Precalculus: Concepts Through Functions, a Unit Circle Approach to Trigonometry 9780321931047, 1292058749, 9781292058740, 5535575615, 0321931041
This edition features the exact same content as the traditional text in a convenient, three-hole- punched, loose-leaf ve
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English Pages 1200 [1191] Year 2013
Cover......Page 1
Title......Page 2
Copyright......Page 3
Contents......Page 6
To the Student......Page 16
Preface to the Instructor......Page 17
Prepare for Class ‘‘Read the Book’’......Page 22
Practice ‘‘Work the Problems’’......Page 23
Review ‘‘Study for Quizzes and Tests’’......Page 24
Resources for Success......Page 25
Applications Index......Page 26
F Foundations: A Prelude to Functions......Page 34
F.1 The Distance and Midpoint Formulas......Page 35
Use the Distance Formula......Page 36
Use the Midpoint Formula......Page 38
Graph Equations by Plotting Points......Page 42
Find Intercepts from a Graph......Page 44
Test an Equation for Symmetry......Page 45
Know How to Graph Key Equations......Page 47
Calculate and Interpret the Slope of a Line......Page 52
Find the Equation of a Vertical Line......Page 55
Use the Point-Slope Form of a Line; Identify Horizontal Lines......Page 56
Write the Equation of a Line in Slope-Intercept Form......Page 57
Identify the Slope and y-Intercept of a Line from Its Equation......Page 58
Graph Lines Written in General Form Using Intercepts......Page 59
Find Equations of Parallel Lines......Page 60
Find Equations of Perpendicular Lines......Page 61
Write the Standard Form of the Equation of a Circle......Page 67
Graph a Circle......Page 68
Work with the General Form of the Equation of a Circle......Page 69
Chapter Project......Page 74
1 Functions and Their Graphs......Page 75
Determine Whether a Relation Represents a Function......Page 76
Find the Value of a Function......Page 79
Find the Domain of a Function Defined by an Equation......Page 82
Formthe Sum, Difference, Product, and Quotient of Two Functions......Page 84
1.2 The Graph of a Function......Page 89
Identify the Graph of a Function......Page 90
Obtain Information from or about the Graph of a Function......Page 91
Determine Even and Odd Functions from a Graph......Page 99
Identify Even and Odd Functions from the Equation......Page 100
Use a Graph to Determine Where a Function is Increasing, Decreasing, or Constant......Page 101
Use a Graph to Locate Local Maxima and Local Minima......Page 102
Use a Graph to Locate the Absolute Maximum and the Absolute Minimum......Page 103
Use a Graphing Utility to Approximate Local Maxima and Local Minima and to Determine Where a Function is Increasing or Decreasing......Page 104
Find the Average Rate of Change of a Function......Page 105
Graph the Functions Listed in the Library of Functions......Page 111
Graph Piecewise-defined Functions......Page 116
Graph Functions Using Vertical and Horizontal Shifts......Page 122
Graph Functions Using Compressions and Stretches......Page 125
Graph Functions Using Reflections about the x-Axis and the y-Axis......Page 127
Build and Analyze Functions......Page 134
Construct a Model Using Direct Variation......Page 139
Construct a Model Using Joint or Combined Variation......Page 140
Chapter Review......Page 144
Chapter Test......Page 148
Chapter Projects......Page 149
2 Linear and Quadratic Functions......Page 151
Use Average Rate of Change to Identify Linear Functions......Page 152
Determine Whether a Linear Function Is Increasing, Decreasing or Constant......Page 155
Find the Zero of a Linear Function......Page 156
Build Linear Models from Verbal Descriptions......Page 157
Draw and Interpret Scatter Diagrams......Page 163
Distinguish between Linear and Nonlinear Relations......Page 164
Use a Graphing Utility to Find the Line of Best Fit......Page 165
2.3 Quadratic Functions and Their Zeros......Page 170
Find the Zeros of a Quadratic Function Using the Square Root Method......Page 171
Find the Zeros of a Quadratic Function by Completing the Square......Page 173
Find the Zeros of a Quadratic Function Using the Quadratic Formula......Page 174
Find the Point of Intersection of Two Functions......Page 176
Solve Equations That Are Quadratic in Form......Page 177
2.4 Properties of Quadratic Functions......Page 181
Graph a Quadratic Function Using Transformations......Page 182
Identify the Vertexand Axis of Symmetry of a Quadratic Function......Page 184
Graph a Quadratic Function Using Its Vertex, Axis, and Intercepts......Page 185
Find a Quadratic Function Given Its Vertex and One Other Point......Page 188
Find the Maximum or Minimum Value of a Quadratic Function......Page 189
Solve Inequalities Involving a Quadratic Function......Page 193
2.6 Building Quadratic Models from Verbal Descriptions and from Data......Page 197
Build Quadratic Models from Verbal Descriptions......Page 198
Build Quadratic Models from Data......Page 202
Find the Complex Zeros of a Quadratic Function......Page 208
2.8 Equations and Inequalities Involving the Absolute Value Function......Page 211
Solve Absolute Value Inequalities......Page 212
Chapter Review......Page 217
Chapter Test......Page 220
Cumulative Review......Page 221
Chapter Projects......Page 222
3 Polynomial and Rational Functions......Page 224
Identify Polynomial Functions and Their Degree......Page 225
Graph Polynomial Functions Using Transformations......Page 229
Identify the Real Zeros of a Polynomial Function and Their Multiplicity......Page 230
Analyze the Graph of a Polynomial Function......Page 237
Build Cubic Models from Data......Page 239
3.2 The Real Zeros of a Polynomial Function......Page 245
Use the Remainder and Factor Theorems......Page 246
Use Descartes’ Rule of Signs to Determine the Number of Positive and the Number of Negative Real Zeros of a Polynomial Function......Page 248
Use the Rational Zeros Theorem to List the Potential Rational Zeros of a Polynomial Function......Page 249
Find the Real Zeros ofa Polynomial Function......Page 250
Solve Polynomial Equations......Page 252
Use the Theorem for Bounds on Zeros......Page 253
Use the Intermediate Value Theorem......Page 254
3.3 Complex Zeros; Fundamental Theorem of Algebra......Page 259
Use the Conjugate Pairs Theorem......Page 260
Find a Polynomial Function with Specified Zeros......Page 261
Find the Complex Zeros of a Polynomial Function......Page 262
3.4 Properties of Rational Functions......Page 265
Find the Domain of a Rational Function......Page 266
Find the Vertical Asymptotes of a Rational Function......Page 269
Find the Horizontal or Oblique Asymptote of a Rational Function......Page 270
Analyze the Graph of a Rational Function......Page 276
Solve Applied Problems Involving Rational Functions......Page 287
3.6 Polynomial and Rational Inequalities......Page 291
Solve Polynomial Inequalities......Page 292
Solve Rational Inequalities......Page 293
Chapter Review......Page 299
Cumulative Review......Page 303
Chapter Projects......Page 304
4 Exponential and Logarithmic Functions......Page 306
Form a Composite Function......Page 307
Find the Domain of a Composite Function......Page 308
Determine Whether a Function Is One-to-One......Page 315
Determine the Inverse of a Function Defined by a Map or a Set of Ordered Pairs......Page 317
Obtain the Graph ofthe Inverse Function from the Graph of the Function......Page 319
Find the Inverse of a Function Defined by an Equation......Page 320
Evaluate Exponential Functions......Page 327
Graph Exponential Functions......Page 331
Define the Number e......Page 334
Solve Exponential Equations......Page 336
4.4 Logarithmic Functions......Page 344
Evaluate Logarithmic Expressions......Page 345
Graph Logarithmic Functions......Page 346
Solve Logarithmic Equations......Page 351
Work with Properties of Logarithms......Page 357
Write a Logarithmic Expression as a Single Logarithm......Page 360
Evaluate a Logarithm Whose Base Is Neither 10 Nor e......Page 361
Graph a Logarithmic Function Whose Base Is Neither 10 Nor e......Page 363
Solve Logarithmic Equations......Page 366
Solve Exponential Equations......Page 368
Solve Logarithmic and Exponential Equations Using a Graphing Utility......Page 369
4.7 Financial Models......Page 372
Determine the Future Value of a Lump Sum of Money......Page 373
Calculate Effective Rates of Return......Page 376
Determine the Present Value of a Lump Sum of Money......Page 377
Determine the Rate of Interest or the Time Required to Double a Lump Sum of Money......Page 378
Find Equations of Populations That Obey the Law of Uninhibited Growth......Page 382
Find Equations of Populations That Obey the Law of Decay......Page 384
Use Newton’s Law of Cooling......Page 385
Use Logistic Models......Page 387
4.9 Building Exponential, Logarithmic, and Logistic Models from Data......Page 392
Build an Exponential Model from Data......Page 393
Build a Logarithmic Model from Data......Page 394
Build a Logistic Model from Data......Page 395
Chapter Review......Page 400
Chapter Test......Page 405
Cumulative Review......Page 406
Chapter Projects......Page 407
5 Trigonometric Functions......Page 408
5.1 Angles and Their Measures......Page 409
Convert between Decimals and Degrees, Minutes, Seconds Measures for Angles......Page 411
Find the Length if an Arc of a Circle......Page 412
Convert from Degreesto Radians and from Radians to Degrees......Page 413
Find the Area of a Sector of a Circle......Page 416
Find the Linear Speed of an Object Traveling in Circular Motion......Page 417
5.2 Trigonometric Functions: Unit Circle Approach......Page 423
Find the Exact Values of the Trigonometric Functions Using a Point on the Unit Circle......Page 425
Find the Exact Values of the Trigonometric Functions of Quadrantal Angles......Page 426
Find the Exact Values of the Trigonometric Functions......Page 428
Find the Exact Values of the Trigonometric Functions......Page 429
Find the Exact Values of the Trigonometric Functions for Integer Multiples......Page 431
Use a Circle of Radius r to Evaluate the Trigonometric Functions......Page 433
Determine the Domain and the Range of the Trigonometric Functions......Page 440
Determine the Period of the Trigonometric Functions......Page 442
Find the Values of the Trigonometric Functions Using Fundamental Identities......Page 444
Find the ExactValues of the Trigonometric Functions of an Angle......Page 446
Use Even-Odd Properties to Find the Exact Values of the Trigonometric Functions......Page 449
5.4 Graphs of the Sine and Cosine Functions......Page 453
Graph Functions of the Form y = A sin......Page 454
Graph Functions of the Form y = A cos......Page 456
Determine the Amplitude and Period of Sinusoidal Functions......Page 457
Graph Sinusoidal Functions Using Key Points......Page 459
Find an Equation for a Sinusoidal Graph......Page 463
5.5 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions......Page 468
Graph Functions of the Form......Page 470
Graph Functions of the Form......Page 473
Graph Sinusoidal Functions of the Form......Page 476
Build Sinusoidal Models from Data......Page 480
Chapter Review......Page 487
Cumulative Review......Page 493
Chapter Projects......Page 494
6 Analytic Trigonometry......Page 496
6.1 The Inverse Sine, Cosine, and Tangent Functions......Page 497
Find the Exact Value of an Inverse Sine Function......Page 498
Find an Approximate Value of an Inverse Sine Function......Page 499
Use Properties of Inverse Functions to Find Exact Values of Certain Composite Functions......Page 500
Find the Inverse Function of a Trigonometric Function......Page 505
Solve Equations Involving Inverse Trigonometric Functions......Page 506
6.2 The Inverse Trigonometric Functions (Continued)......Page 509
Find the Exact Value of Expressions Involving the Inverse Sine, Cosine, and Tangent Functions......Page 510
Define the Inverse Secant, Cosecant, and Cotangent Functions......Page 511
Write a Trigonometric Expression as an Algebraic Expression......Page 512
Solve Equations Involving a Single Trigonometric Function......Page 515
Solve Trigonometric Equations Quadratic in Form......Page 518
Solve Trigonometric Equations Using Fundamental Identities......Page 519
Solve Trigonometric Equations Using a Graphing Utility......Page 520
6.4 Trigonometric Identities......Page 524
Establish Identities......Page 526
6.5 Sum and Difference Formulas......Page 532
Use Sum and Difference Formulas to Find Exact Values......Page 533
Use Sum and Difference Formulas to Establish Identities......Page 534
Use Sum and Difference Formulas Involving Inverse Trigonometric Functions......Page 538
Solve Trigonometric Equations Linear in Sine and Cosine......Page 539
6.6 Double-angle and Half-angle Formulas......Page 544
Use Double-angle Formulas to Establish Identities......Page 545
Use Half-angle Formulas to Find Exact Values......Page 548
Express Products as Sums......Page 554
Express Sums as Products......Page 556
Chapter Review......Page 558
Cumulative Review......Page 562
Chapter Projects......Page 563
7 Applications of Trigonometric Functions......Page 564
Find the Value of Trigonometric Functions of Acute Angles Using Right Triangles......Page 565
Solve Right Triangles......Page 567
Solve Applied Problems......Page 568
7.2 The Law of Sines......Page 577
Solve SSA Triangles......Page 579
Solve Applied Problems......Page 582
7.3 The Law of Cosines......Page 588
Solve SAS Triangles......Page 589
Solve Applied Problems......Page 590
7.4 Area of a Triangle......Page 594
Find the Area of SAS Triangles......Page 595
Find the Area of SSS Triangles......Page 596
Build a Model for an Object in Simple Harmonic Motion......Page 601
Analyze Simple Harmonic Motion......Page 603
Analyze an Object in Damped Motion......Page 604
Graph the Sum of Two Functions......Page 605
Chapter Review......Page 610
Chapter Test......Page 613
Chapter Projects......Page 614
8 Polar Coordinates; Vectors......Page 616
Plot Points Using Polar Coordinates......Page 617
Convert from Polar Coordinates to Rectangular Coordinates......Page 619
Convert from Rectangular Coordinates to Polar Coordinates......Page 621
Transform Equations between Polar and Rectangular Forms......Page 623
8.2 Polar Equations and Graphs......Page 626
Identify and Graph Polar Equations by Converting to Rectangular Equations......Page 627
Test Polar Equations for Symmetry......Page 630
Graph Polar Equations by Plotting Points......Page 631
Plot Points in the Complex Plane......Page 641
Convert a Complex Number between Rectangular Form and Polar Form......Page 642
Find Products and Quotients of Complex Numbers in Polar Form......Page 643
Use De Moivre’s Theorem......Page 644
Find Complex Roots......Page 645
8.4 Vectors......Page 649
Graph Vectors......Page 651
Find a Position Vector......Page 652
Add and Subtract Vectors Algebraically......Page 653
Find a Scalar Multiple and the Magnitude of a Vector......Page 654
Find a Vector from Its Direction and Magnitude......Page 655
Model with Vectors......Page 656
Find the Dot Product of Two Vectors......Page 663
Find the Angle between Two Vectors......Page 664
Determine Whether Two Vectors Are Orthogonal......Page 665
Decompose a Vector into Two Orthogonal Vectors......Page 666
Compute Work......Page 668
8.6 Vectors in Space......Page 671
Find Position Vectors in Space......Page 672
Perform Operations on Vectors......Page 673
Find the Dot Product......Page 674
Find the Angle between Two Vectors......Page 675
Find the Direction Angles of a Vector......Page 676
Find the Cross Product of Two Vectors......Page 680
Know Algebraic Properties of the Cross Product......Page 682
Find aVector Orthogonal to Two Given Vectors......Page 683
Find the Area of a Parallelogram......Page 684
Chapter Review......Page 686
Chapter Test......Page 689
Chapter Projects......Page 690
9 Analytic Geometry......Page 692
Know the Names of the Conics......Page 693
Analyze Parabolas with Vertex at the Origin......Page 694
Analyze Parabolas with Vertex at (h, k)......Page 697
Solve Applied Problems Involving Parabolas......Page 698
9.3 The Ellipse......Page 702
Analyze Ellipses with Center at the Origin......Page 703
Analyze Ellipses with Center at (h, k)......Page 707
Solve Applied Problems Involving Ellipses......Page 708
9.4 The Hyperbola......Page 712
Analyze Hyperbolas with Center at the Origin......Page 713
Find Asymptotes of Hyperbola......Page 717
Analyze Hyperbolas with Center at (h, k)......Page 719
Solve Applied Problems Involving Hyperbolas......Page 720
Identify a Conic......Page 725
Use a Rotation of Axes to Transform Equations......Page 726
Analyze an Equation Using Rotation of Axes......Page 728
Identify Conics without a Rotation of Axes......Page 730
Analyze and Graph Polar Equations of Conics......Page 732
Convert the Polar Equation of a Conic to a Rectangular Equation......Page 736
9.7 Plane Curves and Parametric Equations......Page 738
Find a Rectangular Equation for a Curve Defined Parametrically......Page 739
Use Time as a Parameter in Parametric Equations......Page 741
Find Parametric Equations for Curves Defined by Rectangular Equations......Page 744
Chapter Review......Page 750
Cumulative Review......Page 753
Chapter Projects......Page 754
10 Systems of Equations and Inequalities......Page 755
10.1 Systems of Linear Equations: Substitution and Elimination......Page 756
Solve Systems of Equations by Substitution......Page 758
Solve Systems of Equations by Elimination......Page 759
Identify Inconsistent Systems of Equations Containing Two Variables......Page 761
Solve Systems of Three Equations Containing Three Variables......Page 762
Identify Inconsistent Systems of Equations Containing Three Variables......Page 764
Express the Solution of a System of Dependent Equations Containing Three Variables......Page 765
Write the Augmented Matrix of a System of Linear Equations......Page 771
Write the System of Equations from the Augmented Matrix......Page 772
Perform Row Operationson a Matrix......Page 773
Solve a System of Linear Equations Using Matrices......Page 774
10.3 Systems of Linear Equations: Determinants......Page 785
Use Cramer’s Rule to Solve a System of Two Equations Containing Two Variables......Page 786
Evaluate 3 by 3 Determinants......Page 789
Use Cramer’s Rule to Solve a System of Three Equations Containing Three Variables......Page 790
Know Properties of Determinants......Page 792
10.4 Matrix Algebra......Page 795
Find the Sum and Difference of Two Matrices......Page 797
Find Scalar Multiples of a Matrix......Page 799
Find the Product of Two Matrices......Page 800
Find the Inverse of a Matrix......Page 804
Solve a System of Linear Equations Using an Inverse Matrix......Page 808
10.5 Partial Fraction Decomposition......Page 814
Decompose P/Q, Where Q Has Only Nonrepeated Linear Factors......Page 815
Decompose P/Q, Where Q Has Repeated Linear Factors......Page 817
Decompose P/Q, Where Q Has a Nonrepeated Irreducible Quadratic Factor......Page 819
Decompose P/Q, Where Q Has a Repeated Irreducible Quadratic Factor......Page 820
Solve a System of Nonlinear Equations Using Substitution......Page 822
Solve a System of Nonlinear Equations Using Elimination......Page 823
Graph an Inequality......Page 831
Graph a System of Inequalities......Page 833
Set up a Linear Programming Problem......Page 839
Solve a Linear Programming Problem......Page 840
Chapter Review......Page 846
Chapter Test......Page 849
Cumulative Review......Page 850
Chapter Projects......Page 851
11 Sequences; Induction; the Binomial Theorem......Page 852
Write the First Several Terms of a Sequence......Page 853
Write the Terms of a Sequence Defined by a Recursive Formula......Page 856
Use Summation Notation......Page 857
Find the Sum of a Sequence......Page 858
Determine Whether a Sequence Is Arithmetic......Page 863
Find a Formula for an Arithmetic Sequence......Page 864
Find the Sum of an Arithmetic Sequence......Page 865
Determine Whether a Sequence Is Geometric......Page 869
Find a Formula for a Geometric Sequence......Page 870
Find the Sum of a Geometric Sequence......Page 871
Determine Whether a Geometric Series Converges or Diverges......Page 872
Solve Annuity Problems......Page 875
Prove Statements Using Mathematical Induction......Page 880
11.5 The Binomial Theorem......Page 884
Evaluate n/j......Page 885
Use the Binomial Theorem......Page 887
Chapter Review......Page 891
Cumulative Review......Page 893
Chapter Projects......Page 894
12 Counting and Probability......Page 895
Count the Number of Elements in a Set......Page 896
Solve Counting Problems Using the Multiplication Principle......Page 898
12.2 Permutations and Combinations......Page 901
Solve Counting Problems Using Permutations Involving n Distinct Objects......Page 902
Solve Counting Problems Using Combinations......Page 904
Solve Counting Problems Using Permutations Involving n Nondistinct Objects......Page 906
Construct Probability Models......Page 910
Compute Probabilities of Equally Likely Outcomes......Page 912
Find Probabilities of the Union of Two Events......Page 914
Use the Complement Rule to Find Probabilities......Page 915
Chapter Review......Page 920
Chapter Test......Page 922
Chapter Projects......Page 923
13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function......Page 924
Find a Limit Using a Table......Page 925
Find a Limit Using a Graph......Page 927
13.2 Algebra Techniques for Finding Limits......Page 930
Find the Limit of a Sum, a Difference, and a Product......Page 931
Find the Limit ofa Polynomial......Page 932
Find the Limit of a Power or a Root......Page 933
Find the Limit of a Quotient......Page 934
Find the Limit of an Average Rate of Change......Page 935
Find the One-sided Limits of a Function......Page 937
Determine Whether a FunctionIs Continuous......Page 939
Find an Equation of the Tangent Line to Graph a Function......Page 944
Find the Derivative of a Function......Page 946
Find the Instantaneous Speed of a Particle......Page 947
13.5 The Area Problem; The Integral......Page 951
Approximate the Area Under the Graph of a Function......Page 952
Approximate Integrals Using a Graphing Utility......Page 956
Chapter Review......Page 957
Chapter Test......Page 960
Chapter Projects......Page 961
Work with Sets......Page 962
Graph Inequalities......Page 965
Find Distance on the Real Number Line......Page 966
Evaluate Algebraic Expressions......Page 967
Use the Laws of Exponents......Page 968
Evaluate Square Roots......Page 970
Use a Calculator to Evaluate Exponents......Page 971
Use the Pythagorean Theorem and Its Converse......Page 974
Know Geometry Formulas......Page 976
Understand Congruent Triangles and Similar Triangles......Page 977
Recognize Monomials......Page 983
Recognize Polynomials......Page 984
Add and Subtract Polynomials......Page 985
Multiply Polynomials......Page 986
Know Formulas for Special Products......Page 987
Divide Polynomials Using Long Division......Page 988
Work with Polynomials in Two Variables......Page 991
A.4 Factoring Polynomials......Page 993
Factoring the Difference of Two Squares and the Sum and Difference of Two Cubes......Page 994
Factor a Second-Degree Polynomial......Page 995
Factor by Grouping......Page 997
Factor a Second-Degree Polynomial......Page 998
Complete the Square......Page 999
Divide Polynomials Using Synthetic Division......Page 1002
Reduce a Rational Expression to Lowest Terms......Page 1006
Multiply and Divide Rational Expressions......Page 1007
Add and Subtract Rational Expressions......Page 1008
Use the Least Common Multiple Method......Page 1009
Simplify Complex Rational Expressions......Page 1011
Work with nth Roots......Page 1016
Simplify Radicals......Page 1017
Rationalize Denominators......Page 1018
Simplify Expressions with Rational Exponents......Page 1019
A.8 Solving Equations......Page 1024
Solve Linear Equations......Page 1025
Solve Rational Equations......Page 1027
Solve Equations by Factoring......Page 1028
Solve Radical Equations......Page 1029
Translate Verbal Descriptions into Mathematical Expressions......Page 1033
Solve Interest Problems......Page 1034
Solve Mixture Problems......Page 1035
Solve Uniform Motion Problems......Page 1036
Solve Constant Rate Job Problems......Page 1038
Use Interval Notation......Page 1042
Use Properties of Inequalities......Page 1043
Solve Inequalities......Page 1045
Solve Combined Inequalities......Page 1046
A.11 Complex Numbers......Page 1050
Add, Subtract, Multiply, and Divide Complex Numbers......Page 1051
B.1 The Viewing Rectangle......Page 1058
B.2 Using a Graphing Utility to Graph Equations......Page 1060
B.3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry......Page 1062
B.4 Using a Graphing Utility to Solve Equations......Page 1063
B.5 Square Screens......Page 1065
B.7 Using a Graphing Utility to Solve Systems of Linear Equations......Page 1066
B.9 Using a Graphing Utility to Graph Parametric Equations......Page 1068
Answers......Page 1070
Photo Credits......Page 1176
Index......Page 1178
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GLOBAL EDITION
Precalculus
Concepts Through Functions A Unit Circle Approach to Trigonometry THIRD EDITION
.JDIBFM4VMMJWBOr.JDIBFM4VMMJWBO***
Precalculus Concepts Through Functions
A Unit Circle Approach To Trigonometry
Third Edition Global Edition Michael Sullivan Chicago State University
Michael Sullivan, lll Joliet Junior College
Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montréal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
Editor in Chief: Anne Kelly Acquisitions Editor: Dawn Murrin Assistant Editor: Joseph Colella Senior Managing Editor: Karen Wernholm Associate Managing Editor: Tamela Ambush Senior Production Project Manager: Peggy McMahon Digital Assets Manager: Marianne Groth Associate Media Producer: Marielle Guiney Head, Learning Asset Acquisition, Global Edition: Laura Dent Assistant Acquisitions Editor, Global Edition: Aditee Agarwal Asistant Project Editor, Global Edition: Mrithyunjayan Nilayamgode Associate Print & Media Editor, Global Edition: Anuprova Dey Chowdhuri** Senior Manufacturing Controller, Production, Global Edition: Trudy Kimber
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Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsonglobaleditions.com © Pearson Education Limited 2015 The rights of Michael Sullivan and Michael Sullivan, III to be identified as the authors of this work has been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. Authorized adaptation from the United States edition, entitled Precalculus: Concepts through Functions, A Unit Circle Approach to Trigonometry, 3rd edition, ISBN 978-0-321-93104-7, by Michael Sullivan and Michael Sullivan, III, published by Pearson Education © 2015. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a license permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners. The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. ISBN 10: 1-292-05874-9 ISBN 13: 978-1-292-05874-0 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library 10 9 8 7 6 5 4 3 2 1
14 13 12 11 10
Typeset in Times Ten by Cenveo® Publisher Services Printed and bound by Courier Kendallville in The United States of America
For Michael S., Kevin, and Marissa (Sullivan) Shannon, Patrick, and Ryan (Murphy) Maeve, Sean, and Nolan (Sullivan) Kaleigh, Billy, and Timmy (O’Hara) The Next Generation
Contents
F
To the Student
15
Preface to the Instructor
17
Prepare for Class ‘‘Read the Book’’
21
Practice ‘‘Work the Problems’’
22
Review ‘‘Study for Quizzes and Tests’’
23
Resources for Success
24
Applications Index
25
Foundations: A Prelude to Functions F.1 The Distance and Midpoint Formulas
33 34
6TFUIF%JTUBODF'PSNVMBr6TFUIF.JEQPJOU'PSNVMB
F. 2 Graphs of Equations in Two Variables; Intercepts; Symmetry
41
(SBQI&RVBUJPOTCZ1MPUUJOH1PJOUTr'JOE*OUFSDFQUTGSPNB(SBQIr'JOE *OUFSDFQUTGSPNBO&RVBUJPOr5FTUBO&RVBUJPOGPS4ZNNFUSZr,OPX)PX to Graph Key Equations
F. 3 Lines
51
$BMDVMBUFBOE*OUFSQSFUUIF4MPQFPGB-JOFr(SBQI-JOFT(JWFOB1PJOU BOEUIF4MPQFr'JOEUIF&RVBUJPOPGB7FSUJDBM-JOFr6TFUIF1PJOU4MPQF 'PSNPGB-JOF *EFOUJGZ)PSJ[POUBM-JOFTr'JOEUIF&RVBUJPOPGB-JOF (JWFO5XP1PJOUTr8SJUFUIF&RVBUJPOPGB-JOFJO4MPQF*OUFSDFQU'PSN r*EFOUJGZUIF4MPQFBOEy*OUFSDFQUPGB-JOFGSPN*UT&RVBUJPOr(SBQI -JOFT8SJUUFOJO(FOFSBM'PSN6TJOH*OUFSDFQUTr'JOE&RVBUJPOTPG 1BSBMMFM-JOFTr'JOE&RVBUJPOTPG1FSQFOEJDVMBS-JOFT
F. 4 Circles
66
8SJUFUIF4UBOEBSE'PSNPGUIF&RVBUJPOPGB$JSDMFr(SBQIB$JSDMF r8PSLXJUIUIF(FOFSBM'PSNPGUIF&RVBUJPOPGB$JSDMF
Chapter Project
1
Functions and Their Graphs 1.1 Functions
73
74 75
%FUFSNJOF8IFUIFSB3FMBUJPO3FQSFTFOUTB'VODUJPOr'JOEUIF7BMVFPGB 'VODUJPOr'JOEUIF%PNBJOPGB'VODUJPO%FGJOFECZBO&RVBUJPOr'PSN the Sum, Difference, Product, and Quotient of Two Functions
1.2 The Graph of a Function
88
*EFOUJGZUIF(SBQIPGB'VODUJPOr0CUBJO*OGPSNBUJPOGSPNPSBCPVUUIF Graph of a Function
1.3 Properties of Functions
98
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5
6
CONTENTS
1.4 Library of Functions; Piecewise-defined Functions
110
(SBQIUIF'VODUJPOT-JTUFEJOUIF-JCSBSZPG'VODUJPOTr(SBQI Piecewise-defined Functions
1.5 Graphing Techniques: Transformations
121
(SBQI'VODUJPOT6TJOH7FSUJDBMBOE)PSJ[POUBM4IJGUTr(SBQI'VODUJPOT 6TJOH$PNQSFTTJPOTBOE4USFUDIFTr(SBQI'VODUJPOT6TJOH3FGMFDUJPOT about the x-Axis and the y-Axis
1.6 Mathematical Models: Building Functions
133
Build and Analyze Functions
1.7 Building Mathematical Models Using Variation
138
$POTUSVDUB.PEFM6TJOH%JSFDU7BSJBUJPOr$POTUSVDUB.PEFM6TJOH *OWFSTF7BSJBUJPOr$POTUSVDUB.PEFM6TJOH+PJOUPS$PNCJOFE7BSJBUJPO
2
Chapter Review
143
Chapter Test
147
Chapter Projects
148
Linear and Quadratic Functions 2.1 Properties of Linear Functions and Linear Models
150 151
(SBQI-JOFBS'VODUJPOTr6TF"WFSBHF3BUFPG$IBOHFUP*EFOUJGZ-JOFBS 'VODUJPOTr%FUFSNJOF8IFUIFSB-JOFBS'VODUJPO*T*ODSFBTJOH %FDSFBTJOHPS$POTUBOUr'JOEUIF;FSPPGB-JOFBS'VODUJPOr#VJME-JOFBS Models from Verbal Descriptions
2.2 Building Linear Models from Data
162
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2.3 Quadratic Functions and Their Zeros
169
'JOEUIF;FSPTPGB2VBESBUJD'VODUJPOCZ'BDUPSJOHr'JOEUIF;FSPTPGB 2VBESBUJD'VODUJPO6TJOHUIF4RVBSF3PPU.FUIPEr'JOEUIF;FSPTPGB 2VBESBUJD'VODUJPOCZ$PNQMFUJOHUIF4RVBSFr'JOEUIF;FSPTPGB2VBESBUJD 'VODUJPO6TJOHUIF2VBESBUJD'PSNVMBr'JOEUIF1PJOUPG*OUFSTFDUJPOPG 5XP'VODUJPOTr4PMWF&RVBUJPOT5IBU"SF2VBESBUJDJO'PSN
2.4 Properties of Quadratic Functions
180
(SBQIB2VBESBUJD'VODUJPO6TJOH5SBOTGPSNBUJPOTr*EFOUJGZUIF7FSUFY BOE"YJTPG4ZNNFUSZPGB2VBESBUJD'VODUJPOr(SBQIB2VBESBUJD 'VODUJPO6TJOH*UT7FSUFY "YJT BOE*OUFSDFQUTr'JOEB2VBESBUJD 'VODUJPO(JWFO*UT7FSUFYBOE0OF0UIFS1PJOUr'JOEUIF.BYJNVNPS Minimum Value of a Quadratic Function
2.5 Inequalities Involving Quadratic Functions
192
Solve Inequalities Involving a Quadratic Function
2.6 Building Quadratic Models from Verbal Descriptions and from Data
196
#VJME2VBESBUJD.PEFMTGSPN7FSCBM%FTDSJQUJPOTr#VJME2VBESBUJD Models from Data
2.7 Complex Zeros of a Quadratic Function
207
'JOEUIF$PNQMFY;FSPTPGB2VBESBUJD'VODUJPO
2.8 Equations and Inequalities Involving the Absolute Value Function
210
4PMWF"CTPMVUF7BMVF&RVBUJPOTr4PMWF"CTPMVUF7BMVF*OFRVBMJUJFT
Chapter Review
216
Chapter Test
219
CONTENTS
3
Cumulative Review
220
Chapter Projects
221
Polynomial and Rational Functions
223
3.1 Polynomial Functions and Models
224
*EFOUJGZ1PMZOPNJBM'VODUJPOTBOE5IFJS%FHSFFr(SBQI1PMZOPNJBM 'VODUJPOT6TJOH5SBOTGPSNBUJPOTr*EFOUJGZUIF3FBM;FSPTPGB1PMZOPNJBM 'VODUJPOBOE5IFJS.VMUJQMJDJUZr"OBMZ[FUIF(SBQIPGB1PMZOPNJBM 'VODUJPOr#VJME$VCJD.PEFMTGSPN%BUB
3.2 The Real Zeros of a Polynomial Function
244
6TFUIF3FNBJOEFSBOE'BDUPS5IFPSFNTr6TF%FTDBSUFT3VMFPG4JHOTUP %FUFSNJOFUIF/VNCFSPG1PTJUJWFBOEUIF/VNCFSPG/FHBUJWF3FBM;FSPT PGB1PMZOPNJBM'VODUJPOr6TFUIF3BUJPOBM;FSPT5IFPSFNUP-JTUUIF 1PUFOUJBM3BUJPOBM;FSPTPGB1PMZOPNJBM'VODUJPOr'JOEUIF3FBM;FSPTPG B1PMZOPNJBM'VODUJPOr4PMWF1PMZOPNJBM&RVBUJPOTr6TFUIF5IFPSFNGPS #PVOETPO;FSPTr6TFUIF*OUFSNFEJBUF7BMVF5IFPSFN
3.3 Complex Zeros; Fundamental Theorem of Algebra
258
6TFUIF$POKVHBUF1BJST5IFPSFNr'JOEB1PMZOPNJBM'VODUJPOXJUI 4QFDJGJFE;FSPTr'JOEUIF$PNQMFY;FSPTPGB1PMZOPNJBM'VODUJPO
3.4 Properties of Rational Functions
264
'JOEUIF%PNBJOPGB3BUJPOBM'VODUJPOr'JOEUIF7FSUJDBM"TZNQUPUFT PGB3BUJPOBM'VODUJPOr'JOEUIF)PSJ[POUBMPS0CMJRVF"TZNQUPUFPGB Rational Function
3.5 The Graph of a Rational Function
275
"OBMZ[FUIF(SBQIPGB3BUJPOBM'VODUJPOr4PMWF"QQMJFE1SPCMFNT Involving Rational Functions
3.6 Polynomial and Rational Inequalities
290
4PMWF1PMZOPNJBM*OFRVBMJUJFTr4PMWF3BUJPOBM*OFRVBMJUJFT
4
Chapter Review
298
Chapter Test
302
Cumulative Review
302
Chapter Projects
303
Exponential and Logarithmic Functions 4.1 Composite Functions
305 306
'PSNB$PNQPTJUF'VODUJPOr'JOEUIF%PNBJOPGB$PNQPTJUF'VODUJPO
4.2 One-to-One Functions; Inverse Functions
314
%FUFSNJOF8IFUIFSB'VODUJPO*T0OFUP0OFr%FUFSNJOFUIF*OWFSTFPGB 'VODUJPO%FGJOFECZB.BQPSB4FUPG0SEFSFE1BJSTr0CUBJOUIF(SBQIPG UIF*OWFSTF'VODUJPOGSPNUIF(SBQIPGUIF'VODUJPOr'JOEUIF*OWFSTFPGB Function Defined by an Equation
4.3 Exponential Functions
326
&WBMVBUF&YQPOFOUJBM'VODUJPOTr(SBQI&YQPOFOUJBM'VODUJPOTr%FGJOF the Number er4PMWF&YQPOFOUJBM&RVBUJPOT
4.4 Logarithmic Functions Change Exponential Statements to Logarithmic Statements and Logarithmic 4UBUFNFOUTUP&YQPOFOUJBM4UBUFNFOUTr&WBMVBUF-PHBSJUINJD&YQSFTTJPOT r%FUFSNJOFUIF%PNBJOPGB-PHBSJUINJD'VODUJPOr(SBQI-PHBSJUINJD 'VODUJPOTr4PMWF-PHBSJUINJD&RVBUJPOT
343
7
8
CONTENTS
4.5 Properties of Logarithms
356
8PSLXJUI1SPQFSUJFTPG-PHBSJUINTr8SJUFB-PHBSJUINJD&YQSFTTJPOBT B4VNPS%JGGFSFODFPG-PHBSJUINTr8SJUFB-PHBSJUINJD&YQSFTTJPOBTB 4JOHMF-PHBSJUINr&WBMVBUFB-PHBSJUIN8IPTF#BTF*T/FJUIFS/PSe r(SBQIB-PHBSJUINJD'VODUJPO8IPTF#BTF*T/FJUIFS/PSe
4.6 Logarithmic and Exponential Equations
365
4PMWF-PHBSJUINJD&RVBUJPOTr4PMWF&YQPOFOUJBM&RVBUJPOTr4PMWF Logarithmic and Exponential Equations Using a Graphing Utility
4.7 Financial Models
371
%FUFSNJOFUIF'VUVSF7BMVFPGB-VNQ4VNPG.POFZr$BMDVMBUF&GGFDUJWF 3BUFTPG3FUVSOr%FUFSNJOFUIF1SFTFOU7BMVFPGB-VNQ4VNPG.POFZ r%FUFSNJOFUIF3BUFPG*OUFSFTUPSUIF5JNF3FRVJSFEUP%PVCMFB-VNQ Sum of Money
4.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models
381
'JOE&RVBUJPOTPG1PQVMBUJPOT5IBU0CFZUIF-BXPG6OJOIJCJUFE(SPXUI r'JOE&RVBUJPOTPG1PQVMBUJPOT5IBU0CFZUIF-BXPG%FDBZr6TF /FXUPOT-BXPG$PPMJOHr6TF-PHJTUJD.PEFMT
4.9 Building Exponential, Logarithmic, and Logistic Models from Data
391
#VJMEBO&YQPOFOUJBM.PEFMGSPN%BUBr#VJMEB-PHBSJUINJD.PEFMGSPN %BUBr#VJMEB-PHJTUJD.PEFMGSPN%BUB
5
Chapter Review
399
Chapter Test
404
Cumulative Review
405
Chapter Projects
406
Trigonometric Functions
407
5.1 Angles and Their Measures
408
Convert between Decimals and Degrees, Minutes, Seconds Measures for "OHMFTr'JOEUIF-FOHUIJGBO"SDPGB$JSDMFr$POWFSUGSPN%FHSFFT UP3BEJBOTBOEGSPN3BEJBOTUP%FHSFFTr'JOEUIF"SFBPGB4FDUPSPGB $JSDMFr'JOEUIF-JOFBS4QFFEPGBO0CKFDU5SBWFMJOHJO$JSDVMBS.PUJPO
5.2 Trigonometric Functions: Unit Circle Approach
422
Find the Exact Values of the Trigonometric Functions Using a Point on the 6OJU$JSDMFr'JOEUIF&YBDU7BMVFTPGUIF5SJHPOPNFUSJD'VODUJPOTPG 2VBESBOUBM"OHMFTr'JOEUIF&YBDU7BMVFTPGUIF5SJHPOPNFUSJD Functions of p/4 = 45°r'JOEUIF&YBDU7BMVFTPGUIF5SJHPOPNFUSJD Functions of p/6 = 30° and p/3 = 60°r'JOEUIF&YBDU7BMVFTPGUIF Trigonometric Functions for Integer Multiples of p/6 = 30°, p/4 = 45°, and p/3 = 60°r6TFB$BMDVMBUPSUP"QQSPYJNBUFUIF7BMVFPGB5SJHPOPNFUSJD 'VODUJPOr6TFB$JSDMFPG3BEJVTr to Evaluate the Trigonometric Functions
5.3 Properties of the Trigonometric Functions
439
Determine the Domain and the Range of the Trigonometric Functions r%FUFSNJOFUIF1FSJPEPGUIF5SJHPOPNFUSJD'VODUJPOTr%FUFSNJOFUIF4JHOT PGUIF5SJHPOPNFUSJD'VODUJPOTJOB(JWFO2VBESBOUr'JOEUIF7BMVFTPG UIF5SJHPOPNFUSJD'VODUJPOT6TJOH'VOEBNFOUBM*EFOUJUJFTr'JOEUIF&YBDU 7BMVFTPGUIF5SJHPOPNFUSJD'VODUJPOTPGBO"OHMF(JWFO0OFPGUIF 'VODUJPOTBOEUIF2VBESBOUPGUIF"OHMFr6TF&WFO0EE1SPQFSUJFTUP Find the Exact Values of the Trigonometric Functions
5.4 Graphs of the Sine and Cosine Functions Graph Functions of the Form y = A sin (vx) Using Transformations r(SBQI'VODUJPOTPGUIF'PSNy = A cos (vx) Using Transformations
452
CONTENTS
r%FUFSNJOFUIF"NQMJUVEFBOE1FSJPEPG4JOVTPJEBM'VODUJPOTr(SBQI 4JOVTPJEBM'VODUJPOT6TJOH,FZ1PJOUTr'JOEBO&RVBUJPOGPSB4JOVTPJEBM Graph
5.5 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
467
Graph Functions of the Form y = A tan(vx) + B and y = A cot (vx) + B r(SBQI'VODUJPOTPGUIF'PSNy = A csc (vx) + B and y = A sec (vx) + B
5.6 Phase Shift; Sinusoidal Curve Fitting
475
Graph Sinusoidal Functions of the Form y = A sin (vx - f) + B r#VJME4JOVTPJEBM.PEFMTGSPN%BUB
6
Chapter Review
486
Chapter Test
492
Cumulative Review
492
Chapter Projects
493
Analytic Trigonometry 6.1 The Inverse Sine, Cosine, and Tangent Functions
495 496
'JOEUIF&YBDU7BMVFPGBO*OWFSTF4JOF'VODUJPOr'JOEBO"QQSPYJNBUF 7BMVFPGBO*OWFSTF4JOF'VODUJPOr6TF1SPQFSUJFTPG*OWFSTF'VODUJPOT UP'JOE&YBDU7BMVFTPG$FSUBJO$PNQPTJUF'VODUJPOTr'JOEUIF*OWFSTF 'VODUJPOPGB5SJHPOPNFUSJD'VODUJPOr4PMWF&RVBUJPOT*OWPMWJOH*OWFSTF Trigonometric Functions
6.2 The Inverse Trigonometric Functions (Continued)
508
Find the Exact Value of Expressions Involving the Inverse Sine, Cosine, BOE5BOHFOU'VODUJPOTr%FGJOFUIF*OWFSTF4FDBOU $PTFDBOU BOE $PUBOHFOU'VODUJPOTr6TFB$BMDVMBUPSUP&WBMVBUFTFD -1 x, csc -1 x, and cot -1 xr8SJUFB5SJHPOPNFUSJD&YQSFTTJPOBTBO"MHFCSBJD&YQSFTTJPO
6.3 Trigonometric Equations
514
4PMWF&RVBUJPOT*OWPMWJOHB4JOHMF5SJHPOPNFUSJD'VODUJPOr4PMWF 5SJHPOPNFUSJD&RVBUJPOT6TJOHB$BMDVMBUPSr4PMWF5SJHPOPNFUSJD &RVBUJPOT2VBESBUJDJO'PSNr4PMWF5SJHPOPNFUSJD&RVBUJPOT6TJOH 'VOEBNFOUBM*EFOUJUJFTr4PMWF5SJHPOPNFUSJD&RVBUJPOT6TJOHB Graphing Utility
6.4 Trigonometric Identities
523
6TF"MHFCSBUP4JNQMJGZ5SJHPOPNFUSJD&YQSFTTJPOTr&TUBCMJTI*EFOUJUJFT
6.5 Sum and Difference Formulas
531
6TF4VNBOE%JGGFSFODF'PSNVMBTUP'JOE&YBDU7BMVFTr6TF4VNBOE %JGGFSFODF'PSNVMBTUP&TUBCMJTI*EFOUJUJFTr6TF4VNBOE%JGGFSFODF 'PSNVMBT*OWPMWJOH*OWFSTF5SJHPOPNFUSJD'VODUJPOTr4PMWF Trigonometric Equations Linear in Sine and Cosine
6.6 Double-angle and Half-angle Formulas
543
6TF%PVCMFBOHMF'PSNVMBTUP'JOE&YBDU7BMVFTr6TF%PVCMFBOHMF 'PSNVMBTUP&TUBCMJTI*EFOUJUJFTr6TF)BMGBOHMF'PSNVMBTUP'JOE&YBDU Values
6.7 Product-to-Sum and Sum-to-Product Formulas
553
&YQSFTT1SPEVDUTBT4VNTr&YQSFTT4VNTBT1SPEVDUT
Chapter Review
557
Chapter Test
561
Cumulative Review
561
Chapter Projects
562
9
10
CONTENTS
7
Applications of Trigonometric Functions
563
7.1 Right Triangle Trigonometry; Applications
564
Find the Value of Trigonometric Functions of Acute Angles Using Right 5SJBOHMFTr6TFUIF$PNQMFNFOUBSZ"OHMF5IFPSFNr4PMWF3JHIU5SJBOHMFT r4PMWF"QQMJFE1SPCMFNT
7.2 The Law of Sines
576
4PMWF4""PS"4"5SJBOHMFTr4PMWF44"5SJBOHMFTr4PMWF"QQMJFE Problems
7.3 The Law of Cosines
587
4PMWF4"45SJBOHMFTr4PMWF4445SJBOHMFTr4PMWF"QQMJFE1SPCMFNT
7.4 Area of a Triangle
593
'JOEUIF"SFBPG4"45SJBOHMFTr'JOEUIF"SFBPG4445SJBOHMFT
7.5 Simple Harmonic Motion; Damped Motion; Combining Waves
600
#VJMEB.PEFMGPSBO0CKFDUJO4JNQMF)BSNPOJD.PUJPOr"OBMZ[F4JNQMF )BSNPOJD.PUJPOr"OBMZ[FBO0CKFDUJO%BNQFE.PUJPOr(SBQIUIF4VN of Two Functions
8
Chapter Review
609
Chapter Test
612
Cumulative Review
613
Chapter Projects
613
Polar Coordinates; Vectors 8.1 Polar Coordinates
615 616
1MPU1PJOUT6TJOH1PMBS$PPSEJOBUFTr$POWFSUGSPN1PMBS$PPSEJOBUFTUP 3FDUBOHVMBS$PPSEJOBUFTr$POWFSUGSPN3FDUBOHVMBS$PPSEJOBUFTUP1PMBS $PPSEJOBUFTr5SBOTGPSN&RVBUJPOTCFUXFFO1PMBSBOE3FDUBOHVMBS'PSNT
8.2 Polar Equations and Graphs
625
Identify and Graph Polar Equations by Converting to Rectangular &RVBUJPOTr5FTU1PMBS&RVBUJPOTGPS4ZNNFUSZr(SBQI1PMBS&RVBUJPOT by Plotting Points
8.3 The Complex Plane; De Moivre’s Theorem
640
1MPU1PJOUTJOUIF$PNQMFY1MBOFr$POWFSUB$PNQMFY/VNCFSCFUXFFO 3FDUBOHVMBS'PSNBOE1PMBS'PSNr'JOE1SPEVDUTBOE2VPUJFOUTPG $PNQMFY/VNCFSTJO1PMBS'PSNr6TF%F.PJWSFT5IFPSFNr'JOE Complex Roots
8.4 Vectors
648
(SBQI7FDUPSTr'JOEB1PTJUJPO7FDUPSr"EEBOE4VCUSBDU7FDUPST "MHFCSBJDBMMZr'JOEB4DBMBS.VMUJQMFBOEUIF.BHOJUVEFPGB7FDUPSr'JOE B6OJU7FDUPSr'JOEB7FDUPSGSPN*UT%JSFDUJPOBOE.BHOJUVEFr.PEFM with Vectors
8.5 The Dot Product
662
'JOEUIF%PU1SPEVDUPG5XP7FDUPSTr'JOEUIF"OHMFCFUXFFO5XP 7FDUPSTr%FUFSNJOF8IFUIFS5XP7FDUPST"SF1BSBMMFMr%FUFSNJOF 8IFUIFS5XP7FDUPST"SF0SUIPHPOBMr%FDPNQPTFB7FDUPSJOUP5XP 0SUIPHPOBM7FDUPSTr$PNQVUF8PSL
8.6 Vectors in Space 'JOEUIF%JTUBODFCFUXFFO5XP1PJOUTJO4QBDFr'JOE1PTJUJPO 7FDUPSTJO4QBDFr1FSGPSN0QFSBUJPOTPO7FDUPSTr'JOEUIF%PU 1SPEVDUr'JOEUIF"OHMFCFUXFFO5XP7FDUPSTr'JOEUIF%JSFDUJPO Angles of a Vector
670
CONTENTS
8.7 The Cross Product
11
679
'JOEUIF$SPTT1SPEVDUPG5XP7FDUPSTr,OPX"MHFCSBJD1SPQFSUJFTPGUIF $SPTT1SPEVDUr,OPX(FPNFUSJD1SPQFSUJFTPGUIF$SPTT1SPEVDUr'JOEB 7FDUPS0SUIPHPOBMUP5XP(JWFO7FDUPSTr'JOEUIF"SFBPGB Parallelogram
9
Chapter Review
685
Chapter Test
688
Cumulative Review
689
Chapter Projects
689
Analytic Geometry
691
9.1 Conics
692
Know the Names of the Conics
9.2 The Parabola
693
"OBMZ[F1BSBCPMBTXJUI7FSUFYBUUIF0SJHJOr"OBMZ[F1BSBCPMBTXJUI Vertex at (h, k r4PMWF"QQMJFE1SPCMFNT*OWPMWJOH1BSBCPMBT
9.3 The Ellipse
701
"OBMZ[F&MMJQTFTXJUI$FOUFSBUUIF0SJHJOr"OBMZ[F&MMJQTFTXJUI$FOUFSBU (h, k r4PMWF"QQMJFE1SPCMFNT*OWPMWJOH&MMJQTFT
9.4 The Hyperbola
711
"OBMZ[F)ZQFSCPMBTXJUI$FOUFSBUUIF0SJHJOr'JOE"TZNQUPUFTPG )ZQFSCPMBr"OBMZ[F)ZQFSCPMBTXJUI$FOUFSBU h, k r4PMWF"QQMJFE Problems Involving Hyperbolas
9.5 Rotation of Axes; General Form of a Conic
724
*EFOUJGZB$POJDr6TFB3PUBUJPOPG"YFTUP5SBOTGPSN&RVBUJPOT r"OBMZ[FBO&RVBUJPO6TJOH3PUBUJPOPG"YFTr*EFOUJGZ$POJDTXJUIPVUB Rotation of Axes
9.6 Polar Equations of Conics
731
"OBMZ[FBOE(SBQI1PMBS&RVBUJPOTPG$POJDTr$POWFSUUIF1PMBS&RVBUJPO of a Conic to a Rectangular Equation
9.7 Plane Curves and Parametric Equations
737
(SBQI1BSBNFUSJD&RVBUJPOTr'JOEB3FDUBOHVMBS&RVBUJPOGPSB$VSWF %FGJOFE1BSBNFUSJDBMMZr6TF5JNFBTB1BSBNFUFSJO1BSBNFUSJD &RVBUJPOTr'JOE1BSBNFUSJD&RVBUJPOTGPS$VSWFT%FGJOFECZ3FDUBOHVMBS Equations
10
Chapter Review
749
Chapter Test
752
Cumulative Review
752
Chapter Projects
753
Systems of Equations and Inequalities
754
10.1 Systems of Linear Equations: Substitution and Elimination 4PMWF4ZTUFNTPG&RVBUJPOTCZ4VCTUJUVUJPOr4PMWF4ZTUFNTPG&RVBUJPOT CZ&MJNJOBUJPOr*EFOUJGZ*ODPOTJTUFOU4ZTUFNTPG&RVBUJPOT$POUBJOJOH 5XP7BSJBCMFTr&YQSFTTUIF4PMVUJPOPGB4ZTUFNPG%FQFOEFOU&RVBUJPOT $POUBJOJOH5XP7BSJBCMFTr4PMWF4ZTUFNTPG5ISFF&RVBUJPOT$POUBJOJOH 5ISFF7BSJBCMFTr*EFOUJGZ*ODPOTJTUFOU4ZTUFNTPG&RVBUJPOT$POUBJOJOH 5ISFF7BSJBCMFTr&YQSFTTUIF4PMVUJPOPGB4ZTUFNPG%FQFOEFOU&RVBUJPOT Containing Three Variables
755
12
CONTENTS
10.2 Systems of Linear Equations: Matrices
770
8SJUFUIF"VHNFOUFE.BUSJYPGB4ZTUFNPG-JOFBS&RVBUJPOTr8SJUFUIF 4ZTUFNPG&RVBUJPOTGSPNUIF"VHNFOUFE.BUSJYr1FSGPSN3PX0QFSBUJPOT POB.BUSJYr4PMWFB4ZTUFNPG-JOFBS&RVBUJPOT6TJOH.BUSJDFT
10.3 Systems of Linear Equations: Determinants
784
&WBMVBUFCZ%FUFSNJOBOUTr6TF$SBNFST3VMFUP4PMWFB4ZTUFNPG 5XP&RVBUJPOT$POUBJOJOH5XP7BSJBCMFTr&WBMVBUFCZ%FUFSNJOBOUT r6TF$SBNFST3VMFUP4PMWFB4ZTUFNPG5ISFF&RVBUJPOT$POUBJOJOH 5ISFF7BSJBCMFTr,OPX1SPQFSUJFTPG%FUFSNJOBOUT
10.4 Matrix Algebra
794
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10.5 Partial Fraction Decomposition
813
Decompose P/Q, Where Q)BT0OMZ/POSFQFBUFE-JOFBS'BDUPST r%FDPNQPTFP/Q, Where Q)BT3FQFBUFE-JOFBS'BDUPSTr%FDPNQPTF P/Q, Where Q Has a Nonrepeated Irreducible Quadratic Factor r%FDPNQPTFP/Q, Where Q Has a Repeated Irreducible Quadratic Factor
10.6 Systems of Nonlinear Equations
821
4PMWFB4ZTUFNPG/POMJOFBS&RVBUJPOT6TJOH4VCTUJUVUJPOr4PMWFB4ZTUFN of Nonlinear Equations Using Elimination
10.7 Systems of Inequalities
830
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10.8 Linear Programming
838
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11
Chapter Review
845
Chapter Test
848
Cumulative Review
849
Chapter Projects
850
Sequences; Induction; the Binomial Theorem 11.1 Sequences
851 852
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11.2 Arithmetic Sequences
862
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11.3 Geometric Sequences; Geometric Series
868
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11.4 Mathematical Induction
879
Prove Statements Using Mathematical Induction
11.5 The Binomial Theorem
883
n Evaluate a b r6TFUIF#JOPNJBM5IFPSFN j
Chapter Review
890
Chapter Test
892
Cumulative Review
892
Chapter Projects
893
CONTENTS
12
Counting and Probability 12.1 Counting
13
894 895
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12.2 Permutations and Combinations
900
Solve Counting Problems Using Permutations Involving n%JTUJODU0CKFDUT r4PMWF$PVOUJOH1SPCMFNT6TJOH$PNCJOBUJPOTr4PMWF$PVOUJOH1SPCMFNT Using Permutations Involving n/POEJTUJODU0CKFDUT
12.3 Probability
909
$POTUSVDU1SPCBCJMJUZ.PEFMTr$PNQVUF1SPCBCJMJUJFTPG&RVBMMZ-JLFMZ 0VUDPNFTr'JOE1SPCBCJMJUJFTPGUIF6OJPOPG5XP&WFOUTr6TFUIF Complement Rule to Find Probabilities
13
Chapter Review
919
Chapter Test
921
Cumulative Review
922
Chapter Projects
922
A Preview of Calculus: The Limit, Derivative, and Integral of a Function 13.1 Finding Limits Using Tables and Graphs
923 924
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13.2 Algebra Techniques for Finding Limits
929
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13.3 One-sided Limits; Continuous Functions
936
'JOEUIF0OFTJEFE-JNJUTPGB'VODUJPOr%FUFSNJOF8IFUIFSB'VODUJPO Is Continuous
13.4 The Tangent Problem; The Derivative
943
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13.5 The Area Problem; The Integral
950
"QQSPYJNBUFUIF"SFB6OEFSUIF(SBQIPGB'VODUJPOr"QQSPYJNBUF Integrals Using a Graphing Utility
A
Chapter Review
956
Chapter Test
959
Chapter Projects
960
Review A.1 Algebra Essentials
A1 A1
8PSLXJUI4FUTr(SBQI*OFRVBMJUJFTr'JOE%JTUBODFPOUIF3FBM/VNCFS -JOFr&WBMVBUF"MHFCSBJD&YQSFTTJPOTr%FUFSNJOFUIF%PNBJOPGB 7BSJBCMFr6TFUIF-BXTPG&YQPOFOUTr&WBMVBUF4RVBSF3PPUTr6TFB Calculator to Evaluate Exponents
A.2 Geometry Essentials 6TFUIF1ZUIBHPSFBO5IFPSFNBOE*UT$POWFSTFr,OPX(FPNFUSZ 'PSNVMBTr6OEFSTUBOE$POHSVFOU5SJBOHMFTBOE4JNJMBS5SJBOHMFT
A13
14
CONTENTS
A.3 Polynomials
A22
3FDPHOJ[F.POPNJBMTr3FDPHOJ[F1PMZOPNJBMTr"EEBOE4VCUSBDU 1PMZOPNJBMTr.VMUJQMZ1PMZOPNJBMTr,OPX'PSNVMBTGPS4QFDJBM1SPEVDUT r%JWJEF1PMZOPNJBMT6TJOH-POH%JWJTJPOr8PSLXJUI1PMZOPNJBMTJO5XP Variables
A.4 Factoring Polynomials
A32
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A.5 Synthetic Division
A41
Divide Polynomials Using Synthetic Division
A.6 Rational Expressions
A45
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A.7 nth Roots; Rational Exponents
A55
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A.8 Solving Equations
A63
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A.9 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
A72
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A.10 Interval Notation; Solving Inequalities
A81
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A.11 Complex Numbers
A89
Add, Subtract, Multiply, and Divide Complex Numbers
B
Graphing Utilities
B1
B.1 The Viewing Rectangle
B1
B.2 Using a Graphing Utility to Graph Equations
B3
B.3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry
B5
B.4 Using a Graphing Utility to Solve Equations
B6
B.5 Square Screens
B8
B.6 Using a Graphing Utility to Graph Inequalities
B9
B.7 Using a Graphing Utility to Solve Systems of Linear Equations
B9
B.8 Using a Graphing Utility to Graph a Polar Equation
B11
B.9 Using a Graphing Utility to Graph Parametric Equations
B11
Answers
AN1
Photo Credits Index
C1 I1
To the Student As you begin, you may feel anxious about the number of theorems, definitions, QSPDFEVSFT BOEFRVBUJPOT:PVNBZXPOEFSJGZPVDBOMFBSOJUBMMJOUJNF%POUXPSSZ 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 TLJMMTZPVOFFEUPCFTVDDFTTGVM)FSFTIPXZPVDBOVTFUIFCPPLUPZPVSCFOFGJU
Read Carefully 8IFOZPVHFUCVTZ JUTFBTZUPTLJQSFBEJOHBOEHPSJHIUUPUIFQSPCMFNT%POU 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) TPZPVDBOBTLRVFTUJPOTBCPVUBOZUIJOHZPVEJEOUVOEFSTUBOE:PVMMCFBNB[FEBU IPXNVDINPSFZPVMMHFUPVUPGDMBTTJGZPVEPUIJT
Use the Features We 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 (just 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. 5PNBLFUIJTFBTJFS XFWFQSPWJEFEBCSJFGHVJEFUPHFUUJOHUIFNPTUGSPNUIJT book. Refer to the “Prepare for Class,” “Practice,” and “Review” on pages 21–23. Spend fifteen minutes reviewing the guide and familiarizing yourself 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 us, through Pearson Education, with any questions, suggestions, or comments that would improve this text. We look forward to hearing from you, and good luck with all of your studies.
Best Wishes! Michael Sullivan Michael Sullivan, III
15
Preface to the Instructor
A
s professors at both an urban university and a community college, Michael Sullivan and Michael Sullivan, III, are aware of the varied needs of Precalculus students, ranging from those who have little mathematical background and a fear of mathematics courses, to those having a strong mathematical education and a high level of motivation. For some of your students, this will be their last course in mathematics, whereas others will further their mathematical education. This text is written for both groups. As a teacher, and as an author of precalculus, engineering calculus, finite mathematics, and business calculus texts, Michael Sullivan understands what students must know if they are to be focused and successful in upperlevel math courses. However, as a father of four, he also understands the realities of college life. As an author of BEFWFMPQNFOUBMNBUIFNBUJDTTFSJFT .JDIBFMTDPBVUIPS and son, Michael Sullivan, III, understands the trepidations and skills students bring to the Precalculus course. Michael, III also believes in the value of technology as a tool for learning that enhances understanding without sacrificing math skills. Together, both authors have taken great pains to ensure that the text contains solid, studentfriendly examples and problems, as well as a clear and seamless writing style. A tremendous benefit of authoring a successful series is the broad-based feedback we receive from teachers and students. We are sincerely grateful for their support. Virtually every change in this edition is the result of their thoughtful comments and suggestions. We are sincerely grateful for this support and hope that we have been able to take these ideas and, building upon a successful first edition, make this series an even better tool for learning and teaching. We continue to encourage you to share with us your experiences teaching from this text.
this course. By introducing functions early in the course, students are less likely to develop bad habits. Another advantage of the early introduction of functions is that the discussion of equations and inequalities can focus around the concept of a function. For example, rather than asking students to solve an equation such as 2x2 + 5x + 2 = 0, we ask students to find the zeros of f1x2 = 2x2 + 5x + 2 or solve f1x2 = 0 when f1x2 = 2x2 + 5x + 2. While the technique used to solve this type of problem is the same, the fact that the problem looks different to the student means the student is less apt to say, i0I *BMSFBEZIBWFTFFOUIJTQSPCMFNCFGPSF BOE*LOPX how to solve it.” In addition, in Calculus students are going to be asked to solve equations such as f ′1x2 = 0, so solving f1x2 = 0 is a logical prerequisite skill to practice in Precalculus. Another advantage to solving equations through the eyes of a function is that the properties of functions can be included in the solution. For example, the linear function f 1x2 = 2x - 3 has one real zero because the function f is increasing on its domain.
About This Book
r Retain Your Knowledge This new category of problems in the exercise set are based on the article “To Retain New Learning, Do the Math” published in the Edurati Review in which author Kevin Washburn suggests that “the more students are required to recall new content or skills, the better their memory will be.” It is frustrating when students cannot recall skills learned earlier in the course. To alleviate this recall problem, we have created “Retain Your Knowledge” problems. These are problems considered to be “final exam material” that students must complete to maintain their skills. All the answers to these problems appear in the back of the book. r Guided Lecture Notes Ideal for online, emporium/redesign courses, inverted classrooms or traditional lecture classrooms. These lecture notes assist students in taking thorough, organized, and understandable notes as they watch the Author in Action videos by asking students to complete definitions, procedures, and examples based
This book utilizes a functions approach to Precalculus. Functions are introduced early (Chapter 1) in various formats: maps, tables, sets of ordered pairs, equations, and HSBQIT0VSBQQSPBDIUPGVODUJPOTJMMVTUSBUFTUIFTZNCPMJD numeric, graphic, and verbal representations of functions. This allows students to make connections between the visual representation of a function and its algebraic representation. It is our belief that students need to “hit the ground running” so that they do not become complacent in their studies. After all, it is highly likely that students have been exposed to solving equations and inequalities prior to entering this class. By spending precious time reviewing these concepts, students are likely to think of the course as a rehash of material learned in other courses and say to themTFMWFT i*LOPXUIJTNBUFSJBM TP*EPOUIBWFUPTUVEZu5IJT may result in the students developing poor study habits for
16
Features in the Third Edition Rather than provide a list of new features here, that information can be found on pages 21–23. This places the new features in their proper context, as building blocks of an overall learning system that has been carefully crafted over the years to help students get the most out of the time they put into studying. Please take the time to review the features listed on pages 21–23 and to discuss them with your students at the beginning of your DPVSTF0VSFYQFSJFODFIBTCFFOUIBUXIFOTUVEFOUTVUJMJ[F these features, they are more successful in the course.
New to the Third Edition
PREFACE
on the content of the videos and book. In addition, experience suggests that students learn by doing and understanding the why/how of the concept or property. Therefore, many sections will have an exploration activity to motivate student learning. These explorations will introduce the topic and/or connect it somehow to either a real world application or previous section. For example, when teaching about the vertical line test in Section 1.2, after the theorem statement, the notes ask the students to explain why the vertical line test works by using the definition of a function. This helps students process the information at a higher level of understanding. r Chapter Projects, which apply the concepts of each chapter to a real-world situation, have been enhanced to give students an up-to-the-minute experience. Many projects are new and Internet-based, requiring the student to research information online in order to solve problems. r Exercise Sets at the end of each section remain classified according to purpose. The “Are You Prepared?” exercises have been expanded to better serve the student who needs a just-in-time review of concepts utilized in the section. The Concepts and Vocabulary exercises have been updated. These fill-in-the-blank and True/False problems have been written to serve as reading quizzes. Skill BuildingFYFSDJTFTEFWFMPQUIFTUVEFOUTDPNQVUBtional skills and are often grouped by objective. Mixed Practice exercises have been added where appropriate. These problems offer a comprehensive assessment of the skills learned in the section by asking problems that relate to more than one objective. Sometimes these require information from previous sections so students must utilize skills learned throughout the course. Applications and Extension problems have been updated and many new problems involving sourced information and data have been added to bring relevance and timeliness to the exercises. The Explaining Concepts: Discussion and Writing exercises have been updated and reworded to stimulate discussion of concepts in online discussion forums. These can also be used to spark classroom discussion. r The Chapter Review now includes answers to all the problems. We have created a separate review worksheet for each chapter to help students review and practice key skills to prepare for exams. The worksheets can CFEPXOMPBEFEGSPNUIF*OTUSVDUPST3FTPVSDF$FOUFS
17
Changes in the Third Edition r CONTENT r Chapter 2, Section 4 A new objective “Find a quadratic function given its vertex and one point” has been added. r Chapter 2, Section 5 A new example was added to illustrate that quadratic inequalities may have the empty set or all real numbers as a solution. r Chapter 3, Sections 1 and 4 The content related to describing the behavior of the graph of a polynomial or rational function near a zero has been removed. r Chapter 3, Section 4 Content has been added that discusses the role of multiplicity and behavior of the graph of rational function as the graph approaches a vertical asymptote. r ORGANIZATION r Chapter 3, Sections 5 and 6 Section 5, The Real Zeros of a Polynomial Function and Section 6, Complex Zeros, Fundamental Theorem of Algebra have been moved to Sections 2 and 3, respectively. This was done in response to reviewer requests that “everything involving polynomials” be located sequentially. Skipping the new Sections 2 and 3 and proceeding to Section 4 Properties of Rational Functions can be done without loss of continuity.
Using this Book Effectively and Efficiently with Your Syllabus To meet the varied needs of diverse syllabi, this book contains more content than is likely to be covered in a typical Precalculus course. As the chart illustrates, this book has been organized with flexibility of use in mind. Even within a given chapter, certain sections are optional and can be omitted without loss of continuity. See the detail following the flow chart. F
APPENDIX A
APPENDIX B
1
2
3
4
5
9.1-9.4
6
10
11
12
9.5-9.7
13 7
8.1-8.3
8.4-8.7
Foundations A Prelude to Functions Quick coverage of this chapter, which is mainly review material, will enable you to get to Chapter 1, Functions and Their Graphs, earlier.
18
PREFACE
Chapter 1 Functions and Their Graphs Perhaps the most important chapter. Sections 1.6 and 1.7 are optional.
Chapter 10 Systems of Equations and Inequalities Sections 10.2–10.7 may be covered in any order. Section 10.8 requires Section 10.7.
Chapter 2 Linear and Quadratic Functions Topic selection depends on your syllabus. Sections 2.2, 2.6, and 2.7 may be omitted without a loss of continuity.
Chapter 11 Sequences; Induction; the Binomial Theorem There are three independent parts: Sections 11.1–11.3, Section 11.4, and Section 11.5.
Chapter 3 Polynomial and Rational Functions Topic selection depends on your syllabus. Section 3.6 is optional. Chapter 4 Exponential and Logarithmic Functions Sections 4.1–4.6 follow in sequence. Sections 4.7–4.9 are optional. Chapter 5 Trigonometric Functions The sections follow in sequence. Section 5.6 is optional.
Chapter 12 Counting and Probability The sections follow in sequence. Chapter 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function If time permits, coverage of this chapter will provide your students with a beneficial head-start in calculus. The sections follow in sequence.
Chapter 7 Applications of Trigonometric Functions Sections 7.4 and 7.5 may be omitted in a brief course.
Appendix A Review This review material may be covered at the start of a course or used as a just-in-time review. Specific references to this material occur throughout the text to assist in the review process.
Chapter 8 Polar Coordinates; Vectors Sections 8.1–8.3 and Sections 8.4–8.7 are independent and may be covered separately.
Appendix B Graphing Utilities Reference is made to these sections at the appropriate place in the text.
Chapter 6 Analytic Trigonometry Sections 6.2 and 6.7 may be omitted in a brief course.
Chapter 9 Analytic Geometry Sections 9.1–9.4 follow in sequence. Sections 9.5, 9.6, and 9.7, are independent of each other, but each requires Sections 9.1–9.4.
Acknowledgments Third Edition 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. Don is remembered for his extensive contributions to publishing and mathematics. Thanks are due to the following people for their assistance and encouragement to the preparation of this edition: r From Pearson Education: Anne Kelly for her substantial contributions, ideas, and enthusiasm; Peggy Lucas, who is a huge fan and works tirelessly to get the word out; Dawn Murrin, for her unmatched talent at getting the details right; Peggy McMahon for her organizational skills and leadership in overseeing production; Chris Hoag for her continued support and genuine interest; Greg Tobin for his leadership and commitment to excellence; and the Pearson Math and Science Sales team, for their continued confidence and personal support of our books. r Bob Walters, Production Manager, who passed away after a long and valiant battle fighting lung disease. He was an old and dear friend—a true professional in every sense of the word. r Accuracy checkers: C. Brad Davis, who read the entire manuscript and accuracy checked answers. His attention to detail is amazing; Timothy Britt, for creating the Solutions Manuals and accuracy checking answers. r Michael Sullivan, III would like to thank his colleagues at Joliet Junior College for their support and feedback. Finally, we offer our grateful thanks to the dedicated users and reviewers of our books, whose collective insights form the backbone of each textbook revision. 0VSMJTUPGJOEFCUFEOFTTKVTUHSPXTBOEHSPXT"OE JGXFWFGPSHPUUFOBOZPOF please accept our apology. Thank you all. Gary Amara—South Maine Community College Richard Andrews—Florida A&M University Jay Araas—Sheridan College Jessica Bernards—Portland Community college Rebecca Berthiaume—Edison State College Susan Bradley—Angelina College Michael Brook—University of Delaware Tim Chappell—Penn Valley Community College Christine Cole—Moorpark College Alicia Collins—Mesa Community College Rebecca Cosner—Spokane Community College Jerry DeGroot—Purdue North Central Joanna DelMonaco—Middlesex Community College Stephanie Deacon—Liberty University Jerrett Dumouchel—Florida Community College at Jacksonville Vaden Fitton—North Virginia Community College Carrie Rose Gibson—North Idaho College Nina Girard—University of Pittsburgh at Johnstown Mary Beth Grayson—Liberty University Scott Greenleaf—South Maine Community College Donna Harbin—University of Hawaii-Maui Celeste Hernandez—Richland College Gloria P. Hernandez—Louisiana State University at Eunice .BSJU[B+JNFOF[;FMKBL-PT"OHFMFT)BSCPS$PMMFHF Glenn Johnson—Middlesex Community College
19
20
ACKNOWLEDGMENTS
Susitha Karunaratne—Purdue University North Central Debra Kopcso—Louisiana State University Yelena Kravchuk—University of Alabama at Birmingham Mary Krohn—Butler University Lynn Marecek—Santa Ana College James McLaughlin—West Chester University ,BUIMFFO.JSBOEB46/:BU0ME8FTUCVSZ Chris Mirbaha—The Community College of Baltimore County Brigette M. Myers—Stanly Community College Karla Neal—Louisiana State University Denise Nunley—Maricopa Community Colleges -FUJDJB0SPQFTB6OJWFSTJUZPG.JBNJ Laura Pyzdrowski—West Virginia University Mike Rosenthal—Florida International University Phoebe Rouse—Louisiana State University Brenda Santistevan—Salt Lake Community College Catherine Sausville—George Mason University Ingrid Scott—Montgomery College Charlotte Smedberg—University of Tampa Leslie Soltis—Mercyhurst College Katrina Staley—North Carolina Agricultural and Technical State University Sonya Stephens—Florida A&M University John Sumner—University of Tampa Steve Szabo—Eastern Kentucky University Marilyn Toscano—University of Wisconsin, Superior Timothy L. Warkentin—Cloud County Community College Hayat Weiss—Middlesex Community College Larissa Williamson—University of Florida 4IBSZO;JBT+BNFTUPXO$PNNVOJUZ$PMMFHF Michael Sullivan Chicago State University
Michael Sullivan, III Joliet Junior College
Global Edition Pearson would like to thank and acknowledge the following people for their work on the Global Edition: r Contributor: r Sunila Sharma, Delhi University r Reviewers: r Yosum Kurtulmaz, Bilkent University r Mohd. Hasan Shahid, Jamia Millia Islamia
Feature
Description
Benefit
Page
Every Chapter Opener begins with...
Chapter Opening Article & Project
Each chapter begins with a current article and ends with a related project. The article describes a real situation.
The Article describes a real situation. The 305, 406 Project lets you apply what you learned to solve a related problem.
NEW! Internet Based Projects
The projects allow for the integration of spreadsheet technology that students will need to be a productive member of the workforce.
The projects allow the opportunity for students to collaborate and use mathematics to deal with issues that come up in their lives.
305, 406
Every Section begins with...
Each section begins with a list of objectives. Objectives also appear in the text where the objective is covered.
These focus your studying by emphasizing what’s most important and where to find it.
326
Preparing for this Section
Most sections begin with a list of key concepts to review with page numbers.
Ever forget what you’ve learned? This feature highlights previously learned material to be used in this section. Review it, and you’ll always be prepared to move forward.
326
Now Work the ‘Are You Prepared?’ Problems
Problems that assess whether you have the Not sure you need the Preparing for This 326, 337 prerequisite knowledge for the upcoming Section review? Work the ‘Are You section. Prepared?’ problems. If you get one wrong, you’ll know exactly what you need to review and where to review it!
Now Work
These follow most examples and direct you to a related exercise.
Learning Objectives 2 Sections contain...
PROBLEMS
WARNING
Warnings are provided in the text.
We learn best by doing. You’ll solidify 333, 338 your understanding of examples if you try a similar problem right away, to be sure you understand what you’ve just read. These point out common mistakes and help you to avoid them.
360 232, 347
These represent graphing utility activities to foreshadow a concept or solidify a concept just presented.
You will obtain a deeper and more intuitive understanding of theorems and definition.
These provide alternative descriptions of select definitions and theorems.
Does math ever look foreign to you? This feature translates math into plain English.
CALCULUS
These appear next to information essential for the study of calculus.
Pay attention–if you spend extra time now, 102, 334 you’ll do better later!
SHOWCASE EXAMPLES
These examples provide “how-to” instruction by offering a guided, step-by-step approach to solving a problem.
With each step presented on the left and the mathematics displayed on the right, students can immediately see how each step is employed.
These are examples and problems that require you to build a mathematical model from either a verbal description or data. The homework Model It! problems are marked by purple headings.
351, 379 It is rare for a problem to come in the form, “Solve the following equation”. Rather, the equation must be developed based on an explanation of the problem. These problems require you to develop models that will allow you to describe the problem mathematically and suggest a solution to the problem.
Exploration and Seeing the Concept
In Words
Model It! Examples and Problems
343
236
21
Feature
Description
Benefit
Page
‘Are You Prepared?’ Problems
These assess your retention of the prerequisite material you’ll need. Answers are given at the end of the section exercises. This feature is related to the Preparing for This Section feature.
326, 337 Do you always remember what you’ve learned? Working these problems is the best way to find out. If you get one wrong, you’ll know exactly what you need to review and where to review it!
Concepts and Vocabulary
These short-answer questions, mainly Fillin-the-Blank and True/False items, assess your understanding of key definitions and concepts in the current section.
It is difficult to learn math without knowing the language of mathematics. These problems test your understanding of the formulas and vocabulary.
Skill Building
Correlated to section examples, these problems provide straightforward practice.
It’s important to dig in and develop your skills. These problems provide you with ample practice to do so.
337–339
Mixed Practice
These problems offer comprehensive assessment of the skills learned in the section by asking problems that relate to more than one concept or objective. These problems may also require you to utilize skills learned in previous sections.
Learning mathematics is a building process. Many concepts are interrelated. These problems help you see how mathematics builds on itself and also see how the concepts tie together.
339–340
Applications and Extensions
These problems allow you to apply your skills to real-world problems. They also allow you to extend concepts learned in the section.
You will see that the material learned within the section has many uses in everyday life.
340–342
Discussion and Writing
“Discussion and Writing” problems are colored red. These support class discussion, verbalization of mathematical ideas, and writing and research projects.
To verbalize an idea, or to describe it clearly in writing, shows real understanding. These problems nurture that understanding. Many are challenging but you’ll get out what you put in.
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NEW! Retain Your Knowledge
These problems allow you to practice content learned earlier in the course.
The ability to remember how to solve all the different problems learned throughout the course is difficult. These help you remember.
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Now Work
Many examples refer you to a related homework problem. These related problems are marked by a pencil and orange numbers.
If you get stuck while working problems, look for the closest Now Work problem and refer back to the related example to see if it helps.
336, 339
Chapter Review Problems
Every chapter concludes with a comprehensive list of exercises to practice. Use the list of objectives to determine the objective and examples that correspond to the problems.
401–404 Work these problems to verify you understand all the skills and concepts of the chapter. Think of it as a comprehensive review of the chapter.
PROBLEMS
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Feature
Description
Benefit
Page
Chapter Review at the end of each chapter contains...
Things to Know
A detailed list of important theorems, formulas, and definitions from the chapter.
Review these and you’ll know the most 399–400 important material in the chapter!
You Should Be able to...
Contains a complete list of objectives by section, examples that illustrate the objective, and practice exercises that test your understanding of the objective.
Do the recommended exercises and you’ll have mastery over the key material. If you get something wrong, review the suggested page numbers and try again.
Review Exercises
These provide comprehensive review Practice makes perfect. These problems 401–404 and practice of key skills, matched to the combine exercises from all sections, giving you a comprehensive review in Learning Objectives for each section. one place.
Chapter Test
About 15-20 problems that can be taken as Be prepared. Take the sample practice 404–405 a Chapter Test. Be sure to take the Chapter test under test conditions. This will get you ready for your instructor’s test. If you Test under test conditions—no notes! get a problem wrong, you can watch the Chapter Test Prep Video.
Cumulative Review
These problem sets appear at the end of each chapter, beginning with Chapter 2. They combine problems from previous chapters, providing an ongoing cumulative review.
These are really important. They will ensure that you are not forgetting anything as you go. These will go a long way toward keeping you primed for the final exam.
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Chapter Project
The Chapter Project applies to what you’ve learned in the chapter. Additional projects are available on the Instructor’s Resource Center (IRC).
The Project gives you an opportunity to apply what you’ve learned in the chapter to the opening article.If your instructor allows, these make excellent opportunities to work in a group, which is often the best way of learning math.
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In selected chapters, a web-based project The projects allow the opportunity for students to collaborate and use mathematics is given. to deal with issues that come up in their lives.
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NEW! Internet Based Projects
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Resources for Success Instructor Resources
Student Resources
Additional resources can be downloaded from www.pearsonglobaleditions.com/sullivan.
Additional resources to help student success:
TestGen®
Chapter Test Prep Videos
®
TestGen (www.pearsonglobaleditions.com/sullivan)
enables instructors to build, edit, print, and administer tests using a computerized bank of questions developed to cover all the objectives of the text.
Students can watch instructors work through step-by-step solutions to all chapter test exercises from the textbook. These are available on YouTube.
PowerPoint® Lecture Slides Fully editable slides that correlate to the textbook.
Instructor Solutions Manual Includes fully worked solutions to all textbook exercises.
Mini Lecture Notes Includes additional examples and helpful teaching tips, by section.
Online Chapter Projects Additional projects that let students apply what was learned in the chapter.
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Algebra Review Four Chapters of Intermediate Algebra review. Perfect for a slower-paced course or for individual review.
Applications Index Acoustics
Astronomy
amplifying sound, 403 loudness of sound, 355, 405 loudspeaker, 607 tuning fork, 607 whispering galleries, 708
angle of elevation of Sun, 573 distances of planets from Sun, 861 planetary orbits, 707–08, 711 Earth, 710 Jupiter, 710 Mars, 710 Mercury, 737 Pluto, 711 radius of the Moon, 437
Aerodynamics modeling aircraft motion, 689–90
Aeronautics Challenger disaster, 390
Aviation
Agriculture
modeling aircraft motion, 689–90 orbital launches, 767
farm management, 843 farm workers in U.S., 390–91 field enclosure, 828 grazing area for cow, 598 minimizing cost, 843 removing stump, 661
Biology
bearing of aircraft, 571, 574 flight time and ticket price, 167 frequent flyer miles, 584–85 holding pattern, 521 intersection point of two planes, 134–35 parking at O’Hare International Airport, 118 revising a flight plan, 592 speed and direction of aircraft, 656, 660
age versus total cholesterol, 398 alcohol and driving, 351, 356 bacterial growth, 382–83, 395 E-coli, 109, 152–53 blood pressure, 521 blood types, 899 bone length, 219 cricket chirping, 206 gestation period, 215 healing of wounds, 340, 355 maternal age versus Down syndrome, 168 muscle force, 661 yeast biomass as function of time, 394–95
Archaeology
Business
age of ancient tools, 384 age of fossil, 388, 389 age of tree, 389 date of prehistoric man’s death, 403
advertising, 167, 219 automobile production, 312, 783 blending coffee, A78 car rentals, 159 checkout lines, 918 clothing store, 920 cookie orders, 848 cost of can, 286–87, 289 of commodity, 313 of manufacturing, 243, 297, 837, A13, A78 marginal, 191, 218 minimizing, 218, 843, 848 of production, 108, 313, 810–11, 848 of theater ticket per student, 297 of transporting goods, 119 weekly, 239 cost equation, 63, 142 cost function, 160 average, 92 demand for candy, 142 for jeans, 167 for PCs, 396 demand equation, 218, 303 discounts, 313
Air travel
Architecture brick staircase, 867, 891 Burj Khalifa building, A14–A15 floor design, 865–66, 891 football stadium seating, 867 mosaic design, 867, 891 Norman window, 204, A20 parabolic arch, 204 racetrack design, 710 special window, 204 stadium construction, 867 window design, 204
Area. See also Geometry of Bermuda Triangle, 598 under a curve, 507 of isosceles triangle, 552 of sector of circle, 415–16, 419 of segment of circle, 610
Art fine decorative pieces, 437
drive-thru rate at Burger King, 336 at Citibank, 340, 355 earnings of young adults, 754 equipment depreciation, 877 ethanol production, 395 expense computation, A79 Jiffy Lube’s car arrival rate, 340–41, 355 managing a meat market, 844 milk production, 396 mixing candy, A78 mixing nuts, A78 new-car markup, A88 orange juice production, 783 personal computer price and demand, 396 precision ball bearings, A13 presale order, 767 product design, 844 production scheduling, 843 product promotion, 64 profit, 811 cigar company, 132 maximizing, 841–42, 843–44 profit function, 88, 191–92 rate of return on, 379 restaurant management, 767 revenue, 191, 195-64, A78 advertising and, 167 airline, 844 from calculator sales, 181 of clothing store, 799–800 daily, 192 from digital music, 132 instantaneous rate of change of, 950, 958 maximizing, 191–92, 203 monthly, 191–92 from seating, 878 theater, 768 revenue equation, 142 salary, 867 gross, 87 increases in, 877, 891 sales commission on, A88 of movie theater ticket, 755, 759–60, 767 net, 40 salvage value, 403 straight-line depreciation, 156–57, 160 supply and demand, 157–58, 160 tax, 297 toy truck manufacturing, 837 transporting goods, 837 truck rentals, 63, 161 unemployment, 921 wages of car salesperson, 63
25
26
Applications Index
Calculus
Computers and computing
area under a curve, 507 carrying a ladder around a corner, 522 maximizing rain gutter construction, 552 projectile motion, 522 Simpson’s rule, 204
Dell PCs, 396 graphics, 659, 812–13 households without personal computers, 390 JPEG image compression, 812 laser printers, A79 LCD monitors, 218 social media, 812 three-click rule for web design, 812 website map, 811 Word users, 390
Carpentry. See also Construction pitch, 65
Chemistry alpha particles, 723 decomposition reactions, 389 drug concentration, 288 gas laws, 142 pH, 354 purity of gold, A79 radioactive decay, 388, 389, 396, 404 radioactivity from Chernobyl, 389 reactions, 204 salt solutions, A79 sugar molecules, A80 volume of gas, A88
Combinatorics airport codes, 901 binary codes, 920 birthday permutations, 903, 907, 914–15, 918, 920 blouses and skirts combinations, 899 book arrangements, 907 box stacking, 907 code formation, 907 combination locks, 908 committee formation, 905, 907, 908, 920 Senate committees, 908 flag arrangement, 906, 920 letter codes, 901, 921 letter combinations, 920 license plate possibilities, 907, 920, 921 lining people up, 902, 907 number formation, 899, 907, 908, 921 objects selection, 908 seating arrangements, 920 shirts and ties combinations, 899 telephone numbers, 920 two-symbol codewords, 898 word formation, 905–06, 907, 921
Communications cell phone plan, 74 cell phone service, 118, 148, 161 cell phone towers, 397 installing cable TV, 137 long distance comparing phone companies, 218 international call plan, 161 phone charges, 160 satellite dish, 698, 700 spreading of rumors, 340, 355 Touch-Tone phones, 556–57, 608
Construction of box, 828 closed, 147 open, 137 of brick staircase, 891 of can, 301 of coffee can, A80 of cylindrical tube, 828 of enclosures around garden, A79 around pond, A79 maximizing area of, 198–99, 203 of fencing, 198–99, 203, 828 minimum cost for, 289 of flashlight, 700 of headlight, 700 of highway, 574, 585, 611 installing cable TV, 137 of open box, 179 pitch of roof, 575 of rain gutter, 204, 430, 552, 565–66 of ramp, 584 access ramp, 64 of rectangular field enclosure, 203 sidewalk area, 438 of stadium, 204, 867 of steel drum, 289 of swimming pool, A21 of swing set, 593 of tent, 597 TV dish, 700 vent pipe installation, 710
Crime income and, 398
Cryptography matrices in, 811
Decorating Christmas tree, A15–A16
Demographics birth rate mother’s age and, 206 of unmarried women, 191 diversity index, 354–55 life expectancy, A88 marital status, 900 mosquito colony growth, 388–89
population. See Population rabbit colony growth, 860
Design of awning, 585–86 of box with minimum surface area, 289 of fine decorative pieces, 437 of Little League Field, 421 of water sprinkler, 419
Direction of aircraft, 660 compass heading, 660 for crossing a river, 660, 661 of fireworks display, 722 of lightning strikes, 722 of motorboat, 660 of swimmer, 688
Distance Bermuda Triangle, A21 bicycle riding, 97 from Chicago to Honolulu, 508 circumference of Earth, 421 between cities, 414–15, 419 between Earth and Mercury, 586 between Earth and Venus, 586 from Earth to a star, 573–74 of explosion, 723 height of aircraft, 584, 586 of bouncing ball, 877, 891 of bridge, 584 of building, 574, 575, 576 of cloud, 569 of Eiffel Tower, 573 of embankment, 574 of Ferris Wheel rider, 521 of Great Pyramid of Cheops, 586, A21 of helicopter, 582, 611 of hot-air balloon, 574 of Lincoln’s caricature on Mt. Rushmore, 574 of monument, 574 of mountain, 581, 584 of statue on building, 569–70 of tree, 437 of Willis Tower, 574 from home, 97 from Honolulu to Melbourne, Australia, 508 of hot-air balloon to airport, 612 from intersection, 40 from intersection, 40, 136 length of guy wire, 592 of mountain trail, 574 of ski lift, 584 limiting magnitude of telescope, 403 to the Moon, 584 nautical miles, 420 pendulum swings, 873, 877 to plateau, 573
Applications Index
across pond, 573 range of airplane, A80 reach of ladder, 573 of rotating beacon, 474–75 at sea, 581–82, 585 of ship from shore, 573 to shore, 585, 610 between skyscrapers, 575 sound to measure, A71 to tower, 586 traveled by wheel, A20 between two moving vehicles, 40 toward intersection, 136 between two objects, 574 viewing, 437 visibility of Gibb’s Hill Lighthouse beam, 570–71, 576, A22 visual, A21 walking, 97 of water tower to building, 900 width of gorge, 572 of Mississippi River, 575 of river, 568, 610
Economics Consumer Price Index (CPI), 380 demand equations, 303 earnings of young adults, 754 federal deficit, 379, 403 income versus crime rate, 398 inflation, 379–80 IS-LM model in, 768 marginal propensity to consume, 878 multiplier, 878 participation rate, 88 personal computer price and demand, 396 poverty rates, 301 poverty threshold, 40 relative income of child, 811 unemployment, 921
Education admission probabilities, 921 age distribution of community college, 921 college costs, 379, 877 college tuition and fees, 810 degrees awarded, 897 doctoral degrees awarded, 918 faculty composition, 918 funding a college education, 403 grade computation, A88 grade-point average and video games, 167 IQ tests, A88 learning curve, 341, 355 maximum level achieved, 850 multiple-choice test, 907 Spring break, 843 student loan, 148 interest on, 810 true/false test, 907
Electricity alternating current (ac), 491, 542 alternating current (ac) circuits, 466, 484 alternating current (ac) generators, 466 charging a capacitor, 608 cost of, 116–17 current in RC circuit, 341 current in RL circuit, 341, 355 impedance, A95 Kirchhoff’s Rules, 768, 783 parallel circuits, A95 resistance in, 274 rates for, 63–64, A88 resistance, 142, 143, 274, A52, A54 due to a conductor, 148 voltage foreign, A13 U.S., A13
Electronics loudspeakers, 607 microphones, 50 sawtooth curve, 552, 608
Energy ethanol production, 395 heat loss through wall, 140 through window, 147 nuclear power plant, 722–23 solar, 50, 668, 700 thermostat control, 131–32
Engineering bridges clearance, 466 Golden Gate, 200–01 parabolic arch, 218, 700–01 semielliptical arch, 710, 751 suspension, 204, 700 crushing load, A71 drive wheel, 611 Gateway Arch (St. Louis), 701 grade of road, 65 horsepower, 142 lean of Leaning Tower of Pisa, 585 maximum weight supportable by pine, 139 moment of inertia, 557 piston engines, 436–37 product of inertia, 552 road system, 624 robotic arm, 678 rods and pistons, 593 rod tolerance, 215 safe load for a beam, 143 searchlight, 530, 700, 751 whispering galleries, 710
Entertainment banquet hall rental, 843 cable subscribers, 398 Demon Roller Coaster customer rate, 341
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movie theater, 507 theater revenues, 768
Environment endangered species population, 340 lake pollution control laws, 860 oil leakage, 312
Exercise and fitness. See also Sports heartbeats during exercise, 153–54 for weight loss, A88
Finance. See also Investment(s) balancing a checkbook, A13 bills in wallet, 921 calculator sales revenue, 181 clothes shopping, 849 college costs, 379, 877 computer system purchase, 379 cost of car rental, 119 of driving a car, 63 of electricity, 116–17 of fast food, 767 minimizing, 218, 289 of natural gas, 118–19 of RV rental, 220 of tattoo, 685 of trans-Atlantic travel, 87, 95–96 of triangular lot, 597 cost equation, 142 cost function, 160 cost minimization, 191 credit cards balance on, 820 debt, 860 interest on, 379 minimum payments for, 119–20 payment, 860 demand equation, 203, 220 depreciation, 340, 399 of car, 371, 406 division of money, A73–A74, A78 electricity rates, 63–64 federal income tax, A88 financial planning, 767, 780, 783–84, 834–35, 836, 838, 844, A73–A74, A78 foreign exchange, 313 future value of money, 243 gross salary, 87 international call plan, 161 life cycle hypothesis, 205 loans, A78 car, 860 interest on, 148, 810, A73 repayment of, 379 student, 810 mortgages fees, 119 interest rates on, 379 payments, 138, 141, 147 second, 379 national debt, 108–09 price appreciation of homes, 379
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Applications Index
prices demand vs., 218 of fast food, 769 for soda and hot dog combinations, 161 refunds, 767 rents and square footage, 205 revenue equation, 142 revenue maximization, 191, 197–98 rich man’s promise, 878 salary calculation, 313 salary options, 878–79 saving for a car, 379 for a home, 877 savings accounts interest, 379 sinking fund, 877 taxes, 160 e-filing returns, 119 federal income, 119, 325 luxury, 160 used-car purchase, 379 water bills, A88
Food and nutrition animal, 844 calories in fast foods, 76–77 candy, 166 colored candies, 909–10, 921 cooler contents, 921 cooling time of pizza, 389 fast food, 767, 769 Girl Scout cookies, 918 hospital diet, 768, 783 “light” foods, A88 milk production, 396 number of possible meals, 897–98 pig roasts, 390 raisins, 166 warming time of Beer stein, 389
Forestry wood product classification, 388
Games die rolling, 910–11, 912, 921 grains of wheat on a chess board, 878 Powerball, 921
Gardens and gardening. See also Landscaping enclosure for, A79
Geography area of Bermuda Triangle, 598 area of lake, 597, 611 grade of a mountain trail, 829 inclination of hill, 669 inclination of mountain trail, 568–69, 610 width of a river, 568
Geology earthquakes, 355–56
Geometry
Health. See also Medicine
angle between two lines, 542 balloon volume, 312 circle area of, 597, A78 area of sector of, 415–16, 419 circumference of, A7, A12, A78 equation of, 794 inscribed, 135–36, 599 length of chord of, 593 radius of, 827 collinear points, 793 cone volume, 142, 313 cube length of edge of, 257 surface area of, A13 volume of, A13 cylinder inscribing in cone, 137 inscribing in sphere, 136 volume of, 142, 313 Descartes’s method of equal roots, 828 equation of line, 793 ladder angle, 612 polygon area of, 794 number of sides of, 179 quadrilateral area, 612 rectangle area of, 87, 134, 218, 420, 711, A12 dimensions of, 218, 827 inscribed in semicircle, 136, 553 perimeter of, A12 semicircle inscribed in, 136 semicircle area, 597, 612 sphere surface area of, A12 volume of, A12 square area of, A20, A78 perimeter of, A78 surface area of balloon, 312 of cube, A13 of sphere, A12 triangle area of, 597, 612, 794, A12 circumscribing, 587 equilateral, A12 inscribed in circle, 136 isosceles, 87, 827, 828 Pascal’s, 860 perimeter of, A12 right, 572 sides of, 613 volume of paralleliped, 684
breast cancer survival rate, 396 cigarette use among teens, 64 expenditures on, 88 ideal body weight, 325 life cycle hypothesis, 205 pancreatic cancer survival rate, 340
Government federal deficit, 403 federal income tax, 88, 119, 325, A88 first-class mail charge, 120 national debt, 108–09 stimulus package (2009), 379
Home improvement. See also Construction painting a house, 769 painting a room, 475
Investment(s) annuity, 874–75, 877 in bonds, 844 EE Series, 379 Treasuries, 783, 784, 834–35, 836, 838 Treasury notes vs. Treasury bonds, 780 zero-coupon, 376, 380 in CDs, 375, 844 compound interest on, 372–75, 379, 467, 929 diversified, 768–69 division among instruments, A78 doubling of, 377, 380 in fixed-income securities, 844 433(K), 877, 891 growth rate for, 379 IRA, 379, 874–75, 877 in mutual fund, 392–93 return on, 379, 843, 844 in stock appreciation, 379 beta, 150, 221–22 NASDAQ stocks, 907 NYSE stocks, 907 portfolios of, 900 price of, 878 time to reach goal, 379, 380 tripling of, 377, 380
Landscaping. See also Gardens and gardening pond enclosure, 218 removing stump, 661 tree cutting, 584, 783 watering lawn, 419
Law and law enforcement motor vehicle thefts, 918 violent crimes, 88
Leisure and recreation cable TV, 137 centrifugal force ride, 419 community skating rink, 148 Ferris wheel, 71, 420, 521, 586, 607 gondola, 419 swing displacement, 613 video games and grade-point average, 167
Applications Index
Marketing. See Business Measurement optical methods of, 530 of rainfall, 668
Mechanics. See Physics Medicine. See also Health blood pressure, 521 breast cancer survival rate, 396 drug concentration, 108, 288 drug medication, 340, 355 healing of wounds, 340, 355 pancreatic cancer, 340 spreading of disease, 404
Meteorology weather balloon height and atmospheric pressure, 393–94
Miscellaneous bending wire, 828 biorhythms, 466 carrying a ladder around a corner, 474, 522 citrus ladders, 867 cross-sectional area of beam, 87–88, 95 curve fitting, 768, 782, 847 drafting error, 40 pet ownership, 918 rescue at sea, 581–82 rooms in housing units, 87 surface area of balloon, 312 surveillance satellites, 575–76 volume of balloon, 312 window dimensions, 179 wire enclosure area, 136
Mixtures. See also Chemistry blending coffees, 837, 848, A74–A75, A78 blending teas, A78 cement, A80 mixed nuts, 767, 837, 848, A78 mixing candy, A78 solution, 767 water and antifreeze, A79
Motion. See also Physics catching a train, 752 on a circle, 419 of Ferris Wheel rider, 521 of golf ball, 95, 522 minute hand of clock, 418 objects approaching intersection, 748 of pendulum, 608 revolutions of circular disk, A20 simulating, 742–43 tortoise and the hare race, 827 uniform, 136, 748, A75–A77, A78–A79
Motor vehicles
Pets
alcohol and driving, 351, 356 approaching intersection, 748 automobile production, 312, 783 automobile theft, 918 average car speed, A80 brake repair with tune-up, 921 braking load, 669, 688 crankshafts, 585 depreciation of, 305, 371, 399, 406 distance between, 437 with Global Positioning System (GPS), 403 loans for, 860 miles per gallon, 205–06 new-car markup, A88 RV rental cost, 220 spin balancing tires, 420 stopping distance, 88, 191, 325 used-car purchase, 379 windshield wiper, 419
dog roaming area, 420
Music iPod storage capacity for, 161 revenues from, 132
Navigation avoiding a tropical storm, 592 bearing, 571, 591 of aircraft, 571, 574 of ship, 574, 611 charting a course, 661 commercial, 584–85 compass heading, 660 correct direction for crossing river, 660 error in correcting, 589–90, 611 time lost due to, 585 rescue at sea, 581–82, 584 revising a flight plan, 592
Oceanography tides, 485
Optics angle of incidence, 522–23 angle of refraction, 522–23 bending light, 523 index of refraction, 522–23 intensity of light, 142 laser beam, 573 laser projection, 552 lensmaker’s equation, A54 light obliterated through glass, 340 magnitude of telescope, 403 measurements using, 530 mirrors, 723 reflecting telescope, 700
Pediatrics height vs. head circumference, 196, 325
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Pharmacy vitamin intake, 768, 784
Photography camera distance for full-body shot, 574
Physics angle of elevation of Sun, 573 bouncing balls, 891 braking load, 669, 688 damped motion, 603–04 direction of aircraft, 660 Doppler effect, 289 falling objects, 141 force, 659, A78 muscle, 661 resultant, 659 of wind on a window, 140, 142 gravity, 274, 297 on Earth, 87, 325 on Jupiter, 87 harmonic motion, 602, 607, 611 heat loss through a wall, 140 heat transfer, 522 horsepower, 142 inclination of mountain trail, 568–69 inclination of ramp, 661 intensity of light, 142 kinetic energy, 143, A78 maximum weight supportable by pine, 139 moment of inertia, 557 motion of object, 602, 744 Newton’s law, 141 pendulum motion, 419, 607, 608, 873, A62, A71 period, 132, 326 simple pendulum, 141 pressure, 142, A78 product of inertia, 552 projectile motion, 181, 199–200, 203–04, 436, 437–38, 522, 547, 552, 557, 741–42, 747, 748, 749, 752 artillery, 513 hit object, 748 thrown object, 747 rate of change average, 960 instantaneous, 946, 949 safe load for a beam, 143 simulating motion, 742–43 sound to measure distance, A71 static equilibrium, 657, 660, 661, 688, 689 static friction, 661 stopping distance, 191 stress of materials, 143 stretching a spring, 142 tension, 657, 660, 688, 689, 883 thrown object, 195, 205, 655, 947–48, 949 truck pulls, 660
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Applications Index
uniform motion, 136, 748, 752, A75–A77, A78–A79 velocity down inclined planes, A62 vertically propelled object, 179, 195 vibrating string, 142 weight, 142, 147, 656, 660 effect of elevation on, 96 work, 678, A78
Play wagon pulling, 659, 666, 667
Population. See also Demographics bacterial, 388, 389, 390, 395 decline in, 389 of divorced people, 201–02 E-coli growth, 109, 152–53 of endangered species, 340, 390 of fruit fly, 387 as function of age, 87 growth in, 388, 389 insect, 274, 388 of rabbit colony, 860 of trout, 860 of United States, 370, 396–97, 893 of world, 371, 397, 403, 851 future of, 960
Probability checkout lines, 918 classroom composition, 918 “Deal or No Deal” TV show, 894 exponential, 336, 340, 355 gender composition of 35-child family, 912 household annual income, 918 Monty Hall Game, 922 Poisson, 341 “Price is Right” games, 918 of shared birthdays in room of n people, 391 tossing a fair coin, 909, 911 of winning a lottery, 919
Psychometrics IQ tests, 215
Pyrotechnics fireworks display, 722
Rate. See also Speed of car, 419 catching a bus, 747–48 catching a train, 747 current of stream, 768 of emptying oil tankers, A80 a pool, A80 a tub, A80 to keep up with the Sun, 420 miles per gallon, 205–06 revolutions per minute of bicycle wheels, 419 of pulleys, 421
speed average, A80 of current, A78–A79 of motorboat, A78–A79 of moving walkways, A79 of plane, A80
Real estate commission schedule, A88 cost of triangular lot, 597 ground area covered by building, 597–98 price appreciation of homes, 379 rents and square footage, 205 valuing a home, 33, 73
Recreation bungee jumping, 297 Demon Roller Coaster customer rate, 341 online gambling, 918
Security
basketball, 908 free throws, 95, 575 granny shots, 95 biathlon, A80 bungee jumping, 297 exacta betting, 921 football, 710, 752, 908, A79 golf, 918 distance to the green, 591 putts, 398–99 sand bunkers, 513 hammer throw, 492 Olympic heroes, A80 pool shots, 576 races, 825, 827–28, A80 relay runners, 920 swimming, 613, 688 tennis, A79
Statistics. See Probability
security cameras, 573
Surveys
Seismology
of appliance purchases, 899 data analysis, 896, 899 stock portfolios, 899 of summer session attendance, 899 of TV sets in a house, 918
calibrating instruments, 751
Sequences. See also Combinatorics ceramic tile floor design, 865–66 Drury Lane Theater, 867 Fibonacci, 860 football stadium seating, 867 seats in amphitheater, 867
Speed of aircraft, 660 angular, 419, 491 of current, 420, 848 of cyclists moving in opposite directions, A80 as function of time, 97, 136 of glider, 610 ground, 660 instantaneous of ball, 947–48, 949, 958 on the Moon, 949–50 linear, 416–17 on Earth, 419, 420 of Moon, 420 of motorboat, 660, A76–A77 revolutions per minute of pulley, 420 of rotation of lighthouse beacons, 491 of swimmer, 688 of truck, 573 of wheel pulling cable cars, 420 wind, 767
Sports baseball, 748–49, 908, 920 diamond, 39 dimensions of home plate, 597 field, 592 Little League, 40, 421 on-base percentage, 162–63 stadium, 592 World Series, 908
Technology. See also Computers and computing Blu-ray drive, 419 DVD drive, 419 iPod storage capacity for music, 161
Temperature of air parcel, 867 body, A13 conversion of, 313, 325 cooling time of pizza, 389 cricket chirping and, 206 measuring, 64, 132 after midnight, 243 monthly, 484–85, 491 of portable heater, 403 relationship between scales, 132 sinusoidal function from, 480–81 of skillet, 403 warming time of Beer stein, 389 wind chill factor, 404
Tests and testing IQ, A88
Time for Beer stein to warm, 389 for block to slide down inclined plane, 436 Ferris Wheel rider height as function of, 521 to go from an island to a town, 137 hours of daylight, 482–83, 485–86, 506–07 for pizza to cool, 389 of sunrise, 420, 507 of trip, 437, 451
Applications Index
Transportation. See also Air travel; Motor vehicles de-icing salt, 513 Falls Incline Railway, 574
Travel. See also Air travel; Navigation bearing, 611 drivers stopped by the police, 405 driving to school, 142 parking at O’Hare International Airport, 118
Volume of box, 669 of gasoline in tank, A62 of water in cone, 137
artillery, 513
weather satellites, 71 wind chill, 120, 404
Weather
Work, 667, 678
atmospheric pressure, 340, 355 avoiding a tropical storm, 592 cooling air, 867 hurricanes, 242, 484 lightning strikes, 719–20, 722 rainfall measurement, 668 relative humidity, 341
computing, 667, 668, 688 constant rate jobs, 848 pulling a wagon, 666, 667 ramp angle, 669 wheel barrow push, 659 working together to do a job, A77, A79
Weapons
31
Foundations: A Prelude to Functions How to Value a House
F
Two things to consider in valuing a home are, first, how does it compare to similar homes that have sold recently? Is the asking price fair? And second, what value do you place on the advertised features and amenities? Yes, other people might value them highly, but do you? Zestimate home valuation, RealEstateABC.com, and Reply.com are among the many algorithmic (generated by a computer model) starting points in figuring out the value of a home. They show you how the home is priced relative to other homes in the area, but you need to add in all the things that only someone who has seen the house knows. You can do that using My Estimator, and then you create your own estimate and see how it stacks up against the asking price.
Looking at “Comps” Knowing whether an asking price is fair will be important when you’re ready to make an offer on a house. It will be even more important when your mortgage lender hires an appraiser to determine whether the house is worth the loan you’re after. Check with your agent, Zillow.com, propertyshark.com, or other websites to see recent sales of homes in the area that are similar, or comparable, to what you’re looking for. Print them out and keep these “comps” in a three-ring binder; you’ll be referring to them quite a bit. Note that “recent sales” usually means within the last six months. A sales price from a year ago may bear little or no relation to what is going on in your area right now. In fact, some lenders will not accept comps older than three months. Market activity also determines how easy or difficult it is to find accurate comps. In a “hot” or busy market, with sales happening all the time, you’re likely to have lots of comps to choose from. In a less active market, finding reasonable comps becomes harder. And if the home you’re looking at has special design features, finding a comparable property is harder still. It’s also necessary to know what’s going on in a given sub-segment. Maybe large, high-end homes are selling like hotcakes, but owners of smaller houses are staying put, or vice versa. Source: http://realestate.yahoo.com/Homevalues/How_to_Value_a_House.html
—See the Internet-based Chapter Project—
Here we connect algebra and geometry using the rectangular coordinate system. In the 1600s, algebra had developed to the point that René Descartes (1596–1650) and Pierre de Fermat (1601–1665) were able to use rectangular coordinates to translate geometry problems into algebra problems, and vice versa. This allowed both geometers and algebraists to gain new insights into their subjects, which had been thought to be separate but now were seen as connected.
Intercepts; Symmetry F.3 Lines F.4 Circles
Chapter Project
33
34
CHAPTER F Foundations: A Prelude to Functions
F.1 The Distance and Midpoint Formulas PREPARING FOR THIS SECTION Before getting started, review the following: r "MHFCSB&TTFOUJBMT "QQFOEJY" 4FDUJPO" pp. A1–A10)
r (FPNFUSZ&TTFOUJBMT "QQFOEJY" 4FDUJPO" pp. A13–A19)
Now Work the ‘Are You Prepared?’ problems on page 38.
OBJECTIVES 1 Use the Distance Formula (p. 35) 2 Use the Midpoint Formula (p. 37)
Rectangular Coordinates
Figure 1 y 4 2 –4
–2
O
2
4
x
–2 –4
Figure 2 y 4 3 (–3, 1) 1 –4 3 (–2, –3)
3
(3, 2) 2
O 3
x 4 2 (3, –2)
2
Figure 3 y Quadrant II x < 0, y > 0
Quadrant I x > 0, y > 0
Quadrant III x < 0, y < 0
Quadrant IV x > 0, y < 0
x
A point on the real number line is located by a single real number called the coordinate of the point. For work in a two-dimensional plane, points are located by using two numbers. Begin with two real number lines located in the same plane: one horizontal and the other vertical. The horizontal line is called the x-axis, the vertical line the y-axis, and the point of intersection the origin O. See Figure 1. Assign coordinates to every point on these number lines using a convenient scale. Recall that the scale of a number line is the distance between 0 and 1. In mathematics, we usually use the same scale on each axis, but in applications, a different scale is often used. The origin O has a value of 0 on both the x-axis and the y-axis. Points on the x-axis to the right of O 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 O are associated with positive real numbers, and those below O are associated with negative real numbers. In Figure 1, the x-axis and y-axis are labeled as x and y, respectively, and an arrow at the end of each axis is used 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 1x, y2 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 7 0, and if P is to the left of the y-axis, then x 6 0); and let y denote the signed distance of P from the x-axis. The ordered pair 1x, y2 , 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 1 - 3, 12 , go 3 units along the x-axis to the left of O and then go straight up 1 unit. We plot this point by placing a dot at this location. See Figure 2, in which the points with coordinates 1 - 3, 12 , 1 - 2, - 32 , 13, - 22 , and 13, 22 are plotted. The origin has coordinates 10, 02 . Any point on the x-axis has coordinates of the form 1x, 02 , and any point on the y-axis has coordinates of the form 10, y2 . If 1x, y2 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 1x, y2 by writing P = 1x, y2 . Usually, we will simply say, “the point 1x, y2 ” rather than “the point whose coordinates are 1x, y2 .” 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 negative. Points on the coordinate axes belong to no quadrant.
Now Work
PROBLEM
11
*Named after René Descartes (1596–1650), a French mathematician, philosopher, and theologian.
SECTION F.1 The Distance and Midpoint Formulas
Figure 4
35
COMMENT On a graphing calculator, you can set the scale on each axis. Once this has been done, you obtain the viewing rectangle. See Figure 4 for a typical viewing rectangle. You should now read Section B.1, The Viewing Rectangle, in Appendix B. ■
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.
EXAMPLE 1 Solution
Finding the Distance between Two Points
Find the distance d between the points 11, 32 and 15, 62 .
First plot the points 11, 32 and 15, 62 and connect them with a straight line. See Figure 5(a). To find the length d, begin by drawing a horizontal line from 11, 32 to 15, 32 and a vertical line from 15, 32 to 15, 62 , forming a right triangle, as shown in Figure 5(b). One leg of the triangle is of length 4 (since 0 5 - 1 0 = 4), and the other is of length 3 (since 0 6 - 3 0 = 3). By the Pythagorean Theorem, the square of the distance d that we seek is d 2 = 42 + 32 = 16 + 9 = 25 d = 225 = 5
Figure 5
y 6
y 6
(5, 6)
(5, 6) d
d 3
3 (1, 3)
0
(1, 3) 4 (5, 3)
0
6 x
3
3
3
6 x
(b)
(a)
r
The distance formula provides a straightforward method for computing the distance between two points.
THEOREM
Distance Formula The distance between two points P1 = 1x1 , y1 2 and P2 = 1x2 , y2 2 , denoted by d 1P1 , P2 2 , is
In Words To compute the distance between two points, find the difference of the x-coordinates, square it, and add this to the square of the difference of the y-coordinates. The square root of this sum is the distance.
Figure 6
d1P1 , P2 2 = 2 1x2 - x1 2 2 + 1y2 - y1 2 2
(1)
Proof of the Distance Formula Let 1x1 , y1 2 denote the coordinates of point P1 and let 1x2 , y2 2 denote the coordinates of point P2 . Assume that the line joining P1 and P2 is neither horizontal nor vertical. Refer to Figure 6(a). The coordinates of P3 are 1x2 , y1 2 . The horizontal distance from P1 to P3 is the absolute value of the difference of the x-coordinates, 0 x2 - x1 0 . The vertical distance from P3 to P2 is the
y
y P2 (x2, y2)
y2 y1
P1 (x1, y1)
P3 (x2, y1) x2
x1 (a)
x
P2 (x2, y2)
y2
d(P1, P2)
y1 P1 (x1, y1)
⏐y2 y1⏐
⏐x2 x1⏐ x2
x1 (b)
P3 (x2, y1) x
36
CHAPTER F Foundations: A Prelude to Functions
absolute value of the difference of the y-coordinates, 0 y2 - y1 0 . See Figure 6(b). The distance d 1P1 , P2 2 that we seek is the length of the hypotenuse of the right triangle, so, by the Pythagorean Theorem, it follows that 3 d 1P1 , P2 2 4 2 = 0 x2 - x1 0 2 + 0 y2 - y1 0 2 = 1x2 - x1 2 2 + 1y2 - y1 2 2 d 1P1 , P2 2 = 2 1x2 - x1 2 2 + 1y2 - y1 2 2
Now, if the line joining P1 and P2 is horizontal, then the y-coordinate of P1 equals the y-coordinate of P2; that is, y1 = y2 . Refer to Figure 7(a). In this case, the distance formula (1) still works, because, for y1 = y2 , it reduces to d 1P1 , P2 2 = 2 1x2 - x1 2 2 + 02 = 2 1x2 - x1 2 2 = 0 x2 - x1 0
Figure 7
y
y y2 P1 (x1, y1)
y1
d (P1, P2)
P2 (x2, y1)
⏐y2 y1⏐ d (P1, P2) P1 (x1, y1)
y1
⏐x2 x1⏐ x2
x1
P2 (x1, y2)
x1
x
(a)
x
(b)
A similar argument holds if the line joining P1 and P2 is vertical. See Figure 7(b).
EX A MPL E 2
Solution
■
Using the Distance Formula
Find the distance d between the points 1 - 3, 52 and (3, 2). Use the distance formula, equation (1), with P1 = (x1, y1) = ( - 3, 5) and P2 = (x2, y2) = (3, 2). Then d = 2[3 - 1 - 32]2 + 12 - 52 2 = = = =
Now Work
PROBLEMS
15
AND
262 + 1 - 32 2 236 + 9 245 325 ≈ 6.71
r
19
The distance between two points P1 = 1x1, y1 2 and P2 = 1x2, y2 2 is never a negative number. Furthermore, the distance between two points is 0 only when the points are identical—that is, when x1 = x2 and y1 = y2. Also, because 1x2 - x1 2 2 = 1x1 - x2 2 2 and 1y2 - y1 2 2 = 1y1 - y2 2 2, it makes no difference whether the distance is computed from P1 to P2 or from P2 to P1; that is, d 1P1 , P2 2 = d 1P2 , P1 2 . 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.
EX A MPL E 3
Using Algebra to Solve Geometry Problems
Consider the three points A = 1 - 2, 12 , B = 12, 32 , and C = 13, 12 . (a) (b) (c) (d)
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.
SECTION F.1 The Distance and Midpoint Formulas
Solution
37
(a) Figure 8 shows the points A, B, C and the triangle ABC. (b) To find the length of each side of the triangle, use the distance formula, equation (1).
d 1A, B2 = 2 3 2 - 1 - 22 4 2 + 13 - 12 2 = 216 + 4 = 220 = 225 d 1B, C2 = 2 13 - 22 2 + 11 - 32 2 = 21 + 4 = 25 d 1A, C2 = 2 3 3 - 1 - 22 4 2 + 11 - 12 2 = 225 + 0 = 5
(c) If the triangle is a right triangle, then the sum of the squares of the lengths of two of the sides will equal the square of the length of the third side. (Why is this sufficient?) Looking at Figure 8, it seems reasonable to conjecture that the right angle is at vertex B. We shall check to see whether
Figure 8 y B = (2, 3)
3 A = (–2, 1)
3 d 1A, B2 4 2 + 3 d 1B, C2 4 2 = 3 d 1A, C2 4 2
C = (3, 1)
–3
Using the results from part (b) yields
x
3
3 d 1A, B2 4 2 + 3 d 1B, C2 4 2 = 1 225 2 + 1 25 2 2 = 20 + 5 = 25 = 3 d 1A, C2 4 2 2
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 =
1 1 1Base2 1Height2 = 1 225 2 1 25 2 = 5 square units 2 2
Now Work
29
2 Use the Midpoint Formula
Figure 9 y P2 = (x 2, y2) y2 M = (x, y) y
y1
PROBLEM
r
y – y1
x – x1 P1 = (x1, y1) x1
y2 – y x2 – x B = (x 2, y)
A = (x, y1) x
x2
x
THEOREM
In Words To find the midpoint of a line segment, average the x-coordinates of the endpoints, and average the y-coordinates of the endpoints.
We now derive a formula for the coordinates of the midpoint of a line segment. Let P1 = 1x1 , y1 2 and P2 = 1x2 , y2 2 be the endpoints of a line segment, and let M = 1x, y2 be the point on the line segment that is the same distance from P1 as it is from P2 . See Figure 9. The triangles P1 AM and MBP2 are congruent. [Do you see why? Angle AP1 M = angle BMP2 ,* angle P1 MA = angle MP2 B, and d 1P1 , M2 = d 1M, P2 2 is given. Thus we have angle–side–angle.] Hence, corresponding sides are equal in length. That is, x - x1 = x2 - x 2x = x1 + x2 x1 + x2 x = 2
and
y - y1 = y2 - y 2y = y1 + y2 y1 + y2 y = 2
Midpoint Formula The midpoint M = 1x, y2 of the line segment from P1 = 1x1 , y1 2 to P2 = 1x2 , y2 2 is M = 1x, y2 = ¢
x1 + x2 y1 + y2 , ≤ 2 2
(2)
*A postulate from geometry states that the transversal P1 P2 forms congruent corresponding angles with the parallel line segments P1 A and MB.
38
CHAPTER F Foundations: A Prelude to Functions
Finding the Midpoint of a Line Segment
EX A MPL E 4
Find the midpoint of the line segment from P1 = 1 - 5, 52 to P2 = 13, 12 . Plot the points P1 and P2 and the midpoint. Apply the midpoint formula (2) using x1 = - 5, y1 = 5, x2 = 3, and y2 = 1. Then the coordinates 1x, y2 of the midpoint M are
Solution
Figure 10 y P1 (–5, 5)
x =
5
M (–1, 3)
That is, M = 1 - 1, 32 . See Figure 10.
P2 (3, 1)
–5
5
y1 + y2 x1 + x2 -5 + 3 5 + 1 = = - 1 and y = = = 3 2 2 2 2
Now Work
x
PROBLEM
r
35
F.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. On the real number line the origin is assigned the number 0 . (p. A4) 2. If - 3 and 5 are the coordinates of two points on the real number line, the distance between these points is 8 . (p. A6) 3. If 3 and 4 are the legs of a right triangle, the hypotenuse is 5 . (pp. A13–A14)
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. A14) 112 + 602 = 612 5. State the formula for the area A of a triangle whose base is b and whose altitude is h. (p. A15) A = 1 bh 2 6. State the three cases for which two triangles are congruent. (p. A16) ASA, SSS, SAS
Concepts and Vocabulary 7. If 1x, y2 are the coordinates of a point P in the xy-plane, then x is called the x-coordinate or abscissa of P, and y y-coordinate or ordinate is the of P. 8. The coordinate axes divide the xy-plane into four sections quadrants called .
9. The distance d between two points P1 = (x1, y1) and P2 = (x2, y2) is d = 2(x2 - x1)2 + (y2 - y1)2. 10. If three distinct points P, Q, and R all lie on a line, and if midpoint d1P, Q2 = d1Q, R2 , then Q is called the of the line segment from P to R.
Skill Building In Problems 11 and 12, plot each point in the xy-plane. Tell in which quadrant or on what coordinate axis each point lies.
*11. (a) A = 1 - 3, 22
*12. (a) A = 11, 42
(d) D = 16, 52
(b) B = 16, 02
(b) B = 1 - 3, - 42
(e) E = 10, - 32
(c) C = 1 - 2, - 22
(d) D = 14, 12 (e) E = 10, 12
(c) C = 1 - 3, 42
(f) F = 16, - 32
(f) F = 1 - 3, 02
*13. Plot the points 12, 02 , 12, - 32 , 12, 42 , 12, 12 , and 12, - 12 . Describe the set of all points of the form 12, y2 , where y is a real number.
*14. Plot the points 10, 32 , 11, 32 , 1 - 2, 32 , 15, 32 , and 1 - 4, 32 . Describe the set of all points of the form 1x, 32 , where x is a real number.
In Problems 15–28, find the distance d1P1 , P2 2 between the points P1 and P2 . 15. 25
16. 25
y 2 P = (2, 1) 2 P1 = (0, 0) –2
–1
2
x
17. 210
y P2 = (–2, 1) 2 P = (0, 0) 1 –2
–1
2
x
P2 (–2, 2)
–2
18. 210
y 2
P1 (1, 1) 2 x
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
y P1 = (–1, 1) 2 –2
–1
P2 = (2, 2)
2
x
SECTION F.1 The Distance and Midpoint Formulas
19. P1 = 13, - 42; P2 = 15, 42 21. P1 = 1 - 3, 22; P2 = 16, 02
2217 285
23. P1 = 14, - 22; P2 = 1 - 2, - 52
325
25. P1 = 1 - 0.2, 0.32; P2 = 12.3, 1.12 27. P1 = 1a, b2; P2 = 10, 02
26.89 ≈ 2.62
2a2 + b2
39
20. P1 = 1 - 1, 02; P2 = 12, 42 5 22. P1 = 12, - 32; P2 = 14, 22 229 24. P1 = 1 - 4, - 32; P2 = 16, 22 525 26. P1 = 11.2, 2.32; P2 = 1 - 0.3, 1.12 28. P1 = 1a, a2; P2 = 10, 02
23.69 ≈ 1.92
a 22
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 = 1 - 2, 52; B = 11, 32; C = 1 - 1, 02
*30. A = 1 - 2, 52; B = 112, 32; C = 110, - 112
*33. A = 14, - 32; B = 10, - 32; C = 14, 22
*34. A = 14, - 32; B = 14, 12; C = 12, 12
*31. A = 1 - 5, 32; B = 16, 02; C = 15, 52
*32. A = 1 - 6, 32; B = 13, - 52; C = 1 - 1, 52
In Problems 35–44, find the midpoint of the line segment joining the points P1 and P2 . 35. P1 = 13, - 42; P2 = 15, 42 37. P1 = 1 - 3, 22; P2 = 16, 02
(4, 0)
39. P1 = 14, - 22; P2 = 1 - 2, - 52
3 a , 1b 2
41. P1 = 1 - 0.2, 0.32; P2 = 12.3, 1.12 43. P1 = 1a, b2; P2 = 10, 02 a a, b b 2 2
7 a1, - b 2 (1.05, 0.7)
36. P1 = 1 - 2, 02; P2 = 12, 42 38. P1 = 12, - 32; P2 = 14, 22
40. P1 = 1 - 4, - 32; P2 = 12, 22
1 a3, - b 2
42. P1 = 11.2, 2.32; P2 = 1 - 0.3, 1.12 44. P1 = 1a, a2; P2 = 10, 02 a a, a b 2 2
(0, 2)
1 a - 1, - b 2 (0.45, 1.7)
Applications and Extensions 45. Find all points having an x-coordinate of 2 whose distance from the point 1 - 2, - 12 is 5. (2, 2); (2, - 4) 46. Find all points having a y-coordinate of - 3 whose distance from the point 11, 22 is 13. (13, - 3); ( - 11, - 3) 47. Find all points on the x-axis that are 5 units from the point 14, - 32 . (0, 0); (8, 0) 48. Find all points on the y-axis that are 5 units from the point 14, 42 . (0, 1); (0, 7) 49. 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 = 10, 02, B = 16, 02 , and C = 14, 42 . 217; 225; 229
C
Median Midpoint
A
B
50. Geometry An equilateral triangle is one in which all three sides are of equal length. If two vertices of an equilateral triangle are 10, 42 and 10, 02, find the third vertex. How many of these triangles are possible? Two triangles are possible; vertex: 1 223, 2 2 or 1 - 223, 2 2 s
a 23 a *52. Geometry Verify that the points (0, 0), (a, 0), and a , b 2 2 are the vertices of an equilateral triangle. Then show that the midpoints of the three sides are the vertices of a second equilateral triangle (refer to Problem 50). In Problems 53–56, find the length of each side of the triangle determined by the three points P1 , 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.) *53. P1 = 12, 12; P2 = 1 - 4, 12; P3 = 1 - 4, - 32
*54. P1 = 1 - 1, 42; P2 = 16, 22; P3 = 14, - 52 *55. P1 = 1 - 2,- 12; P2 = 10, 72; P3 = 13, 22 *56. P1 = 17, 22; P2 = 1 - 4, 02; P3 = 14, 62
57. 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)? 9022 ft ≈ 127.28 ft 2nd base
s
90 ft
s
51. 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. [Hint: Use (0, 0), (0, s), (s, 0), and (s, s) as the vertices of the square.] s s a , b 2 2
3rd base
Pitching rubber 1st base
90 ft
Home plate
40
CHAPTER F Foundations: A Prelude to Functions
58. Little League Baseball The layout of a Little League playing field is a square 60 feet on a side. How far is it directly from home plate to second base (the diagonal of the square)? Source: Little League Baseball, Official Regulations and Playing Rules, 2012. 6022 ft ≈ 84.85 ft
(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 49. ≈1.285 units Source: www.uwgb.edu/DutchS/structge/s100.htm 63. (a) (2.65, 1.6) y
59. Baseball Refer to Problem 57. Overlay a rectangular coordinate system on a major league baseball diamond so that the origin is at home plate, the positive x-axis 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.
1.5 1.3
*(a) What are the coordinates of first base, second base, and third base? Use feet as the unit of measurement.
(c) If the center fielder is located at 1300, 3002 , how far is it from there to third base? 302149 ft ≈ 366.20 ft
60. Little League Baseball Refer to Problem 58. Overlay a rectangular coordinate system on a Little League baseball diamond so that the origin is at home plate, the positive x-axis 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 1180, 202 , how far is it from there to second base? 40210 ft ≈ 126.49 ft
(c) If the center fielder is located at 1220, 2202 , how far is it from there to third base? 202185 ft ≈ 272.03 ft 61. Distance between Moving Objects A Ford Focus and a Mack truck leave an intersection at the same time. The Focus 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. d = 50t mi 62. 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 intersection (see the figure). Find an expression for the distance d (measured in feet) from the balloon to the intersection t seconds later. d = 210,000 + 484t 2 ft
15 mph 100 ft
(2.6, 1.5) (1.4, 1.3) 1.4
2.6 2.7
x
64. Net Sales The figure illustrates how net sales of Wal-Mart Stores, Inc., grew from 2008 through 2012. Use the midpoint formula to estimate the net sales of Wal-Mart Stores, Inc., in 2010. How does your result compare to the reported value of $405 billion? $409 billion Source: Wal-Mart Stores, Inc., 2012 Annual Report Wal-Mart Stores, Inc. Net sales ($ billions)
Net sales ($ billions)
(b) If the right fielder is located at 1310, 152 , how far is it from there to second base? 522161 ft ≈ 232.43 ft
(2.7, 1.7)
1.7
500 450 400 374 350 300 250 200 150 100 50 0 2008
444
2009
2010 Year
2011
2012
65. Poverty Threshold Poverty thresholds are determined by the U.S. Census Bureau. A poverty threshold represents the minimum annual household income for a family not to be considered poor. In 2004, the poverty threshold for a family of four with two children under the age of 18 years was $19,157. In 2012, the poverty threshold for a family of four with two children under the age of 18 years was $23,283. Assuming poverty thresholds increase in a straight-line fashion, use the midpoint formula to estimate the poverty threshold of a family of four with two children under the age of 18 in 2008. How does your result compare to the actual poverty threshold in 2008 of $21,834? $21,220 Source: U.S. Census Bureau
66. Horizontal and Vertical Shifts Suppose that A = 12, 52 are the coordinates of a point in the xy-plane. (a) Find the coordinates of the point if A is shifted 3 units to the right and 2 units down. (5, 3) (b) Find the coordinates of the point if A is shifted 2 units to the left and 8 units up. (0, 13)
63. Drafting Error When a draftsman draws three lines that are to intersect at one point, the lines may not intersect as intended 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.
67. Completing a Line Segment Plot the points A = 1 - 1, 82 and M = 12, 32 in the xy-plane. If M is the midpoint of a line segment AB, find the coordinates of B. (5, - 2)
SECTION F.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
41
Discussion and Writing the terms coordinate axes, ordered pair, coordinates, plot, x-coordinate, and y-coordinate.
68. Write a paragraph that describes a Cartesian plane. Then write a second paragraph that describes how to plot points in the Cartesian plane. Your paragraphs should include
‘Are You Prepared?’ Answers 1. 0
2. 8
3. 5
4. 112 + 602 = 612
5. A =
1 bh 2
6. Angle–side–angle; side–side–side; side–angle–side
F.2 Graphs of Equations in Two Variables; Intercepts; Symmetry PREPARING FOR THIS SECTION Before getting started, review the following: r 4PMWJOH&RVBUJPOT "QQFOEJY" 4FDUJPO" QQ"m"
Now Work the ‘Are You Prepared?’ problems on page 48.
OBJECTIVES 1 2 3 4 5
Graph Equations by Plotting Points (p. 41) Find Intercepts from a Graph (p. 43) Find Intercepts from an Equation (p. 44) Test an Equation for Symmetry (p. 44) Know How to Graph Key Equations (p. 46)
1 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: x2 + y 2 = 5
2x - y = 6
y = 2x + 5
x2 = y
The first of these, x2 + y2 = 5, is satisfied for x = 1, y = 2, since 12 + 22 = 1 + 4 = 5. Other choices of x and y, such as x = - 1, y = - 2, also satisfy this equation. It is not satisfied for x = 2 and y = 3, since 22 + 32 = 4 + 9 = 13 ≠ 5. The graph of an equation in two variables x and y consists of the set of points in the xy-plane whose coordinates 1x, y2 satisfy the equation.
EXAM PL E 1
Determining Whether a Point Is on the Graph of an Equation Determine if the following points are on the graph of the equation 2x - y = 6. (a) 12, 32
Solution
(b) 12, - 22
(a) For the point 12, 32, check to see whether x = 2, y = 3 satisfies the equation 2x - y = 6. 2x - y = 2122 - 3 = 4 - 3 = 1 ≠ 6
The equation is not satisfied, so the point 12, 32 is not on the graph.
42
CHAPTER F Foundations: A Prelude to Functions
(b) For the point 12, - 22, 2x - y = 2122 - 1 - 22 = 4 + 2 = 6 The equation is satisfied, so the point 12, - 22 is on the graph.
Now Work
EX AM PL E 2
r
9
PROBLEM
How to Graph an Equation by Plotting Points (SBQIUIFFRVBUJPOy = - 2x + 3
Step-by-Step Solution Step 1: Find points (x, y) that satisfy the equation. To determine these points, choose values of x and use the equation to find the corresponding values for y. See Table 1.
Step 2: Plot the points found in the table as shown in Figure 11(a). Now connect the points to obtain the graph of the equation (a line), as shown in Figure 11(b).
Table 1
y = −2x + 3
(x, y)
-2
- 2( - 2) + 3 = 7
( - 2, 7)
-1
- 2( - 1) + 3 = 5
( - 1, 5)
0
- 2(0) + 3 = 3
(0, 3)
1
- 2(1) + 3 = 1
(1, 1)
2
- 2(2) + 3 = - 1
(2, - 1)
Figure 11 y 8
y 8 (–2, 7) (–1, 5)
(–2, 7)
6 4
–4
6
(–1, 5) (0, 3)
2
(0, 3) 2
(1, 1) 2
–2
4 x (2, –1)
–4
(1, 1) 2
–2
4 x (2, –1)
–2
–2 (a)
EX A MPL E 3
x
r
(b)
Graphing an Equation by Plotting Points (SBQIUIFFRVBUJPOy = x2
Solution
Table 2
Table 2 provides several points on the graph. Plotting these points and connecting them with a smooth curve gives the graph (a parabola) shown in Figure 12.
x
y = x2
(x, y)
-4
16
( - 4, 16)
-3
9
( - 3, 9)
-2
4
( - 2, 4)
-1
1
( - 1, 1)
0
0
(0, 0)
1
1
(1, 1)
2
4
(2, 4)
3
9
(3, 9)
4
16
(4, 16)
Figure 12 y 20 (– 4, 16) (–3, 9)
(4, 16)
15 10
(3, 9)
5 (–2, 4) (2, 4) (1, 1) (–1, 1) (0, 0) –4 4
x
r
SECTION F.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
43
The graphs of the equations shown in Figures 11 and 12 do not show all the points that are on the graph. For example, in Figure 11 the point 120, - 372 is a part of the graph of y = - 2x + 3, but it is not shown. Since the graph of y = - 2x + 3 could be extended out as far as we please, we use arrows to indicate that the pattern shown continues. It is important 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 actually there. This is referred to as a complete graph. One way to obtain a complete graph of an equation is to plot a sufficient number of points on the graph for a pattern to become 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 example, 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 plotting points. Shortly, we shall investigate various techniques that will enable us to graph an equation without plotting so many points. COMMENT Another way to obtain the graph of an equation is to use a graphing utility. Read Section B.2, Using a Graphing Utility to Graph Equations, in Appendix B. ■ Figure 13 Graph 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 Graph crosses x-axis
2 Find Intercepts from a Graph x
Graph touches x-axis
Intercepts
EXAM PL E 4
The points, if any, at which a graph crosses or touches the coordinate axes are called the intercepts. See Figure 13. The x-coordinate of a point at which the graph crosses 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.
Finding Intercepts from a Graph Find the intercepts of the graph in Figure 14. What are its x-intercepts? What are its y-intercepts?
Solution Figure 14
1 - 3, 02,
y 4
3 a , 0b , 2
4 a0, - b , 3
10, - 3.52,
14.5, 02
3 4 The x-intercepts are - 3, , and 4.5; the y-intercepts are - 3.5, - , and 3. 2 3
(4.5, 0) 5 x 4
(0, –3 ) (0, 3.5)
10, 32,
(0, 3)
( 3–2 , 0) 4 (3, 0)
The intercepts of the graph are the points
r
In Example 4, note the following usage: If the type of intercept (x- versus y-) is not specified, then report the intercept as an ordered pair. However, if the type of intercept is specified, then report only the coordinate of the specified intercept. For x-intercepts, report the x-coordinate of the intercept; for y-intercepts, report the y-coordinate of the intercept.
Now Work
PROBLEM
37(A)
44
CHAPTER F Foundations: A Prelude to Functions
3 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 0.
Procedure for Finding Intercepts 1. To find the x-intercept(s), if any, of the graph of an equation, let y = 0 in the equation and solve for x. 2. To find the y-intercept(s), if any, of the graph of an equation, let x = 0 in the equation and solve for y.
Because the x-intercepts of the graph of an equation are those x-values for which y = 0, they are also called the zeros (or roots) of the equation.
EX A MPL E 5
Finding Intercepts from an Equation Find the x-intercept(s) and the y-intercept(s) of the graph of y = x2 - 4.
Solution
To find the x-intercept(s), let y = 0 and obtain the equation
x + 2 = 0 x = -2
x2 - 4 1x + 22 1x - 22 or x - 2 or x
= = = =
0 0 0 2
Factor. Zero-Product Property Solve.
The equation has two solutions, - 2 and 2. The x-intercepts (the zeros) are - 2 and 2. To find the y-intercept(s), let x = 0 in the equation. y = x2 - 4 = 02 - 4 = -4 The y-intercept is - 4.
Now Work
PROBLEM
19
r
COMMENT For many equations, finding intercepts may not be so easy. In such cases, a graphing utility can be used. Read the first part of Section B.3, Using a Graphing Utility to Locate Intercepts and Check for Symmetry, in Appendix B to find out how a graphing utility locates intercepts. ■
4 Test an Equation for Symmetry Another helpful tool for graphing equations by hand involves symmetry, particularly symmetry with respect to the x-axis, the y-axis, and the origin. Symmetry often occurs in nature. Consider the picture of the butterfly. Do you see the symmetry?
DEFINITION
A graph is said to be symmetric with respect to the x-axis if, for every point 1x, y2 on the graph, the point 1x, - y2 is also on the graph. A graph is said to be symmetric with respect to the y-axis if, for every point 1x, y2 on the graph, the point 1 - x, y2 is also on the graph. A graph is said to be symmetric with respect to the origin if, for every point 1x, y2 on the graph, the point 1 - x, - y2 is also on the graph.
SECTION F.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
45
Figure 15 illustrates the definition. Note that when a graph is symmetric with respect to the x-axis, the part of the graph above the x-axis is a reflection or mirror image of the part below it, and vice versa. When a graph is symmetric with respect to the y-axis, 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. Symmetry with respect to the origin may be viewed in two ways: 1. As a reflection about the y-axis, followed by a reflection about the x-axis 2. As a projection along a line through the origin so that the distances from the origin are equal
Figure 15
y
y
y (–x1, y1)
(x2, y2) (x1, y1)
(x3, y3)
(x1, –y1)
x (x3, –y3)
(–x2, y2)
(x2, y2)
(x2, –y2)
x
(x2, y2)
x
(–x2, –y2) (–x1, –y1)
Symmetry with respect to the x-axis
EXAM PL E 6
(x1, y1)
(x1, y1)
Symmetry with respect to the y-axis
Symmetry with respect to the origin
Symmetric Points
(a) If a graph is symmetric with respect to the x-axis, and the point 14, 22 is on the graph, then the point 14, - 22 is also on the graph. (b) If a graph is symmetric with respect to the y-axis, and the point 14, 22 is on the graph, then the point 1 - 4, 22 is also on the graph. (c) If a graph is symmetric with respect to the origin, and the point 14, 22 is on the graph, then the point 1 - 4, - 22 is also on the graph.
Now Work
r
PROBLEM
27
When the graph of an equation is symmetric with respect to the x-axis, the y-axis, or the origin, the number of required points 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 graphing an equation, it is wise to determine whether any symmetry exists. 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
Replace y by - y in the equation. 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. 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. If an equivalent equation results, the graph of the equation is symmetric with respect to the origin.
46
CHAPTER F Foundations: A Prelude to Functions
Testing an Equation for Symmetry
EX A MPL E 7
Test y =
Solution
x-Axis:
y-Axis:
Origin:
4x2 for symmetry. x + 1 2
To test for symmetry with respect to the x-axis, replace y by - y. Since 4x2 4x2 -y = 2 is not equivalent to y = 2 , the graph of the equation x + 1 x + 1 is not symmetric with respect to the x-axis. To test for symmetry with respect to the y-axis, replace x by - x. Since 41 - x2 2 4x2 4x2 y = = is equivalent to y = , the graph of the 1 - x2 2 + 1 x2 + 1 x2 + 1 equation is symmetric with respect to the y-axis. To test for symmetry with respect to the origin, replace x by - x and y by - y. 41 - x2 2
-y =
1 - x2 2 + 1
Replace x by - x and y by - y.
-y =
4x2 x2 + 1
Simplify.
y = -
4x2 x2 + 1
Multiply both sides by - 1.
Since the result is not equivalent to the original equation, the graph of the 4x2 equation y = 2 is not symmetric with respect to the origin. x + 1
Figure 16
r
5
−5
5
Seeing the Concept Figure 16 shows the graph of y = with respect to the y-axis?
4x2 x + 1 2
using a graphing utility. Do you see the symmetry
−5
Now Work
PROBLEM
57
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 they will be used later. The first of these is y = x3.
EX A MPL E 8
Graphing the Equation y = x 3 by Finding Intercepts and Checking for Symmetry (SBQI UIF FRVBUJPO y = x3 by plotting points. Find any intercepts and check for symmetry first.
Solution
First, find the intercepts. When x = 0, then y = 0; and when y = 0, then x = 0. The origin 10, 02 is the only intercept. Now test for symmetry. x-Axis: y-Axis: Origin:
Replace y by - y. 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 = 1 - x2 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 = 1 - x2 3 = - x3 is equivalent to y = x3 (multiply both sides by - 1), the graph is symmetric with respect to the origin.
To graph y = x3, use the equation to obtain several points on the graph. Because of the symmetry, we only need to locate points on the graph for which x Ú 0.
SECTION F.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
47
See Table 3. 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. Figure 17 shows the graph.
Table 3
x
y = x3
(x, y)
0
0
(0, 0)
1
1
(1, 1)
2
8
(2, 8)
3
27
(3, 27)
Figure 17 y
(0, 0) –6
(1, 1) 6
(– 1, – 1)
(– 2, – 8)
EXAM PL E 9
(2, 8)
8
–8
x
r
Graphing the Equation x = y 2 B (SBQIUIFFRVBUJPOx = y2. Find any intercepts and check for symmetry first. C (SBQIx = y2, y Ú 0.
Solution
Table 4 y
x = y2
(x, y)
0
0
(0, 0)
1
1
(1, 1)
2
4
(4, 2)
3
9
(9, 3)
Figure 20 6
(a) The lone intercept is 10, 02 . The graph is symmetric with respect to the x-axis since x = 1 - y2 2 is equivalent to x = y2. The graph is not symmetric with respect to the y-axis or the origin. To graph x = y2, use the equation to obtain several points on the graph. Because the equation is solved for x, it is easier to assign values to y and use the equation to determine the corresponding values of x. Because of the symmetry, start by finding points whose y-coordinates are non-negative. Then use the symmetry to find additional points on the graph. See Table 4. For example, since 11, 12 is on the graph, so is 11, - 12. Since 14, 22 is on the graph, so is 14, - 22, and so on. Plot these points and connect them with a smooth curve to obtain Figure 18. (b) If we restrict y so that y Ú 0, the equation x = y2, y Ú 0, may be written equivalently as y = 1x. The portion of the graph of x = y2 in quadrant I is therefore the graph of y = 1x. See Figure 19. Figure 18
Y1 x
Figure 19
y 6 (9, 3) (1, 1)
(4, 2)
(1, 1)
(0, 0)
2
10
6
y 6
Y2 x
EX AM PL E 10
2 (1, 1)
(9, 3)
(0, 0) 10 x
5 (4, 2)
(9, 3)
–2
5
10 x
r
COMMENT To see the graph of the equation x = y2 on a graphing calculator, graph two equations: Y1 = 1x and Y2 = - 1x. See Figure 20. We discuss why in Chapter 1. ■
Graphing the Equation y =
1 x
1 . Find any intercepts and check for symmetry first. x Check for intercepts first. If we let x = 0, we obtain 0 in the denominator, which makes y undefined. We conclude that there is no y-intercept. If we let y = 0, 1 we get the equation = 0, which has no solution. We conclude that there is no x 1 x-intercept. The graph of y = does not cross or touch the coordinate axes. x (SBQIUIFFRVBUJPOy =
Solution
(4, 2)
48
CHAPTER F Foundations: A Prelude to Functions
Next check for symmetry:
Table 5 1 y = x
x 1 10 1 3 1 2 1
(x, y) a
1
1 , 10b 10 1 a , 3b 3 1 a , 2b 2 11, 12
1 2 1 3 1 10
1 a2, b 2 1 a3, b 3 1 a10, b 10
10 3 2
2 3 10
y-Axis: Origin:
Use the equation to form Table 5, and obtain some points on the graph. Because of the symmetry, we only find points (x, y) for which x is positive. From 1 Table 5 we infer that if x is a large and positive number, then y = is a positive x number close to 0. We also infer that if x is a positive number close to 0, then 1 y = is a large and positive number. Armed with this information, we can graph x the equation. 1 Figure 21 illustrates some of these points and the graph of y = . Observe how x the absence of intercepts and the existence of symmetry with respect to the origin are utilized.
y
Figure 21
3
(––12 , 2) (1, 1) 3
r
(2, ––12 ) x
3
(2, ––12 )
(1, 1)
( ––12 , 2)
1 1 , which is not equivalent to y = . x x 1 1 1 Replacing x by - x yields y = = - , which is not equivalent to y = . -x x x 1 Replacing x by - x and y by - y yields - y = - , which is equivalent to x 1 y = . The graph is symmetric with respect to the origin. x Replacing y by - y yields - y =
x-Axis:
1 COMMENT Refer to Example 2 in Appendix B, Section B.3, for the graph of y = found x using a graphing utility. ■
3
F.2 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. Solve the equation 2(x + 3) - 1 = - 7. (pp. A64–A66)
5 - 66 2. Solve the equation x 2 - 4x - 12 = 0. (pp. A67–A68)
5 - 2, 66
Concepts and Vocabulary 3. The points, if any, at which a graph crosses or touches the coordinate axes are called intercepts .
6. True or False To find the y-intercept(s) of the graph of an equation, let x = 0 and solve for y. True
4. Because the x-intercepts of the graph of an equation are those x-values for which y = 0, they are also called zeros roots or .
7. True or False The graph of an equation in two variables x and y consists of the set of points in the xy-plane whose coordinates (x, y) satisfy the equation. True
5. If, for every point 1x, y2 on the graph of an equation, the point 1 - x, y2 is also on the graph, then the graph is y-axis symmetric with respect to the .
8. True or False If a graph is symmetric with respect to the x-axis, then it cannot be symmetric with respect to the y-axis. False
Skill Building In Problems 9–14, tell whether the given points are on the graph of the equation. 9. Equation: y = x4 - 1x Points: 10, 02; 11, 12; 1 - 1, 02 12. Equation: y = x + 1 3
Points: 11, 22; 10, 12; 1 - 1, 02 (0, 1); 1 - 1, 02
10. Equation: y = x3 - 21x
11. Equation: y2 = x2 + 9
13. Equation: x2 + y2 = 4
14. Equation: x2 + 4y2 = 4
Points: 10, 02; 11, 12; 11, - 12 10, 02; 11, - 12 Points: 10, 32; 13, 02; 1 - 3, 02 (0, 3)
(0, 0)
Points: 10, 22; 1 - 2, 22; 1 22, 22 2 10, 22; 1 22,22 2
1 Points: 10, 12; 12, 02; a2, b 2 (0, 1); (2, 0) In Problems 15–26, find the intercepts and graph each equation by plotting points. Be sure to label the intercepts. *15. y = x + 2
*16. y = x - 6
*17. y = 2x + 8
*18. y = 3x - 9
*19. y = x - 1
*20. y = x - 9
*21. y = - x + 4
*22. y = - x2 + 1
*23. 2x + 3y = 6
*24. 5x + 2y = 10
*25. 9x2 + 4y = 36
*26. 4x2 + y = 4
2
2
2
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
49
SECTION F.2 Graphs of Equations in Two Variables; Intercepts; Symmetry
In Problems 27–36, 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. *27. 13, 42
*28. 15, 32
*29. 1 - 2, 12
*30. 14, - 22
*31. 15, - 22
*32. 1 - 1, - 12
*33. 1 - 3, - 42
*34. 14, 02
*35. 10, - 32
*36. 1 - 3, 02
In Problems 37–48, 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. 37. 38. 39. 40. y y y y 3
3
–3
–3
(a) 1 - 1, 02; 11, 02 (b) x-axis, y-axis, origin 41.
3x
–3
y 3
45.
y 3
4
(a) 1 - 2, 02, 10, - 32, 12, 02 (b) y-axis 44.
y
x
3x
6
6
6 40
(a) 1 - 4, 02, 10, 02, 14, 02
(b) x-axis, y-axis, origin
(b) origin
(b) origin 47.
y 3
4
48.
8
2 3x
3 x
(a) 1 - 2, 02, 10, 22, 12, 02, 10, - 22 (a) 1 - 2, 02, 10, 02, 12, 02 46.
3
3
40
3
3x –3
(a) (0, 0) (b) x-axis
x
6
–3
–3
–– 2
–1
p p (a) a - , 0b, 10, 12, a , 0b 2 2 (b) y-axis 43. y
(a) (0, 1) (b) no symmetry 42.
y 3
– – –– 2
3x
–3
3x
–3
4
1
4
2
4
3x
3
4
8 3
3
(a) x-intercepts: 3 - 2, 14 y-intercept: 0 (b) no symmetry
(a) x-intercepts: 3 - 1, 24 y-intercept: 0 (b) no symmetry
(a) no intercepts (b) origin
(a) no intercepts (b) x-axis
In Problems 49–52, draw a complete graph so that it has the type of symmetry indicated. *49. y-axis
*50. x-axis y
y
5
9
(0, 2)
*51. Origin
(0, 0)
–9
, 2) (–– 2
5x
–5 9x (2, –5)
–9
4
(5, 3)
(–4, 0)
*52. y-axis
y
(2, 2)
(, 0) x
(0, 0)
–2
–4
–5
(0, –9)
*53. y2 = x + 4
*54. y2 = x + 9
3 *55. y = 2 x
*56.
*57. y = x4 - 8x2 - 9
*58. y = x4 - 2x2 - 8
*59. 9x2 + 4y2 = 36
*60. 4x2 + y2 = 4
*61. y = x3 + x2 - 9x - 9
*62. y = x3 + 2x2 - 4x - 8
*63. y = |x| - 4
*64.
y = |x 0 - 2
3x *65. y = 2 x + 9
x2 - 4 *66. y = 2x
- x3 *67. y = 2 x - 9
*68.
y =
5 y = 2 x
*71. y = 1x
*72.
y =
x4 + 1 2x5
In Problems 69–72, draw a quick sketch of each equation. *70. x = y2
3x
–3
In Problems 53–68, list the intercepts and test for symmetry.
*69. y = x3
y (0, 4) 4
1 x
50
CHAPTER F Foundations: A Prelude to Functions
73. If 13, b2 is a point on the graph of y = 4x + 1, what is b?
13
75. If 1a, 42 is a point on the graph of y = x2 + 3x, what is a? - 4 or 1
74. If 1 - 2, b2 is a point on the graph of 2x + 3y = 2, what is b? 2
76. If 1a, - 52 is a point on the graph of y = x2 + 6x, what is a? - 5 or - 1
Mixed Practice In Problems 77–84, (a) find the intercepts of the graph of each equation, (b) test each equation for symmetry with respect to the x-axis, the y-axis, and the origin, and (c) graph each equation by plotting points. Be sure to label the intercepts on the graph and use any symmetry to assist in drawing the graph. *77. y = x2 - 5
*78. y = x2 - 8
*79. x - y2 = - 9
*80. x + y2 = 4
*81. x + y = 9
*82. x + y = 16
*83. y = x - 4x
*84. y = x3 - x
2
2
2
2
3
Applications and Extensions 85. (JWFO UIBU UIF QPJOU JT PO UIF HSBQI PG BO FRVBUJPO that is symmetric with respect to the origin, what other point is on the graph? 1 - 1, - 22 86. 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. - 6
90. 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 16y2 = 120x - 225.
87. 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. 4 88. If the graph of an equation is symmetric with respect to the x-axis, and 2 is a y-intercept, name another y-intercept. - 2 89. Microphones In studios and on stages, cardioid microphones 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 cardioid pattern is given by the equation 1x2 + y2 - x2 2 = x2 + y2. (a) Find the intercepts of the graph of the equation. (b) Test for symmetry with respect to the x-axis, the y-axis, and the origin. x-axis symmetry 89. (a) (0, 0), (2, 0), (0, 1), 10, - 12
(a) Find the intercepts of the graph of the equation. a
15 , 0b 8 (b) Test for symmetry with respect to the x-axis, the y-axis, and the origin. x-axis symmetry Source: U.S. Department of Energy
Discussion and Writing In Problem 91, use a graphing utility.
91. B (SBQIy = 2x2 , y = x, y = 0 x 0 , and y = 1 1x2 2, noting which graphs are the same. same: y = 2x2 and y = |x| (b) Explain why the graphs of y = 2x2 and y = 0 x 0 are the same. 2x2 = 0 x 0 (c) Explain why the graphs of y = x and y = 1 1x2 are not the same. 2
(d) Explain why the graphs of y = 2x2 and y = x are not the same. y Ú 0 for y = 2x2 92. Explain what is meant by a complete graph. 93. 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. 91. (c) y = 1 2x2 2 has domain [0, q 2 ; y = x has domain 1 - q, q 2
‘Are You Prepared?’ Answers 1. 5 - 66
2. 5 - 2, 66
94. Make up an equation with the intercepts 12, 02, 14, 02, and 10, 12. Compare your equation with a friend’s equation. Comment on any similarities. 95. Draw a graph that contains the points 1 - 2, - 12, 10, 12, 11, 32, and 13, 52. Compare your graph with those of other students. Are most of the graphs almost straight lines? How many are “curved”? Discuss the various ways that these points might be connected.
96. 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. 97. Draw a graph that contains the points ( - 2, 5), ( - 1, 3), and 10, 22 that is symmetric with respect to the y-axis. Compare your graph with those of other students; comment on any similarities. Can a graph contain these points and be symmetric with respect to the x-axis? the origin? Why or why not?
SECTION F.3 Lines
51
F.3 Lines OBJECTIVES 1 2 3 4 5 6 7 8 9 10
Instructor Note For many students, this material is review. Consider your audience when setting the pace.
Calculate and Interpret the Slope of a Line (p. 51) Graph Lines Given a Point and the Slope (p. 54) Find the Equation of a Vertical Line (p. 54) Use the Point–Slope Form of a Line; Identify Horizontal Lines (p. 55) Find the Equation of a Line Given Two Points (p. 56) Write the Equation of a Line in Slope–Intercept Form (p. 56) Identify the Slope and y-Intercept of a Line from Its Equation (p. 57) Graph Lines Written in General Form Using Intercepts (p. 58) Find Equations of Parallel Lines (p. 59) Find Equations of Perpendicular Lines (p. 60)
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 Figure 22 Line
Rise
Consider the staircase illustrated in Figure 22. 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.
Run
DEFINITION
Let P = (x1 , y1 2 and Q = (x2 , y2 2 be two distinct points. If x1 ≠ x2 , the slope m of the nonvertical line L containing P and Q is defined by the formula m =
y2 - y1 x2 - x1
x1 ≠ x2
(1)
If x1 = x2 , L is a vertical line and the slope m of L is undefined (since this results in division by 0).
Figure 23(a) provides an illustration of the slope of a nonvertical line; Figure 23(b) illustrates a vertical line. Figure 23 L Q = (x 2, y2)
L
y2
Q = (x 1, y2)
y1
P = (x 1, y1)
Rise = y2 – y1
P = (x 1, y1) y1
y2
Run = x2 – x1 x1
(a) Slope of L is m =
x2 y2 – y1 _______ x2 – x1
x1
(b) Slope is undefined; L is vertical
52
CHAPTER F Foundations: A Prelude to Functions
As Figure 23(a) illustrates, the slope m of a nonvertical line may be viewed as
In Words
The symbol ∆ is the Greek letter delta. In mathematics, ∆ is read ∆y “change in,” so is read “change ∆x in y divided by change in x.”
m =
y2 - y1 Rise = x2 - x1 Run
or as m =
Change in y y2 - y1 ∆y = = x2 - x1 Change in x ∆x
That is, the slope m of a nonvertical line measures the amount y changes when x ∆y changes from x1 to x2. The expression is called the average rate of change of y ∆x with respect to x. 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 24 for justification.) Figure 24 Triangles ABC and PQR are similar (equal angles), so ratios of corresponding sides are equal. Then y2 - y1 Slope using P and Q = = x2 - x1 d(B, C) = Slope using A and B d(A, C)
y
Q = (x 2, y2) y2 – y1
P = (x 1, y1) B
x2 – x1
R x
A
C
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. 2. The slope of a line may be computed from P = (x1 , y1 2 to Q = (x2 , y2 2 or from Q to P because y2 - y1 y1 - y2 = x2 - x1 x1 - x2
EX A MPL E 1
Finding and Interpreting the Slope of a Line Given Two Points
The slope m of the line containing the points 11, 22 and 15, - 32 may be computed as m =
-3 - 2 -5 5 = = 5 - 1 4 4
or as m =
2 - 1 - 32 5 5 = = 1 - 5 -4 4
For every 4-unit change in x, y will change by - 5 units. 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
Now Work
PROBLEMS
11
AND
r
17
To get a better idea of the meaning of the slope m of a line L, consider the following example.
EX A MPL E 2
Finding the Slopes of Various Lines Containing the Same Point (2, 3) Compute the slopes of the lines L1 , L2 , L3 , and L4 containing the following pairs of QPJOUT(SBQIBMMGPVSMJOFTPOUIFTBNFTFUPGDPPSEJOBUFBYFT L1 : L2 : L3 : L4 :
P P P P
= = = =
12, 32 12, 32 12, 32 12, 32
Q1 Q2 Q3 Q4
= = = =
1 - 1, - 22 13, - 12 15, 32 12, 52
SECTION F.3 Lines
Solution
Let m1 , m2 , m3 , and m4 denote the slopes of the lines L1 , L2 , L3 , and L4 , respectively. Then -2 - 3 -5 5 A rise of 5 divided by a run of 3 = = -1 - 2 -3 3 -1 - 3 -4 m2 = = = -4 3 - 2 1 3 - 3 0 m3 = = = 0 5 - 2 3 m4 is undefined because x1 = x2 = 2
Figure 25 L2
m1 =
L4
L1
y 5
Q4 (2, 5) P (2, 3)
m3 0
Q3 (5, 3)
5
m1
5– 3
L3
5 x Q2 (3, 1)
Q1 (1, 2)
53
3 m4 undefined
The graphs of these lines are given in Figure 25. Figure 25 illustrates the following facts:
m2 4
r
1. When the slope of a line is positive, the line slants upward from left to right 1L1 2. 2. When the slope of a line is negative, the line slants downward from left to right 1L2 2. 3. When the slope is 0, the line is horizontal 1L3 2. 4. When the slope is undefined, the line is vertical 1L4 2.
Seeing the Concept On the same screen, graph the following equations: Y1 = 0
Slope of line is 0.
Y2 =
1 x 4 1 Y3 = x 2
1 Slope of line is . 4 1 Slope of line is . 2
Y2 14 x
Y4 = x
Slope of line is 1.
3
Y5 = 2x
Slope of line is 2.
Y6 = 6x
Slope of line is 6.
Figure 26 Y6 6x Y5 2x Y4 x 2 Y3 12 x 3
Y1 0
See Figure 26.
2
Seeing the Concept On the same screen, graph the following equations:
Figure 27 Y6 6x Y5 2x Y4 x
2
Y3 12 x Y2 14 x 3
3
Y1 0 2
Y1 = 0
Slope of line is 0.
1 Y2 = - x 4 1 Y3 = - x 2 Y4 = - x
1 Slope of line is - . 4 1 Slope of line is - . 2 Slope of line is - 1.
Y5 = - 2x
Slope of line is - 2.
Y6 = - 6x
Slope of line is - 6.
See Figure 27.
Figures 26 and 27 illustrate that the closer the line is to the vertical position, the greater the absolute value of the slope.
54
CHAPTER F Foundations: A Prelude to Functions
2 Graph Lines Given a Point and the Slope EX A MPL E 3
Graphing a Line Given a Point and the Slope
Draw a graph of the line that contains the point 13, 22 and has a slope of: (a)
Solution Figure 28 y 6 (7, 5) Rise = 3
3 4
(b) -
Rise 3 . The slope means that for every horizontal movement (run) of Run 4 4 units to the right, there will be a vertical movement (rise) of 3 units. Start at the given point, 13, 22 and move 4 units to the right and 3 units up arriving at the point 17, 52. Drawing the line through this point and the point 13, 22 gives the graph. See Figure 28. (b) The fact that the slope is (a) Slope =
(3, 2)
-
Run = 4 –2
10 x
5
4 -4 Rise = = 5 5 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). Start at the given point 13, 22 and move 5 units to the right and then 4 units down, arriving at the point 18, - 22. Drawing the line through these points gives the graph. See Figure 29.
Figure 29 y
Alternatively, consider
(–2, 6) 6
-
Rise = 4 Rise = – 4 10 x
–2
4 4 Rise = = 5 -5 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 approach leads to the point 1 - 2, 62, which is also on the graph of the line in Figure 29.
(3, 2) Run = 5 Run = –5 –2
4 5
r
(8, –2)
Now Work
PROBLEM
23
3 Find the Equation of a Vertical Line EX A MPL E 4
Graphing a Line (SBQIUIFFRVBUJPOx = 3
Solution
To graph x = 3, find all points 1x, y2 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 30.
Figure 30
y 4 (3, 3) (3, 2) (3, 1) 1 1
(3, 0) (3, 1)
5 x
r
SECTION F.3 Lines
55
Example 4 suggests 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.
COMMENT In order for an equation to be graphed using a graphing utility, the equation must be expressed in the form y = 5expression in x6 . But x = 3 cannot be put in this form. To overcome this, most graphing utilities have special commands for drawing vertical lines. LINE, PLOT, and VERT are among the more common ones. Consult your manual to determine the correct methodology for your graphing utility. ■ Figure 31
4 Use the Point–Slope Form of a Line; Identify Horizontal Lines
y
L (x, y )
Now let L be a nonvertical line with slope m that contains the point 1x1 , y1 2. See Figure 31. Any other point 1x, y2 on L gives
y – y1
(x 1, y1)
m =
x – x1
y - y1 x - x1
or y - y1 = m1x - x1 2
x
THEOREM
Point–Slope Form of an Equation of a Line An equation of a nonvertical line with slope m that contains the point 1x1 , y1 2 is y - y1 = m1x - x1 2
EXAM PL E 5 Figure 32 y 6
(2, 6) Rise 4
(1, 2)
Using the Point–Slope Form of a Line
An equation of the line with slope 4 that contains the point 11, 22 can be found by using the point–slope form with m = 4, x1 = 1, and y1 = 2. y - y1 = m1x - x1 2 y - 2 = 41x - 12
Run 1 2
y = 4x - 2 5 x
2
EXAM PL E 6
Solution
m = 4, x1 = 1, y1 = 2 Solve for y.
r
See Figure 32 for the graph.
Finding the Equation of a Horizontal Line
Find an equation of the horizontal line containing the point 13, 22. Because all the y-values are equal on a horizontal line, the slope of a horizontal line is 0. To get an equation, use the point–slope form with m = 0, x1 = 3, and y1 = 2. y - y1 = m1x - x1 2
Figure 33 y
y - 2 = 0 # 1x - 32
4 (3, 2)
–1
(2)
1
3
m = 0, x1 = 3, and y1 = 2
y - 2 = 0 y = 2
5 x
See Figure 33 for the graph.
r
56
CHAPTER F Foundations: A Prelude to Functions
Example 6 suggests 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 Points (JWFOUXPQPJOUT XFVTFUIFTMPQFGPSNVMBBOEUIFQPJOUmTMPQFGPSNPGBMJOFUP find the equation.
Finding an Equation of a Line Given Two Points
EX A MPL E 7
Find an equation of the line containing the points 12, 32 and 1 - 4, 52.(SBQIUIF line.
Solution
First compute the slope of the line with 1x1, y1 2 = (2, 3) and 1x2, y2 2 = ( - 4, 5). m =
5 - 3 2 1 = = -4 - 2 -6 3
m =
y2 - y1 x2 - x1
1 Use the point 1x1, y1 2 = 12, 32 and the slope m = - to get the point–slope form 3 of the equation of the line.
Figure 34 y (–4, 5)
y - 3 = -
(2, 3)
1 1x - 22 3
y - y1 = m(x - x1)
2 –4
–2
10
x
r
See Figure 34 for the graph.
In the solution to Example 7, we could have used the other point, 1 - 4, 52 , instead of the point 12, 32. The equation that results, although it looks different, is equivalent to the equation that we obtained in the example. (Try it for yourself.)
Now Work
PROBLEM
37
6 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, both the slope m of the line and a point 10, b2 on the line are known; then use the point–slope form, equation (2), to obtain the following equation: y - b = m1x - 02
THEOREM
or y = mx + b
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
Now Work
PROBLEMS
45
AND
51 (EXPRESS
SLOPE–INTERCEPT FORM)
(3)
ANSWERS IN
SECTION F.3 Lines
Figure 35 y = mx + 2
Seeing the Concept
Y4 3x 2 Y2 x 2
Y5 3x 2 Y3 x 2 4
57
To see the role that the slope m plays, graph the following lines on the same square screen.
Y1 2 6
6
Y1 = 2
m = 0
Y2 = x + 2
m = 1
Y3 = - x + 2
m = -1
Y4 = 3x + 2
m = 3
Y5 = - 3x + 2
m = -3
See Figure 35. What do you conclude about the lines y = mx + 2?
4
Seeing the Concept Figure 36 y = 2x + b Y4 2x 4 4
To see the role of the y-intercept b, graph the following lines on the same square screen.
Y2 2x 1 Y1 2x Y3 2x 1 Y5 2x 4
6
Y1 = 2x
b = 0
Y2 = 2x + 1
b = 1
Y3 = 2x - 1
b = -1
Y4 = 2x + 4
b = 4
Y5 = 2x - 4
b = -4
6
See Figure 36. What do you conclude about the lines y = 2x + b?
4
7 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 m and y-intercept b of the line. For example, suppose that the equation of a line is y = - 2x + 3 Compare this equation to y = mx + b. y = - 2x + 3 y =
c
c
mx + b
The slope of this line is - 2 and its y-intercept is 3.
Now Work
EXAM PL E 8
PROBLEM
71
Finding the Slope and y-Intercept Find the slope m and y-intercept b of the equation 2x + 4y = 8. (SBQI UIF equation.
Solution
To obtain the slope and y-intercept, write the equation in slope–intercept form by solving for y. 2x + 4y = 8 4y = - 2x + 8 1 y = - x + 2 2
Figure 37 2x + 4y = 8 y 4 (0, 2) –3
2 1 (2, 1) 3
x
y = mx + b
1 The coefficient of x, - , is the slope, and the constant, 2, is the yJOUFSDFQU(SBQI 2 1 the line with the y-intercept 2 and the slope - . Starting at the point 10, 22, go to 2 the right 2 units and then down 1 unit to the point 12, 12. Draw the line through these points. See Figure 37.
r
Now Work
PROBLEM
77
58
CHAPTER F Foundations: A Prelude to Functions
8 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
(4)
where A, B, and C are real numbers and A and B are not both 0. If B = 0 in equation (4), then A ≠ 0 and the graph of the equation is a vertical C line: x = . If B ≠ 0 in equation (4), then we can solve the equation for y and write A the equation in slope–intercept form as we did in Example 8. Another approach to graphing equation (4) would be to find its intercepts. Remember, the intercepts of the graph of an equation are the points where the graph crosses or touches a coordinate axis.
EX A MPL E 9
Graphing an Equation in General Form Using Its Intercepts (SBQIUIFFRVBUJPO2x + 4y = 8 by finding its intercepts.
Solution
To obtain the x-intercept, let y = 0 in the equation and solve for x. 2x + 4y = 8 2x + 4(0) = 8
Let y = 0.
2x = 8 x = 4
Divide both sides by 2.
The x-intercept is 4 and the point 14, 02 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
Figure 38
y = 2
y 4
(4, 0) 3
Divide both sides by 4.
The y-intercept is 2 and the point 10, 22 is on the graph of the equation. Plot the points 14, 02 and 10, 22 and draw the line through the points. See Figure 38.
(0, 2)
–3
Let x = 0.
x
r
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 + 0#y = a
A = 1, B = 0, C = a
A horizontal line whose equation is y = b can be written in the general form
*Some books use the term standard form.
0#x + 1#y = b
A = 0, B = 1, C = b
SECTION F.3 Lines
59
Lines that are neither vertical nor horizontal have general equations of the form Ax + By = C
A ≠ 0 and 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 39
9 Find Equations of Parallel Lines
y
Rise Run
Rise Run x
THEOREM
When two lines (in the same plane) do not intersect (that is, they have no points in common), they are parallel. Look at Figure 39. There we have drawn two lines and have constructed two right triangles by drawing sides parallel to the coordinate axes. These lines are parallel if and only if the right triangles are similar. (Do you see why? Two angles are equal.) And the triangles are similar if and only if the ratios of corresponding sides are equal.
Criteria for Parallel 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 actually 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.
EX AM PL E 10
Showing That Two Lines Are Parallel Show that the lines given by the following equations are parallel. L1 : 2x + 3y = 6,
Solution Figure 40
To determine whether these lines have equal slopes and different y-intercepts, write each equation in slope–intercept form. L1 : 2x + 3y = 6
y 5
5 x L1 3
L2 : 4x + 6y = 0
3y = - 2x + 6 y = -
5
L2 : 4x + 6y = 0
L2
EX AM PL E 11
Solution
6y = - 4x
2 x + 2 3
2 Slope = - ; y@intercept = 2 3
y = -
2 x 3
2 Slope = - ; y@intercept = 0 3
2 Because these lines have the same slope, - , but different y-intercepts, the lines are 3 parallel. See Figure 40.
r
Finding a Line That Is Parallel to a Given Line
Find an equation for the line that contains the point 12, - 32 and is parallel to the line 2x + y = 6. Since the two lines are to be parallel, the slope of the line being sought equals the slope of the line 2x + y = 6. Begin by writing the equation of the line 2x + y = 6 in slope–intercept form. 2x + y = 6 y = - 2x + 6
60
CHAPTER F Foundations: A Prelude to Functions
The slope is - 2. Since the line being sought contains the point 12, - 32, use the point–slope form to obtain
Figure 41 y
y - y1 y - 1 - 32 y + 3 y 2x + y
6
6
6 x 2x y 6 (2, 3) 5
m1x - x1 2 - 21x - 22 - 2x + 4 - 2x + 1 1
Point–slope form m = - 2, x1 = 2, y1 = - 3 Simplify. Slope–intercept form General form
This line is parallel to the line 2x + y = 6 and contains the point 12, - 32 . See Figure 41.
r
Now Work
2x y 1
= = = = =
PROBLEM
59
10 Find Equations of Perpendicular Lines When two lines intersect at a right angle (90°), they are perpendicular. See Figure 42.
THEOREM
Criterion for Perpendicular Lines Two nonvertical lines are perpendicular if and only if the product of their slopes is - 1.
Figure 42 y
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.
90° x
In Problem 128, you are asked to prove the “if” part of the theorem: If two nonvertical lines have slopes whose product is - 1, then the lines are perpendicular.
Figure 43 y Slope m2 A = (1, m2) Slope m1
Rise = m 2 x
Run = 1 O
1
Rise = m1
B = (1, m1)
Proof Let m1 and m2 denote the slopes of the two lines. There is no loss in generality (that is, neither the angle nor the slopes are affected) if we situate the lines so that they meet at the origin. See Figure 43. The point A = 11, m2 2 is on the line having slope m2 , and the point B = 11, m1 2 is on the line having slope m1 . (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 3 d 1O, A2 4 2 + 3 d 1O, B2 4 2 = 3 d 1A, B2 4 2
(5)
By the distance formula, we can write the squares of these distances as
3 d 1O, A2 4 2 = 11 - 02 2 + 1m2 - 02 2 = 1 + m22 3 d 1O, B2 4 2 = 11 - 02 2 + 1m1 - 02 2 = 1 + m21 3 d 1A, B2 4 2 = 11 - 12 2 + 1m2 - m1 2 2 = m22 - 2m1 m2 + m21
Using these facts in equation (5), we get
11 + m22 2 + 11 + m21 2 = m22 - 2m1 m2 + m21
which, upon simplification, can be written as m1 m2 = - 1 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 m1 m2 = - 1 means that m1 and m2 are 1 1 negative reciprocals of each other; that is, either m1 = or m2 = . m2 m1
SECTION F.3 Lines
61
Finding the Slope of a Line Perpendicular to Another Line
EX AM PL E 12
3 2 If a line has slope , any line having slope - is perpendicular to it. 2 3
Finding the Equation of a Line Perpendicular to a Given Line
EX AM PL E 13
Find an equation of the line that contains the point 11, - 22 and is perpendicular to the line x + 3y = 6.(SBQIUIFUXPMJOFT
Solution
First write the equation of the given line in slope–intercept form to find its slope. x + 3y = 6 3y = - x + 6 1 y = - x + 2 3
y x 3y 6
Proceed to solve for y. Place in the form y = mx + b.
1 The given line has slope - . Any line perpendicular to this line will have slope 3. 3 Because the point 11, - 22 is on this line with slope 3, use the point–slope form of the equation of a line.
Figure 44 y 3x 5
6
y - y1 = m1x - x1 2 y - 1 - 22 = 31x - 12
4 2 x
2
2
4
Point–slope form m = 3, x1 = 1, y1 = - 2
To obtain other forms of the equation, proceed as follows:
6
y + 2 y + 2 y 3x - y
(1, 2)
2
r
4
= = = =
31x - 12 3x - 3 3x - 5 5
Simplify. Slope–intercept form General form
r
Figure 44 shows the graphs.
Now Work
PROBLEM
65
F.3 Assess Your Understanding Concepts and Vocabulary 1. The slope of a vertical line is horizontal line is 0 .
7. Two nonvertical lines have slopes m1 and m2, respectively. The lines are parallel if m1 = m2 and the y-intercepts are unequal; the lines are perpendicular if m1m2 = - 1 .
undefined ; the slope of a
2. For the line 2x + 3y = 6, the x-intercept is y-intercept is 2 .
3
and the
8. The lines y = 2x + 3 and y = ax + 5 are parallel if 2 . a =
3. A horizontal line is given by an equation of the form y = b , where b is the y-intercept .
9. The lines y = 2x - 1 and y = ax + 2 are perpendicular if a = −½ .
4. True or False Vertical lines have an undefined slope. True
10. True or False Perpendicular lines have slopes that are reciprocals of one another. False
5. True or False The slope of the line 2y = 3x + 5 is 3. False 6. True or False The point 11, 22 is on the line 2x + y = 4. True
Skill Building In Problems 11–14, (a) find the slope of the line and (b) interpret the slope. *11.
*12.
y 2
(–2, 1) 2
(2, 1)
(0, 0) –2
–1
*13.
y
*14.
y (–2, 2)
2
(1, 1)
y (–1, 1)
2
(2, 2)
(0, 0) 2
x
–2
–1
2
x
–2
–1
2
x
–2
–1
2
In Problems 15–22, plot each pair of points and determine the slope of the line containing them. Graph the line. *15. 12, 32; 14, 02
*19. 1 - 3, - 12; 12, - 12
*16. 14, 22; 13, 42
*20. 14, 22; 1 - 5, 22
*17. 1 - 2, 32; 12, 12
*21. 1 - 1, 22; 1 - 1, - 22
*18. 1 - 1, 12; 12, 32
*22. 12, 02; 12, 22
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
x
62
CHAPTER F Foundations: A Prelude to Functions
In Problems 23–30, graph the line containing the point P and having slope m. *23. P = 11, 22; m = 3
*27. P = 1 - 1, 32; m = 0
3 2 *26. P = 11, 32; m = 4 5 *29. P = 10, 32 ; slope undefined *30. P = 1 - 2, 02; slope undefined
*24. P = 12, 12; m = 4
*25. P = 12, 42; m = -
*28. P = 12, - 42; m = 0
In Problems 31–36, the slope and a point on a line are given. Use this information to locate three additional points on the line. Answers may vary. [Hint: It is not necessary to find the equation of the line. See Example 2.] 3 *31. Slope 4; point 11, 22 *33. Slope - ; point 12, - 42 *32. Slope 2; point 1 - 2, 32 2 4 *36. Slope - 1; point 14, 12 *35. Slope - 2; point 1 - 2, - 32 *34. Slope ; point 1 - 3, 22 3 In Problems 37–44, find an equation of the line L. *37.
*38.
y 2
(2, 1)
*39. L
y
L
(–2, 1) 2
(–1, 3)
(0, 0) –2
–1
x
y
*41.
3
–2
*42.
–1
3
(3, 3)
–2
*43.
y
3
–2
L
(1, 2)
–1
x
2
–1 L
3 x
L is parallel to y = 2x
–1
L
(2, 2)
2
–1
x
y 3
*44.
y 3 (1, 2) L
L y = 2x
2
(–1, 1)
x
2
y
(1, 1)
(0, 0) 2
*40.
y
3 x y = –x
L is parallel to y = –x
–1 y = 2x
3 x
L is perpendicular to y = 2x
(–1, 1) –3 L
1 x 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 prefer. *45. Slope = 3; containing the point 1 - 2, 32 2 *47. Slope = - ; containing the point 11, - 12 3 *49. Containing the points 11, 32 and 1 - 1, 22
*46. Slope = 2; containing the point 14, - 32 1 *48. Slope = ; containing the point 13, 12 2 *50. Containing the points 1 - 3, 42 and 12, 52
*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 12, 42 57. Horizontal; containing the point 1 - 3, 22
y = 2
*59. Parallel to the line y = 2x; containing the point 1 - 1, 22
*61. Parallel to the line 2x - y = - 2; containing the point 10, 02
*63. Parallel to the line x = 5; containing the point 14, 22
1 65. Perpendicular to the line y = x + 4; containing the point 2 11, - 22 2x + y = 0 or y = - 2x 67. Perpendicular to the line 2x + y = 2; containing the point 1 - 3, 02 x - 2y = - 3 or y = 1x + 3 2 2 *69. Perpendicular to the line x = 8; containing the point 13, 42
*56. Slope undefined; containing the point 13, 82
*58. Vertical; containing the point 14, - 52
*60. Parallel to the line y = - 3x; containing the point 1 - 1, 22
*62. Parallel to the line x - 2y = - 5; containing the point 10, 02 64. Parallel to the line y = 5; containing the point 14, 22
y = 2
66. Perpendicular to the line y = 2x - 3; containing the point 11, - 22 x + 2y = - 3 or y = - 1x - 3 2 2 68. Perpendicular to the line x - 2y = - 5; containing the point 10, 42 2x + y = 4 or y = - 2x + 4
*70. Perpendicular to the line y = 8; containing the point 13, 42
SECTION F.3 Lines
63
In Problems 71–90, find the slope and y-intercept of each line. Graph the line. *71. y = 2x + 3 *76. y = 2x + *81. x + y = 1
1 2
*86. x = 2
*72. y = - 3x + 4
*73.
1 y = x - 1 2
*74.
1 x + y = 2 3
*75. y =
1 x + 2 2
*77. x + 2y = 4
*78. - x + 3y = 6
*79. 2x - 3y = 6
*80. 3x + 2y = 6
*82. x - y = 2
*83. x = - 4
*84. y = - 1
*85. y = 5
*87. y - x = 0
*88. x + y = 0
*89. 2y - 3x = 0
*90. 3x + 2y = 0
In Problems 91–100, (a) find the intercepts of the graph of each equation and (b) graph the equation. *91. 2x + 3y = 6
*92. 3x - 2y = 6
1 1 *97. x + y = 1 2 3 *100. - 0.3x + 0.4y = 1.2
*96. 5x + 3y = 18
*95. 7x + 2y = 21
*94. 6x - 4y = 24
*93. - 4x + 5y = 40
*99. 0.2x - 0.5y = 1
*98. x -
2 y = 4 3
102. Find an equation of the y-axis. x = 0
101. Find an equation of the x-axis. y = 0
In Problems 103–106, the equations of two lines are given. Determine whether the lines are parallel, perpendicular, or neither. 1 103. y = 2x - 3 105. y = 4x + 5 Neither 104. y = x - 3 106. y = - 2x + 3 Neither 2 y = 2x + 4 Parallel y = - 4x + 2 1 y = - 2x + 4 y = - x + 2 2 Perpendicular In Problems 107–110, match each graph with the correct equation: (a) y = x 107. (b)
(b) y = 2x 108. (c)
8
6
6
(c) y = 4
12
8
109. (d)
12
4
x 2
(d) y = 4x 110. (a)
8
3
3
8
2
3
3
2
Applications and Extensions *111. Geometry Use slopes to show that the triangle whose vertices are 1 - 2, 52 , 11, 32 , and 1 - 1, 02 is a right triangle. *112. Geometry Use slopes to show that the quadrilateral whose vertices are 11, - 12 , 14, 12 , 12, 22 , and 15, 42 is a parallelogram. *113. Geometry Use slopes to show that the quadrilateral whose vertices are 1 - 1, 02 , 12, 32 , 11, - 22 , and 14, 12 is a rectangle. *114. Geometry Use slopes and the distance formula to show that the quadrilateral whose vertices are 10, 02 , 11, 32 , 14, 22 , and 13, - 12 is a square. 115. Truck Rentals A truck rental company rents a moving truck for one day by charging $39 plus $0.60 per mile. Write a linear 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? C = 0.60x + 39; $105.00; $177.00 116. 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 machinery. The variable costs of operating a business are the costs that change with the level of output. Variable costs include 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 manufactured. What is the cost of manufacturing 400 pairs of jeans? 740 pairs? C = 8x + 500; $3700; $6420 117. Cost of Driving a Car The annual fixed costs for owning a small sedan are $6735, assuming the car is completely paid for. The cost to drive the car is approximately $0.45 per mile. Write a linear equation that relates the cost C and the number x of miles driven annually. C = 0.45x + 6735 Source: AAA, April 2012 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 represents Dan’s weekly salary S when he has sales that generate a profit of x dollars. S = 0.05x + 375 119. Electricity Rates in Illinois Commonwealth Edison Company supplies electricity to residential customers for a monthly customer charge of $13.04 plus 10.62 cents per kilowatt-hour for up to 800 kilowatt-hours (kw-hr).
64
CHAPTER F Foundations: A Prelude to Functions
*123. 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 Platform 30"
Ramp x
(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 … 800. C = 0.1062x + 13.04 * C (SBQIUIJTFRVBUJPO (c) What is the monthly charge for using 200 kilowatthours? $34.28 (d) What is the monthly charge for using 500 kilowatthours? $66.14 *(e) Interpret the slope of the line. Source: Commonwealth Edison Company, January 2013. 120. Electricity Rates in Florida Florida Power & Light Company supplies electricity to residential customers for a monthly customer charge of $7.00 plus 8.55 cents per kilowatt-hour for up to 1000 kilowatt-hours (kw-hr). (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. C = 0.0855x + 7.00 * C (SBQIUIJTFRVBUJPO (c) What is the monthly charge for using 200 kilowatthours? $24.10 (d) What is the monthly charge for using 500 kilowatthours? $49.75 *(e) Interpret the slope of the line. Source: Florida Power & Light Company, January 2013. *121. Measuring Temperature The relationship between Celsius (°C) and Fahrenheit (°F) degrees of measuring temperature is linear. Find a linear equation relating °C and °F if 0°C corresponds to 32°F and 100°C corresponds to 212°F. Use the equation to find the Celsius measure of 70°F. *122. Measuring Temperature The Kelvin (K) scale for measuring temperature is obtained by adding 273 to the Celsius temperature. (a) Write a linear equation relating K and °C. (b) Write a linear equation relating K and °F (see Problem 121).
(a) Write a linear equation that relates the height y of the ramp above the floor to the horizontal distance x from the platform. (b) Find and interpret the x-intercept of the graph of your equation. (c) Design requirements stipulate that the maximum run be 30 feet and that the maximum slope be a drop of 1 inch for every 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/wood_ramps.php 124. Cigarette Use A report in the Child Trends DataBase indicated that in 2000, 20.6% of twelfth grade students reported daily use of cigarettes. In 2011, 10.3% of twelfth grade students reported daily use of cigarettes. *(a) Write a linear equation that relates the percent y of twelfth grade students who smoke cigarettes daily to the number x of years after 2000. *(b) Find the intercepts of the graph of your equation. *(c) Do these intercepts have any meaningful interpretation? (d) Use your equation to predict the percent for the year 2025. Is this result reasonable? - 2.8%; No Source: www.childtrends.org/databank 125. Product Promotion A cereal company finds that the number of people who will buy one of its products in the first month that the product is introduced is linearly related to the amount of money it spends on advertising. If it spends $40,000 on advertising, 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 expenditure on advertising is needed to sell 300,000 boxes of cereal? $80,000 *(c) Interpret the slope.
*126. Show that the line containing the points 1a, b2 and 1b, a2 , a ≠ b, is perpendicular to the line y = x. Also show that the midpoint of (a, b) and 1b, a2 lies on the line y = x. *127. 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 member of the family? *128. Prove that if two nonvertical lines have slopes whose product is - 1, then the lines are perpendicular. [Hint: Refer to Figure 43 and use the converse of the Pythagorean Theorem.]
SECTION F.3 Lines
65
Discussion and Writing 129. Which of the following equations might have the graph shown? (More than one answer is possible.) (a) 2x + 3y = 6 (b) - 2x + 3y = 6
y
(c) 3x - 4y = - 12 (d) x - y = 1 (e) x - y = - 1
x
(f) y = 3x - 5 (g) y = 2x + 3 (h) y = - 3x + 3
133. m Is for Slope The accepted symbol used to denote the slope of a line is the letter m. Investigate the origin of this symbolism. Begin by consulting a French dictionary and looking up the French word monter. Write a brief essay on your findings. 134. Grade of a Road The term grade is used to describe the inclination of a road. How is this term related to the notion 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.
(b), (c), (e), (g)
130. Which of the following equations might have the graph shown? (More than one answer is possible.) (a) 2x + 3y = 6 (b) 2x - 3y = 6
Steep 7% Grade
(c) 3x + 4y = 12
y
(d) x - y = 1 (e) x - y = - 1
x
(f) y = - 2x - 1 (g) y = -
1 x + 10 2
(h) y = x + 4
(a), (c), (g)
131. The figure shows the graph of two parallel lines. Which of the following pairs of equations might have such a graph? (c) (a) x - 2y = 3 x + 2y = 7 (b) x + y = 2 x + y = -1
135. Carpentry Carpenters use the term pitch to describe the steepness of staircases and roofs. How is pitch related to slope? Investigate typical pitches used for stairs and for roofs. Write a brief essay on your findings. 136. Can the equation of every line be written in slope–intercept form? Why? No, consider vertical lines. *137. Does every line have exactly one x-intercept and one y-intercept? Are there any lines that have no intercepts?
y
138. What can you say about two lines that have equal slopes and equal y-intercepts? They are the same.
(c) x - y = - 2 x - y = 1
x
(d) x - y = - 2 2x - 2y = - 4
139. What can you say about two lines with the same x-intercept and the same y-intercept? Assume that the x-intercept is not 0. They are the same line.
(e) x + 2y = 2 x + 2y = - 1
140. If two distinct lines have the same slope but different x-intercepts, can they have the same y-intercept? No
132. The figure shows the graph of two perpendicular lines. Which of the following pairs of equations might have such a graph? (d) (a) y - 2x = 2 y + 2x = - 1 (b) y - 2x = 0 2y + x = 0 (c) 2y - x = 2 2y + x = - 2 (d) y - 2x = 2 x + 2y = - 1 (e) 2x + y = - 2 2y + x = - 2
y
x
*141. If two distinct lines have the same y-intercept but different slopes, can they have the same x-intercept? 142. 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. *143. What Went Wrong? A student is asked to find the slope of the line joining 1 - 3, 22 and 11, - 42 . He states that the 3 slope is . Is he correct? If not, what went wrong? 2
66
CHAPTER F Foundations: A Prelude to Functions
F.4 Circles PREPARING FOR THIS SECTION Before getting started, review the following: r $PNQMFUJOHUIF4RVBSF "QQFOEJY" 4FDUJPO" pp. A38–A39)
r 4PMWJOH&RVBUJPOT "QQFOEJY" 4FDUJPO" pp. A63–A69)
Now Work the ‘Are You Prepared?’ problems on page 70.
OBJECTIVES 1 Write the Standard Form of the Equation of a Circle (p. 66) 2 Graph a Circle (p. 67) 3 Work with the General Form of the Equation of a Circle (p. 68)
1 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 following geometric statement that defines a circle.
DEFINITION
Figure 45 shows the graph of a circle. To find the equation, let 1x, y2 represent the coordinates of any point on a circle with radius r and center 1h, k). Then the distance between the points 1x, y2 and 1h, k2 must always equal r. That is, by the distance formula,
Figure 45 y
A circle is a set of points in the xy-plane that are a fixed distance r from a fixed point 1h, k2. The fixed distance r is called the radius, and the fixed point 1h, k2 is called the center of the circle.
(x, y) r
2 1x - h2 2 + 1y - k2 2 = r
(h, k ) x
or, equivalently, 1x - h2 2 + 1y - k2 2 = r 2
DEFINITION
The standard form of an equation of a circle with radius r and center 1h, k2 is 1x - h2 2 + 1y - k2 2 = r 2
THEOREM
(1)
The standard form of an equation of a circle of radius r with center at the origin 10, 02 is x2 + y 2 = r 2
DEFINITION
If the radius r = 1, the circle whose center is at the origin is called the unit circle and has the equation x2 + y 2 = 1
See Figure 46. Note that the graph of the unit circle is symmetric with respect to the x-axis, the y-axis, and the origin.
SECTION F.4 Circles
Figure 46 Unit circle x2 + y2 = 1
67
y 1
1
1x
(0,0) 1
EXAM PL E 1
Solution
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 1 - 3, 62.
Substitute the values r = 5, h = - 3, and k = 6, into equation (1). 1x - h2 2 + 1y - k2 2 = r 2
(x - ( - 3))2 + (y - 6)2 = 52
1x + 32 2 + 1y - 62 2 = 25
Now Work
PROBLEM
7
r
2 Graph a Circle
The graph of any equation of the form 1x - h)2 + (y - k)2 = r 2 is that of a circle with radius r and center 1h, k2.
EXAM PL E 2
Solution
Graphing a Circle
(SBQIUIFFRVBUJPO 1x + 32 2 + 1y - 22 2 = 16 Since the equation is in the form of equation (1), its graph is a circle. To graph the equation, compare the given equation to the standard form of the equation of a circle. The comparison yields information about the circle. 1x + 32 2 + 1y - 22 2 = 16 1x - 1 - 32 2 2 + 1y - 22 2 = 42
Figure 47 (–3, 6)
y 6
c
–10
(–3, 2)
c
1x - h2 + 1y - k2 = r 2
4 (–7, 2)
c
2
(1, 2) 2 x
–5 (–3, –2)
We see that h = - 3, k = 2, and r = 4. The circle has center 1 - 3, 22 and a radius of 4 units. To graph this circle, first plot the center 1 - 3, 22. Since the radius is 4, locate four points on the circle by plotting points 4 units to the left, to the right, and up and down from the center. Then use these four points as guides to obtain the graph. See Figure 47.
Now Work
EXAM PL E 3
Solution
2
PROBLEMS
23(A)
r
AND
(B)
Finding the Intercepts of a Circle
For the circle 1x + 32 2 + 1y - 22 2 = 16, find the intercepts, if any, of its graph. This is the equation discussed and graphed in Example 2. To find the x-intercepts, if any, let y = 0 and solve for x. Then 1x + 32 2 + 1y - 22 2 = 16
In Words
The symbol { is read “plus or minus.” It means to add and subtract the quantity following the { symbol. For example, 5 { 2 means “5 - 2 = 3 or 5 + 2 = 7.”
1x + 32 2 + 10 - 22 2 = 16 1x + 32 + 4 = 16 2
1x + 32 2 = 12 x + 3 = { 212 x = - 3 { 223
y = 0 Simplify. Subtract 4 from both sides. Solve for x + 3. Solve for x
The x-intercepts are - 3 - 223 ≈ - 6.46 and - 3 + 223 ≈ 0.46.
68
CHAPTER F Foundations: A Prelude to Functions
To find the y-intercepts, if any, let x = 0 and solve for y. Then 1x + 32 2 + 1y 10 + 32 2 + 1y 9 + 1y 1y y
22 2 22 2 22 2 22 2 - 2 y
= = = = = =
16 16 16 7 { 27 2 { 27
x = 0
Solve for y - 2. Solve for y.
The y-intercepts are 2 - 27 ≈ - 0.65 and 2 + 27 ≈ 4.65. Look back at Figure 47 to verify the approximate locations of the intercepts.
Now Work
PROBLEM
23(C)
r
3 Work with the General Form of the Equation of a Circle Eliminate the parentheses from the standard form of the equation of the circle given in Example 3 to obtain 1x + 32 2 + 1y - 22 2 = 16 x + 6x + 9 + y2 - 4y + 4 = 16 2
which simplifies to x2 + y2 + 6x - 4y - 3 = 0 It can be shown that any equation of the form x2 + y2 + ax + by + c = 0 has a graph that is a circle, or has a graph that is a point, or has no graph at all. For example, the graph of the equation x2 + y2 = 0 is the single point 10, 02 . The equation x2 + y2 + 5 = 0, or x2 + y2 = - 5, has no graph, because sums of squares of real numbers are never negative.
DEFINITION
When its graph is a circle, the equation x2 + y2 + ax + by + c = 0 is the general form of the equation of a circle.
If an equation of a circle is in the general form, the method of completing the square can be used to put the equation in standard form so that its center and radius can be identified.
EX A MPL E 4
Graphing a Circle Whose Equation Is in General Form (SBQIUIFFRVBUJPOx2 + y2 + 4x - 6y + 12 = 0
Solution
(SPVQUIFUFSNTJOWPMWJOHx, group the terms involving y, and put the constant on the right side of the equation. The result is 1x2 + 4x2 + 1y2 - 6y2 = - 12 Next, complete the square of each expression in parentheses. Remember that any number added on the left side of the equation must be added on the right.
SECTION F.4 Circles
1x2 + 4x + 42 + 1y2 - 6y + 92 = - 12 + 4 + 9
Figure 48
c
y
(2, 4)
c
6
(1, 3)
4
1 (3, 3)
69
6
4 a b = 4 2 2
a
-6 2 b = 9 2
1x + 22 2 + 1y - 32 2 = 1
(2, 3) (2, 2)
3
1 x
This equation is the standard form of the equation of a circle with radius 1 and center 1 - 2, 32. To graph the equation, use the center 1 - 2, 32 and the radius 1. See Figure 48.
Now Work
EXAM PL E 5
Solution Figure 49
Factor
PROBLEM
r
27
Finding the General Equation of a Circle
Find the general equation of the circle whose center is 11, - 22 and whose graph contains the point 14, - 22. To find the equation of a circle, we need to know its center and its radius. Here, the center is 11, - 22. Since the point 14, - 22 is on the graph, the radius r will equal the distance from 14, - 22 to the center 11, - 22. See Figure 49. Thus,
y 3
r = 2 14 - 12 2 + 3 - 2 - 1 - 22 4 2 = 29 = 3 r (1, 2)
x
5 (4, 2)
5
The standard form of the equation of the circle is
1x - 12 2 + 1y + 22 2 = 9
Eliminate the parentheses and rearrange the terms to get the general equation x2 + y2 - 2x + 4y - 4 = 0
Now Work
PROBLEM
13
COMMENT Example 6 requires information from Section B.5, Square Screens, in Appendix B.
EXAM PL E 6
r ■
Using a Graphing Utility to Graph a Circle (SBQIUIFFRVBUJPOx2 + y2 = 4
Solution Figure 50 2
This is the equation of a circle with center at the origin and radius 2. To graph this equation, first solve for y. x2 + y 2 = 4
Y1 = 4 − x 2
y 2 = 4 - x2 y = { 24 - x2
−3
3
−2
Y2 = − 4 − x 2
Subtract x 2 from each side. Solve for y.
There are two equations to graph on the same square screen: Y1 = 24 - x2 and Y2 = - 24 - x2. (Your circle will appear oval if you do not use a square screen.) See Figure 50. The graph is “disconnected” because of the resolution of the calculator.
r
Overview The discussion in Sections F.3 and F.4 about lines and circles dealt with two main types of problems that can be generalized as follows: 1. (JWFOBOFRVBUJPO DMBTTJGZJUBOEHSBQIJU 2. (JWFOBHSBQI PSJOGPSNBUJPOBCPVUBHSBQI àOEJUTFRVBUJPO This text deals with both types of problems. We shall study various equations, classify them, and graph them. The second type of problem is usually more difficult to solve than the first.
70
CHAPTER F Foundations: A Prelude to Functions
F.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. 2. Solve the equation 1x - 2)2 = 9. (pp. A63–A69)
1. To complete the square of x2 + 10x, you would (add/ subtract) the number 25 . (pp. A38–A39) add
{ - 1, 5}
Concepts and Vocabulary 3. True or False Every equation of the form
5. The standard form of an equation of a circle with radius r and center (h, k) is 1x - h2 2 + 1y - k2 2 = r 2.
x + y + ax + by + c = 0 2
2
6. The circle whose equation is given by x2 + y2 = 1 is called unit the circle.
has a circle as its graph. False radius is the distance from the center 4. For a circle, the to any point on the circle.
Skill Building In Problems 7–10, find the center and radius of each circle. Write the standard form of the equation. 8.
7. y
9. y
y
(4, 2)
(2, 3) (h, k)
(h, k)
(1, 2) (0, 1)
10. y
(2, 1) (0, 1) (1, 2)
x (1, 0)
center: (2, 1); radius: 2 1x - 22 2 + 1y - 12 2 = 4
x
center: (1, 2); radius: 2
1x - 12 2 + 1y - 22 2 = 4
x
x
3 5 center: a , 2b ; radius: 2 2 9 5 2 ax - b + 1y - 22 2 = 2 4
center: (1, 2); radius: 22 1x - 12 2 + 1y - 22 2 = 2
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 1h, k2. Graph each circle. *11. r = 2; *15. r = 5; *19. r =
1 ; 2
1h, k2 = 10, 02
*12. r = 3;
1h, k2 = 14, - 32 *16. r = 4;
1h, k2 = 10, 02
*13. r = 2;
1h, k2 = 12, - 32 *17. r = 4;
1 1h, k2 = a , 0b 2
*20. r =
1 ; 2
1h, k2 = 10, 22 *14. r = 3;
1h, k2 = 1 - 2, 12 *18. r = 7;
1h, k2 = 11, 02
1h, k2 = 1 - 5, - 22
1 1h, k2 = a0, - b 2
In Problems 21–34, (a) find the center 1h, k2 and radius r of each circle; (b) graph each circle; (c) find the intercepts, if any. *21. x2 + y2 = 4
*22. x2 + 1y - 12 2 = 1
*23. 21x - 32 2 + 2y2 = 8
*24. 31x + 12 2 + 31y - 12 2 = 6
*25. x2 + y2 - 2x - 4y - 4 = 0
*26. x2 + y2 + 4x + 2y - 20 = 0
*27. x2 + y2 + 4x - 4y - 1 = 0
*28. x2 + y2 - 6x + 2y + 9 = 0
*29. x2 + y2 - x + 2y + 1 = 0
1 = 0 2 2 2 *33. 2x + 8x + 2y = 0
*31. 2x2 + 2y2 - 12x + 8y - 24 = 0
*32. 2x2 + 2y2 + 8x + 7 = 0
*30. x2 + y2 + x + y -
*34. 3x2 + 3y2 - 12y = 0
In Problems 35– 42, find the standard form of the equation of each circle. 36. Center 11, 02 and containing the point 1 - 3, 22 35. Center at the origin and containing the point 1 - 2, 32 x2 + y2 = 13 1x - 12 2 + y2 = 20 38. Center 1 - 3, 12 and tangent to the y-axis 37. Center 12, 32 and tangent to the x-axis 1x - 22 2 + 1y - 32 2 = 9 1x + 32 2 + 1y - 12 2 = 9 40. With endpoints of a diameter at (4, 3) and (0, 1) 39. With endpoints of a diameter at (1, 4) and ( - 3, 2) 1x - 22 2 + 1y - 22 2 = 5 1x + 12 2 + 1y - 32 2 = 5 41. Center ( - 1, 3) and tangent to the line y = 2 1x + 12 2 + 1y - 32 2 = 1
42. Center (4, - 2) and tangent to the line x = 1 1x - 42 2 + 1y + 22 2 = 9
* Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
SECTION F.4 Circles
In Problems 43–46, match each graph with the correct equation. (a) 1x - 32 2 + 1y + 32 2 = 9 43. (c)
44. (d)
4
6
(b) 1x + 12 2 + 1y - 22 2 = 4
45. (b)
6
9
6
9
6
4
(c) 1x - 12 2 + 1y + 22 2 = 4
(d) 1x + 32 2 + 1y - 32 2 = 9 46. (a)
4
6
71
6
9
6
9
4
6
Applications and Extensions 47. Find the area of the square in the figure. 18 units2 y
x2 y2 9
x
and a diameter of 520 feet, 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. Source: LasVegas Review Journal x2 + 1y - 2902 2 = 67,600 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 + y2 + 2x + 4y - 4091 = 0. A weather satellite circles 0.6 unit above Earth with the center of its circular orbit at the center of Earth. Find the equation for the orbit of the satellite on this map.
48. Find the area of the blue shaded region in the figure, assuming the quadrilateral inside the circle is a square. 36p - 72 units2 y
r
x 2 y 2 36
x
x2 + y2 + 2x + 4y - 4168.16 = 0 *52. The tangent line to a circle may be defined as the line that intersects the circle in a single point called the point of tangency. See the figure. y
49. Ferris Wheel The original Ferris wheel was built in 1893 by 1JUUTCVSHI 1FOOTZMWBOJB CSJEHFCVJMEFS (FPSHF 8 'FSSJT The Ferris wheel was originally built for the 1893 World’s Fair in Chicago, and it was 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 x2 + 1y - 1392 2 = 15,625 50. Ferris Wheel Opening in 2014 in Las Vegas, The High Roller observation wheel has a maximum height of 550 feet
r x
If the equation of the circle is x2 + y2 = r 2 and the equation of the tangent line is y = mx + b, show that:
(a) r 2 11 + m2 2 = b2 [Hint: The quadratic equation x2 + 1mx + b2 2 = r 2 has exactly one solution.] - r 2m r 2 (b) The point of tangency is ¢ , ≤. b b (c) The tangent line is perpendicular to the line containing the center of the circle and the point of tangency.
53. The Greek Method 5IF (SFFL NFUIPE GPS GJOEJOH UIF 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 tangent line are perpendicular (see Problem 52). Use this method to find an equation of the tangent line to the circle x2 + y2 = 9 at the point 11, 2122. 22x + 4y = 922
72
CHAPTER F Foundations: A Prelude to Functions
54. 6TF UIF (SFFL NFUIPE EFTDSJCFE JO 1SPCMFN to find an equation of the tangent line to the circle x2 + y2 - 4x + 6y + 4 = 0 at the point 13, 212 - 32. 22 x + 4y = 1122 - 12 55. Refer to Problem 52. The line x - 2y + 4 = 0 is tangent to a circle at 10, 22. The line y = 2x - 7 is tangent to the same circle at 13, - 12. Find the center of the circle. (1, 0)
56. Find an equation of the line containing the centers of the two circles x2 + y2 - 4x + 6y + 4 = 0 x + 5y = - 13
and x2 + y2 + 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? y = 2 58. If the circumference of a circle is 6p, what is its radius? 3
Discussion and Writing 59. Which of the following equations might have the graph shown? (More than one answer is possible.) (b), (c), (e), (g) (a) 1x - 22 2 + 1y + 32 2 = 13 (b) 1x - 22 + 1y - 22 = 8 2
2
(a) 1x - 22 2 + y2 = 3
y
(b) 1x + 22 2 + y2 = 3
(c) 1x - 22 2 + 1y - 32 2 = 13 x
(f) x + y + 4x - 2y = 0 2
2
(e) x2 + y2 + 10x + 16 = 0
x
(f) x2 + y2 + 10x - 2y = 1
(g) x2 + y2 - 9x - 4y = 0
(g) x2 + y2 + 9x + 10 = 0
(h) x + y - 4x - 4y = 4 2
(d) 1x + 22 2 + y2 = 4
2
(e) x2 + y2 - 4x - 9y = 0
y
(c) x2 + 1y - 22 2 = 3
(d) 1x + 22 + 1y - 22 = 8 2
60. Which of the following equations might have the graph shown? (More than one answer is possible.) (b), (e), (g)
2
(h) x2 + y2 - 9x - 10 = 0
61. Explain how the center and radius of a circle can be used to graph a circle.
62. What Went Wrong? A student stated that the center and radius of the graph whose equation is 1x + 32 2 + 1y - 22 2 = 16 are (3, - 2) and 4, respectively. Why is this incorrect? The center is 1 - 3, 22 .
‘Are You Prepared?’ Answers 1. add; 25
2. 5 - 1, 56
Chapter Project
73
Chapter Project The graph below, called a scatter diagram, shows the points (291.5, 268), (320, 305), . . . , (368, 385) in a Cartesian plane. From the graph, it appears that the data follow a linear relation.
Sale Price ($ thousands)
Zestimate vs. Sale Price in Oak Park, IL
Internet-based Project Determining the Selling Price of a Home Determining how much to pay for a home is one of the more difficult decisions that must be made when purchasing a home. There are many factors that play a role in a home’s value. Location, size, number of bedrooms, number of bathrooms, lot size, and building materials are just a few. Fortunately, the website Zillow.com has developed its own formula for predicting the selling price of a home. This information is a great tool for predicting the actual sale price. For example, the data below show the “zestimate”—the selling price of a home as predicted by the folks at Zillow and the actual selling price of the home, for homes, in Oak Park, Illinois.
380 360 340 320 300 280 300 320 340 360 Zestimate ($ thousands)
1. Imagine drawing a line through the data that appears to fit the data well. Do you believe the slope of the line would be positive, negative, or close to zero? Why? 2. Pick two points from the scatter diagram. Treat the zestimate as the value of x and treat the sale price as the corresponding value of y. Find the equation of the line through the two points you selected. 3. Interpret the slope of the line. 4. Use your equation to predict the selling price of a home whose zestimate is $335,000.
Zestimate ($ thousands)
Sale Price ($ thousands)
291.5
268
320
305
371.5
375
303.5
283
351.5
350
314
275
(b) Select two points from the scatter diagram and find the equation of the line through the points.
332.5
356
(c) Interpret the slope.
295
300
313
285
368
385
(d) Find a home from the Zillow website that interests you under the “Make Me Move” option for which a zestimate is available. Use your equation to predict the sale price based on the zestimate.
5. Do you believe it would be a good idea to use the equation you found in part 2 if the zestimate were $950,000? Why or why not? 6. $IPPTF B MPDBUJPO JO XIJDI ZPV XPVME MJLF UP MJWF (P UP www.zillow.com and randomly select at least ten homes that have recently sold. (a) Draw a scatter diagram of your data.
1
Functions and Their Graphs Choosing a Cellular Telephone Plan Most consumers choose a cellular telephone provider first and then select lect an appropriate plan from that provider. What type of plan to select depends ds on your use of the phone. For example, is text messaging important? How many minutes do you plan to use the phone? Do you desire a data plan to browse owse the Web? The mathematics learned in this chapter can help you decide cide which plan is best suited for your particular needs.
—See the Internet-based Chapter Project—
In this chapter, we look at a special type of equation involving two variables called a function. This chapter deals with what a function is, how to graph functions, properties of functions, and how functions are used in applications. The word function apparently was introduced by René Descartes in 1637. For him, a function was simply any positive integral power of a variable x. Gottfried Wilhelm Leibniz (1646–1716), who always emphasized the geometric side of mathematics, used the word function to denote any quantity associated with a curve, such as the coordinates of a point on the curve. Leonhard Euler (1707–1783) employed the word to mean any equation or formula involving variables and constants. His idea of a function is similar to the one most often seen in courses that precede calculus. Later, the use of functions in investigating heat flow equations led to a very broad definition that originated with Lejeune Dirichlet (1805–1859), which describes a function as a correspondence between two sets. That is the definition used in this text.
74
Functions The Graph of a Function Properties of Functions Library of Functions; Piecewise-defined Functions 1.5 Graphing Techniques: Transformations 1.6 Mathematical Models: Building Functions 1.7 Building Mathematical Models Using Variation Chapter Review Chapter Test Chapter Projects
SECTION 1.1 Functions
75
1.1 Functions PREPARING FOR THIS SECTION Before getting started, review the following: r *OUFSWBMT "QQFOEJY" 4FDUJPO" QQ"m"
r 4PMWJOH*OFRVBMJUJFT "QQFOEJY" 4FDUJPO" QQ"m"
r &WBMVBUJOH"MHFCSBJD&YQSFTTJPOT %PNBJOPGB 7BSJBCMF "QQFOEJY" 4FDUJPO" QQ"m"
Now Work the ‘Are You Prepared?’ problems on page 85.
OBJECTIVES 1 2 3 4
Determine Whether a Relation Represents a Function (p. 75) Find the Value of a Function (p. 78) Find the Domain of a Function Defined by an Equation (p. 81) Form the Sum, Difference, Product, and Quotient of Two Functions (p. 83)
1 Determine Whether a Relation Represents a Function
Figure 1 y 5 y 3x 1 (1, 2) 4
2 (0, 1)
2
4
x
5
EXAM PL E 1
Often there are situations where the value of 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 S y. There are a number of ways to express relations between two sets. For example, the equation y = 3x - 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. This relation, expressed as a graph is shown in Figure 1. In addition to being expressed in equations and graphs, relations can be expressed through a technique called mapping. A map illustrates a relation as a set of inputs with an arrow drawn from each element in the set of inputs to the corresponding element in the set of outputs. Ordered pairs can be used to represent x S y as 1x, y2.
Maps 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 (Source: www.house.gov). The relation might be named “number of representatives.”
Figure 2 State Alaska Arizona California Colorado Florida North Dakota
Number of Representatives 1 7 9 27 53
In this relation, Alaska corresponds to 1, Arizona corresponds to 9, and so on. Using ordered pairs, this relation would be expressed as 5 (Alaska, 1), (Arizona, 9), (California, 53), (Colorado, ), (Florida, 2), (North Dakota, 1) 6
r
One of the most important concepts in algebra is the function. A function is a special type of relation. To understand the idea behind a function, let’s revisit the relation presented in Example 1. If we were to ask, “How many representatives does
76
CHAPTER 1 Functions and Their Graphs
Figure 3 Person
Phone Number
Dan
555 – 2345
Gizmo
549 – 9402 930 – 3956
Colleen
555 – 8294
Phoebe
839 – 9013
Alaska have?,” you would respond “1.” In fact, each input state corresponds to a single output number of representatives. Let’s consider a second relation, one that involves a correspondence between four people and their phone numbers. See Figure 3. Notice that Colleen has two telephone numbers. There is no single answer to the question “What is Colleen’s phone number?” -FUTMPPLBUPOFNPSFSFMBUJPO'JHVSFJTBSFMBUJPOUIBUTIPXTBDPSSFTQPOEFODF between type of animal 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 PGBSBCCJU XFXPVMEBMMSFTQPOEiZFBSTu Figure 4 Animal
Life Expectancy
Dog
11
Duck
10
Rabbit
7
/PUJDF UIBU UIF SFMBUJPOT QSFTFOUFE JO 'JHVSFT BOE IBWF TPNFUIJOH JO common. What is it? In both of these relations, each input corresponds to exactly one output. This leads to the definition of a function.
DEFINITION
Figure 5
y
x
Range
X Domain
Y
EX A MPL E 2
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. The set X is called the domain of the function. For each element x in X, the corresponding 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 function. See Figure 5. 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. For example, consider the function y = x2. Since x2 Ú for all real numbers x, the range of y = x2 is 5 y y Ú 6 , which is a subset of the set of all real numbers, Y. Not all relations between two sets are functions. The next example shows how to determine whether a relation is a function.
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. B 4FF'JHVSF'PSUIJTSFMBUJPO UIFEPNBJOSFQSFTFOUTUIFOVNCFSPGDBMPSJFTJO a fast-food sandwich, and the range represents the fat content (in grams).
Figure 6 Source: Each company’s Web site.
Calories
Fat
(Wendy’s 1/4-lb Single) 600
34
(Burger King Whopper) 653
35
(Culvers, Deluxe Single) 711
46
(McDonald's Big Mac) 550
29
(Five Guys Hamburger) 700
43
*The sets X and Y will usually be sets of real numbers, in 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 (proposed by Lejeune Dirichlet), X and Y can be any two sets.
SECTION 1.1 Functions
77
C 4FF'JHVSF'PSUIJTSFMBUJPO UIFEPNBJOSFQSFTFOUTHBTPMJOFTUBUJPOTJO)BSSJT County, Texas and the range represents the price per gallon of regular unleaded HBTPMJOFJO'FCSVBSZ D 4FF'JHVSF'PSUIJTSFMBUJPO UIFEPNBJOSFQSFTFOUTUIFXFJHIU JODBSBUT PG pear-cut diamonds, and the range represents the price (in dollars). Figure 8 Source: Used with permission of Diamonds.com
Figure 7 Gas Station Mobil Shell
Price per Gallon
Carats
Price
$3.76
0.70
$1529
$3.69
0.71
$1575
0.75
$1765
0.78
$1798
$3.71
Texaco
$1952
Citgo
Solution
B 5IF SFMBUJPO JO 'JHVSF JT B GVODUJPO CFDBVTF FBDI FMFNFOU JO UIF EPNBJO corresponds to exactly one element in the range. The domain of the function is 5 55, , 53, , 116 , and the range of the function is 5 29, 3, 35, 3, 6 . C 5IF SFMBUJPO JO 'JHVSF JT B GVODUJPO CFDBVTF FBDI FMFNFOU JO UIF EPNBJO corresponds to exactly one element in the range. The domain of the function is 5 Mobil, Shell, Texaco, Citgo 6 . The range of the function is 5 3.9, 6 . Notice that it is okay for more than one element in the domain to correspond UP UIF TBNF FMFNFOU JO UIF SBOHF 4IFMM BOE $JUHP CPUI TFMM HBT GPS B gallon). D 5IFSFMBUJPOJO'JHVSFJTOPUBGVODUJPOCFDBVTFFBDIFMFNFOUJOUIFEPNBJO EPFTOPUDPSSFTQPOEUPFYBDUMZPOFFMFNFOUJOUIFSBOHF*GBDBSBUEJBNPOE is chosen from the domain, a single price cannot be assigned to it.
r
Now Work
In Words For a function, no input has more than one output. The domain of a function is the set of all inputs; the range is the set of all outputs.
EXAM PL E 3
PROBLEM
15
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 QSFEJDUBCJMJUZ -PPL CBDL BU 'JHVSF *G BTLFE i)PX NBOZ HSBNT PG GBU BSF JO B DBMPSJF TBOEXJDI u XF DPVME VTF UIF DPSSFTQPOEFODF UP BOTXFS iu /PX DPOTJEFS'JHVSF*GBTLFE i8IBUJTUIFQSJDFPGBDBSBUEJBNPOE uXFDPVME OPUHJWFBTJOHMFSFTQPOTFCFDBVTFUXPPVUQVUTSFTVMUGSPNUIFTJOHMFJOQVUiu 'PSUIJTSFBTPO UIFSFMBUJPOJO'JHVSFJTOPUBGVODUJPO We may also think of a function as a set of ordered pairs 1x, y2 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) 5 (1, ), (2, 5), (3, ), (, ) 6 (b) 5 (1, ), (2, ), (3, 5), (, 1) 6 (c) 5 ( - 3, 9), ( - 2, ), (, ), (1, 1), ( - 3, ) 6
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 5 1, 2, 3, 6 , and its range is 5 , 5, , 6 .
78
CHAPTER 1 Functions and Their Graphs
(b) 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 5 1, 2, 3, 6 , and its range is 5 , 5, 16 . (c) This relation is not a function because there are two ordered pairs, 1 - 3, 92 and 1 - 3, ), that have the same first element and different second elements.
r
In Example 3(b), notice that 1 and 2 in the domain both have the same image in the range. This does not violate the definition of a function; two different first elements 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).
Now Work
PROBLEM
19
Up to now we have shown how to identify when a relation is a function for relations defined by mappings (Example 2) and ordered pairs (Example 3). But relations can also be expressed as equations. The circumstances under which equations are functions are discussed next. 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.
EX A MPL E 4
Determining Whether an Equation Is a Function Determine whether the equation y = 2x - 5 defines y as a function of x.
Solution
The equation tells us to take an input x, multiply it by 2, and then subtract 5. 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.
r
EX A MPL E 5
Determining Whether an Equation Is a Function Determine whether the equation x2 + y2 = 1 defines y as a function of x.
Solution
To determine whether the equation x2 + y2 = 1, which defines the unit circle, is a function, solve the equation for y. x2 + y 2 = 1 y 2 = 1 - x2 y = { 21 - x2
Instructor Note: We postpone the vertical-line test until the next section when we introduce the graph of a function.
For values of x between - 1 and 1, two values of y result. For example, if x = , then y = {1, so two different outputs result from the same input. This means that the equation x2 + y2 = 1 does not define a function.
r
Now Work
PROBLEM
33
2 Find the Value of a Function Functions are often denoted by letters such, as f, F, g, G, and others. If f is a function, then for each number x in its domain, the corresponding image in the range is designated by the symbol f 1x2, read as “f of x” or as “f at x.” We refer to f1x2 as the value of f at the number x; f1x2 is the number that results when x is given and the function f is applied; f 1x2 is the output corresponding to x or the image of x; f1x2 does not mean “f times x.” For example, the function HJWFOJO&YBNQMFNBZCFXSJUUFOBTy = f1x2 = 2x - 5. Then f 112 = - 3 and f 132 = 1.
SECTION 1.1 Functions
79
Figure 9 illustrates some other functions. Notice that, in every function, for each x in the domain there is one value in the range. Figure 9 1 5 f (1) 5 f (21)
1 21 0
0 5 f (0)
2
2 5 f ( 2)
22
2 1–2 5 F (22)
21
21 5 F (21)
f(x) 5 x 2
x Domain
Domain
0
0 5 g(0)
0
1
1 5 g(1)
22
2 5 g(2)
3 5 G(0) 5 G(22) 5 G(3)
3
2 5 g(4)
4
g(x) 5 x
x Domain
G(x) 5 3
x
Range
Domain
(c) g(x) 5 x
Range (d) G (x) 5 3
Sometimes it is helpful to think of a function f as a machine that receives as JOQVUBOVNCFSGSPNUIFEPNBJO NBOJQVMBUFTJU BOEPVUQVUTBWBMVF4FF'JHVSF The restrictions on this input/output machine are as follows:
Figure 10 Input x
x
Range 1 (b) F (x) 5 – x
2
2
5 F(4)
1 F(x ) 5 – x
x
Range (a) f (x) 5 x
1– 4
4
1. It accepts only numbers from the domain of the function. 2. For each input, there is exactly one output (which may be repeated for different inputs).
f
Output y f(x)
EXAM PL E 6
Solution
For a function y = f1x2, 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 symbols can be used to represent the independent and dependent variables. For example, if f is the cube function, then f can be given by f1x2 = x3 or f1t2 = t 3 or f1z2 = z3. All three functions are the same. Each says to cube the independent 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 f1x2 = x3, then f tells us to cube the argument. Thus, f122 means to cube 2, f1a2 means to cube the number a, and f1x + h2 means to cube the quantity x + h.
Finding Values of a Function
For the function f defined by f 1x2 = 2x2 - 3x, evaluate (a) f132
(b) f1x2 + f132
(c) 3f1x2
(e) - f1x2
(f) f13x2
(g) f1x + 32
(d) f1 - x2 f1x + h2 - f1x2 (h) h
(a) Substitute 3 for x in the equation for f , f1x2 = 2x2 - 3x, to get f 132 = 2132 2 - 3132 = 1 - 9 = 9
The image of 3 is 9.
h ≠
80
CHAPTER 1 Functions and Their Graphs
(b) f1x2 + f132 = 12x2 - 3x2 + 192 = 2x2 - 3x + 9 (c) Multiply the equation for f by 3. 3f1x2 = 312x2 - 3x2 = x2 - 9x (d) Substitute - x for x in the equation for f and simplify. f1 - x2 = 21 - x2 2 - 31 - x2 = 2x2 + 3x
Notice the use of parentheses here.
(e) - f1x2 = - 12x2 - 3x2 = - 2x2 + 3x (f) Substitute 3x for x in the equation for f and simplify. f 13x2 = 213x2 2 - 313x2 = 219x2 2 - 9x = 1x2 - 9x (g) Substitute x + 3 for x in the equation for f and simplify. f1x + 32 = 21x + 32 2 - 31x + 32 = 21x2 + x + 92 - 3x - 9 = 2x2 + 12x + 1 - 3x - 9 = 2x2 + 9x + 9 (h)
f1x + h2 - f1x2 3 21x + h2 2 - 31x + h2 4 - 3 2x2 - 3x4 = h h c f(x + h) = 2(x + h)2 - 3(x + h)
21x2 + 2xh + h2 2 - 3x - 3h - 2x2 + 3x Simplify. h 2x2 + xh + 2h2 - 3h - 2x2 Distribute and combine = like terms. h =
xh + 2h2 - 3h h h(x + 2h - 3) = h =
= x + 2h - 3
Combine like terms. Factor out h.
Divide out the h’s.
r
Notice in this example that f 1x + 32 ≠ f 1x2 + f132 , f 1 - x2 ≠ - f1x2 , and 3f1x2 ≠ f13x2. The expression in part (h) is called the difference quotient of f, an important expression in calculus.
Now Work
PROBLEMS
39
AND
75
Most calculators have special keys that allow you to find the value of certain commonly used functions. For example, you should be able to find the square function f1x2 = x2, the square root function f 1x2 = 1x, the reciprocal function 1 f1x2 = = x -1, and many others that will be discussed later in this text (such as x ln x and log x 7FSJGZUIFSFTVMUTPG&YBNQMF XIJDIGPMMPXT POZPVSDBMDVMBUPS
EX A MPL E 7
Finding Values of a Function on a Calculator (a) f1x2 = x2 1 (b) F1x2 = x (c) g1x2 = 1x
f11.232 = 1.232 = 1.5225 1 F11.232 = ≈ .13215 1.23 g11.232 = 11.23 ≈ 1.115552
r
SECTION 1.1 Functions
81
COMMENT Graphing calculators can be used to evaluate any function. Figure 11 shows the SFTVMUPCUBJOFEJO&YBNQMF B POB5*1MVTHSBQIJOHDBMDVMBUPSXJUIUIFGVODUJPOUPCF evaluated, f(x) = 2x2 - 3x, in Y1. Figure 11
■
COMMENT The explicit form of a function is the form required by a graphing calculator. ■
Implicit 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 = f1x2 and say that the function is given explicitly. For example, Implicit Form 3x + y = 5 x2 - y = xy =
Explicit Form y = f 1x2 = - 3x + 5 y = f 1x2 = x2 - y = f 1x2 = x
SUMMARY Important Facts about Functions (a) For each x in the domain of a function f, there is exactly one image f1x2 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 (rule) that we use to get from an x in the domain to f1x2 in the range. (c) If y = f1x2, 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 Domain of a Function Defined by an Equation Often the domain of a function f is not specified; instead, only the equation defining 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 1x2 is a real number. The domain of a function f is the same as the domain of the variable x in the expression f1x2.
EXAM PL E 8
Finding the Domain of a Function Find the domain of each of the following functions.
Solution
(a) f1x2 = x2 + 5x
(b) g1x2 =
3x x -
(c) h 1t2 = 2 - 3t
(d) F1x2 =
23x + 12 x - 5
2
(a) The function says to square a number and then add five times the number. Since these operations can be performed on any real number, the domain of f is the set of all real numbers. (b) The function g says to divide 3x by x2 - .4JODFEJWJTJPOCZJTOPUEFàOFE UIF denominator x2 - DBOOFWFSCF TPx can never equal - 2 or 2. The domain of the function g is 5 x 0 x ≠ - 2, x ≠ 26 .
82
CHAPTER 1 Functions and Their Graphs
In Words The domain of g found in Example 8(b) is {x 0 x ≠ - 2, x ≠ 2}. This notation is read, “The domain of the function g is the set of all real numbers x such that x does not equal - 2 and x does not equal 2.”
(c) The function h says to take the square root of - 3t. But only nonnegative numbers have real square roots, so the expression under the square root (the radicand) must be nonnegative (greater than or equal to zero). This requires that - 3t Ú - 3t Ú - t … 3 The domain of h is e t ` t …
f or the interval a - q , d . 3 3
(d) The function F says to take the square root of 3x + 12 and divide this result by x - 5. This requires that 3x + 12 Ú , so x Ú - , and that x - 5 ≠ , so x ≠ 5. Combining these two restrictions, the domain of F is {x 0 x Ú - , x ≠ 5}.
r
The following steps may prove helpful for finding the domain of a function that is defined by an equation and whose domain is a subset of the real numbers.
Finding the Domain of a Function Defined by an Equation 1. Start with the domain as the set of real numbers. 2. If the equation has a denominator, exclude any numbers that give a zero denominator. 3. If the equation has a radical of even index, exclude any numbers that cause the expression inside the radical (the radicand) to be negative. That is, solve radicand Ú .
Now Work
PROBLEM
51
If x is in the domain of a function f, we shall say that f is defined at x, or f 1 x2 exists. If x is not in the domain of f, we say that f is not defined at x, or f 1 x2 does x not exist. For example, if f1x2 = 2 , then f12 exists, but f112 and f1 - 12 do x - 1 not exist. (Do you see why?) We will say more about finding the range when we look at the graph of a function in the next section. When a function is defined by an equation, it can be 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 functions are used in applications, the domain may be restricted by physical or geometric considerations. For example, the domain of the function f defined by f1x2 = 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 UPUIFQPTJUJWFSFBMOVNCFST TJODFUIFMFOHUIPGBTJEFDBOOFWFSCF or negative.
EX A MPL E 9
Finding the Domain in an Application Express the area of a circle as a function of its radius. Find the domain.
Solution Figure 12 A
r
See Figure 12. The formula for the area A of a circle of radius r is A = pr 2. Using r to represent the independent variable and A to represent the dependent variable, the function expressing this relationship is A(r) = pr 2
In this setting, the domain is 5 r 0 r 7 6 . (Do you see why?)
r
SECTION 1.1 Functions
83
Observe, in the solution to Example 9, that the symbol A is used 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
PROBLEM
89
4 Form the Sum, Difference, Product, and Quotient of Two Functions Next we introduce some operations on functions. Functions, like numbers, can be added, subtracted, multiplied, and divided. For example, if f1x2 = x2 + 9 and g1x2 = 3x + 5, then f 1x2 + g1x2 = 1x2 + 92 + 13x + 52 = x2 + 3x + 1
The new function y = x2 + 3x + 1 is called the sum function f + g. Similarly, f 1x2 # g1x2 = 1x2 + 92 13x + 52 = 3x3 + 5x2 + 2x + 5
The new function y = 3x3 + 5x2 + 2x + 5 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 1 f + g2 1x2 = f1x2 + g1x2
In Words
Remember, the symbol x stands for intersection. It means you should find the elements that are common to two sets.
DEFINITION
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 ∩ domain of g.
The difference f − g is the function defined by 1f - g2 1x2 = f1x2 - g1x2
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 ∩ domain of g.
DEFINITION
The product f ~ g is the function defined by 1f # g2 1x2 = f 1x2 # g1x2
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 ∩ domain of g.
DEFINITION
The quotient
f is the function defined by g f 1x2 f a b 1x2 = g g1x2
g1x2 ≠
84
CHAPTER 1 Functions and Their Graphs
f consists of the numbers x for which g1x2 ≠ and that are in the g domains of both f and g. That is, f domain of = 5 x 0 g(x) ≠ 6 ∩ domain of f ∩ domain of g g The domain of
EX AM PL E 10
Operations on Functions Let f and g be two functions defined as f 1x2 =
1 x + 2
and g1x2 =
x x - 1
Find the following, and determine the domain in each case. (a) 1f + g2 1x2
Solution
(b) 1f - g2 1x2
(c) 1f # g2 1x2
f (d) a b (x2 g
The domain of f is 5 x 0 x ≠ - 26 and the domain of g is 5 x 0 x ≠ 16 . (a) 1f + g2 1x2 = f 1x2 + g1x2 = =
1 x + x + 2 x - 1
x1x + 22 x - 1 x2 + 3x - 1 + = 1x + 22 1x - 12 1x + 22 1x - 12 1x + 22 1x - 12
The domain of f + g consists of those numbers x that are in the domains of both f and g. Therefore, the domain of f + g is 5 x 0 x ≠ - 2, x ≠ 16 . (b) 1f - g2 1x2 = f1x2 - g1x2 = =
1 x x + 2 x - 1
x1x + 22 - 1x2 + x + 12 x - 1 = 1x + 22 1x - 12 1x + 22 1x - 12 1x + 22 1x - 12
The domain of f - g consists of those numbers x that are in the domains of both f and g. Therefore, the domain of f - g is 5 x 0 x ≠ - 2, x ≠ 16 . (c) 1f # g2 1x2 = f1x2 # g1x2 =
x 1 # x = x + 2 x - 1 1x + 22 1x - 12
The domain of f # g consists of those numbers x that are in the domains of both f and g. Therefore, the domain of f # g is 5 x 0 x ≠ - 2, x ≠ 16 . 1 f 1x2 f x + 2 1 #x - 1 x - 1 (d) a b 1x2 = = = = g x x g1x2 x + 2 x1x + 22 x - 1 f consists of the numbers x for which g(x) ≠ and that are in g the domains of both f and g. Since g(x) = when x = ,XFFYDMVEFBTXFMM f as - 2 and 1 from the domain. The domain of is 5 x 0 x ≠ - 2, x ≠ , x ≠ 16 . g
The domain of
r
Now Work
PROBLEM
63
In calculus, it is sometimes helpful to view a complicated function as the sum, difference, product, or quotient of simpler functions. For example, F1x2 = x2 + 1x is the sum of f1x2 = x2 and g(x) = 1x. x2 - 1 H1x2 = 2 is the quotient of f1x2 = x2 - 1 and g1x2 = x2 + 1. x + 1
SECTION 1.1 Functions
85
SUMMARY A relation between two sets of real numbers so that each number x in the first set, the domain, has corresponding to it exactly one number y in the second set. A set of ordered pairs 1x, y2 or (x, f (x)) in which no first element is paired with two different 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 = f1x2. 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. y = f1x2 f is a symbol for the function. x is the independent variable or argument. y is the dependent variable. f1x2 is the value of the function at x, or the image of x.
Function
Unspecified domain
Function notation
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.
1. The inequality - 1 6 x 6 3 can be written in interval notation as ( - 1, 32 . QQ"m"
1 is 2. If x = - 2, the value of the expression 3x2 - 5x + x 21.5 . QQ"m"
x - 3 3. The domain of the variable in the expression x + is {x 0 x ≠ - } . Q"
4. Solve the inequality 3 - 2x 7 5. Graph the solution set. QQ"m" {x 0 x 6 - 1}
Concepts and Vocabulary 5. If f is a function defined by the equation y = f 1x2, then x is called the independent variable and y is the dependent variable. 6. The set of all images of the elements in the domain of a function is called the range .
7. If the domain of f is all real numbers in the interval 3, 4 and the domain of g is all real numbers in the interval 3 - 2, 54, then the domain of f + g is all real numbers in the interval [, 5] . f consists of numbers x for which 8. The domain of g g(x2 ≠ UIBUBSFJOUIFEPNBJOTPGCPUI f and g .
9. If f 1x2 = x + 1 = x3 - 1x + 12.
and
g(x) = x3,
then
(g - f )(x)
10. True or False Every relation is a function. False
11. True or False The domain of 1f # g2 1x2 consists of the numbers x that are in the domains of both f and g. True 12. True or False The independent variable is sometimes referred to as the argument of the function. True 13. True or False If no domain is specified for a function f, then the domain of f is taken to be the set of real numbers. False x2 - 14. True or False The domain of the function f 1x2 = is x 5 x 0 x ≠ {26. False
Skill Building In Problems 15–26, determine whether each relation represents a function. For each function, state the domain and range. 15.
Person Elvis
Birthday Jan. 8
Function; Domain: {Elvis, Colleen, Kaleigh, Marissa}; 3BOHF\+BO .BS 4FQU^
16.
Father Bob
Colleen
Daughter Beth Diane
Kaleigh
Mar. 15
John
Linda
Marissa
Sept. 17
Chuck
Marcia
Not a function
86
CHAPTER 1 Functions and Their Graphs
17. Hours Worked
Not a function
Salary
*18.
$200
20 Hours
$300 30 Hours
$350
40 Hours
$425
*19. 5 (2, ), ( - 3, ), (, 9), (2, 1) 6
Level of Education
Average Income
Less than 9th grade 9th–12th grade High School Graduate Some College College Graduate
$18,120 $23,251 $36,055
*20. 5( - 2, 5), ( - 1, 3), (3, ), (, 12) 6
*22. 5 (, - 2), (1, 3), (2, 3), (3, ) 6
*25. 5( - 2, ), ( - 1, 1), (, ), (1, 1) 6
*23. 5 ( - 2, ), ( - 2, ), (, 3), (3, ) 6
*26. 5 ( - 2, 1), ( - 1, ), (, 3), (1, ) 6
$45,810 $67,165
*21. 5 (1, 3), (2, 3), (3, 3), (, 3) 6
*24. 5( - , ), ( - 3, 3), ( - 2, 2), ( - 1, 1), ( - , ) 6
In Problems 27–38, determine whether the equation defines y as a function of x.
31. y2 = - x2
29. y =
32. y = { 21 - 2x
No
3x - 1 36. y = x + 2
35. y = 2x2 - 3x + Yes
30. y = 0 x 0
1 Yes x 33. x = y2 No
28. y = x3 Yes
27. y = x2 Yes
No
37. 2x2 + 3y2 = 1
Yes
Yes
34. x + y2 = 1
No
38. x2 - y2 = 1
No
No
In Problems 39– 46, find the following for each function: (a) f 12
(b) f 112
(c) f 1 - 12
(d) f 1 - x2
(e) - f 1x2
*39. f 1x2 = 3x2 + 2x -
*40. f 1x2 = - 2x2 + x - 1
*43. f 1x2 = 0 x 0 +
*44. f 1x2 = 2x2 + x
(f) f 1x + 12
*41. f 1x2 =
(g) f 12x2
(h) f 1x + h2 *42. f 1x2 =
x x + 1 2x + 1 *45. f 1x2 = 3x - 5 2
x2 - 1 x +
*46. f 1x2 = 1-
1 1x + 22 2
In Problems 47–62, find the domain of each function. *47. f 1x2 = - 5x + *51. g1x2 =
*48. f 1x2 = x2 + 2
x x - 1
*52. h1x2 =
2
55. h1x2 = 23x - 12 58. f 1x2 = 61. P 1t2 =
x 2x - 2t - 3t - 21
2x x - 2
{x 0 x Ú } 56. G1x2 = 21 - x
{x 0 x 7 }
59. p1x2 =
2 Ax - 1
x2 x *50. f 1x2 = x2 + 1 x2 + 1 x - 2 x + 53. F1x2 = 3 {x 0 x ≠ } *54. G1x2 = 3 x + x x - x 57. f 1x2 = {x 0 x 7 9} 2x - 9
*49. f 1x2 =
{x 0 x … 1} {x 0 x 7 1}
{t 0 t Ú , t ≠ }
60. q1x2 = 2- x - 2 62. h1z2 =
2z + 3 z - 2
{x 0 x … - 2}
{z 0 z Ú - 3, z ≠ 2}
In Problems 63–72, for the given functions f and g, find the following. For parts (a)–(d), also find the domain. (a) 1f + g2 1x2
(b) 1f - g2 1x2
(c) 1f # g2 1x2
(e) 1f + g2 132
(f) 1f - g2 12
(g) 1 f # g)(22
f (d) a b 1x2 g f (h) a b 112 g
*63. f 1x2 = 3x + ; g1x2 = 2x - 3
*64. f 1x2 = 2x + 1; g1x2 = 3x - 2
*67. f1x2 = 1x; g1x2 = 3x - 5
*68. f 1x2 = 0 x 0 ; g1x2 = x
*65. f 1x2 = x - 1; g1x2 = 2x2
1 1 *69. f 1x2 = 1 + ; g1x2 = x x 2x + 3 x *71. f 1x2 = ; g1x2 = 3x - 2 3x - 2 1 73. Given f 1x2 = 3x + 1 and 1f + g2 1x2 = - x, find the 2 function g. g1x2 = 5 - x 2
*66. f 1x2 = 2x2 + 3; g(x) = x3 + 1 *70. f 1x2 = 2x - 1; g1x2 = 2 - x *72. f 1x2 = 2x + 1; g1x2 =
2 x
f 1 x + 1 , find the and a b 1x2 = 2 x g x - x x - 1 function g. g(x) = x + 1
74. Given f 1x2 =
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
SECTION 1.1 Functions
In Problems 75– 82, find the difference quotient of f ; that is, find 75. f 1x2 = x + 3 *79. f 1x2 =
1 x2
76. f 1x2 = - 3x + 1 *80. f 1x2 =
f 1x + h2 - f 1x2
-3
1 x + 3
h
87
, h ≠ , for each function. Be sure to simplify.
