 5-horsepower (hp) pump can empty a pool in 5 hours. A smaller, 2-hp pump empties the same pool in 8 hours. The pumps are used together to begin emptying this pool. After two hours, the 2-hp pump breaks down. How long will it take the larger pump to empty the pool? 51. A Biathlon Suppose that you have entered an 87-mile biathlon that consists of a run and a bicycle race. During your run, your average velocity is 6 miles per hour, and during your bicycle race, your average velocity is 25 miles per hour. You finish the race in 5 hours. What is the distance of the run? What is the distance of the bicycle race?
52.Cyclists Two cyclists leave a city at the same time, one going east and the other going west. The westbound cyclist bikes 5 mph faster than the eastbound cyclist. After 6 hours they are 246 miles apart. How fast is each cyclist riding? 53.COmlJaring Olympic Heroes In the 1984 Olympics, C. Lewis of the United States won the gold medal in the 100-meter race with a time of 9.99 seconds. I n the 1 896 Olympics, Thomas Burke, also of the United States, won the gold medal in the 100-meter race in 12.0 seconds. If they ran in the same race repeating their respective times, by how many meters would Lewis beat B urke? 54. Constructing a Coffee Can A 39-ounce can of Hills Bros. ® coffee requires 188.5 square inches of aluminum. If its height is 7 inches, what is its radius? [Hint: The surface area S of a right cylinder is S 2 71"r 2 + 2 m"h , where r is the radius and h is the height.] =
T 1
7 in.
Discussion and Writing 55. Critical Thinking You are the manager of a clothing store and have just purchased 100 dress shirts for $20.00 each. After 1 month of selling the shirts at the regular price, you plan to have a sale giving 40% off the original selling price. However, you still want to make a profit of $4 on each shirt at the sale price. What should you price the shirts at initially to ensure this? If, instead of 40% off at the sale, you give 50% off, by how much is your profit reduced? 56 . Critical Thinking Make up a word problem that requires solving a linear equation as part of its solution. Exchange prob lems with a friend. Write a critique of your friend's problem. 57. Critical Thinking Without solving, explain what is wrong with the following mixture problem: How many liters of 25% ethanol should be added to 20 liters of 48% ethanol to obtain a solution of 58% ethanol? Now go through an algebraic so lution. What happens?
58 . COllll)llting Average Speed I n going from Chicago to Atlanta, a car averages 45 miles per hour, and in going from Atlanta to Miami, it averages 55 miles per hour. If Atlanta is halfway between Chicago and Miami, what is the average speed from Chicago to Miami? Discuss an intuitive solution. Write a paragraph defending your intuitive solution. Then solve the problem algebraically. Is your intuitive solution the same as the algebraic one? If not, find the flaw. 59 . Speed of a Plane On a recent flight from Phoenix to Kansas City, a distance of 919 nautical miles, the plane arrived 20 min utes early. On leaving the aircraft, I asked the captain, "What was our tail wind?" He replied, "I don't know, but our ground speed was 550 knots." How can you determine if enough in formation is provided to find the tail wind? If possible, find the tail wind. (1 knot = 1 nautical mile per hour)
CHAPTER REVIEW Quadratic formula (pp. 102 and US)
Things to Know Ifax2
][
+
bx + c =
b2 - 4ac <
If b2
0, a
-b ± Yb2 - 4ac
0, then x = ------2a 0, there are no real solutions. *-
- 4ac > 0, there are two distinct real solutions.
Discriminant (pp. 103 and 116)
b2 4ac = 0, there is one repeated real solution. If b2 - 4ac < 0, there are no real solutions, but there are two distinct complex solutions that are not real; the complex solutions If
-
are conjugates of each other.
Chapter Review
Interval notation (p. 125) {xla:sx:Sb} [a , bJ a, b {xla:sx < b} ) [
(a, b] (a,b)
{xla < x:Sb} {xla < x < b}
[a, 00 ) (a, 00 ) ( -oo,a] ( -oo,a)
{xix � a} {xix> a} {xix:sa} {xix < a} 00 00 ( - , ) All real numbers
Properties of inequalities
Addition property (p. 126)
If a < b, then a + e < b + e If a> b, then a + e> b + e
Multiplication properties (p. 127)
(a) If a < b and if e> If a < b and if e < 1 If a> 0, then -> ° a 1 If -> 0, then a> ° a
Reciprocal properties (p. 128)
0, then ae < be (b) If a> b and if e> 0, then ae> be If a> b and if e < 0, then ae < be 0, then ae> be 1 If a < 0, then - < ° a 1 If - < 0, then a < ° a
Absolute value
If lui = a, a> 0, then u = -a or u = a (p. 135) a> 0, then -a:su:sa (p. 136) If lui � a, a> 0, then u:s -a or u � a (p. 137) If lui:s a,
Objectives Section
1.1 2
3
1.2 2 3
4
5
1.3 2
1.4 2 3
1.5 2 3
4
1 .6 2
1.7
1
2
3
4 5
--------�
You should be able to ...
Review Exercises
Solve a linear equation (p. 88) Solve equations that lead to linear equations (p. 90) Solve applied problems involving linear equations (p. 92)
1-6,11-12 7, 8 86,105
Solve a quadratic equation by factoring (p. 97) Know how to complete the square (p. 99) Solve a quadratic equation by completing the square (p. 100) Solve a quadratic equation using the quadratic formula (p. 102) Solve applied problems involving quadratic equations (p. 105)
10, 13, 14, 33-36 61-64 9, 10, 13-16, 19, 20, 33-36 9,10, 13-16, 19, 20,33-36 88, 94, 99,100,104, 106
Add, subtract, multiply, and divide complex numbers (p. 1 10) Solve quadratic equations in the complex number system (p. 1 14)
65-74 75-82
Solve radical equations (p. 1 18) Solve equations quadratic i n form (p. 119) Solve equations by factoring (p. 121)
17, 1 8,23-30,37, 38 21 , 22, 3 1 , 32 43--46
Use interval notation (p. 125) Use properties of inequalities (p. 1 26) Solve inequalities (p. 128) Solve combined inequalities (p. 129)
47-60 47-60 47, 48 49-52
Solve equations involving absolute value (p. 135) Solve i nequalities involving absolute value (p. 135)
39--42 53-60
Translate verbal descriptions into mathematical expressions (p. 140) Solve interest problems (p. 141) Solve mixture problems (p. 142) Solve uniform motion problems (p. 143) Solve constant rate job problems (p. 145)
83, 84 85, 86 86, 97, 98, 101 87,89-93,109 95,96, 103, 107
150
CHAPTER 1
Equations and Inequalities
Review Exercises In Problems 1-46, find all the real solution5; if any, of each equation. (Where they appeaJ; a, b, m, and n are positive constants.) x x 3.-2(5- 3x) + 8=4 + 5x 2'4- 2=4 1.2-3=8 4- 2x 1 1 3x x 6.-- + "6 = 2x 5. = 4. (6 - 3x) - 2 ( 1 + x) = 6x 3 4 3 12 3 4x- 5 6 x 9.x ( l - x) = 6 8.---=2 x,,=7.--=- x,,= 1 7 3 - 7x x- 1 5 1 - 3x x+6 1 12.--=-- + 2" 11.1:. - =�- � 10. x(l + x) = 6 3 4 2 3 4 6 -
19.
17.
25. \Y2x + 3 =2
-Vx--=s = 3
21.x4-5x2 + 4=0
33. x2 + m2=2mx + (nxf n,,= 1 , n
27. "Vx+l
29.2XI/2 - 3=0
30.3xl/4- 2=0
+
32. 6x-1 - 5x- I/2 + 1 =0 ,,=
34. b2 x2 + 2ax=x2 + a2 b,,= 1, b ,,= -1 2 1 1 36.-- + --=- x,,= 0, x,,= m, x x x-n x-m
-1
yx2- 3x + 9 + 2=0
38.
Yx2 + 3x +
7-
41.[2- 3x[ + 2=9
40.[3x- 1[ =5
yx2 + 3x + 9=2
45.2x3 + 5x2- 8x- 20=0
44.5x4=9x3
43.2x3= 3x2
�= V2x + 1
24.�=x- 2
26. �3x + 1 = -1
31.x-6 - 7x-3 - 8 = 0
39.[2x + 3[=7
18.�=3
�=2
23. V2x - 3 + x = 3
22.3x4 + 4x2 + 1 =0
37.yx2 + 3x + 7-
15.2x + 3=4x2
20.3x2- X + 1=0
x(x + 1 ) + 2=0
-
-
14.x(2 - x) =3(x- 4)
6x=4x2
28. �
-
(r 1:.)
13. ( x- 1 ) (2x + 3 ) =3 16. 1 +
-
,,=
n
42. [1 - 2x[ + 1 =4 46.3x3 + 5x2 - 3x- 5=0
In Problems 47-60, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set. 2x + 3 5- x 2x - 3 x 49. -9 :5 --- :5 7 47.--- + 2 :5 48.-- :5 6x- 4 -4 3 2 5 5- 3x 3-3x 2x - 2 52. -3 :5 --- :5 6 51.2 < -- < 6 50.-4 < -- < 6 12 2 3 1 1 55.[2x- 5[ � 9 56. [3x + 1[ � 10 53.[3x + 4[ < 2" 54. [1- 2x[ < :3 57.2 + [2 - 3x[
In Problems 61.x2
+
:5
61-64,
4
58.
1
� + [ 2X ; [
:5
1
59. 1 - [2 - 3x[ <-4
[
[
2X - 1 60. 1 - -- <-2 3
what number should be added to complete the square of each expression? 62.x2 - lOx
6x
?
63 . x2 - ix 3
64.r
4 5
+-x
In Problems 65-74, use the complex number system and write each expression in the standard form a + bi. 65. (6 + 3i) - (2- 4i ) 66. (8 - 3i) + ( -6 + 2i) 67.4(3- i) + 3 (-5 + 2i ) 68.2(1 + i) - 3(2- 3i) 3 69.__ 3+i
4 70. --. 2-{
73.( 2 + 3i) 3
In Problems 75-82, solve each equation in the complex number system. 77.2X2 + x-2=0 75.x2 + x + 1 =0 76.x2 - x + 1 =0 79.x2 + 3=x
80. 2x2 + 1=2x
83.Translate the following statement into a mathematical expression : The perimeter p of a rectangle is the sum of two times the length I and two times the width w.
81.x( 1-x)=6
74. (3- 2i)3 78.3x2- 2x- 1 = 0
82.x ( l + x) =2
84.Translate the following statement into a mathematical expres sion:The total cost C of manufacturing x bicycles in one day is $50,000 plus $95 times the number of bicycles manufactured.
Cha pter Review
85 . Banking A bank lends out $9000 at 7 % simple interest. At the end of 1 year, how much interest is owed on the loan? 86. Financial Planning Steve, a recent retiree, requires $5000 per year in extra income. He has $70,000 to invest and can in vest in A-rated bonds paying 8% per year or in a certificate of deposit (CD) paying 5% per year. How much money should be invested in each to realize exactly $5000 in interest per year? 87. Lightning and Thunder A flash of lightning is seen, and the resulting thunderclap is heard 3 seconds later. If the speed of sound averages 1 100 feet per second, how far away is the storm?
15 1
per second and 5 meters per second, respectively. How long is it until the bees meet for the first time? How long is it until they meet for the second time? 93 . Physics : Uni form Motion A Metra commuter train leaves Union Station in Chicago at 12 noon. Two hours later, an Amtrak train leaves on the same track, traveling at an aver age speed that is 50 miles per hour faster than the Metra train. At 3 PM the Amtrak train is 10 miles behind the commuter train. How fast is each going? 94. Physics An object is thrown down from the top of a build ing 1280 feet tall with an initial velocity of 32 feet per sec ond. The distance s (in feet) of the object from the ground after t seconds is s 1 280 - 32t - 1 6t2 =
88. Physics : Intensity o f Light The intensity I (in candlepower) 900 . . . a f a certam lIght source obeys the equatIOn I = ? , where x-
x
is the distance (in meters) from the light. Over what range of distances can an object be placed from this light source so that the range of intensity of light is from 1600 to 3600 can dlepower, inclusive? 89.Extent of Search and Rescue A search plane has a cruising speed of 250 miles per hour and carries enough fuel for at most 5 hours of flying. If there is a wind that averages 30 miles per hour and the direction of the search is with the wind one way and against it the other, how far can the search plane travel before it has to turn back? 90. Extent of Search and Rescue If the search plane described in Problem 89 is able to add a supplementary fuel tank that allows for an additional 2 hours of flying, how much farther can the plane extend its search? 91 . Rescue at Sea A life raft, set adrift from a sinking ship 150 miles offshore, travels directly toward a Coast Guard station at the rate of 5 miles per hour. At the time that the raft is set adrift, a rescue helicopter is dispatched from the Coast Guard station. If the helicopter's average speed is 90 miles per hour, how long will it take the helicopter to reach the life raft? �- 90 milhr
5 milhr --
/
�---- 1 50 mi ---� .. : I
92 . Physics : Uni form Motion Two bees leave two locations 1 50 meters apart and fly, without stopping, back and forth between these two locations at average speeds of 3 meters
(a) When will the object strike ground? (b) What is the height of the object after 4 seconds? 95. Working Together to Get a Job Done Clarissa and Shawna, working together, can paint the exterior of a house in 6 days. Clarissa by herself can complete this job in 5 days less than Shawna. How long will it take Clarissa to complete the job by herself? 96.Emptying a Tank Two pumps of different sizes, working to gether, can empty a fuel tank in 5 hours. The larger pump can empty this tank in 4 hours less than the smaller. lf the larger pump is out of order, how long will it take the smaller one to do the job alone? 97.Chemistry: Salt Solutions How much water should be added to 64 ounces of a 10% salt solution to make a 2% salt solution? 98 . Chemistry: Salt Solutions How much water must be evap orated from 64 ounces of a 2 % salt solution to make a 10% salt solution? 99.Geometry The hypotenuse of a right triangle measures 13 centimeters. Find the lengths of the legs if their sum is 17 centimeters. 100. Geometry The diagonal of a rectangle measures 10 inches. If the length is 2 inches more than the width, find the di mensions of the rectangle. 101. Chemistry : Mixing Acids A laboratory has 60 cubic cen timeters (cm3 ) of a solution that is 40% HCI acid. How many cubic centimeters of a 1 5 % solution of HCI acid should be mixed with the 60 cm3 of 40% acid to obtain a solution of 25 % HCI? How much of the 25% solution is there? 102. Framing a Painting An artist has 50 inches of oak trim to frame a painting. The frame is to have a border 3 inches wide surrounding the painting. (a) If the painting is square, what are its dimensions? What are the dimensions of the frame? (b) If the painting is rectangular with a length twice its width, what are the dimensions of the painting? What are the dimensions of the frame? 103.Using Two Pumps An 8-horsepower (hp) pump can fill a tank in 8 hours. A smaller, 3-hp pump fills the same tank in 12 hours. The pumps are used together to begin filling this tank. After four hours, the 8-hp pump breaks down. How long will it take the smaller pump to fill the tank? 104. Pleasing Proportion One formula stating the relationship between the length I and width w of a rectangle of "pleasing proportion" is 12 w(1 + w). How should a 4 foot by 8 foot sheet of plasterboard be cut so that the result is a rectangle of "pleasing proportion" with a width of 4 feet? =
152
CHAPTER 1
Equations and Inequalities
105. F inance An inheritance of $900,000 is to be divided among Scott, Alice, and Tricia in the following manner: Alice is to receive of what Scott gets, while Tricia gets of what Scott gets. How much does each receive? 106. B us iness: Determ in ing the Cos t of a Char te r A group of 20 senior citizens can charter a bus for a one-day excursion trip for $15 per person. The charter company agrees to re duce the price of each ticket by 10¢ for each additional pas senger in excess of 20 who goes on the trip, up to a maximum of 44 passengers (the capacity of the bus). If the final bill from the charter company was $482.40, how many seniors went on the trip, and how much did each pay?
%
they race again, this time with Todd lining up 5 meters be hind the start. (a) Assuming that Todd and Scott run at the same pace as before, does the second race end in a tie? (b) If not, who wins? (c) By how many meters does he win? (d) How far back should Todd start so that the race ends in a tie? After running the race a second time, Scott, to even things up, suggests to Todd that he (Scott) line up 5 meters in front of the start. (e) Assuming again that they run at the same pace as in the first race, does the third race result in a tie? (f) If not, who wins? (g) By how many meters? (h) How far ahead should Scott start so that the race ends in a tie?
�
107. U til iz ing Copying Machines A new copying machine can do a certain job in 1 hour less than an older copier. Together they can do this job in 72 minutes. How long would it take the older copier by itself to do the job? 108. In a 100-meter race, Todd crosses the finish line 5 meters ahead of Scott. To even things up, Todd suggests to Scott that
109. Phys ics: Uniform Mo tion A man is walking at an average speed of 4 miles per hour alongside a railroad track. A freight train, going in the same direction at an average speed of 30 miles per hour, requires 5 seconds to pass the man. How long is the freight train? Give your answer in feet. 30 mi/hr
t
=
0 - 5 sec - t
=
5
CHAPTER TEST In Problems 1-7, find all the real solutions, if any, of each equation. 2x x 5 1. - = 2 . x(x - 1) = 6 3 "2 12 S.
12x - 31 + 7 = 10
6 . 3x3
+
2x2 - 12x - 8 = 0
3.
X4
- 3x2 - 4 = 0
7 . 3x2
- X
+ 1
=
4 . \I2x-=5 + 2 = 4
0
In Problems 8-10, solve each inequality. Express your answer using interval notation. Graph the solution set. 3x - 4 2 :::; 6 8. -3 :::; 9. 13x + 4 1 < 8 10. 2 + 12x - 5 1 � 9 -
11 . Write � in the standard form a + bi . 3-i 12. Solve the equation 4x2 - 4x + 5 = 0 in the complex number system. 13.
B lending Coffee A coffee house has 20 pounds of a coffee that sells for $4 per pound. How many pounds of a coffee that sells for $8 per pound should be mixed with the 20 pounds of $4-per-pound coffee to obtain a blend that will sell for $5 per pound? How much of the $5-per-pound coffee is there to sell?
Chapter Projects
1 S3
CHAPTER PROJECTS
I. Home Mortgages While you may not be in the market for a home right now, it is probably an event that will occur for you within the next few years. The formula below gives the monthly payment P required to pay off a loan L at an an nual rate of interest r, expressed as a decimal, but usually given as a percent. The time t, measured in months, is the duration of the loan, so a IS-year mortgage requires t = 1 2 X IS = 1 80 monthly payments.
P =
L
On (a) (b) 1.
l
r 12
1
-
(1 ....c.12. ) +
_{
l
P = month ly payment L = loan a m ount r = annua l rate of Interest,
(1 )
expressed as a decima l
t
=
length o f loan, In months
May 12, 2006, average interest rates were: 6.S8% on 30-year mortgages 6.17% on IS-year mortgages For each of these loans, calculate the monthly payment for a loan of $200,000.
2. Then compute the total amount paid over the term of the loan. 3. Finally, calculate the interest paid. On May S, 2006, average interest rates were: (a) 6.S9% on 30-year mortgages (b) 6.22 % on IS-year mortgages 4. For each of these loans, calculate the monthly payment for a loan of $200,000. s. Then compute the total amount paid over the term of the loan. 6 . Finally, calculate the interest paid. 7. Solve equation (1 ) for L. 8. If you can afford to pay $ 1000 per month for a mortgage payment, calculate the amount you can borrow on May S, 2006. (a) On a 30-year mortgage (b) On a IS-year mortgage 9. Repeat Problem 8 for the rates on May 12, 2006. 10 . Check with your local lending institution for current rates for 30-year and IS-year mortgages. Calculate how much you can borrow with a $'1000 per month payment. 11. Repeat Problem 10 if you can afford a payment of $ 1 300 per month. 12 . Do you think that the interest rate plays an important role in determining how much you can afford to pay for a house? 1 3. Comment on the two types of mortgages: 30-year and I S-year. Which would you take? Why?
The following project is also available on the Instructor's Resource Center (IRC): I I. Project at Motorola How Many Cellular Phones Can I Make? An industrial engineer uses a model involving equations to be sure production levels meet customer demand.
Graphs The First Modern Olympics: Athens, 1896
The birth of the modern Olympic Games
By John Gettings-"I hereby proclaim the opening of the first Inter national Olympic Games at Athens." With these words on April 6, 1 896, King George I of Greece welcomed the crowd that had gathered in the newly reconstructed Panathenean Stadium to the modern-day Olympic Summer Games. The event was the idea of B aron Pierre de Coubertin of France who traveled the world to gather support for his dream to have nations come together and overcome national disputes, all in the name of sport. The program for the Games included track and field, fencing, weightlifting, rifle and pistol shooting, tennis, cycling, swimming, gym nastics, and wrestling. Although 14 nations participated, most of the athletes were Greek. The Games reached their high point on Day 1 1 with the first modern-day marathon. The idea to hold an event to commemorate the Ancient Olympic games was suggested by a friend of de Coubertin and was met with great anticipation. The race was run from Marathon to Athens (estimated at 22-26 miles), watched by more than 100,000 people and won by a Greek runner, Spiridon Louis. Source: Fact Monsler™ Database, © 2006 Pearson Education, Inc.
- See the Chapter Project-
Source: Adapted from John Gettings, "The First Modern Olympics: Athens, 1896," accessed at www.infoplease.com. ©2006 Pearson Education, Inc. Reprinted with permission.
A Look Back
I n Cha pter R we reviewed algebra essentia l s and geometry essentials. I n Cha pter 1 we stu d ied equations i n one va riable.
A Look Ahead
Here we con nect a lgebra and geometry u s i ng the recta n g u l a r coord i nate system and use it to graph eq uations in two va riables. The idea of using a system of rectan gular coord i n ates d ates back to a ncient ti mes, when such a system was used for s u r veying a n d city p l a n n ing. Apollo n i u s of Perga, i n 200 BC, used a form of recta ngular coord i nates i n his work o n con ics, although this use d oes not sta n d out as clearly a s i t does i n modern treatments. Sporadic u s e o f recta n g u l a r coord i nates continued u ntil the 1600s. By that ti me, algebra has developed sufficiently so that Rene Descartes (1596- 1650) a n d Pierre d e Fermat ( 1 60 1 -1665) could take the crucial step, which was the use of recta n g u l a r coord i nates to translate geometry problems i nto algebra problems, and vice versa. This step was i m porta nt for two reasons. Fi rst, it allowed both geometers a n d algebraists to g a i n new insights i nto their subjects, wh ich previously had been regarded as sepa rate, but now were seen to be con nected i n m a ny i m portant ways. Second, these insights made the development of ca lculus possible, wh ich g reatly enlarg ed the n u mber of a reas in wh ich mathematics cou l d be a p p l ied and made possible a much deeper understanding of these a reas.
Outline
2.1 The Distance and Midpoint Formulas
2.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
2.3 Lines
2.4 Circles
2.5 Variation
Cha pter Review Chapter Test Cumulative Review Cha pter Project
1 55
1 S6
CHAPTER 2
Graphs
2.1 The Distance and Midpoint Formulas PREPARING FOR THIS SECTION
Before getting started, review the following:
Algebra Essentials (Chapter R, Section R.2, pp. 17-26)
•
•
Geometry Essentials (Chapter R, Section R.3, pp. 30-35)
Now Work the 'Are You Prepared? , problems on page 1 60. OBJECTIVES
1 U se the Distance Form u l a (p. 1 57) 2
Use the Mid point Formula (p. 1 59)
Rectangular Coordinates
Figure 1 Y
4
-
2
-2
4
2
0-
2
4
x
-4
Figure 2
y
(-3, 1 ) 1 {e
3
4
3
�
[ 0
3 (-2 , -3j�
-4
(3 e , 2) 2 � x 3 (3, -2)
�
J
�
Figure 3
Quadrant
y
II
Q u a d rant
I
x< O, y > O
x > O, y > O
III x< 0, y< 0
Quadrant IV
Quadrant
x> O , y < O
We locate a point on the real number line by assigning it a single real number, called the coordinate of the point. For work in a two-dimensional plane, we locate points by using two numbers. We begin with two real number lines located in the same plane: one horizontal and the other vertical. We call the horizontal line the x-axis, the vertical line the y-axis, and the point of intersection the origin O. See Figure 1. We assign coordi nates to every point on these number lines using a convenient scale. We usually use the same scale on each axis. In applications, however, different scales appropriate to the application may be used. The origin 0 has a value of 0 on both the x-axis and y-axis. Points on the x-axis to the right of 0 are associated with positive real numbers, and those to the left of o are associated with negative real numbers. Points on the y-axis above 0 are asso ciated with positive real numbers, and those below 0 are associated with negative real numbers. In Figure 1, the x-axis and y-axis are labeled as x and y, respectively, and we have used an arrow at the end of each axis to denote the positive direction. The coordinate system described here is called a rectangular or Cartesian* coordinate system. The plane formed by the x-axis and y-axis is sometimes called the xy-plane, and the x-axis and y-axis are referred to as the coordinate axes. Any point P in the xy-plane can be located by using an ordered pair (x, y) of real numbers. Let x denote the signed distance of P from the y-axis (signed means that, if P is to the right of the y-axis, then x > 0, and if P is to the left of the y-axis, then x < 0); and let y denote the signed distance of P from the x-axis. The ordered pair (x, y), also called the coordinates of P, then gives us enough information to locate the point P in the plane. For example, to locate the point whose coordinates are ( - 3, 1 ) , go 3 units along the x-axis to the left of 0 and then go straight up 1 unit. We plot this point by plac ing a dot at this location. See Figure 2, in which the points with coordinates ( -3, 1 ) , ( -2, -3 ) , (3, - 2), and (3, 2) are plotted. The origin has coordinates (0, 0). Any point on the x-axis has coordinates of the form (x, 0), and any point on the y-axis has coordinates of the form (0, y). If (x, y) are the coordinates of a point P, then x is called the x-coordinate, or abscissa, of P and y is the y-coordinate, or ordinate, of P. We identify the point P by its coordinates (x, y) by writing P ( x, y). Usually, we will simply say "the point (x, y ) " rather than "the point whose coordinates are (x, y ) . " The coordinate axes divide the xy-plane into four sections called quadrants, as shown in Figure 3. In quadrant I, both the x-coordinate and the y-coordinate of all points are positive; in quadrant II, x is negative and y is positive; in quadrant III, both x and y are negative; and in quadrant IV, x is positive and y is negativel Points on the coordinate axes belong to no quadrant. =
x
1;21!:1 = = ....
Now Work P R O B L E M 1 1
':' Named after Rene Descartes (1596-1650), a French mathematician, philosopher, and theologian.
SECTION 2.1
I
The Distance and M i d point Formulas
157
CO M M ENT On a graph ing calculator, you can set the scale on each axis. Once th is has been done,
you obtain the viewing recta ngle. See Figure 4 for a typical v i ewing rectangle. You should now read
_
Section 1, The Viewing Rectangle, in the Appendix.
Figure 4
1
Use the Distance Formula
If the same units of measurement, such as inches, centimeters, and so on, are used for both the x-axis and y-axis, then all distances in the xy-plane can be measured using this unit of measurement. E XA M P L E 1
F i n d i n g the Distance betwee n Two Poi nts
Find the distance d between the points ( 1 , 3 ) and (5, 6 ) . S o l ution
First w e plot the points ( 1 , 3 ) and (5, 6 ) and connect them with a straight line. See Figure Sea). We are looking for the length d. We begin by drawing a horizontal line from ( 1 , 3 ) to (5, 3 ) and a vertical line from (5, 3 ) to (5, 6), forming a right triangle, as shown in Figure S(b). One leg of the triangle is of length 4 (since Is 1 1 = 4), and the other is of length 3 (since 16 - 31 = 3). By the Pythagorean Theorem, the square of the distance d that we seek is d2 = 42 + 32 = 16 + 9 = 25 -
d = V2S = S
� .LJ
Figure 5
( 1 , 3) 4 ( 5 , 3)
6 x
3
3 (b)
(a)
6
x •
The distance formula provides a straightforward method for computing the distance between two points. THEOREM
r
To
r
ference
r
r
( '
r
The distance between two points PI = (Xl , Y I ) and P2 = (X2 ' Yz ), denoted by d ( PI , P2 ), is
I n Word s
r
compute
the
distance
between two points, find the dif of the
Distance Formula
x-coordinates,
square it, and add t h i s to the square of the difference of the y-coordinates. The square r oot of this sum is the distance.
d(PI , P2 ) = V(X2
-
xd
+ ( Y2
-
yd
(1)
I�
�--------------------------------�
Proof of the Distance Formula Let (Xl , Y l ) denote the coordinates o f point PI and let ( X2 ' Y2 ) denote the coordinates of point P2 ' Assume that the line joining PI and P2 is neither horizontal nor vertical. Refer to Figure 6(a). The coordinates of P3 are (X2 ' Y l ) . The horizontal distance from P I to P3 is the absolute value of
1 58
CHAPTER 2
Graphs
the difference of the x-coordinates, I X2 - xI I . The vertical distance from P3 to Pz is the absolute value of the difference of the y-coordinates, IY2 - Y l i . See Fig ure 6(b). The distance d( P l , P2 ) that we seek is the length of the hypotenuse of the right triangle, so, by the Pythagorean Theorem, it follows that 2 2 [ d( P l , P2 ) f = I X2 - xl l + IY2 - Y l l 2 = (X2 - x d + (Yz - Yl ) d( P l , P2 ) =
Figure 6
V(X2 - Xl? + (Y2 - yd
y
L--L
-L
__ __ __ __
�x
__ __
(b)
(a)
Now, if the line joining PI and P2 is horizontal, then the y-coordinate of PI equals the y-coordinate of P2 ; that is, Y l = Y2 . Refer to Figure 7(a). In this case, the distance formula (1) still works, because, for Yl = Yz , it reduces to d( P 1 , P2 ) = Figure 7
Y Y1
P1 =
(X1 ' Y1) •
V(X2 - X l ? + 02 = V( X2 - x d = I X2 - XI I Y2
y
d(P1, P2)
•
Y1 x
I Y2 � Y1 1 d(P� , P2) T P1 - (X1 , Y1)
r�
I,\ , Y,I
x1
(a)
x
(b)
A similar argument holds if the line joining PI and P2 is vertical. See Figure 7(b). The distance formula is valid in all cases. • E XA M P L E 2
U s i n g the Distance Formula
Find the distance d between the points ( - 4 5 ) and (3, 2). ,
Solution
Using the distance formula, equation (1), the solution is obtained as follows : 2 2 d = \1[3 - ( 4 ) f + (2 - 5 ? = \17 + ( - 3 ) -
=
, ;!I,I!
>-
V49+9 = v5s
�
7.62 •
Now Work P R O B L E M S 1 5 A N D 1 9
The distance between two points PI = (X l , Y l ) and P2 = ( X2 ' Y2 ) is never a neg ative number. Furthermore, the distance between two points is 0 only when the points are identical, that is, when Xl = X2 and Y l = Y2 . Also, because (X2 - Xl ) 2 = (Xl - X2 ) 2 and ( Yz - Y l ) 2 = ( Yl - Y2?, it makes no difference whether the dis tance is computed from PI to P2 or from P 2 to PI; that is, d(Pl , P z ) d( P2 , PI). The introduction to this chapter mentioned that rectangular coordinates enable us to translate geometry problems into algebra problems, and vice versa. The next example shows how algebra (the distance formula) can be used to solve geometry problems. =
SECTION 2.1
E XA M P L E 3
(a) (b) (c) (d)
(-2, 1 ) eo::- ____ C (3, 1) -+-
,
,
Plot each point and form the triangle ABC. Find the length of each side of the triangle. Verify that the triangle is a right triangle. Find the area of the triangle.
d(A, C)
=
3
( - 2 1 ) B = (2, 3 ) , and C = (3, 1 ) .
d(A, B) = V[ 2 - ( -2) f + (3 2 2 d ( B, C) = V(3 - 2) + ( 1 - 3 )
y
-3
=
(a) Points A, B, C and triangle ABC are plotted in Figure 8. (b) We use the distance formula, equation ( 1).
S o l ution
A=
1 S9
U s i ng Algebra to Solve G eometry P ro blems
Consider the three points A
Figure 8
The Distance and M i d point Formulas
x
=
V[3 - ( -2) f +
1 ) 2 v16+4 = V20 = 2Vs Vl+4 = Vs =
(1 - 1? = V25+O = 5 =
( c) To show that the triangle is a right triangle, we need to show that the sum of the squares of the lengths of two of the sides equals the square of the length of the third side. (Why is this sufficient?) Looking at Figure 8, it seems rea sonable to conjecture that the right angle is at vertex B. We shall check to see whether [ d(A, B) f + [ d(B, C) f [ d(A, C) f We find that 2 2 [ d(A, B) f + [ d(B, C) f ( 2 Vs ) + ( Vs ) = 20 + 5 = 25 = [ d ( A, C) f =
=
so it follows from the converse of the Pythagorean Theorem that triangle ABC is a right triangle. (d) Because the right angle is at vertex B, the sides AB and BC form the base and height of the triangle. Its area is Area m:::=:;;;;"..2
=
�
(Base ) ( Height) =
� ( 2 Vs) ( Vs )
=
5 square units •
Now Work P R O B L E M 2 9
Use the Midpoint Formula
We now derive a formula for the coordinates of the midpoint of a line segment. Let P I = ( Xl , Y 1 ) and P2 (X2 ' Y2) be the endpoints of a line segment, and let M = ( x, y) be the point on the line segment that is the same distance from P l as it is from P2 . See Figure 9. The triangles Pl AM and MBP2 are congruent. Do you see why? d(PJ , M) = d(M, P2) is given; L AP1 M L BMP2 ,* and L P1 MA L MP2B. So, we have angle-side-angle. Because triangles P l AM and MBP2 are congruent, corresponding sides are equal in length. That is,
Figure 9 Y
=
=
y
and x
x
2x = X l + X2 X l + X2 X= 2
Y - Yl
=
=
Y2
-Y
+ Y2 Yl + Y2 y= 2
2y =
Yl
':' A postulate from geometry states that the transversal P I P2 forms congruent corresponding angles with the parallel line segments P I A and M B.
160
CHAPTER 2
Graphs
THEOREM r
r r r
r
In
Word s
=
( x, y) of the line segment from
M
ment, average the x-coordinates the endpoints.
L-
+ l ( X, y ) _- X +2 Xz ' Y I 2 Yz
(
( X l , YI) to
)
Find the midpoint of a line segment from PI = ( -5, 5 ) to points P I and Pz and their midpoint. Check your answer.
y (-5 , 5 )
-
F i n d i ng the M idpoint of a Line Segment
Sol ution
Figure 1 0
=
_
=
(2)
I
��
__ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __
E XA M P L E 4
P1
The midpoint M Pz = (Xz , Yz) is
fo find the midpoint o f a line seg and average the y-coordinates of
PI
M i d point Formula
We apply the midpoint formula (2) using X l = -5, Y l = 5, Xz the coordinates ( x, y) of the midpoint M are x
5
=
Xl
+
2
Xz
-
5 + 3 2
_ _ _ _
=
- 1 and
Y
=
Yl
Pz =
= (3, 1 ) . Plot the
3, and Yz = 1 . Then
+ Y2 = 5 + 1 2
2
--
=3
That is, M = ( - 1, 3). See Figure 10. 5
-5
x
Check:
Beca u se M is the d(P1, M) = d( M, P2 ):
mid poi nt, we check the a n swer by verifyi ng that
d(Pl, M ) V[ -1 - (-5)J2 + (3 - 5)2 \11 6 + 4 V20 d( M, P2 ) V[3 (-1 ) J 2 + (1 - 3) 2 \116 + 4 = V20 =
=
=
"l' =--
-
=
=
•
Now Work P R O B L E M 3 5
2 . 1 Assess Your Understanding
'Are You Prepared?' A nswers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1 . On the real number line the origin is assigned the number (p. 17-26) 2.If -3 and 5 are the coordinates of two points on the real num ber line, the distance between these points is . (p. 17-26) 3. If 3 and 4 are the legs of a right triangle, the hypotenuse is . (pp. 30-35) __.
4. Use the converse of the Pythagorean Theorem to show that a triangle whose sides are of lengths 11, 60, and 61 is a right triangle. (p. 30-35)
__
__
Concep�s and Vocabulary 5. If (x, y) are the coordinates of a point P in the xy-plane, then x is called the of P and y is the of P 6. The coordinate axes divide the xy-plane into four sections called 7. If three distinct points P, Q, and R all lie on a line and if d(P, Q) = d(Q, R ) , then Q is called the of the line segment from P t a R. __
__
__
The distance between two points is some times a negative number. 9. True or False The point ( - 1 , 4) lies in quadrant IV of the Cartesian plane. 10. True or False The midpoint of a line segment is found by av eraging the x-coordinates and averaging the y-coordinates of the endpoints. 8. True or False
SECTION 2.1
The Distance a nd Midpoint Formulas
161
Skill Building In Problems 1 1 and 12, plot each point in the xy-plane. Tell in which quadrant or on what coordinate axis each point lies. 11.
(a) A = ( -3, 2) (b) B = (6, 0) (c) C = ( -2, -2)
12. (a) A ( 1 , 4) (b) B ( -3, -4) (c) C = ( -3, 4)
(d) D = (6, 5 ) (e) E = (0, -3) (f) F = (6, -3)
(d) D (4, 1 ) (e) E (0, 1 ) (f) F = ( -3, 0)
=
=
=
=
13. Plot the points (2, 0), (2, -3), (2, 4 ) , (2, 1 ) , and (2, - 1 ) . Describe the set of all points of the form (2, y ) , where y is a real number. 14.
Plot the points (0, 3), ( 1 , 3), ( -2, 3), (5, 3 ) , and ( -4, 3). Describe the set of all points of the form ( x, 3 ) , where x is a real number.
In Problems 15-28, find the distance d ( P ] , P 2 ) between the points PI and P2 . 17. 15. Y 16. Y.
-2
2
P1 = (0, 0 )
1 9. P I
=
-1
P2 = (2, 1 )
2
�
P2 = (-
-2
x
-1
P = (0 ' 0 )
-/ 1 I I
2
I
-2 I
x
(3, -4); P 2 = (5, 4)
20.
PI
/1
22. PI = (2, -3); P2
23. PI = (4, -3); P2
24. P I
25.
PI
27. P I
(6, 4)
Y
-2 I
�I -1
I
=
(2, 2) 2
I
x
= ( - 1 , 0); P2 = (2, 4)
21. PI = (-3, 2); P2 = (6, 0) =
(1 , 1)
I . 2 x =
"' -:1 P2
18.
=
= ( -4, -3); P 2
(4, 2) =
(6, 2)
= ( -0.2, 0.3 ); P 2 = (2.3, 1 . 1 )
26.
PI = ( 1. 2, 2.3 ); P2 = ( -0.3, 1 . 1 )
= (a, b); P 2 = (0, 0)
28.
PI
= ( a , a ) ; P2
=
(0, 0)
In Problems 29-34, plot each point and form the triangle ABC. Verify that the triangle is a right triangle. Find its area. 29. A = ( -2, 5 ) ; B = ( 1 , 3 ) ; C = ( - 1 , 0) 30. A = ( -2, 5); B = ( 12, 3 ) ; C = ( 10, - 1 1 ) 31. A = ( -5, 3); B = (6, 0); C = (5, 5 )
32. A = ( -6, 3); B = (3, -5); C = ( - 1 , 5)
33. A = (4, -3); B = (0, -3); C = (4, 2)
34.
A = (4, -3); B = (4, 1 ) ; C = (2, 1 )
In Problems 35-44, find the midpoint of the line segment joining the points PI and P2 · 36. PI = ( -2, 0); P 2 = (2, 4) 35. P I = (3, - 4); P2 = (5, 4) 37.
PI
= ( -3, 2); P2
=
(6, 0)
39. PI = (4, -3); P2 = (6, 1 )
38.
PI
40. P I
= (2, -3); P2 =
=
( -4, -3); P 2
(4, 2) =
(2, 2)
41.
P I = ( -0.2, 0.3) ; P2 = (2.3, 1 . 1 )
42.
PI = ( 1 .2, 2.3) ; P 2 = ( -0.3, 1 . 1 )
43.
PI = ( a , b ) ; P2 = (0, 0)
44.
P I = (a, a ) ; P2 = (0, 0)
The midpoint of the line segment from PI to P2 is (5, -4). If P2 = (7, -2), what is PI?
Applications and Extensions 45.
Find all points having an x-coordinate of 2 whose distance from the point ( -2, - 1 ) is 5.
50.
46.
Find all points having a y-coordinate of -3 whose distance from the point ( 1 , 2 ) is 13.
47.
Find all points on the x-axis that are 5 units from the point (4, -3).
51 . Geometry The medians of a triangle are the line segments from each vertex to the midpoint of the opposite side (see the figure). Find the lengths of the medians of the triangle with vertices at A (0, 0), B = (6, 0), and C = (4, 4).
48.
Find all points on the y-axis that are 5 units from the point (4, 4).
49.
The midpoint of the line segment from Pi ta P2 is (-1, 4). If p] ( -3, 6), what is P2? =
=
C
A
B
162
CHAPTER 2
Graphs
52.
Geometry An equilateral triangle is one in which all three sides are of equal length. If two vertices of an equilateral tri angle are (0, 4) and (0, 0), find the third vertex. How many of these triangles are possible?
54.
Geometry Verify that the points (0, 0), (a, 0), and
53.
Geometry Find the midpoint of each diagonal of a square with side of length s. Draw the conclusion that the diagonals of a square intersect at their midpoints. [Hin t: Use (0, 0), (0, s), (s, 0), and (s, s) as the vertices of the square.]
(%' V;a ) are the vertices of an equilateral triangle. Then show that the mid
points of the three sides are the vertices of a second equilateral triangle (refer to Problem 52).
In Problems 55-58, find the length of each side of the triangle determined by the three points PI , P2 , and P3 . State whether the triangle is an isosceles triangle, a right triangle, neither of these, or both. (An isosceles triangle is one in which at least two of the sides are of equal length.) 55. 56. 57.
Pl = (2, 1 ) ; P2 = ( -4, 1 ) ; P3 = ( -4, -3) p] = ( - 1, 4); P = (6, 2 ) ; P3 = (4, -5) 2 Pl = ( -2, - 1 ) ; P2 = (0, 7 ) ; P3 = (3, 2)
58.
PI = (7, 2 ) ; P2 = ( -4, 0); P3 = (4, 6)
59.
Baseball A major league baseball "diamond" is actually a square, 90 feet on a side (see the figure). What is the distance directly from home plate to second base (the diagonal of the square )?
lies in the direction from home plate to first base, and the positive y-axis lies in the direction from home plate to third base. (a) What are the coordinates of first base, second base, and third base? Use feet as the unit of measurement. (b) If the right fielder is located at ( 1 80, 20) , how far is it from the right fielder to second base? (c) If the center fielder is located at ( 220, 220), how far is it from the center fielder to third base? 63.
Distance between Moving Objects A Dodge Neon and a Mack truck leave an intersection at the same time. The Neon heads east at an average speed of 30 miles per hour, while the truck heads south at an average speed of 40 miles per hour. Find an expression for their distance apart d (in miles) at the end of t hours.
64.
Distance of a Moving Object from a Fixed Point A hot-air balloon, headed due east at an average speed of 15 miles per hour and at a constant altitude of 100 feet, passes over an in tersection (see the figure). Find an expression for the distance d (measured in feet) from the balloon to the intersection t seconds later.
65.
Draft ing Error When a draftsman draws three lines that are to intersect at one point, the lines may not intersect as in tended and subsequently will form an error triangle. If this error triangle is long and thin, one estimate for the location of the desired point is the midpoint of the shortest side. The figure shows one such error triangle. Source: www. uwgb.eduldutchsISTRUCTGElsIOO.htm
2 n d base
60.
Li ttle League Baseball The layout of a Little League playing field is a square, 60 feet on a side: How far is it di rectly from home plate to second base (the diagonal of the square )? Source: Uttle League Baseball, Official Regulations and Play ing Rules, 2006.
61.
Baseball Refer to Problem 59. Overlay a rectangular coor dinate system on a major league baseball diamond so that the origin is at home plate, the positive x-axis lies in the di rection from home plate to first base, and the positive y-axis lies in the direction from home plate to third base. (a) What are the coordinates of first base, second base, and third base? Use feet as the unit of measurement. (b) If the right fielder is located at (310, 1 5 ) , how far is it from the right fielder to second base? (c) If the center fielder is located at (300, 300), how far is it from the center fielder to third base?
62.
Little League Baseball Refer to Problem 60. Overlay a rec tangular coordinate system on a Little League baseball dia mond so that the origin is at home plate, the positive x-axis
SECTION 2.2
Graphs of Equations in Two Variables; Intercepts; Symmetry
1 63
Inc., in 2004. How does your result compare to the re ported value of $256 billion? Source: Wal-Mart Stores, Inc., 2006 Annual Report
y
1.7 1. 5 1. 3
Wal-Mart Stores, Inc.
1 .4
2.6 2.7
Net sales (in $ billions)
� , :� i 350 1 · ··· · ·· ····· ·········· ·· ·········· ···· ·· .. 2. i · . . . . . . . . . ... . ... . . .. . . . · . . . . . . . . . · . . . .. . . . .. . . ... . . .. . . . ... . � 300 1 � f.-"' � 2 50 r ;;;:;,; :::::;,..;;;;- --� 200 I L _ � �................................................................................................, � 1 50 1 i � 1 00 1 ······ · ········· ······························ .. . . . . . . . J � 5 0 1· · ····· ········ ············ ··················· .. .. . ! . . .. . . . .. . . . . . . . . .. . . . . . . . .. . . . . . . . . ........ ) 1 ····· · · · ·· · · ·· · · ·· · · · ··· · · · · ·· · · · ·· · ··· · · ·· · ·· 0
x
........ ...................................
en
(a) Find an estimate for the desired intersection point. (b) Find the length of the median for the midpoint found in part (a). See Problem 51.
.. .
....
........................................................................................................
66.
Ne t Sa les The figure illustrates how net sales of Wal-Mart Stores, Inc., have grown from 2002 through 2006. Use the mid point formula to estimate the net sales of Wal-Mart Stores,
'"
..... ... . ...... .. . ........ ............ ...
..... ..
2002
'Are You Prepared?' A nswers 2. 8
1. 0
3. 5
2003
2004 Year
...... .........
2005
.
...
....
2006
4. 1 12 + 602 = 121 + 3600 = 3721 = 61 2
2.2 Graphs of Equations i n Two Variables; Intercepts; Symmetry PREPARING FOR THIS SECTION •
Before getting started, review the following:
Solving Equations (Section 1.1, pp. 86-93) Now Work the 'Are You Prepared' problems OBJECTIVES
on
page 1 7 1 .
1 Gra p h Eq uations by Plotting Points (p. 163)
2 Find I ntercepts from a G raph (p. 165) 3
1
Find I ntercepts from an Eq uation (p. 166)
4
Test an Eq uation for Symmetry with Respect to the x-Axis, the y-Axis, a n d the Origin (p. 167)
5
Know How to Gra ph Key Equations (p. 169)
Graph Equations by Plotting Points
An equation in two variables, say x and y, is a statement in which two expressions involving x and y are equal. The expressions are called the sides of the equation. Since an equation is a statement, it may be true or false, depending on the value of the variables. Any values of x and y that result in a true statement are said to satisfy the equation. For example, the following are all equations in two variables x and y: 2x - y = 6
The first of these, x2 + l = 5, is satisfied for x = 1, y = 2, since 1 2 + 22 = 1 + 4 = 5 . Other choices of x and y, such as x = - 1 , y = -2, also satisfy this equa tion. It is not satisfied for x = 2 and y = 3, since 22 + 32 = 4 + 9 = 13 "* 5. The graph o f a n equation i n two variables x and y consists of the set of points in the xy-plane whose coordinates ( x, y) satisfy the equation. Graphs play an important role in helping us visualize the relationships that exist between two variable quantities. Table 1 shows the average price of regular y
=
2x + 5
1 64
CHAPTER 2
Graphs
Table 1 Adjusted average price (in dollars) of gasoline in Cal ifornia , 1 978-2005.
Price
Year
1 979
1 .9829
1 993
1 981
2.4977
Year 1 978
1 980 1 982 1 983
1 984
1 .5651
2.4929
1 994
2 . 1 795
1 996
1 .831 0
1 998
1 .8782
1 985
1 .7540
1 987
1 .3274
1 986 1 988 1 989 1 990 1 991
1 992
1 995 1 997
1 999
1 .3459
2000
1 .31 1 1
2002
1 .4656
2004
1 .3589
1 .4973
2001
2003 2005
Figure 1 1
Price
1 .3969
1 .5 1 89
�
3
.
Cal ifornia gasoline prices based on 2005 dollars, adj usted for inflation, 1 978-2005
0
1 .5 ... .
� 2 .5
�m 1 . 0
1 .4709
,
- ··"·"············································.......................................
....
� 2.0
; -;;;- 5 .. .. .. .. . . . .. ! . .g O. LJ. -'..L..J L...L -'-'--' L...L -'-:':--' L...L � � OO O N � � OO O N � � OO O N v ID �
1 .4669 1 .5397
0
1 .5329
I
· ·
f
··································· ················...................................................... · · ····· ···················· ···················· ······
. ..
.... . ...... .. .... .......... ... ...........
0
r- cc CO CO C:O OO (J") Q") (J") (J") Q) o o o o C) O') O') O') O') Q) m m m O") m o o o o 'T'"""" T""" ,.- .,.- r- r- r- .,-- 'T'"""" T""" T"'"" N N N N
1 .3246
Year Source: Statistical Abstract of the
1 .5270
1 .8249
United States.
1 .7536
1 .5955 1 .8950
2.1 521
2.4730
gasoline in California based on 2005 dollars (adjusted for inflation) for the years 1978-2005. In Figure 1 1 , we graph the data from the table, using the year along the x-axis and the price along the y-axis. From the graph, we can see the price was high est in 1980, 1981, and 2005 at about $2.50 per gallon. EXAM P L E 1
Determi n i n g Whether a Point Is on the G raph of an E quation
Determine if the following points are on the graph of the equation 2x - y (b) (2, -2) (a) (2, 3 ) Solution
=
6.
(a) For the point (2, 3), we check to see if x = 2, y = 3 satisfies the equation 2x - y = 6. 2x - y 2(2) - 3 = 4 - 3 = 1 *- 6 The equation is not satisfied, so the point (2, 3) is not on the graph of 2x - y 6. (b) For the point ( 2 , -2), we have 2x - y = 2(2) - ( -2) = 4 + 2 = 6 The equation is satisfied, so the point (2, -2) is on the graph of 2x - y = 6 . =
=
�,
E XA M P L E 2
'.>-
•
Now Work P R O B l E M 1 1
G raphi n g an E quation by P l otti ng Poi nts
Graph the equation: y = 2x + 5 Solution
Figure 1 2
We want to find all points (x, y) that satisfy the equation. To locate some of these points (and get an idea of the pattern of the graph), we assign some numbers to x and find corresponding values for y. If
x = o x = 1
x = -5
25 x
x = 10
Then
y = 2 ( 0) + 5
Point on Graph = 5
Y = 2( 1 ) + 5 = 7
Y = 2( -5 ) +
5 = -5
Y = 2 ( 1 0) + 5 = 25
( 0, 5 ) ( 1 , 7)
( - 5, - 5 ) ( 1 0, 25 )
By plotting these points and then connecting them, we obtain the graph o f the equa tion (a line), as shown in Figure 12.
•
SECTION 2.2
E XA M P L E 3
Graphs of Equations in Two Variables; I ntercepts; Symmetry
1 65
G rap h i n g an E q u ation by P l otti n g P o i nts
Graph the equation: y
x2
=
Table 2 provides several points on the graph. In Figure 13 we plot these points and connect them with a smooth curve to obtain the graph (a parabola).
Solution
Table 2
x
Y = J?
(x, y)
-4
16
( - 4, 1 6)
-3
9
(-3, 9)
-2
4
( - 2, 4)
a
a
(0, 0)
2
4
(2, 4)
3
9
(3, 9)
4
16
(4, 1 6 )
-1
Figure 1 3
(- 1 , 1 )
(1 , 1 )
•
The graphs of the equations shown in Figures 12 and 13 do not show all points. For example, in Figure 12, the point (20, 45 ) is a part of the graph of y 2x + 5, but it is not shown. Since the graph of y = 2x + 5 could be extended out as far as we please, we use arrows to indicate that the pattern shown continues. It is impor tant when illustrating a graph to present enough of the graph so that any viewer of the illustration will "see" the rest of it as an obvious continuation of what is actual ly there. This is referred to as a complete graph. One way to obtain a complete graph of an equation is to plot a sufficient num ber of points on the graph until a pattern becomes evident. Then these points are connected with a smooth curve following the suggested pattern. But how many points are sufficient? Sometimes knowledge about the equation tells us. For exam ple, we will learn in the next section that, if an equation is of the form y = mx + b, then its graph is a line. In this case, only two points are needed to obtain the graph. One purpose of this book is to investigate the properties of equations in order to decide whether a graph is complete. Sometimes we shall graph equations by plot ting points. Shortly, we shall investigate various techniques that will enable us to graph an equation without plotting so many points. =
CO M M ENT Another way to obtain the graph of an equation
is to
use a graphing util ity. Read Sec _
tion 2, Using a Graphing Utility to Graph Equations, i n the Appendix.
Figure 1 4 G raph crosses y-axis
Two techniques that sometimes reduce the number of points required to graph an equation involve finding intercepts and checking for symmetry.
y
�
G raph
2 x
G raph touches x-axis
E XA M P L E 4
Find Intercepts from a Graph
The points, if any, at which a graph crosses or touches the coordinate axes are called the intercepts. See Figure 14. The x-coordinate of a point at which the graph cross es or touches the x-axis is an x-intercept, and the y-coordinate of a point at which the graph crosses or touches the y-axis is a y-intercept. For a graph to be complete, all its intercepts must be displayed. F i n d i n g I ntercepts from a G raph
Find the intercepts of the graph in Figure 15 shown on p. x-intercepts? What are its y-intercepts?
166.
What are its
1 66
CHAPTER 2
Figure 1 5
y 4
Graphs
The intercepts of the graph are the points
Sol ution
( -3, 0), (0, 3), (0, 3)
(0, -3.5 ) , (4.5, 0)
3 4 . . The x-mtercepts are -3, 2 ' and 4.5; the y-mtercepts are -3.5, - "3 ' and 3.
•
In Example 4, you should notice the following usage: If we do not specify the type of intercept (x- versus y-), then we report the intercept as an ordered pair. However, if we specify the type of intercept, then we only report the coordinate of the intercept. For x-intercepts, we report the x-coordinate of the intercept; for y-intercepts, we report the y-coordinate of the intercept.
( 0 , - 3.5)
� ... ....". -
3
� �
(%, O} (o, -�}
COM M ENT For many equations, finding
Find Intercepts from an Equation
The intercepts of a graph can be found from its equation by using the fact that points on the x-axis have y-coordinates equal to 0 and points on the y-axis have x-coordinates equal to
O.
intercepts m a y not be so easy. In such
Procedure for Finding Intercepts 1.
cases, a graphing utility can be used. Read the first part of Section 3, Using
2.
a Graphing Utility to Locate Intercepts and Check for Symmetry, in the Appen dix, to find out how a graph ing utility _
locates intercepts.
EXAM P L E 5
Now Work P R O B L E M 3 9 ( a )
To find the x-intercept(s), if any, of the graph of an equation, let y = 0 in the equation and solve for x, where x is a real number. To find the y-intercept(s), if any, of the graph of an equation, let x = 0 in the equation and solve for y, where y is a real number.
F i n d i n g I ntercepts from an E qu ation
Find the x-intercept(s) and the y-intercept(s) of the graph of y y = x2 - 4 by plotting points. Solution
To find the x-intercept(s), we let y
=
=
x2 - 4. Then graph
0 and obtain the equation
x2 - 4 = 0 y = x2 - 4 with Y = 0 (x + 2)(x - 2) 0 Factor. x+2=0 x - 2 = 0 Zero-Product Property or x -2 or Solve. X = 2 The equation has two solutions, -2 and 2. The x-intercepts are -2 and 2. To find the y-intercept(s), we let x = 0 in the equation. y = x2 - 4 = 02 - 4 = -4 The y-intercept is -4. Since x2 2:: 0 for all x, we deduce from the equation y x2 - 4 that y 2:: -4 for all x. This information, the intercepts, and the points from Table 3 enable us to graph y = x2 - 4. See Figure 16. =
=
=
Ta ble 3
x
y = x2 - 4
(x, y)
-3
5
( - 3, 5)
-1
-3
(- 1 , - 3)
-3
( 1 , - 3)
3
5
(3, 5)
Figure 1 6
5 x
• &:m!J:= =--
Now Work
PROBLEM 2 1
SECTION 2.2
4
Graphs of Equations in Two Variables; Intercepts; Symmetry
167
Test an Equation for Symmetry with Respect to the x-Axis, the y-Axis, and the Origin
We just saw the role that intercepts play in obtaining key points on the graph of an equation. Another helpful tool for graphing equations involves symmetry, particu larly symmetry with respect to the x-axis, the y-axis, and the origin. DEFINITION
A graph is said to be symmetric with respect to the x-axis if, for every point ( x, y) on the graph, the point (x, - y) is also on the graph.
-.J
Figure 17 illustrates the definition. When a graph is symmetric with respect to the x-axis, notice that the part of the graph above the x-axis is a reflection or mirror image of the part below it, and vice versa. Figure 1 7 Symmetry with respect to the x-axis
P o i nts Symm etric with Respect to the x-Axis
EXAM P L E 6
If a graph is symmetric with respect to the x-axis and the point (3, 2) is on the graph, then the point (3, 2 ) is also on the graph. -
DEFINITION
Figure 1 8
A graph is said to be symmetric with respect to the y-axis if, for every point ( x, y) on the graph, the point ( -x, y) is also on the graph.
Symmetry with respect to the y-axis y
(-x, y)
( x, y)
x
E XA M P L E 7
•
-.J
Figure 18 illustrates the definition. When a graph is symmetric with respect to the y-axis, notice that the part of the graph to the right of the y-axis is a reflection of the part to the left of it, and vice versa.
P o i nts Sym m etric with Respect to the y-Axi s
If a graph is symmetric with respect to the y-axis and the point (5, 8) is on the graph, then the point ( -5, 8) is also on the graph. DEFINITION
•
A graph is said to be symmetric with respect to the origin if, for every point (x, y) on the graph, the point ( -x, - y) is also on the graph.
Figure 1 9 Sym metry with respect to the origin y ( x, y)
Figure 19 illustrates the definition. Notice that symmetry with respect to the ori gin may be viewed in three ways : 1.
x ( - x, -y)
-.J
2.
3.
As a reflection about the y-axis, followed by a reflection about the x-axis As a projection along a line through the origin so that the distances from the origin are equal As half of a complete revolution about the origin
1 68
CHAPTER 2
Graphs
EXAM P L E 8
Poi nts Sym m etric with Respect to the Origin
If a graph is symmetric with respect to the origin and the point (4, 2) is on the graph, then the point ( - 4, -2) is also on the graph. .... ,Ii .
>-
•
Now Work P R O B L E M S 2 9 A N D 3 9 ( b )
When the graph of an equation is symmetric with respect to a coordinate axis or the origin, the number of points that you need to plot in order to see the pattern is reduced. For example, if the graph of an equation is symmetric with respect to the y-axis, then, once points to the right of the y-axis are plotted, an equal number of points on the graph can be obtained by reflecting them about the y-axis. Because of this, before we graph an equation, we first want to determine whether it has any symmetry. The following tests are used for this purpose. Tests for Symmetry To test the graph of an equation for symmetry with respect to the
x-Axis y-Axis Origin
E XA M P L E 9
Testin g an Equation for Symm etry
Test y Solution
Replace y by -y in the equation and simplify. If an equivalent equation results, the graph of the equation is symmetric with respect to the x-axis. Replace x by -x in the equation and simplify. If an equivalent equation results, the graph of the equation is symmetric with respect to the y-axis. Replace x by - x and y by -y in the equation and simplify. If an equivalent equation results, the graph of the equation is symmetric with respect to the origin.
=
4x2 --- for symmetry. x2 + 1
x-Axis: To test for symmetry with respect to the x-axis, replace y by -y. Since 4x2 4x2 -Y = - - is not equivalent to y = ---, the graph of the equation x2 + 1 x2 + 1 is not symmetric with respect to the x-axis. y-Axis: To test for symmetry with respect to the y-axis, replace x by -x. Since 4( -x f 4x2 4x2 = --- is equivalent to y = --- , the graph of the Y = 2 ( -x ) + 1 x 2 + 1 x2 + 1 equation is symmetric with respect to the y-axis. Origin: To test for symmetry with respect to the origin, replace x by -x and y by -y o 4( - X ) 2 -y = Replace x by -x and y by -yo ( -x)2 + 1 -
-Y =
4x2 x2 + 1
Sim pl ify.
4x2 M u ltiply both sides by 1 x2 + 1 Since the result is not equivalent to the original equation, the graph of the 4x2 equation y = --- is not symmetric with respect to the origin. x2 + 1 • y
=
-
-
.
SECTION 2.2
Graphs of Equations in Two Variables; Intercepts; Symmetry
169
Seeing the Concept
4x2 Figure 20 shows the g raph of y = -2-- using a graphing utility. Do you see the symmetry with rex + 1 spect to the y-axis?
Figure 20
5
5
-5 Iiml::==:- '- Now Work P R O B l E M S 9 5
Know How to Graph Key Equations
The next three examples use intercepts, symmetry, and point plotting to obtain the graphs of key equations. It is important to know the graphs of these key equations because we use them later. The first of these is y = x3. EXAM P L E 1 0
G raphi n g the E q u ation y = x3 by F i n d i ng I ntercepts, C hecking for Symmetry, and P l otting Points
Graph the equation y = x3 by plotting points. Find any intercepts and check for symmetry first. Solution
First, we seek the intercepts. When x = 0, then y 0; and when y = 0, then x = O. The origin (0, 0) is the only intercept. Now we test for symmetry. =
Replace y by -yo Since -y x3 is not equivalent to y x3, the graph is not symmetric with respect to the x-axis. Replace x by -x. Since y = ( - x)3 = - x3 is not equivalent to y x3, the graph is not symmetric with respect to the y-axis. Replace x by -x and y by -y. Since -y = ( - x ) 3 - x3 is equivalent to y = x3 (multiply both sides by - 1), the graph is symmetric with respect to the origin.
x-Axis:
=
y-Axis:
=
=
Origin:
=
To graph y = x3, we use the equation to obtain several points on the graph. Be cause of the symmetry, we only need to locate points on the graph for which x ;:::: O. See Table 4. Since (1, 1 ) is on the graph, and the graph is symmetric with respect to the origin, the point ( - 1 , -1) is also on the graph. Plot the points from Table 4. Figure 21 shows the graph.
Table 4
x
y=r
(x, y)
a
a
(0, 0)
1
(1, 1 )
2
8
(2, 8)
3
27
(3, 27)
Y
Figure 2 1
8
(2 , 8)
6
-6
(-2, -8)
-8
x
•
1 70
CHAPTER 2
Gra phs G raphing the E q u ation x = y
E XA M P L E 1 1
Graph the equation x =
y 6
Figure 22 x = y2
(9, 3)
-2 ( 1 , -1 )
Figure 23 Y = \IX
Y 6
�
r.�. il
(9, -3)
If we restrict y so that y � 0, the equation x = l, y � 0, may be written equiv alently as y = \IX. The portion of the graph of x = l in quadrant I is therefore the graph of y = \IX . See Figure 23. Y, = Vx a nd 6
-6
1 10
-3 -2 2 3 1
1
10
y = x 1
10
3 2 1
-2 -3 1
10
(x, y)
1 , 10 10
( ) G, 3 ) G' 2 ) ( 2,D ( 3,�) ( �)
(1 , 1 )
1 0,
1
=
.
Vx
.
x
=
y2 on a g ra ph i n g calculator, you wi l l need to graph
We discuss why in Chapter 3. See Fig u re 24.
�
-�
10
\
G raph ing the E q u ation y
Graph the equation y
Solution
Y2
-----
---
-2
Table 5
-
•
CO M M ENr To see the graph of the equation
two equations:
Figure 24
x
l. Find any intercepts and check for symmetry first.
The lone intercept is (0, 0). The graph is symmetric with respect to the x-axis. (Do you see why? Replace y by - y.) Figure 22 shows the graph.
Solution
E XA M P L E 1 2
2
=
•
x 1
1.-. Find any intercepts and check for symmetry first. x
We check for intercepts first. If we let x = 0, we obtain ° in the denominator, which makes y undefined. We conclude that there is no y-intercept. If we let y = 0, we get the equation
1.-x = 0, which has no solution. We conclude that there is no
x-intercept. The graph of y =
1.-x does not cross or touch the coordinate axes.
Next we check for symmetry:
- -x1 ,
' · y b y - y YIelds y x-Axis: Rep I acmg
=
W
' h IS h IC " not eqUivalent to y =
-x1 .
1 1 y-Axis: Replacing x by -x yields y = - = - - , which is not equivalent to x -x 1 y = �. Origin: Replacing x by -x and y by -y yields - y _ 1.-, which is equivalent to x 1 y = . The graph is symmetric with respect to the origin. x
-
=
Now, we set up Table 5, listing several points on the graph. Because of the sym metry with respect to the origin, we use only positive values of x. From Table 5 we
SECTION 2.2
infer that if x is a large and positive number, then y
Figure 25 y
3
(-}, 2 )
(-2 -.1.) 2 (-1 . - 1 ) •
(-t. -2)
=
1 71
1.x is a positive number close
to O. We also infer that if x is a positive number close to 0, then
y =
1. is a large x
and positive number. Armed with this information, we can graph the equation. Figure 25 illustrates some of these points and the graph of y
x
-3 3
f f f f
Graphs of Equations in Two Variables; I ntercepts; Symmetry
=
1.. Observe how x
the absence of intercepts and the existence of symmetry with respect to the origin were utilized.
•
COM M ENT Refer to Example 3 in the Appendix, Section 3, for the graph of y =
-3
graphing utility.
1
x
using a •
2.2 Assess Your Understanding
'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1. Solve the equation 2(x + 3) 2. Solve the equation x2 - 9 0. (p. 86-93) 1 = -7. (pp. 86-93)
-
=
Concepts and Vocabulary 3. The points, if any, at which a graph crosses or touches the coordinate axes are called
7. If the graph of an equation is symmetric with respect to the origin and (3, -4) is a point on the graph, then is also a point on the graph.
_ _ _ _
_ _ _ _
4. The x-intercepts of the graph of an equation are those x-values for which
8.
_ _ _ _ _
5. If for every point (x, y ) on the graph of an equation the point ( - x, y) is also on the graph, then the graph is symmetric with respect to the
9.
_ _ _ _ _
6. If the graph of an equation is symmetric with respect to the yaxis and -4 is an x-intercept of this graph, then is also an x-intercept.
True or False
equation, let x
To find the y-intercepts of the graph of an and solve for y.
= °
True or False The y-coordinate of a point at which the graph crosses or touches the x-axis is an x-intercept.
If a graph is symmetric with respect to the x-axis, then it cannot be symmetric with respect to the y-axis.
10. True or False
_ _ _ _
Skill Building In Problems ll-I6, determine which of the given points are on the graph of the equation. 12. Equation: y x3 - 2 Vx 11. Equation: y = X4 Vx Points: (0, 0) ; ( 1 , 1 ); ( - 1 , 0) Points: (0, 0); ( 1 , 1 ); ( 1, - 1 )
-
14.
=
=
15. Equation: x2 + / 4 Points: (0, 2 ) ; ( -2, 2); (V2, V2)
Equation: i = x + 1 Points: ( 1 , 2 ); (0, 1 ); ( - 1 , 0)
13. Equation: / = x2 + 9 Points: (0, 3); ( 3 , 0 ) ; ( - 3 , 0 ) 16. Equation: x2 + 4/
=
4
(D
Points: (0, 1 ) ; (2, 0); 2,
In Problems 17-28, find the intercepts and graph each equation by plot/ing points. Be sure to label the intercepts. 20. Y 3x - 9 19. y = 2 x + 8 17. y = x + 2 18. y = x - 6 21. Y =
22. y
x2 - 1
25. 2x + 3y
=
6
=
x2 - 9
26. 5 x + 2y
=
23. y
=
-x2 + 4
27. 9x2 + 4y = 36
10
=
24.
Y =
-x2 + 1
28. 4x2 + y
=
4
In Problems 29-38, plot each point. Then plot the point that is symmetric to it with respect to (a) the x-axis; (b) the y-axis; (c) the origin. 33. (5, -2) 32. (4, -2 ) 29. (3, 4) 30. (5, 3 ) 31. ( -2, 1 ) 34. (-1, - 1 )
35. ( -3, -4)
36. (4, 0)
37. (0, -3)
38. ( -3, 0)
1 72
CHAPTER 2
Graphs
In Problems 39-50, the graph of an equation is given. (a) Find the intercepts. (b) Indicate whether the graph is symmetric with respect to the x-axis, the y-axis, or the origin.
40.
39.
41.
-3
43.
Y+
44.
y
3
�
-4 45.
3
3x
-3 1
'
-6
3 x
-2
48.
8
-2 3
/
8
49.
1 /,
50.
\I�
-2
y-axis
52.
-8
x-axis
-8
53.
y
y 4
y
(0, 0\
-5
59. x
2
x +
4
+ y - 9
=0
63. Y
= x3 - 27
67. Y
=
3x 2
x
+
9
56.
/=
x + 9
60.
x2 -
Y
64. Y =
68. y
X4
-
4
- 1
57. y
=
0
2 x - 4 = 2:;:-
In Problems 71-74, dra w a quick sketch of each equation. 71. y = x
3
72. x = /
75.
If (3, b ) is a point on the graph of y =
4x +
77.
If ( a, 4) is a point on the graph of y =
x
2
+
54. y -a xis
('IT,
0)
( 2 , 2)
x
3X
-3
-2
= -0X 2
+ 4/
65. Y
=
x2
69. y
=
73. Y
= Vx
61. 9x
-
58. y
= 36
x
_
62.
= -0'X
4x2
66. y
9
70. Y =
74. Y
/=4
+
3x - 4
-x3 2
14 '�-
/
-4
In Problems 55-70, list the intercepts and test for symmetry.
/=
+ -4
Origin
-5
55.
x
4
-4
2
In Problems 51-54, draw a complete graph so that it has the type of symmetry indicated. 51.
6
-3
W \\
�
46.
-3
3'
6
,] 47.
3 x
-3
3x
4-
-3
42.
y
= x2
+ 4
X4
+ 1
5 2x
=�
1 , what is b?
76.
If ( -2, b) is a point on the graph of 2x
3 x , what is a?
78.
If (a, -5) is a point on the graph of y
1
+
=
3 y = 2, what is b ? x2
+ 6x,
what is a?
SECTION 2.3
Lines
1 73
Applications and Extensions 79.
80.
81.
82. 83.
Given that the point ( 1 , 2) is on the graph of an equation that is symmetric with respect to the origin, what other point is on the graph? If the graph of an equation is symmetric with respect to the y-axis and 6 is an x-intercept of this graph, name another x-intercept. If the graph of an equation is symmetric with respect to the origin and -4 is an x-intercept of this graph, name another x-intercept. If the graph of an equation is symmetric with respect to the x-axis and 2 is a y-intercept, name another y-intercept. Micro phones In studios and on stages, cardioid micro phones are often preferred for the richness they add to voices and for their ability to reduce the level of sound from the sides and rear of the microphone. Suppose one such cardiod pattern is given by the equation (x2 + l - xf = x2 + i. (a) Find the intercepts of the graph of the equation. (b) Test for symmetry with respect to the x-axis, y-axis, and origin.
84.
Solar Energy The solar electric generating systems at Kramer Junction, California, use parabolic troughs to heat a heat-transfer fluid to a high temperature. This fluid is used to generate steam that drives a power conversion system to produce electricity. For troughs 7.5 feet wide, an equation for the cross-section is 161 120x - 225. =
(a) Find the intercepts of the graph of the equation. (b) Test for symmetry with respect to the x-axis, y-axis, and origin. Source: u.s. D epartment of Energy
Source: www.notaviva.com
Discussion and Writing lxi, and y = ( Vx )2, not y = 0 , y = x, y ing which graphs are the same. (b) Explain why the graphs of y 0 and y = Ixl are the same. (c) Explain why the graphs of y x and y = ( Vx)2 are not the same. (d) Explain why the graphs of y 0 and y x are not the same. Explain what is meant by a complete graph. Draw a graph of an equation that contains two x-intercepts; at one the graph crosses the x-axis, and at the other the graph touches the x-axis.
In Problem 85, y o u m ay u s e a graphing utility, but i t is n o t required.
85. (a) G raph
=
88.
=
89.
=
=
86. 87.
=
90.
Make up an equation with the intercepts (2, 0), (4, 0), and (0, 1 ). Compare your equation with a friend's equation. Com ment on any similarities. Draw a graph that contains the points ( -2, - 1 ) , (0, 1 ) , ( 1 , 3 ) , and (3, 5 ) . Compare your graph with those of other students. Are most of the graphs almost straight lines? How many are "curved"? D iscuss the various ways that these points might be connected. An equation is being tested for symmetry with respect to the x-axis, the y-axis, and the origin. Explain why, if two of these symmetries are present, the remaining one must also be present.
'Are You Prepared ?' Answers 1.
{ -6 }
2.3
2.
{-3,3}
Lines OBJECTIVES
1
Calcu late a n d I nterpret the S lope of a Li n e (p. 1 74)
2
G raph Lines Given a Poi nt a n d the Slope (p. 1 76)
4
Use the Point-Slope Form of a Line; Identify Horizontal Li n es (p. 1 78)
6
Write the Equation of a Line in Slope-I ntercept Form (p. 1 79)
3
5
7
8 9
Find the Equation of a Vertical Line (p. 1 7 7) Find the Equation of a Line Given Two Poi nts (p. 179)
I dentify the Slope a n d y-Intercept of a Line from Its Eq uation (p. 180) Graph Li n es Written in General Form Using I ntercepts (p. 1 8 1) Find Equations of Para l lel Lines (p. 1 82)
1 0 Find Equations of Perpendicular Li n es (p. 1 83)
1 74
CHAPTER 2
Graphs
In this section we study a certain type of equation that contains two variables, called a linear equation, and its graph, a line. 1
Calculate and Interpret the Slope of a Line
Consider the staircase illustrated in Figure 26. Each step contains exactly the same horizontal run and the same vertical rise. The ratio of the rise to the run, called the slope, is a numerical measure of the steepness of the staircase. For example, if the run is increased and the rise remains the same, the staircase becomes less steep. If the run is kept the same, but the rise is increased, the staircase becomes more steep. This important characteristic of a line is best defined using rectangular coordinates.
Figure 26
DEFINITION
Let
P
slope
=
m
formula
(Xl , Y l ) and Q = (X2 ' Y2 ) be two distinct points. If Xl "* X2 , the of the nonvertical line L containing P and Q is defined by the ( 1)
Y2 - Y l X2 - X l If X l = X2 , L is a vertical results in division by 0).
line
and the slope
In
of L is
undefined
..J
(since this
Figure 27(a) provides an illustration of the slope of a nonvertical line; Figure 27(b) illustrates a vertical line.
L
Figure 27
(a) Slope of L is m
=
y _) -y 2
x 2
(b)
1
Slope is u ndefined;
L is vertical
As Figure 27(a) illustrates, the slope m of a nonvertical line may be viewed as n1 =
Y2 - Yl
Rise Run
Two comments about computing the slope of a nonvertical line may prove helpful: 1.
Any two distinct points on the line can be used to compute the slope of the line. (See Figure 28 for justification.)
Triangles ABC and PQR are similar (equ a l ang les), so ratios of correspond ing sides are proportional. Then
Figure 28
.
y
Y2 - Yl
Slope uSing P and Q = -- =
d(B, C) d(A, C)
-
=
X2 - Xl
Slope using A and B
x
SECTION 2.3
2.
Lines
1 75
The slope of a line may be computed from P = (X l , Y1 ) to Q = (X2 , Yz) or from Q to P because Y2 - YI x2 - X I Finally, we can also express the slope 111 of a nonvertical line as 111 =
Yz - Y1
Change in Y Change in X
�Y �X
That is, the slope 111 of a non vertical line L measures the amount that Y changes as �Y X changes from Xl to X2 ' The expression is called the average rate of change �x of Y with respect to x. Since any two distinct points can be used to compute the slope of a line, the average rate of change of a line is always the same number. EXAM P L E 1
F i n d i n g and I nterpreting the Slope of a L i n e G iven Two Points
The slope 111 of the line containing the points ( 1 , 2) and (5, - 3) may be computed as In
=
2 - ( -3 ) -3 - 2 -5 5 = - = - - or as n1. = 5 - 1 4 4 1 -5
5 -4
5 4
For every 4-unit change in x, Y will change by -5 UIilts. That is, if X increases by 4 units, then y will decrease by 5 units. The average rate of change of y with respect 5 . to X IS - - . 4 • Wi
*--
Now Work
P R O B L E M S
1 1 AND 1 7
To get a better idea of the meaning of the slope following example. E XA M P L E 2
111
of a line L, consider the
F i n d ing the S lopes of Various Lines Contai n ing the Same Point (2, 3)
Compute the slopes of the lines L1 , L2 , L3 , and L4 containing the following pairs of points. Graph all four lines on the same set of coordinate axes. L1 : L2 : L3: L4: Solution
Figure 29
P = (2, 3)
P = (2, 3)
P = (2, 3) P = (2, 3)
Ql Q2 Q3 Q4
= ( - 1, -2) = (3, - 1 )
= (5 3 ) = (2, 5 ) ,
Let 111 1 , 1112 , 1113 , and 1114 denote the slopes of the lines L I , L� , L3 , and L.. , respectively. Then -2 - 3 -5 5 A rise of 5 divided by a - 1 - 2 -3 3 - 1 3 -4 In" = = - = -4 3-2 1 3 -3 0 In, = =-=0 J 5 -2 3 111-+ is undefined because XI = X2 = 2 n1' 1 =
run of 3
-
---
--
m
1
=
� 3
m4
u n defined
m2 = -4
The graphs of these lines are given in Figure 29.
•
1 76
CHAPTER 2
Graphs
Figure 29 illustrates the following facts: 1.
2.
3. 4.
Figure 30
2
j
'�II Y6
= Ys
6x =
.f "
-3
. -. t-
=
x
..... Y3
........
�::::-:: �--:;.-:::O;�
.-- ....-:'--;j ... ./ /j ....
.:.
2x
Y4
./" '"
�
-2
"
./
3 Y1
� �
Seeing the Concept
=
Y1 = Y2
x
0
Y3
\2
4
1 = -x
1 = -x 2
Y4 = x See Figure 30.
�'. �
0
On the same screen, graph the following equations:
= x
-6x -2x -x \
= = =
Y6 Y4
=
I
Figure 31 Ys
Y2
When the slope of a line is positive, the line slants upward from left to right (Ll) ' When the slope of a line is negative, the line slants downward from left to right ( L2 ) ' When the slope is 0 , the line is horizontal ( L3)' When the slope is undefined, the line is vertical (L4) '
Ys
= 2x
Y6
= 6x
Slope of l ine is
O.
4 1
S lope of l ine is - .
Slope of l ine is
1
"2 '
Slope of l i n e is 1 .
Slope of line is 2. Slope of line is 6.
Seeing the Concept 0
On the same screen, graph the following equations: Y1
=
4 1 1
Slope of line is
O.
4 1
Y2
= - -x
S lope of line is - - .
Y3
= - -x 2
Slope of line is
- "2 '
Y4
= -x
Slope of l i ne is
-1 .
Ys
=
- 2x
Slope of l i ne is - 2 ,
Y6
= - 6x
Slope of l i ne is - 6,
See Figure 3 1 .
1
Figures 30 and 31 illustrate that the closer the line is to the vertical position, the greater the magnitude of the slope. 2
Graph Lines Given a Point and the Slope
The next example illustrates how the slope of a line can be used to graph the line. EXAM P L E 3
G raphing a Line Give n a Point and a Slope
Draw a graph of the line that contains the point (3, 2) and has a slope of: (a) Solution
3 4
4 (b) - 5
-
Rise . 3 . . The fact that the slope 1S - means that for every honzontal Run 4 movement (run) of 4 units to the right there will be a vertical movement (rise) of 3 units. Look at Figure 32. If we start at the given point (3, 2 ) and move 4 units to the right and 3 units up, we reach the point (7, 5 ) . By drawing the line through this point and the point (3, 2), we have the graph.
(a) Slope
= -
SECTION 2.3
y 6
Figure 32
Lines
1 77
(b) The fact that the slope is
(3 , 2��= Run
5
=
4
4 5
10
-4 5
Rise Run
means that for every horizontal movement of 5 units to the right there will be a corresponding vertical movement of -4 units (a downward movement). If we start at the given point (3, 2) and move 5 units to the right and then 4 units down, we arrive at the point ( 8, -2). By drawing the line through these points, we have the graph. See Figure 33. Alternatively, we can set
x
4 5
Figure 33
4 -5
Rise Run
so that for every horizontal movement of -5 units (a movement to the left) there will be a corresponding vertical movement of 4 units (upward). This ap proach brings us to the point ( -2, 6 ) , which is also on the graph shown in Figure 33. • ".".
3
.;>-
Now Work P R O B L E M 2 3
Find the Equation of a Vertical Line
Now that we have discussed the slope of a line, we are ready to derive equations of lines. As we shall see, there are several forms of the equation of a line. Let's start with an example. EXAM P L E 4
G raphing a Line
Graph the equation : x = 3 Solution
To graph x = 3, recall that we are looking for all points (x, y) in the plane for which x = 3. No matter what y-coordinate is used, the corresponding x-coordinate always equals 3. Consequently, the graph of the equation x = 3 is a vertical line with x-intercept 3 and undefined slope. See Figure 34. y 4
Figure 34
-1
-1
(3, 3) (3, 2 ) (3, 1 )
(3, 0) (3, -1 )
5
x
As suggested by Example 4, we have the following result: THEOREM
Equation of a Vertical Line
A vertical line is given by an equation of the form x=a where a is the x-intercept.
•
1 78
CHAPTER 2
Graphs
{ expression in x } . But x = 3 cannot be put in this form. To overcome this, most graph
COMM ENT To graph an equation using a g ra ph i n g util ity, we need to express the equation in the
form y =
ing utilities have special commands for drawing vertical l i nes. DRAW, LINE, PLOT, and VERT a re
a mong the more common ones. Consult your m a n u a l to determ ine the correct methodology for your •
g ra p h i n g util ity.
4
Figure 3S Y
L
( Xl , Yl ) See
Use the Point-Slope Form of a Line; I dentify Horizontal Lines
Now let L be a nonvertical line with slope m and containing the point Figure 35. For any other point ( X , y) on L, we have m=
Y - Yl
X - Xl
---
or Y - Y l = m(x -
Xl
'
)
x
THEOREM
Point-Slope Form of an Equation of a Line
An equation of a nonvertical line with slope m that contains the point ( X l , Y l ) is Y - Yl = m(x - Xl )
(2)
I�
� -----------------�
Using the Poi nt-Slope Form of a Line
EXAM P L E 5
An equation of the line with slope 4 and containing the point ( 1 , 2) can be found by using the point-slope form with m = 4, x l = 1, and Yl = 2.
Figure 36
( 1 , 2)
10
-2
Y - YI = m(x - X l ) Y - 2 = 4(x - 1 ) Y = 4x - 2 x
=
1, Y1 = 2
Solve for y.
See Figure 36 for the graph. I II ,'
EXAM P L E 6
m = 4, X1
•
;",... Now Work P R O B L E M 4 5
F i n d i n g the E q u ation of a Horizontal Line
Find an equation of the horizontal line containing the point (3, 2 ) . Solution
Figure 37 y 4 ( 3 , 2)
-1
3
5 x
Because all the y-values are equal on a horizontal line, the slope of a horizontal line is O. To get an equation, we use the point-slope form with m = 0, Xl = 3, and Yl = 2. Y - Yl = m(x - X l ) Y - 2 = O ' (X - 3 ) y-2=0 Y=2
See Figure 37 for the graph.
m = 0, X1 = 3, a n d Y1 = 2
•
SECTION 2.3
Lines
1 79
As suggested by Example 6, we have the following result: THEOREM
Equation of a Horizontal Line
A horizontal line is given by an equation of the form y=b where b is the y-intercept. 5
Find the Equation of a Line Given Two Poi n ts
We use the slope formula and the point-slope form of a line to find the equation of a line given two points. EXAM P L E 7
F i n d ing an Equation of a Line G iven Two Points
Find an equation of the line containing the points (2, 3 ) and ( -4, 5 ) . Graph the line. We first compute the slope of the line.
Solution
n1
Figure 38
=
5 -3 -4 - 2
-6
2
---
We use the point (2, 3) and the slope m = equation of the line.
1
1
3
_ l. to get the point-slope form of the 3
y - 3 = - - (x - 2) 3 See Figure 38 for the graph.
•
In the solution to Example 7, we could have used the other point, ( -4, 5), instead of the point (2, 3 ) . The equation that results, although it looks different, is equivalent to the equation that we obtained in the example. (Try it for yourself.) '1'= ;
6
'''''''
Now Work P R O B L E M 3 7
Write the Equation of a Line in Slope-Intercept Form
Another useful equation of a line is obtained when the slope m and y-intercept b are known. In this event, we know both the slope m of the line and a point (0, b) on the line; then we may use the point-slope form, equation (2), to obtain the follow ing equation: y - b = m(x - 0 ) o r y = mx + b THEOREM
Slope-Intercept Form of an Equation of a Line
An equation of a line with slope m and y-intercept b is y = mx + b
(3)
I�
�----------------------------------� = =t<1I!l;;
Now Work P R O B L E M S 1 ( E X P R E S S A N S W E R IN SLOPE-INTERCEPT FORM)
1 80
Graphs
CHAPTER 2
= Y4 3x 2 Y3 Y5= -X 3x 2 = X '.... ..\ l" . /\"\ . I ..... I \\ . ."..... .. Y
Figure 39
-
=
-6
."
,/
+
2
.../
,/
/'
Figure 40
+
mx + 2
\, 4 ,/
....'-!
.
..
.
... .
l
-4
.. ... ....
, .
� Seeing the Concept
2
To see the role that the slope m plays, graph the following lines on the same screen.
=X Y3 = Y,
2x + b
I
I
/ 11/ j , ''1; / I /.1 / '
/ .//�I I'I
'l
1/" .' /
'
I
1:,...0:'
+1
Y2
+ 2
=
3x + 2
- 3x + 2
See Figure 39. What do you conclude a bout the lines y Ys
..
= Y4 = 2X ,/, /Y1=3Y5=2x==2X2x2x-- 1 I·!GJ j fllII/ y
2
-x + 2
Y4
.
.
=
Y2
.
I.
+4 4
-6
Y2
' /
.. ..
..
+ +
=
,
y.
Y2Y3 == Y,
=
Y4
=
Ys
=
2x
2x + 2x -
1 1
2x + 4 2x
-
See Figure 40. What do you conclude a bout the lines y
7
mx + 2?
Seeing the Concept
6
-4
=
To see the role of the y-intercept b, graph the following lines on the same screen.
4
1
=
= 4
2x + b?
Identify the Slope and y-Intercept of a Line from Its Equation
When the equation of a line is written in slope-intercept form, it is easy to find the slope rn and y-intercept b of the line. For example, suppose that the equation of a line is y = -2x + 7 Compare it to y = rnx + b. y = -2x + 7 y
=
i
i
mx + b
The slope of this line is -2 and its y-intercept is 7. U!J ! � ,.-
EXAM P L E 8
Now Work P R O B L E M 7 1
F i n d i n g the Slope and y-I nte rcept
Find the slope equation. Solution
Figure 41
x
rn
and y-intercept b of the equation 2x + 4y = 8. Graph the
To obtain the slope and y-intercept, we write the equation in slope-intercept form by solving for y. 2x + 4y = 8 4y = -2x + 8 1 Y = - 2" x + 2 y mx + b The coefficient of x, -
�,
=
is the slope, and the y-intercept is 2. We can graph the line
using the fact that the y-intercept is 2 and the slope is -
�.
Then, starting at the point
(0, 2 ) , go to the right 2 units and then down 1 unit to the point (2, 1 ) . See Figure 41 . �-
Now Work P R O B L E M 7 7
•
SECTION 2.3
8
Lines
181
Graph Lines Written in General Form Using Intercepts
Refer to Example 8. The form of the equation of the line 2x + 4y = 8, is called the general form. DEFINITION
The equation of a line is in
general form*
when it is written as
Ax + By = C where A, B, and C are real numbers and A and B are not both
O.
(4)
When we want to graph an equation that is written in general form, we can solve the equation for y and write the equation in slope-intercept form as we did in Example 8. Another approach to graphing the equation would be to find its inter cepts. Remember, the intercepts of the graph of an equation are the points where the graph crosses or touches a coordinate axis. EXAM P L E 9
G raph ing an Equation in General Form Using Its I ntercepts
Graph the equation 2x + 4y = 8 by finding its intercepts. Solution
To obtain the x-intercept, let y = 0 in the equation and solve for x. 2x + 4y 2x + 4(0) 2x x
= = = =
8 8 8 4
Let y
= o.
Divide both sides by 2.
The x-intercept is 4 and the point (4, 0) is on the graph of the equation. To obtain the y-intercept, let x = 0 in the equation and solve for y. 2x + 4y = 8 2(0) + 4y = 8 4y 8 y=2
Figure 42
L et x
=
o.
=
x
Divide both sides by 4.
The y-intercept is 2 and the point (0, 2 ) is on the graph of the equation. We plot the points (4, 0) and (0, 2) on a Cartesian plane and draw a line through the points. See Figure 42.
•
� = = -
Now Work P R O B L E M 9 1
Every line has an equation that is equivalent to an equation written in general form. For example, a vertical line whose equation is x=a can be written in the general form 1 x + O·y = a A = A horizontal line whose equation is y=b can be written in the general form o x + 1·y = b A = '
.
1, B = 0, C
.
0, B
Some books use the term standard form.
=a
= 1, C =
b
182
CHAPTER 2
Graphs
Lines that are neither vertical nor horizontal have general equations of the form Ax + By =
C
A
=/= 0 a n d B =/= 0
Because the equation of every line can be written in general form, any equation equivalent to equation (4) is called a linear equation. Figure 43
9
y
Fin d Equations of Parallel Lines
When two lines (in the plane) do not intersect (that is, they have no points in common), they are said to be parallel. Look at Figure 43. There we have drawn two parallel lines and have constructed two right triangles by drawing sides paral lel to the coordinate axes. The right triangles are similar. (Do you see why? Two angles are equal.) Because the triangles are similar, the ratios of corresponding sides are equal. This suggests the following result: THEOREM
Criterion for Para l lel Lines
�
Two nonvertical lines are parallel if and only if their slopes are equal and they have different y-intercepts.
The use of the words "if and only if" in the preceding theorem means that actu ally two statements are being made, one the converse of the other. If two nonvertical lines are parallel, then their slopes are equal and they have different y-intercepts. If two nonvertical lines have equal slopes and they have different y-intercepts, then they are parallel.
E XA M P L E 1 0
S howing That Two Lines Are Paral lel
Show that the lines given by the following equations are parallel: Solution
Lr : 2x + 3y = 6, To determine whether these lines have equal slopes and different y-intercepts, we write each equation in slope-intercept form: L1 : 2x + 3y = 6 3y = -2x + 6 2 y = - "3 x + 2
Figure 44
Slope =
L 2 : 4x + 6y = 0 6y = -4x 2 y = - -x 3
-�; y-intercept = 2
Because these lines have the same slope, are parallel. See Figure 44.
Slope =
-
-�; y-intercept = 0
�, but different y-intercepts, the lines 3
•
E XA M P L E 1 1
F i nd i ng a Line That Is Paral lel to a G iven Line
Find an equation for the line that contains the point (2, -3 ) and is parallel to the line 2x + y = 6.
SECTION 2.3
Solution
Figure 45
1 83
Since the two lines are to be parallel, the slope of the line that we seek equals the slope of the line 2x + y = 6. We begin by writing the equation of the line 2x + y = 6 in slope-intercept form. 2x
Y
Lines
+ Y = y =
6
-2x
+
6
The slope is -2. Since the line that we seek also has slope -2 and contains the point (2, -3), we use the point-slope form to obtain its equation. Y 6 x
-6
y
- Y1 - ( -3 ) Y + 3
=
=
Y =
2x 2x+ Y= 1
=
+ y =
m(x - Xl) -2(x - 2 ) -2x + 4 -2x + 1 1
This line is parallel to the line 2x Figure 45. == l!')!;:;: --
Now Work
+ y =
Point-slope form
m = -2,X1 = 2'Y1 =
-3
Simplify. Slope-intercept form General form
6 and contains the point (2, -3).
See •
P R O B L EM S 9
10 Find Equations of Perpendicular Lines
When two lines intersect at a right angle (90°), they are said to be perpendicular. See Figure 46. Figure 46
Y
x
The following result gives a condition, in terms of their slopes, for two lines to be perpendicular. THEOREM
Criterion for Perpend icular Lines
Two nonvertical lines are perpendicular if and only if the product of their slopes is -1.
-.J
Here we shall prove the "only if" part of the statement: If two nonvertical lines are perpendicular, then the product of their slopes is -1. In Problem 128 you are asked to prove the " if" part of the theorem; that is: If two nonvertical lines have slopes whose product is -1 , then the lines are perpendicular.
1 84
CHAPTER 2
Figure 47
Graphs
y
Slope
m2
Proof Let ml and m2 denote the slopes of the two lines. There is no loss in gen erality (that is, neither the angle nor the slopes are affected) if we situate the lines so that they meet at the origin. See Figure 47. The point A = (1, m2) is on the line having slope m2, and the point B = (1, ml) is on the line having slope mI' (Do you see why this must be true?) Suppose that the lines are perpendicular. Then triangle OAB is a right triangle. As a result of the Pythagorean Theorem, it follows that
[d(O, A)]2 + [d(O, B)f = [dCA,B)f
(5)
Using the distance formula, we can write the squares of these distances as [d(O,A)f = (1- o? + (m2 - 0)2 = 1 + m� [d(o,B)f = (1 - 0)2 + (ml - 0)2 = 1 + mi [dCA,B)f = (1 - 1? + (m2 - m1? = m� - 2mIm2 + my Using these facts in equation (5), we get
(1 + m�) + (1 + mi) = m� - 2mIm2 + my which, upon simplification, can be written as If the lines are perpendicular, the product of their slopes is-1.
•
You may find it easier to remember the condition for two nonvertical lines to be perpendicular by observing that the equality m]m2 = -1 means that ml and m2 . . 1 1 are negatIve recIproca . . . Is 0f eac h other; that IS, eIther ml = - - or m2 = ml m2 -
E XA M P L E
E XA M P L E
12
Finding the Slope of a Line Perpendicular to Another Line
3 2 If a line has slope "2' any line having slope- 3' is perpendicular to it. 13
•
Finding the Equation of a Line Perpendicular to a Given Line Find an equation of the line that contains the point (1, -2) and is perpendicular to the line x + 3Y = 6. Graph the two lines.
Solution
We first write the equation of the given line in slope-intercept form to find its slope.
x + 3y = 6 3y = -x + 6 1
y = -3' x + 2 The given line has slope
-
Proceed to solve for y. Place
in
the form y = mx + b.
�. Any line perpendicular to this line will have slope
3.
Because we require the point (1, -2) to be on this line with slope 3, we use the point-slope form of the equation of a line.
y - YI = m(x - Xl) Y- (-2) = 3(x - 1)
Point-slope form m = 3,x, = 1,y, =
-2
SECTION 2.3
Lines
1 85
To obtain other forms of the equation, we proceed as follows:
Figure 48
y + 2 =
=
y + 2
3x
-
Figure 48 shows the graphs.
y =
y =
Now Work P R O B L E M
to= -==
3 ( x - 1) 3x - 3 3x - 5 5
Simplify. Slope-intercept form General
form •
65
WARNING Be sure to use a square screen when you graph perpendicular lines. Otherwise, the angle
!§,!
between the two lines will appear distorted. A discussion of square screens is given in section 5 of the Appendix. _
2.3 Assess Your Understanding
Concepts and Vocabulary 1. The slope of a vertical line is zontal line is =
____
; the slope of a hori-
7. Two nonvertical lines have slopes m[ and m 2 , respectively. The lines are parallel if are and the unequal; the lines are perpendicular if
___ _
2. For the line 2x + 3y the y-intercept is
6, the x-intercept is ____ and
___ _
___ _
8. The lines y= 2x + 3 and y= ax + 5 are parallel if a =
3. A horizontal line is given by a n equation of the form , where b is the ____
9. The lines y=2x - 1 and y a=
____
4.
True or False
Vertical lines have an undefined slope.
5.
True or False
The slope of the line 2y = 3x + 5 is 3.
6.
True or False
The point ( 1 , 2) is on the line 2x + y = 4.
10.
=
ax + 2 are perpendicular if
True or False Perpendicular lines have slopes that are rec iprocals of one another.
Skill Building In Problems 11-14, (a) find the slope of the line a n d (b) interpret the slope. 13. 12. 11. y y 2
(-2, 1 ) 2
( 0,0) -2
y
14.
-1
In Problems 15-22, plot each pair of points and determine the slope of the line containing them. Graph the line. 15. (2, 3 ) ; (4, 0 )
16. ( 4, 2); (3, 4)
17. ( -2, 3 ) ; (2, 1 )
18. ( - 1 , 1 ); (2, 3 )
19. ( -3, - 1 ) ; (2, - 1 )
20. (4, 2); ( -5 , 2)
21. ( - 1 , 2); ( - 1 , -2)
22. (2, 0 ) ; (2, 2)
In Problems 23-30, grap h the line containing the point P and having slope m.
'- 23.
P = ( 1 , 2); m = 3
27. P = (-1 , 3 ) ;m
In Problems may vary.
31 -36,
=
24. P = (2, 1); m = 4 0
28. P
=
(2,
-4 ) ; m
= 0
25. P = (2, 4); m =
29. P = (0, 3 ) ; slope undefined
[Hint: It is not necessary to find the equation of the line. See Example
4
4
30. P = (-2, 0 ) ; slope undefined
the slope and a point on a line are given. Use this information to locate three additional points on the line. Answers
31. Slope 4; point ( 1 , 2) 34. Slope
2 26. P = ( 1 " 3 ) ' m = - 5
-�
. '3; pomt ( -3, 2)
3.]
-�;
point (2, -4)
32. Slope 2; point ( -2, 3 )
33. Slope
35. Slope -2; point ( -2, -3)
36. Slope -1; point (4, 1 )
1 86
CHAPTER 2
In Problems
jtnd an equation of the line L.
37-44,
a
37.
Graphs
38. 10;\
L
Y (-2,1) 2
39. L
�
(-1, ) 3 3
( 0, 0 )
-1
41.
.J
Y4
.J
3 y=2x L is parallel to y
=
42.
2x
,"-y.\.
(1,1)
�
Y.\.
43.
-2
L'"
.J
(2,2) _L 2
-1
x
Y
44.
y=2x L is perpendicular to y=2x
L is parallel to y=-x
Y
40.
y=-x L is perpendicular to y=-x
In Problems 45-70, find an equation for the line with the given properties. Express your answer using either the general form or the slope -intercept form of the equation of a line, whichever you prefe r. 45. Slope = 3 ; containing the point ( -2 , 3 ) 47. Slope
=
-�;
46. Slope
containing the point (1, - 1 )
=
48. Slope =
2 ; containing the point ( 4 , -3 )
�;
containing the point ( 3 , 1)
49. Containing the points (1, 3 ) and ( -1, 2 )
50. Containing the points (- 3 , 4) and (2, 5 )
51. Slope = - 3 ; y-intercept = 3
52. Slope = - 2 ; y-intercept = - 2
53. x-intercept=2 ; y-intercept = -1
54. x-intercept = - 4 ; y-intercept=4
55. Slope undefined; containing the point ( 2 , 4)
56. Slope undefined; containing the point ( 3 , 8)
57. Horizontal; containing the point (- 3 , 2 )
58. Vertical; containing the point (4, - 5 )
'- 59.
Parallel to the line y = 2x ; containing the point ( -1, 2)
61. Parallel to the line 2x
-y=
- 2 ; containing the point (0, 0)
60. Parallel to the line y= -3x; containing the point ( - 1, 2) 62. Parallel to the line x - 2y = -5; containing the point (0, 0)
63. Parallel to the line x = 5 ; containing the point (4, 2 )
64. Parallel to the line y = 5; containing the point ( 4 , 2 )
65. Perpendicular t o the line y = 1:. x + 4; containing the 2 point ( 1 , -2)
66. Perpendicular t o the line y = 2 x - 3 ; containing the point (1, -2)
67. Perpendicular to the line 2x + y = 2 ; containing the point (- 3 , 0)
68. Perpendicular to the line x - 2y= -5; containing the point (0, 4)
69. Perpendicular to the line x=8; containing the point ( 3 , 4)
70. Perpendicular to the line y = 8; containing the point ( 3 , 4)
In Problems
71-90,
,71. Y = 2x + 3
1 76. Y = 2x +" 2
find the slope and y-intercept of each line. Graph the line. 72. y = -3x + 4
" 77. x + 2y = 4
1 Y = x -I "2
1 74. " x + y = 2 3
1 75. Y = " x + 2 2
78. - x + 3y = 6
79. 2x - 3y = 6
80. 3x + 2y = 6
73.
81. x + y=1
82. x - y = 2
83. x = -4
84. y = -1
85. y =5
86. x = 2
87. Y - x=0
88. x + y = 0
89. 2y - 3x=0
90. 3x + 2y = 0
SECTION 2.3
In Problems
91-100,
92. 3x - 2y = 6
93. - 4x + 5y = 40
94. 6x - 4y=24
95. 7x + 2y = 21
96. 5x + 3y = 18
1 1 97. - x + - y = 1 2 3
2 98. x - - y = 4 3
99. 0.2x - 0.5y = 1
101. Find an equation of the x-axis. 103-106,
103. y = 2x - 3
1 87
(a) find the intercepts of the graph of each equation and (b) graph the equation.
91. 2x + 3y=6
In Problems
Lines
100. -0.3x + O.4y = 1.2
102. Find an equation of the y-axis.
the equations of two lines are given. Determine if the lines are parallel, perpendicular, or neither. 104. y =
y=2x + 4
y
=
1 2:x - 3 - 2x + 4
105. y = 4x + 5
106. y = -2x + 3
y = - 4x + 2
1 y = - 2:x + 2
In Problems 1 0 7-110, write an equation of each line. Express your answer using either the general form or the slope-intercept form of the equation of a line, whichever you prefer. 107.
108.
2
�
""
-4
109.
-2
,."
'-"....
3
-3
....----. .. .
---
2
1----:---
�
2
110. 3
-2
-3 '-..
3
-2
Applications and Extensions 111. Geometry Use slopes to show that the triangle whose ver tices are ( - 2, 5 ) , (1,3), and ( -1,0) is a right triangle. 112. Geometry Use slopes to show that the quadrilateral whose vertices are (1, -1), ( 4,1), (2, 2 ) , and (5, 4) is a parallelo gram. 113. Geometry Use slopes to show that the quadrilateral whose vertices are ( -1, 0), ( 2, 3), (1, - 2 ) , and ( 4, 1) is a rectangle. 114. Geometry Use slopes and the distance formula to show that the quadrilateral whose vertices are (0, 0) , (1, 3 ) , ( 4,2 ) , and ( 3, -1) is a square. 115. Truck Rentals A truck rental company rents a moving truck for one day by charging $29 plus $0.20 per mile. Write a lin ear equation that relates the cost C, in dollars, of renting the truck to the number x of miles driven. What is the cost of renting the truck if the truck is driven 110 miles? 230 miles? 116. Cost Equation The fixed costs of operating a business are the costs incurred regardless of the level of production. Fixed costs include rent, fixed salaries, and costs of leasing ma chinery. The variable costs of operating a business are the costs that change with the level of output. Variable costs in clude raw materials, hourly wages, and electricity. Suppose that a manufacturer of jeans has fixed daily costs of $500 and variable costs of $8 for each pair of jeans manufactured. Write a linear equation that relates the daily cost C, in dollars, of manufacturing the jeans to the number x of jeans manufac tured. What is the cost of manufacturing 400 pairs of jeans? 740 pairs? 117. Cost of Sunday Home Delivery The cost to the Chicago Tribune for Sunday home delivery is approximately $0.53 per newspaper with fixed costs of $1,070,000. Write a linear
equation that relates the cost C and the number x of copies delivered. Source: Chicago Tribune, 2002. 118. Wages of a Car Salesperson Dan receives $375 per week for selling new and used cars at a car dealership in Oak Lawn, Illinois. In addition, he receives 5% of the profit on any sales that he generates. Write a linear equation that relates Dan's weekly salary S when he has sales that generate a profit of x dollars. 119. Electricity Rates in Illinois Commonwealth Edison Com pany supplies electricity to residential customers for a month ly customer charge of $7.58 plus 8.275 cents per kilowatt-hour for up to 400 kilowatt-hours.
f (a) Write a linear equation that relates the monthly charge C, in dollars, to the number'x of kilowatt-hours used in a month, 0 ::; x ::; 400. (b) Graph this equation.
1 88
CHAPTER 2
G ra p h s
(c) What is the monthly charge for using 100 k ilowatt hours? (d) What is the monthly charge for using 300 kilowatt hours? (e) Interpret the slope of the line. Source: Commonwealth Edison Company, April, 2006.
120. Electricity Rates in Florida Florida Power & Light Company supplies electricity to residential customers for a monthly customer charge of $5.17 plus 1 0.07 cents per kilowatt-hour for up to 1000 kilowatt-hours. (a) Write a linear equation that relates the monthly charge C, in dollars, to the number x of kilowatt-hours used in a month, 0 :=; x :=; 1000. (b) Graph this equation. (c) What is the monthly charge for using 200 kilowatt hours? (d) What is the monthly charge for using 500 kilowatt hours? (e) Interpret the slope of the line. Source:
Florida Power & Light Company, April, 2006.
121. Measuring Temperature The relationship between Celsius (0C) and Fahrenheit (OF) degrees of measuring temperature is linear. Find a linear equation relating °C and of if O°C cor responds to 32°F and 100°C corresponds to 212°F. Use the equation to find the Celsius measure of 70°F. U2. Measuring Temperature The Kelvin (K) scale for measur ing temperature is obtained by adding 273 to the Celsius temperature. (a) Write a linear equation relating K and dc. (b) Write a linear equation relating K and of (see Prob lem 121). U3. Access Ramp A wooden access ramp is being built to reach a platform that sits 30 inches above the floor. The ramp drops 2 inches for every 25-inch run. y
(c) Design requirements stipulate that the maximum run be 30 feet and that the maximum slope be a drop of 1 inch for each 12 inches of run. Will this ramp meet the requirements? Explain. (d) What slopes could be used to obtain the 30-inch rise and still meet design requirements? Source: www.adaptiveaccess. com/woodJamps.php 124. Cigarette Use A study by the Partnership for a Drug-Free America indicated that, in 1998, 42% of teens in grades 7 through 12 had recently used cigarettes. A similar study in 2005 indicated that 22% of such teens had recently used cigarettes. (a) Write a linear equation that relates the percent of teens y to the number of years after 1998, x. (b) Find the intercepts of the graph of your equation. (c) Do the intercepts have any meaningful interpretation? (d) Use your equation to predict the percent for the year 2019. Is this result reasonable? Source:
www.drugfree. org
125. Product Promotion A cereal company finds that the num ber of people who will buy one of its products in the first month that it is introduced is linearly related to the amount of money it spends on advertising. If it spends $40,000 on ad vertising, then 100,000 boxes of cereal will be sold, and if it spends $60,000, then 200,000 boxes will be sold. (a) Write a linear equation that relates the amount A spent on advertising to the number x of boxes the company aims to sell. (b) How much advertising is needed to sell 300,000 boxes of cereal? (c) Interpret the slope. 126. Show that the line containing the points ( a, b) and (b, a), a "* b, is perpendicular to the line y x. Also show that the midpoint of ( a, b) and (b, a) lies on the line y = x. =
U7. The equation 2x - y = C defines a family of lines, one line for each value of C. On one set of coordinate axes, graph the members of the family when C = -4, C = 0, and C = 2. Can you draw a conclusion from the graph about each mem ber of the family?
Platform x
(a) Write a linear equation that relates the height y above the floor to the horizontal distance x from the platform. (b) Find and interpret the x-intercept of the graph of your equation.
128. Prove that if two non vertical lines have slopes whose product is - 1 then the lines are perpendicular. [Hint: Refer to Figure 47 and use the converse of the Pythagorean Theorem.]
Discussion and Writing U9.
Which of the following equations might have the graph shown? (More than one answer is possible.) (a) 2x + 3y = 6 y (b) -2x + 3y = 6 (c) 3x - 4y = -12 (d) x - y = 1 (e) x - y = - 1 (f) y = 3x - 5 (g) y
=
2x + 3
(h) y = -3x + 3
x
+ '
130. Which of the following equations might h ave the graph shown? (More than one answer is possible.) (a) 2x + 3y = 6 (b) 2x - 3y = 6 (c) 3x + 4y = 12 (d) x y = 1 (e) x - y -1 (f) y = -2x - 1 -
=
(g) y (h) y
=
1
-"2 x + 10
= x
+
4
x
SECTION 2.4
131. The figure shows the graph of two parallel lines. Which of the following pairs of equations might have such a graph? (a) x - 2y = 3 x + 2y = 7 (b) x + y = 2
Circles
1 89
of slope of a line? Is a 4% grade very steep? Investigate the grades of some mountainous roads and determine their slopes. Write a brief essay on your findings.
x + Y = -1 (c) x - y = - 2 x - y = 1 (d) x - y = -2 2x - 2y = -4 (e) x + 2y = 2 x + 2y = - 1 132.
The figure shows t h e graph o f two perpendicular lines. Which of the following pairs of equations might have such a graph?
=2 -1 = 0 = 0 =2 = -2 =2 = -1 = -2 2y + x = -2
(a) y - 2x y + 2x (b) y - 2x 2y + x (c) 2y - x 2y + x (d) y - 2x x + 2y (e) 2x + y
133.
134.
135.
=
136.
137. 138. 139.
140.
11 1
is for Slope The accepted symbol used to denote the slope of a line is the letter m. Investigate the origin of this symbol ism. Begin by consulting a French dictionary and looking up the French word monter. Write a brief essay on your findings.
141. 142.
Grade of a Road The term grade is used to describe the inclination of a road. How does this term relate to the notion
Carpentry Carpenters use the term pitch to describe the steepness of staircases and roofs. How does pitch relate to slope? Investigate typical pitches used for stairs and for roofs. Write a brief essay on your findings. Can the equation of every line be written in slope-intercept form? Why? Does every line have exactly one x-intercept and one y-intercept? Are there any lines that have no intercepts? What can you say about two lines that have equal slopes and equal y-intercepts? What can you say about two lines with the same x-intercept and the same y-intercept? Assume that the x-intercept is not O. If two distinct lines have the same slope, but different x intercepts, can they have the same y-intercept? If two distinct lines have the same y-intercept, but different slopes, can they have the same x-intercept? Which form of the equation of a line do you prefer to use? Justify your position with an example that shows that your choice is better than another. Have reasons.
2.4 Circles PREPARING FOR THIS SECTION •
Before getting started, review the following:
Completing the Square (Section 1.2, pp. 99-100) Now Work the 'Are You
•
Square Root Method (Section 1 .2, pp. 98-99)
Prepared?' problems on page 1 93 .
OBJECTIVES
1 Write t h e Standard Form o f t h e Equation o f a Ci rcle ( p . 189)
2 3
1
Gra ph a Circle (p. 191) Work with the General Form of the Equation of a Circle (p. 192)
Write the Standard Form of the Equation of a Circle
One advantage of a coordinate system is that it enables us to translate a geometric statement into an algebraic statement, and vice versa. Consider, for example, the fol lowing geometric statement that defines a circle. DEFINITION
A circle is a set of points in the xy-plane that are a fixed distance r from a fixed point (h, k). The fixed distance r is called the radius, and the fixed point (h, k) is called the center of the circle.
--1
1 90
C H A PTER 2
Figure 49
Y
Graphs
Figure 49 shows the graph of a circle. To find the equation, we let ( x, y ) repre sent the coordinates of any point on a circle with radius r and center ( h, k). Then the distance between the points ( x, y ) and ( h, k) must always equal r. That is, by the distance formula
(X, Y)
x
or, equivalently,
( x - h? + ( y - k)2 =r2
DEFINITION
The standard form of an equation of a circle with radius r and center (h, k) is
( x - hf+(y - k?=r2
(1)
I
�
�----------------------------------�
THEOREM
The standard form of an equation of a circle of radius r with center at the origin ( 0, 0) is
X_2_+ y 2 =r2
�
______________________
DEFINITION
__
__
�
__________________________
If the radius r = 1 , the circle whose center is at the origin is called the and has the equation
unit
circle
�+;= 1
I�
�------------------�
See Figure 50. Notice that the graph of the unit circle is symmetric with respect to the x-axis, the y-axis, and the origin. Figure SO
y
Unit circle X2 + i = 1
x
-1 -1
EXA M P L E
1
Writing the Standard Form of the Equation of a Circle Write the standard form of the equation of the circle with radius 5 and center ( -3, 6 ) .
Solution
Using the form o f equation ( 1 ) and substituting the values r = 5, h = - 3 , and k = 6, we have ( x - hf + ( y - k) 2 =r2 ( x + 3 ) 2 + ( y - 6? = 25 -
- Now Work P R O B L E M
7
•
SECTION 2.4
2
Circles
1 91
Graph a Circle
The graph of any equation in the form of equation (1) is that of a circle with radius r and center (h, k).
E XA M P L E
2
Graphing a Circle Graph the equation: (x + 3f + ( y - 2? = 1 6
Solution Figure 5 1
(-7, 2 )
(1,2 ) 2 x
-10
(-3 , -2)
Since the equation is i n the form of equation ( 1 ) , its graph i s a circle. To graph the equation, we first compare the given equation to the standard form of the equation of a circle. The comparison yields information about the circle. (x + 3? + (y - 2? = 1 6 (x - ( -3) ? + ( y - 2f = 42 i i i (x - h)2 + (y - k)2 = ,.2 We see thath = -3, k = 2, and r = 4. The circle has center ( -3,2) and a ra dius of 4 units. To graph this circle, we first plot the center ( -3, 2 ) . Since the radius is 4, we can locate four points on the circle by plotting points 4 units to the left, to the right, up, and down from the center. These four points can then be used as guides to obtain the graph. See Figure 51 . lI!l!C==-
EXA M P L E
3
Now Work P R O B L E M S 23 (a)
•
AND (b)
Finding the Intercepts of a Circle For the circle (x + 3? + (y - 2? = 16, find the intercepts, if any, of its graph.
Solution
This is the equation discussed and graphed in Example 2. To find the x-intercepts, if any, let y = O. Then (x + 3? + (y - 2? (x + 3 ? + ( 0 - 2 ? (x + 3)2 + 4 (x + 3)2 X + 3 x
= = = = = =
16 16 16
y = 0
12
Simplify.
Simplify.
Apply the S quare Root Method. ± Vi2 -3 ± 2 \13 Solve for x.
The x-intercepts are -3 - 2\13 � - 6.46 and -3 + 2\13 To find the y-intercepts, if any, we let x = O. Then (x + 3)2 + (y (0 + 3f + (y 9 + (y (y
- 2? - 2)2 - 2)2 - 2)2 y-2 y
�
0.46.
= 16 = 16 = 16 =7 = ±V7 = 2± V7
The y-intercepts are 2 - V7 � -0.65 and 2 + V7 � 4.65. Look back at Figure 51 to verify the approximate locations of the intercepts. ·fI'
,- Now Work P R O B L E M 23 (e)
•
CHAPTER 2
1 92
Graphs
3
Work with the General Form of the Equation of a Circle
If we eliminate the parentheses from the standard form of the equation of the cir cle given in Example 2, we get (x + 3? + ( y - 2? = 16 x2 + 6x + 9 + l - 4y + 4 = 16
which we find, upon simplifying, is equivalent to x2 + l + 6x - 4y - 3
=
(2)
0
It can be shown that any equation of the form x2 + l + ax + by +
c = °
has a graph that is a circle, or a point, or has no graph at all. For example, the graph of the equation x2 + l = ° is the single point ( 0, 0). The equation x2 + l + 5 = 0, or x2 + l = -5, has no graph, because sums of squares of real numbers are never negative. DEFINITION
When its graph is a circle, the equation x2 + l + ax + by + is referred to as the Il'l'l:: ==�
c =
0
general form of the equation of a circle.
Now Work P R O B L E M
1 3
If an equation of a circle is in the general form, we use the method of complet ing the square to put the equation in standard form so that we can identify its cen ter and radius.
E XA M P L E 4
Graphing a Circle Whose Equation Is in General Form Graph the equation x2 + l + 4x - 6y + 12
Solution
We complete the square in both x and y to put the equation in standard form. Group the expression involving x, group the expression involving y, and put the constant on the right side of the equation. The result is ( x2 + 4x ) +
Figure 52
y
(-2,4)
@
4
1
(-3 ,3)
(l- 6y )
=
- 12
Next, complete the square of each expression in parentheses. Remember that any number added on the left side of the equation must also be added on the right. ( x2 + 4x + 4 ) + (l - 6y + 9 ) = - 12 + 4 + 9
------L
(-1,3)
(iY
'
(-2,3)
=
4
�
(-26Y
=
9
( x + 2 ) 2 + (y - 3? = 1
(-2,2)
-3
= °
1 x
Factor.
We recognize this equation as the standard form of the equation of a circle with ra dius 1 and center ( -2, 3 ) . To graph the equation use the center ( -2, 3 ) and the radius 1. See Figure 52 . I'JI='> -
Now Work P R O B L E M 2 7
•
SECTION 2.4
=I
Graph the equation: x2
2
\
I ---....;
\
',-----...../. \ \ -2 Y2
=
l =4
x2 + l = 4 l = 4 - x2
.... \, "/
(..
+
This is the equation of a circle with center at the origin and radius 2. To graph this equation, we must first solve for y.
Solution /---
-3
Using a Graphing Utility to Graph a Circle
EXA M P L E 5
Figure 53
3
-
1 93
Ci rcles
y =
±
'\14
_
x2
Subtract ;Z from each side. Apply the Square Root Method to solve for y
There are two equations to graph: first, we graph Y1 = '\14 - x2 and then Y = - '\14 - x2 on the same square screen. (Your circle will appear oval if you 2 do not use a square screen.) See Figure 53.
..)4 - X2
•
2.4 Assess Your Understanding
Are You Prepared? Answers are given at the end of these exercises. If you. get a wrong answel; read the pages listed in red. 1.
To complete the square of x2 + lOx , you would (addl subtracl) the number . (pp. 99-100)
2.
__
Use the Square Root Method to solve the equation (x - 2)2 = 9. (pp. 98-99)
Concepts and Vocabulary 3. True or False
5. True or False
Every equation of the form x2 + i + ax + by + c = 0
6. True o r False
has a circle as its graph. 4.
For a circle, the point on the circle.
__
The radius of the circle x2 + i
=9
is 3.
The center of the circle (x + 3 ) 2 + (y - 2 ) 2 =13 is (3, -2).
is the distance from the center to any
Skill Building In Problems 7.
Y
7-10,
find the center and radius of each circle. Write the standard form of the equation. Y
8.
9.
x
Y
10. Y
x
x
x
In Problems 11-20, write the standard form of the equation and the general form of the equation of each circle of radius r and center (h,k) . Graph each circle. 11.
r
=
2;
15. I' = 5;
19. 1'
�,
r
3;
(h,k) = (0, 0)
12.
( h ,k) = ( 4,
16. I' = 4;
(h,k)=
-3 )
=
(h,k)
=
(0, 0)
(h,k) = (2, -3)
G ) ,O
= 2;
(h,k)
= ( 0,2 )
17. I ' = 4;
( h,k)
=
( -2,1)
�
(h,k)
=
(o,-D
13. I '
20.
r
=
;
14. I' =
3;
18. 1' =7;
(h,k)
=
(1,0)
( h,k) =( -5,-2)
In Problems 21-34, (a ) find (he center (h,k) and radius r of each circle; (b) graph each circle; (c) find the intercepts, if any. 2 2 2 2 23. 2(x - 3 ) + 2 1 = 8 22. x + (y - 1) = 1 21. x + l = 4 24.
3(x + 1 ) 2
+
3 ( y - 1) 2
=6
25.
x2 + i - 2x - 4y - 4 = 0
26.
x2 + i +
4x
+ 2Y - 20
=
0
1 94
CHAPTER 2
27. x2 +
Graphs
l + 4x - 4y - 1 = 0
28. x2 +
1 2
l - 6x + 2Y + 9 = 0
31. 2X2 + 2i - 12x + 8y
30. x2 + i + x + y - - = 0
-
29. x2 +
l - x + 2y + 1 = 0
32. 2X2 + 2i + 8x + 7 = 0
24 = 0
34. 3x2 + 3i - 12y = 0
33. 2x2 + 8x + 2i = 0
In Problems 35-42, find the standard form of the equation of each circle. 35. Center at the origin and containing the point ( -2, 3)
36. Center ( 1 , 0) and containing the point ( -3, 2)
37. Center (2, 3 ) and tangent to the x-axis
38. Center ( -3, 1 ) and tangent to the y-axis
39. With endpoints of a diameter at (1, 4) and (- 3,2)
40. With endpoints of a diameter at (4, 3) and (0, 1)
41. Center ( - 1 , 3) and tangent to the line y = 2
42. Center (4, -2) and tangent to the line x = 1
,� In Problems
43-46,
(a) (x - 3)2 + (y + 3) 2 = 9 43.
4
6 � � -6
match each graph with the correct equation.
-6� 6 ------v-
(b) (x + 1 )2 + (y - 2)2 = 4 44.
-4
-9
(c) (x 45.
9
I? + (y + 2)2 = 4
-6+6
(d) (x + 3 ) 2 + (y - 3) 2 = 9
4
-4
-6
Applications and Extensions 50. Ferris Wheel In 2006, the star of Nanchang (in the Jiangxi province) opened as the world's largest Ferris wheel. It has a maximum height of 1 60 meters and a diameter of 153 me ters, with one full rotation taking approximately 30 minutes. Find an equation for the wheel if the center of the wheel is on the y-axis.
47. Find the area of the square in the figure.
x
Source:
AsiaOne Travel
48. Find the area of the shaded region in the figure assuming the quadrilateral inside the circle is a square.
x
49. Ferris Wheel The original Ferris wheel was built in 1893 by Pittsburg, Pennsylvania, bridge builder George W. Ferris. The Ferris wheel was originally built for the 1893 World's Fair in Chicago, but was also later reconstructed for the 1904 World's Fair in St. Louis. It had a maximum height of 264 feet and a wheel diameter of 250 feet. Find an equation for the wheel if the center of the wheel is on the y-axis. Source: inventors. about. com
Note: Two other even larger Ferris wheels are reportedly to be completed in Asia by 2008 in time for the 2008 summer Olympics 51. Weather Satellites Earth is represented on a map of a portion of the solar system so that its surface is the circle with equation x2 + i + 2x + 4y - 4091 = O. A weather
SECTION 2.5
satellite circles 0.6 unit above Earth with the center of its cir cular orbit at the center of Earth. Find the equation for the orbit of the satellite on this map.
Variation
1 95
[Hint: The quadratic equation x2 + (mx + b)2 = r2 has exactly one solution.]
(b) The point of tangency is
( -,:m, :) '
-
(c) The tangent line is perpendicular to the line containing the center of the circle and the point of tangency. The Greek method for finding the equation of the tangent line to a circle uses the fact that at any point on a circle the lines containing the center and the tan gent line are perpendicular (see Problem 52). Use this method to find an equation of the tangent line to the circle x2 + i = 9 at the point ( 1 , 2\1'2 ) .
53. The Greek Method
52. The tangent line to a circle may be defined as the line that in tersects the circle in a single point, called the point of tan gency. See the figure. y
54. U s e the Greek method described in Problem 5 3 to
find an equation of the tangent line to the circle x2 + i - 4x + 6y + 4 = O at the point (3, 2\1'2 - 3). 55. Refer to Problem 52. The line x - 2y + 4 = 0 is tangent to a circle at (0, 2 ) . The line y = 2x - 7 is tangent to the same
circle at (3, - 1 ) . Find the center of the circle. 56. Find an equation of the line containing the centers of the two
circles
x
and
m
If the equation of the circle is x2 + i = r2 and the equation of the tangent line is y = x + b, show that: (a) r2( 1 + 2) = b2
m
_-2 + I - 4x + 6y + 4 = 0 � + I + 6x
+
4Y
+
9= 0
57. If a circle of radius 2 is made to roll along the x-axis, what is
an equation for the path of the center of the circle? 58. If the circumference of a circle is 67T, what is its radius?
Discussion and Writing 59. Which of the following equations might have the graph
-
shown? (More than one answer is possible.) (a) (x 2)2 + (y + 3)2 = 1 3 ( b ) (x - 2)2 + ( y - 2 f = 8 (c) (x - 2f + (y - 3 f = 13 (d) (x + 2)2 + (y - 2 f = 8 (e) x2 + i - 4x - 9y = 0 (f) x2 + i + 4x 2y = 0 (g) x2 + i - 9x - 4y = 0 (h) x2 + i - 4x 4y = 4
-
61.
60. Which of the following equations might have the graph
shown? (More than one answer is possible.) (a) (x - 2)2 + i = 3 (b) (x + 2)2 + i = 3 y (c) x2 + (y - 2f = 3 (d) (x + 2)2 + i = 4 (e) x2 + i + lOx + 16 0 (f) x2 + i + lOx - 2y = 1 (g) x2 + i + 9x + 1 0 = 0 (h) x2 + i - 9x - 1 0 = 0 =
x
Explain how the center and radius of a circle can be used to graph a circle.
Are You Prepared? Answers 1.
add;25
2. { - 1 , 5 }
'"2 .5 Variation OBJECTIVES
1 Construct a Model Using Di rect Va riation (p. 196)
2
3
Construct a Model Using I nverse Variation (p. 197) Construct a Model Using Joint or Com bi ned Variation (p. 197)
x
1 96
C H A PTER 2
Graphs
When a mathematical model is developed for a real-world problem, it often involves relationships between quantities that are expressed in terms of proportionality: Force is proportional to acceleration. When an ideal gas is held at a constant temperature, pressure and volume are inversely proportional. The force of attraction between two heavenly bodies is inversely proportional to the square of the distance between them. Revenue is directly proportional to sales. Each of these statements illustrates the idea of variation, or how one quantity varies in relation to another quantity. Quantities may vary directly, inversely, or jointly. 1 DEFINITION
Construct a Model Using Direct Variation
Let x and y denote two quantities. Then y varies directly with x, or y is x, if there is a nonzero number k such that
directly proportional to
y = kx
Figure S4
The number k is called the constant of proportionality.
Y = kx;k > a,x 2:: a
The graph in Figure 54 illustrates the relationship between y and x if y varies directly with x and Ie > 0, X :2: O. Note that the constant of proportionality is, in fact, the slope of the line. If we know that two quantities vary directly, then knowing the value of each quantity in one instance enables us to write a formula that is true in all cases.
x
E XA M P LE
1
Solution
P
E iii' 600 a. >-
�
::2:
o
800
The monthly payment p on a mortgage varies directly with the amount borrowed B. If the monthly payment on a 30-year mortgage is $6.65 for every $1000 borrowed, find a formula that relates the monthly payment p to the amount borrowed B for a mortgage with these terms. Then find the monthly payment p when the amount borrowed B is $120,000. Because p varies directly with B, we know that
for some constant k. Because
Figure SS
c '"
Mortgage Payments
p =
6.65 (120 000,798)
Since p
=
kB, we have
In particular, when B
400
=
Ie
=
=
p
=
kB
6.65 when B
=
k( 1000) 0.00665 Solve for k. p
=
0.00665B
$120,000, we find that p = 0.00665( $120,000)
200
1000, it follows that
=
$798
Figure 55 illustrates the relationship between the monthly payment
-;;of--'-----;'4o;;-'--:8::'::o:--;-;12�0--;1-=-60::--..,B a mou n t borrowed B. Amount borrowed (OOO's) == 1l'I!liI: -
Now Work P R O B L E M S
3 AND 2 1
p
and the •
SECTION 2.5
Figure 56
y =
;k >
k
x
-
y
2 0, x
> °
DEFINITION
Variation
1 97
Construct a Model Using Inverse Variation
Let x and y denote two quantities. Then y varies inversely with x, or y is inversely proportional to x, if there is a nonzero constant k such that =
I
�
�-------- ---------y - - (- ----------------�
�
The graph in Figure 56 illustrates the relationship between y and x if y varies inversely with x and k > 0, x > O.
x
M aximum Weight That Can Be Supported by a Piece of Pine
EXA M P L E 2
See Figure 57. The maximum weight W that can be safely supported by a 2-inch by 4-inch piece of pine varies inversely with its length t. Experiments indicate that the maximum weight that a 10-foot-long 2-by-4 piece of pine can support is 500 pounds. Write a general formula relating the maximum weight W (in pounds) to length l (in feet). Find the maximum weight Wthat can be safely supported by a length of 25 feet.
Figure 57
Solution Because Wvaries inversely with I, we know that =
W
k 1
-
for some constant k. Because W = 500 when l 500
k S'mce W
/
=
= =
k 10 5000
k
5000 l
In particular, the maximum weight Wthat can be safely supported by a piece of pine 25 feet in length is
W
5000 W = 2s
600 500
=
200 pounds
Figure 58 illustrates the relationship between the weight Wand the length l.
400 300 200
li'l!IOl=� ....
Now Work P R O B L E M
100 o
1 0, we have
/' we have
W=
Figure 58
=
5
10
15
20
25
3
3 1
•
Construct a Model Using Joint Variation or Combined Variation
When a variable quantity Q is proportional to the product of two or more other variables, we say that Q varies jointly with these quantities. Finally, combinations of direct and/or inverse variation may occur. This is usually referred to as combined variation.
Let's look at an example.
1 98
CHAPTER 2
Graphs
EXA M P L E
3
Loss of Heat Through a Wall The loss of heat through a wall varies jointly with the area of the wall and the dif ference between the inside and outside temperatures and varies inversely with the thickness of the wall. Write an equation that relates these quantities.
Solution
We begin by assigning symbols to represent the quantities: L = Heat loss A = Area of wall
T = Temperature difference d = Thickness of wall
Then AT L = k d
where k is the constant of proportionality.
•
In direct or inverse variation, the quantities that vary may be raised to powers. For example, in the early seventeenth century, Johannes Kepler (1571-1630) discov ered that the square of the period of revolution T around the Sun varies directly with the cube of its mean distance a from the Sun. That is, T2 = ka3 , where k is the constant of proportionality.
EXAM P L E 4
Force of the Wind on a Window The force F of the wind on a flat surface positioned at a right angle to the direc tion of the wind varies jointly with the area A of the surface and the square of the speed v of the wind. A wind of 30 miles per hour blowing on a window measuring 4 feet by 5 feet has a force of 150 pounds. See Figure 59. What is the force on a window measuring 3 feet by 4 feet caused by a wind of 50 miles per hour?
Figure 59
Solution
Since F varies jointly with A and v2, we have
�\. �
where k is the constant of proportionality. We are told that F = 150 when A 4· 5 20 and v = 30. Then we have
� Wind �
=
=
150
=
F = kAv2, F
k(20 ) ( 900)
=
150, A
=
20, v = 30
1 k =120 Since F = kAv2; we have
F
=
1 - Av-? 1 20
For a wind of 50 miles per hour blowing on a window whose area is A = 3· 4 = 12 square feet, the force F is
F 1,1
,,_-
=
1 ( 12 ) ( 2500) = 250 pounds 120
Now Work P R O B L E M 3 9
•
SECTION 2.5
Variation
1 99
2.5 Assess Your Understanding
Concepts and Vocabulary 1. If x and y are two quantities, then y is directly proportional
to x if there is a nonzero number k such that
2.
___ _
True or False
k is a constant.
If y varies directly with x, then y
=
'5:.-, where x
Skill Building In Problems 3-14, write a general formula t o describe each variation. 3.
4. 5.
6. 7. 8. 9. 10.
1 1. 12. 13.
14.
y varies directly with x; y = 2 when x = 1 0 v varies directly with t; v = 16 when t = 2 A varies directly with x2 ; A = 47i when x = 2 V varies directly with x3; V 367i when x = 3 F varies inversely with d2 ; F = 1 0 when d = 5 y varies inversely with \IX; y = 4 when x = 9 z varies directly with the sum of the squares of x and y; z = 5 when x = 3 and y = 4 T varies jointly with the cube root of x and the square of d; T = 1 8 when x 8 and d 3 9 and d 4 M varies directly with the square of d and inversely with the square root of x; M = 24 when x z varies directly with the sum of the cube of x and the square of y; z = 1 when x = 2 and y 3 The square of T varies directly with the cube of a and inversely with the square of d; T 2 when a = 2 and d The cube of z varies directly with the sum of the squares of x and y; z = 2 when x = 9 and y = 4 =
=
=
=
=
=
=
=
4
Appl ications and Extensions In Problems
15-20,
write an equation that relates the quantities.
The volume V of a sphere varies directly with 47i . the cube of its radius r. The constant of proportionality is 3 Geometry The square of the length of the hypotenuse c of a right triangle varies jointly with the sum of the squares of the lengths of its legs a and b. The constant of proportional ity is 1 .
$ 10 00 borrowed, find a linear equation that relates the monthly payment p to the amount borrowed B for a mort gage with the same terms. Then find the monthly payment p when the amount borrowed B is $ 145,000.
15. Geometry
16.
22. Mortgage Payments The monthly payment p on a mortgage
varies directly with the amount borrowed B. lf the monthly payment on a IS-year mortgage is $8.99 for every $1000 bor rowed, find a linear equation that relates the monthly pay ment p to the amount borrowed B for a mortgage with the same terms. Then find the monthly payment p when the amount borrowed B is $1 75,000.
The area A of a triangle varies jointly with the lengths of the base b and the height h. The constant of 1 proportionality is "2 '
17. Geometry
The perimeter p of a rectangle varies jointly with the sum of the lengths of its sides I and w . The constant of proportionality is 2.
18.
Geometry
19.
Physics: Newton's Law The force F (in newtons) of attrac tion between two bodies varies jointly with their masses m and M (in kilograms) and inversely with the square of the distance d (in meters) between them. The constant of pro portionality is G = 6.67 X 10- 1 1 .
TI1e period of a pendulum is the time required for one oscillation; the pendulum is usually re ferred to as simple when the angle made to the vertical is less than 5°. The period T of a simple pendulum (in seconds) varies directly with the square root of its length l (in feet). . 27i . Th e constant 0f proportJOna I'Ity IS ;;::;: . v 32 Mortgage Payments The monthly payment p on a mort gage varies directly with the amount borrowed B. If the monthly payment on a 30-year mortgage is $6.49 for every
20. Physics: Simple Pendulum
_
21.
23.
Physics: Falling Objects The distance s that an object falls is directly proportional to the square of the time t of the fall. If an object faBs 16 feet in 1 second, how far will it faB in 3 sec onds? How long will it take an object to fall 64 feet?
The velocity v of a falling object is directly proportional to the time t of the fall. If, after 2 sec onds, the velocity of the object is 64 feet per second, what will its veloci ty be after 3 seconds?
24. Physics: Falling Objects
25.
Physics: Stretching a Spring The elongation E of a spring balance varies directly with the applied weight W (see the figure). If E = 3 when W = 20, find E when W = 15.
T E
1
200
CHAPTER 2
Graphs
26. Physics: Vibrating String The rate of vibration of a string
under constant tension varies inversely with the length of the string. If a string is 48 inches long and vibrates 256 times per second, what is the length of a string that vibrates 576 times per second?
constant of proportionality is 1T. See the figure. Write an equa tion for V.
T 1 h
At the corner Shell station, the rev enue R varies directly with the number g of gallons of gaso line sold. If the revenue is $47 AO when the number of gallons sold is 12, find a linear equation that relates revenue R to the number g of gallons of gasoline. Then find the revenue R when the number of gallons of gasoline sold is 10.5.
27. Revenue Equation
28. Cost Equation The cost C of chocolate-covered almonds
varies directly with the number A of pounds of almonds pur chased. If the cost is $23.75 when the number of pounds of chocolate-covered almonds purchased is 5, find a linear equa tion that relates the cost C to the number A of pounds of al monds purchased. Then find the cost C when the number of pounds of almonds purchased is 3.5. Suppose that the demand D for candy at the movie theater is inversely related to the price p . (a) When the price of candy is $2.75 per bag, the theater sells 156 bags of candy. Express the demand for candy in terms of its price. (b) Determine the number of bags of candy that will be sold if the price is raised to $3 a bag.
The volume V of a right circular cone varies jointly with the square of its radius r and its height h. The 1T constant of proportionality is -. See the figure. Write an 3 equation for V.
36. Geometry
I 1 h
29. Demand
The time t that it takes to get to school varies inversely with your average speed s. (a) Suppose that it takes you 40 minutes to get to school when your average speed is 30 miles per hour. Express the driving time to school in terms of average speed. (b) Suppose that your average speed to school is 40 miles per hour. How long will it take you to get to school?
30. Driving to School
The volume of a gas V held at a constant tem perature in a closed container varies inversely with its pres sure P. If the volume of a gas is 600 cubic centimeters (cm3) when the pressure is 1 50 millimeters of mercury (mm Hg), find the volume when the pressure is 200 mm Hg.
31. P ressure
The current i in a circuit is inversely propor tional to its resistance Z measured in ohms. Suppose that when the current in a circuit is 30 amperes the resistance is 8 ohms. Find the current in the same circuit when the resis tance is 10 ohms.
32. Resistance
33. Weight The weight of an object above the surface of Earth
varies inversely with the square of the distance from the cen ter of Earth. If Maria weighs 125 pounds when she is on the surface of Earth (3960 miles from the center) , determine Maria's weight if she is at the top of Mount McKinley (3.8 miles from the surface of Earth). 34. Intensity of Light The intensity 1 of light (measured in foot
candles) varies inversely with the square of the distance from the bulb. Suppose that the intensity of a 100-watt light bulb at a distance of 2 meters is 0.075 foot-candle. Determine the intensity of the bulb at a distance of 5 meters.
The volume V of a right circular cylinder varies jointly with the square of its radius r and its height h. The
35. Geometry
The weight of a body above the surface of Earth varies inversely with the square of the distance from the center of Earth. If a certain body weighs 55 pounds when it is 3960 miles from the center of Earth, how much will it weigh when it is 3965 miles from the center?
37. Weight of a Body
38. Force of the Wind on a Window The force exerted by the
wind on a plane surface varies jointly with the area of the surface and the square of the velocity of the wind. If the force on an area of 20 square feet is 1 1 pounds when the wind ve locity is 22 miles per hour, find the force on a surface area of 47. 125 square feet when the wind velocity is 36.5 miles per hour. The horsepower (hp) that a shaft can safely transmit varies jointly with its speed (in revolutions per minute, rpm) and the cube of its diameter. If a shaft of a certain material 2 inches in diameter can transmit 36 hp at 75 rpm, what diameter must the shaft have in order to trans mit 45 hp at 125 rpm?
39. Horsepower
40. Chemistry: Gas Laws The volume V of an ideal gas varies directly with the temperature T and inversely with the pres sure P . Write an equation relating V, T,and P using k as the
constant of proportionality. If a cylinder contains oxygen at a temperature of 300 K and a pressure of 1 5 atmospheres in a volume of 100 liters, what is the constant of proportion ality k? If a piston is lowered into the cylinder, decreasing the volume occupied by the gas to 80 liters and raising the temperature to 310 K, what is the gas pressure?
41. Physics: Kinetic Energy The kinetic energy K of a moving
object varies jointly with its mass m and the square of its ve locity v . If an object weighing 25 kilograms and moving with a velocity of 10 meters per second has a kinetic energy of 1250 joules, find its kinetic energy when the velocity is 15 me ters per second.
Chapter Review
The electrical resistance of a wire varies directly with the length of the wire and inversely with the square of the diameter of the wire. If a wire 432 feet long and 4 millimeters in diameter has a resistance of 1 .24 ohms, find the length of a wire of the same material whose resistance is 1 .44 ohms and whose diameter is 3 millimeters. 43. Measuring the Stress of Materials The stress in the mater ial of a pipe subject to internal pressure varies jointly with the internal pressure and the internal diameter of the pipe and inversely with the thickness of the pipe. The stress is 100 pounds per square inch when the diameter is 5 inches, 42. Electrical Resistance of a Wire
201
the thickness is 0.75 inch, and the internal pressure is 25 pounds per square inch. Find the stress when the internal pressure is 40 pounds per square inch if the diameter is 8 inch es and the thickness is 0.50 inch. 44. Safe Load for a Beam The maximum safe load for a hori zontal rectangular beam varies jointly with the width of the beam and the square of the thickness of the beam and in versely with its length. If an 8-foot beam will support up to 750 pounds when the beam is 4 inches wide and 2 inches thick, what is the maximum safe load in a similar beam 10 feet long, 6 inches wide, and 2 inches thick?
Discussion and Writing 45. In the early 17th century, Johannes Kepler discovered that
the square of the period T varies directly with the cube of its mean distance a from the Sun. Go to the library and research this law and Kepler's other two laws. Write a brief paper about these laws and Kepler's place in history.
46. Using a situation that has not been discussed in the text, write
47. Using a situation that has not been discussed in the text, write
a real-world problem that you think involves two variables that vary inversely. Exchange your problem with another stu dent's to solve and critique.
48. Using a situation that has not been discussed in the text, write
a real-world problem that you think involves three variables that vary jointly. Exchange your problem with another stu dent's to solve and critique.
a real-world problem that you think involves two variables that vary directly. Exchange your problem with another stu dent's to solve and critique.
CHAPTER REVIEW
Things to Know Formulas
Distance formula (p. 157) Midpoint formula (p. 160) Slope (p. 174) Parallel lines (p. 182) Perpendicular lines (p. 1 83) Direct variation (p. 196) Inverse variation (p. 197) Equations of Lines and Circles
Vertical line (p. 177) Horizontal line (p. 179)
Equal slopes (117. 1 = 117.2 ) and different y-intercepts (b 1 Product of slopes is - 1 ( m1 117.2 = -1 ) y = kx k y= x
=F
b2 )
•
x = a; a is the x-intercept y = b; b is the y-intercept
y - Y I = m(x - X l ) ; m is the slope of the line, (x] , Y I ) is a point on the line Slope-intercept form of the equation of a line (p. 179) Y = mx + b; m is the slope of the line, b is the y-intercept Ax + By = C; A, B not both 0 General form of the equation of a line (p. 181) ( x - hf + (y - k f = ,2; r is the radius of the circle, (h, k ) is the Standard form of the equation of a circle (p. 1 90) center of the circle Equation of the unit circle (p. 190) X2 + l = 1 Point-slope form of the equation of a line (p. 178)
General form of the equation of a circle (p. 192)
2 x
+
l+
ax
+ by +
c
= 0, with restrictions on a, b, and c
Objectives --------. Section
2.1
2
You should be able to . . .
Review Exercises
Use the distance formula (p. 157) Use the midpoint formula (p. 1 59)
1(a)-6(a), 48, 49(a), 50 1(b )-6(b), 50 (continued)
202
CHAPTER 2
Graphs
Section
You should be able to . . .
Review Exercises
2.2
Graph equations by plotting points (p. 163) Find intercepts from a graph (p. 165) Find intercepts from an equation (p. 166) Test an equation for symmetry with respect to the x-axis, the y-axis, and the origin (p. 167) Know how to graph key equations (p. 169)
7 8 9-16 9-16 45-46
Calculate and interpret the slope of a line (p. 174) Graph lines given a point and the slope (p. 176) Find the equation of a vertical line (p. 177) Use the point-slope form of a line; identify horizontal lines (p. 178) Find the equation of a line given two points (p. 179) Write the equation of a line in slope-intercept form (p. 1 79) Identify the slope and y-intercept of a line from its equation (p. 180) Graph lines written in general form using intercepts (p. 181) Find equations of parallel lines (p. 182) Find equations of perpendicular lines (p. 183)
l(c)-6(c), I (d)-6(d), 49(b), 51 47 29 27, 28 30-32 27, 28, 30-36 37-40 41-44 33, 34 35, 36
2
Write the standard form of the equation of a circle (p. 189) Graph a circle (p. 191) Work with the general form of the equation of a circle (p. 192)
1 7-20 21-26 23-26
2
Construct a model using direct variation (p. 196) Construct a model using inverse variation (p. 197) Construct a model using joint or combined variation (p. 197)
52, 53, 55 54 55
2
3
4 5
2.3 2 3
4 5
6
7 8
9
10
2.4
3
2.5
3
Review Exercises In Problems 1-6, find the following for each pair of points: (a) The distance between the points (b) The midpoint of the line segment connecting the points (c) The slope of the line containing the points (d) Interpret the slope found in part (c) 1. (0, 0); (4, 2)
2. (0, 0) ; ( -4, 6)
3. ( 1 , - 1 ) ; ( -2, 3 )
4. ( -2, 2 ) ; ( 1 , 4)
5 . (4, -4); (4, 8)
6. (-3, 4); (2, 4)
In Problems 9. 2x
=
9-16,
3/
7. Graph y
=
x2 + 4 by plotting points.
8. List the intercepts of the given graph.
x
list the intercepts and test for symmetry with respect to the x-axis, the y-axis, and the origin. 10. y = 5x 11. x2 + 41 16 12. 9x2 - l 14. y
=
=
x3 -
15. x2 + x +
X
l + 2y
=
=
16. x2 + 4x +
0
9
l - 2y
=
0
In Problems 1 7-20, find the standard form of the equation of the circle whose center and radius are given. 1 7. ( h,
k)
=
In Problems 21. x2
+
24. x2 +
( -2, 3 ) ; r
21-26,
(y - If
=
18. ( h, k)
4
=
(3, 4) ; r
=
19. (h, k)
4
=
( -1 , -2 ) ; r
=
1
20. ( h, k )
=
(2, -4) ; r
find the center and radius of each circle. Graph each circle. Find the intercepts, if any, of each circle. 23. x2 + l - 2x + 4y - 4 22. (x + 2)2 + l 9 = 4
l + 4x - 4y - 1
=
=
0
25. 3x2 + 31 - 6x
+
12y
=
0
26. 2x2 + 21 - 4x
=
0
=
0
=
3
C h a pter Review
203
In Problems 27-36, find an equation of the line having the given characteristics. Express your answer using either the general form or the slope-intercept form of the equation of a line, whichever you prefa 27. Slope
=
-2; containing the point (3, - 1 )
28. Slope
containing the point ( -3, 4)
29. Vertical;
31. y-intercept
=
34. Parallel to the line x + y
=
=
0; containing the point ( -5, 4)
30. x-intercept
-2; containing the point (5, -3)
33. Parallel to the line 2x - 3y
=
=
2; containing the point (4, -5)
32. Containing the points (3, -4) and (2, 1 )
-4; containing the point ( -5, 3)
2; containing the point (1, -3)
35. Perpendicular to the line x + y
=
36. Perpendicular to the line 3x - y
2; containing the point (4, - 3)
=
-4; containing the point ( -2, 4)
In Problems 3 7-40, find the slope and y-intercept of each line. Graph the line, labeling any intercepts. 37. 4x
-
5y
=
38. 3x + 4y
-20
=
1 2
In Problems 41-44, find the intercepts and graph each line. 41. 2x - 3y = 12 42. x - 2y 8
43
=
45. Sketch a graph of y 46. Sketch a graph of y
=
=
•
=
1 6
3
40. - '4 x
- -
+
1
2' y
=
0
2
52. Mortgage Payments
Yx. 2
= (3, 4 ) , B = ( 1 , 1 ) , and C ( -2, 3) are the vertices of an isosceles triangle.
48. Show that the points A
= ( -2, 0 ) , B = ( -4, 4), and C (8, 5 ) are the vertices of a right triangle in two ways: (a) By using the converse of the Pythagorean Theorem (b) By using the slopes of the lines joining the vertices
49. Show that the points A
=
R
At the corner Esso station, the revenue varies directly with the number g of gallons of gasoline sold. If the revenue is $46.67 when the number of gaUons sold is 13, find an equation that relates revenue R to the number g of gallons of gasoline. Then find the revenue R when the number of gallons of gasoline sold is 1 1 .2.
53. Revenue Function
=
50. The endpoints of the diameter of a circle are ( -3, 2) and
54. Weight of a Body The weight of a body varies inversely with
(5, -6) . Find the center and radius of the circle. Write the standard equation of this circle. that the points A = (2, 5 ) , B (8, - 1 ) lie on a line by using slopes.
51. Show =
1 1 + -y 2" 3
-,
=
The monthly payment p on a mortgage varies directly with the amount borrowed B. If the monthly payment on a 30-year mortgage is $854.00 when $ 130,000 is borrowed, find an equation that relates the monthly payment p to the amount borrowed B for a mortgage with the same terms. Then find the monthly payment p when the amount borrowed B is $165,000.
x3.
47. Graph the line with slope '3 containing the point ( 1 , 2 ) .
C
1 3
39. -x - - y
12
=
(6, 1 ) ,
the square of its distance from the center of Earth. Assuming that the radius of Earth is 3960 miles, how much would a man weigh at an altitude of 1 mile above Earth's surface if he weighs 200 pounds on Earth's surface?
and
55. Kepler's Third Law of Planetary Motion Kepler's Third Law of Planetary Motion states that the square of the period of revolu tion T of a planet varies directly with the cube of its mean distance a from the Sun. If the mean distance of Earth from the Sun is
93 million miles, what is the mean distance of the planet Mercury from the Sun, given that Mercury has a "year" of 88 days?
.
/
_
Mercury T 88 da�s
_ - - /
=
,
"
---
_ _ - -
/
/
/
Earth days
T = 365
204
C H A PTER 2
Graphs
56. Create four problems that you might be asked to do given the two points ( -3, 4) and (6, 1 ) . Each problem should involve a dif
ferent concept. Be sure that your directions are clearly stated.
57. Describe each of the following graphs in the xy-plane. Give justification. (a) x = 0 (b) y = 0 (c) x + Y = 0 (d) xy = 0 (e) x2 +
l=0
58. Suppose that you have a rectangular field that requires watering. Your watering system consists of an arm of variable length that
rotates so that the watering pattern is a circle. Decide where to position the arm and what length it should be so that the entire field is watered most efficiently. When does it become desirable to use more than one arm? [Hint: Use a rectangular coordinate system positioned as shown in the figures. Write equations for the circle(s) swept out by the watering arm(s).] y
y
Rectangular field, one arm
Square field
Rectangular field, two arms
CHAPTER TEST In Problems
1 -3,
use PI = ( - 1 , 3 ) and P = (5, - 1 ).
1. Find the distance from PI to P .
2
2. Find the midpoint of the line segment joining p] and P .
2
3. (a) Find the slope of the line containing PI and P . 4.
(b) Interpret this slope. Graph y = x2 - 9 by plotting points.
2
2
5. Sketch the graph of l = x.
6. List the intercepts and test for symmetry: x2 + y = 9.
7. Write the slope-intercept form of the line with slope -2 con
taining the point (3, -4). Graph the line. 8. Write the general form of the circle with center (4, -3) and radius 5.
9. Find the center and radius of the circle x2 + l + 4x - 2y - 4 = O. Graph this circle.
10. For the line 2x + 3 y = 6, find a line parallel to it containing
the point ( 1 , - 1 ) . Also find a line perpendicular to it con taining the point (0, 3). 1 1 . Resistance due t o a Conductor The resistance (in ohms) of a circular conductor varies directly with the length of the conductor and inversely with the square of the radius of the conductor. If 50 feet of wire with a radius of 6 X 10-3 inch has a resistance of 10 ohms, what would be the resistance of 1 00 feet of the same wire if the radius is increased to 7 X 10-3 inch?
C U M U LATIVE REVI EW In Problems
1 -8,
find the real solution (s) of each equation. X - 12 = 0 x2 2x - 2 = 0 6. � = 3
1. 3x - 5 = 0
2. x2 -
3. 2x2 - 5x - 3 = 0 5. x2 + 2x + 5 = 0
4.
7. Ix - 21
=
1
8.
-
Yx2 + 4x = 2
In Problems 9 and 1 0, solve each equation in the complex number system. 9. x2 = - 9
In Problems
1 1-14,
10. x2 - 2x + 5 = 0
solve each inequality. Graph the solution set.
7 12. - 1 < x + 4 < 5 14. 12 + xl > 3 1 15. Find the distance between the points P = ( - 1 , 3 ) and Q = (4, -2). Find the midpoint of the line segment from P to Q.
11. 2x - 3
:s;
13. Ix - 21
:s;
16. Which of the following points are on the graph of
y = x3 - 3x + 1 ? (a) ( -2, - 1 )
(b) (2, 3 )
(c) (3, 1 )
17. Sketch the graph of y = x3. 18. Find the equation of the line containing the points ( - 1 , 4)
and (2, -2). Express your answer in slope-intercept form. 19. Find the equation of the line perpendicular to the
line
y = 2x + 1 and containing the point (3, 5 ) . Express your an swer in slope-intercept form and graph the line. 20. Graph the equation x 2 + l 4x + 8y - 5 O. -
=
Chapter Project
205
CHAPTER PROJ ECT 1. Treating year as the independent variable and the winning
2.
3.
4. Predicting Olympic Performance Measurements of human performance over time sometimes follow a strong linear re lationship for reasonably short periods. In 2004 the Summer Olympic Games returned to Greece, the home of both the ancient Olympics and the first modern Olympics. The fol lowing data represent the winning times (in hours) for men and women in the Olympic marathon. Year
Men
Women
1 984
2 .1 6
2.41
1 988
2. 1 8
2.43
1 992
2.22
2.54
1 996
2.2 1
2.43
2000
2.1 7
2.39
SOl/ree: www.hiekoksports. col11lhistorylo!mtan.dfshtml
5.
6.
value as the dependent variable, find linear equations re lating these variables (separately for men and women) us ing the data for the years 1992 and 1996. Compare the equations and comment on any similarities or differences. Interpret the slopes in your equations from part 1. Do the y-intercepts have a reasonable interpretation? Why or why not? Use your lines to predict the winning time in the 2004 Olympics. Compare your predictions to the actual results (2. 18 hours for men and 2.44 hours for women). How well did your equations do in predicting the winning times? Repeat parts 1 to 3 using the data for the years 1996 and 2000. How do your results compare? Would your equations be useful in predicting the winning marathon times in the 2104 Summer Olympics? Why or why not? Pick your favorite Winter Olympics event and find the winning value (that is distance, time, or the like) in two Winter Olympics prior to 2006. Repeat parts 1 to 3 using your selected event and years and compare to the actual results of the 2006 Winter Olympics in Torino, Italy.
Functions a nd Their Graphs
u.s. Consumers Make Wireless Top Choice
July 1 1 , 200S-WASHINGTON,
D.C.,
eTIA-The Wireless Association
President and CEO Steve Largent highlights the Federal Communi cations Commission (FCC) report on local telephone competition, which indicates for the first time there are more wireless subscribers in the United States than wireline service lines. According to the FCC report, as of December 3 1 , 2004 there were 181.1 million wireless subscribers in the U.S., compared to a combined 178 million incum bent local exchange carrier switched access lines and competitive local exchange carrier switched access lines. Also according to the report, there was a 1 5 % increase in wireless subscribers for the 12-month period ending December 31, 2004. "This significant milestone reflects consumers' approval of a vi brant, ultracompetitive industry that lets us communicate how we want, where we want and with whom we want," said Largent. "Wireless is so popular because it lets people communicate on their terms. And it's not j ust about talking anymore. You can browse the web, take pictures and video, download music, play games or conduct business. Wireless can satisfy a lot of different communication needs and desires and that's extremely popular to millions of consumers." Source: www. CTIA. org
- See the Chapter Project -
A Look Back
So far, we have developed tech n i q u es for g ra p h i ng equations containing two vari ables.
A Look Ahead
In this chapter, we look at a specia l type of eq uation i nvolving two va riables ca l led a
function. This chapter deals with what a fu nction is, how to g raph functions, proper
ties of functions, and how fu nctions a re used in appl ications. The word function a p pa rently was introduced by Rene Descartes in 1637. For h i m, a function s i m ply meant a ny positive integral power of a variable x . Gottfried Wilhelm Leibniz ( 1646- 1 7 1 6), who always emphasized the g eo metric side of mathematics, used the word function to denote a ny q u a ntity a ssociated with a cu rve, such as the coordi nates of a point on the cu rve. Leon h a rd Euler ( 1707-1 783) employed the word to mean any eq uation or form u l a i nvolving va riables and constants. His idea of a func tion is s i m i l a r to the one m ost often seen in cou rses that precede ca lculus. Later, the use of functions i n investigating heat flow equations led to a very broad defi n ition, due to Lejeune Dirichlet ( 1805-1859), which describes a function as a correspon dence between two sets. It is h i s defi n ition that we use here.
Outline 3.1
Functions
3.2 The Graph of a Function
3.3 Properties of Functions
3.4 Library of Functions; Piecewise-defined Functions
3 .5 Graphing Techniques: Transformations 3.6 Mathematical Models: Building Functions
Chapter Review Chapter Test Cumulative Review Chapter Projects
207
CHAPTER 3
208
Functions a n d Their Graphs
3 . 1 Functions PREPARING FOR THIS SECTION •
•
Before getting started, review the following:
Intervals (Section 1 .5, pp. 125-126) Evaluating Algebraic Expressions, Domain of a Variable (Chapter R, Section R.2, pp. 20-21 )
"NOW Work
•
Solving Inequalities (Section 1 .5, pp. 128-13 1 )
the 'Are You Prepared?' problems on page 2 1 9.
OBJECTIVES
1 Determine Whether a Relation Represents a Function (p. 208)
2
3
4
1
Find the Va lue of a Fu nction (p. 2 12) Find the Domain of a Function (p. 2 15) Form the Sum, Difference, Product, and Quotient ofTwo Functions (p. 2 17)
Determine Whether a Relation Represents a Function
We often see situations where one variable is somehow linked to the value of another variable. For example, an individual's level of education is linked to annual income. Engine size is linked to gas mileage. When the value of one variable is related to the value of a second variable, we have a relation. A relation is a correspondence between two sets. If x and y are two elements in these sets and if a relation exists between x and y, then we say that x corresponds to y or that y depends on x, and we write x � y. We have a number of ways to express relations between two sets. For example, the equation y = 3 x - 1 shows a relation between x and y. It says that if we take some number x, multiply it by 3, and then subtract 1 we obtain the corresponding value of y. In this sense, x serves as the input to the relation and y is the output of the relation. We can also express this relation as a graph as shown in Figure 1. Not only can a relation be expressed through an equation or graph, but we can also express a relation through a technique called mapping. A map illustrates a rela tion by using a set of inputs and drawing arrows to the corresponding element in the set of outputs. Ordered pairs can be used to represent x � y as ( x, y ) . We illustrate these two concepts in Example 1 .
Figure 1
-4
EXA M P L E
1
Figure 2
M aps and Ordered Pairs as Relations Figure 2 shows a relation between states and the number of representatives each state has in the House of Representatives. The relation might be named "number of representatives." State
N umber of Representatives
Alaska Arizona California Colorado Florida --+-?""-=----==:::.....d-_+_ North Dakota
_::::�-.o:::::�/;:Z:>--==F:
7 8 25 53
In this relation, Alaska corresponds to 7, Arizona corresponds to 8, and so on. Us ing ordered pairs, this relation would be expressed as { (Alaska, 7 ) , (Arizona, 8), (California, 53 ) , (Colorado, 7), (Florida, 25) , (North D akota, I ) }
•
We now present one of the most important concepts in algebra - the function. A function is a special type of relation. To understand the idea behind a function,
SECTION 3.1
Figure 3
Functions
209
let's revisit the relation presented in Example 1. If we were to ask, "How many rep resentatives does Alaska have?," you would respond 7. In other words, each input state corresponds to a single output number of representatives. Let's consider a second relation where we have a correspondence between four people and their phone numbers. See Figure 3. Notice that Colleen has two tele phone numbers. If asked, "What is Colleen's phone number?", you cannot assign a single number to her. Person
Phone number
Dan
555 - 2345
Gizmo
549 - 9402
Colleen
... 930 - 3956 555 - 8294
Phoebe
839 - 901 3
Let's look at one more relation. Figure 4 is a relation that shows a correspon dence between animals and life expectancy. If asked to determine the life expectancy of a dog, we would all respond "11 years." If asked to determine the life expectancy of a rabbit, we would all respond "7 years." Figure 4
Animal
Life Expectancy
Dog
11
Duck
10
Rabbit
7
Notice that the relations presented in Figures 2 and 4 have something in com mon. What is it? The common link between these two relations is that each input corresponds to exactly one output. This leads to the definition of a function. DEFINITION
Let X and Y be two nonempty sets.'" A function from X into Y is a relation that associates with each element of X exactly one element of Y.
.J
The set X is called the domain of the function. For each element x in X, the cor responding element y in Y is called the value of the function at x, or the image of x. The set of all images of the elements in the domain is called the range of the func tion. See Figure 5. Figure 5
':' The sets X and Y will usually b e sets o f real numbers, i n which case a (real) function results. The two sets can also be sets of complex numbers, and then we have defined a complex function. In the broad definition (due to Lejeune Dirichlet), X and Y can be any two sets.
21 0
CHAPTER 3
Functions a n d Their Graphs
Since there may be some elements in Y that are not the image of some x in X, it follows that the range of a function may be a subset of Y, as shown in Figure 5. Not all relations between two sets are functions. The next example shows how to determine whether a relation is a function or not.
EXAM P L E 2
Determining Whether a Relation Represents a Function Determine which of the following relations represent a function. If the relation is a function, then state its domain and range. (a) See Figure 6. For this relation, the domain represents the level of education and the range represents the unemployment rate.
Figure 6 United State:,; 2006
Level of Education
Unemployment Rate
No High School Diploma
8.5%
High School Diploma
5.0%
Some College
4.2%
College Graduate
2.7%
SOURCE: Statistical A bstract
of the
(b) See Figure 7. For this relation, the domain represents the number of calories in a sandwich from a fast-food restaurant and the range represents the fat con tent (in grams). Figure 7 SOURCE: Each company's Web site
Calories
Fat
(Wendy's Single) 4 1 0
19
(Wendy's B i g Bacon Classic) 580
29
(Burger King Whopper) 540
24
(Burger King Chicken Sandwich) 750 (McDonald's Big Mac) 600 (McDonald's McChicken) 430
�
�
-.....
33
� 23
(c) See Figure S. For this relation, the domain represents the weight (in carats) of pear-cut diamonds and the range represents the price (in dollars).
Solution
(a) The relation in Figure 6 is a function because each element in the domain corresponds to exactly one element in the range. The domain of the relation is { No High School Diploma, High School Diploma, Some College, College Graduate } and the range of the relation is { S. 5%, 5.0 % , 4.2 % , 2.7%}. (b) The relation in Figure 7 is a function because each element in the domain cor responds to exactly one element in the range. The domain of the relation is {410, 5S0, 540, 750, 600, 430 } . The range of the relation is { 19, 29, 24, 33, 23} .
SECTION 3.1
Functions
21 1
Notice that it is okay for more than one element in the domain to correspond to the same element in the range (McDonald's and Burger King's chicken sand wich both have 23 grams of fat). (c) The relation in Figure 8 is not a function because each element in the domain does not correspond to exactly one element in the range. If a 0.86-carat dia mond is chosen from the domain, a single price cannot be assigned to it. �==�·- Now Work P R O B L E M 1 5 r r
r r
r
r
r
r
In
Word s
In
Word s
For a function, no input has more than one output.
For a function, the domain is the set of inputs, and the range is . the set of outputs.
EXA M P L E 3
•
The idea behind a function is its predictability. If the input is known, we can use the function to determine the output. With "nonfunctions," we don't have this pre dictability. Look back at Figure 7. The inputs are {410, 580, 540, 750, 600, 430 } . The correspondence is num ber offat grams, and the outputs are { 19, 29, 24, 33, 23 } . If asked, "How many grams of fat are in a 410-calorie sandwich?", we can use the cor respondence to answer " 19." Now consider Figure 8. If asked, "What is the price of a 0.86-carat diamond?", we could not give a single response because two outputs result from the single input "0.86". For this reason, the relation in Figure 8 is not a function. We may also think of a function as a set of ordered pairs ( x , y) in which no ordered pairs have the same first element and different second elements. The set of all first elements x is the domain of the function, and the set of all second elements y is its range. Each element x in the domain corresponds to exactly one element y in the range.
Determining Whether a Relation Represents a Function Determine whether each relation represents a function. If it is a function, state the domain and range. (a) { ( I , 4) , (2, 5 ) , (3, 6 ) , (4, 7 ) } (b) { (1 , 4) , (2, 4) , (3, 5 ) , (6, 10) } (c) { ( -3, 9), (-2, 4) , (0, 0), ( 1,1 ), (-3, 8) }
Solution
(a) This relation is a function because there are no ordered pairs with the same first element and different second elements. The domain of this function is { I, 2, 3, 4} , and its range is { 4, 5, 6, 7 } . ( b ) This relation i s a function because there are n o ordered pairs with the same first element and different second elements. The domain of this function is { I , 2, 3, 6 } , and its range is { 4, 5, 10} . (c) This relation is not a function because there are two ordered pairs, (-3, 9) and (-3, 8), that have the same first element and different second elements.
•
In Example 3(b), notice that 1 and 2 in the domain each have the same image in the range. This does not violate the definition of a function; two different first ele ments can have the same second element. A violation of the definition occurs when two ordered pairs have the same first element and different second elements, as in Example 3(c). li!I!i:: = =
... -.
Now Work P R O B L E M 1 9
Up to now we have shown how to identify when a relation is a function for rela tions defined by mappings (Example 2) and ordered pairs (Example 3). We know that relations can also be expressed as equations. We discuss next the circumstances under which equations are functions.
21 2
CHAPTER 3
Functions a n d Their Graphs
To determine whether an equation, where y depends on x, is a function, it is often easiest to solve the equation for y. If any value of x in the domain corresponds to more than one y, the equation does not define a function; otherwise, it does define a function.
Determining Whether an Equation Is a Function
EXAM P L E 4
Determine if the equation y
=
2x
-5
defines y as a function of x.
The equation tells us to take an input x, multiply it by 2, and then subtract s . For any input x, these operations yield only one output y. For example, if x = 1 , then y = 2(1) 5 = -3. If x = 3, then y = 2 ( 3 ) 5 = 1 . For this reason, the equation is a function.
Solution
-
-
•
Determining Whether an Equation Is a Function
EXA M P LE 5
Determine if the equation x2 + l = 1 defines y as a function of x.
To determine whether the equation x2 + l = 1 , which defines the unit circle, is a function, we need to solve the equation for y .
Solution
x2 + l
=
1
l= 1 Y =
-
x2
± -v1=7
For values of x between - 1 and 1 , two values of y result. For example, if x 0, then y = ± 1 , so two different outputs result from the same input. This means that the equation x2 + l = 1 does not define a function. =
-
m=
Figure 9
x Domain
2
_ f(x) = x2
Range
(a)
f(x) = x2
•
Now Work P R O B L E M 3 3
Find the Value of a Function
Functions are often denoted by letters such as f, P, g, G, and others. If f is a func tion, then for each number x in its domain the corresponding image in the range is designated by the symbol f(x), read as "f of x" or as "f at x." We refer to f(x) as the value of f at the number X; f(x) is the number that results when x is given and the function f is applied; f(x) is the output corresponding to x or the image of x; f(x) does not mean "f times x." For example, the function given in Example 4 may be written as y
=
f(x)
=
2x
- S.
Then f
(%)
=
-2.
Figure 9 illustrates some other functions. Notice that, in every function, for each x in the domain there is one value in the range.
�-"r -3 = G(O) = G( - 2) = G(3) X Domain
(b)
_ F(x) = xl F(x) = �
Range
x Domain
- g(x) = -Yx
Range
(c)
g(x) = -Yx
x - G(x) = 3 Domain
Range (d)
G(x) = 3
SECTION 3 . 1
Figure 1 0 Input x
Functions
21 3
Sometimes it is helpful to think of a function f as a machine that receives as input a number from the domain, manipulates it, and outputs the value. See Figure 10. The restrictions on this input/output machine are as follows:
�
1.
2.
Output Y = fIx)
EXA M P L E 6
It only accepts numbers from the domain of the function. For each input, there is exactly one output (which may be repeated for differ ent inputs).
For a function y = f(x), the variable x is called the independent variable, because it can be assigned any of the permissible numbers from the domain. The variable y is called the dependent variable, because its value depends on x. Any symbol can be used to represent the independent and dependent variables. For example, if f is the cube function, then f can be given by f (x) = x3 or f (t) = t3 or f ez) = Z3. All three functions are the same. Each tells us to cube the indepen dent variable to get the output. In practice, the symbols used for the independent and dependent variables are based on common usage, such as using C for cost in business. The independent variable is also called the argument of the function. Thinking of the independent variable as an argument can sometimes make it easier to find the value of a function. For example, if f is the function defined by f(x) = x3, then f tells us to cube the argument. Thus, f(2) means to cube 2, f(a) means to cube the number a, and f(x + h) means to cube the quantity x + h.
F inding Values of a Function For the function f defined by f(x)
Solution
=
(a) f(3) (d) f e -x)
(b) f(x) + f ( 3 ) (e) -f(x)
(g) f(x + 3 )
(h)
f(x
+
2X2 - 3x, evaluate
l
h - f(x)
h
*
(c) 3f(x) (f) f(3x) 0
(a) We substitute 3 for x in the equation for f, f(x) = 2X2 - 3x, to get f(3 )
=
2(3? - 3 (3 )
=
18 - 9
(b) f(x) + f(3) = (2x2 - 3x) + (9) = 2X2 - 3x + 9 (c) We multiply the equation for f by 3 . 3f(x)
=
3(2x2 - 3 x )
=
=
9
6x2 - 9x
(d) We substitute -x for x in the equation for f and simplify. f e -x ) =
=
2 ( -x )2 - 3 ( -x ) = 2X2
+
3x
_ (2X2 - 3x) = -2x2 + 3x (f) We substitute 3x for x in the equation for f and simplify. (e) -f(x)
f(3x )
=
2(3x )2 - 3(3x )
=
2(9x2) - 9x = 1 8x2 - 9x
(g) We substitute x + 3 for x in the equation for f and simplify. f(x + 3 )
=
=
= =
2(x + 3? - 3 (x + 3 ) 2(x2 + 6x + 9) - 3x - 9 2x2 2X2
+ +
12x + 18 - 3x - 9 9x + 9
Notice the use of parentheses here.
214
CHAPTER 3
Functions a n d Their Graphs
(h)
+
f(x
[ 2(x + h ) 2 - 3(x + h ) ] - [ 2X2 - 3x] h
h) - f(x) h =
f(x + h)
l' 2(x + h) 2 - 3(x + h)
2 ( x2
+
2xh + h2) - 3x - 3h - 2x2 + 3x Simplify. h
2X2 + 4xh + 2h2 - 3h - 2x2 Distribute and combine like terms. h
=
4xh + 2h2 - 3h Combine like terms. h h ( 4x + 2h - 3) Factor out h. h 4x + 2h - 3 Divide out the h's.
•
Notice in this example that f(x + 3 ) =F f(x) + f ( 3 ) , f e -x) =F -f(x), and 3f(x) =F f(3x). The expression in part (h) is called the difference quotient of f, an important expression in calculus. � :::: _ ::l -
Now Work P R O B L E M S 3 9 AND 7 3
Most calculators have special keys that enable you to find the value of certain commonly used functions. For example, you should be able to find the square func tion f(x) = x2, the square root function f(x) = Vx, the reciprocal function
f(x)
=
! x
=
x-I, and many others that will be discussed later in this book (such as
In x and log x). Verify the results of Example 7, which follows, on your calculator.
EXA M P LE 7
Finding Values of a Function on a Calculator (a) f(x) x2 f ( 1.234) 1.2342 1.522756 =
(b) F (x) (c) g(x)
=
=
=
�
F ( 1.234)
=
=
�
�
1 . 34 Vx g(1.234) = v'1 .234
0.8103727715 �
1 . 1 10855526
•
Graphing calculators can be used to evaluate any function that you wish. Figure I) shows the result obtained in Example 6(a) on a TI-84 Plus graphing calculator with the function COM M E NT
to be evaluated, f(x) Figure 1 1
Plotl
P lot2
Plot3
'·.Y 1 ElZ"; 2 -3�< .... Ii ;:: = I...I � = '·S ... = '·.Y � = ,..v Ii = .... Y 7 =
=
11
2x2 - 3x, in Yl ' Y1 (3)
9
•..•
•
SECTION 3.1
Functions
21 5
I m plicit Form of a Function
In general, when a function f is defined by an equation in x and y, we say that the function f is given implicitly . If it is possible to solve the equation for y in terms of x, then we write y = f(x) and say that the function is given explicitly. For example, Implicit Form
3x
1:1 tion is the form The explicit form of a func re uired by a graphing CO M M ENT
q
x2
_
calculator.
+
y=
- Y
5
Explicit Form
y f(x) = -3x + y = f(x) = x2 - 6 4 y = f(x) =. x =
=6
xy = 4
5
We list next a summary of some important facts to remember about a func tion f. S U M M A RY
I m porta nt Facts about Functions
(a) For each x in a domain of a function f, there is exactly one image f(x) in the range; however, an element in the range can result from more than one x in the domain. (b) f is the symbol that we use to denote the function. It is symbolic of the equation that we use to get from an x in the domain to f (x) in the range. (c) If y = f(x), then x is called the independent variable or argument of f, and y is called the dependent variable or the value of f at x. 3
Find the Dom ain of a Function
Often the domain of a function f is not specified; instead, only the equation defin ing the function is given. In such cases, we agree that the domain of f is the largest set of real numbers for which the value f( x) is a real number. The domain of a func tion f is the same as the domain of the variable x in the expression f(x).
EXA M P L E
8
Finding the Domai n of a Function Find the domain of each of the following functions: (a) f(x)
Solution
=
-
3x (b) g(x) = X2 - 4
x2 + Sx
(c) h ( t)
=
v'4=3t
(a) The function tells us to square a number and then add five times the number. Since these operations can be performed on any real number, we conclude that the domain of f is the set of all real numbers. (b) The function g tells us to divide 3x by x2 - 4. Since division by 0 is not de fined, the denominator x2 - 4 can never be 0, so x can never equal -2 or 2. The domain of the function g is { x i x =1= -2, x =1= 2 } . (c) The function h tells u s to take the square root of 4 - 3t. But only nonnegative numbers have real square roots, so the expression under the square root must be nonnegative (greater than or equal to zero). This requires that
{ �}
The domain of h is t I t
:s
4 - 3t 2: 0 -3t 2: -4 4 t :S 3 or the interval
( � ]. - 00,
•
21 6
CHAPTER 3
Functions a n d Their Graphs
For the functions that we will encounter in this book, the following steps may prove helpful for finding the domain of a function whose domain is a subset of real numbers and is defined by an equation. Start with the set of real numbers l. If the equation has a denominator, exclude any numbers that glve a zero denominator. 2. If the equation has a radical of even index, exclude any numbers that cause the expression inside the radical to be negative. ",.' m==> _
Now Work P R O B L E M 5 1
If x is in the domain of a function f, we shall say that f is defined at x, orf(x) If x is not in the domain of f, we say that f is not defined at x, or f(x) does
exists.
not exist .
For example, if f(x)
=
� - 1 r
,
then f(O) exists, but f ( l ) and f ( - l ) do
not exist. (Do you see why?) We have not said much about finding the range of a function. The reason is that when a function is defined by an equation it is often difficult to find the range.* Therefore, we shall usually be content to find just the domain of a function when the function is defined by an equation. We shall express the domain of a function using inequalities, interval notation, set notation, or words, whichever is most convenient. When we use functions in applications, the domain may be restricted by physi cal or geometric considerations. For example, the domain of the function f defined by f(x) x2 is the set of all real numbers. However, if f is used to obtain the area of a square when the length x of a side is known, then we must restrict the domain of f to the positive real numbers, since the length of a side can never be 0 or negative. =
EXA M P L E
9
Finding the Domain in an Application Express the area of a circle as a function of its radius. Find the domain.
Figure 1 2
Solution
G
See Figure 12. We know that the formula for the area A of a circle of radius r is 7Tr2. If we use r to represent the independent variable and A to represent the A dependent variable, the function expressing this relationship is =
A(r)
=
7TT2
In this setting, the domain is { r l r > O } . (Do you see why?)
•
Observe in the solution to Example 9 that we used the symbol A in two ways: It is used to name the function, and it is used to symbolize the dependent variable. This double use is common in applications and should not cause any difficulty. �-
Now Work P R O B L E M 8 7
;, In Section 6.2 we discuss a way to find the range for a special class of functions.
SECTION 3.1
4
Functions
217
Form the Sum, Differen ce, Product, and Quotien t of Two Functions
Next we introduce some operations on functions. We shall see that functions, like numbers, can be added, subtracted, multiplied, and divided. For example, if f(x) = x2 + 9 and g(x) = 3x + 5, then f(x)
+
g(x) = (x2
The new function y = x2
+
3x
+
+
f(x) · g (x) = ( x2
+
9)
+
+
(3x + 5) = x2
3x
+
14 is called the sum function f
9) (3x
+
5 ) = 3x3
+
5x2
+
27x
14 +
+
g. Similarly, 45
The new function y = 3x3 + 5x2 + 27x + 45 is called the product function f · g. The general definitions are given next. DEFINITION
If f and g are functions: The
sum f + g
is the function defined by (.f
+
�------
g) (x) = f(x)
+
I
g(x)
------------��
The domain of f + g consists of the numbers x that are in the domains of both f and g. That is, domain of f + g = domain of f n domain of g. DEFINITION
The difference f -
g
is the function defined by ( .f - g) (x) = f(x) - g(x)
I
----��
�--------------
The domain of f - g consists of the numbers x that are in the domains of both f and g. That is, domain of f - g = domain of f n domain of g. DEFINITION
The product f· g is the function defined by (.f . g ) ( x )
�------
I
f(x) . g(x)
------------�� =
The domain of f . g consists of the numbers x that are in the domains of both f and g. That is, domain of f . g = domain of f n domain of g. DEFINITION
The
quotient
!..g is the function defined by
(f}X) ���� =
g(x) * 0
L---____
I
_
_
�
The domain of L consists of the numbers x for which g( x) * 0 that are in the g domains of both f and g. That is, domain of L = { x J g(x) * O} g
n
domain of f n domain of g.
21 8
CHAPTER 3
Functions a n d Their Graphs
EXAM PLE
10
Operations on Functions Let I and g be two functions defined as I(x)
=
--
--
1 x and g(x) = x- 1 x+2
Find the following, and determine the domain in each case.
Solution
(a) (f + g ) ( x )
(b) (f - g ) ( x )
(c) (f · g) (x)
(d)
(?)c X )
The domain of I is { x i x '" -2} and the domain of g is { x i x '" I } . (a) (f + g ) ( x )
=
I(x) + g(x) =
x 1 + x+2 x- I
x(x + 2) x- I + = (x + 2 ) ( x - 1 ) (x + 2) (x - 1 )
x
=
2
+ 3x - 1 (x + 2 ) ( x - 1 )
------
The domain of I + g consists o f those numbers x that are in the domains of both I and g. Therefore, the domain of I + g is { x i x '" -2, x '" I } . (b) (f - g ) ( x )
=
I(x) - g(x)
=
-- -1 x x+2 x- I
x- I (x + 2 ) ( x - 1)
X
- (x2 + + 1 ) ( x + 2) (x - 1 )
x(x + 2) (x + 2) (x - 1)
The domain of I - g consists of those numbers x that are in the domains of both I and g. Therefore, the domain of I - g is { x i x '" -2, x '" I } . (c) (f . g ) (x)
=
I(x ) · g(x )
1 =
x+2
.x
x - I
=
x (x + 2)(x - 1 )
The domain of I , g consists of those numbers x that are i n the domains of both I and g. Therefore, the domain of I ' g is { x i x '" -2, x '" I } . (d)
(g) I
(x)
=
I(x) g(x)
=
1 x- I 1 x+2 = ' xx+2 x x- I
=
x - I x(x + 2)
The domain of L consists of the numbers x for which g(x) '" ° that are in the g domains of both I and g. Since g(x) ° when x 0, we exclude ° as well as =
=
-2 and 1 from the domain. The domain of L is { x i x '" -2, x '" 0, x '" I } . g �
Now Work P R O B L E M 6 1
•
In calculus , it is sometimes helpful to view a complicated function as the sum, difference, product, or quotient of simpler functions. For example, F(x) H(x)
=
=
+ \IX is the sum of I(x) x2 and g(x) \IX . x - 1 - is the quotient of I(x) x2 1 and g(x) 2 x + 1
2 x
-
2
=
=
-
=
=
x2 +
l.
SECTION 3.1
Functions
21 9
SUMMARY
We list here some of the important vocabulary introduced in this section, with a brief description of each term. A relation between two sets of real numbers so that each number x in the first set, the do main, has corresponding to it exactly one number y in the second set. A set of ordered pairs (x, y) or (x, f( x ) ) in which no first element is paired with two dif ferent second elements. The range is the set of y values of the function that are the images of the x values in the domain. A function f may be defined implicitly by an equation involving x and y or explicitly by writing y = f(x) . Numbers for which f(x) 0 are the zeros of f.
Function
=
Unspecified domain
If a function f is defined by an equation and no domain is specified, then the domain will be taken to be the largest set of real numbers for which the equation defines a real number.
Function notation
y = f(x)
f is a symbol for the function. x is the independent variable or argument. y is the dependent variable. f(x) is the value of the function at x, or the image of x.
3 . 1 Assess Your Understanding
'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1 . The inequality - 1 < x < 3 can be written in interval nota tion as . (pp. 125-] 26)
2. If x
=
__
__
1 -2, the value of the expression 3x2 - 5x + - is x . (pp. 20-21)
3. The domain of the variable in the expression
x \s x + 4
__
.
(pp. 20-2 1 ) 4. Solve the inequality: 3 - 2x > 5. G raph the solution set. (pp. ] 28-131)
Concepts and Vocabu lary S.
If f is a function defined by the equation y = f(x), then x is called the variable and y is the variable. 6. The set of all images of the elements in the domain of a func tion is called the 7. If the domain of f is all real numbers in the interval [0, 7] and the domain of g is all real numbers in the interval [ -2, 5], the domain of f + g is all real numbers in the interval . __
__
__
8. The domain of L consists of numbers x for which g(x) g o
9.
that are in the domains of both If f(x) = x + 1 and g(x) x3 ,
__
and
__
.
Every relation is a function.
1 0.
True or False
1 1.
True or False The domain of (f . g) (x) consists of the nUlll bel's x that are in the domains of both f and g.
1 2.
True or False The independent variable is sometimes referred to as the argument of the function.
1 3.
True o r False If no domain is specified for a function f, then the domain of f is taken to be the set of real numbers.
1 4.
True or False
__
=
{xix
#-
±2}.
The domain of the function f(x) =
= x3 - (x + l ) .
then
Skill Building In Problems 15.
15-26,
Person Elvis
determine whether each relation represents a function. For each function, state the domain and range. Birthday
__ Jan. 8
Colieen ---
16.
Father
Bob _
Kaleigh
Mar. 1 5
John
Marissa
Sept. 1 7
Chuck
Daughter Beth
I--t->-- Diane Linda
Marcia
--
x2
4
x
is
220
17.
CHAPTER 3
Functions and T h e i r Gra p h s
Average I ncome
$200
Less than 9th grade
$1 8,1 20
$300
9th· 1 2th grade
$23,251
$350
High School Graduate
$36,055
Some College
$45,8 1 0
College Graduate
$67, 1 65
Salary
20 Hours
-r.--
30 Hours
18.
Level of Education
Hours Worked
$425
40 Hours
1 9. { (2, 6), ( -3, 6 ) , (4, 9 ) , (2, 10) }
20. { ( -2, 5 ) , ( -1 , 3), (3, 7 ) , (4, 12) }
21. { ( I, 3 ) , (2, 3 ) , (3, 3 ) , (4, 3 ) }
22. { (O, -2) , ( 1 , 3 ) , (2, 3 ) , (3, 7 ) }
23. { ( -2, 4), ( -2, 6) , (0, 3 ) , (3, 7) }
24. { ( -4, 4), ( - 3, 3 ) , ( -2, 2 ) , ( - 1 , 1 ) , ( - 4, 0 ) }
25. { ( -2, 4), ( -1,
26. { ( -2, 16), ( - 1 , 4), (0, 3 ) , ( 1 , 4) }
1 ) , (0, 0), ( 1 , I ) }
In Problems 27-38, determine whether the equation defines y as a function of x.
27.
Y
=
- x2 35. y = 2 x2 - 3x + 4 31. l = 4
In Problems
39-46,
(a) f(O)
+
+
43. f(x) = I x l In Problems <
51. g(x) = 55.
36. y =
f( l)
(c) f( - 1 )
r
=
_ 2 x2
f( x + 1 ) x 41. f(x) = X +1 2x + 1 45. x = --
(e) -f(x)
+x-I
(f)
2
f( )
find the domain of each function.
+l=1
-
57. f ( x )
Jx 2- I
=
h)
-
x+4
46. f ( x )
=
1 -
(x 2
1
+ 2)
2
x -x2 + 1 x+4 54. G(x) = -,:--XO - 4x
x
50. f(x) =
x2
x-2
56. G(x) = �
+
x2 1 42. f(x) = -
+1 53. F(x) = -,-XO + x
2x 52. h ( x ) = 2 x -4
(h) f(x
(g) f(2x)
3x - 5
49. f(x) =
48. f(x) = x2 + 2
h(x) = V3x - 12
59. p(x) =
(d) f( -x)
44. f(x) = �
- 16
34. x
38. x2 - 41 = 1
4
-?=--
x
- x+2
40. f ( x )
+4
30. y = I x l
3x - 1
2x - 4
4 7-60,
47. f(x) = -5x
32. y = ± �
find the following for each function:
(b)
39. f(x) = 3x2
1
29. y = x 33. x = l
28. y = x3
x2
, ;---;:. - 9 V
4
58. f(x) =
x
, r---:
x
vx - 4
60. q(x) = �
1n Problems 61-70, for the given functions f and g, find the following. For parts (a)- (d) , also find the domain.
+ g) (x) + g ) (3 ) 61. f(x) = 3x + 4; (a) (f
(b) ( f - g ) ( x )
(c) (f . g) (x)
(d)
(e) (f
(f) (f - g ) ( 4 )
(g) (f . g) (2)
(h)
67. f ( x ) 69. f(x)
=
0;
= 1 + =
g(x)
1
=
64. f ( x )
1
function g.
= 3x - 2
g(x) = 4x3
+1
68. f(x) = �; g(x) = �
4x -- ; g(x) = --
71. Given f(x) = 3x
=
g(x )
66. f(x) = I x l ; g(x) = x
3x - 5
- ; g(x) = x x
2x + 3 3x - 2
+ 1; 2 x2 + 3;
62. f(x) = 2x
g(x) = 2x - 3
63. f(x) = x - I ; g(x) = 2x2 65. f ( x )
(f}X ) (f) ( l)
3x - 2
+ 1 and + (f
g) (x)
= 6 - .!.x, find the 2
70. f(x) = Vx+l; g ( x ) = � x f 1 x . 72. Given f ( x ) = and - (x) = -? x g r
-:- ( )
+ 1 find the function --x ,
g.
SECTION 3.1
(f. In Problems
73-80,
. 73. f(x)
=
4x + 3
74. f ( x )
=
77. f(x)
=
3x2 - 2x + 6
78. f ( x )
=
find the difference quotient of f; that is, find
f(x
+
h) - f(x) , h 1.
7'"
-3x + 1
75. f(x) = x2
4x2 + 5 x - 7
79. f(x) = x3 - 2
- X
Functions
221
0, for each function. Be sure to simplify.
76. f ( x ) = x2
+4
80. f ( x ) =
x
+
1
+
5x - 1
-
3
Applications and Extensions 81. 82. 83.
If
f(x) = 2 x3 value of A ? I f f(x) of B?
I f f (x)
.
I f f(x)
+
Ax2 + 4x - 5 and f(2) = 5, what is the
(a) What is the height of the rock when x 1 second? x 1.1 seconds? x = 1.2 seconds? x = 1.3 seconds? (b) When is the height of the rock 15 meters? When is it 10 meters? When is it 5 meters? (c) When does the rock strike the ground? =
=
= 3x2 - Bx + 4 and f ( - 1 ) = 1 2, what i s the value =
3x + 8 . and f(O) = 2, what IS the value of A ? 2x - A 2x - B
94.
1
. and f (2) = 2 ' what IS the value of B? 3x + 4 2x - A 85. If f(x) = and f(4) 0, what is the value of A ? x - 3
84.
=
.
H(x)
x - B , f(2) = 0 a n d f( l ) is undefined, what are x - A the values of A and B?
=
Geometry Express the area A of a rectangle as a function of the length x if the length of the rectangle is twice its width.
88.
Geometry Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides.
89.
Constructing Functions Express the gross salary G of a per son who earns $ 1 0 per hour as a function of the number x of hours worked.
90.
Constructing Functions Tiffany, a commissioned salesper son, earns $ 1 00 base pay plus $10 per item sold. Express her gross salary G as a function of the number x of items sold.
91.
Population as a Function of Age
95.
The function
P e a ) = 0.015a2 - 4.962a
+
I0
Census Bureau (a) Identify the dependent and independent variable. (b) Evaluate P(20). Provide a verbal explanation of the meaning of P(20).
(c) Evaluate P(O). Provide a verbal explanation of the meaning of P(O). Number of Rooms The function
2
represents the number N of housing units (in millions) in 2005 that have r rooms, where r is an integer and 2 S r S 9. Source: U s.
Census Bureau (a) Identify the dependent and independent variable. (b) Evaluate N(3) . Provide a verbal explanation of the meaning of N(3) .
20 meters on Earth, the height H (in meters) after x seconds Effect o f Gravity on Earth
If a rock falls from a height of
is approximately
H (x)
=
20 - 4 . 9 x2
+
36,000 x
-
where x is the ground speed (airspeed ± wind ). (a) What is the cost per passenger for quiescent (no wind) conditions? (b) What is the cost per passenger with a head wind of 50 miles per hour? (c) What is the cost per passenger with a tail wind of 100 miles per hour? (d) What is the cost per passenger with a head wind of 100 miles per hour?
Source: Us.
N ( r ) = - 1.441' + 14.52r - 14.96
Cost of Trans-Atlantic Travel A Boeing 747 crosses the Atlantic Ocean (3000 m iles) with an airspeed of 500 miles per hour. The cost C (in dollars) per passenger is given by
x C ( x ) = 100 + -
290.580
represents the population P (in millions) of Americans in 2005 that are a years of age or older.
93.
20 - 13x2
=
" 87.
92.
=
(a) What is the height of the rock when x = 1 second? x 1 . 1 seconds? x = 1 .2 seconds? (b) When is the height of the rock 15 meters? When is it 10 meters? When is it 5 meters? (c) When does the rock strike the ground?
Where is f not defined?
I f f(x)
If a rock falls from a height of
x seconds is approximately
=
86.
EtTect of Gravity on Jupiter
20 meters on the planet Jupiter, its height H (in meters) after
96.
Cross-sectional Area The cross-sectional area of a beam cut from a log with radius 1 foot is given by the function A ( x ) = 4x \ll - x2 , where x represents the length, i n feet, of half the base of the beam. See the figure. Determine the cross-sectional area of the beam if the length of half the base of the beam is as follows: (a) One-third of a foot (b) One-half of a foot (c) Two-thirds of a foot
A(X) = 4X�
222
CHAPTER 3
Functions a n d Their Graphs
97. Economics The participation rate is the number of people
in the labor force divided by the civilian population (excludes military). Let L(x) represent the size of the labor force in year x and P (x) represent the civilian population in year x. Determine a function that represents the participation rate R as a function of x. Suppose that Vex) represents the number of vio lent crimes committed in year x and P(x) represents the number of property crimes committed in year x. Determine a function T that represents the combined total of violent crimes and property crimes in year x.
98. Crimes
Suppose that P(x) represents the percentage of income spent on health care in year x and I (x) represents income in year x. Determine a function H that represents total health care expenditures in year x.
99. Health Care
Suppose that 1 (x) represents the income of an individual in year x before taxes and T(x) represents the in dividual's tax bill in year x. Determine a function N that rep resents the individual's net income (income after taxes) in year x.
1 00. Income Tax
Suppose that the revenue R, in dollars, from selling x cell phones, in hundreds, is R(x) = - 1 .2x2 + 220x. The cost C, in dollars, of selling x cell phones is C(x) 0.05x3 - 2x2 + 65x + 500. (a) Find the profit function, P(x) = R( x ) - C(x). (b) Find the profit if x = 15 hundred cell phones are sold. (c) Interpret P(15). 1 02. Profit Function Suppose that the revenue R, in dollars, from selling x clocks is R(x) = 30x. The cost C, in dollars, of sell ing x clocks is C(x) = 0.lx2 + 7x + 400. (a) Find the profit function, P(x) = R(x) - C(x). (b) Find the profit if x = 30 clocks are sold. (c) Interpret P(30). 1 03. Some functions f have the property that f(a + b) = f ( a ) + feb) for all real numbers a and b. Which of the fol lowing functions have this property? (b) g (x) = x2 (a) hex) 2x 1 (c) F(x) = 5x - 2 (d) G( x ) = x
101. Profit Function
=
=
Discussion and Writing 1 04. Are the functions f(x)
same? Explain.
=
x - I and g (x) =
x2 - 1 the x+ 1
--
'Are You Prepared?' Answers 2 . 2 1 .5
1. ( - 1 , 3 )
3. { x i x
*-
105. Investigate when, historically, the use of the function nota tion y = f(x) first appeared.
-4}
4.
{xix
<
-I}
-1
o
3.2 The Graph of a Function PREPARING FOR THIS SECTION •
Before getting started, review the following:
Graphs of Equations (Section 2.2, pp. 163-165) Now Work
the 'Are You Prepared?' problems
OBJECTIVES
on page 226.
•
Intercepts (Section 2.2, pp. 1 65-167)
1 Identify the Graph of a Function (p. 223)
2
O btain I nformation from o r a bout the Graph of a Function (p. 224)
In applications, a graph often demonstrates more clearly the relationship between two variables than, say, an equation or table would. For example, Table 1 shows the average price of gasoline in California, adjusted for inflation (based on 2005 dollars) for the years 1978-2005. If we plot these data and then connect the points, we obtain Figure 13. Table 1
Year
Price
Year
Price
Year
Price
Year
Price
1 979
1 .9829
1 986
1 .3459
1 993
1 .5 1 89
2000
1 .8249
2.4977
1 988
1 .3 1 1 1
1 995
1 .4669
2002
1 .8782
1 990
1 978
1 980
2.4929
1 982
2.1 795
1 98 1
1 983
1 984
1 985
1 .565 1
1 .83 1 0
Source: Statistical A bstract
1 987
of the
1 .7540
1 .3274
1 992
1 994
1 989
1 .3589
1 996
1 99 1
1 .4973
1 998
United States.
1 .4656
1 997
1 .3969 1 .4709
1 999
200 1
1 .5397
2003
1 .3246
2005
1 .5329
2004
1 .5270 1 .7536
1 .5955
1 .8950
2.1521
2.4730
SECTION 3.2
Figure 1 3
California Gasoline Prices Based on 2005 Dollars, Adjusted for Inflation, 1 978-2005
3.0 ,g
c
'"
:;;
c:n
223
The Graph of a Function
2.5
2.0
� 1 .5 "'-
"0 :::0 '"
.�
OJ
Cl...
1 .0
0.5
0.0
00 ..... m
0 00 m
N 00 m
.... 00 m
c.o 00 m
00 00 m
Year
SOURCE: Statistical Abstact of the United States.
COM M E NT When we select a viewing window to graph a function on a calcu lator, the values of Xmin, Xmax give the domain that we wish to view, while Ymin, Ymax give the range that we wish to view. These settings usua l ly do not rep resent the actual domain and range of the function. _
We can see from the graph that the price of gasoline (adjusted for inflation) rose rapidly from 1978 to 1981 and was falling from 2000 to 2002. The graph also shows that the lowest price occurred in 1998. To learn information such as this from an equation requires that some calculations be made. Look again at Figure 13. The graph shows that for each date on the horizontal axis there is only one price on the vertical axis. The graph represents a function, although the exact rule for getting from date to price is not given. When a function is defined by an equation in x and y, the graph of the function is the graph of the equation, that is, the set of points (x, y) in the xy-plane that sat isfies the equation. 1
Identify the Graph of a Function
Not every collection of points in the xy-plane represents the graph of a function. Remember, for a function, each number x in the domain has exactly one image y in the range. This means that the graph of a function cannot contain two points with the same x-coordinate and different y-coordinates. Therefore, the graph of a func tion must satisfy the following vertical-line test. THEOREM
Vertical-line Test
A set of points in the xy-plane is the graph of a function if and only if every vertical line intersects the graph in at most one point.
--1
In other words, if any vertical line intersects a graph at more than one point, the graph is not the graph of a function.
EXA M P L E 1
Identifying the G raph of a Function Which of the graphs in Figure 14 are graphs of functions?
Figure 1 4
Y
3
-3
I
(a) y x2 =
3x
(b)
-3 y = x3
1 x
-1
4X
-4
(c)
X = y2
(d)
t1 I
x 2 + y2
=
1
224
CHAPTER 3
Functions a n d T h e i r G ra p h s
The graphs in Figures 14(a) and 14(b) are graphs of functions, because every vertical line intersects each graph in at most one point. The graphs in Figures 14(c) and 1 4(d) are not graphs of functions, because there is a vertical line that intersects each graph in more than one point.
Solution
•
" "'I
2
-
Now Work P R O B L E M 1 5
Obtain Information from or about the Graph of a Function
If (x, y) is a point on the graph of a function f, then y is the value of f at x; that is, y f(x). The next example illustrates how to obtain information about a function if its graph is given. =
EXA M P L E 2 Figure 1 5
Y
( 4'JT, 4 )
(2 'JT,4)
4 2
Obtaining Information from the G raph of a Function Let f be the function whose graph is given in Figure 1 5. (The graph of f might rep resent the distance that the bob of a pendulum is from its at-rest position. Negative values of y mean that the pendulum is to the left of the at-rest position, and posi tive values of y mean that the pendulum is to the right of the at-rest position.) (a) What are f ( O ) , f
( 'JT, - 4 )
(3'JT, -4)
e;)
, and f(37T ) ?
(b) What i s the domain of f? (c) What is the range of f? (d) List the intercepts. (Recall that these are the points, if any, where the graph crosses or touches the coordinate axes.) (e) How often does the line y 2 intersect the graph? (f) For what values of x does f(x) - 4? (g) For what values of x is f(x) > O? =
=
Solution
(a) Since (0, 4) is on the graph of f, the y-coordinate 4 is the value of f at the 37T x-coordinate 0; that is, f(O) 4. In a similar way, we find that when x , then y
=
0, so f
e;)
=
=
=
0. When x
=
37T, then y
=
-4, so f(37T)
=
-4.
2
(b) To determine the domain of f, we notice that the points on the graph of f have x-coordinates between ° and 47T, inclusive; and for each number x between ° and 47T, there is a point (x, f(x) ) on the graph. The domain of f is {x I ° ::; x ::; 47T} or the interval [0, 47T ] . (c) The points on the graph all have y-coordinates between - 4 and 4, inclusive; and for each such number y, there is at least one number x in the domain. The range of f is {yl -4 ::; Y ::; 4} or the interval [ -4, 4] . (d) The in tercepts are
(e) If we draw the horizontal line y 2 on the graph in Figure 1 5, we find that it intersects the graph four times. (f) Since (7T, -4) and (37T, -4) are the only points on the graph for which y f(x) -4, we have f(x) -4 when x 7T and x 37T. =
=
=
=
=
=
(g) To determine where f(x) > 0, we look at Figure 1 5 and determine the x-values from ° to 47T for which the y-coordinate is positive. This occurs
SECTION 3 . 2
T h e G r a p h o f a Function
7T 7T ] [ 7T ) ( 37T 57T ) (72,4 -7T -37T -57T -77T ,. 47T
on 0, 2
u
for 0 ::S X <
2' 2
2
or
2
u
225
. Using inequality notation, f(x) > 0
< x< 2 '
or
2
<
A
::S
•
•
When the graph of a function is given, its domain may be viewed as the shadow created by the graph on the x-axis by vertical beams of light. Its range can be viewed as the shadow created by the graph on the y-axis by horizontal beams of light. Try this technique with the graph given in Figure
15.
, ====- -
EXA M P L E 3
N o w W o r k P R O B L E M S 9 AND 1 3
Obtaining I nformation about the G raph of a Function Consider the function: f(x) =
x+ 1 x+2
(a) Is the point ( 1 ' �) on the graph of f? (b) If x = 2, what is f(x)? What point is on the graph of f? (c) If f(x) = 2, what is x? What point is on the graph of f? (d) What are the x-intercepts of the graph of f (if any)? What point(s) are on the graph of f?
Solution
(a) When x = 1, then
x+ 1 x +2 1 + 1 2 = f( l ) = 1 + 2 "3
f(x) =
The point ( 1 , (b) If x = 2, then
�)
is on the graph of f; the point x x 2 f(2) = 2
f(x) =
( �)
The point 2,
+ + + +
(
1,
�) is not.
1 2 1 = 2 "4
3
is on the graph of f.
(c) If f(x) = 2, then f(x) x+ 1 -x+2 x+ 1 x+ 1 X
=2 =2 = 2 ( x + 2 ) Multiply both sides by x + Remove parentheses. = 2x + Solve for x. =
-3
4
2.
If f(x) = 2, then x = -3. The point ( -3, 2 ) is on the graph of f· (d) The x-intercepts of the graph off are the real solutions of the equation f(x) = o. x+ 1 . . 1 IS 2 = 0 , IS X = - 1 , so The only real solution of the equation f(x) = x+ the only x-intercept. Since f( - 1 ) = 0, the point ( - 1 , 0) is on the graph of f.
-
== .... "15: �
Now Work P R O B L E M 2 5
•
226
CHAPTER 3
Functions a n d Their Graphs
Average Cost Function
E XA M P L E 4
The average cost C of manufacturing x computers per day is given by the function 20,000 C(x) 0.56x2 - 34.39x + 1212.57 + x =
--
Determine the average cost of manufacturing:
1f?1
�
(a) (b) (c) (d) (e)
30 computers in a day 40 computers in a day 50 computers in a day Graph the function C = C(x), 0 < x :::; 80. Create a TAB LE with TblStart 1 and t:, Tbl the average cost? =
=
1 . Which value of x minimizes
(a) The average cost of manufacturing x = 30 computers is 20 000 C (30) 0.S6(30) 2 - 34.39( 30) + 1212.S7 + ---tJ-
Solution
_
=
=
(b) The average cost of manufacturing x
C ( 40) = 0.S6( 40) 2 - 34.39(40)
_
=
+
$13S1 .S4
40 computers is 20 000 1212.S7 + � = $1232.97
(c) The average cost of manufacturing x = SO computers is 20 000 C( SO) 0.S6( SO) 2 - 34.39( SO) + 1212.S7 + � = $1293.07 =
[iP.:!! (d) See Figure 16 for the graph of C C(x). � (e) With the function C = C(x) in Y1 , we create Table 2. We scroll down until we =
Figure 1 6
find a value of x for which Y 1 is smallest. Table 3 shows that manufacturing x = 41 computers minimizes the average cost at $1231 .74 per computer. 1 2 3 It � .. 7
o
O �============� 8 0 .,I,I!
SU MMARY Graph of a function Vertical Line Test
Table 3
Table 2
4000
,, -
x
21179 111'16 77B1.1 6(1B'I �(I�'1.6 '13�9.7 3B�".'1
m
1./ 1 S . 56X 2 -34 . 39X ...
X
3B 39 '1(1 '11 '12 '1 3 '1'1
\-' 1
12'1(1.7 123�.9
Mea 12 :;:;::' 2 123'1.'1 123B.1
\-' 1 = 1 23 1 . 74487805
•
Now Work P R O B L E M 3 1
The collection of points (x, y) that satisfies the equation y = f(x). A collection of points is the graph of a function provided that every vertical line intersects the graph in at most one point.
3.2 Assess Your Understanding
'Are You Prepared?' A nswers are given at the end of these exercises. If you get a wrong answel; read the pages listed in 1.
The intercepts of the equation (pp. 1 65-167)
x2
+ 4/
=
16 are
__ .
2.
True or False
equation x
=
red.
The point ( -2 , - 6) i s on the graph of the 2y - 2. (pp. 163-1 65)
SECTION 3.2
T h e G ra p h o f a Function
227
Concepts and Voca bulary 3. A set of points in the xy-plane is the graph of a function if and only if every line intersects the graph in at most one point.
6. True or False
4. If the point (5, -3 ) i s a point on the graph of f, then f( ) = _
_.
5. Find a so that the point ( - 1 , 2 ) is on the graph of f(x) ax2 + 4.
A function can h ave more than one
y-intercept.
__
7. 8.
True or False The graph of a function y crosses the y-axis. True or False
y
=
f(x) always
The y-intercept of the graph of the function
= f(x), whose domain is all real numbers, is f(O).
=
Skill Building ' 9.
Use the given graph o f the function f t o answer parts (a)-(n).
1 0. Use the given graph of the function f to answer parts (a)-(n). y
( 2, 4)
4 ( 6,0)
x
( - 6, - 3 )
(2, -2)
(a) Find f ( O ) and f( - 6 ) .
(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (1) (m) (n)
(b) Find f(6) and f( l1 ) . (c) (d) (e) ([) (g) (h)
Is f(3) positive or negative? Is f( -4) positive or negative? For what values of x i s f(x) = O? For what values of x is f ( x ) > O? What is the domain of f ? What is the range of f ?
(i) What are the x-intercepts? (j ) What is the y-intercept? (k) How often does the line y
=
�
intersect the graph?
(I) How often does the line x 5 intersect the graph? (m) For what values of x does f ( x ) = 3? (n) For what values of x does f ( x ) = -2? =
In Problems
1 1 -22,
Find f(O) and f ( 6 ) . Find f(2) a n d f( - 2 ) . Is f(3) positive or negative? ]s f( - 1 ) positive or negative? For what values of x is f(x) O? For what values of x is f(x) < O? What is the domain of f? What is the range of f? What are the x-intercepts? What is the y-intercept? How often does the line y = - 1 i ntersect the graph? How often does the line x = 1 intersect the graph? For what value of x does f(x) = 3? For what value of x does f(x) -2? =
=
determine whether the graph is that of a function by using the verlical-line lest. If it i5; use the graph to find:
(a) The domain and range (b) The intercepts, if any (c) A ny symmetry with respect to the x-axis, the y-axis, or the origin
-3 -3 15.
Y+
- 3 1-
�
3x
y
16.
3x
y
13.
12.
11.
3X -3
I�
I
�'\
3x -3
y
18.
3
-3
- 'IT
� )(
y
17.
3
-3
- ---
y
14.
" ,k 4
-4
4x
-4
228
CHAPTER 3
y
3
19.
Functions a n d Their Graphs
y
3
20.
(1 , 2)
y
21.
22.
3X
-3 -3
-3
-3
-1
(2, -3 )
In Problems 23-28, answer the questions about the given function. = 2x2 - x - I (a) Is the point ( - 1, 2) on the graph of f? (b) Ifx = -2, what is f(x)? What point is on the graph off? (c) If f(x) = - 1 , what is x? What point(s) are on the graph of f? (d) What is the domain of f? (e) List the x-intercepts, if any, of the graph of f. (f) List the y-intercept, if there is one, of the graph of f. 24. f(x) = -3x2 + 5x (a) Is the point ( - 1 , 2) on the graph of f? (b) If x = -2, what is f(x)? What point is on the graph of f? (c) H f(x) = -2, what is x? What point(s) are on the graph of f? (d) What is the domain of f? (e) List the x-intercepts, if any, of the graph of f. (f) List the y-intercept, if there is one, of the graph of f. x+2 25. f(x) = -x-6 (a) Is the point (3, 14) on the graph of f? (b) If x = 4, what is f(x)? What point is on the graph of f? (c) If f(x) = 2, what is x? What point(s) are on the graph of f? (d) What is the domain of f? (e) List the x-intercepts, if any, of the graph of f. (f) List the y-intercept, if there is one, of the graph of f. x2 + 2 26. f(x) = x+4
23. f ( x)
--
( �) on the graph of
(a) Is the point 1 ,
f?