*77. f 1x2 = x2 - x +
*78. f 1x2 = 3x2 - 2x +
*81. f 1x2 = 2x
*82. f 1x2 = 2x + 1
[Hint: Rationalize the numerator.]
Applications and Extensions 83. If f 1x2 = 2x3 + Ax2 + x - 5 and f 122 = 5, what is the value of A? - 2 84. If f 1x2 = 3x2 - Bx + and f 1 - 12 = 12, what is the value of B? 5 3x + and f 12 = 2, what is the value of A? - 85. If f 1x2 = 2x - A 2x - B 1 86. If f 1x2 = and f 122 = , what is the value of B? - 1 3x + 2 2x - A and f 12 = , what is the value of A? 87. If f 1x2 = x - 3 Where is f not defined? A = ; undefined at x = 3 x - B 88. If f 1x2 = , f 122 = , and f 112 is undefined, what x - A are the values of A and B? A = 1, B = 2
*(b) When is the height of the rock 15 meters? When is it NFUFST 8IFOJTJUNFUFST (c) When does the rock strike the ground? TFD 96. Effect of Gravity on Jupiter If a rock falls from a height of NFUFSTPOUIFQMBOFU+VQJUFS JUTIFJHIUH (in meters) after x seconds is approximately H1x2 = 2 - 13x2 *(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 NFUFST 8IFOJTJUNFUFST (c) When does the rock strike the ground? TFD
*89. 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. *90. Geometry Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides. 91. Constructing Functions Express the gross salary G of a QFSTPOXIPFBSOTQFSIPVSBTBGVODUJPOPGUIFOVNCFS x of hours worked. G(x) = 1x *92. Constructing Functions Tiffany, a commissioned sales QFSTPO FBSOTCBTFQBZQMVTQFSJUFNTPME&YQSFTT her gross salary G as a function of the number x of items sold. 93. Population as a Function of Age The function P(a) = .a2 - 3.92a + 31.9 represents the population P (in millions) of Americans that were aZFBSTPGBHFPSPMEFSJO Source: U.S. Census Bureau *(a) Identify the dependent and independent variables. *(b) Evaluate P 1SPWJEF B WFSCBM FYQMBOBUJPO PG UIF meaning of P *(c) Evaluate P 1SPWJEF B WFSCBM FYQMBOBUJPO PG UIF meaning of P 94. Number of Rooms The function N(r) = - 2.r 2 + 22.91r - 3. represents the number N of housing units (in millions) in UIBUIBEr rooms, where r is an integer and 2 … r … 9. Source: U.S. Census Bureau *(a) Identify the dependent and independent variables. *(b) Evaluate N 1SPWJEF B WFSCBM FYQMBOBUJPO PG UIF meaning of N(3). 95. Effect of Gravity on Earth If a rock falls from a height of NFUFSTPO&BSUI UIFIFJHIUH (in meters) after x seconds is approximately H1x2 = 2 - .9x2 *(a) What is the height of the rock when x = 1 second? x = 1.1 seconds? x = 1.2 seconds? x = 1.3 seconds?
97. Cost of Trans-Atlantic Travel " #PFJOH DSPTTFT UIF "UMBOUJD0DFBO NJMFT XJUIBOBJSTQFFEPGNJMFT per hour. The cost C (in dollars) per passenger is given by C 1x2 = 1 +
3, x + 1 x
where x is the ground speed (airspeed { wind). (a) What is the cost per passenger for quiescent (no wind) conditions? $222 C 8IBUJTUIFDPTUQFSQBTTFOHFSXJUIBIFBEXJOEPG miles per hour? $225 D 8IBUJTUIFDPTUQFSQBTTFOHFSXJUIBUBJMXJOEPG miles per hour? E 8IBUJTUIFDPTUQFSQBTTFOHFSXJUIBIFBEXJOEPG miles per hour? 98. Cross-sectional Area The cross-sectional area of a beam cut from a log with radius 1 foot is given by the function A1x2 = x21 - x2 , where x represents the length, in feet, of half the base of the beam. See the figure on the next page. 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 GU2 (b) One-half of a foot GU2 (c) Two-thirds of a foot 1.99 ft2
88
CHAPTER 1 Functions and Their Graphs
The cost C, in dollars, of selling x cell phones, in hundreds, is C 1x2 = .5x3 - 2x2 + 5x + 5. *(a) Find the profit function, P 1x2 = R 1x2 - C 1x2. *(b) Find the profit if x = 15 hundred cell phones are sold. *(c) Interpret P(15).
A(x ) ⫽ 4x 公1 ⫺ x 2 1
x
*99. Economics The participation rate is the number of people in the labor force divided by the civilian population (excludes military). Let L 1x2 represent the size of the labor force in year x and P 1x2 represent the civilian population in year x. Determine a function that represents the participation rate R as a function of x. 100. Crimes Suppose that V 1x2 represents the number of violent crimes committed in year x and P 1x2 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. T(x) = V 1x2 + P 1x2
101. Health Care Suppose that P 1x2 represents the percentage of income spent on health care in year x and I 1x2 represents income in year x. Determine a function H that represents total health care expenditures in year x. H(x) = P(x) # I(x)
102. Income Tax Suppose that I 1x2 represents the income of an individual in year x before taxes and T 1x2 represents the individual’s tax bill in year x. Determine a function N that represents the individual’s net income (income after taxes) in year x. N(x) = I(x) - T(x) 103. Profit Function Suppose that the revenue R, in dollars, from selling x cell phones, in hundreds, is R 1x2 = - 1.2x2 + 22x.
Discussion and Writing x2 - 1 *107. Are the functions f 1x2 = x - 1 and g1x2 = the x + 1 same? Explain. 108. Investigate when, historically, the use of the function notation y = f 1x2 first appeared.
‘Are You Prepared?’ Answers 1. 1 - 1, 32
2. 21.5
104. Profit Function Suppose that the revenue R, in dollars, from selling x clocks is R 1x2 = 3x. The cost C, in dollars, of selling x clocks is C 1x2 = .1x2 + x + . *(a) Find the profit function, P 1x2 = R 1x2 - C 1x2. (b) Find the profit if x = 3 clocks are sold. *(c) Interpret P 105. Stopping Distance When the driver of a vehicle observes an impediment, the total stopping distance involves both the reaction distance (the distance the vehicle travels while the driver moves his or her foot to the brake pedal) and the braking distance (the distance the vehicle travels once the brakes are applied). For a car traveling at a speed of v miles per hour, the reaction distance R, in feet, can be estimated by R 1v2 = 2.2v. Suppose that the braking distance B, in feet, for a car is given by B 1v2 = .5v2 + .v - 15. *(a) Find the stopping distance function D(v) = R(v) + B(v). (b) Find the stopping distance if the car is traveling at a TQFFEPGNQI 321 feet *(c) Interpret D
106. Some functions f have the property that f 1a + b2 = f 1a2 + f 1b2 for all real numbers a and b. Which of the following functions have this property? Only h(x) = 2x (a) h1x2 = 2x (b) g1x2 = x2 1 (c) F 1x2 = 5x - 2 (d) G1x2 = x 109. Find a function H that multiplies a number x by 3 and then subtracts the cube of x and divides the result by your age. 3x - x3 H(x) = age
3. 5 x 0 x ≠ - 6
4. 5x 0 x 6 - 16
⫺1
0
1
1.2 The Graph of a Function PREPARING FOR THIS SECTION Before getting started, review the following: r (SBQITPG&RVBUJPOT 'PVOEBUJPOT4FDUJPO QQm
r *OUFSDFQUT 'PVOEBUJPOT4FDUJPO QQm
Now Work the ‘Are You Prepared?’ problems on page 93.
OBJECTIVES 1 Identify the Graph of a Function (p. 89) 2 Obtain Information from or about the Graph of a Function (p. 90)
In applications, a graph often demonstrates more clearly the relationship between two variables than, say, an equation or table would. For example, Table 1 on the next page shows the average price of gasoline at a particular gas station in Texas (for the ZFBSTmBEKVTUFEGPSJOGMBUJPO CBTFEPOEPMMBST *GXFQMPUUIFTFEBUB and then connect the points, we obtain Figure 13 (also on the next page).
SECTION 1.2 The Graph of a Function
Table 1
Year
Price
Year
Price
Year
Price
1983
3.10
1993
1.93
2003
2.13
1984
2.79
1994
1.94
2004
2.36
1985
2.64
1995
1.98
2005
2.92
1986
2.00
1996
2.02
2006
3.13
1987
2.12
1997
1.94
2007
3.31
1988
2.00
1998
1.70
2008
3.61
1989
2.01
1999
1.83
2009
2.70
1990
2.25
2000
2.19
2010
3.08
1991
2.07
2001
2.02
2011
3.78
1992
2.02
2002
1.99
2012
3.84
89
Source: http://www.randomuseless.info/gasprice/gasprice.html
Figure 13 Average retail price of gasoline (2012 dollars) 4.50 Price (dollars per gallon)
4.00 3.50 3.00 2.50 2.00 1.50 1.00
2012
2010
2008
2006
2004
2002
2000
1998
1996
1994
1992
1990
1988
1986
0.00
1984
0.50
Source: http://www.randomuseless.info/gasprice/gasprice.html
We can see from the graph that the price of gasoline (adjusted for inflation) fell GSPNUP TUBZFESPVHIMZUIFTBNFGSPNUP BOESPTFSBQJEMZGSPN UP5IFHSBQIBMTPTIPXTUIBUUIFMPXFTUQSJDFPDDVSSFEJO5PMFBSO 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, it is the set of points 1x, y2 in the xy-plane that satisfy the equation.
1 Identify the Graph of a Function In Words If any vertical line intersects a graph at more than one point, the graph is not the graph of a function.
THEOREM
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 function must satisfy the following vertical-line test.
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.
EXAM PL E 1
Identifying the Graph of a Function 8IJDIPGUIFHSBQITJO'JHVSFPOUIFOFYUQBHFBSFHSBQITPGGVODUJPOT
90
CHAPTER 1 Functions and Their Graphs y 6
Figure 14
y 4
y
y 3
1 (1, 1)
4 3
3x (a) y x 2
Solution
4x 4
(1, 1)
6 x
1 1
3
(b) y x 3
1 x
(c) x y 2
(d) x 2 y 2 1
5IFHSBQITJO'JHVSFT B BOE C BSFHSBQITPGGVODUJPOT CFDBVTFFWFSZWFSUJDBM MJOFJOUFSTFDUTFBDIHSBQIJOBUNPTUPOFQPJOU5IFHSBQITJO'JHVSFT D BOE E are not graphs of functions, because there is a vertical line that intersects each graph JONPSFUIBOPOFQPJOU/PUJDFJO'JHVSF D UIBUUIFJOQVUDPSSFTQPOETUPUXP outputs, - 1 and 1. This is why the graph does not represent a function.
Now Work
r
PROBLEM
15
2 Obtain Information from or about the Graph of a Function
If 1x, y2 is a point on the graph of a function f, then y is the value of f at x; that is, y = f1x2 . Also if y = f 1x2, then 1x, y2 is a point on the graph of f. For example, if 1 - 2, 2 is on the graph of f, then f1 - 22 = , and if f152 = , then the point 15, 2 is on the graph of y = f1x2 . The next example illustrates how to obtain information about a function if its graph is given.
Obtaining Information from the Graph of a Function
EX A MPL E 2 Figure 15 y 4 2
⫺2 ⫺4
(4, 4)
(2, 4)
(0, 4)
(5––2, 0)
(––2 , 0)
(7––2, 0)
(3––2, 0) (, ⫺4)
x
(3, ⫺4)
Solution
Let f be the function whose graph is given in Figure 15. (The graph of f might represent the distance y that the bob of a pendulum is from its at-rest position at time x. Negative values of y mean that the pendulum is to the left of the at-rest position, and positive values of y mean that the pendulum is to the right of the at-rest position.) 3p (a) What are f 12, f a b , and f13p2? 2 (b) What is 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 many times does the line y = 2 intersect the graph? (f) For what values of x does f1x2 = - ? (g) For what values of x is f1x2 7 ? (a) Since 1, 2 is on the graph of f, the yDPPSEJOBUF JT UIF WBMVF PG f at the 3p xDPPSEJOBUF UIBUJT f12 = . In a similar way, when x = , then y = , so 2 3p f a b = . When x = 3p, then y = - , so f13p2 = - . 2 (b) To determine the domain of f, notice that the points on the graph of f have xDPPSEJOBUFTCFUXFFOBOEp, inclusive; and for each number xCFUXFFOBOE p, there is a point (x, f(x)) on the graph. The domain of f is 5 x 0 … x … p6 or the interval 3 , p4 . (c) The points on the graph all have y-coordinates between - BOE JODMVTJWF and for each such number y, there is at least one corresponding number x in the domain. The range of f is 5 y 0 - … y … 6 or the interval 3 - , 4 . (d) The intercepts are the points p 3p 5p (, ), a , b , a , b , a , b , and 2 2 2
a
p , b 2
SECTION 1.2 The Graph of a Function
91
(e) Draw the horizontal line y = 2 on the graph in Figure 15. Notice that the line intersects the graph four times. (f) Since 1p, - 2 and 13p, - 2 are the only points on the graph for which y = f1x2 = - , we have f 1x2 = - when x = p and x = 3p. (g) To determine where f1x2 7 , look at Figure 15 and determine the xWBMVFT GSPN UP p for which the y-coordinate is positive. This occurs on 3p 5p p p b ∪ a , b ∪ a , p d . Using inequality notation, f1x2 7 for 2 2 2 2 p 3p 5p p … x 6 or 6 x 6 or 6 x … p. 2 2 2 2 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. c ,
r
The Zeros of a Function If f1r2 = for a real number r, then r is called a real zero of f . For example, the zeros of the function f1x2 = x2 - are - 2 and 2 since f1 - 22 = and f122 = . We can identify the zeros of a function from its graph. To see how, we must remember that y = f1x2 means that the point 1x, y2 is on the graph of f . So, if r is a zero of f , then f1r2 = , which means the point 1r, 2 is on the graph of f . That is, r is an x@intercept. We conclude that the x@intercepts of the graph of a function are also the zeros of the function.
EXAM PL E 3
Finding the Zeros of a Function from Its Graph Find the zeros of the function f whose graph is shown in Figure 15.
Solution
The zeros of a function are the x@intercepts of the graph of the function. Because the p 3p 5p p x@intercepts of the function shown in Figure 15 are , , , and , the zeros of 2 2 2 2 p 3p 5p p f are , , , and . 2 2 2 2
Now Work
EXAM PL E 4
PROBLEMS
9
AND
13
r
Obtaining Information about the Graph of a Function Consider the function: f1x2 =
x + 1 x + 2
(a) Find the domain of f. 1 (b) Is the point a1, b on the graph of f ? 2 (c) If x = 2, what is f1x2 ? What point is on the graph of f ? (d) If f1x2 = 2, what is x? What point is on the graph of f ? (e) What are the x-intercepts of the graph of f (if any)? What point(s) are on the graph of f ?
Solution
(a) The domain of f is {x 0 x ≠ - 2}, since x = - 2SFTVMUTJOEJWJTJPOCZ (b) When x = 1, f112 =
1 + 1 2 = 1 + 2 3
2 1 The point a1, b is on the graph of f; the point a1, b is not. 3 2 (c) If x = 2, then 2 + 1 3 f122 = = 2 + 2 3 The point ¢ 2, ≤ is on the graph of f.
92
CHAPTER 1 Functions and Their Graphs
(d) If f1x2 = 2, then x x x x
+ + + +
1 2 1 1 x
= 2 = 2(x + 22 = 2x + = -3
Multiply both sides by x + 2. Distribute. Solve for x.
If f1x2 = 2, then x = - 3. The point 1 - 3, 22 is on the graph of f. (e) The x-intercepts of the graph of f are the real solutions of the equation f1x2 = that are in the domain of f. x + 1 = x + 2 x + 1 = x = -1
Multiply both sides by x + 2. Solve for x.
x + 1 = is x = - 1, so - 1 is the x + 2 only x-intercept. Since f 1 - 12 = , the point ( - 1 JTPOUIFHSBQIPGf.
The only real solution of the equation f1x2 =
Now Work
EX A MPL E 5
r
PROBLEM
25
Average Cost Function The average cost C (per computer) of manufacturing x computers per day is given by the function C(x) = .5x2 - 3.39x + 1212.5 +
2, x
Determine the average cost of manufacturing: B DPNQVUFSTJOBEBZ C DPNQVUFSTJOBEBZ D DPNQVUFSTJOBEBZ (d) Graph the function C = C(x), 6 x … . (e) Create a TABLE with TblStart = 1 and ∆Tbl = 1. Which value of x minimizes the average cost?
Solution
(a) The average cost of manufacturing x = 3 computers is 2, = $1351.5 3 (b) The average cost of manufacturing x = computers is C(3) = .5(3)2 - 3.39(3) + 1212.5 +
2, = $1232.9 (c) The average cost of manufacturing x = 5 computers is C() = .5()2 - 3.39() + 1212.5 +
C(5) = .5(5)2 - 3.39(5) + 1212.5 +
2, = $1293. 5
(d) 4FF'JHVSFGPSUIFHSBQIPGC = C 1x2. (e) With the function C = C 1x2 in Y1 , we create Table 2. We scroll down until we find a value of x for which Y1 is smallest. Table 3 shows that manufacturing x = 1DPNQVUFSTNJOJNJ[FTUIFBWFSBHFDPTUBUQFSDPNQVUFS Figure 16
Table 2
4000
0
80 0
Now Work
PROBLEM
31
Table 3
r
SECTION 1.2 The Graph of a Function
93
SUMMARY Graph of a Function Vertical-Line Test
The collection of points 1x, y2 that satisfies the equation y = f1x2. A collection of points is the graph of a function provided that every vertical line intersects the graph in at most one point.
1.2 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. The intercepts of the equation x2 + y2 = 1 ( - , ), (, ), (, - 2), (, 22 . Q
2. True or False The point 1 - 2, - 2 is on the graph of the equation x = 2y - 2. QQm False
are
Concepts and Vocabulary 3. 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.
6. True or False A function can have more than one yintercept. False
5. Find a such that the point 1 - 1, 22 is on the graph of f 1x2 = ax2 + . a = - 2
8. True or False The y-intercept of the graph of the function y = f 1x2, whose domain is all real numbers, is f 12. True
7. True or False The graph of a function y = f 1x2 always crosses the y-axis. False
4. If the point 15, - 32 is a point on the graph of f, then f( 5 ) = -3 .
Skill Building 9. Use the given graph of the function f to answer parts B m P y (0, 3) 4
10. Use the given graph of the function f to answer parts B m P y
(2, 4) 4
(4, 3)
(5, 3)
(–4, 2)
(10, 0) (11, 1)
(–2, 1) 2
(–3, 0)
(4, 0) 5
–5
(6, 0)
(–5, –2) (–6, –3)
–3
11 x
–2 (0, 0) –2
(8, – 2)
Find f 12 and f 1 - 2. f 12 = 3; f 1 - 2 = - 3 Find f 12 and f 1112. f 12 = ; f 1112 = 1 Is f 132 positive or negative? 1PTJUJWF Is f 1 - 2 positive or negative? Negative For what values of x is f 1x2 = ? - 3, , and 1 For what values of x is f 1x2 7 ? - 3 6 x 6 ; 1 6 x … 11 What is the domain of f ? 5x 0 - … x … 116 What is the range of f ? 5y 0 - 3 … y … 6 What are the x-intercepts? - 3, , and 1 What is the y-intercept? 3 1 How many times does the line y = intersect the 2 graph? Three times (l) How many times does the line x = 5 intersect the graph? Once (m) For what values of x does f 1x2 = 3? BOE (n) For what values of x does f 1x2 = - 2? - 5 and (o) What are the zeros of f? ¢
(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k)
–4
2
4
(6, 0) 6
x
(2, –2)
Find f 12 and f 12. f 12 = ; f 12 = Find f 122 and f 1 - 22. f 122 = - 2; f 1 - 22 = 1 Is f 132 positive or negative? Negative Is f 1 - 12 positive or negative? 1PTJUJWF For what values of x is f 1x2 = ? BOE For what values of x is f 1x2 6 ? 6 x 6 What is the domain of f ? 5 x 0 - … x … 6 What is the range of ƒ? 5y 0 - 2 … y … 36 What are the x-intercepts? BOE What is the y-intercept? How many times does the line y = - 1 intersect the graph? Twice (l) How many times does the line x = 1 intersect the graph? Once (m) For what value of x does f 1x2 = 3? 5 (n) For what value of x does f 1x2 = - 2? 2 (o) What are the zeros of f?
(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k)
94
CHAPTER 1 Functions and Their Graphs
In Problems 11–22, determine whether the graph is that of a function by using the vertical-line test. If it is, use the graph to find: (a) The domain and range *11.
(b) The intercepts, if any *12.
y 3
⫺3
*16.
y 3
⫺3
*19.
y (⫺1, 2) 3
*20.
3 ⫺3
y 3
1 –– 2
⫺
x
*21.
y 3
*18.
–– 2
y 4
(3, 2)
3x
⫺3
x
4x 4
*22.
y
y
9
3x
(4, 3)
4
(1–2 , 5)
4
6 ⫺3
⫺ –– 2 ⫺1
3
⫺3
(1, 2)
⫺–– 2 ⫺1
3
3x
x ⫺3
⫺
*17.
y 3
⫺3
y
1
3x
⫺3
3x
*14.
y
⫺3
⫺3
*15.
*13.
y 3
⫺3
3x
(c) Any symmetry with respect to the x-axis, the y-axis, or the origin
3 ⫺1 ⫺3
3 x (2, ⫺3)
⫺3
3 x
In Problems 23–28, answer the questions about the given function. 23. f 1x2 = 2x2 - x - 1 (a) Is the point 1 - 1, 22 on the graph of ƒ? Yes (b) If x = - 2, what is f 1x2? What point is on the graph of ƒ? f ( - 2) = 9; ( - 2, 9) *(c) If f 1x2 = - 1, what is x? What point(s) is (are) on the graph of ƒ? 1 (d) What is the domain of ƒ? All real numbers - ,1 (e) List the x-intercepts, if any, of the graph of f. 2 (f) List the y-intercept, if there is one, of the graph of f. - 1 1 (g) What are the zeros of f ? - , 1 2 24. f 1x2 = - 3x2 + 5x (a) Is the point 1 - 1, 22 on the graph of ƒ? No (b) If x = - 2, what is f 1x2? What point is on the graph of ƒ? f ( - 2) = - 22; ( - 2, - 22) *(c) If f 1x2 = - 2, what is x? What point(s) is (are) on the graph of ƒ? 5 (d) What is the domain of ƒ? All real numbers , (e) List the x-intercepts, if any, of the graph of f. 3 (f) List the y-intercept, if there is one, of the graph of f. 5 (g) What are the zeros of f ? , 3 x + 2 25. f 1x2 = x - (a) Is the point 13, 12 on the graph of ƒ? No *(b) If x = , what is f 1x2? What point is on the graph of ƒ? (c) If f 1x2 = 2, what is x? What point(s) is (are) on the graph of ƒ? 1; (1, 22 (d) What is the domain of ƒ? {x x ≠ } (e) List the x-intercepts, if any, of the graph of f. - 2 1 (f) List the y-intercept, if there is one, of the graph of f. 3 (g) What are the zeros of f ? - 2
x2 + 2 x + 3 (a) Is the point a1, b on the graph of ƒ? Yes 5 *(b) If x = , what is f 1x2? What point is on the graph of ƒ? 1 *(c) If f 1x2 = , what is x? What point(s) is (are) on the graph of ƒ? 2
26. f 1x2 =
What is the domain of ƒ? {x 0 x ≠ - } List the x-intercepts, if any, of the graph of f. None 1 List the y-intercept, if there is one, of the graph of f. 2 What are the zeros of f ? None 2x2 27. f 1x2 = x + 1 (a) Is the point 1 - 1, 12 on the graph of ƒ? Yes *(b) If x = 2, what is f 1x2? What point is on the graph of ƒ? (c) If f 1x2 = 1, what is x? What point(s) is (are) on the graph of ƒ? - 1, 1; ( - 1, 1), (1, 12 (d) What is the domain of ƒ? All real numbers (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. (g) What are the zeros of f ? 2x 28. f 1x2 = x - 2 1 2 (a) Is the point a , - b on the graph of ƒ? Yes 2 3 *(b) If x = , what is f 1x2? What point is on the graph of ƒ? (c) If f 1x2 = 1, what is x? What point(s) is (are) on the graph of ƒ? - 2; ( - 2, 12 (d) What is the domain of ƒ? {x 0 x ≠ 2} (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. (g) What are the zeros of f ? (d) (e) (f) (g)
* Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
SECTION 1.2 The Graph of a Function
95
Applications and Extensions 29. Free-throw Shots "DDPSEJOH UP QIZTJDJTU 1FUFS #SBODB[JP the key to a successful foul shot in basketball lies in the arc of the shot. Brancazio determined the optimal angle of UIF BSD GSPN UIF GSFFUISPX MJOF UP CF EFHSFFT5IF BSD also depends on the velocity with which the ball is shot. If BQMBZFSTIPPUTBGPVMTIPU SFMFBTJOHUIFCBMMBUBEFHSFF BOHMFGSPNBQPTJUJPOGFFUBCPWFUIFGMPPS UIFOUIFQBUIPG the ball can be modeled by the function h(x) = -
where x is the horizontal distance that the golf ball has traveled.
x2 + x + v2
where h is the height of the ball above the floor, x is the 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. Suppose a player shoots a ball with an initial velocity PGGFFUQFSTFDPOE (a) Determine the height of the ball after it has traveled GFFUJOGSPOUPGUIFGPVMMJOF ≈1. ft *(b) Determine h(12). What does this value represent? ≈9.9 ft *(c) Find additional points, and graph the path of the basketball. * E 5IF DFOUFS PG UIF IPPQ JT GFFU BCPWF UIF GMPPS BOE 15 feet in front of the foul line. Will the ball go through the hoop? Why or why not? If not, with what initial velocity must the ball be shot in order for the ball to go through the hoop? Source: The Physics of Foul Shots, Discover, Vol. 21, No. 10, October 2000 30. Granny Shots The last player in the NBA to use an underhand foul shot (a “granny” shot) was Hall of Fame forward Rick #BSSZ XIP SFUJSFE JO #BSSZ CFMJFWFT UIBU DVSSFOU 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 GPVM TIPU SFMFBTJOH UIF CBMM BU B EFHSFF BOHMF GSPN B position 3.5 feet above the floor, then the path of the ball can 13x2 be modeled by the function h(x) = - 2 + 2.x + 3.5, v where h is the height of the ball above the floor, x is the 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. B 5IF DFOUFS PG UIF IPPQ JT GFFU BCPWF UIF GMPPS BOE 15 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. GUTFD *(b) Write the function for the path of the ball using the velocity found in part (a). *(c) Determine h(9). What does this value represent? GU *(d) Find additional points, and graph the path of the basketball. Source: The Physics of Foul Shots, Discover, Vol. 21, No. 10, October 2000 31. Motion of a Golf Ball A golf ball is hit with an initial WFMPDJUZPGGFFUQFSTFDPOEBUBOJODMJOBUJPOPGUPUIF horizontal. In physics, it is established that the height h of the golf ball is given by the function h1x2 =
- 32x2 + x 132
(a) Determine the height of the golf ball after it has traveled GFFU ≈1. ft * C 8IBUJTUIFIFJHIUBGUFSJUIBTUSBWFMFEGFFU *(c) What is h *OUFSQSFUUIJTWBMVF ≈2.3 (d) How far was the golf ball hit? ≈52.13 ft *(e) Use a graphing utility to graph the function h = h(x2. *(f) Use a graphing utility to determine the distance that the CBMMIBTUSBWFMFEXIFOUIFIFJHIUPGUIFCBMMJTGFFU (g) Create a TABLE with TblStart = and ∆Tbl = 25. To the nearest 25 feet, how far does the ball travel before it reaches a maximum height? What is the maximum height? 25 ft; 131. ft (h) Adjust the value of ∆Tbl until you determine the distance, to within 1 foot, that the ball travels before it reaches the maximum height. GU 32. 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) = x21 - x2 , where x represents the length, in feet, of half the base of the beam. See the figure. A(x ) ⫽ 4x 公1 ⫺ x 2 1
x
(a) Find the domain of A. 5x 0 6 x 6 16 *(b) Use a graphing utility to graph the function A = A(x2. (c) Create a TABLE with TblStart = and ∆Tbl = .1. Which value of x in the domain found in part (a) maximizes the cross-sectional area? What should be the length of the base of the beam to maximize the crosssectional area? GU GU 33. Cost of Trans-Atlantic Travel " #PFJOH DSPTTFT UIF "UMBOUJD0DFBO NJMFT XJUIBOBJSTQFFEPGNJMFT per hour. The cost C (in dollars) per passenger is given by C 1x2 = 1 +
3, x + 1 x
CHAPTER 1 Functions and Their Graphs
where x is the ground speed 1airspeed { wind). B 8IBUJTUIFDPTUXIFOUIFHSPVOETQFFEJTNJMFTQFS IPVS NJMFTQFSIPVS (b) Find the domain of C. {x 0 x 7 } *(c) Use a graphing utility to graph the function C = C(x). *(d) Create a TABLE with TblStart = and ∆Tbl = 5. F 5P UIF OFBSFTU NJMFT QFS IPVS XIBU HSPVOE TQFFE minimizes the cost per passenger? NJIS 34. Effect of Elevation on Weight If an object weighs m pounds at sea level, then its weight W (in pounds) at a height of h miles above sea level is given approximately by 2 W(h) = ma b + h B *G"NZXFJHITQPVOETBUTFBMFWFM IPXNVDIXJMM TIFXFJHIPO1JLFT1FBL XIJDIJT GFFUBCPWFTFB level? ≈119. lb *(b) Use a graphing utility to graph the function W = W(h2. Use m = 12 pounds. *(c) Create a TABLE with TblStart = and ∆Tbl = .5 to see how the weight W varies as hDIBOHFTGSPNUPNJMFT *(d) At what height will Amy weigh 119.95 pounds? (e) Does your answer to part (d) seem reasonable? Explain. 35. The graph of two functions, f and g, is illustrated. Use the HSBQIUPBOTXFSQBSUT B m G y
C (100, 280000) 250,000
Cost (dollars per day)
96
150,000
100,000
50,000
(50, 51000) (30, 32000)
(0, 5000)
(10, 19000) 10
20
30
40 50 60 70 80 Number of computers
90 100 q
37. Reading and Interpreting Graphs Let C be the function whose graph is given below. This graph represents the cost C of using m anytime cell phone minutes in a month for a five-person family plan.
yg(x) C 2000
(2, 2) (4, 1) 2 2
(6, 1) (6, 0) x yf(x)
4
(3, 2)
(5, 2)
(4, 3)
4
(a) 1 f + g)(22 (c) 1 f - g)(2
3 -1 2
(14400, 1822)
(b) 1 f + g)(2 - 2 (d) 1g - ƒ2 12 1 f 1 (f) a b(2 g 3
36. Reading and Interpreting Graphs Let C be the function whose graph is given in the next column. This graph represents the cost C of manufacturing q computers in a day. *(a) Determine C *OUFSQSFUUIJTWBMVF *(b) Determine C *OUFSQSFUUIJTWBMVF *(c) Determine C *OUFSQSFUUIJTWBMVF *(d) What is the domain of C? What does this domain imply in terms of daily production? 5q 0 … q … 16 (e) Describe the shape of the graph. (f) 5IFQPJOU JTDBMMFEBOinflection point. Describe the behavior of the graph around the inflection point.
1500 Cost (dollars)
2 (2, 1)
(e) 1 f # g)(22
200,000
1000
500 (2000, 210) (0, 80)
(1000, 80) 5000
10000 Number of minutes
15000 m
Determine C *OUFSQSFUUIJTWBMVF Determine C *OUFSQSFUUIJTWBMVF Determine C *OUFSQSFUUIJTWBMVF What is the domain of C? What does this domain imply in terms of the number of anytime minutes? (e) Describe the shape of the graph.
*(a) *(b) *(c) *(d)
Discussion and Writing 38. 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? *39. How many x-intercepts can the graph of a function have? How many y-intercepts can the graph of a function have? *40. Is a graph that consists of a single point the graph of a function? Can you write the equation of such a function? 41. Match each of the following functions with the graph on the next page that best describes the situation. (a) The cost of building a house as a function of its square footage III C 5IFIFJHIUPGBOFHHESPQQFEGSPNBGPPUCVJMEJOHBTBGVODUJPOPGUJNF IV (c) The height of a human as a function of time I (d) The demand for Big Macs as a function of price V (e) The height of a child on a swing as a function of time? II
SECTION 1.2 The Graph of a Function y
y
y
y
x
x
y
x
x
x
(III)
(II)
(I)
97
(V)
(IV)
42. Match each of the following functions with the graph that best describes the situation. (a) The temperature of a bowl of soup as a function of time II (b) The number of hours of daylight per day over a 2-year period V (c) The population of Texas as a function of time IV (d) The distance traveled by a car going at a constant velocity as a function of time III F 5IFIFJHIUPGBHPMGCBMMIJUXJUIBJSPOBTBGVODUJPOPGUJNF I y
y
y
x
y
x
x
(II)
(I)
*44. Consider the following scenario: Jayne enjoys riding her bicycle through the woods. At the forest preserve, she gets POIFSCJDZDMFBOESJEFTVQBGPPUJODMJOFJONJOVUFT She then travels down the incline in 3 minutes. The next GFFUJTMFWFMUFSSBJOBOETIFDPWFSTUIFEJTUBODFJO NJOVUFT4IFSFTUTGPSNJOVUFT+BZOFUIFOUSBWFMT GFFUJONJOVUFT%SBXBHSBQIPG+BZOFTEJTUBODFUSBWFMFE (in feet) as a function of time. 45. The following sketch represents the distance d (in miles) that Kevin was from home as a function of time t (in hours). Answer the questions by referring to the graph. In parts B m H
IPXNBOZIPVSTFMBQTFEBOEIPXGBSXBT,FWJO from home during this time? d (t ) (2.5, 3)
(2.8, 0)
(3.9, 2.8)
(3, 0)
(4.2, 2.8)
(5.3, 0)
t
x (V)
*(d) From t = 2. to t = 3 *(e) From t = 3 to t = 3.9 *(f) From t = 3.9 to t = .2 *(g) From t = .2 to t = 5.3 *(h) What is the farthest distance that Kevin was from home? (i) How many times did Kevin return home? Twice 46. The following sketch represents the speed v (in miles per hour) of Michael’s car as a function of time t (in minutes). v (t ) (7, 50)
(2, 30)
(8, 38)
(4, 30)
(4.2, 0)
(7.4, 50)
(7.6, 38)
(6, 0)
(9.1, 0)
t
*(a) Over what interval of time was Michael traveling fastest? *(b) Over what interval(s) of time was Michael’s speed zero? * D 8IBUXBT.JDIBFMTTQFFECFUXFFOBOENJOVUFT * E 8IBUXBT.JDIBFMTTQFFECFUXFFOBOENJOVUFT * F 8IBUXBT.JDIBFMTTQFFECFUXFFOBOENJOVUFT *(f) When was Michael’s speed constant? *47. Draw the graph of a function whose domain is and whose range is 5x 0 - 3 … x … , x ≠ 56 5 y 0 - 1 … y … 2, y ≠ 6. What point(s) in the rectangle - 3 … x … , - 1 … y … 2 cannot be on the graph? Compare your graph with those of other students. What differences do you see? *48. Is there a function whose graph is symmetric with respect to the x-axis? Explain.
*(a) From t = to t = 2 *(b) From t = 2 to t = 2.5 *(c) From t = 2.5 to t = 2.
49. Explain why the vertical-line test works.
‘Are You Prepared?’ Answers 1. 1 - , ), 1, ), 1, - 2), 1, 22
x (IV)
(III)
*43. Consider the following scenario: Barbara decides to take a walk. She leaves home, walks 2 blocks in 5 minutes at a constant 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 minute to find her keys and lock the door. Barbara walks 5 blocks in 15 minutes and then decides to jog home. It UBLFTIFSNJOVUFTUPHFUIPNF%SBXBHSBQIPG#BSCBSBT distance from home (in blocks) as a function of time.
(2, 3)
y
2. False
98
CHAPTER 1 Functions and Their Graphs
1.3 Properties of Functions PREPARING FOR THIS SECTION Before getting started, review the following: r *OUFSWBMT "QQFOEJY" 4FDUJPO" QQ"m"
r *OUFSDFQUT 'PVOEBUJPOT 4FDUJPO QQm
r 4MPQFPGB-JOF 'PVOEBUJPOT 4FDUJPO QQm
r 1PJOUm4MPQF'PSNPGB-JOF 'PVOEBUJPOT 4FDUJPO p. 55) r 4ZNNFUSZ 'PVOEBUJPOT 4FDUJPO QQm
Now Work the ‘Are You Prepared?’ problems on page 106.
OBJECTIVES 1 Determine Even and Odd Functions from a Graph (p. 98) 2 Identify Even and Odd Functions from the Equation (p. 99) 3 Use a Graph to Determine Where a Function is Increasing, Decreasing, or Constant (p. 100) 4 Use a Graph to Locate Local Maxima and Local Minima (p. 101) 5 Use a Graph to Locate the Absolute Maximum and the Absolute Minimum (p. 102) 6 Use a Graphing Utility to Approximate Local Maxima and Local Minima and to Determine Where a Function is Increasing or Decreasing (p. 103) 7 Find the Average Rate of Change of a Function (p. 104)
To obtain the graph of a function y = f1x2 , it is often helpful to know certain properties 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 1x, y2 is on the graph of f , the point 1 - x, y2 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 1 - x2 = f 1x2 A function f is odd if and only if, whenever the point 1x, y2 is on the graph of f , the point 1 - x, - y2 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 f1 - x2 = - f 1x2 3FGFSUPQBHF XIFSFUIFUFTUTGPSTZNNFUSZBSFMJTUFE5IFGPMMPXJOHSFTVMUT 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.
SECTION 1.3 Properties of Functions
EXAM PL E 1
99
Determining Even and Odd Functions from the Graph %FUFSNJOFXIFUIFSFBDIHSBQIHJWFOJO'JHVSFJTUIFHSBQIPGBOFWFOGVODUJPO an odd function, or a function that is neither even nor odd.
Figure 17
y
y
x
x
x
(b)
(a)
Solution
y
(c)
B 5IF HSBQI JO 'JHVSF B JT UIBU PG BO FWFO GVODUJPO CFDBVTF UIF HSBQI JT symmetric with respect to the y-axis. C 5IF GVODUJPO XIPTF HSBQI JT HJWFO JO 'JHVSF C JT OFJUIFS FWFO OPS PEE because the graph is neither symmetric with respect to the y-axis nor symmetric with respect to the origin. D 5IFGVODUJPOXIPTFHSBQIJTHJWFOJO'JHVSF D JTPEE CFDBVTFJUTHSBQIJT symmetric with respect to the origin.
r
Now Work
PROBLEMS
21(a), (b),
AND
(d)
2 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.
EXAM PL E 2
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, with respect to the origin, or neither. (a) f1x2 = x2 - 5 (c) h 1x2 = 5x3 - x
Solution
(b) g1x2 = x3 - 1 (d) F1x2 = 0 x 0
(a) To determine whether f is even, odd, or neither, replace x by - x in f1x2 = x2 - 5. Then f1 - x2 = 1 - x2 2 - 5 = x2 - 5 = f 1x2 Since f1 - x2 = f1x2, the function is even, and the graph of f is symmetric with respect to the y-axis. (b) Replace x by - x in g1x2 = x3 - 1. g1 - x2 = 1 - x2 3 - 1 = - x3 - 1 Since g1 - x2 ≠ g1x2 and g1 - x2 ≠ - g1x2 = - 1x3 - 12 = - x3 + 1, the function is neither even nor odd. The graph of g is not symmetric with respect to the y-axis, nor is it symmetric with respect to the origin. (c) Replace x by - x in h 1x2 = 5x3 - x. h 1 - x2 = 51 - x2 3 - 1 - x2 = - 5x3 + x = - 15x3 - x2 = - h 1x2 Since h 1 - x2 = - h 1x2 , h is an odd function, and the graph of h is symmetric with respect to the origin.
100
CHAPTER 1 Functions and Their Graphs
(d) Replace x by - x in F1x2 = 0 x 0 . Then
Figure 18 y
F1 - x2 = 0 - x 0 = 0 - 1 0 # 0 x 0 = 0 x 0 = F1x2
5 (0, 4)
(⫺6, 0)
(3, 4) y = f (x ) (6, 1)
(⫺2, 0) ⫺4
Now Work
6 x
0
(⫺4, ⫺2)
Since F1 - x2 = F1x2, F is an even function, and the graph of F is symmetric with respect to the y-axis.
⫺2
r
PROBLEM
33
3 Use a Graph to Determine Where a Function Is Increasing, Decreasing, or Constant $POTJEFSUIFHSBQIHJWFOJO'JHVSF-PPLGSPNMFGUUPSJHIUBMPOHUIFHSBQIPGUIF function, and note that parts of the graph are going up, parts are going down, and parts are horizontal. In such cases, the function is described as increasing, decreasing, or constant, respectively.
Determining Where a Function Is Increasing, Decreasing, or Constant from Its Graph
EX A MPL E 3
Determine the values of xGPSXIJDIUIFGVODUJPOJO'JHVSFJTJODSFBTJOH8IFSFJT it decreasing? Where is it constant?
Solution
To determine 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 xDPPSEJOBUFT5IF GVODUJPO XIPTF HSBQI JT HJWFO JO 'JHVSF JT increasing on the open interval 1 - , 2, or for - 6 x 6 . The function is decreasing on the open intervals 1 - , - 2 and 13, 2, or for - 6 x 6 - and 3 6 x 6 . The function is constant on the open interval 1, 32, or for 6 x 6 3.
WARNING Describe the behavior of a graph in terms of its x-values. Do OPU TBZ UIF HSBQI JO 'JHVSF JT increasing from the point 1 - , - 22 to the point 1, 2. Rather, say it is increasing on the interval ( - , ). j
DEFINITIONS
A function f is increasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f1x1 2 6 f 1x2 2. A function f is decreasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f1x1 2 7 f 1x2 2. A function f is constant on an open interval I if, for all choices of x in I, the values of f1x2 are equal.
In Words If a function is decreasing, then, as the values of x get bigger, the values of the function get smaller. If a function is increasing, then, as the values of x get bigger, the values of the function also get bigger. If a function is constant, then, as the values of x get bigger, the values of the function remain unchanged. Figure 19
r
More precise definitions follow:
Figure 19 illustrates the definitions. Graphs are read like a book—from left to right. Thus, 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. y
y
y
f (x 1)
f (x 2) f (x 1)
x1
x2
x
f (x 1)
f (x 2)
x1
x2
x
x1
(b) For x 1 < x 2 in l, f(x 1) > f (x 2); f is decreasing on I.
(a) For x 1 < x 2 in l, f (x 1) < f (x 2); f is increasing on I.
Now Work
PROBLEMS
x2
l
l
l
f(x 2)
11, 13, 15,
(c) For all x in I, the values of f are equal; f is constant on I. AND
21(C)
*The open interval 1a, b2 consists of all real numbers x for which a 6 x 6 b.
x
SECTION 1.3 Properties of Functions
101
4 Use a Graph to Locate Local Maxima and Local Minima Suppose f is a function defined on an open interval I containing c. If the value of f at c is greater than or equal to the values of f on I, then f has a local maximum at c 4FF'JHVSF B If the value of f at c is less than or equal to the values of f on I, then f has a local minimum at c4FF'JHVSF C Figure 20
y
f(c)
y
f has a local maximum at c. (c, f(c))
c l
f(c)
x
(a)
DEFINITIONS
(c, f(c))
c l
f has a local minimum at c.
x
(b)
A function f has a local maximum at c if there is an open interval I containing c so that, for all x in I, f1x2 … f1c2. We call f1c2 a local maximum value of f. A function f has a local minimum at c if there is an open interval I containing c so that, for all x in I, f 1x2 Ú f1c2. We call f1c2 a local minimum value of f. If f has a local maximum at c, then the value of f at c is greater than or equal to the values of f near c. If f has a local minimum at c, then the value of f at c is less than or equal to the values of f near c. The word local is used to suggest that it is only near c, not necessarily over the entire domain, that the value f1c2 has these properties.
EXAM PL E 4 Figure 21 y
2
y ⫽ f(x)
Figure 21 shows the graph of a function f.
(1, 2)
(–1, 1) –2
3
Finding Local Maxima and Local Minima from the Graph of a Function and Determining Where the Function Is Increasing, Decreasing, or Constant
x
Solution
WARNING The y-value is the local maximum value or local minimum value, and it occurs at some x-value. For example, in Figure 21, we say f has a local maximum at 1 and the local maximum value is 2. j
(a) At what value(s) of x, if any, does f have a local maximum? List the local maximum values. (b) At what value(s) of x, if any, does f have a local minimum? List the local minimum values. (c) Find the intervals on which f is increasing. Find the intervals on which f is decreasing. 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, we have f1x2 … f112. The local maximum value is f112 = 2. (b) f has local minima at - 1 and at 3. The local minimum values are f1 - 12 = 1 and f132 = . (c) 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, 12 and 13, q 2 , or for - 1 6 x 6 1 and x 7 3. The function is decreasing for all values of x less than - 1 and for all values of x *Some texts use the term relative instead of local.
102
CHAPTER 1 Functions and Their Graphs
between 1 and 3. That is, the function is decreasing on the intervals 1 - q , - 12 and 11, 32 , or for x 6 - 1 and 1 6 x 6 3.
r
Now Work
PROBLEMS
17
AND
19
5 Use a Graph to Locate the Absolute Maximum and the Absolute Minimum Look at the graph of the function f given in Figure 22. The domain of f is the closed interval 3 a, b4 . Also, the largest value of f is f1u2 and the smallest value of f is f1v2 . These are called, respectively, the absolute maximum and the absolute minimum of f on 3 a, b4 .
Figure 22 y (u, f(u)) y = f (x)
DEFINITION
Let f denote a function defined on some interval I. If there is a number u in I for which f 1x2 … f1u2 for all x in I, then f1u2 is the absolute maximum of f on I, and we say the absolute maximum of f occurs at u.
(b, f (b))
(a, f(a))
(y, f (y)) a
u
x
y b
If there is a number v in I for which f1x2 Ú f1v2 for all x in I, then f1v2 is the absolute minimum of f on I, and we say the absolute minimum of f occurs at v.
domain: [a, b] for all x in [a, b], f(x) ≤ f (u) for all x in [a, b], f(x) ≥ f (y) absolute maximum: f(u) absolute minimum: f(y)
The absolute maximum and absolute minimum of a function f are sometimes called the extreme values of f on I. The absolute maximum or absolute minimum of a function f may not exist. Let’s look at some examples.
Finding the Absolute Maximum and the Absolute Minimum from the Graph of a Function
EX A MPL E 5
For each graph of a function y = f1x2 in Figure 23, find the absolute maximum and the absolute minimum, if they exist. Also, find any local maxima or local minima. Figure 23 y
y
y
y
y (3, 6) 6
6
6
(5, 5)
6
6
4
4
(5, 4) 4
4
4
(4, 4)
(5, 3)
(4, 3)
(0, 3)
(1, 2) (0, 1) 1
(3, 1) 3
5
x
1
3
(b)
(a)
Solution
WARNING A function may have an absolute maximum or an absolute minimum at an endpoint, but not a local maximum or a local minimum. Why? Local maxima and local minima are found over some open interval I, and this interval cannot be created around an endpoint. ■
(1, 1) (2, 1) 5
x
2
2
2
2
2
(1, 4)
1
3
(c)
(2, 2)
(0, 0) 5
x
1
3
(d)
5
x
1
3
5
(e)
(a) The function f whose graph is given in Figure 23(a) has the closed interval BTJUTEPNBJO5IFMBSHFTUWBMVFPGf is f132 = , the absolute maximum. The smallest value of f is f 12 = 1, the absolute minimum. The function has a MPDBMNBYJNVNPGBUx = 3BOEBMPDBMNJOJNVNPGBUx = . (b) The function f whose graph is given in Figure 23(b) has the domain 5 x|1 … x … 5, x ≠ 36 . Note that we exclude 3 from the domain because of the “hole” at (3, 1). The largest value of f on its domain is f152 = 3, the absolute maximum. There is no absolute minimum. Do you see why? As you trace the graph, getting closer to the point (3, 1), there is no single smallest value. [As soon as you claim a smallest value, we can trace closer to (3, 1) and get a smaller value!] The function has no local maxima or minima.
x
SECTION 1.3 Properties of Functions
103
(c) The function f whose graph is given in Figure 23(c) has the interval 3 , 54 as its domain. The absolute maximum of f is f152 = . The absolute minimum is 1. Notice that the absolute minimum 1 occurs at any number in the interval 3 1, 24 . The function has a local minimum of 1 at every x in the interval 3 1, 24 , but it has no local maxima. (d) The graph of the function f given in Figure 23(d) has the interval 3 , q 2 as its domain. The function has no absolute maximum; the absolute minimum is f12 = . The function has no local maxima or local minima. (e) The graph of the function f in Figure 23(e) has the domain 5 x 0 1 6 x 6 5, x ≠ 26 . The function f has no absolute maximum and no absolute minimum. Do you see why? The function has a local maximum of 3 at x = , but no local minima.
r
In calculus, there is a theorem with conditions that guarantee a function will have an absolute maximum and an absolute minimum.
THEOREM
Extreme Value Theorem If f is a continuous function* whose domain is a closed interval 3 a, b 4 , then f has an absolute maximum and an absolute minimum on 3 a, b4 . The absolute maximum (minimum) can be found by selecting the largest (smallest) value of f from the following list: 1. The value of f at any local maxima and local minima of f in 1a, b2 . 2. The value of f at each endpoint of 3 a, b4 —that is, f (a) and f (b). 3. The value of f on any interval in 3 a, b4 on which f is constant.
For example, the graph of the function f given in Figure 23(a) is continuous on the closed interval 3 , 54 . The Extreme Value Theorem guarantees that f has extreme values on 3 , 54 . To find them, we list 1. The value of f at the local extrema: f (3) = , f () = 2. The value of f at the endpoints: f () = 1, f (5) = 5 5IFMBSHFTUPGUIFTF JTUIFBCTPMVUFNBYJNVN UIFTNBMMFTUPGUIFTF JTUIF absolute minimum.
Now Work
PROBLEM
45
6 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 minimum usually requires calculus. However, a graphing utility may be used to approximate these values by using the MAXIMUM and MINIMUM features.
EXAM PL E 6
Using a Graphing Utility to Approximate Local Maxima and Minima and to Determine Where a Function Is Increasing or Decreasing (a) Use a graphing utility to graph f1x2 = x3 - 12x + 5 for - 2 6 x 6 2. Approximate where f has a local maximum and where f has a local minimum. (b) Determine where f is increasing and where it is decreasing.
*Although a precise definition requires calculus, we’ll agree for now that a continuous function is one whose graph has no gaps or holes and can be traced without lifting the pencil from the paper.
104
CHAPTER 1 Functions and Their Graphs
Solution
(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 6 x 6 2. The MAXIMUM and MINIMUM commands require us to first determine the open interval I. The graphing utility will then approximate the maximum or minimum value in the interval. Using MAXIMUM, we find that the local maximum is 11.53 and that it occurs at x = - .2, rounded to two decimal places. See Figure B 6TJOH.*/*.6. XFGJOEUIBUUIFMPDBMNJOJNVNJT - 1.53 and that it occurs at x = .2,SPVOEFEUPUXPEFDJNBMQMBDFT4FF'JHVSF C
Figure 24
30
30
⫺2
⫺2
2
2 ⫺10 (b)
⫺10 (a)
C -PPLJOHBU'JHVSFT B BOE C
XFTFFUIBUUIFHSBQIPG f is increasing from x = - 2 to x = - .2 and from x = .2 to x = 2, so f is increasing on the intervals 1 - 2, - .22 and 1.2, 22 , or for - 2 6 x 6 - .2 and .2 6 x 6 2. The graph is decreasing from x = - .2 to x = .2, so f is decreasing on the interval 1 - .2, .22 , or for - .2 6 x 6 .2.
r
Now Work
PROBLEM
53
7 Find the Average Rate of Change of a Function In Foundations, Section 3, we said that the slope of a line could be interpreted as the average rate of change. To find the average rate of change of a function between any two points on its graph, calculate the slope of the line containing the two points.
DEFINITION In Words
The symbol ∆ is the Greek capital letter delta and is read “change in.”
If a and b, a ≠ b, are in the domain of a function y = f1x2, the average rate of change of f from a to b is defined as Average rate of change =
f1b2 - f1a2 ∆y = ∆x b - a
a ≠ b
(1)
The symbol ∆y in equation (1) is the “change in y,” and ∆x is the “change in x.” The average rate of change of f is the change in y divided by the change in x.
EX A MPL E 7
Finding the Average Rate of Change Find the average rate of change of f1x2 = 3x2: B 'SPNUP C 'SPNUP D 'SPNUP
Solution
(a) The average rate of change of f1x2 = 3x2 from 1 to 3 is f132 - f112 ∆y 2 - 3 2 = = = = 12 ∆x 3 - 1 3 - 1 2 (b) The average rate of change of f1x2 = 3x2 from 1 to 5 is f152 - f112 ∆y 5 - 3 2 = = = = 1 ∆x 5 - 1 5 - 1 (c) The average rate of change of f1x2 = 3x2GSPNUPJT f12 - f112 ∆y 1 - 3 1 = = = = 2 ∆x - 1 - 1
r
SECTION 1.3 Properties of Functions
See Figure 25 for a graph of f1x2 = 3x2. The function f is increasing for x 7 . The fact that the average rate of change is positive for any x1, x2, x1 ≠ x2, in the interval 11, 2 indicates that the graph is increasing on 1 6 x 6 . Further, the average rate of change is consistently getting larger for 1 6 x 6 , which indicates that the graph is increasing at an increasing rate.
Figure 25 y 160
Now Work
(7, 147) 120
Average rate of change 5 24
80
Average rate of change 5 18 (3, 27)
(1, 3) 2
4
PROBLEM
61
The Secant Line (5, 75)
40
(0, 0)
105
The average rate of change of a function has an important geometric interpretation. Look at the graph of y = f1x2 JO'JHVSF5XPQPJOUTBSFMBCFMFEPOUIFHSBQI (a, f(a)) and (b, f(b)). The line containing these two points is called the secant line; its slope is msec =
Average rate of change 5 12 6
f 1b2 - f1a2 f1a + h2 - f 1a2 = b - a h
x
Figure 26
y
y f (x ) Secant line (b, f (b )) (a h, f (a h)) f (b ) f (a ) f (a h) f (a )
(a, f (a ))
Instructor Note: You may want to draw the connection between msec f(x + h) - f(x) and . h
THEOREM
bah
a
b ah
x
Slope of the Secant Line The average rate of change of a function f 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 PL E 8
Finding 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 1. (b) Find an equation of the secant line containing 1 - 2, g( - 2)2 and 11, g(1)2 . (c) Using a graphing utility, draw the graph of g and the secant line obtained in part (b) on the same screen.
Solution
(a) The average rate of change of g1x2 = 3x2 - 2x + 3 from - 2 to 1 is g112 - g1 - 22 1 - 1 - 22 - 19 = 3 15 = = -5 3
Average rate of change =
g(1) = 3(1)2 - 2(1) + 3 = 4 g(- 2) = 3(- 2)2 - 2(- 2) + 3 = 19
(b) The slope of the secant line containing 1 - 2, g( - 2)) = ( - 2, 19) and (1, g(1)) = 11, 2 is msec = - 56TFUIFQPJOUmTMPQFGPSNUPàOEBOFRVBUJPOPG the secant line.
106
CHAPTER 1 Functions and Their Graphs
y - y1 y - 19 y - 19 y
Figure 27 24
= = = =
msec 1x - x1 2 - 5(x - ( - 2)) - 5x - 1 - 5x + 9
Point–slope form of the secant line x1 = - 2, y1 = g(- 2) = 19, m sec = - 5 Distribute. Slope–intercept form of the secant line
D 'JHVSFTIPXTUIFHSBQIPGg along with the secant line y = - 5x + 9. ⫺3
3
Now Work
⫺4
r
67
PROBLEM
1.3 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. The interval 12, 52 can be written as the inequality 2 6 x 6 5 . QQ"m"
4. 8SJUFUIFQPJOUmTMPQFGPSNPGUIFMJOFXJUITMPQFDPOUBJOJOH the point 13, - 22. (p. 55) y + 2 = 5(x - 3)
2. The slope of the line containing the points 1 - 2, 32 and 13, 2 is 1 . QQm
5. The intercepts of the equation ( - 3, ), (3, ), (, - 9) . QQm
y = x2 - 9
are
3. Test the equation y = 5x2 - 1 for symmetry with respect to the x-axis, the y-axis, and the origin. QQm y-axis
Concepts and Vocabulary 9. True or False A function f has a local maximum at c if there is an open interval I containing c so that for all x in I, f 1x2 … f 1c2. True
6. A function f is increasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 6 f 1x2 2. 7. A(n) even function f is one for which f 1 - x2 = f 1x2 for every x in the domain of ƒ; a(n) odd function f is one for which f 1 - x2 = - f 1x2 for every x in the domain of f.
10. True or False Even functions have graphs that are symmetric with respect to the origin. False
8. True or False A function f is decreasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 7 f 1x2 2. True
Skill Building In Problems 11–20, use the graph of the function f given.
y
11. Is f increasing on the interval 1 - , - 2)? Yes 12. Is f decreasing on the interval 1 - , - )?
13. Is f increasing on the interval 12, 1)?
10
No
(2, 10)
(⫺2, 6)
No
14. Is f decreasing on the interval 12, 5)? Yes
(5, 0)
(⫺5, 0)
15. List the interval(s) on which f is increasing. ( - , - 2), (, 2), (5, q ) ⫺10 16. List the interval(s) on which f is decreasing. ( - q , - ), ( - 2, ), (2, 5) 17. Is there a local maximum value at 2? If yes, what is it? :FT
⫺5
(0, 0)
(⫺8, ⫺4)
10 x
5
⫺6
18. Is there a local maximum value at 5? If yes, what is it? No 19. List the number(s) at which f has a local maximum. What are the local maximum values? 20. List the number(s) at which f has a local minimum. What are the local minimum values?
- 2, 2; , 1 - , , 5; - , ,
In Problems 21–28, the graph of a function is given. Use the graph to find: (a) The intercepts, if any (b) The domain and range (c) The intervals on which the function is increasing, decreasing, or constant (d) Whether the function is even, odd, or neither *21.
*22.
y 4 (⫺4, 2)
(⫺3, 3)
y
(3, 3)
*23.
3 (0, 3)
*24.
y 3
y 3
(0, 2)
(4, 2)
(0, 1) ⫺4 (⫺2, 0)
(2, 0)
4x
⫺3
(⫺1, 0)
(1, 0)
3 x
⫺3
3 x
⫺3
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
(1, 0)
3x
107
SECTION 1.3 Properties of Functions
*25.
*26.
y 2
*27.
y 2
(––2 , 1)
( – 3, 2)
(0, 1)
*28.
y 3
1 0, – 2
( ) (3, 1)
( –1, 2) ⫺
⫺–– 2
–– 2
⫺
x
⫺ 2
(⫺, ⫺1)
(⫺ ––2 , ⫺1)
2
x
1 –, 0 3
( )
(, ⫺1)
(2, 2)
(⫺2, 1)
(⫺2.3, 0)
(0, 1)
⫺3
3 x (1, –1) (2, –1)
–3
⫺2
⫺2
y 3
3 x
(⫺3, ⫺2)
–3
(3, 0)
⫺2
In Problems 29–32, the graph of a function f is given. Use the graph to find: (a) The numbers, if any, at which f has a local maximum value. What are the local maximum values? (b) The numbers, if any, at which f has a local minimum value. What are the local minimum values? *29.
*30.
y 4
*31.
y
*32.
y
(
3 (0, 3)
⫺4 (⫺2, 0)
1
(0, 2) ⫺
4x
(2, 0)
⫺3 (⫺1, 0)
(1, 0)
3 x
–– , 2
⫺–– 2
1)
–– 2
(⫺ ––2 , ⫺1)
y 2
⫺ π2
⫺π
x
π 2
π x
(⫺π, ⫺1)
⫺1
(π, ⫺1) ⫺2
In Problems 33–44, determine algebraically whether each function is even, odd, or neither. 33. f 1x2 = x3
37. F 1x2 = 2x
34. f 1x2 = 2x - x2
Odd
3
38. G1x2 = 1x
Odd
x2 + 3 41. g1x2 = 2 x - 1
Even
Neither
x 42. h1x2 = 2 x - 1
Even
35. g1x2 = - 3x2 - 5
39. f 1x2 = x + 0 x 0
- x3 43. h1x2 = 3x2 - 9
Odd
36. h1x2 = 3x3 + 5
Even
40. f 1x2 = 22x + 1 3
Neither
2x 44. F 1x2 = 0x0
Odd
Neither
2
Even
Odd
In Problems 45–52, for each graph of a function y = f 1x2, find the absolute maximum and the absolute minimum, if they exist. Identify any local maxima or local minima. *45.
*46.
y
*47.
y
(1, 4)
4 2
2
(2, 2) (5, 1) 1
*49.
3
1
*50.
2 (0, 1)
(1, 1) 1
x
5
*51.
3
(3, 2) (0, 0)
1
1
1
x
3
1
3
x
x
(1, 3) 2
(1, 1)
(4, 1)
1
3
y
(3, 2)
2
1
1
*52.
y
(2, 4)
2 (0, 2)
x
5
(2, 0) 2 (1, 1)
(1, 3)
(4, 3)
(5, 0) 3
(2, 4)
4
2
y 4 (1, 3)
(2, 3)
(3, 4)
4
(0, 2)
x
5
y
(0, 3)
(1, 1)
y 4
(4, 4)
4
(3, 3)
*48.
y
(0, 2)
(3, 1)
x
3
1
(2, 2) 2
1 (2, 0) 3
x
2
(1, 3)
In Problems 53–60, use a graphing utility to graph each function over the indicated interval and approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places. 1 - 2, 22
*53. f 1x2 = x3 - 3x + 2
*55. f 1x2 = x - x 5
3
1 - 2, 22
*59. f 1x2 = .25x + .3x - .9x + 3 3
*56. f1x2 = x - x
*57. f 1x2 = - .2x3 - .x2 + x -
*54. f 1x2 = x3 - 3x2 + 5
2
1 - , 2
1 - 3, 22
61. Find the average rate of change of f 1x2 = - 2x2 + . B 'SPNUP - (b) From 1 to 3 - D 'SPNUP - 1
2
1 - 1, 32
1 - 2, 22
*58. f1x2 = - .x3 + .x2 + 3x - 2
1 - , 52
*60. f 1x2 = - .x - .5x + .x - 2
3
2
1 - 3, 22
62. Find the average rate of change of f 1x2 = - x3 + 1. B 'SPNUP - (b) From 1 to 3 - 13 (c) From - 1 to 1 - 1
108
CHAPTER 1 Functions and Their Graphs
63. Find the average rate of change of g1x2 = x3 - 2x + 1. (a) From - 3 to - 2 1 (b) From - 1 to 1 - 1 (c) From 1 to 3 11
67. g(x) = x2 - 2 (a) Find the average rate of change from - 2 to 1. - 1 (b) Find an equation of the secant line containing 1 - 2, g( - 2)2 and 11, g(1)2. y = - x
64. Find the average rate of change of h1x2 = x2 - 2x + 3. (a) From - 1 to 1 - 2 C 'SPNUP (c) From 2 to 5 5
68. g(x) = x2 + 1 (a) Find the average rate of change from - 1 to 2. 1 (b) Find an equation of the secant line containing 1 - 1, g( - 12 and 12, g(2)2. y = x + 3
65. f 1x2 = 5x - 2 (a) Find the average rate of change from 1 to 3. 5 (b) Find an equation of the secant line containing 11, f 112 2 and 13, f 132 2. y = 5x - 2
66. f 1x2 = - x + 1 (a) Find the average rate of change from 2 to 5. - (b) Find an equation of the secant line containing 12, f 122 2 and 15, f 152 2. y = - x + 1
69. h(x) = x2 - 2x B 'JOEUIFBWFSBHFSBUFPGDIBOHFGSPNUP (b) Find an equation of the secant line containing 12, h(2)2 and 1, h()2. y = x - 70. h(x) = - 2x2 + x B 'JOEUIFBWFSBHFSBUFPGDIBOHFGSPNUP - 5 (b) Find an equation of the secant line containing 1, h()2 and 13, h(3)2. y = - 5x
Mixed Practice 71. g(x) = x3 - 2x (a) Determine whether g is even, odd, or neither. Odd (b) There is a local minimum value of - 5 at 3. Determine a local maximum value. BUx = - 3
72. f 1x2 = - x3 + 12x (a) Determine whether f is even, odd, or neither. Odd C 5IFSFJTBMPDBMNBYJNVNWBMVFPGBU%FUFSNJOF a local minimum value. - 1 at x = - 2
73. F 1x2 = - x + x2 + (a) Determine whether F is even, odd, or neither. Even C 5IFSFJTBMPDBMNBYJNVNWBMVFPGBUx = 2. Determine a second local maximum value. BU x = - 2
(c) Suppose the area under the graph of F between x = and x = 3 that is bounded below by the xBYJT JT square units. Using the result from part (a), determine the area under the graph of F between x = - 3 and x = bounded below by the x-axis. TRVOJUT 74. G1x2 = - x + 32x2 + 1 (a) Determine whether G is even, odd, or neither. Even C 5IFSFJTBMPDBMNBYJNVNWBMVFPGBUx = . Determine a second local maximum value. BUx = - (c) Suppose the area under the graph of G between x = and x = that is bounded below by the xBYJTJT square units. Using the result from part (a), determine the area under the graph of G between x = - and x = bounded below by the x-axis. TRVOJUT
Applications and Extensions 75. Minimum Average Cost The average cost per hour in dollars, C, of producing x riding lawn mowers can be modeled by the function C(x) = .3x2 + 21x - 251 +
25 x
*(a) Use a graphing utility to graph C = C(x2 . (b) Determine the number of riding lawn mowers to produce in order to minimize average cost. NPXFSTIS (c) What is the minimum average cost? $239/mower 76. Medicine Concentration The concentration C of a medication in the bloodstream t hours after being administered is modeled by the function C(t) = - .2t + .39t 3 - .25t 2 + .t + .5 (a) After how many hours will the concentration be highest? IS (b) A woman nursing a child must wait until the DPODFOUSBUJPO JT CFMPX CFGPSF TIF DBO GFFE IFS PS him. After taking the medication, how long must she wait before feeding her child? IS 77. National Debt The size of the total debt owed by the United States federal government continues to grow. In fact, according to the Department of the Treasury, the debt per QFSTPOMJWJOHJOUIF6OJUFE4UBUFTJTBQQSPYJNBUFMZ PS PWFS QFS 64 IPVTFIPME 5IF GPMMPXJOH EBUB
SFQSFTFOU UIF 64 EFCU GPS UIF ZFBST m 4JODF UIF debt D depends on the year y, and each input corresponds to exactly one output, the debt is a function of the year. Thus D(y), represents the debt for each year y. Year
Debt (billions of dollars)
Year
Debt (billions of dollars)
2001
5807
2007
9008
2002
6228
2008
10,025
2003
6783
2009
11,910
2004
7379
2010
13,562
2005
7933
2011
14,790
2006
8507
2012
16,066
Source: www.treasurydirect.gov
* B 1MPUUIFQPJOUT
BOETPPOJOB Cartesian plane. * C %SBX B MJOF TFHNFOU GSPN UIF QPJOU UP 8IBU EPFT UIF TMPQF PG UIJT MJOF TFHNFOU represent? D 'JOEUIFBWFSBHFSBUFPGDIBOHFPGUIFEFCUGSPNUP CJMMJPOQFSZFBS E 'JOEUIFBWFSBHFSBUFPGDIBOHFPGUIFEFCUGSPNUP CJMMJPOQFSZFBS
SECTION 1.3 Properties of Functions
F 'JOEUIFBWFSBHFSBUFPGDIBOHFPGUIFEFCUGSPNUP $1252 billion per year *(f) What is happening to the average rate of change as time passes?
*(e) What is happening to the average rate of change as time passes? 79. For the function f1x2 = x2, compute each average rate of change: B 'SPNUP 1 C 'SPNUP D 'SPNUP E 'SPNUP F 'SPNUP *(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? 5IFZBSFHFUUJOHDMPTFSUP.
78. E.coli Growth A strain of E.coli#FVSFD"JTQMBDFE JOUP B OVUSJFOU CSPUI BU $FMTJVT BOE BMMPXFE UP HSPX The data shown in the table 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. Thus P(t) represents the population at time t. Time (hours), t
Population (grams), P
0
0.09
2.5
0.18
3.5
0.26
4.5
0.35
6
0.50
109
80. For the function f 1x2 = x2, compute each average rate of change: (a) From 1 to 2 3 (b) From 1 to 1.5 2.5 (c) From 1 to 1.1 2.1 E 'SPNUP F 'SPNUP *(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? They are getting closer to 2.
* B 1MPU UIF QPJOUT
BOE TP PO JO B Cartesian plane. * C %SBXBMJOFTFHNFOUGSPNUIFQPJOU UP What does the slope of this line segment represent? (c) Find the average rate of change of the population from UPIPVST HIS (d) Find the average rate of change of the population from UPIPVST HIS
Problems 81–88 require the following discussion of a secant line. The slope of the secant line containing the two points 1x, f (x)2 and (x + h, f (x + h)) on the graph of a function y = f 1x2 may be given as msec =
f 1x + h2 - f 1x2 1x + h2 - x
=
f 1x + h2 - f 1x2 h
h ≠
In calculus, this expression is called the difference quotient of f. (a) Express the slope of the secant line of each function in terms of x and h. Be sure to simplify your answer. (b) Find msec for h = .5, and at x = 1. What value does msec approach as h approaches ? (c) Find the equation for the secant line at x = 1 with h = .1. (d) Use a graphing utility to graph f and the secant line found in part (c) on the same viewing window.
*81. f 1x2 = 2x + 5
*82. f 1x2 = - 3x + 2
*83. f 1x2 = x2 + 2x
*84. f 1x2 = 2x2 + x
*85. f 1x2 = 2x2 - 3x + 1
*86. f 1x2 = - x2 + 3x - 2
*87. f 1x2 =
*88. f 1x2 =
1 x
1 x2
Discussion and Writing 89. Draw the graph of a function that has the following properties: domain: all real numbers; range: all real numbers; intercepts: 1, - 32 and 13, ); a local maximum value of - 2 is at - 1; a local minimum value of - is at 2. Compare your graph with those of others. Comment on any differences. 90. 3FEP1SPCMFNXJUIUIFGPMMPXJOHBEEJUJPOBMJOGPSNBUJPO increasing on 1 - q , - 1), (2, q ); decreasing on 1 - 1, 22. Again compare your graph with others and comment on any differences. 91. How many x-intercepts can a function defined on an interval have if it is increasing on that interval? Explain. At most one
92. Suppose that a friend of yours does not understand the JEFB PG JODSFBTJOH BOE EFDSFBTJOH GVODUJPOT 1SPWJEF BO explanation, complete with graphs, that clarifies the idea. *93. Can a function be both even and odd? Explain.
94. Using a graphing utility, graph y = 5 on the interval 1 - 3, 32. Use MAXIMUM to find the local maximum values on 1 - 3, 32. Comment on the result provided by the calculator.
*95. A function f has a positive average rate of change on the interval 3 2, 5 4 . Is f increasing on 3 2, 5 4 ? Explain. *96. Show that a constant function f(x) = b has an average SBUFPGDIBOHFPG$PNQVUFUIFBWFSBHFSBUFPGDIBOHFPG y = 2 - x2 on the interval 3 - 2, 24. Explain how this can happen.
110
CHAPTER 1 Functions and Their Graphs
‘Are You Prepared?’ Answers 1. 2 6 x 6 5
2. 1
4. y + 2 = 5(x - 32
3. symmetric with respect to the y-axis
5. 1 - 3, ), (3, ), (, - 92
1.4 Library of Functions; Piecewise-defined Functions PREPARING FOR THIS SECTION Before getting started, review the following: r *OUFSDFQUT 'PVOEBUJPOT 4FDUJPO QQm
r (SBQITPG,FZ&RVBUJPOT 'PVOEBUJPOT 4FDUJPO &YBNQMF Q &YBNQMF Q &YBNQMF Q &YBNQMF Q
Now Work the ‘Are You Prepared?’ problems on page 117.
OBJECTIVES 1 Graph the Functions Listed in the Library of Functions (p. 110) 2 Graph Piecewise-defined Functions (p. 115)
1 Graph the Functions Listed in the Library of Functions Figure 28
First we introduce a few more functions, beginning with the square root function. 0OQBHF XFHSBQIFEUIFFRVBUJPOy = 1x'JHVSFTIPXTBHSBQIPGUIF function f1x2 = 1x. Based on the graph, we have the following properties:
y 6 (1, 1)
(4, 2)
(9, 3)
Properties of f1 x 2 = 1x
(0, 0) –2
5
10 x
EX A MPL E 1
Solution
1. The domain and the range are the set of nonnegative real numbers. 2. The x-intercept of the graph of f 1x2 = 1x JT 5IF y-intercept of the graph of f1x2 = 1xJTBMTP 3. The function is neither even nor odd. 4. The function is increasing on the interval 1, q 2. 5. 5IFGVODUJPOIBTBOBCTPMVUFNJOJNVNPGBUx = .
Graphing the Cube Root Function
3 (a) Determine whether f 1x2 = 2 x 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. 3 (b) Determine the intercepts, if any, of the graph of f1x2 = 2 x. 3 (c) Graph f1x2 = 2x.
(a) Because 3 3 -x = - 2 x = - f1x2 f1 - x2 = 2
the function is odd. The graph of f is symmetric with respect to the origin. 3 (b) The y-intercept is f12 = 2 = . The x-intercept is found by solving the equation f1x2 = . f1x2 = 3 2 x = x =
The xJOUFSDFQUJTBMTP
3
f (x) = 2x Cube both sides of the equation.
111
SECTION 1.4 Library of Functions; Piecewise-defined Functions
D 6TFUIFGVODUJPOUPGPSN5BCMFBOEPCUBJOTPNFQPJOUTPOUIFHSBQI#FDBVTF of the symmetry with respect to the origin, we find only points 1x, y2 for which 3 x Ú . Figure 29 shows the graph of f1x2 = 2 x.
Table 4
3
Figure 29
x
y =f1x2 = 2x
(x, y)
0
0
10, 02
1 8
1 2
1 1 a , b 8 2
1
1
11, 12
2
22 ≈ 1.26
3 12, 222
8
2
3
y 3 3
(1, 1)
( 1–8, 1–2)
(2, 2 )
( 1–8 , 1–2)
3
3
x
(0, 0)
18, 22
3
(1, 1)
(2, 2 )
r
3
From the results of Example 1 and Figure 29, we have the following properties of the cube root function.
3 Properties of f1 x 2 = 2x
1. The domain and the range are the set of all real numbers. 3 2. The x-intercept of the graph of f 1x2 = 2 x JT 5IF y-intercept of the 3 graph of f1x2 = 2xJTBMTP 3. The graph is symmetric with respect to the origin. The function is odd. 4. The function is increasing on the interval 1 - q , q 2. 5. The function does not have any local minima or any local maxima.
EXAM PL E 2
Solution
Graphing the Absolute Value Function
(a) Determine whether f1x2 = 0 x 0 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 f1x2 = 0 x 0 . (c) Graph f1x2 = 0 x 0 . (a) Because
f1 - x2 = 0 - x 0 = 0 x 0 = f 1x2
the function is even. The graph of f is symmetric with respect to the y-axis. (b) The y-intercept is f12 = 0 0 = . The x-intercept is found by solving the equation f1x2 = or 0 x 0 = . So the xJOUFSDFQUJT (c) Use the function to form Table 5 and obtain some points on the graph. Because of the symmetry with respect to the y-axis, we need to find only points 1x, y2 for which x Ú .'JHVSFTIPXTUIFHSBQIPGf1x2 = 0 x 0 . Figure 30
Table 5
x
y = f(x) = ∣ x ∣
1x, y 2
0
0
(0, 0)
1
1
(1, 1)
2
2
(2, 2)
3
3
(3, 3)
y 3
(3, 3) (2, 2) (1, 1)
3 2 1
(3, 3) (2, 2)
2 1
1
(1, 1) 1 2 (0, 0)
3
x
r
112
CHAPTER 1 Functions and Their Graphs
Instructor Note: Emphasize the importance of knowing the properties of the key functions presented here. Being able to visualize basic functions will be helpful later with topics such as nonlinear systems and graphing using transformations.
'SPNUIFSFTVMUTPG&YBNQMFBOE'JHVSF XFIBWFUIFGPMMPXJOHQSPQFSUJFT of the absolute value function.
Properties of f(x)= ∣ x ∣
1. The domain is the set of all real numbers. The range of f is 5 y 0 y Ú 6 . 2. The x-intercept of the graph of f 1x2 = 0 x 0 JT5IFy-intercept of the graph of f1x2 = 0 x 0 JTBMTP 3. The graph is symmetric with respect to the y-axis. The function is even. 4. The function is decreasing on the interval 1 - q , 2. It is increasing on the interval 1, q 2. 5. 5IFGVODUJPOIBTBOBCTPMVUFNJOJNVNPGBUx = .
Seeing the Concept
Graph y = 0 x 0 on a square screen and compare what you see with Figure 30. Note that some graphing calculators use abs 1x2 for absolute value.
Below is a list of the key functions that we have discussed. In going through this list, pay special attention to the properties of each function, particularly to the shape of each graph. Knowing these graphs, along with key points on each graph, will lay the foundation for further graphing techniques.
Figure 31 Constant Function
Constant Function
y
f1x2 = b
f(x) = b
b is a real number
(0,b) x
Figure 32 Identity Function
See Figure 31. 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.
Identity Function
f(x) = x
y 3
f1x2 = x (1, 1)
–3 (–1, –1)
(0, 0)
3 x
See Figure 32. The domain and the range of the identity function are the set of all real numbers. Its graph is a line whose slope is 1 and whose yJOUFSDFQU JT 5IF MJOF consists of all points for which the x-coordinate equals the y-coordinate. The identity function is an odd function that is increasing over its domain. Note that the graph bisects quadrants I and III.
Square Function f1x2 = x2
SECTION 1.4 Library of Functions; Piecewise-defined Functions
See Figure 33. The domain of the square function 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 vertex is at 1, 2, which is also the only intercept. The square function is an even function that is decreasing on the interval 1 - q , 2 and increasing on the interval 1, q 2 .
Figure 33 Square Function f (x ) =
y (–2, 4)
x2
(2, 4)
4
(–1, 1)
(1, 1) 4 x
(0, 0)
–4
113
Cube Function f1x2 = x3
Figure 34 Cube Function y 4
f (x ) = x 3
4FF'JHVSF The domain and the range of the cube function are the set of all real numbers. The intercept of the graph is at 1, 2. The cube function is odd and is increasing on the interval 1 - q , q 2.
(1, 1) ⫺4 (⫺1, ⫺1)
(0, 0)
x
4
⫺4
Figure 35 Square Root Function y
f(x) =
2
⫺1
Square Root Function
x
f1x2 = 1x
(1, 1)
(4, 2) 5 x
(0, 0)
Figure 36 Cube Root Function y 3
f (x ) =
3
See Figure 35. The domain and the range of the square root function are the set of nonnegative real numbers. The intercept of the graph is at 1, 2. The square root function is neither even nor odd and is increasing on the interval 1, q 2.
Cube Root Function
x
3 f1x2 = 2 x
3
(1, 1)
(⫺ 1–8,⫺ 1–2)
(2, 2 )
( 1–8 , 1–2)
⫺3
3 x (0, 0) 3
(⫺1, ⫺1)
(⫺2,⫺ 2 )
⫺3
Figure 37 Reciprocal Function y 2
Reciprocal Function
1– , 2
( 2) f (x ) =
(⫺2, ⫺ 1–2 )
4FF'JHVSF The domain and the range of the cube root function are the set of all real numbers. The intercept of the graph is at 1, 2. The cube root function is an odd function that is increasing on the interval 1 - q , q 2.
1 –– x
f1x2 =
1 x
(1, 1)
⫺2
2 x
(⫺1, ⫺1) ⫺2
1 3FGFS UP &YBNQMF QBHF GPS B EJTDVTTJPO PG UIF FRVBUJPO y = . See x 'JHVSF 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 1 - q , 2 and 1, q 2 and is an odd function.
114
CHAPTER 1 Functions and Their Graphs
Figure 38 Absolute Value Function y
f(x) = ⏐x ⏐
Absolute Value Function f1x2 = 0 x 0
3 (2, 2)
(⫺2, 2)
(1, 1)
(⫺1, 1) ⫺3
(0, 0)
3 x
Instructor Note: Students often apply the greatest integer function by just dropping the decimal portion of a number. Be sure to show them why this won’t work for negative arguments.
DEFINITION
4FF'JHVSF 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 1, 2 . If x Ú , then f1x2 = x, and this part of the graph of f is the line y = x; if x 6 , then f1x2 = - x, and this part of the graph of f is the line y = - x. The absolute value function is an even function; it is decreasing on the interval 1 - q , 2 and increasing on the interval 1, q 2. The notation int1x2 stands for the largest integer less than or equal to x. For example, 1 3 int112 = 1, int12.52 = 2, inta b = , inta - b = - 1, int1p2 = 3 2 This type of correspondence occurs frequently enough in mathematics that we give it a name.
Greatest Integer Function f1x2 = int1x2* = greatest integer less than or equal to x
Table 6 x
y =f(x) =int(x)
-1
-1
( - 1, - 1)
1 2
-1
1 a - , - 1b 2
1 4
-1
1 a - , - 1b 4
0
0
(0, 0)
0
1 a , b 4
0
1 a , b 2
0
3 a , b 4
-
1 4 1 2 3 4
(x, y)
We obtain the graph of f 1x2 = int1x2 CZ QMPUUJOH TFWFSBM QPJOUT 4FF 5BCMF For values of x, - 1 … x 6 , the value of f 1x2 = int1x2 is - 1; for values of x, … x 6 1, the value of f JT4FF'JHVSFGPSUIFHSBQI Figure 39 Greatest Integer Function y 4 2 ⫺2
2
4
x
⫺3
The domain of the greatest integer function is the set of all real numbers; its range is the set of integers. The yJOUFSDFQU PG UIF HSBQI JT 5IF x-intercepts lie in the interval 3 , 12. The greatest integer function is neither even nor odd. It is constant on every interval of the form 3 k, k + 12, for k an integer. In Figure 39, a solid dot is used to indicate, for example, that at x = 1 the value of f is f112 = 1; BOPQFODJSDMFJTVTFEUPJMMVTUSBUFUIBUUIFGVODUJPOEPFTOPUBTTVNFUIFWBMVFPG at x = 1. Although a precise definition requires the idea of a limit, discussed in calculus, in a rough sense, a function is said to be continuous if its graph has no gaps or holes and can be drawn without lifting a pencil from the paper on which the graph is drawn. We contrast this with a discontinuous function. A function is discontinuous if its graph has gaps or holes so that its graph cannot be drawn without lifting a pencil from the paper. *Some books use the notation f 1x2 = 3 x 4 instead of int 1x2.
SECTION 1.4 Library of Functions; Piecewise-defined Functions
Figure 40 f 1x2 = int 1x2 6
⫺2
6
From the graph of the greatest integer function, we can see why it is also called a step function. At x = , x = {1, x = { 2, and so on, this function is discontinuous because, 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 at x = 3 and to the immediate right of x = 3, the y-coordinates of the points on the graph are 3. Consequently, the graph has gaps in it. 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 may be necessary to be in the dot mode. This is to prevent the utility from “connecting the dots” when f 1x2 changes from POFJOUFHFSWBMVFUPUIFOFYU4FF'JHVSF/PUFUIBUTPNFHSBQIJOHVUJMJUJFTXJMMEJTQMBZUIF gaps even when in “connected” mode. ■
⫺2 (a) Connected mode 6
⫺2
115
6 ⫺2 (b) Dot mode
The functions 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 f1x2 = x2, you should see in your mind’s eye a picture like Figure 33.
Now Work
PROBLEMS
9
THROUGH
16
2 Graph Piecewise-defined Functions Sometimes a function is defined using different equations on different parts of its domain. For example, the absolute value function f1x2 = 0 x 0 is defined by two equations: f 1x2 = x if x Ú and f1x2 = - x if x 6 . For convenience, these equations are generally combined into one expression as f1x2 = 0 x 0 = e
x if x Ú - x if x 6
When a function is defined by different equations on different parts of its domain, it is called a piecewise-defined function.
EXAM PL E 3
Analyzing a Piecewise-defined Function The function f is defined as - 2x + 1 if - 3 … x 6 1 f1x2 = c 2 if x = 1 x2 if x 7 1 (a) Find f1 - 22, f112, and f122. (c) Locate any intercepts. (e) Use the graph to find the range of f.
Solution
(b) Determine the domain of f. (d) Graph f . (f) Is f continuous on its domain?
(a) To find f1 - 22, observe that when x = - 2, the equation for f is given by f1x2 = - 2x + 1, so f 1 - 22 = - 2( - 2) + 1 = 5 When x = 1, the equation for f is f1x2 = 2. That is, f112 = 2 When x = 2, the equation for f is f1x2 = x2, so f122 = 22 = (b) To find the domain of f, look at its definition. Since f is defined for all x greater than or equal to - 3, the domain of f is {x 0 x Ú - 3}, or the interval 3 - 3, q 2 . (c) The y-intercept of the graph of the function is f 12. Because the equation for f when x = is f1x2 = - 2x + 1, the y-intercept is f12 = - 212 + 1 = 1.
116
CHAPTER 1 Functions and Their Graphs
The x-intercepts of the graph of a function f are the real solutions to the equation f1x2 = . To find the x-intercepts of f, solve f1x2 = for each “piece” of the function, and then determine whether the values of x, if any, satisfy the condition that defines the piece. f1x2 = - 2x + 1 = - 3 … x 6 1 - 2x = - 1 1 x = 2
Figure 41 y 8
4
(1,2)
(2,4)
(0,1)
( 1–2 , 0)
(1, ⫺1)
4
x
f1x2 = x2 = x 7 1 x =
1 The first potential x-intercept, x = , satisfies the condition - 3 … x 6 1, so 2 1 x = is an x-intercept. The second potential x-intercept, x = , does not satisfy 2 1 the condition x 7 1, so x = is not an x-intercept. The only x-intercept is . The 2 1 JOUFSDFQUTBSF BOE a , b . 2 (d) To graph f , graph each “piece.” First graph the line y = - 2x + 1 and keep only the part for which - 3 … x 6 1. Then plot the point 11, 22 because, when x = 1, f1x2 = 2. Finally, graph the parabola y = x2 and keep only the part for which x 7 1.4FF'JHVSF (e) From the graph, we conclude that the range of f is 5 y y 7 - 16 , or the interval 1 - 1, q 2. (f) The function f is not continuous because there is a “jump” in the graph at x = 1.
Now Work
EX A MPL E 4
f 1x2 = 2 = x = 1 No solution
PROBLEM
29
r
Cost of Electricity *O UIF TQSJOH PG %VLF &OFSHZ 1SPHSFTT TVQQMJFE FMFDUSJDJUZ UP SFTJEFODFT JO 4PVUI $BSPMJOB GPS B NPOUIMZ DVTUPNFS DIBSHF PG QMVT a QFS LJMPXBUU IPVS L8IS GPSUIFGJSTUL8ISTVQQMJFEJOUIFNPOUIBOEaQFSL8ISGPS BMMVTBHFPWFSL8ISJOUIFNPOUI B 8IBUJTUIFDIBSHFGPSVTJOHL8ISJOBNPOUI C 8IBUJTUIFDIBSHFGPSVTJOHL8ISJOBNPOUI (c) If C is the monthly charge for x kWhr, develop a model relating the monthly charge and kilowatt-hours used. That is, express C as a function of x. Source: Duke Energy Progress, 2013
Solution
B 'PSL8IS UIFDIBSHFJTQMVT9.5c = $.95 per kWhr. That is, Charge = $.5 + $.95132 = $35.1 C 'PSL8IS UIFDIBSHFJTQMVTcQFSL8ISGPSUIFàSTUL8IS QMVTcQFSL8ISGPSUIFJOFYDFTTPG5IBUJT Charge = $.5 + $.9512 + $.512 = $12.9 (c) Let x represent the number of kilowatt-hours used. If … x … , then the monthly charge C (in dollars) can be found by multiplying x UJNFT BOEBEEJOHUIFNPOUIMZDVTUPNFSDIBSHFPG5IVT JG … x … , then C 1x2 = .95x + .5. For x 7 , the charge is .95() + .5 + .51x - 2 , since 1x - 2 FRVBMTUIFVTBHFJOFYDFTTPGL8IS XIJDIDPTUT per kWhr. That is, if x 7 , then C 1x2 = .9512 + .5 + .51x - 2 = . + .5 + .5x - . = .5x + 1.5
117
SECTION 1.4 Library of Functions; Piecewise-defined Functions
The rule for computing C follows two equations:
Figure 42 C
C 1x2 = e
.95x + .5 if … x … .5x + 1.5 if x 7
The Model
Charge (dollars)
180 (1500, 142.90)
r
4FF'JHVSFGPSUIFHSBQI
120 60
(800, 82.98)
6.50 (300, 35.18)
0
400
x
800 1200 Usage (kWhr)
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.
1. Sketch the graph of y = 1x. Q
2. Sketch the graph of y =
3. List the intercepts of the equation y = x3 - . Q
1, - 2, 12, 2
1 . QQm
x
Concepts and Vocabulary 4. The function f 1x2 = x2 is decreasing on the interval 1 - q , 2 .
7. True or False The cube root function is odd and is decreasing on the interval 1 - q , q 2. False
5. When functions are defined by more than one equation, they are called piecewise-defined functions.
8. True or False The domain and the range of the reciprocal function are the set of all real numbers. False
6. True or False The cube function is odd and is increasing on the interval 1 - q , q 2. True
Skill Building In Problems 9 –16, match each graph to its function. A. Constant function B. Identity function E. Square root function F. Reciprocal function
C. Square function G. Absolute value function
D. Cube function H. Cube root function
9.
C
10.
A
11.
E
12.
G
13.
B
14.
D
15.
F
16.
H
In Problems 17–24, sketch the graph of each function. Be sure to label three points on the graph. *17. f 1x2 = x
*18. f 1x2 = x2
*19. f 1x2 = x3
*20. f 1x2 = 1x
*21. f 1x2 =
*22. f 1x2 = 0 x 0
3 *23. f 1x2 = 2 x
*24. f 1x2 = 3
1 x
x2 25. If f 1x2 = c 2 2x + 1 find: (a) f 1 - 22
27. If f 1x2 = e
if x 6 if x = if x 7
(b) f 12
2x - x3 - 2
2
if - 1 … x … 2 if 2 6 x … 3
find: (a) f 12 - (b) f 112 - 2
(c) f 122
- 3x 26. If f 1x2 = c 2x2 + 1 5
(c) f 122 (d) f 132 25
find: (a) f 1 - 22
28. If f 1x2 = e
if x 6 - 1 if x = - 1 if x 7 - 1
(b) f 1 - 12
x3 3x + 2
find: (a) f 1 - 12 - 1
if - 2 … x 6 1 if 1 … x …
(c) f 12
1
(b) f 12 (c) f 112 5 (d) f 132 11
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
118
CHAPTER 1 Functions and Their Graphs
In Problems 29–40: (a) Find the domain of each function.
(b) Locate any intercepts.
(d) Based on the graph, find the range.
(e) Is f continuous on its domain?
if x ≠ if x =
*29. f 1x2 = b
2x 1
*32. f1x2 = b
x + 3 - 2x - 3
1 + x *35. f1x2 = b 2 x *38. f 1x2 = b
2 - x 2x
*30. f 1x2 = b
if x ≠ if x =
3x
x + 3 *33. f 1x2 = c 5 -x + 2
if x 6 - 2 if x Ú - 2
1 *36. f 1x2 = c x 3 2 x
if x 6 if x Ú
(c) Graph each function. *31. f 1x2 = b
- 2x + 3 3x - 2
if x 6 1 if x Ú 1
if - 2 … x 6 1 2x + 5 if x = 1 *34. f 1x2 = c - 3 if x 7 1 - 5x if x 6
*37. f 1x2 = b
if x Ú
if - 3 … x 6 1 *39. f 1x2 = 2 int 1x2 if x Ú 1
0x0 x3
if - 3 … x 6 if x = if x 7
if - 2 … x 6 if x Ú
*40. f 1x2 = int 12x2
In Problems 41–44, the graph of a piecewise-defined function is given. Write a definition for each function. *41.
*42.
y
*43.
y 2
2
(2, 2) (1, 1)
(⫺1, 1) ⫺2
⫺2
2 x
(0, 0)
y (0, 2)
2 (2, 1)
(2, 1)
(⫺1, 1)
*44.
y
2 x
(0, 0)
⫺2
(⫺1, 0) (0, 0)
(2, 0) x
(1, 1)
⫺2
2 x
(⫺1, ⫺1)
45. If f 1x2 = int 12x2, find (a) f 11.22
2
(b) f 11.2
3
(c) f 1 - 1.2
-
x 46. If f 1x2 = int a b, find 2 (a) f 11.22 (b) f 11.2
(c) f 1 - 1.2
-1
Applications and Extensions 47. Cell Phone Service 4QSJOU 1$4 PGGFST B NPOUIMZ DFMMVMBS QIPOF QMBO GPS *U JODMVEFT BOZUJNF NJOVUFT BOE DIBSHFT QFS NJOVUF GPS BEEJUJPOBM NJOVUFT 5IF following function is used to compute the monthly cost for a subscriber: C 1x2 = b
39.99 .5x - 12.51
if … x … 5 if x 7 5
where x is the number of anytime minutes used. Compute the monthly cost of the cellular phone for use of the following numbers of anytime minutes: B $39.99 C D Source: Sprint PCS 48. Parking at O’Hare International Airport The short-term OPNPSFUIBOIPVST QBSLJOHGFFF (in dollars) for parking x hours on a weekday at O’Hare International Airport’s main parking garage can be modeled by the function 2 F 1x2 = e 1 5 int 1x + 12 + 2 51
if if 1 if 3 if if 9
6 6 6 6 …
x x x x x
… … … 6 …
1 3 9 2
Determine the fee for parking in the short-term parking garage for (a) 2 hours C IPVST (c) 15 hours $51 E IPVSTBOENJOVUFT Source: O’Hare International Airport 49. Cost of Natural Gas *O.BSDI 1FPQMFT&OFSHZIBEUIF following rate schedule for natural gas usage in single-family residences: Monthly service charge $22.25 1FSUIFSNTFSWJDFDIBSHF 'JSTUUIFSNT UIFSN 0WFSUIFSNT UIFSN (BTDIBSHF UIFSN * B 8IBUJTUIFDIBSHFGPSVTJOHUIFSNTJOBNPOUI * C 8IBUJTUIFDIBSHFGPSVTJOHUIFSNTJOBNPOUI *(c) Develop a function that models the monthly charge C for x therms of gas. *(d) Graph the function found in part (c). Source: Peoples Energy, Chicago, Illinois, 2013
SECTION 1.4 Library of Functions; Piecewise-defined Functions
50. Cost of Natural Gas *O'FCSVBSZ -BDMFEF(BTIBEUIF following rate schedule for natural gas usage in single-family residences: .POUIMZDVTUPNFSDIBSHF Distribution charge 'JSTUUIFSNT UIFSN 0WFSUIFSNT UIFSN (BTTVQQMZDIBSHF UIFSN
119
* B 8IBUJTUIFDIBSHFGPSVTJOHUIFSNTJOBNPOUI * C 8IBUJTUIFDIBSHFGPSVTJOHUIFSNTJOBNPOUI *(c) Develop a model that gives the monthly charge C for x therms of gas. *(d) Graph the function found in part (c). Source: Laclede Gas, 2013 *51. Federal Income Tax 3FGFSUPUIF5BY3BUF4DIFEVMFT If x equals taxable income and y equals the tax due, construct a function y = f 1x2 for Schedule X.
*52. Federal Income Tax 3FGFSUPUIFUBYSBUFTDIFEVMFT*Gx equals taxable income and y equals the tax due, construct a function y = f 1x2 for Schedule Y-1. 2013 Tax Rate Schedules Schedule X-Single If Taxable Income Is Over
But Not Over
The Tax Is This Amount
Schedule Y-l - Married Filing Jointly or Qualified Widow(er) Plus This %
Of the Excess Over
If Taxable Income Is Over
But Not Over
The Tax Is This Amount
Plus This %
Of the Excess Over
$0
$8,925
$0
+
10%
$0
$0
$17,850
$0
+
10%
$0
8,925
36,250
892.50
+
15%
8,925
17,850
72,500
1,785
+
15%
17,850
36,250
87,850
4,991.25
+
25%
36,250
72,500
146,400
9,982.50
+
25%
72,500
87,850
183,250
17,891.25
+
28%
87,850
146,400
223,050
28,457.50
+
28%
146,400
183,250
398,350
44,603.25
+
33%
183,250
223,050
398,350
49,919.50
+
33%
223,050
398,350
400,000
115,586.25
+
35%
398,350
398,350
450,000
107,768.50
+
35%
398,350
400,000
–
116,163.75
+
39.6%
400,000
450,000
–
125,846
+
39.6%
450,000
Source: Internal Revenue Service
53. Cost of Transporting Goods A trucking company transports HPPET CFUXFFO $IJDBHP BOE /FX:PSL B EJTUBODF PG miles. The company’s policy is to charge, for each pound, QFSNJMFGPSUIFGJSTUNJMFT QFSNJMFGPSUIF OFYUNJMFT QFSNJMFGPSUIFOFYUNJMFT BOEOP DIBSHFGPSUIFSFNBJOJOHNJMFT *(a) Graph the relationship between the cost of transportation JOEPMMBSTBOENJMFBHFPWFSUIFFOUJSFNJMFSPVUF (b) Find the cost as a function of mileage for hauls between BOENJMFTGSPN$IJDBHP C 1x2 = 1 + .x (c) Find the cost as a function of mileage for hauls between BOENJMFTGSPN$IJDBHP C 1x2 = + .25x *54. Car Rental Costs An economy car rented in Florida from Enterprise® PO B XFFLMZ CBTJT DPTUT QFS XFFL &YUSB EBZTDPTUQFSEBZVOUJMUIFEBZSBUFFYDFFETUIFXFFLMZ 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 function of the number x of days used, where … x … 1. Graph this function. Source: enterprise.com 55. Mortgage Fees Fannie Mae charges a loan-level price BEKVTUNFOU --1" POBMMNPSUHBHFT XIJDISFQSFTFOUTBGFF homebuyers seeking a loan must pay. The rate paid depends on the credit score of the borrower, the amount borrowed, and the loan-to-value (LTV) ratio. The LTV ratio is the ratio of amount borrowed to appraised value of the home. For FYBNQMF BIPNFCVZFSXIPXJTIFTUPCPSSPX XJUI BDSFEJUTDPSFPGBOEBO-57SBUJPPGXJMMQBZ
PG PS5IFUBCMFTIPXTUIF--1"GPS WBSJPVTDSFEJUTDPSFTBOEBO-57SBUJPPG
Credit Score
Loan-Level Price Adjustment Rate
… 659
3.00%
660–679
2.50%
680–699
1.75%
700–719
1%
720–739
0.5%
Ú 740
0.25%
Source: Fannie Mae.
*(a) Construct a function C = C(s), where C is the loan-level QSJDFBEKVTUNFOU --1" BOEs is the credit score of an JOEJWJEVBMXIPXJTIFTUPCPSSPX XJUIBO LTV ratio. C 8IBUJTUIF--1"POB MPBOXJUIBO-57 SBUJPGPSBCPSSPXFSXIPTFDSFEJUTDPSFJT D 8IBUJTUIF--1"POB MPBOXJUIBO-57 SBUJPGPSBCPSSPXFSXIPTFDSFEJUTDPSFJT *56. 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 depends on the total amount owed. One such credit card
120
CHAPTER 1 Functions and Their Graphs
DPNQBOZVTFTUIFGPMMPXJOHSVMFT'PSBCJMMPGMFTTUIBO UIFFOUJSFBNPVOUJTEVF'PSBCJMMPGBUMFBTUCVUMFTT UIBO UIF NJOJNVN EVF JT " NJOJNVN PG JT EVFPOBCJMMPGBUMFBTUCVUMFTTUIBO BNJOJNVN PGJTEVFPOBCJMMPGBUMFBTUCVUMFTTUIBO BOEBNJOJNVNPGJTEVFPOCJMMTPGPSNPSF'JOE the function f that describes the minimum payment due on a bill of x dollars. Graph f.
B "OBJSUFNQFSBUVSFPG$BOEBXJOETQFFEPGNFUFS per second 1m/sec2 $ C "O BJS UFNQFSBUVSF PG $ BOE B XJOE TQFFE PG 5 m/sec °C D "O BJS UFNQFSBUVSF PG $ BOE B XJOE TQFFE PG 15 m/sec - 3°C E "O BJS UFNQFSBUVSF PG $ BOE B XJOE TQFFE PG 25 m/sec - °C *(e) Explain the physical meaning of the equation corresponding to … v 6 1.9. *(f) Explain the physical meaning of the equation corresponding to v 7 2.
57. Wind Chill The wind chill factor represents the equivalent 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 t W = d 33 -
11.5 + 11v - v2 133 - t2
22. 33 - 1.595133 - t2
… v 6 1.9 1.9 … v … 2 v 7 2
where v represents the wind speed (in meters per second) and tSFQSFTFOUTUIFBJSUFNQFSBUVSF $ $PNQVUFUIFXJOE chill for the following:
*58. Wind Chill 3FEP1SPCMFN B m E GPSBOBJSUFNQFSBUVSF of - 1°C. *59. First-class Mail *O UIF 64 1PTUBM 4FSWJDF DIBSHFE QPTUBHFGPSGJSTUDMBTTNBJMSFUBJMGMBUT TVDIBTBO CZ FOWFMPQF XFJHIJOH VQ UP PVODF QMVT GPS each additional ounce up to 13 ounces. First-class rates do not apply to flats weighing more than 13 ounces. Develop a model that relates C, the first-class postage charged, for a flat weighing x ounces. Graph the function. Source: United States Postal Service
Discussion and Writing *65. Exploration Graph y = x3. Then on the same screen graph y = 1x - 12 3 + 2. Could you have predicted the result?
In Problems 60–67, use a graphing utility. *60. Exploration Graph y = x . Then on the same screen graph y = x2 + 2, followed by y = x2 + , followed by y = x2 - 2. What pattern do you observe? Can you predict the graph of y = x2 - ? Of y = x2 + 5? 2
*66. Exploration Graph y = x2, y = x, and y = x on the same screen. What do you notice is the same about each graph? What do you notice is different?
*61. Exploration Graph y = x2. Then on the same screen graph y = 1x - 22 2, followed by y = 1x - 2 2, followed by y = 1x + 22 2. What pattern do you observe? Can you predict the graph of y = 1x + 2 2? Of y = 1x - 52 2?
*67. Exploration Graph y = x3, y = x5, and y = x on the same screen. What do you notice is the same about each graph? What do you notice is different? *68. Consider the equation
*62. Exploration Graph y = 0 x 0 . Then on the same screen graph 1 y = 2 0 x 0 , followed by y = 0 x 0 , followed by y = 0 x 0 . 2 What pattern do you observe? Can you predict the graph of 1 y = 0 x 0 ? Of y = 5 0 x 0 ? *63. Exploration Graph y = x2. Then on the same screen graph y = - x2. What pattern do you observe? Now try y = 0 x 0 and y = - 0 x 0 . What do you conclude?
y = b
1
if 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?
69. Define some functions that pass through 1, 2 and 11, 12 and are increasing for x Ú . Begin your list with y = 1x, y = x, and y = x2. Can you propose a general result about such functions?
*64. Exploration Graph y = 1x. Then on the same screen graph y = 1- x. What pattern do you observe? Now try y = 2x + 1 and y = 21 - x2 + 1. What do you conclude?
‘Are You Prepared?’ Answers 1. y
2.
2 (1, 1)
y 2
(4, 2)
3. 1, - 2, 12, 2 (1, 1) 2 x
(0, 0)
4
x
(⫺1, ⫺1)
SECTION 1.5 Graphing Techniques: Transformations
121
1.5 Graphing Techniques: Transformations OBJECTIVES 1 Graph Functions Using Vertical and Horizontal Shifts (p. 121) 2 Graph Functions Using Compressions and Stretches (p. 124) 3 Graph Functions Using Reflections about the x-Axis and the y-Axis (p. 126)
At this stage, if you were asked to graph any of the functions defined by 1 3 y = x, y = x2, y = x3, y = 1x, y = 2 x, y = , or y = 0 x 0 , your response should x be “Yes, I recognize these functions and know the general shapes of their graphs.” *GUIJTJTOPUZPVSBOTXFS SFWJFXUIFQSFWJPVTTFDUJPO 'JHVSFTUISPVHI
Sometimes we are asked to graph a function that is “almost” like one that we already know how to graph. In this section, we develop techniques for graphing such functions. Collectively, these techniques are referred to as transformations.
1 Graph Functions Using Vertical and Horizontal Shifts EXAM PL E 1
Vertical Shift Up Use the graph of f1x2 = x2 to obtain the graph of g1x2 = x2 + 3. Find the domain and range of g.
Solution
Table 7
Begin by obtaining some points on the graphs of f and g. For example, when x = , then y = f12 = and y = g12 = 3. When x = 1, then y = f112 = 1 and y = g112 = .5BCMFMJTUTUIFTFBOEBGFXPUIFSQPJOUTPOFBDIHSBQI/PUJDFUIBU 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,FYDFQUUIBUJUJTTIJGUFEWFSUJDBMMZVQVOJUT4FF'JHVSF
Figure 43
x
y = f1 x2 = x2
y = g1x2 = x2 + 3
-2
4
7
(⫺2, 7)
-1
1
4
(⫺1, 4)
0
0
3
1
1
4
2
4
7
y = x2 + 3 y (2, 7) (1, 4) 5 (⫺2, 4)
Up 3 units
(2, 4) (0, 3) y = x 2
(⫺1, 1) ⫺3
(1, 1) (0, 0)
3
x
The domain of g is all real numbers, or ( - q , q ). The range of g is [3, q ).
EXAM PL E 2
r
Vertical Shift Down Use the graph of f1x2 = x2 to obtain the graph of g1x2 = x2 - . Find the domain and range of g.
Solution
5BCMFPOUIFOFYUQBHFMJTUTTPNFQPJOUTPOUIFHSBQITPGf and g. Notice that each y-coordinate of gJTVOJUTMFTTUIBOUIFDPSSFTQPOEJOHy-coordinate of f. To obtain the graph of g from the graph of f TVCUSBDUGSPNFBDIy-coordinate on the graph of f. The graph of g is identical to that of f, except that it is shifted down VOJUT4FF'JHVSFon the next page.
122
CHAPTER 1 Functions and Their Graphs
Table 8
y = f1 x2 = x2
x
Figure 44
y = g1x2 = x2 − 4
-2
4
0
-1
1
-3
0
0
-4
1
1
-3
2
4
0
y (–2, 4)
y = x2
4
(2, 4)
Down 4 units
Down 4 units (2, 0) 4 x
(⫺2, 0) (0, 0)
y = x2 ⫺ 4 ⫺5
(0, ⫺4)
The domain of g is all real numbers, or 1 - q , q 2 . The range of g is [ - , q ).
r
Note that a vertical shift affects only the range of a function, not the domain. For example, the range of f1x2 = x2 is 3 , q 2 . In Example 1 the range of g is 3 3, q 2 , whereas in Example 2 the range of g is 3 - , q 2 . The domain for all three functions is all real numbers. We are led to the following conclusions: If a positive real number k is added to the output of a function y = f1x2, the graph of the new function y = f1x2 + k is the graph of f shifted vertically up k units. If a positive real number k is subtracted from the output of a function y = f1x2, the graph of the new function y = f1x2 - k is the graph of f shifted vertically down k units.
Now Work
PROBLEM
39
Horizontal Shift to the Right
EX A MPL E 3
Use the graph of f1x2 = 2x to obtain the graph of g1x2 = 1x - 2. Find the domain and range of g.
Solution
Table 9
The function g1x2 = 1x - 2 is basically a square root function. Table 9 lists some points on the graphs of f and g. Note that when f1x2 = then x = , and when g1x2 = , then x = 2. Also, when f1x2 = 2, then x = , and when g1x2 = 2, then x = . Notice that the x-coordinates on the graph of g are 2 units larger than the corresponding x-coordinates on the graph of f for any given y-coordinate. We conclude that the graph of g is identical to that of f, except that it is shifted horizontally 2 units to the right. See 'JHVSF y = f1 x2
y = g1x2
x
= 2x
x
= 2x − 2
0
0
2
0
1
1
3
1
4
2
6
2
9
3
11
3
Figure 45
y 5
y = "x
Right 2 units (4, 2)
y = "x − 2
(6, 2) (0, 0)
(2, 0)
9
x
Right 2 units
The domain of g is 3 2, q 2 and the range is 3 , q 2 .
EX A MPL E 4
r
Horizontal Shift to the Left Use the graph of f1x2 = 1x to obtain the graph of g1x2 = 1x + . Find the domain and range of g.
Solution
Again, the function g1x2 = 1x + is basically a square root function. Its graph is the same as that of f,FYDFQUUIBUJUJTTIJGUFEIPSJ[POUBMMZVOJUTUPUIFMFGU4FF'JHVSF
123
SECTION 1.5 Graphing Techniques: Transformations
Figure 46
y 5
y = "x + 4
Left 4 units
(0, 2) (4, 2)
(−4, 0) −5
(0, 0)
y = "x x
5
Left 4 units
The domain of g is 3 - , q 2 and the range is 3 , q 2 .
Now Work
PROBLEM
r
43
Note that a horizontal shift affects only the domain of a function, not the range. For example, the domain of f1x2 = 1x is 3 , q 2 . In Example 3 the domain of g is 3 2, q 2 XIFSFBTJO&YBNQMFUIFEPNBJOPGg is 3 - , q 2 . The range for all three functions is 3 , q 2 . We are led to the following conclusion.
In Words
For y = f1x - h2, add h to each x-coordinate on the graph of y = f1x2. For y = f1x + h2, subtract h from each x-coordinate on the graph of y = f1x2.
EXAM PL E 5
Solution
Figure 47
If the argument x of a function f is replaced by x - h, h 7 , the graph of the new function y = f1x - h2 is the graph of f shifted horizontally right h units. If the argument x of a function f is replaced by x + h, h 7 , the graph of the new function y = f1x + h2 is the graph of f shifted horizontally left h units. Observe the distinction between vertical and horizontal shifts. The graph of f1x2 = x3 + 2 is obtained by shifting the graph of y = x3 up 2 units, because we evaluate the cube function first and then add 2. The graph of g1x2 = 1x + 22 3 is obtained by shifting the graph of y = x3 left 2 units, because we add 2 to x before we evaluate the cube function. Vertical and horizontal shifts are sometimes combined.
Combining Vertical and Horizontal Shifts
Graph the function f1x2 = 0 x + 3 0 - 5. Find the domain and range of f. We graph f in steps. First, note that the rule for f is basically an absolute value function, so begin with the graph of y = 0 x 0 BTTIPXOJO'JHVSF B /FYU UPHFUUIFHSBQIPG y = 0 x + 3 0 , shift the graph of y = 0 x 0 IPSJ[POUBMMZVOJUTUPUIFMFGU4FF'JHVSF C Finally, to get the graph of y = 0 x + 3 0 - 5, shift the graph of y = 0 x + 3 0 vertically EPXOVOJUT4FF'JHVSF D /PUFUIFQPJOUTQMPUUFEPOFBDIHSBQI6TJOHLFZQPJOUT can be helpful in keeping track of the transformation that has taken place.
y
y
y
5
5
5 (−1, 2)
(−2, 2)
(2, 2) (0, 0)
(−5, 2) 5
x
(−3, 0)
2
x
(−1, −3)
(−5, −3)
y = 0x 0 (a)
Replace x by x + 3; Horizontal shift left 3 units
y = 0 x + 30 (b)
Subtract 5: Vertical shift down 5 units
x
2
(−3, −5) y = 0 x + 30 − 5 (c)
The domain of f is all real numbers, or 1 - q , q 2 . The range of f is 3 - 5, q 2 .
Check: Graph Y1 = f 1x2 = 0 x + 3 0 - 5BOEDPNQBSFUIFHSBQIUP'JHVSF D
r
124
CHAPTER 1 Functions and Their Graphs
In Example 5, if the vertical shift had been done first, followed by the horizontal shift, the final graph would have been the same. Try it for yourself.
Now Work
PROBLEMS
45
AND
75
2 Graph Functions Using Compressions and Stretches EX A MPL E 6
Vertical Stretch Use the graph of f1x2 = 2x to obtain the graph of g1x2 = 22x.
Solution
Table 10
EX A MPL E 7
To see the relationship between the graphs of f and g XFGPSN5BCMF MJTUJOHQPJOUT on each graph. For each x, the y-coordinate of a point on the graph of g is 2 times as large as the corresponding y-coordinate on the graph of f. The graph of f1x2 = 2x is vertically stretched by a factor of 2 to obtain the graph of g1x2 = 22x [for example, 11, 12 is on the graph of f, but 11, 22 is on the graph of g>4FF'JHVSF
x
y = f1x2
y = g1x2
= 2x
= 22x
0
0
0
1
1
2
4
2
4
9
3
6
y = 2"x
Figure 48 y 5 (1, 2)
y = "x
(4, 4) (4, 2)
(0, 0)
10 x
5
(1, 1)
r
Vertical Compression 1 Use the graph of f1x2 = 0 x 0 to obtain the graph of g1x2 = 0 x 0 . 2
Solution
Table 11
1 as large as the 2 corresponding y-coordinate on the graph of f. The graph of f1x2 = 0 x 0 is vertically 1 1 compressed by a factor of to obtain the graph of g1x2 = 0 x 0 . For example, 12, 22 2 2 is on the graph of f, but 12, 12 is on the graph of g4FF5BCMFBOE'JHVSF
For each x, the y-coordinate of a point on the graph of g is
y = f1 x2 x -2
In Words
For y = af1x2 , the factor a is “outside” the function, so it affects the y-coordinates. Multiply each y-coordinate on the graph of y = f1x2 by a.
= 0x0 2
-1
1
0
0
1
1
2
2
Figure 49
y = g1x2 1 = 0x0 2
y
y =⏐x⏐
4
1 1 2 0 1 2 1
y= (2, 2)
(2, 2)
(– 2, 1) 4
1– x ⏐ ⏐ 2
(2, 1) (0, 0)
4 x
r
When the right side of a function y = f1x2 is multiplied by a positive number a, the graph of the new function y = af1x2 is obtained by multiplying each y-coordinate on the graph of y = f1x2 by a. The new graph is a vertically compressed (if 6 a 6 1) or a vertically stretched (if a 7 1) version of the graph of y = f1x2.
Now Work
PROBLEM
47
What happens if the argument x of a function y = f1x2 is multiplied by a positive number a, creating a new function y = f1ax2? To find the answer, look at the following Exploration.
SECTION 1.5 Graphing Techniques: Transformations
125
Exploration On the same screen, graph each of the following functions:
Figure 50 3 Y2 =
2x
Y1 =
x
Y3 =
–x 2
4
0 0
1 1 x Y3 = f a x b = x = 2 A2 A2 Create a table of values to explore the relation between the x- and y-coordinates of each function. Y1 = f(x) = 1x
Y2 = f(2x) = 12x
Result You should have obtained the graphs in Figure 50. Look at Table 12(a). Notice that (1, 1), (4, 2), and (9, 3) are points on the graph of Y1 = 1x. Also, (0.5, 1), (2, 2), and (4.5, 3) are points on the graph of 1 Y2 = 22x. For a given y-coordinate, the x-coordinate on the graph of Y2 is of the x-coordinate on Y1. 2
Table 12
We conclude that the graph of Y2 = 22x is obtained by multiplying the x-coordinate of each point on 1 the graph of Y1 = 1x by . The graph of Y2 = 22x is the graph of Y1 = 1x compressed horizontally. 2 Look at Table 12(b). Notice that (1, 1), (4, 2), and (9, 3) are points on the graph of Y1 = 1x. Also x . For a given y-coordinate, notice that (2, 1), (8, 2), and (18, 3) are points on the graph of Y3 = A2 the x-coordinate on the graph of Y3 is 2 times the x-coordinate on Y1. We conclude that the graph of x is obtained by multiplying the x-coordinate of each point on the graph of Y1 = 1x by 2. The Y3 = A2 x is the graph of Y1 = 1x stretched horizontally. graph of Y3 = A2
Based on the Exploration, we have the following result:
In Words
For y = f (ax), the factor a is “inside” the function, so it affects the x-coordinates. Multiply each x-coordinate on the graph of 1 y = f (x) by . a
If the argument x of a function y = f1x2 is multiplied by a positive number a, then the graph of the new function y = f 1ax2 is obtained by multiplying each 1 x-coordinate of y = f 1x2 by . A horizontal compression results if a 7 1, and a a horizontal stretch results if 6 a 6 1. Let’s look at an example.
EXAM PL E 8
Graphing Using Stretches and Compressions
The graph of y = f 1x2 is given in Figure 51. Use this graph to find the graphs of (a) y = 2f1x2
Solution
(a) The graph of y = 2f1x2 is obtained by multiplying each y-coordinate of y = f1x2 by 2. See Figure 52. (b) The graph of y = f 13x2 is obtained from the graph of y = f1x2 by multiplying 1 each x-coordinate of y = f1x2 by . See Figure 53. 3 Figure 52
Figure 51
1
1
( 2 , 1(
3 2 5 3 2 2 3 , 1 2
(
Figure 53
( 2 , 2(
(
y
( 52 , 2(
2
( 52 , 1(
y f(x)
2
y 3
y
(b) y = f13x2
2 1
1
x
( 6 , 1( ( 56 , 1(
2
3 2
2 52 3
1
(
3
(
3 , 2 2
2
PROBLEMS
y f(3x)
69(e)
AND
(g)
3
x
2 3
( 2 , 1 (
y 2f(x)
Now Work
2
x
1
r
126
CHAPTER 1 Functions and Their Graphs
3 Graph Functions Using Reflections about the x-Axis and the y-Axis Reflection about the x-Axis
EX A MPL E 9
Graph the function f1x2 = - x2. Find the domain and range of f.
Solution
Begin with the graph of y = x2,BTTIPXOJOCMBDLJO'JHVSF'PSFBDIQPJOU1x, y2 on the graph of y = x2, the point 1x, - y2 is on the graph of y = - x2, as indicated in Table 13. Draw the graph of y = - x2 by reflecting the graph of y = x2 about the xBYJT4FF'JHVSF x
y = x2
y = −x2
-2
4
-4
-1
1
-1
0
0
0
1
1
-1
2
4
-4
Table 13 Figure 54 y 4
(⫺2, 4)
(⫺1, 1) ⫺4
y = x2 (2, 4)
The domain of f is all real numbers, or 1 - q , q 2 . The range of f is 1 - q , 4 .
(1, 1)
(–1, –1)
(1, – 1)
(–2, –4)
–4
4
r
x
When the right side of the function y = f 1x2 is multiplied by - 1, the graph of the new function y = - f1x2 is the reflection about the x-axis of the graph of the function y = f1x2.
(2, – 4) y = –x 2
Now Work
EX AM PL E 10
PROBLEM
51
Reflection about the y-Axis Graph the function f1x2 = 1- x. Find the domain and range of f.
Solution
To get the graph of f1x2 = 1- x, begin with the graph of y = 1x, as shown in Figure 55. For each point 1x, y2 on the graph of y = 1x, the point 1 - x, y2 is on the graph of y = 1- x. Obtain the graph of y = 1- x by reflecting the graph of y = 1x about the y-axis. See Figure 55.
Figure 55
y 4 y=
y=
–x ( – 1, 1) –5
x
(4, 2)
( – 4, 2) (0, 0)
(1, 1) 5
x
The domain of f is 1 - q , 4 . The range of f is 3 , q 2.
In Words
For y = - f (x), multiply each y-coordinate on the graph of y = f1x2 by - 1. For y = f1 - x2 , multiply each x-coordinate by - 1.
When the graph of the function y = f 1x2 is known, the graph of the new function y = f1 - x2 is the reflection about the y-axis of the graph of the function y = f1x2.
r
SECTION 1.5 Graphing Techniques: Transformations
127
SUMMARY OF GRAPHING TECHNIQUES To Graph:
Draw the Graph of f and:
Functional Change to f(x)
Vertical shifts
y = f1x2 + k, k 7 y = f1x2 - k, k 7
Raise the graph of f by k units. Lower the graph of f by k units.
Horizontal shifts
y = f1x + h2 , h 7 y = f1x - h2, h 7
Shift the graph of f to the left h units. Shift the graph of f to the right h units.
Compressing or stretching
Add k to f 1x2. Subtract k from f1x2 .
Multiply each y-coordinate of y = f 1x2 by a. Stretch the graph of f vertically if a 7 1. Compress the graph of f vertically if 6 a 6 1. 1 Multiply each x-coordinate of y = f 1x2 by . a Stretch the graph of f horizontally if 6 a 6 1. Compress the graph of f horizontally if a 7 1.
Reflection about the x-axis
y = af 1x2, a 7 y = f1ax2 , a 7
y = - f 1x2
y = f1 - x2
Reflect the graph of f about the y-axis.
Solution
Replace x by ax. Multiply f1x2 by - 1.
Reflection about the y-axis
Instructor Note: Encourage students to write the current function after each transformation, and remind them that transformations are cumulative.
Multiply f1x2 by a.
Reflect the graph of f about the x-axis.
EX AM PL E 11
Replace x by x + h. Replace x by x - h.
Replace x by - x.
Determining the Function Obtained from a Series of Transformations Find the function that is finally graphed after the following three transformations are applied to the graph of y = 0 x 0 . 1. Shift left 2 units 2. Shift up 3 units 3. Reflect about the y-axis 1. Shift left 2 units: Replace x by x + 2. 2. Shift up 3 units: Add 3. 3. Reflect about the y-axis: Replace x by - x.
Now Work
PROBLEM
y = 0x + 20 y = 0x + 20 + 3 y = 0 -x + 20 + 3
27
r
The examples that follow combine some of the procedures outlined in this section to get the required graph.
EX AM PL E 12
Combining Graphing Procedures Graph the function f1x2 =
3 + 1. Find the domain and range of f . x - 2
128
CHAPTER 1 Functions and Their Graphs
Solution
1 It is helpful to write f as f(x) = 3 a b + 1. Now use the following steps to x - 2 obtain the graph of f : STEP 1: y =
1 x
Reciprocal function
1 3 STEP 2: y = 3 # a b = x x
1 Multiply by 3; vertical stretch of the graph of y = by x a factor of 3.
3 x - 2 3 STEP 4: y = + 1 x - 2
Replace x by x - 2; horizontal shift to the right 2 units.
STEP 3: y =
Add 1; vertical shift up 1 unit.
4FF'JHVSF Figure 56 y 4
y 4 (1, 1)
24
(1, 3)
(3, 3)
3 2, – 2
( )
(2, 12–) 4 x
y 4
y 4
24
(3, 4)
(4, 3–2 )
4 x
x
4
(21, 21)
(4, 5–2 ) 4 x
24 (1, 22)
(21, 23)
24 1 x
D y 5 ––
Multiply by 3; Vertical stretch
(1, 23) 24
24
24 3 x
E y 5 ––
Replace x by x 2 2; Horizontal shift right 2 units
3 x –2
F y 5 –––
Add 1; Vertical shift up 1 unit
3 x –2
G y 5 ––– 1 1
1 is {x 0 x ≠ } and its range is {y 0 y ≠ }. Because we x shifted right 2 units and up 1 unit to obtain f , the domain of f is {x 0 x ≠ 2} and its The domain of y =
range is {y 0 y ≠ 1}.
Hint: Although the order in which transformations are performed can be altered, you may consider using the following order for consistency: 1. Reflections 2. Compressions and stretches 3. Shifts ■
EX AM PL E 13
r
Other orderings of the steps shown in Example 12 would also result in the graph of f. For example, try this one: STEP 1: y =
1 x
1 x - 2 3 STEP 3: y = x - 2 3 STEP 4: y = + 1 x - 2 STEP 2: y =
Reciprocal function Replace x by x - 2; horizontal shift to the right 2 units. 1 Multiply by 3; vertical stretch of the graph of y = x - 2 by a factor of 3. Add 1; vertical shift up 1 unit.
Combining Graphing Procedures Graph the function f1x2 = 21 - x + 2. Find the domain and range of f .
Solution
Because horizontal shifts require the form x - h, begin by rewriting f1x2 as f1x2 = 21 - x + 2 = 2 - 1x - 12 + 2. Now use the following steps:
129
SECTION 1.5 Graphing Techniques: Transformations
STEP 1: y = 1x
Square root function
STEP 2: y = 2 - x
Replace x by - x; reflect about the y-axis.
STEP 3: y = 2 - (x - 1) = 21 - x
Replace x by x - 1; horizontal shift to the right 1 unit.
STEP 4: y = 21 - x + 2
Add 2; vertical shift up 2 units.
4FF'JHVSF Figure 57
(1, 1)
(4, 2)
(a) y
(4, 2)
y (3, 4) 5 (0, 3)
(3, 2)
(1, 2)
(0, 1) 5 x 5
(0, 0)
5
y 5
y 5
y 5
(1, 1) (0, 0)
5 x 5
(b) y x
x Replace x by x; Reflect about y-axis
5 x 5
(1, 0)
Replace x by x 1; (c) y Horizontal shift right 1 unit
5 x
(x 1) Add 2; (d) y x 1 Vertical shift up 2 units 1x
1x2
The domain of f is ( - q , 1] and the range is [2, q ).
Now Work
PROBLEM
r
61
1.5 Assess Your Understanding Concepts and Vocabulary 4. True or False The graph of y = - f 1x2 is the reflection about the x-axis of the graph of y = f 1x2. True
1. Suppose that the graph of a function f is known. Then the graph of y = f 1x - 22 may be obtained by a(n) horizontal shift of the graph of f to the right
5. True or False To obtain the graph of f(x) = 2x + 2, shift the graph of y = 2x horizontally to the right 2 units. False
a distance of 2 units. 2. Suppose that the graph of a function f is known. Then the graph of y = f 1 - x2 may be obtained by a reflection y about the -axis of the graph of the function y = f 1x2.
6. True or False To obtain the graph of f(x) = x3 + 5, shift the graph of y = x3 vertically up 5 units. True
3. Suppose that the graph of a function g is known. The graph vertical of y = g(x) + 2 may be obtained by a up shift of the graph of g a distance of 2 units.
Skill Building In Problems 7–18, match each graph to one of the following functions: A. y = x2 + 2
E. y = 1x - 22
F. y = - 1x + 22
2
I. y = 2x2 7.
C. y = 0 x 0 + 2
B. y = - x2 + 2
B
8.
E
y 3
9.
3 x
3
3 x
L. y = - 2 0 x 0 H
y 1 3
3
H. y = - 0 x + 2 0
K. y = 2 0 x 0
J. y = - 2x2 y 3
D. y = - 0 x 0 + 2
G. y = 0 x - 2 0
2
10.
D
y 3
3x 3
3 x
130
CHAPTER 1 Functions and Their Graphs
11.
I
y 3
12.
A
y 5
13.
3 x
3
L
y 3
3
14.
3 x 6
3
3
15.
F
y 4
16.
3 x
3
4
17.
y 4
G
4 x
4
6 x 4
3
J
y 3
3
4 x
4
1
3 x
C
y 8
18.
K
y 3
3 x
3
4
3
In Problems 19–26, write the function whose graph is the graph of y = x3, but is: 19. 4IJGUFEUPUIFSJHIUVOJUT y = (x - )3
20. 4IJGUFEUPUIFMFGUVOJUTy = (x + )3
21. 4IJGUFEVQVOJUT y = x3 +
22. 4IJGUFEEPXOVOJUT y = x3 -
23. Reflected about the y-axis
y = -x
3
25. 7FSUJDBMMZTUSFUDIFECZBGBDUPSPG y = x3
24. Reflected about the x-axis
y = - x3
1 3 1 3 26. )PSJ[POUBMMZTUSFUDIFECZBGBDUPSPG y = a xb = x
In Problems 27–30, find the function that is finally graphed after each of the following transformations is applied to the graph of y = 1x in the order stated. 27. (1) Shift up 2 units (2) Reflect about the x-axis (3) Reflect about the y-axis
y = - ( 2- x + 2)
29. (1) Reflect about the x-axis (2) Shift up 2 units (3) Shift left 3 units y = - 2x + 3 + 2
28. (1) Reflect about the x-axis (2) Shift right 3 units (3) Shift down 2 units y = - 2x - 3 - 2 30. (1) Shift up 2 units (2) Reflect about the y-axis (3) Shift left 3 units y = 2- (x + 3) + 2 = 2- x - 3 + 2
31. If 13, 2 is a point on the graph of y = f 1x2, which of the following points must be on the graph of y = - f 1x2? (c) (a) 1, 32 (b) 1, - 32 (c) 13, - 2 (d) 1 - 3, 2
32. If 13, 2 is a point on the graph of y = f 1x2, which of the following points must be on the graph of y = f 1 - x2? (d) (a) 1, 32 (b) 1, - 32 (c) 13, - 2 (d) 1 - 3, 2
33. If 11, 32 is a point on the graph of y = f 1x2, which of the following points must be on the graph of y = 2f 1x2? (c) (a) 11, 32 (b) 12, 32
34. If 1, 22 is a point on the graph of y = f 1x2, which of the following points must be on the graph of y = f(2x)? (c) (a) 1, 12 (b) 1, 22 (c) 12, - 22 (d) 1, 2
(c) 11, 2
1 (d) a , 3b 2 35. Suppose that the x-intercepts of the graph of y = f 1x2 are - 5 and 3. *(a) What are the x-intercepts of the graph of y = f 1x + 22? *(b) What are the x-intercepts of the graph of y = f 1x - 22? *(c) What are the x-intercepts of the graph of y = f 1x2? *(d) What are the x-intercepts of the graph of y = f 1 - x2? 37. Suppose that the function y = f 1x2 is increasing on the interval 1 - 1, 52. (a) Over what interval is the graph of y = f 1x + 22 increasing? ( - 3, 32 (b) Over what interval is the graph of y = f 1x - 52 increasing? (, 12 *(c) What can be said about the graph of y = - f 1x2? *(d) What can be said about the graph of y = f 1 - x2?
36. Suppose that the x-intercepts of the graph of y = f 1x2 are - and 1. *(a) What are the x-intercepts of the graph of y = f 1x + 2? *(b) What are the x-intercepts of the graph of y = f 1x - 32? *(c) What are the x-intercepts of the graph of y = 2f 1x2? *(d) What are the x-intercepts of the graph of y = f 1 - x2? 38. Suppose that the function y = f 1x2 is decreasing on the interval 1 - 2, 2. (a) Over what interval is the graph of y = f 1x + 22 decreasing? ( - , 52 (b) Over what interval is the graph of y = f 1x - 52 decreasing? (3, 122 *(c) What can be said about the graph of y = - f 1x2? *(d) What can be said about the graph of y = f 1 - x2?
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
SECTION 1.5 Graphing Techniques: Transformations
131
In Problems 39–68, 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. Be sure to show at least three key points. Find the domain and the range of each function. *40. f 1x2 = x2 +
*39. f 1x2 = x2 - 1 *42. g1x2 = x - 1
*45. f 1x2 = 1x - 12 3 + 2 *48. g1x2 =
*41. g1x2 = x3 + 1
*43. h1x2 = 2x + 2
3
*44. h1x2 = 2x + 1
*46. f 1x2 = 1x + 22 3 - 3
1 1x 2
*49. h1x2 =
3 *51. f 1x2 = - 2 x
*47. g1x2 = 1x
1 2x
3 *50. h1x2 = 22x
*52. f 1x2 = - 1x
3 *53. g1x2 = 2 -x
*55. h1x2 = - x3 + 2
*56. h1x2 =
*57. f 1x2 = 21x + 12 2 - 3
*58. f 1x2 = 31x - 22 2 + 1
*59. g1x2 = 22x - 2 + 1
*60. g1x2 = 3 0 x + 1 0 - 3
*61. h1x2 = 1- x - 2
*62. h1x2 =
*63. f 1x2 = - 1x + 12 3 - 1
*64. f 1x2 = - 2x - 1
*65. g1x2 = 2 0 1 - x 0
*54. g1x2 =
1 -x
*67. h1x2 = 2 int 1x - 12
*66. g1x2 = 22 - x
1 + 2 -x + 2 x
*68. h1x2 = int 1 - x2
In Problems 69–72, 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 1x2 = f 1x2 + 3 (e) Q1x2 = 69.
1 f 1x2 2
y 4 (0, 2)
(b) G1x2 = f 1x + 22
(c) P 1x2 = - f 1x2
(f) g1x2 = f 1 - x2
(g) h1x2 = f 12x2 *71.
y 4
*70. (2, 2)
1
(2, 2)
2 (4, 0)
4 (4, 2)
2
x
4 2 (4, 2)
2
y
4 x
2
2 (2, 2)
(4, 2)
π
(d) H1x2 = f 1x + 12 - 2
π –2
π– 2
1 π ( –2 , 1)
y
*72.
(π–2 , 1)
1
π
x
π
π –2
(π, 1)
π– 1
2
π
x
(π, 1)
Mixed Practice In Problems 73–82, complete the square of each quadratic expression. Then graph each function using the technique of shifting. (If necessary, refer to Appendix A, Section A.4 to review completing the square.)
*73. f 1x2 = x2 + 2x
*77. f 1x2 = x + x + 1 2
*81. f 1x2 = - 3x2 - 12x - 1
*74. f 1x2 = x2 - x
*78. f 1x2 = x - x + 1 2
*82. f 1x2 = - 2x2 - 12x - 13
83. (a) Graph f(x) = x - 3 - 3 using transformations. (b) Find the area of the region bounded by f and the x-axis that lies below the x-axis.
*75. f 1x2 = x2 - x + 1
*79. f 1x2 = 2x - 12x + 19 2
*76. f 1x2 = x2 + x + 2
*80. f 1x2 = 3x2 + x + 1
84. (a) Graph f(x) = - 2 x - + using transformations. (b) Find the area of the region bounded by f and the x-axis that lies above the x-axis.
Applications and Extensions
*86. 3FQFBU1SPCMFNGPSUIFGBNJMZPGQBSBCPMBTy = x2 + c. 87. Thermostat Control Energy conservation experts estimate UIBU IPNFPXOFST DBO TBWF UP PO XJOUFS IFBUJOH CJMMT CZ QSPHSBNNJOH UIFJS UIFSNPTUBUT UP EFHSFFT lower while sleeping. In the given graph, the temperature T (in degrees Fahrenheit) of a home is given as a function of time t JOIPVSTBGUFSNJEOJHIU PWFSBIPVSQFSJPE
T 80 Temperature (°F )
*85. The equation y = 1x - c2 2 defines a family of parabolas, one parabola for each value of c. On one set of coordinate axes, graph the members of the family for c = , c = 3, and c = - 2.
76 72 68 64 60 56 0
t 4 8 12 16 20 24 Time (hours after midnight)
132
CHAPTER 1 Functions and Their Graphs
(a) At what temperature is the thermostat set during daytime hours? At what temperature is the thermostat set overnight? ' ' *(b) The homeowner reprograms the thermostat to y = T 1t2 - 2. Explain how this affects the temperature in the house. Graph this new function. *(c) The homeowner reprograms the thermostat to y = T 1t + 12. Explain how this affects the temperature in the house. Graph this new function. Source: Roger Albright, 547 Ways to Be Fuel Smart, 2000 88. Digital Music Revenues The total projected worldwide digital music revenues R, in millions of dollars, for the years UISPVHIDBOCFFTUJNBUFECZUIFGVODUJPO R 1x2 = 2.x + 3x + 3 2
where xJTUIFOVNCFSPGZFBSTBGUFS *(a) Find R 12, R 132, and R 152 and explain what each value represents.* r(x) = 2. x2 + 15. x + 35. (b) Find r = R 1x - 22.
*(c) Find r 122, r 152, and r 12 and explain what each value represents. *(d) In the model r, what does x represent? (e) Would there be an advantage in using the model r when estimating the projected revenues for a given year instead of the model R? Source: IFPI Digital Music Report, 2013
*89. Temperature Measurements The relationship between UIF $FMTJVT $ BOE 'BISFOIFJU ' TDBMFT GPS NFBTVSJOH temperature is given by the equation F =
*(a) Use a graphing utility to graph the function T = T 1l2. *(b) Now graph the functions T = T 1l + 12, T = T 1l + 22, and T = T 1l + 32. (c) Discuss how adding to the length l changes the period T. *(d) Now graph the functions T = T 12l2, T = T 13l2, and T = T 1l2. (e) Discuss how multiplying the length l by factors of 2, 3, BOEDIBOHFTUIFQFSJPET. 91. Cigar Company Profits The daily profits of a cigar company from selling x cigars are given by p1x2 = - .5x2 + 1x - 2 The government wishes to impose a tax on cigars (sometimes called a sin tax) that gives the company the option of either QBZJOHBGMBUUBYPG QFSEBZPSBUBYPGPOQSPGJUT 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 = p1x2 - 1, and Y2 = 11 - .12p1x2. (b) Based on the graph, which option would you select? Why? UBY *(c) Using the terminology learned in this section, describe each graph in terms of the graph of p1x2. (d) Suppose that the government offered the options of a GMBUUBYPGPSBUBYPGPOQSPGJUT8IJDIXPVME you select? Why? UBY 92. The graph of a function f is illustrated in the figure. *(a) Draw the graph of y = 0 f 1x2 0 . *(b) Draw the graph of y = f 1 0 x 0 2.
9 C + 32 5
5IF SFMBUJPOTIJQ CFUXFFO UIF $FMTJVT $ BOE ,FMWJO , 9 scales is K = C + 23. Graph the equation F = C + 32 5 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. Period of a Pendulum The period T (in seconds) of a simple pendulum is a function of its length l (in feet) defined by the equation T = 2p
y 2 (1, 1) 3 (2, 1)
(2, 0) 3 x (1, 1) 2
93. The graph of a function f is illustrated in the figure. *(a) Draw the graph of y = 0 f 1x2 0 . *(b) Draw the graph of y = f 1 0 x 0 2.
l Ag
y 2
where g ≈ 32.2 feet per second per second is the acceleration of gravity.
(1, 1)
(22, 0) 23
(2, 0) 3 x
(21, 21) (0, 21) 22
94. Suppose (1, 3) is a point on the graph of y = f 1x2 . (a) What point is on the graph of y = f 1x + 32 - 5? 1 - 2, - 22 (b) What point is on the graph of y = - 2f 1x - 22 + 1? (3, - 5) (c) What point is on the graph of y = f 12x + 32? ( - 1, 3)
95. Suppose 1 - 3, 52 is a point on the graph of y = g1x2 . (a) What point is on the graph of y = g1x + 12 - 3? ( - , 2) (b) What point is on the graph of y = - 3g1x - 2 + 3? (1, - 12) (c) What point is on the graph of y = g13x + 92? ( - , 5)
SECTION 1.6 Mathematical Models: Building Functions
133
Discussion and Writing *96. Suppose that the graph of a function f is known. Explain how the graph of y = f 1x2 differs from the graph of y = f 1x2. *97. Suppose that the graph of a function f is known. Explain how the graph of y = f 1x2 - 2 differs from the graph of y = f 1x - 22. 98. The area under the curve y = 1x bounded below by the 1 square units. Using the x-axis and on the right by x = is 3
ideas presented in this section, what do you think is the area under the curve of y = 1- x bounded below by the x-axis and on the left by x = - ? Justify your answer. *99. Explain how the range of the function f1x2 = x2 compares to the range of g1x2 = f 1x2 + k. *100. Explain how the domain of g1x2 = 2x compares to the domain of g1x - k2, where k Ú . 1 sq. units 3
1.6 Mathematical Models: Building Functions OBJECTIVE 1 Build and Analyze Functions (p. 133)
1 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 language of mathematics. This is done by assigning symbols to represent the independent and dependent variables and then finding the function or rule that relates these variables.
EXAM PL E 1
Finding the Distance from the Origin to a Point on a Graph Let P = 1x, y2 be a point on the graph of y = x2 - 1.
(a) Express the distance d from P to the origin O as a function of x. (b) What is d if x = ? (c) What is d if x = 1? 22 (d) What is d if x = ? 2 (e) Use a graphing utility to graph the function d = d 1x2, x Ú . 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.]
Solution
Figure 58
B 'JHVSFJMMVTUSBUFTUIFHSBQIPGy = x2 - 1. The distance d from P to O is
y
d = 2 1x - 2 2 + 1y - 2 2 = 2x2 + y2
2 y ⫽ x2 ⫺ 1 1 P⫽ (x, y) (0, 0) d 1 ⫺1 2 x
Since P is a point on the graph of y = x2 - 1, substitute x2 - 1 for y. Then d 1x2 = 2x2 + 1x2 - 12 2 = 2x - x2 + 1 The distance d is expressed as a function of x. (b) If x = , the distance d is
⫺1
d 12 = 2 - 2 + 1 = 21 = 1 (c) If x = 1, the distance d is
Figure 59
d 112 = 21 - 12 + 1 = 1
2
22 , the distance d is 2 22 22 22 2 1 1 23 da b = a b - a b + 1 = - + 1 = 2 2 2 B 2 B 2
(d) If x =
2
0 0
(e) Figure 59 shows the graph of Y1 = 2x - x2 + 1. Using the MINIMUM feature on a graphing utility, we find that when x ≈ .1, the value of d is
134
CHAPTER 1 Functions and Their Graphs
smallest. The local minimum is d ≈ . rounded to two decimal places. Since d 1x2 is even, it follows by symmetry that when x ≈ - .1, the value of d is also a local minimum. Since 1 {.12 2 - 1 ≈ - .5, the points 1 - .1, - .52 and 1.1, - .52 on the graph of y = x2 - 1 are closest to the origin.
r
Now Work
EX A MPL E 2
PROBLEM
1
Area of a Rectangle 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 'JHVSF (a) (b) (c) (d)
Solution
Figure 60 y 30 20
(x, y) y ⫽ 25 ⫺ x 2
10 ⫺1
1 2 3 4 5
x
Express the area A of the rectangle as a function of x. What is the domain of A? Graph A = A 1x2. 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 1x2 = x125 - x2 2 = 25x - x3. (b) Since 1x, y2 is in quadrant I, we have x 7 . Also, y = 25 - x2 7 , which implies that x2 6 25, so - 5 6 x 6 5. Combining these restrictions, we have the domain of A as 5 x 6 x 6 56 , or 1, 52 using interval notation. (c) 4FF'JHVSFGPSUIFHSBQIPGA = A 1x2. (d) 6TJOH ."9*.6. XF GJOE UIBU UIF NBYJNVN BSFB JT TRVBSF VOJUT BU x = 2.9VOJUT FBDISPVOEFEUPUXPEFDJNBMQMBDFT4FF'JHVSF Figure 61
(0,0)
Figure 62
50
50
0
5
0
0
Now Work
EX A MPL E 3
5 0
PROBLEM
7
r
Close Call? Suppose two planes flying at the same altitude are headed toward each other. One QMBOFJTGMZJOHEVFTPVUIBUBHSPVOETQFFEPGNJMFTQFSIPVSBOEJTNJMFT from the potential intersection point of the planes. The other plane is flying due XFTUXJUIBHSPVOETQFFEPGNJMFTQFSIPVSBOEJTNJMFTGSPNUIFQPUFOUJBM JOUFSTFDUJPOQPJOUPGUIFQMBOFT4FF'JHVSF (a) Build a model that expresses the distance d between the planes as a function of time t. (b) Use a graphing utility to graph d = d 1t2 . How close do the planes come to each other? At what time are the planes closest?
Solution
B 3FGFSUP'JHVSF5IFEJTUBODFd between the two planes is the hypotenuse of a right triangle. At any time t the length of the north/south leg of the triangle is - t. At any time t, the length of the east/west leg of the triangle is - 25t. Using UIF1ZUIBHPSFBO5IFPSFN UIFTRVBSFPGUIFEJTUBODFCFUXFFOUIFUXPQMBOFTJT d 2 = 1 - t2 2 + 1 - 25t2 2
Therefore, the distance between the two planes as a function of time is given by the model d 1t2 = 2 1 - t2 2 + 1 - 25t2 2
SECTION 1.6 Mathematical Models: Building Functions
(b) 'JHVSF B TIPXT UIF HSBQI PG d = d 1t2 . Using MINIMUM, the minimum EJTUBODFCFUXFFOUIFQMBOFTJTNJMFT BOEUIFUJNFBUXIJDIUIFQMBOFTBSF DMPTFTUJTBGUFSIPVST FBDISPVOEFEUPUXPEFDJNBMQMBDFT4FF'JHVSF C
Figure 63 N Plane
135
400 mph
Figure 64 500 600 miles
d Plane 250 mph E 400 miles
0
2
⫺50
Now Work
r
(b)
(a) PROBLEM
19
1.6 Assess Your Understanding Applications and Extensions 1. Let P = 1x, y2 be a point on the graph of y = x2 - . (a) Express the distance d from P to the origin as a function of x. d(x) = 2x - 15x2 + (b) What is d if x = ? (c) What is d if x = 1? 25 ≈ . *(d) Use a graphing utility to graph d = d1x2. (e) For what values of x is d smallest? ≈ - 2. or ≈2. 2. Let P = 1x, y2 be a point on the graph of y = x2 - . (a) Express the distance d from P to the point 1, - 12 as a function of x. d1x2 = 2x - 13x2 + 9 (b) What is d if x = ? (c) What is d if x = - 1? 23 ≈ . *(d) Use a graphing utility to graph d = d1x2. (e) For what values of x is d smallest? ≈ - 2.55 or ≈2.55
3. Let P = 1x, y2 be a point on the graph of y = 1x. (a) Express the distance d from P to the point 11, 2 as a function of x. d1x2 = 2x2 - x + 1 *(b) Use a graphing utility to graph d = d1x2. (c) For what values of x is d smallest? .5 1 4. Let P = 1x, y2 be a point on the graph of y = . x *(a) Express the distance d from P to the origin as a function of x. *(b) Use a graphing utility to graph d = d1x2. (c) For what values of x is d smallest? 1 or - 1 5. A right triangle has one vertex on the graph of y = x , x 7 , at 1x, y2, another at the origin, and the third on the positive y-axis at 1, y2, as shown in the figure. Express the area A of 1 the triangle as a function of x. A1x2 = x 2 3 3
y
(0, y)
y⫽x
(x, y)
*6. A right triangle has one vertex on the graph of y = 9 - x2, x 7 , at 1x, y2, another at the origin, and the third on the positive x-axis at 1x, 2. 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 = 1 - x2, another at the origin, a third on the positive y-axis, and the fourth on the positive x-axis. See the figure. (a) Express the area A of the rectangle as a function of x. A(x) = x(1 - x2) (b) What is the domain of A? {x 6 x 6 } *(c) Graph A = A1x2. For what value of x is A largest? y 16
y ⫽ 16 ⫺ x 2 (x, y)
8
4
(0,0)
x
8. A rectangle is inscribed in a semicircle of radius 2. See the figure. Let P = 1x, y2 be the point in quadrant I that is a vertex of the rectangle and is on the circle. y y ⫽ 4 ⫺ x2
⫺2
P ⫽ (x, y )
2
x
(a) Express the area A of the rectangle as a function of x. A(x) = 2x2 - x2 (b) Express the perimeter p of the rectangle as a function of x. p(x) = x + 22 - x2 *(c) Graph A = A1x2. For what value of x is A largest? *(d) Graph p = p1x2. For what value of x is p largest?
9. A rectangle is inscribed in a circle of radius 2. See the figure on the next page. Let P = 1x, y2 be the point in quadrant I x (0, 0) that is a vertex of the rectangle and is on the circle. (a) Express the area A of the rectangle as a function of x. A(x) = x2 - x2 *Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
136
CHAPTER 1 Functions and Their Graphs y 2
P ⫽ (x, y)
⫺2
x
2 ⫺2
16. Geometry An equilateral triangle is inscribed in a circle of radius r. See the figure. Express the circumference C of the circle as a function of the length x of a side of the triangle. x2 [Hint: First show that r 2 = .] C 1x2 = 223 px 3 3 x
x r
2 2 x ⫹y ⫽4
(b) Express the perimeter p of the rectangle as a function of x. p(x) = x + 2 - x2 *(c) Graph A = A1x2. For what value of x is A largest? *(d) Graph p = p1x2. For what value of x is p largest? 10. A circle of radius r is inscribed in a square. See the figure. r
(a) Express the area A of the square as a function of the radius r of the circle. A(r) = r 2 (b) Express the perimeter p of the square as a function of r. p(r) = r 11. Geometry " XJSF NFUFST MPOH JT UP CF DVU JOUP UXP pieces. One piece will be shaped as a square, and the other piece will be shaped as a circle. See the figure.
p 23 2 bx 3 *17. Geometry An equilateral triangle is inscribed in a circle of radius r 4FF UIF GJHVSF JO 1SPCMFN &YQSFTT UIF BSFB A within the circle, but outside the triangle, as a function of the length x of a side of the triangle. 17. A1x2 = a
x
18. Uniform Motion Two cars leave an intersection at the TBNFUJNF0OFJTIFBEFETPVUIBUBDPOTUBOUTQFFEPGNJMFT per hour, and the other is headed west at a constant speed PG NJMFT QFS IPVS TFF UIF GJHVSF #VJME B NPEFM UIBU expresses the distance d between the cars as a function of the time t. [Hint: At t = , the cars leave the intersection.] d(t) = 5t N W
E S
x 4x 10 m 10 ⫺ 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? {x 6 x 6 2.5} *(c) Graph A = A1x2. For what value of x is A smallest? 12. Geometry "XJSFNFUFSTMPOHJTUPCFDVUJOUPUXPQJFDFT 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? *(c) Graph A = A1x2. 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 C of the circle as a function of x. C(x) = x *(b) Express the area A 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 p of the square as a function of x. p(x) = x *(b) Express the area A of the square as a function of x.
d
19. Uniform Motion Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a DPOTUBOUTQFFEPGNJMFTQFSIPVS"UUIFTBNFUJNF UIF other car is 3 miles east of the intersection and is moving at BDPOTUBOUTQFFEPGNJMFTQFSIPVS (a) Build a model that expresses the distance d between the cars as a function of time t. d(t) = 225t 2 - 3t + 13 [Hint: At t = , the cars are 2 miles south and 3 miles east of the intersection, respectively.] *(b) Use a graphing utility to graph d = d1t2. For what value of t is d smallest? 20. Inscribing a Cylinder in a Sphere Inscribe a right circular cylinder of height h and radius r in a sphere of fixed radius R. See the illustration. Express the volume V of the cylinder h2 as a function of h. 2 V 1h2 = phaR b [Hint: V = pr 2 h. Note also the right triangle.] r
15. Geometry 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. R r
(a) Express the area A of the rectangle as a function of the radius r of the semicircle. A(r) = 2r 2 (b) Express the perimeter p of the rectangle as a function of r. 15. (b) p1r2 = r
Sphere
h
SECTION 1.6 Mathematical Models: Building Functions
21. Inscribing a Cylinder in a Cone 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 2 pH1R - r2r V of the cylinder as a function of r. V 1r2 = 2 R [Hint: V = pr h. Note also the similar triangles.] r
H
137
*(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, build a model that expresses 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 ? 5 x … x … 126 (c) How long will it take to travel from the island to town if UIFQFSTPOMBOETUIFCPBUNJMFTGSPNP? IS E )PXMPOHXJMMJUUBLFJGUIFQFSTPOMBOETUIFCPBUNJMFT from P? 3.55 hr 24. Filling a Conical Tank Water is poured into a container JOUIFTIBQFPGBSJHIUDJSDVMBSDPOFXJUISBEJVTGFFUBOE IFJHIUGFFU4FFUIFGJHVSF&YQSFTTUIFWPMVNFV of the water in the cone as a function of the height h of the water.
h
R
[Hint: The volume V of a cone of radius r and height h is p 1 V = pr 2 h.] V(h) = h3 3
Cone
22. Installing Cable TV 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 connection box for the cable is located 5 miles down the road. See the figure.
4 r 16
House
h
Stream 2 mi
Box 5 mi
x
25. Constructing an Open Box An open box with a square base JTUPCFNBEFGSPNBTRVBSFQJFDFPGDBSECPBSEJODIFTPO a side by cutting out a square from each corner and turning up the sides. See the figure.
x
* B *GUIFJOTUBMMBUJPODPTUJTQFSNJMFBMPOHUIFSPBEBOE QFSNJMFPGGUIFSPBE CVJMEBNPEFMUIBUFYQSFTTFT 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(1) = $33.5 (c) Compute the cost if x = 3 miles. C(3) = $39.9 *(d) Graph the function C = C 1x2. Use TRACE to see how the cost C varies as xDIBOHFTGSPNUP (e) What value of x results in the least cost? x ≈ 2.9 mi 23. Time Required to Go from an Island to a Town 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.
d2 Town
P
12 ⫺ x
2 mi
x 12 mi d1
Island
x
x
x
24 in.
x
x x
x 24 in.
(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. V(x) = x(2 - 2x)2 (b) What is the volume if a 3-inch square is cut out? 92 in.3 D 8IBUJTUIFWPMVNFJGBJODITRVBSFJTDVUPVU 1 in.3 *(d) Graph V = V 1x2. For what value of x is V largest? 26. Constructing an Open Box An open box with a square base JTSFRVJSFEUPIBWFBWPMVNFPGDVCJDGFFU *(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? 1 ft 2 (c) How much material is required for a base 2 feet by 2 feet? 2 ft 2 *(d) Use a graphing utility to graph A = A1x2. For what value of x is A smallest?
138
CHAPTER 1 Functions and Their Graphs
1.7 Building Mathematical Models Using Variation OBJECTIVES 1 Construct a Model Using Direct Variation (p. 138) 2 Construct a Model Using Inverse Variation (p. 139) 3 Construct a Model Using Joint or Combined Variation (p. 139)
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 objects is inversely proportional to the square of the distance between them. Revenue is directly proportional to sales.
Instructor Note: The material in this section is optional and can be skipped without loss of continuity.
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 Construct a Model Using Direct Variation DEFINITION
Let x and y denote two quantities. Then y varies directly with x, or y is directly proportional to x, if there is a nonzero number k such that y = kx
Figure 65 y
The number k is called the constant of proportionality. 5IFHSBQIJO'JHVSFJMMVTUSBUFTUIFSFMBUJPOTIJQCFUXFFOy and x if y varies directly with x and k 7 , x Ú . Note that the constant of proportionality is, in fact, the slope of the line. If two quantities vary directly, then knowing the value of each quantity in one instance enables us to write a model that is true in all cases.
x
Mortgage Payments
EX A MPL E 1
The monthly payment p on a mortgage varies directly with the amount B borrowed. If UIFNPOUIMZQBZNFOUPOBZFBSNPSUHBHFJTGPSFWFSZCPSSPXFE GJOEB model that relates the monthly payment p to the amount B borrowed for a mortgage with these terms. Then find the monthly payment p when the amount BCPSSPXFEJT
Solution
Because p varies directly with B, we know that p = kB for some constant k. Because p = .5 when B = 1, it follows that
Figure 66
.5 = k 112 k = .5
Monthly payment
p 800
(120, 798)
600
Since p = kB, we have p = .5B
400
The model
In particular, when B = $12,, we find that
200 0
Solve for k.
p = .51$12,2 = $9 40 80 120 160 Amount borrowed (000's)
B
'JHVSFJMMVTUSBUFTUIFSFMBUJPOTIJQCFUXFFOUIFNPOUIMZQBZNFOUp and the amount B borrowed.
Now Work
PROBLEMS
3
AND
21
r
SECTION 1.7 Building Mathematical Models Using Variation
139
2 Construct a Model Using Inverse Variation DEFINITION Figure 67
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
y
y =
k x
5IFHSBQIJO'JHVSFJMMVTUSBUFTUIFSFMBUJPOTIJQCFUXFFOy and x if y varies inversely with x and k 7 , x 7 .
x
EXAM PL E 2
Maximum Weight That Can Be Supported by a Piece of Pine 4FF'JHVSF5IFNBYJNVNXFJHIUW that can be safely supported by a 2-inch CZJODIQJFDFPGQJOFWBSJFTJOWFSTFMZXJUIJUTMFOHUIl. Experiments indicate that UIFNBYJNVNXFJHIUUIBUBGPPUMPOHCZQJFDFPGQJOFDBOTVQQPSU JT pounds. Write a model relating the maximum weight W (in pounds) to length l (in feet). Find the maximum weight W that can be safely supported by a length of 25 feet.
Solution
Because W varies inversely with l, we know that
Figure 68
W =
k l
for some constant k. Because W = 5 when l = 1, we have k 1 k = 5
5 =
Since W =
k , we have l W =
5 l
The Model
In particular, the maximum weight W that can be safely supported by a piece of pine 25 feet in length is W = Figure 69
5 = 2 pounds 25
'JHVSFJMMVTUSBUFTUIFSFMBUJPOTIJQCFUXFFOUIFXFJHIUW and the length l.
W 600
Now Work
(10, 500)
500 400
31
3 Construct a Model Using Joint or Combined Variation
300 200
(25, 200)
100 0
PROBLEM
r
5
10
15
20
25
l
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. Combinations of direct and/or inverse variation may also occur. This is usually referred to as combined variation. Let’s look at an example.
140
CHAPTER 1 Functions and Their Graphs
EX A MPL 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 difference between the inside and outside temperatures, and inversely with the thickness of the wall. Write an equation that relates these quantities.
Solution
Begin by assigning symbols to represent the quantities: L = Heat loss
T = Temperature difference d = Thickness of wall
A = Area of wall Then
L = k
AT d
where k is the constant of proportionality.
r
In direct or inverse variation, the quantities that vary may be raised to powers. 'PS FYBNQMF JO UIF FBSMZ TFWFOUFFOUI DFOUVSZ +PIBOOFT ,FQMFS m discovered that the square of the period of revolution T of a planet around the Sun varies directly with the cube of its mean distance a from the Sun. That is, T 2 = ka3, where k is the constant of proportionality.
EX A MPL 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 direction of the wind varies jointly with the area A of the surface and the square of the speed vPGUIFXJOE"XJOEPGNJMFTQFSIPVSCMPXJOHPOBXJOEPXNFBTVSJOHGFFUCZ GFFUIBTBGPSDFPGQPVOET4FF'JHVSF8IBUGPSDFEPFTBXJOEPGNJMFT QFSIPVSDBVTFPOBXJOEPXNFBTVSJOHGFFUCZGFFU
Solution
Since F varies jointly with A and v2, we have
Figure 70
F = kAv2 Wind
where k is the constant of proportionality. We are told that F = 15 pounds when A = # 5 = 2 feet 2 and v = 3 miles per hour. This gives 15 = k 122 192 1 k = 12
F = kAv 2, F = 150, A = 20, v = 30
Since F = kAv2, we have F =
1 Av2 12
'PSBXJOEPGNJMFTQFSIPVSCMPXJOHPOBXJOEPXXIPTFBSFBJTA = 3 # = 12 square feet, the force F is F =
Now Work
1 1122 1252 = 25 pounds 12
PROBLEM
39
r
141
SECTION 1.7 Building Mathematical Models Using Variation
1.7 Assess Your Understanding Concepts and Vocabulary k 2. True or False If y varies directly with x, then y = , where k x is a constant. False
1. If x and y are two quantities, then y is directly proportional to x if there is a nonzero number k such that y = kx.
Skill Building In Problems 3–14, write a general formula to describe each variation. 1 3. y varies directly with x; y = 2 when x = 1 y = x 5 5. A varies directly with x2; A = p when x = 2
A = px2
7. F varies inversely with d 2; F = 1 when d = 5
F =
25 d2
4. v varies directly with t; v = 1 when t = 2
v = t
6. V varies directly with x3; V = 3p when x = 3
V =
y varies inversely with 1x; y = when x = 9
y =
8.
9. z varies directly with the sum of the squares of x and y; z = 5 when x = 3 and y =
1 z = (x2 + y2 2 5
10. T varies jointly with the cube root of x and the square of d; T = 1 when x = and d = 3
12. z varies directly with the sum of the cube of x and the square of y; z = 1 when x = 2 and y = 3
M =
1 3 z = (x + y2 2 1
13. 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 = 14. The cube of z varies directly with the sum of the squares of x and y; z = 2 when x = 9 and y =
Applications and Extensions
p 3 r 3
19. F = . * 1 - 11 a
2x
3 T = d2 2 x
11. M varies directly with the square of d and inversely with the square root of x; M = 2 when x = 9 and d =
15. V =
p 3 x 3 12
z3 =
9d 2 22x
T2 =
2 (x + y2 2 9
a3 d2
mM b d2
In Problems 15–20, write an equation that relates the quantities. 15. Geometry The volume V of a sphere varies directly with the p cube of its radius r. The constant of proportionality is . 3 16. 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 proportionality is 1. c 2 = a2 + b2
21. Mortgage Payments The monthly payment p on a mortgage varies directly with the amount B borrowed. If the monthly QBZNFOU PO B ZFBS NPSUHBHF JT GPS FWFSZ borrowed, find a model that relates the monthly payment p to the amount B borrowed for a mortgage with the same terms. Then find the monthly payment p when the amount B CPSSPXFEJT p = .9B; $91.5
17. Geometry The area A of a triangle varies jointly with the lengths of the base b and the height h. The constant of 1 1 proportionality is . A = bh 2 2 18. Geometry The perimeter p of a rectangle varies jointly with the sum of the lengths of its sides l and w. The constant of proportionality is 2. p = 21l + w2
22. Mortgage Payments The monthly payment p on a mortgage varies directly with the amount B borrowed. If the monthly QBZNFOU PO B ZFBS NPSUHBHF JT GPS FWFSZ borrowed, find a model that relates the monthly payment p to the amount B borrowed for a mortgage with the same terms. Then find the monthly payment p when the amount B CPSSPXFEJT p = .99B; $153.25
19. Physics: Newton’s Law The force F (in newtons) of attraction between two objects 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 proportionality is G = . * 1 - 11.
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 BOPCKFDUGBMMTGFFUJOTFDPOE IPXGBSXJMMJUGBMMJOTFDPOET )PXMPOHXJMMJUUBLFBOPCKFDUUPGBMMGFFU GU TFD
20. Physics: Simple Pendulum The period of a pendulum is the time required for one oscillation; the pendulum is usually referred to as simple when the angle made to the vertical is MFTTUIBO5IFQFSJPET of a simple pendulum (in seconds) varies directly with the square root of its length l (in feet). 2p 2p The constant of proportionality is . T = 2l 232 232
24. Physics: Falling Objects The velocity v of a falling object is directly proportional to the time t of the fall. If, after TFDPOET UIF WFMPDJUZ PG UIF PCKFDU JT GFFU QFS TFDPOE what will its velocity be after 3 seconds? GUT
142
CHAPTER 1 Functions and Their Graphs
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 = 2, find E when W = 15. 2.25
29. (a) D =
29 p
E
34. Intensity of Light The intensity I of light (measured in footcandles) varies inversely with the square of the distance from UIFCVMC4VQQPTFUIBUUIFJOUFOTJUZPGBXBUUMJHIUCVMCBUB EJTUBODFPGNFUFSTJTGPPUDBOEMF%FUFSNJOFUIFJOUFOTJUZ of the bulb at a distance of 5 meters. GPPUDBOEMF 35. Geometry The volume V of a right circular cylinder varies jointly with the square of its radius r and its height h. The constant of proportionality is p. See the figure. Write an equation for V. V = pr 2h
W 12 r s 26. Physics: Vibrating String The rate of vibration of a string under constant tension varies inversely with the length of the h TUSJOH*GBTUSJOHJTJODIFTMPOHBOEWJCSBUFTUJNFTQFS TFDPOE XIBUJTUIFMFOHUIPGBTUSJOHUIBUWJCSBUFTUJNFT per second? 26. in. 27. Revenue Equation At the corner Shell station, the revenue 3 R varies directly with the number g of gallons of gasoline TPME*GUIFSFWFOVFJTXIFOUIFOVNCFSPGHBMMPOTTPME 36. Geometry The volume V of a right circular cone varies is 12, find a model that relates revenue R to the number g of jointly with the square of its radius r and its height h. The gallons of gasoline sold. Then find the revenue R when the p constant of proportionality is . See the figure. Write an OVNCFSPGHBMMPOTPGHBTPMJOFTPMEJT R = 3.95g p 2 3 equation for V. V = r h 28. Cost Equation The cost C of chocolate-covered almonds 3 varies directly with the number A of pounds of almonds QVSDIBTFE*GUIFDPTUJTXIFOUIFOVNCFSPGQPVOET of chocolate-covered almonds purchased is 5, find a model that relates the cost C to the number A of pounds of almonds purchased. Then find the cost C when the number of pounds h of almonds purchased is 3.5. C = .5A
30. (a) t =
29. Demand Suppose that the demand D for candy at the movie theater is inversely related to the price p. B 8IFO UIF QSJDF PG DBOEZ JT QFS CBH UIF UIFBUFS TFMMTCBHTPGDBOEZ&YQSFTTUIFEFNBOEGPSDBOEZJO 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. CBHT 30. Driving to School The time t that it takes to get to school varies inversely with your average speed s. B 4VQQPTF UIBU JU UBLFT ZPV NJOVUFT UP HFU UP TDIPPM XIFOZPVSBWFSBHFTQFFEJTNJMFTQFSIPVS&YQSFTT the driving time to school in terms of average speed. C 4VQQPTFUIBUZPVSBWFSBHFTQFFEUPTDIPPMJTNJMFTQFS hour. How long will it take you to get to school? NJO 31. Pressure The volume V of a gas held at a constant temperature in a closed container varies inversely with its pressure P. If UIF WPMVNF PG B HBT JT DVCJD DFOUJNFUFST 1cm3 2 when UIFQSFTTVSFJTNJMMJNFUFSTPGNFSDVSZ NN)H
GJOEUIF WPMVNFXIFOUIFQSFTTVSFJTNN)H DN3 32. Resistance The current I in a circuit is inversely proportional to its resistance Z measured in ohms. Suppose that when the DVSSFOUJOBDJSDVJUJTBNQFSFT UIFSFTJTUBODFJTPINT Find the current in the same circuit when the resistance is PINT " 33. Weight The weight of an object above the surface of Earth varies inversely with the square of the distance from the center of Earth. If Maria weighs 125 pounds when she is on UIFTVSGBDFPG&BSUI NJMFTGSPNUIFDFOUFS
EFUFSNJOF Maria’s weight when she is at the top of Mount McKinley NJMFTGSPNUIFTVSGBDFPG&BSUI MC
r
37. Weight of a Body 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 JUJTNJMFTGSPNUIFDFOUFSPG&BSUI IPXNVDIXJMMJU XFJHIXIFOJUJTNJMFTGSPNUIFDFOUFS MC 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 PO BO BSFB PG TRVBSF GFFU JT QPVOET XIFO UIF XJOE velocity is 22 miles per hour, find the force on a surface area PGTRVBSFGFFUXIFOUIFXJOEWFMPDJUZJTNJMFTQFS hour. MC 39. Horsepower 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 DFSUBJO NBUFSJBM JODIFT JO EJBNFUFS DBO USBOTNJU IQ BU SQN XIBU EJBNFUFS NVTU UIF TIBGU IBWF JO PSEFS UP 3 USBOTNJUIQBUSQN 2 ≈ 1.2 in. 40. Chemistry: Gas Laws The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P. Write a model relating V, T, and P using k as the constant of proportionality. If a cylinder contains PYZHFO BU B UFNQFSBUVSF PG , BOE B QSFTTVSF PG BUNPTQIFSFTJOBWPMVNFPGMJUFST XIBUJTUIFDPOTUBOU of proportionality k? If a piston is lowered into the DZMJOEFS EFDSFBTJOHUIFWPMVNFPDDVQJFECZUIFHBTUP MJUFSTBOESBJTJOHUIFUFNQFSBUVSFUP, XIBUJTUIFHBT kT pressure? V = ; k = 5 BUN P
143
Chapter Review
41. Physics: Kinetic Energy The kinetic energy K of a moving object varies jointly with its mass m and the square of its velocity v. If an object weighing 25 kilograms and moving XJUIBWFMPDJUZPGNFUFSTQFSTFDPOEIBTBLJOFUJDFOFSHZ PG KPVMFT GJOE JUT LJOFUJD FOFSHZ XIFO UIF WFMPDJUZ JT 15 meters per second. KPVMFT 42. Electrical Resistance of a Wire The electrical resistance of a wire varies directly with the length of the wire and inversely XJUI UIF TRVBSF PG UIF EJBNFUFS PG UIF XJSF *G B XJSF GFFUMPOHBOENJMMJNFUFSTJOEJBNFUFSIBTBSFTJTUBODFPG PINT GJOE UIF MFOHUI PG B XJSF PG UIF TBNF NBUFSJBM XIPTF SFTJTUBODF JT PINT BOE XIPTF EJBNFUFS JT millimeters. GU 43. Measuring the Stress of Materials The stress in the material of a pipe subject to internal pressure varies jointly with the
internal pressure and the internal diameter of the pipe and JOWFSTFMZ XJUI UIF UIJDLOFTT PG UIF QJQF 5IF TUSFTT JT pounds per square inch (c) when the diameter is 5 inches, UIF UIJDLOFTT JT JODI BOE UIF JOUFSOBM QSFTTVSF JT pounds per square inch. Find the stress when the internal QSFTTVSF JT QPVOET QFS TRVBSF JODI JG UIF EJBNFUFS JT JODIFTBOEUIFUIJDLOFTTJTJODI QTJ 44. Safe Load for a Beam The maximum safe load for a horizontal rectangular beam varies jointly with the width of the beam and the square of the thickness of the beam and JOWFSTFMZXJUIJUTMFOHUI*GBOGPPUCFBNXJMMTVQQPSUVQ UPQPVOETXIFOUIFCFBNJTJODIFTXJEFBOEJODIFT thick, what is the maximum safe load in a similar beam GFFUMPOH JODIFTXJEF BOEJODIFT UIJDL MC
Discussion and Writing 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 student to solve and critique.
45. In the early seventeenth century, Johannes Kepler discovered that the square of the period T for a planet to orbit the Sun 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 a real-world problem that you think involves two variables that vary directly. Exchange your problem with another student 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 student to solve and critique.
Chapter Review Library of Functions Constant function (p. 112) f 1x2 = b The graph is a horizontal line with y-intercept b.
Identity function (p. 112) f 1x2 = x The graph is a line with slope 1 and yJOUFSDFQU
y
y 3
y f (x ) = b
Square function (pp. 112–113) f 1x2 = x2 The graph is a parabola with intercept at 1, 2. ( – 2, 4)
(2, 4)
4
(0,b) (1, 1) x
–3 ( – 1, – 1)
(0, 0)
(– 1, 1)
3 x
Square root function (pp. 110 and 113) f 1x2 = 1x
y
y
4
2 (1, 1)
⫺4 (⫺1, ⫺1)
(0, 0)
4
x
⫺1
4 x
(0, 0)
–4
Cube function (p. 113) f 1x2 = x3
(1, 1)
Cube root function (pp. 111 and 113) 3 f 1x2 = 2 x y 3
(1, 1)
(0, 0)
(4, 2) 5 x
3
(1, 1)
(⫺ 1–8,⫺ 1–2)
(2, 2 )
( 1–8 , 1–2)
⫺3
3 x (0, 0)
⫺4
3
(⫺1, ⫺1)
(⫺2,⫺ 2 )
⫺3
144
CHAPTER 1 Functions and Their Graphs
Reciprocal function (p. 113) 1 f1x2 = x
Absolute value function (pp. 112 and 114) f 1x2 = 0 x 0
y
y
2
3
⫺2
(⫺1, 1) 2 x
⫺3
y 4 (2, 2)
(⫺2, 2) (1, 1)
Greatest integer function (p. 114) f 1x2 = int 1x2
(0, 0)
2
(1, 1) 3 x
⫺2
(⫺1, ⫺1)
2
4
x
⫺3 ⫺2
Things to Know Function (pp. 75–78)
Function notation (pp. 78–81)
A relation between two sets such 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 1x, y2 in which no first element is paired with two different second elements. y = f 1x2
f is a symbol for the function.
x is the argument, or independent variable.
y is the dependent variable.
Difference quotient of f (pp. 80 and 109) Domain (pp. 81–83) Vertical-line test (p. 89) Even function f (p. 98) Odd function f (p. 98)
f 1x2 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 1x2. f 1x + h2 - f 1x2
h ≠ h If unspecified, the domain of a function f defined by an equation is the largest set of real numbers for which f 1x2 is a real number. 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 1 - x2 = f 1x2 for every x in the domain ( - x must also be in the domain).
f 1 - x2 = - f 1x2 for every x in the domain ( - x must also be in the domain).
Increasing function (p. 100)
A function f is increasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 6 f 1x2 2.
Decreasing function (p. 100)
A function f is decreasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 7 f 1x2 2.
Constant function (p. 100)
A function f is constant on an open interval I if, for all choices of x in I, the values of f 1x2 are equal.
Local maximum (p. 101)
A function f has a local maximum at c if there is an open interval I containing c so that, for all x in I, f 1x2 … f 1c2.
Local minimum (p. 101)
A function f has a local minimum at c if there is an open interval I containing c so that, for all x in I, f 1x2 Ú f 1c2.
Absolute maximum and Absolute minimum (p. 102)
Let f denote a function defined on some interval I.
If there is a number u in I for which f 1x2 … f 1u2 for all x in I, then f 1u2 is the absolute maximum of f on I and we say the absolute maximum of f occurs at u. If there is a number v in I for which f 1x2 Ú f 1v2 , for all x in I, then f 1v2 is the absolute minimum of f on I and we say the absolute minimum of f occurs at v.
Average rate of change of a function (p. 104)
Direct variation (p. 138) Inverse variation (p. 139)
The average rate of change of f from a to b is f 1b2 - f 1a2 ∆y = ∆x b - a y = kx, k ≠ k y = ,k ≠ x
a ≠ b
Chapter Review
145
Objectives Section
You should be able to...
Examples
Review Exercises
Determine whether a relation represents a function (p. 75) Find the value of a function (p. 78) Find the domain of a function defined by an equation (p. 81) Form the sum, difference, product, and quotient of two functions (p. 83)
1–5 6, 7 8, 9 10
1, 2 3–5, 15, 39 6–11 12–14
Identify the graph of a function (p. 89) Obtain information from or about the graph of a function (p. 90)
1 2–5
27, 28 16(a)–(e), 17(a), 17(e), 17(g)
1 2
17(f) 18–21
3 4
17(b) 17(c)
5
17(d)
2
Determine even and odd functions from a graph (p. 98) Identify even and odd functions from the equation (p. 99) Use a graph to determine where a function is increasing, decreasing, or constant (p. 100) Use a graph to locate local maxima and local minima (p. 101) Use a graph to locate the absolute maximum and the absolute minimum (p. 102) Use a graphing utility to approximate local maxima and local minima and to determine where a function is increasing or decreasing (p. 103) Find the average rate of change of a function (p. 104) Graph the functions listed in the library of functions (p. 110) Graph piecewise-defined functions (p. 115)
6 7, 8 1, 2 3, 4
22, 23, 40(d), 41(b) 24–26 29, 30 37, 38
1.5
1
Graph functions using vertical and horizontal shifts (p. 121)
1–5, 11, 12, 13
2 3
Graph functions using compressions and stretches (p. 124) Graph functions using reflections about the x-axis or y-axis (p. 126)
6–8, 12 9–11, 13
16(f), 31, 33, 34, 35, 36 16(g), 32, 36 16(h), 32, 34, 36
1.6
1
Build and analyze functions (p. 133)
1–3
40, 41
1.7
1
Construct a model using direct variation (p. 138) Construct a model using inverse variation (p. 139) Construct a model using joint or combined variation (p. 139)
1 2 3,4
42 43 44
1.1
1
1.2
1
1.3
1
4
6
1.4
7
2 3 4
2
2 3
5
1
2 3
Review Exercises In Problems 1 and 2, determine whether each relation represents a function. For each function, state the domain and range. 2. 5 15, 32, 12, 52, 15, - 32 6 Not a function
1. 5 13, 22, 1 - 2, 52, 11, 22 6 Function; domain { - 2, 1, 3}; range {2, 5}
In Problems 3–5, find the following for each function: (a) f 132 (b) f 1 - 32 (c) f 1 - x2 *3. f 1x2 =
3x x2 - 1
(d) - f 1x2
*4. f 1x2 = 2x2 - 3x
(e) f 1x - 12 *5. f 1x2 =
In Problems 6–11, find the domain of each function. x - 1 {x 0 x ≠ 2, x ≠ - 2} x2 - 4 x *9. f 1x2 = 2 x + 2x - 3 6. f 1x2 =
7. f 1x2 = 22 - x 10. f 1x2 =
1x x2 - 1
{x x … 2} {x 0 x Ú 0, x ≠ 1}
8. g(x) = 11. g(x) =
7 - x x2
0x0 x
{x x ≠ 0}
x 1x + 8
In Problems 12–14, find f + g, f - g, f # g, and
5x x 7 - 86
f for each pair of functions. State the domain of each of these functions. g x + 1 1 *13. f 1x2 = 3x2 + x + 1; g(x) = 3x *14. f 1x2 = ; g(x) = *12. f 1x2 = 2 - x; g(x) = 3x + 1 x - 1 x f 1x + h2 f 1x2 15. Find the difference quotient of f 1x2 = - 2x2 + x + 1; that is, find , h ≠ 0. - 4x + 1 - 2h h
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
(f) f 13x2
2
146
CHAPTER 1 Functions and Their Graphs
16. Use the graph of the function f shown to: *(a) Find the domain and the range of f. (b) List the intercepts. (0, 0) (c) Find f 1 - 22. - 1 (d) Find the value(s) of x for which f 1x2 = - 3. (e) Solve f 1x2 7 0. {x 0 6 x … 3} *(f) Graph y = f 1x - 32. 1 *(g) Graph y = f a xb. 2 *(h) Graph y = - f 1x2.
17. Use the graph of the function f shown to find: *(a) The domain and the range of f . *(b) The intervals on which f is increasing, decreasing, or constant. *(c) The local minimum values and local maximum values. *(d) The absolute maximum and absolute minimum. (e) Whether the graph is symmetric with respect to the x-axis, the y-axis, or the origin. No symmetry (f) Whether the function is even, odd, or neither. Neither *(g) The intercepts, if any.
-4
y 4
y 4
(3, 3)
(3, 0)
(⫺2, 1) ⫺5 (⫺2, ⫺1)
(0, 0)
⫺6 (⫺4,⫺3) (⫺3, 0)
x
5
(4, 3)
6 x (2, ⫺1)
⫺4
(⫺4, ⫺3)
⫺4
In Problems 18–21, determine (algebraically) whether the given function is even, odd, or neither. 18. f 1x2 = 2x5 - x
19. g(x) =
Odd
4 + x2 1 + x4
21. f 1x2 =
*20. G(x) = x2 - 3x + 4
Even
x 1 + x2
Odd
In Problems 22 and 23, use a graphing utility to graph each function over the indicated interval. Approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. *22. f 1x2 = 2x3 - 5x + 1
1 - 3, 32
*23. f 1x2 = 2x4 - 5x3 + 2x + 1
24. Find the average rate of change of f 1x2 = 8x - x. (a) From 1 to 2 23 (b) From 0 to 1 7
1 - 2, 32
2
(c) From 2 to 4
47
In Problems 25 and 26, find the average rate of change from 3 to 5 for each function f. Be sure to simplify. 25. f 1x2 = 2x - 3
26. f 1x2 = x3 - 5x
2
44
In Problems 27 and 28, is the graph shown the graph of a function? 27. No
28. Yes
y
y
x
x
In Problems 29 and 30, sketch the graph of each function. Be sure to label at least three points. *29. f 1x2 = 0 x 0
*30. f 1x2 = 1x
In Problems 31–36, 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. *31. F 1x2 = 0 x 0 - 4
*32. g1x2 = - 2 0 x 0
*34. f 1x2 = 21 - x
*35. h(x) = (x - 1) + 2
In Problems 37 and 38, (a) Find the domain of each function. (d) Based on the graph, find the range. *37. f1x2 = b
3x x + 1
x *38. f 1x2 = c 1 3x
*33. h(x) = 2x - 1 *36. g(x) = - 2(x + 2)3 - 8
2
if - 2 6 x … 1 if x 7 1 if - 4 … x 6 0 if x = 0 if x 7 0
(b) Locate any intercepts. (e) Is f continuous on its domain?
(c) Graph each function.
39. A function f is defined by f 1x2 =
Ax + 5 6x - 2
If f 112 = 4, find A. A = 11
42. p =
427 B; $1083.92 65,000
Chapter Test
40. Constructing a Closed Box A closed box with a square base is required to have a volume of 10 cubic feet. *(a) Build a model that expresses 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? 42 ft 2 (c) How much material is required for a base 2 feet by 2 feet? 28 ft 2 *(d) Graph A = A1x2. For what value of x is A smallest? 41. Area of a Rectangle 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. ≈12.17 sq. units
147
terms. Then find the monthly payment p when the amount borrowed is $165,000 43. Weight of a Body The weight of a body varies inversely with 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? 199.9 lb 44. Heat Loss The amount of heat transferred per hour through a glass window varies jointly with the surface area of the window and the difference in temperature between the areas separated by the glass. A window with a surface area of 7.5 square feet loses 135 BTU per hour when the temperature difference is 40°F. How much heat is lost per hour for a similar window with a surface area of 12 square feet when the temperature difference is 35°F? 189 BTU
42. Mortgage Payments The monthly payment p on a mortgage varies directly with the amount B borrowed. If the monthly payment on a 30-year mortgage is $854.00 when $130,000 is borrowed, find a model that relates the monthly payment p to the amount B borrowed for a mortgage with the same
Chapter Test Prep Videos include step-by-step solutions to all chapter test exercises and can be found on this text’s Channel. (search “SullivanPrecalcUC3e”)
Chapter Test
1. Determine whether each relation represents a function. For each function, state the domain and the range. *(a) 5 12, 52, 14, 62, 16, 72, 18, 82 6 Function (b) 5 11, 32, 14, - 22, 1 - 3, 52, 11, 72 6 Not a function (c) Not a function y 6
In Problems 2– 4, find the domain of each function and evaluate each function at x = - 1. x + 2 *3. g1x2 = *2. f 1x2 = 24 - 5x 0x + 20 x - 4 *4. h1x2 = 2 x + 5x - 36 5. Using the graph of the function f : y
4
4 (1, 3)
2
(0, 2)
x ⫺4
⫺2
2
4
⫺2
(5, 3)
x 4
2
⫺4
(d)
(2, 0)
(2, 0)
4
4
(5, 2) (3, 3)
y
*(a) Find the domain and the range of f. (b) List the intercepts. (0, 2), ( - 2, 0), (2, 0) (c) Find f 112. f 112 = 3 (d) For what value(s) of x does f 1x2 = - 3? - 5 and 3 *(e) Solve f 1x2 6 0.
6 4 2 x 4
2
2
4
2
Function; domain: all real numbers; range: {y 0 y Ú 2}
*6. Use a graphing utility to graph the function f 1x2 = - x4 + 2x3 + 4x2 - 2 on the interval 1 - 5, 52. Approximate any local maximum values and local minimum values rounded to two decimal places. Determine where the function is increasing and where it is decreasing. 2x + 1 7. Consider the function g(x) = b x - 4 *(a) Graph the function. (b) List the intercepts. (0, - 4), (4, 0) (c) Find g1 - 52. g( - 5) = - 9 (d) Find g122. g(2) = - 2
if x 6 - 1 if x Ú - 1
148
CHAPTER 1 Functions and Their Graphs
8. For the function f 1x2 = 3x2 - 2x + , find the average rate of change of fGSPNUP 19 9. For the functions f 1x2 = 2x2 + 1 and g(x) = 3x - 2, find the following and simplify: (a) f - g (f - g)(x) = 2x2 - 3x + 3 (b) f # g (f # g)(x) = x3 - x2 + 3x - 2 (c) f 1x + h2 - f 1x2 xh + 2h2
12. A community skating rink is in the shape of a rectangle with semicircles attached at the ends. The length of the rectangle JTGFFUMFTTUIBOUXJDFUIFXJEUI5IFUIJDLOFTTPGUIFJDFJT JODI *(a) Build a model that expresses the ice volume, V, as a function of the width, x. * C )PXNVDIJDFJTJOUIFSJOLJGUIFXJEUIJTGFFU
10. Graph each function using the techniques of shifting, compressing or stretching, and reflections. Start with the graph of the basic function and show all stages. *(a) h1x2 = - 21x + 12 3 + 3 *(b) g1x2 = 0 x + 0 + 2 11. The variable interest rate on a student loan changes each July 1 based on the bank prime loan rate. For the years m UIJT SBUF DBO CF BQQSPYJNBUFE CZ UIF NPEFM r 1x2 = - .115x2 + 1.13x + 5.23, 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 during this time period. During which year was the interest rate the highest? PDDVSSJOHJO(x ≈ 5) C 6TF UIF NPEFM UP FTUJNBUF UIF SBUF JO %PFT UIJT value seem reasonable? - 1.33; No Source: U.S. Federal Reserve
13. The resistance (in ohms) of a circular conductor varies directly with the length of the conductor and inversely with UIFTRVBSFPGUIFSBEJVTPGUIFDPOEVDUPS*GGFFUPGXJSF with a radius of * 1 - 3JODIIBTBSFTJTUBODFPGPINT XIBUXPVMECFUIFSFTJTUBODFPGGFFUPGUIFTBNFXJSFJG the radius were increased to * 1 - 3 inch? PINT
Chapter Projects 4. 4VQQPTF ZPV FYQFDU UP VTF BOZUJNF NJOVUFT XJUI VOMJNJUFEUFYUJOHBOE.#PGEBUB8IBUXPVMECFUIF monthly cost of each plan you are considering? 5. Build a model that describes the monthly cost C as a function of the number of anytime minutes used m, BTTVNJOH VOMJNJUFE UFYUJOH BOE .# PG EBUB FBDI month for each plan you are considering. 6. (SBQIFBDIGVODUJPOGSPN1SPCMFN 7. Based on your particular usage, which plan is best for you?
I.
Internet-based Project Choosing a Cellular Telephone Plan Collect information from your family, friends, or consumer agencies such as Consumer Reports. Then decide on a cellular telephone provider, choosing the company that you feel offers the best service. Once you have selected a service provider, research the various types of individual plans offered by the company by visiting the provider’s website. 1. 4VQQPTFZPVFYQFDUUPVTFBOZUJNFNJOVUFTXJUIPVU a texting or data plan. What would be the monthly cost of each plan you are considering?
8. Now, develop an Excel spreadsheet to analyze the various plans you are considering. Suppose you want a plan that PGGFST BOZUJNF NJOVUFT XJUI BEEJUJPOBM NJOVUFT DPTUJOH QFS NJOVUF UIBU DPTUT QFS NPOUI In addition, you want unlimited texting, which costs an BEEJUJPOBM QFS NPOUI BOE B EBUB QMBO UIBU PGGFST up to 25 MB of data each month, with each additional .#DPTUJOH#FDBVTFDFMMVMBSUFMFQIPOFQMBOTDPTU structure is based on piecewise-defined functions, we need “if-then” statements within Excel to analyze the cost of the plan. Use the Excel spreadsheet on the next page as a guide in developing your worksheet. Enter into your spreadsheet a variety of possible minutes and data used to help arrive at a decision regarding which plan is best for you. 9. Write a paragraph supporting the choice in plans that best meets your needs.
2. 4VQQPTF ZPV FYQFDU UP VTF BOZUJNF NJOVUFT XJUI unlimited texting, but no data plan. What would be the monthly cost of each plan you are considering?
10. How are “if/then” loops similar to a piecewise-defined function?
3. 4VQQPTF ZPV FYQFDU UP VTF BOZUJNF NJOVUFT XJUI unlimited texting and an unlimited data plan. What would be the monthly cost of each plan you are considering?
Citation: &YDFM ¥ .JDSPTPGU $PSQPSBUJPO 6TFE XJUI permission from Microsoft.
Chapter Projects
A 1 2 Monthly fee 3 Allotted number of anytime minutes 4 5 6 7 8 9 10 11 12 13 14 15 16
Number of anytime minutes used Cost per additional minute Monthly cost of text messaging Monthly cost of data plan Allotted data per month (MB) Data used Cost per additional MB of data
B $
C
149
D
39.99 700 700
$ $ $
0.40 20.00 9.99 25 30
$
0.20
Cost of phone minutes Cost of data
=IF(B42; g1x2 =
3 x + 2 2
2 x + 1
In Problems 21–30, find the domain of the composite function f ∘ g. 21. f 1x2 =
2 3 ; g1x2 = x - 1 x
5 x x ≠ 0, x ≠ 26
22. f 1x2 =
1 2 ; g1x2 = x + 3 x
%VFUPTQBDFSFTUSJDUJPOT BOTXFSTUPUIFTFFYFSDJTFTNBZCFGPVOEJOUIF"OTXFSTJOUIFCBDLPGUIFCPPL
e x x ≠ 0, x ≠
2 f 3
312
CHAPTER 4 Exponential and Logarithmic Functions
23. f 1x2 =
x 4 ; g1x2 = x - 1 x
5 x x≠ - 4, x ≠ 06
25. f 1x2 = 1x ; g1x2 = 2x + 3
3 e x x Ú - f 2 27. f 1x2 = x2 + 1; g1x2 = 2x - 1 5 x x Ú 16
29. f 1x2 =
x - 6 ; g1x2 = 2x x - 2
5 x 0 x Ú 0, x ≠ 46
24. f 1x2 =
x 2 ; g1x2 = x + 3 x
2 e x x≠ - , x≠0 f 3
26. f 1x2 = x - 2; g1x2 = 21 - x
28. f 1x2 = x2 + 4; g1x2 = 2x - 2 30. f 1x2 = 2x; g1x2 =
x x - 3
5x x … 16 5x x Ú 26
5x 0 x … 0 or x 7 36
In Problems 31–46, for the given functions f and g, find: B f ∘ g C g ∘ f D f ∘ f E g ∘ g State the domain of each composite function. *31. f 1x2 = 2x + 3; g1x2 = 3x *33. f 1x2 = 3x + 1; g1x2 = x *35. f 1x2 = x2; 3 *37. f 1x2 = x x *39. f 1x2 = x *41. f 1x2 = 1x;
*32. f 1x2 = - x; g1x2 = 2x - 4
*34. f 1x2 = x + 1; g1x2 = x2 + 4
2
*36. f 1x2 = x2 + 1; g1x2 = 2x2 + 3 1 2 *38. f 1x2 = ; g1x2 = x + 3 x x 2 ; g1x2 = *40. f 1x2 = x + 3 x *42. f 1x2 = 2x - 2 ; g1x2 = 1 - 2x
g1x2 = x2 + 4 2 ; g1x2 = 1 x 1
4 x g1x2 = 2x + 3
; g1x2 = -
*44. f 1x2 = x2 + 4 ; g1x2 = 2x - 2
*43. f1x2 = x2 + 1; g1x2 = 2x - 1 *45. f 1x2 =
x - 5 x + 2 ; g1x2 = x + 1 x - 3
*46. f 1x2 =
In Problems 47–54, show that 1f ∘ g2 1x2 = 1g ∘ f2 1x2 = x.
2x - 1 x + 4 ; g1x2 = x - 2 2x - 5
*47. f 1x2 = 2x; g1x2 =
1 x 2 3 *49. f 1x2 = x3; g1x2 = 1x
*48. f 1x2 = 4x; g1x2 =
1 x 4 *50. f 1x2 = x + 5; g1x2 = x - 5
1 1x + 62 2 1 *53. f 1x2 = ax + b; g1x2 = 1x - b2 a
*52. f 1x2 = 4 - 3x; g1x2 =
*51. f 1x2 = 2x - 6; g1x2 =
a ≠ 0
*54. f 1x2 =
1 4 - x
3
1 1 ; g1x2 = x x
In Problems 55–60, find functions f and g so that f ∘ g = H. For 55–60, other answers are possible. 55. H x = 12x + 32 4 57. H x = 2x2 + 1 59. H x = 0 2x + 1 0
f 1x2 = x4; g1x2 = 2x + 3
f 1x2 = 2x ; g1x2 = x2 + 1
f 1x2 = |x|; g1x2 = 2x + 1
56. H1x2 = 11 + x2 2
58. H x = 21 - x2
60. H x = 0 2x + 3 0 2
3
f 1x2 = x3; g1x2 = 1 + x2
f 1x2 = 2x ; g1x2 = 1 - x2 f 1x2 = |x|; g1x2 = 2x2 + 3
Applications and Extensions 61. If f 1x2 = 2x3 - 3x2 + 4x - 1 and g1x2 = 2, find 1f ∘ g2 1x2 and 1g ∘ f2 1x2. 1f ∘ g2 1x2 = 11; 1g ∘ f2 1x2 = 2 x + 1 62. If f 1x2 = , find 1f ∘ f2 1x2. 1f ∘ f2 1x2 = x, x ≠ 1 x - 1 2 63. If f 1x2 = 2x + 5 and g1x2 = 3x + a, find a so that the graph of f ∘ g crosses the y-axis at 23. - 3, 3
64. If f 1x2 = 3x2 - and g1x2 = 2x + a, find a so that the graph of f ∘ g crosses the yBYJTBU - 5, 5 In Problems 65 and 66, use the functions f and g to find: (a) f ∘ g (b) g ∘ f (c) the domain of f ∘ g and of g ∘ f (d) the conditions for which f ∘ g = g ∘ f
*65. f 1x2 = ax + b; g1x2 = cx + d ax + b ; g1x2 = mx *66. f 1x2 = cx + d 67. Surface Area of a Balloon The surface area S JO TRVBSF NFUFST PGBIPUBJSCBMMPPOJTHJWFOCZ S r = 4pr 2
where rJTUIFSBEJVTPGUIFCBMMPPO JONFUFST *GUIFSBEJVT r is increasing with time t JO TFDPOET BDDPSEJOH UP UIF 2 formula r 1t2 = t 3, t Ú 0, find the surface area S of the 3 16 6 balloon as a function of the time t. S t = pt 9 *68. Volume of a Balloon The volume V JO DVCJD NFUFST PG UIF IPUBJS CBMMPPO EFTDSJCFE JO 1SPCMFN JT HJWFO CZ 4 V 1r2 = pr 3. If the radius r is the same function of t as in 3 1SPCMFN GJOEUIFWPMVNFV as a function of the time t. 69. Automobile Production The number N of cars produced at a certain factory in one day after t hours of operation is given by N1t2 = 100t - 5t 2, 0 … t … 10. If the cost C JO EPMMBST PG QSPEVDJOH N cars is C 1N2 = 15,000 + 000N, find the cost C as a function of the time t of operation of the factory. C t = 15,000 + 00,000t - 40,000t 2 70. Environmental Concerns The spread of oil leaking from a tanker is in the shape of a circle. If the radius r JOGFFU PGUIF spread after t hours is r 1t2 = 2001t, find the area A of the oil slick as a function of the time t. A t = 40,000pt
SECTION 4.1 Composite Functions
*71. Production Cost The price p, in dollars, of a certain product and the quantity x sold obey the demand equation p = -
1 x + 100 0 … x … 400 4
Suppose that the cost C, in dollars, of producing x units is C =
1x + 600 25
Assuming that all items produced are sold, find the cost C as a function of the price p. [Hint: Solve for x in the demand equation and then form the composite.] *72. Cost of a Commodity The price p, in dollars, of a certain commodity and the quantity x sold obey the demand equation p = -
1 x + 200 0 … x … 1000 5
Suppose that the cost C, in dollars, of producing x units is C =
1x + 400 10
Assuming that all items produced are sold, find the cost C as a function of the price p. 73. Volume of a Cylinder The volume V of a right circular cylinder of height h and radius r is V = pr 2h. If the height is twice the radius, express the volume V as a function of r. V r = 2pr 3
313
* E 8IBUJTg1f 110002 2 Interpret this result. 9,11.51942 yen 5 76. Temperature Conversion The function C 1F2 = 1F - 322 9 converts a temperature in degrees Fahrenheit, F, to a temperature in degrees Celsius, C. The function K C = C + 23 converts a temperature in degrees Celsius to a temperature in kelvins, K. * B 'JOEBGVODUJPOUIBUDPOWFSUTBUFNQFSBUVSFJOEFHSFFT Fahrenheit to a temperature in kelvins. C %FUFSNJOFEFHSFFT'BISFOIFJUJOLFMWJOT , 77. Discounts The manufacturer of a computer is offering two discounts on last year’s model computer. The first discount is a $200 rebate and the second discount is 20% off the regular price, p. B 8SJUFBGVODUJPOf that represents the sale price if only the rebate applies. f 1p2 = p - 200 C 8SJUFBGVODUJPOg that represents the sale price if only the 20% discount applies. g1p2 = 0.p * D 'JOE f ∘ g and g ∘ f 8IBUEPFTFBDIPGUIFTFGVODUJPOT SFQSFTFOU 8IJDIDPNCJOBUJPOPGEJTDPVOUTSFQSFTFOUTB CFUUFSEFBMGPSUIFDPOTVNFS 8IZ *78. Taxes 4VQQPTF UIBU ZPV XPSL GPS QFS IPVS 8SJUF B function that represents gross salary G as a function of hours worked h. Your employer is required to withhold taxes GFEFSBM JODPNF UBY 4PDJBM 4FDVSJUZ .FEJDBSF GSPN ZPVS paycheck. Suppose your employer withholds 20% of your JODPNFGPSUBYFT8SJUFBGVODUJPOUIBUSFQSFTFOUTOFUTBMBSZ N as a function of gross salary G. Find and interpret N ∘ G.
74. Volume of a Cone The volume V of a right circular cone 1 is V = pr 2h. If the height is twice the radius, express the 79. 3 2 volume V as a function of r. V r = pr 3 3 75. Foreign Exchange Traders often buy foreign currency 80. in the hope of making money when the currency’s value DIBOHFT 'PS FYBNQMF PO "QSJM POF 64 EPMMBS *81. DPVMEQVSDIBTFFVSPT BOEPOFFVSPDPVMEQVSDIBTF ZFO-FU f 1x2 represent the number of euros you *82. can buy with x dollars, and let g1x2 represent the number of yen you can buy with x euros. * B 'JOEBGVODUJPOUIBUSFMBUFTEPMMBSTUPFVSPT 83. C 'JOEBGVODUJPOUIBUSFMBUFTFVSPTUPZFO g1x2 = 12.4594 x D 6TF UIF SFTVMUT PG QBSUT B BOE C UP GJOE B function that relates dollars to yen. That is, find 1g ∘ f2 x = g1f 1x2 2. g1f 1x2 2 = 9.1151942x
Suppose that f x = x3 + x2 - 16x - 16 and g x = x2 - 4. Find the zeros of f ∘ g x . - 222, - 23, 0, 222, 23 Suppose that f x = 2x3 - 3x2 - x + 12 and g x = x + 5. Find the zeros of f ∘ g x . - , - , - 3 2 If f and g are odd functions, show that the composite function f ∘ g is also odd. If f is an odd function and g is an even function, show that the composite functions f ∘ g and g ∘ f are both even. Let f x = ax + b and g x = bx + a, where a and b are integers. If f 1 = and f g 20
- g f 20
= - 14, find the product of a and b.* 15
Retain Your Knowledge Problems 84–87 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. *84. Given f x = 3x + and g x = x - 5, find 1f + g2 x , f 1f - g2 x , 1f # g2 x , and a b x . State the domain of g each. 85. Find the real zeros of f 1x2 = 2x - 52x + 2.
1 ,4 4
*86. 6TFBHSBQIJOHVUJMJUZUPHSBQI f x = - x3 + 4x - 2 over the interval - 3, 3 . Approximate any local maxima and MPDBM NJOJNB %FUFSNJOF XIFSF UIF GVODUJPO JT JODSFBTJOH and where it is decreasing. x2 + 6x + 5 *87. Find the domain of R x = . Find any x - 3 horizontal, vertical, or oblique asymptotes.
‘Are You Prepared?’ Answers 1. - 21
2. 4 - 1x2
$PVSUFTZPGUIF+PMJFU+VOJPS$PMMFHF .BUIFNBUJDT%FQBSUNFOU
3. 5x 0 x ≠ - 5, x ≠ 56
314
CHAPTER 4 Exponential and Logarithmic Functions
4.2 One-to-One Functions; Inverse Functions PREPARING FOR THIS SECTION Before getting started, review the following: r 'VODUJPOT 4FDUJPO QQm
r *ODSFBTJOH%FDSFBTJOH'VODUJPOT 4FDUJPO QQm
r 3BUJPOBM&YQSFTTJPOT "QQFOEJY" 4FDUJPO" QQ"m"
Now Work the ‘Are You Prepared?’ problems on page 322.
OBJECTIVES 1 Determine Whether a Function Is One-to-One (p. 314) 2 Determine the Inverse of a Function Defined by a Map or a Set of Ordered Pairs (p. 316) 3 Obtain the Graph of the Inverse Function from the Graph of the Function (p. 318) 4 Find the Inverse of a Function Defined by an Equation (p. 319)
1 Determine Whether a Function Is One-to-One 4FDUJPOQSFTFOUFEGPVSEJGGFSFOUXBZTUPSFQSFTFOUBGVODUJPO BNBQ BTFUPG ordered pairs, BHSBQI BOE BOFRVBUJPO'PSFYBNQMF 'JHVSFTBOEJMMVTUSBUF two different functions represented as mappings. The function in Figure 6 shows the DPSSFTQPOEFODF CFUXFFO TUBUFT BOE UIFJS QPQVMBUJPOT JO NJMMJPOT 5IF GVODUJPO JO 'JHVSFTIPXTBDPSSFTQPOEFODFCFUXFFOBOJNBMTBOEMJGFFYQFDUBODJFT JOZFBST Figure 6 State
Figure 7 Population (in millions)
Animal
Indiana
6.5
Dog
Washington
6.9
Cat
South Dakota
0.8
Duck
North Carolina
9.8
Lion
Oklahoma
3.8
Pig
Life Expectancy (in years) 11
10
Rabbit
7
4VQQPTF TFWFSBM QFPQMF BSF BTLFE UP OBNF B TUBUF UIBU IBT B QPQVMBUJPO PG NJMMJPOCBTFEPOUIFGVODUJPOJO'JHVSF&WFSZPOFXJMMSFTQPOEi4PVUI%BLPUBu/PX if the same people are asked to name an animal whose life expectancy is 11 years CBTFEPOUIFGVODUJPOJO'JHVSF TPNFNBZSFTQPOEiEPHu XIJMFPUIFSTNBZSFTQPOE iDBUu8IBUJTUIFEJGGFSFODFCFUXFFOUIFGVODUJPOTJO'JHVSFTBOE *O'JHVSF OP UXPFMFNFOUTJOUIFEPNBJODPSSFTQPOEUPUIFTBNFFMFNFOUJOUIFSBOHF*O'JHVSF UIJTJTOPUUIFDBTF%JGGFSFOUFMFNFOUTJOUIFEPNBJODPSSFTQPOEUPUIFTBNFFMFNFOU in the range. Functions such as the one in Figure 6 are given a special name.
DEFINITION
In Words A function is not one-to-one if two different inputs correspond to the same output.
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 x1 and x2 are any two different inputs of a function f, then f is one-to-one if f1x1 2 ≠ f1x2 2 . 1VU BOPUIFS XBZ B GVODUJPO 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 elements in the domain correspond to the same element in the range. So the GVODUJPOJO'JHVSFJTOPUPOFUPPOFCFDBVTFUXPEJGGFSFOUFMFNFOUTJOUIFEPNBJO dog and cat CPUIDPSSFTQPOEUP'JHVSFJMMVTUSBUFTUIFEJTUJODUJPOBNPOHPOFUPPOF functions, functions that are not one-to-one, and relations that are not functions.
SECTION 4.2 One-to-One Functions; Inverse Functions
315
Figure 8 y1
x1
y2
x2
Domain
EXAM PL E 1
y3
x3
Range
Domain
(a) One-to-one function: Each x in the domain has one and only one image in the range.
x1
y1 y2
x2
y3
x3
y1
x1
x3
Range
Domain
(b) Not a one-to-one function: y1 is the image of both x 1 and x 2.
y3
Range
(c) Not a function: x 1 has two images, y1 and y2.
Determining Whether a Function Is One-to-One %FUFSNJOFXIFUIFSUIFGPMMPXJOHGVODUJPOTBSFPOFUPPOF B 'PSUIFGPMMPXJOHGVODUJPO UIFEPNBJOSFQSFTFOUTUIFBHFPGGJWFNBMFT BOEUIF SBOHFSFQSFTFOUTUIFJS)%- HPPE DIPMFTUFSPMTDPSFT mgdL . Age
HDL Cholesterol
38
57
42
54
46
34
55
38
61
C { - 2, 6 , - 1, 3 , 0, 2 , 1, 5 , 2, }
Solution
B 5IFGVODUJPOJTOPUPOFUPPOFCFDBVTFUIFSFBSFUXPEJGGFSFOUJOQVUT BOE UIBUDPSSFTQPOEUPUIFTBNFPVUQVU C 5IF GVODUJPO JT POFUPPOF CFDBVTF UIFSF BSF OP UXP EJTUJODU JOQVUT UIBU correspond to the same output.
Now Work
PROBLEMS
11
AND
15
r
For functions defined by an equation y = f 1x2 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.
THEOREM Figure 9 f 1x1 2 = f 1x2 2 = h and x1 ≠ x2; f is not a one-to-one function.
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 y ⫽ f (x) (x 1, h)
(x 2, h) h
x1
x2
y⫽h x
EXAM PL E 2
The reason why this test works can be seen in Figure 9, where the horizontal line y = h intersects the graph at two distinct points, 1x1 , h2 and 1x2 , h2 . Since h is the image of both x1 , and x2 , and x1 ≠ x2 , f is not one-to-one. Based on Figure 9, the horizontal-line test can be stated 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.
Using the Horizontal-line Test For each function, use its graph to determine whether the function is one-to-one. B f1x2 = x2
Solution
C g1x2 = x3
B 'JHVSF B POUIFOFYUQBHFJMMVTUSBUFTUIFIPSJ[POUBMMJOFUFTUGPS f1x2 = x2. The horizontal line y = 1 intersects the graph of f twice, at 11, 12 and at 1 - 1, 12 , so f is not one-to-one.
316
CHAPTER 4 Exponential and Logarithmic Functions
C 'JHVSF C JMMVTUSBUFT UIF IPSJ[POUBMMJOF UFTU GPS g1x2 = x3. Because every horizontal line intersects the graph of g exactly once, it follows that g is one-to-one. Figure 10
y
y5x2
y
(1, 1)
23
y51 23
3 x
23
3 x
23
(a) A horizontal line intersects the graph twice; f is not one-to-one.
Now Work
3
3
3 (21, 1)
y5x
PROBLEM
(b) Every horizontal line intersects the graph exactly once; g is one-to-one.
r
19
Look more closely at the one-to-one function g1x2 = x3. This function is an JODSFBTJOHGVODUJPO#FDBVTFBOJODSFBTJOH PSEFDSFBTJOH GVODUJPOXJMMBMXBZTIBWF different y-values for unequal x-values, it follows that a function that is increasing PSEFDSFBTJOH PWFSJUTEPNBJOJTBMTPBPOFUPPOFGVODUJPO
THEOREM
A function that is increasing on an interval I is a one-to-one function on I. A function that is decreasing on an interval I is a one-to-one function on I.
2 Determine the Inverse of a Function Defined by a Map or a Set of Ordered Pairs DEFINITION
In Words Suppose that f is a one-to-one function such that the input 5 corresponds to the output 10. In the inverse function f -1, the input 10 will correspond to the output 5.
EX AMPL E 3
Suppose that f is a one-to-one function. Then, corresponding to each x in the domain of f, there is exactly one yJOUIFSBOHF CFDBVTF f JTBGVODUJPO BOE corresponding to each y in the range of f, there is exactly one x in the domain CFDBVTFf JTPOFUPPOF 5IFDPSSFTQPOEFODFGSPNUIFSBOHFPGf back to the domain of f is called the inverse function of f. The symbol f -1 is used to denote the inverse of f. 8F XJMM EJTDVTT IPX UP GJOE JOWFSTFT GPS BMM GPVS SFQSFTFOUBUJPOT PG GVODUJPOT NBQT TFUT PG PSEFSFE QBJST HSBQIT BOE FRVBUJPOT8F CFHJO XJUI finding inverses of functions represented by maps or sets of ordered pairs.
Finding the Inverse of a Function Defined by a Map Find the inverse of the following function. Let the domain of the function represent DFSUBJOTUBUFT BOEMFUUIFSBOHFSFQSFTFOUUIFTUBUFTQPQVMBUJPOT JONJMMJPOT 'JOE the domain and the range of the inverse function. State
Population (in millions)
Oklahoma
3.8
Indiana
6.5
Washington
6.9
South Dakota
0.8
North Carolina
9.8
SECTION 4.2 One-to-One Functions; Inverse Functions
Solution
317
The function is one-to-one. To find the inverse function, interchange the elements in the domain with the elements in the range. For example, the function receives as input Indiana and outputs 6.5 million. So the inverse receives as input 6.5 million and outputs Indiana. The inverse function is shown next. Population (in millions)
State
3.8
Oklahoma
6.5
Indiana
6.9
Washington
0.8
South Dakota
9.8
North Carolina
The domain of the inverse function is {3., 6.5, 6.9, 0., 9.}. The range of the inverse function is {Oklahoma, Indiana,8BTIJOHUPO 4PVUI%BLPUB /PSUI$BSPMJOB^
r
If the function f is a set of ordered pairs 1x, y2 , then the inverse of f , denoted f -1, is the set of ordered pairs 1y, x2 .
Finding the Inverse of a Function Defined by a Set of Ordered Pairs
EXAM PL E 4
Find the inverse of the following one-to-one function: {1 - 3, - 22, 1 - 2, - 2, 1 - 1, - 12, 10, 02, 11, 12, 12, 2, 13, 22} State the domain and the range of the function and its inverse.
Solution
The inverse of the given function is found by interchanging the entries in each ordered pair and so is given by {1 - 2, - 32, 1 - , - 22, 1 - 1, - 12, 10, 02, 11, 12, 1, 22, 12, 32 } The domain of the function is { - 3, - 2, - 1, 0, 1, 2, 3}. The range of the function is {- 2, - , - 1, 0, 1, , 2}. The domain of the inverse function is {- 2, - , - 1, 0, 1, , 2}. The range of the inverse function is {- 3, - 2, - 1, 0, 1, 2, 3}.
r
Now Work
Figure 11 Domain of f
Range of f f
f Range of f ⫺1
PROBLEMS
25
AND
29
3FNFNCFS JG f is a one-to-one function, it has an inverse function, f -1. See Figure 11. The results of Example 4 and Figure 11 suggest two facts about a one-to-one function f and its inverse f -1.
⫺1
Domain of f
⫺1
WARNING Be careful! f -1 is a symbol for the inverse function of f. The - 1 used in f -1 is not an exponent. That is, f -1 does not mean the reciprocal 1 . j of f; f -1 x is not equal to f x
%omain of f = 3ange of f -1 3ange of f = %omain of f -1
Look again at Figure 11 to visualize the relationship. Starting with x, applying f, and then applying f -1 gets x back again. Starting with x, applying f -1, and then applying f, gets x back again. To put it simply, what f does, f -1 undoes, and vice versa. See the illustration that follows. Input x from domain of f Input x from domain of f - 1
Apply f Apply f -1
f1x2 f - 1 1x2
Apply f -1
f - 1 1f 1x2 2 = x
Apply f
f 1 f - 1 1x2 2 = x
318
CHAPTER 4 Exponential and Logarithmic Functions
In other words, f -1 1f1x2 2 = x, where x is in the domain of f
f1f -1 1x2 2 = x, where x is in the domain of f -1
Figure 12 f f −1
x
f(x ) = 2x
f −1(2x) = 1–2 (2x) = x
Consider the function f1x2 = 2x, which multiplies the argument x by 2. The inverse function f -1 undoes whatever f does. So the inverse function of f is 1 f -1 1x2 = x, which divides the argument by 2. For example, f132 = 2132 = 6 2 1 and f - 1 162 = 162 = 3, so f - 1 undoes what f did. This is verified by showing that 2 1 1 1 f -1 1 f1x2 2 = f -1 12x2 = 12x2 = x and f1 f -1 1x2 2 = f a xb = 2a xb = x 2 2 2 See Figure 12.
EX A MPL E 5
Verifying Inverse Functions
3 B 7FSJGZUIBUUIFJOWFSTFPGg1x2 = x3 is g -1 1x2 = 1 x.
C 7FSJGZUIBUUIFJOWFSTFPGf1x2 = 2x + 3 is f -1 1x2 =
Solution
3 3 B g -1 1g1x2 2 = g -1 1x3 2 = 2 x = x
1 1x - 32 . 2
for all x in the domain of g
g1g 1x2 2 = g1 2x2 = 1 2x2 = x for all x in the domain of g -1 for all x in the 1 1 C f -1 1 f1x2 2 = f -1 12x + 32 = 3 12x + 32 - 34 = 12x2 = x domain of f 2 2 1 1 for all x in the f1 f -1 1x2 2 = f a 1x - 32 b = 2c 1x - 32 d + 3 = 1x - 32 + 3 = x domain of f -1 2 2 -1
3
3
3
r
EX A MPL E 6
Verifying Inverse Functions
1 1 is f -1 1x2 = + 1. For what values of x is x x - 1 f -1 1 f1x2 2 = x For what values of x is f1 f -1 1x2 2 = x
7FSJGZUIBUUIFJOWFSTFPGf1x2 =
Solution
The domain of f is {x x ≠ 1} and the domain of f -1 is {x x ≠ 0}. Now f -1 1 f1x2 2 = f -1 a
1 b = x - 1
1 + 1 = 1x - 12 + 1 = x provided x ≠ 1 1 x - 1 1 1 1 f1 f -1 1x2 2 = f a + 1b = = = x provided x ≠ 0 x 1 1 a + 1b - 1 x x
Now Work
Figure 13 y
y⫽x
(a, b)
b a
(b, a) a
b
x
PROBLEMS
33
AND
r
37
3 Obtain the Graph of the Inverse Function from the Graph of the Function
Suppose that 1a, b2 is a point on the graph of a one-to-one function f defined by y = f1x2 . Then b = f1a2. This means that a = f -1 1b2, so 1b, a2 is a point on the graph of the inverse function f -1. The relationship between the point 1a, b2 on f and the point 1b, a2 on f -1 is shown in Figure 13. The line segment with endpoints 1a, b2 and 1b, a2 is perpendicular to the line y = x and is bisected by the line y = x. %PZPVTFFXIZ *UGPMMPXTUIBUUIFQPJOU 1b, a2 on f -1 is the reflection about the line y = x of the point 1a, b2 on f.
SECTION 4.2 One-to-One Functions; Inverse Functions
THEOREM
319
The graph of a one-to-one function f and the graph of its inverse f -1 are symmetric with respect to the line y = x.
Figure 14 illustrates this result. Once the graph of f is known, the graph of f -1 may be obtained by reflecting the graph of f about the line y = x. Figure 14
y
y ⫽ f (x)
y⫽x
(a 3, b 3) y ⫽ f ⫺1(x)
(a 2, b 2)
(b 3, a 3) x
(a 1, b 1) (b 2, a 2) (b 1, a 1)
EXAMPL E 7
Graphing the Inverse Function 5IFHSBQIJO'JHVSF B JTUIBUPGBPOFUPPOFGVODUJPOy = f1x2.%SBXUIFHSBQI of its inverse.
Solution
Begin by adding the graph of y = x UP 'JHVSF B 4JODF UIF QPJOUT 1 - 2, - 12, 1 - 1, 02, and 12, 12 are on the graph of f, the points 1 - 1, - 22, 10, - 12, and 11, 22 must be on the graph of f -1.,FFQJOHJONJOEUIBUUIFHSBQIPG f -1 is the reflection about the line y = x of the graph of f, draw f -1.4FF'JHVSF C
Figure 15
y 3
y 3
y⫽x (1, 2)
Exploration Simultaneously graph Y1 = x, Y2 = x3, 3 and Y3 = 2x on a square screen with - 3 … x … 3. What do you observe about the graphs of Y2 = x3, its inverse 3 Y3 = 2x, and the line Y1 = x? Repeat this experiment by simultaneously graphing Y1 = x, 1 Y2 = 2x + 3, and Y3 = (x - 3) on a 2 square screen with - 6 … x … 3. Do you see the symmetry of the graph of Y2 and its inverse Y3 with respect to the line Y1 = x?
y ⫽ f (x) (⫺1, 0) ⫺3
y ⫽ f(x) (2, 1) 3 x
(⫺2, ⫺1)
⫺3 (⫺2, ⫺1)
3 x (0, ⫺1) (⫺1, ⫺2) ⫺3
⫺3
y ⫽ f ⫺1(x)
(b)
(a)
Now Work
(2, 1)
(⫺1, 0)
PROBLEM
r
43
4 Find the Inverse of a Function Defined by an Equation The fact that the graphs of a one-to-one function f and its inverse function f -1 are symmetric with respect to the line y = x tells us more. It says that we can obtain f -1 by interchanging the roles of x and y in f. Look again at Figure 14. If f is defined by the equation y = f1x2 then f -1 is defined by the equation x = f1y2
320
CHAPTER 4 Exponential and Logarithmic Functions
The equation x = f 1y2 defines f -1 implicitly. Solving this equation for y yields the explicit form of f -1, that is, y = f -1 x
Let’s use this procedure to find the inverse of f1x2 = 2x + 3. #FDBVTF f is a linear function and is increasing, f JTPOFUPPOFBOEUIVTIBTBOJOWFSTFGVODUJPO
How to Find the Inverse Function
EX A MPL E 8
Find the inverse of f1x2 = 2x + 3. Graph f and f -1 on the same coordinate axes.
Step-by-Step Solution
3FQMBDF f1x2 with y in f 1x2 = 2x + 3 and obtain y = 2x + 3. Now interchange the variables x and y to obtain
Step 1 Replace f(x) with y. In y = f(x), interchange the variables x and y to obtain x = f(y). This equation defines the inverse function f -1 implicitly.
x = 2y + 3 This equation defines the inverse f -1 implicitly. To find the explicit form of the inverse, solve x = 2y + 3 for y.
Step 2 If possible, solve the implicit equation for y in terms of x to obtain the explicit form of f -1, y = f -1(x).
x = 2y + 3 2y + 3 = x 2y = x - 3 1 y = 1x - 32 2
Reflexive Property; If a = b, then b = a. Subtract 3 from both sides. 1 Multiply both sides by . 2
The explicit form of the inverse f - 1 is f -1 1x2 =
1 1x - 32 2
4FF&YBNQMF C GPSWFSJGJDBUJPOUIBUf and f - 1 are inverses. 1 The graphs of f1x2 = 2x + 3 and its inverse f -1 1x2 = 1x - 32 are shown in 2 Figure 16. Note the symmetry of the graphs with respect to the line y = x.
Step 3 Check the result by showing that f -1(f(x)) = x and f(f -1(x)) = x.
r
Figure 16 y
Procedure for Finding the Inverse of a One-to-One Function STEP 1: In y = f 1x2, interchange the variables x and y to obtain
f(x) 2x 3 5
yx f
5
x = f1y2
1
(x) 1–2(x 3) 5
x
This equation defines the inverse function f -1 implicitly. STEP 2: If possible, solve the implicit equation for y in terms of x to obtain the explicit form of f -1: y = f -1 1x2
5
EX A MPL E 9
STEP 3: Check the result by showing that
f -1 1 f x2 2 = x and f1 f -1 1x2 2 = x
Finding the Inverse Function The function f 1x2 =
2x + 1 x - 1
x ≠ 1
is one-to-one. Find its inverse and check the result.
SECTION 4.2 One-to-One Functions; Inverse Functions
Solution
321
STEP 1: 3FQMBDFf1x2 with y and interchange the variables x and y in y =
2x + 1 x - 1
x =
2y + 1 y - 1
to obtain
STEP 2: Solve for y. x = x1y - 12 xy - x xy - 2y 1x - 22y
= = = =
y =
2y + 1 y - 1 2y + 1 2y + 1 x + 1 x + 1
Multiply both sides by y - 1. Apply the Distributive Property. Subtract 2y from both sides; add x to both sides. Factor.
x + 1 x - 2
Divide by x - 2.
The inverse is f -1 1x2 = STEP 3:
x + 1 , x ≠ 2 x - 2
Replace y by f -1(x).
Check:
2x + 1 + 1 2x + 1 x - 1 2x + 1 + x - 1 3x f -1 1 f1x2 2 = f -1 a b = = = = x, x ≠ 1 x - 1 2x + 1 2x + 1 - 21x - 12 3 - 2 x - 1 f1 f -1 1x2 2 = f a
x + 1 b = x - 2
x + 1 b + 1 21x + 12 + x - 2 x - 2 3x = = = x, x ≠ 2 x + 1 x + 1 - 1x - 22 3 - 1 x - 2
2a
r
Exploration
2x + 1 x + 1 In Example 9, it was found that, if f(x) = , then f-1(x) = . Compare the vertical and x - 1 x - 2 -1 horizontal asymptotes of f and f . Result The vertical asymptote of f is x = 1, and the horizontal asymptote is y = 2. The vertical asymptote of f-1 is x = 2, and the horizontal asymptote is y = 1.
Now Work
PROBLEMS
51
AND
65
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.
EX AM PL E 10
Solution
Finding the Inverse of a Domain-restricted Function Find the inverse of y = f 1x2 = x2 if x Ú 0. Graph f and f - 1.
The function y = x2 JT OPU POFUPPOF )PXFWFS restricting the domain of this function to x Ú 0, as indicated, yields a new function that is increasing and, therefore, is one-to-one. As a result, the function defined by y = f1x2 = x2, x Ú 0, has an inverse function, f -1. Follow the steps given previously to find f -1.
322
CHAPTER 4 Exponential and Logarithmic Functions
STEP 1: In the equation y = x2, x Ú 0, interchange the variables x and y. The result is
Figure 17 y 2
f (x ) x 2, x 0
x = y2
yx f 1(x ) x
2
x
y Ú 0
5IJTFRVBUJPOEFGJOFT JNQMJDJUMZ UIFJOWFSTFGVODUJPO
STEP 2: Solve for y to get the explicit form of the inverse. Since y Ú 0, only one solution for y is obtained: y = 1x. So f -1 1x2 = 1x. STEP 3:
Check:
f -1 1 f1x2 2 = f -1 1x2 2 = 2x2 = 0 x 0 = x since x Ú 0
f 1 f -1 1x2 2 = f 1 1x2 = 1 1x2 2 = x
'JHVSFJMMVTUSBUFTUIFHSBQITPG f1x2 = x2, x Ú 0, and f -1 1x2 = 1x. Note that the domain of f = range of f - 1 = [0, q and the domain of f - 1 = range of f = [0, q .
r
SUMMARY 1. If a function f is one-to-one, then it has an inverse function f -1. 2. %PNBJOPGf = 3ange of f -1;3BOHFPGf = %omain of f -1. 3. To verify that f -1 is the inverse of f, show that f -1 1 f1x2 2 = x for every x in the domain of f and that f1 f -1 1x2 2 = x for every x in the domain of f -1. 4. The graphs of f and f -1 are symmetric with respect to the line y = x.
4.2 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. Is the set of ordered pairs { 11, 32, 12, 32, 1 - 1, 22} a GVODUJPO 8IZPSXIZOPU QQm yes
2. 8IFSF JT UIF GVODUJPO f 1x2 = x2 JODSFBTJOH 8IFSF JT JU EFDSFBTJOH Q 10, q 2; 1 - q , 02 x + 5 QQm
3. 8IBUJTUIFEPNBJOPGf 1x2 = 2 x + 3x - 1 {x 0 x ≠ - 6, x ≠ 3}
1 + 1 x 4. Simplify: QQ"m"
1 1 x2 x , x ≠ 0, x ≠ - 1, x ≠ 1 1-x
Concepts and Vocabulary 5. If x1 and x2 are any two different inputs of a function f, then f is one-to-one if f 1x1 2 ≠ f 1x2 2 . 6. If every horizontal line intersects the graph of a function f at one@to@one function. no more than one point, f JTB O 7. If f is a one-to-one function and f 132 = , then f -1 12 = 3 .
8. If f -1 denotes the inverse of a function f, then the graphs of f and f -1 are symmetric with respect to the line y = x . 9. If the domain of a one-to-one function f is 34, q 2, then the 34, q 2 . range of its inverse, f -1, is 10. True or False If f and g are inverse functions, the domain of f is the same as the range of g. True
Skill Building In Problems 11–18, determine whether the function is one-to-one. 11.
Domain
Range
20 Hours 25 Hours
one-to-one
12.
Domain
Range
$200
Bob
Karla
$300
Dave
Debra
30 Hours
$350
John
Dawn
40 Hours
$425
Chuck
Phoebe
one-to-one
SECTION 4.2 One-to-One Functions; Inverse Functions
13.
Domain
Range
20 Hours
$200
not one-to-one
14.
Domain
25 Hours
Karla
Dave
Debra Phoebe
$350
John
40 Hours
$425
Chuck
15. { 12, 62, 1 - 3, 62, 14, 92, 11, 102} 17. { 10, 02, 11, 12, 12, 162, 13, 12}
16. { 1 - 2, 52, 1 - 1, 32, 13, 2, 14, 122}
not one-to-one
18. { 11, 22, 12, 2, 13, 12, 14, 322}
one-to-one
not one-to-one
Range
Bob
30 Hours
323
one-to-one
one-to-one
In Problems 19–24, the graph of a function f is given. Use the horizontal-line test to determine whether f is one-to-one. 19.
one-to-one
y 3
⫺3
20.
⫺3
3 x
⫺3
22.
one-to-one
y 3
21.
y 3
⫺3
3 x
3 x
⫺3
⫺3
not one-to-one 23.
y
not one-to-one
one-to-one
y 3
24.
y 3
not one-to-one
2 ⫺3
3 ⫺3
x
⫺3
3 x
3 x
In Problems 25–32, find the inverse of each one-to-one function. State the domain and the range of each inverse function. *25. Location
Annual Precipitation (inches)
*26.
Domestic Gross (in millions)
Title
Atlanta, GA
49.7
Avatar
$761
Boston, MA
43.8
Titanic
$659
Las Vegas, NV
4.2
Marvel’s The Avengers
$623
Miami, FL
61.9
The Dark Knight
$535
Los Angeles, CA
12.8
Star Wars: Episode One – The Phantom Menace
$475
Source: currentresults.com
*27. Age 30 40
Source: boxofficemojo.com Monthly Cost of Life Insurance $10.55 $12.89 $19.29
45 Source: acequotes.com
*29. { 1 - 3, 52, 1 - 2, 92, 1 - 1, 22, 10, 112, 11, - 52}
*31. { 1 - 2, 12, 1 - 3, 22, 1 - 10, 02, 11, 92, 12, 42}
*28.
State
Unemployment Rate
Virginia Nevada Tennessee Texas
5.6% 9.7% 7.7% 6.3%
Source: United States Bureau of Labor Statistics, Jan. 2013
*30. { 1 - 2, 22, 1 - 1, 62, 10, 2, 11, - 32, 12, 92}
*32. { 1 - 2, - 2, 1 - 1, - 12, 10, 02, 11, 12, 12, 2}
In Problems 33–42, verify that the functions f and g are inverses of each other by showing that f 1 g1x2 2 = x and g 1 f 1x2 2 = x. Give any values of x that need to be excluded from the domain of f and the domain of g. 1 1x - 42 3 x *35. f 1x2 = 4x - ; g1x2 = + 2 4
*33. f 1x2 = 3x + 4; g1x2 =
*34. f 1x2 = 3 - 2x; g1x2 = *36. f 1x2 = 2x + 6; g1x2 =
%VFUPTQBDFSFTUSJDUJPOT BOTXFSTUPUIFTFFYFSDJTFTNBZCFGPVOEJOUIF"OTXFSTJOUIFCBDLPGUIFCPPL
1 1x - 32 2
1 x - 3 2
324
CHAPTER 4 Exponential and Logarithmic Functions
3 *37. f 1x2 = x3 - ; g1x2 = 2 x +
*38. f 1x2 = 1x - 22 2, x Ú 2; g1x2 = 1x + 2
*39. f 1x2 =
*40. f 1x2 = x; g1x2 = x
1 1 ; g1x2 = x x 4x - 3 2x + 3 ; g1x2 = *41. f 1x2 = x + 4 2 - x
*42. f 1x2 =
x - 5 3x + 5 ; g1x2 = 2x + 3 1 - 2x
In Problems 43–48, the graph of a one-to-one function f is given. Draw the graph of the inverse function f -1. For convenience (and as a hint), the graph of y = x is also given. *43.
y 3
*44.
y⫽x
y⫽x
y 3
y 3
y⫽x
(1, 2)
(0, 1)
(2, 1–2)
(⫺1, 0) ⫺3
⫺3
3 x
(1, 0)
(2, 1) ⫺3 (⫺1, ⫺1)
3 x
(0, ⫺1)
(⫺2, ⫺2)
(⫺2, ⫺2)
⫺3
⫺3
*46.
*45.
y⫽x
y 3
*47.
y 3
3 x
⫺3 y⫽x
*48.
y 2
y⫽x
(⫺2, 1) ⫺3
⫺3
3 x (1, ⫺1) ⫺3
⫺2
3 x
2 x
⫺2
⫺3
In Problems 49–60, the function f is one-to-one. Find its inverse and check your answer. Graph f, f -1, and y = x on the same coordinate axes. *49. f 1x2 = 3x
f -1 1x2 =
*51. f 1x2 = 4x + 2 *53. f 1x2 = x3 - 1
*50. f 1x2 = - 4x
1 x 3
f -1 1x2 =
x 1 4 2 3 -1 f 1x2 = 1x + 1
*55. f 1x2 = x2 + 4, x Ú 0 f -1 1x2 = 2x - 4, x Ú 4 4 4 *57. f 1x2 = f -1 1x2 = x x 1 2x + 1 f -1 1x2 = *59. f 1x2 = x - 2 x
1 f -1 1x2 = - x 4 1 - x -1 *52. f 1x2 = 1 - 3x f 1x2 = 3 3 *54. f 1x2 = x3 + 1 f -1 1x2 = 1 x - 1
*56. f 1x2 = x2 + 9, x Ú 0 f -1 1x2 = 2x - 9, x Ú 9 3 3 *58. f 1x2 = f -1 1x2 = x x 4 4 *60. f 1x2 = f -1 1x2 = - 2 x + 2 x
In Problems 61–72, the function f is one-to-one. Find its inverse and check your answer. *61. f 1x2 = *63. *65. *67. *69. *71.
2 3 + x 3x f 1x2 = x + 2 2x f 1x2 = 3x - 1 3x + 4 f 1x2 = 2x - 3 2x + 3 f 1x2 = x + 2 x2 - 4 f 1x2 = , 2x2
2 - 3x x 2x f -1 1x2 = x - 3 x f -1 1x2 = 3x - 2 3x + 4 f -1 1x2 = 2x - 3 - 2x + 3 f -1 1x2 = x - 2 f -1 1x2 =
x 7 0
f -1 1x2 =
2 11 - 2x
*62. f 1x2 = *64. f 1x2 = *66. f 1x2 = *68. f 1x2 = *70. f 1x2 = *72. f 1x2 =
4 4 f -1 1x2 = 2 2 - x x 2x x f -1 1x2 = x - 1 x + 2 3x + 1 -1 f -1 1x2 = x x + 3 2x - 3 3 + 4x f -1 1x2 = x + 4 2 - x - 3x - 4 2x - 4 f -1 1x2 = x - 2 x + 3 x2 + 3 3 , x 7 0 f -1 1x2 = A 3x - 1 3x2
Applications and Extensions 73. 6TFUIFHSBQIPGy = f 1x2 HJWFOJO1SPCMFNUPFWBMVBUF the following: B f 1 - 12 0 C f 112 2 D f -1 112 0 E f -1 122 1
74. 6TFUIFHSBQIPGy = f 1x2 HJWFOJO1SPCMFNUPFWBMVBUFUIF following: 1 B f 122 C f 112 0 D f -1 102 1 E f -1 1 - 12 0 2
SECTION 4.2 One-to-One Functions; Inverse Functions
75. If f 12 = 13 and f is one-to-one, what is f -1 1132
-5 76. If g1 - 52 = 3 and g is one-to-one, what is g -1 132 *77. The domain of a one-to-one function f is [5, q 2, and its range is [ - 2, q 2. State the domain and the range of f -1. *78. The domain of a one-to-one function f is [0, q 2, and its range is [5, q 2. State the domain and the range of f -1. *79. The domain of a one-to-one function g is 1 - q , 04 , and its range is [0, q 2. State the domain and the range of g -1. *80. The domain of a one-to-one function g is [0, 15], and its SBOHFJT 4UBUFUIFEPNBJOBOEUIFSBOHFPGg -1.
*81. A function y = f 1x2JTJODSFBTJOHPOUIFJOUFSWBM 8IBU conclusions can you draw about the graph of y = f -1 1x2
82. A function y = f 1x2 JT EFDSFBTJOH PO UIF JOUFSWBM 8IBU DPODMVTJPOT DBO ZPV ESBX BCPVU UIF HSBQI PG y = f -1 1x2 %FDSFBTJOHPOUIFJOUFSWBM 1 f 152, f 102 2
325
*83. Find the inverse of the linear function f 1x2 = mx + b, m ≠ 0 *84. Find the inverse of the function f 1x2 = 2r 2 - x2, 0 … x … r 85. A function f has an inverse function. If the graph of f lies in quadrant I, in which quadrant does the graph of f -1MJF I 86. A function f has an inverse function. If the graph of f lies in quadrant II, in which quadrant does the graph of f -1MJF *7
*87. The function f 1x2 = 0 x 0 is not one-to-one. Find a suitable restriction on the domain of f so that the new function that results is one-to-one. Then find the inverse of f.
*88. The function f 1x2 = x4 is not one-to-one. Find a suitable restriction on the domain of f so that the new function that results is one-to-one. Then find the inverse of f.
In applications, the symbols used for the independent and dependent variables are often based on common usage. So, rather than using y = f 1x2 to represent a function, an applied problem might use C = C 1q2 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 inverses of applied problems. Rather, the inverse of a function such as C = C 1q2 will be q = q1C2. So C = C 1q2 is a function that represents the cost C as a function of the output q, and q = q1C2 is a function that represents the output q as a function of the cost C. Problems 89–92 illustrate this idea. * B &YQSFTT UIF UFNQFSBUVSF JO EFHSFFT $FMTJVT C as a 89. Vehicle Stopping Distance Taking into account reaction time, function of the temperature in degrees Fahrenheit F. the distance d JOGFFU UIBUBDBSSFRVJSFTUPDPNFUPBDPNQMFUF * C 7FSJGZ UIBU C = C 1F2 is the inverse of F = F 1C2 by stop while traveling r miles per hour is given by the function showing that C 1F 1C2 = C and F 1C 1F2 2 = F. d1r2 = 6.9r - 90.39 D 8IBU JT UIF UFNQFSBUVSF JO EFHSFFT $FMTJVT JG JU JT EFHSFFT'BISFOIFJU 21.1°C * B &YQSFTT UIF TQFFE r at which the car is traveling as a function of the distance d required to come to a complete stop. * C 7FSJGZ UIBU r = r 1d2 is the inverse of d = d1r2 by showing that r 1d1r2 2 = r and d1r 1d2 2 = d. D 1SFEJDUUIFTQFFEUIBUBDBSXBTUSBWFMJOHJGUIFEJTUBODF required to stop was 300 feet. 56 mph 90. Height and Head Circumference The head circumference C of a child is related to the height HPGUIFDIJME CPUIJO JODIFT UISPVHIUIFGVODUJPO H1C2 = 2.15C - 10.53 * B &YQSFTTUIFIFBEDJSDVNGFSFODFC as a function of height H. * C 7FSJGZUIBUC = C 1H2 is the inverse of H = H1C2 by showing that H1C 1H2 2 = H and C 1H1C2 2 = C. D 1SFEJDU UIF IFBE DJSDVNGFSFODF PG B DIJME XIP JT inches tall. JO 91. Ideal Body Weight One model for the ideal body weight W GPSNFO JOLJMPHSBNT BTBGVODUJPOPGIFJHIUh JOJODIFT JT given by the function W 1h2 = 50 + 2.31h - 602
B 8IBUJTUIFJEFBMXFJHIUPGBGPPUNBMF LH * C &YQSFTTUIFIFJHIUh as a function of weight W. * D 7FSJGZUIBUh = h1W2 is the inverse of W = W 1h2 by showing that h1W 1h2 2 = h and W 1h1W2 2 = W. E 8IBUJTUIFIFJHIUPGBNBMFXIPJTBUIJTJEFBMXFJHIUPG LJMPHSBNT JO [Note: The ideal body weight W GPS XPNFO JO LJMPHSBNT BTBGVODUJPOPGIFJHIUh JOJODIFT JTHJWFOCZ W h = 45.5 + 2.3 h - 60 .] 9 92. Temperature Conversion The function F 1C2 = C + 32 5 converts a temperature from C degrees Celsius to F degrees Fahrenheit.