(b) If x =
0,
3 X
-3
what is f ( x ) ? What point is on the graph of f?
(c) If f(x) = .!. , what is x? What point(s) are on the graph 2 of f? (d) What is the domain of f? (e) List the x-intercepts, if any, of the graph of f. (£) List the y-intercept, if there is one, of the graph of f. 2X2 27. f(x) = X4 + 1 (a) Is the point ( - 1 , 1 ) on the graph of f? (b) If x = 2, what is f(x)? What point is on the graph of f? (c) If f(x) = 1, what is x? What point(s) are on the graph of f? (d) What is the domain of f? (e) List the x-intercepts, if any, of the graph of f. (f) List the y-intercept, if there is one, of the graph of f. 2x 28. f(x) = x-2 -
(a) Is the point
(.!.,2 -�)3 on the graph of
f?
(b) If x = 4, what is I(x)? What point is on the graph of f? (c) If f(x) = 1 , what is x? What point(s) are on the graph of I? (d) What is the domain of f? (e) List the x-intercepts, if any, of the graph of f. (f) List the y-intercept, if there is one, of the graph of f.
Applications and Extensions According to physicist Peter Brancazio, the key to a successful foul shot in basketball lies in the arc of the shot. Brancazio determined the optimal angle of the arc from the free-throw line to be 45 degrees. The arc also de pends on the velocity with which the ball is shot. If a player shoots a foul shot, releasing the ball at a 45-degree angle from a position 6 feet above the floor, then the path of the ball can be modeled by the quadratic function
29. Free-throw Shots
hex) =
44x2 v
2
- -
+x +
6
where h is the height of the ball above the floor, x is the for ward distance of the ball in front of the foul line, and v is the initial velocity with which the ball is shot in feet per second.
Suppose a player shoots a ball with an initial velocity of 28 feet per second. (a) Determine the height of the ball after it has traveled 8 feet in front of the foul line. (b) Determine the height of the ball after it has traveled 12 feet in front of the foul line. (c) Find additional points and graph the path of the bas ketball. (d) The center of the hoop is 10 feet above the floor and 15 feet in front of the foul line. Will the ball go through the hoop? Why or why not? If not, with what initial ve locity must the ball be shot in order for the ball to go through the hoop? Source: The Physics of Foul Shots, Discover, Vo!' 21, No. 10, October 2000
SECTION 3.2
The last player in the NBA to use an under hand foul shot (a "granny" shot) was Hall of Fame forward Rick Barry who retired in Barry believes that current NBA players could increase their free-throw percentage if they were to use an underhand shot. Since underhand shots are released from a lower position, the angle of the shot must be increased. If a player shoots an underhand foul shot, re leasing the ball at a 70-degree angle from a position feet above the floor, then the path of the ball can be modeled
A(x) 4x�,
3.5 136x2 2.7x 3.5, by the quadratic function h e x ) where h is the height of the ball above the floor, x is the ?- + --
+
v-
forward distance of the ball in front of the foul line, and v is the initial velocity with which the ball is shot in feet per second. (a) The center of the hoop is feet above the floor and feet in front of the foul line. Determine the initial velocity with which the ball must be shot in order for the ball to go through the hoop. (b) Write the function for the path of the ball using the velocity found in part (a). (c) Determine the height of the ball after it has traveled feet in front of the foul line. (d) Find additional points and graph the path of the basketball. Source: The Physics of Foul Shots, D iscover, Vol. 21, No. 10,
(a) Find the domain of A . / (b) Use a graphing utility to graph the function
10
15
1.
!Jill
October 2000
A golf ball is hit with an initial ve locity of feet per second at an inclination of to the horizontal In physics, it is established that the height h of the golf ball is given by the function
130
45°
- 3 ? x2 x hex) ---=-:,130=
x
+
0
x
=
(3000 C C(x) 100 + 10x + 36,000 =
500
-
x
where x is the ground speed (airspeed ± wind). (a) Use a graphing utility to graph the function = (b) Create a TAB LE with TblStart = 0 and 11 Tbl 50. (c) To the nearest miles per hour, what ground speed minimizes the cost per passenger? 34. Etlect of Elevation on Weight If an object weighs In pounds at sea level, then its weight (in pounds) at a height of h miles above sea level is given approximately by
C C(x).
50
=
W
( 40004000 h )2 (a) If Amy weighs 120 pounds at sea level, how much will she weigh on Pike's Peak, which is 14,110 feet above sea level? Iil (b) Use a graphing utility to graph the function W W(h). Use 120 pounds. (c) Create a Table with TblStart O and l1Tbl 0. 5 to see how the weight W varies as h changes from 0 to 5 miles. (d) At what height will Amy weigh 119. 9 5 pounds? W(h)
where is the horizontal distance that the golf ball has traveled.
=
A A(x). Tbl 0.1 for
(c) Create a TABLE with TblStart = and 11 o ::; x ::; Which value of maximizes the cross sectional area? What should be the length of the base of the beam to maximize the cross-sectional area? 33. Cost of Trans-Atlantic Travel A B oeing 747 crosses the Atlantic Ocean miles) with an airspeed of miles per hour. The cost (in dollars) per passenger is given by
9
31. Motion of a Golf Ball
1 x
cut from a log with radius foot is given by the function = where represents the length, in feet, of half the base of the beam. See the figure.
1980.
-
229
32. Cross-sectional Area The cross-sectional area of a beam
30. Granny Shots
=
The Graph of a Function
m
+
= In
=
=
=
=
(e) Does your answer to part (d) seem reasonable? Explain.
35. The graph of two functions, f and g, is illustrated. Use the
(a) Determine the height of the golf ball after it has traveled feet. (b) What is the height after it has traveled feet? (c) What is the height after it has traveled feet? (d) How far was the golf ball hit? ;.� (e) Use a graphing utility to graph the function ( I") Use a graphing utility to determine the distance that the ball has traveled when the height of the ball is feet. (g) Create a TABLE with TblStart = and 11 Tbl = To the nearest feet, how far does the ball travel before it reaches a maximum height? What is the maximum height? (h) Adjust the value of I1Tbl until you determine the dis tance, to within 1 foot, that the ball travels before it reaches a maximum height.
100
graph to answer parts (a)-(t).
300 500
h hex). 90 25. =
25
0
-2
(a) (f (c) (f
+ -
-4
g) (2)
g )(6)
( e ) (f . g ) (2)
x y= � x) (5, -2)
( 3, - 2
(4, -3 )
(b) (f (d) ( g (f)
+ -
g)
(4)
f)(6)
(f) (4)
230
CHAPTER 3
Functions a n d Their Graphs
Discussion and Writing 36. Describe how you would proceed to find the domain and range of a function if you were given its graph. How would your strategy change if you were given the equation defining the function instead of its graph? 37. How many x-intercepts can the graph of a function have? How many y-intercepts can the graph of a function have? 38. Is a graph that consists of a single point the graph of a function? Can you write the equation of such a function? 39. Match each of the following functions with the graph that best describes the situation. (a) The cost of building a house as a function of its square footage (b) The height of an egg dropped from a 300-foot building as a function of time (c) The height of a human as a function of time (d) The demand for Big M acs as a function of price (e) The height of a child on a swing as a function of time
y
y
x
x
(V)
(IV)
(III)
(II)
(I)
40. Match each o f the following functions with t h e graph that best describes t h e situation. (a) The temperature of a bowl of soup as a function of time (b) The number of hours of daylight per day over a 2-year period ( c) The population of Florida as a function of time (d) The distance travelled by a car going at a constant velocity as a function of time (e) The height of a golf ball hit with a 7-iron as a function of time
y
y
x (I)
(II)
x
d(t)
41. Consider the following scenario: Barbara decides to take a walk. She leaves home, walks 2 blocks in 5 minutes at a con stant speed, and realizes that she forgot to lock the door. So Barbara runs home in 1 minute. While at her doorstep, it takes her 1 min ute to find her keys and lock the door. Barbara walks 5 blocks in 1 5 minutes and then decides to jog home. It takes her 7 minutes to get home. Draw a graph of Barbara's distance from home (in blocks) as a function of time. 42. Consider the following scenario: Jayne enjoys riding her bi cycle through the woods. At the forest preserve, she gets on her bicycle and rides up a 2000-foot incline in 10 minutes. She then travels down the incline in 3 minutes. The next 5000 feet is level terrain and she covers the distance in 20 minutes. She rests for 15 minutes. Jayne then travels 10,000 feet in 30 min utes. Draw a graph of Jayne's distance traveled (in feet) as a function of time. 43. The following sketch represents the distance d (in miles) that Kevin is from home as a function of time t (in hours). An swer the questions based on the graph. In parts (a)-(g) , how many hours elapsed and how far was Kevin from home dur ing this time?
(V)
( IV)
(III)
(2, 3)
(2.8, 0)
t From t From t From t From t From t From t
(a) From (b) (c) (d) (e) (f) (g)
= = = = = = =
____.... (4.2, 2.8)
(2.5, 3)
(3, 0)
(5.3, 0)
t=2 2 to t = 2.5 2.5 to t = 2.8 2.8 to t = 3 3 to t = 3.9 3.9 to t = 4.2 4.2 to t = 5 .3 0 to
(h) What is the farthest distance that Kevin is from home? (i) How many times did Kevin return h ome?
SECTION 3.3 v (in miles per hour) of Michael's car as a function of time t (in minutes).
44. The following sketch represents the speed v(t)
(7.4 , 50)
(7, 50)
(8, 38)
(9.1 , 0)
(a) Over what interval of time is Michael traveling fastest? (b) Over what interval(s) of time is Michael's speed zero?
Properties of Functions
231
(c) What is Michael's speed between 0 and 2 minutes? (d) What is Michael's speed between 4.2 and 6 minutes? (e) What is Michael's speed between 7 and 7.4 minutes? ([) When is Michael's speed constant? 45. Draw the
graph of a function whose domain is {xl -3 ::; x ::; 8, x i= 5} and whose range is { y l - 1 ::; Y ::; 2, y i= O} . What point(s) in the rectangle -3 ::; x ::; 8, -1 ::; y ::; 2 cannot be on the graph? Com pare your graph with those of other students. What differ ences do you see?
46. Is there a function whose graph is symmetric with respect to
the x-axis? Explain.
'Are You Prepared?' Answers 1.
( -4, 0), (4, 0 ) , (0, -2), (0, 2)
2 . False
3 .3 Properties of Functions PREPARING FOR THIS SECTION •
• •
Before getting started, review the following:
Intervals (Section 1 .5, pp. 125-126) Intercepts (Section 2.2, pp. 1 65-167) Slope of a Line (Section 2.3, pp. 174-176)
• •
Point-Slope Form of a Line (Section 2.3, pp. 176-179) Symmetry (Section 2.2, pp. 167-168)
Now Work the 'Are You Prepared?' problems on page 238. OBJECTIVES
1 Determ ine Even and Odd Fu nctions from a G ra p h (p. 231)
2
3
4
Identify Even and Odd Functions from the Eq u ation (p. 232) Use a G ra p h to Determine Where a Function I s I ncreasi ng, Decreasing, or Consta nt (p, 233) U se a Gra ph to Locate Local M axima and Local M i nima (p. 234)
• 5 Use a Graphing Utility to Approximate Local Maxima a n d Local 6
M inima and to Determine Where a Fu nction Is I ncreasing or Decreasing ( p. 235) Find the Average Rate of Change of a Function (p. 236)
It is easiest to obtain the graph of a function y = f ( x ) by knowing certain proper ties that the function has and the impact of these properties on the way that the graph will look. 1
Determine Even and Odd Functions from a Graph
The words even and odd, when applied to a function f, describe the symmetry that exists for the graph of the function. A function f is even, if and only if, whenever the point (x, y) is on the graph of f then the point ( -x, y ) is also on the graph. Using function notation, we define an even function as follows: DEFINITION
A function f is even if, for every number x in its domain, the number -x is also in the domain and f ( -x )
f(x)
I�
�----------------------� =
232
CHAPTER 3
Functions a n d Their Graphs
A function f is odd, if and only if, whenever the point (x, y) is on the graph of f then the point ( - x, - y) is also on the graph. Using function notation, we define an odd function as follows: DEFINITION
A function f is odd if, for every number x in its domain, the number - x is also in the domain and
f( -x)
-f(x)
I
�---------------=
------��
Refer to page 1 68, where the tests for symmetry are listed. The following results are then evident. THEOREM
A function is even if and only if its graph is symmetric with respect to the y-axis. A function is odd if and only if its graph is symmetric with respect to the origin.
�
EXA M P L E
1
Determining Even and Odd Functions from the G raph Determine whether each graph given in Figure 17 is the graph of an even function, an odd function, or a function that is neither even nor odd.
Figure 1 7
Y
x
(b)
(a)
(a) The graph in Figure 17(a) is that of an even function, because the graph is symmetric with respect to the y-axis. (b) The function whose graph is given in Figure 17(b) is neither even nor odd, be cause the graph is neither symmetric with respect to the y-axis nor symmetric with respect to the origin. (c) The function whose graph is given in Figure 17(c) is odd, because its graph is symmetric with respect to the origin.
Solution
lI'I!l: = :::ZOo-
2
EXA M P L E 2
(c)
Now Work P R O B L E M S 2 1 ( a ) , ( b ) , A ND ( d )
•
Identify Even and Odd Functions from the Equation
In the next example, we use algebraic techniques to verify whether a given function is even, odd, or neither.
Identifying Even and Odd Functions Algebraically Determine whether each of the following functions is even, odd, or neither. Then determine whether the graph is symmetric with respect to the y-axis, the origin, or neither. ( a ) f ( x ) x2 - 5 (b) g(x) x3 - 1 (c) h(x) 5x3 - X (d) F(x) Ixl =
=
=
=
S ECTION 3.3
Properties of Functions
233
(a) To determine whether f is even, odd, or neither, we replace x by -x f (x ) x2 - 5. Then f ( -x) = ( -x )2 - 5 x2 - 5 f (x) Since fe -x) f(x), we conclude that f is an even function, and the graph is symmetric with respect to the y-axis. (b) We replace x by -x in g(x) x3 - 1. Then g( - x) = ( -x? - 1 = -x3 - 1 Since g(-x ) =f. g ( x ) and g(-x ) =f. -g(x) - ( x3 - 1 ) -x3 + 1 , we con clude that g is neither even nor odd. The graph is not symmetric with respect to the y-axis nor is it symmetric with respect to the origin. (c) We replace x by -x in hex) = 5x3 - x. Then he-x) 5 ( -x? - ( -x) = -5x3 + - (5x3 - x) -hex) Since he-x) = -hex), h is an odd function, and the graph of h is symmetric with respect to the origin. (d) We replace x by -x in F(x) = I xl. Then F( -x) I -xl = l - l l ' lxl Ixl F(x) Since F( -x) = F(x), Fis an even function, and the graph of Fis symmetric with respect to the y-axis.
Solution
111
=
=
=
=
=
=
=
=
X =
=
=
=
=
•
..=-�--
3
Now Work P R O B l E M 3 3
Use a Graph to Determine Where a Function Is Increasing, Decreasing, or Constant
Consider the graph given in Figure 18. If you look from left to right along the graph of the function, you will notice that parts of the graph are rising, parts are falling, and parts are horizontal. In such cases, the function is described as increasing, decreasing, or constant, respectively. Figure 1 8
y
(-4, -2)
Determ i n i n g Where a F unction Is Inc reasing, D ecreasing,
E XA M P L E 3
or Constant from Its G raph
Where is the function in Figure constant? S o l ution
WARNING W e describe the behavior of a graph in terms of its x-values. Do not
say the graph in Figure 18 is increasing from the point
(-4, -2)
to
(0,4).
18
increasing? Where is it decreasing? Where is it
To answer the question of where a function is increasing, where it is decreasing, and where it is constant, we use strict inequalities involving the independent variable x, or we use open intervals'" of x-coordinates. T he graph in Figure 18 is rising (increasing) from the point (-4, -2) to the point (0,4), so we conclude that it is increasing on the open interval (-4,0) or for -4 < x < O. The graph is falling (decreasing) from the point (-6,0) to the point ( -4, -2) and from the point (3, 4 ) to the point (6, 1). We conclude that the graph is decreasing on the open intervals
Rather, say it is increasing on the inter
val
(-4,0).
•
':' The open interval (a, b) consists of all real numbers x for which a
< x <
b.
234
CHAPTER 3
Functions and Their Graphs
(- 6, -4) and ( 3,6) or for -6 < x < -4 and 3 < x < 6. The graph is constant on the open interval ( 0, 3) or for 0 < x < 3.
•
More precise definitions follow: DEFINITIONS
A function / is increasing on an open interval I if, for any choice of Xl and X2 in I, with Xl < X2, we have /(X1) < /(X2)'
-.J
A function / is decreasing on an open interval I if, for any choice of Xl and X2 in I, with Xl < X2, we have /(X1) > /(X2)'
-.J
A function / is constant on an open interval I if, for all choices of X in I, the values le x) are equal.
-.J
Figure 19 illustrates the definitions. The graph of an increasing function goes up from left to right, the graph of a decreasing function goes down from left to right, and the graph of a constant function remains at a fixed height. Figure 19
y
y
y
I I I I
I I I I
: t(X2) x
X2
X1
I I
I I
:t(X1) :t(X2)
t(x1:) X1
x
X1
x
·1
t(X1)> t(x2;)
t(X1)< t(X2);
tis decreasing on 1
tis increasing on 1 . , I'l'>l::m:l.....,..
4
(c) For all x in I, the values of tare equal; tis constant on 1
(b) For x1< X2 in I,
(a) For X1< x2in I,
Now Work P R O B L E M S
1 1
I
1 3, 1 5
I
AND 2 1
Use a Graph to Locate Local Maxima and Local Minima
When the graph of a function is increasing to the left of X = c and decreasing to the right of x c,then at c the value of / is largest. This value is called a local maxi mum of f. See Figure 20(a). When the graph of a function is decreasing to the left of x = c and is increasing to the right of x = c,then at c the value of / is the smallest. This value is called a local minimum of f. See Figure 20(b). =
Figure 20
(c)
Y
t(e)
y
(e, t(e))
-71\ e
t(e)
x
--
V I (e, I I I
t(e))
e
increasing decreasing
decreasing increasing
The local maximum is f(e) and occurs at X= e.
The local minimum is t ( e) and occurs at X= e.
(a)
(b)
x
SECTION 3.3
235
Properties of Functions
A function f has a local maximum at c if there is an open interval I containing c so that, for all x not equal to c in I, f(x) < f(c). We call f(c) a local
DEFINITIONS
maximum off
A function f has a local minimum at c if there is an open interval I containing c so that, for all x not equal to c in I,f(x) > f(c). We call f(c) a local minimum off
�
If f has a local maximum at c, then the value of f at c is greater than the values of f near c. If f has a local minimum at c,then the value of f at c is less than the val ues of f near c. The word local is used to suggest that it is only near c that the value f (c) is largest or smallest. E XA M P L E 4
F i n d i n g Local M axima and Local M i n i m a from the G raph of a F un ction and Determi n i ng Where the Function Is Increasing, Decreasing, o r Constant
Figure 2 1
Figure 21 shows the graph of a function f. (a) At what number(s), if any, does f have a local maximum? (b) What are the local maxima? (c) At what number(s), if any, does f have a local minimum? (d) What are the local minima? (e) List the intervals on which f is increasing. List the intervals on which f is decreasing.
x
Solution
The domain of f is the set of real numbers. (a) f has a local maximum at 1, since for all x close to 1, x =f. 1, we have f(x) < f(l ) . (b) The local maximum is f(l ) 2. (c) f has a local minimum at -1 and at 3. (d) The local minima are f( -1) = 1 and f(3) = O. (e) The function whose graph is given in Figure 21 is increasing for all values of x between -1 and 1 and for all values of x greater than 3. That is, the function is increasing on the intervals (-1, 1) and (3, 00 ) or for - 1 < x < 1 and > 3. The function is decreasing for all values of x less than -1 and for all values of x between 1 and 3. That is, the function is decreasing on the intervals ( - , -1) and (1,3) or for x < -1 and 1 < x < 3. =
WARNING The y-value i s the local max
x
imum or local minim um and it occurs at some x-value. For Figure local maximum is
2
21,
we say the
00
and that the local
maximum occurs at x =
1.
•
•
I;!l!>: == ::> '-
I
5
Now Work
PRO B L EMS 1 7 AND 1 9
Use a Graphing Utility to Approximate Local Maxima and Local Minima and to Determine Where a Function Is Increasing or Decreasing
To locate the exact value at which a function f has a local maximum or a local min imum usually requires calculus. However, a graphing utility may be used to approx imate these values by using the MAXIMUM and MINIMUM features. E XA M P L E 5
U s i n g a G raph ing Util ity to Approxim ate Local M axima and M in i m a and to Determine Where a Functio n Is Increasing or Decreasing
(a) Use a graphing utility to graph f(x) = 6x3 - 12x + 5 for -2 < x < 2. Ap proximate where f has a local maximum and where f has a local minimum. (b) Determine where f is increasing and where it is decreasing.
236
CHAPTER 3
Functions and Their Graphs
(a) Graphing utilities have a feature that finds the maximum or minimum point of a graph within a given interval. Graph the function f for -2 < x < 2. See Figure 22(a). Using MAXIMUM, we find that the local maximum is 11 .53 and it occurs at x -0.82, rounded to two decimal places. See Figure 22(b). Using MINIMUM, we find that the local minimum is - 1 .53 and it occurs at x = 0.82, rounded to two decimal places.
Solution
=
Figure 22
30
30
-2
,r"�o;.-......_ I
�
"(J iriIUf"i1 1-:= -.B16'19'11
l /
.........
--
I
/
.......
,l.----
2
"
-2
�
"i if"iluf"il 1-:=.B16'19'1S:
'�=11_S:"19n
-1 0 (a)
-
.......
;11:".
)
2
','= -1.S:�19n
10
(b)
(b) Looking at Figures 22(a) and (b), we see that the graph of f is increasing from x -2 to x -0.82 and from x = 0.82 to x 2, so f is increasing on the in tervals ( -2, -0.82 ) and (0.82, 2) or for -2 < x < -0.82 and 0.82 < x < 2. The graph is decreasing from x = -0.82 to x 0.82, so f is decreasing on the interval ( -0.82, 0.82) for -0.82 < x < 0.82. =
=
=
=
or
....... Now Work
' I!.
6
•
PRO B LEM 4 5
Find the Average Rate of Change of a Function
In Section 2.3, we said that the slope of a line could be interpreted as the average rate of change of the line. To find the average rate of change of a function between any two points on its graph, we calculate the slope of the line containing the two points. DEFINITION
If a and b, a #- b, are in the domain of a function y of change of f from a to b is defined as
=
f(x), the average rate
ily f(b) - f(a) Average rate of change = - = --'---'----'--'ilx b-a
a#- b
(1 )
�
�------�
The symbol ily in (1) is the "change in y" and ilx is the "change in x." The aver age rate of change of f is the change in y divided by the change in x. E XA M P L E 6
F i n d i n g the Average Rate of Change
Find the average rate of change of f(x) = 3x2: (a) From 1 to 3 Solution
(b) From 1 to 5
(c) From 1 to 7
(a) The average rate of change of f(x) = 3x2 from 1 to 3 is ily ilx
f(3 ) f (I ) 27 - 3 = 3-1 3-1
_ -_ _ _ _
_ _
(b) The average rate of change of f(x) ily ilx
f(5) - f(l ) 5 - 1
=
=
=
24 = 12 2
3x2 from 1 to 5 is 75 - 3 5 - 1
=
72 4
=
18
SECTION 3.3
Figure 23
(c) The average rate of change of f(x)
Y
�y �x
1 60
120
=
Properties o f Functions
237
3x2 from 1 to 7 is
f(7) - f(l) 7 1 147 - 3 144 = 7 1 6 -
=
=
24
-
•
80
See Figure 23 for a graph of f(x) = 3x2 . The function f is increasing for x > O. The fact that the average rates of change are getting larger indicates that the graph is getting steeper; that is, it is increasing at an increasing rate.
40
:>I!
-
Now Work
PRO B L E M S 3
x
(0, 0)
The Secant Line
The average rate of change of a function has an important geometric interpretation. Look at the graph of y = f(x) in Figure 24. We have labeled two points on the graph: (a,f(a)) and (b,f (b) ). The line containing these two points is called the secant line; its slope is feb) - f(a) 111sec = b -
Figure 24
y
y=
f(x)
b
a
THEOREM
a
x
Slope of the Secant Line
The average rate of change of a function from a to b equals the slope of the secant line containing the two points (a,f(a)) and (b, f(b)) on its graph. EXAM P L E 7
.-J
F i n d i n g the Equation of a Secant Line
Suppose that g(x)
=
3x2 - 2x + 3.
(a) Find the average rate of change of g from -2 to l. (b) Find an equation of the secant line containing ( -2, g( 2 ) ) and ( 1, g(l)). -
S o l ution
(a) The average rate of change of g(x) Average rate of change
=
3x2
g(l) - g( - 2 ) 1 (-2) 4 19 3 15 -- = -5 3 _
-
=
=
-
2x + 3 from -2 to 1 is
g(l) 3(12) - 2(1) + 3 = 4 g(-2) 3(-2)2 - 2(-2) + 3 =
=
= 19
238
CHAPTER 3
Functions and Their Graphs
(b) The slope of the secant line containing (-2, g(-2)) (-2, 19) and (l, g(l) ) (1, 4) is msec -5. We use the point-slope form to find an equation of the secant line. =
=
=
Y Y Y
-
Yl
19 19 Y
k �IIImI_ "" '-
=
=
Insec ( x - Xl) -5 ( x - ( -2 ) )
- 5x - 10 = -5x + 9
X,
=
-2'YI
=
g(-2)
=
19,
msec
=-5
Simplify.
=
Now Work P R O B L E M
Point-slope form of the secant line
Slope-intercept form of the secant line
•
59
3.3 Assess Your Understanding 'Are You Prepa red?' Answers are given at the end of these exercises. Ifyou get a wrong answel; read the pages listed in 1.
2.
The interval (2, 5) can be written as the inequality . (pp.125-L26) The slope of the line containing the points ( -2, 3 ) and (3, 8) is . (pp. 174-176) Test the equation y = 5x2 1 for symmetry with respect to the x-axis, the y-axis, and the origin. (pp. L67- 168) __
__
3.
red.
Write the point-slope form of the line with slope 5 contain ing the point (3, -2). (pp. L76-179) 2 . 5. The intercepts of the equation y = x - 9 are (pp. 165-167)
4.
__
-
Concepts a n d Vocabu lary 6.
7.
A function f is on an open interval I if, for any choice of x, and X2 in I, with X, < X2, we have f( x,) < f( X2)' A(n) function f is one for which f( -x) = f( x) for every x in the domain of f; (an) function f is one for which f( -x) = -f(x) for every x in the domain of f. True or False A function f is decreasing on an open inter val! if, for any choice of Xl and X2 in I, with x, < X2 , we have f( x,) > f( X2)' __
__
__
8.
A function f has a local maximum at c if there is an open interval! containing c so that, for all X not equal to c in I,f( x) < f( c). True or False Even functions have graphs that are sym metric with respect to the origin.
9. True or False
10.
Skil l Bu ilding !n Problems 11-20, use the graph o f the f�lI1ction f given.
(-8 , -4)
. 11. 13.
Is f increasing on the interval ( -8, -2)?
12.
Is f decreasing on the interval (-8, -4)?
Is f increasing on the interval (2, IO) ?
14.
Is f decreasing on the interval (2, 5)?
16.
List the interval(s) on which f is decreasing.
18.
Is there a local maximum at 5? If yes, what is it?
15. List the interval(s) o n which f is increasing. . 17.
-6
Is there a local maximum at 2?
If yes,
what is it?
19.
List the numbers at which f has a local maximum. What are these local maxima?
20.
List the numbers at which f has a local minimum. What are these local minima?
SECTION 3.3
Properties of Functions
239
I n Problems 21-28, the graph o f a function i s given. Use the graph t o find: (a) The intercepts, if any (b) The domain and range (c) The intervals on which it is increasing, decreasing, or constanl (d) Whether it is even, odd, or neither '21.
y
(-2,0) (2,0)
-4
,,_:l.
4
2
(33 , )
23.
y
26.
2
(t,1)
3x
-3
(- t,
-
-2
1)
3
(2,2)
3x
1 / (2 (3,0) (1,-1) ,
-1)
y
28.
3
(0,1) ('IT,
-3
y
27.
-3 (-'IT, - 1 )
y
24.
3
-3(-1,0)(1,0) 3x
x
y
y
(-3,3)
22.
4
-
1)
(-3,-2)
-3
-2
In Problems 29-32, the graph of a function f is given. Use the graph to find: (a) The n umbers, if any, at which f has a local maxim um. What are these local maxima? (b) The n umbers, if any, at which f has a local minim um. What are these local minima?
y
29.
30.
y
31.
-4
(-2,0) (2,0)
4
-3(-1,0)(1,0)
x
y
32.
2
4
2
(t,1)
3x
(-re, -1)
(- t,-1 ) -2
(re,
-2
-1)
In Problems 33-44, determine algebraically whether each function is even, odd, or neither. 33. f(x) = 4x3 37. F(x) =
41. g(x)
llirl
-\YX 1
=?
34. f(x) =
2X4 -
38. G(x) =
\IX
42. h(x)
[
=
x
x2
-
2
x
-I
3
36. h(x) = 3x3 + 5
35. g(x) = - x 2 - 5 39. f(x) = x +
43. h(x)
=
40. f(x) =
Ixl
-x3
-? .:LC - 9
-�
44. F(x) =
\12 x2
+ 1
2x
N
In Problems 45-52, use a graphing utility to graph each function over the indicated interval and approximate any local maxima and local minima. Determine where the function is increasing and where it is decreasing. Ro und answers 10 two decimal places.
,.45.
'
f(x) = x' - 3x +
47. f(x) = x5 - x3 49. f(x)
=
'
2 (-2,2)
(-2,2)
-0.2x3 - 0.6x2 + 4 x -
?
46. f(x) = x' - 3[ + 5 48. f(x)
6
51. f(x) = 0.25x4 + 0.3x3 - 0.9x2 +
(-6,4)
3 ( -3,2)
=
x4 - x2
(-1,3)
( -2,2)
50. f(x) = -0.4x3 + 0.6x2 + 3x
-2
52. f(x) = -0.4x4 - 0.5x3 + 0.8x2 -
(-4,5)
2 ( -3,2)
240
CHAPTER 3
Functions and Their Graphs
, 53. Find the average rate of change of f(x) = _2X2 + 4 (a) From 0 to 2 (b) From 1 to 3 (c) From 1 to 4 54. Find the average rate of change of f(x) = -x3 + 1 (a) From 0 to 2 (b) From 1 to 3 (c) From -1 to 1 55. Find the average rate of change of g(x) = x3 - 2x + 1 (a) From -3 to -2 (b) From -1 to 1 (c) From 1 to 3 56. Find the average rate of change of h(x) = x2 - 2x + 3 (a) From -1 to 1 (b) From 0 to 2 (c) From 2 to 5 57. f(x) = 5x - 2 (a) Find the average rate of change from 1 to 3. (b) Find an equation of the secant line containing (1,f(l) ) and (3,/(3)).
58.
.59.
f(x) = -4x
+1 (a) Find the average rate of change from 2 to 5. (b) Find an equation of the secant line containing and (5,/ (5)).
(2,f(2))
g(x) = x2 - 2
(a) Find the average rate of change from -2 to l. (b) Find an equation of the secant line containing (-2,g( -2)) and (l,g(l)). 60. g(x) = x2 + 1 (a) Find the average rate of change from -1 to 2. (b) Find an equation of the secant line containing (-1, g( - 1)) and (2,g(2) ).
61.
h(x) = x2 - 2x
(a) Find the average rate of change from 2 to 4. (b) Find an equation of the secant line containing (2,h(2)) and (4, h( 4) ). 62. h(x) = -2x2 + x (a) Find the average rate of change from 0 to 3. (b) Find an equation of the secant line containing (0, h(O)) and (3, h(3) ).
Appl ications a n d Extensions 63.
An open box with a square base is to be made from a square piece of cardboard 24 inches on a side by cutting out a square from each corner and turning up the sides. See the figure. Constructing an Open Box
x x
'I ,1
24in.
x x
x
1 � 24in. - 1
r•.:
.,?
(a) Express the volume V of the box as a function of the length x of the side of the square cut from each corner. (b) What is the volume if a 3-inch square is cut out? (c) What is the volume if a lO-inch square is cut out? (d) Graph V = V(x). For what value of x is V largest?
(a) Use a graphing utility t o graph s = s(t). (b) Determine the time at which height is maximum. (c) What is the maximum height? On July 1,2004, the Cassini ":;: 66. Maximum Height of a Ball probe became the first spacecraft to orbit the planet Saturn. Although Saturn is about 764 times the size of Earth, it has a very similar gravitational force. The height s of an object thrown upward from Saturn's surface with an initial velocity of 100 feet per second is given as a function of time t (in sec onds) by s(t) = -17.28t2 + lOOt. (a) Use a graphing utility to graph s = s(t). (b) Determine the time at which height is a maximum. (c) What is the maximum height? (d) The same object thrown from the surface of Earth would have a height given by s(t) = -16t2 + lOOt. De termine the maximum height of the object on Earth and compare this to your result from part (c). ,,1 67. Minimum Average Cost The average cost per hour in dol lars of producing x riding lawn mowers is given by
-C(x) = 0.3x2 + 21x - 251 + 2500 x Use a graphing utility to graph C = C(x)
64. Constructing an Open Box An open box with a square base is required to have a volume of 10 cubic feet. (a) Express the amount A of material used to make such a box as a function of the length x of a side of the square base. (b) How much material is required for a base 1 foot by 1 foot? (c) How much material is required for a base 2 feet by 2 feet? ;it (d) Use a graphing utility to graph A = A (x). For what .. value of x is A smallest?
(a) (b) Determine the number of riding lawn mowers to pro duce in order to minimize average cost. (c) What is the minimum average cost? ii.r 68. Medicine Concentration The concentration C of a medica tion in the bloodstream t hours after being administered is given by C(t) = -0.002X4 + 0.039t3 - 0.285t2 + 0.766t + 0.085
65. Maximum Height of a Ball The height s of a ball (in feet) thrown with an initial velocity of 80 feet per second from an initial height of 6 feet is given as a function of the time t (in seconds) by s(t) = -16t2 + 80t + 6
(a) After how many hours will the concentration be highest? (b) A woman nursing a child must wait until the concen tration is below 0.5 before she can feed him. After tak ing the medication, how long must she wait before feed ing her child?
--
SECTION 3.3
69. E-coli Growth A strain of E-coli Beu 397-recA441 is placed into a nutrient broth at 30° Celsius and allowed to grow. The data shown below are collected. The population is measured in grams and the time in hours. Since population P depends on time t and each input corresponds to exactly one output, we can say that population is a function of time; so pet ) rep resents the population at time t. ( a ) Find the average rate o f change o f the population from o to 2.5 hours. (b) Find the average rate of change of the population from 4.5 to 6 hours. (c) What is happening to the average rate of change as time passes?
70.
Time (hoursl.
t
Population (gramsl. P
o
0.09
2.5
0.18
3.5
0.26
4.5
0.35
6
0.50
241
(d) What is happening to the average rate of change as time passes?
�------------� o�
Properties of Functions
Year
Percentage of returns e-filed
1998
20.7
1999
23.5
2000
27.6
2001
30.7
2002
35.6
2003
40.2
2004
46.5
2005
51.1
2006
57.1
SOURCE: Internal Revenue Service
For the function f(x) = x2, compute each average rate of change: (a) From 0 to 1 (d) From 0 to 0.01 (b) From 0 to 0.5 (e) From 0 to 0.001 (c) From 0 to 0.1 ij> (f) Use a graphing utility to graph each of the secant lines along with f. (g) What do you think is happening to the secant lines? (h) What is happening to the slopes of the secant lines? Is there some number that they are getting closer to? What is that number? 72. For the function f(x) = x2, compute each average rate of change: (d) From 1 to 1.01 (a) From 1 to 2 (e) From 1 to 1.001 (b) From 1 to 1.5 (c) From 1 to 1.1 hi? (0 Use a graphing utility to graph each of the secant lines along with f. (g) What do you think is happening to the secant lines? (h) What is happening to the slopes of the secant lines? Is there some number that they are getting closer to? What is that number? 71.
e-Filing Tax Returns The Internal Revenue Service Restructuring and Reform Act (RRA) was signed into law by President Bill Clinton in 1 998. A major objective of the RRA was to promote electronic filing of tax returns. The data in the table show the percentage of individual income tax re turns filed electronically for filing years 1 998-2006. Since the percentage P of returns filed electronically depends on the fil ing year y and each input corresponds to exactly one output, the percentage of returns filed electronically is a function of the filing year; so P(y) represents the percentage of returns filed electronically for filing year y. (a) Find the average rate of change of the percentage of e-filed returns from 1998 to 2000. (b) Find the average rate of change of the percentage of e-filed returns from 2001 to 2003. (c) Find the average rate of change of the percentage of e-filed returns from 2004 to 2006.
(f. Problems 73-80 require the following discussion of a secant line. The slope of the secant line containing the two points (x,f(x ) ) and (x + h,f(x + h)) on the graph of a function y = f(x) may be given as f(x + h) - f(x) (x + h) - x
f(x
+
h ) - f(x) h
h*O
In calculus, this expression is called the difference quotient of f. (a) Express the slope ofthe secant line of each function in terms ofx and h. Be sure to simplify your answer. (b) Find msecfor h = 0.5,0.1, and 0.01 at x = 1. What value does mscc approach as h approaches O? (c) Find the equation for the secant line at x = 1 with h = 0.01. ro.: (d) Use a graphing utility to graph f and the secant line found in part (c) on the same viewing window. 73. f(x) = 2x + 5 74. f(x) = -3x + 2 76. f(x) 75. f(x) x2 + 2 x =
77.
f(x)
=
2x2 - 3x
+1
78.
f(x)
=
-x2
+
3x
- 2
79.
f(x)
=
� x
80.
f(x)
=
2x 2
1 =-
x2
+
X
242
C H A PTER 3
Functions and Their Graphs
Discussion a n d Writing 81.
82.
83.
Draw the graph of a function that has the following proper ties: domain: all real numbers; range: all real numbers; inter cepts: (0,-3) and (3,0); a local maximum of -2 is at -1; a local minimum of -6 is at 2. Compare your graph with oth ers. Comment on any differences. Redo Problem 81 with the following additional information: increasing on (-00,-1),(2,00); decreasing on (-1,2). Again compare your graph with others and comment on any differences. How many x-intercepts can a function defined on an interval have if it is increasing on that interval? Explain.
84.
85.
L;TI 86.
Suppose that a friend of yours does not understand the idea of increasing and decreasing functions. Provide an explana tion, complete with graphs, that clarifies the idea. Can a function be both even and odd? Explain. Using a graphing utility, graph y = 5 on the interval (-3,3). Use MAXIMUM to find the local maxima on ( 3,3). Com ment on the result provided by the calculator. -
'Are You Prepa red?' Answers 1.2<x<5
2.
1
3.
symmetric with respect to the y-axis
4.
Y
+2
=
5(x - 3)
5.
(-3,0),(3,0), (0,-9)
3.4 Library of Functions; Piecewise-defined Functions Before getting started, review th.e following:
PREPARING FOR THIS SECTION •
Intercepts ( Section 2.2, pp. 165-167) Now Work the 'Are You
•
Prepared?, problems
Graphs of Key Equations ( Section 2.2: Example 3, p. 165; Example 10, p. 169; Example 11, p. 170; Example 12, p. 170)
on page 248.
1 Graph the Functions Listed in the Library of Functions (p.244)
OBJECTIVES
2 Graph Piecewise-defined Functions (p. 247)
Figure 25
We now introduce a few more functions to add to our list of important functions. We begin with the square root function. On page 170, we graphed the equation y = \IX. Figure 25 shows a graph of the function f(x) \IX. Based on the graph, we have the following properties:
y 6
=
-2
5
10
x
Properties of fIx) 1.
2.
3.
4.
5.
E XA M P L E 1
=
Vii
The domain and the range are the set of nonnegative real numbers. The x-intercept of the graph of f(x) \IX is O. The y-intercept of the graph of f(x) = \IX is also O. The function is neither even nor odd. It is increasing on the interval (0, ) It has a minimum value of 0 at x = O. =
00
.
G raph i n g the C u be Root F u n ction
(a) Determine whether f(x) Vx is even, odd, or neither. State whether the graph of f is symmetric with respect to the y-axis or symmetric with respect to the origin. (b) Determine the intercepts, if any, of the graph of f(x) Vx. (c) Graph f(x) Vx. =
=
=
Soluti o n
(a) Because f( -x) -\Y=X -Vx -f(x) the function is odd. The graph of f is symmetric with respect to the origin. =
=
=
SECTION 3.4
Library o f Functions; Piecewise-defined Functions
243
(b) The y-intercept is f(O) = \YO = O. The x-intercept is found by solving the equa tion f(x) O. f(x) = 0 -0X = 0 f(x) = -\YX x = 0 Cube both sides of the equation. The x-intercept is also O. ( c) We use the function to form Table 4 and obtain some points on the graph. Be cause of the symmetry with respect to the origin, we need to find only points (x, y) for which x 2:: O. Figure 26 shows the graph of f(x) -0X. =
=
Table 4
Y =
x
f(x)
0 -1 8
=
�
0 -1 2 V2
2 8
""
Figure 26
(x, y)
(0,0)
y 3
G,�)
(1, 1) (2, V2) (8,2)
1.26
2
3
-3
x
(-1, -1 ) -3 •
From the results of Example 1 and Figure 26, we have the following properties of the cube root function. Properties of f(x) 1.
2.
3.
4. 5.
EXAM P LE 2
=
Vx
The domain and the range are the set of all real numbers. The x-intercept of the graph of f(x) -0X is O. The y-intercept of the graph of f(x) = -0X is also O. The graph is symmetric with respect to the origin. The function is odd. It is increasing on the interval (- 00, 00). It does not have a local minimum or a local maximum. =
G raph i ng the Absolute Val ue F unction
(a) Determine whether f(x) = Ixl is even, odd, or neither. State whether the graph of f is symmetric with respect to the y-axis or symmetric with respect to the origin. (b) Determine the intercepts, if any, of the graph of f(x) = Ixl. (c) Graph f(x) = Ixl . Solution
(a) B ecause fe - x)
=
=
I -xl I x l = f (x )
the function is even. The graph of f is symmetric with respect to the y-axis. (b) The y-intercept is f(O) = 101 O. The x-intercept is found by solving the equa tion f(x) 0 or Ixl O. So the x-intercept is O. ( c) We use the function to form Table 5 and obtain some points on the graph. Be cause of the symmetry with respect to the y-axis, we need to find only points (x, y) for which x 2:: O. Figure 27 shows the graph of f ( x ) = Ixl· =
=
=
244
CHAPTER 3
Functions and Their Graphs
Table 5
y
x
=
f(x)
=
o
o
Figure 27
(x, y)
Ixl
2
2
(0, 0) ( 1, 1) (2,2)
3
3
(3,3)
-3 - 2 -1
3
x
-1 •
From the results of Example 2 and Figure 27, we have the following properties of the absolute value function. Properties of f(x) 1.
2.
3.
4.
5.
=
Ixl
The domain is the set of all real numbers. The range of f is { y l y � O}. The x-intercept of the graph of f(x ) = I x l is O. The y-intercept of the graph of f(x) I xl is also O. The graph is symmetric with respect to the y-axis. The function is even. It is decreasing on the interval ( -00, 0). It is increasing on the interval (0, 00). It has a local minimum of 0 at x = O. =
Seeing the Concept
I l
Graph y = x on a square screen and com pare what you see with Figure
27. Note that some graph
ing calculator s use abs(x) for absolute value.
1
Graph the Functions Listed in the Library of Functions
We now provide a summary of the key functions that we have encountered. In going through this list, pay special attention to the properties of each function, particularly to the shape of each graph. Knowing these graphs will lay the foundation for later graphing techniques. Constant Function
f(x)
=
b
b is
a real number
Figure 28 Constant Function
y
(O,b)
f(x)
=
See Figure 28. The domain of a constant function is the set of all real numbers; its range is the set consisting of a single number b. Its graph is a horizontal line whose y-intercept is b. The constant function is an even function whose graph is constant over its domain.
b
x
Identity Function
Figure 29 Identity Function f(x)
=
x
f(x)
=
x
See Figure 29. The domain and the range of the identity function are the set of all real num bers. Its graph is a line whose slope is m = 1 and whose y-intercept is O. The line consists of all points for which the x-coordinate equals the y-coordinate. The identi ty function is an odd function that is increasing over its domain. Note that the graph bisects quadrants I and III.
SECTION 3.4
Figure 30
Library of Functions; Piecewise-defined Functions
245
Square Function
Square Function
f(x)
f ( x)
X2
=
(2, 4)
(-2, 4)
4 x
-4
=
x2
See Figure 30. The domain of the square function f is the set of all real numbers; its range is the set of nonnegative real numbers. The graph of this function is a parabola whose i ntercept is at (0, 0). The square function is an even function that is decreasing on the interval ( -00, 0) and increasing on the interval (0, ) (0
.
Cube Function
Figure 31 Cube Function
f (x) f(x)
=
x3
4
x
-4
(0
.
Square Root Function
Figure 32 Square Root Function
f(x)
=
Vx
See Figure 32. The domain and the range of the square root function are the set of nonnega tive real numbers. The intercept of the graph is at (0, 0). The square root function is neither even nor odd and is increasing on the interval (0, )
-JX
2
-1
x3
See Figure 3l. The domain and the range of the cube function are the set of all real numbers. The i ntercept of the graph is at (0, 0). The cube function is odd and is increasing on the i nterval ( -00, )
f(x)
Y
=
(0
5 x
(0, 0)
Cube Root Function
f (x)
Figure 33
.
-0X
See Figure 33. The domain and the range of the cube root function are the set of all real num bers. The intercept of the graph is at (0,0). The cube root function is an odd func tion that is increasing on the interval ( -00,00 ) .
Cube Root Function
Y 3
Reciprocal Function 3 x
-3
3 ( -2,-'V2)
f (x)
( -1 , - 1) -3
Figure 34 Reciprocal Function
Y 2
f(x) = 1.. x
1
x
1 Refer to Example 12, page 170, for a discussi on of the equation y = -. See x Figure 34. The domain and the range of the reciprocal function are the set of all nonzero real numbers. The graph has no intercepts. The reciprocal function is decreasing on the intervals ( -00, 0 ) and ( 0, 00 ) and is an odd function. Absolute Value Function
2 x
-2
f (x) See Figure 35 on page 246.
Ixl
246
CHAPTER 3
Functions and Their Graphs
Figure 35
The domain of the absolute value function is the set of all real numbers; its range is the set of nonnegative real numbers. The intercept of the graph is at (0, 0). If x 2: 0, then I(x) = x, and the graph of 1 is part of the line y = x; if x < 0, then I(x) = -x, and the graph of 1 is part of the line y = -x. The absolute value func tion is an even function; it is decreasing on the interval (-00, 0) and increasing on the interval (0, 00). The notation int( x) stands for the largest integer less than or equal to x. For example,
Absolute Value Function
Y
fIx) I x I =
int ( l )
=
1, int(2.5) = 2, int
(�) = 0, int ( -%) = - 1,
int(7T) = 3
This type of correspondence occurs frequently enough in mathematics that we give it a name. DEFINITION
Greatest Integer Function
I(x) Table 6 Y =
x
-1 2 1 -4 0
-
4 -1 2 3 4
=
fIx) int(x)
-1 -1
-1 0 0 0 0
(x,y) (-1,
=
Figure 36
greatest integer less than or equal to x
Y 4
Greatest Integer Function
2
-1
e-o
e-o
2
-2
4
x
e-o
(0,0)
G,O) G,O) G,O)
int(x) *
We obtain the graph of I(x) = int(x) by plotting several points. See Table 6. For values of x, - 1 ::; x< 0, the value of I(x) = int(x) is - 1; for values of x, a ::; x < 1, the value of 1 is O. See Figure 36 for the graph.
- 1)
(-�, - 1 ) ( -�, )
=
-3
The domain of the greatest integer function is the set of all real numbers; its range is the set of integers. The y-intercept of the graph is O. The x-intercepts lie in the interval [0, 1 ) . The greatest integer function is neither even nor odd. It is constant on every interval of the form [k, k + 1 ) , for k an integer. In Figure 36, we use a solid dot to indicate, for example, that at x = 1 the value of 1 is 1( 1) = 1; we use an open circle to illustrate that the function does not assume the value of a at x = l. From the graph of the greatest integer function, we can see why it is also called a step function. At x = 0, x = ±1, x = ±2, and so on, this function exhibits what is called a discontinuity; that is, at integer values, the graph suddenly "steps" from one value to another without taking on any of the intermediate values. For example, to the immediate left of x = 3, the y-coordinates of the points on the graph are 2, and to the immediate right of x = 3, the y-coordinates of the points on the graph are 3. COMMENT When graphing a function using a graphing utility, you can choose either the connected
mode, in which points plotted on the screen are connected, making the graph appear without
any breaks, or the dot mode. in which only the points plotted appear. When graphing the greatest
integer function with a graphing utility. it is necessary to be in the dot mode. This is to prevent the utility from "connecting the dots" when Figure
37.
f(x)
changes from one integer value to the next. See
'" Some books use the notation fIx ) = [x l instead of int(x ) .
SECTION 3.4
Figure 37
((x)
6
= i nt(x)
Library of Functions; Piecewise-defined Functions
247
6
I
/
-2
1----1'-2
�:
6
-2 (b) Dot mode
(a) Connected mode
•
The functions that we have discussed so far are basic. Whenever you encounter one of them, you should see a mental picture of its graph. For example, if you encounter the function f(x) = x2, you should see in your mind's eye a picture like Figure 30. "'!I'W:=="-' Now Work 2
T
PROBlEMS 9
HR0UGH 1 6
Graph Piecewise-defined Functions
Sometimes a function is defined differently on different parts of its domain. For example, the absolute value function f(x) Ixl is actually defined by two equa tions: f(x) = x if x 2: 0 and f(x) = -x if x < O. For convenience, we generally combine these equations into one expression as =
f(x)
=
Ixl
=
{ -xX if'f Xx 1
0 <0 2:
When functions are defined by more than one equation, they are called piecewise defined functions. Let 's look at another example of a piecewise-defined function. E XA M P L E 3
A n alyzi n g a Piecewise-defined Function
The function f is defined as f(x)
=
{ �2
-x + 1 if - 1 :S x < 1 if x 1 if x > 1
(a) Find f(O), f ( 1 ) , and f(2). (c) Graph f. S o l ution
=
(b) Determine the domain of f. (d) Use the graph to find the range of f.
(a) To find f(O), we observe that when f(x) -x + 1 . So we have
x
0 t he equation for f is given by
=
=
f(O) When x
=
=
-0 + 1
1, the equation for f is f(x) f(1)
When x
Figure 38
5 );
(-1 , 2)
,\1
(0 , 1 ) �' ,
-3
(2 4)
•
/
3
22
2 =
=
x2. So 4
(b) To find the domain of f, we look at its definition. We conclude that the domain of f is { x i x 2: - 1 } , or the interval [ - 1 , ) (c) To graph f, we graph "each piece." First we graph the line y -x + 1 and keep only the part for which -1 :S x < 1 . Then we plot the point ( 1 , 2) be cause, when x = 1 , f(x) = 2. Finally, we graph the parabola y = x2 and keep only the part for which x > 1. See Figure 38. .
=
/
-x+ 1
=
1
2. Thus,
(0
(1 , 2
, Y=
2, the equation for f is f(x) f(2)
Y = x2
Y
=
=
=
=
x
248
CHAPTER 3
Functions and Their Graphs
(d) From the graph, we conclude that the range of f is { y l y > O } , or the interval (0,00 ) .
•
Now Work P R O B L E M
1l'l
EXAM PLE 4
29
Cost of E lectricity
In May 2006, Commonwealth Edison Company supplied electricity to residences for a monthly customer charge of $7.58 plus 8.275¢ per kilowatt-hour (kWhr) for the first 400 kWhr supplied in the month and 6.208¢ per kWhr for all usage over 400 kWhr in the month. (a) What is the charge for using 300 kWhr in a month? (b) What is the charge for using 700 kWhr in a month? (c) If C is the monthly charge for x kWhr,express C as a function of x. Source: Commonwealth Edison Co., Chicago, Illinois, 2006.
Sol ution
(a) For 300 kWhr, the charge is $7.58 plus 8.275¢ Charge
=
$7.58
+
=
$0.08275(300)
$0.08275 per kWhr. That is, =
$32.41
(b) For 700 kWhr, the charge is $7.58 plus 8.275¢ per kWhr for the first 400 kWhr plus 6.208¢ per kWhr for the 300 kWhr in excess of 400. That is, Charge (c)
=
$7.58 + $0.08275(400) + $0.06208(300)
=
$59.30
If 0 ::; x ::; 400, the monthly charge C (in dollars) can be found by multiply ing x times $0.08275 and adding the monthly customer charge of $7.58. So, if o ::; x ::; 400, then C(x) 0.08275x + 7.58. For x > 400, the charge is 0.08275 (400) + 7.58 + 0.06208(x - 400) , since - 400 equals the usage in excess of 400 kWhr, which costs $0.06208 per kWhr. That is, if x > 400, then =
x
Figure 39 �
C(x)
80 1++H1++H+H++++l-I++HI++H+H+H++I++H-H
� 60 1++t11++H+H++++l-1-H o en
"0
-;; 40
�
u
..c
7
=
= =
.58 +++l-1+l+1+l+1+I±iw-H-i'ffi-!-H-H+H
1n+1++H��I++H�TItt��-H
The rule for computing C follows two equations: _ C(x) -
20 l+t1tft1.!1'H+ttt l
1 00 200 300 400 500 600 700
Usage (kWhr)
0.08275(400) + 7.58 + 0.06208(x - 400) 40.68 + 0.06208(x - 400) 0.06208x + 15.848 0.08275X { 0.06208x
+ 7.58 if 0 ::; x ::; 400 + 1 5.848 if x > 400
See Figure 39 for the graph. •
3.4 Assess Your Understanding
Answers are given at the end ofthese exercises. [{you get a wrong answel; read the pages listed in red. 3. List the intercepts of the equation y = - 8. (pp. 1 65-1 67) Sketch the graph of y \IX . (p. l70) 1 Sketch the graph of y - . (p. 170) x
'Are You Prepa red?' 1. 2.
x3
=
=
Concepts and Vocabulary
4. The function f(x) = x2 is decreasing on the interval __. S. When functions are defined by more than one equation, they functions. are called 6. True or False The cube function is odd and is increasing on the interval ( - (X), (0 ) .
The cube root function is odd and is de creasing on the interval ( - (X) , (0 ) . True or False The domain and the range o f the reciprocal function are the set of all real numbers.
7. True o r False
8.
SECTION 3.4
Library of Functions; Piecewise-defined Functions
249
Ski l l Bu ilding In Problems
9-16,
match each graph to its function.
A. ConSLanl function D. Cube function G. Absolute value function
"· 13.
V /
C.
B. Identity function E. Square root function H. Cube root fitnction
F
. 10.
Square function Reciprocal funClion
11.
12.
/
i�
15.
V
16.
�
1
_ _ _ - - - _ _ - _ _ _ _ _
In Problems
1 7-24,
\1
sketch the graph of each function. Be sure to label three points on the graph.
17. f(x) = x 1 21. f(x) = x
{
x2 2 2x + 1 find : (a) f ( -2)
25. If f(x) =
18. f(x) = x2
19. f(x) = x3
22. f(x) = Ix l
23. f(x) = -0X
if x < 0 if x = O if x > 0 (b) f ( O )
26. If f(x) = (c) f(2)
In Problems
28. If f(x) =
-
38. f(x) =
+
1
{ X3
3x + 2
if x < - 1 if x = - 1 if x > - 1
(c ) f( O )
(b) f( -1)
if -2 � x < 1 if l � x � 4 (c) f( l ) (b) f( O )
{ 2X 1
X
{I
+
2
if x of. a if x = 0
30. f(x) =
3- 3 if
x < -2 if x 2: -2
x
{2 - x �
{3X if
33. f(x) =
x of. 0 if x = 0
4
r J {� 5
-x
+
if x < 0 if x 2: 0
3
if �x< 1 if x > 1
39. f(x) = 2 int(x)
-3
�
VX
31. f(x) =
if -2 � x < 1 if x = 1 if x > 1
2
if x < 0 if x
34. f(x) =
37. f(x) =
2: a
{ 2X � .)x
r-3 x
+ ?
-
+
5 - x
(0, 0)
�
(2, 1 )
( 2, 1 )
(-1 , 1 )
y
2
2
-2
( d) f (3 )
(0, 0)
-2 ( - 1 , - 1)
2
x
-2
(0, 0)
_
5
x< 1 x 2: 1
-3
if �x< if x = a if x > a
if -2 � x < 0 if x > a
{ Ix l x3
40. f(x) = int(2x)
U
y
�
y
3
(b) Locate any intercepts. (d) Based on the graph,find the range.
In Problems 41-44, the graph of a piecewise-defined function is given. Write a definition for each function. U
3
29-40:
{x + 32. f(x) = -2x 35. f(x) =
�
2x2
find: (a) f( - 1 )
( d) f( 3 )
(a) Find the domain of each function. (c) Graph each function.
29. f(x) =
0
24. f(x) =
find: (a) f( -2)
if - 1 � x � 2 { 2X - 4 if 2 < x � 3 x3 2 find: (a) f( O ) (b) f( l ) (c) f(2)
27. If f(x) =
{-3l:
20. f(x) = \IX
-2
a
250 45.
Functions and Their Graphs
CHAPTER 3
=
If f(x)
int(2x), find (a) f( 1.2),
46. If f(x) = in
{1} find
(a) f ( 1 .2),
(b) f( 1 .6),
(c) f( - 1 .8).
(b) f ( 1 .6),
(c) f( - 1 .8).
Appl ications a n d Extensions 47.
Sprint PCS offers a monthly cellular phone plan for $35. It includes 300 anytime minutes and charges $0.40 per minute for additional minutes. The follow ing function is used to compute the monthly cost for a sub scriber: Cell Phone Service
C(x)
=
{
35 0.40x - 85
49.
if 0 < x :$ 300 if x > 300
where x is the number of anytime minutes used. Compute the monthly cost of the cellular phone for use of the follow ing anytime minutes: (c) 301 (a) 200 (b) 365 Source: 48.
Sprint PCS
The short-term (no more than 24 hours) parking fee F (in dollars) for park ing x hours at O'Hare International Airport's main parking garage can be modeled by the function
Source:
Parking at O'Hare International Airport
F(x)
=
{�
int(X + 1 ) + 1 50
if 0 if 3 if 9
< < :$
50.
x :$ 3 x<9 x :$ 24
Determine the fee for parking in the short-term parking garage for (a) 2 hours (b) 7 hours (c) 15 hours (d) 8 hours and 24 minutes Source:
51.
In May 2006, Peoples Energy had the following rate schedule for natural gas usage in single-family residences: Monthly service charge $9.45 Per therm service charge 1 st 50 therms $0.36375/therm Over 50 therms $0.1 1445/therm Gas charge $0.7958/therm (a) What is the charge for using 50 therms in a month? (b) What is the charge for using 500 therms in a month? (c) Construct a function that relates the monthly charge C for x therms of gas. (d) Graph this function.
Cost of Natural Gas
O'Hare Intemational Airport
Peoples Energy, Chicago, Illinois, 2006
In May 2006, Nicor Gas had the fol lowing rate schedule for natural gas usage in single-family residences: $8.85 Monthly customer charge Distribution charge $0.1 557/therm 1st 20 therms $0.0663/therm Next 30 therms $0.051 9/therm Over 50 therms $0.66/therm Gas supply charge (a) What is the charge for using 40 therms in a month? (b) What is the charge for using 202 therms in a month? (c) Construct a function that gives the monthly charge C for x therms of gas. (d) Graph this function.
Cost of Natural Gas
Source:
Nicor Gas, Aurora, Illinois, 2006
Federal Income Tax Two 2006 Tax Rate Schedules are given in the accompanying table. I f x equals taxable income and y equals the tax due, construct a function y = f(x) for Schedule X. REVISED 2006 TAX RATE SCHEDULES
Schedule X- Single If Taxable Income Is over
But Not over
Schedule Y-l -Married Filing jointly or qualifying Widow(er) Plus This
The Tax Is This Amount
%
Of the Excess over
If Taxable Income Is over
But Not over
The Tax Is This Amount
Plus This %
Of the Excess over
0
7550
0.00
+
1 0%
$0
$0
1 5,1 00
0.00
+
1 0%
$0
7550
30,650
755.00
+
1 5%
7550
1 5,1 00
61,300
1 51 0.00
+
1 5%
1 5,100
30,650
74,200
4220.00
+
25%
30,650
61,300
1 23,700
8440.00
+
25%
61 ,300
74,200
1 54,800
1 5,1 07.50
+
28%
74,200
1 23,700
1 88,450
24,040.00
+
28%
1 23,700
1 54,800
336,550
37,675.50
+
33%
1 54,800
1 88,450
336,550
42,1 76.00
+
33%
1 88,450
97,653.00
+
35%
336,550
336,550
-
91 ,043.00
+
35%
336,550
336,550
Source: Internal Revenue Service
52.
Refer to the revised 2006 tax rate schedules. If x equals taxable income and y equals the tax due, construct a function y = f(x) for Schedule Y-1 .
Federal Income Tax
53.
A trucking company trans ports goods between Chicago and New York, a distance of 960 miles. The company's policy is to charge, for each pound,
Cost of Transporting Goods
SECTION 3.4
54.
55.
56.
$0.50 per mile for the first 100 miles, $0.40 per mile for the next 300 miles, $0.25 per mile for the next 400 miles, and no charge for the remaining 160 miles. (a) Graph the relationship between the cost of transporta tion in dollars and mileage over the entire 960-mile route. (b) Find the cost as a function of mileage for hauls between 100 and 400 miles from Chicago. (c) Find the cost as a function of mileage for hauls between 400 and 800 miles from Chicago. Car Rental Costs An economy car rented in Florida from National Car Rental® on a weekly basis costs $95 per week. Extra days cost $24 per day until the day rate exceeds the weekly rate, in which case the weekly rate applies. Also, any part of a day used counts as a full day. Find the cost C of renting an economy car as a piecewise-defined function of the num ber x of days used, where 7 :S x :S 14. Graph this function. Minimum Payments for Credit Cards Holders of credit cards issued by banks, department stores, oil companies, and so on, receive bills each month that state minimum amounts that must be paid by a certain due date. The minimum due de pends on the total amount owed. One such credit card com pany uses the following rules: For a bill of less than $10, the entire amount is due. For a bill of at least $10 but less than $500, the minimum due is $10. A minimum of $30 is due on a bill of at least $500 but less than $1000, a minimum of $50 is due on a bill of at least $1000 but less than $1500, and a minimum of $70 is due on bills of $ 1500 or more. Find the function f that describes the minimum payment due on a bill of x dollars. Graph f. Interest Payments for Credit Cards Refer to Problem 55. The card holder may pay any amount between the minimum due and the total owed. The organization issuing the card charges the card holder interest of 1 . 5 % per month for the first $1000 owed and 1 % per month on any unpaid balance over $1000. Find the function g that gives the amount of in terest charged per month on a balance of x dollars. Graph g.
Library of Functions; Piecewise-defined Functions
251
Wind Chill The wind chill factor represents the equiva lent air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed. One formula for computing the equivalent temperature is
57.
W
5S.
59.
=
{
t
O :s V < 1.79 ( 10.45 + 10VV - v) (33 - t ) 33 1 .79 :S v :S 20 22.04 33 - 1 .5958(33 - t ) v > 20
where v represents the wind speed (in meters per second) and t represents the air temperature (0C). Compute the wind chill for the following: (a) An air temperature of 1 0°C and a wind speed of 1 me ter per second ( m/sec) (b) An air temperature of 10°C and a wind speed of 5 m/sec (c) An air temperature of lOoC and a wind speed of 1 5 m/sec (d) An air temperature of lOoC and a wind speed of 25 m/sec (e) Explain the physical meaning of the equation corre sponding to 0 :S V < 1.79. (f) Explain the physical meaning of the equation corre sponding to v > 20. Wind Chill Redo Problem 57(a)-(d) for an air temperature of -10°C. First-class Mail In 2006 the U.S. Postal Service charged $0.39 postage for a first-class letter weighing up to 1 ounce, plus $0.24 for each additional ounce up to 13 ounces. First-class rates do not apply to letters weighing more than 13 ounces. Write a piecewise-defined function C for the first-class postage charged for a letter weighing x ounces. Graph the function. Source:
United States Postal Service
Discussion a n d Writing
� In Problems 60-67, use a graphing utility. 60. Exploration Graph y = x2. Then on the same screen graph y = x2 + 2, followed by y = x2 + 4, followed by y = x2 - 2. What pattern do you observe? Can you predict the graph of y = x2 - 4? Of y = x2 + 5? 61. Exploration Graph y = x2 Then on the same screen graph y = (x - 2)2, followed by y = (x - 4)2, followed by y = (x + 2f What pattern do you observe? Can you predict the graph of y = (x + 4)2? Of y = (x - Sf? 62. Exploration Graph y = I xl. Then on the same screen graph 1 y = 21xl, followed by y = 41xl, followed by y = 2: lxl. What
63.
64.
pattern do you observe? Can you predict the graph of y = � Ixl ? Of y = 51xl? 4 Exploration Graph y x2. Then on the same screen graph y = -x2. What pattern do you observe? Now try y = Ixl and y = - Ixl. What do you conclude?
65.
y = 2x
+ 1 and y
=
2( -x)
+
1 . What do you conclude?
Graph y = x2, Y = X4, and y = x6 on the same screen. What do you notice is the same about each graph? What do you notice that is different?
66.
Exploration
67.
E xploration
6S.
Consider the equation
Graph y = x3, Y = xS, and y = x7 on the same screen. What do you notice is the same about each graph? What do you notice that is different?
y=
{�
i f x is rational if x is irrational
Is this a function? What is its domain? What is its range? What is its y-intercept, if any? What are its x-intercepts, if any? Is it even, odd, or neither? How would you describe its graph?
=
E xploration Graph y = \IX. Then on the same screen graph y = V=X. What pattern do you observe? Now try
Graph y = x3 Then on the same screen graph (x - 1 )3 + 2. Could you have predicted the result?
Exploration
y=
69.
Define some functions that pass through (0, 0) and ( 1 , 1 ) and are increasing for x 2: O. Begin your list with y = \IX, y = x, and y = x2. Can you propose a general result about such functions?
252
CHAPTER 3
Functions and Their Graphs
'Are You Prepa red' Answers 2.
1. y
2
~
3.
(0,-8), (2, 0)
(1 , 1 ) 2
x
(-1 , -1 )
x
OBJECTIVES
1 Graph Functions Using Vertical and Horizontal Shifts (p. 252) 2 Graph Functions Using Compressions and Stretches (p.255) 3 Graph Functions Using Reflections about the x-Axis and the y-Axis
(p.258)
At this stage, if you were asked to graph any of the functions defined by y = x, y = x2 , Y = x3 , Y = Vx, Y = \Y.X, y = I x!, or y 1.. , your response should be, x "Yes,I recognize these functions and know the general shapes of their graphs." (If this is not your answer, review the previous section, Figures 29 through 35.) Sometimes we are asked to graph a function that is "almost" like one that we already know how to graph. In this section,we look at some of these functions and develop techniques for graphing them. Collectively, these techniques are referred to as transformations. =
1 EXAM P L E 1
Graph Functions Using Vertical and Horizontal Shifts Vertical Shift U p
Use the graph of f(x) = x2 to obtain the graph of g(x) Solution
=
x2 + 3.
We begin by obtaining some points on the graphs of f and g. For example, when x = 0, then y f(O) = 0 and y = g(O) = 3. When x = 1, then y = f(l) = 1 and y = g ( l ) 4. Table 7 lists these and a few other points on each graph. Notice that each y-coordinate of a point on the graph of g is 3 units larger than the y-coordinate of the corresponding point on the graph of f. We conclude that the graph of g is identical to that of f, except that it is shifted vertically up 3 units. See Figure 40. =
=
Table 7 x
Figure 40 y = fIx) = J?
-2 -1 0
4
2
4
0
y = g(x) = J? + 3
7 4 3 4 7
-3 •
SECTION 3.5
I
Graphing Techniques: Transformations
253
Exploration
Figure 41
On the same screen, graph each of the fol lowing functions: Y1 = x2
Y1
=
Y = x2 + 2 2 Y3 = x2 - 2
X2 _ - 6 1----������� 6
\
.
.
\
Result Fig u re 41 illustrates the g raphs. You should have observed a general pattern. With Y1 = x2 on the screen, the g raph of Y2 = x2 + 2 is identical to that of Y1 = x2, except that it is shifted vertically u p 2 units. The g raph of Y3 = x2 - 2 is identical to that of Y1 = x2, except that it is shifted vertically down 2 u nits.
We are led to the following conclusions: If a positive real number k is added to the right side of a function y = f(x), the graph of the new function y = f (x) + k is the graph of f shifted verti cally up k units. If a positive real number k is subtracted from the right side of a function y = f (x), the graph of the new function y = f (x) - k is the graph of f shifted vertically down k units. Let's look at another example. EXAM P L E 2
Solutio n
Vertical S hift Down
Use the graph of f(x) = x2 to obtain the graph of g(x) = x2 - 4.
Table 8 lists some points on the graphs of f and g. Notice that each y-coordinate of g is 4 units less than the corresponding y-coordinate of f. The graph of g is identical to that of f, except that it is shifted down 4 units. See Figure 42. Figure 42
Table 8 X
y= =
-2 -1 0
4
2
4
0
f(x) x2
y = =
g(x) x2 -
0 -3 -4 -3 0
4
Down 4 units
Y = X2 (2, 4)
1 -5
(0, - 4) •
_ �_ C!lll
E XA M P L E 3
Now Work P R O B L E M 3 5
H orizontal Shift to the Right
Use the graph of f(x) = x2 to obtain the graph of g(x) = (x - 2f Solution
The function g ( x) = (x - 2 f is basically a square function. Table 9 lists some points on the graphs of f and g. Note that when f(x) = 0 then x = 0, and when g(x) = 0, then x = 2. Also, when f(x) = 4, then x = -2 or 2, and when g(x) = 4, then x = 0 or 4. We conclude that the graph of g is identical to that of f, except that it is shifted 2 units to the right. See Figure 43.
254
CHAPTER 3
Functions and Their Graphs
Table 9
Figure 43
x
g(x)
= (x
4 0 4 16
-2 0 2 4
;J
Y =
y = f(x) =�
-
Y
2)2
16 4 0 4
Y = x2
Y = (x- 2)2
( - 2, 4)
( 4, 4)
R ight 2 units
(0, 0)
(2, 0)
4
x •
Exploration
Figure 44
On the same screen, graph each of the following functions:
x2
y1 =\
Yl Y2 Y3
\ 6
x2 = (x - 3 / = (x + 2) 2
=
Result Figure 44 ill ustrates the graphs. You should observe the fol lowing pattern. With the graph of Y1 x2 on the screen, the graph of Y2 = (x - 3)2 is identical to that of Y1 x2, except that it is shifted horizontally to the right 3 u nits. The graph of Y3 = (x + 2) 2 is identical to that of Y1 x2, except that it is shifted horizonta l ly to the left 2 u n its.
=
=
=
We are led to the following conclusion. If the argument x of a function f is replaced by x h, h > 0, the graph of the new function y f(x h ) is the graph of f shifted horizontally right h units. If the argument x of a function f is replaced by x + h, h > 0, the graph of the new function y = f(x + h ) is the graph of f shifted horizontally left h units. -
=
E XA M P L E 4
Horizontal S h ift to the Left
Use the graph of f(x) Solution
-
=
x2 to obtain the graph of g(x)
=
(x + 4f
Again, the function g(x) = (x + 4) 2 is basically a square function. Its graph is the same as that of f , except that it is shifted horizontally 4 units to the left. See Figure 45. Figure 45
Y
Y = (X t 4) 2
5
( - 6 , 4)
(2, 4)
x Left 4 units • = =� -
Now Work P R O B L E M 3 9
Vertical and horizontal shifts are sometimes combined.
SECTION 3.5
E XA M P L E 5
Graphing Techniques: Transformations
255
C o m b i n i n g Vertical and H orizontal Sh ifts
Graph the function: f(x)
=
(x
+
3? - 5
We graph f in steps. First,we note that the rule for f is basically a square function, so we begin with the graph of y x2 as shown in Figure 46(a). Next, to get the graph of y (x + 3?, we shift the graph of y = x2 horizontally 3 units to the left. See Figure 46(b). Finally, to get t he graph of y (x + 3? - 5,we shift the graph of y (x + 3)2 vertically down 5 units. See Figure 46(c). Note the points plotted on each graph. Using key points can be helpful in keeping track of the transformation that has taken place.
Solution
=
=
=
=
Figure 46
Y 5
( -2, 1 ) -5
5 x
5 x -5
5
( -2, -4) -5 2 y = ( x + 3)
Replace x b y x + 3 ; Horizontal shift left 3 units
(b)
(a)
I
Check: Graph
=
2
3)2 -
•
(e)
•
5 and compare the g raph to Figu re 46(c).
Now Work P R O B L E M 4 1
Graph Functions Using Compressions and Stretches Vertical Stretch
Use the graph of f(x) Solution
(x +
2 Y = (x + 3) - 5
Subtract 5; Vertical shift down 5 units
In Example 5, if the vertical shift had been done first, followed by the horizon tal shift, the final graph would have been the same. Try it for yourself. ,,�-
E XA M P L E 6
f(x)
Y1 =
Vertex ( - 3, - 5)
=
Ixl to obtain the graph of g(x)
=
2 1 xl·
To see t he relationship between the graphs of f and g, we form Table 10, listing points on each graph. For each the y-coordinate of a point on t he graph of g is 2 times as large as the corresponding y-coordinate on t he graph of .f. The graph of f (x) = Ixl is vertically stretched by a factor of 2 to obtain the graph of g( ) 2 1 xl [for example, ( 1, 1 ) is on the graph of f, but ( 1 , 2) is on the graph of g]. See Figure 47. x,
x
Table 1 0
x -2
Figure 47 Y =
fIx) I xl
= 2
y =
=
g(x) 21xl
4 2
-1
0
=
0
0 2
2
2
4 •
256
C H A PTER 3
Functions and Their Graphs
E XA M P L E 7
Vertical Compression
Use the graph of I(x) Solution
=
.
Ixl to obtam the graph of g(x)
1
=
"2 lxl.
1 For each x, the y-coordinate of a point on the graph of g is "2 as large as the corresponding y-coordinate on the graph of f. The graph of I(x)
�
compressed by a factor of to obtain the graph of g(x)
=
=
Ixl is vertically
� Ixl [for example, (2, 2)
is on the graph of I,but (2, 1 ) is on the graph of g.]. See Table 1 1 and Figure 48. Table 1 1 y = ((x) = I xl
x
2
-2 -1 o
o
2
2 •
When the right side of a function y I(x) is multiplied by a positive number a, the graph of the new function y = al(x) is obtained by multiplying each y-coordinate on the graph of y = I(x) by a. The new graph is a vertically com pressed (if 0 < a < 1 ) or a vertically stretched (if a > 1 ) version of the graph of y I(x). =
=
""
..... Now Work P R O B l EM
43
What happens if the argument x of a function y = I(x) is multiplied by a posi tive number a, creating a new function y = I(ax)? To find the answer, we look at the following Exploration. Exploration On the same screen, g raph each of the fol lowing functions: Y1
Y 2
Result
of Y1
Figure 49
=
=
((x)
=
= ((2x)
x2
= (2X) 2
You should have obtained the g raphs shown in Figure 49. The graph of Y = (2x)2 is the graph 2 x2 compressed horizontally. Look also at Table 1 2(a).
Table 1 2 X '1 1 (0 (0 .� .,� 1 1 , '1 '1 iii B Ii '1 iii 2�6 Y , a ( 2X ) 2 (a)
M
(0 1 '1 i ii Ii '1 2�1i 1(02'1
(b)
SECTION 3.5
257
Graphing Techniques: Transformations
Notice i n Table 1 2(a) that (1, 1 ), (2, 4), (4, 1 6), and ( 1 6, 256) are points on the graph of Y1 = x2• Also, (0.5, 1 ), ( 1 , 4). (2, 1 6), and (8, 256) are points on the graph of Y2 (2X)2. For equal y-coordinates, the =
1
x-coordinate on the graph of Y2 is - the x-coordi nate of Y1 . The graph of Y2 2 1 2 tiplying the x-coordi nate of each point on the graph of Y1 = x by "2 ' The graph of Y3
=
2
(�x) i s the graph of Y1
(2X)2 is obtained by mul-
x2 stretched horizontal ly. Look a t Table 1 2(b). Notice
and (4, 1 6) are points on the graph of Y1 x2. Also, (1 , 0.25), (2, 1 ) , (4, 4), 2 and (8, 1 6) are points on the graph of Y3 = (]"x) . For equal y-coordinates, the x-coordinate on the 2 graph of Y3 is 2 times the x-coordinate on Y1 . The graph of Y3 = '2 x is obtained by mUltiplying the that (0.5, 0.25), (1 , 1 ),
(2, 4),
=
=
=
x-coordinate of each point on the graph of Y1
(1 )2
2
x by a factor of 2.
=
If the argument x of a function y = f(x) is multiplied by a positive number the graph of the new function y = f(ax ) is obtained by multiplying each x-coordinate of y = f(x) by � . A horizontal compression results if > 1, and a,
a
a
a horizontal stretch occurs if 0 <
a
< 1.
Let's look a t a n example. E XA M P L E 8
G raphi n g Using Stretches and C o m p ressio n s
The graph of y (a) y Figure 5 0
=
=
f(x) is given in Figure 50. Use this graph to find the graphs of:
(b) y
3f( x )
=
f(3x)
Y
rr
-1
'2
1T
31T - (3; , 1)
�
V
21T fu!
2
x
-
Y = f( xl
S o l ution
(a) The graph of y 3f( x ) is obtained by multiplying each y-coordinate of y by a factor of 3. See Figure 51(a). =
(b) The graph of y
=
f(3x) is obtained from the graph of y
each x-coordinate of y Figure 51
Y
(� , 3)
( 52 , 3)
=
1[
2 (:IT 1 )/ ( 561T ' 1 )
-1
I
1T
-3 y=
3 f(x) (a)
..-=>� -
21T
31T
X
�1
(� , - 1 )
-2
f(x) by multiplying
y
x
-1
f(x)
� . See Figure 51(b).
f(x) by a factor Of
J'
'2
=
=
Now Work P R O B L E M S 6 S ( e )
3
Y = f( x)
(b) AND
•
(9)
258
CHAPTER 3
Functions and Their Graphs
3
Graph Functions Using Reflections about the x-Axis and the y-Axis
EXAM PLE 9
Reflection about the x-Axis
Graph the function: f (x) Sol ution
=
-
x2
We begin with the graph of y = x2 , as shown in black in Figure 52. For each point (x, y) on the graph of y x2 , the point (x, -y) is on the graph of y = -x2, as indicated in Table 13. We can draw the graph of y = x2 by reflecting the graph of y = x2 about the x-axis. See Figure 52. =
-
Figure 52
Y=
Table 1 3
x2
Y
-2 -1 0
(2, 4)
-4
x
4
= 4 0 4
2
x
y = - J? -4 -1 0 -1 -4
J?
•
When the right side of the function y f(x) is multiplied by - 1 , the graph of the new function y f (x) is the reflection about the x-axis of the graph of the function y f(x). =
=
-
=
Il!I!
E XA M P L E 1 0
i;> -
Now Work P R O B L EM 4 7
Reflecti on about the y-Axi s
Graph the function: f(x) Solution
=
v=x
First, notice that the domain of f consists of all real numbers x for which -x ;::: 0 or, equivalently,x ::; O. To get the graph of f(x) = V=X , we begin with the graph of y = \IX, as shown in Figure 53. For each point (x, y ) on the graph of y \IX , t h e point ( -x, y) i s on the graph o f y V=X . We obtain the graph of y = v=x by reflecting the graph of y = \IX about the y-axis. See Figure 53. =
=
Figure 53 Y
4
•
When the graph of the function y = f(x) is known,the graph of the new func tion y = f( - x) is the reflection about the y-axis of the graph of the function y = f(x).
SECTION 3.5
Graphing Techniques: Transformat ions
259
SUMMARY O F GRAPH ING TECHNIQUES To Graph:
Draw the Graph of f and:
Functional Change to f(x)
Raise the graph of f by k units. Lower the graph of f by k units.
Add k to f(x). Subtract k from f(x).
Shift the graph of f to the left h units. Shift the graph of f to the right h units.
Replace x by x + h . Replace x by x - h .
Vertical shifts
y = f(x) + k, k > 0 y = f(x) k, k > 0 -
Horizontal shifts
y = f(x + h ), h > 0 y = f(x - h ) , h > 0 Compressing or stretching
y = af(x), a > 0
y
= f(ax), a > 0
Multiply f(x) by a. Multiply each y-coordinate of y = f(x) by a. Stretch the graph of f vertically if a > l . Compress the graph of f vertically i f 0 < a < l . Multiply each x-coordinate o f y = f(x) by .l. Replace x by ax. a Stretch the graph of f horizontally if 0 < a < l . Compress the graph of f horizontally i f a > l .
Reflection about the x-axis
y = -f(x)
Reflect the graph of f about the x-axis.
Multiply f(x) by - l .
Reflect the graph of f about the y-axis.
Replace x by -x.
Reflection about the y-axis
y = f(- x)
The examples that follow combine some of the procedures outlined in this sec tion to get the required graph. EXAM P L E 1 1
Determi n i n g the F unction Obtai ned from a Series of T ransformations
Find the function that is finally graphed after the following three transformations are applied to the graph of y = lxi1. 2.
3. S o l ution
1.
2.
3.
Shift left 2 units. Shift up 3 units. Reflect about the y-axis. Shift left 2 units: Replace x by x + 2. Shift up 3 units: Add 3. Reflect about the y-axis: Replace x by -x.
'!'I!:==-- Now Work E XA M P L E 1 2
PROBlEM
•
27
C o m b i n i ng G raphi n g P roced u res
. Graph the functlOn: f(x) = S o l ution
y = Ix + 2 1 y = Ix + 2 1 + 3 y = I -x + 2 1 + 3
3 + x-2
--
1
We use the following steps to obtain the graph of f: 1
STEP 1:
y=-
Reciprocal function.
STEP 2:
3 y=x
Multiply by
x
3; vertical stretch of the graph of y
=
1
-
x
by a factor of 3.
260
C H A PTER 3
Functions and Their Graphs
STEP 3:
y =
STEP 4:
y =
3 x - 2 3 -- + 1 x - 2
--
Replace
x
by
x -
2; horizontal shift to the right 2 units,
Add 1 ; vertical shift up 1 unit,
See Figure 54. Figure 54
Y
Y
4
Y
Y
4
4
4
- - - - - - -
4
x
-4
4
x
1
1 1 1 (1 , -3)
(a )
1 Y = -x
3 (b) Y = x
x
Y = x-2 -
Add 1 ; Vertical shift u p 1 unit
1 1 1
-4 M ultiply by 3; Vertical stretch
4
Replace x by x - 2; Horizontal shift right 2 u n its
-
-4
-4
(c)
3
1
(3, 4) 1 1
1 1
1
- - - - - - - -
1
1 4 1 ( 1 , -2) 1 1 1 1
x
I
3
(d) Y =x-2 -+
1 •
Other ordering of the steps shown in Example 12 would also result in the graph of f. For example, try this one: STEP
1: y =
STEP 2:
y =
STEP 3:
y =
STEP 4:
y =
1 x
-
Reciprocal function
1 x - 2 3 - x - 2 3 -- + 1 x - 2
-
Replace
x
by
x -
2; horizontal shif t to the right 2 units,
1 M u lti p ly by 3; ve rti ca l stretch of the g ra p h of y = -x - 2 by a factor of 3, Add 1; vertical shif t u p 1 unit,
C o m b i n i n g G raphing Proce d u res
EXAM P L E 1 3
Graph the function: f(x)
=
�
+ 2
Because horizontal shifts require the form x - h, we begin by rewriting f(x) as f(x) = � + 2 = V- (x - 1 ) + 2. Now use the following steps:
Sol ution
STEP 1:
y =
STEP 2:
y =
STEP 3:
y =
STEP 4:
y =
\IX
Square root function
�
V-(x - 1 )
�+
=
�
2
Re p l ace
x
by
- x;
Replace
x
by x
-
reflect about the y-ax is,
1; horizontal shift to the right 1 unit,
Add 2; vertical shift u p 2 units,
See Figure 55. Figure 5 5
Y
-5
5
5
5
(a) Y = ..JX
Y
Y
5
x
-5
Replace x by - x; Reflect about y-axis 'I.'!
5
(b)
x
-5
y = -r=x Replace x by x - 1 ; (c) Horizontal shift right 1 u n it
"- Now Work P R O B L E M 5 7
(1, 0)
Y=
5
x
-5
--.! - (x - 1 ) Add 2; Vertical shift up 2 units
�
= -ff-=-x
5
(d) Y = ..f1="X+
x
2 •
SECTION 3.5
Graphing Techniq ues: Transformations
261
3.5 Assess Your Understanding Concepts a n d Vocabulary 1. Suppose that the graph of a function
4.
f is known. Then the graph of y f(x - 2) may be obtained by a(n) shift of the graph of f to the a distance of 2 units. 2. Suppose that the graph of a function f is known. Then the graph of y f( - x) may be obtained by a reflection about the -axis of the graph of the function y f(x). 3 . Suppose that the x-intercepts of the graph of y = f(x) are -2, 1, and 5. The x-intercepts of y = f(x + 3) are , , and =
True or False The graph of y = -f(x) is the reflection about the x-axis of the graph of y = f(x). 5. True o r False To obtain the graph of y = f ( x + 2) - 3, shift the graph of y = f(x) horizontally to the right 2 units and vertically down 3 units. 6. True or False Suppose that the x-intercepts of the graph of y f( x) are -3 and 2. Then the x-intercepts of the graph of y = 2f(x) are -3 and 2.
__
__
=
=
__
=
__
__
__ .
Ski l l Bui ld i n g
In Problems 7-18, match each graph to one of the following functions: A. y = x2 + 2 B. y -x2 + 2 E. y = (x - 2)2 F y - ( x + 2f 1. y = 2x2 _2x2 1. y =
=
=
7.
8.
� 3
-3
x
-3
3
G. K.
y = Ix l + 2 y Ix - 21 y = 2 1 xl
D. Y = -I x l + 2 H. y = - I x + 2 1 L. y = -2 1 xl
=
9.
10.
y
1
y 3
-3
12.
11.
c.
13.
x
3
3
-3
15.
16.
y
4
4
x
-1
3 x
3
3 x
-3
-4
18.
4
-4
-3
In Problems 19-26, write the function whose graph is the graph of y 19. Shifted to the right 4 units 21. Shifted up 4 units
x
-4
17.
Y
6
x -6
-3
x
14.
y
-3
3
x
3 x
-3
-4
-3
x3, but is: 20. Shifted to the left 4 units
=
22. Shifted down 4 units
23. Reflected about the y-axis
24. Reflected about the x-axis
25. Vertically stretched by a factor of 4
26. Horizontally stretched by a factor of 4
In Problems 2 7-30, find the function that is finally graphed after the following transformations are applied to the graph of y = 27. (1) Shift up 2 units (2) Reflect about the x-axis (3) Reflect about the y-axis
28. ( 1 ) Reflect about the x-axis (2) Shift right 3 units (3) Shift down 2 units
29. ( 1 ) Reflect about the x-axis (2) Shift up 2 units (3) Shift left 3 units
30. ( 1 ) Shift up 2 units (2) Reflect about the y-axis (3) Shift left 3 units
Yx .
262
CHAPTER 3
Functions and Their Graphs
31. If (3,0) is a point on the graph of y = f(x), which of the fol lowing points must be on the graph of y = -f(x)? ( a ) (0,3) ( b) (0, -3) (d) ( -3,0) (c ) (3,0)
32. If (3,0) is a point on the graph of y = f(x), which of the fol lowing points must be on the graph of y = f( -x)? ( a) (0,3) (b ) (0, -3) (d) (-3,0) (c) (3,0)
33. If (0,3) is a point on the graph of y = f(x), which of the fol lowing points must be on the graph of y = 2f(x)? (b) (0,2) ( a) (0,3) (d) (6,0) (c ) (0,6)
34. If
(3,0) is a point on the graph of y = f(x), which of the following points must be on the graph of y = f(x)? ( a) ( c)
G, o) (d) G, o)
(3,0)
�
(b)
(o,D
In Problems 35-64, graph each function using the techniques of shifting, compressing, stretching, and/or reflecting. Start with the graph of the basic function (for example, y = x2) and show all stages. 36. f(x) = x2 + 4 37. g(x) x3 + 1 . 35. f(x) = x2 - 1 =
38.
g(x)
41.
f(x)
=
x3 - 1
., 39.
hex) f(x)
=
Vx"-=2
=
Vx+l
" 43.
hex) g(x)
=
4 Yx
46.
hex)
=
3\YX
49.
g(x)
=
�
40.
42.
44.
(x - 1 )3 + 2 g(x) = "21 Yx
47.
f(x)
48.
(x + 2)3 - 3 1 hex) = 2x f(x) = - Yx
51.
hex)
=
-x3
2
52.
hex)
=
54.
f(x)
=
3(x - 2)2 + 1
55.
g(x)
=
'" 57.
hex)
=
v=x
58.
hex)
=
-4 �
61.
x+2 g(x) = 211 - x l
2 int ( x - 1 )
64.
hex)
=
=
=
45.
- \YX
=
53.
1 x f(x) = 2(x
56.
g(x)
=
Ix + 1 1 - 3
59.
f(x)
=
-(x + 1 )3 - 1
60.
f(x)
=
62.
g(x)
=
4�
63.
hex)
=
50.
g(x)
= --
+
1f - 3
+
-2
1 -x
-
+
2
Vx"-=2
+
1
4
-
int ( -x )
In Problems 65-68, the graph of a function f is illustrated. Use the graph of f as the first step toward graphing each of the following functions: (a) F(x) = f(x) + 3 (b) G(x) = f(x + 2) (c) P(x) = -f(x) (d) H (x) = f(x + 1 ) - 2 (e) Q(x) = "21 f(x) (f) g(x) = fe-x) (g) hex) = f(2x) 65.
66.
y 4
(0, 2) I
y 4
(2, 2) (4, 0)
x ( 4, - 2)
( - 4 , -2)
67.
y
68.
y
( -n, - 1 )
( n, - 1 )
S ECTION 3.5
Graphing Techniques: Transformations
263
Appl ications a n d Exten sions
69. Suppose that the x-intercepts of the graph of y = f(x) are -5 and 3. (a) What are the x-intercepts of the graph of y = f(x + 2)? (b) What are the x-intercepts of the graph of y = f(x - 2)? (c) What are the x-intercepts of the graph of y = 4f(x)? (d) What are the x-intercepts of the graph of y f e - x) ?
70. Suppose that the x-intercepts of the graph of y = f(x) are -8 and 1. (a) What are the x-intercepts of the graph o f y = f(x + 4)? (b) What are the x-intercepts of the graph of y = f(x - 3)? (c) What are the x-intercepts of the graph of y = 2f(x)? (d) What are the x-intercepts of the graph of y = f( -x)?
71. Suppose that the function y = f(x) is increasing on the interval ( - 1 , 5 ) . ( a ) Over what interval is the graph o f y = f(x + 2) increasing? (b) Over what interval is the graph of y = f(x - 5 ) increasing? (c) What can be said about the graph of y = -f(x)? (d) What can be said about the graph of y = f( -x)? 73. 111e graph of a function f is illustrated in the figure. (a) Draw the graph of y = If(x) l . (b) Draw the graph of y = f( l x l ).
72. Suppose that the function y = f(x) is decreasing on the interval ( -2, 7). (a) Over what interval is the graph of y = f(x + 2) decreasing? (b) Over what interval is the graph of y = f(x - 5) decreasing? (c) What can be said about the graph of y = -f(x)? (d) What can be said about the graph of y = fe-x)? 74. The graph of a function f is illustrated in the figure. (a) Draw the graph of y = If ( x) l . (b) Draw the graph of y = I ( l x l ).
=
y 2
y 2
3
-3
( - 2, - 1 )
x
(-1 , - 1 ) -2
( -1 , -1 ) -2
In Problems 75-84, complete the square of each quadratic expression. Then graph each function using the technique of shifting. (If necessary, refer to Section 1 . 2 to review completing the square.) 75. f(x) = x2 + 2x 77. f(x) = x2 - 8x + 1 76. f(x) = x2 - 6x 78. f(x) = x2 + 4x + 2 79. f(x) = x2 + X + 1 80. f(x) = x2 - X + 1 81. f(x) 2x2 - 12x + 19 82. f(x) = 3x2 + 6x + 1 83. f(x) = -3x2 - 12x - 17 84. f(x) = - 2 x - 12x - 13 =
2
85. The equation y = (x - C)2 defines a family ofparab 0 las, one parabola for each value of c. On one set of coordinate axes, graph the members of the family for c = 0, c = 3, and c = -2. 86. Repeat Problem 85 for the family of parabolas y = x2 + c. 87. Thermostat Control Energy conservation experts estimate that homeowners can save 5 to 10 percent on winter heating bills by programming their thermostats 5 to 10 degrees lower while sleeping. In the given graph, the temperature T (in degrees Fahrenheit) of a home is given as a function of time t (in hours after midnight) over a 24-hour period. T
88. Digital Music Revenues The total projected worldwide dig ital music revenues R, in millions of dollars, for the years 2005 through 2010 can be estimated by the function
R(x) = 170.7x2 + 1373x +
80
LL 76 � 72 �
�
(b) 111e homeowner reprograms the thermostat to y = T(t) - 2. Explain how this affects the temperature in the house. Graph this new function. (c) The homeowner reprograms the thermostat to y = T(t + 1). Explain how this affects the temperature in the house. Graph this new function. Source: Roger Albright, 54 7 Ways to be Fuel Smart, 2000.
68
� E 64
� 60
4 8 1 2 16 20 24 Time (hours after midnight)
(a) At what temperature is the thermostat set during day time hours? At what temperature is the thermostat set overnight?
1080
where x is the number of years after 2005. (a) Find R(O), R(3), and R(S) and explain what each value represents. (b) Find r = R(x - 5). (c) Find 1'(5), 1'(8), and 1'(10) and explain what each value represents. (d) In the model l', what does x represent? (e) Would there be an advantage in using the model I' when estimating the projected revenues for a given year instead of the model R? Source:
eMarketel:com, May 2006
264 89.
CHAPTER 3
Functions and Their Graphs
where g '" 32.2 feet per second per second is the accelera tion of gravity. (a) Use a graphing utility to graph the function T T(l). (b) Now graph the functions T = T(l + 1), T = T(l + 2), and T = T(l + 3). (c) Discuss how adding to the length I changes the period T (d) Now graph the functions T T(2l), T = T(31), and
The relationship between the Celsius (oq and Fahrenheit (OF) scales for measuring tem perature is given by the equation
Temperature Measurements
F
sC + 9
=
=
32
The relationship between the Celsius (oq and Kelvin (K) scales is
K
=
C
+
273. Graph the equation F =
�C +
using degrees Fahrenheit on the y-axis and degrees Celsius on the x-axis. Use the techniques introduced in this section to obtain the graph showing the relationship between Kelvin and Fahrenheit temperatures. , 90.
=
32
T = T(41).
i't 91.
(e) Discuss how multiplying the length l by factors of 2, 3, and 4 changes the period T Cigar Company Profits The daily profits of a cigar company from selling x cigars are given by p(x) = -0.05x2
The period T (in seconds) of a sim ple pendulum is a function of its length I (in feet) defined by the equation Period of a Pendulum
+
100x - 2000
The government wishes to impose a tax on cigars (sometimes called a sin tax) that gives the company the option of either paying a flat tax of $10,000 per day or a tax of 10% on profits. As chief financial officer (CFO) of the company, you need to decide which tax is the better option for the company. (a) On the same screen, graph Y1 = p(x) - 10,000 and Yz = ( 1 - 0. 1 0)p( x). (b) Based on the graph, which option would you select? Why? ( c) Using the terminology learned in this section, describe each graph in terms of the graph of p(x). (d) Suppose that the government offered the options of a flat tax of $4800 or a tax of 10% on profits. Which would you select? Why?
T = 27T'V(l g
Discussion and Writing 92.
Suppose that the graph of a function f is known. Explain how the graph of y = 4f(x) differs from the graph of y = f( 4x).
OBJECTIVE 1
E XA M P LE 1
1
Build and Analyze Functions (p.264)
Build and Analyze Functions
Real-world problems often result in mathematical models that involve functions. These functions need to be constructed or built based on the information given. In building functions, we must be able to translate the verbal description into the lan guage of mathematics. We do this by assigning symbols to represent the indepen dent and dependent variables and then by finding the function or rule that relates these variables. F i n d i n g the D istance from the O rigin to a Point on a G raph
Let
P =
(x, y ) be a point on the graph of y
O?I? v,:?
=
x2 - 1.
(a) Express the distance d from P to the origin 0 as a function of x. (b) What is d if x = ( c) What is d if x = (d) What is d if x
=
(e) Use a graphing utility to graph the function d = d(x), x 2': O. Rounded to two decimal places, find the value(s) of x at which d has a local minimum. [This gives the point(s) on the graph of y = x2 - 1 closest to the origin.]
SECTION 3.6
Solution
(a) Figure 56 illustrates the graph of y = x2 - 1. The distance d from P to
Y
is
Vx2
(x2 - 1 )2 = VX4 - x2 + 1 We have expressed the distance d as a function of x. (b) If x = 0, the distance d is d(O) = VI = 1 ( c) If x = 1 , the distance d is
y = x2 - 1
2
0
d = V(x - O? + ( y - O? = Vx2 + i Since P is a point on the graph of y = x2 - 1 , we substitute x2 - 1 for y. Then
Figure 56 2
265
Mathematical Models: Building Functions
d(x)
x
=
+
d ( l ) = \/1 - 1
(d) If x = Figure 57
+
1 =1
-v: , the distance d is d (;) = �(;y - (;y + 1 = �� - � + 1 = V;
(e) Figure 57 shows the graph of Y = Vx4 - x2 + 1 . Using the MINIMUM fea ture on a graphing utility, we find that when x 0.71 the value of d is small est. The local minimum is d 0.87 rounded to two decimal places. [Since d(x) is even, by symmetry, it follows that when x -0.71 the value of d is also a local minimum.]
2
1
;:::j
;:::j
;:::j
•
0 �======� 2 o
E XA M P L E 2
Y 30
-1
Area of a Rectangle
(a) (b) (c) (d)
\�
Now Work P R O B L E M 1
;
A rectangle has one corner in quadrant I on the graph of y = 25 - x2, another at the origin, a third on the positive y-axis, and the fourth on the positive x-axis. See Figure 58.
Figure 58
10
L'! Ji>
2
(0,0)
x
S o l ution
Express the area A of the rectangle as a function of x. What is the domain of A? Graph A = A(x). For what value of x is the area largest?
(a) The area A of the rectangle is A xy, where y = 25 - x2. Substituting this expression for y, we obtain A(x) = x(25 - x2) = 25x - x3 . (b) Since (x, y) is in quadrant I, we have x > O. Also, y = 25 - x2 > 0, which im plies that x2 < 25, so -5 < x < 5. Combining these restrictions, we have the domain of A as { x i O < x < 5 } , or (0, 5 ) using interval notation. (c) See Figure 59 for the graph of A = A (x). (d) Using MAXIMUM, we find that the maximum area is 48.11 at x = 2.89, each rounded to two decimal places. See Figure 60. =
Figure 60
Figure 59
50
50
.
....-. -.....
,/
/
.'
o
1//' o
,= -
. ...,;-
.......
....\
\
\
Now Work P R O B L E M 7
5
o
f, /\ \ f /1
\ HoJxif'ilu�'1 l-:=2:.8867�2:1 _ V= '18.112:52:2: � 5
o
•
266
CHAPTER 3
Functions and Their Graphs
E XA M P L E 3
M aking a P l aypen
'"
A manufacturer of children's playpens makes a square model that can be opened at one corner and attached at right angles to a wall or, perhaps, the side of a house. If each side is 3 feet in length, the open configuration doubles the available area in which the child can play from 9 square feet to 18 square feet. See Figure 6l. Now suppose that we place hinges at the outer corners to allow for a configu ration like the one shown in Figure 62. Figure 6 1
Figure 6 2
(a) Express the area A of the configuration shown in Figure 62 as a function of the distance x between the two parallel sides. (b) Find the domain of A. (c) Find A if x = 5. (d) Graph A = A( x). For what value of x is the area largest? What is the maximum area? Sol ution
(a) Refer to Figure 62. The area A that we seek consists of the area of a rectangle (with width 3 and length x) and the area of an isosceles triangle (with base x and two equal sides of length 3). The height h of the triangle may be found using the Pythagorean Theorem. h2 +
(�y h2 h
=
=
=
32 32
-
( X2 )2 -
_ 9 -
x2 4
-
A(x)
=
36 - x2 4
---
2
area of rectangle + area of triangle 3x +
-
l Y36 - x2
The area A enclosed by the playpen is A =
_
xV36 - x2 4
=
� (� V36 - X2 )
3x + x
Now the area A is expressed as a function of x. (b) To find the domain of A, we note first that x > 0, since x is a length. Also, the expression under the square root must be positive, so 36 - x2 > ° x2 < 36 -6 < x < 6 Combining these restrictions, we find that the domain of A is ° < x < 6, or (0, 6) using interval notation. * Adapted from Proceedings, Summer Conference for College Teachers on Applied Mathematics (University of Missouri, Rolla), 197 1 .
SECTION 3.6
Figure 63
(c) If x
=
Mathematical Models: Building Functions
267
5, the area is A(5)
=
3 ( 5 ) + � V36 - ( 5 )2 4
�
19.15 square feet
If the length of the playpen is 5 feet, its area is 19.15 square feet. �I (d) See Figure 63. The maximum area is about 19.82 square feet, obtained when x � is about 5.58 feet. •
3.6 Assess Your Understanding Appl ications a n d Extensions
Let P = (x, y) be a point on the graph of y = x2 8. (a) Express the distance d from P to the origin as a function of x. (b) What is d if x = O? (c) What is d if x 1 ? / (d) Use a graphing utility to graph d = d(x). (e) For what values of x is d smallest? 2. Let P = (x, y) be a point on the graph of y = x2 8. (a) Express the distance d from P to the point (0, - 1 ) as a function of x. (b) What is d if x = O? (c) What is d if x = - 1 ? " � (d) Use a graphing utility to graph d d(x). (c) For what values of x is d smallest? 3. Let P (x, y) be a point on the graph of y Vi . ( a ) Express the distance d from P t o the point ( 1 , 0) a s a function of x. } (b) Use a graphing utility to graph d = d(x). ( c) For what values of x is d smallest? . 1 4. Let P = (x, y) be a pomt on the graph of y = - . 1.
-
=
third on the positive x-axis at (x, 0). Express the area A of the triangle as a function of x. 7. A rectangle has one corner in quadrant I on the graph of y = 16 - x2, another at the origin, a third on the positive y axis, and the fourth on the positive x-axis. See the figure.