*93. Income Taxes The function
T 1g2 = 4991.25 + 0.251g - 36,2502
represents the 2013 federal income tax T JO EPMMBST EVF for a “single” filer whose modified adjusted gross income is g dollars, where 36,250 … g … ,50. B 8IBUJTUIFEPNBJOPGUIFGVODUJPOT C (JWFOUIBUUIFUBYEVFT is an increasing linear function of modified adjusted gross income g, find the range of the function T. D 'JOE BEKVTUFE HSPTT JODPNF g as a function of federal income tax T8IBUBSFUIFEPNBJOBOEUIFSBOHFPGUIJT GVODUJPO *94. Income Taxes The function
T 1g2 = 15 + 0.151g - 1,502
represents the 2013 federal income tax T JOEPMMBST EVFGPS a “married filing jointly” filer whose modified adjusted gross income is g dollars, where 1,50 … g … 2,500. B 8IBUJTUIFEPNBJOPGUIFGVODUJPOT C (JWFOUIBUUIFUBYEVFT is an increasing linear function of modified adjusted gross income g, find the range of the function T. D 'JOEBEKVTUFEHSPTTJODPNFg as a function of federal income tax T8IBUBSFUIFEPNBJOBOEUIFSBOHFPGUIJTGVODUJPO 95. Gravity on Earth If a rock falls from a height of 100 meters on Earth, the height H JONFUFST BGUFSt seconds is approximately H 1t2 = 100 - 4.9t 2
B In general, quadratic functions are not one-to-one. However, the function HJTPOFUPPOF8IZ t Ú 0 * C 'JOEUIFJOWFSTFPGH and verify your result. D )PXMPOHXJMMJUUBLFBSPDLUPGBMMNFUFST 2.02 s
326
CHAPTER 4 Exponential and Logarithmic Functions
96. Period of a Pendulum The period T (in seconds) of a simple pendulum as a function of its length l (in feet) is given by
97. Given f 1x2 =
l T 1l2 = 2p A 32.2
ax + b cx + d
find f -1 1x2. If c ≠ 0, under what conditions on a, b, c, and d - dx + b ; f = f -1if a = - d is f = f -1? f - 1 1x2 = cx - a
*(a) Express the length l as a function of the period T. (b) How long is a pendulum whose period is 3 seconds? 7.34 ft
Discussion and Writing *98. 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. 99. Draw the graph of a one-to-one function that contains the points ( - 2, - 32, 10, 02 , and 11, 52 . Now draw the graph of its inverse. Compare your graph to those of other students. Discuss any similarities. What differences do you see? *100. 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.
101. Is every odd function one-to-one? Explain. No
*102. Suppose that C 1g2 represents the cost C, in dollars, of manufacturing g cars. Explain what C -1 1800,0002 represents. 103. Explain why the horizontal-line test can be used to identify one-to-one functions from a graph. 104. Explain why a function must be one-to-one in order to have an inverse that is a function. Use the function y = x2 to support your explanation.
[Hint: Use a piecewise-defined function.]
Retain Your Knowledge Problems 105–108 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 6x2 - 11x - 2 105. If f 1x2 = 3x2 - 7x, find f 1x + h2 - f 1x2 6xh + 3h2 - 7h *108. Find the domain of R(x) = . Find any 2x2 - x - 6 *106. Use the techniques of shifting, compressing or stretching, horizontal, vertical, or oblique asymptotes. and reflections to graph f(x) = - x + 2 0 + 3. 3 Domain: {x 0 x ≠ - , x ≠ 2}; *107. Find the zeros of the quadratic function f(x) = 3x2 + 5x + 1. 2 3 What are the x-intercepts, if any, of the graph of the function? Vertical asymptote: x = - ; Horizontal asymptote: y = 3 2
‘Are You Prepared?’ Answers 1. Yes; for each input x there is one output y. 2. Increasing on 10, q 2; decreasing on 1 - q , 02
3. {x x ≠ - 6, x ≠ 3} x , x ≠ 0, x ≠ - 1, x ≠ 1 4. 1 - x
4.3 Exponential Functions PREPARING FOR THIS SECTION Before getting started, review the following: r &YQPOFOUT "QQFOEJY" 4FDUJPO" QQ"m" BOE4FDUJPO" QQ"m"
r (SBQIJOH5FDIOJRVFT5SBOTGPSNBUJPOT 4FDUJPO QQm
r -JOFBS&RVBUJPOT "QQFOEJY" 4FDUJPO" QQ"m"
r r r r
"WFSBHF3BUFPG$IBOHF 4FDUJPO QQm
2VBESBUJD'VODUJPOT 4FDUJPO QQm
-JOFBS 'VODUJPOT 4FDUJPO QQm
)PSJ[POUBM"TZNQUPUFT 4FDUJPO QQm
Now Work the ‘Are You Prepared?’ problems on page 337.
OBJECTIVES 1 2 3 4
Evaluate Exponential Functions (p. 326) Graph Exponential Functions (p. 330) Define the Number e (p. 333) Solve Exponential Equations (p. 335)
1 Evaluate Exponential Functions "QQFOEJY" 4FDUJPO"HJWFTBEFGJOJUJPOGPSSBJTJOHBSFBMOVNCFSa to a rational power. That discussion provides meaning to expressions of the form
SECTION 4.3 Exponential Functions
327
ar where the base a is a positive real number and the exponent r is a rational number. But what is the meaning of ax, where the base a is a positive real number and the exponent x JT BO JSSBUJPOBM OVNCFS "MUIPVHI B SJHPSPVT EFGJOJUJPO SFRVJSFT methods discussed in calculus, the basis for the definition is easy to follow: Select a rational number rUIBUJTGPSNFECZUSVODBUJOH SFNPWJOH BMMCVUBGJOJUFOVNCFSPG digits from the irrational number x. Then it is reasonable to expect that ax ≈ ar For example, take the irrational number p = 3.14159 c . Then an approximation to ap is ap ≈ a3.14 where the digits after the hundredths position have been removed from the value for p. A better approximation would be ap ≈ a3.14159 where the digits after the hundred-thousandths position have been removed. Continuing in this way, approximations to ap can be obtained to any desired degree of accuracy. .PTUDBMDVMBUPSTIBWFBO xy key or a caret key for working with exponents. To evaluate expressions of the form ax, enter the base a, then press the xy key PSUIF LFZ
FOUFSUIFFYQPOFOUx, and press = PS ENTE3
^
^
EXAM PL E 1
Using a Calculator to Evaluate Powers of 2 6TJOHBDBMDVMBUPS FWBMVBUF B 21.4
Solution Figure 18
C 21.41
D 21.414
E 21.4142
F 222
'JHVSFTIPXTUIFTPMVUJPOTUPQBSUT B BOE F VTJOHB5*1MVTHSBQIJOHDBMDVMBUPS B 21.4 ≈ 2.63901522 D 21.414 ≈ 2.6644965 F 222 ≈ 2.665144143
Now Work
PROBLEM
C 21.41 ≈ 2.653162 E 21.4142 ≈ 2.66511909
r
15
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 7 0 and b 7 0, then as # at = as + t 1s = 1
1as 2 = ast 1ab2 s = as # bs s 1 1 a-s = s = a b a0 = 1 a a t
(1)
Introduction to Exponential Growth Suppose a function f has the following two properties:
Table 1 x
f(x)
0
5
1
10
2
20
3
40
4
80
1. The value of f doubles with every 1-unit increase in the independent variable x. 2. The value of f at x = 0 is 5, so f 102 = 5. Table 1 shows values of the function f for x = 0, 1, 2, 3, and 4. Let’s find an equation y = f1x2 that describes this function f. The key fact is that the value of f doubles for every 1-unit increase in x. f102 = 5 f112 = 2f 102 = 2 # 5 = 5 # 21 f122 = 2f 112 = 215 # 22 = 5 # 22
Double the value of f at 0 to get the value at 1. Double the value of f at 1 to get the value at 2.
328
CHAPTER 4 Exponential and Logarithmic Functions
f 132 = 2f 122 = 215 # 22 2 = 5 # 23 f142 = 2f 132 = 215 # 23 2 = 5 # 24 The pattern leads to f1x2 = 2f1x - 12 = 215 # 2x-1 2 = 5 # 2x
DEFINITION
An exponential function is a function of the form f1x2 = Cax
WARNING It is important to distinguish a power function, g x = axn, n Ú 2, an integer, from an exponential function, f x = C # ax, a ≠ 1, a 7 0. In a power function, the base is a variable and the exponent is a constant. In an exponential function, the base is a constant and the exponent is a variable. j
where a is a positive real number a 7 0 , a ≠ 1, and C ≠ 0 is a real number. The domain of f is the set of all real numbers. The base a is the growth factor, and because f 0 = Ca0 = C, C is called the initial value. In the definition of an exponential function, the base a = 1 is excluded because this function is simply the constant function f1x2 = C # 1x = C. Bases that are negative also are excluded; otherwise, many values of x would have to be excluded 3 1 and x = . 2 = 2 - 2, from the domain, such as x = 2 4 4 4 1 - 32 34 = 2 1 - 32 3 = 2 - 2, and so on, are not defined in the set of real OVNCFST>'JOBMMZ USBOTGPSNBUJPOT WFSUJDBMTIJGUT IPSJ[POUBMTIJGUT SFGMFDUJPOT BOE TPPO PGBGVODUJPOPGUIFGPSN f 1x2 = Cax also represent exponential functions. Some examples of exponential functions are f 1x2 = 2x
1 x F1x2 = a b + 5 3
G1x2 = 2 # 3x - 3
For each function, note that the base of the exponential expression is a constant and the exponent contains a variable. In the function, f1x2 = 5 # 2x, notice that the ratio of consecutive outputs is constant for 1-unit increases in the input. This ratio equals the constant 2, the base of the exponential function. In other words, f112 5 # 21 = = 2 f102 5
f122 5 # 22 = = 2 f112 5 # 21
f132 5 # 23 = = 2 and so on f122 5 # 22
This leads to the following result.
THEOREM
For an exponential function f1x2 = Cax, where a 7 0 and a ≠ 1, if x is any real number, then f1x + 12 = a f1x2
In Words For 1-unit changes in the input x of an exponential function f(x) = C # a x , the ratio of consecutive outputs is the constant a.
or f1x + 12 = af1x2
Proof f1x + 12 Cax + 1 = = ax + 1 - x = a1 = a f1x2 Cax
EX AMPL E 2
■
Identifying Linear or Exponential Functions %FUFSNJOFXIFUIFSUIFHJWFOGVODUJPOJTMJOFBS FYQPOFOUJBM PSOFJUIFS'PSUIPTF that are linear, find a linear function that models the data. For those that are exponential, find an exponential function that models the data.
SECTION 4.3 Exponential Functions
B
Solution
x
y
-1
C
x
y
5
-1
0
2
1
-1
2 3
D
x
y
32
-1
2
0
16
0
4
1
8
1
7
-4
2
4
2
11
-7
3
2
3
16
329
For each function, compute the average rate of change of y with respect to x and the ratio of consecutive outputs. If the average rate of change is constant, then the function is linear. If the ratio of consecutive outputs is constant, then the function is exponential.
Table 2 (a) x
y
-1
5
Average Rate of Change
Ratio of Consecutive Outputs
∆y
2 5
∆x
(b)
=
2 - 5 = -3 0 - ( - 1)
-1 - 2 = -3 1 - 0
1 -1 = 2 2
-1
- 4 - ( - 1) = -3 2 - 1
-4 = 4 -1
2
-4
- 7 - ( - 4) = -3 3 - 2
7 -7 = -4 4
3
-7
Average Rate of Change
Ratio of Consecutive Outputs
0
2
1
x
y
-1
32
∆y ∆x
0
16
1
8
2
4
3
2
=
16 - 32 = - 16 0 - ( - 1)
16 1 = 32 2
-8
8 1 = 16 2
-4
4 1 = 8 2
-2
1 2 = 4 2
(c) x
y
Average Rate of Change
-1
2
∆y ∆x
0
4
1
7
2 3
4 - 2 = 2 0 - ( - 1)
2
3
7 4
4
11 7
5
16 11
11 16
=
Ratio of Consecutive Outputs
330
CHAPTER 4 Exponential and Logarithmic Functions
B 4FF5BCMF B 5IFBWFSBHFSBUFPGDIBOHFGPSFWFSZVOJUJODSFBTFJOx is - 3. Therefore, the function is a linear function. In a linear function the average rate of change is the slope m, so m = - 3. The y-intercept b is the value of the function at x = 0, so b = 2. The linear function that models the data is f1x2 = mx + b = - 3x + 2. C 4FF 5BCMF C 'PS UIJT GVODUJPO UIF BWFSBHF SBUF PG DIBOHF GSPN - 1 to 0 is - 16, and the average rate of change from 0 to 1 is - . Because the average rate of change is not constant, the function is not a linear function. The ratio of consecutive outputs for a 1-unit increase in the inputs is 1 a constant, . Because the ratio of consecutive outputs is constant, the function 2 1 is an exponential function with growth factor a = . The initial value of the 2 exponential function is C = 16. Therefore, the exponential function that models 1 x the data is g x = Cax = 16 # a b . 2 D 4FF5BCMF D 'PSUIJTGVODUJPO UIFBWFSBHFSBUFPGDIBOHFGSPN - 1 to 0 is 2, and the average rate of change from 0 to 1 is 3. Because the average rate of change is not constant, the function is not a linear function. The ratio of consecutive outputs from - 1 to 0 is 2, and the ratio of consecutive outputs from 0 to 1 is . Because the ratio of consecutive outputs is not a constant, the function 4 is not an exponential function.
r
Now Work
PROBLEM
25
2 Graph Exponential Functions If we know how to graph an exponential function of the form f x = ax, then we DBOVTFUSBOTGPSNBUJPOT TIJGUJOH TUSFUDIJOH BOETPPO UPPCUBJOUIFHSBQIPGBOZ exponential function. First, let’s graph the exponential function f 1x2 = 2x.
EX A MPL E 3
Solution Table 3 x
f(x) = 2x
−10
2-10 ≈ 0.00098 1 2-3 = 8 1 2-2 = 4 1 2-1 = 2
−3 −2 −1 0
20 = 1
1
21 = 2
2
22 = 4
3
23 = 8
10
Graphing an Exponential Function
Graph the exponential function: f 1x2 = 2x The domain of f1x2 = 2x is the set of all real numbers. Begin by locating some points on the graph of f1x2 = 2x, as listed in Table 3. Because 2x 7 0 for all x, the range of f is 0, q . Therefore, the graph has no x-intercepts, and in fact, the graph will lie above the x-axis for all x. As Table 3 indicates, the y-intercept is 1. Table 3 also indicates that as x S - q , the value of f1x2 = 2x gets closer and closer to 0. Thus, the x-axis 1y = 02 is a horizontal asymptote to the graph as x S - q . This provides the end behavior for x large and negative. To determine the end behavior for x large and positive, look again at Table 3. As x S q , f1x2 = 2x grows very quickly, causing the graph of f 1x2 = 2x to rise very rapidly. It is apparent that f is an increasing function and, hence, is one-to-one. 6TJOH BMM UIJT JOGPSNBUJPO QMPU TPNF PG UIF QPJOUT GSPN5BCMF BOE DPOOFDU them with a smooth, continuous curve, as shown in Figure 19. Figure 19 y
(2, 4)
210 = 1024 3
(–2, 1–4 ) (–1, 1–2 ) (–3, 1–8 ) y=0
(1, 2) (0, 1) 3
x
r
SECTION 4.3 Exponential Functions
331
Graphs that look like the one in Figure 19 occur very frequently in a variety of situations. For example, the graph in Figure 20 illustrates the number of Facebook subscribers by year from 2004 to 2012. One might conclude from this graph that the number of Facebook subscribers is growing exponentially.
Subscribers (in millions)
Figure 20
1100 1000 900 800 700 600 500 400 300 200 100 0 2004 2005 2006 2007 2008 2009 2010 2011 2012 Year
Later in this chapter, more will be said about situations that lead to exponential growth. For now, let’s continue to explore properties of exponential functions. The graph of f1x2 = 2x in Figure 19 is typical of all exponential functions of the form f1x2 = ax with a 7 1. Such functions are increasing functions and hence are one-to-one. Their graphs lie above the x-axis, pass through the point 10, 12 , and thereafter rise rapidly as x S q . As x S - q , the x-axis 1y = 02 is a horizontal asymptote. There are no vertical asymptotes. Finally, the graphs are smooth and continuous with no corners or gaps. Figure 21 illustrates the graphs of two more exponential functions whose bases are larger than 1. Notice that the larger the base, the steeper the graph is when x 7 0, and when x 6 0, the larger the base, the closer the graph of the equation is to the x-axis.
Figure 21 y
y=6
x
(1, 6)
6
y = 3x 3 1– 3
(–1, ) y⫽0
–3
(1, 3)
Seeing the Concept Graph Y1 = 2x and compare what you see to Figure 19. Clear the screen, graph Y1 = 3x and Y2 = 6x , and compare what you see to Figure 21. Clear the screen and graph Y1 = 10x and Y2 = 100x.
(0, 1)
(–1, 1–6 )
3
x
The following display summarizes information about f1x2 = ax, a 7 1.
Properties of the Exponential Function f(x) = ax, a + 1
Figure 22 y
(1, a)
(−1, a1 ) y⫽0
(0, 1) x
1. The domain is the set of all real numbers, or - q , q using interval notation; the range is the set of positive real numbers, or 0, q using interval notation. 2. There are no x-intercepts; the y-intercept is 1. 3. The x-axis 1y = 02 is a horizontal asymptote as x S - q 3 lim ax = 0 4 . xS - q 4. f1x2 = ax, a 7 1, is an increasing function and is one-to-one. 1 5. The graph of f contains the points 10, 12, 11, a2, and a - 1, b . a 6. The graph of f is smooth and continuous, with no corners or gaps. See Figure 22.
Now consider f1x2 = ax when 0 6 a 6 1.
332
CHAPTER 4 Exponential and Logarithmic Functions
EX A MPL E 4
Graphing an Exponential Function 1 x Graph the exponential function: f 1x2 = a b 2
Solution Table 4 x
1 x f(x) = a b 2
−10
1 - 10 = 1024 a b 2
−3
1 -3 = 8 a b 2
−2
1 -2 = 4 a b 2
−1
1 -1 a b = 2 2
0
1 0 a b = 1 2
1 x The domain of f1x2 = a b consists of all real numbers. As before, locate some 2 1 x points on the graph as shown in Table 4. Because, a b 7 0 for all x, the range of f is 2 the interval 10, q 2. The graph lies above the x-axis and has no x-intercepts. The 1 x y-intercept is 1. As x S - q , f1x2 = a b grows very quickly. As x S q , the 2 values of f1x2 approach 0. The x-axis 1y = 02 is a horizontal asymptote as x S q . It is apparent that f is a decreasing function and hence is one-to-one. Figure 23 illustrates the graph.
Figure 23
y 6
(–2, 4)
1
1
1 1 a b = 2 2
2
1 1 2 a b = 2 4
3
1 1 3 a b = 2 8
10
1 10 a b ≈ 0.00098 2
3 (–1, 2) (0, 1)
(1, 1–2)
(2, 1–4 ) (3, 1– ) 8
–3
3
x y=0
r
1 x The graph of y = a b also can be obtained from the graph of y = 2x 2 1 x using transformations. The graph of y = a b = 2-x is a reflection about the y-axis 2 of the graph of y = 2x SFQMBDFx by - x 4FF'JHVSFT B BOE C
y
Figure 24
y y = 2x
6
6 1 x
y = ( –2) (2, 4)
(–2, 4)
3
(–2, 1–4 ) (–1, 1–2 ) (–3, 1–8 )
3 (1, 2)
Using a graphing utility, simultaneously graph: 1 x (a) Y1 = 3x, Y2 = a b 3 1 x (b) Y1 = 6x, Y2 = a b 6
1 x Conclude that the graph of Y2 = a b , a for a 7 0, is the reflection about the y-axis of the graph of Y1 = ax.
(0, 1)
(0, 1)
y=0
Seeing the Concept
(–1, 2)
3 (a) y = 2x
x
(1, 1–2)
(2, 1–4 ) (3, 1– ) 8
–3
Replace x by −x ; Reflect about the y-axis.
3
x y=0
1 x
(b) y = 2−x = ( –2)
1 x The graph of f 1x2 = a b in Figure 23 is typical of all exponential functions of 2 the form f x = ax with 0 6 a 6 1. Such functions are decreasing and one-to-one. Their graphs lie above the x-axis and pass through the point 10, 12. The graphs rise rapidly as x S - q . As x S q , the x-axis 1y = 02 is a horizontal asymptote. There are no vertical asymptotes. Finally, the graphs are smooth and continuous, with no corners or gaps.
SECTION 4.3 Exponential Functions
Figure 25 illustrates the graphs of two more exponential functions whose bases are between 0 and 1. Notice that the smaller base results in a graph that is steeper when x 6 0.8IFOx 7 0, the graph of the equation with the smaller base is closer to the x-axis. The following display summarizes information about the function f1x2 = ax, 0 6 a 6 1.
Figure 25 y=(
1– x 6 y
)
6
(–1, 6)
333
y = ( 1– ) x 3
3
(–1, 3)
(0, 1)
(1, 1–6 )
–3
Properties of the Exponential Function f(x) = ax, 0 * a * 1
(1, 1–3) 3
x y⫽0
Figure 26 y
(–1, 1–a) (0, 1)
(1, a) x y⫽0
EXAM PL E 5
1. The domain is the set of all real numbers, or 1 - q , q 2 using interval notation; the range is the set of positive real numbers, or 10, q 2 using interval notation. 2. There are no x-intercepts; the y-intercept is 1. 3. The x-axis 1y = 02 is a horizontal asymptote as x S q 3 lim ax = 0 4 . x Sq 4. f1x2 = ax, 0 6 a 6 1, is a decreasing function and is one-to-one. 1 5. The graph of f contains the points a - 1, b , 10, 12, and 11, a2 . a 6. The graph of f is smooth and continuous, with no corners or gaps. See Figure 26.
Graphing Exponential Functions Using Transformations Graph f1x2 = 2-x - 3 and determine the domain, range, and horizontal asymptote of f.
Solution
Begin with the graph of y = 2x.'JHVSFTIPXTUIFTUBHFT
Figure 27
y
y
y
10
10
10
(3, 8)
(23, 8)
(23, 5) (2, 4)
(21, 1–2)
(22, 4) (21, 2) (0, 1)
(1, 2) (0, 1)
y50
3
x
23
(1, 1–2 ) 1
(22, 1) x y50
(21, 21) (0, 22) 24
(a) y 5 2 x
Replace x by 2 x ; reflect about y-axis.
(b) y 5 22x
Subtract 3; shift down 3 units.
(
x 2 5– 1, 2 ) 2 y 5 23
(c) y 5 22x 2 3
"T'JHVSF D JMMVTUSBUFT UIFEPNBJOPGf1x2 = 2-x - 3 is the interval 1 - q , q 2 and the range is the interval 1 - 3, q 2. The horizontal asymptote of f is the line y = - 3.
r
Now Work
PROBLEM
41
3 Define the Number e .BOZ QSPCMFNT UIBU PDDVS JO OBUVSF SFRVJSF UIF VTF PG BO FYQPOFOUJBM GVODUJPO whose base is a certain irrational number, symbolized by the letter e.
334
CHAPTER 4 Exponential and Logarithmic Functions
One way of arriving at this important number e is given next.
DEFINITION
The number e is defined as the number that the expression a1 +
1 n b n
(2)
approaches as n S q . In calculus, this is expressed, using limit notation, as e = lim a1 + n Sq
1 n b n
5BCMF JMMVTUSBUFT XIBU IBQQFOT UP UIF EFGJOJOH FYQSFTTJPO BT n takes on increasingly large values. The last number in the right column in the table is correct to nine decimal places. Therefore, e = 2.1212c3FNFNCFS UIFUISFFEPUT indicate that the decimal places continue. Because these decimal places continue but do not repeat, e is an irrational number. The number e is often expressed as a decimal rounded to a specific number of places. For example, e ≈ 2.12 is rounded to five decimal places. The exponential function f1x2 = e x, 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 e x or exp 1x2 , which may be used to evaluate the exponential function for a given value of x.*
Table 6
Table 5 1 n
n
y
1 n
a1 +
1 b n
x
ex
-2
e - 2 ≈ 0.14
-1
e - 1 ≈ 0.37
1
1
2
2
2
0.5
1.5
2.25
0
e0 ≈ 1
5
0.2
1.2
2.48832
1
e1 ≈ 2.72
10
0.1
1.1
2.59374246
2
100
0.01
1.01
2.704813829
e2 ≈ 7.39
1,000
0.001
1.001
2.716923932
10,000
0.0001
1.0001
2.718145927
100,000
0.00001
1.00001
2.718268237
0.000001
1.000001
2.718280469
-9
2.718281827
1,000,000
Figure 28
1 +
n
1,000,000,000
10
-9
1 + 10
(2, e 2)
Now use your calculator to approximate e x for x = - 2, x = - 1, x = 0, x = 1, and x = 2. See Table 6. The graph of the exponential function f1x2 = e x is given in 'JHVSF4JODF2 6 e 6 3, the graph of y = e x lies between the graphs of y = 2x and y = 3x.%PZPVTFFXIZ 3FGFSUP'JHVSFTBOE
6
3
1– e
(–1, ) y⫽0
(–2, e1–2)
Seeing the Concept
(1, e)
Graph Y1 = ex and compare what you see to Figure 28. Use eVALUEate or TABLE to verify the points on the graph shown in Figure 28. Now graph Y2 = 2x and Y3 = 3x on the same screen as Y1 = ex. Notice that the graph of Y1 = ex lies between these two graphs.
(0, 1) 0
3
x
EX A MPL E 6
Graphing Exponential Functions Using Transformations Graph f1x2 = - e x - 3 and determine the domain, range, and horizontal asymptote of f.
Solution
Begin with the graph of y = e x. Figure 29 shows the stages. *If your calculator does not have one of these keys, refer to your owner’s manual.
SECTION 4.3 Exponential Functions
Figure 29
y
(0, 21)
(–1, – 1–e )
6
y50
(1, 2e)
23
3
x
3
(–2, – e1–2)
(2, e 2 )
y
y
y50
y50
( 1, – e1–2) 23
(2, – 1–e )
x
(3, 21) (4, 2e)
(1, e) 26
26
(–1, 1–e )
335
(2, 2e 2 )
(0, 1) 3
(–2, e1–2) (a) y 5 e x
(5, 2e 2 )
x (b) y 5 2e x
Multiply by −1; Reflect about the x-axis.
Replace x by x − 3; Shift right 3 units.
(c) y 5 2e x23
"T'JHVSF D JMMVTUSBUFT UIFEPNBJOPG f 1x2 = - e x - 3 is the interval 1 - q , q 2, and the range is the interval 1 - q , 02. The horizontal asymptote is the line y = 0.
Now Work
PROBLEM
r
53
4 Solve Exponential Equations Equations that involve terms of the form ax, a 7 0, a ≠ 1, are referred to as exponential equations. Such equations can sometimes be solved by appropriately BQQMZJOHUIF-BXTPG&YQPOFOUTBOEQSPQFSUZ If au = av, then u = v.
In Words When two exponential expressions with the same base are equal, then their exponents are equal.
EXAM PL E 7
(3)
1SPQFSUZ JTBDPOTFRVFODFPGUIFGBDUUIBUFYQPOFOUJBMGVODUJPOTBSFPOFUPPOF 5PVTFQSPQFSUZ
FBDITJEFPGUIFFRVBMJUZNVTUCFXSJUUFOXJUIUIFTBNFCBTF
Solving Exponential Equations Solve each exponential equation. B 3x + 1 = 1 C 42x - 1 = x + 3
Solution
B #FDBVTF1 = 34, write the equation as 3x + 1 = 1 = 34 Now the expressions on each side of the equation have the same base, 3. Set the exponents equal to each other to obtain x + 1 = 4 x = 3 The solution set is {3}. C 42x - 1 = x + 3 2 3 122 2 2x - 1 = 123 2 x + 3 4 = 2 ; 8 = 2 r s rs 2 2x - 1
3 x + 3
1a 2 = a 2 = 2 2 2x - 1 = 3 x + 3 If a u = a v, then u = v. 4x - 2 = 3x + 9 x = 11 The solution set is {11}.
Now Work
PROBLEMS
63
AND
73
r
336
CHAPTER 4 Exponential and Logarithmic Functions
EX A MPL E 8
Solving an Exponential Equation Solve: e -x = 1e x 2 2
Solution
2
#
1 e3
6TFUIF-BXTPG&YQPOFOUTàSTUUPHFUBTJOHMFFYQSFTTJPOXJUIUIFCBTFe on the right side. 1e x 2
2
#
1 = e 2x # e -3 = e 2x - 3 e3
As a result, 2
e -x - x2 2 x + 2x - 3 1x + 32 1x - 12 x = - 3 or x
= = = = =
e 2x - 3 2x - 3 Apply property (3). Place the quadratic equation in standard form. 0 Factor. 0 Use the Zero-Product Property. 1
The solution set is 5 - 3, 16 .
Now Work
EX A MPL E 9
PROBLEM
r 79
Exponential Probability Between 9:00 pm and 10:00 pmDBSTBSSJWFBU#VSHFS,JOHTESJWFUISVBUUIFSBUFPG DBSTQFSIPVS DBSQFSNJOVUF 5IFGPMMPXJOHGPSNVMBGSPNTUBUJTUJDTDBOCF used to determine the probability that a car will arrive within t minutes of 9:00 pm. F1t2 = 1 - e -0.2t B %FUFSNJOFUIFQSPCBCJMJUZUIBUBDBSXJMMBSSJWFXJUIJONJOVUFTPGpm UIBUJT before 9:05 pm C %FUFSNJOF UIF QSPCBCJMJUZ UIBU B DBS XJMM BSSJWF XJUIJO NJOVUFT PG pm CFGPSFpm D Graph F using your graphing utility. E 8IBU WBMVF EPFT F approach as t increases without bound in the positive EJSFDUJPO
Solution
B 5IFQSPCBCJMJUZUIBUBDBSXJMMBSSJWFXJUIJONJOVUFTJTGPVOECZFWBMVBUJOHF t
at t = 5. F 5 = 1 - e -0.2 5 ≈ 0.63212 c
Use a calculator. Figure 30
'JHVSFTIPXTUIJTDBMDVMBUJPOPOB5*1MVTHSBQIJOHDBMDVMBUPS5IFSFJTB 63% probability that a car will arrive within 5 minutes. C 5IFQSPCBCJMJUZUIBUBDBSXJMMBSSJWFXJUIJONJOVUFTJTGPVOECZFWBMVBUJOH F t at t = 30. F 30 = 1 - e -0.2 30 ≈ 0.995
c
Use a calculator. Figure 31
5IFSFJTBQSPCBCJMJUZUIBUBDBSXJMMBSSJWFXJUIJONJOVUFT D See Figure 31 for the graph of F. E "TUJNFQBTTFT UIFQSPCBCJMJUZUIBUBDBSXJMMBSSJWFJODSFBTFT5IFWBMVFUIBUF 1 approaches can be found by letting t S q . Since e -0.2t = 0.2t , it follows that e e -0.2t S 0 as t S q . Therefore, F approaches 1 as t gets large. The algebraic analysis is confirmed by Figure 31.
1
0
30 0
Now Work
r
PROBLEM
109
SECTION 4.3 Exponential Functions
337
SUMMARY Properties of the Exponential Function %PNBJOUIFJOUFSWBM - q , q ; range: the interval 0, q
x-intercepts: none; y-intercept: 1 Horizontal asymptote: x-axis y = 0 as x S - q Increasing; one-to-one; smooth; continuous See Figure 22 for a typical graph. %PNBJOUIFJOUFSWBM - q , q ; range: the interval 0, q
x-intercepts: none; y-intercept: 1 Horizontal asymptote: x-axis y = 0 as x S q %FDSFBTJOH POFUPPOF TNPPUI DPOUJOVPVT See Figure 26 for a typical graph.
f1x2 = ax, a 7 1
f1x2 = ax, 0 6 a 6 1
If au = av, then u = v.
4.3 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 1. 43 = 64 ; 2>3 = 4 ; 3-2 = . QQ "m" BOE 4. Find the average rate of change of f 1x2 = 3x - 5 from 9 QQ"m"
x = 0 to x = 4. QQm m 3 2x 2 2. Solve: x + 3x = 4 QQ"m" { - 4, 1} has y = 2 as a 5. True or False The function f 1x2 = x - 3 3. True or False To graph y = 1x - 22 3, shift the graph of horizontal asymptote. QQm True y = x3 to the left 2 units. QQm False
Concepts and Vocabulary 6. " O exponential function is a function of the form f 1x2 = Cax, where a 7 0, a ≠ 1, and C ≠ 0 are real numbers. The base a is the growth factor and C is the initial value. f 1x + 12 = a . 7. For an exponential function f 1x2 = Cax, f 1x2 8. True or False The domain of the exponential function f 1x2 = ax, where a 7 0 and a ≠ 1, is the set of all real numbers. True 9. True or False The range of the exponential function f x = ax, where a 7 0 and a ≠ 1, is the set of all real numbers. False
10. True or False The graph of the exponential function f 1x2 = ax, where a 7 0 and a ≠ 1,has no x-intercept. True
*11. The graph of every exponential function f 1x2 = ax, where a 7 0 and a ≠ 1, passes through three points: , . , and 12. If the graph of the exponential function f 1x2 = ax, where a 7 0 and a ≠ 1, is decreasing, then a must be less than 1 . 13. If 3x = 34, then x =
4 . 1 x 14. True or False The graphs of y = 3x and y = a b are 3 identical. False
Skill Building In Problems 15–24, approximate each number using a calculator. Express your answer rounded to three decimal places. *15. B 32.2 *17. B 23.14
C 32.23 C 23.141
D 32.236 D 23.1415
E 315 E 2p
*16. B 51. *18. B 22.
C 51.3 C 22.1
D 51.32 D 22.1
E 513 E 2e
*19. B 3.12.
C 3.142.1
D 3.1412.1
E pe
*20. B 2.3.1
C 2.13.14
D 2.13.141
E e p
21. e
1.2
3.320
22. e
-1.3
23. e
-0.5
%VFUPTQBDFSFTUSJDUJPOT BOTXFSTUPUIFTFFYFSDJTFTNBZCFGPVOEJOUIF"OTXFSTJOUIFCBDLPGUIFCPPL
24. e
2.1
338
CHAPTER 4 Exponential and Logarithmic Functions
In Problems 25–32, determine whether the given function is linear, exponential, or neither. For those that are linear functions, find a linear function that models the data; for those that are exponential, find an exponential function that models the data. 25.
26.
x
f1 x 2
-1
3
-1
2
0
6
0
5
1
12
1
8
2
18
2
11
3
30
27.
g1x2
3
x
H 1x 2
-1
30.
x
g1x2
2
-1
6
0
4
0
1
1
6
1
0
2
8
2
3
3
10
3
10
Linear; H1x2 = 2x + 4
x
f1x 2
1 4 1
-1
3 2 3
1
4
1
6
2
16
2
12
3
64
3
24
-1 0
14
28.
H 1 x2
x
Linear; g1x2 = 3x + 5
Neither 29.
x
0
Exponential; f 1x2 = 3 # 2x
Exponential; H1x2 = 4x 31.
x
F 1x 2
-1
2 3 1 3 2 9 4 27 8
0 1 2
Neither 3
32.
x
F 1x 2
-1
1 2 1 4 1 8 1 16 1 32
0 1 2 3
3 x Exponential; F x = a b 2
Exponential; F x =
1# 1 x a b 4 2
In Problems 33–40, the graph of an exponential function is given. Match each graph to one of the following functions. B y = 3x C y = 3-x D y = - 3x E y = - 3-x G y = 3x - 1
F y = 3x - 1 33.
B
y 3
H y = 31 - x
34.
F
y 3
35.
I y = 1 - 3x %
y 1
y⫽0 2x
⫺2 y⫽0 ⫺2
y⫽0 ⫺2
2x ⫺1
37.
A
y 3
H
y 1
⫺2
2x
38.
⫺3
C
y
⫺3
39.
E 40.
y 3
1
y
G
3
2x
2x ⫺1
y⫽1
2x ⫺1
y⫽0 ⫺2 y⫽0 ⫺2
36.
⫺2 y ⫽ ⫺1
⫺3
2x ⫺1
⫺2
2x
y⫽0
⫺1
In Problems 41–52, use transformations to graph each function. Determine the domain, range, and horizontal asymptote of each function. *42. f 1x2 = 3x - 2
*41. f 1x2 = 2x + 1 *45. f 1x2 = 3
#
1 a b 2
x
*49. f 1x2 = 2 + 4x-1
*43. f 1x2 = 3x - 1
*44. f 1x2 = 2x + 2
x
*47. f 1x2 = 3-x - 2
*48. f 1x2 = - 3x + 1
*50. f 1x2 = 1 - 2x + 3
*51. f 1x2 = 2 + 3x>2
*52. f 1x2 = 1 - 2-x>3
*46. f 1x2 = 4
#
1 a b 3
In Problems 53–60, begin with the graph of y = e x [Figure 28] and use transformations to graph each function. Determine the domain, range, and horizontal asymptote of each function. *53. f 1x2 = e -x
*57. f 1x2 = 5 - e -x
*54. f 1x2 = - e x
*58. f 1x2 = 9 - 3e -x
*55. f 1x2 = e x + 2
*59. f 1x2 = 2 - e -x>2
*56. f 1x2 = e x - 1
*60. f 1x2 = - 3e 2x
SECTION 4.3 Exponential Functions
339
In Problems 61–80, solve each equation. 61. x = 3
5 36
62. 5x = 5-6
1 x 1 5 26 65. a b = 5 25 3 69. 3x = 9x 5 - 22, 0, 226 2
73. 3x
-
= 22x
1 1 x 66. a b = 536 4 64 2 1 70. 4x = 2x e 0, f 2 2 74. 5x + = 1252x 52, 46
5 - 1, 6 5 - 46
77. e x = e 3x +
5 - 66
78. e 3x = e 2 - x 1 49 1 4
81. If 4x = , what does 4 - 2x equal 83. If 3 - x = 2, what does 32x equal
1 e f 2
5 - 46
63. 2-x = 16
64. 3-x = 1
1 68. 5x + 3 = 5 - 46 5 72. 9-x + 15 = 2x 566
75. 4x # 2x = 162
76. 92x # 2x = 3-1
2
5 - 4, 26
2
1 2 51, 26 80. e 4 x # e x = e 12 e2 1 82. If 2x = 3, what does 4 - x equal 9 1 -x 3x 84. If 5 = 3, what does 5 equal 2 2
79. e x = e 3x
#
In Problems 85–88, determine the exponential function whose graph is given. 85.
f 1x2 = 3x
y
86.
20
16
16
12
y⫽0
87.
(–1, 1–3 ) –3
y (–1, – 1–6 ) –1 –10
–2
5 - 6, 26
12 (2, 9)
8
(0, 1)
4
–1 –2
(–1, 1–5 )
(1, 3) 1
2
3 x
–3
f 1x2 = - 6x
(0, –1) 1
1 e - 1, - f 3
f 1x2 = 5x
y
20
8
5 - 46
3 67. 22x - 1 = 4 e f 2 71. -x + 14 = 16x 566
2 3 x (1, –6)
y⫽0
88.
–2
4
–1 –2
y 2
(1, 5) (0, 1) 1
2
3 x
y⫽0
f 1x2 = - e x (0, –1) 3 x
y⫽0
(1, –e) (–1, – 1–e ) –4
–20 –8
–30 (2, –36) –40
*89. Find an exponential function with horizontal asymptote y = 2 whose graph contains the points 10, 32 and 11, 52 .
(2, –e 2)
–12
*90. Find an exponential function with horizontal asymptote y = - 3 whose graph contains the points 10, - 22 and 1 - 2, 12 .
Mixed Practice *91. Suppose that f 1x2 = 2x. B 8IBUJTf 142 8IBUQPJOUJTPOUIFHSBQIPGf 1 C *G f 1x2 = , what is x 8IBU QPJOU JT PO UIF HSBQI 16 of f
*92. Suppose that f 1x2 = 3x. B 8IBUJTf 142 8IBUQPJOUJTPOUIFHSBQIPGf 1 C *G f 1x2 = , what is x 8IBU QPJOU JT PO UIF HSBQI 9 of f
93. Suppose that g1x2 = 4x + 2. * B 8IBUJTg1 - 12 8IBUQPJOUJTPOUIFHSBQIPGg C *G g1x2 = 66, what is x 8IBU QPJOU JT PO UIF HSBQI of g 3; 13, 662
94. Suppose that g1x2 = 5x - 3. * B 8IBUJTg1 - 12 8IBUQPJOUJTPOUIFHSBQIPGg C *Gg1x2 = 122, what is x 8IBUQPJOUJTPOUIFHSBQI of g 3; 13, 1222
1 x 95. Suppose that H x = a b - 4. 2 * B 8IBUJTH1 - 62 8IBUQPJOUJTPOUIFHSBQIPGH C *G H1x2 = 12, what is x 8IBU QPJOU JT PO UIF graph of H - 4; 1 - 4, 122 D 'JOEUIF[FSPPGH. - 2
1 x 96. Suppose that F 1x2 = a b - 3. 3 * B 8IBUJTF 1 - 52 8IBUQPJOUJTPOUIFHSBQIPGF? C *G F 1x2 = 24, what is x 8IBU QPJOU JT PO UIF graph of F - 3; 1 - 3, 242 D 'JOEUIF[FSPPGF. - 1
340
CHAPTER 4 Exponential and Logarithmic Functions
In Problems 97–100, graph each function. Based on the graph, state the domain and the range, and find any intercepts. *97. f 1x2 = e
e -x ex
*99. f 1x2 = e
- ex - e -x
if x 6 0 if x Ú 0 if x 6 0 if x Ú 0
*98. f 1x2 = e *100. f 1x2 = e
ex e -x - e -x - ex
if x 6 0 if x Ú 0 if x 6 0 if x Ú 0
Applications and Extensions 101. 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 p1n2 = 10010.92 n B 8IBUQFSDFOUPGMJHIUXJMMQBTTUISPVHIQBOFT C 8IBUQFSDFOUPGMJHIUXJMMQBTTUISPVHIQBOFT * D &YQMBJOUIFNFBOJOHPGUIFCBTFJOUIJTQSPCMFN 102. Atmospheric Pressure The atmospheric pressure p on a balloon or plane decreases with increasing height. This pressure, measured in millimeters of mercury, is related to the height h JOLJMPNFUFST BCPWFTFBMFWFMCZUIFGVODUJPO p1h2 = 60e -0.145h B 'JOEUIFBUNPTQIFSJDQSFTTVSFBUBIFJHIUPGLJMPNFUFST PWFSBNJMF 56.6 mmHg C 8IBU JT JU BU B IFJHIU PG LJMPNFUFST PWFS GFFU 1.2 mmHg *103. Depreciation The price p, in dollars, of a Honda Civic EX-L Sedan that is x years old is modeled by p1x2 = 22,265 0.90 x B )PXNVDITIPVMEBZFBSPME$JWJD&9-4FEBODPTU C )PXNVDITIPVMEBZFBSPME$JWJD&9-4FEBODPTU D Explain the meaning of the base 0.90 in this problem. 104. Healing of Wounds The normal healing of wounds can be modeled by an exponential function. If A0 represents the original area of the wound and if A equals the area of the wound, then the function
B 8IBUJTUIFTJ[FPGUIFJOJUJBMQPQVMBUJPOPGUIFTQFDJFT 30 C "DDPSEJOHUPUIFNPEFM XIBUXJMMCFUIFQPQVMBUJPOPG UIFTQFDJFTJOZFBST 60 D "DDPSEJOHUPUIFNPEFM XIBUXJMMCFUIFQPQVMBUJPOPG UIFTQFDJFTJOZFBST 120 E "DDPSEJOHUPUIFNPEFM XIBUXJMMCFUIFQPQVMBUJPOPG UIFTQFDJFTJOZFBST 241 F 8IBU JT IBQQFOJOH UP UIF QPQVMBUJPO FWFSZ ZFBST %PVCMJOH 107. Drug Medication The function D1h2 = 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 has been administered. How many milligrams will be QSFTFOUBGUFSIPVS "GUFSIPVST 3.35 mg; 0.45 mg 108. Spreading of Rumors A model for the number N of people in a college community who have heard a certain rumor is N = P 11 - e -0.15d 2
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 IBWFIFBSEUIFSVNPSBGUFSEBZT 362 students 109. Exponential Probability Between 12:00 pm and 1:00 pm, cars arrive at Citibank’s drive-thru at the rate of 6 cars QFSIPVS DBSQFSNJOVUF 5IFGPMMPXJOHGPSNVMBGSPN probability can be used to determine the probability that a car will arrive within t minutes of 12:00 pm:
A1n2 = A0 e -0.35n
F1t2 = 1 - e -0.1t
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. B *GIFBMJOHJTUBLJOHQMBDF IPXMBSHFXJMMUIFBSFBPGUIF XPVOECFBGUFSEBZT 34.99 mm2 C )PXMBSHFXJMMJUCFBGUFSEBZT 3.02 mm2
B %FUFSNJOFUIFQSPCBCJMJUZUIBUBDBSXJMMBSSJWFXJUIJO 10 minutes of 12:00 pm UIBUJT CFGPSFpm 0.632 C %FUFSNJOFUIFQSPCBCJMJUZUIBUBDBSXJMMBSSJWFXJUIJO 40 minutes of 12:00 pm CFGPSFpm D 8IBUWBMVFEPFTF approach as t becomes unbounded in UIFQPTJUJWFEJSFDUJPO 1 * E Graph F using a graphing utility. F 6TJOH*/5&34&$5 EFUFSNJOFIPXNBOZNJOVUFTBSF needed for the probability to reach 50%. NJO
105. Advanced-Stage Pancreatic Cancer The percentage of patients P who have survived t years after initial diagnosis of advanced-stage pancreatic cancer is modeled by the function P t = 100 0.3 t Source: Cancer Treatment Centers of America B "DDPSEJOHUPUIFNPEFM XIBUQFSDFOUPGQBUJFOUTTVSWJWF ZFBSBGUFSJOJUJBMEJBHOPTJT 30% C 8IBU QFSDFOU PG QBUJFOUT TVSWJWF ZFBST BGUFS JOJUJBM EJBHOPTJT 9% * D Explain the meaning of the base, 0.3, in the context of this problem. 106. Endangered Species In a protected environment, the population P of an endangered species recovers over time t according to the model P t = 30 1.149 t
110. Exponential Probability Between 5:00 pm and 6:00 pm, cars BSSJWFBU+JGGZ-VCFBUUIFSBUFPGDBSTQFSIPVS DBS QFS NJOVUF 5IF GPMMPXJOH GPSNVMB GSPN QSPCBCJMJUZ DBO be used to determine the probability that a car will arrive within t minutes of 5:00 pm: F 1t2 = 1 - e -0.15t
B %FUFSNJOFUIFQSPCBCJMJUZUIBUBDBSXJMMBSSJWFXJUIJO 15 minutes of 5:00 pm UIBUJT CFGPSFpm C %FUFSNJOFUIFQSPCBCJMJUZUIBUBDBSXJMMBSSJWFXJUIJO 30 minutes of 5:00 pm CFGPSFpm D 8IBUWBMVFEPFTF approach as t becomes unbounded in UIFQPTJUJWFEJSFDUJPO 1
SECTION 4.3 Exponential Functions
* E Graph F using a graphing utility. F 6TJOH*/5&34&$5 EFUFSNJOFIPXNBOZNJOVUFTBSF needed for the probability to reach 60%. ≈6 min 111. Poisson Probability Between 5:00 pm and 6:00 pm, cars BSSJWF BU .D%POBMET ESJWFUISV BU UIF SBUF PG DBST QFS hour. The following formula from probability can be used to determine the probability that x cars will arrive between 5:00 pm and 6:00 pm.
where x! = x
#
1x - 12
#
#g#3#2#1
B %FUFSNJOF UIF QSPCBCJMJUZ UIBU x = 15 cars will arrive between 5:00 pm and 6:00 pm. 0.0516 C %FUFSNJOF UIF QSPCBCJMJUZ UIBU x = 20 cars will arrive between 5:00 pm and 6:00 pm. 112. Poisson Probability 1FPQMF FOUFS B MJOF GPS UIF Demon Roller Coaster at the rate of 4 per minute. The following formula from probability can be used to determine the probability that x people will arrive within the next minute.
where x! = x
#
1x - 12
4xe -4 x!
# x - 2 # g # 3 # 2 # 1
B %FUFSNJOFUIFQSPCBCJMJUZUIBUx = 5 people will arrive within the next minute. 0.1563 C %FUFSNJOFUIFQSPCBCJMJUZUIBUx = people will arrive within the next minute. 113. Relative Humidity The relative humidity is the ratio FYQSFTTFE BT B QFSDFOU PG UIF BNPVOU PG XBUFS WBQPS JO the air to the maximum amount that it can hold at a specific temperature. The relative humidity, R, is found using the following formula: 4221
I =
E 31 - e - R>L t 4 R I
20xe -20 x!
1x - 22
P 1x2 =
115. Current in a RL Circuit The equation governing the amount of current I JOBNQFSFT BGUFSUJNFt JOTFDPOET JO a single RL circuit consisting of a resistance R JOPINT
BO inductance L JOIFOSZT
BOEBOFMFDUSPNPUJWFGPSDFE JO WPMUT JT
R ⫹
P 1x2 =
E
* B *GE = 120 volts, R = 10 ohms, and L = 5 henrys, how much current I1 JT GMPXJOH BGUFS TFDPOE "GUFS TFDPOE "GUFSTFDPOE C 8IBUJTUIFNBYJNVNDVSSFOU 12 amp * D (SBQI UIJT GVODUJPO I = I1 t , measuring I along the y-axis and t along the x-axis. * E *GE = 120 volts, R = 5 ohms, and L = 10 henrys, how much current I2 JT GMPXJOH BGUFS TFDPOE "GUFS TFDPOE "GUFSTFDPOE F 8IBUJTUIFNBYJNVNDVSSFOU 24 amp * G (SBQI UIF GVODUJPO I = I2 1t2 on the same coordinate axes as I1 1t2. 116. Current in a RC Circuit The equation governing the amount of current I JOBNQFSFT BGUFSUJNFt JONJDSPTFDPOET JOB single RC circuit consisting of a resistance R JO PINT
B capacitance C JONJDSPGBSBET
BOEBOFMFDUSPNPUJWFGPSDF E JOWPMUT JT I =
L1t2 = 500 1 - e -0.0061t
approximates the number of words L that the student will have learned after t minutes. B )PXNBOZXPSETXJMMUIFTUVEFOUIBWFMFBSOFEBGUFS NJOVUFT XPSET C )PXNBOZXPSETXJMMUIFTUVEFOUIBWFMFBSOFEBGUFS NJOVUFT 153 words
E -t> RC
e R I
4221
114. Learning Curve Suppose that a student has 500 vocabulary words to learn. If the student learns 15 words after 5 minutes, the function
L
⫺
R = 10 T + 459.4 - D + 459.4 + 2
R ⫹
where TJTUIFBJSUFNQFSBUVSF JO' BOED is the dew point UFNQFSBUVSF JO' B %FUFSNJOFUIFSFMBUJWFIVNJEJUZJGUIFBJSUFNQFSBUVSF is 50° Fahrenheit and the dew point temperature is 41° Fahrenheit. C %FUFSNJOFUIFSFMBUJWFIVNJEJUZJGUIFBJSUFNQFSBUVSF JT'BISFOIFJUBOEUIFEFXQPJOUUFNQFSBUVSFJT Fahrenheit. D 8IBUJTUIFSFMBUJWFIVNJEJUZJGUIFBJSUFNQFSBUVSFBOE UIFEFXQPJOUUFNQFSBUVSFBSFUIFTBNF 100%
341
E
⫺
C
* B *G E = 120 volts, R = 2000 ohms, and C = 1.0 microfarad, how much current I1 is flowing initially t = 0 "GUFS NJDSPTFDPOET "GUFS NJDSPTFDPOET C 8IBUJTUIFNBYJNVNDVSSFOU 0.06 amp * D (SBQI UIF GVODUJPO I = I1 1t2, measuring I along the y-axis and t along the x-axis. * E *G E = 120 volts, R = 1000 ohms, and C = 2.0 microfarads, how much current I2 JT GMPXJOH JOJUJBMMZ "GUFSNJDSPTFDPOET "GUFSNJDSPTFDPOET F 8IBUJTUIFNBYJNVNDVSSFOU 0.12 amp * G (SBQI UIF GVODUJPO I = I2 1t2 on the same coordinate axes as I1 1t2.
117. If f is an exponential function of the form f 1x2 = C # ax with growth factor 3 and f 162 = 12, what is f 12 36
342
CHAPTER 4 Exponential and Logarithmic Functions
*118. Another Formula for e 6TF B DBMDVMBUPS UP DPNQVUF UIF values of 2 +
Problems 124 and 125 provide definitions for two other transcendental functions. *124. The hyperbolic sine function, designated by sinh x, is defined as
1 1 1 + + g + 2! 3! n!
for n = 4, 6, , and 10. Compare each result with e. [Hint: 1! = 1, 2! = 2 # 1, 3! = 3 # 2 # 1, n! = n1n - 12
#g#
sinh x =
B 4IPXUIBUf 1x2 = sinh x is an odd function. C Graph f 1x2 = sinh x using a graphing utility.
132 122 112.]
*119. Another Formula for e 6TF B DBMDVMBUPS UP DPNQVUF UIF various values of the expression. Compare the values to e.
*125. The hyperbolic cosine function, designated by cosh x, is defined as
2 + 1 1 + 1 2 + 2 3 + 3 4 + 4 etc.
cosh x =
f 1x + h2 - f 1x2 h
= ax
#a
- 1 h
1cosh x2 2 - 1sinh x2 2 = 1 *126. Historical Problem 1JFSSF conjectured that the function
h ≠ 0
*121. If f 1x2 = ax, show that f 1A + B2 = f 1A2 *122. If f 1x2 = ax, show that f 1 - x2 =
1 x 1e + e -x 2 2
B 4IPXUIBUf 1x2 = cosh x is an even function. C Graph f 1x2 = cosh x using a graphing utility. D 3FGFSUP1SPCMFN4IPXUIBU GPSFWFSZx,
*120. Difference Quotient If f 1x2 = ax, show that h
1 x 1e - e -x 2 2
EF
'FSNBU
m
f1x2 = 212 2 + 1 x
# f 1B2.
for x = 1, 2, 3, c , would always have a value equal to a prime number. But Leonhard Euler 1109132 showed that this formula fails for x = 5.6TFBDBMDVMBUPSUPEFUFSNJOF the prime numbers produced by f for x = 1, 2, 3, 4. Then show that f 152 = 641 * 6,00,41, which is not prime.
1 . f 1x2
*123. If f 1x2 = ax, show that f 1ax2 = 3f 1x2 4 a.
Discussion and Writing 127. The bacteria in a 4-liter container double every minute. After 60 minutes the container is full. How long did it take UPGJMMIBMGUIFDPOUBJOFS 59 minutes 128. Explain in your own words what the number eJT1SPWJEFBU least two applications that use this number. 129. %PZPVUIJOLUIBUUIFSFJTBQPXFSGVODUJPOUIBUJODSFBTFT more rapidly than an exponential function whose base is HSFBUFSUIBO &YQMBJO
130. As the base a of an exponential function f 1x2 = ax, where a 7 1, increases, what happens to the behavior of its graph for x 7 0 8IBU IBQQFOT UP UIF CFIBWJPS PG JUT HSBQI GPS x 6 0 1 x 131. The graphs of y = a-x and y = a b BSFJEFOUJDBM8IZ a
Retain Your Knowledge Problems 132–135 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 132. Solve the inequality: x3 + 5x2 … 4x + 20 *135. - q , - 5] h [ - 2, 2] x + 1 133. Solve the inequality: Ú 1 2, q
x - 2 134. %FUFSNJOF XIFUIFS UIF GVODUJPO JT MJOFBS PS OPOMJOFBS *G linear, determine the equation that defines y = f x . 1 Linear; y = - x - 5 x y = f(x) 3 -6
-3
-3
-4
0
-5
3
-6
6
-7
Consider the quadratic function f x = x2 + 2x - 3. B (SBQI f by determining whether its graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. C %FUFSNJOFUIFEPNBJOBOEUIFSBOHFPGf. D %FUFSNJOF XIFSF f is increasing and where it is decreasing.
‘Are You Prepared?’ Answers 1. 64; 4;
1 9
2. { - 4, 1}
3. False
4. 3
5. True
SECTION 4.4 Logarithmic Functions
343
4.4 Logarithmic Functions PREPARING FOR THIS SECTION Before getting started, review the following: r 4PMWF-JOFBS*OFRVBMJUJFT "QQFOEJY" 4FDUJPO" QQ"m"
r 4PMWF2VBESBUJD*OFRVBMJUJFT 4FDUJPO QQm
r 1PMZOPNJBMBOE3BUJPOBM*OFRVBMJUJFT 4FDUJPO QQm
r 4PMWF-JOFBS&RVBUJPOT "QQFOEJY" 4FDUJPO" QQ"m"
Now Work the ‘Are You Prepared?’ problems on page 352.
OBJECTIVES 1 Change Exponential Statements to Logarithmic Statements and Logarithmic Statements to Exponential Statements (p. 344) 2 Evaluate Logarithmic Expressions (p. 344) 3 Determine the Domain of a Logarithmic Function (p. 345) 4 Graph Logarithmic Functions (p. 345) 5 Solve Logarithmic Equations (p. 350)
3FDBMMUIBUBPOFUPPOFGVODUJPOy = f1x2 has an inverse function that is defined JNQMJDJUMZ CZ UIF FRVBUJPO x = f 1y2. In particular, the exponential function y = f1x2 = a x, where a 7 0 and a ≠ 1, is one-to-one and, hence, has an inverse function that is defined implicitly by the equation x = ay, a 7 0, a ≠ 1 This inverse function is so important that it is given a name, the logarithmic function.
DEFINITION
The logarithmic function to the base a, where a 7 0 and a ≠ 1, is denoted by y = log a x SFBEBTiy is the logarithm to the base a of xu BOEJTEFGJOFECZ y = log a x if and only if x = ay
In Words When you read logax, think to yourself “a raised to what power gives me x? ”
The domain of the logarithmic function y = log a x is x 7 0.
As this definition illustrates, a logarithm is a name for a certain exponent. So log ax represents the exponent to which a must be raised to obtain x.
EXAM PL E 1
Relating Logarithms to Exponents B *Gy = log 3 x, then x = 3y. For example, the logarithmic statement 4 = log 3 1 is equivalent to the exponential statement 1 = 34. 1 C *G y = log 5 x, then x = 5y. For example, - 1 = log 5 a b is equivalent to 5 1 = 5-1. 5
r
344
CHAPTER 4 Exponential and Logarithmic Functions
1 Change Exponential Statements to Logarithmic Statements and Logarithmic Statements to Exponential Statements The definition of a logarithm can be used to convert from exponential form to logarithmic form, and vice versa, as the following two examples illustrate.
EX A MPL E 2
Changing Exponential Statements to Logarithmic Statements Change each exponential statement to an equivalent statement involving a logarithm. B 1.23 = m C e b = 9 D a4 = 24
Solution
6TFUIFGBDUUIBUy = log a x and x = ay, where a 7 0 and a ≠ 1, are equivalent. B *G1.23 = m, then 3 = log 1.2 m. C *Ge b = 9, then b = log e 9. D *Ga4 = 24, then 4 = log a 24.
Now Work
EX A MPL E 3
PROBLEM
r
9
Changing Logarithmic Statements to Exponential Statements Change each logarithmic statement to an equivalent statement involving an exponent. B log a 4 = 5 C log e b = - 3 D log 3 5 = c
Solution
B *Glog a 4 = 5, then a5 = 4. C *Glog e b = - 3, then e -3 = b. D *Glog 3 5 = c, then 3c = 5.
Now Work
PROBLEM
r
17
2 Evaluate Logarithmic Expressions To find the exact value of a logarithm, write the logarithm in exponential notation using the fact that y = log a x is equivalent to ay = x, and use the fact that if au = ay, then u = y.
EX A MPL E 4
Finding the Exact Value of a Logarithmic Expression Find the exact value of: B log 2 16
Solution
C log 3
1 2
B 5P FWBMVBUF log 216, think “2 SBJTFE UP XIBU QPXFS ZJFMET u y = log 2 16 2y = 16 2y = 24 y = 4
Change to exponential form. 16 = 24 Equate exponents.
1 , think “3 2 1 raised to what power yields u 2 1 y = log 3 2
C 5P FWBMVBUF log 3
3y =
1 2
3y = 3-3
Therefore, log 2 16 = 4.
y = -3 Therefore, log 3
Now Work
PROBLEM
25
Change to exponential form. 1 1 = 3 = 3-3 27 3 Equate exponents.
1 = - 3. 2
r
SECTION 4.4 Logarithmic Functions
345
3 Determine the Domain of a Logarithmic Function The logarithmic function y = log a x has been defined as the inverse of the exponential function y = a x. That is, if f1x2 = a x, then f -1 1x2 = log a x. Based on the discussion given in Section 4.2 on inverse functions, for a function f and its inverse f -1, %omain of f -1 = 3ange of f and 3ange of f -1 = %omain of f Consequently, it follows that %omain of the logarithmic function = 3ange of the exponential function = 10, q 2 3ange of the logarithmic function = %omain of the exponential function = 1 - q , q 2
The next box summarizes some properties of the logarithmic function:
y = log a x %omain:
1defining equation: x = ay 2 10, q 2 3ange: 1 - q , q 2
The domain of a logarithmic function consists of the positive real numbers, so the argument of a logarithmic function must be greater than zero.
EXAMPL E 5
Finding the Domain of a Logarithmic Function Find the domain of each logarithmic function.
B F1x2 = log 2 1x + 32 1 + x C g1x2 = log 5 a b 1 - x D h 1x2 = log 1>2 0 x 0
Solution
B 5IFEPNBJOPGF consists of all x for which x + 3 7 0, that is, x 7 - 3.6TJOH interval notation, the domain of F is 1 - 3, q 2 . C 5IFEPNBJOPGg is restricted to 1 + x 7 0 1 - x Solve this inequality to find that the domain of g consists of all x between - 1 and 1, that is, - 1 6 x 6 1 or, using interval notation, 1 - 1, 12. D 4JODF 0 x 0 7 0, provided that x ≠ 0, the domain of h consists of all real numbers except zero or, using interval notation, 1 - q , 02 ∪ 10, q 2.
r
Now Work
PROBLEMS
39
AND
45
4 Graph Logarithmic Functions Because exponential functions and logarithmic functions are inverses of each other, the graph of the logarithmic function y = log a x is the reflection about the line y = x of the graph of the exponential function y = a x, as shown in Figure 32 on the following page.
346
CHAPTER 4 Exponential and Logarithmic Functions
Figure 32
y
y ⫽ ax ( ⫺1,
1– a
y
y⫽x
3
y ⫽ ax y ⫽ x
3 (1, a)
)
(a, 1) (⫺1, 1–a
(1, a)
(0, 1) ⫺3
(1, 0)
3 x
1– a,
⫺1) y ⫽ loga x
(
(0, 1) )
⫺3
y ⫽ loga x (a, 1)
(1, 0)
3 x
( 1–a ,⫺1)
⫺3
⫺3
(a) 0 ⬍ a ⬍ 1
(b) a ⬎ 1
For example, to graph y = log 2 x, graph y = 2x and reflect it about the line 1 x y = x. See Figure 33. To graph y = log 1>3 x, graph y = a b and reflect it about 3 the line y = x. See Figure 34.
Figure 33
Figure 34 y 2
(⫺1, 12 )
y⫽ (1, 2)
2x
y⫽ 1
y⫽x
y
(⫺1, 3)
3
y ⫽ log2x
(0, 1)
(2, 1)
⫺2
x
(3)
(0, 1) ⫺3
(1, 0) 2 x
( 13 , 1) (1, 13 ) 3 x (1, 0) (3, ⫺1) y ⫽ log1/3x
( 12 , ⫺1) ⫺3
⫺2
Now Work
y⫽x
PROBLEM
59
The graphs of y = log a xJO'JHVSFT B BOE C MFBEUPUIFGPMMPXJOHQSPQFSUJFT
Properties of the Logarithmic Function f1 x 2 = loga x
1. The domain is the set of positive real numbers, or 10, q 2 using interval notation; the range is the set of all real numbers, or 1 - q , q 2 using interval notation. 2. The x-intercept of the graph is 1. There is no y-intercept. 3. The y-axis x = 0 is a vertical asymptote of the graph. 4. A logarithmic function is decreasing if 0 6 a 6 1 and is increasing if a 7 1. 1 5. The graph of f contains the points 11, 02, 1a, 12, and a , - 1b . a 6. The graph is smooth and continuous, with no corners or gaps.
In Words
y = logex is written y = ln x.
If the base of a logarithmic function is the number e, the result is the natural logarithm function. This function occurs so frequently in applications that it is given a special symbol, ln GSPNUIF-BUJO logarithmus naturalis 5IBUJT
SECTION 4.4 Logarithmic Functions
y = ln x if and only if x = e y
347
(1)
Because y = ln x and the exponential function y = e x are inverse functions, the graph of y = ln x can be obtained by reflecting the graph of y = e x about the line y = x. See Figure 35. Other points on the graph of f 1x2 = ln x can be found by using a calculator with an ln LFZ4FF5BCMF Figure 35
y
Table 7
5
x
ln x
1 2
- 0.69
2
0.69
3
1.10
ye x
yx
(1, e) ( 0, 1) (1, 1–e )
Seeing the Concept Graph Y1 = ex and Y2 = ln x on the same square screen. Use eVALUEate to verify the points on the graph given in Figure 35. Do you see the symmetry of the two graphs with respect to the line y = x?
y 0 3
( 1, 0) 1
( 1–e ,1)
Graphing a Logarithmic Function and Its Inverse
B 'JOEUIFEPNBJOPGUIFMPHBSJUINJDGVODUJPOf 1x2 = - ln 1x - 22. C (SBQIf. D 'SPNUIFHSBQI EFUFSNJOFUIFSBOHFBOEWFSUJDBMBTZNQUPUFPGf. E 'JOEf -1, the inverse of f. F 6TF f -1 to confirm the range of fGPVOEJOQBSU D 'SPNUIFEPNBJOPGf, find the range of f - 1. G (SBQIf -1. B 5IFEPNBJOPGf consists of all x for which x - 2 7 0, or, equivalently, x 7 2. The domain of f is {x 0 x 7 2} or 2, q in interval notation. C 5PPCUBJOUIFHSBQIPGy = - ln 1x - 22, begin with the graph of y = ln x and use transformations. See Figure 36.
Solution
y
3
3 x0
x0
1
1
(e, 1) (1, 0)
1
(
1– , e
D y In x
x2
y
y
3
1
3 x
x0
E XAM PL E 6
Figure 36
yIn x
(e, 1)
3 x
1
1)
( 1–e , 1)
1
Multiply by 1; reflect about the x-axis.
E y In x
(3, 0) 3 x
(1, 0)
( 1–e 2,1)
1 1 1
(e, 1) Replace x by x 2; shift right 2 units.
3
5
x
(e2, 1)
F y In (x 2)
D 5IFSBOHFPGf1x2 = - ln 1x - 22 is the set of all real numbers. The vertical asymptote is x = 2.L t 4 R
to measure the amount L learned at time t. The number A represents the amount to be learned, and the number k measures the rate of learning. Suppose that a student has an amount A of 200 vocabulary words to learn. A psychologist determines that the student had learned 20 vocabulary words after 5 minutes. ln 0.9 B %FUFSNJOFUIFSBUFPGMFBSOJOHk. - 5 ≈ 0.0211 C "QQSPYJNBUFMZIPXNBOZXPSETXJMMUIFTUVEFOUIBWF MFBSOFEBGUFSNJOVUFT XPSET D "GUFSNJOVUFT 54 words E )PX MPOH EPFT JU UBLF GPS UIF TUVEFOU UP MFBSO XPSET NJO
Loudness of Sound Problems 127–130 use the following discussion: The loudness L x , measured in decibels (dB), of a sound of x intensity x, measured in watts per square meter, is defined as L x = 10 log , where I0 = 10-12 watt per square meter is the least intense I0 sound that a human ear can detect. Determine the loudness, in decibels, of each of the following sounds. 130. %JFTFM USVDL USBWFMJOH NJMFT QFS IPVS GFFU BXBZ 127. Normal conversation: intensity of x = 10- watt per square intensity 10 times that of a passenger car traveling 50 miles meter. 50 dB QFSIPVSGFFUBXBZXIPTFMPVEOFTTJTEFDJCFMT E# -1 128. Amplified rock music: intensity of 10 watt per square meter. 110 dB 129. Heavy city traffic: intensity of x = 10-3 watt per square meter. 90 dB The Richter Scale Problems 131 and 132 use the following discussion: The Richter scale is one way of converting seismographic readings into numbers that provide an easy reference for measuring the magnitude M of an earthquake. All earthquakes are compared to a zero-level earthquake whose seismographic reading measures 0.001 millimeter at a distance of 100 kilometers from the epicenter. An earthquake whose seismographic reading measures x millimeters has magnitude M x , given by M x = log¢
x ≤ x0
356
CHAPTER 4 Exponential and Logarithmic Functions
where x0 = 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 .FYJDP $JUZ JO TFJTNPHSBQIJDSFBEJOHPG NJMMJNFUFSTLJMPNFUFST from the center 132. Magnitude of an Earthquake San Francisco in 1906: seismographic reading of 50,119 millimeters 100 kilometers from the center 133. Alcohol and Driving The concentration of alcohol in a person’s bloodstream is measurable. Suppose that the relative risk R of having an accident while driving a car can be modeled by an equation of the form R = e kx where x is the percent of concentration of alcohol in the bloodstream and k is a constant.
* B 4VQQPTF UIBU B DPODFOUSBUJPO PG BMDPIPM JO UIF bloodstream of 0.03 percent results in a relative risk of an accident of 1.4. Find the constant k in the equation. C 6TJOH UIJT WBMVF PG k, what is the relative risk if the DPODFOUSBUJPOJTQFSDFOU D 6TJOHUIFTBNFWBMVFPGk, what concentration of alcohol DPSSFTQPOETUPBSFMBUJWFSJTLPG 0.41% E *G UIF MBX BTTFSUT UIBU BOZPOF XJUI B SFMBUJWF SJTL PG having an accident of 5 or more should not have driving privileges, at what concentration of alcohol in the bloodstream should a driver be arrested and charged XJUI%6* 0.14% F Compare this situation with that of Example 10. If you XFSF B MBXNBLFS XIJDI TJUVBUJPO XPVME ZPV TVQQPSU Give your reasons.
Discussion and Writing 134. Is there any function of the form y = xa, 0 6 a 6 1, that increases more slowly than a logarithmic function whose CBTFJTHSFBUFSUIBO &YQMBJO *135. In the definition of the logarithmic function, the base a is not BMMPXFEUPFRVBM8IZ 136. Critical Thinking In buying a new car, one consideration might be how well the price of the car holds up over time. %JGGFSFOU NBLFT PG DBST IBWF EJGGFSFOU EFQSFDJBUJPO SBUFT One way to compute a depreciation rate for a car is given here. Suppose that the current prices of a certain automobile are as shown in the table.
Age in Years New
1
2
3
4
5
$38,000
$36,600
$32,400
$28,750
$25,400
$21,200
6TF UIF GPSNVMB New = Old e Rt to find R, the annual depreciation rate, for a specific time t8IFO NJHIU CF UIF CFTUUJNFUPUSBEFJOUIFDBS $POTVMUUIF/"%" iCMVFu book and compare two like models that you are interested JO8IJDIIBTUIFCFUUFSEFQSFDJBUJPOSBUF
Retain Your Knowledge Problems 137–140 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. *139. 6TFUIF*OUFSNFEJBUF7BMVF5IFPSFNUPTIPXUIBUUIFGVODUJPO *137. Find the real zeros of g x = 4x4 - 3x2 + 9.8IBUBSF f x = 4x3 - 2x2 - has a real zero in the interval [1, 2]. the x-intercepts of the graph of g 138. Find the average rate of change of f x = 9x from
*140. A complex polynomial function f of degree 4 has the zeros 1 to 1. 12 - 1, 2, and 3 - i.'JOEUIFSFNBJOJOH[FSP T PGf. Then find a 2 polynomial function with real coefficients that has the zeros.
‘Are You Prepared?’ Answers 1. B {x 0 x … 3} C {x 0 x 6 - 2 or x 7 3}
2. {x 0 x 6 - 4 or x 7 1}
3. {3}
4.5 Properties of Logarithms OBJECTIVES 1 Work with the Properties of Logarithms (p. 356) 2 Write a Logarithmic Expression as a Sum or Difference of Logarithms (p. 359) 3 Write a Logarithmic Expression as a Single Logarithm (p. 359) 4 Evaluate a Logarithm Whose Base Is Neither 10 Nor e (p. 360) 5 Graph a Logarithmic Function Whose Base Is Neither 10 Nor e (p. 362)
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.
SECTION 4.5 Properties of Logarithms
EXAMPL E 1
Establishing Properties of Logarithms B 4IPXUIBUlog a 1 = 0.
Solution
357
C 4IPXUIBUlog a a = 1.
B 5IJT GBDU XBT FTUBCMJTIFE FBSMJFS XIJMF HSBQIJOH y = log a x TFF 'JHVSF PO QBHF 5PTIPXUIFSFTVMUBMHFCSBJDBMMZ MFUy = log a 1. Then y ay ay y log a 1
= = = = =
log a 1 1 a0 0 0
= = = = =
log a a a a1 1 1
Change to an exponential statement. a 0 = 1 since a 7 0, a ≠ 1 Solve for y. y = loga 1
C -FUy = log a a. Then y ay ay y log a a
Change to an exponential statement. a = a1 Solve for y. y = loga a
r
To summarize: log a 1 = 0
THEOREM
log a a = 1
Properties of Logarithms In the properties given next, M and a are positive real numbers, a ≠ 1, and r is any real number. The number log a M is the exponent to which a must be raised to obtain M. That is, aloga M = M
(1)
The logarithm to the base a of a raised to a power equals that power. That is, log a ar = r
(2)
The proofs use the fact that y = ax and y = log a x are inverses.
Proof of Property (1) For inverse functions, f 1f -1 1x2 2 = x for all x in the domain of f - 1 6TFf1x2 = ax and f -1 1x2 = log a x to find f1f -1 1x2 2 = aloga x = x for x 7 0 Now let x = M to obtain aloga M = M, where M 7 0.
■
Proof of Property (2) For inverse functions, f -1 1f1x2 2 = x for all x in the domain of f 6TFf1x2 = ax and f -1 1x2 = log a x to find f -1 1f1x2 2 = log a ax = x for all real numbers x Now let x = r to obtain log a ar = r, where r is any real number.
■
358
CHAPTER 4 Exponential and Logarithmic Functions
EX A MPL E 2
Using Properties (1) and (2) B 2log2 p = p C log 0.2 0.2-22 = - 22 D ln e kt = kt
Now Work
PROBLEM
r
15
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 ≠ 1, and r is any real number.
The Log of a Product Equals the Sum of the Logs log a 1MN2 = log a M + log a N
(3)
The Log of a Quotient Equals the Difference of the Logs log a a
M b = log a M - log a N N
(4)
The Log of a Power Equals the Product of the Power and the Log log a Mr = r log a M
(5)
ax = e x ln a
(6)
8FTIBMMEFSJWFQSPQFSUJFT
BOE BOEMFBWFUIFEFSJWBUJPOPGQSPQFSUZ BTBOFYFSDJTF TFF1SPCMFN
Proof of Property (3) Let A = log a M and let B = log a N. These expressions are equivalent to the exponential expressions aA = M and aB = N Now
log a 1MN = log a 1aA aB 2 = log a aA + B = A + B = log a M + log a N
Law of Exponents Property (2) of logarithms ■
Proof of Property (5) Let A = log a M. This expression is equivalent to aA = M Now
log a Mr = log a 1aA 2 = log a arA = rA = r log a M r
Law of Exponents Property (2) of logarithms ■
Proof of Property (6) 1SPQFSUZ
XJUIa = e, yields e ln M = M Now let M = axBOEBQQMZQSPQFSUZ
e ln a = e x ln a = 1e ln a 2 x = ax x
Now Work
PROBLEM
19
■
SECTION 4.5 Properties of Logarithms
359
2 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 calculus problems.
Writing a Logarithmic Expression as a Sum of Logarithms
EXAM PL E 3
8SJUFlog a 1x2x2 + 12, x 7 0, as a sum of logarithms. Express all powers as factors. log a 1x2x2 + 12 = log a x + log a 2x2 + 1 1>2 = log a x + log a 1x2 + 12 1 = log a x + log a 1x2 + 12 2
Solution
loga 1M # N2 = loga M + loga N
loga M r = r loga M
r
Writing a Logarithmic Expression as a Difference of Logarithms
EXAM PL E 4
8SJUF ln
x2 , 1x - 12 3
x 7 1
as a difference of logarithms. Express all powers as factors.
Solution
ln
x2 = ln x2 - ln 1x - 12 3 = 2 ln x - 3 ln1x - 12 1x - 12 3 c c M loga a b = loga M - loga N N
r
logaM r = r logaM
Writing a Logarithmic Expression as a Sum and Difference of Logarithms
EXAM PL E 5
8SJUF log a
2x2 + 1 , x 7 0 x3 1x + 12 4
as a sum and difference of logarithms. Express all powers as factors.
Solution
log a
WARNING *O VTJOH QSPQFSUJFT UISPVHI
CFDBSFGVMBCPVUUIFWBMVFT that the variable may assume. For example, the domain of the variable for log a x is x 7 0 and for log a x - 1 is x 7 1. If these functions are added, the domain is x 7 1. That is, the equality log a x + log a x - 1 = log a[x x - 1 ] is true only for x 7 1. j
2x2 + 1 = log a 2x2 + 1 - log a 3 x3 1x + 12 4 4 x3 1x + 12 4 = log a 2x2 + 1 - 3 log a x3 + log a 1x + 12 4 4
= log a 1x + 12 - log a x - log a 1x + 12 1 = log a 1x2 + 12 - 3 log a x - 4 log a 1x + 12 2 2
Now Work
PROBLEM
1>2
3
Property (4) Property (3)
4
Property (5)
r
51
3 Write a Logarithmic Expression as a Single Logarithm "OPUIFS VTF PG QSPQFSUJFT UISPVHI JT UP XSJUF TVNT BOEPS EJGGFSFODFT PG logarithms with the same base as a single logarithm. This skill will be needed to solve certain logarithmic equations discussed in the next section.
EXAM PL E 6
Writing Expressions as a Single Logarithm 8SJUFFBDIPGUIFGPMMPXJOHBTBTJOHMFMPHBSJUIN 2 ln - ln152 - 1
3 D log a x + log a 9 + log a 1x2 + 12 - log a 5 B log a + 4 log a 3
C
360
CHAPTER 4 Exponential and Logarithmic Functions
Solution
B log a + 4 log a 3 = log a + log a 34 r logaM = logaM r = log a + log a 1 logaM + logaN = loga 1M # N2 = log a 1 # 12 = log a 56 2 C ln - ln 152 - 12 = ln 2>3 - ln 125 - 12 r logaM = logaM r 3 3 82/3 = 1 1 82 2 = 22 = 4 = ln 4 - ln 24 4 M = ln a b logaM - logaN = loga a b N 24 1 = ln a b 6 = ln 1 - ln 6 = - ln 6 ln1 = 0
D log a x + log a 9 + log a 1x2 + 12 - log a 5 = log a 19x2 + log a 1x2 + 12 - log a 5 = log a 3 9x1x2 + 12 4 - log a 5 = log a J
9x1x2 + 12 R 5
r
WARNING A common error made by some students is to express the logarithm of a sum as the sum of logarithms. log a 1M + N2
Correct statement
is not equal to
log a 1MN2 = log a M + log a N
log a M + log a N Property (3)
Another common error is to express the difference of logarithms as the quotient of logarithms. log a M - log a N is not equal to
log a M log a N
M Correct statement log a M - log a N = log a a b Property (4) N A third common error is to express a logarithm raised to a power as the product of the power times the logarithm. 1log a M2 r is not equal to r log a M
Correct statement
Now Work
log a Mr = r log a M PROBLEMS
57
AND
j
Property (5)
63
Two other important properties of logarithms are consequences of the fact that the logarithmic function y = log a x is a one-to-one function.
THEOREM
Properties of Logarithms In the following properties, M, N, and a are positive real numbers, a ≠ 1. If M = N, then log a M = log a N. If log a M = log a N, then M = N.
(7) (8)
1SPQFSUZ JT VTFE BT GPMMPXT 4UBSUJOH XJUI UIF FRVBUJPO M = N, “take the logarithm of both sides” to obtain log a M = log a N. 1SPQFSUJFT BOE BSF VTFGVM GPS TPMWJOH exponential and logarithmic equations, a topic discussed in the next section.
4 Evaluate a Logarithm Whose Base Is Neither 10 Nor e Logarithms to the base 10, common logarithms, were used to facilitate arithmetic DPNQVUBUJPOTCFGPSFUIFXJEFTQSFBEVTFPGDBMDVMBUPST 4FFUIF)JTUPSJDBM'FBUVSF BU UIF FOE PG UIJT TFDUJPO /BUVSBM MPHBSJUINTUIBU JT MPHBSJUINT XIPTF CBTF JT
SECTION 4.5 Properties of Logarithms
361
the number e remain very important because they arise frequently in the study of natural phenomena. Common logarithms are usually abbreviated by writing log, with the base understood to be 10, just as natural logarithms are abbreviated by ln, with the base understood to be e. .PTU DBMDVMBUPST IBWF CPUI log and ln keys to calculate the common logarithm and the natural logarithm of a number, respectively. Let’s look at an example to see how to approximate logarithms having a base other than 10 or e.
EXAMPL E 7
Approximating a Logarithm Whose Base Is Neither 10 Nor e Approximate log 2 .3PVOEUIFBOTXFSUPGPVSEFDJNBMQMBDFT
Solution
3FNFNCFS log 2 NFBOT i SBJTFE UP XIBU FYQPOFOU FRVBMT u -FU y = log 2 . Then 2y = . Because 22 = 4 and 23 = , the value of log 2 is between 2 and 3. 2y = ln 2y = ln y ln 2 = ln ln y = ln 2 y ≈ 2.04
Property (7) Property (5) Exact value
r
Approximate value rounded to four decimal places
&YBNQMFTIPXTIPXUPBQQSPYJNBUFBMPHBSJUINXIPTFCBTFJTCZDIBOHJOH to logarithms involving the base e. In general, the Change-of-Base Formula is used.
THEOREM
Change-of-Base Formula If a ≠ 1, b ≠ 1, and M are positive real numbers, then log a M =
log b M log b a
(9)
Proof Let y = log a M. Then ay = M log b ay = log b M
Property (7)
y log b a = log b M log b M y = log b a log b M log a M = log b a
Property (5) Solve for y.
r
y = loga M
Because calculators have keys only for log and ln , in practice, the Change-of-Base Formula uses either b = 10 or b = e. That is, log a M =
EXAMPL E 8
log M log a
and log a M =
ln M ln a
Using the Change-of-Base Formula Approximate: B log 5 9 3PVOEBOTXFSTUPGPVSEFDJNBMQMBDFT
C log 22 25
(10)
362
CHAPTER 4 Exponential and Logarithmic Functions
Solution
1 log 5 log 25 log 9 2 1.94939000 = C log 22 25 = B log 5 9 = ≈ 1 log 5 0.69900043 log 22 log 2 2 ≈ 2.9 log 5 = ≈ 2.3219 log 2 or or ln 9 4.46363 ≈ log 5 9 = ln 5 1.60943912 1 ln 5 ln 25 2 ≈ 2.9 log 22 25 = = 1 ln 22 ln 2 2 ln 5 = ≈ 2.3219 ln 2
r
Now Work
PROBLEMS
23
AND
71
5 Graph a Logarithmic Function Whose Base Is Neither 10 Nor e The Change-of-Base Formula also can be used to graph logarithmic functions whose base is neither 10 nor e.
EX AMPL E 9
Graphing a Logarithmic Function Whose Base Is Neither 10 Nor e 6TFBHSBQIJOHVUJMJUZUPHSBQIy = log 2 x.
Figure 42
Solution
2
0
2
4
Because graphing utilities have only logarithms with the base 10 or the base e, use the Change-of-Base Formula to express y = log 2 x in terms of logarithms with base log x ln x 10 or base e. Graph either y = or y = to obtain the graph of y = log 2 x. ln 2 log 2 See Figure 42. log x ln x Check: 7FSJGZUIBUy = and y = result in the same graph by graphing ln 2 log 2 each on the same screen.
Now Work
PROBLEM
79
SUMMARY Properties of Logarithms In the list that follows, a, b, M, N, and r are real numbers. Also, a 7 0, a ≠ 1, b 7 0, b ≠ 1, M 7 0, and N 7 0. Definition
y = log a x means x = ay
Properties of logarithms
log a 1 = 0; log a a = 1 aloga M = M; log a ar = r log a 1MN2 = log a M + log a N M log a a b = log a M - log a N N log b M log a M = log b a
log a Mr = r log a M ax = e x ln a If M = N, then log a M = log a N.
Change-of-Base Formula
If log a M = log a N, then M = N.
r
363
SECTION 4.5 Properties of Logarithms
Historical Feature
L
tables, published in 1614, listed what would now be called natural logarithms of sines and were rather difficult to use. A London professor, Henry Briggs, became interested in the tables and visited Napier. In their conversations, they developed the idea of common logarithms, which were published in 1617. Their importance for calculation was immediately recognized, and by 1650 they were being printed as far away as China. They remained an important calculation tool until the advent of the inexpensive handheld calculator about 1972, which has decreased their calculational, but not their theoretical, importance. A side effect of the invention of logarithms was the popularization of the decimal system of notation for real numbers.
ogarithms were invented about 1590 by John Napier (1550–1617) and Joost Bürgi (1552–1632), working independently. Napier, whose work had the greater influence, was a Scottish lord, a secretive man whose neighbors were inclined to believe him to be in league with the devil. His approach to logarithms was very John Napier different from ours; it was based on the relationship (1550–1617) between arithmetic and geometric sequences, discussed in a later chapter, and not on the inverse function relationship of logarithms to exponential functions (described in Section 4.4). Napier’s
4.5 Assess Your Understanding Concepts and Vocabulary 1. log a1 =
2. log aa =
0
5. log a 1MN2 =
log aM
+
3. alogaM =
1
M 6. log a a b = log aM N 8. If log a x = log a6, then x =
log aN
7. log aM = r
r log a M log 5 , then M = 9. If log M = log 5
4. log aar =
M
log aN 6 .
10. True or False ln1x + 32 - ln12x2 =
.
11. True or False log 2 13x4 2 = 4 log 2 13x2
12. True or False
False
ln = 2 ln 4
r
ln1x + 32
False
ln12x2
False
Skill Building In Problems 13–28, use properties of logarithms to find the exact value of each expression. Do not use a calculator. 13. log 3 31 log2
17. 2
14. log 2 2-13
18. e
ln
15. ln e -4
- 13
19. log 2 + log 4
16. ln e 22
-4
22
20. log 6 9 + log 6 4
1
2
22. log 16 - log 2 1 23. log 2 6 # log 6 3 24. log 3 # log 9 2 21. log 6 1 - log 6 3 1 log3 5 - log3 4 5 log5 6 + log5 loge2 16 42 27. e 4 28. e loge2 9 3 26. 5 25. 3 4 In Problems 29–36, suppose that ln 2 = a and ln 3 = b. Use properties of logarithms to write each logarithm in terms of a and b. 2 29. ln 6 a + b 30. ln a - b 31. ln 1.5 b - a 32. ln 0.5 - a 3 1 2 1 5 35. ln 2 6 1a + b2 36. ln 4 1a - b2 33. MO 3a 34. MO 3b 5 A3 4 In Problems 37–56, write each expression as a sum and/or difference of logarithms. Express powers as factors. x 38. log 3 log 3 x - 2 39. log 2 z3 3 log 2z 40. log x5 5 log x 37. log 5 125x2 2 + log 5 x 9 e x 42. ln 1 - ln x 43. ln x ln x - x 44. ln1xe x 2 ln x + x 41. ln1ex2 1 + ln x x e a *45. log a 1u2 v3 2 u 7 0, v 7 0 *46. log 2 a 2 b a 7 0, b 7 0 *47. ln1x2 21 - x2 0 6 x 6 1 *48. ln1x21 + x2 2 x 7 0 b 3 2 x1x + 22 x3 2 x + 1 x3 2x + 1 *49. log 2 ¢ b x 7 1 *51. logJ ≤ x 7 3 *50. log 5 a 2 R x 7 0 *52. logJ R x 7 2 2 x - 3 x - 1 1x + 32 1x - 22 2 *53. lnJ
x2 - x - 2 R 1x + 42 2
1>3
x 7 2*54. lnJ
1x - 42 2 x - 1 2
2>3
R
x 7 4 *55. ln
5x21 + 3x 1x - 42 3
x 7 4
*56. lnc
3 5x2 2 1 - x d 41x + 12 2
0 6 x 6 1
In Problems 57–70, write each expression as a single logarithm. 57. 3 log 5 u + 4 log 5 v
log 5 1u3v4 2
1 1 60. log 2 a b + log 2 ¢ 2 ≤ x x *63. lna
log 2 a
1 b x3
58. 2 log 3 u - log 3 v
log 3 a
u2 b v
*61. log 4 1x2 - 12 - 5 log 4 1x + 12
59. log 3 2x - log 3 x3
log 3 a
1 x5>2
b
*62. log1x2 + 3x + 22 - 2 log1x + 12
x x2 + x + 6 x + 1 x2 + 2x - 3 4 ≤ - log¢ b + lna b - ln1x2 - 12*64. log¢ ≤ *65. log 2 23x - 2 - log 2 a b + log 2 4 2 x - 1 x x + 2 x x - 4
%VFUPTQBDFSFTUSJDUJPOT BOTXFSTUPUIFTFFYFSDJTFTNBZCFGPVOEJOUIF"OTXFSTJOUIFCBDLPGUIFCPPL
364
CHAPTER 4 Exponential and Logarithmic Functions
3 *66. 21 log 3 1 x + log 3 19x2 2 - log 3 9
*67. 2 log a 15x3 2 -
*69. 2 log 2 1x + 12 - log 2 1x + 32 - log 2 1x - 12
1 1 1 *68. log1x3 + 12 + log1x2 + 12 log a 12x + 32 2 3 2 *70. 3 log 5 13x + 12 - 2 log 5 12x - 12 - log 5 x
In Problems 71–78, use the Change-of-Base Formula and a calculator to evaluate each logarithm. Round your answer to three decimal places. 71. log 3 21
72. log 5 1
73. log 1>3 1
75. log 12
5.615
76. log 15
77. log p e
- 3.0
74. log 1>2 15
- 3.90
78. log p 22
0.303
In Problems 79–84, graph each function using a graphing utility and the Change-of-Base Formula.
*79. y = log 4 x * 80. y = log 5 x *81. y = log 2 1x + 22 * 82. y = log 4 1x - 32 * 83. y = log x - 1 1x + 12 *84. y = log x + 2 1x - 22
Mixed Practice 86. If f 1x2 = log 2 x, g1x2 = 2x, and h1x2 = 4x, find: * B 1f ∘ g2 1x2.8IBUJTUIFEPNBJOPGf ∘ g * C 1g ∘ f2 1x2.8IBUJTUIFEPNBJOPGg ∘ f D 1f ∘ g2 132 3 * E 1f ∘ h2 1x2 8IBUJTUIFEPNBJOPGf ∘ h F 1f ∘ h2 12 5
85. If f 1x2 = ln x, g1x2 = e x, and h1x2 = x2, find: * B 1f ∘ g2 1x2.8IBUJTUIFEPNBJOPGf ∘ g * C 1g ∘ f2 1x2.8IBUJTUIFEPNBJOPGg ∘ f D 1f ∘ g2 152 5 * E 1f ∘ h2 1x2 8IBUJTUIFEPNBJOPGf ∘ h F 1f ∘ h2 1e2 2
Applications and Extensions In Problems 87–96, express y as a function of x. The constant C is a positive number. 87. ln y = ln x + ln C
89. ln y = ln x + ln1x + 12 + ln C 91. ln y = 3x + ln C
88. ln y = ln1x + C2
y = Cx y = Cx1x + 12
y = x + C
90. ln y = 2 ln x - ln1x + 12 + ln C
y = Ce 3x
92. ln y = - 2x + ln C
93. ln1y - 32 = - 4x + ln C y = Ce - 4x + 3 1 1 *95. 3 ln y = ln12x + 12 - ln1x + 42 + ln C 2 3 97. Find the value of log 2 3 # log 3 4 # log 4 5 # log 5 6 # log 6 # log . 3
99. Find the value of log 2 3 # log 3 4 # g # log n 1n + 12 # log n + 1 2. 1 *101. Show that log a 1x + 2x2 - 12 + log a 1x - 2x2 - 12 = 0.
y =
y = Ce - 2x
Cx2 x + 1
94. ln1y + 42 = 5x + ln C y = Ce 5x - 4 1 1 *96. 2 ln y = - ln x + ln1x2 + 12 + ln C 2 3 98. Find the value of log 2 4 # log 4 6 # log 6 . 3
100. Find the value of log 2 2 # log 2 4 # g # log 2 2n.
n!
*102. Show that log a 1 2x + 2x - 12 + log a 1 2x - 2x - 12 = 0. *103. Show that ln11 + e 2x 2 = 2x + ln11 + e -2x 2. f 1x + h2 - f 1x2 h 1>h = log a a1 + b , h ≠ 0. *104. Difference Quotient If f 1x2 = log a x, show that h x *106. If f 1x2 = log a x, show that f 1AB2 = f 1A2 + f 1B2. *105. If f 1x2 = log a x, show that - f 1x2 = log 1>a x. 1 * 107. If f 1x2 = log a x, show that f a b = - f 1x2. x M *109. Show that log a a b = log a M - log a N, where a, M, and N N are positive real numbers and a ≠ 1.
*108. If f 1x2 = log a x, show that f 1xa 2 = af 1x2. *110. Show that log a a
1 b = - log a N, where a and N are N positive real numbers and a ≠ 1.
Discussion and Writing 111. Graph Y1 = log1x2 2 and Y2 = 2 log1x2 using a graphing VUJMJUZ "SF UIFZ FRVJWBMFOU 8IBU NJHIU BDDPVOU GPS BOZ EJGGFSFODFTJOUIFUXPGVODUJPOT
112. 8SJUFBOFYBNQMFUIBUJMMVTUSBUFTXIZ 1log a x2 ≠ r log a x.
113. 8 SJUFBOFYBNQMFUIBUJMMVTUSBUFTXIZ log 2 1x + y2 ≠ log 2 x + log 2 y. 114. %PFT3log31-52 = - 5 8IZPSXIZOPU
r
Retain Your Knowledge Problems 115–118 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. *115. 6TF B HSBQIJOH VUJMJUZ UP TPMWF x3 - 3x2 - 4x + = 0. 3PVOEBOTXFSTUPUXPEFDJNBMQMBDFT { - 1., 1.29, 3.49} 116. Find the real zeros of f 1x2 = 5x5 + 44x4 + 116x3 + 95x2 - 4x - 4
1 - 5 { 221 - 2, , 5 2
*117. 8JUIPVUTPMWJOH EFUFSNJOFUIFDIBSBDUFSPGUIFTPMVUJPOPG the quadratic equation 4x2 - 2x + 49 = 0 in the complex system. *118. Graph f x = 22 - x using the techniques of shifting, compressing or stretching, and reflections. State the domain and the range of f.
365
SECTION 4.6 Logarithmic and Exponential Equations
4.6 Logarithmic and Exponential Equations PREPARING FOR THIS SECTION Before getting started, review the following: r 4PMWJOH&RVBUJPOT6TJOHB(SBQIJOH6UJMJUZ "QQFOEJY# 4FDUJPO# QQ#m#
r 4PMWJOH2VBESBUJD&RVBUJPOT 4FDUJPO QQm
r 4PMWJOH&RVBUJPOT2VBESBUJDJO'PSN 4FDUJPO QQm
Now Work the ‘Are You Prepared?’ problems on page 369.
OBJECTIVES 1 Solve Logarithmic Equations (p. 365) 2 Solve Exponential Equations (p. 367) 3 Solve Logarithmic and Exponential Equations Using a Graphing Utility (p. 368)
1 Solve Logarithmic Equations In Section 4.4 logarithmic equations were solved by changing a logarithmic expression to an exponential expression. That is, they were solved by using the definition of a logarithm: y = log a x is equivalent to x = ay
a 7 0, a ≠ 1
For example, to solve the equation log 2 11 - 2x2 = 3, write the logarithmic equation as an equivalent exponential equation 1 - 2x = 23 and solve for x. log 2 11 - 2x2 = 3
1 - 2x = 23 - 2x = x = -
Change to an exponential equation. Simplify.
2
Solve.
Check: log 2 11 - 2x2 = log 2 a1 - 2a - b b = log 2 11 + 2 = log 2 = 3 2
23 =
'PS NPTU MPHBSJUINJD FRVBUJPOT TPNF NBOJQVMBUJPO PG UIF FRVBUJPO VTVBMMZ VTJOH QSPQFSUJFT PG MPHBSJUINT JT SFRVJSFE UP PCUBJO B TPMVUJPO "MTP UP BWPJE extraneous solutions with logarithmic equations, it is wise to determine the domain of the variable first. Let’s begin with an example of a logarithmic equation that requires using the fact that a logarithmic function is a one-to-one function. If log a M = log a N, then M = N
EXAM PL E 1
M, N, and a are positive and a ≠ 1.
Solving a Logarithmic Equation Solve: 2 log 5 x = log 5 9
Solution
The domain of the variable in this equation is x 7 0. Note that each logarithm is to the same base, 5. Then find the exact solution as follows: 2 log 5 x = log 5 9 loga M r = r loga M log 5 x2 = log 5 9 If loga M = loga N, then M = N. x2 = 9 x = 3 or x = - 3 3FDBMM UIBU UIF EPNBJO PG UIF WBSJBCMF JT x 7 0. Therefore, - 3 is extraneous and must be discarded.
366
CHAPTER 4 Exponential and Logarithmic Functions
Check: 2 log 5 3 ≟ log 5 9 log 5 32 ≟ log 5 9 log 5 9 = log 5 9
r loga M = loga M r
The solution set is 5 36 .
Now Work
r
PROBLEM
13
Often one or more properties of logarithms are needed to rewrite the equation as a single logarithm. In the next example, the log of a product property is used.
EX A MPL E 2
Solution
WARNING A negative solution is not automatically extraneous. Check whether the potential solution causes the argument of any logarithmic expression in the equation to be negative. j
Solving a Logarithmic Equation
Solve: log 5 1x + 62 + log 5 1x + 22 = 1 The domain of the variable requires that x + 6 7 0 and x + 2 7 0, so x 7 - 6 and x 7 - 2. This means any solution must satisfy x 7 - 2. To obtain an exact solution, express the left side as a single logarithm. Then change the equation to an equivalent exponential equation. log 5 1x + 62 + log 5 1x + 22 log 5 3 1x + 62 1x + 22 4 1x + 62 1x + 22 x2 + x + 12 x2 + x + 1x + 2 1x + 12 x = - or x
1 1 51 5 0 0 -1
loga M + loga N = loga(MN) Change to an exponential equation. Simplify. Place the quadratic equation in standard form. Factor. Zero-Product Property
Only x = - 1 satisfies the restriction that x 7 - 2, so x = - is extraneous. The solution set is 5 - 16 , which you should check.
Now Work
EX A MPL E 3
= = = = = = =
PROBLEM
r
21
Solving a Logarithmic Equation Solve: ln x = ln x + 6 - ln x - 4
Solution
The domain of the variable requires that x 7 0, x + 6 7 0, and x - 4 7 0. As a result, the domain of the variable here is x 7 4. Begin the solution using the log of a difference property. ln x = ln x + 6 - ln x - 4
x + 6 ln x = ln a b x - 4 x + 6 x = x - 4 x x - 4 = x + 6 x2 - 4x = x + 6 x2 - 5x - 6 = 0 x - 6 x + 1 = 0 x = 6 or x = - 1
M In M - ln N = ln a b N If ln M = ln N, then M = N. Multiply both sides by x - 4. Simplify. Place the quadratic equation in standard form. Factor. Zero-Product Property
Because the domain of the variable is x 7 4, discard - 1 as extraneous. The solution set is {6}, which you should check.
Now Work
PROBLEM
31
r
SECTION 4.6 Logarithmic and Exponential Equations
367
2 Solve Exponential Equations In Sections 4.3 and 4.4, exponential equations were solved algebraically by expressing each side of the equation using the same base. That is, they were solved by using the one-to-one property of the exponential function: If au = av, then u = v
a 7 0, a ≠ 1
For example, to solve the exponential equation 42x + 1 = 16, notice that 16 = 42 and apply the property above to obtain the equation 2x + 1 = 2, and the solution is 1 x = . 2 Not all exponential equations can be readily expressed so that each side of the equation has the same base. For such equations, algebraic techniques often can be used to obtain exact solutions. In the next example two exponential equations are solved by changing the exponential expression to a logarithmic expression.
EXAMPL E 4
Solving Exponential Equations 4PMWF B 2x = 5
Solution
C # 3x = 5
B #FDBVTFDBOOPUCFXSJUUFOBTBOJOUFHFSQPXFSPG 22 = 4 and 23 = , write the exponential equation as the equivalent logarithmic equation. 2x = 5 x = log 2 5 =
c
ln 5 ln 2
Change@of@Base Formula (9), Section 4 .5
Alternatively, the equation 2x = 5 can be solved by taking the natural logarithm PSDPNNPOMPHBSJUIN PGFBDITJEF5BLJOHUIFOBUVSBMMPHBSJUINZJFMET 2x = 5 ln 2x = ln 5 x ln 2 = ln 5 ln 5 x = ln 2 ≈ 2.322 The solution set is e C # 3x = 5 5 3x =
Exact solution Approximate solution
ln 5 f. ln 2
5 ln a b ln 3
Exact solution
≈ - 0.42
Now Work
In M r = r ln M
Solve for 3x.
5 x = log 3 a b =
The solution set is
If M = N, then ln M = ln N.
W
Approximate solution
5 ln a b ¶ . ln 3
PROBLEM
43
r
368
CHAPTER 4 Exponential and Logarithmic Functions
EX A MPL E 5
Solving an Exponential Equation Solve: 5x - 2 = 33x + 2
Solution
#FDBVTFUIFCBTFTBSFEJGGFSFOU àSTUBQQMZQSPQFSUZ
4FDUJPO UBLFUIFOBUVSBM MPHBSJUINPGFBDITJEF
BOEUIFOVTFBQSPQFSUZPGMPHBSJUINT5IFSFTVMUJTBOFRVBUJPO in x that can be solved. 5x - 2 = 33x + 2 ln 5x - 2 = ln 33x + 2
If M = N, ln M = ln N.
1x - 22 ln 5 = 13x + 22 ln 3
1ln 52 x - 2 ln 5 = 13 ln 32x + 2 ln 3
NOTE: Because of the properties of logarithms, exact solutions involving logarithms often can be expressed in multiple ways. For example, the solution to 5x - 2 = 33x + 2 from Example 5 can be expressed equivalently 2 ln 15 ln 225 as or among others. ln 5 - ln 27 ln( 5/27) j Do you see why?
1ln 52x - 13 ln 32x = 2 ln 3 + 2 ln 5
1ln 5 - 3 ln 32x = 21ln 3 + ln 52 21ln 3 + ln 52 x = ln 5 - 3 ln 3 ≈ - 3.212
The solution set is e
Now Work
ln M r = r ln M Distribute. Place terms involving x on the left. Factor. Exact solution Approximate solution
21ln 3 + ln 52 f. ln 5 - 3 ln 3
PROBLEM
r
53
The next example deals with an exponential equation that is quadratic in form.
EX A MPL E 6
Solving an Exponential Equation That Is Quadratic in Form Solve: 4x - 2x - 12 = 0
Solution
Note that 4x = 22 x = 2 2x = 2x 2, so the equation is quadratic in form and can be written as 2x 2 - 2x - 12 = 0
Let u = 2x; then u 2 - u - 12 = 0.
Now factor as usual. 12x - 42 12x + 32 = 0 2 - 4 = 0 or 2x + 3 = 0 2x = 4 2x = - 3 x
(u - 4)(u + 3) = 0 u - 4 = 0 or u + 3 = 0 u = 2x = 4
u = 2x = - 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 7 0 for all x. The only solution is 2. The solution set is 5 26 .
r
Now Work
PROBLEM
61
3 Solve Logarithmic and Exponential Equations Using a Graphing 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, a graphing utility can be used to approximate the solution.
369
SECTION 4.6 Logarithmic and Exponential Equations
Solving Equations Using a Graphing Utility
EXAM PL E 7
Solve: x + e x = 2 &YQSFTTUIFTPMVUJPO T SPVOEFEUPUXPEFDJNBMQMBDFT
Figure 43
Solution
4
Y1 x e x
r
Y2 2
0
The solution is found by graphing Y1 = x + e x and Y2 = 2. Since Y1 is an increasing GVODUJPO EPZPVLOPXXIZ
UIFSFJTPOMZPOFQPJOUPGJOUFSTFDUJPOGPSY1 and Y2 . Figure 43 shows the graphs of Y1 and Y2 .6TJOHUIF*/5&34&$5DPNNBOESFWFBMT that the solution is 0.44, rounded to two decimal places.
Now Work
1
PROBLEM
71
0
4.6 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. 5 - 3, 106
1. Solve x2 - x - 30 = 0. QQm
2. Solve 1x + 32 - 41x + 32 + 3 = 0. QQm 5 - 2, 06 2
3. "QQSPYJNBUFUIFTPMVUJPO T UPx3 = x2 - 5 using a graphing utility. QQ#m# 5 - 1.436 4. "QQSPYJNBUF UIF TPMVUJPO T UP x3 - 2x + 2 = 0 using a graphing utility. QQ#m# 5 - 1.6
Skill Building In Problems 5–40, solve each logarithmic equation. Express any irrational solution in exact form and as a decimal rounded to three decimal places. 16 5. log 4 x = 2 {16} 6. log 1x + 62 = 1 {4} 7. log 2 15x2 = 4 e f 5 10 8. log 3 13x - 12 = 2 e f 9. log 4 1x + 22 = log 4 {6} 10. log 5 12x + 32 = log 5 3 {0} 3 1 1 1 11. log 3 x = 2 log 3 2 {16} 12. - 2 log 4 x = log 4 9 e f 13. 3 log 2 x = - log 2 2 e f 2 3 3 14. 2 log 5 x = 3 log 5 4 {} 15. 3 log 2 1x - 12 + log 2 4 = 5 {3} 16. 2 log 3 1x + 42 - log 3 9 = 2 {5} 21 17. log x + log1x + 152 = 2 {5} 18. log x + log 1x - 212 = 2 {25} 19. log12x + 12 = 1 + log1x - 22 e f 15 21. log 2 1x + 2 + log 2 1x + 2 = 1 { - 6} 22. log 6 1x + 42 + log 6 1x + 32 = 1 { - 1} 20. log12x2 - log1x - 32 = 1 e f 4 23. log 1x + 62 = 1 - log 1x + 42 { - 2} 24. log 5 1x + 32 = 1 - log 5 1x - 12 5 26 *25. ln x + ln1x + 22 = 4 *27. log 3 x + 1 + log 3 1x + 42 = 2
*26. ln1x + 12 - ln x = 2
29. log 1>3 1x2 + x2 - log 1>3 1x2 - x2 = - 1 5 26
*28. log 2 x + 1 + log 2 1x + 2 = 3
30. log 4 1x2 - 92 - log 4 1x + 32 = 3
{6} 9 31. log a 1x - 12 - log a 1x + 62 = log a 1x - 22 - log a 1x + 32 e f 32. log a x + log a 1x - 22 = log a 1x + 42 {4} 2 33. 2 log 5 x - 3 - log 5 = log 5 2 \^ 34. log 3 x - 2 log 3 5 = log 3 x + 1 - 2 log 3 10 35. 2 log 6 x + 2 = 3 log 6 2 + log 6 4
5 - 2 + 4226
3 36. 3 log x - log 2 = 2 log 4 5 42 26 1 3 38. log x - 1 = log 2 5 1 + 226 3 40. ln x - 32 ln x + 2 = 0 5e, e 4 6
37. 2 log 13 x + 2 = log 13 4x + 5 - 23, 236 1 39. log 3 x 2 - 5 log 3 x = 6 e , 29 f 3
1 e f 3
In Problems 41–68, solve each exponential equation. Express any irrational solution in exact form and as a decimal rounded to three decimal places. 41. 2x - 5 =
42. 5-x = 25
{}
*45. -x = 1.2 53. 1.2x = 10.52 -x
{0}
*61. 16 + 4
- 3 = 0
*65. 3 # 4x + 4 # 2x + = 0
58. 32x + 3x - 2 = 0 x+1
*62. 9 - 3
*48. 0.3140.2x 2 = 0.2 4 1-x *52. a b = 5x 3 *56. e x + 3 = px
x
3 *51. a b = 1 - x 5 *55. p1 - x = e x
*54. 0.31 + x = 1.2x - 1 x
*44. 3x = 14
*47. 5123x 2 =
*50. 2x + 1 = 51 - 2x
*57. 22x + 2x - 12 = 0 x+1
*43. 2x = 10
*46. 2-x = 1.5
*49. 31 - 2x = 4x
x
{ - 2}
+ 1 = 0
{0}
*66. 2 # 49x + 11 # x + 5 = 0
59. 32x + 3x + 1 - 4 = 0
*63. 25 - # 5 = - 16 x
x
*67. 4x - 10 # 4-x = 3
{0}
60. 22x + 2x + 2 - 12 = 0
*64. 36x - 6 # 6x = - 9
*68. 3x - 14 # 3-x = 5
In Problems 69–82, use a graphing utility to solve each equation. Express your answer rounded to two decimal places. 69. log 5 1x + 12 - log 4 1x - 22 = 1
71. e x = - x
{ - 0.5}
{2.9}
72. e 2x = x + 2
70. log 2 1x - 12 - log 6 1x + 22 = 2
{ - 1.9, 0.45} 73. e x = x2
{ - 0.0}
{12.15}
74. e x = x3
%VFUPTQBDFSFTUSJDUJPOT BOTXFSTUPUIFTFFYFSDJTFTNBZCFGPVOEJOUIF"OTXFSTJOUIFCBDLPGUIFCPPL
{1.6, 4.54}
{1}
370
CHAPTER 4 Exponential and Logarithmic Functions
75. ln x = - x
{0.5}
79. e x + ln x = 4
{1.32}
76. ln12x2 = - x + 2 80. e x - ln x = 4
{1.16}
77. ln x = x3 - 1
{0.05, 1.4} 81. e -x = ln x
{0.39, 1}
{1.31}
78. ln x = - x2 82. e -x = - ln x
{0.65} {0.5}
Mixed Practice In Problems 83–96, solve each equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places. 83. log 9 1x - 52 = log 3 1x + 12 {2, 3} [Hint: Change log 9 1x - 52 to base 3.]
84. log 2 1x + 12 - log 4 x = 1
86. log 16 x + log 4 x + log 2 x =
87. log 9 x + 3 log 3 x = 14 {1} 1 90. log 2 xlog2 x = 4 e 4, f 4
{1} * 85. log 2 13x + 22 - log 4 x = 3
1 e , 4f 88. 21log 4x2 2 + 3log x = log 2 16 16 2 e x + e -x 2 3 e - 1, f 91. = 1 {0} 89. 1 222 2 - x = 2x 3 2 [Hint:.VMUJQMZFBDITJEFCZe x.] {ln13 + 2222, ln 3 - 222 } e x - e -x e x + e -x e x - e -x 92. = 3 93. = 2 94. = -2 ≈{1.63, - 1.63} 2 2 2 {ln1 - 2 + 252} ≈ { - 1.444} {ln12 + 252} ≈ {1.444} 96. log 2 x + log 6 x = 3 *95. log 5 x + log 3 x = 1 # 3 ln 2 ln 6 [Hint:6TFUIF$IBOHFPG#BTF'PSNVMB> 5 e ln 12 6 ≈ {4.49} 101.* B (SBQIf 1x2 = 3x and g x = 10 on the same Cartesian 97. f 1x2 = log 2 1x + 32 and g1x2 = log 2 13x + 12. B 4PMWFf 1x2 = 3.8IBUQPJOUJTPOUIFHSBQIPGf {5}; 5, 3
plane. C 4PMWFg1x2 = 4.8IBUQPJOUJTPOUIFHSBQIPGg {5}; 5, 4
* C 4IBEF UIF SFHJPO CPVOEFE CZ UIF y-axis, f 1x2 = 3x, and g x = 10POUIFHSBQIESBXOJOQBSU B D 4PMWF f 1x2 = g1x2.%PUIFHSBQITPGf and gJOUFSTFDU D 4PMWFf 1x2 = g1x2 and label the point of intersection *GTP XIFSF {1}; yes; 11, 22 POUIFHSBQIESBXOJOQBSU B {log 310} E 4PMWF 1f + g2 1x2 = . {5} 1 F 4PMWF 1f - g2 1x2 = 2. e- f 102.* B (SBQIf 1x2 = 2x and g x = 12 on the same Cartesian 11 plane. 98. f 1x2 = log 3 1x + 52 and g1x2 = log 3 1x - 12. {4}; 14, 22 C 4IBEF UIF SFHJPO CPVOEFE CZ UIF y-axis, f 1x2 = 2x, * B 4PMWFf 1x2 = 2.8IBUQPJOUJTPOUIFHSBQIPGf and g x = 12POUIFHSBQIESBXOJOQBSU B C 4PMWFg1x2 = 3.8IBUQPJOUJTPOUIFHSBQIPGg {2}; 12, 32 D 4PMWFf 1x2 = g1x2 and label the point of intersection D 4PMWFf 1x2 = g1x2.%PUIFHSBQITPGf and gJOUFSTFDU POUIFHSBQIESBXOJOQBSU B {log 212} *GTP XIFSF No real solution; no E 4PMWF 1f + g2 1x2 = 3. {4} *103. B (SBQI f 1x2 = 2x + 1 and g x = 2-x + 2 on the same e f F 4PMWF 1f - g2 1x2 = 2. Cartesian plane. 4 x+1 x+2 C 4IBEFUIFSFHJPOCPVOEFECZUIFy-axis, f 1x2 = 2x + 1, and g x = 2 , on the same 99.* B (SBQI f 1x2 = 3 -x + 2 and g x = 2 POUIFHSBQIESBXOJOQBSU B Cartesian plane. D 4PMWFf 1x2 = g1x2 and label the point of intersection C 'JOEUIFQPJOU T PGJOUFSTFDUJPOPGUIFHSBQITPGf and POUIFHSBQIESBXOJOQBSU B g by solving f 1x2 = g1x2. 3PVOE BOTXFST UP UISFF *104. B (SBQI f 1x2 = 3-x + 1 and g1x2 = 3x - 2 on the same decimal places. Label any intersection points on the Cartesian plane. HSBQIESBXOJOQBSU B
C 4IBEFUIFSFHJPOCPVOEFECZUIFy-axis, f 1x2 = 3-x + 1, D #BTFEPOUIFHSBQI TPMWFf 1x2 7 g1x2. , q
and g1x2 = 3x - 2POUIFHSBQIESBXJOQBSU B 100.* B (SBQI f 1x2 = 5x - 1 and g x = 2x + 1, on the same D 4PMWFf 1x2 = g1x2 and label the point of intersection Cartesian plane. POUIFHSBQIESBXOJOQBSU B 3 C 'JOEUIFQPJOU T PGJOUFSTFDUJPOPGUIFHSBQITPGf and g e f 105.* B (SBQIf 1x2 = 2x - 4. 2 by solving f 1x2 = g1x2. Label any intersection points C 'JOEUIF[FSPPGf. 2 POUIFHSBQIESBXOJOQBSU B
D #BTFEPOUIFHSBQI TPMWFf 1x2 6 0. 1 - q , 22 q D #BTFEPOUIFHSBQI TPMWFf 1x2 7 g1x2.
106.* B (SBQIg1x2 = 3x - 9. C 'JOEUIF[FSPPGg. 2 D #BTFEPOUIFHSBQI TPMWFg1x2 7 0. 12, q 2 {16}
Applications and Extensions 107. A Population Model 5IFSFTJEFOUQPQVMBUJPOPGUIF6OJUFE States in 2013 was 316 million people and was growing at B SBUF PG QFS ZFBS "TTVNJOH UIBU UIJT HSPXUI SBUF continues, the model P 1t = 31611.00 t - 2013 represents the population P JONJMMJPOTPGQFPQMF JOZFBSt. B "DDPSEJOHUPUIJTNPEFM XIFOXJMMUIFQPQVMBUJPOPGUIF 6OJUFE4UBUFTCFNJMMJPOQFPQMF C "DDPSEJOHUPUIJTNPEFM XIFOXJMMUIFQPQVMBUJPOPGUIF 6OJUFE4UBUFTCFNJMMJPOQFPQMF Source: U.S. Census Bureau
SECTION 4.7 Financial Models
371
108. A Population Model The population of the world in 2013 XBTCJMMJPOQFPQMFBOEXBTHSPXJOHBUBSBUFPGQFS year. Assuming that this growth rate continues, the model P 1t2 = .1311.0112 t - 2013 represents the population P JO CJMMJPOTPGQFPQMF JOZFBSt. B "DDPSEJOHUPUIJTNPEFM XIFOXJMMUIFQPQVMBUJPOPGUIF XPSMECFCJMMJPOQFPQMF 2044 C "DDPSEJOHUPUIJTNPEFM XIFOXJMMUIFQPQVMBUJPOPGUIF XPSMECFCJMMJPOQFPQMF 2062 Source: U.S. Census Bureau 109. Depreciation The value V of a Chevy Cruze LT that is t years old can be modeled by V 1t2 = 1,0010.42 t. B "DDPSEJOH UP UIF NPEFM XIFO XJMM UIF DBS CF XPSUI After 4.2 years C "DDPSEJOH UP UIF NPEFM XIFO XJMM UIF DBS CF XPSUI After 6.5 years D "DDPSEJOH UP UIF NPEFM XIFO XJMM UIF DBS CF XPSUI "GUFSZFBST Source: Kelley Blue Book
110. Depreciation The value V of a Honda Civic LX that is t years old can be modeled by V 1t2 = 1,95510.9052 t. B "DDPSEJOH UP UIF NPEFM XIFO XJMM UIF DBS CF XPSUI "GUFSZFBST C "DDPSEJOH UP UIF NPEFM XIFO XJMM UIF DBS CF XPSUI After 6.4 years D "DDPSEJOH UP UIF NPEFM XIFO XJMM UIF DBS CF XPSUI After 9.3 years Source: Kelley Blue Book
Discussion and Writing 111. Fill in a reason for each step in the following two solutions. Solve: log 3 1x - 12 2 = 2
Solution A
Solution B
log 3 1x - 1 2 = 2
1x - 12 2 = 32 = 9 ______ 1x - 12 = { 3 ______ x - 1 = - 3 or x - 1 = 3 ______
log 3 1x - 12 2 = 2 2 log 3 1x - 12 = 2 ______ log 3 1x - 12 = 1 ______
x - 1 = 31 = 3 ______ x = - 2 or x = 4 ______ x = 4 ______ Both solutions given in Solution A check. Explain what caused the solution x = - 2 to be lost in Solution B.
Retain Your Knowledge Problems 112–115 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. x + 5 x and g x = , find f ∘ g. Then find *112. Solve: 4x3 + 3x2 - 25x + 6 = 0. *114. For f x = x - 2 x - 3 113. Find the domain of the domain of f ∘ g. f 1x2 = 2x + 3 + 2x - 1. {x x Ú 1} 115. %FUFSNJOF XIFUIFS UIF GVODUJPO { 10, - 42, 12, - 22, 14, 02, 16, 22} is one-to-one. one-to-one
‘Are You Prepared?’ Answers 1. { - 3, 10}
2. { - 2, 0}
3. { - 1.43}
4. { - 1.}
4.7 Financial Models PREPARING FOR THIS SECTION Before getting started, review the following: r 4JNQMF*OUFSFTU "QQFOEJY" 4FDUJPO" QQ"m"
Now Work the ‘Are You Prepared?’ problems on page 381.
OBJECTIVES 1 2 3 4
Determine the Future Value of a Lump Sum of Money (p. 372) Calculate Effective Rates of Return (p. 375) Determine the Present Value of a Lump Sum of Money (p. 376) Determine the Rate of Interest or the Time Required to Double a Lump Sum of Money (p. 377)
372
CHAPTER 4 Exponential and Logarithmic Functions
1 Determine the Future Value of a Lump Sum of Money * *OUFSFTUJTNPOFZQBJEGPSUIFVTFPGNPOFZ5IFUPUBMBNPVOUCPSSPXFE XIFUIFS by an individual from a bank in the form of a loan or by a bank from an individual in b UIFGPSNPGBTBWJOHT BDDPVOU JTDBMMFEUIFprincipal. The rate of interest, expressed as a percent, is the amount charged for the use of the principal for a given period of UJNF VTVBMMZPOBZFBSMZ UIBUJT QFSBOOVN CBTJT
THEOREM
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 I = Prt
(1)
*OUFSFTUDIBSHFEBDDPSEJOHUPGPSNVMB JTDBMMFEsimple interest. In problems involving interest, the term payment period is defined as follows: Annually: Semiannually: Quarterly:
Once per year Twice per year Four times per year
Monthly: 12 times per year Daily: 365 times per year*
8IFO UIF JOUFSFTU EVF BU UIF FOE PG B QBZNFOU QFSJPE JT BEEFE UP UIF QSJODJQBM TP UIBU the interest computed at the end of the next payment period is based on this new principal amount 1old principal + interest , the interest is said to have been compounded. Compound interest is interest paid on the principal and on previously earned interest.
EX A MPL E 1
Computing Compound Interest A credit union pays interest of 2% per annum compounded quarterly on a certain savings plan. If $1000 is deposited in such a plan and the interest is left to accumulate, IPXNVDIJTJOUIFBDDPVOUBGUFSZFBS
Solution
6TFUIFTJNQMFJOUFSFTUGPSNVMB I = Prt. The principal P is $1000 and the rate of 1 interest is 2, = 0.02. After the first quarter of a year, the time t is year, so the 4 interest earned is 1 I = Prt = 1$10002 10.022 a b = $5 4 The new principal is P + I = $1000 + $5 = $1005. At the end of the second quarter, the interest on this principal is 1 I = 1$10052 10.022 a b = $5.03 4 At the end of the third quarter, the interest on the new principal of $1005 + $5.03 = $1010.03 is 1 I = 1$1010.032 10.022 a b = $5.05 4 Finally, after the fourth quarter, the interest is 1 I = 1$1015.02 10.022 a b = $5.0 4 After 1 year the account contains $1015.0 + $5.0 = $1020.16. The pattern of the calculations performed in Example 1 leads to a general formula for compound interest. For this purpose, let P represent the principal to be invested at a per annum interest rate r that is compounded n times per year, so the time of each 1 compounding period is years. 'PS DPNQVUJOH QVSQPTFT r is expressed as a n EFDJNBM 5IFJOUFSFTUFBSOFEBGUFSFBDIDPNQPVOEJOHQFSJPEJTHJWFOCZGPSNVMB
r
.PTUCBOLTVTFBEBZiZFBSu8IZEPZPVUIJOLUIFZEP
SECTION 4.7 Financial Models
Interest = principal * rate * time = P # r
#
1 = P n
#
373
r a b n
The amount A after one compounding period is A = P + P
#
r a b = P n
#
a1 +
r b n
After two compounding periods, the amount A, based on the new principal r P # a1 + b , is n A = P # a1 +
r r r r 2 r r b + P # a1 + b a b = P # a1 + b a1 + b = P # a1 + b n n n n n n
5
5 New principal
c
Factor out P # 11 + nr 2
Interest on new principal
After three compounding periods, the amount A is A = P
#
a1 +
r 2 b + P n
#
r 2 r b a b = P n n
a1 +
#
a1 +
r 2 b n
#
a1 +
r b = P n
#
a1 +
r 3 b n
Continuing this way, after nDPNQPVOEJOHQFSJPET ZFBS
UIFBNPVOUA is A = P
#
r n b n
a1 +
Because t years will contain n # t compounding periods, the amount after t years is A = P
THEOREM
Exploration To see the effects of compounding interest monthly on an initial deposit r 12x b with of $1, graph Y1 = a1 + 12 r = 0.06 and r = 0.12 for 0 … x … 30. What is the future value of $1 in 30 years when the interest rate per annum is r = 0.06 (6%)? What is the future value of $1 in 30 years when the interest rate per annum is r = 0.12 (12%)? Does doubling the interest rate double the future value?
NOTE The results shown here differ from those in Example 1 due to rounding. j
EXAM PL E 2
#
a1 +
r nt b n
Compound Interest Formula The amount A after t years due to a principal P invested at an annual interest rate r compounded n times per year is A = P
#
a1 +
r nt b n
(2)
For example, to rework Example 1, use P = $1000, r = 0.02, n = 4 RVBSUFSMZ DPNQPVOEJOH
BOEt = 1 year to obtain
#
= P # a1 +
r nt 0.02 4 1 b A = $1020.15 b = 1000a1 + n 4 *OFRVBUJPO
UIFBNPVOUA is typically referred to as the future value of the account, and P is called the present value.
Now Work
PROBLEM
7
Comparing Investments Using Different Compounding Periods Investing $1000 at an annual rate of 10% compounded annually, semiannually, quarterly, monthly, and daily will yield the following amounts after 1 year: Annual compounding 1n = 12:
Semiannual compounding 1n = 22: 2VBSUFSMZDPNQPVOEJOH 1n = 42:
A = P # 11 + r2 = 1$10002 11 + 0.102 = $1100.00 r 2 A = P # a1 + b 2 = 1$10002 11 + 0.052 2 = $1102.50 r 4 A = P # a1 + b 4 = 1$10002 11 + 0.0252 4 = $1103.1
374
CHAPTER 4 Exponential and Logarithmic Functions
.POUIMZDPNQPVOEJOH 1n = 122:
A = P
#
a1 +
r 12 b 12
= 1$10002 a1 + %BJMZDPNQPVOEJOH 1n = 3652:
A = P
#
a1 +
0.10 12 b = $1104.1 12
r 365 b 365
= 1$10002 a1 +
0.10 365 b = $1105.16 365
r From Example 2, note that the effect of compounding more frequently is that the amount after 1 year is higher: $1000 compounded 4 times a year at 10% results JO DPNQPVOEFE UJNFT B ZFBS BU SFTVMUT JO BOE $1000 compounded 365 times a year at 10% results in $1105.16. This leads to the GPMMPXJOHRVFTUJPO8IBUXPVMEIBQQFOUPUIFBNPVOUBGUFSZFBSJGUIFOVNCFSPG UJNFTUIBUUIFJOUFSFTUJTDPNQPVOEFEXFSFJODSFBTFEXJUIPVUCPVOE Let’s find the answer. Suppose that P is the principal, r is the per annum interest rate, and n is the number of times that the interest is compounded each year. The amount after 1 year is A = P
#
a1 +
r n b n
3FXSJUFUIJTFYQSFTTJPOBTGPMMPXT A = P
#
r n a1 + b = P n
#
n
1 £1 + ≥ = P n r
#
n>r
1 C £1 + ≥ n r
r
S = P c h =
#
c a1 +
r
1 h b d (3) h
n r
Now suppose that the number n of times that the interest is compounded per year n gets larger and larger; that is, suppose that n S q . Then h = S q , and the r FYQSFTTJPOJOCSBDLFUTJOFRVBUJPO FRVBMTe. That is, A S Pe r. r n 5BCMFDPNQBSFT a1 + b , for large values of n, to e r for r = 0.05, r = 0.10, n r n r = 0.15, and r = 1. The larger that n gets, the closer a1 + b gets to e r. No matter n how frequent the compounding, the amount after 1 year has the definite ceiling Pe r.
Table 8
(1 nr )n n = 100
n = 1000
n = 10,000
er
r = 0.05
1.0512580
1.0512698
1.0512710
1.0512711
r = 0.10
1.1051157
1.1051654
1.1051704
1.1051709
r = 0.15
1.1617037
1.1618212
1.1618329
1.1618342
r = 1
2.7048138
2.7169239
2.7181459
2.7182818
8IFOJOUFSFTUJTDPNQPVOEFETPUIBUUIFBNPVOUBGUFSZFBSJTPe r, the interest is said to be compounded continuously.
THEOREM
Continuous Compounding The amount A after t years due to a principal P invested at an annual interest rate r compounded continuously is A = Pe rt
(4)
SECTION 4.7 Financial Models
EXAMPL E 3
375
Using Continuous Compounding The amount A that results from investing a principal P of $1000 at an annual rate r of 10% compounded continuously for a time t of 1 year is A = $1000e 0.10 = 1$10002 11.10512 = $1105.1
Now Work
PROBLEM
r
13
2 Calculate Effective Rates of Return Suppose that you have $1000 and a bank offers to pay 3% annual interest on a TBWJOHTBDDPVOUXJUIJOUFSFTUDPNQPVOEFENPOUIMZ8IBUBOOVBMJOUFSFTUSBUFNVTU be earned for you to have the same amount at the end of the year as if the interest IBE CFFO DPNQPVOEFE BOOVBMMZ PODF QFS ZFBS 5P BOTXFS UIJT RVFTUJPO GJSTU determine the value of the $1000 in the account that earns 3% compounded monthly. A = +1000a1 +
0.03 12 b 12
r n Use A = P a1 + b with P = $1000, r = 0.03, n = 12. n
= $1030.42 4PUIFJOUFSFTUFBSOFEJT6TJOHI = Prt with t = 1, I = +30.42, and P = $1000, the annual simple interest rate is 0.03042 = 3.042,. This interest rate is known as the effective rate of interest. The effective rate of interest is the equivalent annual simple interest rate that would yield the same amount as compounding n times per year, or continuously, after 1 year.
THEOREM
Effective Rate of Interest The effective rate of interest re of an investment earning an annual interest rate r is given by r n Compounding n times per year: re = a1 + b - 1 n r Continuous compounding: re = e - 1
EXAMPL E 4
Computing the Effective Rate of Interest—Which Is the Best Deal? 4VQQPTFZPVXBOUUPCVZBZFBSDFSUJGJDBUFPGEFQPTJU $% :PVWJTJUUISFFCBOLT UP EFUFSNJOF UIFJS $% SBUFT"NFSJDBO &YQSFTT PGGFST ZPV BOOVBM JOUFSFTU compounded monthly, and First Internet Bank offers you 2.20% compounded RVBSUFSMZ %JTDPWFS PGGFST DPNQPVOEFE EBJMZ %FUFSNJOF XIJDI CBOL JT offering the best deal.
Solution
The bank that offers the best deal is the one with the highest effective interest rate. American Express
First Internet Bank
0.022 4 0.0215 12 b - 1 re = a1 + b - 1 re = a1 + 4 12 ≈ 1.0211 - 1 ≈ 1.0221 - 1
Discover
0.0212 365 b - 1 365 ≈ 1.02143 - 1
re = a1 +
= 0.0211
= 0.0221
= 0.02143
= 2.11,
= 2.21,
= 2.143,
The effective rate of interest is highest for First Internet Bank, so First Internet Bank is offering the best deal.
r
Now Work
PROBLEM
23
376
CHAPTER 4 Exponential and Logarithmic Functions
3 Determine the Present Value of a Lump Sum of Money 8IFO QFPQMF JO GJOBODF TQFBL PG UIF iUJNF WBMVF PG NPOFZ u UIFZ BSF VTVBMMZ referring to the present value of money. The present value of A dollars to be received at a future date is the principal that you would need to invest now so that it will grow to A dollars in the specified time period. The present value of money to be received at a future date is always less than the amount to be received, since the amount to CF SFDFJWFE XJMM FRVBM UIF QSFTFOU WBMVF NPOFZ JOWFTUFE OPX plus the interest accrued over the time period. 5IF DPNQPVOE JOUFSFTU GPSNVMB JT VTFE UP EFWFMPQ B GPSNVMB GPS QSFTFOU value. If P is the present value of A dollars to be received after t years at a per annum interest rate r compounded nUJNFTQFSZFBS UIFO CZGPSNVMB
A = P
#
a1 +
To solve for P, divide both sides by a1 + A a1 +
THEOREM
r nt b n
r nt b n
r nt b . The result is n
= P or P = A # a1 +
r -nt b n
Present Value Formulas The present value P of A dollars to be received after t years, assuming a per annum interest rate r compounded n times per year, is P = A
#
a1 +
r -nt b n
(5)
If the interest is compounded continuously, then P = Ae -rt
(6)
5PEFSJWFGPSNVMB
TPMWFGPSNVMB GPSP.
EX AMPL E 5
Computing the Value of a Zero-Coupon Bond "[FSPDPVQPO OPOJOUFSFTUCFBSJOH CPOEDBOCFSFEFFNFEJOZFBSTGPS How much should you be willing to pay for it now if you want a return of B DPNQPVOEFENPOUIMZ C DPNQPVOEFEDPOUJOVPVTMZ
Solution
B 5P GJOE UIF QSFTFOU WBMVF PG VTF GPSNVMB XJUI A = $1000, n = 12, r = 0.0, and t = 10. P = A
#
a1 +
r -nt 0.0 -121102 = $1000a1 + b = $450.52 b n 12
'PSBSFUVSOPGDPNQPVOEFENPOUIMZ QBZGPSUIFCPOE C )FSFVTFGPSNVMB XJUIA = $1000, r = 0.0, and t = 10. P = Ae -rt = $1000e -10.021102 = $496.59 'PSBSFUVSOPGDPNQPVOEFEDPOUJOVPVTMZ QBZGPSUIFCPOE
Now Work
PROBLEM
15
r
SECTION 4.7 Financial Models
377
4 Determine the Rate of Interest or the Time Required to Double a Lump Sum of Money EXAM PL E 6
Rate of Interest Required to Double an Investment 8IBUBOOVBMSBUFPGJOUFSFTUDPNQPVOEFEBOOVBMMZJTOFFEFEJOPSEFSUPEPVCMFBO JOWFTUNFOUJOZFBST
Solution
If P is the principal and P is to double, the amount A will be 2P6TFUIFDPNQPVOE interest formula with n = 1 and t = 5 to find r.
# 2P = P # A = P
r nt b n 11 + r2 5 a1 +
2 = 11 + r2
5
A = 2P, n = 1, t = 5 Divide both sides by P.
5
1 + r = 22
Take the fifth root of each side.
5
r = 22 - 1 ≈ 1.1469 - 1 = 0.1469 5IFBOOVBMSBUFPGJOUFSFTUOFFEFEUPEPVCMFUIFQSJODJQBMJOZFBSTJT
Now Work
EXAM PL E 7
PROBLEM
r
31
Time Required to Double or Triple an Investment B )PX MPOH XJMM JU UBLF GPS BO JOWFTUNFOU UP EPVCMF JO WBMVF JG JU FBSOT DPNQPVOEFEDPOUJOVPVTMZ C )PXMPOHXJMMJUUBLFUPUSJQMFBUUIJTSBUF
Solution
B *GP is the initial investment and P is to double, the amount A will be 2P.6TF GPSNVMB GPSDPOUJOVPVTMZDPNQPVOEFEJOUFSFTUXJUIr = 0.05. Then = = = =
Pe rt Pe 0.05t e 0.05t ln 2 ln 2 t = ≈ 13.6 0.05
A 2P 2 0.05t
A = 2P, r = 0.05 Divide out the P’s. Rewrite as a logarithm. Solve for t.
It will take about 14 years to double the investment. C 5PUSJQMFUIFJOWFTUNFOU MFUA = 3PJOGPSNVMB = = = =
Pe rt Pe 0.05t e 0.05t ln 3 ln 3 t = ≈ 21.9 0.05
A 3P 3 0.05t
A = 3P, r = 0.05 Divide out the P’s. Rewrite as a logarithm. Solve for t.
It will take about 22 years to triple the investment.
Now Work
PROBLEM
35
r
378
CHAPTER 4 Exponential and Logarithmic Functions
4.7 Assess Your Understanding ‘Are You Prepared?’
Answers are given at the end of these exercises. If you get a wrong answer, read the page listed in red.
1. 8IBUJTUIFJOUFSFTUEVFJGJTCPSSPXFEGPSNPOUITBUB TJNQMFJOUFSFTUSBUFPGQFSBOOVN QQ"m" $15
2. If you borrow $5000 and, after 9 months, pay off the loan in the amount of $5500, what per annum rate of interest was DIBSHFE QQ"m" 131% 3
Concepts and Vocabulary 3. 5IFUPUBMBNPVOUCPSSPXFE XIFUIFSCZBOJOEJWJEVBMGSPN a bank in the form of a loan or by a bank from an individual JOUIFGPSNPGBTBWJOHTBDDPVOU JTDBMMFEUIF principal.
5. In working problems involving interest, if the payment period of the interest is quarterly, then interest is paid 4 times per year.
4. 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 I = Prt . Interest charged according to this formula is called simple interest.
6. The effective rate of interest is the equivalent annual simple interest rate that would yield the same amount as compounding n times per year, or continuously, after 1 year.
Skill Building In Problems 7–14, find the amount that results from each investment. 7. $100 invested at 4% compounded quarterly after a period of 2 years
11. $600 invested at 5% compounded daily after a period of 3 years
8. $50 invested at 6% compounded monthly after a period of 3 years
12. JOWFTUFE BU DPNQPVOEFE EBJMZ BGUFS B QFSJPE PG 2 years
9. JOWFTUFEBUDPNQPVOEFERVBSUFSMZBGUFSBQFSJPEPG 1 2 years $609.50 2 10. $300 invested at 12% compounded monthly after a period 1 of 1 years 2
13. $1000 invested at 11% compounded continuously after a period of 2 years 14. JOWFTUFE BU DPNQPVOEFE DPOUJOVPVTMZ BGUFS B period of 3 years
In Problems 15–22, find the principal needed now to get each amount; that is, find the present value. 15. To get $100 after 2 years at 6% compounded monthly
19. To get $600 after quarterly $554.09
2
years
at
4%
compounded
16. 5PHFUBGUFSZFBSTBUDPNQPVOEFERVBSUFSMZ $59.14 20. To get $300 after 4 years at 3% compounded daily 1 17. To get $1000 after 2 years at 6% compounded daily 1 2 *21. 5PHFUBGUFS3 years at 9% compounded continuously 4 1 *18. 5PHFUBGUFS3 yearsBUDPNQPVOEFENPOUIMZ 1 2 *22. 5PHFUBGUFS2 yearsBUDPNQPVOEFEDPOUJOVPVTMZ 2 In Problems 23–26, find the effective rate of interest. 23. For 5% compounded quarterly 5.095% 24. For 6% compounded monthly
25. For 5% compounded continuously 26. For 6% compounded continuously
In Problems 27–30, determine the rate that represents the better deal. 1 *27. 6% compounded quarterly or 6 , compounded annually 4 1 *28. 9% compounded quarterly or 9 , compounded annually 4 *29. DPNQPVOEFENPOUIMZPSDPNQPVOEFEEBJMZ *30. DPNQPVOEFETFNJBOOVBMMZPSDPNQPVOEFEEBJMZ 31. 8IBU SBUF PG JOUFSFTU DPNQPVOEFE BOOVBMMZ JT SFRVJSFE UP EPVCMFBOJOWFTUNFOUJOZFBST 25.992% 32. 8IBU SBUF PG JOUFSFTU DPNQPVOEFE BOOVBMMZ JT SFRVJSFE UP EPVCMFBOJOWFTUNFOUJOZFBST 12.246% 33. 8IBU SBUF PG JOUFSFTU DPNQPVOEFE BOOVBMMZ JT SFRVJSFE UP USJQMFBOJOWFTUNFOUJOZFBST 34. 8IBU SBUF PG JOUFSFTU DPNQPVOEFE BOOVBMMZ JT SFRVJSFE UP USJQMFBOJOWFTUNFOUJOZFBST 11.612%
35.* B )PX MPOH EPFT JU UBLF GPS BO JOWFTUNFOU UP EPVCMF JO WBMVFJGJUJTJOWFTUFEBUDPNQPVOEFENPOUIMZ C )PX MPOH EPFT JU UBLF JG UIF JOUFSFTU JT DPNQPVOEFE DPOUJOVPVTMZ ≈.66 yr 36.* B )PX MPOH EPFT JU UBLF GPS BO JOWFTUNFOU UP USJQMF JO WBMVFJGJUJTJOWFTUFEBUDPNQPVOEFENPOUIMZ C )PX MPOH EPFT JU UBLF JG UIF JOUFSFTU JT DPNQPVOEFE DPOUJOVPVTMZ ≈1.31 yr 37. 8IBU SBUF PG JOUFSFTU DPNQPVOEFE RVBSUFSMZ XJMM ZJFME BO FGGFDUJWFJOUFSFTUSBUFPG 38. 8IBU SBUF PG JOUFSFTU DPNQPVOEFE DPOUJOVPVTMZ XJMM ZJFME BOFGGFDUJWFJOUFSFTUSBUFPG
%VFUPTQBDFSFTUSJDUJPOT BOTXFSTUPUIFTFFYFSDJTFTNBZCFGPVOEJOUIF"OTXFSTJOUIFCBDLPGUIFCPPL
SECTION 4.7 Financial Models
379
Applications and Extensions *51. Comparing IRA Investments 8JMM JOWFTUT JO IJT *3" JO B CPOE USVTU UIBU QBZT JOUFSFTU DPNQPVOEFE TFNJBOOVBMMZ )JT GSJFOE )FOSZ JOWFTUT JO IJT *3" 1 in a certificate of deposit that pays , compounded 2 DPOUJOVPVTMZ8IP IBT NPSF NPOFZ BGUFS ZFBST 8JMM PS )FOSZ
39. Time Required to Reach a Goal If Tanisha has $100 to JOWFTUBUQFSBOOVNDPNQPVOEFENPOUIMZ IPXMPOHXJMM JUCFCFGPSFTIFIBT *GUIFDPNQPVOEJOHJTDPOUJOVPVT IPXMPOHXJMMJUCF ZS ZS 40. Time Required to Reach a Goal If Angela has $100 to invest at 10% per annum compounded monthly, how long will it be CFGPSFTIFIBT *GUIFDPNQPVOEJOHJTDPOUJOVPVT IPX MPOHXJMMJUCF ZS ZS
*52. Comparing Two Alternatives Suppose that April has access to an investment that will pay 10% interest compounded DPOUJOVPVTMZ8IJDIJTCFUUFSUPCFHJWFOOPXTPUIBU she can take advantage of this investment opportunity or to CFHJWFOBGUFSZFBST
*41. Time Required to Reach a Goal How many years will it UBLFGPSBOJOJUJBMJOWFTUNFOUPG UPHSPXUP Assume a rate of interest of 6% compounded continuously. *42. Time Required to Reach a Goal How many years will it UBLFGPSBOJOJUJBMJOWFTUNFOUPG UPHSPXUP "TTVNFBSBUFPGJOUFSFTUPGDPNQPVOEFEDPOUJOVPVTMZ
53. College Costs The average annual cost of college at 4-year QSJWBUF DPMMFHFT XBT JO UIF m BDBEFNJD year. This was a 4.2% increase from the previous year. Source: The College Board B *GUIFDPTUPGDPMMFHFJODSFBTFTCZFBDIZFBS XIBU will be the average cost of college at a 4-year private DPMMFHFGPSUIFmBDBEFNJDZFBS $60,933 C $PMMFHF TBWJOHT QMBOT TVDI BT B QMBO BMMPX individuals to put money aside now to help pay for college later. If one such plan offers a rate of 4% compounded continuously, how much should be put in a college savings plan in 2015 to pay for 1 year of the cost of college at a 4-year private college for an incoming GSFTINBOJO $33,441
43. Price Appreciation of Homes 8IBU XJMM B condominium cost 5 years from now if the price appreciation for condos over that period averages 3% compounded BOOVBMMZ $104,335 44. Credit Card Interest A department store charges 1.25% per month on the unpaid balance for customers with charge BDDPVOUT JOUFSFTU JT DPNQPVOEFE NPOUIMZ " DVTUPNFS DIBSHFTBOEEPFTOPUQBZIFSCJMMGPSNPOUIT8IBUJT UIFCJMMBUUIBUUJNF 45. Saving for a Car Jerome will be buying a used car for $15,000 in 3 years. How much money should he ask his parents for now so that, if he invests it at 5% compounded continuously, IFXJMMIBWFFOPVHIUPCVZUIFDBS $12,910.62
*54. Analyzing Interest Rates on a Mortgage Colleen and Bill have just purchased a house for $650,000, with the seller holding a second mortgage of $100,000. They promise to pay the seller $100,000 plus all accrued interest 5 years from now. The seller offers them three interest options on the second mortgage: B 4JNQMFJOUFSFTUBUQFSBOOVN 1 C 11 , interest compounded monthly 2 1 D 11 , interest compounded continuously 4 8IJDI PQUJPO JT CFTU 5IBU JT XIJDI SFTVMUT JO QBZJOH UIF MFBTUJOUFSFTUPOUIFMPBO
46. Paying off a Loan John requires $3000 in 6 months to pay off a loan that has no prepayment privileges. If he has the $3000 now, how much of it should he save in an account paying 3% compounded monthly so that in 6 months he will IBWFFYBDUMZ $2955.39 47. Return on a Stock George contemplates the purchase of 100 shares of a stock selling for $15 per share. The stock pays no dividends. The history of the stock indicates that it should grow at an annual rate of 15% per year. How much should UIFTIBSFTPGTUPDLCFXPSUIJOZFBST ≈$301 48. Return on an Investment A business purchased for $650,000 JOJTTPMEJOGPS 8IBUJTUIFBOOVBMSBUF PGSFUVSOGPSUIJTJOWFTUNFOU 9.35%
55. 2009 Federal Stimulus Package *O'FCSVBSZ 1SFTJEFOU 0CBNBTJHOFEJOUPMBXBCJMMJPOGFEFSBMTUJNVMVTQBDLBHF At that time, 20-year Series EE bonds had a fixed rate of 1.3% compounded semiannually. If the federal government financed the stimulus through EE bonds, how much would JUIBWFUPQBZCBDLJO )PXNVDIJOUFSFTUXBTQBJEUP GJOBODFUIFTUJNVMVT ≈$1020 billion; $233 billion Source: U.S. Treasury Department
*49. Comparing Savings Plans Jim places $1000 in a bank account that pays 5.6% compounded continuously. After 1 year, will he have enough money to buy a computer TZTUFNUIBUDPTUT *GBOPUIFSCBOLXJMMQBZ+JN DPNQPVOEFENPOUIMZ JTUIJTBCFUUFSEFBM 50. Savings Plans 0O+BOVBSZ ,JNQMBDFTJOBDFSUJGJDBUF PG EFQPTJU UIBU QBZT DPNQPVOEFE DPOUJOVPVTMZ BOE NBUVSFT JO NPOUIT5IFO ,JN QMBDFT UIF BOE UIF interest in a passbook account that pays 5.25% compounded NPOUIMZ)PXNVDIEPFT,JNIBWFJOUIFQBTTCPPLBDDPVOU PO.BZ $1021.60
56. Per Capita Federal Debt In 2013, the federal debt was about USJMMJPO*O UIF64QPQVMBUJPOXBTBCPVUNJMMJPO "TTVNJOHUIBUUIFGFEFSBMEFCUJTJODSFBTJOHBCPVUQFS ZFBS BOE UIF 64 QPQVMBUJPO JT JODSFBTJOH BCPVU QFS ZFBS EFUFSNJOF UIF QFS DBQJUB EFCU UPUBM EFCU EJWJEFE CZ QPQVMBUJPO JOSPVOEFEUPUIFOFBSFTUEPMMBS $90,646
Inflation Problems 57–62 require the following discussion. Inflation is a term used to describe the erosion of the purchasing power of money. For example, if the annual inflation rate is 3%, then $1000 worth of purchasing power now will have only $970 worth of purchasing power in 1 year because 3% of the original $1000 (0.03 * 1000 = 30) has been eroded due to inflation. In general, if the rate of inflation averages r per annum over n years, the amount A that $P will purchase after n years is A = P where r is expressed as a decimal.
#
11 - r2 n
380
CHAPTER 4 Exponential and Logarithmic Functions
57. Inflation If the inflation rate averages 3%, how much will QVSDIBTFJOZFBST $940.90
60. Inflation If the amount that $1000 will purchase is only $930 BGUFSZFBST XIBUXBTUIFBWFSBHFJOGMBUJPOSBUF 3.56%
58. Inflation If the inflation rate averages 2%, how much will QVSDIBTFJOZFBST $941.19
61. Inflation If the average inflation rate is 2%, how long is it VOUJMQVSDIBTJOHQPXFSJTDVUJOIBMG 34.31 yr
59. Inflation If the amount that $1000 will purchase is only $950 BGUFSZFBST XIBUXBTUIFBWFSBHFJOGMBUJPOSBUF 2.53%
62. Inflation If the average inflation rate is 4%, how long is it VOUJMQVSDIBTJOHQPXFSJTDVUJOIBMG ZS
Problems 63–66 involve zero-coupon bonds. A zero-coupon bond is a bond that is sold now at a discount and will pay its face value at the time when it matures; no interest payments are made. 63. Zero-Coupon Bonds A zero-coupon bond can be redeemed in 20 years for $10,000. How much should you be willing to pay for it now if you want a return of: B DPNQPVOEFENPOUIMZ $1364.62 C DPNQPVOEFEDPOUJOVPVTMZ $1353.35 64. Zero-Coupon Bonds A child’s grandparents are considering buying a $40,000 face-value, zero-coupon bond at her birth TP UIBU TIF XJMM IBWF NPOFZ GPS IFS DPMMFHF FEVDBUJPO ZFBSTMBUFS*GUIFZXBOUBSBUFPGSFUVSOPGDPNQPVOEFE BOOVBMMZ XIBUTIPVMEUIFZQBZGPSUIFCPOE *65. Zero-Coupon Bonds How much should a $10,000 facevalue, zero-coupon bond, maturing in 10 years, be sold for OPXJGJUTSBUFPGSFUVSOJTUPCFDPNQPVOEFEBOOVBMMZ 66. Zero-Coupon Bonds *G 1BU QBZT GPS B GBDFWBMVF [FSPDPVQPOCPOEUIBUNBUVSFTJOZFBST XIBU JTIJTBOOVBMSBUFPGSFUVSO 67. Time to Double or Triple an Investment The formula ln m
t =
n lna1 +
r b n
can be used to find the number of years t required to multiply an investment m times when r is the per annum interest rate compounded n times a year. * B )PXNBOZZFBSTXJMMJUUBLFUPEPVCMFUIFWBMVFPGBO *3"UIBUDPNQPVOETBOOVBMMZBUUIFSBUFPG C )PX NBOZ ZFBST XJMM JU UBLF UP USJQMF UIF WBMVF PG B savings account that compounds quarterly at an annual SBUFPG ZS * D (JWFBEFSJWBUJPOPGUIJTGPSNVMB 68. Time to Reach an Investment Goal The formula t =
ln A - ln P r
can be used to find the number of years t required for an investment P to grow to a value A when compounded continuously at an annual rate r. B )PXMPOHXJMMJUUBLFUPJODSFBTFBOJOJUJBMJOWFTUNFOUPG UPBUBOBOOVBMSBUFPG ZS C 8IBUBOOVBMSBUFJTSFRVJSFEUPJODSFBTFUIFWBMVFPGB *3"UP JOZFBST * D (JWFBEFSJWBUJPOPGUIJTGPSNVMB
Problems 69–72 require the following discussion. The Consumer Price Index (CPI) indicates the relative change in price over time for a fixed basket of goods and services. It is a cost-of-living index that helps measure the effect of inflation on the cost of goods and services. The CPI uses the base period 1982–1984 for comparison (the CPI for this period is 100). The CPI for January 2013 was 230.28. This means that $100 in the period 1982–1984 had the same purchasing power as $230.28 in January 2013. In general, if the rate of inflation averages r percent per annum over n years, then the CPI index after n years is C1I = C1I 0 a1 +
r n b 100
where CPI0 is the CPI index at the beginning of the n-year period. Source: U.S. Bureau of Labor Statistics 69. Consumer Price Index B 5IF $1* XBT GPS BOE GPS Assuming that annual inflation remained constant for this time period, determine the average annual inflation rate. C 6TJOHUIFJOGMBUJPOSBUFGSPNQBSU B
JOXIBUZFBSXJMM UIF$1*SFBDI In 2023 70. Consumer Price Index *G UIF DVSSFOU $1* JT BOE UIF BWFSBHFBOOVBMJOGMBUJPOSBUFJT XIBUXJMMCFUIF$1* JOZFBST
71. Consumer Price Index If the average annual inflation rate is IPXMPOHXJMMJUUBLFGPSUIF$1*JOEFYUPEPVCMF " EPVCMJOHPGUIF$1*JOEFYNFBOTQVSDIBTJOHQPXFSJTDVUJO IBMG ZS 72. Consumer Price Index5IFCBTFQFSJPEGPSUIF$1*DIBOHFE JO6OEFSUIFQSFWJPVTXFJHIUBOEJUFNTUSVDUVSF UIF $1*GPSXBT*GUIFBWFSBHFBOOVBMJOGMBUJPOSBUF XBT XIBU ZFBS XBT VTFE BT UIF CBTF QFSJPE GPS UIF $1*
Discussion and Writing 73. Explain in your own words what the term compound interest NFBOT8IBUEPFTcontinuous compoundingNFBO 74. Explain in your own words the meaning of present value. 75. Critical Thinking You have just contracted to buy a house and will seek financing in the amount of $100,000. You go to several banks. Bank 1 will lend you $100,000
BU UIF SBUF PG BNPSUJ[FE PWFS ZFBST XJUI B MPBO PSJHJOBUJPO GFF PG #BOL XJMM MFOE ZPV BU UIF SBUF PG BNPSUJ[FEPWFSZFBST XJUI BMPBO origination fee of 1.5%. Bank 3 will lend you $100,000 at the rate of 9.125% amortized over 30 years with no loan origination fee. Bank 4 will lend you $100,000 at the rate of BNPSUJ[FE PWFS ZFBST XJUI OP MPBO PSJHJOBUJPO
SECTION 4.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models
GFF 8IJDI MPBO XPVME ZPV UBLF 8IZ #F TVSF UP IBWF TPVOE SFBTPOT GPS ZPVS DIPJDF 6TF UIF JOGPSNBUJPO JO UIF table to assist you. If the amount of the monthly payment EJEOPUNBUUFSUPZPV XIJDIMPBOXPVMEZPVUBLF "HBJO have sound reasons for your choice. Compare your final EFDJTJPOXJUIPUIFSTJOUIFDMBTT%JTDVTT
381
Monthly Payment
Loan Origination Fee
Bank 1
$786.70
$1,750.00
Bank 2
$977.42
$1,500.00
Bank 3
$813.63
$0.00
Bank 4
$992.08
$0.00
Retain Your Knowledge Problems 76–79 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. x 76. Find the remainder R when f x = 6x3 + 3x2 + 2x - 11 is *78. The function f x = is one-to-one. Find f -1. x 2 divided by g x = x - 1. Is g a factor of f R = 0; yes 79. Solve: log 2 x + 3 = 2 log 2 x - 3 {6} *77. Find the real zeros of f x = x5 - x4 - 15x3 - 21x2 - 16x - 20. Then write f in factored form.
‘Are You Prepared?’ Answers 1 2. 13 % 3
1. $15
4.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models OBJECTIVES 1 Find Equations of Populations That Obey the Law of Uninhibited Growth (p. 381) 2 Find Equations of Populations That Obey the Law of Decay (p. 383) 3 Use Newton’s Law of Cooling (p. 384) 4 Use Logistic Models (p. 386)
1 Find Equations of Populations That Obey the Law of Uninhibited Growth Figure 44
.BOZ OBUVSBM QIFOPNFOB IBWF CFFO GPVOE UP GPMMPX UIF MBX UIBU BO BNPVOU A varies with time t according to the function
A
A 1t2 = A0e kt A0 t (a) A(t ) ⫽ A0 e k t , k ⬎ 0 A A0
t (b) A(t ) ⫽ A0 e kt , k ⬍ 0
(1)
Here A0 is the original amount 1t = 02 and k ≠ 0 is a constant. If k 7 0, UIFO FRVBUJPO TUBUFT UIBU UIF BNPVOU A is increasing over time; if k 6 0, the amount A is decreasing over time. In either case, when an amount A WBSJFTPWFSUJNFBDDPSEJOHUPFRVBUJPO
JUJTTBJEUPGPMMPXUIFexponential law, or the law of uninhibited growth 1k 7 02 or decay 1k 6 02. See Figure 44. 'PS FYBNQMF BT TFFO JO 4FDUJPO DPOUJOVPVTMZ DPNQPVOEFE JOUFSFTU was shown to follow the law of uninhibited growth. In this section, additional phenomena that follow the exponential law will be studied. Cell division is the growth process of many living organisms, such as amoebas, plants, and human skin cells. Based on an ideal situation in which no cells die and no by-products are produced, the number of cells present at a given time follows the law of uninhibited growth. Actually, however, after enough time has passed, growth at an exponential rate will cease due to the influence of factors such as lack of living space and dwindling food supply. The law of uninhibited growth accurately models only the early stages of the cell division process. The cell division process begins with a culture containing N0 cells. Each cell in the culture grows for a certain period of time and then divides into two identical
382
CHAPTER 4 Exponential and Logarithmic Functions
cells. Assume that the time needed for each cell to divide in two is constant and does not change as the number of cells increases. These new cells then grow, and eventually each divides in two, and so on.
Uninhibited Growth of Cells A model that gives the number N of cells in a culture after a time t has passed JOUIFFBSMZTUBHFTPGHSPXUI JT N 1t2 = N0e kt
k 7 0
(2)
where N0 is the initial number of cells and k is a positive constant that represents the growth rate of the cells. 6TJOHGPSNVMB UPNPEFMUIFHSPXUIPGDFMMTFNQMPZTBGVODUJPOUIBUZJFMET positive real numbers, even though the number of cells being counted must be an integer. This is a common practice in many applications.
EX A MPL E 1
Bacterial Growth A colony of bacteria grows according to the law of uninhibited growth according to the function N 1t2 = 100e 0.045t, where N is measured in grams and t is measured in days. B %FUFSNJOFUIFJOJUJBMBNPVOUPGCBDUFSJB C 8IBUJTUIFHSPXUISBUFPGUIFCBDUFSJB D 8IBUJTUIFQPQVMBUJPOBGUFSEBZT E )PXMPOHXJMMJUUBLFGPSUIFQPQVMBUJPOUPSFBDIHSBNT F 8IBUJTUIFEPVCMJOHUJNFGPSUIFQPQVMBUJPO
Solution
B 5IFJOJUJBMBNPVOUPGCBDUFSJB N0 , is obtained when t = 0, so N0 = N 102 = 100e 0.045102 = 100 grams
C $PNQBSFN 1t2 = 100e 0.045t to N 1t2 = N0e kt. The value of k, 0.045, indicates a growth rate of 4.5%. D 5IFQPQVMBUJPOBGUFSEBZTJTN 152 = 100e 0.045152 ≈ 125.2 grams. E 5P àOE IPX MPOH JU UBLFT GPS UIF QPQVMBUJPO UP SFBDI HSBNT TPMWF UIF equation N 1t2 = 140. 100e 0.045t = 140 e 0.045t = 1.4 0.045t = ln 1.4 ln 1.4 t = 0.045 ≈ .5 days
Divide both sides of the equation by 100. Rewrite as a logarithm. Divide both sides of the equation by 0.045.
5IFQPQVMBUJPOSFBDIFTHSBNTJOBCPVUEBZT F 5IFQPQVMBUJPOEPVCMFTXIFON 1t2 = 200 grams, so the doubling time can be found by solving the equation 200 = 100e 0.045t for t. 200 = 100e 0.045t 2 = e 0.045t ln 2 = 0.045t ln 2 t = 0.045 ≈ 15.4 days
Divide both sides of the equation by 100. Rewrite as a logarithm. Divide both sides of the equation by 0.045.
The population doubles approximately every 15.4 days.
Now Work
PROBLEM
1
r
SECTION 4.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models
EXAMPL E 2
383
Bacterial Growth A colony of bacteria increases according to the law of uninhibited growth. B *GN is the number of cells and t is the time in hours, express N as a function of t. C *G UIF OVNCFS PG CBDUFSJB EPVCMFT JO IPVST GJOE UIF GVODUJPO UIBU HJWFT UIF number of cells in the culture. D )PXMPOHXJMMJUUBLFGPSUIFTJ[FPGUIFDPMPOZUPUSJQMF E )PXMPOHXJMMJUUBLFGPSUIFQPQVMBUJPOUPEPVCMFBTFDPOEUJNF UIBUJT JODSFBTF GPVSUJNFT
Solution
B 6TJOHGPSNVMB
UIFOVNCFSN of cells at time t is N 1t2 = N0e kt
where N0 is the initial number of bacteria present and k is a positive number. C 5PàOEUIFHSPXUISBUFk, note that the number of cells doubles in 3 hours. Hence But N 132 = N0e
N 132 = 2N0
k132
, so
N0e k132 = 2N0 e 3k = 2 Divide both sides by N0 3k = ln 2 Write the exponential equation as a logarithm. 1 k = ln 2 ≈ 0.23105 3 The function that models this growth process is therefore N 1t2 = N0e 0.23105t D 5IFUJNFt needed for the size of the colony to triple requires that N = 3N0 . Substitute 3N0 for N to get 3N0 = N0e 0.23105t Divide both sides by N0 3 = e 0.23105t Write the exponential equation as a logarithm. 0.23105t = ln 3 ln 3 t = ≈ 4.55 hours 0.23105 *UXJMMUBLFBCPVUIPVSTPSIPVST NJOVUFTGPSUIFTJ[FPGUIFDPMPOZ to triple. E *GBQPQVMBUJPOEPVCMFTJOIPVST JUXJMMEPVCMFBTFDPOEUJNFJONPSFIPVST for a total time of 6 hours.
r
2 Find Equations of Populations That Obey the Law of Decay 3BEJPBDUJWFNBUFSJBMTGPMMPXUIFMBXPGVOJOIJCJUFEEFDBZ
Uninhibited Radioactive Decay The amount A of a radioactive material present at time t is given by A 1t2 = A0e kt
k 6 0
(3)
where A0 is the original amount of radioactive material and k is a negative number that represents the rate of decay. All radioactive substances have a specific half-life, which is the time required for half of the radioactive substance to decay. Carbon dating uses the fact that all living PSHBOJTNTDPOUBJOUXPLJOETPGDBSCPO DBSCPO BTUBCMFDBSCPO BOEDBSCPO
384
CHAPTER 4 Exponential and Logarithmic Functions
BSBEJPBDUJWFDBSCPOXJUIBIBMGMJGFPGZFBST 8IJMFBOPSHBOJTNJTMJWJOH UIF ratio of carbon 12 to carbon 14 is constant. But when an organism dies, the original amount of carbon 12 present remains unchanged, whereas the amount of carbon 14 begins to decrease. This change in the amount of carbon 14 present relative to the amount of carbon 12 present makes it possible to calculate when the organism died.
EX AMPL E 3
Estimating the Age of Ancient Tools Traces of burned wood along with ancient stone tools in an archeological dig in Chile XFSFGPVOEUPDPOUBJOBQQSPYJNBUFMZPGUIFPSJHJOBMBNPVOUPGDBSCPO*GUIF IBMGMJGFPGDBSCPOJTZFBST BQQSPYJNBUFMZXIFOXBTUIFUSFFDVUBOECVSOFE
Solution
6TJOHGPSNVMB
UIFBNPVOUA of carbon 14 present at time t is A 1t2 = A0e kt
where A0 is the original amount of carbon 14 present and kJTBOFHBUJWFOVNCFS8F first seek the number k5PàOEJU XFVTFUIFGBDUUIBUBGUFSZFBST IBMGPGUIF 1 original amount of carbon 14 remains, so A 15302 = A0 . Then 2 1 A = A0e k15302 2 0 1 Divide both sides of the equation by A 0. = e 530k 2 1 Rewrite as a logarithm. 530k = ln 2 1 1 k = ln ≈ - 0.00012096 530 2 'PSNVMB
UIFSFGPSF CFDPNFT A 1t2 = A0e - 0.00012096t
If the amount APGDBSCPOOPXQSFTFOUJTPGUIFPSJHJOBMBNPVOU JU follows that 0.016A0 = A0e - 0.00012096t Divide both sides of the equation by A 0. 0.016 = e - 0.00012096t Rewrite as a logarithm. - 0.00012096t = ln 0.016 ln 0.016 t = ≈ 33,30 years - 0.00012096 5IF USFF XBT DVU BOE CVSOFE BCPVU ZFBST BHP 4PNF BSDIFPMPHJTUT VTF UIJT conclusion to argue that humans lived in the Americas nearly 34,000 years ago, much earlier than is generally accepted.
Now Work
r
PROBLEM
3
3 Use Newton’s Law of Cooling Newton’s Law of Cooling* states that the temperature of a heated object decreases exponentially over time toward the temperature of the surrounding medium.
Newton’s Law of Cooling The temperature u of a heated object at a given time t can be modeled by the following function: u 1t2 = T + 1u 0 - T2e kt
k 6 0
(4)
where T is the constant temperature of the surrounding medium, u 0 is the initial temperature of the heated object, and k is a negative constant. /BNFEBGUFS4JS*TBBD/FXUPO m
POFPGUIFDPGPVOEFSTPGDBMDVMVT
SECTION 4.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models
EXAM PL E 4
385
Using Newton’s Law of Cooling "OPCKFDUJTIFBUFEUP$ EFHSFFT$FMTJVT BOEJTUIFOBMMPXFEUPDPPMJOBSPPN whose air temperature is 30°C. B *GUIFUFNQFSBUVSFPGUIFPCKFDUJT$BGUFSNJOVUFT XIFOXJMMJUTUFNQFSBUVSF CF$ C %FUFSNJOFUIFFMBQTFEUJNFCFGPSFUIFUFNQFSBUVSFPGUIFPCKFDUJT$ D 8IBUEPZPVOPUJDFBCPVUUIFUFNQFSBUVSFBTUJNFQBTTFT
Solution
B 6TJOHGPSNVMB XJUIT = 30 and u 0 = 100, the temperature u t JOEFHSFFT $FMTJVT PGUIFPCKFDUBUUJNFt JONJOVUFT JT u 1t2 = 30 + 1100 - 302e kt = 30 + 0e kt
where k is a negative constant. To find k, use the fact that u = 0 when t = 5. Then u 1t2 = 30 + 0e kt 0 = 30 + 0e k152 50 = 0e 5k 50 e 5k = 0 5 5k = ln 1 5 k = ln ≈ - 0.063 5
u(5) = 80 Simplify. Solve for e 5k. Take ln of both sides. Solve for k.
'PSNVMB
UIFSFGPSF CFDPNFT u t = 30 + 0e - 0.063t
(5)
Find t when u = 50°C. 50 = 30 + 0e - 0.063t 20 = 0e - 0.063t 20 e - 0.063t = 0 2 - 0.063t = ln 2 ln t = ≈ 1.6 minutes - 0.063
Simplify.
Take ln of both sides.
Solve for t.
5IFUFNQFSBUVSFPGUIFPCKFDUXJMMCF$BGUFSBCPVUNJOVUFT PSNJOVUFT 36 seconds. C 6TFFRVBUJPO UPàOEt when u = 35°C. 35 = 30 + 0e - 0.063t 5 = 0e - 0.063t 5 e - 0.063t = 0 5 - 0.063t = ln 0 5 ln 0 t = ≈ 39.2 minutes - 0.063
Simplify.
Take ln of both sides.
Solve for t.
The object will reach a temperature of 35°C after about 39.2 minutes. D -PPLBUFRVBUJPO "Tt increases, the exponent - 0.063t becomes unbounded in the negative direction. As a result, the value of e - 0.063t approaches zero, so the
386
CHAPTER 4 Exponential and Logarithmic Functions
value of u, the temperature of the object, approaches 30°C, the air temperature of the room.
r
Now Work
PROBLEM
13
4 Use Logistic Models
The exponential growth model A 1t2 = A0e kt, k 7 0, assumes uninhibited growth, NFBOJOHUIBUUIFWBMVFPGUIFGVODUJPOHSPXTXJUIPVUMJNJU3FDBMMUIBUDFMMEJWJTJPO could be modeled using this function, assuming that no cells die and no by-products are produced. However, cell division eventually is limited by factors such as living space and food supply. The logistic model, given next, can describe situations where the growth or decay of the dependent variable is limited.
Logistic Model In a logistic model, the population P after time t is given by the function P1t2 =
c 1 + ae -bt
(6)
where a, b, and c are constants with a 7 0 and c 7 0. The model is a growth model if b 7 0; the model is a decay model if b 6 0. The number c is called the carrying capacity GPSHSPXUINPEFMT CFDBVTFUIF value P1t2 approaches c as t approaches infinity; that is, lim P1t2 = c. The number t Sq 0 b 0 is the growth rate for b 7 0 and the decay rate for b 6 0.'JHVSF B TIPXT UIFHSBQIPGBUZQJDBMMPHJTUJDHSPXUIGVODUJPO BOE'JHVSF C TIPXTUIFHSBQIPG a typical logistic decay function. Figure 45
P(t )
yc
P(t )
yc
(0, P(0))
1– c 2
1– c 2
Inflection point
Inflection point
(0, P(0)) t
t (a)
(b)
From the figures, the following properties of logistic growth functions emerge. Properties of the Logistic Model, Equation (6) 1. The domain is the set of all real numbers. The range is the interval 10, c2, where c is the carrying capacity. 2. There are no x-intercepts; the y-intercept is P102. 3. There are two horizontal asymptotes: y = 0 and y = c. 4. P1t2 is an increasing function if b 7 0 and a decreasing function if b 6 0. 1 5. There is an inflection point where P1t2 equals of the carrying capacity. 2 The inflection point is the point on the graph where the graph changes from being curved upward to being curved downward for growth functions, and the point where the graph changes from being curved downward to being curved upward for decay functions. 6. The graph is smooth and continuous, with no corners or gaps.
SECTION 4.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models
EXAM PL E 5
387
Fruit Fly Population 'SVJUGMJFTBSFQMBDFEJOBIBMGQJOUNJMLCPUUMFXJUIBCBOBOB GPSGPPE BOEZFBTU QMBOUT GPS GPPE BOE UP QSPWJEF B TUJNVMVT UP MBZ FHHT 4VQQPTF UIBU UIF GSVJU GMZ population after t days is given by 230 P 1t2 = 1 + 56.5e -0.3t B 4UBUFUIFDBSSZJOHDBQBDJUZBOEUIFHSPXUISBUF C %FUFSNJOFUIFJOJUJBMQPQVMBUJPO D 8IBUJTUIFQPQVMBUJPOBGUFSEBZT E )PXMPOHEPFTJUUBLFGPSUIFQPQVMBUJPOUPSFBDI F 6TFBHSBQIJOHVUJMJUZUPEFUFSNJOFIPXMPOHJUUBLFTGPSUIFQPQVMBUJPOUPSFBDI one-half of the carrying capacity.
Solution
230 . The carrying capacity of the half-pint 1 bottle is 230 fruit flies. The growth rate is 0 b 0 = 0 0.3 0 = 3, per day. C 5PàOEUIFJOJUJBMOVNCFSPGGSVJUáJFTJOUIFIBMGQJOUCPUUMF FWBMVBUFP102.
B "T t S q , e -0.3t S 0 and P1t2 S
P102 =
230
1 + 56.5e -0.3102 230 = 1 + 56.5 = 4
Thus, initially, there were 4 fruit flies in the half-pint bottle. D 5PàOEUIFOVNCFSPGGSVJUáJFTJOUIFIBMGQJOUCPUUMFBGUFSEBZT FWBMVBUFP152. P152 =
230
≈ 23 fruit flies 1 + 56.5e -0.3152 After 5 days, there are approximately 23 fruit flies in the bottle. E 5PEFUFSNJOFXIFOUIFQPQVMBUJPOPGGSVJUáJFTXJMMCF TPMWFUIFFRVBUJPO P1t2 = 10. 230 1 + 56.5e -0.3t 230 1.2 0.2 0.0049 ln 10.00492 t
Figure 46
= 10 = = = = = ≈
1011 + 56.5e -0.3t 2 1 + 56.5e -0.3t 56.5e -0.3t e -0.3t - 0.3t 14.4 days
Divide both sides by 180. Subtract 1 from both sides. Divide both sides by 56.5. Rewrite as a logarithmic expression. Divide both sides by - 0.37.
230
250
Y1 5 1 1 56.5e20.37x
Y2 5 115 0
25
*U XJMM UBLF BQQSPYJNBUFMZ EBZT EBZT IPVST GPS UIF QPQVMBUJPO UP SFBDIGSVJUáJFT F One-half of the carrying capacity is 115 fruit flies. Solve P1t2 = 115 by graphing 230 Y1 = and Y2 = 115BOEVTJOH*/5&34&$54FF'JHVSF5IF 1 + 56.5e -0.3x population will reach one-half of the carrying capacity in about 10.9 days EBZT IPVST
r
250
Look back at Figure 46. Notice the point where the graph reaches 115 fruit flies POFIBMGPGUIFDBSSZJOHDBQBDJUZ 5IFHSBQIDIBOHFTGSPNCFJOHDVSWFEVQXBSE UPCFJOHDVSWFEEPXOXBSE6TJOHUIFMBOHVBHFPGDBMDVMVT UIFHSBQIDIBOHFTGSPN increasing at an increasing rate to increasing at a decreasing rate. For any logistic growth function, when the population reaches one-half the carrying capacity, the population growth starts to slow down.
Now Work
PROBLEM
23
388
CHAPTER 4 Exponential and Logarithmic Functions
EX A MPL E 6
Exploration On the same viewing rectangle, graph Y1 =
500 1 + 24e-0.03x
and Y2 =
500
Wood Products The EFISCEN wood product model classifies wood products according to their life-span. 5IFSFBSFGPVSDMBTTJGJDBUJPOTTIPSU ZFBS
NFEJVNTIPSU ZFBST
NFEJVNMPOH ZFBST
BOEMPOH ZFBST #BTFEPOEBUBPCUBJOFEGSPNUIF&VSPQFBO'PSFTU Institute, the percentage of remaining wood products after t years for wood products XJUIMPOHMJGFTQBOT TVDIBTUIPTFVTFEJOUIFCVJMEJOHJOEVTUSZ JTHJWFOCZ
1 + 24e-0.08x
What effect does the growth rate 0 b 0 have on the logistic growth function?
P1t2 =
100.3952 1 + 0.0316e 0.051t
B 8IBUJTUIFEFDBZSBUF C 8IBUJTUIFQFSDFOUBHFPGSFNBJOJOHXPPEQSPEVDUTBGUFSZFBST D )PXMPOHEPFTJUUBLFGPSUIFQFSDFOUBHFPGSFNBJOJOHXPPEQSPEVDUTUPSFBDI E Explain why the numerator given in the model is reasonable.
Solution
B 5IFEFDBZSBUFJT 0 b 0 = 0 - 0.051 0 = 5.1,. C &WBMVBUFP1102. 100.3952 ≈ 95.0 P1102 = 1 + 0.0316e 0.0511102 So 95% of long-life-span wood products remain after 10 years. D 4PMWFUIFFRVBUJPOP1t2 = 50. 100.3952 1 + 0.0316e 0.051t 100.3952 2.009 1.009 31.956 ln 131.9562 t
= 50 = = = = = ≈
5011 + 0.0316e 0.051t 2 1 + 0.0316e 0.051t 0.0316e 0.051t e 0.051t 0.051t 59.6 years
Divide both sides by 50. Subtract 1 from both sides. Divide both sides by 0.0316. Rewrite as a logarithmic expression. Divide both sides by 0.0581.
It will take approximately 59.6 years for the percentage of long-life-span wood products remaining to reach 50%. E The numerator of 100.3952 is reasonable because the maximum percentage of wood products remaining that is possible is 100%.
Now Work
r
PROBLEM
27
4.8 Assess Your Understanding Applications and Extensions 1. Growth of an Insect Population The size P of a certain insect population at time t JOEBZT PCFZTUIFNPEFMP 1t2 = 500e 0.02t. B %FUFSNJOFUIFOVNCFSPGJOTFDUTBUt = 0 days. 500 C 8IBUJTUIFHSPXUISBUFPGUIFJOTFDUQPQVMBUJPO 2% D 8IBUJTUIFQPQVMBUJPOBGUFSEBZT ≈611 E 8IFOXJMMUIFJOTFDUQPQVMBUJPOSFBDI ≈23.5 days F 8IFOXJMMUIFJOTFDUQPQVMBUJPOEPVCMF ≈34. days 2. Growth of Bacteria The number N of bacteria present in a culture at time t JOIPVST PCFZTUIFNPEFMN1t2 = 1000e 0.01t. * B %FUFSNJOFUIFOVNCFSPGCBDUFSJBBUt = 0 hours. C 8IBUJTUIFHSPXUISBUFPGUIFCBDUFSJB 1% D 8IBUJTUIFQPQVMBUJPOBGUFSIPVST ≈1041 * E 8IFOXJMMUIFOVNCFSPGCBDUFSJBSFBDI F 8IFOXJMMUIFOVNCFSPGCBDUFSJBEPVCMF ≈69.3 h 3. Radioactive Decay Strontium-90 is a radioactive material that decays according to the function A1t2 = A0e -0.0244t, where A0 is the initial amount present and A is the amount
present at time t JO ZFBST "TTVNF UIBU B TDJFOUJTU IBT B sample of 500 grams of strontium-90. B 8IBUJTUIFEFDBZSBUFPGTUSPOUJVN - 2.44, C )PXNVDITUSPOUJVNJTMFGUBGUFSZFBST ≈391. g D 8IFOXJMMHSBNTPGTUSPOUJVNCFMFGU ≈9.1 yr E 8IBUJTUIFIBMGMJGFPGTUSPOUJVN ZS 4. Radioactive Decay Iodine-131 is a radioactive material that decays according to the function A1t2 = A0e -0.0t, where A0 is the initial amount present and A is the amount present at time t JOEBZT "TTVNFUIBUBTDJFOUJTUIBTBTBNQMFPG 100 grams of iodine-131. B 8IBUJTUIFEFDBZSBUFPGJPEJOF - .% C )PXNVDIJPEJOFJTMFGUBGUFSEBZT ≈45. g D 8IFOXJMMHSBNTPGJPEJOFCFMFGU ≈ 4.1 days E 8IBUJTUIFIBMGMJGFPGJPEJOF ≈ EBZT 5. Growth of a Colony of Mosquitoes The population of a colony of mosquitoes obeys the law of uninhibited growth.
%VFUPTQBDFSFTUSJDUJPOT BOTXFSTUPUIFTFFYFSDJTFTNBZCFGPVOEJOUIF"OTXFSTJOUIFCBDLPGUIFCPPL
SECTION 4.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models
B *GN is the population of the colony and t is the time in days, express N as a function of t. N1t2 = N0e kt * C *GUIFSFBSFNPTRVJUPFTJOJUJBMMZBOEUIFSFBSF BGUFSEBZ XIBUJTUIFTJ[FPGUIFDPMPOZBGUFSEBZT * D )PXMPOHJTJUVOUJMUIFSFBSF NPTRVJUPFT 6. Bacterial Growth A culture of bacteria obeys the law of uninhibited growth. * B *GN is the number of bacteria in the culture and t is the * time in hours, express N as a function of t. C *G CBDUFSJB BSF QSFTFOU JOJUJBMMZ BOE UIFSF BSF after 1 hour, how many will be present in the culture BGUFSIPVST ≈5243 * D )PXMPOHJTJUVOUJMUIFSFBSF CBDUFSJB 7. Population Growth The population of a southern city follows the exponential law. B *G N is the population of the city and t is the time in years, express N as a function of t. N1t2 = N0e kt C *G UIF QPQVMBUJPO EPVCMFE JO TJ[F PWFS BO NPOUI period and the current population is 10,000, what will UIFQPQVMBUJPOCFZFBSTGSPNOPX 8. Population Decline The population of a midwestern city follows the exponential law. B *GN is the population of the city and t is the time in years, express N as a function of t. N1t2 = N0e kt, k 6 0 C *G UIF QPQVMBUJPO EFDSFBTFE GSPN UP GSPN UP XIBU XBT UIF QPQVMBUJPO JO 9. Radioactive Decay The half-life of radium is 1690 years. If 10 grams is present now, how much will be present in ZFBST H *10. Radioactive Decay The half-life of radioactive potassium is 1.3 billion years. If 10 grams is present now, how much will be QSFTFOUJOZFBST *OZFBST 11. Estimating the Age of a Tree A piece of charcoal is found UPDPOUBJOPGUIFDBSCPOUIBUJUPSJHJOBMMZIBE8IFO EJE UIF USFF GSPN XIJDI UIF DIBSDPBM DBNF EJF 6TF years as the half-life of carbon-14. 9953 yr ago 12. Estimating the Age of a Fossil "GPTTJMJ[FEMFBGDPOUBJOT PGJUTOPSNBMBNPVOUPGDBSCPO)PXPMEJTUIFGPTTJM 6TF ZFBSTBTUIFIBMGMJGFPGDBSCPO 2949 yr old 13. Cooling Time of a Pizza A pizza baked at 450°F is removed from the oven at 5:00 pm and placed in a room that is a DPOTUBOU'"GUFSNJOVUFT UIFQJ[[BJTBU' B "UXIBUUJNFDBOZPVCFHJOFBUJOHUIFQJ[[BJGZPVXBOU JUTUFNQFSBUVSFUPCF' ≈5:1 pm C %FUFSNJOF UIF UJNF UIBU OFFET UP FMBQTF CFGPSF UIF pizza is 160°F. ≈14.3 min * D 8IBU EP ZPV OPUJDF BCPVU UIF UFNQFSBUVSF BT UJNF QBTTFT
14. Newton’s Law of Cooling " UIFSNPNFUFS SFBEJOH ' JT placed in a refrigerator where the temperature is a constant '
389
B *GUIFUIFSNPNFUFSSFBET'BGUFSNJOVUFT XIBUXJMM JUSFBEBGUFSNJOVUFT 45.41°F C )PX MPOH XJMM JU UBLF CFGPSF UIF UIFSNPNFUFS SFBET ' ≈16.2 min D %FUFSNJOF UIF UJNF UIBU NVTU FMBQTF CFGPSF UIF thermometer reads 45°F. ≈.26 min * E 8IBUEPZPVOPUJDFBCPVUUIFUFNQFSBUVSFBTUJNFQBTT 15. Newton’s Law of Heating " UIFSNPNFUFS SFBEJOH $ is brought into a room with a constant temperature of 35°C. If the thermometer reads 15°C after 3 minutes, what XJMM JU SFBE BGUFS CFJOH JO UIF SPPN GPS NJOVUFT 'PS NJOVUFT $ $ [Hint: You need to construct a formula similar to FRVBUJPO > 16. Warming Time of a Beer Stein A beer stein has a UFNQFSBUVSFPG'*UJTQMBDFEJOBSPPNXJUIBDPOTUBOU UFNQFSBUVSF PG '"GUFS NJOVUFT UIF UFNQFSBUVSF PG UIFTUFJOIBTSJTFOUP'8IBUXJMMUIFUFNQFSBUVSFPGUIF TUFJO CF BGUFS NJOVUFT )PX MPOH XJMM JU UBLF UIF TUFJO UP SFBDI B UFNQFSBUVSF PG ' 4FF UIF IJOU HJWFO GPS 1SPCMFN ' NJO 17. Decomposition of Chlorine in a Pool 6OEFSDFSUBJOXBUFS DPOEJUJPOT UIF GSFF DIMPSJOF IZQPDIMPSPVT BDJE )0$M in a swimming pool decomposes according to the law of uninhibited decay. After shocking his pool, Ben tested the water and found the amount of free chlorine to be 2.5 parts QFSNJMMJPO QQN 5XFOUZGPVSIPVSTMBUFS #FOUFTUFEUIF water again and found the amount of free chlorine to be QQN8IBU XJMM CF UIF SFBEJOH BGUFS EBZT UIBU JT IPVST 8IFOUIFDIMPSJOFMFWFMSFBDIFTQQN #FONVTU shock the pool again. How long can Ben go before he must TIPDLUIFQPPMBHBJO QQN EBZT PSI 18. Decomposition of Dinitrogen Pentoxide At 45°C, dinitrogen QFOUPYJEF N2O 5 EFDPNQPTFT JOUP OJUSPVT EJPYJEF NO 2 BOEPYZHFO O 2 BDDPSEJOHUPUIFMBXPGVOJOIJCJUFEEFDBZ An initial amount of 0.25 mole of dinitrogen pentoxide EFDPNQPTFT UP NPMF JO NJOVUFT )PX NVDI EJOJUSPHFO QFOUPYJEF XJMM SFNBJO BGUFS NJOVUFT )PX long will it take until 0.01 mole of dinitrogen pentoxide SFNBJOT NPM NJO 19. Decomposition of Sucrose 3FBDUJOHXJUIXBUFSJOBOBDJEJD TPMVUJPO BU $ TVDSPTF C 12H 22O 11 EFDPNQPTFT JOUP HMVDPTF C 6H 12O 6 BOE GSVDUPTF C 6H 12O 6 BDDPSEJOH UP the law of uninhibited decay. An initial amount of 0.40 mole of sucrose decomposes to 0.36 mole in 30 minutes. How NVDI TVDSPTF XJMM SFNBJO BGUFS IPVST )PX MPOH XJMM JU UBLFVOUJMNPMFPGTVDSPTFSFNBJOT NPM I 20. Decomposition of Salt in Water 4BMU /B$M EFDPNQPTFTJO water into sodium 1Na + 2 and chloride 1Cl- 2 ions according to the law of uninhibited decay. If the initial amount of salt is 25 kilograms and, after 10 hours, 15 kilograms of salt is left, IPXNVDITBMUJTMFGUBGUFSEBZ )PXMPOHEPFTJUUBLFVOUJM 1 LJMPHSBNPGTBMUJTMFGU LH I PSEBZT 2 21. Radioactivity from Chernobyl After the release of radioactive material into the atmosphere from a nuclear QPXFS QMBOU BU $IFSOPCZM 6LSBJOF JO UIF IBZ JO "VTUSJBXBTDPOUBNJOBUFECZJPEJOF IBMGMJGFEBZT *G it is safe to feed the hay to cows when 10% of the iodine-131 remains, how long did the farmers need to wait to use this IBZ 26.6 days *Author’s Note: Surprisingly, the chemical formulas for glucose and fructose are the same: This is not a typo.
390
CHAPTER 4 Exponential and Logarithmic Functions
22. Pig Roasts The hotel Bora-Bora is having a pig roast. At noon, the chef put the pig in a large earthen oven. The pig’s PSJHJOBMUFNQFSBUVSFXBT'"Upm the chef checked the pig’s temperature and was upset because it had reached only 100°F. If the oven’s temperature remains a constant 325°F, at what time may the hotel serve its guests, assuming UIBUQPSLJTEPOFXIFOJUSFBDIFT' ≈ 1.
B 8IBU JT UIF QSFEJDUFE OVNCFS PG FSPEFE PS MFBLZ QSJNBSZ0SJOHTBUBUFNQFSBUVSFPG' ≈0 C 8IBU JT UIF QSFEJDUFE OVNCFS PG FSPEFE PS MFBLZ QSJNBSZ0SJOHTBUBUFNQFSBUVSFPG' ≈1 D 8IBU JT UIF QSFEJDUFE OVNCFS PG FSPEFE PS MFBLZ QSJNBSZ0SJOHTBUBUFNQFSBUVSFPG' ≈5 * E Graph the equation. At what temperature is the QSFEJDUFEOVNCFSPGFSPEFEPSMFBLZ0SJOHT Source: Linda Tappin, “Analyzing Data Relating to the Challenger Disaster,” .BUIFNBUJDT 5FBDIFS Vol. 87, No. 6, September 1994, pp. 423–426.
23. Population of a Bacteria Culture The logistic growth model P 1t2 =
1000 1 + 32.33e -0.439t SFQSFTFOUT UIF QPQVMBUJPO JO HSBNT PG B CBDUFSJVN BGUFS t hours. * B %FUFSNJOFUIFDBSSZJOHDBQBDJUZPGUIFFOWJSPONFOU C 8IBUJTUIFHSPXUISBUFPGUIFCBDUFSJB 43.9% D %FUFSNJOFUIFJOJUJBMQPQVMBUJPOTJ[F 30 g E 8IBUJTUIFQPQVMBUJPOBGUFSIPVST 616.6 g F 8IFOXJMMUIFQPQVMBUJPOCFHSBNT I G )PX MPOH EPFT JU UBLF GPS UIF QPQVMBUJPO UP SFBDI POFIBMGUIFDBSSZJOHDBQBDJUZ ≈.9 h 24. Population of an Endangered Species Often environmentalists capture an endangered species and transport the species to a controlled environment where the species can produce offspring and regenerate its population. Suppose that six "NFSJDBOCBMEFBHMFTBSFDBQUVSFE USBOTQPSUFEUP.POUBOB and set free. Based on experience, the environmentalists expect the population to grow according to the model P 1t2 =
500
1 + 2.33e -0.162t where t is measured in years.
26. Word Users According to a survey by Olsten Staffing Services, the percentage of companies reporting usage of .JDSPTPGU8PSEtZFBSTTJODFJTHJWFOCZ P 1t2 =
99.44 1 + 3.014e -0.99t
B 8IBUJTUIFHSPXUISBUFJOUIFQFSDFOUBHFPG.JDSPTPGU 8PSEVTFST * C 6TFBHSBQIJOHVUJMJUZUPHSBQIP = P 1t2. D 8IBU XBT UIF QFSDFOUBHF PG .JDSPTPGU 8PSE VTFST JO E %VSJOHXIBUZFBSEJEUIFQFSDFOUBHFPG.JDSPTPGU8PSE VTFSTSFBDI * F Explain why the numerator given in the model is SFBTPOBCMF8IBUEPFTJUJNQMZ 27. Home Computers The logistic model P 1t2 =
* B %FUFSNJOFUIFDBSSZJOHDBQBDJUZPGUIFFOWJSPONFOU C 8IBUJTUIFHSPXUISBUFPGUIFCBMEFBHMF 16.2% D 8IBUJTUIFQPQVMBUJPOBGUFSZFBST ≈ 10 eagles E 8IFOXJMMUIFQPQVMBUJPOCFFBHMFT ZS F )PX MPOH EPFT JU UBLF GPS UIF QPQVMBUJPO UP SFBDI POFIBMGPGUIFDBSSZJOHDBQBDJUZ yr 25. The Challenger Disaster After the Challenger disaster in BTUVEZXBTNBEFPGUIFMBVODIFTUIBUQSFDFEFEUIF fatal flight. A mathematical model was developed involving the relationship between the Fahrenheit temperature x around the O-rings and the number y of eroded or leaky primary O-rings. The model stated that y =
6
1 + e -15.05 - 0.1156x2 where the number 6 indicates the 6 primary O-rings on the spacecraft.
95.4993 1 + 0.0405e 0.196t
represents the percentage of households that do not own a personal computer tZFBSTTJODF * B &WBMVBUFBOEJOUFSQSFUP 102. P 102 = 91. * C 6TFBHSBQIJOHVUJMJUZUPHSBQIP = P 1t2. D 8IBUQFSDFOUBHFPGIPVTFIPMETEJEOPUPXOBQFSTPOBM DPNQVUFSJO E *OXIBUZFBSEJEUIFQFSDFOUBHFPGIPVTFIPMETUIBUEP OPUPXOBQFSTPOBMDPNQVUFSSFBDI 2011 Source: U.S. Department of Commerce 28. Farmers The logistic model W 1t2 =
14,656,24 1 + 0.059e 0.05t
SFQSFTFOUTUIFOVNCFSPGGBSNXPSLFSTJOUIF6OJUFE4UBUFT t years after 1910. * B &WBMVBUFBOEJOUFSQSFUW 102. W 102 = 13,39,05 * C 6TFBHSBQIJOHVUJMJUZUPHSBQIW = W 1t2. D )PX NBOZ GBSN XPSLFST XFSF UIFSF JO UIF 6OJUFE 4UBUFTJO
SECTION 4.9 Building Exponential, Logarithmic, and Logistic Models from Data
E 8IFO EJE UIF OVNCFS PG GBSN XPSLFST JO UIF 6OJUFE 4UBUFTSFBDI 1946 * F According to this model, what happens to the number of GBSNXPSLFSTJOUIF6OJUFE4UBUFTBTt approaches q Based on this result, do you think that it is reasonable to use this model to predict the number of farm workers in UIF6OJUFE4UBUFTJO 8IZ Source: U.S. Department of Agriculture 29. Birthdays The logistic model P 1n2 =
113.319
391
models the probability that in a room of n people, no two people share the same birthday. * B 6TFBHSBQIJOHVUJMJUZUPHSBQIP = P 1n2. C *OBSPPNPGn = 15 people, what is the probability that OPUXPTIBSFUIFTBNFCJSUIEBZ PS D )PX NBOZ QFPQMF NVTU CF JO B SPPN CFGPSF UIF probability that no two people share the same birthday GBMMTCFMPX 50 people * E 8IBUIBQQFOTUPUIFQSPCBCJMJUZBTnJODSFBTFT &YQMBJO what this result means.
1 + 0.115e 0.0912n
Retain Your Knowledge Problems 30–33 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. x2 2y 30. Find the equation of the linear function f that passes *32. 8SJUF UIF MPHBSJUINJD FYQSFTTJPO ln a b as the sum z through the points 4, 1 and , - 5 . and/or difference of logarithms. Express powers as factors. 31. %FUFSNJOF XIFUIFS UIF HSBQIT PG UIF MJOFBS GVODUJPOT 10 3 1 33. 3BUJPOBMJ[FUIFEFOPNJOBUPSPG 3 . 22 5 are parallel, 3 f x = 5x - 1 and g x = x + 1 225 5 30. f x = - x + perpendicular, or neither. Neither 2
4.9 Building Exponential, Logarithmic, and Logistic Models from Data PREPARING FOR THIS SECTION Before getting started, review the following: r #VJMEJOH-JOFBS.PEFMTGSPN%BUB 4FDUJPO QQm
r #VJMEJOH$VCJD.PEFMTGSPN%BUB 4FDUJPO QQm
OBJECTIVES
r #VJMEJOH2VBESBUJD.PEFMTGSPN%BUB 4FDUJPO QQm
1 Build an Exponential Model from Data (p. 392) 2 Build a Logarithmic Model from Data (p. 393) 3 Build a Logistic Model from Data (p. 394)
Finding the linear function of best fit 1y = ax + b2 for a set of data was discussed in Section 2.2. Likewise, finding the quadratic function of best fit 1y = ax2 + bx + c2 and finding the cubic function of best fit 1y = ax3 + bx2 + cx + d2 were discussed in Sections 2.6 and 3.1, respectively. In this section we discuss how to use a graphing utility to find equations of best fit that describe the relation between two variables when the relation is thought to be exponential 1y = abx 2, logarithmic 1y = a + b ln x2 , or logistic c ¢y = ≤. As before, a scatter diagram of the data is drawn to help 1 + ae -bx determine the appropriate model to use. 'JHVSF TIPXT TDBUUFS EJBHSBNT UIBU XJMM UZQJDBMMZ CF PCTFSWFE GPS UIF UISFF models. Below each scatter diagram are any restrictions on the values of the parameters.
Figure 47 y
y
x
y
y
x
y
x
x
y ab x, a 0, b 1
y ab x, 0 b 1, a 0
y a b In x, a 0, b 0
Exponential
Exponential
Logarithmic
y a b In x, a 0, b 0 y Logarithmic
x c
bx , a 0, b 0, c 0
1 ae
Logistic
392
CHAPTER 4 Exponential and Logarithmic Functions
.PTUHSBQIJOHVUJMJUJFTIBWF3&(SFTTJPOPQUJPOTUIBUGJUEBUBUPBTQFDJGJDUZQF of curve. Once the data have been entered and a scatter diagram obtained, the type PGDVSWFUIBUZPVXBOUUPGJUUPUIFEBUBJTTFMFDUFE5IFOUIBU3&(SFTTJPOPQUJPOJT used to obtain the curve of best fit of the type selected. The correlation coefficient r will appear only if the model can be written as a linear expression. As it turns out, r will appear for the linear, power, exponential, and logarithmic models, since these models can be written as a linear expression. 3FNFNCFS UIFDMPTFS 0 r 0 is to 1, the better the fit.
1 Build an Exponential Model from Data 8FTBXJO4FDUJPOUIBUUIFGVUVSFWBMVFPGNPOFZCFIBWFTFYQPOFOUJBMMZ BOEXF TBXJO4FDUJPOUIBUHSPXUIBOEEFDBZNPEFMTBMTPCFIBWFFYQPOFOUJBMMZ5IFOFYU example shows how data can lead to an exponential model.
Table 9
EX A MPL E 1
Year, x
Account Value, y
0
20,000
1
21,516
2
23,355
3
24,885
4
27,484
5
30,053
6
32,622
Solution
Fitting an Exponential Function to Data .BSJBIEFQPTJUFE JOUPBXFMMEJWFSTJGJFENVUVBMGVOEZFBSTBHP5IFEBUB JO5BCMFSFQSFTFOUUIFWBMVFPGUIFBDDPVOUFBDIZFBSGPSUIFMBTUZFBST B 6TJOHBHSBQIJOHVUJMJUZ ESBXBTDBUUFSEJBHSBNXJUIZFBSBTUIFJOEFQFOEFOU variable. C 6TJOHBHSBQIJOHVUJMJUZ CVJMEBOFYQPOFOUJBMNPEFMGSPNUIFEBUB D &YQSFTTUIFGVODUJPOGPVOEJOQBSU C JOUIFGPSNA = A0 e kt. E (SBQIUIFFYQPOFOUJBMGVODUJPOGPVOEJOQBSU C PS D POUIFTDBUUFSEJBHSBN F 6TJOHUIFTPMVUJPOUPQBSU C PS D
QSFEJDUUIFWBMVFPGUIFBDDPVOUBGUFSZFBST G Interpret the value of kGPVOEJOQBSU D B &OUFSUIFEBUBJOUPUIFHSBQIJOHVUJMJUZBOEESBXUIFTDBUUFSEJBHSBNBTTIPXO JO'JHVSF C "HSBQIJOHVUJMJUZàUTUIFEBUBJO5BCMFUPBOFYQPOFOUJBMNPEFMPGUIFGPSN y = abx VTJOH UIF &91POFOUJBM 3&(SFTTJPO PQUJPO 'JHVSF TIPXT UIBU y = abx = 19,20.4311.05562 x. Notice that 0 r 0 = 0.999, which is close to 1, indicating a good fit. Figure 48
Figure 49
40,000
⫺1
7 0
D 5PFYQSFTTy = abx in the form A = A0 e kt, where x = t and y = A, proceed as follows: abx = A0 e kt, x = t If x = t = 0, then a = A0 . This leads to a = A0, bx = e kt t bx = 1e k 2 b = ek x = t x x Because y = ab = 19,20.4311.05562 , this means that a = 19,20.43 and b = 1.0556. a = A0 = 19,20.43 and b = e k = 1.0556 To find k, rewrite e k = 1.0556 as a logarithm to obtain k = ln 11.05562 ≈ 0.0210 As a result, A = A0 e kt = 19,20.43e 0.0210t.
SECTION 4.9 Building Exponential, Logarithmic, and Logistic Models from Data
393
E 4FF'JHVSFGPSUIFHSBQIPGUIFFYQPOFOUJBMGVODUJPOPGCFTUàU Figure 50 40,000
⫺1
7 0
F -FUt = 10JOUIFGVODUJPOGPVOEJOQBSU D 5IFQSFEJDUFEWBMVFPGUIFBDDPVOU after 10 years is A = A0e kt = 19,20.43e 0.02101102 ≈ +45,04 G The value of k = 0.0210 = .210% represents the growth rate of the account. It represents the rate of interest earned, assuming the account is growing continuously.
r
Now Work
PROBLEM
1
2 Build a Logarithmic Model from Data .BOZSFMBUJPOTCFUXFFOWBSJBCMFTEPOPUGPMMPXBOFYQPOFOUJBMNPEFM JOTUFBE UIF independent variable is related to the dependent variable using a logarithmic model.
Table 10
EXAM PL E 2
Atmospheric Pressure, p
Height, h
760
0
740
0.184
725
0.328
700
0.565
650
1.079
630
1.291
600
1.634
580
1.862
550
2.235
Fitting a Logarithmic Function to Data Jodi, a meteorologist, is interested in finding a function that explains the relation CFUXFFO UIF IFJHIU PG B XFBUIFS CBMMPPO JO LJMPNFUFST BOE UIF BUNPTQIFSJD QSFTTVSF NFBTVSFEJONJMMJNFUFSTPGNFSDVSZ POUIFCBMMPPO4IFDPMMFDUTUIFEBUB shown in Table 10. B 6TJOH B HSBQIJOH VUJMJUZ ESBX B TDBUUFS EJBHSBN PG UIF EBUB XJUI BUNPTQIFSJD pressure as the independent variable. C *UJTLOPXOUIBUUIFSFMBUJPOCFUXFFOBUNPTQIFSJDQSFTTVSFBOEIFJHIUGPMMPXTB MPHBSJUINJDNPEFM6TJOHBHSBQIJOHVUJMJUZ CVJMEBMPHBSJUINJDNPEFMGSPNUIF data. D %SBXUIFMPHBSJUINJDGVODUJPOGPVOEJOQBSU C POUIFTDBUUFSEJBHSBN E 6TFUIFGVODUJPOGPVOEJOQBSU C UPQSFEJDUUIFIFJHIUPGUIFXFBUIFSCBMMPPO if the atmospheric pressure is 560 millimeters of mercury.
Solution
Figure 51 2.4
B &OUFS UIF EBUB JOUP UIF HSBQIJOH VUJMJUZ BOE ESBX UIF TDBUUFS EJBHSBN 4FF Figure 51. C "HSBQIJOHVUJMJUZàUTUIFEBUBJO5BCMFUPBMPHBSJUINJDGVODUJPOPGUIFGPSN y = a + b ln xCZVTJOHUIF-0(BSJUIN3&(SFTTJPOPQUJPO4FF'JHVSFPO the next page. The logarithmic model from the data is h 1p2 = 45.63 - 6.9025 ln p
525 ⫺0.2
775
where h is the height of the weather balloon and p is the atmospheric pressure. Notice that 0 r 0 is close to 1, indicating a good fit. D 'JHVSF TIPXT UIF HSBQI PG h 1p2 = 45.63 - 6.9025 ln p on the scatter diagram.
394
CHAPTER 4 Exponential and Logarithmic Functions
Figure 53
Figure 52
2.4
775
525 ⫺0.2
E 6TJOH UIF GVODUJPO GPVOE JO QBSU C
+PEJ QSFEJDUT UIF IFJHIU PG UIF XFBUIFS balloon when the atmospheric pressure is 560 to be h 15602 = 45.63 - 6.9025 ln 560 ≈ 2.10 kilometers
Now Work
PROBLEM
r
5
3 Build a Logistic Model from Data Logistic growth models can be used to model situations for which the value of the EFQFOEFOUWBSJBCMFJTMJNJUFE.BOZSFBMXPSMETJUVBUJPOTDPOGPSNUPUIJTTDFOBSJP For example, the population of the human race is limited by the availability of OBUVSBM SFTPVSDFT TVDI BT GPPE BOE TIFMUFS 8IFO UIF WBMVF PG UIF EFQFOEFOU variable is limited, a logistic growth model is often appropriate.
EX A MPL E 3
Fitting a Logistic Function to Data The data in Table 11 represent the amount of yeast biomass in a culture after t hours.
Table 11
700
20 0
Figure 55
Yeast Biomass
Time (hours)
Yeast Biomass
0
9.6
10
513.3
1
18.3
11
559.7
2
29.0
12
594.8
3
47.2
13
629.4
4
71.1
14
640.8
5
119.1
15
651.1
6
174.6
16
655.9
7
257.3
17
659.6
8
350.7
18
661.8
9
441.0
Source:5PS$BSMTPO ¾CFS(FTDIXJOEJHLFJUVOE(SÕTTFEFS)FGFWFSNFISVOHJO8ÛS[F #JPDIFNJTDIF ;FJUTDISJGU #E QQm
Figure 54
⫺2
Time (hours)
Solution
B 6TJOH B HSBQIJOH VUJMJUZ ESBX B TDBUUFS EJBHSBN PG UIF EBUB XJUI UJNF BT UIF independent variable. C 6TJOHBHSBQIJOHVUJMJUZ CVJMEBMPHJTUJDNPEFMGSPNUIFEBUB D 6TJOH B HSBQIJOH VUJMJUZ HSBQI UIF GVODUJPO GPVOE JO QBSU C PO UIF TDBUUFS diagram. E 8IBUJTUIFQSFEJDUFEDBSSZJOHDBQBDJUZPGUIFDVMUVSF F 6TFUIFGVODUJPOGPVOEJOQBSU C UPQSFEJDUUIFQPQVMBUJPOPGUIFDVMUVSFBU t = 19 hours. B 4FF'JHVSFGPSBTDBUUFSEJBHSBNPGUIFEBUB C " HSBQIJOH VUJMJUZ àUT B MPHJTUJD HSPXUI NPEFM PG UIF GPSN y =
c by 1 + ae -bx using the LOGISTIC regression option. See Figure 55. The logistic model from the data is 663.0 y = 1 + 1.6e -0.540x where y is the amount of yeast biomass in the culture and x is the time.
SECTION 4.9 Building Exponential, Logarithmic, and Logistic Models from Data
395
D 4FF'JHVSFGPSUIFHSBQIPGUIFMPHJTUJDNPEFM Figure 56 700
⫺2
20 0
E #BTFEPOUIFMPHJTUJDHSPXUINPEFMGPVOEJOQBSU C
UIFDBSSZJOHDBQBDJUZPG the culture is 663. F 6TJOHUIFMPHJTUJDHSPXUINPEFMGPVOEJOQBSU C
UIFQSFEJDUFEBNPVOUPGZFBTU biomass at t = 19 hours is y =
Now Work
663.0 1 + 1.6e -0.540 19
PROBLEM
≈ 661.5
r
7
4.9 Assess Your Understanding Applications and Extensions 1. Biology A strain of E. coli#FVSFD"JTQMBDFEJOUP a nutrient broth at 30° Celsius and allowed to grow. The following data are collected. Theory states that the number of bacteria in the petri dish will initially grow according to the law of uninhibited growth. The population is measured using an optical device in which the amount of light that passes through the petri dish is measured.
2. Ethanol Production The data in the table below represent FUIBOPM QSPEVDUJPO JO CJMMJPOT PG HBMMPOT JO UIF 6OJUFE States from 2000 to 2012.
Year
Ethanol Produced (billion gallons)
2000 (x = 0)
1.6
2007 (x = 7)
Population, y
2001 (x = 1)
1.8
2008 (x = 8)
9.0
0
0.09
2002 (x = 2)
2.1
2009 (x = 9)
10.6
2.5
0.18
2003 (x = 3)
2.8
2010 (x = 10)
13.2
3.5
0.26
2004 (x = 4)
3.4
2011 (x = 11)
13.9
4.5
0.35
2005 (x = 5)
3.9
2012 (x = 12)
13.3
6
0.50
2006 (x = 6)
4.9
Time (hours), x
Source:%S1PMMZ-BWFSZ +PMJFU Junior College
* B %SBXBTDBUUFSEJBHSBNUSFBUJOHUJNFBTUIFJOEFQFOEFOU variable. C 6TJOH B HSBQIJOH VUJMJUZ CVJME BO FYQPOFOUJBM NPEFM from the data. y = 0.090311.3342 x D &YQSFTT UIF GVODUJPO GPVOE JO QBSU C JO UIF GPSN N1t2 = N0 e kt. N t = 0.0903e 0.2915t * E (SBQIUIFFYQPOFOUJBMGVODUJPOGPVOEJOQBSU C PS D on the scatter diagram. F 6TF UIF FYQPOFOUJBM GVODUJPO GSPN QBSU C PS D UP predict the population at x = hours. 0.69 G 6TF UIF FYQPOFOUJBM GVODUJPO GSPN QBSU C PS D UP QSFEJDUXIFOUIFQPQVMBUJPOXJMMSFBDI I
Year
Ethanol Produced (billion gallons) 6.5
Source:3FOFXBCMF'VFMT"TTPDJBUJPO
* B 6TJOH B HSBQIJOH VUJMJUZ ESBX B TDBUUFS EJBHSBN PG the data using 0 for 2000, 1 for 2001, and so on as the independent variable. C 6TJOH B HSBQIJOH VUJMJUZ CVJME BO FYQPOFOUJBM NPEFM from the data. y = 1.5256 1.2236 x D &YQSFTT UIF GVODUJPO GPVOE JO QBSU C JO UIF GPSN A t = A0e kt. A t = 1.5256e 0.201t * E (SBQIUIFFYQPOFOUJBMGVODUJPOGPVOEJOQBSU C PS D on the scatter diagram F 6TFUIFNPEFMUPQSFEJDUUIFBNPVOUPGFUIBOPMUIBUXJMM be produced in 2015. 31.5 billion gallons * G Interpret the meaning of k in the function found in QBSU D
%VFUPTQBDFSFTUSJDUJPOT BOTXFSTUPUIFTFFYFSDJTFTNBZCFGPVOEJOUIF"OTXFSTJOUIFCBDLPGUIFCPPL
396
CHAPTER 4 Exponential and Logarithmic Functions
3. Advanced-Stage Breast Cancer The data in the table below represent the percentage of patients who have survived after diagnosis of advanced-stage breast cancer at 6-month intervals of time. Time after Diagnosis (years)
5. Milk Production The data in the table below represent the OVNCFS PG EBJSZ GBSNT JO UIPVTBOET BOE UIF BNPVOU PG NJMLQSPEVDFE JOCJMMJPOTPGQPVOET JOUIF6OJUFE4UBUFT for various years.
Percentage Surviving
Dairy Farms (thousands)
Milk Produced (billion pounds)
0.5
95.7
Year
1
83.6
1980
334
128
1.5
74.0
1985
269
143
2
58.6
1990
193
148
2.5
47.4
1995
140
155
3
41.9
2000
105
167
33.6
2005
78
177
2010
63
193
3.5 Source: Cancer Treatment Centers of America
Source: Statistical Abstract of the United States, 2012 * B 6TJOH B HSBQIJOH VUJMJUZ ESBX B TDBUUFS EJBHSBN PG UIF data, with time after diagnosis as the independent variable. C 6TJOH B HSBQIJOH VUJMJUZ CVJME BO FYQPOFOUJBM NPEFM from the data. y = 11.226 0.013 x D &YQSFTT UIF GVODUJPO GPVOE JO QBSU C JO UIF GPSN A t = A0e kt. A t = 11.226e -0354t * E (SBQIUIFFYQPOFOUJBMGVODUJPOGPVOEJOQBSU C PS D on the scatter diagram. F 8IBUQFSDFOUBHFPGQBUJFOUTEJBHOPTFEXJUIBEWBODFE stage cancer are expected to survive for 4 years after JOJUJBMEJBHOPTJT 2.% * G Interpret the meaning of k in the function found in part D
* B 6TJOHBHSBQIJOHVUJMJUZ ESBXBTDBUUFSEJBHSBNPGUIF data with the number of dairy farms as the independent variable. C 6TJOHBHSBQIJOHVUJMJUZ CVJMEBMPHBSJUINJDNPEFMGSPN the data. y = 330.0549 - 34.500 ln x * D (SBQIUIFMPHBSJUINJDGVODUJPOGPVOEJOQBSU C POUIF scatter diagram. E *O UIFSF XFSF UIPVTBOE EBJSZ GBSNT JO UIF 6OJUFE4UBUFT6TFUIFGVODUJPOJOQBSU C UPQSFEJDUUIF BNPVOUPGNJMLQSPEVDFEJO CJMMJPOQPVOET F 5IF BDUVBM BNPVOU PG NJML QSPEVDFE JO XBT CJMMJPO QPVOET )PX EPFT ZPVS QSFEJDUJPO JO QBSU E DPNQBSFUPUIJT 6OEFSCZCJMMJPOQPVOET
4. Chemistry A chemist has a 100-gram sample of a radioactive material. He records the amount of radioactive material FWFSZXFFLGPSXFFLTBOEPCUBJOTUIFGPMMPXJOHEBUB
6. Economics and Marketing The following data represent the QSJDFBOERVBOUJUZEFNBOEFEGPS%FMMQFSTPOBMDPNQVUFST in a recent year.
Week
Weight (in grams)
0
100.0
Price ($/Computer)
Quantity Demanded
2300
152 159
1
88.3
2000
2
75.9
1700
164
3
69.4
1500
171
4
59.1
1300
176
5
51.8
1200
180
6
45.5
1000
189
* B 6TJOH B HSBQIJOH VUJMJUZ ESBX B TDBUUFS EJBHSBN XJUI week as the independent variable. C 6TJOH B HSBQIJOH VUJMJUZ CVJME BO FYQPOFOUJBM NPEFM from the data. y = 100.32610.692 x D &YQSFTT UIF GVODUJPO GPVOE JO QBSU C JO UIF GPSN A1t2 = A0 e kt. A1t2 = 100.326e - 0.1314t * E (SBQIUIFFYQPOFOUJBMGVODUJPOGPVOEJOQBSU C PS D on the scatter diagram. F 'SPNUIFSFTVMUGPVOEJOQBSU C
EFUFSNJOFUIFIBMGMJGF of the radioactive material. 5.3 weeks G )PX NVDI SBEJPBDUJWF NBUFSJBM XJMM CF MFGU BGUFS XFFLT 0.14 g * H 8IFOXJMMUIFSFCFHSBNTPGSBEJPBDUJWFNBUFSJBM
* B 6TJOHBHSBQIJOHVUJMJUZ ESBXBTDBUUFSEJBHSBNPGUIF data with price as the dependent variable. C 6TJOHBHSBQIJOHVUJMJUZ CVJMEBMPHBSJUINJDNPEFMGSPN the data. y = 32,41.02 - 600.96 ln x * D 6TJOHBHSBQIJOHVUJMJUZ ESBXUIFMPHBSJUINJDGVODUJPO GPVOEJOQBSU C POUIFTDBUUFSEJBHSBN E 6TFUIFGVODUJPOGPVOEJOQBSU C UPQSFEJDUUIFOVNCFS PG%FMMQFSTPOBMDPNQVUFSTUIBUXJMMCFEFNBOEFEJGUIF price is $1650. ≈16 computers 7. Population Model The following data represent the QPQVMBUJPOPGUIF6OJUFE4UBUFT"OFDPMPHJTUJTJOUFSFTUFEJO CVJMEJOHBNPEFMUIBUEFTDSJCFTUIFQPQVMBUJPOPGUIF6OJUFE States.
SECTION 4.9 Building Exponential, Logarithmic, and Logistic Models from Data
397
8. Population Model The following data represent the world population. An ecologist is interested in building a model that describes the world population.
Year
Population
1900
76,212,168
1910
92,228,496
1920
106,021,537
Year
Population (billions)
1930
123,202,624
2001
6.17
1940
132,164,569
2002
6.24
1950
151,325,798
2003
6.32
1960
179,323,175
2004
6.40
1970
203,302,031
2005
6.47
1980
226,542,203
2006
6.55
1990
248,709,873
2007
6.63
2000
281,421,906
2008
6.71
2010
308,745,538
2009
6.79
2010
6.86
2011
6.94
2012
7.02
Source: U.S. Census Bureau
* B 6TJOHBHSBQIJOHVUJMJUZ ESBXBTDBUUFSEJBHSBNPGUIF data using years since 1900 as the independent variable and population as the dependent variable. * C 6TJOHBHSBQIJOHVUJMJUZ CVJMEBMPHJTUJDNPEFMGSPNUIF data. * D 6TJOHBHSBQIJOHVUJMJUZ ESBXUIFGVODUJPOGPVOEJOQBSU C POUIFTDBUUFSEJBHSBN E #BTFE PO UIF GVODUJPO GPVOE JO QBSU C
XIBU JT UIF DBSSZJOHDBQBDJUZPGUIF6OJUFE4UBUFT F 6TF UIF GVODUJPO GPVOE JO QBSU C UP QSFEJDU UIF QPQVMBUJPOPGUIF6OJUFE4UBUFTJO ≈315,203,2 * G 8IFOXJMMUIF6OJUFE4UBUFTQPQVMBUJPOCF H $PNQBSF BDUVBM 64 $FOTVT GJHVSFT UP UIF QSFEJDUJPOT GPVOEJOQBSUT F BOE G %JTDVTTBOZEJGGFSFODFT
Source: U.S. Census Bureau
* B 6TJOHBHSBQIJOHVUJMJUZ ESBXBTDBUUFSEJBHSBNPGUIF data using years since 2000 as the independent variable and population as the dependent variable. * C 6TJOHBHSBQIJOHVUJMJUZ CVJMEBMPHJTUJDNPEFMGSPNUIFEBUB * D 6TJOHBHSBQIJOHVUJMJUZ ESBXUIFGVODUJPOGPVOEJOQBSU C POUIFTDBUUFSEJBHSBN E #BTFE PO UIF GVODUJPO GPVOE JO QBSU C
XIBU JT UIF DBSSZJOHDBQBDJUZPGUIFXPSME CJMMJPOQFPQMF F 6TF UIF GVODUJPO GPVOE JO QBSU C UP QSFEJDU UIF population of the world in 2020. ≈.65 billion G 8IFOXJMMXPSMEQPQVMBUJPOCFCJMMJPO About 2051
9. Cell Phone Towers 5IFGPMMPXJOHEBUBSFQSFTFOUUIFOVNCFSPGDFMMTJUFTJOTFSWJDFJOUIF6OJUFE4UBUFTGSPNUPBUUIF end of June each year. Year
Cell Sites (thousands)
Year
Cell Sites (thousands)
1985 (x = 1)
0.6
1999 (x = 15)
74.2
1986 (x = 2)
1.2
2000 (x = 16)
95.7
1987 (x = 3)
1.7
2001 (x = 17)
114.1
1988 (x = 4)
2.8
2002 (x = 18)
131.4
1989 (x = 5)
3.6
2003 (x = 19)
147.7
1990 (x = 6)
4.8
2004 (x = 20)
174.4
1991 (x = 7)
6.7
2005 (x = 21)
178.0
1992 (x = 8)
8.9
2006 (x = 22)
197.6
1993 (x = 9)
11.6
2007 (x = 23)
210.4
1994 (x = 10)
14.7
2008 (x = 24)
220.5
1995 (x = 11)
19.8
2009 (x = 25)
245.9
1996 (x = 12)
24.8
2010 (x = 26)
251.6
1997 (x = 13)
38.7
2011 (x = 27)
256.9
1998 (x = 14)
57.7
2012 (x = 28)
285.6
Source:¥$5*"5IF8JSFMFTT"TTPDJBUJPO®"MM3JHIUT3FTFSWFE
* B 6TJOHBHSBQIJOHVUJMJUZ ESBXBTDBUUFSEJBHSBNPGUIFEBUBVTJOHGPS GPS BOETPPOBTUIFJOEFQFOEFOUWBSJBCMF and number of cell sites as the dependent variable. * C 6TJOHBHSBQIJOHVUJMJUZ CVJMEBMPHJTUJDNPEFMGSPNUIFEBUB * D (SBQIUIFMPHJTUJDGVODUJPOGPVOEJOQBSU C POUIFTDBUUFSEJBHSBN E 8IBUJTUIFQSFEJDUFEDBSSZJOHDBQBDJUZGPSDFMMTJUFTJOUIF6OJUFE4UBUFT UIPVTBOE F 6TFUIFNPEFMUPQSFEJDUUIFOVNCFSPGDFMMTJUFTJOUIF6OJUFE4UBUFTBUUIFFOEPG+VOF UIPVTBOE
398
CHAPTER 4 Exponential and Logarithmic Functions
10. Cable Rates The following date represents the average monthly rate charged for a basic cable television subscription JOUIF6OJUFE4UBUFTGSPNUP"NBSLFUSFTFBSDIFS believes that external factors, such as the growth of satellite television Internet programming, have affected the cost of basic cable. She is interested in building a model that will describe the average monthly cost of basic cable. * B 6TJOH B HSBQIJOH VUJMJUZ ESBX B TDBUUFS EJBHSBN PG UIF EBUB VTJOH GPS GPS BOE TP PO BT UIF independent variable and average monthly rate as the dependent variable. * C 6TJOHBHSBQIJOHVUJMJUZ CVJMEBMPHJTUJDNPEFMGSPNUIF data. * D (SBQI UIF MPHJTUJD GVODUJPO GPVOE JO QBSU C PO UIF scatter diagram. E #BTFE PO UIF NPEFM GPVOE JO QBSU C XIBU JT UIF maximum possible average monthly rate for basic DBCMF F 6TFUIFNPEFMUPQSFEJDUUIFBWFSBHFSBUFGPSCBTJDDBCMF JO
Average Monthly Rate (dollars)
Year 1980 (x = 0)
7.69
1985 (x = 5)
9.73
1990 (x = 10)
16.78
1995 (x = 15)
23.07
1997 (x = 17)
26.48
1998 (x = 18)
27.81
1999 (x = 19)
28.92
2000 (x = 20)
30.37
2001 (x = 21)
32.87
2002 (x = 22)
34.71
2003 (x = 23)
36.59
2004 (x = 24)
38.14
2005 (x = 25)
39.63
2006 (x = 26)
41.17
2007 (x = 27)
42.72
2008 (x = 28)
44.28
2009 (x = 29)
46.13
2010 (x = 30)
47.89
Source: Statistical Abstract of the United States, 2012
Mixed Practice 11. Age versus Total Cholesterol The following data represent the age and average total cholesterol for adult males at various ages.
Age
Total Cholesterol
27
189
40
205
50
215
60
210
70
210
80
194
* B 6TJOHBHSBQIJOHVUJMJUZ ESBXBTDBUUFSEJBHSBNPGUIF data using age, x, as the independent variable and total cholesterol, y, as the dependent variable. C #BTFEPOUIFTDBUUFSEJBHSBNESBXOJOQBSU B
EFDJEF PO B NPEFM MJOFBS RVBESBUJD DVCJD FYQPOFOUJBM MPHBSJUINJD PSMPHJTUJD UIBUZPVUIJOLCFTUEFTDSJCFTUIF relation between age and total cholesterol. Be sure to justify your choice of model. 2uadratic with a 6 0 * D 6TJOHBHSBQIJOHVUJMJUZ GJOEUIFNPEFMPGCFTUGJU * E 6TJOHBHSBQIJOHVUJMJUZ ESBXUIFNPEFMPGCFTUGJUPO UIFTDBUUFSEJBHSBNESBXOJOQBSU B F 6TF ZPVS NPEFM UP QSFEJDU UIF UPUBM DIPMFTUFSPM PG B 35-year-old male. 201 12. Income versus Crime Rate The following data represent QSPQFSUZ DSJNF SBUF BHBJOTU JOEJWJEVBMT DSJNFT QFS IPVTFIPMET BOEUIFJSIPVTFIPMEJODPNF JOEPMMBST JOUIF 6OJUFE4UBUFTJO * B 6TJOHBHSBQIJOHVUJMJUZ ESBXBTDBUUFSEJBHSBNPGUIF data using income, x, as the independent variable and crime rate, y, as the dependent variable.
C #BTFEPOUIFTDBUUFSEJBHSBNESBXOJOQBSU B
EFDJEF PO B NPEFM MJOFBS RVBESBUJD DVCJD FYQPOFOUJBM MPHBSJUINJD PS MPHJTUJD UIBU ZPV UIJOL CFTU EFTDSJCFT the relation between income and crime rate. Be sure to justify your choice of model. Logarithmic * D 6TJOHBHSBQIJOHVUJMJUZ GJOEUIFNPEFMPGCFTUGJU * E 6TJOHBHSBQIJOHVUJMJUZ ESBXUIFNPEFMPGCFTUGJUPO UIFTDBUUFSEJBHSBNESBXOJOQBSU B F 6TF ZPVS NPEFM UP QSFEJDU UIF DSJNF SBUF PG B IPVTFIPME whose income is $55,000. 121.4 crimes per 1000 households Income Level
Property Crime Rate
5000
201.1
11,250
157.0
20,000
141.6
30,000
134.1
42,500
139.7
62,500
120.0
Source: Statistical Abstract of the United States, 2012
13. Golfing The data on the following page represent the expected percentage of putts that will be made by QSPGFTTJPOBMHPMGFSTPOUIF1("5PVSEFQFOEJOHPOEJTUBODF For example, it is expected that 99.3% of 2-foot putts will be made. * B 6TJOHBHSBQIJOHVUJMJUZ ESBXBTDBUUFSEJBHSBNPGUIF data with distance as the independent variable. C #BTFEPOUIFTDBUUFSEJBHSBNESBXOJOQBSU B
EFDJEF PO B NPEFM MJOFBS RVBESBUJD DVCJD FYQPOFOUJBM MPHBSJUINJD PSMPHJTUJD UIBUZPVUIJOLCFTUEFTDSJCFTUIF relation between distance and expected percentage. Be sure to justify your choice of model. Exponential
Chapter Review
114.. Depreciation of a Chevrolet Impala The following data 14 represent the asking price and age of a Chevrolet Impala SS.
Distance (feet)
Expected Percentage
Distance (feet)
Expected Percentage
2
99.3
14
25.0
Age
Asking Price
3
94.8
15
22.0
1
$27,417
4
85.8
16
20.0
1
$26,750
5
74.7
17
19.0
2
$22,995
6
64.7
18
17.0
2
$23,195
7
55.6
19
16.0
3
$17,999
8
48.5
20
14.0
4
$16,995
9
43.4
21
13.0
4
$16,490
10
38.3
22
12.0
11
34.2
23
11.0
12
30.1
24
11.0
13
27.0
25
10.0
Source: TheSandTrap.com
* D 6TJOHBHSBQIJOHVUJMJUZ GJOEUIFNPEFMPGCFTUGJU * E (SBQI UIF GVODUJPO GPVOE JO QBSU D PO UIF TDBUUFS diagram. F 6TF UIF GVODUJPO GPVOE JO QBSU D UP QSFEJDU XIBU percentage of 30-foot putts will be made. 5.1%
399
Source: cars.com
* B 6TJOHBHSBQIJOHVUJMJUZ ESBXBTDBUUFSEJBHSBNPGUIF data using age as the independent variable and asking price as the dependent variable. C #BTFEPOUIFTDBUUFSEJBHSBNESBXOJOQBSU B
EFDJEF PO B NPEFM MJOFBS RVBESBUJD DVCJD FYQPOFOUJBM MPHBSJUINJD PSMPHJTUJD UIBUZPVUIJOLCFTUEFTDSJCFTUIF relation between age and asking price. Be sure to justify your choice of model. Exponential * D 6TJOHBHSBQIJOHVUJMJUZ GJOEUIFNPEFMPGCFTUGJU * E 6TJOHBHSBQIJOHVUJMJUZ ESBXUIFNPEFMPGCFTUGJUPO UIFTDBUUFSEJBHSBNESBXOJOQBSU B F 6TFZPVSNPEFMUPQSFEJDUUIFBTLJOHQSJDFPGB$IFWSPMFU Impala SS that is 5 years old.
Retain Your Knowledge Problems 15–18 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 3 322 15. Construct a polynomial function that might have the graph 16. 3BUJPOBMJ[FUIFEFOPNJOBUPSPG . 2 22 TIPXO .PSFUIBOPOFBOTXFSJTQPTTJCMF
6TFUIF1ZUIBHPSFBO5IFPSFNUPGJOEUIFFYBDUMFOHUIPG 17. 1 y f x = x + 3 x + 1 2 x - 2
the unlabeled side in the given right triangle. 23 3 4 2 2 –4
–2
2
4 x
–2 1 –4
*18. Graph the equation x - 3 2 + y2 = 25.
Chapter Review Things to Know Composite function (p. 306)
f ∘ g x = f g x
; The domain of f ∘ g is the set of all numbers x in the domain of g for which g1x2 is in the domain of f.
One-to-one function f (p. 314)
A function for which any two different inputs in the domain correspond to two different outputs in the range For any choice of elements x1 , x2 in the domain of f, if x1 ≠ x2 , then f 1x1 2 ≠ f 1x2 2.
Horizontal-line test (p. 315)
If every horizontal line intersects the graph of a function f in at most one point, f is one-to-one.
400
CHAPTER 4 Exponential and Logarithmic Functions
Inverse function f −1 of f (pp. 316–319)
%PNBJOPGf = range of f -1; range of f = domain of f -1 f -1 1f 1x2 2 = x for all x in the domain of f
f 1f -1 1x2 2 = x for all x in the domain of f -1
The graphs of f and f -1 are symmetric with respect to the line y = x. Properties of the exponential function (pp. 328, 331, 333)
f 1x2 = Cax, a 7 1, C 7 0
%PNBJOUIFJOUFSWBM 1 - q , q 2 3BOHFUIFJOUFSWBM 10, q 2
x-intercepts: none; y-intercept: C
Horizontal asymptote: x-axis 1y = 02 as x S - q Increasing; one-to-one; smooth; continuous See Figure 22 for a typical graph.
f 1x2 = Cax, 0 6 a 6 1, C 7 0 %PNBJOUIFJOUFSWBM 1 - q , q 2 3BOHFUIFJOUFSWBM 10, q 2
x-intercepts: none; y-intercept: C
Horizontal asymptote: x-axis 1y = 02 as x S q
%FDSFBTJOH POFUPPOF TNPPUI DPOUJOVPVT See Figure 26 for a typical graph.
Number e (p. 334)
7BMVFBQQSPBDIFECZUIFFYQSFTTJPOa1 +
Property of exponents (p. 335)
If au = av, then u = v.
Properties of the logarithmic function (pp. 343, 345, 346)
f 1x2 = log a x, a 7 1
1y = log a x means x = a 2 y
1 n 1 n b as n S q ; that is, lim a1 + b = e. n Sq n n
%PNBJOUIFJOUFSWBM 10, q 2
3BOHFUIFJOUFSWBM 1 - q , q 2 x-intercept: 1; y-intercept: none 7FSUJDBMBTZNQUPUFx = 0 yBYJT
Increasing; one-to-one; smooth; continuous
f 1x2 = log a x, 0 6 a 6 1 1y = log a x means x = ay 2
4FF'JHVSF C GPSBUZQJDBMHSBQI
%PNBJOUIFJOUFSWBM 10, q 2 3BOHFUIFJOUFSWBM 1 - q , q 2
x-intercept: 1; y-intercept: none
7FSUJDBMBTZNQUPUFx = 0 yBYJT
%FDSFBTJOH POFUPPOF TNPPUI DPOUJOVPVT
4FF'JHVSF B GPSBUZQJDBMHSBQI
Natural logarithm (p. 347)
y = ln x means x = e .
Properties of logarithms (pp. 357, 358, 360)
log a1 = 0
y
log a a = 1 aloga M = M
log a ar = r
M log a 1MN2 = log a M + log a N log a a b = log a M - log a N N r log a M = r log a M x If M = N, then log a M = log a N a = e x ln a
Formulas
If log a M = log a N, then M = N
Change-of-Base Formula (p. 361)
log a M =
log b M
Continuous compounding (p. 374)
log b a r nt A = P # a1 + b n A = Pe rt
Effective rate of interest (p. 375)
Compounding n times per year: re = a1 +
Present Value Formulas (p. 376)
Continuous compounding: re = e r - 1 r -nt P = A # a1 + b or P = Ae -rt n A1t2 = A0 e kt
Compound Interest Formula (p. 373)
Growth and decay (p. 382, 383) Newton’s Law of Cooling (p. 384) Logistic model (p. 386)
u1t2 = T + u 0 - T e kt k 6 0 c P 1t2 = 1 + ae -bt
r n b - 1 n
Chapter Review
401
Objectives Section
You should be able to . . .
Example(s)
Review Exercises
1, 2, 4, 5 2–4
1–7 5–7
1, 2 3, 4
8(a), 9 8(b)
3 Obtain the graph of the inverse function from the graph
7
9
of the function (p. 318) 4 Find the inverse of a function defined by an equation (p. 319)
8, 9, 10
10–13 14(a), (c), 47(a) 31–33 35, 36, 39, 41
4.1
1 Form a composite function (p. 306)
4.2
1 Determine whether a function is one-to-one (p. 314)
2 Find the domain of a composite function (p. 307) 2 Determine the inverse of a function defined by a map or
a set of ordered pairs (p. 316)
4.3
1 Evaluate exponential functions (p. 326)
4 Solve exponential equations (p. 335)
1 3–6 p. 334 7, 8
4.4
1 Change exponential statements to logarithmic statements
2, 3
15, 16
and logarithmic statements to exponential statements (p. 344) 2 Evaluate logarithmic expressions (p. 344)
4
14(b), (d), 19, 46(b), 48(a), 49 17, 18, 34(a) 34(b), 46(a) 37, 40, 46(c), 48(b) 20, 21
2 Graph exponential functions (p. 330) 3 Define the number e (p. 333)
3 Determine the domain of a logarithmic function (p. 345)
5 6, 7 8, 9
4.5
1 Work with the properties of logarithms (p. 356)
4.6
1 Solve logarithmic equations (p. 365)
4.7
1 Determine the future value of a lump sum of money (p. 372)
4 Graph logarithmic functions (p. 345) 5 Solve logarithmic equations (p. 350) 2 Write a logarithmic expression as a sum or difference of logarithms (p. 359) 3 Write a logarithmic expression as a single logarithm (p. 359) 4 Evaluate a logarithm whose base is neither 10 nor e (p. 360) 5 Graph a logarithmic function whose base is neither 10 nor e (p. 362) 2 Solve exponential equations (p. 367) 3 Solve logarithmic and exponential equations using a
1, 2 3–5 6 7, 8 9
22–25 26–28 29 30
1–3 4–6 7
37, 43 38, 42, 44, 45 35–45
1–3 4 5 6, 7
50, 55 50 51 50
graphing utility (p. 368) 2 Calculate effective rates of return (p. 375) 3 Determine the present value of a lump sum of money (p. 376) 4 Determine the rate of interest or the time required to double
a lump sum of money (p. 377) 4.8
1 Find equations of populations that obey the law
1, 2
54
of uninhibited growth (p. 381) 2 Find equations of populations that obey the law of decay (p. 383) 3 Use Newton’s Law of Cooling (p. 384) 4 Use logistic models (p. 386)
3 4 5, 6
52 53 56
1 2 3
57 58 59
1 Build an exponential model from data (p. 392)
4.9
2 Build a logarithmic model from data (p. 393) 3 Build a logistic model from data (p. 394)
Review Exercises 1. Evaluate each expression using the graphs of y = f 1x2 and y = g1x2 shown in the figure. y
(–5, 4) (–8, 2) (–2, 1)
(7, 6)
6 4 2
–10 –8 –6 –4 –2 –2 (0, –3) (–8, –2) –4 (–5, –4) –6
y = f (x) (4, 3) (10, 3) (2, 0) (7, 0) 2
4
(2, –4)
6
8 10 x (10, –2) y = g(x)
(0, –6) (4, –6)
(a) (b) (c) (d)
1g ∘ f2 1 - 82 - 4 1f ∘ g2 1 - 82 1 1g ∘ g2 172 - 6 1g ∘ f2 1 - 52 - 6
In Problems 2–4, for the given functions f and g find: (a) (b) (c) (d)
1f ∘ g2 132 - 65 1g ∘ f2 1 - 32 665 1f ∘ f2 122 23 1g ∘ g2 1 - 22 2
*2. f 1x2 = 3 - 4x; g1x2 = 3x2 - 10
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
402
CHAPTER 4 Exponential and Logarithmic Functions
*3. f 1x2 = 2x + 2: g1x2 = 2x2 + 1
*9. State why the graph of the function is one-to-one. Then draw the graph of the inverse function f -1. For convenience (and as a hint), the graph of y = x is also given.