-
=
=
=
;� 8.
y
x
(a) Express the distance d from P to the origin as a func tion of x. tIl (b) Use a graphing utility to graph d d(x). ( c ) For what values of x is d smallest? 5. A right triangle has one vertex on the graph of y = x3, X > 0, at (x, y), another at the origin, and the third on the positive y-axis at (0, y), as shown in the figure. Ex press the area A of the triangle as a function of x.
(a) Express the area A of the rectangle as a function of x. (b) What is the domain of A ? (c) Graph A = A(x). For what value of x is A largest? A rectangle is inscribed in a semicircle of radius 2. See the figure. Let P = ( x, y) be the point in quadrant I that is a vertex of the rectangle and is on the circle.
y = .J4
x2 x
=
•
9.
(a) Express the area A of the rectangle as a function of x. (b) Express the perimeter p of the rectangle as a function of x. ( c ) Graph A = A (x). For what value of x is A largest? (d) Graph p = p( x). For what value of x is p largest? A rectangle is inscribed in a circle of radius 2. See the figure. Let P = (x, y) be the point in quadrant I that is a vertex of the rectangle and is on the circle.
x -
6. A right triangle has one vertex on the graph of y = 9 - x2, X > 0, at (x, y), another at the origin, and the
-
2
268
Ij 10.
C H A PTER 3
Functions and Their Graphs
(a) Express the area A of the rectangle as a function of x. (b) Express the perimeter p of the rectangle as a function of x. (c) Graph A = A (x ) . For what value of x is A largest? (d) Graph p = p( x ) . For what value of x is p largest? A circle of radius r is inscribed in a square. See the figure.
16.
x2 [Hint: First show that r2 = 3 ']
17.
1 1.
(a) Express the area A of the square as a function of the radius r of the circle. (b) Express the perimeter p of the square as a function of r. A wire 1 0 meters long is to be cut into two pieces. One piece will be shaped as a square, and the other piece will be shaped as a circle. See the figure.
I 1
10
An equilateral triangle is inscribed in a circle of radius r. See the figure in Problem 16. Express the area A within the cir cle, but outside the triangle, as a function of the length x of a side of the triangle. Two cars leave an intersection at the same time. One is head ed south at a constant speed of 30 miles per hour, and the other is headed west at a constant speed of 40 miles per hour (see the figure). Express the distance d between the cars as a function of the time t. [Hint: At t = 0, the cars leave the intersection.]
o
m
10
18.
[Q] x
4x
An equilateral triangle is inscribed in a circle of radius r. See the figure. Express the circumference C of the circle as a func tion of the length x of a side of the triangle.
-
4x
(a) Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the square. (b) What is the domain of A ? .r (c) Graph A = A ( x ) . For what value of x is A smallest? 12. A wire 10 meters long is to be cut into two pieces. One piece will be shaped as an equilateral triangle, and the other piece will be shaped as a circle. (a) Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the equilateral triangle. (b) What is the domain of A ? , (e) Graph A = A ( x ) . For what value of x is A smallest? 13. A wire of length x is bent into the shape of a circle. (a) Express the circumference of the circle as a function of x. (b) Express the area of the circle as a function of x. 14. A wire of length x is bent into the shape of a square. (a) Express the perimeter of the square as a function of x. (b) Express the area of the square as a function of x. 15. A semicircle of radius r is inscribed in a rectangle so that the diameter of the semicircle is the length of the rectangle. See the figure.
19.
1� 20.
Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a constant speed of 30 miles per hour. At the same time, the other car is 3 miles east of the intersection and is moving at a constant speed of 40 miles per hour. (a) Express the distance d between the cars as a function of time t. [ Hint: At t = 0, the cars are 2 miles south and 3 miles east of the intersection, respectively.] (b) Use a graphing utility to graph d = d(t). For what value of t is d smallest? Inscribing a Cylinder in a Sphere Inscribe a right circular cylinder of height h and radius r in a sphere af fixed radius R. See the illustration. Express the volume V of the cylinder as a function of h. [Hint: V = 7Tr2h. Note also the right triangle.] r
T h
(a) Express the area A of the rectangle as a function of the radius r of the semicircle. (b) Express the perimeter p of the rectangle as a function of r.
Sphere
1
Chapter Review
Inscribe a right circular cylinder of height h and radius r in a cone of fixed radius R and fixed height H. See the illustration. Express the volume V of the cylinder as a function of r. [Hint: V = 7Tr2h. Note also the similar triangles.)
21. Inscribing a Cylinder in a Cone
An island is 2 miles from the nearest point P on a straight shoreline. A town is 12 miles down the shore from P. See the illustration.
23. Time Required to Go from an Island to a Town
'E N
I t
\R
Cone
MetroMedia Cable is asked to provide service to a customer whose house is located 2 miles from the road along which the cable is buried. The nearest con nection box for the cable is located 5 miles down the road. See the figure.
269
"
12-x�x ", ".,,-" 1 2Jni<:o-�--""".--'Z.� :. I
,,"d1
: "�"
I v"
Island *. "'", · .... �ri
22. Installing Cable TV
�
(a) If the installation cost is $500 per mile along the road and $700 per mile off the road, express the total cost C of installation as a function of the distance x (in miles) from the connection box to the point where the cable installation turns off the road. Give the domain. (b) Compute the cost if x = 1 mile. (c) Compute the cost if x = 3 miles. (d) Graph the function C = C(x). Use TRACE to see how the cost C varies as x changes from 0 to 5. (e) What value of x results in the least cost?
(a) If a person can row a boat at an average speed of 3 miles per hour and the same person can walk 5 miles per hour, express the time T that it takes to go from the island to town as a function of the distance x from P to where the person lands the boat. (b) What is the domain of T? (c) How long will it take to travel from the island to town if the person lands the boat 4 miles from P? (d) How long will it take if the person lands the boat 8 miles from P? 24. Filling a Conical Tank Water is poured into a container in the shape of a right circular cone with radius 4 feet and height 16 feet. See the figure. Express the volume V of the water in the cone as a function of the height h of the water. [Hint: The volume V of a cone of radius r and height h is V
1
= -m·2h.) 3
CHAPTER REVIEW Library of Functions
Identity function (p. 244)
Constant function (p. 244)
[(x) = b The graph is a horizontal line with y-intercept b. y
(O, b)
f(x) = b x
[(x) = x The graph is a line with slope 1 and y-intercept O.
270
CHAPTER 3
Functions and Their Graphs
Xl
Square root function (p. 245)
Cube function (p. 245)
Square function (p. 245)
f(x) = vX
f(x) = x3
f(x) = The graph is a parabola with intercept at (0, 0).
y
y
4
Reciprocal function (p. 245)
Cube root function (p. 245)
f(x) = rx
1 . f(x) = x
y 3
y
2
3 x
-3
(-2,-12)
5 x
(0,0)
-4
4 x
-4
-1
x
-2
2 x
�
-2
(- 1 , - 1 ) -3
Greatest integer function (p. 246)
Absolute value function (pp. 245-246)
f(x) = int(x)
f(x) = Ixl y
-3
(0,0)
-2
3 x
e-{) -3
2
4
x
Things to Know
Function (pp. 208-212)
Function notation (pp. 212-215)
Difference quotient of f (pp. 214 and 241) Domain (pp. 209 and 215)
A relation between two sets so that each element x in the first set, the domain, has corresponding to it exactly one element y in the second set. The range is the set of y values of the function for the x values in the domain. A function can also be characterized as a set of ordered pairs (x, y) in which no first element is paired with two different second elements. y f(x) f is a symbol for the function. x is the argument, or independent variable. y is the dependent variable. f(x) is the value of the function at x, or the image of x. A function f may be defined implicitly by an equation involving x and y or explicitly by writing y = f(x). =
h) - f(x) h *- 0 h If unspecified, the domain of a function f is the largest set of real numbers for which f(x) is a real number. f(x
+
Chapter Review
Vertical-line test (p. 223) Even function f (p. 231) Odd function f (p. 232) Increasing function (p. 234) Decreasing function (p. 234) Constant function (p. 234) Local maximum (p. 235) Local minimum (p. 235) A,'erage rate o f change o f a function (p. 236)
271
A set of points in the plane is the graph of a function if and only if every vertical line intersects the graph in at most one point. f( - x) = f (x ) for every x in the domain ( - x must also be in the domain). f( -x ) = -f(x) for every x in the domain ( x must also be in the domain). A function f is increasing on an open interval 1 if, for any choice of XI an d X2 in 1, wi th XI < X2, we have f(xl ) < f(X2 ) ' A function f is decreasing on an open interval I if, for any choice of XI an d X2 in 1, wi th XI < X2, we have f(xl ) > f(X2 ) ' A function f is constant on an open interval f if, for all choices of x in I, the values of f(x)are equal. A function f has a local maximum at c if there is an open interval I con taining c so that,for all x not equal to c in l,f (x) < fCc). A function f has a local minimum at c if there is an open interval! containing c so that, for all x not equal to c in!, f(x) > f(c). The average rate of change of f from a to b is -
tl y tlx
feb) - f(a) b-a
a"* b
Objectives --------� Section
You should be a ble to ...
Review Exercises
3.1
Determine whether a relation represents a function (p. 208) Find the value of a fun ction (p. 212) Find the domain of a function (p. 215) Form the sum, difference, product, and quotient of two functions (p. 217)
1,2 3-8, 23, 24, 7 1 , 72, 73 9-16 17-22
Identify the graph of a function (p. 223) Obtain information from or about the graph of a function (p. 224)
47-50 25(a)-(e), 26(a)-(e), 27(a), 27(d) 27(f), 28(a), 28(d), 28(f)
Determine even and odd functions from a graph (p. 231 ) Iden tify even and odd functions from the equation (p. 232) Use a graph to determine where a function is increasing, decreasing, or constan t (p. 233) Use a graph to locate local maxi ma and local minima (p. 234) Use a graphing utility to approximate local maxima and local minima and to determjne where a function is increasing or decreasing (p. 235) Find the average rate of change of a function (p. 236)
27(e), 28(e) 29-36 27(b), 28(b) 27(c), 28(c)
Graph the functions listed in the library of functions (p. 244) G raph piecewise-defined functions (p. 247)
5 1 -54 67-70
Graph functions using vertical and horizontal shifts (p. 252) Graph functions using compressions and stretches (p. 255) Graph functions using reflections about the x-axis and the y-axis (p. 258)
25(f), 26(f), 26(g) 55, 56, 59-66 25(g), 26(h), 57, 58, 65, 66 25(h), 57, 61, 62, 66
Build and an alyze function s (p. 264)
74-76
2
3
4
3.2 2
3.3 2
3
�
4 5
6
3.4 2
3.5 2
3
3.6
37-40, 75(e), 76(b) 41-46
Review Exercises
In Problems 1 and 2, determine whether each relation represents a function. For each function, state the domain and range. 2. {( 4, - 1 ), (2, 1 ) , (4, 2) } 1. { ( -1, 0), (2, 3 ) , (4, 0)} In Problems 3-8, find the following for each function: (c) f( -x ) (b) f( - 2 ) (a) f(2) x2 3x 4. f(x) = 3. f(x) = 1 x +1 r- 1
--
(d) -f(x)
(e) f(x
-
2)
5 . f(x) =
(f) f(2x)
�
272
CHAPTE R 3
6. f(x )
= Ix2 - 41
In Problems
9. f(x ) 13. h(x )
Functions and Their Graphs
=
=
7. f(x )
20. f(x )
=
=
x2 - 4 x2
find the domain of each function. 3x2 x 10. f(x ) = -2x 2 x -9
9-16,
=� x 15. f(x ) = x2 + 2x
_
Ixl 14. g(x ) = �
\IX
Ixl
+
2 - x; g(x ) 3x; g(x )
8. f(x )
11. f(x)
-
In Problems 1 7-22, find f
17. f(x)
=
= x2 x3 9 _
12.
_
f(x)
16. F(x )
3
= Vx+2 =
= 3x + 1
= 2x - 1; g(x ) = 2x + 1 1 x+ 1 21. f(x) --; g(x ) = x x- 1
= 1 + x + x2
=
f(x
+
25. Using the graph of the function f shown: y 4
19. f(x)
�;
-
=
-
X
+
4
26. Using the graph of the function g shown: ( -5,1) .
y 3 (-1,1) ...
-5
x
5
(a) (b) (c) (d) (e) (f)
h) - f(x ) h "* 0 h 24. f(x ) 3x2 - 2x
(3,3)
x
5 (3, -3)
-4
Fi nd the domain and the range of f . List the intercepts. Find f( -2). For what value of x does f(x ) -3? Solve f(x) > O. Graph y f(x - 3).
=
= (g) Graph y = f G x ) . (h) Graph y = -f(x) .
(a) (b) (c) (d) (e) (f) (g) (h)
Find the domain and the range of g. Find g(- 1 ) . List the intercepts. For what value of x does g(x ) -3? Solve g(x ) > O. Graph y g(x - 2). Graph y g(x ) + l. Graph y 2g(x ) .
===
=
In Problems 27 and 28, use the graph of the fitnction f to find: (a) The domain and the range of f . (b) The intervals on which f is increasing, decreasing, or constant. (c) The local minima and local maxima. (d) Whether the graph is symmetric with respect to the x-axis, the y-axis, o r the origin. (e) Whether the function is even, odd, or neither. (j) The intercepts, if any. 27. 28. y (-1,1 ) -5
3
(4,0)
5 x
(3, -3)
-
4
= 3x2 + X + 1 ; g(x ) = 3x 3 1 22. f(x) = -- g(x ) = x - .)
18. f(x )
= -2x2 + X + 1
(-4, -3)
_
g, f - g, f . g, and L for each pair of functions. State the domain of each one. g
In Problems 23 and 24, find the difference quotient of each function f ; that is, flnd
23. f( x)
x2
1 3x
6 x
Chapter Review
In Problems 29-36, determine (algebraically) whether the given function is even, odd, or neitha 4 + x2 1 1 29. f(x) = x3 - 4x 30. g(x) = --4 31. hex) = 4 + 2" + 1 1 +x x x
33. G ( x) = 1 - x + x3 ;,�
34. H (x) = I + x + x2
35. f(x)
32. F(x)= � 1 + x2 3 36. g(x) = -X
x + x2
_ = _
l
273
In Problems 37-40, use a graphing utility to graph each function over the indicated interval. Approximate any local maxima and local minima. Determine where the function is increasing and where it is decreasing. 37. f(x) = 2x3 - Sx + I (-3,3) 38. f(x) = -x3 + 3x - 5 (-3,3)
2X4 - Sx3
2x + 1 (-2,3)
40. f(x)
=
-x4 + 3x3 - 4x
In Problems 41 and 42, find the average rate of change of f: ( c) From 2 to 4 (b) From ° to 1 (a) From 1 to 2 = 2 x 8x 41. f(x) 42. f(x)
=
2x3 + X
39. f(x)
=
+
+
3 ( -2, 3 )
In Problems 43-46, find the average rate of change from 2 to 3 for each function f. Be sure to simplify. 43. f(x) = 2 - Sx 44. f(x) = 2x + 7 46. f(x) = x2 - 3x + 2 45. f(x) = 3x - 4x2
2
In Problems 47-50, is the graph shown the graph of a function? 47. 48. y y
x
49.
50.
y
x
x
x
In Problems 51-54, sketch the graph of each function. Be sure to label at least three points. 51. f(x) = Ixl 52. f(x) = \YX 53. f(x) = �
54. f(x) = ..!.. x
In Problems 55-66, graph each function using the techniques of shifting, compressing or stretching, and reflections. Identify any intercepts on the graph. State the domain and, based on the graph, find the range. I 57. g(x) = -21xl 55. F(x) = Ixl - 4 56. f(x) = Ixl + 4 58. g(x) = 21xl
60. hex) = vx - 1
63. hex) = (x - If
+
64. h e x ) = (x + 2f -
2
In Problems 67-70, (a) Find the domain of each function. ( c ) Graph each functiol1. if -2 < x :s; 1 67. f(x) = if x > 1 x+ 1
{3X
69. f(x) =
{
1 3x \:
if -4 :s; x < 0 if x = 0 if x > 0
71. A function f is defined by
If f(l) = 4, find A .
Ax + 5 f(x) = -6x - 2
3
61. f(x)
=
62. f (x) = -Vx + 3
�
66. g(x) = -2(x + 2)3 - 8
65. g(x) = 3(x - 1)3 + 1
(b) Locate any intercepts. ( d ) Based on the graph, find the range. i f -3 < x < 0 x-1 68. f ( x) = if x :2: 0 3x - 1
{
70. f(x) =
{X2
2x - 1
if -2 :s; x :s; 2 if x > 2
72. A function g is defined by g(x)
If g( -1) = 0, fi nd A.
A
=
x
8 +? r
274
C H A PTER 3
73. Spheres
Functions and Their Graphs
The volume
V of a
r
sphere of radius is V
r
=
�'7Tr3;
the surface area S of this sphere is S 4'7T 2 . If the radius doubles, how does the volume change? How does the surface area change? 1 74. Page Design A page with dimensions of 8 inches by 2" 11 inches has a border of uniform width x surrounding the printed matter of the page, as shown in the figure. =
8� in.
Cd) Graph the function A = A(x). (c) Use TRACE to determine what margin should be used to obtain areas of 70 square inches and 50 square inches. 75. Material Needed to Make a Drum A steel drum in the shape of a right circular cylinder is required to have a volume of 1 00 cubic feet. (a) Express the amount A of material required to make the drum as a function of the radius of the cylinder. (b) How much material is required if the drum is of radius 3 feet? (c) How much material is required if the drum is of radius 4 feet? (d) How much material is required if the drum is of radius 5 feet? ?� (e) Graph A = A ( ). For what value of is A smallest? Hint: The volume V and surface area S of a right circular cylinder are S = 2m.2 + :,�
r
r
where
(a) Express the area A of the printed part of the page as a function of the width x of the border. (b) Give the domain and the range of A . (c) Find the area of printed page for borders of widths 1 inch, 1.2 inches, and 1 .5 inches.
r
h is the height and r is the radius
2'7Trh
76. A rectangle has one vertex in quadrant I on the graph of
y = 10 x2, another at the origin, one on the positive x-axis, and one on the positive y-axis. (a) Express the area A of the rectangle as a function of x. (b ) Find the largest area A that can be enclosed by the rectangle. -
'��
CHAPTER TEST 1. Determine whether each relation represents a function. For
each function, state the domain and range. (a) {(2 , 5 ), (4, 6), (6, 7), (8, 8) } (b) { (1, 3), (4, 2) ( -3 5) ( 1, 7 )} (c) -
,
,
,
In Problems 2-4, find the domain of each function and evaluate each function at x = - 1 . 2. f(x) � x + 2 3. g (x) x+2 I-I x-4 4. hex) = 2 X + 5x 36 5. Using the graph of the function f: =
=
--- -
y
-4
-2
4 (1 , 3)
y
(d)
6
( - 5 , -3)
2 -4
-2
2 -2
4
x
(a) (b) (c) (d) (e)
-4
(3, - 3)
Find the domain and the range of f. List the intercepts. Find f ( l). For what value(s) of x does f(x) = -3 ? Solve f(x) < O.
Cumulative Review
LJ
6. Use
a graphing utility to graph the function f(x) -x4 + 2x3 + 4x2 - 2 on the interval ( -5, 5). Ap proximate any local maxima and local minima rounded to two decimal places. Determine where the f unction is in creasing and where it is decreasing. =
{
. 2X + 1 . 7. Consider the function g( x) = x-4
8. 9.
if x if x
<
:2!
-1 -1
(a) Graph the function. (b) List the intercepts. (c) Find g( -5). (d) Find g(2). For the function f(x) 3x2 - 2x + 4, f ind the average rate of change of f from 3 to 4. For the functions f(x) = 2x2 + 1 and g(x) 3x - 2, find the following and simplify: (a) f - g (b) f · g (c) f(x + 11) f(x) Graph each function using the techniques of shifting, com pressing or stretching, and reflections. Start with the graph of the basic function and show all stages. (a) h(x) -2(x + 1 )3 + 3 (b) g(x) Ix + 41 + The variable interest rate on a studen t loan changes each July 1 based on the bank prime loan rate. For the years 1992-2004, this rate can be approximated by the model =
=
r(x) = -0.1 15x2 + 1. 183x + 5.623, where x is the number of years since 1992 and r is the interest rate as a percent. - (a ) Use a graphing utility to estimate the highest rate dur ing this time period. During which year was the interest rate the highest? (b) Use the model to estimate the rate in 2010. Does this value seem reasonable? Source: u.s. Federal Reserve 12. A community skating rink is in the shape of a rectangle with semicircles attached at the ends. The length of the rectangle is 20 feet less than twice the width. The thickness of the ice is 0.75 inch. (a) Write the ice volume, V, as a function of the width, x. (b) How much ice is in the rink if the width is 90 feet?
-
10.
2
=
=
11.
275
o
o
o
o
CUMULATIVE REVIEW In Problems 1-6, find the real solutions o f each equation. 1. 3x - 8 = 10
-
X=0
4. 6x2 - 5x + 1 = 0
3 . x2 - 8x - 9 = 0 S.
2. 3x2
12x + 31 = 4
6.�=2
In Problems 7-9, solve each inequality. G ra ph the solution set. 8. 12x - 51 < 3 7. 2 - 3x > 6
9. 14x
+
11 :2! 7
= (-2, -3) to P 2 = (3, -5). (b) What is the midpoint of the line segment from P1 to P ? 2 (c) What is the slope of the line containing the points PI to P2 ?
10. (a) Find the distance from PI
In Problems 11-14, graph each equation. 11. 3x - 2y = 12 13. x2
+ (y - 3)2
=
16
15. For the equa tion 3x2 - 4 y = 12, find the intercepts and check for symmetry. In Problems 17-19, graph each function. 1 18. f(x) = 17. f(x) (x + 2)2 - 3 x =
12.
x
14. y
=i =
V;
16. Find the slope-intercept form of the equa tion of the line con taining the points ( - 2,4) and (6, 8).
19. f(x) =
{2 - x Ixl
2
if x ::5 if x > 2
276
CHAPTER 3
Functions and Their Graphs
CHAPTER PROJECTS I.
This month, as you were paying your bills, you noticed tha t your cell phone service contract had expired. Since cell phone numbers are now portable, you decided to investigate different companies. Since you regularly travel outside your local calling area, you decide to look only at plans with no roaming charges. Here are your findings: Cell Phone Service
Anytime Minutes Included
Charge for Each Extra Minute
Mobile-to-Mobile Minutes
National Long Distance
Nights and Weekends
Plan A 1 : $39.99
450 with rollover
$0.45
Unlimited
Included
5000 (9 PM-6 AM)
Plan A2: $59.99
900 with rollover
$0.40
Unlimited
Included
Unlimited (9 PM-6 AM)
600
$0.40
None
Included
Unlimited (9 PM-7 AM)
1 000
$0.40
None
Included
Unlimited (9 PM-7 AM)
55 1 -1050, $5 for each
Unlimited
Included
Unlimited (7 PM-7 AM)
Unlimited
Included
Unlimited (7 PM-7 AM)
Company A:
Company B: Plan 81: $39.99 Plan 82: $49.99
Company C: Plan ( 1 : $59.99
550
50 minutes; Above 1 050, $0. 1 0 each minute
Plan C2: $69.99
800
80 1 - 1 300, $5 for each 50 minutes; Above 1 300, $0. 1 0 each minute
Based on rates from the Web sites of the companies Cingular, T Mobile and Sprint PCS for the zip code 76201 on May 20, 2006. (www.cingular.com , www. t-mobile. com, www.sprintpcs.com)
Source:
Each plan requires a 2-year contract. 1.
Determine the total cost of each plan for the life of the con tract, assuming that you stay within the allotted anytime min utes provided by each contract. 2. If you expect to use 400 anytime, 200 mobile-to-mobile, and 4500 night and weekend minutes per month, which plan pro vides the best deal? If you expect to use 400 anytime, 200 mo bile-to-mobile, and 5500 night and weekend minutes per month, which plan provides the best deal? If you expect to use 500 anytime, and 1 000 mobile-to-mobile, and 2000 night and weekend minutes per month, which plan provides the best deal? 3. Ignoring any mobile-to-mobile and night and weekend usage, if you expect to use 850 anytime minutes each month, which option provides the best deal? What if you use 1050 anytime minutes per month?
4. Each monthly charge includes a specific number of peak time
minutes in the monthly fee. Write a function for each option, where C is the monthly cost and x is the number of anytime minutes used. 5. Graph each function from part 4. 6. For each of the companies A, B, and C, determine the aver age price per minute for each plan, based on no extra min utes used. For each company, which plan is better? 7. Looking at the three plans that you found to be the best for compa nies A, B, and C, in part 6, which of these three seems to be the best deal? 8. Based on your own cell phone usage, which plan would be the best for you?
The following projects are available on the Instructor's Resource Center (IRC): Wireless Internet Service Use functions and their graphs to analyze the total cost of various wireless In ternet service plans. III. Cost of Cable When government regulations and customer preference influence the path of a new cable line, the Pythagorean Theorem can be used to assess the cost of installation. IV. Oil Spill Functions are used to analyze the size and spread of an oil spill from a leaking tanker.
II. Project at Motorola:
Linear and Quadratic Functions Valuing Hewlett-Packard 17 MAY 2006-With the market focused on Hewlett-Packard Co. (HPQ) taking market share from Dell Inc. (DELL), the casual observer might be fooled into mistaking it as a computer hardware company. It's not; while the Compaq acquisition of 2002 expanded the company's hardware business, 53.4 percent of H PQ's profits still come from its imaging and printing division. When we value stocks, we use the capital asset pricing model to price shares, which ex amines the volatility of the stock through beta and then projects how much earnings growth would be needed to justify the current stock price. Historically, HPQ has a beta of 2.05. The computer peripherals industry average, though, is 1.98. But the industry average beta for the computer hardware companies is 1 .58. A weighted average beta for the two industries seems appropriate for a company that walks the line between them. In this case, a beta weighted between the computer peripher als and computer hardware industries by the percentages of HPQ's profits yields an expected beta of 1 .8. If we use this beta in a capital-asset-pricing model, assuming that Hewlett-Packard will maintain a price-to-earnings to-growth ratio of 1 .3 and that full fiscal year 2006 earnings per share will be at the low end of management's guidance, $2.04, then a new position taken at the current price would require an annualized, per-share earnings growth rate of 9.9% over the next five years to break even. The average long-term growth rate among the 12 analysts who provide one is 12.3 % for Hewlett-Packard; this is well above what our analysis indicates that we require, implying a strong undervaluation even after the price spike. Source: Adapted from Paul DeMartino, "Valuing Hewlett-Packard," May 17, 2006, www.reuters.com . A dapted with permission.
-See the Chapter Project-
A Look Back Up to now, o u r d i scussion has focused on g raphs of equations and the idea of a function. We have learned how to graph equations using the point-plotting method, intercepts, and the tests for symm etry. In addition, we have learned what a function is and how to identi fy whether a relation represents a function. We also in troduced properties of functions, such as domain/range, increasing/decreasing, even/odd, and average rate of change.
A Look Ahead Going forward, we w i l l l ook at classes of functions. In this chapter, we focus on linear and quad ratic functions and their properties and appl ications.
Outline
4.1 linear Functions and Their Properties 4.2 Building linear Functions from Data
4.3 Quadratic Functions and Their Properties 4.4 Quadratic Models; Building Quadratic Functions from Data
4.5 Inequalities Involving Quadratic Functions
Chapter Review Chapter Test Cumulative Review Chapter Projects
277
278
CHAPTER 4
Linear and Quadratic Functions
4.1 Linear Functions and Their Properties PREPARING FOR THIS SECTION •
• •
Before getting started, review the following: •
Lines (Section 2.3, pp. 173-185) Graphs of Equations in Two Variables; Intercepts; Symmetry (Section 2.2, pp. 163-171 ) Linear Equations (Section 1.1, pp. 86-93)
•
Now Work the 'Are You Prepared?' problems on page 284.
•
Functions (Section 3.1, pp. 208-219) The Graph of a Function (Section 3.2, pp. 222-226) Properties of Functions (Section 3.3, pp. 231-238)
OBJECTIVES 1 Gra ph Linear Functions (p. 278)
2 Use Average Rate of Change to Identify Linear Functions (p. 278) 3 Determ ine Whether a Linear Function is Increasing, Decreasing,
o r Constant (p. 281)
4 Work with Applications of Linear Functions (p. 282)
1
Graph Linear Functions
In Section 2.3 we discussed lines. In particular, we developed the slope-intercept form of the equation of a line y = mx + b. When we write the slope-intercept form of a line using function notation, we have a linear function.
DEFINITION
A
linear function
is a function of the form
f(x) = mx + b The graph of a linear function is a line with slope m and y-intercept b. Its domain is the set of all real numbers.
E XAM P L E 1 Figure
1 Y
G raphing a Linear Function Graph the linear function:
(0 , 7)
.J
f(x) = -3x + 7
This is a linear function with slope m = -3 and y-intercept b = 7. To graph this function, we start by plotting the point (0, 7), the y-intercept, and use the slope to find an additional point by moving right 1 unit and down 3 units. See Figure 1 .
Solution
•
Alternatively, we could have found an additional point by evaluating the func tion at some x =F- O. For x = 1 , we find f(l) = -3 ( 1 ) + 7 = 4 and obtain the point ( 1 , 4 ) on the graph.
5 x
I '-
2
Now Work
P R O B L EMS 1 3 (a ) AND (b )
Use Average Rate of Change to Identify Linear Functions
Look at Table 1, which shows certain values of the independent variable x and cor responding values of the dependent variable y for the function f(x) = -3x + 7. Notice that as the value of the independent variable, x, increases by 1 the value of the dependent variable y decreases by 3. That is, the average rate of change of y with respect to x is a constant, -3.
SECTION 4.1
Table
1 x
y=f(x)=-3x+7
-2
Linear Functions and Their P roperties
279
!l.y Average Rate of Change =!l.x
13 10- 13
-----:-----,--
-1 -( - 2)
-3 1
= - =
-3
10
-1
7
- 10
-3
--- = - =
0-(-1) o
1
-3
7
-3 4
-3 2
-3 -2
3
It is not a coincidence that the average rate of change of the linear function
� f(x) = - 3x + 7 is the slope of the linear function. That is, y = m = -3. The fol �x lowing theorem states this fact. THEOREM
Average Rate of Change of a Linear Function
Linear functions have a constant average rate of change. That is, the average rate of change of a linear function f(x) = mx + b is
I
�y =m �x
_.J
� -------'---
Proof
The average rate of change of f(x) = mx + b from X l to x2 , X l
�y �x
=1=
X2 , is
(mx2 + b) - (mxl + b) ! ( X2 ) - ! ( Xl) X 2 - Xl X2 Xl mX2 - mXl m(x2 - X l) = nl x2 - Xl x2 - X l
•
B ased on the theorem just proved, the average rate of change of the function
2 2 g(x) = - - x + 5 is - -. 5
:!I!l
Now Work
5
PROBLEM
1 3(c)
As it turns out, only linear functions have a constant average rate of change. Because of this, we can use the average rate of change to determine whether a par ticular function is linear or not. Functions that are not linear are said to be nonlinear.
E XA M P L E 2
U sing the Average Rate of Change to Identify Linear Functions (a) A strain of E-coli Beu 397-recA441 is placed into a Petri dish at 30° Celsius and allowed to grow. The data shown in Table 2 are collected. The population is measured in grams and the time in hours. Plot the ordered pairs (x, y) in the Cartesian plane and use the average rate of change to determine whether the function is linear.
280
CHAPTER 4
Linear and Quadratic Functions
(b) The data in Table 3 represent the maximum number of heartbeats that a healthy individual should have during a IS-second interval of time while exercising for different ages. Plot the ordered pairs (x, y ) in the Cartesian plane and use the average rate of change to determine whether the function is linear. Table 3
Table 2
r
j�
l�
Age, x
Maximum Number of Heart Beats, y
(x,y)
(0,0.09)
20
50
(20,50)
(1,0.12)
30
47.5
(30,47.5)
Time (hours), x
Population (grams), y
(x, y)
0
0.09 0.12
�
2
0.16
(2,0.16)
40
45
(40,45)
3
0.22
(3,0.22)
50
42.5
(50,42.5)
4
0.29
(4,0.29)
60
40
(60,40)
5
0.39
(5,0.39)
70
37.5
(70,37.5)
Source: American Heart Association
We compute the average rate of change of each function. If the average rate of change is constant, the function is linear. If the average rate of change is not constant, the function is nonlinear.
Solution
(a) Figure 2 shows the points listed in Table 2 plotted in the Cartesian plane. Notice that it is impossible to draw a straight line that contains all the points. Table 4 displays the average rate of change of population. Table 4
Figure 2 >-
y
Time (hours), x
a
c::
�
:::J Cl.
a 0..
•
0.3 0.2 0.1
Ily
Average Rate of Change =
e;;- OA E � 9
Population (grams), y
•
0
•
•
2
•
0
Ilx
0.09 0.12 - 0.09
•
1
-
0
=
0.03
0. 1 2 0.04
3
4
Time (hou rs), x
5
X
2
0. 1 6 0.06
3
0.22 0.07
4
0.29 0. 1 0
5
0.39
Because the average rate of change is not constant, we know that the func tion is not linear. In fact, because the average rate of change is increasing as the value of the independent variable increases, we say that the function is in creasing at an increasing rate. So not only is the population increasing over time, but it is also growing more rapidly as time passes. (b) Figure 3 shows the points listed in Table 3 plotted in the Cartesian plane. We can see that the data in Figure 3 lie on a straight line. Table 5 contains the av erage rate of change of the maximum number of heartbeats. The average rate of change of the heartbeat data is constant, -0.25 beats per year, so the func tion is linear.
SECTION 4.1
Figure 3
Table
y
Linear Functions an d Their Properties
5
50
Age, x
Maximum Number of Heartbeats, y
20
50
281
Average Rate of !!.y Change !!.x =-
(/)
� Q)
co
47.5 - 50
45 � Q)
30 - 20
I
30
47.5
40
45
50
42.5
60
40
70
37.5
=
-0.25
-0.25
40 20
30
40
50
60
x
70
-0.25
Age
-0.25
-0.25
�
3
=--
Now Work
• PROBLEM 21
Determine Whether a Linear Function Is Increasing, Decreasing, or Constant
Look back at the Seeing the Concept on page 176. We know that when the slope m of a linear function is positive ( m > 0 ) the line slants upward from left to right. When the slope m of a linear function is negative ( m < 0 ) , the line slants downward from left to right. When the slope m of a linear function is zero ( m = 0 ) , the line is horizontal. Based on these results, we have the following theorem. THEOREM
Increasing, Decreasing, and Constant Linear Functions
A linear function f (x) = mx + b is increasing over its domain if its slope, m, is positive. It is decreasing over its domain if its slope, m, is negative. It is constant over its domain if its slope, m, is zero.
--1
E XAM P LE 3
Determining Whether a Linear F unction I s Increasing, Decreasing, or Constant Determine whether the following linear functions are increasing, decreasing, or constant. (a) f ( x )
5x - 2 3 (c) set ) = - t - 4 4
Solution
(b) g (x) = -2x + 8
=
(d) h ( z)
=
7
(a) For the linear function f(x) = 5x - 2, the slope is 5, which is positive. The function f is increasing on the interval ( -00, (0 ) .
(b) For the linear function g( x) = -2x + 8 , the slope is -2, which is negative. The function g is decreasing on the interval ( -00,00 ) .
- 4 , the slope is 2, which is positive. The func4 4 tion s is increasing on the interval ( 00, (0 ) .
( c) For the linear function s ( t) =
2t
-
(d) We can write the linear function h as h ( z) = Oz + 7. Because the slope is 0, the function h is constant on the interval ( -00, 00 ) . ,.
1_
Now Work
PROBLEM 1
3(d)
•
282
CHAPTER 4
Linear and Quadratic Functions
4
Work with Applications of Linear Functions
There are many applications of linear functions. Let's look at one from accounting.
E XA M P LE 4
Straight-line Depreciation Book value is the value of an asset that a company uses to create its balance sheet. Some companies depreciate their assets using straight-line depreciation so that the value of the asset declines by a fixed amount each year. The amount of the decline depends on the useful life that the company places on the asset. Suppose that a com pany just purchased a fleet of new cars for its sales force at a cost of $28,000 per car. The company chooses to depreciate each vehicle using the straight-line method over $28,000 7 years. This means that each car will depreciate by = $4000 per year. 7 (a) Write a linear function that expresses the book value V of each car as a function of its age, x. (b) Graph the linear function. (c) What is the book value of each car after 3 years? (d) Interpret the slope. (e) When will the book value of each car be $8000? [Hint: Solve the equation Vex) = 8000.]
Solution
(a) If we let Vex) represent the value of each car after x years, then V(O) represents the original value of each car, so V (O ) = $28,000. The y-intercept of the linear function is $28,000. Because each car depreciates by $4000 per year, the slope of the linear function is -4000. The linear function that represents the book value V of each car after x years is Vex) = -4000x + 28,000 (b) Figure 4 shows the graph of V.
Figure 4
(c) The book value of each car after 3 years is 28,000
V ( 3 ) = -4000 ( 3 ) + 28,000
24,000 _
�
� <0 >
= $16,000
20,000
(d) Since the slope of V (x) = -4000x + 28,000 is -4000, the average rate of change of book value is -$4000/year. So for each additional year that passes the book value of the car decreases by $4000.
16,000
g 12,000 en
(e) To find when the book value is $8000, we solve the equation
8000
Vex)
4000 2 3 4 5 6 7 Age of vehicle (years)
=
8000
-4000x + 28,000 = 8000 -4000x = -20,000 x =
-20,000 = 5 -4000
Su btract 28,000 from each side. Divide by -4000.
The car will have a book value of $8000 when it is 5 years old. ""=>-
Now Work
•
P R O B LE M 4 7
Next we look at an application from economics.
E XA M P L E 5
Supply and Demand The quantity supplied of a good is the amount of a product that a company is will ing to make available for sale at a given price. The quantity demanded of a good is the amount of a product that consumers are willing to purchase at a given price.
SECTION 4.1
Linear Functions an d Their Properties
283
Suppose that the quantity supplied, S, and quantity demanded, D, of cellular tele phones each month are given by the following functions: S(p) D(p)
60p - 900
=
=
- 1 5p + 2850
where p is the price (in dollars) of the telephone. (a) The equilibrium price of a product is defined as the price at which quantity sup plied equals quantity demanded. That is, the equilibrium price is the price at which S(p) = D(p). Find the equilibrium price of cellular telephones. What is the equil ibrium quantity, the amount demanded (or supplied) at the equilibrium price? (b) Determine the prices for which quantity supplied is greater than quantity de manded. That is, solve the inequality S(p) > D ( p ) . (c) Graph S = S(p), D = D(p) and label the equilibrium price.
Solution
(a) To find the equilibrium price, we solve the equation S(p) = D ( p ) . 60p - 900
=
- 1 5p + 2850
60p
=
- 1 5p + 3750
S(p) = 60p - 900; D(p) = -15p + 2850 Add 900 to each side.
75 p = 3750 p
=
Add
15p to each side.
Divide each side by 75.
50
The equilibrium price is $50 per cellular phone. To find the equilibrium quan tity, we evaluate either S(P) or D(P) at p = 50. S(50)
=
60(50) - 900
=
2100
The equilibrium quantity is 2100 cellular phones. At a price of $50 per phone, the company will produce and sell 2100 phones each month and have no short ages or excess inventory.
(b) We solve the inequality S(p) > D ( p ) .
60p - 900 > - 15p + 2850 60p > -15p + 3750 75p > 3750
S(p) > D(p) Add 900 to each side. Add
15p to each side.
Divide each side by 75.
p > 50
If the company charges more than $50 per phone, quantity supplied will exceed quantity demanded. In this case the company will have excess phones in in ventory. (c) Figure 5 shows the graphs of S point labeled. Figure
S, D
5 _"0 "0 '" '" "0 :.= C '"
C. C.E (/)0 .2:'»
::::l '"
:.;::::::;-� c � '" C ::::l '" a::::l a
=
S(p) and D = D(p) with the equilibrium
S= S(p)
3000 2000 1 000
50 Price ($) ==�
Now Work
P R O B L EM 4
1
•
284
CHAPTER 4
Linear and Quadratic Functions
4. 1 Assess Your Understa n ding JAre You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red.
1. Graph y
2 x - 3. (pp. 1 73- ( 85) 2. Find the slope of the line joining the points (2, 5) and ( -1, 3 ) . (pp. 1 73-1 74) =
3. Find the average rate of change of f(x)
=
2 to 4. (pp. 23 1 -238)
3x2 - 2, from
=
4. Solve: 60x - 900 5. If f(x) 6.
=
- 1 5x +
2850. (pp. 86-93)
x2 - 4, find f( -2). (pp. 2 08-2 1 9)
True or False The graph of the function f(x) creasing on the interval (0, oc ) . (pp. 23 1-238)
=
x2 is in
Concepts and Vocabulary
7. For the graph of the linear function f(x)
and b is the
=
8. For the graph of the linear function H ( z )
slope is 9.
__
and the y-intercept is
__
.
mx + b, m is the =
-4z + 3, the
If the slope m of the graph of a linear function is function is increasing over its domain.
__
, the
True or False The slope of a line is the average rate of change of the linear function. 11. True or False If the average rate of change of a linear func-
10.
tion 12.
iS�, then if
True or False
y
increases by 3 , x will increase by 2.
The average rate of change of f(x)
is 8.
2x + 8
=
Skill Building
In Problems 13-20, a linear fitnction is given. (a) Determine the slope and y-intercept of each function. (b) Use the slope and y-intercept to graph the linear function. (c) Determine the average rate of change of each function. (d) Determine whether the linear function is increasing, decreasing, or constant. 15. hex) = 14. g(x) = 5x - 4 13. f(x) = 2 x + 3 17. f( X )
=
�
X
18. hex)
-3
=
2 --x 3
+
4
19. F(x)
=
- 3x +
16. p( x)
4
20. G(x)
4
=
=
-x
+ 6
-2
In Problems 21 -28, determine whether the given function is linear or nonlinear. If it is linear, determine the slope. 21.
x
22.
x
y =f (x) 1 /4
-2
4
-2
0
-2
0
2
-8
-1
25.
y =f (x)
x
-1
-5
y =f (x)
2
26.
x
-2
-26
-2
0
2
0
-1
2
-4 -2
-10
-1
2
23.
2
4
y = f (x) -4
-3.5
-3
- 2.5
-2
-8
-2
-4
0
0
4
-2 0 2
27.
y =f (x)
y = f (x)
-1
1 /2
x -2
-1
0 2
24.
x
x
-1
-3
2
0
y =f (x)
28.
x
0 8
12
y = f (x)
8
-2
0
8
0
4
2
16
8
8
8
-1
9
Applications and Extensions = 4x - 1 and g(x) = - 2 x + 5. (a) So lve f(x) = O . (b) Solve f(x) > O. (c) Sol ve f(x) = g(x). (d) Solve f(x) :s g(x). (e) Graph y = f(x) and y = g(x) and label the point that represents the solution to the equation f(x) = g(x).
29. Suppose that f(x)
= 3 x + 5 and g ( x ) = - 2 x + 15. (a) Solve f(x) = O . (b) Solve f(x) < O. (c) Solve f(x) = g(x). (d) Solve f(x) 2: g(x). (e) Graph y = f(x) and y = g(x) and label the point that represents the solution to the equation f(x) = g(x).
30. Suppose that f(x)
SECTION 4.1
31. In parts (a)-(f), use the following figure.
Linea r Functions an d Their Properties
(a) Solve the equation: f(x) = g(x). (b) Solve the inequality: g(x) :5 f(x)
Y
<
285
h ex).
36. In parts (a) and (b), use the following figure.
Y= h(x)
x
(a) Solve f(x) 50. (b) Solve f(x) 80. (c) SOlve f(x) = O. (d) Solve f(x) > 50. (e) Solve f(x) :5 80. (f) Solve 0 < f(x) < 80. 32. In parts (a)-(f), use the following figure. =
x
=
Y
Y= g (x)
Y= g (x)
Y= f(x)
(a) Solve the equation: f(x) = g(x). (b) Solve the inequality: g(x) < f(x)
:5
hex).
The cost C, in dollars, of renting a moving truck for a day is given by the function C(x) = 0.25x + 35, where x is the number of miles driven. (a) What is the cost if you drive x 40 miles? (b) If the cost of renting the moving truck is $80, how many miles did you drive? (c) Suppose that you want the cost to be no more than $100. What is the maximum number of miles that you can drive?
37. Car Rentals
=
x
(a) Solve g(x) 20. (b) Solve g(x) = 60. (c) Solve g(x) = O. (d) Solve g(x) > 20. (e) Solve g(x) :5 60. (f) Solve 0 < g(x) < 60. 33. In parts (a) and (b) use the following figure. =
Y = f(x)
Y
tional calls on a certain cellular phone plan is given by the function C(x) = 0.38x + 5, where x is the number of min utes used.
x
(a) Solve the equation: f(x) = g(x). (b) Solve the inequality: f(x) > g(x). 34. In parts (a) and (b), use the following figure. Y
38. Phone Charges The monthly cost C, in dollars, for interna
(a) What is the cost if you talk on the phone for x = 50 minutes? (b) Suppose that your monthly bill is $29.32. How many minutes did you use the phone? (c) Suppose that you budget yourself $60 per month for the phone. What is the maximum number of minutes that you can talk? The average monthly benefit B, in dol lars, for individuals on disability is given by the function B(t) = 19.25t + 585.72 where t is the number of years since January 1 , 1 990. (a) What was the average monthly benefit in 2000 (t = 10)? (b) In what year will the average monthly benefit be $893.72? [Hint: use a table with 6t = 1 .] (c) In what year will the average monthly benefit exceed $1000?
39. Disability Benefits
Y= f(x)
x
(a) Solve the equation: f(x) = g(x). (b) Solve the inequality: f(x) :5 g(x). 35. In parts (a) and (b), use the following figure.
40. Health Expenditures The total private health expenditures
Y = f(x)
Y= h (x)
x Y= g (x)
H, in billions of dollars, is given by the function H ( t ) = 26t + 411 where t is the number of years since January 1 , 1 990. (a) What were total private health expenditures in 2000 (t = 1O) ? (b) In what year will total private health expenditures be $879 billion? [Hint: use a table with 6t = 1.] (c) In what year will total private health expenditures exceed $1 trillion ($1,000 billion)?
286
CHAPTER 4
Linear and Quadratic Functions
41. Supply and Demand Suppose that the quantity supplied S
and quantity demanded D of T-shirts at a concert are given by the following functions: S(p) = -200 + SOp D(p) = 1 000 - 25p where p is the price of a T-shirt. (a) Find the equilibrium price for T-shirts at this concert. What is the equilibrium quantity? (b) Determine the prices for which quantity demanded is greater than quantity supplied. (c) What do you think will eventually happen to the price of T-shirts if quantity demanded is greater than quan tity supplied? 42. Supply and Demand Suppose that the quantity supplied S and quantity demanded D of hot dogs at a baseball game are given by the following functions: S(p) = -2000 + 3000p D(p) = 10,000 - 1 000p where p is the price of a hot dog. (a) Find the equilibrium price for hot dogs at the baseball game. What is the equilibrium quantity? (b) Determine the prices for which quantity demanded is less than quantity supplied. (c) What do you think will eventually happen to the price of hot dogs if quantity demanded is less than quantity supplied? 43. Taxes The function T(x) = 0.15(x - 7300) + 730 repre sents the tax bill T of a single person whose adjusted gross in come is x dollars for income between $7300 and $29,700, inclusive in 2005. Source: Internal Revenue Service (a) What is the domain of this linear function? (b) What is a single filer's tax bill if adjusted gross income is $18,000? (c) Which variable is independent and which is dependent? (d) Graph the linear function over the domain specified in part (a). (e) What is a single filer's adjusted gross income if the tax bill is $2860? 44. Luxury Tax In 2002, major league baseball signed a labor agreement with the players. In this agreement, any team whose payroll exceeds $128 million in 2005 will have to pay a luxury tax of 22.5% (for first-time offenses). The linear func tion T(p) = 0.225( p 128) describes the luxury tax T of a team whose payroll is p (in millions of dollars). (a) What is the implied domain of this linear function? (b) What is the luxury tax for a team whose payroll is $160 million? (c) Graph the linear function. (d) What is the payroll of a team that pays a lUxury tax of $11.7 million? -
The point at which a company 's profits equal zero is called the company 's break-even point. For Problems 45 and 46, let R represent a company 's revenue, let C represent the company's costs, and let x represent the number of units produced and sold each day. (a) Find the flrm 's break-even point; that is,find x so that R = C. (b) Find the values ofx such that R (x) > C(x). This represents the number of units that the company must sell to earn a profit.
= 8x C(x) = 4.5x + 1 7,500 46. R(x) = 1 2x C(x) = lOx + 15,000
45. R(x)
Suppose that a company has just purchased a new computer for $3000. The company chooses to depreciate the computer using the straight-line method over 3 years. (a) Write a linear function that expresses the book value V of the computer as a function of its age x. (b) Graph the linear function. (c) What is the book value of the computer after 2 years? (d) When will the computer have a book value of $2000?
47. Straight-line Depreciation
Suppose that a company has just purchased a new machine for its manufacturing facility for $ 120,000. The company chooses to depreciate the machine using the straight-line method over 10 years. (a) Write a linear function that expresses the book value V of the machine as a function of its age x. (b) Graph the linear function. (c) What is the book value of the machine after 4 years? (d) When will the machine have a book value of $72,000?
48. Straight-line Depreciation
The simplest cost function is the linear cost function, C( x) = mx + b, where the y-intercept b represents the fixed costs of operating a business and the slope m rep resents the cost of each item produced. Suppose that a small bicycle manufacturer has daily fixed costs of $1800 and each bicycle costs $90 to manufacture. (a) Write a linear function that expresses the cost C of man ufacturing x bicycles in a day. (b) Graph the linear function. (c) What is the cost of manufacturing 14 bicycles in a day? (d) How many bicycles could be manufactured for $3780?
49. Cost Function
Refer to Problem 49. Suppose that the land lord of the building increases the bicycle manufacturer's rent by $100 per month. (a) Assuming that the manufacturer is open for business 20 days per month, what are the new daily fixed costs? (b) Write a linear function that expresses the cost C of man ufacturing x bicycles in a day with the higher rent. (c) Graph the linear function. (d) What is the cost of manufacturing 14 bicycles in a day? (e) How many bicycles could be manufactured for $3780?
50. Cost Function
51. Truck Rentals A truck rental company rents a truck for one
day by charging $29 plus $0.07 per mile. (a) Write a linear function that relates the cost C, in dollars, of renting the truck to the number x of miles driven. (b) What is the cost of renting the truck if the truck is dri ven 110 miles? 230 miles? A phone company offers a domestic long distance package by charging $5 plus $0.05 per minute. (a) Write a linear function that relates the cost C, in dol lars, of talking x minutes. (b) What is the cost of talking 1 05 minutes? 180 minutes?
52. Long Dishll1ce
SECTION 4.2
Bui lding Linear Functions from Data
287
Discussion and Writing 53.
Which of the following functions might have the graph shown? (More than one answer is possible.) Y (a) f ( x ) = 2x - 7 (b) g ( x ) = -3x + 4 (c) H ( x ) = 5 (d) F ( x ) = 3x + 4
shown? (More than one answer is possible.) (a) f ( x ) = 3x + 1 Y (b) g ( x ) = -2x + 3 (c) H ( x ) = 3 (d) F ( x ) = -4x - 1 2 (e) G( ,x: ) = - - x + 3
x
1 (e) G ( x ) = 2 x + 2
55.
54. Which of the following functions might have the graph
3
Under what circumstances is a linear function f ( x ) = mx + b odd? Can a linear function ever be even?
tAre You Prepared?, Answers
1.
2.
Y
.3 2
4. {50}
3. 36
5. 0
6. True
2 x
4.2 Building Linear Functions from Data 1 Draw and Interpret Scatter Diagrams (p. 287)
OBJECTIVES
2
Disting u i s h between Linear and Nonl inear Relations (p. 288)
• 3 Use a G raph ing Util ity to Find the Line of Best Fit (p. 289)
1
Draw and Interpret Scatter Diagrams
In Section 4.1 , we built linear functions from verbal descriptions. Linear functions can also be constructed by fitting a linear function to data. The first step is to plot the ordered pairs using rectangular coordinates. The resulting graph is called a scatter diagram.
E XA M P L E 1
Drawing and I nterpreting a Scatter Diagram The data listed in Table 6 represent the apparent temperature versus the relative humidity in a room whose actual temperature is 72° Fahrenheit.
Table 6
Relative Humidity (%), x
Apparent Temperature of, y
(x, y)
°
64
(0, 64)
10
65
( 1 0, 65)
20
67
(20, 67)
30
68
(30, 68)
40
70
(40, 70)
50
71
(50, 7 1 )
60
72
(60, 72)
70
73
(70, 73)
80
74
(80, 74)
90
75
(90, 75)
76
( 1 00, 76)
1 00
288
CHAPTER 4
Linear and Quadratic Functions
�
(a) Draw a scatter diagram by hand treating relative humidity as the independent variable. (b) Use a graphing utility to draw a scatter diagram.* (c) Describe what happens to the apparent temperature as the relative humidity increases. (a) To draw a scatter diagram by hand, we plot the ordered pairs listed in Table 6, with relative humidity as the x-coordinate and apparent temperature as the y-coordinate. See Figure 6(a). Notice that the points in a scatter diagram are not connected.
Solution
(b) Figure 6(b) shows a scatter diagram using a graphing utility. 80 ,..----, 78 r----------------------1 ---� � 75 r-----·--��--- 1 - ---------� 74 �---.
Figure 6
78
m
E 72 r----------.---�-------1
� 70 �--------·-------1 58 • � 56 f_-------------------J · � 54 �----------------------1 5 c
�
�f-?------�
-10 .
o 1 0 20 30 40 50 60 70 80 90 1 00 Relative Humidity (a)
a
a
a
a
a
a
a
a
a
a
110
52 (b)
(c) We see from the scatter diagrams that, as the relative humidity increases, the apparent temperature increases. 'I"
2
w, """'""
Now Work
PROBLEM 9
•
(a )
Distinguish between Linear and N onlinear Relations
Scatter diagrams are used to help us see the type of relation that exists between two variables. In this text, we will discuss a variety of different relations that may exist between two variables. For now, we concentrate on distinguishing between linear and nonlinear relations. See Figure 7. Figure 7
. .
.
(a) Linear Y = mx + b, m > 0
E XA M P L E 2
(b) Linear Y = mx + b, m < 0
(c) Nonlinear
.
' .
'
.
.
(e) Nonlinear
(d) Nonlinear
Distinguishing between Linear and Nonlinear Relations Determine whether the relation between the two variables in Figure 8 is linear or nonlinear.
Figure 8 ' .
.
(a)
(b)
' .
(c)
'" Consult your owner's manual for the proper keystrokes.
.
' .
.
.
.
.
(d)
SECTION 4.2
(a) Linear
Solution
(b) Nonlinear
�- - N o w Work
B ui l d i ng Linear Functions from Data
( c) Nonlinear
289
(d) Nonlinear •
PROBLEM 3
In this section we will study data whose scatter diagrams imply that a linear relation exists between the two variables. Suppose that the scatter diagram of a set of data appears to be linearly related as in Figure 7(a) or (b). We might wish to find an equation of a line that relates the two variables. One way to obtain an equation for such data is to draw a line through two points on the scatter diagram and determine the equation of the line.
EXAM P L E 3
Finding an Equation for Linearly Related Data Using the data in Table 6 from Example 1 , (a) Select two points and find an equation of the line containing the points. (b) Graph the line on the scatter diagram obtained in Example l (a). (a) Select two points, say ( 1 0, 65) and (70, 73). The slope of the line joining the points (10, 65) and (70, 73) is
Solution
n1 =
The equation o f the line with slope
Figure 9
.a Q)
�
E
� C
�
�
73 - 65 70 - 10
80 78 /' 76 /74 / 72 � 70 / 68 V 66 64 / 62 ? o 1 0 20 3D 40 50 60 70 80 90 1 DO Relative Humidity
Y
- 65
� and passing through ( 10, 65) is found 2 ' Xl = 10, and 15
= m (x - X l ) =
Y =
2 15
1
using the point-slope form with m = Y - YI
8 60
2 (x - 10) 15 191 2 -x + 15 3
Yl =
65 .
POint- slope form m
y
2
=-
15'
= mx
x,
=
1 0, y,
=
65
+ b
(b) Figure 9 shows the scatter diagram from Example l (a) with the graph of the line. • = = � �� >= :::;:;:z
II
E XA M P L E 4
3
Select two other points and complete the solution. Add the line obtained to Figure 9.
Now Work P R O B L E M S 9 ( b ) AND ( c )
Use a Graphing Utility to Find the Line of Best Fit
The line obtained in Example 3 depends on the selection of points, which will vary from person to person. So the line that we found might be different from the line you found. Although the line we found in Example 3 appears to fit the data well, there may be a line that "fits it better." Do you think your line fits the data better? Is there a line of best fit? As it turns out, there is a method for finding the line that best fits linearly related data (called the line of best fit) . '"
Finding the Line of Best Fit Using the data in Table 6 from Example
1:
(a) Find the line o f best fit using a graphing utility. (b) Graph the line of best fit on the scatter diagram obtained in Example l (b). , . We s h a l l n o t discuss t h e underlying Ill athematics of l i nes of best f i t in t h i s book.
290
CHAPTER 4
Linear and Quadratic Functions
(c) Interpret the slope of the line of best fit. (d) Use the line of best fit to predict the apparent temperature of a room whose actual temperature is nOF and relative humidity is 45 % . Solution Figure
10
L i nRe9 '::I = ax+b a= . 1 2090'30909 b=64 . 4090909 1 t-' <: = . 9:3:32 1 22905 r·= . 9940886734
Figure
(a) Graphing utilities contain built-in programs that find the line of best fit for a collection of points in a scatter diagram. (Look in your owner's manual under Linear Regression or Line of Best Fit for details on how to execute the program.) Upon executing the LINear REGression program, we obtain the results shown in Figure 10. The output that the utility provides shows us the equation y ax + b, where a is the slope of the line and b is the y-intercept. The line of best fit that relates relative humidity to apparent temperature may be expressed as the line y = 0.121x + 64.409. =
(b) Figure 11 shows the graph of the line of best fit, along with the scatter diagram.
(c) The slope of the line of best fit is 0.121, which means that, for every 1 % in crease in the relative humidity, apparent room temperature increases 0.121°F.
11 78
(d) Letting x = 45 in the equation of the line of best fit, we obtain y 0.121 (45) + 64.409 � 70°F, which is the apparent temperature in the room . =
= =(,i!Ir::: ::
-10
.
. . . . . . . 110
62
•
Now Work P R O B L E M S 9 ( d ) AND ( e )
Does the line of best fit appear to be a good fit? In other words, does the line appear to accurately describe the relation between temperature and relative humidity? And just how "good" is this line of best fit? The answers are given by what is called the correlation coefficient. Look again at Figure 10. The last line of output is r = 0.994. This number, called the correlation coefficient, r, - 1 ::; r ::; 1, is a mea sure of the strength of the linear relation that exists between two variables. The clos er that I r I is to 1 , the more perfect the linear relationship is. If r is close to 0, there is little or no linear relationship between the variables. A negative value of r, r < 0, indicates that as x increases y decreases; a positive value of r, r > 0, indicates that as x increases y does also. The data given in Example 1 , having a correlation coeffi cient of 0.994, are indicative of a strong l inear relationship with positive slope.
4.2 Assess Your Understa n d i n g Concepts and Vocabulary
1. A ( n )
is used to help us to see the type of relation, if any, that may exist between two variables.
2.
__ __
True or False The correlation coefficient is a measure of the strength of a linear relation between two variables and must lie between -1 and 1, inclusive.
Skill Building
In Problems 3-8, examine the scatter diagram and determine whether the type of relation that may exist is linear or nonlinear.
, 3.
y 35 30 25 20 15 10 5
. . '
.
. .'
4. Y 14 12 10
.,
-2
.
7. 25 .,pP'.
20
.
,po .
8.
• a a
•
•
0
B
35
• •
•
.
•
30
• •
12
o 2 4 6 8 10121416 x
0 5 1 0 1 5 20 2530 3540 x
5
•
6 4 2
a
.---.----,
8
.
50
6.
22
5.
0
0
10
0
•
0
B
•
B
45
SECTION 4.2
In Problems 9-14 (a) Draw a scatter diagram. (b) Select two points from the scatter diagram and find the equation of the line containing the points selected. * 9.
12.
x
I
4
3
Y
4
x
-2
7
6
-1
7
Y
5
6
6
7
10
12
0
2
3
9
8 14
10.
16
13.
2
0
1
x
3
Y
0
x
- 20
Y
1 00
I,:�
5
7
2
3 -17 1 20
Building Linear Functions from Data
291
(c) Graph the line found in part (b) on the scatter diagram. (Ii) Use a graphing utility to find the line of best fit. (e) Use a graphing utility to draw the scatter diagram an.d graph the line of best fit on it.
9
11
13
6
9
11
-15
-14
-10
118
1 30
1
11. 14.
140
1
x
-2
Y
-4
0
x
-30
- 27
Y
-1
10
0
12
4
2
5
- 25
-20
- 14
13
13
18
Applications and Extensions
15. Consumption and Disposable Income An economist wants to estimate a line that relates personal consumption ex penditures C and disposable income 1. Both C and I are in thousands of dollars. She interviews eight heads of house holds for families of size 3 and obtains the data shown below. Let I represent the independent variable and C the depen dent variable. (a) Draw a scatter diagram. (b) Find a line that fits the data. * (c) Interpret the slope. The slope of this line is called the
Hershey's Milk Chocol ate®
marginal propensity to consume.
�
(d) Predict the consumption of a family whose disposable income is $42,000. (e) Use a graphing utility to find the line of best fit to the data. What is the correlation coefficient?
It �
1 (000)
e (OOO)
20
16
20
18
18
13
27
21
36
27
37
26
45
36
50
39
Calories, y
44.28
230
Nestle's Crunch®
44.84
230
Butterfinger®
61.30
270
B a by Ruth®
66.45
280
Almond J oy®
47.33
220
Twix® (with Ca ramel)
58.00
280
Snickers®
61.12
280
Heath®
39.52
210
Source: Megan Pocius, Student at Joliet Junior College
(c) Select two points and find an equation of the line con taining the points.* (d) Graph the line on the scatter diagram drawn in part (a). (e) Predict the number of calories in a candy bar that weighs 62.3 grams. (f) Interpret the slope of the line found in part (c).
18. Raisins The following data represent the weight w (in grams) of a box of raisins and the number N of raisins in the box.
Number of Raisins, N 42.3
The same economist as in Problem 15 wants to estimate a line that relates savings S and disposable income I. Let S = I C be the dependent vari able and I the independent variable. (a) Draw a scatter diagram. (b) Find a line that fits the data. ':' (c) Interpret the slope. The slope of this line is called the
16. Marginal P ropensity to save
-
marginal propensity to save.
(d) Predict the savings of a family whose income is $42,000. (c) Use a graphing utility to find the line of best fit. What is the correlation coefficient? 17. Candy The following data represent the weight (in grams) of various candy bars and the corresponding number of calories. (a) Draw a scatter diagram of the data treating weight as the independent variable. (b) What type of relation appears to exist between the weight of a candy bar and the number of calories? * Answers
Weight, x
87
42.7
91
42.8
93
42.4
87
42.6
89
42.4
90
42.3
82
42.5
86
42.7
86
42.5
86
Source: Jennifer Maxwell, Student at Joliet Junior College
(a) Does the relation defined by the set of ordered pairs (w, N) represent a function? (b) Draw a scatter diagram of the data treating weight as the independent variable. (c) Select two points and find the equation of the line con taining the points.*
will vary. We will use the first and the last data points in the answer section.
292
�
1 9.
CHAPTEH 4
Linear and Quadratic Functions
(d) Graph the line on the scatter diagram drawn in part (b). (e) Express the relationship found in part (c) using func tion notation. (f) Predict the number of raisins in a box that weighs 42.5 grams. (g) Interpret the slope of the line found in part (c). Height versus Head Circumference A pediatrician wanted to estimate a linear function that relates a child's height, H, to his or her head circumference, C. She randomly selects nine children from her practice, measures their height and head circumference, and obtains the data shown below. Let H represent the independent variable and C the dependent variable. (a) Use a graphing utility to draw a scatter diagram. (b) Use a graphing utility to find the line of best fit to the data. Express the solution using function notation. (c) Interpret the slope. (d) Predict the head circumference of a child that is 26 inches tall. ( e) Predict the height of a child whose head circumference is 1 7.4 inches.
,'�
Height, H (inches)
Head Ci rcumference, C (inches)
25.25
16.4
25.75
16.9
25
16.9
27.75
17.6
26.5
17.3
27
17.5
26.75
17.3
26.75
1 7.5
27.5
17.5
Source: Denise Slucki, Student at Joliet Junior College
8 '0 Gestation Period versus Life Expectancy ,� .
_
.
A researcher would like to estimate the linear function relating the gesta tion period of an animal, G, and its life expectancy, L. She collects the following data. ..... . ..---
Y "",
Animal
Gestatio n (or incubation) Period, G (days)
Cat
63
Chicken
22
Dog
Duck
Goat Lion
Pig
11 10 12
1 08
10
18
8
115
10
Ra bbit
31
7
S q uirrel
44
9
Source: Tllne Almanac 2000
20
60
22
57
23
56
23
53
27
52
29
49
30
44
-!o �
7.5
28
Demand (Pairs of Jeans Sold per Day), 0
�i
11
63
Price (S/Pair), p
(a) Does the relation defined by the set of ordered pairs (p, D) represent a function? (b) Draw a scatter diagram of the data. " (c) Using a graphing utility, find the line of best fit relating price and quantity demanded. What is the correlation coefficient? (d) Interpret the slope. (e) Express the relationship found in part (c) using func tion notation. (f) What is the domain of the function? (g) How many jeans will be demanded if the price is $28 a pair? 22. Advertising and Sales Revenue A marketing firm wishes to find a function that relates the sales S of a product and A , the amount spent o n advertising the product. The data are obtained from past experience. Advertising and sales are measured in thousands of dollars.
Life Expectancy, L (years)
151
Para keet
Let G represent the independent variable and L the depen dent variable. (a) Use a graphing utility to draw a scatter diagram. (b) Use a graphing utility to find the line of best fit to the data. Express the solution using function notation. (c) Interpret the slope. (d) Predict the life expectancy of an animal whose gesta tion period is 89 days. 21. Demand for Jeans The marketing manager at Levi-Strauss wishes to find a function that relates the demand D for men's jeans and p, the price of the jeans. The following data were ob tained based on a price history of the jeans.
;,�
Advertising Expenditures, A
Sales, S
20
335
22
339
22.5
338
24
343
24
341
27
350
28.3
351
(a) Does the relation defined by the set of ordered pairs (A , S) represent a function? (b) Draw a scatter diagram of the data. (c) Using a graphing utility, find the line of best fit relating advertising expenditures and sales. What is the correla tion coefficient?
SECTION 4.3
(d) (e) (f) (g)
Quadratic Functions an d Their P roperties
Interpret the slope. Express the relationship found in part (c) using function notation. What is the domain of the function? Predict sales if advertising expenditures are $25,000.
Discussion and Writing 23.
Maternal Age \'erSlIS Down Syndrome A biologist would like to know how the age of the mother affects the incidence rate of Down syndrome. The following data represent the age of the mother and the incidence rate of Down syndrome per 1000 pregnancies. Draw a scatter diagram treating age of the mother as the independent variable. Explain why it would not make sense to find the line of best fit for these data. '�.' 24. Find the line of best fit for the ordered pairs (1, 5) and (3, 8). ,,' What is the correlation coefficient for these data? Why is this result reasonable? 25. What does a correlation coefficient of 0 imply?
Incidence of Down Syndrome. y 33
2.4
34
3.1
35
4
36
5
37
6.7
38
8.3
39
10
40
1 3.3
41
1 6.7
42
22.2
43
28.6
44
33.3
45
50
Source: E . B. Hook, P. K. Cross, and D. M. Schreinemachers, " Ch romosomal abnormality rates at amniocentesis and in
live-born infants," Journal of the American Medical
Association, 249(15),
PREPARING FOR THIS SECTION •
•
1983, pp. 2034-2038.
Before getting started, review the following:
Intercepts (Section 2.2, pp. 165-166) Graphing Techniques: Transformations (Section 3.5, pp. 252-260)
• •
Completing the Square (Section 1.2, pp. 99-100) Quadratic Equations (Section 1 .2, pp. 100-106)
N o w Work the 'Are You Prepared?' problems on page 302.
OBJECTIVES
1 G raph a Quadratic Fu nction Using Transformations (p. 295)
2
Identify the Vertex and Axis of Sym m etry of a Quadratic Fu nction (p. 297)
3 G raph a Quad ratic Fu nction Using Its Vertex, Axis, and I ntercepts
(p. 297)
4 Find the Maxi m u m or M i n i m u m Va l u e of a Q u a d ratic Function (p. 3 0 1 )
Quadratic Functions
Here are some examples of quadratic functions.
F (x)
=
3x2 - 5x + 1
g(x)
=
- 6x2 + 1
H (x)
=
2 1 -x2 + -x 2 3
293
294
CHAPTE R 4
Linear and Quadratic Functions
DEFINITION
A quadratic function
is a function of the form
f(x)
=
ax2
+
bx
+
c
where a, b, and c are real numbers and a *- O. The domain of a quadratic func tion is the set of all real numbers.
We may also say that a quadratic function is a function defined by a second degree polynomial in one variable. Many applications require a knowledge of quadratic functions. For example, suppose that Texas Instruments collects the data shown in Table 7, which relate the number of calculators sold at the price p ( in dollars ) per calculator. Since the price of a product determines the quantity that will be purchased, we treat price as the independent variable. The relationship between the number x of calculators sold and the price p per calculator is given by the linear equation
x
=
21,000 - 150 p
Table 7 Price per Calculator, p (Dol larsl 60
Number of Calculators, x 1 2,000
65
1 1 ,250
70
1 0,500
75
9,750
80
9,000
85
8,250
90
7,500
Then the revenue R derived from selling x calculators at the price p per calcu lator is equal to the unit selling price p of the calculator times the number x of units actually sold. That is,
R R (p )
=
=
=
xp
( 21 , 000 - 150p ) p -150l + 21 ,OOOp
x =
21,000 - 150p
So the revenue R is a quadratic function of the price p. Figure 12 illustrates the graph of this revenue function, whose domain is 0 ::; p ::; 140, since both x and p must be nonnegative. In the next section we shall determine the price p that max imizes revenue. R
Figure 1 2
800,000 I
·····················································..............................................................................................................................................................
700,000
..
en
600,000
..
'0
'
�o 500 000 "
-;- 400 , 000 ..... :::> c:
g;
w
a:
.
300,000 .. . 200,000 ... 1 00,000 . ... o
.
.
14
28
42
56
70
84
98
1 1 2 1 26 1 40
Price p per calculator (dollars)
p
SECTION 4.3
Figure 1 3 Path of a cannon ball
Quadratic Functions an d Their Properties
295
A second situation in which a quadratic function appears involves the motion of a projectile. Based on Newton's Second Law of Motion (force equals mass times acceleration, F = mal, it can be shown that, ignoring air resistance, the path of a projectile propelled upward at an inclination to the horizontal is the graph of a qua dratic function. See Figure 13 for an illustration. In the next section we shall analyze the path of a projectile.
1
Graph a Quadratic Function Using Transformations We know how to graph the square function f(x) = x2 . Figure 14 shows the graph
of three functions of the form f(x) = ax2 , a
=
1:.,
and a = 3. 2 Notice that the larger the value of a, the "narrower" the graph is, and the smaller the value of a, the "wider" the graph is. Figure 15 shows the graphs of f(x) ax2 for a < O. Notice that these graphs are reflections about the x-axis of the graphs in Figure 14. Based on the results of these two figures, we can draw some general conclusions about the graph of f(x) = ax2 . First, as l a l increases, the graph becomes "taller" (a vertical stretch), and as l a l gets closer to zero, the graph gets "shorter" (a vertical compression). Second, if a is positive, the graph opens "up," and if a is negative, the graph opens "down."
> 0,
for a
=
1, a
=
Figure 14
y
2 l / f(X) = X2 f(x) � X2
f(x)
t
=
Figure 1 5
3x
y
4
,
=
-4
4
-4
x
\
-4
'",- f(x)
f(x)
Figure 1 6 Graphs of a quadratic fu nction , f( x) = a>? + bx + c, a -:f- 0
Axis of symmetry
Vertex is highest point
Vertex is lowest point
Axis of symmetry
(a) Opens up a>O
(b) Opens down
a
E XA M P L E 1
3 2
= - x
=
-x2
The graphs in Figures 14 and 15 are typical of the graphs of all quadratic func tions, which we call parabolas. ' Refer to Figure 16, where two parabolas are pic tured. The one on the left opens up and has a lowest point; the one on the right opens down and has a highest point. The lowest or highest point of a parabola is called the vertex. The vertical line passing through the vertex in each parabola in Figure 16 is called the axis of symmetry (usually abbreviated to axis) of the parabola. Because the parabola is symmetric about its axis, the axis of symmetry of a parabo la can be used to find additional points on the parabola. The parabolas shown in Figure 16 are the graphs of a quadratic function f(x) = ax2 + bx + c , a -=1= O. Notice that the coordinate axes are not included in the figure. Depending on the values of a, b, and c , the axes could be placed any where. The important fact is that the shape of the graph of a quadratic function will look like one of the parabolas in Figure 16. In the following example, we use techniques from Section 3.5 to graph a quadratic function f(x) = ax2 + bx + c , a -=1= O. In so doing, we shall complete the square and write the function f in the form f(x) = a ( x - hf + k.
G raphing a Q uadratic Function U sing Transformations Graph the function f(x)
=
2x2 + 8x +
5.
Find the vertex and axis of symmetry.
" We shall study parabolas using a geometric definition later in this book.
296
CHAPTER 4
Linear and Q u a d ratic Functions
Solution
We begin by completing the square on the right side.
f (x) = 2 x2 + 8x + 5 = 2 (x2 + 4x) + 5
Factor out the 2 from 2XZ + 8x .
Complete the square of XZ + 4x by adding 4, Notice that the factor of 2 requ ires that 8 be added and subtracted,
= 2 (x2 + 4x + 4) + 5 - 8 = 2 (x + 2 )2 - 3
The graph of f can be obtained in three stages, as shown in Figure 17. Now com pare this graph to the graph in Figure 16(a). The graph of f(x) = 2 x2 + 8x + 5 is a parabola that opens up and has its vertex (lowest point) at ( - 2 , - 3 ) Its axis of symmetry is the line x = - 2 . .
Axis of Y Symmetry3
Figure 1 7
x = -2 x
3 ( - 1 , -1 )
(a) Y = x 2
-3
-3
-3 M ultiply by 2; Vertical stretch
Replace x by x + 2; Shift left 2 units
(b) y = 2 x2
Vertex
�
(c) y = 2 (x + 2) 2
x
Subtract 3; Shift down 3 units
( - 2, -3) (d) y = 2 (x + 2)L 3 •
&1 !l! =
Now Work P R O B l E M 2 7
The method used in Example 1 can be used to graph any quadratic function =I- 0, as follows:
f(x) = ax2 + bx + c, a f(x)
=
ax2 + bx +
(
C
) = a x2 + � x +
Factor out a from aXZ + bx,
C
( b2 ) b = a x,- + - x + -,' + a 4a-
( b )2 + b2 ( x + -b )2 + ---b2
=a x+ =a
C
-
2a
C
2a
4ac 4a
4a
( 2)
b - a -, 4a-
tl
Com plete the square by adding 2' 4a Look closely at th is step! Factor, tl
4a
c - - = c · -
4a
4a
-
tl
4ac - tl
4a
4a
- = ---
Based on these results, we conclude the following: If h = -
� 2a
and
f(x)
=
k =
4ac - b2 then 4a '
ax2 + bx + c = a(x - h ? + k
( 1)
The graph of f(x) = a(x - h)2 + k is the parabola y = ax2 shifted horizon tally h units (replace x by x - h) and vertically k units (add k). As a result, the ver tex is at (h, k), and the graph opens up if a > ° and down if a < 0. The axis of sym metry is the vertical line x = h.
SECTION 4.3
f
For example, compare equation ( 1 ) with the solution given in Example (x) = 2 ( x + 2 )2 - 3 2 ( x - ( -2) ? + ( - 3 ) = a(x h ) 2 + Ie =
-
We conclude that a = 2, so the graph opens up. Also, we find that Ie = -3, so its vertex is at ( -2, - 3 ) . 2
297
Q u a d ratic Functions an d Their Properties
h
1.
-2 and
=
Identify the Vertex and Axis of Symmetry of a Quadratic Function
We do not need to complete the square to obtain the vertex. In almost every case, it is easier to obtain the vertex of a quadratic function f by remembering that its
-�. f (- �) .
x-coordinate is
h
=
- �. That is' Ie
=
.
2a
2a
The y-coordinate Ie can then be found by evaluating
f
at
2a
We summarize these remarks as follows:
f (-�,f(-�))
Properties of the Graph of a Quadratic Function
( x)
Vertex
=
2a
=
2a
ax2 + bx +
c
a *" O
Axis of symmetry: the line x =
-� 2a
(2)
Parabola opens up if a >
0; the vertex is a minimum point. Parabola opens down if a < 0; the vertex is a maximum point. E XAM P L E 2
Locating the Vertex without G raphing
f
Without graphing, locate the vertex and axis of symmetry of the parabola defined by ( x ) = -3x2 + 6x + 1 . Does it open up or down? For this quadratic function, a = -3, b vertex is b h = --
Solution
2a
=
6, and
c
=
1.
The x-coordinate of the
= -- = l
6 -6
The y-coordinate of the vertex is Ie
=f
( : ) f(l) -
a
=
=
-3 + 6 + 1
=
4
The vertex is located at the point ( 1 , 4). The axis of symmetry is the line x Because a = - 3 < 0, the parabola opens down. 3
=
1. •
Graph a Quadratic Function Using Its Vertex, Axis, and Intercepts
f
The information we gathered in Example 2, together with the location of the inter cepts, usually provides enough information to graph ( x ) = ax2 + bx + c, a *" O.
f
The y-intercept is the value of at x
=
f
0; that is, the y-intercept is ( O )
=
c.
The x-intercepts, if there are any, are found by solving the quadratic equation ax2 + bx + c = 0
298
CHAPTER 4
Linear and Quadratic Functions
This equation has two, one, or no real solutions, depending on whether the discrim inant b2 - 4ac is positive, 0, or negative. Depending on the value of the discrimi nant, the graph of 1 has x-intercepts, as follows: The x-Intercepts of a Quadratic Function
ax2 + bx + c has places. ax2 + bx + c has
- 4ac > 0, the graph of I(x) = two distinct x-intercepts so it crosses the x-axis in two 2. If the discriminant b2 - 4ac 0, the graph of 1 ( x ) one x-intercept so it touches the x-axis at its vertex. 3. If the discriminant b2 - 4ac < 0, the graph of I ( x ) = no x-intercept so it does not cross or touch the x-axis. 1. If the discriminant b2
=
=
a x2
+
bx
+
c has
Figure 18 illustrates these possibilities for parabolas that open up. ((x)
Figure 1 8
=
ax2 + bx + c, a >
0
Axis of symmetry x = - JL
Axis of symmetry x = _JL
Axis of symmetry x = _ JL
2a
2a
2a
\)
x-intercept (_ JL f (- JL))
2a 2a ' (a) bL 4ac > 0
(c) bL 4ac < 0
(b) bL 4ac = 0
No x-intercepts
One x-intercept
Two x-intercepts
E XAM P L E 3
x
b 0) (- 2a '
G raphing a Quadratic F unction Using Its Vertex, Axis, and I ntercepts (a) Use the information from Example 2 and the locations of the intercepts to graph I ( x ) = -3x2 + 6x + 1 . (b) Determine the domain and the range o f f (c) Determine where 1 is increasing and where it is decreasing.
Solution
(a) In Example 2, we found the vertex to be at (1, 4) and the axis of symmetry to be x = 1. The y-intercept is found by letting x = O. The y-intercept is 1 (0) = 1 . The x-intercepts are found b y solving the equation I ( x ) = O. This results in the equation -3x2 + 6x +
1
= a
a =
- 3,
b
=
6,
-6
+
C =
1
The discriminant b2 - 4ac = ( 6 ? - 4( - 3) ( 1 ) = 3 6 + 12 = 48 > 0, so the equation has two real solutions and the graph has two x-intercepts. Using the quadratic formula, we find that x = Figure 1 9
Axis of symmetry x =l
2a
-6 + V48 -6
4 \1'3 ;:::::: -0.15 -6
and
y
x = (2, 1 )
�4
-b + Yb2 - 4ac
x
(2.1 5, 0)
-b - Yb2 - 4ac
2a
-6 - V48 -6
-6 - 4 \1'3 ;:::::: 2.15 -6
The x-intercepts are approximately -0.15 and 2.15. The graph is illustrated in Figure 19. Notice how we used the y-intercept and the axis of symmetry, x = 1 , to obtain the additional point (2, 1 ) on the graph.
SECTION 4.3
Quadratic Functions and Their Properties
299
(b) The domain of f is the set of all real numbers. B ased on the graph, the range of f is the interval ( - 00 , 4]. (c) The function f is increasing on the interval ( - 00, 1) and decreasing on the interval ( 1 , 00 ) . •
'I,j
=___
_'l!>: = ="- '-
Graph the function in Example
3 by completing the square and using
transformations. Which method do you prefer?
Now Work P R O B L E M 3 5
If the graph of a quadratic function has only one x-intercept or no x-intercepts, it is usually necessary to plot an additional point to obtain the graph.
EXAM P L E 4
G raphing a Quadratic F unction Using Its Vertex, Axis, and I ntercepts (a) Graph f(x) = x2 - 6x + 9 by determining whether the graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range of f (c) Determine where f is increasing and where it is decreasing.
Solution
-6 b h = - - = - -- = 3 2( 1 ) 2a
Figure 20
y
(a) For f(x) = x2 - 6x + 9, we have a = 1 , b = -6, and c = 9. Since a = 1 > 0, the parabola opens up. The x-coordinate of the vertex is
Axis of symmetry
x= 3
The y-coordinate of the vertex is
k = f(3) = ( 3 ? - 6(3)
+
9 =0
So the vertex is at (3, 0). The axis of symmetry is the line x = 3. The y-intercept is f(O) = 9. Since the vertex (3, 0) lies on the x-axis, the graph touches the x-axis at the x-intercept. By using the axis of symmetry and the y-intercept at (0, 9), we can locate the additional point (6, 9) on the graph. See Figure 20. (b) The domain of f is the set of all real numbers. B ased on the graph, the range of f is the interval [0, 00 ) . (3, 0)
x
(c) The function f i s decreasing o n the interval ( - 00, 3 ) and increasing on the interval (3, 00 ) . •
Now Wor k P R O B L E M 4 3
E XA M P L E 5
G raphing a Quad ratic F unction Using Its Vertex, Axis, and I ntercepts (a) Graph f(x) = 2x2 + X + 1 by determining whether the graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range of f (c) Determine where f is increasing and where it is decreasing.
Solution
(a) For f(x) = 2 x2 + X + 1, we have a = 2, b = 1, and c = l . Since a = 2 > 0, the parabola opens up. The x-coordinate of the vertex is
b 1 17. = - - = - 4 2a
300
CHAPTER 4
Linear and Quadratic Functions
The y-coordinate of the vertex is NGrE In Example 5, since the vertex is above the x-axis and the parabola opens up, we can conclude that the graph of the quadratic function will have no • x-intercepts.
So the vertex is at
(-±, i}
The axis of symmetry is the line x
-
=
-
±. The
y-intercept is f(O) = 1. The x-intercept(s), if any, obey the equation 2x2 + x + 1 = 0. Since the discriminant b2 4ac = ( 1 ? - 4(2) ( 1 ) = -7 < 0, this equation has no real solutions, and therefore the graph has no x-intercepts.
Figure 2 1
We use the point (0, 1) and the axis of symmetry x = tional point
(-�, ) [i, ) (-±, )
_
1:. to locate the addi4
on the graph. See Figure 2 1 .
1
(b) The domain of f is the set of all real numbers. Based on the graph, the range 00
of f is the interval
.
(c) The function f is decreasing on the interval -1
the interval
x
00
(-oo,-±)
and is increasing on
.
• 1l!I!lI _ _ ",,- '-
Now Work P R O B l E M 4 7
Given the vertex (h, k) and one additional point on the graph of a qua dratic function f(x) = ax2 + bx + c , a *' 0, we can use
f(x) = a(x - h ) 2
+
(3 )
k
to obtain the quadratic function.
E XA M P L E 6
Finding the Quadratic F unction G iven Its Vertex and One Other Point
-
Determine the quadratic function whose vertex is ( 1 , -5) and whose y-intercept is 3. The graph of the parabola is shown in Figure 22.
Solution Figure 22
The vertex is ( 1 , - 5 ) , so h = 1 and k = -5. Substitute these values into equation (3).
f(x) = a(x
-
h? f(x) = a(x - 1 ) 2
y
-
+
k
Equation (3)
5
h
=
1, k
=
To determine the value of a, we use the fact that f(O) =
f(x) = a(x
-8
-3
(the y-intercept).
- 1) 2 - 5
-3 = a(O - 1 ? -3 = a - 5 (1 , -5 )
-5
-
5 x=
G, y
= f( G ) =
-3
a=2 The quadratic function whose graph is shown in Figure 22 is
f(x)
=
a(x
- h?
+
k = 2(x - 1 ?
- 5 = 2x2 - 4x - 3 •
� =:::I> '-
Now Work
PROBlEM 53
SECTION 4.3
SUMMA RY Option STEP STEP
Option
1
Steps for Graphing a Quadratic Function f(x)
+
bx +
c,
a
2
STEP
2: Determine the vertex
STEP
3: Determine the axis of symmetry, x = -
(-:a ' ( : ) ) f -
a
.
(a
>
0) or down
(a <
301
* 0
=
2: Graph the function in stages using transformations.
1: Determine whether the parabola open up
a(x
- h )2
+ k.
0).
a
b . 2
4: Determine the y-intercept, f(O), and the x-intercepts, if any. (a) If b2 - 4ac > 0, the graph of the quadratic function has two x-intercepts, which are found by solving the equation + bx + c = O. (b) If b2 - 4 c = 0, the vertex is the x-intercept.
(c) If b2 -
STEP 5: STEP
a>?
1: Complete the square in x to write the quadratic function in the form f(x)
STEP
STEP
=
Q u a d ratic Functions and Their Properties
a4ac
0, there are no x-intercepts.
Determine an additional point by using the y-intercept and the axis of symmetry.
6: Plot the points and draw the graph.
4
Find the Maximum or M inimum Value of a Quadratic Function
The graph of a quadratic function f (x) = is a parabola with vertex at the graph if
a< (a < (a
E XA M P L E 7
+ bx
+
( :a ,f(-:a ) ) (-:a ) ( -:a ) -
a
c
*
>
0
. This vertex is the highest point on
0 and the lowest point on the graph if
highest point lowest point
ax2
a
> O. If the vertex is the
0), then f
is the
maximum value
of f If the vertex is the
0), then f
is the
minimum value
of f
F i nding the M aximum or M i n i m u m Value of a Quad ratic Function Determine whether the quadratic function f(x)
=
x2 - 4x - 5
has a maximum or minimum value. Then find the maximum or minimum value.
Solution
We compare f(x) = x2 - 4x - 5 to f(x) = ax2 + bx + c . We conclude that a = 1, b = -4, and c = -5. Since a > 0, the graph of f opens up, so the vertex is a minimum point. The minimum value occurs at x
-4 = - - = - -- = - = 2 a = 1, b
The minimum value is f
&
(-:a )
b 2a i
= f(2)
=
2( 1 )
=
4 2
-4
22 - 4(2) - 5 = 4 - 8 - 5
N o w Work P R O B L E M 6 1
=
-9 •
302
C H A PTER 4
Linear and Quadratic Functions
4.3 Assess You r Understa nding JAre You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. List the intercepts of the equation y
=
x2 - 9. (pp. 165-1 66) + 7 x - 4 = O.
3. To complete the square of x2 - 5x, you add the number
2. Find the real solutions of the equation 2x2
(pp. 1 00-106)
__ .
(pp. 99-100) = (x - 4)2, you shift the graph of y a distance of units. (pp. 252-260)
4. To graph y __
__
=
x2 to the
Concepts and Vocabulary
5. The graph of a quadratic function is called a(n)
__
.
The y-coordinate of the vertex of f(x) = -x2 + 4x + 5 is f(2) .
9.
True or False
6. The vertical line passing through the vertex of a parabola is
called the 7. The x-coordinate o f the vertex o f f ( x ) = ax2 + b x + c, a #- 0, is . 8. True or False The graph of f(x) = 2x2 + 3x - 4 opens up.
True or False If the discriminant b2 - 4ac = 0, the graph of f(x) = ax2 + bx + c, a #- 0, will touch the x-axis at its vertex.
10.
__
S kill Building
In Problems 1 1-18, match each graph to one the following functions. 13. f ( x) = x2 - 2x + 1 12. f ( x) = -x2 - 1 11. f ( x) = x2 - 1 15. f(x) = x2 - 2x + 2 16. f(x) = x2 + 2x 17. f(x) = x2 - 2x A B e y y Y 2
2
3
2 x
E
y 1
(-1, - 1I ) -2 F
2 x
-2
2 x
-2 G
y
= x2 + 2x + 1 18. f(x) = x2 + 2x + 2 D y 14. f(x)
(0, - 1 )
2 x
-2 -1 H
y
y 2
-2 -1
(1 , 0)
3 x
-2
-1
3 x
-1
x
-2 -2
(1 , - 1 )
In Problems 1 9-34, graph the function f by starting with the graph of y = x2 and using transformations (shifting, compressing, stretching, and/or reflection). [Hint: If necessary, write fin the form f(x) = a(x - hl + k.] 1 1 19. f(x) = - x2 20. f(x) = 2x2 21. f(x) = X2 - 2 22. f(x) = 2X2 - 3 "4 4 1 23. f ( x) = x2 + 2 24. f(x) = 2x2 + 4 25. f(x) = X2 + 1 26. f(x) = _2x2 - 2 "4 27. f(x) = x2 + 4x + 2 28. f ( x) = x2 - 6x - 1 29. f(x) = 2X2 - 4x + 1 30. f(x) = 3x2 + 6x
�
31. f(x)
= -x2 - 2x
32. f(x)
=
_2X2 + 6x + 2
33. f (x)
1 = x2 + x - I 2 -
34. f( x)
2 = - x2 3
+
4 -x - I 3
In Problems 35-52, (a) graph each quadratic function by determining whether its graph apens up ar down and by finding its vertex, axis af symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range af the functian. (c) Determine where the functian is increasing and where it is decreasing. 35. f(x) = x2 + 2x 37. f(x) = -x2 - 6x 36. f(x) = x2 - 4x 38. f(x) = -x2 + 4x 39. f(x) = 2x2 - 8x 40. f(x) = 3x2 + 18x 41. f(x) = x2 + 2 x - 8 43. f ( x) = x2 + 2x + 1 42. f(x) = x2 - 2x - 3 44. f ( x) = x2 + 6x + 9 46. f(x) = 4x2 - 2x + 1 45. f ( x) = 2x2 - X + 2
SECTION 4.3
= 2x2 + 2x - 3 50. f(x) = 2x2 + 5x + 3
. 47. f(x)
-
= -3x2 + 3x - 2 51. f(x) = - 4x2 - 6x + 2
48. f(x)
49. f(x) 52. f(x)
In Problems 53-58, determine the quadratic function whose graph is given. 53.
y
54.
= 3x2 + 6x + 2 = 3x2 - 8x + 2
55.
y
303
Quadratic Functions and Their Properties
Vertex: (-3, 5)
\
Y 6
1
-1
56.
2
3
4
x
5 x
57.
58.
Vertex: (-2, 6)
Y 6
-1 x -4
Vertex: (1 , -3)
In Problems 59-66, determine, without graphing, whether the given quadratic function has a maximum value or a minimum value and then find the value. 59. f(x) = 2X2 + 12x 60. f(x) = _2x2 + 12x 61. f(x) = 2x2 + 12x - 3 2 62. f(x) = 4x - 8x + 3 63. f(x) = -x2 + lax - 4 64. f(x) = _2x2 + 8x + 3 65. f(x) = -3x2 + 12x + 1 66. f(x) = 4x2 - 4x Applications and Extensions = ax2 + bx + c has ver tex at (0, 2) and passes through the point (1, 8). Find a, b, and c.
67. The graph of the function f(x)
= ax2 + bx + c has ver tex at ( 1 , 4) and passes through the point ( - 1 , - 8 ) . Find a, b, and c.
68. The graph of the function f(x)
In Problems 69-74, for the given functions f and g, (a) Graph f and g on the same Cartesian plane. (b) Solve f(x) = g(x). (c) Use the result ofpart (b) to label the points of intersection of the graphs off and g (d) Shade the region for which f(x) > g(x), that is, the region below f and above g 69. f(x) = 2x - 1; g(x) = x2 - 4 70. f(x) = - 2x - 1 ; g(x) = x2 9 2 = 2 71. f(x) = - x + 4; g(x) - 2x + 1 72. f(x) = -x + 9; g(x) = 2x + 1 73. f(x) = -x2 + 5x; g(x) = x2 + 3x - 4 74. f(x) = -x2 + 7x - 6; g(x) = x2 + X - 6 -
Answer Problems 75 and 76 using the following: A quadratic function of the form f(x) = ax2 + bx + c with b2 - 4ac > a may a/so be written in the form f(x) = a(x - r1 ) (x - r2 ) , where r1 and r2 are the x-intercepts of the graph of the quadratic function. 76. (a) Find a quadratic function whose x-intercepts are -5 and 75. (a) Find a quadratic function whose x-intercepts are -3 and 3 with a = 1; a = 2; a = -2; a = 5. 1 with a = 1; a = 2; a = -2; a = 5. (b) How does the value of a affect the intercepts? (b) How does the value of a affect the intercepts? (c) How does the value of a affect the axis of sym (c) How does the value of a affect the axis of sym metry? metry? (d) How does the value of a affect the vertex? (d) How does the value of a affect the vertex? (e) Compare the x-coordinate of the vertex with the mid (e) Compare the x-coordinate of the vertex with the mid point of the x-intercepts. What might you conclude? point of the x-intercepts. What might you conclude?
304
CHAPTER 4
Linear and Quadratic Functions
77. Suppose that f(x) = x2 + 4x - 21
(a) What is the vertex of f? (b) What are the x-intercepts of the graph of f? (c) Solve f(x) = -21 for x. What points are on the graph of f? (d) Use the information obtained in parts (a)-(c) to graph f(x) = x2 + 4x - 2l. 78. Suppose that f(x) = x2 + 2x - 8 (a) What is the vertex of f? (b) What are the x-intercepts of the graph of f? (c) Solve f(x) = -8 for x. What points are on the graph off? (d) Use the information obtained in parts (a)-(c) to graph f(x) = x2 + 2x - 8. 79. Find the point on the line y = x that is closest to the point (3, 1 ) . [Hint: Express the distance d from the point to the line as a function of x, and then find the minimum value of [d(x)f 80. Find the point on the line y = x + 1 that is closest to the point (4, 1 ) . 81. Maximizing Revenue Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is p dollars, the revenue R (in dollars) is R(p) -4p2 + 4000p What unit price should be established for the dryer to maxi mize revenue? What is the maximum revenue? 82. Maximizing Revenue The John Deere company has found that the revcnue, in dollars, from sales of heavy-duty tractors is a function of the unit price p, in dollars, that it charges. If the revenue R is 1 R(p) - Z p2 + 1900p what unit price p should be charged to maximize revenue? What is the maximum revenue? 83. Minimizing Marginal Cost The marginal cost of a product can be thought of as the cost of producing one additional unit of output. For example, if the marginal cost of producing the 50th product is $6.20, it cost $6.20 to increase production from 49 to 50 units of output. Suppose the marginal cost C (in dollars) to produce x thousand mp3 players is given by the function C(x) = x2 - 140x + 7400 (a) How many players should be produced to minimize the marginal cost? (b) What is the minimum marginal cost? 84. Minimizing Marginal Cost (See Problem 83.) The margin al cost C (in dollars) of manufacturing x cell phones (in thou sands) is given by C(x) = 5x2 - 200x + 4000. (a) How many cell phones should be manufactured to min imize the marginal cost? (b) What is the minimum marginal cost? 85. Hunting The function H(x) = -3.24x2 + 242.1x - 738.4 models the number of individuals whose age is x and engage in hunting activities. (a) What is the age at which there are the most hunters? Approximately how many hunters are this age? =
=
(b) Are the number of hunters increasing or decreasing for individuals who are between 40 and 45 years of age? Source: National Sporting Goods Association 86. Advanced Degrees The function P(x) -0.008x2 + 0.815x - 9.983 models the percentage of the U.S. population whose age is x that have earned an advanced degree (more than a bache lor's degree) in March 2003. (a) What is the age for which the highest percentage of Americans have earned an advanced degree? What is the highest percentage? (b) Is the percentage of Americans who have earned an ad vanced degree increasing or decreasing for individuals between the ages of 40 and 50? Source: Us. Census Bureau 87. Male Murder V ictims The function M(x) = l.00x2 - 136.74x + 4764.89 models the number of male murder victims who are x years of age (20 :=; x < 90). (a) Use the model to approximate the number of male murder victims who are x = 23 years of age. (b) At what age is the number of male murder victims 1456? Describe what happens to the number of male murder (c) victims as age increases from 20 to 65. Source: Federal Bureau of Investigation 88. Health Care Expenditures The function H(x) 0.004x2 - 0.197x + 5.406 models the percentage of total income that an individual who is x years of age spends on health care. (a) Use the model to approximate the percentage of total income an individual who is x 45 years of age spends on health care. (b) At what age is the percentage of income spent on health care 1O%? ;;r . . (c) Using a graphing utility, graph H = H(x). (d) Based on the graph drawn in part (c), describe what happens to the percentage of income spent on health care as individuals age. Source: Bureau of Labor Statistics 89. Business The monthly revenue R achieved by selling x wristwatches is figured to be R(x) = 75x - 0.2x2. The monthly cost C of selling x wristwatches is C(x) 32x + 1750. (a) How many wristwatches must the firm sell to maximize revenue? What is the maximum revenue? (b) Profit is given as P(x) = R(x) - C(x). What is the profit function? (c) How many wristwatches must the firm sell to maximize profit? What is the maximum profit? (d) Provide a reasonable explanation as to why the answers found in parts (a) and (c) differ. Explain why a qua dratic function is a reasonable model for revenue. 90. Business The daily revenue R achieved by selling x boxes of candy is figured to be R(x) = 9.5x - 0.04x2. The daily cost C of selling x boxes of candy is C (x) 1.25x + 250. =
=
=
=
=
SECTION 4.4
(a) How many boxes of candy must the firm sell to maxi mize revenue? What is the maximum revenue? (b) Profit is given as P(x) = R(x) - C(x). What is the profit function? (c) How many boxes of candy must the firm sell to maxi mize profit? What is the maximum profit?
91.
Quadratic Models; Building Q u a d ratic Functions from Data
305
(d) Provide a reasonable explanation as to why the answers found in parts (a) and (c) differ. Explain why a qua dratic function is a reasonable model for revenue. Let f(x ) = ax2 + bx + c, where a, b, and c are odd integers. If x is an integer, show that f(x) must be an odd integer. [Hint: x is either an even integer or an odd integer.]
Discussion and Writing 92.
93.
94.
95.
Make up a quadratic function that opens down and has only one x-intercept. Compare yours with others in the class. What are the similarities? What are the differences? On one set of coordinate axes, graph the family of parabolas f(x) = x2 + 2x + c for c = -3, c = 0, and c = 1. Describe the characteristics of a member of this family. On one set of coordinate axes, graph the family of parabolas f(x) = x2 + bx + 1 for b = -4, b = 0, and b = 4. De scribe the general characteristics of this family.
96. 97. 98.
State the circumstances that cause the graph of a quadratic function f(x) = ax2 + bx + c to have no x-intercepts. Why does the graph of a quadratic function open up if a > 0 and down if a < O? Can a quadratic function have a range of ( - 00, oo )? Justify your answer. What are the possibilities for the number of times the graphs of two different quadratic functions intersect?
'Are You Prepared?' Answers
1. (0, -9), ( -3, 0), (3, 0)
2.
{-4, �}
25 4
3. -
4.
right; 4
4.4 Quadratic Models; Build ing Quadratic Functions from Data PREPARING FOR THIS SECTION •
Before getting started, review the following:
Problem Solving (Section 1.7, pp. 139-145)
•
Building Linear Functions from Data (Section 4.2, pp. 287-290)
'\. N o w Work the 'Are You Prepared?' problems on page 3 1 0. OBJECTIVES
1 Solve Applied Problems I nvolving Quad ratic Fu nctions (p. 305)
� 2
Use a G ra p h i ng Util ity to Find the Quadratic Function of Best Fit (p. 309)
In this section we will first discuss models that lead to a quadratic function when a verbal description of the problem is given. We end the section by fitting a quadratic function to data, which is another form of modeling. When a mathematical model leads to a quadratic function, the properties of the graph of the quadratic function can provide important information about the model. In particular, we can use the quadratic function to determine the maximum or minimum value of the function. The fact that the graph of a quadratic function has a maximum or minimum value enables us to answer questions involving optim ization, that is, finding the maximum or minimum values in models involving qua dratic functions. 1
S olve Applied Problems Involving Quadratic Functions
In economics, revenue R, in dollars, is defined as the amount of money received from the sale of an item and is equal to the unit selling price p, in dollars, of the item times the number x of units actually sold. That is,
R
=
xp
In economics, the Law of Demand states that p and x are related: As one increases, the other decreases. The equation that relates p and x is called the demand
equation.
306
CHAPTER 4
Linear and Quadratic Functions
EXAM P L E 1
M aximizing Revenue The marketing department at Texas Instruments has found that, when certain cal culators are sold at a price of p dollars per unit, the number x of calculators sold is given by the demand equation x = 21,000 - 150p (a) (b) (c) (d) (e)
Solution
Express the revenue R as a function of the price p. What urnt price should be used to maximize revenue? If this price is charged, what is the maximum revenue? How many units are sold at this price? Graph R.
(a) The revenue R is R R
=
xp
=
=
xp, where x = 21,000 - 150p. (21 ,000 - 150p) p
=
- 150p2
+
21,000p
(b) The function R is a quadratic function with a = - 1 50, b = 21 ,000, and c = O. Because a < 0, the vertex is the highest point on the parabola. The revenue R is therefore a maximum when the price p is
_ _ � _ _ 21,000 _ _ 21 ,000 - $ 70.00 _
p -
2a
2( - 150)
-300
i
a = - 1 50, b = 21,000
( c) The maximum revenue R is R (70) = - 150(70) 2 + 21,000(70)
=
$735,000
(d) The number of calculators sold is given by the demand equation x = 21,000 - 150p. At a price of p = $70, x calculators are sold.
=
21,000 - 150(70)
=
10,500
(e) See Figure 23 for the graph of R. R 800 , 000 I ·
Figure 23
··· ············································································:�c···cc·�·�·�·�·:·········........................................,
700,000 ... _
�
i o
�
�
600,000 .... . . 500 '000 ... 400,000 ... . ... 300,000 ... . .. 200 , 000 ... . . . ................................ . .. .
14
28
42
....... . . ... ... . .. .... . .. .... ...
56
Now Work
E XA M P LE 2
70
84
p
98
Price p per calculator (dollars)
•
PROBLEM 3
M aximizing the Area E nclosed by a F ence A farmer has 2000 yards of fence to enclose a rectangular field. What are the di mensions of the rectangle that encloses the most area?
Solution
Figure 24 illustrates the situation. The available fence represents the perimeter of the rectangle. If x is the length and w is the width, then 2x
+
2w = 2000
The area A of the rectangle is
A
=
xw
(1)
SECTION 4.4
Q uad ratic Models; Building Q uadratic Fu nctions from Data
307
To express A in terms of a single variable, we solve equation (1) for w and substi tute the result in A = xw. Then A involves only the variable x. [You could also solve equation (1) for x and express A in terms of w alone. Try it!]
Figure 24
x
2x
w
2w 2w
+
=
=
w
x
=
2000 2000 - 2x 2000 - 2x 2
Equation (1)
=
Solve for w.
1000 - x
Then the area A is
A
xw
=
=
x( lOOO - x)
=
-x2 + 1000x
Now, A is a quadratic function of x. =
A ( x)
1000x
+
a = - 1, b = 1000, C = 0
Figure 25 shows a graph of A(x) = -x2 + 1000x. Since a < 0, the vertex is a maximum point on the graph of A . The maximum value occurs at
Figure 2S A
-x2
(500, 250000)
x (1 000, 0) -'--'---'*- ---'--�,..:.... 500 1 000 x
=
b
-2a
=
1000 - -2(-1)
=
500
The maximum value of A is
( :a)
A -
=
A (500)
=
-5002 + 1000(500)
=
-250,000 + 500,000
=
250,000
The largest rectangle that can be enclosed by 2000 yards of fence has an area of 250,000 square yards. Its dimensions are 500 yards by 500 yards.
•
= = =-
E XA M P L E 3
Now Work P R O B L E M 7
Analyzing the Motion of a Projectil e A projectile i s fired from a cliff 500 feet above t h e water a t a n inclination o f 45° to the horizontal, with a muzzle velocity of 400 feet per second. In physics, it is estab lished that the height h of the proj ectile above the water is given by
h(x)
=
-32x2 + x + 500 (400?
where x is the horizontal distance of the proj ectile from the base of the cliff. See Figure 26. Figure 26
h (x)
2500 2000 1 500 /' 1 000 5° V4 500
-
"9-
t-.....
....... -
1 000 2000
""
�
3000 4000 5000
x
(a) Find the maximum height of the projectile. (b) How far from the base of the cliff will the projectile strike the water?
Solution
(a) The height of the projectile is given by a quadratic function.
h (x)
=
-32x2 + X + 500 (400) -
-7
=
-1 --x2 + X + 500 5000
308
CHAPTER 4
Linear and Quadratic Functions
We are looking for the maximum value of h. Since a < 0, the maximum value is obtained at the vertex. We compute x = - !!..- = 2a
2
(
)
=
=
- 1250 + 2500 + 500
1
1 __ 5000
5000 = 2500 2
The maximum height of the projectile is h(2500) =
-1 (2500) 2 + 2500 5000
+
500
=
1750 ft
(b) The projectile will strike the water when the height is zero. To find the distance x traveled, we need to solve the equation hex) =
1/ Seeing the Concept
We find the discriminant first.
Graph
h (x) = o :s; Use
-1 --
5000
x2 +
x :s; 5 5 00
MAXIMUM
X
-1 x2 + X + 500 = 0 5000
b2 - 4ac
+ 500
= 12 - 4
( )
2 ( 500) = 1 . 4 5000
Then
to find the maxi m u m
height o f t h e projectile, a n d u s e ROOT o r
x =
ZERO t o find t h e distance from t h e base
-b
±
Vb2
- 4ac
2a
(
-1
±
2 -
of the cliff to where it strikes the water. Compare you r results with those ob
ViA
1 5000
)
�
{
-458 5458
We discard the negative solution and find that the projectile will strike the water at a distance of about 5458 feet from the base of the cliff.
tained in Example 3.
• �
EXAM P L E 4
Now Work P R O B L E M 1 1
The Golden Gate Bridge The Golden Gate Bridge, a suspension bridge, spans the entrance to San Francisco Bay. Its 746-foot-tall towers are 4200 feet apart. The bridge is suspended from two huge cables more than 3 feet in diameter; the 90-foot-wide roadway is 220 feet above the water. The cables are parabolic in shape':' and touch the road surface at the center of the bridge. Find the height of the cable above the road at a distance of 1000 feet from the center.
Solution
Figure 27
See Figure 27. We begin by choosing the placement of the coordinate axes so that the x-axis coincides with the road surface and the origin coincides with the center of the bridge. As a result, the twin towers will be vertical (height 746 - 220 = 526 feet above the road) and located 2100 feet from the center. Also, the cable, which has the shape of a parabola, will extend from the towers, open up, and have its vertex at (0, 0). The choice of placement of the axes enables us to identify the equation of the parabola as y = ax2 , a > O. We can also see that the points ( -2100, 526) and (2100, 526) are on the graph. ( - 21 00, 526)
y
(21 00, 526)
':' A cable suspended from two towers is in the shape of a catenary, but when a horizontal roadway is
suspended from the cable, the cable takes the shape of a parabola.
SECTION 4.4
Quadratic Models; B u i l d i n g Quadratic Functions from Data
309
B ased on these facts, we can find the value of a in y = ax2 . y =
526
ax2
=
a(2100) 2
x = 2100, y = 526
526 (2100) 2
a=
---
The equation of the parabola is therefore
526 2 x (2100 ) 2
=
Y
The height of the cable when x = 1000 is
526 ( 1000) 2 (2100) 2
y =
�
1 19.3 feet
The cable is 1 19.3 feet above the road at a distance of 1000 feet from the center of the bridge. • Q) =:;:>-
2
Now Work P R O B L E M 1 3
Use a Graphing Utility to Find the Qua dratic Function of Best Fit
In Section 4.2 we found the line of best fit for data that appeared to be linearly related. It was noted that data may also follow a nonlinear relation. Figures 28(a) and (b) show scatter diagrams of data that follow a quadratic relation. Figure 28 ..
.
.
.
. •
.
•
"0 • ° 0
• ° 0 .
0 •
.
•
• . .
y
E XA M P L E 5
= ax2
+
bx + (a)
c,
a>0
y
.
° 0
•
:' •
.
' .
. .
.
'
"0 "
.
.
= ax2 + bx + (b)
c,
a<0
Fittin g a Quad ratic F unction to Data A farmer collected the data given in Table 8 (on page 310), which shows crop yields Y for various amounts x of fertilizer used. (a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) Use a graphing utility to find the quadratic function of best fit to these data. (c) Use the function found in part (b) to determine the optimal amount of fertil izer to apply. (d) Use the function found in part (b) to predict crop yield when the optimal amount of fertilizer is applied. (e) Draw the quadratic function of best fit on the scatter diagram.
Solution
(a) Figure 29 shows the scatter diagram, from which it appears that the data follow a quadratic relation, with a < O.
31 0
CHAPTER 4
Linear and Quadratic Functions
Table 8
Fertilizer, x (Pounds/100 ft21
(b) Upon executing the QUADratic REGression program, we obtain the results shown in Figure 30. The output that the utility provides shows us the equation y = ax2 + bx + c. The quadratic function of best fit is
Yield, Y (Bushelsl
0
4
2
0
6
3
5
10
4
5 10
12
6
10
10
7
15
15
8
15
17
9
20
18
10
20
21
11
25
20
12
25
21
13
30
21
14
30
22
15
35
21
16
35
20
17
40
19
18
40
19
Y(x)
=
-0.0171x2 + 1 .0765x + 3.8939 Figure 30
Figure 29
25
a a a a
a B a a a
B B
Qu.:adRe9
':I=a:x: ;:: +b�<+c. a= - . 0 1 7 1 2 1 2 1 2 1 b= 1 . 0765 1 5 1 52 c.=3 . 8939393'34
a
-5 �======� 45 a
where x represents the amount of fertilizer used and Y represents crop yield.
( c) B ased on the quadratic function of best fit, the optimal amount of fertilizer to apply is
x =
1 .0765 b = 2( -0.0171 ) 2a
- -
>=:::;
31.5 pounds of fertilizer per 100 square feet
(d) We evaluate the function Y(x) for x = 31.5.
Y ( 3 1 .5 )
=
-0.0171 (31.5)2 + 1.0765 (31.5) + 3.8939
>=:::;
20.8 bushels
If we apply 31.5 pounds of fertilizer per 100 square feet, the crop yield will be 20.8 bushels according to the quadratic function of best fit. (e) Figure 31 shows the graph of the quadratic function found in part (b) drawn
Figure 3 1
on the scatter diagram.
25
.y� /.�
Look again at Figure 30. Notice that the output given by the graphing calculator does not include r, the correlation coefficient. Recall that the correlation coefficient is a measure of the strength of a linear relation that exists between two variables. The graphing calculator does not provide an indication of how well the function fits the data in terms of r since a quadratic function cannot be expressed as a linear function.
a ...a.-i-iI--
11'__�
...d
}r /a
-5
•
45 a
i.lm�
Now Work P R O B L E M 2 7
4.4 Assess You r U nderstan d i n g 'Are You Prepared?' A nswers are given a t the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Translate the following sentence to a mathematical equation: The total revenue R from selling x hot dogs is $3 times the number
� 2.
of hot dogs sold. (pp. 139-1 45) Use a graphing utility to find the line of best fit for the following data: (pp. 287-290)
x y
I
3 10
5 13
5 12
6 15
7 16
8 19
Applications and Extensions . 3.
The price p (in dollars) and the quan tity x sold of a certain product obey the demand equation
Demand E(luation
P =
1 6
- -x
+ 100
a :S X :S 600
(a) Express the revenue R as a function of x. (Remember, R = xp.) (b) What is the revenue if 200 units are sold?
(c) What quantity x maximizes revenue? What is the max imum revenue? (d) What price should the company charge to maximize revenue? 4. Demand Equation The price p (in dollars) and the quantity x sold of a certain product obey the demand equation 1 P = - - x + 1 00 a :S X :S 300 3
SECTION 4.4
(a) Express the revenue R as a function of x. (b) What is the revenue if 100 units are sold? (c) What quantity x maximizes revenue? What is the max imum revenue? (d) What price should the company charge to maximize revenue? s. Demand Equation The price p (in dollars) and the quanti ty x sold of a certain product obey the demand equation x = -5p + 100, 0 :5 P :5 20 (a) Express the revenue R as a function of x. (b) What is the revenue if 15 units are sold? (c) What quantity x maximizes revenue? What is the max imum revenue? (d) What price should the company charge to maximize revenue? 6. Demand Equation The price p (in dollars) and the quanti ty x sold of a certain product obey the demand equation x = -20p + 500, 0 :5 P :5 25 (a) Express the revenue R as a function of x. (b) What is the revenue if 20 units are sold? (c) What quantity x maximizes revenue? What is the max imum revenue? (d) What price should the company charge to maximize revenue? 7. Enclosing a Rectangular Field D avid has 400 yards of fenc ing and wishes to enclose a rectangular area. (a) Express the area A of the rectangle as a function of the width w of the rectangle. (b) For what value of w is the area largest? (c) What is the maximum area? 8. Enclosing a Rectangular Field Beth has 3000 feet of fenc ing available to enclose a rectangular field. (a) Express the area A of the rectangle as a function of x, where x is the length of the rectangle. (b) For what value of x is the area largest? (c) What is the maximum area? 9. Enclosing the Most Area with a Fence A farmer with
4000 meters of fencing wants to enclose a rectangular plot that borders on a river. If the farmer does not fence the side along the river, what is the largest area that can be enclosed? (See the figure.)
x
A farmer with 2000 meters of fencing wants to enclose a rectangular plot that bor ders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed? 11. Anal)'zing the Motion of a Projectile A projectile is fired from a cliff 200 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 50 feet per second. The height h of the projectile above the water is given by -32x2 hex) = -, + x + 200 (50)-
10. Enclosing the Most Area with a Fence
Quadratic Models; Building Quad ratic Functions from Data
31 1
where x is the horizontal distance of the projectile from the face of the cliff. (a) At what horizontal distance from the face of the cliff is the height of the projectile a maximum? (b) Find the maximum height of the projectile. (c) At what horizontal distance from the face of the cliff will the projectile strike the water? � (d) Using a graphing utility, graph the function h, o :5 X :5 200. (e) Use a graphing utility to verify the solutions found in parts (b) and (c). (f) When the height of the projectile is 100 feet above the water, how far is it from the cliff? 12. Analyzing the Motion of a Projectile A projectile is fired at an inclination of 45° to the horizontal, with a muzzle velocity of 100 feet per second. The height h of the projectile is given by -32x2 hex) = + x ( 100) 2 where x is the horizontal distance of the projectile from the firing point. (a) At what horizontal distance from the firing point is the height of the projectile a maximum? (b) Find the maximum height of the projectile. (c) At what horizontal distance from the firing point will the projectile strike the ground? � (d) Using a graphing utility, graph the function h, o :5 X :5 350. (e) Use a graphing utility to verify the results obtained in parts (b) and (c). (f) When the height of the projectile is 50 feet above the ground, how far has it traveled horizontally? 13. Suspension Bridge A suspension bridge with weight uni formly distributed along its length has twin towers that extend 75 meters above the road surface and are 400 meters apart. The cables are parabolic in shape and are suspended from the tops of the towers. The cables touch the road surface at the center of the bridge. Find the height of the cables at a point 100 meters from the center. (Assume that the road is level.) 14. Architecture A parabolic arch has a span of 120 feet and a maximum height of 25 feet. Choose suitable rectangular co ordinate axes and find the equation of the parabola. Then calculate the height of the arch at points 10 feet, 20 feet, and 40 feet from the center. 15. Constructing Rain Gutters A rain gutter is to be made of aluminum sheets that are 12 inches wide by turning up the edges 90°. See the illustration. What depth will provide maximum cross-sectional area and hence allow the most water to flow?
31 2
CHAPTER 4
Linear and Quadratic Functions
A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle. See the figure. If the perimeter of the window is 20 feet, what dimensions will admit the most light (maximize the area)? [Hint: Circumference of a circle = 2 7fr ; area of a circle = 7f r2, where r is the radius of the circle.]
16. Norman Windows
catastrophes in U.S. history (as of 2005). For the United States Automobile Association (USAA) and its affiliates, the total cost of claims for catastrophic losses, in millions, can be approximated by C(x) = 34.87x2 - 98.1x + 258.3, where x = 0 for 1999, x = 1 for 2000, x = 2 for 2001, and so on. (a) Estimate the total cost of claims for the year 2003. (b) According to the model, during which year were cata strophic loss claims at a minimum? ( c) Would C(x) be useful for predicting total catastrophic loss claims for the year 2015? Why or why not? Source: USAA Report to Members 2005 A self-catalytic chemical reaction re sults in the formation of a compound that causes the forma tion ratio to increase. If the reaction rate V is given by
21. Chemical Reactions
Vex) A track and field playing area is in the shape of a rectangle with semicircles at each end. See the figure. The inside perimeter of the track is to be 1500 meters. What should the dimensions of the rectangle be so that the area of the rectangle is a maximum?
17. Constructing a Stadium
=
kx(a - x),
where k is a positive constant, a is the initial amount of the compound, and x is the variable amount of the compound, for what value of x is the reaction rate a maximum?
f,. 22. Calculus: Siml)SOn'S Rule The figure shows the graph of y = ax2 + bx + c. Suppose that the points ( -h, Yo) , (0, Yl ) , and ( h , Y2 ) are o n the graph. It can b e shown that the area en closed by the parabola, the x-axis, and the lines x = -h and x = h is h Area = (2ah2 + 6c) 3
-
Show that this area may also be given by
y
A special window has the shape of a rectan gle surmounted by an equilateral triangle. See the figure. If the perimeter of the window is 16 feet, what dimensions will admit the most light? [Hint: Area of an equilateral triangle = x2, where x is the length of a side of the triangle.]
18. Architecture
(�)
x
f,. 23. Use the result obtained in Problem 22 to find the area en closed by f (x) = 5x2 + 8, the x-axis, and the lines x = -1 and x = 1 .
-
An accepted relationship between stopping distance, d (in feet), and the speed of a car, v (in mph), is d = 1 . 1 v + 0.06v2 on dry, level concrete. (a) How many feet will it take a car traveling 45 mph to stop on dry, level concrete? (b) If an accident occurs 200 feet ahead of you, what is the maximum speed you should be traveling to avoid being involved? (c) What might the term 1.1v represent? Source: www2.nsta.orgIEnergylfn_braking.html
19. Stopping Distance
The years 1999 to 2005 were particularly costly for insurance companies, with 7 of the 10 most costly
20. Insurance Claims
f,. 24. Use the result obtained in Problem 2 2 t o find the area en closed by f (x) = 2x2 + 8, the x-axis, and the lines x = - 2 and x = 2.
f,. 25. Use the result obtained in Problem 22 to find the area en closed by f(x) = x2 + 3x + 5, the x-axis, and the lines x = -4 and x = 4.
f,. 26. Use the result obtained in Problem 22 to find the area en closed by f (x) = -x2 + X + 4, the x-axis, and the lines x = -1 and x = 1 . A n individual's income varies with his or her age. The following table shows the median in come I of individuals of different age groups within the United States for 2003. For each age group, let the class midpoint represent the independent variable x. For the class
27. Life Cycle Hypothesis
SECTION 4.4
"65 years and older," we will assume that the class midpoint is 69.5. Age (yearsl 1 5-24
Class Midpoint, x
Median Income (Sl, I
1 9.5
8,614
25-34
29.5
26,21 2
35-44
39.5
30,9 14
45-54
49.5
32,583
55-64
59.5
28,068
65 a n d older
69.5
1 4,664
"�
29.
Quadratic Models; Building Quadratic Functions from Data
(d) Use a graphing utility to verify that the function given in part (b) is the quadratic function of best fit. (e) With a graphing utility, draw a scatter diagram of the data and then graph the quadratic function of best fit on the scatter diagram. Height of a Ball A shot-putter throws a ball at an inclina tion of 45° to the horizontal. The following data represent the height h of the ball at the instant that it has traveled x feet horizontally.
Distance, x
Source: U.S. Census Bureau, 2003 Annual Social and Economic Supplement
(a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) The quadratic function of best fit to these data is I ( x ) = -34.3x2 + 3157x - 39,115
Use this function to determine the age at which an in dividual can expect to earn the most income. (c) Use the function to predict the peak income earned. � (d) Use a graphing utility to verify that the function given in part (b) is the quadratic function of best fit. (e) With a graphing utility, draw a scatter diagram of the data and then graph the quadratic function of best fit on the scatter diagram. 28. Life Cycle Hypothesis An individual's income varies with his or her age. The following table shows the median income I of individuals of different age groups within the United States for 2004. For each age group, let the class mid point represent the independent variable x. For the class "65 years and older," we will assume that the class midpoint is 69.5.
Age (yearsl 1 5-24 25-34
Class Midpoint, x
Median Income (Sl, I
1 9.5
8,782
29.5
26,642
�
Height, h
20
25
40
40
60
55
80
65
1 00
71
1 20
77
1 40
77
1 60
75
1 80
71
200
64
(a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) The quadratic function of best fit to these data is hex) = -O.0037x2 + 1 .03x + 5.7 Use this function to determine the horizontal distance the ball will travel before it reaches its maximum height. (c) Use the function to find the maximum height of the ball. (d) Use a graphing utility to verify that the function given in part (b) is the quadratic function of best fit. (e) With a graphing utility, draw a scatter diagram of the data and then graph the quadratic function of best fit on the scatter diagram. An engineer collects data showing the speed s of a Ford Taurus and its average miles per gallon, M. See the table.
30. Miles per Gallon
35-44
39.5
31 ,629
45-54
49.5
32,908
Speed, s
Miles per Gal lon, M
55-64
59.5
28,5 1 8
30
18
6 5 a n d older
69.5
1 5, 1 93
35
20
40
23
40
25
45
25
50
28
Source: U.S. Census B ureau, 2004 Annual Social and Economic Supplement
(a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) The quadratic function of best fit to these data is I ( x ) = -34.5x2 + 3186x - 39,335 Use this function to determine the age at which an in dividual can expect to earn the most income. (c) Use the function to predict the peak income earned.
31 3
55
30
60
29
65
26
65
25
70
25
31 4
CHAPTER 4
Linear and Quadratic Functions
(c) Use the function to predict miles per gallon for a speed of 63 miles per hour. ;,; (d) Use a graphing utility to verify that the function given in part (b) is the quadratic function of best fit. (e) With a graphing utility, draw a scatter diagram of the data and then graph the quadratic function of best fit on the scatter diagram.
(a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) The quadratic function of best fit to these data is M(s) = -0.017s2 + 1 .93s - 25.34
Use this function to determine the speed that maximizes miles per gallon. Discussion and Writing 31.
Refer to Example 1 on page 306. Notice that if the price charged for the calculators is $0 or $140 the revenue is $0. It
is easy to explain why revenue would be $0 if the price charged is $0, but how can revenue be $0 if the price charged is $140?
'Are You Prepared?' Answers
1. R
=
2.
3x
Y =
l.7826x + 4.0652
4.5 Ineq ualities Involving Quadratic Functions PREPARING FOR THIS SECTION •
Before getting started, review the following: •
Solve Inequalities (Section 1 .5, p. 131)
Use Interval Notation (Section 1 .5, pp. 125-126)
Now Work the 'Are You Prepared? , problem on page 3 1 6.
OBJECTIVE
1
1
Solve Inequal ities I nvolving a Q u a d ratic Function (p. 314)
Solve Inequalities Involving a Quadratic Function
In this section we solve inequalities that involve quadratic functions. We will accom plish this by using their graphs. For example, to solve the inequality
a *- O ax2 + bx + c > 0 2 we graph the function f( x ) = ax + bx + c and, from the graph, determine where it is above the x-axis, that is, where f( x ) > O. To solve the inequality ax2 + bx + c < 0, a *- 0, we graph the function f(x) = ax2 + bx + c and deter mine where the graph is below the x-axis. If the inequality is not strict, we include the x-intercepts in the solution. Let's look at an example.
EXA M P L E 1
Solving an I n equal ity
Solve the inequality x2 - 4x - 12 :5 0 and graph the solution set. Solution
We graph the function f(x) = x2 - 4x - 12. The intercepts are y-intercept:
x-intercepts ( if any ) :
f(O) = - 12
x2 - 4x - 12 = 0
( x - 6 ) (x + 2 ) x - 6 = 0 or x + 2 x = 6 or x
=
=
=
0 0 -2
Eva luate f at O. Solve f(x) = O. Factor. Apply the Zero-Product Property.
The y-intercept is - 12; the x-intercepts are -2 and 6. The vertex is at x = -
b
2a
S e e Figure 32 for the graph.
=
-4
- 2 = 2. Since f(2) = - 16, the vertex is (2, -16).
SECTION 4.5
I [ -4 -2
0
2
4
1 I I 6 8
•
x
31 5
y
Figure 32
Figure 33
I nequalities Involving Quadratic Functions
The graph is below the x-axis for -2 < x < 6. Since the original inequality is not strict, we include the x-intercepts. The solution set is { x 1 - 2 ::5 X ::5 6} or, us ing interval notation, [-2, 6]. See Figure 33 for the graph of the solution set. • I
EXAM P L E 2
Now Work P R O B L E M 9
Solving an I neq uality
Solve the inequality 2X2 < x + 10 and graph the solution set. Solution
Method 1
We rearrange the inequality so that 0 is on the right side. 2x2 < x + 10 2x2 - X - 10 < 0
Subtract x +
10 from both sides.
This inequality is equivalent to the one that we wish to solve. Next we graph the function f(x) = 2x2 - X - 10. The intercepts are
f(O)
y-intercept:
=
-10
Eva luate f at O.
x-intercepts (if any) : 2x2 - x - 1 0 = 0 ( 2x - 5 ) (x + 2 ) 0
Solve f(x) = O. Factor.
=
2x - 5
=
Figure 34
y 4
x
0 or x 5
=
-
2
or
x
+ 2 =
=
0
Apply the Zero-Product Property.
-2
5 . . - 10 ; th e x-lI1tercepts IS are - 2 an d 2" Th e y-mtercept ' TIle vertex
(�,
)
.
IS
at x
=
-
b 2a
=
-1 -4
=
()
l 1 S mce ' f 4 4'
=
-10 . 125, the vertex is
- 10.125 . See Figure 34 for the graph. The graph is below the x-axis between x
is strict, the solution set is
{
x1 -2 < x <
%}
=
-2 and x
=
%.
Since the inequality
or, using interval notation,
( %). -2,
31 6
CHAPTER 4
Linear and Quadratic Functions
Method 2 If f(x) = 2x2 and g(x) = x + 10, the inequality that we wish to solve is f(x) < g(x) . We graph the functions f(x) 2X2 and g(x) = x + 10. See Figure 35. The graphs intersect where f(x) = g ( x ) . Then =
Figure
35 g(x)
=
2X2
x+ 10
=
f(x)
X + 10
2x2 - X - 10 = 0 (2x - 5 ) (x + 2) = 0 2x - 5 = 0 or x + 2 5 x = - or x 2 �
Figure
36 I
-4
I
( -2
I
0
g(x)
Factor. =
0
=
-2
The graphs intersect at the points ( -2, 8) and 2
=
Apply the Zero-Product Property.
(%' ;) 2
. The inequality that we
wish to solve is f(x) < g(x), so we seek where the graph of f is below that of g. This happens between the points of intersection. Since the inequality is strict, the
4 x
solution set is
I) I I 2 4
{
X I -2
%}
< x <
or, using interval notation,
( ,%) -
2
.
See Figure 36 for the graph of the solution set. • �
Now Work P R O B L E M S 5 AN D 1 3
4.5 Assess You r Understa n d i n g
-
'Are You Prepare d ?' Answers are given a t the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Solve the inequality -3x
2 < 7 (p. 1 31 )
2. Write ( -2, 7] using inequality notation. (pp. 1 25-1 26)
Skill Building
In Problems 3-6, use the figure to solve each inequality.
Y = f(x)
3.
4.
Y
6.
(2 , 0) 3
(a) f(x) > 0 (b) f(x) :s 0
x
In Problems 7-22, solve each inequality. 7. x2 - 3x - 1 0 < 0 8. x2 11. x2 - 9 15.
<
0
2X2 < 5x
+
(a) g(x) 2: f(x) (b) f(x) > g(x)
(a) g(x) < 0 (b) g(x) 2: 0 +
3x - 10
>
0
12. x2 - 1 < 0
3
19. 4x2 + 9 < 6x
. 9. x2 - 4x > 0 . 13.
x2 +
X
-
> 12
16. 6x2 < 6 + 5x
17. x(x
20. 25x2 + 16 < 40x
21. 6(x2 - 1 ) > 5x
23. What is the domain of the function lex)
=
Vx2 - 167
7) > 8
(a) f(x) < g(x) (b) f(x) 2: g(x) 10. x2 + 8x > 0 14. x2
+
7x < -12
18. x(x + 1 ) > 20 22. 2 (2x2 - 3x) > -9
24. What is the domain of the function f(x)
=
Vx - 3x2 7
SECTION 4.5
In Problems 25-32, use the given functions f and g. (a) Solve f(x) = O. (b) Solve g(x) = O. (e) Solve g(x) � o. (f) Solve f(x) > g(x). 25. f( x ) = x2 - 1 26. f(x) = - x2 + 3 g(x) = 3x + 3 g(x) = -3x + 3 29. f(x) = x2 - 4 30. f ( x) = x2 - 2x + 1 g(x) = -x2 + 4 g(x) = -x2 + 1
I nequalities Involving Quadratic Functions
(c) Solve f(x) = g(x). (g) Solve f(x) � l . 27. f(x) = -x2 + 1 g(x) = 4x + 1 31. f(x) = x2 X - 2 2 g(x) = x + X 2 -
-
(d) Solve f(x)
>
31 7
O.
28. f(x) = -x2 + 4
g(x) = -x - 2 32. f( x ) = -x2 - X + 1 g(x) = -x2 + X + 6
Applications and Extensions
A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is set) = 80t - 16t2
33. Physics
I 1
I j
where x= y = v = g=
96 ft
s = 80t - 1 6t 2
(a) At what prices p is revenue zero? (b) For what range of prices will revenue exceed $1 ,200,000? 37. Artillery A projectile fired from the point (0, 0) at an angle to the positive x-axis has a trajectory given by
(a) At what time t will the ball strike the ground? (b) For what time t is the ball more than 96 feet above the ground? 34. Physics A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is set) = 96t - 1 6t2 . (a) At what time t will the ball strike the ground? (b) For what time t is the ball more than 128 feet above the ground? Revenue Suppose that the manufacturer of a gas clothes 35. dryer has found that, when the unit price is p dollars, the revenue R (in dollars) is R(p) = 4p2 + 4000p -
(a) At what prices p is revenue zero? (b) For what range of prices will revenue exceed $800,000? 36. Revenue The John Deere company has found that the revenue from sales of heavy-duty tractors is a function of the unit price p, in dollars, that it charges. If the revenue R, in dollars, is 1 R(p) = - 2P? + 1900p
horizontal distance in meters height in meters initial muzzle velocity in meters per second (m/sec) acceleration due to gravity = 9.81 meters per second squared (m/sec2) c > 0 is a constant determined by the angle of elevation. A howitzer fires an artillery round with a muzzle velocity of 897 m/sec. (a) If the round must clear a hill 200 meters high at a dis tance of 2000 meters in front of the howitzer, what c val ues are permitted in the trajectory equation? (b) If the goal in part (a) is to hit a target on the ground 75 kilometers away, is it possible to do so? If so, for what values of c? If not, what is the maximum distance the round will travel? Source: www.answers.com 38. Runaway Car Using Hooke's Law, we can show that the work done in compressing a spring a distance of x feet from its 1 at-rest position is W = 2 kx2, where k is a stiffness constant depending on the spring. It can also be shown that the work done by a body in motion before it comes to rest is given by w W = -v2, where w = weight of the object (lb), g = acceler2g ation due to gravity (32.2 ftlsec2), and v = object's velocity (in ftlsec). A parking garage has a spring shock absorber at the end of a ramp to stop runaway cars. The spring has a stiffness constant k = 9450 Ib/ft and must be able to stop a 4000-lb car traveling at 25 mph. What is the least compression required of the spring? Express your answer using feet to the nearest tenth. [Hint: Solve W > Hi, x � 0]. Source: www.sciforums.com -
Discussion and Writing
39. Show that the inequality (x - 4)2
solution. 40. Show that the inequality (x that is not a solution.
-
2)2
>
�
0 has exactly one
0 has one real number
'Are You Prepared?' Answers
1. {x I x
>
-3 } or ( -3,
00
)
2. - 2
<
x
�
7
41. Explain why the inequality x2 + x + 1
>
bers as the solution set. 42. Explain why the inequality x2 - x + 1 set as solution set.
0 has all real num
<
0 has the empty
318
CHAPTER 4
Linear and Quadratic Functions
CHAPTER R EV I EW Things to Know
Linear function (p. 278)
f ( x)
=
Average rate of change = In The graph is a line with slope m and y-intercept b.
mx + b
Quadratic function (pp. 293-301)
f ( x)
=
ax2 + bx + c, a
i=
The graph is a parabola that opens up if a
0
. AXIs 0 f symmetry: x
=
>
0 and opens down if a
<
O.
b 2a
- -
y-intercept f (O) = c x-intercept(s): If any, found by finding the real solutions of the equation ax2 + bx + c = 0 Objectives ------� Section
You should be a ble to . . .
Review Exercises
Graph linear functions (p. 278) Use average rate of change to identify linear functions (p. 278) Determine whether a linear function is increasing, decreasing, or constant (p. 281 ) Work with applications of linear functions (p. 282)
1 (a)-6(a), 1 (b)-6(b) 7, 8 1 (c)-6( c) 35, 36
Draw and interpret scatter diagrams (p. 287) Distinguish between linear and nonlinear relations (p. 288) Use a graphing utility to find the line of best fit (p. 289)
45(a), 46(a) 45(b), 46(a) 45(c)
Graph a quadratic function using transformations (p. 295) Identify the vertex and axis of symmetry of a quadratic function (p. 297) Graph a quadratic function using its vertex, axis, and intercepts (p. 297) Find the maximum or minimum value of a quadratic function (p. 301)
9-14 15-24 15-24 25-30, 37--42
4.4
Solve applied problems involving quadratic functions (p. 305) Use a graphing utility to find the quadratic function of best fit (p. 309)
37-44 46
4.5
Solve inequalities involving a quadratic function (p. 314)
31-34
4.1 2
3
4
4.2
�
2 3
4.3 2
3
4
Review Exercises
In Problems 1-6: (a) Determine the slope and y-intercept of each linear fimction. (b) Graph each function. Label the intercepts. (c) Determine whether the function is increasing, decreasing, or constant. 1. f (x ) = 2x - 5 2. g (x ) = -4x + 7
1 4. F(x) = - 3 x + 1
5.
G (x )
=
3. h ( x)
4
6.
H(x)
= =
4 x S
-
6
-3
In Problems 7 and 8, determine whether the function is linear or nonlineal: If the function is lineal; state its slope. 7.
x
y
=
fIx)
8.
x
y
=
g (x)
-1
-2
-1
-3
0
3
0
4
2
13
2
6
3
18
3
8
7
31 9
Chapter Review
In Problems 9.
12.
9-14, graph each quadratic function using transformations (shifting , compressing , stretching, and/or r eflecting ). f(x) = (x - 2 f + 2 10. f(x) = (x + 1 ) 2 - 4 11. f(x) - ( x - 4) 2 =
f(x)
=
(x - I f - 3
13.
f(x)
=
2(x + 1) 2 + 4
14.
f(x) = -3 (x + 2 f + 1
In Problems 15-24, (a ) graph each quadratic function by determining whether its graph opens u p or down and by finding its vertex , axis of symmetry, y-intercept , and x -intercepts, if any. (b ) Determine the do main and the range of the function. (c ) Deter mine where the func tion is increasing and where it is decreasin g. 15.
f ( x)
19.
f(x)
23.
f(x)
=
=
=
1
(x - 2 f + 2
16.
f ( x)
(x + 1 f - 4
17.
f(x) = "4x2 - 1 6
-4x2 + 4x
20.
f(x) = 9x2 - 6x + 3
21.
f(x) =
24.
f ( x) = - 2x2 -
=
3x2 + 4x - 1
� x2
3x + 1
+
X
18.
1 2 f(x) - - "2 x +2
22.
f ( x) = -x2 +
_
X
1 +2
+ 4
In Problems 25-30, determine whether the given quadratic function has a maximum value or a minimu m value, and then find the value. 25. f(x) = 3x2 - 6x + 4 26. f(x) = 2x2 + 8x + 5 27. f(x) = -x2 + 8x - 4 28.
f(x)
=
-x2 - lOx - 3
29.
f(x) = -3x2 + 12x + 4
In Problems 31-34, solve each quadratic ine quality. 31. x2 + 6x - 16 < 0 32. 3x2 - 2x - 1
2:
0
Marissa must decide be tween one of two companies as her long-distance phone provider. Company A charges a monthly fee of $7.00 plus $0.06 per minute, while Company B does not have a monthly fee, but charges $0.08 per m inute. (a) Find a linear function that relates cost, C, to total min utes on the phone, x, for each company. (b) Determine the number of minutes x for which the bi.1I from Company A wil l equal the bill from Company B. (c) Over what interval of minutes x will the bill from Com pany B be less than the bill from Company A? Sales Commissions Bill was just offered a sales position for a computer company. His salary would be $15 ,000 per year plus 1 % of his total annual sales. (a) Find a linear function that relates Bill's annual salary, S, to his total annual sales, x. (b) In 2005, Bill had total annual sales of $1 ,000 ,000. What was Bill's salary? (c) What would Bill have to sell to earn $100 ,000? (d) Determine the sales required of Bill for his salary to exceed $1 50 ,000. Demand Equation The price p (in dollars) and the quantity x sold of a certain product obey the demand equation
33.
37.
p =-
38.
39.
1
10
x + 150
0
:S
x
:S
2:
14x
+
5
34.
4x2
<
13x - 3
A special window in the shape of a rectangle with semicircles at each end is to be constructed so that the outside dimensions are 1 00 feet in length. See the illustration. Find the dimensions of the rectangle that maximizes its area.
40. A rchitecture
Callaway Golf Company has determined that the marginal cost C of manufacturing x Big Bertha golf clubs may be expressed by the quadratic function
41. Minimizing Marginal Cost
1 500
(a) Express the revenue R as a function of x. (b) What is the revenue if 1 00 units are sold? (c) What quantity x maximizes revenue? What is the max imum reven ue? (d) What price should the company charge to maximize revenue? Landscaping A landscape engineer has 200 feet of border to enclose a rectangular pond. What dimensions will result in the largest pond? Enclosing the Most Area with a Fence A farmer with 10,000 meters of fencing wants to enclose a rectangular field and then divide it into two plots with a fence parallel to one of the
3x2
f(x) = _2x2 + 4
sides. See the figure. What is the largest area that can be enclosed?
35. Comparing Phone Companies
36.
30.
C(x) = 4. 9x2 - 617.4x + 1 9,600
42.
(a) How many clubs should be manufactured to minimize the marginal cost? (b) At this level of production, what is the marginal cost? Violent Crimes The function
V(t) = 5.0 P - 87.3t + 1761.1 models the number V ( i n thousands) of violent crimes com mitted in the United States t years after 1 995 , based on data obtained from the Federal Bureau of Investigation. So t 0 represents 1 995 , t = 1 represents 1 996 , and so on. =
320
; J 43.
CHAPTER 4
Linear and Quadratic Functions
(a) Determine the year in which the fewest violent crimes were committed. (b) Approximately how many violent crimes were commit ted during this year? (c) Using a graphing utility, graph V = Vet). Were the num ber of violent crimes increasing or decreasing during the years 1995 to 2003? A rectangle has one vertex on the line y = 10 - x, x > 0, another at the origin, one on the positive x-axis, and one on the positive y-axis. Express the area A of the rectangle as a function of x. Find the largest area A that can be enclosed by the rectangle.
[ij
(a) Draw a scatter diagram of the data treating length of the right humerus as the independent variable. (b) Based on the scatter diagram, do you think that there is a linear relation between the length of the right humerus and the length of the right tibia? (c) Use a graphing utility to find the line of best fit relat ing length of the right humerus and length of the right tibia. (d) Predict the length of the right tibia on a rat whose right humerus is 26.5 millimeters (mm). A small manufacturing firm collected the fol lowing data on advertising expenditures A (in thousands of dollars) and total revenue R (in thousands of dollars).
46. Ad"ertising
A horizontal bridge is in the shape of a parabolic arch. Given the information shown in the fig ure, what is the height h of the arch 2 feet from shore?
44. Parabolic Arch Bridge
Research performed at NASA, led by Dr. Emily R. Morey-Holton, measured the lengths of the right humerus and right tibia in 1 1 rats that were sent to space on Spacelab Life Sciences 2. The following data were collected.
45. Bone Length
Right Humerus (mm).x
(mm),y
24.80
36.05
24.59
35.57
24.59
35.57 34.58
23.81
34.20
24.87
34.73
25.90
37.38
26.11
37.96
26.63
37.46
26.31
37.75
26.84
38.50
Total Revenue
20
$6101
22
$6222
25
$6350
25
$6378
27
$6453
28
$6423
29
$6360
31
$6231
(a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) The quadratic function of best fit to these data is R(A) = -7.76A2 + 411.88A + 942.72
Right Tibia
24.29
Advertising
�
Use this function to determine the optimal level of advertising. (c) Use the function to predict the total revenue when the optimal level of advertising is spent. (d) Use a graphing utility to verify that the function given in part (b) is the quadratic function of best fit. (e) Use a graphing utility to draw a scatter diagram of the data and then graph the quadratic function of best fit on the scatter diagram.
Source: NASA Life Sciences Data Archive
CHAPTER TEST 1. For the linear function f(x) = -4x + 3 , (a) Find t h e slope a n d y-intercept. (b) Determine whether f is increasing, decreasing, or constant. (c) Graphf.
[n Problems 2 and 3, fin d the intercepts ofeach quadrat ic funct ion. 2. f(x) = 3x2 - 2x - 8 3. G(x) _2x2 + 4x + 1 =
Cumulative Review
4.
Given that f(x) = x2 + 3x andg(x)
5.
Graph f(x) = (x - 3 ) 2 - 2 using transformations.
6.
=
5x
321
3, solve f(x) = g ( x ) . Graph each function and label the points of i ntersection.
+
For the quadratic function f(x) = 3 x 2 - 12x + 4, (a) Determine whether the graph opens up or down. (b) Determine the vertex. (c) Determine the axis of symmetry. (d) Determine the intercepts. (e) Use the information from parts (a)-(d) to graphj.
7.
Determine whetherf (x) = - 2x 2 + 1 2x + 3 has a maximum or minimum value. Then find the maximum or minimum value.
8.
Solve x 2
-
lOx + 24
;:0::
O.
The weekly rental cost of a 20-foot recreational vehicle is $129.50 plus $0.1 5 per mile. (a) Find a linear function that expresses the cost C as a function of miles driven m. (b) What is the rental cost if 860 miles are driven? (c) How many miles were driven if the rental cost is $213.80?
9 . R V Rental
10. The variable interest rate on a student loan changes each July 1 based on the bank prime loan rate. For the years 1 992 to 2004, this rate can be approximated by the model
r ex) = -0. 1 1 5x2 + 1. 183x
+
5 .623
where x is the number of years since 1992 and r is the interest rate as a percent. Source: u.s. Federal Reserve (a) During which year was the interest rate the highest? Determine the highest rate during this time period. (b) Use the model to estimate the rate in 2010. Does this value seem reasonable?
CUMULATIVE REVIEW 1.
Find the distance between the points P ( - 1 , 3 ) and Q ( 4,-2). Find the midpoint of the line segment P to Q.
10. Is the following graph the graph of a function?
=
=
2. Which of the following points are on the graph of y x3 - 3x + I? (a) ( -2, - 1 ) (b) (2,3) (c) (3, 1 ) =
3.
Solve the inequality 5x + 3
4.
Find the equation of the line containing the points ( -1, 4) and (2, -2). Express your answer in slope-intercept form and graph the line.
5.
Find the equation of the line perpendicular to the line y = 2x + 1 and containing the point (3, 5 ) . Express your answer in slope-intercept form and graph the line.
6.
;:0::
0 and graph the solution set.
Graph the equation x2 + i - 4x + 8y - 5
=
Does the following relation represent a function? {( -3, 8),(1, 3 ) , (2, 5 ) , (3, 8)}.
8.
For the functionf defined by f(x) (a) f(2 ) (b) f(x) + f(2) (c) f( -x) (d) -f (x) (e) f (x + 2 ) f(x + h ) - f(x) (f) h h *" 0
9.
Find the domain of h(z) =
3z - 1 6z _ 7 ·
x 11. Consider the function f (x) = _ _ .
x + 4
(a) Is the point
O.
7.
=
x
( �) 1,
on the graph off?
x -2, what is f(x)? What point is on the graph off? (c) If f(x) = 2, what is x? What point is on the graph off?
(b)
x2 - 4x + 1 , find:
If
=
x2
even, odd, or neither? 12. Is the function f(x ) = _ 2x + 1 __
IiJ
13. Approximate the local maxima and local minima of f(x) = x3 - 5x + I on ( - 4,4) . Determine where the func tion is increasing and where it is decreasing.
14. Iff(x) = 3x + 5 andg(x) ( a ) Solve f(x) = g ( x ) . (b) Solve f (x) > g(x ).
=
2x + 1 ,
322 15.
CHAPTER 4
Linear and Quadratic Functions
For the graph of the functionf,
y 4
(4,3 )
5
x
(a) Find the domain and the range of f. (b) Find the intercepts. (c) Is the graph off symmetric with respect to the x-axis, the y-axis, or the origin? (d) Findf(2). (e) For what value(s) of x isf(x) = 3? (f) Solvef(x) < O. (g) Graph y = f (x) + 2 (h) Graph y = fe-x). (i) Graph y 2f (x). U) Isf even, odd, or neither? (k) Find the interval(s) on whichf is increasing. =
-4
CHAPTER PROJECTS 1.
Find the weekly closing price of your favorite stock and the S&P 500 for 20 weeks. One good source is on the In ternet at Ifinance.yahoo. com.
2. Compute the rate of return by computing the weekly per centage change in the closing price of your stock and the weekly percentage change in the S&P 500 using the fol lowing formula: P2 - PI Weekly % change PI =
where PI is last week's price and P2 is this week's price. 3. Using a graphing utility, find the line of best fit, treating the weekly rate of return of the S&P 500 as the indepen dent variable and the weekly rate of return of your stock as the dependent variable. 4. What is the beta of your stock? I.
The beta, {3, of a stock represents the rel ative risk of a stock compared with a market basket of stocks, such as Standard and Poor's 500 Index of stocks. Beta is com puted by finding the slope of the line of best fit between the rate of return of the stock and rate of return of the S&P 500. The rates of return are computed on a weekly basis.
Analyzing a Stock
5.
Compare your result with that of the Value Line Invest ment Survey found in your library. What might account for any differences?
The following projects are available on the Instructor's Resource Center (IRC): II.
III.
Cannons A battery commander uses the weight of a missile, its initial velocity, and the position of its gun to determine where the missile will travel. First and Second Differences
Finite differences provide a numerical method that is used to estimate the graph of an unknown
function. IV.
CBL Experiment
Computer simulation is used to study the physical properties of a bouncing ball.
Polynomial and Rational Functions We Don't Want to Be Number One Again - in Tree Pollen LEXINGTON, Ky. (March 29, 2006)-Were your allergies especially bothersome last spring? It's no wonder. In the spring of 2005 , Lexington, Ky., was the tree pollen capital of the U.S. No allergy sufferer is rooting for the Bluegrass to earn that title again in 2006. Either way, allergy sufferers should be prepared. The spring allergy season is here. If you can't stop sneezing, blowing your nose, and rubbing your itchy eyes, you may have allergic rhinitis, as do 40 million other Amer icans. The pollens of trees such as maple, oak and walnut trigger much of the allergic rhinitis, commonly referred to as hay fever, in the early spring. Pollinating grasses and weeds are responsible for summer hay fever. Perennial allergic rhinitis can be triggered any time of year by indoor allergens, such as animal dander, mold or dust mites. "Spring can be the start of misery for many allergy sufferers be cause they have multiple pollen allergies," said Dr. Beth Miller, director of the UK Asthma, Allergy and Sinus Clinics. Source: Adapted from Melanie Jackson, "We Don't Want to Be Number One Again-In Tree Pollen," University of Kentucky News, March 29, 2006,
hllP://news. uky. edu.
-See the Chapter Project-
A Look Back In Chapter 3, we began our discussion of functions.We defined domain and range and independent and dependent variables; we found the value of a function and graphed functions. We continued our study of functions by listing the properties that a function might have, like being even or odd, and we created a library of func tions, naming key f unctions and listing their properties, including their graphs. In Chapter 4, we discussed linear functions and quadratiC functions, which be long to the class of polynomial functions.
A Look Ahead In this chapter, we look at two general classes of functions, polynomial functions and rational functions, and examine their properties. Polynomial functions are ar guably the simplest expressions in a lgebra. For this reason, they are often used to approximate other, more complicated functions. Rational functions are ratios of polynomial functions.
Outline 5.1 Polynomial Functions and Models 5.2 Properties of Rational Functions 5.3 The Graph of a Rational Function 5.4 Polynomial and Rational Inequalities 5.5 The Real Zeros of a Polynomial Function 5.6 Complex Zeros; Fundamental Theorem of Algebra Chapter Review Chapter Test Cumulative Review Chapter Projects
323
324
CHAPTER 5
Polynomial and Rational Functions
-
5.1 Polynomial Functions and Models PREPARING FOR THIS SECTION •
•
•
Before getting started, review the following:
Polynomials (Chapter R, Section R.4, pp. 39-47) Using a Graphing Utility to Approximate Local Maxima and Local Minima (Section 3.3, pp. 235-236) Intercepts of a Function (Section 3.2, pp. 224-225) Now Work the 'Are You
•
•
Graphing Techniques: Transformations (Section 3.5, pp. 252-260) Intercepts (Section 2.2, pp. 165-166)
Prepared?, problems on page 339.
OBJECTIVES 1 Identify Polynomial Functions and Their Degree (p. 324)
1
2
Graph Polynomial Functions Using Transformations (p. 328)
3
Identify the Real Zeros of a Polynomial Function and Their Multiplicity (p.329)
4
Analyze the Graph of a Polynomial Function (p. 336)
Identify Polynomial Functions and Their Degree
In Chapter 4 we studied the linear function f(x) f(x)
=
alx
+
=
mx + b, which we can write as
ao
and we studied the quadratic function f(x) = ax2 + bx + c, a write as f(x) = a2x2 + alx + ao , a 2 i= 0
i=
0, which we can
Each of these functions is an example of a polynomial function.
DEFINITION
A polynomial function is a function of the form (1)
where an, an-l , .. . , aI , ao are real numbers and n is a nonnegative integer. The domain of a polynomial function is the set of all real numbers.
-.J
A polynomial function is a function whose rule is given by a polynomial in one variable. The degree of a polynomial function is the largest power of x that appears. The zero polynomial function f(x) = 0 is not assigned a degree. Polynomial functions are among the simplest expressions in algebra. They are easy to evaluate: only addition and repeated multiplication are required. Because of this, they are often used to approximate other, more complicated functions. In this section, we investigate properties of this important class of functions. E XA M P L E 1
Identifying Polynom ial Functions
Determine which of the following are polynomial functions. For those that are, state the degree; for those that are not, tell why not. x2 - 2 ( a ) f(x) = 2 - 3x4 ( b ) g(x) = \IX ( c) h ex) = -3-
(el) F (x )
Solution
=
0
( e ) G(x)
=
8
( f) H(x )
( a ) f is a polynomial function of degree 4 . ( b ) g is not a polynomial function because g (x )
1
=
\IX
=
X
I
= Xl,
raised to the .2 power, which is not a nonnegative integer.
-
1
-2x3 ( x
-
1f
so the variable x is
SECTION 5.1
( c)
325
Polynomial Functions and Models
h
is not a polynomial function. It is the ratio of two polynomials, and the poly nomial in the denominator is of positive degree. (d) F is the zero polynomial function; it is not assigned a degree. (e) G is a nonzero constant function. It is a polynomial function of degree 0 since G(x) = 8 = 8xo.
(f) H(x) = -2x3(x - If = -2x3(x2 2x + 1) = -2x5 + 4X4 - 2x3. So H is a polynomial function of degree 5. Do you see a way to find the degree of H without multiplying out? -
1l!l!lI==__
Now Work
• PRO B l EMS 1 1
A ND 1 5
We have already discussed in detail polynomial functions of degrees 0, 1 , and 2. See Table 1 for a summary of the properties of the graphs of these polynomial functions. Table
1
Form
Degree No degree
{(xl
=
o
{(xl
=
Q o,
{(xl
=
Q ,X
0 Qo
+
of. 0
Qo,
Q,
Graph
Name
7'= 0
2
Zero function
The x-axis
Constant function
Horizontal line with y-intercept
Linear function
Nonvertical, nonhorizontal line with slope Q, and y-intercept Qo
Quadratic function
Parabola: Graph opens up if graph opens down if
Qo
Q2 >
Q2 <
0;
0
One objective of this section is to analyze the graph of a polynomial function. If you take a course in calculus, you will learn that the graph of every polynomial function is both smooth and continuous. By smooth, we mean that the graph contains no sharp corners or cusps; by continuous, we mean that the graph has no gaps or holes and can be drawn without lifting pencil from paper. See Figures l(a) and (b). Figure
1
y
y Cusp
x (b) Cannot be the graph of a
(a) Graph of a polynomial function: smooth, continuous
polynomial fun ction
Power Functions
We begin the analysis of the graph of a polynomial function by discussing power functions, a special kind of polynomial function.
DEFINITION
A power function of degree n is a monomial of the form f(x)
where a is a real number, a
oj:.
=
axil
0, and n > 0 is an integer.
(2)
326 t r r
r
CHAPTER 5
Polynomial and Rational Functions
Examples of power functions are
In Words
single monomial.
f(x) = -5x2
f(x) = 3x
A power function is defined by a
f(x) = -5x4
f(x) = 8x3 degree 3
degree 2
degree 1
degree 4
The graph of a power function of degree 1 , f(x) = ax, is a straight line, with slope a, that passes through the origin. The graph of a power function of degree 2, f(x) = ax2, is a parabola, with vertex at the origin, that opens up if a > 0 and down if a < O. If we know how to graph a power function of the form f(x) = x", a compres sion or stretch and, perhaps, a reflection about the x-axis will enable us to obtain the graph of g(x) = ax". Consequently, we shall concentrate on graphing power func tions of the form f (x) = x". We begin with power functions of even degree of the form f (x) = XII, n 2:: 2 and n even. The domain of f is the set of all real numbers, and the range is the set of nonnegative real numbers. Such a power function is an even function (do you see why?) , so its graph is symmetric with respect to the y-axis. Its graph always contains the origin and the points ( - 1 , 1) and ( 1 , 1). If n = 2, the graph is the familiar parabola y = x2 that opens up, with vertex at the origin. If n 2:: 4, the graph of f(x) = x", n even, will be closer to the x-axis than the parabola y = x2 if - 1 < x < 1 and farther from the x-axis than the parabola y = x2 if x < -1 or if x > 1 . Figure 2(a) illustrates this conclusion. Figure 2(b) shows the graphs of y = x4 and y = x8 for comparison. f(x) = x" n2 4 n even
Figure 2
Y=
x8 Y
4 Y= X
4
2
(-1 ,1)
-3
3
x
-3
(1, 1 ) (0,0)
(a)
3
x
(b)
From Figure 2, we can see tha t as n increases the graph of f (x) = XII, n 2:: 2 and n even, tends to flatten out near the origin and to increase very rapidly when x is far from O. For large n, it may appear that the graph coincides with the x-axis near the origin, but it does not; the graph actually touches the x-axis only at the origin (see Table 2). Also, for large n, it may appear that for x < -1 or for x > 1 the graph is vertical, but it is not; it is only increasing very rapidly in these intervals. If the graphs were enlarged many times, these distinctions would be clear. Table 2
x =
{ (x)
=
{ (x)
=
{ (x)
=
x =
0.1
x8 x20
1 0-8 1 0-20
x40
1 0-40
0.3
x =
0.5
0.0000656
0.0039063
3.487 ' 1 0-'1 1 .2 1 6. 1 0-21
0.000001 9.095' 1 0-13
Seeing the Concept Graph Y, X4, Y2 = xB, and Y3 X'2 usi n g the viewing rectangle -2 :5 X :5 2, -4 :5 Y :5 1 6. Then g raph each again usi n g the viewing rectangle -1 :5 x :5 1 ,0 :5 Y :5 1 . See Figure 3.TRACE along one =
=
of the graphs to confirm that for x close to 0 the graph is above the x-axis a n d that for x is i ncreasi ng.
>
0 the graph
SECTION 5.1
Polynomial Functions and Models
327
Figure 3
'\..11' II'
... III -2
�
/1)
. j.' .., '
�II \,\.
2 -1
-4
�1��
�
I�
..�... . ... .
,.//{
..
_
o
(a)
..
(b)
Properties of Power Functions, ((xl
=
X',
n Is an Even Integer
1. f is an even function, so its graph is symmetric with respect to the y-axis. 2. The domain is the set of all real numbers. The range is the set of non negative real numbers. 3. The graph always contains the points ( - 1 , 1), (0,0), and ( 1 , 1). 4. As the exponent n increases in magnitude, the graph becomes more ver tical when x < - l or x > 1 ; but for x near the origin, the graph tends to flatten out and lie closer to the x-axis. Now we consider power functions of odd degree of the form f(x) = XII, n 2: 3 and n odd. The domain and the range of f are the set of real numbers. Such a power function is an odd function (do you see why?), so its graph is symmetric with respect to the origin. Its graph always contains the origin and the points ( - 1 , - 1 ) and (1, 1). The graph of f(x) = x" when n = 3 has been shown several times and is repeated in Figure 4. If n 2: 5, the graph of f(x) = x", n odd, will be closer to the x-axis than that of y = x3 if - 1 < x < 1 and farther from the x-axis than that of y = x 3 if x < - 1 or if x > 1. Figure 4 also illustrates this conclusion. Figure 5 shows the graph of y = x S and the graph of y = x 9 for further comparison. n
Figure 5
Y= x n? 5 n odd
Figure 4
-3
3 x
-3
3 x
It appears that each graph coincides with the x-axis near the origin, but it does not; each graph actually crosses the x-axis at the origin. Also, it appears that as x increases the graph becomes vertical, but it does not; each graph is increasing very rapidly.
328
CHAPTER 5
Polynomial and Ratio nal Functions
�I Seeing the Concept eJ Graph Yl = x3, Y2 = x7, a n d Y3 = Xll using the viewing rectangle -2 :oS X :oS
2, - 1 6 :oS Y :oS 1 6. Then graph each again using the viewing rectangle -1 :oS x :oS 1, -1 :oS Y :oS 1 . See Figure 6. TRACE along one of the graphs to confirm that the graph is increasing and crosses the x-axis at the origin.
Figure 6
-1
-16
(b)
(a)
To summarize: Properties of Power Functions, ((x)
=
x",
n Is an Odd Integer
1. f is an odd function, so its graph is symmetric with respect to the origin. The domain and the range are the set of all real numbers. 3. The graph always contains the points (-1 , -1), (0, 0), and (1,1). 4. As the exponent n increases in magnitude, the graph becomes more vertical when x < -1 or x > 1; but for x near the origin, the graph tends to flatten out and lie closer to the x-axis.
2.
2
Graph Polynomial Functions Using Transformations
The methods of shifting, compression, stretching, and reflection studied in Section 3.5, when used with the facts just presented, will enable us to graph polyno mial functions that are transformations of power functions. E XA M P L E 2
Graphing Polynomial Functions Using Transformations
Graph: f(x) Solution
=
1 - x5
Figure 7 shows the required stages.
Figure 7
2
-2
x
(-1,-1) -2 Multi ply by -1; ref lect about x-a xis
(a) y = XS
EXAM P L E 3
Add 1 ;
shi ft up
(b)
s Y = -x
1 unit
(c)
y = -x s + 1
Graphing Polynomial Functions Using Transformatio n s
Graph: f(x)
=
1
2 (x - 1)4
= 1 -xs
•
SECTION 5.1
329
Figure 8 shows the required stages.
Solution Figu re
Polynomial Functions and Models
8
Y 2
-2 -2
-2
-2
(a) Y= x4 � == ....
3
x
(1 , 0) 2
Replace x by x - 1; shift right 1 unit
�
(b) y=(x-1)4
Now Work
PRO B
L
-2
�;
Multiply by co mpression by a factor o f
(c) Y= � (X_1)4
9
EMS 2 3 A N D 2
•
Identify the Real Zeros of a Polynomial Function and Their Multiplicity
Figure 9 shows the graph of a polynomial function with four x-intercepts. Notice that at the x-intercepts the graph must either cross the x-axis or touch the x-axis. Consequently, between consecutive x-intercepts the graph is either above the x-axis or below the x-axis. We will make use of this property of the graph of a polynomial function shortly. Figure 9
Y - Above---+ x-axis
x-axis ___
Below x-axis
x-axis
__
x-axis
Below x-axis
If a polynomial function f is factored completely, it is easy to solve the equation f(x) = 0 using the Zero-Product Property and locate the x-intercepts of the graph . For example, if f(x) = (x - 1)2(x + 3) , then the solutions of the equation f(x)
=
(x - l)\x + 3)
=
0
are identified as 1 and - 3. Based on this result, we make the following observations: DEFINITION
If f is a function and r is a real number for which f(r) a real zero of f·
=
0, then r is called
.J
As a consequence of this definition, the following statements are equivalent. 1. r is a real zero of a polynomial function f 2. r is an x-intercept of the graph of f 3. x - r is a factor of f So the real zeros of a polynomial function are the x-intercepts of its graph, and they are found by solving the equation f(x) = O.
330
CHAPTER 5
Polynomial and Rational Functions
F i nding a Polynom ial from Its Zeros
E XA M P L E 4
Solution Figure
10 40
.
.. .---.
l
,/'
"
l 6
- 50
(a) Find a polynomial of degree 3 whose zeros are -3,2, and 5. (b) Use a graphing utility to graph the polynomial found in part (a) to verify your result. (a) If r is a real zero of a polynomial [, then x - r is a factor of f. This means that x - ( -3) = x + 3, x - 2, and x - 5 are factors of f. As a result, any polynomial of the form [(x) = a(x + 3)(x - 2)(x - 5) where a is any nonzero real number, qualifies. The value of a causes a stretch, compression, or reflection, but does not affect the x-intercepts. Do you know why? We choose to graph [ with a = 1. Then [(x) = (x + 3)(x - 2)(x - 5) = x3 - 4x2 - llx + 3 0 Figure 10 shows the graph of f. Notice that the x-intercepts are -3, 2, and 5.
I (b)
•
Seeing the Concept Graph the function found in Example 4 for
a=
2 and
of f? How does the value of a affect the graph of f?
lJ'l!l:=iOlIII ""II: .- Now Work
a=
-1. Does the value of
a affect the
zeros
PRO B L E M 3 7
If the same factor x -
occurs more than once,
r
r
is called a repeated, or
multiple, zero of f. More precisely, we have the following definition.
If (x - r ) 111 is a factor of a polynomial [ and ( x then r is called a zero of multiplicity m off· *
DEFINITION E XA M P L E 5
-
r
) I11+1 is not a factor of [,
.J
Identifyi ng Zeros and Thei r M u lti p l icities
For the polynomial [(x)
=
( �y
5 (x - 2)(x + 3? X -
2 is a zero of multiplicity 1 because the exponent on the factor x - 2 is 1. -3 is a zero of multiplicity 2 because the exponent on the factor x + 3 is 2.
�
is a zero of multiplicity 4 because the exponent on the factor x -
�
Now Work
PRO B L EM 4 5 (a)
�
is 4. •
Suppose that it is possible to factor completely a polynomial function and, as a result, locate all the x-intercepts of its graph (the real zeros of the function). These x-intercepts then divide the x-axis into open intervals and, on each such interval, the graph of the polynomial will be either above or below the x-axis. Let's look at an example. E XA M P L E 6
Graphing a Polynomial Using Its x-Intercepts
For the polynomial: [(x)
=
x2(x - 2)
(a) Find the x- and y-intercepts of the graph of f. (b) Use the x-intercepts to find the intervals on which the graph of [ is above the x-axis and the intervals on which the graph of [ is below the x-axis. ':' Some books use the terms lIIultiple root and root of lIIultiplicity
m.
SECTION 5.1
Polynomial Functions and Models
331
(c) Locate other points on the graph and connect all the points plotted with a smooth, continuous curve. Solution
(a) The y-intercept is f (O)
=
02(0 - 2) = O. The x-intercepts satisfy the equation f(x) =x2(x - 2) = 0
from which we find x2=0 or x - 2=0 x =O or x=2 The x-intercepts are 0 and 2. (b) The two x-intercepts divide the x-axis into three intervals: (0, 2)
( - 00, 0)
( 2,
(0
)
Since the graph of f crosses or touches the x-axis only at x = 0 and x = 2, it follows that the graph of f is either above the x-axis [I(x) > 0] or below the x-axis [I(x) < 0] on each of these three intervals. To see where the graph lies, we only need to pick a number in each interval, evaluate f there, and see whether the value is positive (above the x-axis) or negative (below the x-axis). See Table 3. (c) In constructing Table 3, we obtained three additional points on the graph: ( - 1 , -3), (1, -1), and (3, 9). Figure 11 illustrates these points, the intercepts, and a smooth, continuous curve (the graph of f) connecting them. Table 3
Figure
.2
.0 Interval Number chosen
-1
Value of f
f(-1)
Location of graph
Below x-axis
Point on graph
=
-3
(-1,-3)
f(1)
3
f(3)
= -1
Below x-axis
(1, -1)
(3,9)
(2,00)
(0,2)
(-00,0)
• x
11
=9
Above x-axis
(3,9)
-2
4
x
•
Look again at Table 3. Since the graph of f(x ) = x2 (x - 2) is below the x-axis on both sides of 0, the graph of f touches the x-axis at x = 0, a zero of multiplicity 2. Since the graph of f is below the x-axis for x < 2 and above the x-axis for x > 2, the graph of f crosses the x-axis at x = 2, a zero of multip licity 1. This suggests the following results: If r Is a Zero of Even Multiplicity
The sign of f (x) does not change from one side to the other side of r. If r Is a Zero of Odd Multiplicity
The sign of f(x) changes from one side to the other side of r. 'I!=>-
Now Work
PRO B L EM 4 5 (b)
The graph of f touches the x-axis at r.
The graph of f crosses the x-axis at r.
332
C H A PTER 5
Polynomial and Rational Functions
Behavior Near a Zero
We have just learned how the multiplicity of a zero can be used to determine whether the graph of a function touches or crosses the x-axis at the zero. However, we can learn more about the behavior of the graph near its zeros than just whether the graph crosses or touches the x-axis. Consider the function f(x) = x2(x - 2) whose graph is drawn in Figure 1 1 . The zeros of f are 0 and 2. Table 4 shows the values of f(x) = x2(x - 2) and y = -2x2 for x near O. Figure 12 shows the points ( -0.1 , -0.021) , ( -0.05, -0.0051), and so on, that are on the graph of f(x) = x2(x - 2) along with the graph of y = _2x2 on the same Cartesian plane. From the table and graph, we can see that the points on the graph of f(x) = x2(x - 2) and the points on the graph of y = -2x2 are indistinguishable near x = O. So y = _2x2 describes the behavior of the graph of f(x) = x2(x - 2) near x = O. Table 4
Figure
fIx) = �(x - 2)
x
y=-�
-0.1
-0.021
-0.02
-0.05
-0.005 1 25
-0.005
-0.03
-0.00 1 827
-0.00 1 8
-0.01
-0.000201
-0.0002
0
0
12 Y 0.005 x
0
0.01
-0.0001 99
-0.0002
0.03
-0.00 1 773
-0.001 8
0.05
-0.004875
-0.005
0.1
-0.01 9
-0.02
But how did we know that the function f(x) = x\x - 2) behaves like -2x2 when x is close to O? In other words, where did y = -2x2 come from? Because the zero, 0, comes from the factor x2, we evaluate all factors in the function f at 0 with the exception of x2. y =
f(x)
=
�
=
x2(x - 2) x2(0 - 2) _2x2
The factor the factor
J'- g ives rise to the J'- a nd let x = 0 in
zero, so we kee p the rema i ni ng
factors to fi nd the behavio r near O.
This tells us that the graph of f(x) = x2(x - 2) will behave like the graph of y = _2x2 near x = O. Now let's discuss the behavior of f(x) = x2(x - 2) near x = 2, the other zero. Because the zero, 2, comes from the factor x - 2, we evaluate all factors of the func tion f at 2 with the exception of x - 2. f(x)
=
�
=
x2(x - 2) 22(x - 2) 4(x - 2)
T h e factor
x
-
2 gives rise t o the zero, so we
keep the factor x
-
2 a nd let x = 2 in the
rema i n ing factors to find the behavior near 2.
So the graph of f(x) = x2(x - 2) will behave like the graph of y = 4(x - 2) near x = 2. Table 5 verifies that f(x) = x2(x - 2) and y = 4(x - 2 ) have similar values for x near 2. Figure 13 shows the points ( 1 .9, -0.361 ) , ( 1 99 -0.0396 ) , and so on, that are on the graph of f(x) = x2(x - 2) along with the graph of y = 4( x - 2) on the same Cartesian plane. We can see that the points on the graph of f(x) = x2(x - 2) and the points on the graph of y = 4(x - 2) are indistin guishable near x = 2. So Y = 4(x - 2 ) , a line with slope 4, describes the behavior of the graph of f(x) = x2(x - 2) near x = 2. .
,
SECTION 5.1
Table 5
f(x)
x
1.9
Figure
14
=
xl(x - 2)
y
=
4(x - 2)
-0.361
-0.4
1.99
-0.0396
-0.04
1.999
-0.003996
-0.004
2
0
0
2.001
0.004004
0.004
2.01
0.0404
0.04
2.1
0.441
0.4
Polynomial Functions and Models
Figure
13
333
y •
x
2.2
Figure 14 illustrates how we would use this information to begin to graph f(x) = x2(x - 2). (2,0)
1
Behaves 1 ke -2 y=-2x near x= 0 -4
3 x
Behaves l i ke Y= 4(x- 2) near x= 2
By determining the multiplicity of a real zero, we determine whether the graph crosses or touches the x-axis at the zero. By determining the behavior of the graph near the real zero, we determine how the graph touches or crosses the x-axis. =:>-
Now Work
PRO B l EM
45( c )
Turning Points
Look again at Figure 11. We cannot be sure just how low the graph actually goes between x = 0 and x = 2. But we do know that somewhere in the interval (0, 2) the graph of f must change direction (from decreasing to increasing). The points at which a graph changes direction are called turning points. In calculus, such points are called local maxima or local minima, and techniques for locating them are given. So we shall not ask for the location of turning points in our graphs. Instead, we will use the following result from calculus, which tells us the maximum number of turn ing points that the graph of a polynomial function can have. THEOREM
Turning Points
If f is a polynomial function of degree n, then f has at most n - 1 turning points. If the graph of a polynomial function f has n - 1 turning points, the degree of f is at least n.
.J
For example, the graph of f(x) = x\x - 2) shown in Figure 11 is the graph of a polynomial of degree 3 and has 3 - 1 = 2 turning points: one at (0, 0) and the other somewhere between x = 0 and x = 2.
l�.·1 Exploration lM A g raphing utility can be used to locate the turning points of a g raph. Graph MINIMUM to find the location of the turning point for 0
Figure 15
=- Now Work
PRO B L EM
4 5(d)
<
x
<
2. See Figure 1 5.
Y1
=
xl(x
-
2). Use
334
CHAPTER 5
Polynomial and Rational Functions
Identifying the Graph of a Polynomial F u n ction
E XA M P L E 7
Which of the graphs in Figure 16 could be the graph of a polynomial function? For those that could, list the real zeros and state the least degree the polynomial can have. For those that could not, say why not. Figure
16
y 3
y
x -2
(a) Sol ution
(d)
(c)
(b)
(a) The graph in Figure 16(a) cannot be the graph of a polynomial function because of the gap that occurs at x = -1. Remember, the graph of a polynomial function is continuous-no gaps or holes. (b) The graph in Figure 16(b) could be the graph of a polynomial function. It has three real zeros, at -2, at 1, and at 2. Since the graph has two turning points, the degree of the polynomial function must be at least 3. (c) The graph in Figure 16(c) cannot be the graph of a polynomial function because of the cusp at x = 1. Remember, the graph of a polynomial function is smooth. (d) The graph in Figure 16( d) could be the graph of a polynomial function. It has two real zeros, at -2 and at 1. Since the graph has three turning points, the degree of the polynomial function is at least 4. !'l
-
•
Now Work PRO B L E M 5 7
End Behavior
One last remark about Figure 11. For very large values of x, either positive or neg ative, the graph of f(x) = x2(x - 2) looks like the graph of y = x3. To see why, we write f in the form f (x)
=
x2 (x - 2 )
=
x3 - 2x2
=
( �)
x3 1
-
Now, for large values of x, either positive or negative, the term � is close to 0, so x for large values of x f (x)
=
x3 - 2x2
=
( �)
x3 1
-
�
x3
The behavior of the graph of a function for large values of x, either positive or negative, is referred to as its end behavior. THEOREM
End Behavior
For large values of x, either positive or negative, the graph of the polynomial
resembles the graph of the power function y L-____________________
= __ a" x_ _
n
�I�
_ ____________________
SECTION 5.1
Polynomial Functions and Models
335
Look back at Figures 2 and 4. Based on the above theorem and the previous dis cussion on power functions, the end behavior of a polynomial can only be of four types. See Figure 17. Figure
17
E n d Behavior of ((x) = onX'
\
y
+
0n_ 1 X'- 1
+ ... +
y
I
0lX
X
(a)
y
I
x
I n ? 2 even; an > 0
+ 00
\
y
X
X
I
\
\ (d)
(e)
(b)
n ? 2 even; an < O
n ? 3 odd; an > 0
n ? 3 odd; an < 0
For example, if I(x) = _2x4 + x3 + 4x2 - 7x + 1, the graph of 1 will resem ble the graph of the power function y = _2X4 for large I x i . The graph of 1 will look like Figure 17(b) for large I x l . �==-- Now Work P RO B L E M 4 5 ( e )
Identifying the Graph of a Polynomial F u nction
E XA M P L E 8
Which of the graphs in Figure 18 could be the graph of I(x) Figure
=
x4 + Sx3 + Sx2 - Sx
18
-
6? y
X
(a)
Solution
(e)
(b)
(d)
The y -intercept of lis 1(0) = -6. We can eliminate the graph in Figure 18(a), whose y-intercept is positive. We don ' t have any methods for finding the x-intercepts of I, so we move on to investigate the turning points of each graph. Since 1 is of degree 4, the graph of 1 has at most 3 turning points. We eliminate the graph in Figure 18( c) since that graph has 5 turning pain ts. Now we look at end behavior. For large values of x, the graph of I will behave like the graph of y = x4. This eliminates the graph in Figure 18( d), whose end be havior is like the graph of y = -X4. Only the graph in Figure 18(b) could be (and, in fact, is) the graph of I(x) = x4 + Sx3 + x2 - Sx - 6. I/T.=;;;;�;;ol Now Work PRO B L E M 6
•
1
336
CHAPTER 5
SUMMARY
Polynomial and Rational Functions
G raph of a Polynomial Function fix)
=
Q,y< n
+
Qn _ ,X n
-,
Degree of the polynomial f: n Maximum number of turning points: n - 1 At a zero of even multiplicity: The graph of f touches the x-axis. At a zero of odd multiplicity: The graph of f crosses the x-axis. Between zeros, the graph of f is either above or below the x-axis. End behavior: For large lxi, the graph of [ behaves like the graph of y
4 E XA M P LE 9
+ ... +
=
Q,X + Qo Q n
* 0
al1xl1.
Analyze the Graph of a Polynomial Function Analyzing the Graph of a Polynomial F u n ction
For the polynomial f(x) = x3 + x2 - 12x: (a) Find the x- and y-intercepts of the graph of f. (b) Determine whether the graph crosses or touches the x-axis at each x-intercept. (c) End behavior: find the power function that the graph of f resembles for large values of I x l . (d) Determine the maximum number of turning points on the graph of f. (e) Determine the behavior of the graph of [ near each x-intercept. (f) Put all the information together to obtain the graph of .f. Solution
(a) The y-intercept is [(0) = O. To find the x-intercepts, if any, we first factor .f. [( x )
=
=
=
x3 + x2 - 12x x(x2 + X - 12)
x(x + 4) (x - 3)
We find the x-intercepts by solving the equation
(b) ( c) (d) (e) Near -4: f(x) [(x) Near 0: Near 3:
[(x)
[(x) x
=
x
=
=
x(x + 4) (x - 3) = 0 0 or x + 4 = 0 or 0 or x = -4 or
x-3 x
=
=
0
Zero-Product Property
3
Solve.
The x-intercepts, or real zeros of [, are -4, 0, and 3. Since each real zero of f is of multiplicity 1, the graph of f will cross the x-axis at each x-intercept. End behavior: the graph of [ resembles that of the power function y x3 for large values of I x l . The graph o f [ will contain a t most two turning points. The three x-intercepts are -4, 0, and 3. =
=
= =
x(x + 4) (x - 3) � -4(x + 4) ( -4 - 3) 28(x + 4) x(x + 4) (x - 3) ::::: x(O + 4) (0 - 3) = - 12x x(x + 4) (x - 3 ) ::::: 3 (3 + 4) (x - 3) = 21(x - 3) =
A line with slope
28
A line with slope
-12
A line with slope
21
(f) Figure 1 9(a) illustrates the information obtained from parts (a) , (b), (c), and (e). The graph of [is given in Figure 19(b). Notice that we evaluated [at -5, -2, 1, and 4 to help establish the scale on the y-axis.
SECTION 5.1
Figure
337
Polynomial Functions and Mode l s
19 Y
Y
Near x -4 the graph behaves like Y = 28(x + 4), a line with slope 28 =
Near x 0 the graph =
\
x
3
-5
x
Near x = 3 the graph behaves like Y = 21 (x - 3), a line with slope 21
End behavior: 3 resembles Y= x
I (b)
(a)
•
� Exploration I:iI!f G ra p h Yj = >? + x? -
1 2x. Compa re what you see with Figure 1 9(b). Use MAXIMUM/MINIMUM to locate the two turning points.
L.'l'l:=::::o:- Now Work P R O B L E M 6 7 E XA M P L E 1 0
Analyzing the Graph of a Polynom ial F u n ction
Follow the instructions of Example 9 for the following polynomial: f(x) (a) The y-intercept is f(O)
Solution
=
=
x2(x - 4 ) ( x
+
1)
O. The x-intercepts satisfy the equation
f(x)
=
x2(x - 4 ) ( x
+
1) = 0
So x2 0 or x - 4 x x = 0 or =
(b) (c) (d) (e)
=
=
=
0 or x + 1 0 x = 1 4 or =
-
The x-intercepts are - 1 , 0, and 4. The intercept 0 is a zero of multiplicity 2, so the graph of f will touch the x-axis at 0; 4 and - 1 are zeros of multiplicity 1, so the graph of f will cross the x-axis at 4 and - l . End behavior: the graph of [" resembles that of the power function y = X4 for large values of I x l . The graph of f will contain at most three turning points. The three x-intercepts are - 1 , 0, and 4.
Near - I : f(x) = x2( x - 4 ) ( x + 1) Near 0: f(x) x2( x - 4)(x + 1) Near 4: f(x)
=
x2( x - 4 ) ( x + 1 )
� � �
( - I f( - I - 4) (x + 1)
=
-5(x
x2(0 - 4 ) (0 + 1) = -4x2 42(x - 4 ) (4 + 1 ) = 80(x - 4)
+
1)
A l ine with slope - 5 A parabola opening down A l in e with slope
80
(f) Figure 20(a) illustrates the information obtained from parts (a), (b), (c), and (e). 1
The graph of f is given in Figure 20(b) . Notice that we evaluated f at -2' - 2' 2, and 5 to help establish the scale on the y-axis.
338
CHAPTER 5
Polynomial and Rational Functions
Figure 20
y
I
End behavior: 4 Resembles y = x
Near x = 0 the graph behaves like y = -4} a parabola opening down
End behavior: 4 Resembles y = x
\
Near x = 4 the graph behaves like y = 80(x - 4) a line with slope 80 (4, 0)
2
-2
5
3
x
Near x = -1 the graph behaves like y = -5(x + 1 ) a line with slope -5
x
- 40
(b)
(a)
Exploration Graph Yl = x2 (x
- 4)(x + 1 ). Compare
•
what you see with F i g u re 20(b). Use MAXIMUM/MINIMUM to
locate the two turning points besides (0, 0).
SUMMARY
Steps for Ana lyzing the Graph of a Polynomial
To analyze the graph of a polynomial function
y
=
1 (x), follow these steps:
1: (a) Find the y-intercept by letting x = 0 and finding the value of 1(0). (b) Find the x-intercepts, if any, by solving the equation I(x) = O. STEP 2: Determine whether the graph of 1 crosses or touches the x-axis at each x-intercept. STEP 3: End behavior: find the power function that the graph of 1 resembles for large values of I x i . STEP 4: Determine the maximum number of turning points on the graph of I. STEP 5: Determine the behavior of the graph of 1 near each x-intercept. STEP 6: Put all the information together to obtain the graph of f. This usually requires finding additional points on the graph. STEP
'1'1
,-
Now Work PRO B L E M 7 7
For polynomial functions that have non integer coefficients and for polynomials that are not easily factored, we utilize the graphing utility early in the analysis of the graph. This is because the amount of information that can be obtained from algebraic analysis is limited.
II
E XA M P L E 1 1
Using a Graphing Utility to Analyze the Graph of a Polynomial F unction
For the polynomial I(x)
=
x3 + 2.48x2 - 4.3155x + 2.484406:
(a) Find the degree of the polynomial. Determine the end behavior; that is, find the power function that the graph of 1 resembles for large values of I x l . (b) Graph 1 using a graphing utility.
SECTION 5 . 1
Polynomial Functio ns and Models
339
(c) Find the x- and y-intercepts of the graph. (d) Use a TABLE to find points on the graph around each x-intercept. Determine on which intervals the graph is above and below the x-axis. ( e) Determine the local maxima and local minima, if any exist, rounded to two dec imal places. That is, locate any turning points. (f) Use the information obtained in parts (a) to (e) to draw a complete graph of I by hand. Be sure to label the intercepts, turning points, and the points ob tained in part (d). (g) Find the domain of f. Use the graph to find the range of I. (h) Use the graph to determine where I is increasing and decreasing. Solution Figure 2 1 15
Table 6 ::-:: � -3.D "3:.6 -3.'1
-�.2 -3 -2.8
X= - 4
V1 -'1.S;''1 -.177S �.S(lS 6.S219 D.!'<:12 10.7S1 12.(lS9
(a) The degree of the polynomial is 3. End behavior: the graph of I resembles that of the power function y = x3 for large values of lx l . (b) See Figure 21 for the graph of f. (c) The y-intercept is 1(0) = 2.484406. In Examples 9 and 10 we could easily fac tor I(x) to find the x-intercepts. However, it is not readily apparent how I(x) factors in this example. Therefore, we use a graphing utility's ZERO (or ROOT) feature and find the lone x-intercept to be -3.79, rounded to two decimal places. (d) Table 6 shows values of x around the x-intercept. The points ( -4, -4.57 ) and ( - 3, 10.75) are on the graph. The graph is below the x-axis on the interval ( - 00 , -3.79) and above the x-axis on the interval ( -3.79, 00 ) . (e) From the graph we see that it has two turning points: one between - 3 and -2, the other between 0 and 1 . Rounded to two decimal places, the local maximum is 13.36 and occurs at x = -2.28; the local minimum is 1 and occurs at x = 0.63. The turning points are ( -2.28, 13.36) and (0.63, 1 ) . (f) Figure 22 shows a graph of I drawn by hand using the information obtained in parts (a) to (e). Figure 22
(-3,1 0.75) ------
-5 End behavior: Resembles y
=
x3
(g) The domain and the range of I are the set of all real numbers. (h) Based on the graph, I is decreasing on the interval ( -2.28, 0.63 ) and is in creasing on the intervals ( - 00 , -2.28) and (0.63, 00 ) . il! q
5.1
I 7"-
Now Work P RO B L E M
• 89
Assess You r U nderstan di ng
fAre You Prepa red ?' Answers are given a t the end of these exercises. If you get a wrong answel; read the pages listed in red. 1.
The intercepts of the equation 9x2 + 4y = 36 are . ( pp. 1 65- ] 66) ____
2. True or False The expression 4x3 - 3.6x2 - V2 is a poly nomial. (pp. 39--47 )
3. To graph y = x2 - 4, you would shift the graph of y = x2 a distance of units. (pp. 25 2-2 60 ) 4.
True or False
The x-intercepts of the graph of a function
y = f(x) are the real solutions of the equation f(x) = ( pp. 224-225 )
O.
340
C H A PTER 5
Polynom ial and Rational Functions
Concepts and Voca bulary 5.
The graph of every polynomial function is both and
8.
_ _ _ _
_ _ _ _
6.
7.
A real number r for which f e r ) of the function f.
0 is called a(n)
In Problems why not.
., 1 1.
f(x)
14.
hex)
17.
g(x)
20.
F(x)
1 1-22,
=
4x + x3
12.
f(x)
3 - 2x
, 15.
X?/2 - x2 + 2
=
x2 - 5 x-'
,= --
In Problems
23-36,
=
=
1 -
18.
hex)
=
vx( vx -
21.
G(x) = 2(x - 1 )2(x2 + 1 )
.
use transformations of the graph of y
=
1
-x
X4 or y
1)
=
f(x)
f(x) = 3 xs
29.
f(x)
f(x)
33.
f(x)
36.
f(x)
27.
f(x)
2 X4 1
28.
31.
f(x)
=
( x - 1 )5 + 2
32.
35.
f(x)
=
4 - (x - 2 ) 5
=
(x + 2 ) 4 - 3
1 - x2 2
16.
f(x)
x(x - 1 )
19.
F(x)
=
=
, 1 5x4 - 7rX' + 2
26.
f ( x)
-x5
30.
f(x)
=
2 (x + 1 ) 4 + 1
34.
f(x)
=
3 - (x + 2) 4
=
=
=
=
X4 + 2 -x4 1 - (x - 1 )) - 2
2
form a polynomial whose real zeros and degree are given. Answers will vary depending on the choice of a leading 38.
Zeros: -2, 2, 3;
40.
Zeros: -4, 0, 2; degree 3
41.
Zeros: -4, - 1 , 2, 3; degree 4
43.
Zeros: - 1 , multiplicity 1;
Zeros:
g(x) =
x5 - 3
=
1 , 3; degree 3
. 37.
13.
x5 to graph each function.
25.
24.
-1,
True or False End behavior: the graph of the function f(x) = 3 x4 - 6x2 + 2 x + 5 resembles y X4 for large val ues of I x l .
f (x)
f(x) = (x + 1 )4
37-44,
10.
5x2 + 4X4
23.
[n Problems coefficient.
True or False The x-intercepts of the graph of a polynomial function are called turning points.
=
f ( x ) = (x - 2 ) 5
=
x2( x - 3 ) (x + 4) has
determine which functions are polynomial functions. For those that are, state the degree. For those that are not, tell
1
=
=
9. _ _ _ _
If r is a real zero of even multiplicity of a function f, the graph the x-axis at r. of f
Skill B u i l d ing
"
=
True or False The graph of f(x) exactly three x-intercepts.
3, multiplicity 2; degree 3
degree 3
44.
39.
Zeros: - 3, 0, 4 ;
42.
Zeros: - 3, - 1 , 2, 5; degree 4
Zeros: -2, multiplicity 2;
degree 3
4, mUltiplicity 1 ;
degree 3
In Problems 45-56, for each polynomial function:
(a) List each real zero and its multiplicity. (b) Determine whether the graph crosses or touches the x-axis at each x-intercept. (c) Determine the behavior of the graph near each x-intercept. (d) Determine the maximum number of turning points on the graph. (e) Determine the end behavior; that is, find the power function that the graph off resembles for large values of I x l .
'- 45.
f(x)
=
3 (x - 7 ) (x
+
3)2
48.
f(x) = 2 ( x - 3 ) ( x + 4)3
51.
f(x)
54.
f(x) = -2 (x2
=
(x - 5 ) 3 ( x + 4)2 +
3 )3
46.
52.
f(x)
f(x)
=
=
4(x + 4 ) ( x + 3 )3
(x + V3/(x - 2 ) 4
47.
f(x)
=
4(x2 + l ) (x - 2) 3
50.
f(x)
=
(x - �y(X - 1 )3
53.
f(x)
=
3 (x2 + 8 ) ( x2 + 9) 2
56.
f(x)
=
4x(x2 - 3 )
SECTION 5.1
Polynomial Functions and Models
341
In Problems 5 7-60, identify which of the graphs could be the graph of a polynomial function. For those that could, list the real zeros and state the least degree the polynomial can have. For those that could not, say why not. 57.
Y+
58.
Y.
59.
\
y+
60.
Y
x 4 x
-4
-2
-4
2
4 x
-2
-2
-4
-2
2
4
x
-2 -4
In Problems possible.) 61.
61-64,
decide which of the polynomial functions in the list might have the given graph. (More than one answer may be
Y
62.
x
(a) y (b) y (c) y
= = =
-4x(x - l ) (x - 2 ) x2 (x - 1 ) 2 (x - 2 ) 3x(x - 1 ) (x - 2)
63.
(d) y (e) y (f)
=
=
Y =
(a ) y (b) y ( c) y
x(x - 1 ) 2 (x - 2? x3 (x - 1 ) (x - 2) -x( 1 - x ) (x - 2)
Y
(r I o'pdx = =
=
2x3 (x - l ) (x - 2) 2 x2 (x - 1 ) (x - 2 ) x3 (x - 1 ) 2 (x - 2 )
64.
3
x
( c)
=
Y =
(d) y
=
(e) y
=
(f) y
=
In Problems
=
=
x
3
-2
(b) y
(e) y (f) y
x2 (x - 1 ) 2 (x - 2) 2 5x(x - 1 ) 2 (x - 2) -2x(x - 1 ) 2 (2 - x)
2
-2
=
=
Y
2
(a ) y
(d) y
-2
1 ? - (x- - 1 ) (x - 2) 2 1 - - (x2 + 1 ) (x - 2 ) 2
(a) y
=
(b) y
=
(x2 - 1 ) 1 -
(c)
- I (x-? - 1 ) ?-(x - 2) Z
(d) y
=
(x - 1 ?(x +
( e) y
=
- (x - 1 ?(x - 2 ) ( x + 1 )
( �)
( x2 D (x2 - 1 ) (2 - x) +
Y =
(f) y
- (x - 1 ) (x - 2 ) (x + 1 )
=
1 - (x2 - 1 ) (x 2 ) (x + 1 ) Z 1 - Z (x2 + l ) (x - 2 ) (x + 1 ) -
1
- ( x + l )- ? (x - l ) (x - 2 ) Z
1)( 1 - �)
( D (x - 1 ?(x
_ X2 +
+ 1 ) (x - 2 )
65-88:
(a) Find the x- and y-intercepts of each polynomial function f. (b) Determine whether the graph of f crosses o r touches the x-axis a t each x-intercept. (c) End behavior:find the power function that the graph of f resembles for large values of Ix!. (d) Determine the maximum number of turning points on the graph of f. (e) Determine the behavior of the graph off near each x-intercept. (f) Put all the information together to obtain the graph off (You may need to locate additional points on the graph.) x2 (x - 3 ) 66. f(x) = (x - 2) 3 " 67. f(x) 65. f(x) = (x - 1 ) 2 =
68.
f(x)
=
x(x + 2 ) 2
69.
f(x)
=
6x3 (x + 4)
70.
f(x)
=
5x(x - 1 ) 3
342
C HAPTER 5
Polynomial and Rational Functions
�
71. f( x)
=
-4x2 ( x + 2 )
72. f(x) = - x\x + 4)
73. f(x) = (x - l ) ( x - 2 ) ( x
74. f(x)
=
( x + l ) ( x + 4) (x - 3 )
75. f(x) = 4x - x3
76. f(x)
78. f(x) = x2 (x - 3 ) ( x + 4 )
79. f(x) = (x + 2 ) 2 ( x - 2 ) 2
(x + 1 ) 3 (x - 3 )
81. f(x) = (x - 1 ) 2 (x - 3 ) ( x + 1 )
82. f(x)
(x + 2 ? ( x - 4?
84. f ( x)
. 77. f(x) = x2 ( x - 2 ) ( x + 2 ) 80. f(x) 83. f(x)
=
=
86. f(x) = x2 ( x2 + l ) ( x
+
=
+
=
4)
x - x3
(x + 1 ) 2 (x - 3) ( x - 1 )
85. f(x) = x2 (x - 2 ) (x2 + 3 )
(x - 2 ) 2 (x + 2 ) ( x + 4 )
87. f(x) = -x2 (X2 - l ) (x
4)
=
+
88. f(x) = -x2 ( x2 - 4 ) ( x - 5 )
1)
, - In Problems 89-98, for each polynomial function f:
(a) Find the degree of the polynomial. D etermine the end behavior: that is, find the power function that the graph of f resembles for large values of l x l . (b) Graph f using a graphing utility. (c) Find the x- and y-intercepts of the graph, rounded to two decimal places. (d) Use TA BLE to find points on the graph around each x-intercept. Determine on which intervals the graph is above and below the x-axis. (e) Determine the local maxima and local minima, if any exist, rounded to two decimal places. That is, locate any turning points. (f) Use the information obtained in parts (a) to (e) to draw a complete graph of f by hand. Be sure to label the intercepts, turning points, and the points obtained in part (d). (g) Find the domain ofI Use the graph to find the range of r (h) Use the graph to determine where f is increasing and where f is decreasing 90. f(x) = x3 - 0.8x2 - 4.6656x + 3.73248 89. f(x) = x3 + 0.2x2 - l.5876x - 0.31752 91. f(x) = x3 + 2 . 56x2 - 3.31x + 0.89 93. f(x) 95. f(x)
=
=
x4 - 2.5x2 + 0.5625
94. f(x)
2 X4 - 7TX3 + V5x - 4
96. f(x)
=
- 1.2x4 + 0.5x2 - v3 x + 2
98. f(x)
=
7TX5
97. f(x) = - 2x5 - V2 x2 99.
92. f(x) = x3 - 2.91x2 - 7.668x - 3.8151
X
- V2
=
X4 - 1 8.5x2 + 50.2619
+
7TX4 + v3 x + 1
Hurricanes In 2005, hurricane Katrina struck the Gulf Coast of the United States, killing 1 289 people and causing an estimated $200 billion in damage. The following data represent the number of major hurricane strikes in the US. (category 3, 4, or 5) each decade from 192 1 to 2000. (a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) The cubic function of best fit to these data is
H(x) = 0.16x3 - 2.32x2 + 9.33x - 2.21
(c) (d) (e)
(f)
Decade,x
Major Hurricanes Striking United States, H
1 92 1 -1 930, 1
5
1 93 1 -1940, 2
8
1 941-1 950, 3
10
1 951-1 960, 4 8 Use this function to predict the number of major hurricanes that struck the United States between 1961 and 1 970. 1 961-1 970, 5 6 Use a graphing utility to verify that the function given in part (b) is the cubic func 1 971 -1980, 6 4 tion of best fit. 1 981-1990, 7 5 With a graphing utility, draw a scatter diagram of the data and then graph the cu 1991-2000, 8 5 bic function of best fit on the scatter diagram. Source: National Oceanic & Concern has risen about the increase in the number and intensity of hurricanes, Atmospheric Administration but some scientists believe this is just a natural fluctuation that could last another decade or two. Use your model to predict the number of major hurricanes that will strike the United States between 2001 and 2010. Does your result appear to agree with what these scientists believe? From 2001 -2005, five major hurricanes had struck the United States. Does this support or contradict your prediction in part (e)?
SECTION 5.1
100.
The following data represent the number of active duty military personnel (in millions) in the four major branches of the U.S. military for the years 1 998-2005, where 1 represents 1 998, 2 represents 1 999, and so on. Active Duty
101.
Temperature
Polynomial Functions and Models
The following data represent the temperature
T (OFahrenheit) in Kansas City, Missouri, x hours after mid
night on May 15, 2005.
ours after M idnight, x Year,x
Temperature (OF), T
Active Duty Personnel, N
6
44.1
1998, 1
1 .41
9
51.1
1999, 2
1 .39
12
57.9
2000, 3
1.38
15
63.0
2001, 4
1.39
18
63.0
2002, 5
1.41
21
59.0
2003, 6
1.42
24
54.0
2004, 7
1.41
2005, 8
1.39
Source: infoplease.com
(a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) The cubic function of best fit to these data is N(x) = - 0.001 46x3 + 0.01 98x2 - 0.0742x + 1 .4671 Use this function to predict the number of active duty military personnel in 2002. , (c) Use a graphing utility to verify that the function given in part (b) is the cubic function of best fit. (d) With a graphing utility, draw a scatter diagram of the data and then graph the cubic function of best fit on the scatter diagram. ( e) Do you think that the function given in part (b) will be useful in predicting the number of active duty military personnel in 2010? Explain.
343
45.0
Source: The Weather Underground
(a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) Find the average rate of change in temperature from 9 A M to 1 2 noon. (c) What is the average rate of change in temperature from 3 PM to 6 PM? (d) The cubic function of best fit to these data is T( x ) = -0.0103x3 + 0.32x2 - 1 .37x + 45.39 Use this function to predict the temperature at 5 P M . Use a graphing utility to verify that the function given in part (b) is the cubic function of best fit. ( I') With a graphing utility, draw a scatter diagram of the data and then graph the cubic function of best fit on the scatter diagram. (g) Interpret the y-intercept.
lfl (e)
Discussion a n d Writing
1 02. Can the graph of a polynomial function have no y-intercept?
Can it have no x-intercepts? Explain. 103. Write a few paragraphs that provide a general strategy
for graphing a polynomial function. Be sure to mention the following: degree, intercepts, end behavior, and turning points. 104. Make up a polynomial that has the following characteristics:
crosses the x-axis at - 1 and 4, touches the x-axis at 0 and 2, and is above the x-axis between 0 and 2. Give your polyno mial to a fellow classmate and ask for a written critique of your polynomial. 105. Make up two polynomials, not of the same degree, with the
following characteristics: crosses the x-axis at - 2 , touches the x-axis at 1, and is above the x-axis between -2 and 1 . Give your polynomials to a fellow classmate and ask for a written critique of your polynomials.
107. Which of the following statements are true regarding the graph of the cubic polynomial [(x) x3 + bx2 + ex + d? =
(Give reasons for your conclusions.) (a) It intersects the y-axis in one and only one point. (b) It intersects the x-axis in at most three points. (c) It intersects the x-axis at least once. (d) For I x l very large, it behaves like the graph of y = x3 . (e) It is symmetric with respect to the origin. (f) It passes through the origin.
108. The illustration shows the graph of a polynomial function.
(a) Is the degree of the polynomial even or odd? (b) Is the leading coefficient positive or negative? (c) Is the function even, odd, or neither?
y
106. The graph of a polynomial function is always smooth and
continuous. Name a function studied earlier that is smooth and not continuous. Name one that is continuous, but not smooth.
x
344
C H A PTER 5
Polynomial and Rational Functions
(d) Why is x2 necessarily a factor of the polynomial? (e) What is the minimum degree of the polynomial? (f) Formulate five different polynomials whose graphs could look like the one shown. Compare yours to those of other students. What similarities do you see? What differences?
109. Design a polynomial function with the following characteris tics: degree 6; four distinct real zeros, one of multiplicity 3 ; y intercept 3; behaves like y -5 x6 for large values of I x l. Is thjs polynomial unique? Compare your polynomial with those of other students. What terms will be the same as everyone else's? Add some more characteristics, such as symmetry or naming the real zeros. How does this modify the polynomial? =
'Are You Prepared?' Answers
1. ( -2 , 0) , ( 2 , 0 ) , ( 0 , 9)
2. True
3. down; 4
4. True
S.2 Properties of Rational Functions PREPARING FOR THIS SECTION •
•
Before getting started, review the following:
Rational Expressions (Chapter R, Section R.7, pp. 61-68) Polynomial Division (Chapter R, Section R.4, pp. 44-47) Now Work
• •
Graph of f(x) 1:. (Section 2.2, Example 12, x pp. 1 70-1 7 1 ) Graphing Techniques: Transformations (Section 3.5, pp. 252-260) =
the 'Are You Prepared?' problems on page 352.
Find the Domain of a Rational Fu nction (p. 344)
OBJECTIVES 2
Find the Vertical Asymptotes of a Rational Fu nction (p. 347)
3
Find the Horizontal or Oblique Asymptotes of a Rational Fu nction (p. 348)
Ratios of integers are called rational numbers. Similarly, ratios of polynomial func tions are called rational functions. Examples of rational functions are F(x)
DEFINITION
=
x3
7
r
-4
G(x)
=
� x4 - 1
A rational function is a function of the form R(x)
=
p (x) q(x)
where p and q are polynomial functions and q is not the zero polynomial. The domain of a rational function is the set of all real numbers except those for which the denominator q is O.
-.J
1 E XA M P L E 1
Find the Domain of a Rational Function F i nd i ng the Domain of a Rational F u nction
2x2 - 4 is the set of all real numbers x except -5; that x + 5 is, the domain is {xix *' - 5 } .
(a) The domain of R(x)
=
( b ) The domain o f R ( x )
=
_2_1
-
is the set of all real numbers x except -2 and
x - 4 2; that is, the domain is { x i x *' -2, x
(c) The domain of R ( x)
=
x3
2 -
-
x + 1
*'
2}.
is the set of all real numbers.
SECTION 5.2
Properties of Rational Functions
345
3 -_- is the set of all real numbers. x2 + 2 x2 - 1 (e) The domain of R (x) = is the set of all real numbers x except 1; that is, x - I the domain is { x i x "* I } . (d) The domain of R (x)
=
•
It is important to observe that the functions R (x)
=
x2 - 1 x- I
-- and f ( x )
are not equal, since the domain of R is { x i x of all real numbers.
"*
=
x + 1
I } and the domain of f is the set
���� ·- Now Work PRO B L E M 1 3
If R ( x)
=
:�;�
is a rational function and if p and q have no common factors,
then the rational function R is said to be in lowest terms. For a rational function R(x)
=
p(x)
-- in lowest terms, the real zeros, if any, of the numerator are the
( x) x-intercepts of the graph of R and so will play a major role in the graph of R. The real zeros of the denominator of R [that is, the numbers x, if any, for which q ( x) = 0], although not in the domain of R, also play a major role in the graph of R. We will discuss this role shortly. 1 We have already discussed the properties of the rational function f(x) = - . x ( Refer to Example 12, pages 170-171). The next rational function that we take up is 1 H (x) x-7 · E XA M P L E 2
q
1
Graphing y = 2" x
Analyze the graph of: H(x) Sol ution Table 7
x
2
1 1 00 1 0,000
2 1 00 1 0,000
H(x)
=
1
J?
4 1 0,000
The domain of H (x)
=
1
= -
x2
� is the set of all real numbers ;c
4 1 0,000
1 00,000,000
except O. The graph has
no y-intercept, because x can never equal O. The graph has no x-intercept because the equation H (x) = 0 has no solution. Therefore, the graph of H will not cross either of the coordinate axes. Because H ( -x)
1 ( -x)2
1 x-
= -- = ? =
H (x)
H is an even function, so its graph is symmetric with respect to the y-axis. Table 7 shows the behavior of H(x)
1 00,000,000
x
=
�x for selected positive numbers x. (We
will use symmetry to obtain the graph of H when x < 0). From the first three rows of Table 7, we see that, as the values of x approach (get closer to) 0, the values of IA H ( x ) become larger and larger positive numbers. When this happens, we say that � H is unbounded in the positive direction. We symbolize this by writing H � 00 (read as "H approaches infinity" ) . In calculus, limits are used to convey these ideas. There we use the symbolism lim H (x) = 00, read "the limit of H (x) as x X -' O approaches zero equals infinity," to mean that H ( x ) � 00 as x � O. Look at the last four rows of Table 7 . As x � 00 , the values of H(x) approach 0 (the end behavior of the graph). In calculus, this is symbolized by writing
346
C H A PTER 5
Polynomial a nd Rational Functions
lim H (x) = O. Figure 23 shows the graph. Notice the use of red dashed lines to convey the ideas discussed above. x-+oo
Figure 23 1
H(x)
= :?
y 5
: x= 0 l
0 , 4)
3 x Y=
y= 0 - 3
0 •
Sometimes transformations (shifting, compressing, stretching, and reflection) can be used to graph a rational function. E XA M P L E 3
Using Transformations to Graph a Rational F unction
Graph the rational function: R ( x ) Solution
Figure 24
=
1
(x - 2 )
2 +
1
First, we take note of the fact that the domain o f R is the set of all real numbers See Figure 24 for the except x = 2. To graph R, we start with the graph of y = x steps. •
�.
x= 2
: x= 0
x= 2
IJ
y= 1
y= 0 - 2
3
(a) y= x2 1
x
- - - - - .., - - - - - - - - I I
y= O Replace x by x - 2; shift right 2 units
(b) y
5
x
Add 1 ; shift u p 1 unit
1
(c)
= (x - 2)2
1
Y = (x 2)2 _
+
1
1;..=== - Now Work P R O B L E M 3 1
Asymptotes
In Figure 24(c), notice that as the values of x become more negative, that is, as x becomes unbounded in the negative direction (x ---,> - 00 , read as "x approaches negative infinity"), the values R(x) approach 1 . In fact, we can conclude the follow ing from Figure 24(c): 1. As x ---,> - 00 , the values R(x) approach 1 . 2. As x approaches 2, the values R(x) ---,> 00 . 3. As x ---,> 00 , the values R(x) approach 1 .
= 1J
[ lim R(x) x--> -OO
[ lim R(x) x-->2
[ lim R(x) x-->oo
=
ooJ
= 1J
SECTION 5.2
Properties of Rational Functions
347
This behavior of the graph is depicted by the vertical line x = 2 and the horizon tal line y = 1. These lines are called asymptotes of the graph, which we define as follows: Let R denote a function .
DEFINITION
If, as x � - 00 or as x � 00 , the values of R ( x) approach some fixed number L, then the line y = L is a horizontal asymptote of the graph of R. [Refer to Figures 2S(a) and (b).] If, as x approaches some number c , the values I R ( x) I � 00, then the line x = c is a vertical asymptote of the graph of R. The graph of R never intersects a vertical asymptote. [Refer to Figures 2S(c) and (d).]