*4. f 1x2 = 2x2 - 1; g1x2 = 3x - 5 In Problems 5–7, find f ∘ g, g ∘ f, f ∘ f, and g ∘ g for each pair of functions. State the domain of each composite function.
y 4
*5. f 1x2 = 2 - x; g1x2 = 3x + 1
y=x (3, 3)
*6. f 1x2 = 2x - 1; g1x2 = x2 + 3x + 1 *7. f 1x2 =
1 x + 1 ; g1x2 = x - 1 x
(2, 0) 4 x
–4
8. For the function { 13, 22, 15, 102, 12, 52, 17, 32}
(0, –2) (–1, –3) –4
(a) Verify that the function is one-to-one. (b) Find the inverse of the given function. (a) one-to-one (b) {(2, 3), (10, 5), (5, 2), (3, 7)}
In Problems 10–13, the function f is one-to-one. Find the inverse of each function and check your answer. State the domain and the range of f and f - 1 . 2x + 3 2x + 3 f - 1 1x2 = 5x - 2 5x - 2 *12. f 1x2 = 2x - 2 f - 1 1x2 = x2 + 2, x Ú 2
1 x + 1 f - 1 1x2 = x - 1 x *13. f 1x2 = x1>3 + 1 f - 1 1x2 = 1x - 12 3 *11. f 1x2 =
*10. f 1x2 =
14. If f 1x2 = 4x and g1x2 = log 2 x, evaluate each of the following. (a) f 122
1 1 (d) ga b - 3 64 8 15. Convert 35 = z to an equivalent statement involving a logarithm. log 3 z = 5 16
(b) g1162
4
(c) ƒ( - 32
16. Convert log3 z = 7 to an equivalent statement involving an exponent. 37 = z In Problems 17 and 18, find the domain of each logarithmic function. 2 18. H1x2 = log 2 1x2 - 3x + 22 a , qb 3 In Problems 19–21, evaluate each expression. Do not use a calculator. 17. f 1x2 = log13x - 22
19. log 3 27
20. ln e 22
3
22
( - q , 1) ∪ (2, q )
21. 5log5 0.3
0.3
In Problems 22–25, write each expression as the sum and/or difference of logarithms. Express powers as factors. xy 23. log 2 1a2 2b2 4, a 7 0, b 7 0 8 log 2 a + 2 log 2 6 *22. log 5 ¢ 2 ≤, x 7 0, y 7 0, z 7 0 z *24. lna
2x2 + 1 y 1x
2
b, x 7 0, y 7 0
25. ln¢
2x + 3 ≤ , x 7 2 2 ln(2x + 3) - 2 ln(x - 1) - 2 ln(x - 2) x2 - 3x + 2
In Problems 26–28, write each expression as a single logarithm. 3
26.
1 log 2 x3 - 3 log 2 1x2 + 12, x 7 0 2
28.
1 1 1 ln1x2 + 12 - 4 ln - 3ln1x - 42 + ln x4 2 2 2
x2 1 log 2 2 27. 3 ln x2 - ln 3 - 43ln1x2 + 32 - ln1x4 3 2 (x + 1) lnc
162x2 + 1 2x1x - 42
ln
x8 13(x2 + 3)4
d
29. Use the Change-of-Base Formula and a calculator to evaluate log 4 19 . Round your answer to three decimal places. 2.124 *30. Graph y = log 3 x using a graphing utility and the Change-of-Base Formula. In Problems 31–34, use the given function f to: (a) Find the domain of f. (b) Graph f. (c) From the graph, determine the range and any asymptotes of f. (d) Find f -1. (e) Find the domain and the range of f -1 . (f) Graph f -1 . *31. f 1x2 = 2x - 3
*32. f 1x2 = 1 + 3 - x
*33. f 1x2 = 3e x - 2
*34. f 1x2 =
1 ln1x + 32 2
In Problems 35–45, solve each equation. Express any irrational solution in exact form and as a decimal rounded to 3 decimal places. 5 1 2 35. 53x + 7 = 25 e - f *36. 3x + x = 23 37. log x 64 = - 3 e f 3 4 2 39. 252x = 5x - 12 5 - 2, 66 40. log 3 2x - 2 = 2 {83} *38. 5x = 3x + 2 1 x2 # 5x x# 41. 8 = 4 2 e , -3f 43. log 7 1x + 22 + log 7 1x - 42 = 1 {5} 42. 2 5 = 10x 516 2 *45. 9x + 4 # 3x - 3 = 0 44. e 1 - x = 5 {1 - ln 5} ≈ { - 0.609}
Chapter Review
*46. Suppose that f 1x2 = log 2 1x - 22 + 1. B (SBQIf. C 8IBUJTf 1 62 8IBUQPJOUJTPOUIFHSBQIPGf? D 4PMWFf 1x2 = 4.8IBUQPJOUJTPOUIFHSBQIPGf E #BTFEPOUIFHSBQIESBXOJOQBSU B
TPMWFf 1x2 7 0. F 'JOEf -1 1x2. Graph f -1 on the same Cartesian plane as f. 47. Amplifying Sound An amplifier’s power output P JOXBUUT is related to its decibel voltage gain d by the formula P = 25e 0.1d
B 'JOE UIF QPXFS PVUQVU GPS B EFDJCFM WPMUBHF HBJO PG decibels. 8 C 'PS B QPXFS PVUQVU PG XBUUT XIBU JT UIF EFDJCFM WPMUBHFHBJO 6.9 dB *48. Limiting Magnitude of a Telescope A telescope is limited in its usefulness by the brightness of the star that it is aimed at and by the diameter of its lens. One measure of a star’s brightness is its magnitude; the dimmer the star, the larger its magnitude. A formula for the limiting magnitude L of a UFMFTDPQFUIBUJT UIFNBHOJUVEFPGUIFEJNNFTUTUBSUIBUJU DBOCFVTFEUPWJFXJTHJWFOCZ L = 9 + 5.1 log d where dJTUIFEJBNFUFS JOJODIFT PGUIFMFOT B 8IBUJTUIFMJNJUJOHNBHOJUVEFPGBJODIUFMFTDPQF C 8IBUEJBNFUFSJTSFRVJSFEUPWJFXBTUBSPGNBHOJUVEF 49. Salvage Value The number of years n for a piece of machinery to depreciate to a known salvage value can be found using the formula log s - log i n = log 1 - d
where s is the salvage value of the machinery, i is its initial value, and d is the annual rate of depreciation. B )PXNBOZZFBSTXJMMJUUBLFGPSBQJFDFPGNBDIJOFSZUP decline in value from $90,000 to $10,000 if the annual SBUFPGEFQSFDJBUJPOJT ZS C )PXNBOZZFBSTXJMMJUUBLFGPSBQJFDFPGNBDIJOFSZUP lose half of its value if the annual rate of depreciation is ZS 50. Funding a College Education A child’s grandparents QVSDIBTF B CPOE GVOE UIBU NBUVSFT JO ZFBST UP be used for her college education. The bond fund pays 4% interest compounded semiannually. How much will the CPOEGVOECFXPSUIBUNBUVSJUZ 8IBUJTUIFFGGFDUJWFSBUFPG JOUFSFTU )PXMPOHXJMMJUUBLFUIFCPOEUPEPVCMFJOWBMVF VOEFSUIFTFUFSNT ZS 51. Funding a College Education A child’s grandparents wish to QVSDIBTFBCPOEUIBUNBUVSFTJOZFBSTUPCFVTFEGPSIFS college education. The bond pays 4% interest compounded semiannually. How much should they pay so that the bond XJMMCFXPSUI BUNBUVSJUZ 52. When Did a Prehistoric Man Die? The bones of a prehistoric NBO GPVOE JO UIF EFTFSU PG /FX .FYJDP DPOUBJO approximately 5% of the original amount of carbon-14. If UIFIBMGMJGFPGDBSCPOJTZFBST BQQSPYJNBUFMZIPX MPOHBHPEJEUIFNBOEJF ≈24,65 yr ago
403
53. Temperature of a Skillet A skillet is removed from an oven where the temperature is 450°F and placed in a room whose UFNQFSBUVSFJT'"GUFSNJOVUFT UIFUFNQFSBUVSFPGUIF skillet is 400°F. How long will it be until its temperature is ' 55.22 min, or 55 min, 13 sec 54. World Population The annual growth rate of the world’s population in 2012 was 1.1, = 0.011. The population PG UIF XPSME JO XBT -FUUJOH t = 0 represent 2012, predict the world’s population in the year Source: U.S. Census Bureau *55. Federal Deficit In fiscal year 2012, the federal deficit was $1.1 trillion. At that time, 10-year Treasury notes were QBZJOHJOUFSFTUQFSBOOVN*GUIFGFEFSBMHPWFSONFOU financed this deficit through 10-year notes, how much would JUIBWFUPQBZCBDLJO )PXNVDIXBTUIFJOUFSFTU Source: U.S. Treasury Department 56. Logistic Growth The logistic growth model 0. P 1t2 = 1 + 1.6e -0.16t represents the proportion of new cars with a global QPTJUJPOJOH TZTUFN (14 -FU t = 0 represent 2006, t = 1 SFQSFTFOU BOETPPO B 8IBUQSPQPSUJPOPGOFXDBSTJOIBEB(14 0.3 C %FUFSNJOF UIF NBYJNVN QSPQPSUJPO PG OFX DBST UIBU IBWFB(14 * D 6TJOHBHSBQIJOHVUJMJUZ HSBQIP = P 1t2. E 8IFOXJMMPGOFXDBSTIBWFB(14 In 2026 57. CBL Experiment The following data were collected by placing a temperature probe in a portable heater, removing the probe, and then recording the temperature of the probe over time. Time (s)
Temperature (°F)
0
165.07
1
164.77
2
163.99
3
163.22
4
162.82
5
161.96
6
161.20
7
160.45
8
159.35
9
158.61
10
157.89
11
156.83
12
156.11
13
155.08
14
154.40
15
153.72
According to Newton’s Law of Cooling, these data should follow an exponential model. * B 6TJOHBHSBQIJOHVUJMJUZ ESBXBTDBUUFSEJBHSBNGPSUIFEBUB C 6TJOH B HSBQIJOH VUJMJUZ CVJME BO FYQPOFOUJBM NPEFM from the data. y = 165.310.99512 x
404
CHAPTER 4 Exponential and Logarithmic Functions
*(c) Graph the exponential function found in part (b) on the scatter diagram. (d) Predict how long it will take for the probe to reach a temperature of 110°F. ≈83 s
cold. Those who come in contact with someone who has this cold will themselves catch the cold. The following data represent the number of people in the small town who have caught the cold after t days.
58. Wind Chill Factor The data below represent the wind speed (mph) and wind chill factor at an air temperature of 15°F. Days, t Wind Speed (mph)
Wind Chill Factor (°F)
5
7
10
3
15
0
20
⫺2
25
⫺4
30
⫺5
35 ⫺7 Source: U.S. National Weather Service
*(a) Using a graphing utility, draw a scatter diagram with wind speed as the independent variable. (b) Using a graphing utility, build a logarithmic model from the data. y = 18.921 - 7.096 ln x *(c) Using a graphing utility, draw the logarithmic function found in part (b) on the scatter diagram. (d) Use the function found in part (b) to predict the wind chill factor if the air temperature is 15°F and the wind speed is 23 mph. ≈ - 3°F 59. Spreading of a Disease Jack and Diane live in a small town of 50 people. Unfortunately, both Jack and Diane have a
1. Given f 1x2 =
5
2. Determine whether the function is one-to-one. (a) y = 4x2 + 3 not one-to-one (b) y = 2x + 3 - 5 one-to-one 2 and check your answer. *3. Find the inverse of f 1x2 = 3x - 5 State the domain and the range of f and f -1. 4. If the point 13, - 52 is on the graph of a one-to-one function f, what point must be on the graph of f -1? 1 - 5, 32
In Problems 5–7, solve each equation. 5. 3x = 243
{5}
6. log b16 = 2
{4}
7. log 5 x = 4
{625}
In Problems 8–11, use a calculator to evaluate each expression. Round your answer to three decimal places. 8. e + 2
22.086
3
9. log 20
1.301
10. log 321
2.771
11. ln 133
4.890
2 4 8 14 22 30 37 42 44
*(a) Using a graphing utility, draw a scatter diagram of the data. Comment on the type of relation that appears to exist between the day and the number of people with a cold. *(b) Using a graphing utility, build a logistic model from the data. *(c) Graph the function found in part (b) on the scatter diagram. (d) According to the function found in part (b), what is the maximum number of people who will catch the cold? In reality, what is the maximum number of people who could catch the cold? ≈47; 50 *(e) Sometime between the second and third day, 10 people in the town had a cold. According to the model found in part (b), when did 10 people have a cold? *(f) How long will it take for 46 people to catch the cold?
Chapter Test Prep Videos include step-by-step solutions to all chapter test exercises and can be found on this text’s Channel. (search “SullivanPrecalcUC3e”)
Chapter Test x + 2 and g1x2 = 2x + 5, find: x - 2 (b) 1g ∘ f2 1 - 22 *(a) f ∘ g and state its domain (c) 1f ∘ g2 1 - 22 - 3
Number of People with Cold, C
0 1 2 3 4 5 6 7 8
In Problems 12 and 13, use the given function f to: (a) Find the domain of f. (b) Graph f. (c) From the graph, determine the range and any asymptotes of f. (d) Find f - 1 , the inverse of f. (e) Find the domain and the range of f - 1 . (f) Graph f - 1 .
* 12. f 1x2 = 4x + 1 - 2 *13. f 1x2 = 1 - log 5 1x - 22 In Problems 14–19, solve each equation. 14. 5x + 2 = 125 * 16. 8 - 2e
-x
{1}
= 4
* 18. 7x + 3 = e x
15. log1x + 92 = 2
{91}
* 17. log1x + 32 = log1x + 62 2
*19. log 2 1x - 42 + log 2 1x + 42 = 3
4x3 ≤ as the sum and/or difference of x - 3x - 18 logarithms. Express powers as factors. 21. A 50-mg sample of a radioactive substance decays to 34 mg after 30 days. How long will it take for there to be 2 mg remaining? ≈250.39 days
* 20. Write log 2 ¢
2
22. (a) If $1000 is invested at 5% compounded monthly, how much is there after 8 months? $1033.82 (b) If you want to have $1000 in 9 months, how much do you need to place in a savings account now that pays 5% compounded quarterly? $963.42 (c) How long does it take to double your money if you can invest it at 6% compounded annually? 11.9 yr
Cumulative Review
23. The decibel level, D, of sound is given by the equation I D = 10 log¢ ≤, where I is the intensity of the sound and I0 I0 = 10-12 watt per square meter. B *G UIF TIPVU PG B TJOHMF QFSTPO NFBTVSFT EFDJCFMT how loud will the sound be if two people shout at the TBNFUJNF 5IBUJT IPXMPVEXPVMEUIFTPVOECFJGUIF JOUFOTJUZEPVCMFE ≈3 dB
405
C 5IFQBJOUISFTIPMEGPSTPVOEJTEFDJCFMT*GUIF"UIFOT 0MZNQJD4UBEJVN 0MZNQJBLP4UBEJP"UIJOBTA4QZSPT -PVJT DBO TFBU QFPQMF IPX NBOZ QFPQMF JO UIF crowd need to shout at the same time for the resulting TPVOEMFWFMUPNFFUPSFYDFFEUIFQBJOUISFTIPME *HOPSF BOZQPTTJCMFTPVOEEBNQFOJOH 31,623 people
Cumulative Review 1. B *TUIFGPMMPXJOHHSBQIUIFHSBQIPGBGVODUJPO *GJUJT JT UIFGVODUJPOPOFUPPOF Yes; no y 4
*9. Graph f1x2 = 31x + 12 3 - 2 using transformations. 2 *10. B (JWFO UIBU f 1x2 = x2 + 2 and g1x2 = find x - 3, f 1g1x2 2 BOETUBUFJUTEPNBJO8IBUJTf 1g152 2 C *G f 1x2 = x + 2 and g1x2 = log 2 x, find 1f 1g1x2 2 and TUBUFJUTEPNBJO8IBUJTf 1g1142 2
4 x
–4
–4
* C "TTVNJOHUIFHSBQIJTBGVODUJPO XIBUUZQFPGGVODUJPO NJHIUJUCF QPMZOPNJBM FYQPOFOUJBM BOETPPO 8IZ 2. For the function f1x2 = 2x2 - 3x + 1, find the following: B f 132 10 * C f 1 - x2 * D f 1x + h2 3. %FUFSNJOF XIJDI PG UIF GPMMPXJOH QPJOUT JT PS BSF PO UIF graph of x2 + y2 = 1. 0OMZ C JTPOUIFHSBQI 1 1 1 23 B a , b C a , b 2 2 2 2 4. Solve the equation 31x - 22 = 41x + 52. { - 26} *5. Graph the line 2x - 4y = 16.
6.* B (SBQIUIFRVBESBUJDGVODUJPO f 1x2 = - x2 + 2x - 3 by determining whether its graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and xJOUFSDFQU T
JGBOZ C 4PMWFf 1x2 … 0. All real numbers; 1 - q , q 2 7. %FUFSNJOF UIF RVBESBUJD GVODUJPO XIPTF HSBQI JT HJWFO JO the figure. f 1x2 = 21x - 42 2 - = 2x2 - 16x + 24
*11. For the polynomial function f 1x2 = 4x3 + 9x2 - 30x - : B 'JOEUIFSFBM[FSPTPGf. C %FUFSNJOFUIFJOUFSDFQUTPGUIFHSBQIPGf. D 6TFBHSBQIJOHVUJMJUZUPBQQSPYJNBUFUIFMPDBMNBYJNB and local minima. E %SBX B DPNQMFUF HSBQI PG f. Be sure to label the intercepts and turning points. *12. For the function g1x2 = 3x + 2: B (SBQIg using transformations. State the domain, range, and horizontal asymptote of g. C %FUFSNJOFUIFJOWFSTFPGg. State the domain, range, and vertical asymptote of g -1. g - 1 1x2 = log 3 1x - 22 D 0OUIFTBNFHSBQIBTg, graph g -1. *13. Solve the equation: 4x - 3 = 2x
*14. Solve the equation: log 3 1x + 12 + log 3 12x - 32 = log 9 9 15. Suppose that f 1x2 = log 3 1x + 22. Solve: B f 1x2 = 0 { - 1} C f 1x2 7 0 - 1, q
D f 1x2 = 3 {25}
16. Data Analysis The following data represent the percent of all drivers by age who have been stopped by the police for any reason within the past year. The median age represents the midpoint of the upper and lower limit for the age range.
y 50
Age Range (0, 24)
–2
4 –10
8
x
Vertex: (4, –8)
8. Is the graph that of a polynomial, exponential, or logarithmic GVODUJPO %FUFSNJOF UIF function whose graph is given. Exponential; f 1x2 = 2 # 3x
y 6
(1, 6)
4
(–1, 2–3) –4
2
–2
(0, 2) 2
–2
4 x
Median Age, x
Percent Stopped, y
16–19
17.5
18.2
20–29
24.5
16.8
30–39
34.5
11.3
40–49
44.5
9.4
50–59
54.5
7.7
≥60
69.5
3.8
* B 6TJOHZPVSHSBQIJOHVUJMJUZ ESBXBTDBUUFSEJBHSBNPGUIF data treating median age, x, as the independent variable. * C %FUFSNJOF B NPEFM UIBU ZPV GFFM CFTU EFTDSJCFT UIF relation between median age and percent stopped. You may choose from among linear, quadratic, cubic, exponential, logarithmic, and logistic models. D 1SPWJEFBKVTUJGJDBUJPOGPSUIFNPEFMUIBUZPVTFMFDUFEJO QBSU C Highest value of 0 r 0
406
CHAPTER 4 Exponential and Logarithmic Functions
Chapter Projects diagram. To do this in Excel, click on any data point in the scatter diagram. Now click the Layout menu, TFMFDU5SFOEMJOFXJUIJOUIF"OBMZTJTSFHJPO TFMFDU.PSF Trendline Options. Select the Exponential radio button BOE TFMFDU %JTQMBZ &RVBUJPO PO $IBSU 4FF 'JHVSF .PWFUIF5SFOEMJOF0QUJPOTXJOEPXPGGUPUIFTJEF BOE you will see the exponential function of best fit displayed PO UIF TDBUUFS EJBHSBN %P ZPV UIJOL UIF GVODUJPO accurately describes the relation between age of the car BOETVHHFTUFESFUBJMQSJDF Figure 57
Internet-based Project I.
Depreciation of Cars ,FMMFZ #MVF #PPL JT BO PGGJDJBM HVJEF that provides the current retail price of cars. You can access the ,FMMFZ#MVF#PPLBUZPVSMJCSBSZPSPOMJOFBUwww.kbb.com. 1.
Identify three cars that you are considering purchasing, BOE GJOE UIF ,FMMFZ #MVF #PPL WBMVF PG UIF DBST GPS CSBOE OFX
BOE ZFBST PG BHF 0OMJOF UIF WBMVF PG UIF DBS DBO CF GPVOE CZ TFMFDUJOH 6TFE $BST UIFO6TFE$BS7BMVFT&OUFSUIFZFBS NBLF BOENPEFM of the car you are selecting. To be consistent, we will assume the cars will be driven 12,000 miles per year, so a 1-year-old car will have 12,000 miles, a 2-year-old car will have 24,000 miles, and so on. Choose the same options for each year, and finally determine the suggested retail price for cars that are in Excellent, Good, and Fair shape. :PV TIPVME IBWF B UPUBM PG PCTFSWBUJPOT POF GPS B brand new car, 3 for a 1-year-old car, 3 for a 2-year-old DBS BOETPPO
2.
%SBX B TDBUUFS EJBHSBN PG UIF EBUB XJUI BHF BT UIF independent variable and value as the dependent variable using Excel, a TI-graphing calculator, or some other spreadsheet. The Chapter 2 project describes how to draw a scatter diagram in Excel.
3.
%FUFSNJOF UIF FYQPOFOUJBM GVODUJPO PG CFTU GJU (SBQI the exponential function of best fit on the scatter
4.
The exponential function of best fit is of the form y = Ce rx, where y is the suggested retail value of the car and xJTUIFBHFPGUIFDBS JOZFBST 8IBUEPFTUIFWBMVF of C SFQSFTFOU 8IBU EPFT UIF WBMVF PG r SFQSFTFOU 8IBUJTUIFEFQSFDJBUJPOSBUFGPSFBDIDBSUIBUZPVBSF DPOTJEFSJOH
5.
8SJUF B SFQPSU EFUBJMJOH XIJDI DBS ZPV XPVME QVSDIBTF based on the depreciation rate you found for each car.
The following projects are available on the Instructor’s Resource Center (IRC): II. Hot Coffee A fast-food restaurant wants a special container to hold coffee. The restaurant wishes the container to quickly cool the coffee from 200° to 130°F and keep the liquid between 110° and 130°F as long as possible. The restaurant has three containers to TFMFDUGSPN8IJDIPOFTIPVMECFQVSDIBTFE III. Project at Motorola Thermal Fatigue of Solder Connections 1SPEVDU SFMJBCJMJUZ JT B NBKPS DPODFSO PG B NBOVGBDUVSFS )FSF B logarithmic transformation is used to simplify the analysis of a cell phone’s ability to withstand temperature change. Citation:&YDFM¥.JDSPTPGU$PSQPSBUJPO6TFEXJUIQFSNJTTJPOGSPN.JDSPTPGU
5
Trigonometric Functions Length of Day Revisited The length of a day depends on the day of the year as well as on the latitude of the location. Latitude gives the location of a point on Earth north or south of the equator. In Chapter 3 we found a model that describes the relation between the length of day and latitude for a specific day of the year. In the Internet Project at the end of this chapter, we will find a model that describes the relation between the length of day and day of the year for a specific latitude.
—See the Internet-based Chapter Project I—
In this chapter we define the trigonometric functions, six functions that have wide application. We shall talk about their domain and range, see how to find values for them, graph them, and develop a list of their properties. There are two widely accepted approaches to the development of the trigonometric functions: one uses right triangles; the other uses circles, especially the unit circle. In this book, we develop the trigonometric functions using the unit circle. In Section 7.1, we present right triangle trigonometry.
5.3 5.4 5.5 5.6
Angles and Their Measure Trigonometric Functions: Unit Circle Approach Properties 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 Chapter Review Chapter Test Cumulative Review Chapter Projects
407
408
CHAPTER 5 Trigonometric Functions
5.1 Angles and Their Measure PREPARING FOR THIS SECTION Before getting started, review the following: r $JSDVNGFSFODFBOE"SFBPGB$JSDMF "QQFOEJY" 4FDUJPO" Q"
r 6OJGPSN.PUJPO "QQFOEJY" 4FDUJPO" QQ"m"
Now Work the ‘Are You Prepared?’ problems on page 417.
OBJECTIVES 1 Convert between Decimals and Degrees, Minutes, Seconds Measures for Angles (p. 410) 2 Find the Length of an Arc of a Circle (p. 411) 3 Convert from Degrees to Radians and from Radians to Degrees (p. 412) 4 Find the Area of a Sector of a Circle (p. 415) 5 Find the Linear Speed of an Object Traveling in Circular Motion (p. 416)
Figure 1 V
Ray
Line
"ray, or half-line, is that portion of a line that starts at a point V on the line and FYUFOETJOEFGJOJUFMZJOPOFEJSFDUJPO5IFTUBSUJOHQPJOUV of a ray is called its vertex. 4FF'JHVSF *GUXPSBZTBSFESBXOXJUIBDPNNPOWFSUFY UIFZGPSNBOangle. We call one ray of an angle the initial side and the other the terminal side. The angle formed is identified by showing the direction and amount of rotation from the initial side to the terminal side. If the rotation is in the counterclockwise direction, the angle is positive; if the rotation is clockwise, the angle is negative.4FF'JHVSF
Figure 2 ide al s
al min
Vertex Initial side
e
al min
Vertex
e sid
Ter
Ter

n mi Ter ␣
sid
Initial side
␥
Vertex
Initial side
Counterclockwise rotation Positive angle
Clockwise rotation Negative angle
Counterclockwise rotation Positive angle
(a)
(b)
(c)
Lowercase Greek letters, such as a BMQIB
b CFUB
g HBNNB
BOEu UIFUB
BSF PGUFOVTFEUPEFOPUFBOHMFT/PUJDFJO'JHVSF B UIBUUIFBOHMFa is positive because the direction of the rotation from the initial side to the terminal side is counterclockwise. The angle bJO'JHVSF C JTOFHBUJWFCFDBVTFUIFSPUBUJPOJTDMPDLXJTF5IFBOHMFg in 'JHVSF D JT QPTJUJWF /PUJDF UIBU UIF BOHMF a JO 'JHVSF B BOE UIF BOHMF g in 'JHVSF D IBWF UIF TBNF JOJUJBM TJEF BOE UIF TBNF UFSNJOBM TJEF )PXFWFS UIF measures of a and g are unequal, because the amount of rotation required to go from the initial side to the terminal side is greater for angle g than for angle a. "O BOHMF u is said to be in standard position JG JUT WFSUFY JT BU UIF PSJHJO PG B rectangular coordinate system and its initial side coincides with the positive xBYJT See Figure 3. y
Figure 3
Terminal side Vertex
y
Initial side
x Terminal side
(a) is in standard position; is positive
Vertex Initial side x
(b) is in standard position; is negative
SECTION 5.1 Angles and Their Measure
409
When an angle u is in standard position, the terminal side will lie either in a quadrant, in which case we say that u lies in that quadrant, or the terminal side will lie on the xBYJTPSUIFyBYJT JOXIJDIDBTFXFTBZUIBUu is a quadrantal angle. For FYBNQMF UIFBOHMFuJO'JHVSF B MJFTJORVBESBOU** UIFBOHMFuJO'JHVSF C MJFT in quadrant IV, and the angle uJO'JHVSF D JTBRVBESBOUBMBOHMF y
Figure 4
y
y
x
(a) lies in quadrant II
x
x
(b) lies in quadrant IV
(c) is a quadrantal angle
"OHMFT BSF NFBTVSFE CZ EFUFSNJOJOH UIF BNPVOU PG SPUBUJPO OFFEFE GPS UIF initial side to become coincident with the terminal side. The two commonly used measures for angles are degrees and radians.
Degrees HISTORICAL NOTE One counterclockwise rotation is 360° due to the Babylonian year, which had 360 days. j
Figure 5
5IFBOHMFGPSNFECZSPUBUJOHUIFJOJUJBMTJEFFYBDUMZPODFJOUIFDPVOUFSDMPDLXJTF EJSFDUJPOVOUJMJUDPJODJEFTXJUIJUTFMG SFWPMVUJPO JTTBJEUPNFBTVSFEFHSFFT BCCSFWJBUFEOne degree, 1°, is SFWPMVUJPO"right angle is an angle that 3 NFBTVSFT PS revolution; a straight angle JT BO BOHMF UIBU NFBTVSFT PS 4 SFWPMVUJPO 4FF 'JHVSF "T 'JHVSF C TIPXT JU JT DVTUPNBSZ UP JOEJDBUF a right angle by using the symbol . y
y
y
Terminal side
Terminal side x
Initial side Vertex
x
Vertex Initial side (b) right angle, 1–4 revolution counterclockwise, 90°
(a) 1 revolution counterclockwise, 360°
Terminal side Vertex
x
Initial side 1
(c) straight angle, –2 revolution counterclockwise, 180°
It is also customary to refer to an angle that measures u degrees as an angle of u degrees.
EXAM PL E 1
Drawing an Angle Draw each angle in standard position. B C - ° D E B "OBOHMFPGJT of a right angle. 4FF'JHVSF Figure 6
revolution in 4 UIFDMPDLXJTFEJSFDUJPO4FF'JHVSF
C "OBOHMFPG - ° is
Figure 7
y Te rm in al sid e
Solution
Vertex
Initial side x
Terminal side
45°
Vertex Initial side
y
x
90°
CHAPTER 5 Trigonometric Functions
D "OBOHMFPGDPOTJTUTPGB SPUBUJPOUISPVHIGPMMPXFECZB SPUBUJPOUISPVHI4FF'JHVSF
E "OBOHMFPGDPOTJTUTPGSFWPMVUJPO GPMMPXFECZBSPUBUJPOUISPVHI 4FF'JHVSF
Figure 8
Figure 9
y
y
225° Initial side x
405°
sid e
Vertex
Te rm in al sid e
410
al rm
in
x
Initial side
Vertex
Te
r
Now Work
PROBLEM
11
1 Convert between Decimals and Degrees, Minutes, Seconds Measures for Angles "MUIPVHITVCEJWJTJPOTPGBEFHSFFNBZCFFYQSFTTFEVTJOHEFDJNBMT UIFOPUJPOPG minutes and seconds may also be used. One minute, denoted by 1′, is defined as degree. One second, denoted by 1″, is defined as minute, or, equivalently, EFHSFF"OBOHMFPG TBZ EFHSFFT NJOVUFT TFDPOETJTXSJUUFODPNQBDUMZ 3 as 3°4′″. To summarize: counterclockwise revolution = 3° ° = ′ ′ = ″
(1)
It is sometimes necessary to convert from the degree, minute, second notation 1D°.′S″2 to a decimal form, and vice versa.
EX A MPL E 2
Converting between Degrees, Minutes, Seconds, and Decimal Forms B $POWFSU °′″ to a decimal in degrees. Round the answer to four decimal places. C $POWFSUUPUIFEFHSFF NJOVUF TFDPOE D°.′S″ OPUBUJPO3PVOEUIF answer to the nearest second.
Solution
B #FDBVTF′ = a
5 = b and ″ = a b = a
#
5 b , convert as follows:
a
°′″ = ° + ′ + ″
= ° + # ′ + # ″
COMMENT Graphing calculators BOETPNFTDJFOUJGJDDBMDVMBUPST IBWF the ability to convert from degrees, minutes, seconds to decimal form, and vice versa. Consult your owner’s manual. ■
= ° +
#
a
5 b +
≈ ° + .° + .° = .°
#
#
5 b
Convert minutes and seconds to degrees.
SECTION 5.1 Angles and Their Measure
411
C #FDBVTF° = ′ and ′ = ″, proceed as follows: .° = = = = = = = = ≈
Now Work
° + .° ° + . # ° ° + 1.2 1′2 ° + .3′ ° + ′ + .3′ ° + ′ + .3 # ′ ° + ′ + 1.32 1″2 ° + ′ + .″ °′″
PROBLEMS
23
AND
Convert fraction of degree to minutes;1° = 60′.
Convert fraction of minute to seconds; 1′ = 60″. Round to the nearest second.
r
29
In NBOZ BQQMJDBUJPOT TVDI BT EFTDSJCJOH UIF FYBDU MPDBUJPO PG B TUBS PS UIF precise position of a ship at sea, angles measured in degrees, minutes, and even seconds are used. For calculation purposes, these are transformed to decimal form. In other applications, especially those in calculus, angles are measured using radians.
Radians "central angleJTBQPTJUJWFBOHMFXIPTFWFSUFYJTBUUIFDFOUFSPGBDJSDMF5IFSBZT PGBDFOUSBMBOHMFTVCUFOE JOUFSTFDU BOBSDPOUIFDJSDMF*GUIFSBEJVTPGUIFDJSDMFJT r and the length of the arc subtended by the central angle is also r, then the measure of the angle is 1 radian.4FF'JHVSF B
Ter mi
Ter
nal si
de
mi nal sid e
Figure 10
r
1 radian
3
r r
1 Initial side
Initial side
1 1 radian
(a)
3
(b)
'PSBDJSDMFPGSBEJVT UIFSBZTPGBDFOUSBMBOHMFXJUINFBTVSFSBEJBOTVCUFOE BOBSDPGMFOHUI'PSBDJSDMFPGSBEJVT UIFSBZTPGBDFOUSBMBOHMFXJUINFBTVSF SBEJBOTVCUFOEBOBSDPGMFOHUI4FF'JHVSF C Figure 11 u s = u1 s1
2 Find the Length of an Arc of a Circle s
1 s1
r
Now consider a circle of radius r and two central angles, u and u, measured in radians. Suppose that these central angles subtend arcs of lengths s and s, SFTQFDUJWFMZ BT TIPXO JO 'JHVSF 'SPN HFPNFUSZ UIF SBUJP PG UIF NFBTVSFT PG the angles equals the ratio of the corresponding lengths of the arcs subtended by these angles; that is, u s = s u
(2)
412
CHAPTER 5 Trigonometric Functions
Suppose that u = SBEJBO3FGFSBHBJOUP'JHVSF B 5IFMFOHUIs of the arc subtended by the central angle u = radian equals the radius r of the circle. Then s = r TPFRVBUJPO SFEVDFTUP u s = r
THEOREM
or s = ru
(3)
Arc Length For a circle of radius r, a central angle of u radians subtends an arc whose length s is (4)
NOTE Formulas must be consistent with regard to the units used. In equation (4), we write s = ru To see the units, however, go back to equation (3) and write s length units u radians = 1 radian r length units s length units = r length units
u radians 1 radian
The radians divide out, leaving s length units = 1r length units2u
s = ru
where u appears to be “dimensionless” but, in fact, is measured in radians. So, in using the formula s = r u, the dimension for u is radians, and any convenient unit of length (such as inches or meters) may be used for s and r. j
EX A MPL E 3
Finding the Length of an Arc of a Circle 'JOEUIFMFOHUIPGUIFBSDPGBDJSDMFPGSBEJVTNFUFSTTVCUFOEFECZBDFOUSBMBOHMF PGSBEJBO
Solution
6TFFRVBUJPO XJUIr = meters and u = .. The length s of the arc is
r
s = ru = 1.2 = . meter
Now Work
PROBLEM
71
3 Convert from Degrees to Radians and from Radians to Degrees Figure 12 1 revolution = 2p radians s 2r 1 revolution
With two ways to measure angles, it is important to be able to convert from one to the other. Consider a circle of radius r"DFOUSBMBOHMFPG SFWPMVUJPOXJMMTVCUFOEBO BSDFRVBMUPUIFDJSDVNGFSFODFPGUIFDJSDMF 'JHVSF #FDBVTFUIFDJSDVNGFSFODFPG a circle of radius r equals pr, we substitute pr for sJOFRVBUJPO UPGJOEUIBU GPS an angle uPGSFWPMVUJPO s = ru
r
pr = ru u = p radians
u = 1 revolution; s = 2pr Solve for u.
From this, we have
SFWPMVUJPO = p radians
(5)
SECTION 5.1 Angles and Their Measure
413
4JODFSFWPMVUJPO = 3°, we have 3° = p radians %JWJEJOHCPUITJEFTCZZJFMET ° = p radians
(6)
%JWJEFCPUITJEFTPGFRVBUJPO CZ5IFO degree =
p radian
%JWJEFCPUITJEFTPG CZp. Then degrees = radian p We have the following two conversion formulas:* degree =
EXAM PL E 4
p radian
radian =
degrees p
(7)
Converting from Degrees to Radians Convert each angle in degrees to radians. B C D - 4° E F
Solution
B ° = # degree =
#
p p radian = radians 3 p p C ° = # ° = # radian = radians p p D - 4° = - 4 # radian = - radian 4 p p E ° = # radian = radians p F ° = # radian ≈ . radians
r
&YBNQMF QBSUT B m E
JMMVTUSBUFT UIBU BOHMFT UIBU BSFiOJDFu GSBDUJPOT PG B SFWPMVUJPOBSFFYQSFTTFEJOSBEJBONFBTVSFBTGSBDUJPOBMNVMUJQMFTPGp, rather than as p EFDJNBMT'PSFYBNQMF BSJHIUBOHMF BTJO&YBNQMF E
JTMFGUJOUIFGPSN radians, p 3.4 XIJDI JT FYBDU SBUIFS UIBO VTJOH UIF BQQSPYJNBUJPO ≈ = . radians. 8IFOUIFGSBDUJPOTBSFOPUiOJDF uVTFUIFEFDJNBMBQQSPYJNBUJPOPGUIFBOHMF BT JO&YBNQMF F
Now Work
EXAM PL E 5
PROBLEMS
35
AND
61
Converting Radians to Degrees Convert each angle in radians to degrees. 3p 3p p radian C radians D radians 4 p E radians F SBEJBOT 3 B
*
Some students prefer instead to use the proportion
given and solve for the measurement sought.
Degree °
=
Radian . Then substitute for what is p
414
CHAPTER 5 Trigonometric Functions
Solution
p p # p # radian = radian = degrees = 3° p 3p # 3p radians = C degrees = ° p 3p # 3p radians = D degrees = - 3° p 4 4 p # p radians = E degrees = 4° p 3 3 F 4 radians = 4 # degrees ≈ .° p B
Now Work
PROBLEM
r
47
5BCMFMJTUTUIFEFHSFFBOESBEJBONFBTVSFTPGTPNFDPNNPOMZFODPVOUFSFE angles. You should learn to feel equally comfortable using degree or radian measure for these angles.
Table 1
EX A MPL E 6
Degrees
0°
30° p 6
45° p 4
60° p 3
90° p 2
120° 2p 3
135° 3p 4
150° 5p 6
180°
Radians
0
Degrees
210°
225°
240°
270°
300°
315°
330°
360°
Radians
7p 6
5p 4
4p 3
3p 2
5p 3
7p 4
11p 6
2p
p
Finding the Distance between Two Cities The latitude of a location L is the measure of the angle formed by a ray drawn from the center of Earth to the equator and a ray drawn from the center of Earth to L4FF'JHVSF B 4JPVY'BMMT 4PVUI%BLPUB JTEVFOPSUIPG%BMMBT 5FYBT'JOE UIFEJTUBODFCFUXFFO4JPVY'BMMT ¿OPSUIMBUJUVEF BOE%BMMBT ¿ north MBUJUVEF 4FF'JHVSF C "TTVNFUIBUUIFSBEJVTPG&BSUIJTNJMFT
Figure 13
North Pole
North Pole
L
Sioux Falls 43°33' Dallas
θ° Equator
South Pole
South Pole (a)
Solution
Equator
32°46'
(b)
5IFNFBTVSFPGUIFDFOUSBMBOHMFCFUXFFOUIFUXPDJUJFTJT¿ - ¿ =¿. 6TF FRVBUJPO
s = ru #VU SFNFNCFS UP àSTU DPOWFSU UIF BOHMF PG ¿ to radians. u = °4′ ≈ .33° = .33 c 47′ = 47a
1 b° 60
#
p radian ≈ . radian
SECTION 5.1 Angles and Their Measure
415
6TF u = . radian and r = 3 miles JO FRVBUJPO 5IF EJTUBODF CFUXFFO the two cities is s = ru = 3 # . ≈ 44 miles
NOTE If the measure of an angle is given as 5, it is understood to mean 5 radians; if the measure of an angle is given as 5°, it means 5 degrees.
j
r
When an angle is measured in degrees, the degree symbol will always be shown. )PXFWFS XIFO BO BOHMF JT NFBTVSFE JO SBEJBOT UIF VTVBM QSBDUJDF JT UP PNJU UIF p word radians. Thus, if the measure of an angle is given as , it is understood to mean p radian.
Now Work
PROBLEM
107
4 Find the Area of a Sector of a Circle Consider a circle of radius r. Suppose that u, measured in radians, is a central angle PGUIJTDJSDMF4FF'JHVSF8FTFFLBGPSNVMBGPSUIFBSFBAPGUIFTFDUPS TIPXO JOCMVF GPSNFECZUIFBOHMFu. Now consider a circle of radius r and two central angles u and u, both measured JO SBEJBOT 4FF 'JHVSF 'SPN HFPNFUSZ UIF SBUJP PG UIF NFBTVSFT PG UIF BOHMFT equals the ratio of the corresponding areas of the sectors formed by these angles. That is,
Figure 14 A θ r
u A = u A
Figure 15 u A = u1 A1
Suppose that u = p radians. Then A = area of the circle = pr . Solving for A, we find A
A = A
θ
u u = pr = r u u p c
r
A1
A1 = pr2
θ1
u1 = 2p
THEOREM
Area of a Sector The area A of the sector of a circle of radius r formed by a central angle of u radians is A =
EXAM PL E 7
r u
(8)
Finding the Area of a Sector of a Circle 'JOE UIF BSFB PG UIF TFDUPS PG B DJSDMF PG SBEJVT GFFU GPSNFE CZ BO BOHMF PG Round the answer to two decimal places.
Solution
6TFFRVBUJPO XJUIr = feet and u = 3° = u must be in radians.] A =
p SBEJBO - 2 3 - 2 = = C
r
*UJTJOUFSFTUJOHUPDPNQBSFUIFBOTXFSGPVOEJO&YBNQMF B XJUIUIFBOTXFS UP&YBNQMFPG4FDUJPO5IFSFJUXBTGPVOEUIBU cos
p = cos ° = 1 2 + 2 2
#BTFEPOUIJTBOEUIFSFTVMUPG&YBNQMF B
UIJTNFBOTUIBU 1 2 + 2 2
and
3 + 2
BSFFRVBM 4JODFFBDIFYQSFTTJPOJTQPTJUJWF ZPVDBOWFSJGZUIJTFRVBMJUZCZTRVBSJOH FBDIFYQSFTTJPO 5XPWFSZEJGGFSFOUMPPLJOH ZFUDPSSFDU BOTXFSTDBOCFPCUBJOFE depending on the approach taken to solve a problem.
Now Work
EX A MPL E 7
PROBLEM
19
Finding Exact Values Using Half-angle Formulas p If cos a = - , p 6 a 6 ,GJOEUIFFYBDUWBMVFPG a a a B sin C cos D tan
Solution
'JSTU PCTFSWFUIBUJGp 6 a 6
p p a p a , then 6 6 . As a result, lies in quadrant II.
a a B #FDBVTF lies in quadrant II, sin 7 , so use the + TJHOJOGPSNVMB B UPHFU - a- b a - cos a sin = = B R 2 = = = = H A 2
SECTION 6.6 Double-angle and Half-angle Formulas
549
a a C #FDBVTF lies in quadrant II, cos 6 , so use the - TJHOJOGPSNVMB C UPHFU + a- b a + cos a cos = = B R 2 = = = R 2 a a D #FDBVTF lies in quadrant II, tan 6 , so use the - TJHOJOGPSNVMB D UPHFU
a - cos a tan = = B + cos a
- a- b = = - + a- b b b
r
"OPUIFSXBZUPTPMWF&YBNQMF D JTUPVTFUIFSFTVMUTPGQBSUT B BOE C
a tan =
Now Work
PROBLEM
a 2 = - = a 2 cos sin
7(C), (D),
AND
(F)
a that does not contain + and - signs, making it NPSFVTFGVMUIBOGPSNVMB D 5PEFSJWFJU VTFUIFGPSNVMBT There is a formula for tan
- cos a = sin
a
Formula (9)
and a a a sin a = sin c a b d = sin cos
Double-angle Formula
Then - cos a = sin a
a
a a = = tan a a a sin cos cos sin
sin
Because it also can be shown that - cos a sin a = sin a + cos a UIJTSFTVMUTJOUIFGPMMPXJOHUXP)BMGBOHMF'PSNVMBT
Half-angle Formulas for tan
tan
A 2
a - cos a sin a = = sin a + cos a
(11)
550
CHAPTER 6 Analytic Trigonometry
8JUIUIJTGPSNVMB UIFTPMVUJPOUP&YBNQMF D DBOCFPCUBJOFEBTGPMMPXT cos a = -
p
p 6 a 6
sin a = - 2 - cos a = -
B
-
= = B
5IFO CZFRVBUJPO
- a- b a - cos a tan = = = = - sin a
6.6 Assess Your Understanding Concepts and Vocabulary 1. cos 1u2 = cos u - sin u = cos u - = - sin u.
4. True or False tan1u2 =
- cos u u 2. sin = 3. tan
tan u - tan u
True
5. True or False sin1u2 IBTUXPFRVJWBMFOUGPSNT
u - cos u = sin u
sin u cos u and sin u - cos u
'BMTF
6. True or False tan1u2 + tan1u2 = tan1u2
'BMTF
Skill Building In Problems 7–18, use the information given about the angle u, … u 6 p, to find the exact value of: u u u (a) sin1u2 (b) cos 1u2 (c) sin (d) cos (e) tan1u2 (f) tan p p p *7. sin u = , 6 u 6 *8. cos u = , 6 u 6 *9. tan u = , p 6 u 6 *10. tan u =
p , p 6 u 6
*11. cos u = -
2 ,
p 6 u 6 p
*12. sin u = -
2 ,
p 6 u 6 p
*13. sec u = , sin u 7
*14. csc u = - 2, cos u 6
*15. cot u = - , sec u 6
*16. sec u = , csc u 6
*17. tan u = - , sin u 6
*18. cot u = , cos u 6
In Problems 19–28, use the Half-angle Formulas to find the exact value of each expression. 19. sin .°
3 - 2
23. DPT 27. sina -
p b
3 + 2 -
20. DPT
3 + 2
24. TJO -
3 - 2
3 - 2
21. tan
p
*25. sec
p
28. cos a -
- 2
p b
22. tan
p
*26. csc
p
3 - 2
In Problems 29–40, use the figures to evaluate each function given that f1x2 = sin x, g1x2 = cos x, and h1x2 = tan x. y
y x2 ⫹ y2 ⫽ 1
x2 ⫹ y2 ⫽ 5
(x, 2)
␣
x
x (⫺1–4 , y)
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
- + 2
551
SECTION 6.6 Double-angle and Half-angle Formulas
29. f 1u2
-
u 31. ga b
-
33. h1u2
u 34. ha b
+ 2
35. g1a2
a 37. ƒa b
2
a 38. ga b
-
2
a 39. ha b
-
- cos 1u2 + cos 1u2.
*41. 4IPXUIBUsin u =
u *32. ƒa b
31 - 22
30. g1u2
2
36. f 1a2
2
40. h1a2
-
2
*42. 4IPXUIBUsin1u2 = 1cos u2 1 sin u - sin u2.
*43. Develop a formula for cos 1u2 as a third-degree polynomial in the variable cos u.
*44. Develop a formula for cos 1u2 as a fourth-degree polynomial in the variable cos u.
*45. 'JOEBOFYQSFTTJPOGPSsin1u2 as a fifth-degree polynomial *46. 'JOEBOFYQSFTTJPOGPScos 1u2 as a fifth-degree polynomial in the variable sin u. in the variable cos u. In Problems 47–68, establish each identity. cot u - cot u - tan u *47. cos u - sin u = cos 1u2 *48. = cos 1u2 *49. cot 1u2 = cot u + tan u cot u *50. cot 1u2 =
1cot u - tan u2
*53. cos 1u2 - sin 1u2 = cos 1u2 *56. sin u cos u =
3 - cos 1u2 4
y sec y + *59. cot = sec y - *62. *64.
*51. sec1u2 =
sec u - sec u
*54. 1 sin u cos u2 1 - sin u2 = sin1u2 *57. sec
*55.
u = + cos u
*63.
cos u + sin u cos u - sin u = tan1u2 cos u - sin u cos u + sin u
cos 1u2
sec u csc u
+ sin1u2
*58. csc
=
cot u - cot u +
u = - cos u
u *61. cos u = u + tan - tan
y *60. tan = csc y - cot y
sin u + cos u sin1u2 = sin u + cos u
sin1u2 sin u
-
cos 1u2 cos u
=
tan u - tan u - tan u *67. ln 0 sin u 0 = 1ln 0 - cos 1u2 0 - ln 2
*65. tan1u2 =
*66. tan u + tan1u + °2 + tan1u + °2 = tan1u2 *68. ln 0 cos u 0 =
*52. csc1u2 =
1ln 0 + cos 1u2 0 - ln 2
In Problems 69–78, solve each equation on the interval … u 6 p. *69. cos 1u2 + sin u =
*70. cos 1u2 = - sin u
p p p p e , , , f
72. sin1u2 = cos u
75. - sin u = cos 1u2
76. cos 1u2 + cos u + =
p p e , , p, f
e ,
p p , f
*74. cos 1u2 + cos 1u2 =
*73. sin1u2 + sin1u2 =
No real solution
77. tan1u2 + sin u =
71. cos 1u2 = cos u
78. tan1u2 + cos u =
e
p p , f
p p p p e , , , f
Mixed Practice In Problems 79–90, find the exact value of each expression. 79. sina sin- b
2
83. tanc cos- a - b d 87. sin a cos- b
80. sinc sin-
2 d
2
81. cos a sin- b
82. cos a cos- b
84. tana tan- b
85. sina cos- b
86. cos c tan- a - b d
-
88. cos a sin- b
89. seca tan- b
90. cscc sin- a - b d
-
In Problems 91–93, find the real zeros of each trigonometric function on the interval … u 6 p. p p p p 91. f 1x2 = sin1x2 - sin x , , p, 92. f 1x2 = cos 1x2 + cos x , p, 93. f 1x2 = cos 1x2 + sin x
p p ,
552
CHAPTER 6 Analytic Trigonometry
Applications and Extensions 94. Constructing a Rain Gutter A rain gutter is to be constructed PG BMVNJOVN TIFFUT JODIFT XJEF "GUFS NBSLJOH PGG B MFOHUI PG JODIFT GSPN FBDI FEHF UIF CVJMEFS CFOET UIJT length up at an angle u.4FFUIFJMMVTUSBUJPO5IFBSFBA of the opening as a function of u is given by A1u2 = sin u 1cos u + 2
R
° 6 u 6 °
45°
θ θ
θ
4 in
4 in
* B 4IPXUIBU
4 in
12 in
R 1u2 =
B *O DBMDVMVT ZPV XJMM CF BTLFE UP GJOE UIF BOHMF u that maximizes A by solving the equation
C *O DBMDVMVT ZPV XJMM CF BTLFE UP GJOE UIF BOHMF u that maximizes R by solving the equation
cos 1u2 + cos u = , ° 6 u 6 °
sin1u2 + cos 1u2 =
4PMWFUIJTFRVBUJPOGPSu.
* C What is the maximum area A of the opening?
p , or .° D 8IBU JT UIF NBYJNVN EJTUBODF R if v = feet per second? - 2 ≈ . ft
* D (SBQI A = A1u2, ° … u … °, and find the angle u that maximizes the area A. Also find the maximum area. $PNQBSFUIFSFTVMUTUPUIFBOTXFSGPVOEFBSMJFS
* B 4IPXUIBUUIFQSPKFDUFEJNBHFXJEUIJTHJWFOCZ u W = D tan
4PMWFUIJTFRVBUJPOGPSu.
* E (SBQIR = R 1u2, ° … u … °, and find the angle u that maximizes the distance R. Also find the maximum distance. Use v = GFFU QFS TFDPOE $PNQBSF UIF results with the answers found earlier.
95. Laser Projection In a laser projection system, the optical angle or scanning angle u is related to the throw distance D from the scanner to the screen and the projected image width W by the equation W D = csc u - cot u
v 2 3sin1u2 - cos 1u2 - 4
*98. Sawtooth Curve An oscilloscope often displays a sawtooth curve. This curve can be approximated by sinusoidal curves of varying periods and amplitudes. A first approximation to the sawtooth curve is given by y =
C 'JOEUIFPQUJDBMBOHMFJGUIFUISPXEJTUBODFJTGFFUBOE UIFQSPKFDUFEJNBHFXJEUIJTGFFU u = .° Source: Pangolin Laser Systems, Inc.
sin1px2 + sin1px2
4IPXUIBUy = sin1px2 cos 1px2.
V1
2B. Gm.V
Trig
TVline
OH1
*96. Product of Inertia The product of inertia for an area about inclined axes is given by the formula Iuy = Ix sin u cos u - Iy sin u cos u + Ixy 1cos u - sin u2 50mv
Obase1
4IPXUIBUUIJTJTFRVJWBMFOUUP Iuy =
Ix - Iy
sin1u2 + Ixy cos 1u2
Source: Adapted from Hibbeler, Engineering Mechanics: Statics, UIFE 1SFOUJDF)BMM¥. 97. Projectile Motion An object is propelled upward at an angle u, ° 6 u 6 °, to the horizontal with an initial velocity of v feet per second from the base of a plane that makes an angle of ° XJUI UIF IPSJ[POUBM 4FF UIF JMMVTUSBUJPO *G BJS resistance is ignored, the distance R that it travels up the inclined plane is given by the function
*99. Area of an Isosceles Triangle 4IPX UIBU UIF BSFB A of an isosceles triangle whose equal sides are of length s, and where u is the angle between them, is A =
3 cos 1pt2,
0 … t … 3
(b) At what times t does the graph of V touch the graph of y = e -t>3? When does the graph of V touch the graph of y = - e -t>3? (c) When is the voltage V between - 0.4 and 0.4 volt? *56. The Sawtooth Curve An oscilloscope often displays a sawtooth curve. This curve can be approximated by sinusoidal curves of varying periods and amplitudes. (a) Use a graphing utility to graph the following function, which can be used to approximate the sawtooth curve. 1 1 f 1x2 = sin12px2 + sin 14px2, 2 4
and y = sin12pht2
0 … x … 4
y = sin32p18522t4 + sin32p112092t4 Use a graphing utility to graph the sound emitted by touching 7. Touch-Tone phone
1
2
3
697 cycles/sec
4
5
6
770 cycles/sec
7
8
9
852 cycles/sec
*
0
#
941 cycles/sec
(b) A better approximation to the sawtooth curve is given by f 1x2 =
1 1 1 sin12px2 + sin 14px2 + sin 18px2 2 4 8
Use a graphing utility to graph this function for 0 … x … 4 and compare the result to the graph obtained in part (a). (c) A third and even better approximation to the sawtooth curve is given by 1 1 1 1 f 1x2 = sin12px2 + sin 14px2 + sin 18px2 + sin11px2 2 4 8 1
V1
2B. Gm.V
50mv
Trig
TVline
OH1
Obase1
1209 cycles/sec
1477 cycles/sec
1336 cycles/sec
*58. Use a graphing utility to graph the sound emitted by the * key on a Touch-Tone phone. See Problem 57. 59. CBL Experiment Pendulum motion is analyzed to estimate simple harmonic motion. A plot is generated with the position of the pendulum over time. The graph is used to find a sinusoidal curve of the form y = A cos 3B 1x - C2 4 + D. %FUFSNJOF UIF BNQMJUVEF QFSJPE BOE GSFRVFODZ "DUJWJUZ 3FBM8PSME.BUIXJUIUIF$#-4ZTUFN
60. CBL Experiment The sound from a tuning fork is collected PWFSUJNF%FUFSNJOFUIFBNQMJUVEF GSFRVFODZ BOEQFSJPE of the graph. A model of the form y = A cos 3B 1x - C2 4 is fitted to the data. (Activity 23, Real-World Math with the CBL System.)
Discussion and Writing 61. Use
a
graphing utility to graph the function sin x f 1x2 = , x 7 0. Based on the graph, what do you x sin x conjecture about the value of for x close to 0? x 62. Use a graphing utility to graph y = x sin x, y = x2 sin x, and y = x3 sin x for x 7 0. What patterns do you observe?
1 1 63. Use a graphing utility to graph y = sin x, y = 2 sin x, x x 1 and y = 3 sin x for x 7 0. What patterns do you observe? x 64. How would you explain to a friend what simple harmonic motion is? How would you explain damped motion?
Chapter Review
609
Retain Your Knowledge Problems 65–68 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your xy3 4x - 3 mind so that you are better prepared for the final exam. 65. f - 1(x) = 66. log 7 a b x - 1 x + y x - 3 4 p 65. The function f(x) = , x ≠ 4, is one-to-one. Find its 68. Given cos a = , 0 6 a 6 , find the exact value of: x - 4 5 2 inverse function. a 3210 a 210 a 1 66. Write as a single logarithm: log 7 x + 3 log 7 y - log 7 (x + y) (a) cos (b) sin (c) tan 2 10 2 10 2 3 67. Solve: log (x + 1) + log (x - 2) = 1 {4}
‘Are You Prepared?’ Answer 1. A = 5 T =
p 2
Chapter Review Things to Know Formulas Law of Sines (p. 577)
sin B sin C sin A = = a b c
Law of Cosines (p. 587–588)
c 2 = a2 + b2 - 2ab cos C b2 = a2 + c 2 - 2ac cos B a2 = b2 + c 2 - 2bc cos A
Area of a triangle (pp. 594–595)
K =
1 1 1 1 bh K = ab sin C K = bc sin A K = ac sin B 2 2 2 2
K = 2s 1s - a2 1s - b2 1s - c2
where s =
1 1a + b + c2 2
Objectives You should be able to N
Example(s)
Review Exercises
7.1
1
Find the value of trigonometric functions of acute BOHMFTVTJOHSJHIUUSJBOHMFT Q
1, 2
1, 2, 27
2
6TFUIFDPNQMFNFOUBSZBOHMFUIFPSFN Q
3
3–5
3
4PMWFSJHIUUSJBOHMFT Q
4, 5
4
4PMWFBQQMJFEQSPCMFNT Q
m
m m
7.2
1
Solve SAA or ASA triangles (p. 578)
1, 2
8, 19
2
Solve SSA triangles (p. 578)
3–5
3
Solve applied problems (p. 581)
32, 33
7.3
1
Solve SAS triangles (p. 588)
1
11, 15, 20
2
Solve SSS triangles (p. 589)
2
13, 14, 17
3
Solve applied problems (p. 589)
3
34
7.4
1
Find the area of SAS triangles (p. 594)
1
2
Find the area of SSS triangles (p. 595)
2
23, 24
7.5
1
#VJMEBNPEFMGPSBOPCKFDUJOTJNQMFIBSNPOJDNPUJPO Q
1
39
2
"OBMZ[FTJNQMFIBSNPOJDNPUJPO Q
2
40, 41
3
"OBMZ[FBOPCKFDUJOEBNQFENPUJPO Q
3
42, 43
4
(SBQIUIFTVNPGUXPGVODUJPOT Q
4, 5
44
Section
610
CHAPTER 7 Applications of Trigonometric Functions
Review Exercises In Problems 1 and 2, find the exact value of the six trigonometric functions of the angle u in each figure. *2.
*1.
4
4
2
3
In Problems 3–5, find the exact value of each expression. Do not use a calculator. 3. cos 62° - sin 28°
4.
0
tan 70° cot 20°
5. cos2 40° + cos2 50°
1
1
In Problems 6 and 7, solve each triangle. *6.
10
A
20° a
*7.
5
b
A 2
B a
In Problems 8–20, find the remaining angle(s) and side(s) of each triangle, if it (they) exists. If no triangle exists, say “No triangle.” *8. A = 30°, B = 75°, b = 5
*9. A = 100°, a = 5, c = 2
*10. a = 3, c = 1, C = 110°
*11. a = 3, c = 1, B = 100°
*12. a = 4, b = 6, A = 50°
*13. a = 2, b = 3, c = 1
*14. a = 6, b = 8, c = 4
*15. a = 1, b = 3, C = 40°
*16. a = 3, b = 5, C = 40°
1 4 *17. a = 1, b = , c = 2 3 *20. c = 5, b = 4, A = 70°
*18. a = 3, A = 10°, b = 4
*19. a = 4, A = 20°, B = 100°
In Problems 21–25, find the area of each triangle. 21. a = 2, b = 3, C = 40° 24. a = 4, b = 2, c = 5
1.93
3.80
22. a = 7, c = 10, B = 60°
30.31 23. a = 4, b = 3, c = 5
25. B = 40°, C = 60°, b = 4
6
10.57
26. Area of a Segment Find the area of the segment of a circle whose radius is 8 inches formed by a central angle of 30°. 0.76 in2 27. Geometry The hypotenuse of a right triangle is 7 feet. If one leg is 4 feet, find the degree measure of each angle. 34.85° and 55.15° 28. Finding the Width of a River Find the distance from A to C across the river illustrated in the figure. 23.32 ft
1454 ft 5¯ Lake Michigan
1 mi
30. Finding the Speed of a Glider From a glider 200 feet above the ground, two sightings of a stationary object directly in front are taken 1 minute apart (see the figure). What is the speed of the glider? 132.55 ft/min
A
25¯ C
50 ft
40¯
B 10¯
29. Finding the Distance to Shore The Willis Tower in Chicago is 1454 feet tall and is situated about 1 mile inland from the shore of Lake Michigan, as indicated in the figure on the top right. An observer in a pleasure boat on the lake directly in front of the Willis Tower looks at the top of the tower and measures the angle of elevation as 5°. How far offshore is the boat? 2.15 mi
200 ft
31. Finding the Grade of a Mountain Trail A straight trail with a uniform inclination leads from a hotel, elevation 5000 feet, to a lake in a valley, elevation 4100 feet. The length of the trail is 4100 feet. What is the inclination (grade) of the trail? 12.7°
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
Chapter Review
32. Finding the Height of a Helicopter Two observers simultaneously measure the angle of elevation of a IFMJDPQUFS 0OF BOHMF JT NFBTVSFE BT UIF PUIFS BT (see the figure). If the observers are 100 feet apart and the helicopter lies over the line joining them, how high is the helicopter? 29.97 ft
611
35. Approximating the Area of a Lake To approximate the area of a lake, Cindy walks around the perimeter of the lake, taking the measurements shown in the illustration. Using UIJTUFDIOJRVF XIBUJTUIFBQQSPYJNBUFBSFBPGUIFMBLF [Hint: Use the Law of Cosines on the three triangles shown and then find the sum of their areas.] TRGU
100'
70' 100°
50°
50' 125' 50'
33. Constructing a Highway A highway whose primary directions are north–south is being constructed along the west coast of Florida. Near Naples, a bay obstructs the straight path of the road. Since the cost of a bridge is prohibitive, engineers decide to go around the bay. The illustration shows the path that they decide on and the measurements taken. What is the length of highway needed to go around the bay? NJ
36. Finding the Bearing of a Ship The Majesty leaves the Port at #PTUPOGPS#FSNVEBXJUIBCFBSJOHPG4&BUBOBWFSBHF TQFFE PG LOPUT"GUFS IPVS UIF TIJQ UVSOT UPXBSE the southwest. After 2 hours at an average speed of 20 knots, what is the bearing of the ship from Boston? 4& 37. Drive Wheels of an Engine The drive wheel of an engine is 13 inches in diameter, and the pulley on the rotary pump is 5 inches in diameter. If the shafts of the drive wheel and the QVMMFZBSFGFFUBQBSU XIBUMFOHUIPGCFMUJTSFRVJSFEUPKPJO them as shown in the figure? JO
120° Gulf
3 mi Clam Bay
1– 4
mi
1– 4
mi
6.5 in.
2.5 in.
115° 2 ft.
34. Correcting a Navigational Error A sailboat leaves St. Thomas bound for an island in the British West Indies, 200 miles away. Maintaining a constant speed of 18 miles per hour, but encountering heavy crosswinds and strong currents, the crew finds, after 4 hours, that the sailboat is off DPVSTFCZ (a) 131.8 mi (a) How far is the sailboat from the island at this time? (b) Through what angle should the sailboat turn to correct its course? (c) How much time has been added to the trip because of this? (Assume that the speed remains at 18 miles per hour.) 0.21 hr British West Indies
St. Thomas
200 mi
15°
38. Rework Problem 37 if the belt is crossed, as shown in the figure. JO
6.5 in.
2.5 in.
2 ft.
39. An object attached to a coiled spring is pulled down a distance a = 3 units from its rest position and then released. Assuming that the motion is simple harmonic with period T = 4 seconds, develop a model that relates the displacement d of the object from its rest position after t seconds. Also assume that the positive direction of the motion is up. p d = - 3 cos a tb 2
612
CHAPTER 7 Applications of Trigonometric Functions
In Problems 40 and 41, the displacement d (in feet) that an object travels in time t (in seconds) is given. (a) Describe the motion of the object. (b) What is the maximum displacement from its rest position? (c) What is the time required for one oscillation? (d) What is the frequency? *40. d = 6 sin 12t2
*41. d = - 2 cos 1pt2
*42. An object of mass m = 40 grams attached to a coiled spring with damping factor b = 0.75 gram/second is pulled down a distance a = 15 centimeters from its rest position and then released. Assume that the positive direction of the motion is up and the period is T = 5 seconds under simple harmonic motion. (a) Develop a model that relates the displacement d of the object from its rest position after t seconds. (b) Graph the equation found in part (a) for 5 oscillations. *43. The displacement d (in meters) of the bob of a pendulum of mass m (in kilograms) from its rest position at time t (in seconds) is given as d = - 15e -0.6t>40 cos ¢ (a) (b) (c) (d) (e)
2p 2 0.36 b t≤ C 5 1600 a
Describe the motion of the object. What is the initial displacement of the bob? That is, what is the displacement at t = 0? Graph the motion using a graphing utility. What is the displacement of the bob at the start of the second oscillation? What happens to the displacement of the bob as time increases without bound?
*44. Use the method of adding y-coordinates to graph y = 2 sin x + cos 12x2 .
Chapter Test Prep Videos include step-by-step solutions to all chapter test exercises and can be found on this text’s Channel. (search “SullivanPrecalcUC3e”)
Chapter Test
2. Find the exact value of sin 40° - cos 50°. 0
*1. Find the exact value of the six trigonometric functions of the angle u in the figure. 3 6
In Problems 3–5, use the given information to determine the three remaining parts of each triangle. *3.
*4.
C 17
b a
*5.
12
C
5
41⬚
22⬚
8
A
c
B 10
B
52⬚
C
19
In Problems 6–8, solve each triangle. *6. A = 55°, C = 20°, a = 4
*7. a = 3, b = 7, A = 40°
*9. Find the area of the triangle described in Problem 8.
*8. a = 8, b = 4, C = 70°
[Hint: Triangle ABC is a right triangle.]
*10. Find the area of the triangle described in Problem 5.
B
11. A 12-foot ladder leans against a building. The top of the ladder leans against the wall 10.5 feet from the ground. What is the angle formed by the ground and the ladder? 61.0° 12. A hot-air balloon is flying at a height of 600 feet and is directly above the Marshall Space Flight Center in Huntsville, Alabama. The pilot of the balloon looks down at the airport that is known to be 5 miles from the Marshall Space Flight Center. What is the angle of depression from the balloon to the airport? 1.3° 13. Find the area of the shaded region enclosed in a semicircle of diameter 8 centimeters. The length of the chord AB is 6 centimeters. 9.26 cm2
6 A
C
8
14. Find the area of the quadrilateral shown. 54.15 square units 5 72
11
7
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
8
Chapter Projects
15. Madison wants to swim across Lake William from the fishing lodge (A) to the boat ramp (B), but she wants to know the distance first. Highway 20 goes right past the boat ramp and County Road 3 goes to the lodge. The two roads intersect at point (C), 4.2 miles from the ramp and 3.5 miles from the lodge. Madison uses a transit to measure the angle of intersection of the two roads to be 32°. How far will she need to swim? ≈2.23 mi
613
B
A
5 40⬚
Lake William
B
O
3.5
6x
3
5x 32⬚
CR
iles 4.2 m
m ile s
20 Hwy
A
17. The area of the triangle shown below is 542TRVBSFVOJUT find the lengths of the sides. 15, 18, 21
7x
C
16. Given that △OAB is an isosceles triangle and the shaded sector is a semicircle, find the area of the entire region. &YQSFTT ZPVS BOTXFS BT B EFDJNBM SPVOEFE UP UXP places. TRVOJUT
18. Logan is playing on her swing. One full swing (front to back UPGSPOU UBLFTTFDPOETBOEBUUIFQFBLPGIFSTXJOHTIFJT at an angle of 42° with the vertical. If her swing is 5 feet long, and we ignore all resistive forces, build a model that relates her horizontal displacement (from the rest position) after time t. pt d = 5(sin 42°) sina b 3
Cumulative Review *1. 'JOEUIFSFBMTPMVUJPOT JGBOZ PGUIFFRVBUJPO3x2 + 1 = 4x. *2. 'JOE BO FRVBUJPO GPS UIF DJSDMF XJUI DFOUFS BU UIF QPJOU 1 - 5, 12 and radius 3. Graph this circle. *3. Determine the domain of the function f 1x2 = 2x2 - 3x - 4 *4. Graph the function y = 3 sin 1px2.
*5. Graph the function y = - 2 cos 12x - p2. 3p 6 u 6 2p, find the exact value of: 2 *(b) cos u *(c) sin12u2 3 1 1 *(e) sina u b *(f) cos a u b 5 2 2
6. If tan u = - 2 and *(a) sin u (d) cos 12u2
*7. Graph each of the following functions on the interval 30, 44: (a) y = e x (b) y = sin x (c) y = e x sin x (d) y = 2x + sin x *8. Sketch the graph of each of the following functions: (a) y = x (b) y = x2 (c) y = 1x (d) y = x3
(e) y = e x (g) y = sin x (i) y = tan x
(f) y = ln x (h) y = cos x
*9. Solve the triangle in which side a is 20, side c is 15, and angle CJT *10. *OUIFDPNQMFYOVNCFSTZTUFN TPMWFUIFFRVBUJPO 3x5 - 10x4 + 21x3 - 42x2 + 3x - 8 = 0 *11. Analyze the graph of the rational function R 1x2 =
2x2 - 7x - 4 x2 + 2x - 15
*12. Solve 3x = 12. Round your answer to two decimal places. 13. Solve log 3 1x + 82 + log 3 x = 2.
{1}
14. Suppose that f 1x2 = 4x + 5 and g1x2 = x2 + 5x - 24. (b) Solve f 1x2 = 13. {2} *(a) Solve f 1x2 = 0. *(c) Solve f 1x2 = g1x2. *(d) Solve f 1x2 7 0. *(e) Solve g1x2 … 0. *(f) Graph y = f 1x2. *(g) Graph y = g1x2.
Chapter Projects I.
Spherical Trigonometry When the distance between two MPDBUJPOTPOUIFTVSGBDFPG&BSUIJTTNBMM UIFEJTUBODFDBOCF computed in statutory miles. Using this assumption, the Law of Sines and the Law of Cosines can be used to approximate EJTUBODFT BOE BOHMFT )PXFWFS &BSUI JT B TQIFSF TP BT UIF distance between two points on its surface increases, the linear distance gets less accurate because of curvature. Under
UIJTDJSDVNTUBODF UIFDVSWBUVSFPG&BSUINVTUCFDPOTJEFSFE when using the Law of Sines and the Law of Cosines. 1. Draw a spherical triangle and label the vertices A, B, and C. Then connect each vertex by a radius to the center O of the sphere. Now draw tangent lines to the sides a and b of the triangle that go through C&YUFOEUIFMJOFTOA and
614
CHAPTER 7 Applications of Trigonometric Functions
OB to intersect the tangent lines at P and Q, respectively. See the figure. List the plane right triangles. Determine the measures of the central angles. C a
b O
Q
B
A
c
2. Apply the Law of Cosines to triangles OPQ and CPQ to find two expressions for the length of PQ. 3. Subtract the expressions in part (2) from each other. Solve for the term containing cos c. 4. Use the Pythagorean Theorem to find another value for OQ2 - CQ2 and OP 2 - CP 2. Now solve for cos c. 5. Replacing the ratios in part (4) by the cosines of the sides of the spherical triangle, you should now have the Law of Cosines for spherical triangles: cos c = cos a cos b + sin a sin b cos C Source: 'PSUIFTQIFSJDBM-BXPG$PTJOFT TFF Mathematics from the Birth of Numbers by Jan Gullberg. W. W. Norton $P 1VCMJTIFST QQm.
P
II. The Lewis and Clark Expedition Lewis and Clark followed several rivers in their trek from what is now Great Falls, Montana, to the Pacific coast. First, they went down the Missouri and Jefferson rivers from Great Falls to Lemhi, Idaho. Because the two cities are at different longitudes and EJGGFSFOUMBUJUVEFT UIFDVSWBUVSFPG&BSUINVTUCFBDDPVOUFE for when computing the distance that they traveled. Assume UIBUUIFSBEJVTPG&BSUIJTNJMFT
angles are found by using the differences in the latitudes and longitudes of the towns. See the diagram.) Then find the length of the arc joining the two towns. (Recall s = ru.) 2. From Lemhi, they went up the Bitteroot River and the Snake River to what is now Lewiston and Clarkston on the border of Idaho and Washington. Although this is not really a side to a triangle, make a side that goes from Lemhi to Lewiston and Clarkston. If Lewiston and Clarkston are at about 4.5°N 117.0°W, find the distance from Lemhi using the Law of Cosines for a spherical triangle and the arc length. 3. How far did the explorers travel just to get that far?
1. Great Falls is at approximately 47.5°N and 111.3°W. Lemhi is at approximately 45.5°N and 113.5°W. (Assume that the rivers flow straight from Great Falls to Lemhi on UIF TVSGBDF PG &BSUI 5IJT MJOF JT DBMMFE B HFPEFTJD MJOF Apply the Law of Cosines for a spherical triangle to find the angle between Great Falls and Lemhi. (The central North b a Lemhi
c
Great Falls
4. Draw a plane triangle connecting the three towns. If the distance from Lewiston to Great Falls is 282 miles, and if the angle at Great Falls is 42° and the angle at Lewiston is 48.5°, find the distance from Great Falls to Lemhi and from Lemhi to Lewiston. How do these distances compare with the ones computed in parts (1) and (2)? Source: 'PS -FXJT BOE $MBSL &YQFEJUJPO American Journey: The Quest for Liberty to 1877, Texas Edition. Prentice Hall, 1992, p. 345. Source: For map coordinates: National Geographic Atlas of the World, published by National Geographic Society, 1981, pp. 74–75.The following projects are available at the Instructor’s Resource Center (IRC): III. Project at Motorola: How Can You Build or Analyze a Vibration Profile? Fourier functions not only are important to analyze vibrations, but they are also what a mathematician would call interesting. Complete the project to see why. IV. Leaning Tower of Pisa Trigonometry is used to analyze the apparent height and tilt of the Leaning Tower of Pisa. V. Locating Lost Treasure Clever treasure seekers who know the Law of Sines are able to find a buried treasure efficiently.
South
VI. Jacob’s Field Angles of elevation and the Law of Sines are used to determine the height of the stadium wall and the distance from home plate to the top of the wall.
Polar Coordinates; Vectors How Airplanes Fly Essentially there are 4 aerodynamic forces that act on an airplane in flight; these are lift, drag, thrust and weight (i.e. gravity). In simple terms, drag is the resistance of air molecules hitting the airplane (the backward force), thrust is the power of the airplane’s engine (the forward force), lift is the upward force and weight is the downward force. So for airplanes to fly and stay airborne, the thrust must be greater than the drag and the lift must be greater than the weight (so as you can see, drag opposes thrust and lift opposes weight). This is certainly the case when an airplane takes off or climbs. However, when it is in straight and level flight the opposing forces of lift and weight are balanced. During a descent, weight exceeds lift and to slow an airplane drag has to overcome thrust. Thrust is generated by the airplane’s engine (propeller or jet), weight is created by the natural force of gravity acting upon the airplane and drag comes from friction as the plane moves through air molecules. Drag is also a reaction to lift, and this lift must be generated by the airplane flight. This is done by the wings of the airplane. . . The generation of lift is a widely discussed and sometimes disputed theory, but there are some key factors that nobody argues. A cross section of a typical airplane wing will show the top surface to be more curved than the bottom surface. This shaped profile is called an ‘airfoil’ (or ‘aerofoil’) and the shape exists because it’s long been proven (since the dawn of flight) that an airfoil generates significantly more lift than opposing drag i.e. it’s very efficient at generating lift. During flight air naturally flows over and beneath the wing and is deflected upwards over the top surface and downwards beneath the lower surface. Any difference in deflection causes a difference in air pressure (‘pressure gradient’) and because of the airfoil shape the pressure of the deflected air is lower above the airfoil than below it. As a result the wing is ‘pushed’ upwards by the higher pressure beneath or, you can argue, it is ‘sucked’ upwards by the lower pressure above.
8
Source: Pete Carpenter. How Airplanes Fly—the basic principles of flight.
http://www.rc-airplane-world.com/how-airplanes-fly.html, accessed July 2013. © rc-airplane-world.com
—See Chapter Project I—
Outline
This chapter is in two parts: Polar Coordinates, Sections 8.1–8.3, and Vectors, Sections 8.4–8.7. They are independent of each other and may be covered in either order. Sections 8.1–8.3: In Foundations we introduced rectangular coordinates x and y and discussed the graph of an equation in two variables involving x and y. In Sections 8.1 and 8.2, we introduce polar coordinates, an alternative to rectangular coordinates, and discuss graphing equations that involve polar coordinates. In Section 4.3, we discussed raising a real number to a real power. In Section 8.3, we extend this idea by raising a complex number to a real power. As it turns out, polar coordinates are useful for the discussion. Sections 8.4–8.7: We have seen in many chapters that we are often required to solve an equation to obtain a solution to applied problems. In the last four sections of this chapter, we develop the notion of a vector and show how it can be used to model applied problems in physics and engineering.
8.1 8.2 8.3 8.4 8.5 8.6 8.7
Polar Coordinates Polar Equations and Graphs The Complex Plane; De Moivre’s Theorem Vectors The Dot Product Vectors in Space The Cross Product Chapter Review Chapter Test Cumulative Review Chapter Projects
615
616
CHAPTER 8 Polar Coordinates; Vectors
8.1 Polar Coordinates PREPARING FOR THIS SECTION Before getting started, review the following: r *OWFSTF5BOHFOU'VODUJPO 4FDUJPO QQm
r $PNQMFUJOHUIF4RVBSF "QQFOEJY" 4FDUJPO" pp. A38–A39)
r 3FDUBOHVMBS$PPSEJOBUFT 'PVOEBUJPOT 4FDUJPO pp. 34–35) r %FGJOJUJPOPGUIF5SJHPOPNFUSJD'VODUJPOT (Section 5.2, pp. 422–433)
Now Work the ‘Are You Prepared?’ problems on page 623. OBJECTIVES 1 2 3 4
Figure 1 y
Polar axis O Pole
x
Plot Points Using Polar Coordinates (p. 616) Convert from Polar Coordinates to Rectangular Coordinates (p. 618) Convert from Rectangular Coordinates to Polar Coordinates (p. 620) Transform Equations between Polar and Rectangular Forms (p. 622)
So far, we have always used a system of rectangular coordinates to plot points in the plane. Now we are ready to describe another system, called polar coordinates. In many instances, polar coordinates offer certain advantages over rectangular coordinates. In a rectangular coordinate system, you will recall, a point in the plane is represented by an ordered pair of numbers 1x, y2, where x and y equal the signed distances of the point from the y-axis and from the x-axis, respectively. In a polar coordinate system, we select a point, called the pole, and then a ray with vertex at the pole, called the polar axis. See Figure 1. Comparing the rectangular and polar coordinate systems, notice that the origin in rectangular coordinates coincides with the pole in polar coordinates, and the positive x-axis in rectangular coordinates coincides with the polar axis in polar coordinates.
1 Plot Points Using Polar Coordinates A point P in a polar coordinate system is represented by an ordered pair of numbers 1r, u2. If r 7 0, then r is the distance of the point from the pole; u is an angle (in degrees or radians) formed by the polar axis and a ray from the pole through the point. We call the ordered pair 1r, u2 the polar coordinates of the point. See Figure 2. p As an example, suppose that a point P has polar coordinates a2, b . Locate P 4 p by first drawing an angle of radian, placing its vertex at the pole and its initial side 4 along the polar axis. Then go out a distance of 2 units along the terminal side of the angle to reach the point P. See Figure 3. Figure 2
Figure 3 P (r, )
r O Pole
2 Polar axis
– 4
O Pole
P (2, –4 ) Polar axis
In using polar coordinates 1r, u2, it is possible for r to be negative. When this happens, instead of the point being on the terminal side of u, it is on the ray from the pole extending in the direction opposite the terminal side of u at a distance 0 r 0 units from the pole. See Figure 4 for an illustration. 2p For example, to plot the point a - 3, b , use the ray in the opposite direction 3 2p of and go out 0 - 3 0 = 3 units along that ray. See Figure 5. 3
SECTION 8.1 Polar Coordinates
Figure 4
617
Figure 5
2 –– 3
O
O
–– ) (3, 2 3
⏐r⏐
P (r, ), r 0
EXAM PL E 1
Plotting Points Using Polar Coordinates Plot the points with the following polar coordinates: (a) a3,
Solution
5p b 3
(b) a2, -
p b 4
(c) 13, 02
(d) a - 2,
p b 4
'JHVSFTIPXTUIF points. Figure 6 5–– 3
– 4
O
–
O
( 2,
( 3, 5––3 ) (a)
–)
O
4
(2, –4 )
4
(c)
(b)
Now Work
O
(3, 0)
PROBLEMS
9, 17,
AND
(d)
r
27
3FDBMM UIBU BO BOHMF NFBTVSFE DPVOUFSDMPDLXJTF JT QPTJUJWF BOE BO BOHMF measured clockwise is negative. This convention has some interesting consequences related to polar coordinates.
EXAM PL E 2
Finding Several Polar Coordinates of a Single Point
Figure 7
p Consider again the point P with polar coordinates a2, b , as shown in Figure 7(a). 4 7p p 9p Because , , and all have the same terminal side, this point P could also 4 4 4 9p 7p b or a2, b , as shown have been located by using the polar coordinates a2, 4 4 p in Figures 7(b) and (c). The point a2, b can also be represented by the polar 4 5p b . See Figure 7(d). coordinates a - 2, 4
2
–
P (2, –4 )
2
4
O
9–– 4
P (2, 9––– 4 )
O
2 7––
P ( 2, 7––– 4 )
2
5 –– 4
O
–
) P (2, 5–– 4
4
O
4
(a)
(b)
(c)
(d)
r
618
CHAPTER 8 Polar Coordinates; Vectors
Finding Other Polar Coordinates of a Given Point
EX A MPL E 3
Plot the point P with polar coordinates a3,
p b , and find other polar coordinates
(a) r 7 0, 2p … u 6 4p (c) r 7 0, - 2p … u 6 0
(b) r 6 0, 0 … u 6 2p
1r, u2 of this same point for which:
Solution P (3,
6
p b is plotted in Figure 8.
p p (a) Add 1 revolution 12p radians2 to the angle to get P = a3, + 2pb = 13p a3, b . See Figure 9.
Figure 8 –
The point a3,
)
– 6
O
1 p revolution 1p radians2 to the angle and replace 3 by - 3 to get 2 7p p b . See Figure 10. P = a - 3, + pb = a - 3,
(b) Add
p p 11p (c) Subtract 2p from the angle to get P = a3, - 2pb = a3, b . See Figure 11. Figure 10
Figure 9
7–– 6
) ––– P (3, 13 6 13 ––– 6
O
Figure 11 P (3,
7–– 6
)
11 ––– 6
––– P (3, 11 6 )
O
O
r
These examples show a major difference between rectangular coordinates and polar coordinates. A point has exactly one pair of rectangular coordinates; however, a point has infinitely many pairs of polar coordinates.
SUMMARY A point with polar coordinates 1r, u2 , u in radians, can also be represented by either of the following: 1r, u + 2pk2
or
1 - r, u + p + 2pk2
k any integer
The polar coordinates of the pole are 10, u2, where u can be any angle.
Now Work
PROBLEM
31
2 Convert from Polar Coordinates to Rectangular Coordinates Sometimes it is necessary to convert coordinates or equations in rectangular form to polar form, and vice versa. To do this, recall that the origin in rectangular coordinates is the pole in polar coordinates and that the positive x-axis in rectangular coordinates is the polar axis in polar coordinates.
THEOREM
Conversion from Polar Coordinates to Rectangular Coordinates If P is a point with polar coordinates 1r, u2 u2,, the rectangular coordinates 1x, y2 of P are given by x = r cos u
y = r sin u
(1)
SECTION 8.1 Polar Coordinates
Figure 12 y
P r
y
x
O
x
619
Proof Suppose that P has the polar coordinates 1r, u2. We seek the rectangular coordinates 1x, y2 of P3FGFSUP'JHVSF If r = 0, then, regardless of u, the point P is the pole, for which the rectangular coordinates are 10, 02. Formula (1) is valid for r = 0. If r 7 0, the point P is on the terminal side of u, and r = d 1O, P2 = 2x2 + y2 . Since y x cos u = sin u = r r we have x = r cos u
y = r sin u
If r 6 0 and u is in radians, then the point P = 1r, u2 can be represented as 1 - r, p + u2, where - r 7 0. Since cos 1p + u2 = - cos u =
sin 1p + u2 = - sin u =
x -r
y -r
we have x = r cos u
EXAM PL E 4
y = r sin u
■
Converting from Polar Coordinates to Rectangular Coordinates Find the rectangular coordinates of the points with the following polar coordinates: (a) a,
Solution
p b
(b) a - 4, -
p b 4
Use formula (1): x = r cos u and y = r sin u. p p b plotted. Notice that a, b lies in quadrant I of the rectangular coordinate system. So the x-coordinate and the y-coordinate should p both be positive. With r = and u = , we have
(a) Figure 13(a) shows a, Figure 13 y 3
6
(6, –6 )
–
p 23 = # = 323 2 p 1 y = r sin u = sin = # = 3 2
6
3 3
x = r cos u = cos
x
(a) y
(4, –4 )
p The rectangular coordinates of the point a, b are 1 323, 3 2 , which lies in quadrant I, as expected.
2 2
– 4
2 2
x
4
(b)
COMMENT Most calculators have the capability of converting from polar coordinates to rectangular coordinates. Consult your owner’s manual for the proper keystrokes. Since in most cases this procedure is tedious, you will find that using formula (1) is faster. ■
p p b plotted. Notice that a - 4, - b lies in quadrant II 4 4 p of the rectangular coordinate system. With r = - 4 and u = - , we have 4
(b) Figure 13(b) shows a - 4, -
x = r cos u = - 4 cos a y = r sin u = - 4 sin a -
p b = -4 4
#
22 = - 222 2
p 22 b = - 4a b = 222 4 2
p The rectangular coordinates of the point a - 4, - b are 1 - 222, 222 2 , which 4 lies in quadrant II, as expected.
r
Now Work
PROBLEMS
39
AND
51
620
CHAPTER 8 Polar Coordinates; Vectors
3 Convert from Rectangular Coordinates to Polar Coordinates
Converting from rectangular coordinates 1x, y2 to polar coordinates 1r, u2 is a little more complicated. Notice that each solution begins by plotting the given rectangular coordinates.
How to Convert from Rectangular Coordinates to Polar Coordinates with the Point on a Coordinate Axis
EX A MPL E 5
Find polar coordinates of a point whose rectangular coordinates are 10, 32 .
Step-by-Step Solution
Figure 14
y
Plot the point 10, 32 in a rectangular coordinate system. See Figure 14. The point lies on the positive y-axis.
(x, y) (0, 3)
Step 1 Plot the point (x, y) and note the quadrant the point lies in or the coordinate axis the point lies on.
3
– 2
x
The point (0, 3) lies on the y-axis a distance of 3 units from the origin (pole), so r = 3.
Step 2 Determine the distance r from the origin to the point.
p with the polar axis. 2 p Polar coordinates for this point can be given by a3, b . Other possible 2 p 5p representations include a - 3, - b and a3, b. 2 2
Step 3 Determine u.
A ray with vertex at the pole through (0, 3) forms an angle u =
COMMENT Most graphing calculators have the capability of converting from rectangular coordinates to polar coordinates. Consult your owner’s manual for the proper keystrokes. ■
r
Figure 15 shows polar coordinates of points that lie on either the x-axis or the y-axis. In each illustration, a 7 0.
Figure 15 y
a
(x, y) (a, 0) (r, θ) (a, 0) a
–
(x, y) (a, 0) (r, ) (a, )
2
x
y
y
y (x, y) (0, a ) –) (r, ) (a, 2
x
a
3 ––– 2
x
x (x, y) (0, a) (r, ) (a, 3–– 2 )
(a) (x, y) (a, 0), a 0
(b) (x, y) (0, a), a 0
Now Work
EX A MPL E 6
PROBLEM
(c) (x, y) ( a, 0), a 0
a
(d) (x, y) (0, a), a 0
55
How to Convert from Rectangular Coordinates to Polar Coordinates with the Point in a Quadrant
Find the polar coordinates of a point whose rectangular coordinates are 12, - 22 .
Step-by-Step Solution Step 1 Plot the point (x, y) and note the quadrant the point lies in or the coordinate axis the point lies on.
Plot the point 12, - 22 in a rectangular DPPSEJOBUFTZTUFN4FF'JHVSF The point lies in quadrant IV.
Figure 16
y 1 1 1 2
2
x
r (x, y) (2, 2)
SECTION 8.1 Polar Coordinates
621
r = 2x2 + y2 = 2 122 2 + 1 - 22 2 = 28 = 222
Step 2 Determine the distance r from the origin to the point using r = 2x 2 + y 2.
y y p p 6 u 6 . Since 12, - 22 lies , so u = tan - 1 , x x 2 2 p in quadrant IV, we know that 6 u 6 0. As a result, 2
Step 3 Determine u.
Find u by recalling that tan u =
u = tan-1
y -2 p b = tan-1 1 - 12 = = tan-1 a x 2 4
A set of polar coordinates for the point 12, - 22 is a222, representations include a222,
EXAM PL E 7
p b . Other possible 4
7p 3p b and a - 222, b. 4 4
r
Converting from Rectangular Coordinates to Polar Coordinates Find polar coordinates of a point whose rectangular coordinates are
Solution
1 - 1, - 23 2 .
STEP 1: See Figure 17. The point lies in quadrant III. STEP 2: The distance r from the origin to the point 1 - 1, - 23 2 is
Figure 17
r = 3 1 - 12 2 +
y u
1 - 23 2 2
= 24 = 2
y - 23 p p = tan - 1 23, 6 a 6 . = tan - 1 x -1 2 2 Since the point 1 - 1, - 23 2 lies in quadrant III, and the inverse tangent function gives an angle in quadrant I, add p to the result to obtain an angle in quadrant III.
STEP 3: To find u, use a = tan - 1
x r (x, y) 5 (21, 2 3)
u = p + tan-1 a
- 23 p 4p b = p + tan-1 23 = p + = -1 3 3
4p A set of polar coordinates for this point is a2, b . Other possible 3 p 2p representations include a - 2, b and a2, b. 3 3
r
Figure 18 shows how to find polar coordinates of a point that lies in a quadrant when its rectangular coordinates 1x, y2 are given.
Figure 18 y
y (x, y )
(x, y ) r
r
y
y
x
x
x r
r
(x, y )
(a) r x 2 y 2 tan1 y x
(b) r x 2 y 2 tan1 y x
(x, y)
(c) r x 2 y 2 tan1 y x
(d) r x 2 y 2 tan1 y x
Based on the preceding discussion, we have the formulas r 2 = x2 + y 2
tan u =
y x
if x ≠ 0
(2)
x
622
CHAPTER 8 Polar Coordinates; Vectors
To use formula (2) effectively, follow these steps:
Steps for Converting from Rectangular to Polar Coordinates STEP 1: Always plot the point 1x, y2 GJSTU BTTIPXOJO&YBNQMFT BOE GJSTU BT TIPXO JO &YBNQMFT BOE Note the quadrant the point lies in or the coordinate axis the point lies on. STEP 2: If x = 0 or y = 0, use your illustration to find r. If x ≠ 0 and y ≠ 0, then r = 2x 2 2 + y 2. STEP 3: Find u. If x = 0 or y = 0, use your illustration to find u. If x ≠ 0 and y ≠ 0, note the quadrant in which the point lies. u = tan-1
Quadrant I or IV:
y x
Quadrant II or III: u = p + tan-1
Now Work
PROBLEM
y x
59
4 Transform Equations between Polar and Rectangular Forms Formulas (1) and (2) may also be used to transform equations from polar form to rectangular form, and vice versa. Two common techniques for transforming an equation from polar form to rectangular form are 1. Multiplying both sides of the equation by r 2. Squaring both sides of the equation
EX A MPL E 8
Transforming an Equation from Polar to Rectangular Form Transform the equation r = cos u from polar coordinates to rectangular coordinates, and identify the graph.
Solution
Multiplying each side by r makes it easier to apply formulas (1) and (2). r = cos u r 2 = r cos u x + y = x 2
2
Multiply each side by r. r 2 = x 2 + y 2; x = r cos u
This is the equation of a circle. Proceed to complete the square to obtain the standard form of the equation. x2 + y2 = x
1x2 - x2 + y2 = 0
1x2 - x + 92 + y2 = 9 1x - 32 + y = 9 2
2
General form Complete the square in x. Factor.
This is the standard form of the equation of a circle with center 13, 02 and radius 3.
Now Work
EX A MPL E 9
PROBLEM
r
75
Transforming an Equation from Rectangular to Polar Form Transform the equation 4xy = 9 from rectangular coordinates to polar coordinates.
Solution
Use formula (1): x = r cos u and y = r sin u. 4xy = 9 41r cos u2 1r sin u2 = 9 x = r cos u, y = r sin u 4r 2 cos u sin u = 9
SECTION 8.1 Polar Coordinates
623
This is the polar form of the equation. It can be simplified as follows: 2r 2 12 sin u cos u2 = 9 Factor out 2r 2.
2r 2 sin 12u2 = 9 Double-angle Formula
Now Work
PROBLEM
r
69
8.1 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. Plot the point whose rectangular coordinates are 13, - 12. What quadrant does the point lie in? (pp. 34–35)
3. If P = 1a, b2 is a point on the terminal side of the angle u at a distance r from the origin, then tan u = . (p. 433) b p a -1 . (pp. 502–504) 4. tan 1 - 12 = 4
2. To complete the square of x2 + x, add 9 . (pp. A38–A39)
Concepts and Vocabulary 7. True or False The polar coordinates of a point are unique.
5. The origin in rectangular coordinates coincides with the pole in polar coordinates; the positive x-axis in rectangular coordinates coincides with the polar axis in polar coordinates.
8. If P is a point with polar coordinates (r, u), the rectangular coordinates (x, y) of P are given by x = r cos u and y = r sin u.
6. True or False In the polar coordinates (r, u), r can be negative. True
7. False
Skill Building In Problems 9–16, match each point in polar coordinates with either A, B, C, or D on the graph. 9. a2, 13. a2,
11p b
5p b
10. a - 2, -
A
14. a - 2,
B
p b
5p b
B D
11. a - 2,
p b
15. a - 2,
7p b
C A
12. a2,
7p b
16. a2,
11p b
B
C
A
2
D
π 6
D
C
In Problems 17–30, plot each point given in polar coordinates. p b
*17. 13, 90°2
*18. 14, 270°2
*19. 1 - 2, 02
*20. 1 - 3, p2
*21. a,
*22. a5,
*23. 1 - 2, 135°2
*24. 1 - 3, 120°2
*25. a4, -
*26. a2, -
*28. a - 3, -
*29. 1 - 2, - p2
*30. a - 3, -
5p b 3
*27. a - 1, -
p b 3
3p b 4
2p b 3
5p b 4
p b 2
In Problems 31–38, plot each point given in polar coordinates, and find other polar coordinates 1r, u2 of the point for which: (a) r 7 0,
- 2p … u 6 0
2p b *31. a5, 3 *35. a1,
p b 2
(b) r 6 0, 0 … u 6 2p
(c) r 7 0, 2p … u 6 4p
3p *32. a4, b 4
*33. 1 - 2, 3p2
*36. 12, p2
*37. a - 3, -
*34. 1 - 3, 4p2
p b 4
*38. a - 2, -
2p b 3
In Problems 39–54, the polar coordinates of a point are given. Find the rectangular coordinates of each point. 39. a3,
p b 2
43. 1, 150°2
(0, 3)
( - 323, 3)
40. a4,
3p b 2
44. 15, 300°2
(0, - 4) 5 523 a ,b 2 2
41. 1 - 2, 02 45. a - 2,
3p b 4
( - 2, 0)
1 22, - 22 2
42. 1 - 3, p2 46. a - 2,
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
2p b 3
(3, 0)
1 1, - 23 2
624
CHAPTER 8 Polar Coordinates; Vectors
47. a - 1, -
p b 3
51. 17.5, 110°2
1 23 a- , b 2 2
48. a - 3, -
3p b 4
( - 2.57, 7.05) 52. 1 - 3.1, 182°2
a
322 322 , b 2 2
(3.10, 0.11)
49. 1 - 2, - 180°2 53. 1.3, 3.82
50. 1 - 3, - 90°2
(2, 0)
54. 18.1, 5.22
( - 4.98, - 3.85)
(0, 3) (3.79, - 7.1)
In Problems 55–66, the rectangular coordinates of a point are given. Find polar coordinates for each point. p p 55. 13, 02 (3, 0) 56. 10, 22 a2, b 58. 10, - 22 a2, - b 57. 1 - 1, 02 (1, p) 2 2 p 59. 11, - 12 a 22, - p b 62. 1 - 2, - 223 2 a4, - 2p b 60. 1 - 3, 32 a322, 3p b 61. 1 23, 1 2 a2, b 4 3 4 65. 18.3, 4.22 (9.30, 0.47) 66. 1 - 2.3, 0.22 (2.31, 3.05) 63. 11.3, - 2.12 (2.47, - 1.02) 64. 1 - 0.8, - 2.12 (2.25, 4.35) In Problems 67–74, the letters x and y represent rectangular coordinates. Write each equation using polar coordinates 1r, u2. *67. 2x2 + 2y2 = 3 71. 2xy = 1
68. x2 + y2 = x
r 2 sin 12u2 = 1 *72. 4x2 y = 1
r = cos u
*69. x2 = 4y 73. x = 4
*70. y2 = 2x r cos u = 4
74. y = - 3
r sin u = - 3
In Problems 75–82, the letters r and u represent polar coordinates. Write each equation using rectangular coordinates 1x, y2. *75. r = cos u 79. r = 2
*76. r = sin u + 1
x2 + y 2 = 4
80. r = 4
x2 + y2 = 1
*77. r 2 = cos u 81. r =
*78. r = sin u - cos u
4 1 - cos u
3 3 - cos u
*82. r =
Applications and Extensions 83. Chicago In Chicago, the road system is set up like a Cartesian plane, where streets are indicated by the number of blocks they are from Madison Street and State Street. 'PS FYBNQMF 8SJHMFZ 'JFME JO $IJDBHP JT MPDBUFE BU West Addison, which is 10 blocks west of State Street and CMPDLTOPSUIPG.BEJTPO4USFFU5SFBUUIFJOUFSTFDUJPOPG Madison Street and State Street as the origin of a coordinate system, with east being the positive x-axis. (a) Write the location of Wrigley Field using rectangular coordinates. ( - 10, 3) *(b) Write the location of Wrigley Field using polar coordinates. Use the east direction for the polar axis. Express u in degrees.
City of Chicago, Illinois
1 mile 1 km N Wrigley Field 1060 West Addison Madison Street
d =
2r 21
+
r 22
- 2r1 r2 cos 1u2 - u1 2
State Street
(c) U.S. Cellular Field, home of the White Sox, is located at 35th and Princeton, which is 3 blocks west of State Street and 35 blocks south of Madison. Write the location of U.S. Cellular Field using rectangular coordinates. ( - 3, - 35) *(d) Write the location of U.S. Cellular Field using polar coordinates. Use the east direction for the polar axis. Express u in degrees. *84. Show that the formula for the distance d between two points P1 = 1r1 , u1 2 and P2 = 1r2 , u2 2 is
Addison Street
Addison Street
U.S. Cellular Field 35th and Princeton 35th Street
35th Street
Explaining Concepts: Discussion and Writing 85. In converting from polar coordinates to rectangular coordinates, what formulas will you use? 86. Explain how you proceed to convert from rectangular coordinates to polar coordinates.
87. Is the street system in your town based on a rectangular coordinate system, a polar coordinate system, or some other system? Explain.
SECTION 8.2 Polar Equations and Graphs
625
Retain Your Knowledge Problems 88–91 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 19 88. Solve: log 4 1x + 32 - log 4 1x - 12 = 2 e 15 f 89. 6TF%FTDBSUFT3VMFPG4JHOTUPEFUFSNJOFUIFQPTTJCMFOVNCFSPGQPTJUJWFPSOFHBUJWF real zeros for the function f 1x2 = - 2x3 + x2 - 7x - 8. 2 or 0 positive real zeros; 1 negative real zero 1 5 9 90. Find the midpoint of the line segment connecting the points 1 - 3, 72 and a , 2b. a - , b 2 4 2 91. Given that the point (3, 8) is on the graph of y = f 1x2 , what is the corresponding point on the graph of y = - 2f 1x + 32 + 5? (0, - 11)
‘Are You Prepared?’ Answers y
1.
;quadrant IV
2. 9
3.
2 2 2
b a
4. -
p 4
2 4 x (3, 1)
8.2 Polar Equations and Graphs PREPARING FOR THIS SECTION Before getting started, review the following: r %JGGFSFODF'PSNVMBTGPS4JOFBOE$PTJOF 4FDUJPO QQBOE
r 7BMVFPGUIF4JOFBOE$PTJOF'VODUJPOTBU$FSUBJO Angles (Section 5.2, pp. 425–432)
r 4ZNNFUSZ 'PVOEBUJPOT 4FDUJPO QQm
r $JSDMFT 'PVOEBUJPOT 4FDUJPO QQm
r &WFOm0EE1SPQFSUJFTPG5SJHPOPNFUSJD'VODUJPOT (Section 5.3, pp. 448–449) Now Work the ‘Are You Prepared?’ problems on page 637.
OBJECTIVES 1 Identify and Graph Polar Equations by Converting to Rectangular Equations (p. 626) 2 Test Polar Equations for Symmetry (p. 629) 3 Graph Polar Equations by Plotting Points (p. 630)
Just as a rectangular grid may be used to plot points given by rectangular coordinates, as in Figure 19(a), a grid consisting of concentric circles (with centers at the pole) and rays (with vertices at the pole) can be used to plot points given by polar coordinates, as shown in Figure 19(b). We shall use such polar grids to graph polar equations. Figure 19
y 4
3 2 1 B (3, 1) 4 3 2 1 O
– 2
3 –– 4
( )
A (1, 2) 1
2
3 4
P 2, –4 x
2 3 4
(
O
r1
5
)
Q 4, –– 4
5
–– 4
(a) Rectangular grid
–4
r3 r5
3 –– 2
(b) Polar grid
7 –– 4
0
626
CHAPTER 8 Polar Coordinates; Vectors
DEFINITION
An equation whose variables are polar coordinates is called a polar equation. The graph of a polar equation consists of all points whose polar coordinates satisfy the equation.
1 Identify and Graph Polar Equations by Converting to Rectangular Equations One method that can be used to graph a polar equation is to convert the equation to rectangular coordinates. In the following discussion, 1x, y2 represents the rectangular coordinates of a point P, and 1r, u2 represents the polar coordinates of the point P.
EX A MPL E 1
Identifying and Graphing a Polar Equation (Circle) Identify and graph the equation: r = 3
Solution
Convert the polar equation to a rectangular equation. r = 3 r2 = 9
Square both sides.
x + y = 9 r2 = x2 + y 2 2
2
The graph of r = 3 is a circle, with center at the pole and radius 3. See Figure 20. Figure 20
y
r = 3 or x2 + y2 = 9
–2
3–– 4
O
5 –– 4
Now Work
EX A MPL E 2
PROBLEM
Solution
Convert the polar equation to a rectangular equation.
x 4 O 1 2 3 4 5 0
p 4
tan u = tan
– 4
y = 1 x y = x
–
p 4
u =
– 2
r
Identifying and Graphing a Polar Equation (Line)
y
3–– 4
7–– 4
13
Figure 21
p or y = x 4
x 1 2 3 4 5 0
3 –– 2
Identify and graph the equation: u = u =
–4
p 4
Take the tangent of both sides.
tan u =
y p ; tan = 1 x 4
p p is a line passing through the pole making an angle of with the 4 4 polar axis. See Figure 21. The graph of u =
5–– 4
3–– 2
7–– 4
r
Now Work
PROBLEM
15
SECTION 8.2 Polar Equations and Graphs
EXAM PL E 3
627
Identifying and Graphing a Polar Equation (Horizontal Line) Identify and graph the equation: r sin u = 2
Solution
Since y = r sin u, we can write the equation as
Figure 22 r sin u = 2 or y = 2 y
y = 2
– 2
3–– 4
Thus, the graph of r sin u = 2 is a horizontal line 2 units above the pole. See Figure 22.
x 1 2 3 4 5 O 0
COMMENT A graphing utility can be used to graph polar equations. 3FBE 6TJOH B (SBQIJOH 6UJMJUZ UP Graph a Polar Equation, Appendix B, Section B.8. ■
EXAM PL E 4
– 4
5–– 4
3–– 2
7–– 4
r
Identifying and Graphing a Polar Equation (Vertical Line) Identify and graph the equation: r cos u = - 3
Solution
Since x = r cos u, we can write the equation as
Figure 23 r cos u = - 3 or x = - 3 y
x = -3
Thus, the graph of r cos u = - 3 is a vertical line 3 units to the left of the pole. See Figure 23.
⫽ 3–– 4
⫽
– ⫽ 2
O
– ⫽ 4
x 1 2 3 4 5 ⫽0
⫽ 7–– 4
⫽ 5–– 4 ⫽ 3–– 2
r Based on Examples 3 and 4, we are led to the following results. (The proofs are left as exercises. See Problems 81 and 82.)
THEOREM
Let a be a nonzero real number. Then the graph of the equation r sin u = a is a horizontal line a units above the pole if a 7 0 and 0 a 0 units below the pole if a 6 0. The graph of the equation r cos u = a is a vertical line a units to the right of the pole if a 7 0 and 0 a 0 units to the left of the pole if a 6 0.
Now Work
PROBLEM
19
628
CHAPTER 8 Polar Coordinates; Vectors
Identifying and Graphing a Polar Equation (Circle)
EX A MPL E 5
Identify and graph the equation: r = 4 sin u
Solution
To transform the equation to rectangular coordinates, multiply each side by r.
Figure 24 r = 4 sin u or x2 + (y - 2)2 = 4
r 2 = 4r sin u
y
Now use the facts that r 2 = x2 + y2 and y = r sin u. Then
– ⫽ 2
⫽ 3–– 4
– ⫽ 4
x2 + y2 = 4y
x2 + 1y2 - 4y2 = 0
x2 + 1y2 - 4y + 42 = 4
x
⫽
O
x2 + 1y - 22 2 = 4
1 2 3 4 5 ⫽0
⫽ 5–– 4
Complete the square in y. Factor.
This is the standard equation of a circle with center at 10, 22 in rectangular coordinates and radius 2. See Figure 24.
⫽ 7–– 4
r
⫽ 3–– 2
Identifying and Graphing a Polar Equation (Circle)
EX A MPL E 6
Identify and graph the equation: r = - 2 cos u
Solution Figure 25 2 r = - 2 cos u or (x + 1) + y2 = 1
Proceed as in Example 5. r 2 = - 2r cos u Multiply both sides by r.
y ⫽ 3–– 4
– ⫽ 2
x2 + y2 = - 2x – ⫽ 4
x2 + 2x + y2 = 0
1x2 + 2x + 12 + y2 = 1 1x + 12 + y = 1 2
x
⫽
O
r 2 = x 2 + y 2; x = r cos u
2
Complete the square in x Factor.
1 2 3 4 5 ⫽0
This is the standard equation of a circle with center at 1 - 1, 02 in rectangular coordinates and radius 1. See Figure 25.
⫽ 7–– 4
⫽ 5–– 4 ⫽ 3–– 2
r
Exploration
Using a square screen, graph r1 = sin u, r2 = 2 sin u, and r3 = 3 sin u. Do you see the pattern? Clear the screen and graph r1 = - sin u, r2 = - 2 sin u, and r3 = - 3 sin u. Do you see the pattern? Clear the screen and graph r1 = cos u, r2 = 2 cos u, and r3 = 3 cos u. Do you see the pattern? Clear the screen and graph r1 = - cos u, r2 = - 2 cos u, and r3 = - 3 cos u. Do you see the pattern?
#BTFE PO &YBNQMFT BOE BOE UIF QSFDFEJOH &YQMPSBUJPO XF BSF MFE UP UIF GPMMPXJOHSFTVMUT 5IFQSPPGTBSFMFGUBTFYFSDJTFT4FF1SPCMFNTm
THEOREM
Let a be a positive real number. Then Equation (a) r (b) r (c) r (d) r
= = = =
2a sin u - 2a sin u 2a cos u - 2a cos u
Description
Circle: radius a; center at 10, a2 in rectangular coordinates Circle: radius a; center at 10, - a2 in rectangular coordinates Circle: radius a; center at 1a, 02 in rectangular coordinates Circle: radius a; center at 1 - a, 02 in rectangular coordinates
Each circle passes through the pole.
Now Work
PROBLEM
21
SECTION 8.2 Polar Equations and Graphs
629
The method of converting a polar equation to an identifiable rectangular equation to obtain the graph is not always helpful, nor is it always necessary. Usually, a table is created that lists several points on the graph. By checking for symmetry, it may be possible to reduce the number of points needed to draw the graph.
2 Test Polar Equations for Symmetry
In polar coordinates, the points 1r, u2 and 1r, - u2 are symmetric with respect to the polar axis (the xBYJT 4FF'JHVSF B 5IFQPJOUT1r, u2 and 1r, p - u2 are symmetric p with respect to the line u = (the yBYJT 4FF 'JHVSF C 5IF QPJOUT 1r, u2 and 2 1 - r, u2 BSFTZNNFUSJDXJUISFTQFDUUPUIFQPMF UIFPSJHJO 4FF'JHVSF D
Figure 26 y 2
3–– 4
– 4
3–– 4
(r, )
O
5–– 4
y
y
–
x
1 2 3 4 5 (r, )
3–– 2
0
O
3–– 4
– 4
(r, )
(r, )
1 2 3 4 5
– 2
x 0
O
5–– 4
3–– 2
7–– 4
(b) Points symmetric with respect to the line –– 2
5–– 4
– 4
(r, )
(r, )
7–– 4
(a) Points symmetric with respect to the polar axis
– 2
x
1 2 3 4 5
0
(r, )
3–– 2
7–– 4
(c) Points symmetric with respect to the pole
The following tests are a consequence of these observations.
THEOREM
Tests T Te sts for Symmetry Symmetry with Respect to the Polar Axis ((x (x-Axis) -Axis) In a polar equation, replace u by - u. If an equivalent equation results, the graph is symmetric with respect to the polar axis. P Symmetry with Respect to the Line U = ((y-Axis) (y -Axis) 2 In a polar equation, replace u by p - u. If an equivalent equation results, p the graph is symmetric with respect to the line u = . 2 Symmetry with Respect to the Pole (Origin) In a polar equation, replace r by - r. If an equivalent equation results, the graph is symmetric with respect to the pole.
The three tests for symmetry given here are sufficient conditions for symmetry, but they are not necessary conditions. That is, an equation may fail these tests and p still have a graph that is symmetric with respect to the polar axis, the line u = , 2 or the pole. For example, the graph of r = sin 12u2 turns out to be symmetric with p respect to the polar axis, the line u = , and the pole, but all three tests given here 2 fail. See also Problems 87–89.
630
CHAPTER 8 Polar Coordinates; Vectors
3 Graph Polar Equations by Plotting Points EX A MPL E 7
Graphing a Polar Equation (Cardioid) Graph the equation: r = 1 - sin u
Solution Table 1 U p 2 p 3
-
p 6
Check for symmetry first. Polar Axis: 3FQMBDFu by - u. The result is r = 1 - sin 1 - u2 = 1 + sin u
r = 1 − sin U 1 - ( - 1) = 2 23 b ≈ 1.87 1 - a2 1 3 1 - a- b = 2 2
0 p 6
1 - 0 = 1 1 1 1 = 2 2
p 3
1 -
p 2
1 - 1 = 0
sin (- u) = - sin u
The test fails, so the graph may or may not be symmetric with respect to the polar axis. The Line U =
P : 3FQMBDFu by p - u. The result is 2
r = 1 - sin 1p - u2 = 1 - 1sin p cos u - cos p sin u2
= 1 - 3 0 # cos u - 1 - 12 sin u4 = 1 - sin u
23 ≈ 0.13 2
The test is satisfied, so the graph is symmetric with respect to the line u =
p . 2
The Pole: 3FQMBDFr by - r. Then the result is - r = 1 - sin u, so r = - 1 + sin u. The test fails, so the graph may or may not be symmetric with respect to the pole. Next, identify points on the graph by assigning values to the angle u and calculating the corresponding values of r. Due to the symmetry with respect to the line p p p u = , just assign values to u from - to , as given in Table 1. 2 2 2 Now plot the points 1r, u2 from Table 1 and trace out the graph, beginning at p p the point a2, - b and ending at the point a0, b . Then reflect this portion of the 2 2 p graph about the line u = (the y-axis) to obtain the complete graph. See Figure 27. 2 Figure 27 r = 1 - sin u
y ⫽ 3–– 4
– ⫽ 2
(0.13, –3 ) ⫽
– ⫽ 4
( ––12 , –– 6 ) (1, 0) 1 2
O
3
x
⫽0
( ––2 , ⫺––6 )
Exploration Graph r1 = 1 + sin u. Clear the screen and graph r1 = 1 - cos u. Clear the screen and graph r1 = 1 + cos u. Do you see a pattern?
DEFINITION
⫽ 5–– 4
–
(2, ⫺ 2 ) ⫽ 3––
(1.87, ⫺ –3 ) ⫽ 7–– 4
r
2
The curve in Figure 27 is an example of a cardioid (a heart-shaped curve). Cardioids are characterized by equations of the form r = a11 + cos u2
r = a11 + sin u2
r = a11 - cos u2
r = a11 - sin u2
where a 7 0. The graph of a cardioid passes through the pole.
Now Work
PROBLEM
37
SECTION 8.2 Polar Equations and Graphs
EXAM PL E 8
631
Graphing a Polar Equation (Limaçon without an Inner Loop) Graph the equation: r = 3 + 2 cos u
Solution
Check for symmetry first. Polar Axis: 3FQMBDFu by - u. The result is r = 3 + 2 cos 1 - u2 = 3 + 2 cos u
cos (- u) = cos u
The test is satisfied, so the graph is symmetric with respect to the polar axis. P : 3FQMBDFu by p - u. The result is 2
The Line U =
Table 2 U
r = 3 + 2 cos U
0
3 + 2(1) = 5
p 6
23 b ≈ 4.73 3 + 2a 2
p 3
1 3 + 2a b = 4 2
p 2
3 + 2(0) = 3
2p 3
1 3 + 2a - b = 2 2
5p 6
3 + 2a -
p
3 + 2( - 1) = 1
23 b ≈ 1.27 2
r = 3 + 2 cos 1p - u2 = 3 + 21cos p cos u + sin p sin u2 = 3 - 2 cos u The test fails, so the graph may or may not be symmetric with respect to the p line u = . 2 The Pole: 3FQMBDFr by - r. The test fails, so the graph may or may not be symmetric with respect to the pole. Next, identify points on the graph by assigning values to the angle u and calculating the corresponding values of r. Due to the symmetry with respect to the polar axis, just assign values to u from 0 to p, as given in Table 2. Now plot the points 1r, u2 from Table 2 and trace out the graph, beginning at the point 15, 02 and ending at the point 11, p2. Then reflect this portion of the graph about the polar axis (the x-axis) to obtain the complete graph. See Figure 28.
Figure 28 r = 3 + 2 cos u
y ⫽ 3–– 4
– ⫽ 2
(4, –3 )
(3, –2 ) (2, 2––3)
⫽
(1.27, 5––6 ) (1, ) O
⫽ 5–– 4
Exploration Graph r1 = 3 - 2 cos u. Clear the screen and graph r1 = 3 + 2 sin u. Clear the screen and graph r1 = 3 - 2 sin u. Do you see a pattern?
DEFINITION
– ⫽ 4
(4.73, –6 )
(5, 0) x 1 2 3 4 5 ⫽0
⫽ 3–– 2
⫽ 7–– 4
r
The curve in Figure 28 is an example of a limaçon (a French word for snail) without an inner loop. Limaçons without an inner loop are characterized by equations of the form r = a + b cos u
r = a + b sin u
r = a - b cos u
r = a - b sin u
where a 7 0, b 7 0, and a 7 b. The graph of a limaÇon without an inner loop does not pass through the pole.
Now Work
PROBLEM
43
632
CHAPTER 8 Polar Coordinates; Vectors
EX A MPL E 9
Graphing a Polar Equation (Limaçon with an Inner Loop) Graph the equation: r = 1 + 2 cos u
Solution
First, check for symmetry. Polar Axis: 3FQMBDFu by - u. The result is r = 1 + 2 cos 1 - u2 = 1 + 2 cos u
Table 3 U
r = 1 + 2 cos U
0
1 + 2(1) = 3
p 6
1 + 2a
p 3
1 1 + 2a b = 2 2
p 2
1 + 2(0) = 1
2p 3
1 1 + 2a - b = 0 2
5p 6
1 + 2a -
p
1 + 2( - 1) = - 1
23 b ≈ 2.73 2
The test is satisfied, so the graph is symmetric with respect to the polar axis. The Line U =
P : 3FQMBDFu by p - u. The result is 2
r = 1 + 2 cos 1p - u2 = 1 + 21cos p cos u + sin p sin u2 = 1 - 2 cos u The test fails, so the graph may or may not be symmetric with respect to the line p u = . 2
23 b ≈ - 0.73 2
The Pole: 3FQMBDFr by - r. The test fails, so the graph may or may not be symmetric with respect to the pole. Next, identify points on the graph of r = 1 + 2 cos u by assigning values to the angle u and calculating the corresponding values of r. Due to the symmetry with respect to the polar axis just assign values to u from 0 to p, as given in Table 3. Now plot the points 1r, u2 from Table 3, beginning at 13, 02 and ending at 1 - 1, p2. See Figure 29(a). Finally, reflect this portion of the graph about the polar axis (the x-axis) to obtain the complete graph. See Figure 29(b).
Figure 29
y
y
– ⫽ 2
⫽3–– 4
(2, –
– 3
(1, 2 ) (0, 2––3 )
⫽
5 ⫺0.73, –– 6
(
Exploration Graph r1 = 1 - 2 cos u. Clear the screen and graph r1 = 1 + 2 sin u. Clear the screen and graph r1 = 1 - 2 sin u. Do you see a pattern?
DEFINITION
⫽5–– 4
– ⫽ 4
– ⫽ 2
⫽3–– 4
(2, –3 ) (1, –2 ) (0, 2––3 )
) ( 2.73, – ) 6 2
)
(3, 0) 4
x ⫽0
⫽
(⫺1, )
⫽3–– 2
5 ⫺0.73, –– 6
(
⫽ 7–– 4
⫽5–– 4
– ⫽ 4
( 2.73, –6 ) 2
(3, 0) 4
x ⫽0
) (⫺1, )
⫽3–– 2
⫽ 7–– 4
(b) r ⫽ 1 ⫹ 2 cos
(a)
The curve in Figure 29(b) is an example of a limaçon with an inner loop.
Limaçons with an inner loop are characterized by equations of the form r = a + b cos u
r = a + b sin u
r = a - b cos u
r = a - b sin u
where a 7 0, b 7 0, and a 6 b. The graph of a limaçon with an inner loop will pass through the pole twice.
Now Work
PROBLEM
45
r
SECTION 8.2 Polar Equations and Graphs
EX AM PL E 10
Solution
633
Graphing a Polar Equation (Rose) Graph the equation: r = 2 cos 12u2 Check for symmetry. Polar Axis: 3FQMBDFu by - u. The result is r = 2 cos 3 21 - u2 4 = 2 cos 12u2 The test is satisfied, so the graph is symmetric with respect to the polar axis. P : 3FQMBDFu by p - u. The result is 2
The Line U =
r = 2 cos 3 21p - u2 4 = 2 cos 12p - 2u2 = 2 cos 12u2 p . 2 The Pole: Since the graph is symmetric with respect to both the polar axis and the p line u = , it must be symmetric with respect to the pole. 2 The test is satisfied, so the graph is symmetric with respect to the line u =
Table 4 U
r = 2 cos(2U)
0
2(1) = 2
p 6
1 2a b = 1 2
p 4
2(0) = 0
p 3
1 2a - b = - 1 2
p 2
2( - 1) = - 2
Next, construct Table 4. Due to the symmetry with respect to the polar axis, the p p line u = , and the pole, just consider values of u from 0 to . 2 2 Plot and connect these points as shown in Figure 30(a). Finally, because of symmetry, reflect this portion of the graph first about the polar axis (the x-axis) and p then about the line u = (the y-axis) to obtain the complete graph. See Figure 30(b). 2
Figure 30
y 2
⫽3–– 4
Exploration
Graph r1 = 2 cos 14u2 ; clear the screen and graph r1 = 2 cos 16u2 . How many petals did each of these graphs have? Clear the screen and graph, in order, each on a clear screen, r1 = 2 cos 13u2 , r1 = 2 cos 15u2 , and r1 = 2 cos 17u2 . What do you notice about the number of petals?
(0,
⫽
y
– ⫽
⫺1, –
(
3
(1, – 4
– ⫽ 4
– 6
)
)
(⫺2, ⫽5–– 4
– 2
x ⫽0
)
⫽3–– 2
– ⫽ 4
(1, –6 )
(2, 0)
1 2 3 4 5
)
– ⫽ 2
⫽3–– 4
(2, 0) x 2 3 4 5 ⫽0
⫽
(⫺1, –3 )
(⫺2, –2 )
⫽5–– 4
⫽ 7–– 4
⫽3–– 2
⫽ 7–– 4
(b) r ⫽ 2 cos (2)
(a)
The curve in Figure 30(b) is called a rose with four petals.
DEFINITION
Rose curves are characterized by equations of the form r = a cos 1nu2,
r = a sin 1nu2,
a ≠ 0
and have graphs that are rose shaped. If n ≠ 0 is even, the rose has 2n petals; if n ≠ {1 is odd, the rose has n petals.
Now Work
PROBLEM
49
r
634
CHAPTER 8 Polar Coordinates; Vectors
EX AM PL E 11
Solution Table 5 U
r2 = 4 sin(2U)
r
0
4(0) = 0
0
p 6
4a
p 4
4(1) = 4
{2
p 3
23 b = 223 4a 2
{ 1.9
p 2
4(0) = 0
23 b = 223 2
{ 1.9
0
Graphing a Polar Equation (Lemniscate) Graph the equation: r 2 = 4 sin 12u2
We leave it to you to verify that the graph is symmetric with respect to the pole. Because of the symmetry with respect to the pole, consider only those values of u between u = 0 and u = p. Note that there are no points on the graph for p 6 u 6 p (quadrant II), since sin 12u2 6 0 for such values. Table 5 lists points on 2 p the graph for values of u = 0 through u = . The points from Table 5 where r Ú 0 2 are plotted in Figure 31(a). The remaining points on the graph may be obtained by using symmetry. Figure 31(b) shows the final graph drawn.
Figure 31 y
⫽3–– 4
⫽
⫽5–– 4
y
– ⫽ 2
(1.9,–3 )
(0, 0)
⫽3–– 2
1
– ⫽ 4 –
(2, 4 ) (1.9,–6 ) 2
x ⫽0
⫽
(1.9,–3 )
(⫺ 1.9, –6 ) (⫺ 2, –4 ) ⫽5–– 4
⫽ 7–– 4
– ⫽ 2
⫽3–– 4
– ⫽ 4
(2, –4 ) (1.9,–6 )
(0, 0)1
(⫺ 1.9, –3 ) ⫽3–– 2
2
⫽ 7–– 4
(b) r 2 ⫽ 4 sin (2)
(a)
x ⫽0
r
The curve in Figure 31(b) is an example of a lemniscate (from the Greek word ribbon).
DEFINITION
Lemniscates are characterized by equations of the form r 2 = a2 sin 12u2
r 2 = a2 cos 12u2
where a ≠ 0, and have graphs that are propeller shaped.
Now Work
EX AM PL E 12
PROBLEM
53
Graphing a Polar Equation (Spiral) Graph the equation: r = e u>5
Solution
p 2 fail. Furthermore, there is no number u for which r = 0, so the graph does not pass through the pole. Observe that r is positive for all u, r increases as u increases, r S 0 as u S - q , and r S q as u S q . With the help of a calculator, the values JO5BCMFDBOCFPCUBJOFE4FF'JHVSF
The tests for symmetry with respect to the pole, the polar axis, and the line u =
635
SECTION 8.2 Polar Equations and Graphs
Figure 32 r = eu/5
Table 6
-
U>5
U
r = e
3p 2
0.39
-p p 2 p 4
y
0.53
– ⫽ 2
⫽3–– 4
(1.37, –2 ) (1.17, –4 )
0.73
(1.87, )
⫽
(1, 0) (3.51, 2) x 2 4 ⫽0
0.85
0 p 4 p 2
1
p
1.87
3p 2
2.57
2p
3.51
– ⫽ 4
1.17
⫽5–– 4
(2.57, 3––2 ) ⫽3–– 2
⫽ 7–– 4
r
1.37
The curve in Figure 32 is called a logarithmic spiral, since its equation may be written as u = 5 ln r and it spirals infinitely both toward the pole and away from it.
Classification of Polar Equations The equations of some lines and circles in polar coordinates and their corresponding equations in rectangular coordinates are given in Table 7. Also included are the names and graphs of a few of the more frequently encountered polar equations.
Table 7 Lines Description
Line passing through the pole making an angle a with the polar axis
Rectangular equation
y = (tan a)x
Polar equation
u = a
Typical graph
Vertical line
y
Horizontal line
x = a
y = b
r cos u = a
r sin u = b
y
y
x
x
x
Circles Description
Center at the pole, radius a
Passing through the pole, p tangent to the line u = , 2 center on the polar axis, radius a
Passing through the pole, tangent to the polar axis, p center on the line u = , 2 radius a
Rectangular equation
x2 + y2 = a2, a 7 0
x2 + y2 = { 2ax, a 7 0
x2 + y2 = { 2ay, a 7 0
Polar equation
r = a, a 7 0
r = { 2a cos u, a 7 0
r = { 2a sin u, a 7 0
Typical graph
y
y
y
a a
a x
x
x
636
CHAPTER 8 Polar Coordinates; Vectors
Table 7 (Continued) Other Equations Name
Cardioid
Limaçon without inner loop
Limaçon with inner loop
Polar equations
r = a { a cos u, a 7 0
r = a { b cos u, 0 6 b 6 a
r = a { b cos u, 0 6 a 6 b
r = a { a sin u, a 7 0
r = a { b sin u, 0 6 b 6 a
r = a { b sin u, 0 6 a 6 b
Typical graph
y
y
x
y
x
x
Name
Lemniscate
Rose with three petals
Rose with four petals
Polar equations
r2 = a2 cos(2u), a ≠ 0
r = a sin(3u), a 7 0
r = a sin(2u), a 7 0
r2 = a2 sin(2u), a ≠ 0
r = a cos(3u), a 7 0
r = a cos(2u), a 7 0
Typical graph
y
y
x
y
x
x
Sketching Quickly If a polar equation involves only a sine (or cosine) function, you can quickly obtain a sketch of its graph by making use of Table 7, periodicity, and a short table.
EX AM PL E 13
Sketching the Graph of a Polar Equation Quickly Graph the equation: r = 2 + 2 sin u
Solution
You should recognize the polar equation: Its graph is a cardioid. The period of sin u is 2p, so form a table using 0 … u … 2p, compute r, plot the points 1r, u2 , and sketch the graph of a cardioid as u varies from 0 to 2p. See Table 8 and Figure 33.
Table 8 U
r = 2 + 2 sin U
0 p 2
2 + 2(0) = 2 2 + 2(1) = 4
p
2 + 2(0) = 2
3p 2
2 + 2( - 1) = 0
2p
2 + 2(0) = 2
Figure 33 r = 2 + 2 sin u y
3–– 4
(2, )
5–– 4
– 2
(4, ––2 ) – 4
(2, 0) 1 2 3 4 5
x 0
(0, 3––2) 3–– 2
7–– 4
r
SECTION 8.2 Polar Equations and Graphs
637
Calculus Comment For those of you who are planning to study calculus, a comment about one important role of polar equations is in order. In rectangular coordinates, the equation x2 + y2 = 1, whose graph is the unit circle, is not the graph of a function. In fact, it requires two functions to obtain the graph of the unit circle: y1 = 21 - x2
Upper semicircle
y2 = - 21 - x2
Lower semicircle
In polar coordinates, the equation r = 1, whose graph is also the unit circle, does define a function. For each choice of u, there is only one corresponding value of r—that is, r = 1. Since many problems in calculus require the use of functions, the opportunity to express relations that are nonfunctions in rectangular coordinates as functions in polar coordinates becomes extremely useful. Note also that the vertical-line test for functions is valid only for equations in rectangular coordinates.
Historical Feature
P
olar coordinates seem to have been invented by Jakob Bernoulli (1654–1705) in about 1691, although, as with most such ideas, earlier traces of the notion exist. Early users of calculus remained committed to rectangular coordinates, and polar coordinates did not become widely used until the early
1800s. Even then, it was mostly geometers who used them for describing odd curves. Finally, about the mid-1800s, applied mathematicians realized the tremendous simplification that polar coordinates make possible in the description of objects with circular or cylindrical symmetry. From then on, their use became widespread.
Jakob Bernoulli (1654–1705)
8.2 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. If the rectangular coordinates of a point are 14, - 2, the point symmetric to it with respect to the origin is 1 - 4, 2 . QQm
2. The difference formula for cosine is cos 1A - B2 = ______. (p. 531) cos A cos B + sin A sin B
3. The standard equation of a circle with center at 1 - 2, 52 and radius 3 is ______. QQm 1x + 22 2 + 1y - 52 2 = 9
4. Is the sine function even, odd, or neither? Q
5. sin
5p = ______. (pp. 430–432) 4
-
22 2
6. cos
2p = ______. (pp. 430–432) 3
-
1 2
Odd
Concepts and Vocabulary 7. An equation whose variables are polar coordinates is called a(n) polar equation. 8. True or False The tests for symmetry in polar coordinates are necessary but not sufficient. False 9. To test whether the graph of a polar equation may be symmetric with respect to the polar axis, replace u by - u .
10. To test whether the graph of a polar equation may be symmetric p with respect to the line u = , replace u by p - u . 2 11. True or False A cardioid passes through the pole. True 12. 3PTF DVSWFT BSF DIBSBDUFSJ[FE CZ FRVBUJPOT PG UIF GPSN r = a cos (n u) or r = a sin (n u), a ≠ 0. If n ≠ 0 is even, the rose has 2n petals; if n ≠ {1 is odd, the rose has n petals.
638
CHAPTER 8 Polar Coordinates; Vectors
Skill Building In Problems 13–28, transform each polar equation to an equation in rectangular coordinates. Then identify and graph the equation. *13. r = 4
*14. r = 2
*15. u =
p 3
*16. u = -
p 4
*17. r sin u = 4
*18. r cos u = 4
*19. r cos u = - 2
*20. r sin u = - 2
*21. r = 2 cos u
*22. r = 2 sin u
*23. r = - 4 sin u
*24. r = - 4 cos u
*25. r sec u = 4
*26. r csc u = 8
*27. r csc u = - 2
*28. r sec u = - 4
In Problems 29–36, match each of the graphs (A) through (H) to one of the following polar equations. 29. r = 2
30. u =
E
33. r = 1 + cos u
3–– 4
5–– 4
3–– 4
5–– 4
– 4
x 0
2
O
3–– 2
31. r = 2 cos u
A
34. r = 2 sin u
H
y – 2
p 4
3–– 4
G
y – 2
O
5–– 4
7–– 4
3p 35. u = 4
1
3–– 2
– 4
D
3–– 4
x 0
3
7–– 4
F
y – 2
O
5–– 4
32. r cos u = 2
B
36. r sin u = 2
C
3–– 4
– 4
x 0
2
O
5–– 4
7–– 4
3–– 2
y – 2
2
3–– 2
(A)
(B)
(C)
(D)
y – 2
y – 2
y – 2
y – 2
O
1
3–– 2
– 4
3
x 0
7–– 4
3–– 4
5–– 4
(E)
O
1
3–– 2
3–– 4
– 4
3
x 0
7–– 4
5–– 4
O
3–– 2
– 4
2
4
3–– 4
x 0
7–– 4
5–– 4
O
3–– 2
(G)
(F)
– 4
x 0
4
7–– 4
– 4
x 2 0
7–– 4
(H)
In Problems 37–60, identify and graph each polar equation. *37. r = 2 + 2 cos u
*38. r = 1 + sin u
*39. r = 3 - 3 sin u
*40. r = 2 - 2 cos u
*41. r = 2 + sin u
*42. r = 2 - cos u
*43. r = 4 - 2 cos u
*44. r = 4 + 2 sin u
*45. r = 1 + 2 sin u
*46. r = 1 - 2 sin u
*47. r = 2 - 3 cos u
*48. r = 2 + 4 cos u
*53. r = 9 cos 12u2
*54. r = sin12u2
*55. r = 2
*58. r = 3 + cos u
*59. r = 1 - 3 cos u
*49. r = 3 cos 12u2 2
*57. r = 1 - cos u
*50. r = 2 sin 13u2 2
*51. r = 4 sin 15u2 u
*52. r = 3 cos 14u2 *56. r = 3u
*60. r = 4 cos 13u2
Mixed Practice In Problems 61–66, graph each pair of polar equations on the same polar grid. Find the polar coordinates of the point(s) of intersection. *61. r = 8 cos u; r = 2 sec u
*62. r = 8 sin u; r = 4 csc u
*63. r = sin u; r = 1 + cos u
*64. r = 3; r = 2 + 2 cos u
*65. r = 1 + sin u; r = 1 + cos u
*66. r = 1 + cos u; r = 3 cos u
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
SECTION 8.2 Polar Equations and Graphs
639
Applications and Extensions In Problems 67–70, the polar equation for each graph is either r = a + b cos u or r = a + b sin u, a 7 0. Select the correct equation and find the values of a and b. 67.
r = 3 + 3 cos u
y
68.
2
3–– 4
(
3, – 2
)
–
0 2 4 6 8 10
5–– 4
2
3–– 4
–4 x
(6, 0)
0
7–– 4
(3, –2 )
(
2
3–– 4
)
7–– 4
0 1 2 3 4 5
5–– 4
70. 3–– 4
x
0
*77. r = csc u - 2, 0 6 u 6 p (conchoid) *79. r = tan u,
-
p p 6 u 6 (kappa curve) 2 2
–2
5–– 4
3–– 2
*75. r = u, u Ú 0 (spiral of Archimedes)
(5, –2 )
–4
(1, 0) 0 1 2 3 4 5
7–– 4
In Problems 71–80, graph each polar equation. 2 *71. r = (parabola) 1 - cos u 1 *73. r = (ellipse) 3 - 2 cos u
r = 1 + 4 sin u
y
–4
(4, 0)
0
3–– 2
r = 4 + sin u
–2
x 0 2 4 6 8 10
5–– 4
y
5, –
–4
(6, )
3–– 2
69.
r = 3 - 3 cos u
y
–
x
0
7–– 4 3–– 2
2 (hyperbola) 1 - 2 cos u 1 *74. r = (parabola) 1 - cos u 3 *76. r = (reciprocal spiral) u *72. r =
*78. r = sin u tan u (cissoid) *80. r = cos
u 2
*81. Show that the graph of the equation r sin u = a is a horizontal line a units above the pole if a Ú 0 and 0 a 0 units below the pole if a 6 0.
*82. Show that the graph of the equation r cos u = a is a vertical line a units to the right of the pole if a Ú 0 and 0 a 0 units to the left of the pole if a 6 0.
*83. Show that the graph of the equation r = 2a sin u, a 7 0, is a circle of radius a with center at 10, a2 in rectangular coordinates.
*84. Show that the graph of the equation r = - 2a sin u, a 7 0, is a circle of radius a with center at 10, - a2 in rectangular coordinates.
*85. Show that the graph of the equation r = 2a cos u, a 7 0, is a circle of radius a with center at 1a, 02 in rectangular coordinates.
*86. Show that the graph of the equation r = - 2a cos u, a 7 0, is a circle of radius a with center at 1 - a, 02 in rectangular coordinates.
Discussion and Writing 87. &YQMBJOXIZUIFGPMMPXJOHUFTUGPSTZNNFUSZJTWBMJE3FQMBDF r by - r and u by - u in a polar equation. If an equivalent equation results, the graph is symmetric with respect to the p line u = (y-axis). 2 * B 4IPXUIBUUIFUFTUPOQBHFGBJMTGPSr 2 = cos u, yet this new test works. * C 4IPXUIBUUIFUFTUPOQBHFXPSLTGPSr 2 = sin u, yet this new test fails.
88. Write down two different tests for symmetry with respect to the polar axis. Find examples in which one test works and the other fails. Which test do you prefer to use? Justify your answer. 89. 5IFUFTUTGPSTZNNFUSZHJWFOPOQBHFBSFTVGGJDJFOU CVU not necessary. Explain what this means. 90. Explain why the vertical-line test used to identify functions in rectangular coordinates does not work for equations expressed in polar coordinates.
640
CHAPTER 8 Polar Coordinates; Vectors
Retain Your Knowledge Problems 91–94 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 91. Using the given graph of y = f 1x2 , find the absolute maximum and the absolute minimum, if it exists. Absolute maximum: f(4) = 1; y Absolute minimum; none 4
(4, 1) 26
(1, 21)
O
6 x
(25, 22)
7p radians to degrees. 420° 3 93. Determine the amplitude and period of y = -2 sin (5x) without graphing. 94. Find any asymptotes for the graph of x + 3 . R 1x2 = 2 x - x - 12 Horizontal asymptote: y = 0; Vertical asymptote: x = 4 92. Convert
93. Amplitude = 2; Period =
2p 5
(4, 25) (23, 26) 28
‘Are You Prepared?’ Answers 1. 1 - 4, 2
2. cos A cos B + sin A sin B
3. 1x + 22 2 + 1y - 52 2 = 9
4. Odd
5. -
22 2
6. -
1 2
8.3 The Complex Plane; De Moivre’s Theorem PREPARING FOR THIS SECTION Before getting started, review the following: r $PNQMFY/VNCFST "QQFOEJY" 4FDUJPO" pp. A89–A94) r 7BMVFPGUIF4JOFBOE$PTJOF'VODUJPOTBU$FSUBJO Angles (Section 5.2, pp. 425–432)
r 4VNBOE%JGGFSFODF'PSNVMBTGPS4JOFBOE$PTJOF 4FDUJPO QQBOE
Now Work the ‘Are You Prepared?’ problems on page 646. OBJECTIVES 1 Plot Points in the Complex Plane (p. 640) 2 Convert a Complex Number between Rectangular Form and Polar Form (p. 641) 3 Find Products and Quotients of Complex Numbers in Polar Form (p. 642) 4 Use De Moivre’s Theorem (p. 643) 5 Find Complex Roots (p. 644)
1 Plot Points in the Complex Plane Figure 34 Imaginary axis
z x yi
y
O
x
Real axis
Complex numbers are discussed in Appendix A, Section A.11. In that discussion, we were not prepared to give a geometric interpretation of a complex number. Now we are ready. A complex number z = x + yi can be interpreted geometrically as the point 1x, y2 in the xy-plane. Each point in the plane corresponds to a complex number, and conversely, each complex number corresponds to a point in the plane. The collection of such points is referred to as the complex plane. The x-axis is referred to as the real axis, because any point that lies on the real axis is of the form z = x + 0i = x, a real number. The y-axis is called the imaginary axis, because any point that lies on it is of the form z = 0 + yi = yi, a pure imaginary number. See Figure 34.
SECTION 8.3 The Complex Plane; De Moivre’s Theorem
641
Plotting a Point in the Complex Plane
EXAM PL E 1
Plot the point corresponding to z = 13 - i in the complex plane.
Solution
The point corresponding to z = 23 - i has the rectangular coordinates 1 13, - 12 . This point, located in quadrant IV, is plotted in Figure 35.
Figure 35 Imaginary axis 2 2
O 2
0z0 = 2 2x2 + y2
Figure 36 Imaginary axis 2
y z x yi x y ⏐z ⏐ Real axis x O
z 3i
r
Let z = x + yi be a complex number. The magnitude or modulus of z, denoted by 0 z 0 , is defined as the distance from the origin to the point 1x, y2. That is,
DEFINITION
2
Real axis
2
(1)
4FF'JHVSFGPSBOJMMVTUSBUJPO This definition for 0 z 0 is consistent with the definition for the absolute value of a real number: If z = x + yi is real, then z = x + 0i and
0 z 0 = 2x2 + 02 = 2x2 = 0 x 0
For this reason, the magnitude of z is sometimes called the absolute value of z. 3FDBMMUIBUJGz = x + yi, then its conjugate, denoted by z, is z = x - yi. Since zz = x2 + y2, which is a nonnegative real number, it follows from equation (1) that the magnitude of z can be written as
0 z 0 = 2zz
(2)
2 Convert a Complex Number between Rectangular Form and Polar Form When a complex number is written in the standard form z = x + yi, we say that it is in rectangular, or Cartesian, form, because 1x, y2 are the rectangular coordinates of the corresponding point in the complex plane. Suppose that 1r, u2 are the polar coordinates of this point. Then x = r cos u,
z = x + yi = 1rr cos u2 + 1rr sin u2i = r 1cos u + i sin u2
Figure 37 Imaginary axis
z r x
y Real axis
z x yi r (cos i sin ), r ≥ 0, 0 ≤ 2
(3)
If r Ú 0 and 0 … u 6 2p, the complex number z = x + yi may be written in polar form as
DEFINITION
O
y = r sin u
(4)
See Figure 37. If z = r 1cos u + i sin u2 is the polar form of a complex number,* the angle u, 0 … u 6 2p, is called the argument of z. Also, because r Ú 0, we have r = 2x2 + y2 . From equation (1), it follows that the magnitude of z = r 1cos u + i sin u2 is
0z0 = r Some texts abbreviate the polar form using z = r(cos u + i sin u) = r cis u.
*
642
CHAPTER 8 Polar Coordinates; Vectors
EX A MPL E 2
Writing a Complex Number in Polar Form Write an expression for z = 23 - i in polar form.
Solution
The point, located in quadrant IV, is plotted in Figure 35. Because x = 23 and y = - 1, it follows that r = 2x2 + y2 = 3 1 23 2 + 1 - 12 2 = 24 = 2 2
so sin u =
y -1 = , r 2
cos u =
23 x = , r 2
0 … u 6 2p
The angle u, 0 … u 6 2p, that satisfies both equations is u = and r = 2, the polar form of z = 23 - i is z = r 1cos u + i sin u2 = 2acos
Now Work
EX A MPL E 3
PROBLEM
11p 11p . With u =
11p 11p + i sin b
r
11
Plotting a Point in the Complex Plane and Converting from Polar to Rectangular Form Plot the point corresponding to z = 21cos 30° + i sin 30°2 in the complex plane, and write an expression for z in rectangular form.
Solution Figure 38
z = 21cos 30° + i sin 30°2 = 2a
Imaginary axis 2
z 2(cos 30° i sin 30°) 2
O
To plot the complex number z = 21cos 30° + i sin 30°2, plot the point whose polar coordinates are 1r, u2 = 12, 30°2, as shown in Figure 38. In rectangular form,
30° 2
Real axis
2
Now Work
PROBLEM
23 1 + ib = 23 + i 2 2
r
23
3 Find Products and Quotients of Complex Numbers in Polar Form The polar form of a complex number provides an alternative method for finding products and quotients of complex numbers.
THEOREM In Words The magnitude of a complex number z is r and its argument is u, so when z = r (cos u + i sin u) the magnitude of the product (quotient) of two complex numbers equals the product (quotient) of their magnitudes; the argument of the product (quotient) of two complex numbers is determined by the sum (difference) of their arguments.
Let z1 = r1 1cos u1 + i sin u1 2 and z2 = r2 1cos u2 + i sin u2 2 be two complex numbers. Then z1 z2 = r1 r2 3 cos 1u1 + u2 2 + i sin 1u1 + u2 2 4
(5)
If z2 ≠ 0, then z1 r1 = 3 cos 1u1 - u2 2 + i sin 1u1 - u2 2 4 z2 r2
(6)
Proof 8F XJMM QSPWF GPSNVMB 5IF QSPPG PG GPSNVMB JT MFGU BT BO FYFSDJTF TFF1SPCMFN z1 z2 = 3 r1 1cos u1 + i sin u1 2 4 3 r2 1cos u2 + i sin u2 2 4 = r1 r2 3 1cos u1 + i sin u1 2 1cos u2 + i sin u2 2 4
= r1 r2 3 1cos u1 cos u2 - sin u1 sin u2 2 + i1sin u1 cos u2 + cos u1 sin u2 2 4
= r1 r2 3 cos 1u1 + u2 2 + i sin 1u1 + u2 2 4
Let’s look at an example of how this theorem can be used.
■
SECTION 8.3 The Complex Plane; De Moivre’s Theorem
EXAM PL E 4
643
Finding Products and Quotients of Complex Numbers in Polar Form If z = 31cos 20° + i sin 20°2 and w = 51cos 100° + i sin 100°2, find the following (leave your answers in polar form): (a) zw
Solution
z w
(b)
(a) zw = 3 31cos 20° + i sin 20°2 4 3 51cos 100° + i sin 100°2 4 = 13 # 52 3 cos 120° + 100°2 + i sin 120° + 100°2 4
Apply equation (5).
= 151cos 120° + i sin 120°2 (b)
31cos 20° + i sin 20°2 z = w 51cos 100° + i sin 100°2 3 3 cos 120° - 100°2 + i sin 120° - 100°2 4 5 3 = 3 cos 1 - 80°2 + i sin 1 - 80°2 4 5 3 = 1cos 280° + i sin 280°2 5 =
Now Work
PROBLEM
Apply equation (6).
The argument must lie between 0° and 360°.
r
33
4 Use De Moivre’s Theorem %F .PJWSFT 5IFPSFN TUBUFE CZ "CSBIBN %F .PJWSF m JO CVU already known to many people by 1710, is important for the following reason: The fundamental processes of algebra are the four operations of addition, subtraction, multiplication, and division, together with powers and the extraction of roots. De Moivre’s Theorem allows these latter fundamental algebraic operations to be applied to complex numbers. De Moivre’s Theorem, in its most basic form, is a formula for raising a complex number z to the power n, where n Ú 1 is a positive integer. Let’s try to conjecture the form of the result. Let z = r 1cos u + i sin u2 be a complex number. Then, based on equation (5), we have n = 2: z2 = r 2 3 cos 12u2 + i sin 12u2 4
n = 3: z3 = z2 # z
Equation (5)
= 5 r 2 3 cos 12u2 + i sin 12u2 4 6 3 r 1cos u + i sin u2 4 = r 3 3 cos 13u2 + i sin 13u2 4
n = 4: z = z 4
3
#z
Equation (5)
= 5 r 3 3 cos 13u2 + i sin 13u2 4 6 3 r 1cos u + i sin u2 4 = r 4 3 cos 14u2 + i sin 14u2 4
Equation (5)
Do you see the pattern?
THEOREM
De Moivre’s Theorem If z = r 1cos u + i sin u2 is a complex number, then zn = r n 3 cos 1nu2 + i sin 1nu2 4 where n Ú 1 is a positive integer.
(7)
644
CHAPTER 8 Polar Coordinates; Vectors
The proof of De Moivre’s Theorem requires mathematical induction (which is not discussed until Section 11.4), so it is omitted here. The theorem is actually true for all integers, n:PVBSFBTLFEUPQSPWFUIJTJO1SPCMFN
EX A MPL E 5
Solution
Using De Moivre’s Theorem
Write 3 21cos 20° + i sin 20°2 4 3 in the standard form a + bi. 3 21cos 20° + i sin 20°2 4 3 = 23 3 cos 13 # 20°2 + i sin 13 # 20°2 4 Apply De Moivre’s Theorem. = 81cos 0° + i sin 0°2 = 8a
Now Work
EX A MPL E 6
Solution
PROBLEM
1 23 + ib = 4 + 423i 2 2
r
41
Using De Moivre’s Theorem
Write 11 + i2 5 in the standard form a + bi. To apply De Moivre’s Theorem, first write the complex number in polar form. Since the magnitude of 1 + i is 212 + 12 = 22, begin by writing 1 + i = 22 a
1 22
+
1 22
ib = 22 acos
p p + i sin b 4 4
Now 11 + i2 5 = c 22 acos =
NOTE In the solution of Example 6, the approach used in Example 2 could also be used to write 1 + i in polar form. j
p p 5 + i sin b d 4 4
1 22 2 5 c cos a5 #
p p b + i sin a5 # b d 4 4
= 422 acos
5p 5p + i sin b 4 4
= 422 c -
1 22
+ a-
1 22
b i d = - 4 - 4i
r
5 Find Complex Roots Let w be a given complex number, and let n Ú 2 denote a positive integer. Any complex number z that satisfies the equation zn = w is called a complex nth root of w. In keeping with previous usage, if n = 2, the solutions of the equation z2 = w are called complex square roots of w, and if n = 3, the solutions of the equation z3 = w are called complex cube roots of w.
THEOREM
Finding Complex Roots Let w = r 1cos u0 + i sin u0 2 be a complex number, and let n Ú 2 be an integer. If w ≠ 0, there are n distinct complex nth roots of w, given by the formula n
zk = 1r Jcos ¢
u0 u0 2kp 2kp + ≤ + i sin ¢ + ≤R n n n n
where k = 0, 1, 2, c , n - 1.
(8)
SECTION 8.3 The Complex Plane; De Moivre’s Theorem
645
Proof (Outline) We will not prove this result in its entirety. Instead, we shall show only that each zk in equation (8) satisfies the equation znk = w, proving that each zk is a complex nth root of w. znk
n u0 u0 2kp 2kp = b 1r Jcos ¢ + ≤ + i sin ¢ + ≤R r n n n n n
= 1 1r2 n b cos Jn ¢ n
u0 u0 2kp 2kp + ≤ R + i sin Jn ¢ + ≤ R r Apply De Moivre’s Theorem. n n n n
= r 3 cos 1u0 + 2kp2 + i sin 1u0 + 2kp2 4
Simplify.
= r 1cos u0 + i sin u0 2 = w
The Periodic Property
Thus each zk , k = 0, 1, c , n - 1, is a complex nth root of w. To complete the proof, we would need to show that each zk , k = 0, 1, c , n - 1, is, in fact, distinct and that there are no complex nth roots of w other than those given by equation (8). ■
EXAM PL E 7
Finding Complex Cube Roots Find the complex cube roots of - 1 + 23i. Leave your answers in polar form, with the argument in degrees.
Solution
First, express - 1 + 23i in polar form using degrees. - 1 + 23i = 2 a -
WARNING Most graphing utilities will provide only the answer z0 to the calculation 1 - 1 + 23i 2 ^ 11>32. The paragraph following Example 7 explains how to obtain z1 and z2 j from z0.
1 23 + ib = 21cos 120° + i sin 120°2 2 2
The three complex cube roots of - 1 + 23i = 21cos 120° + i sin 120°2 are 3 zk = 2 2 c cos a
120° 30°k 120° 30°k + b + i sin a + bd 3 3 3 3
3 = 2 23 cos 140° + 120°k2 + i sin 140° + 120°k2 4
k = 0, 1, 2
Therefore, z0 = 22 3 cos 140° + 120° # 02 + i sin 140° + 120° # 02 4 = 221cos 40° + i sin 40°2 3
3
3 3 z1 = 2 2 3 cos 140° + 120° # 12 + i sin 140° + 120° # 12 4 = 2 21cos 10° + i sin 10°2 3 3 z2 = 2 2 3 cos 140° + 120° # 22 + i sin 140° + 120° # 22 4 = 2 21cos 280° + i sin 280°2
r Notice that each of the three complex roots of - 1 + 23i has the same 3 magnitude, 2 2. This means that the points corresponding to each cube root lie the same distance from the origin; that is, the three points lie on a circle with center at 3 the origin and radius 2 2. Furthermore, the arguments of these cube roots are 40°, 30° BOE UIFEJGGFSFODFPGDPOTFDVUJWFQBJSTCFJOH 120° = . This means 3 that the three points are equally spaced on the circle, as shown in Figure 39 on the next page. These results are not coincidental. In fact, you are asked to show that these results hold for complex nUISPPUTJO1SPCMFNTUISPVHI
646
CHAPTER 8 Polar Coordinates; Vectors
Figure 39
Imaginary axis 2 2
2
3
2
x y ( 2) z1
3
2(cos 160° i sin 160°)
1
z0
120° 2
1
1
2(cos 40° i sin 40°)
40°
O 120°
3
1
2
Real axis
120° z2
3
2(cos 280° i sin 280°)
2
Now Work
PROBLEM
53
Historical Feature
T
John Wallis
he Babylonians, Greeks, and Arabs considered square roots of negative quantities to be impossible and equations with complex solutions to be unsolvable. The first hint that there was some connection between real solutions of equations and complex numbers came when Girolamo Cardano (1501–1576) and Tartaglia (1499–1557) found real roots of cubic equations by taking cube roots of complex quantities. For centuries
thereafter, mathematicians worked with complex numbers without much belief in their actual existence. In 1673, John Wallis appears to have been the first to suggest the graphical representation of complex numbers, a truly significant idea that was not pursued further until about 1800. Several people, including Karl Friedrich Gauss (1777–1855), then rediscovered the idea, and the graphical representation helped to establish complex numbers as equal members of the number family. In practical applications, complex numbers have found their greatest uses in the study of alternating current, where they are a commonplace tool, and in the field of subatomic physics.
Historical Problems 1. The quadratic formula works perfectly well if the coefficients of a quadratic equation are complex numbers. Solve the following. [Hint: The answers are “nice.”] (a) z2 - (2 + 5i )z - 3 + 5i = 0
1 + 4i, 1 + i
(b) z2 - (1 + i )z - 2 - i = 0
- 1, 2 + i
8.3 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. The conjugate of - 4 - 3i is - 4 + 3i . (p. A91) 2. The sum formula for the sine function is sin1A + B2 = _______ . (p. 534) sin A cos B + cos A sin B
Concepts and Vocabulary
3. The sum formula for the cosine function is cos 1A + B2 = _______. (p. 531) cos A cos B - sin A sin B 4. sin 120° = _______; cos 240° = _______ . (pp. 430–432) 1 23 2 2
5. In the complex plane, the x-axis is referred to as the real axis and the y-axis is called the imaginary axis.
8. If z = r(cos u + i sin u) is a complex number, then zn = r n 3 cos 1 nu 2 + i sin 1 nu 2 4 .
6. When a complex number z is written in the polar form z = r 1cos u + i sin u2 , the nonnegative number r is the magnitude or modulus of z, and the angle u, 0 … u 6 2p, is the argument of z.
9. Every nonzero complex number has exactly three distinct cube roots.
7. Let z1 = r1 1cos u1 + i sin u1 2 and z2 = r2 1cos u2 + i sin u2 2 be two complex numbers. Then z1z2 = r1r2 3 cos 1u1 + u2 2 + i sin 1u1 + u2 2 4 .
10. True or False The polar form of a nonzero complex number is unique. True
Skill Building In Problems 11–22, plot each complex number in the complex plane and write it in polar form. Express the argument in degrees. *11. 1 + i
*12. - 1 + i
*13. 23 - i
*14. 1 - 23i
*15. - 3i
*16. - 2
*17. 4 - 4i
*18. 923 + 9i
*19. 3 - 4i
*20. 2 + 23i
*21. - 2 + 3i
*22. 25 - i
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
647
SECTION 8.3 The Complex Plane; De Moivre’s Theorem
In Problems 23–32, write each complex number in rectangular form. 23. 21cos 120° + i sin 120°2
7p 7p + i sin b 222 - 222i 4 4 3p p p 3p 27. 3acos + i sin b - 3i 28. 4acos + i sin b 4i 2 2 2 2 30. 0.41cos 200° + i sin 200°2 - 0.37 - 0.137i
- 1 + 23i 24. 31cos 210° + i sin 210°2
5p 5p + i sin b - 23 + i *29. 0.21cos 100° + i sin 100°2 26. 2acos
31. 2acos
p p + i sin b 18 18
1.970 + 0.347i 32. 3acos
p p + i sin b 10 10
-
323 3 - i 2 2
25. 4acos
2.853 + 0.927i
z . Leave your answers in polar form. w *34. z = cos 120° + i sin 120° *33. z = 21cos 40° + i sin 40°2 w = cos 100° + i sin 100° w = 41cos 20° + i sin 20°2
In Problems 33–40, find zw and
*35. z = 31cos 130° + i sin 130°2 w = 41cos 270° + i sin 270°2
p p + i sin b 8 8 p p + i sin b w = 2acos 10 10
*36. z = 21cos 80° + i sin 80°2 w = 1cos 200° + i sin 200°2
*37. z = 2acos
*39. z = 2 + 2i w = 23 - i
*40. z = 1 - i w = 1 - 23i
3p 3p + i sin b 8 8 9p 9p + i sin b w = 2acos 1 1
*38. z = 4acos
In Problems 41–52, write each expression in the standard form a + bi. 42. 331cos 80° + i sin 80°2 4 3
*41. 341cos 40° + i sin 40°2 4 3
3 23 1cos 10°
*44. c 22 acos
5p 5p 4 + i sin bd 1 1
*45.
*47. c 25 acos
3p 3p 4 + i sin bd 1 1
*48. c 23 acos
50.
1 23
- i2
- 4
51.
1 22
- i2
+ i sin 10°2 4
-
27 2723 i 2 2
p 5 p + i sin b d 10 10
32i
5 1 46. c 1cos 72° + i sin 72°2 d 2
5p 5p + i sin bd 18 18
43. c 2acos
49. 11 - i2 5
- 23 + 14.142i
52.
11
1 32
- 4 + 4i
- 25i 2
8
- 124 - 28.217i
In Problems 53–60, find all the complex roots. Leave your answers in polar form with the argument in degrees. *53. The complex cube roots of 1 + i
*54. The complex fourth roots of 23 - i
*55. The complex fourth roots of 4 - 423i
*56. The complex cube roots of - 8 - 8i
*57. The complex fourth roots of - 1i
*58. The complex cube roots of - 8
*59. The complex fifth roots of i
*60. The complex fifth roots of - i
Applications and Extensions *61. Find the four complex fourth roots of unity (1) and plot them. *62. Find the six complex sixth roots of unity (1) and plot them. *63. Show that each complex nth root of a nonzero complex number w has the same magnitude. *64. 6TF UIF SFTVMU PG 1SPCMFN UP ESBX UIF DPODMVTJPO UIBU each complex nth root lies on a circle with center at the origin. What is the radius of this circle? *65. 3FGFSUP1SPCMFN4IPXUIBUUIFDPNQMFYnth roots of a nonzero complex number w are equally spaced on the circle. *66. 1SPWFGPSNVMB *67. Prove that De Moivre’s Theorem is true for all integers, n, by assuming it is true for integers n Ú 1 and then showing it is true for 0 and for negative integers. Hint: Multiply the numerator and the denominator by the conjugate of the denominator and use even-odd properties. *68. Mandelbrot Sets (a) Consider the expression an = 1an - 1 2 2 + z, where z is some complex number (called the seed) and a0 = z. Compute a1 1=a20 + z2, a2 1=a21 + z2, a3 1=a22 + z2, a4 , a5, and a for the following seeds: z1 = 0.1 - 0.4i, z2 = 0.5 + 0.8i, z3 = - 0.9 + 0.7i, z4 = - 1.1 + 0.1i, z5 = 0 - 1.3i, and z = 1 + 1i.
(b) The dark portion of the graph represents the set of all values z = x + yi that are in the Mandelbrot set. Determine which complex numbers in part (a) are in this set by plotting them on the graph. Do the complex numbers that are not in the Mandelbrot set have any common characteristics regarding the values of a found in part (a)? (c) Compute 0 z 0 = 2x2 + y2 for each of the complex numbers in part (a). Now compute 0 a 0 for each of the complex numbers in part (a). For which complex numbers is 0 a 0 … 0 z 0 and 0 z 0 … 2? Conclude that the criterion for a complex number to be in the Mandelbrot set is that 0 an 0 … 0 z 0 and 0 z 0 … 2. Imaginary axis y 1
Real axis 1 x
–2
–1
648
CHAPTER 8 Polar Coordinates; Vectors
Retain Your Knowledge Problems 69–72 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 69. Find the area of the triangle with a = 8, b = 11, and C = 113°. ≈40.50 70. Determine whether f(x) = 5x2 - 12x + 4 has a maximum value or a minimum value, and then find the value. 4 71. Convert 240° to radians. Express your answer as a multiple of p. p 3 3 3 1 72. Simplify: 224x2y5 2y23x2y2 70. Minimum; f a b = 5 5
‘Are You Prepared?’ Answers 1. - 4 + 3i
2. sin A cos B + cos A sin B
3. cos A cos B - sin A sin B
4.
1 13 ;2 2
8.4 Vectors OBJECTIVES 1 2 3 4 5 6 7
Graph Vectors (p. 650) Find a Position Vector (p. 651) Add and Subtract Vectors Algebraically (p. 652) Find a Scalar Multiple and the Magnitude of a Vector (p. 653) Find a Unit Vector (p. 654) Find a Vector from Its Direction and Magnitude (p. 654) Model with Vectors (p. 655)
In simple terms, a vector (derived from the Latin vehere, meaning “to carry”) is a quantity that has both magnitude and direction. It is customary to represent a vector by using an arrow. The length of the arrow represents the magnitude of the vector, and the arrowhead indicates the direction of the vector. Many quantities in physics can be represented by vectors. For example, the velocity of an aircraft can be represented by an arrow that points in the direction of movement; the length of the arrow represents speed. If the aircraft speeds up, we lengthen the arrow; if the aircraft changes direction, we introduce an arrow in the new direction. See Figure 40. Based on this representation, it is not surprising that vectors and directed line segments are somehow related.
Figure 40
Geometric Vectors If P and Q are two distinct points in the xy-plane, there is exactly one line containing both P and Q [Figure 41(a)]. The points on that part of the line that joins P to Q, including P and Q, form what is called the line segment PQ [Figure 41(b)]. Ordering the points so that they proceed from P to Q, we have a> directed line segment from P to > Q, or a geometric vector, which we denote by PQ . In a directed line segment PQ , P is called the initial point and Q the terminal point, as indicated in Figure 41(c). Figure 41
Q
Q
Q
Terminal point
P (a) Line containing P and Q
P (b) Line segment PQ
Initial point
P
(c) Directed line segment PQ
SECTION 8.4 Vectors
649
>
Figure 42 U Q S T P R
Figure 43
The magnitude of the directed line segment PQ is the distance from the point > P to the point Q; that is, it is the length of the line segment. The direction of PQ is from P to Q. If a vector v> * has the same magnitude and the same direction as the directed line segment PQ , we write > v = PQ The vector v whose magnitude is 0 is called the zero vector, 0. The zero vector is assigned no direction. Two vectors v and w are equal, written v = w if they have the same magnitude and the same direction. For example, the three vectors shown in Figure 42 have the same magnitude and the same direction, so they are equal, even though they have different initial points and different terminal points. As a result, it is useful to think of a vector simply as an arrow, keeping in mind that two arrows (vectors) are equal if they have the same direction and the same magnitude (length).
Adding Vectors Geometrically Terminal point of w vw
w
v Initial point of v
The sum v + w of two vectors is defined as follows: Position the vectors v and w so that the terminal point of v coincides with the initial point of w, as shown in Figure 43. The vector v + w is then the unique vector whose initial point coincides with the initial point of v and whose terminal point coincides with the terminal point of w. Vector addition is commutative. That is, if v and w are any two vectors, then v + w = w + v
Figure 44 v v w w w v
w
Figure 44 illustrates this fact. (The commutative property is another way of saying that opposite sides of a parallelogram are equal and parallel.) Vector addition is also associative. That is, if u, v, and w are vectors, then u + 1v + w2 = 1u + v2 + w
v
Figure 45 illustrates the associative property for vectors. Figure 45
1u + v2 + w = u + 1v + w2 vw
u
The zero vector 0 has the property that
w
v
v + 0 = 0 + v = v
uv
for any vector v. If v is a vector, then - v is the vector that has the same magnitude as v but whose direction is opposite to v BTTIPXOJO'JHVSF Furthermore,
Figure 46 v
v
v + 1 - v2 = 0 If v and w are two vectors, then the difference v - w is defined as v - w = v + 1 - w2
Figure 47
Figure 47 illustrates the relationships among v, w, v + w, and v - w.
w
v
v w
w
v
w v
Multiplying Vectors by Numbers Geometrically When dealing with vectors, real numbers are referred to as scalars. Scalars are quantities that have only magnitude. Examples of scalar quantities from physics are temperature, speed, and time. We now define how to multiply a vector by a scalar. *
Boldface letters are used to denote vectors, to distinguish them from numbers. For handwritten work, S an arrow is placed over the letter to signify a vector. For example, a vector is written by hand as v .
650
CHAPTER 8 Polar Coordinates; Vectors
DEFINITION
If a is a scalar and v is a vector, the scalar multiple av is defined as follows: 1.. If a 7 0, av is the vector whose magnitude is a times the magnitude of v and whose direction is the same as v. 2.. If a 6 0, av is the vector whose magnitude is 0 a 0 times the magnitude of v and whose direction is opposite that of v. 3. If a = 0 or if v = 0, then av = 0.
Figure 48
2v v
1v
See Figure 48 for some illustrations. For example, if a is the acceleration of an object of mass m due to a force F being exerted on it, then, by Newton’s second law of motion, F = ma. Here, ma is the product of the scalar m and the vector a. Scalar multiples have the following properties: 0v = 0
1v = v
1a + b2v = av + bv
- 1v = - v a1v + w2 = av + aw
a 1bv2 = 1ab2v
1 Graph Vectors EX A MPL E 1
Graphing Vectors Use the vectors illustrated in Figure 49 to graph each of the following vectors: (a) v - w
Solution
(c) 2v - w + u
Figure 50 shows each graph. Figure 50
Figure 49
v
(b) 2v + 3w
w
u
u
2v w u
w
2v
w
3w
vw v
2v 2v 3w (b) 2v 3w
(a) v w
Now Work
PROBLEMS
9
AND
(c) 2v w u
11
r
Magnitude of Vectors
The symbol 7 v 7 is used to represent the magnitude of a vector v. Since 7 v 7 equals the length of a directed line segment, it follows that 7 v 7 has the following properties:
THEOREM
Properties of 7 v 7 If v is a vector and if a is a scalar, then
(a) 7 v 7 Ú 0 (c) 7 - v 7 = 7 v 7
(b) 7 v 7 = 0 if and only if v = 0 (d) 7 av 7 = 0 a 0 7 v 7
Property (a) is a consequence of the fact that distance is a nonnegative number. > Property (b) follows because the length of the directed line segment PQ is
SECTION 8.4 Vectors
651
positive unless P and Q are the same point, in which case the length is 0. Property (c) follows because the length of the line segment PQ equals the length of the line segment QP. Property (d) is a direct consequence of the definition of a scalar multiple.
DEFINITION
A vector u for which 7 u 7 = 1 is called a unit vector.
2 Find a Position Vector To compute the magnitude and direction of a vector, an algebraic way of representing vectors is needed.
DEFINITION
An algebraic vector v is represented as v = 8a, b b99
Figure 51 y
,b
v
a =
where a and b are real numbers (scalars) called the components of the vector v.
P (a, b)
x
THEOREM
A rectangular coordinate system is used to represent algebraic vectors in the plane. If v = 8a, b9 is an algebraic vector whose initial point is at the origin, then v is called a position vector. See Figure 51. Notice that the terminal point of the position vector v = 8a, b9 is P = 1a, b2. The next result states that any vector whose initial point is not at the origin is equal to a unique position vector. Suppose that v is a vector with initial point P1 = 1x1 ,> y1 2, not necessarily the origin, and terminal point P2 = 1x2 , y2 2. If v = P1P2 , then v is equal to the position vector v = 8x2 - x1 , y2 - y1 9
In Words An algebraic vector represents “driving directions” to get from the initial point to the terminal point of a vector. So if v = 85, 4 9 , travel 5 units right and 4 units up from the initial point to arrive at the terminal point.
(1)
To see why this is true, look at Figure 52. Figure 52 v = 8a, b 9 = 8x2 - x1, y2 - y1 9 y P2 (x2, y2) P (a, b)
b
b
v
A O
a
a
v P1 (x1, y1)
y2 y1 Q
x2 x1
x
Triangle OPA and triangle P1P2Q are congruent. [Do you see why? The line segments have the same magnitude, so d 1O, P2 = d 1P1 , P2 2; and they have the same direction, so ∠POA = ∠P2P1Q. Since the triangles are right triangles, we have angle–side–angle.] It follows that corresponding sides are equal. As a result, x2 - x1 = a and y2 - y1 = b, so v may be written as v = 8a, b9 = 8x2 - x1 , y2 - y1 9
Because of this result, any algebraic vector can be replaced by a unique position vector, and vice versa. This flexibility is one of the main reasons for the wide use of vectors.
652
CHAPTER 8 Polar Coordinates; Vectors
Finding a Position Vector
EX A MPL E 2
>
Find the position vector of the vector v = P1P2 , if P1 = 1 - 1, 22 and P2 = 14, 2 .
Solution
By equation (1), the position vector equal to v is
v = 84 - 1 - 12, - 29 = 85, 49
r
See Figure 53.
Figure 53 y
Two position vectors v and w are equal if and only if the terminal point of v is the same as the terminal point of w. This leads to the following result:
P2 (4, 6)
5
(5, 4)
THEOREM
v
P1 (1, 2)
5
O
x
Equality of Vectors Two vectors v and w are equal if and only if their corresponding components are equal. That is, If v = 8a1 , b1 9
v = w if and only if a1 = a2 and b1 = b2 .
then
Figure 54
and w = 8a2 , b2 9
We now present an alternative representation of a vector in the plane that is common in the physical sciences. Let i denote the unit vector whose direction is along the positive x-axis; let j denote the unit vector whose direction is along the positive y-axis. Then i = 81, 09 and j = 80, 19, as shown in Figure 54. Any vector v = 8a, b9 can be written using the unit vectors i and j as follows:
y (0, 1) j
v = 8a, b 9 = a81, 09 + b 80, 19 = ai + bj i
(1, 0)
x
The quantities a and b are called the horizontal and vertical components of v, respectively. For example, if v = 85, 49 = 5i + 4j, then 5 is the horizontal component and 4 is the vertical component.
Now Work
PROBLEM
29
3 Add and Subtract Vectors Algebraically The sum, difference, scalar multiple, and magnitude of algebraic vectors are defined in terms of their components.
DEFINITION
Let v = a1i + b1j = 8a1 , b1 9 and w = a2 i + b2 j = 8a2 , b2 9 be two vectors, and let a be a scalar. Then v + w = 1a1 + a2 2i + 1b1 + b2 2j = 8a1 + a2 , b1 + b2 9 v - w = 1a1 - a2 2i + 1b1 - b2 2j = 8a1 - a2 , b1 - b2 9 av = 1aa1 2i + 1ab1 2j = 8aa1 , ab1 9
In Words To add two vectors, add corresponding components. To subtract two vectors, subtract corresponding components.
7v7 = 2 2a21 + b21
(2) (3) (4) (5)
These definitions are compatible with the geometric definitions given earlier in this section. See Figure 55.
653
SECTION 8.4 Vectors
Figure 55 y (a2, b2)
w
(a1, b1)
w
b2
v
b2
y
y
(a1 a2, b1 b2)
v
b1 O
a1 a2
v b1
(a1, b1) x
a2
a1
x
a1
(b) Illustration of property (4), 0
O
P1 (a1, b1) v a1
b1 x
(c) Illustration of property (5): || v || Distance from O to P1 || v || a 21 b 21
Adding and Subtracting Vectors
If v = 2i + 3j = 82, 39 and w = 3i - 4j = 83, - 49, find: (a) v + w
Solution
v O
(a) Illustration of property (2)
EXAM PL E 3
b1
b1
(a1, b1)
(b) v - w
(a) v + w = 12i + 3j2 + 13i - 4j2 = 12 + 32 i + 13 - 42j = 5i - j or
v + w = 82, 39 + 83, - 49 = 82 + 3, 3 + 1 - 42 9 = 85, - 19
(b) v - w = 12i + 3j2 - 13i - 4j2 = 12 - 32i + 3 3 - 1 - 42 4 j = - i + 7j or
v - w = 82, 39 - 83, - 49 = 82 - 3, 3 - 1 - 42 9 = 8 - 1, 79
r
4 Find a Scalar Multiple and the Magnitude of a Vector EXAM PL E 4
Finding Scalar Multiples and Magnitudes of Vectors If v = 2i + 3j = 82, 39 and w = 3i - 4j = 83, - 49, find: (a) 3v
Solution
(b) 2v - 3w
(c) 7 v 7
(a) 3v = 312i + 3j2 = i + 9j or
3v = 382, 39 = 8, 99
(b) 2v - 3w = 212i + 3j2 - 313i - 4j2 = 4i + j - 9i + 12j = - 5i + 18j or 2v - 3w = 282, 39 - 383, - 49 = 84, 9 - 89, - 129 = 84 - 9, - 1 - 122 9 = 8 - 5, 189
(c) 7 v 7 = 7 2i + 3j 7 = 222 + 32 = 213
Now Work
PROBLEMS
35
AND
r 41
For the remainder of the section, we will express a vector v in the form ai + bj.
654
CHAPTER 8 Polar Coordinates; Vectors
5 Find a Unit Vector
3FDBMMUIBUBVOJUWFDUPSu is a vector for which 7 u 7 = 1. In many applications, it is useful to be able to find a unit vector u that has the same direction as a given vector v.
THEOREM
Unit Vector in the Direction of v For any nonzero vector v, the vector u =
v 7v7
(6)
is a unit vector that has the same direction as v.
Proof Let v = ai + bj. Then 7 v 7 = 2a2 + b2 and u =
ai + bj v a b = = i + j 2 2 2 2 2 7v7 2a + b 2a + b 2a + b2
The vector u is in the same direction as v, since 7 v 7 7 0. Furthermore,
7u7 =
a2 b2 a 2 + b2 + = = 1 B a 2 + b2 B a 2 + b2 a 2 + b2
That is, u is a unit vector in the direction of v.
■
As a consequence of this theorem, if u is a unit vector in the same direction as a vector v, then v may be expressed as v = 7v7u
(7)
This way of expressing a vector is useful in many applications.
EX A MPL E 5
Finding a Unit Vector Find a unit vector in the same direction as v = 4i - 3j.
Solution
Find 7 v 7 first.
7 v 7 = 7 4i - 3j 7 = 21 + 9 = 5
Now multiply v by the scalar
1 1 = . A unit vector in the same direction as v is 5 7v7 v
7v7
=
4i - 3j 4 3 = i - j 5 5 5
Check: This vector is indeed a unit vector, because 4 2 3 2 1 9 25 a b + a- b = + = = 1 5 5 25 25 25
Now Work
PROBLEM
r
51
6 Find a Vector from Its Direction and Magnitude If a vector represents the speed and direction of an object, it is called a velocity vector. If a vector represents the direction and amount of a force acting on an object, it is called a force vector. In many applications, a vector is described in terms of its
SECTION 8.4 Vectors
Figure 56 y 1 v j
u
(cos , sin )
x
1
i
655
magnitude and direction, rather than in terms of its components. For example, a ball thrown with an initial speed of 25 miles per hour at an angle of 30° to the horizontal is a velocity vector. Suppose that we are given the magnitude 7 v 7 of a nonzero vector v and the direction angle a, 0° … a 6 30°, between v and i. To express v in terms of 7 v 7 and a, first find the unit vector u having the same direction as v. -PPLBU'JHVSF5IFDPPSEJOBUFTPGUIFUFSNJOBMQPJOUPGu are 1cos a, sin a2. Then u = cos a i + sin a j and, from (7), v = 7 v 7 1cos ai + sin aj2
(8)
where a is the direction angle between v and i.
Writing a Vector When Its Magnitude and Direction Are Given
EXAM PL E 6
A ball is thrown with an initial speed of 25 miles per hour in a direction that makes an angle of 30° with the positive x-axis. Express the velocity vector v in terms of i and j. What is the initial speed in the horizontal direction? What is the initial speed in the vertical direction?
Solution
The magnitude of v is 7 v 7 = 25 miles per hour, and the angle between the direction of v and i, the positive x-axis, is a = 30°. By equation (8),
v = 7 v 7 1cos ai + sin aj2 = 251cos 30°i + sin 30°j2 = 25a
Figure 57 y 12.5
v = 25(cos 30°i + sin 30°j)
12.5 j 25
x
21.65
The initial speed of the ball in the horizontal direction is the horizontal
2523 ≈ 21.5 miles per hour. The initial speed in the vertical 2 25 direction is the vertical component of v, = 12.5 miles per hour. See Figure 57. 2 component of v,
21.65 i
30°
23 1 2523 25 i + jb = i + j 2 2 2 2
r
Now Work
EXAM PL E 7
PROBLEM
57
Finding the Direction Angle of a Vector Find the direction angle a of v = 4i - 4j.
Solution
See Figure 58. The direction angle a of v = 4i - 4j can be found by solving
Figure 58
tan a =
4
x
-4 = -1 4
Because 0° … a 6 30°, the direction angle is a = 315°.
v 4i 4j
Now Work
PROBLEM
r
63
(4, 4)
4
7 Model with Vectors Figure 59 Resultant F1 + F2
F1
F2
Because forces can be represented by vectors, two forces “combine” the way that vectors “add.” If F1 and F2 are two forces simultaneously acting on an object, the vector sum F1 + F2 is the resultant force. The resultant force produces the same effect on the object as that obtained when the two forces F1 and F2 act on the object. See Figure 59.
656
CHAPTER 8 Polar Coordinates; Vectors
EX A MPL E 8
A Boeing 737 aircraft maintains a constant airspeed of 500 miles per hour headed due south. The jet stream is 80 miles per hour in the northeasterly direction.
N W
E S
Orlando
Wind
Naples
Finding the Actual Speed and Direction of an Aircraft
Miami
Solution Figure 60 N
(a) Express the velocity va of the 737 relative to the air and the velocity vw of the jet stream in terms of i and j. (b) Find the velocity of the 737 relative to the ground. (c) Find the actual speed and direction of the 737 relative to the ground. (a) Set up a coordinate system in which north (N) is along the positive y-axis. See 'JHVSF5IFWFMPDJUZPGUIFSFMBUJWFUPUIFBJSJTva = - 500j. The velocity of the jet stream vw has magnitude 80 and direction NE (northeast), so the angle between vw and i is 45°. Express vw in terms of i and j as
y vw W
500 x
E
vw = 801cos 45° i + sin 45° j2 = 80a
22 22 i + jb = 4022 1i + j2 2 2
(b) The velocity of the 737 relative to the ground vg is va 500j
vg = va + vw = - 500j + 4022 1i + j2 = 4022i +
vg
500
1 4022
- 500 2 j
(c) The actual speed of the 737 is
S
7 vg 7 = 3 1 4022 2 2 + 1 4022 - 500 2 2 ≈ 447 miles per hour To find the actual direction of the 737 relative to the ground. Determine the direction angle of vg. The direction angle is found by solving tan a =
4022 - 500 4022
Then a ≈ 277.3°. The 737 is traveling S7.3°E.
Now Work
EX A MPL E 9
PROBLEM
r
77
Finding the Weight of a Piano Two movers require a magnitude of force of 300 pounds to push a piano up a ramp inclined at an angle 20° from the horizontal. How much does the piano weigh?
Solution Figure 61
Let F1 represent the force of gravity, F2 represent the force required to move the piano up the ramp, and F3 represent the force of the piano against the ramp. See 'JHVSF5IFBOHMFCFUXFFOUIFHSPVOEBOEUIFSBNQJTUIFTBNFBTUIFBOHMFCFUXFFO F1 and F3 because triangles ABC and BDE are similar, so ∠BAC = ∠DBE = 20°. To find the magnitude of F1, calculate sin 20° =
7 F1 7 =
B A
7 F2 7 300 = 7 F1 7 7 F1 7 300 lb ≈ 877 lb sin 20°
F3
20° C F1 D
20° E
F2
The piano weighs approximately 877 pounds.
r
An object is said to be in static equilibrium if (1) the object is at rest and (2) the sum of all forces acting on the object is zero—that is, if the resultant force is 0.
SECTION 8.4 Vectors
EX AM PL E 10
657
An Object in Static Equilibrium A box of supplies that weighs 1200 pounds is suspended by two cables attached to UIFDFJMJOH BTTIPXOJO'JHVSF8IBUBSFUIFUFOTJPOTJOUIFUXPDBCMFT
Solution
Figure 62
30°
F1 = 7 F1 7 1cos 150°i + sin 150°j2 = 7 F1 7 a -
45° 30°
%SBXBGPSDFEJBHSBNVTJOHUIFWFDUPSTBTTIPXOJO'JHVSF5IFUFOTJPOTJOUIF cables are the magnitudes 7 F1 7 and 7 F2 7 of the force vectors F1 and F2 . The magnitude of the force vector F3 equals 1200 pounds, the weight of the box. Now write each force vector in terms of the unit vectors i and j. For F1 and F2 , use equation (8). 3FNFNCFSUIBUa is the angle between the vector and the positive x-axis.
45°
F2 = 7 F2 7 1cos 45°i + sin 45°j2 = 7 F2 7 a
1200 pounds
23 1 23 1 7 F1 7 i + 7 F1 7 j i + jb = 2 2 2 2
22 22 22 22 7 F2 7 i + 7 F2 7 j i + jb = 2 2 2 2
F3 = - 1200j
Figure 63
For static equilibrium, the sum of the force vectors must equal zero. y
F1
F2 150°
30°
F1 + F2 + F3 = -
23 1 22 22 7 F1 7 i + 7 F1 7 j + 7 F2 7 i + 7 F2 7 j - 1200j = 0 2 2 2 2
45° x
The i component and j component must each equal zero. This results in the two equations
F3
23 22 7 F1 7 + 7 F2 7 = 0 2 2
(9)
1 22 7 F1 7 + 7 F2 7 - 1200 = 0 2 2
(10)
-
Solve equation (9) for 7 F2 7 to obtain
7 F2 7 =
23 22
7 F1 7
(11)
Substituting into equation (10) and solving for 7 F1 7 yields 1 22 23 7 F1 7 + 7 F1 7 b - 1200 = 0 a 2 2 22 1 23 7F 7 + 7 F1 7 - 1200 = 0 2 1 2 1 + 23 7 F1 7 = 1200 2 2400 7 F1 7 = ≈ 878.5 pounds 1 + 23 Substituting this value into equation (11) gives 7 F2 7 .
7 F2 7 =
23 22
7 F1 7 =
23 #
2400
22 1 + 23
≈ 1075.9 pounds
The left cable has tension of approximately 878.5 pounds, and the right cable has tension of approximately 1075.9 pounds.
Now Work
r
PROBLEM
85
658
CHAPTER 8 Polar Coordinates; Vectors
Historical Feature
T
he history of vectors is surprisingly complicated for such a natural concept. In the xy-plane, complex numbers do a good job of imitating vectors. About 1840, mathematicians became interested in finding a system that would do for three dimensions what the complex numbers do for two dimensions. Hermann Grassmann (1809–1877), in Germany, Josiah Gibbs and William Rowan Hamilton (1805–1865), in (1839 –1903) Ireland, both attempted to find solutions. Hamilton’s system was the quaternions, which are best thought of as a real number plus a vector, and do for four dimensions what complex numbers do for two dimensions. In this system the order of multiplication matters; that is, ab ≠ ba. Also, two
products of vectors emerged, the scalar product (or dot product) and the vector product (or cross product). Grassmann’s abstract style, although easily read today, was almost impenetrable during the nineteenth century, and only a few of his ideas were appreciated. Among those few were the same scalar and vector products that Hamilton had found. About 1880, the American physicist Josiah Willard Gibbs (1839–1903) worked out an algebra involving only the simplest concepts: the vectors and the two products. He then added some calculus, and the resulting system was simple, flexible, and well adapted to expressing a large number of physical laws. This system remains in use essentially unchanged. Hamilton’s and Grassmann’s more extensive systems each gave birth to much interesting mathematics, but little of this mathematics is seen at elementary levels.
8.4 Assess Your Understanding Concepts and Vocabulary 1. A vector is a quantity that has both magnitude and direction.
5. If v = ai + bj, then a is called the horizontal component of v and b is called the vertical component of v.
0 . 3. A vector u for which 7 u 7 = 1 is called a(n) unit vector.
6. If F1 and F2 are two forces simultaneously acting on an object, the vector sum F1 + F2 is called the resultant force.
4. If v = 6a, b7 is an algebraic vector whose initial point is the origin, then v is called a(n) position vector.
7. True or False Force is an example of a vector. True
2. If v is a vector, then v + 1 - v2 =
8. True or False Mass is an example of a vector. False
Skill Building In Problems 9–16, use the vectors in the figure at the right to graph each of the following vectors. *9. v + w
*10. u + v
*11. 3v
*12. 4w
*13. v - w
*14. u - v
*15. 3v + u - 2w
*16. 2u - 3v + w
w u
In Problems 17–24, use the figure at the right. Determine whether each statement given is true or false. 17. A + B = F True 19. C = D - E + F 21. E + D = G + H
18. K + G = F
20. G + H + E = D True
False
22. H - C = G - F
False
23. A + B + K + G = 0 True 25. If 7 v 7 = 4, what is 7 3v 7 ?
v
False False
B
24. A + B + C + H + G = 0 True 26. If 7 v 7 = 2, what is 7 - 4v 7 ?
12
A
8
F G
27. P = 10, 02; Q = 13, 42
v = 3i + 4j v = 2i + 4j
31. P = 1 - 2, - 12; Q = 1, - 22 33. P = 11, 02; Q = 10, 12
v = 8i - j
v = -i + j
E
28. P = 10, 02; Q = 1 - 3, - 52 30. P = 1 - 3, 22; Q = 1, 52
32. P = 1 - 1, 42; Q = 1, 22 34. P = 11, 12; Q = 12, 22
v = - 3i - 5j v = 9i + 3j v = 7i - 2j
v = i + j
In Problems 35–40, find 7 v 7 . 35. v = 3i - 4j
5
36. v = - 5i + 12j
13
38. v = - i - j
22
39. v = - 2i + 3j
213
H D
In Problems 27–34, the vector v has initial point P and terminal point Q. Write v in the form ai + bj; that is, find its position vector. 29. P = 13, 22; Q = 15, 2
C
K
37. v = i - j 40. v = i + 2j
22 2210
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
SECTION 8.4 Vectors
In Problems 41–46, find each quantity if v = 3i - 5j and w = - 2i + 3j. 41. 2v + 3w
-j
44. 7 v + w 7
42. 3v - 2w
45. 7 v 7 - 7 w 7
25
13i - 21j 234 - 213
43. 7 v - w 7
46. 7 v 7 + 7 w 7
659
289 234 + 213
In Problems 47–52, find the unit vector in the same direction as v. 47. v = 5i
48. v = - 3j
i
50. v = - 5i + 12j
-
5 12 i + j 13 13
-j 22 22 i j 2 2
51. v = i - j
4 3 i - j 5 5 225 25 i j 5 5
49. v = 3i - 4j
*53. Find a vector v whose magnitude is 4 and whose component in the i direction is twice the component in the j direction.
52. v = 2i - j
*54. Find a vector v whose magnitude is 3 and whose component in the i direction is equal to the component in the j direction. 56. If P = 1 - 3, 12 and Q = 1x, 42, > find all numbers x such that the vector represented by PQ has length 5. { - 7, 1}
55. If v = 2i - j and w = xi + 3j, find all numbers x for which 7 v + w 7 = 5. { - 2 + 221, - 2 - 221}
In Problems 57–62, write the vector v in the form ai + bj, given its magnitude 7 v 7 and the angle a it makes with the positive x-axis. *57. 7 v 7 = 5, a = 0°
*58. 7 v 7 = 8,
*60. 7 v 7 = 3, a = 240°
a = 45°
*61. 7 v 7 = 25, a = 330°
*59. 7 v 7 = 14, a = 120° *62. 7 v 7 = 15, a = 315°
In Problems 63–70, find the direction angle of v for each vector. 63. v = 3i + 3j
45°
64. v = i + 23 j
67. v = 4i - 2j
333.4°
68. v = i - 4j
65. v = - 323i + 3j 69. v = - i - 5j
150°
258.7°
66. v = - 5i - 5j
225°
70. v = - i + 3j
108.4°
Applications and Extensions 71. Computer Graphics The field of computer graphics utilizes vectors to compute translations of points. For example, if the point 1 - 3, 22 is to be translated by v = 85, 29 , then the new location will be u′ = u + v = 8 - 3, 29 + 85, 29 = 82, 49. As illustrated in the figure, the point 1 - 3, 22 is translated to 12, 42 by v. Source: Phil Dadd. Vectors and Matrices: A Primer.
www.gamedev.net/reference/articles/article1832.asp
(a) Determine the new coordinates of 13, - 12 if it is translated by v = 8 - 4, 59. 1 - 1, 42 *(b) Illustrate this translation graphically.
*73. Force Vectors A child pulls a wagon with a force of 40 pounds. The handle of the wagon makes an angle of 30° with the ground. Express the force vector F in terms of i and j. 74. Force Vectors A man pushes a wheelbarrow up an incline of 20° with a force of 100 pounds. Express the force vector F in terms of i and j. F = 1001cos 20°i + sin 20°j2 *75. Resultant Force Two forces of magnitude 40 newtons (N) BOE/BDUPOBOPCKFDUBUBOHMFTPG30° and - 45° with the positive x-axis, as shown in the figure. Find the direction and magnitude of the resultant force; that is, find F1 + F2 . y
y
F1
40 N
5
(2, 4)
30°
v
45°
(5, 2) (3, 2)
x
u' u
v
5
5
F2
60 N
x
5
72. Computer Graphics 3FGFS UP 1SPCMFN 5IF QPJOUT 1 - 3, 02, 1 - 1, - 22, 13, 12 , and 11, 32 are the vertices of a parallelogram ABCD. *(a) Find the new vertices of a parallelogram A′B′C′D′ if it is translated by v = 83, - 29. *(b) Find the new vertices of a parallelogram A′B′C′D′ if it 1 is translated by - v. 2
*76. Resultant Force Two forces of magnitude 30 newtons (N) and 70 N act on an object at angles of 45° and 120° with the positive x-axis, as shown in the figure. Find the direction and magnitude of the resultant force; that is, find F1 + F2 .
F2
70 N y
F1
30 N 120° 45° x
660
CHAPTER 8 Polar Coordinates; Vectors
77. Finding the Actual Speed and Direction of an Aircraft A Boeing 747 jumbo jet maintains a constant airspeed of 550 miles per hour (mph) headed due north. The jet stream is 100 mph in the northeasterly direction. *(a) Express the velocity va of the 747 relative to the air and the velocity vw of the jet stream in terms of i and j. *(b) Find the velocity of the 747 relative to the ground. (c) Find the actual speed and direction of the 747 relative to the ground. NQI /& 78. Finding the Actual Speed and Direction of an Aircraft An Airbus A320 jet maintains a constant airspeed of 500 mph headed due west. The jet stream is 100 mph in the southeasterly direction. *(a) Express the velocity va of the A320 relative to the air and the velocity vw of the jet stream in terms of i and j. *(b) Find the velocity of the A320 relative to the ground. (c) Find the actual speed and direction of the A320 relative to the ground. NQI 48
85. Static Equilibrium A weight of 1000 pounds is suspended from two cables as shown in the figure. What are the tensions in the two cables? 3JHIUMC -FGUMC
25°
1000 pounds
86. Static Equilibrium A weight of 800 pounds is suspended from two cables, as shown in the figure. What are the tensions in the two cables? 3JHIUMC -FGUMC
*79. Ground Speed and Direction of an Airplane An airplane has an airspeed of 500 kilometers per hour (km/h) bearing /&5IFXJOEWFMPDJUZJTLNIJOUIFEJSFDUJPO/8 Find the resultant vector representing the path of the plane relative to the ground. What is the ground speed of the plane? What is its direction? *80. Ground Speed and Direction of an Airplane An airplane IBTBOBJSTQFFEPGLNICFBSJOH4&5IFXJOEWFMPDJUZ is 40 km/h in the direction S45°E. Find the resultant vector representing the path of the plane relative to the ground. What is the ground speed of the plane? What is its direction? 81. Weight of a Boat A magnitude of 700 pounds of force is required to hold a boat and its trailer in place on a ramp whose incline is 10° to the horizontal. What is the combined weight of the boat and its trailer? ≈4031 lb 82. Weight of a Car A magnitude of 1200 pounds of force is required to prevent a car from rolling down a hill whose incline is 15° to the horizontal. What is the weight of the car? ≈43 lb *83. Correct Direction for Crossing a River A river has a constant current of 3 km/h. At what angle to a boat dock should a motorboat capable of maintaining a constant speed of 20 km/h be headed in order to reach a point directly 1 opposite the dock? If the river is kilometer wide, how long 2 will it take to cross?
40°
35°
50°
800 pounds
87. Static Equilibrium A tightrope walker located at a certain point deflects the rope as indicated in the figure. If the weight of the tightrope walker is 150 pounds, how much tension is in each part of the rope? 3JHIUMC -FGUMC
4.2°
3.7°
150 pounds
88. Static Equilibrium 3FQFBU 1SPCMFN JG UIF BOHMF PO UIF left is 3.8°, the angle on the right is 2.°, and the weight of the tightrope walker is 135 pounds. 3JHIUMC MFGUMC 89. Truck Pull At a county fair truck pull, two pickup trucks are attached to the back end of a monster truck as illustrated in the figure. One of the pickups pulls with a force of 2000 pounds, and the other pulls with a force of 3000 pounds. There is an angle of 45° between them. With how much force must the monster truck pull in order to remain unmoved?
Current
Boat Direction of boat due to current
84. Finding the Correct Compass Heading The pilot of an aircraft wishes to head directly east but is faced with a wind speed of 40 mph from the northwest. If the pilot maintains an airspeed of 250 mph, what compass heading should be maintained to head directly east? What is the actual speed of the aircraft? /& NQI
[Hint: Find the resultant force of the two trucks.]
0 lb 200 45˚ 300 0 lb
MC
SECTION 8.4 Vectors
90. Removing a Stump A farmer wishes to remove a stump from a field by pulling it out with his tractor. Having removed many stumps before, he estimates that he will OFFEUPOT QPVOET PGGPSDFUPSFNPWFUIFTUVNQ However, his tractor is only capable of pulling with a force of 7000 pounds, so he asks his neighbor to help. His neighbor’s tractor can pull with a force of 5500 pounds. They attach the two tractors to the stump with a 40° angle between the forces as shown in the figure. * B "TTVNJOHUIFGBSNFSTFTUJNBUFPGBOFFEFEUPOGPSDF is correct, will the farmer be successful in removing the stump? Explain. No *(b) Had the farmer arranged the tractors with a 25° angle between the forces, would he have been successful in removing the stump? Explain. Yes
661
94. Inclined Ramp A 2-pound weight is attached to a 3-pound weight by a rope that passes over an ideal pulley. The smaller weight hangs vertically, while the larger weight sits on a frictionless inclined ramp with angle u. The rope exerts a tension force T on both weights along the direction of the rope. Find the angle measure for u that is needed to keep UIFMBSHFSXFJHIUGSPNTMJEJOHEPXOUIFSBNQ3PVOEZPVS answer to the nearest tenth of a degree. 41.8°
T
T
3 pounds
2 pounds
u
b
l 00
55 40˚
7000 lb
91. Charting a Course A helicopter pilot needs to travel to a regional airport 25 miles away. She flies at an actual heading PG/&XJUIBOBJSTQFFEPGNQI BOEUIFSFJTBXJOE blowing directly east at 20 mph. (a) Determine the compass heading that the pilot needs to reach her destination. N 7.05°E (b) How long will it take her to reach her destination? 3PVOEUPUIFOFBSFTUNJOVUF 12 min 92. Crossing a River A captain needs to pilot a boat across a river that is 2 km wide. The current in the river is 2 km/h and the speed of the boat in still water is 10 km/h. The desired landing point on the other side is 1 km upstream. (a) Determine the direction in which the captain should aim the boat. "CPVUMFGUPGBQFSQFOEJDVMBSUPUIFTIPSF (b) How long will the trip take? 0.25 h, or 15 min 93. Static Friction A 20-pound box sits at rest on a horizontal surface, and there is friction between the box and the surface. One side of the surface is raised slowly to create a ramp. The friction force f opposes the direction of motion and is proportional to the normal force FN exerted by the surface on the box. The proportionality constant is called the coefficient of friction, m. When the angle of the ramp, u, reaches 20°, the box begins to slide. Find the value of m to two decimal places. m = 0.3
95. Inclined Ramp A box sitting on a horizontal surface is attached to a second box sitting on an inclined ramp by a rope that passes over an ideal pulley. The rope exerts a tension force T on both weights along the direction of the rope, and the coefficient of friction between the surface BOE CPYFT JT 4FF 1SPCMFNT BOE *G UIF CPY PO the right weighs 100 pounds and the angle of the ramp is 35°, how much must the box on the left weigh for the TZTUFNUPCFJOTUBUJDFRVJMJCSJVN 3PVOE ZPVSBOTXFSUP two decimal places. MC FN 1 F1
T
T F2
W1
96. Muscle Force Two muscles exert force on a bone at the same point. The first muscle exerts a force of 800 N at a 10° angle with the bone. The second muscle exerts a force of 710 N at a 35° angle with the bone. What are the direction and magnitude of the resulting force on the bone? 1474.3 N at an angle of about 21.7° *97. Static Equilibrium Show on the following graph the force needed for the object at P to be in static equilibrium.
F2 P F3 F4
f
20 pounds
u
35°
W2
F1
FN
FN 2
662
CHAPTER 8 Polar Coordinates; Vectors
Explaining Concepts: Discussion and Writing 98. Explain in your own words what a vector is. Give an example of a vector.
100. Explain the difference between an algebraic vector and a position vector.
99. Write a brief paragraph comparing the algebra of complex numbers and the algebra of vectors.
Retain Your Knowledge Problems 101–104 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 3 101. Solve: 2 x - 2 = 3.
5296
102. Factor - 3x + 12x + 3x completely. 3
2
- 3x(x + 2)(x - )
*103. Find the amplitude, period, and phase shift of y = 1 104. Find the exact value of tan c cos - 1 a b d . 2
23
3 cos (x + 3p). Graph the function, showing at least two periods. 2 p p 3 Amplitude = ; Period = ; Phase shift = 2 3 2
8.5 The Dot Product PREPARING FOR THIS SECTION Before getting started, review the following: r -BXPG$PTJOFT 4FDUJPO Q
Now Work the ‘Are You Prepared?’ problem on page 668. OBJECTIVES 1 2 3 4 5 6
Find the Dot Product of Two Vectors (p. 662) Find the Angle between Two Vectors (p. 663) Determine Whether Two Vectors Are Parallel (p. 664) Determine Whether Two Vectors Are Orthogonal (p. 664) Decompose a Vector into Two Orthogonal Vectors (p. 665) Compute Work (p. 667)
1 Find the Dot Product of Two Vectors The definition for a product of two vectors is somewhat unexpected. However, such a product has meaning in many geometric and physical applications.
DEFINITION
If v = a1 i + b1 j and w = a2 i + b2 j are two vectors, the dot product v # w is defined as v # w = a1 a2 + b1 b2
EX A MPL E 1
Finding Dot Products If v = 2i - 3j and w = 5i + 3j, find: (a) v # w
Solution
(1)
(b) w # v
(c) v # v
(a) v # w = 2152 + 1 - 323 = 1 (c) v # v = 2122 + 1 - 32 1 - 32 = 13 (e) 7 v 7 = 222 + 1 - 32 2 = 213
(d) w # w
(e) 7 v 7
(f) 7 w 7
(b) w # v = 5122 + 31 - 32 = 1 (d) w # w = 5152 + 3132 = 34 (f) 7 w 7 = 252 + 32 = 234
r
663
SECTION 8.5 The Dot Product
COMMENT A scalar multiple av is a vector. A dot product u # v is a scalar (real number). ■
THEOREM
Since the dot product v # w of two vectors v and w is a real number (scalar), it is sometimes referred to as the scalar product. The results obtained in Example 1 suggest some general properties.
Properties of the Dot Product If u, v, and w are vectors, then Commutative Property u#v = v#u
(2)
u # 1v + w2 = u # v + u # w
(3)
v # v = 7v72
(4)
Distributive Property
0#v = 0
(5)
Proof We prove properties (2) and (4) here and leave properties (3) and (5) as FYFSDJTFT TFF1SPCMFNTBOE To prove property (2), let u = a1 i + b1 j and v = a2 i + b2 j. Then u # v = a1 a2 + b1 b2 = a2 a1 + b2 b1 = v # u To prove property (4), let v = ai + bj. Then
v # v = a 2 + b2 = 7 v 7 2
■
2 Find the Angle between Two Vectors
Figure 64 uv
u A
v
One use of the dot product is to calculate the angle between two vectors. Let u and v be two vectors with the same initial point A. Then the vectors u, v, and u - v form a triangle. The angle u at vertex A of the triangle is the angle between the vectors u and v4FF'JHVSF8FXJTIUPGJOEBGPSNVMBGPSDBMDVMBUJOH the angle u. The sides of the triangle have lengths 7 v 7 , 7 u 7 , and 7 u - v 7 , and u is the included angle between the sides of length 7 v 7 and 7 u 7 . The Law of Cosines (Section 7.3) can be used to find the cosine of the included angle.
7 u - v 7 2 = 7 u 7 2 + 7 v 7 2 - 2 7 u 7 7 v 7 cos u
Now use property (4) to rewrite this equation in terms of dot products. 1u - v2 # 1u - v2 = u # u + v # v - 2 7 u 7 7 v 7 cos u
(6)
5IFOBQQMZUIFEJTUSJCVUJWFQSPQFSUZ UXJDFPOUIFMFGUTJEFPG UPPCUBJO 1u - v2 # 1u - v2 = u # 1u - v2 - v # 1u - v2
= u#u - u#v - v#u + v#v = u#u + v#v - 2u#v
c
Property (2)
$PNCJOJOHFRVBUJPOT BOE HJWFT u # u + v # v - 2 7 u 7 7 v 7 cos u = u # u + v # v - 2 u # v
7 u 7 7 v 7 cos u = u # v cos u =
u#v 7u7 7v7
(7)
664
CHAPTER 8 Polar Coordinates; Vectors
THEOREM
Angle between Vectors If u and v are two nonzero vectors, the angle u, 0 … u … p, between u and v is determined by the formula cos u =
EX A MPL E 2
u#v 7u7 7v7
(8)
Finding the Angle between Two Vectors Find the angle u between u = 4i - 3j and v = 2i + 5j.
Solution
Find u # v, 7 u 7 , and 7 v 7 .
Figure 65
u # v = 4122 + 1 - 32 152 = - 7
7 u 7 = 242 + 1 - 32 2 = 5
y
7 v 7 = 222 + 52 = 229
v 2i 5j
By formula (8), if u is the angle between u and v, then
u 4i 3j
x
cos u =
u#v -7 = ≈ - 0.2 7u7 7v7 5229
Thus, u ≈ cos-1 1 - 0.22 ≈ 105°.4FF'JHVSF
Now Work
PROBLEMS
7(a)
AND
r
(b)
3 Determine Whether Two Vectors Are Parallel Two vectors v and w are said to be parallel if there is a nonzero scalar a so that v = aw. In this case, the angle u between v and w is 0 or p.
EX A MPL E 3
Determining Whether Vectors Are Parallel The vectors v = 3i - j and w = i - 2j are parallel, since v = since cos u =
1 w. Furthermore, 2
v#w 18 + 2 20 = = = 1 7v7 7w7 210 240 2400
the angle u between v and w is 0.
r
4 Determine Whether Two Vectors Are Orthogonal Figure 66 v is orthogonal to w.
w
v
p If the angle u between two nonzero vectors v and w is , the vectors v and w are 2 * called orthogonal. 4FF'JHVSF p Since cos = 0, it follows from formula (8) that if v and w are orthogonal, then 2 v # w = 0. On the other hand, if v # w = 0, then v = 0 or w = 0 or cos u = 0. If cos u = 0, p then u = , and v and w are orthogonal. If v or w is the zero vector, then, since the 2 zero vector has no specific direction, we adopt the convention that the zero vector is orthogonal to every vector. *
Orthogonal, perpendicular, and normal are all terms that mean “meet at a right angle.” It is customary to refer to two vectors as being orthogonal, to two lines as being perpendicular, and to a line and a plane or a vector and a plane as being normal.
SECTION 8.5 The Dot Product
THEOREM
665
Two vectors v and w are orthogonal if and only if v#w = 0
EXAM PL E 4 Figure 67
Determining Whether Two Vectors Are Orthogonal The vectors
y
v = 2i - j and w = 3i + j are orthogonal, since
w 3i 6j
v#w = - = 0
x
r
4FF'JHVSF
v 2i j
Now Work
PROBLEM
7(c)
5 Decompose a Vector into Two Orthogonal Vectors Figure 68 F1
F
F2
Figure 69 v2
v
P
v1
w (a)
v
v2
v1 P
w (b)
In the last section, we discussed how to add two vectors to find the resultant vector. Now, we discuss the reverse problem, decomposing a vector into the sum of two components. In many physical applications, it is necessary to find “how much” of a vector JT BQQMJFE JO B HJWFO EJSFDUJPO -PPL BU 'JHVSF 5IF GPSDF F due to gravity is pulling straight down (toward the center of Earth) on the block. To study the effect of gravity on the block, it is necessary to determine how much of F is actually pushing the block down the incline 1F1 2 and how much is pressing the block against the incline 1F2 2, at a right angle to the incline. Knowing the decomposition of F often enables us to determine when friction (the force holding the block in place on the incline) is overcome and the block slides down the incline. Suppose that v and w are two nonzero vectors with the same initial point P. We seek to decompose v into two vectors: v1 , which is parallel to w, and v2 , which is orthogonal to w 4FF 'JHVSF B BOE C 5IF WFDUPS v1 is called the vector projection of v onto w. The vector v1 is obtained as follows: From the terminal point of v, drop a perpendicular to the line containing w. The vector v1 is the vector from P to the foot of this perpendicular. The vector v2 is given by v2 = v - v1 . Note that v = v1 + v2 , the vector v1 is parallel to w, and the vector v2 is orthogonal to w. This is the decomposition of v that was sought. Now we seek a formula for v1 that is based on a knowledge of the vectors v and w. Since v = v1 + v2 , we have v # w = 1v1 + v2 2 # w = v1 # w + v2 # w
(9)
Since v2 is orthogonal to w, we have v2 # w = 0. Since v1 is parallel to w, we have v1 = aw for some scalar a. Equation (9) can be written as v # w = aw # w = a 7 w 7 2 v1 = aw ; v2 # w = 0 a =
v#w 7w72
Then v1 = aw =
v#w w 7w72
666
CHAPTER 8 Polar Coordinates; Vectors
THEOREM
If v and w are two nonzero vectors, the vector projection of v onto w is v1 =
v#w w 7w72
(10)
The decomposition of v into v1 and v2 , where v1 is parallel to w and v2 is orthogonal to w, is v#w w 7w72
v1 =
EX A MPL E 5
v2 = v - v1
(11)
Decomposing a Vector into Two Orthogonal Vectors Find the vector projection of v = i + 3j onto w = i + j. Decompose v into two vectors, v1 and v2 , where v1 is parallel to w and v2 is orthogonal to w.
Solution
Use formulas (10) and (11).
Figure 70
v1 =
y v i 3j
v#w 1 + 3 w = w = 2w = 21i + j2 2 7w7 1 22 2 2
v2 = v - v1 = 1i + 3j2 - 21i + j2 = - i + j
v2 i j
r
See Figure 70.
v1 2(i j) wij x
EX A MPL E 6
Now Work
PROBLEM
19
Finding the Force Required to Hold a Wagon on a Hill A wagon with two small children as occupants that weighs 100 pounds is on a hill with a grade of 20°. What is the magnitude of the force that is required to keep the wagon from rolling down the hill?
Solution Figure 71
See Figure 71. We wish to find the magnitude of the force v that is acting to cause the wagon to roll down the hill. A force with the same magnitude in the opposite direction of v will keep the wagon from rolling down the hill. The force of gravity is orthogonal to the level ground, so the force of the wagon due to gravity can be represented by the vector Fg = - 100j
v
w 20°
Fg
Determine the vector projection of Fg onto w, which is the force parallel to the hill. The vector w is given by w = cos 20°i + sin 20°j The vector projection of Fg onto w is v = =
Fg # w
7w72
w - 100(sin 20°)
1 2cos2 20°
+ sin2 20° 2 2
(cos 20°i + sin 20°j)
≈ - 34.2(cos 20°i + sin 20°j) The magnitude of v is 34.2 pounds, so the magnitude of the force required to keep the wagon from rolling down the hill is 34.2 pounds.
r
SECTION 8.5 The Dot Product
Figure 72
667
6 Compute Work A
F θ
B
A
In elementary physics, the work W done by a constant force F in moving an object from a point A to a point B is defined as
>
W = 1magnitude of force2 1distance2 = 7 F 7 7 AB 7
Work is commonly measured in foot-pounds or in newton-meters (joules). In this definition, it is assumed that the force F is applied along the line of motion. If the constant force F is not along the line of motion, but instead is at an angle u to the direction of the motion, as illustrated in Figure 72, then the work W done by F in moving an object from A to B is defined as W = F # AB
>
(12)
This definition is compatible with the force-times-distance definition, since
>
W = 1amount of force in the direction of AB 2 1distance2
>
= 7 projection of F on AB 7 7 AB 7 =
F # AB
7 AB
>
>
72
> > > 7 AB 7 7 AB 7 = F # AB
c Use formula 1102.
EXAM PL E 7
Computing Work Figure 73(a) shows a girl pulling a wagon with a force of 50 pounds. How much work is done in moving the wagon 100 feet if the handle makes an angle of 30° with the ground?
Figure 73
y
F 50(sin 30°)j
F
50 30° 30° (0, 0)
(100, 0) x
(b)
(a)
Solution
50(cos 30°)i
Position the vectors in a coordinate system in such a way that the wagon is moved from> 10, 02 to 1100, 02. The motion is from A = 10, 02 to B = 1100, 02, so AB = 100i. The force vector F, as shown in Figure 73(b), is F = 501cos 30°i + sin 30°j2 = 50a
23 1 i + jb = 25 1 23i + j 2 2 2
By formula (12), the work done is
W = F # AB = 25 1 23i + j 2 # 100i = 250023 foot@pounds
>
Now Work
PROBLEM
r
25
Historical Feature
W
e stated in an earlier Historical Feature that complex numbers were used as vectors in the plane before the general notion of a vector was clarified. Suppose that we make the correspondence Vector 4 Complex number ai + bj 4 a + bi ci + dj 4 c + di
Show that (ai + bj) # (ci + dj) = real part [(a + bi)(c + di)] This is how the dot product was found originally. The imaginary part is also interesting. It is called a determinant (see Section 10.3) and represents the area of the parallelogram whose edges are the vectors. This is close to some of Hermann Grassmann’s ideas and is also connected with the scalar triple product of three-dimensional vectors.*
668
CHAPTER 8 Polar Coordinates; Vectors
8.5 Assess Your Understanding ‘Are You Prepared?’ The answer is given at the end of these exercises. If you get the wrong answer, read the page listed in red. 1. In a triangle with sides a, b, c and angles A, B, C, the Law of Cosines states that ________. (p. 587) c 2 = a2 + b2 - 2ab cos C
Concepts and Vocabulary 2. If v = a1i + b1j and w = a2i + b2j are two vectors, then the dot product is defined as v # w = a1a2 + b1b2. 3. If v # w = 0, then the two vectors v and w are orthogonal .
5. True or False Given two nonzero vectors v and w, it is always possible to decompose v into two vectors, one parallel to w and the other perpendicular to w. True
4. If v = 3w, then the two vectors v and w are parallel .
6. True or False Work is a physical example of a vector. False
Skill Building In Problems 7–16, (a) find the dot product v # w; (b) find the angle between v and w; (c) state whether the vectors are parallel, orthogonal, or neither. *7. v = i - j, w = i + j
*8. v = i + j, w = - i + j
*9. v = 2i + j, w = i - 2j
*10. v = 2i + 2j, w = i + 2j
*11. v = 23i - j, w = i + j
*12. v = i + 23j, w = i - j
*13. v = 3i + 4j, w = - i - 8j
*14. v = 3i - 4j, w = 9i - 12j
*15. v = 4i, w = j
*16. v = i, w = - 3j 17. Find a so that the vectors v = i - aj 18. Find b so that the vectors v = i + j and w = 2i + 3j are orthogonal. 2 and w = i + bj are orthogonal. - 1 3 In Problems 19–24, decompose v into two vectors v1 and v2 , where v1 is parallel to w and v2 is orthogonal to w. *19. v = 2i - 3j, w = i - j
*20. v = - 3i + 2j, w = 2i + j
*21. v = i - j, w = - i - 2j
*22. v = 2i - j, w = i - 2j
*23. v = 3i + j, w = - 2i - j
*24. v = i - 3j, w = 4i - j
Applications and Extensions 25. Computing Work Find the work done by a force of 3 pounds acting in the direction 0° to the horizontal in moving an PCKFDUGFFUGSPN 10, 02 to 1, 02. 9 ft@lb 26. Computing Work A wagon is pulled horizontally by exerting a force of 20 pounds on the handle at an angle of 30° with the horizontal. How much work is done in moving the wagon 100 feet? 100023 ft@lb ≈ 1732 ft@lb 27. Solar Energy The amount of energy collected by a solar panel depends on the intensity of the sun’s rays and the area of the panel. Let the vector I represent the intensity, in watts per square centimeter, having the direction of the sun’s rays. Let the vector A represent the area, in square centimeters, whose direction is the orientation of a solar panel. See the figure. The total number of watts collected by the panel is given by W = 0 I # A 0 .
*(c) If the solar panel is to collect the maximum number of watts, what must be true about I and A? 28. Rainfall Measurement Let the vector R represent the amount of rainfall, in inches, whose direction is the inclination of the rain to a rain gauge. Let the vector A represent the area, in square inches, whose direction is the orientation of the opening of the rain gauge. See the figure. The volume of rain collected in the gauge, in cubic inches, is given by V = 0 R # A 0 , even when the rain falls in a slanted direction or the gauge is not perfectly vertical. R A 9 8 7 6 5 4 3 2 1
I A
Suppose that I = 8 - 0.02, - 0.019 and A = 8300, 4009 . *(a) Find 7 I 7 and 7 A 7 and interpret the meaning of each. *(b) Compute W and interpret its meaning.
Suppose that R = 80.75, - 1.759 and A = 80.3, 19. *(a) Find 7 R 7 and 7 A 7 and interpret the meaning of each. *(b) Compute V and interpret its meaning. *(c) If the gauge is to collect the maximum volume of rain, what must be true about R and A?
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
SECTION 8.5 The Dot Product
*29. Braking Load A Toyota Sienna with a gross weight of 5300 pounds is parked on a street with an 8° grade. See the figure. Find the magnitude of the force required to keep the Sienna from rolling down the hill. What is the magnitude of the force perpendicular to the hill? MCT MCT
669
34. Given vectors u = xi + 2j and v = 7i − 3j, find x so that the angle between the vectors is 75°. State the answer correct to three decimal places. 1.574 35. Find the acute angle that a constant unit force vector makes with the positive x-axis if the work done by the force in moving a particle from 10, 02 to 14, 02 equals 2. *36. Prove the distributive property:
u # 1v + w2 = u # v + u # w
Weight 5300 pounds
30. Braking Load A Chevrolet Silverado with a gross weight of 4500 pounds is parked on a street with a 10° grade. Find the magnitude of the force required to keep the Silverado from rolling down the hill. What is the magnitude of the force perpendicular to the hill? MCT MCT 31. Ramp Angle Billy and Timmy are using a ramp to load furniture into a truck. While rolling a 250-pound piano up the ramp, they discover that the truck is too full of other furniture for the piano to fit. Timmy holds the piano in place on the ramp while Billy repositions other items to make room for it in the truck. If the angle of inclination of the ramp is 20°, how many pounds of force must Timmy exert to hold the piano in position? 85.5 lb
*37. Prove property (5), 0 # v = 0.
*38. If v is a unit vector and the angle between v and i is a, show that v = cos ai + sin aj. *39. Suppose that v and w are unit vectors. If the angle between v and i is a and that between w and i is b, use the idea of the dot product v # w to prove that cos 1a - b2 = cos a cos b + sin a sin b
*40. Show that the projection of v onto i is 1v # i2i. Then show that we can always write a vector v as v = 1v # i2i + 1v # j2j
*41. (a) If u and v have the same magnitude, show that u + v and u - v are orthogonal. (b) Use this to prove that an angle inscribed in a semicircle is a right angle (see the figure).
u v
20° 250 lb
32. Incline Angle A bulldozer exerts 1000 pounds of force to prevent a 5000-pound boulder from rolling down a hill. Determine the angle of inclination of the hill. 11.5° 33. Given vectors u = i + 5j and v = 4i + yj, find y so that the angle between the vectors is 114°. State the answer correct to three decimal places.+ −2.833
v
*42. Let v and w denote two nonzero vectors. Show that the v#w vector v - aw is orthogonal to w if a = . 7w72 *43. Let v and w denote two nonzero vectors. Show that the vectors 7 w 7 v + 7 v 7 w and 7 w 7 v - 7 v 7 w are orthogonal. *44. In the definition of work given in this > section, what is the work done if F is orthogonal to AB ? *45. Prove the polarization identity,
7 u + v 7 2 - 7 u - v 7 2 = 41u # v2
Discussion and Writing 46. Create an application (different from any found in the text) that requires a dot product.
Retain Your Knowledge Problems 47–50 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 47. Find the average rate of change of f(x) = x3 - 5x2 + 27 from −3 to 2. 12 49. (1 - sin 2u)(1 + tan 2u) = ( cos 2u)( sec 2u) 1 p 9 = cos 2u # 48. Find the exact value of 5 cos 0° + 2 tan . Do not use a calculator. cos 2u 4 2 = 1 49. Establish the identity: (1 - sin2 u)(1 + tan2 u) = 1 50. Volume of a Box An open-top box is made from a sheet of metal by cutting squares from each corner and folding up the sides. The sheet has a length of 19 inches and a width of 13 inches. If x is the length of one side of the square to be cut out, write a function, V(x), for the volume of the box in terms of x. V(x) = x(19 - 2x)(13 - 2x) or V(x) = 4x3 - 4x2 + 247x
‘Are You Prepared?’ Answer 1. c 2 = a2 + b2 - 2ab cos C +
Courtesy of the Joliet Junior College Mathematics Department
670
CHAPTER 8 Polar Coordinates; Vectors
8.6 Vectors in Space PREPARING FOR THIS SECTION Before getting started, review the following: r %JTUBODF'PSNVMB 'PVOEBUJPOT 4FDUJPO Q
Now Work the ‘Are You Prepared?’ problem on page 677.
OBJECTIVES 1 2 3 4 5 6
Figure 74
Rectangular Coordinates in Space z 4 2 2
2 2 4 x
Find the Distance between Two Points in Space (p. 671) Find Position Vectors in Space (p. 671) Perform Operations on Vectors (p. 672) Find the Dot Product (p. 673) Find the Angle between Two Vectors (p. 674) Find the Direction Angles of a Vector (p. 675)
O 2
2
4
y
In the plane, each point is associated with an ordered pair of real numbers. In space, each point is associated with an ordered triple of real numbers. Through a fixed point, called the origin O, draw three mutually perpendicular lines: the x-axis, the y-axis, and the z-axis. On each of these axes, select an appropriate scale and the positive direction. See Figure 74. The direction chosen for the positive z-axis in Figure 74 makes the system right-handed. This conforms to the right-hand rule, which states that if the index finger of the right hand points in the direction of the positive x-axis and the middle finger points in the direction of the positive y-axis, then the thumb will point in the direction of the positive z-axis. See Figure 75. Associate with each point P an ordered triple 1x, y, z2 of real numbers, the coordinates of P. For example, the point 12, 3, 42 is located by starting at the origin and moving 2 units along the positive x-axis, 3 units in the direction of the positive y-axis, and 4 units in the direction of the positive zBYJT4FF'JHVSF Figure 75
Figure 76 z 8
z
6 4
O
2
y
(0, 0, 4)
(2, 3, 4)
x
(0, 3, 0) 2 4
(2, 0, 0) 2
4
y
(2, 3, 0)
x
'JHVSF BMTP TIPXT UIF MPDBUJPO PG UIF QPJOUT 12, 0, 02, 10, 3, 02, 10, 0, 42, and 12, 3, 02 . Points of the form 1x, 0, 02 lie on the x-axis, and points of the forms 10, y, 02 and 10, 0, z2 lie on the y-axis and z-axis, respectively. Points of the form 1x, y, 02 lie in a plane called the xy-plane. Its equation is z = 0. Similarly, points of the form 1x, 0, z2 lie in the xz-plane (equation y = 0), and points of the form 10, y, z2 lie in the yz-plane (equation x = 0). See Figure 77(a). By extension of these ideas, all points obeying the equation z = 3 will lie in a plane parallel to and 3 units above the xy-plane. The equation y = 4 represents a plane parallel to the xz-plane and 4 units to the right of the plane y = 0. See Figure 77(b).
SECTION 8.6 Vectors in Space
Figure 77
671
z z
Plane z 3
3 y0 xz-plane
Plane y 4
x0 yz-plane z0 xy-plane
4
y
y
x
x (a)
Now Work
(b) PROBLEM
9
1 Find the Distance between Two Points in Space The formula for the distance between two points in space is an extension of the Distance Formula for points in the plane given in Foundations, Section 1.
THEOREM
Distance Formula in Space If P1 = 1x1 , y1 , z1 2 and P2 = 1x2 , y2 , z2 2 are two points in space, the distance d from P1 to P2 is d = 2 1x2 - x1 2 2 + 1y2 - y1 2 2 + 1z2 - z1 2 2
(1)
The proof, which we omit, utilizes a double application of the Pythagorean Theorem.
EXAM PL E 1
Solution
Using the Distance Formula
Find the distance from P1 = 1 - 1, 3, 22 to P2 = 14, - 2, 52. d = 2 3 4 - 1 - 12 4 2 + 3 - 2 - 34 2 + 3 5 - 24 2 = 225 + 25 + 9 = 259
Now Work
15
2 Find Position Vectors in Space
Figure 78
To represent vectors in space, we introduce the unit vectors i, j, and k whose directions are along the positive x-axis, the positive y-axis, and the positive z-axis, respectively. If v is a vector with initial point at the origin O and terminal point at P = 1a, b, c2, then we can represent v in terms of the vectors i, j, and k as
z
P (a, b, c) k O i
PROBLEM
r
j
v = ai + bj + ck
v ai bj ck y
x
THEOREM
See Figure 78. The scalars a, b, and c are called the components of the vector v = ai + bj + ck, with a being the component in the direction i, b the component in the direction j, and c the component in the direction k. A vector whose initial point is at the origin is called a position vector. The next result states that any vector whose initial point is not at the origin is equal to a unique position vector. Suppose that v is a vector with initial point P1 = 1x1 , y1 , z1 2, not necessarily > the origin, and terminal point P2 = 1x2 , y2 , z2 2. If v = P1P2 , then v is equal to the position vector v = 1x2 - x1 2i + 1y2 - y1 2j + 1z2 - z1 2k
(2)
672
CHAPTER 8 Polar Coordinates; Vectors
Figure 79 illustrates this result. Figure 79
z P1 (x1, y1, z1)
O
P2 (x2, y2, z2)
v P1P2 (x2 x1)i (y2 y1)j (z2 z1)k y
x
EX A MPL E 2
Solution
Finding a Position Vector
>
Find the position vector of the vector v = P1 P2 if P1 = 1 - 1, 2, 32 and P2 = 14, , 22. By equation (2), the position vector equal to v is v = 3 4 - 1 - 12 4 i + 1 - 22j + 12 - 32k = 5i + 4j - k
Now Work
PROBLEM
r
29
3 Perform Operations on Vectors Equality, addition, subtraction, scalar product, and magnitude can be defined in terms of the components of a vector.
DEFINITION
Let v = a1 i + b1 j + c1 k and w = a2 i + b2 j + c2 k be two vectors, and let a be a scalar. Then v = w if and only iff
a1 = a2 , b1 = b2 , and c1 = c2
v + w = 1a1 + a2 2i + 1b1 + b2 2j + 1c1 + c2 2k
v - w = 1a1 - a2 2i + 1b1 - b2 2j + 1c1 - c2 2k av = 1aa1 2i + 1ab1 2j + 1ac1 2k
‘v‘ = 2 2a21 + b21 + c 21
These definitions are compatible with the geometric definitions given in Section 8.4 for vectors in a plane.
EX A MPL E 3
Adding and Subtracting Vectors If v = 2i + 3j - 2k and w = 3i - 4j + 5k, find: (a) v + w
Solution
(b) v - w
(a) v + w = 12i + 3j - 2k2 + 13i - 4j + 5k2
= 12 + 32i + 13 - 42j + 1 - 2 + 52k = 5i - j + 3k
(b) v - w = 12i + 3j - 2k2 - 13i - 4j + 5k2
= 12 - 32i + 3 3 - 1 - 42 4 j + 3 - 2 - 54 k = - i + 7j - 7k
r
SECTION 8.6 Vectors in Space
EXAM PL E 4
673
Finding Scalar Products and Magnitudes If v = 2i + 3j - 2k and w = 3i - 4j + 5k, find: (b) 2v - 3w
(a) 3v
Solution
(c) ‘v‘
(a) 3v = 312i + 3j - 2k2 = i + 9j - k (b) 2v - 3w = 212i + 3j - 2k2 - 313i - 4j + 5k2 = 4i + j - 4k - 9i + 12j - 15k = - 5i + 18j - 19k
(c) ‘v‘ = ‘2i + 3j - 2k ‘ = 222 + 32 + 1 - 22 2 = 217
Now Work
PROBLEMS
33
AND
r
39
3FDBMMUIBUBVOJUWFDUPSu is one for which ‘u ‘ = 1. In many applications, it is useful to be able to find a unit vector u that has the same direction as a given vector v.
THEOREM
Unit Vector in the Direction of v For any nonzero vector v, the vector u =
v ‘v‘
is a unit vector that has the same direction as v.
As a consequence of this theorem, if u is a unit vector in the same direction as a vector v, then v may be expressed as v = ‘v‘u
EXAM PL E 5
Finding a Unit Vector Find the unit vector in the same direction as v = 2i - 3j - k.
Solution
Find ‘v‘ first. ‘v‘ = ‘2i - 3j - k ‘ = 24 + 9 + 3 = 249 = 7 Now multiply v by the scalar u =
Now Work
1 1 = . The result is the unit vector 7 ‘v‘
2i - 3j - k v 2 3 = i - j - k = 7 7 7 7 ‘v‘
PROBLEM
r
47
4 Find the Dot Product The definition of dot product in space is an extension of the definition given for vectors in a plane.
DEFINITION
If v = a1 i + b1 j + c1 k and w = a2 i + b2 j + c2 k are two vectors, the dot product v ~ w is defined as v # w = a1 a2 + b1 b2 + c1 c2
(3)
674
CHAPTER 8 Polar Coordinates; Vectors
EX A MPL E 6
Finding Dot Products If v = 2i - 3j + k and w = 5i + 3j - k, find: (a) v ~ w (d) w ~ w
Solution
(a) (b) (c) (d)
(b) w ~ v (e) ‘v‘
v ~ w = 2152 + 1 - 323 + 1 - 12 = - 5 w ~ v = 5122 + 31 - 32 + 1 - 12 12 = - 5 v ~ v = 2122 + 1 - 32 1 - 32 + 12 = 49 w ~ w = 5152 + 3132 + 1 - 12 1 - 12 = 35
(c) v ~ v (f) ‘ w ‘
(e) ‘v‘ = 222 + 1 - 32 2 + 2 = 249 = 7
(f) ‘w ‘ = 252 + 32 + 1 - 12 2 = 235
The dot product in space has the same properties as the dot product in a plane.
THEOREM
r
Properties of the Dot Product If u, v, and w are vectors, then
Commutative Property u~v = v~u
Distributive Property u ~ 1v + w2 = u ~ v + u ~ w v ~ v = ‘v‘ 2 0~v = 0
5 Find the Angle between Two Vectors The angle u between two vectors in space follows the same formula as for two vectors in a plane.
THEOREM
Angle between Vectors If u and v are two nonzero vectors, the angle u, 0 … u … p, between u and v is determined by the formula cos u =
EX A MPL E 7
u~v ‘u ‘ ‘v‘
Finding the Angle between Two Vectors Find the angle u between u = 2i - 3j + k and v = 2i + 5j - k.
Solution
Compute the quantities u ~ v, ‘u ‘, and ‘v‘. u ~ v = 2122 + 1 - 32 152 + 1 - 12 = - 17 ‘u ‘ = 222 + 1 - 32 2 + 2 = 249 = 7 ‘v‘ = 222 + 52 + 1 - 12 2 = 230
(4)
SECTION 8.6 Vectors in Space
675
By formula (4), if u is the angle between u and v, then cos u =
u~v - 17 ≈ - 0.443 = ‘u ‘ ‘v‘ 7230
Thus, u ≈ cos -1 1 - 0.4432 ≈ 11.3°.
Now Work
PROBLEM
r
51
6 Find the Direction Angles of a Vector A nonzero vector v in space can be described by specifying its magnitude and its three direction angles a, b, and g. These direction angles are defined as a = the angle between v and i, the positive x@axis, 0 … a … p b = the angle between v and j, the positive y@axis, 0 … b … p g = the angle between v and k, the positive z@axis, 0 … g … p See Figure 80. Figure 80
z C 5 (0, 0, c)
␥ P 5 (a, b, c)
v  ␣
B 5 (0, b, 0)
A 5 (a, 0, 0)
y
x 0 … ␣ … , 0 …  … , 0 … ␥ …
Our first goal is to find expressions for a, b, and g in terms of the components of a vector. Let v = ai + bj + ck denote a nonzero vector. The angle a between v and i, the positive x-axis, obeys cos a =
v~i a = ‘v‘ ‘i‘ ‘v‘
Similarly, cos b =
b ‘v‘
and cos g =
c ‘v‘
Since ‘ v‘ = 2a2 + b2 + c 2 , the following result is obtained.
THEOREM
Direction Angles If v = ai + bj b + ck is a nonzero vector in space, the direction angles a, b, and g obey cos a =
a 2a + b + c 2 c 2
2
2
=
a ‘v‘
c cos g = = ‘v‘ 2a2 + b2 + c 2 2
cos b =
b 2a + b + c 2 2
2
2
=
b ‘v‘ (5)
676
CHAPTER 8 Polar Coordinates; Vectors
The numbers cos a, cos b, and cos g are called the direction cosines of the vector v.
EX A MPL E 8
Finding the Direction Angles of a Vector Find the direction angles of v = - 3i + 2j - k.
Solution
‘v‘ = 2 1 - 32 2 + 22 + 1 - 2 2 = 249 = 7 Using the formulas in equation (5), we have -3 7 a ≈ 115.4°
2 7 b ≈ 73.4°
cos a =
THEOREM
cos b =
- 7 g ≈ 149.0°
cos g =
r
Property of the Direction Cosines If a, b, and g are the direction angles of a nonzero vector v in space, then cos2 a + cos2 b + cos2 g = 1
(6)
The proof is a direct consequence of the equations in (5). #BTFE PO FRVBUJPO
XIFO UXP EJSFDUJPO DPTJOFT BSF LOPXO UIF UIJSE JT determined up to its sign. Knowing two direction cosines is not sufficient to uniquely determine the direction of a vector in space.
EX A MPL E 9
Finding a Direction Angle of a Vector
p p with the positive x-axis, an angle of b = 3 3 with the positive y-axis, and an acute angle g with the positive z-axis. Find g. The vector v makes an angle of a =
Solution
#ZFRVBUJPO
XFIBWF p p cos2 a b + cos2 a b + cos2 g = 1 3 3
0 6 g 6
p 2
1 2 1 2 a b + a b + cos2 g = 1 2 2 cos2 g = cos g =
22 2 g =
Since g must be acute, g =
1 2
or cos g = p 4
or g =
22 2
3p 4
p . 4
r
The direction cosines of a vector give information about only the direction of the vector; they provide no information about its magnitude. For example, any p vector that is parallel to the xy-plane and makes an angle of radian with the 4 positive x-axis and y-axis has direction cosines cos a =
22 2
cos b =
22 2
cos g = 0
However, if the direction angles and the magnitude of a vector are known, the vector is uniquely determined.
SECTION 8.6 Vectors in Space
677
Writing a Vector in Terms of Its Magnitude and Direction Cosines
EX AM PL E 10
Show that any nonzero vector v in space can be written in terms of its magnitude and direction cosines as v = ‘v‘ 3 1cos a2i + 1cos b2j + 1cos g2k 4
(7)
Let v = ai + bj + ck. From the equations in (5), note that
Solution
a = ‘v‘ cos a
b = ‘v‘ cos b
c = ‘v‘ cos g
Substituting gives
v = ai + bj + ck = ‘v‘ 1cos a2i + ‘v‘ 1cos b2j + ‘v‘ 1cos g2k = ‘v‘ 3 1cos a2i + 1cos b2j + 1cos g2k 4
Now Work
PROBLEM
59
r
Example 10 shows that the direction cosines of a vector v are also the components of the unit vector in the direction of v.
8.6 Assess Your Understanding ‘Are You Prepared?’ The answer is given at the end of these exercises. If you get the wrong answer, read the page listed in red. 1. The distance d from P1 = 1x1 , y1 2 to P2 = 1x2 , y2 2 is d = _______. (p. 35)
21x2 - x1 2 2 + 1y2 - y1 2 2
Concepts and Vocabulary 2. In space, points of the form 1x, y, 02 lie in a plane called the xy@plane.
5. True or False In space, the dot product of two vectors is a positive number. False
3. If v = ai + bj + ck is a vector in space, the scalars a, b, c are called the components of v.
6. True or False A vector in space may be described by specifying its magnitude and its direction angles. True
4. The squares of the direction cosines of a vector in space add up to 1 .
Skill Building In Problems 7–14, describe the set of points 1x, y, z2 defined by the equation(s). *7. y = 0
*8. x = 0
*11. x = - 4
*12. z = - 3
*9. z = 2 *13. x = 1 and y = 2
In Problems 15–20, find the distance from P1 to P2 . 15. P1 = 10, 0, 02 and P2 = 14, 1, 22
221
17. P1 = 1 - 1, 2, - 32 and P2 = 10, - 2, 12
19. P1 = 14, - 2, - 22 and P2 = 13, 2, 12
233 22
*10. y = 3 *14. x = 3 and z = 1
16. P1 = 10, 0, 02 and P2 = 11, - 2, 32
18. P1 = 1 - 2, 2, 32 and P2 = 14, 0, - 32
214
20. P1 = 12, - 3, - 32 and P2 = 14, 1, - 12
2219 22
In Problems 21–26, opposite vertices of a rectangular box whose edges are parallel to the coordinate axes are given. List the coordinates of the other six vertices of the box. *21. 10, 0, 02;
*24. 15, , 12;
12, 1, 32
*22. 10, 0, 02;
13, 8, 22
*25. 1 - 1, 0, 22;
14, 2, 22
14, 2, 52
*23. 11, 2, 32;
13, 4, 52
*26. 1 - 2, - 3, 02;
1 - , 7, 12
In Problems 27–32, the vector v has initial point P and terminal point Q. Write v in the form ai + bj + ck; that is, find its position vector. 27. P = 10, 0, 02; Q = 13, 4, - 12
29. P = 13, 2, - 12; Q = 15, , 02
v = 3i + 4j - k v = 2i + 4j + k
31. P = 1 - 2, - 1, 42; Q = 1, - 2, 42
v = 8i - j
28. P = 10, 0, 02; Q = 1 - 3, - 5, 42
30. P = 1 - 3, 2, 02; Q = 1, 5, - 12
32. P = 1 - 1, 4, - 22; Q = 1, 2, 22
v = - 3i - 5j + 4k v = 9i + 3j - k v = 7i - 2j + 4k
678
CHAPTER 8 Polar Coordinates; Vectors
In Problems 33–38, find ‘v‘. 33. v = 3i - j - 2k 36. v = - i - j + k
7
34. v = - i + 12j + 4k
14
23
37. v = - 2i + 3j - 3k
222
35. v = i - j + k
23
38. v = i + 2j - 2k
2211
In Problems 39–44, find each quantity if v = 3i - 5j + 2k and w = - 2i + 3j - 2k. 39. 2v + 3w
- j - 2k
40. 3v - 2w
42. ‘v + w ‘
25
43. ‘v‘ - ‘w ‘
13i - 21j + 10k
41. ‘v - w ‘
238 - 217
44. ‘v‘ + ‘w ‘
46. v = - 3j
i
*48. v = - i + 12j + 4k
-j
49. v = i + j + k
23 23 23 i + j + k 3 3 3
In Problems 51–58, find the dot product v ~ w and the angle between v and w. 51. v = i - j, w = i + j + k
v # w = 0; u = 90°
53. v = 2i + j - 3k, w = i + 2j + 2k 55. v = 3i - j + 2k, w = i + j - k
v#w
52. v = i + j, w = - i + j - k
v # w = 0; u = 90°
= - 2; u ≈ 100.3° 54. v = 2i + 2j - k, w = i + 2j + 3k
v # w = 0; u = 90°
57. v = 3i + 4j + k, w = i + 8j + 2k
238 + 217
2 3 i - j - k 7 7 7 47. v = 3i - j - 2k 2 2 2 50. v = 2i - j + k i j + k 3
In Problems 45–50, find the unit vector in the same direction as v. 45. v = 5i
2105
v#w
= 52; u = 0°
56. v = i + 3j + 2k, w = i - j + k
v # w = 3; u ≈ 74.5°
v # w = 0; u = 90°
58. v = 3i - 4j + k, w = i - 8j + 2k
v # w = 52; u = 0°
In Problems 59 – 66, find the direction angles of each vector. Write each vector in the form of equation (7). *59. v = 3i - j - 2k
*60. v = - i + 12j + 4k
*61. v = i + j + k
*62. v = i - j - k
*63. v = i + j
*64. v = j + k
*65. v = 3i - 5j + 2k
*66. v = 2i + 3j - 4k
Applications and Extensions 67. Robotic Arm Consider the double-jointed robotic arm shown in the figure. Let the lower arm be modeled by a = 82, 3, 49, the middle arm be modeled by b = 81, - 1, 39, and the upper arm be modeled by c = 84, - 1, - 29, where units are in feet.
In Problems 69 and 70, find the equation of a sphere with radius r and center P0 .
69. r = 1; P0 = 13, 1, 12 (x - 3)2 + (y - 1)2 + (z - 1)2 = 1
70. r = 2; P0 = 11, 2, 22 (x - 1)2 + (y - 2)2 + (z - 2)2 = 4 In Problems 71–76, find the radius and center of each sphere. [Hint: Complete the square in each variable.]
c b
71. x2 + y2 + z2 + 2x - 2y = 2
r = 2; ( - 1, 1, 0)
72. x + y + z + 2x - 2z = - 1 2
2
r = 1; ( - 1, 0, 1)
2
73. x2 + y2 + z2 - 4x + 4y + 2z = 0
r = 3; (2, - 2, - 1)
74. x + y + z - 4x = 0
r = 2; (2, 0, 0) 322 r = ; 12, 0, - 12 2 2 2 2 76. 3x + 3y + 3z + x - y = 3 r = 23; ( - 1, 1, 0) 2
a
2
2
75. 2x2 + 2y2 + 2z2 - 8x + 4z = - 1
*(a) Find a vector d that represents the position of the hand. *(b) Determine the distance of the hand from the origin. *68. The Sphere In space, the collection of all points that are the same distance from some fixed point is called a sphere. See the illustration. The constant distance is called the radius, and the fixed point is the center of the sphere. Show that the equation of a sphere with center at 1x0 , y0 , z0 2 and radius r is 1x - x0 2 2 + 1y - y0 2 2 + 1z - z0 2 2 = r 2
[Hint: Use the Distance Formula (1).]
The work W done by a constant force F in moving an object from a point A > in space to a point B in space is defined as W = F ~ AB . Use this definition in Problems 77–79. 77. Work Find the work done by a force of 3 newtons acting in the direction 2i + j + 2k in moving an object 2 meters from 10, 0, 02 to 10, 2, 02. 2 newton@meters = 2 joules 78. Work Find the work done by a force of 1 newton acting in the direction 2i + 2j + k in moving an object 3 meters from 10, 0, 02 to 11, 2, 22. 79. Work Find the work done in moving an object along a vector u = 3i + 2j - 5k if the applied force is F = 2i - j - k. Use meters for distance and newtons for force.
z
9 newton@meters = 9 joules P ⫽ (x, y, z ) r
78.
P0 ⫽ (x0, y0, z0)
8 8 newton@meters = joules 3 3
y x
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
SECTION 8.7 The Cross Product
679
Retain Your Knowledge Problems 80–83 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 80. Solve:
e x 2 6 x …
3 Ú 5 x - 2
13 f 5
or
a2,
13 d 5
81. Given f(x) = 2x - 3 and g(x) = x2 + x - 1, find (f ∘ g)(x). 2x2 + 2x - 5 1 82. Find the exact value of sin 80° cos 50° − cos 80° sin 50°. 2 83. Solve the triangle. c = 325 ≈ .71; A ≈ 2.°; B ≈ 3.4° B c
3
A 6
‘Are You Prepared?’ Answer 1. 21x2 - x1 2 2 + 1y2 - y1 2 2
8.7 The Cross Product OBJECTIVES 1 2 3 4 5
Find the Cross Product of Two Vectors (p. 679) Know Algebraic Properties of the Cross Product (p. 681) Know Geometric Properties of the Cross Product (p. 682) Find a Vector Orthogonal to Two Given Vectors (p. 682) Find the Area of a Parallelogram (p. 683)
1 Find the Cross Product of Two Vectors For vectors in space, and only for vectors in space, a second product of two vectors is defined, called the cross product. The cross product of two vectors in space is also a vector that has applications in both geometry and physics.
DEFINITION
If v = a1 i + b1 j + c1 k and w = a2 i + b2 j + c2 k are two vectors in space, the cross product v * w is defined as the vector v * w = 1b1 c2 - b2 c1 2i - 1a1 c2 - a2 c1 22jj + 1a1 b2 - a2 b1 2k
(1)
Notice that the cross product v * w of two vectors is a vector. Because of this, it is sometimes referred to as the vector product.
EXAM PL E 1
Finding a Cross Product Using Equation (1) If v = 2i + 3j + 5k and w = i + 2j + 3k, find v * w.
Solution
v * w = 13 # 3 - 2 # 52i - 12 # 3 - 1 # 52j + 12 # 2 - 1 # 32k Equation (1) = 19 - 102i - 1 - 52j + 14 - 32k
= -i - j + k
r
680
CHAPTER 8 Polar Coordinates; Vectors
Determinants* may be used as an aid in computing cross products. A 2 by 2 determinant, symbolized by `
a1 a2
b1 ` b2
has the value a1 b2 - a2 b1; that is, `
a1 a2
b1 ` = a1 b2 - a2 b1 b2
A 3 by 3 determinant has the value A 3 a1
B b1 b2
a2
EX A MPL E 2
C b c1 3 = ` 1 b2 c2
c1 a `A - ` 1 c2 a2
c1 a `B + ` 1 c2 a2
b1 `C b2
Evaluating Determinants 2 3 ` = 2#2 1 2 A B C (b) 3 2 3 5 3 = 1 2 3 = (a) `
- 1#3 = 4 - 3 = 1 `
3 5 2 5 2 3 `A - ` `B + ` `C 2 3 1 3 1 2
19 - 102A - 1 - 52B + 14 - 32 C
r
= -A - B + C
Now Work
PROBLEM
7
The cross product of the vectors v = a1 i + b1 j + c1 k and w = a2 i + b2 j + c2 k, that is, v * w = 1b1 c2 - b2 c1 2i - 1a1 c2 - a2 c1 2j + 1a1 b2 - a2 b1 2k
may be written symbolically using determinants as i v * w = 3 a1 a2
EX A MPL E 3
j b1 b2
k b c1 3 = ` 1 b2 c2
c1 a `i - ` 1 c2 a2
c1 a `j + ` 1 c2 a2
b1 `k b2
Using Determinants to Find Cross Products If v = 2i + 3j + 5k and w = i + 2j + 3k, find: (a) v * w
Solution
(b) w * v
(c) v * v
(d) w * w
i j k 3 5 2 5 2 3 (a) v * w = 3 2 3 5 3 = ` `i - ` `j + ` ` k = -i - j + k 2 3 1 3 1 2 1 2 3 i j k 2 3 1 3 1 2 3 (b) w * v = 1 2 3 3 = ` `i - ` `j + ` `k = i + j - k 3 5 2 5 2 3 2 3 5 i j k 3 5 2 5 2 3 (c) v * v = 3 2 3 5 3 = ` `i - ` `j + ` ` k = 0i - 0j + 0k = 0 3 5 2 5 2 3 2 3 5 *
Determinants are discussed in detail in Section 10.3.
681
SECTION 8.7 The Cross Product
i j k 3 (d) w * w = 1 2 3 3 1 2 3 = `
2 3 1 3 1 2 `i - ` `j + ` ` k = 0i - 0j + 0k = 0 2 3 1 3 1 2
Now Work
PROBLEM
r
15
2 Know Algebraic Properties of the Cross Product Notice in Examples 3(a) and (b) that v * w and w * v are negatives of one another. From Examples 3(c) and (d), one might conjecture that the cross product of a vector with itself is the zero vector. These and other algebraic properties of the cross product are given next.
THEOREM
Algebraic Properties of the Cross Product If u, v, and w are vectors in space and if a is a scalar, then u * u = 0
(2)
u * v = - 1v * u2
(3)
a1u a 1u * v2 = 1au2 * v = u * 1av2
(4)
u * 1v + w2 = 1u * v2 + 1u * w2
(5)
Proof We will prove properties (2) and (4) here and leave properties (3) and (5) as FYFSDJTFT TFF1SPCMFNTBOE To prove property (2), let u = a1 i + b1 j + c1 k. Then i u * u = 3 a1 a1
j b1 b1
k b c1 3 = ` 1 b1 c1
c1 a `i - ` 1 c1 a1
c1 a `j + ` 1 c1 a1
b1 `k b1
= 0i - 0j + 0k = 0 To prove property (4), let u = a1 i + b1 j + c1 k and v = a2 i + b2 j + c2 k. Then a1u * v2 = a3 1b1 c2 - b2 c1 2i - 1a1 c2 - a2 c1 2j + 1a1 b2 - a2 b1 2k 4 c
Apply (1).
= a1b1 c2 - b2 c1 2i - a1a1 c2 - a2 c1 2j + a1a1 b2 - a2 b1 2k
(6)
Since au = aa1 i + ab1 j + ac1 k, we have 1au2 * v = 1ab1 c2 - b2ac1 2i - 1aa1 c2 - a2ac1 2j + 1aa1 b2 - a2ab1 2k = a1b1 c2 - b2 c1 2i - a1a1 c2 - a2 c1 2j + a1a1 b2 - a2 b1 2k
(7)
#BTFEPOFRVBUJPOT BOE
UIFGJSTUQBSUPGQSPQFSUZ GPMMPXT5IFTFDPOEQBSU can be proved in like fashion. ■
Now Work
PROBLEM
17
682
CHAPTER 8 Polar Coordinates; Vectors
3 Know Geometric Properties of the Cross Product THEOREM
Geometric Properties of the Cross Product Let u and v be vectors in space. u * v is orthogonal to both u and v.
(8)
7 u * v 7 = 7 u 7 7 v 7 sin u,
(9)
where u is the angle between u and v.
7 u * v 7 is the area of the parallelogram
having u ≠ 0 and v ≠ 0 as adjacent sides.
(10)
u * v = 0 if and only if u and v are parallel.
(11)
Proof of Property (8) Let u = a1 i + b1 j + c1 k and v = a2 i + b2 j + c2 k. Then
Figure 81
u * v = 1b1 c2 - b2 c1 2i - 1a1 c2 - a2 c1 2j + 1a1 b2 - a2 b1 2k Now compute the dot product u ~ 1u * v2.
u
u ~ 1u * v2 = 1a1 i + b1 j + c1 k2 # 3 1b1 c2 - b2 c1 2i - 1a1 c2 - a2 c1 2j + 1a1 b2 - a2 b1 2k 4 = a1 1b1 c2 - b2 c1 2 - b1 1a1 c2 - a2 c1 2 + c1 1a1 b2 - a2 b1 2 = 0
v
Since two vectors are orthogonal if their dot product is zero, it follows that u and u * v are orthogonal. Similarly, v ~ 1u * v2 = 0, so v and u * v are orthogonal. ■ Figure 82
4 Find a Vector Orthogonal to Two Given Vectors As long as the vectors u and v are not parallel, they will form a plane in space. See Figure 81. Based on property (8), the vector u * v is normal to this plane. As Figure 81 illustrates, there are essentially (without regard to magnitude) two vectors normal to the plane containing u and v. It can be shown that the vector u * v is the one determined by the thumb of the right hand when the other fingers of the right hand are cupped so that they point in a direction from u to v. See Figure 82.*
u
v
Finding a Vector Orthogonal to Two Given Vectors
EX A MPL E 4
Find a vector that is orthogonal to u = 3i - 2j + k and v = - i + 3j - k.
Solution
Based on property (8), such a vector is u * v.
u * v = 3
i 3 -1
j -2 3
k 1 3 = 12 - 32 i - 3 - 3 - 1 - 12 4 j + 19 - 22k = - i + 2j + 7k -1
The vector - i + 2j + 7k is orthogonal to both u and v. Check: Two vectors are orthogonal if their dot product is zero.
u ~ 1 - i + 2j + 7k2 = 13i - 2j + k2 ~ 1 - i + 2j + 7k2 = - 3 - 4 + 7 = 0 v ~ 1 - i + 2j + 7k2 = 1 - i + 3j - k2 ~ 1 - i + 2j + 7k2 = 1 + - 7 = 0
Now Work
PROBLEM
41
5IFQSPPGPGQSPQFSUZ JTMFGUBTBOFYFSDJTF4FF1SPCMFN *
This is a consequence of using a “right-handed” coordinate system.
r
SECTION 8.7 The Cross Product
Figure 83
683
5 Find the Area of a Parallelogram Proof of Property (10) Suppose that u and v are adjacent sides of a parallelogram. See Figure 83. Then the lengths of these sides are 7 u 7 and 7 v 7 . If u is the angle between u and v, then the height of the parallelogram is 7 v 7 sin u and its area is
v
Area of parallelogram = Base * Height = 7 u 7 3 7 v 7 sin u4 = 7 u * v 7
θ
c
u
Property (9)
j
Finding the Area of a Parallelogram
EXAM PL E 5
Find the area of the parallelogram whose vertices are P1 = 10, 0, 02, P2 = 13, - 2, 12, P3 = 1 - 1, 3, - 12, and P4 = 12, 1, 02.
Solution
Two adjacent sides of this parallelogram are
>
>
u = P1 P2 = 3i - 2j + k and v = P1 P3 = - i + 3j - k
WARNING Not all pairs of vertices> give rise to a side. For example, P1P4 is a diagonal > of the > parallelogram > since> P1P3 + P>3P4 = P1P4 . Also, P1P3 and P2P4 are not adjacent sides; they are parallel sides. ■
Since u * v = - i + 2j + 7k (Example 4), the area of the parallelogram is
Area of parallelogram = 7 u * v 7 = 21 + 4 + 49 = 254 = 32 square units
Now Work
PROBLEM
r
49
Proof of Property (11) The proof requires two parts. If u and v are parallel, then there is a scalar a such that u = av. Then u * v = 1av2 * v = a 1v * v2 = 0 c
c
Property (4)
Property (2)
If u * v = 0, then, by property (9), we have
7 u * v 7 = 7 u 7 7 v 7 sin u = 0
Since u ≠ 0 and v ≠ 0, we must have sin u = 0, so u = 0 or u = p. In either case, since u is the angle between u and v, then u and v are parallel. j
8.7 Assess Your Understanding Concepts and Vocabulary 1. True or False If u and v are parallel vectors, then u * v = 0. True 2. True or False For any vector v, v * v = 0. True
5. True or False 7 u * v 7 = 7 u 7 7 v 7 cos u, where u is the angle between u and v. False
3. True or False If u and v are vectors, then u * v + v * u = 0. True 4. True or False u * v is a vector that is parallel to both u and v. False
6. True or False The area of the parallelogram having u and v as adjacent sides is the magnitude of the cross product of u and v. True
Skill Building In Problems 7–14, find the value of each determinant. 7. `
3 4 ` 1 2
2
A B 11. 3 2 1 1 3
C 4 3 1
- 11A + 2B + 5C
8. `
-2 2
5 ` -3
A B 12. 3 0 2 3 1
-4
C 4 3 3
2A + 12B - C
9. `
-2
5 ` -1
A B 13. 3 - 1 3 5 0
4 C 53 -2
- A + 23B - 15C
10. `
-4 0 ` 5 3
A 14. 3 1 0
B -2 2
- 12 C -3 3 -2
10A + 2B + 2C
684
CHAPTER 8 Polar Coordinates; Vectors
In Problems 15–22, find (a) v * w, (b) w * v, (c) w * w, and (d) v * v. *15. v = 2i - 3j + k w = 3i - 2j - k
*16. v = - i + 3j + 2k w = 3i - 2j - k
*17. v = i + j w = 2i + j + k
*18. v = i - 4j + 2k w = 3i + 2j + k
*19. v = 2i - j + 2k w = j - k
*20. v = 3i + j + 3k w = i - k
*21. v = i - j - k w = 4i - 3k
*22. v = 2i - 3j w = 3j - 2k
In Problems 23–44, use the given vectors u, v, and w to find each expression. u = 2i - 3j + k 23. u * v
- 9i - 7j - 3k
26. w * v
- 7i - 11j + k
29. 13u2 * v
32. 1 - 3v2 * w
35. u ~ 1v * w2
38. 1v * u2 ~ w
- 27i - 21j - 9k - 21i - 33j + 18k - 25
v = - 3i + 3j + 2k
24. v * w
7i + 11j - k
27. v * v
0
30. v * 14w2
33. u ~ 1u * v2
36. 1u * v2 ~ w
25. v * u 28. w * w
28i + 44j - 24k
39. u * 1v * v2
25
w = i + j + 3k
0
9i + 7j + 3k 0
31. u * 12v2
34. v ~ 1v * w2
- 18i - 14j - k 0
37. v ~ 1u * w2
- 25
40. 1w * w2 * v
0
25 0
*41. Find a vector orthogonal to both u and v.
*42. Find a vector orthogonal to both u and w.
*43. Find a vector orthogonal to both u and i + j.
*44. Find a vector orthogonal to both u and j + k.
>
>
In Problems 45–48, find the area of the parallelogram with one corner at P1 and adjacent sides P1 P2 and P1 P3 . 45. P1 = 10, 0, 02, P2 = 11, 2, 32, P3 = 1 - 2, 3, 02
47. P1 = 11, 2, 02, P2 = 1 - 2, 3, 42, P3 = 10, - 2, 32
21 2555
46. P1 = 10, 0, 02, P2 = 12, 3, 12, P3 = 1 - 2, 4, 12
2213
48. P1 = 1 - 2, 0, 22, P2 = 12, 1, - 12, P3 = 12, - 1, 22 2217
In Problems 49–52, find the area of the parallelogram with vertices P1 , P2 , P3 , and P4 . 49. P1 = 11, 1, 22, P2 = 11, 2, 32, P3 = 1 - 2, 3, 02, P4 = 1 - 2, 4, 12 234
51. P1 = 11, 2, - 12, P2 = 14, 2, - 32, P3 = 1, - 5, 22, P4 = 19, - 5, 02 2998
50. P1 = 12, 1, 12, P2 = 12, 3, 12, P3 = 1 - 2, 4, 12, P4 = 1 - 2, , 12 8
52. P1 = 1 - 1, 1, 12, P2 = 1 - 1, 2, 22, P3 = 1 - 3, 4, - 52, P4 = 1 - 3, 5, - 42 289
Applications and Extensions *53. Find a unit vector normal to the plane containing v = i + 3j - 2k and w = - 2i + j + 3k. *54. Find a unit vector normal to the plane containing v = 2i + 3j - k and w = - 2i - 4j - 3k. 55. Volume of a Parallelepiped A parallelepiped is a prism whose faces are all parallelograms. Let A, B, and C be the vectors that define the parallelepiped shown in the figure. The volume V of the parallelepiped is given by the formula V = 0 (A * B) # C 0 .
C
*56. Volume of a Parallelepiped 3FGFS UP 1SPCMFN 'JOE the volume of a parallelepiped whose defining vectors are A = 81, 0, 9, B = 82, 3, - 89, and C = 88, - 5, 9. *57. Prove for vectors u and v that
7 u * v 7 2 = 7 u 7 2 7 v 7 2 - 1u ~ v2 2
[Hint: Proceed as in the proof of property (4), computing first the left side and then the right side.] *58. Show that if u and v are orthogonal, then
7u * v7 = 7u7 7v7
*59. Show that if u and v are orthogonal unit vectors, then u * v is also a unit vector.
B
*60. Prove property (3). *61. Prove property (5). A
Find the volume of a parallelepiped if the defining vectors are A = 3i - 2j + 4k, B = 2i + j - 2k, and C = 3i - j - 2k. 48 cubic units
*62. Prove property (9). [Hint: Use the result of Problem 57 and the fact that if u is the angle between u and v, then u ~ v = 7 u 7 7 v 7 cos u.]
Discussion and Writing 63. If u ~ v = 0 and u * v = 0, what, if anything, can you conclude about u and v? *Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
Chapter Review
685
Retain Your Knowledge Problems 64–67 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 64. Find the exact value of cos - 1 a
1 22
b.
p 4
65. Find two pairs of polar coordinates (r, u), one with r 7 0 and the other with r 6 0, for the point with rectangular coordinates ( - 8, - 15). Express u in radians. 117, 4.222, 1 - 17, 1.082 66. Tattoo Cost For custom tattoos, a tattoo shop charges $150 for the first hour and $90 per hour for each additional hour of work. If the total cost for a tattoo is $780, how many hours of custom work were required? 8 hours 67. Use properties of logarithms to write log 4
2x 1 as a sum or difference of logarithms. Express powers as factors. log 4 x - 3 log 4 z 3 2 z
Chapter Review Things to Know Polar Coordinates (pp. 616–623) x = r cos u, y = r sin u
3FMBUJPOTIJQCFUXFFOQPMBS coordinates 1r, u2 and rectangular coordinates 1x, y2 QQBOE
r 2 = x2 + y2, tan u =
1PMBSGPSNPGBDPNQMFYOVNCFS Q
If z = x + yi, then z = r 1cos u + i sin u2,
y , x ≠ 0 x
where r = 0 z 0 = 2x2 + y2 , sin u =
y x , cos u = , 0 … u 6 2p. r r
%F.PJWSFT5IFPSFN Q
If z = r 1cos u + i sin u2, then zn = r n 3cos 1nu2 + i sin1nu2 4, where n Ú 1 is a positive integer.
nth root of a complex number w = r 1cos u0 + i sin u0 2 Q
u0 u0 2kp 2kp + b + i sin a + b d , k = 0, c , n - 1, n n n n where n Ú 2 is an integer zk = 1r c cos a n
Vectors (pp. 648–657)
Quantity having magnitude and direction; equivalent to a directed line ¡ segment PQ
1PTJUJPOWFDUPS QQBOE
Vector whose initial point is at the origin
6OJUWFDUPS QQBOE
Vector whose magnitude is 1
%PUQSPEVDU QQBOE
If v = a1 i + b1 j and w = a2 i + b2 j, then v # w = a1a2 + b1b2. If v = a1 i + b1 j + c1 k and w = a2 i + b2 j + c2 k, then v # w = a1a2 + b1b2 + c1c2.
Angle u between two nonzero vectors u and v QQBOE
Direction angles of a WFDUPSJOTQBDF Q
cos u =
u#v
7u7 7v 7
If v = ai + bj + ck, then v = 7 v 7 3 1cos a2i + 1cos b2j + 1cos g2k4, where cos a =
a
7v7
, cos b =
b
7v7
, and cos g =
c . 7v7
$SPTTQSPEVDU Q
If v = a1 i + b1 j + c1 k and w = a2 i + b2 j + c2 k,
"SFBPGBQBSBMMFMPHSBN Q
7 u * v 7 = 7 u 7 7 v 7 sin u, where u is the angle between the two adjacent sides u and v.
then v * w = 3b1 c2 - b2 c1 4i - 3a1 c2 - a2 c1 4j + 3a1 b2 - a2 b1 4k.
686
CHAPTER 8 Polar Coordinates; Vectors
Objectives Section
You should be able to…
8.1 8.2
1
8.3
2
3
8.4 8.5 8.6 8.7
4
2 3 4 1
3 1 2
5 1 2 3 4 5 6 7 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5
Example(s)
Plot points using polar coordinates (p. 616) Convert from polar coordinates to rectangular coordinates (p. 618) Convert from rectangular coordinates to polar coordinates (p. 620) Transform equations between polar and rectangular forms (p. 622) Identify and graph polar equations by converting to rectangular equations (p. 626) Test polar equations for symmetry (p. 629) Graph polar equations by plotting points (p. 630) Plot points in the complex plane (p. 640) Convert a complex number between rectangular form and polar form (p. 641) Find products and quotients of complex numbers in polar form (p. 642) Use De Moivre’s Theorem (p. 643) Find complex roots (p. 644) Graph vectors (p. 650) Find a position vector (p. 651) Add and subtract vectors algebraically (p. 652) Find a scalar multiple and the magnitude of a vector (p. 653) Find a unit vector (p. 654) Find a vector from its direction and magnitude (p. 654) Model with vectors (p. 655) Find the dot product of two vectors (p. 662) Find the angle between two vectors (p. 663) Determine whether two vectors are parallel (p. 664) Determine whether two vectors are orthogonal (p. 664) Decompose a vector into two orthogonal vectors (p. 665) Compute work (p. 667) Find the distance between two points in space (p. 671) Find position vectors in space (p. 671) Perform operations on vectors (p. 672) Find the dot product (p. 673) Find the angle between two vectors (p. 674) Find the direction angles of a vector (p. 675) Find the cross product of two vectors (p. 679) Know algebraic properties of the cross product (p. 681) Know geometric properties of the cross product (p. 682) Find a vector orthogonal to two given vectors (p. 682) Find the area of a parallelogram (p. 683)
Review Exercises
1–3 4 5–7 8, 9 1– 6
1–3 1–3 4–6 7(a)–10(a) 7(b)–10(b)
7–10 7–13 1 2, 3
11–13 11–13 16–18 14–18
4
19–21
5, 6 7 1 2 3 4 5 6, 7 8–10 1 2 3 4 5, 6 7 1 2 3–5 6 7 8–10 1–3 1 3 4 5
22–25 26 27, 28 29, 30 31 29, 30, 32–34 35 36, 37 59, 60 46, 47 46, 47 50–52 50–52 53, 54, 62 61 38 39 40- 42 48, 49 48, 49 55 43, 44 57, 58 56 45 56
Review Exercises In Problems 1–3, plot each point given in polar coordinates, and find its rectangular coordinates. *1. a3,
p b 6
*2. a - 2,
4p b 3
*3. a - 3, -
p b 2
In Problems 4–6, the rectangular coordinates of a point are given. Find two pairs of polar coordinates 1r, u2 for each point, one with r 7 0 and the other with r 6 0. Express u in radians. *4. 1 - 8, - 82
*5. 1 - 13, 12
*6. 1 - 5, 02
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
Chapter Review
687
In Problems 7–10, the variables r and u represent polar coordinates. (a) Write each polar equation as an equation in rectangular coordinates 1x, y2. (b) Identify the equation and graph it. *7. r = 2 sin u p *9. u = 4
*8. r = 5 *10. r 2 + 4r sin u - 8r cos u = 5
In Problems 11–13, sketch the graph of each polar equation. Be sure to test for symmetry. *11. r = 4 cos u
*12. r = 3 - 3 sin u
*13. r = 4 - cos u
In Problems 14 and 15, write each complex number in polar form. Express each argument in degrees. 14. - 6 + 4i
21cos 146.3° + i sin 146.3°2
15. 4 - 3i
51cos 323.1° + i sin 323.1°2
In Problems 16–18, write each complex number in the standard form a + bi, and plot each in the complex plane. *16. 21cos 150° + i sin 150°2
- 23 + i
*18. 0.11cos 350° + i sin 350°2 In Problems 19–21, find zw and
*17. 3acos
2p 2p + i sin b 3 3
-
3 323 i + 2 2
0.10 - 0.02i z . Leave your answers in polar form. w 5p 5p + i sin b 6 6
*20. z = 3acos
*19. z = 21cos 40° + i sin 40°2 w = - 21cos 10° + i sin 10°2
w = 4acos
*21. z = 61cos 325° + i sin 325°2 w = 1cos 35° + i sin 35°2
p p + i sin b 6 6
In Problems 22–25, write each expression in the standard form a + bi. 22. 321cos 10° + i sin 10°2 4 6 24.
11
+ i28
7p 7p 3 b + i sin a b d 12 12 4 25. 18 - 6i2 - 8432 - 5376i
23. c 25 acos a
32 + 3213i
16
5
5 5 - 5 i A2 A2
*26. Find all the complex fifth roots of 32.
*27. u + v
v
u
In Problems 27 and 28, use the figure to the right to graph each of the following: *28. 2u + 3v
¡ In Problems 29 and 30, the vector v is represented by the directed line segment PQ . Write v in the form ai + bj and find 7 v 7 . 29. P = 11, - 22; Q = 13, - 62
30. P = 10, - 22; Q = 1 - 1, 12
v = 2i - 4j; 225
v = - i + 3j; 210
In Problems 31–35, use the vectors v = - 2i + j and w = 4i - 3j to find: 31. v + w 2i - 2j 33. 7 v 7 25
32. 4v - 3w - 20i + 13j
34. 7 v 7 + 7 w 7 25 + 5 ≈ 7.24
*36. Find the vector v in the xy-plane with magnitude 3 if the direction angle of v is 60°.
*35. Find a unit vector in the same direction as v. 37. Find the direction angle between i and v = - i + 23 j.
120° *38. Find the distance from P1 = 11, 3, - 22 to P2 = 14, - 2, 12.
*39. A vector v has initial point P = 11, 3, - 22 and terminal point Q = 14, - 2, 12. Write v in the form v = ai + bj + ck.
In Problems 40–45, use the vectors v = 3i + j - 2k and w = - 3i + 2j - k to find each expression. 40. 4v - 3w 43. v * w
41. 7 v - w 7 238
21i - 2j - 5k
44. v # (v * w)
3i + 9j + 9k
42. 7 v 7 - 7 w 7
In Problems 46–49, find the dot product v ~ w and the angle between v and w. 46. v = - 2i + j, w = 4i - 3j
v # w = - 11; u ≈ 169.7°
48. v = i + j + k, w = i - j + k
v # w = 1; u ≈ 70.5°
0
*45. Find a unit vector orthogonal to both v and w.
0
47. v = i - 3j, w = - i + j
v # w = - 4; u ≈ 153.4°
49. v = 4i - j + 2k, w = i - 2j - 3k
v # w = 0; u ≈ 90°
In Problems 50–52, determine whether v and w are parallel, orthogonal, or neither. *50. v = 2i + 3j; w = - 4i - 6j
*51. v = - 2i + 2j; w = - 3i + 2j
*52. v = 3i - 2j; w = 4i + 6j
688
CHAPTER 8 Polar Coordinates; Vectors
In Problems 53 and 54, decompose v into two vectors, one parallel to w and the other orthogonal to w. *53. v = 2i + j; w = - 4i + 3j
*54. v = 2i + 3j; w = 3i + j
56. Find the area of the parallelogram with vertices P1 = 11, 1, 12, P2 = 12, 3, 42, P3 = 16, 5, 22, and P4 = 17, 7, 52. 2283
*55. Find the direction angles of the vector v = 3i - 4j + 2k.
58. Suppose that u = 3v. What is u * v?
57. If u * v = 2i - 3j + k, what is v * u? - 2i + 3j - k
59. Actual Speed and Direction of a Swimmer A swimmer can maintain a constant speed of 5 miles per hour. If the swimmer heads directly across a river that has a current moving at the rate of 2 miles per hour, what is the actual speed of the swimmer? (See the figure.) If the river is 1 mile wide, how far downstream will the swimmer end up from the point directly across the river from the starting point? 229 ≈ 5.39 mph; 0.4 mi
0
60. Static Equilibrium A weight of 2000 pounds is suspended from two cables, as shown in the figure. What are the tensions in the two cables? Left: 1843.21 lb; right: 1630.41 lb
40°
30°
2000 pounds Current
61. Computing Work Find the work done by a force of 5 pounds acting in the direction 60° to the horizontal in moving an object 20 feet from 10, 02 to 120, 02. 50 ft@lb
Swimmer's direction
*62. Braking Load A moving van with a gross weight of 8000 pounds is parked on a street with a 5° grade. Find the magnitude of the force required to keep the van from rolling down the hill. What is the magnitude of the force perpendicular to the hill? 697.2 lbs; 7969.6 lbs
Direction of swimmer due to current
Chapter Test Prep Videos include step-by-step solutions to all chapter test exercises and can be found on this text’s Channel. (search “SullivanPrecalcUC3e”)
Chapter Test In Problems 1–3, plot each point given in polar coordinates. *1. a2,
3p b 4
*2. a3, -
p b 6
*3. a - 4,
p b 3
4. Convert 1 2, 223 2 from rectangular coordinates to polar coordinates 1r, u2, where r 7 0 and 0 … u 6 2p.
a4,
p b 3
In Problems 5–7, convert the polar equation to a rectangular equation. Graph the equation. *7. r sin2 u + 8 sin u = r p In Problems 8 and 9, test the polar equation for symmetry with respect to the pole, the polar axis, and the line u = . 2 *9. r = 5 sin u cos2 u *8. r 2 cos u = 5 *5. r = 7
x2 + y2 = 49
*6. tan u = 3
y = 3x
8y = x2
In Problems 10–12, perform the given operation, where z = 21cos 85° + i sin 85°2 and w = 31cos 22° + i sin 22°2. Write your answer in polar form. w 3 10. z # w 6(cos 107° + i sin 107°) (cos 297° + i sin 297°) 12. w 5 343(cos 297° + i sin 297°) 11. z 2 *13. Find all the complex cube roots of - 8 + 823i. Then plot them in the complex plane.
1 322, 722 2 and P2 = 1 822, 222 2 . ¡ 14. Find the position vector v equal toP1P2 . 8522, - 5229
In Problems 14–18, P1 = 15. Find 7 v 7 . 10
16. Find the unit vector in the direction of v. h 22, - 22 i 2 2
17. Find the direction angle of v. 315° 18. Decompose v into its vertical components. v = 522i - 522j
and
horizontal
Chapter Projects
In Problems 19–22, v1 = 84, 9, v2 = 8 - 3, - 9, v3 = 8 - 8, 49, and v4 = 810, 159. 19. Find the vector v1 + 2v2 - v3. 20. Which two vectors are parallel?
8, - 109
v1 and v4
21. Which two vectors are orthogonal?
v2 and v3
22. Find the angle between the vectors v1 and v2 .
172.87°
In Problems 23–25, use the vectors u = 2i - 3j + k and v = - i + 3j + 2k. 23. Find u * v.
- 9i - 5j + 3k
689
*24. Find the direction angles for u. 25. Find the area of the parallelogram that has u and v as adjacent sides. 2115 26. A 1200-pound chandelier is to be suspended over a large ballroom; the chandelier will be hung on two cables of FRVBMMFOHUIXIPTFFOETXJMMCFBUUBDIFEUPUIFDFJMJOH feet apart. The chandelier will be free-hanging so that the ends of the cable will make equal angles with the ceiling. If UIFUPQPGUIFDIBOEFMJFSJTUPCFGFFUGSPNUIFDFJMJOH what is the minimum tension each cable must be able to endure? ≈70.82 lb
Cumulative Review 2
1. Find the real solutions, if any, of the equation e x
-9
= 1. { - 3, 3}
*2. Find an equation for the line containing the origin that makes an angle of 30° with the positive x-axis.
*3. Find an equation for the circle with center at the point 10, 12 and radius 3. Graph this circle. *4. What is the domain of the function f 1x2 = ln11 - 2x2?
*5. Test the equation x2 + y3 = 2x4 for symmetry with respect to the x-axis, the y-axis, and the origin. *6. Graph the function y = 0 ln x 0 .
*7. Graph the function y = 0 sin x 0 .
*8. Graph the function y = sin 0 x 0 .
1 p 9. Find the exact value of sin-1 a - b. 2 *10. Graph the equations x = 3 and y = 4 on the same set of rectangular axes. p *11. Graph the equations r = 2 and u = on the same set of 3 polar axes. *12. What are the amplitude and period of y = - 4 cos(px)?
Chapter Projects Lift
Drag
Thrust
Weight
I.
Modeling Aircraft Motion Four aerodynamic forces act on an airplane in flight: lift, weight, thrust, and drag. While an aircraft is in flight, these four forces continuously battle each other. Weight opposes lift, and drag opposes thrust. See the figure. In balanced flight at constant speed, both the lift and weight are equal and the thrust and drag are equal. 1.
What will happen to the aircraft if the lift is held constant while the weight is decreased (say, from burning off fuel)?
2.
What will happen to the aircraft if the lift is decreased while the weight is held constant?
3.
What will happen to the aircraft if the thrust is increased while the drag is held constant?
4.
What will happen to the aircraft if the drag is increased while the thrust is held constant?
Source: www.aeromuseum.org/eduHowtoFly.html
In 1903 the Wright brothers made the first controlled powered flight. The weight of their plane was approximately 700 pounds (lb). Newton’s Second Law of Motion states that force = mass * acceleration 1F = ma2 . If the mass is measured in kilograms (kg) and acceleration in meters per second squared (m/sec2), then the force will be measured in newtons (N). 3Note: 1 N = 1 kg # m/sec2.4 5.
If 1 kg = 2.205 lb, convert the weight of the Wright brothers’ plane to kilograms.
6.
If acceleration due to gravity is a = 9.80 m/sec2, determine the force due to weight on the Wright brothers’ plane.
690
CHAPTER 8 Polar Coordinates; Vectors
7.
What must be true about the lift force of the Wright brothers’ plane for it to get off the ground?
8.
The weight of a fully loaded Cessna 170B is 2200 lb. What lift force is required to get this plane off the ground?
9.
The maximum gross weight of a Boeing 747 is 255,000 lb. What lift force is required to get this jet off the ground?
The following projects are available at the Instructors’ Resource Center (IRC): II. Project at Motorola Signal Fades due to Interference Complex trigonometric functions are used to ensure that a cellphone has optimal reception as the user travels up and down an elevator. III. Compound Interest The effect of continuously compounded interest is analyzed using polar coordinates. IV. Complex Equations Analysis of complex equations illustrates the connections between complex and real equations. At times, using complex equations is more efficient for proving mathematical theorems.
9
Analytic Geometry The Orbit of the Hale–Bopp Comet The orbits of the Hale–Bopp comet and Earth can be modeled using ellipses, the subject of Section 9.3. The Internet-based Project at the end of this chapter explores the possibility of the Hale–Bopp comet colliding with Earth.
—See the Internet-based Chapter Project I—
In this chapter, geometric definitions are given for the conics, and the distance formula and rectangular coordinates are used to obtain their equations. Historically, Apollonius (200 BC) was among the first to study conics and discover some of their interesting properties. Today, conics are still studied because of their many uses. Paraboloids of revolution (parabolas rotated about their axes of symmetry) are used as signal collectors (the satellite dishes used with radar and dish TV, for example), as solar energy collectors, and as reflectors (telescopes, light projection, and so on). The planets circle the Sun in approximately elliptical orbits. Elliptical surfaces can be used to reflect signals such as light and sound from one place to another. A third conic, the hyperbola, can be used to determine the location of lightning strikes. The Greeks used the methods of Euclidean geometry to study conics. However, we shall use the more powerful methods of analytic geometry, which employs both algebra and geometry, for our study of conics.
Conics The Parabola The Ellipse The Hyperbola Rotation of Axes; General Form of a Conic Polar Equations of Conics Plane Curves and Parametric Equations Chapter Review Chapter Test Cumulative Review Chapter Projects
691
692
CHAPTER 9 Analytic Geometry
9.1 Conics OBJECTIVE 1 Know the Names of the Conics (p. 692)
1 Know the Names of the Conics The word conic derives from the word cone, which is a geometric figure that can be constructed in the following way: Let a and g be two distinct lines that intersect at a point V. Keep the line a fixed. Now rotate the line g about a, while maintaining the same angle between a and g. The collection of points swept out (generated) by the line g is called a (right circular) cone. See Figure 1. The fixed line a is called the axis of the cone; the point V is its vertex; the lines that pass through V and make the same angle with a as g are generators of the cone. Each generator is a line that lies entirely on the cone. The cone consists of two parts, called nappes, that intersect at the vertex. Figure 1
Axis, a
Generators Vertex, V g
Conics, an abbreviation for conic sections, are curves that result from the intersection of a right circular cone and a plane. The conics we shall study arise when the plane does not contain the vertex, as shown in Figure 2. These conics are circles when the plane is perpendicular to the axis of the cone and intersects each generator; ellipses when the plane is tilted slightly so that it intersects each generator, but intersects only one nappe of the cone; parabolas when the plane is tilted farther so that it is parallel to one (and only one) generator and intersects only one nappe of the cone; and hyperbolas when the plane intersects both nappes. If the plane does contain the vertex, the intersection of the plane and the cone is a point, a line, or a pair of intersecting lines. These are usually called degenerate conics. Figure 2
Axis
Axis
Axis
Axis
Generator
(a) Circle
(b) Ellipse
(c) Parabola
(d) Hyperbola
Conic sections are used in modeling many different applications. For example, parabolas are used in describing satellite dishes and telescopes (see Figures 14 and 15 on page 698). Ellipses are used to model the orbits of planets and whispering chambers (see pages 707 and 708). And hyperbolas are used to locate lightning strikes and model nuclear cooling towers (see Problems 76 and 77 in Section 9.4).
SECTION 9.2 The Parabola
693
9.2 The Parabola PREPARING FOR THIS SECTION Before getting started, review the following: r %JTUBODF'PSNVMB 'PVOEBUJPOT 4FDUJPO Q
r 4ZNNFUSZ 'PVOEBUJPOT 4FDUJPO QQm
r 4RVBSF3PPU.FUIPE 4FDUJPO QQm
r $PNQMFUJOHUIF4RVBSF "QQFOEJY" 4FDUJPO" pp. A38–A39) r (SBQIJOH5FDIOJRVFT5SBOTGPSNBUJPOT 4FDUJPO pp. 121–129)
Now Work the ‘Are You Prepared?’ problems on page 699.
OBJECTIVES 1 Analyze Parabolas with Vertex at the Origin (p. 693) 2 Analyze Parabolas with Vertex at (h, k) (p. 696) 3 Solve Applied Problems Involving Parabolas (p. 697)
Section 2.4 taught us that the graph of a quadratic function is a parabola. This section gives a geometric definition of a parabola and uses it to obtain an equation.
DEFINITION
A parabola is the collection of all points P in the plane that are the same distance d from a fixed point F as they are from a fixed line D. The point F is called the focus of the parabola, and the line D is its directrix. As a result, a parabola is the set of points P for which d 1F, P2 = d 1P, D2
Figure 3 P d(P, D)
Axis of symmetry d(F, P) F
2a
a
a
V
(1)
Figure 3 shows a parabola (in blue). The line through the focus F and perpendicular to the directrix D is called the axis of symmetry of the parabola. The point of intersection of the parabola with its axis of symmetry is called the vertex V. Because the vertex V lies on the parabola, it must satisfy equation (1): d 1F, V2 = d 1V, D2. The vertex is midway between the focus and the directrix. We shall let a equal the distance d 1F, V2 from F to V. Now we are ready to derive an equation for a parabola. To do this, we use a rectangular system of coordinates positioned so that the vertex V, focus F, and directrix D of the parabola are conveniently located.
Directrix D
1 Analyze Parabolas with Vertex at the Origin
If we choose to locate the vertex V at the origin 10, 02, we can conveniently position the focus F on either the x-axis or the y-axis. First, consider the case where the focus F is on the positive x-axis, as shown in Figure 4. Because the distance from F to V is a, the coordinates of F will be 1a, 02 with a 7 0. Similarly, because the distance from V to the directrix D is also a, and because D must be perpendicular to the x-axis (since the x-axis is the axis of symmetry), the equation of the directrix D must be x = - a. Now, if P = 1x, y2 is any point on the parabola, P must obey equation (1):
Figure 4 D: x ⫽ ⫺a y
d(P, D ) (⫺a, y)
P ⫽ (x, y )
d 1F, P2 = d 1P, D2
d (F, P ) V (0, 0)
F ⫽ (a, 0)
x
2 1x - a2 + 1y - 02 2 = 2(x - ( - a))2 + (y - y)2 2
1x - a2 2 + y2 = 1x + a2 2
x - 2ax + a + y = x + 2ax + a 2
2
2
2
y2 = 4ax
Use the Distance Formula. Square both sides.
2
Square binomials. Simplify.
694
CHAPTER 9 Analytic Geometry
THEOREM
Equation of a Parabola: Vertex at (0, 0), Focus at (a, 0), a + 0
The equation of a parabola with vertex at 10, 02, focus at 1a, 02, and directrix x = - a, a 7 0, is y2 = 4ax
(2)
3FDBMMUIBUa is the distance from the vertex to the focus of a parabola. When graphing the parabola y2 = 4ax, it is helpful to determine the “opening” by finding the points that lie directly above or below the focus 1a, 02 . This is done by letting x = a in y2 = 4ax, so y2 = 4a1a2 = 4a2, or y = {2a. The line segment joining these two points, 1a, 2a2 and 1a, - 2a2, is called the latus rectum; its length is 4a.
EX A MPL E 1
Solution
Figure 5
Finding the Equation of a Parabola and Graphing It
Find an equation of the parabola with vertex at 10, 02 and focus at 13, 02.(SBQI the equation.
The distance from the vertex 10, 02 to the focus 13, 02 is a = 3. Based on equation (2), the equation of this parabola is
y
D : x 3
6
y2 = 4ax
(3, 6)
y2 = 12x Latus rectum V (0, 0)
6
F (3, 0)
a = 3
To graph this parabola, find the two points that determine the latus rectum by letting x = 3. Then 6x
y2 = 12x = 12132 = 36 y = {6
6
(3, 6)
The points 13, 62 and 13, - 62 determine the latus rectum. These points help in graphing the parabola because they determine the “opening.” See Figure 5.
Now Work COMMENT To graph the parabola y2 = 12x discussed in Example 1, graph the two functions Y1 = 212x and Y2 = - 212x . %P UIJT BOE compare what you see with Figure 5. ■
Figure 6
EX A MPL E 2
D: x 2
V
F (2, 0) x 5
(2, 4) 5
PROBLEM
19
By reversing the steps used to obtain equation (2), it follows that the graph of an equation of the form of equation (2), y2 = 4ax, is a parabola; its vertex is at 10, 02 , its focus is at 1a, 02, its directrix is the line x = - a, and its axis of symmetry is the x-axis. For the remainder of this section, the direction “Analyze the equation” will mean to find the vertex, focus, and directrix of the parabola and graph it.
Analyzing the Equation of a Parabola Solution The equation y2 = 8x is of the form y2 = 4ax, where 4a = 8, so a = 2.
Latus rectum
(0, 0)
r
Analyze the equation: y2 = 8x
y 5 (2, 4)
5
Solve for y.
Consequently, the graph of the equation is a parabola with vertex at 10, 02 and focus on the positive x-axis at (a, 0) = 12, 02. The directrix is the vertical line x = - 2. The two points that determine the latus rectum are obtained by letting x = 2. Then y2 = 16, so y = {4. The points 12, - 42 and 12, 42 determine the latus rectum. See Figure 6 for the graph.
r
SECTION 9.2 The Parabola
695
3FDBMM UIBU XF BSSJWFE BU FRVBUJPO BGUFS QMBDJOH UIF GPDVT PO UIF QPTJUJWF x-axis. If the focus is placed on the negative x-axis, positive y-axis, or negative y-axis, a different form of the equation for the parabola results. The four forms of the equation of a parabola with vertex at 10, 02 and focus on a coordinate axis a distance a from 10, 02 are given in Table 1, and their graphs are given in Figure 7. Notice that each graph is symmetric with respect to its axis of symmetry.
Table 1
Equations of a Parabola: Vertex at (0, 0); Focus on an Axis; a + 0 Vertex
Focus
Directrix
Equation
Description
(0, 0)
(a, 0)
x = -a
y = 4ax
Axis of symmetry is the x-axis, opens right
(0, 0)
( - a, 0)
x = a
y = - 4ax
Axis of symmetry is the x-axis, opens left
(0, 0)
(0, a)
y = -a
x = 4ay
Axis of symmetry is the y-axis, opens up
(0, 0)
(0, - a)
y = a
x = - 4ay
Axis of symmetry is the y-axis, opens down
2 2 2 2
Figure 7 D: x a y
D: x a
y
y
y F (0, a)
F (a, 0)
V
D: y a
V
V
V x
x
x
x F (a, 0)
D: y a F (0, a)
(b) y 2 4ax
(a) y 2 4ax
Analyze the equation: x2 = - 12y
Figure 8
Solution
y 6
D: y 3 V 6
(0, 0) (6, 3)
F (0, 3)
Figure 9 y
V (0, 0) 6
PROBLEM
39
Find the equation of the parabola with focus at 10, 42 and directrix the line y = - 4. (SBQIUIFFRVBUJPO
Solution
F (0,4)
Now Work
Finding the Equation of a Parabola
10
(8, 4)
The equation x2 = - 12y is of the form x2 = - 4ay, with a = 3. Consequently, the graph of the equation is a parabola with vertex at 10, 02, focus at 10, - 32 , and directrix the line y = 3. The parabola opens down, and its axis of symmetry is the y-axis. To obtain the points defining the latus rectum, let y = - 3. Then x2 = 36, so x = {6. The points 1 - 6, - 32 and 16, - 32 determine the latus rectum. See Figure 8 for the graph.
r
x (6, 3)
EXAM PL E 4
10
(d) x 2 4ay
Analyzing the Equation of a Parabola
EXAM PL E 3
6
(c) x 2 4ay
A parabola whose focus is at 10, 42 and whose directrix is the horizontal line y = - 4 will have its vertex at 10, 02. %P ZPV TFF XIZ 5IF WFSUFY JT NJEXBZ CFUXFFO the focus and the directrix.) Since the focus is on the positive y-axis at 10, 42, the equation of this parabola is of the form x2 = 4ay, with a = 4; that is, x2 = 4ay = 4142y = 16y
(8, 4)
c
10 D: y 4
x
a = 4
Letting y = 4 yields x2 = 64, so x = {8. The points (8, 4) and ( - 8, 4) determine the latus rectum. Figure 9 shows the graph of x2 = 16y.
r
696
CHAPTER 9 Analytic Geometry
Finding the Equation of a Parabola
EX A MPL E 5
Find the equation of a parabola with vertex at 10, 02 if its axis of symmetry is the 1 x-axis and its graph contains the point a - , 2b . Find its focus and directrix, and 2 graph the equation.
Solution
The vertex is at the origin, the axis of symmetry is the x-axis, and the graph contains a point in the second quadrant, so the parabola opens to the left. From Table 1, note that the form of the equation is y2 = - 4ax 1 1 Because the point a - , 2b is on the parabola, the coordinates x = - , y = 2 must 2 2 1 satisfy y2 = - 4ax. Substituting x = - and y = 2 into this equation leads to 2
Figure 10 y
D: x 2
5
1 4 = - 4aa - b 2 a = 2
(2, 4) 1 –2 ,
(
2)
V 5 F (2, 0) (0, 0)
5
x
1 y 2 = - 4ax; x = - , y = 2 2
The equation of the parabola is y2 = - 4122x = - 8x
(2, 4)
The focus is at 1 - 2, 02 and the directrix is the line x = 2. Letting x = - 2 gives y2 = 16, so y = {4. The points 1 - 2, 42 and 1 - 2, - 42 determine the latus rectum. See Figure 10.
5
r
Now Work
PROBLEM
27
2 Analyze Parabolas with Vertex at (h, k) If a parabola with vertex at the origin and axis of symmetry along a coordinate axis is shifted horizontally h units and then vertically k units, the result is a parabola with vertex at 1h, k2 and axis of symmetry parallel to a coordinate axis. The equations of such parabolas have the same forms as those in Table 1, but x is replaced by x - h (the horizontal shift) and y is replaced by y - k (the vertical shift). Table 2 gives the forms of the equations of such parabolas. Figures 11(a)–(d) illustrate the graphs for h 7 0, k 7 0. NOTE It is not recommended that Table 2 be memorized. Rather use the ideas of transformations (shift horizontally h units, vertically k units), along with the fact that a represents the distance from the vertex to the focus, to determine the various components of a parabola. It is also helpful to understand that parabolas of the form “x 2 = ” will open up or down, while parabolas of the form “y 2 = ” will open left or right. j
Table 2
Equations of a Parabola: Vertex at (h, k); Axis of Symmetry Parallel to a Coordinate Axis; a + 0 Vertex
Focus
Directrix
Equation
(h, k)
(h + a, k)
x = h - a
( y - k) = 4a(x - h)
Axis of symmetry is parallel to the x-axis, opens right
(h, k)
(h - a, k)
x = h + a
( y - k)2 = - 4a(x - h)
Axis of symmetry is parallel to the x-axis, opens left
(h, k)
(h, k + a)
y = k - a
(x - h)2 = 4a( y - k)
Axis of symmetry is parallel to the y-axis, opens up
(h, k)
(h, k - a)
y = k + a
(x - h)2 = - 4a( y - k)
Axis of symmetry is parallel to the y-axis, opens down
2
Description
SECTION 9.2 The Parabola
Figure 11
Axis of symmetry x5h D: x 5 h 2 a
y
Axis of symmetry y5k
V 5 (h, k) F 5 (h 1 a, k)
y
D: x 5 h 1 a
y
Axis of symmetry x5h
F 5 (h, k 1 a)
Axis of V5 symmetry (h, k) y5k
y D: y 5 k 1 a V 5 (h, k)
V 5 (h, k)
F 5 (h 2 a, k)
x
x
x
x
D: y 5 k 2 a
EXAM PL E 6
Solution Figure 12 D: x 4
F 5 (h, k 2 a) 2 (d) (x 2 h) 5 24a(y 2 k)
Finding the Equation of a Parabola,Vertex Not at the Origin
Find an equation of the parabola with vertex at 1 - 2, 32 and focus at 10, 32 .(SBQI the equation.
The vertex 1 - 2, 32 and focus 10, 32 both lie on the horizontal line y = 3 (the axis of symmetry). The distance a from the vertex 1 - 2, 32 to the focus 10, 32 is a = 2. Also, because the focus lies to the right of the vertex, the parabola opens to the right. Consequently, the form of the equation is 1y - k2 2 = 4a1x - h2
y
where 1h, k2 = 1 - 2, 32 and a = 2. Therefore, the equation is
8 (0, 7) Axis of symmetry y3
V (2, 3) F (0, 3) 6
(c) (x 2 h)2 5 4a(y 2 k)
(b) (y 2 k)2 5 24a(x 2 h)
(a) (y 2 k)2 5 4a(x 2 h)
697
6x
(0, 1)
1y - 32 2 = 4 # 2[x - ( - 2)] 1y - 32 2 = 81x + 22
To find the points that define the latus rectum, let x = 0, so that 1y - 32 2 = 16. Then y - 3 = {4, so y = - 1 or y = 7. The points 10, - 12 and 10, 72 determine the latus rectum; the line x = - 4 is the directrix. See Figure 12.
r
Now Work
PROBLEM
29
4
Polynomial equations define parabolas whenever they involve two variables that are quadratic in one variable and linear in the other.
EXAM PL E 7
Analyzing the Equation of a Parabola Analyze the equation: x2 + 4x - 4y = 0
Solution
x2 + 4x - 4y x2 + 4x x2 + 4x + 4 1x + 22 2
Figure 13 Axis of symmetry x 2
y 4
(4, 0) F (2, 0) 4 V (2, 1) 3
To analyze the equation x2 + 4x - 4y = 0, complete the square involving the variable x.
(0, 0)
4 x
D: y 2
= = = =
0 4y 4y + 4 41y + 12
Isolate the terms involving x on the left side. Complete the square on the left side. Factor.
This equation is of the form 1x - h2 2 = 4a1y - k2, with h = - 2, k = - 1, and a = 1. The graph is a parabola with vertex at 1h, k2 = 1 - 2, - 12 that opens up. The focus is a = 1 unit above the vertex at 1 - 2, 02, and the directrix is the line y = - 2. See Figure 13.
r
Now Work
PROBLEM
47
3 Solve Applied Problems Involving Parabolas P Parabolas find their way into many applications. For example, as discussed in Section 2.6, suspension bridges have cables in the shape of a parabola. Another S property of parabolas that is used in applications is their reflecting property.
698
CHAPTER 9 Analytic Geometry
Suppose that a mirror is shaped like a paraboloid of revolution, a surface formed by rotating a parabola about its axis of symmetry. If a light (or any other emitting source) is placed at the focus of the parabola, all the rays emanating from the light reflect off the mirror in lines parallel to the axis of symmetry. This principle is used in the design of searchlights, flashlights, certain automobile headlights, and other such devices. See Figure 14. Conversely, suppose that rays of light (or other signals) emanate from a distant source so that they are essentially parallel. When these rays strike the surface of a parabolic mirror whose axis of symmetry is parallel to these rays, they are reflected to a single point at the focus. This principle is used in the design of some solar energy devices, satellite dishes, and the mirrors used in some types of telescopes. See Figure 15. Figure 14 Searchlight
Figure 15 Telescope Rays
ht
of lig
Light at focus
EX A MPL E 8
Satellite Dish A satellite dish is shaped like a paraboloid of revolution. The signals that emanate from a satellite strike the surface of the dish and are reflected to a single point, where the receiver is located. If the dish is 8 feet across at its opening and 3 feet deep at its DFOUFS BUXIBUQPTJUJPOTIPVMEUIFSFDFJWFSCFQMBDFE 5IBUJT XIFSFJTUIFGPDVT
Solution
'JHVSF B TIPXTUIFTBUFMMJUFEJTI%SBXUIFQBSBCPMBVTFEUPGPSNUIFEJTIPOB rectangular coordinate system so that the vertex of the parabola is at the origin and its focus is on the positive y-axis. See Figure 16(b).
Figure 16 y
8'
8'
(4, 3)
4
3' 2 4
USA Cable
2
(4, 3) F (0, a)
0
3' 2
4 x
(b)
(a)
The form of the equation of the parabola is x2 = 4ay
and its focus is at 10, a2. Since 14, 32 is a point on the graph, this gives 42 = 4a132 4 a = 3
x 2 = 4ay ; x = 4, y = 3 Solve for a.
1 The receiver should be located 1 feet (1 foot, 4 inches) from the base of the dish, 3 along its axis of symmetry.
Now Work
r
PROBLEM
63
SECTION 9.2 The Parabola
699
9.2 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. The formula for the distance d from P1 = 1x1 , y1 2 to 2 2 P2 = 1x2 , y2 2 is d = 2(x2 - x1) + (y2 - y1) . (p. 35) 2. To complete the square of x2 - 4x, add
4. The point that is symmetric with respect to the x-axis to the point 1 - 2, 52 is ( - 2, - 5 ). (pp. 44–45)
5. To graph y = 1x - 32 2 + 1, shift the graph of y = x2 to the right 3 units and then up 1 unit. (pp. 121–123)
4 . (pp. A38–A39)
3. 6TF UIF 4RVBSF 3PPU .FUIPE UP GJOE UIF SFBM TPMVUJPOT PG 1x + 42 2 = 9. (pp. 170–171) { - 7, - 1}
Concepts and Vocabulary 6. A(n) parabola is the collection of all points in the plane such that the distance from each point to a fixed point equals its distance to a fixed line. Answer Problems 7–10 using the figure. 7. If a (a) (b) (c) (d)
7 0, the equation of the parabola is of the form 1y - k2 2 = 4a1x - h2 1y - k2 2 = - 4a1x - h2 1x - h2 2 = 4a1y - k2 1x - h2 2 = - 4a1y - k2
8. The coordinates of the vertex are
(3, 2)
y
(c)
F
V (3, 2)
.
9. If a = 4, then the coordinates of the focus are 10. If a = 4, then the equation of the directrix is
(3, 6)
x
.
y = -2 .
D
Skill Building In Problems 11–18, the graph of a parabola is given. Match each graph to its equation. (A) y2 = 4x (C) y2 = - 4x (E) 1y - 12 2 = 41x - 12 2 2 (B) x = 4y % x = - 4y (F) 1x + 12 2 = 41y + 12 11.
(
12.
B
13.
14.
E
y
y
y
2
3
2
(2, 1)
( 1y - 12 2 = - 41x - 12 (H) 1x + 12 2 = - 41y + 12 % y 2
(1, 1) (1, 1)
2
2
2 x 2
2
15.
16.
H
2 x
1
18.
C
F
y 2
y
y 2
2
2 x
(2, 1) 2
2
17.
A
y
2
2 x
2
(1, 2)
(1, 1) 2
2 x
2
2 x
2
2 (1, 2)
2
2 x 2
3 (1, 1)
1 x 2
In Problems 19–36, find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.
*19. Focus at 14, 02; vertex at 10, 02
* 21. Focus at 10, - 32; vertex at 10, 02
y2 = 16x x = - 12y 2
* 23. Focus at 1 - 2, 02; directrix the line x = 2 y2 = - 8x 1 * 25. %JSFDUSJYUIFMJOFy = - ; vertex at 10, 02 x2 = 2y 2 * 27. Vertex at 10, 02; axis of symmetry the y-axis; containing the point 12, 32 *29. Vertex at 12, - 32; focus at 12, - 52
* 31. Vertex at 1 - 1, - 22; focus at 10, - 22
1x - 22 2 = - 81y + 32
* 33. Focus at 1 - 3, 42; directrix the line y = 2
* 20. Focus at 10, 22; vertex at 10, 02
* 22. Focus at 1 - 4, 02; vertex at 10, 02
x2 = 8y y2 = - 16x
* 24. Focus at 10, - 12; directrix the line y = 1 x2 = - 4y 1 * 26. %JSFDUSJYUIFMJOFx = - ; vertex at 10, 02 y2 = 2x 2 *28. Vertex at 10, 02; axis of symmetry the x-axis; containing the point 12, 32 *30. Vertex at 14, - 22; focus at 16, - 22
1y + 22 2 = 41x + 12 *32. Vertex at 13, 02; focus at 13, - 22
1y + 22 2 = 81x - 42
1x - 32 2 = - 8y
1x + 32 2 = 41y - 32 *34. Focus at 12, 42; directrix the line x = - 4 1y - 42 2 = 121x + 12
* 35. Focus at 1 - 3, - 22; directrix the line x = 1
*36. Focus at 1 - 4, 42; directrix the line y = - 2
%VFUPTQBDFSFTUSJDUJPOT BOTXFSTUPUIFTFFYFSDJTFTNBZCFGPVOEJOUIF"OTXFSTJOUIFCBDLPGUIFCPPL
700
CHAPTER 9 Analytic Geometry
In Problems 37–54, find the vertex, focus, and directrix of each parabola. Graph the equation. *37. x2 = 4y
*38. y2 = 8x
*41. 1y - 22 2 = 81x + 12
*39. y2 = - 16x
*42. 1x + 42 2 = 161y + 22 *46. 1x - 22 = 41y - 32
*45. 1y + 32 = 81x - 22 2
2
*49. x2 + 8x = 4y - 8
*50. y2 - 2y = 8x - 1
*53. x - 4x = y + 4
*54. y + 12y = - x + 1
2
*40. x2 = - 4y
*43. 1x - 32 2 = - 1y + 12
*44. 1y + 12 2 = - 41x - 22
*47. y - 4y + 4x + 4 = 0
*48. x2 + 6x - 4y + 1 = 0
*51. y2 + 2y - x = 0
*52. x2 - 4x = 2y
2
2
In Problems 55–62, write an equation for each parabola. y
*55.
*56.
2
2
(1, 2)
y
*58.
2
2
(2, 1)
2
x
2
2
(2, 0)
2
2
x
*60.
2
(0, 1)
(0, 1) 2
2 2
2
y
*62.
2 (0, 1) (1, 0)
2
x
x
(0, 1)
2 (0, 1)
2
2 2
y
*61.
y
(2, 2)
2
x
(1, 0) 2
2
y
2
(1, 2)
(2, 1)
(0, 1)
*59.
y
*57.
y
x
(1, 1)
2
(2, 0)
2
x
2
x
2
2
Applications and Extensions *63. Satellite Dish A satellite dish is shaped like a paraboloid of revolution. The signals that emanate from a satellite strike the surface of the dish and are reflected to a single point, where the receiver is located. If the dish is 10 feet across at its opening and 4 feet deep at its center, at what position TIPVMEUIFSFDFJWFSCFQMBDFE *64. Constructing a TV Dish A cable TV receiving dish is in the shape of a paraboloid of revolution. Find the location of the receiver, which is placed at the focus, if the dish is 6 feet across at its opening and 2 feet deep. *65. Constructing a Flashlight The reflector of a flashlight is in the shape of a paraboloid of revolution. Its diameter is 4 inches and its depth is 1 inch. How far from the vertex should the light bulb be placed so that the rays will be SFGMFDUFEQBSBMMFMUPUIFBYJT 66. Constructing a Headlight A sealed-beam headlight is in the shape of a paraboloid of revolution. The bulb, which is placed at the focus, is 1 inch from the vertex. If the depth is to be 2 inches, XIBUJTUIFEJBNFUFSPGUIFIFBEMJHIUBUJUTPQFOJOH 422 in. 67. Suspension Bridge The cables of a suspension bridge are in the shape of a parabola, as shown in the figure. The towers supporting the cable are 600 feet apart and 80 feet high. If the cables touch the road surface midway between the towers, what is the height of the cable from the road at a QPJOUGFFUGSPNUIFDFOUFSPGUIFCSJEHF 20 ft
400 feet apart and 100 feet high. If the cables are at a height of 10 feet midway between the towers, what is the height of the DBCMFBUBQPJOUGFFUGSPNUIFDFOUFSPGUIFCSJEHF 15.625 ft 69. Searchlight A searchlight is shaped like a paraboloid of revolution. If the light source is located 2 feet from the base along the axis of symmetry and the opening is 5 feet across, IPXEFFQTIPVMEUIFTFBSDIMJHIUCF 0.78125 ft 70. Searchlight A searchlight is shaped like a paraboloid of revolution. If the light source is located 2 feet from the base along the axis of symmetry and the depth of the searchlight JTGFFU XIBUTIPVMEUIFXJEUIPGUIFPQFOJOHCF 822 ft *71. Solar Heat A mirror is shaped like a paraboloid of revolution and is used to concentrate the rays of the sun at its focus, creating a heat source. See the figure. If the mirror is 20 feet across at its opening and is 6 feet deep, where will UIFIFBUTPVSDFCFDPODFOUSBUFE Sun’s rays
20'
6'
600 ft
*72. Reflecting Telescope A reflecting telescope contains a mirror shaped like a paraboloid of revolution. If the mirror is 4 inches across at its opening and is 3 inches deep, where XJMMUIFDPMMFDUFEMJHIUCFDPODFOUSBUFE
68. Suspension Bridge The cables of a suspension bridge are in the shape of a parabola. The towers supporting the cable are
*73. Parabolic Arch Bridge A bridge is built in the shape of a parabolic arch. The bridge has a span of 120 feet and a maximum height of 25 feet. See the illustration. Choose a
80 ft ? 150 ft
SECTION 9.3 The Ellipse
701
(c) %PUIFEBUBTVQQPSUUIFOPUJPOUIBUUIF"SDIJTJOUIF TIBQFPGBQBSBCPMB No Source: gatewayarch.com
suitable rectangular coordinate system and find the height of the arch at distances of 10, 30, and 50 feet from the center.
*76. Show that an equation of the form Ax2 + Ey = 0,
74. Parabolic Arch Bridge A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch a distance of 40 feet from the center is to be 10 feet. Find the height of the arch at its center. 27.78 ft 75. Gateway Arch 5IF (BUFXBZ "SDI JO 4U -PVJT JT PGUFO mistaken to be parabolic in shape. In fact, it is a catenary, which has a more complicated formula than a parabola. The Arch is 630 feet high and 630 feet wide at its base. *(a) Find the equation of a parabola with the same dimensions. Let x equal the horizontal distance from the center of the arc. (b) The following table gives the height of the Arch at various widths; find the corresponding heights for the parabola found in (a). 119.7 ft, 267.3 ft, 479.4 ft Width (ft)
A ≠ 0, E ≠ 0
is the equation of a parabola with vertex at 10, 02 and axis of symmetry the y-axis. Find its focus and directrix.
25 ft 120 ft
*77. Show that an equation of the form Cy2 + Dx = 0,
is the equation of a parabola with vertex at 10, 02 and axis of symmetry the x-axis. Find its focus and directrix. *78. Show that the graph of an equation of the form Ax2 + Dx + Ey + F = 0,
100
478
312.5
308
525
A ≠ 0
(a) Is a parabola if E ≠ 0. (b) Is a vertical line if E = 0 and D2 - 4AF = 0. (c) Is two vertical lines if E = 0 and D2 - 4AF 7 0. (d) Contains no points if E = 0 and D2 - 4AF 6 0. *79. Show that the graph of an equation of the form Cy2 + Dx + Ey + F = 0,
Height (ft)
567
C ≠ 0, D ≠ 0
C ≠ 0
(a) Is a parabola if D ≠ 0. (b) Is a horizontal line if D = 0 and E 2 - 4CF = 0. (c) Is two horizontal lines if D = 0 and E 2 - 4CF 7 0. (d) Contains no points if D = 0 and E 2 - 4CF 6 0.
Retain Your Knowledge Problems 80–83 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 3 2210 83. Find the exact value: tan c cos - 1 a - b d *80. For x = 9y2 - 36, list the intercepts and test for symmetry. 7 3 81. Solve: 4x + 1 = 8x - 1 5 56 5 p 6 u 6 p, find the exact value of *82. (JWFO tan u = - , 8 2 each of the remaining trigonometric functions.
‘Are You Prepared?’ Answers 1. 21x2 - x1 2 2 + 1y2 - y1 2 2
2. 4
3. x + 4 = {3; 5 - 7, - 16
4. 1 - 2, - 52
5. 3; up
9.3 The Ellipse PREPARING FOR THIS SECTION Before getting started, review the following: r %JTUBODF'PSNVMB 'PVOEBUJPOT 4FDUJPO Q
r $PNQMFUJOHUIF4RVBSF "QQFOEJY" 4FDUJPO" pp. A38–A39) r *OUFSDFQUT 'PVOEBUJPOT 4FDUJPO QQm
r 4ZNNFUSZ 'PVOEBUJPOT 4FDUJPO QQm
r $JSDMFT 'PVOEBUJPOT 4FDUJPO QQm
r (SBQIJOH5FDIOJRVFT5SBOTGPSNBUJPOT 4FDUJPO pp. 121–129)
Now Work the ‘Are You Prepared?’ problems on page 708.
OBJECTIVES 1 Analyze Ellipses with Center at the Origin (p. 702) 2 Analyze Ellipses with Center at (h, k) (p. 706) 3 Solve Applied Problems Involving Ellipses (p. 707)
702
CHAPTER 9 Analytic Geometry
DEFINITION
Figure 17
Minor axis Major axis
P
Center V1
F2
V2
F1
Figure 18
An ellipse is the collection of all points in the plane the sum of whose distances from two fixed points, called the foci, is a constant. The definition contains within it a physical means for drawing an ellipse. Find a piece of string (the length of this string is the constant referred to in the definition). Then take two thumbtacks (the foci) and stick them into a piece of cardboard so that the distance between them is less than the length of the string. Now attach the ends of the string to the thumbtacks and, using the point of a pencil, pull the string taut. See Figure 17. Keeping the string taut, rotate the pencil around the two thumbtacks. The pencil traces out an ellipse, as shown in Figure 17. In Figure 17, the foci are labeled F1 and F2 . The line containing the foci is called the major axis. The midpoint of the line segment joining the foci is the center of the ellipse. The line through the center and perpendicular to the major axis is the minor axis. The two points of intersection of the ellipse and the major axis are the vertices, V1 and V2 , of the ellipse. The distance from one vertex to the other is the length of the major axis. The ellipse is symmetric with respect to its major axis, with respect to its minor axis, and with respect to its center.
y
1 Analyze Ellipses with Center at the Origin
P ⫽ (x, y ) d(F1, P )
d(F2, P )
F1 ⫽ (⫺c, 0)
F2 ⫽ (c, 0)
x
With these ideas in mind, we are ready to find the equation of an ellipse in a rectangular coordinate system. First, place the center of the ellipse at the origin. Second, position the ellipse so that its major axis coincides with a coordinate axis, say the x-axis, as shown in Figure 18. If c is the distance from the center to a focus, one focus is at F1 = 1 - c, 02 and the other at F2 = 1c, 02. As we shall see, it is convenient to let 2a denote the constant distance referred to in the definition. Then, if P = 1x, y2 is any point on the ellipse, d1F1 , P2 + d 1F2 , P2 = 2a
2 1x - ( - c)2 2 + (y - 0)2 + 2 1x - c2 2 + (y - 0)2 = 2a 2 1x + c2 2 + y2 = 2a - 2 1x - c2 2 + y2
The sum of the distances from P to the foci equals a constant, 2a. Use the Distance Formula. Isolate one radical.
1x + c2 2 + y2 = 4a2 - 4a2 1x - c2 2 + y2 Square both sides. + 1x - c2 2 + y2
x2 + 2cx + c 2 + y2 = 4a2 - 4a2 1x - c2 2 + y2 Square binomials. + x2 - 2cx + c 2 + y2
4cx - 4a2 = - 4a2 1x - c2 2 + y2 cx - a = - a2 1x - c2 + y 2
2
Simplify; isolate the radical.
2
Divide each side by 4.
1cx - a 2 = a 3 1x - c2 + y 4 2 2
2
2
2
Square both sides again.
c x - 2a cx + a = a 1x - 2cx + c + y 2 2 2
2
4
2
2
1c 2 - a2 2x2 - a2 y2 = a2 c 2 - a4 1a2 - c 2 2x2 + a2 y2 = a2 1a2 - c 2 2
2
2
Square binomials. Rearrange the terms. Multiply each side by - 1; factor a 2 on the right side.
(1)
To obtain points on the ellipse off the x-axis, it must be that a 7 c. To see why, look again at Figure 18. Then d 1F1 , P2 + d 1F2 , P2 7 d 1F1 , F2 2 2a 7 2c a 7 c
The sum of the lengths of two sides of a triangle is greater than the length of the third side. d(F1 , P ) + d(F2 , P ) = 2a, d(F1 , F2) = 2c
SECTION 9.3 The Ellipse
703
Because a 7 c 7 0, this means a2 7 c 2, so a2 - c 2 7 0. Let b2 = a2 - c 2, b 7 0. Then a 7 b and equation (1) can be written as b2 x2 + a2 y2 = a2 b2 y2 x2 + = 1 a2 b2
Divide each side by a 2 b 2.
As you can verify, the graph of this equation has symmetry with respect to the x-axis, the y-axis, and the origin. Because the major axis is the x-axis, find the vertices of this ellipse by letting x2 y = 0. The vertices satisfy the equation 2 = 1, the solutions of which are x = {a. a Consequently, the vertices of this ellipse are V1 = 1 - a, 02 and V2 = 1a, 02. The y-intercepts of the ellipse, found by letting x = 0, have coordinates 10, - b2 and 10, b2 . These four intercepts, 1a, 02, 1 - a, 02, 10, b2, and 10, - b2, are used to graph the ellipse.
THEOREM
Equation of an Ellipse: Center at (0, 0); Major Axis along the x-Axis An equation of the ellipse with center at 10, 02, foci at 1 - c, 02 and 1c, 02 , and vertices at 1 - a, 02 and 1a, 02 is y2 x2 + = 1, a2 b2
where a 7 b 7 0 and b2 = a2 - c 2
(2)
The major axis is the x-axis. See Figure 19. Figure 19
y (0, b) V1 ⫽ (⫺a, 0)
a V2 ⫽ (a, 0) c x F1 ⫽ (⫺c, 0) F2 ⫽ (c, 0) b
(0, ⫺b)
Notice in Figure 19 the right triangle formed by the points 10, 02 , 1c, 02, and 10, b2. Because b2 = a2 - c 2 (or b2 + c 2 = a2), the distance from the focus at 1c, 02 to the point 10, b2 is a. This can be seen another way. Look at the two right triangles in Figure 19. They BSFDPOHSVFOU%PZPVTFFXIZ #FDBVTFUIFTVNPGUIFEJTUBODFTGSPNUIFGPDJUPB point on the ellipse is 2a, it follows that the distance from (c, 0) to (0, b) is a.
EXAM PL E 1
Finding an Equation of an Ellipse
y
The ellipse has its center at the origin, and because the given focus and vertex lie on the x-axis, the major axis is the x-axis. The distance from the center, 10, 02 , to one of the foci, 13, 02, is c = 3. The distance from the center, 10, 02, to one of the vertices, 1 - 4, 02 , is a = 4. From equation (2), it follows that
Figure 20
Solution
5 (0, 7 )
F1 ⫽ (⫺3, 0)
F2 ⫽ (3, 0)
⫺5
5 x
V1 ⫽ (⫺4, 0)
(0, ⫺ 7 ) ⫺5
Find an equation of the ellipse with center at the origin, one focus at 13, 02, and a vertex at 1 - 4, 02.(SBQIUIFFRVBUJPO
b2 = a2 - c 2 = 16 - 9 = 7 so an equation of the ellipse is
V2 ⫽ (4, 0)
y2 x2 + = 1 16 7 Figure 20 shows the graph.
r
704
CHAPTER 9 Analytic Geometry
In Figure 20 the intercepts of the equation are used to graph the ellipse. Following this practice will make it easier for you to obtain an accurate graph of an ellipse. COMMENT The intercepts of the ellipse also provide information about how to set the viewing rectangle for graphing an ellipse. To graph the ellipse y2 x2 + = 1 16 7 discussed in Example 1, we set the viewing rectangle using a square screen that includes the intercepts, perhaps - 4.5 … x … 4.5, - 3 … y … 3. Then we proceed to solve the equation for y: y2 x2 + = 1 16 7 y2 x2 = 1 7 16 Figure 21
y2 = 7¢1 -
(
x2 3 Y1 7 1 16
) y = {
4.5
4.5
(
)
x2 ≤ 16
7¢1 -
x2 from each side. 16
Multiply both sides by 7. x2 ≤ 16
Take the square root of each side.
Now graph the two functions Y1 =
3 x2 Y2 7 1 16
B
Subtract
C
7¢1 -
x2 x2 ≤ and Y2 = - 7¢1 ≤ 16 C 16
Figure 21 shows the result.
Now Work
■
PROBLEM
27
An equation of the form of equation (2), with a 7 b, is the equation of an ellipse with center at the origin, foci on the x-axis at 1 - c, 02 and 1c, 02 , where c 2 = a2 - b2, and major axis along the x-axis. For the remainder of this section, the direction “Analyze the equation” will mean to find the center, major axis, foci, and vertices of the ellipse and graph it.
EX A MPL E 2
Analyzing the Equation of an Ellipse Analyze the equation:
Solution
y2 x2 + = 1 25 9
The given equation is of the form of equation (2), with a2 = 25 and b2 = 9. The equation is that of an ellipse with center 10, 02 and major axis along the x-axis. The vertices are at 1 {a, 02 = 1 {5, 02. Because b2 = a2 - c 2, this means c 2 = a2 - b2 = 25 - 9 = 16
The foci are at 1 {c, 02 = 1 {4, 02. Figure 22 shows the graph. Figure 22
y 6 (0, 3) V1 (5, 0) F (4, 0) 1
F 2 (4, 0) V2 (5, 0) 6 x
6
(0, 3)
Now Work
PROBLEM
17
r
SECTION 9.3 The Ellipse
705
If the major axis of an ellipse with center at 10, 02 lies on the y-axis, the foci are at 10, - c2 and 10, c2. Using the same steps as before, the definition of an ellipse leads to the following result:
THEOREM
y F 2 (0, c) c
(b, 0) x
F 1 (0, c) V 1 (0, a)
Analyzing the Equation of an Ellipse
EXAM PL E 3
Analyze the equation: 9x2 + y2 = 9
Figure 24
Solution y 3 V2 (0, 3)
To put the equation in proper form, divide each side by 9. x2 +
F2 (0, 2 2)
(1, 0)
3
EXAM PL E 4
Solution Figure 25 y 3 V 2 (0, 3)
PROBLEM
21
Finding an Equation of an Ellipse
Find an equation of the ellipse having one focus at 10, 22 and vertices at 10, - 32 and 10, 32.(SBQIUIFFRVBUJPO By plotting the given focus and vertices, note that the major axis is the y-axis. Because the vertices are at 10, - 32 and 10, 32, the center of this ellipse is at their midpoint, the origin. The distance from the center, 10, 02, to one of the foci, 10, 22, is c = 2. The distance from the center, 10, 02, to one of the vertices, 10, 32, is a = 3. So b2 = a2 - c 2 = 9 - 4 = 5. The form of the equation of this ellipse is given by equation (3). y2 x2 + = 1 b2 a2
( 5 , 0) 3 x
3
F 1 (0, 2)
The larger denominator, 9, is in the y2@term so, based on equation (3), this is the equation of an ellipse with center at the origin and major axis along the y-axis. Also, a2 = 9, b2 = 1, and c 2 = a2 - b2 = 9 - 1 = 8. The vertices are at 10, {a2 = 10, {32, and the foci are at 10, {c2 = 10, {2222. The x-intercepts are at 1 {b, 02 = 1 {1, 02 . Figure 24 shows the graph.
Now Work
3 V (0, 3) 1
( 5 , 0)
x
y2 = 1 9
r
F1 (0, 2 2)
F 2 (0, 2)
(3)
Figure 23 illustrates the graph of such an ellipse. Again, notice the right triangle formed by the points at 10, 02, 1b, 02, and 10, c2 , so that a2 = b2 + c 2 (or b2 = a2 - c 2). Look closely at equations (2) and (3). Although they may look alike, there is a difference! In equation (2), the larger number, a2, is in the denominator of the x2@term, so the major axis of the ellipse is along the x-axis. In equation (3), the larger number, a2, is in the denominator of the y2@term, so the major axis is along the y-axis.
a b
where a 7 b 7 0 and b2 = a2 - c 2
The major axis is the y-axis.
V 2 (0, a )
3 (1, 0)
An equation of the ellipse with center at 10, 02, foci at 10, - c2 and 10, c2 , and vertices at 10, - a2 and 10, a2 is y2 x2 + = 1, b2 a2
Figure 23
(b, 0)
Equation of an Ellipse: Center at (0, 0); Major Axis along the y-Axis
y2 x2 + = 1 5 9 3 V1 (0, 3)
r
Figure 25 shows the graph.
Now Work
PROBLEM
29
706
CHAPTER 9 Analytic Geometry
The circle may be considered a special kind of ellipse. To see why, let a = b in equation (2) or (3). Then y2 x2 + = 1 a2 a2 x 2 + y 2 = a2 This is the equation of a circle with center at the origin and radius a. The value of c is c 2 = a 2 - b2 = 0 c
a = b
This indicates that the closer the two foci of an ellipse are to the center, the more the ellipse will look like a circle.
2 Analyze Ellipses with Center at (h, k) If an ellipse with center at the origin and major axis coinciding with a coordinate axis is shifted horizontally h units and then vertically k units, the result is an ellipse with center at 1h, k2 and major axis parallel to a coordinate axis. The equations of such ellipses have the same forms as those given in equations (2) and (3), except that x is replaced by x - h (the horizontal shift) and y is replaced by y - k (the vertical shift). Table 3 gives the forms of the equations of such ellipses, and Figure 26 shows their graphs.
Table 3 NOTE It is not recommended that Table 3 be memorized. Rather, use the ideas of transformations (horizontally h units, vertically k units), along with the fact that a represents the distance from the center to the vertices, c represents the distance from the center to the foci, and b 2 = a 2 - c 2 (or c 2 = a 2 - b 2).
Equations of an Ellipse: Center at (h, k); Major Axis Parallel to a Coordinate Axis Center
Major Axis
Foci
Vertices
(h, k)
Parallel to the x-axis
(h + c, k)
(h + a, k)
(h - c, k)
(h - a, k)
(h, k + c)
(h, k + a)
(h, k - c)
(h, k - a)
(h, k)
Parallel to the y-axis
Equation (y - k)2 (x - h)2 + = 1, a2 b2 2 a 7 b 7 0 and b = a2 - c2 (y - k)2 (x - h)2 + = 1, 2 b a2 2 a 7 b 7 0 and b = a2 - c2
j
y
Figure 26 y
Major axis (h, k a) (h, k c)
(h c, k) Major axis (h a, k)
(h c, k) (h , k) (h , k)
(h a, k) x
(x h)2 (y k)2 (a) –––––– –––––– 1 a2 b2
EX A MPL E 5
Solution
x (h, k a)
(h, k c)
(x h)2 (y k)2 (b) –––––– –––––– 1 b2 a2
Finding an Equation of an Ellipse, Center Not at the Origin
Find an equation for the ellipse with center at 12, - 32, one focus at 13, - 32 , and one vertex at 15, - 32.(SBQIUIFFRVBUJPO
The center is at 1h, k2 = 12, - 32, so h = 2 and k = - 3. Plot the center, focus, and vertex, and note that the points all lie on the line y = - 3. Hence the major axis
SECTION 9.3 The Ellipse
707
is parallel to the x-axis. The distance from the center 12, - 32 to a focus 13, - 32 is c = 1; the distance from the center 12, - 32 to a vertex 15, - 32 is a = 3. Then b2 = a2 - c 2 = 9 - 1 = 8. The form of the equation is 1x - h2 2
+
1y - k2 2
= 1, where h = 2, k = - 3, a = 3, b = 2 22 a2 b2 1x - 22 2 1y + 32 2 + = 1 9 8
To graph the equation, use the center 1h, k2 = 12, - 32 to locate the vertices. The major axis is parallel to the x-axis, so the vertices are a = 3 units left and right of the center 12, - 32. Therefore, the vertices are V1 = 12 - 3, - 32 = 1 - 1, - 32
Figure 27 y 2
V1 (1, 3)
Since c = 1 and the major axis is parallel to the x-axis, the foci are 1 unit left and right of the center. Therefore, the foci are
(2, 3 2 2 )
F1 = 12 - 1, - 32 = 11, - 32
6 x
2 F1
F2
V2 (5, 3)
(2, 3 2 2 )
- 222 2
and
1 2, - 3
+ 222 2
r
Figure 27 shows the graph.
Now Work
EXAM PL E 6
and F2 = 12 + 1, - 32 = 13, - 32
Finally, we use the value of b = 222 to find the two points above and below the center.
1 2, - 3
(2, 3)
and V2 = 12 + 3, - 32 = 15, - 32
PROBLEM
55
Analyzing the Equation of an Ellipse Analyze the equation: 4x2 + y2 - 8x + 4y + 4 = 0
Solution
Complete the squares in x and in y. 4x2 + y2 - 8x + 4y + 4 = 0 4x2 - 8x + y2 + 4y = - 4 41x2 - 2x2 + 1y2 + 4y2 = - 4
Figure 28 y (1, 0) x (1, 2 3 ) (0, 2)
4
(1, 2)
(1, 4)
(2, 2) (1, 2 3 )
41x2 - 2x + 12 + 1y2 + 4y + 42 = - 4 + 4 + 4 41x - 12 2 + 1y + 22 2 = 4 1y + 22 2 1x - 12 2 + = 1 4
Group like variables; place the constant on the right side. Factor out 4 from the first two terms. Complete each square. Factor. Divide each side by 4.
This is the equation of an ellipse with center at 11, - 22 and major axis parallel to the y-axis. Since a2 = 4 and b2 = 1, we have c 2 = a2 - b2 = 4 - 1 = 3. The vertices are at 1h, k { a2 = 11, - 2 { 22 , or 11, 02 and 11, - 42. The foci are at 1h, k { c2 = 11, - 2 { 232 , or 11, - 2 - 232 and 11, - 2 + 232. Figure 28 shows the graph.
r
Now Work
PROBLEM
47
3 Solve Applied Problems Involving Ellipses Ellipses are found in many applications in science and engineering. For example, the orbits of the planets around the Sun are elliptical, with the Sun’s position at a focus. See Figure 29 on the following page.
708
CHAPTER 9 Analytic Geometry
Figure 29
Venus
Jupiter
Mars Earth
Asteroids
Stone and concrete bridges are often shaped as semielliptical arches. Elliptical gears are used in machinery when a variable rate of motion is required. Ellipses also have an interesting reflection property. If a source of light (or sound) is placed at one focus, the waves transmitted by the source reflect off the ellipse and concentrate at the other focus. This is the principle behind whispering galleries, which are rooms designed with elliptical ceilings. A person standing at one focus of the ellipse can whisper and be heard by a person standing at the other focus, because all the sound waves that reach the ceiling are reflected to the other person.
EX A MPL E 7
A Whispering Gallery 5IF XIJTQFSJOH HBMMFSZ JO UIF .VTFVN PG 4DJFODF BOE *OEVTUSZ JO $IJDBHP JT 47.3 feet long. The distance from the center of the room to the foci is 20.3 feet. Find BOFRVBUJPOUIBUEFTDSJCFTUIFTIBQFPGUIFSPPN)PXIJHIJTUIFSPPNBUJUTDFOUFS Source: Chicago Museum of Science and Industry Web site; www.msichicago.org
Solution
Set up a rectangular coordinate system so that the center of the ellipse is at the origin and the major axis is along the x-axis. The equation of the ellipse is y2 x2 + = 1 a2 b2 Since the length of the room is 47.3 feet, the distance from the center of the room 47.3 to each vertex (the end of the room) is = 23.65 feet, so a = 23.65 feet. The 2 distance from the center of the room to each focus is c = 20.3 feet. See Figure 30. Because b2 = a2 - c 2, this means b2 = 23.652 - 20.32 = 147.2325. An equation that describes the shape of the room is given by
Figure 30
y2 x2 + = 1 147.2325 23.652
y 15 (0, 12.1) (⫺23.65, 0) ⫺25 (⫺20.3, 0)
(23.65, 0)
The height of the room at its center is b = 2147.2325 ≈ 12.1 feet.
25 x (20.3, 0)
Now Work
PROBLEM
r
71
9.3 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. The distance d from P1 = 12, - 52 to P2 = 14, - 22 is d = 213 . (p. 35) 9 2. To complete the square of x2 - 3x, add 4 . (pp. A38–A39) 3. The intercepts of the equation y2 = 16 - 4x2 are ( - 2, 0), (2, 0), (0, - 4), (0, 4) . (pp. 43–44) 4. The point that is symmetric with respect to the y-axis to the point 1 - 2, 52 is (2, 5) . (pp. 44–45)
5. To graph y = 1x + 12 2 - 4, shift the graph of y = x2 to the (left/right) left 1 unit(s) and then (up/down) down 4 unit(s). (pp. 121–123)
6. The standard equation of a circle with center at 12, - 32 and radius 1 is . (pp. 66–67) (x - 2)2 + (y + 3)2 = 1
709
SECTION 9.3 The Ellipse
Concepts and Vocabulary 7. A(n) ellipse is the collection of all points in the plane the sum of whose distances from two fixed points is a constant. 8. For an ellipse, the foci lie on a line called the major axis. y2 x2 9. For the ellipse + = 1, the vertices are the points 4 25 (0, - 5) and (0, 5) . 10. For the ellipse value of b is
11. If the center of an ellipse is 12, - 32, the major axis is parallel to the x-axis, and the distance from the center of the ellipse to its vertices is a = 4 units, then the coordinates of the vertices are ( - 2, - 3) and (6, - 3) .
12. If the foci of an ellipse are 1 - 4, 42 and 16, 42 , then the coordinates of the center of the ellipse are (1, 4) .
y2 x2 + = 1, the value of a is 5 , the 25 9 3 , and the major axis is the x -axis.
Skill Building In Problems 13–16, the graph of an ellipse is given. Match each graph to its equation. y2 y2 x2 x2 (A) + y2 = 1 (B) x2 + = 1 (C) + = 1 4 4 16 4 13. C
14. %
y 4
15. B
y
%
y2 x2 + = 1 4 16
16. A
y 3
y 3
2 ⫺2
2
⫺4
x
⫺3
4 x
⫺2
⫺4
⫺3
3 x
3x ⫺3
⫺3
In Problems 17–26, find the vertices and foci of each ellipse. Graph each equation. y2 y2 y2 x2 x2 x2 + = 1 *18. + = 1 *19. + = 1 *17. 25 4 9 4 9 25 *21. 4x2 + y2 = 16
*22. x2 + 9y2 = 18
*25. x2 + y2 = 16
*26. x2 + y2 = 4
*20. x2 +
*23. 4y2 + x2 = 8
In Problems 27–38, find an equation for each ellipse. Graph the equation. *27. Center at 10, 02; focus at 13, 02; vertex at 15, 02
y2 = 1 16
*24. 4y2 + 9x2 = 36
*28. Center at 10, 02; focus at 1 - 1, 02; vertex at 13, 02
*29. Center at 10, 02; focus at 10, - 42; vertex at 10, 52
*30. Center at 10, 02; focus at 10, 12; vertex at 10, - 22
*33. Focus at 1 - 4, 02; vertices at 1 {5, 02
*34. Focus at 10, - 42; vertices at 10, {82
*37. Center at 10, 02; vertex at 10, 42; b = 1
*38. Vertices at 1 {5, 02; c = 2
*31. Foci at 1 {2, 02; length of the major axis is 6
*32. Foci at 10, {22; length of the major axis is 8
*35. Foci at 10, {32; x-intercepts are {2
*36. Vertices at 1 {4, 02; y-intercepts are {1
In Problems 39–42, write an equation for each ellipse. *39. (⫺1, 1)
*40.
y 3
*41.
y 3
*42.
y 3
y 3 (0, 1)
⫺3
3 x
⫺3
⫺3
(⫺1, ⫺1)
3 x
⫺3
⫺3
(1, 0)
3 x
⫺3
⫺3
3 x
⫺3
In Problems 43–54, analyze each equation; that is, find the center, foci, and vertices of each ellipse. Graph each equation. 1x - 32 2 1y + 12 2 1x + 42 2 1y + 22 2 *43. + = 1 *44. + = 1 *45. 1x + 52 2 + 41y - 42 2 = 16 4 9 9 4 *46. 91x - 32 2 + 1y + 22 2 = 18 *47. x2 + 4x + 4y2 - 8y + 4 = 0 *48. x2 + 3y2 - 12y + 9 = 0 *49. 2x2 + 3y2 - 8x + 6y + 5 = 0
*50. 4x2 + 3y2 + 8x - 6y = 5
*51. 9x2 + 4y2 - 18x + 16y - 11 = 0
*52. x2 + 9y2 + 6x - 18y + 9 = 0
*53. 4x2 + y2 + 4y = 0
*54. 9x2 + y2 - 18x = 0
%VFUPTQBDFSFTUSJDUJPOT BOTXFSTUPUIFTFFYFSDJTFTNBZCFGPVOEJOUIF"OTXFSTJOUIFCBDLPGUIFCPPL
710
CHAPTER 9 Analytic Geometry
In Problems 55–64, find an equation for each ellipse. Graph the equation. *55. Center at 12, - 22; vertex at 17, - 22; focus at 14, - 22
*56. Center at 1 - 3, 12; vertex at 1 - 3, 32; focus at 1 - 3, 02
*57. Vertices at 14, 32 and 14, 92; focus at 14, 82
*58. Foci at 11, 22 and 1 - 3, 22; vertex at 1 - 4, 22
*61. Center at 11, 22; focus at 14, 22; contains the point 11, 32
*62. Center at 11, 22; focus at 11, 42; contains the point 12, 22
*59. Foci at 15, 12 and 1 - 1, 12; length of the major axis is 8
*63. Center at 11, 22; vertex at 14, 22; contains the point 11, 52
*60. Vertices at 12, 52 and 12, - 12; c = 2
*64. Center at 11, 22; vertex at 11, 42; contains the point 11 + 23, 32
In Problems 65–68, graph each function. Be sure to label all the intercepts. [Hint: Notice that each function is half an ellipse.] *66. f 1x2 = 29 - 9x2
*65. f 1x2 = 216 - 4x2
*67. f 1x2 = - 264 - 16x2
*68. f 1x2 = - 24 - 4x2
Applications and Extensions 69. Semielliptical Arch Bridge An arch in the shape of the upper half of an ellipse is used to support a bridge that is to span a river 20 meters wide. The center of the arch is 6 meters above the center of the river. See the figure. Write an equation for the ellipse in which the x-axis coincides with the water level and the y-axis passes through the center of the arch. y2 x2 + = 1 100 36
74. Semielliptical Arch Bridge A bridge is to be built in the shape of a semielliptical arch and is to have a span of 100 feet. The height of the arch, at a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center. 16.67 ft 75. Racetrack Design Consult the figure. A racetrack is in the shape of an ellipse 100 feet long and 50 feet wide. What is UIFXJEUIGFFUGSPNBWFSUFY 30 ft
10 ft 6m
?
100 ft 50 ft
20 m
*70. Semielliptical Arch Bridge The arch of a bridge is a semiellipse with a horizontal major axis. The span is 30 feet, and the top of the arch is 10 feet above the major axis. The roadway is horizontal and is 2 feet above the top of the arch. Find the vertical distance from the roadway to the arch at 5-foot intervals along the roadway. 71. Whispering Gallery A hall 100 feet in length is to be designed as a whispering gallery. If the foci are located 25 feet from the DFOUFS IPXIJHIXJMMUIFDFJMJOHCFBUUIFDFOUFS 43.3 ft 72. Whispering Gallery Jim, standing at one focus of a whispering gallery, is 6 feet from the nearest wall. His friend is standing at the other focus, 100 feet away. What is the MFOHUIPGUIJTXIJTQFSJOHHBMMFSZ )PXIJHIJTJUTFMMJQUJDBM DFJMJOHBUUIFDFOUFS 112 ft; 25.2 ft 73. Semielliptical Arch Bridge A bridge is built in the shape of a semielliptical arch. The bridge has a span of 120 feet and a maximum height of 25 feet. Choose a suitable rectangular coordinate system and find the height of the arch at distances of 10, 30, and 50 feet from the center. 24.65 ft, 21.65 ft, 13.82 ft
76. Semielliptical Arch Bridge An arch for a bridge over a highway is in the form of half an ellipse. The top of the arch is 20 feet above the ground level (the major axis). The highway has four lanes, each 12 feet wide; a center safety strip 8 feet wide; and two side strips, each 4 feet wide. What should the span of the bridge be (the length of its major axis) if the IFJHIUGFFUGSPNUIFDFOUFSJTUPCFGFFU 73.69 ft *77. Installing a Vent Pipe A homeowner is putting in a fireplace that has a 4-inch-radius vent pipe. He needs to cut an elliptical hole in his roof to accommodate the pipe. If the pitch of his roof 5 is BSJTFPG SVOPG
XIBUBSFUIFEJNFOTJPOTPGUIFIPMF 4 Source: www.doe.virginia.gov 78. Volume of a Football A football is in the shape of a prolate spheroid, which is simply a solid obtained by rotating an y2 x2 ellipse a 2 + 2 = 1b about its major axis. An inflated NFL a b football averages 11.125 inches in length and 28.25 inches in center circumference. If the volume of a prolate spheroid 4 is pab2 IPXNVDIBJSEPFTUIFGPPUCBMMDPOUBJO /FHMFDU 3 material thickness). ≈471 in.3 Source: www.answerbag.com
In Problems 79–83, use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion of a planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from the Sun is the length of the semimajor axis of the elliptical orbit. See the illustration. *79. Earth The mean distance of Earth from the Sun is 93 million miles. If the aphelion of Earth is 94.5 million NJMFT XIBUJTUIFQFSJIFMJPO 8SJUFBOFRVBUJPOGPSUIFPSCJU of Earth around the Sun. *80. Mars 5IFNFBOEJTUBODFPG.BSTGSPNUIF4VOJTNJMMJPO NJMFT*GUIFQFSJIFMJPOPG.BSTJTNJMMJPONJMFT XIBUJT
Mean distance Aphelion Center
Perihelion
Major axis
Sun
UIFBQIFMJPO 8SJUFBOFRVBUJPOGPSUIFPSCJUPG.BSTBCPVU the Sun. *81. Jupiter The aphelion of Jupiter is 507 million miles. If the distance from the center of its elliptical orbit to the Sun is 23.2 NJMMJPONJMFT XIBUJTUIFQFSJIFMJPO 8IBUJTUIFNFBOEJTUBODF Write an equation for the orbit of Jupiter around the Sun.
SECTION 9.4 The Hyperbola
711
y-coordinate. Determine the exact y-coordinate of the point M on D that is equidistant from points B and C.† 525 - 4
*82. Pluto The perihelion of Pluto is 4551 million miles, and the distance from the center of its elliptical orbit to the Sun is 897.5 million miles. Find the aphelion of Pluto. What is the mean distance of Pluto from the Sun? Write an equation for the orbit of Pluto about the Sun.
*86. Show that an equation of the form Ax2 + Cy2 + F = 0,
A ≠ 0, C ≠ 0, F ≠ 0
where A and C are of the same sign and F is of opposite sign, (a) is the equation of an ellipse with center at 10, 02 if A ≠ C. (b) is the equation of a circle with center 10, 02 if A = C.
83. Elliptical Orbit A planet orbits a star in an elliptical orbit with the star located at one foci. The perihelion of the planet c is 5 million miles. The eccentricity e of a conic section is e = . a If the eccentricity of the orbit is 0.75, find the aphelion of the planet.† 35 million mi
*87. Show that the graph of an equation of the form Ax2 + Cy2 + Dx + Ey + F = 0,
84. A rectangle is inscribed in an ellipse with major axis of length 14 meters and minor axis of length 4 meters. Find the maximum area of a rectangle inscribed in the ellipse. Round your answer to two decimal places.† 28 m2
A ≠ 0, C ≠ 0
where A and C are of the same sign, D2 E2 (a) is an ellipse if + - F is the same sign as A. 4A 4C 2 2 D E (b) is a point if + - F = 0. 4A 4C D2 E2 (c) contains no points if + - F is of opposite sign to A. 4A 4C
85. Let D be the line given by the equation x + 5 = 0. Let E be the conic section given by the equation x2 + 5y2 = 20. Let the point C be the vertex of E with the smaller x-coordinate, and let B be the endpoint of the minor axis of E with the larger
Discussion and Writing c 88. The eccentricity e of an ellipse is defined as the number , where a is the distance of a vertex from the center and c is the distance a of a focus from the center. Because a 7 c, it follows that e 6 1. Write a brief paragraph about the general shape of each of the following ellipses. Be sure to justify your conclusions. (a) Eccentricity close to 0 (b) Eccentricity = 0.5 (c) Eccentricity close to 1
Retain Your knowledge Problems 89–92 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. *89. Find the zeros of the quadratic function f(x) = (x - 5)2 - 12. What are the x-intercepts, if any, of the graph of the function? 2x - 3 *90. Find the domain of the rational function f(x) = . x - 5 Find any horizontal, vertical, or oblique asymptotes.
92. Solve the right triangle shown. b ≈ 10.94, c ≈ 17.77, B = 38° 528 c
91. Find the work done by a force of 80 pounds acting in the direction of 50° to the horizontal in moving an object 12 feet from (0, 0) to (12, 0). Round to one decimal place. 617.1 ft-lb
b
B 14
‘Are You Prepared?’ Answers 1. 213
2.
9 4
3. 1 - 2, 02, 12, 02, 10, - 42, 10, 42
4. 12, 52
5. left; 1; down: 4
6. 1x - 22 2 + 1y + 32 2 = 1
9.4 The Hyperbola PREPARING FOR THIS SECTION Before getting started, review the following: r %JTUBODF'PSNVMB 'PVOEBUJPOT 4FDUJPO Q
r $PNQMFUJOHUIF4RVBSF "QQFOEJY" 4FDUJPO" pp. A38–A39) r *OUFSDFQUT 'PVOEBUJPOT 4FDUJPO QQm
r 4ZNNFUSZ 'PVOEBUJPOT 4FDUJPO QQm
r "TZNQUPUFT 4FDUJPO QQm
r (SBQIJOH5FDIOJRVFT5SBOTGPSNBUJPOT 4FDUJPO pp. 121–129) r 4RVBSF3PPU.FUIPE 4FDUJPO QQm
Now Work the ‘Are You Prepared?’ problems on page 721.
OBJECTIVES 1 2 3 4 †
Analyze Hyperbolas with Center at the Origin (p. 712) Find the Asymptotes of a Hyperbola (p. 716) Analyze Hyperbolas with Center at (h, k) (p. 718) Solve Applied Problems Involving Hyperbolas (p. 719)
Courtesy off the C h Joliet J li JJunior i C College ll M Mathematics h i D Department
712
CHAPTER 9 Analytic Geometry
DEFINITION
A hyperbola is the collection of all points in the plane the difference of whose distances from two fixed points, called the foci, is a constant.
Figure 31
Conjugate axis
Transverse axis V2
V1
F2
Figure 31 illustrates a hyperbola with foci F1 and F2 . The line containing the foci is called the transverse axis. The midpoint of the line segment joining the foci is the center of the hyperbola. The line through the center and perpendicular to the transverse axis is the conjugate axis. The hyperbola consists of two separate curves, called branches, that are symmetric with respect to the transverse axis, conjugate axis, and center. The two points of intersection of the hyperbola and the transverse axis are the vertices, V1 and V2 , of the hyperbola.
Center
F1
1 Analyze Hyperbolas with Center at the Origin
Figure 32
d 1F1, P2 - d 1F2, P2 = {2a y d(F 1, P ) Transverse axis F 1 ⫽ (⫺c, 0)
P ⫽ (x, y ) d (F 2 , P ) F 2 ⫽ (c, 0) x
With these ideas in mind, we are now ready to find the equation of a hyperbola in the rectangular coordinate system. First, place the center at the origin. Next, position the hyperbola so that its transverse axis coincides with a coordinate axis. Suppose that the transverse axis coincides with the x-axis, as shown in Figure 32. If c is the distance from the center to a focus, one focus is at F1 = 1 - c, 02 and the other at F2 = 1c, 02 . Now we let the constant difference of the distances from any point P = 1x, y2 on the hyperbola to the foci F1 and F2 be denoted by {2a. (If P is on the right branch, the + sign is used; if P is on the left branch, the - sign is used.) The coordinates of P must satisfy the equation d 1F1 , P2 - d 1F2 , P2 = {2a 2 1x - 1 - c2 2 2 + y2 - 2 1x - c2 2 + y2 = {2a 2 1x + c2 2 + y2 = {2a + 2 1x - c2 2 + y2 1x + c2 2 + y2 = 4a2 { 4a2 1x - c2 2 + y2
Difference of the distances from P to the foci equals {2a. Use the Distance Formula. Isolate one radical. Square both sides.
+ 1x - c2 2 + y2 Next, remove the parentheses. x2 + 2cx + c 2 + y2 = 4a2 { 4a2 1x - c2 2 + y2 + x2 - 2cx + c 2 + y2 4cx - 4a2 = {4a2 1x - c2 2 + y2 cx - a = {a2 1x - c2 + y 2
2
2
1cx - a2 2 = a2 3 1x - c2 2 + y2 4 2
c 2 x2 - 2ca2 x + a4 = a2 1x2 - 2cx + c 2 + y2 2 c 2 x 2 + a4 = a2 x 2 + a2 c 2 + a2 y 2 1c - a 2x - a y = a c - a 2
2
2
2 2
2 2
4
1c 2 - a2 2x2 - a2 y2 = a2 1c 2 - a2 2
Simplify; isolate the radical. Divide each side by 4. Square both sides. Simplify. Distribute and simplify. Rearrange terms. Factor a 2 on the right side.
(1)
To obtain points on the hyperbola off the x-axis, it must be that a 6 c. To see why, look again at Figure 32. d 1F1 , P2 6 d 1F2 , P2 + d 1F1 , F2 2
d 1F1 , P2 - d 1F2 , P2 6 d 1F1 , F2 2 2a 6 2c a 6 c
Use triangle F1 P F2 .
P is on the right branch, so d(F1 , P) - d(F2 , P) = 2a; d1F1, F2 2 = 2c.
SECTION 9.4 The Hyperbola
713
Because a 6 c, this means a2 6 c 2, so c 2 - a2 7 0. Let b2 = c 2 - a2, b 7 0. Then equation (1) can be written as b2 x2 - a2 y2 = a2 b2 y2 x2 = 1 a2 b2
Divide each side by a 2 b 2.
To find the vertices of the hyperbola defined by this equation, let y = 0. x2 The vertices satisfy the equation 2 = 1, the solutions of which are x = {a. a Consequently, the vertices of the hyperbola are V1 = 1 - a, 02 and V2 = 1a, 02. Notice that the distance from the center 10, 02 to either vertex is a.
THEOREM
Equation of a Hyperbola: Center at (0, 0); Transverse Axis along the x-Axis An equation of the hyperbola with center at 10, 02, foci at 1 - c, 02 and 1c, 02, and vertices at 1 - a, 02 and 1a, 02 is y2 x2 = 1, a2 b2
Figure 33
where b2 = c 2 - a2
(2)
y
The transverse axis is the x-axis. V 1 ⫽ (⫺a, 0) Transverse axis F 1 ⫽ (⫺c, 0)
V 2 ⫽ (a, 0) F 2 ⫽ (c, 0) x
EXAMPL E 1
Solution
See Figure 33. As you can verify, the hyperbola defined by equation (2) is symmetric with respect to the x-axis, y-axis, and origin. To find the y-intercepts, if y2 any, let x = 0 in equation (2). This results in the equation 2 = - 1, which has no b real solution, so the hyperbola defined by equation (2) has no y-intercepts. In fact, y2 x2 x2 since 2 - 1 = 2 Ú 0, it follows that 2 Ú 1. There are no points on the graph for a b a - a 6 x 6 a.
Finding and Graphing an Equation of a Hyperbola
Find an equation of the hyperbola with center at the origin, one focus at 13, 02, and one vertex at 1 - 2, 02.(SBQIUIFFRVBUJPO The hyperbola has its center at the origin. Plot the center, focus, and vertex. Since they all lie on the x-axis, the transverse axis coincides with the x-axis. One focus is at 1c, 02 = 13, 02, so c = 3. One vertex is at 1 - a, 02 = 1 - 2, 02, so a = 2. From equation (2), it follows that b2 = c 2 - a2 = 9 - 4 = 5, so an equation of the hyperbola is y2 x2 = 1 4 5 To graph a hyperbola, it is helpful to locate and plot other points on the graph. For example, to find the points above and below the foci, let x = {3. Then y2 x2 = 1 4 5 1 {32 2 y2 = 1 4 5 y2 9 = 1 4 5 y2 5 = 5 4
x = {3
714
CHAPTER 9 Analytic Geometry
Figure 34
y2 =
y 5
(3, 5–2)
(3, 5–2)
V1 (2, 0)
y = {
V 2 (2, 0)
5 F 1 (3, 0)
25 4
5 x F 2 (3, 0)
(3, 5–2) (3, 5–2)
5 2
5 5 The points above and below the foci are a {3, b and a {3, - b . These points 2 2 determine the “opening” of the hyperbola. See Figure 34.
r
5
COMMENT To graph the hyperbola
y2 x2 = 1 discussed in Example 1, the two functions 4 5
x2 x2 - 1 and Y2 = - 25 - 1 NVTU CF HSBQIFE %P UIJT BOE DPNQBSF UIF B4 B4 ■ result with Figure 34. Y1 = 25
Now Work
PROBLEM
19
An equation of the form of equation (2) is the equation of a hyperbola with center at the origin; foci on the x-axis at 1 - c, 02 and 1c, 02 , where c 2 = a2 + b2; and transverse axis along the x-axis. For the next two examples, the direction “Analyze the equation” will mean to find the center, transverse axis, vertices, and foci of the hyperbola and graph it.
EX A MPL E 2
Analyzing the Equation of a Hyperbola Analyze the equation:
Solution
y2 x2 = 1 16 4
The given equation is of the form of equation (2), with a2 = 16 and b2 = 4. The graph of the equation is a hyperbola with center at 10, 02 and transverse axis along the x-axis. Also, c 2 = a2 + b2 = 16 + 4 = 20. The vertices are at 1 {a, 02 = 1 {4, 02, and the foci are at 1 {c, 02 = 1 {225, 02. To locate the points on the graph above and below the foci, let x = {225. Then y2 x2 = 1 16 4
1 {225 2 2 16
-
y2 = 1 4
x = {2 25
y2 20 = 1 16 4 y2 5 = 1 4 4
Figure 35 y 4 (– 2 5 , 1) (2 5 , 1) V 1 = (–4, 0) V = (4, 0) 2 –5
F1 = (–2 5 , 0) (–2 5 , –1) –4
5 x F2 = (2 5 , 0) (2 5 , –1)
y2 1 = 4 4 y = {1 The points above and below the foci are 1 {225, 12 and 1 {225, - 12. See Figure 35.
r
SECTION 9.4 The Hyperbola
THEOREM
715
Equation of a Hyperbola: Center at (0, 0); Transverse Axis along the y-Axis An equation of the hyperbola with center at 10, 02, foci at 10, - c2 and 10, c2 , and vertices at 10, - a2 and 10, a2 is y2 a2
-
x2 = 1, b2
where b2 = c 2 - a2
(3)
The transverse axis is the y-axis.
Figure 36 y2 x2 = 1, b 2 = c 2 - a 2 a2 b2 y F 2 (0, c ) V 2 (0, a) x V 1 (0, a )
F 1 (0, c )
EXAM PL E 3
Figure 36 shows the graph of a typical hyperbola defined by equation (3). y2 x2 An equation of the form of equation (2), 2 - 2 = 1, is the equation of a a b hyperbola with center at the origin; foci on the x-axis at 1 - c, 02 and 1c, 02, where b2 = c 2 - a2 (or c 2 = a2 + b2); and transverse axis along the x-axis. y2 x2 An equation of the form of equation (3), 2 - 2 = 1, is the equation of a a b hyperbola with center at the origin; foci on the y-axis at 10, - c2 and 10, c2 , where b2 = c 2 - a2 (or c 2 = a2 + b2); and transverse axis along the y-axis. Notice the difference in the forms of equations (2) and (3). When the y2@term is subtracted from the x2@term, the transverse axis is along the x-axis. When the x2@term is subtracted from the y2@term, the transverse axis is along the y-axis.
Analyzing the Equation of a Hyperbola Analyze the equation: y2 - 4x2 = 4
Solution
To put the equation in proper form, divide each side by 4: y2 - x2 = 1 4 Since the x2@term is subtracted from the y2@term, the equation is that of a hyperbola with center at the origin and transverse axis along the y-axis. Also, comparing the above equation to equation (3), note that a2 = 4, b2 = 1, and c 2 = a2 + b2 = 5. The vertices are at 10, {a2 = 10, {22, and the foci are at 10, {c2 = 10, { 252. To locate other points on the graph, let x = {2. Then
Figure 37 y 5 (2, 2 5 )
(– 2, 2 5 )
F 2 = (0, 5 ) V 2 = (0, 2) 5 x
–5 V 1 = (0, – 2)
F 1 = (0, – 5 ) (–2, –2 5 )
(2, – 2 5 ) –5
y2 - 4x2 = 4 y2 - 41 {22 2 = 4
x = {2
y2 - 16 = 4 y2 = 20 y = {225 Four other points on the graph are 1 {2, 2252 and 1 {2, - 2252. See Figure 37.
r
EXAM PL E 4
Solution
Finding an Equation of a Hyperbola
Find an equation of the hyperbola that has one vertex at 10, 22 and foci at 10, - 32 and 10, 32.(SBQIUIFFRVBUJPO
Because the foci are at 10, - 32 and 10, 32, the center of the hyperbola, which is at their midpoint, is the origin. Because the foci lie on the y-axis, the transverse
716
CHAPTER 9 Analytic Geometry
axis is along the y-axis. The given information also reveals that c = 3, a = 2, and b2 = c 2 - a2 = 9 - 4 = 5. The form of the equation of the hyperbola is given by equation (3):
Figure 38 y 5 F = (0, 3) 2
(– 52 , 3)
y2
( 52 , 3)
a
V 2 = (0, 2) V 1 = (0, – 2)
(– 52 , –3)
( 52 , –3) –5
F 1 = (0, – 3)
-
x2 = 1 b2
y2 x2 = 1 4 5
5 x
–5
2
Let y = {3 to obtain points on the graph on either side of each focus. See Figure 38.
Now Work
PROBLEM
r
21
Look at the equations of the hyperbolas in Examples 2 and 4. For the hyperbola in Example 2, a2 = 16 and b2 = 4, so a 7 b; for the hyperbola in Example 4, a2 = 4 and b2 = 5, so a 6 b. This indicates that for hyperbolas, there are no requirements involving the relative sizes of a and b. Contrast this situation to the case of an ellipse, in which the relative sizes of a and b dictate which axis is the major axis. Hyperbolas have another feature to distinguish them from ellipses and parabolas: Hyperbolas have asymptotes.
2 Find the Asymptotes of a Hyperbola 3FDBMMGSPN4FDUJPOUIBUBIPSJ[POUBMPSPCMJRVFBTZNQUPUFPGBHSBQIJTBMJOF with the property that the distance from the line to points on the graph approaches 0 as x S - q or as x S q . The asymptotes provide information about the end behavior of the graph of a hyperbola.
THEOREM
Asymptotes of a Hyperbola The hyperbola
y2 x2 = 1 has the two oblique asymptotes a2 b2 y =
b b x and y = - x a a
(4)
Proof Begin by solving for y in the equation of the hyperbola. y2 x2 = 1 a2 b2 y2 x2 = - 1 b2 a2 y 2 = b2 ¢
x2 - 1≤ a2
Because x ≠ 0, the right side can be rearranged in the form y2 =
b2 x 2 a2 ¢1 - 2 ≤ 2 a x
y = {
bx a2 1 - 2 a B x
a2 approaches 0, so the expression under the x2 bx radical approaches 1. So, as x S - q or as x S q , the value of y approaches { ; a that is, the graph of the hyperbola approaches the lines
Now, as x S - q or as x S q , the term
b b y = - x and y = x a a These lines are oblique asymptotes of the hyperbola.
■
SECTION 9.4 The Hyperbola
717
The asymptotes of a hyperbola are not part of the hyperbola, but they do serve as a guide for graphing a hyperbola. For example, suppose that we want to graph the equation
Figure 39 y2 x2 = 1 a2 b2 y y
y2 x2 = 1 a2 b2
b– x a
(0, b) V2 (a, 0) x (0, b) V1 (a, 0)
y b–a x
THEOREM
Begin by plotting the vertices 1 - a, 02 and 1a, 02. Then plot the points 10, - b2 and 10, b2 and use these four points to construct a rectangle, as shown in Figure 39. b b The diagonals of this rectangle have slopes and - , and their extensions are the a a b b asymptotes y = x and y = - x of the hyperbola. If we graph the asymptotes, a a we can use them to establish the “opening” of the hyperbola and avoid plotting other points.
Asymptotes of a Hyperbola The hyperbola
y2 a2
-
x2 = 1 has the two oblique asymptotes b2 y =
a a x and y = - x b b
(5)
You are asked to prove this result in Problem 84. For the remainder of this section, the direction “Analyze the equation” will mean to find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola and graph it.
EXAM PL E 5 Figure 40 y 5 22x
Analyzing the Equation of a Hyperbola Analyze the equation:
y 5
y 5 2x
F2 5 (0, 5) V2 5 (0, 2) 25
5 x V1 5 (0, 2 2) F1 5 (0, 2 5) 25
y2 - x2 = 1 4
Solution Because the x2@term is subtracted from the y2@term, the equation is of the form of equation (3) and is a hyperbola with center at the origin and transverse axis along the y-axis. Also, comparing this equation to equation (3), note that a2 = 4, b2 = 1, and c 2 = a2 + b2 = 5. The vertices are at 10, {a2 = 10, {22, and the foci are at 10, {c2 = 10, { 252. Using equation (5) with a = 2 and b = 1, the a a asymptotes are the lines y = x = 2x and y = - x = - 2x. Form the rectangle b b containing the points 10, {a2 = 10, {22 and 1 {b, 02 = 1 {1, 02. The extensions of the diagonals of this rectangle are the asymptotes. Now graph the rectangle, the asymptotes, and the hyperbola. See Figure 40.
r
EXAM PL E 6
Analyzing the Equation of a Hyperbola Analyze the equation: 9x2 - 4y2 = 36
Solution
%JWJEFFBDITJEFPGUIFFRVBUJPOCZUPQVUUIFFRVBUJPOJOQSPQFSGPSN y2 x2 = 1 4 9 Now analyze the equation. The center of the hyperbola is the origin. Because the x2@term is first in the equation, the transverse axis is along the x-axis, and the vertices and foci will lie on the x-axis. Using equation (2), note that a2 = 4, b2 = 9, and c 2 = a2 + b2 = 13. The vertices are a = 2 units left and right of the center
718
CHAPTER 9 Analytic Geometry
at 1 {a, 02 = 1 {2, 02, the foci are c = 213 units left and right of the center at 1 {c, 02 = 1 { 213, 02, and the asymptotes have the equations
Figure 41 y 3– x
y
y 3– x
2
2
5
y =
(0, 3) V1 (2, 0)
V2 (2, 0)
5
5
x
To graph the hyperbola, form the rectangle containing the points 1 {a, 02 and 10, {b2 , that is, 1 - 2, 02, 12, 02, 10, - 32, and 10, 32. The extensions of the diagonals of this rectangle are the asymptotes. See Figure 41 for the graph.
r
F 2 ( 13, 0)
F 1 ( 13, 0)
3 b 3 b x = x and y = - x = - x a a 2 2
Now Work
(0, 3)
31
PROBLEM
5
3 Analyze Hyperbolas with Center at (h, k) If a hyperbola with center at the origin and transverse axis coinciding with a coordinate axis is shifted horizontally h units and then vertically k units, the result is a hyperbola with center at 1h, k2 and transverse axis parallel to a coordinate axis. The equations of such hyperbolas have the same forms as those given in equations (2) and (3), except that x is replaced by x - h (the horizontal shift) and y is replaced by y - k (the vertical shift). Table 4 gives the forms of the equations of such hyperbolas. See Figure 42 for typical graphs.
Table 4
Equation of a Hyperbola: Center at (h, k); Transverse Axis Parallel to a Coordinate Axis Center
Transverse Axis
Foci
Vertices
Equation
Asymptotes
(h, k)
Parallel to the x-axis
(h { c, k)
(h { a, k)
(y - k) (x - h) = 1, b2 = c2 - a2 a2 b2
b y - k = { (x - h) a
(h, k)
Parallel to the y-axis
(h, k { c)
(h, k { a)
(y - k)2
a y - k = { (x - h) b
2
a
Figure 42 NOTE It is not recommended that Table 4 be memorized. Rather, use the ideas of transformations (shift horizontally h units, vertically k units), along with the fact that a represents the distance from the center to the vertices, c represents the distance from the center to the foci, and j b 2 = c 2 - a 2 (or c 2 = a 2 + b 2).
(x - h)2 = 1, b2 = c2 - a2 b2
y
Transverse F1 V1 axis
Transverse axis F2
(h, k)
V2
V2 F 2
(h, k ) x
V1
x
F1
2
Solution
-
y
2
(y k) (x h ) (a) ––––––– –––––– 1 a2 b2
EX A MPL E 7
2
2
(y k )2 (x h)2 (b) ––––––– –––––– 1 2 a b2
Finding an Equation of a Hyperbola, Center Not at the Origin
Find an equation for the hyperbola with center at 11, - 22, one focus at 14, - 22, and one vertex at 13, - 22.(SBQIUIFFRVBUJPO
The center is at 1h, k2 = 11, - 22, so h = 1 and k = - 2. Because the center, focus, and vertex all lie on the line y = - 2, the transverse axis is parallel to the x-axis. The distance from the center 11, - 22 to the focus 14, - 22 is c = 3; the distance from
SECTION 9.4 The Hyperbola
Figure 43
the center 11, - 22 to the vertex 13, - 22 is a = 2. Then b2 = c 2 - a2 = 9 - 4 = 5. The equation is
y 6
1x - h2 2 a2
(1, 2 5)
6 x V2 (3, 2) (1, 2) F2 (4, 2)
1
6
-
1y - k2 2 b2
= 1
1x - 12 2 1y + 22 2 = 1 4 5
V1 (1, 2) 6 Transverse axis F (2, 2)
719
r
See Figure 43.
Now Work
(1, 2 5)
PROBLEM
41
Analyzing the Equation of a Hyperbola
EXAM PL E 8
Analyze the equation: - x2 + 4y2 - 2x - 16y + 11 = 0
Solution
Complete the squares in x and in y. - x2 + 4y2 - 2x - 16y + 11 = 0
- 1x2 + 2x2 + 41y2 - 4y2 = - 11
Figure 44
- 1x + 2x + 12 + 41y - 4y + 42 = - 11 - 1 + 16
Transverse axis y F2 (1, 2 5 ) V2 (1, 3) 5
(3, 2)
2
- 1x + 12 2 + 41y - 22 2 = 4 1y - 22 2 -
(1, 2)
5 5 x F1 (1, 2
V1 (1, 1)
5)
2
1x + 12 2 = 1 4
Group terms. Complete each square. Factor. Divide each side by 4.
This is the equation of a hyperbola with center at 1 - 1, 22 and transverse axis parallel to the y-axis. Also, a2 = 1 and b2 = 4, so c 2 = a2 + b2 = 5. Because the transverse axis is parallel to the y-axis, the vertices and foci are located a and c units above and below the center, respectively. The vertices are at 1h, k { a2 = 1 - 1, 2 { 12, or 1 - 1, 12 and 1 - 1, 32. The foci are at 1h, k { c2 = ( - 1, 2 { 25). The asymptotes 1 1 are y - 2 = 1x + 12 and y - 2 = - 1x + 12. Figure 44 shows the graph. 2 2
r
Now Work
PROBLEM
55
4 Solve Applied Problems Involving Hyperbolas Figure 45
S
O3 O1
O2
EXAM PL E 9
L Look at Figure 45. Suppose that three microphones are located at points O1 , O2 , aand O3 (the foci of the two hyperbolas). In addition, suppose that a gun is fired at S, and the microphone at O1 records the gunshot 1 second after the microphone at O2 . Because sound travels at about 1100 feet per second, we conclude that the microphone at O1 is 1100 feet farther from the gunshot than O2 . We can model this situation by saying that S lies on a branch of a hyperbola with foci at O1 and O2 . %PZPVTFFXIZ 5IFEJGGFSFODFPGUIFEJTUBODFTGSPNS to O1 and from S to O2 is the constant 1100.) If the third microphone, at O3, records the gunshot 2 seconds after O1 , then S lies on a branch of a second hyperbola with foci at O1 and O3 . In this case, the constant difference is 2200. The intersection of the two hyperbolas identifies the location of S.
Lightning Strikes Suppose that two people standing 1 mile apart both see a flash of lightning. After a period of time, the person standing at point A hears the thunder. One second later, the person standing at point B hears the thunder. If the person at B is due west of the person at A, and the lightning strike is known to occur due north of the person standing at point A XIFSFEJEUIFMJHIUOJOHTUSJLFPDDVS
720
CHAPTER 9 Analytic Geometry
Solution
See Figure 46, in which the ordered pair 1x, y2 represents the location of the lightning strike. Sound travels at 1100 feet per second, so the person at point A is 1100 feet closer to the lightning strike than the person at point B. Since the difference of the distance from 1x, y2 to B and the distance from 1x, y2 to A is the constant 1100, the point 1x, y2 lies on a hyperbola whose foci are at A and B.
Figure 46
North (x, y)
East B ⫽ (⫺2640, 0) (⫺a, o)
(a, o)
A ⫽ (2640, 0)
1 mile ⫽ 5280 feet
An equation of the hyperbola is y2 x2 = 1 a2 b2 where 2a = 1100, so a = 550. Because the distance between the two people is 1 mile (5280 feet) and each person is at a focus of the hyperbola, this means 2c = 5280 5280 c = = 2640 2 Because b2 = c 2 - a2 = 26402 - 5502 = 6,667,100, the equation of the hyperbola that describes the location of the lightning strike is y2 x2 = 1 6,667,100 5502 3FGFSUP'JHVSF#FDBVTFUIFMJHIUOJOHTUSJLFPDDVSSFEEVFOPSUIPGUIFJOEJWJEVBM at the point A = 12640, 02, let x = 2640 and solve the resulting equation. y2 26402 = 1 6,667,100 5502 y2 = - 22.04 6,667,100 y2 = 146,942,884 y = 12,122
Subtract
26402 from both sides. 5502
Multiply both sides by - 6,667,100. y 7 0 since the lightning strike occurred in quadrant I.
Check: The difference between the distance from 12640, 12,1222 to the person at the point B = 1 - 2640, 02 and the distance from 12640, 12,1222 to the person at the point A = 12640, 02 should be 1100. From the distance formula, the difference in the distances is 3 3 2640 - 1 - 26402 4 2 + 112,122 - 02 2 - 3 12640 - 26402 2 + 112,122 - 02 2 = 1100 as required. The lightning strike occurred 12,122 feet north of the person standing at point A.
Now Work
PROBLEM
75
r
SECTION 9.4 The Hyperbola
721
9.4 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.
1. The distance d from P1 = 13, - 42 to P2 = 1 - 2, 12 is d = 522 . (p. 35) 25 2 2. To complete the square of x + 5x, add 4 . (pp. A38–A39)
5. To graph y = 1x - 52 3 - 4, shift the graph of y = x3 to the (left/right) 5 unit(s) and then (up/down) 4 unit(s). (pp. 121–123) right; down
y2 = 9 + 4x2.
6. Find the vertical asymptotes, if any, and the horizontal or x2 - 9 . (pp. 268–271) oblique asymptote, if any, of y = 2 x - 4 Vertical: x = - 2, x = 2; horizontal: y = 1
3. Find the intercepts of the equation (p. 44) (0, - 3), (0, 3)
4. True or False The equation y2 = 9 + x2 is symmetric with respect to the x-axis, the y-axis, and the origin. (pp. 44–46) True
Concepts and Vocabulary 7. A(n) hyperbola is the collection of points in the plane the difference of whose distances from two fixed points is a constant. 8. For a hyperbola, the foci lie on a line called the transverse axis . Answer Problems 9–11 using the figure. y
Transverse axis
9. The equation of the hyperbola is of the form (b) 1x - h2 2 1y - k2 2 = 1 (a) 2 a b2 1y - k2 2 1x - h2 2 (b) = 1 2 a b2 10. If the center of the hyperbola is (2, 1) and a = 3, then the coordinates of the vertices are (2, 4) and (2, - 2) .
F2
11. If the center of the hyperbola is (2, 1) and c = 5, then the coordinates of the foci are (2, 6) and (2, - 4) .
V2
12. In a hyperbola, if a = 3 and c = 5, then b = 4 . y2 x2 13. For the hyperbola = 1, the value of a is 2 , the 4 9 value of b is 3 , and the transverse axis is the x -axis. y2 x2 14. For the hyperbola = 1, the asymptotes are 16 81 4 4 y = x y = - x 9 . 9 and
(h, k ) V1
x
F1
Skill Building In Problems 15–18, the graph of a hyperbola is given. Match each graph to its equation. y2 y2 x2 x2 (A) - y2 = 1 (B) x2 = 1 (C) - x2 = 1 (D) y2 = 1 4 4 4 4 15. B
16. C
y 3
⫺3
3 x
17. A
y 4
⫺4
⫺3
4 x
18. D
y 4
⫺4
⫺4
4x ⫺4
y 3
⫺3
3x ⫺3
In Problems 19–28, find an equation for the hyperbola described. Graph the equation. *19. Center at 10, 02; focus at 13, 02; vertex at 11, 02
*21. Center at 10, 02; focus at 10, - 62; vertex at 10, 42
*23. Foci at 1 - 5, 02 and 15, 02; vertex at 13, 02
*25. Vertices at 10, - 62 and 10, 62; asymptote the line y = 2x *27. Foci at 1 - 4, 02 and 14, 02; asymptote the line y = - x
*20. Center at 10, 02; focus at 10, 52; vertex at 10, 32
*22. Center at 10, 02; focus at 1 - 3, 02; vertex at 12, 02 *24. Focus at 10, 62; vertices at 10, - 22 and 10, 22
*26. Vertices at 1 - 4, 02 and 14, 02; asymptote the line y = 2x
*28. Foci at 10, - 22 and 10, 22; asymptote the line y = - x
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
722
CHAPTER 9 Analytic Geometry
In Problems 29–36, find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation. *29.
y2 x2 = 1 25 9
*30.
*33. y2 - 9x2 = 9
y2 x2 = 1 16 4
*34. x2 - y2 = 4
*31. 4x2 - y2 = 16
*32. 4y2 - x2 = 16
*35. y2 - x2 = 25
*36. 2x2 - y2 = 4
In Problems 37–40, write an equation for each hyperbola. *37.
y 3
y ⫽ ⫺x
y⫽x
⫺3
*38.
3x
y 3
⫺3
⫺3
⫺3
y⫽x
*39. y ⫽ ⫺2x
3x
⫺5
y 10
y ⫽ 2x
5 x
*40.
y ⫽ ⫺2x y 5
⫺5
⫺10
y ⫽ ⫺x
y ⫽ 2x
5 x
⫺5
In Problems 41–48, find an equation for the hyperbola described. Graph the equation. *41. Center at 14, - 12; focus at 17, - 12; vertex at 16, - 12
*42. Center at 1 - 3, 12; focus at 1 - 3, 62; vertex at 1 - 3, 42
*45. Foci at 13, 72 and 17, 72; vertex at 16, 72
*46. Focus at 1 - 4, 02 ; vertices at 1 - 4, 42 and 1 - 4, 22
*43. Center at 1 - 3, - 42; focus at 1 - 3, - 82; vertex at 1 - 3, - 22 *47. Vertices at 1 - 1, - 12 and 13, - 12; asymptote the line 3 y + 1 = 1x - 12 2
*44. Center at 11, 42; focus at 1 - 2, 42; vertex at 10, 42
*48. Vertices at 11, - 32 and 11, 12; asymptote the line 3 y + 1 = 1x - 12 2
In Problems 49–62, find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation. 1x - 22 2 1y + 32 2 1y + 32 2 1x - 22 2 = 1 *50. = 1 *51. 1y - 22 2 - 41x + 22 2 = 4 *49. 4 9 4 9 *52. 1x + 42 2 - 91y - 32 2 = 9
*53. 1x + 12 2 - 1y + 22 2 = 4
*54. 1y - 32 2 - 1x + 22 2 = 4
*55. x - y - 2x - 2y - 1 = 0
*56. y - x - 4y + 4x - 1 = 0
*57. y2 - 4x2 - 4y - 8x - 4 = 0
*58. 2x2 - y2 + 4x + 4y - 4 = 0
*59. 4x2 - y2 - 24x - 4y + 16 = 0
*60. 2y2 - x2 + 2x + 8y + 3 = 0
*61. y - 4x - 16x - 2y - 19 = 0
*62. x - 3y + 8x - 6y + 4 = 0
2
2
2
2
2
2
2
2
In Problems 63–66, graph each function. Be sure to label any intercepts. [Hint: Notice that each function is half a hyperbola.] *63. f 1x2 = 216 + 4x2
*64. f 1x2 = - 29 + 9x2
*65. f 1x2 = - 2- 25 + x2
*66. f 1x2 = 2- 1 + x2
Mixed Practice In Problems 67–74, analyze each conic. (x - 3)2 y2 = 1 *67. 4 25 *70. y2 = - 12(x + 1)
= 1 16 4 *71. 25x2 + 9y2 - 250x + 400 = 0
*73. x2 - 6x - 8y - 31 = 0
*74. 9x2 - y2 - 18x - 8y - 88 = 0
*68.
(y + 2)2
-
(x - 2)2
*69. x2 = 16(y - 3) *72. x2 + 36y2 - 2x + 288y + 541 = 0
Applications and Extensions 75. Fireworks Display Suppose that two people standing 2 miles apart both see the burst from a fireworks display. After a period of time the first person, standing at point A, hears the burst. One second later the second person, standing at point B, hears the burst. If the person at point B is due west of the person at point A, and if the display is known to occur due north of the person at point A, where EJEUIFGJSFXPSLTEJTQMBZPDDVS 50,138 ft north of point A 76. Lightning Strikes Suppose that two people standing 1 mile apart both see a flash of lightning. After a period of time
the first person, standing at point A, hears the thunder. Two seconds later the second person, standing at point B, hears the thunder. If the person at point B is due west of the person at point A, and if the lightning strike is known to occur due north of the person standing at point A, where did UIFMJHIUOJOHTUSJLFPDDVS 5236 ft north of point A 77. Nuclear Power Plant Some nuclear power plants utilize “natural draft” cooling towers in the shape of a hyperboloid, a solid obtained by rotating a hyperbola about its conjugate axis. Suppose that such a cooling tower has a base diameter
SECTION 9.4 The Hyperbola
of 400 feet and the diameter at its narrowest point, 360 feet above the ground, is 200 feet. If the diameter at the top of UIFUPXFSJTGFFU IPXUBMMJTUIFUPXFS 592.4 ft Source: Bay Area Air Quality Management District 78. An Explosion Two recording devices are set 2400 feet apart, with the device at point A to the west of the device at point B. At a point between the devices, 300 feet from point B, a small amount of explosive is detonated. The recording devices record the time until the sound reaches each. How far directly north of point B should a second explosion be done so that the measured time difference recorded by the EFWJDFTJTUIFTBNFBTUIBUGPSUIFGJSTUEFUPOBUJPO 700 ft *79. Rutherford’s Experiment *O .BZ &SOFTU 3VUIFSGPSE published a paper in Philosophical Magazine. In this article, he described the motion of alpha particles as they are shot at a piece of gold foil 0.00004 cm thick. Before conducting this FYQFSJNFOU 3VUIFSGPSE FYQFDUFE UIBU UIF BMQIB QBSUJDMFT would shoot through the foil just as a bullet would shoot through snow. Instead, a small fraction of the alpha particles bounced off the foil. This led to the conclusion that the nucleus of an atom is dense, while the remainder of the atom is sparse. Only the density of the nucleus could cause the alpha particles to deviate from their path. The figure shows a diagram from 3VUIFSGPSET QBQFS UIBU JOEJDBUFT UIBU UIF EFGMFDUFE BMQIB particles follow the path of one branch of a hyperbola. y
723
same point. The rays are collected by the parabolic mirror, reflected toward the (common) focus, and thus are reflected by the hyperbolic mirror through the opening to its front focus, where the eyepiece is located. If the equation of the hyperbola y2 x2 is = 1 and the focal length (distance from the 9 16 vertex to the focus) of the parabola is 6, find the equation of the parabola. Source: www.enchantedlearning.com *81. The eccentricity e of a hyperbola is defined as the number c , where a is the distance of a vertex from the center, and a c is the distance of a focus from the center. Because c 7 a, it follows that e 7 1. %FTDSJCF UIF HFOFSBM TIBQF PG B hyperbola whose eccentricity is close to 1. What is the shape if eJTWFSZMBSHF 82. A hyperbola for which a = b is called an equilateral hyperbola. Find the eccentricity e of an equilateral hyperbola. 22 [Note: The eccentricity of a hyperbola is defined in Problem 81.] *83. Two hyperbolas that have the same set of asymptotes are called conjugate. Show that the hyperbolas x2 x2 - y2 = 1 and y2 = 1 4 4 BSF DPOKVHBUF (SBQI FBDI IZQFSCPMB PO UIF TBNF TFU PG coordinate axes. *84. Prove that the hyperbola y2
x2 = 1 a b2 has the two oblique asymptotes a a y = x and y = - x b b *85. Show that the graph of an equation of the form
45⬚
2
x
(a) Find an equation of the asymptotes under this scenario. (b) If the vertex of the path of the alpha particles is 10 cm from the center of the hyperbola, find a model that describes the path of the particle. *80. Hyperbolic Mirrors Hyperbolas have interesting reflective properties that make them useful for lenses and mirrors. For example, if a ray of light strikes a convex hyperbolic mirror on a line that would (theoretically) pass through its rear focus, it is reflected through the front focus. This property, and that of the parabola, were used to develop the Cassegrain telescope in 1672. The focus of the parabolic mirror and the rear focus of the hyperbolic mirror are the
-
Ax2 + Cy2 + F = 0
A ≠ 0, C ≠ 0, F ≠ 0
where A and C are opposite in sign, is a hyperbola with center at 10, 02. *86. Show that the graph of an equation of the form Ax2 + Cy2 + Dx + Ey + F = 0
A ≠ 0, C ≠ 0
where A and C are opposite in sign, (a) is a hyperbola if
E2 D2 + - F ≠ 0. 4A 4C
(b) is two intersecting lines if
D2 E2 + - F = 0. 4A 4C
Retain Your Knowledge Problems 87–90 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 3 89. Find the rectangular coordinates of the point with the polar *87. For y = cos (6x + 3p), find the amplitude, period, and 2 p phase shift. Then graph the function, showing at least two coordinates a12, - b. (6, - 623) 3 periods. *90. Transform the polar equation r = 6 sin u to an equation 88. Solve the triangle described: a = 7, b = 10, and C = 100°. in rectangular coordinates. Then identify and graph the c ≈ 13.16, A ≈ 31.6°, B ≈ 48.4°, equation.
‘Are You Prepared?’ Answers 1. 522
2.
25 4
3. 10, - 32, 10, 32
4. True
5. right; 5; down; 4
6. Vertical: x = - 2, x = 2; horizontal: y = 1
724
CHAPTER 9 Analytic Geometry
9.5 Rotation of Axes; General Form of a Conic PREPARING FOR THIS SECTION Before getting started, review the following: r %PVCMFBOHMF'PSNVMBTGPS4JOFBOE$PTJOF 4FDUJPO 6.6, p. 544)
r 4VN'PSNVMBTGPS4JOFBOE$PTJOF 4FDUJPO pp. 531 and 534) r )BMGBOHMF'PSNVMBTGPS4JOFBOE$PTJOF 4FDUJPO 6.6, p. 547) Now Work the ‘Are You Prepared?’ problems on page 730.
OBJECTIVES 1 2 3 4
Identify a Conic (p. 724) Use a Rotation of Axes to Transform Equations (p. 725) Analyze an Equation Using a Rotation of Axes (p. 727) Identify Conics without a Rotation of Axes (p. 729)
In this section, we show that the graph of a general second-degree polynomial containing two variables x and y—that is, an equation of the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
(1)
where A, B, and C are not simultaneously 0—is a conic.We shall not concern ourselves here with the degenerate cases of equation (1), such as x2 + y2 = 0, whose graph is a single point 10, 02; or x2 + 3y2 + 3 = 0, whose graph contains no points; or x2 - 4y2 = 0, whose graph is two lines, x - 2y = 0 and x + 2y = 0. We begin with the case where B = 0. In this case, the term containing xy is not present, so equation (1) has the form Ax2 + Cy2 + Dx + Ey + F = 0 where either A ≠ 0 or C ≠ 0.
1 Identify a Conic We have already discussed the procedure for identifying the graph of this kind of equation; complete the squares of the quadratic expressions in x or y, or both. Once this has been done, the conic can be identified by comparing it to one of the forms studied in Sections 9.2 through 9.4. In fact, though, the conic can be identified directly from the equation without completing the squares.
THEOREM
Identifying Conics without Completing the Squares Excluding degenerate cases, the equation Ax2 + Cy2 + Dx + Ey + F = 0
(2)
where A and C cannot both equal zero: B %FGJOFTBQBSBCPMBJGAC = 0. C %FGJOFTBOFMMJQTF PSBDJSDMF JGAC 7 0. D %FGJOFTBIZQFSCPMBJGAC 6 0.
Proof (a) If AC = 0, then either A = 0 or C = 0, but not both, so the form of equation (2) is either Ax2 + Dx + Ey + F = 0,
A ≠ 0
Cy2 + Dx + Ey + F = 0,
C ≠ 0
or
SECTION 9.5 Rotation of Axes; General Form of a Conic
725
Using the results of Problems 78 and 79 in Exercise 9.2, it follows that, except for the degenerate cases, the equation is a parabola. (b) If AC 7 0, then A and C are of the same sign. Using the results of Problem 87 in Exercise 9.3, except for the degenerate cases, the equation is an ellipse. (c) If AC 6 0, then A and C are of opposite signs. Using the results of Problem 86 in Exercise 9.4, except for the degenerate cases, the equation is a hyperbola. ■ We will not be concerned with the degenerate cases of equation (2). However, in practice, you should be alert to the possibility of degeneracy.
EXAM PL E 1
Identifying a Conic without Completing the Squares Identify the graph of each equation without completing the squares. (a) 3x2 + 6y2 + 6x - 12y = 0 (c) y2 - 2x + 4 = 0
Solution
(b) 2x2 - 3y2 + 6y + 4 = 0
(a) Compare the given equation to equation (2) and conclude that A = 3 and C = 6. Since AC = 18 7 0, the equation defines an ellipse. (b) Here A = 2 and C = - 3, so AC = - 6 6 0. The equation defines a hyperbola. (c) Here A = 0 and C = 1, so AC = 0. The equation defines a parabola.
r
Now Work
PROBLEM
11
Although we can now identify the type of conic represented by any equation of the form of equation (2) without completing the squares, we still need to complete the squares if we desire additional information about the conic, such as its graph.
2 Use a Rotation of Axes to Transform Equations
Figure 47 y′
y θ
x′ θ x
O
Now let’s turn our attention to equations of the form of equation (1), where B ≠ 0. To discuss this case, we introduce a new procedure: rotation of axes. In a rotation of axes, the origin remains fixed while the x-axis and y-axis are rotated through an angle u to a new position; the new positions of the x-axis and the y-axis are denoted by x′ and y′, respectively, as shown in Figure 47(a). Now look at Figure 47(b). There the point P has the coordinates 1x, y2 relative to the xy-plane, while the same point P has coordinates 1x′, y′2 relative to the x′y′@plane. We seek relationships that will enable us to express x and y in terms of x′, y′, and u. As Figure 47(b) shows, r denotes the distance from the origin O to the point P, and a denotes the angle between the positive x′@axis and the ray from O through P. Then, using the definitions of sine and cosine, we have x′ = r cos a y′ = r sin a x = r cos 1u + a2 y = r sin 1u + a2
(a)
Now y′
x = r cos 1u + a2
y r α O
x
P ⫽ (x, y) ⫽ (x ′, y ′) y′ x′ x′ θ
y
= r 1cos u cos a - sin u sin a2
Apply the Sum Formula for cosine.
= x′ cos u - y′ sin u
By equation (3)
= 1r cos a2 1cos u2 - 1r sin a2 1sin u2
x
Similarly, (b)
y = r sin 1u + a2
= r 1sin u cos a + cos u sin a2
Apply the Sum Formula for sine.
= x′ sin u + y′ cos u
By equation (3)
(3) (4)
726
CHAPTER 9 Analytic Geometry
THEOREM
Rotation Formulas If the x- and y-axes are rotated through an angle u, the coordinates 1x, y2 of a point P relative to the xy-plane and the coordinates 1x′, y′2 of the same point relative to the new x′@ and y′@axes are related by the formulas x = x′ cos u - y′ sin u
y = x′ sin u + y′ cos u
(5)
Rotating Axes
EX A MPL E 2
Express the equation xy = 1 in terms of new x′y′@coordinates by rotating the axes through a 45°BOHMF%JTDVTTUIFOFXFRVBUJPO
Solution
x′
y
( 2 , 0)
1
45° ⫺1
(⫺ 2 , 0)
22 22 22 - y′ = 1x′ - y′2 2 2 2
y = x′ sin 45° + y′ cos 45° = x′
22 22 22 + y′ = 1x′ + y′2 2 2 2
c
2
⫺2
x = x′ cos 45° - y′ sin 45° = x′
Substituting these expressions for x and y in xy = 1 gives
Figure 48 y′
Let u = 45° in equation (5). Then
1 ⫺1 ⫺2
2
x
22 22 1x′ - y′2 d c 1x′ + y′2 d = 1 2 2 1 1x′2 - y′2 2 = 1 2 y′2 x′2 = 1 2 2
This is the equation of a hyperbola with center at 10, 02 and transverse axis along the x′@axis. The vertices are at 1 { 22, 02 on the x′@axis; the asymptotes are y′ = x′ and y′ = - x′ (which correspond to the original x- and y-axes). See Figure 48 for the graph.
r
As Example 2 illustrates, a rotation of axes through an appropriate angle can transform a second-degree equation in x and y containing an xy-term into one in x′ and y′ in which no x′ y′@term appears. In fact, we will show that a rotation of axes through an appropriate angle transforms any equation of the form of equation (1) into an equation in x′ and y′ without an x′ y′@term. To find the formula for choosing an appropriate angle u through which to rotate the axes, begin with equation (1), Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
B ≠ 0
Next rotate through an angle u using the rotation formulas (5).
A 1x′ cos u - y′ sin u2 + B 1x′ cos u - y′ sin u2 1x′ sin u + y′ cos u2 2 + C 1x′ sin u + y′ cos u2 + D1x′ cos u - y′ sin u2 + E 1x′ sin u + y′ cos u2 + F = 0 2
Expanding and collecting like terms gives 1A cos2 u + B sin u cos u + C sin2 u2x′2 + + + +
3 B 1cos2 u - sin2 u2 + 21C - A2 1sin u cos u2 4 x′ y′ 1A sin2 u - B sin u cos u + C cos2 u2y′2 1D cos u + E sin u2x′ (6) 1 - D sin u + E cos u2y′ + F = 0
In equation (6), the coefficient of x′ y′ is
B 1cos2 u - sin2 u2 + 21C - A2 1sin u cos u2
SECTION 9.5 Rotation of Axes; General Form of a Conic
727
Because we want to eliminate the x′ y′@term, we select an angle u so that this coefficient is 0. B 1cos2 u - sin2 u2 + 21C - A2 1sin u cos u2 = 0 Double-angle B cos 12u2 + 1C - A2 sin 12u2 = 0 B cos 12u2 = 1A - C2 sin 12u2 Formulas A - C cot12u2 = B ≠ 0 B
THEOREM
To transform the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
B ≠ 0
into an equation in x′ and y′ without an x′ y′@term, rotate the axes through an angle u that satisfies the equation cot12u2 =
A - C B
(7)
WARNING Be careful if you use a calculator to solve equation (7). 1. If cot 12u2 = 0, then 2u = 90° and u = 45°.
2. If cot 12u2 ≠ 0, first find cos 12u2. Then use the inverse cosine function key(s) to obtain 2u, 0° 6 2u 6 180°. Finally, divide by 2 to obtain the j correct acute angle u.
Equation (7) has an infinite number of solutions for u. We shall adopt the convention of choosing the acute angle u that satisfies (7). There are two possibilities: If cot12u2 Ú 0, then 0° 6 2u … 90°, so 0° 6 u … 45°. If cot12u2 6 0, then 90° 6 2u 6 180°, so 45° 6 u 6 90°. Each of these results in a counterclockwise rotation of the axes through an acute angle u.*
3 Analyze an Equation Using a Rotation of Axes For the remainder of this section, the direction “Analyze the equation” will mean to transform the given equation so that it contains no xy-term and to graph the equation.
EXAM PL E 3
Analyzing an Equation Using a Rotation of Axes Analyze the equation: x2 + 23xy + 2y2 - 10 = 0
Solution
Because an xy-term is present, the axes must be rotated. Using A = 1, B = 23, and C = 2 in equation (7), the appropriate acute angle u through which to rotate the axes satisfies the equation cot12u2 =
A - C -1 23 = = B 3 23
0° 6 2u 6 180°
23 , this means 2u = 120°, so u = 60°. Using u = 60° in the 3 rotation formulas (5) yields Because cot12u2 = -
x = x′cos 60° - y′sin 60° =
1 23 1 x′ y′ = 1 x′ - 23y′ 2 2 2 2
y = x′sin 60° + y′cos 60° =
23 1 1 x′ + y′ = 1 23x′ + y′ 2 2 2 2
A - C eliminates the x′ y′@term. However, the B final form of the transformed equation may be different (but equivalent), depending on the angle chosen. *Any rotation through an angle u that satisfies cot 12u2 =
728
CHAPTER 9 Analytic Geometry
Substituting these values into the original equation and simplifying give 1 1 x′ - 23y′ 2 2 4
x2 + 23xy + 2y2 - 10 = 0 1 1 1 + 23 c 1 x′ - 23y′ 2 d c 1 23x′ + y′ 2 d + 2c 1 23x′ + y′ 2 2 d = 10 2 2 4
.VMUJQMZCPUITJEFTCZBOEFYQBOEUPPCUBJO x′2 - 223x′ y′ + 3y′2 + 23 1 23x′2 - 2x′ y′ - 23y′2 2 + 2 1 3x′2 + 223x′ y′ + y′2 2 = 40 10x′2 + 2y′2 = 40 x′ y′2 x′2 + = 1 4 20
Figure 49 y y′
(2, 0)
(0, 2 5 )
60° (⫺2, 0)
x (0, ⫺2 5 )
This is the equation of an ellipse with center at 10, 02 and major axis along the y′@axis. The vertices are at 10, {2252 on the y′@axis. See Figure 49 for the graph.
r
Now Work
PROBLEM
31
In Example 3, the acute angle u through which to rotate the axes was easy to find because of the numbers used in the given equation. In general, the equation A - C cot12u2 = will not have such a “nice” solution. As the next example shows, B we can still find the appropriate rotation formulas without using a calculator approximation by applying Half-angle Formulas.
EX A MPL E 4
Analyzing an Equation Using a Rotation of Axes Analyze the equation: 4x2 - 4xy + y2 + 525x + 5 = 0
Solution
Letting A = 4, B = - 4, and C = 1 in equation (7), the appropriate angle u through which to rotate the axes satisfies A - C 3 3 = = B -4 4 To use the rotation formulas (5), we need to know the values of sin u and cos u. Since we seek an acute angle u, we know that sin u 7 0 and cos u 7 0. Use the Half-angle Formulas in the form cot12u2 =
sin u =
1 - cos 12u2 B 2
cos u =
1 + cos 12u2 B 2
3 Now we need to find the value of cos 12u2 . Since cot12u2 = - , then 4 3 90° 6 2u 6 180° %PZPVLOPXXIZ
TPcos 12u2 = - . Then 5 1 - cos 12u2 sin u = = B 2 R
3 1 - a- b 5 4 2 225 = = = 2 B5 5 25
1 + cos 12u2 cos u = = B 2 R
3 1 + a- b 5 1 1 25 = = = 2 B5 5 25
With these values, the rotation formulas (5) are x =
25 225 25 x′ y′ = 1x′ - 2y′2 5 5 5
y =
225 25 25 x′ + y′ = 12x′ + y′2 5 5 5
SECTION 9.5 Rotation of Axes; General Form of a Conic
729
Substituting these values in the original equation and simplifying give 4x2 - 4xy + y2 + 525x + 5 = 0 4c
2 25 25 25 1x′ - 2y′2 d - 4c 1x′ - 2y′2 d c 12x′ + y′2 d 5 5 5
+ c
2 25 25 12x′ + y′2 d + 525 c 1x′ - 2y′2 d = - 5 5 5
.VMUJQMZCPUITJEFTCZBOEFYQBOEUPPCUBJO Figure 50 y
x′
y′ (0, 1)
63.4° x
41x′2 - 4x′ y′ + 4y′2 2 - 412x′2 - 3x′ y′ - 2y′2 2 + 4x′2 + 4x′y′ + y′2 + 251x′ - 2y′2 25y′2 - 50y′ + 25x′ y′2 - 2y′ + x′ y′2 - 2y′ + 1 2 1y′ - 12
= = = = =
- 25 - 25 -1 - x′ - x′
Combine like terms. Divide by 25. Complete the square in y′.
This is the equation of a parabola with vertex at 10, 12 in the x′y′@plane. The axis of 225 , and find symmetry is parallel to the x′@axis. Use a calculator to solve sin u = 5 that u ≈ 63.4°. See Figure 50 for the graph.
r
Now Work
PROBLEM
37
4 Identify Conics without a Rotation of Axes Suppose that we are required only to identify (rather than analyze) the graph of an equation of the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
B ≠ 0
(8)
Applying the rotation formulas (5) to this equation gives an equation of the form A′ x′2 + B′ x′ y′ + C′ y′2 + D′ x′ + E′ y′ + F′ = 0
(9)
where A′, B′, C′, D′, E′, and F′ can be expressed in terms of A, B, C, D, E, F and the angle u of rotation (see Problem 53). It can be shown that the value of B2 - 4AC in equation (8) and the value of B′2 - 4A′ C′ in equation (9) are equal no matter what angle u of rotation is chosen (see Problem 55). In particular, if the angle u of rotation satisfies equation (7), then B′ = 0 in equation (9), and B2 - 4AC = - 4A′ C′. Because equation (9) then has the form of equation (2), A′ x′2 + C′ y′2 + D′ x′ + E′ y′ + F′ = 0 we can identify its graph without completing the squares, as we did in the beginning of this section. In fact, now we can identify the conic described by any equation of the form of equation (8) without a rotation of axes.
THEOREM
Identifying Conics without a Rotation of Axes Except for degenerate cases, the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 B %FGJOFTBQBSBCPMBJGB2 - 4AC = 0. C %FGJOFTBOFMMJQTF PSBDJSDMF JGB2 - 4AC 6 0. D %FGJOFTBIZQFSCPMBJGB2 - 4AC 7 0. You are asked to prove this theorem in Problem 56.
730
CHAPTER 9 Analytic Geometry
Identifying a Conic without a Rotation of Axes
EX A MPL E 5
Identify the graph of the equation: 8x2 - 12xy + 17y2 - 425x - 225y - 15 = 0
Solution
Here A = 8, B = - 12, and C = 17, so B2 - 4AC = - 400. Since B2 - 4AC 6 0, the equation defines an ellipse.
Now Work
r
PROBLEM
43
9.5 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. The sum formula for the sine function is sin1A + B2 = sin A cos B + cos A sin B. (p. 534) 2. 5IF %PVCMFBOHMF 'PSNVMB GPS UIF TJOF GVODUJPO JT sin12u2 = 2 sin u cos u. (p. 544)
3. If u is acute, the Half-angle Formula for the sine function is 1- cos u u . (p. 547) sin = A 2 2 4. If u is acute, the Half-angle Formula for the cosine function 1 + cos u u is cos = A . (p. 547) 2 2
Concepts and Vocabulary *5. To transform the equation Ax + Bxy + Cy + Dx + Ey + F = 0, 2
2
B ≠ 0
into one in x′ and y′ without an x′y′@term, rotate the axes through an acute angle u that satisfies the equation. 6. Except for degenerate cases, the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 defines a(n) parabola if B2 - 4AC = 0.
8. True or False The equation ax2 + 6y2 - 12y = 0 defines an ellipse if a 7 0. True 9. True or False The equation 3x2 + Bxy + 12y2 = 10 defines a parabola if B = - 12. True 10. True or False To eliminate the xy-term from the equation x2 - 2xy + y2 - 2x + 3y + 5 = 0, rotate the axes through an angle u, where cot u = B2 - 4AC. False
7. Except for degenerate cases, the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 defines an ellipse if B2 - 4AC 6 0.
Skill Building In Problems 11–20, identify the graph of each equation without completing the squares. 11. x2 + 4x + y + 3 = 0
Parabola
13. 6x + 3y - 12x + 6y = 0 2
2
15. 3x - 2y + 6x + 4 = 0 2
2
17. 2y2 - x2 - y + x = 0
Parabola
Ellipse
14. 2x2 + y2 - 8x + 4y + 2 = 0
Hyperbola
16. 4x - 3y - 8x + 6y + 1 = 0
Hyperbola
19. x2 + y2 - 8x + 4y = 0
12. 2y2 - 3y + 3x = 0
Circle
2
2
18. y2 - 8x2 - 2x - y = 0
Ellipse Hyperbola
Hyperbola
20. 2x2 + 2y2 - 8x + 8y = 0
Circle
In Problems 21–30, determine the appropriate rotation formulas to use so that the new equation contains no xy-term. *21. x2 + 4xy + y2 - 3 = 0
*22. x2 - 4xy + y2 - 3 = 0
*23. 5x2 + 6xy + 5y2 - 8 = 0
*24. 3x2 - 10xy + 3y2 - 32 = 0
*25. 13x - 623xy + 7y - 16 = 0
*26. 11x2 + 1023xy + y2 - 4 = 0
*27. 4x2 - 4xy + y2 - 825x - 1625y = 0
*28. x2 + 4xy + 4y2 + 525y + 5 = 0
*29. 25x - 36xy + 40y - 12213x - 8213y = 0
*30. 34x2 - 24xy + 41y2 - 25 = 0
2
2
2
2
In Problems 31–42, rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation. Refer to Problems 21–30 for Problems 31–40. *31. x2 + 4xy + y2 - 3 = 0
*32. x2 - 4xy + y2 - 3 = 0
*33. 5x2 + 6xy + 5y2 - 8 = 0
*34. 3x2 - 10xy + 3y2 - 32 = 0
*35. 13x - 623xy + 7y - 16 = 0
*36. 11x2 + 1023xy + y2 - 4 = 0
*37. 4x2 - 4xy + y2 - 825x - 1625y = 0
*38. x2 + 4xy + 4y2 + 525y + 5 = 0
*39. 25x - 36xy + 40y - 12213x - 8213y = 0
*40. 34x2 - 24xy + 41y2 - 25 = 0
*41. 16x2 + 24xy + 9y2 - 130x + 90y = 0
*42. 16x2 + 24xy + 9y2 - 60x + 80y = 0
2
2
2
2
%VFUPTQBDFSFTUSJDUJPOT BOTXFSTUPUIFTFFYFSDJTFTNBZCFGPVOEJOUIF"OTXFSTJOUIFCBDLPGUIFCPPL
SECTION 9.6 Polar Equations of Conics
731
In Problems 43–52, identify the graph of each equation without applying a rotation of axes. 43. x2 + 3xy - 2y2 + 3x + 2y + 5 = 0 45. x2 - 7xy + 3y2 - y - 10 = 0
Hyperbola
47. 9x2 + 12xy + 4y2 - x - y = 0
Parabola
49. 10x2 - 12xy + 4y2 - x - y - 10 = 0 51. 3x2 - 2xy + y2 + 4x + 2y - 1 = 0
44. 2x2 - 3xy + 4y2 + 2x + 3y - 5 = 0
Hyperbola
46. 2x2 - 3xy + 2y2 - 4x - 2 = 0
48. 10x2 + 12xy + 4y2 - x - y + 10 = 0 50. 4x2 + 12xy + 9y2 - x - y = 0
Ellipse
Ellipse
Parabola
52. 3x2 + 2xy + y2 + 4x - 2y + 10 = 0
Ellipse
Ellipse
Ellipse
Ellipse
Applications and Extensions In Problems 53–56, apply the rotation formulas (5) to
*56. Prove that, except for degenerate cases, the equation
Ax + Bxy + Cy + Dx + Ey + F = 0 2
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
2
B %FGJOFTBQBSBCPMBJGB2 - 4AC = 0. C %FGJOFTBOFMMJQTF PSBDJSDMF JGB2 - 4AC 6 0. D %FGJOFTBIZQFSCPMBJGB2 - 4AC 7 0.
to obtain the equation A′ x′2 + B′ x′ y′ + C′ y′2 + D′ x′ + E′ y′ + F′ = 0 *53. Express A′, B′, C′, D′, E′, and F′ in terms of A, B, C, D, E, F, and the angle u of rotation. [Hint:3FGFSUPFRVBUJPO > *54. Show that A + C = A′ + C′, and thus show that A + C is invariant; that is, its value does not change under a rotation of axes. *55. 3FGFSUP1SPCMFN4IPXUIBUB2 - 4AC is invariant.
*57. Use the rotation formulas (5) to show that distance is invariant under a rotation of axes. That is, show that the distance from P1 = 1x1 , y1 2 to P2 = 1x2 , y2 2 in the xy-plane equals the distance from P1 = 1x1= , y1= 2 to P2 = 1x2= , y2= 2 in the x′ y′@plane. *58. Show that the graph of the equation x1>2 + y1>2 = a1>2 is part of the graph of a parabola.
Discussion and Writing 59. Formulate a strategy for discussing and graphing an equation of the form
60. Explain how your strategy presented in Problem 59 changes if the equation is of the following form:
Ax2 + Cy2 + Dx + Ey + F = 0
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
Retain Your Knowledge Problems 61–64 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 61. Solve the triangle described: a = 7, b = 9, and c = 11. 62. Find the area of the triangle described: a = 14, b = 11, and C = 30°. 38.5 61. A ≈ 39.4°, B ≈ 54.7°, C ≈ 85.9°
63. Transform the equation xy = 1 from rectangular coordinates to polar coordinates. r 2 cos u sin u = 1 64. Write the complex number 2 - 5i in polar form. 229( cos 291.8° + i sin 291.8°)
‘Are You Prepared?’ Answers 1. sin A cos B + cos A sin B
2. 2 sin u cos u
3.
1 - cos u B 2
4.
1 + cos u B 2
9.6 Polar Equations of Conics PREPARING FOR THIS SECTION Before getting started, review the following: r 1PMBS$PPSEJOBUFT 4FDUJPO QQm
Now Work the ‘Are You Prepared?’ problems on page 736.
OBJECTIVES 1 Analyze and Graph Polar Equations of Conics (p. 731) 2 Convert the Polar Equation of a Conic to a Rectangular Equation (p. 735)
1 Analyze and Graph Polar Equations of Conics Sections 9.2 through 9.4 gave separate definitions for the parabola, ellipse, and hyperbola based on geometric properties and the distance formula. This section presents an alternative definition that simultaneously defines all these conics. As
732
CHAPTER 9 Analytic Geometry
XFTIBMMTFF UIJTBQQSPBDIJTXFMMTVJUFEUPQPMBSDPPSEJOBUFSFQSFTFOUBUJPO 3FGFS to Section 8.1.)
DEFINITION
Let D denote a fixed line called the directrix; let F denote a fixed point called the focus, which is not on D; and let e be a fixed positive number called the eccentricity. A conic is the set of points P in the plane such that the ratio of the distance from F to P to the distance from D to P equals e. That is, a conic is the collection of points P for which d 1F, P2 = e d 1D, P2
(1)
If e = 1, the conic is a parabola. If e 6 1, the conic is an ellipse. If e 7 1, the conic is a hyperbola. Observe that if e = 1, the definition of a parabola in equation (1) is exactly the same as the definition used earlier in Section 9.2. In the case of an ellipse, the major axis is a line through the focus perpendicular to the directrix. In the case of a hyperbola, the transverse axis is a line through the focus perpendicular to the directrix. For both an ellipse and a hyperbola, the eccentricity e satisfies
e =
Figure 51 Directrix D
P ⫽ (r, θ) d(D, P ) r
p Pole O (Focus F )
θ Q
Polar axis
c a
(2)
where c is the distance from the center to the focus and a is the distance from the center to a vertex. Just as we did earlier using rectangular coordinates, we derive equations for the conics in polar coordinates by choosing a convenient position for the focus F and the directrix D. The focus F is positioned at the pole, and the directrix D is either parallel or perpendicular to the polar axis. Suppose that we start with the directrix D perpendicular to the polar axis at a distance p units to the left of the pole (the focus F). See Figure 51. If P = 1r, u2 is any point on the conic, then, by equation (1), d 1F, P2 = e or d 1F, P2 = e # d 1D, P2 d 1D, P2
(3)
Now use the point Q obtained by dropping the perpendicular from P to the polar axis to calculate d 1D, P2. d 1D, P2 = p + d 1O, Q2 = p + r cos u Using this expression and the fact that d 1F, P2 = d 1O, P2 = r in equation (3) gives e # d 1D, P2 e1p + r cos u2 ep + er cos u ep ep ep r = 1 - e cos u
d 1F, P2 r r r - er cos u r 11 - e cos u2
= = = = =
SECTION 9.6 Polar Equations of Conics
THEOREM
733
Polar Equation of a Conic; Focus at the Pole; Directrix Perpendicular to the Polar Axis a Distance p to the Left of the Pole The polar equation of a conic with focus at the pole and directrix perpendicular to the polar axis at a distance p to the left of the pole is r =
ep 1 - e cos u
(4)
where e is the eccentricity of the conic.
EXAM PL E 1
Analyzing and Graphing the Polar Equation of a Conic 4 2 - cos u The given equation is not quite in the form of equation (4), since the first term in the EFOPNJOBUPSJTJOTUFBEPG%JWJEFUIFOVNFSBUPSBOEEFOPNJOBUPSCZUPPCUBJO ep 2 r = r = 1 1 - e cos u 1 - cos u 2 This equation is in the form of equation (4), with Analyze and graph the equation: r =
Solution
e =
1 2
and
ep = 2
Then
Figure 52 Directrix
4 3 3
( , π) 4– 3
F
4– 3
( , 0)
(4, 0)
Polar axis
1 p = 2, so p = 4 2 1 Hence the conic is an ellipse, because e = 6 1. One focus is at the pole, and 2 the directrix is perpendicular to the polar axis, a distance of p = 4 units to the left of the pole. It follows that the major axis is along the polar axis. To find the vertices, let 4 u = 0 and u = p. The vertices of the ellipse are 14, 02 and a , pb . 3 4 The midpoint of the vertices, a , 0b in polar coordinates, is the center of the ellipse. 3 4 2
xS1
x S -1 x S -3
xS4
xSp
x S -1
xS2
xS0
1 x
xS2
*35. lim f 1x2, f 1x2 = b
x2 if x Ú 0 2x if x 6 0
*36. lim f 1x2, f 1x2 = b
x - 1 if x 6 0 3x - 1 if x Ú 0
*37. lim f 1x2, f 1x2 = b
3x if x … 1 x + 1 if x 7 1
*38. lim f 1x2, f 1x2 = b
x2 if x … 2 2x - 1 if x 7 2
*40. lim f 1x2, f 1x2 = b
1 if x 6 0 - 1 if x 7 0
*42. lim f 1x2, f 1x2 = b
ex if x 7 0 1 - x if x … 0
xS0
xS1
xS0
xS2
x if x 6 0 *39. lim f 1x2, f 1x2 = c 1 if x = 0 xS0 3x if x 7 0 *41. lim f 1x2, f 1x2 = b xS0
xS0
sin x if x … 0 x2 if x 7 0
xS0
1 x2
In Problems 43–48, use a graphing utility to find the indicated limit rounded to two decimal places. x3 - x2 + x - 1 x S 1 x4 - x3 + 2x - 2
0.67
44. lim
x3 - x2 + 3x - 3 xS1 x2 + 3x - 4
0.8
47. lim
43. lim
46. lim
x3 + x2 + 3x + 3 x S -1 x4 + x3 + 2x + 2
4
45. lim
x3 - 2x2 + 4x - 8 xS2 x2 + x - 6
x3 + 2x2 + x x S -1 x4 + x3 + 2x + 2
0
48. lim
x3 - 3x2 + 4x - 12 x S 3 x4 - 3x3 + x - 3
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
1.6 0.46
SECTION 13.2 Algebra Techniques for Finding Limits
929
Retain Your Knowledge Problems 49–52 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 49. Let A12, - 32 and B 16, - 112 CFQPJOUTJOBQMBOF'JOEUIF distance between the points and the midpoint of the line segment connecting the points. d = 42; M = (4, - 7)
*50. 'JOE UIF DFOUFS GPDJ BOE WFSUJDFT PG UIF FMMJQTF 1x - 22 2 1y + 12 2 + = 1. 9 13
‘Are You Prepared?’ Answers
51. Logan invests $4000 at an annual interest rate of 6%. How much money will she have after 10 years if interest is compounded continuously? $7288.48 52. Assuming r 7 0 and 0 … u 6 2p, find the polar coordinates of the point whose rectangular coordinates are 1 - 2, 2232. 2p a4, b 3
2. f 102 = 0
1. 4FF'JHVSFPOQBHF
13.2 Algebra Techniques for Finding Limits OBJECTIVES 1 2 3 4 5
Find the Limit of a Sum, a Difference, and a Product (p. 930) Find the Limit of a Polynomial (p. 931) Find the Limit of a Power or a Root (p. 932) Find the Limit of a Quotient (p. 933) Find the Limit of an Average Rate of Change (p. 934)
We mentioned in the previous section that algebra can sometimes be used to find the exact value of a limit. This is accomplished by developing two formulas involving limits and several properties of limits.
THEOREM
Two Formulas: lim b and lim x xSc
xSc
Limit of a Constant
In Words
'PSUIFDPOTUBOUGVODUJPOf1x2 = b,
The limit of a constant is the constant.
lim f1x2 = lim b = b
(1)
xSc
xSc
where c is any number.
Limit of x 'PSUIFJEFOUJUZGVODUJPOf1x2 = x,
In Words The limit of x as x approaches c is c.
lim f1x2 = lim x = c
xSc
(2)
xSc
where c is any number. We use graphs to establish formulas (1) and (2). Since the graph of a constant function is a horizontal line, it follows that, no matter how close x is to c, the corresponding value of f1x2 equals b. That is, lim b = b.4FF'JHVSF xSc 4FF'JHVSF'PSBOZDIPJDFPGc, as x gets closer to c, the corresponding value of f1x2 is x, which is just as close to c. That is, lim x = c. xSc
Figure 4
(0, b)
y
Figure 5
y f(x ) ⫽ x
f(x) b c
x
c (c, c)
c
x
930
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
EX A MPL E 1
Using Formulas (1) and (2) (a) lim =
(b) lim x = 3
xS3
(c) lim ( - 8) = - 8
xS3
Now Work
PROBLEM
(c)
xS0
lim x = -
x S -1>2
7
1 2
r
'PSNVMBT BOE
XIFO VTFE XJUI UIF QSPQFSUJFT UIBU GPMMPX FOBCMF VT UP evaluate limits of more complicated functions.
1 Find the Limit of a Sum, a Difference, and a Product In the following properties, we assume that f and g are two functions for which both lim f1x2 and lim g1x2 exist. xSc
THEOREM
xSc
Limit of a Sum lim 3 f 1x2 + g1x2 4 = lim f1x2 + lim g1x2
xSc
In Words
xSc
(3)
xSc
The limit of the sum of two functions equals the sum of their limits.
EX A MPL E 2
Finding the Limit of a Sum 'JOE
Solution
lim 1x + 42
x S -3
The limit we seek is the sum of two functions: f1x2 = x and g1x2 = 4.'SPNGPSNVlas (1) and (2), we know that lim f1x2 = lim x = - 3 and x S -3
x S -3
lim g1x2 = lim 4 = 4
x S -3
x S -3
'SPNGPSNVMB
JUGPMMPXTUIBU
lim 1x + 42 = lim x + lim 4 = - 3 + 4 = 1 x S -3
x S -3
THEOREM
r
x S -3
Limit of a Difference lim 3 f 1x2 - g1x2 4 = lim f1x2 - lim g1x2 xSc
xSc
In Words
xSc
(4)
The limit of the difference of two functions equals the difference of their limits.
EX A MPL E 3
Finding the Limit of a Difference 'JOE lim 16 - x2 xS4
Solution
The limit we seek is the difference of two functions: f 1x2 = 6 and g1x2 = x.'SPN formulas (1) and (2), we know that lim f1x2 = lim 6 = 6 and
xS4
xS4
lim g1x2 = lim x = 4
xS4
xS4
'SPNGPSNVMB
JUGPMMPXTUIBU
lim 16 - x2 = lim 6 - lim x = 6 - 4 = 2
xS4
THEOREM
xS4
Limit of a Product lim 3 f 1x2 # g1x2 4 = 3 lim f1x2 4 3 lim g1x2 4
xSc
In Words The limit of the product of two functions equals the product of their limits.
EX A MPL E 4
r
xS4
Finding the Limit of a Product 'JOE
lim 1 - 4x2
x S -
xSc
xSc
(5)
SECTION 13.2 Algebra Techniques for Finding Limits
Solution
931
The limit we seek is the product of two functions: f1x2 = - 4 and g1x2 = x.'SPN formulas (1) and (2), we know that lim f1x2 = lim 1 - 42 = - 4 and
x S -
lim g1x2 = lim x = -
x S -
x S -
x S -
'SPNGPSNVMB
JUGPMMPXTUIBU
lim 1 - 4x2 = 3 lim 1 - 42 4 3 lim x4 = 1 - 42 1 - 2 = 20
x S -
EXAM PL E 5
x S -
Finding Limits Using Algebraic Properties 'JOE B lim 13x - 2
(b) lim 1x2 2
x S -2
Solution
r
x S -
xS2
(a) lim 13x - 2 = lim 13x2 - lim = 3 lim 34 3 lim x4 - lim x S -2
x S -2
x S -2
x S -2
x S -2
x S -2
= 132 1 - 22 - = - 6 - = - 11
(b) lim 1x2 2 = 3 lim 4 3 lim x2 4 = lim 1x # x2 = 3 lim x4 3 lim x4 xS2
xS2
Now Work
xS2
xS2
xS2
xS2
= # 2 # 2 = 20
r
15
PROBLEM
Notice in the solution to part (b) that lim 1x2 2 = # 22. xS2
THEOREM
Limit of a Monomial If n Ú 1 is a positive integer and a is a constant, then lim 1axn 2 = ac n
(6)
xSc
for any number c.
Proof
lim (axn) = [lim a][lim xn] = a[lim (x · x · x · . . . · x)]
x
c
x
x
c
c
x
c
n factors
= a[lim x][lim x][lim x] . . . [lim x] x
c
x
c
x c n factors
x
c
= a · c · c · c · . . . · c = acn n factors
EXAM PL E 6
■
Finding the Limit of a Monomial 'JOE lim 1 - 4x3 2 xS2
Solution
lim 1 - 4x3 2 = - 4 # 23 = - 4 # 8 = - 32
r
xS2
2 Find the Limit of a Polynomial Since a polynomial is a sum of monomials, we can use formula (6) and the repeated use of formula (3) to obtain the following result:
THEOREM
Limit of a Polynomial If P is a polynomial function, then lim P1x2 = P1c2
xSc
for any number c.
(7)
932
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
Proof If P is a polynomial function—that is, if P1x2 = an xn + an - 1 xn - 1 + g + a1 x + a0 then
In Words
lim P1x2 = lim 3 an xn + an - 1 xn - 1 + g + a1 x + a0 4 S
To find the limit of a polynomial as x approaches c, all we need to do is evaluate the polynomial at c.
xSc
x
c
x
c
= lim 1an xn 2 + lim 1an - 1 xn - 1 2 + g + lim 1a1 x2 + lim a0 S S S S x
= an c + an - 1 c n
c
n-1
x
c
x
c
+ g + a1 c + a0
= P1c2
EX A MPL E 7
■
Finding the Limit of a Polynomial 'JOE lim 3 x4 - 6x3 + 3x2 + 4x - 24 xS2
Solution
lim 3 x4 - 6x3 + 3x2 + 4x - 24 = # 24 - 6 # 23 + 3 # 22 + 4 # 2 - 2
xS2
= # 16 - 6 # 8 + 3 # 4 + 8 - 2
r
= 80 - 48 + 12 + 6 = 0
Now Work
PROBLEM
17
3 Find the Limit of a Power or a Root THEOREM
Limit of a Power or Root If lim f1x2 exists and if n Ú 2 is a positive integer, then xSc
lim 3 f 1x2 4 n = 3 lim f1x2 4 n
(8)
xSc
xSc
and n
n lim f1x2 lim 2f1x2 = 5 xSc
(9)
xSc
n
n lim f1x2 In formula (9), we require that both 2f1x2 and 5 be defined. xSc
Look carefully at equations (8) and (9) and compare each side.
EX A MPL E 8
Finding the Limit of a Power or a Root 'JOE B lim 13x - 2 4 xS1
Solution
(b) lim 2x2 + 8 xS0
(c) lim 1x3 - x + 32
4>3
x S -1
(a) lim 13x - 2 4 = 3 lim 13x - 2 4 4 = 1 - 22 4 = 16 xS1
xS1
(x2 + 8) = 28 = 222 (b) lim 2x2 + 8 = 5 lim xS0 xS0 (c)
lim 1x3 - x + 32
x S -1
4>3
3 lim (x3 - x + 3)4 = 5 x S -1 3 3 3 [ lim (x3 - x + 3)]4 = 2 = 5 1 - 12 4 = 2 1 = 1 x S -1
Now Work
PROBLEM
27
r
SECTION 13.2 Algebra Techniques for Finding Limits
933
4 Find the Limit of a Quotient THEOREM
Limit of a Quotient lim c
In Words The limit of the quotient of two functions equals the quotient of their limits, provided that the limit of the denominator is not zero.
EXAM PL E 9
xSc
(10)
xSc
provided that lim g1x2 ≠ 0. xSc
Finding the Limit of a Quotient 'JOE lim
xS1
Solution
lim f1x2 f 1x2 xSc d = g1x2 lim g1x2
x3 - x + 2 3x + 4
The limit we seek involves the quotient of two functions: f1x2 = x3 - x + 2 and g1x2 = 3x + 4.'JSTU XFàOEUIFMJNJUPGUIFEFOPNJOBUPSg1x2. lim g1x2 = lim 13x + 42 = 7 xS1
xS1
Since the limit of the denominator is not zero, we can proceed to use formula (10). x3 - x + 2 lim = xS1 3x + 4
Now Work
PROBLEM
lim 1x3 - x + 22
xS1
lim 13x + 42
xS1
=
6 7
r
25
When the limit of the denominator is zero, formula (10) cannot be used. In such cases, other strategies need to be used. Let’s look at two examples.
EX AM PL E 10
Finding the Limit of a Quotient 'JOE (a) lim
xS3
Solution
x2 - x - 6 x2 - 9
(b) lim
xS0
x - sin x x
(a) The limit of the denominator equals zero, so formula (10) cannot be used. Instead, notice that the expression can be factored as 1x - 32 1x + 22 x2 - x - 6 = 2 1x - 32 1x + 32 x - 9 When we compute a limit as x approaches 3, we are interested in the values of the function when x is close to 3 but unequal to 3. Since x ≠ 3, we can divide out the 1x - 32 >s.'PSNVMB DBOUIFOCFVTFE lim
xS3
lim 1x + 22 1x - 32 1x + 22 x2 - x - 6 xS3 = = = lim 2 S x 3 1x 32 1x + 32 lim 1x + 32 6 x - 9 xS3
(b) Again, the limit of the denominator is zero. In this situation, perform the indicated operation and divide by x. lim
xS0
x - sin x x x sin x sin x - lim = lim c d = lim = - 1 = 4 xS0 x xS0 x xS0 x x x c c Limit of a difference
Refer to Example 3, Section 13.1
r
934
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
EX AM PL E 11
Finding Limits Using Algebraic Properties 'JOE lim
xS2
Solution
x3 - 2x2 + 4x - 8 x4 - 2x3 + x - 2
The limit of the denominator is zero, so formula (10) cannot be used. We factor the expression. x2 1x - 22 + 41x - 22 1x2 + 42 1x - 22 x3 - 2x2 + 4x - 8 = = x3 1x - 22 + 11x - 22 1x3 + 12 1x - 22 x4 - 2x3 + x - 2 c
Factor by grouping
Then lim
xS2
1x2 + 42 1x - 22 x3 - 2x2 + 4x - 8 8 = lim = 3 4 3 x S 2 1x + 12 1x - 22 9 x - 2x + x - 2
r
which is exact.
Compare the exact solution above with the approximate solution found in Example 6 of Section 13.1.
5 Find the Limit of an Average Rate of Change EX AM PL E 12
Finding the Limit of an Average Rate of Change 'JOEUIFMJNJUBTx approaches 2 of the average rate of change of the function f1x2 = x2 + 3x from 2 to x.
Solution
The average rate of change of f from 2 to x is
f 1x2 - f 122 1x2 + 3x2 - 10 1x + 2 1x - 22 ∆y = = = ∆x x - 2 x - 2 x - 2
The limit of the average rate of change is lim
xS2
f1x2 - f122 1x2 + 3x2 - 10 1x + 2 1x - 22 = lim = lim = 7 xS2 xS2 x - 2 x - 2 x - 2
r
SUMMARY To find exact values for lim f1x2 , try the following: xSc
1. If f is a polynomial function, lim f1x2 = f1c2 . (formula (7)) xSc 2. If f is a polynomial raised to a power or is the root of a polynomial, use formula (8) or (9) with formula (7). 3. If f is a quotient and the limit of the denominator is not zero, use the fact that the limit of a quotient is the quotient of the limits. (formula (10)) 4. If f is a quotient and the limit of the denominator is zero, use other techniques, such as factoring.
13.2 Assess Your Understanding Concepts and Vocabulary 1. The limit of the product of two functions equals the product of their limits. 2. lim b = b . xSc
3. lim x = xSc
c .
4. True or False The limit of a polynomial function as x BQQSPBDIFTFRVBMTUIFWBMVFPGUIFQPMZOPNJBMBU True 5. True or False 5IFMJNJUPGBSBUJPOBMGVODUJPOBUFRVBMTUIF WBMVFPGUIFSBUJPOBMGVODUJPOBU 'BMTF 6. True or False The limit of a quotient equals the quotient of the limits. 'BMTF
SECTION 13.2 Algebra Techniques for Finding Limits
935
Skill Building In Problems 7–42, find each limit algebraically. 7. lim xS1
8. lim 1 - 32
-3
xS1
11. lim 1x2
12. lim 1 - 3x2
- 10
x S -2
15. lim 13x + 22
xS4
19. lim 1x4 - 3x2 + 6x - 92
-1
21. lim 1x2 + 12
22. lim 13x - 42 2
3
8
xS1
25. lim
xS0
x2 - 4 x2 + 4
x2 - 4 x S 2 x2 - 2x
29. lim
x3 - 1 xS1 x - 1
33. lim
-1
26. lim
2
30. lim
xS2
3x + 4 x2 + x
x4 - 1 xS1 x - 1
2 3
38. lim
x3 - x2 + 3x - 3 xS1 x2 + 3x - 4
4
41. lim
40. lim
lim 18x - 7x3 + 8x2 + x - 42
xS0
x2 - x - 12 x S -3 x2 - 9
35.
lim
x S -1
1x + 12 2 x - 1 2
x S -1
x2 + x - 6 x S -3 x2 + 2x - 3
7 6
32. lim
x3 - 8 x S 2 x2 - 4
36. lim
0
x3 + x2 + 3x + 3 x S -1 x4 + x3 + 2x + 2
4
39. lim
x3 + 2x2 + x x S -1 x + x3 + 2x + 2
0
42. lim
1
28. lim 12x + 12 >3
32
lim
28
2
24. lim 21 - 2x
3
31.
4
xS2
x S -1
xS2
- 4
18. lim 18x2 - 42
8
27. lim 13x - 22 >2
4
x3 - x2 + x - 1 x S 1 x4 - x3 + 2x - 2
37. lim
x S -3
x S -1
xS1
1 2
34. lim
14. lim 12x3 2
80
23. lim 2x + 4
4
3
x2 + x x S -1 x2 - 1
x S -3
lim 13x2 - x2
17. 20.
xS2
3
xS2
- 13
xS3
xS1
13. lim 1x4 2
-3
10. lim x
4
xS4
- 12
16. lim 12 - x2
8
xS2
9. lim x
x3 - 2x2 + 4x - 8 xS2 x2 + x - 6 x3 - 3x2 + 4x - 12 x S 3 x4 - 3x3 + x - 3
-1 4
3
8 13 28
In Problems 43–52, find the limit as x approaches c of the average rate of change of each function from c to x. 43. c = 2; f 1x2 = x - 3
46. c = 3; f 1x2 = x
3
44. c = - 2; f 1x2 = 4 - 3x
47. c = - 1; f 1x2 = x + 2x 2
27
49. c = 0; f 1x2 = 3x3 - 2x2 + 4 51. c = 1; f 1x2 =
1 x
45. c = 3; f 1x2 = x2
-3
48. c = - 1; f 1x2 = 2x - 3x
0
50. c = 0; f 1x2 = 4x3 - x + 8
0
52. c = 1; f 1x2 =
-1
6 2
1 x2
-7
-
-2
In Problems 53–56, use the properties of limits and the facts that lim
xS0
sin x = 1 x
lim
xS0
cos x - 1 = 0 x
lim sin x = 0
xS0
lim cos x = 1
xS0
where x is in radians, to find each limit. 53. lim
xS0
tan x x
3 sin x + cos x - 1 xS0 4x
55. lim
54. lim
1
sin12x2
2 x [Hint: 6TFB%PVCMFBOHMF'PSNVMB> xS0
3 4
56. lim
xS0
sin2 x + sin x1cos x - 12 x2
1
Retain Your Knowledge Problems 57–60 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 3 - x 23 p -1 . 60° or rad 59. Give the exact value of sin-1 *57. Graph the function f 1x2 = x3 + x2 + 1. 58. g (x) = x - 2 2 3 2x + 3 60. Use the Binomial Theorem to expand 1x + 22 4. 58. 'JOEUIFJOWFSTFPGUIFGVODUJPOg1x2 = . x + 1 x4 + 8x3 + 24x2 + 32x + 16
936
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
13.3 One-sided Limits; Continuous Functions PREPARING FOR THIS SECTION Before getting started, review the following: r r r r
1JFDFXJTFEFGJOFE'VODUJPOT 4FDUJPO QQm
-JCSBSZPG'VODUJPOT 4FDUJPO QQm
1PMZOPNJBM'VODUJPOT 4FDUJPO QQm
1SPQFSUJFTPG3BUJPOBM'VODUJPOT 4FDUJPO QQm
r 5IF(SBQIPGB3BUJPOBM'VODUJPO 4FDUJPO QQm
r 1SPQFSUJFTPGUIF&YQPOFOUJBM'VODUJPO 4FDUJPO pp. 331 and 333) r 1SPQFSUJFTPGUIF-PHBSJUINJD'VODUJPO 4FDUJPO p. 346) r 1SPQFSUJFTPGUIF5SJHPOPNFUSJD'VODUJPOT 4FDUJPO QQBOE BOE4FDUJPO QQm
Now Work the ‘Are You Prepared?’ problems on page 940.
OBJECTIVES 1 Find the One-sided Limits of a Function (p. 936) 2 Determine Whether a Function Is Continuous (p. 938)
1 Find the One-sided Limits of a Function Earlier we described lim f1x2 = N by saying that as x gets closer to c but remains xSc unequal to c, the corresponding values of f1x2 get closer to N. Whether we use a numerical argument or the graph of the function f, the variable x can get closer to c in only two ways: by approaching c from the left, through numbers less than c, or by approaching c from the right, through numbers greater than c. If we approach c from only one side, we have a one-sided limit. The notation
lim f1x2 = L
x S c-
is called the left limit. It is read “the limit of f1x2 as x approaches c from the left equals L” and may be described by the following statement:
In Words
x S c - means x is approaching the value of c from the left, so x 6 c.
As x gets closer to c but remains less than c, the corresponding value of f1x2 gets closer to L.
The notation x S c - is used to remind us that x is less than c. The notation
lim f1x2 = R
x S c+
is called the right limit. It is read “the limit of f1x2 as x approaches c from the right equals R” and may be described by the following statement:
In Words
x S c + means x is approaching the value of c from the right, so x 7 c.
As x gets closer to c, but remains greater than c, the corresponding value of f1x2 gets closer to R.
The notation x S c + is used to remind us that x is greater than c.
SECTION 13.3 One-sided Limits; Continuous Functions
937
'JHVSFJMMVTUSBUFTMFGUBOESJHIUMJNJUT Figure 6
y
y
L
R
c
x
x
c
x
x
lim
lim
x
x
(b)
(a)
The left and right limits can be used to determine whether lim f1x2 exists. See xSc 'JHVSF Figure 7
y R
y LR
L c
c
x
lim f(x) lim f(x)
lim f(x) lim f(x)
x c
x
x c
x c
x c
(b)
(a)
"T'JHVSF B JMMVTUSBUFT lim f1x2 exists and equals the common value of the xSc left limit and the right limit 1L = R2.*O'JHVSF C
XFTFFUIBUlim f1x2 does not xSc exist because L ≠ R. This leads us to the following result.
THEOREM
Suppose that lim- f1x2 = L and lim+ f1x2 = R. Then lim f1x2 exists if and xSc xSc xSc only if L = R.'VSUIFSNPSF JGL = R, then lim f1x2 = L = R. xSc
Collectively, the left and right limits of a function are called one-sided limits of the function.
Finding One-sided Limits of a Function
EXAM PL E 1
'PSUIFGVODUJPO 2x - 1 if x 6 2 f1x2 = c 1 if x = 2 x - 2 if x 7 2 find: (a) lim- f1x2 xS2
Solution
Figure 8 y
(b) lim+ f1x2
(c) lim f1x2
xS2
xS2
'JHVSFTIPXTUIFHSBQIPGf. (a) To find lim- f (x), look at the values of f when x is close to 2 but less than 2. xS2
Since f1x2 = 2x - 1 for such numbers, we conclude that lim f1x2 = lim- 12x - 12 = 3
2
x S 2-
(2, 1)
xS2
(b) To find lim+ f1x2, look at the values of f when x is close to 2 but greater than 2. xS2
22
2 22
4
x
Since f1x2 = x - 2 for such numbers, we conclude that lim f1x2 = lim+ 1x - 22 = 0
x S 2+
xS2
(c) Since the left and right limits are unequal, lim f1x2 does not exist. xS2
Now Work
PROBLEMS
21
AND
35
r
938
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
2 Determine Whether a Function Is Continuous We have observed that f1c2, the value of the function f at c, plays no role in determining the one-sided limits of f at c. What is the role of the value of a function at c and its one-sided limits at c? Let’s look at some of the possibilities. See 'JHVSF Figure 9
y y
y
y 5 f(x) f(c)
c x lim f(x) 5 lim f(x), so lim f(x) exists;
x
c2
x
c1
x
c
x
lim f(x) 5 f(c)
x
c
x
y 5 f(x)
y 5 f(x)
f(c) c x lim f(x) 5 lim f(x), so lim f(x) exists; c2
x
c1
x
c x lim f(x) 5 lim f(x), so lim f(x) exists; c⫺
x
c
lim f(x) fi f(c)
x
c⫹
x
c
(a)
(b)
y
(c)
y
y 5 f(x)
y y 5 f(x)
y 5 f(x) f(c) c x lim f(x) fi lim f(x), so lim f(x) does not exist;
x
2
c
x
1
c
x
c
f(c) is defined
c x lim f(x) fi lim f(x), so lim f(x) does not exist;
x
c
⫺
c
f(c) is not defined
x
c
⫹
x
c
f(c) is not defined
c lim f(x) 5 f(c) fi lim f(x),
x
c
x
c
x
⫹
so lim f(x) does not exist; x
(e)
(d)
⫺
c
f(c) is defined
(f)
Much earlier in this text, we said that a function f was continuous if its graph DPVME CF ESBXO XJUIPVU MJGUJOH QFODJM GSPN QBQFS 'JHVSF SFWFBMT UIBU UIF POMZ HSBQI UIBU IBT UIJT DIBSBDUFSJTUJD JT UIF HSBQI JO 'JHVSF B
GPS XIJDI UIF POF sided limits at c each exist and are equal to the value of f at c. This leads us to the following definition.
DEFINITION
A function f is continuous at c if 1. f is defined at c; that is, c is in the domain of f so that f1c2 equals a number. 2. lim- f1x2 = f 1c2 xSc
3. lim+ f1x2 = f1c2 xSc
In other words, a function f is continuous at c if lim f1x2 = f1c2
xSc
If f is not continuous at c, we say that f is discontinuous at c. Each function XIPTFHSBQIBQQFBSTJO'JHVSFT C UP G JTEJTDPOUJOVPVTBUc. Look again at formula (7) on page 931. Based on (7), we conclude that a polynomial function is continuous at every number. Look at formula (10) on page 933. We conclude that a rational function is continuous at every number, except any at which it is not defined. At numbers where a rational function is not defined, either a hole appears in the graph or else an asymptote appears.
Now Work
PROBLEM
27
SECTION 13.3 One-sided Limits; Continuous Functions
939
Determining the Numbers at Which a Rational Function Is Continuous
EXAM PL E 2
(a) Determine the numbers at which the rational function R 1x2 =
x - 2 x - 6x + 8 2
is continuous. (b) Use limits to analyze the graph of R near 2 and near 4. (c) Graph R.
Solution
(a) Since R 1x2 =
x - 2 , the domain of R is 5 x x ≠ 2, x ≠ 46 . 1x - 22 1x - 42
R is a rational function, so it is defined at every number except 2 and 4. We conclude that R is discontinuous at both 2 and 4. (Condition 1 of the definition is violated.) (b) To determine the behavior of the graph near 2 and near 4, look at lim R 1x2 and xS2 lim R 1x2. xS4
'PSlim R 1x2, we have xS2
x - 2 1 1 = lim = xS2 x - 4 1x - 22 1x - 42 2
lim R 1x2 = lim
xS2
xS2
1 As x gets closer to 2, the graph of R gets closer to - . Since R is not defined at 2, 2 1 the graph will have a hole at a2, - b . 2 'PSlim R 1x2, we have xS4
lim R 1x2 = lim
xS4
x4
x S 4+
Since 0 R 1x2 0 S q for x close to 4, the graph of R will have a vertical asymptote at x = 4. (c) It is easiest to graph R by observing that
2 1
2
12
x - 2 1 = lim xS4 x - 4 1x - 22 1x - 42
1 If x 6 4 and x is getting closer to 4, the value of is negative and is becomx - 4 ing unbounded; that is, lim- R 1x2 = - q . xS4 1 If x 7 4 and x is getting closer to 4, the value of is positive and is becomx 4 ing unbounded; that is, lim R 1x2 = q .
Figure 10 y
xS4
2
4
6
x
if x ≠ 2, then R 1x2 =
1
1 x - 2 = 1x - 22 1x - 42 x - 4
1 Therefore, the graph of R is the graph of y = shifted to the right 4 units with x 1 a hole at a2, - b .4FF'JHVSF. 2
r
Now Work
PROBLEM
73
The exponential, logarithmic, sine, and cosine functions are continuous at every number in their domain. The tangent, cotangent, secant, and cosecant functions are continuous except at numbers for which they are not defined, where asymptotes occur. The square root function and absolute value function are continuous at every number in their domain. The function f 1x2 = int 1x2 is continuous except for x = an integer, where a jump occurs in the graph. Piecewise-defined functions require special attention.
940
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
Determining Where a Piecewise-defined Function Is Continuous
EX A MPL E 3
Determine the numbers at which the following function is continuous. x2 if x … 0 f1x2 = c x + 1 if 0 6 x 6 2 - x if 2 … x …
Solution
The “pieces” of f —that is, y = x2, y = x + 1, and y = - x—are each continuous for every number since they are polynomials. In other words, when we graph the pieces, we will not lift our pencil. When we graph the function f, however, we have to be careful, because the pieces change at x = 0 and at x = 2. So the numbers we need to investigate for f are x = 0 and x = 2. At x = 0:
f102 = 02 = 0 lim f1x2 = lim- x2 = 0
Figure 11
3
x S 0+
xS0
Since lim+ f1x2 ≠ f102, f is not continuous at x = 0. xS0
(2, 3)
At x = 2:
2
2 (0, 0)
xS0
lim f1x2 = lim+ 1x + 12 = 1
y 4
x S 0-
2
4
6
f122 = - 2 = 3
lim f1x2 = lim- 1x + 12 = 3
x S 2-
xS2
x S 2+
xS2
lim f1x2 = lim+ 1 - x2 = 3
x
So f is continuous at x = 2. The graph of f,HJWFOJO'JHVSF EFNPOTUSBUFTUIFDPODMVTJPOTESBXOBCPWF
2
Now Work
PROBLEMS
53
AND
61
r
SUMMARY Library of Functions: Continuity Properties Function
Domain
Property
Polynomial function
All real numbers
Continuous at every number in the domain
P1x2 3BUJPOBMGVODUJPOR 1x2 = , Q1x2 P, Q are polynomials
5 x Q1x2 ≠ 06
Continuous at every number in the domain Hole or vertical asymptote where R is undefined
Exponential function
All real numbers
Continuous at every number in the domain
Logarithmic function
Positive real numbers
Continuous at every number in the domain
Sine and cosine functions
All real numbers
Continuous at every number in the domain
Tangent and secant functions
All real numbers, except p odd integer multiples of 2 All real numbers, except integer multiples of p
Continuous at every number in the domain p Vertical asymptotes at odd integer multiples of 2 Continuous at every number in the domain Vertical asymptotes at integer multiples of p
Cotangent and cosecant functions
13.3 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.
if x … 0 x2 1. 'PSUIFGVODUJPOf 1x2 = c x + 1 if 0 6 x 6 2, - x if 2 … x … find f 102 and f 122. QQm f(0) = 0; f(2) = 3
2. What are the domain and range of f 1x2 = ln x? (p. 346)
3. True or False The exponential function f 1x2 = e x is increasing on the interval 1 - q , q 2. (p. 331) True
941
SECTION 13.3 One-sided Limits; Continuous Functions
6. True or False Every polynomial function has a graph that can be traced without lifting pencil from paper. QQm
True
4. Name the trigonometric functions that have asymptotes. QQm m Secant, cosecant, tangent, cotangent 5. True or False Some rational functions have holes in their graph. QQm True
Concepts and Vocabulary 10. True or False 'PSBOZGVODUJPOf, lim- f 1x2 = lim+ f 1x2. 'BMTF
7. If we approach c from only one side, then we have a(n) one@sided limit. lim f 1x2 = R is used to describe the fact 8. The notation x S c + that as x gets closer to c but remains greater than c, the value of f 1x2 gets closer to R. 9. If lim f 1x2 = f 1c2, then f is continuous at xSc
xSc
xSc
11. True or False If f is continuous at c, then lim+ f 1x2 = f 1c2. True xSc
12. True or False Every polynomial function is continuous at every real number. True
c .
Skill Building In Problems 13–32, use the accompanying graph of y = f 1x2.
y
*13. What is the domain of f?
4
*14. What is the range of f?
15. 'JOEUIFx-intercept(s), if any, of f. - 8, - , - 3 17. 'JOEf 1 - 82 and f 1 - 42. 0; 2
16. 'JOEUIFy-intercept(s), if any, of f.
19. 'JOE lim - f 1x2.
q
20. 'JOE lim + f 1x2.
-q
21. 'JOE lim - f 1x2.
2
22. 'JOE lim + f 1x2.
-2
x S -6 x S -4
23. 'JOE lim- f 1x2.
18. 'JOEf 122 and f 162. x S -6 x S -4
24. 'JOE lim+ f 1x2.
1
xS2
xS2
3, 2
26. Does lim f 1x2 exist? If it does, what is it? Yes; 3 xS0
xS1
37.
lim + sin x
x S p>2
41. lim x S -1
x2 - 1 x3 + 1
0
35. lim- 12x3 + x2
38. lim- 13 cos x2
-3
39. lim+
xSp
-
2 3
32. Is f DPOUJOVPVTBU Yes
34. lim- 14 - 2x2 xS2
1
29. Is f continuous at 0? Yes
No
31. Is f continuous at 4? No
In Problems 33–44, find the one-sided limit. 33. lim+ 12x + 32
42. lim+ xS0
x3 - x2 x4 + x2
6
-4
28. Is f continuous at - 4?
30. Is f continuous at 2? No
4
2
xS4
No
(6, 2)
2
25. Does lim f 1x2 exist? If it does, what is it? Yes; 0 27. Is f continuous at - 6?
(2, 3)
3
xS1
xS2
-1
43.
x2 - 4 x - 2
lim +
x S -2
36.
7
40. lim-
4
x2 + x - 2 x2 + 2x
lim 13x2 - 82
xS1
3 2
44.
4
x S -2+
x3 - x x - 1
lim -
x S -4
2
x2 + x - 12 x2 + 4x
In Problems 45–60, determine whether f is continuous at c. 46. f 1x2 = 3x2 - 6x + c = - 3 Yes
45. f 1x2 = x3 - 3x2 + 2x - 6 c = 2 Yes 47. f 1x2 =
x2 + x - 6
50. f 1x2 =
x - 6 x + 6
c = 3 Yes c = -6
x3 + 3x 53. f 1x2 = c x2 - 3x 1 x3 + 3x 55. f 1x2 = c x2 - 3x -1
No
48. f 1x2 =
x3 - 8 x2 + 4
51. f 1x2 =
x3 + 3x x2 - 3x
if x ≠ 0 if x = 0
c = 0
No
if x ≠ 0 if x = 0
c = 0
Yes
c = 2 Yes c = 0
No
x2 - 6x 54. f 1x2 = c x2 + 6x -2 x2 - 6x 56. f 1x2 = c x2 + 6x -1
49. f 1x2 =
x + 3 x - 3
52. f 1x2 =
x2 - 6x x2 + 6x
c = 3 c = 0
if x ≠ 0 if x = 0
c = 0
No
c = 0
Yes
if x ≠ 0 if x = 0
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
No No
7 4
x
942
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
x3 - 1 x2 - 1 57. f 1x2 = e 2 3 x + 1
x2 - 2x x - 2 58. f 1x2 = e 2 x - 4 x - 1
if x 6 1 if x = 1
c = 1
No
if x 7 1
2e x 2 59. f 1x2 = d 3 x + 2x2 x2
if x 6 0 if x = 0
c = 0
if x 6 2 if x = 2
if x 7 0
No
if x 7 2
3 cos x 3 60. f 1x2 = d 3 x + 3x2 x2
Yes
c = 2
if x 6 0 if x = 0
c = 0
Yes
if x 7 0
In Problems 61–72, find the numbers at which f is continuous. At which numbers is f discontinuous? *61. f 1x2 = 2x + 3
*62. f 1x2 = 4 - 3x
*63. f 1x2 = 3x2 + x
*64. f 1x2 = - 3x3 + 7
*65. f 1x2 = 4 sin x
*66. f 1x2 = - 2 cos x
*67. f 1x2 = 2 tan x
*68. f 1x2 = 4 csc x
*69. f 1x2 =
2x + x2 - 4
*70. f 1x2 =
x - 4 x2 - 9 2
*71. f 1x2 =
x - 3 ln x
*72. f 1x2 =
ln x x - 3
In Problems 73–76, discuss whether R is continuous at c. Use limits to analyze the graph of R at c. Graph R. *73. R 1x2 =
x - 1 , c = - 1 and c = 1 x2 - 1
*74. R 1x2 =
3x + 6 , c = - 2 and c = 2 x2 - 4
*75. R 1x2 =
x2 + x , c = - 1 and c = 1 x2 - 1
*76. R 1x2 =
x2 + 4x , c = - 4 and c = 4 x2 - 16
In Problems 77–82, determine where each rational function is undefined. Determine whether an asymptote or a hole appears at such numbers. *77. R 1x2 =
x3 - x2 + x - 1 x4 - x3 + 2x - 2
*78. R 1x2 =
x3 + x2 + 3x + 3 x4 + x3 + 2x + 2
*79. R 1x2 =
x3 - 2x2 + 4x - 8 x2 + x - 6
*80. R 1x2 =
x3 - x2 + 3x - 3 x2 + 3x - 4
*81. R 1x2 =
x3 + 2x2 + x x + x3 + 2x + 2
*82. R 1x2 =
x3 - 3x2 + 4x - 12 x4 - 3x3 + x - 3
4
For Problems 83–88, use a graphing utility to graph the functions R given in Problems 77–82. Verify the solutions found above.*
Discussion and Writing 89. Name three functions that are continuous at every real number.
90. $SFBUFBGVODUJPOUIBUJTOPUDPOUJOVPVTBUUIFOVNCFS
Retain Your Knowledge Problems 91–94 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 91. 'JOEBOZWFSUJDBMPSIPSJ[POUBMBTZNQUPUFTGPSUIFHSBQIPG 3x - 4 f 1x2 = . Vertical: x = 4; horizontal: y = 3 x - 4 92. Evaluate P(, 3) 60
93. Write ln x + 2ln y - 4ln z as a single natural logarithm. *94. Write the augmented matrix for the following system: 3x + y + 2z = 4
c
93. lna
x + 2z = y - 3z = - 2
‘Are You Prepared?’ Answers 1. f 102 = 0; f 122 = 3
2. Domain: 5x x 7 06; range 5y - q 6 y 6 q 6
3. True
4. Secant, cosecant, tangent, cotangent
5. True
6. True
xy2 z4
b
SECTION 13.4 The Tangent Problem; The Derivative
943
13.4 The Tangent Problem; The Derivative PREPARING FOR THIS SECTION Before getting started, review the following: r 1PJOUm4MPQF'PSNPGB-JOF 'PVOEBUJPOT 4FDUJPO Q
r "WFSBHF3BUFPG$IBOHF 4FDUJPO QQm
Now Work the ‘Are You Prepared?’ problems on page 916.
OBJECTIVES 1 2 3 4
Find an Equation of the Tangent Line to the Graph of a Function (p. 943) Find the Derivative of a Function (p. 945) Find Instantaneous Rates of Change (p. 946) Find the Instantaneous Speed of a Particle (p. 946)
The Tangent Problem One question that motivated the development of calculus was a geometry problem, the tangent problem. This problem asks, “What is the slope of the tangent line to the graph of a function y = f1x2 at a point PPOJUTHSBQI u4FF'JHVSF We first need to define what we mean by a tangent line. In high school geometry, the tangent line to a circle at a point is defined as the line that intersects the circle at FYBDUMZUIBUPOFQPJOU-PPLBU'JHVSF/PUJDFUIBUUIFUBOHFOUMJOFKVTUUPVDIFT the graph of the circle. Figure 12
Figure 13 Tangent line
y
P y f(x) Tangent line to f at P
P
x
5IJTEFGJOJUJPO IPXFWFS EPFTOPUXPSLJOHFOFSBM-PPLBU'JHVSF5IFMJOFT L1 and L2 intersect the graph at only one point P, but neither just touches the graph at P. Additionally, the tangent line LT TIPXOJO'JHVSFUPVDIFTUIFHSBQIPG f at P but also intersects the graph elsewhere. So how should we define the tangent line to the graph of f at a point P? Figure 14
Figure 15 y
y
L1
L2 LT P
P x
x
1 Find an Equation of the Tangent Line to the Graph of a Function The tangent line LT to the graph of a function y = f1x2 at a point P necessarily contains the point P. To find an equation for LT VTJOHUIFQPJOUmTMPQFGPSNPGUIF equation of a line, we need to find the slope mtan of the tangent line.
944
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
Figure 16 y Q (x, f(x)) y f(x)
Q1 (x1, f(x1)) LT P (c, f(c)) c
x
Suppose that the coordinates of the point P are 1c, f1c2 2. Locate another point Q = 1x, f1x2 2 on the graph of f. The line containing P and Q is a secant line. 3FGFSUP4FDUJPO 5IFTMPQFmsec of the secant line is msec =
f 1x2 - f 1c2 x - c
/PXMPPLBU'JHVSF As we move along the graph of f from Q toward P, we obtain a succession of secant lines. The closer we get to P, the closer the secant line is to the tangent line LT . The limiting position of these secant lines is the tangent line LT . Therefore, the limiting value of the slopes of these secant lines equals the slope of the tangent line. Also, as we move from Q toward P, the values of x get closer to c. Therefore, mtan = lim msec = lim xSc
DEFINITION
xSc
f1x2 - f1c2 x - c
The tangent line to the graph of a function y = f1x2 at a point P = 1c, f1c2 2 on its graph is defined as the line containing the point P whose slope is mtan = lim
xSc
f1x2 - f1c2 x - c
(1)
provided that this limit exists.
THEOREM
If mtan exists, an equation of the tangent line is y - f 1c2 = mtan 1x - c2
(2)
Finding an Equation of the Tangent Line
EX A MPL E 1
Solution
x2 1 'JOEBOFRVBUJPOPGUIFUBOHFOUMJOFUPUIFHSBQIPG f(x) = at the point a1, b . 4 4 Graph f and the tangent line. 1 The tangent line contains the point a1, b . The slope of the tangent line to the 4 x2 1 graph of f(x) = at a1, b is 4 4
mtan
x2 1 f1x2 - f112 1x - 12 1x + 12 4 4 = lim = lim = lim xS1 xS1 x - 1 xS1 x - 1 4 1x - 12 = lim
xS1
Figure 17
x + 1 1 = 4 2
An equation of the tangent line is
y y 1 2
x2 4
y -
1 4
( 1, ) y 12 x 14 1 2
1
1 1 = 1x - 12 4 2 y =
x
'JHVSFTIPXTUIFHSBQIPGy =
Now Work
PROBLEM
y - f(c) = mtan(x - c)
1 1 x 2 4 x2 1 and the tangent line at a1, b . 4 4 11
r
SECTION 13.4 The Tangent Problem; The Derivative
945
2 Find the Derivative of a Function The limit in formula (1) has an important generalization: it is called the derivative of f at c.
DEFINITION
Let y = f1x2 denote a function f. If c is a number in the domain of f, then the derivative of f at c, denoted by f′(c) and read “f prime of c,” is defined as f′1c2 = lim
xSc
f1x2 - f1c2 x - c
(3)
provided that this limit exists.
EXAM PL E 2
Finding the Derivative of a Function 'JOEUIFEFSJWBUJWFPGf1x2 = 2x2 - x at 2. That is, find f′(2).
Solution
Since f122 = 2122 2 - 122 = - 2, we have f1x2 - f122 12x2 - x2 - 1 - 22 12x - 12 1x - 22 2x2 - x + 2 = = = x - 2 x - 2 x - 2 x - 2 The derivative of f at 2 is f′122 = lim
xS2
Now Work
f1x2 - f122 12x - 12 1x - 22 = lim = 3 xS2 x - 2 x - 2
PROBLEM
r
21
Example 2 provides a way of finding the derivative at 2 analytically. Graphing utilities have built-in procedures to approximate the derivative of a function at any number c. Consult your owner’s manual for the appropriate keystrokes.
EXAM PL E 3
Finding the Derivative of a Function Using a Graphing Utility Use a graphing utility to find the derivative of f1x2 = 2x2 - x at 2. That is, find f′122.
Solution
'JHVSFTIPXTUIFTPMVUJPOVTJOHB5*1MVT Figure 18 graphing calculator. As shown, f′122 = 3.
Now Work
EXAM PL E 4
PROBLEM
33
r
Finding the Derivative of a Function 'JOEUIFEFSJWBUJWFPGf1x2 = x2 at c. That is, find f′(c).
Solution
Since f1c2 = c 2, we have
f1x2 - f1c2 1x + c2 1x - c2 x2 - c 2 = = x - c x - c x - c
The derivative of f at c is f′1c2 = lim
xSc
f1x2 - f1c2 1x + c2 1x - c2 = 2c = lim xSc x - c x - c
r
946
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
As Example 4 illustrates, the derivative of f 1x2 = x2 exists and equals 2c for any number c. In other words, the derivative is itself a function, and using x for the independent variable, we can write f′(x) = 2x. The function f′ is called the derivative function of f or the derivative of f . We also say that f is differentiable. The instruction “differentiate f ” means “find the derivative of f.”
3 Find Instantaneous Rates of Change The average rate of change of a function f from c to x is f1x2 - f1c2 ∆y = x - c ∆x The limit as x approaches c of the average rate of change of f, based on formula (3), is the derivative of f at c.
DEFINITION
The derivative of f at c is also called the instantaneous rate of change of f with respect to x at c. That is, ¢
EX A MPL E 5
f1x2 - f1c2 Instantaneous rate of ≤ = f′1c2 = lim change of f with respect to x at c xSc x - c
(4)
Finding the Instantaneous Rate of Change The volume V of a right circular cone of height h = 6 feet and radius r feet is 1 V = V 1r2 = pr 2h = 2pr 2. If r is increasing, find the instantaneous rate of change 3 of the volume V with respect to the radius r at r = 3.
Solution
The instantaneous rate of change of V with respect to r at r = 3 is the derivative V′(3). V′132 = lim
rS3
V 1r2 - V 132 2p1r 2 - 92 2pr 2 - 18p = lim = lim rS3 rS3 r - 3 r - 3 r - 3
= lim 3 2p1r + 32 4 = 12p rS3
At the instant r = 3 feet, the volume of the cone is increasing with respect to r at a rate of 12p ≈ 37.699 cubic feet per 1-foot change in the radius.
r
Now Work
PROBLEM
43
4 Find the Instantaneous Speed of a Particle If s = f1t2 denotes the position of a particle at time t, then the average speed of the particle from c to t is f 1t2 - f1c2 Change in position ∆s = = Change in time ∆t t - c
DEFINITION
(5)
The limit as t approaches cPGUIFFYQSFTTJPOJOGPSNVMB JTUIFinstantaneous speed of the particle at c or the velocity of the particle at c. That is, a
f1t2 - f1c2 Instantaneous speed of b = f′1c2 = lim a particle at time c tSc t - c
(6)
SECTION 13.4 The Tangent Problem; The Derivative
EXAM PL E 6
947
Finding the Instantaneous Speed of a Particle In physics it is shown that the height s of a ball thrown straight up with an initial speed of 80 feet per second (ft/sec) from a rooftop 96 feet high is s = s 1t2 = - 16t 2 + 80t + 96
Figure 19
where t is the elapsed time that the ball is in the air. The ball misses the rooftop on JUTXBZEPXOBOEFWFOUVBMMZTUSJLFTUIFHSPVOE4FF'JHVSF Roof top
96 ft
Solution
(a) (b) (c) (d) (e) (f) (g)
When does the ball strike the ground? That is, how long is the ball in the air? At what time t will the ball pass the rooftop on its way down? What is the average speed of the ball from t = 0 to t = 2? What is the instantaneous speed of the ball at time t 0? What is the instantaneous speed of the ball at t = 2? When is the instantaneous speed of the ball equal to zero? What is the instantaneous speed of the ball as it passes the rooftop on the way down? (h) What is the instantaneous speed of the ball when it strikes the ground? (a) The ball strikes the ground when s = s 1t2 = 0. - 16t 2 + 80t + 96 t 2 - t - 6 1t - 62 1t + 12 t = 6 or t
= = = =
0 0 0 -1
Discard the solution t = - 1. The ball strikes the ground after 6 sec. (b) The ball passes the rooftop when s = s 1t2 = 96. - 16t 2 + 80t + 96 t 2 - t t1t - 2 t = 0 or t
= = = =
96 0 0
Discard the solution t = 0. The ball passes the rooftop on the way down BGUFSTFD (c) The average speed of the ball from t = 0 to t = 2 is s 122 - s 102 ∆s 192 - 96 = = = 48 ft/sec ∆t 2 - 0 2
(d) The instantaneous speed of the ball at time t 0 is the derivative s′1t 0 2; that is, s 1t2 - s 1t 0 2 t - t0 1 - 16t 2 + 80t + 962 - 1 - 16t 20 + 80t 0 + 962 lim t S t0 t - t0 - 163 t 2 - t 20 - t + t 0 4 lim t S t0 t - t0 - 163 1t + t 0 2 1t - t 0 2 - 1t - t 0 2 4 lim t S t0 t - t0 - 163 1t + t 0 - 2 1t - t 0 2 4 lim = lim 3 - 161t + t 0 - 2 4 t S t0 t S t0 t - t0 = - 1612t 0 - 2 ft/sec
s′1t 0 2 = lim
t S t0
= = = =
3FQMBDFt0 by t. The instantaneous speed of the ball at time t is s′1t2 = - 1612t - 2 ft/sec
948
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
(e) At t = 2 sec, the instantaneous speed of the ball is
s′ 122 = - 1612 # 2 - 2 = - 161 - 12 = 16 ft/sec
(f) The instantaneous speed of the ball is zero when s′1t2 = 0 - 1612t - 2 = 0 t = = 2. sec 2 (g) The ball passes the rooftop on the way down when t = . The instantaneous speed at t = is s′12 = - 16110 - 2 = - 80 ft/sec
Exploration Determine the vertex of the quadratic function given in Example 6. What do you conclude about the velocity when s(t) is a maximum?
At t = sec, the ball is traveling - 80 ft/sec. When the instantaneous rate of change is negative, it means that the direction of the object is downward. The ball is traveling 80 ft/sec in the downward direction when t = sec. (h) The ball strikes the ground when t = 6. The instantaneous speed when t = 6 is s′162 = - 16112 - 2 = - 112 ft/sec The speed of the ball at t = 6 sec is - 112 ft/sec. Again, the negative value implies that the ball is traveling downward.
r
SUMMARY The derivative of a function y = f1x2 at c is defined as f′1c2 = lim
xSc
f1x2 - f1c2 x - c
In geometry, f′(c) equals the slope of the tangent line to the graph of f at the point 1c, f1c2 2. In physics, f′(c) equals the instantaneous speed (velocity) of a particle at time c, where s = f 1t2 is the position of the particle at time t. In applications, if two variables are related by the function y = f1x2, then f′(c) equals the instantaneous rate of change of f with respect to x at c.
13.4 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. 1. 'JOE BO FRVBUJPO PG UIF MJOF XJUI TMPQF DPOUBJOJOH UIF point 12, - 42. Q y = x - 14
2. True or False The average rate of change of a function f from a to b is f 1b2 + f 1a2 b + a
QQm 'BMTF
Concepts and Vocabulary 3. If lim
f 1x2 - f 1c2
exists, it equals the slope of the x - c tangent line to the graph of f at the point 1c, f 1c2 2. f 1x2 - f 1c2 4. If lim exists, it is called the derivative of f at c. xSc x - c xSc
5. If s = f 1t2 denotes the position of a particle at time t, the derivative f′1c2 is the velocity of the particle at c.
6. True or False The tangent line to a function is the limiting position of a secant line. True 7. True or False The slope of the tangent line to the graph of f at 1c, f 1c2 2 is the derivative of f at c. True 8. True or False The velocity of a particle whose position at time t is s 1t2 is the derivative s′ 1t2. True
949
SECTION 13.4 The Tangent Problem; The Derivative
Skill Building In Problems 9–20, find the slope of the tangent line to the graph of f at the given point. Graph f and the tangent line. *9. f 1x2 = 3x + at 11, 82
*12. f 1x2 = 3 - x
2
*10. f 1x2 = - 2x + 1 at 1 - 1, 32
at 11, 22
*13. f 1x2 = 3x
*15. f 1x2 = 2x2 + x at 11, 32
*18. f 1x2 = - 2x + x - 3 at 11, - 42 2
2
*11. f 1x2 = x2 + 2 at 1 - 1, 32
at 12, 122
*14. f 1x2 = - 4x2 at 1 - 2, - 162
*16. f 1x2 = 3x2 - x at 10, 02
*17. f 1x2 = x2 - 2x + 3 at 1 - 1, 62
*19. f 1x2 = x + x at 12, 102
*20. f 1x2 = x3 - x2 at 11, 02
3
In Problems 21–32, find the derivative of each function at the given number. 21. f 1x2 = - 4x + at 3
24. f 1x2 = 2x + 1 at - 1 2
27. f1x2 = x + 4x at - 1 3
22. f 1x2 = - 4 + 3x at 1
-4
25. f 1x2 = 2x + 3x at 1
-4
2
28. f 1x2 = 2x - x 3
7
30. f1x2 = x3 - 2x2 + x at - 1
2
31. f 1x2 = sin x at 0
8
at 2
23. f 1x2 = x2 - 3 at 0
3
26. f 1x2 = 3x2 - 4x at 2
7
0 8
29. f 1x2 = x + x - 2x at 1 3
20
2
32. f 1x2 = cos x at 0
1
3
0
In Problems 33–42, use a graphing utility to find the derivative of each function at the given number. 33. f 1x2 = 3x3 - 6x2 + 2 at - 2
-x + 1 at 8 x2 + x + 7 p *37. f 1x2 = x sin x at 3 p 2 *40. f 1x2 = x sin x at 4 35. f 1x2 =
3
34. f 1x2 = - x4 + 6x2 - 10 at
60
- x + 9x + 3 x3 + x2 - 6 p *38. f 1x2 = x sin x at 4
- 0.88777696
4
36. f 1x2 =
*41. f 1x2 = e x sin x at 2
at - 3
- 2440
*39. f 1x2 = x2 sin x at
p 3
*42. f 1x2 = e -x sin x at 2
Applications and Extensions 43. Instantaneous Rate of Change The volume V of a right circular cylinder of height 3 feet and radius r feet is V = V 1r2 = 3pr 2.'JOEUIFJOTUBOUBOFPVTSBUFPGDIBOHFPG the volume with respect to the radius r at r = 3. 18p ft 3 >ft 44. Instantaneous Rate of Change The surface area S of a sphere of radius r feet is S = S1r2 = 4pr 2. 'JOE UIF instantaneous rate of change of the surface area with respect to the radius r at r = 2. 16p ft 2 >ft 45. Instantaneous Rate of Change The volume V of a sphere of 4 radius r feet is V = V 1r2 = pr 3.'JOEUIFJOTUBOUBOFPVT 3 rate of change of the volume with respect to the radius r at r = 2. 16p ft 3 >ft *46. Instantaneous Rate of Change The volume V of a cube of side x meters is V = V 1x2 = x3.'JOEUIFJOTUBOUBOFPVTSBUF of change of the volume with respect to the side x at x = 3. 47. Instantaneous Speed of a Ball In physics it is shown that the height s of a ball thrown straight up with an initial speed of 96 ft/sec from ground level is s = s 1t2 = - 16t 2 + 96t where t is the elapsed time that the ball is in the air. (a) When does the ball strike the ground? That is, how long is the ball in the air? 6 seconds (b) What is the average speed of the ball from t = 0 to t = 2? 64 feet per second *(c) What is the instantaneous speed of the ball at time t? *(d) What is the instantaneous speed of the ball at t = 2? (e) When is the instantaneous speed of the ball equal to zero? 3 seconds (f) How high is the ball when its instantaneous speed equals zero? 144 feet (g) What is the instantaneous speed of the ball when it strikes the ground? - 96 feet per second
48. Instantaneous Speed of a Ball In physics it is shown that the height s of a ball thrown straight down with an initial speed of 48 ft/sec from a rooftop 160 feet high is s = s 1t2 = - 16t 2 - 48t + 160 where t is the elapsed time that the ball is in the air. (a) When does the ball strike the ground? That is, how long is the ball in the air? 2 seconds (b) What is the average speed of the ball from t = 0 to t = 1? - 64 feet per second *(c) What is the instantaneous speed of the ball at time t? *(d) What is the instantaneous speed of the ball at t = 1? (e) What is the instantaneous speed of the ball when it strikes the ground? - 112 feet per second 4 . Instantaneous Speed on the Moon An astronaut throws a 49 49. ball down into a crater on the moon. The height s (in feet) of the ball from the bottom of the crater after t seconds is given in the following table:
Time, t (in seconds)
Distance, s (in feet)
0
1000
1
987
2
969
3
945
4
917
5
883
6
843
7
800
8
749
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
950
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
* B 'JOEUIFBWFSBHFTQFFEGSPNt = 1 to t = 4 seconds. * C 'JOEUIFBWFSBHFTQFFEGSPNt = 1 to t = 3 seconds. * D 'JOEUIFBWFSBHFTQFFEGSPNt = 1 to t = 2 seconds. *(d) Using a graphing utility, find the quadratic function of best fit. *(e) Using the function found in part (d), determine the instantaneous speed at t = 1 second. 50. Instantaneous Rate of Change The data to the right represent the total revenue R (in dollars) received from selling x bicycles at Tunney’s Bicycle Shop. B 'JOEUIFBWFSBHFSBUFPGDIBOHFJOSFWFOVFGSPNx = 2 to x = 10 bicycles. $249.28 per bicycle C 'JOEUIFBWFSBHFSBUFPGDIBOHFJOSFWFOVFGSPNx = 2 to x = 102 bicycles. ≈$329.87 per bicycle D 'JOEUIFBWFSBHFSBUFPGDIBOHFJOSFWFOVFGSPNx = 2 to x = 60 bicycles. ≈$48.71 per bicycle (d) Using a graphing utility, find the quadratic function of best fit. R(x) = - 1.22x2 + 97.791x + 7944.40
Number of Bicycles, x
Total Revenue, R (in dollars)
0
0
25
28,000
60
45,000
102
53,400
150
59,160
190
62,360
223
64,835
249
66,525
(e) Using the function found in part (d), determine the instantaneous rate of change of revenue at x = 2 bicycles. ≈$21.69 per bicycle
Retain Your Knowledge Problems 51–54 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 51. 'JOEUIFWFSUFYBOEGPDVTPGUIFQBSBCPMB 54. 'JOEUIFBSFBPGUIFHJWFOUSJBOHMF SPVOEFEUPUXPEFDJNBM x2 - 2x - 2y + 7 = 0. places. 23.67 sq. units B y = x + 4 52. Solve the system by substitution: e 2 9 x - y = - 2 (2, 6), ( - 1, 3) 6
53. Evaluate: C(, 3)
10
7 51. Vertex: (1, 3); focus: a1, b 2
C A
13
‘Are You Prepared?’ Answers 1. y = x - 14
2. 'BMTF
13.5 The Area Problem; The Integral PREPARING FOR THIS SECTION Before getting started, review the following: r (FPNFUSZ'PSNVMBT "QQFOEJY" 4FDUJPO" QQ"m"
r 4VNNBUJPO/PUBUJPO 4FDUJPO QQm
Now Work the ‘Are You Prepared?’ problems on page 955.
OBJECTIVES 1 Approximate the Area under the Graph of a Function (p. 951) 2 Approximate Integrals Using a Graphing Utility (p. 955)
The Area Problem The development of the integral, like that of the derivative, was originally motivated to a large extent by a problem in geometry: the area problem. Figure 20 y
y f(x)
Area A a
b
Area A area under the graph of f from a to b
Area Problem Suppose that y = f1x2 is a function whose domain is a closed interval 3 a, b 4 . We assume that f1x2 Ú 0 for all x in 3 a, b4 . 'JOE UIF BSFB FODMPTFE CZ UIF graph of f, the x-axis, and the vertical lines x = a and x = b.
x
'JHVSFJMMVTUSBUFTUIFBSFBQSPCMFN8FSFGFSUPUIFBSFBA shown as the area under the graph of f from a to b.
SECTION 13.5 The Area Problem; The Integral
951
'PSBDPOTUBOUGVODUJPOf 1x2 = k and for a linear function f1x2 = mx + B, we DBOTPMWFUIFBSFBQSPCMFNVTJOHGPSNVMBTGSPNHFPNFUSZ4FF'JHVSFT B BOE C Figure 21
y
y
f(x) mx B f(x) k
k
f(b) k
A a
b
A
f(a) x
a
b
x
ba A area of trapezoid
ba A area of rectangle height base k (b a)
1
2 [f (b) f (a)](b a)
(a)
(b)
'PSNPTUPUIFSGVODUJPOT OPGPSNVMBTGSPNHFPNFUSZBSFBWBJMBCMF We begin by discussing a way to approximate the area under the graph of a function f from a to b.
1 Approximate the Area under the Graph of a Function We use rectangles to approximate the area under the graph of a function f. We do this by partitioning or dividing the interval 3 a, b4 into subintervals of equal length. On each subinterval, we form a rectangle whose base is the length of the subinterval and whose height is f 1u2 for some number u in the subinterval. Look at 'JHVSF Figure 22
y
y y f(x)
y f(x)
f(u2)
f (u1)
f(u1) a u1
b
u2 ba 2
ba 2
2 subintervals (a)
x
a
u1 ba 4
f (u2) u2
ba ba 4 4 4 subintervals (b)
f(u3)
f(u4) b u4 x
u3
ba 4
*O'JHVSF B
UIFJOUFSWBM 3 a, b 4 is partitioned into two subintervals, each of b - a length , and the number u is chosen as the left endpoint of each subinterval. 2 *O'JHVSF C
UIFJOUFSWBM 3 a, b4 is partitioned into four subintervals, each of b - a length , and the number u is chosen as the right endpoint of each subinterval. 4 The area A under f from a to b is approximated by adding the areas of the rectangles formed by the partition. 6TJOH'JHVSF B
Area A ≈ area of first rectangle + area of second rectangle b - a b - a = f 1u 1 2 + f1u 2 2 2 2
952
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
6TJOH'JHVSF C
Area A ≈ area of first rectangle + area of second rectangle + area of third rectangle + area of fourth rectangle b - a b - a b - a b - a = f 1u 1 2 + f1u 2 2 + f1u 3 2 + f1u 4 2 4 4 4 4 In approximating the area under the graph of a function f from a to b, the choice of the number uJOFBDITVCJOUFSWBMJTBSCJUSBSZ'PSDPOWFOJFODF XFTIBMMBMXBZTQJDL u as either the left endpoint of each subinterval or the right endpoint. The choice of how many subintervals to use is also arbitrary. In general, the more subintervals used, the better the approximation will be. Let’s look at a specific example.
EX A MPL E 1
Approximating the Area under the Graph of f(x) = 2x from 0 to 1 Approximate the area A under the graph of f1x2 = 2x from 0 to 1 as follows:
Figure 23 y f (x ) ⫽ 2x
2
1
( 2)
f 1 f (0) 0
x
1
1 2
Solution
1 2
1 2
2 subintervals; u’s are left endpoints (a) y f (x ) ⫽ 2x
2
1
f (1)
( 2)
f 1
0 1 2
x
1
1 2
(a) Partition 3 0, 14 into two subintervals of equal length and choose u as the left endpoint. (b) Partition 3 0, 14 into two subintervals of equal length and choose u as the right endpoint. (c) Partition 3 0, 14 into four subintervals of equal length and choose u as the left endpoint. (d) Partition 3 0, 14 into four subintervals of equal length and choose u as the right endpoint. F $PNQBSFUIFBQQSPYJNBUJPOTGPVOEJOQBSUT B m E XJUIUIFBDUVBMBSFB 1 (a) Partition 3 0, 14 into two subintervals, each of length , and choose u as the left 2 FOEQPJOU4FF'JHVSF B 5IFBSFBA is approximated as 1 1 1 A ≈ f102 a b + f a b a b 2 2 2 1 1 = 102 a b + 112 a b 2 2 1 = = 0. 2
1 (b) Partition 3 0, 14 into two subintervals, each of length , and choose u as the right 2 FOEQPJOU4FF'JHVSF C 5IFBSFBA is approximated as 1 1 1 A ≈ f a b a b + f112 a b 2 2 2
1 2
2 subintervals; u’s are right endpoints
1 1 = 112 a b + 122 a b 2 2
(b) y f (x ) ⫽ 2x
2
1 f (0) 0
1 4
1 2
3 4
1 1 f(0) = 2 # 0 = 0; f a b = 2 # = 1 2 2
1
x
( 4) f ( 12) f ( 34)
f 1
4 subintervals; u’s are left endpoints (c)
=
3 = 1. 2
1 (c) Partition 3 0, 14 into four subintervals, each of length , and choose u as the left 4 FOEQPJOU4FF'JHVSF D 5IFBSFBA is approximated as 1 1 1 1 1 3 1 A ≈ f102 a b + f a b a b + f a b a b + f a b a b 4 4 4 2 4 4 4 1 1 1 1 3 1 = 102 a b + a b a b + 112 a b + a b a b 4 2 4 4 2 4 =
3 = 0.7 4
SECTION 13.5 The Area Problem; The Integral
(d) Partition 3 0, 14 into four subintervals, each of length
Figure 23 (continued) y f (x ) ⫽ 2x
2
1
953
1 , and choose u as the 4 SJHIUFOEQPJOU4FF'JHVSF E 5IFBSFBA is approximated as 1 1 1 1 3 1 1 A ≈ f a b a b + f a b a b + f a b a b + f112 a b 4 4 2 4 4 4 4
f (1)
1 1 1 3 1 1 = a b a b + 112 a b + a b a b + 122 a b 2 4 4 2 4 4 0
1 4
1 2
3 4
x
1
= 1.2 4 (e) The actual area under the graph of f1x2 = 2x from 0 to 1 is the area of a right triangle whose base is of length 1 and whose height is 2. The actual area A is therefore =
( 4) f ( 12) f ( 34)
f 1
4 subintervals; u’s are right endpoints (d)
A =
1 1 base * height = a b 112 122 = 1 2 2
r
Now look at Table 7, which shows the approximations to the area under the graph of f1x2 = 2x from 0 to 1 for n = 2, 4, 10, and 100 subintervals. Notice that the approximations to the actual area improve as the number of subintervals increases.
Table 7
Using left endpoints:
n Area
2 0.5
4 0.75
10 0.9
100 0.99
Using right endpoints:
n Area
2 1.5
4 1.25
10 1.1
100 1.01
You are asked to confirm the entries in Table 7 in Problem 31. 5IFSF JT BOPUIFS VTFGVM PCTFSWBUJPO BCPVU &YBNQMF -PPL BHBJO BU 'JHVSFT B m E BOEBU5BCMF4JODFUIFHSBQIPG f1x2 = 2x is increasing on 3 0, 14 , the choice of u as the left endpoint gives a lower-bound estimate to the actual area, while choosing u as the right endpoint gives an upper-bound estimate. Do you see why?
Now Work
EXAM PL E 2
PROBLEM
9
Approximating the Area under the Graph of f(x) = x 2 Approximate the area under the graph of f1x2 = x2GSPNUPBTGPMMPXT (a) Using four subintervals of equal length (b) Using eight subintervals of equal length In each case, choose the number u to be the left endpoint of each subinterval.