.-J
Figure 25
y
x= c
y --------- y = L
--------- y= L
x
x
(a) End behavior:
(b) End behavior:
x � 00, the values of R(x) approach L [ symbolized by )i..rll R(x) =
As
As x � the values of R(x) approach L [ symbolized by x�rl1 R(x) = L]. That is, the points on the g raph of R are getting closer to the line y = L; Y = L is a horizontal asymptote. - 00 ,
L]. That is, the points on the g raph of R are getting closer to the line y = L; Y = L is a horizontal asymptote.
li t I I I
y
x= c
y
it I I I
X
X
I I
(c) As x approaches c, the values of I R(x) I � 00 [ for x < c, this is symbolized by �!!;lc R(x) = 00; for x > c, this is sym bolized by V..Tc + R(x) = 00] That is, the points on the g raph of R are getting closer to the l ine x = c ; x = C is a vertical asym ptote.
)1
(d) As x approaches c, the values of I R(x) I � 00
[ for x < c, this is symbolized by �!!;lc R(x) = - 00; for x> c, this is symbolized by �!Ilc+ R(x) = 00]. That is, the points on the graph of R are getting closer to the line x = c ; x = C is a vertical asymptote.
A horizontal asymptote, when it occurs, describes the end behavior of the graph as x � 00 or as x � - 00 . The graph of a function may intersect a horizontal asymptote.
A vertical asymptote, when it occurs, describes the behavior of the graph when
x is close to some number
Figure 26
c.
The graph of a function will never intersect a vertical
asymptote.
Y
/
/
/
/
/
/
/
/
There is a third possibility. If, as x � - 00 or as x � 00, the value of a rational function R ( x) approaches a linear expression ax + b, a =1= 0, then the line y = ax + b , a =1= 0, is an oblique asymptote of R. Figure 26 shows an oblique asymptote. An oblique asymptote, when it occurs, describes the end behavior of the graph . The graph of a function may intersect an oblique asymptote.
/
X
1iI!l!l:l==-- Now Work P RO B L E M 2 3 2
Find the Vertical Asymptotes of a Rational Function
( x) p . Iowest terms, are q( x) located at the real zeros of the denominator of q ( x). Suppose that is a real zero of q, so x - is a factor of q. As x approaches symbolized as x � r, the val ues of x - approach 0, causing the ratio to become unbounded, that is, IR ( x) I � 00 . Based on the definition, we conclude that the line x is a vertical
. If ' I functIOn ' R(x) The vertlca asymptotes o ' a ra tIOna r
asymptote .
r
=
, m
r
r,
= r
348
C H A PTER 5
P o l yn o mi a l and Rational Functions
Locating Vertical Asymptotes
THEOREM
(x ) , in lowest terms, will have a vertical asymp q(x) tote x = r if r is a real zero of the denominator q, That is, if x - r is a factor p (x) . of the denominator q of a rational function R(x) = -- , 1ll lowest terms, R q (x ) will have the vertical asymptote x = r . A rational function R ( x )
WARN I NG I f a rational function is not in lowest terms, an a pp l ication of this theorem will res u lt i n a n incorrect l i sting
p
.J
_
of vertical asym ptotes,
=
F i n d i ng Vertical Asymptotes
E XA M P L E 4
Find the vertical asymptotes, if any, of the graph of each rational function. (a) R(x)
=
( C) H ( x )
WARN I N G I n Exam p l e 4(a ) , the vertical
=
- 2 and x
=
2, Do
not say that the vertical asym ptotes
are -2 a n d 2,
x2
(d) G(x)
x2 + 1
=
x2 - 9 ---,:-- - x2 + 4x - 21
(a) R is in lowest terms and the zeros of the denominator x2 - 4 are -2 and 2 . The lines x = - 2 and x = 2 are the vertical asymptotes of the graph of R. (b) F is in lowest terms and the only zero of the denominator is 1. The line x = 1 is the vertical asymptote of the graph of F.
Solution asym ptotes are x
=
x + 3 (b) F (x) = x _ 1
x x2 _ 4
_
(c) H is in lowest terms and the denominator has no real zeros, because the equa tion x2 + 1 has no real solutions. The graph of H has no vertical asymptotes.
(d) Factor G( x) to determine if it is in lowest terms. G (x )
=
x2 - 9 ----:--- x2 + 4x - 21
(x + 3 ) (x - 3 ) (x + 7 ) (x - 3 )
x + 3 x + 7
x *- 3
The only zero of the denominator of G(x) in lowest terms is -7. The line x = -7 is the only vertical asymptote of the graph of G.
•
As Example 4 points out, rational functions can have no vertical asymptotes, one vertical asymptote, or more than one vertical asymptote. However, the graph of a rational function will never intersect any of i ts vertical asymptotes. (Do you know why?)
i�1 Exploration i!mll Graph each of the fol lowing ratio n a l functions: R(x)
1
1
� R(x) = _ 1 ) 2 (x
R(x) =
1
1 )3
R(x)
=
1 4 _ (x 1 )
(x x = 1 , What happens to the va l u e of R(x) as x a p p roaches 1 from the right side of the vertical asym ptote; that is, what is l i m R(x)? What happens to the val ue of R(x) as x- l + X a pproaches 1 from the l eft side of the vertical asym ptote; that is, what is l i m R(x)? How does the m u lx� , tiplicity of the zero in the denomi nator affect the graph of R? =
_
Each has the vertical asym ptote
I.i!l !�
Now Work P R O B L E M 4 5 ( F I N D T H E A S Y M P TO T E S , I F A N Y . )
3
V
E RT I CA L
Find the Horizontal or Oblique Asymptotes of a Rational Function
The procedure for finding horizontal and oblique asymptotes is somewhat more involved. To find such asymptotes, we need to know how the values of a function behave as x --7 - 00 or as x --7 00 .
SECTION 5 . 2
Properties of Rational Functions
349
If a rational function R ( x ) is proper, that is, if the degree of the numerator is less than the degree of the denominator, then as x � - 00 or as x � 00 the value of R ( x ) approaches O. Consequently, the line y = 0 ( the x-axis) is a horizontal asymp tote of the graph.
THEOREM
E XA M P L E 5
If a rational function is proper, the line y = 0 is a horizontal asymptote of its graph.
-.J
F i n d i n g H orizontal Asym ptotes
Find the horizontal asymptotes, if any, of the graph of R( x ) = Solution
"x: _-__ 1 2_ ----:_ 4x2 + x + 1
Since the degree of the numerator, 1 , is less than the degree of the denominator, 2, the rational function R is proper. We conclude that the line y = 0 is a horizontal asymptote of the graph of R.
•
To see why y = 0 is a horizontal asymptote of the function R in Example 5, we need to i nvestigate the behavior of R as x � - 00 and x � 00 . When Ixl is unbounded, the numerator of R, which is x - 12, can be approximated by the power function y = x, while the denominator of R, which is 4x2 + X + 1, can be approximated by the power function y = 4x2. Applying these ideas to R( x ) , we find R(x) =
x - 12 2 4x + X + 1 For
Ixl
7
1 x �O = 4x l' 4x2
unbou nded
As
x --> -00
or
x --> 00
This shows that the line y = 0 is a horizontal asymptote of the graph of R. p(x) If a rational function R ( x ) = is improper, that is, if the degree of the q(x) numerator is greater than or equal to the degree of the denominator, we must use long division to write the rational function as the sum of a polynomial f ( x ) plus a rex) . proper ratlona ' Th at IS, . I functIOn ' ' we wrIte q(x) R(x) =
rex) p(x) = f(x) + q(x) q(x)
r e x) rex) where f(x) is a polynomial and -- is a proper rational function. Since -- is (x) q q(x ) rex) proper, -- � 0 a s x � - 00 o r a s x � 00 . A s a result, q(x) p(x) R ( x ) = - � f(x) q(x)
as x �
- 00 or as x � 00
The possibilities are listed next.
1. If f ( x ) = b, a constant, the line y = b is a horizontal asymptote of the graph of R. 2. If f ( x ) = ax + b, a oF 0, the line y = ax + b is an oblique asymptote of the graph of R. 3. In all other cases, the graph of R approaches the graph of f, and there are no horizontal or oblique asymptotes. The following examples demonstrate these conclusions.
350
C H A PTER 5
Polynomial and Rational Functions
EXAM P L E 6
F i nding H o rizontal o r Oblique Asymptotes
Find the horizontal or oblique asymptotes, if any, of the graph of
3x4 - x2 H(x) = , x� - x2 + 1 Solution
Since the degree of the numerator, 4, is larger than the degree of the denominator, 3, the rational function H is improper. To find any horizontal or oblique asymptotes, we use long division.
3x + 3 b x3 - x2 + 1 x4 + 3x 3x4 - 3x3 3x3 - x2 - 3x 3x3 - 3x2 +3 2x2 - 3x - 3 As a result,
H(x) = As x ----;.
- 00
2 -,v2 - 3x - 3 3 x4 - x2 = 3x + 3 + x3 - x2 + 1 x3 - x2 + 1
or as x ----;. 00 ,
A s x ----;. - 00 o r a s x ----;. 00 , we have H(x) ----;. 3 x + 3. We conclude that the graph of the rational function H has an oblique asymptote y = 3x + 3. •
E XA M P L E 7
F i nding H orizontal or O b l i q u e Asymptotes
Find the horizontal or oblique asymptotes, if any, of the graph of
R (x) = Solution
8x2 - X + 2 4x2 - 1
Since the degree of the numerator, 2, equals the degree of the denominator, 2, the rational function R is improper. To find any horizontal or oblique asymptotes, we use long division.
2 ) 4x2 - 1 8x2 - X + 2 8x2 -2 -x+4 As a result,
R( x ) = Then, as x ----;.
- 00
8X2 - X + 2 -x + 4 = 2 + ---::,--4x2 - 1 4x2 - 1
or as x ----;. 00 ,
-x + 4 4x2 - 1
-1 -x ----;. 0 = 4x2 4x
- � ::--
As x ----;. - 00 or as x ----;. 00 , we have R(x) ----;. 2. We conclude that y = 2 is a hori zontal asymptote of the graph. •
SECTION 5 .2
Properties of Rational Functions
3S1
In Example 7, we note that the quotient 2 obtained by long division is the quo tient of the leading coefficients of the numerator polynomial and the denominator polynomial
(�). This means that we can avoid the long division process for rational
functions whose numerator and denominator are of the same degree and conclude that the quotient of the leading coefficients will give us the horizontal asymptote. & .$ >-
•
Now Work PRO B L E M S 4 1 A N D 4 3
F i n d i n g Horizontal or O b l i q u e Asymptotes
E XA M P L E 8
Find the horizontal or oblique asymptotes, if any, of the graph of 2x5 - x3 + 2 G ( x) = ----x3 - 1 Since the degree of the numerator, S, is larger than the degree of the denominator, 3, the rational function G is improper. To find any horizontal or oblique asymptotes, we use long division.
Solution
x3 - 1 hx5 - x3 + 2x5 - 2x2 -x3 + 2x2 + + -x3 2x2 +
2 2 1 1
As a result, G( x) =
2x5 - x3 + 2 x' - 1 0
=
2x2 - 1
+
2x2 + 1 x3 - 1
---
Then, as x --i> - 00 or as x --i> 00 , 2X2 + 1 2x2 2 ::::; 0 = - --i> 0 x x3 - 1 x '
---
-
As x --i> - 00 or as x --i> 00, we have G (x) --i> 2x2 - 1. We conclude that, for large values of lxi, the graph of G approaches the graph of y = 2x2 - 1 . That is, the graph of G will look like the graph of y = 2x2 - 1 as x --i> - 00 or x --i> 00 . Since y = 2X2 - 1 is not a linear function, G has no horizontal or oblique asymptotes. •
We now summarize the procedure for finding horizontal and oblique asymptotes. SUMMARY
Finding Horizo nta l and Obl ique Asym ptotes of a Rational Function R
Consider the rational function R(x) =
p (x) q(x)
=
x" + a,,_ l x"- 1 + . . . + a l x + ao " bm xm + bm_ 1 ,·'C'I/-l + . . . + b1x + bo a
in which the degree of the numerator is n. and the degree of the denominator is m . 1. If the degree of the numerator is less than the degree of the denominator, R is a proper rational function, and the graph of R will have the horizontal asymptote y = 0 (the x-axis) .
352
Pol y n omia l and Rational Functions
CHAPTER 5
2. If the degree of the numerator is greater than or equal to the degree of the denominator, then R is improper.
Here long division is used. (a) If the degree of the numerator equals the degree of the denominator, the quotient obtained will be the an
number - , and the line bm
y
an
= - is a horizontal asymptote. bm
(b) If the degree of the numerator is one more than the degree of the denominator, the quotient obtained is of the form a x + b (a polynomial of degree 1), and the line y = ax + b is an oblique asymptote. (c) If the degree of the numerator is two or more than the degree of the denominator, the quotient obtained is a polynomial of degree 2 or higher, and R has neither a horizontal nor an oblique asymptote. In this case, for Ixl unbounded, the graph of R will behave like the graph of the quotient. Note: The graph of a rational function either has one horizontal or one oblique asymptote or else has no horizontal and no oblique asymptote. It cannot have both a horizontal asymptote and an oblique asymptote.
5.2 Assess You r U ndersta nding 'Are You P repared?, Answers are given a t the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. True or False The quotient of two polynomial expressions is a rational expression. (pp. 6 1-68) 2. What is the quotient and remainder when 3x4 - x2 is divid ed by x3 - x2 + 1 . ( pp. 44-47)
3. Graph y =
1 -
x
. (pp. 1 70-171 )
4. Graph y = 2(x + 1) 2 - 3 using transformations.
(pp. 2S2-260)
Concepts and Voca bulary 5•
· The hne
6. The line 7.
__
.
IS
__
x3 - 1 3+ 1
.
a honzontal asymptote of R ( x ) = --- ' X
is a vertical asymptote of R(x) =
: : �.
9. True or False If an asymptote is neither horizontal nor ver tical, it is called oblique.
10.
For a rational function R, if the degree of the numerator is less than the degree of the denominator, then R is . __
True or False If the degree of the numerator of a rational function equals the degree of the denominator, then the ratio of the leading coefficients gives rise to the horizontal asymptote.
8. True or False The domain of every rational function is the set of all real numbers. Skill B u i l d i n g
In Problems 11-22, find the domain of each rational function. 4x Sx2 11. R(x) = 12 R( x ) = x-3 3+x . --
14. G(x) =
6
(x + 3) (4 - x) x
17. R(x) = -- X3 - 8
20. G(x) =
x - 3 + 1
4 X
3x(x - 1 ) 15. F ( x) = ---'-----'--2x2 - Sx - 3 x
13.
H(x) =
-4x2 (x - 2)(x + 4) -x( 1 - x)
16. Q ( x) = ---:-------'---
3x2 + Sx - 2 3x2 + X x2 + 4
18. R(x) = -4- x - I
19. H(x) =
3(x2 - X - 6) 21. R(x) = -----'4(x2 - 9)
22. F (x) = ---'-----'--
-
2 ( x2 - 4)
3(x2 + 4x + 4)
SECTION 5.2
In Problems 23-28, use the graph shown to find: (a) The domain and range ofeach function (d) Vertical asymptotes, if any y
23.
(b) The intercepts, if any (e) Oblique asymptotes, if any
4 x
-4
25.
�!
-- l�
y
3 x
I I I I I I
-3
35 3
(c) Horizontal asymptotes, if any
24.
4
Properties of Rational Functions
-3
1 -3
-4
I
26.
27. y
) � : l y
:I
3
I
I I
I I
I I
_ _ _ -.-l _ _ _ _ -3
3 x
In Problems
29-40, 1
35. R(x ) = 38. F(x ) =
3 x
2
=
33. H ( x ) =
-
x2
+
-
L___ 3 x
-3
30. Q(x )
x
=
_ _ _
x
graph each rational function using transformations.
29. F(x ) = 2 + 32. R(x )
28.
-1 4x
+
1
1
x +
4
36. R(x ) = 39. R (x )
=
1
3+? x-
31. R(x )
-2
=
3x
x
+
42. R(x )
4 +
-x2 1 44. G(x ) = x2 _ 5x + 6 47. Q(x )
=
50. R(x ) =
5 - x2
=
45. T(x ) =
�
48. F(x ) =
6x2 + X + 1 2 3x2 - 5x - 2
51. G ( x )
=
2
x+ 1
34. G(x ) =
1 + 1 x-I
37. G(x ) = 1 +
x?- - 4 x2
40. R(x ) =
--
In Problems 41-52, find the vertical, horizontal, and oblique asymptotes, if any, of each rational function.
41. R(x ) =
1 (x - I f (x
+
2f 2
(x _ 3 ) 2
x -4 x
3x + 5 x -6
43. H ( x ) =
x3 - 8 2 x _ 5x + 6
x3 _ 1
46. P(x ) =
4x5 x3 _
x4
-2X2
2x 3
+
1
+ 4x2
x3 - 1
x - x2
49. R(x )
=
52. F(x ) =
3x4
x3
1
+
4
+ 3x
x - I
x
_
x3
354
C H APTER 5
Polynomial and Rational Functions
Appl ications a n d Extensions
53.
(a) Let Rl = 10 ohms, and graph Rtot as a function of R2 . (b) Find and interpret any asymptotes of the graph obtained in part (a). (c) If R2 = 2�, what value of R I will yield an Rtot of 1 7 ohms?
In physics, it is established that the acceleration due Gravity to gravity, g (in meters/sec2 ), at a height h meters above sea level is given by 3.99
g(h) = (6.374
X
10 1 4 106 + h ) 2
X
where 6.374 X 106 is the radius of Earth in meters. (a) What is the acceleration due to gravity at sea level? (b) The Sears Tower in Chicago, Illinois, is 443 meters tall. What is the acceleration due to gravity at the top of the Sears Tower? (c) The peak of Mount Everest is 8848 meters above sea level. What is the acceleration due to gravity on the peak of Mount Everest? (d) Find the horizontal asymptote of g( h ) . ( e ) Solve g ( h ) O . How d o you interpret your answer? 54. Population Model A rare species of insect was discovered in the Amazon Rain Forest. To protect the species, environ mentalists declare the insect endangered and transplant the insect into a protected area. The population P of the insect ( months after being transplanted is
Source: en. wikipedia.org/wiki/SeriesJmd_paralleCcircuits
gs, 56.
2
is an efficient method for finding the x-intercepts (or real zeros) of a function, such as p(x). The steps below outline Newton's Method STEP 1 : Select an initial value Xo that is somewhat close to the x-intercept being sought. STEP 2: Find values for x using the relation p(X,,) X,,+ I = X" - ' - n = 1 , 2, . . . p ( x" )
50( 1 + 0.5t) (2 + O.Olt)
until you get two consecutive values Xk and Xk+ 1 that agree to whatever decimal place accuracy you desire. STEP 3: The approximate zero will be Xk+ l ' Consider the polynomial p(x) = x3 7x - 40. (a) Evaluate p (5) and p( -3). (b) What might we conclude about a zero of p . Explain. (c) Use Newton's Method to approximate an x-intercept, 1', -3 < r < 5, of p(x) to four decimal places. .- (d) Use a graphing utility to graph p(x) and verify your an swer in part (c). (e) Using a graphing utility, evaluate per) to verify your result.
(a) How many insects were discovered? In other words,
what was the population when t = O? (b) What will the population be after 5 years? (c) Determine the horizontal asymptote of p e t ) . What is the largest population that the protected area can sustain? 55. R esistance in Parallel Circuits From Ohm's law for circuits, it follows that the total resistance R tot of two components hooked in parallel is given by the equation Rtot
=
RI R 2 RI + R
"
Newton's Method
=
pet ) =
I n calculus you will learn that, if p(x) = a"x + a,,_lx,,- I + . . . + ajX + ao is a polynomial, then the derivative of p(x) is p ' (x) = na"x,,- I + ( n - 1 )a,,_jx,,-2 + . . . + 2a x + al
Newton's Method
-
---
2
where R l and R 2 are the individual resistances. Discussion a n d Writing
57. If the graph of a rational function R has the vertical asymp tote x = 4, the factor x 4 must be present in the denomi
59. Can the graph of a rational function have both a horizontal
and an oblique asymptote? Explain.
-
nator of R. Explain why. 58. If the graph of a rational function R has the horizontal asymptote y = 2, the degree of the numerator of R equals the degree of the denominator of R . Explain why.
60. Make up a rational function that has y
2x + 1 as an oblique asymptote. Explain the methodology that you used. =
'Are You Prepa red ?' An swers 1.
True
2. Quotient: 3x + 3; Remainder: 2x2 - 3x - 3
3.
4.
y 3
-
3 x
3 (0, - 1 )
SECTION 5.3
The Graph of a Rational Function
355
5.3 The Graph of a Rational Function PREPARING FOR THIS SECTION •
Before getting started, review the following:
Intercepts (Section 2.2, pp. 1 65-166) Now Work the 'Are You
Prepared?' problems on page 366.
OBJECTIVES 1 Ana lyze the G raph
of a
Rational Fu nction (p. 355)
2 Solve Appl ied Problems I nvolving Rational F u n ctions (p. 365)
1
Analyze the Graph of a Rational Function
We commented earlier that calculus provides the tools required to graph a polyno mial function accurately. The same holds true for rational functions. However, we can gather together quite a bit of information about their graphs to get an idea of the general shape and position of the graph. In the examples that follow, we will analyze the graph of a rational function by applying the following steps:
Analyzing the Graph of a Rational Function R
STEP 1: Factor the numerator and denominator of R and find its domain. If o is in the domain, find the y-intercept, R(O), and plot it.
STEP 2: Write R in lowest terms as
STEP 3:
STEP 4:
STEP 5:
STEP 6: STEP
E XA M P L E 1
7:
�i��
and find the real zeros of the numer
ator; that is, find the real solutions of the equation p(x) 0, if any. These are the x-intercepts of the graph. D etermine the behavior of the graph of R near each x-intercept, using the same procedure as for polynomial functions. Plot each x-intercept and indicate the behavior of the graph near it. ( x) With R written in lowest terms as p , find the real zeros of the deq( x) nominator; that is, find the real solutions of the equation q(x) = 0, if any. These determine the vertical asymptotes of the graph. Graph each vertical asymptote using a dashed line. Locate any horizontal or oblique asymptotes using the procedure given in the previous section. Graph the asymptotes using a dashed line. D etermine the points, if any, at which the graph of R intersects these asymptotes. Plot any such points. Using the real zeros of the numerator and the denominator of the given equation for R, divide the x-axis into intervals and determine where the graph is above the x-axis and where it is below the x-axis by choosing a number in each interval and evaluating R there. Plot the points found. Analyze the behavior of the graph of R near each asymptote and in dicate this behavior on the graph. Put all the information together to obtain the graph of R. =
A n alyzi ng the G raph of a Ration al Functio n
Analyze the graph of the rational function:
R(x)
=
x - I x- - 4
-?-
356
CHAPTER 5
Polynomial a nd Rational Functions
Solution
STEP 1: We factor the numerator and denominator of R, obtaining
R(x ) The domain of R is { x i x
"*
x - 1 (x + 2) (x - 2)
- ----
-2, x
"*
2 } . The y-intercept is
-1 1 R(O) = - = 4 -4
( �) .
Plot the point 0,
STEP 2: R is in lowest terms. The real zero of the numerator satisfies the equation
x - 1 = O. The only x-intercept is 1 . Near 1 :
x - 1 (x + 2) (x - 2)
R(x ) =
�
x - 1 1 = - 3(x - 1 ) (1 + 2) (1 - 2)
Plot the point (1, 0) and indicate a line with slope -
�
there.
STEP 3: R is in lowest terms. The real zeros of the denominator are the real solutions
of the equation (x + 2) (x - 2) = 0, that is, -2 and 2. The graph of R has two vertical asymptotes: the lines x = -2 and x = 2. Graph each of these asymptotes using dashed lines. STEP 4: The degree of the numerator is less than the degree of the denominator, so R is proper and the line y = 0 (the x-axis) is a horizontal asymptote of the graph. Indicate this line by graphing y = 0 using a dashed line. To determine if the graph of R intersects the horizontal asymptote, we solve the equation R( x) = o. x - 1 -= 0 x2 - 4
Multiply both sides by
x - 1 = 0 x = 1
:l- 4.
The only solution is x = 1 , so the graph of R intersects the horizontal asymptote at ( 1 , 0). We have already plotted this point. STEP 5: The zero of the numerator, 1, and the zeros of the denominator, -2 and 2, divide the x-axis into four intervals:
( -00, -2)
(2,
( 1 , 2)
( -2, 1 )
(0
)
Now construct Table 8.
-----------. ----- ----- .2 -----
Table 8 r-
_-:2
�
Interval
( -00, -2)
(-2, 1)
Number chosen
-3
0
Value of R
R(-3)
Location of graph
Below x-axis
Point on graph
( - 3, -0.8)
=
-0.8
R(O)
-::-
(2, 00)
(1, 2) 3
=
� 4
Above x-axis
( O,�)
---,
, x
3
2
R(%) � =
-
Below x-axis
e- -�) 2'
7
R(3)
=
0.4
Above x-axis (3, 0.4)
Plot the points from Table 8. You should now have Figure 27(a). STEP 6: Next, we determine the behavior of the graph near the asymptotes. • Since the x-axis is a horizontal asymptote and the graph lies below the x-axis for x < -2, we can sketch a portion of the graph by placing a small arrow to the far left and under the x-axis.
357
SECTION 5.3 The Graph of a Rational Function
Since the line x = -2 is a vertical asymptote and the graph lies below the x-axis for x < -2, we continue by placing an arrow well below the x-axis and approaching the line x = - 2 on the left. • Since the graph is above the x-axis for -2 < x < 1 and x = -2 is a ver tical asymptote, the graph will continue on the right of x = -2 at the top. Similar explanations account for the other arrows shown in Figure 27(b). STEP 7: Figure 27(c) shows the complete graph. •
Figure 27
X=-2
Y
x=2 I I I I I I I I
3
-3 (-3, -0.8)
(0, �)
•
(1,0) : •
(�2' _f) 7
X=-2
i\
(3,0.4) •
3
xY=
0 -.... -3 (-3,-0.8) •
I I I I I I I I I I I I I
I I I I I
(1,0) :
(0, �)
•
I
(�,-n :
(a)
(3,0.4) 3 XY=0
1,
.-
1-3, -0.8)
�
li I
(b)
�. lim
x=2
3
I I I
-3
Y
X= -2
it
3
V
-3
x=2
Y
.Jl 3
l
0
X Y=
-3 (c)
•
Exploration x-l Graph R(x) = � x-
-4
Result The analysis just completed in Example 1 helps us to set the viewing rectangle to obtain a com x-l -2-- in connected mode, and Figure 28(b) shows x -4 it in dot mode. Notice in Figure 28(a) that the g ra ph has vertical lines at x = -2 and x = 2. This is due plete g ra ph. Figure 28(a) shows the graph of R(x) =
to the fact that, when the g ra phing utility is in connected mode, it will connect the dots between con secutive pixels. We know that the graph of R does not cross the lines x = -2 and x = 2, since R is not defined at x = -2 or x = 2. So, when graphing rational functions, dot mode should be used to avoid extraneous vertical lines that are not part of the g raph. See Figure 28(b).
Figure 28
-4
l
---,\
I
4
4 '\
"--.,-,. 4
4 -4
-4
Dot mode
Connected mode
(b)
(a)
"'1'1
EXAM P L E 2
¥ 5 .-
Now Work
PRO B L E M
7
A n alyzi ng the G raph of a Rational Function
Analyze the graph of the rational function: Solution
STEP 1: The domain of R is { x i x
=1=
R(x)
x2
-
1
= --
x
o} . B ecause x cannot equal 0, there is no y-intercept. Now factor R to obtain (x + l ) ( x - 1) R(x) x STEP 2: R is in lowest terms.
358
CHAPTER 5
Polynomial a n d Rational Functions
Solving the equation R(x) = 0, we find the graph has two x-intercepts: - 1 and 1 . (x + l ) (x - 1) (x + 1)( - 1 - 1 ) � = 2(x + 1) Near -1: R(x) = x -1 (x + 1) (x - 1) (1 + l) (x - 1) � = 2 ( x - 1) Near 1 : R(x) = 1 x
Plot the point ( - 1, 0) and indicate a line with slope 2 there. Plot the point ( 1, 0) and indicate a line with slope 2 there. STEP 3: R is in lowest terms, so the graph of R has the line x = 0 (the y-axis) as a vertical asymptote. Graph x = 0 using a dashed line. STEP 4: The rational function R is improper, since the degree of the numerator, 2, is larger than the degree of the denominator, 1. To find any horizontal or oblique asymptotes, we use long division. x
-1 x i s a n oblique asymptote o f the graph.
The quotient is x , s o the line y = Graph y = x using a dashed line. To determine whether the graph of R intersects the asymptote y we solve the equation R(x) = X. R(x)
=
x2 - 1 -X
=
=
x,
x
x2 - 1 = x2 -1 = 0
Im possible
---
x2 - 1 . ' We conclud e th at t h e equatIOn = x h as no soi utlOn, so t he x graph of R does not intersect the line y = X. STEP 5: The zeros of the numerator are -1 and 1; the zero of the denominator is O. We use these values to divide the x-axis into four intervals: ( - 1, 0)
( - 00 , -1) Table 9
(0, 1)
( 1, 00 )
Now we construct Table 9.
.
-1
Interval
(-00, - 1)
Number chosen
-2
Value of R
R(-2)
Location of graph
Below x-axis
Point on graph
( -1, 0) -
=
( D 2 - ,-
3
--
2
.0
'x
(0, 1)
(1,00)
-
2
1
1
-
2
2
( -D = % (_ � i)
G) = - %
R
R
Above x-axis
Below x-axis
U _ i)
2'2
2' 2
R(2)
=%
Above x-axis
(D 2
,
Plot the points from Table 9. You should now have Figure 29(a) . STEP 6: • Since the graph of R is below the x-axis for x < - 1 and is above the x-axis for x > 1, and since the graph of R does not intersect the oblique asymptote y = x, the graph of R will approach the line y = x as shown in Figure 29(b). • Since the graph of R is above the x-axis for - 1 < x < 0, the graph of R will approach the vertical asymptote x 0 at the top to the left of x = O. =
SECTION 5.3 The Graph of a Figure 29
x= O
Rational Function
x= O
y
x=O
y 3 /
/ //
//
/ // e
e
y
y =x / //
y= x /
( 2 3) ,2 3
359
/ // x
/
/
//
/
/
( 2, 23)
e
x
3
-3
x
(12' _;1)2
-3
(a)
.[� ;.;I �
(b)
Seeing the Concept x2 - 1 Graph R(x) = -- and compare what x you see with Figu re 29(c). Could you have predicted from the g raph that y = x is an oblique asym ptote? Graph y = x and ZOOM-OUT. What do you observe?
EXA M P L E 3
(c)
Since the graph of R is below the x-axis for 0 < x < 1 , the graph of R will approach the vertical asymptote x = 0 at the bottom to the right of x = O. See Figure 29 (b ) . STEP 7: The complete graph is given in Figure 29 ( c) . •
•
Iii-
:;.>-
Now Work
PROBlEM 1 5
Analyzing the G raph of a Rational Function
Analyze the graph of the rational function: Solution
R( x )
=
4 + 1 -- X2
X
STEP 1: R is completely factored. The domain of R is
y-intercept.
"*
O } . There is no
4
+ 1 = 0 has no real solutions, there are no x-intercepts. STEP 3: R is in lowest terms, so x = 0 (the y-axis) is a vertical asymptote of R. Graph the line x = 0 using dashes. STEP 4: The rational function R is improper. To find any horizontal or oblique asymptotes, we use long division.
STEP 2: R is in lowest terms. Since
X
{x i x
1 The quotient is x2 , so the graph has no horizontal or oblique asymptotes. However, the graph of R will approach the graph of y = x2 as x � -00 and as x � 00. The graph of R does not intersect y = x2 . Graph y = x2 using dashes. STEP 5: The numerator has no zeros, and the denominator has one zero at O. We divide the x-axis into the two intervals
( -00, 0) and construct Table 10.
( 0, 00 )
360
CHAPTER 5 Polynomial and Rational Functions o
Table 10
.
. )(
Interval
(-00,0)
(0,00)
Number chosen
-1
1
Value of R
R(-l)
Location of graph
Above x-axis
Above x-axis
Point on graph
(-1, 2)
( 1 , 2)
R( 1 )
2
=
=
2
Plot the points ( - 1 ,2 ) and ( 1 , 2 ) . STEP 6: • Since the graph of R is above the x-axis and does not intersect y = x2, we place arrows above y = x2 as shown in Figure30(a). • Also,since the graph of R is above the x-axis,it will approach the verti cal asymptote x = 0 at the top to the left of x = 0 and at the top to the right of x = O. See Figure30(a) . STEP 7: Figure30(b) shows the complete graph. Figure 30 x=O
x=O
Y
Y
6
\
Graph R(x)
=
\
\ \
I �I Seeing the Concept
I\lm
t
X4 + 1
-2
- and compare what
x you see with Figure 30(b). Use MINIMUM to find the two turning points. Enter Y2 = x2 and ZOOM-OUT. What do you see?
\
\
,
.
\
,
-3
/
/
.
(1, 2)
(-1,2) \
,
6
2 Y= x
,
/
/
/
/
1/
I I
I
/ / / /
/ /
,
(-1,2)
/
x
3
\
\
EXAM P L E 4
Now Work
"
3
x
(b)
P R O BLE M 1
•
3
A n alyzi ng the Graph of a Ration al Function
Analyze the graph of the rational function: Solution
,
-3
(a)
""= -
2 Y= x
I I
R(x )
=
3x2 -3x x2 + X - 12
STEP 1: We factor R to get
R(x)
=
3x(x - 1 ) (x + 4 ) ( x - 3)
The domain of R is { x i x -=I- -4, x -=I- 3}. The y-intercept is R(O) = O. Plot the point (0,0). STEP 2: R is in lowest terms. Since the real solutions of the equation3x( x - I ) = 0 are x = 0 and x = 1, the graph has two x-intercepts, 0 and 1. We determine the behavior of the graph of R near each x-intercept. Near 0:
R( x )
=
3x( x - 1 ) (x + 4 ) ( x - 3)
-,----'� -,-----'
Near 1:
R(x)
=
3x(x - 1 ) ( x + 4 ) ( x - 3)
�
3x(0 - 1 ) (0 + 4) (0 - 3)
=
1 -x 4
3( 1 ) (x - 1 ) (1 + 4) ( 1 - 3)
=
-
3 (x 10
1)
SECTION 5.3
The Graph of a Rational Function
36 1
Plot the point (0, 0) and show a line with slope � there. 4 3 Plot the point (1, 0) and show a line with slope there. 10 STEP 3: R is in lowest terms. Since the real solutions of the equation (x + 4 ) ( x - 3) = 0 are x = -4 and x = 3, the graph of R has two verti cal asymptotes, the lines x = -4 and x 3. Plot these lines using dashes. STEP 4: Since the degree of the numerator equals the degree of the denominator, the graph has a horizontal asymptote. To find it, we either use long division or form the quotient of the leading coefficient of the numerator, 3, and the leading coefficient of the denominator, 1. The graph of R has the horizon tal asymptote y = 3. To find out whether the graph of R intersects the asymptote, we solve the equation R(x) = 3. =
3x2 - 3x ==3 x2 + x - 12 3x2 - 3x = 3x2 + 3x - 36 -6x = -36 x 6
R(x)
=
The graph intersects the line y = 3 only at x 6, and (6, 3) is a point on the graph of R. Plot the point (6, 3 ) and the line y = 3 using dashes. STEP 5: The zeros of the numerator, 0 and 1 , and the zeros of the denominator, -4 and 3, divide the x-axis into five intervals: ( -00, -4) (-4, 0 ) (0, 1 ) ( 1 , 3 ) (3, (0 ) =
Construct Table 1 1 .
Table 11
-4
.0
.
.3
Interval
(-00, -4)
( -4, 0)
Number chosen
-5
-2
Value of R
R(-5) = 1 1 .25
R(-2) = -1.8
R
Location of graph
Above x-axis
Below x-axis
Above x-axis
Point on graph
( - 5, 1 1 .25)
(-2, - 1 .8)
(0, 1 ) 2
G)
=
1 15
G �) '1
• x
( 1 , 3)
(3, 00)
2
4
R(2) = - 1
R(4) = 4.5
Below x-axis
Above x-axis
(2, - 1 )
(4, 4.5)
Plot the points from Table 1 1 . Figure 31(a) shows the graph we have so far. STEP 6: • Since the graph of R is above the x-axis for x < -4 and only crosses the line y = 3 at (6, 3 ) , as x approaches -00 the graph of R will approach the horizontal asymptote y = 3 from above. • The graph of R will approach the vertical asymptote x = -4 at the top to the left of x = -4 and at the bottom to the right of x -4. • The graph of R will approach the vertical asymptote x = 3 at the bot tom to the left of x 3 and at the top to the right of x 3. • We do not know whether the graph of R crosses or touches the line y 3 at (6, 3 ) . To see whether the graph, in fact, crosses or touches the line y = 3, we plot an additional point to the right of (6, 3 ) . We use x = 7 =
=
=
=
to find R(7)
=
�� < 3. The graph crosses
y =
3 at x
=
6. Because (6, 3)
is the only point where the graph of R intersects the asymptote y = 3, the graph must approach the line y = 3 from below as x � 00. See Figure 31(b). STEP 7: The complete graph is shown in Figure 3 1 (c).
362
CHAPTER 5
Polynomial a nd Rational Functions
Figure 31
x= -4 (-5,11.25)-
I I I I
10
I
-----...1 - - (0,6)" -5
I
_
1(-2,-1.8) I I I I I
x=-4 t (-5,11.25)-
x=3
Y
1.2'15.1..) I
-( --t------ -I \ I
(4,4.5) _ (6,3)
-(2 -1) : (1,0) I
I
-----
(a)
'i
I I
:It
2'15 I
.1..)1 (4,4.5) _ (6,3)
( 6)
\
_
I
if
.
-(2,-1) I
(1,0)
I
Ii
(7,�) 22
5
Y= 3
x
+
-10 (b)
ii� '
x= 3
Y 10
"
:
-
---------:: - - (0, 0) -5
10
1(-2,-1.8)
x = -4
(-5 11 25)
--
I
-5
I I I
-10
x= 3
___ ...1 6:- " -I-(1 -t--------
Y=3
x
5
Y
I
11 45) ( 2' 15):4 :
---1-- - - -- (6,3) - Y=3 -
,
I
(1,0)
(7�)
I 5 ,-1) ' 22 :
x
-10 (c)
II
Exploration
= -,2:-----3 x 2 - 3x
x +x - 1 2
Result Figure 32 shows the graph in connected mode, and Figure 33(a) shows it in dot mode. Neither graph displays clearly the behavior between the two x-intercepts, 0 and 1. Nor do they clearly display the fact that the graph crosses the horizonta l asymptote at (6, 3).To see these parts better, we graph R for -1 oS x oS 2 and - 1 oS Y oS 0.5 [Figure 33(b)] and for 4 oS X oS 60 and 2.5 oS Y oS 3.5. [Figure 34(b)].
Figure 32
Figure 33
10
-----/
�
-10
Graph R(x)
•
/. I
l
-10
\
�
10
:
'
---""",/
10
-10
0.5
.
-1 10
.'"
\\
-1
-10
Connected mode
�
/
/,..
Dot mode
---
10
/
-10
(b)
(a)
Figure 34
3.5
'
.
,
�
10
-10 Dot mode
(a)
2
-
4. 2.5
.
.
.
(b)
.
y= 3
60
SECTION 5.3 The Graph of a Rational Function
363
The new g raphs reflect the behavior produced by the analysis. Furthermore, we observe two turning points, one between 0 and 1 and the other to the right of 4. Rounded to two decimal places, these turn ing points a re (0.52,0.07) and ( 1 1 .48,2.75).
E XA M P L E 5
A n alyzi ng the G raph of a Ratio nal Function with a Hole
R( x )
Analyze the graph of the rational function: Soluti o n
2x2 - 5x + 2 x2 - 4
=
STEP 1: We factor R and obtain
R(x) The domain of R is { x i x
( �) .
=
=1=
( 2x - l ) (x - 2 ) ( x + 2 ) (x - 2 ) -2, x
=1=
2 } . The y-intercept i s R(O)
=
-�.
-
Plot the point 0, STEP 2: In lowest terms,
R(x) =
2x - 1 x + 2
x
=1=
-2
1
.
The graph has one x-mtercept: "2' Near
Plot the point
1 i
(�, ) °
R(x) =
2x - 1 x + 2
�
2x - 1 2 = - (2x - 1 ) 5 1 - + 2 2
showing a line with slope
�.
STEP 3: Look at R in lowest terms. The graph has one vertical asymptote, x = -2, since x + 2 is the only factor of the denominator of R(x) in lowest terms.
Remember though, the rational function is undefined at both x 2 and x = -2. Graph the line x = -2 using dashes. STEP 4: Since the degree of the numerator equals the degree of the denominator, the graph has a horizontal asymptote. To find it, we either use long division or form the quotient of the leading coefficient of the numerator, 2, and the leading coefficient of the denominator, 1 . The graph of R has the horizon tal asymptote y 2. Graph the line y = 2 using dashes. To find out whether the graph of R intersects the horizontal asymptote y = 2, we solve the equation R ( x ) = 2. =
=
R(x)
=
2x - 1 = 2 x + 2 2x - 1 = 2 ( x + 2 ) 2x - 1 = 2x + 4 -1 = 4
Im possible
The graph does not intersect the line y = 2. STEP 5: Look at the given expression for R. The zeros of the numerator and denominator, -2, .!, and 2, divide the x-axis into four intervals: 2
( -00, -2) Construct Table 12.
(2, (0)
364
CHAPTER 5 Polynomial and Rational Functions
.
Table 12
-2
.
( l)
(-00,-2)
Number chosen
-3
Value of R
R(-3)
Location of graph
Above x-axis
Below x-axis
Point on graph
(-3,7)
(-1, -3)
-1
R(-1)
=
G)
• x
2 '
-2,
Interval
=7
.2
1/2
R(l)
-3
(2,00) 3 1
=-
R(3)
3
=1
Above x-axis
Above x-axis
(1'�)
(3,1)
Plot these points. STEP 6: • From Table 12 we know the graph of R is above the x-axis for x < -2. From Step 4 we know the graph of R does not intersect the asymptote y = 2. Therefore, the graph of R will approach y = 2 from above as x � -00 and will approach the vertical asymptote x = -2 at the top from the left. • Since the graph of R is below the x-axis for -2 < x < �, the graph of R 2 will approach x - 2 at the bottom from the right. =
•
Finally, since the graph of R is above the x-axis for x >
�
and does not
intersect the horizontal asymptote y = 2, the graph of R will approach 2 from below as x � 00. See Figure35 ( a) . STEP 7: See Figure35 ( b ) for the complete graph. Since R is not defined at 2, there y =
( %).
is a hole at the point 2, Figure 35
(-3,7).
/1II x=-2 I I
I I I I I
y 8 6
:
I
4
I
I I I I
x=-2
Y 8
6
4
(11) I 1 2 ----------- i io-:::: l)- 2 \---- - (3.�- y 2 -----------,----I (0, -�) �, -4 -3 -2 -1 -4 -3 \1 2 3 x -2 0,0) /
:
' 3
=
"
2
Hole at
(1 1) (2, �)
3
x
1(-1,\-3)-4
(a)
�
Exploration
Graph R ( x )
=
2X2
( %}
the hole at 2,
-
2
x
5x + 2 -
4
.
(b)
•
NOTE The coordinates of the hole were obta ined by evaluating R in lowest terms at 2. R in lowest Do you see
TRACE along the graph.
Did you obtain an ERROR at x = 2 ? Are you convinced that an algebraic analysis of a rational function is required in order to accurately interpret the graph obtained with a g ra phing utility?
terms is
--
2x - 1 , x+2
which, at
x = 2,
As Example 5 shows,the
is
2(2) - 1 3 = -. 2+2 4
real zeros of the denominator of a rational function
give rise to either vertical asymptotes or holes on the graph. m>:: -==:a
Now Work
•
PROBLEM
33
We now discuss the problem of finding a rational function from its graph.
SECTION 5.3 EXAM P L E 6
The Graph of a Rational Function
365
Constructing a Rational Function from Its G raph
Find a rational function that might have the graph shown in Figure 36. Figure 36
x= -5 I 1 1
y 10
x= 2 I I I I I 1 1 1
:
5 1 --------1---1
-15
T
---------
-10
y= 2
15 x
10
-5
-10
The numerator of a rational function R( x)
Solution
=
��;?
in lowest terms determines the
x-intercepts of its graph. The graph shown in Figure 36 has x-intercepts -2 (even multiplicity; graph touches the x-axis) and 5 (odd multiplicity; graph crosses the x-axis). So one possibility for the numerator is p(x) = (x + 2)2(x - 5 ) . The denominator o f a rational function i n lowest terms determines the vertical asymptotes of its graph. The vertical asymptotes of the graph are x = -5 and x = 2. Since R( x) approaches 00 to the left of x = -5 and R(x) approaches - 00 to the right of x = -5, then x + 5 is a factor of odd multiplicity in q(x). Also, because R(x) approaches -00 on both sides of x = 2, then x - 2 is a factor of even mul tiplicity in q(x ) . A possibility for the denominator is q(x) = (x + 5 ) (x - 2 f So far w e have R(x)
-15
(x
+
2)2(x - 5 )
(x + 5 ) (x - 2 ) 2
'
The horizontal asymptote o f the graph given i n Figure 36 i s y = 2, s o w e know that the degree of the numerator must equal the degree of the denominator and the
Figure 37
J
=
quotient of leading coefficients must be
5
I(
I
-5
,------
R( x ) -
10
I
- -
2 ( x + 2)2(x - 5 )
----'--
----'-'--
---'--
( x + 5 ) (x - 2 ) 2
•
R on a g ra p h i n g util ity. S i n ce Fig u re 37 looks s i m i l a r to Fig u re 36, we have fou n d a rational fu nction R for the g ra p h in Fig u re 36.
Check: Figure 37 shows the g ra p h of r.i!i�
2 EXAM P L E 7
i. This leads to
Now Work
PROBLEM
45
Solve Applied Problems Involving Rational Functions F ind ing the Least Cost of a Can
Reynolds M etal Company manufactures aluminum cans in the shape of a right circular cylinder with a capacity of 500 cubic centimeters
(� )
liter ' The top and
bottom of the can are made of a special aluminum alloy that costs 0.05¢ per square centimeter. The sides of the can are made of material that costs 0.02¢ per square centimeter.
366
CHAPTER 5
Polynomial a nd Rational Functions
\?!
(a) Express the cost C of material for the can as a function of the radius r of the can. (b) Find any vertical asymptotes. Discuss the cost of the can near any vertical asymptotes. C (r ). (c) Use a graphing utility to graph the function C (d) What value of r will result in the least cost? (e) What is this least cost? =
(a) Figure 38 illustrates the components of a can in the shape of a right circular cylinder. Notice that the material required to produce a cylindrical can of height h and radius r consists of a rectangle of area 27Trh and two circles, each of area 7Tr 2 . The total cost C (in cents) of manufacturing the can is therefore
Solution
Figure 38
rd r
.
.
Top Area
h h
=
Tir
2
C
La_ t--:eral Surface Area = 2Tirh
=
----'
L__
Area
=
=
Tir2
Cost of the top and bottom + Cost of the side 2(7Tr2) (0.05) + (2 'ITI'h) (0.02) "---v---'
�
"---v---'
Total area
Cost/unit
of top and
area
=
Cost/unit
area of
area
side
bottom
Bottom
�
Total
O.lO'TTr2 + O.04'TTrh
But we have the additional restriction that the height h and radius r must be chosen so that the volume V of the can is 500 cubic centimeters. Since V = 7Tr 2 h, we have 500 500 = 7TI, 2h so h = -
7Tr 2
Substituting this expression for h, the cost C , in cents, as a function of the radius r is
500 20 0.107Tr 3 + 20 0.047Tr-) = 0. 107Tr 2 + - = ----r r 7Tr (b) The only vertical asymptote is r = O. As the radius r of the can gets closer to 0, the cost C of the can gets higher. (c) See Figure 39 for the graph of C = C ( r ) . (d) Using the M INI M UM command, the cost is least for a radius of about 3.17 cen
Figure 39
C (r )
=
0.107Tr2
+
timeters. (c) The least cost is C (3.17 ) I'll
- Now Work
'
�
9.47¢.
•
PROBLEM S5
5.3 Assess Your Understanding
'Are You Prepared?' The answer is given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1.
The intercepts of the equation y =
- 1 -)--
x are x- - 4
____
. (pp. 1 65- 1 66)
Concepts and Vocabulary 2. If the numerator and the denominator of a rational function
have no common factors, the rational function is
____
3. True or False
The graph of a polynomial function some times has a hole.
4. True or False
5. True or False
The graph of a rational function sometimes in tersects a vertical asymptote.
6. True or False
The graph of a rational function sometimes
has a hole.
The graph of a rational function never inter sects a horizontal asymptote.
Skill Building In Problems 7. R(x)
=
7-44,
follow Steps
x + 1 x( A.v + 4)
1
through
7
on page 355 to analyze the graph of each function. 8.
R(x) =
x ( x - 1)( x + 2)
9.
R(x) =
3x + 3 2" _, + 4
SECTION 5.3 2x + 4 x - I
10. R(x)
=
13. P( x)
=
16. G(x)
=
-
19. G( x)
=
- -
22. R (x)
=
--
25. F(x )
=
--
28. R ( x )
-
31. R(x )
_
=
4
X + x2 + 1 ----:=--" \: 2 - 1
=
40. f(x)
=
43. f (x)
=
45.
=
20. G( x)
-4
(x + 1 ) ( x2 - 9)
x2 - 3x x + 2
---
4-
2
3 -
- -
X
4
12. R ( x )
4
X - 1 -? x- - 4
,
. 15. H ( x )
23. H (x)
=
=
3x ? x- - 1
- -
6 r- x - 6
=
18. R ( x )
=
21. R ( x )
=
x3? - 1 r- 9
- -
X - 12 x2 - 4
x2 +
3 ? (x - 1 ) ( r - 4)
x2 - 1
4 - X
- 16
x2 + X - 12 x - 4
=
x2 + 3x + 2 x-I
27. R ( x )
=
29. F(x)
=
x2 + X - 12 x + 2
30. G ( x )
=
---
32. R ( x )
=
(x - 1 ) (x + 2 ) ( x - 3 ) ? x(x - 4)-
33. R ( x )
=
x2 + X - 12 ? x- - x - 6
x;: :-2 + 3 x l0 ? r + 8x + 15
35. R(x )
=
6x2 - 7x - 3 2r - 7x + 6
36 . R ( x ) -
8x2 + 26x + 15 2x2 - x - 1 5
x2 + 5x + 6 x +j
3 8. R( x
39. f ( x )
=
1 x +x
9 2x + x
41. f ( x )
42. f(x)
=
x2 - X - 1 2 x + ) _
x(x - 1)2 ,
(x + 3 ) "
,.,
x +
1 x
3"
45-48,
)
=
=
367
-?:---
=
26. F ( x)
_
-=? ------
x2 + x
X
+
- 30 6
1 ? r +x
44. f(x)
)[-': " l
In Problems
14. Q(x)
�v3 + 1
-
37. R(x)
=
"
x- + 2x x X2 - 4
_ _ _ _ _ _ _ _
34. R ( x ) -
11. R ( x )
The Graph of a Rationa[ Function
=
2x +
x2 - x - 12 x + 1
2x2 +
16
X
9 x
3"
: '�1
find a rational function that might have the given graph. (More than one answer might be possible.)
'�-
46.
---�---- ---f--- y=1 -3 3 x
x 3
x= -3
48.
47.
I
Y
10 8
6
-4
-3
4
-2 x=2
-15
-10
-5
x= 4 I
[ [ [ [ [ [
�-------------y=3 [ [
[
-2 [
[5 [ [
[ -8
[
[
[ -4 [ [ -6
[ [ [ [
1
0
15
20 x
368
CHAPTER 5
Polynomial and Rational Functions
Applications and Extensions
tf0, lLYx
The concentration C of a certain drug in a patient's bloodstream I hours after injection is given by
49. Drug Concentration
C( I )
=
212
1
+ 1
(a) Find the horizontal asymptote of C ( I ) . What happens to the concentration of the drug as I increases? c, (b) Using your graphing utility, graph C ( t ) . , (c) Determine the time at which the concentration is highest. 50. Drug Concentration The concentration C of a certain drug in a patient's bloodstream t minutes after injection is given by C( I )
fil
=
SOt 25
+ 12
(a) Find the horizontal asymptote of C ( I ) . What happens to the concentration of the drug as t increases? (b) Using our graphing utility, graph C ( t ) . . . � (c) Determme the time at which the concentration IS highest.
A rectangular area adjacent to a river is to be fenced in; no fence is needed on the river side. The en closed area is to be 1000 square feet. Fencing for the side par allel to the river is $5 per linear foot, and fencing for the other two sides is $8 per linear foot; the four corner posts are $25 apiece. Let x be the length of one of the sides perpendicular to the river. (a) Write a function C(x ) that describes the cost of the project. (b) What is the domain of C? 6tl (c) Use a graphing utility to graph C(x) . (d) Find the dimensions of the cheapest enclosure. Source: www.uncwil. edulcourseslmalhlllhbIPandRirationa/1 rational.hlml 52. Doppler Effect The Doppler effect (named after Christian Doppler) is the change in the pitch (frequency) of the sound from a source ( s ) as heard by an observer ( 0 ) when one or both are in motion. If we assume both the source and the ob server are moving in the same direction, the relationship is
x
(a) Express the surface area S of the box as a function of x. 'J; (11) Using a graphing utility, graph the function found in part (a). (c) What is the minimum amount of cardboard that can be used to construct the box? (d) What are the dimensions of the box that minimize the surface area? (e) Why might UPS be interested in designing a box that minimizes the surface area? United Parcel Service has con tracted you to design an open box with a square base that has a volume of 5000 cubic inches. See the illustration.
54. Minimizing Surface Area
r' lLYx
51. Minimum Cost
f' f' fa
where
v va
Vs
= = = = =
=
fa
C = ::)
perceived pitch by the observer actual pitch of the source speed of sound in air (assume 772.4 mph) speed of the observer speed of the source
Suppose you are traveling down the road at 45 mph and you hear an ambulance (with siren) coming toward you from the rear. The actual pitch of the siren is 600 hertz (Hz). (a) Write a function f'(v\) that describes this scenario. (b) If f' 620 Hz, find the speed of the ambulance. ;'1 (c) Use a graphing utility to graph the function. (d) Verify your answer from part (b). Source: www. kettering. edul-drusselll 53. Minimizing Surface Area United Parcel Service has con tracted you to design a closed box with a square base that has a volume of 10,000 cubic inches. See the illustration. =
x
(a) Express the surface area S of the box as a function of x. Using a graphing utility, graph the function found in part (a). (c) What is the minimum amount of cardboard that can be used to construct the box? (d) What are the dimensions of the box that minimize the surface area? (e) Why might UPS be interested in designing a box that minimizes the surface area?
lMl (b)
A can in the shape of a right circular cylinder is required to have a volume of 500 cubic centimeters. The top and bottom are made of material that costs 6¢ per square centimeter, while the sides are made of material that costs 4¢ per square centimeter. (a) Express the total cost C of the material as a function of the radius r of the cylinder. ( Refer to Figure 38.) C ( r ) . For what value of r is the cost C a i ,;? (b) Graph C " minimum?
55. Cost of a Can
=
A steel drum in the shape of a right circular cylinder is required to have a vol ume of 100 cubic feet.
56. Material Needed to Make a Drum
SECTION 5.4
(a) Express the amount A of material required to make the drum as a function of the radius r of the cylinder. (b) How much material is required if the drum's radius is 3 feet? (c) How much material is required if the drum's radius is 4 feet?
Polynomial and Rational Inequalities
369
(d) How much material is required if the drum's radius is 5 feet? 'g� (e) Graph A = A(r). For what value of r is A smallest?
Discussion and Writing 57. Graph each of the following functions:
x2-1 x-I X4 1 y=- x-I y=
x=
-
y= y=
60. Create a rational function that has the following characteris
tics: crosses the x-axis at 2; touches the x-axis at -1; one ver tical asymptote at x = -5 and another at x = 6; and one horizontal asymptote, y= 3. Compare your function to a fellow classmate's. How do they differ? What are their similarities?
x3- 1
x -I x5 - 1
-- x-I
1 a vertical asymptote? Why not? What is happening x" - 1 . for x I ? What do you conjecture about y= , x-I n 2: 1 an integer, for x = I ? Is
---
=
58. Graph each of the following functions: y=
- x-I x2
y=
- x-I X4
x6
y=-
x-I
y=
xS -
61. Create a rational function that has the following characteris
tics: crosses the x-axis at 3; touches the x-axis at - 2; one ver tical asymptote, x = 1; and one horizontal asymptote, y= 2. Give your rational function to a fellow classmate and ask for a written critique of your rational function. 62. Create a rational function with the following characteristics:
x-I
What similarities do you see? What differences? 59. Write a few paragraphs that provide a general strategy for
graphing a rational function. Be sure to mention the follow ing: proper, improper, intercepts, and asymptotes.
three real zeros, one of multiplicity 2;y-intercept 1; vertical as ymptotes x = - 2 and x = 3; oblique asymptote y = 2x + l. Is this rational function unique? Compare your function with those of other students. What will be the same as everyone else's? Add some more characteristics, such as symmetry or naming the real zeros. How does this modify the rational function?
'Are You Prepared?' Answers 1.
( �), 0,
(1, 0)
5.4 Polynomial and Rational Inequalities PREPARING FOR THIS SECTION •
Before getting started, review the following:
Solving Inequalities (Section 1 .5, pp. 124--131) Now Work
the 'Are You Prepared?' problem o n page 373.
OBJECTIVES
1 Solve Polynomial Inequal ities (p. 370)
2 Solve Rational I neq u a l ities (p. 372)
In this section we solve inequalities that involve polynomials of degree 3 and higher, as well as some that involve rational expressions. To solve such inequalities, we use the information obtained in the previous three sections about the graph of polyno mial and rational functions. The general idea follows: Suppose that the polynomial or rational inequality is in one of the forms
f(x) < 0
f(x) > 0
f(x)
:5
0
f(x)
2:
0
Locate the zeros of f if f is a polynomial function, and locate the zeros of the nu merator and the denominator if f is a rational function. If we use these zeros to di vide the real number line into intervals, we know that on each interval the graph of f is either above the x-axis [f(x) > 0 ] or below the x-axis [ f(x) < 0]. In other words, we have found the solution of the inequality.
370
CHAPTER 5
Polynomial and Rational Functions
The following steps provide more detail.
Steps for Solving Polynomial and Rational Inequalities STEP 1: Write the inequality so that a polynomial or rational expression f is on
the left side and zero is on the right side in one of the following forms:
f(x) > 0
2:
f(x)
f(x) < 0
0
$
f(x)
0
For rational expressions, be sure that the left side is written as a single quotient and find its domain. STEP 2: Determine the real numbers at which the expression f on the left side equals zero and, if the expression is rational, the real numbers at which the expression f on the left side is undefined. STEP 3: Use the numbers found in Step 2 to separate the real number line into intervals. STEP 4: Select a number in each interval and evaluate f at the number. (a) If the value of f is positive, then f (x) > 0 for all numbers x in the interval. (b) If the value of f is negative, then f (x) < 0 for all numbers x in the interval. If the inequality is not strict, include the solutions of f (x) 0 in the solution set. =
1 EXAM P L E 1
Solve Polynomial Ineq ualities Solving a Polynomial Inequality
Solve the inequality x4 Solution
$
4x2 , and graph the solution set .
STEP 1: Rearrange the inequality s o that 0 is o n the right side.
x4 $ 4x2 X4 - 4x2 $ 0
Subtract 41 from both sides of the inequality.
This inequality is equivalent to the one that we wish to solve. STEP 2: Find the real zeros of
x4 - 4x2
x
=
=
f(x)
o.
X4 - 4x2 x2 ( x2 - 4) x2 (x + 2 ) ( x - 2 ) 0 or x -2 or x =
=
=
=
=
=
x4 - 4x2 by solving the equation 0 0
Factor. Factor.
0 2
Use the Zero-Product Property and solve.
STEP 3: We use these zeros of f to separate the real number line into four intervals:
( -00, -2)
( -2, 0)
(0, 2 )
(2,
(0
STEP 4: Select a number in each interval and evaluate f(x) mine if f(x) is positive or negative. See Table 13. • -2
Table 1 3 Interval
( -00, -2)
Number chosen
-3
Value of f
f(-3)
Conclusion
Positive
=
45
• 0
( -2,0) -1
f(-1)
=
Negative
• 2 (0, 2) 1
-3
f(1)
=
x (2, 00) 3
=
)
-3
Negative
f(3) = 45 Positive
X4 - 4x2 to deter
SECTION 5.4
I [
-2
1
2
0
4
6
8
•
x
•
� !==-
E XA M P L E 2
371
B ased on Table 13, we know that f(x) < 0 for all x in the intervals ( -2, 0 ) U (0, 2 ) , that is, for all x such that -2 < x < 0 or 0 < x < 2. How ever, because the original inequality is not strict, numbers x that satisfy the equation f(x) = x4 - 4x2 = 0 are also solutions of the inequality X4 :::; 4x2 . Thus, we include -2, 0, and 2. The solution set of the given inequality is { x l -2 :::; x :::; 2} or, using interval notation, [ -2, 2 J . Figure 40 shows the graph of the solution set.
Figure 40 -4
Polynomial and Rational Inequal ities
Now Work
5 AND 1 3
PROBlEMS
Solving a Po lynomial I nequality
Solve the inequality X4 > x, and graph the solution set. Solution
STEP 1: Rearrange the inequality so that 0 is on the right side.
X4 > X 4 x - X > 0
S ubtract x from both sides of the inequa lity.
This inequality is equivalent to the one that we wish to solve. STEP 2: Find the real zeros of f(x) = X4 - x by solving X4 - x = O. X4 - X = 0 x(x3 - 1 ) = 0
Factor out x.
x( x - I ) (x2 + X + 1 ) = 0
Factor the difference of two cu bes.
x = 0
or x - I = 0 or x2 + x + 1 = 0
x = 0
or
x = 1
Set each factor equal to zero a n d solve. The equation x2 + x + 1 0 ha s no real solutions. (Do you see why?) =
STEP 3: We use these zeros 0 and 1 to separate the real number line into three
intervals: ( 1 , 00 )
(0, 1 )
( - 00 , 0 )
STEP 4: We choose a number in each interval and evaluate f(x) = x4 - x to deter
mine if f(x) is positive or negative. See Table 14. Table 1 4
.0 (- 00,0)
(0, 1 )
(1,00)
Number chosen
-1
-
2
Value of f
f(- l )
Conclusion
Positive
Interval
Figure 41 I
I
-2 -1
) ( 0
1
• x
1
2
=
f
2
G)
=
Negative
-
7 16
f(2)
=
14
Positive
B ased on Table 14, we know that f(x) > 0 for all x in ( - 00 , 0) U ( 1 , 00 ) , that is, for all numbers x for which x < 0 or x > 1 . Because the orig inal inequality is strict, the solution set of the given inequality is {x i x < 0 or x > I j or, using interval notation, ( - 00 , 0) U ( 1 , 00 ) . Figure 41 shows the graph of the solution set. •
2
� - Now Work
.
PROBLEM
17
372
CHAPTER 5
Polynomial and Rational Functions
2
Solve Rational Inequalities Solving a Rational I nequal ity
E XA M P L E 3
(x + 3 ) ( 2 - x) > 0, and graph the solution set. (x - 1 ) 2
Solve the inequality Solution
STEP 1: The domain of the variable x i s { x I x
form with 0 on the right side.
=f.
I } . The inequality is already in a
(x + 3 ) ( 2 - x)
2 . The real zeros of the numerator of f are -3 (x - 1 ) and 2 ; the real zero of the denominator is l . STEP 3: We use the zeros -3, 1, and 2 , found i n Step 2, to separate the real number line into four intervals:
STEP 2 : Let f(x)
=
(
-
00
,
-3)
( -3, 1 )
(2,
(1, 2)
(0
STEP 4: Select a number in each interval and evaluate f(x)
)
=
(x + 3 ) (2 - x) (x - I f
determine if f(x) is positive or negative. See Table 15. Table
Figure 42 (
-4 -3
2 -1
-
.
lS
@)
a 1 2 3
Interval
(-00, -3)
( -3, 1 )
( 1 , 2)
(2, (0)
Number chosen
-4
0
-
3 2
3
Value of f
f( -4)
Conclusion
Negative
6
- -
flO)
25
=
f(V =
6
Positive
Positive
9
f(3)
=
)(
3
- -
2
Negative
=
•
Now Work
PROBLEM 2 1
Solving a Rational Inequality
Solve the inequality Solution
=
.
Based on Table 15, we know that f(x) > O for all x in ( -3, 1 ) U ( 1 , 2), that is, for all x such that -3 < x < 1 or 1 < x < 2. Because the original inequality is strict, the solution set of the given inequality is { x l -3 < x < 2, x =f. I } or, using interval notation, ( -3, 1 ) U ( 1 , 2 ) . Figure 42 shows the graph of the solution set. Notice the hole at x 1 to indicate that 1 is to be excluded. m=
E XA M P L E 4
.2
-3
to
4x + 5 x + 2
:2:
3, and graph the solution set.
STEP 1: The domain of the variable x is { x i x
=f.
- 2 } . Rearrange the inequality so that 0 is on the right side. Then express the left side as a single quotient.
--( --) -------
4x + 5 - 3 :2: 0 x + 2
4x + 5 X + 2 --- - 3 x + 2 x + 2
:2: 0
4x + 5 - 3x - 6 :2: 0 x + 2 x - I -- :2: 0 x + 2
Su btract 3 from both sides of the inequality. Multiply -3 by
--. x + 2 x + 2
Write as a single quotient. Combine like terms.
The domain of the variable is { x I x
=f.
-2}.
SECTION 5.4
Polynomial and Rational Inequal ities
373
x - I --. The zero of the numerator of f is 1 , and the zero of the x + 2 denominator is -2.
STEP 2: Let f(x)
=
STEP 3: We use the zeros found in Step 2, -2 and
into three intervals: ( - 00 , -2)
1,
( -2, 1 )
to separate the real number line (1,
00
)
STEP 4: Select a number in each interval and evaluate f(x)
determine if f( x ) is positive or negative. See Table 16. Table
16
-2
• x
.
Interval
( -00,-2)
Number chosen
-3
Value of f
f(-3)
Conclusion
Positive
=
( 1 , (0 )
( - 2, 1 )
0
4
4x + 5 - 3 to x + 2
2
f(O) =
1
f(2)
- -
2
1
= 4"
Positive
Negative
B ased on Table 16, we know that f ( x ) > 0 for all x in ( - 00 , -2) U ( 1 , 00 ) , that is, for all x such that x < -2 or x > 1 . Because the original x - I inequality is not strict, numbers x that satisfy the equation f(x) = -- = 0 x + 2 x - I = 0 only if x are also solutions of the inequality. Since 1, we x + 2 conclude that the solution set is { x i x < - 2 or x 2: I } or, using interval notation, ( - 00 , -2) U [ 1 , 00 ) . =
Figure 43 I.. I I ) -4
-2
I
[
I
0
! ); ,
2
4
Figure 43 shows the graph of the solution set. = :::cD1l'l"i:
Now Work
•
PROBLEM 29
5.4 Assess Your Understanding
'Are You Prepared?' The answer is given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1. Solve the inequality: 3 - 4x > 5. Graph the solution set. ( pp. 124- 1 3 1 )
Concepts and Vocabulary 2.
True
or
False
The first step in solving the inequality x2 + 4x
2::
-4 is to factor the expression x2 + 4x.
Skill Building In Problems 3-40, solve each inequality. 3. (x - 5)2(x + 2) < 0 4. (x - 5 ) ( x + 2f > 0 7. x3 - 9x :::; 0
11. (x - 1 ) ( x2 +
X
+ 4)
2::
0
12. (x + 2) (x2
(x - 2)2 x2 - I
2::
-
X4
< 9x2
+ 1)
-
26 .
(x + 5)2 x2 - 4
2::
0
. 13. (x
2::
23.
0
- l ) (x - 2 ) ( x - 3) :::; 0
16. x3 + 2x2 - 3x > 0 19.
X 3 22. -- > 0 x + 1
0
X
15. x3 - 2x2 - 3x > 0 18.
25.
10. 3x3 < -15x2
9. 2x3 > -8x2
14. (x + l ) (x + 2 ) ( x + 3) :::; 0
x + 1 21. -- > 0 x - I
6. x3 + 8x2 < 0
5. x3 - 4x2 > 0
X4
20. x3 >
> 1
(x - 1 ) (x + 1 ) x 6 x
27. 6 x - ) < �
:::; 0
24.
1
(x - 3 ) (x + 2) x - I
12 28. x + - < 7 x
:::; 0
374
CHAPTER 5
Polynomial and Rational Functions
x +2 � 1 x - 4 5 3 34. -- > -x - 3 x + 1
x +4 � 1 x - 2 2 1 33. -- < -x -2 3x - 9 29.
37.
x2( 3 + x ) ( x + 4) (x + 5 ) (x - 1 )
�
0
38.
x -4 1 2x + 4 1 3 36. -- > -x + 2 x + 1
3x - 5 �2 x + 2 2x + 5 x + 1 35. -- > -x + 1 x - I
30.
31.
x(x2 + l ) (x - 2) (x - l ) (x + 1 )
�
0
39.
(3 - x)3(2x + 1 ) 3
x - I
32.
--- �
(2 - x )3 (3x - 2)
40. -----:3:--- < 0
< 0
X + 1
Application and Extensions 41. For what positive numbers will the cube of a number exceed
four times its square? 42. For what positive numbers will the cube of a number be less than the number? 43. What is the domain of the function f(x) = � ? 44. What is the domain of the function f(x) = 45. What is the domain of the function f(x) = 46. What is the domain of the function f ( x ) =
V� - 3�?
) )
x x x x
+ +
2 ? 4 1 ? 4
In Problems 47-50, determine where the graph of f is below the graph of g by solving the inequality f ( x ) � g ( x ) . Graph f and g togethel: 47. f(x) = X4 - 1 48. f ( x) = X4 - 1 g(x)
=
_2x2 + 2
g(x) = x - I
49. f(x) = X4 - 4
50. f ( x ) = X4
g(x) = 3x2
g(x)
=2
- x2
Suppose that the daily cost C of manufac turing bicycles is given by C(x) = 80x + 5000. Then the 80x + 5000 -. . . average dally cost C IS given by C(x) = . How x many bicycles must be produced each day for the average cost to be no more than $100? 52. Average Cost See Problem 5 1 . Suppose that the govern ment imposes a $1000 per day tax on the bicycle manufac turer so that the daily cost C of manufacturing x bicycles is now given by C(x) = 80x + 6000. Now the average daily 80x + 6000 . How many bicycles cost C is given by C(x) = x must be produced each day for the average cost to be no more than $100? 53. Bungee "umping Originating on Pentecost Island in the Pa cific, the practice of a person j umping from a high place har nessed to a flexible attachment was introduced to Western culture in 1 979 by the Oxford University Dangerous Sport 51. Average Cost
Club. One important parameter to know before attempting a bungee j ump is the amount the cord will stretch at the bot tom of the fall. The stiffness of the cord is related to the amount of stretch by the equation K
where
=
2W(S + L) -'---:,--: "":" S2
-
W = weight of the jumper (pounds) K = cord's stiffness
(pounds per foot)
L = free length of the cord (feet)
S = stretch (feet) (a) A ISO-pound person plans to j ump off a ledge attached to a cord of length 42 feet. If the stiffness of the cord is no less than 16 pounds per foot, how much will the cord stretch? (b) If safety requirements will not permit the jumper to get any closer than 3 feet to the ground, what is the mini mum height required for the ledge in part (a)?
Source: American
Institute of Physics, Physics News Update, No. 150, November 5, 1993
According to Newton's Law of uni versal gravitation, the attractive force F between two bodies is given by m 1m F = G .2 2 , where m 1 , m2 are the masses of the two bodies
54. Gravitational Force
r
=
distance between the Two bodies
G = gravitational constant 6.6742 newtons meter2 kilogram- 2
X
10- 1 1
Suppose an object is traveling directly from Earth to the moon. The mass of the Earth is 5.9742 X 1024 kilograms, the mass of the moon is 7.349 X 1022 kilograms and the mean dis tance from Earth to the moon is 384,400 kilometers. For an object between Earth and the moon, how far from Earth is the force on the object due to the moon greater than the force on the object due to Earth? Source: www.solarviews. com;en. wikipedia. org
Discussion and Writing 55. Make up an inequality that has no solution. Make up one that
has exactly one solution. 56. The inequality X4 + 1 < -5 has no solution. Explain why.
. . x +4 57. A student attempted to solve the Inequality -- � 0 by x -3 multiplying both sides of the inequality by x - 3 to get
1} "
'Are You Prepared?, 1.
{i
x x < - 2"
-2
x + 4 � O. This led to a solution of {xix � -4J. Is the stu dent correct? Explain. 58. Write
a rational inequality whose solution set is {xl -3 < x � 5 J .
SECTION 5.5
The Real Zeros of a Polynomial Function
375
5.5 The Real Zeros of a Polynomial Function PREPARING FOR THIS SECTION •
• •
Before getting started, review the following:
Evaluating Functions (Section 3 . 1 , pp. 212-214) Factoring Polynomials (Chapter R, Section R.5, pp. 49-55) Synthetic Division (Chapter R, Section R.6, pp. 57-60) Now Work
the 'Are You Prepared?' problems on page
OBJECTIVES
• •
Polynomial Division (Chapter R, Section R.4, pp. 44-47) Quadratic Formula (Section 1 .2, pp. 102-104)
386.
1 Use the Remainder and Factor Theorems (p. 375) 2
Use Descartes' Rule of Signs to Determine the N u mber of Positive and the N u m ber of Negative Rea l Zeros of a Polynomia l Fu nction (p. 378)
3 U se the Rational Zeros Theorem to List the Potential Rational Zeros of
a Polynomial Fu nction (p. 379)
4 Find the Real Zeros of a Polynom i a l Function (p. 38 1 ) 5 Solve Polynomial Equations (p. 382)
6 Use the Theorem for Bounds on Zeros (p. 383) 7
Use the I ntermed iate Va l u e Theorem (p. 384)
In this section, we discuss techniques that can be used to find the real zeros of a polynomial function. Recall that if r is a real zero of a polynomial function f then f(r) = 0, r is an x-intercept of the graph of f, and r is a solution of the equation f(x) = 0. For polynomial and rational functions, we have seen the importance of the real zeros for graphing. In most cases, however, the real zeros of a polynomial function are difficult to find using algebraic methods. No nice formulas like the quadratic formula are available to help us find zeros for polynomials of degree 3 or higher. Formulas do exist for solving any third- or fourth-degree polynomial equation, but they are some what complicated. No general formulas exist for polynomial equations of degree 5 or higher. Refer to the Historical Feature at the end of this section for more information. 1
Use the Remainder and Factor Theorems
When we divide one polynomial (the dividend) by another (the divisor), we obtain a quotient polynomial and a remainder, the remainder being either the zero poly nomial or a polynomial whose degree is less than the degree of the divisor. To check our work, we verify that ( Quotient) (Divisor) + Remainder
=
Dividend
This checking routine is the basis for a famous theorem called the division algo which we now state without proof.
rithm* for polynomials,
THEOREM
Division Algorithm for Polynomials
If f(x) and g(x) denote polynomial functions and if g(x) is not the zero poly nomial, there are unique polynomial functions q(x) and r(x) such that
r(x) f(x) - q (x) + or f(x) g(x) g(x)
-
-
i
=
q(x)g(x) + r(x) i
i
dividend quotient divisor
i
(1)
remainder
where r( x ) is either the zero polynomial or a polynomial of degree less than that of g(x).
--.J
,', A systematic process in which certain steps are repeated a fin i te number of times is called an algorithm.
For example, long division is an algorithm.
376
CHAPTER 5
Polynomial a n d Rational Functions
In equation (1), f(x) is the dividend, g(x) is the divisor, q(x) is the and r ex ) is the remainder. If the divisor g( x) is a first-degree polynomial of the form
quotient,
c a real number
g(x) = x - c,
then the remainder r ex) is either the zero polynomial or a polynomial of degree O. As a result, for such divisors, the remainder is some number, say R, and we may write
f(x)
=
(x - c) q (x) + R
(2)
This equation is an identity in x and is true for all real numbers x. Suppose that x = c. Then equation (2) becomes
f(c) = (c - c)q(c) + R f(c) = R Substitute f(c) for R in equation (2) to obtain
f(x) We have now proved the REMAINDER THEOREM
E XA M P L E 1
=
(x - c) q (x) + f(c)
Remainder Theorem.
Let f be a polynomial function. If f(x) is divided by x - c, then the remainder is f( c).
--1
U s i ng the Remainder Theorem
Find the remainder if f(x) = x3 - 4x2 (a) x - 3 Solutio n
( 3)
-
5 is divided by
(b) x + 2
(a) We could use long division or synthetic division, but it is easier to use the Re mainder Theorem, which says that the remainder is f(3).
f(3)
=
(3)3 - 4(3) 2 - 5
=
27 - 3 6 - 5 = - 14
The remainder is -14. (b) To find the remainder when f(x) is divided by x ate f( -2).
f(-2) = ( -2) 3 - 4( -2) 2 - 5
=
-8
-
+
2
=
16 - 5
x - ( -2), we evalu =
-29
The remainder is -29.
•
Compare the method used in Example l (a) with the method used in Example 1 of Chapter R, Section R.6. Which method do you prefer? Give reasons.
� COM MENT
A graphing utility provides another way to find the value of a function using the eVALUEate feature. Consult your manual for details. Then check the results of Exam ple 1. •
An important and useful consequence of the Remainder Theorem is the
Factor
Theorem.
FACTOR THEOREM
Let f be a polynomial function. Then x - c is a factor of f (x) if and only if f(c) = o.
--1
The Factor Theorem actually consists of two separate statements: 1. If f(c)
=
0, then x - c is a factor of f(x). 2 . If x - c i s a factor o f f(x), then f(c) = o .
SECTION 5.5
The Real Zeros o f a Polynomial Function
377
The proof requires two parts. Proof
1. Suppose that
2.
f(c) = O. Then, by equation (3), we have
f(x) = (x - c) q (x) for some polynomial q (x). That is, x - c is a factor o f f(x). Suppose that x - c is a factor of f(x). Then there is a polynomial function q such that
f(x) = (x - c) q (x) Replacing x by c, we find that
f(c) = (c - c)q(c) = O · q (c) = 0 •
This completes the proof.
One use of the Factor Theorem is to determine whether a polynomial has a par ticular factor. EXAM P L E 2
U s i n g the Factor Theorem
Use the Factor Theorem to determine whether the function
f (x) = 2x3 - x2 + 2x - 3 has the factor (a) x - I Soluti o n
(b) x + 3
The Factor Theorem states that if f(c) = 0 then x - c is a factor. (a) Because x - I is of the form x - c with c choose to use substitution.
f( l )
=
=
1, we find the value of f( l ) . We
2 ( 1 ) 3 - ( 1 )2 + 2 ( 1 ) - 3 = 2 - 1 + 2 - 3 = 0
By the Factor Theorem, x - I is a factor of f(x). (b) To test the factor x + 3, w e first need t o write i t in the form x - c. Since x + 3 = x - ( - 3 ) , we find the value of f( - 3 ) . We choose to use synthetic division. -3 )2 - 1 2 -6 21 2 - 7 23
-3 -69 -72
Because f( -3) = -72 * 0, we conclude from the Factor Theorem that x - ( -3) x + 3 is not a factor of f(x). =
..m:.: = =-
Now Work
•
PROBLEM
1 1
In Example 2(a), we found that x - I was a factor of f. To write f in factored form, we use long division or synthetic division. Using synthetic division, we find that 1 )2 - 1 2
2
The quotient is q (x) 2x2 write f in factored form as =
+
1
2 1 3
-3 3 0
X + 3 with a remainder of 0, as expected. We can
f (x) = 2 x3 - x2 + 2x - 3 = (x - 1 ) ( 2x2 +
X
+ 3)
378
CHAPTER 5
Polynomial and Rational Functions
The next theorem concerns the number of real zeros that a polynomial function may have. In counting the zeros of a polynomial, we count each zero as many times as its multiplicity. THEOREM
N u m ber of Real Zeros A
polynomial function cannot have more real zeros than its degree.
-.J
Proof The proof is based on the Factor Theorem. If I' is a real zero of a polynomial function f , then fe r ) a and, hence, x - I' is a factor of f(x). Each real zero cor responds to a factor of degree Because f cannot have more first-degree factors • than its degree, the result follows.
1.
=
2
Use Descartes' Rule of Signs to Determine the Number of Positive and the Number of Negative Real Zeros of a Polynomial Function
Descartes' Rule of Signs provides information about the number and location of the real zeros of a polynomial function written in standard form (descending powers of x). It requires that we count the number of variations in the sign of the coeffi cients of f(x) and f( -x). For example, the following polynomial function has two variations in the signs of the coefficients.
f(x)
=
=
-3x7 + 4X4 + 3x2 - 2x - 1 -3x7 + Ox6 + Ox5 + 4X4 + Ox3 + 3x2 - 2x -
�
�
1
+ to -
- to +
Notice that we ignored the zero coefficients in Ox6, Ox5, and Ox3 in counting the number of variations in the sign of f(x). Replacing x by -x, we get f( -x )
=
=
- 3( -x ) 7 + 4( -X )4 + 3( -X )2 - 2( -x ) 3x7 + 4X4 + 3x2 + 2x - 1 �
-1
+ to -
which has one variation in sign. THEOREM
Descartes' Rule of S i gns
Let f denote a polynomial function written in standard form. The number of positive real zeros of f either equals the number of variations in the sign of the nonzero coefficients of f(x) or else equals that number less an even integer. The number of negative real zeros of f either equals the number of variations in the sign of the nonzero coefficients of f( -x) or else equals that number less an even integer. We shall not prove Descartes' Rule of Signs. Let's see how it is used. EXAM P L E 3
U s i n g the Number of Real Zeros Theorem and Descartes' Rule of Signs
Discuss the real zeros of f (x) Solution
=
3x6 - 4x4 + 3x3 + 2x2 - X - 3.
Because the polynomial is of degree 6, by the Number of Real Zeros Theorem there are at most six real zeros. Since there are three variations in the sign of the nonzero
SECTION 5.5
The Real Zeros of a Polynomial Function
379
coefficients of f ( x ) , by Descartes' Rule of Signs we expect either three or one pos itive real zeros. To continue, we look at f( -x ) . f ( -x )
=
3x6 - 4X4 - 3x3 + 2x2 + X - 3
There are three variations in sign, so we expect either three or one negative real zeros. Equivalently, we now know that the graph of f has either three or one pos itive x-intercepts and three or one negative x-intercepts.
•
1i::!l!:O: :::; ;Z: =--=
3
Now Work
21
PROBLEM
Use the Rational Zeros Theorem to List the Potential Rational Zeros of a Polynomial Function
The next result, called the Rational Zeros Theorem, provides information about the rational zeros of a polynomial with integer coefficients. THEOREM
Rational Zeros Theorem
Let f be a polynomial function of degree 1 or higher of the form an
=I- 0,
ao
=I- °
where each coefficient is an integer. If E , in lowest terms, is a rational zero of
q
f, then p must be a factor of ao , and q must be a factor of EXA M P L E 4
an '
.J
L isti ng Potential Rational Zeros
List the potential rational zeros of
f (x) Solution
± 1 , ±2, ±3, ± 6 ±1, ±2
q:
r
r r
r
r
r
2x3 + llx2 - 7 x - 6
Because f has integer coefficients, we may use the Rational Zeros Theorem. First, we list all the integers p that are factors of the constant term ao = -6 and all the integers q that are factors of the leading coefficient a3 = 2. p:
r
=
Factors of - 6
Factors of 2
Now we form all possible ratios E .
q
In Words
For the polynomial function f(x) = 2>1' + 11><" - 7x - 6, we know 5 is not a zero, because 5 is not in the list of potential rational zeros. However, -1 may or may not be a zero.
p
q
3 1 ± 1 ± 2 ±3 ±6, ± "2 ' ± "2
If f has a rational zero, it will be found in this list, whicb contains 12 possibilities.
•
= = Ii<'l!t: ___
Now Work
PROBLEM
33
Be sure that you understand what the Rational Zeros Theorem says: For a poly nomial with integer coefficients, if there is a rational zero, it is one of those listed. It may be the case that the function does not have any rational zeros. Long division, synthetic division, or substitution can be used to test each poten tial rational zero to determine whether it is indeed a zero. To make the work easier, integers are usually tested first. Let 's continue this example.
380
CHAPTER 5
Polynomial a n d Rational Functions
E XA M P L E 5
F i n d i n g the Rational Zeros of a Polynom ial Fu nction
Continue working with Example 4 to find the rational zeros of
[( x)
=
2x3 + 1 1x2 - 7 x - 6
Write [ in factored form. Solution
We gather all the information that we can about the zeros. STEP 1: There are at most three real zeros.
STEP 2: By Descartes ' Rule of Signs, there is one positive real zero. Also, because
[(-x) = -2x3 + 1 1x2 + 7x - 6 there are two negative zeros or no negative zeros. STEP 3: Now we use the list of potential rational zeros obtained in Example
4:
1 3 ± 1 , ±2, ±3, ±6, ± 2" ± 2' We choose to test the potential rational zero 1 using substitution.
[( 1 ) = 2 ( 1 ) 3 + 1 1 ( 1 ? - 7 ( 1 ) - 6 = 2 + 1 1 - 7 - 6
=
0
Since [ ( 1 ) = 0, 1 is a zero and x - I is a factor of f. We can use long divi sion or synthetic division to factor f.
[(x) = 2x3 =
+
1 1 x2 - 7x - 6
(x - 1 ) (2x2 + 13x + 6 )
Now any solution o f the equation 2x2 + 1 3 x + 6 = 0 will b e a zero o f f. Because of this, we call the equation 2X2 + 13x + 6 = 0 a depressed equa tion of [. Since the degree of the depressed equation of [ is less than that of the original polynomial, it is easier to work with the depressed equation to find the zeros of [. STEP 4: The depressed equation 2x2 + 13x + 6 = 0 is a quadratic equation with discriminant b2 - 4ac = 169 - 48 = 121 > O. The equation has two real solutions, which can be found by factoring.
2X2 + 13x + 6 = (2x + l ) (x
+
6)
=
0
2x + 1 = 0 or x + 6 0 1 x= x = -6 2 =
- -
1
The zeros of [ are -6, - 2" and 1 . We know from Step 3 that [(x) = 2x3 + 11x2 - 7 x - 6 can b e written a s [(x) = ( x - 1 ) (2x2 + 13x + 6 ) . Because 2X2 + 13x + 6 = (2x + 1 ) (x + 6), we write [ in factored form as
[(x) = (x - 1 ) (2x + 1) (x
+
6)
•
Notice that the three zeros of [ found in this example are among those given in the list of potential rational zeros in Example 4.
SECTION 5.5 4
T h e Real Zeros o f a Polynomial Function
381
Find the Real Zeros of a Polynomial Function
To obtain information about the real zeros of a polynomial function, follow these steps:
Steps for Finding the Real Zeros of a Polynomial Function STEP 1: Use the degree of the polynomial to determine the maximum num
ber of real zeros.
STEP 2: Use Descartes' Rule of Signs to determine the possible number of
positive zeros and negative zeros. STEP 3: (a) If the polynomial has integer coefficients, use the Rational Zeros Theorem to identify those rational numbers that potentially could be zeros. (b) Use substitution, synthetic division, or long division to test each po tential rational zero. Each time that a zero (and thus a factor) is found, repeat Step 3 on the depressed equation. STEP 4: I n attempting to find the zeros, remember to use (if possible) the fac toring techniques that you already know (special products, factoring by grouping, and so on). EXA M P L E 6
F i n d i ng the Real Zeros of a Polynomial F u n ction
Find the real zeros of f(x) = x5 - 5x4 + 12x3 - 24x2 tored form. Solution
+
32x - 1 6. Write f in fac
We gather all the information that we can about the zeros. STEP 1: There are at most five real zeros. STEP 2: By Descartes' Rule of Signs, there are five, three, or one positive zeros.
There are no negative zeros because
f( -x)
=
-xs - 5x4 - 12x3 - 24x2 - 32x - 16
has no variation i n sign. STEP 3: Because the leading coefficient as 1 and there are no negative zeros, the potential rational zeros are the integers 1, 2, 4, 8, and 16, the positive factors of the constant term, 16. We test the potential rational zero 1 first, using syn thetic division. =
1 h -5 12 -24 32 - 16 8 - 16 16 1 -4 1 -4 8 -16 16 0 The remainder is f( l ) 0, so 1 is a zero and x - I is a factor of f. Using the entries in the bottom row of the synthetic division, we can begin to factor f. f(x) x5 - 5x4 + 12x3 - 24x2 + 32x - 16 =
=
=
( x - 1 ) (x4 - 4x3
+
8x2 - 16x + 16 )
We now work with the first depressed equation:
Ql (X)
=
X4 - 4x3
+
8x2 - 16x
REPEAT STEP 3: The potential rational zeros of
ql
1 first, since it may be a repeated zero of f.
1h -4 1
+
16
=
0
are still 1 , 2, 4, 8, and 16. We test
8 - 16 16 1 -3 5 -11 -3 5 - 11 5
382
CHAPTER 5
Polynomial and Rational Functions
Since the remainder is 5, 1 is not a repeated zero. We try 2 next.
2h -4 8 -16 16 2 -4 8 -16 4 -8 1 -2 o The remainder is f(2) = 0, so 2 is a zero and x - 2 is a factor of f. Again using the bottom row, we find
f(x) = x5 - 5x4 + 12x3 - 24x2 + 32x - 16 = (x - 1 ) (x - 2 ) (x3 - 2x2 + 4x - 8) The remaining zeros satisfy the new depressed equation
Q (X) = x3 - 2X2 + 4x - 8 = 0 2
Notice that Q (X) can be factored using grouping. (Alternatively, you could 2 repeat Step 3 and check the potential rational zero 2.) Then
x3 - 2x2 + 4x - 8 = 0 x2 (x - 2 ) + 4(x - 2 ) = 0 (x2 + 4) (x - 2) = 0 x2 + 4 = 0 or x - 2 = 0 x=2 Since x2 + 4 = 0 has no real solutions, the real zeros of f are 1 and 2, the lat ter being a zero of multiplicity 2. The factored form of f is
f(x) = x5 - 5x4 + 12x3 - 24x2 + 32x - 16 = (x - l ) (x - 2)2(X2 + 4) """ 5 E XA M P L E 7
'" .. - Now Work
PROBLEM 45
Solve Polynomial Equations Solving a Polynomial Equation
Find the real solutions of the equation: Solution
•
x5 - 5x4 + 12x3 - 24x2 + 32x - 16 = 0
The real solutions of this equation are the real zeros of the polynomial function
f(x) = x5 - 5x4 + 12x3 - 24x2 + 32x - 16 Using the result of Example 6, the real zeros of f are 1 and 2. These are the real solutions of the equation
x5 - 5x4 + 12x3 - 24x2 + 32x - 16 = 0 1<11 = __
Now Work
•
PROBLEM 57
In Example 6, the quadratic factor x2 + 4 that appears in the factored form of f is called irreducible, because the polynomial x2 + 4 cannot be factored over the real numbers. In general, we say that a quadratic factor ax2 + bx + c is irreducible if it cannot be factored over the real numbers, that is, if it is prime over the real numbers. Refer to Examples 5 and 6. The polynomial function of Example 5 has three real zeros, and its factored form contains three linear factors. The polynomial func tion of Example 6 has two distinct real zeros, and its factored form contains two dis tinct linear factors and one irreducible quadratic factor.
SECTION 5.5
THEOREM
The Real Zeros o f a Polynomial Function
383
Every polynomial function with real coefficients can be uniquely factored into a product of linear factors and/or irreducible quadratic factors.
...J
We shall prove this result in Section 5.6, and, in fact, we shall draw several addi tional conclusions about the zeros of a polynomial function. One conclusion is worth noting now. If a polynomial with real coefficients is of odd degree, it must contain at least one linear factor. (Do you see why?) This means that it must have at least one real zero. THEOREM
A polynomial function with real coefficients of odd degree has at least one real zero. 6
Use the Theorem for Bounds on Zeros
The search for the real zeros of a polynomial function can be reduced somewhat if bounds on the zeros are found. A number M is a bound on the zeros of a polyno mial if every zero lies between - M and M, inclusive. That is, M is a bound on the zeros of a polynomial f if - M ::;
THEOREM
::; M
any real zero of f
Bounds on Zeros
Let f denote a polynomial function whose leading coefficient is 1 .
f(x) A bound
M
=
x" + all _ l xn- 1 + . . .
+
a l x + aD
on the real zeros of f is the smaller of the two numbers
where Max{ } means "choose the largest entry in { } ."
...J
An example will help to make the theorem clear. E XA M P L E 8
U s i n g the Theorem for F i n d i n g Bounds on Zeros
Find a bound on the real zeros of each polynomial.
Solution
(a) f(x)
=
(b) g(x)
=
x5 + 3x3 - 9x2 + 5 4x5 - 2x3 + 2x2 + 1
(a) The leading coefficient of f is 1.
f(x)
=
x5 + 3x3 - 9x2 + 5
a4 = Q, a3
=
3, a2
=
- 9, al
=
Q, ao
=
5
We evaluate the two expressions in (4).
Max { l , l aol
+
la l l
+ ... +
l an - l l }
1 + Max{ l aol , l a l l , · · · , l an - l l }
=
= =
=
Max { 1 , 1 5 1 + 1 0 1 + 1 -9 1 + 1 3 1 + 1 0 1 } Max { l , 17 } = 17 1 + Max { ISI , 1 0 1 , 1 -9 1 , 131, 1 0 1 } 1 +9
=
10
The smaller of the two numbers, 10 , is the bound. Every real zero of f lies be tween - 10 and 10.
384
C HAPTER 5
Polynomial and Rational Functions
(b) First we write g so that it is the product of a constant times a polynomial whose leading coefficient is 1 .
-
g ( x ) = 4x5
2x3 +
2X2
( . . () .
+ 1
= 4 x5 - � x3 + � x2 + �
2
2
0
4
)
-, 1
1
Next we evaluate the two expressIOns In 4 wIth a = , a 3 = a2 = 4 2 2' 1 and ao = 4 ' al =
0,
{ I � I 101 I � I I-�I I I } { �} � { I � I I I I � I, 1-�1 , 101}
M a x { l , l ao l + l al l + . . . + l an-I I } = M ax 1 ,
= M ax l 1 + M ax{ l ao l , l al l , . . . , l all-l l }
I� COMMENT ..
'4
= 1 + M ax 1
The bounds on the zeros of a polynomial provide good choices for set
+
+
=
+
+ O
4
, O,
3
= 1 +-=-
2
ting Xm in and Xmax of the viewing rectangle. With these choices, all the x-intercepts of the gra ph can be seen. _
The smaller of the two numbers,
5
5
between - - and - .
>
=
7
4
-
2
% , is the bound. Every real zero of g lies
4
Now Work
•
PROBLEM 8 1
Use the Intermediate Value Theorem
The next result, called the Intermediate Value Theorem, is based on the fact that the graph of a polynomial function is continuous; that is, it contains no "holes" or "gaps." THEOREM
I ntermediate Value Theorem
Let f denote a polynomial function. If a < b and if f ( a ) and f (b) are of opposite sign, there is at least one real zero of f between a and b.
-.J
Although the proof of this result requires advanced methods in calculus, it is easy to "see" why the result is true. Look at Figure 44. Figure 44 If ((0) < ° a nd ((b) >
zero between
a
and b.
0, there is a
Y
Y=
f(b)
f(x)
T
f(b)
t a
b
t
x
f(a) f(a)
E XA M P L E 9
1
U s i n g the I ntermediate Val ue Theorem to Locate Real Zeros
Show that f(x) = x5 - x3 - 1 has a zero between 1 and 2.
Solution
We evaluate f at 1 and at 2.
f( l ) =
-1
and f(2) = 23
SECTION 5.5
The Real Zeros of a Polynomial
Function
385
B ecause / ( 1 ) < 0 and /(2) > 0, it follows from the Intermediate Value Theorem that / has a zero between 1 and 2.
•
, "I!l: = = -
Now Work
PROBLEM 89
Let's look at the polynomial / of Example 9 more closely. B ased on Descartes' Rule of Signs, / has exactly one positive real zero. Based on the Rational Zeros Theorem, 1 is the only potential positive rational zero. Since / ( 1 ) =F- 0, we conclude that the zero between 1 and 2 is irrational. We can use the Intermediate Value Theorem to approximate it. The steps to use follow:
Approximating the Real Zeros of a Polynomial Function STEP 1: Find two consecutive integers a and a + 1 such that / has a zero between them.
STEP 2: Divide the interval [a, a + 1 ] into 10 equal subintervals. STEP 3: Evaluate / at each endpoint of the subintervals until the Intermedi
ate Value Theorem applies; this interval then contains a zero.
STEP 4: Repeat the process starting at Step 2 until the desired accuracy is
achieved.
EXAM P L E 1 0
Approximating the Real Zeros of a Polynomial F u nction
Find the positive zero of /(x)
;]
Solution Exploration We examine the polynom i a l ( g iven in Example
1 0. The
Theorem on Bounds of Zeros tells us that ,,;
this
zero
ZERO
or
ROOT,
we find
to be 1 .24 rounded to two deci
mal places. Correct to two decimal places,the zero is 1 .23.
Figure 4S
-2
4
I
J
�T
..-�
/' 2';:1'(0
�
=
/(1.1)
=
-4
The TABLE feature of a
graphing calculator makes the computa tions in the solution to Exam ple 10 a lot • easier.
-
x3 - 1 correct to two decimal places.
x5 - x3 - 1
-1
/ ( 1 .2 )
=
-0.72049
/ ( 1 .3 )
=
-0.23968 0.51593
We can stop here and conclude that the zero is between 1 .2 and 1 .3. Now we divide the interval [ 1 .2, 1.3] into 10 equal subintervals and proceed to evaluate / at each endpoint.
/( 1.21 )
l-:=1. 2�:6�(l�7 ',1:(1
� COMMENT
/ ( 1 .0)
/ ( 1 .20) 2
=
/(x)
2 (see Fig
ure 45), we see that ( has exactly one x
intercept. Using
x5
From Example 9 we know that the positive zero is between 1 and 2. We divide the interval [ 1 , 2 ] into 10 equal subintervals: [ 1 , 1 . 1 ] , [ 1 .1, 1 . 2 ] , [ 1 .2, 1 .3 ] , [ 1 .3, 1.4 ] , [ 1 .4, 1.5J, [ 1 .5, 1.6], [ 1 .6, 1.7], [ 1 .7, 1 . 8 ] , [ 1 .8, 1 .9 ] , [ 1 .9, 2 ] ' Now w e find the value of / at each endpoint until the Intermediate Value Theorem applies.
every zero is between - 2 and 2. If we graph ( using -2 ,,; x
=
/ ( 1 .22)
= �
�
/ ( 1 .23 )
�
- 0.0455613
-0.1778185 / ( 1 .24)
�
0.025001
-0.23968 -0.1131398
We conclude that the zero lies between 1 .23 and 1 .24, and so, correct to two deci mal places, the zero is 1 .23.
•
�::;:s =-:> -
Now Work
PROBLEM
1 0 1
There are many other numerical techniques for approximating the zeros of a polynomial. The one outlined in Example 10 (a variation of the bisection method) has the advantages that it will always work, that it can be programmed rather easily on a computer, and each time it is used another decimal place of accuracy is achieved. See Problem 1 15 for the bisection method, which places the zero in a succession of inter vals, with each new interval being half the length of the preceding one.
386
CHAPTER 5 Polynomial and Rational Functions
ormulas for the sol ution of third- and fou rth-degree polynomial equations exist, and, while not very practical, they do have a n in teresting history. In the 1 500s in Ita ly, mathematical contests were a popular pastime, and persons possessing methods for solving problems kept them se cret. (Solutions that were published were already common knowledge.) Niccolo of Brescia ( 1 499- 1 557), commonly referred to as Tartaglia ("the stammerer"), had the secret for solving cubic (third-degree) equations, which gave him a decided advantage i n the contests. Girolamo Cardano ( 1 50 1 - 1 576) found out that Tartag lia had the secret, and, being inter ested in cubics, he requested it from Tartaglia. The reluctant Tartaglia hesitated for some time, but finally, swearing Cardano to secrecy with midnight oaths by candlelig ht, told him the secret. Cardano then pub-
F
H i storica l Pro b l e m s
Probl ia-Cardano tion andemsshowdevel why iotpis thenot alTartagl together practical.solution of the cubic equa l + by + cy d + px + q b y px q -q. -p, -p -q -p + -q. 1 )q2 p 3 1 -8
1. Show that the general cubic equation
+
can be transformed into an equation of the form ;(3 =
by using the s ubstitution
2. In
x -
=
=
3
the equation x3 + = 0, replace x by H + K. Let + 3HK and show that H3 + K3 =
3. Based on Problem 2, we have the two equations
3HK
Solve for K in 3HK show that
=
=
and substitute i nto H3
=
H
H3 + K3
and
�+
=
2
4
+
K3
=
Ars Magna.
4. Use the solution for H from Problem 3 and the equation
H3 + K3
=
-q
to show that
0 0
-.
=
Ars Magna
lished the solution in his book ( 1 545), g iving Tartaglia the credit but rather compromising the secrecy. Ta rtaglia exploded into bit ter recrimi nations, and each wrote pamphlets that reflected on the other's mathematics, moral character, and a ncestry. The quartic (fourth-deg ree) equation was solved by Cardano's stu dent Lodovico Ferrari, and this solution a lso was included, with credit and this time with permission, in the Attempts were made to solve the fifth-degree equation in similar ways, all of which failed. I n the early 1 800s, P. Ruffini, Niels Abel, and Evariste Galois all found ways to show that it is not possible to solve fifth-degree equations by formula, but the proofs requ i red the intro duction of new methods. Galois's methods eventually developed into a large part of modern a lgebra.
Then
27
[Hint: Look for an eq uation that is quadratic i n form.]
K
=
1 �q )� �; -
+
5. Use the results from Problems 2 to 4 to show that the solution of
x3 +
px q -q p -q p - 1--+ M, 2 3 1 M, 2 3 +
=
0 is
x -
- +- + 4 27
2
-
2
-
- +4 27
6. Use the result of Problem 5 to solve the equation x3 - 6x
-9
=
O.
7. Use a calculator and the result of Problem 5 to solve the equation
x3 + 3x - 1 4
=
O.
8. Use the methods of this chapter to solve the equation x3 + 3x - 1 4 = O.
5.5 Assess Your Understanding
'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Find f( - 1 ) if f ( x )
2x2 - x. (pp. 2 1 0- 2 1 4) 2. (pp. 49-55) 2. Factor the expression 6x2 + x =
-
3. Find the quotient and remainder if 3x4 - 52 + 7x
-
4 is
divided by x - 3. (pp. 44-47 or 57-60) 4. Solve the equation x2 + x - 3 = O. (pp. 1 0 2 -1 04)
Concepts and Vocabulary 5. In the process of polynomial division, ( Divisor) ( Quotient) +
9. True or False
The only potential rational zeros of 2x5 - x3+ x2 - X + 1 are ± 1, ± 2. 10. True or False If f is a polynomial function of degree 4 and if f(2) = 5, then f(x)
6. When a polynomial function f is divided by x - c, the remainder is 7. If a function f, whose domain is all real numbers, is even and if 4 is a zero of f, then is also a zero. 8. True or False Every polynomial function of degree 3 with real coefficients has exactly three real zeros. _ _ _ _ _
=
f(x) x - 2
=
p(x) +
5 x - 2
where p ( x ) is a polynomial of degree 3.