Solution Figure 24 y f (x ) ⫽ x 2
25
B 4FF 'JHVSF 6TJOH GPVS TVCJOUFSWBMT PG FRVBM MFOHUI UIF JOUFSWBM 3 1, 4 is - 1 = 1 as follows: partitioned into subintervals of length 4 3 2, 34 3 3, 44 3 4, 4 3 1, 24 Each of these subintervals is of length 1. Choosing u as the left endpoint of each subinterval, the area A under the graph of f1x2 = x2 is approximated by Area A ≈ f112 112 + f122 112 + f132 112 + f142 112 = 1 + 4 + 9 + 16 = 30
15
5 0
1
2
3
4
5
f (1) f (2) f (3) f (4) 4 subintervals; each of length 1
x
C 4FF 'JHVSF PO UIF OFYU QBHF 6TJOH FJHIU TVCJOUFSWBMT PG FRVBM MFOHUI UIF - 1 interval 3 1, 4 is partitioned into subintervals of length = 0. as follows: 8 3 1, 1.4
3 1., 24
3 2, 2.4
3 2., 34
3 3, 3.4
3 3., 44
3 4, 4.4
3 4., 4
954
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
&BDIPGUIFTFTVCJOUFSWBMTJTPGMFOHUI$IPPTJOHu as the left endpoint of each subinterval, the area A under the graph of f 1x2 = x2 is approximated by
Figure 25 y
15
Area A ≈ f112 10.2 + f 11.2 10.2 + f122 10.2 + f 12.2 10.2
f (4.5) f (x ) ⫽ x 2
25
+ f132 10.2 + f 13.2 10.2 + f142 10.2 + f14.2 10.2
f (3.5) f (2.5) f (1.5)
= 3 f112 + f 11.2 + f122 + f12.2 + f 132 + f13.2 + f142 + f 14.2 4 10.2
= 3 1 + 2.2 + 4 + 6.2 + 9 + 12.2 + 16 + 20.24 10.2
5
= 3.
0
1
3
5
x
In general, we approximate the area under the graph of a function y = f1x2 from a to b as follows:
f (1) f (2) f (3) f (4) 8 subintervals; each of length 1/2
1. Partition the interval 3 a, b4 into n subintervals of equal length. The length ∆x of each subinterval is then
Figure 26 y f(u3) f(u2) f(u1)
a u1 u2 u3 Dx Dx Dx
r
f (un)
∆x =
y 5 f(x)
un b x Dx
b - a n
2. In each of these subintervals, pick a number u and evaluate the function f at each u. This results in n numbers u 1 , u 2 , c , u n and n functional values f1u 1 2, f1u 2 2, c , f1u n 2. 3. 'PSN n rectangles with base equal to ∆x, the length of each subinterval, and with height equal to the functional value f1u i 2, i = 1, 2, c , n.4FF'JHVSF 4. Add up the areas of the n rectangles. A1 + A2 + g + An = f 1u 1 2 ∆x + f 1u 2 2 ∆x + g + f1u n 2 ∆x n
= a f1u i 2 ∆x i=1
This number is the approximation to the area under the graph of f from a to b.
Definition of Area We have observed that the larger the number n of subintervals used, the better the approximation to the area. If we let n become unbounded, we obtain the exact area under the graph of f from a to b.
DEFINITION
Area under a Graph Let f denote a function whose domain is a closed interval 3 a, b4 and suppose f1x2 Ú 0 on 3 a, b 4 . Partition 3 a, b4 into n subintervals, each of length b - a ∆x = . In each subinterval, pick a number u i , i = 1, 2, c , n, and evaluate n f1u i 2.'PSNUIFQSPEVDUTf1u i 2 ∆x and add them up, obtaining the sum n
a f1u i 2 ∆x
i=1
If the limit of this sum exists as n S q , that is, n
if
lim f1u i 2 ∆x n Sq a i=1
exists
it is defined as the area under the graph of f from a to b. If this limit exists, it is denoted by b
La
f1x2 dx
which is read as “the integral from a to b of f1x2.”
SECTION 13.5 The Area Problem; The Integral
955
2 Approximate Integrals Using a Graphing Utility Using a Graphing Utility to Approximate an Integral
EXAM PL E 3
Use a graphing utility to approximate the area under the graph of f1x2 = x2 from 1 UP5IBUJT FWBMVBUFUIFJOUFHSBM
L1
Solution
Figure 27
x2 dx
'JHVSF TIPXT UIF SFTVMU VTJOH B 5* 1MVT DBMDVMBUPS $POTVMU ZPVS PXOFST manual for the proper keystrokes. 124 The value of the integral is . 3 In calculus, techniques are given for evaluating integrals to obtain exact answers.
r
13.5 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. 4
1. The formula for the area A of a rectangle of length l and width w is A = lw . Q"
2. a 12k + 12 = 24 . QQm
k=1
Concepts and Vocabulary 3. The integral from a to b of f 1x2 is denoted by 1a f(x) dx . b
4. The area under the graph of f from a to b is denoted by b 1a f(x) dx .
Skill Building In Problems 5 and 6, refer to the illustration. The interval 31, 34 is partitioned into two subintervals 31, 24 and 32, 34. y y ⫽ f(x)
f(3) ⫽ 4
f(2) ⫽ 2 f(1) ⫽ 1
A
x 1 2 3 f(1) ⫽ 1, f(2) ⫽ 2, f(3) ⫽ 4
5. Approximate the area A, choosing u as the left endpoint of each subinterval. 3 6. Approximate the area A, choosing u as the right endpoint of each subinterval. 6
In Problems 7 and 8, refer to the illustration. The interval 30, 84 is partitioned into four subintervals 30, 24, 32, 44, 34, 64, and 36, 84. y 10
5
y ⫽ f(x)
7. Approximate the area A, choosing u as the left endpoint of each subinterval. 8. Approximate the area A, choosing u as the right endpoint of each subinterval. 38
*9. The function f 1x2 = 3x is defined on the interval 30, 64. *(a) Graph f. *O C m F
BQQSPYJNBUFUIFBSFBA under f from 0 to 6 as follows: *(b) Partition 30, 64 into three subintervals of equal length and choose u as the left endpoint of each subinterval. *(c) Partition 30, 64 into three subintervals of equal length and choose u as the right endpoint of each subinterval. *(d) Partition 30, 64 into six subintervals of equal length and choose u as the left endpoint of each subinterval. *(e) Partition 30, 64 into six subintervals of equal length and choose u as the right endpoint of each subinterval. *(f) What is the actual area A?
*10. 3FQFBU1SPCMFNGPSf 1x2 = 4x.
11. The function f 1x2 = - 3x + 9 is defined on the interval 30, 34. *(a) Graph f. *O C m F
BQQSPYJNBUFUIFBSFBA under f from 0 to 3 as follows: *(b) Partition 30, 34 into three subintervals of equal length and choose u as the left endpoint of each subinterval. *(c) Partition 30, 34 into three subintervals of equal length and choose u as the right endpoint of each subinterval. *(d) Partition 30, 34 into six subintervals of equal length and choose u as the left endpoint of each subinterval.
2 4 6 8 x *Due to space restrictions, answers to these exercises may be found in the Answers f(0) ⫽ 10, f(2) ⫽ 6, f(4) ⫽ 7, f(6) ⫽ 5, f(8) ⫽ 1 in the back of the book.
956
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
*(e) Partition 30, 34 into six subintervals of equal length and choose u as the right endpoint of each subinterval. *(f) What is the actual area A?
*12. 3FQFBU1SPCMFNGPSf 1x2 = - 2x + 8.
In Problems 13–22, a function f is defined over an interval 3a, b4. (a) Graph f, indicating the area A under f from a to b. (b) Approximate the area A by partitioning 3a, b4 into four subintervals of equal length and choosing u as the left endpoint of each subinterval. (c) Approximate the area A by partitioning 3a, b4 into eight subintervals of equal length and choosing u as the left endpoint of each subinterval. (d) Express the area A as an integral. (e) Use a graphing utility to approximate the integral. *13. f 1x2 = x2 + 2,
30, 44
*15. f 1x2 = x3, 30, 44 1 *17. f 1x2 = , 31, 4 x *19. f 1x2 = e x, 3 - 1, 34 *21. f 1x2 = sin x,
30, p4
*14. f 1x2 = x2 - 4,
*16. f 1x2 = x3,
32, 64
31, 4
*18. f 1x2 = 1x,
30, 44
*20. f 1x2 = ln x,
33, 74
*22. f 1x2 = cos x,
c 0,
In Problems 23–30, an integral is given. (a) What area does the integral represent? (b) Provide a graph that illustrates this area. (c) Use a graphing utility to approximate this area.
p d 2
4
*23.
L0
*25.
L2
13x + 12 dx
*24.
1x2 - 12 dx
*26.
p>2
*27.
1 - 2x + 72 dx
4
116 - x2 2 dx
L1 L0
p>4
sin x dx
L0
3
*28.
2
L-p>4
cos x dx
2e
e x dx
ln x dx Le L0 *31. Confirm the entries in Table 7. [Hint: 3FWJFX UIF GPSNVMB GPS UIF TVN PG BO BSJUINFUJD sequence.] *29.
*30.
32. Consider the function f 1x2 = 21 - x2 whose domain is the interval 3 - 1, 14. *(a) Graph f. *(b) Approximate the area under the graph of f from - 1 to 1 by dividing 3 - 1, 14 into five subintervals, each of equal length. *(c) Approximate the area under the graph of f from - 1 to 1 by dividing 3 - 1, 14 into ten subintervals, each of equal length. *(d) Express the area as an integral. *(e) Evaluate the integral using a graphing utility. *(f) What is the actual area?
Retain Your Knowledge Problems 33–36 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. f 1x + h2 - f 1x2 2 19 22 2 and *33. Graph the function f 1x2 = log 2 x. d 35. if f 1x2 = 2x + 3x + 1, find 34. c h 43 0 4 simplify. 4x + 2h + 3 1 2 6 0 16 34. If A = c d and B = c d , find AB. 36. 'JOEUIFQBSUJBMGSBDUJPOEFDPNQPTJUJPOPG . 3 4 7 8 1 1x - 22 1x + 22 2 36.
1 4 1 x - 2 x + 2 (x + 2)2
‘Are You Prepared?’ Answers 1. A = lw
2. 24
Chapter Review Things to Know Limit (p. 924)
lim f 1x2 = N
xSc
Limit formulas (p. 929)
As x gets closer to c, x ≠ c, the value of f gets closer to N.
lim b = b
The limit of a constant is the constant.
lim x = c
The limit of x as x approaches c is c.
xSc xSc
Review Exercises
Limit properties (pp. 930, 932, 933)
lim 3f 1x2 + g1x2 4 = lim f 1x2 + lim g1x2
xSc
xSc
xSc
xSc
xSc
xSc
The limit of a sum equals the sum of the limits.
lim 3f 1x2 - g1x2 4 = lim f 1x2 - lim g1x2 lim 3f 1x2 # g1x2 4
xSc
lim c
xSc
f 1x2 g1x2
=
lim f 1x2
d =
The limit of a difference equals the difference of the limits.
lim f 1x2 # lim g1x2
xSc
xSc
lim g1x2
The limit of a product equals the product of the limits.
xSc
1lim g1x2 ≠ 02
The limit of a quotient equals the quotient of the limits, provided that the limit of the denominator is not zero.
xSc
xSc
957
lim 3f 1x2 4 n = [lim f 1x2]n
Provided lim f(x) exists, n Ú 2 an integer.
lim 2f 1x2 = 2 lim f(x)
Provided 2f(x) and 2 lim f(x) are both defined, n Ú 2 an integer.
xSc
xSc
xSc
n
n
n
xSc
xSc
n
xSc
Limit of a polynomial (p. 931)
Derivative of a function (p. 945)
lim P 1x2 = P 1c2, where P is a polynomial
f′ 1c2 = lim
xSc
f 1x2 - f 1c2 x - c
xSc
Continuous function (p. 938)
, provided that the limit exists
Area under a graph (p. 954) b
lim f 1x2 = f 1c2
xSc
La
n
f 1x2 dx = lim a f 1u i 2 ∆x, provided that the limit exists n Sq i=1
Objectives Section
You should be able to N
Example(s)
Review Exercises
13.1
1 'JOEBMJNJUVTJOHBUBCMF Q
m
m
2 'JOEBMJNJUVTJOHBHSBQI Q
m
m
13.2
1 'JOEUIFMJNJUPGBTVN BEJGGFSFODF BOEBQSPEVDU Q
m
1
2 'JOEUIFMJNJUPGBQPMZOPNJBM Q
7
1, 2
3 'JOEUIFMJNJUPGBQPXFSPSBSPPU Q
8
4 'JOEUIFMJNJUPGBRVPUJFOU Q
m
m
5 'JOEUIFMJNJUPGBOBWFSBHFSBUFPGDIBOHF Q
12
m
13.3
1 'JOEUIFPOFTJEFEMJNJUTPGBGVODUJPO Q
1
m
2 Determine whether a function is continuous (p. 938)
2, 3
m m
13.4
1 'JOEBOFRVBUJPOPGUIFUBOHFOUMJOFUPUIFHSBQIPGBGVODUJPO Q
1
m
2 'JOEUIFEFSJWBUJWFPGBGVODUJPO Q
m
m
3 'JOEJOTUBOUBOFPVTSBUFTPGDIBOHF Q
39
4 'JOEUIFJOTUBOUBOFPVTTQFFEPGBQBSUJDMF Q
6
38
1 "QQSPYJNBUFUIFBSFBVOEFSUIFHSBQIPGBGVODUJPO Q
1, 2
m
2 "QQSPYJNBUFJOUFHSBMTVTJOHBHSBQIJOHVUJMJUZ Q
3
41(e), 42(e), 43(c), 44(c)
Review Exercises In Problems 1–11, find the limit. 1. lim 1x3 - 2x2 + x - 42 xS3
5. lim 1x2 + x + 32 3>2 xS1
x2 - 1 x S -1 x3 - 1
9. lim
2. lim 1x3 - x2 + 2 3
8
x S -2
- 343
x + x + 2 1 4 x2 - 9 x3 - 8 10. lim 3 x S 2 x - 2x2 + 4x - 8 2
27
0
3. lim 2x2 + 7 xS3
4. lim 2x2 - 4
4
x S -3
x - 4 4 x - 2 4 3 x - 2x + x - 2 11. lim 3 x S 2 x - 2x2 + 2x - 4 2
6. lim
7. lim
x S -1
8. lim
xS2
3 2
xS3
1
x - x + 6 x2 - 8x + 1 2
-
1 2
3 2
In Problems 12–15, determine whether f is continuous at c. 12. f 1x2 = x - 2x3 + 6 c = 2 x2 - 9 14. f 1x2 = c x - 3 6
Continuous
if x ≠ 3 if x = 3
c = 3
Continuous
13. f 1x2 =
x2 - 4 x + 2
x2 - 9 15. f 1x2 = c x - 3 -6
c = -2
No
if x ≠ 3 if x = 3
c = 3
Not continuous
958
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
In Problems 16–27, use the accompanying graph of y = f 1x2. 16. What is the domain of f?
y
{x - 6 … x 6 2 or 2 6 x 6 or 6 x … 6}
4
17. What is the range of f? All real numbers
2
18. 'JOEUIFx-intercept(s), if any, of f. 1, 6 2
19. 'JOEUIFy-intercept(s), if any, of f. 4 20. 'JOEf 1 - 62 and f 1 - 42. 21. 'JOE lim - f 1x2.
4
22. 'JOE lim + f 1x2.
-2
x S -4 x S -4
4
6
x
f( - 6) = 2; f( - 4) = 1 23. 'JOE lim- f 1x2. xS2
-q
24. 'JOE lim+ f 1x2. xS2
q
25. Does lim f 1x2 exist? If it does, what is it? Does not exist 26. Is f continuous at 0? No 27. Is f continuous at 4? Yes xS0 x + 4 is continuous at c = - 4 and c = 4. Use limits to analyze the graph of R at c. *28. Discuss whether R 1x2 = 2 x - 16 x3 - 2x2 + 4x - 8 is undefined. Determine whether an asymptote or a hole *29. Determine where the rational function R 1x2 = x2 - 11x + 18 appears at such numbers. In Problems 30–32, find the slope of the tangent line to the graph of f at the given point. Graph f and the tangent line. *30. f 1x2 = 2x2 + 8x at 11, 102
*31. f 1x2 = x2 + 2x - 3 at 1 - 1, - 42
*32. f 1x2 = x3 + x2 at 12, 122
In Problems 33–35, find the derivative of each function at the number indicated. *33. f 1x2 = - 4x2 + at 3
34. f 1x2 = x2 - 3x at 0
- 24
35. f 1x2 = 2x2 + 3x + 2 at 1
-3
7
In Problems 36 and 37, find the derivative of each function at the number indicated using a graphing utility. p 36. f 1x2 = 4x4 - 3x3 + 6x - 9 at - 2 - 18 37. f 1x2 = x3 tan x at 0.66621763 6
3
39. Instantaneous Rate of Change The following data represent the revenue R (in dollars) received from selling x wristwatches at Wilk’s Watch Shop.
12
where t is the elapsed time that the ball is in the air. The ball misses the rooftop on its way down and eventually strikes the ground. (a) When does the ball strike the ground? That is, how long is the ball in the air? 7 seconds (b) At what time t will the ball pass the rooftop on its way down? 6 seconds (c) What is the average speed of the ball from t = 0 to t = 2? 64 feet per second *(d) What is the instantaneous speed of the ball at time t? *(e) What is the instantaneous speed of the ball at t = 2? (f) When is the instantaneous speed of the ball equal to zero? At t = 3 sec (g) What is the instantaneous speed of the ball as it passes the rooftop on the way down? −96 feet per second (h) What is the instantaneous speed of the ball when it strikes the ground? −128 feet per second
6
s = s 1t2 = - 16t + 96t + 112 2
9
38. Instantaneous Speed of a Ball In physics it is shown that the height s of a ball thrown straight up with an initial speed of 96 ft/sec from a rooftop 112 feet high is
Wristwatches, x
Revenue, R
0
0
25
2340
40
3600
50
4375
90
6975
130
8775
160
9600
200
10,000
220
9900
250
9375
* B 'JOEUIFBWFSBHFSBUFPGDIBOHFPGSFWFOVFGSPNx = 2 to x = 130 wristwatches. * C 'JOEUIFBWFSBHFSBUFPGDIBOHFPGSFWFOVFGSPNx = 2 to x = 90 wristwatches. * D 'JOEUIFBWFSBHFSBUFPGDIBOHFPGSFWFOVFGSPNx = 2 to x = 0 wristwatches. *(d) Using a graphing utility, find the quadratic function of best fit. *(e) Using the function found in part (d), determine the instantaneous rate of change of revenue at x = 2 wristwatches.
Chapter Test
40. The function f 1x2 = 2x + 3 is defined on the interval 30, 44. *(a) Graph f. In (b)–(e), approximate the area A under f from x = 0 to x = 4 as follows: *(b) Partition 30, 44 into four subintervals of equal length and choose u as the left endpoint of each subinterval. *(c) Partition 30, 44 into four subintervals of equal length and choose u as the right endpoint of each subinterval. *(d) Partition 30, 44 into eight subintervals of equal length and choose u as the left endpoint of each subinterval. *(e) Partition 30, 44 into eight subintervals of equal length and choose u as the right endpoint of each subinterval. *(f) What is the actual area A? 28
959
In Problems 41 and 42, a function f is defined over an interval 3a, b4. (a) Graph f, indicating the area A under f from a to b. (b) Approximate the area A by partitioning 3a, b4 into three subintervals of equal length and choosing u as the left endpoint of each subinterval. (c) Approximate the area A by partitioning 3a, b4 into six subintervals of equal length and choosing u as the left endpoint of each subinterval. (d) Express the area A as an integral. (e) Use a graphing utility to approximate the integral. *41. f 1x2 = 4 - x2, *42. f 1x2 =
1 , x2
3 - 1, 24
31, 44
In Problems 43 and 44, an integral is given. (a) What area does the integral represent? (b) Provide a graph that illustrates this area. (c) Use a graphing utility to approximate this area. 3
43.
L-1
1
19 - x2 2 dx
44.
L-1
e x dx
Chapter Test Prep Videos include step-by-step solutions to all chapter test exercises and can be found on this text’s Channel (search “SullivanPrecalcUC3e”).
Chapter Test
0x - 20
In Problems 1–6, find each limit. 1. lim
xS3
4. lim
1 - x2
x S -1
+ 3x - 5 2
x2 - 4x - 5 x3 + 1
-5
2.
-2
lim
xS2 +
5. lim
xS5
3x - 6
3 13x2 1x
x2 - 9 if x … 4 f 1x2 = c x + 3 kx + 5 if x 7 4
In Problems 8–12, use the accompanying graph of y = f 1x2. y
4
9. Find lim- f 1x2 xS3
10. Find lim f 1x2 xS - 2
x
-3 5 2
*11. Does lim f 1x2 exist? If so, what is it? If not, explain why not. xS1
135
6.
lim 27 - 3x
5
tan x 1 + cos2 x
2 3
x S -6
limp
xS
4
*12. Determine whether f is continuous at each of the following numbers. If it is not, explain why not. (a) x = - 2 (b) x = 1 (c) x = 3 (d) x = 4 *13. Determine where the rational function R 1x2 =
x3 + 6x2 - 4x - 24 x2 + 5x - 14
is undefined. Determine whether an asymptote or a hole appears at such numbers.
4
xS3
3.
- 22 2 4
7. Determine the value for k that will make the function continuous at c = 4. - 1
8. Find lim+ f 1x2
1 3
14. For the function f 1x2 = 4x2 - 11x - 3: (a) Find the derivative of f at x = 2. 5 (b) Find the equation of the tangent line to the graph of f at the point 12, - 92 . y = 5x - 19 *(c) Graph f and the tangent line.
15. The function f 1x2 = 216 - x2 is defined on the interval 30, 44 . *(a) Graph f. *(b) Partition 30, 44 into eight subintervals of equal length and choose u as the left endpoint of each subinterval. Use the partition to approximate the area under the graph of f and above the x-axis from x = 0 to x = 4. (c) Find the exact area of the region and compare it to the approximation in part (b). 4p ≈ 12.566
960
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function
16. Write the integral that represents the shaded area. Do not attempt to evaluate. 4 ( - x2 + x + 3) dx L1 y
17. A particle is moving along a straight line according to some position function s 1t2 . The distance (in feet) of the particle, s, from its starting point after t seconds is given in the table. 'JOE UIF BWFSBHF SBUF PG DIBOHF PG EJTUBODF GSPN t = 3 to t = 6 seconds. 1 3 ft/sec 3 t s
8 f (x) = –x 2 + 5x + 3 4
x 4
8
0
0
1
2.5
2
14
3
31
4
49
5
89
6
137
7
173
8
240
Chapter Projects 2. Graph Y1 = f 1t2, where f 1t2 represents the logistic growth function of best fit found in part (a). 3. Determine the instantaneous rate of growth of population in 1960 using the numerical derivative function on your graphing utility. 4. Use the result from part (c) to predict the population in 1961. What was the actual population in 1961? 5. Determine the instantaneous growth of population in 1970, 1980, 1990, 2000, and 2010. What is happening to the instantaneous growth rate as time passes? Is Malthus’ contention of a geometric growth rate accurate?
I.
World Population Thomas Malthus believed that “population, when unchecked, increases in a geometrical progression of such nature as to double itself every twenty-five years.” However, the growth of population is limited because the resources available to us are limited in supply. If Malthus’s conjecture were true, geometric growth of the world’s population would imply that Pt = r + 1, where r is the growth rate Pt - 1 1. Using world population data and a graphing utility, find the logistic growth function of best fit, treating the year as the independent variable. Let t = 0 SFQSFTFOU t = 1 SFQSFTFOU BOETPPO VOUJMZPVIBWFFOUFSFEBMMUIFZFBST and the corresponding populations up to the current year.
6. Using the numerical derivative function on your graphing utility, graph Y2 = f′(t), where f′(t) represents the derivative of f 1t2 with respect to time. Y2 is the growth rate of the population at any time t. 7. Using the MAXIMUM function on your graphing utility, determine the year in which the growth rate of the population is largest. What is happening to the growth rate JOUIFZFBSTGPMMPXJOHUIFNBYJNVN 'JOEUIJTQPJOUPOUIF graph of Y1 = f 1t2.
8. Evaluate lim f 1t2. This limiting value is the carrying t Sq
capacity of Earth. What is the carrying capacity of Earth? 9. What do you think will happen if the population of Earth exceeds the carrying capacity? Do you think that agricultural output will continue to increase at the same rate as population growth? What effect will urban sprawl have on agricultural output?
The following projects are available on the Instructor’s Resource Center (IRC). II. Project at Motorola: Curing Rates Engineers at Motorola use calculus to find the curing rate of a sealant. III. Finding the Profit-maximizing Level of Output A manufacturer uses calculus to maximize profit.
Appendix
A
Review Outline A.1 A.2 A.3 A.4 A.5 A.6
Algebra Essentials Geometry Essentials Polynomials Factoring Polynomials Synthetic Division Rational Expressions
nth Roots; Rational Exponents Solving Equations Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications A.10 Interval Notation; Solving Inequalities A.11 Complex Numbers A.7 A.8 A.9
A.1 Algebra Essentials PREPARING FOR THIS BOOK Before getting started, read “To the Student” at the beginning of this text on page xxiv. OBJECTIVES 1 2 3 4 5 6 7 8
Work with Sets (p. A1) Graph Inequalities (p. A4) Find Distance on the Real Number Line (p. A5) Evaluate Algebraic Expressions (p. A6) Determine the Domain of a Variable (p. A7) Use the Laws of Exponents (p. A7) Evaluate Square Roots (p. A9) Use a Calculator to Evaluate Exponents (p. A10)
1 Work 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 determine 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 ∅. 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 = 5 0, 1, 2, 3, 4, 5, 6, 7, 8, 96
In this notation, the braces 5 6 are used to enclose the objects, or 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 =
5
x c
x is a digit6
c
c
c
5
c
Read as “D is the set of all x such that x is a digit”
A1
A2
APPENDIX A Review
EX A MPL E 1
Using Set-Builder Notation and the Roster Method
(a) E = 5 x 0 x is an even digit6 = 5 0, 2, 4, 6, 86 (b) O = 5 x 0 x is an odd digit6 = 5 1, 3, 5, 7, 96
r
Because the elements of a set are distinct, we never repeat elements. For example, we would never write 5 1, 2, 3, 26 ; the correct listing is 5 1, 2, 36 . Because a set is a collection, the order in which the elements are listed does not matter. Thus, 5 1, 2, 36 , 5 1, 3, 26 , 5 2, 1, 36 , and so on, all represent the same set. If every element of a set A is also an element of a set B, then A is a subset of B, which is denoted A ⊆ B. If two sets A and B have the same elements, then A equals B, which is denoted A = B. For example, 5 1, 2, 36 ⊆ 5 1, 2, 3, 4, 56 and 5 1, 2, 36 = 5 2, 3, 16 . If A and B are sets, the intersection of A with B, denoted A ∩ B, is the set consisting of elements that belong to both A and B. The union of A with B, denoted A ∪ B, is the set consisting of elements that belong to either set A or set B or to both.
DEFINITION
EX A MPL E 2
Finding the Intersection and Union of Sets
Let A = 5 1, 3, 5, 86 , B = 5 3, 5, 76 , and C = 5 2, 4, 6, 86 . Find: (a) A ∩ B
Solution
(b) A ∪ B
(c) B ∩ 1A ∪ C2
(a) A ∩ B = 5 1, 3, 5, 86 ∩ 5 3, 5, 76 = 5 3, 56 (b) A ∪ B = 5 1, 3, 5, 86 ∪ 5 3, 5, 76 = 5 1, 3, 5, 7, 86 (c) B ∩ 1A ∪ C2 = 5 3, 5, 76 ∩ 1 5 1, 3, 5, 86 ∪ 5 2, 4, 6, 86 2 = 5 3, 5, 76 ∩ 5 1, 2, 3, 4, 5, 6, 86 = 5 3, 56
Now Work
PROBLEM
r
13
When working with sets, it is common practice to designate a universal set U, the set consisting of all the elements to be considered. With a universal set designated, elements of the universal set not found in a given set can be considered.
DEFINITION
EX A MPL E 3
If A is a set, then the complement of A, denoted A, is the set consisting of all the elements in the universal set that are not in A.*
Finding the Complement of a Set
If the universal set is U = 5 1, 2, 3, 4, 5, 6, 7, 8, 96 and if A = 5 1, 3, 5, 7, 96 , then A = 5 2, 4, 6, 86 .
r
Figure 1 Universal set
It follows from the definition of complement that A ∪ A = U and A ∩ A = ∅. Do you see why?
Now Work
B A C
PROBLEM
17
It is often helpful to draw pictures of sets. Such pictures, called Venn diagrams, represent sets as circles enclosed in a rectangle. The rectangle represents the universal set. Such diagrams often make it easier to visualize various relationships among sets. See Figure 1. Some books use the notation A′ or A c for the complement of A.
*
SECTION A.1 Algebra Essentials
A3
The Venn diagram in Figure 2(a) illustrates that A ⊆ B. The Venn diagram in Figure 2(b) illustrates that A and B have no elements in common—that is, A ∩ B = ∅. The sets A and B in Figure 2(b) are disjoint. Figure 2
Universal set
Universal set
B A
A
(a) A ⊆ B subset
B
(b) A ∩ B = ∅ disjoint sets
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
Universal set
Universal set
B
A
(a) A B intersection
A
B
A
(c) A complement
(b) A B union
Real Numbers Real numbers are represented by symbols such as
25, 0,
Figure 4 p =
C d
C d
- 3,
1 , 2
5 - , 0.125, 4
22, p,
3 2 - 2, 0.666 c
The set of counting numbers, or natural numbers, contains the numbers in the set 5 1, 2, 3, 4, c 6 . (The three dots, called an ellipsis, indicate that the pattern continues indefinitely.) The set of integers contains the numbers in the set 5 c , - 3, - 2, - 1, 0, 1, 2, 3, c 6 . A rational number is a number that can be a expressed as a quotient of two integers, where the integer b cannot be 0. Examples b 3 5 0 2 a of rational numbers are , , , and - . Since = a for any integer a, every 4 2 4 3 1 integer is also a rational number. Real numbers that are not rational are called irrational. Examples of irrational numbers are 22 and p (the Greek letter pi), which equals the constant ratio of the circumference to the diameter of a circle. See Figure 4. Real numbers can be represented as decimals. Rational real numbers have decimal representations that either terminate or are nonterminating with repeating 3 2 blocks of digits. For example, = 0.75, which terminates; and = 0.666c, 4 3 in which the digit 6 repeats indefinitely. Irrational real numbers have decimal representations that neither repeat nor terminate. For example, 22 = 1.414213c and p = 3.14159. c In practice, the decimal representation of an irrational number is given as an approximation. We use the symbol ≈ (read as “approximately equal to”) to write 22 ≈ 1.4142 and p ≈ 3.1416.
A4
APPENDIX A Review
Two frequently used properties of real numbers are given next. Suppose that a, b, and c are real numbers.
Distributive Property a # 1b + c2 = ab + ac
Zero-Product Property
In Words If a product equals 0, then one or both of the factors is 0.
If ab = 0, then either a = 0 or b = 0 or both equal 0. The Distributive Property can be used to remove parentheses: 2(x + 3) = 2x + 2 # 3 = 2x + 6
The Zero-Product Property will be used to solve equations (Section A.8). For example, if 2x = 0, then 2 = 0 or x = 0. Since 2 ≠ 0, it follows that x = 0.
The Real Number Line
Figure 5 Real number line 2 units Scale 1 unit O ⫺3
⫺2
⫺1 ⫺1–2 0 1–2 1 2 2
3
DEFINITION
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 0. See Figure 5. The point 1 unit to the right of O corresponds to the number 1. The distance between 0 and 1 determines the scale of the number line. For example, the point associated with the number 2 is twice as far from O 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 5 also shows 1 1 the points associated with the rational numbers - and and the points associated 2 2 with the irrational numbers 12 and p. The 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.
Figure 6 O ⫺3
⫺2 ⫺ 3–2 ⫺1⫺ 1–2 0 1–2 1 3–2 2 Negative real numbers
Zero
Positive real numbers
Figure 7 a
(a) a ⬍ b
3
b
The real number line consists of three classes of real numbers, as shown in Figure 6. 1. The negative real numbers are the coordinates of points to the left of the origin O. 2. 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.
Now Work a b (b) a ⫽ b
b
(c) a ⬎ b
PROBLEM
21
2 Graph Inequalities
a
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 7. If a is to the left of b, then “a is less than b,” which is denoted a 6 b. If a is to the right of b, then “a is greater than b,” which is denoted a 7 b. If a is at the same location as b, then a = b. If a is either less than or equal to b, then write a … b.
SECTION A.1 Algebra Essentials
A5
Similarly, a Ú b means that a is either greater than or equal to b. Collectively, the symbols 6 , 7 , … , and Ú are called inequality symbols. Note that a 6 b and b 7 a mean the same thing. It does not matter whether we write 2 6 3 or 3 7 2. Furthermore, if a 6 b or if b 7 a, then the difference b - a is positive. Do you see why? An inequality is a statement in which two expressions are related by an inequality symbol. The expressions are referred to as the sides of the inequality. Statements of the form a 6 b or b 7 a are called strict inequalities, whereas statements of the form a … b or b Ú a are called nonstrict inequalities. The following conclusions are based on the discussion so far. a 7 0 is equivalent to a is positive a 6 0 is equivalent to a is negative The inequality a 7 0 is sometimes read as “a is positive.” If a Ú 0, then either a 7 0 or a = 0, and this is read as “a is nonnegative.”
Now Work
PROBLEMS
25
AND
35
Graphing Inequalities
EXAM PL E 4
(a) On the real number line, graph all numbers x for which x 7 4. (b) On the real number line, graph all numbers x for which x … 5.
Solution
Figure 8 –2 –1
0
1
2
3
4
5
6
7
r
Figure 9 ⫺2 ⫺1
(a) See Figure 8. Note that we use a left parenthesis to indicate that the number 4 is not part of the graph. (b) See Figure 9. Note that we use a right bracket to indicate that the number 5 is part of the graph.
Now Work 0
1
2
3
4
5
6
PROBLEM
41
7
3 Find Distance on the Real Number Line Figure 10 4 units
3 units
⫺5 ⫺4 ⫺3 ⫺2 ⫺1
0
1
2
3
4
DEFINITION
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 0. See Figure 10. 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 0 a 0 , is defined by the rules
0 a 0 = a if a Ú 0
and
0 a 0 = - a if a 6 0
For example, because - 4 6 0, the second rule must be used to get 0 - 4 0 = - 1 - 42 = 4.
EXAM PL E 5
Computing Absolute Value (a) 0 8 0 = 8
(b) 0 0 0 = 0
(c) 0 - 15 0 = - 1 - 152 = 15
r
Look again at Figure 10. The distance from - 4 to 3 is 7 units. This distance is the difference 3 - 1 - 42 , obtained by subtracting the smaller coordinate from
A6
APPENDIX A Review
the larger. However, because 0 3 - 1 - 42 0 = 0 7 0 = 7 and 0 - 4 - 3 0 = 0 - 7 0 = 7, absolute value can be used to calculate the distance between two points without being concerned about which is smaller.
DEFINITION
If P and Q are two points on a real number line with coordinates a and b, respectively, then the distance between P and Q, denoted by d 1P, Q2, is d 1P, Q2 = 0 b - a 0 Since 0 b - a 0 = 0 a - b 0 , it follows that d 1P, Q2 = d 1Q, P2.
EX A MPL E 6
Finding Distance on a Number Line Let P, Q, and R be points on a real number line with coordinates - 5, 7, and - 3, respectively. Find the distance (a) between P and Q
Solution
(b) between Q and R
See Figure 11. Figure 11 P
R
Q
⫺5 ⫺4 ⫺3 ⫺2 ⫺1
0
1
2
3
4
5
6
7
d (P, Q) ⫽ ⏐7 ⫺ (⫺5)⏐ ⫽ 12 d (Q, R) ⫽ ⏐ ⫺3 ⫺ 7 ⏐ ⫽ 10
(a) d 1P, Q2 = 0 7 - 1 - 52 0 = 0 12 0 = 12 (b) d 1Q, R2 = 0 - 3 - 7 0 = 0 - 10 0 = 10
Now Work
PROBLEM
r
47
4 Evaluate Algebraic Expressions In algebra, letters such as x, y, a, b, and c are used to represent numbers. If the letter used represents any number from a given set of numbers, it is called a variable. A constant is either a fixed number, such as 5 or 13, or a letter that represents a fixed (possibly unspecified) number. Constants and variables are combined using the operations of addition, subtraction, multiplication, and division to form algebraic expressions. Examples of algebraic expressions include x + 3
3 1 - t
7x - 2y
To evaluate an algebraic expression, substitute a numerical value for each variable.
EX A MPL E 7
Evaluating an Algebraic Expression Evaluate each expression if x = 3 and y = - 1. 3y (d) 0 - 4x + y 0 2 - 2x (a) Substitute 3 for x and - 1 for y in the expression x + 3y. (a) x + 3y
Solution
(b) 5xy
(c)
x + 3y = 3 + 31 - 12 = 3 + 1 - 32 = 0
c
x = 3, y = - 1
SECTION A.1 Algebra Essentials
A7
(b) If x = 3 and y = - 1, then 5xy = 5132 1 - 12 = - 15 (c) If x = 3 and y = - 1, then 31 - 12 3y -3 -3 3 = = = = 2 - 2x 2 - 2132 2 - 6 -4 4 (d) If x = 3 and y = - 1, then
0 - 4x + y 0 = 0 - 4132 + 1 - 12 0 = 0 - 12 + 1 - 12 0 = 0 - 13 0 = 13 Now Work
PROBLEMS
49
AND
r
57
5 Determine the Domain of a Variable 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 = pr 2, the variable r is restricted to the positive real numbers (since a radius cannot be 0 or negative). In the expression 1 , the variable x cannot take on the value 0, since division by 0 is not defined. x
DEFINITION
EXAM PL E 8
The set of values that a variable may assume is called the domain of the variable.
Finding the Domain of a Variable The domain of the variable x in the expression 5 x - 2 is 5 x 0 x ≠ 26 , since if x = 2, the denominator becomes 0, which is not defined.
EXAM PL E 9
r
Circumference of a Circle In the formula for the circumference C of a circle of radius r, which is C = 2pr the domain of the variable r, representing the radius of the circle, is the set of positive real numbers, {r 0 r 7 0}. The domain of the variable C, representing the circumference of the circle, is also the set of positive real numbers, {C 0 C 7 0}.
r
In describing the domain of a variable, either set notation or words may be used, whichever is more convenient.
Now Work
PROBLEM
67
6 Use the Laws of Exponents Integer exponents provide a shorthand notation for representing repeated multiplications of a real number. For example, 23 = 2 # 2 # 2 = 8
34 = 3 # 3 # 3 # 3 = 81
A8
APPENDIX A Review
DEFINITION
If a is a real number and n is a positive integer, then the symbol an represents the product of n factors of a. That is, an = a # a # g # a
c
(1)
n factors
WARNING Be careful with negatives and exponents. - 24 = - 1 # 24 = - 16 whereas ( - 2)4 = ( - 2)( - 2)( - 2)( - 2) = 16 䊏
In the definition, it is understood that a1 = a. In addition, a2 = a # a, a3 = a # a # a, and so on. In the expression an, a is called the base and n is called the exponent, or power. an is read as “a raised to the power n” or as “a to the nth power.” Usually, a2 is read as “a squared” and a3 is read as “a cubed.” In working with exponents, the operation of raising to a power is performed before any other operation. Here are some 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, 1 - 22 4 = 1 - 22 1 - 22 1 - 22 1 - 22 = 16
DEFINITION
12 + 32 2 = 52 = 25
If a ≠ 0, then a0 = 1
DEFINITION
If a ≠ 0 and if n is a positive integer, then 1 an
a -n =
Whenever you encounter a negative exponent, think “reciprocal.”
EX AM PL E 10
Evaluating Expressions Containing Negative Exponents (a) 2-3 =
1 1 = 3 8 2
(b) x -4 =
Now Work
PROBLEMS
1 -2 (c) a b = 5
1 x4
85
AND
1 1 a b 5
2
=
1 = 25 1 25
105
r
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 aman = am + n am 1 = am - n = n - m an a
1am 2 n = amn if a ≠ 0
1ab2 n = anbn a n an a b = n b b
if b ≠ 0
SECTION A.1 Algebra Essentials
EX AM PL E 11
A9
Using the Laws of Exponents Write each expression so that all exponents are positive. (a)
Solution
(a)
x5 y -2 x3 y x5 y -2 x3 y
(b) ¢
x ≠ 0, y ≠ 0
=
(b) ¢
x -3 -2 ≤ x ≠ 0, y ≠ 0 3y -1
x2 x5 # y -2 5 - 3 # -2 - 1 2 -3 2# 1 y = x y = x = = x x3 y y3 y3
1x -3 2 -2 x -3 -2 x6 x6 9x6 ≤ = = = = 1 2 y2 3y -1 13y -1 2 -2 3-2 1y -1 2 -2 y 9
Now Work
PROBLEMS
87
AND
r
97
7 Evaluate Square Roots In Words
136 means “give me the nonnegative number whose square is 36.”
DEFINITION
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 1 - 62 2 = 36, the numbers 6 and - 6 are square roots of 36. The symbol 1 , called a radical sign, is used to denote the principal, or nonnegative, square root. For example, 136 = 6. If a is a nonnegative real number, the nonnegative number b, such that b2 = a, is the principal square root of a and is denoted by b = 1a.
The following comments are noteworthy: 1. Negative numbers do not have square roots (in the real number system), because the square of any real number is nonnegative. For example, 1- 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 = 0. That is, 10 = 0. 3. The principal square root of a positive number is positive. 4. If c Ú 0, then 1 1c2 2 = c. For example, 1 122 2 = 2 and 1 132 2 = 3.
EX AM PL E 12
Evaluating Square Roots (a) 164 = 8
(b)
1 1 = A 16 4
(c)
1 11.4 2 2
r
= 1.4
Examples 12(a) and (b) are examples of square roots of perfect squares, since 1 1 2 64 = 82 and = a b . 16 4 Consider the expression 2a2. Since a2 Ú 0, the principal square root of a2 is defined whether a 7 0 or a 6 0. However, since the principal square root is nonnegative, an absolute value is needed to ensure the nonnegative result. That is, 2a2 = 0 a 0
EX AM PL E 13
where a is any real number
Simplifying Expressions Using Equation (2) (a) 2 12.32 2 = 0 2.3 0 = 2.3
Now Work
(b) 2 1 - 2.32 2 = 0 - 2.3 0 = 2.3
(2)
(c) 2x2 = 0 x 0
r PROBLEM
93
A10
APPENDIX A Review
Calculators Calculators are finite machines. As a result, they are incapable of displaying decimals that contain a large number of digits. For example, some calculators are capable of displaying only eight digits. When a number requires more than eight digits, the calculator either truncates or rounds. To see how your calculator handles decimals, 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 ln, log, sin, cos, tan, xy, inv, and so on. Graphing calculators have all the capabilities of scientific calculators and contain a screen on which graphs can be displayed. In this book, the graphing calculator is optional. The symbol is shown whenever a graphing calculator is used.
8 Use a Calculator to Evaluate Exponents Your calculator has a caret key, ^ , which is used for computations involving exponents.
Exponents on a Graphing Calculator
EX AM PL E 14
Evaluate: 12.32 5
Solution
r
Figure 12 shows the result using a TI-84 graphing calculator. Figure 12
Now Work
PROBLEM
123
A.1 Assess Your Understanding Concepts and Vocabulary 1. A(n) variable is a letter used in algebra to represent any number from a given set of numbers.
5. True or False The product of two negative real numbers is always greater than zero. True
2. On the real number line, the real number zero is the coordinate of the origin .
6. True or False The distance between two distinct points on the real number line is always greater than zero. True
3. An inequality of the form a 7 b is called a(n) strict inequality.
7. True or False The absolute value of a real number is always greater than zero. False
4. In the expression 24, the number 2 is called the base and 4 is called the exponent or power .
8. True or False To multiply two expressions having the same base, retain the base and multiply the exponents. False
Skill Building In Problems 9–20, use U = universal set = 5 0, 1, 2, 3, 4, 5, 6, 7, 8, 96, A = 51, 3, 4, 5, 96 , B = 52, 4, 6, 7, 86 , and C = 51, 3, 4, 66 to find each set. *9. A ∪ B
13. 1A ∪ B2 ∩ C *17. A ∩ B
51, 3, 4, 66
10. A ∪ C
5 1, 3, 4, 5, 6, 9 6
18. B ∪ C
5 0, 5, 96
14. 1A ∩ B2 ∪ C
51, 3, 4, 66
11. A ∩ B 15. A
5 46
50, 2, 6, 7, 86
*19. A ∪ B
3 5 *21. On the real number line, label the points with coordinates 0, 1, - 1, , - 2.5, , and 0.25. 2 4 2 3 1 *22. On the real number line, label the points with coordinates 0, - 2, 2, - 1.5, , , and . 2 3 3 *Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
12. A ∩ C 16. C
51, 3, 46
50, 2, 5, 7, 8, 96
20. B ∩ C
50, 5, 96
A11
SECTION A.1 Algebra Essentials
In Problems 23–32, replace the question mark by 6, 7, or =, whichever is correct. 1 5 23. ? 0 7 24. 5 ? 6 6 25. - 1 ? - 2 7 26. - 3 ? 2 2 28. 22 ? 1.41
7
29.
1 ? 0.5 2
=
30.
1 ? 0.33 3
7
6
2 ? 0.67 3
31.
27. p ? 3.14
6
32.
7
1 ? 0.25 4
=
In Problems 33–38, write each statement as an inequality. 33. x is positive
x 7 0
36. y is greater than - 5
z 6 0
34. z is negative y 7 -5
35. x is less than 2 x … 1
37. x is less than or equal to 1
x 6 2 x Ú 2
38. x is greater than or equal to 2
In Problems 39–42, graph the numbers x on the real number line. *39. x Ú - 2
*40. x 6 4
*41. x 7 - 1
*42. x … 7
In Problems 43–48, use the given real number line to compute each distance. A
43. d1C, D2
1
44. d1C, A2
B
C
D
⫺4 ⫺3 ⫺2 ⫺1
0
1
45. d1D, E2
3
2
E 2
3
4
5
46. d1C, E2
6
47. d1A, E2
3
48. d1D, B2
6
2
In Problems 49–56, evaluate each expression if x = - 2 and y = 3. 49. x + 2y
53.
2x x - y
50. 3x + y
4 4 5
54.
x + y x - y
-3 -
51. 5xy + 2
1 5
55.
In Problems 57–66, find the value of each expression if x = 3 and y = - 2. 57. 0 x + y 0
62.
0y0 y
1
58. 0 x - y 0
5
63. 0 4x - 5y 0
-1
22
3x + 2y 2 + y
- 28
52. - 2x + xy
0
56.
59. 0 x 0 + 0 y 0
5
60. 0 x 0 - 0 y 0
64. 0 3x + 2y 0
5
65. 0 0 4x 0 - 0 5y 0 0
61.
1
2x - 3 y
0x0 x
-
-2 7 3
1
66. 3 0 x 0 + 2 0 y 0
2
13
In Problems 67–74, determine which of the value(s) (a) through (d), if any, must be excluded from the domain of the variable in each expression. (a) x = 3
(b) x = 1
(c) x = 0
(d) x = - 1
67.
x2 - 1 x
x = 0
68.
x2 + 1 x
x = 0
71.
x2 x + 1
None
72.
x3 x - 1
x = 1, x = - 1
2
2
x x - 9
69.
x = 3
2
x2 + 5x - 10 x3 - x
*73.
70.
x x + 9
74.
- 9x2 - x + 1 x3 + x
78.
x - 2 x - 6
None
2
x = 0
In Problems 75–78, determine the domain of the variable x in each expression. 75.
4 x - 5
5x x ≠ 56
76.
-6 x + 4
5 x x ≠ - 46
77.
x x + 4
5x x ≠ - 46
5x x ≠ 66
5 In Problems 79–82, use the formula C = 1F - 322 for converting degrees Fahrenheit into degrees Celsius to find the Celsius measure 9 of each Fahrenheit temperature. 79. F = 32 °
80. F = 212 °
0°C
81. F = 77 °
100°C
82. F = - 4°
25°C
- 20°C
In Problems 83–94, simplify each expression. 83. 1 - 42 2
16
89. 13-2 2 -1
9
84. - 42
- 16
90. 12-1 2 -3
8
85. 4-2 91. 225
1 16 5
86. - 4-2
-
92. 236
6
1 16
87. 3-6 # 34
88. 4-2 # 43
1 9
93. 21 - 42 2
4
4
94. 21 - 32 2
3
A12
APPENDIX A Review
In Problems 95–104, simplify each expression. Express the answer so that all exponents are positive. Whenever an exponent is 0 or negative, assume that the base is not 0. y3 x2 y 3 x 1 x4 3 2 2 96. 1 - 4x2 2 -1 - 2 97. 1x2 y -1 2 98. 1x -1 y2 99. 95. 18x3 2 64x6 2 3 y 4x x xy4 y 100.
x -2 y xy
2
1 x3y
*101.
1 - 22 3 x4 1yz2 2 2
*102.
3
3 xy z
4x -2 1yz2 -1
103. ¢
3 4
2x y
In Problems 105–116, find the value of each expression if x = 2 and y = - 1. 3 105. 2xy -1 - 4 106. - 3x -1 y 107. x2 + y2 2 109. 1xy2 2
110. 1x + y2 2
4
111. 2x2
1
3x -1 4y
-1
25
114. 2x2 + 2y2
3
16662 4
12222 4
?
16x2 9y2
104. ¢
112.
2
115. xy
216x6 125y6
4
1 2x 2 2
116. yx
2
1
10; 0 8; 44
120. What is the value of 10.12 3 1202 3 ?
81
5x -2 -3 ≤ 6y -2
108. x2 y2
118. Find the value of the expression 4x3 + 3x2 - x + 2 if x = 1. What is the value if x = 2? 119. What is the value of
-2
5
1 2 117. Find the value of the expression 2x3 - 3x2 + 5x - 4 if x = 2. What is the value if x = 1?
113. 2x2 + y2
≤
8
In Problems 121–128, use a calculator to evaluate each expression. Round your answer to three decimal places. 121. 18.22 6 125. 1 - 2.82 6
122. 13.72 5
304,006.671
126. - 12.82 6
481.890
693.440 - 481.890
123. 16.12 -3
124. 12.22 -5
0.004
127. 1 - 8.112 -4
0.019
128. - 18.112 -4
0.000
- 0.000
Applications and Extensions In Problems 129–138, express each statement as an equation involving the indicated variables. 129. Area of a Rectangle The area A of a rectangle is the product of its length l and its width w. A = lw
133. Area of an Equilateral Triangle The area A of an 13 equilateral triangle is times the square of the length x of 4 one side.
l
A = A
w x
x
x
130. Perimeter of a Rectangle The perimeter P of a rectangle is twice the sum of its length l and its width w. P = 21l + w2 131. Circumference of a Circle The circumference C of a circle is the product of p and its diameter d. C = pd C
13 2 x 4
134. Perimeter of an Equilateral Triangle The perimeter P of an equilateral triangle is 3 times the length x of one side. P = 3x 4 135. Volume of a Sphere The volume V of a sphere is times p 3 times the cube of the radius r.
d
V = 132. Area of a Triangle The area A of a triangle is one-half the product of its base b and its height h. A =
4 3 pr 3
r
1 bh 2
h b
136. Surface Area of a Sphere The surface area S of a sphere is 4 times p times the square of the radius r. S = 4pr 2
SECTION A.2 Geometry Essentials
137. Volume of a Cube The volume V of a cube is the cube of the length x of a side. V = x3
(a) Show that a voltage of 108 volts is acceptable. 2 … 5 *(b) Show that a voltage of 104 volts is not acceptable. 144. Foreign Voltage In some countries, normal household voltage is 220 volts. It is acceptable for the actual voltage x to differ from normal by at most 8 volts. A formula that describes this is
x
0 x - 220 0 … 8
x
x
A13
138. Surface Area of a Cube The surface area S of a cube is 6 times the square of the length x of a side. S = 6x2 139. Manufacturing Cost The weekly production cost C of manufacturing x watches is given by the formula C = 4000 + 2x, where the variable C is in dollars. (a) What is the cost of producing 1000 watches? $6000 (b) What is the cost of producing 2000 watches? $8000 140. Balancing a Checkbook At the beginning of the month, Mike had a balance of $210 in his checking account. During the next month, he deposited $80, wrote a check for $120, made another deposit of $25, and wrote two checks: one for $60 and the other for $32. He was also assessed a monthly service charge of $5. What was his balance at the end of the month? $98 In Problems 141 and 142, write an inequality using an absolute value to describe each statement.
0x - 40 Ú 6 142. x is more than 5 units from 2. 0 x - 2 0 7 5
(a) Show that a voltage of 214 volts is acceptable. 6 … 8 *(b) Show that a voltage of 209 volts is not acceptable. 145. Making Precision Ball Bearings The FireBall Company manufactures ball bearings for precision equipment. One of its products is a ball bearing with a stated radius of 3 centimeters (cm). Only ball bearings with a radius within 0.01 cm of this stated radius are acceptable. If x is the radius of a ball bearing, a formula describing this situation is
0 x - 3 0 … 0.01
(a) Is a ball bearing of radius x = 2.999 acceptable? Yes (b) Is a ball bearing of radius x = 2.89 acceptable? No 146. Body Temperature Normal human body temperature is 98.6°F. A temperature x that differs from normal by at least 1.5°F is considered unhealthy. A formula that describes this is
0 x - 98.6 0 Ú 1.5
141. x is at least 6 units from 4.
143. U.S. Voltage In the United States, normal household voltage is 110 volts. It is acceptable for the actual voltage x to differ from normal by at most 5 volts. A formula that describes this is
0 x - 110 0 … 5
*(a) Show that a temperature of 97°F is unhealthy. *(b) Show that a temperature of 100°F is not unhealthy. 1 *147. Does equal 0.333? If not, which is larger? By how much? 3 *148. Does
2 equal 0.666? If not, which is larger? By how much? 3
Discussion and Writing 149. Is there a positive real number “closest” to 0? No 150. Number Game I’m thinking of a number! It lies between 1 and 10; its square is rational and lies between 1 and 10. The number is larger than p. 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. 3.14, 3.15, or 3.16
151. Write a brief paragraph that illustrates the similarities and differences between “less than” 1 6 2 and “less than or equal to” 1 … 2. 152. Give a reason why the statement 5 6 8 is true.
A.2 Geometry Essentials OBJECTIVES 1 Use the Pythagorean Theorem and Its Converse (p. A13) 2 Know Geometry Formulas (p. A15) 3 Understand Congruent Triangles and Similar Triangles (p. A16)
1 Use the Pythagorean Theorem and Its Converse
Figure 13 Hypotenuse c
b Leg 90°
a Leg
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 13, c represents the length of the hypotenuse, and a and b represent the lengths of the legs. Note the use of the symbol Pythagorean Theorem is stated next.
to show the 90° angle. The
A14
APPENDIX A Review
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 13, c 2 = a 2 + b2
(1)
A proof of the Pythagorean Theorem is given at the end of this section.
EX A MPL E 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, use the Pythagorean Theorem with a = 4 and b = 3 to find the length c of the hypotenuse. From equation (1), c 2 = a 2 + b2 c 2 = 42 + 32 = 16 + 9 = 25 c = 125 = 5
Now Work P R O B L E M 13 The converse of the Pythagorean Theorem is also true.
CONVERSE OF THE PYTHAGOREAN THEOREM
r
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.
EX A MPL E 2
Verifying That a Triangle Is a Right Triangle Show that a triangle whose sides are of lengths 5, 12, and 13 is a right triangle. Identify the hypotenuse.
Solution
Figure 14
Square the lengths of the sides. 52 = 25
13 5 90° 12
EX A MPL E 3
122 = 144
132 = 169
Notice that the sum of the first two squares (25 and 144) equals the third square (169). That is, because 52 + 122 = 132, the triangle is a right triangle. The longest side, 13, is the hypotenuse. See Figure 14.
Now Work
PROBLEM
21
r
Applying the Pythagorean Theorem The tallest building in the world is Burj Khalifa in Dubai, United Arab Emirates, at 2717 feet and 163 floors. The observation deck is 1483 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. Source: Council on Tall Buildings and Urban Habitat
Solution From the center of Earth, draw two radii: one through Burj Khalifa and the other to the farthest point a person can see from the observation deck. See Figure 15. Apply the Pythagorean Theorem to the right triangle. 1483 Since 1 mile = 5280 feet, 1483 feet = mile. Thus 5280 1483 2 d 2 + 139602 2 = a3960 + b 5280 1483 2 d 2 = a3960 + b - 139602 2 ≈ 2224.58 5280 d ≈ 47.17 A person can see more than 47 miles from the observation deck.
SECTION A.2 Geometry Essentials
A15
Figure 15 1483 ft
d
3960 mi
r Now Work
PROBLEM
53
2 Know Geometry Formulas Certain formulas from geometry are useful in solving algebra problems. For a rectangle of length l and width w, w
Area = lw
Perimeter = 2l + 2w
l
For a triangle with base b and altitude h, h
Area = b
1 bh 2
For a circle of radius r (diameter d = 2r), d
r
Area = pr 2
Circumference = 2pr = pd
For a closed rectangular box of length l, width w, and height h, h w
l
Volume = lwh
Surface area = 2lh + 2wh + 2lw
For a sphere of radius r, r
Volume =
4 3 pr 3
Surface area = 4pr 2
For a right circular cylinder of height h and radius r,
r
h
Volume = pr 2 h
Now Work
EXAM PL E 4
PROBLEM
Surface area = 2pr 2 + 2prh 29
Using Geometry Formulas 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?
A16
APPENDIX A Review
Solution Figure 16 4
See Figure 16. The amount of copper required equals the shaded area. This area is the sum of the areas 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. Area = Area of triangle + Area of semicircle =
l6
1 1 1 1 bh + pr 2 = 142 162 + p # 22 2 2 2 2
b = 4; h = 6; r = 2
= 12 + 2p ≈ 18.28 cm2
r
About 18.28 cm2 of copper is required.
Now Work
PROBLEM
47
3 Understand Congruent Triangles and Similar Triangles In Words Two triangles are congruent if they have the same size and shape.
DEFINITION
Throughout the text we will make reference to triangles, beginning here, 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 pair of corresponding angles have the same measure and each pair of corresponding sides are the same length. In Figure 17, corresponding angles are equal and the lengths of the corresponding sides are equal: a = d, b = e, and c = f . Hence these triangles are congruent. Figure 17 Congruent Triangles 100⬚
a 30⬚
50⬚
100⬚
d
b
e 50⬚
30⬚
c
f
Actually, 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 Congruent Triangles 1. Angle–Side–Angle Case Two triangles are congruent if the measures of two of the angles are equal and the lengths of the corresponding sides between the two angles are equal. For example, in Figure 18(a), the two triangles are congruent because the measures of 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 18(b), the two triangles are congruent because the lengths of 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 measures of the angles between the two sides are the same. For example, in Figure 18(c), the two triangles are congruent because the lengths of two sides and the measure of the included angle are equal.
A17
SECTION A.2 Geometry Essentials
Figure 18 15
15 20
80⬚
80⬚
20
40⬚
8
40⬚
7
10
10 40⬚
40⬚
8
8
7
8 (b)
(a)
(c)
See the following definition to contrast congruent triangles with similar triangles.
DEFINITION
In Words Two triangles are similar if they have the same shape, but (possibly) different sizes.
Two triangles are similar if the measures of the corresponding angles are equal and the lengths of the corresponding sides are proportional. For example, the triangles in Figure 19 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 f d e constant: = = = 2. a c b Figure 19
80
d ⫽ 2a a
80
30
e ⫽ 2b
b 30
70
70 f ⫽ 2c
c
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.
Determining Similar Triangles 1. Angle–Angle Case Two triangles are similar if two of the corresponding angles have equal measures. For example, in Figure 20(a), the two triangles are similar because two angles have equal measures. 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 20(b), the two triangles are similar because 10 5 6 1 = = = 30 15 18 3 3. Side–Angle–Side Case Two triangles are similar if two corresponding sides are proportional and the angles between the two sides have equal measure. For example, in Figure 20(c), the two triangles are similar because 4 12 2 = = and the angles between the sides have equal measure. 6 18 3 Figure 20 15 80⬚ 80⬚
30
5 10
18
(a)
120⬚
120⬚
6
35⬚
35⬚
18
12
6
4 (b)
(c)
A18
APPENDIX A Review
EX A MPL E 5
Using Similar Triangles Given that the triangles in Figure 21 are similar, find the missing length x and the angles A, B, and C. Figure 21 60⬚
6
x
30⬚
B
3
C
5 A
90⬚
Solution
Because the triangles are similar, corresponding angles are equal, so A = 90°, B 3 and C = 30°. Also, the corresponding sides are proportional. That is, = 5 solve this equation for x. 3 5 3 5x # 5 3x x
6 x
=
= 5x #
6 x
= 30 = 10
Multiply both sides by 5x. Simplify. Divide both sides by 3.
r
The missing length is 10 units.
Now Work
= 60°, 6 . We x
PROBLEM
41
Proof of the Pythagorean Theorem Begin with a square, each side of length a + b. In this square, form four right triangles, each having legs equal in length to a and b. See Figure 22. 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 22 indicates a square with an area equal to c 2. Figure 22 b
a
1
Area = –2 ab
1
Area = –2 ab
a
c c
b
Area = c 2 c
b
c
1
Area = –2 ab a
a
1
Area = –2 ab
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 ab) plus the area of the square with side c. That is, 2
Figure 23 x b
a (a)
1a + b2 2 =
1 1 1 1 ab + ab + ab + ab + c 2 2 2 2 2 a2 + 2ab + b2 = 2ab + c 2 a 2 + b2 = c 2 The proof is complete.
c
b
a 2
2
(b) c = a + b
2
■
Proof of the Converse of the Pythagorean Theorem Begin with two triangles: one a right triangle with legs a and b and the other a triangle with sides a, b, and c for which c 2 = a2 + b2. See Figure 23. By the Pythagorean Theorem, the length x of the third side of the first triangle is x 2 = a 2 + b2
SECTION A.2 Geometry Essentials
A19
But c 2 = a2 + b2. Hence, x2 = c 2 x = c The two triangles have sides with the same length and are therefore congruent. This means corresponding angles are equal, so the angle opposite side c of the second triangle equals 90°. The proof is complete. ■
A.2 Assess Your Understanding Concepts and Vocabulary 9. True or False The triangles shown are similar. True
1. A(n) right triangle is one that contains an angle of 90 degrees. The longest side is called the hypotenuse . 2. For a triangle with base b and altitude h, a formula for the 1 A = bh area A is . 2
25⬚
25⬚
3. The formula for the circumference C of a circle of radius r is C = 2pr . 100⬚
4. Two triangles are similar 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. True
100⬚
10. True or False The triangles shown are similar. False
6. True or False The triangle with sides of lengths 6, 8, and 10 is a right triangle. True 4 7. True or False The volume of a sphere of radius r is pr 2. 3 False
4
8. True or False The triangles shown are congruent. True
3 120⬚ 120⬚
10
3
2 30 30 29 29 10
Skill Building In Problems 11–16, the lengths of the legs of a right triangle are given. Find the hypotenuse. 11. a = 5, b = 12 14. a = 4, b = 3
13 5
12. a = 6, b = 8 15. a = 7, b = 24
10 25
13. a = 10, b = 24
26
16. a = 14, b = 48
50
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 Yes; 5
18. 6, 8, 10 Yes; 10
19. 4, 5, 6
No
20. 2, 2, 3
No
21. 7, 24, 25 Yes; 25
22. 10, 24, 26 Yes; 26
23. 6, 4, 3
No
24. 5, 4, 7
No
2
25. Find the area A of a rectangle with length 4 inches and width 2 inches. 8 in
26. Find the area A of a rectangle with length 9 centimeters and width 4 centimeters. 36 cm2 27. Find the area A of a triangle with height 4 inches and base 2 inches. 4 in2 28. Find the area A of a triangle with height 9 centimeters and base 4 centimeters. 18 cm2 29. Find the area A and circumference C of a circle of radius 5 meters. A = 25p m2; C = 10p m 30. Find the area A and circumference C of a circle of radius 2 feet. A = 4p ft 2; C = 4p ft
A20
APPENDIX A Review
*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. 256 p cm3; S = 64p cm2 33. Find the volume V and surface area S of a sphere of radius 4 centimeters. V = 3 3 34. Find the volume V and surface area S of a sphere of radius 3 feet. V = 36p ft ; S = 36p ft 2 *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. In Problems 37–40, find the area of the shaded region. 37.
38.
2
39.
2
40. 2
2 2
2
p square units
4- p square units
2
2
2p - 4 square units
2p square units
In Problems 41–44, each pair of triangles is similar. Find the missing length x and the missing angles A, B, and C. *41. 60⬚
*42.
2
90⬚
30°
16
*43.
*44. 20 60⬚
75°
4
125⬚
50⬚
95⬚
12
30⬚
10
50 75°
45
25⬚
x
B
5⬚ x A
A
B
8
6
A
C 30
C
A
8 B
B
x C x C
Applications and Extensions 45. How many feet does a wheel with a diameter of 16 inches travel after four revolutions? About 16.8 ft *46. How many revolutions will a circular disk with a diameter of 4 feet have completed after it has rolled 20 feet? 47. In the figure shown, ABCD is a square, with each side of length 6 feet. The width of the border (shaded portion) between the outer square EFGH and ABCD is 2 feet. Find the area of the border. 64 ft2
48. Refer to the figure. Square ABCD has an area of 100 square feet; square BEFG has an area of 16 square feet. What is the area of the triangle CGF? 12 ft2 A G
D E
E F
C
F A
B 6 ft
D H
B
2 ft C G
49. Architecture A Norman window consists of a rectangle surmounted by a semicircle. Find the area of the Norman window shown in the illustration. How much wood frame is needed to enclose the window? 24 + 2p ≈ 30.28 ft 2; 16 + 2p ≈ 22.28 ft
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
6'
4'
SECTION A.2 Geometry Essentials
50. Construction A circular swimming pool that is 20 feet in diameter 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?
3' 20'
69 p ≈ 216.77 ft 2; 26 p ≈ 81.68 ft 51. How Tall Is the Great Pyramid? The ancient Greek philosopher Thales of Miletus is reported on one occasion to have visited Egypt and calculated the height of the Great Pyramid of Cheops by means of shadow reckoning. Thales knew that each side of the base of the pyramid was 252 paces and that his own height was 2 paces. He measured the length of the pyramid’s shadow to be 114 paces and determined the length of his shadow to be 3 paces. See the illustration. Using similar triangles, determine the height of the Great Pyramid in terms of the number of paces. 160 paces*
A21
52. The Bermuda Triangle Karen is doing research on the Bermuda Triangle, which she defines roughly by Hamilton, Bermuda; San Juan, Puerto Rico; and Fort Lauderdale, Florida. On her atlas Karen measures the straight-line distances from Hamilton to Fort Lauderdale, Fort Lauderdale to San Juan, and San Juan to Hamilton to be approximately 57 millimeters (mm), 58 mm, and 53.5 mm, respectively. If the actual distance from Fort Lauderdale to San Juan is 1046 miles, approximate the actual distances from San Juan to Hamilton and from Hamilton to Fort Lauderdale.
Source: Reprinted with permission from Red River Press, Inc. Winnipeg, Canada. *San Juan to Hamilton: ≈ 965 mi; Hamilton to Fort Lauderdale: ≈ 1028 mi
Source: Diggins, Julia E., illustrations by Corydon Bell, String, Straightedge and Shadow: The Story of Geometry, 2003 Whole Spirit Press, http://wholespiritpress.com.
In Problems 53–55, use the facts that the radius of Earth is 3960 miles and 1 mile = 5280 feet. 53. How Far Can You See? The conning tower of the U.S.S. Silversides, a World War II submarine now permanently stationed in Muskegon, Michigan, is approximately 20 feet above sea level. How far can you see from the conning tower? About 5.477 mi* 54. How Far Can You See? A person who is 6 feet tall is standing on the beach in Fort Lauderdale, Florida, and looks out onto the Atlantic Ocean. Suddenly, a ship appears on the horizon. How far is the ship from shore? About 3.0 mi*
Discussion and Writing 57. You have 1000 feet of flexible pool siding and wish to construct a swimming pool. Experiment with rectangular-shaped pools with perimeters of 1000 feet. How do their areas vary? What is the shape of the rectangle with the largest area? Now compute the area enclosed by a circular pool with a perimeter (circumference) of 1000 feet. What would be your choice of shape for the pool? If rectangular, what is your preference for dimensions? Justify your choice. If your only consideration is to have a pool that encloses the most area, what shape should you use?
*55. How Far Can You See? The deck of a destroyer is 100 feet above sea level. How far can a person see from the deck? How far can a person see from the bridge, which is 150 feet above sea level? *56. Suppose that m and n are positive integers with m 7 n. If a = m2 - n2, b = 2mn, and c = m2 + n2, show that a, b, and c are the lengths of the sides of a right triangle. (This formula can be used to find the sides of a right triangle that are integers, such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triples.)
A22
APPENDIX A Review
58. The Gibb’s Hill Lighthouse, Southampton, Bermuda, in operation since 1846, stands 117 feet high on a hill 245 feet high, so its beam of light is 362 feet above sea level. A brochure states that the light itself can be seen on the horizon about 26 miles from the lighthouse. Verify the accuracy of this information. The brochure further states ships 40 miles away can see the light and that planes flying at 10,000 feet can see it 120 miles away. Verify the accuracy of these statements. What assumption did the brochure make about the height of the ship?
120 miles 40 miles
A.3 Polynomials OBJECTIVES 1 2 3 4 5 6 7
Recognize Monomials (p. A22) Recognize Polynomials (p. A23) Add and Subtract Polynomials (p. A24) Multiply Polynomials (p. A25) Know Formulas for Special Products (p. A26) Divide Polynomials Using Long Division (p. A27) Work with Polynomials in Two Variables (p. A30)
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 shall use the letters at the beginning 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 constants, even though the constants a and b are unspecified. As you will find out, the context usually makes the intended meaning clear.
1 Recognize Monomials 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 integers are the integers 0, 1, 2, 3, …. 䊏
axk where a is a constant, x is a variable, and k Ú 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.
EX AMPL E 1
Examples of Monomials Monomial
Coefficient
Degree
6
2
(b) - 22x3
- 22
3
(c) 3
3
0
Since 3 = 3 # 1 = 3x 0, x ≠ 0
(d) - 5x
-5
1
Since - 5x = - 5x 1
(e) x4
1
4
(a) 6x
2
Since x 4 = 1 # x 4
r
SECTION A.3 Polynomials
EXAM PL E 2
A23
Examples of Nonmonomial Expressions 1 1 (a) 3x1>2 is not a monomial, since the exponent of the variable x is , and is not a 2 2 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.
r
Now Work
PROBLEM
7
2 Recognize Polynomials 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. Like terms may be added or subtracted using the Distributive Property. For example, 2x2 + 5x2 = 12 + 52x2 = 7x2 and 8x3 - 5x3 = 18 - 52x3 = 3x3
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 a trinomial. For example, x2 - 2 is a binomial x3 - 3x + 5 is a trinomial 2x2 + 5x2 + 2 = 7x2 + 2 is a binomial
DEFINITION
A polynomial in one variable is an algebraic expression of the form an xn + an - 1 xn - 1 + g + a1 x + a0
(1)
where an , an - 1 , c , a1 , a0 are constants,* called the coefficients of the polynomial, n Ú 0 is an integer, and x is a variable. If an ≠ 0, it is called the leading coefficient, and n is called the degree of the polynomial.
In Words A polynomial is a sum of monomials.
The monomials that make up a polynomial are called its terms. If all the coefficients 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.
EXAM PL E 3
Examples of Polynomials Polynomial
Coefficients
Degree
- 8x + 4x - 6x + 2
- 8, 4, - 6, 2
3
3, 0, - 5
2
1, - 2, 8
2
5, 22
1
3
0
0
No degree
3
2
3x2 - 5 = 3x2 + 0 # x + 1 - 52
8 - 2x + x = 1 # x + 1 - 22x + 8 2
2
5x + 22 = 5x1 + 22
3 = 3 # 1 = 3 # x0 0
*The notation an is read as “a sub n.” The number n is called a subscript and should not be confused with an exponent. Subscripts are used to distinguish one constant from another when a large or undetermined number of constants are required.
r
A24
APPENDIX A Review
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 - 2y2 + y - 3 is a polynomial (in y) of degree 3. z5 + p is a polynomial (in z) of degree 5. Algebraic expressions such as 1 x
and
x2 + 1 x + 5
1 are not polynomials. The first is not a polynomial because = x -1 has an exponent x 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
PROBLEM
17
3 Add and Subtract Polynomials Polynomials are added and subtracted by combining like terms.
EX A MPL E 4
Adding Polynomials Find the sum of the polynomials: 8x3 - 2x2 + 6x - 2 and 3x4 - 2x3 + x2 + x
Solution
The sum can be found using a horizontal or vertical format. Horizontal Addition: The idea here is to group the like terms and then combine them. 18x3 - 2x2 + 6x - 22 + 13x4 - 2x3 + x2 + x2 = 3x4 + 18x3 - 2x3 2 + 1 - 2x2 + x2 2 + 16x + x2 - 2 = 3x4 + 6x3 - x2 + 7x - 2
Vertical Addition: The idea here is to vertically line up the like terms in each polynomial and then add the coefficients. x4
x3
+
x2
x1
x0
8x - 2x + 6x - 2 3x4 - 2x3 + x2 + x 3x4 + 6x3 - x2 + 7x - 2 3
2
r
Subtraction of polynomials can be performed horizontally or vertically as well.
EX A MPL E 5
Solution
Subtracting Polynomials
Find the difference: 13x4 - 4x3 + 6x2 - 12 - 12x4 - 8x2 - 6x + 52 Horizontal Subtraction: 13x4 - 4x3 + 6x2 - 12 - 12x4 - 8x2 - 6x + 52 = 3x4 - 4x3 + 6x2 - 1 + 1 - 2x4 + 8x2 + 6x - 52
w Be sure to change the sign of each term in the second polynomial.
= 13x4 - 2x4 2 + 1 - 4x3 2 + 16x2 + 8x2 2 + 6x + 1 - 1 - 52 c Group like terms.
= x4 - 4x3 + 14x2 + 6x - 6
A25
SECTION A.3 Polynomials
Vertical Subtraction: We line up like terms, change the sign of each coefficient of the second polynomial, and add vertically. x4
-
x3
x2
x1
x0
3x - 4x + 6x - 1 = 3 2x4 - 8x2 - 6x + 54 = + 4
3
2
x4
x3
x2
x1
x0
3x - 4x + 6x - 1 - 2x4 + 8x2 + 6x - 5 x4 - 4x3 + 14x2 + 6x - 6 4
3
2
r
The choice of which of these methods to use for adding and subtracting polynomials is left to you. To save space, we most often use the horizontal format.
Now Work
PROBLEM
29
COMMENT Vertical subtraction will be used when we divide polynomials.
■
4 Multiply Polynomials Two monomials may be multiplied using the Laws of Exponents and the Commutative and Associative Properties. For example, 12x3 2 # 15x4 2 = 12 # 52 # 1x3 # x4 2 = 10x3 + 4 = 10x7 Products of polynomials are found by repeated use of the Distributive Property and the Laws of Exponents. Again, you have a choice of a horizontal or vertical format.
EXAM PL E 6
Solution
Multiplying Polynomials
Find the product: 12x + 52 1x2 - x + 22 Horizontal Multiplication: 12x + 52 1x2 - x + 22 = 2x1x2 - x + 22 + 51x2 - x + 22 c Distributive Property
= 12x # x2 - 2x # x + 2x # 22 + 15 # x2 - 5 # x + 5 # 22 c
Distributive Property
= 12x3 - 2x2 + 4x2 + 15x2 - 5x + 102 c
Law of Exponents
= 2x3 + 3x2 - x + 10 c Combine like terms.
Vertical Multiplication: The idea here is very much like multiplying a two-digit number by a three-digit number. x2 -
x 2x 5x2 - 5x 3 1 + 2 2x - 2x2 + 4x 2x3 + 3x2 - x
Now Work
+ 2 + 5 + 10
This line is 5(x 2 - x + 2). This line is 2x(x 2 - x + 2).
+ 10
Sum of the above two lines
r PROBLEM
45
A26
APPENDIX A Review
5 Know Formulas for Special Products Certain products, called special products, occur frequently in algebra. They can be calculated using the FOIL (First, Outer, Inner, Last) method of multiplying two binomials. Outer First
(ax + b)(cx + d) = ax(cx + d) + b(cx + d) b
Inner
Last b
Outer
= ax # cx + ax # d
b
First
b
Inner Last
+ b # cx + b # d
= acx2 + adx + bcx + bd = acx2 + (ad + bc)x + bd
EX A MPL E 7
Using FOIL F
(a) (b) (c) (d) (e)
O
I
L
1x - 32 1x + 32 = x2 + 3x - 3x - 9 = x2 - 9 1x + 22 2 = 1x + 22 1x + 22 = x2 + 2x + 2x + 4 1x - 32 2 = 1x - 32 1x - 32 = x2 - 3x - 3x + 9 1x + 32 1x + 12 = x2 + x + 3x + 3 = x2 + 4x + 12x + 12 13x + 42 = 6x2 + 8x + 3x + 4 = 6x2 +
Now Work
PROBLEMS
47
AND
= x2 + 4x + 4 = x2 - 6x + 9 3 11x + 4
r
55
Some products have been given special names because of their form. The following special products are based on Examples 7(a), (b), and (c).
Difference of Two Squares 1a - b2 1a + b2 = a2 - b2
(2)
Squares of Binomials, or Perfect Squares 1a + b2 2 = a2 + 2ab + b2
(3a)
1a - b2 2 = a2 - 2ab + b2
EX A MPL E 8
(3b)
Using Special Product Formulas
(a) 1x - 52 1x + 52 = x2 - 52 = x2 - 25 (b) 1x + 72 2 = x2 + 2 # 7 # x + 72 = x2 + 14x + 49 (c) 12x + 12 2 = 12x2 2 + 2 # 1 # 2x + 12 = 4x2 + 4x + 1
Difference of two squares Square of a binomial Note that we used 2x in place of a in formula (3a).
(d) 13x - 42 2 = 13x2 2 - 2 # 4 # 3x + 42 = 9x2 - 24x + 16 Replace a by 3x in formula (3b).
Now Work
PROBLEMS
65, 67,
AND
69
Let’s look at some more examples that lead to general formulas.
r
SECTION A.3 Polynomials
EXAM PL E 9
A27
Cubing a Binomial
(a) 1x + 22 3 = 1x + 22 1x + 22 2 = 1x + 22 1x2 + 4x + 42
Formula (3a)
= x1x + 4x + 42 + 21x + 4x + 42 = 1x3 + 4x2 + 4x2 + 12x2 + 8x + 82 = x3 + 6x2 + 12x + 8 2
2
(b) 1x - 12 3 = 1x - 12 1x - 12 2 = 1x - 12 1x2 - 2x + 12
Formula (3b)
= x1x - 2x + 12 - 11x - 2x + 12 = 1x3 - 2x2 + x2 + 1 - x2 + 2x - 12 = x3 - 3x2 + 3x - 1 2
2
r
Cubes of Binomials, or Perfect Cubes 1a + b2 3 = a3 + 3a2b + 3ab2 + b3 1a - b2 3 = a3 - 3a2b + 3ab2 - b3
Now Work
EX AM PL E 10
EX AM PL E 11
PROBLEM
(4a) (4b)
85
Forming the Difference of Two Cubes
1x - 12 1x2 + x + 12 = x1x2 + x + 12 - 11x2 + x + 12 = x3 + x2 + x - x2 - x - 1 = x3 - 1
r
Forming the Sum of Two Cubes
1x + 22 1x2 - 2x + 42 = x1x2 - 2x + 42 + 21x2 - 2x + 42 = x3 - 2x2 + 4x + 2x2 - 4x + 8 = x3 + 8
r
Examples 10 and 11 lead to two more special products.
Difference of Two Cubes 1a - b2 1a2 + ab + b2 2 = a3 - b3
(5)
1a + b2 1a2 - ab + b2 2 = a3 + b3
(6)
Sum of Two Cubes
6 Divide Polynomials Using Long Division The procedure for dividing two polynomials is similar to the procedure for dividing two integers.
A28
APPENDIX A Review
EX AM PL E 12
Dividing Two Integers Divide 842 by 15.
Solution Divisor S
Thus,
56 15) 842 75 92 90 2
d Quotient d Dividend
d 5 # 15 (subtract) d 6 # 15 (subtract) d Remainder
842 2 = 56 + . 15 15
r
In the long-division process detailed in Example 12, the number 15 is called the divisor, the number 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. 1Quotient2 1Divisor2 + Remainder = Dividend For example, check the results obtained in Example 12 as follows: 1562 1152 + 2 = 840 + 2 = 842
To divide two polynomials, first write each polynomial in standard form. The process then follows a pattern similar to that of Example 12. The next example illustrates the procedure.
EX AM PL E 13
Dividing Two Polynomials Find the quotient and the remainder when 3x3 + 4x2 + x + 7 is divided by x2 + 1
Solution NOTE Remember, a polynomial is in standard form when the terms are written according to descending degree. 䊏
Each polynomial is in standard form. The dividend is 3x3 + 4x2 + x + 7, and the divisor is x2 + 1. STEP 1: Divide the leading term of the dividend, 3x3, by the leading term of the divisor, x2. Enter the result, 3x, over the term 3x3, as follows: 3x x2
+ 1) 3x3 + 4x2 + x + 7
STEP 2: Multiply 3x by x2 + 1, and enter the result below the dividend.
x2
3x + 1) 3x3 + 4x2 + x + 7 3x3 + 3x
d 3x # (x 2 + 1) = 3x 3 + 3x
c Align the 3x term under the x to make the next step easier.
STEP 3: Subtract and bring down the remaining terms. 3x x2 + 1) 3x3 + 4x2 + x + 7 3x3 + 3x 4x2 - 2x + 7
d Subtract (change the signs and add). d Bring down the 4x 2 and the 7.
SECTION A.3 Polynomials
A29
STEP 4: Repeat Steps 1–3 using 4x2 - 2x + 7 as the dividend.
COMMENT If the degree of the divisor is greater than the degree of the dividend, the process ends. 䊏
3x + 4 x2 + 1) 3x3 + 4x2 + x 3x3 + 3x 4x2 - 2x 4x2 - 2x
d
+ 7 + 7 + 4 + 3
d Divide 4x 2 by x 2 to get 4. d Multiply x 2 + 1 by 4; subtract.
Since x2 does not divide into - 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: 1Quotient2 1Divisor2 + Remainder = 13x + 42 1x2 + 12 + 1 - 2x + 32
= 3x3 + 3x + 4x2 + 4 + 1 - 2x + 32
= 3x3 + 4x2 + x + 7 = Dividend Then 3x3 + 4x2 + x + 7 - 2x + 3 = 3x + 4 + 2 2 x + 1 x + 1
r
The next example combines the steps involved in long division.
EX AM PL E 14
Dividing Two Polynomials Find the quotient and the remainder when x4 - 3x3 + 2x - 5 is divided by x2 - x + 1
Solution
In setting up this division problem, it is necessary to leave a space for the missing x2 term in the dividend.
Divisor S Subtract S Subtract S Subtract S
x2 - 2x - 3 x - x + 1) x4 - 3x3 x4 - x3 + x2 - 2x3 - x2 - 2x3 + 2x2 - 3x2 - 3x2 2
+ 2x - 5 + + +
2x 2x 4x 3x x
d Quotient d Dividend
- 5 - 5 - 3 - 2
d Remainder
Check: 1Quotient2 1Divisor2 + Remainder
= 1x2 - 2x - 32 1x2 - x + 12 + 1x - 22 = x4 - x3 + x2 - 2x3 + 2x2 - 2x - 3x2 + 3x - 3 + x - 2 = x4 - 3x3 + 2x - 5 = Dividend
As a result, x4 - 3x3 + 2x - 5 x - 2 = x2 - 2x - 3 + 2 2 x - x + 1 x - x + 1
r
A30
APPENDIX A Review
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.
Now Work
PROBLEM
93
7 Work with Polynomials in Two Variables A monomial in two variables x and y has the form axn ym, where a is a constant, x and y are variables, and n and m are nonnegative integers. The degree of a monomial is the sum of the powers of the variables. For example, 2xy3, x2 y2, and x3 y are all monomials that have 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.
Examples of Polynomials in Two Variables
EX AM PL E 15
3x2 + 2x3 y + 5 Two variables, degree is 4.
px3 - y2
x4 + 4x3 y - xy3 + y4
Two variables, degree is 3.
Two variables, degree is 4.
r
Multiplying polynomials in two variables is handled in the same way as multiplying polynomials in one variable.
Using a Special Product Formula
EX AM PL E 16
To multiply 12x - y2 2, use the Square of a Binomial formula (3b) with 2x instead of a and with y instead of b. 12x - y2 2 = 12x2 2 - 2 # y # 2x + y2
r
= 4x2 - 4xy + y2
Now Work
PROBLEM
79
A.3 Assess Your Understanding Concepts and Vocabulary 1. The polynomial 3x4 - 2x3 + 13x2 - 5 is of degree The leading coefficient is 3 . 2. 1x2 - 42 1x2 + 42 =
x4 - 16.
3. 1x - 22 1x2 + 2x + 42 =
x3 - 8 .
4
.
4. True or False 4x -2 is a monomial of degree - 2.
False
5. True or False The degree of the product of two nonzero polynomials equals the sum of their degrees. True 6. True or False
1x + a2 1x2 + ax + a2 = x3 + a3.
False
SECTION A.3 Polynomials
A31
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. 8 *7. 2x3 *8. - 4x2 *9. *10. - 2x -3 *11. - 2xy2 x 2x2 8x *12. 5x2 y3 *13. *14. - 3 *15. x2 + y2 *16. 3x2 + 4 y y In Problems 17–26, tell whether the expression is a polynomial. If it is, give its degree. If it is not, state why not. 17. 3x2 - 5 Yes; 2 *22.
3 + 2 x
18. 1 - 4x Yes; 1
19. 5 Yes; 0
20. - p Yes; 0
23. 2y3 - 22 Yes; 3
24. 10z2 + z Yes; 2
*25.
*21. 3x2 -
x2 + 5 x3 - 1
5 x
3x3 + 2x - 1 x2 + x + 1
*26.
In Problems 27–46, add, subtract, or multiply, as indicated. Express your answer as a single polynomial in standard form. 27. (x2 + 4x + 5) + (3x - 3)
x2 + 7x + 2
28. (x3 + 3x2 + 2) + (x2 - 4x + 4)
29. (x3 - 2x2 + 5x + 10) - (2x2 - 4x + 3) 31. (6x + x + x) + (5x - x + 3x ) 5
3
4
3
2
33. (x2 - 3x + 1) + 2(3x2 + x - 4) 2
3
x3 - 4x2 + 9x + 7 30. (x2 - 3x - 4) - (x3 - 3x2 + x + 5)
6x + 5x + 3x + x 5
4
3
3
38. (x2 + 1) - (4x2 + 5) + (x2 + x - 2)
3
40. 8(1 - y ) + 4(1 + y + y + y )
2
3
2
x3 + x2 - 4x
42. 4x2(x3 - x + 2)
43. - 2x (4x + 5)
- 8x - 10x
44. 5x (3x - 4)
3
5
*45. (x + 1)(x2 + 2x - 4)
- 7x2 - 3x
2x2 - 4x + 6
2
41. x(x2 + x - 4) 2
5
36. 8(4x - 3x - 1) - 6(4x + 8x - 2) 3
2
3
x3 + 3x2 - 2x - 4
- x3 + 4x2 - 4x - 9
10x + 3x3 - 10x2 + 6
2
34. - 2(x2 + x + 1) + ( - 5x2 - x + 2)
15y - 27y + 30
2
2
2
37. (x2 - x + 2) + (2x2 - 3x + 5) - (x2 + 1) 39. 9(y - 3y + 4) - 6(1 - y )
5
- 2x + 18x - 18
2
2
32. (10x - 8x ) + (3x - 2x + 6)
2
7x2 - x - 7
35. 6(x + x - 3) - 4(2x - 3x ) 3
x3 + 4x2 - 4x + 6
8x3 - 24x2 - 48x + 4 - 2x2 + x - 6
- 4y + 4y2 + 4y + 12
3
3
4x5 - 4x3 + 8x2
15x - 20x3 4
46. (2x - 3)(x2 + x + 1)
2x3 - x2 - x - 3
In Problems 47–64, multiply the polynomials using the FOIL method. Express your answer as a single polynomial in standard form. 47. (x + 2)(x + 4)
x2 + 6x + 8 6x + 5x + 1
50. (3x + 1)(2x + 1) 53. (x - 3)(x - 2)
2
x2 - 5x + 6 6x - 10x - 4
56. (2x - 4)(3x + 1)
2
59. ( - x - 2)( - 2x - 4)
2x2 + 8x + 8
62. (2x + 3y)(x - y)
2
2x + xy - 3y
48. (x + 3)(x + 5)
x2 + 8x + 15
49. (2x + 5)(x + 2)
51. (x - 4)(x + 2)
x - 2x - 8
52. (x + 4)(x - 2)
54. (x - 5)(x - 1)
x2 - 6x + 5
55. (2x + 3)(x - 2)
2
x2 + 2x - 8 2x2 - x - 6
57. ( - 2x + 3)(x - 4)
- 2x + 11x - 12
58. ( - 3x - 1)(x + 1) - 3x2 - 4x - 1
60. ( - 2x - 3)(3 - x)
2x2 - 3x - 9
61. (x - 2y)(x + y)
2
*63. ( - 2x - 3y)(3x + 2y)
2
2x2 + 9x + 10
x2 - xy - 2y2
*64. (x - 3y)( - 2x + y)
In Problems 65–88, multiply the polynomials using the special product formulas. Express your answer as a single polynomial in standard form. 65. (x - 7)(x + 7)
x2 - 49 9x - 4
68. (3x + 2)(3x - 2) 71. (x - 4)
x - 8x + 16
2
2
25x - 9
74. (5x - 3)(5x + 3) 77. (x + y)(x - y)
x - y 2
83. (x - 2y)2
9x - 16y 2
x + 8x + 16
70. (x + 5)
72. (x - 5)
2
x - 10x + 25
73. (3x + 4)(3x - 4)
2
2
4x - 12x + 9 x + 2xy + y
2
87. (2x + 1)
2
x - 9y 2
2
3
82. (x - y)
2
4x2 + 12xy + 9y2 8x + 12x + 6x + 1 3
2
2
2
9x2 - 16
9x - 24x + 16 2
79. (3x + y)(3x - y)
2
4x2 - 9
x + 10x + 25
2
76. (3x - 4)
2
84. (2x + 3y)2
x + 3x + 3x + 1 3
69. (x + 4)
81. (x + y)
2
67. (2x + 3)(2x - 3)
2
78. (x + 3y)(x - 3y)
2
x2 - 4xy + 4y2
3
x2 - 1
2
75. (2x - 3)
2
80. (3x + 4y)(3x - 4y) 86. (x + 1)
66. (x - 1)(x + 1)
2
9x2 - y2
x - 2xy + y2
2
2
85. (x - 2)3
x3 - 6x2 + 12x - 8
88. (3x - 2) 27x3 - 54x2 + 36x - 8 3
In Problems 89–104, find the quotient and the remainder. Check your work by verifying that (Quotient)(Divisor) + Remainder = Dividend 89. 4x - 3x + x + 1 divided by x + 2 3
2
91. 4x - 3x + x + 1 divided by x 3
2
4x2 - 11x + 23; R: - 45 90. 3x3 - x2 + x - 2 divided by x + 2
4x - 3; R: x + 1
2
93. 5x - 3x + x + 1 divided by x + 2 4
2
5x - 13; R: x + 27
2
2
95. 4x - 3x + x + 1 divided by 2x - 1 5
2
2x ; R: - x + x + 1
3
2
2
*97. 2x4 - 3x3 + x + 1 divided by 2x2 + x + 1 99. - 4x + x - 4 divided by x - 1 3
2
101. 1 - x2 + x4 divided by x2 + x + 1 103. x - a divided by x - a 3
3
- 4x - 3x - 3; R: - 7 2
x2 - x - 1; R: 2x + 2
x + ax + a ; R: 0 2
2
92. 3x - x + x - 2 divided by x 3
2
3x2 - 7x + 15; R: - 32
3x - 1; R: x - 2
2
94. 5x - x + x - 2 divided by x + 2 4
2
2
96. 3x - x + x - 2 divided by 3x - 1 5
2
3
5x2 - 11; R: x + 20 x2; R: x - 2
*98. 3x4 - x3 + x - 2 divided by 3x2 + x + 1 *100. - 3x4 - 2x - 1 divided by x - 1 102. 1 - x2 + x4 divided by x2 - x + 1 104. x - a divided by x - a 5
5
x2 + x - 1; R: - 2x + 2
x + ax + a2x2 + a3x + a4; R: 0
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
4
3
A32
APPENDIX A Review
Discussion and Writing 105. Explain why the degree of the product of two nonzero polynomials equals the sum of their degrees. 106. Explain why the degree of the sum of two polynomials of different degrees equals the larger of their degrees. 107. Give a careful statement about the degree of the sum of two polynomials of the same degree.
108. Do you prefer adding two polynomials using the horizontal method or the vertical method? Write a brief position paper defending your choice. 109. Do you prefer to memorize the rule for the square of a binomial (x + a)2 or to use FOIL to obtain the product? Write a brief position paper defending your choice.
A.4 Factoring Polynomials OBJECTIVES 1 Factor the Difference of Two Squares and the Sum and Difference of Two Cubes (p. A33) 2 Factor Perfect Squares (p. A34) 3 Factor a Second-Degree Polynomial: x2 + Bx + C (p. A34) 4 Factor by Grouping (p. A36) 5 Factor a Second-Degree Polynomial: Ax2 + Bx + C, A ≠ 1 (p. A37) 6 Complete the Square (p. A38)
Consider the following product:
12x + 32 1x - 42 = 2x2 - 5x - 12
COMMENT Over the real numbers, 4 3x + 4 factors into 3ax + b . It is 3 4 the noninteger that causes 3x + 4 3 to be prime over the integers. ■
EX A MPL E 1
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. 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 polynomials (excluding 1 and - 1), then the polynomial is prime. When a polynomial has been written as a product consisting only of prime factors, it is factored completely. Examples of prime polynomials (over the integers) are 2, 3, 5, x, x + 1, x - 1, 3x + 4, x2 + 4 The first factor to look for in a factoring problem is a common monomial factor present in each term of the polynomial. If one is present, use the Distributive Property to factor it out. Continue factoring out monomial factors until none are left.
Identifying Common Monomial Factors
Polynomial
Common Monomial Factor
Remaining Factor
Factored Form
2x + 4
2
x + 2
2x + 4 = 21x + 22
3x - 6
3
x - 2
3x - 6 = 31x - 22
2x2 - 4x + 8
2
x2 - 2x + 4
2x2 - 4x + 8 = 21x2 - 2x + 42
8x - 12
4
2x - 3
8x - 12 = 412x - 32
x + x
x
x + 1
x2 + x = x1x + 12
x3 - 3x2
x2
x - 3
x3 - 3x2 = x2 1x - 32
6x2 + 9x
3x
2x + 3
6x2 + 9x = 3x12x + 32
2
r
SECTION A.4 Factoring Polynomials
A33
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 polynomial of degree 2 or higher. (Do you see why?)
Now Work
PROBLEM
5
1 Factor the Difference of Two Squares and the Sum and Difference of Two Cubes 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. Difference of Two Squares
a2 - b2 = 1a - b2 1a + b2
Perfect Squares
a2 + 2ab + b2 = 1a + b2 2 a2 - 2ab + b2 = 1a - b2 2
Sum of Two Cubes
a3 + b3 = 1a + b2 1a2 - ab + b2 2
Difference of Two Cubes
EXAM PL E 2
a3 - b3 = 1a - b2 1a2 + ab + b2 2
Factoring the Difference of Two Squares Factor completely: x2 - 4
Solution
Note that x2 - 4 is the difference of two squares, x2 and 22. x2 - 4 = 1x - 22 1x + 22
r EXAM PL E 3
Factoring the Difference of Two Cubes Factor completely: x3 - 1
Solution
Because x3 - 1 is the difference of two cubes, x3 and 13, x3 - 1 = (x - 1)(x2 + x + 1)
r EXAM PL E 4
Factoring the Sum of Two Cubes Factor completely: x3 + 8
Solution
Because x3 + 8 is the sum of two cubes, x3 and 23, x3 + 8 = 1x + 22 1x2 - 2x + 42
r EXAM PL E 5
Factoring the Difference of Two Squares Factor completely: x4 - 16
Solution
Because x4 - 16 is the difference of two squares, x4 = 1x2 2 and 16 = 42, 2
x4 - 16 = 1x2 - 42 1x2 + 42 But x2 - 4 is also the difference of two squares, so x4 - 16 = 1x2 - 42 1x2 + 42 = 1x - 22 1x + 22 1x2 + 42
Now Work
r PROBLEMS
15 AND 33
A34
APPENDIX A Review
2 Factor Perfect Squares 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.
EX A MPL E 6
Factoring Perfect Squares Factor completely: x2 + 6x + 9
Solution
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, the trinomial is a perfect square. x2 + 6x + 9 = 1x + 32 2
EX A MPL E 7
r
Factoring Perfect Squares Factor completely: 9x2 - 6x + 1
Solution
The first term, 9x2 = 13x2 2, and the third term, 1 = 12, are perfect squares. Because the middle term, - 6x, is - 2 times the product of 3x and 1, the trinomial is a perfect square. 9x2 - 6x + 1 = 13x - 12 2
EX A MPL E 8
r
Factoring Perfect Squares Factor completely: 25x2 + 30x + 9
Solution
The first term, 25x2 = (5x)2, and the third term, 9 = 32, are perfect squares. Because the middle term, 30x, is twice the product of 5x and 3, the trinomial is a perfect square. 25x2 + 30x + 9 = 15x + 32 2
Now Work
PROBLEMS
25
AND
99
r
If a trinomial is not a perfect square, it may be possible to factor it using the technique discussed next.
3 Factor a Second-Degree Polynomial: x 2 + 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, consider (x + 3)(x + 4) = x2 + 7x + 12 The factors of x2 + 7x + 12 are x + 3 and x + 4. Note the following: x2 + 7x + 12 = (x + 3)(x + 4) 12 is the product of 3 and 4. 7 is the sum of 3 and 4.
In general, if x2 + Bx + C = 1x + a2 1x + b2 = x2 + (a + b)x + ab, then ab = C and a + b = B. To factor a second-degree polynomial x2 + Bx + C, find integers whose product 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 = 1x + a2 1x + b2
SECTION A.4 Factoring Polynomials
EXAM PL E 9
A35
Factoring Trinomials Factor completely: x2 + 7x + 10
Solution
First determine all integers whose product is 10, and then compute their sums. Integers whose product is 10 Sum
1, 10
- 1, - 10
2, 5
- 2, - 5
11
- 11
7
-7
The integers 2 and 5 have a product of 10 and add up to 7, the coefficient of the middle term. As a result, x2 + 7x + 10 = 1x + 22 1x + 52
EX AM PL E 10
r
Factoring Trinomials Factor completely: x2 - 6x + 8
Solution
First determine all integers whose product is 8, and then compute each sum. Integers whose product is 8 Sum
1, 8
- 1, - 8
2, 4
- 2, - 4
9
-9
6
-6
Since - 6 is the coefficient of the middle term,
x2 - 6x + 8 = 1x - 22 1x - 42
r EX AM PL E 11
Factoring Trinomials Factor completely: x2 - x - 12
Solution
First determine all integers whose product is - 12, and then compute each sum. Integers whose product is − 12 Sum
1, - 12
- 1, 12
2, - 6
- 2, 6
3, - 4
- 3, 4
11
-4
4
-1
1
- 11
Since - 1 is the coefficient of the middle term,
x2 - x - 12 = 1x + 32 1x - 42
r EX AM PL E 12
Factoring Trinomials Factor completely: x2 + 4x - 12
Solution
The integers - 2 and 6 have a product of - 12 and have the sum 4, so x2 + 4x - 12 = 1x - 22 1x + 62
r
To avoid errors in factoring, always check your answer by multiplying it out to see whether the result equals the original expression. When none of the possibilities works, the polynomial is prime.
A36
APPENDIX A Review
EX AM PL E 13
Identifying Prime Polynomials Show that x2 + 9 is prime.
Solution
First list the integers whose product is 9, and then compute their sums. Integers whose product is 9
1, 9
- 1, - 9
3, 3
- 3, - 3
Sum
10
- 10
6
-6
Since the coefficient of the middle term in x2 + 9 = x2 + 0x + 9 is 0, and none of the sums equals 0, this means that x2 + 9 is prime. Example 13 demonstrates a more general result:
THEOREM
r
Any polynomial of the form x2 + a2, a real, is prime.
Now Work
PROBLEMS
39
AND
83
4 Factor by Grouping Sometimes a common factor does not occur in every term of the polynomial but does occur 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.
EX AM PL E 14
Solution
Factoring by Grouping
Factor completely by grouping: 1x2 + 22x + 1x2 + 22 # 3 Note the common factor x2 + 2. Applying the Distributive Property yields 1x2 + 22x + 1x2 + 22 # 3 = 1x2 + 22 1x + 32 Since x2 + 2 and x + 3 are prime, the factorization is complete.
r
The next example shows a factoring problem that occurs in calculus.
EX AM PL E 15
Solution
Factoring by Grouping
Factor completely by grouping: 31x - 12 2 1x + 22 4 + 41x - 12 3 1x + 22 3
Here, 1x - 12 2 1x + 22 3 is a common factor of both 31x - 12 2 1x + 22 4 and 41x - 12 3 1x + 22 3. As a result,
31x - 12 2 1x + 22 4 + 41x - 12 3 1x + 22 3 = 1x - 12 2 1x + 22 3 3 31x + 22 + 41x - 124 = 1x - 12 2 1x + 22 3 3 3x + 6 + 4x - 44 = 1x - 12 2 1x + 22 3 17x + 22
r EX AM PL E 16
Factoring by Grouping Factor completely by grouping: x3 - 4x2 + 2x - 8
Solution
To see whether 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, factor
SECTION A.4 Factoring Polynomials
A37
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: x3 - 4x2 + 2x - 8 = 1x3 - 4x2 2 + 12x - 82 = x2 1x - 42 + 21x - 42 = 1x - 42 1x2 + 22
r
Since x2 + 2 and x - 4 are prime, the factorization is complete.
Now Work
PROBLEMS
51
AND
127
5 Factor a Second-Degree Polynomial: Ax 2 + Bx + C, A 3 1 To factor a second-degree polynomial Ax 2 + Bx + C, when A ≠ 1 and A, B, and C have no common factors, follow these steps:
Steps for Factoring Ax 2 + Bx + C, when A 3 1 and A, B, and C Have No Common Factors STEP 1: Find the value of AC. STEP 2: Find integers whose product is AC that add up to B. That is, find a and b such that ab = AC and a + b = B. STEP 3: Write Ax2 + Bx + C = Ax2 + ax + bx + C. STEP 4: Factor this last expression by grouping.
EX AM PL E 17
Factoring Trinomials Factor completely: 2x2 + 5x + 3
Solution
Comparing 2x2 + 5x + 3 to Ax2 + Bx + C, note that A = 2, B = 5, and C = 3.
STEP 1: The value of AC is 2 # 3 = 6. STEP 2: Determine the integers whose product is AC = 6, and compute their sums. Integers whose product is 6 Sum
1, 6
- 1, - 6
2, 3
- 2, - 3
7
-7
5
-5
STEP 3: The integers whose product is 6 that add up to B ⫽ 5 are 2 and 3.
2x2 ⫹ 5x ⫹ 3 ⫽ 2x2 ⫹ 2x ⫹ 3x ⫹ 3 STEP 4: Factor by grouping.
2x2 + 2x + 3x + 3 = 12x2 + 2x2 + 13x + 32 = 2x1x + 12 + 31x + 12 = 1x + 12 12x + 32
As a result,
2x2 + 5x + 3 = 1x + 12 12x + 32
r EX AM PL E 18
Factoring Trinomials Factor completely: 2x2 - x - 6
Solution
Comparing 2x2 - x - 6 to Ax2 + Bx + C, note that A = 2, B = - 1, and C = - 6.
A38
APPENDIX A Review
STEP 1: The value of AC is 2 # 1 - 62 = - 12. STEP 2: Determine the integers whose product is AC = - 12, and compute their sums. Integers whose product is − 12 Sum
1, - 12
- 1, 12
2, - 6
- 2, 6
3, - 4
- 3, 4
- 11
11
-4
4
-1
1
STEP 3: The integers whose product is ⫺12 that add up to B ⫽ ⫺1 are ⫺4 and 3. 2x2 ⫺ x ⫺ 6 ⫽ 2x2 ⫺ 4x ⫹ 3x ⫺ 6 STEP 4: Factor by grouping. 2x2 - 4x + 3x - 6 = 12x2 - 4x2 + 13x - 62 = 2x1x - 22 + 31x - 22 = 1x - 22 12x + 32 As a result,
Now Work
2x2 - x - 6 = 1x - 22 12x + 32 PROBLEM
57
r
SUMMARY Type of Polynomial Any polynomial
Method Look for common monomial factors. (Always do this first!)
Example
Binomials of degree 2 or higher
Check for a special product: Difference of two squares, a2 - b2 Difference of two cubes, a3 - b3 Sum of two cubes, a3 + b3
x2 - 16 = 1x - 42 1x + 42 x3 - 64 = 1x - 42 1x2 + 4x + 162 x3 + 27 = 1x + 32 1x2 - 3x + 92
Trinomials of degree 2 Four or more terms
Check for a perfect square, 1a { b2 2 Factoring x2 + Bx + C (p. A34) Factoring Ax2 + Bx + C (p. A37) Grouping
x2 + 8x + 16 = 1x + 42 2 x2 - 10x + 25 = 1x - 52 2 x2 - x - 2 = 1x - 22 1x + 12 6x2 + x - 1 = 12x + 12 13x - 12 2x3 - 3x2 + 4x - 6 = 12x - 32 1x2 + 22
6x2 + 9x = 3x12x + 32
6 Complete the Square The idea behind completing the square in one variable is to “adjust” an expression of the form x2 + bx to make it a perfect square. Perfect squares are trinomials of the form x2 + 2ax + a2 = 1x + a2 2
or
x2 - 2ax + a2 = 1x - a2 2
For example, x2 + 6x + 9 is a perfect square because x2 + 6x + 9 = 1x + 32 2. And p2 - 12p + 36 is a perfect square because p2 - 12p + 36 = 1p - 62 2. So how do we “adjust” x2 + bx to make it a perfect square? We do it by adding a number. For example, to make x2 + 6x a perfect square, add 9. But how do we know to add 9? Divide the coefficient on the first-degree term, 6, by 2, and then square the result to obtain 9. This approach works in general.
SECTION A.4 Factoring Polynomials
A39
Completing the Square 1 2 and then square the result. That is, determine the value of b in x2 + bx and 1 2 compute a b b . 2 Identify the coefficient of the first-degree term. Multiply this coefficient by
EX AM PL E 19
Completing the Square Determine the number that must be added to each expression to complete the square. Then factor the expression. Start
Add
Result
Factored Form
y2 + 8y
a
1# 2 8b = 16 2
y2 + 8y + 16
1y + 42 2
x2 + 12x
a
2 1# 12b = 36 2
x2 + 12x + 36
1x + 62 2
a2 - 20a
a
2 1# 1 - 202 b = 100 2
a2 - 20a + 100
1a - 102 2
p2 - 5p
a
2 1# 25 1 - 52 b = 2 4
p2 - 5p +
ap -
25 4
5 2 b 2
r Note that the factored form of a perfect square is either
Figure 24
y
y
4
Area = y 2
Area = 4y
Now Work
PROBLEM
or
b 2 b 2 x2 - bx + a b = ax - b 2 2
69
Are you wondering why making an expression a perfect square is called “completing the square”? Look at the square in Figure 24. Its area is 1y + 42 2. The yellow area is y2 and each orange area is 4y (for a total area of 8y). The sum of these areas is y2 + 8y. To complete the square, we need to add the area of the green region: 4 # 4 = 16. As a result, y2 + 8y + 16 = 1y + 42 2.
Area = 4y
4
b 2 b 2 x2 + bx + a b = ax + b 2 2
A.4 Assess Your Understanding Concepts and Vocabulary 1. If factored completely, 3x3 - 12x = 3x(x - 2)(x + 2).
3. True or False The polynomial x2 + 4 is prime. True
2. If a polynomial cannot be written as the product of two other polynomials (excluding 1 and - 1), then the polynomial is said to be prime .
4. True or False 3x3 - 2x2 - 6x + 4 = (3x - 2)(x2 + 2). False
Skill Building In Problems 5–14, factor each polynomial by removing the common monomial factor. 5. 3x + 6
3(x + 2)
*9. x + x + x 3
2
13. 3x2 y - 6xy2 + 12xy
6. 7x - 14
7(x - 2)
10. x - x + x 3
3xy(x - 2y + 4)
2
x(x - x + 1) 2
7. ax2 + a 11. 2x - 2x 2
a(x2 + 1) 2x(x - 1)
14. 60x2 y - 48xy2 + 72x3 y
8. ax - a 12. 3x - 3x 2
a(x - 1) 3x(x - 1)
12xy(5x - 4y + 6x2)
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
A40
APPENDIX A Review
In Problems 15–22, factor the difference of two squares. 15. x2 - 1
(x + 1)(x - 1)
19. x2 - 16
16. x2 - 4
(x + 4)(x - 4)
(x + 2)(x - 2)
20. x2 - 25
17. 4x2 - 1
(x + 5)(x - 5)
(2x + 1)(2x - 1)
21. 25x2 - 4
18. 9x2 - 1
(5x + 2)(5x - 2)
(3x + 1)(3x - 1)
22. 36x2 - 9 9(2x + 1)(2x - 1)
In Problems 23–32, factor the perfect squares. 23. x2 + 2x + 1
(x + 1)2
24. x2 - 4x + 4
(x - 2)2
27. x2 - 10x + 25
(x - 5)2 28. x2 + 10x + 25
31. 16x + 8x + 1
(4x + 1) 32. 25x + 10x + 1
2
2
25. x2 + 4x + 4
(x + 5)2
29. 4x2 + 4x + 1
(5x + 1)
2
(x + 2)2
26. x2 - 2x + 1
(2x + 1)2
(x - 1)2
30. 9x2 + 6x + 1
(3x + 1)2
2
In Problems 33–38, factor the sum or difference of two cubes. *33. x3 - 27
*34. x3 + 125
*35. x3 + 27
*36. 27 - 8x3
*37. 8x3 + 27
*38. 64 - 27x3
In Problems 39–50, factor each polynomial. 39. x2 + 5x + 6
(x + 2)(x + 3)
40. x2 + 6x + 8
42. x + 9x + 8
(x + 1)(x + 8)
43. x + 7x + 10
2
45. x - 10x + 16
(x - 8)(x - 2)
2
48. x2 - 2x - 8
(x + 2)(x + 4)
2
(x + 5)(x + 2)
46. x - 17x + 16 49. x2 + 7x - 8
(x + 6)(x + 1)
44. x2 + 11x + 10
(x - 16)(x - 1)
2
(x - 4)(x + 2)
41. x2 + 7x + 6
(x + 8)(x - 1)
(x + 1)(x + 10)
47. x - 7x - 8
(x - 8)(x + 1)
50. x2 + 2x - 8
(x - 2)(x + 4)
2
In Problems 51–56, factor by grouping. 51. 2x2 + 4x + 3x + 6 53. 2x2 - 4x + x - 2
(x + 2)(2x + 3)
52. 3x2 - 3x + 2x - 2
(x - 2)(2x + 1)
55. 6x2 + 9x + 4x + 6
54. 3x2 + 6x - x - 2
(2x + 3)(3x + 2)
(x - 1)(3x + 2) (x + 2)(3x - 1)
56. 9x2 - 6x + 3x - 2
(3x - 2)(3x + 1)
In Problems 57–68, factor each polynomial. *57. 3x2 + 4x + 1
*58. 2x2 + 3x + 1
*59. 2z2 + 5z + 3
*60. 6z2 + 5z + 1
*61. 3x + 2x - 8
*62. 3x + 10x + 8
*63. 3x - 2x - 8
*64. 3x2 - 10x + 8
*65. 3x2 + 14x + 8
*66. 3x2 - 14x + 8
*67. 3x2 + 10x - 8
*68. 3x2 - 10x - 8
2
2
2
In Problems 69–74, determine the number that should be added to complete the square of each expression. Then factor each expression. 69. x2 + 10x 72. x2 - 4x
25; (x + 5)2
70. p2 + 14p
4; (x - 2)2
73. x2 -
1 x 2
49; (p + 7)2
71. y2 - 6y
1 1 2 ; ¢x - ≤ 16 4
74. x2 +
1 x 3
9; (y - 3)2 1 1 2 ; ¢x + ≤ 36 6
Mixed Practice In Problems 75–122, factor each polynomial completely. If the polynomial cannot be factored, say it is prime. 75. x2 - 36
(x + 6)(x - 6)
76. x2 - 9
(x + 3)(x - 3)
*77. 2 - 8x2
*78. 3 - 27x2
*81. x - 10x + 21
*82. x2 - 6x + 8
*79. x + 11x + 10
*80. x + 5x + 4
*83. 4x2 - 8x + 32
*84. 3x2 - 12x + 15
*85. x2 + 4x + 16
*86. x2 + 12x + 36
*87. 15 + 2x - x
*88. 14 + 6x - x
*89. 3x - 12x - 36
*90. x3 + 8x2 - 20x
2
2
2
2
2
2
*91. y4 + 11y3 + 30y2
*92. 3y3 - 18y2 - 48y
*93. 4x2 + 12x + 9
*94. 9x2 - 12x + 4
*95. 6x + 8x + 2
*96. 8x + 6x - 2
*97. x - 81
*98. x4 - 1
*99. x6 - 2x3 + 1
*100. x6 + 2x3 + 1
*101. x7 - x5
*103. 16x + 24x + 9
*104. 9x - 24x + 16
*105. 5 + 16x - 16x
*107. 4y2 - 16y + 15
*108. 9y2 + 9y - 4
*109. 1 - 8x2 - 9x4
2
2
2
4
2
*102. x8 - x5 *106. 5 + 11x - 16x2
2
*110. 4 - 14x2 - 8x4
*111. x(x + 3) - 6(x + 3)
*112. 5(3x - 7) + x(3x - 7)
*114. (x - 1) - 2(x - 1)
*115. (3x - 2) - 27
*117. 3(x + 10x + 25) - 4(x + 5)
*118. 7(x - 6x + 9) + 5(x - 3)
*119. x + 2x - x - 2
*120. x - 3x - x + 3
*121. x - x + x - 1
*122. x4 + x3 + x + 1
2
2
3
2
3
2
4
3
*113. (x + 2) - 5(x + 2) 2
116. (5x + 1)3 - 1 3
5x(25x2 + 15x + 3)
2
Applications and Extensions In Problems 123–132, expressions that occur in calculus are given. Factor each expression completely. *123. 2(3x + 4)2 + (2x + 3) # 2(3x + 4) # 3 125. 2x(2x + 5) + x
2
#2
2x(3x + 5)
*127. 2(x + 3)(x - 2)3 + (x + 3)2 # 3(x - 2)2
*124. 5(2x + 1)2 + (5x - 6) # 2(2x + 1) # 2 126. 3x2(8x - 3) + x3 # 8
x2(32x - 9)
*128. 4(x + 5)3(x - 1)2 + (x + 5)4 # 2(x - 1)
SECTION A.5 Synthetic Division
129. (4x - 3)2 + x # 2(4x - 3) # 4
3(4x - 3)(4x - 1)
130. 3x2(3x + 4)2 + x3 # 2(3x + 4) # 3
3x2(3x + 4)(5x + 4)
* 131. 2(3x - 5) # 3(2x + 1)3 + (3x - 5)2 # 3(2x + 1)2 # 2
*132. 3(4x + 5)2 # 4(5x + 1)2 + (4x + 5)3 # 2(5x + 1) # 5
* 133. Show that x + 4 is prime.
*134. Show that x2 + x + 1 is prime.
2
A41
Discussion and Writing 135. Make up a polynomial that factors into a perfect square.
136. Explain to a fellow student what you look for first when presented with a factoring problem. What do you do next?
A.5 Synthetic Division OBJECTIVE 1 Divide Polynomials Using Synthetic Division (p. A41)
1 Divide Polynomials Using Synthetic 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, first consider long division for dividing the polynomial 2x3 - x2 + 3 by x - 3. d Quotient 2x2 + 5x + 15 3 2 x - 3) 2x - x + 3 2x3 - 6x2 5x2 5x2 - 15x 15x + 3 15x - 45 48 d Remainder
Check: 1Divisor2 # 1Quotient2 + Remainder = 1x - 32 12x2 + 5x + 152 + 48 = 2x3 + 5x2 + 15x - 6x2 - 15x - 45 + 48 = 2x3 - x2 + 3 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. Removing these terms gives the following display: 2x2 + 5x + 15 x - 3) 2x3 - x2 + 3 2 - 6x 5x2 - 15x 15x - 45 48
APPENDIX A Review
Most of the x’s that appear in this process can also be removed, provided that care is used in positioning each coefficient. In this regard, use 0 as the coefficient of x in the dividend, because that power of x is missing. This gives 2x2 + 5x + 15 x - 3) 2 - 1 0 - 6 5 - 15 15
3
- 45 48 To make this display more compact, move the lines up until the numbers in blue align horizontally. 2x2 + 5x + 15 x - 3) 2 - 1 0 3 - 6 - 15 - 45 5 15 48 ○
Row 1 Row 2 Row 3 Row 4
Because the leading coefficient of the divisor is always 1, the leading coefficient of the dividend will also be the leading coefficient of the quotient. 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 the process can be compressed to three rows, with the bottom row containing both the coefficients of the quotient and the remainder. - 1 0 3 - 6 - 15 - 45 5 15 48
x - 3) 2 2
Row 1 Row 2 (subtract) 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, change the sign of each entry and add. With this modification, the display will look like this: x - 3) 2 2
- 1 6 5
0 15 15
3 45 48
Row 1 Row 2 (add) Row 3
Note that the entries in row 2 are three times the prior entries in row 3. As the last modification to the display, replace the x - 3 with 3. The entries in row 3 give the quotient and the remainder, as shown next. 3) 2
0 15 15
3 45 48
d
2
- 1 6 5 e Quotient
Row 1 Row 2 (add) Row 3 Remainder
e
d
A42
2x2 + 5x + 15 48
Let’s go through an example step by step.
EX A MPL E 1
Using Synthetic Division to Find the Quotient and Remainder Use synthetic division to find the quotient and remainder when x3 - 4x2 - 5 is divided by x - 3
SECTION A.5 Synthetic Division
Solution
A43
STEP 1: Write the dividend in descending powers of x. Then copy the coefficients, remembering to insert a 0 for any missing powers of x. -4 0
1
-5
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, insert 3 to the left of the division symbol. -4 0
3) 1
-5
Row 1
STEP 3: Bring the 1 down two rows, and enter it in row 3. -4 0
3) 1 T 1
-5
Row 1 Row 2 Row 3
STEP 4: Multiply the latest entry in row 3 by 3, and place the result in row 2, one column over to the right. 3) 1
-5
1
Row 1 Row 2 Row 3
c
x3
-4 0 3
STEP 5: Add the entry in row 2 to the entry above it in row 1, and enter the sum in row 3. -4 0 3 x3 1 -1
-5
3) 1
Row 1
c
Row 2 Row 3
STEP 6: Repeat Steps 4 and 5 until no more entries are available in row 1. -5 -9 x3 - 14 c
c
c
3) 1 - 4 0 3 -3 x3 x3 1 -1 -3
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
Remainder = - 14
Check: 1Divisor2 1Quotient2 + Remainder
= 1x - 32 1x2 - x - 32 + 1 - 142 = 1x3 - x2 - 3x - 3x2 + 3x + 92 + 1 - 142 = x3 - 4x2 - 5 = Dividend
Let’s do an example in which all seven steps are combined.
EXAM PL E 2
r
Using Synthetic Division to Verify a Factor Use synthetic division to show that x + 3 is a factor of 2x5 + 5x4 - 2x3 + 2x2 - 2x + 3
Solution
If the remainder is 0, then the divisor is a factor of the dividend, and the quotient is the other factor. The divisor is x + 3 = x - ( - 3), so 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. - 3) 2 2
5 -6 -1
-2 3 1
2 -3 -1
-2 3 1
3 -3 0
Row 1 Row 2 Row 3
A44
APPENDIX A Review
The remainder is 0, so Dividend = 1Divisor2 1Quotient2 + Remainder
2x + 5x - 2x3 + 2x2 - 2x + 3 = 1x + 32 12x4 - x3 + x2 - x + 12 5
4
Hence, x + 3 is a factor of 2x5 + 5x4 - 2x3 + 2x2 - 2x + 3.
r As Example 2 illustrates, the remainder after division gives information about whether the divisor is or is not a factor. More will come about this in Chapter 3.
Now Work
PROBLEMS
7
AND
17
A.5 Assess Your Understanding Concepts and Vocabulary 1. To check division, ensure that the expression being divided—the dividend—equals the product of the quotient and the divisor plus the remainder . 2. To divide 2x3 - 5x + 1 by x + 3 using synthetic division, the first step is to write - 3
)2 0 - 5 1 . 3. True or False In using synthetic division, the divisor is always a polynomial of degree 1, whose leading coefficient is 1. True 4. True or False - 2) 5 5
3 2 - 10 14 - 7 16
1 means - 32 - 31
- 31 5x3 + 3x2 + 2x + 1 = 5x2 - 7x + 16 + . True x + 2 x + 2
Skill Building In Problems 5–16, use synthetic division to find the quotient and remainder when: *5. x3 - x2 + 2x + 4 *7. 3x + 2x - x + 3 3
2
*9. x - 4x + x 5
3
*13. 0.1x + 0.2x *15. x5 - 1
is divided by x - 3
is divided by x + 3
*11. 4x6 - 3x4 + x2 + 5 3
is divided by x - 2
is divided by x + 1.1
is divided by x + 1
*8. - 4x3 + 2x2 - x + 1 *10. x + x + 2 4
is divided by x - 1
is divided by x - 1
*6. x3 + 2x2 - 3x + 1
*12. x5 + 5x3 - 10 *14. 0.1x - 0.2 2
*16. x5 + 1
is divided by x + 2
is divided by x - 2
2
is divided by x + 1
is divided by x + 2.1
is divided by x + 1
In Problems 17–26, use synthetic division to determine whether x - c is a factor of the given polynomial. 17. 4x3 - 3x2 - 8x + 4; x - 2
No
18. - 4x3 + 5x2 + 8; x + 3
No
19. 3x - 6x - 5x + 10; x - 2 Yes
20. 4x - 15x - 4; x - 2 Yes
21. 3x6 + 82x3 + 27; x + 3 Yes
22. 2x6 - 18x4 + x2 - 9; x + 3 Yes
4
3
23. 4x6 - 64x4 + x2 - 15; x + 4 25. 2x4 - x3 + 2x - 1; x -
1 2
No
Yes
4
2
24. x6 - 16x4 + x2 - 16; x + 4 Yes 26. 3x4 + x3 - 3x + 1; x +
1 3
No
Applications and Extensions 27. Find the sum of a, b, c, and d if d x3 - 2x2 + 3x + 5 = ax2 + bx + c + x + 2 x + 2
-9
Discussion and Writing 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 difference to you in choosing? Give reasons. *Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
SECTION A.6 Rational Expressions
A45
A.6 Rational Expressions OBJECTIVES 1 2 3 4 5
Reduce a Rational Expression to Lowest Terms (p. A45) Multiply and Divide Rational Expressions (p. A46) Add and Subtract Rational Expressions (p. A47) Use the Least Common Multiple Method (p. A48) Simplify Complex Rational Expressions (p. A50)
1 Reduce a Rational Expression to Lowest Terms The quotient of two polynomials forms a result called a rational expression. Some examples of rational expressions are (a)
x3 + 1 x
(b)
3x2 + x - 2 x2 + 5
(c)
x x2 - 1
(d)
xy2 (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. Rational expressions are described in the same manner as rational numbers. In expression (a), the polynomial x3 + 1 is the numerator, and x is the denominator. When the numerator and denominator of a rational expression contain no common factors (except 1 and - 1), the rational expression is reduced to lowest terms, or simplified. The polynomial in the denominator of a rational expression cannot be equal x3 + 1 to 0 because division by 0 is not defined. For example, for the expression , x x cannot take on the value 0. The domain of the variable x is 5 x 0 x ≠ 06 . A rational expression is reduced to lowest terms by factoring the numerator and the denominator completely and dividing out any common factors using the Reduction Property: ac a = bc b
EXAM PL E 1
b ≠ 0, c ≠ 0
(1)
Reducing a Rational Expression to Lowest Terms Reduce to lowest terms:
Solution
if
x2 + 4x + 4 x2 + 3x + 2
Begin by factoring the numerator and the denominator. x2 + 4x + 4 = (x + 2)(x + 2) x2 + 3x + 2 = (x + 2)(x + 1)
WARNING Apply the Reduction Property only to rational expressions written in factored form. Be sure to divide out only common factors, not common terms! 䊏
Since a common factor, x + 2, appears, the original expression is not in lowest terms. To reduce it to lowest terms, use the Reduction Property: (x + 2) (x + 2) x2 + 4x + 4 x + 2 = = 2 (x + 2) (x + 1) x + 1 x + 3x + 2
x ≠ - 2, - 1
r EXAM PL E 2
Reducing Rational Expressions to Lowest Terms Reduce each rational expression to lowest terms. (a)
x3 - 8 x3 - 2x2
(b)
8 - 2x x2 - x - 12
A46
APPENDIX A Review
Solution
(a)
(x - 2) (x2 + 2x + 4) x3 - 8 x2 + 2x + 4 = = x3 - 2x2 x2 (x - 2) x2
(b)
2( - 1) (x - 4) 2(4 - x) 8 - 2x -2 = = = (x - 4)(x + 3) (x - 4) (x + 3) x + 3 x - x - 12
x ≠ 0, 2
2
Now Work
PROBLEM
x ≠ - 3, 4
r
5
2 Multiply and Divide Rational Expressions The rules for multiplying and dividing rational expressions are the same as the rules a c for multiplying and dividing rational numbers. If and , b ≠ 0, d ≠ 0, are two b d rational expressions, then a#c ac = b d bd
if
a b a d ad = # = c c b bc d
if
b ≠ 0, d ≠ 0
b ≠ 0, c ≠ 0, d ≠ 0
(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 divided out. Leave your answer in factored form.
EX A MPL E 3
Multiplying and Dividing Rational Expressions Perform the indicated operation and simplify the result. Leave your answer in factored form. x + 3 2 2 x - 2x + 1 # 4x + 4 x2 - 4 (a) (b) x3 + x x2 + x - 2 x2 - x - 12 x3 - 8
Solution
(a)
4(x2 + 1) (x - 1)2 x2 - 2x + 1 # 4x2 + 4 # = x3 + x x2 + x - 2 x(x2 + 1) (x + 2)(x - 1) = =
(x - 1) 2 (4) (x2 + 12 x (x2 + 12 (x + 2) (x - 12 4(x - 1) x(x + 2)
x ≠ - 2, 0, 1
x + 3 x + 3 # x3 - 8 x2 - 4 = 2 (b) 2 x - 4 x2 - x - 12 x - x - 12 3 x - 8 =
2 x + 3 # (x - 2)(x + 2x + 4) (x - 2)(x + 2) (x - 4)(x + 3)
=
(x + 32 (x - 2) (x2 + 2x + 4) (x - 2) (x + 2)(x - 4) (x + 32
=
x2 + 2x + 4 (x + 2)(x - 4)
Now Work
PROBLEMS
17
x ≠ - 3, - 2, 2, 4 AND
25
r
SECTION A.6 Rational Expressions
A47
3 Add and Subtract Rational Expressions In Words To add (or subtract) two rational expressions with the same denominator, keep the common denominator and add (or subtract) the numerators.
The rules for adding and subtracting rational expressions are the same as the rules for adding and subtracting rational numbers. If the denominators of two rational expressions to be added (or subtracted) are equal, then add (or subtract) the numerators and keep the common denominator.
If
a c and are two rational expressions, then b b a c a + c + = b b b
EXAM PL E 4
a c a - c = b b 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 factored form.
Solution
(a)
2x2 - 4 x + 3 + 2x + 5 2x + 5
(a)
(2x2 - 4) + (x + 3) 2x2 - 4 x + 3 + = 2x + 5 2x + 5 2x + 5
x ≠ -
= (b)
(b)
x 3x + 2 x - 3 x - 3
x ≠ 3
(2x - 1)(x + 1) 2x2 + x - 1 = 2x + 5 2x + 5
x - (3x + 2) x 3x + 2 x - 3x - 2 = = x - 3 x - 3 x - 3 x - 3 =
EXAM PL E 5
5 2
- 2(x + 1) - 2x - 2 = x - 3 x - 3
r
Adding Rational Expressions Whose Denominators Are Additive Inverses of Each Other Perform the indicated operation and simplify the result. Leave your answer in factored 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 2x 5 2x 5 2x -5 + = + = + x - 3 3 - x x - 3 - (x - 3) x - 3 x - 3
c
c
a -a = -b b
3 - x = - (x - 3)
=
Now Work
PROBLEMS
37
AND
43
2x + ( - 5) 2x - 5 = x - 3 x - 3
r
A48
APPENDIX A Review
If the denominators of two rational expressions to be added or subtracted are not equal, the general formulas for adding and subtracting rational expressions can be used. 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
EX A MPL E 6
if
b ≠ 0, d ≠ 0
(5a)
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 factored form. x - 3 x x2 1 (a) + x ≠ - 4, 2 (b) 2 x ≠ - 2, 0, 2 x x + 4 x - 2 x - 4
Solution
(a)
x x - 3#x - 2 x + 4# x x - 3 + = + x + 4 x - 2 x + 4 x - 2 x + 4 x - 2 æ (5a)
(b)
=
(x - 3)(x - 2) + (x + 4)(x) (x + 4)(x - 2)
=
x2 - 5x + 6 + x2 + 4x 2x2 - x + 6 = (x + 4)(x - 2) (x + 4)(x - 2)
x2(x) - (x2 - 4)(1) x2 x2 # x x2 - 4 # 1 1 = = x x2 - 4 x2 - 4 x x2 - 4 x (x2 - 4)(x) æ (5b)
=
x3 - x2 + 4 x (x - 2)(x + 2)
Now Work
PROBLEM
r
47
4 Use the Least Common Multiple Method If the denominators of two rational expressions to be added (or subtracted) have common factors, it usually is better not to use the general rules given by equations (5a) and (5b). Instead, just as with fractions, the least common multiple (LCM) method is used. The LCM method uses the polynomial of least degree that has each denominator polynomial as a factor.
The LCM Method for Adding or Subtracting 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 denominators 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 consider an example that requires only Steps 1 and 2.
SECTION A.6 Rational Expressions
EXAM PL E 7
A49
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 - 1)(x + 1)3
Solution
STEP 1: The polynomials are already factored completely as x(x - 1)2(x + 1) and 4 (x - 1)(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.) x(x - 1)2(x + 1) 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 - 1, 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
r
Notice that the LCM in Example 7 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.
Now Work
EXAM PL E 8
PROBLEM
53
Using the Least Common Multiple to Add Rational Expressions Perform the indicated operation and simplify the result. Leave your answer in factored form. x 2x - 3 + 2 x + 3x + 2 x - 1 2
Solution
x ≠ - 2, - 1, 1
STEP 1: Factor completely the polynomials in the denominators. x2 + 3x + 2 = (x + 2)(x + 1) x2 - 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(x - 1) x x x #x - 1 = = = (x + 2)(x + 1) (x + 2)(x + 1) x 1 (x + 2)(x + 1)(x - 1) x + 3x + 2 2
æ
Multiply numerator and denominator by x − 1 to get the LCM in the denominator.
2x - 3 2x - 3 2x - 3 # x + 2 = (2x - 3)(x + 2) = = 2 (x 1)(x + 1) (x 1)(x + 1) x + 2 (x - 1)(x + 1)(x + 2) x - 1 æ
Multiply numerator and denominator by x + 2 to get the LCM in the denominator.
A50
APPENDIX A Review
STEP 4: Now add by using equation (4). (2x - 3)(x + 2) x(x - 1) x 2x - 3 + + 2 = (x + 2)(x + 1)(x - 1) (x + 2)(x + 1)(x - 1) x + 3x + 2 x - 1 2
EX A MPL E 9
=
(x2 - x) + (2x2 + x - 6) (x + 2)(x + 1)(x - 1)
=
3(x2 - 2) 3x2 - 6 = (x + 2)(x + 1)(x - 1) (x + 2)(x + 1)(x - 1)
r
Using the Least Common Multiple to Subtract Rational Expressions Perform the indicated operation and simplify the result. Leave your answer in factored form. 3 x + 4 x ≠ - 1, 0 - 2 2 x + x x + 2x + 1
Solution
STEP 1: Factor completely the polynomials in the denominators. x2 + x = x (x + 1) x2 + 2x + 1 = (x + 1)2 STEP 2: The LCM is x(x + 1)2. STEP 3: Write each rational expression using the LCM as the denominator. 3 3 3 # x + 1 = 3(x + 1)2 = = x(x + 1) x(x + 1) x + 1 x + x x(x + 1) 2
x(x + 4) x + 4 x + 4 #x x + 4 = = = 2 2 x x + 2x + 1 (x + 1) (x + 1) x(x + 1)2 2
STEP 4: Subtract, using equation (4). x(x + 4) 3(x + 1) 3 x + 4 - 2 = 2 2 x + x x + 2x + 1 x(x + 1) x(x + 1)2 3(x + 1) - x(x + 4) = x(x + 1)2
Now Work
PROBLEM
=
3x + 3 - x2 - 4x x(x + 1)2
=
- x2 - x + 3 x(x + 1)2
r
63
5 Simplify Complex 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 expression.* For example, 1 x 1 1 x
1 +
x2 - 3 x - 4 x - 3 - 1 x + 2 2
and
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 accomplished in either of two ways. *Some texts use the term complex fraction.
SECTION A.6 Rational Expressions
A51
Simplifying a Complex Rational Expression METHOD 1: Treat the numerator and denominator of the complex rational expression separately, performing whatever operations are indicated and simplifying the results. Follow this by simplifying the resulting 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. Both methods are shown 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.
EX AM PL E 10
Simplifying a Complex Rational Expression
Simplify:
Solution
1 3 + x 2 x + 3 4
x ≠ - 3, 0
Method 1: First perform the indicated operation in the numerator, and then divide. 1 3 1#x + 2#3 x + 6 + # x 2 2 x 2x x + 6# 4 = = = x + 3 x + 3 x + 3 2x x + 3 æ æ 4 4 4 Rule for adding quotients
=
Rule for dividing quotients
(x + 6) # 4 2(x + 6) 2 # 2 # (x + 6) = = # # # # 2 x (x + 3) 2 x (x + 3) x(x + 3)
æ Rule for multiplying quotients
Method 2: The rational expressions that appear in the complex rational expression are 1 , 2
3 , x
x + 3 4
The LCM of their denominators is 4x. Multiply the numerator and denominator of the complex rational expression by 4x and then simplify. 1 3 1 3 1 3 4x # a + b 4x # + 4x # + x x x 2 2 2 = = # 4x (x + 3) x + 3 x + 3 4x # a b 4 4 4 æ
æ Use the Distributive Multiply the Property in the numerator. numerator and denominator by 4x.
2 # 2x # =
1 3 + 4x # 2(x + 6) x 2 2x + 12 = = # 4 x (x + 3) x(x + 3) x(x + 3) 4 æ Simplify.
æ Factor.
r
A52
APPENDIX A Review
EX AM PL E 11
Simplifying a Complex Rational Expression
Simplify:
Solution
x2 + 2 x - 4 2x - 2 - 1 x
x ≠ 0, 2, 4
Here Method 1 is used. 2(x - 4) x2 x2 x2 + 2x - 8 + 2 + x - 4 x - 4 x - 4 x - 4 = = 2x - 2 2x - 2 2x - 2 - x x - 1 x x x x (x + 4)(x - 2) (x + 4) (x - 2) x - 4 = = x - 2 x - 4 x (x + 4) # x x(x + 4) = = x - 4 x - 4
Now Work
PROBLEM
#
x x - 2
r
73
Application EX AM PL E 12
Solving an Application in Electricity An electrical circuit contains two resistors connected in parallel, as shown in Figure 25. If their resistances are R1 and R2 ohms, respectively, their combined resistance R is given by the formula
Figure 25 R1
R = R2
1 1 1 + R1 R2
Express R as a rational expression; that is, simplify the right-hand side of this formula. Evaluate the rational expression if R1 = 6 ohms and R2 = 10 ohms.
Solution
1 Using Method 2, consider 1 as the fraction . Then the rational expressions in the 1 complex rational expression are 1 , 1
1 , R1
1 R2
The LCM of the denominators is R1R2 . Multiply the numerator and denominator of the complex rational expression by R1R2 and simplify. 1 1 1 + R1 R2
=
1 # R1R2 a
1 1 # + b R1R2 R1 R2
=
R1R2 1 # 1 # R1R2 + RR R1 R2 1 2
=
R1R2 R2 + R1
Thus, R = If R1 = 6 and R2 = 10, then R =
R1R2 R2 + R1
6 # 10 60 15 = = ohms 10 + 6 16 4
r
SECTION A.6 Rational Expressions
A53
A.6 Assess Your Understanding Concepts and Vocabulary 2x3 - 4x is reduced x - 2
1. When the numerator and denominator of a rational expression contain no common factors (except 1 and - 1), the rational expression is in lowest terms.
3. True or False The rational expression to lowest terms. True
2. LCM is an abbreviation for least common multiple .
4. True or False The LCM of 2x3 + 6x2 and 6x4 + 4x3 is 4x3(x + 1). False
Skill Building In Problems 5–16, reduce each rational expression to lowest terms. 5.
3x + 9 x2 - 9
9.
24x2 12x2 - 6x
13.
3 x - 3
6.
4x 2x - 1
x2 + 4x - 5 x2 - 2x + 1
x + 5 x - 1
4x2 + 8x 12x + 24
x 3
10.
x2 + 4x + 4 x2 - 4
14.
x - x2 x + x - 2 2
7. x + 2 x - 2
11.
x x + 2
15.
-
x2 - 2x 3x - 6
x 3
8.
y2 - 25
y + 5
2y - 8y - 10
2(y + 1)
12.
- (x + 7)
16.
2
x2 + 5x - 14 2 - x
15x2 + 24x 3x2
5x + 8 x
3y2 - y - 2
y - 1
3y + 5y + 2
y + 1
2x2 + 5x - 3 1 - 2x
- (x + 3)
2
In Problems 17–34, perform the indicated operation and simplify the result. Leave your answer in factored form. 17.
3x + 6 # x 5x2 x2 - 4
21.
12 4x - 8 # - 3x 12 - 6x
3 5x(x - 2)
18.
3 # x2 2x 6x + 10
8 3x
22.
8x x - 1 27. 10x x + 1
4 5(x - 1)
x2 + 7x + 12 x2 - 7x + 12 31. 2 x + x - 12 x2 - x - 12
28.
(x + 3)2 (x - 3)2
*19.
6x - 27 # 2 5x 4x - 18
6x x2 - 4 25. 3x - 9 2x + 4
x2 + x - 6 # x2 - 25 *24. 2 x + 4x - 5 x2 + 2x - 15
2
3x 4(3x + 5)
x - 2 4x x - 4x + 4 12x 2
x2 + 7x + 6 x2 + x - 6 32. 2 x + 5x - 6 x2 + 5x + 6
4x2 # x3 - 64 2x x - 16
3 5x
*23.
4x (x - 2)(x - 3)
26.
12 # x3 + 1 x + x 4x - 2
*20.
2
2
x2 - 3x - 10 # x2 + 4x - 21 x2 + 2x - 35 x2 + 9x + 14 12x 5x + 20
3(x - 4)
2
4x x - 16
5x
2
4 - x 4 + x 29. 4x x2 - 16
3 x - 2
-
(x - 4)2
2x2 - x - 28 3x2 - x - 2 33. (x - 2)(x - 1) 4x2 + 16x + 7 3x2 + 11x + 6 (x + 1)(x + 2)
30.
4x
3 + x 3 - x x - 9 9x3 2
-
9x3 (x - 3)2
9x2 + 3x - 2 12x2 + 5x - 2 *34. (x - 1)(2x + 1) 9x2 - 6x + 1 8x2 - 10x - 3 (x - 4)(x + 3)
In Problems 35–52, perform the indicated operation and simplify the result. Leave your answer in factored form. x + 5 2
35.
5 x + 2 2
38.
3x2 9 2x - 1 2x - 1
3(x2 - 3)
41.
2x - 4 3x + 5 2x - 1 2x - 1
x + 9 2x - 1
44.
x 6 x - 1 1 - x
47.
2x - 3 x + x + 1 x - 1
50.
2x + 1 2x - 3 x - 1 x + 1
2x - 1
x + 6 x - 1 3x - 2x - 3 (x + 1)(x - 1)
36.
3 6 x x
39.
x + 1 2x - 3 + x - 3 x - 3
42.
5x - 4 x + 1 3x + 4 3x + 4
45.
4 2 x - 1 x + 2
(x - 1)(x + 2)
48.
3x 2x + x - 4 x + 3
(x - 4)(x + 3)
51.
1 x + x x2 - 4
2
-
2 (x - 1)(x + 1)
-
3 x 3x - 2 x - 3 4x - 5 3x + 4 2(x + 5) x(5x + 1) 2(x2 - 2) x(x - 2)(x + 2)
*37.
x2 4 2x - 3 2x - 3
40.
2x - 5 x + 4 + 3x + 2 3x + 2
43.
4 x + x - 2 2 - x
4 - x x - 2
46.
2 5 x + 5 x - 5
-
49.
x - 3 x + 4 x + 2 x - 2
*52.
x - 1 x + 2 x + 1 x3
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
3x - 1 3x + 2
3x + 35 (x - 5)(x + 5) - (11x + 2)
(x + 2)(x - 2)
A54
APPENDIX A Review
In Problems 53–60, find the LCM of the given polynomials. *53. x2 - 4, x2 - x - 2
*54. x2 - x - 12, x2 - 8x + 16
*55. x3 - x, x2 - x
*56. 3x2 - 27, 2x2 - x - 15
*57. 4x3 - 4x2 + x, 2x3 - x2, x3
*58. x - 3, x2 + 3x, x3 - 9x
*59. x3 - x, x3 - 2x2 + x, x3 - 1
60. x2 + 4x + 4, x3 + 2x2, (x + 2)3
x2(x + 2)3
In Problems 61–72, perform the indicated operations and simplify the result. Leave your answer in factored form. x x - 2 x2 - 7x + 6 x - 2x - 24 3x 3x2 - 4x + 4 x - 4 64. - 2 x - 1 x - 2x + 1 (x - 1)2
*61.
*67.
2x + 3 x + 4 - 2 x2 - x - 2 x + 2x - 8
*70.
x 2 x + 1 + - 3 x (x - 1)2 x - x2
x x + 1 - 2 x - 3 x + 5x - 24 2 3 + *65. (x - 1)2(x + 1) (x - 1)(x + 1)2
4x 2 - 2 x2 - 4 x + x - 6 6 2 *66. (x + 2)2(x - 1) (x + 2)(x - 1)2
*62.
*68.
71.
*63.
2x - 3 x - 2 x2 + 8x + 7 (x + 1)2 -1 x(x + h)
1 1 1 a - b h x + h x
3 1 2 + 3 - 2 x x + x x - x2
*69.
72.
- (2x + h)
1 1 1 J - 2R h (x + h)2 x
x2(x + h)2
In Problems 73–84, perform the indicated operations and simplify the result. Leave your answer in factored form. 1 x 73. 1 1 x 1 +
1 x2 74. 1 3 - 2 x 4 +
x + 1 x - 1
x - 3 x + 4 x - 2 x + 1 *77. x + 1
81. 1 -
1 1 1 x
x + 1 x 75. x - 1 3 + x + 1 2 -
4x2 + 1 3x2 - 1
x - 2 x x + 1 x - 2 *78. x + 3
-1 x - 1
82. 1 -
1 1 1 1 - x
x x + 1 76. x - 1 2 x 1 -
(x - 1)(x + 1) 2x(2x + 1)
x - 2 x - 1 + x + 2 x + 1 *79. x 2x - 3 x + 1 x 1 x
83.
2(x - 1) - 1 + 3 3(x - 1)
-1
+ 2
*80.
3x - 1 2x + 1
84.
x (x + 1)2
2x + 5 x x x - 3 (x + 1)2 x2 x - 3 x + 3 4(x + 2) - 1 - 3 3(x + 2)
-1
- 1
3x + 2 x - 1
In Problems 85–92, expressions that occur in calculus are given. Reduce each expression to lowest terms. 85.
88.
91.
(2x + 3) # 3 - (3x - 5) # 2
19 (3x - 5)2
86.
x2 + 4 (x + 2)2(x - 2)2
89.
(3x - 5)2 x # 2x - (x2 - 4) # 1 (x2 - 4)
2
(x2 + 1) # 3 - (3x + 4) # 2x (x2 + 1)
2
-
(4x + 1) # 5 - (5x - 2) # 4 (5x - 2)2
13 (5x - 2)2
(3x + 1) # 2x - x2 # 3
x(3x + 2)
(3x + 1)2
(3x + 1)2
(x + 3)(3x - 1) (x2 + 1)2
92.
87.
90.
(x2 + 9) # 2 - (2x - 5) # 2x (x2 + 9)
2
x # 2x - (x2 + 1) # 1 (x + 1) 2
(x + 1)(x - 1)
2
(x2 + 1)2
(2x - 5) # 3x2 - x3 # 2
x2(4x - 15)
(2x - 5)2
(2x - 5)2
-
2(x2 - 5x - 9) (x2 + 9)2
Applications and Extensions 93. The Lensmaker’s Equation The focal length f of a lens with index of refraction n is 1 1 1 = (n - 1) J + R f R1 R2 where R1 and R2 are the radii of curvature of the front and back surfaces of the lens. Express f as a rational expression. Evaluate the rational expression for n = 1.5, R1 = 0.1 meter, and R2 = 0.2 meter. f =
R1 # R2 2 m ; (n - 1)(R1 + R2) 15
94. Electrical Circuits An electrical circuit contains three resistors connected in parallel. If their resistances are R1 , R2 , and R3 ohms, respectively, their combined resistance R is given by the formula 1 1 1 1 + + = R R1 R2 R3 Express R as a rational expression. Evaluate R for R1 = 5 ohms, R2 = 4 ohms, and R3 = 10 ohms. R =
R1R2R3 20 ; ohms R1R2 + R1R3 + R2R3 11
SECTION A.7 nth Roots; Rational Exponents
A55
Discussion and Writing 95. The following expressions are called continued fractions: 1 +
1 , 1 + x
1 1 1 + x
1
, 1 + 1 +
1
, 1 +
1
1 +
1 x
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 method to add two rational expressions. Give two examples of adding two rational expressions, one in which you use the LCM and the other in which you do not.
97. Which of the two methods given in the text for simplifying complex rational expressions do you prefer? Write a brief paragraph stating the reasons for your choice.
A.7 nth Roots; Rational Exponents PREPARING FOR THIS SECTION Before getting started, review the following: r &YQPOFOUT 4RVBSF3PPUT "QQFOEJY" 4FDUJPO" QQ"m"
Now Work the ‘Are You Prepared?’ problems on page A60.
OBJECTIVES 1 2 3 4
Work with nth Roots (p. A55) Simplify Radicals (p. A56) Rationalize Denominators (p. A57) Simplify Expressions with Rational Exponents (p. A58)
1 Work with nth Roots DEFINITION
The principal nth root of a real number a, n Ú 2 an integer, symbolized by n 2a, is defined as follows: n
1a = b means a = bn
In Words
n
The symbol 1a means “give me the number that, when raised to the power n, equals a.”
where a Ú 0 and b Ú 0 if n is even, and a, b are any real numbers if n is odd.
n
Notice that if a is negative and n is even, then 2a is not defined. When it is defined, the principal nth root of a number is unique. n The symbol 2a for the principal nth root of a is called a radical; the integer n is 2 called the index, and a is called the radicand. If the index of a radical is 2, then 2 a is called the square root of a, and the index 2 is omitted to simply write 2a. If the 3 index is 3, then 2 a is called the cube root of a.
EXAM PL E 1
Simplifying Principal nth Roots 3 3 3 (a) 2 8 = 2 2 = 2
(c)
The number whose cube is 8 is 2.
4 1 = 4 a1 b = 1 A 16 2 B 2 4
3 3 (b) 2 - 64 = 2 1 - 42 3 = - 4 6 (d) 2 1 - 22 6 = 0 - 2 0 = 2
r
A56
APPENDIX A Review
These are examples of perfect roots, since each simplifies to a rational number. Notice the absolute value in Example 1(d). If n is even, then the principal nth root must be nonnegative. In general, if n Ú 2 is an integer and a is a real number, then n
2a n = a
2a = 0 a 0 n
Now Work
n
PROBLEM
if n Ú 3 is odd
(1a)
if n Ú 2 is even
(1b)
7
Radicals provide a way of representing many irrational real numbers. For example, there is no rational number whose square is 2. In terms of radicals, it can be said that 12 is the positive number whose square is 2.
EX A MPL E 2
Using a Calculator to Approximate Roots 5 Use a calculator to approximate 1 16.
Solution Figure 26
Figure 26 shows the result using a TI-84 Plus graphing calculator.
Now Work
PROBLEM
r
101
2 Simplify Radicals Let n Ú 2 and m Ú 2 denote positive integers, and let a and b represent real numbers. Assuming that all radicals are defined, the following properties are useful.
Properties of Radicals n
n
n
2ab = 2a 2b
(2a)
n
2a n a = n Ab 2b
2am = ( 2aa22 m n
n
(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.
EX AMPL E 3
Simplifying Radicals
(a) 232 = 216 # 2 = 216 # 22 = 422 c
c
Factor out 16, (2a) a perfect square.
3 3 3 3 3 3# 3 3 (b) 2 16 = 2 8#2 = 2 8# 2 2 = 2 2 22 = 22 2
c
c
Factor out 8, a perfect cube.
(2a)
SECTION A.7 nth Roots; Rational Exponents
A57
3 3 3 (c) 2 - 16x4 = 2 - 8 # 2 # x3 # x = 2 1 - 8x3 2 12x2
c
c
Factor perfect cubes inside radical
Group perfect cubes.
3 3 3 3 = 2 1 - 2x2 3 # 2x = 2 1 - 2x2 3 # 2 2x = - 2x2 2x
c
(2a)
(d)
2x 16x5 24 x4 x = 4 = 4 a b A 81 A 34 B 3 4
Now Work
PROBLEMS
4
#x =
2x 4 # 4 2x 4 2 4 b 2x = ` ` 2x = x 2 x 3 3 3 B 4
a
11 AND 17
r
Two or more radicals can be added or subtracted, provided that they have the same index and the same radicand. Such radicals are called like radicals.
EXAM PL E 4
Combining Like Radicals
(a) - 8212 + 23 = - 824 # 3 + 23
= - 8 # 2423 + 23 = - 1623 + 123 = - 1523
3 3 3 3 3 3 3 3 3 (b) 2 8x4 + 2 - x + 42 27x = 2 2xx + 2 - 1 # x + 42 3 x
3 3 3 3 3 3# 3 = 2 12x2 3 # 2 x + 2 -1 # 2 x + 42 3 2x 3 3 3 = 2x2 x - 1# 2 x + 122 x 3 = 12x + 112 2 x
Now Work
PROBLEM
r
33
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 denominator contains no radicals. For example: If a Denominator Contains the Factor
Multiply by
To Obtain a Denominator Free of Radicals
13
13
1 23 2 2
13 + 1
13 - 1
12 - 3
12 + 3
15 - 13
15 + 13
1 25 2 2
3 24
3 22
3 3 3 # 22 24 = 28 = 2
1 23 2
2
1 22 2 2
= 3 - 12 = 3 - 1 = 2 - 32 = 2 - 9 = - 7 -
1 23 2 2
= 5 - 3 = 2
In rationalizing the denominator of a quotient, be sure to multiply both the numerator and the denominator by the expression.
EXAM PL E 5
Rationalizing Denominators Rationalize the denominator of each expression: (a)
1 23
(b)
5 3
424
(c)
22 23 - 322
A58
APPENDIX A Review
Solution
(a) The denominator contains the factor 13, so multiply the numerator and denominator by 13 to obtain 1 23
=
# 23
1
23 23
23
1 23 2
=
2
=
23 3
3 (b) The denominator contains the factor 2 4, so multiply the numerator and 3 denominator by 22 to obtain
5 3 42 4
=
5
# 23 2 3
3 42 4 22
=
3 52 2 3 42 8
=
3 52 2 8
(c) The denominator contains the factor 13 - 312, so multiply the numerator and denominator by 13 + 312 to obtain 22 23 - 322
=
=
Now Work
22
# 23 + 322
23 - 322 23 + 322
2223 + 3 1 22 2
2
=
3 - 18
PROBLEM
=
22 1 23 + 322 2
1 23 2 2
-
1 322 2 2
26 + 6 6 + 26 = - 15 15
r
47
4 Simplify Expressions with Rational Exponents Radicals are used to define rational exponents.
DEFINITION
If a is a real number and n Ú 2 is an integer, then n
a1>n = 2a
(3)
n
provided that 2a exists.
n
Note that if n is even and a 6 0, then 2a and a1>n do not exist.
EX A MPL E 6
Writing Expressions Containing Fractional Exponents as Radicals (a) 41>2 = 24 = 2 3 (c) 1 - 272 1>3 = 2 - 27 = - 3
DEFINITION
(b) 81>2 = 28 = 222 3 3 (d) 161>3 = 2 16 = 22 2
r
If a is a real number and m and n are integers containing no common factors, with n Ú 2, then am>n = 2am = 1 2aa22 m n
n
(4)
n
provided that 2a exists. Two comments about equation (4) should be noted: m must be in lowest terms, and n must be positive. n n n 2. In simplifying the rational expression am>n, either 2am or 1 2a2 m may be used, the choice depending on which is easier to simplify. Generally, taking the root first, n as in 1 2a2 m, is easier. 1. The exponent
SECTION A.7 nth Roots; Rational Exponents
EXAM PL E 7
Using Equation (4)
1 24 2 3
(a) 43>2 =
(c) 1322 -2>5 =
5 12 32 2 -2
Now Work
4 3 12 - 82
(b) 1 - 82 4>3 =
= 23 = 8
PROBLEM
= 1 - 22 4 = 16
1 225 2 3
1 (d) 256>4 = 253>2 = 4
= 2-2 =
A59
= 53 = 125
r
55
It can be shown that the Laws of Exponents (page A8) hold for rational exponents. The next example illustrates using the Law of Exponents to simplify.
EXAM PL E 8
Simplifying Expressions Containing Rational Exponents Simplify each expression. Express your answer in such a way that only positive exponents occur. Assume that the variables are positive.
Solution
(a) 1x2>3 y2 1x -2 y2
1>2
(a) 1x2>3 y21x -2 y2
1>2
(b) ¢
2x1>3 y2>3
≤
= 1x2>3 y231x -2 2
-3
(c) ¢
1>2 1>2
y
4
= x2>3 yx -1 y1>2
(a m)n = a mn
= x -1>3 y3>2
am # an = am + n
=
(c) ¢
2x1>3 y2>3
≤
-3
9x2 y1>3 x1>3 y
≤
= ¢ 1>2
y2>3 2x
= ¢
Now Work
x1>3 y
y3>2
a -n =
x1>3 3
≤ = 1>3 9x2 - 11>32 y1 - 11>32
1y2>3 2
3
12x1>3 2 ≤
1>2
PROBLEM
3
= ¢
=
y2
23 1x1>3 2
9x5>3 y2>3
≤
≤
1>2
(ab)n = a nb n
= 1x2>3 # x -1 2 1y # y1>2 2
(b) ¢
9x2 y1>3
1>2
3
=
=
1 an
y2 8x
91>2 1x5>3 2 1y2>3 2
1>2
1>2
=
3x5>6
r
y1>3
71
The next two examples illustrate some algebra needed for certain calculus problems.
EXAM PL E 9
Writing an Expression as a Single Quotient Write the following expression as a single quotient in which only positive exponents appear. 1x2 + 12
Solution
1x2 + 12
1>2
+ x#
1>2
+ x#
1 2 1x + 12 -1>2 # 2x 2
1 2 x2 - 1>2 # 1>2 1x + 12 2 x = 1x2 + 12 + 1>2 2 1x2 + 12 = = =
Now Work
PROBLEM
77
1x2 + 12
1>2
1x2 + 12
1x2 + 12
1>2
+ x2
1>2
1x2 + 12 + x2 1x2 + 12
1>2
2x2 + 1
1x2 + 12
1>2
r
A60
APPENDIX A Review
Factoring an Expression Containing Rational Exponents
EX AM PL E 10
4 Factor: x1>3 12x + 12 + 2x4>3 3 Begin by writing 2x4>3 as a fraction with 3 as the denominator.
Solution
4x1>3(2x + 1) 4x1>3(2x + 1) + 6x4>3 4 1>3 6x4>3 x 12x + 12 + 2x4>3 = + = 3 3 3 c 3 Add the two fractions.
2x1>3[2(2x + 1) + 3x] 2x1>3(7x + 2) = = 3 3 c
c
2 and x1>3 are common factors.
Simplify.
Now Work
PROBLEM
r
89
Historical Note
T
he radical sign, 2 , was first used in print by Christoff Rudolff in 1525. It is thought to be the manuscript form of the letter r (for the Latin word radix = root), although this is not quite conclusively confirmed. It took a long time for 2 to become the standard symbol for a square root and much longer to standardize 4 5 3 2, 2, 2 and so on. The indexes of the root were placed in every conceivable position, with
3
4
all being variants for 28. The notation 2 216 was popular for 216. By the 1700s, the index had settled where we now put it. The bar on top of the present radical symbol, as follows,
2a2 + 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,
3
28,
2 3 8, and
ab + c = a(b + c)
28 3
A.7 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages in red. 1. ( - 3)2 =
; - 32 =
9
-9
2. 216 =
(p. A8)
; 2( - 4)2 =
4
4
(p. A9)
Concepts and Vocabulary n
3. In the symbol 2a, the integer n is called the 3 5. We call 2a the
cube
root
index
5 4. True or False 2 - 32 = - 2 True
.
4 6. True or False 2 ( - 3)4 = - 3
of a.
False
Skill Building In Problems 7–42, simplify each expression. Assume that all variables are positive when they appear. 3 7. 227
11. 28
4 8. 2 16
3
4 12 8 15. 2 x y
19. 236x 23.
1 25 29 2
2
35.
20. 29x 3
24.
1523
5
223 *32. + 1
*36.
17.
3x 2x 4
4
x9 y 7
B xy
3
3 10. 2 -1
-2 4
3
x2y
18. 6x2x
1 326 2 1 222 2
30013
25.
225
29. - 218 + 228
1 25 - 2 2 1 25 1 1x + 25 2 2
+ 32
3
3
33. 522 - 2254 3
-1
14. 248x
2
3
4
- 2x2x
21. 23x 212x
2
28. 625 - 425
722
1 23 + 3 2 1 23 - 1 2 1 1x - 1 2 2 x - 22x
13. 2 - 8x
x2y
1 23 210 2 3
3
322
5 10 5 16. 2 x y
62x
3
3
12. 254
x3y2
27. 322 + 422 31.
3
222
3 9. 2 -8
2
3
*37. 216x4 - 22x
3
- 22
3xy2
3
B 81x y
4 2
4 2x1 3x
1 3x
22. 25x 220x3
1223 26. 22
5
10x2
1 528 2 1 - 323 2
30. 2212 - 3227 3
3
34. 9224 - 281 4
4
*38. 232x + 22x5
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
- 3026 - 523
3 152 3
A61
SECTION A.7 nth Roots; Rational Exponents
39. 28x3 - 3250x
(2x - 15) 12x
3 3 3 41. 2 16x4 y - 3x2 2xy + 52 - 2xy4
40. 3x29y + 4225y 3 - (x + 5y) 1 2xy
(9x + 20) 1y
3 42. 8xy - 225x2 y2 + 2 8x3 y3
5xy
In Problems 43–54, rationalize the denominator of each expression. Assume that all variables are positive when they appear. 43.
47.
51.
22 2
1 22
44.
23
(5 + 12) 13
5 - 22
23
3 524 2
5 3
22
48.
52.
2
223 3
23
45.
22
( 27 - 2) 12
27 + 2
3
-2 3
29
-
3 21 3 3
49.
*53.
- 23
215 5
-
25
2 - 25
46.
825 - 19 41
2 + 325
50.
2x + h - 1x
*54.
2x + h + 1x
- 23
-
28
26 4 9 - 523 3
23 - 1 223 + 3
2x + h + 2x - h 2x + h - 2x - h
In Problems 55–66, simplify each expression. 55. 82>3
56. 43>2
4
59. 163>2
60. 253>2
64
9 3>2 63. a b 8
64. a
2722 32
57. ( - 27)1>3
8
27 2>3 b 8
8 -3>2 65. a b 9
9 4
58. 163>4
1 27
61. 9-3>2
125
-3
8 1 64
62. 16 -3>2 66. a
2722 32
8 -2>3 b 27
9 4
In Problems 67–74, simplify each expression. Express your answer in such a way that only positive exponents occur.Assume that the variables are positive. 67. x3>4 x1>3 x -1>2 71.
(x2 y)
1>3
68. x2>3 x1>2 x -1>4
x7>12
(xy2)
2>3
x2>3y
x2>3 y2>3
72.
(xy)1>4(x2 y2)
x11>12 1>2
(x2 y)3>4
69. (x3 y6)
y1>2
73.
x1>4
1>3
xy2
(16x2 y -1>3)
70. (x4y8)
3>4
8x5>4
(xy2)1>4
74.
y3>4
3>4
x3y6
(4x -1 y1>3)
3>2
(xy)3>2
8 x3y
Applications and Extensions In Problems 75–88, expressions that occur in calculus are given. Write each expression as a single quotient in which only positive exponents and/or radicals appear. 75.
x
+ 2(1 + x)1>2
(1 + x)1>2
x 7 -1
3x + 2
1 77. 2x(x2 + 1)1>2 + x2 # (x2 + 1)-1>2 # 2x 2 *79. 24x + 3 #
83.
(x + 4)1>2 - 2x(x + 4)-1>2 x + 4 x2
85.
- (x - 1)
1>2
x 1 + x2
87.
2
(x2 - 1)
22x
2
x 7 5
*80.
x 7 -4
+ x1>2
x 7 0
3 2 8x + 1 3
32(x - 2)
2
+
82.
3>2
4 - x (x + 4)
3x + 1 2x1>2
84.
3>2
x ≠ -1
3 2 x - 2 3
242(8x + 1)
2x2 + 1 - x #
2 + x 2(1 + x)
2x1>2
4x + 3 3(x + 1)2>3
x ≠ 2, x ≠ -
2
1 8
2x
22x2 + 1 x + 1 2
1 (x2 + 1)3>2
(9 - x2)1>2 + x2(9 - x2)-1>2
-3 6 x 6 3
9 - x2
9 (9 - x2)3>2
1>2
x 6 - 1 or x 7 1
2
- 2x2x
(1 + x2)
x 7 -1
1 + x
1 78. (x + 1)1>3 + x # (x + 1)-2>3 3
(x2 + 1)1>2
1
221 + x 1 + x
81.
x(3x2 + 2)
1 1 + 2x - 5 # 21x - 5 514x + 3
21 + x - x #
76.
(1 + x)1>2
x 7 0
1 - 3x2 22x(1 + x2)2
1 x (x - 1) 2
2
1>2
86.
(x2 + 4)1>2 - x2(x2 + 4)-1>2
4
x + 4
(x + 4)3>2
2
2x(1 - x2)1>3 +
88.
2 3 x (1 - x2)-2>3 3
(1 - x2)2>3
2
x ≠ - 1, x ≠ 1
2x(3 - 2x2) 3(1 - x2)4>3
A62
APPENDIX A Review
In Problems 89–98, expressions that occur in calculus are given. Factor each expression. Express your answer in such a way that only positive exponents occur. 3 4 1 1 3>2 4>3 1>3 89. (x + 1) + x # (x + 1)1>2 x Ú - 1 (5x + 2)(x + 1)1>2 90. (x2 + 4) + x # (x2 + 4) # 2x (x2 + 4)1>3(11x2 + 12) 2 2 3 3 *91. 6x1>2(x2 + x) - 8x3>2 - 8x1>2 93. 3(x2 + 4)
4>3
+ x # 4(x2 + 4)
1>3
92. 6x1>2(2x + 3) + x3>2 # 8 x Ú 0
x Ú 0
# 2x
*95. 4(3x + 5)1>3(2x + 3)3>2 + 3(3x + 5)4>3(2x + 3)1>2 x Ú 97. 3x -1>2 +
3 1>2 x 2
x 7 0
94. 2x(3x + 4)4>3 + x2 # 4(3x + 4)1>3
(x2 + 4)1>3(11x2 + 12) 3 2
2x(3x + 4)1>3(5x + 4)
*96. 6(6x + 1)1>3(4x - 3)3>2 + 6(6x + 1)4>3(4x - 3)1>2 x Ú
3(x + 2) 2x
2x1>2(10x + 9)
98. 8x1>3 - 4x -2>3
1>2
x ≠ 0
3 4
4(2x - 1) x2>3
In Problems 99–106, use a calculator to approximate each radical. Round your answer to two decimal places. 99. 22 103.
100. 27
1.41
2 + 23 3 - 25
4.89
104.
3 101. 2 4
2.65
25 - 2 22 + 4
3 102. 2 -5
1.59
3
105.
0.04
325 - 22 23
2.15
106.
- 1.71
3 223 - 2 4
22
1.33
107. Calculating the Amount of Gasoline in a Tank 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
96 - 0.608 Ah
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?
15,660.4 gal
390.7 gal
108. Inclined Planes The final velocity y of an object in feet per second (ft>sec) after it slides down a frictionless inclined plane of height h feet is y = 264h + y20 where y0 is the initial velocity (in ft/sec) of the object. v0
(a) What is the final velocity y of an object that slides down a frictionless inclined plane of height 4 feet? Assume that the initial velocity is 0. 16 ft/sec (b) What is the final velocity y of an object that slides down a frictionless inclined plane of height 16 feet? Assume that the initial velocity is 0. 32 ft/sec (c) What is the final velocity y of an object that slides down a frictionless inclined plane of height 2 feet with an initial velocity of 4 ft/sec ? 12 ft/sec
h v
Problems 109–112 require the following information. Period of a Pendulum The period T, in seconds, of a pendulum of length l, in feet, may be approximated using the formula T = 2p
l A 32
In Problems 109–112, express your answer both as a square root and as a decimal. 109. Find the period T of a pendulum whose length is 64 feet. 222p ≈ 8.89 sec 110. Find the period T of a pendulum whose length is 16 feet. p22 ≈ 4.44 sec p23 111. Find the period T of a pendulum whose length is 8 inches. ≈ 0.91 sec 6 112. Find the period T of a pendulum whose length is 4 inches. p26 ≈ 0.64 sec 12
Discussion and Writing 113. Give an example to show that 2a2 is not equal to a. Use it to explain why 2a2 = 0 a 0 .
‘Are You Prepared?’ Answers 1. 9; - 9
2. 4; 4
SECTION A.8 Solving Equations
A63
A.8 Solving Equations PREPARING FOR THIS SECTION Before getting started, review the following: r 4RVBSF3PPUT "QQFOEJY" 4FDUJPO" QQ"m"
r nth Roots; Rational Exponents (Appendix A, Section A.7, pp. A55–A60)
r 'BDUPSJOH1PMZOPNJBMT "QQFOEJY" 4FDUJPO" pp. A32–A38) r ;FSP1SPEVDU1SPQFSUZ "QQFOEJY" 4FDUJPO" p. A4) r 3BUJPOBM&YQSFTTJPOT "QQFOEJY" 4FDUJPO" pp. A45–A50) Now Work the ‘Are You Prepared?’ problems on page A70.
OBJECTIVES 1 2 3 4
Solve Linear Equations (p. A64) Solve Rational Equations (p. A66) Solve Equations by Factoring (p. A67) Solve Radical Equations (p. A68)
An equation in one variable is a statement in which two expressions, at least one containing the variable, are set 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 variable. Unless otherwise restricted, the admissible values of the variable are those in the domain of the variable. Those 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
2x2 + 9 = 5
The first of these statements, x + 5 = 9, is true when x = 4 and false for any other choice of x. Thus 4 is a solution of the equation x + 5 = 9. Also, 4 satisfies the equation x + 5 = 9, because, when 4 is substituted for x, a true statement 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, set notation is used to write the solution of an equation. This set is called the solution set of the equation. For example, the solution set of the equation x2 - 9 = 0 is 5 - 3, 36 . Some equations have no real solution. For example, x2 + 9 = 5 has no real solution, because there is no real number whose square, when added to 9, equals 5. An equation that is satisfied for every choice 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 equations requires that a series of equivalent equations be developed from the original equation until an obvious solution results. For example, all the following equations are equivalent, because each has only the solution x = 5. 2x + 3 = 13 2x = 10 x = 5
A64
APPENDIX A Review
The question, though, is “How do I obtain an equivalent equation?” In general, there are five ways to do so.
Procedures That Result in Equivalent Equations 1. Interchange the two sides of the equation: 3 = x is equivalent to x = 3 2. Simplify the sides of the equation by combining like terms, eliminating parentheses, and so on: x + 2 + 6 = 2x + 31x + 12 is equivalent to x + 8 = 5x + 3 3. Add or subtract the same expression on both sides of the equation: 3x - 5 = 4 is equivalent to 13x - 52 + 5 = 4 + 5 4. Multiply or divide both sides of the equation by the same nonzero expression: 3x 6 = x ≠ 1 x - 1 x - 1 3x # 6 # is equivalent to 1x - 12 = 1x - 12 x - 1 x - 1 5. If one side of the equation is 0 and the other side can be factored, use the Zero-Product Property and set each factor equal to 0. x1x - 32 = 0 is equivalent to x = 0 or x - 3 = 0
WARNING Squaring both sides of an equation does not necessarily lead to an equivalent equation. 䊏
Whenever it is possible to solve an equation in your head, do so. For example: The solution of 2x = 8 is x = 4. The solution of 3x - 15 = 0 is x = 5.
Now Work
PROBLEM
9
Some specific types of equations that can be solved algebraically to obtain exact solutions are now introduced, starting with linear equations.
1 Solve Linear Equations Linear equations are equations such as 3x + 12 = 0
3 1 x - = 0 4 5
0.62x - 0.3 = 0
A linear equation in one variable is equivalent to an equation of the form ax + b = 0 where a and b are real numbers and a ≠ 0. Sometimes a linear equation is called a first-degree equation, because the left side is a polynomial in x of degree 1.
SECTION A.8 Solving Equations
EXAM PL E 1
A65
Solving a Linear Equation Solve the equation: 31x - 22 = 51x - 12 31x - 22 = 51x - 12
Solution
3x - 6 = 5x - 5 3x - 6 - 5x = 5x - 5 - 5x - 2x - 6 = - 5
Subtract 5x from each side. Simplify.
- 2x - 6 + 6 = - 5 + 6 - 2x = 1 -2x 1 = -2 -2 x = -
Use the Distributive Property.
Add 6 to each side. Simplify. Divide each side by - 2.
1 2
Simplify.
1 in the expression on the left side of the equation and simplify. 2 1 Let x = - in the expression on the right side of the equation and simplify. 2 If the two expressions are equal, the solution checks.
Check: Let x = -
31x - 22 = 3a -
1 5 15 - 2b = 3a - b = 2 2 2
51x - 12 = 5a -
1 3 15 - 1b = 5a - b = 2 2 2
1 Since the two expressions are equal, the solution x = - checks. The solution 2 1 set is e - f. 2
r
Now Work
PROBLEM
15
The next example illustrates the solution of an equation that does not appear to be linear but leads to a linear equation upon simplification.
EXAM PL E 2
Solving an Equation That Leads to a Linear Equation Solve the equation: 12x - 12 1x - 12 = 1x - 52 12x - 52
Solution
12x - 12 1x - 12 = 1x - 52 12x - 52 2x2 - 3x + 1 = 2x2 - 15x + 25 2x - 3x + 1 - 2x - 3x + 1 - 3x + 1 - 1 - 3x - 3x + 15x 12x 2
2
= = = = = =
2x - 15x - 15x - 15x - 15x 24 2
12x 24 = 12 12 x = 2
15x + 25 - 2x + 25 + 25 - 1 + 24 + 24 + 15x
Multiply and combine like terms. 2
Subtract 2x2 from each side. Simplify. Subtract 1 from each side. Simplify. Add 15x to each side. Simplify. Divide each side by 12. Simplify.
A66
APPENDIX A Review
Check: 12x - 12 1x - 12 = 12 # 2 - 12 12 - 12 = 132 112 = 3
1x - 52 12x - 52 = 12 - 52 12 # 2 - 52 = 1 - 32 1 - 12 = 3
Since the two expressions are equal, the solution checks. The solution set is 5 26 .
Now Work
PROBLEM
r
25
2 Solve Rational Equations A rational equation is an equation that contains a rational expression. Examples of rational equations are 3 2 = + 7 and x + 1 x - 1
x - 5 3 = x - 4 x + 2
To solve a rational equation, multiply both sides of the equation by the least common multiple of the denominators of the rational expressions that make up the rational equation.
Solving a Rational Equation
EX A MPL E 3
Solve the equation:
Solution
3 1 7 = + x - 2 x - 1 1x - 12 1x - 22
First, note that the domain of the variable is 5 x 0 x ≠ 1, x ≠ 26 . Clear the equation of rational expressions by multiplying both sides by the least common multiple of the denominators of the three rational expressions, 1x - 12 1x - 22.
3 1 7 = + x - 2 x - 1 1x - 12 1x - 22 1x - 12 1x - 22
3 1 7 = 1x - 12 1x - 22 c + d x - 2 x - 1 1x - 12 1x - 22 3x - 3 = 1x - 12 1x - 22
1 7 + 1x - 12 1x - 22 x - 1 1x - 12 1x - 22
3x - 3 = 1x - 22 + 7
Use the Distributive Property on each side; divide out common factors on the right. Rewrite the equation.
3x - 3 = x + 5 2x = 8
Combine like terms. Add 3 to each side; subtract x from each side.
x = 4
Check:
Multiply both sides by (x - 1)(x - 2); divide out common factors on the left.
Divide by 2.
3 3 3 = = x - 2 4 - 2 2 1 7 1 7 1 7 2 7 9 3 + = + = + # = + = = x - 1 1x - 12 1x - 22 4 - 1 14 - 12 14 - 22 3 3 2 6 6 6 2 Since the two expressions are equal, x = 4 checks. The solution set is 5 46 .
Now Work
PROBLEM
35
r
SECTION A.8 Solving Equations
A67
Sometimes the process of creating equivalent equations leads to apparent solutions that are not solutions of the original equation. These are called extraneous solutions.
EXAM PL E 4
A Rational Equation with No Solution Solve the equation:
Solution
3x 3 + 2 = x - 1 x - 1
First, note that the domain of the variable is 5 x 0 x ≠ 16 . Since the two rational expressions in the equation have the same denominator, x - 1, multiply both sides by x - 1 to clear the rational expressions from the equation. The resulting equation is equivalent to the original equation for x ≠ 1. 3x 3 + 2 = x - 1 x - 1 a
3x x - 1
#
3x 3 + 2b # 1x - 12 = x - 1 x - 1
#
1x - 12
1x - 12 + 2 # 1x - 12 = 3
Multiply both sides by x - 1; divide out common factors on the right. Use the Distributive Property on the left side; divide out common factors on the left.
3x + 2x - 2 = 3
Simplify.
5x - 2 = 3
Combine like terms.
5x = 5
Add 2 to each side.
x = 1
Divide both sides by 5.
The solution appears to be 1. But recall that x = 1 is not in the domain of the variable, so x = 1 is an extraneous solution. The equation has no solution. The solution set is ∅.
r
Now Work
PROBLEM
37
3 Solve Equations by Factoring Many equations can be solved by factoring and then using the Zero-Product Property.
EXAM PL E 5
Solving Equations by Factoring Solve the equations: (a) x2 = 4x
Solution
(b) x3 - x2 - 4x + 4 = 0
(a) Begin by collecting all terms on one side. This results in 0 on one side and an expression to be factored on the other. x2 = 4x x2 - 4x = 0 x(x - 4) = 0 x = 0 or x - 4 = 0 x = 4
Factor. Apply the Zero-Product Property.
A68
APPENDIX A Review
x = 0: 02 = 4 # 0 x = 4: 42 = 4 # 4
Check:
So 0 is a solution. So 4 is a solution.
The solution set is {0, 4}. (b) Group the terms of x3 - x2 - 4x + 4 = 0 as follows: 1x3 - x2 2 - 14x - 42 = 0
Factor out x2 from the first grouping and 4 from the second. x2 1x - 12 - 41x - 12 = 0
This reveals the common factor 1x - 12, so 1x2 - 42 1x - 12 = 0
1x - 22 1x + 22 1x - 12 = 0 x - 2 = 0 or
Factor again.
x + 2 = 0 or
x = 2
x - 1 = 0
x = -2
Set each factor equal to 0.
x = 1 Solve.
1 - 22 3 - 1 - 22 2 - 41 - 22 + 4 = - 8 - 4 + 8 + 4 = 0
Check: x = - 2:
−2 is a solution.
x = 1: 1 - 1 - 4112 + 4 = 1 - 1 - 4 + 4 = 0
1 is a solution.
x = 2: 23 - 22 - 4122 + 4 = 8 - 4 - 8 + 4 = 0
2 is a solution.
3
2
The solution set is 5 - 2, 1, 26 .
Now Work
PROBLEM
r
41
4 Solve Radical 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 changes 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. Recall that these are called extraneous solutions. In radical equations, extraneous solutions may occur when the index of the radical is even. Therefore, it is important to check all answers when working with radical equations.
EX A MPL E 6
Solving a Radical Equation 3
Find the real solutions of the equation: 22x - 4 - 2 = 0
Solution
The equation contains a radical whose index is 3. Isolate it on the left side: 3
22x - 4 - 2 = 0 3 22x - 4 = 2 Because the index of the radical is 3, raise each side to the third power and solve. 3 12 2x
- 42 2x - 4 2x x 3
= = = =
23 8 12 6
Raise each side to the power 3. Simplify. Add 4 to both sides. Divide both sides by 2.
3 3 3 Check: 2 2162 - 4 - 2 = 2 12 - 4 - 2 = 2 8 - 2 = 2 - 2 = 0
The solution set is 5 66 .
Now Work
PROBLEM
r 45
SECTION A.8 Solving Equations
EXAM PL E 7
A69
Solving a Radical Equation Find the real solutions of the equation: 2x - 1 = x - 7
Solution
Square both sides since the index of a square root is 2. 2x - 1 = 2 - 12 = x - 1 = 2 x - 15x + 50 = 1x - 102 1x - 52 = x = 10 or x =
1 2x
Check: x = 10: x = 5:
x - 7 1x - 72 2 Square both sides. 2 x - 14x + 49 Remove parentheses. 0 Put in standard form. 0 Factor. 5 Apply the Zero-Product Property and solve.
2x - 1 = 210 - 1 = 29 = 3 and x - 7 = 10 - 7 = 3 2x - 1 = 25 - 1 = 24 = 2 and x - 7 = 5 - 7 = - 2
The apparent solution x = 5 is extraneous; the only solution of the equation is x = 10. The solution set is 5 106 .
Now Work
r
PROBLEM
51
Sometimes, it is necessary to raise each side to a power more than once in order to solve a radical equation.
EXAM PL E 8
Solving a Radical Equation Find the real solutions of the equation: 22x + 3 - 2x + 2 = 2
Solution
First, isolate the more complicated radical expression (in this case, 22x + 3) on the left side: 22x + 3 = 2x + 2 + 2 Now square both sides (the index of the radical is 2).
1 22x
+ 32 = 2
2x + 3 =
1 2x 1 2x
+ 2 + 222
+ 2 2 + 42x + 2 + 4 Remove parentheses. 2
2x + 3 = x + 2 + 42x + 2 + 4
Simplify.
2x + 3 = x + 6 + 42x + 2
Combine like terms.
The equation still contains a radical. Isolate the remaining radical on the right side and again square both sides. x - 3 1x - 32 2 x2 - 6x + 9 x2 - 22x - 23 1x - 232 1x + 12 x = 23 or x Check: x = 23: x = - 1:
= = = = = =
42x + 2 161x + 22 16x + 32 0 0 -1
Isolate the radical on the right side. Square both sides. Remove parentheses. Put in standard form. Factor. Apply the Zero-Product Property and solve.
221232 + 3 - 223 + 2 = 249 - 225 = 7 - 5 = 2 221 - 12 + 3 - 2 - 1 + 2 = 21 - 21 = 1 - 1 = 0
The apparent solution x = - 1 is extraneous; the only solution is x = 23. The solution set is 5 236 .
Now Work
r
PROBLEM
57
A70
APPENDIX A Review
A.8 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 the least common multiple of the denominators of 5 3 . (p. A49) and 2 2 x - 4 x - 3x + 2 1x - 22 1x + 22 1x - 12
3. Factor: (a) 2x2 - 10x (p. A32)
2x1x - 52
(b) x3 + 3x2 - 4x - 12 (p. A36)
2. The solution set of the equation 1x - 32 13x + 52 = 0 is . (p. A4) 5 e 3, - f 3
3 3 4. (a) 2 x =
(b)
4 12 x
x
- 324 =
. (p. A56) x - 3
1x + 22 1x - 22 1x + 32 . (p. A56)
Concepts and Vocabulary 5. Two equations that have the same solution set are called equivalent equations .
7. When an apparent solution does not satisfy the original equation, it is called a(n) extraneous solution.
6. An equation that is satisfied for every choice of the variable for which both sides are defined is called a(n) identity .
8. True or False Some equations have no solution. True
Skill Building In Problems 9–66, solve each equation. 9. 2x - 3 = 5
10. 3x + 4 = - 8
{4}
13. 6 - x = 2x + 9
14. 3 - 2x = 2 - x
{91}
21. 0.9t = 0.4 + 0.1t
22. 0.9t = 1 + t
{0.5}
25. (x + 7)(x - 1) = (x + 1)2 28. w(4 - w 2) = 8 - w 3 31. t 3 - 9t = 0
{2}
-3 -2 = x + 4 x + 1
37.
x 2 + 3 = x - 2 x - 2
{910} ∅
19.
3 1 x - 4 = x 2 4
23.
4 2 + = 3 y y
26. (x + 2)(x - 3) = (x - 3)2
35.
2 3 10 = + x - 2 x + 5 (x + 5)(x - 2)
38.
2x -6 = - 2 x + 3 x + 3
43. 2x3 + 4 = x2 + 8x
1 e - 2, , 2 f 2
44. 3x3 + 4x2 = 27x + 36 47. 23t + 4 = - 6
3 49. 2 1 - 2x - 3 = 0
52. 212 - x = x
{913}
{3}
{91, 1} 4 e - 3, - , 3 f 3
No real solution
59. 13x + 12 1>2 = 4
64.
5 4 -3 + = 5z - 11 2z - 3 5 - z
62. 12x + 12 1>3 = - 1 {2}
No real solution {3}
54. 2 + 212 - 2x = x
{18}
{2}
48. 25t + 3 = - 2
{8}
65.
{92}
{91}
x + 3 -3 x - 2 = 2 x2 - 1 x - x x + x
{96}
{4}
57. 23x + 1 - 2x - 1 = 2 60. 13x - 52 1>2 = 2
{5}
{94, - 1, 1}
{1}
53. 3 + 23x + 1 = x
58. 23x - 5 - 2x + 7 = 2 61. 15x - 22 1>3 = 2
45. 22t - 1 = 1
51. 215 - 2x = x
56. 23x + 7 + 2x + 2 = 1
{95, 0, 4}
42. x3 + 4x2 - x - 4 = 0
{0}
{91, 3}
{3}
1 1 1 1 e- f + = 3 2x + 3 x - 1 (2x + 3)(x - 1)
3 50. 2 1 - 2x - 1 = 0
55. 22x + 3 - 2x + 1 = 1
3 f 10
{21}
39. x3 + x2 - 20x = 0
∅
41. x3 + x2 - x - 1 = 0
36.
e
{0, 4}
2 3 = 33. 2x - 3 x + 5 {6}
7 e f 5
{98}
4 5 - 5 = y 2y
27. z(z2 + 1) = 3 + z3
{3}
27 f 4
1 x = 5 2
20. 1 24.
{2}
5 - 22, 0, 226
{97, 0, 1}
{0}
{916}
e
2 9 x = 3 2
{ - 18} 16. 3(2 - x) = 2x - 1
30. x3 = 4x2
40. x3 + 6x2 - 7x = 0
46. 23t + 4 = 2
12.
{0, 9}
32. 4z3 - 8z = 0
{93, 0, 3}
34.
{92}
{910}
29. x2 = 9x
{2}
5 e f 4
15. 2(3 + 2x) = 3(x - 4)
{1}
17. 8x - (2x + 1) = 3x - 10 18. 5 - (2x - 1) = 10 { - 3}
5 1 x = 3 12
11.
{94}
{1, 5}
{3} e-
11 f 6
63.
2 3 5 + = y + 3 y - 4 y + 6
66.
x + 1 x + 4 -3 - 2 = 2 x2 + 2x x + x x + 3x + 2
{97}
SECTION A.8 Solving Equations
A71
Problems 67–72, list some formulas that occur in applications. Solve each formula for the indicated variable. R1R2 1 1 A - P 1 + for R R = 68. Finance A = P 11 + rt2 for r r = 67. Electricity = R R1 R2 R1 + R2 Pt 69. Mechanics F =
mv2 for R R
71. Mathematics S =
R =
a for r 1 - r
mv2 F
r =
S - a S
70. Chemistry PV = nRT for T
T =
72. Mechanics v = - gt + v0 for t
t =
PV nR v0 - v g
Applications and Extensions 73. Physics: Using Sound to Measure Distance The distance to the surface of the water in a well can sometimes be found by dropping an object into the well and measuring the time elapsed until a sound is heard. If t 1 is the time (measured in seconds) that it takes for the object to strike the water, then t 1 will obey the equation s = 16t 21 , where s is the distance 1s (measured in feet). It follows that t 1 = . Suppose that t 2 4 is the time that it takes for the sound of the impact to reach your ears. Because sound travels at a speed of approximately 1100 feet per second, the time t 2 for the sound to travel
s . See the illustration. Now 1100 t 1 + t 2 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 the distance s will be t 2 =
Total time elapsed =
s 1s + 4 1100
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. ≈229.94 ft 74. Crushing Load A civil engineer relates the thickness T, in inches, and the height H, in feet, of a square wooden pillar LH 2 . B 25 If a square wooden pillar is 4 inches thick and 10 feet high, what is its crushing load? 64 tons
to its crushing load L, in tons, using the model T = Sound waves: s t2 ⫽ –––– 1100
Falling object: s t1 ⫽ –– 4
4
75. Foucault’s Pendulum The period of a pendulum is the time it takes the pendulum to make one full swing back and forth. The period T, in seconds, is given by the formula l , where l is the length, in feet, of the pendulum. A 32 In 1851, Jean-Bernard-Leon 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 16.5 seconds. What was its length? ≈221 ft T = 2p
Discussion and Writing 76. Make up a radical equation that has no solution. 77. Make up a radical equation that has an extraneous solution. *78. 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? *79. What Went Wrong? On an exam, Jane solved the equation 22x + 3 - x = 0 and wrote that the solution set was { - 1, 3}. Jane received 3 out of 5 points for the problem. Jane asks you why she received 3 out of 5 points. Provide an explanation. 80. Which of the following pairs of equations are equivalent? Explain. (b) (a) x2 = 9; x = 3
(b) x = 29; x = 3
(c) 1x - 12 1x - 22 = 1x - 12 2; x - 2 = x - 1 *81. The equation 5 8 + x + 3 = x + 3 x + 3 has no solution, yet when we go through the process of solving it we obtain x = - 3. Write a brief paragraph to explain what causes this to happen. 82. Make up an equation that has no solution and give it to a fellow student to solve. Ask the student to write a critique of your equation.
‘Are You Prepared?’ Answers 1. (x - 2)(x + 2)(x - 1)
3. (a) 2x(x - 5)
(b) (x + 2)(x - 2)(x + 3)
4. (a) x (b) x - 3 5 2. e - , 3 f 3 *Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
A72
APPENDIX A Review
A.9 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications OBJECTIVES 1 2 3 4 5
The icon is a Model It! icon. It indicates that the discussion or problem involves modeling.
Translate Verbal Descriptions into Mathematical Expressions (p. A72) Solve Interest Problems (p. A73) Solve Mixture Problems (p. A74) Solve Uniform Motion Problems (p. A75) Solve Constant Rate Job Problems (p. A77)
Applied (word) problems do not come in the form “Solve the equation. c” Instead, they supply information using words, a verbal description of the real problem. Solving applied problems requires the ability to translate the verbal description into the language of mathematics. This can be done 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. The equation or inequality that represents the relationship among the variables is called a model. Any solution to the mathematical problem must be checked against the mathematical problem, the verbal description, and the real problem. See Figure 27 for an illustration of the modeling process.
Figure 27 Real problem
Verbal description
Language of mathematics
Mathematical problem Check
Check
Check Solution
1 Translate Verbal Descriptions into Mathematical Expressions Let’s look at a few examples that will help with translating certain words into mathematical symbols.
EX A MPL E 1
Translating Verbal Descriptions into Mathematical Expressions (a) The (average) speed of an object equals the distance traveled divided by the time required. d Translation: If r is the speed, d the distance, and t the time, then r = . 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 - x.
r
Now Work
PROBLEM
7
Always check the units used to measure the variables of an applied problem. In Example 1(a), if r is measured in miles per hour, then the distance d 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.
SECTION A.9 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
A73
Steps for Solving Applied Problems STEP 1: Read the problem carefully, perhaps two or three times. Pay particular 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 mathematical 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, creating 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, congratulations! If it does not agree, try again.
2 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 I = Prt
(1)
Interest charged according to formula (1) is called simple interest. When using formula (1), be sure to express r as a decimal. For example, if the rate of interest is 4%, then r = 0.04.
EXAM PL E 2
Finance: Computing Interest 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?
Solution
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 1 the rate of interest 19, = 0.092 times the time in years, : 2 1 Interest charged = I = Prt = 15002 10.092 a b = +22.50 2 After 6 months, Juanita will owe what she borrowed plus the interest: +500 + +22.50 = +522.50
EXAM PL E 3
r
Financial Planning Candy has $70,000 to invest and wants an annual return of $2800, which requires an overall rate of return of 4%. She can invest in a safe, government-insured certificate of deposit, but it pays only 2%. To obtain 4%, she agrees to invest some of her
A74
APPENDIX A Review
money in noninsured corporate bonds paying 7%. How much should be placed in each investment to achieve her goal?
Solution
STEP 1: The question is asking for two dollar amounts: the principal to invest in the corporate bonds and the principal to invest in the certificate of deposit. STEP 2: Let b represent the amount (in dollars) to be invested in the bonds. Then 70,000 - b is the amount that will be invested in the certificate. (Do you see why?) STEP 3: Now set up Table 1:
Table 1
Principal ($)
Rate
Time (yr)
Interest ($)
Bonds
b
7% = 0.07
1
0.07b
Certificate
70,000 - b
2% = 0.02
1
0.02(70,000 - b)
Total
70,000
4% = 0.04
1
0.04(70,000) = 2800
The total interest from the investments is equal to 0.04 170,0002 = 2800, which leads to the equation 0.07b + 0.02170,000 - b2 = 2800 (Note that the units are consistent: the unit is dollars on each side.) STEP 4: 0.07b + 1400 - 0.02b = 2800 0.05b = 1400 Simplify. b = 28,000 Divide both sides by 0.05. Candy should place $28,000 in the bonds and +70,000 - +28,000 = +42,000 in the certificate.
STEP 5: The interest on the bonds after 1 year is 0.07 1+28,0002 = +1960; the interest on the certificate after 1 year is 0.02 1 +42,0002 = +840. The total annual interest is $2800, the required amount.
r
Now Work
PROBLEM
17
3 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.
EX A MPL E 4
Blending Coffees The manager of a local coffee shop decides to experiment with a new blend of coffee. 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 between selling the new blend and selling the other types. How many pounds of the B grade Colombian coffee and how many pounds of the A grade Arabica coffee are required?
Solution
STEP 1: The question is asking how many pounds of Colombian coffee and how many pounds of Arabica coffee are needed to make 100 pounds of the mixture. STEP 2: Let c represent the number of pounds of the B grade Colombian coffee. Then 100 - c equals the number of pounds of the A grade Arabica coffee. STEP 3: See Figure 28.
SECTION A.9 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
Figure 28
$5 per pound
$10 per pound
⫹
B Grade Colombian c pounds
⫹
A75
$7 per pound
⫽
A Grade Arabica 100 − c pounds
Blend ⫽
100 pounds
There is to be no difference in revenue between selling the A and B grades separately and the blend. This leads to the following equation: e
Price per pound Pounds of Price per pound Pounds of Price per pound Pounds of fe f + e fe f = e fe f of B grade B grade of A grade A grade of blend blend # # (100 - c) = # c + +10 +7 100 +5 STEP 4: Solve the equation. 5c + 101100 - c2 5c + 1000 - 10c - 5c c
= = = =
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.
STEP 5: The 60 pounds of B grade coffee would sell for 1+52 1602 = +300, and the 40 pounds of A grade coffee would sell for 1+102 1402 = +400; the total revenue, $700, equals the revenue obtained from selling the blend, as desired.
r
Now Work
PROBLEM
21
4 Solve Uniform Motion Problems Objects that move at a constant speed are said to be in uniform motion. When the average speed of an object is known, it can be interpreted as that object’s constant speed. For example, a bicyclist traveling at an average speed of 25 miles per hour can be modeled as being in uniform motion with a constant speed of 25 miles per hour.
Uniform Motion Formula If an object moves at an average speed (rate) r, the distance d covered in time t is given by the formula d = rt
(2)
That is, Distance = Rate # Time.
EXAM PL E 5
Physics: Uniform Motion Tanya, who is a long-distance runner, runs at an average speed of 8 miles per hour (mi>hr). Two hours after Tanya leaves your house, you leave in your Honda and
A76
APPENDIX A Review
follow the same route. If your average speed is 40 mi>hr, how long will it be before you catch up to Tanya? How far will each of you be from your home?
Solution
Refer to Figure 29. 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.
Figure 29
t⫽0
2 hr
Time t
Time t
t⫽0
Set up Table 2:
Table 2
Rate (mi/hr)
Time (hr)
Distance (mi)
Tanya
8
t + 2
8(t + 2)
Honda
40
t
40t
The distance traveled is the same for both, which leads to the following equation: 81t + 22 = 40t 8t + 16 = 40t 32t = 16 1 t = hour 2 It will take the Honda
1 hour to catch up to Tanya. Each will have gone 20 miles. 2
1 Check: In 2.5 hours, Tanya travels a distance of 12.52 182 = 20 miles. In hour, 2 1 the Honda travels a distance of a b 1402 = 20 miles. 2
r
EX A MPL E 6
Physics: Uniform Motion A motorboat heads upstream a distance of 24 miles on a river whose current is running at 3 miles per hour (mi/hr). The trip up and back takes 6 hours. Assuming that the motorboat maintained a constant speed relative to the water, what was its speed?
Solution
See Figure 30. Use r to represent the constant speed of the motorboat relative to the water. Then the true speed going upstream is r - 3 mi/hr, and the true speed going Distance downstream is r + 3 mi/hr. Since Distance = Rate * Time, then Time = . Rate Set up Table 3:
Figure 30 24 miles r ⫺ 3 mi/hr r ⫹ 3 mi/hr
Rate (mi/hr)
Table 3
Distance (mi)
Distance Time = (hr) Rate
Upstream
r - 3
24
24 r - 3
Downstream
r + 3
24
24 r + 3
SECTION A.9 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
A77
The total time up and back is 6 hours, which gives the equation 24 24 + = 6 r - 3 r + 3 241r + 32 + 241r - 32 = 6 1r - 32 1r + 32
Add the quotients on the left.
48r = 6 r2 - 9
Simplify.
48r = 61r 2 - 92 6r 2 - 48r - 54 r 2 - 8r - 9 1r - 92 1r + 12 r = 9 or r
= = = =
0 0 0 -1
Multiply both sides by r 2 − 9. Place in standard form. Divide by 6. Factor. Apply the Zero-Product Property and solve.
Discard the solution r = - 1 mi/hr, and conclude that the speed of the motorboat relative to the water is 9 mi/hr.
Now Work
r
PROBLEM
27
5 Solve Constant Rate Job Problems This section involves jobs that are performed at a constant rate. The 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 Thus, if a job takes 4 hours, then ¼ of the job is done in 1 hour.
EXAM PL E 7
Solution Table 4 Hours to Do Job
Part of Job Done in 1 Hour
Danny
4
1 4
Mike
6
1 6
Together
t
1 t
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 help Danny. Assuming no gain or loss of efficiency, when will they finish if they work together? Can they make the golf date? 1 1 Set up Table 4. In 1 hour, Danny does of the job, and in 1 hour, Mike does of the 4 6 job. Let t be the time (in hours) that it takes them to do the job together. In 1 hour, 1 then, of the job is completed. Reason as follows: t a
Part done by Danny Part done by Mike Part done together b + a b = a b in 1 hour in 1 hour in 1 hour
From Table 4, 1 1 + 4 6 3 2 + 12 12 5 12 5t
= = = =
t =
1 t 1 t 1 t 12 12 5
The model LCD = 12 (left side)
Multiply both sides by 12t. Divide both sides by 5.
12 Working together, Danny and Mike can do the job in hours, or 2 hours, 5 24 minutes. They should make the golf date, since they will finish at 12:24 pm.
Now Work
r
PROBLEM
33
A78
APPENDIX A Review
A.9 Assess Your Understanding Concepts and Vocabulary 1. The process of using variables to represent unknown quantities and then finding relationships that involve these variables is referred to as mathematical modeling .
5. True or False If an object moves at an average speed r, the distance d covered in time t is given by the formula d = rt. True
2. The money paid for the use of money is interest .
6. 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. 100 - x
3. Objects that move at a constant rate are said to be in uniform motion . 4. True or False The amount charged for the use of principal for a given period of time is called the rate of interest. False
Applications and Extensions In Problems 7–16, translate each sentence into a mathematical equation. Be sure to identify the meaning of all symbols. *7. Geometry The area of a circle is the product of the number p and the square of the radius. A = pr 2 *8. Geometry The circumference of a circle is the product of the number p and twice the radius. C = 2pr *9. Geometry The area of a square is the square of the length of a side. A = s2 *10. Geometry The perimeter of a square is four times the length of a side. P = 4s *11. Physics Force equals the product of mass and acceleration. *12. Physics Pressure is force per unit area. P = F>A *13. Physics Work equals force times distance. W = Fd *14. Physics Kinetic energy is one-half the product of the mass and the square of the velocity. *15. Business The total variable cost C of manufacturing x dishwashers is $150 per dishwasher times the number of dishwashers manufactured. C = 150x *16. Business The total revenue R derived from selling x dishwashers is $250 per dishwasher times the number of dishwashers sold. R = 250x *17. Financial Planning Betsy, a recent retiree, requires $6000 per year in extra income. She has $50,000 to invest and can invest in B-rated bonds paying 15% per year or in a certificate of deposit (CD) paying 7% per year. How much money should Betsy invest in each to realize exactly $6000 in interest per year? $31,250 in bonds; $18,750 in CDs 18. Financial Planning After 2 years, Betsy (see Problem 17) finds that she will now require $7000 per year. Assuming that the remaining information is the same, how should the money be reinvested? $43,750 in bonds; $6250 in CDs
of the new blend. The selling price of the new blend is to be $4.50 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 Earl Grey tea and of the Orange Pekoe tea are required? Earl Grey: 75 lb; Orange Pekoe: 25 lb 22. Business: Blending Coffee A coffee manufacturer wants to market a new blend of coffee that sells for $3.90 per pound by mixing two coffees that sell for $2.75 and $5 per pound, respectively. What amounts of each coffee should be blended to obtain the desired mixture? [Hint: Assume that the total weight of the desired blend is 100 pounds.] Coffee I: 49 lb; Coffee II: 51 lb 23. Business: Mixing Nuts A nut store normally sells cashews for $9.00 per pound and almonds for $3.50 per pound. But at 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 60 pounds of almonds with some cashews and sell the mixture for $7.50 per pound. How many pounds of cashews should be mixed with the almonds to ensure no change in the profit? 160 lb of cashews 24. Business: Mixing Candy A candy store sells boxes of candy containing caramels and cremes. Each box sells for $12.50 and holds 30 pieces of candy (all pieces are the same size). If the caramels cost $0.25 to produce and the cremes cost $0.45 to produce, how many of each should be in a box to yield a profit of $3? 20 caramels and 10 cremes 25. Physics: Uniform Motion A motorboat can maintain a constant speed of 16 miles per hour relative to the water. The boat makes a trip upstream to a certain point in 20 minutes; the return trip takes 15 minutes. What is the speed of the current? See the figure. 2.286 mi/hr
19. Banking A bank loaned out $12,000, part of it at the rate of 8% per year and the rest at the rate of 18% per year. If the interest received totaled $1000, how much was loaned at 8%? $11,600 20. Banking Wendy, a loan officer at a bank, has $1,000,000 to lend and is required to obtain an average return of 18% per year. If she can lend at the rate of 19% or at the rate of 16%, how much can she lend at the 16% rate and still meet her requirement? $333,333.33 21. Blending Teas The manager of a store that specializes in selling tea decides to experiment with a new blend. She will mix some Earl Grey tea that sells for $5 per pound with some Orange Pekoe tea that sells for $3 per pound to get 100 pounds
26. Physics: Uniform Motion A motorboat heads upstream on a river that has a current of 3 miles per hour. The trip upstream takes 5 hours, and the return trip takes 2.5 hours. What is the speed of the motorboat? (Assume that the motorboat maintains a constant speed relative to the water.) 9 mi/hr
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
SECTION A.9 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications
40 30 TE DB
20
20 10
31. Tennis A regulation doubles tennis court has an area of 2808 square feet. If it is 6 feet longer than twice its width, determine the dimensions of the court. 78 ft by 36 ft Source: United States Tennis Association
30
30. High-Speed Walkways Toronto’s Pearson International Airport has a high-speed version of a moving walkway. If Liam walks while riding this moving walkway, he can travel 280 meters in 60 seconds less time than if he stands still on the moving walkway. If Liam walks at a normal rate of 1.5 meters per second, what is the speed of the high-speed walkway? 2 m/sec
[Hint: At time t = 0, the defensive back is 5 yards behind the tight end.]
40
29. Moving Walkways The speed of a moving walkway is typically about 2.5 feet per second. Walking on such a moving walkway, it takes Karen a total of 40 seconds to travel 50 feet with the movement of the walkway and then back again against the movement of the walkway. What is Karen’s normal walking speed? 4.05 ft/sec Source: Answers.com
37. Football A tight end can run the 100-yard dash in 12 seconds. 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 15-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? tight end’s 45-yd line
10
*28. Physics: Uniform Motion Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 am, each heading for Wildwood. One car’s average speed is 10 miles per hour more than the other’s. The faster car arrives at 1 Wildwood at 11:00 am, hour before the other car. What 2 was the average speed of each car? How far did each travel?
36. 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? 25.83 ft (d) Which pond has the greatest area? the circular one
SOUTH
27. Physics: Uniform Motion A motorboat maintained a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning. The total time for the trip was 1.5 hours. Use this information to find the speed of the current. 5 mi/hr
A79
*32. Laser Printers It takes an HP 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? Source: Hewlett-Packard
38. Computing Business Expense Therese, an outside salesperson, 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 deduct as a business expense? 20,000 mi
33. Working Together on a Job Trent can deliver his newspapers 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 worked together? 12 min
*39. Mixing Water and Antifreeze How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60% antifreeze?
34. Working Together on a Job Patrick, 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? 15 hr 35. 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? 10 ft by 5 ft (b) What is the area of the garden? 50 sq ft * (c) If the length and width of the garden are to be the same, what will be the dimensions of the garden? * (d) What will be the area of the square garden? 2 ft 2 ft
40. Mixing Water and Antifreeze The cooling system of a certain foreign-made car has a capacity of 15 liters. If the system is filled with a mixture that is 40% antifreeze, how much of this mixture should be drained and replaced by pure antifreeze so that the system is filled with a solution that is 60% antifreeze? 5 L *41. Chemistry: Salt Solutions How much water must be evaporated from 32 ounces of a 4% salt solution to make a 6% salt solution? 42. Chemistry: Salt Solutions How much water must be evaporated from 240 gallons of a 3% salt solution to produce a 5% salt solution? 96 gal *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, 18-karat 18 12 gold is , or 75% pure gold; 12-karat gold is , or 50% 24 24 pure gold; and so on. How much 12-karat gold should be mixed with pure gold to obtain 60 grams of 16-karat gold?
A80
APPENDIX A Review
44. Chemistry: Sugar Molecules A sugar molecule has twice as many atoms of hydrogen as of oxygen and one more atom of carbon than of oxygen. If a sugar molecule has a total of 45 atoms, how many are oxygen? How many are hydrogen? 11 oxygen; 22 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. Dan
Mike
Start 1– 4
1– 2
mi
mi
3– 4
mi
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 speed remains constant.) 742.5 mi 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 is the latest the auxiliary pump can be started so that the tanker is emptied by noon? 9:45 am
faucets closed and the stopper removed, the tub will empty in 20 minutes. How long will it take for the tub to fill if both faucets are open and the stopper is removed? 1 hr 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? 1hr, 45 min *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 speed is 6 miles per hour, and during your bicycle race, your average speed 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. Comparing Olympic Heroes In the 2012 Olympics, Usain Bolt of Jamaica won the gold medal in the 100-meter race with a time of 9.69 seconds. In the 1896 Olympics, Thomas Burke, 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 Bolt beat Burke? 19.25 m 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 = 2pr 2 + 2prh, where r is the radius and h is the height.] 3 in.
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? 5 lb 49. Emptying a Tub A bathroom tub will fill in 15 minutes with both faucets open and the stopper in place. With both
7 in.
39 oz.
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? $ 40; no profit 56. Critical Thinking Make up a word problem that requires solving a linear equation as part of its solution. Exchange problems 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 solution. What happens?
58. Computing Average Speed In 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. 49.5 mi/hr *59. Speed of a Plane On a recent flight from Phoenix to Kansas City, a distance of 919 nautical miles, the plane arrived 20 minutes 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 whether enough information is provided to find the tail wind? If possible, find the tail wind. (1 knot = 1 nautical mile per hour)
SECTION A.10 Interval Notation; Solving Inequalities
A81
A.10 Interval Notation; Solving Inequalities PREPARING FOR THIS SECTION Before getting started, review the following: r "MHFCSB&TTFOUJBMT "QQFOEJY" 4FDUJPO" ƭQQ"m"
Now Work the ‘Are You Prepared?’ problems on page A86.
OBJECTIVES 1 2 3 4
Use Interval Notation (p. A81) Use Properties of Inequalities (p. A82) Solve Inequalities (p. A84) Solve Combined Inequalities (p. A85)
Suppose that a and b are two real numbers and a 6 b. The notation a 6 x 6 b means that x is a number between a and b. The expression a 6 x 6 b is equivalent to the two inequalities a 6 x and x 6 b. Similarly, the expression a … x … b is equivalent to the two inequalities a … x and x … b. The remaining two possibilities, a … x 6 b and a 6 x … b, are defined similarly. Although it is acceptable to write 3 Ú x Ú 2, it is preferable to reverse the inequality symbols and write instead 2 … x … 3 so that the values go from smaller to larger, reading left to right. A statement such as 2 … x … 1 is false because there is no number x for which 2 … x and x … 1. Finally, never mix inequality symbols, as in 2 … x Ú 3.
1 Use Interval Notation Let a and b represent two real numbers with a 6 b.
DEFINITION
In Words The notation [a, b] represents all real numbers between a and b, inclusive. The notation (a, b) represents all real numbers between a and b, not including either a or b.
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 6 x 6 b. The half-open, or half-closed, intervals are (a, b], consisting of all real numbers x for which a 6 x … b, and [a, b), consisting of all real numbers x for which a … x 6 b.
In each of these definitions, a is called the left endpoint and b the right endpoint of the interval. The symbol q (read as “infinity”) is not a real number, but a notational device used to indicate unboundedness in the positive direction. The symbol - q (read as “negative infinity”) also is not a real number, but a notational device used to indicate unboundedness in the negative direction. The symbols q and - q , are used to define five other kinds of intervals: [a, ˆ ) Consists of all real numbers x for which x Ú (a, ˆ ) Consists of all real numbers x for which x 7 ( − ˆ , a] Consists of all real numbers x for which x ( − ˆ , a) Consists of all real numbers x for which x ( − ˆ , ˆ ) Consists of all real numbers x
a a … a 6 a
Note that q and - q are never included as endpoints, since neither is a real number. Table 5 on the following page summarizes interval notation, corresponding inequality notation, and their graphs.
A82
APPENDIX A Review
Table 5
EX A MPL E 1
Interval
Inequality
Graph
The open interval (a, b)
a⬍x⬍b
a
b
The closed interval [a, b]
aⱕxⱕb
a
b
The half-open interval [a, b)
aⱕx⬍b
a
b
The half-open interval (a, b]
a⬍xⱕb
a
b
The interval [a, ⬁)
xⱖa
a
The interval (a, ⬁)
x⬎a
a
The interval (⫺⬁, a]
xⱕa
The interval (⫺⬁, a)
x⬍a
The interval (⫺⬁, ⬁)
All real numbers
a a
Writing Inequalities Using Interval Notation Write each inequality using interval notation. (a) 1 … x … 3
Solution
(b) - 4 6 x 6 0
(c) x 7 5
(d) x … 1
(a) 1 … x … 3 describes all real numbers x between 1 and 3, inclusive, which is 3 1, 34 in interval notation. (b) In interval notation, - 4 6 x 6 0 is written 1 - 4, 02. (c) x 7 5 consists of all real numbers x greater than 5, which in interval notation, is 15, q 2 . (d) In interval notation, x … 1 is written 1 - q , 14 .
r
EX A MPL E 2
Writing Intervals Using Inequality Notation Write each interval as an inequality involving x. (a) 3 1, 42
Solution
(a) (b) (c) (d)
(b) 12, q 2
(c) 3 2, 34
(d) 1 - q , - 34
3 1, 42 consists of all real numbers x for which 1 … x 6 4. 12, q 2 consists of all real numbers x for which x 7 2. 3 2, 34 consists of all real numbers x for which 2 … x … 3. 1 - q , - 34 consists of all real numbers x for which x … - 3.
Now Work
PROBLEMS
11, 23,
AND
r
31
2 Use Properties of Inequalities 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 0. For any real number a, the value of a2 is 0 or positive; that is, a2 is nonnegative. This is called the nonnegative property.
Nonnegative Property
In Words The square of a real number is never negative.
For any real number a, a2 Ú 0
(1)
SECTION A.10 Interval Notation; Solving Inequalities
A83
If the same number is added to both sides of an inequality, an equivalent inequality results. For example, since 3 6 5, then 3 + 4 6 5 + 4 or 7 6 9. This is called the addition property of inequalities.
Addition Property of Inequalities For real numbers a, b, and c, If a 6 b, then a + c 6 b + c.
(2a)
If a 7 b, then a + c 7 b + c.
(2b)
The addition property states that the sense, or direction, of an inequality remains unchanged when the same number is added to each side. Now let’s see what happens when each side of an inequality is multiplied by a nonzero number. Begin with 3 6 7 and multiply each side by 2. The numbers 6 and 14 that result obey the inequality 6 6 14. Now start with 9 7 2 and multiply each side by - 4. The numbers - 36 and - 8 that result obey the inequality - 36 6 - 8. Note that the effect of multiplying both sides of 9 7 2 by the negative number - 4 is that the direction of the inequality symbol is reversed. This leads to the following general multiplication properties for inequalities:
Multiplication Properties for Inequalities For real numbers a, b, and c:
In Words Multiplying by a negative number reverses the inequality.
If a 6 b and if c 7 0, then ac 6 bc. If a 6 b and if c 6 0, then ac 7 bc.
(3a)
If a 7 b and if c 7 0, then ac 7 bc. If a 7 b and if c 6 0, then ac 6 bc.
(3b)
The multiplication properties state that the sense, or direction, of an inequality remains the same when each side is multiplied by a positive real number, whereas the direction is reversed when each side is multiplied by a negative real number.
EXAM PL E 3
Multiplication Property of Inequalities 1 1 (a) If 2x 6 6, then 12x2 6 162 , or x 6 3. 2 2 x x (b) If 7 12, then - 3a b 6 - 31122 , or x 6 - 36. -3 -3 - 4x -8 7 , or x 7 2. -4 -4 (d) If - x 7 8, then 1 - 12 1 - x2 6 1 - 12 182 , or x 6 - 8.
(c) If - 4x 6 - 8, then
Now Work
PROBLEM
r
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.
A84
APPENDIX A Review
Reciprocal Property for Inequalities If a 7 0, then
1 7 0. a
If
1 7 0, then a 7 0. a
(4a)
If a 6 0, then
1 6 0. a
If
1 6 0, then a 6 0. a
(4b)
3 Solve Inequalities An inequality in one variable is a statement involving two expressions, at least one containing the variable, separated by one of the inequality symbols 6 , … , 7 , or Ú . To solve an inequality means to find all values of the variable for which the statement is true. These values are solutions of the inequality. For example, the following are all inequalities involving one variable x: x + 1 x + 5 6 8 2x - 3 Ú 4 x2 - 1 … 3 7 0 x - 2 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 6 3, is obtained. Equivalent inequalities are obtained by applying some of the same properties as those used to find equivalent equations. The addition property and the multiplication properties form the basis for the following procedures.
Procedures That Leave the Inequality Symbol Unchanged 1. Simplify both sides of the inequality by combining like terms and eliminating parentheses: is equivalent to
1x + 22 + 6 7 2x + 51x + 12 x + 8 7 7x + 5
2. Add or subtract the same expression on both sides of the inequality: is equivalent to
3x - 5 6 4 13x - 52 + 5 6 4 + 5
3. Multiply or divide both sides of the inequality by the same positive expression: 4x 7 16 is equivalent to
4x 16 7 4 4
Procedures That Reverse the Sense or Direction of the Inequality Symbol 1. Interchange the two sides of the inequality: 3 6 x is equivalent to x 7 3 2. Multiply or divide both sides of the inequality by the same negative expression: - 2x 7 6 is equivalent to
- 2x 6 6 -2 -2
As the examples that follow illustrate, inequalities can be solved by using many of the same steps that would be used to solve equations. In writing the solution of an inequality, either set notation or interval notation may be used, whichever is more convenient.
SECTION A.10 Interval Notation; Solving Inequalities
EXAM PL E 4
A85
Solving an Inequality Solve the inequality 4x + 7 Ú 2x - 3, and graph the solution set. 4x + 7 Ú 2x - 3
Solution
4x + 7 - 7 Ú 2x - 3 - 7 4x Ú 2x - 10 4x - 2x Ú 2x - 10 - 2x
⫺6
⫺5
⫺4
⫺3
⫺2
⫺1
Simplify. Subtract 2x from both sides.
2x Ú - 10
Simplify.
2x - 10 Ú 2 2
Divide both sides by 2. (The direction of the inequality symbol is unchanged.)
x Ú -5 Figure 31
Subtract 7 from both sides.
Simplify.
The solution set is 5 x 0 x Ú - 56 or, using interval notation, all numbers in the interval 3 - 5, q 2. See Figure 31 for the graph.
r
Now Work
PROBLEM
57
4 Solve Combined Inequalities EXAM PL E 5
Solving Combined Inequalities Solve the inequality - 5 6 3x - 2 6 1, and graph the solution set.
Solution
Recall that the inequality - 5 6 3x - 2 6 1 is equivalent to the two inequalities - 5 6 3x - 2 and 3x - 2 6 1 Solve each of these inequalities separately. - 5 6 3x - 2 - 5 + 2 6 3x - 2 + 2 - 3 6 3x -3 3x 6 3 3 -1 6 x
3x - 2 6 1 Add 2 to both sides.
3x - 2 + 2 6 1 + 2
Simplify. Divide both sides by 3. Simplify.
3x 6 3 3x 3 6 3 3 x 6 1
The solution set of the original pair of inequalities consists of all x for which - 1 6 x and x 6 1 Figure 32 ⫺3
⫺2
⫺1
0
1
2
This may be written more compactly as 5 x - 1 6 x 6 16 . In interval notation, the solution is 1 - 1, 12. See Figure 32 for the graph.
r
A86
APPENDIX A Review
Observe in the preceding process that solving the two inequalities required exactly the same steps. A shortcut to solving the original inequality algebraically is to deal with the two inequalities at the same time, as follows: -5 -5 + 2 -3 -3 3 -1
Now Work
EX A MPL E 6
Solution
6 3x - 2 6 1 6 3x - 2 + 2 6 1 + 2 6 3x 6 3 3x 3 6 6 3 3 6 x 6 1 PROBLEM
Add 2 to each part. Simplify. Divide each part by 3. Simplify.
73
Using the Reciprocal Property to Solve an Inequality
Solve the inequality 14x - 12 -1 7 0, and graph the solution set. Recall that 14x - 12 -1 = then a 7 0. Hence,
1 1 . The Reciprocal Property states that when 7 0, a 4x - 1
14x - 12 -1 1 4x - 1 4x - 1 4x
7 0 7 0
7 0 Reciprocal Property 7 1 1 x 7 4
The solution set is b x 0 x 7
Figure 33 0
1 – 4
1
1 1 r —that is, all x in the interval a , q b . Figure 33 4 4
r
illustrates the graph.
Now Work
PROBLEM
83
A.10 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 Ú - 2. (pp. A4–A5)
2. Graph the inequality: x 6 1. (pp. A4–A5)
Concepts and Vocabulary 3. If each side of an inequality is multiplied by a(n) negative number, then the direction of the inequality symbol is reversed. 4. A(n) closed interval , denoted 3a, b4, consists of all real numbers x for which a … x … b.
5. The Multiplication Property states that the sense, or direction, of an inequality remains the same if each side is multiplied by a positive number, whereas the direction is reversed if each side is multiplied by a negative number.
In Problems 6–9, determine whether the statement is True or False if a 6 b and c 6 0. 6. a + c 6 b + c
True
7. a - c 6 b - c
True
8. ac 7 bc
10. True or False The square of any real number is always nonnegative. True
True
9.
b a 6 c c
False
A87
SECTION A.10 Interval Notation; Solving Inequalities
Skill Building In Problems 11–16, express the graph shown in blue using interval notation. Also express each as an inequality involving x. *11.
*12.
–1
0
1
2
3
–2
–1
0
1
2
*14.
*15.
–2
–1
0
1
2
–1
0
1
2
3
*13.
–1
0
1
2
3
–1
0
1
2
3
*16.
In Problems 17–22, an inequality is given. Write the inequality obtained by: (a) Adding 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. 3 6 5
*18. 2 7 1
*19. 4 7 - 3
*20. - 3 7 - 5
*21. 2x + 1 6 2
*22. 1 - 2x 7 5
In Problems 23–30, write each inequality using interval notation, and illustrate each inequality using the real number line. *23. 0 … x … 4 *27. x Ú 4
*24. - 1 6 x 6 5
[0, 4]
34, q )
*28. x … 5
( - 1, 5)
( - q , 54
*25. 4 … x 6 6
*26. - 2 6 x 6 0
[4, 6)
( - q , - 4)
*29. x 6 - 4
*30. x 7 1
( - 2, 0)
(1, q )
In Problems 31–38, write each interval as an inequality involving x, and illustrate each inequality using the real number line. *31. 32, 54
*35. 34, q 2
*32. 11, 22
2 … x … 5
*36. 1 - q , 24
x Ú 4
*33. 1 - 3, - 22
1 6 x 6 2
*37. 1 - q , - 32
x … 2
*34. 30, 12
-3 6 x 6 -2
0 … x 6 1
*38. 1 - 8, q 2
x 6 -3
x 7 -8
In Problems 39–52, fill in the blank with the correct inequality symbol. 39. If x 6 5, then x - 5
6
0.
40. If x 6 - 4, then x + 4
42. If x 7 6, then x - 6
7
0.
43. If x Ú - 4, then 3x
45. If x 7 6, then - 2x
6
- 12.
Ú
12.
49. If 2x 7 6, then x
1 x … 3, then x 2
Ú
- 6.
52. If -
0.
Ú
46. If x 7 - 2, then - 4x
48. If x … - 4, then - 3x 51. If -
6
- 12. 6
7
1 x 7 1, then x 4
8. 3.
6
41. If x 7 - 4, then x + 4 44. If x … 3, then 2x
7 …
6.
47. If x Ú 5, then - 4x
…
50. If 3x … 12, then x
…
- 4.
In Problems 53–88, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set. *53. x + 1 6 5
*54. x - 6 6 1
*55. 1 - 2x … 3
*56. 2 - 3x … 5
*57. 3x - 7 7 2
*58. 2x + 5 7 1
*59. 3x - 1 Ú 3 + x
*60. 2x - 2 Ú 3 + x
*61. - 21x + 32 6 8
*62. - 311 - x2 6 12
*63. 4 - 311 - x2 … 3
*64. 8 - 412 - x2 … - 2x
1 1x - 42 7 x + 8 2 x x *68. Ú 2 + 3 6
*65.
*71. - 5 … 4 - 3x … 2 *74. 0 6
3x + 2 6 4 2
*66. 3x + 4 7
1 1x - 22 3
*70. 4 … 2x + 2 … 10
*72. - 3 … 3 - 2x … 9
*73. - 3 6
*75. 1 6 1 -
1 x 6 4 2
*80. x19x - 52 … 13x - 12 2
*81.
*83. 14x + 22 -1 6 0
*84. 12x - 12 -1 7 0
1 x + 1 3 … 6 2 3 4
*87. 0 6 12x - 42 -1 6
2x - 1 6 0 4
*76. 0 6 1 -
*78. 1x - 12 1x + 12 7 1x - 32 1x + 42
4 2 6 x 3
x x Ú 1 2 4
*69. 0 … 2x - 6 … 4
*77. 1x + 22 1x - 32 7 1x - 12 1x + 12
*86. 0 6
*67.
1 x + 1 2 6 … 3 2 3
*85. 0 6 1 2
1 x 6 1 3
*79. x14x + 32 … 12x + 12 2 *82.
2 3 6 x 5
*88. 0 6 13x + 62 -1 6
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
1 3
0.
- 20. 4.
A88
APPENDIX A Review
Applications and Extensions 89. What is the domain of the variable in the expression 2 3x + 6 ? {x x Ú - 2} 90. What is the domain of the variable in the expression 2 8 + 2x ? {x x Ú - 4} 91. A young adult may be defined as someone older than 21 and less than 30 years of age. Express this statement using inequalities. 21 6 age 6 30 92. Middle-aged may be defined as being 40 or more and less than 60. Express this statement using inequalities. 40 … age 6 60 93. Life Expectancy According to the National Center for Health Statistics, an average 30-year-old male in 2008 could expect to live at least 47.3 more years, and an average 30-year-old female in 2008 could expect to live at least 51.6 more years. (a) To what age could an average 30-year-old male expect to live? Express your answer as an inequality. male Ú 77.3 yr (b) To what age could an average 30-year-old female expect to live? Express your answer as an inequality. female Ú 81.6 yr (c) Who could expect to live longer, a male or a female? By how many years? female; 4.3 yr Source: National Vital Statistics Reports, Vol. 61, No. 3, September 2012.
JAN 2008
AUG 2059
APR 2055
94. 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? 1600 cm3 to 2400 cm3, inclusive *95. 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. Assuming that the complex will sell at some price between $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 the selling price? *96. Sales Commission A used-car salesperson is paid a commission 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 salesperson expect the commission to vary? *97. Federal Tax Withholding The percentage method of withholding for federal income tax (2013) states that a single person whose weekly wages, after subtracting withholding allowances, are over $739, but not over $1732, shall have $95.95 plus 25% of the excess over $739 withheld. Over
what range does the amount withheld vary if the weekly wages vary from $800 to $1000, inclusive? Source: Employer’s Tax Guide. Department of the Treasury, Internal Revenue Service, Publication 2013. *98. Exercising Sue wants to lose weight. For healthy weight loss, the American College of Sports Medicine (ACSM) recommends 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? *99. Electricity Rates Commonwealth Edison Company’s charge for electricity in January 2013 is 10.62¢ per kilowatt-hour. In addition, each monthly bill contains a customer charge of $13.04. If last year’s bills ranged from a low of $82.07 to a high of $278.54, over what range did usage vary (in kilowatthours)? Source: Commonwealth Edison Co., January 2013. * 100. Water Bills The Village of Oak Lawn charges homeowners $52.74 per quarter-year, plus $5.37 per 1000 gallons for water usage in excess of 10,000 gallons. In 2013 one homeowner’s quarterly bill ranged from a high of $165.51 to a low of $101.07. Over what range did water usage vary? Source: Village of Oak Lawn, Illinois, January 2013. * 101. Markup of a New Car The markup over dealer’s cost of a new car ranges from 12% to 18%. If the sticker price is $18,000, over what range will the dealer’s cost vary? * 102. 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 1.96s above the average, where s (sigma, a number called the standard deviation) depends on the nature of the test. If s = 12 for this test and there is (in principle) no upper limit to the score possible on the test, write the interval of possible test scores of the people in the top 2.5%. 103. 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. at least a 74 (b) What score do you need if the fifth test counts double? at least a 77 What do I need to get a B?
82 68 89 87
* 104. “Light” Foods For food products to be labeled “light,” the U.S. Food and Drug Administration requires that the altered product must either contain at least one-third calories of the regular product or contain at least one-half less fat than the regular product. If a serving of Miracle Whip® Light contains 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®?
SECTION A.11 Complex Numbers
* 105. Arithmetic Mean If a 6 b, show that a 6
a + b 6 b. 2
* 109. Harmonic Mean For 0 6 a 6 b, let h be defined by 1 1 1 1 = a + b h 2 a b
a + b The number is called the arithmetic mean of a and b. 2 * 106. Refer to Problem 105. Show that the arithmetic mean of a and b is equidistant from a and b. * 107. Geometric Mean If 0 6 a 6 b, show that a 6 1 ab 6 b. The number 1 ab is called the geometric mean of a and b. * 108. Refer to Problems 105 and 107. Show that the geometric mean of a and b is less than the arithmetic mean of a and b.
A89
Show that a 6 h 6 b. The number h is called the harmonic mean of a and b. * 110. Refer to Problems 105, 107, and 109. Show that the harmonic mean of a and b equals the geometric mean squared divided by the arithmetic mean. * 111. Another Reciprocal Property Prove that if 0 6 a 6 b, then 1 1 6 . 0 6 b a
Discussion and Writing 112. Make up an inequality that has no solution. Make up one that has exactly one solution. * 113. The inequality x 2 + 1 6 - 5 has no real solution. Explain why. 114. 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.
115. How would you explain to a fellow student the underlying reason for the multiplication properties for inequalities (page A83)? That is, the sense (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.
‘Are You Prepared?’ Answers 1.
⫺4
⫺2
2.
0
–1
0
1
A.11 Complex Numbers PREPARING FOR THIS SECTION Before getting started, review the following: r $MBTTJGJDBUJPOPG/VNCFST "QQFOEJY" Section A.1, p. A3)
r 3BUJPOBMJ[JOH%FOPNJOBUPST "QQFOEJY" Section A.7, pp. A57–A58)
Now Work the ‘Are You Prepared?’ problems on page A94.
OBJECTIVE 1 Add, Subtract, Multiply, and Divide Complex Numbers (p. A90)
Complex Numbers One property of a real number is that its square is nonnegative (greater than or equal to 0). For example, there is no real number x for which x2 = - 1 The introduction of the imaginary unit will remedy this situation.
DEFINITION
The imaginary unit, denoted by i, is the number whose square is - 1; that is, i2 = - 1 This should not be a surprise. 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 and , the rational numbers. If our 2 3 universe were to consist only of rational numbers, there would be no x whose square
A90
APPENDIX A Review
equals 2. That is, there would be no number x for which x2 = 2. To remedy this, 3 mathematicians introduced numbers such as 2 2 and 2 5, the irrational numbers. Recall that the real numbers consist of the rational numbers and the irrational numbers. Now, if our universe were to consist only of real numbers, there would be no number x whose square is - 1. To remedy this, mathematicians introduced a number i, whose square is - 1. In the progression outlined, each time that a situation was encountered that was unsuitable, a new number system was introduced to remedy the situation. And each new number system contained the earlier number system as a subset. The number system that results from introducing the number i is called the complex number system.
DEFINITION
Complex numbers are numbers of the form a + bi, 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. For example, the complex number - 5 + 6i has the real part - 5 and the imaginary part 6. When a complex number is written in the form a + bi, where a and b are real numbers, it is in standard form. However, if the imaginary part of a complex number is negative, such as in the complex number 3 + 1 - 22i, then it is written in the form 3 - 2i instead. Also, the complex number a + 0i is usually written simply 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, Subtract, Multiply, and Divide Complex Numbers Equality, addition, subtraction, and multiplication of complex numbers are defined so as to preserve the familiar rules of algebra for real numbers. Thus, two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.
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 imaginary parts.
Sum of Complex Numbers 1a + bi2 + 1c + di2 = 1a + c2 + 1b + d2i
(2)
To subtract two complex numbers, use this rule:
Difference of Complex Numbers 1a + bi2 - 1c + di2 = 1a - c2 + 1b - d2i
EX A MPL E 1
(3)
Adding and Subtracting Complex Numbers
(a) 13 + 5i2 + 1 - 2 + 3i2 = 3 3 + 1 - 22 4 + 15 + 32i = 1 + 8i (b) 16 + 4i2 - 13 + 6i2 = 16 - 32 + 14 - 62 i = 3 + 1 - 22i = 3 - 2i
Now Work
PROBLEM
13
Products of complex numbers are calculated as illustrated in Example 2.
r
SECTION A.11 Complex Numbers
EXAM PL E 2
A91
Multiplying Complex Numbers
15 + 3i2 # 12 + 7i2 = 512 + 7i2 + 3i12 + 7i2 = 10 + 35i + 6i + 21i 2 = 10 + 41i + 211 - 12 = - 11 + 41i
Distributive Property Distributive Property i 2 =−1
r
Based on the procedure of Example 2, the product of two complex numbers is defined by the following formula:
Product of Complex Numbers 1a + bi2 # 1c + di2 = 1ac - bd2 + 1ad + bc2i
(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 binomials, as in Example 2, remembering that i 2 = - 1. For example, 12i2 12i2 = 4i 2 = - 4
12 + i2 11 - i2 = 2 - 2i + i - i 2 = 3 - i
Now Work
PROBLEM
19
Algebraic properties for addition and multiplication, such as the Commutative, Associative, and Distributive Properties, hold for complex numbers. However, the property that every nonzero complex number has a multiplicative inverse, or reciprocal, requires a closer look.
Conjugates DEFINITION
If z = a + bi is a complex number, then its conjugate, denoted by z, is defined as z = a + bi = a - bi For example, 2 + 3i = 2 - 3i and - 6 - 2i = - 6 + 2i.
EXAM PL E 3
Multiplying a Complex Number by Its Conjugate Find the product of the complex number z = 3 + 4i and its conjugate z.
Solution
Since z = 3 - 4i,
zz = 13 + 4i2 13 - 4i2 = 9 - 12i + 12i - 16i 2 = 9 + 16 = 25
r
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 zz = a2 + b2
(5)
Proof If z = a + bi, then
zz = 1a + bi2 1a - bi2 = a2 - abi + abi - 1bi2 2 = a2 - b2 i 2 = a2 + b2
■
A92
APPENDIX A Review
To express the reciprocal of a nonzero complex number z in standard form, 1 multiply the numerator and denominator of by its conjugate z. That is, if z z = a + bi is a nonzero complex number, then 1 1 z a - bi a b 1 z = = # = = = 2 - 2 i z z z a + bi zz c a2 + b2 a + b2 a + b2 Use (5).
EX A MPL E 4
Writing the Reciprocal of a Complex Number in Standard Form 1 in standard form a + bi; that is, find the reciprocal of 3 + 4i. 3 + 4i 1 Multiply the numerator and denominator of by the conjugate of 3 + 4i, the 3 + 4i complex number 3 - 4i. The result is Write
Solution
1 1 # 3 - 4i 3 - 4i 3 4 = = = i 3 + 4i 3 + 4i 3 - 4i 9 + 16 25 25
r
To express the quotient of two complex numbers in standard form, multiply the numerator and denominator of the quotient by the conjugate of the denominator.
EX A MPL E 5
Writing the Quotient of Complex Numbers in Standard Form Write each of the following in standard form. (a)
Solution
(a)
(b)
1 + 4i 5 - 12i
2 - 3i 4 - 3i
1 + 4i 1 + 4i # 5 + 12i 5 + 12i + 20i + 48i 2 = = 5 - 12i 5 - 12i 5 + 12i 25 + 144 - 43 + 32i 43 32 = = + i 169 169 169 2 - 3i 2 - 3i # 4 + 3i 8 + 6i - 12i - 9i 2 17 - 6i 17 6 = = = = i 4 - 3i 4 - 3i 4 + 3i 16 + 9 25 25 25
Now Work
EX A MPL E 6
(b)
PROBLEM
27
r
Writing Other Expressions in Standard Form If z = 2 - 3i and w = 5 + 2i, write each of the following expressions in standard form. z (a) (c) z + z (b) z + w w
Solution
(a)
12 - 3i2 15 - 2i2 z 10 - 4i - 15i + 6i 2 z#w = = # = w w w 15 + 2i2 15 - 2i2 25 + 4 4 - 19i 4 19 = = i 29 29 29
(b) z + w = 12 - 3i2 + 15 + 2i2 = 7 - i = 7 + i (c) z + z = 12 - 3i2 + 12 + 3i2 = 4
r
The conjugate of a complex number has certain general properties that will be useful later. For a real number a = a + 0i, the conjugate is a = a + 0i = a - 0i = a.
SECTION A.11 Complex Numbers
THEOREM
A93
The conjugate of a real number is the real number itself. Other properties that are direct consequences of the definition of the conjugate 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)
The conjugate of the sum of two complex numbers equals the sum of their conjugates. z + w = z + w
(7)
The conjugate of the product of two complex numbers equals the product of their conjugates. z#w = z#w
(8)
The proofs of equations (6), (7), and (8) are left as exercises. See Problems 60–62.
Powers of i The powers of i follow a pattern that is useful to know.
i5 = i4 # i = 1 # i = i
i1 = i i2 = - 1
i6 = i4 # i2 = - 1
i 4 = i 2 # i 2 = 1 - 12 1 - 12 = 1
i8 = i4 # i4 = 1
i3 = i2 # i = - i
i7 = i4 # i3 = - i
And so on. The powers of i repeat with every fourth power.
EXAM PL E 7
Evaluating Powers of i (a) i 27 = i 24 # i 3 = 1i 4 2
# i3 = 16 # i3 = - i 25 1i 4 2 # i = 125 # i = i 6
(b) i 101 = i 100 # i 1 =
EXAM PL E 8
Solution
r
Writing the Power of a Complex Number in Standard Form Write 12 + i2 3 in standard form.
Use the special product formula for 1a + b2 3. 1a + b2 3 = a3 + 3a2 b + 3ab2 + b3
Using this special product formula,
NOTE: If you did not remember the special product formula for (a ⫹ b)3, you could find (2 ⫹ i)3 by simplifying (2 ⫹ i)2 (2 ⫹ i ). 䊏
12 + i2 3 = 23 + 3 # 22 # i + 3 # 2 # i 2 + i 3 = 8 + 61 - 12 + 12i + 1 - i2 = 2 + 11i
Now Work
PROBLEMS
33
AND
41
r
A94
APPENDIX A Review
Square Roots of Negative Numbers Because i 2 = - 1, the square root of a negative number can be defined.
DEFINITION
If N is a positive real number, then the principal square root of −N, denoted by 2 −N, −N N , is 2 - N = 2Ni 2Ni 2N where i is the imaginary unit and i 2 = - 1.
EX A MPL E 9
Evaluating the Square Root of a Negative Number (a) 2 - 1 = 21i = i
(b) 2 - 16 = 216i = 4i
r
(c) 2 - 8 = 28i = 222i
Now Work Problem 47
A.11 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. Name the integers and the rational numbers in the set 6 e - 3, 0, 22, , p f. (p. A3) 5 2. True or False Rational numbers and irrational numbers are in the set of real numbers. (p. A3) True
3. Rationalize the denominator of 312 - 232
3 2 + 23
. (pp. A57–A58)
Concepts and Vocabulary 4. In the complex number 5 + 2i, the number 5 is called the real part; the number 2 is called the imaginary part; the number i is called the imaginary unit .