Skill Building In Problems I I -20, use the Remainder Theorem to find the remainder when f(x) is divided by x - c. Then use the Factor Theorem to determine whether x - c is a factor of f ( x ) . 11 .
f ( x ) = 4x3 - 3x2
-
8x+ 4; x - 2
12. f(x)
=
-4x3 + 5x2
+
8; x + 3
SECTION 5.5 13. f(x) = 3x4 - 6x3 - 5x + 10; 15. f(x)
=
14. f(x) = 4X4 - 15x2 - 4;
x - 2
3x6 + 82x3 + 27; x + 3
17. f(x) = 4x6 - 64x4 + x2 - 15; 19. f(x) = 2x4 - x3 + 2x - 1 ;
The Real Zeros of a Polynomial Function
x + 4
1 x - 2"
387
x - 2
16. f(x)
=
2x6 - 18x4 + x2 - 9; x + 3
18. f (x)
=
x6 - 16x4 + x2 - 16; x + 4
20. f(x) = 3x4 + x3 - 3x + 1 ;
x +
�
In Problems 21-32, tell the maximum number of real zeros that each polynomial function may have. Then use Descartes ' Rule of Signs to determine how many positive and how many negative zeros each polynomial function may have. Do not attempt to find the zeros. 21. f(x) = -4x7 + x3 - x2 + 2 24. f(x)
= -3x5 + 4X4 + 2
27. f ( x) = -x4 + x2 - 1
22. f(x) = 5x4 + 2x2 - 6x - 5
23. f ( x ) = 2x6 - 3x2 - X + 1
25. f (x) = 3x3 - 2x2 + X + 2
26. f ( x)
28. f(x) = x4 + 5x3 - 2
29. f(x) = x5 + X4 + x2 + X + 1
31. f (x) = x6 - 1
32. f ( x)
=
=
-x3 - x2 + X + 1
x6 + 1
In Problems 33-44, list the potential rational zeros of each polynomial function. Do not attempt to find the zeros. . 33. f(x) = 3x4 - 3x3 + x2 - X + 1 36. f ( x) = 2x5 - X4 - x2 + 1 39. f(x)
= 6x4 - x2 + 9
42. f(x)
=
34. f(x) = x5 - X4 + 2X2 + 3
35. f(x) = x5 - 6x2 + 9x - 3
37. f(x) = - 4x3 - x2 + X + 2
38. f ( x ) = 6x4 - x2 + 2
40. f(x)
3x5 - x2 + 2x + 18
=
41. f ( x ) = 2x5 - x3 + 2x2 + 12
-4x3 + x2 + X + 6
43. f(x) = 6x4 + 2x3 - x2 + 20
44. f ( x )
=
- 6x3 - x2 + X + 10
In Problems 45-56, use the Rational Zeros Theorem to find all the real zeros of each polynomial function. Use the zeros to factor f over the real numbers. 45. f(x)
= x3 + 2x2 - 5x - 6
46. f(x)
=
x3 + 8x2 + llx - 20
47. f(x) = 2x3 - x2 + 2x - 1
48. f(x) = 2x3 + x2 + 2x + 1
49. f(x) = 2x3 - 4x2 - lOx + 20
50. f(x) = 3x3 + 6x2 - 15x - 30
51. f ( x) = 2X4 + x3 - 7 x2 - 3x + 3
52. f ( x) = 2X4 - x3 - 5x2 + 2x + 2
53. f ( x) = X4 + x3 - 3x2 - X + 2
54. f(x) = x4 - x3 - 6x2 + 4x + 8
55. f(x) = 4X4 + 5x3 + 9x2 + lOx + 2
56. f(x) = 3x4 + 4x3 + 7x2 + 8x + 2
In Problems
5 7-68,
solve each equation in the real number system
.
. 57. x4 - x3 + 2x2 - 4x - 8 = 0
58. 2x3 + 3x2 + 2x + 3
=
0
59. 3x3 + 4x2 - 7x + 2 = 0
60. 2x3 - 3x2 - 3x - 5 = 0
61. 3x3 - x2 - 15x + 5 = 0
62. 2x3 - l lx2 + lOx + 8
63. X4 + 4x3 + 2x2 - X + 6 = 0
64. X4 - 2x3 + lOx2 - 18x + 9
,
65. XO -
8 2 x2 + x + 1 "3 "3
=
0
67. 2x4 - 19x3 + 57x2 - 64x + 20 = 0
=
0 =
0
3 2
66. x3 + -x2 + 3x - 2 = 0 68. 2x4 + x3 - 24x2 + 20x + 16
=
0
388
CHAPTER 5
In Problems
69-80,
Polynomial and Rational Functions
graph each polynomial function.
69. f(x) = x3 + 2x2 - 5x - 6 72. f(x)
=
70. f(x)
=
76. f(x)
=
80. f(x) = 4x5 + 12x4 - X - 3
In Problems 81-88, find bounds on the real zeros of each polynomial function . '- 81. f(x) = X4 - 3x2 - 4 82. f(x) 83. f ( x)
84. f ( x)
X4 + x3 - x - I
85. f(x) = 3x4 + 3x3 - x2 - 12x - 12 87. f(x)
=
In Problems 89. f(x)
=
4x5 - X4 + 2x3 - 2X2
+
2x3 - x2 + 2x - 1
77. f(x) = X4 + x3 - 3x2 - X + 2
4X4 + 15x2 - 4
79. f(x) = 4x5 - 8x4 - X + 2
78. f(x) = X4 - x3 - 6x2 + 4x + 8
=
74. f(x) = X4 - 3x2 - 4
73. f ( x) = X4 + x2 - 2
2x3 + x2 + 2x + 1
75. f(x) = 4X4 + 7x2 - 2
=
71. f(x)
x3 + 8x2 + 1 1x - 20
= =
X4 - 5x2 - 36 X4 - x3 + x - I
86. f(x) = 3x4 - 3x3 - 5x2 + 27x - 36
88. f(x) = 4x5 + x4 + x3 + x2 - 2x - 2
x - I
use the Intermediate Value Theorem to show that each polynomial function has a zero in the given interval. 90. f ( x ) = x4 + 8x3 - x2 + 2 ; [ - 1 , OJ 8x4 - 2x2 + 5x - 1; [O, l J
89-94,
91. f ( x ) = 2x3 + 6x2 - 8x + 2;
[ -5, -4J
93. f ( x ) = x5 - X4 + 7x3 - 7x2 - 18x + 18;
[ 1.4, 1.5J
92. f(x)
=
3x3 - LOx + 9;
94. f(x)
=
xS - 3x4 - 2x3
[ -3, -2J +
6x2 + X + 2;
[ 1. 7 , 1.8J
In Problems 95-98, each equation has a solution r in the interval indicated. Use the method of Example 1 0 to approximate this solution correct 10 two decimal places. 95. 8x4 - 2x2 + 5x - 1
=
0; O :s r :s
97. 2x3 + 6x2 - 8x + 2 = 0;
1
96. x 4 + 8x3 - x 2 + 2 = 0;
- 5 :s r :s - 4
98. 3x3 - LOx + 9
=
0;
- 1 :s r :s 0
-3 :s r :s -2
In Problems 99- 1 02, each polynomial function has exactly one positive zero. Use the method of Example correct to two decimal places. 99. f(x) = x3 + x2 + X - 4
10
to approximate the zero
100. f(x) = 2X4 + x2 - 1
101. f(x) = 2X4 - 3x3 - 4x2 - 8
102. f(x) = 3x3 - 2x2 - 20
Applications and Extensions 103. Find k such that f(x) = x3 - kx2 + kx + 2 has the
factor x - 2.
104. Find Ie such that f(x) = X4 - kx3 + kx2 + 1 has the
factor x
+
2.
105. What is the remainder when f(x) = 2x20 - 8x l0 + X - 2
is divided by x - I?
106. What is the remainder when f(x) = - 3X 1 7 + x9 - XS + 2x
is divided by x + I?
107. Use the Factor Theorem to prove that x -
x" -
"
e
for any positive integer n.
c
is a factor of
108. Use the Factor Theorem to prove that x + c is a factor of
x" +
e
"
if n
�
1 is an odd integer.
109. One solution of the equation x3 - 8x2 + 16x - 3
=
0 is 3.
Find the sum of the remaining solutions. 110. One solution of the equation x3 + 5x2 + 5x - 2
=
0 is -2.
Find the sum of the remaining solutions. What is the length of the edge of a cube if, after a slice 1 inch thick is cut from one side, the volume remain ing is 294 cubic inches?
111. Geometry
What is the length of the edge of a cube if its volume could be doubled by an increase of 6 centimeters in one edge, an increase of 12 centimeters in a second edge, and a decrease of 4 centimeters in the third edge?
112. Geometry
113. Let f(x) be a polynomial function whose coefficients are in
tegers. Suppose that r is a real zero of f and that the leading coefficient of f is 1. Use the Rational Zeros Theorem to show that r is either an integer or an irrational number.
114. Prove the Rational Zeros Theorem.
[Hint: Let E , where p and q have no common factors except q 1 and -1, be a zero of the polynomial function f(x) = aw'(l1 + al_lx"-1 + . . . + a j X + ao
whose coefficients are all integers. Show that a p" + a,, _ I P"- lq + . . . + a 1 pq"- l + aoq" = 0 " Now, because p is a factor of the first n terms of this equa tion,p must also be a factor of the term aoqll. Since p is not a factor of q (why?),p must be a factor of ao. Similarly, q must be a factor of awJ 115. Bisection Method for Approximating Zeros of a Function f
We begin with two consecutive integers, a and a + 1 , such that f(a) and f(a + 1 ) are of opposite sign. Evaluate f at the midpoint Yi1 1 of a and a + 1. If f(l11l ) = 0, then 111 1 is the zero of f, and we are finished. Otherwise,f ( m l ) is of oppo site sign to either f(a) or f(a + 1 ) . Suppose that it is f(a) and f(m l ) that are of opposite sign. Now evaluate f at the midpoint /112 of a and 111 1 ' Repeat this process until the
SECTION 5.6
desired degree of accuracy is obtained. Note that each itera tion places the zero in an interval whose length is half that of the previous interval. Use the bisection method to solve Prob lems 95-102.
[Hint: The process ends when both endpoints agree to the desired number of decimal places.]
Discussion and Writing 1
-
1
-
+ 3x2 - 6x + 7? Explain.
11 8.
Is
1 17. IS " a zero of f(x) = 4x' - 5x2 - 3x + 1 ? Explain.
1 19.
Is
1 1 6. Is
3" a zero of f(x) = 2x' 3
� �
a zero of f(x) = 2x6 - 5x4 + x3 a zero of f(x) = x7
+
'Are You Prepared?' Answers 1.
2. ( 3x + 2) (2x - 1 )
3
389
Complex Zeros; Fundamental Theorem of Algebra
3. Quotient: 3x3 + 4x2 + 12x + 43; Remainder: 125
4.
X
+ I? Explain.
6x5 - X4 + X + 2? Explain.
{ -I
-
2
VT3 -1
'
+
2
VT3
}
5.6 Complex Zeros; Fundamental Theorem of Algebra PREPARING FOR THIS SECTION •
Before getting started, review the following:
Complex Numbers (Section 1.3, pp. 109-1 1 4) Now Work
•
Quadratic Equations in the Complex Number System (Section 1 .3, pp. 1 14-1 16)
the 'Are You Prepared?' problems on page 393.
OBJECTIVES
1 Use the Conjugate Pairs Theorem (p. 390) 2 Find a Polynomial Fu nction with Specified Zeros (p. 3 9 1 ) 3 Find the Com plex Zeros of a Polynom i a l (p. 392)
In Section 1.2, we found the real solutions of a quadratic equation. That is, we found the real zeros of a polynomial function of degree 2. Then, in Section 1.3 we found the solutions of a quadratic equation in the complex number system. These are referred to as the complex zeros of a polynomial function of degree 2. In Section 5.5 we found the real zeros of polynomial functions of degree 3 or higher. In this section we will find the complex zeros of polynomial functions of degree3 or higher. DEFINITION
A variable in the complex number system is referred to as a complex vari A complex polynomial function f of degree n is a function of the form
able.
(1 )
where all , all - I , . . . , a1 , ao are complex numbers, all * 0, n is a nonnegative in teger, and x is a complex variable. As before, an is called the leading coeffi cient of f. A complex number r is called a complex zero of f if f(r) 0. =
.-I
In most of our work the coefficients in ( 1 ) will be real numbers. We have learned that some quadratic equations have no real solutions, but that in the complex number system every quadratic equation has a solution, either real or complex. The next result, proved by Karl Friedrich Gauss ( 1777-1855) when he was 22 years old,* gives an extension to complex polynomials. In fact, this result is so important and useful that it has become known as the Fundamental Theorem of Algebra. ,;, In all, G a uss gave four d i fferent proofs of this theorem, the first one in 1 799 being the subject of his doctoral dissertation.
390
C HAPTER 5
Polynomial and Rational Functions
FUNDAMENTAL THEOREM OF ALGEBRA
Every complex polynomial function f(x) of degree �� �
n
::::::
1 has at least one
�
We shall not prove this result, as the proof is beyond the scope of this book. However, using the Fundamental Theorem of Algebra and the Factor Theorem, we can prove the following result: THEOREM
Every complex polynomial function f(x) of degree n linear factors (not necessarily distinct) of the form
n
=
f(x)
an (x - r l ) ( x - r ) · 2
. . .
•
::::::
1 can be factored into
(x - rn )
(2)
where an , r1 , r , . . . , rn are complex numbers. That is, every complex polyno 2 mial function of degree n :::::: 1 has exactly n complex zeros, some of which may repeat.
�
Proof
Let
By the Fundamental Theorem of Algebra, f has at least one zero, say r1 . Then, by the Factor Theorem, x - r 1 is a factor, and where q1 (x) is a complex polynomial of degree n - 1 whose leading coefficient is an . Repeating this argument n times, we arrive at
(x - r l ) (x - r2 ) · . . . . (x - rn )qn(x) where qn( x) is a complex polynomial of degree n - n = 0 whose leading coefficient is an . That is, qn( x) anxo = a,P and so f(x) = an (x - r1 ) (x - r ) · . . . ( x - rn ) 2 f(x)
=
=
.
We conclude that every complex polynomial function f(x) of degree n actly n (not necessarily distinct) zeros. 1
::::::
1 has ex
•
Use the Conjugate Pairs Theorem
We can use the Fundamental Theorem of Algebra to obtain valuable information about the complex zeros of polynomials whose coefficients are real numbers.
CONJUGATE PAIRS THEOREM
Let f(x) be a polynomial whose coefficients are real numbers. If r a zero of f, the complex conjugate r = a - bi is also a zero of f.
=
a
+
bi is
�
In other words, for polynomials whose coefficients are real numbers, the com plex zeros occur in conjugate pairs. This result should not be all that surprising since the complex zeros of a quadratic function occurred in conjugate pairs. Proof
Let
where an ' an-I , · . . , a1 , ao are real numbers and an then f(r) = f(a + bi) 0, so =
anr
n
+
an _l ri1 - 1 + . . . + a I r
=f.
+
O. If r
ao
=
=
0
a
+
bi is a zero of f,
SECTION 5.6
Complex Zeros; Fundamental Theorem o f Al gebra
We take the conjugate of both sides to get 0 an yn + an _ 1 yn - 1 + . . . + a[ y + aa all yn + an - I Y" 1 + . . . + a l Y + aa 0 anCr) n + an _ l (n n l + . . . + all' + aa = 0 a n ( n n + a - J ( n n - 1 + . . . + all' + ao ° n
391
=
=
=
The conjugate of a sum equa ls the sum of the conjugates (see Section 1.3). The conjugate of a product equals the product of the conjugates. The conj ugate of a rea l n u m ber equals the real number.
This last equation states that f ( n = 0; that is, I' = a - bi is a zero of f.
•
The importance of this result should be clear. Once we know that, say, 3 + 4i is a zero of a polynomial with real coefficients, then we know that 3 - 4i is also a zero. This result has an important corollary. A polynomial f of odd degree with real coefficients has at least one real zero.-.J
C O ROLLARY
Proof Because complex zeros occur as conjugate pairs in a polynomial with real coefficients, there will always be an even number of zeros that are not real numbers. Consequently, since f is of odd degree, one of its zeros has to be a real number. •
For example, the polynomial f(x) x5 - 3x4 + 4x3 - 5 has at least one zero that is a real number, since f is of degree 5 (odd) and has real coefficients. =
E XA M P L E 1
U s i n g the Conjugate Pairs Theorem
A polynomial f of degree 5 whose coefficients are real numbers has the zeros 1, 5i, a n d 1 + i. Find the remaining two zeros.
Since f has coefficients that are real numbers, complex zeros appear as conjugate pairs. It follows that -5i, the conjugate of 5i, and 1 - i, the conjugate of 1 + i, are the two remaining zeros.
Solution
•
2 E XA M P L E 2
Find a Polynomial Function with Specified Zeros F i n d i n g a P olynom ial F unction Whose Zeros Are G iven
Find a polynomial f of degree 4 whose coefficients are real numbers that has the zeros 1 , 1, and -4 + i. Solution
Since -4 + i is a zero, by the Conjugate Pairs Theorem, -4 - i must also be a zero of f. Because of the Factor Theorem, if f(c) = 0, then x - c is a factor of f(x). So we can now write f as
f(x) = a(x - l ) ( x - l ) [ x - ( -4 + i) ] [ x - ( - 4 - i) ] where a is any real number. Then
f ( x) = a(x - l ) ( x - l ) [ x - ( -4 + i) ] [ x - ( -4 - i) ] = a( x2 - 2x + 1) [ x2 - ( -4 + i)x - ( -4 - i)x + ( -4 + i ) ( - 4 - i)] = a(x2 - 2x
+
1 ) (x2 + 4x - ix + 4x + ix + 16
+
4i - 4i - p )
= a(x2 - 2x + 1 ) ( x2 + 8x + 17 ) = a (x4 + 8x3 + 17x2 - 2x3 - 16x2 - 34x + x 2 + 8x = a ( x4 + 6x� + 2x-? - 26x + 17) 0
+
17) •
392
I
CHAPTER 5 Polynomial and Rational Functions
Exploration
Graph the function ( found i n Example 2 for Q
Figure 46
Result A quick analysis of the polynomial ( tells us what to expect:
=
1 . Does the value of a affect the zeros of ( ? How does
the value of a affect the graph of (? What information a bout ( is sufficient to uniq uely determine a? 50
At most three turning points.
For large I xl . the graph will behave like y
=
x4 .
A repeated real zero at 1 , so the graph will touch the x-axis at 1 . The only x-intercept i s at 1 ; the y-intercept is 1 7.
-5
Figure 46 shows the complete graph. (Do you see why?The graph has exactly three turning points.) The value of a causes a stretch or compression; a reflection also occurs if a < O. The zeros a re not affected. If a ny point other than an x-intercept on the graph of ( is known, then a can be determined. For example, if (2,3) is on the graph, then f( 2 ) 3 a ( 3 7 ) , so a 3/37. Why won't an x-intercept work?
3
=
=
=
Now we can prove the theorem we conjectured in Section 5.5. THEOREM
Every polynomial function with real coefficients can be uniquely factored over the real numbers into a product of linear factors and/or irreducible quadratic factors. Proof Every complex polynomial f of degree n has exactly n zeros and can be factored into a product of n linear factors. If its coefficients are real, those zeros that are complex numbers will always occur as conjugate pairs. As a result, if r = a + bi is a complex zero, then so is r = a - bi. Consequently, when the linear factors x - r and x - r of f are multiplied, we have
(x - r) (x - r) = x2 - ( r + r) x + rr = x2 - 2ax + a2 + b2 This second-degree polynomial has real coefficients and is irreducible (over the real numbers). Thus, the factors of fare either linear or irreducible quadratic factors. • ]1\
3 E XA M P L E 3
.",- Now Work
PROBLEM 1 7
Find the Complex Zeros of a Polynomial F i n d i n g the Complex Zeros of a Polynom ial
Find the complex zeros of the polynomial function
f(x)
=
3 x4
+ 5x3 + 25x2 + 45x - 18
Write f in factored form. Solution
STEP 1: The degree of f is
4. So f will have four complex zeros.
STEP 2: Descartes' Rule of Signs provides information about the real zeros. For this
polynomial, there is one positive real zero. There are three or one negative real zeros because
f(-x) = 3x4 - 5x3 + 25x2 - 45x - 18 has three variations in sign. STEP 3: The Rational Zeros Theorem provides information about the potential
rational zeros of polynomials with integer coefficients. For this polynomial (which has integer coefficients) , the potential rational zeros are
1 ±-
3'
2 ±-
3'
± 1 , ±2, ± 3 , ±6, ± 9, ± 1 8
SECTION 5.6
Complex Zeros; Fundamental Theorem of Algebra
25 45 -18 78 3 8 33 3 8 33 78 60
393
We test 1 first: 1 )3 5
1 is not a zero.
45 - 1 8 5 25 -3 -2 -23 -22 2 23 22 -40 3
- 1 )3
We test -1:
2)3
5 25 45 6 22 94 3 11 47 13 9
We test 2:
- 18
278 260
2 is not a zero.
45 -18 25 2 -54 18 27 -9 0
-2)3
5 -6 3 -1 Since f( -2) = 0, then -2
We test -2:
- 1 is not a zero.
is a zero and x
depressed equation is
3x3 - x2
+
27x - 9
=
+
2 is a factor of f. The
0
STEP 4: We factor the depressed equation by grouping.
3x3 - x2 + 27x - 9 = 0 x2(3x - 1 ) + 9(3x - 1 ) = 0 (x2
+
9 ) (3x - 1)
or =
-3i, x
=
3i
The four complex zeros of The factored form o f f is
f(x)
fare { 3x4
=
=
� = =-
Now Work
0
3x - 1 = 0 1 x=3 1 x=3 or
or
x
=
3(x
-3i, 3i, -2 , 0
25x-?
-
l- and
9 from
Factor out the common factor 3x - 1 .
Apply the Zero-Product Property.
�}.
+
5xJ
+
3i) (x - 3i ) (x
PROBLEM
+
-
Factor l- from 3/J 27x 9.
+
45x - 18 +
( �)
2) x -
•
33
5.6 Assess Your Understanding
'Are You Prepared?' Answers are given at the end of these exercises. If you. get the wrong answel; read the pages listed in red. 1.
Find the sum and the product of the complex numbers 3 - 21 and -3 + 51. (pp. 109- 1 1 4)
2. In the complex number system, solve x2 + 2x + 2 = O.
(pp. 1 l 4--1 16)
Concepts and Vocabulary 3. Every polynomial function of odd degree with real coeffi-
cients will have at least
real zeroes). 4. If 3 + 41 is a zero of a polynomial function of degree 5 with real coefficients, then so is __ __
5. True o r False
A polynomial function of degree n with real coefficients h as exactly n complex zeros. At most n of them are real zeros.
6. True or False
A polynomial function of degree 4 with real coefficients could have -3, 2 + i, 2 - i, and -3 + 5i as its zeros.
394
CHAPTER 5
Polynomial a nd Rational Functions
Skill Building In Problems
7-1 6,
information is given about a polynomial f(x) whose coefficients are real numbers. Find the remaining zeros of f.
7. Degree 3;
zeros: 3, 4 - i
8. Degree 3;
zeros: 4, 3 + i
9. Degree 4;
zeros: i, 1 + i
10. Degree 4;
zeros: 1 , 2, 2 + i
11. Degree 5;
zeros: 1 , i, 2i
12. Degree 5;
zeros: 0, 1 , 2, i
13. Degree 4;
zeros: i, 2, -2
14. Degree 4;
zeros: 2 - i, -i
15. Degree 6;
zeros: 2, 2 + i, -3 - i, 0
16. Degree 6;
zeros: i, 3 - 2i, -2 + i
In Problems 1 7-22, form a polynomial f(x) with real coefficients having the given degree and zeros. Answers will vary depending on the choice of leading coefficient. 17. Degree 4;
zeros: 3 + 2i; 4, multiplicity 2
18. Degree 4;
zeros: i, 1 + 2i
19. Degree 5 ;
zeros: 2 ; -i; 1 + i
20. Degree 6;
zeros: i, 4 - i; 2 + i
21. Degree 4;
zeros: 3, multiplicity 2; -i
22. Degree 5;
zeros: 1 , multiplicity 3; 1 + i
In Problems 23-30, use the given zero to find the remaining zeros of each function. 23. f(x)
= .2 - 4� + 4x - 16;
24. g(x)
zero: 2i
25. f(x)
= 2x4 + 5.2 + 5� + 20x - 12;
27. h(x)
=
29. h( x)
= 3Xi + 2x4 + 15.2 + 10� - 528x - 352;
In Problems
X4 - 9.2 + 21� + 21x - 130;
31-40,
=
.2 + 3� + 25x + 75;
26. h(x) = 3x4 + 5.2 + 25� + 45x - 18;
zero: -2i zero: 3 - 2i zero: -4i
x4 - 7.2 + 14� - 38x - 60;
zero: 3i
28. f(x)
=
30. g(x)
= 2Xi - 3x4 - 5.2 - 15� - 207x + 108;
zero: 1 + 3i zero: 3i
find the complex zeros of each polynomial function. Write f in factored form.
31. f(x)
=
.2
33. f(x)
=
.2 - 8� + 25x - 26
34. f(x)
=
35. f(x)
=
X4 + 5� + 4
36. f(x)
= x4 + 13� + 36
38. f(x)
=
32. f(x) = X4 - 1
- 1
37. f(x) = x4 + 2.2 + 22� + 50x - 75 39. f(x)
zero: -5i
= 3x4 - .2 - 9� + 159x - 52
.2 + 13� + 57x + 85
x4 + 3.2 - 19� + 27x - 252
40. f(x) = 2X4 + .2 - 35x2 - 1 13x + 65
Discussion and Writing In Problems 41 and 42, explain why the facts given are contradictory. 41. f(x) is a polynomial of degree 3 whose coefficients are real
numbers; its zeros are 4 + i, 4 - i, and 2 + i. 42. f(x) is a polynomial of degree 3 whose coefficients are real
numbers; its zeros are 2, i, and 3 + i. 43. f(x) is a polynomial of degree 4 whose coefficients are real
numbers; three of its zeros are 2, 1 + 2i, and 1 - 2i. Explain why the remaining zero must be a real number. 44. f(x) is a polynomial of degree 4 whose coefficients are real
numbers; two of its zeros are -3 and 4 - i. Explain why one of the remaining zeros must be a real number. Write down one of the missing zeros.
'Are You Prepared?' Answers 1. Sum: 3i; product: 1 + 21i
2. -1 - i, -1 + i
CHAPTER R EVI EW
Things to Know Power function (pp. 325-328)
f(x)
=
x",
n
�
2 even
Domain: all real numbers Range: nonnegative real numbers Passes through ( - 1 , 1 ) , (0, 0 ) , ( 1 , 1 ) Even function Decreasing on ( - 00, 0), increasing on (0, 00 )
C hapter Review
f(x) = xn,
n ::=::
3 odd
Domain: all real numbers
395
Range: all real numbers
Passes through ( - 1 , - 1 ) , (0, 0), ( 1 , 1 ) Odd function Increasing on ( - (X),
(X)
)
Polynomial function (pp. 324, 328-339) + ... +
a\ x
+
Domain: aa ,
an "* 0
all real numbers
At most n - 1 turning points End behavior:
Real Zeros of a polynomial function f (p. 329)
Behaves like y = aw'ClI for large I xl
Real Numbers for which f(x) = 0; the real zeros of f are the x-intercepts of the graph of f.
Rational function (p. 344-352)
R(x)
=
p(x) q(x)
Domain:
p, q are polynomial functions.
{x l q(x) "* O}
Vertical asymptotes: With R(x) in lowest terms, if q (r) then x = r is a vertical asymptote.
=
0 for some real number,
Horizontal or oblique asymptotes: See the summary on pages 35 1-352. Remainder Theorem (p. 376)
If a polynomial function f(x) is divided by x -
c,
then the remainder is f(c).
Factor Theorem (p. 376)
x - c is a factor of a polynomial function f ( x ) if and only if f(c) = O.
Descartes' Rule of Signs (p. 378)
Let f denote a polynomial function written in standard form. The number of positive zeros of f either equals the number of variations in sign of the nonzero coefficients of f(x) or else equals that number less some even integer. The number of negative zeros of f either equals the number of variations in sign of the nonzero coefficients of f( - x) or else equals that number less some even integer.
Rational Zeros Theorem (p. 379)
Let f be a polynomial function of degree 1 or higher of the form f(x) = all XII + alI _ \ xlI - \ + . . . + a l x + aa all "* 0, aa "* 0 where each coefficient is an integer. lf E , in lowest terms, is a rational zero of .t, then q p must be a factor of aa, and q must be a factor of all '
Intermediate Value Theorem (p. 384)
Let f be a polynomial function. If a < b and f( a) and f( b) are of opposite sign, then there is at least one real zero of f between a and b.
Fundamental Theorem of Algebra (p. 390)
Every complex polynomial function f(x) of degree complex zero.
Conjugate Pairs Theorem (p. 390)
Let f(x) be a polynomial whose coefficients are real numbers. If r = a of f, then its complex conj ugate r = a - bi is also a zero of f.
n ::=::
1 has at least one +
bi is a zero
Objectives --------� Section
5.1
2
3
4
5.2
2
3
5.3 5.4
2 2
You should be able to . . .
Review Exercises
Identify polynomial functions and their degree (p. 324) Graph polynomial functions using transformations (p. 328) Identify the real zeros of a polynomial function and their multiplicity (p. 329) Analyze the graph of a polynomial function (p. 336)
1-4 5-10 11-18 (a) and (b) 1 1-18
Find the domain of a rational function (p. 344) Find the vertical asymptotes of a rational function (p. 347) Find the horizontal or oblique asymptotes of a rational function (p. 348)
1 9-22 19-22 19-22
Analyze the graph of a rational function (p. 355) Solve applied problems involving rational functions (p. 365)
23-34 89
Solve polynomial inequalities (p. 370) Solve rational inequalities (p. 372)
35 , 36 37-44
396
CHAPTER 5
5.5
45-50
Use the Remainder and Factor Theorems (p. 375) Use Descartes' Rule of Signs to determine the number of positive and the number of negative real zeros of a polynomial function (p. 378)
2 3 4
5· 6
7
5.6
Polynomial a nd Rational Functions
2
3
5 1 , 52
Use the Rational Zeros Theorem to list the potential rational zeros of a polynomial function (p. 379) Find the real zeros of a polynomial function (p. 381 ) Solve polynomial equations (p. 382) Use the Theorem for Bounds on Zeros (p. 383) Use the Intermediate Value Theorem (p. 384)
53-60 55-60 61-64 65-68 69-76
Use the Conjugate Pairs Theorem (p. 390) Find a polynomial function with specified zeros (p. 391) Find the complex zeros of a polynomial (p. 392)
77-80 77-80 81-88
Review Exercises In Problems not. 1.
f(x)
=
1-4,
determine which functions are polynomial functions. For those that are, state the degree. For those that are not, tell why 3x5 + 1 2x-
2. f ( x ) = -
4x5 - 3x2 + 5x - 2
3. f(x)
=
3x2 + 5xl/2 - 1
4. f(x) = 3
[n Problems 5-10, graph each function using transformations (shifting, compressing, stretching, and reflection). Show all the stages. 6. f(x) = -x3 + 3
5. f(x) = (x + 2?
8. f(x) = (x - 1 )4 - 2
9. f(x) = (x - 1 )4 + 2
7. f(x)
=
10. f(x)
=
- (x - 1 )4 ( 1 - x)3
In Problems 1 1-18: (a) Find the x- and y-intercepts of each polynomial function f. (b) Determine whether the graph of f touches or crosses the x-axis at each x-intercept. (c) End behavior: find the power function that the graph off resembles for large values of Ix!. (d) Determine the maximum number of turning points of the graph of .f. (e) Determine the behavior of the graph off near each x-intercept. (f) Put all the information together to obtain the graph off (You may need to locate additional points on the graph.) 13. f(x) = (x - 2)2(x + 4) 12. f(x) = x(x - 2 ) ( x - 4) 1 1 . f(x) = x(x + 2 ) ( x + 4) 1 4. f(x)
=
15. f(x) = -2x3 + 4x2
(x - 2 ) (x + 4f
In Problems 19.
1 9-22,
16. f(x) = -4x3 + 4x
18. f(x) = ( x - 4 ) ( x + 2)2(x - 2 )
17. f(x) = (x - 1 f( x + 3) ( x + 1 )
find the domain of each rational function. Find any horizontal, vertical, o r oblique asymptotes.
x + 2 R ( x ) = -x2 - 9
x2 + 4 x - 2
20. R(x) = --
21. R(x) =
x2 + 3x + 2 (x + 2 ) 2
22. R(x)
=
x3
,-
x' - 1
In Problems 23-34, discuss each rational function following the seven steps given on page 355. 2x - 6 x
23. R(x) = -27. R(x)
=
31. R(x)
=
24. R(x)
�
x + x - 6 x - x - 6 2x4 -:: (x - 1 )2
=
28. R(x) = 32. R(x) =
4 - x x
--
x2 - 6: + 9 xx4 - , x- - 9
25. H ( x ) =
x + 2 x(x - 2 )
29. F(x) =
x3 --X2 - 4 x2 - 4
39.
2x - 6 1 - X
43.
x2 - 8x + 12 x2 16 _
<
36. x3 + 4x2 2: X + 4
3 - 2x 2x + 5
2
40. --- 2: 2 >
0
30. F(x) =
33. G (x) = -2::---- X
In Problems 35-44, solve each inequality. Graph the solution set. 35. x3 + x2 < 4x + 4
26. H ( x) =
- X
- 2
6 x + 3
37. -- 2: 1 41.
44.
(x - 2 ) ( x - 1 ) x - 3 x(x2
+ X
- 2)
x2 + 9x + 20
34. F (x) =
38.
2: 0 ::;
0
42.
1
-2 3x
_
x
_
x2
1
3x3 (x - 1 ) 2 ( x - 1 )2
-'--'
x2 - 1
<
x + 1 x(x - 5 )
1 ::;
0
Chapter Review
In Problems 45-48, find the remainder R when f ( x ) is divided by g ( x ) . Is g a faclOr of f? 45. f(x)
=
8x3 - 3x2 + X + 4;
g(x)
47. f(x) = X4 - 2x3 + 15x - 2;
=
g(x)
x - I
=
x + 2
46. f(x)
=
2x3 + 8x2 - 5x + 5;
48. f(x)
=
X4 - x2 + 2x + 2;
g(x) g(x)
=
=
397
x - 2
x + 1
49. Find the value of f(x) = 12x6 - 8x4 + 1 at x = 4.
50. Find the value of f(x) = - 16x3 + 1 8x2 - X + 2 at x = -2.
In Problems 51 and 52, use Descartes ' Rule of Signs to determine how many positive and negative zeros each polynomial function may have. Do not attempt to find the zeros. 51. f(x)
=
12x8 - x7 + 8x4 - 2x3 +
X
+ 3
52. f(x)
- 6x5 + X4 + 5x3 + X
=
+
1
53. List all the potential rational zeros of f ( x ) = 12x8 - x7 + 6x4 - x3 + X - 3. 54. List all the potential rational zeros of f(x) = - 6x5 + X4 + 2x3 - X + 1.
In Problems 55-60, use the Rational Zeros Theorem 1 0 find all the real zeros of each polynomial function. Use the zeros 1 0 faclOr f over the real numbers. -
55. f(x)
=
x3 - 3x2 - 6x + 8
56. f(x)
57. f(x)
=
4x3 + 4x2 - 7x + 2
58. f(x) = 4x3 - 4x2 - 7x - 2
59. f(x)
=
X4
In Problems
-
6 1--64,
63. 2 X4 + 7 x3 + x2 - 7 x - 3 65. f(x) = x3 - x2 - 4x + 2
In Problems 69. f(x)
=
=
=
In Problems 65-68, find bounds
=
66. f(x)
=
x3 + x2 - l Ox - 5
68. f(x)
=
3x3 - 7x2 - 6x + 14
use the Intermediate Value Theorem to show that each polynomial has a zero in the given interval.
3x3 - x - I;
73-76,
- 6=0
the real zeros of each polynomial function.
2x3 - 7x2 - lOx + 35 69-72,
X
64. 2 X4 + 7x3 - 5x2 - 28x - 12 = 0
0
10
8
60. f(x) = X4 + 6x3 + l lx2 + 12x + 18 62. 3x4 + 3x3 - 1 7x 2 +
0
70. f(x)
[0, 1 ]
71. f(x) = 8x4 - 4x3 - 2x - 1 ;
In Problems
20
x3 - x2 - l Ox
solve each equation in the real number system.
61. 2 X4 + 2x3 - l l x2 + X - 6
67. f(x)
+
4x3 + 9x2 - 20x
=
[0, 1 ]
=
2x3
-
x2 - 3 ;
[1, 2]
72. f(x) = 3x4 + 4x3 - 8x - 2;
[ 1, 2]
each polynomial has exactly one positive zero. Approximate the zero correct to two decimal places. -
73. f(x)
=
x3 - X
2
75. f(x)
=
8x4 - 4x3 - 2x - 1
74. f(x)
=
2x3 - x2 - 3
76. f(x)
=
3x4 + 4x3 - 8x - 2
In Problems 77-80, information is given about a complex polynomial f(x) whose coefficients are real numbers. Find the remaining zeros of f. Then find a polynomial with real coefficients that has the zeros. 77. Degree 3;
zeros: 4 + i, 6
78. Degree 3;
zeros: 3 + 4i , 5
79. Degree 4;
zeros: i, 1 + i
80. Degree 4;
zeros: 1 , 2, 1 + i
In Problems 81-88, find the complex zeros of each polynomial function f(x). Write f in factored form. 81. f(x) = x3 - 3x2 - 6x + 8 83. f(x)
=
4x3 + 4x2 - 7x + 2
85. f(x) = X4 - 4x3 + 9x2 - 20x + 20 87. f(x)
=
2X4 + 2x3 - l l x2 + X
-
6
A can in the shape of a right circular cylin der is required to have a volume of 250 cubic centimeters. (a) Express the amount A of material to make the can as a function of the radius r of the cylinder. (b) How much material is required if the can is of radius 3 centimeters?
89. Making a Can
82. f(x)
=
x3 - x2 - l Ox - 8
84. f(x)
=
4x3 - 4x2 - 7x - 2
86. f(x) = X4 + 6x3 + l l x2 + 12x + 1 8
88. f(x) = 3x4 + 3 x3 - 1 7x 2 +
X
- 6
(c) How much material is required if the can is of radius 5 centimeters? ( d ) Graph A = A ( r ) . For what value of r is A smallest?
398
CHAPTER 5
Polynomial a n d Rational Functions
CHAPTER TEST 1. Graph f(x)
=
(x - 3)4 2 using transformations. 2. For the polynomial function g(x) = 20 + 5 2 - 28x - 15, (a) Determine the maximum number of real zeros that the function may have. (b) Find bounds to the zeros of the function. (c) List the potential rational zeros. (d) Determine the real zeros of g. Factor g over the reals. (e) Find the x- and y-intercepts of the graph of g. (f) Determine whether the graph crosses or touches the x-axis at each x-intercept. (g) Find the power function that the graph of g resembles for large values of Ixl. (h) Determine the behavior of the graph of g near each x-intercept. (i) Put all the information together to obtain the graph of g. -
3 . Find the complex zeros of f(x) = 0 4. Solve 30 + 2x - 1 = 8x2
-
-
42 + 25x - 100.
4 in the complex number
system.
In Problems 5 and 6, find the domain of each function. Find any horizontal, vertical, or oblique asymptotes.
22 - 14x + 24 2 + 2x 3 6. rex) = x+l 2 + 6x - 40 7. Sketch the graph of the function in Problem 6. Label all in tercepts, vertical asymptotes, horizontal asymptotes, and oblique asymptotes. -
5. g (x) =
In Problems conditions.
8
and
9,
write a function that meets the given
8. Fourth-degree polynomial with real coefficients;
-2, 0, 3 + i
zeros:
9. Rational function; asymptotes: y = 2, x = 4;
domain: {xix
-:f-
4, x
-:f-
9}
10. Use the Intermediate Value Theorem to show that the func
tion f(x) = _2x2 - 3x interval [0, 4]. x+2 11. Solve: <2 x-3
+
8 has at least one real zero on the
--
CUMU LATIVE REVI EW 1. Find the distance between the points P = ( 1 , 3 )
Q
=
and
( -4, 2).
2. Solve the inequality 2
2: X
and graph the solution set. 3. Solve the inequality 2 - 3x < 4 and graph the solution set. 4. Find a linear function with slope -3 that contains the point ( -1, 4). Graph the function.
5. Find the equation of the line parallel to the line y = 2x + 1
and containing the point (3, 5). Express your answer in slope-intercept form and graph the line.
6. Graph the equation y = 0.
7. Does the relation { (3 , 6), ( 1 , 3 ) , (2 , 5 ) , (3, 8) }
function? Why or why not? 8. Solve the equation 0 - 62 9. Solve the inequality 3x + 2 tion set.
+
:=;
represent a
8x = O.
+
4x
+
(a) f(3) (c) -f(x) (e)
f(x + It)
h
-
f(x)
=
2 + 5x - 2, find (b) f e -x) (d) f(3x)
h -:f- O
15. Answer the following questions regarding the function
f(x) (a) (b) (c) (d)
=
x+5 x - I
What is the domain of f? Is the point (2, 6) on the graph of f? If x = 3, what is f(x)? What point is on the graph of f? If f(x) = 9, what is x? What point is on the graph of f?
16. Graph the function f(x) = -3x + 7.
5x - 1 and graph the solu
10. Find the center and radius of the circle
2
14. For the function f(x)
y - 2y - 4 = O. Graph the circle.
11. For the equation y = 0 - 9x, determine the intercepts and
test for symmetry.
12. Find an equation of the line perpendicular to 3x - 2y = 7
17. Graph f(x) = 22 - 4x + 1
by determining whether its graph opens up or down and by finding its vertex, axis of sym metry, y-intercept, and x-intercepts, if any.
18. Find the average rate of change of f(x) = x2 + 3x + 1 from
1 to 2. Use this result to find the equation of the secant line containing ( 1 , f( I ) ) and (2, f(2 ) ) .
19. I n parts ( a ) t o (f), use the following graph.
that contains the point ( 1 , 5 ) .
13. Is the following graph the graph o f a function? Why o r why
not? y
-7
x
7x
399
Cha pter Projects
(a) Determine the intercepts. (b) Based on the graph, tell whether the graph is symmetric with respect to the x-axis, the y-axis, and/or the origin. (c) Based on the graph, tell whether the function is even, odd, or neither. (d) List the intervals on which f is increasing. List the inter vals on which f is decreasing. (e) List the numbers, if any, at which f has a local maxi mum. What are these local maxima? (f) List the numbers, if any, at which f has a local mini mum. What are these local minima? 20. Determine algebraically whether the function f(x) is even, odd, or neither. 21. For the function f( x) =
(a) (b) (c) (d)
{
=
_
22. Graph the function f(x)
+
1
+ 4
-3(x
+
1 )2
+
5
mations. 23. Suppose that f(x) = �-? - 5x + 1 and g(x) + (a) Find f g and state its domain.
using transfor = -4x
-
7.
(b) Find !.. and state its domain. g 24. Demand Equation The price p (in dollars) and the quantity x sold of a certain product obey the demand equation P =
1 � 1O
- - :c
+ 150 ,
o :s
x
:s 1500
(a) Express the revenue R as a function of x. (b) What is the revenue if 100 units are sold? (c) What quantity x maximizes revenue? What is the maxi mum revenue? (d) What price should the company charge to maximize revenue?
� � - 9
2x 3x
=
if -3 < x < 2 if x 2:: 2
Find the domain of f. Locate any intercepts. Graph the function. Based on the graph, find the range.
C HAPTER PROJ E CTS
I. Weed Pollen The table shows the tree pollen count for Lexington, Kentucky, from April 1, 2006, to May 16, 2006, during the prime season for tree pollen. Date
Count
4/2
43
4/4
73
4/6
138
4/1 5
264
4/1 7
4/1
4/3
4/5 4/7
4/8
4/9
Date
Count
Date
Count
Date
4/1 1
207
4/20
924
4/30
4/1 3
1 058
4/22
82
1 832
4/25
724
5/4
4/27
110
5/6
76
4/1 0
114
4/1 2
140
4/1 4
981
4/1 6
2058
4/1 8
1 1 88
1 53
203
19
903
1 387
4/1 9 4/21
4/23
4/26
4/28
1 1 62 1 030
1514 1 02
320
Count
Date
Count
1 64
5/9
91
4/29
359
5/1
65
5/2
5/1 0
69
5/1 1
181
5/13
300
5/15
5/3
30
5/5
95
5/7
5/8
91
5/12
63
1 66
80
1 53
71
5/1 4
63
5/1 6
9
13
Source: NatIOnal A llergy Bureau at the Amencan Academy of Allergy, Asthma an d I m munology webSIte. Used with permission.
400
CHAPTER 5
Polynomial a nd Rational Functions
1. Let the independent variable D represent the date, where
D = 1 on 4/1 , D 2 on 4/2, D = 3 on 4/3, and D = 46 on 5/16. Let the dependent variable P represent the num ber of grains per cubic meter of weed pollen. Draw a scat ter diagram of the data using your graphing utility and by hand on graph paper. =
2. In the hand-drawn scatter diagram, sketch a smooth curve
that fits the data. Try to have the curve pass through as many of the points as closely as possible. How many turn ing points are there? What does this tell you about the de gree of the polynomial that could be fit to the data? If the ends of the graph on the left and on the right are extended downward, how many real zeros are indicated by the graph? How many complex? Why? 3. Use a graphing utility to determine the quartic function
of best fit. Graph it on your graphing utility. Does it look
like what you expected from your hand-drawn graph? Explain. 4. Use a graphing utility to determine the cubic function of best fit. Graph it on your graphing utility. Does it look like what you expected from your hand-drawn graph? Explain. s. Use a graphing utility to determine the quadratic function of best fit. Graph it on your graphing utility. Does it look like what you expected from your hand-drawn graph? Explain. 6. Which of the three functions seems to fit the best? Explain your reasoning. 7. Investigate weed pollen counts at other times of the year. ( Go to the National Allergy Bureau at www.aaaai.org.) Would the polynomial you found in part (3) work for other 6-week periods? Why or why not? What about for the same period of time in other years?
The following project is available at the Instructor's Resource Center (IRC):
II.
Theory of Equations The coefficients of a polynomial function can be found if its zeros are known, an advantage of using poly nomials in modeling.
Exponentia l and Logarithmic Functions The McDonald's Scalding Coffee Case
April 3 , 1 996-There
is a lot of hype about the McDonald's scalding coffee case. No one is in favor of frivolous cases or outlandish results; however, it is important to understand some points that were not re ported in most of the stories about the case. McDonald's coffee was not only hot, it was scalding, capable of almost instantaneous de struction of skin, flesh, and muscle. Plaintiff's expert, a scholar in thermodynamics applied to human skin burns, testified that liquids at 180 degrees will cause a full thick ness burn to human skin in two to seven seconds. Other testimony showed that, as the temperature decreases toward degrees, the extent of the burn relative to that temperature decreases exponen tialJy. Thus, if (the) spill had involved coffee at ISS degrees, the liquid would have cooled and given her time to avoid a serious burn.
ISS
SOl/rce: A TLA fact sheet © 1 995, 1996 Consumer Auorneys of CA. Used with
permission of the Association of Trial Lawyers of America.
- See the Chapter Project-
A Look Back
U ntil now, our study of fu nctions has concentrated on polynom ial and ration a l func tions. These fu nction s belong to the class of algebraic functions, that is, fu nctions that can be expressed i n terms of s u m s, d i fferences, products, q u otients, powers, or roots of polynom i als. F u n ctions that a re not a lg ebraic a re termed transcendental (they tra nscend, or go beyond, a lgebraic fu nctions).
A Look Ahead
In this chapter, we study two tra nscendenta l fu nctions: the exponential fu nction and the logarith mic fu nction. These fu n ctions occur freq uently i n a wide variety of applications, such as biology, chemistry, economics, and psychology. The cha pter beg i n s with a d i scussion of com posite, one-to-one, a n d i nverse
functions, concepts needed to see the relati o n s h i p between exponential a n d logarithmic fu nctions.
Outline
6.1 Composite Functions
6.2 One-to-One Functions; Inverse Functions 6.3 Exponential Functions
6.4 logarithmic Functions
6.5 Properties of logarithms
6.6 logarith mic and Exponential Equations 6.7 Compound Interest
6.8 Exponential Growth and Decay Models; Newton's law; logistic Growth
and Decay Models
6.9 Building Exponential, logarithmic, and logistic Models from Data Chapter Review Chapter Test Cumulative Review Cha pter Projects
40 1
402
CHAPTER 6
Exponential and Logarithmic Functions
PREPARING FOR THIS SECTION •
Before getting started, review the following:
Finding Values of a Function (Section 3.1, pp. 212-214)
'\..Now Work
•
Domain of a Function (Section 3.1, 215-216)
pp.
209-2 1 1 and
the 'Are You Prepared?, problems on page 407.
OBJECTIVES
1
Figure 1
1 Form a Composite Function (p. 402)
2 Find the Domain of a Composite Fu n ction (p. 403)
Form a Composite Function
Suppose that an oil tanker is leaking oil and you want to determine the area of the circular oil patch around the ship. See Figure 1. It is determined that the oil is leak ing from the tanker in such a way that the radius of the circular patch of oil around the ship is increasing at a rate of 3 feet per minute. Therefore, the radius r of the oil patch at any time t, in minutes, is given by r(t) = 3t. So after 20 minutes the radius of the oil patch is r(20) = 3(20) = 60 feet. The area A of a circle as a function of the radius r is given by A(r) = 7Tr2 . The area of the circular patch of oil after 20 minutes is A ( 60) = 7T( 60) 2 = 36007T square feet. Notice that 60 = r(20), so A ( 60) = A(r(20 ) ) . The argument of the function A is itself a function! In general, we can find the area of the oil patch as a function of time t by eval uating A(r(t ) ) . The function A(r(t ) ) is a special type of function called a composite
function.
As another example, consider the function y = (2x + 3 f If we write f(u) = u2 and u g(x) = 2x + 3, then, by a substitution process, we can obtain the original function: y = f(u) = f(g(x ) ) (2x + 3 f I n general, suppose that f and g are two functions and that x i s a number in the domain of g. By evaluating g at x, we get g(x). If g(x) is in the domain of f, then we may evaluate f at g(x) and obtain the expression f(g(x ) ) . The correspondence from x to f(g(x) ) is called a composite function f o g. y =
DEFINITION
=
=
Given two functions f and g, the composite function, denoted by f o g (read as "f composed with g"), is defined by
(f
0
g) (x)
=
f(g(x) )
The domain of f o g is the set of all numbers x in the domain of g such that g(x) is in the domain of f.
.J
Look carefully at Figure 2. Only those x ' s in the domain of g for which g( x) is in the domain of f can be in the domain of f o g. The reason is that if g( x) is not in the domain of f then f (g( x ) ) is not defined. Because of this, the domain of f o g is a subset of the domain of g; the range of f o g is a subset of the range of f. Figure 2
Domain o f 9
Range of 9
to 9
SECTION 6.1
403
Com posite Functions
Figure 3 provides a second illustration of the definition. Here x is the input to the function g, yielding g(x). Then g(x) is the input to the function f, yielding f ( g ( x ) ) . Notice that the "inside" function g in f ( g ( x ) ) is done first. Figu re 3
I N PUT x
g(x)
-
-
-
OUTPUT f(g(x))
Let's look at some examples. EXAM P L E 1
Evaluating a Composite F u nction
Suppose that f(x) (a) (f
(a) (f
Solution
a
g)(l)
a
g)(l )
=
2x2 - 3 and g(x)
(b) (g
a
r
f(g( l )
=
g(x) g(1 )
(b) ( g
(c ) (f (d) ( g
a
a
a
f) ( 1 )
g(f( l ) )
=
f) ( -2 ) g) ( - 1 )
=
=
f(x) f(1)
= =
f) ( 1 )
= =
f(4)
=
4x. Find:
(c) (f =
1
a
f) ( -2)
2 . 42 - 3
=
4x f(x) = 22- - 3 4 =
g( - l )
=
1 22- - 3 -1
f(f( -2)
=
1
1
4 · (- 1 )
1
g)(-1)
=
-4
g(x) = 4x
f(5 )
=
a
29
=
2 . 52 - 3
=
47
' f( - 2) = 2(-2)2 - 3 = 5
g(g( - 1 ) )
(d) (g
g( -4)
=
4 · ( -4 )
=
-16
g(-1) = -4
Figure 4 29
� lIm
COMM ENT Graphing
calculators can be used to evaluate com posite functions! Let Y2 g(x) 4x. Then, using a TI-84 Plus g raphing calculator, (f g ) ( 1 ) i s found as shown in Figure 4 . Notice that this i s the result obtained in Example 1(a). _ Yl =
f(x) = 22- - 3 and
I.. = = :::::::a
2 EXAM P L E 2
•
Now Work
=
=
0
PROBLEM 1 1
Find the Domain of a Composite Function F i n d i n g a Composite F u nction and Its Domain
Suppose that f ( x ) Find:
=
x2 + 3x - 1 and g(x)
(a) f o g
=
2x + 3.
(b) g o f
Then find the domain of each composite function. Solution
The domain of f and the domain of g are the set of all real numbers. (a) (f
a
g ) (x)
=
=
f(g(x) 4x2
+
=
f(2x + 3 )
=
1
( 2x + 3 ? + 3 (2x + 3 ) - 1
f(x) = x2 + 3x - 1
12x + 9 + 6x + 9 - 1
=
4x2
+
1 8x + 17
Since the domains of both f and g are the set of all real numbers, the domain of f o g is the set of all real numbers. ':' Consu l t your owner's manual for the appropriate keystrokes.
404
C HAPTER 6
Exponential and Logarithmic Functions
( b ) (g
0
f) (x)
=
=
g(f(x))
t 2(x2 + 3x - 1 )
g( x2 + 3x - 1 )
= 2x2 + 6x - 2 + 3
=
2x + 3
=
g(x)
+3
2x2 + 6x + 1
Since the domains of both f and g are the set of all real numbers, the domain of f o g is the set of all real numbers.
•
(f
0
1. 2.
Look back at Figure 2. In determining the domain of the composite function g) (x) f(g(x) ) , keep the following two thoughts in mind about the input x. =
g(x) must be defined so that any x not in the domain of g must be excluded. f(g( x ) ) must be defined so that any x for which g(x) is not in the domain of f must be excluded. f og
F i n d i n g the Domain of
E XA M P L E 3
Find the domain of (f Solution
0
g) (x) if f(x) =
4 1 _ and g(x) = __. x+2 x-I
_
For (f g ) (x) = f(g(x) ) , we first note that the domain of g is {xix "* I } , so we exclude 1 from the domain of f o g. Next we note that the domain of f is { x i x "* -2 } , which means that g(x) cannot equal - 2 . We solve the equation g(x) = -2 to determine what values of x to exclude. 0
4
--
x-I
=
g(x) = -2
-2
= -2(x - 1 ) 4 = -2x + 2 2x = -2 x = -1
4
We also exclude -1 from the domain of f o g. The domain of f o g is
{ x l x "* - 1 , x "* I } .
Check: For
x
=
1 , g(x)
=
d e f·med .
For x defined. ' I '=>-
=
-1, g( - 1 )
=
4 _ is not defined, so (f x-I
_
4 2
=
-2, and (f
g)(-1)
0
0
=
g) (x)
=
f(g(x ) ) is not
f(g( - 1 ) )
=
. not f( -2) IS •
Now Work
PROBLEM 2 1
F i nding a Composite Function and Its Domain
EXAM P L E 4
Suppose that f(x) = Find:
1 and g(x) x+2
--
( a) f o g
=
4
--
x-I (b ) f f 0
Then find the domain of each composite function. Solution
( a ) (f
0
The domain of f is {xix "* -2} and the domain of g is { x i x "* I } .
g)(x)
=
f(g(x ) )
=
C � 1)
f �
f{x)
=
1 4 -- + 2 i x - I i
= 1
--
x + 2
4
x - I + 2(x - 1 )
x - I 2x + 2
x - 1
M u lti p ly by --.
x - 1
In Example 3, we found the domain of f o g to be {xix "* - 1 , x "* I } .
x-I 2(x + 1 )
SECTION 6.1
Composite Functions
405
We could also find the domain of fog by first looking at the domain of g: {x I x =F- 1}. We exclude 1 from the domain of fog as a result. Then we look at fog and notice that x cannot equal -1, since x = -1 results in division by O. So we also exclude -1 from the domain of fog. Therefore, the domain of fog is { xix =F- -1, x =F- 1}. (b )
(f 0 f) (x) = f(f (x) ) = f x � = 2 i
(
)
1
1 f(x) = - x+2
The domain of f
1
--+ \':
'+2
2
i
x+ 2 1+ 2(x+ 2)
x+ 2 2x+ 5
x+2 x+2
M u ltiply by --.
0 f consists of those x in the domain of f, {x I x =F- -2} , for which 1
f(x) =-- =F- -2 x+ 2
-- = -2 1
x+2 1 = -2(x + 2) 1 = -2x - 4 2x = -5 x =
or, equivalently,
5
2
--
5 x =F- - 2
0 f is x i x =F- -%, x =F- - 2 . also find the domain of f 0 f by
The domain of f
}
{
recognizing that -2 is not in the We could domain of f and so should be excluded from the domain of f 0 f. Then, looking at
f 0 f, we see that x cannot equal -%. Do you see why? Therefore, the domain of
f 0 f is x I x =F- - % , x =F- -2
{
}.
• L'!l!l: =="" .-
Now Work P R O B L E M S 3 3 A N D 3 5
Look back at Example 2, which illustrates that, in general, Sometimes fog does equal g 0 f, as shown in the next example.
EXAMPLE 5
Showing That Two Composite Functions Are Equal
If f(x)
= 3x - 4 and g(x)
1
3 (x+4), show that (f 0 g) (x) = (g 0 f) (x) =x =
for every x in the domain of fog and Solution
fog =F- g 0 f.
(f 0 g) (x) = f(g(x)) x 4 =f
(;) x 4 = 3( ; ) - 4 = x+4 - 4=x
g 0 f.
1 g(x) = -(x+ 4) 3 Substitute
g(x)
=
x+ 4 3
--
into the ru le for
f, f(x)
=
3x - 4.
406
1 1 � �
CHAPTER 6
Exponential and Logarithmic Functions
Seeing the Concept
(g f)(x) =g(f(x) ) =g(3x - 4) 1 ="3 [(3x - 4) + 4J 0
Using a g raphing calculator, let
Yl = f(x) = 3x
-
4
Y2 = g(x) = "3(x+ 4) Y3 = fog, Y4 = 9 f. 1
f(x) = 3x
-
S ubstitute
4
f(x)
into the rule for
g, g(x) = 3" (x + 4). 1
1 =- (3x) =X
0
Using the viewin g window 3 $ x $ 3, 2 $ Y $ 2, graph only Y3 a n d Y4. What do you see?TRACE to verify that Y3 = Y4.
3
-
We conclude that (f
-
0
g)(x) = (g f)(x) = x. 0
•
In Section 6.2, we shall see that there is an important relationship between func tions f and g for which (f g) (x) = (g f)(x) = x. 0
£w="...
0
Now Work PROBLEM 45
Ca l c u l u s Appl ication
Some techniques in calculus require that we be able to determine the components of a composite function. For example, the function H (x) = v:x+l is the compo sition of the functions f and g, where f(x) = \IX and g(x) = x + 1, because
H(x) = (f g)(x) =f(g(x)) 0
=
f(x + 1 )
=
v:x+l.
Finding the Components of a C omposite Function
E XAMPLE 6
f and g such that fog = H if H(x) = (x2+ 1 )50. The function H takes x2+ 1 and raises it to the power 50. A natural way to decompose H is to raise the function g( x) = x2+ 1 to the power 50. If we let f(x) = x50 and g(x) = x2+ 1, then Find functions
S o l ution
g <29 -.!_�(g(X)) 2 (x x g(x) x Figure 5
•__
=
H
+
1
•
=
f(�2
=
H(x) (x2 =
+
+
+
1�0
(f g)(x) =f(g(x)) =f(x2+ 1 ) = (x2+ 1)50 = H(x) 0
1)
1 )50
See Figure
5. •
Other functions f and g may be found for which example, if f(x) = x2 and g(x) = (x2+ 1 ) 25, then
(f
fog =H in Example 6. For
g)(x) =f(g(x) ) = f( (x2 + 1 ?5 ) = [ (x2+ 1 ?5F = (x2+ 1 ) 50 Although the functions f and g found as a solution to Example 6 are not unique, there is usually a "natural" selection for f and g that comes to mind first. E XAMPLE 7
0
F i n d i ng the Components of a C omposite Function
Find functions Solution
1 . x+ 1 f(x) =x1 and g(x) = x+ 1,
f and g such that fog = H if H(x) =
Here H is the reciprocal of we find that
g(x) =x + 1.
If we let
(f g)(x) = f(g(x) ) = f(x+ 1 ) = 0
--
-
1 = H(x) x+ 1
--
• I.i!l=::::=> -
Now Work PROB L EM 5 3
SECTION 6.1
Composite Functions
407
6.1 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.
1. Find f (3 ) if f ( x ) .
-4x
=
.
2. Find f (3x) If f ( x )
=
+ 5x. (pp. 212-214)
2
x2 - 1 . . 3 F'Ind th e d omam 0f th e functIOn f ( x ) = -2.-- . - 25 x (pp. 215-216) •
.
2x2. (pp. 212-214)
4 -
Concepts and Vocabulary
4. If f(x) 5.
=
= x + 1 and g(x) f (g(x»
TrueorFalse
=
x3 , t hen
__
f( x ) · g(x) .
= (x + 1)3
6.
The domain of the composite function g) (x) is the same as the domain of g(x) .
True or False
(f
0
Skill Building
In Problems 7 and 8, evaluate each expression using the values given in the table. 7. , x
f(x)
g(x)
(a ) ( f 8. ,
0
-2
-1
0
2
3
-7
-5
-3
-1
3
5
5
8
3
0
-1
0
3
8
(b) (f
g ) (l )
x
f(x)
g(x)
(a ) ( f
-3
g) ( - 1 )
(c) (g
-3
-2
-1
0
11
9
7
5
-3
-8 0
0
0
g ) (2 )
f ) ( -1)
(d ) (g
0
f) (0)
(e ) (g
(b) g(f ( O» (d ) f (g(4»
f)(2)
(d ) (g
(e) (g
f) ( 3 )
0
y=g(x)
�
�A I
-2
�
In Problems 11-20, for the given functions f and g, find: (b ) ( g
13. f (x) = 4x2 - 3; 15. f(x) = v'X; 1 7. f(x) = I xl;
=
g(x)
g(x) g(x)
3 19. f ( x ) = -- ; x + 1
0
=
=
3 -
x2 + 1
g(x)
=
,� 3r: x
g(x) shown below.
(6,5) (7,5)
r
(5,4)
(1,4)
1 ? x2
(c ) ( f
0
f)(1)
1
2
'7" /, 74
(5'1
.. (8,4)
1)
6
1
I
II
8 x
----y -- = f(x) \ y(2, -2)
(1, -1)
(d) (g g ) ( O ) 12. f ( x ) = 3x + 2; 0
14. f (x)
=
16. f(x) =
2x 1
=
f)(2)
3x2 + 1
I
I'\..
-2 �
0
(f) ( f
g ) (l )
(-1,3) (-1,1)
(a) (f g) (4) 1 1 . f( x ) = 2x; g ( x )
=
0
y
(b) g(f ( 5» ( d ) f (g( 2»
1 0. (a ) g(f ( l» (c) f (g( O»
(f) ( f
g)( -2)
0
-8
-3
In Problems 9 and 10, evaluate each expression using the graphs of y = f( x ) and y 9. (a ) g(f ( - 1 » ( c) f (g( - 1 »
0
-1
3 0
(c) (g
0
3
2
0
(b) ( f
g ) (l )
0
2x2;
g(x)
=
2x2
- 1
g ( x ) = 1 - 3x2
-Vx+l; g ( x )
1 8. f(x) = I x - 21;
g(x)
20. f( x ) = x3/2; g ( x )
=
=
=
X
3x 3
x2 + 2
2 + 1
0
f) ( 3 )
f) ( - 1 )
408
CHAPTER 6
Exponential and Logarithmic Functions
In Problems 21-28, find the domain of the composite function fog. 3 22. f(x) 21. f(x) = __; g(x) = � x -I x =
25. f(x)
= Vx;
27. f(x)
= xl+ 1;
g(x)
=
= 2x+3
26. f(x)
=
x - 2; g(x)
28. f(x)
=
Xl+4; g(x)
=
-x; g(x)
�
(b) g
0
f
(c) f
0
f
=
30. f(x)
= Xl
32. f(x)
= x2+4
34. f(x)
g(x)
� x
=� =�
(d) gog
State the domain of each composite function. 29. f(x) = 2x+3; g(x) 3x
= 3x+ 1;
g(x)
x =_ _; x+ 3
In Problems 29-44, for the given functions f and g , find:
31. f(x)
= --x2
24. f(x)
g(x) =
(a) fog
g(x)
--
= - �4
x ; g(x) x - I
23. f(x)
1 = x+ ; 3
= x+ 1; = x2+ 1;
= 2x - 4
g(x)
= x2+4
33. f(x)
=
xl; g(x)
'.35. f(x)
=
3 ; g(x) x - I
= �2
36. f(x)
1 = --; x+3
37. f(x)
=
x ; g(x) x -I
=-�4
38. f(x)
=
39. f(x)
= Vx;
2x+ 3
40. f(x)
=�;
41. f(x)
= x2 +
42. f(x)
= x2+4;
g(x) =
�
43. f(x)
-5 = xx+1 ;
44. f(x)
- 1 = 2x ; x 2
g(x) =
x +4 2x - 5
<
g(x)
=
1; g(x) =
--
�
= xx+2 -3
g(x)
--
In Problems 45-52, show that (f 1 45. f(x) = 2x; g(x) = 2"x 48. f(x) 51.
f (x)
= x+ 5; = ax+b;
g(x)
=x
g(x)
0
-
g )(x)
=(g
5
1 = -(x - b) a
0
H(x)
56. H(x)
=
_
46. f(x)
= 4x;
49. f(x)
= 2x - 6;
54.
=�
=
2x2+3
= --x2 = �2
g(x)
=
1 - 2x
f)(x) = x. g(x)
= 41 x
g(x)
a"* 0
(2x+ 3)4
g(x)
x ; g(x) x+ 3
=
1
2"(x+6)
52. f(x)
=
-x1 ;
g(x)
(IS, In Problems 53-58, find functions f and g so that fog � 53. ,, .
g(x)
= H. H(x) = (1 +x2) 3
57. H(x)
= 12x+ 11
x3 ; g(x)
47. f(x)
=
50. f(x)
=4
= -\YX
- 3x; g(x)
=
1 3(4 - x)
=-x1 55. H(x)
=�
58. H(x)
=
12x2+ 31
Applications and Extensions
59. If f(x) find (f 61.
= 2x3 0
- 3x2+ 4x - 1 and g(x) g )(x) and (g f )(x).
= 2,
60. If f(x)
0
f(x) = 2x2 + 5 and g(x) = 3x+a , find a so that the graph of fog crosses the y-axis at 23.
11
1 = xx+ , find (f - I --
0
f )(x).
62. If f(x) = 3x2 - 7 and g(x) = 2x+a , find a so that the graph of fog crosses the y-axis at 68.
In Problems 63 and 64, use the functions f and g to find: (a) fog (c) the domain of fog and of g
63. f(x)
=
ax+b; g(x)
(b) g f (d) the conditions for which fog 0
0
= cx+d
f
64. f(x)
=g
0
f
+b = ax ; cx+d --
g(x)
=
mx
SECTION 6.2
65. Surface Area of a Balloon The surface area S (in square meters) of a hot-air balloon is given by
?:
loon as a function of the time t. 66. Volume of a Balloon The volume V (in cubic meters) of the hot-air balloon described in Problem 65 is given by
V(r) =
Yx C = 10
67. Automobile Production The numberN of cars produced at a certain factory in 1 day after t hours of operation is given byN(t) = l OOt - 5 t , 0::; t::; 10. If the cost C (in dollars) of producing N cars is C(N) = 15,000 + S OOON, find the cost C as a function of the time t of operation of the factory.
2
r
68. Environmental Concerns The spread of oil leaking from a tanker is in the shape of a circle. If the radius (in feet) of the spread after t hours is t) = 200Vt, find the area A of the oil slick as a function of the time t.
r(
= - -x + 100 0::; x::; 400
h
600
The volume Vof a right circular cone is
�7Tr2h. If the height is twice the radius, express the vol .) ume V as a function of r. V=
73. Foreign Exchange Traders often buy foreign currency in hope of making money when the currency's value changes. For example, on February 17, 2006, one U.S. dollar could pur chase 0.8382 Euros, and one Euro could purchase 140.9687 yen. Let f(x) represent the number of Euros you can buy with x dollars, and let g(x) represent the number of yen you can buy with x Euros.
(a) Find a function that relates dollars to Euros. (b) Find a function that relates Euros to yen . (c) Use the results of parts (a) and (b) to find a function that relates dollars to yen. That is, find (g f)(x) = 0
g(f(x)) . (d) What i s g(f(1000)) ?
74. Temperature Conversion The function C(F) = "9(F - 32) converts a temperature in degrees Fahrenheit, F, to a tem perature in degrees Celsius, C. The function K(C) C + 273, converts a temperature in degrees Celsius to a temperature in kelvins, K. =
Assuming that all items produced are sold, find the cost Cas a function of the price p. [Hint: Solve for x in the demand equation and then form the composite.] 70. Cost of a Commodity The price p, in dollars, of a certain commodity and the quantity x sold obey the demand equation p
7Tr2h.
5
Suppose that the cost C, in dollars, of producing x units is +
r
72. Volume of a Cone
1
4
Vx C = 25
400
71. Volume of a Cylinder The volume V of a right circular cylin der of height and radius is V = If the height is twice the radius, express the volume Vas a function of I:
69. Production Cost The price p, in dollars, of a certain prod uct and the quantity x sold obey the demand equation p
+
Assuming that all items produced are sold, find the cost Cas a function of the price p.
�7Tr3. If the radius r is the same function of t as i n
Problem 65, find t h e volume Vas a function o f the time t.
1
= -"5 x + 200 0::; x::; 1000
(a) Find a function that converts a temperature in degrees Fahrenheit to a temperature in kelvins. (b) Determine 80 degrees Fahrenheit in kelvins. 75. If f and g are odd functions, show that the composite func tion fog is also odd. 76. If f is an odd function and g is an even function, show that the composite functions f og and g f are both even. 0
IAre You Prepared?1 Answers
1. -21 :; 1"-- '
2. 4 - 18x2
•
�
�
3. {xix
�.'-� \ > ..2".;",
.�,�
*
-5,x
...,.. �'O_�;"�:''''''' '''
*
5}
• "'���"";�-"'', qr;.-.'"f:"'-Oc!:1i'-::;i! ;'� ��<.4,... �,
:;;-
�. >
::,6.2 . One-to_� Ori'e'�Eunctioris; hiverse··Eunctions� �'.rl�'"k,"� ' , "',�<�.",-�"�,,j:;� _::-ff<'"'�.:''!i�'''''''' _ ,.,. '" ;...
,
. � . "- n�.
"
PREPARING FOR THIS SECTION •
409
Suppose that the cost C, in dollars, of producing x units is
S(r) = 47Tr2
where r is the radius of the balloon (in meters). If the radius r is increasing 2with, time t (in seconds) according to the . formula r(t) = -;:; to, t 0, fmd the surface area S of the bal .)
One-to-One Functions; Inverse Functions
"
�
�«
<.�
' -
�
Before getting started , review the following: •
Functions (Section 3.1, pp. 208-219)
Increasing/Decreasing Functions (Section 3.3, pp. 233-234)
Now Work the 'Are You Prepared?' problems on page 419.
OBJECTIVES 1
Determine Whether a Function Is One-to-One (p.410)
2 Determine the Inverse of a Function Defined by a Map or a Set of
Ordered Pairs (p.412) 3 Obtain the Graph of the Inverse Function from the Graph of the
Function (p.414) 4 Find the Inverse of a Function Defined by an Equation (p.416)
410
CHAPTER 6
Exponential a n d Logarithmic Functions
1
Determine Whether a Function Is One-to-One
In Section 3.1, we presented four different ways to represent a function: as (1) a map, (2) a set of ordered pairs, (3) a graph, and (4) an equation. For example, Figures 6 and 7 illustrate two different functions represented as mappings. The function in Figure 6 shows the correspondence between states and their population (in mil lions). The function in Figure 7 shows a correspondence between animals and life expectancy (in years). Figure 6
Population (in millions)
State
Figure 7 Animal
Life Expectancy (in years)
Indiana
6.2
Dog --+----+-- 1 1
Washington
6.1
Cat
South Dakota
0.8
Duck
North Carolina
8.3
Lion
Tennessee
5.8
Pig
-..,----1-_
10
Rabbit ---i---+-__ 7
Suppose we asked a group of people to name the state that has a population of
0.8 million based on the function in Figure 6. Everyone in the group would respond
South D akota. Now, if we asked the same group of people to name the animal whose life expectancy is 1 1 years based on the function in Figure 7, some would respond dog, while others would respond cat. What is the difference between the functions in Figures 6 and 7? In Figure 6, we can see that each element in the domain corresponds to exactly one element in the range. In Figure 7, this is not the case: two different elements in the domain correspond to the same element in the range. We give functions such as the one in Figure 6 a special name. DEFINITION
A function is one-to-one if any two different inputs in the domain correspond to two different outputs in the range. That is, if Xl and Xz are two different inputs of a function f, then f(Xl ) =F f(xz ) ·
.-J
r r r r
r
In Words
A function is not one-to-one if two different inputs correspond to the same output.
Put another way, a function f is one-to-one if no y in the range is the image of more than one x in the domain. A function is not one-to-one if two different ele ments in the domain correspond to the same element in the range. So the function in Figure 7 is not one-to-one because two different elements in the domain, dog and cat , both correspond to 1 1 . Figure 8 illustrates the distinction of one-to-one func tions, non-one-to-one functions, and nonfunctions.
Figure 8
Domain
Range
(a) One-to-one function: Each x in the domain has one and only one image in the range
Domain
Range
(b) Not a one-to-one function: Y1 is the image of both x1 and x2 .
(c) Not a function:
x1 has two images, Y1 and h
SECTION 6.2
EXAMPLE 1
41 1
One-to-One Functions; Inverse Functions
Determ i n i ng Whether a Fu nction I s One-to-On e
Determine whether the following functions are one-to-one. (a) For the following function, the domain represents the age of five males and the range represents their HDL (good) cholesterol (mg/dL). Age
HDL Cholesterol
38
57
42
54
46
34
55
38
�
61 -
(b) {( -2, 6), ( - 1 , 3), (0, 2), ( 1 , 5), (2, 8)} Solution
(a) The function is not one-to-one because there are two different inputs, 55 and 61, that correspond to the same output, 38. (b) The function is one-to-one because there are no two distinct inputs that corre spond to the same output. •
"'/0\
""-
Now Work P R O B L E M S 9 A N D 1 3
For functions defined by an equation y = f(x) and for which the graph of f is known, there is a simple test, called the horizontal-line test, to determine whether f is one-to-one.
Figure 9
THEOREM
Horizontal-line Test
If every horizontal line intersects the graph of a function f in at most one point, then f is one-to-one.
y
...J
((Xl) = ((X2) = h and Xl -:f- X2; ( is n ot a one-to-one function.
EXAMPLE 2
The reason that this test works can be seen in Figure 9, where the horizontal line y = h intersects the graph at two distinct points, (Xl , h) and (X2' h). Since h is the image of both Xl and x2, Xl -:f- X2, f is not one-to-one. Based on Figure 9, we can state the horizontal-line test in another way: If the graph of any horizontal line intersects the graph of a function f at more than one point, then f is not one-to-one.
U s i ng the Horizontal-l i n e Test
For each function, use its graph to determine whether the function is one-to-one. (a) Solution
f(x) = x2
(b)
g(x) = x3
2
(a) Figure 1 0(a) on page 412 illustrates the horizontal-line test for f(x) = x . The horizontal line y = 1 intersects the graph of f twice, at ( 1, 1) and at ( - 1 , 1), so f is not one-to-one. (b) Figure 10(b) illustrates the horizontal-line test for g( x) = x 3 . Because every horizontal line intersects the graph of g exactly once, it follows that g is one to-one.
41 2
CHAPTER 6
Exponential and Logarithmic Functions
Figure 1 0
3 x
-3
-3
(a) A horizontal line intersects the graph twice; is not one-to-one
f
�-
(b) Every horizontal line intersects the graph exactly once; 9 is one-to-one
•
Now Work P R O B L E M 17
Let's look more closely at the one-to-one function g( x) = x3 . This function is an increasing function. Because an increasing (or decreasing) function will always have different y-values for unequal x-values, it follows that a function that is increas ing (or decreasing) over its domain is also a one-to-one function. A function that is increasing on an interval I is a one-to-one function on
THEOREM
1.
A function that is decreasing on an interval I is a one-to-one function on 1.
2
.-J
Determine the Inverse of a Fu nction Defi ned by a Ma p or a Set of Ordered Pa irs
We begin with the definition. DEFINITION
Suppose f is a one-to-one function. Then, to each x in the domain of f, there is exactly one y in the range (because f is a function); and to each y in the range of f, there is exactly one x in the domain (because f is one-to-one). The correspondence from the range of f back to the domain of f is called the inverse function of f.
.-J
Recall that we have a variety of ways of representing functions. We will discuss how to find inverses for all four representations of functions: (1) maps, (2) sets of ordered pairs (3) graphs, and (4) equations. We begin with finding inverses of func tions represented by maps or sets of ordered pairs. EXAMPLE 3
Finding the Inverse of a Function Defined by a Map
Find the inverse of the following function. Let the domain of the function repre sent certain states, and let the range represent the state's population (in millions). State the domain and the range of the inverse function. State
Population (in millions)
Indiana
6.2
Washington
6.1
South Dakota
0.8
North Carolina
8.3
Tennessee
5.8
SECTION 6.2
Solution
One-to-One Functions; Inverse Functions
41 3
The function is one-to-one. To find the inverse function, we interchange the elements in the domain with the elements in the range. For example, the function receives as input Indiana and outputs 6.2 million. So the inverse receives as input 6.2 million and outputs Indiana. The inverse function is shown next. Population (in millions)
State
6.2
Indiana
6.1
Washington
0.8
South Dakota
8.3
North Carolina
5.8
Tennessee
The domain of the inverse function is { 6.2 , 6.1 , O .S , S .3, 5.S }. The range of the in verse function is {Indiana, Washington, South D akota, North Carolina, Tennessee} . •
If the function is a set of ordered pairs ( x, y), then the inverse of of ordered pairs (y, x ) .
f
EXA M P LE 4
f is the set
Finding the I nverse of a Fu nction Defin ed by a Set of Ordered Pairs
Find the inverse of the following one-to-one function:
{ ( -3 , -27), ( -2, -S ) , ( - 1 , - 1 ) , ( 0, 0), ( 1 , 1 ) , (2 , S ) , (3 , 27)} Solution
The inverse of the given function is found by interchanging the entries in each ordered pair and so is given by
{ ( -27 , -3) , ( -S , -2) , ( - 1 , - 1 ) , (0, 0), ( 1 , 1 ) , (S , 2) , (27 , 3)} P"
Figure 1 1
Range of,
Domain of,
D>-
Now Work P R O B L E M S 2 3 A N D 2 7
f
Remember, if is a one-to-one function, it has an inverse function. We use the symbol r l to denote the inverse of f. Figure 1 1 illustrates this definition. Based on Figure 11 , two facts are now apparent about a one-to-one function and its inverse rl.
f
Domain of
1
,1
1 Domain of,-
Range of,-
f = Range of r1
Range of f = Domain of rI
f,
Look again at Figure 1 1 to visualize the relationship. If we start with x, apply and then applyrI, we get x back again. If we start with x, apply rI, and then apply f, we get the number x back again. To put it simply, what does, rl undoes, and vice versa. See the illustration that follows.
f
WARNING Be ca reful!
f-1
is a symbol
for the inverse function of f. The -1 used i n
f-1
is not a n exponent. That is,
f-1 does not mea n the reCiprocal of f; 1 f-\x) is not equal to • f(x)'
•
I
Inpu t x
I
I Input x I
Apply f
-1
Apply f
)
-1
Apply f
I f(x) I
) rl(x)
I
I
Apply f
) rl(f(x) )
I
)
=
xI
I f(f-l ( X) ) = x I
41 4
CHAPTER 6
Exponential and Logarithmic Functions
In other words,
= x where x is in the domain of f
rl(f(X»
f(f-l(x» = x where x is in the domain of f -1
Consider the one-to-one function f(x) = 2x, which multiplies the argument x by 2. The inverse function f -1 undoes whatever f does. So the inverse function of f
is rl(x) = %x, which divides the argument by 2. For example, f(3) = 2(3) = 6 Figure
and r I ( 6) = "2(6) = 3, so r1 undoes what f did. We can verify this by showing that
1
12
f
X
r\f"(x) = rl(2x) =
� f(X)=2X
j
% (2X) = x
and f(r1 (x))
=f
f-\x) = -x 2
r-\2x) = � ( 2x)= x
See Figure 12.
2
X
(� ) =j (� ) = X
x
f(x) = 2x
1
Verifyi ng I nverse Functions
EXAMPLE 5
(a) We verify that the inverse of g(x)
g-l(g(x))
g(g-I(X))
= x3 is g-l(x)
= g-1(x3) = # = x = g(\YX) = (\YX? = x
1 = r1(2x+ 3) = "2[(2x + 3)
.nrl(x»
=f
(
-
.!. . ( X-3» = 2 1:. (x - 3) 2 2
) [
]
\YX by showing that
for all x in the domain of g
for all x in the domain of g-I
(b) We verify that the inverse of f(x) = 2x that
r1(f (x»
=
+
3 is rl(x) = .!..(x-3) by showing 2
1
3J = "2(2x) +
3
for all x in the domain of f
=x
for all x in the domain of r1
= (x - 3)+ 3 = x
•
Verifying I nverse F unctions
EXAMPLE 6
_ is rl(x) = 1:.x
Verify that the inverse of f(x) = _1
x -I
r1(f(x» = x? For what values of x is f(r1(x» The domain of f is {xix
Solution
rl(f(x» f(r1(x»)
"'m
Figure 1 3
y =x
Y b
3
t
= rl
(
_ l _ x-I
(� )
a
b
x
I}
1. For what values of x is
= x?
and the domain of r1 is {xix
) = _ 11_ + 1 = x - I + 1 =
=f +1 =
x-I
1 1 -+ 1-1 x
1 =-=x 1 x
x
O}. Now
oj.
provided x
provided x
oj.
oj.
1
0 •
i - Now Work P R O B L E M 3 1
O bta i n the Gra p h of the Inverse Function fro m the Gra p h of t h e Function
a
oj.
+
Suppose that ( a, b) is a point on the graph of a one-to-one function f defined by y = f(x). Then b = f( a ) . This means that a = rl(b), so (b, a ) is a point on the graph of the inverse function ri. The relationship between the point ( a, b) on f and the point (b, a ) on rl is shown in Figure 13. The line segment containing ( a, b) and
SECTION 6.2
One-to-One Functions; Inverse Functions
41 5
(b, a ) is perpendicular to the line y = x and is bisected by the line y = x. (Do you see why?) It follows that the point (b, a ) on r1 is the reflection about the line y = x of the point ( a, b) on f. THEOREM
The graph of a function j and the graph of its inverse j-l are symmetric with respect to the line y = x.
-.J
Figure 14 illustrates this result. Notice that, once the graph of j is known, the graph of r1 may be obtained by reflecting the graph of j about the line y = x. Figure
14
Y = f(x)
Y
Seeing the Concept
= X, Y2 =
3 x , and Y3 = Vx on a square screen with 3 � x � 3. What do 3 you observe about the graphs of Y2 = x , its inverse Y3 = Vx, and the line Y1 = x? Repeat this Simultaneously graph Y1
-
experiment by simultaneously graphing Y1 with
-6 � x � 3. = x?
=
X,
Y2
=
2x + 3,
and Y3
=
1
"2(x
-
3) on a square screen
Do you see the symmetry of the graph of Y2 and its inverse Y3 with respect to the
line Y1
EXAMPLE 7
Solution
G raphi ng the I nverse Function
The graph in Figure lS(a) is that of a one-to-one function y = j ( x). Draw the graph of its inverse.
We begin by adding the graph of y = x to Figure lS(a). Since the points ( -2, - 1 ) , ( - 1 , 0 ) , and (2, 1 ) are on the graph of j, we know that the points (-1 , -2), (0, - 1 ) , and ( 1 , 2) must be on the graph of rl. Keeping in mil1d that the graph of rJ is the reflection about the line y = x of the graph of j, we can draw ri. See Figure lS(b). Figu re
15
Y 3
Y 3 (1 , 2) Y = f(x)
(2, 1 ) 3 x
-3
( -2 - 1 ) ,
Y=x
(2, 1 ) 3 x
-3
(-2, - 1 )
-3
(a)
1 Y = f- (x)
(b) •
==".. -
Now Work P R O B L E M 4 1
416
CHAPTER 6
Exponential a n d Logarithmic Functions
4
F i n d the Inverse of a Function Defi ned by an Equation
The fact that the graphs of a one-to-one function I and its inverse function 1-1 are symmetric with respect to the line y = x tells us more. It says that we can obtain 1-1 by interchanging the roles of x and y in I. Look again at Figure 14. If I is defined by the equation
y = I(x) then r1 is defined by the equation
x = I(y) The equation x = I(y) defines r1 implicitly. If we can solve this equation for y , we will have the explicit form of rl , that is, y = rl(x) Let's use this procedure t o find the inverse o f I(x) = 2x+ 3. (Since I i s a lin
ear function and is increasing, we know that I is one-to-one and so has an inverse function.) E X A M P LE 8
F i n d i ng the I nverse Fu nction
Find the inverse of I(x) = 2x+ 3. Also find the domain and range of I and Graph I and I-Ion the same coordinate axes.
Solution
r1 .
y = 2x + 3 , interchange the variables x and y. The result, x = 2y+ 3 that defines the inverse r1 implicitly. To find the explicit form, we
In the equation is an equation solve for y.
2y+ 3 = x 2y = x - 3 y = "21 (x - 3) The explicit form of the inverse r1 is therefore 1 rJ (x) = "2 (x - 3) which w e verified i n Example 5 (b). Next we find
= Range of I =
Domain ofl The graphs of I(x) Figure
Figure 16
=( Domain of r1 = Range ofr1
= 2x+ 3 and
its inverse
- 00 , 00
)
( -00,00 )
rl (x)
=
� (x - 3)
are shown in
16. Note the symmetry of the graphs with respect to the line y = x. y
•
SECTION 6.2
One-to-One Functions; Inverse Functions
41 7
We now outline the steps to follow for finding the inverse of a one-to-one function. Procedure for Finding the Inverse of a One-to-One Function STEP
1: In y= f(x), interchange the variables x and y to obtain
x= f(y)
STEP
STEP
This equation defines the inverse function r1 implicitly.
2: If possible, solve the implicit equation for y in terms of x to obtain
the explicit form of rl:
= r1(x)
3: Check the result by showing that
r1(f(x))
EXA M P LE 9
y
=x
and
f(r1(x))= x
Finding the I nverse Function
The function
f(x)
+1 = 2xx-I
x =t- 1
is one-to-one. Find its inverse and check the result. Solut i o n
STEP
1: Interchange the variables x and y in
y =
2x + 1 x-I
x=
2y + 1 y- 1
to obtain
STEP
2: Solve for y.
2y + 1 y- 1 x(y -1) 2y + 1 xy -x 2y + 1 xy - 2y = x + 1 (x -2)y x + 1 x+1 y x-2 x=
= =
= =
--
The inverse is
f-l(X) STEP
3:
M u ltiply both sides by
-
1.
Apply the Distributive Property. Subtract
2y from both sides; add x to both sides.
Factor. Divide by x
-
2.
+1 _X X =t- 2 x-2
Replace
y by f-1 (x).
Check:
2x+1 +1 2X + 1 x-I r1(f(x))= rl = x -I 2x+l _2 x -I
(
f(r1(x))
y
)
= f ( :� � �) =
--
X+ 2 --1 +1 -2 + 1 - 1 x-2
(�r ) --
3x 2x + 1+x-I =x = 2x+I-2(x-l) = 3 2(x + 1)+ x -2 x + 1 - (x-2)
3x ==x 3
x =t-l
x=t-2 •
41 8
CHAPTER 6
Exponential and Logarithmic Functions
Exploration
In Example 9, we found that, if {(x) = --, then (-'(x) = --, Compare the vertical and hori2x+ 1
x+ 1
x- 1
x- 2
zontal asymptotes of { and {-'. What did you find? Are you su rprised?
Result You should have found that the vertical asymptote of { is x = 1, and the horizontal asymptote , is y = 2. The vertical asymptote of (- is x = 2, and the horizontal asymptote is y = 1 . "",\::::
: : >- Now Work
P R O B L E M 49
We said in Chapter 3 that finding the range of a function f is not easy. However, if f is one-to-one, we can find its range by finding the domain of the inverse func tion r1. EXAM P LE 10
Finding the Range of a Function
Find the domain and range of
f(x) Solution
1 = 2xx+ - 1
The domain of f is {xix *' 1 }. To find the range of f, we first find the inverse rl . B ased on Example 9, we have
The domain of r1 is {xix *' 2}, so the range of f is {yly "!IIi
>-
*'
2}.
•
Now Work P R O B L E M 6 3
If a function is not one-to-one, it has no inverse function. Sometimes, though, an appropriate restriction on the domain of such a function will yield a new function that is one-to-one. Then the function defined on the restricted domain has an inverse function. Let's look at an example of this common practice. EXAMPLE 1 1
Finding the I nverse of a Domain-restricted Function
Find the inverse of y
Solution
=
STEP
Y
2
f(x)
= x2 if x 2: 0.
The function y x2 is not one-to-one. [Refer to Example 2(a).] However, if we restrict the domain of this function to x 2: 0, as indicated, we have a new function that is increasing and therefore is one-to-one. As a result, the function defined by y f(x) x2, X 2: 0, has an inverse function, rl . We follow the steps given previously to find rl.
=
Figure
=
=
1: In the equation y
=
x2, X
2:
0, interchange the variables x and y. The result is
x= l
This equation defines (implicitly) the inverse function.
17
f(x) = x2, X2: 0
STEP
Y= x
x
2: We solve for y to get the explicit form of the inverse. Since y 2:
solution for y is obtained: y
= Yx. So rl (x)
=
Yx.
0, only one
= rl(x2) # Ixl x since x 2: ° = f(Yx) = (Yxf x Figure 17 illustrates the graphs of f(x) = x2, X 2: 0, and r1(x) Yx. STEP
2
y 2: 0
3:
Check:
r1(f(x) ) f(rl(x) )
=
=
=
=
=
•
SECTION 6.2
One-to-One Functions; Inverse Functions
41 9
SUMMARY 1. If a function I is one-to-one, then it has an inverse function 2. Domain of I
=
Range of rl; Range of I
=
1-1.
Domain of rl.
r1(f(x) ) = x for every x in the domain of I and l(r1(x) ) = x for every x in the domain of rl . 4. The graphs of I and r1 are symmetric with respect to the line y = x. 5. To find the range of a one-to-one function I, find the domain of its inverse function ri. 3. To verify that r1 is the inverse of I, show that
6.2 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of tfuse exercises. If you get a wrong answel; read the pages listed in red. x + 5 1. Is the set of ordered pairs { ( I,3 ) ,( 2,3 ) ,( - 1,2)}a function? ?(pp. 215-216) 3. What is the domain of f ( x ) x2 + 3x - 18 =
Why or why not? (pp. 208-210)
2. Where is the function f(x) decreasing? (pp. 233-234 )
=
x2 increasing? Where is it
Concepts and Vocabulary
4. If every horizontal line intersects the graph of a function f at no more than one point,f is a(n) function.
7.
True orFalse If f and g are inverse functions,the domain of f is the same as the domain of g.
8.
True orFalse If f and g are i nverse functions,their graphs are symmetric with respect to the line y = x.
__
5. If rl denotes the inverse of a function f, then the graphs of f and rl are symmetric with respect to the line . __
6. If the domain of a one-to-one function f is [4,00 ) , the range of its inverse,f -I, is . __
Skill Building In Problems 9-16, determine whether the function is one-to-one.
9.
11.
Domain
Range
20 Hou rs
$200
25 Hou rs
$300
30 Hours
$350
40 Hou rs
$425
Domain
Range
20 Hours
10.
Domain Bob
Range -� Karla
t---
-
Dave -�
� Debra
John
12.
Dawn
Chuck
Phoebe
Domain
Range
...j.- $200
Bob
25 Hours -
Karla
Dave
30 Hours
$350
John
40 Hours
$425
Chuck
Debra ....j.- Phoebe r-
13. { ( 2,6 ) ,( - 3,6) , ( 4,9 ) ,( 1,10)}
14. {( - 2,5 ) ,( - 1,3 ) , (3,7 ) ,( 4,12)}
15. {C O,0),( 1,1), ( 2,1 6 ) ,(3 , 8 1 ) }
16. { ( I,2) ,( 2,8) , (3,18) ,( 4,3 2)}
In Problems 17-22, the graph of a function f is given. Use the horizontal-line test to determine whether f is one-to-one.
17.
18.
19.
y
3
3 x
-3
-3
3 x
-3
-3
3 x
-3
-3
420
CHAPTER 6
20.
Exponential and Logarithmic Functions
21.
Y
x
"
22.
-3
Y 3
3 x
-3
In Problems 23-30, find the inverse of each one-to-one function. State the domain and the range of each inverse function. 23.
Annual Rainfall (inches)
Location Mt Waialeale, Hawaii
24.
460.00
Star Wars
$461
Star Wars: Episode One - The Phantom Menace
$431
E. T. the Extra Terrestrial
$400
Jurassic Park
$357
Forrest Gump
$330
Monrovia, Liberia
202 01
Pago Pago, American Samoa
196.46
Moulmein, Bu rma
191.02
Lae, Papua New Guinea
182.87
Source: Information Please Almanac
25.
Domestic Gross (in millions)
Title
Source: Information Please Almanac
Age
Monthly Cost of Life Insurance
30
$7.09
26.
$8.40
40
$11.29
45
State
Unemployment Rate
Virginia
11%
Nevada
5.5%
Tennessee
5.1%
Texas
6.3%
Source: United
Source: eterm.com
.27. {( -3, 5), (-2, 9), (-1, 2), ( 0, ll), (1, -5)}
States Statistical Abstract
28. {( -2, 2), ( - 1 ,6), (0, 8), (1, -3), (2, 9)}
29. { (-2, 1), (-3,2), ( - 10,0) , (1, 9), (2, 4)}
30. {( -2, -8), ( - 1 , -1), ( 0,0) , (1, 1), (2, 8)}
In Problems 31-40, verify that the functions f and g are inverses of each other by showing that f (g (x)) = x and g (f (x)) = x. Give any values of x that need to be excluded. 1 l .31. f (x) = 3x + 4; g (x) = 3 (x - 4) 32. f (x) = 3 - 2x; g (x) = _ (x - 3) 2 x
33. f (x)
= 4x - 8; g (x) = "4
35. f(x)
= x3 - 8; g (x) = �
1 37. f (x) = - ; x 39. f (x)
=
g (x)
2x+ 3 x+4
-- ;
=
+
2
1
-
x
4x - 3 g (x) = -2 - x
1 2
34. f (x)
= 2x
36. f (x)
= (x - 2j2, x
38. f (x)
= x; g (x) = x
40. f (x)
=
+
6; g (x) = -x - 3 2:
2; g (x) = vX + 2
x -5 ; g (x) 2x + 3
--
3x + 5 = -1 - 2x
In Problems 41-46, the graph of a one-to-one function f is given. Draw the graph of the inverse function rl. For convenience (and as a hint), the graph of y = x is also given. 41.
Y 3
Y= x
42.
Y 3
y=X
43.
y 3
(1,2 )
Y=X
(2,1) 3 x
-3
3 x
-3
3 x
-3 ( -1, -1)
-3
SECTION 6.2
44.
Y =X
Y 3
45.
One-to-One Functions; Inverse Functions
421
46.
Y=X
( - 2,1) 3 x (1 , -1)
-3
3 x
-3
-3
-3
In Problems 47-58, the [unction [ is one-to-one. F ind its inverse and check your answeJ: State the domain and the range o[ [ and rl . Graph [, [-I , and y = x on the same coordinate axes. 49. [(x) = 4x+ 2 48. [ (x) = -4x 47. [(x) = 3x 50. [(x)
= 1 - 3x
51. [(x)
= x3 - 1
52. [( x)
=
x3+1
53. [( x)
= x2
54. [( x)
= x2+ 9 x � o
55. [ (x)
=
-
56. [(x)
=
57. [(x)
=
58. [ (x)
=
--
+
4 x�o
3 x
- -
1 x-2
--
4 x
4 x+2
In Problems 59-70, the [unction [ is one-to-one. F ind its inverse and check your answeJ: State the domain o[[ and find its range using rl . 4 3x 2 .,9. [(x) = 61. [ (x) = 60. [(x) 3+x x+2 2-x _
--
62. [(x)
=
65. [(x)
=
68. [(x)
=
=
2x x - I
'-,, 63. [(x)
- -
=
-
--
2x .)x -
64. [(x)
=
- --
67. [(x)
=
2x+ 3 x+2
70. [(x)
=
-
1
3x+4 2x - 3
66. [(x)
=
2x - 3 x+4
-3x - 4 x-2
69. [(x)
=
x2 - 4 x> 2x-,
--
3x+ 1 x
� -
°
x2 + 3 x> 3x-,
°
Applications and Extensions
71. Use the graph of y the following:
(a) (b) (c) (d)
[( - 1) [(1) rl (l) rl (2)
=
[ (x) given in Problem 41 to eval uate
79. A function y = [(x) is increasing o n the interval (0, 5). What conclusions can you draw about the graph of y = rl (x) ?
= [ (x) given in Problem 42 to evaluate
(c) rl (O) (d) rl ( - l )
= 13 and [ is one-to-one, what is r l ( B)?
74. If g (-5)
=
80. A function y [(x) is decreasing on the interval (0,5). What conclusions can you draw about the graph of y = rl (x) ? =
72. Use the graph of y the following: (a) [(2) (b) [(1)
73. If [(7)
78. The domain of a one-to-one function g is [0, 15] , and its range is (0, 8). State the domain and the range of g-I.
3 and g is one-to-one,what is g-I (3)?
75. The domain of a one-to-one function [is [5,00 ) , and its range is [-2, 00). State the domain and the range of rl . 76. The domain of a one-to-one function [ is [0, 00 ) ,and its range is [5, 00 ) . State the domain and the range of r l .
77. The domain o f a one-to-one function g is the set o f all real n umbers, and its range is [0,00). State the domain and the range of g-I .
81. Find the inverse of the linear function
[(x) = mx+b
In "* °
82. Find the inverse of the function
[(x) = Vr2 - x2 O :=;
X
:=;
r
83. A function [ has an inverse function. If the graph of [ lies in quadrant I, in which q uadrant does the graph of r 1 lie? 84. A function [ has an inverse function. If the graph of [ lies in quadrant II, in which quadrant does the graph of r 1 lie?
85. The function [(x) = I x l is not one-to-one. Find a suitable re striction on the domain of [ so that the new function that re sults is one-to-one. Then find the inverse off. 86. The function [(x) = x4 is not one-to-one. Find a suitable re striction on the domain of [ so that the new function that results is one-to-one. Then find the inverse of f.
422
C HAPTER 6
Exponential and Logarithmic Functions
In applications, the symbols used for the independent and dependent variables are often based on common usage. So, rather than using y = f(x ) to represent a function, an applied problem might use C C(q) to represent the cost C of manufacturing q units of a good since, in economics, q is used for output. Because of this, the inverse notation f-1 used in a pure mathematics problem is not used when finding in verses of applied problems. Rathel; the inverse of a function such as C C( q) will be q = q( C). So C = C( q) is a function that represents the cost C as a function of the output q, while q = q( C) is a jil.l1ction that represents the output q as a function of the cost C. Problems 87-90 illustrate this idea. =
=
87. Vehicle Stopping Distance Taking into account reaction time, the distance d (in feet) that a car requires to come to a complete stop while traveling r miles per hour is given by the function d ( r)
= 6.97r - 90.39
(a) Express the speed r at which the car is traveling as a function of the distance d required to come to a com plete stop. (b) Verify that r = r e d ) is the inverse of d = d ( r ) by show ing that r ( d ( r ) ) r and d ( r ( d) ) = d. (c) Predict the speed that a car was traveling if the distance required to stop was 300 feet. =
88. Height and Head Circumference The head circumference C of a child is related to the height H of the child (both in inches) through the function
H (C) = 2.15C - 1 0. 53 (a) Express the head circumference C as a function of height H. (b) Verify that C = C ( H ) is the inverse of H H(C) by showing that H ( C ( H ) ) = H and C ( H ( C ) ) C. (c) Predict the head circumference of a child who is 26 inches tall. =
=
89. Ideal Body Weight The ideal body weight W for men (in kilograms) as a function of height h (in inches) is given by the function W(h) SO + 2.3(h - 60 ) =
(a) What is the ideal weight of a 6-foot male? (b) Express the height h as a function of weight W. (c) Verify that h = h e W ) is the inverse of W = W ( h ) by W. showing that h ( W ( h ) ) = h and W ( h ( W ) ) (d) What is the height of a male who is at his ideal weight of 80 kilograms? =
Note: The ideal body weight W for women (in kilograms)
as a function of height h (in inches) is given by W ( h) 45.5 + 2.3 ( h - 60) . =
9 90. Temperature Con"ersion The function F(C) = s C + 32 converts a temperature from C degrees Celsius to F degrees Fahrenheit. (a) Express the temperature in degrees Celsius C as a func tion of the temperature in degrees Fahrenheit F. (b) Verify that C C ( F ) is the inverse of F F ( C) by showing that C ( F(C) ) = C and F ( C ( F ) ) F. (c) What is the temperature in degrees Celsius if it is 70 de grees Fahrenheit? =
91. Income Taxes
The function
T ( g ) = 4220 + 0.25(g - 30,650 ) represents the 2006 federal income tax T (in dollars) due for a "single" filer whose adjusted gross income is g dollars, where 30,650 � g � 74,200. (a) What is the domain of the function T? (b) Given that the tax due T is an increasing linear func tion of adjusted gross income g, find the range of the function T. (c) Find adjusted gross income g as a function of federal in come tax T. What are the domain and the range of this function? 92. Income Taxes
The function
T ( g ) = 1510 + 0. 15(g - 15,100 ) represents the 2006 federal income tax T (in dollars) due for a "married filing jointly" filer whose adjusted gross income is g dollars, where 15,100 � g � 61 ,300. (a) What is the domain of the function T? (b) Given that the tax due T is an increasing linear function of adjusted gross income g, find the range of the func tion T. (c) Find adjusted gross income g as a function of federal in come Lax T. What are the domain and the range of this function?
93. Gra"ity on Earth If a rock falls from a height of 100 meters on Earth, the height H (in meters) after t seconds is approx imately H ( t ) = 100 - 4.9t2
(a) In general, quadratic functions are not one-to-one. How ever, the function H(t) is one-to-one. Why? (b) Find the inverse of H and verify your result. ( c) How long will it take a rock to fall 80 meters? 94. Period of a Pendululll The period T (in seconds) of a sim ple pendulum as a function of its length I (in feet) is given by
�32.2
T ( l ) = 27T
(a) Express the length I as a function of the period T. (b) How long is a pendulum whose period is 3 seconds? 95. Given f(x) =
=
=
I
find r1 ( x ) . If c is f r l ?
oF
ax + b cx + d
0, under what conditions on a, b, c, and d
=
Discussion and Writing
96. Can a one-to-one function and its inverse be equal? What must be true about the graph of f for this to happen? Give some examples to support your conclusion.