6. True or False The conjugate of 2 + 5i is - 2 - 5i.
False
7. True or False All real numbers are complex numbers. True 8. True or False The product of a complex number and its conjugate is a nonnegative real number. True
5. i 2 = - 1 ; i 3 = - i ; i 4 = 1
Skill Building In Problems 9–46, write each expression in the standard form a + bi. 9. 12 - 3i2 + 16 + 8i2
12. 13 - 4i2 - 1 - 3 - 4i2 15. 312 - 6i2
27.
6 - i 1 + i
30. a
33. i 23 36. i
-23
37
5 12 + i 13 13 5 7 - i 2 2
1 2 23 - ib 2 2 -i
13. 12 - 5i2 - 18 + 6i2
1 23 i 2 2
- 6 - 11i
- 8 - 32i
11. 1 - 3 + 2i2 - 14 - 4i2
14. 1 - 8 + 4i2 - 12 - 2i2 17. 2i12 - 3i2
20. 15 + 3i2 12 - i2
22. 1 - 3 + i2 13 + i2
- 10
23.
10 3 - 4i
26.
2 - i - 2i
25.
2 + i i
1 - 2i
28.
2 + 3i 1 - i
-
31. 11 + i2 2
6
- 5
1 23 2 + ib 2 2
2i
32. 11 - i2 2
-6 3 - 4i
13 + i
1 + i 2
29. a
40. 4i 3 - 2i 2 + 1
- 10 + 6i
6 8 + i 5 5
1 5 + i 2 2
-1
- 7 + 6i
6 + 4i
10 - 5i
37. i - 10i
- 4 + 7i
19. 13 - 4i2 12 + i2
34. i 14
i
39. 6i 3 - 4i 5
10. 14 + 5i2 + 1 - 8 + 2i2
16. - 412 + 8i2
- 12 - 9i
21. 1 - 6 + i2 1 - 6 - i2 13 5 - 12i
6
6 - 18i
18. 3i1 - 3 + 4i2
24.
8 + 5i
35. i -15
i
38. 4 + i
3
- 2i
4 - i
-
1 23 + i 2 2
SECTION A.11 Complex Numbers
41. 11 + i2 3
44. 2i 11 + i 2 4
2
42. 13i2 4 + 1
- 2 + 2i
45. i
0
6
+ i
4
43. i 7 11 + i 2 2
82
+ i + 1 2
46. i
0
+ i
7
0
+ i + i
5
A95
3
0
In Problems 47–52, perform the indicated operations and express your answer in the standard form a + bi. 47. 2- 4
48. 2- 9
2i
50. 2- 64
51. 213 + 4i2 14i - 32
8i
49. 2- 25
3i
5i
52. 214 + 3i2 13i - 42
5i
5i
Applications and Extensions In Problems 53–56, z = 3 - 4i and w = 8 + 3i. Write each expression in the standard form a + bi. 53. z + z
6
54. w - w
55. zz
6i
57. 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, V through the elements. That is, Z = . If the voltage across I a circuit element is 18 + i volts and the current through the element is 3 - 4i amperes, determine the impedance. 2 + 3i *58. Parallel Circuits In an ac circuit with two parallel pathways, the total impedance Z, in ohms, satisfies the formula 1 1 1 = + , where Z1 is the impedance of the first Z Z1 Z2
56. z - w
25
- 5 + 7i
pathway and Z2 is the impedance of the second pathway. Determine the total impedance if the impedances of the two pathways are Z1 = 2 + i ohms and Z2 = 4 - 3i ohms. *59. Use z = a + bi to show that z + z = 2a and that z - z = 2bi. *60. Use z = a + bi to show that z = z. *61. Use z = a + bi z + w = z + w.
and
w = c + di
to
show
that
*62. Use z = a + bi and w = c + di to show that z # w = z # w.
Discussion and Writing 63. Explain to a friend how you would add two complex numbers and how you would multiply two complex numbers. Explain any differences between the two explanations. 64. Write a brief paragraph that compares the method used to rationalize denominators and the method used to write the quotient of two complex numbers in standard form.
66. Explain how the method of multiplying two complex numbers is related to multiplying two binomials. *67. What Went Wrong? A student multiplied 2- 9 and 2- 9 as follows:
65. Use an Internet search engine to investigate the origins of complex numbers. Write a paragraph describing what you find, and present it to the class.
2- 9
# 2- 9 =
= 281 = 9 The instructor marked the problem incorrect. Why?
‘Are You Prepared?’ Answers 6 1. Integers: 5 - 3, 06; rational numbers: e - 3, 0, f 5
2. True
2( - 9)( - 9)
3. 3(2 - 13)
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
Appendix
B
Graphing Utilities Outline B.1 The Viewing Rectangle B.2 Using a Graphing Utility to Graph Equations B.3 Using a Graphing Utility to Locate Intercepts
and Check for Symmetry B.4 Using a Graphing Utility to Solve Equations B.5 Square Screens
B.6 Using a Graphing Utility to Graph
B.9 Using a Graphing Utility to Graph Parametric
Inequalities B.7 Using a Graphing Utility to Solve Systems of Linear Equations B.8 Using a Graphing Utility to Graph a Polar Equation
Equations
B.1 The Viewing Rectangle Figure 1 y = 2x
Figure 2
All graphing utilities—that is, all graphing calculators and all computer software graphing packages—graph equations by plotting points on a screen. The screen itself actually consists of small rectangles called pixels. The more pixels the screen has, the better the resolution. Most graphing calculators have 2048 pixels per square inch; most computer screens have 4096 to 8192 pixels per square inch. When a point to be plotted lies inside a pixel, the pixel is turned on (lights up). The graph of an equation is a collection of pixels. Figure 1 shows how the graph of y = 2x looks on a TI-84 Plus graphing calculator. The screen of a graphing utility will display the coordinate axes of a rectangular coordinate system. However, the scale must be set on each axis. The smallest and largest values of x and y to be included in the graph must also be set. This is called setting the viewing rectangle or viewing window. Figure 2 illustrates a typical viewing window. To select the viewing window, values must be given to the following expressions: Xmin: Xmax: Xscl: Ymin: Ymax: Yscl:
the smallest value of x the largest value of x the number of units per tick mark on the x-axis the smallest value of y the largest value of y the number of units per tick mark on the y-axis
Figure 3 illustrates these settings and their relation to the Cartesian coordinate system.
Figure 3
y Ymax Yscl
x
Xmin
Xmax Xscl
Ymin
B1
B2
APPENDIX B Graphing Utilities
If the scale used on each axis is known, the minimum and maximum values of x and y shown on the screen can be determined by counting the tick marks. Look again at Figure 2 on the previous page. For a scale of 1 on each axis, the minimum and maximum values of x are - 10 and 10, respectively, the minimum and maximum values of y are also - 10 and 10. If the scale is 2 on each axis, then the minimum and maximum values of x are - 20 and 20, respectively, and the minimum and maximum values of y are - 20 and 20, respectively. Conversely, if the minimum and maximum values of x and y are known, then the scales can be found by counting the tick marks displayed. This text follows the practice of showing the minimum and maximum values of x and y in illustrations so that the reader will know how the viewing window was set. See Figure 4. Figure 4
4
⫺3
3
means
Xmin = - 3 Xmax = 3 Xscl = 1
Ymin = - 4 Ymax = 4 Yscl = 2
⫺4
Finding the Coordinates of a Point Shown on a Graphing Utility Screen
EX A MPL E 1 Figure 5
Find the coordinates of the point shown in Figure 5. Assume that the coordinates are integers.
4
Solution First note that the viewing window used in Figure 5 is
⫺3
3
Xmin = - 3 Xmax = 3 Xscl = 1
⫺4
Ymin = - 4 Ymax = 4 Yscl = 2
The point shown is 2 tick units to the left on the horizontal axis 1scale = 12 and 1 tick unit up on the vertical axis 1scale = 22. The coordinates of the point shown are 1 - 2, 22.
r
B.1 Exercises In Problems 1–4, determine the coordinates of the points shown. Tell in which quadrant each point lies. Assume that the coordinates are integers. 1.
2.
10
⫺5
5
⫺10 ( - 1, 4); II
3.
10
⫺5
⫺5
5
4.
5
⫺10
5
⫺10 (3, 4); I
10
10
⫺10 ( - 6, - 4); III
⫺5 (3, 1); I
In Problems 5–10, determine the viewing window used. *5.
*6.
4
⫺6
6
*7.
2
⫺3
3
3 ⫺6
⫺4
⫺2
6 ⫺1
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
SECTION B.2 Using a Graphing Utility to Graph Equations
*8.
*9.
4 ⫺9
*10.
10
B3
8
9
⫺12
3
9
⫺22
⫺10 4
2
In Problems 11–16, select a setting such that each of the given points will lie within the viewing rectangle. *11. 1 - 10, 52, 13, - 22, 14, - 12
*14. 1 - 80, 602, 120, - 302, 1 - 20, - 402
*12. 15, 02, 16, 82, 1 - 2, - 32
*15. 10, 02, 1100, 52, 15, 1502
*13. 140, 202, 1 - 20, - 802, 110, 402 *16. 10, - 12, 1100, 502, 1 - 10, 302
B.2 Using a Graphing Utility to Graph Equations From Examples 2 and 3 in Foundations, Section 2, recall that a graph can be obtained by plotting points in a rectangular coordinate system and connecting them. Graphing utilities perform these same steps when graphing an equation. For example, the TI-84 Plus determines 95 evenly spaced input values,* starting at Xmin and ending at Xmax, uses the equation to determine the output values, plots these points on the screen, and finally (if in the connected mode) draws a line between consecutive points. To graph an equation in two variables x and y using a graphing utility requires that the equation be written in the form y = 5 expression in x6 . If the original equation is not in this form, replace it by equivalent equations until the form y = 5 expression in x6 is obtained.
Steps for Graphing an Equation Using a Graphing Utility STEP 1: Solve the equation for y in terms of x. STEP 2: Get into the graphing mode of the graphing utility. The screen will usually display Y1 = , prompting the user to enter the expression involving x that was found in Step 1. (Consult your manual for the correct way to enter the expression. For example, y = x2 might be entered as x ¿ 2 or as x*x or as x xY 2). STEP 3: Select the viewing window. Without prior knowledge about the behavior of the graph of the equation, it is common to select the standard viewing window** initially. The viewing window is then adjusted based on the graph that appears. In this text the standard viewing window is Xmin = - 10 Xmax = 10 Xscl = 1
Ymin = - 10 Ymax = 10 Yscl = 1
STEP 4: Graph. STEP 5: Adjust the viewing window until a complete graph is obtained.
*These input values depend on the values of Xmin and Xmax. For example, if Xmin = - 10 and Xmax = 10, 10 - 1 - 102 then the first input value will be - 10 and the next input value will be - 10 + = - 9.7872, 94 and so on. **Some graphing utilities have a ZOOM-STANDARD feature that automatically sets the viewing window to the standard viewing window and graphs the equation.
B4
APPENDIX B Graphing Utilities
Graphing an Equation on a Graphing Utility
EX A MPL E 1
Graph the equation: 6x2 + 3y = 36
Solution
STEP 1: Solve for y in terms of x. 6x2 + 3y = 36 3y = - 6x2 + 36 Subtract 6x2 from both sides of the equation. y = - 2x2 + 12 Divide both sides of the equation by 3 and simplify. STEP 2: STEP 3: STEP 4: STEP 5:
From the Y1 = screen, enter the expression - 2x2 + 12 after the prompt. Set the viewing window to the standard viewing window. Graph. The screen should look like Figure 6. The graph of y = - 2x2 + 12 is not complete. The value of Ymax must be increased so that the top portion of the graph is visible. Increasing the value of Ymax to 12 gives the graph in Figure 7. The graph is now complete. Figure 6
Figure 7 10
⫺10
Figure 8
12
10
⫺10
10
12
⫺10 ⫺4
⫺10
r
4
Look again at Figure 7. Although a complete graph is shown, the graph might be improved by adjusting the values of Xmin and Xmax. Figure 8 shows the graph of y = - 2x2 + 12 using Xmin = - 4 and Xmax = 4. Do you think this is a better choice for the viewing window?
⫺10
EX A MPL E 2
Creating a Table and Graphing an Equation Create a table and graph the equation: y = x3
Solution
Most graphing utilities have the capability of creating a table of values for an equation. (Check your manual to see if your graphing utility has this capability.) Table 1 illustrates a table of values for y = x3 on a TI-84 Plus. See Figure 9 for the graph. Figure 9
Table 1
10
⫺3
3
⫺10
B.2 Exercises In Problems 1–16, graph each equation using the following viewing windows: (a) Xmin Xmax Xscl Ymin Ymax Yscl
= = = = = =
-5 5 1 -4 4 1
(b) Xmin Xmax Xscl Ymin Ymax Yscl
= = = = = =
- 10 10 2 -8 8 2
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
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SECTION B.3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry
*1. y = x + 2
*2. y = x - 2
*3. y = - x + 2
B5
*4. y = - x - 2
*5. y = 2x + 2
*6. y = 2x - 2
*7. y = - 2x + 2
*8. y = - 2x - 2
*9. y = x2 + 2
*10. y = x2 - 2
*11. y = - x2 + 2
*12. y = - x2 - 2
*13. 3x + 2y = 6
*14. 3x - 2y = 6
*15. - 3x + 2y = 6
*16. - 3x - 2y = 6
*17–32. For each of the above equations, create a table, - 3 … x … 3, and list points on the graph.
B.3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry Value and Zero (or Root) Most graphing utilities have an eVALUEate feature that, given a value of x, determines the value of y for an equation. This feature can be used to evaluate an equation at x = 0 to determine the y-intercept. Most graphing utilities also have a ZERO (or ROOT) feature that can be used to determine the x-intercept(s) of an equation.
Finding Intercepts Using a Graphing Utility
EXAM PL E 1
Use a graphing utility to find the intercepts of the equation y = x3 - 8.
Solution
Figure 10(a) shows the graph of y = x3 - 8.
Figure 10
10
−5
5
−20 (a)
(b)
(c)
The eVALUEate feature of a TI-84 Plus graphing calculator accepts as input a value of x and determines the value of y. If we let x = 0, we find that the y-intercept is - 8. See Figure 10(b). The ZERO feature of a TI-84 Plus is used to find the x-intercept(s). See Figure 10(c). The x-intercept is 2.
EXAM PL E 2
1 Graphing the Equation y = x Graph the equation y = and symmetry.
Solution
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1 . Based on the graph, infer information about intercepts x
Figure 11 shows the graph. Infer from the graph that there are no intercepts. Also infer that symmetry with respect to the origin is a possibility. The TABLE feature on a graphing utility can provide further evidence of symmetry with respect to the origin. Using a TABLE, observe that for any ordered pair 1x, y2 , the ordered pair 1 - x, - y2 is also a point on the graph.
Figure 11
Y1 ⫽ 1x
4
⫺3
3
⫺4
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B6
APPENDIX B Graphing Utilities
B.3 Exercises In Problems 1–6, use ZERO (or ROOT) to approximate the smaller of the two x-intercepts of each equation. Express the answer rounded to two decimal places. 1. y = x2 + 4x + 2 4. y = 3x + 5x + 1 2
- 3.41
2. y = x2 + 4x - 3
- 1.43
- 4.65
5. y = 2x - 3x - 1 2
- 0.28
3. y = 2x2 + 4x + 1
- 1.71
6. y = 2x - 4x - 1
- 0.22
2
In Problems 7–12, use ZERO (or ROOT) to approximate the positive x-intercepts of each equation. Express each answer rounded to two decimal places. 7. y = x3 + 3.2x2 - 16.83x - 5.31
8. y = x3 + 3.2x2 - 7.25x - 6.3
3.00
9. y = x4 - 1.4x3 - 33.71x2 + 23.94x + 292.41 11. y = x + 19.5x - 1021x + 1000.5 3
2
4.50
1.00; 23.00
2.00
10. y = x4 + 1.2x3 - 7.46x2 - 4.692x + 15.2881 12. y = x + 14.2x - 4.8x - 12.4 3
2
1.70
1.07
B.4 Using a Graphing Utility to Solve Equations For many equations, there are no algebraic techniques that lead to a solution. For such equations, a graphing utility can often be used to investigate possible solutions. When a graphing utility is used to solve an equation, approximate solutions usually are obtained. Unless otherwise stated, this text follows the practice of giving approximate solutions rounded to two decimal places. The ZERO (or ROOT) feature of a graphing utility can be used to find the solutions of an equation when one side of the equation is 0. In using this feature to solve equations, make use of the fact that the x-intercepts (or zeros) of the graph of an equation are found by letting y = 0 and solving the equation for x. Solving an equation for x when one side of the equation is 0 is equivalent to finding where the graph of the corresponding equation crosses or touches the x-axis.
Using ZERO (or ROOT) to Approximate Solutions of an Equation
EX A MPL E 1
Find the solution(s) of the equation x2 - 6x + 7 = 0. Round answers to two decimal places. The solutions of the equation x2 - 6x + 7 = 0 are the same as the x-intercepts of the graph of Y1 = x2 - 6x + 7. Begin by graphing the equation. See Figure 12(a). From the graph there appear to be two x-intercepts (solutions to the equation): one between 1 and 2, the other between 4 and 5.
Solution
Figure 12
8
8
⫺1
8
7 ⫺1 ⫺2
7 ⫺2
(a)
⫺1
7 ⫺2
(b)
(c)
Using the ZERO (or ROOT) feature of the graphing utility, determine that the x-intercepts, and thus the solutions to the equation, are x = 1.59 and x = 4.41, rounded to two decimal places. See Figures 12(b) and (c).
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A second method for solving equations using a graphing utility involves the INTERSECT feature of the graphing utility. This feature is used most effectively when one side of the equation is not 0.
SECTION B.4 Using a Graphing Utility to Solve Equations
EXAM PL E 2
B7
Using INTERSECT to Approximate Solutions of an Equation Find the solution(s) to the equation 31x - 22 = 51x - 12. Round answers to two decimal places.
Solution
Begin by graphing each side of the equation as follows: graph Y1 = 31x - 22 and Y2 = 51x - 12. See Figure 13(a).
Figure 13
5 ⫺4
5 4
⫺15 (a)
⫺4
4
⫺15 (b)
At the point of intersection of the graphs, the values of the y-coordinates are the same. Conclude that the x-coordinate of the point of intersection represents the solution to the equation. Do you see why? The INTERSECT feature on a graphing utility determines the point of intersection of the graphs. Using this feature, find that the graphs intersect at 1 - 0.5, - 7.52. See Figure 13(b). The solution of the equation is therefore x = - 0.5.
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SUMMARY Here are the steps to follow for approximating solutions of equations. Steps for Approximating Solutions of Equations Using ZERO (or ROOT) STEP 1: Write the equation in the form 5 expression in x6 = 0. STEP 2: Graph Y1 = 5 expression in x6 . Be sure that the graph is complete. That is, be sure that all the intercepts are shown on the screen. STEP 3: Use ZERO (or ROOT) to determine each x-intercept of the graph. Steps for Approximating Solutions of Equations Using INTERSECT STEP 1: Graph Y1 = 5 expression in x on the left side of the equation6 . Graph Y2 = 5 expression in x on the right side of the equation6 . STEP 2: Use INTERSECT to determine each x-coordinate of the point(s) of intersection, if any. Be sure that the graphs are complete. That is, be sure that all the points of intersection are shown on the screen.
EXAM PL E 3
Solving a Radical Equation 3 Find the real solutions of the equation 2 2x - 4 - 2 = 0.
Solution
3 Figure 14 shows the graph of the equation Y1 = 2 2x - 4 - 2. From the graph, note that one x-intercept is near 6. Using ZERO (or ROOT), find that the x-intercept is 6. The only solution is x = 6.
Figure 14
1 ⫺1
10
⫺4
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B8
APPENDIX B Graphing Utilities
B.5 Square Screens Figure 15 4
(4, 4)
⫺4
4
(⫺4, ⫺4)
Most graphing utilities have a rectangular screen. Because of this, using the same settings for both x and y will result in a distorted view. For example, Figure 15 shows the graph of the line y = x connecting the points 1 - 4, - 42 and 14, 42. The line is expected to bisect the first and third quadrants, but it doesn’t. The selections for Xmin, Xmax, Ymin, and Ymax must be adjusted so that a square screen results. On most graphing utilities, this is accomplished by setting the ratio of x to y at 3∶2.* For example, if Xmin = - 6 Xmax = 6
⫺4
Ymin = - 4 Ymax = 4
then the ratio of x to y is 6 - 1 - 62 Xmax - Xmin 12 3 = = = Ymax - Ymin 4 - 1 - 42 8 2 for a ratio of 3∶2, resulting in a square screen.
Examples of Viewing Rectangles That Result in Square Screens
EX A MPL E 1
Figure 16 4
⫺6
(4, 4)
6
(⫺4, ⫺4)
⫺4
(a) Xmin Xmax Xscl Ymin Ymax Yscl
= -3 = 3 = 1 = -2 = 2 = 1
(b) Xmin Xmax Xscl Ymin Ymax Yscl
= -6 = 6 = 1 = -4 = 4 = 1
(c) Xmin Xmax Xscl Ymin Ymax Yscl
= - 12 = 12 = 1 = -8 = 8 = 2
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Figure 16 shows the graph of the line y = x on a square screen using the viewing rectangle given in part (b). Notice that the line now bisects the first and third quadrants. Compare this illustration to Figure 15.
B.5 Exercises In Problems 1–8, determine which of the given viewing rectangles result in a square screen. 1. Xmin Xmax Xscl Ymin Ymax Yscl
= - 3 Yes = 3 = 2 = -2 = 2 = 2
2. Xmin Xmax Xscl Ymin Ymax Yscl
= - 5 No = 5 = 1 = -4 = 4 = 1
3. Xmin Xmax Xscl Ymin Ymax Yscl
= 0 Yes = 9 = 3 = -2 = 4 = 2
4. Xmin Xmax Xscl Ymin Ymax Yscl
= - 6 Yes = 6 = 1 = -4 = 4 = 2
5. Xmin Xmax Xscl Ymin Ymax Yscl
= - 6 No = 6 = 1 = -2 = 2 = 0.5
6. Xmin Xmax Xscl Ymin Ymax Yscl
= - 6 Yes = 6 = 2 = -4 = 4 = 1
7. Xmin Xmax Xscl Ymin Ymax Yscl
= 0 Yes = 9 = 1 = -2 = 4 = 1
8. Xmin Xmax Xscl Ymin Ymax Yscl
= - 6 Yes = 6 = 2 = -4 = 4 = 2
*9. If Xmin = - 4, Xmax = 8, and Xscl = 1, how should Ymin, Ymax, and Yscl be selected so that the viewing rectangle contains the point 14, 82 and the screen is square? *10. If Xmin = - 6, Xmax = 12, and Xscl = 2, how should Ymin, Ymax, and Yscl be selected so that the viewing rectangle contains the point 14, 82 and the screen is square? *Some graphing utilities have a built-in function that automatically squares the screen. For example, the TI-84 has a ZSquare function that does this. Some graphing utilities require a ratio other than 3 ∶ 2 to square the screen. For example, the HP 48G requires the ratio of x to y to be 2 ∶ 1 for a square screen. Consult your manual.
*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.
SECTION B.7 Using a Graphing Utility to Solve Systems of Linear Equations
B9
B.6 Using a Graphing Utility to Graph Inequalities EXAM PL E 1
Graphing an Inequality Using a Graphing Utility Use a graphing utility to graph: 3x + y - 6 … 0
Solution
Begin by graphing the equation 3x + y - 6 = 0 1Y1 = - 3x + 62. See Figure 17. As with graphing by hand, test points selected from each region and determine whether they satisfy the inequality. To test the point 1 - 1, 22, for example, enter 31 - 12 + 2 - 6 … 0. See Figure 18(a). The 1 that appears indicates that the statement entered (the inequality) is true. When the point 15, 52 is tested, a 0 appears, indicating that the statement entered is false. Thus 1 - 1, 22 is a part of the graph of the inequality, and 15, 52 is not. Figure 18(b) shows the graph of the inequality on a TI-84 Plus.*
Figure 17
Figure 18
10
⫺2
10
⫺2
6
6
Y1 ⫽ ⫺3x ⫹ 6 ⫺10
⫺10 (a)
(b)
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The steps to follow to graph an inequality using a graphing utility are given next.
Steps for Graphing an Inequality Using a Graphing Utility STEP 1: Replace the inequality symbol by an equal sign, solve the equation for y, and graph the equation. STEP 2: In each region, select a test point P and determine whether the coordinates of P satisfy the inequality. (a) If the test point satisfies the inequality, then so do all the points in the region. Indicate this by using the graphing utility to shade the region. (b) If the coordinates of P do not satisfy the inequality, then none of the points in that region do.
B.7 Using a Graphing Utility to Solve Systems of Linear Equations Most graphing utilities have the capability to put the augmented matrix of a system of linear equations in row echelon form. The next example, Example 6 from Section 10.2, demonstrates this feature using a TI-84 Plus graphing calculator.
*Consult your owner’s manual for shading techniques.
B10
APPENDIX B Graphing Utilities
EX A MPL E 1
Solving a System of Linear Equations Using a Graphing Utility x - y + z = 8 Solve: c 2x + 3y - z = - 2 3x - 2y - 9z = 9
Solution
(1) (2) (3)
The augmented matrix of the system is -1 3 -2
1 C2 3
1 8 3 -1 -2S -9 9
Enter this matrix into the graphing utility and name it A. See Figure 19(a). Use the REF (row echelon form) command on matrix A. The results are shown in Figure 19(b). Figure 19
(a)
(b)
The system of equations represented by the matrix in row echelon form is 1
-
2 3
E0
1
0
0
-3 15 13 1
x -
3 5
-
24 U 13 1
e
2 y 3 y +
3z =
3
15 24 z = 13 13 z = 1
(1) (2) (3)
Using z = 1, back-substitute to get x d
2 y 3
3112 =
x -
3 (1)
15 24 (2) ¡ y + 112 = 13 13 Simplify.
d
2 y = 3 y =
6 - 39 = -3 13
(1) (2)
Solve the second equation for y, to find that y = - 3. Back-substitute y = - 3 into 2 x - y = 6, to find that x = 4. The solution of the system is x = 4, y = - 3, z = 1. 3
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Figure 20
Notice that the row echelon form of the augmented matrix using the graphing utility differs from the row echelon form in Chapter 10 (p. 746), yet both matrices provide the same solution! This is because the two solutions used different row operations to obtain the row echelon form. In all likelihood, the two solutions parted ways in Step 4 of the algebraic solution, where fractions were avoided by interchanging rows 2 and 3. Most graphing utilities also have the ability to put a matrix in reduced row echelon form. Figure 20 shows the reduced row echelon form of the augmented matrix from Example 1 using the RREF command on a TI-84 Plus graphing calculator. From this command, note that the solution of the system is x = 4, y = - 3, z = 1.
SECTION B.9 Using a Graphing Utility to Graph Parametric Equations
B11
B.8 Using a Graphing Utility to Graph a Polar Equation Most graphing utilities require the following steps in order to obtain the graph of a polar equation. Be sure to be in POLar mode.
Graphing a Polar Equation Using a Graphing Utility STEP 1: Set the mode to POLar. Solve the equation for r in terms of u. STEP 2: Select the viewing rectangle in polar mode. Besides setting Xmin, Xmax, Xscl, and so forth, the viewing rectangle in polar mode requires setting the minimum and maximum values for u and an increment setting for u (ustep). In addition, a square screen and radian measure should be used. STEP 3: Enter the expression involving u that you found in Step 1. (Consult your manual for the correct way to enter the expression.) STEP 4: Graph.
Graphing a Polar Equation Using a Graphing Utility
EXAM PL E 1
Use a graphing utility to graph the polar equation r sin u = 2.
Solution
STEP 1: Solve the equation for r in terms of u. r sin u = 2 r =
2 sin u
STEP 2: From the POLar mode, select the viewing rectangle. umin = 0 umax = 2p p ustep = 24
6
9
STEP 3: Enter the expression STEP 4: Graph.
⫺6
Ymin = - 6 Ymax = 6
Xscl = 1
Yscl = 1
ustep determines the number of points that the graphing utility will plot. For example, p p 2p 3p if ustep is , the graphing utility will evaluate r at u = 01umin2, , , , and 24 24 24 24 so forth, up to 2p1umax2. The smaller ustep is, the more points the graphing utility will plot. The reader is encouraged to experiment with different values for umin, umax, and ustep to see how the graph is affected.
Figure 21
⫺9
Xmin = - 9 Xmax = 9
2 after the prompt r1 = . sin u
The graph is shown in Figure 21.
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B.9 Using a Graphing Utility to Graph Parametric Equations Most graphing utilities have the capability of graphing parametric equations. The following steps are usually required to obtain the graph of parametric equations. Check your owner’s manual to see how yours works.
B12
APPENDIX B Graphing Utilities
Graphing Parametric Equations Using a Graphing Utility STEP 1: Set the mode to PARametric. Enter x1t2 and y1t2. STEP 2: Select the viewing window. In addition to setting Xmin, Xmax, Xscl, and so on, the viewing window in parametric mode requires setting minimum and maximum values for the parameter t and an increment setting for t (Tstep). STEP 3: Graph.
EX A MPL E 1
Graphing a Curve Defined by Parametric Equations Using a Graphing Utility Graph the curve defined by the parametric equations x = 3t 2,
Solution
5
0
15
-2 … t … 2
STEP 1: Enter the equations x1t2 = 3t 2, y1t2 = 2t with the graphing utility in PARametric mode. STEP 2: Select the viewing window. The interval is - 2 … t … 2, so select the following square viewing window: Tmin = - 2 Tmax = 2 Tstep = 0.1
Figure 22
y = 2t,
Xmin = 0 Xmax = 15 Xscl = 1
Ymin = - 5 Ymax = 5 Yscl = 1
Choose Tmin = - 2 and Tmax = 2 because - 2 … t … 2. Finally, the choice for Tstep will determine the number of points that the graphing utility will plot. For example, with Tstep at 0.1, the graphing utility will evaluate x and y at t = - 2, - 1.9, - 1.8, and so on. The smaller the Tstep, the more points the graphing utility will plot. The reader is encouraged to experiment with different values of Tstep to see how the graph is affected. STEP 3: Graph. Note the direction in which the graph is drawn. This direction shows the orientation of the curve.
⫺5
The graph shown in Figure 22 is complete.
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Exploration Graph the following parametric equations using a graphing utility with Xmin = 0, Xmax = 15, Ymin = - 5, Ymax = 5, and Tstep = 0.1. 1. x =
3t 2 , y = t, 4
-4 … t … 4
2. x = 3t2 + 12t + 12, y = 2t + 4, 3
3. x = 3t2>3, y = 22t ,
-4 … t … 0
-8 … t … 8
Compare these graphs to the graph in Figure 22. Conclude that parametric equations defining a curve are not unique; that is, different parametric equations can represent the same graph.
Exploration 3y2
4x 4x aY1 = and Y2 = b with Xmin = 0, Xmax = 15, 4 A3 A3 Ymin = - 5, Ymax = 5. Compare this graph with Figure 22. Why do the graphs differ?
In FUNCtion mode, graph x =
Answers CHAPTER F Foundations: A Prelude to Functions F.1 Assess Your Understanding (page 38) 7. x-coordinate or abscissa; y-coordinate or ordinate 8. quadrants 9. 2 1x2 - x1 2 2 + 1y2 - y1 2 2 10. midpoint 29. d 1A, B2 = 213 11. (a) Quadrant II (b) Positive x-axis 13. The points will be on a 15. 25 (c) Quadrant III (d) Quadrant I vertical line that is 2 units 17. 210 d 1B, C2 = 213 (e) Negative y-axis (f) Quadrant IV to the right of the y-axis. 19. 2 217 d 1A, C2 = 226 (2, 4) y 21. 285 y 1 213 2 2 + 1 213 2 2 = 1 226 2 2 5 D (6, 5) (2, 1) 6 23. 3 25 13 (2, 0) Area = square units A (3, 2) 25. 26.89 ≈ 2.62 2 B (6, 0) 2 2 5 x 27. 2a + b y 7 x C (2, 2) (2, 1) (2, 3)
E (0, 3)
A (2, 5) 5
F (6, 3)
C (1, 0)
31. d 1A, B2 = 2130 d 1B, C2 = 226 d 1A, C2 = 2 226 1 226 2 2 + 1 2 226 2 2 = Area = 26 square units y 10 A (5, 3)
33. d 1A, B2 = 4 d 1A, C2 = 5 d 1B, C2 = 241 42 + 52 = 16 + 25 = 1 241 2 2 Area = 10 square units
1 2130 2 2
y 5 C (5, 5) B (6, 0) 10 x
B (0, 3)
C (4, 2) 5 x A (4, 3)
B (1, 3) 5 x
3 7 a b 35. (4, 0) 37. a , 1 b 39. a1, - b 41. (1.05, 0.7) 43. a , b 2 2 2 2 45. (2, 2); 12, - 42 47. (0, 0); (8, 0) 49. 217; 2 25; 229 s s 51. a , b 53. d 1P1 , P2 2 = 6; d 1P2 , P3 2 = 4; d 1P1 , P3 2 = 2 213; 2 2 right triangle 55. d 1P1 , P2 2 = 2 217; d 1P2 , P3 2 = 234; d 1P1 , P3 2 = 234; isosceles right triangle 57. 90 22 ≈ 127.28 ft 59. (a) (90, 0), (90, 90), (0, 90) (b) 5 22161 ≈ 232.43 ft (c) 30 2149 ≈ 366.20 ft 61. d = 50t mi 63. (a) (2.65, 1.6) (b) ≈1.285 units 65. $21,220; a slight underestimate 67. 15,- 22
F.2 Assess Your Understanding (page 48)
3. intercepts 4. zeros; roots 5. y-axis 6. True 7. True 8. False 9. (0, 0) is on the graph. 11. (0, 3) is on the graph. 13. (0, 2) and 1 22 , 22 2 are on the graph. 15. 1 - 2, 02, 10, 22 17. 1 - 4, 02, 10, 82 19. 1 - 1, 02, 11, 02, 10, - 12 21. 1 - 2, 02, 12, 02, 10, 42 23. 13, 02, 10, 22 yx2
y 5
y 10
y 5
y 2x 8 (0, 8)
(2, 0)
(0, 2)
(4, 0) 10 x
5 x
y 5 (0, 4)
y x2 1 (2, 0)
(1, 0) 5 x
(1, 0)
y 10 (0, 9) (2, 0)
(c) (3, 4)
y 5 (2, 1) (a) (2, 1)
5 x
43. (a) ( - 2, 0), (0, 0), (2, 0) (b) Symmetric with respect to the origin
31. (b) (2, 1) 5 x (c) (2, 1)
(c) (5, 2)
y 5
p p , 0 b , 10, 12, a , 0 b 2 2 (b) Symmetric with respect to the y-axis
39. (a) a -
45. (a) x-intercepts: 3 - 2, 1 4 ; y-intercept: 0 (b) No symmetry
33. (a) (5, 2)
5 (3, 0)
y 5 (a) (3, 4)
(5, 2)
x
(c) (3, 4) 5 x
5 x (b) (5, 2)
(a) (3, 4)
37. (a) ( - 1, 0), (1, 0) (b) Symmetric with respect to the x-axis, the y-axis, and the origin
y 5 (a) (0, 3) (c) (0, 3) (0, 3) (b) (0, 3)
29.
y 5 (b) (3, 4) (3, 4) 5 x
(2, 0) 10 x 9x 2 4y 36
35.
27.
2x 3y 6
(0, 2)
(2, 0) 5 x
y x 2 4
(0, 1)
25. 1 - 2, 02, 12, 02, 10, 92
y 5
(3, 4)
(b) (3, 4)
41. (a) (0, 0) (b) Symmetric with respect to the x-axis
47. (a) No intercepts (b) Symmetric with respect to the origin
AN1
AN2
ANSWERS Section F.2
49.
51.
y 4 5 x
(2, 5) (0, 9)
69.
(, 0) (0, 0) 5 x
, 2 2
71.
y 5 (1, 1) 5 x
(1, 1)
(, 0)
(2, 5)
53. 1 - 4, 02, 10, - 22, 10, 22; symmetric with respect to the x-axis 55. (0, 0); symmetric with respect to the origin 57. 10, - 92, 13, 02, 1 - 3, 02; symmetric with respect to the y-axis 59. 1 - 2, 02, 12, 02, 10, - 32, 10, 32; symmetric with respect to the x-axis, y-axis, and origin 61. 10, - 92, 13, 02, 1 - 3, 02, 1 - 1, 02 ; no symmetry 63. 10, - 42, 14, 02, 1 - 4, 02 ; symmetric with respect to the y-axis 65. (0, 0); symmetric with respect to the origin 67. (0, 0); symmetric with respect to the origin
y 5 , 2 2
(0, 0)
73. b = 13 75. - 4 or 1
y (1, 1) 5 (4, 2) (0, 0)
5 x
77. (a) 10, - 52, 1 - 25, 02, 1 25, 02 (b) Symmetric with respect to the y-axis (c) y 5 (兹5, 0)
79. (a) 1 - 9, 02, 10, - 32, 10, 32 (b) Symmetric with respect to the x-axis (c) y 5
(兹5, 0)
(2, 1)
(0, 3)
5 x (2, 1)
81. (a) 10, 32, 10, - 32, 1 - 3, 02, 13, 02 83. (a) 10, 02, 1 - 2, 02, 12, 02 (b) Symmetric with respect to the x-axis, (b) Symmetric with respect y-axis, and origin to the origin (c) (c) y y 5
5
(0, 3)
(1, 3) (2, 0)
(3, 0) 5 x
(3, 0)
(0, 0)
(0, 3)
(2, 0) 5 x
1 x
(9, 0)
(1, 4) (0, 5) (1, 4)
(0, 3)
85. 1 - 1, - 22 87. 4 89. (a) 10, 02, 12, 02, 10, 12, 10, - 12 (b) Symmetric with respect to the x-axis 91. (a) y = 2x2 and y = x have the same graph. (b) 2x2 = x (c) x Ú 0 for y = 1 2x2 2, while x can be any real number for y = x (d) y Ú 0 for y = 2x2
(1, 3)
F.3 Assess Your Understanding (page 61) 1 1. undefined; 0 2. 3; 2 3. y = b; y-intercept 4. True 5. False 6. True 7. m1 = m2; y-intercepts; m1m2 = - 1 8. 2 9. 10. False 2 1 1 (b) If x increases by 2 units, y will increase by 1 unit. 13. (a) Slope = - (b) If x increases by 3 units, y will decrease by 1 unit. 11. (a) Slope = 2 3 21. Slope undefined 3 19. Slope = 0 1 15. Slope = 17. Slope = y 2 y 2 y 8
5
(3, 1) (2,1)
x
(4, 0)
(1, 2)
5
(2, 1)
(2, 3)
(2, 3)
5
5
y 5
x
5 x
(1, 2)
8 x
23.
25.
y 8 (2, 5) 3
P (1, 2)
27.
y 8 3
P (2, 4) 4
1 5 x
29.
y 5
y 5 P (0, 3)
P (1, 3) 5 x
(6, 1) 7 x
5 x
1 37. x - 2y = 0 or y = x 2 1 5 39. x + y = 2 or y = - x + 2 41. 2x - y = 3 or y = 2x - 3 43. x + 2y = 5 or y = - x + 45. 3x - y = - 9 or y = 3x + 9 2 2 2 1 1 5 1 47. 2x + 3y = - 1 or y = - x 49. x - 2y = - 5 or y = x + 51. 3x + y = 3 or y = - 3x + 3 53. x - 2y = 2 or y = x - 1 3 3 2 2 2 55. x = 2; no slope–intercept form 57. y = 2 59. 2x - y = - 4 or y = 2x + 4 61. 2x - y = 0 or y = 2x 1 3 63. x = 4; no slope–intercept form 65. 2x + y = 0 or y = - 2x 67. x - 2y = - 3 or y = x + 69. y = 4 2 2 1 1 75. Slope = ; y@intercept = 2 77. Slope = - ; y@intercept = 2 71. Slope = 2; y@intercept = 3 73. Slope = 2; y@intercept = - 2 2 2 31. (2, 6); (3, 10); (4, 14)
33. (4, - 7); (6, - 10); (8, - 13)
y 5
y 8 (0, 3)
35. ( - 1, - 5); (0, - 7); (1, - 9)
(1, 5) (0, 2)
(1, 0) 5 x
y 5 (0, 2)
y 5 (0, 2)
(2, 3)
5 5 x
(4, 0)
x
5 x
79. Slope =
2 ; y@intercept = - 2 3
y 5
5 x
83. Slope undefined; no y-intercept y 5
y 2.5 (0, 1)
(3, 0) (0, 2)
81. Slope = - 1; y@intercept = 1
(1, 0) 2.5 x
85. Slope = 0; y@intercept = 5 y 8 (0, 5)
(4, 0) 5 x
5 x
AN3
ANSWERS Section F.4 87. Slope = 1; y@intercept = 0
89. Slope =
y 5
3 ; y@intercept = 0 2
(0, 0)
(1, 1)
(0, 2)
(0, 0)
5 x
5 x
(10, 0)
5
8 x
x
(3, 0)
97. (a) x-intercept: 2; y-intercept: 3 (b) y
101. y = 0 103. Parallel 105. Neither 107. (b) 109. (d)
99. (a) x-intercept: 5; y-intercept: - 2 (b) y
5
12
10
(0, 8)
(2, 3)
95. (a) x-intercept: 3; 21 y-intercept: 2 (b) y 21 0, 2
y 5
(b)
y 5 (2, 2)
93. (a) x-intercept: - 10; y-intercept: 8 y (b)
91. (a) x-intercept: 3; y-intercept: 2
5
(0, 3)
(2, 0) 5 x
(3, 0) 5 x
(5, 0) 6 x (0, 2)
Cost (dollars)
3 2 111. P1 = ( - 2, 5), P2 = (1, 3), m1 = - ; P2 = (1, 3), P3 = ( - 1, 0), m2 = ; because m1m2 = - 1, the lines are perpendicular and the points 3 2 ( - 2, 5), (1, 3), and ( - 1, 0) are the vertices of a right triangle. 113. P1 = ( - 1, 0), P2 = (2, 3), m = 1; P3 = (1, - 2), P4 = (4, 1), m = 1; P1 = ( - 1, 0), P3 = (1, - 2), m = - 1; P2 = (2, 3), P4 = (4, 1), m = - 1; opposite sides are parallel, and adjacent sides are perpendicular; the points are the vertices of a rectangle. 115. C = 0.60x + 39; +105.00; +177.00 117. C = 0.45x + 6735 2 5 119. (a) C = 0.1062x + 13.04, 0 … x … 800 121. °C = (°F - 32); approximately 21.1°C 123. (a) y = - x + 30 (b) x-intercept: 375; The ramp 9 25 (b) C = 0.1062x + 13.04 meets the floor 375 in. (31.25 ft) from the base of the platform. (c) The ramp does not meet design 0 ≤ x ≤ 800 C (800, 98.00) requirements. It has a run of 31.25 ft. (d) The only slope possible for the ramp to comply with the 100 requirement is for it to drop 1 in. for every 12-in. run. 80 60 (0, 13.04) 1 40 125. (a) A = x + 20,000 (b) +80,000 (c) Each additional box sold requires an additional +0.20 in 5 20 y 2x y 0 advertising. 127. All have the same slope, 2; the lines are parallel. 0 100 300 500 700 900 x 6 (0, 4) (0, 0)
kW-hr
(c) +34.28 (d) +66.14 (e) Each additional kilowatt-hour used adds +0.1062 to the bill. 129. (b), (c), (e), (g)
131. (c)
5 x (0, 2)
2 x y 4
137. No; no
2x y 2
3 139. They are the same line. 141. Yes, if the y-intercept is 0. 143. No, the slope should be negative, - . 2
F.4 Assess Your Understanding (page 70) 4. radius 5. 1x - h2 2 + 1y - k2 2 = r 2 6. unit 9 5 3 5 2 9. Center a , 2 b ; radius = ; ax - b + 1y - 22 2 = 2 2 2 4
3. False
11. x2 + y2 = 4; x2 + y 2 - 4 = 0
13. x2 + 1y - 22 2 = 4; x2 + y2 - 4y = 0
y 5
y 5
7. Center (2, 1); radius = 2; 1x - 22 2 + 1y - 12 2 = 4
1 2 1 17. 1x + 22 2 + 1y - 12 2 = 16; 19. ax - b + y2 = ; 2 4 x2 + y2 + 4x - 2y - 11 = 0 x2 + y 2 - x = 0
15. 1x - 42 2 + 1y + 32 2 = 25; x2 + y2 - 8x + 6y = 0 y 2
(0, 2)
y (2, 1) 6
(0, 0)
y 2.5
9 x
5 x
(4, 3)
5 x
4 x 1, 0 2
21. (a) 1h, k2 = 10, 02; r = 2 (b) y
23. (a) 1h, k2 = 13, 02; r = 2 (b) y 5
5
(c) 1 {2, 02; 10, {22 1 1 29. (a) 1h, k2 = a , - 1 b ; r = 2 2 (b) y
(2, 2) 6
6 x
5 x
5 x (1, 2)
4 x
(c) 1 1 { 25 , 0 2 ; 1 0, 2 { 2 22 2
(c) (1, 0); (5, 0) 31. (a) 1h, k2 = 13, - 22; r = 5 (b) y 3
2.5
33. (a) 1h, k2 = 1 - 2, 02; r = 2 (b) y (2, 0)
8
5
x 2.5 x
(c) 10, - 12 51. x2 + y2 + 2x + 4y - 4168.16 = 0
27. (a) 1h, k2 = 1 - 2, 22; r = 3 (b) y
5
(3, 0)
(0, 0)
1, 1 2
25. (a) 1h, k2 = 11, 22; r = 3 (b) y
(3, 2)
5 x
(c) 1 3 { 221 , 0 2 ; 10, - 62, 10, 22 53. 22x + 4y = 9 22
2.5 x
(c) 10, 02, 1 - 4, 02
55. (1, 0) 57. y = 2
59. (b), (c), (e), (g)
(c)
1 -2
{ 25 , 0 2 ; 1 0, 2 { 25 2
35. x2 + y2 = 13 37. 1x - 22 2 + 1y - 32 2 = 9 39. 1x + 12 2 + 1y - 32 2 = 5 41. 1x + 12 2 + 1y - 32 2 = 1 43. (c) 45. (b) 47. 18 units2 49. x2 + (y - 139)2 = 15,625
AN4
ANSWERS Section 1.1
CHAPTER 1 Functions and Their Graphs 1.1 Assess Your Understanding (page 85) 5. independent; dependent 6. range 7. [0, 5] 8. ≠; ƒ; g 9. 1g - f 2 1x2 10. F 11. T 12. T 13. F 14. F 15. Function; Domain: 5Elvis, Colleen, Kaleigh, Marissa 6 ; Range: 5January 8, March 15, September 17 6 17. Not a function 19. Not a function 21. Function; Domain: 51, 2, 3, 4 6; Range: 53 6 23. Not a function 25. Function; Domain: 5 - 2, - 1, 0, 1 6; Range: 50, 1, 4 6 27. Function 29. Function 31. Not a function 33. Not a function 35. Function 37. Not a function 39. (a) - 4 (b) 1 (c) - 3 (d) 3x2 - 2x - 4 1 -x 1 (c) (d) 2 (e) - 3x2 - 2x + 4 (f) 3x2 + 8x + 1 (g) 12x2 + 4x - 4 (h) 3x2 + 6xh + 3h2 + 2x + 2h - 4 41. (a) 0 (b) 2 2 x + 1 -x x + 1 2x x + h (e) 2 (f) 2 (g) 2 (h) 2 43. (a) 4 (b) 5 (c) 5 (d) x + 4 (e) - x - 4 (f) x + 1 + 4 x + 1 x + 2x + 2 4x + 1 x + 2xh + h2 + 1 3 1 2x - 1 - 2x - 1 2x + 3 4x + 1 2x + 2h + 1 1 (b) (c) (d) (e) (f) (g) (h) (g) 2 x + 4 (h) x + h + 4 45. (a) 5 2 8 3x + 5 3x - 5 3x - 2 6x - 5 3x + 3h - 5 47. All real numbers 49. All real numbers 51. 5x x ≠ - 4, x ≠ 4 6 53. 5x x ≠ 0 6 55. 5x x Ú 4 6 57. 5x x 7 9 6 59. 5x x 7 1 6 61. 5t t Ú 4, t ≠ 7 6 63. (a) 1f + g2 1x2 = 5x + 1; All real numbers (b) 1f - g2 1x2 = x + 7; All real numbers (c) 1ƒ # g2 1x2 = 6x2 - x - 12; f 3x + 4 3 All real numbers (d) a b 1x2 = ; e x ` x ≠ f (e) 16 (f) 11 (g) 10 (h) - 7 65. (a) 1f + g2 1x2 = 2x2 + x - 1; All real numbers g 2x - 3 2 f x - 1 (b) 1f - g2 1x2 = - 2x2 + x - 1; All real numbers (c) 1ƒ # g2 1x2 = 2x3 - 2x2; All real numbers (d) a b 1x2 = ; 5x x ≠ 0 6 g 2x2 (e) 20 (f) - 29 (g) 8 (h) 0 67. (a) 1f + g2 1x2 = 1x + 3x - 5; 5x x Ú 0 6 (b) 1 f - g2 1x2 = 1x - 3x + 5; 5x x Ú 0 6 f 1x 5 1 ; e x ` x Ú 0, x ≠ f (e) 23 + 4 (f) - 5 (g) 22 (h) (c) 1ƒ # g2 1x2 = 3x 1x - 5 1x; 5x x Ú 0 6 (d) a b 1x2 = g 3x - 5 3 2 2 1 1 + 2 ; 5x x ≠ 0 6 69. (a) 1f + g2 1x2 = 1 + ; 5x x ≠ 0 6 (b) 1f - g2 1x2 = 1; 5x x ≠ 0 6 (c) 1ƒ # g2 1x2 = x x x f 5 3 6x + 3 2 - 2x + 3 2 (f) 1 (g) (h) 2 71. (a) 1f + g2 1x2 = ; e x ` x ≠ f (b) 1f - g2 1x2 = ; ex ` x ≠ f (d) a b 1x2 = x + 1; 5x x ≠ 0 6 (e) g 3 4 3x - 2 3 3x - 2 3 2 f 8x + 12x 2 7 2x + 3 2 1 7 5 (c) 1ƒ # g2 1x2 = ; e x ` x ≠ f (d) a b 1x2 = ; e x ` x ≠ 0, x ≠ f (e) 3 (f) (g) (h) 73. g 1x2 = 5 - x 3 g 4x 3 2 2 4 2 13x - 22 2
- 12x + h2 1 7 1 85. A = - 4 87. A = 8; undefined at x = 3 89. A1x2 = x2 81. 83. A = 2 2 x2 1x + h2 2 2x + h + 2x 91. G1x2 = 10x 93. (a) P is the dependent variable; a is the independent variable. (b) P1202 = 243.706 million; There were 243.706 million Americans 20 years of age or older in 2011. (c) P102 = 317.946 million; There were 317.946 million Americans in 2011. 95. (a) 15.1 m, 14.071 m, 12.944 m, L1x2 101. H1x2 = P1x2 # I 1x2 11.719 m (b) 1.01 sec, 1.43 sec, 1.75 sec (c) 2.02 sec 97. (a) $222 (b) $225 (c) $220 (d) $230 99. R1x2 = P1x2 3 2 103. (a) P1x2 = - 0.05x + 0.8x + 155x - 500 (b) P1152 = $1836.25 (c) When 15 hundred cellphones are sold, the profit is $1836.25. 75. 4
77. 2x + h - 1
79.
105. (a) D1v2 = 0.05v2 + 2.6v - 15
(b) 321 feet
(c) The car will need 321 feet to stop once the impediment is observed.
(x + 1)(x - 1) x2 - 1 107. No. g(x) = = = x - 1, x ≠ - 1. x + 1 x + 1
109. H1x2 =
3x - x3 your age
1.2 Assess Your Understanding (page 93) 3. vertical 4. 5; - 3 5. a = - 2 6. F 7. F 8. T 9. (a) f 102 = 3; f 1 - 62 = - 3 (b) f 162 = 0; f 1112 = 1 (c) Positive (d) Negative (e) - 3, 6, and 10 (f) - 3 6 x 6 6; 10 6 x … 11 (g) 5x - 6 … x … 11 6 (h) 5y - 3 … y … 4 6 (i) - 3, 6, 10 (j) 3 (k) Three times (l) Once (m) 0, 4 (n) - 5, 8 (o) - 3, 6, 10 11. Not a function 13. Function (a) Domain: 5x - p … x … p 6; Range: 5y - 1 … y … 1 6 p p (b) a - , 0 b , a , 0 b , 10, 12 (c) y-axis 15. Not a function 17. Function (a) Domain: 5x 0 6 x 6 3 6; Range: {y|y 6 2} (b) 11, 02 (c) None 2 2 19. Function (a) Domain: all real numbers; Range: 5y y … 2 6 (b) 1 - 3, 02, 13, 02, 10, 22 (c) y-axis 21. Function (a) Domain: all real 1 1 numbers; Range: 5y y Ú - 3 6 (b) 11, 02, 13, 02, 10, 92 (c) None 23. (a) Yes (b) f 1 - 22 = 9; 1 - 2, 92 (c) 0, ; 10, - 12, a , - 1 b 2 2 1 1 1 (d) All real numbers (e) - , 1 (f) - 1 (g) - , 1 25. (a) No (b) f 142 = - 3; 14, - 32 (c) 14; 114, 22 (d) 5x x ≠ 6 6 (e) - 2 (f) (g) - 2 2 2 3 8 8 27. (a) Yes (b) f 122 = ; a2, b (c) - 1, 1; 1 - 1, 12, 11, 12 (d) All real numbers (e) 0 (f) 0 (g) 0 17 17 29. (a) Approximately 10.4 ft high (b) h 1122 ≈ 9.9 represents the height of the ball, in feet, after it has traveled 12 feet in front of the foul line. (c) h (d) The ball will not go through (8, 10.4) (12, 9.9) 10
(15, 8.4)
5 (0, 6)
(20, 3.6)
the hoop; h 1152 ≈ 8.4 ft . If v = 30 ft/sec, h 1152 = 10 ft .
31. (a) About 81.07 ft (b) About 129.59 ft (c) h 15002 ≈ 26.63 represents the height of the golf ball, in feet, after it has traveled a horizontal distance of 500 feet. (d) About 528.13 ft (e) 150 (f) About 115.07 ft and 413.05 ft (g) 275 ft; maximum height shown in the table is 131.8 ft (h) 264 ft
(22.6, 0) 0
5 10 15 20 25
x
0 0
550
ANSWERS Section 1.3 (b) 5x x 7 0 6
33. (a) $223; $220 (c) 500
1 3 37. (a) $80; it costs $80 if you use 0 minutes. (b) $80; it costs $80 if you use 1000 minutes. (c) $210; it costs $210 if you use 2000 minutes. (d) {m|0 … m … 14,400}. There are at most 14,400 anytime minutes in a month. 39. The x-intercepts can number anywhere from 0 to infinitely many. There is at most one y-intercept. 41. (a) III (b) IV (c) I (e) 600 mi/hr (d) V (e) II 45. (a) 2 hr elapsed during which Kevin was between 0 and 3 mi from home (b) 0.5 hr elapsed during which Kevin was 3 mi from home (c) 0.3 hr elapsed during which Kevin was between 0 and 3 mi from home (d) 0.2 hr elapsed during which Kevin was 0 mi from home (e) 0.9 hr elapsed during which Kevin was between 0 and 2.8 mi from home (f) 0.3 hr elapsed during which Kevin was 2.8 mi from home (g) 1.1 hr elapsed during which Kevin was between 0 and 2.8 mi from home (h) 3 mi (i) Twice 47. No points whose x-coordinate is 5 or whose y-coordinate is 0 can be on the graph.
1000 0 y Distance (blocks)
43.
(22, 5) 5 4 3 (5, 2) 2 (7, 0) 1 0
(6, 0)
35. (a) 3 (b) - 2
(d)
0
AN5
(29, 0)
x 10 20 Time (minutes)
(c) - 1
(d) 1
(e) 2
(f) -
1.3 Assess Your Understanding (page 106) 6. increasing 7. even; odd 8. T 9. T 10. F 11. Yes 13. No 15. 1 - 8, - 22; 10, 22; 15, q 2 17. Yes; 10 19. - 2, 2; 6, 10 21. (a) 1 - 2, 02, 10, 32, 12, 02 (b) Domain: 5x - 4 … x … 4 6 or [ - 4, 4]; Range: 5y 0 … y … 3 6 or [0, 3] (c) Increasing on 1 - 2, 02 and 12, 42 ; Decreasing on 1 - 4, - 22 and 10, 22 (d) Even 23. (a) 10, 12 (b) Domain: all real numbers; Range: 5y y 7 0 6 or 10, q 2 (c) Increasing on 1 - q , q 2 (d) Neither 25. (a) 1 - p, 02, 10, 02, 1p, 02 (b) Domain: 5x - p … x … p 6 or [ - p, p]; Range: 5y - 1 … y … 1 6 or [ - 1, 1] p p 1 1 5 p p (c) Increasing on a - , b ; Decreasing on a - p, - b and a , p b (d) Odd 27. (a) a0, b , a , 0 b , a , 0 b (b) Domain: 5x - 3 … x … 3 6 or 2 2 2 2 2 3 2 [ - 3, 3]; Range: 5y - 1 … y … 2 6 or [ - 1, 2] (c) Increasing on 12, 32 ; Decreasing on 1 - 1, 12; Constant on 1 - 3, - 12 and 11, 22 (d) Neither p p 29. (a) 0; 3 (b) - 2, 2; 0, 0 31. (a) ; 1 (b) - ; - 1 33. Odd 35. Even 37. Odd 39. Neither 41. Even 43. Odd 45. Absolute maximum: 2 2 f 112 = 4; absolute minimum: f 152 = 1; local maximum: f 132 = 3; local minimum: f 122 = 2 47. Absolute maximum: f 132 = 4; absolute minimum: f 112 = 1; local maximum: f 132 = 4; local minimum: f 112 = 1 49. Absolute maximum: none; absolute minimum: f 102 = 0; local maximum: f 122 = 3; local minima: f 102 = 0 and f 132 = 2 51. Absolute maximum: none; absolute minimum: none; local maximum: none; local minimum: none 55. 0 59. 53. 0.5 4 57. 8 6 2
2
4
2
3
2 0.5
0
Increasing: 1 - 2, - 0.772, 10.77, 22 Decreasing: 1 - 0.77, 0.772 Local maximum: 1 - 0.77, 0.192 Local minimum: 10.77, - 0.192
Increasing: 1 - 2, - 12, 11, 22 Decreasing: 1 - 1, 12 Local maximum: 1 - 1, 42 Local minimum: 11, 02
2
20
0
Increasing: 1 - 3.77, 1.772 Decreasing: 1 - 6, - 3.772, 11.77, 42 Local maximum: 11.77, - 1.912 Local minimum: 1 - 3.77, - 18.892
Increasing: 1 - 1.87, 02, 10.97, 22 Decreasing: 1 - 3, - 1.872, 10, 0.972 Local maximum: 10, 32 Local minima: 1 - 1.87, 0.952, 10.97, 2.652
75. (a)
77. (a), (b)
2500
0
40 0
(b) 10 riding lawn mowers/hr (c) $239/mower
Debt (trillions of dollars)
61. (a) - 4 (b) - 8 (c) - 10 63. (a) 17 (b) - 1 (c) 11 65. (a) 5 (b) y = 5x - 2 67. (a) - 1 (b) y = - x 69. (a) 4 (b) y = 4x - 8 71. (a) Odd (b) Local maximum value: 54 at x = - 3 73. (a) Even (b) Local maximum value: 24 at x = - 2 79. (a) 1 (b) 0.5 (c) 0.1 (d) 0.01 (e) 0.001 (f) y x2
D 14 12 10 8 6 4 2000
2
(c) 47.4 sq. units
yx y 0.5x y 0.1x
2005 2010 Year
y
The slope represents the average rate of change of the debt from 2001 to 2006. (c) $575.5 billion (d) $759 billion (e) $1252 billion (f) The average rate of change is increasing over time.
3
3
y 0.01x y 0.001x
2
(g) They are getting closer to the tangent line at 10, 02 . (h) They are getting closer to 0.
AN6
ANSWERS Section 1.3
81. (a) 2 (b) 2; 2; 2; 2 (c) y = 2x + 5 (d)
83. (a) 2x + h + 2 (b) 4.5; 4.1; 4.01; 4 (c) y = 4.01x - 1.01 (d) 10
10
10
1 1x + h2x 2 10 100 (b) - ; - ; ; -1 3 11 101 100 201 (c) y = x + 101 101 (d) 2
87. (a) -
85. (a) 4x + 2h - 3 (b) 2; 1.2; 1.02; 1 (c) y = 1.02x - 1.02 (d) 10
10 2 5
4
3
10
2
2
4
5 2
91. At most one
93. Yes; the function f 1x2 = 0 is both even and odd. 95. Not necessarily. It just means f 152 7 f 122.
1.4 Assess Your Understanding (page 117) 4. 1 - q , 02 y 17.
5. piecewise-defined
6. T 7. F 19.
8. F y 10
10 (4, 4)
(0, 0)
11. E
13. B
15. F 21.
5 x
(1, 1)
(2, 8)
25. (a) 4 (b) 2 (c) 5 27. (a) - 4 29. (a) All real numbers (b) 10, 12 (c) y 2.5 (0, 1)
(b) - 2
(c) 0
(1, 2)
(d) 5y y ≠ 0 6; 1 - q , 02 h 10, q 2 (e) Discontinuous at x = 0
35. (a) All real numbers (b) 1 - 1, 02, 10, 02 (c) y 2.5 (0, 1) (1, 0)
23. 1 2, 2
5
(1, 1) (0, 0) 5 x (1, 1)
33. (a) 5x x Ú - 2 6; 3 - 2, q 2 (b) 10, 32, 12, 02 (c) y 5 (0, 3) (2, 1)
(2, 4)
(0, 3)
(1, 1) 5 x
(d) 5y y Ú 1 6; 3 1, q 2 (e) Continuous
37. (a) 5x x Ú - 2 6; 3 - 2, q 2 (b) (0, 0) (c) y
2.5 x
(ⴚ2, 2)
(1, 5) (1, 4) (1, 1) (2, 0) 5 x
(d) 5y y 6 4 or y = 5 6; 1 - q , 42 h 55 6 (e) Discontinuous at x = 1
39. (a) All real numbers (b) 1x, 02 for 0 … x 6 1 (c) y 5
2.5 (1, 1)
y 5
5 x
(d) 25 31. (a) All real numbers (b) 10, 32 (c) y
2.5 x
(1, 2)
y 5 (1, 1)
(2, 8)
(0, 0)
10 x
(4, 4)
9. C
(1, 1) 5 x
2.5 x
(0, 0)
(d) All real numbers (e) Discontinuous at x = 0
(d) 5y y Ú 0 6; 3 0, q 2 (e) Continuous
(d) Set of even integers (e) Discontinuous at 5x x is an integer 6
0.10x 892.50 + 0.15 1x - 89252 4991.25 + 0.25 1x - 36,2502 51. f 1x2 = g 17,891.25 + 0.28 1x - 87,8502 44,603.25 + 0.33 1x - 183,2502 115,586.25 + 0.35 1x - 398,3502 116,163.75 + 0.396 1x - 400,002
if 0 … x … 8925 if 8925 6 x … 36,250 if 36,250 6 x … 87,850 if 87,850 6 x … 183,250 if 183,250 6 x … 398,350 if 398,350 6 x … 400,000 if x 7 400,000
Charge (dollars)
- x if - 1 … x … 0 -x 41. f 1x2 = e 1 (Other answers are possible.) 43. f 1x2 = e if 0 6 x … 2 -x + 2 2x 45. (a) 2 (b) 3 (c) - 4 47. (a) $39.99 (b) $46.74 (c) $40.44 0.65183x + 22.25 if 0 … x … 50 49. (a) $54.84 (b) $284.46 (c) C1x2 = e (d) 0.51026x + 29.3285 if x 7 50
if x … 0 (Other answers are possible.) if 0 6 x … 2
C 100
50
(100, 80.35) (50, 54.84) (0, 22.25)
0
50 100 x Usage (therms)
ANSWERS Section 1.5 y 275 (800, 270)
9000 if 7500 if 5250 if 55. (a) C1s2 = g 3000 if 1500 if 750 if
(960, 270)
200 (400, 170) 100 (100, 50)
(0, 0)
500 1000 x Distance (miles)
(b) C1x2 = 10 + 0.4x
s … 660 680 700 720 s Ú
659 … s … s … s … s 740
… … … …
679 699 719 739
(b) $1500
(c) $7500
(c) C1x2 = 70 + 0.25x
57. (a) 10 °C (b) 4 °C (c) - 3°C (d) - 4°C (e) The wind chill is equal to the air temperature. (f) At wind speed greater than 20 m/s, the wind chill factor depends only on the air temperature.
61. Each graph is that of y = x2, but shifted horizontally. If y = 1x - k2 2, k 7 0, the shift is right k units; if y = 1x + k2 2, k 7 0, the shift is left k units. 59. C1x2 = 63. The graph of y = - f 1x2 is the reflection about the x-axis of the graph of y = f 1x2. 3 3 65. Yes. The graph of y = 1x - 12 + 2 is the graph of y = x shifted right 1 unit and up 2 units. 67. They all have the same general shape. All three go through the points 1 - 1, - 12, 10, 02 , and 11, 12 . As the exponent increases, the steepness of the curve increases (except near x = 0).
i
0.92 1.12 1.32 1.52 1.72 1.92 2.12 2.32 2.52 2.72 2.92 3.12 3.32
0 6 x … 1 1 6 x … 2 2 6 x … 3 3 6 x … 4 4 6 x … 5 5 6 x … 6 6 6 x … 7 7 6 x … 8 8 6 x … 9 9 6 x … 10 10 6 x … 11 11 6 x … 12 12 6 x … 13
Postage (dollars)
Cost (dollars)
53. (a)
AN7
C 3.32 2.92 2.52 2.12 1.72 1.32 0.92 0
2 4 6 8 10 12 x Weight (ounces)
1.5 Assess Your Understanding (page 129) 1. horizontal; right 2. y 3. vertical; up 4. T 5. F 6. T 7. B 9. H 11. I 13. L 15. F 17. G 19. y = 1x - 42 3 21. y = x3 + 4 23. y = - x3 25. y = 4x3 27. y = - 1 2 - x + 22 29. y = - 2x + 3 + 2 31. (c) 33. (c) 35. (a) - 7 and 1 (b) - 3 and 5 (c) - 5 and 3 (d) - 3 and 5 37. (a) 1 - 3, 32 (b) 14, 102 (c) Decreasing on 1 - 1, 52 (d) Decreasing on 1 - 5, 12 y y 39. 45. 47. y 43. 41. y y 5
2.5 (1, 0) 2.5 x
(1, 0)
(21, 1)
(1, 2)
(0, 1)
(0, 1)
23 (22, 0)
5 x
(1, 0)
(0, 1)
5
(2, 2) 3
x
9
(2, 3) (1, 2) 5 x
(4, 8) (1, 4) (0, 0) 8 x
Domain: 1 - q, q2; Range: [ - 1, q2
Domain: 1 - q, q2; Range: 1 - q, q2
Domain: [ - 2, q2; Range: [0, q2
Domain: 1 - q, q2; Range: 1 - q, q2
Domain: [0, q2; Range: [0, q2
49.
51.
53.
55.
57.
y 2.5
1, 1 2 1,
1,
1 2
y (1, 1) 10 (8, 2) 10
1 , 1 2
y 5 (1, 3) (0, 2)
10
x (8, 2) (1, 1)
2.5 x
1 2
y (1, 1) 10 (8, 2) x (8, 2) (1, 1)
(1, 1) 5 x
Domain: 1 - q, q2; Domain: 1 - q, 02 h 10, q2 Range: 1 - q, q2 Range: 1 - q, 02 h 10, q2
Domain: 1 - q, q2; Range: 1 - q, q2
Domain: 1 - q, q2; Range: 1 - q, q2
59.
63.
65.
61.
y 8 (6, 5)
y 5 (4, 0)
(3, 3) (2, 1) 8 x
y 5
(1, 1)
5 x
(1, 5)
5 x (0, 1)
(2, 1) (1, 3)
Domain: 1 - q, q2; Range: [ - 3, q2
67.
y 5 (0, 2)
(2, 0) 5 x (0, 2)
(1, 1)
y 6
(3, 5)
y 5
(2, 2) (1, 0) 5 x
5 x
(0, 2)
Domain: [2, q2; Range: [1, q2
Domain: 1 - q, 0]; Range: [ - 2, q2
Domain: 1 - q, q2; Range: 1 - q, q2
Domain: 1 - q, q2; Range: [0, q2
Domain: 1 - q, q2; Range: 5 y y is an even integer 6
69. (a) F1x2 = f 1x2 + 3
(b) G1x2 = f 1x + 22
(c) P1x2 = - f 1x2
(d) H1x2 = f 1x + 12 - 2
(e) Q 1x2 =
y 7 (0, 5) (4, 1)
(2, 5)
y 5 (2, 2)
(4, 3) 5 x
(6, 2)
y 5 (0, 2) (2, 0) 3 x
(4, 2) (0, 2)
(4, 0) 5 x (2, 2)
y 3 (1, 0) (1, 0) 4 x (3, 2) (5, 4)
1 f 1x2 2
y 5 (0, 1) (4, 1)
(2, 1) (4, 0) 5 x
AN8
ANSWERS Section 1.5
(f) g 1x2 = f 1 - x2
(g) h 1x2 = f 12x2
y 5 (2, 2) (4, 0)
y 5 (0, 2)
(0, 2)
5 x (4, 2)
y 2.5
2 x
(4, 15)
c 2
y 9
1 2 3 b + 2 4
3 7 , 2 4
c0
y (2, 3) 5 1 7 , 2 4
1 3 , 2 4
F 320 256 192 128 64
(100, 212)
2
2
x
83. (a)
y 5 (6, 0) (0, 0)
7
x
(3, −3)
(2, 5) (3, 8)
(b) 9 square units
(c) The time at which the temperature adjusts between the daytime and overnight settings is moved to 1 hr sooner. It begins warming up at 5:00 am instead of 6:00 am, and it begins cooling down at 8:00 pm instead of 9:00 pm. T
76 72 68 64 60 5 10 15 20 25 t Time (hours after midnight)
76 72 68 64 60 0
45,000
Y2
5 10 15 20 25 t Time (hours after midnight)
Y1
(373, 212)
(b)
y 2.5 (1, 1) (1, 1)
(0, 0) 5 x
(1, 1)
5 x
0
K
15,000
93. (a)
(2, 0)
y 1
91. (a)
(273, 32) 0 280 310 350
(0, 32) 0 20 40 60 80 100 C
y 5
, 0 2
Temperature (F)
Temperature (F)
F 320 256 192 128 64
73. f 1x2 = 1x + 12 2 - 1
(1, 8)
0
89.
, 1 2
81. f 1x2 = - 3 1x + 22 2 - 5
(4, 3)
T 5 x
x
, 1 4
, 1 4
87. (a) 72°F; 65°F (b) The temperature decreases by 2° to 70°F during the day and 63°F overnight.
c3
x
(3, 1) 5 x
(0, 1) 2.5 x
y 2.5
x
79. f 1x2 = 2 1x - 32 2 + 1
y 4
7.5
y , 1 2.5 2
( 2, 0) 2 x
( 2, 0)
,0 2
, 1 2
77. f 1x2 = ax +
(8, 1) 9 x
2, 1 2
(g) h 1x2 = f 12x2
y 2.5
1 , 2 2
y (0, 1)
,1 2
, 1 2 2 x
1, 3 2
75. f 1x2 = 1x - 42 2 - 15
(f) g 1x2 = f 1 - x2
1 f 1x2 2
(, 0)
( 1, 2)
(c) P1x2 = - f 1x2
2, 1 2
1, 1 2
( 1, 2)
(, 3) x
(, 3)
(2, 2)
(1, 2)
y 2.5 (2, 0)
(1, 2) (2, 0) 5 x
(e) Q 1x2 =
y 1
(b) G1x2 = f 1x + 22
, 4 2
,2 y 2 5
(d) H1x2 = f 1x + 12 - 2
85.
71. (a) F1x2 = f 1x2 + 3
y 2.5 (1, 1) (1, 1) 2
2
x
2000
(b) 10% tax (c) Y1 is the graph of p(x) shifted down vertically 10,000 units. Y2 is the graph of p(x) vertically compressed by a factor of 0.9. (d) 10% tax
95. (a) 1 - 4, 22 (b) 11, - 122 (c) 1 - 4, 52 97. The graph of y = f 1x2 - 2 is the graph of y = f 1x2 shifted down 2 units. The graph of y = f 1x - 22 is the graph of y = f 1x2 shifted right 2 units. 99. The range of f 1x2 = x2 is [0, q 2. The graph of g 1x2 = f 1x2 + k is the graph of f shifted up k units if k 7 0 and down 0 k 0 units if k 6 0, so the range of g is [k, q 2 .
ANSWERS Review Exercises
AN9
1.6 Assess Your Understanding (page 135) 3. (a) d 1x2 = 2x2 - x + 1
1. (a) d 1x2 = 2x4 - 15x2 + 64 (b) d 102 = 8 (c) d 112 = 250 ≈ 7.07 (d)
(b)
1 4 x 2 7. (a) A1x2 = x 116 - x2 2 (b) Domain: 5x 0 6 x 6 4 6 (c) The area is largest when x ≈ 2.31.
5. A1x2 =
2
40
30
0
2 0
10
10
(c) d is smallest when x = 0.5.
5
0
4 0
(e) d is smallest when x ≈ - 2.74 or x ≈ 2.74
25 - 20x + 4x2 p (b) Domain: 5x 0 6 x 6 2.5 6 (c) A is smallest when x ≈ 1.40 m.
(b) p 1x2 = 4x + 4 24 - x2 (d) p is largest when x ≈ 1.41.
9. (a) A1x2 = 4x 24 - x2 (c) A is largest when x ≈ 1.41.
11. (a) A1x2 = x2 +
12
10
8
0
0
2
2 0
0
0
2.5 0
13. (a) C1x2 = x
(b) A1x2 =
x2 4p
15. (a) A1r2 = 2r 2
19. (a) d 1t2 = 22500t 2 - 360t + 13 (b) d is smallest when t ≈ 0.07 hr.
(b) p 1r2 = 6r
21. V1r2 =
17. A1x2 = a
pH1R - r2r 2
p 23 2 bx 3 4
25. (a) V1x2 = x 124 - 2x2 2 (b) 972 in .3 (c) 160 in .3 (d) V is largest when x = 4.
R
12 - x 2x2 + 4 + 5 3 (b) 5x 0 … x … 12 6 (c) 3.09 hr (d) 3.55 hr
23. (a) T1x2 =
5
0
1100
0.2 0 0
12 0
1.7 Assess Your Understanding (page 141) 1 250 1 9d 2 8a3 4p 3 1 x 5. A = px2 7. F = 2 9. z = 1x2 + y2 2 11. M = 13. T 2 = 2 15. V = r 17. A = bh 5 5 3 2 d d 2 2x mM 429 (b) 143 bags 19. F = 6.67 * 10 - 11 a 2 b 21. p = 0.00649B; $941.05 23. 144 ft; 2 s 25. 2.25 27. R = 3.95g; $41.48 29. (a) D = p d 3 3 2 31. 450 cm 33. 124.76 lb 35. V = pr h 37. 54.86 lb 39. 26 ≈ 1.82 in . 41. 2812.5 joules 43. 384 psi 1. y = kx
2. F 3. y =
Review Exercises (page 145) 1. Function; domain 5 - 2, 1, 3 6 , range 52, 5 6 (f) (c)
9x 9x2 - 1 7 - x2 x
2
4. (a) 0 (d)
x2 - 7 x
2
10. 5x 0 x Ú 0, x ≠ 1 6
(c) 2x2 + 3x
(b) 3 12 (e)
- x2 + 2x + 6 x - 2x + 1 2
11. {x 0 x 7 - 8}
(f)
(b) - 9>8
2. Not a function
3. (a) 9>8
(d) - 2x2 - 3x
(e) 2x2 - 5x + 4
7 - 9x2 9x2
6. 5x 0 x ≠ 2, x ≠ - 2 6
12. 1f + g2 1x2 = 2x + 3; Domain: all real numbers 1f - g2 1x2 = - 4x + 1; Domain: all real numbers 1ƒ # g2 1x2 = - 3x2 + 5x + 2; Domain: all real numbers f 2 - x 1 a b 1x2 = ; Domain: e x ` x ≠ - f g 3x + 1 3
(c) -
3x x2 - 1
(f) {3 2x2 - x
3x 3 (e) x + 1 x2 - 1 2 2 5. (a) (b) 9 9
(d) -
7. 5x 0 x … 2 6 8. 5x 0 x ≠ 0 6
9. 5x 0 x ≠ - 3, x ≠ 1 6
13. 1f + g2 1x2 = 3x2 + 4x + 1; Domain: all real numbers 1f - g2 1x2 = 3x2 - 2x + 1; Domain: all real numbers 1ƒ # g2 1x2 = 9x3 + 3x2 + 3x; Domain: all real numbers f 3x2 + x + 1 a b 1x2 = ; Domain: 5x x ≠ 0 6 g 3x
AN10
ANSWERS Review Exercises
x2 + 1 ; Domain: 5x x ≠ 0, x ≠ 1 6 x 1x - 12
1f - g2 1x2 = 1ƒ # g2 1x2 =
15. - 4x + 1 - 2h 16. (a) Domain: 5x - 4 … x … 3 6; Range: 5y - 3 … y … 3 6 (b) 10, 02 (c) - 1 (d) - 4 (e) 5x 0 6 x … 3 6 (g) (h) (f) y y
x2 + 2x - 1 ; Domain: 5x x ≠ 0, x ≠ 1 6 x 1x - 12
14. 1f + g2 1x2 =
5
x + 1 ; Domain: 5x x ≠ 0, x ≠ 1 6 x 1x - 12
(1, 3)
x 1x + 12 f a b 1x2 = ; Domain: 5x x ≠ 0, x ≠ 1 6 g x - 1
23.
20
3
y 2
33. (0, 0) 5 x
(4, 0)
(2, 4)
5 x (2, 2)
(2, 4)
Intercepts: 1 - 4, 02, 14, 02, 10, - 42 Domain: all real numbers Range: 5y y Ú - 4 6 or 3 - 4, q 2
Intercept: 10, 02 Domain: all real numbers Range: 5y y … 0 6 or 1 - q , 0 4
35.
36.
y (0, 3)
5
(2, 3)
y 30 (3.6, 0)
(1, 2) 5 x
3 (0, 24)
(3, 6) (2, 8)
Intercepts: 10, - 242 , 3 1 - 2- 2 4 , 02 or about 1 - 3 .6 , 02 Domain: all real numbers Range: all real numbers
Intercept: 10, 32 Domain: all real numbers Range: 5y y Ú 2 6 or 32, q 2
39. A = 11
40 40. (a) A1x2 = 2x + x (d) 100
0
(b) 42 ft
7 0
(1, 1)
5 x
(4, 2) (0, 0) 5 x
34.
y 5 (2, 1)
A is smallest when x ≈ 2 .15 ft .
2
(c) 28 ft
2
y 5
(5, 2) (3, 2)
(1, 0)
7 x
(0, 1) (1, 0) 5 x
Intercept: 11, 02 Domain: 5x x Ú 1 6 or 3 1, q 2 Range: 5y y Ú 0 6 or 3 0, q 2
Intercepts: 10, 12, 11, 02 Domain: 5x x … 1 6 or 1 - q , 1 4 Range: 5y y Ú 0 6 or 30, q 2
37. (a) 5x x 7 - 2 6 or 1 - 2, q 2 (b) 10, 02 (c) y (1, 3)
38. (a) 5x x Ú - 4 6 or [ - 4, q 2 (b) 10, 12 (c) y
5
(1, 10)
2
5 (4, 4)
Local maximum value: 1.53 at x = 0.41 Local minima values: 0.54 at x = - 0.34 and - 3.56 at x = 1.80 Increasing: 1 - 0.34, 0.412; 11.80, 32 Decreasing: 1 - 2, - 0.342; 10.41, 1.802
32.
y 5
(2, 2) (0, 4)
26. 44 27. No 30. y
4
Local maximum value: 4.04 at x = - 0.91 Local minimum value: - 2.04 at x = 0.91 Increasing: 1 - 3, - 0.912; 10.91, 32 Decreasing: 1 - 0.91, 0.912
(4, 0)
y 5
(0, 0)
3
20
5 x (3, 3)
(c) 47 25. 2
(4, 4) 2
(2, 1) (0, 0)
10 x
(8, 3)
24. (a) 23 (b) 7 28. Yes 29.
40
3
31.
(0, 0) (4, 1)
5 x (1, 1)
y (4, 3) 5
(6, 3)
(d) Absolute maximum: f 142 = 3 Absolute minimum: none (e) No symmetry (f) Neither (g) x-intercepts: - 3, 0, 3; y-intercept: 0 18. Odd 19. Even 20. Neither 21. Odd
17. (a) Domain: 5x x … 4 6 or 1 - q , 4]; Range: 5y y … 3 6 or 1 - q , 3] (b) Increasing on 1 - q , - 22 and 12, 42 ; Decreasing on 1 - 2, 22 (c) Local maximum value is 1 and occurs at x = - 2. Local minimum value is - 1 and occurs at x = 2. 22.
5
(6, 3) (3, 0)
5
(2, 3) (1, 2) 5 x
(2, 6)
(d) 5y y 7 - 6 6 or 1 - 6, q 2 (e) Discontinuous at x = 1
(0, 1) (4, 4)
(1, 3) 5 x
(d) 5y - 4 … y 6 0 or y 7 0 6 or 3 - 4, 02 h 10, q 2 (e) Discontinuous at x = 0
41. (a) A1x2 = 10x - x3 (b) The largest area that can be enclosed by the rectangle is approximately 12.17 square units. 427 42. P = B; $1083.92 65,000 43. 199.9 lb 44. 189 BTU
ANSWERS Section 2.1
AN11
Chapter Test (page 147) 1. (a) Function; Domain: 52, 4, 6, 8 6; Range: 55, 6, 7, 8 6 (b) Not a function (c) Not a function (d) Function; Domain: all real numbers; Range: 4 1 5y y Ú 2 6 2. Domain: e x ` x … f ; f 1 - 12 = 3 3. Domain: 5x x ≠ - 2 6; g 1 - 12 = 1 4. Domain: 5x x ≠ - 9, x ≠ 4 6; h 1 - 12 = 5 8 5. (a) Domain: 5x - 5 … x … 5 6; Range: 5y - 3 … y … 3 6 (b) 10, 22 , 1 - 2, 02, and 12, 02 (c) f 112 = 3 (d) x = - 5 and x = 3 (e) 5x - 5 … x 6 - 2 or 2 6 x … 5 6 or [ - 5, - 22 h 12, 5] 6. Local maximum values: f 1 - 0.852 ≈ - 0.86; f 12.352 ≈ 15.55; local minimum value: f 102 = - 2; the function is increasing on the intervals 1 - 5, - 0.852 and 10, 2.352 and decreasing on the intervals 1 - 0.85, 02 and 12.35, 52 . y 10. (a) (b) 8. 19 y 7. (a) (b) 10, - 42, 14, 02 y 3 10 10 9. (a) 1 f - g2 1x2 = 2x2 - 3x + 3 (c) g 1 - 52 = - 9 (2, 5) (0, 1) (b) 1 ƒ # g2 1x2 = 6x3 - 4x2 + 3x - 2 (6, 4) 5 x (1, 3) (2, 4) (d) g 122 = - 2 2.5 x 10 x (c) f 1x + h2 - f 1x2 = 4xh + 2h2 (4, 2) 11. (a) 8.67% occurring in 1997 1x ≈ 52 (b) The model predicts that the interest rate will be - 10.343,. This is not reasonable. 5x px2 x2 + (b) 1297.61 ft 3 13. 14.69 ohms 12. (a) V1x2 = 8 4 64
CHAPTER 2 Linear and Quadratic Functions 2.1 Assess Your Understanding (page 158) 7. slope; y-intercept 8. - 4; 3 13. (a) m = 2; b = 3 (b) y
9. positive
12. False 1 17. (a) m = ; b = - 3 4 (b) y
8
(0, 3) 1
(1, 5) 2
(0, 4) 1
(d) Increasing
(c) - 3 25. (a) 8 (b)
y
3 (1, 1)
(0, 3)
(d) Decreasing
5 (0, 4) 9 (2, 0) 5 x
(8, 0)
(c)
1 4
21. (a) 4 (b) y
5 (0, 4) (4, 2) 1 4
(4, 0)
1
9 x
8
5 x
x
(0, 8)
(c) 0 (d) Constant
(d) Increasing
27. Linear; f 1x2 = - 3x - 2 1 1 1 35. (a) (b) e x ` x 7 f or a , q b 29. Nonlinear 4 4 4 (c) 1 (d) 5x x … 1 6 or 1 - q , 1 4 31. Nonlinear y (e) 33. Linear; f 1x2 = 8
y
16 (0, 10)
19. (a) m = 0; b = 4 (b) y
5
8 x
8 x
23. (a) 2 (b)
11. False
15. (a) m = - 3; b = 4 (b) y
8
(c) 2
10. True
x
4
(1, 3)
y f (x)
5 x y g (x)
37. (a) 40 (b) 88 (c) - 40 (d) 5x x 7 40 6 or 140, q 2 (e) 5x x … 88 6 or ( - q , 88] (f) 5x - 40 6 x 6 88 6 or 1 - 40, 882 39. (a) - 4 (b) 5x x 6 - 4 6 or 1 - q , - 42 41. (a) - 6 (b) 5x - 6 … x 6 5 6 or 3 - 6, 52 43. (a) $59 (b) 180 mi (c) 300 mi (d) 5x x Ú 0 6 or 30, q ) (e) The cost of renting the car for a day increases $0.35 for each mile driven, or there is a charge of $0.35 per mile to rent the car in addition to a fixed charge of $45. (f) It costs $45 to rent the car if 0 miles are driven, or there is a fixed charge of $45 to rent the car, in addition to a charge that depends on mileage. 45. (a) $24; 600 T-shirts (b) $0 … p 6 +24 (c) The price will increase.
5,000 4,000 3,000 2,000 1,000
51. (a) V(x) = - 1000x + 3000 (b) 5x 0 … x … 3 6 or 3 0, 3 4 (c) V(x) 3000
(36,250, 4991.25)
2000 1000 0
(8925, 892.50)
0 12,000 30,000 x Adjusted Gross Income ($)
(e) $27,600 (f) For each additional dollar of taxable income between $8925 and $36,250, the tax bill of a single person in 2013 increased by $0.15.
3000 2000 1000 x 0 4 8 12 Number of Bicycles
1 2 3 x Age (years)
(d) $1000 (e) After 1 year 55. (a) C (x) = 0.89x + 31.95 (b) $129.85; $236.65
53. (a) C(x) = 90x + 1800 (b) C(x) Cost (dollars)
49. (a) x = 5000 (b) x 7 5000
Book Value (dollars)
Tax Bill ($)
47. (a) 5x 8925 … x … 36,250 6 or 3 8925, 36,250 4 (b) $2553.75 (c) The independent variable is adjusted gross income, x. The dependent variable is the tax bill, T. (d) T (20,000, 2553.75)
(c) $3060 (d) 22 bicycles
ANSWERS Section 2.1 n
(b) Since each input (memory) corresponds to a single output (number of songs), we know that number of songs is a function of memory. Also, because the average rate of change is a constant 218.75 songs per gigabyte, the function is linear. (c) n(m) = 218.75 m (d) {m 0 m Ú 0} or [0, q )
16,000 12,000 8,000 4,000 0 16 32 48 64 m Memory (gigabytes)
61. b = 0; yes, f 1x2 = b
59. (d), (e)
63.
y
64. 6
(e)
Number of Songs
57. (a)
Number of Songs
AN12
n 16,000 12,000 8,000 4,000 0 16 32 48 64 m Memory (gigabytes)
(f) If memory increases by 1 GB, then the number of songs increases by 218.75.
65. 7
y
66.
2 10 x
(−2, 4)
5 (4, 3)
(−1, 1) (2, −5)
(1, 2) 5
x
2.2 Assess Your Understanding (page 165) 3. scatter diagram 4. decrease; 0.008 11. (a) 20
0
5. Linear relation, m 7 0
7. Linear relation, m 6 0 (c) 20
10
0
0
10
13. (a)
3
6
3
3
25
25
0
0 90
(b) Answers will vary. Using ( - 20, 100) and ( - 10, 140), y = 4x + 180.
150
25
0 90
38 46 54 62 70 x Weight (grams)
(d)
Number of Calories
(d) y = 3.8613x + 180.2920
(c) Answers will vary. Using the points (39.52, 210) and (66.45, 280), y = 2.599x + 107.288.
y 280 260 240 220 200
(e)
150
90
3
6
(d) y = 2.2x + 1.2 (c)
150
Number of Calories
6
6
9 1 (b) Answers will vary. Using ( - 2, - 4) and (2, 5), y = x + . 4 2 15. (a)
(e)
6
3
3
10 0
(d) y = 2.0357x - 2.3571 (c)
6
20
0
0
(b) Answers will vary. Using (4, 6) and (8, 14), y = 2x - 2.
17. (a)
9. Nonlinear relation (e)
(e) 269 calories (f) If the weight of a candy bar is increased by 1 gram, the number of calories will increase by 2.599, on average.
y 280 260 240 220 200 38 46 54 62 70 x Weight (grams)
(b) Linear with positive slope 19. (a) The independent variable is the number of hours spent playing video games, and the dependent variable is cumulative grade-point average, because we are using number of hours playing video games to predict (or explain) cumulative grade-point average.
(b)
4
1
13 0
(c) G(h) = - 0.0942h + 3.2763 (d) If the number of hours playing video games in a week increases by 1 hour, the cumulative grade-point average decreases 0.09, on average. (e) 2.52 (f) Approximately 9.3 hours
ANSWERS Section 2.3 (c) D = - 1.3355p + 86.1974 (d) If the price increases $1, the quantity sold per day decreases by about 1.34 pairs of jeans, on average. (e) D1p2 = - 1.3355p + 86.1974 (f) 5p 0 6 p … 64 6 (g) About 49 pairs
D 60 55 50 45 40 18 22 26 30 34 p Price (dollars/pair)
23.
y Incidence Rate (per 1000)
Demand (pairs of jeans sold per day)
21. (a) No (b)
AN13
50 40 30 20 10 x 40 35 Age of Mother
No, the data do not follow a linear pattern. 25. No linear relation 27. 34.8 hours; A student whose GPA is 0 spends 34.8 hours each week playing video games.; G(0) = 3.28; The average GPA of a student who does not play video games is 3.28. 28. 2x + y = 3 or y = - 2x + 3 29. {x x ≠ - 5, x ≠ 5} 30. (g - f )(x) = x2 - 8x + 12 31. y = (x + 3)2 - 4
2.3 Assess Your Understanding (page 177) - b { 2b2 - 4ac 11. Zeros: 0, 9; x-intercepts: 0, 9 2a 1 1 13. Zeros: - 5, 5; x-intercepts: - 5, 5 15. Zeros: - 3, 2; x-intercepts: - 3, 2 17. Zeros: - , 3; x-intercepts: - , 3 19. Zeros: - 4, 4; x-intercepts: - 4, 4 2 2 3 3 21. Zeros: - 6, - 2; x-intercepts: - 6, - 2 23. Zero: ; x-intercept: 25. Zeros: - 2 22, 2 22; x-intercepts: - 2 22, 2 22 2 2 7. repeated; multiplicity 2
8. discriminant; negative 9. 0, 1, 2
27. Zeros: - 1, 3; x-intercepts: - 1, 3
29. Zeros:
10. x =
- 3 - 4 22 - 3 + 4 22 - 3 - 4 22 - 3 + 4 22 , ; x-intercepts: , 2 2 2 2
1 3 1 3 31. Zeros: - 2 - 2 23, - 2 + 2 23; x-intercepts: - 2 - 2 23, - 2 + 2 23 33. Zeros: - , ; x-intercepts: - , 4 4 4 4 - 1 - 27 - 1 + 27 - 1 - 27 - 1 + 27 35. Zeros: , ; x-intercepts: , 37. Zeros: 2 - 22, 2 + 22; x-intercepts: 2 - 22, 2 + 22 6 6 6 6 3 3 39. Zeros: 2 - 25, 2 + 25; x-intercepts: 2 - 25, 2 + 25 41. Zeros: 1, ; x-intercepts: 1, 43. No real zeros; no x-intercepts 2 2 - 1 - 25 - 1 + 25 - 1 - 25 - 1 + 25 - 2 - 210 - 2 + 210 45. Zeros: , ; x-intercepts: , 47. Zeros: , ; 4 4 4 4 2 2 - 2 - 210 - 2 + 210 1 1 1 1 1 , 49. Zero: ; x-intercept: 51. 5 - 6, 0 6; 1 - 6, 32, (0, 3) 53. e - 1, - f ; 1 - 1, - 12, a - , b x-intercepts: 2 2 3 3 2 2 2 55. 5 - 3, 5 6; 1 - 3, 132, (5, 21) 57. Zeros: - 2 22, 2 22; x-intercepts: - 2 22, 2 22 59. Zeros: - 2, - 1, 1, 2; x-intercepts: - 2, - 1, 1, 2 1 1 61. Zeros: - 1, 1; x-intercepts: - 1, 1 63. Zeros: - 2, 1; x-intercepts: - 2, 1 65. Zeros: - 6, - 5; x-intercepts: - 6, - 5 67. Zero: - ; x-intercept: 3 3 3 3 69. Zeros: - , 2; x-intercepts: - , 2 71. Zeros: 0, 16; x-intercepts: 0, 16 73. Zero: 16; x-intercept: 16 75. Zeros: - 5 22, 5 22; 2 2 1 1 3 5 3 5 1 2 1 2 79. Zeros: - , ; x-intercepts: - , 81. Zeros: - , ; x-intercepts: - , x-intercepts: - 5 22, 5 22 77. Zero: ; x-intercept: 4 4 5 2 5 2 2 3 2 3 - 22 + 2 - 22 - 2 - 22 + 2 - 22 - 2 83. Zeros: , ; x-intercepts: , 2 2 2 2 - 1 - 217 - 1 + 217 - 1 - 217 - 1 + 217 85. Zeros: , ; x-intercepts: , 2 2 2 2 87. (a)
(b) - 1, 3
y 5 (1, 0) (0, 3)
(3, 0) 5 x (2, 3) (1, 4)
1 2 1 25 2 10 93. e - , f ; a - , b , a , - b 4 3 4 16 3 9
89. (a)
(b) - 2, - 6
y 40
(6, 0)
91. (a)
y (3, 6) 6 (4, 3) x 3 (3 兹2, 0) (3 兹2, 0)
(b) 3 - 22, 3 + 22
(2, 3)
(2, 0) 4 x
(4, 8)
95. { - 8}; ( - 8, 180)
5 5 45 97. e f ; a , - b 3 3 88
99. (a) 92 - 213, - 2 + 213 (b) 5 (c) - 7, - 4, 1, 2 101. 11 ft by 13 ft 103. 4 ft by 4 ft 105. (a) The ball strikes the ground after 6 sec. (b) The ball passes the top of the building on its way down after 5 sec. 107. 20 consecutive integers - b + 2b2 - 4ac - b - 2b2 - 4ac - 2b b 1 1 + = = 111. k = or k = 2a 2a 2a a 2 2 b { 2 1 - b2 2 - 4ac - b { 2b2 - 4ac b { 2b2 - 4ac - b { 2b2 - 4ac 2 2 113. ax + bx + c = 0, x = ; ax - bx + c = 0, x = = = 2a 2a 2a 2a 115. (b) y 122. Domain: { - 3, - 1, 1, 3}; Range: {2, 4}; 121. Function 6 3 123. a - 4, b 6 2 x 124. (1, 4) 109.
AN14
ANSWERS Section 2.4
2.4 Assess Your Understanding (page 189) 7. -
5. parabola 6. axis or axis of symmetry 19.
21.
y 5 (4, 2)
29. f 1x2 = 2 1x + 12 2 - 4
23.
(0, 1) 5 x
(1, 1)
31. (a)
y 5
(2, 0)
5 x (0, 2)
39. (a)
(2, 1)
5 x
(2, 1) 5 x (1, 1)
(4, 2) (3, 1) (2, 2)
(2, 0)
5 x
5 x
35. (a)
y (2, 0) 5 x
1
(0, 0) 2 x
(6, 0)
(b) Domain: 1 - q , q 2 Range: 1 - q , 9] (c) Increasing: 1 - q , - 32 Decreasing: 1 - 3, q 2
1 4
1, 5 2 2
y 2
1, 2 2
5 x (0, 3)
1 , 15 4 8
(0, 8) (1, 9) x 1
x 3
41. (a)
y 5
x
(0, 0)
(1, 1)
y 9
(3, 9)
(1, 1)
(0, 2)
y 5
5 x
(0, 2)
x 1
27. f 1x2 = - 1x + 12 2 + 1
(0, 0)
(0, 1)
(1, 0)
17. H y 5
x 1
y 5
15. G
(4, 0)
(b) Domain: 1 - q , q 2 Range: [ - 1, q 2 (c) Decreasing: 1 - q , - 12 Increasing: 1 - 1, q 2
37. (a)
13. F
25. f(x) = (x + 2)2 - 2
33. (a)
y 2
(1, 1) (2, 2) (1, 4)
10. True 11. C y 5
(0, 2)
(3, 1) (2, 2)
5 x (2, 1)
(2, 1) (0, 2)
8. True 9. True
y 5 (4, 2)
(4, 2)
b 2a
(1, 3)
(b) Domain: 1 - q , q 2 Range: [ - 9, q 2 (c) Decreasing: 1 - q , - 12 Increasing: 1 - 1, q 2 43. (a)
3 17 , 4 4
y 5 (0, 2)
(1.78, 0)
2.5 x (0.28, 0)
3 x x
1 2
x
3 4
(b) Domain: 1 - q , q 2 (b) Domain: 1 - q , q 2 (b) Domain: 1 - q , q 2 15 5 17 Range: c , q b Range: a - q , - d d Range: a - q , 8 2 4 1 1 3 (c) Decreasing: a - q , b (c) Increasing: a - q , b (c) Increasing: a - q , - b 4 2 4 1 1 3 Increasing: a , q b Decreasing: a , q b Decreasing: a - , q b 4 2 4 2 2 2 2 2 45. f 1x2 = 1x + 12 - 2 = x + 2x - 1 47. f 1x2 = - 1x + 32 + 5 = - x - 6x - 4 49. f 1x2 = 2 1x - 12 - 3 = 2x2 - 4x - 1 51. Minimum value; - 18 53. Minimum value; - 21 55. Maximum value; 21 57. Maximum value; 13 y y y 65. (a) y 61. (a) 63. (a) 59. (a) (3, 2) (b) Domain: 1 - q , q 2 Range: [0, q 2 (c) Decreasing: 1 - q , - 12 Increasing: ( - 1, q)
2
(3, 0)
4
(5, 0)
(b) Domain: 1 - q , q 2 Range: [ - 16, q 2 (c) Decreasing: 1 - q , 12 Increasing: 11, q 2
67. (a)
(b) Domain: 1 - q , q 2 Range: 1 - q , q 2 (c) Increasing: 1 - q , q 2
69. (a)
y 10
1 ,0 2
(0, 4) 12 (10, 0) x
(b) Domain: 1 - q , q 2 Range: 1 - q , q 2 (c) Decreasing: 1 - q , q 2
(1, 1)
1 ,1 2
(2, 0)
(b) Domain: 1 - q , q 2 Range: 1 - q , 2] (c) Increasing: 1 - q , 32 Decreasing: 13, q 2
71. a = 6, b = 0, c = 2 73. (a), (c), (d)
y 2 5 x
y 6
(0, 1)
(b) Domain: 1 - q , q 2 Range: 1 - q , 0] 1 (c) Increasing: a - q , - b 2 1 Decreasing: a - , q b 2
(0, 1)
(6, 16)
(0, 16)
(0, 5)
8
7 x
5 x (1, 16)
(4, 0)
2
5, 0 2
7 x
5 x 1 7 , 4 8
(b) Domain: 1 - q , q 2 7 Range: c , q b 8 1 (c) Decreasing: a - q , - b 4 1 Increasing: a - , q b 4 75. (a), (c), (d) y (1, 3) 1
(3, 5)
(1, 3)
(b) 5 - 1, 3 6
5 x
5 x (3, 5)
(b) 5 - 1, 3 6
ANSWERS Section 2.6 79. (a) a = 1: f 1x2 = 1x + 32 1x - 12 = x2 + 2x - 3 a = 2: f 1x2 = 2 1x + 32 1x - 12 = 2x2 + 4x - 6 a = - 2: f 1x2 = - 2 1x + 32 1x - 12 = - 2x2 - 4x + 6 a = 5: f 1x2 = 5 1x + 32 1x - 12 = 5x2 + 10x - 15 (b) The value of a does not affect the x-intercepts, but it changes the y-intercept by a factor of a. (c) The value of a does not affect the axis of symmetry. It is x = - 1 for all values of a. (d) The value of a does not affect the x-coordinate of the vertex. However, the y-coordinate of the vertex is multiplied by a. (e) The mean of the x-intercepts is the x-coordinate of the vertex.
77. (a), (c), (d) y 10
(2, 6) 10 x
(1, 6)
(b) 5 - 1, 2 6
AN15
81. (a) 1 - 2, - 252 (b) - 7, 3 (c) - 4, 0; 1 - 4, - 212, 10, - 212 y (d) 12 (7, 0) (4, 21) (2, 25)
(3, 0) 10 x (0, 21)
83. $500; $1,000,000 85. (a) 70,000 mp3 players (b) $2500 87. (a) 171 ft (b) 49 mph (c) Reaction time 89. (a) 187 or 188 watches; $7031.20 (b) P1x2 = - 0.2x2 + 43x - 1750 (c) 107 or 108 watches; $561.20 91. f(x2 = 2 1x + 42 1x - 22 93. If x is even, then ax2 and bx 2 2 are even and ax + bx is even, which means that ax + bx + c is odd. If x is odd, then ax2 and bx are odd and ax2 + bx is even, which means that ax2 + bx + c is odd. In either case, f(x) is odd. 95. 97. b2 - 4ac 6 0 99. No y 5
y x 2 2x 1 y x 2 2x 3 5 x
y x 2 2x
101. Symmetric with respect to the x-axis, the y-axis, and the origin. 102. {x x … 4} or ( - q , 4] 103. Center (5, - 2); radius = 3 104. Zeros: 3 - 25, 3 + 25; x-intercepts: 3 - 25, 3 + 25
2.5 Assess Your Understanding (page 194) 3. (a) 5x x 6 - 2 or x 7 2 6; 1 - q , - 22 h 12. q 2 (b) 5x - 2 … x … 2 6; [ - 2, 2] 5. (a) 5x - 2 … x … 1 6; [ - 2, 1] (b) 5x x 6 - 2 or x 7 1 6; 1 - q , - 22 h 11, q 2 7. 5x - 2 … x … 5 6; [ - 2, 5] 9. 5x x 6 0 or x 7 4 6; 1 - q , 02 h 14, q 2 11. 5x x 6 - 4 or x 7 3 6; 1 - q , - 42 h 13, q 2 13. { - 3} 15. e x ` -
1 1 … x … 3 f ; c - , 3 d 17. 5x x 6 - 1 or x 7 8 6; 1 - q , - 12 h 18, q 2 2 2 19. No real solution 21. All real numbers; 1 - q , q 2 23. 5x x … - 4 or x Ú 4 6; 1 - q , - 4] h [4, q ) 25. (a) 5 - 1, 1 6 (b) 5 - 1 6 (c) 5 - 1, 4 6 (d) 5x x 6 - 1 or x 7 1 6; 1 - q , - 12 h 11, q 2 (e) 5x x … - 1 6 ; 1 - q , - 1] (f) 5x x 6 - 1 or x 7 4 6; 1 - q , - 12 h 14, q 2 1 (g) 5 x x … - 22or x Ú 22 6 ; 1 - q , - 22 4 h 3 2 2, q 2 27. (a) 5 - 1, 1 6 (b) e - f (c) 5 - 4, 0 6 (d) 5x - 1 6 x 6 1 6; 1 - 1, 12 4 1 1 (e) e x ` x … - f or a - q , - d (f) 5x - 4 6 x 6 0 6; 1 - 4, 02 (g) 50 6 29. (a) 5 - 2, 2 6 (b) 5 - 2, 2 6 (c) 5 - 2, 2 6 4 4 (d) 5x x 6 - 2 or x 7 2 6; 1 - q , - 22 h 12, q 2 (e) 5x x … - 2 or x Ú 2 6; 1 - q , - 2] h [2, q ) (f) 5x x 6 - 2 or x 7 2 6; 1 - q , - 22 h 12, q 2 (g) 5 x x … - 25or x Ú 25 6 ; 1 - q , - 25 4 h 3 25, q 2 31. (a) 5 - 1, 2 6 (b) 5 - 2, 1 6 (c) 50 6 (d) 5x x 6 - 1 or x 7 2 6; 1 - q , - 12 h 12, q 2 (e) 5x - 2 … x … 1 6; [ - 2, 1] 1 - 213 1 + 213 1 - 213 1 + 213 or x Ú f; a - q, d h c , qb 2 2 2 2 33. (a) 5 sec (b) The ball is more than 96 ft above the ground for time t between 2 and 3 sec, 2 6 t 6 3. 35. (a) $0, $1000 (b) The revenue is more than $800,000 for prices between $276.39 and $723.61, +276.39 6 p 6 +723.61. 41. When the inequality is not strict. 42. {x x … 5} 43. Odd 44. (a) 9 45. (a) 18 y (b) (f) 5x x 6 0 6; 1 - q , 02
(g) e x ` x …
5 (9, 0) 5
x 24
28 16
(0, −6)
(b) C(H) = 0.3734H + 7.3268 (c) If height increases by 1 in., then head circumference increases by about 0.3734 in., on average. (d) Approximately 17.0 in. (e) Approximately 26.98 in.
2.6 Assess Your Understanding (page 202) 1 2 1 x + 100x (b) 5x 0 … x … 600 6 (c) $13,333.33 (d) 300; $15,000 (e) $50 5. (a) R1x2 = - x2 + 20x (b) $255 6 5 (d) $10 (e) Between $8 and $12 7. (a) A1w2 = - w 2 + 200w (b) A is largest when w = 100 yd. (c) 10,000 yd2 9. 2,000,000 m2
3. (a) R1x2 = (c) 50; $500
AN16 11. (a) (d)
ANSWERS Section 2.6
625 ≈ 39 ft 16
(b)
7025 ≈ 219.5 ft 32
(c) About 170 ft 13. 18.75 m 15. (a) 3 in. (b) Between 2 in. and 4 in. 750 ≈ 238.73 m by 375 m 17. p a 38 248 19. x = 21. 23. 2 3 3
220
(f) When the height is 100 ft, the projectile is about 135.7 ft from the cliff. 0
200 0
25. (a)
50,000
15
75
27. (a)
50,000
(e)
15
0
75 0
The data appear to follow a quadratic relation with a 6 0. (b) I(x) = - 43.335x2 + 4184.883x - 54, 062.439 (c) About 48.3 years of age (d) Approximately $46,972 32. 15i
33. 13
34. (x + 6)2 + y2 = 7
35.
29. (a)
1800
400 1000
1100
125
10
45 0
The data appear to follow a linear relation with positive slope. (b) R(x) = 0.953x + 704.186 (c) $1514
The data appear to follow a quadratic relation with a 6 0. (b) B(a) = - 0.483a2 - 26.356a - 251.342 (c) 79.4 births per 1000 population
- 4 - 231 - 4 + 231 , 5 5
2.7 Assess Your Understanding (page 209) 7. 2 + 3i 8. True 9. - 2i, 2i
11. - 4, 4
y 7
13. 3 - 2i, 3 + 2i y 4
(0, 4)
(4, 0)
(4, 0) 10 x
(0, 13)
y 16
18 (0, 10)
(6, 10)
(3, 4)
5 x
17. 2 - 23, 2 + 23
19. -
y
1 1 1 1 - i, - + i 2 2 2 2
21. -
1 23 1 23 i, - + i 2 2 2 2
y 8
6
(3, 1) 8 x
8 x
(0, 16)
(0.27, 0)
15. 3 - i, 3 + i
y
23.
4 - 3 22 4 + 3 22 , 2 2 y 10
y 8
(3.73, 0) 7 x
(2, 9)
(0, 1) (1, 1)
(2, 3)
(0, 1) 5 x
1 1 , 2 2
25. Two complex solutions that are conjugates of each other
(1, 1)
(0.12, 0)
(0, 1)
1 3 , 2 4
27. Two unequal real solutions
5 x
3
(4, 1) x (4.12, 0)
29. A repeated real solution 31. - 2, 2, - 2i, 2i
1 23 1 23 3x + 2 33. - i, - + i, - 1 - 23i, - 1 + 23i, 1, 2 35. (g - f)(x) = ; Domain: {x x ≠ - 1, x ≠ 0} 2 2 2 2 x(x + 1) 36. (a) Domain: [ - 3, 3]; Range: [ - 2, 2] (b) ( - 3, 0), (0, 0), (3, 0) (c) Symmetric with respect to the origin (d) The relation is a function. 600 25 38. y = 2 37. x −4
4
−25
Local maximum: (0, 0) Local minima: ( - 2.12, - 20.25), (2.12, - 20.25) Increasing: ( - 2.12, 0), (2.12, 4) Decreasing: ( - 4, - 2.12), (0, 2.12)
2.8 Assess Your Understanding (page 213) 7. - a; a 8. - a 6 u 6 a 9. … 10. True 11. False 12. False 13. (a) 5 - 9, 3 6 (b) 5x - 9 … x … 3 6; [ - 9, 3] (c) 5x x 6 - 9 or x 7 3 6 ; 1 - q , - 92 h 13, q 2 15. (a) 5 - 2, 3 6 (b) 5x x … - 2 or x Ú 3 6; 1 - q , - 2] h [3, q 2 (c) 5x - 2 6 x 6 3 6; 1 - 2, 32 17. 5 - 6, 6 6
AN17
ANSWERS Review Exercises 19. 5 - 4, 1 6
21. e - 1,
3 1 1 f 23. 5 - 4, 4 6 25. e - , f 2 2 2 35. 5 - 1, 3, 1 - 22i, 1 + 22i 6 37. 5 - 2, - 1, 0, 1 6 39. 5x - 6 6 x 6 6}; 1 - 6, 62
27. e -
27 27 , f 2 2
41. 5x x 6 - 4 or x 7 4}; 1 - q , - 42 h (4, q )
29. e -
36 24 , f 5 5
31. No real solution
43. 5x - 4 6 x 6 4}; 1 - 4, 42 4
4
47. 5x 1 6 x 6 3}; (1, 3)
49. e t ` -
4
45. 5x x 6 - 4 or x 7 4}; 1 - q , - 42 h (4, q )
4 4
51. 5x x … 1 or x Ú 5}; 1 - q, 1 4 h [5, q 2
2 2 … t … 2 f; c - , 2 d 3 3
33. 5 - 3, 3 6
4
53. e x ` - 1 6 x 6
3 3 f ; a - 1, b 2 2
3
1
55. 5x x 6 - 1 or x 7 2 6; 1 - q , - 12 h 12, q 2 1
2 3
1
2
59. e x `
57. No real solution; ∅
65. (a) { - 2, 6}
19 17 19 17 6 x 6 f; a , b 6 6 6 6
2
19 6
17 6
3 2
61. {x - 2 6 x 6 4}; ( - 2, 4)
0
2
1 63. (a) e - , 1 f 5
1
5
4
(b) e x ` -
1 1 1 1 6 x 6 1 f ; a - , 1 b (c) e x ` x … - or x Ú 1 f ; a - q , - d h 3 1, q ) 5 5 5 5 (b) {x x … - 2 or x Ú 6}; ( - q , - 2 4 h 3 6, q ) (c) {x - 2 6 x 6 6}; ( - 2, 6) 67. x - 10 6 2; 5x 8 6 x 6 12 6
69. 2x + 1 7 5; 5x x 6 - 3 or x 7 2 6 71. Between 5.6995 in. and 5.7005 in. 73. IQs less than 71 or greater than 129 75. 5x + 1 + 7 = 5 is equivalent to 5x + 1 = - 2. Because u Ú 0 for any real number u, the equation has no real solution. 77. 2x - 1 is greater than or equal to 0 for 1 all real numbers x. The only real number x such that 2x - 1 … 0 is the value of x for which 2x - 1 = 0, which is . 2 78. f( - 4) = 15 79. (0, q ) −2 −1 0
1 2
3
4
5
6
80. 17 + 7i 81. (a) (0, 0), (4, 0) (b) Domain: [ - 2, 5]; Range: [ - 2, 4] (c) Increasing: (3, 5); Decreasing: ( - 2, 1); Constant: (1, 3) (d) Neither
7
Review Exercises (page 217) 1. (a) m = 2; b = - 5 (b) 5 y 2
2
2. (a) m = ,0
(b)
5 x (0, 5)
(1, 3)
(0, 4)
(4, 4) 5 x
(5, 2)
5. Linear; y = 5x + 3
13. { - 1, 7}
1 5 1 5 8. Zeros: - , ; x-intercepts: - , 9. Zeros: 1, 5; x-intercepts: 1, 5 3 2 3 2 5 5 5 5 1 1 11. Zeros: 1 + ,1 ; x-intercepts: 1 + ,1 12. Zeros: - , 1; x-intercepts: - , 1 A3 A3 A3 A3 2 2
6. Nonlinear
7. Zeros: - 5, 3; x-intercepts: - 5, 3
y 10
y 20
(2, 7) (7, 16) (4, 5) 8 x
1 x
(c) Domain: 1 - q , q 2 Range: 5y y = 4 6 (d) 0 (e) Constant
(c) Domain and Range: 1 - q , q 2 4 (e) Increasing (d) 5
14. { - 2, 2}
(0, 9)
(7, 0)
(0, 6)
1 1 10. Zero: - ; x-intercept: 3 3
(3, 0)
(3, 4)
8 x
(c) Domain and Range: 1 - q , q 2 (d) 2 (e) Increasing
(1, 16)
y 18 (0, 14)
5
15 ,0 2
y 2
4. Zero: - 7; y-intercept: 14
3. (a) m = 0; b = 4 (b) y
4 ; b = -6 5
(2, 9)
4 x (0, 5)
15. Zeros: - 12, 12; x-intercepts: - 12, 12 16. Zero: - 1; x-intercept: - 1 17. Zeros: - 3, 5; x-intercepts: - 3, 5 1 1 1 1 18. Zeros: - , ; x-intercepts: - , 2 6 2 6 y 19. 8 (1, 3) (3, 3) (2, 2) 8 x
20.
21.
y 5
(2, 6)
(4, 0) (3, 1)
y 9
8 x (5, 1)
(0, 6)
(1, 4) 5 x
AN18
ANSWERS Review Exercises
22. (a)
23. (a)
y (0, 6)
24. (a)
y 2
8 (ⴚ8, 0)
(4, 6)
y 2.5
(8, 0) 10 x
(0, 0) (2, 2) 8 x
(0, 1) 2.5 x ⴚ2.5 1 3
1 2
xⴝⴚ
(b) Domain: 1 - q , q 2 Range: ( - q , 1] 1 (c) Increasing: a - q , b 2 1 Decreasing: a , q b 2
(b) Domain: 1 - q , q 2 1 Range: c , q b 2 1 (c) Decreasing: a - q , - b 3 1 Increasing: a - , q b 3
14 28. Maximum value; 12 29. Maximum value; 5 5 2 30. f(x) = - 3(x - 2) - 4 = - 3x2 + 12x - 16 31. f(x) = (x + 1)2 + 2 = x2 + 2x + 3 1 1 32. 5x 0 - 5 6 x 6 2 6; 1 - 5, 22 33. e x ` x … - 3 or x Ú - f ; a - q , - 3 d h c - , q ≤ 2 2
(0.22, 0) 5 x (0, ⴚ1)
7 2 ⴚ ,ⴚ ⴚ5 3 3 2 xⴝⴚ 3
34. - 2 22i, 2 22i
(b) Domain: 1 - q , q 2 7 Range: c - , q b 3
y 5 (ⴚ1ⴚÏ5, 0) (0, 8) (ⴚ1, ⴚ5)
5 x
3 9 38. e - , f 5 5
7 3 7 3 6 x 6 - f; a - , - b 2 6 2 6
22 22 i, 1 + i 2 2
37. -
1 1 22 22 i, - + i 2 2 2 2
y 2 (ⴚ1ⴙÏ5, 0) 5 x
y 9 (ⴚ1, 3)
5 x (1, ⴚ1)
(0, ⴚ3)
(0, 3)
1 ⴚ ,2 2
(2, ⴚ3)
(0, ⴚ4)
3 x
39. 51, 5 6
41. {x x … - 2 or x Ú 7}; ( - q , - 2 4 h 3 7, q ) ⴚ2
7 ⴚ 6
36. 1 -
35. - 1 - 25, - 1 + 25
y 18
2 (c) Decreasing: a - q , - b 3 2 Increasing: a - , q b 3
3 ⴚ 2
(1, 0) 2.5 x
27. Minimum value;
y
40. e x ` -
xⴝ
(b) Domain: 1 - q , q 2 Range: [ - 16, q 2 (c) Decreasing: ( - q , 0) Increasing: (0, q )
(b) Domain: 1 - q , q 2 Range: 32, q ) (c) Decreasing: ( - q , 2) Increasing: (2, q )
(ⴚ1.55, 0)
y 1 1 ⴚ , 3 2
(0, ⴚ16)
xⴝ2
26. (a)
25. (a)
1, 1 2
42. e x ` 0 … x …
4 4 f ; c 0, d 3 3
0
7
43. e x ` x 6 - 1 or x 7
7 f; 3
7 ( - q , - 1) h a , q b 3
4 3
ⴚ1
7 3
120
52. (a)
80 40 0
100 200 300 Price (dollars)
p
(b) Since each input (price) corresponds to a (e) single output (quantity demanded), we know that quantity demanded is a function of price. Also, because the average rate of change is a constant - 0.4 24-in. LCD monitor per dollar, the function is linear. (c) q( p) = - 0.4p + 160 (d) 5p 0 … p … 400 6 or 3 0, 400 4 53. (a) Quadratic, a 6 0
Tibia (mm)
y 38 37 36 35 34 24 25 26 27 28 x Humerus (mm)
(b) Yes; the two variables appear to be linearly related. (c) y = 1.3902x + 1.114 (d) 38.0 mm
Quantity Demanded
q
Total Revenue (thousands of dollars)
51. (a)
Quantity Demanded
44. (a) Company A: C1x2 = 0.06x + 7; Company B: C1x2 = 0.08x (b) 350 min (c) 0 … x 6 350 45. (a) R(p) = - 10p2 + 1500p (b) {p 0 6 p … 150} (c) $75 (d) $56,250 (e) 750 units (f) Between $70 and $80 46. The width is 6 in. and the length is 8 in. 47. (a) 63 clubs (b) $151.90 48. 50 by 50 ft 49. 25 square units 50. 3.6 ft q 120 80 40 0
100 200 300 Price (dollars)
p
(f ) If price increases by $1, quantity demanded of 24-in. LCD monitors decreases by 0.4 monitor. (g) q-intercept: When the price is $0, 160 24-in. LCD monitors will be demanded. p-intercept: There will be 0 24-in. LCD monitors demanded when the price is $400. (d)
6500
R 6500 6400 6300 6200 6100 25 30 A 20 Advertising (thousands of dollars)
(b) R(A) = - 7.76 A2 + 411.88A + 942.72; $26.5 thousand (c) $6408 thousand
15 6000
35
AN19
ANSWERS Cumulative Review
Chapter Test (page 219) 1. (a) Slope: - 4; y-intercept: 3 (b) Decreasing (c) y
2. Linear; y = f(x) = - 5x + 2
4 3. - , 2 3
7.
y 9
y 21
(0, 3)
2 - 26 2 + 26 , 2 2
6. - 3, 0
5. { - 1, 3}
5
4.
(0, 7)
(6, 7)
(3, 18) (4, ⴚ1)
5 x (1, ⴚ1)
(0, 0) 4 x (ⴚ1, ⴚ2)
(ⴚ3, 0)
8. (a) Opens up (b) 12, - 82 (c) x = 2
(e)
(2, ⴚ1)
(f) Domain: All real numbers; ( - q, q) Range: 5y y Ú - 8 6; [ - 8, q2 (g) Decreasing: ( - q, 2) ; Increasing: 12, q2
y 10 (0, 4)
8 x (3, ⴚ2)
(4, 4)
10 x (3.63, 0) (2, ⴚ8) xⴝ2
6 - 2 26 6 + 2 26 , ; 3 3 y-intercept: 4
(0.37, 0)
(d) x-intercepts:
9. (a) Opens down (b) 11, - 32 (c) x = 1 (d) No x-intercepts: y-intercept: - 5
(e)
(f) Domain: All real numbers; ( - q, q) Range: {y y … - 3}; ( - q, - 3] (g) Increasing: ( - q, 1); Decreasing: 11, q2
y 2 5 x (1, ⴚ3) (0, ⴚ5)
(2, ⴚ5)
xⴝ1
10. f 1x2 = 2x2 - 4x - 30
16. {x x … - 5 or x Ú 2}; ( - q , - 5 4 h 3 2, q )
15. {x - 11 6 x 6 5}; ( - 11, 5) ⴚ11
0
26 26 i, - 1 + i 2 2 17. (a) C1m2 = 0.15m + 129.50 (b) $258.50 (c) 562 miles
11. Maximum value; 21 12. {x x … 4 or x Ú 6}; ( - q , 4 4 h 3 6, q )
ⴚ5
5
1 2 x + 1000x 18. (a) R1x2 = 10 10 19. (a) ⴚ3
0
(b) $384,000
13. - 1 -
7 14. e - 3, f 3
2
(c) 5000 units; $2,500,000
(d) $500 (b) y = - 4.234x - 2.362
15
(c) y = 1.993x 2 + 0.289x + 2.503
3
ⴚ3
3
ⴚ15
0
Quadratic that opens up
Linear with negative slope
Cumulative Review (page 220) 1. 5 22; (1.5, 0.5)
2. 1 - 2, - 12 and (2, 3) are on the graph.
3 3 3. e x ` x Ú - f or c - , q b 5 5 ⴚ
3 5
4. y = - 2x + 2 (ⴚ1, 4)
y 5
6. 1x - 22 2 + 1y + 42 2 = 25
1 13 5. y = - x + 2 2
y 10
y 9 5 x (2, ⴚ2)
(2, 1)
(3, 5) (ⴚ3, ⴚ4) 5 x
7. Yes, a function 8. (a) - 3
(b) x2 - 4x - 2
(c) x2 + 4x + 1
(d) - x2 + 4x - 1
10 x (7, ⴚ4) (2, ⴚ4) (2, ⴚ9)
(e) x2 - 3
(f) 2x + h - 4
9. e z ` z ≠
7 f 6
10. Yes, a function 11. (a) No (b) - 1; 1 - 2, - 12 is on the graph. (c) - 8; 1 - 8, 22 is on the graph. 12. Neither 13. Local maximum is 5.30 and occurs at x = - 1.29. Local minimum is - 3.30 and occurs at x = 1.29. Increasing: 1 - 4, - 1.292 or (1.29, 4) Decreasing: 1 - 1.29, 1.292 14. (a) - 4 (b) 5x x 7 - 4 6 or 1 - 4, q 2 15. (a) Domain: 5x - 4 … x … 4 6; Range: 5y - 1 … y … 3 6 (b) 1 - 1, 02, 10, - 12, 11, 02 (d) 1 (e) - 4 and 4 (f ) 5x - 1 6 x 6 1 6 y (g) (h) (i) y y (ⴚ4, 5) (4, 5) 5 (2, 3) (ⴚ2, 3) (1, 2) (ⴚ1, 2) 5 x (0, 1)
(ⴚ4, 3) (ⴚ2, 1) (ⴚ1, 0)
5
(0, ⴚ1)
( j) Even
(k) (0, 4)
(4, 3) (2, 1) 5 x (1, 0)
10 (ⴚ4, 6) (ⴚ2, 2) (ⴚ1, 0) (0, ⴚ2)
(c) y-axis
(4, 6) (2, 2) 5 x (1, 0)
AN20
ANSWERS Section 3.1
CHAPTER 3: Polynomial and Rational Functions 3.1 Assess Your Understanding (page 239) 7. smooth; continuous 8. touches 9. ( - 1, 1); (0, 0); (1, 1) 10. r is a real zero of f; r is an x-intercept of the graph of f; x - r is a factor of f. 11. turning points 12. y = 3x4 13. q ; - q 14. As x increases in the positive direction, f(x) decreases without bound. 1 1 1 1 15. Yes; degree 3; f(x) = x3 + 4x; leading term: x3; constant term: 0 17. Yes; degree 2; g(x) = - x2 + ; leading term: - x2; constant term: 2 2 2 2 3 1 1 power. 23. Yes; degree 4; F (x) = 5x4 - px3 + ; leading term: 5x4; constant term: . 2 2 2 25. Yes; degree 4; G(x) = 2x4 - 4x3 + 4x2 - 4x + 2; leading term: 2x4; constant term: 2 19. No; x is raised to the - 1 power.
27.
29.
y 5 (0, 1) (ⴚ2, 1) 5 x (ⴚ1, 0)
35.
31.
y 5
5 x (1, ⴚ2)
37. (ⴚ2, 3) (ⴚ1, 1)
(2, 3) (1, 2) 5 x
1,
(0, 3)
y 6 (1, 5)
1 2
(2, 4) (3, 3)
5 x
5 x
y 5 (ⴚ1, 1) (0, 0)
5 x
(0, 0)
39.
y 5
33.
y 5
1 ⴚ1, 2
(0, ⴚ3) (ⴚ1, ⴚ4)
y 5 (0, 1)
21. No; x is raised to the
5 x (1, ⴚ1)
41. f(x) = x3 - 3x2 - x + 3 for a = 1 43. f(x) = x3 - x2 - 12x for a = 1 45. f(x) = x4 - 15x2 + 10x + 24 for a = 1 47. f(x) = x3 - 5x2 + 3x + 9 for a = 1
49. (a) 7, multiplicity 1; - 3, multiplicity 2
(b) Graph touches the x-axis at - 3 and crosses it at 7. (c) 2 (d) y = 3x3 1 5 51. (a) 2, multiplicity 3 (b) Graph crosses the x-axis at 2. (c) 4 (d) y = 4x 53. (a) - , multiplicity 2; - 4, multiplicity 3 2 1 (b) Graph touches the x-axis at - and crosses it at - 4. (c) 4 (d) y = - 2x5 55. (a) 5, multiplicity 3; - 4, multiplicity 2 2 (b) Graph touches the x-axis at - 4 and crosses it at 5. (c) 4 (d) y = x5 57. (a) No real zeros (b) Graph neither crosses nor touches the x-axis. (c) 5
(d) y = 3x6
(c) 3
(d) y = - 2x
59. (a) 0, multiplicity 2; - 22, 22, multiplicity 1 61. Could be; zeros: - 1, 1, 2; Least degree is 3
4
65. f(x) = x(x - 1)(x - 2)
(4, 16) (3, 0) 5 x (2, ⴚ4)
79. Step 1: y = - x3 Step 2: x-intercepts: - 4, 1; y-intercept: 16 Step 3: - 4: multiplicity 2, touches; 1: multiplicity 1, crosses Step 4: At most 2 turning points (−2, 12) y Step 5: 25 (−5, 6) −6
(−4, 0)
(0, 16) (1, 0) 4 x
69. f(x) = 0.2(x + 4)(x + 1)2(x - 3)
75. Step 1: y = x3 Step 2: x-intercepts: - 4, 2; y-intercept: 16 Step 3: - 4: multiplicity 1, crosses; 2: multiplicity 2, touches Step 4: At most 2 turning points Step 5: y (ⴚ2, 32) 40
24
(0, 0) (ⴚ1, ⴚ4)
63. Cannot be the graph of a polynomial; gap at x = - 1
1 67. f(x) = - (x + 1)(x - 1)2 (x - 2) 2
73. Step 1: y = x3 Step 2: x-intercepts: 0, 3; y-intercept: 0 Step 3: 0: multiplicity 2, touches; 3: multiplicity 1, crosses Step 4: At most 2 turning points y Step 5:
(b) Graph touches the x-axis at 0 and crosses it at - 22 and 22.
(ⴚ4, 0)
(4, 32) (0, 16) (2, 0) 5 x (ⴚ5, ⴚ49)
81. Step 1: y = x3 Step 2: x-intercepts: - 4, - 1, 2; y-intercept: - 8 Step 3: - 4, - 1, 2: multiplicity 1, crosses Step 4: At most 2 turning points Step 5: y (0, ⴚ8) 50 (ⴚ2, 8) (ⴚ4, 0) (ⴚ5, ⴚ28)
−25
77. Step 1: y = - 2x4 Step 2: x-intercepts: - 2, 2; y-intercept: 32 Step 3: - 2: multiplicity 1, crosses; 2: multiplicity 3, crosses Step 4: At most 3 turning points Step 5: (ⴚ1, 54) y (0, 32) (ⴚ2, 0)
(ⴚ3, ⴚ250)
(2, 0) 5 x (3, ⴚ10) ⴚ240
83. Step 1: y = x4 Step 2: x-intercepts: - 2, 0, 2; y-intercept: 0 Step 3: - 2, 2: multiplicity 1, crosses; 0, multiplicity 2, touches Step 4: At most 3 turning points Step 5: y 90
(3, 28) 5 x (2, 0) (1, ⴚ10)
(ⴚ1, 0)
71. f(x) = - x(x + 3)2(x - 3)2
(ⴚ3, 45) (1, ⴚ3) (ⴚ2, 0) (ⴚ1, ⴚ3)
(3, 45) (0, 0) (2, 0) 5 x
ANSWERS Section 3.1 85. Step 1: y = x4 Step 2: x-intercepts: - 1, 2; y-intercept: 4 Step 3: - 1, 2: multiplicity 2, touches Step 4: At most 3 turning points y Step 5: (ⴚ2, 16)
20
(0, 4)
2
(3, 16) (1, 4)
(ⴚ1, 0)
87. Step 1: y = x3 Step 2: x-intercepts: 0, 1, 2; y-intercept: 0 Step 3: 0, 1, 2: multiplicity 1, crosses Step 4: At most 2 turning points y Step 5: Q52 , 15 8R
(1, 0)
89. Step 1: y = x4 Step 2: x-intercepts: - 3, - 1, 0; y-intercept: 0 Step 3: - 1, - 3: multiplicity 1, crosses; 0: multiplicity 2, touches Step 4: At most 3 turning points y Step 5: 4
(2, 0) 3 x
(0, 0) −2
(2, 0) 5 x
AN21
(−3, 0) (−1, 0) −2
Q− 12 , − 15 R 8
2 x (0, 0)
(−2, −4) −6
91. Step 1: y = x4 Step 2: x-intercepts: - 4, - 2, 2; y-intercept: 32 Step 3: - 2, - 4: multiplicity 1, crosses; 2: multiplicity 2, touches Step 4: At most 3 turning points y Step 5:
93. Step 1: y = 5x4 Step 2: x-intercepts: - 3, - 2, 0, 2; y-intercept: 0 Step 3: - 3, - 2, 0, 2: multiplicity 1, crosses Step 4: At most 3 turning points (−1, 30) y Step 5: 40
200
(ⴚ4, 0) (ⴚ3, ⴚ25)
(3, 35) (2, 0) 5 x
97. Step 1: y = x3 Step 2:
−60
99. Step 1: y = x3 Step 2:
2
ⴚ2
2
160
(0, 0)
(−3, 0) −3 (−2, 0)
(0, 32)
(ⴚ2, 0)
95. Step 1: y = x5 Step 2: x-intercepts: 0, 2; y-intercept: 0 Step 3: 2: multiplicity 1, crosses; 0: multiplicity 2, touches Step 4: At most 4 turning points y Step 5:
(2, 0) 2 x (0, 0) (ⴚ1, ⴚ12) (1, −60)
101. Step 1: y = x4 Step 2:
10
ⴚ4
2
3
ⴚ2
2
ⴚ10
ⴚ2
(3, 108) (1, ⴚ4) (2, 0) 5 x
ⴚ2
Step 3: x-intercepts: - 1.26, - 0.20, 1.26 y-intercept: - 0.31752 Step 4:
Step 3: x-intercepts: - 3.56, 0.50 y-intercept: 0.89 Step 4:
Step 3: x-intercepts: - 1.5, - 0.5, 0.5, 1.5 y-intercept: 0.5625 Step 4:
Step 5: ( - 0.80, 0.57); (0.66, - 0.99) y Step 6:
Step 5: ( - 2.21, 9.91); (0.50, 0) y Step 6:
Step 5: ( - 1.12, - 1); (1.12, - 1); (0, 0.56) Step 6: (0, 0.56)
(ⴚ0.5, 0.40) 2.5 (ⴚ0.80, 0.57) (ⴚ1.26, 0) (ⴚ1.5, ⴚ0.86) (0, ⴚ0.32)
(ⴚ0.20, 0) (1.5, 1.13) 2.5 x (1.26, 0) (0.66, ⴚ0.99)
Step 7: Domain: ( - q , q ); Range: ( - q , q ) Step 8: Increasing on ( - q , - 0.80) and (0.66, q ) Decreasing on ( - 0.80, 0.66)
(ⴚ2.21, 9.91)
10 (ⴚ1, 5.76)
(ⴚ3.56, 0) (0, 0.89) (ⴚ3.9, ⴚ6.58)
(0.5, 0)
1 x (1, 1.14)
Step 7: Domain: ( - q , q ); Range: ( - q , q ) Step 8: Increasing on ( - q , - 2.21) and (0.50, q ) Decreasing on ( - 2.21, 0.50)
(ⴚ0.5, 0) (ⴚ1.75, 2.29) (ⴚ1.5, 0)
y 5
(0.5, 0) (1.75, 2.29) (1.5, 0)
2.5 x (ⴚ1.12, ⴚ1) (1.12, ⴚ1) (ⴚ1, ⴚ0.94) (1, ⴚ0.94)
Step 7: Domain: ( - q , q ); Range: [ - 1, q ) Step 8: Increasing on ( - 1.12, 0) and (1.12, q ) Decreasing on ( - q , - 1.12) and (0, 1.12)
AN22
ANSWERS Section 3.1
103. Step 1: y = 2x4 Step 2:
Step 5: ( - 0.42, - 4.64) Step 6: y
6
(ⴚ1.25, 4.22)
ⴚ2
5
(1.75, 1.83) (1.62, 0) 2 x
(ⴚ1.07, 0)
2
(ⴚ0.42, ⴚ4.64) ⴚ5
(0, ⴚ4)
107. f(x) = x(x + 4)(x - 3) Step 1: y = x3 Step 2: x-intercepts: - 4, 0, 3; y-intercept: 0 Step 3: - 4, 0, 3: multiplicity 1, crosses Step 4: At most 2 turning points y Step 5:
700 (ⴚ7, 630) (ⴚ1, 30)
(4, 32) (3, 0) 5 x (0, 0) (2, ⴚ12)
(ⴚ2, 0) (ⴚ6, 0) (ⴚ4, ⴚ192)
111. f(x) = - x2 (x + 1)2 (x - 1) Step 1: y = - x5 Step 2: x-intercepts: - 1, 0, 1; y-intercept: 0 Step 3: 1: multiplicity 1, crosses; - 1, 0: multiplicity 2, touches Step 4: At most 4 turning points 113. f(x) = 3(x + 3)(x - 1)(x - 4)
(ⴚ0.54, 0.10) (0, 0)
117. (a) - 3, 2
121. (a)
2 4 6 8 10
T 66 60 54 48 42 36
x
0
The relation appears to be cubic. (b) H(x) = 0.3948x3 - 5.9563x2 + 26.1965x - 7.4127 (c) Approximately 24 (d) 35
0
10
(3, 270) (2, 0) 3 x (1, ⴚ42)
(0.74, 0.43) (1, 0) 2.5 x (1.2, ⴚ1.39392)
(b) - 6, - 1 (b) Approximately 4.67ºF per hour. (c) Approximately - 2.33°F per hour
6 12 18 24 x
The relation appears to be cubic. (d) T(x) = - 0.02357x3 + 0.809x2 - 6.2893x + 52.0714; 63.2°F (e)
66
0
0
y 2.5 (ⴚ1.5, 1.40625) (ⴚ1, 0)
115. f(x) = - 2(x + 5)2 (x - 2)(x - 4)
H
(0, 0)
Step 5:
50 40 30 20 10 0
(ⴚ2, 0)
109. f(x) = 2x(x + 6)(x - 2)(x + 2) Step 1: y = 2x4 Step 2: x-intercepts: - 6, - 2, 0, 2; y-intercept: 0 Step 3: - 6, - 2, 0, 2: multiplicity 1, crosses Step 4: At most 3 turning points y Step 5:
50 (ⴚ2, 20) (ⴚ4, 0)
119. (a)
20 (0, 0) (1, 3) (2, 0) 2.5 x (ⴚ1, ⴚ3) (3, ⴚ15)
(ⴚ3, 15)
Step 7: Domain: ( - q , q ); Range: [ - 4.64, q ) Step 8: Increasing on ( - 0.42, q ) Decreasing on ( - q , - 0.42)
Step 3: x-intercepts: - 1.07, 1.62; y-intercept: - 4 Step 4:
(ⴚ5, ⴚ40)
105. f(x) = - x(x + 2)(x - 2) Step 1: y = - x3 Step 2: x-intercepts: - 2, 0, 2; y-intercept: 0 Step 3: - 2, 0, 2: multiplicity 1, crosses Step 4: At most 2 turning points y Step 5:
26 0
(e) Approximately 54; no
(f) The predicted temperature at midnight is 52.1°F.
Account value of deposit 1
Account value of deposit 2
5
c
e
123. (a) T(r) = 500(1 + r)(1 + r) + 500(1 + r) + 500 Deposit 3
= 500(1 + 2r + r 2) + 500 + 500r + 500 = 500 + 1000r + 500r 2 + 500 + 500r + 500 = 500r 2 + 1500r + 1500 (b) $2155.06 2 11 - 2 - 27 - 2 + 27 4 129. (a)–(d) 132. y = - x 133. 5x x ≠ - 5 6 134. , 135. e - , 2 f 5 5 2 2 5
AN23
ANSWERS Section 3.2
Historical Problems (page 255) b 3 b 2 b b + b ax - b + c ax - b + d = 0 3 3 3
ax -
1. 2
3
2
3
bx b 2b x b bc + bx2 + + cx + d = 0 3 27 3 9 3
x3 - bx2 +
x3 + ac Let p = c -
b2 2b3 bc bx + a + db = 0 3 27 3
(H + K)3 + p(H + K) + q H 3 + 3H 2K + 3HK2 + K3 + pH + pK + q Let 3HK H 3 - pH - pK + K3 + pH + pK + q = 0, H 3 + K3
= = = =
0 0 - p. -q
2b3 bc b2 and q = + d. Then x3 + px + q = 0. 3 27 3
3HK = - p p K = 3H p 3 b = -q H3 + a 3H
3.
2.
H3 -
p3 27H 3
4. H 3 + K3 = - q K3 = - q - H 3 K3 = - q - c K3 =
= -q
27H 6 - p3 = - 27qH 3
K =
27H 6 + 27qH 3 - p3 = 0 H3 = H3 = H3 = H =
2 # 27
-q 2 -q 2 3
{
-q
B 2
B4 +
+
q2 A4
4(27)p
+
B 22(272) q2
x =
3
27 q
{
2
2
3
-q
B 2
2 q2
B4 -
+ q2
A4
q2
+
B4
p3 27
d
p3 27 +
p3 27
3
-q
B 2
+
q2 B4
+
p3 27
+
3
-q
B 2
-
q2 B4
+
p3 27
(Note that if we had used the negative root in 3, the result would have been the same.) 6. x = 3 7. x = 2 8. x = 2
2 (27 )
p3 27 +
2
-
+
5. x = H + K
- 27q { 2(27q)2 - 4(27)( - p3) 2 2
-q
-q
p3 27
Choose the positive root for now.
3.2 Assess Your Understanding (page 255) 7. - 4 8. F 9. 0 10. T 11. R = f(2) = 8; no 13. R = f(2) = 0; yes 15. R = f( - 3) = 0; yes 1 21. 7; 3 or 1 positive; 2 or 0 negative 23. 6; 2 or 0 positive; 2 or 0 negative 17. R = f( - 4) = 1; no 19. R = f a b = 0; yes 2 25. 3; 2 or 0 positive; 1 negative 27. 4; 2 or 0 positive; 2 or 0 negative 29. 5; 0 positive; 3 or 1 negative 31. 6; 1 positive; 1 negative 1 1 1 1 1 1 3 9 33. {1, { 35. {1, {3 37. {1, {2, { , { 39. {1, {3, {9, { , { , { , { , { 3 4 2 2 3 6 2 2 1 3 1 5 1 2 4 5 10 20 1 5 41. {1, {2, {3, {4, {6, {12, { , { 43. {1, {2, {4, {5, {10, {20, { , { , { , { , { , { , { , { , { , { 2 2 2 2 3 3 3 3 3 3 6 6 1 1 2 45. - 3, - 1, 2; f(x) = (x + 3)(x + 1)(x - 2) 47. ; f(x) = 2 ax b (x + 1) 49. 2, 25, - 25; f(x) = 2(x - 2)(x - 25)(x + 25) 2 2 1 1 51. - 1, , 23, - 23; f(x) = 2(x + 1) ax b (x - 23)(x + 23) 53. 1, multiplicity 2; - 2, - 1; f(x) = (x + 2)(x + 1)(x - 1)2 2 2 6. f(c)
5. Remainder; dividend
55. - 1, 65. e -
1 1 ; f(x) = 4(x + 1) ax + b (x2 + 2) 4 4
1 f 3
67. e
1 , 2, 5 f 2
77. LB = - 3; UB = 3 85. 0.21 87. - 4.04 93.
y 30 (ⴚ2, 4) (ⴚ3, 0) (ⴚ1, 0) (ⴚ4, ⴚ18)
69. LB = - 2; UB = 2
79. f(0) = - 1; f(1) = 10 89. 1.15 91. 2.53 95.
(3, 24)
57. { - 1, 2}
97.
5 x 1 ,0 2
1 , 25, - 25 f 3
63. { - 3, - 2}
75. LB = - 1; UB = 1
83. f(1.4) = - 0.17536; f(1.5) = 1.40625
99.
y 2.5 (ⴚ1, 0)
61. e
73. LB = - 2; UB = 2
81. f( - 5) = - 58; f( - 4) = 2
y 5
(0, ⴚ1)
2 , - 1 + 22, - 1 - 22 f 3
71. LB = - 1; UB = 1
(1, 2) (2, 0) 5 x (0, ⴚ6)
59. e
(1, 0) 2.5 x (0, ⴚ2)
y 1 ⴚ ,0 2
2
1 ,0 2 2.5 x (0, ⴚ2)
AN24
ANSWERS Section 3.2
101.
103.
y 16 (2, 12)
(1, 0)
兹2 ,0 2
y (0, 2) 4
兹2 ,0 2
105. - 8, - 4, -
7 3
107. k = 5
109. - 7
111. If f(x) = xn - c n, then f(c) = c n - c n = 0, so x - c is a factor of f. 113. 5 115. 7 in. 117. All the potential rational zeros are integers, so (1, 3) (1, 9) (1.5, 1.5625) r either is an integer or is not a rational zero (and is therefore irrational). 119. 0.215 1 2 121. No; by the Rational Zeros Theorem, is not a potential rational zero. 123. No; by the Rational Zeros Theorem, is not a potential rational zero. 3 3 2 3 127. ( - 3, 2) and (5, q ) 124. y = x 125. [3, 8) 126. (0, - 2 23), (0, 2 23), (4, 0) 5 5 (0, 2) (1, 0) 5 x
(2, 0)
(2, 0) 5 x
3.3 Assess Your Understanding (page 262) 4. 3 - 4i
3. one
5. T
7. 4 + i
6. F
9. - i, 1 - i
17. f(x) = x - 14x + 77x - 200x + 208; a = 1 4
3
21. f(x) = x4 - 6x3 + 10x2 - 6x + 9; a = 1 31. 1, -
11. - i, - 2i
19. f(x) = x - 4x
2
5
4
+ 7x
1 25. 2i, - 3, 2
23. - 2i, 4
13. - i 3
15. 2 - i, - 3 + i
- 8x + 6x - 4; a = 1 2
27. 3 + 2i, - 2, 5
29. 4i, - 211, 211, -
2 3
1 23 1 23 1 23 1 23 i, - + i; f(x) = (x - 1) ax + + i b ax + ib 2 2 2 2 2 2 2 2
33. 2, 3 - 2i, 3 + 2i; f(x) = (x - 2)(x - 3 + 2i)(x - 3 - 2i)
35. - i, i, - 2i, 2i; f(x) = (x + i)(x - i)(x + 2i)(x - 2i)
37. - 5i, 5i, - 3, 1; f(x) = (x + 5i)(x - 5i)(x + 3)(x - 1)
1 1 39. - 4, , 2 - 3i, 2 + 3i; f(x) = 3(x + 4) ax - b (x - 2 + 3i)(x - 2 - 3i) 3 3
43. (a) f(x) = (x2 - 22x + 1)(x2 + 22x + 1)
22 22 22 22 22 22 22 22 i, + i, i, + i 2 2 2 2 2 2 2 2
(b) -
41. 130
45. Zeros that are complex numbers must occur in conjugate pairs; or a polynomial with real coefficients of odd degree must have at least one real zero. 47. If the remaining zero were a complex number, its conjugate would also be a zero, creating a polynomial of degree 5. 50. { - 22} 49. y 7 51. 6x3 - 13x2 - 13x + 20 52. A = 9p ft 2 ( ≈28.274 ft 2); C = 6p ft ( ≈18.850 ft) −2
10 x
−5
3.4 Assess Your Understanding (page 272) 5. F
6. horizontal asymptote
7. vertical asymptote
13. All real numbers except 3; 5x x ≠ 3 6
8. proper
9. T
10. F
11. y = 0
15. All real numbers except 2 and - 4; 5x x ≠ 2, x ≠ - 4 6
12. T
1 1 and 3; e x x ≠ - , x ≠ 3 f 19. All real numbers except 2; 5x x ≠ 2 6 21. All real numbers 2 2 23. All real numbers except - 3 and 3; 5x x ≠ - 3, x ≠ 3 6 25. (a) Domain: 5x x ≠ 2 6; Range: 5y y ≠ 1 6 (b) (0, 0) (c) y = 1 (d) x = 2 (e) None 27. (a) Domain: 5x x ≠ 0 6; Range: all real numbers (b) ( - 1, 0), (1, 0) (c) None (d) x = 0 (e) y = 2x 29. (a) Domain: 5x x ≠ - 2, x ≠ 2 6; Range: 5y y … 0, y 7 1 6 (b) (0, 0) (c) y = 1 (d) x = - 2, x = 2 (e) None 17. All real numbers except -
31.
33.
y 8
35.
y 10
(ⴚ2, 2) yⴝ0
(1, 3) yⴝ2 5 x
(ⴚ1, 1)
37.
y 5
yⴝ0
(0, 1)
(1, ⴚ1)
(2, 1)
xⴝ1
(ⴚ3, ⴚ1)
5
39.
y 2 4 x (ⴚ1, ⴚ1)
y 10
yⴝ0
(2, 3) x ⴝ ⴚ2
x ⴝ ⴚ1
y 3 yⴝ1 5 x (2, 0)
(ⴚ2, 0)
x
5 x
41.
xⴝ3
(4, 3)
yⴝ1 9 x
xⴝ0
xⴝ0
43. Vertical asymptote: x = - 4; horizontal asymptote: y = 3
45. Vertical asymptote: x = 3; oblique asymptote: y = x + 5 1 2 47. Vertical asymptotes: x = 1, x = - 1; horizontal asymptote: y = 0 49. Vertical asymptote: x = - ; horizontal asymptote: y = 3 3 51. Vertical asymptote: none; oblique asymptote: y = 2x - 1 53. Vertical asymptote: x = 0; no horizontal or oblique asymptote 55. (a)
57. (a)
x = −3 y 4 (−2, −1) 4 x y = −2
−6 (−4, −3) −6
(b) Domain: {x x ≠ - 3}; Range: {y y ≠ - 2} (c) Vertical asymptote: x = - 3; Horizontal asymptote: y = - 2
59. (a)
y x=1 6
y x=1 7 (0, 5) (2, 5)
y=3 (0, 1) −5
(2, 1) 5 x
−6
(b) Domain: {x x ≠ 1}; Range: {y y 6 3} (c) Vertical asymptote: x = 1; Horizontal asymptote: y = 3
y=1 6x
−6 −3
(b) Domain: {x x ≠ 1}; Range: {y y 7 1} (c) Vertical asymptote: x = 1; Horizontal asymptote: y = 1
ANSWERS Section 3.5 61. (a) 9.8208 m/sec2 63. (a)
(b) 9.8195 m/sec2
(c) 9.7936 m/sec2
(e) ∅
(d) h-axis
65. (a) R(x) = 2 +
Rtot
(b)
10
y x=1 10 (2, 7)
5 0
5 1 = 5¢ ≤ + 2 x - 1 x - 1
5 10 15 20 25 R2
(b) Horizontal: Rtot = 10; as the resistance of R2 increases without bound, the total resistance approaches 10 ohms, the resistance R1. (c) R1 ≈ 103.5 ohms 71. x = 5
72. e -
4 r 19
y=2 5 x
y=2 −5 (0, −3) −10
(c) Vertical asymptote: x = 1; Horizontal asymptote: y = 2
74. ( - 3, 11), (2, - 4)
73. x-axis symmetry
3.5 Assess Your Understanding (page 287) 3. in lowest terms 4. vertical 5. True 6. True 7. 1. Domain: { x x ≠ 0, x ≠ - 4} 2. R is in lowest terms 4. R is in lowest terms; vertical asymptotes: x = 0, x = - 4 4 1 6. Interval
(, 4)
(4, 1)
Number Chosen
5
2
Value of R
R(5)
45
Location of Graph Below x-axis
(
9. 1. R(x) =
)
3(x + 1) 2(x + 2)
R
2,
1
( ) 1 2
2 7
R(1)
2 5
Below x-axis
Above x-axis
(
(
( )
)
)
2. R is in lowest terms
4. R is in lowest terms; vertical asymptote: x = - 2 2 1 6. Interval
(, 2)
(2, 1)
Number Chosen
3
2
Value of R
R(3) 3
R 2
1 4 4 x
5. Horizontal asymptote: y =
3 ; x-intercept: - 1 4
3 , not intersected 2 7.
x 2 y 5 (3, 3)
(1, )
3
0,
3
3 2
R(0)
3 4
Location of Graph Above x-axis
Below x-axis
Above x-axis
Point on Graph
(
(0, )
)
3, 3 2 2
3 4
y
0
( )
y0
1 2 , 2 7
(1, 0)
3. y-intercept:
2 5
1,
4 5, 5
2 1, 5
1 2 2,7
; Domain: 5x x ≠ - 2 6
(3, 3)
x 4 y 2.5
Above x-axis 1 2, 4
4 5, 5
Point on Graph
1
2
R(2)
0
(0, )
(1, 0)
1 4
3. no y-intercept; x-intercept: - 1 5. Horizontal asymptote: y = 0, intersected at ( - 1, 0) 7. x0
3 2
5 x (1, 0)
3 3 , 2 2
3 4
3 3 ; Domain: {x x ≠ - 2, x ≠ 2} 2. R is in lowest terms 3. y-intercept: - ; no x-intercept (x + 2)(x - 2) 4 4. R is in lowest terms; vertical asymptotes: x = 2, x = - 2 5. Horizontal asymptote: y = 0, not intersected 2 2 6. 7. x 2 y x 2
11. 1. R(x) =
Interval
(, 2)
(2, 2)
(2, )
Number Chosen
3
0
3
Value of R
R(3) 5
R(0) 34
R(3)
3
Location of Graph Above x-axis
(3, ) 3 5
Point on Graph
13. 1. P(x) =
Below x-axis
Above x-axis
(3, )
3 4
1x2 + x + 12 1x2 - x + 12 (x + 1)(x - 1)
0,
3 4
2
3,
3 5 y0
3 5
; Domain: {x x ≠ - 1, x ≠ 1}
4. P is in lowest terms; vertical asymptotes: x = - 1, x = 1 1 1 6. Interval
(, 1)
(1, 1)
(1, )
Number Chosen
2
0
2
Value of P
P(2) 7
P(0) 1
P(2) 7
Location of Graph Above x-axis
Below x-axis
Above x-axis
Point on Graph
(0, 1)
(2, 7)
(2, 7)
3 5
5 x
3 5
(0, )
3,
2. P is in lowest terms
3. y-intercept: - 1; no x-intercept
5. No horizontal or oblique asymptote 7.
y xⴝ1 (ⴚ2, 7)
6
(2, 7)
(0, ⴚ1)
x ⴝ ⴚ1
5 x
AN25
AN26
ANSWERS Section 3.5 (x - 1)(x2 + x + 1)
15. 1. H(x) =
(x + 3)(x - 3)
1 3. y-intercept: ; x-intercept: 1 9 1 1 5. Oblique asymptote: y = x, intersected at a , b 9 9 7. (4, 9) y
; Domain: {x x ≠ - 3, x ≠ 3}
2. H is in lowest terms
4. H is in lowest terms; vertical asymptotes: x = 3, x = - 3 6.
3
1
3
0,
Interval
(, 3)
(3, 1)
(1, 3)
(3, )
Number Chosen
4
0
2
4
Value of H
H(4) ⬇ 9.3
H(0) 9
H(2) 1.4
H(4) 9
Above x-axis
Below x-axis
Above x-axis
(0, )
(2, 1.4)
(4, 9)
Location of Graph Below x-axis Point on Graph
(4, 9.3)
1
1 9
1 9
8
yx (1, 0) 10 x (2, 1.4)
(4, 9.3)
x 3 x 3
x2 ; Domain: {x ≠ - 3, x ≠ 2} 2. R is in lowest terms 3. y-intercept: 0; x-intercept: 0 (x + 3)(x - 2) 4. R is in lowest terms; vertical asymptotes: x = 2, x = - 3 5. Horizontal asymptote: y = 1, intersected at (6, 1) 3 0 2 6. 7.
17. 1. R(x) =
Interval Number Chosen Value of R
(, 3)
(3, 0)
(0, 2)
(2, )
6
1
1
3
R(6) 1.5
R(1) 6
R(1) 0.25
R(3) 1.5
1
Location of Graph Above x-axis
Below x-axis
Below x-axis
Above x-axis
Point on Graph
(1, )
(1, 0.25)
(3, 1.5)
(6, 1.5)
1 6
x ; Domain: {x x ≠ - 2, x ≠ 2} (x + 2)(x - 2) 4. G is in lowest terms; vertical asymptotes: x = - 2, x = 2 2 0 6.
19. 1. G(x) =
(2, 0)
(0, 2)
(2, )
Number Chosen
3
1
1
3
G(3)
Location of Graph Below x-axis Point on Graph
(
)
3
3, 5
1
1 3
10 x (0, 0) (1, 0.25) x 3 x 2
3. y-intercept: 0; x-intercept: 0
1, 3
G(1)
Above x-axis
Below x-axis
Above x-axis
(1, )
(1, )
(3, )
G(3) 5
3,
1 3 (0, 0)
3 5
2 5 x
3,
3 5
1 3
(3, 1.5)
y
G(1) 3 1 3
1 6
2
5. Horizontal asymptote: y = 0, intersected at (0, 0) 7.
2
(, 2) 35
1,
2. G is in lowest terms
Interval
Value of G
y
(6, 1.5) y1
3 5
y0
1, x 2 x 2
1 3
3 3 ; Domain: {x x ≠ 1, x ≠ - 2, x ≠ 2} 2. R is in lowest terms 3. y-intercept: ; no x-intercept (x - 1)(x + 2)(x - 2) 4 4. R is in lowest terms; vertical asymptotes: x = - 2, x = 1, x = 2 5. Horizontal asymptote: y = 0, not intersected 2 1 2 6. 7. x 2 y
21. 1. R(x) =
Interval
(, 2)
(2, 1)
Number Chosen
3
0
Value of R
3 R(3) 20
R(0)
Location of Graph Below x-axis Point on Graph
(3, ) 3 20
3 4
0,
(1, 2)
(2, )
1.5
3
R(1.5) 24 7
R(3) 10
Above x-axis
Below x-axis
Above x-axis
(0, )
(1.5, )
(3, )
3 4
24 7
3,
3
3 4
5
3, 5 x
3 20
x1 x2
3 10
3 10
y0 3 24 , 2 7
(x + 1)(x - 1)
1 ; Domain: {x x ≠ - 2, x ≠ 2} 2. H is in lowest terms 3. y-intercept: ; x-intercepts: - 1, 1 16 (x + 4)(x + 2)(x - 2) 4. H is in lowest terms; vertical asymptotes: x = - 2, x = 2 5. Horizontal asymptote: y = 0, intersected at ( - 1, 0) and (1, 0) 1 2 6. 2 1 7. 1
23. 1. H(x) =
2
(2, 1)
(1, 1)
3
1.5
0
H(3) ⬇ 0.12
H(1.5) ⬇ 0.11 H(0) 16
Location of Graph Above x-axis
Below x-axis
Point on Graph
(1.5, 0.11)
Interval
(, 2)
Number Chosen Value of H
25. 1. F(x) =
(3, 0.12)
(1, 2)
(2, )
1.5
3
H(1.5) ⬇ 0.11
H(3) ⬇ 0.12
Above x-axis
Below x-axis
Above x-axis
(0, 161 )
(1.5, 0.11)
(3, 0.12)
1
(x + 1)(x - 4)
; Domain: {x x ≠ - 2} x + 2 4. F is in lowest terms; vertical asymptote: x = - 2 2 1 6.
2. F is in lowest terms
(, 2)
(2, 1)
(1, 4)
(4, )
Number Chosen
3
1.5
0
5
Value of F
F(3) 14
F(1.5) 5.5
F(0) 2
F(5) ⬇ 0.86
Above x-axis
Below x-axis
Above x-axis
(1.5, 5.5)
(0, 2)
(5, 0.86)
Point on Graph
(3, 14)
0,
16
(3, 0.12)
(3, 0.12) y0 5 x (1, 0) (1.5, 0.11) x 2 x 2
(1, 0) (1.5, 0.11)
3. y-intercept: - 2; x-intercepts: - 1, 4
5. Oblique asymptote: y = x - 5, not intersected 4 7.
Interval
Location of Graph Below x-axis
y 0.3
y
12 (1.5, 5.5) (4, 0)
yx5 (5, 0.86) 10 x (0, 2)
(1, 0)
(3, 14) x 2
AN27
ANSWERS Section 3.5
27. 1. R(x) =
(x + 4)(x - 3)
; Domain: {x x ≠ 4} x - 4 4. R is in lowest terms; vertical asymptote: x = 4 4 3 6. Interval
(, 4)
Number Chosen
5
Value of R
R(5) 9
8
Location of Graph Below x-axis Point on Graph
29. 1. F(x) =
(5, ) 8 9
2. R is in lowest terms
5. Oblique asymptote: y = x + 5, not intersected 4 7.
(4, 3)
(3, 4)
(4, )
0
3.5
5
R(0) 3
R(3.5) 7.5
R(5) 18
Above x-axis
Below x-axis
Above x-axis
(0, 3)
(3.5, 7.5)
(5, 18)
(x + 4)(x - 3)
; Domain: {x x ≠ - 2} x + 2 4. F is in lowest terms; vertical asymptote: x = - 2 6. 4 2 Interval
(, 4)
Number Chosen
5
Value of F
F (5) –3
8
Location of Graph Below x-axis Point on Graph
(5, ) 8– 3
3. y-intercept: 3; x-intercepts: - 4, 3
2. F is in lowest terms
(2, 3)
(3, )
3
0
4
F (3) 6
F (0) 6
F (4) –3
3. y-intercept: - 6; x-intercepts: - 4, 3
Above x-axis
Below x-axis
Above x-axis
(3, 6)
(0, 6)
(4, )
4
ⴚ5, ⴚ
(0, 1)
(1, )
Number Chosen
4
1
1– 2
2
Value of R
R(4) 100
R (1) 0.5
R
( ) ⬇ 0.003
R(2) 0.016
Location of Graph Above x-axis
Below x-axis
Above x-axis
Above x-axis
Point on Graph
(1, 0.5)
(
(2, 0.016)
Interval
(ⴚⴥ, ⴚ4)
Number Chosen
ⴚ5
Value of R
R(ⴚ5) ⴝ
1– 3
Location of Graph Above x-axis
35. 1. R(x) =
)
0.003
; Domain: {x x ≠ - 2, x ≠ 3}
7 4. Vertical asymptote: x = - 2; hole at a3, b 5 ⴚ4 ⴚ2 6.
Point on Graph
1– 2
(ⴚ5, ) 1– 3
(3x + 1)(2x - 3) (x - 2)(2x - 3)
x ⴝ ⴚ2
0
4
R(ⴚ3) ⴝ ⴚ1
R(0) ⴝ 2
R(4) ⴝ –3
Below x-axis
Above x-axis
Above x-axis
(ⴚ3, ⴚ1)
(0, 2)
(4, )
Number Chosen
1
0
Value of R
R(1) 23
R(0) 2
Location of Graph Above x-axis Point on Graph
37. 1. R(x) =
(1, ) 2 3
2. In lowest terms, R(x) =
(0, 0)
(1, 0)
0.01 (1, 0) 1.25 x
Enlarged view
y
1 3
(ⴚ4, 0)
7 5 3, 5 (0, 2) yⴝ1 10 x
(ⴚ3, ⴚ1)
4,
x ⴝ ⴚ2
3x + 1 x - 2
4 3
1 1 3. y-intercept: - ; x-intercept: 2 3
7.
2 (2, )
1.7
6
R(1.7) ⬇ 20.3
R(6) 4.75
Below x-axis
Below x-axis
Above x-axis
(0, )
(1.7, 20.3)
(6, 4.75)
1 2
yⴝ1 10 x
5. Horizontal asymptote: y = 3, not intersected
3 2,
1
y
3. y-intercept: 2; x-intercept: - 4
4– 3
( 2)
)
x + 4 x + 2
4
3 , x ≠ 2f 2
3 4. Vertical asymptote: x = 2; hole at a , - 11 b 2 1 3 6. 3 2
(
See enlarged view at right. 10
x ⴝ ⴚ3
ⴚ5,
ⴚ3
(, )
(0, 0)
7. (3, ⴥ)
Interval
7.
3 (ⴚ2, 3)
1 3 3 , 2
10 x (3, 0) (0, ⴚ6)
5. Horizontal asymptote: y = 1, not intersected
(ⴚ4, ⴚ2)
; Domain: e x x ≠
1 3
2. In lowest terms, R(x) =
yⴝxⴚ1 4 4, 3
10
8 3
y
(3, 0)
(x + 2)(x - 3)
y
4– 3
(, 3)
(x + 4)(x - 3)
x4
(ⴚ3, 6) (ⴚ4, 0)
Interval
33. 1. R(x) =
8 9
5,
(3, 0) 10 x (3.5, 7.5)
5. Oblique asymptote: y = x - 1, not intersected 3 7.
(4, 2)
1– 2,
(5, 18) yx5
(0, 3)
(4, 0)
31. 1. Domain: {x x ≠ - 3} 2. R is in lowest terms 3. y-intercept: 0; x-intercepts: 0, 1 4. Vertical asymptote: x = - 3 5. Horizontal asymptote: y = 1, not intersected 1 3 0 6.
(4, 100)
y 21
ⴚ1,
2 3
y xⴝ2 8 (6, 4.75)
yⴝ3 ⴚ
10 x 3 , ⴚ11 2
1 ,0 3
0, ⴚ
1 2
(x + 3)(x + 2)
; Domain: {x x ≠ - 3} 2. In lowest terms, R(x) = x + 2 3. y-intercept: 2; x-intercept: - 2 x + 3 4. Vertical asymptote: none; hole at ( - 3, - 1) 5. Oblique asymptote: y = x + 2 intersected at all points except x = - 3 3 2 6. 7. y Interval
(, 3)
(3, 2)
Number Chosen
4
2
Value of R
R(4) 2
Location of Graph Below x-axis Point on Graph
(4, 2)
5
R
0
( ) 5 2
1 2
R(0) 2
Below x-axis
Above x-axis
(
(0, 2)
)
5, 1 2 2
5
(2, ) (ⴚ2, 0) (ⴚ3, ⴚ1) (ⴚ4, ⴚ2)
(0, 2) 5 x 5 1 ⴚ ,ⴚ 2 2
AN28
ANSWERS Section 3.5
39. 1. H(x) =
- 3(x - 2) (x - 2)(x + 2)
; Domain: {x x ≠ - 2, x ≠ 2}
3 4. Vertical asymptote: x = - 2; hole at ¢ 2, - ≤ 4 2 2 6. (2, 2)
(2, )
Number Chosen
3
0
3 3
H(3) 5
Above x-axis
Below x-axis
Below x-axis
Point on Graph
(3, 3)
(0 , )
(3 , )
3 2
x - 4 ; Domain: {x x ≠ 1} 2. In lowest terms, F(x) = x - 1 (x - 1)2 4. Vertical asymptote: x = 1 5. Horizontal asymptote: y = 1; not intersected 1 4 6. Interval
(, 1)
(1, 4)
(4, )
Number Chosen
0
2
5
x Q0, − 32R
F(5)
3. y-intercept: 4; x-intercept: 4
7.
1 4
Value of F
F(0) 4
F(2) 2
Location of Graph
Above x-axis
Below x-axis
Above x-axis
(2, 2)
(5 , )
(0, 4)
Q3, − 35R
3 5
(x - 1)(x - 4)
Point on Graph
x = −2 y Q2, − 34R 10 (−3, 3) y=0 −7
3
H(0) 2
Location of Graph
41. 1. F(x) =
3 3. y-intercept: - ; no x-intercept 2
7.
(, 2)
H(3) 3
-3 x + 2
5. Horizontal asymptote: y = 0; not intersected
Interval
Value of H
2. In lowest terms, H(x) =
y x=1 5 Q5, 14R (0, 4) y=1 −5
5 x (4, 0) (2, −2)
−5
1 4
x
; Domain: {x x ≠ - 2} 2. G is in lowest terms 3. y-intercept: 0; x-intercept: 0 (x + 2)2 4. Vertical asymptote: x = - 2 5. Horizontal asymptote: y = 0; intersected at (0, 0) 2 0 6. 7.
43. 1. G(x) =
Interval
(, 2)
(2, 0)
(0, )
Number Chosen
3
1
1 G(1)
x = −2
y Q1, 19R
3 −7 y=0 1 9
Value of G
G(3) 3
G(1) 1
Location of Graph
Below x-axis
Below x-axis
Above x-axis
Point on Graph
(3, 3)
(1, 1)
(1 , )
3 x (0, 0) (−1, −1) −7
(−3, −3)
1 9
x2 + 1 ; Domain: {x x ≠ 0} 2. f is in lowest terms 3. no y-intercept; no x-intercepts x 4. f is in lowest terms; vertical asymptote: x = 0 5. Oblique asymptote: y = x, not intersected 0 6. 7.
45. 1. f(x) =
Interval
(, 0)
(0, )
Number Chosen
1
1
Value of f
f(1) 2
Location of Graph Below x-axis Point on Graph
(1, 2)
x0 y 5 (1, 2) yx 5 x (1, 2)
f(1) 2 Above x-axis (1, 2)
(x + 1)(x2 - x + 1) x3 + 1 = ; Domain: {x x ≠ 0} 2. f is in lowest terms 3. no y-intercept; x-intercept: - 1 x x 4. f is in lowest terms; vertical asymptote: x = 0 5. No horizontal or oblique asymptote 1 0 6. 7. y 7
47. 1. f(x) =
Interval Number Chosen Value of f
(, 1)
(1, 0)
2
1 2
f(2)
7 2
Location of Graph Above x-axis Point on Graph
49. 1. f(x) =
(2, ) 7 2
f
(0, )
2,
5
2
1
( ) 1 2
7 4
f (1) 2
Below x-axis
Above x-axis
(
(1, 2)
)
1, 7 2 4
(1, 0) 1 7 , 2 4
x0
x4 + 1
; Domain: {x x ≠ 0} 2. f is in lowest terms 3. no y-intercept; no x-intercepts x3 4. f is in lowest terms; vertical asymptote: x = 0 5. Oblique asymptote: y = x, not intersected 0 6. 7. Interval
(, 0)
(0, )
Number Chosen
1
1
Value of f
f(1) 2
f(1) 2
Location of Graph Below x-axis Point on Graph
(1, 2)
Above x-axis (1, 2)
(1, 2) 5 x
yx
y 5
(1, 2) 5 x (1, 2) x0
ANSWERS Section 3.6 53. Minimum value: 1.89 at x = 0.79
51. Minimum value: 2.00 at x = 1.00 5
55. Minimum value: 1.75 at x = 1.32
14
0
10
0
5 0
4
0
0
57. One possibility: R(x) =
x2 - 4
59. One possibility: R(x) =
61. (a) t-axis; C(t) S 0 (b) 0.4
4 b 3
(x + 1)2 (x - 2)2
63. (a) C(x) = 16x +
65. (a) S(x) = 2x2 +
5000 + 100 x
(b)
(b) x 7 0 (c) 10,000
0
5 0
(x - 1)(x - 3) ax2 +
x2
12
40,000 x
10,000
0
0
60 0
(c) 0.71 h after injection
0
300
(c) 2784.95 in.2 (d) 21.54 in. * 21.54 in. * 21.54 in. (e) To minimize the cost of materials needed for construction
0
(d) Approximately 17.7 ft by 56.5 ft (longer side parallel to river) 67. (a) C(r) = 12pr 2 +
4000 r
x2 + x - 12 x - x - 6 2
6000
(b)
69. (a) R(x) = e
7 5
0
10 0
The cost is smallest when r = 3.76 cm.
6x2 - 7x - 3
if x ≠ 3 (b) R(x) = e if x = 3
2x2 - 7x + 6 - 11
4. F
5. (a) 5x 0 0 6 x 6 1 or x 7 2 6; (0, 1) h (2, q )
7. (a) 5x 0 - 1 6 x 6 0 or x 7 1 6; ( - 1, 0) h (1, q ) 9. 5x 0 x 6 0 or 0 6 x 6 3 6; ( - q , 0) h (0, 3)
11. 5x 0 x Ú - 4 6; 3 - 4, q )
31. 5x 0 x 6 - 1 or x 7 1 6; ( - q , - 1) h (1, q )
35. 5x 0 x … - 1 or 0 6 x … 1 6; ( - q , - 1] h (0, 1 4
13. 5x 0 x … - 2 or x Ú 2 6; ( - q , - 2 4 h 3 2, q )
17. 5x 0 - 2 6 x … - 1 6; ( - 2, - 1]
23. 5x 0 - 4 6 x 6 0 or x 7 0 6; ( - 4, 0) h (0, q )
27. 5x 0 - 1 6 x 6 0 or x 7 3 6; ( - 1, 0) h (3, q )
39. 5x 0 x 6 2 6; ( - q , 2)
(b) 5x 0 x … 0 or 1 … x … 2 6; ( - q , 0 4 h [1, 2]
(b) 5x 0 x 6 - 1 or 0 … x 6 1 6; ( - q , - 1) h 3 0, 1)
15. 5x 0 - 4 6 x 6 - 1 or x 7 0 6; ( - 4, - 1) h (0, q ) 21. 5x 0 x 7 4 6; (4, q )
19. 5x 0 x 6 - 2 6; ( - q , - 2)
25. 5x 0 x … 1 or 2 … x … 3 6; ( - q , 1] h 3 2, 3 4
29. 5x 0 x 6 - 1 or x 7 1 6; ( - q , - 1) h (1, q )
33. 5x 0 x 6 - 1 or x 7 1 6; ( - q , - 1) h (1, q )
37. 5x 0 x 6 - 1 or x 7 1 6; ( - q , - 1) h (1, q )
41. 5x 0 - 2 6 x … 9 6; ( - 2, 9 4
43. 5x 0 x 6 2 or 3 6 x 6 5 6; ( - q , 2) h (3, 5)
45. 5x 0 x 6 - 5 or - 4 … x … - 3 or x = 0 or x 7 1 6; ( - q , - 5) h 3 - 4, - 3 4 h 50 6 h (1, q ) 47. e x ` -
1 1 6 x 6 1 or x 7 3 f ; a - , 1 b h (3, q ) 2 2
49. 5x 0 - 1 6 x 6 3 or x 7 5 6; ( - 1, 3) h (5, q )
1 1 55. 5x 0 x 6 2 6 ; ( - q , 2) r; (- q, - 44 h c , q b 53. 5x 0 x 6 3 or x Ú 7 6 ; ( - q , 3) h [7, q ) 2 2 2 3 2 3 57. b x ` x 6 - or 0 6 x 6 r ; a - q , - b h a0, b 59. 5x 0 x … - 3 or 0 … x … 3 6 ; ( - q , - 3 4 h 3 0, 3 4 3 2 3 2
51. b x ` x … - 4 or x Ú
if x ≠
3 2
if x =
3 2
71. No. Each function is a quotient of polynomials, but it is not written in lowest terms Each function is undefined for x = 1; each graph has a hole at x = 1. 77. If there is a common factor between the numerator and the denominator, and the factor yields a real zero, then the graph will have a hole. 1 17 79. e - f 80. 81. ≈ - 0.164 78. 4x3 - 5x2 + 2x - 2 10 2
3.6 Assess Your Understanding (page 295) 3. T
AN29
AN30
ANSWERS Section 3.6 63. (a)
1 61. (a) - 4, - , 3 2 (b) f(x) = (x + 4)2(2x + 1)(x - 3) (c) (1.87, 60.54)
y 10 (6, 0) y1 0 ,
y
75
1 ,0 2 (0, 48)
15 2
65. (a) 9 10 , 4
(1, 0) 10 x
3 2
x2
(3, 0) 5 x
(4, 0)
4,
y y x 3 10 28 3, 5 (4, 0) (0, 2) 10 x (1, 0) x 2
(b) 3 - 4, - 2) h 3 - 1, 3) h (3, q )
(b) ( - q , - 6] h [1, 2 ) h (2, q )
(1.74, 185.98)
1 (d) a - , 3 b 2 67. ¢ - 3,
-4 4 ≤ h ¢ , 3≤ 3 3
75.
69. 5x 0 x 7 4 6 ; (4, q ) 77.
y f(x) x4 1 2.5 (1, 0) 2.5 x
(1, 0)
71. 5x 0 x … - 2 or x Ú 2 6 ; ( - q , - 2 4 h 3 2, q )
y g(x) 3x 2 32 (2, 12)
(2, 12)
2.5 x f (x) x4 4
g(x) 2x2 2
ƒ(x) … g(x) if - 1 … x … 1
ƒ(x) … g(x) if - 2 … x … 2
73. 5x 0 x 6 - 4 or x Ú 2 6 ; ( - q , - 4) h 32, q )
79. Produce at least 250 bicycles 81. (a) The stretch is less than 39 ft. (b) The ledge should be at least 84 ft above the ground for a 150-lb jumper. 4 83. At least 50 students must attend. 88. J , q ≤ 3 2 2 2 4 89. y = 2x 90. x - x - 4 91. 3x y (x + 2y)(2x - 3y) 3
Review Exercises (page 299) 1. Polynomial of degree 7 y 5.
2. Neither 6.
3. Rational y 4 (1, 0)
15 (2, 0)
(0, 8)
(0, 1)
2 x
(4, 8)
18 7 x
(2, 1)
(4, 0) (5, 15)
20
(1, 2) (2, 3) 3 x
(0, 3)
8. 1. y = x3 2. x-intercepts: - 4, - 2, 0; y-intercept: 0 3. - 4, - 2, 0, multiplicity 1, crosses 4. 2 y 5. (2, 0) (3, 3)
4. Polynomial of degree 0 y 7.
(1, 15) (0, 0) 2 x (1, 3)
9. 1. y = x3 2. x-intercepts: - 4, 2; y-intercept: 16 3. - 4 multiplicity 1, crosses; 2, multiplicity 2, touches 4. 2 y (0, 16) 5. (2, 32) (3, 7) (2, 0) 10 x
(4, 0) (5, 49)
10. 1. y = - 2x3 2. f(x) = - 2x2 (x - 2) x-intercepts: 0, 2; y-intercept: 0 3. 0, multiplicity 2, touches: 2, multiplicity 1, crosses 4. 2 y 5. 20 (1, 2) (2, 0) 6 x (0, 0)
(1, 6)
(3, 18)
12. R = 0; g is a factor of f.
60
11. 1. y = x4 2. x-intercepts: - 3, - 1, 1; y-intercept: 3 3. - 3, - 1, multiplicity 1, crosses; 1, multiplicity 2, touches 4. 3 y 5. (4, 75)
80
(2, 15) (1, 0) 5 x (2, 9) (0, 3) (1, 0)
(3, 0)
13. R = - 5x + 10; g is not a factor of f.. 14. f(4) = 47,105 15. 5, 3, or 1 positive; 3 or 1 negative 1 3 1 1 3 1 1 16. 1 positive; 2 or 0 negative 17. {1, {3, { , { , { , { , { , { , { 18. - 2, 1, 4; f(x) = (x + 2)(x - 1)(x - 4) 2 2 3 4 4 6 12 2 1 1 1 19. , multiplicity 2; - 2; f(x) = 4 ax - b (x + 2) 20. 2, multiplicity 2; f(x) = (x - 2)2 (x2 + 5) 21. 5 - 3, 2 6 22. e - 3, - 1, - , 1 f 2 2 2
ANSWERS Review Exercises 23. LB: - 2, UB; 3
24. LB: - 5, UB: 5
25. f(0) = - 1; f(2) = 5
26. f(0) = - 1; f(1) = 1
28. 6 + i, 2 - 5i; f(x) = x5 - 19x4 + 162x3 - 838x2 + 2561x - 3219 30. - 2,
1 1 (multiplicity 2); f(x) = 4(x + 2) ax - b 2 2
32. - 3, 2, -
2
27. 2 - 3i; f(x) = x3 - 8x2 + 29x - 52
29. - 2, 1, 4; f(x) = (x + 2)(x - 1)(x - 4)
31. 2 (multiplicity 2), - 25i, 25i; f(x) = (x + 25i)(x - 25i)(x - 2)2
22 22 22 22 i, i; f(x) = 2(x + 3)(x - 2) ax + i b ax ib 2 2 2 2
33. Domain: 5x 0 x 34. Domain: 5x x 35. Domain: 5x x 2(x 36. 1. R(x) =
≠ - 3 6 ; horizontal asymptote: y = 0; vertical asymptotes: x = - 3 ≠ 2, x ≠ - 2 6 ; horizontal asymptote: y = 2; vertical asymptote: x = 2, x = - 2 ≠ - 2}; horizontal asymptote: y = 1; vertical asymptote: x = - 2 - 3) ; Domain: {x x ≠ 0} 2. R is in lowest terms 3. no y-intercept; x-intercept: 3 x 4. R is in lowest terms; vertical asymptote: x = 0 5. Horizontal asymptote: y = 2; not intersected 0 3 6. 7. Interval
(, 0)
(0, 3)
(3, )
Number Chosen
2
1
4
R(2) 5
R(1) 4
R(4)
Below x-axis
Above x-axis
Point on Graph
(1, 4)
(4, )
(2, 5)
10 (2, 5)
3. no y-intercept; x-intercept: - 2 5. Horizontal asymptote: y = 0; intersected at ( - 2, 0) x0 2 7. y 1,
Interval
(, 2)
(2, 0)
(0, 2)
(2, )
Number Chosen
3
1
1
3
Value of H
H(3) –– 15
H(1) 1–3
H(1) 3
H(3) 5–3
Above x-axis
Below x-axis
Above x-axis
(1, )
(1, 3)
(3, )
1
Location of Graph Below x-axis Point on Graph
(3, ) 1 –– 15
1– 3
1 2
x0
1– 2
37. 1. Domain: {x x ≠ 0, x ≠ 2} 2. H is in lowest terms 4. H is in lowest terms; vertical asymptotes: x = 0, x = 2 2 0 6.
4,
y2 (3, 0) x (1, 4)
10
1 –2
Location of Graph Above x-axis
Value of R
y
1 3
5
3,
5 3
(2, 0) 5 x 1 3, 15
y0
(1, 3) x2
5– 3
(x + 3)(x - 2)
; Domain: {x x ≠ - 2, x ≠ 3} 2. R is in lowest terms 3. y-intercept: 1; x-intercepts: - 3, 2 (x - 3)(x + 2) 4. R is in lowest terms; vertical asymptotes: x = - 2, x = 3 5. Horizontal asymptote: y = 1; intersected at (0, 1) 3 2 2 3 6. 7. y (0, 1)
38. 1. R(x) =
(3, 2)
Interval
(, 3)
Number Chosen
4
Value of R
R(4) –– 7
5
5– 2
R(0) 1
R
Below x-axis
Above x-axis
Below x-axis
Above x-axis
(
(0, 1)
(
(4, )
5
3 –– 7
4,
0
( )
(4, )
(3, )
(2, 3)
9 R –2 –– 11
–2 3
Location of Graph Above x-axis Point on Graph
(2, 2)
5– 2,
9
)
–– 11
4
( ) 5– 2
5– 2,
11 –– 9
)
11
–– 9
4,
5
3 7
y1 5 x (2, 0)
(3, 0)
7
R (4) –– 3
7 –– 3
7 3
5 9 5 11 , , 2 11 2 9 x 2 x 3
x3 ; Domain: {x x ≠ - 2, x ≠ 2} 2. F is in lowest terms 3. y-intercept: 0; x-intercept: 0 (x + 2)(x - 2) 4. F is in lowest terms; vertical asymptotes: x = - 2, x = 2 5. Oblique asymptote: y = x; intersected at (0, 0) 2 6. 0 7. 2 x 2
39. 1. F(x) =
Interval
(, 2)
(2, 0)
(0, 2)
(2, )
Number Chosen
3
1
1
3
Value of F
F(3) –– 5
F(1 ) 1–3
F(1) 1–3
–– F(3) 27 5
Above x-axis
Below x-axis
Above x-axis
27
Location of Graph Below x-axis Point on Graph
(3, ) 27 –– 5
(1, )
(1 , )
1– 3
1– 3
40. 1. Domain: {x x ≠ 1} 2. R is in lowest terms 4. R is in lowest terms; vertical asymptote: x = 1 0 1 6. (, 0)
(0, 1)
(1, )
2
1– 2
2
Value of R
R(2)
Location of Graph Above x-axis Point on Graph
(2, ) 32 –– 9
( )
R
1– 2
R(2) 32
Above x-axis
Above x-axis
(
(2, 32)
1– 2
,
1– 2
)
yx
4 10 x 1 3
1,
(3, ) 27 – 5
27 3, 5
7.
Number Chosen
1– 2
1 3 (0, 0)
1,
27 5
3,
x2
3. y-intercept: 0; x-intercept: 0 5. No oblique or horizontal asymptote
Interval
32 –– 9
y
y 40 (2, 32) 1 1 , 2 2
32 2, 9 (0, 0) x1
5 x
AN31
AN32
ANSWERS Review Exercises (x + 2)(x - 2)
41. 1. G(x) =
(x + 1)(x - 2)
; Domain: {x x ≠ - 1, x ≠ 2}
4 4. Vertical asymptote: x = - 1; hole at a2, b 3 1 2 6. Interval
(, 2)
Number Chosen
3
Value of G
G(3)
(2, 1) 3
Location of Graph Above x-axis
(3, ) 1– 2
Point on Graph
3 G (3) –4
Below x-axis
Above x-axis
Above x-axis
(
(0, 2)
(3, )
3– 2 , 1
(b) ( - 3, 2) ∪ (2, q )
43. (a) y = 0.25
(b) x = - 2, x = 2
)
5– 4
(c) ( - 3, - 2) ∪ ( - 1, 2)
44. {x x 6 - 2 or - 1 6 x 6 2}; ( - q , - 2) ∪ ( - 1, 2)
3
2
2
(d) f(x) = (x - 2)2(x + 3) (d) ( - q , - 3] ∪ ( - 2, - 1] ∪ (2, q )
3
(b) 223.22 cm2 (d) 1000
4x2 - 16
2
1
48. {x x 6 - 4 or 2 6 x 6 4 or x 7 6}; ( - q , - 4) h (2, 4) ∪ (6, q )
3 4
49. (a) A(r) = 2pr 2 +
x2 + 4x + 3
(e) R(x) =
46. {x x 6 1 or x 7 2}; ( - q , 1) ∪ (2, q )
45. {x - 3 6 x … 3}; ( - 3, 3]
47. {x 1 … x … 2 or x 7 3}; [1, 2] ∪ (3, q ) 1
4 5 2, 3 (0, 2) y1 5 x (2, 0) 5 3, 3 4 , 1 2 x 1
5
(c) ( - q , - 3] ∪ 52 6
y 1 2
3,
G (0) 2
42. (a) 5 - 3, 2 6
2 1
7. (2, )
0 3
3. y-intercept: 2; x-intercept: - 2
2
G –2 1
( )
x + 2 x + 1
5. Horizontal asymptote: y = 1, not intersected
(1, 2)
–2 1– 2
2. In lowest terms, G(x) =
50. (a)
500 r (c) 257.08 cm2
2
4 6
(c)
17
−1
17
−1
23 10
0
23 10
Data appear to follow a cubic relation. (b) p(t) = 0.0029t 3 - 0.0723t 2 + 0.2990t + 14.0082; 17.9%
8 0
A is smallest when r ≈ 3.41 cm. 53. (a) Even
(b) Positive
(d) The graph touches the x-axis at x = 0 but does not cross it there.
(c) Even
(e) 8
Chapter Test (page 302) 1.
(4, 1) 8 x (3, 2) (2, 1)
p
1 3 5 15 : { , {1, { , { , {3, {5, { , {15 q 2 2 2 2 1 (c) - 5, - , 3; g(x) = (x + 5)(2x + 1)(x - 3) 2 1 (d) y-intercept: - 15; x-intercepts: - 5, - , 3 2 1 (e) Crosses at - 5, - , 3 (f) y = 2x3 2
2. (a) 3
y 7
(b)
(g) (2, 45) y (3, 60) 60
1 ,0 2
(3, 0) 5 x
(5, 0)
(2, 35) (0, 15) (1, 36)
5 - 261 5 + 261 , r 5. Domain: {x x ≠ - 10, x ≠ 4}; asymptotes: x = - 10, y = 2 6 6 6. Domain: {x x ≠ - 1}; asymptotes: x = - 1, y = x + 1 yx1 7. y 3. 4, - 5i, 5i
4. b 1,
5 (3, 0)
(1, 0) 5 x (0, 3)
x 1
8. Answers may vary. One possibility is f(x) = x4 - 4x3 - 2x2 + 20x. 9. Answers may vary. One possibility is r(x) =
2(x - 9)(x - 1) (x - 4)(x - 9)
.
10. f(0) = 8; f(4) = - 36 Since f(0) = 8 7 0 and f(4) = - 36 6 0, the Intermediate Value Theorem guarantees that there is at least one real zero between 0 and 4. 11. 5x x 6 3 or x 7 8}; ( - q , 3) ∪ (8, q )
AN33
ANSWERS Section 4.1
Cumulative Review (page 302) 1. 226
4. f(x) = - 3x + 1
3. {x - 1 6 x 6 4}; ( - 1, 4)
2. { x x … 0 or x Ú 1}; ( - q , 0] or [1, q )
y 1
4
5
(1, 4)
1
0
5 x
5. y = 2x - 1 3 3 7. Not a function; 3 has two images. 8. {0, 2, 4} 9. e x ` x Ú f ; J , q b 0 1 2 3 2 2 11. x-intercepts: - 3, 0, 3; y-intercept: 0; symmetric with respect to the origin 10. Center: ( - 2, 1); radius: 3 2 17 y 12. y = - x + 3 3 (2, 4) 5 13. Not a function; it fails the vertical-line test. (5, 1) (1, 1) 2 x (2, 1) 14. (a) 22 (b) x2 - 5x - 2 (c) - x2 - 5x + 2 (d) 9x2 + 15x - 2 (2, 2)
15. (a) {x x ≠ 1}
(b) No; (2, 7) is on the graph.
y 6
6.
(3, 5) 5 x
y 10
(1, 1)
(2, 8) (1, 1) 10 x
(2, 8)
(e) 2x + h + 5
(c) 4; (3, 4) is on the graph.
7 7 d) ; a , 9 b is on the graph. 4 4
(e) Rational 16.
17.
y (0, 7)
y
8
6 7 ,0 3
(0, 1)
兹2 1 ,0 2 3 x (1, 1)
兹2 1 ,0 2 x1
8 x
21. (a) Domain: {x x 7 - 3} or ( - 3, q ) 1 (b) x-intercept: - ; y-intercept: 1 2 (c) y 1 ,0 2
5
18. 6; y = 6x - 1 19. (a) x-intercepts: - 5, - 1, 5; y-intercept: - 3 (b) No symmetry (c) Neither (d) Increasing: ( - q , - 3) and (2, q ); decreasing: ( - 3, 2) (e) Local maximum is 5 and occurs at x = - 3. (e) Local minimum is - 6 and occurs at x = 2. 20. Odd
22.
23. (a) (f + g)(x) = x2 - 9x - 6; domain: all real numbers
y 8
(b) a
(1, 5) (2, 2)
(0, 2) 2 x
(2, 5)
g
b (x) =
24. (a) R(x) = -
(0, 1)
(3, 5)
f
(b) $14,000
5 x (2, 2)
x2 - 5x + 1 7 ; domain: e x ` x ≠ - f - 4x - 7 4
1 2 x + 150x 10 (c) 750; $56,250
(d) $75
(d) Range: {y y 6 5} or ( - q , 5)
CHAPTER 4 Exponential and Logarithmic Functions 4.1 Assess Your Understanding (page 311) 4. composite function; f 1g 1x2 2
5. False 6. False 7. (a) - 1 (b) - 1 (c) 8 (d) 0 (e) 8 (f) - 7 9. (a) 4 (b) 5 (c) - 1 (d) - 2 11. (a) 98 163 3 1 1 1 (b) 49 (c) 4 (d) 4 13. (a) 97 (b) (c) 1 (d) 15. (a) 2 22 (b) 2 22 (c) 1 (d) 0 17. (a) (b) (c) 1 (d) 2 2 17 5 2 3 6 3 19. (a) (b) 1 (c) (d) 0 21. {x x ≠ 0, x ≠ 2} 23. {x x ≠ - 4, x ≠ 0} 25. e x ` x Ú - f 27. {x x Ú 1} 29. {x x Ú 0, x ≠ 4} 3 5 2 2 4 + 1 31. (a) 1f ∘ g2 1x2 = 6x + 3; all real numbers (b) 1g ∘ f2 1x2 = 6x + 9; all real numbers (c) 1f ∘ f2 1x2 = 4x + 9; all real numbers (d) 1g ∘ g2 1x2 = 9x; all real numbers
37. (a) 1f ∘ g2 1x2 =
3x ; {x x ≠ 0, x ≠ 2} 2 - x
(d) 1g ∘ g2 1x2 = x; {x x ≠ 0} (c) 1f ∘ f2 1x2 = x; {x x ≠ 1}
33. (a) 1f ∘ g2 1x2 = 3x2 + 1; all real numbers (b) 1g ∘ f2 1x2 = 9x2 + 6x + 1; all real numbers (c) 1f ∘ f2 1x2 = 9x + 4; all real numbers (d) 1g ∘ g2 1x2 = x4; all real numbers (b) 1g ∘ f2 1x2 =
39. (a) 1f ∘ g2 1x2 =
2 1x - 12
4 ; {x x ≠ - 4, x ≠ 0} 4 + x
(d) 1g ∘ g2 1x2 = x; {x x ≠ 0}
(b) 1g ∘ f2 1x2 = 2 2x + 3; {x x Ú 0}
3
; {x x ≠ 1}
35. (a) (f ∘ g2(x2 = x4 + 8x2 + 16; all real numbers (b) (g ∘ f2(x2 = x4 + 4; all real numbers (c) 1f ∘ f2 1x2 = x4; all real numbers (d) 1g ∘ g2 1x2 = x4 + 8x2 + 20; all real numbers
(c) 1f ∘ f2 1x2 =
3 1x - 12
; {x x ≠ 1, x ≠ 4} 4 - x - 4 1x - 12 (b) 1g ∘ f2 1x2 = ; {x x ≠ 0, x ≠ 1} x
3 41. (a) 1f ∘ g2 1x2 = 22x + 3; e x ` x Ú - f 2
4 (c) 1f ∘ f2 1x2 = 2 x; {x x Ú 0}
(d) 1g ∘ g2 1x2 = 4x + 9; all real numbers
AN34
ANSWERS Section 4.1
43. (a) 1f ∘ g2 1x2 = x; {x x Ú 1}
(b) 1g ∘ f2 1x2 = x ; all real numbers
(d) 1g ∘ g2 1x2 = 2 1x - 1 - 1; {x x Ú 2}
(c) 1f ∘ f2 1x2 = x4 + 2x2 + 2; all real numbers
4x - 17 1 ; e x ` x ≠ 3; x ≠ f 2x - 1 2 2x + 5 (c) 1f ∘ f2 1x2 = ; {x x ≠ - 1; x ≠ 2} x - 2
45. (a) 1f ∘ g2 1x2 = -
(b) 1g ∘ f2 1x2 = -
3x - 3 ; {x x ≠ - 4; x ≠ - 1} 2x + 8
(d) 1g ∘ g2 1x2 = -
3x - 4 11 ; ex ` x ≠ ; x ≠ 3f 2x - 11 2
1 1 1 47. 1f ∘ g2 1x2 = f 1g 1x2 2 = ƒ a x b = 2 a x b = x; 1g ∘ f2 1x2 = g 1f 1x2 2 = g 12x2 = 12x2 = x 2 2 2
3 3 3 3 49. 1f ∘ g2 1x2 = f 1g 1x2 2 = f 1 1 x2 = 1 1 x2 3 = x; 1g ∘ f2 1x2 = g 1f 1x2 2 = g 1x3 2 = 2 x = x 1 1 51. 1f ∘ g2 1x2 = f 1g 1x2 2 = f a 1x + 62 b = 2 c 1x + 62 d - 6 = x + 6 - 6 = x; 2 2
1g ∘ f2 1x2 = g 1f 1x2 2 = g 12x - 62 =
1 1 12x - 6 + 62 = 12x2 = x 2 2 1 1 1 53. 1f ∘ g2 1x2 = f 1g 1x2 2 = ƒ a 1x - b2 b = a c 1x - b2 d + b = x; 1g ∘ f2 1x2 = g 1f 1x2 2 = g 1ax + b2 = 1ax + b - b2 = x a a a
55. f 1x2 = x4; g 1x2 = 2x + 3 (Other answers are possible.)
57. f 1x2 = 2x; g 1x2 = x2 + 1 (Other answers are possible.)
59. f 1x 2 = x ; g 1x 2 = 2x + 1 (Other answers are possible.) 65. (a) 1f ∘ g2(x2 = acx + ad + b
61. 1f ∘ g2 1x2 = 11; 1g ∘ f2 1x2 = 2
63. - 3, 3
(b) 1g ∘ f2 1x2 = acx + bc + d (c) The domains of both f ∘ g and g ∘ f are all real numbers. 16 6 pt 69. C1t2 = 15,000 + 800,000t - 40,000t 2 (d) f ∘ g = g ∘ f when ad + b = bc + d 67. S 1t2 = 9 2 2100 - p 71. C1p2 = + 600, 0 … p … 100 73. V1r2 = 2pr 3 25 75. (a) f 1x2 = 0.7643x (b) g 1x2 = 128.4594x (c) g 1f 1x2 2 = g 10.7643x2 = 98.18151942x (d) g 1f 110002 2 = 98,181.51942; On April 18, 2013, one thousand dollars could purchase 98,181.51942 yen. 77. (a) f 1p2 = p - 200 (b) g 1p2 = 0.8p (c) 1f ∘ g2 1p2 = 0.8p - 200; 1g ∘ f2 1p2 = 0.8p - 160; The 20% discount followed by the $200 rebate is the better deal. 79. - 2 22, - 23, 0, 23, 2 22 81. f and g are odd functions, so f 1 - x2 = - f 1x2 and g 1 - x2 = - g 1x2. Then 1f ∘ g2 1 - x2 = f 1g 1 - x2 2 = f 1 - g 1x2 2 = - f 1g 1x2 2 = - 1f ∘ g2 1x2. So f ∘ g is also odd. 83. 15 84. (f + g)(x) = 4x + 3; Domain: all real numbers (f - g)(x) = 2x + 13; Domain: all real numbers
(f # g)(x) = 3x2 - 7x - 40; Domain: all real numbers 3x + 8 ; Domain: {x x ≠ 5} ¢ ≤(x) = g x - 5
86.
87. Domain: {x x ≠ 3} Vertical asymptote: x = 3 Oblique asymptote: y = x + 9
10
−3
3
f
−10
1 85. , 4 4
Local minimum: - 5.08 at x = - 1.15 Local maximum: 1.08 at x = 1.15 Decreasing: ( - 3, - 1.15); (1.15, 3) Increasing: ( - 1.15, 1.15)
4.2 Assess Your Understanding (Page 322) 5. ƒ(x1) ≠ ƒ(x2) 6. one-to-one 7. 3 8. y = x 9. [4, q 2 10. True 11. one-to-one 13. not one-to-one 15. not one-to-one 17. one-to-one 19. one-to-one 21. not one-to-one 23. one-to-one 25.
Annual Rainfall (inches) 49.7 43.8 4.2 61.9 12.8
27. Location Atlanta, Georgia Boston, Massachusetts Las Vegas, Nevada Miami, Florida Los Angeles, California
Monthly Cost of Life Insurance
Age
$10.55 $12.89 $19.29
30 40 45
Domain: {$10.55, $12.89, $19.29} Range: 530, 40, 45 6
Domain: {49.7, 43.8, 4.2, 61.9, 12.8} Range: {Atlanta, Boston, Las Vegas, Miami, Los Angeles} 29. 5 15, - 32, 19, - 22, 12, - 12, 111, 02, 1 - 5, 12 6 Domain: 55, 9, 2, 11, - 5 6 Range: 5 - 3, - 2, - 1, 0, 1 6
31. 5 11, - 2 2, 12, - 32, 10, - 102, 19, 12, 14, 22 6 Domain: 51, 2, 0, 9, 4 6 Range: 5 - 2, - 3, - 10, 1, 2 6
ANSWERS Section 4.2 1 1 33. f 1g 1x2 2 = f a 1x - 42 b = 3 c 1x - 42 d + 4 = 1x - 42 + 4 = x 3 3 1 1 g 1f 1x2 2 = g 13x + 42 = 3 13x + 42 - 4 4 = 13x2 = x 3 3
35. f 1g 1x 2 2 = f a
37. f 1g 1x2 2 = f 1 2x + 82 = 1 2x + 82 3 - 8 = 1x + 82 - 8 = x
1 39. f 1g 1x2 2 = f a b = x
3
x x + 2 b = 4 c + 2 d - 8 = 1x + 82 - 8 = x 4 4
3 3 3 g 1f 1x2 2 = g 1x3 - 82 = 2 1x3 - 82 + 8 = 2 x = x
1 = x; x ≠ 0 1 a b x
1 g 1f 1x 2 2 = g a b = x
4x - 3 41. f 1g 1x2 2 = ƒ a b = 2 - x
2x + 3 g 1f 1x2 2 = g a b = x + 4 43.
4x - 3 b + 3 2 14x - 32 + 3 12 - x2 2 - x 5x = = x, x ≠ 2 = 4x - 3 4x - 3 + 4 12 - x2 5 + 4 2 - x
2x + 3 b - 3 4 12x + 32 - 3 1x + 42 x + 4 5x = = x, x ≠ - 4 = 2x + 3 2 1x + 42 - 12x + 32 5 2 x + 4
4a
y 2.5 (1, 2)
(2, 1) f 1 2.5 x
(2, 2)
f
47.
yx
y 2.5
yx
2.5 x
(1, 1)
2.5 x
f 1
(1, 0)
1
1 = x, x ≠ 0 1 a b x
2a
45.
yx
y 2.5
4x - 8 + 2 = 1x - 22 + 2 = x 4
g 1f 1x2 2 = g 14x - 82 =
3
AN35
f 1
(0, 1)
49. f - 1 1x2 =
51. f -1 1x2 =
1 x 3
1 1 f 1f - 1 1x2 2 = ƒ a x b = 3 a x b = x 3 3 1 -1 -1 f 1f 1x2 2 = f 13x2 = 13x2 = x 3
3 f 1f -1 1x2 2 = f 1 2 x + 12
f 1f - 1 1x2 2 = f a
x 1 x 1 - b = 4a - b + 2 4 2 4 2 = 1x - 22 + 2 = x 4x + 2 1 -1 f 1f 1x2 2 = f - 1 14x + 22 = 4 2 1 1 = x = ax + b 2 2
yx
y 5
3 53. f -1 1x2 = 2 x + 1
x 1 4 2
f(x) 3x 5 x
f -1 1f 1x2 2 = f -1 1x3 - 12
3 = 2 1x3 - 12 + 1 = x
f
1
3
y 5
yx
(x) 兹x 1
yx
y 5
1 f 1(x) x 3
3 = 12 x + 12 3 - 1 = x
5 x f (x) x 3 1
f (x) 4x 2 5 x f 1(x)
55. f - 1 1x) = 2x - 4, x Ú 4 f 1f
-1
1x)) = f 1 2x - 4) = 1 2x - 4) + 4 = x 2
f - 1 1f 1x)) = f - 1 1x2 + 4) = 2 1x2 + 4) - 4 = 2x2 = x, x Ú 0 y 8
f(x) x 2 4, xⱖ0
f
57. f - 1 1x2 = ƒ 1f
-1
4 x
y 5
4 1x2 2 = ƒ a b = x
4
= x
4 a b x
4 f - 1 1 ƒ 1x2 2 = f - 1 a b = x
yx 1
1 x 2 4
(x) 兹x 4
4 4 a b x
8 x
2x + 1 x 2x + 1 1 x f 1f -1 1x2 2 = f a b = = = x x 2x + 1 12x + 12 - 2x - 2 x
59. f
f
-1
-1
1x2 =
1f 1x2 2 = f
-1
1 a b = x - 2
2a
1 b + 1 2 + 1x - 22 x - 2 = = x 1 1 x - 2
f 1(x)
2x 1 x
x0 y x2 yx 5 y2 y0
5 x
f(x)
1 x2
= x
yx 4 f(x) f 1(x) x 5 x
AN36
ANSWERS Section 4.2
2 - 3x x 2 - 3x f 1f -1 1x2 2 = f a b = x
61. f -1 1x2 =
f
65. f
-1
-1
1f 1x2 2 = f
1x2 =
f 1f
-1
-1
63. f 2 2x 2x = = x = 2 - 3x 3x + 2 - 3x 2 3 + x
2 a b = 3 + x
- 2x x - 3
1x2 =
2 - 3a
2 b 2 13 + x2 - 3 # 2 3 + x 2x = = = x 2 2 2 3 + x
=
- 2x b x - 3 - 2x + 2 x - 3 3a
- 2x f 1f -1 1x2 2 = f a b = x - 3 3 1 - 2x2
- 2x + 2 1x - 32
x 1x2 2 = f a b = 3x - 2
2a 3a
x b 3x - 2
=
x b - 1 3x - 2
- 2 13x2
3x - 3 1x + 22
- 6x = x -6
=
- 2a
3x b = f -1 1f 1x2 2 = f -1 a x + 2
x 3x - 2
=
-1
3x b x + 2
3x - 3 x + 2 =
- 6x = x -6
2x 2x = = x 3x - 13x - 22 2
2x f -1 1f 1x2 2 = f -1 a b = 3x - 1
2x 3x - 1
2x b - 2 3x - 1 2x 2x = = = x 6x - 2 13x - 12 2
67. f
-1
1x2 =
3x + 4 2x - 3
69. f
3x + 4 f 1f -1 1x2 2 = f a b = 2x - 3
=
f
-1
-1
1x2 =
3a
3x + 4 b + 4 2x - 3
2a
3x + 4 b - 3 2x - 3
3 13x + 42 + 4 12x - 32
=
2 13x + 42 - 3 12x - 32
1f 1x2 2 = f
=
71. f
3a
-1
3x + 4 b = a 2x - 3
3a 2a
17x = x 17
=
f
17x = x 17
2 21 - 2x
4 - 4 4 - 4 11 - 2x2 1 2x f 1f -1 1x2 2 = f a b = = 4 2#4 21 - 2x 2# 1 - 2x 2
8x = = x 8 f
-1
1f 1x2 2 = f
-1
a
x2 - 4 2x2
b =
2 B
1 - 2a
= 2x2 = x, since x 7 0
x2 - 4 2x2
- 2x + 3 x - 2
=
3x + 4 b - 3 2x - 3
2 13x + 42 - 3 12x - 32
1x2 =
- 2x + 3 f 1f -1 1x2 2 = f a b = x - 2
3x + 4 b + 4 2x - 3
3 13x + 42 + 4 12x - 32
-1
b
=
2 4 B x2
-1
2a
- 2x + 3 b + 3 x - 2 - 2x + 3 + 2 x - 2
2 1 - 2x + 32 + 3 1x - 22 - 2x + 3 + 2 1x - 22
=
-x = x -1
2x + 3 b + 3 x + 2 1f 1x2 2 = f 2x + 3 - 2 x + 2 - 2 12x + 32 + 3 1x + 22 -x = = = x 2x + 3 - 2 1x + 22 -1 -1
2x + 3 b = a x + 2
- 2a
73. (a) 0 (b) 2 (c) 0 (d) 1 75. 7 77. Domain of f -1: [ - 2, q 2; range of f -1: [5, q 2 79. Domain of g -1: [0, q 2; range of g -1: ( - q , 0] 81. Increasing on the interval 1f 102, f 152 2 1 83. f -1 1x2 = 1x - b2, m ≠ 0 85. Quadrant I m
87. Possible answer: f 1x2 = x , x Ú 0, is one-to-one; f -1 1x2 = x, x Ú 0 d + 90.39 89. (a) r 1d2 = 6.97 6.97r - 90.39 + 90.39 6.97r (b) r 1d 1r2 2 = = = r 6.97 6.97 d 1r 1d2 2 = 6.97 a (c) 56 miles per hour
d + 90.39 b - 90.39 = d + 90.39 - 90.39 = d 6.97
AN37
ANSWERS Section 4.3
93. (a) 5g 36,250 … g … 87,850 6 (b) 5T 4991.25 … T … 17,891.25 6 T - 4991.25 (c) g 1T2 = + 36,250; 0.25 Domain: 5T 4991.25 … T … 17,891.25 6; Range: 5g 36,250 … g … 87,850 6 95. (a) t represents time, so t Ú 0. 100 - H H - 100 = (c) 2.02 seconds (b) t 1H2 = A 4.9 A - 4.9 - dx + b 97. f -1 1x2 = ; f = f -1 if a = - d cx - a
91. (a) 77.6 kg W + 88 W - 50 + 60 = (b) h 1W2 = 2.3 2.3 50 + 2.3 1h - 602 + 88 2.3h (c) h 1W1h2 2 = = = h 2.3 2.3 W1h 1W2 2 = 50 + 2.3 a
W + 88 - 60 b 2.3
= 50 + W + 88 - 138 = W (d) 73 inches 101. No 105. 6xh + 3h2 - 7h y 5
106.
- 5 - 213 - 5 + 213 , ; 6 6
107. Zeros:
- 5 - 213 - 5 + 213 , 6 6
x-intercepts:
−4
3 x
3 3 108. Domain: e x ` x ≠ - , x ≠ 2 f ; Vertical asymptote: x = - ; 2 2 Horizontal asymptote: y = 3
−5
4.3 Assess Your Understanding (page 337) 1 10. True 11. a - 1, b ; (0, 1); (1, a) 12. 1 a 15. (a) 11.212 (b) 11.587 (c) 11.664 (d) 11.665 17. (a) 8.815 (b) 8.821 (c) 8.824 (d) 8.825 19. (a) 21.217 (b) 22.217 (c) 22.440 (d) 22.459 21. 3.320 23. 0.427 25. Neither 27. Exponential; H1x2 = 4x 3 x 29. Linear; H1x2 = 2x + 4 31. Exponential; f 1x2 = - a b 33. B 35. D 37. A 39. E 2 6. exponential function; growth factor; initial value 7. a
41.
43.
y 9 1,
3 2
y 5
45. (2, 3)
(1, 3) (0, 2)
0,
2.5 x
49.
(2, 6)
Domain: All real numbers Range: 5y y 7 2 6 or 12, q 2 Horizontal asymptote: y = 2 57. y5
(2, e2)
7 3 (0, 3)
(2, 5)
Domain: All real numbers Range: 5y y 7 2 6 or 12, q 2 Horizontal asymptote: y = 2
(0, 4)
(2, e 2) (1, e)
y 8
Domain: All real numbers Range: 5y y 6 2 6 or 1 - q , 22 Horizontal asymptote: y = 2
(1, e) (0, 1) 5 x
Domain: 1 - q , q 2 Range: [1, q 2 Intercept: (0, 1)
2,
1,
1 e2
y 3
1 e (0, 1)
(0, 1) 2.5 x
y 2
1,
1 e
y0 5 x 1 2, 2 e
Domain: 1 - q , q 2 Range: [ - 1, 02 Intercept: 10, - 12
y0
5 3
Domain: All real numbers Range: 5y y 7 - 2 6 or 1 - 2, q 2 Horizontal asymptote: y = - 2 55. (0, e2)
1 1, e
y 8
(1, e) 5 x
1 3, e
Domain: All real numbers Range: 5y y 7 0 6 or 10, q 2 Horizontal asymptote: y = 0 63. 5 - 4 6
95. (a) 60; 1 - 6, 602
(2, 1) y05 x
Domain: All real numbers Range: 5y y 7 0 6 or 10, q 2 Horizontal asymptote: y = 0
3 67. e f 69. 5 - 22 , 0, 22 6 2 75. 5 - 4, 2 6 77. 5 - 4 6 79. 51, 2 6
65. 52 6
87. f 1x2 = - 6 x
91. (a) 16; (4, 16) (b) - 4; a - 4,
99. (2, e 2)
y0
71. 56 6 73. 5 - 1, 7 6 1 1 81. 83. 85. f 1x2 = 3x 49 4
y2
5 x
97.
(1, 1)
y 8
(1, e) (0, 1) y0
61. 53 6
y 5
5 x
Domain: All real numbers Range: 5y y 6 5 6 or 1 - q , 52 Horizontal asymptote: y = 5
3 2
Domain: All real numbers Range: 5y y 7 0 6 or 10, q 2 Horizontal asymptote: y = 0 53.
(0, 1)
1,
y 10
14. False
1,
y2 2.5 x
59.
y 8
1 3
y 8 2,
(1, 3) y2 6 x
(0, 3)
13. 4
47.
y 7
2.5 x
Domain: All real numbers Range: 5y y 7 0 6 or 10, q 2 Horizontal asymptote: y = 0 51.
y 8 33 1, 16
(1, 6)
(1, 1) y0 2.5 x
y1
Domain: All real numbers Range: 5y y 7 1 6 or 11, q 2 Horizontal asymptote: y = 1
8. True 9. False
89. f 1x2 = 3x + 2
1 9 9 b 93. (a) ; a - 1, b 16 4 4 (b) - 4; 1 - 4, 122 (c) - 2
(b) 3; (3, 66)
101. (a) 74% (b) 47% (c) Each pane allows only 97% of light to pass through. 103. (a) $16,231 (b) $8626 (c) As each year passes, the sedan is worth 90% of its value the previous year. 105. (a) 30% (b) 9% (c) As each year passes, only 30% of the previous survivors survive again. 107. 3.35 mg; 0.45 mg 109. (a) 0.632 (b) 0.982 (c) 1 (d) y 1 e0.1t 1 (e) About 7 min
0
40 0
AN38
ANSWERS Section 4.3 117. 36
111. (a) 0.0516 (b) 0.0888 113. (a) 70.95% (b) 72.62% (c) 100% 115. (a) 5.41 amp, 7.59 amp, 10.38 amp (b) 12 amp (d) 3.34 amp, 5.31 amp, 9.44 amp (e) 24 amp (c), (f) I 30
119.
Final Denominator 1 + 1
Value of Expression
Compare Value to e ? 2.718281828 2.5 6 e
2.5
I2 (t) 24(1 e0.5t )
2 + 2
2.8
2.8 7 e
I1 (t) 12(1 e2t )
3 + 3
2.7
2.7 6 e
5 t
121. f 1A + B2 = aA + B = aA # aB = f 1A2 # f 1B2 123. f 1ax2 = aax = 1ax 2 a = 3 f 1x2 4 a 1 -x 1e + e -1 - x2 2 2 1 = 1e - x + e x 2 2 1 x = 1e + e - x 2 2 = f 1x2
125. (a) f 1 - x2 =
(b)
y
6
4 + 4
2.721649485
2.721649485 7 e
5 + 5
2.717770035
2.717770035 6 e
6 + 6
2.718348855
2.718348855 7 e
(c) 1cosh x2 2 - 1sinh x2 2 2 2 1 1 = c 1e x + e - x 2 d - c 1e x - e - x 2 d 2 2 1 = 3 e 2x + 2 + e - 2x - e 2x + 2 - e - 2x 4 4 1 = 142 = 1 4
1 x (e ex ) 2
6
6 1
1 132. ( - q , - 5] ∪ [ - 2, 2] 133. (2, q ) 134. Linear; y = - x - 5 3 y (b) Domain: ( - q , q ) 5 Range: [ - 4, q ) (c) Decreasing: ( - q , - 1) (1, 0) 5 x Increasing: ( - 1, q )
127. 59 minutes 135. (a) (3, 0) (2, 3)
(0, 3) (1, 4) x 1
4.4 Assess Your Understanding (page 352) 4. 5x x 7 0 6 or 10, q 2 17. 23 = 8
19. a6 = 3
1 5. a , - 1 b , 11, 02, 1a, 12 a
21. 3x = 2
23. e x = 4
25. 0
6. 1
7. False
27. 2
29. - 4
59.
y f(x) 3 x yx (1, 3) 5 (3, 1) 1 1, (0, 1) f 1 (x) log 3 x 3
61.
5 (1, 2)
5 x (1, 0)
47. 5x x Ú 1 6; [1, q 2
1 2 y
63. B
1 1, 2
65. D
51. 30.099 53. 2.303 67. A
55. - 53.991
69. E
(3, 0) 2 x
x 4
75. (a) Domain: 10, q 2 (b) y 2
5
(0, 3)
x0
8
(5, 2) 8 x
x0
(c) Range: 1 - q , q 2 Vertical asymptote: x = 0 1 (d) f - 1 1x2 = e x + 3 2 (e) Domain of f -1:1 - q , q 2 Range of f -1: 10, q 2 y (f) 8
y0
7 x
y0
(c) Range: 1 - q , q 2 Vertical asymptote: x = 4 (d) f - 1 1x2 = 10x - 2 + 4 (e) Domain of f - 1: 1 - q , q 2 Range of f -1: 14, q 2 y (f) 8 (2, 5)
1 3, 2
(2, 1)
y 4
2.5
8 1, 3 2
(c) Range: 1 - q , q 2 Vertical asymptote: x = 0 (d) f - 1 1x2 = e x - 2 (e) Domain of f -1: 1 - q , q 2 Range of f -1: 10, q 2 (f) y
5 x
77. (a) Domain: 14, q 2 (b) y x 4
x
(1, 2) 5 x
57. 22
5
(c) Range: 1 - q , q 2 Vertical asymptote: x = - 4 73. (a) Domain: 10, q 2 y (b)
15. x = ln 8
(d) f - 1 1x2 = e x - 4 (e) Domain of f - 1: 1 - q , q 2 Range of f -1: 1 - 4, q 2 y (f)
5
x f 1 (x) log 1/2 x (2, 1)
1 , 1 3
49. 0.511
71. (a) Domain: 1 - 4, q 2 y (b)
1 ,1 yx 2
5
13. x = log 27.2
31.
x
f (x)
11. 2 = log a 1.6
1 1 33. 4 35. 37. 5x x 7 3 6; 13, q 2 2 2 41. 5x x 7 10 6; 110, q 2 43. 5x x 7 - 1 6; 1 - 1, q 2
39. All real numbers except 0; 5x x ≠ 0 6; ( - q , 0) ∪ (0, q )
45. 5x x 6 - 1 or x 7 0 6; 1 - q , - 12 ∪ 10, q 2
9. 2 = log 3 9
8. True
y4
5 x 3 x
ANSWERS Section 4.5 81. (a) Domain: 1 - 2, q 2 (b) x 2 y
79. (a) Domain: 10, q 2 (b) y 2.5
5,
1 2
(1, 3)
5 (1, 4) 5 x y 3
(c) Range: 1 - q , q 2 Vertical asymptote: x = - 2 (d) f - 1 1x2 = 3x - 3 - 2 (e) Domain of f - 1: 1 - q , q 2 Range of f -1: 1 - 2, q 2 y (f)
(c) Range: 1 - q , q 2 Vertical asymptote: x = 0 1 # 2x 10 (d) f - 1 1x2 = 2 -1 (e) Domain of f : 1 - q , q 2 Range of f -1: 10, q 2 (f) y
(c) Range: 1 - 3, q 2 Horizontal asymptote: y = - 3 (d) f - 1 1x2 = ln 1x + 32 - 2 (e) Domain of f - 1: 1 - 3, q 2 Range of f -1: 1 - q , q 2 y (f) 5
5 (4, 1)
8
y0
(2, 2)
5 x
x0
0,
83. (a) Domain: 1 - q , q 2 y (b)
5
8 x
1, 0 2
AN39
1, 5 2
1 2
(2, 2)
5 x (3, 1)
y 2
x 3
5 x
85. (a) Domain: 1 - q , q 2 y (b)
87. 59 6
7 89. e f 2
91. 52 6
95. 53 6
97. 52 6
99. e
9 (6, 8)
(0, 5) y4
8 x
(c) Range: 14, q 2 Horizontal asymptote: y = 4 (d) f - 1 1x2 = 3 log 2 1x - 42 (e) Domain of f - 1:14, q 2 Range of f -1: 1 - q , q 2 (f) y 8
5 x
101. e
ln 8 - 5 f 2
105. 5 - 1 6
103.
107. e 5 ln
109. e 2 - log
5
113.
93. 55 6
y 5 (1, 0)
ln 10 f 3
5 x
(1, 0)
- 2 22 , 2 22 6
x0
Domain: 5x x ≠ 0 6 Range: 1 - q , q 2 Intercepts: 1 - 1, 02, (1, 0)
7 f 5
5 f 2
1 1 111. (a) e x ` x 7 - f ; a - , q b 2 2
(8, 6)
(b) 2; (40, 2) (c) 121; (121, 3) (d) 4
(5, 0) 10 x x4 y 2.5
(1, 0) 5 x x0
Domain: 5x x 7 0 6 Range: 5y y Ú 0 6 Intercept: (1, 0)
117. (a) 1 (b) 2 (c) 3 (d) It increases. (e) 0.000316 (f) 3.981 * 10-8 119. (a) 5.97 km (b) 0.90 km 121. (a) 6.93 min (b) 16.09 min 123. h ≈ 2.29, so the time between injections is about 2 h, 17 min.
1 1 1 1 137. Zeros: - 3, - , , 3; x-intercepts: - 3, - , , 3 2 2 2 2
125. 0.2695 s 0.8959 s 2.0 1.6 1.2 0.8 0.4
(0.8959, 1)
(0.2695, 0.5) 0 0.4 1.2 2.0 Seconds
139. f(1) = - 5; f(2) = 17
138. 12
127. 50 decibels (dB) 129. 90 dB 131. 8.1 133. (a) k ≈ 11.216 (b) 6.73 (c) 0.41% (d) 0.14% 135. Because y = log 1 x means 1y = 1 = x, which cannot be true for x ≠ 1
y Amperes
115.
x
140. 3 + i; f(x) = x4 - 7x3 + 14x2 + 2x - 20; a = 1
4.5 Assess Your Understanding (page 363) 5. log a M; log a N 6. log a M; log a N 7. r log a M 8. 6 9. 7 10. False 11. False 12. False 13. 71 15. - 4 17. 7 5 1 19. 1 21. 1 23. 3 25. 27. 4 29. a + b 31. b - a 33. 3a 35. 1a + b2 37. 2 + log 5 x 39. 3 log 2 z 41. 1 + ln x 43. ln x - x 4 5 1 45. 2 log a u + 3 log av 47. 2 ln x + ln 11 - x2 49. 3 log 2 x - log 2 1x - 32 51. log x + log 1x + 22 - 2 log 1x + 32 2 1 1 2 1 1 x - 1 53. ln 1x - 22 + ln 1x + 12 - ln 1x + 42 55. ln 5 + ln x + ln 11 + 3x2 - 3 ln 1x - 42 57. log 5 u3v4 59. log 3 a 5/2 b 61. log 4 c d 3 3 3 2 1x + 12 4 x 1. 0
2. 1
3. M 4. r
63. - 2 ln 1x - 12
65. log 2[x 13x - 22 4]
67. log a a
25x6 22x + 3
b
69. log 2 c
1x + 12 2
1x + 32 1x - 12
d
71. 2.771
73. - 3.880
75. 5.615
77. 0.874
AN40 79. y =
ANSWERS Section 4.5
log x
81. y =
log 4
log 1x + 22
83. y =
log 2
2
4
2
0
5
2
log 1x - 12
4
3
1
log 1x + 12
5
4
2
85. (a) 1f ∘ g2 1x2 = x; 5x x is any real number 6 or 1 - q , q 2 (b) 1g ∘ f2 1x2 = x; 5x x 7 0 6 or 10, q 2 (c) 5 (d) 1f ∘ h2 1x2 = ln x2; {x x ≠ 0} or ( - q , 0) ∪ (0, q ) (e) 2
89. y = Cx 1x + 12
87. y = Cx 95. y =
3 2 C12x + 12 1>6
1x + 42 1>9
97. 3
91. y = Ce 3x
93. y = Ce -4x + 3
99. 1
101. log a 1x + 2x2 - 12 + log a 1x - 2x2 - 12 = log a 3 1x + 2x2 - 12 1x - 2x2 - 12 4 = log a[x2 - 1x2 - 12 4 = log a 1 = 0
103. ln 11 + e 2x 2 = ln[e 2x 1e -2x + 12 4 = ln e 2x + ln 1e -2x + 12 = 2x + ln 11 + e -2x 2 1 -y 105. y = f 1x2 = log a x; ay = x implies ay = a b = x, so - y = log 1>a x = - f 1x2. a 1 1 107. f 1x2 = log a x; f a b = log a = log a 1 - log a x = - f 1x2 x x M -1 -1 109. log a = log a 1M # N -1 2 = log a M + log a N -1 = log a M - log a N, since aloga N = N -1 implies a -loga N = N; that is, log a N = - log a N -1. N 1 - 5 - 221 - 5 + 221 116. - 2, , , 5 2 2
115. { - 1.78, 1.29, 3.49} 118.
y 10 −10
117. A repeated real solution (double root)
10 x −10
Domain: {x x … 2} or ( - q , 2] Range: {y y Ú 0} or [0, q )
4.6 Assess Your Understanding (page 369)
16 1 21 f 9. 56 6 11. 516 6 13. e f 15. 53 6 17. 55 6 19. e f 21. 5 - 6 6 23. 5 - 2 6 25. 5 - 1 + 21 + e 4 6 ≈ 56.456 6 5 3 8 - 5 + 3 25 1 9 27. e f ≈ {0.854} 29. 52 6 31. e f 33. {7} 35. { - 2 + 4 22} 37. { - 23, 23} 39. e , 729 f 41. 58 6 2 2 3 8 ln ln 10 ln 1.2 8 5 1 43. {log 2 10} = e f ≈ {3.322} 45. { - log 81.2} = e f ≈ { - 0.088} 47. e log 2 f = d t ≈ 50.226 6 ln 2 ln 8 3 5 3 ln 2 5. 516 6
49. e
7. e
ln 3 f ≈ 50.307 6 2 ln 3 + ln 4
51. e
61. 5log 4 1 - 2 + 272 6 ≈ 5 - 0.315 6 71. { - 0.57}
87. {81}
73. { - 0.70}
2 89. e - 1, f 3
97. (a) 55 6 ; (5, 3)
75. {0.57}
91. 50 6
(b) 55 6 ; (5, 4)
93.
ln 7 f ≈ 51.356 6 ln 0.6 + ln 7
63. 5log 5 4 6 ≈ 50.861 6
77. {0.39}
5 ln 1 2
79. {1.32}
99. (a), (b)
(d) 55 6
83. {2, 3}
y 18
95. c e (e) e -
g(x) 2 2.5 x
(c) 5x x 7 0.710 6 or 10.710, q 2
g(x) 10
85. e
57. e
ln 3 f ≈ 51.585 6 ln 2
67. 5log 4 5 6 ≈ 51.161 6
69. {2.79}
26 + 8 210 26 - 8 210 , f ≈ {5.7, 0.078} 9 9
ln 15
s ≈ {1.921}
1 f 11 103. (a), (b), (c) y
f (x) 3x
5 x 2
g(x) 2
x2
(0.710, 6.541)
ln p f ≈ 50.534 6 1 + ln p
ln 5 # ln 3
101. (a), (b), (c)
y f(x) 3x 1 18
55. e
65. No real solution
81. {1.31}
+ 25 2 6 ≈ 51.444 6
(c) 51 6; yes, at (1, 2)
53. 50 6
(log3 10, 10) 3
x
f (x) 2x 1
1 , 2兹2 2 5 x
59. 50 6
ANSWERS Section 4.9 105. (a)
107. (a) 2048 (b) 2068 109. (a) After 4.2 yr (b) After 6.5 yr (c) After 12.8 yr 1 112. e - 3, , 2 f 113. {x x Ú 1} or [1, q ) 4 x + 5 114. (f ∘ g)(x) = ; {x x ≠ 3, x ≠ 11} 115. One-to-one - x + 11
y 5
f(x) 2x 4
AN41
5 x y 4
(c) 5x x 6 2 6or 1 - q , 22
(b) 2
4.7 Assess Your Understanding (page 378) 3. principal 4. I; Prt; simple interest
5. 4
6. effective rate of interest 7. $108.29 9. $609.50 11. $697.09 13. $1246.08 15. $88.72 1 17. $860.72 19. $554.09 21. $59.71 23. 5.095% 25. 5.127% 27. 6 , compounded annually 29. 9% compounded monthly 31. 25.992% 4 33. 24.573% 35. (a) About 8.69 yr (b) About 8.66 yr 37. 6.823% 39. 5.09 yr; 5.07 yr 41. 15.27 yr or 15 yr, 3 mo 43. $104,335 45. $12,910.62 47. About $30.17 per share or $3017 49. Not quite. Jim will have $1057.60. The second bank gives a better deal, since Jim will have $1060.62 after 1 yr. 51. Will has $11,632.73; Henry has $10,947.89. 53. (a) $60,933 (b) $33,441 55. About $1020 billion; about $233 billion 57. $940.90 59. 2.53% 61. 34.31 yr 63. (a) $1364.62 (b) $1353.35 65. $4631.93 67. (a) 6.12 yr (b) 18.45 yr 69. (a) 2.47% (b) In 2023 71. 22.7 yr 76. R = 0; yes r nt 2x (c) mP = Pa1 + b 77. - 2, 5; f(x) = (x + 2)2(x - 5)(x2 + 1) 78. f -1(x) = n x - 1 r nt m = a1 + b n r nt r ln m = ln a1 + b = nt ln a1 + b n n ln m t = r n ln a1 + b n
79. {6}
4.8 Assess Your Understanding (page 388) 1. (a) 500 insects (b) 0.02 = 2, (c) About 611 insects (d) After about 23.5 days (e) After about 34.7 days 3. (a) - 0.0244 = - 2.44, (b) About 391.7 g (c) After about 9.1 yr (d) 28.4 yr 5. (a) N(t) = N0 e kt (b) 5832 (c) 3.9 days 7. (a) N(t) = N0 e kt (b) 25,198 9. 9.797 g 11. 9953 yr ago 13. (a) About 5:18 pm (b) About 14.3 min (c) The temperature of the pizza approaches 70°F. 15 18.63°C; 25.07°C 17. 1.7 ppm; 7.17 days or 172 h 19. 0.26 mol; 6.58 h or 395 min 21. 26.6 days 23. (a) 1000 g (b) 43.9% (c) 30 g (d) 616.6 g (e) After 9.85 h (f) About 7.9 h 25. (a) 9.23 * 10-3, or about 0 (b) 0.81, or about 1 (c) 5.01, or about 5 (d) 57.91°, 43.99°, 30.07°
27. (a) In 1984, 91.8% of households did not own a personal computer. (b) 100
29. (a)
120
6
0
100 0
0 0
40 0
100 0
(c) 70.6%
3 30. f(x) = - x + 7 2
31. Neither
32. 2 ln x +
1 ln y - ln z 2
(d) During 2011
(b) 0.78 or 78% (c) 50 people (d) As n increases, the probability decreases.
3 33. 2 2 5
4.9 Assess Your Understanding (page 395) 1. (a)
(d)
1
1
7 0
(b) y = 0.0903 11.33842 x (c) N1t2 = 0.0903e 0.2915t
3. (a)
1
1
7 0
(e) 0.69 (f) After about 7.26 hr
(d)
100
0
4 20
(b) y = 118.7226(0.7013)x (c) A(t) = 118.7226e -0.3548t
100
0
4 20
(e) 28.7% (f) k = - 0.3548 = - 35.48% is the exponential decay rate. It represents the rate at which the percentage of patients surviving advanced-stage breast cancer is decreasing.
AN42 5. (a)
ANSWERS Section 4.9 (c)
210
20 110
20 110
380
(b) y = 330.0549 - 34.5008 ln x
(c)
0
30
13. (a)
286.2055
(d)
110
(d)
225
90
20
(b) Quadratic with a 6 0 because of the “upside down U-shape” of the data (c) y = - 0.0311x2 + 3.4444x + 118.2493
110
90
3 22 2
(e) 201
1 (x + 3)(x + 1)2(x - 2) 3
15. f(x) =
0
30 0
17. 23
30
18.
y 10 −10
0
(b) Exponential (c) y = 115.5779(0.9012)x = 115.5779e -0.1040x
225
175
16.
0
120
(d) 762,176,844 (e) Approximately 315,203,288 (f) 2023
1 + 8.7428e - 0.0162x
175
(d) 286.2 thousand cell sites (e) 281.3 thousand cell sites
1 + 226.8644e -0.2873x
762,176,844.4
320,000,000
10 70,00,000
120
20
30 0
0
(b) y =
(b) y =
11. (a)
300
(c)
320,000,000
10 70,000,000
380
(d) 185 billion pounds (e) The prediction was under by 5 billion pounds.
9. (a) 300
0
7. (a)
210
10 x −10
(e) 5.1%
Review Exercises (page 401) 1. (a) - 4 (b) 1 (c) - 6 (d) - 6 2. (a) - 65 (b) 665 (c) 23 (d) 2 3. (a) 121 4. (a) 115 (b) 3 18 - 5 (c) 12 (d) - 38 5. 1f ∘ g2 1x2 = 1 - 3x, all real numbers; 1g ∘ f2 1x2 = 7 - 3x, all real numbers; 1f ∘ f2 1x2 = x, all real numbers; 1g ∘ g2 1x2 = 9x + 4, all real numbers
(b) 3 (c) 2 (d) 163
6. 1f ∘ g2 1x2 = 2x2 + 3x, 5x 0 x Ú 0 6 h 5x 0 x … - 3 6 1g ∘ f2 1x2 = x + 3 2x - 1, 5x 0 x Ú 1 6
1f ∘ f2 1x2 = 2 1x - 1 - 1, 5x 0 x Ú 2 6 1g ∘ g2 1x2 = x4 + 6x3 + 14x2 + 15x + 5, all real numbers 1 + x x - 1 , 5x x ≠ 0, x ≠ 1 6; 1g ∘ f2 1x2 = , 5x x ≠ - 1, x ≠ 1 6; 1f ∘ f2 1x2 = x, 5x x ≠ 1 6; 1g ∘ g2 1x2 = x, 5x x ≠ 0 6 7. 1f ∘ g2 1x2 = 1 - x x + 1 8. (a) one-to-one (b) 5 12, 32, 110, 52, 15, 22, 13, 72 6 9. The graph passes the y yx 5 horizontal line test. (0, 2) (2, 0)
(3, 3) 5 x
(3, 1)
10. f -1 1x2 =
2x + 3 5x - 2
f 1f -1 1x2 2 =
f
-1
1f 1x2 2 =
2a
11. f
2x + 3 b + 3 5x - 2
2x + 3 5a b - 2 5x - 2 2a
2x + 3 b + 3 5x - 2
2x + 3 5a b - 2 5x - 2
Domain of f = range of f
-1
Range of f = domain of f
-1
=
=
2(2x + 3) + 3(5x - 2) 5(2x + 3) - 2(5x - 2)
2(2x + 3) + 3(5x - 2) 5(2x + 3) - 2(5x - 2)
2 5 2 = all real numbers except 5
= all real numbers except
=
19x = x 19
=
19x = x 19
-1
1x2 =
x + 1 x
f 1f -1 1x2 2 =
1 x = = x x + 1 x + 1 - x - 1 x 1 + 1 x - 1 1 + x - 1 -1 f 1f 1x2 2 = = = x 1 1 x - 1
Domain of f = range of f
-1
= all real numbers except 1
Range of f = domain of f
-1
= all real numbers except 0
ANSWERS Review Exercises 12. f -1 1x2 = x2 + 2, x Ú 0 f 1f f
-1
-1
13. ƒ - 1 1x2 = 1x - 12 3
f 1f -1 1x2 2 = 1 1x - 12 3 2 1>3 + 1 = x f -1 1f 1x2 2 = 1x1>3 + 1 - 12 3 = x Domain of f = range of f -1 = 1 - q , q 2
1x2 2 = 2x + 2 - 2 = x = x, x Ú 0 2
1f 1x2 2 = 1 2x - 22 2 + 2 = x
Domain of f = range of f
-1
Range of f = domain of f
-1
= 3 2, q 2
Range of f = domain of f -1 = 1 - q , q 2 1 14. (a) 16 (b) 4 (c) (d) - 3 15. log 3 z = 5 64
= 3 0, q 2
18. 5x x 6 1 or x 7 2 6; 1 - q , 12 ∪ 12, q 2
AN43
19. 3
20. 22
21. 0.3
22. log 5 x + log 5 y - 2 log 5 z
1 1 24. ln 1x2 + 12 + ln y - ln x 25. 2 ln 12x + 32 - 2 ln 1x - 12 - 2 ln 1x - 22 2 2
26. log 2
x
16. 37 = z
(x + 1)
3
2 2 f; a , q b 3 3
23. 8 log 2 a + 2 log 2 b
3 2
2
17. e x ` x 7
27. ln
x8 13(x + 3) 2
4
28. ln c
16 2x2 + 1 2x 1x - 42
d
29. 2.124 30.