97. Draw the graph o f a one-to-one function that contains the points ( -2, -3 ) , (0, 0), and ( 1 , 5 ) . Now draw the graph of its inverse. Compare your graph to those of other students. Dis cuss any similarities. What differences do you see?
SECTION 6.3
98. Give an example of a function whose domain is the set of real numbers and that is neither increasing nor decreasing on its domain, but is one-to-one. [Hint: Use a piecewise-defined function.]
Exponential Functions
423
99. Is every odd function one-to-one? Explain. 100. Suppose C(g) represents the cost C, in dollars, of manufac turing g cars. Explain what C-1 (800,000) represents.
'Are You Prepared?' Answers
3. {xix
1. Yes; for each input x there is one output y .
2. Increasing on (0, 00 ) ; decreasing on ( - 00 , 0)
*"
-
6, x
3}
*"
6.3 Exponentia l Functions PREPARING FOR THIS SECT ION
• •
Before getting started, review the following: •
Exponents (Section R.2, pp. 21-24, and Section R.8, pp. 75-76) Graphing Techniques: Transformations (Section 3.5, pp. 252-260) Now Work the 'Are You Prepared?, problems on page
OBJECT IVES 1
• •
Average Rate of Change (Section 3.3, pp. 236-238) Solving Equations (Section 1 . 1 , pp. 86-92) Horizontal Asymptotes (Section 5.2, pp. 346-352)
432.
Evaluate Exponential Functions (p. 423)
2 Graph Exponential Functions (p. 425) 3 Defin e the Number e (p.429) 4 Solve Exponential Equations (p.431)
1
Eva l ua te Expo nential Fu nctions
In Chapter R, Section R.8, we give a definition for raising a real number a to a ratio nal power. Based on that discussion, we gave meaning to expressions of the form
a
r
where the base a is a positive real number and the exponent r is a rational number. B ut what is the meaning of aX, where the b ase a is a positive real number and the exponent x is an irrational number? Although a rigorous definition requires methods discussed in calculus, the basis for the definition is easy to follow: Select a rational number r that is formed by truncating (removing) all but a finite number of digits from the irrational number x. Then it is reasonable to expect that For example, take the irrational number tion to a7T is
7T =
3.141 5 9 . . . . Then an approxima
where the digits after the hundredths position have been removed from the value for 7T. A better approximation would be
a7T
:::::
3 14159 a .
where the digits after the hundred-thousandths position have been removed. Con . tinuing in this way, we can obtain approximations to aT to any desired degree of accuracy. Most calculators have an key or a caret key for working with expo
[2]
nents. To evaluate expressions of the form key (or the
[] key), enter the exponent
X
a ,
x,
�I
enter the base
and press
a,
then press the
I 1 (or I enter I)· =
[2]
424
CHAPTER 6
Exponential and Logarithmic Functions Using a Calcu l ator to Evaluate Powers of 2
E XA M P LE 1
Using a calculator, evaluate: Solution
(a) 21 .4 � 2.639015822 ( c) 2 1 .41 4 � 2.66474965 (e) 2° � 2.665144143 �j ! =---
(e) 20
(d) 21 .4142
(c) 21 .414
(b) 2 1 .41
(a) 2 1 .4
(b) 2 1.41 � 2.657371628 (d) 2 1.4142 � 2.665119089
•
Now Work P R O B L E M 1 1
It can be shown that the familiar laws for rational exponents hold for real exponents. THEOREM
Laws of Exponents
If
s,
t, a, and b are real numbers with a
> 0 and
b
> 0, then
(aby
a-S = � as
1S = 1
=
=
as . bS
(!a ) S
( 1)
�------�
�
We are now ready for the following definition: DEFINITION
An
exponential function
is a function of the form
f(x) where a is a positive real number set of all real numbers.
WARNING It is important to distinguish a power function, g(x)
=
x",
n �
2, an
integer, from an exponential function, f(x)
=
aX, a *- 1, a
>
O.
In
a power
function, the base i s a variable and the exponent is a constant. In a n exponen tial function, the base i s a constant and the exponent is a variable.
_
(a
=
aX
> 0) and
a
1 . The domain of
*-
f is the
We exclude the base a = 1 because this function is simply the constant func tion f (x) = 1x = 1. We also need to exclude bases that are negative; otherwise, we
�
�.
would have to exclude many values of x from the domain, such as x = and x = [Recall that ( _2 ) 1/2 = vC2, ( _3 )3/4 = �, and s o on, are not defined in the set of real numbers.] Some examples of exponential functions are =
f(x)
=
�
F (x)
2x
=
(3)X 1
Notice for each function that the base is a constant and the exponent is a variable. You may wonder what role the base a plays in the exponential function f(x) = aX. We use the following Exploration to find out.
Exploration
(a) Evaluate ((x) (b) Eval u ate g(x)
=
=
2X at x
=
-
2, - 1 , 0, 1, 2, and 3.
3x + 2 at x
=
-
2, - 1, 0, 1, 2, and 3.
(c) Comment on the pattern that exists in the val ues of ( and g. Result
(a) Table 1 shows the values of ((x) (b) Table 2 shows the val ues of g(x)
= =
2x for x
=
-
2, - 1, 0, 1 , 2, and 3.
3x + 2 for x
=
-
2, - 1 , 0, 1 , 2, and 3.
SECTION 6.3
Table
x
1
f(x)
2
f(- 2) =
1
=
r2
Table 2
r
x
1
1
2
2
4
3
8
=
3x +
425
2
g(- 2) = 3 ( - 2) + 2 = -4
-1
2
0
g(x)
-2
1 1 = - = 4 22
Exponential Functions
-1
0
2
1
5
2
8
3
11
(c) In Table 1 we notice that each va lue of the exponential function f(x) = aX = Y could be found by m Ultiplying the previous value of the function by the base, a = 2. For example, f( - 1 ) = 2 · f( - 2) = 2 '
1 1 = 2' "4
f(O) = 2 · f( - 1 ) = 2 · 2 = 1 , 1
f(1 ) = 2 · f(O) = 2 · 1 = 2
and so on. Put another way, the ratio of consecutive outputs is constant for u n it increases in the input. The constant equals the value of the base a of the exponential function. For example, for the function f(x) = 2x, we notice that 1
2 f(- 1 ) -- = - = 2' f(- 2) 1
---
f( 1 ) 2 - = - = 2' f(O) 1
f(x + 1 ) f(x)
y+ 1 = - = 2 Y
4
and so on. From Table 2 we see that the function g(x) = 3x + 2 does not have the ratio of consecutive outputs that a re constant because it is not exponential. For example, --
g(- 1 )
g(-2)
g(1 ) 5 -1 1 = - = - i' - = -4 4 g(O) 2
Instead, because g(x) = 3x + 2 is a linear function, for unit increases in the input, the outputs increase by a fixed amount equal to the value of the slope, 3.
The results of the Exploration lead to the following result. THEOREM r
r
r
r
r
In Words
For
f(x)
an
=
For an exponential function f(x) then
=
f(x
exponential
X a , a
+
1)
f(x)
'----
> 0,
a
I
= a
-----------.l..J
f u n ction
aX, for l-unit c h anges in
Proof
the input x, the ratio of consecu
f(x
tive outputs is the constant a.
+
1)
•
f(x) �1iiIi = ill:lO "'"
2
EXAM P L E 2
Now Work P R O B L E M 2 1
G r a p h Exponential Fu nctions
First, we graph the exponential function f(x)
=
2x.
G raphing an Exponential Function
Graph the exponential function: Solution
=F 1, if x is any real number,
f ( x ) = 2x
The domain of f(x) = 2x consists of all real numbers. We begin by locating some points on the graph of f(x) = 2x, as listed in Table 3.
426
CHAPTER 6
Exponential and Logarithmic Functions
Table 3
Figure 1 8
x r 1 0 '" 0.00098
-10 -3
r3
=
�
-2
r2
=
�
-1
r1
=
�
o
20
=
1
21
=
2
2
22
=
4
3
23
=
8
10
2 10
=
1 024
8
4 2
Since 2x > ° for all x, the range of f i s the interval (0, 00 ) . From this, we con clude that the graph has no x-intercepts, and, in fact, the graph will lie above the x-axis. As Table 3 indicates, the y-intercept is Table 3 also indicates that as x � - 00 the value of f (x) = 2x gets closer and closer to 0. We conclude that the x-axis is a horizontal asymptote to the graph as x � - 00. This gives us the end behavior of the graph for x large and negative. To determine the end behavior for x large and positive, look again at Table 3. As x � 00 , f(x) 2x grows very quickly, causing the graph of f(x) = 2x to rise very rapidly. It is apparent that f is an increasing function and so is one-to-one. Using all this information, we plot some of the points from Table 3 and con nect them with a smooth, continuous curve, as shown in Figure 18.
l.
=
•
As we shall see, graphs that look like the one in Figure 18 occur very frequently in a variety of situations. For example, look at the graph in Figure 19, which illus trates the number of cellular telephone subscribers at the end of each year from 1 985 to 2005. We might conclude from this graph that the number of cellular tele phone subscribers is growing exponentially. Figure
Number of Cellular Phone Subscribers at Year End
19
_
220 200
'E 180 c 160 =
� 140
(l) .D
.'" 120 C>
'0 � .D E z
::::l
�
'
___ __ ______ __ __ __ __ __ ____ __ __ __ ___ __ __ _ ______ ____
.--_._---_.--.• -._--_. _---
�-----------------------------------------Ir� r_
80
-,--
40 0
U') eo
�
co eo
�
reo
�
eo eo
�
==�L::iDoDO cr> eo
�
----
,,---
--- - --- ---- --------------. _--- --
60 20
Ir-
0 cr>
�
0; �
N cr>
�
C') cr>
�
cr> cr>
'"
U') cr> cr>
co cr> cr>
r- . ,: " --
rcr>
�
eo cr>
�
cr> cr>
�
0 0 0 N
0 N
c;
N 0 0 N
C') 0 0 N
'" 0 0 N
U') 0 0 N
Year Source: ©2006 CTIA-The Wireless Association®. All Rights Reserved.
We shall have more to say about situations that lead to exponential growth later in this chapter. For now, we continue to seek properties of the exponential functions. The graph of f(x) = 2x in Figure 18 is typical of all exponential functions that have a base larger than Such functions are increasing functions and hence are one-to-one. Their graphs lie above the x-axis, pass through the point ( 0, 1 ) , and thereafter rise rapidly as x � 00 . As x � - 00 , the x-axis (y = 0) is a horizontal
l.
SECTION 6.3
Figure 20
Exponential Functions
427
asymptote. There are no vertical asymptotes. Finally, the graphs are smooth and con tinuous, with no corners or gaps. Figure 20 illustrates the graphs of two more exponential functions whose bases are larger than 1 . Notice that for the larger base the graph is steeper when x > 0 and is closer to the x-axis when x < O .
l�1 I!m
Seeing the Concept
Gra p h Y, 2x and compare what you see to Figure 1 8. Clear the screen and g raph Y, 3x and Y2 = 6x = = a n d compare what you see to Figure 20. Clear the screen a n d graph Y, 1 0x and Y2 1 00x. What = viewing recta ngle seems to work best? =
-3
Y=0
The following display summarizes the information that we have about f(x) = > 1.
aX, a
Properties of the Exponential Function f(x)
((x)
4. f(x)
=
aX, a
= 0) i s a horizontal asymptote a s x � - (X) .
> 1, i s an increasing function and is one-to-one.
( �} -1,
5. The graph of f contains the points y= o
x
E XA M P L E 3
Solution
f(x)
-2 -1 o
2 3 10
=
(1"2)-10
(�r
=
1 024
Gr3 Gr2 Gr' GY GY = � GY � GY � G) =
8
= 4 =
=
2
a).
See Figure 21.
aX
when 0
<
a<
G raphing an Exponential F u nction
Graph the exponential function: Table 4
(0, 1 ) , and ( 1,
6 . The graph o f f i s smooth and continuous, with n o corners or gaps.
Now w e consider f ( x ) =
-3
1
2. There are no x-intercepts; the y-intercept is 1 .
3. The x-axis ( y
-10
a >
real numbers.
if, a > 1
x
a" ,
1. The domain is the set of all real numbers; the range is the set of positive
Figure 2 1 =
=
The domain of f(x)
=
(�r
f(x) =
1.
(l)X "2
consists of all real numbers. As before, we locate some
points on the graph by creating Table 4. Since
(�r
> 0 for all x, the range of f is
the interval ( 0, (0 ) . The graph lies above the x-axis and so has no x-intercepts. The
y-intercept is L As x � - (X) , f(x) =
(l)X "2
grows very quickly. As x �
00 ,
the values
of f(x) approach O . The x-axis (y = 0) is a horizontal asymptote as x � 00 . It is apparent that f is a decreasing function and so is one-to-one. Figure 22 illustrates the graph. Figure 22
1
=
=
10 ""
0.0009 8
-3
3
x Y=0 •
428
CHAPTER 6
Exponential and Logarithmic Functions
We could have obtained the graph of y transformations. The graph of y = the graph of y
=
(�r
=
=
(�r
from the graph of y
TX is a reflection
=
2x using
about the y-axis of
2x (replace x by -x). See Figures 23(a) and 23(b).
Figure 23
(a) y = 2 X
l� mlI
Replace x by Reflect about the y-axis
Yj
y=
GY
(b) Yj
W
(-1 , 6) y=
G) x
-3
=
3
x y= 0
Y = 2-x =
=
(�Y,
for
a>
=
=
=
(�Y
(�r
in Figure 22 is typical of all exponential functions
that have a base between ° and 1. Such functions are decreasing and one-to-one. Their graphs lie above the x-axis and pass through the point (0, 1 ) . The graphs rise rapidly as x -7 - 00 . As x -7 00 , the x-axis (y = 0) is a horizontal asymptote. There are no vertical asymptotes. Finally, the graphs are smooth and continuous, with no corners or gaps. Figure 24 illustrates the graphs of two more exponential functions whose bases are between ° and 1. Notice that the smaller base results in a graph that is steeper when x < 0. When x > 0, the graph of the equation with the smaller base is closer to the x-axis. The following display summarizes the information that we have about the func tion f(x) = aX, ° < a < 1 .
1.
Figure 25
6x, Y2
0, i s the reflection about the y-axis of the g raph of
Properties o f the Exponential Function f(x)
{(x)
(W
«.
The graph of f(x) =
X y
(b)
Using a g raphing utility, simultaneously g raph:
Conclude that the graph of Y2
24
x;
Seeing the Concept (a) Yj = 3x, Y2 =
Figure
-
=
tr,
0 < a < 1
The domain is the set of all real numbers; the range is the set of positive real numbers.
2. There are no x-intercepts; the y-intercept is 1 .
aX, 0 < a < 1
3. The x-axis ( y 4. f(x)
=
=
0 ) is a horizontal asymptote as x -7 00 .
aX, ° < a <
1, is a decreasing function and is one-to-one.
5. The graph of f contains the points x y= 0
( �) -1,
, (0, 1 ) , and ( 1 , a).
6. The graph o f f i s smooth and continuous, with no corners o r gaps.
See Figure 25.
SECTION 6.3
E XA M P L E 4
Exponential Functions
429
G raph i n g Exponential F u nctions U s i n g Transfo rmations
Graph f ( x) of f.
=
TX -
3 and determine the domain, range, and horizontal asymptote
We begin with the graph of
Solution
Figure 26
y
= 2x. Figure 26 shows the stages.
y
y
Y 10
x
Xy= O
-3
y= O
-4 Replace x by - x; reflect about y-axis
(a) y = 2x
Subtract 3; shift down 3 units
(b) y = 2-x
y = -3
(c) y = 2 -x - 3
As Figure 26(c) illustrates, the domain of f (x) = TX - 3 is the interval ( - 00 , 00 ) and the range is the interval ( -3, (0 ) . The horizontal asymptote of f is the line y = - 3.
•
Si!I!lO = IIIII> I;;Ii -
3
Now Work P R O B L E M 3 7
Defi n e the N u mber
e
As we shall see shortly, many problems that occur in nature require the use of an exponential function whose base is a certain irrational number, symbolized by the letter e . Let's look at one way of arriving at this important number e .
DEFINITION
The number
e
is defined as the number that the expression
( I )"
( 2)
1 +n
approaches as
n �
00 . In calculus, this is expressed using limit notation as
e
=
lim
,, --> 00
(
n
1 + I
)
"
I
-.J
Table 5, on page 430, illustrates what happens to the defining expression (2) as takes on increasingly large values. The last number in the right column in the table is correct to nine decimal places and is the same as the entry given for e on your cal culator (if expressed correctly to nine decimal places). The exponential function f(x) = eX, whose base is the number e , occurs with such frequency in applications that it is usually referred to as the exponential function. Indeed, most calculators have the key or exp(x) which may be used to
n
[2]
1
evaluate the exponential function for a given value of x . *
I,
':' I f your calculator does n o t have one of these keys, refer t o y o u r Owner's Manual.
430
CHAPTER 6
Table 5
Exponential and Logarithmic Functions
n
(
+ n 1
n 2
1
)
-2
0.1 4
-1
0.37
0.5
1 .5
2.25
0.2
1 .2
2.48832
10
0.1
1 .1
2.59374246
0.01
1 .01
2.70481 3829
a
2.72
0.001
1 .001
2.7 1 6923932
1 0,000
0.0001
1 .0001
2.7 1 8 1 45927
1 00,000
0.00001
1 .00001
2.71 8268237
0.000001
1 .000001
2.7 1 8280469
1 0- 9
1 + 1 0- 9
2.7 1 8281 827
1 ,000,000 1 ,000,000,000
y
e"
x
2
5
1 ,000
Table 6
"
2
1 00
Figure 27 y = e"
+ n
1
2
eX
7.39
Now use your calculator to approximate for x -2, x = - 1 , x = 0, x = 1 , and x = 2 , as we have done to create Table 6 . The graph of the exponential function f(x) = is given in Figure 27. Since 2 < < 3, the graph of y = lies between the graphs of y = 2 and y Y. Do you see why? (Refer to Figures 18 and 20.)
(2, 7.39)
eX
I� . --:· I �
x
e
=
=
eX
Seeing the Concept
Gra ph Y1 = e" and compare what you see to Fig u re 27. Use eVALUEate or TABLE to ver ify the points on the g raph shown in Figure 27. Now graph Y2 = 2x and Y3 = 3x on the same screen as Y1 e". Notice that the graph of Y1 = eX l ies between these two graphs. =
Y= 0
3
(-2, 0.1 4)
x
E XA M P L E 5
G raphing Exponential F unctions Using Transformations
Graph f(x) of f. Solution
=
_ex-3
and determine the domain, range, and horizontal asymptote
We begin with the graph of
Figure 28
y
y
=
eX.
Figure 28 shows the stages. y
y= o
3
(2, 7.39) ( - 2 , - 0 . 1 4) ( -1, -0.37)
6
x
Y= 0
x
-3
3
-I--..J---,---,I
..-1_
_ _ __
Y= 0
-6 (2, -7.39 )
__
3
(-2, 0.1 4 )
(a) y = eX
x
Multiply by -1; Reflect about x-axis.
(b) y = - ex
Replace x by x - 3; Shift right 3 units.
(c) Y = _ ex-3
_ex-3
As Figure 28(c) illustrates, the domain of f(x) = is the interval ( - 00 , 00 ) , and the range is the interval ( - 00 , 0). The horizontal asymptote is the line y = O . rr �
-
•
Now Work P R O B L E M 4 9
SECTION 6.3
4 r
r , r r
In Words
E XA M P L E 6
If
a
U
=
V a ,
then
u
=
v
( 3)
Property (3) is a consequence of the fact that exponential functions are one-to one. To use property (3), each side of the equality must be written with the same base. Solving an Exponential Eq uation
Solve: Solution
431
Equations that involve terms of the form aX , a > 0, a "* 1 , are often referred to as exponential equations. Such equations can sometimes be solved by appropriately applying the Laws of Exponents and property (3):
sions with the sa m e base are equal.
Functions
So lve Exponential Equations
When two exponential expres equal, then their exponents are
Exponential
3x +1 = 81
Since 81 = 34 , we can write the equation as 3x + 1 = 81 = 34 Now we have the same base, 3, on each side, so we can set the exponents equal to each other to obtain x + 1 = 4 x = 3 The solution set is (3). L"!J1!
E XA M P L E 7
Now Work P R O B L E M 5 9
Solving a n Exponential Equation
Solve: Solution
•
e-x
2
=
(ex) 2 .
�e�
We use the Laws of Exponents first to get the base e on the right side. (ex ) 2 . As a result,
�e� = e2x • e-3 = e2x -3
e-X2 = e2x -3 -x2 = 2x - 3
x2 + 2x - 3 = 0 (x + 3 ) (x - 1 ) = 0 x = -3 or x = 1
A p ply Property ( 3 ) . Place th e quadratic equation i n standard form . Factor. U se the Zero-Prod uct Property.
The solution set is { -3, 1 } .
•
Many applications involve the exponential function. Let's look at one. E XA M P L E 8
Exponential Probabil ity
B etween 9:00 PM and 10:00 PM cars arrive at Burger King's drive-thru at the rate of 12 cars per hour (0.2 car per minute ) . The following formula from statistics can be used to determine the probability that a car will arrive within t minutes of 9:00 PM.
F(t)
=
1 - e-O.2t
( a) Determine the probability that a car will arrive within 5 minutes of 9 PM ( that is, before 9:05
PM).
( b ) Determine the probability that a car will arrive within 30 minutes of 9 ( before 9:30 PM).
PM
432
CHAPTER 6
Exponential and Logarithmic Functions
I:l
(c) What value does F approach as t becomes unbounded in the positive direction? (d ) Graph F using your graphing utility.
5
(a) The probability that a car will arrive within F(t) at t = 5 .
Solution
F( 5 )
=
1
- e-O. 2(5)
;::::
i
minutes is found by evaluating
0.63212
U se a ca lc ulator.
We conclude that there is a 63 % probability that a car will arrive within 5 minutes. (b) The probability that a car will arrive within 30 minutes is found by evaluating F(t) at t = 30.
F(30)
=
1 - e-O. 2(30)
Figure 29
;::::
i
0.9975
Use a ca lcu lator.
1
There is a 99.75 % probability that a car will arrive within 30 minutes. (c) As time passes, the probability that a car will arrive increases. The value that
1
t ---? 00. Since e-O.2t = 02' it follows that e.t e-O . 2t ---? 0 as t ---? 00. We conclude that F approaches 1 as t gets large.
approaches can be found by letting
I:l
O �======� 30
o
(d)
S e e Figure 2 9 for the graph o f F.
�-� ..... - Now Work
SUMMARY f(x)
X a ,
=
a
X a ,
=
•
P R O B L EM 1 0 1
Properties of the Exponential Function
>
1
Domain: the interval ( - 00, 00 ) ; range: the interval x-intercepts: none; y-intercept: 1 Horizontal asymptote: x-axis ( y
f (x)
F
a
°<
=
0)
as x ---? - 00
Increasing; one-to-one; smooth; continuous See Figure 21 for a typical graph.
<1
Domain: the interval
( - 00 , 00 ) ;
Range: the interval
x-intercepts: none; y-intercept: 1 Horizontal asymptote: x-axis (y
=
0)
(0, 00 )
as x ---?
(0, 00 ) .
00
Decreasing; one-to-one; smooth; continuous If
"
a
V
a ,
=
See Figure 25 for a typical graph.
then
u = v.
6.3 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red.
1 . 43 = pp. 75-76)
__
2. Solve: 3.
=
82/3
5x - 2
=
; T2
__
=
=
__
.
( pp. 21 -24 and
3 (pp. 86-92 ) To graph y (x - 2 ) 3 , shift the graph of 3 x to the left 2 units. (pp. 252-260)
True
y
;
or False
=
4. Find the average rate of change of f ( x) = 3 x - 5 from x = 0 to x = 4. (pp. 236-238 ) 2x � has y = 2 as a 5. True or False The function f ( x) = -x horizontal asymptote. (pp. 346-352) j
SECTION 6.3
Concepts and Vocabulary
,
6. The graph of every exponential function f(x ) = aX, a > 0, a =ft 1, passes through three points: , and
8. lf 3 x
7. If the graph of the exponential function f(x) = aX, a > 0, a =ft 1 , is decreasing, then a must be less than .
10. True
__
=
=
34 , then x
9. True or False identical.
__
or
__
433
Exponential Functions
()
.
The graphs of y
=
Y
and y
1 X
-
=
3
are
The range of the exponential function
False
f(x) = a'r, a > 0, a =ft 1, is the set of all real numbers.
__
Skill Building In Problems 11-20, approximate each number using a calculator. Express your answer rounded to three decimal places. 0 22 (d) 3 (c) 32236 (b) 3223 (d) (c) 5 17 3 2 (b) 5 17 3 12. (a) 5 17 . 11. (a) 3 . 23.14
13. (a)
(b)
15. (a) 3 . 1 2.7
23 141
(b) 3 . 1 42 71
23. 1 4 1 5
(d)
2 '"
14. (a)
22.7
(b)
22.71
(c) 3.141 2 718 (d)
7re
16. (a)
2.73 1
(b)
2.7 1 3 1 4
(c)
19. e-O.85
(d) (c)
2.71 83. 1 4 1
50
2e
(d) err
20. e2.1
In Problems 21-28, determine whether the given function is exponential or not. For those that are exponential functions, identify the value of the base. [Hint: Look at the ratio of consecutive values. ]
x
g(x)
3
-1
2
6
0
5
x
f(x)
-1
°' 21.
0
22.
3
g(x)
-1
3
-1
6
2
0
0
26.
3
27.
F(x)
x
2
-1
0
14
x
f(x)
24.
4
3
0
1
4
3
2
16
2
3
64
11
3
30
x
25.
2
18
H(x)
x
-1
8
12 2
23.
x
H(x)
-1
2
0
4
0
9
2
28.
2
3
2
8
2
12
3
10
3
10
3
24
3
-
x
F(x)
27 8
1
-1
6
6
4
2
0
4 8
2
16
3
32
In Problems 29-36, the graph of an exponential function is given. Match each graph to one of the following functions. C. y = - 3x A. y = 3x D. y = B. Y = TX E. y
=
3x -
29.
1
F.
Y =
3"- 1
30.
-2
-1
G. y
y
31.
y= o
-2
-2 -1
=
H. Y =
3 J -x
-
r'
1 - Y
32. y= o
� 2X
2X
434
CHAPTER 6
y
33.
3
y= O
-2
Exponential and Logarithmic Functions
y
34.
)
y= o
35.
-2
36.
2x 2x
2x
-1
-2
-3
2x
-1
y= O
In Problems 37-48, use transformations to graph each function. Determine the domain, range, and horizontal asymptote of each function. 37. f(x)
= 2.1 + 1
41. f(x)
=
3·
45. f(x)
=
38. f(x)
or1
-2 4· Gr - 2.1+ 3
= Y
42. f(x)
=
2 + 4.1-
46. f(x)
=
5 - e-X
54. f(x)
1
39. f(x)
=
43. f ( x )
= r'
47. f(x)
=
3x- 1
-2
2
+
2
- e-xl2
3xI2
40. f(x)
=
2x + 2
44. f(x)
=
-3.1 + 1
48. f(x)
= 1
- Txl3
In Problems 49-56, begin with the graph of y = eX [Figure 27J and use transformations to graph each function. Determine the domain, range, and horizontal asymptote of each function. 52. f ( x) = eX - 1 51. f(x) = ex+ 2 50. f(x) = - ex " 49. f(x) e-X =
53. f(x)
=
In Problems 57-76, solve each equation. 57. 7x
61.
69.
= 73
58. 5 .1
Gr
(l)X 62. -
1
25
4 2 66. 4.1
=
=
=
'+
70. 5 .1
272.1 x e +S
y.2_7 =
73. e X
=
3
8
9 - 3e-x
5-6 =
1
-
64
1252.1
=
1 , what i s x ? What point is o n the graph o ff? 16
79. Suppose that g(x)
-
=
.2
-
=
(b) If H ( x ) =
_
of H?
83. If 4.1
=
85. If TX
( I ) ' - 2.
3 ' "2
e-
=
82. Suppose that F(x)
,
2, what does 3 2.1 equal?
86. If 5-x
-3
-2
=
=
(I )' +
-2 · :3
3, what does 4-x equal?
=
3, what does 53.1 equal? y
88.
20
20
16
16
12
12
�)
-1
(-1 , -2
2
3 x
e l2
=
8 (-1 ,
=
T1
1.
( a ) What is F ( - I ) ? What point i s o n the graph o f F ? ( b ) I f F ( x ) = -53, what i s x ? What point i s o n the graph of F?
In Problems 87-90, determine the exponential function whose graph is given.
y= O
=
%, what is x? What point is on the graph of f? -
: what is x? What point is on the graph
y
X'
27.1
5x - 3. (a) What is g( I ) ? What point is on the graph of g? (b) If g(x) = 122, what is x? What point is on the graph of g?
84. If 2.1
87.
=
76. (e4y . e
"1
80. Suppose that g(x)
4.1 + 2.
1
72. 92x · 27x'
1
(b) If f(x)
7, what does 4-2.1 equal?
=
•
-5
=
(a) What is H( 2 ) ? What point is on the graph of H? 1
e
=
81
=
68. 9-x + 15
1 6.1
= 1 62
3 . .r
TX =
64. 5x +3
4
=
2
60.
7 - 3e2x
=
78. Suppose that f(x) 3x. (a) What is f( 4)? What point is on the graph of f?
(a) What is g ( I ) ? What point is on the graph of g? (b) If g(x) = 66, what is x? What point is on the graph of g? 81. Suppose that H ( x )
=
71. 4.1 · 2x
56. f(x)
16
=
63. 22x - 1
75. eX
77. Suppose that f(x) = 2 ' . (a) What is f ( 4 ) ? What point is on the graph of f?
(b) If f( x )
59. TX
=
67. 8-.1+14
2.1 =
55. f(x)
-3
-2
�) -1 -2
2
3 x
y=O
SECTION 6.3
89.
(-1 ,
(0, -1 )
Y
-
�)
90.
�_"1��-7---:��3�X -1 0
Y=
y
2
0
/
(-1 , - �)
-20
(0, -1 )
3
x
Exponential Functions
435
y= o
-4 -8
-30
(2, -36)
-40
-1 2
In Problems 91-94, graph each function. Based on the graph, state the domain and the range and find any intercepts.
91. f(x) 93. f(x)
{ x ifif if = { - x if =
e-X e
-eX e
x < 0
x
�
92. f(x)
0
x < °
94. f(x)
x � °
{ ifif if = {=::x if
=
eX
-r
e '
x < 0
x � 0 x < 0 x � °
Applications and Extensions
95. Optics If a single pane of glass obliterates 3 % of the light passing through it, the percent p of light that passes through n successive panes is given approximately by the function pen)
100. Spreading of Rumors A model for the number N of people in a college community who have heard a certain rumor is
N
= 100(0.97 )"
=
Between 12:00 PM and 1 :00 PM, cars arrive at Citibank's drive-thru at the rate of 6 cars per hour (0.1 car per minute). The following formula from probability can be used to determine the probability that a car will arrive within t minutes of 1 2:00 PM:
1 0 1. Exponential Probability
-
(a) Find the atmospheric pressure at a height of 2 kilome ters (over a mile). (b) What is it at a height of 10 kilometers (over 30,000 feet)?
F(t)
97. Depreciation The price p, in dollars, of a Honda Civic DX Sedan that is x years old is given by p(x)
= 16,630(0.90Y
(a) How much does a 3-year-old Civic DX Sedan cost? (b) How much does a 9-year-old Civic DX Sedan cost? 98. Healing of Wounds The normal healing of wounds can be modeled by an exponential function. If Ao represents the original area of the wound and if A equals the area of the wound, then the function A(n) = Aoe-O.3Sn
describes the area of a wound after n days following an injury when no infection is present to retard the healing. Suppose that a wound initially had an area of 1 00 square millimeters. (a) If healing is taking place, how large will the area of the wound be after 3 days? (b) How large will it be after 10 days?
6:l
=
1 - e-01 1
(a) Determine the probability that a car will arrive within 10 minutes of 1 2:00 PM (that is, before 12:10 PM). (b) Determine the probability that a car will arrive within 40 minutes of 1 2:00 PM (before 12:40 PM). (c) What value does F approach as t becomes unbounded in the positive direction? (d) Graph F using a graphing utility. (e ) Using TRACE, determine how many minutes are needed for the probability to reach 50%.
F ( t) = 1 - e-0 151
5e-OAh
can be used to find the number of milligrams D of a certain drug that is in a patient's bloodstream h hours after the drug has been administered. How many milligrams will be pre sent after 1 hour? After 6 hours?
=
102. Exponential Probability Between 5:00 PM and 6:00 PM, cars arrive at Jiffy Lube at the rate of 9 cars per hour (0.15 car per minute). The following formula from probability can be used to determine the probability that a car will arrive with in t minutes of 5:00 PM:
99. Drug Medication The function
D(h)
P ( l - e-O.1 5d)
where P is the total population of the community and d is the number of days that have elapsed since the rumor began. In a community of 1000 students, how many students will have heard the rumor after 3 days?
(a) What percent of light will pass through 10 panes? (b) What percent of light will pass through 25 panes? 96. Atmospheric Pressure The atmospheric pressure p on a bal loon or plane decreases with increasing height. This pressure, measured in millimeters of mercury, is related to the height h (in kilometers) above sea level by the function p ( h) 760e 0 14Sh
=
�
(a) Determine the probability that a car will arrive within 15 minutes of 5 :00 PM (that is, before 5:15 PM). (b) Determine the probability that a car will arrive within 30 minutes of 5:00 PM (before 5:30 PM). (c) What value does F approach as t becomes unbounded in the positive direction? (e1 ) Graph F using a graphing utility. (e ) Using TRACE, determine how many minutes are needed for the probability to reach 60 %.
436
CHAPTER 6
Exponential and Logarithmic Functions
103. Poisson Probability Between 5:00 PM and 6:00 PM, cars ar rive at McDonald's drive-thru at the rate of 20 cars per hour. The following formula from probability can be used to de termine the probability that x cars will arrive between 5:00 PM and 6:00 P M .
E
-
[
P(x) where x!
=
x ' (x - 1 ) ' (x
-
2) ·
3 2 1
. . · ·
(a) Determine the probability that x between 5:00 PM and 6:00 P M . (b) Determine the probability that x between 5:00 PM and 6:00 P M .
·
·
1 5 cars will arrive
=
=
20 cars will arrive
104. Poisson Probabilit), People enter a line for the Demon Roller Coaster at the rate of 4 per minute. The following for mula from probability can be used to determine the proba bility that x people will arrive within the next minute.
P(x) where x!
=
x
'
(x
=
1 ) (x
-
'
4xe-4 x.
,
- 2)
(a) Determine the probability that x within the next minute. (b) Determine the probability that x within the next minute.
· . . · ·
3 2 1 ·
·
=
5 people will arrive
=
8 people will arrive
(a) If E = 120 volts, R 10 ohms, and L 5 henrys, how much current I , is flowing after 0.3 second? After 0.5 second? After 1 second? (b) What is the maximum current? (c) Graph this function 1 = 1] (/), measuring 1 along the y-axis and / along the x-axis. (d) If E = 1 20 volts, R = 5 ohms, and L = 10 henrys, how much current 1 2 is flowing after 0.3 second? After 0.5 second? After 1 second? (e) What is the maximum current? (f) Graph this function 1 = 12 ( t ) on the same coordinate axes as l , ( t ) . =
108. Cl\I'rent in a R C Circuit The equation governing the amount of current 1 (in amperes) after time t (in microsec onds) in a single RC circuit consisting of a resistance R (in ohms), a capacitance C (in microfarads), and an electromo tive force E (in volts) is 1
The relative humidity is the ratio (ex pressed as a percent) of the amoun t of water vapor in the air to the maximum amount that it can hold at a specific tem perature. The relative humidity, R, is found using the follow ing formula:
lOS. Relative Humidity
R
=
E e-t/(RC)
R
+
106. Learning Curve Suppose that a student has 500 vocabulary words to learn. If the student learns 1 5 words after 5 minutes, the function =
500 ( 1 -
e-0 006 1 t
)
approximates the number of words L that the student will learn a fter t minutes. (a) How many words will the student learn after 30 minutes? (b) How many words will the student learn after 60 minutes? 107. Current in a RL Circuit The equation governing the amount of current 1 (in amperes) after time / (in seconds) in a single RL circuit consisting of a resistance R (in ohms), an induc tance L (in henrys), and an electromotive force E (in volts) is
1
=
(T+459.4 4221 422 1 ) -D 459.4 + 2
10
where T is the air temperature (in OF) and D is the dew point temperature (in OF). (a) Determine the relative humidity if the air temperature is 50° Fahrenheit and the dew point temperature is 4 1 ° Fahrenheit. (b) Determine the relative humidity if the air temperature is 68° Fahrenheit and the dew point temperature is 59° Fahrenheit. (c) What is the relative humidity if the air temperature and the dew point temperature are the same?
L ( t)
=
=
� ( 1 - e-(R/L)t ] R
(a) If E = 120 volts, R = 2000 ohms, and C = 1.0 micro farad, how much current I , is flowing initially ( t = O)? After 1 000 microseconds? After 3000 mi croseconds? (b) What is the maximum current? (c) Graph this function I = 1J ( t ) , measuring 1 along the y-axis and t along the x-axis. (d) I f E = 120 volts, R 1000 ohms, and C = 2.0 micro farads, how much current 1 2 is flowing initially? Af ter 1000 microseconds? After 3000 microseconds? (e) What is the maximum current? (f) Graph this function 1 = 1 2 ( t ) on the same coordinate axes as 1J ( / ) . =
109. Another Formula for e Use a calculator t o compute the values of 1 1 1 2 + - + - + ..· + 3! n! 2!
for n
=
4, 6, 8, a n d 10. Compare each result with e.
[ Hint: I ! n!
=
=
1, 2! = 2 · 1, 3! = 3 2 ' 1, n ( n - 1 ) · . . ( 3 ) ( 2 ) ( 1 ) .] ·
· ·
SECTION 6.4
110. Another Formula for e Use a calculator to compute the various values of the expression. Compare the values to e.
:f, 111.
2 + 1 1 + 1 2 + 2 3 + 3 4 + 4 etc. Difference Quotient
.• .,
(a) Show that f(x) = sinh x is an odd function. (b) Graph f ( x ) = sinh x using a graphing utility.
1 cosh x = - ( eX + e-" ) 2 0
f (x + h ) - f (x) '--'---'-----'-----'- = a o' a" - 1 h h
• --
•
h "* 0
112. If f(x) = a'\ show that f ( A + B) = f ( A ) ' f ( B ) .
, 113. H f(x) = a" show that f( -x) =
437
116. The hyperbolic cosine function, designated by cosh x, is defined as
If f ( x ) = aX, show that 0
Logarithmic Functions
1 " f(x)
(a) Show that f ( x ) = cosh x is an even function. (b) Graph f(x) = cosh x using a graphing utility. (c) Refer to Problem 1 1 5 . Show that, for every x, (cosh x) 2 - (sinh X) 2 = 1
117. Historical Problem Pierre de Fermat (1601-1665) conjec tured that the function ' f ( x ) = 2(2 ) + 1
for x = 1 , 2, 3, . . . , would always have a value equal to a prime number. But Leonhard Euler (1707-1783) showed that this formula fails for x = 5. Use a calculator to deter mine the prime numbers produced by f for x = 1 , 2, 3, 4. Then show that f( 5 ) 641 X 6,700,4 1 7, which is not prime.
114. If f(x) = a'\ show that f( ax) = [f(x) J". Problems 115 and 116 provide definitions for two other transcendental functions.
=
115. The hyperbolic sine function, designated by sinh x, is de fined as
Discussion and Writing
1 1 8. The bacteria in a 4-liter container double every minute. After 60 minutes the container is full. How long did it take to fill half the container?
1 19. Explain in your own words what the number e is. Provide at least two applications that use this number. UO.
121. As the base a of an exponential function f(x) = aX, a > 1, increases, what happens to the behavior of its graph for x > O? What happens to the behavior of its graph for x < O? 1 22.
The graphs of y = a-x and y =
(�r are identical. Why?
Do you think that there is a power function that increases more rapidly than an exponential function whose base is greater than I? Explain.
'Are You Prepared?' Answers
1. 64; 4;
1
"9
2. ( I }
3. False
4. 3
5. True
6.4 Logarithmic Functions PREPARING FOR THIS SECTION
•
•
Before getting started, review the following:
Solving Inequalities (Section 1 .5 , pp. 128-131 ) Quadratic Inequalities (Section 4.5, pp. 314-316)
•
Polynomial and Rational Inequalities (Section 5.4, pp. 369-373)
Now Work the 'Are You Prepared?' problems on page 446.
OBJECTIVES 1
Change Exponential Expressions to Logarithmic Expressions and Logarithmic Expressions to Exponential Expressions (p.438)
2 Evaluate Logarithmic Expressions (p.438) 3 Determine the Domain of a Logarithmic Function (p.439) 4 Graph Logarithmic Functions (p.440) 5 Solve Logarithmic Equations (p.443)
438
CHAPTER 6
Exponential and Logarithmic Functions
Recall that a one-to-one function y = f(x) has an inverse function that is defined (implicitly) by the equation x = f(y). In particular, the exponential function y f(x) = aX, a > 0, a * 1 , is one-to-one and hence has an inverse function that is defined implicitly by the equation =
a
a > 0,
* 1
This inverse function is so important that it is given a name, the
DEFINITION
The
y
logarithmic function to the base
a,
where a > ° and
a
logarithmic function. * 1,
is denoted by
= loga x (read as "y is the logarithm to the base a of x") and is defined by
y=
loga x
if and only if
The domain of the logarithmic function As this definition illustrates, a
y
=
x = aY
loga x is
x > 0.
.J
logarithm is a name for a certain exponent.
Relating Logarithm s to Exponents
EXAM P L E 1
(a) If
y = log3 x, then x = 3Y•
(b) If y
=
logs x , then x
For example, 4 = log3 81 is equivalent to 81
= 5Y• For example, - 1
=
10gs
(�)
is equivalent to
�
= 34. =
5-1 . •
1
Change Exponential Expressions to Logarit h m i c Expressions and Logarithmic Expressions to Exponentia l Expressions
We can use the definition of a logarithm to convert from exponential form to loga rithmic form, and vice versa, as the following two examples illustrate. C hanging Exponential Expressions to Logarith m i c Expressions
EXAM P L E 2
Change each exponential expression to an equivalent expression involving a logarithm. (c) a4 = 24 (b) eb = 9 (a) 1 . 23 = m We use the fact that y = loga x and (a) If 1 .23 = m , then 3 = 10g1.2 m.
Solution
(c) If a4 U 1 '•
,t .
=
,
-
24, then 4
=
x
=
aY, a > 0, a * 1 , are equivalent. (b) If eb = 9, then b = loge 9.
loga 24.
•
Now Work P R O B L E M 9
C hanging Logarith m i c Expressions to Exponential Expressions
E XA M P L E 3
Change each logarithmic expression to an equivalent expression involving an exponent. (a) log" 4
=
5
(a) If loga 4 =
Solution
(c)
=
5, then as = 4. If log3 5 = c, then 3c = 5.
UJ1':I = ::::r.tl � .,..
2
(b) loge b
(c) log3 5
=
c
(b) If loge b
=
-3, then e-3
-3
=
b. •
Now Work P R O B L E M 1 7
Eva l u ate Loga rithmic Expressi o n s
To find the exact value of a logarithm, we write the logarithm in exponential nota tion and use the fact that if a" = aV then u = v.
SECTION 6.4
E XA M P L E 4
Logarithmic Functions
439
F i nd i ng the Exact Value of a Logarit h m i c Expression
Find the exact value of:
1 b (b) louo J 27 1 (b) y = log3 27
( a) log2 1 6 (a)
Solution
y= 2Y = 2Y = Y
=
log2 1 6 16 24 4
Ch ang e to expone ntial form .
16
= 24
3Y
Equate expon ents.
Therefore, log2 16
1
=27
1
T3
= y=
3Y
= 4.
C h ange to exponential form .
27
-
-3
3
-
Equate exponents.
Therefore, log3 �I"
1
= 33 = 3-3
1 27
=
-3. •
, or-
Now Work P R O B L E M 2 5
Determine the Doma i n of a Logarit h m ic Fu nction
The logarithmic function y = loga x has been defined as the inverse of the expo nential function y = aX. That is, if f(x) = aX, then rl (x) = loga x. Based on the discussion given in Section 6.2 on inverse functions, for a function f and its inverse r\ we have Domain ofr1
= Range of f
and
Range ofr1
=
Domain of f
Consequently, it follows that Domain of the logarithmic function Range of the logarithmic function
=
=
Range of the exponential function
Domain of the exponential function
= (0, 00 ) = ( - 00, 00 )
In the next box, we summarize some properties of the logarithmic function:
y = loga x Domain:
(defining equation:
° < x <
00
x = aY)
Range:
-00
<
y
<
00
The domain of a logarithmic function consists of the positive real numbers, so the argument of a logarithmic function must be greater than zero. EXAM P L E 5
F i nd i ng the Domain of a Logarith m i c F un ction
Find the domain of each logarithmic function. (a) Solution
F (x ) =
log2 (x + 3 )
(b)
g(x) =
logs
(--) 1 + x
1 - x
(a) The domain of F consists of all x for which x + 3 interval notation, the domain of f is ( -3, 00 ) .
(c)
hex) = logl/2 l x l
> 0, that is, x >
-3. Using
(b) The domain of g is restricted to
1 + x 1 - x
-- > 0 Solving this inequality, we find that the domain of g consists of all x between - 1 and 1 , that is, -1 < x < 1 Of, using interval notation, ( -1 , 1 ) .
( c) Since Ixl > 0, provided that x "* 0, the domain of h consists of all real num bers except zero or, using interval notation, ( - 00 , 0) U (0, 00 ). "'Ill!
•
Now Work P R O B L E M S 3 9 A N D 4 5
440
CHAPTER 6
Exponential and Logarithmic Functions
4
G ra p h Log a rith mic Fu nctions
Since exponential functions and logarithmic functions are inverses of each other, the graph of the logarithmic function y = logo x is the reflection about the line y = x of the graph of the exponential function y = a X , as shown in Figure 30.
Figure 30
Y Y = aX
Y = aX y = x
Y
y= x
3 x
-3
-3
-3
-3
(b) a > 1
(a) a < a < 1
For example, to graph y
=
log2 x, graph y
y = x. See Figure 31. To graph y the line y = x. See Figure 32.
=
=
2x
[Ogl/3 x, graph Figure 3 2
Figure 3 1
Y = (�
r
( - 1 , 3)
and reflect it about the line
y =
(l)X 3
and reflect it about
y=x
Y
-3
-3
�====> - Now Work P R O B L E M 5 9
Properties of the Logarithmic Function f(x) 1.
loga x
The domain is the set of positive real numbers; the range is the set of all real numbers.
2. The x-intercept of the graph is
3. The y-axis (x 4.
=
1 . There is no y-intercept.
0) is a vertical asymptote of the graph. A logarithmic function is decreasing if 0 < a < 1 and increasing if a =
5. The graph of f contains the points
( 1 , 0), (a, 1 ), and
(�, -1).
>
1.
6. The graph is smooth and continuous, with n o corners or gaps.
If the base of a logarithmic function is the number e, then we have the natural function occurs so frequently in applications that it is given a special symbol, In (from the Latin, logarithmus naturalis). That is,
logarithm function. This
y
=
In x if and only if x
=
eY
( 1)
Since y = In x and the exponential function y = eX are inverse functions, we can obtain the graph of y = In x by reflecting the graph of y = eX about the line y = x. See Figure 33.
SECTION 6.4
Using a calculator with an f ( x ) = In x. See Table 7. Figure 3 3
Logarithmic Functions
� key, we can obtain other points on the graph of Table 7
Y
5
y= e x
y=x
In x
x 2
� �
441
-0.69
2
0.69
3
1 .1 0
Seeing the Concept Graph Y1 = eX and Y2
I n x on the same square screen. Use eVALUEate to verify the points on the graph given in Figure 33. Do you see the symmetry of the two g raphs with respect to the line y = x? =
E XA M P L E 6
X= 0
Figure 34
I I I
G raph i ng a Logarithmic F u n ction and Its I nverse
(a) (b) ( c) (d) (e) (f) Solution
3 x
y = 0 -3
Find the domain of the logarithmic function f ( x ) = -In ( x - 2 ) . Graph f From the graph, determine the range and vertical asymptote of f Find f - 1 , the inverse of f Use rl to find the range of f Graph f - l .
(a) The domain of f consists of all x for which x - 2 > 0 or, equivalently, x > 2. (b) To obtain the graph of y = -In ( x - 2 ) , we begin with the graph of y = In x and use transformations. See Figure 34.
Y
Y 3
3
x= O
Y 3
x- 2
x= o
x
-1
-1
-1
(a) y = In x
Multi ply by - 1 ; reflect about x-axis
Replace x by x - 2; shift right 2 u n its.
(b) y = - I n x
(c) y = - l n (x - 2)
(c) The range of f(x) = - I n( x - 2 ) is the set of all real numbers. The vertical as ymptote is x = 2. [Do you see why? The original asymptote ( x = 0) is shifted to the right 2 units.] (d) We begin with y - In ( x - 2 ) . The inverse function is defined (implicitly) by the equation =
x = -In ( y - 2 ) We proceed to solve for y. -x = In(y - 2 ) e-x = y - 2 y e -x + 2 =
Isolate the l ogarith m . Change to an exponential expression . Solve for y.
The inverse of f is f -l ( x ) = e-x
+
2.
442
CHAPTER 6
Exponential and Logarithmic Functions
(e) The range of jis the domain of j -I, the set of all real numbers, confirming what we found from the graph of f. (f) To graph rI, we use the graph of j in Figure 34 (c) and reflect it about the line y = x. See Figure 35.
Figure 3S
Y 5
x= 2 Y= X
-1
I I
fix) = : I n (x
-1
Figure 36
I I I
-
-
2)
y =x
• 1.l!J ! �
Y = log
4
x
Now Work P R O B L E M 7 1
If the base of a logarithmic function is the number 10, then we have the common If the base a of the logarithmic function is not indicated, it is understood to be 10. That is,
logarithm function.
y=
x
log x if and only if
x=
lOY
Since y = log x and the exponential function y = lOX are inverse functions, we can obtain the graph of y = log x by reflecting the graph of y = lOX about the line y = x. See Figure 36.
-2 EXAM PLE 7
G raphing a Logarith m i c F un ction and Its I nverse
(a) (b) ( c) (d) ( e) (f)
Find the domain of the logarithmic function j (x) = 3 log (x - 1 ) . Graph f From the graph, determine the range and vertical asymptote of f Find j -l, the inverse of f Use j -1 to find the range of f Graph j -1 .
( a) The domain of j consists of all x for which x - 1 > a or, equivalently, x > l . (b) To obtain the graph of y = 3 loge x - 1 ) , we begin with the graph of y = log x and use transformations. See Figure 37.
Solution
Figure 3 7
2 -2 -2
I I I I
Y X= 1
Y X= 0
2
(10, 1 ) ( 1 , 0) ._-----1 2 (fo, - 1 )
4
6
8
10
1 2 x -2
Y X= 1
(1 1 , 1 ) : (2, 0) ._-----6
10
1 2 x -2
-2
-2 ----+-
Replace x by x - 1 ; horizontal shift right 1 unit (a) Y = log x
8
2
M u ltiply by 3; vertical stretch by a factor of 3.
(b) y = log (x - 1 )
(c) y = 3 log (x - 1 )
SECTION 6.4
Logarithmic Functions
443
(c) The range of f(x) = 3 10g(x - 1 ) is the set of all real numbers. The vertical as ymptote is x = 1 . (d) We begin with y = 3 10g(x - 1 ) . The inverse function i s defined (implicitly) by the equation x
=
3 log(y - 1 )
We proceed to solve for y. x '3 = log (y - 1 ) lOx/3
Isolate th e logarith m .
- 1 Change to an exponential expression . 3 Solve for y. Y = lOx/ + 1 3 The inverse of f is f -1 ( x) = lOx/ + 1 . (e) The range of fis the domain of f -l, the set of all real numbers, confirming what we found from the graph of f. (f) To graph f -1 , we use the graph of f in Figure 37(c) and reflect it about the line y = x. See Figure 38. Figure 38
=
Y
y
•
�==> .5
E XA M P L E 8
So lve Loga rith mic Equations
Equations that contain logarithms are called logarithmic equations. Care must be taken when solving logarithmic equations algebraically. In the expression logo M , remember that a and M are positive and a *' 1 . B e sure to check each apparent solution in the original equation and discard any that are extraneous. Some logarithmic equations can be solved by changing from a logarithmic expression to an exponential expression. Solving a Logarith m i c E q u ation
Solve: S o l ution
Now Work P R O B L E M 7 9
(a) log3 (4x - 7)
=
2
(b) logx 64
=
2
(a) We can obtain an exact solution by changing the logarithm to exponential form. log3 (4x - 7) 4x - 7
=
=
4x - 7
=
4x
=
x
=
2
32
9 16 4
C h a ng e to an exponential expression .
444
CHAPTER 6
Exponential and Logarithmic Fu nctions
Check: log3 ( 4x - 7 )
log3 ( 1 6
=
-
7)
=
i og3 9
=
2
32
= 9
The solution set is (4J. (b) We can obtain an exact solution by changing the logarithm to exponential form. logx 64 x2
=
2
=
64
X = ±
V64
Change to an exponentia l expression .
=
±8
Square Root Method
The base of a logarithm is always positive. As a result, we discard -8. We check the solution 8.
Check: logs 64
=
2
82
=
64
The solution set is (8J.
EXAM PLE 9
U s i n g Logarithms t o Solve Exponential E quations
Solve: Solution
•
e2x =
5
We can obtain an exact solution by changing the exponential equation to logarithmic form. e2 x 5 =
In 5
x
The solution set 1"
EXAM P L E 1 0
E .....
=
= �
2x
In 5 -2 0.805
Change to a loga rith m ic expression using the
fact that if eY = x then y = In
x.
Exact solution A pproxi mate sol ution
. { In 5 }
IS
2 .
•
Now Work P R O B L E M S 8 7 A N D 9 9
Alcohol and D rivi ng
The blood alcohol concentration (BAC) is the amount of alcohol in a person's blood stream. A BAC of 0.04% means that a person has 4 parts alcohol per 10,000 parts blood in the body. Relative risk is defined as the likelihood of one event occurring divided by the likelihood of a second event occurring. For example, if an individ ual with a BAC of 0.02% is 1 .4 times as likely to have a car accident as an indi vidual that has not been drinking, the relative risk of an accident with a BAC of 0.02% is 1 .4. Recent medical research suggests that the relative risk R of having an accident while driving a car can be modeled by the equation R = e
kx
where x is the percent of concentration of alcohol in the bloodstream and k is a constant. (a) Medical research indicates that the relative risk of a person having an accident with a BAC of 0.02% is 1 .4. Find the constant k in the equation. (b) Using this value of k, what is the relative risk if the concentration is 0. 17%? (c) Using this same value of k , what BAC corresponds to a relative risk of 100? (d) If the law asserts that anyone with a relative risk of 5 or more should not have driving privileges, at what concentration of alcohol in the bloodstream should a driver be arrested and charged with a DUI (driving under the influence)? Solution
(a) For a concentration of alcohol in the blood of 0.02% and a relative risk of 1 .4, we let x = 0.02 and R = 1 .4 in the equation and solve for k. R
1.4
k e x k (O.02) = e
=
R
= 1.4; x =
0.02
S E CTION 6.4
0.02k
=
In 1.4 In 1.4 --
445
Change to a logarithm ic expression. �
16.82 Solve for k. 0.02 For a concentration of 0.17 % , we have x = 0.17. Using k = 16.82 in the equa tion, we find the relative risk R to be R = ekx e(16.82)(O.17) = 17.5 For a concentration of alcohol in the blood of 0.17%, the relative risk of an ac cident is about 17.5. That is, a person with a BAC of 0.17% is 17.5 times as likely k
(b)
=
Logarithmic Functions
=
to have a car accident as a person with no alcohol in the bloodstream. risk of 100, we have R = 100. Using k = 16.82 in the equation ( c) For a relative x ek R , we find the concentration x of alcohol in the blood obeys =
NOTE A BAC of 0.30 results in a loss of consciousness in most people. •
100 = e16.82x 16.82x = In 100 In 100 x= 16.82
R =
ekx, R
=
100; k = 1 6.82
Change to a logarithmic expression. �
0.27
Solve for x.
For a concentration of alcohol in the blood of 0.27 % the relative risk of an accident is 100. (d) For a relative risk of 5, we have R 5. Using k = 16.82 in the equation R = ekx, we find the concentration x of alcohol in the bloodstream obeys =
5 = e16.82x 16.82x = In 5 In 5 � 0.096 x 16.82 =
NOTE Most states use 0.08 or 0.10 as the blood alcohol content at which a • DUI citation is g iven.
SUMMARY =
BAC
of 0.096 % or more should be arrested and charged •
Properties of the Logarithmic Function
logax, a> 1 (y = loga x means x aY)
f(x)
A
driver with a with DUI.
=
loga x, 0 < a < 1 (y = loga x means x = aY)
f(x) =
Domain: the interval (0, 00 ); range: the interval ( -00 , 00) x-intercept: 1; y-intercept: none; vertical asymptote: x 0 (y-axis); increasing; one-to-one See Figure 39(a) for a typical graph. Domain: the interval (0, ) range: the interval ( -00, (0) x-intercept: 1; y-intercept: none; vertical asymptote: x = 0 (y-axis); decreasing; one-to-one See Figure 39(b) for a typical graph. =
(0 ;
Figure 39
y 3
Y 3
: X= 0
-3 -3 (a)
-3
I
: X= 0
a>
1
(b)
0<
a<
1
446
CHAPTER 6
6.4
Exponential and Logarithmic Functions
Assess You r Understa nding
'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. x -1 1. Solve the inequality: 3 x - 7 :=; 8 - 2x (pp. 128-13 1) 3. Solve the inequality: > 0 (pp. 369-373) x+4 2 2 . Solve the inequality: x - x - 6 > 0 (pp. 314-316) Concepts and Vocabulary 4. The domain of the logarithmic function f(x) = log" x is
7. True or False 8. True or False
5. The graph of every logarithmic function f(x) = log" x,
a >
0,
y
=
log"
x, then y
The graph of
f(x)
=
aX.
= log"
x, a >
has an x-intercept equal to1 and no y-intercept.
*- 1, passes through three points: __, __, and
a
If
6. If the graph of a logarithmic function f(x) = log" x, 0,
a >
than
a
*-
1,
is increasing, then its base must be larger
Skill Building
In Problems 9-16, change each exponential expression to an equivalent expression in volving a logarithm. 9. 9
13. 2x
=
32
=
10. 16
=
42
11.
14. 3x = 4. 6
7. 2
a2
12. a3 = 2.1
1 .6
=
15. eX = 8
16. e2.2 = M
In Problems 1 7-24, change each logarithmic expression to an equivalent expression involving an exponent.
{�)
17. log28 = 3
18.10g
21. log3 2 = x
22. log 6
2
=
= -2
19. log" 3 = 6
20.10gb 4
x
23. In 4
24. In x = 4
=
x
In Problems 25-36, find the exact value of each logarithm without using a calculatOl:
=
2
(i)
25. log21
26. logs 8
27. logs 25
28.10g
29. log'!216 33. logv'2 4
30. log'!39
31. 1 0g lO ViO
32.10gsV2s
34.10gy'39
3
36. In e
35.1nVe
3
In Problems 37-48, find the domain of each function. 37. f(x) 40. H(x)
=
I n ( x- 3)
38. g ( x) = In (x - 1)
logs
41. f(x)
=
3- 210g4
44. g(x)
=
l_ In _ x- 5
47·f(x)
=
=
x3
(x + ) C� )
43. f(x) = In _l_ 1
46. hex)
=
IOg3
1
(
[1 ] ) - 5
�
8 + 5 In(2x + 3)
42. g(x)
=
45. g(x)
=
48. g(x)
=-
log5
In
l
(X-+x- )
x
In Problems 49-56, use a calculator to evaluate each expression. Round your answer to three decimal places.
49.
5 In]'
53.
----
In 4 + In 2
log4 + log 2
10
50
•
54.
In-
�
51.
--=----=
55.
3
log 15+ log 20 In15+ In 20
57. Find a so that the graph of f( x)
=
log"
x contains the point ( 2, 2).
58. Find a so that the graph of f(x)
=
log"
x contains the point
3
52.
0 . 04
21n 5+ log 50
----=-
56.
log4 - In 2
(�, )
-4 .
In Problems 59-62, graph each fun ction and its inverse on the same Cartesian plane. 59. f(x) = 3'\ r' (x) = log3 x
60. f(x) = 4\rl (x)
=
log4
x
1
2 In3 -0 .1
310g80 - In 5
-=-----.,--
log 5 + In
-
20
0,
a
*- 1,
SECTION 6.4
61. f(x) =
(1)X "2
;r 1(x) = loglx
(l)X;rl(x)
62. f(x) =
� J
=
447
Logarithmic Functions
logp: '
In Problems 63-70, the graph of a logarithmic function is given. Match each graph to on e of the following functions: A. E.
Y = log3X
Y = IOg3x-I
F.
3 rx x -3 Y
63.
=
0
x=O3 -3
= IOg3(X- 1)
y
31- :x -3 Yt
64.
y
= -log3x
D.
G. Y = log3(1 - x)
I I
1
67.
c.
Y = IOg3(-x)
B.
YI
65. = 1
H.
66.
_�o
68.
----'----
69.
5x
-3 1-
-1
-1
y
1
= - log3x
3 �x=O L t � ytx=3 1 Yt
,..
5
-1
-1
Y
Y =-log3(-x)
3 rx= -3 Y
-3 �
x
-3 70.
0
-3
-1
In Problems 71-86, use the given function f to: (a) Find the domain of! (b) Graph! (c) From the graph, determine the range and any asymptotes of! (d) Find r1, the in verse of! (e) Use f-1 to find the range of! (f) Graph rl.
73. f(x) = 2 +Inx
74. f(x)
76. f(x) =-21n (x +1)
77. f(x) = log(x- 4) +2
78. f(x) =- Iogx- 5 2
80. f(x) = loge -2x)
81. f(x) = 3 + log3(X
82. f(x)
84. f(x) = 3ex +2
3 85. f(x) = 2 x/ + 4
86. f(x) =-3 x+1
89. IOg (2x + 1) = 3 2
90. log3(3x- 2) = 2
93.In e = 5
94.In e-2x = 8
71. f(x) = In(x + 4)
72. f(x)
75. f(x) = In(2x)- 3 79. f(x) =
1
"2
log(2x)
83. f(x) = ex+2- 3
=
In(x- 3)
In Prob lems 87-11 0, solve each equation. 87. log3x = 2
88. log5 x = 3
91. log 4 x
92. IOg
2
=
103. log3(X2+ 107. 5eO.2x
=
1)
(�)
=
3
96. logs625 = x
95. IOg464 = x
99. e3x = 10
x
100. e-2x = =2
111. SupposethatG(x)
+
108. 8·102x-7
=
IOg3(2x +
(a) What is thedomain of G?
x + 4) = 2 =
3
1).
(b) What i s G(40)? What point is on the graph o f G? (c) If G(x) = 2, what is x? What point is on the graph of G?
97. log3243 = 2x + 101. e2x+S
3
104. logs(x 2
7
.!.
X
=
=
-In(-x)
=
98. log636 102. e- 2x +1
8
105. IOg 8x = -3 2 109.2 _102-x
1
+ 2)
=
5
1
2- IOg3(X + 1)
= =
106. log3 3x = 110. 4 ex+1
=
5x +3
13 -1
5
112. Supposethat F(x) = IOg (X+ 1)- 3. 2 (a) What is thedomain of F ?
(b) What is F(7)? What point is on the graph of f? (c) If F(x) = -1, what is x? What point is graph of F ?
on
the
448
CHAPTER 6
{ { -In x
Exponential and Logarithmic Functions
{ {ln x
In Problems 113-116, graph each function. Based on the graph, state the domain and the range and find any intercepts. if x :5 - 1 ln (- X) In (-X) if x < 0 114. f (x ) = 113. f(x) = - l n ( -x ) if - 1 < x < 0 In x if x > 0 115. f ( x)
=
In x
if 0 < x < 1 if x 2: 1
116. f (x)
=
- In x
if 0 < x < 1 if x 2: 1
Applications and Extensions
The pH of a chemical solution is given by the
117. Chemistry formula
where [H +] is the concentration of hydrogen ions in moles per liter. Values of pH range from 0 (acidic) to 14 (alkaline). (a) What is the pH of a solution for which [H+] is 0.1? (b) What is the pH of a solution for which [H+] is 0.01? (c) What is the pH of a solution for which [H+] is 0.001? (d) What happens to pH as the hydrogen ion concentration decreases? (e) Determine the hydrogen ion concentration of an orange (pH 3.5 ) . (f) Determine the hydrogen i o n concentration o f human blood (pH = 7.4 ). =
118. Diversity Index Shannon's diversity index is a measure of the diversity of a population. The diversity index is given by the formula H
=
- (PI log PI
+ P2 10g P2 + . . . +
Pl1log PH )
where PI is the proportion of the population that is species 1 , P 2 i s the proportion o f the population that is species 2 , and so on. (a) According to the U.S. Census B ureau, the distribution of race in the United States in 2000 was as follows: Proportion
Race American Indian or Native
0.014
Alaskan
0.041
Asian Black or African American
0.128
Hispanic
0.124
Native Hawaiian or Pacific Islander
0.003
White
0.690
Source: u.s.
Census Bureau
Compute the diversity index of the United States in 2000. (b) The largest value of the diversity index is given by logeS) , where S is the number of categories of Hmax race. Compute Hmax' =
(c) The evenness ratio is given by Ef{
=
H
Hmax
--
, where
0:5 EN:5 1 . If Ef{ = 1 , there is complete evenness. Compute the evenness ratio for the United States. (d) Obtain the distribution of race for the United States in 1 990 from the Census B ureau. Compute Shannon's di versity index. Is the United States becoming more di verse? Why?
119. Atmospheric Pressure TIle atmospheric pressure p on a bal loon or an aircraft decreases with increasing height. This pres sure, measured in millimeters of mercury, is related to the height h (in kilometers) above sea level by the formula O 45h P 760e- .1 (a) Find the height of an aircraft if the atmospheric pres sure is 320 millimeters of mercury. (b) Find the height of a mountain if the atmospheric pres sure is 667 millimeters of mercury. =
120. Healing of Wounds The normal healing of wounds can be modeled by an exponential function. If Ao represents the original area of the wound and if A equals the area of the wound, then the formula A Aoe-0.3511 =
describes the area of a wound after n days following an injury when no infection is present to retard the healing. Suppose that a wound initially had an area of 100 square millimeters. (a) If healing is taking place, after how many days will the wound be one-half its original size? (b) How long before the wound is 10% of its original size?
121. Exponential Probability Between 1 2:00 PM and 1 :00 PM, cars arrive at Citibank's drive-thru at the rate of 6 cars per hour (0.1 car per minute). The following formula from statistics can be used to determine the probability that a car will arrive within t minutes of 12:00 PM. F(t ) 1 e-O.II =
-
(a) Determine how many minutes are needed for the prob ability to reach 50%. (b) Determine how many minutes are needed for the prob ability to reach 80% . (c) I s it possible for the probability t o equal 100% ? Explain.
122. Exponential Probability Between 5 :00 PM and 6:00 PM, cars arrive at Jiffy Lube at the rate of 9 cars per hour (0.15 car per minute). The following formula from statistics can be used to determine the probability that a car will arrive within t minutes of 5:00 PM. F(t ) = 1 - e-O.151 (a) Determine how many minutes are needed for the prob ability to reach 50 % . (b) Determine how many minutes are needed for the prob ability to reach 80% . 123. Drug Medication The formula D
=
5e-0.4h
can be used to find the number of milligrams D of a certain drug that is in a patient's bloodstream h hours after the drug was administered. When the number of milligrams reaches 2, the drug is to be administered again. What is the time between injections?
SECTION 6.4
124. Spreading of Rumors A model for the number N of people in a college community who have heard a certain rumor is N
=
126. Learning Curve
P ( l - e-0.J5d)
449
Psychologists sometimes use the function L(t)
=
A( l - e -kl)
to measure the amount L learned at time t. The number A represents the amount to be learned, and the number k mea sures the rate of learning. Suppose that a student has an amount A of 200 vocabulary words to learn. A psychologist determines that the student learned 20 vocabulary words after 5 minutes. (a) Determine the rate of learning k. (b) Approximately how many words will the student have learned after 10 minutes? (c) After 15 minutes? (d) How long does i t take for the student to learn 1 80 words?
where P is the total population of the community and d is the number of days that have elapsed since the rumor began. In a community of 1000 students, how many days will elapse before 450 students have heard the rumor? 125. Current in a RL Circuit The equation governing the amount of current 1 (in amperes) after time t (in seconds) in a simple RL circuit consisting of a resistance R (in ohms), an induc tance L (in henrys), and an electromotive force E (in volts) is I =
Logarithmic Functions
.£ [ 1 - e -(R/L)I] R
If E = 12 volts, R = 10 ohms, and L = 5 henrys, how long does it take to obtain a current of 0.5 ampere? Of 1.0 am pere? Graph the equation.
-
Problems 127-130 use the following discussion: The loudness L( x), measured in decibels, of a sound of intensity x, x measured in watts per square metel; is defined as L ( x) = 10 log , where 10 = 10- 1 2 watt per square meter is the least intense sound
Loudness of Sound
10
that a human ear can detect. Determine the loudness, in decibels, of each of the following sounds. 3 127. Normal conversation: intensity of x = 10- 7 watt per square 129. Heavy city traffic: intensity of x = 10- watt per square meter. meter. 130. Diesel truck traveling 40 miles per hour 50 feet away: inten 128. Amplified rock music: intensity of 1 0- 1 watt per square meter. sity 10 times that of a passenger car traveling 50 miles per hour 50 feet away whose loudness is 70 decibels.
The Richter Scale Problems 131 and 132 use the following discussion: The Richter scale is one way of converting seismographic readings into numbers the Richter scale that provide an easy reference for measuring the magnitude M of an earthquake. All earthquakes are com pared to a zero-level earthquake whose seismographic reading measures 0.001 millimeter at a distance of 1 00 kilometers from the epicenteJ: An earthquake whose seismogra phic reading measures x millimeters has magnitude M( x), given by M( x)
= 10g
where Xo = 10- 3 is the reading of a zero-level earthquake the same distance from its epicenter. In Problems 131 and 132, determine the magnitude of each earthquake.
131. Magnitude of an Earthquake Mexico City in 1985: seismo graphic reading of 125,892 millimeters 100 kilometers from the center. 132. Magnitude of an Earthquake San Francisco in 1 906: seis mographic reading of 7943 millimeters 100 kilometers from the center. 133. Alcohol and Driving The concentration of alcohol in a per son's bloodstream is measurable. Suppose that the relative risk R of having an accident while driving a car can be mod eled by the equation
where x is the percent of concentration of alcohol in the bloodstream and k is a constant.
( �) Xo
(a) Suppose that a concentration of alcohol in the blood stream of 0.03 percent results in a relative risk of an ac cident of 1 .4. Find the constant k in the equation. (b) Using this value of k, what is the relative risk if the con centration is 0.17 percent? (c) Using the same value of k, what concentration of alco hol corresponds to a relative risk of 100? (d) If the law asserts that anyone with a relative risk of hav ing an accident of 5 or more should not have driving privileges, at what concentration of alcohol in the blood stream should a driver be arrested and charged with a DUI? (e) Compare this situation with that of Example 10. If you were a lawmaker, which situation would you support? Give your reasons.
450
CHAPTER 6
Exponential and Logarithmic Functions
Discussion and Writing
134. Is there any function of the form y x G', O < a < 1 , that in creases more slowly than a logarithmic function whose base is greater than I? Explain. =
Age in Years
2
3
4
5
$32,400
$28,750
$25,400
$21 ,200
New
135. In the definition of the logarithmic function, the base a is not
$38,000
$36,600
allowed to equal 1. Why?
Use the formula New = Old (eRt) to find R, the annual de preciation rate, for a specific time t. When might be the best time to trade in the car? Consult the NADA ("blue") book and compare two like models that you are interested in. Which has the better depreciation rate?
136. Critical Thinking In buying a new car, one consideration might be how well the price of the car holds up over time. Different makes of cars have different depreciation rates. One way to compute a depreciation rate for a car is given here. Suppose that the current prices of a certain Mercedes automobile are as follows: 'Are You Prepared?' Answers
1. x ::; 3 '�'-
3. x < -4 or x > 1
2. x < -2 or x > 3
� ..
6.5 Properties of Logarithms OBJECTIVES 1 Work with the Properties of Logarithms (p.450)
2 Write a Logarithmic Expression as a Sum or Difference of Logarithms (p.452)
3
Write a Logarithmic Expression as a Single Logarithm (p.453)
4 Evaluate Logarithms Whose Base Is Neither 10 nor e (p.454)
1
E XA M PLE 1
Work with the Properties of Logarithms
Logarithms have some very useful properties that can be derived directly from the definition and the laws of exponents. Establishing Properties of Logarithms
(a) Show that loga 1 = o. (b) Show that loga a 1. (a) This fact was established when we graphed y = loga (see Figure 30). To show the result algebraically, let y loga 1 . Then y = loga 1 =
Solution
x
=
Change to an exponential expression. aO = 1
y =o
(b) Let y
=
loga 1 = 0 loga a. Then y = loga a
Solve for y.
y
aY = a aY = a 1 y= 1 loga a = 1
=
loga 1
Change to an exponential expression. a a1 =
Solve for y.
y
=
logaa
To summarize: loga 1
=
0
loga a = 1
•
SECTION 6.S
THEOREM
45 1
Properties of Logarithms
Properties of Logarithms =1=
r
In the properties given next, M and a are positive real numbers, a 1, and is any real number. The number loga M is the exponent to which a must be raised to obtain M. That is, (1)
I
The logarithm to the base a of a raised to a power equals that power. That is, logaaf (2) I --
�----------
= r
----
--------
y
=
--
--
.�
--
�
----
y =
The proof uses the fact that aX and loga x are inverses. Proof of Property (1) For inverse functions, .f(r'(x)) x for all x in the domain of r' Using f(x) aX and r1(x) toga we find f(r'(x)) a'ogax x for x > 0 Now letx M to obtain a'og{/I'vl M, where M > o. Proof of Property (2) For inverse functions, rl(f(x)) x for all x in the domain of f Using f(x) aXand rl(x) !og"x, we find r' (f(x)) logac["' x for all real numbers x Now let x to obtain loga af where r is any real number. =
=
=
X,
=
=
=
=
•
=
=
=
=
= r
E XA M PLE 2
=
= r,
•
Using Properties (1) and (2) 2'og21T
(a) L'Ji
-
=
1T
(b) log0 2 o.rV2
=
-
v2
(c) In
ekl =
kt
•
Now Work PRO B L E M 9
Other useful properties of logarithms are given next. THEOREM
Properties of Logarithms
In the following properties, M, N, and a are positive real numbers, a is any real number.
r
=1= 1,
and
The Log of a Product Equals the S um of the Logs
(3) The Log of a Quotient Equals the Difference of the Logs
(4)
452
CHAPTER 6
Exponential and Logarithmic Functions
The Log of a Power Equals the Prod uct of the Power and the Log
logaMr = r logaM
(5)
I
--------------------------------��
�
We shall derive properties (3) and (5) and leave the derivation of property (4) as an exercise (see Problem 103). Proof of Property (3) Let A = logaM and let B = logaN. These expressions are equivalent to the exponential expressions A a = M and aB = N Now loga(MN) = loga( Aa ) = loga aA+B Law of Exponents a
B
= A + B = logaM
Proof of Property (5)
Now
+ logaN Let A = logaM. This expression is equivalent to A a =M
loga M r = loga (aA) r = loga arA
= rA = r logaM
� == » -
2
Property (2) of logarithms •
Law of Exponents Property (2) of logarithms •
Now Work P R O B L E M 1 3
Write a Logarithmic Expression as a Sum or Difference of Logarithms
Logarithms can be used to transform products into sums, quotients into differences, and powers into factors. Such transformations prove useful in certain types of cal culus problems. EXAMPLE 3
Solution
Writing a Logarithmic Expression as a Sum of Logarithms
Write loga (x�i) ,x> 0, as a sum of logarithms. Express all powers as factors. loga (xP+1) = logax + loga P+1 logA . N) logaM + logaN = logax + loga(x2 + 1) 1 /2 1 = logax + "2 loga(x- + 1) loga loga =
M
?
Mr
= r
M
•
EXAMPLE 4
Writing a Logarithmic Expression as a Difference of Logarithms
Write
x2--:: x> 1 I n -(x 1)3 as a difference of logarithms. Express all powers as factors. -
Properties o f Logarithms
SECTION 6.5
453
In (x x2- 1) 3 Inx2 - In(x - 1)3 2 1nx - 3 1n(x -1)
Solution
=
i
i
=
loga
(�) =
loga M - loga N
loga M'
= r
logaM •
Writing a Logarithmic Expression as a Su m and Difference of Logarithms
E XA M P L E 5
Write as a sum and difference of logarithms. Express all powers as factors. , � loga xyQ"+l 3 (x + 1)4 loga V + 1 -loga[x3 (x + 1)4]
WARNING
I n using properties (3) through (5), be careful about the values that the variab le may assume. For example, the domain of the variable for loga x is x > 0 and for loga(x - 1) it is x > 1 . If we add these func tions, the domain is x > 1. T hat is, the equality loga x + loga(x - 1) = 10gA x(x - 1)]
is true only for x > 1 .
= =
=
3
Solution
=::;;:;;:>
�
•
E XAMPLE 6
Xk
=
Solution
-
Property (4)
loga yQ"+l - [logax3 + loga(x + 1)4] loga(x2 + 1)1/2 - logax3 - log(,(x + 1)4 1 2 loga(x2 + 1) - 3 loga - 4 loga( + 1) X
X
Property (3)
Property (5) •
Now Work PRO B L E M 4 5
Write a Logarithmic Expression as a Single Logarithm
Another use of properties (3) through (5) is to write sums and/or differences of log arithms with the same base as a single logarithm. This skill will be needed to solve certain logarithmic equations discussed in the next section. Writing Expressions as a Single Logarithm
Write each of the following as a single logarithm. (a) loga 7 + 4 logo 3 (b) "32 1n 8 - In(34 - 8) (c) logax + loga 9 + loga(x2 + 1) - loga 5 (a) loga 7 + 4 logo 3 logo 7 + loga 34 logaM logaM' logo 7 + loga 81 logo (7' 81) logo 567 (b) "32 In 8 - In(34 - 8) In 82/3 - In( 81 - 8) rlogaM logaM' In 4 - In 73 In ( 7�) r
=
=
= =
=
=
=
=
=
454
CHAPTER 6
Exponential and Logarithmic Functions
• WARNING
A common error made by some students is to express the logarithm of a sum as the sum of logarithms. loga( M
Correct statement
is not equal to
+ N)
loga( MN)
=
loga M
+
loga M
+
loga N
Property (3)
loga N
Another common error is to express the d ifference of logarithms as the quotient of logarithms.
Correct statement
M-
loga M
loga N
loga
loga N
is not equal to =
(�)
IOga
loga M loga N
Property (4)
A third common error is to express a logarithm ra ised to a power as the product of the power times the logarithm. is not equal to
(Ioga MY Correct statement
�:;;:::::.
-
loga Mr
= r loga M
r
loga M
•
Property (5)
Now Work P R O B L E M 5 1
Two other properties of logarithms that we need to know are consequences of the fact that the logarithmic function loga x is a one-to-one function. y =
THEOREM
Properties of Logarithms =f.
In the following properties, M, N, and a are positive real numbers, a 1. If M N, then loga M loga N. (6) If loga M loga N , then M N . (7) =
=
=
=
�------�� =
When property (6) is used, we start with the equation M N and say "take the logarithm of both sides" to obtain loga M loga N . Properties (6) and (7) are useful for solving exponential and logarithmic equa tions, a topic discussed in the next section. =
4
Evaluate Logarithms Whose Base Is Neither 1 0 nor
e
Logarithms to the base 10, common logarithms, were used to facilitate arithmetic computations before the widespread use of calculators. (See the Historical Feature at the end of this section.) Natural logarithms, that is, logarithms whose base is the number e, remain very important because they arise frequently in the study of nat ural phenomena. Common logarithms are usually abbreviated by writing with the base understood to be 10, just as natural logarithms are abbreviated by with the base understood to be e. Most calculators have both � and � keys to calculate the common loga rithm and natural logarithm of a number. Let's look at an example to see how to approximate logarithms having a base other than 10 or e. log,
In,
S ECTION 6.S
EXAMPLE 7 Solution
Properties o f Logarithms
Approximating a Logarithm Whose Base Is Neither 1 0 nor
455
e
Approximate log2 7. Round the answer to four decimal places. Let log2 7. Then 2Y 7, so 2Y 7 In 2Y In 7 Property (6) y in 2 In 7 Property (5) In 7 Exact va lue In 2 2.8074 Approximate val ue rounded to four y=
=
=
=
=
y= y
�
decima l places
•
Example 7 shows how to approximate a logarithm whose base is 2 by changing to logarithms involving the base In general, we use the e.
THEOREM
Change-or-Base Formula.
Change-of-Base Formula
If a i= 1, b i= 1, and M are positive real numbers, then (8)
I
�
�----------------------------------�
Proof
y=
We derive this formula as follows: Let logoM. Then aY M 10gb aY 10gbM Property (6) 10gb a 10gbM Property (5) 10gbM -- Solve for 10gb a 10gaM 10gbM loga M 10gb a =
=
=
y
y=
y.
=
y =
•
Since calculators have keys only for � and � , in practice, the Change-of Base Formula uses either b 10 or b That is, =
= e.
M and log M In M logaM log -1-og In a =
E XAMPLE 8
Solution
a
a
Using the C hange-of-Base Form ula
Approximate: (a) logs 89 (b ) 10gy'2 Vs Round answers to four decimal places. log 89 1.949390007 2.7889 ( a ) logs 89 log 5 0.6989700043 or 4.48863637 2 .7889 log-) 89 InIn895 1.609437912 =
=
�
�
�
�
__
(9)
456
CHAPTER 6
Exponential and Logarithmic Functions
,r;
b logyz v5
( )
1 log 5 2.3219 log V 5 -2 log 5 12 log V 2 l. log 2 og 2 ,
=
r; ;:::.
,
=
-- =
�
or logyz V 5
,r;
C'l:: == _ -
COMMENT
1 In 5 In V 5 = 2 In 5 2.3219 ;:::. In In V 2 l. In 2 2 2 , r;
=
,
--
=
-
�
Now Work P R O B L E M S , 7 AND 6 5
To graph logarithmic functions when the base is different from
Change-of-Base Formula. For exam ple, to g raph y = log2
0�
C'l::==" -
SUMMARY
•
x,
e
or 10 requi res the In x we would instead g raph y = hl'
•
Now Work P R O B L E M 7 3
Properties of Logarithms
In the list that follows, a, b, M, N, and are real numbers. Also, a> 0, a *- 1, b> 0, b *- 1, M > ° and N > 0. aY y = loga means loga 1 0; toga a 1 logaMr = r logaM 1a 0gaM = M; loga ar = r If M N, then logaM logaN. 10ga(MN) = logaM logaN If logaM = logaN, then M N. lOga( �) logaM -logaN M 1-10gaM -10gb ogb a r
x
Definition
Properties of logarithms
x
=
=
=
=
=
+
=
=
=
Change-of-Base Formula
I-Hs:torical Feature
L John
Napier (1550-1617)
ogarithms were invented about 1 590 by
tion 6.4). Napier's tables, p u blished in 1 6 1 4, listed what would now
John Napier (1 550- 1 6 1 7) and Joost Burgi
be ca lled natural logarithms of sines and were rather difficult to use. A
(1 552- 1 632),
independently.
London professor, Henry Briggs, became interested in the tables and vis
N a pier, whose work had the greater influence,
ited Napier. I n their conversations, they developed the idea of common
working
was a Scottish lord,a secretive man whose neigh
logarithms, which were published in 1 61 7. Their importance for calcula
bors were inclined to believe him to be in leag ue
tion was immediately recognized, and by 1 650 they were being printed
with the devil. His approach to logarithms was
as far away as China. They remained an important calculation tool until
very different from ou rs; it was based on the re-
the advent of the inexpensive handheld calculator about 1972,which has
lationship between arithmetic and geometric
decreased their calculational, but not their theoretical, importance.
sequences, discussed in a later cha pter, and not on the inverse function relationship of logarithms to exponential functions (described in Sec-
A side effect of the invention of logarithms was the popularization
of the decimal system of notation for real numbers.
457
Properties of Logarithms
SECTION 6.5
6.S Assess Yo u r Understanding Concepts and Vocabulary 1. The logarithm of a product equals the logarithms.
logs 7
-- ' 1 ogs 8
2. Hlogs M = 3. loga Mr
=
__
__
of the
5. True or False log2(3x4)
then M =
6. True or False logz16
.
In(x + 3)
In(x + 3) - In(2x) =
4. True or False
=
In(2x)
4 log2 (3x)
In 16
= --
In
2
Skill Building
In Prob lems 7-22, use p rop eriies of logarithms to find the exact value of each expression. Do not use a calculatO/: 13 -4 · 15 8. log2 T 7. log 3371 ', 9. In e 10. In eV" 12. e1n 15. log6 18 - log63
s
16. logs 16 - logs 2
In Problems 23-30, suppose that
In 2
=
a and
In 3 =
',13. logs 2 + logs 4
14. log69 + log64
.,17. log2 6 ·log64
18. log3 8 . logs
9
b . Use properties of logarithms to write each logarithm in terms of a and b.
2
23. I n 6
24. In '3
25. In 1.5
26. In 0 .5
27. In 8
28. In 27
29. In
30. ln
'#6
�
In Prob lems 31-50, write each expression as a sum and/or differen ce of logarithms. Express po wers as factors. x
31. logs(25x)
32. log 3'9
35. In(ex)
36. In�
x
40. IOg2
43. log2 47. In
49. In
[
(--) x
3
x-
3
X2 - x - 2 2 ( x + 4)
5 x V1+3x (x
- 4)
x> 3
J
3
3 1/
44. logs
x e'
38.lnr
37. In( xeX)
(; ) ( ' 3�)
2
41. In(x �)
a> O,b > 0
2
V r+1 2
x -I
x> 1
'
. 45.
[ [ [
log
X(X + 2)
(x
2
48. In -'---2----'-- 5O. In
x>4
JJ
+ 3)
(x - 4)
x>2
0
x -I
<x<1
( �)
46. log
x> 0
2
2/3
5X2� ? 4(x + 1 )-
[
X3�
( x - 2)
x>4
J
x> 0
42. In x
2
J
x>2
O<x<1
In Problems 51-64, write each expression as a single logarithm. 51. 3 logs u + 4 logs
52. 2 log 3U - log 3V
v
2 56. log(x + 3x + 2 )
55. lo�(x2 - 1) - 5 Io�(x + 1 )
( ) ( ) x
\:+1
2
57. In -- + In "-- - In(x - 1 ) x- I
x
58. log
(
2 X +2X - 3 ?
r- 4 ,
61. 2Ioga(5xJ) -
1
z
) ( - log
2 x +7X+6 x+2
loga(2x + 3)
)
- 2log(x + 1)
1
2
62. � log(x3+ 1 ) + ?,log(x +1) .)
1
64. 3 logs(3x + 1 ) - 2 l ogs(2 x
-
- 1)
- logs x
458
CHAPTER 6
Exponential and Logarithmic Functions
In Problems 65-72, use the Change-of-Base Form ula and a calc ulator to eval uate each logarithm. Round yo ur answer to three decimal places.
.65. log3 21 69. logy12 7
� In Problems 73-78, '- 73. Y = log4 x
66. IOg5 1 8
67. logt/3 71
68. logt/2 15
70. logys 8
71. log" e
72. log7/" v2
75. Y = log2( x + 2)
graph each function using a graphing utility and the Change-of-Base Form ula.
74. Y = logs x
77. Y = logx - t( x + 1)
76. Y = log4( x - 3)
78. y = logx+2( x - 2)
Applications and Extensions
2 79. Hf(x) = In x, g( x) = e\ and h(x) = x , find: (a) (f g) ( x) . What is the domain of fog? (b) (g f)(x).What is the domain of go f? (c) (f g)(5) (d) (f h)(x). What is the domain of f h? (e) (f h)(e)
80. Hf(x) = log2 x,g(x) = 2\ and h(x) = 4x, find: (a) (f g)(x). What is the domain of fog? (b) (g f)(x). What is the domain of g of? (c) (f g)(3) (d) (f h)(x).What is the domain of f h? (e) (f h)(8) a
a
a
a
a
a a
a
a
a
a
a
In Problems 81-90, express y as a function of x. The constant C is a positive n umba
81. In y = In x + In C 83. In y = In x + In( x 85. In y = 3x + In C 87. In (y - 3)
=
+
82. In y = In( x + C)
1 ) + InC
84. In y = 21n x - In( x + 1 ) + InC 86. In y = -2x
-4x + InC
88. In (y
+
+
InC
4) = 5x + InC
1 1 89. 31n y = "2ln( 2x + 1 ) - "3 ln( x + 4) + In C
1 1 2 90. 21n y = - "2 ln x + "3ln( x + 1 )
91. Find the value of log2 3·log3 4 ·log4 5 . logs 6· log6 7· log7 8
92. Find the value of log2 4 · log4 6 ·log6 8
93. Find the value of log2 3·log3 4· ... · Iog (n + 1) ·IOg,,+1 2 ,,
95. Show that log,, ( x +
�)
+ log,, ( x
-�) = o. 96. Show that log,, ( Vx + �) + log,, ( Vx - �) = o. 2 2 97. Show that In(l + e x ) = 2x + In( 1 + e- x ). le x
!fi 98. Difference Quotient If f(x) = log" x,show that·
+
101.lf f( x) = IOga x, show that f 103. Show that IOg,,
(�) = log"
(�) = -f(x).
M
InC
94. Find the value of log2 2 . log2 4· ... . log2 2"
(
)
/ h) - f(x) h lh , h = log" 1 + � h
99. If f(x) = log" x, show that -f(x) = IOgl/" x.
+
'1= o.
100. If f(x) = log" x,show that f(AB) = f(A) + feB).
102. If f(x) = log" x, show that f(x") = a f( x) .
- log" N, where a,M, and N
are positive real numbers and a '1= 1 .
104. Show that IOg,,
(�) = - log" N, where a and N are positive
real numbers and a '1= 1.
Discussion and Writing
[;i
2 105. Graph Yt = log( x ) and Y2 = 2 Iog( x) using a graphing utility. Are they equivalent? What might account for any dif ferences in the two functions? 106. Write an example that illustrates why ( log" x)'
'1=
r log" x.
107. Write an example that illustrates why log2( X + y) '1= log2 X + log2 y. 108. Does 310g3(-5) = -5? Why or why not?
Logarithmic and Exponential Equations
SECTION 6.6
459
6.6 Logarithmic and Exponential Equations PREPARING FOR THIS SECTION •
Before getting started, review the following: •
Solving Equations Using a Graphing Utility (Appendix, Section 4, pp. A8-A10)
•
Equations Quadratic in Form (Section 1 .4, pp. 1 1 9-1 21 ) Solving Quadratic Equations (Section 1 . 2, pp. 97-106)
Now Work the 'Are You Prepared?' problems on page 463.
OBJECTIVES 1 Solve Logarithmic Equations (p.459)
2 Solve Exponential Equations (p.461)
fu!j 3
1
Solve Logarithmic and Exponential Equations Using a Graphing Utility (p.462)
Solve Logarithmic Equations
In Section 6.4 we solved logarithmic equations by changing a logarithmic expression to an exponential expression. That is, we used the definition of a logarithm: y=
loga x is equivalent to
x = aY
a>
=F
0, a 1
For example, to solve the equation log2 (1 2x) = 3, we use the equivalent expo nential expression 1 2x = 23 and solve for x. log2( 1 2x) = 3 1 2x = 23 Change to an exponential expression. -
-
-
-
-2x = 7 X=
Simplify.
7 2
--
Solve.
You should check this solution for yourself. For most logarithmic equations, some manipulation of the equation (usually using properties of logarithms) is required to obtain a solution. Also, to avoid extra neous solutions with logarithmic equations, we determine the domain of the vari able first. We begin with an example of a logarithmic equation that requires using the fact that a logarithmic function is a one-to-one function. If logaM logaN, thenM N M, N, and a are positive and a 1 =
EXAMPLE 1 Solution
=F
=
Solvi ng a Logarithmic E quation
Solve: 2 logs x = logs 9 The domain of the variable in this equation is x> O. Because each logarithm is to the same base, 5, we can obtain an exact solution as follows: 2 logs x = logs 9 loga loga logs x2 = logs 9 x2 = 9 If loga M loga N, then N. x = 3 or � Reca ll that the domain of the variable is x > Mr = r =
M
M =
o.
Therefore, -3 is extraneous and we discard it.
460
CHAPTER 6
Exponential and Logarithmic Functions
Check:
2 logs 3
J,
logs 9 logs 32 J, logs 9 logs 9 logs 9 The solution set is {3}.
r
loga M = loga Mr
=
mt=- -
•
Now Work PR O B L E M 1 3
In the next example we employ the log of a product property to solve a loga rithmic equation. EXAMPLE 2
Solving a Logarithmic E quation +
Solution
=
Solve: logs(x + 6) logs(x + 2) 1 The domain of the variable requires that x + 6 > 0 and x + 2 > 0 so x > -6 and x > -2. This means any solution must satisfy x > -2. To obtain an exact solution, we need to express the left side as a single logarithm. Then we will change the expression to exponential form. logs(x + 6) + logs(x + 2) 1 logs[(x + 6)(x + 2)J 1 loga M + loga N = loga (MN) (x + 6)(x + 2) Sl S Change to an exponentia l expression. x2 + 8x 12 S Simplify. x2 + 8x + 7 0 P lace the quad ratic equation in standard form. (x + 7)(x + 1 ) 0 Factor. x -7 or x -1 Zero-Product Property Only x -1 satisfies the restriction that x > -2, so x -7 is extraneous. The so lution set is {-1}, which you should check. = =
=
+
=
= =
WARNING
A negative sol ution is not a utomatica l ly extra neous. You m u st determine whether the potential solu tion causes the a rg u ment of any loga rithmic expression in the equation to be negative. _
=
=
=
Solution
=
•
,"'l!l;: =� �
EXA M P L E 3
=
Now Work PR O B L E M 2 1
Solving a Logarithmic E quation =
Solve: In x + In(x - 4) In(x + 6) The domain of the variable requires that x > 0, x 4 > 0, and x + 6 > O. As a result, the domain of the variable here is x > 4. We begin the solution using the log of a product property. In x + In(x 4) In(x + 6) In[ x(x - 4)J In(x + 6) I n M + I n N I n (MN) x(x - 4) x + 6 If In M In N, then M N. x2 - 4x x 6 Si m pl ify. x2 - Sx - 6 0 P lace the quad ratic equation in sta nda rd form. (x - 6)(x + 1) 0 Factor. 6 or x - 1 Zero-Prod uct Property Since the domain of the variable is x > 4, we discard -1 as extraneous. The solu tion set is 161, which you should check. -
-
=
=
=
=
=
=
X =
+
=
=
=
=
��"-=�- Now Work PR O B L E M 3 1
•
SECTION 6.6
2
Logarithmic and Exponential Equations
461
Solve Exponential Equations
In Sections 6.3 and 6.4, we solved exponential equations algebraically by expressing each side of the equation using the same base. That is, we used the one-to-one prop erty of the exponential function: If aU aV, then a > O, a *- 1 =
u
= v
For example, to solve the exponential equation 42x+1 16, we notice that 16 42 and apply the property above to obtain 2x + 1 2, from which we find �. For most exponential equations, we cannot express each side of the equation using the same base. In such cases, algebraic techniques can sometimes be used to obtain exact solutions. When algebraic techniques cannot be used, we use a graph ing utility to obtain approximate solutions. You should pay particular attention to the form of equations for which exact solutions are obtained. In the next example we solve two exponential equations by changing the expo nential expression to a logarithmic expression. =
=
EXAMPLE 4
x =
Solving an Exponential Equation =
Solve: (a) 2x 5 (b) 8 · 3x 5 (a) Since 5 cannot be written as an integral power of 2, we write the exponential equation as the equivalent logarithmic equation. 21' 5 x log 5 2 Now use the Change of Base Formula (9) on page 455. In 5 Exact solution x = In 2 =
Solution
=
=
=
In 5- } . . set IS. { The solutIOn In 2 (b) 8 · 3x 5 . 5 3"
:::; 2.322
Approximate sol ution
=
= -
8
Solve for Y.
Exact solution
In 3 :::; -0.428
Approximate solution
{ (�) }.
In The solution set is In 3
•
",,,
EXAMPLE 5
;:.. - Now Work P R O B L E
M
35
Solving an Exponential E quation
Solve:
y-2
=
33x+ 2
462
CHAPTER 6
Exponential and Logarithmic Functions
Because the bases are different, we first apply Property (6), Section 6.5 (take the natural logarithm of each side), and then use a property of logarithms. The result is an equation in x that we can solve.
Solution
5x -2 5x - 2
=
33x +2
+
In 33x 2 In ( 2) In 5 (3x + 2) In 3 (ln 5)x - 2 1n 5 = (3 1n 3)x + 2 1n 3 (In 5)x - (3 In 3)x 2 In 3 + 2 in 5 (In 5 3 1n 3)x = 2(1n 3 In 5) 3 + In 5 ) x = 2(in In S - 3 1n 3 -3.212 3 + In 5 ) } The solution set is { 2(In In 5 - 3 In 3 . x
=
=
-
=
+
-
�
If M = N, In
M
= I n N.
In Mr = r In M
Distribute. Place terms i nvolving
x
on the left.
Factor. Exact solution Approxi mate solution
•
"""' - Now Work P R O B L E M 4 5
'l'l
The next example deals with an exponential equation that is quadratic in form. E XA M P L E 6
Solving an Exponential Equation That Is Quadratic in Form
Solve: 4x - 2x - 12 0 We note that 4'" (22 ) X 2(2x) = (2x /, so the equation is quadratic in form, and we can rewrite it as (2x) 2 - 2x - 12 0 Let u = then u2 - u - = Now we can factor as usual. (2X - 4)(2X + 3) 0 (u - 4)(u + 3) = 0 2X - 4 0 or 2x + 3 = 0 u - 4 = 0 or u + 3 = 0 2x 4 2x -3 u= =4 u= = -3 The equation on the left has the solution x 2, since 2x 4 22; the equation on the right has no solution, since 2x 0 for all x. The only solution is 2. The solution set is {2}. =
=
Solution
=
=
12
2x;
o.
=
=
>
.� "'=
3
2X
=
=
=
2x
=
=
- - Now Work P R O B L E M S 3
•
Solve Logarithmic and Exponential Equations Using a G raphing Utility
The algebraic techniques introduced in this section to obtain exact solutions apply only to certain types of logarithmic and exponential equations. Solutions for other types are usually studied in calculus, using numerical methods. For such types, we can use a graphing utility to approximate the solution. EXAMPLE 7
Solving Equations Using a G raphing Utility
Solve: x + eX = 2 Express the solution(s) rounded to two decimal places.
4
- -- ---------- �- .--.. -- . -
=
1
•
.
�---- "'--'"
= X + eX
1
The solution is found by graphing Y and Y 2 2. Since Y is an increasing function (do you know why?), there is only one point of intersection for Y 1 and Y2 . Figure 40 shows the graphs of Y1 and Y 2 . Using the INTERSECT command, the solution is 0.44 rounded to two decimal places.
Solution
Figure 40
463
Logarithmic and Exponential Equations
SECTION 6.6
'rr = ..__ -
Ir.tti:t"�.;:·:ti_:· ... o :-:=.'1'1;?;B�'1'1 �'(=;?; o
Now Work P R O B L E M 6 3
6.6 Assess You r U nde rsta nd i n g 'Are You Prepared?' A nswers are giv en at the end of thes e exercises. I f yo u get a wrong answel; read the pages Lis ted in red. 1.
Solve x2 - 7x - 30 O. (pp. 97-106) 2. Solve ( x + 3 ) 2 - 4 ( x + 3 ) + 3 = O. (pp. 11 9-1 2 1 )
6 3. Approximate the solution(s) to x3
=
= x2 - 5 using a graphing
utility. (pp. AS-AIO) � 4. Approximate the solution(s) to x3 - 2x + 2 graphing utility. (pp. AS-A lO)
0
=
using a
Skill Building
In Pro blems 5-32, s olve each logarithmic eq uation. Express irrarional sol utions in exact form and as a decimal rou nded to 3 decimal places.
8. IOg3 ( 3 x - 1 ) 11.
1
- 100, x 2 bJ
14. 2 IOg5 x
=
=
6. log (x + 6) 9. IOg4( X + 2)
2
2 2 100, bJ
12. -2 IOg4 x
17. log x + log(x + 1 5 )
=
=
. 21. log2 (x
1 - logs ( x + 4 )
26. In (x + 1 ) - I n x
=
24. log5 (x
31. log{[ (x - 1 ) - log{[ (x + 6 )
=
=
+
7)
+
=
=
2
+
19. log(2x 1
=
IOg2 (X + S)
=
32. logll x
2
1 ) = 1 + log(x - 2)
22. log6(X + 4) + log6 (X + 3 ) 25. In x + In(x
+
2)
=
4
28. IOg2 (X + 1 ) + IOg2 (X + 7)
2
30. IOg4(X2 - 9 ) - lo�(x + 3 )
log{[ (x - 2 ) - loga(x + 3 )
=
16. 2 log3 (x + 4) - IOg3 9
5
3 ) = 1 - IOg5 (X - 1 )
-1
= log5 3
. 13. 31og2 X = - log2 27
IOg4 9
27. log3(X + 1 ) + IOg3(X + 4 )
2
29. IOg l/3(X2 + x) - IOg1 /3 (X2 - x)
+
4
10. IOg5(2x + 3)
IOg4 S
18. log x + log (x - 21 )
2
20. log(2x) - log(x - 3 ) = 1 23. logs(x + 6)
=
=
=
7. log2 ( 5 x)
1
15. 3 IOg2 ( X - 1) + IOg2 4
3 IOg5 4
=
=
+ 1 0ga C x
- 2)
=
=
=
1
=3
3
loga(x + 4)
In Problems 33-60, s olve each exponential eq uation. Express irrational sol utions in exact form and as a decimal rounded to 3 decimal places. . 35. 2x = 10 36. 3x = 14 34. 5 - x = 25 33. 2x - 5 = S
37. S-x
=
41. 3 1 - 2x
1.2 =
38.
4x
TX
=
42. 2x + 1
45. 1 .2x (0.5 fx 2 49. 2 x + 2x - 12
1 .5
=
=
46. 0.3 1 + x 50. 32x
0
. 53. 1 6x + 4x+ 1 - 3 = 0 57. 3 · 4x + 4 · 2x + S
=
+ 3"
(_.)"')o )X
=
7 ' -x
44.
47. 7T 1 - x = eX 51. 3 2x + 3x + I - 4
.
58. 2 · 49x + 1 1 · 7x + 5
0
43.
1 72x - 1
- 2 = 0 54. 9x - 3x + 1 + 1 = 0
=
=
5 1 - 2x
40. 0.3 ( 402X )
55. 25x - S · 5x =
0
=
59. 4x - 10 · 4-x
=
=
),-x G3
=
=
5x
48. ex + = 7Tx 2 52. 22x + 2x + - 1 2
0
63. eX
=
67. In x 71.
+
1 ) - IOg4(X - 2 )
56. 36x - 6 · 6x
3
60. 3"' - 14 · Tx
-x
eX + In x
=
62. IOg2 (X - 1 ) - IOg6(X + 2 )
1
64. e2x
-x =
=
=
4
=
2
x + 2
68. I n (2x)
=
72. eX - In x
-x + 2
69. In x
=
73. e-x
4
= =
x3 - 1
In x
=
-16
!.{ In Pro blems 61-74, use a graphing utility to s olve each eq uation. Express yo ur answer rounded to two decimal places.