31. (a) Domain of f:1 - q , q 2 (b) y
3
(c) Range of f: 10, q 2 Horizontal asymptote: y = 0 (d) f -1 1x2 = 3 + log 2 x (e) Domain of f - 1: 10, q 2 Range of f -1:1 - q , q 2
9 1
8 (4, 2) (3, 1) 9 x
3
32. (a) Domain of f: 1 - q , q 2 y (b) (1, 4)
5 (0, 2)
y0
(d) f -1 1x2 = - log 3 1x - 12 (e) Domain of f - 1: 11, q 2 Range of f -1: 1 - q , q 2 y (f)
y1 5 x
33. (a) Domain of f: 1 - q , q 2 y (b)
x0
34. (a) Domain of ƒ: 1 - 3, q 2 (b) y 5
y0 5 x 1,
x
(2, 0) 5 x
3 e
0, x 3
(c) Range of f: 10, q 2 Horizontal asymptote: y = 0 x (d) f -1 1x2 = 2 + ln a b 3 (e) Domain of f - 1: 10, q 2 Range of f -1: 1 - q , q 2 y (f)
x1
(2, 4) (1, 3) 5 x
(2, 3)
(2, 0) 5 (4, 1)
y 5
5
5
(c) Range of f: 11, q 2 Horizontal asymptote: y = 1
(f)
1 ln 3 2
(c) Range of f: 1 - q , q 2 Vertical asymptote: x = - 3 (d) f -1 1x2 = e 2x - 3 (e) Domain of f - 1: 1 - q , q 2 Range of f - 1: 1 - 3, q 2 y (f) 5
5
1 ln 3, 0 2
(0, 2)
(3, 2)
5 x y 3
5 x 3 ,1 e x0
5 35. e - f 3
36. e
1 41. e , - 3 f 2 46. (a), (e)
- 1 - 23 - 1 + 23 , f ≈ 5 - 1.366, 0.366 6 2 2
42. 51 6
43. 55 6
44. 51 - ln 5 6 ≈ 5 - 0.609 6
y f 1 (x) 2 x1 2 14 yx 5 (3, 6) 0, f (x) log2 (x 2) 1 2 (6, 3) 10 x 5 ,0 2
1 37. e f 4
38. e
2 ln 3 f ≈ 54.301 6 ln 5 - ln 3
45. 5log 3 1 - 2 + 27 2 6 = c
(b) 3; (6, 3) (c) 10; (10, 4) 5 5 (d) e x ` x 7 f or a , q b 2 2 (e) f -1 1x2 = 2x - 1 + 2
39. 5 - 2, 6 6
ln 1 - 2 + 27 2 ln 3
40. 583 6
s ≈ 5 - 0.398 6
47. (a) 37.3 W (b) 6.9 dB 48. (a) 11.77 (b) 9.56 in. 49. (a) 9.85 yr (b) 4.27 yr 50. $20,398.87; 4.04%; 17.5 yr 51. $41,668.97 52. 24,765 yr ago 53. 55.22 min or 55 min, 13 sec 54. 7,412,094,029 55. $1.316 trillion; ≈ +216 billion
AN44
ANSWERS Review Exercises
56. (a) 0.3 (b) 0.8 1 (c)
57. (a)
(c)
165.1
58. (a)
165.1
10
0
0 0
0
16 153
25
40
16 10
153
0
(b) y = 165.73(0.9951)x
(d) In 2026 (c)
59. (a)
10
0
50
(d) Approximately 83 s
(b) y = 18.921 - 7.096 ln x
(c)
(e) 2.4 days; during the tenth hour of day 3 (f) 9.5 days
50
40
1
10
1
9 0
(d) ≈ - 3°F
(b) C =
9 0
46.93
(d) About 47 people; 50 people
1 + 21.273e -0.7306t
Chapter Test (page 404) 2x + 7 3 ; domain: e x ` x ≠ - f (b) 1g ∘ f2 1 - 22 = 5 (c) 1f ∘ g2 1 - 22 = - 3 2x + 3 2 2. (a) The function is not one-to-one. (b) The function is one-to-one. 2 + 5x 5 ; domain of f = b x 0 x ≠ r , range of f = 5 y y ≠ 0 6 ; domain of f - 1 = 3. f -1 1x2 = 3x 3 1. (a) f ∘ g =
4. The point 1 - 5, 32 must be on the graph of f -1. ln 21 ≈ 2.771 11. ln 133 ≈ 4.890 10. log 3 21 = ln 3 12. (a) Domain of f: 5x - q 6 x 6 q 6 or 1 - q , q 2 y (b) 16
(1, 1)
(0, 2) 2.5 y 2
15. 591 6
6. {4}
7. {625}
8. e 3 + 2 ≈ 22.086
(e) Domain of f - 1: {x 0 x 7 - 2} or 1 - 2, q 2 Range of f - 1: {y 0 - q 6 y 6 q } or 1 - q , q 2 y (f) 5
x
16. 5 - ln 2 6 ≈ 5 - 0.693 6
17. e
13. (a) Domain of f: 5x x 7 2 6 or (2, q ) (b) y 8 (3, 1)
(e) Domain of f - 1: {x 0 - q 6 x 6 q } or (- q, q) Range of f - 1: 5y y 7 2} or (2, q ) (f) y 8
x
x2
1 - 213 1 + 213 , f ≈ 5 - 1.303, 2.303 6 2 2
(0, 7) (1, 3) y2
(c) Range of f: 5y - q 6 y 6 q 6 or 1 - q , q 2; vertical asymptote: x = 2 (d) f - 1 1x2 = 51 - x + 2
x 2
5 r 3
9. log 20 ≈ 1.301
18 5 x
(1, 1)
≠ 0 6 ; range of f - 1 = b y 0 y ≠
(7, 0)
(2, 0)
(c) Range of f: 5y y 7 - 2 6 or 1 - 2, q 2 ; Horizontal asymptote: y = -2 (d) f - 1 1x2 = log 4 1x + 22 - 1 14. 51 6
5. {5}
5 x x
18. e
8
x
3 ln 7 f ≈ 5 - 6.172 6 1 - ln 7
19. 5 2 26 6 ≈ 54.899 6 20. 2 + 3 log 2 x - log 2 1x - 62 - log 2 1x + 32 21. About 250.39 days 22. (a) $1033.82 (b) $963.42 23. (a) About 83 dB (b) The pain threshold will be exceeded if 31,623 people shout at the same time.
(c) 11.9 yr
Cumulative Review (page 405) 1. (a) Yes; no (b) Polynomial; The graph is smooth and continuous. 2. (a) 10 (b) 2x2 + 3x + 1 1 23 3. a , b is on the graph. 4. { - 26} 2 2 5.
6. (a)
y 10 (8, 0) 10 x (0, 4)
y 10
(0, 3)
(1, 2) 10 x
7. f 1x2 = 2 1x - 42 2 - 8 = 2x2 - 16x + 24 8. Exponential; f 1x2 = 2 # 3x y 9. 10 (0, 1)
(b) 5x - q 6 x 6 q 6
(1, 2)
5 x
(c) 2x2 + 4xh + 2h2 - 3x - 3h + 1
10. (a) f 1g 1x2 2 =
4
1x - 32 2
domain: 5x x ≠ 3 6; 3
+ 2;
(b) f 1g 1x2 2 = log 2 x + 2; domain: 5x x 7 0 6; 2 + log 2 14
AN45
ANSWERS Section 5.2 12. (a),(c)
1 11. (a) Zeros: - 4, - , 2 4
60
2
(4, 0) (0, 8)
x2
(b) 5x x 7 - 1 6 or 1 - 1, q 2 16. (a) 20
5 x (3, 0) g1(x)
yx
Domain g = range g -1 = 1 - q , q 2 Range g = domain g -1 = 12, q 2 Vertical asymptote: x = 2 (b) g -1 1x2 = log 3 1x - 22
4 70
1 ,0 4
3 13. e - f 14. 52 6 2 15. (a) 5 - 1 6
y 5
g(x) y2
1 (b) x-intercepts: - 4, - , 2; y-intercept: - 8 4 (c) Local maximum value of 60.75 occurs at x = - 2.5. Local minimum value of - 25 occurs at x = 1. (d) 5, 3 y
(0, 3)
0
(c) 525 6
80 0
(2, 0) 5 x (1, 25)
(b) Logarithmic; y = 49.293 - 10.563 ln x (c) Highest value of r
CHAPTER 5 Trigonometric Functions 5.1 Assess Your Understanding (page 417) 3. standard position 4. central angle 11. y
13. 30
1 6. ru; r 2u 2
5. radian
15.
y
7. p
s u 8. ; t t
9. T
17.
y
10. F 19. y
y 3 4
x 135
450
x
– 6
x
21.
y 16 3
x
x
p p 3p p 4p 37. 39. 41. p 43. 45. 47. 60° 6 3 3 4 2 49. - 225° 51. 90° 53. 15° 55. - 90° 57. - 30° 59. 0.30 61. - 0.70 63. 2.18 65. 179.91° 67. 114.59° 69. 362.11° 71. 5 m 73. 6 ft p p 75. 0.6 radian 77. ≈ 1.047 in. 79. 25 m2 81. 2 13 ≈ 3.464 ft 83. 0.24 radian 85. ≈ 1.047 in .2 87. s = 2.094 ft; A = 2.094 ft 2 3 3 675p 1075 p ≈ 1060.29 ft 2 97. ≈ 1125.74 in2. 89. s = 14.661 yd; A = 87.965 yd2 91. 3p ≈ 9.42 in; 5p ≈ 15.71 in . 93. 2p ≈ 6.28 m2 95. 2 3 1 1 radian/s; v = cm/s 101. ≈23.2 mph 103. ≈120.6 km>h 105. ≈452.5 rpm 107. ≈359 mi 109. ≈898 mi/h 111. ≈2292 mi/h 99. v = 60 12 3 113. rpm 115. ≈2.86 mi/h 117. ≈31.47 rpm 119. 63 p ft 2 121. ≈1037 mi/h 123. Radius ≈ 3979 mi; circumference ≈ 25,000 mi 4 r1 v2 7 1 = . 134. x = 135. e - 3, f 136. y = - x + 3 - 4 125. v1 = r1v1, v2 = r2v2, and v1 = v2, so r1v1 = r2v2 and r2 v1 3 5 137. Horizontal asymptote: y = 3; Vertical asymptote: x = 7 23. 40.17°
25. 1.03°
27. 9.15°
29. 40°19′12″
31. 18°15′18″
33. 19°59′24″
35.
5.2 Assess Your Understanding (page 434) 7. cosine
8. (0, 1)
9. a
22 22 , b 2 2
1 23 10. a , b 2 2
y x 11. ; r r
12. F
13. sin t =
1 23 23 2 23 ; cos t = ; tan t = ; csc t = 2; sec t = ; cot t = 23 2 2 3 3
2 221 5 2 221 22 221 5 221 22 ; cos t = - ; tan t = ; csc t = ; sec t = - ; cot t = 17. sin t = ; cos t = ; 5 5 2 21 2 21 2 2 1 2 22 22 3 22 tan t = - 1; csc t = 22; sec t = - 22; cot t = - 1 19. sin t = - ; cos t = ; tan t = ; csc t = - 3; sec t = ; cot t = - 2 22 3 3 4 4 1 1 4 23 21. - 1 23. 0 25. - 1 27. 0 29. - 1 31. 1 22 + 1 2 33. 2 35. 37. 26 39. 4 41. 0 43. 2 22 + 45. 1 2 2 3 15. sin t =
47. sin
2p 23 2p 1 2p 2p 2 23 2p 2p 23 = ; cos = - ; tan = - 23; csc = ; sec = - 2; cot = 3 2 3 2 3 3 3 3 3 3
1 23 23 2 23 ; tan 210° = ; csc 210° = - 2; sec 210° = ; cot 210° = 23 49. sin 210° = - ; cos 210° = 2 2 3 3 51. sin
3p 22 3p 22 3p 3p 3p 3p = ; cos = ; tan = - 1; csc = 22; sec = - 22; cot = -1 4 2 4 2 4 4 4 4
53. sin
8p 23 8p 1 8p 8p 2 23 8p 8p 23 = ; cos = - ; tan = - 23; csc = ; sec = - 2; cot = 3 2 3 2 3 3 3 3 3 3
55. sin 405° = 57. sin a -
22 22 ; cos 405° = ; tan 405° = 1; csc 405° = 22; sec 405° = 22; cot 405° = 1 2 2
p p 23 p p p 2 23 p 1 23 b = - ; cos a - b = ; tan a - b = ; csc a - b = - 2; sec a - b = ; cot a - b = - 23 6 2 6 2 6 3 6 6 3 6
59. sin 1 - 135°2 = -
22 22 ; cos 1 - 135°2 = ; tan 1 - 135°2 = 1; csc 1 - 135°2 = - 22; sec 1 - 135°2 = - 22; cot 1 - 135°2 = 1 2 2
x
AN46 61. sin
ANSWERS Section 5.2
5p 5p 5p 5p 5p 5p = 1; cos = 0; tan is undefined; csc = 1; sec is undefined; cot = 0 2 2 2 2 2 2 23 14p 23 14p 1 14p 14p 2 23 14p 14p b = ; cos a b = - ; tan a b = 23; csc a b = ; sec a b = - 2; cot a b = 3 2 3 2 3 3 3 3 3 3 4 5 3 4 5 3 67. 1.07 69. 0.32 71. 3.73 73. 0.84 75. 0.02 77. sin u = ; cos u = - ; tan u = - ; csc u = ; sec u = - ; cot u = 5 5 3 4 3 4
63. sin a 65. 0.47
2 213 213 3 213 3 213 2 ; cos u = ; tan u = - ; csc u = ; sec u = ; cot u = 13 13 2 3 2 3
79. sin u = -
22 22 ; cos u = ; tan u = 1; csc u = - 22; sec u = - 22; cot u = 1 2 2 3 4 3 5 5 4 83. sin u = ; cos u = ; tan u = ; csc u = ; sec u = ; cot u = 85. 0 87. 0 89. - 0.1 5 5 4 3 4 3 81. sin u = -
23 22 22 p 22 113. 115. (a) ;a , b 2 4 2 4 2 11p 5p p 7p 13p 117. Answers may vary. One set of possible answers is ,, , , . 3 3 3 3 3 119. 105. -
103. 23
23 2
107.
1 + 23 2
91. 3
109. -
1 2
111.
93. 5
(b) a
95.
23 2
22 p , b 2 4
97.
1 2
99.
3 4
101.
23 2
p (c) a , - 2 b 4
U
0.5
0.4
0.2
0.1
0.01
0.001
0.0001
0.00001
sin U
0.4794
0.3894
0.1987
0.0998
0.0100
0.0010
0.0001
0.00001
sin U U
0.9589
0.9735
0.9933
0.9983
1.0000
1.0000
1.0000
1.0000
sin u approaches 1 as u approaches 0. u 121. R ≈ 310.56 ft; H ≈ 77.64 ft 123. R ≈ 19,541.95 m; H ≈ 2278.14 m 127. (a) 1.9 hr; 0.57 hr (b) 1.69 hr; 0.75 hr 135. (a) 16.56 ft
(b)
(c) 1.63 hr; 0.86 hr
(d) 1.67 hr; tan 90° is undefined
(c) 67.5°
20
45
90 0
2 2 143. e x x 7 - f or a - , q b 5 5
144. 4 - 3i , - 5
125. (a) 1.20 s (b) 1.12 s
(c) 1.20 s
129. 15.4 ft
131. 71 ft
133. y = -
4 23 7
137. (a) Values estimated to the nearest tenth: sin 1 ≈ 0.8; cos 1 ≈ 0.5; tan 1 ≈ 1.6; csc 1 ≈ 1.3; sec 1 ≈ 2.0; cot 1 ≈ 0.6; actual values to the nearest tenth: sin 1 ≈ 0.8; cos 1 ≈ 0.5; tan 1 ≈ 1.6; csc 1 ≈ 1.2; sec 1 ≈ 1.9; cot 1 ≈ 0.6 (b) Values estimated to the nearest tenth: sin 5.1 ≈ - 0.9; cos 5.1 ≈ 0.4; tan 5.1 ≈ - 2.3; csc 5.1 ≈ - 1.1; sec 5.1 ≈ 2.5; cot 5.1 ≈ - 0.4; actual values to the nearest tenth: sin 5.1 ≈ - 0.9; cos 5.1 ≈ 0.4; tan 5.1 ≈ - 2.4; csc 5.1 ≈ - 1.1; sec 5.1 ≈ 2.6; cot 5.1 ≈ - 0.4
145. R = 134
146. 81p ft 2
5.3 Assess Your Understanding (page 449) 5. 2p; p 23. 22
6. All real numbers except odd multiples of 25.
23 3
27. II
29. IV
31. IV
33. II
p 2
7. [ - 1, 1]
8. T
9. 1
10. F
11.
22 2
22 21. 0 2 1 37. tan u = 2; cot u = ; sec u = 25; 2
13. 1
15. 1
3 4 5 5 35. tan u = - ; cot u = - ; sec u = ; csc u = 4 3 4 3
17. 23
19.
25 23 2 23 22 3 22 39. tan u = ; cot u = 23; sec u = ; csc u = 2 41. tan u = ; cot u = - 2 22; sec u = ; csc u = - 3 2 3 3 4 4 5 12 13 13 5 3 3 5 5 4 43. cos u = - ; tan u = - ; csc u = ; sec u = - ; cot u = 45. sin u = - ; tan u = ; csc u = - ; sec u = - ; cot u = 13 5 12 5 12 5 4 3 4 3 csc u =
47. cos u = -
12 5 13 12 13 ; tan u = - ; csc u = ; sec u = - ; cot u = 13 12 5 12 5
51. cos u = -
25 2 25 3 3 25 25 ; tan u = ; csc u = ; sec u = ; cot u = 3 5 2 5 2
csc u = -
2 23 23 ; cot u = 3 3
csc u = 210; sec u = 81. 0
83. 1
85. - 1
103. All real numbers
87. 0
105. y Ú 1
22 2 22 3 22 ; tan u = - 2 22; csc u = ; sec u = - 3; cot u = 3 4 4
53. sin u = -
23 1 ; cos u = ; tan u = - 23; 2 2
3 4 5 5 4 55. sin u = - ; cos u = - ;csc u = - ; sec u = - ; cot u = 5 5 3 4 3
210 ; cot u = - 3 3 89. 0.9
49. sin u =
91. 9
59. -
93. 0
23 2
61. -
23 3
95. All real numbers
107. Odd; yes; origin
63. 2 65. - 1
67. - 1
97. Odd multiples of
109. Odd; yes; origin
69. p 2
57. sin u =
22 2
210 3 210 ; cos u = ; 10 10
71. 0 73. - 22
99. Odd multiples of
111. Even; yes; y-axis
113. (a) -
1 3
p 2
75.
2 23 3
77. 1 79. 1
101. - 1 … y … 1
(b) 1
AN47
ANSWERS Section 5.4
119. ≈ 15.81 min 121. Let a be a real number and P = 1x, y2 be the point on the unit circle that y 1 a = a. Then y = ax. But x2 + y2 = 1, so x2 + a2x2 = 1. So x = { and y = { ; corresponds to t. Consider the equation tan t = 2 x 21 + a2 21 + a that is, for any real number a, there is a point P = 1x, y2 on the unit circle for which tan t = a. In other words, the range of the tangent function 115. (a) - 2
117. (a) - 4
(b) 6
(b) - 12
is the set of all real numbers. 123. Suppose that there is a number p, 0 6 p 6 2p, for which sin 1u + p2 = sin u for all u. If u = 0, then p p p 3p p sin 10 + p2 = sin p = sin 0 = 0, so p = p. If u = , then sin a + p b = sin a b . But p = p. Thus, sin a b = - 1 = sin a b = 1. This is 2 2 2 2 2 1 ; since cos u has period 2p, so impossible. Therefore, the smallest positive number p for which sin 1u + p2 = sin u for all u is 2p. 125. sec u = cos u does sec u. 127. If P = 1a, b2 is the point on the unit circle corresponding to u, then Q = 1 - a, - b2 is the point on the unit circle corresponding -b b to u + p. Thus, tan 1u + p2 = = = tan u. Suppose that there exists a number p, 0 6 p 6 p, for which tan 1u + p2 = tan u for all u. Then, if -a a u = 0, then tan p = tan 0 = 0. But this means that p is a multiple of p. Since no multiple of p exists in the interval 10, p2, this is a contradiction. Therefore, the period of f 1u2 = tan u is p. 1 1 1 1 a 1 1 129. Let P = 1a, b2 be the point on the unit circle corresponding to u. Then csc u = = ; sec u = = ; and cot u = = = . b sin u a cos u b b/a tan u 2 2 2 2 2 2 2 2 2 2 2 2 2 2 131. 1sin u cos f2 + 1sin u sin f2 + cos u = sin u cos f + sin u sin f + cos u = sin u 1cos f + sin f2 + cos u = sin u + cos u = 1 137. All real numbers
138.
139. E ln 6 + 4 F 140. 9
y
6
(3, 5)
(2, 3)
(4, 3)
-6
6x -6
xⴝ3
5.4 Assess Your Understanding (page 463) p 2
p p p 6. T 7. F 8. T 9. (a) 0 (b) 6 x 6 (c) 1 (d) 0, p, 2p 3 2 2 3p p p 3p 5p p 7p 11p , ; f 1x2 = - 1 for x = - , (f) ,- , , (g) 5xƒ x = kp, k an integer 6 (e) f 1x2 = 1 for x = 2 2 2 2 6 6 6 6 3. 1;
4. 3; p
5. 3;
11. Amplitude = 2; Period = 2p 19. Amplitude = 35.
5 ; Period = 3 3
y (22P, 4) 5 (0, 4)
(2P, 4)
13. Amplitude = 4; Period = p 21. F 37.
23. A
25. H
P 2 ,4 2
y (0, 0) 5
3P ,0 2 25P 2 (2P, 24)
5P x 2 (P, 24)
y (23P, 2) 2.5 (P, 2)
22P
P ,0 2
3 , 10 2
45.
10
(3, 4)
2p, 2
0, 2
53. (3, 4)
P 2 ,5 2
P 2 ,8 4 y
(0, 4)
10 x 3 , 2 2
Domain: 1 - q , q 2 Range: 3 - 2, 10 4
P 2 2
P , 21 4
3P 2 ,5 2
x
(0, 5)
(P, 0) 22P
(0, 3) y 6
(0, 0)
Domain: 1 - q , q 2 Range: 3 - 1, 1 4 49.
P ,5 2
55.
3 5 , 4 3
y (2, 2)
(P, 3)
3 (0, 2) (2, 2)
2
(2P, 3) 3P ,1 2P x 2
3 , 3 2
2
P
Domain: 1 - q , q 2 Range: 3 2, 8 4
x
(1, 8)
9
57. 9 5 , 4 3
2.5
Domain: 1 - q , q 2 5 5 Range: c - , d 3 3
3 , 3 2
Domain: 1 - q , q 2 Range: 3 - 8, 2 4
(3, 0) 5 x
3 5 , 4 3
x
1 , 3 2
y (0, 0) 2.5 5
9 5 , 4 3
P ,0 2
P , 21 4
P , 21 4
Domain: 1 - q , q 2 Range: 3 1, 5 4
3P ,8 4
x
2P
24
(P, 5) P ,2 4 22
P 2 ,1 4
3P ,0 8
x
3P ,1 4
y 1.25
22.5
8
2P
P 2
P 2 ,1 2 22P
P ,5 2
(2P, 5) 3P 2 ,2 4
1 4p ; Period = 2 3
41.
Domain: 1 - q , q 2 Range: 3 - 1, 1 4
5P ,0 4
1 2
P ,0 8 y (0, 1) P ,1 2.5 2
47.
2P
1 2
3P ,0 8
P 2 ,0 8
3P ,0 4
(6, 4) 10 9 , 2 10 2
2
Domain: 1 - q , q 2 1 1 Range: c - , d 2 2
9 , 10 2
2 P 2 ,1 2
P , 24 2
y P 1 , 2.5 2 2 P ,0 4 22P
(3P, 22) (0, 0)
y
39.
Domain: 1 - q , q 2 Range: 3 - 4, 4 4
(2P, 0) (4P, 0)
17. Amplitude =
33. B
2P x
3P 2 , 24 2
Domain: 1 - q , q 2 Range: 3 - 2, 2 4
51.
3P ,4 2
31. A
25
5P x (2P, 22)
29. J
(2P, 0)
Domain: 1 - q , q 2 Range: 3 - 4, 4 4
43.
27. C
15. Amplitude = 6; Period = 2
3 ,0 2
2, (4, 2) 1 6, 2
y 2.5
1 2 (4, 2) 6, 1 2 10 x
(8, 1) (8, 1) 2.5 (0, 1)
Domain: 1 - q , q 2 Range: 3 - 1, 2 4
AN48
ANSWERS Section 5.4
59. y = {3 sin 12x2 71. y = - cos a
4p xb + 1 3
p 73. y = 3 sin a x b 2
81. 1f ∘ g2 1x2 = sin 14x2 y 1.25
P 2 ,0 2
p 63. y = 5 cos a x b 4
61. y = {3 sin 1px2
y 2.25
x
1g ∘ f2 1x2 = 4 sin x
y 1.25
3 sin 12px2 4
67. y =
3 69. y = - sin a x b 2
12 p 1 s 30 Amplitude = 220 amp
y , 1 (, 0) 1.25 2 (2, 1) 2
(0, 0)
(2P, 22)
5 , 兹2 4 2
87. Period =
I 220
x
3 , 0 2 1 15
220 (P, 1)
P, 21 2
x
220
t
(c) I 1t2 = 22 sin 1120pt2 (d) Amplitude = 22 amp 1 s Period = 60
89. (a) Amplitude = 220 V; 1 period = s 60 (b) , (e)
x
3P , 24 2
25
79.
1g ∘ f2 1x2 = cos 1 - 2x2
P, 4 2
2P
85.
(P, 2)
P 22P
2 p
77.
2P x (0, 22)
y 5
75. y = - 4 cos 13x2
83. 1f ∘ g2 1x2 = - 2 cos x
P, 0 2
2P
1 65. y = - 3 cos a x b 2
V
22 22
I
t
1 30
220
[V0 sin 12pft2]2
V 20
R
0
=
R
sin 2 12pft2
V 20
1 and is of the form y = A cos 1vt2 + B, 2f 2 2 2 2 V0 V0 V0 V0 V 20 1 2p then A = and B = . Since = , then v = 4pf . Therefore, P1t2 = cos 14pft2 + = [1 - cos 14pft2]. 2R 2R 2f v 2R 2R 2R 2p p 2p ; emotional potential: v = ; intellectual potential: v = 93. (a) Physical potential: v = 23 14 33 y (b) 100 (c) No (d) Physical potential peaks at 15 days after 20th birthday. 95. 1.25 Emotional potential is 50% at 17 days, with a maximum at 10 days and a minimum at 24 days. Intellectual potential 2P x starts fairly high, drops to a minimum at 13 days, and rises to a maximum at 29 days. 91. (a) P1t2 =
(b) Since the graph of P has amplitude
2R
and period
36 0
5p 1 p 1 p 1 5p 1 3p p 5p 9p , b , a - , b , a , b , a , b 99. Answers may vary. a ,1b, a ,1b, a ,1b, a ,1b 3 2 3 2 3 2 3 2 4 4 4 4 5 7 106. (2 , 5) 107. 10 , 52, a - , 0 b , a - , 0 b 108. E F or ∅ 3 3
97. Answers may vary. a 105. 2x + h - 5
5.5 Assess Your Understanding (page 473) 3. origin; odd multiples of
p 2
4. y-axis; odd multiples of
p 2
5. y = cos x
11. sec x = 1 for x = - 2p, 0, 2p; sec x = - 1 for x = - p, p 17.
y P ,3 4
6
13. -
19.
9. 1 15. -
3p p p 3p ,- , , 2 2 2 2
y P ,4 4
21.
4
x
1 ,1 2 5 x
P 2 , 24 4
25.
2 (0, 2)
x (2P, 21)
Domain: 5x|x ≠ 4kp, k is an integer 6 Range: 1 - q , q 2
Domain: 5x|x does not equal an odd integer 6 Range: 1 - q , q 2 27.
y
(P, 1) 4P
1 , 1 2
Domain: 5x x ≠ kp, k is an integer 6 Range: 1 - q , q 2
kp , k is an odd integer f 2 Range: 1 - q , q 2 y 5
y
2P x
Domain: e x 2 x ≠
23.
7. 0
3p p p 3p ,- , , 2 2 2 2
16
2P P 2 , 23 4
6. T
(2P, 22)
1
(2P, 2) 2P x (P, 22)
Domain: e x 2 x ≠
kp , k is an odd integer f 2 Range: 5y|y … - 2 or y Ú 2 6
P ,3 2
y 15
3P ,3 2 2P
2
3P , 23 2
x P , 23 2
Domain: 5x 0 x ≠ kp, k is an integer 6 Range: 5y 0 y … - 3 or y Ú 3 6
AN49
ANSWERS Section 5.6 29.
31.
y 16
(0, 4)
(22P, 24)
1 ,2 2
(2P, 24)
2 13 p
43.
6 13 p
y
2
y 2.5
P , 21 16
Domain: 5x 0 x ≠ 2pk, k is an odd integer 6 Range: 1 - q , q 2 1g ∘ f2 1x2 = 4 tan x
P ,1 16 P 4
P 2
, 0 2
(0, 0)
(c) ≈0.83
x (, 1)
y 5
2.5
2P
(b)
P
x
53.
25
y 5
3P 2 P 2
0 0
54. 15p - 2q2 2 125p2 + 10pq2 + 4q4 2 1 55. 2log 3(x + 3) + log 3(x - 2) - log 3(x - 1) 2 56. E - 1 , 3 F
57.
x
y 5
(d) ≈9.86 ft
3 , 兹2 4
x 2
1g ∘ f2 1x2 = cot ( - 2x)
y
x
51. (a) L1u2 =
9P , 23 2
3P , 23 2
47. 1f ∘ g2 1x2 = - 2 cot x
P, 4 4
P 2 2 P , 24 2 4
x
3P ,1 2 5P x
Domain: 5x 0 x ≠ 3pk, k is an integer 6 Range: 5y|y … - 3 or y Ú 1 6
(0, 22)
y 5
y 10
9P 2 ,1 2 2
4 3 + cos u sin u = 3sec u + 4 cscu
, 1 4
39.
3 2
x
1
49.
P, 2
5 2P, 2 2
45. 1f ∘ g2 1x2 = tan 14x2
5P x
Domain: 5x 0 x ≠ 2pk, k is an odd integer 6 Range: 1 - q , q 2
5P
3 Domain: e x 2 x ≠ k, k is an odd integer f 4 Range: 5y|y … 1 or y Ú 3 6 41.
y 10
3 ,1 2 x
(P, 2)
(0, 1)
(3P, 0)
37.
2.5
(23P, 2) 1 , 2 2
Domain: 5x 0 x does not equal an integer 6 Range: 5y 0 y … - 2 or y Ú 2 6
(0, 3)
3 ,1 2
2.5 x
y 10
3 , 2 2
Domain:5x 0 x ≠ kp, k is an odd integer 6 Range: 5y 0 y … - 4 or y Ú 4 6 y 8
33.
3 ,2 2
6
5P x
35.
y
y 5 tan x
x
3P 2 P y 5 2cot x 1 2
y 6
26 (0, 23) 26
18 x (4, 22)
5.6 Assess Your Understanding (page 483)
P ,0 2
Phase shift = -
p Phase shift = 6
3P ,4 4
y 5
7. Amplitude = 3 Period = p
5. Amplitude = 2 2p Period = 3
1. phase shift 2. F 3. Amplitude = 4 Period = p p Phase shift = 2
P 2 ,2 6
7P ,4 4
P ,2 2
y 2.5
P 2 ,3 2
2P ,0 3
(0, 0) P , 24 4
3P ,0 2
(P, 0) 5P , 24 4
9. Amplitude = 4 Period = 2 Phase shift = 1 2 2 , 21 2 P
12 2
2
2 , 25 P
y 9
x
2
2 , 25 P
3 2 2 , 29 2 P
2
2 ,8 P
21
2 ,8 P
11
21 1
2 ,2 P
(0, 23)
5P , 22 6
13. Amplitude = 3 Period = p p Phase shift = 4
2 2 2 , 25 P
1 2 2 , 29 2 P
P ,0 3
P , 22 6
11. Amplitude = 3 Period = 2 2 Phase shift = p
2 p
y 1
P 2 ,0 4
P x
x
2P
3
2 ,2 P
y 5
2
P ,3 2
3P ,0 4
y 5
2
P ,0 4
3P ,0 4 x
(P, 23) (0, 23)
x (P, 23)
P ,0 4
1 b d or 2 y = 2 sin 12x - 12
P
x
P
3P ,0 4
15. y = 2 sin c 2 ax -
P ,3 2
(2P, 23)
P ,3 2
p 4
P ,0 4
2 1 17. y = 3 sin c ax + b d or 3 3 2 2 y = 3 sin a x + b 3 9
x
AN50 19.
ANSWERS Section 5.6 3P 2 ,2 16
21.
5P ,2 16
y 5
P 2 ,0 4
P 2
5P 2 , 22 16
P ,0 4 x
7P P x5 x52 8 y 8 5 5P 2 ,3 8
1 s 15
3P x52 8
29. (a)
P (0, 0) x52 4
P 2 , 21 8
x
10 5
10
x
I 120
(c)
(1, 1) 1
x 1 , 1 2
x
1 4
x
3 4
(d) y = 9.46 sin (1.247x + 2.096) + 24.088 (e)
40
H 0
30 2 t 15
3 1 x 4 y 4
1 , 1 2
3P , 21 8
2p 11 y = 8.5 sin c ax b d + 24.5 5 4 or 2p 11p y = 8.5 sin a x b + 24.5 5 10
20
x
3
(b)
H
25.
P ,1 8
(1, 1) P 2
5P x5 8
30
Amplitude = 120 amp 1 Phase shift = s 90
3P 2 ,1 8
7P , 23 8
(0, 0)
27. Period =
3P ,3 8
P y x5 4 2.5
P x
P 2 , 23 8
3P , 22 16
23.
10 10
20 10 5
31. (a)
(b) y = 23.65 sin c
y 80
(c) 50 20
10
x
p p 2p 1x - 42 d + 51.75 or y = 23.65 sin a x b + 51.75 6 6 3
(d) y = 24.25 sin 10.493x - 1.9272 + 51.61 86 (e)
y 80 50
5
10
x 20 5
10
x 0
13 14
33. (a) 11:55 pm (b) y = 3.105 sin c
24p 24p 1x - 8.39582 d + 2.735 or y = 3.105 sin c x - 4.2485 d + 2.735 149 149
2p x - 1.39 b + 12.135 35. (a) y = 1.615 sin a 365 (b) 12.42 h (c) y
2p 37. (a) y = 6.96 sin a x - 1.39 b + 12.41 365 (b) 13.63 h (c) y
20
20
10
10 x 150 280 420
x 150 280 420
(d) The actual hours of sunlight on April 1, 2013, were 12.45 hours. This is close to the predicted amount of 12.42 hours. 41. f -1 1x2 =
2x - 9 4
7 42. e f 3
(c) 2.12 ft
43. 64x2 + 240xy + 225y2
(d) The actual hours of sunlight on April 1, 2013, were 13.47 hours. This is close to the predicted amount of 13.63 hours.
44. 2 213
Review Exercises (page 490) p 1 5 3. 315° 4. - 450° 5. 6. + 2 13 7. - 3 22 - 2 23 8. 3 9. 0 10. 0 11. 1 12. 1 13. 1 10 2 2 12 5 13 3 4 5 5 3 16. cos u = ; tan u = ; csc u = ; sec u = ; cot u = 17. sin u = - ; cos u = - ; csc u = - ; sec u = 5 3 4 3 4 13 13 12 3 4 3 5 4 12 12 13 18. sin u = ; cos u = - ; tan u = - ; csc u = ; cot u = 19. sin u = ; tan u = - ; csc u = ; sec u = 5 5 4 3 3 13 5 12 1.
5p 6
2.
20. cos u =
12 5 13 13 12 ; tan u = - ; csc u = - ; sec u = ; cot u = 13 12 5 12 5
22. sin u = 23. sin u =
21. sin u = -
2 22 1 3 22 22 ; cos u = ; tan u = - 2 22; csc u = ; cot u = 3 3 4 4
25 2 25 1 25 ; cos u = ; tan u = - ; csc u = 25; sec u = 5 5 2 2
14. - 1
15. 1
13 5 ; cot u = 5 12 13 5 - ; cot u = 5 12
210 3 210 210 ; cos u = ; csc u = - 210; sec u = ; cot u = 3 10 10 3
ANSWERS Review Exercises 24.
25.
P, 2 8
y 2.5
P
y 5
26.
P, 3 2 P
x
y 5 P x 2
x
(P, 23) 3P , 22 8
Domain: ( - q , q ) Range: 3 - 3, 3 4
Domain: ( - q , q ) Range: 3 - 2, 2 4 27.
x5
y
P 2
28.
P 2 6
Range: 19 q , q 2 30.
kp , k is an odd integer f 2 Range: 1 - q , q 2
29.
p p + k # , k is an integer f 6 3
Domain: e x 2 x ≠ Range: 1 - q , q 2 31.
y
y 5P 4
P x
x
Domain: e x 2 x ≠
Domain: e x 2 x ≠
3P y x5 4 5
8
AN51
23
p + kp, k is an integer f 4
Domain: e x 2 x ≠
kp , k is an odd integer f 4 Range: {y y … - 4 or y Ú 4} 32.
y
5
x P , 24 2
y 25
5 x
4P x
5P x 7
p Domain: e x 2 x ≠ + kp, k is an integer f 4 Range: {y y … - 1 or y Ú 1} 33. Amplitude = 3; Period = 1
P, 4 6
36. Amplitude = 1 Period = 4p Phase shift = - p y 2.5
P , 24 2
39. y = 5 cos
x 4
p 40. y = - 7sin a x b 4
2 3 37. Amplitude = Period =
1 2
38. Amplitude = Period = 2
4p 3
Phase shift =
2p Phase shift = 3 5P x
P x
3p + k # 3p, k is an integerf 4 Range: 1 - q , q 2
Domain: ex 2 x ≠
Range: [ - 6, 2]
34. Amplitude = 2; Period =
35. Amplitude = 4 2p Period = 3 Phase shift = 0 y 5
Domain: ( - q , q )
y 2 3
y 1.25
12 P
2 3 6 p
x
2P x
41. 0.38
42. 1.02
43. Sine, tangent, cosecant, and cotangent: negative; cosine and secant: positive
23 1 2 1 , cos t = , tan t = - 13, csc t = , sec t = 2; cot t = 2 2 13 13 4 3 4 46. sin t = - , cos t = , tan t = 47. Domain: 5x 0 x ≠ np, n is an integer 6; range: 1 - q , q 2 ; period = p 5 5 3 48. (a) 32.34° (b) 63°10′48″ 3p 49. p ≈ 3.14 ft; ≈ 4.71 ft 2 50. 9p ≈ 28.26 in.; 2p ≈ 6.28 in. 51. Approximately 114.59 revolutions/hr 2 p 52. 0.1 revolution/sec = radian/sec 5 54. (a) y (c) y 1 53. (a) 90 90 15 (b) 220 70 70 1 (c) 50 50 180 x x 5 10 5 10 I (d) p 220 (d) y = 19.81 sin(0.543x - 2.296) + 75.66 (b) y = 20 sin c 1x - 42 d + 75 or 6 100 (e) 2p p 2 t b + 75 y = 20 sin a x 15 6 3 44. III
45. sin t = -
220
0
13 40
AN52 55.
ANSWERS Review Exercises
p 5 2 1 , Ï3 2 2 p 5 (0, 1) 2 2 p 5 1 , Ï3 p 5 2Ï , Ï y P P 2 2 2P 2 2 2 3 P 3 2 2 3P p 5 Ï ,Ï 4 2 2 4 P 5P 6 6 3 1 Ï , 3 1 p5 p 5 2Ï , 2 2 2 2 p 5 (1, 0) 0 P p 5 (21, 0) 1 x Ï3 3 1 p 5 2Ï , 2 , 21 p5 2 2 2 2 11P 7P 6 6 2 2 5P 7P p 5 2Ï , 2Ï 2 2 4 4P 5P 4 3P 2 2 3 2 3 p 5 Ï , 2Ï p 5 2 1 , 2Ï3 2 2 p 5 (0, 21) 2 2 p 5 1 , 2Ï3 2 2
Chapter Test (page 492) 1.
13p 9
2. -
15. - 1.524
20p 9
3.
13p 180
16. 2.747
4. - 22.5°
17.
1 2
9. -
10. -
13 3
6. 135°
7.
sin U
cos U
tan U
sec U
csc U
cot U
+
+
+
+
+
+
U in QI
8. 0
1 2
5. 810°
11. 2
U in QII
+
-
-
-
+
-
U in QIII
-
-
+
-
-
+
U in QIV
-
+
-
+
-
-
12.
3 11 - 122
13. 0.292
2
18. -
14. 0.309
3 5
2 16 5 16 7 7 16 2 16 15 15 3 15 ; tan u = ; csc u = ; sec u = ; cot u = 20. sin u = ; tan u = ; csc u = ; 7 12 5 12 5 3 2 5 3 2 15 5 13 5 5 1146 1 12 13 7 153 sec u = ; cot u = 21. sin u = ; cos u = - ; csc u = ; sec u = - ; cot u = 22. 23. 24. 2 5 13 13 12 5 12 53 146 2
19. cos u = -
25. 11P , 0 2 2
26.
y 5 (24P , 2) (2P, 2)
7P , 0 2
27. y = - 3 sin a3x +
(2P, 3) y (0, 3) 4
28. 78.93 ft 2 29. 143.5 rpm
5P x 2
5P , 0 2
(5P, 22)
P
3p b 4
x
P (2P , 22) 2 , 0
Cumulative Review (page 492) 1 1. e - 1, f 2
2. y - 5 = - 3 1x + 22 or y = - 3x - 1
3. x2 + 1y + 22 2 = 16
2 4. A line; slope ; intercepts (6, 0) and 10, - 42 3
5. A circle; center 11, - 22; radius 3
y 5
6.
y 2
y 8 (2, 3)
(6, 0)
(4, 3) (3, 2) 6 x
5 x 10 x (1, 2)
(0, 4)
7. (a)
(b)
y 4.5
(c)
y 2.5
(1, 1) 2.5 x
2.5 x
(1, 1)
(e)
y 3 (e, 1)
(1, 1)
(0, 0) (1, 1) (0, 0)
(d)
y 4.5 1 1, e
(1, e)
(1, 0) 3
(0, 1) 2.5 x
1 e , 1
8. f -1 1x2 =
1 1x + 22 3
9. - 2
10.
3p 2 ,3 4
p 2 ,0 2 p 2 , 23 4
y p, 3 p ,0 4 5 2 P P x 2 3p, 23 4 (0, 0)
11. 3 -
3 13 2
12. y = 2 13x 2
x
y 3 ,1 2 (0, 0) , 1 2
p 13. y = 3 cos a x b 6
(f)
x
y 2.5
p 2 , 21 4
p, 1 4 (0, 0) P x
ANSWERS Section 6.3 14. (a) f 1x2 = - 3x - 3;
(b) f 1x2 = 1x - 12 2 - 6; 10, - 52 ,
m = - 3; 1 - 1, 02, 10, - 32
(2, 3)
(1, 0)
y 4
(2, 3)
(3.45, 0) 5 x
(1.45, 0)
5 x (0, 3) (1, 6)
15. (a) f 1x2 = y 10
(1, 6)
1 1x + 22 1x - 32 1x - 52 6
(b) R1x2 = y 10 (0, 5) (2, 0)
(0, 5)
(2, 0) (3, 0)
(c) We have that y = 3 when x = - 2 and y = - 6 when x = 1. Both points satisfy y = ae x. Therefore, for 1 - 2, 32 we have 3 = ae -2, which implies that a = 3e 2. But for 11, - 62 we have - 6 = ae 1, which implies that a = - 6e -1. Therefore, there is no exponential function y = ae x that contains 1 - 2, 32 and 11, - 62 .
1 - 16 + 1, 02, 1 16 + 1, 02
y 4
AN53
(5, 0) 6 x
1x + 22 1x - 32 1x - 52 3 1x - 22
(5, 0) x 4 (3, 0)
CHAPTER 6 Analytic Trigonometry 6.1 Assess Your Understanding (page 505) 9. - q 6 x 6 q
7. x = sin y 8. 0 … x … p 29. 0.51
31. - 0.38
33. - 0.12
35. 1.08
4p 37. 5
x- 2 5 Range of f = Domain of f - 1 = 3 - 3, 7 4 p p Range of f -1 = c - , d 2 2
53. f -1 1x2 = sin-1
59. f -1 1x2 =
1 x c sin - 1 a b - 1 d 2 3 Range of f = Domain of f - 1 = 3 - 3, 3 4 1 p 1 p Range of f -1 = c - - , - + d 2 4 2 4
77. (a)
p square units 3
(b)
5p square units 12
10. False
11. True
12. True 13. 0
15. -
p 2
17. 0
19.
p 4
21.
p 3
23.
5p 6
25. 0.10
27. 1.37
p p 1 41. 43. 45. 47. 4 49. Not defined 51. p 8 5 4 1 x 55. f -1 1x2 = cos - 1 a - b 3 2 57. f -1 1x2 = - tan - 1 1x + 32 - 1 Range of f = Domain of f - 1 = 3 - 2, 2 4 Range of f = Domain of f - 1 = 1 - q , q 2 p p p Range of f -1 = c 0, d - 1b Range of f -1 = a - 1 - , 3 2 2 22 1 61. e f 63. e - f 65. 5 23 6 67. 5 - 1 6 2 4 69. (a) 13.92 h, or 13 h, 55 min (b) 12 h (c) 13.85 h, or 13 h, 51 min 71. (a) 13.3 h, or 13 h, 18 min (b) 12 h (c) 13.26 h, or 13 h, 15 min 73. (a) 12 h (b) 12 h (c) 12 h (d) It is 12 h. 75. 3.35 min 3p 39. 8
2 81. c - , 2 d 3
79. 4250 mi
82. The graph passes the horizontal-line test. 83. f -1 1x2 = log 2 1x - 12
1,
1 2
y 5 (0, 1)
1
y
x
3
84. (2x + 1)- 2 (x2 + 3)- 2 (3x2 + 2)
(1, 2) 5 x
6.2 Assess Your Understanding (page 512) 23 22 2 23 3p p 22 13. 2 15. 22 17. 19. 21. 23. 25. 3 2 3 4 3 4 25 214 3 210 p p p p 2p 27. 29. 31. 33. 25 35. 37. 39. 41. 43. 45. 1.32 47. 0.46 49. - 0.34 51. 2.72 53. - 0.73 2 2 10 4 6 2 6 3 5 3p 5 p 1 u 2u2 - 1 2u2 - 1 1 3 67. 69. 71. 73. 75. 77. - 215 55. 2.55 57. 59. 61. 63. 65. u 13 4 4 13 6 0u0 0u0 21 + u2 21 - u2 79. (a) u = 31.89° (b) 54.64 ft in diameter (c) 37.96 ft high 81. (a) u = 22.3° (b) y0 = 2940.23 ft/s 4. x = sec y; Ú 1; 0; p
5. cosine
P
83.
25
6. False
7. True
87. - 5i, 5i, - 2, 2
8. True
9.
22 2
88. Neither
11. -
89.
7p 4
90.
5p in. ≈7.85 in. 2
5 0
6.3 Assess Your Understanding (page 519) p 5p p 5p 7p 11p p 2p 4p 5p p 3p 5p 7p ; 8. e u ` u = + 2pk, u = + 2pk, k is any integer f 9. False 10. False 11. e , f 13. e , , , f 15. e , , , f 6 6 6 6 6 6 3 3 3 3 4 4 4 4 p 7p 11p p 2p 4p 5p 4p 8p 16p 7p 11p 3p 7p 2p 4p 3p 5p 17. e , , f 19. e , , , f 21. e , , f 23. e , f 25. e , f 27. e , f 29. e , f 2 6 6 3 3 3 3 9 9 9 6 6 4 4 3 3 4 4 7.
AN54
ANSWERS Section 6.3
3p 7p 11p p 5p p 5p 13p 17p 25p 29p , f 33. e f 35. e u ` u = + 2kp, u = + 2kp f ; , , , , , 4 4 6 6 6 6 6 6 6 6 6 5p 11p 17p 23p 29p 35p p 3p p 3p 5p 7p 9p 11p 5p + kp f ; , , , , , 39. e u ` u = + 2kp, u = + 2kp f ; , , , , , 37. e u ` u = 6 6 6 6 6 6 6 2 2 2 2 2 2 2 2 p 2p p 2p 4p 5p 7p 8p 8p 10p 8p 10p 20p 22p 32p 34p 41. e u ` u = + kp, u = + kp f ; , , , , , 43. e u ` u = + 4kp, u = + 4kp f ; , , , , , 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 p 2p 4p 3p 45. 50.41, 2.73 6 47. 51.37, 4.51 6 49. 52.69, 3.59 6 51. 51.82, 4.46 6 53. 52.08, 5.22 6 55. 50.73, 2.41 6 57. e , , , f 2 3 3 2 p 7p 11p p 5p p 2p 4p 3p p 5p p 5p p 5p 3p p 59. e , , f 61. e 0, , f 63. e , , , f 65. 5p 6 67. e , f 69. e 0, , p, f 71. e , , f 73. e f 75. 50 6 2 6 6 4 4 2 3 3 2 4 4 3 3 6 6 2 2 p 5p p 2p 4p 5p 77. e , f 79. No real solution 81. { - 1.31, 1.98, 3.84} 83. {0.52} 85. {1.26} 87. { - 1.02, 1.02} 89. {0, 2.15} 91. {0.76, 1.35} 93. , , , 3 3 3 3 3 3 31. e
95. (a) - 2p, - p, 0, p, 2p, 3p, 4p
(b)
5P 3 , 6 2 2
11P 3 , 6 2
11p 7p p 5p 13p 17p ,, , , , f 6 6 6 6 6 6 11p 7p p 5p 13p 17p (d) e x ` 6 x 6 or 6 x 6 or 6 x 6 f 6 6 6 6 6 6 (c) e -
13P 3 , 6 2 17P 3 , 6 2
y 3.75
3P x 7P 3 2 , 6 2 P, 3 6 2
97. (a) e x ` x = -
p + kp, k is any integer f 4
99. (a), (d) P 7 , 12 2
y 7
5P 7 , 12 2 g(x) 5
7 2
(b) -
p p p p 6 x 6 - or a - , - b 2 4 2 4
(b) e
p 5p , f 12 12
(c) e x `
(b) 123.6 m
111. 28.90° 113. Yes; it varies from 1.25 to 1.34 115. 1.47 117. If u is the original angle of incidence and f is the angle sin u = n2 . The angle of incidence of the of refraction, then sin f 1 emerging beam is also f, and the index of refraction is . n2 Thus, u is the angle of refraction of the emerging beam.
90
124. Amplitude: 2 Period: π p Phase shift: 2
9 - 217 9 + 217 , 4 4 1 210 123. tan u = - ; csc u = - 210; sec u = ; cot u = - 3 3 3
y 2.5 3P x
6.4 Assess Your Understanding (page 528) 3. identity; conditional 4. - 1 19. csc u # cos u =
5. 0
6. True 7. False
8. True
9.
1 cos u
11.
1 + sin u cos u
13.
1 sin u cos u
15. 2
17.
3 sin u + 1 sin u + 1
1 # cos u cos u = = cot u 21. 1 + tan2 1 - u2 = 1 + 1 - tan u2 2 = 1 + tan2 u = sec2 u sin u sin u sin u cos u sin2 u + cos2 u 1 1 23. cos u 1tan u + cot u2 = cos ua + b = cos u a b = cos u a b = = csc u cos u sin u cos u sin u cos u sin u sin u 1 25. tan u cot u - cos2 u = tan u # - cos2 u = 1 - cos2 u = sin2 u 27. 1sec u - 12 1sec u + 12 = sec2 u - 1 = tan2 u tan u 1 = 1 29. 1sec u + tan u2 1sec u - tan u2 = sec2 u - tan2 u = 1 31. cos2 u 11 + tan2 u2 = cos2 u sec2 u = cos2 u # cos2 u 2 2 2 2 2 2 33. 1sin u + cos u2 + 1sin u - cos u2 = sin u + 2 sin u cos u + cos u + sin u - 2 sin u cos u + cos u = sin2 u + cos2 u + sin2 u + cos2 u = 1 + 1 = 2 35. sec4 u - sec2 u = sec2 u 1sec2 u - 12 = 11 + tan2 u2 tan2 u = tan4 u + tan2 u 37. cos3x = cos x # cos2 x = cos x 11 - sin2 x2 = cos x - sin2 x cos x
2p 4p , f 3 3
2p 4p 2p 4p 6 x 6 f or a , b 3 3 3 3
0
122.
(b) e
f(x) 5 24 cos x
(c) e x `
0
4P ,2 3 2P x
P x f(x) 5 3 sin(2x) 1 2
109. (a) 30°, 60° 103. (a) 0 s, 0.43 s, 0.86 s (b) 0.21 s (c) (c) 30, 0.03 4 ∪ 30.39, 0.43 4 h 3 0.86, 0.89 4 105. (a) 150 mi (b) 6.06, 8.44, 15.72, 18.11 min 130 (c) Before 6.06 min, between 8.44 and 15.72 min, and after 18.11 min (d) No 107. 2.03, 4.91
g(x) 5 2 cos x 1 3
2P ,2 3
p 5p p 5p 6 x 6 f or a , b 12 12 12 12
121. x = log 6 y
y 5
101. (a), (d)
ANSWERS Section 6.4
39. sec u - tan u =
AN55
1 sin u 1 - sin u # 1 + sin u 1 - sin2 u cos2 u cos u = = = = cos u cos u cos u 1 + sin u cos u 11 + sin u2 cos u 11 + sin u2 1 + sin u
41. 3 sin2 u + 4 cos2 u = 3 sin2 u + 3 cos2 u + cos2 u = 3 1sin2 u + cos2 u2 + cos2 u = 3 + cos2 u 11 + sin u2 11 - sin u2 cos2 u 1 - sin2 u 43. 1 = 1 = 1 = 1 - 11 - sin u2 = sin u 1 + sin u 1 + sin u 1 + sin u 1 cot y + 1 1 1 + sec u sin u 1 + tan y cot y cot y cot y + 1 cos u sin u = 47. + = + tan u = tan u + tan u = 2 tan u 45. = = + tan u = 1 - tan y 1 cot y - 1 cot y - 1 csc u cos u 1 cos u 1 cot y cot y sin u csc u + 1 1 1 + 1 + sin u csc u csc u csc u + 1 49. = = = 1 1 - sin u csc u - 1 csc u - 1 1 csc u csc u 51. 53.
11 - sin y2 2 + cos2 y 2 11 - sin y2 cos y 1 - 2 sin y + sin2 y + cos2 y 2 - 2 sin y 2 1 - sin y + = = = = = = 2 sec y cos y 1 - sin y cos y11 - sin y2 cos y11 - sin y2 cos y11 - sin y2 cos y11 - sin y2 cos y sin u 1 1 1 = = = cos u sin u - cos u sin u - cos u 1 - cot u 1 sin u sin u
55. 1sec u - tan u2 2 = sec2 u - 2 sec u tan u + tan2 u = 11 - sin u2 2
1 - sin u = = 11 - sin u2 11 + sin u2 1 + sin u 57.
cos u 2
-
2 sin u cos u 2
+
sin2 u cos u 2
=
1 - 2 sin u + sin2 u cos u 2
=
11 - sin u2 2 1 - sin2 u
cos u sin u + = 1 - tan u 1 - cot u
cos u sin2 u sin u cos u sin u cos2 u + + = + = sin u cos u cos u - sin u sin u - cos u cos u - sin u sin u - cos u 1 1 cos u sin u cos u sin u 1cos u - sin u2 1cos u + sin u2 cos2 u - sin2 u = = = sin u + cos u cos u - sin u cos u - sin u
59. tan u + 61.
1
sin u11 + sin u2 + cos2 u cos u sin u cos u sin u + sin2 u + cos2 u sin u + 1 1 = + = = = = = sec u 1 + sin u cos u 1 + sin u cos u 11 + sin u2 cos u 11 + sin u2 cos u 11 + sin u2 cos u
tan u + 1sec u - 12 tan u + 1sec u - 12 tan2 u + 2 tan u1sec u - 12 + sec2 u - 2 sec u + 1 tan u + sec u - 1 # = = tan u - sec u + 1 tan u - 1sec u - 12 tan u + 1sec u - 12 tan2 u - 1sec2 u - 2 sec u + 12 =
sec2 u - 1 + 2 tan u1sec u - 12 + sec2 u - 2 sec u + 1
=
2 sec2 u - 2 sec u + 2 tan u1sec u - 12
- 2 + 2 sec u sec2 u - 1 - sec2 u + 2 sec u - 1 2 sec u1sec u - 12 + 2 tan u1sec u - 12 2 1sec u - 12 1sec u + tan u2 = = = tan u + sec u 2 1sec u - 12 2 1sec u - 12
sin2 u - cos2 u sin u cos u sin2 u - cos2 u tan u - cot u cos u sin u cos u sin u = 63. = = = sin2 u - cos2 u tan u + cot u sin u cos u 1 sin2 u + cos2 u + cos u sin u cos u sin u sin u cos u sin2 u - cos2 u tan u - cot u cos u sin u cos u sin u 65. + 1 = + 1 = + 1 = sin2 u - cos2 u + 1 = sin2 u + 11 - cos2 u2 = 2 sin2 u tan u + cot u sin u cos u sin2 u + cos2 u + cos u sin u cos u sin u 1 sin u 1 + sin u + sec u + tan u cos u cos u cos u 1 + sin u # sin u sin u # 1 67. = = = = = tan u sec u cot u + cos u cos u cos u + cos u sin u cos u cos u 11 + sin u2 cos u cos u + cos u sin u sin u 1 - tan2 u
1 - tan2 u + 1 + tan2 u
2 2 = = = 2 cos2 u 1 + tan u 1 + tan2 u 1 + tan2 u sec2 u sec u - csc u sec u csc u 1 1 71. = = = sin u - cos u sec u csc u sec u csc u sec u csc u csc u sec u 2 2 1 1 - cos u sin u sin u 73. sec u - cos u = - cos u = = = sin u # = sin u tan u cos u cos u cos u cos u 1 1 1 + sin u + 1 - sin u 2 2 75. + = = = = 2 sec2 u 1 - sin u 1 + sin u 11 + sin u2 11 - sin u2 1 - sin2 u cos2 u sec u 11 + sin u2 sec u11 + sin u2 1 + sin u sec u sec u # 1 + sin u = = 77. = = 2 1 - sin u 1 - sin u 1 + sin u cos3 u 1 - sin u cos2 u 69.
2
+ 1 =
AN56
79.
ANSWERS Section 6.4
1sec y - tan y2 2 + 1 csc y1sec y - tan y2
=
sec2 y - 2 sec y tan y + tan2 y + 1
1 1 sin y a b sin y cos y cos y 2 11 - sin y2 # sin y = 2 sin y = 2 tan y = cos y 1 - sin y cos y
=
2 sec2 y - 2 sec y tan y 1 1 - sin y a b sin y cos y
2 sin y 2 2 - 2 sin y # sin y cos y cos2 y cos2 y = = 1 - sin y 1 - sin y cos2 y sin y cos y
sin u + cos u sin u - cos u sin u cos u sin2 u + cos2 u 1 = + 1 - 1 + = = = sec u csc u cos u sin u cos u sin u cos u sin u cos u sin u 1sin u + cos u2 1sin2 u - sin u cos u + cos2 u2 sin3 u + cos3 u = = sin2 u + cos2 u - sin u cos u = 1 - sin u cos u 83. sin u + cos u sin u + cos u cos2 u - sin2 u cos2 u - sin2 u cos2 u - sin2 u = = = cos2 u 85. 2 1 - tan2 u sin u cos2 u - sin2 u 1 cos2 u cos2 u 2 2 2 2 [2 cos u - 1sin u + cos2 u2]2 12 cos u - 12 1cos2 u - sin2 u2 2 87. = = = cos2 u - sin2 u = 11 - sin2 u2 - sin2 u = 1 - 2 sin2 u 4 4 2 2 2 2 cos u - sin u 1cos u - sin u2 1cos u + sin u2 cos2 u - sin2 u 81.
89.
11 + sin u2 + cos u 11 + sin u2 + cos u 1 + 2 sin u + sin2 u + 2 11 + sin u2 cos u + cos2 u 1 + sin u + cos u # = = 1 + sin u - cos u 11 + sin u2 - cos u 11 + sin u2 + cos u 1 + 2 sin u + sin2 u - cos2 u = =
1 + 2 sin u + sin2 u + 2 11 + sin u2 1cos u2 + 11 - sin2 u2 1 + 2 sin u + sin u - 11 - sin u2 2
2
2 11 + sin u2 + 2 11 + sin u2 1cos u2 2 sin u11 + sin u2
=
=
2 + 2 sin u + 2 11 + sin u2 1cos u2
2 11 + sin u2 11 + cos u2 2 sin u 11 + sin u2
2 sin u + 2 sin2 u =
1 + cos u sin u
91. 1a sin u + b cos u2 2 + 1a cos u - b sin u2 2 = a2 sin2 u + 2ab sin u cos u + b2 cos2 u + a2 cos2 u - 2ab sin u cos u + b2 sin2 u
= a2 1sin2 u + cos2 u2 + b2 1cos2 u + sin2 u2 = a2 + b2 tan a + tan b tan a + tan b tan a tan b tan a + tan b = = = 1tan a + tan b2 # = tan a tan b 93. tan b + tan a cot a + cot b 1 1 tan a + tan b + tan a tan b tan a tan b
95. 1sin a + cos b2 2 + 1cos b + sin a2 1cos b - sin a2 = 1sin2 a + 2 sin a cos b + cos2 b2 + 1cos2 b - sin2 a2 = 2 cos2 b + 2 sin a cos b = 2 cos b 1cos b + sin a2 = 2 cos b 1sin a + cos b2 97. ln 0 sec u 0 = ln 0 cos u 0 -1 = - ln 0 cos u 0 99. ln 0 1 + cos u 0 + ln 0 1 - cos u 0 = ln 1 0 1 + cos u 0 0 1 - cos u 0 2 = ln 0 1 - cos2 u 0 = ln 0 sin2 u 0 = 2 ln 0 sin u 0 1 1 cos2 x 1 - cos2 x sin2 x sin x - cos x = = = = sin x # = sin x # tan x = f 1x2 101. g 1x2 = sec x - cos x = cos x cos x cos x cos x cos x cos x 1 - sin u cos u 1 - sin u # 1 + sin u cos u # cos u 1 - sin2 u cos2 u = = 103. f 1u2 = cos u 1 + sin u cos u 1 + sin u 1 + sin u cos u cos u 11 + sin u2 cos u 11 + sin u2 cos2 u cos2 u = 0 = g 1u2 = cos u11 + sin u2 cos u 11 + sin u2 p p 105. 216 + 16 tan2 u = 216 21 + tan2 u = 4 2sec2 u = 4 sec u, since sec u 7 0 for - 6 u 6 2 2 1 2 1 2 cos2 u 1 2 - cos2 u 2 107. 1200 sec u (2 sec u - 1) = 1200 a 2 - 1 b = 1200 a 2 b = 1200 ¢ ≤ 2 cos u cos u cos u cos u cos u cos u cos2 u =
1200 (1 + 1 - cos2 u) cos u 3
113. Maximum, 1250 x - 1 114. 1f ∘ g2 1x2 = x - 2
=
1200 (1 + sin2 u)
115. sin u =
cos3 u 5 12 5 13 13 12 ; cos u = - ; tan u = - ; csc u = ; sec u = - ; cot u = 13 13 12 5 12 5
116. -
2 p
6.5 Assess Your Understanding (page 540) 1 1 1 22 - 262 15. - 1 22 + 262 17. 2 - 23 4 4 1 1 1 2 25 11 25 2 25 19. - 1 26 + 222 21. 26 - 22 23. 25. 0 27. 1 29. - 1 31. 33. (a) (b) (c) (d) 2 4 2 2 25 25 5 5 + 12 23 4 - 3 23 - 3 - 4 23 4 + 3 23 25 23 + 48 12 - 5 23 - 5 + 12 23 - 240 + 169 23 35. (a) (b) (c) (d) 37. (a) (b) (c) (d) 10 10 10 39 26 26 26 69 2 22 - 2 22 + 23 - 2 22 + 23 9 - 4 22 1 - 2 26 23 - 2 22 8 22 - 9 23 39. (a) (b) (c) (d) 41. 43. 45. 3 6 6 7 6 6 5 p p p 47. sin a + ub = sin cos u + cos sin u = 1 # cos u + 0 # sin u = cos u 2 2 2 49. sin 1p - u2 = sin p cos u - cos p sin u = 0 # cos u - 1 - 12 sin u = sin u 5. -
6. -
7. False
8. False
9. False
10. True
11.
1 1 26 + 222 4
13.
ANSWERS Section 6.5
AN57
51. sin 1p + u2 = sin p cos u + cos p sin u = 0 # cos u + 1 - 12 sin u = - sin u 0 - tan u 3p tan p - tan u 3p 3p cos u + cos sin u = 1 - 12 cos u + 0 # sin u = - cos u = = - tan u 55. sin a + u b = sin 53. tan 1p - u2 = 1 + tan p tan u 1 + 0 # tan u 2 2 2 57. sin 1a + b2 + sin 1a - b2 = sin a cos b + cos a sin b + sin a cos b - cos a sin b = 2 sin a cos b sin 1a + b2 sin a cos b + cos a sin b sin a cos b cos a sin b 59. = = + = 1 + cot a tan b sin a cos b sin a cos b sin a cos b sin a cos b 61.
cos 1a + b2
=
cos a cos b
cos a cos b - sin a sin b cos a cos b
=
cos a cos b
-
sin a sin b
cos a cos b cos a cos b sin a cos b + cos a sin b
= 1 - tan a tan b
sin a cos b cos a sin b + cos a cos b cos a cos b cos a cos b tan a + tan b 63. = = = = sin a cos b - cos a sin b sin a cos b cos a sin b sin 1a - b2 sin a cos b - cos a sin b tan a - tan b cos a cos b cos a cos b cos a cos b sin 1a + b2
sin a cos b + cos a sin b
cos a cos b - sin a sin b
cos a cos b sin a sin b sin a sin b sin a sin b sin a sin b cot a cot b - 1 = = 65. cot 1a + b2 = = = sin a cos b + cos a sin b sin a cos b cos a sin b sin 1a + b2 sin a cos b + cos a sin b cot b + cot a + sin a sin b sin a sin b sin a sin b cos 1a + b2
cos a cos b - sin a sin b
1 # 1 1 sin a sin b sin a sin b csc a csc b 1 1 67. sec 1a + b2 = = = = = sin a sin b cos a cos b - sin a sin b cos a cos b cos 1a + b2 cos a cos b - sin a sin b cot a cot b - 1 sin a sin b sin a sin b sin a sin b 69. sin 1a - b2 sin 1a + b2 = 1sin a cos b - cos a sin b2 1sin a cos b + cos a sin b2 = sin2 a cos2 b - cos2 a sin2 b = 1sin2 a2 11 - sin2 b2 - 11 - sin2 a2 1sin2 b2 = sin2 a - sin2 b 71. sin 1u + kp2 = sin u cos kp + cos u sin kp = 1sin u2 1 - 12 k + 1cos u2 102 = 1 - 12 k sin u, k any integer 73. 87.
23 2
24 25
75. -
u 21 - y2 - y 21 + u
2
77. -
33 65
79.
63 65
81.
48 + 25 23 39
: - q 6 u 6 q; - 1 … y … 1
89.
83.
4 3
85. u 21 - v 2 - v 21 - u2: - 1 … u … 1; - 1 … v … 1
uy - 21 - u2 21 - y2 y21 - u + u 21 - y 2
2
: - 1 … u … 1; - 1 … y … 1
p 7p f 91. e , 2 6
93. e
p f 4
95. e
11p f 6
p p - ab , cos a - ab = cos b. 2 2 p p p p p p If y Ú 0, then 0 … a … , so a - ab and b both lie on c 0, d . If y 6 0, then … a 6 0, so a - ab and b both lie on a , p d . 2 2 2 2 2 2 p p p - a = b, or a + b = . Either way, cos a - ab = cos b implies 2 2 2 -1
97. Let a = sin
-1
y and b = cos
y.
Then sin a = cos b = y, and since sin a = cos a
1 1 1 p = = cot b, and since tan a = cot a and b = tan-1 y. Because y ≠ 0, a, b ≠ 0. Then tan a = y y tan b 2 p p p p p Because y 7 0, 0 6 a 6 , and so a - ab and b both lie on a0, b . Then cot a - ab = cot b implies - a = 2 2 2 2 2 101. sin 1sin-1 y + cos-1 y2 = sin 1sin-1 y2 cos 1cos-1 y2 + cos 1sin-1 y2 sin 1cos-1 y2 = 1y2 1y2 + 21 - y2 21 - y2 = sin 1x + h2 - sin x cos x sin h - sin x 11 - cos h2 sin x cos h + cos x sin h - sin x sin h 1 103. = = = cos x # - sin x # h h h h 99. Let a = tan-1
105. (a) tan(tan - 1 1 + tan - 1 2 + tan - 1 3) = tan((tan - 1 1 + tan - 1 2) + tan - 1 3) = tan(tan - 1 1) + tan(tan - 1 2)
1 + 2 + 3 1 - 1#2 = = -1 -1 1 + 2 tan(tan 1) + tan(tan 2) #3 1 - 1 - 1#2#3 1 -1 -1 1 - tan(tan 1)tan(tan 2) 1 - tan(tan - 1 1)tan(tan - 1 2)
+ 3
p - ab = cot b. 2 p b, or a = - b. 2 y2 + 1 - y2 = 1 - cos h h - ab , cot a
tan(tan - 1 1 + tan - 1 2) + tan(tan - 1 3) 1 - tan(tan - 1 1 + tan - 1 2) tan(tan - 1 3)
3 + 3 -1 -3 + 3 0 = = = = 0 3 # 1 + 9 10 1 3 -1
p p p (b) From the definition of the inverse tangent function, 0 6 tan - 1 1 6 , 0 6 tan - 1 2 6 , and 0 6 tan - 1 3 6 , 2 2 2 3p so 0 6 tan - 11 + tan - 12 + tan - 13 6 . 2 3p On the interval a0, b , tan u = 0 if and only if u = p. Therefore, from part (a), tan - 11 + tan - 12 + tan - 13 = p. 2 tan u2 - tan u1 m2 - m1 107. tan u = tan 1u2 - u1 2 = = 1 + tan u1 tan u2 1 + m1m2 1 + 1x + 12 1x - 12 1 + tan a tan b 2 1 + x2 - 1 2x2 = 2a b = 2a b = 2a b = = x2 109. 2 cot 1a - b2 = tan 1a - b2 tan a - tan b 1x + 12 - 1x - 12 x + 1 - x + 1 2
AN58
ANSWERS Section 6.5
1 5 112. a - , - b , ( - 5, 1) 3 9
113. 510°
p - ub 2
sin a
p p 111. tan is not defined; tan a - ub = 2 2
p cos a - ub 2
cos u = cot u. sin u
=
9p cm2 ≈14.14 cm2 2
114.
115. sin u = -
2 25 25 25 1 ; cos u = ; csc u = ; sec u = 25; cot u = 5 5 2 2
6.6 Assess Your Understanding (page 550) 1. sin2 u; 2 cos2 u; 2 sin2 u 24 25
9. (a)
11. (a) 13. (a)
7 25
(b) -
2 22 3
4 22 9
2 25 5
(c)
(b)
1 3
(c)
(b) -
7 9
(c)
3 5
(c)
15. (a) -
4 5
(b)
17. (a) -
3 5
(b) -
19.
2. 1 - cos u
32 - 22 2
4 5
C
(c)
C
3. sin u
(d) -
25 5
3 + 26 6
(d)
23 3
(d)
5 + 2 25 10
26 3
310 15 - 252
23. -
(e) -
C
(d)
C
24 7
5. False
(e) - 2 22
4 22 7
5 - 2 25 10
(f)
32 + 23 2
7. (a)
24 25
(b)
7 25
(c)
210 10
(d)
3 210 10
(e)
4 3
(e) -
3 + 26
C 3 - 26
5 + 2 25
(f) (e)
2
25.
(f)
4 5
31.
10
32 + 22
C 5 - 2 25 3 4
(f) -
or 49 + 4 25 or 2 + 25
10 - 210
C 10 + 210
or -
= 12 - 222 32 + 22
45. sin 15u2 = 16 sin5 u - 20 sin3 u + 5 sin u
1 1 - tan2 u = = 49. cot 12u2 = tan 12u2 2 tan u
1
cot u - 1
cot 2 u = 1 b 2a cot u
cot 2 u 2 cot u
1 -
1 1 = = cos 12u2 2 cos2 u - 1
411 - 2 210 1 - 210 or 3 3
27. -
32 - 22 2
= - 1
1cos u cos2 u - sin2 u = 1 + sin 12u2 1 + 2 sin u cos u sin2 u sin u cos u - sin u cos u sin u sin u sin u cot u = = = cos u + sin u cos u sin u cot u + sin u sin u sin u
57. sec2
u = 2
y 59. cot 2 = 2
=
1 u cos2 a b 2
=
2 - sec u 2
sin u
=
sec2 u 2 - sec2 u
sec u 2
- sin u2 1cos u + sin u2
+ cos2 u + 2 sin u cos u
=
1cos u - sin u2 1cos u + sin u2 1sin u + cos u2 1sin u + cos u2
=
cos u - sin u cos u + sin u
- 1 + 1
1 2 = 1 + cos u 1 + cos u 2
1 sec y + 1 1 + 1 1 + cos y sec y sec y sec y + 1 # sec y sec y + 1 = = = = = = 1 cos y 1 cos y 1 sec y 1 sec y sec y 1 sec y - 1 y 1 tan2 a b 1 + cos y sec y sec y 2
1 1 61. = 1 u 1 + 1 + tan2 a b 1 2 sin 13u2
cot 2 u - 1 # cot u cot 2 u - 1 = 2 2 2 cot u cot u
1
u 1 - tan2 a b 2
63.
=
1
53. cos2 12u2 - sin2 12u2 = cos[2 12u2] = cos 14u2 cos 12u2
47. cos4 u - sin4 u = 1cos2 u + sin2 u2 1cos2 u - sin2 u2 = cos 12u2
2
1 2 sec2 u
55.
-
1 -
cos 13u2 cos u
=
+ +
1 3
33.
43. cos 13u2 = 4 cos3 u - 3 cos u
51. sec 12u2 =
(f)
or 45 + 2 26 or 22 + 23
4 7 210 215 35. 37. 39. 3 8 4 3 1 - cos 12u2 2 1 1 1 1 4 2 2 b = [1 - 2 cos 12u2 + cos2 12u2] = - cos 12u2 + cos2 12u2 41. sin u = 1sin u2 = a 2 4 4 2 4 1 1 1 1 + cos 14u2 1 1 1 1 3 1 1 - cos 12u2 + a b = - cos 12u2 + + cos 14u2 = - cos 12u2 + cos 14u2 = 4 2 4 2 4 2 8 8 8 2 8
29. -
24 7
22 2
1 10 + 210 2C 5
(d) -
6. False
(f) - 2
3 - 26 6
(e) -
1 10 - 210 2C 5
21. 1 - 22
4. True
1 + cos u - 11 - cos u2 cos u cos u 1 + cos u 2 cos u # 1 + cos u = = = cos u cos u 1 + cos u + 1 - cos u 1 + cos u 2 cos u 1 + cos u
sin 13u2 cos u - cos 13u2sin u sin u cos u
=
sin 13u - u2
1 12 sin u cos u2 2
=
2 sin 12u2 sin 12u2
= 2
ANSWERS Section 6.7
65. tan 13u2 = tan 1u + 2u2 =
67.
4p 3p 5p p p 2p , , , p, , , f 3 2 3 3 2 3
p 3p , 2 2
= 1 -
0 1 - cos 12u2 0
b
2
1/2
75. No real solution
95. (a) W = 2D(csc u - cot u) = 2Da
(b)
y20
1 - tan2 u = ln 0 sin2 u 0 1/2 = ln 0 sin u 0 77. e 0,
p 5p , p, f 3 3
79.
p 2p 4p 5p 69. e , , , f 3 3 3 3
23 2
81.
7 25
83.
1 cos u 1 - cos u u b = 2D = 2D tan sin u sin u sin u 2
y20 22
cos u 1sin u - cos u2 16 22 = 12 cos u sin u - 2 cos2 u2 32 y20 22 = [sin 12u2 - cos 12u2 - 1] 32
97. (a) R =
2 tan u tan u - tan3 u + 2 tan u 3 tan u - tan3 u 1 - tan2 u = = 2 2 tan u 12 tan u2 1 - tan u - 2 tan u 1 - 3 tan2 u
tan u +
1 - tan u tan 12u2
1 1ln 0 1 - cos 12u2 0 - ln 22 = ln a 2
73. e 0, 93.
tan u + tan 12u2
3p or 67.5° 8
(d)
24 7
85.
71. e 0, 24 25
87.
2p 4p , f 3 3 1 5
89.
25 7
91. 0,
p 5p , p, 3 3
(b) u = 24.45°
u = 67.5° 3p a radians b makes R 8 largest. R = 18.75 ft
20
(c) 32 12 - 222 ≈ 18.75 ft
45
90 0
99. A =
1 1 u u 1 h 1base2 = h a base b = s cos # s sin = s2 sin u 2 2 2 2 2
101. sin 12u2 =
4x 4 + x2
103. -
1 4
a 2 sin a b 2 a 2 tan a b 2
a cos a b 2 a a a 105. = = = = 2 sin a b cos a b = sin a2 # b = sin a 1 2 2 2 1 + z2 2 a 2 a 1 + tan a b sec a b a 2 2 cos2 a b 2 2z
a 2 tan a b 2
107.
y 1.25
2P x
p 22 = 44 - 36 - 22 24 4 p 22 = 44 + 36 + 22 cos 24 4 3 3 111. sin u + sin 1u + 120°2 + sin3 1u + 240°2 = sin3 u + 1sin u cos 120° + cos u sin 120°2 3 + 1sin u cos 240° + cos u sin 240°2 3 109. sin
= sin3 u + a -
3 3 1 23 1 23 sin u + cos u b + a - sin u cos u b 2 2 2 2
1 1 13 23 cos3 u - 9 cos2 u sin u + 3 23 cos u sin2 u - sin3 u2 - 1sin3 u + 3 23 sin2 u cos u + 9 sin u cos2 u + 3 23 cos3 u2 8 8 3 3 9 3 3 3 1 2 3 2 3 = sin u - cos u sin u = 3 sin u - 3 sin u11 - sin u2 4 = 14 sin u - 3 sin u2 = - sin 13u2 (from Example 2) 113. 4 4 4 4 4 2 = sin3 u +
115. y =
1 x - 4 2
y
116.
(3, 16)
117.
23 + 1 2
y (22, 2) (2, 2) 2
118.
(0, 7) (21, 0)
2422
(7, 0)
2 4
x
22 (24, 22) (0, 22)(4, 22)
x
6.7 Assess Your Understanding (page 556) 1 23 1 23 22 1 1 1 a - 1 b 3. - a + 1 b 5. 7. 3 cos 12u2 - cos 16u2 4 9. 3 sin 16u2 + sin 12u2 4 11. 3 cos 12u2 + cos 18u2 4 2 2 2 2 2 2 2 2 1 1 u 13. 3cos u - cos 13u2 4 15. 3 sin 12u2 + sin u4 17. 2 sin u cos 13u2 19. 2 cos 13u2 cos u 21. 2 sin 12u2 cos u 23. 2 sin u sin 2 2 2 2 sin 12u2 cos u sin 14u2 + sin 12u2 2 sin 13u2 cos u sin 13u2 sin u + sin 13u2 = = cos u 27. = = = tan 13u2 25. 2 sin 12u2 2 sin 12u2 cos 14u2 + cos 12u2 2 cos 13u2 cos u cos 13u2 1.
29.
cos u - cos 13u2 sin u + sin 13u2
=
2 sin 12u2 sin u
2 sin 12u2 cos u
=
sin u = tan u cos u
31. sin u 3sin u + sin 13u2 4 = sin u3 2 sin 12u2 cos u4 = cos u 3 2 sin 12u2 sin u 4 = cos u c 2 # 33.
sin 14u2 + sin 18u2
cos 14u2 + cos 18u2
=
2 sin 16u2 cos 12u2
2 cos 16u2 cos 12u2
=
AN59
sin 16u2
cos 16u2
= tan 16u2
1 3 cos u - cos 13u2 4 = cos u 3 cos u - cos 13u2 4 d 2
AN60
ANSWERS Section 6.7
sin 14u2 + sin 18u2
39.
sin 14u2 - sin 18u2
2 sin 16u2 cos 1 - 2u2
=
=
sin 16u2
#
cos 12u2
= tan 16u2 3 - cot 12u2 4 = -
2 sin 1 - 2u2 cos 16u2 cos 16u2 - sin 12u2 a + b a + b a - b a - b 2 sin cos sin cos sin a + sin b a + b a - b 2 2 2 2 # = = = tan 37. cot a - b a + b a + b a - b sin a - sin b 2 2 2 sin cos cos sin 2 2 2 2 35.
sin a + sin b cos a + cos b
2 sin = 2 cos
a + b 2 a + b 2
cos cos
a - b 2 = a - b 2
sin
tan 16u2 tan 12u2
a + b
a + b 2 = tan a + b 2 cos 2
41. 1 + cos 12u2 + cos 14u2 + cos 16u2 = 3 1 + cos 16u2 4 + 3 cos 12u2 + cos 14u2 4 = 2 cos2 13u2 + 2 cos 13u2 cos 1 - u2 = 2 cos 13u2 3cos 13u2 + cos u4 = 2 cos 13u2 3 2 cos 12u2 cos u4 = 4 cos u cos 12u2 cos 13u2 p p 2p 4p 3p 5p p 2p 3p 4p 6p 7p 8p 9p 43. e 0, , , , p, , , f 45. e 0, , , , , p, , , , f 3 2 3 3 2 3 5 5 5 5 5 5 5 5 47. (a) y = 2 sin 11906 pt2 cos 1512 pt2 (b) ymax = 2 (c)
49. Iu = Ix cos2 u + Iy sin2 u - 2Ixy sin u cos u = Ixcos2 u + Iy sin2 u - Ixy sin 2u
2
= 0
0.01
= Iv =
2
cos 2u + 1 1 - cos 2u b + Iy a b - Ixy sin 2u 2 2 Iy Iy Ix Ix cos 2u + + - cos 2u - Ixy sin 2u 2 2 2 2 Ix - Iy Ix + Iy + cos 2u - Ixy sin 2u 2 2 1 - cos 2u cos 2u + 1 Ix sin2 u + Iy cos2 u + 2Ixy sin u cos u = Ix a b + Iy a b + Ixy sin 2u 2 2 Iy Iy Ix Ix - cos 2u + cos 2u + + Ixy sin 2u 2 2 2 2 Ix + Iy Ix - Iy cos 2u + Ixy sin 2u 2 2
= Ix a
= =
51. sin 12a2 + sin 12b2 + sin 12g2 = 2 sin 1a + b2 cos 1a - b2 + sin 12g2 = 2 sin 1a + b2 cos 1a - b2 + 2 sin g cos g = 2 sin 1p - g2 cos 1a - b2 + 2 sin g cos g = 2 sin g cos 1a - b2 + 2 sin g cos g = 2 sin g[cos 1a - b2 + cos g] a - b + g a - b - g p - 2b 2a - p p p cos b = 4 sin g cos cos = 4 sin g cos a - b b cos aa - b = 2 sin ga2 cos 2 2 2 2 2 2
= 4 sin g sin b sin a = 4 sin a sin b sin g
sin 1a - b2 = sin a cos b - cos a sin b
53.
sin 1a + b2 = sin a cos b + cos a sin b
sin 1a - b2 + sin 1a + b2 = 2 sin a cos b 1 sin a cos b = 3 sin 1a + b2 + sin 1a - b2 4 2 a + b a - b a + b a - b a + b a - b 2b 1 2a cos = 2 # c cos a + b + cos a b d = cos + cos = cos a + cos b 55. 2 cos 2 2 2 2 2 2 2 2 2 p p 2 26 57. {7} 58. Amplitude: 5; Period: ; Phase shift: 59. 2 4 7 x + 5 p p 60. f -1 1x2 = sin-1 a b ; Range of f = Domain of f -1 = 3 - 8, - 2 4 ; Range of f -1 = c - , d 3 2 2
Review Exercises (page 559) p 5p p p 3p 3p 3p p p 6. 7. 8. 9. 10. 11. 12. 13. 14. 0.9 15. 0.6 16. 5 17. Not defined 6 3 4 4 8 4 3 7 9 p 2 23 4 4 1 x p p 18. 19. p 20. - 23 21. 22. 23. 24. f -1(x) = sin - 1 a b ; Range of f = Domain of f -1 = [ - 2, 2]; Range of f -1 = c - , d 6 3 5 3 3 2 6 6 0 u 0 25. f -1 1x2 = cos-1 13 - x2; Range of f = Domain of f -1 = 3 2, 4 4 ; Range of f -1 = 3 0, p 4 26. 21 - u2 27. u 2u2 - 1 28. tan u cot u - sin2 u = 1 - sin2 u = cos2 u 29. cos2 u11 + tan2 u2 = cos2 u sec2 u = 1 30. 5 cos2 u + 3 sin2 u = 2 cos2 u + 3 1cos2 u + sin2 u2 = 3 + 2 cos2 u 11 - cos u2 2 + sin2 u 2 11 - cos u2 sin u 1 - 2 cos u + cos2 u + sin2 u 1 - cos u + = = = = 2 csc u 31. sin u 1 - cos u sin u11 - cos u2 sin u 11 - cos u2 sin u 11 - cos u2 cos u cos u cos u 1 1 32. = = = cos u - sin u cos u - sin u sin u 1 - tan u 1 cos u cos u 1.
p 2
2.
p 2
3.
p 4
4. -
p 6
5.
ANSWERS Chapter Test 1 sin u 1 1 # 1 - sin u = = = 1 1 + sin u 1 + sin u 1 - sin u 1 + sin u 1 - cos2 u sin2 u 1 - cos u = = = sin u # 34. sec u - cos u = cos u cos u cos u csc u 33. = 1 + csc u
35.
38.
1 - sin u
=
cos2 u
sin u = sin u tan u cos u
sin u11 - cos2 u2 1 + cos u sin3 u 1 - cos u = sin u 11 + cos u2 # = = csc u 1 - cos u 1 - cos u 1 - cos u cos u sin u cos2 u - sin2 u 1 - 2 sin2 u = = sin u cos u sin u cos u sin u cos u
36. cot u - tan u = 37.
1 - sin u 1 - sin2 u
AN61
sin 1a - b2 cos a cos b
cos 1a - b2 cos a cos b
sin a cos b - cos a sin b
=
cos a cos b
=
cos a cos b + sin a sin b
=
sin a cos b cos a cos b
=
cos a cos b
cos a cos b cos a cos b
-
cos a sin b cos a cos b
= tan a - tan b
sin a sin b
+
= 1 + tan a tan b
cos a cos b
u sin u = sin u 39. 11 + cos u2 atan b = 11 + cos u2 # 2 1 + cos u 2 cos u1cos2 u - sin2 u2 cos u cos 2u cos2 u - sin2 u ba b = = = cot 2 u - 1 2 sin u sin 2u 2 sin u cos u sin2 u sin 12u2 sin 13u2 cos u - sin u cos 13u2 = = 1 41. 1 - 8 sin2 u cos2 u = 1 - 2 12 sin u cos u2 2 = 1 - 2 sin2 12u2 = cos 14u2 42. sin 12u2 sin 12u2 sin 12u2 + sin 14u2 2 sin 13u2 cos 1 - u2 43. = = tan 13u2 cos 12u2 + cos 14u2 2 cos 13u2 cos 1 - u2 40. 2 cot u cot 2u = 2 a
44. 46.
cos 12u2 - cos 14u2
cos 12u2 + cos 14u2 1
47.
23 + 1
48. 1
49. 1
23 - 2 22 6
55. (a) -
63 65
(b)
16 65
- 23 - 2 22 6
(b) (c)
2 cos 13u2 cos 1 - u2
50. 0
22 2 22 12 - 3 27 9 + 4 27 53. (a) (b) 20 20 54. (a)
- 2 sin 13u2 sin 1 - u2
- tan u tan 13u2 =
51.
23 + 1
52.
- tan u tan 13u2 = tan 13u2 tan u - tan u tan 13u2 = 0
45.
1 + 2 26 6 33 65
63 16
24 25
(f) -
119 169
- 23 + 2 22 6
(g)
2 213 13
(h) -
8 22 + 9 23 23
210 10 23 2
(f) -
7 9
p 5p , f 3 3
74. e
p p 3p 3p , , , f 4 2 4 2
23 2 27(5 + 3 217) 1 4 25 1 230 26 33 - 25 13 (d) Not defined (e) (f) (g) (h) 58. 57. (a) 1 (b) 0 (c) 59. 9 9 9 6 6 2493 7 217 - 15 48 + 25 23 22 24 7 p 5p 2p 5p 3p 7p p 3p p 2p 4p 5p 61. 62. 63. 64. e , f 65. e , f 66. e , f 67. e 0, , p, f 68. e , , , f 60. 39 10 25 25 3 3 3 3 4 4 2 2 3 3 3 3 56. (a)
69. 50.25, 2.89 6 77. - 1.11
(b)
70. e 0,
1 - 2 26 6
(e)
2 12
42 - 22 2
23 - 1 3 27 27 12 + 3 27 12 - 3 27 7 2 (c) (d) (e) (f) (g) (h) 20 8 25 9 + 4 27 25 2 22 23 + 2 22 23 - 2 22 4 22 22 - 13 23 + 2 22 7 (c) (d) (g) (h) (e) (f) 6 2 2 C 6 9 1 + 2 26
(d) -
13 - 1
(c)
2p 4p , p, f 3 3
71. e 0,
81. 51.11 6
78. 1.77 79. 1.23 80. 2.90
1 - cos 30° 86. sin 15° = = B 2 T
p 5p , f 6 6
13 2
1 -
(d)
72. e
p p 5p , , f 6 2 6
82. 50.87 6
(e) -
73. e
83. 52.22 6
84. e -
23 f 2
(g)
23 3
(h)
75. e
p , pf 2
85. {0}
2 - 23 22 - 13 = ; 4 2 22 # 23 22 # 1 26 22 26 - 22 sin 15° = sin(45° - 30°) = sin 45° cos 30° - cos 45° sin 30° = = = ; 2 2 2 2 4 4 4 2 2 4 12 232 42 - 23 2 - 23 8 - 4 23 6 - 2 212 + 2 26 - 22 J R = = = = = ¢ ≤ 2 4 4 # 4 16 16 4 2
=
C
Chapter Test (page 561) p p p 7 4 2. 3. 4. 5. 3 6. 7. 0.39 8. 0.78 9. 1.25 10. 0.20 6 4 5 3 3 csc u + cot u csc u + cot u # csc u - cot u csc2 u - cot 2 u 1 11. = = = sec u + tan u sec u + tan u csc u - cot u 1sec u + tan u2 1csc u - cot u2 1sec u + tan u2 1csc u - cot u2 sec u - tan u 1 sec u - tan u # sec u - tan u = = = 1sec u + tan u2 1csc u - cot u2 sec u - tan u csc u - cot u 1sec2 u - tan2 u2 1csc u - cot u2 1.
12. sin u tan u + cos u = sin u #
sin u sin2 u cos2 u sin2 u + cos2 u 1 + cos u = + = = = sec u cos u cos u cos u cos u cos u
76. 0.78
AN62
ANSWERS Chapter Test sin u cos u sin2 u cos2 u sin2 u + cos2 u 1 2 2 + = + = = = = = 2 csc 12u2 cos u sin u sin u cos u sin u cos u sin u cos u sin u cos u 2 sin u cos u sin 12u2
13. tan u + cot u = 14.
sin 1a + b2 tan a + tan b
=
sin a cos b + cos a sin b sin a cos b + cos a sin b sin a cos b + cos a sin b = = sin a cos b sin a cos b + cos a sin b sin b cos a sin b sin a + + cos a cos b cos a cos b cos a cos b cos a cos b
=
sin a cos b + cos a sin b 1
#
cos a cos b sin a cos b + cos a sin b
= cos a cos b
15. sin 13u2 = sin 1u + 2u2 = sin u cos 12u2 + cos u sin 12u2 = sin u # 1cos2 u - sin2 u2 + cos u # 2 sin u cos u = sin u cos2 u - sin3 u + 2 sin u cos2 u = 3 sin u cos2 u - sin3 u = 3 sin u11 - sin2 u2 - sin3 u = 3 sin u - 3 sin3 u - sin3 u = 3 sin u - 4 sin3 u sin u cos u sin2 u - cos2 u 11 - cos2 u2 - cos2 u sin2 u - cos2 u tan u - cot u cos u sin u sin u cos u = = 16. = = 1 - 2 cos2 u = 2 2 2 2 tan u + cot u sin u cos u 1 sin u + cos u sin u + cos u + cos u sin u sin u cos u
2 213 1 25 - 32 25 12 285 2 + 23 20. 21. 22. 5 49 39 4 3p 7p 11p 15p 27. e , , , f 28. 50.285, 3.427 6 29. 50.253, 2.889 6 8 8 8 8 18. 2 + 23
19.
23.
26 2
24.
22 2
25. e
17.
p 2p 4p 5p , , , f 3 3 3 3
1 1 26 + 222 4 26. 50, 1.911, p, 4.373 6
Cumulative Review (page 561) 1. e
- 1 - 213 - 1 + 213 , f 6 6
4.
y 8
2. y + 1 = - 1 1x - 42 or x + y = 3; 6 22; y 5
5. (0, 5)
(6, 5)
6.
3. x-axis symmetry; 10, - 32, 10, 32, 13, 02
y 0.5 2P
(0, 1)
x
6 x y 2
(3, 2) 8 x
y 2.5 y x3
7. (a)
11, 22
(0, 0) (1, 1)
(b)
y 5
1 1, e
(1, 1) 2.5 x y ex
(c)
(1, e) (0, 1)
1,
y 1.25
5 x f 1(x) ln x
P 2 , 21 2
(d)
P 2
P ,1 2
(21, P)
y 3
y 5 sin x
(0, 0)
P
f 21(x) 5 cos21 x
(0, 1) P x y 5 cos x (P, 21)
x
P ,0 2
3
f 1(x) 兹x f 21(x) 5 sin21 x
8. (a) -
2 22 3
(b)
22 4
(c)
4 22 9
(d)
7 9
(e)
C
3 + 2 22 6
11. (a) f 1x2 = 12x - 12 1x - 12 2 1x + 12 2; 1 multiplicity 1; 1 and - 1 multiplicity 2 2 1 (b) 10, - 12; a , 0 b ; 1 - 1, 02; (1, 0) (c) y = 2x5 2 (d) 2
(f) -
C
3 - 2 22 6
25 5
10. (a) -
2 22 3
(b) -
(e) Local minimum value - 1.33 at x = - 0.29 Local minimum value 0 at x = 1 Local maximum value 0 at x = - 1 Local maximum value 0.10 at x = 0.69 (f) 1, 0 y 1.25
2
(1, 0) 2
9.
P 21, 2 2
2 22 3
(c)
7 9
(d)
4 22 9
1 f 2 (b) 5 - 1, 1 6
1 (c) 1 - q , - 12 ∪ a - , q b 2 (d) 1 - q , - 1 4 ∪ 31, q 2
(1, 0)
(g) Increasing: 1 - q , - 12, 1 - 0.29, 0.692, 11, q 2 Decreasing: 1 - 1, - 0.292, 10.69, 12
2
CHAPTER 7 Applications of Trigonometric Functions 7.1 Assess Your Understanding (page 571) 4. False
5. True
6. angle of elevation 7. True
8. False
9. sin u =
5 12 5 12 13 13 ; cos u = ; tan u = ; cot u = ; sec u = ; csc u = 13 13 12 5 12 5
11. sin u =
2 213 3 213 2 3 213 213 ; cos u = ; tan u = ; cot u = ; sec u = ; csc u = 13 13 3 2 3 2
13. sin u =
23 1 23 2 23 ; cos u = ; tan u = 23; cot u = ; sec u = 2; csc u = 2 2 3 3
26 3
12. (a) e - 1, -
1.25 x (0.69, 0.10) (0, 1) (0.29, 1.33)
2
(e)
ANSWERS Section 7.3
15. sin u =
AN63
26 23 22 26 ; cos u = ; tan u = 22; cot u = ; sec u = 23; csc u = 3 3 2 2
25 2 25 1 25 ; cos u = ; tan u = ; cot u = 2; sec u = ; csc u = 25 5 5 2 2 19. 0 21. 1 23. 0 25. 0 27. 1 29. a ≈ 13.74, c ≈ 14.62, A = 70° 31. b ≈ 5.03, c ≈ 7.83, A = 50° 33. a ≈ 0.71, c ≈ 4.06, B = 80° 35. b ≈ 10.72, c ≈ 11.83, B = 65° 37. b ≈ 3.08, a ≈ 8.46, A = 70° 39. c ≈ 5.83, A ≈ 59.0°, B ≈ 31.0° 41. b ≈ 4.58, A ≈ 23.6°, B ≈ 66.4° 43. 23.6° and 66.4° 45. 4.59 in.; 6.55 in. 47. (a) 5.52 in. or 11.83 in. 49. 70.02 ft 51. 985.91 ft 53. 137.37 m 55. 58.8° 57. (a) 111.96 ft/sec or 76.3 mi/hr (b) 82.42 ft/sec or 56.2 mi/hr (c) Under 18.8° 59. (a) 2.4898 * 1013 miles (b) 0.000214° 61. 554.52 ft 63. S76.6°E 65. The embankment is 30.5 m high. 67. 3.83 mi 69. 1978.09 ft 71. 60.27 ft 73. The buildings are 7979 ft apart. 75. 69.0° 77. 38.9° 79. The white ball should hit the top cushion 4.125 ft from the upper left corner. 84. Yes 17. sin u =
85.
26 - 22 32 - 23 or 4 2
87. e
86. 0.236, 0.243, 0.248
p 7p 11p , , f 2 6 6
7.2 Assess Your Understanding (page 583) sin A sin B sin C = = 6. False 7. False 8. ambiguous case 9. a ≈ 3.23, b ≈ 3.55, A = 40° 11. a ≈ 3.25, c ≈ 4.23, B = 45° a b c 13. C = 95°, c ≈ 9.86, a ≈ 6.36 15. A = 40°, a = 2, c ≈ 3.06 17. C = 120°, b ≈ 1.06, c ≈ 2.69 19. A = 100°, a ≈ 5.24, c ≈ 0.92 21. B = 40°, a ≈ 5.64, b ≈ 3.86 23. C = 100°, a ≈ 1.31, b ≈ 1.31 25. One triangle; B ≈ 30.7°, C ≈ 99.3°, c ≈ 3.86 27. One triangle; C ≈ 36.2°, A ≈ 43.8°, a ≈ 3.51 29. No triangle 31. Two triangles; C1 ≈ 30.9°, A1 ≈ 129.1°, a1 ≈ 9.07 or C2 ≈ 149.1°, A2 ≈ 10.9°, a2 ≈ 2.20 33. No triangle 35. Two triangles; A1 ≈ 57.7°, B1 ≈ 97.3°, b1 ≈ 2.35 or A2 ≈ 122.3°, B2 ≈ 32.7°, b2 ≈ 1.28 37. (a) Station Able is about 143.33 mi from the ship. Station Baker is about 135.58 mi from the ship. (b) Approximately 41 min 39. 1490.48 ft 41. 381.69 ft 43. The tree is 39.39 ft high. 45. Adam receives 100.6 more frequent flyer miles. 47. 84.7°; 183.72 ft 49. 2.64 mi 51. 38.5 in. 53. 449.36 ft 55. 187,600,000 km or 101,440,000 km 57. The diameter is 252 ft. 4. oblique 5.
a - b a b sin A sin B sin A - sin B = = = = 59. c c c sin C sin C sin C
2 sin a
A - B A - B A - B A + B p C sin a sin a b cos a b b cos a - b b 2 2 2 2 2 2 = = C C C C C 2 sin cos sin cos cos 2 2 2 2 2
1 sin c 1A - B2 d 2 C cos 2
1 1 1 a - b tan c 1A - B2 d tan c 1A - B2 d tan c 1A - B2 d 2 2 2 a - b c 61. = = = = = a + b a + b C 1 p 1 C cot cos c 1A - B2 d tan a - b tan c 1A + B2 d c 2 2 2 2 2 C sin 2 69.
4 66. e - 3 , - , 3 f 3
(23P, 4) (24P, 0)
67. 3 25 ≈ 6.71 68. -
215 7
y 5 (P, 4) (2P, 0) (4P, 0) 5P x
(22P, 0) (2P, 24)
(3P, 24) (0, 0)
7.3 Assess Your Understanding (page 590) 3. Cosines 4. Sines 5. Cosines 6. False 7. False 8. True 9. b ≈ 2.95, A ≈ 28.7°, C ≈ 106.3° 11. c ≈ 3.75, A ≈ 32.1°, B ≈ 52.9° 13. A ≈ 48.5°, B ≈ 38.6°, C ≈ 92.9° 15. A ≈ 127.2°, B ≈ 32.1°, C ≈ 20.7° 17. c ≈ 2.57, A ≈ 48.6°, B ≈ 91.4° 19. a ≈ 2.99, B ≈ 19.2°, C ≈ 80.8° 21. b ≈ 4.14, A ≈ 43.0°, C ≈ 27.0° 23. c ≈ 1.69, A = 65.0°, B = 65.0° 25. A ≈ 67.4°, B = 90°, C ≈ 22.6° 27. A = 60°, B = 60°, C = 60° 29. A ≈ 33.6°, B ≈ 62.2°, C ≈ 84.3° 31. A ≈ 97.9°, B ≈ 52.4°, C ≈ 29.7° 33. A = 85°, a = 14.56, c = 14.12 35. A = 40.8°, B = 60.6°, C = 78.6° 37. A = 80°, b = 8.74, c = 13.80 39. Two triangles: B1 = 35.4°, C1 = 134.6°, c1 = 12.29; B2 = 144.6°, C2 = 25.4°, c2 = 7.40 41. B = 24.5°, C = 95.5°, a = 10.44 43. 165 yd 45. (a) 26.4° (b) 30.8 h 47. (a) 63.72 ft (b) 66.78 ft (c) 92.8° 49. (a) 492.58 ft (b) 269.26 ft 51. 342.33 ft 53. The footings should be 7.65 ft apart. 1 - cos u 1 - cos u u b 1 d = 2r = 2r sin . 2 B 2 2 u u Since, for any angle in 10, p2, d is strictly less than the length of the arc subtended by u—that is, d 6 ru—then 2r sin 6 ru, or 2 sin 6 u. 2 2 u u u u Since cos 6 1, then, for 0 6 u 6 p, sin u = 2 sin cos 6 2 sin 6 u. If u Ú p, then, since sin u … 1, sin u 6 u. Thus sin u 6 u for all u 7 0. 2 2 2 2
55. Suppose 0 6 u 6 p. Then, by the Law of Cosines, d 2 = r 2 + r 2 - 2r 2 cos u = 4r 2 a
57. sin
C 1 - cos C = = 2 2 R B =
1 -
12s - 2b2 12s - 2a2
B
4ab
a 2 + b2 - c 2 c 2 - 1a - b2 2 1c + a - b2 1c + b - a2 2ab 2ab - a2 - b2 + c 2 = = = 2 4ab 4ab 4ab B B B =
1s - a2 1s - b2
B
ab
AN64 64.
ANSWERS Section 7.3
1 2 ,0 2
y
65. e
(4, 9)
8
5,
y=2 22 1 0, 2 3
11 2
5 x
x=3
ln 3 f ≈ E 3.819 F ln 4 - ln 3
66. 2 26 7 26 7 5 26 sin u = ; csc u = ; sec u = - ; cot u = 7 12 5 12 67. y = - 3 sin 14x2
7.4 Assess Your Understanding (page 596) 1 2. ab sin C 2
1 3. 2s(s - a)(s - b)(s - c); (a + b + c) 4. True 5. 2.83 7. 2.99 9. 14.98 11. 9.56 13. 3.86 15. 1.48 17. 2.82 19. 30 2 a2 sin B sin C 1 1 a sin B b = 27. 0.92 29. 2.27 31. 5.44 33. 9.03 sq ft 35. $5446.38 21. 1.73 23. 19.90 25. K = ab sin C = a sin Ca 2 2 sin A 2 sin A 1 37. The area of home plate is about 216.5 in.2 39. K = r 2 1u + sin u2 41. The ground area is 7517.4 ft2. 2 1 OC # AC 1 1 OC AC = # = sin a cos a 2 2 1 1 2 BC OC 1 1 1 2 # = OB 2 sin b cos b (b) Area ∆OCB = BC OC = OB 2 2 2 OB OB
43. (a) Area ∆OAC =
(c) Area ∆OAB =
BD 1 1 1 BD OA = OB = OB sin 1a + b2 2 2 2 OB
OC
(e) Area ∆OAB = Area ∆OAC + Area ∆OCB 1 1 1 OB sin(a + b) = sin a cos a + OB 2 sin b cos b 2 2 2 1 sin(a + b) = sin a cos a + OB sin b cos b OB sin(a + b) =
cos a 1 = (d) = OB cos b OC
cos b cos a
sin a cos a +
cos a sin b cos b cos b
sin(a + b) = sin a cos b + cos a sin b
OB 45. 31,145 ft 2 47. (a) The perimeter and area are both 36. (b) The perimeter and area are both 60. 1 1 a sin B sin C 49. K = ah = ab sin C 1 h = b sin C = 2 2 sin A A B 1 C C C C + b = 180° - (180° - C) = 90° + , and sin a90° + b = cos a- b = cos , since cosine is an even function. 2 2 2 2 2 2 2 A B B A c sin sin c sin sin 2 2 2 2 Therefore, r = . = C C cos sin a90° + b 2 2 3s - (a + b + c) A B C s - a s - b s - c 3s - 2s s + cot + cot = + + = = = 53. cot 2 2 2 r r r r r r
51. ∠POQ = 180° - a
22 27 214 3 22 3 27 214 ; cos t = ; tan t = ; csc t = ; sec t = ; cot t = 3 3 7 2 7 2 1 1 - sin2 u cos2 u cos u - sin u = = = cos u # = cos u cot u 61. csc u - sin u = sin u sin u sin u sin u 58. Maximum value; 17 59. 1 - q , - 32 h [ - 1 , 32
60. sin t =
7.5 Assess Your Understanding (page 606) 2. simple harmonic; amplitude 3. simple harmonic; damped 4. True 5. d = - 5 cos 1pt2 7. d = - 6 cos 12t2 9. d = - 5 sin 1pt2 2p 3 11. d = - 6 sin 12t2 13. (a) Simple harmonic (b) 5 m (c) sec (d) oscillation/sec 15. (a) Simple harmonic (b) 6 m (c) 2 sec 3 2p 1 1 (d) oscillation/sec 17. (a) Simple harmonic (b) 3 m (c) 4p sec (d) oscillation/sec 2 4p 19. (a) Simple harmonic (b) 2 m (c) 1 sec (d) 1 oscillation/sec
21.
25.
27.
y 3P
y5x y 5 cos x 2P x
23.
y 1.25
P
t
2P
29.
y 3P
y5x y 5 2sin x 3P
x
y 1.25
t
y y 5 cos x 1.25 y 5 sin x
3P
x
ANSWERS Review Exercises 31.
33. (a) f 1x2 =
y 2.5
(b)
3P y 5 sin x
x y 5 sin(2x)
1 3 cos x - cos (3x) 4 2
1 3 cos(6x) + cos(2x) 4 2
35. (a) G(x) = (b)
y 1
1 y 5 cos x 2
1 cos(2x) 2
y5
y 1
2P x 1 y 5 2 cos(3x) 2
2P x
y5
37. (a) H(x) = sin(4x) + sin(2x)
39. (a) d = - 10e -0.7t/50 cos a
(b) y 2
y 5 sin(2x) y 5 sin(4x)
(b)
4p2 0.49 tb B 25 2500
AN65
1 cos(6x) 2
41. (a) d = - 18e -0.6t/60 cos a (b)
10
p2 0.36 tb B 4 3600
18
2P x 0
0
25
ⴚ18
10
43. (a) d = - 5e -0.8t/20 cos a (b)
45. (a) The motion is damped. The bob has mass m = 20 kg with a damping factor of 0.7 kg/s. (b) 20 m leftward (c) 20
4p2 0.64 tb B 9 400
5
0
20
47. (a) The motion is damped. The bob has mass m = 40 kg with a damping factor of 0.6 kg/s. (b) 30 m leftward (c) 30
15 0
0
25
35
5 30
20
(d) 18.33 m leftward
(e) d S 0
51. v = 1040p; d = 0.80 cos(1040pt) 53. v = 880p; d = 0.01 sin(880pt) 55. (a) V
49. (a) The motion is damped. The bob has mass m = 15 kg with a damping factor of 0.9 kg/s. (b) 15 m leftward (c) 15
(d) 28.47 m leftward (e) d S 0 57.
1.25
2.5
0
2
3 t 2.5
0
30
(b) At t = 0, 2; at t = 1, t = 3 (c) During the approximate intervals 0.35 6 t 6 0.67, 1.29 6 t 6 1.75, and 2.19 6 t … 3
15
(d) 12.53 m leftward (e) d S 0 61.
63. y =
1
1 sin x x
y =
1
0
0.1
0 0
1 x3
sin x
0.05
4x - 3 1x2 = x - 1
66. log 7 a
xy3 x + y
b
20.06
67. {4}
3 210 68. (a) 10
7P 0
5P
20.3
65. f
y =
sin x
5P
20.3
-1
1 x2
210 (b) 10
3P
20.015
1 (c) 3
Review Exercises (page 610) 23 4 3 4 3 5 5 1 23 2 23 ; cos u = ; tan u = ; cot u = ; sec u = ; csc u = 2.sin u = ; cos u = ; tan u = 23; cot u = ; sec u = 2; csc u = 5 5 3 4 3 4 2 2 3 3 3. 0 4. 1 5. 1 6. A = 70°, b ≈ 3.42, a ≈ 9.40 7. a ≈ 4.58, A ≈ 66.4°, B ≈ 23.6° 8. C = 75°, b ≈ 2.6, c ≈ 5 9. B ≈ 56.8°, C ≈ 23.2°, b ≈ 4.25 10. No triangle 11. b ≈ 3.32, A ≈ 62.8°, C ≈ 17.2° 12. No triangle 13. No triangle 14. A ≈ 46.6°, B ≈ 104.5°, C ≈ 28.9° 15. c ≈ 2.32, A ≈ 16.1°, B ≈ 123.9° 16. A ≈ 35°, B ≈ 105°, c ≈ 3.3 17. A ≈ 39.6°, B ≈ 18.6°, C ≈ 121.9° 18. Two triangles: B1 ≈ 13.4°, C1 ≈ 156.6°, c1 ≈ 6.86 or B2 ≈ 166.6°, C2 ≈ 3.4°, c2 ≈ 1.02 19. b ≈ 11.52, c ≈ 10.13, C ≈ 60° 20. a ≈ 5.23, B ≈ 46.0°, C ≈ 64.0° 21. 1.93 22. 30.31 23. 6 24. 3.80 25. 10.57 26. 0.76 in2 1. sin u =
AN66
ANSWERS Review Exercises
27. 34.85° and 55.15°
28. 23.32 ft
29. 2.15 mi
30. 132.55 ft/min
31. 12.7°
32. 29.97 ft
33. 6.22 mi
34. (a) 131.8 mi
(b) 23.1°
(c) 0.21 hr
p 37. 76.94 in. 38. 79.69 in. 39. d = - 3 cos a t b 2 1 40. (a) Simple harmonic (b) 6 ft (c) p s (d) oscillation/s p 1 41. (a) Simple harmonic (b) 2 ft (c) 2 s (d) oscillation/s 2
35. 8798.67 sq ft
36. S4.0°E
42. (a) d = - 15e -0.75t>80 cos a (b)
0.5625 4p2 tb 6400 B 25
43. (a) The motion is damped. The bob has mass m = 20 kg with a damping factor of 0.6 kg/s. (b) 15 m leftward (c) 15 (d) 13.92 m leftward (e) d S 0
15
0
25
0
15
44.
y 2
y 5 2 sin x
x 2P y 5 cos(2 x)
25
15
Chapter Test (page 612) 25 2 25 1 25 ; cos u = ; tan u = ; csc u = 25; sec u = ; cot u = 2 2. 0 3. a = 15.88, B ≈ 57.5°, C ≈ 70.5° 5 5 2 2 4. b ≈ 6.85, C = 117°, c ≈ 16.30 5. A ≈ 52.4°, B ≈ 29.7°, C ≈ 97.9° 6. b ≈ 4.72, c ≈ 1.67, B = 105° 7. No triangle 8. c ≈ 7.62, A ≈ 80.5°, B ≈ 29.5° 9. 15.04 square units 10. 19.81 square units 11. 61.0° 12. 1.3° 13. The area of the shaded region is 9.26 cm2. 14. 54.15 square units 15. Madison will have to swim about 2.23 miles. 16. 12.63 square units 17. The lengths of the sides are 15, 18, and 21. pt pt 18. d = 5(sin 42°) sin a b or d ≈ 3.346 sin a b 3 3 1. sin u =
Cumulative Review (page 613) 1 1. e , 1 f 3
2. 1x + 52 2 + 1y - 12 2 = 9
3. 5x x … - 1 or x Ú 4 6
4.
y 5
(5, 1)
5.
y 3
y 2.5 P
2.5 x
x
1 x
6. (a) 7. (a)
2 25 5
(b)
25 5
(c) -
4 5
(d) -
3 5
(e)
(b)
60
B
5 - 15 10
(f) -
B
5 + 15 10 (c)
1.5
(d)
10 0
0
0
4
8. (a)
(b)
y 5
(d)
y 5
5 x
(h)
y 1.25
2P
x
4 0
(e)
y 5
(f)
y 9
5 x 5 x
(g)
0
50
(c)
y 10
4
4
1.5
0
8
5 x
(i)
y 1.25
y 2
2P
x
P
x
y 5
8 x 5 x
9. Two triangles: A1 ≈ 59.0°, B1 ≈ 81.0°, b1 ≈ 23.05 or A2 ≈ 121.0°, B2 ≈ 19.0°, b2 ≈ 7.59 1 10. e - 2i, 2i, , 1, 2 f 3
ANSWERS Section 8.2 12x + 12 1x - 42
11. R1x2 =
1x + 52 1x - 32
; Domain: 5x x ≠ - 5, x ≠ 3 6
1 4 Intercepts: a - , 0 b , (4, 0), a0, b 2 15 No symmetry Vertical asymptotes: x = - 5, x = 3 Horizontal asymptote: y = 2 Intersects: a 12. 52.26 6
y 10
8
AN67
26 ,2 11
10 x 10
8
(4, 0) 1 ,0 2
4
0,
4 15
26 , 2b 11
13. 51 6
5 f 4
14. (a) e -
(e) {x| - 8 … x … 3} or [ - 8, 3]
(b) 52 6
(f)
y 6
(c) e
- 1 - 3 213 - 1 + 3 213 , f 2 2 (g)
(d) e x ` x 7 -
5 5 f or a - , q b 4 4
y 15
(0, 5)
(8, 0)
(3, 0) 10 x
(1.25, 0) 5 x
(0, 24) (2.5, 30.25)
CHAPTER 8 Polar Coordinates; Vectors 8.1 Assess Your Understanding (page 623) 8. r cos u; r sin u
5. pole; polar axis 6. True 7. False 19.
17.
(3, 90)
(2, 0)
9. A
11. C
21.
O
90
13. B
27.
O
25. 135 O
P 6
O
15. A
23.
P 6, 6
4, 2
(2, 135)
2P 3
2
2P 3
O
O
29.
O (22, 2P) 2P
31.
(a) a5, -
5, 2P 3
4p b 3
(b) a - 5,
5p b 3
(c) a5,
8p b 3
O
2P 3
P 3
2
P 3
(a) 12, - 2p2 (b) 1 - 2, p2 (c) 12, 2p2
3P
33.
21, 2
(22, 3P)
O
35.
37.
1, P 2
P 2
23, 2
O
P 4
O
3p b 2 3p (b) a - 1, b 2 5p b (c) a1, 2
2
(a) a1, -
41. 1 - 2, 02
5p b 4 7p b (b) a - 3, 4 11p (c) a3, b 4
1 23 b 47. a - , 2 2
45. 1 22, - 222
49. 12, 02
53. 1 - 4.98, - 3.852
18 b b ≈ 137.36, 105.5°2 (c) 1 - 3, - 352 5 5 9 89. 2 or 0 positive real zeros; 1 negative real zero 90. a - , b 91. (0, - 11) 4 2
19 f 15
(b) a2 2349, 180° + tan
43. 1 - 3 23, 32
p p 55. 13, 02 57. 11, p2 59. a 22, - b 61. a2, b 4 6 3 26 69. r 2 cos2 u - 4r sin u = 0 63. 12.47, - 1.022 65. 19.30, 0.472 67. r 2 = or r = 2 2 1 2 1 71. r 2 sin (2u) = 1 73. r cos u = 4 75. x2 + y2 - x = 0 or ax - b + y2 = 2 4 77. 1x2 + y2 2 3>2 - x = 0 79. x2 + y2 = 4 81. y2 = 8 1x + 22
51. 1 - 2.57, 7.052
P 4
(a) a3, -
83. (a) 1 - 10, 362 88. e
39. 10, 32
-1
a-
(d) a 21234, 180° + tan
-1
a
35 b b ≈ 135.13, 265.1°2 3
8.2 Assess Your Understanding (page 637) 7. polar equation 8. False
9. - u
13. x + y = 16; circle, radius 4, center at pole 2
2
y U5 U5
3P 4
5P 4
P U5 4
U5 3P U5 2
11. True
12. 2n; n
15. y = 23x; line through pole, making an p angle of with polar axis 3
17. y = 4; horizontal line 4 units above the pole y
y
x 1 2 3 4 5 U50
U5P
U5
P 2
10. p - u
7P 4
3P U5 4
P U5 2
x 1 2 3 4 5 U50
U5P
U5
P U5 3 P U5 4
U5
5P 4 U5
3P 2
7P 4
U5
3P 4
U5
U5
P 4
x 1 2 3 4 5 U50
U5P
U5
P 2
U5
5P 4 U5
3P 2
7P 4
AN68
ANSWERS Section 8.2
19. x = - 2; vertical line 2 units to the left of the pole
21. 1x - 12 2 + y2 = 1; circle, radius 1, center 11, 02 in rectangular coordinates
y U5 U5
3P 4
P 4
x 1 2 3 4 5 U50
5P 4
U5
U5
7P 4
3P U5 2
U5
3P 4
U5
U5
U5
P 4
x 1 2 3 4 5 U50
7P U5 4
5P 4 U5
3P U5 4
P 2
4
P 4
x 6 8 10 U 5 0
U5 U5
7P 4
3P U5 4
U5P
U5
3P U5 4
P 2
U5
U5
P 4
7P 4
3P 2
3P U5 4
U5
P U5 4
7P 4
3P U5 4
x 1 2 3 4 5 U50
U5P
U5
5P 4 U5
3P 2
7P 4
U5
3P 4
P 4
U5
5P 4
U5
7P 4
3P U5 2
U5
3P 4
U5
P 2
P 4
x 4 6 8 10 U 5 0
U5P
U5
U5
U5
5P 4
7P 4
3P 2
y P U5 4
7P 4
3P U5 4
U5
U5
P 4
x 1 2 3 4 5 U50
U5P
U5
P 2
5P 4
3P 2
U5
U5
7P 4
U5
P 4
3P 2
55. Spiral U5
y P 2
U5
P 4
x 1 2 3 4 5 U50
U5P
U5
U5
x 1 2 3 4 5 U50
y P U5 4
P 2
49. Rose
53. Lemniscate P 2
U5
U5
U5 U5
35. D
43. Limaçon without inner loop
x 1 2 3 4 5 U50
5P 4
33. H
U5P
3P 2
P U5 2
U5P
3P 2
y P 2
U5
5P 4
y U5
3P 4
3P 2
x 1 2 3 4 5 U50
U5P
7P 4
y
y
51. Rose
U5
x 2 3 4 5 U50
47. Limaçon with inner loop
U5 U5
U5
U5
x 1 2 3 4 5 U50
5P 4
29. E 31. F 37. Cardioid
x 2 3 4 5 U50
U5
5P 4
P 4
P 4
7P U5 4
5P U5 4
y
U5P
1
U5 U5
1
3P 2
45. Limaçon with inner loop U5
U5
P 2 U5
U5P
y
2
5P 4
3P 4
U5
P U5 2
U5
U5
U5P
U5
3P 2
41. Limaçon without inner loop U5
3P 4
7P 4
27. x2 + 1y + 12 2 = 1, y ≠ 0; circle, radius 1, center at 10, - 12 in rectangular coordinates, hole at (0, 0)
y
U5
U5
5P 4
3P 2
39. Cardioid
U5
P U5 4
y
P U5 2
U5P
U5
x 1 2 3 4 5 U50
U5P
y 3P 4
P U5 2
U5
25. 1x - 22 2 + y2 = 4, x ≠ 0; circle, radius 2, center at (2, 0) in rectangular coordinates, hole at (0, 0)
U5
y
y
P 2 U5
U5P
23. x2 + 1y + 22 2 = 4; circle, radius 2, center at 10, - 22 in rectangular coordinates
U5
5P 4 U5
3P 2
7P 4
U5 3P U5 4
U5P
U5
P 2
15 30 45 60
U5
5P 4 U5
3P 2
x U50
7P 4
ANSWERS Historical Problems 57. Cardioid
59. Limaçon with inner loop y
U5
P 2
U5
3P 4
U5
U5P
U5
1
63. 3P 4
U5
P 2
U5
U5P
U5
1
P 4
U5
x U50
2
3P 4
U5
U5P
1
5P 4
U5
U5
3P 4
U5
U5
U5P
U5
1 2 3 3
5P 4
x U50
1
U5 U5
P 4
3P 2
7P 4
U5 U5
3P 4
U5
3P 4
U5
U5
U5P
U5
U5
3P 2
5P 4
U5
1. (a) 1 + 4i, 1 + i
(b) - 1, 2 + i
3P 2
3P U5 4
3P 2
7P 4
U5
U5
5P 4
2p 5
7P 4
3P 2
r = 2a cos u r 2 = 2ar cos u x2 + y2 = 2ax x2 - 2ax + y2 = 0 1x - a2 2 + y2 = a2 Circle, radius a, center at 1a, 02 in rectangular coordinates 85.
(b) r 2 = sin u; r 2 = sin 1p - u2 r 2 = sin u Test works. 1 - r2 2 = sin 1 - u2 r 2 = - sin u Not equivalent; new test fails.
93. Amplitude = 2; Period =
P 4
x 1 2 3 4 5 U50
U5
r = 2a sin u r 2 = 2ar sin u x2 + y2 = 2ay x2 + y2 - 2ay = 0 x2 + 1y - a2 2 = a2 Circle, radius a, center at 10, a2 in rectangular coordinates
P 2 U5
U5P
83.
7P 4
y
P 4
87. (a) r 2 = cos u; r 2 = cos 1p - u2 r 2 = - cos u Not equivalent; test fails. 1 - r2 2 = cos 1 - u2 r 2 = cos u New test works.
91. Absolute maximum: f 142 = 1; absolute minimum: none 92. 420° 94. Horizontal asymptote: y = 0; vertical asymptote: x = 4
Historical Problems (page 646)
U5
5P 4
U5
x 5P U 5 0
P 4
U5
x 1 2 3 4 5 U50
77.
3P
P 4, 2 3
P 2
7P 4 U5
P 2
P
P 4
7P 4
U5
x U50
3P 2
U5
x 1 2 3 4 5 U50
5P 4
3P 4
U5
81. r sin u = a y = a
P 2 U5
U5P
U5
U5
U5
y
79.
U5
y
75.
P 2
3P 2
7P 4
67. r = 3 + 3 cos u 69. r = 4 + sin u y 71.
P 4
2
U5
5P 4
2 2 Ï2 , 5P 2 4
3P U5 2
y
73.
U5
P 4
U5
U5P
7P 4
U5
5P 4 U5
2 1 Ï2 , P 2 4
P 2
(0, P) U5
U5
3P 2
y
65.
P U5 2
P 3
x 2 4 6 8 10 U 5 0
U5P
7P U5 4
5P 4
3P 4
4,
P 2
U5
P 4
x 1 2 3 4 5 U50
U5
y
U5 U5
U5P
3P U5 2
1, U5
3P U5 4
x U50
2
P 2
U5
P 4
7P U5 4
5P 4
y
61.
y
AN69
AN70
ANSWERS Section 8.3
8.3 Assess Your Understanding (page 646) 5. real; imaginary 6. magnitude; modulus; argument 7. r1 r2; u1 + u2; u1 + u2 13. 11. Imaginary Imaginary axis
8. r n; nu; nu
9. three
axis
10. True 15.
Imaginary axis 3
1
1 Real axis
1
1
2
Real axis
3
1
17.
19.
Imaginary axis
3 1cos 270° + i sin 270°2 21.
Imaginary axis
4
Imaginary axis
1
2 2
4
3
2
Real axis
Real axis
3
2 1cos 330° + i sin 330°2
22 1cos 45° + i sin 45°2
3
1
3
Real axis 1
2
2
2
4
4
5 1cos 306.9° + i sin 306.9°2
4 22 1cos 315° + i sin 315°2
2
Real axis
1
213 1cos 123.7° + i sin 123.7°2
23. - 1 + 23i
25. 2 22 - 2 22i 27. - 3i 29. - 0.035 + 0.197i 31. 1.970 + 0.347i z z 1 3 33. zw = 8 1cos 60° + i sin 60°2; = 1cos 20° + i sin 20°2 35. zw = 12 1cos 40° + i sin 40°2; = 1cos 220° + i sin 220°2 w 2 w 4 9p 9p z p p z + i sin b; = cos + i sin 39. zw = 4 22 1cos 15° + i sin 15°2; = 22 1cos 75° + i sin 75°2 37. zw = 4 acos 40 40 w 40 40 w 27 25 22 27 23 25 22 + i 47. + i 49. - 4 + 4i 41. - 32 + 32 23i 43. 32i 45. 2 2 2 2 51. - 23 + 14.142i 6 6 6 53. 2 2 1cos 15° + i sin 15°2, 2 2 1cos 135° + i sin 135°2, 2 2 1cos 255° + i sin 255°2 4 4 4 4 55. 28 1cos 75° + i sin 75°2, 28 1cos 165° + i sin 165°2,28 1cos 255° + i sin 255°2,2 8 1cos 345° + i sin 345°2 57. 2 1cos 67.5° + i sin 67.5°2, 2 1cos 157.5° + i sin 157.5°2, 2 1cos 247.5° + i sin 247.5°2, 2 1cos 337.5° + i sin 337.5°2 59. cos 18° + i sin 18°, cos 90° + i sin 90°, cos 162° + i sin 162°, cos 234° + i sin 234°, cos 306° + i sin 306° 61. 1, i, - 1, - i
n
63. Look at formula (8); zk = 1r for all k.
Imaginary i axis
1
1
65. Look at formula (8). The zk are spaced apart by an angle of
2p . n
Real axis
i
67. Assume the theorem is true for n Ú 1. For n = 0: z0 = r 0 3cos(0 # u) + i sin(0 # u) 4 1 = 1 # 3cos(0) + i sin(0) 4 1 = 1 # 31 + 0 4 1 = 1 True
For negative integers: z-n = (zn)-1 = (r n[cos(nu) + i sin(nu)])-1 with n Ú 1 1 = n r [cos(nu) + i sin(nu)] = = =
1 # cos(nu) - i sin(nu) r n[cos(nu) + i sin(nu)] cos(nu) - i sin(nu) cos(nu) - i sin(nu) r (cos2(nu) + sin2(nu)) cos(nu) - i sin(nu) n
n
sin2 u + cos2 u = 1
r = r -n[cos(nu) - i sin(nu)]
= r -n[cos( - nu) + i sin( - nu)] Thus, De Moivre’s Theorem is true for all integers. 69. ≈40.50
6 16 70. Minimum; f a b = 5 5
4 71. p 3
3 72. 2y 2 3x2y2
cos(- u) = cos(u); sin(- u) = - sin(u)
AN71
ANSWERS Section 8.5
8.4 Assess Your Understanding (page 658) 1. vector 2. 0
3. unit
4. position
9.
5. horizontal; vertical 6. resultant 11.
7. True 8. False 13.
3v
15.
v
u
3v
vw vw
w
w
3v u 2w
2w
v
17. T
19. F 21. F
23. T 25. 12
43. 289
45. 234 - 213
41. - j 55.
5-2
+ 221, - 2 - 221 6
71. (a) 1 - 1, 4 2 (b) (1, 4) y
u (4, 5)
5 v
v u 5 x (3, 1)
107.
y 2.5
27. v = 3i + 4j 47. i
57. v =
29. v = 2i + 4j
3 4 49. i - j 5 5
5 5 23 i + j 2 2
51.
31. v = 8i - j
22 22 i j 2 2
59. v = - 7i + 7 23j
53. v = 61. v =
33. v = - i + j 8 25 4 25 i + j 5 5 25 23 25 i j 2 2
35. 5
37. 22
or v = 63. 45°
73. F = 20 23i + 20 j 75. F = 1 20 23 + 30 22 2 i + 1 20 - 30 22 2 j 77. (a) va = 550j; vw = 50 22i + 50 22j (b) vg = 50 22i + 1550 + 50 222j (c) 7 vg 7 = 624.7 mph; N6.5°E 79. v = 1250 22 - 302i + 1250 22 + 30 232j; 518.8 km/h; N38 .6°E 81. Approximately 4031 lb 83. 8.6° left of direct heading across the river; 1.52 min 85. Tension in right cable: 1000 lb; tension in left cable: 845.2 lb 87. Tension in right part: 1088.4 lb; tension in left part: 1089.1 lb 89. The truck must pull with a force of 4635.2 lb. 91. (a) N7.05°E (b) 12 min 93. m = 0.36 95. 13.68 lb
39. 213
8 25 4 25 i j 5 5
65. 150°
67. 333.4°
69. 258.7°
99. F2 P F1
F3
F4
105. {29} 106. - 3x(x + 2)(x - 6)
108. 23 P 3 x P 3 ,2 3 2
3 p ; Period = 2 3 p Phase shift = 2 Amplitude =
Historical Problem (page 667)
1ai + bj2 # 1ci + dj2 = ac + bd Real part [ 1a + bi2 1c + di2 4 = Real part[ 1a - bi2 1c + di2 4 = Real part[ac + adi - bci - bdi 2] = ac + bd
8.5 Assess Your Understanding (page 668) 9. (a) 0 (b) 90° (c) orthogonal 11. (a) 13 - 1 2 5 5 1 1 19. v1 = i - j, v2 = - i - j (b) 75° (c) neither 13. (a) - 50 (b) 180° (c) parallel 15. (a) 0 (b) 90° (c) orthogonal 17. 3 2 2 2 2 1 2 6 3 14 7 1 2 i + j, v2 = i - j 25. 9 ft-lb 27. (a) 7 I 7 ≈ 0.022; the intensity of the sun’s rays is 21. v1 = - i - j, v2 = i - j 23. v1 = 5 5 5 5 5 5 5 5 approximately 0 .022 W/cm2. 7 A 7 = 500; the area of the solar panel is 500 cm2. (b) W = 10; ten watts of energy is collected. (c) Vectors I and A should be parallel with the solar panels facing the sun. 29. Force required to keep Sienna from rolling down the hill: 737.6 lb; force perpendicular to the hill: 5248.4 lb 31. Timmy must exert 85.5 lb. 33. - 2.833 35. 60° 37. Let v = ai + bj. Then 0 # v = 0a + 0b = 0. 39. v = cos ai + sin aj, 0 … a … p; w = cos bi + sin bj, 0 … b … p. If u is the angle between v and w, then v # w = cos u, since 7 v 7 = 1 and 7 w 7 = 1. Now u = a - b or u = b - a. Since the cosine function is even, v # w = cos(a - b). Also, v # w = cos a cos b + sin a sin b. So cos(a - b) = cos a cos b + sin a sin b. 41. (a) If u = a1i + b1j and v = a2i + b2 j, then, since 7 u 7 = 7 v 7 , a21 + b21 = 7 u 7 2 = 7 v 7 2 = a22 + b22, 1u + v 2 # 1u - v 2 = 1a1 + a2 2 1a1 - a2 2 + 1b1 + b2 2 1b1 - b2 2 = 1a21 + b21 2 - 1a22 + b22 2 = 0. (b) The legs of the angle can be made to correspond to vectors u + v and u - v. 43. 1 7 w 7 v + 7 v 7 w2 # 1 7 w 7 v - 7 v 7 w2 = 7 w 7 2v # v - 7 w 7 7 v 7 v # w + 7 v 7 7 w 7 w # v - 7 v 7 2w # w = 7 w 7 2v # v - 7 v 7 2w # w = 7 w 7 2 7 v 7 2 - 7 v 7 2 7 w 7 2 = 0 2. dot product 3. orthogonal
4. parallel
5. T 6. F
7. (a) 0 (b) 90°
(c) orthogonal
45. 7 u + v 7 2 - 7 u - v 7 2 = 1u + v2 # 1u + v2 - 1u - v2 # 1u - v2 = 1u # u + u # v + v # u + v # v2 - 1u # u - u # v - v # u + v # v2 9 = 2 1u # v2 + 2 1v # u2 = 4 1u # v2 47. 12 48. 2 50. V(x) = x(19 - 2x)(13 - 2x) or V(x) = 4x3 - 64x2 + 247x 49. 11 - sin2u2 11 + tan2u2 = 1cos2u2 1sec2u2 1 = cos2u # cos2u = 1
AN72
ANSWERS Section 8.6
8.6 Assess Your Understanding (page 677) 2. xy-plane 3. components 4. 1 5. F 6. T 7. All points of the form (x, 0, z) 9. All points of the form (x, y, 2) 11. All points of the form ( - 4, y, z) 13. All points of the form (1, 2, z) 15. 221 17. 233 19. 226 21. (2, 0, 0); (2, 1, 0); (0, 1, 0); (2, 0, 3); (0, 1, 3); (0, 0, 3) 23. (1, 4, 3); (3, 2, 3); (3, 4, 3); (3, 2, 5); (1, 4, 5); (1, 2, 5) 25. ( - 1, 2, 2); (4, 0, 2); (4, 2, 2); ( - 1, 2, 5); (4, 0, 5); ( - 1, 0, 5) 27. v = 3i + 4j - k 3 6 2 29. v = 2i + 4j + k 31. v = 8i - j 33. 7 35. 23 37. 222 39. - j - 2k 41. 2105 43. 238 - 217 45. i 47. i - j - k 7 7 7 23 23 23 # # # # 49. i + j + k 51. v w = 0; u = 90° 53. v w = - 2, u ≈ 100.3° 55. v w = 0; u = 90° 57. v w = 52; u = 0° 3 3 3 59. a ≈ 64.6°; b ≈ 149.0°; g ≈ 106.6°; v = 7(cos 64.6°i + cos 149.0°j + cos 106.6°k) 61. a = b = g ≈ 54.7°; v = 23(cos 54.7°i + cos 54.7°j + cos 54.7°k) 63. a = b = 45°; g = 90°; v = 22(cos 45°i + cos 45°j + cos 90°k) 65. a ≈ 60.9°; b ≈ 144.2°; g ≈ 71.1°; v = 238(cos 60.9°i + cos 144.2°j + cos 71.1°k) 67. (a) d = a + b + c = 6 7, 1, 5 7 (b) 8.66 ft 3 22 69. (x - 3)2 + (y - 1)2 + (z - 1)2 = 1 71. Radius = 2, center ( - 1, 1, 0) 73. Radius = 3, center (2, - 2, - 1) 75. Radius = , Center (2, 0, - 1) 2 13 13 1 f or a2, d 81. 2x2 + 2x - 5 82. 77. 2 newton-meters = 2 joules 79. 9 newton-meters = 9 joules 80. e x ` 2 6 x … 5 5 2 83. c = 3 25 ≈ 6.71; A ≈ 26.6°; B ≈ 63.4°
8.7 Assess Your Understanding (page 683) 1. T 2. T 3. T 4. F 5. F 6. T 7. 2 9. 4 11. - 11A + 2B + 5C 13. - 6A + 23B - 15C 15. (a) 5i + 5j + 5k (b) - 5i - 5j - 5k (c) 0 (d) 0 17. (a) i - j - k (b) - i + j + k (c) 0 (d) 0 19. (a) - i + 2j + 2k (b) i - 2j - 2k (c) 0 (d) 0 21. (a) 3i - j + 4k (b) - 3i + j - 4k (c) 0 (d) 0 23. - 9i - 7j - 3k 25. 9i + 7j + 3k 27. 0 29. - 27i - 21j - 9k 31. - 18i - 14j - 6k 33. 0 35. - 25 37. 25 39. 0 41. Any vector of the form c( - 9i - 7j - 3k), where c is a nonzero scalar 43. Any vector of the form c( - i + j + 5k), where c is a nonzero scalar 45. 2166
47. 2555
i 55. 98 cubic units 57. u * v = 3 a1 a2
j b1 b2
49. 234
51. 2998
53.
11 219 219 7 219 i + j + k 57 57 57
or -
11 219 219 7 219 i j k 57 57 57
k c1 3 = (b1c2 - b2c1)i - (a1c2 - a2c1)j + (a1b2 - a2b1)k c2
7 u * v 7 2 = 2(b1c2 - b2c1)2 + (a1c2 - a2c1)2 + (a1b2 - a2b1)2 2 2 7u
72
7u
= b21c 22 - 2b1b2c1c2 + b22c 21 + a21c 22 - 2a1a2c1c2 + a22c 21 + a21b22 - 2a1a2b1b2 + a22b21 =
727
v
a21
72
+ b21 + c 21, 7 v 7 2 = a22 + b22 + c 22
= (a21 + b21 + c 21)(a22 + b22 + c 22) = a21a22 + a21b22 + a21c 22 + b21a22 + b21b22 + b21c 22 + a22c 21 + b22c 21 + c 21c 22
(u # v)2 = (a1a2 + b1b2 + c1c2)2 = (a1a2 + b1b2 + c1c2)(a1a2 + b1b2 + c1c2)
= a21a22 + a1a2b1b2 + a1a2c1c2 + b1b2c1c2 + b1b2a1a2 + b21b22 + b1b2c1c2 + a1a2c1c2 + c 21c 22
7u
= a21a22 + b21b22 + c 21c 22 + 2a1a2b1b2 + 2b1b2c1c2 + 2a1a2c1c2
727
v 7 2 - (u # v)2 = a21b22 + a21c 22 + b21a22 + a22c 21 + b22c 21 + b21c 22 - 2a1a2b1b2 - 2b1b2c1c2 - 2a1a2c1c2, which equals 7 u * v 7 2. 59. By Problem 58, since u and v are orthogonal, 7 u * v 7 = 7 u 7 7 v 7 . If, in addition, u and v are unit vectors, 7 u * v 7 = 1 # 1 = 1. 61. Assume that u = ai + bj + ck, v = di + ej + fk, and w = li + mj + nk. Then u * v = (bf - ec)i - (af - dc)j + (ae - db)k, u * w = 1bn - mc2i - 1an - lc2j + 1am - lb2k, and v + w = 1d + l2i + 1e + m2j + 1f + n2k. Therefore, (u * v) + (u * w) = (bf - ec + bn - mc)i - (af - dc + an - lc)j + (ae - db + am - lb)k and u * 1v + w2 = [b(f + n) - (e + m)c]i - [a(f + n) - (d + l)c]j + [a(e + m) - (d + l)b]k = 1bf - ec + bn - mc2i - 1af - dc + an - lc2j + 1ae - db + am - lb)k, which equals 1u * v2 + 1u * w2. p 1 64. 65. (17, 4.22), ( - 17, 1.08) 66. 8 hours 67. log 4 x - 3log 4 z 4 2
Review Exercises (page 686) 1. a
3 23 3 , b 2 2 3,
P 6
P 6
2. 1 1, 23 2
3. (0, 3) 22,
4P 3
23, 2
4P 3
O
O 2
P 2
P 2
O
4. a8,
5p p b , a - 8, b 4 4
5. a2,
5p p b , a - 2, - b 6 6
6. (5, p), 1 - 5, 02
AN73
ANSWERS Review Exercises 7. (a) x2 + 1y - 12 2 = 1 (b) circle, radius 1, center (0, 1) in rectangular coordinates
8. (a) x2 + y2 = 25 at pole
y U5 U5
3P 4
1
P 4
2
x U50
U5
3P 4
7P 4
U5
P 4
U5
U5P
4
U5
5P 4 U5
U5P
7P 4
U5
13. Limaçon without inner loop; symmetric with respect to the polar axis
U5
U5
3P 4
P 2
P 2 U5
U5
(5, P)
5P 4
U5
0,
U5
P 4
U5
x 5 U50
P 2
3P 2
3P 4
y U5
P 2
U5
P 4
(3, 0) x 2 4 6 8 10 U 5 0
U5P
7P 4
U5
3P 2
U5
5P 4 U5
14. 2 1cos 146.3° + i sin 146.3°2 16. - 23 + i
P 4
3P 2
P 6, 2 2
17. -
18. 0.10 - 0.02i
3 3 23 + i 2 2
Imaginary axis 0.06
Imaginary axis 3
2
7P 4
15. 5 1cos 323.1° + i sin 323.1°2
2
x 4 5 U50
U5 3P U5 2
7P 4
12. Cardioid; symmetric with respect to the line p u = 2
0.10
Real axis
0.02
(3, 0) 1 2
U5
5P 4
P 3
Imaginary axis
2
2
2
Real axis
Real axis
0.06
1
7P 4
4, 3P 2
19. zw = - 1cos 50° + i sin 50°2; 21. zw = 6;
2,
U5
5P 4
2 U5P
x 1 2 3 4 5 U50
U5
1 2 3
U5
4,
7P 4
P 4
P 2
3P 2
y
3P 4
P 2
U5
(4, 0)
x U50
8
U5
P 2
U5P
3P U5 2
y P 2
3P U5 4
11. Circle; radius 2, center at (2, 0) in rectangular coordinates; symmetric with respect to the polar axis
0,
U5
P U5 4
U5
5P 4
y 3P 4
U5
x 1 2 3 4 5 U50
U5
10. (a) 1x - 42 2 + 1y + 22 2 = 25 (b) circle, radius 5, center 14, - 22 in rectangular coordinates
U5
P 2
U5P
3P U5 2
U5
y
U5
U5
U5
5P 4
(b) line through pole, making p an angle of with polar axis 4
y
P 2
U5P
U5
9. (a) x - y = 0
(b) circle, radius 5, center
z = - 1cos 30° + i sin 30°2 w
z = 6 1cos 70° - i sin 70°2 w
22. 32 + 32 23i
20. zw = - 12;
23. 5
5 5 - 5 i A2 A2
z 3 2p 2p = acos + i sin b w 4 3 3 24. 16
25. - 8432 - 5376i
26. 2, 2 1cos 72° + i sin 72°2, 2 1cos 144° + i sin 144°2, 2 1cos 216° + i sin 216°2, 2 1cos 288° + i sin 288°2 or 2, 27.
25 - 1 25 + 1 25 + 1 25 + 15 25 - 15 25 - 15 25 - 1 25 + 15 + i, + i, i, i 4 4 4 4 2 12 2 12 2 12 2 12 v
28.
u uv
2u
3v
29. v = 2i - 4j; 7 v 7 = 2 25 30. v = - i + 3j; 7 v 7 = 210 31. 2i - 2j 32. - 20i + 13j 33. 25 2 25 25 3 3 23 34. 25 + 5 ≈ 7.24 35. i + j 36. v = i + j 37. 120° 38. 243 ≈ 6.56 5 5 2 2 39. v = 3i - 5j + 3k 40. 21i - 2j - 5k 41. 238 42. 0 43. 3i + 9j + 9k 44. 0
219 3 219 3 219 219 3 219 3 219 i + j + k or i j k 19 19 19 19 19 19 # # # 46. v w = - 11; u ≈ 169.7° 47. v w = - 4; u ≈ 153.4° 48. v w = 1; u ≈ 70.5° 49. v # w = 0; u = 90° 4 3 6 8 9 7 21 50. Parallel 51. Neither 52. Orthogonal 53. v1 = i - j; v2 = i + j 54. v1 = 13i + j2; v2 = - i + j 5 5 5 5 10 10 10 55. a ≈ 56.1°; b ≈ 138°; g ≈ 68.2° 56. 2 283 57. - 2i + 3j - k 58. 0 59. 229 ≈ 5.39 mi/h 0.4 mi 60. Left cable: 1843.21 lb; right cable: 1630.41 lb 61 50 ft-lb 62. A force of 697.2 lb is needed to keep the van from rolling down the hill. The magnitude of the force perpendicular to the hill is 7969.6 lb. 2u 3v
45.
AN74
ANSWERS Chapter Test
Chapter Test (page 688) 1–3.
P 2
2, 3P 4
P 3
P 4
p b 3
4. a4,
5. x2 + y2 = 49
6.
y U5
P 6
3P U5 4
P 2 U5
y x
= 3 or y = 3x y U5
P 4
3P U5 4
P 2 U5
P 4
0 P 3, 2 6 24,
x U50
U5P
P 3 U5
U5
5P 4 U5
7. 8y = x2 3P 4
P 2 U5
P 4
x U50
U5P
7P 4
U5
3P 2
U5
5P 4 U5
3P 2
3
z1 2兹2(cos 160 i sin 160) U5
U5
5P 4 U5
Imaginary axis
7P 4
2
3P 2
3
z0 2兹2(cos 40 i sin 40) 40 Real axis
2
3
z2 2兹2(cos 280 i sin 280)
15. 7 v 7 = 10
14. v = 8 5 22, - 5 22 9
19. v1 + 2v2 - v3 = 8 6, - 10 9
22 22 ,i 17. 315° off the positive x-axis 18. v = 5 22i - 5 22j 2 2 20. Vectors v1 and v4 are parallel. 21. Vectors v2 and v3 are orthogonal. 22. 172.87° 23. - 9i - 5j + 3k
24. a ≈ 57.7°, b ≈ 143.3°, g ≈ 74.5°
v
16. u =
7v7
25. 2115
= h
26. The cable must be able to endure a tension of approximately 670.82 lb.
Cumulative Review (page 689) 1. 5 - 3, 3 6
2. y =
23 x 3
3. x2 + 1y - 12 2 = 9
y (0, 1) 5 (0, 4) (3, 1)
6.
y 2
7. 1, 1 e
(1, 0)
10.
y 4.5
8.
y 1.25
4. e x ` x 6
1 1 f or a - q , b 2 2
5. Symmetry with respect to the y-axis
(3, 1) 5 x (0, 2)
y 1.25
9. -
p 6
2P
(e, 1)
2P
x
x
4 x
y
11.
x3 y4
7P 4
p 8. r 2 cos u = 5 is symmetric about the pole, the polar axis, and the line u = . p 2 9. r = 5 sin u cos2 u is symmetric about the line u = . The tests for symmetry about the pole and the polar 2 2 axis fail, so the graph of r = 5 sin u cos u may or may not be symmetric about the pole or the polar axis. w 3 10. z # w = 6 1cos 107° + i sin 107°2 11. = 1cos 297° + i sin 297°2 z 2 5 12. w = 243 1cos 110° + i sin 110°2 3 3 3 13. z0 = 2 22 1cos 40° + i sin 40°2, z1 = 2 22 1cos 160° + i sin 160°2, z2 = 2 2 2 1cos 280° + i sin 280°2
y U5
U5
x U50
U5P
3P U5 4
U5
P 3 P U5 4 r52
4 x U5P
U5
1
5P 4
x U50
2
U5
7P 4
12. Amplitude: 4; period: 2
ANSWERS Section 9.2
AN75
CHAPTER 9 Analytic Geometry 9.2 Assess Your Understanding (page 699) 8. 13, 2)
6. parabola 7. (c)
9. 13, 6)
19. y = 16x 20
V (0, 0)
13. E
15. H
17. C
23. y2 = - 8x
y
(4, 8) F (4, 0) 20 x
(6, 3)
(2, 4) F (2, 0)
10 x (6, 3)
(2, 3) 2, 1 3 3 2.5 x 1 1 D: y F 0, 3 3 V (0, 0)
V (2, 3) 10 x (6, 5)
(2, 3)
D: y 1
2 1 , 3 3
(2, 5)
V (1, 2)
37. Vertex: (0, 0); Focus: (0, 1); Directrix: y = - 1
D: x 1
F (0, 1)
2 x
(2, 1)
V (0, 0)
43. Vertex: 13, - 12; Focus: 5 3 a3, - b ; Directrix: y = 4 4
2.5 x
45. Vertex: 12, - 32; Focus: 14, - 32; Directrix: x = 0 y 2
y 1 V (3, 1) x F 3,
5 x F (0, 2) (0, 4)
5 4
D: x 0
(4, 8) F (4, 0)
D: x 4
V (0, 0) (4, 8)
10 x
47. Vertex: (0, 2); Focus: 1 - 1, 22; Directrix: x = 1 y (1, 4) 8
(4, 1) 8 x V (2, 3)
F (1, 2)
F (4, 3) (4, 7)
y
V (1, 1)
y 2.5
1 8 x 2.5 x
F 3 , 1 4
3 3 , 2 4
77. Cy2 + Dx = 0, C ≠ 0, D ≠ 0 Cy2 = - Dx D y2 = - x C
V (2, 8)
F 2, D: y
33 4
31 4
D: x 3 V (1, 2)
8 (1, 6) F (1, 2) 5 (1, 2) x
49. Vertex: 1 - 4, - 22; Focus: 1 - 4, - 12; Directrix: y = - 3 F (4, 1) y 5
V (0, 2) (6, 1) 2 x
3 31 51. Vertex: 1 - 1, - 12; Focus: a - , - 1 b ; 53. Vertex: 12, - 82; Focus: a2, - b ; 4 4 5 33 Directrix: x = Directrix: y = 4 4
41. Vertex: 1 - 1, 22; Focus: (1, 2); Directrix: x = - 3
D: x 1
(1, 0)
5 D: x 4 3, 1 2 4
8 (5, 4) (1, 4) V (3, 3) D: y 2 2 x
y
y 10
2.5
D: y 1
3
33. 1x + 3)2 = 4 1y - 3) F (3, 4) y
5 (0, 0)
39. Vertex: (0, 0); Focus: 1 - 4, 02; Directrix: x = 4
y
(2, 1) V (1, 2)
3 4
1, 1 2 x 1 D: y 2 2.5
F (2, 5)
35. 1y + 22 2 = - 8 1x + 12 y 2
D: x 2
(2, 2)
2.5
(2, 2) 1 1, 2 V (0, 0)
y 10
4
D: y
5 x
31. 1y + 2)2 = 4 1x + 1)
y y
(3, 6)
D: x 2
V (0, 0) (2, 4)
29. 1x - 2)2 = - 8 1y + 3)
F (3, 2)
1 F 0, 2 y
5
F (0, 3)
4 y 3
(3, 2)
25. x2 = 2y
y
10 V (0, 0) D: y 3
(4, 8)
27. x2 =
11. B
2
y D: x 4
10. y = - 2
21. x = - 12y
2
D: y 3
1 x (2, 1)
V (4, 2)
55. 1y - 12 2 = x
57. 1y - 12 2 = - 1x - 22 1 59. x2 = 4 1y - 12 61. y2 = 1x + 22 2 63. 1.5625 ft from the base of the dish, along the axis of symmetry 65. 1 in. from the vertex, along the axis of symmetry 67. 20 ft 69. 0.78125 ft 71. 4.17 ft from the base, along the axis of symmetry 73. 24.31 ft, 18.75 ft, 7.64 ft 2 2 75. (a) y = x + 630 315 (b) 567 ft: 119.7 ft; 478 ft: 267.3 ft; 308 ft: 479.4 ft (c) No
This is the equation of a parabola with vertex at 10, 02 and axis of symmetry the x-axis. D D The focus is a , 0 b ; the directrix is the line x = . The parabola opens to the right if 4C 4C D D 7 0 and to the left if 6 0. C C
AN76
ANSWERS Section 9.2 (a) If D ≠ 0, then the equation may be written as E 2 - 4CF E 2 D b = - ax b. ay + 2C C 4CD E 2 - 4CF E This is the equation of a parabola with vertex at a ,b 4CD 2C and axis of symmetry parallel to the x-axis. (b)–(d) If D = 0, the graph of the equation contains no points if E 2 - 4CF 6 0, is a single horizontal line if E 2 - 4CF = 0, and is two horizontal lines if E 2 - 4CF 7 0.
79. Cy2 + Dx + Ey + F = 0, C ≠ 0 Cy2 + Ey = - Dx - F D E F y = - x C C C E 2 D F E2 ay + b = - x + 2C C C 4C 2 E 2 D E 2 - 4CF ay + b = - x + 2C C 4C 2 y2 +
80. (0, 2), (0, −2), ( - 36, 0); symmetric with respect to the x-axis
81. {5}
5 289 8 289 289 289 8 ; cos u = ; csc u = ; sec u = ; cot u = 89 89 5 8 5
82. sin u =
83. -
2 210 3
9.3 Assess Your Understanding (page 708) 8. major 9. 10, - 5); 10, 5)
7. ellipse
17. Vertices: 1 - 5, 02, (5, 0)
Foci: 1 - 221, 0), 1 221, 0)
(0, 5)
(5, 0) (兹21, 0)
5 (0, 4) (3, 0) 5 x (0, 4)
(3, 0)
13. C
(0, 2 兹3)
(0, 5)
(3, 0) (4, 0) 5 x
(4, 0)
y
5 (0, 4)
y 5 (0, 4)
(3, 0)
(4, 0)
y 5
35.
3
(5, 0) (0, 3)
y2 x2 + = 1 4 13
(5, 0) x
(2, 0)
(4, 0)
(0, 兹13)
43. Center: 13, - 12; Vertices: 13, - 42, (3, 2); Foci: 1 3, - 1 - 25 2 , 1 3, - 1 + 25 2
45.
(3, 1 兹5 )
(0, 4)
(5, 6)
y 7 (5 2 兹3, 4)
(9, 4)
(1, 4) (5, 4) 1 x
(5 2 兹3, 4)
1x - 22 2
+
1y + 12 2
= 1
3 2 Center: 12, - 12; Vertices: 1 2 - 23 , - 1 2 ; 1 2 + 23 , - 1 2 ; Foci: 11, - 12, 13, - 12 (2, 1)
1x - 12 2
+
(2 兹3, 1) 5 x (3, 1) (2, 1 兹2)
1y + 22 2
= 1
4 9 Center: 11, - 22; Vertices: 11, - 52, 11, 12 ; Foci: 1 1, - 2 - 25 2 , 1 1, - 2 + 25 2 y 5
y 5 (2, 1 兹2)
(2 兹3, 1) (1, 1)
51.
(1, 2 兹5 ) (1, 2) (1, 2) (1, 5)
5 x (3, 2) (1, 2 兹5 )
4
+ 1y - 12 2 = 1
41. 1x - 12 2 +
47.
y2 4
= 1
1x + 22 2
+ 1y - 12 2 = 1 4 Center: 1 - 2, 12; Vertices: 1 - 4, 12, (0, 1);
Foci: 1 - 2 - 23 , 1 2 , 1 - 2 + 23 , 1 2 y 5 (2, 2) (4, 1) 5 (2 兹3 , 1) (2, 1)
53. x2 +
(2 兹3 , 1) (0, 1) x (2, 0)
1y + 22 2
= 1 4 Center: 10, - 22; Vertices:10, - 42, 10, 02 ; Foci: 1 0, - 2 - 23 2 , 1 0, - 2 + 23 2 y
(1, 1)
1x + 12 2
(0, 兹15 )
1y - 42 2 + = 1 16 4 Center: 1 - 5, 42; Vertices: 1 - 9, 42, 1 - 1, 42;
(3, 2)
(3, 4)
39.
(0, 兹5) (3, 0) 5 x (2, 0)
(0, 兹5)
= 1
1x + 52 2
(5, 2)
49.
y2 16
Foci: 1 - 5 - 2 23 , 4 2 , 1 - 5 + 2 23 , 4 2
x (5, 1)
(3, 0)
(3, 0) 5 x
y 5 (0, 兹15 ) (0, 4) (1, 0) (1, 0) 5 x
(2, 0) 5 x (0, 3)
y (3, 1 兹5 ) 5 (1, 1) 5 (3, 1)
37. x2 +
y2 x2 + = 1 9 5 y (2, 0) 5
(0, 5)
y 5 (0, 3) (0, 兹13)
(0, 3)
31.
(0, 4)
(0, 4)
y2 x2 + = 1 25 9
(0, 兹2)
(0, 5)
2
(2 兹2, 0) 5 x ( 兹6, 0)
(兹6, 0)
(0, 4)
(0, 4)
33.
(2 兹2 , 0)
(0, 4)
(3, 0) (5, 0) 4 x
(5, 0)
y 5 (0, 兹2)
y2 x2 + = 1 9 25
29.
y x2 + = 1 8 2 Vertices: 1 - 2 22 , 0 2 , 1 2 22 , 0 2 Foci: 1 - 26 , 0 2 , 1 26 , 0 2
(2, 0) 5 x
(2, 0)
y2 y2 x2 x2 + = 1; Vertices: 1 - 4, 02, 27. + = 1 16 16 25 16 (4, 0), 10, - 42, 10, 4); Focus: (0, 0) y
2
23.
y 5 (0, 4)
(0, 2 兹3)
25.
15. B
2
y x + = 1 4 16 Vertices: 10, - 4), 10, 4) Foci: 10, - 2 23), 10, 2 23)
21.
y
5
(兹21, 0) (5, 0) 5 x (0, 2)
12. 11, 4) 2
19. Vertices: 10, - 52, (0, 5) Foci: 10, - 4), 10, 4)
y (0, 2)
11. 1 - 2, - 3); 16, - 3)
10. 5; 3; x
(0, 0)
(0, 2 兹3)
3 x
(1, 2)
(1, 2) (0, 2)
(0, 2 兹3)
5
(0, 4)
ANSWERS Section 9.4
55.
1x - 22 2 25
+
1y + 22 2 21
y (0, 2) 3
10
x (7, 2) (4, 2) (2, 2)
= 1
9
y 9
59.
(4, 9) (4, 8)
1x - 22 2 y 5
(x - 1)2 9 (1, 2)
(1 兹10, 2) 5 x (1, 2) (1, 1)
+
1y - 2)2 9
= 1
65.
= 1
y (0, 4)
y 6 (1, 5)
6 (2, 0) 2.5 x
(2, 0)
(4, 2)
(2, 2)
7
(2, 1 兹7 )
7 x
63.
1y - 12 2
(2, 1 兹7 ) (5, 1) (6, 1) 7 x (2, 1)
(4, 6)
(4, 3)
+
16 (1, 1) (2, 1)
(4 兹5 , 6)
(4, 4)
+ 1y - 22 2 = 1
(1 兹10, 2)
1y - 62 2
+
5 (4 兹5 , 6)
y (2, 2) 5 (1, 3) (4, 2)
67.
1x - 42 2
6
(2, 2 兹21)
1x - 12 2
57.
(2, 2 兹21)
(3, 2)
61.
= 1
AN77
5 x (1, 1)
y2 x2 + = 1 71. 43.3 ft 73. 24.65 ft, 21.65 ft, 13.82 ft 75. 30 ft 100 36 77. The elliptical hole will have a major axis of length 2 241 in. and a minor axis of length 8 in. y2 x2 = 1 79. 91.5 million mi; + 8646.75 1932 2 y2 x2 81. Perihelion: 460.6 million mi; mean distance: 483.8 million mi; + = 1 233,524.2 1483.82 2
y
69.
(2, 0) 2 (2, 0) 2.5 x
(0, 8)
83. 35 million mi 85. 5 25 - 4 87. Ax2 + Cy2 + Dx + Ey + F = 0 A ≠ 0. C ≠ 0 Ax2 + Dx + Cy2 + Ey = - F D E b = -F Aax2 + x b + Cay2 + A Cy 2 D E 2 D2 E2 Aax + b + Cay + b = -F + + 2A 2C 4A 4C D2 E2 D E (a) If + - F is of the same sign as A (and C), this is the equation of an ellipse with center at a ,b. 4A 4C 2A 2C 2 2 D E D E + - F, the graph is the single point a ,b. (b) If 4A 4C 2A 2C D2 E2 (c) If + - F is of the opposite sign to A (and C), the graph contains no points since, in this case, the left side has a sign opposite that of the 4A 4C right side. 89. Zeros: 5 - 2 23, 5 + 2 23; x-intercepts: 5 - 2 23, 5 + 2 23 90. Domain: {x 0 x ≠ 5}; Horizontal asymptote: y = 2; Vertical asymptote: x = 5 91. 617.1 ft-lb 92. b ≈ 10.94, c ≈ 17.77, B = 38°
9.4 Assess Your Understanding (page 721) 7. hyperbola 8. transverse axis 9. b 10. 12, 42; 12, - 22 11. 12, 6); 12, - 4) 4 4 12. 4 13. 2; 3; x 14. y = - x; y = x 9 9 15. B 17. A
19. x2 -
y2 8
= 1
21.
y (0, 2兹2) 5 V1 (1, 0) V2 (1, 0) F2 (3, 0) F1 (3, 0) 5 x (0, 2兹2) y 2兹2 x
y2 16
-
x2 = 1 20
2 兹5 x 5 V2 (0, 4)
y
(2 兹5, 0)
y F2 (0, 6) 2 兹5 10 y x 5 (2 兹5, 0) 10 x
y 2兹2 x V1 (0, 4)
y2 x2 23. = 1 9 16 y
4 x 3
y2
(0, 4) y 10
V1 (3, 0) F1 (5, 0) (0, 4)
y
4 x 3
V2 (3, 0) 10 x F2 (5, 0)
F2 (0, 3 兹5 ) y 2 x y y 2x 10 V2 (0, 6) (3, 0) (3, 0) 10 x V1 (0, 6) F1 (0, 3 兹5 )
F1 (0, 6)
y2 x2 27. = 1 8 8
x2 25. = 1 36 9
y x V1 (2 兹2, 0) F1 (4, 0)
y 5
(0, 2 兹2 ) yx V2 (2 兹2, 0) 5 x F2 (4, 0) (0, 2 兹2 )
AN78
ANSWERS Section 9.4 2
29.
y x2 = 1 25 9 Center: 10, 02 Transverse axis: x-axis Vertices: 1 - 5, 02, 15, 02
2
31.
Foci: 1 - 234 , 0 2 , 1 234 , 0 2 3 Asymptotes: y = { x 5
y x2 = 1 4 16 Center: 10, 02 Transverse axis: x-axis Vertices: 1 - 2, 02, 12, 02 Foci: 1 - 2 25 , 0 2 , 1 2 25 , 0 2 Asymptotes: y = {2x y 2 x y 5
y 10 (0, 3)
3 y x 5 F2 ( 兹34, 0) 10 x
3 y x 5 F1 (兹34, 0) V1 (5, 0)
(0, 3)
- x2 = 1 9 Center: 10, 02 Transverse axis: y-axis Vertices: 10, - 32, 10, 32 Foci: 1 0, - 210 2 , 1 0, 210 2 Asymptotes: y = {3x y 3x
y 5
F2 (0, 兹10) V2 (0, 3)
(1, 0)
V2 (2, 0) 5 x F2 (2 兹5, 0)
F1 (2 兹5, 0)
y2
y 3 x
(0, 4) y 2x
V1 (2, 0)
V2 (5, 0)
33.
5 x (1, 0)
V1 (0, 3)
F1 (0, 兹10)
(0, 4)
35.
41.
y2
x2 = 1 25 25 Center: 10, 02 Transverse axis: y-axis Vertices: 10, - 52, 10, 52 Foci: 1 0, - 5 22 2 , 1 0, 5 22 2 Asymptotes: y = {x -
1x - 42 2 4
-
1y + 12 2 5
= 1
V2 (0, 5) y x
43.
47.
1x - 12 2 4
-
1y + 12 2
3 y 1 (x1) 2 (1, 1)
9
= 1
1y - 22 2
49.
Foci: 1 - 2, 2 - 25 2 , 1 - 2, 2 + 25 2 Asymptotes: y - 2 = {2 1x + 22 y 2 2(x 2) y y 2 2(x 2) F2 (2, 2 兹5 ) V2 (2, 4) (2, 2) 4 (1, 2) (3, 2) V1 (2, 0)
2 x F1 (2, 2 兹5 )
y4
兹3 (x 3) 3
8 x (3 2 兹3, 4) 兹3 y4 (x 3) 3
1x - 22 2
-
53.
-
y 10 V1 (4, 7)
1y + 22 2
= 1 4 4 Center: 1 - 1, - 22 Transverse axis: parallel to x-axis Vertices: 1 - 3, - 22, 11, - 22
Foci: 1 - 1 - 2 22 , - 2 2 , 1 - 1 + 2 22 , - 2 2 Asymptotes: y + 2 = { 1x + 12 (1, 0) y 4
y2x1
4 x F2 (1 2 兹2, 2) V1 (3, 2) V2 (1, 2) y 2 (x 1) (1, 4) (1, 2)
F1 (1 2 兹2, 2)
= 1
y 7 兹3 (x 5)
V2 (6, 7) F2 (7, 7)
F1 (3, 7)
(5, 7)
1y + 32 2
= 1 4 9 Center: 12, - 32 Transverse axis: parallel to x-axis Vertices: 10, - 32, 14, - 32
1x + 12 2
3
(5, 7 兹3 )
8
(5, 7 兹3 )
Foci: 1 2 - 213 , - 3 2 , 1 2 + 213 , - 3 2 3 Asymptotes: y + 3 = { 1x - 22 2
V2 (3, 1)
4 Center: 1 - 2, 22 Transverse axis: parallel to y-axis Vertices: 1 - 2, 02, 1 - 2, 42
y 6
1y - 72 2
45. 1x - 52 2 -
= 1
F1 (3, 8)
x F2 (1 兹13 , 1)
- 1x + 22 2 = 1
12
V1 (3, 6)
(1, 4)
51.
1x + 32 2
(3 2 兹3, 4)
5
V1 (1, 1)
4
-
V2 (3, 2)
3 y (1, 2) y 1 2 (x1) 4
F1 (1 兹13 , 1)
F1 (0, 5 兹2 )
F2 (3, 0)
兹5 y1 (x 4) 2
(4, 1 兹5 )
1y + 42 2
37. x2 - y2 = 1 y2 x2 39. = 1 36 9
(5, 0) 10 x
V1 (0, 5)
(4, 1) 9 x F2 (7, 1) V2 (6, 1)
V1 (2, 1)
F2 (0, 5 兹2 ) yx
(5, 0)
y (4, 1 兹5 ) 兹5 y1 (x 4) 2 4 F1 (1, 1)
y 10
y (2, 3)
x y 7 兹3 (x 5)
(2, 0)
y3
2
3 (x2) 2
6 x F2 (2 兹13 , 3) V2 (4, 3)
V1 (0, 3) F1 (2 兹13 , 3) (2, 6)
3 y 3 (x2) 2
55. 1x - 12 2 - 1y + 12 2 = 1 Center: 11, - 12 Transverse axis: parallel to x-axis Vertices: 10, - 12, 12, - 12
Foci: 1 1 - 22 , - 1 2 , 1 1 + 22 , - 1 2 Asymptotes: y + 1 = { 1x - 12 y 1 (x 1)
y 2 (1, 0)
F1 (1 兹2 , 1) V1 (0, 1)
y1x1 4 x F2 (1 兹2 , 1)
V2 (2, 1) (1, 2)
ANSWERS Section 9.4
57.
1y - 22 2
- 1x + 12 2 = 1
4 Center: 1 - 1, 22 Transverse axis: parallel to y-axis Vertices: 1 - 1, 02, 1 - 1, 42
59.
Foci: 1 - 1, 2 - 25 2 , 1 - 1, 2 + 25 2 Asymptotes: y - 2 = {2 1x + 12
V1 (1, 0)
F2 (1, 2 兹5 ) V2 (1, 4) (0, 2) 5 x
63.
(3, 6)
yx 10 x (5, 0)
(5, 0) 5 x
1y - 12 2
- 1x + 22 2 = 1 4 Center: 1 - 2, 12 Transverse axis: parallel to y-axis Vertices: 1 - 2, - 12, 1 - 2, 32 Foci: 1 - 2, 1 - 25 2 , 1 - 2, 1 + 25 2 Asymptotes: y - 1 = {2 1x + 22 F2 (2, 1 兹5) y y 1 2(x 2) 6 V2 (2, 3) (2, 1) (1, 1) (3, 1)
y 1 2(x 2)
4 x
V1 (2, 1)
(3, 2)
10 y x
y 2x
(0, 4)
61.
V2 (5, 2)
y
8 y 2x
= 1
V1 (1, 2)
65.
y
1y + 22 2
4 16 Center: 13, - 22 Transverse axis: parallel to x-axis Vertices: 11, - 22, 15, - 22 Foci: 1 3 - 2 25 , - 2 2 , 1 3 + 2 25 , - 2 2 Asymptotes: y + 2 = {2 1x - 32
F1 (1, 2 兹5 )
5
-
y 2 2(x 3) y 2 2(x 3) y (3, 2) 2 x 8 F1 (3 2兹5, 2) F2 (3 2兹5, 2)
y 2 2(x 1)
y 2 2(x 1) y (1, 2) (2, 2)
1x - 32 2
AN79
F1 (2, 1 兹5)
67. Center: 13, 02 Transverse axis: parallel to x-axis Vertices: 11, 0), 15, 0) Foci: 13 - 229, 0), 13 + 229, 0) 5 Asymptotes: y = { 1x - 3) 2 (3, 5)
y 10 (3, 0)
F1 (3 兹29, 0) 10 x
F2 (3 兹29, 0) V2 (1, 0)
69. Vertex: 10, 3); Focus: 10, 7); Directrix: y = - 1
71.
F (0, 7) y
1x - 5)
2
2
y + = 1 9 25 Center: 15, 02 ; Vertices: 15, 5), 15, - 5); Foci: 15, - 4), 15, 4)
73. 1x - 3)2 = 8 1y + 5) Vertex: 13, - 5); Focus: 13, - 3); Directrix: y = - 7 y
8 y V (0, 3) D: y 1
x
2
V1 (5, 5) 5 F1 (5, 4)
10
10 x
F (3, 3)
(5, 0) (8, 0) 9 x F2 (5, 4)
(2, 0)
V1 (5, 0) (3, 5)
V (3, 5) D: y 7
V2 (5, 5)
y2 x2 75. The fireworks display is 50,138 ft north of the person at point A. 77. The tower is 592.4 ft tall. 79. (a) y = {x (b) = 1, x Ú 0 100 100 81. If the eccentricity is close to 1, the “opening” of the hyperbola is very small. As e increases, the opening gets bigger. x2 1 x2 - y2 = 1; asymptotes y = { x 83. y2 1 y 4 4 2 2.5 1 2 (0, 1) y x x 1 2 y2 = 1; asymptotes y = { x (2, 0) (2, 0) 4 2 1.5
(0, 1)
x
x2 y2 1 4
1 y x 2
P, 3 2 2 3
If A and C are opposite in sign and F ≠ 0, this equation may be written as
x2
y2
= 1, F F b a- b A C F F where - and - are opposite in sign. This is the equation of a hyperbola with center 10, 02 . The A C F F transverse axis is the x-axis if 7 0; the transverse axis is the y-axis if 6 0. A A 3 p p y 88. c ≈ 13.16, A ≈ 31.6°, B ≈ 48.4° 87. Amplitude = ; Period = ; Phase shift = P 2 3 2 U5 89. 16, - 6 232 2 y 3P 2 2 P U5 90. x + (y 3) = 9; circle, U5 4 2.5 4 radius 3, center at (0, 3) in P 3 rectangular coordinates x
85. Ax2 + Cy2 + F = 0 Ax2 + Cy2 = - F
a-
+
x 1 2 3 4 5 6 U50
U5P
U5
U5
5P 4 U5
3P 2
7P 4
AN80
ANSWERS Section 9.5
9.5 Assess Your Understanding (page 730) 5. cot 12u2 =
A - C B
17. Hyperbola 19. Circle 25. x =
7. B2 - 4AC 6 0
6. parabola 21. x =
y 2.5 x (1, 0) 2.5 x
y
23. x =
13. Ellipse
15. Hyperbola
22 22 1x′ - y′2, y = 1x′ + y′2 2 2
25 25 1x′ - 2y′2, y = 12x′ + y′2 5 5
y
29. x =
213 213 13x′ - 2y′2, y = 12x′ + 3y′2 13 13
35. u = 60° (see Problem 25) x′2 + y′2 = 1 4 Ellipse Center at 10, 02 Major axis is the x′@axis. Vertices at 1 {2, 02
y
y 2.5 x
2.5 x
(0, 2) 2.5
(2, 0) 5 x
y
2.5 x
(2, 0)
39. u ≈ 34° (see Problem 29) 1x′ - 22 2 + y′2 = 1 4 Ellipse Center at 12, 02 Major axis is the x′@axis. Vertices at 14, 02 and 10, 02
x
(2, 0)
y
x (0, 2)
37. u ≈ 63°(see Problem 27) y′2 = 8x′ Parabola Vertex at 10, 02 Focus at 12, 02 y 5
27. x =
10. False 11. Parabola
33. u = 45° (see Problem 23) y′2 = 1 x′2 + 4 Ellipse Center at 10, 02 Major axis is the y′@axis. Vertices at 10, {22
31. u = 45° (see Problem 21) y′2 = 1 x′2 3 Hyperbola Center at origin Transverse axis is the x′@axis. Vertices at 1 {1, 02
(1, 0)
9. True
22 22 1x′ - y′2, y = 1x′ + y′2 2 2
1 1 1 x′ - 23y′ 2 , y = 1 23x′ + y′ 2 2 2
y
8. True
y 6 (2, 1) x (4, 0) 5 x (2, 1)
41. cot 12u2 =
7 ; 24 3 u = sin-1 a b ≈ 37° 5 1x′ - 12 2 = - 6 ay′ -
1 b 6
Parabola
1 Vertex at a1, b 6 4 Focus at a1, - b 3 y y
5
1,
1 6 x 5 x 1,
4 3
43. Hyperbola 45. Hyperbola 47. Parabola 49. Ellipse 51. Ellipse 53. Refer to equation (6): A′ = A cos2 u + B sin u cos u + C sin2 u B′ = B1cos2 u - sin2 u2 + 2 1C - A2 1sin u cos u2 C′ = A sin2 u - B sin u cos u + C cos2 u D′ = D cos u + E sin u E′ = - D sin u + E cos u F′ = F 55. Use Problem 53 to find B′2 - 4A′C′. After much cancellation, B′2 - 4A′C′ = B2 - 4AC. 57. The distance between P1 and P2 in the x′y′@plane equals 2 1x2 = - x1 = 2 2 + 1y2 = - y1 = 2 2 . Assuming that x′ = x cos u - y sin u and y′ = x sin u + y cos u, then 1x2 = - x1 = 2 2 = 1x2 cos u - y2 sin u - x1 cos u + y1 sin u2 2 = cos2 u1x2 - x1 2 2 - 2 sin u cos u1x2 - x1 2 1y2 - y1 2 + sin2 u 1y2 - y1 2 2, and = = 2 1y2 - y1 2 = 1x2 sin u + y2 cos u - x1 sin u - y1 cos u2 2 = sin2 u 1x2 - x1 2 2 + 2 sin u cos u1x2 - x1 2 1y2 - y1 2 + cos2 u 1y2 - y1 2 2. Therefore, 1x2 = - x1 = 2 2 + 1y2 = - y1 = 2 2 = cos2 u1x2 - x1 2 2 + sin2 u 1x2 - x1 2 2 + sin2 u 1y2 - y1 2 2 + cos2 u 1y2 - y1 2 2 = 1x2 - x1 2 2 1cos2 u + sin2 u2 + 1y2 - y1 2 2 1sin2 u + cos2 u2 = 1x2 - x1 2 2 + 1y2 - y1 2 2. 61. A ≈ 39.4°, B ≈ 54.7°, C ≈ 85.9°
62. 38.5
63. r 2cos u sin u = 1
64. 229(cos 291.8° + i sin 291.8°)
9.6 Assess Your Understanding (page 736) 3. conic; focus; directrix 4. 1; 6 1; 7 1 5. True 6. True 7. Parabola; directrix is perpendicular to the polar axis 1 unit to the right of the pole. 4 9. Hyperbola; directrix is parallel to the polar axis units below the pole. 3 3 11. Ellipse; directrix is perpendicular to the polar axis units to the left of the pole. 2
ANSWERS Section 9.7 15. Ellipse; directrix is parallel to the polar 8 axis units above the pole; vertices are at 3 8 p 3p a , b and a8, b. 7 2 2
13. Parabola; directrix is perpendicular to the polar axis 1 unit to the right of the pole; 1 vertex is at a , 0 b . 2 y Directrix P 2
1,
1 ,0 2
2
2
3P 2
1,
8,
(4, 0)
Polar x axis 4 8 3P (4, P) , Directrix 3 2
(2, P)
Directrix y P 5 3, 2 (23, 0) (1, P) Polar 5 x axis 3P 3, 2
23. Ellipse; directrix is perpendicular to the polar axis 6 units to the left of the pole; vertices are at (6, 0) and 12, p2. Directrix y 5 (2, P)
3,
P 2
(6, 0) 4
Polar x axis
(2, 0)
Polar 5 x axis
3,
3P 2
Directrix
6 3P , 5 2
25. y2 + 2x - 1 = 0
P 6, 2
y 7
4
17. Hyperbola; directrix is perpendicular to the 3 polar axis units to the left of the pole; 2 vertices are at 1 - 3, 02 and 11, p2.
3P 2
y P 2
(2, 0) Polar 5 x axis
21. Ellipse; directrix is parallel to the polar axis 3 units below the pole; vertices are at p 6 3p a6, b and a , b. 2 5 2
19. Ellipse; directrix is parallel to the polar axis 8 units below the pole; vertices are at 8 3p p b. a8, b and a , 2 3 2
8,
8 P , 7 2
Directrix y (2, P) 2
Polar x axis
27. 16x2 + 7y2 + 48y - 64 = 0
29. 3x2 - y2 + 12x + 9 = 0 31. 4x2 + 3y2 - 16y - 64 = 0 1 12 12 39. r = 41. r = 33. 9x2 + 5y2 - 24y - 36 = 0 35. 3x2 + 4y2 - 12x - 36 = 0 37. r = 1 + sin u 5 - 4 cos u 1 - 6 sin u 43. Use d 1D, P2 = p - r cos u in the derivation of equation (a) in Table 5. 45. Use d 1D, P2 = p + r sin u in the derivation of equation (a) in Table 5. p 5p 47. 27.81 48. Amplitude = 4; Period = 10p 49. e , p, f 50. 26 3 3
9.7 Assess Your Understanding (page 746) 2. plane curve; parameter 3. ellipse 7.
4. cycloid 9.
y
5. False
6. True 11.
y 5
13.
y 10
y 5
6 x
3
17.
y 5
x = 3 1y - 12 2
y = x - 8
y = 2x - 2
x - 3y + 1 = 0 15.
10 x
x
5 x
12
19.
y 3
21.
y 5
3 x
y 5
5 x
5 x
5 x
23.
25.
y
y 1
2 (兹2 , 1)
27. x y 29. x y 31. x y 33. x y
back and forth twice 1 x
x
2
x - y = 1 2
y2 x2 + = 1 4 9
y = x3
2y = 2 + x
x + y = 1
2
= = = = = = = =
t or 4t - 1 t or t2 + 1 t or t3 t or t 2/3, t Ú 0
35. x = t + 2, y = t, 0 … t … 5 37. x = 3 cos t, y = 2 sin t, 0 … t … 2p 39. x = 2 cos 1pt2, y = - 3 sin 1pt2, 0 … t … 2 41. x = 2 sin 12pt2, y = 3 cos 12pt2, 0 … t … 1 y y y 43. y (4, 16) 20
(4, 16)
(1, 1) (0, 0) C1
(1, 1) 5 x
16
20
(1, 1)
(1, 1) C2
2.5 x
AN81
(4, 16)
C3
(1, 1) 5 x
20
(4, 16)
C4
(1, 1) 5 x
y2 x2 + = 1 4 9 t + 1 x = 4 y = t x = t3 y = t6 + 1 3 x = 2 t y = t x = t3 y = t 2, t Ú 0
AN82
ANSWERS Section 9.7
45.
47.
7
4
6 5
6
9
6
5
49. (a) x = 3 y = - 16t 2 + 50t + 6 (b) 3.24 s (c) 1.56 s; 45.06 ft (d) 50
53. (a) x = 1145 cos 20°2t y = - 16t 2 + 1145 sin 20°2t + 5 (b) 3.20 s (c) 435.65 ft (d) 1.55 s; 43.43 ft (e) 170
51. (a) Train: x1 = t 2, y1 = 1; Bill: x2 = 5 1t - 52, y2 = 3 (b) Bill won't catch the train. (c) 5
Bill Train 0 0
100
0
440
0
5 0
120
55. (a) x = 140 cos 45°2t y = - 4.9t 2 + 140 sin 45°2t + 300 (b) 11.23 s (c) 317.52 m (d) 2.89 s; 340.82 m (e) 400
57. (a) Camry: x = 40t - 5, y = 0; Impala: x = 0, y = 30t - 4 (d) 0.2 mi; 7.68 min (b) d = 2 140t - 52 2 + 130t - 42 2 (e) Turn axes off to see the graph: (c) 7 4
6 0
6
0.2 0 4
160
320 0
22 22 y t, y = - 16t 2 + y t + 3 (b) Maximum height is 139.1 ft. (c) The ball is 272.25 ft from home plate. 2 0 2 0 (d) Yes; the ball will clear the wall by about 99.5 ft. 61. The orientation is from 1x1 , y1 2 to 1x2 , y2 2.
59. (a) x = 65.
66.
y 6
y 3
y 5 sin
x 2
4 2 2P 26 24 22 22
2
4
x
6 x y 5 2 cos(2x)
24
67. Approximately 2733 miles 68. (a) Simple harmonic (b) 2 ft p (c) s 2 2 (d) oscillation/s p
26
Review Exercises (page 751) 1. Parabola; vertex (0, 0), focus (0, 3), directrix y = - 3 2. Hyperbola; center (0, 0), vertices (3, 0) and ( - 3, 0), foci (5, 0) and ( - 5, 0), asymptotes 4 4 y = x and y = - x 3. Ellipse; center (0, 0), vertices (6, 0) and ( - 6, 0), foci 13 23, 02 and 1 - 3 23, 02 4. Parabola; vertex (3, 1), 3 3 focus (5, 1), directrix x = 1
5. Hyperbola; center (0, 0), vertices 1 22 , 0 2 and 1 - 22 , 0 2 , foci 1 210 , 0 2 and 1 - 210 , 0 2 ,
3 5 7 3 asymptotes y = 2x and y = - 2x 6. Parabola; vertex a - , 3 b , focus a - , b , directrix y = 2 2 2 2 y - 1 x + 1 7. Hyperbola; center ( - 1, 1), vertices (3, 1) and ( - 5, 1), foci ( - 6, 1) and (4, 1), asymptotes = { 4 3
8. Ellipse; center (1, - 3), vertices 11, 2 23 - 32 and 11, - 2 23 - 32 , foci 11,22 - 32 and 111, - 22 - 322 1 1 1 9. Parabola; vertex a , - 1 b , focus a , - 1 b directrix x = 2 2 2
10. Ellipse; center 11, - 12, vertices (1, 2) and 11, - 42, foci 1 1, - 1 + 25 2 and 1 1, - 1 - 25 2
ANSWERS Review Exercises
11. y2 = - 8x
12. y 5
5 x
V (0, 0) (2, 4)
y 2
1x + 12 2
-
(4, 4)
9
1y - 22 2 7
18.
V2 (2, 2) F2 (3, 2) 8 x
(1, 2) y 2 兹7 (x 1) (1, 2 兹7 ) 3
y′2
y2 x2 + = 1 16 7
= 1
y (0, 兹7) 5 V1 (4, 0)
1x - 32 2
-
9
5 x
16.
1x + 42 2
9 Hyperbola Center at the origin Transverse axis is the x′@axis. Vertices at 1 {1, 02 y 5
y
16
1y - 12 2 4
x 9
= 1
= 1
(0, 5) F1 (4, 2) V1 (4, 0)
19. Parabola 20. Ellipse 21. Parabola 22. Hyperbola 23. Ellipse
4 213 x′ 13
26. y′2 = -
Parabola Vertex at the origin Focus on the x′@axis at a -
x
(0, 2) 2
25
(8, 5)
213 , 0b 13
y 5
y x
1y - 52 2
V2 (1, 3)
y′ x′2 + = 1 2 4 Ellipse Center at origin Major axis is the y′@axis. Vertices at 10, {22
(1, 0) 5 x
(1, 0)
+
y V2 (4, 10) 10 F2 (4, 8) (4, 5)
2 x F2 (0, 3)
y (3, 1) y 1 2 (x 3) 5 (3, 3) 3 V2 (6, 1) V1 (0, 1) F2 (3 兹13, 1) F1 (3 兹13, 1) x 5 2 y1 (x 3) (3, 1) 3
25.
F2 (3, 0)
(0, 兹7 )
2
= 1
V2 (4, 0)
F1 (3, 0)
(2, 3 兹3 )
y 2 兹7 (x 1) y (1, 2 兹7 ) 3 12
24. x′2 -
3
(2, 3)
= 1
V1 (4, 2) F1 (5, 2)
1y + 32 2
V1 (3, 3) F1 (4, 3)
F (2, 4)
17.
13.
y 3 兹3 (x 2) y y 3 兹3 (x 2) 2 (2, 3 兹3 )
8 x V (2, 3)
(0, 4)
x2 = 1 12
15. 1x + 22 2 -
14. 1x - 22 2 = - 4 1y + 32
D: y 2
4
-
V2 (0, 2) y 5 F2 (0, 4) 兹3 y x 兹3 y 3 x 3 5 x (2兹3, 0) (2兹3, 0) V1 (0, 2) F1 (0, 4)
D: x 2
(2, 4) F (2, 0)
y2
AN83
y
x
y
5 x
x (0, 2)
27. Parabola; directrix is perpendicular to the polar axis 4 units to the left of the pole; vertex is 12, p2. Directrix y 5
28. Ellipse; directrix is parallel to the polar axis 6 units below the pole; p 3p vertices are a6, b and a2, b. 2 2
P 2 (2, P) Polar 5 x axis 3P 4, 2
y 5
4,
6,
(3, P) 2,
29. Hyperbola; directrix is perpendicular to the polar axis 1 unit to the right of the pole; 2 vertices are a , 0 b and 1 - 2, p2. 3
P 2
2,
2 ,0 3
(3, 0) Polar 5 x axis
3P 2
P 2
2,
y
Directrix (22, P) Polar 3 x axis
1 3P 2
Directrix
30. x = 4 1y + 12 2
32.
31. 3x - y - 8x + 4 = 0 2
2
33.
y (2, 1) 2 (2, 0) 2
x
34.
y (0, 6) 7 (3, 2)
2 (0, 2) (3, 2) 5 x (0, 2)
x + 4y = 2
y
1y - 22 2 x2 + = 1 9 16
(2, 1) 2 (1, 0)
x
1 + y = x
p p t - 4 , y = t, - q 6 t 6 q 36. x = 4 cos a t b , y = 3 sin a t b , 0 … t … 4 35. x = t, y = - 2t + 4, - q 6 t 6 q ; x = -2 2 2 2 y x 1 38. The ellipse + = 1 39. ft, or 3 in. 40. 19.72 ft, 18.86 ft, 14.91 ft 41. 450 ft 16 7 4
37.
y2 x2 = 1 5 4
AN84
ANSWERS Review Exercises 43. (a) x = 180 cos 35°2t y = - 16t 2 + 180 sin 35°2t + 6 (b) 2.9932 s (c) 1.4339 s; 38.9 ft (d) 196.15 ft (e) 50
3 2 t Mary: x2 = 6 1t - 22 2 y1 = 1 y2 = 3 (b) Mary won't catch the train. (c) 5
42. (a) Train: x1 =
0 0
250
100 t8
0
50
Chapter Test (page 752) 1. Hyperbola; Center: 1 - 1, 02; Vertices: 1 - 3, 02 and 11, 02; Foci: 1 - 1 - 213 , 0 2 and 1 - 1 + 213 , 0 2 ; 3 3 Asymptotes: y = - 1x + 12 and y = 1x + 12 2 2 3 1 5 2. Parabola; Vertex: a1, - b ; Focus: a1, b ; Directrix: y = 2 2 2 3. Ellipse; Center: 1 - 1, 12; Foci: 1 - 1 - 23 , 1 2 and
1 - 1 + 23 , 1 2 ; Vertices: 1 - 4, 12 and 12, 12 y2 x2 + = 1 5. 7 16
4. 1x + 12 2 = 6 1y - 32 F (1, 4.5)
y
V (1, 3)
1y - 22 2 4
y 5 V1 (0, 4)
9
(4, 4.5)
6.
(兹7, 0) F1
(2, 4.5) D: y 1.5
F2
5 x
y 8
(兹7, 0) 5 x
1x - 22 2 8 F1
(2 2兹2, 2)
= 1
V1 (2, 4) (2 兹10, 5) (2, 2) (2 2兹2, 2) 8 x
V2 (0, 4)
7. Hyperbola 8. Ellipse
-
F2
V2 (2, 0)
9. Parabola
10. x′2 + 2y′2 = 1. This is the equation of an ellipse with center at 10, 02 in the x′y′@plane. The vertices are at 1 - 1, 02 and 11, 02 in the x′y′@plane.
y 5 y
x
11. Hyperbola; 1x + 22 2 2
y2 3
= 1
1 5 x
12. y = 1 -
B
x + 2 3
y 5 (2, 1)
(1, 0) (10, 1)
13. The microphone should be located
45 x (25, 2)
2 ft (or 8 in.) from the base of the reflector, along its axis of symmetry. 3
Cumulative Review (page 752) 1 2. e - 5, - , 2 f 3. {x - 3 … x … 2} or [ - 3, 2] 4. (a) Domain: 1 - q , q 2; Range: 12, q 2 3 y2 x2 (b) y = log 3 1x - 22; Domain: 12, q 2; Range: 1 - q , q 2 5. (a) 518 6 (b) (2, 18] 6. (a) y = 2x - 2 (b) (x - 2)2 + y2 = 4 (c) + = 1 9 4 2 x p 5p p (d) y = 2(x - 1)2 (e) y2 = 1 (f) y = 4x 7. u = { pk, k is any integer; u = { pk, k is any integer 8. u = 3 12 12 6 3p x2 y 9. r = 8 sin u 10. e x ` x ≠ { pk, k is an integer f 11. 522.5° 6 12. y = + 5 10 4 5 1. - 6x + 5 - 3h
10 x
ANSWERS Section 10.2
AN85
CHAPTER 10 Systems of Equations and Inequalities 10.1 Assess Your Understanding (page 765) 3. inconsistent 4. consistent; independent 5. 13, - 22 1 3 122 - 4 a b = 4 2 122 - 1 - 12 = 5 2 7. e 9. d 5 122 + 2 1 - 12 = 8 1 1 1 122 - 3 a b = 2 2 2 17. x = 6, y = 2; 16, 22 27. x =
3 3 , y = 3; a , 3 b 2 2
33. x =
3 3 , y = 1; a , 1 b 2 2
6. consistent; dependent
3 112 + 3 1 - 12 + 2 122 = 4 3 122 + 3 1 - 22 + 2 122 = 4 4 - 1 = 3 13. c 1 - 1 - 12 2 = 0 15. c 2 - 3 1 - 22 + 2 11. c 1 = 10 142 + 1 = 3 2 1 - 12 - 3 122 = - 8 5 122 - 2 1 - 22 - 3 122 = 8 2
1 1 1 1 , y = - ; a , - b 25. Inconsistent 3 6 3 6 4 - x 29. 5 1x, y2 x = 4 - 2y, y is any real number 6 or b 1x, y2 ` y = , x is any real number r 31. x = 1, y = 1; 11, 12 2
19. x = 3, y = - 6; 13, - 62
35. x = 4, y = 3; 14, 32
43. x = 2, y = - 1, z = 1; 12, - 1, 12
21. x = 8, y = - 4; 18, - 42
37. x =
4 1 4 1 ,y = ; a , b 3 5 3 5
23. x =
39. x =
1 1 1 1 ,y = ; a , b 5 3 5 3
41. x = 8, y = 2, z = 0; 18, 2, 02
47. 5 1x, y, z2 x = 5z - 2, y = 4z - 3; z is any real number 6 49. Inconsistent 1 1 51. x = 1, y = 3, z = - 2; 11, 3, - 22 53. x = - 3, y = , z = 1; a - 3, , 1 b 55. Length 30 ft; width 15 ft 57. There were 20 commercial launches 2 2 and 58 noncommercial launches in 2012. 59. 22.5 lb 61. Smartphone: $325; tablet: $640 63. Average wind speed 25 mph; average airspeed 175 mph 65. 80 $25 sets and 120 $45 sets 67. $9.96 69. Mix 50 mg of first compound with 75 mg of second. 10 65 55 4 5 ,I = ,I = 71. a = , b = - , c = 1 73. Y = 9000, r = 0.06 75. I1 = 3 3 71 2 71 3 71 77. 100 orchestra, 210 main, and 190 balcony seats 79. 1.5 chicken, 1 corn, 2 milk 45. Inconsistent
81. If x = price of hamburgers, y = price of fries, and z = price of colas, then 41 1 x = 2.75 - z, y = + z, +0.60 … z … +0.90. 60 3 There is not sufficient information; see table.
x
$2.13
$2.01
$1.86
y
$0.89
$0.93
$0.98
z
$0.62
$0.74
$0.89
83. It will take Beth 30 hr, Bill 24 hr, and Edie 40 hr. y 87. 6 y=2
(1, 1) 26
6 x
(0, 21)
26
88. (a) 2(2x - 3)3(x3 + 5)(10x3 - 9x2 + 20) 3
1
(b) 7(3x - 5)- 2 (x + 3)- 2 p 90. 2(cos 150° + i sin 150°) 89. 9
10.2 Assess Your Understanding (page 781) 1. matrix 1 5. c 4
2. augmented
-5 2 5 d 3 6
2 7. c 4
3. third; fifth 3 2 6 d -6 -2
112 ; 122
17. e
x - 3y = - 2 2x - 5y = 5
21. c
x - 3y + 2z = - 6 2x - 5y + 3z = - 4 - 3x - 6y + 4z = 6
c
1 0
112 122 ; 132
0.01 9. c 0.13 -3 -2 ` d 1 9 1 C0 0
x = 5 y = -1 Consistent; x = 5, y = - 1 or 15, - 12
25. e
4. True
-3 1 - 15
- 0.03 2 0.06 d 0.10 0.20
19. W
1 11. C 3 1
-1 3 1
x - 3y + 4z = 3 3x - 5y + 6z = 6 - 5x + 3y + 4z = 6
2 -6 -1 3 8S 10 - 12
23. c
1 10 0 3 5S 2 2 112 122 ; 132
1 C0 0
1 13. C 3 5 -3 4 - 12
5x - 3y + z = - 2 2x - 5y + 6z = - 2 - 4x + y + 4z = 6
x = 1 27. c y = 2 0 = 3 Inconsistent
1 -2 3
-1 2 0 3 2S -1 1
1 2 15. D -3 4
4 3 -6 3 -3S 24 21
112 122 ; 132
1 C2 0
7 -5 -9
- 11 2 6 3 -2S 16 2
-1 1 4 -5
- 1 10 2 4 -1 T 0 5 1 0
AN86
ANSWERS Section 10.2 x1 + 4x4 = 2 x1 = 1 x1 + x4 = - 2 31. c x2 + x4 = 2 33. c x2 + x3 + 3x4 = 3 x2 + 2x4 = 2 35. d x3 + 2x4 = 3 0 = 0 x3 - x4 = 0 Consistent: Consistent: 0 = 0 x1 = 2 - 4x4 x1 = 1, x2 = 2 - x4 Consistent: c x3 = 3 - 2x4 c x2 = 3 - x3 - 3x4 x1 = - 2 - x4 x4 is any real number or x3, x4 are any real numbers or x = 2 - 2x4 μ 2 5 1x1, x2, x3, x4 2 x1 = 1, 5 1x1, x2, x3, x4 2 x1 = 2 - 4x4, x3 = x4 x2 = 2 - x4, x3 = 3 - 2x4, x4 is x2 = 3 - x3 - 3x4, x3, x4 are any x4 is any real number or any real number 6 real numbers 6 5 1x1, x2, x3, x4 2 x1 = - 2 - x4, x2 = 2 - 2x4, x3 = x4, x4 is any real number 6
x + 2z = - 1 29. c y - 4z = - 2 0 = 0 Consistent: x = - 1 - 2z c y = - 2 + 4z z is any real number or 5 1x, y, z2 x = - 1 - 2z, y = - 2 + 4z, z is any real number 6
3 1 3 1 , y = ; a , b 41. x = 4 - 2y, y is any real number; 5 1x, y2 x = 4 - 2y, y is any real number 6 2 4 2 4 3 3 4 1 4 1 , y = 1; a , 1 b 45. x = , y = ; a , b 47. x = 8, y = 2, z = 0; 18, 2, 02 49. x = 2, y = - 1, z = 1; 12, - 1, 12 51. Inconsistent 2 2 3 5 3 5 5z - 2, y = 4z - 3, where z is any real number; 5 1x, y, z2 x = 5z - 2, y = 4z - 3, z is any real number 6 55. Inconsistent 1 1 1 2 1 2 1, y = 3, z = - 2; 11, 3, - 22 59. x = - 3, y = , z = 1; a - 3, , 1 b 61. x = , y = , z = 1; a , , 1 b 2 2 3 3 3 3 1, y = 2, z = 0, w = 1; 11, 2, 0, 12 65. y = 0, z = 1 - x, x is any real number; 5 1x, y, z2 y = 0, z = 1 - x, x is any real number 6
37. x = 6, y = 2; 16, 22 43. x = 53. x = 57. x = 63. x =
39. x =
67. x = 2, y = z - 3, z is any real number; 5 1x, y, z2 x = 2, y = z - 3, z is any real number 6
13 7 19 13 7 19 ,y = ,z = ;a , , b 9 18 18 9 18 18 2 8 13 7 8 7 3 7 3 2 7 13 71. x = - z - w, y = - + z + w, where z and w are any real numbers; e 1x, y, z, w2 ` x = - z - w, y = - + z + w, 5 5 5 5 5 5 5 5 5 5 5 5 z and w are any real numbers r
69. x =
75. f 1x2 = 3x3 - 4x2 + 5
73. y = - 2x2 + x + 3
77. 1.5 salmon steak, 2 baked eggs, 1 acorn squash 44 16 28 , I = 2, I3 = ,I = 79. $4000 in Treasury bills, $4000 in Treasury bonds, $2000 in corporate bonds 81. 8 Deltas, 5 Betas, 10 Sigmas 83. I1 = 23 2 23 4 23 85. (a) (b) (c) All the money invested at 7% provides $2100, Amount Invested At Amount Invested At more than what is required. 7% 9% 11% 7% 9% 11%
87.
0
10,000
10,000
12,500
12,500
0
1000
8000
11,000
14,500
8500
2000
2000
6000
12,000
16,500
4500
4000
3000
4000
13,000
18,750
0
6250
4000
2000
14,000
5000
0
15,000
92. {x x is any real nunber} or ( - q , q ) 93. {x 0 - 1 6 x 6 6} or ( - 1, 6) 94. y 9
First Supplement
Second Supplement
Third Supplement
50 mg
75 mg
0 mg
36 mg
76 mg
8 mg
22 mg
77 mg
16 mg
8 mg
78 mg
24 mg
12 35 , 2 2
(23, 5)
13, 78 2 y=2
26
(1, 0)
12 12 , 0 2
6 x
(0, 21) x = 21
95. 2.42
10.3 Assess Your Understanding (page 792) 1. ad - bc
2. 2
5 -3
3. False 4. False 5. False 6. False 7. 22 9. - 2 11. 10 13. - 26 15. x = 6, y = 2; 16, 22
32 -4
19. x = 8, y = - 4; 18, - 42 29. x =
3 3 , y = 1; a , 1 b 2 2
21. x = 4, y = - 2; 14, - 22
31. x =
4 1 4 1 ,y = ;a , b 3 5 3 5
23. Not applicable
= = = =
1 3 1 3 ,y = ;a , b 2 4 2 4
33. x = 1, y = 3, z = - 2; 11, 3, - 22
37. Not applicable 39. x = 0, y = 0, z = 0; (0, 0, 0) 41. Not applicable 57. 1y1 - y2 2x - 1x1 - x2 2y + 1x1y2 - x2y1 2 1y1 - y2 2x + 1x2 - x1 2y 1x2 - x1 2y - 1x2 - x1 2y1 1x2 - x1 2 1y - y1 2
25. x =
43. - 4
45. 12
0 x2y1 - x1y2 1y2 - y1 2x + x2y1 - x1y2 - 1x2 - x1 2y1 1y2 - y1 2x - 1y2 - y1 2x1 y2 - y1 y - y1 = 1x - x1 2 x2 - x1
27. x =
35. x = - 3, y = 47. 8
17. x = 3, y = 2; 13, 22
1 2 1 2 ,y = ;a , b 10 5 10 5
1 1 , z = 1; a - 3, , 1 b 2 2
13 55. 0 or - 9 11 59. The triangle has an area of 5 square units. 49. 8
51. - 5
53.
AN87
ANSWERS Section 10.4 61. 50.5 square units 63. (x - 3)2 + (y + 2)2 = 25 65. If a = 0, we have If b = 0, we have by = s ax = s cx + dy = t cx + dy = t s Since D = ad ≠ 0, then Thus, y = and b a ≠ 0 and d ≠ 0. t - dy tb - ds s x = = Thus, x = and c bc a t - cx ta - cs Using Cramer’s Rule, we get y = = d ad sd - tb tb - sd x = = Using Cramer’s Rule, we get - bc bc sd s - sc s x = = y = = ad a - bc b ta - cs y = ad a11 67. 3 ka21 a31
a12 ka22 a32
If c = 0, we have ax + by = s dy = t Since D = ad ≠ 0, then a ≠ 0 and d ≠ 0. t Thus, y = and d s - by sd - bt x = = a ad Using Cramer’s Rule, we get sd - bt x = ad at t y = = ad d
If d = 0, we have ax + by = s cx = t Since D = - bc ≠ 0, then b ≠ 0 and c ≠ 0. t Thus, x = and c s - ax sc - at y = = b bc Using Cramer’s Rule, we get - tb t x = = - bc c at - sc sc - at y = = - bc bc
a13 ka23 3 = - ka21 1a12a33 - a32a13 2 + ka22 1a11a33 - a31a13 2 - ka23 1a11a32 - a31a12 2 a33 a11 = k[ - a21 1a12a33 - a32a13 2 + a22 1a11a33 - a31a13 2 - a23 1a11a32 - a31a12 2] = k 3 a21 a31
69. 3
a11 + ka21 a21 a31
a12 + ka22 a22 a32
a12 a22 a32
a13 a23 3 a33
a13 + ka23 3 = 1a11 + ka21 2 1a22a33 - a32a23 2 - 1a12 + ka22 2 1a21a33 - a31a23 2 + 1a13 + ka23 2 1a21a32 - a31a22 2 a23 a33 = a11a22a33 - a11a32a23 + ka21a22a33 - ka21a32a23 - a12a21a33 + a12a31a23 - ka22a21a33 + ka22a31a23 + a13a21a32 - a13a31a22 + ka23a21a32 - ka23a31a22 = a11a22a33 - a11a32a23 - a12a21a33 + a12a31a23 + a13a21a32 - a13a31a22 = a11 1a22a33 - a32a23 2 - a12 1a21a33 - a31a23 2 + a13 1a21a32 - a31a22 2 a11 a12 a13 = 3 a21 a22 a23 3 a31 a32 a33 72.
70. v = 9i - 4j; 297 1 5 71. { , { , {1, {2, {5, {10 2 2
73. 0
y 5
5 x (22, 23) (0, 23) (21, 24)
Historical Problem (page 808) 1. (a) 2 - 5i · c
-5 1 d , 1 + 3i · c 2 -3
2 5
3 d 1
(b) c
-5 1 dc 2 -3
2 5
3 17 d = c 1 -1
1 d 17
(c) 17 + i
a b a -b a 2 + b2 0 dc d = c d ; the product is a real number. -b a b a 0 b2 + a 2 ka + lc 3. (a) x = k 1ar + bs2 + l 1cr + ds2 = r 1ka + lc2 + s 1kb + ld2 (b) A = c ma + nc y = m1ar + bs2 + n 1cr + ds2 = r 1ma + nc2 + s 1mb + nd2
(d) 17 + i
2. c
kb + ld d mb + nd
10.4 Assess Your Understanding (page 808) 1. square 9. c
4 -1
- 13 23. C - 18 17
2. True 4 5
-5 d 4 7 10 -7
3. columns; rows 4. False 11. c
- 12 - 14 S 34
0 4
12 8
-2 25. c 2
- 20 d 24
4 1
13. c
2 4
8 d 6
5. inverse -8 7
7 0
5 27. c 9
- 15 d 22
14 d 16
7. True 8. A- 1B 1 14 28 - 9 15. c d 17. C 2 22 4 23 3 0
6. singular
9 29. C 34 47
2 13 S 20
1 31. c -1
-1 d 2
- 14 - 18 S 28 33. £
15 19. C 22 - 11 1 -1
-
5 2§ 3
21 34 7
- 16 - 22 S 22 1
35. D -1
21. c
-
1 a T 2 a
25 4
-9 d 20
AN88
ANSWERS Section 10.4
-3 2 5
3 37. C - 2 -4
1 -1S -2
39. F
1 7 1 7 2 7
5 7 9 7 3 7
3 7 4 - V 7 1 7
41. x = 3, y = 2; 13, 22
43. x = - 5, y = 10; 1 - 5, 102
45. x = 2, y = - 1; 12, - 12
1 1 2 3 2 3 , y = 2; a , 2 b 49. x = - 2, y = 1; 1 - 2, 12 51. x = , y = ; a , b 53. x = - 2, y = 3, z = 5; 1 - 2, 3, 52 2 2 a a a a 1 1 1 1 34 85 12 34 85 12 1 2 1 2 55. x = , y = - , z = 1; a , - , 1 b 57. x = - , y = ,z = ; a - , , b 59. x = , y = 1, z = ; a , 1, b 2 2 2 2 7 7 7 7 7 7 3 3 3 3
47. x =
4 61. c 2 -3 65. C 1 1
-1 1 -7 3 0 5 0
1 -4 2 1
2
5
SF 0
1
2 6 0
0
1
2
0.01 67. C 0.01 - 0.02
1 1 2 3 4 1 0
1 0 dSC 1 2
2 2 1 1 0
0.05 - 0.02 0.01
0 1 0
0 -
1 7
0 1
0
0 1 0S S C 1 1 -3
2 -4 1 1
0
1
1 6
1 V S F 0 6 3 0 7
0 - 0.01 0.01 S 0.03
0.02 - 0.02 69. D 0.02 - 0.02
1 4 1 2
1 2 4
1 S SD
0
0 T 1
5 0 -7 3 0 -1 1
2
5
1
2 6 0
0
0
0 1 0
0 -
1 7
- 0.04 0.05 0.01 0.06
15 63. c 10
3 2 1 2 0
1 1 0S S C0 0 0
0
1
1 6 1 6
1 V 6 11 42
- 0.01 0.03 - 0.04 0.07
0.01 - 0.03 T 0.00 0.06
2 -6 7
1 1 5 3 15 2 0
1 0 d S£ 1 10 5 0 - 12 3 0 14 1
0 1 0
0
§ SD
1
1 0
1 1 5 4 15 2 0 3
0 T 1
1 -1S 3
71. x = 4.57, y = - 6.44, z = - 24.07 or 14.57, - 6.44, - 24.072
73. x = - 1.19, y = 2.46, z = 8.27 or 1 - 1.19, 2.46, 8.272 5 5 75. x = - 5, y = 7; 1 - 5, 72 77. x = - 4, y = 2, z = ; a - 4, 2, b 79. Inconsistent; ∅ 2 2 1 1 1 6 1 1 1 6 81. x = - z + , y = z - , where z is any real number; e 1x, y, z2 2 x = - z + , y = z - , z is any real number f 5 5 5 5 5 5 5 5 6 9 100.00 3591.42 83. (a) A = c d; B = c d (b) AB = c d ; Nikki’s total tuition and fees are $3591.42, and Joe’s total tuition and fees are $4288.56. 332.38 4288.56 3 12 85. (a) c
500 700
350 500
1 87. (a) K -1 = £ - 1 0
500 400 d ; C 350 850 400 0 1 -1
-1 1§ 1
700 500 S 850
15 (b) C 8 S 3
13 (b) M = £ 8 6
(c) c 1 9 21
11,500 d 17,050
20 19 § 14
(d) [0.75
0.80]
(e) $22,265
(c) Math is fun.
89. If D = ad - bc ≠ 0, then a ≠ 0 and d ≠ 0, or b ≠ 0 and c ≠ 0. Assuming the former, a c c
b 2 1 d 0
0 1 d S£ 1 c R1 =
b 1 a 3 a d 0
1 r a 1
2 5 91. (a) B3 = A + A2 + A3 = E 4 2 1
0
§ SD
1 0
1
b 1 a a 4 D c a a
R2 = - cr1 + r2 4 3 2 2 3
5 2 2 3 2
2 5 4 2 1
0
1
b a 4
0
1
T SD 1 R2 =
a r D 2
1 a c D
0 a D
1 T SD 0
d D 4 c 1 D 0
-
b D T a D
b R1 = - r2 + r1 a
3 4 2 U ; Yes, all pages can reach every other page within 3 clicks. (b) Page 3 3 2
567.75 - 2.902 - 1.983 2.335 - 1.75 - 1.167 0.661 0.361 0.565 0.005 - 2.517 - 0.427 3.768 1.307 0.405 1.050 - 0.310 0.050 - 2.123 1.053 - 1.738 - 0.496 0.487 1.164 93. H 0 0.929 - 0.462 0.473 - 3.5 0.565 - 1.657 - 1.744 2.063 - 2.470 - 0.162 - 0.034 0.604 - 1.747 - 0.875 - 1.810 1.211 1.147 - 0.431 1.204 0.652 - 0.292 0.168 0.242
1.092 - 0.433 0.875 - 0.356 0.191 - 0.664 - 0.655 0.851
0.715 - 0.662 - 0.568 - 0.493 X - 1.046 - 0.727 0.985 0.169
ANSWERS Section 10.6 23 1 2 2 95. (a) 13 - 2 23, 3 23 + 22 (b) R - 1 = E 23 1 2 2 0 0 coordinates back to the original coordinates. 101. x6 - 4x5 - 3x4 + 18x3
0 0
U ; This is the rotation matrix needed to get the translated
1
102. v # w = - 5; u = 180°
103. 50, 3 6
2u2 - 1 u
104.
10.5 Assess Your Understanding (page 819) 7. Improper; 1 +
5. Proper
9 x - 4
5 -4 -5 25. + + x + 2 x + 1 1x + 12 2 35.
1 x2 + 4
2x - 1
+
1x2 + 42 2
37.
2 1 1 1 9 3 6 18 45. + 2 + + x x - 3 x + 3 x
x - 4
1 1 - 1x + 42 4 1 4 27. + 2 + x x x2 + 4
-1 2 -1 + + x x - 3 x + 1
-
58. a
22 22 , b; 2 2
3 4
15.
1 -x + 2 x x + 1
3 1 4 4 33. + x + 3 x - 1 8 4 x - 16x 7 7 41. 2 + 43. + 2x + 1 x - 3 1x + 162 2 1x2 + 162 3
51 51 1 1 5 5 5 5 47. x - 2 + 2 ; + ;x - 2 + x - 1 x + 4 x - 1 x + 3x - 4 x + 4 -
-4 4 + x x - 1
1 2 7 7 + 31. 3x - 2 2x + 1
10x - 11
- 11x - 32 - 11 - 10 + ; ; x2 - 4x + 7 x2 + 4x + 4 x + 2 1x + 22 2 1 1 2x3 + x2 - x + 1 1 4 ; + + + 53. x + 1 + x + 1 x - 1 x4 - 2x2 + 1 1x + 12 2 1x
13.
1 1 1 1 4 4 4 4 + + + 23. x - 1 x + 1 1x - 12 2 1x + 12 2
4 -3 -1 + + x - 2 x - 1 1x - 12 2
39.
- 2 1x - 62
1x + 42 1x - 32
1 2 1x + 12 3 3 + 2 29. x + 1 x + 2x + 4
51. x2 - 4x + 7 +
55. 3.85 years 56. - 2 57. 1
11. Improper; 1 +
2
1 1 - 1x + 42 12 12 21. + 2 x - 2 x + 2x + 4
1 3 1 4 4 2 19. + + x + 1 x - 1 1x - 12 2
2 -1 + 17. x - 1 x - 2
22x - 1
9. Improper; 5x +
2
49. x -
x x2 + 1
11 10 x + 2 1x + 22 2
1 3 1 1 4 4 ;x + 1 + + + + x + 1 x - 1 - 12 2 1x + 12 2 1x - 12 2
121, 5P 4 2
5P 4 o
Historical Problem (page 826) x = 6 units, y = 8 units
10.6 Assess Your Understanding (page 826) 5.
7.
y y x2 1 yx1
y (4 兹2, 4 兹2 ) (4 兹2, 4 兹2 ) 10
5 y 兹36 x
(1, 2) (0, 1)
13. x 2 2x y2 0 (2, 0)
9.
y 5
2
10
2.5 x
x 2 y2 4
2
(2, 1) 5 x (1, 2)
5 x
21. No points of intersection yx 9 y 10 2
x2 y2 4 10 x
y 3x 5
x y 5 2
23.
y x2 4
y 20
y 6 x 13
(3, 5) 10 x
x 2y (8, 4) 9 x x y2 2y
5 x y2x
17.
y 5
15.
y 兹x
y 5 (0, 0)
(1, 1)
x y8x
y 5
11.
19.
y (1, 兹3) 5 (0, 2) x 2 y2 4 5 x
(1, 兹3) 2 (0, 2) y x 4
y 5 xy 4 x y 8 (2, 2) 2
2
5 x (2, 2)
AN89
AN90
ANSWERS Section 10.6
25. x = 1, y = 4; x = - 1, y = - 4; x = 2 22 , y = 22; x = - 2 22 , y = - 22 or 11, 42, 1 - 1, - 42, 12 22, 222, 1 - 2 22, - 222 1 2 1 2 27. x = 0, y = 1; x = - , y = - or 10, 12, a - , - b 3 3 3 3
29. x = 0, y = - 1; x =
7 5 7 5 , y = - or 10, - 12, a , - b 2 2 2 2
1 1 4 1 1 4 ; x = , y = or a2, b , a , b 3 2 3 3 2 3 33. x = 3, y = 2; x = 3, y = - 2; x = - 3, y = 2; x = - 3, y = - 2 or 13, 22, 13, - 22, 1 - 3, 22, 1 - 3, - 22
31. x = 2, y =
35. x =
3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 , y = ; x = , y = - ; x = - , y = ; x = - , y = - or a , b , a , - b , a - , b , a - , - b 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
37. x = 22 , y = 2 22; x = - 22 , y = - 2 22 or 1 22, 2 222, 1 - 22, - 2 222 41. x =
39. No real solution exists.
2 210 8 2 210 8 2 210 8 2 210 8 2 210 8 2 210 8 2 210 8 8 2 210 ,y = ;x = - ,y = ;x = ,y = ;x = - ,y = or a , b, a - , b, a , b , a - ,b 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
43. x = 1, y =
1 1 1 1 1 1 1 1 ; x = - 1, y = ; x = 1, y = - ; x = - 1, y = - or a1, b , a - 1, b , a1, - b , a - 1, - b 2 2 2 2 2 2 2 2
45. No real solution exists.
47. x = 23 , y = 23; x = - 23 , y = - 23; x = 2, y = 1; x = - 2, y = - 1 or 1 23, 232 , 1 - 23,- 232, 12, 12, 1 - 2, - 12
49. x = 0, y = - 2; x = 0, y = 1; x = 2, y = - 1 or 10, - 22, 10, 12, 12, - 12 55.
51. x = 2, y = 8 or 12, 82
53. x = 81, y = 3 or 181, 32
57. x = 0.48, y = 0.62 59. x = - 1.65, y = - 0.89 61. x = 0.58, y = 1.86; x = 1.81, y = 1.05; x = 0.58, y = - 1.86; x = 1.81, y = - 1.05 63. x = 2.35, y = 0.85
y x 2 x y 2 3y 2 0 3 (1, 1) 2 x x1
65.
y2 y x 0
67.
y 5 (x 1)2 (y 1)2 5
6 3 , 5 5
(0, 兹 3 2) 5 x x 2y 0 (2, 1)
69.
y 5 (1, 0)
y 5
y
(5, 2) 5 x (x 1)2 (y 2)2 4
y 2 4y x 1 0 (0, 兹 3 2)
4 x3
8 x
(1, 4)
x 2 6x y 2 1 0 (1, 2)
1 1 1 and 77. 5 79. 5 in. by 3 in. 81. 2 cm and 4 cm 83. Tortoise: 7 m/hr, hare: 7 m/hr 2 3 2 P + 2P 2 - 16A P - 2P 2 - 16A ;w = 91. y = 4x - 4 93. y = 2x + 1 85. 12 cm by 18 cm 87. x = 60 ft; y = 30 ft 89. l = 4 4 1 7 - b + 2b2 - 4ac - b - 2b2 - 4ac 95. y = - x + 97. y = 2x - 3 99. r1 = ; r2 = 101. (a) 4.274 ft by 4.274 ft or 0.093 ft by 0.093 ft 3 3 2a 2a 71. 3 and 1; - 1 and - 3
102. e
73. 2 and 2; - 2 and - 2
- 3 - 265 - 3 + 265 , f 7 7
75.
2 103. y = - x - 3 5
104. sin u = -
7 24 7 25 25 ; cos u = - ; tan u = ; csc u = - ; sec u = 25 25 24 7 24
105. ≈15.8°
10.7 Assess Your Understanding (page 835) 7. dashes; solid 8. half-planes y 11.
9. False 13.
10. unbounded 15.
y 5
5 xⱖ0 5 x
8
2x y ⱖ 6
xⱖ4
17.
y
y x y ⬎1 2
2
5
5 x
5 x 8 x
19.
21.
y 5
23.
y 5
xⴙyⴝ2
xy ⱖ 4 5 x
25.
y
y 5
5 2x ⴙ y ⴝ 4
5 x
3x 2y 6
2x y 4 5 x
5 x
y ⱕ x2 1
27.
29.
y 5
y 5
31.
33. No solution
y
2x 4y 0
5
y 5
2x ⴙ y ⴝ ⴚ2
2x 3y 0
5 x 3x 2y 6
6 x x 2y 6
5 x 2x ⴙ y ⴝ 2
2x ⴙ 3y ⴝ 6 5 x 2x ⴙ 3y ⴝ 0
AN91
ANSWERS Section 10.8 35.
37.
y 5
x2 ⴙ y2 ⴝ 9
yⴝx ⴚ4 2
y 5
39.
yⴝxⴚ2
5 x
y 5
5 x
45. Unbounded
y
xⴙyⴝ2 (0, 4)
(0, 0) (3, 0)
x x ⴙ 2y ⴝ 6
x x + y 53. d x y
(1, 0)
x + y x 57. (a) d y y
… Ú … Ú
(b)
(10, 0) 16 x
… … Ú Ú
y
3x ⴙ y ⴝ 12
8
(2, 6) xⴙyⴝ8
(0, 2)
x (4, 0) 2x ⴙ 3y ⴝ 12
9 x (2, 0)
… Ú … … Ú
x y 55. e x + y x - y x
4 6 0 0
2x ⴙ y ⴝ 10
(0, 8) 5 xⴙyⴝ2
24 12 , 7 7
(2, 0)
2x ⴙ y ⴝ 4
x ⴙ 2y ⴝ 10 (0, 5)
49. Bounded y 8
(0, 2)
(2, 0)
y 16
1 2
(0, 4) xⴙyⴝ2 5 x
51. Bounded
0,
5
(2, 2) 8
xy ⴝ 4
47. Bounded
y 8
y ⴝ x2 ⴙ 1
5 x
x 2 ⴙ y 2 ⴝ 16
43. Bounded
(0, 3)
y 5
5 x
xⴙyⴝ3
2x ⴙ y ⴝ 6
41.
y ⴝ x2 ⴚ 4
(5, 0)
20 15 50 0 0
x ⴙ 2y ⴝ 1
50,000 35,000 10,000 0 y
x y 59. (a) d x + 2y 3x + 2y
Ú Ú … …
(b)
x ⴝ 35,000
3x + 2y 2x + 3y 61. (a) d x y
0 0 300 480
(b)
y
80 (0, 50)
40,000 (35,000, 10,000) ⴚ20,000 (35,000, 0)
(40,000, 10,000)
(90, 105) 400
(0, 150)
y ⴝ 10,000 x (50,000, 0) x ⴙ y ⴝ 50,000
U5 3P 4
P 2 U5
U5P
U5
1 6
1 3
U5
5P 4 U5
80
2x ⴙ 3y ⴝ 150 x
160, 0 3
64. f 1 - 12 = - 5; f 122 = 28 2p 4p , f 65. e 0, 3 3
y
U5
(36, 26)
(0, 0)
x ⴙ 2y ⴝ 300
(160, 0)
62. 5 - 1 - 2i, - 1 + 2i 6 1 2 1 ; 63. x2 + ay - b = 6 36 1 1 circle with radius and center a0, b ; 6 6
3x ⴙ 2y ⴝ 160
x
(0, 0)
160 150 0 0
y
400
3x ⴙ 2y ⴝ 480
… … Ú Ú
P 4
x U50
7P 4
3P 2
10.8 Assess Your Understanding (page 842) 1. objective function 2. True 3. Maximum value is 11; minimum value is 3. 5. Maximum value is 65; minimum value is 4. 7. Maximum value is 67; minimum value is 20. 9. The maximum value of z is 12, and it occurs at the point (6, 0). 11. The minimum value of z is 4, and it occurs at the point (2, 0). 13. The maximum value of z is 20, and it occurs at the point (0, 4). 15. The minimum value of z is 8, and it occurs at the point (0, 2). 17. The maximum value of z is 50, and it occurs at the point (10, 0). 19. Produce 8 downhill and 24 cross-country; $1760; $1920, which is the profit when producing 16 downhill and 16 cross-country. 21. Rent 15 rectangular tables and 16 round tables for a minimum cost of $1252. 23. (a) $10,000 in the junk bond and $10,000 in Treasury bills (b) $12,000 in the junk bond and $8000 in Treasury bills 25. Manufacture 10 racing skates and 15 figure skates. 27. 100 lb of ground beef should be mixed with 50 lb of pork. 29. Order 2 metal samples and 4 plastic samples; $34 31. (a) Configure with 10 first-class seats and 120 coach seats. (b) Configure with 15 first-class seats and 120 coach seats. 1 33. e - , 1 f 32 34.
y 3 P
2P x
Domain: 5x 0 x ≠ kp, k is an integer 6 ; range: 1 - q , q 2
35. 89.1 years
36. y = 3x + 7
AN92
ANSWERS Review Exercises
Review Exercises (page 846) 1. x =
3 1 3 1 ,y = or a , - b 4 20 4 20
6. x = 2, y = 3 or 12, 32
2. x = 2, y =
1 1 or a2, b 2 2
3. x = 2, y = - 1 or 12, - 12
4. x = 1, y = 1 or 11, 12
5. Inconsistent
7. Inconsistent
8. x = - 1, y = 2, z = - 3 or 1 - 1, 2, - 32 7 39 9 69 7 39 9 69 9. x = z + ,y = z + , where z is any real number or b 1x, y, z2 ` x = z + ,y = z + , z is any real number r 4 4 8 8 4 4 8 8 11. e
10. Inconsistent
3x + 2y = 8 x + 4y = - 1 -
1 2 17. D 1 6
-1 2 3
T
18. F
5 7 1 7 3 7
9 7 1 7 4 7
12. W
3 7 2 - V 7 1 7
3x + 2y + 5z = - 1 2x + 7z = 3 - 4x + y - 3z = 2
19. Singular
0 13. £ 8 2
1 2§ 8
22. Inconsistent
45. 3x ⴙ 4y ⱕ 12
2, y = 3 or 12, 32 1 9 x + 10 10 x2 + 9
41. x = 0, y = 0; x = - 3, y = 3; x = 3, y = 3 or
47. Bounded y 5
y 8
xⴙyⴝ2 (0, 2)
5 x
5 x
ⴚ2x ⴙ y ⴝ 2
(0, 2)
xⴙyⴝ4 8 x
(0, 0) (3, 0)
49.
50.
y 5
x 2 ⴙ y 2 ⴝ 16
9 (0, 8)
5 x
2x ⴙ y ⴝ 8
(0, 1) x ⴙ 2y ⴝ 2
xⴙyⴝ2
(4, 0)
13 d -4
43. x = 1, y = - 1 or 11, - 12
y ⱕ x2
y
15 -7
4 2 4 2 22, y = - 22; x = - 22, y = 22 or 3 3 3 3
5
48. Bounded
16. c
- 1; x = - 1, y = 2 or 1 - 2, - 12 1 - 1, 22
46. Unbounded
y 5
x
-1 -3§ 0
2 12 4
+ 1, z = - 1, y is any real number 6 29. x = 2, y = - 1 or 12, - 12 30. x = 3 1 2 -3 3 4 10 35. + + 2 36. + - 4 x - 1 x x + 1 x
42. x = 22, y = - 22; x = - 22, y = 22; x =
4 2 4 2 1 22, - 222, 1 - 22, 222, a 22, - 22 b , a - 22, 22 b 3 3 3 3 y 5
7 15. £ 9 -8
1 2 1 13 13 13 13 2 ,y = or a , b 21. x = 9, y = ,z = or a9, , b 5 10 5 10 3 3 3 3 1 7 6 1 7 6 23. x = - , y = - , z = - or a - , - , - b 5 5 5 5 5 5
40. x = 2 22 , y = 22; x = - 2 22 , y = - 22 or 12 22, 222, 1 - 2 22, - 222
44.
- 12 0§ 24
20. x =
24. z = - 1, x = y + 1, where y is any real number or 5 1x, y, z2 0 x = y 25. x = 4, y = 2, z = 3, t = - 2 or 14, 2, 3, - 22 26. 5 27. 67 28. - 100 3 2 + 31. x = - 1, y = 2, z = - 3 or 1 - 1, 2, - 32 32. 15 33. - 8 34. x x 1 1 1 2 4 4 x - 4x + 37. 2 + 38. 2 + 39. x = - 2, y = x - 1 x + 1 x + 4 1x2 + 42 2 x + 1 10, 02, 1 - 3, 32, 13, 32
6 14. £ 18 0
y
2x ⴙ 3y ⴝ 6
xy ⴝ 24
3 3
x
x2 ⴙ y2 ⴝ 25
9 x (2, 0)
51. The maximum value is 32 when x = 0 and y = 8. 52. The minimum value is 3 when x = 1 and y = 0. 53. 9 54. A is any real number, A ≠ 9. 2 1 55. y = - x2 - x + 1 56. Mix 70 lb of $6.00 coffee and 30 lb of $9.00 coffee. 57. Buy 1 small, 5 medium, and 2 large. 3 3 y
x y 58. (a) d 4x + 3y 2x + 3y
Ú Ú … …
0 0 960 576
400 4x ⴙ3y ⴝ 960
(b)
(0, 192)
(192, 64) 400 x 2x ⴙ3y ⴝ 576 (240, 0)
59. Speedboat: 36.67 km/hr; Aguarico River: 3.33 km/hr 60. Bruce: 4 hr; Bryce: 2 hr; Marty: 8 hr 61. Produce 35 gasoline engines and 15 diesel engines; the factory is producing an excess of 15 gasoline engines and 0 diesel engines.
AN93
ANSWERS Cumulative Review
Chapter Test (page 848) 1. x = 3, y = - 1 or 13, - 12 b 1x, y, z2 ` x = - z +
6. c
3. x = - z +
2. Inconsistent
18 17 ,y = z , where z is any real number or 7 7
18 17 ,y = z , z is any real number r 7 7
4. x =
3x + 2y + 4z = - 6 3x + 2y + 4z = - 6 6 1x + 0y + 8z = 2 or c x + 8z = 2 7. C 1 - 2x + 1y + 3z = - 11 - 2x + y + 3z = - 11 5
4 - 11 - 11 S 8. C - 3 12 6
- 19 4 5 S 9. C 1 - 22 -1
-5 -1 5
4 5. C - 2 1
1 1 , y = - 2, z = 0 or a , - 2, 0 b 3 3
10 - 11 26
1 0 0 3 - 25 S -5 10
26 16 2 S 10. c 3 3
17 d - 10
11. D -
2
-1
5 2
3 2
T
3 3 -4 1 1 1 1 12. C - 2 - 2 3 S 13. x = , y = 3 or a , 3 b 14. x = - y + 7, where y is any real number or b 1x, y2 ` x = - y + 7, y is any real number r 2 2 4 4 -4 -5 7 15. x = 1, y = - 2, z = 0 or 11,- 2, 02 16. Inconsistent 17. - 29 18. - 12 19. x = - 2, y = - 5 or 1 - 2, - 52 20. x = 1, y = - 1, z = 4 or 11, - 1, 42 21. 11, - 32 and (1, 3) 22. (3, 4) and (1, 2) 23.
y 15
24. x 2 ⴙ y 2 ⴝ 100
3 -2 + x + 3 1x + 32 2
26. Unbounded y 9
2x ⴚ 3y ⴝ 2
1 1 x 5x 3 3 + + 25. 2 x 1x + 32 1x2 + 32 2
15 x
4x ⴚ 3y ⴝ 0
(4, 2) 9 (8, 0)
x x ⴙ 2y ⴝ 8
27. The maximum value of z is 64, and it occurs at the point (0, 8). 28. Flare jeans cost $24.50, camisoles cost $8.50, and T-shirts cost $6.00.
Cumulative Review (page 849) 1. e 0,
1 f 2
2. 55 6
1 3. e - 1, - , 3 f 2
4. 5 - 2 6
8. Center: 11, - 22; radius = 4
5 5. e f 2
1 f ln 3
7. Odd; symmetric with respect to the origin 10. f -1 1x2 =
9. Domain: all real numbers Range: 5y 0 y 7 1 6 Horizontal asymptote: y = 1
y 3
(2, 2)
yⴝ1 5 x
11. (a)
(b)
y 8
(c)
y
y
(0, 2) (0, 6)
(ⴚ2, 0)
1 1
(ⴚ2, 0) 2 x
(e)
y
(d)
2 (2, 0) x
5 - 2 x
Domain of f: 5x 0 x ≠ - 2 6 Range of f: 5y 0 y ≠ 0 6 Domain of f -1: 5x 0 x ≠ 0 6 Range of f -1: 5y 0 y ≠ - 2 6
y 5
x
3
6. e
2
(1, 1)
(0, 0)
2
(1, 1)
yⴝ0
x
2
(ⴚ1, ⴚ1)
y
2
(0, 0)
x
(1, 1) 2 x
(ⴚ1, ⴚ1)
(0, ⴚ2)
xⴝ0
(f)
y 2 yⴝ0
(g)
x
(1, 0)
4 x
xⴝ0
12. (a)
zero: - 2.28
10
ⴚ10
10
ⴚ10
y 兹2 ,0 ⴚ 2
2
(0, 1) 2
(h)
y
1
0,
y (ⴚ2, 1)
1 兹5 0, ⴚ 5
兹5 5
(i)
x
兹2 ,0 2
(ⴚ1, 0) (ⴚ2, ⴚ1)
1
yⴝ
兹3 x 3
(j)
y 3 (3, 1)
(2, 1) (1, 0) 2 x (2, ⴚ1) 兹3 yⴝⴚ x 3
(b) Local maximum of 7 at x = - 1; local minimum of 3 at x = 1 (c) 1 - q , - 12, 11, q 2
2,
(ⴚ1, 1) 0,
1 4
(1, 0)
1 4
7 x
AN94
ANSWERS Section 11.1
CHAPTER 11 Sequences; Induction; the Binomial Theorem 11.1 Assess Your Understanding (page 858)
5. n(n - 1) # # # 3 # 2 # 1 6. recursive 7. summation 8. True 9. 3,628,800 11. 504 13. 1260 15. s1 = 1, s2 = 2, s3 = 3, 1 1 3 2 5 1 2 2 8 8 19. c1 = 1, c2 = - 4, c3 = 9, c4 = - 16, c5 = 25 21. s1 = , s2 = , s3 = , s4 = , s5 = s4 = 4, s5 = 5 17. a1 = , a2 = , a3 = , a4 = , a5 = 3 2 5 3 7 2 5 7 41 61 1 1 1 1 1 1 2 3 4 5 n 1 , t = - , t4 = ,t = 25. b1 = , b2 = 2 , b3 = 3 , b4 = 4 , b5 = 5 27. an = 29. an = n - 1 23. t 1 = - , t 2 = 6 12 3 20 30 5 42 e n + 1 e e e e 2 31. an = ( - 1)n + 1 33. an = ( - 1)n + 1n 35. a1 = 2, a2 = 5, a3 = 8, a4 = 11, a5 = 14 37. a1 = - 2, a2 = 0, a3 = 3, a4 = 7, a5 = 12 3 1 1 1 43. a1 = 1, a2 = 2, a3 = 2, a4 = 4, a5 = 8 39. a1 = 5, a2 = 10, a3 = 20, a4 = 40, a5 = 80 41. a1 = 3, a2 = , a3 = , a4 = , a5 = 2 2 8 40 45. a1 = A, a2 = A + d, a3 = A + 2d, a4 = A + 3d, a5 = A + 4d 3. sequence 4. True
47. a1 = 22 , a2 = 32 + 12 , a3 = 32 + 22 + 12 , a4 = 42 + 32 + 22 + 12 ,
49. 3 + 4 + g + (n + 2) 55.
a5 = 52 + 42 + 32 + 22 + 12
1 1 1 + + g + n 3 9 3
51.
9 n2 1 + 2 + + g + 2 2 2
53. 1 +
57. ln 2 - ln 3 + ln 4 - g + ( - 1)n ln n
1 1 1 + + g + n 3 9 3 20
59. a k k=1
13 6 n n n+1 k 1 3k 61. a 63. a ( - 1)k a k b 65. a 67. a (a + kd) or a [a + (k - 1)d] 69. 200 71. 820 73. 1110 75. 1560 77. 3570 3 k=1 k + 1 k=0 k=1 k k=0 k=1 79. 44,000 81. $2930 83. $18,058.03 85. 21 pairs 87. Fibonacci sequence 89. (a) 3.630170833 (b) 3.669060828 (c) 3.669296668 (d) 12 91. (a) a1 = 0.4; a2 = 0.7; a3 = 1; a4 = 1.6; a5 = 2.8; a6 = 5.2; a7 = 10; a8 = 19.6 (b) Except for term 5, which has no match, Bode’s formula provides excellent approximations for the mean distances of the planets from the Sun. (c) The mean distance of Ceres from the Sun is approximated by a5 = 2.8, and that of Uranus is a8 = 19.6. (d) a9 = 38.8; a10 = 77.2 (e) Pluto’s distance is approximated by a9, but no term approximates Neptune’s mean distance from the Sun. (f) According to Bode’s Law, the mean orbital distance of Eris from the Sun will be 154 AU. 93. a0 = 2; a5 = 2.236067977; 2.236067977 95. a0 = 4; a5 = 4.582575695; 4.582575695
98. $2654.39
99. 22(cos 225° + i sin 225°)
100. - 5i + 5j + 5k
101. (y - 4)2 = 16(x + 3)
11.2 Assess Your Understanding (page 866) 1. arithmetic 2. False 3. 17 4. True 5. sn - sn - 1 = (n + 4) - [(n - 1) + 4] = n + 4 - (n + 3) = n + 4 - n - 3 = 1, a constant; d = 1; s1 = 5, s2 = 6, s3 = 7, s4 = 8 7. an - an - 1 = (2n - 5) - [2(n - 1) - 5] = 2n - 5 - (2n - 2 - 5) = 2n - 5 - (2n - 7) = 2n - 5 - 2n + 7 = 2, a constant; d = 2; a1 = - 3, a2 = - 1, a3 = 1, a4 = 3 9. cn - cn - 1 = (6 - 2n) - [6 - 2(n - 1)] = 6 - 2n - (6 - 2n + 2) = 6 - 2n - (8 - 2n) = 6 - 2n - 8 + 2n = - 2, a constant; d = - 2; c1 = 4, c2 = 2, c3 = 0, c4 = - 2 1 1 1 1 1 5 1 5 1 1 1 1 1 1 1 1 1 1 - n - a - n + b = - n - a - nb = - n + n = - , a constant; 11. t n - t n - 1 = a - n b - c - (n - 1) d = 2 3 2 3 2 3 2 3 3 2 3 6 3 2 3 6 3 3 1 1 1 1 5 d = - ; t1 = , t2 = - , t3 = - , t4 = 3 6 6 2 6 13. sn - sn - 1 = ln 3n - ln 3n - 1 = n ln 3 - (n - 1) ln 3 = n ln 3 - (n ln 3 - ln 3) = n ln 3 - n ln 3 + ln 3 = ln 3, a constant; d = ln 3; s1 = ln 3, s2 = 2 ln 3, s3 = 3 ln 3, s4 = 4 ln 3 1 83 15. an = 3n - 1; a51 = 152 17. an = 8 - 3n; a51 = - 145 19. an = (n - 1); a51 = 25 21. an = 22n; a51 = 51 22 23. 200 25. - 266 27. 2 2 29. a1 = - 13; d = 3; an = an - 1 + 3; an = - 16 + 3n 31. a1 = - 53; d = 6; an = an - 1 + 6; an = - 59 + 6n n 33. a1 = 28; d = - 2; an = an - 1 - 2; an = 30 - 2n 35. a1 = 25; d = - 2; an = an - 1 - 2; an = 27 - 2n 37. n2 39. (9 + 5n) 41. 1260 2 3 43. 324 45. 30,919 47. 10,036 49. 6080 51. - 1925 53. 15,960 55. 57. 24 terms 59. 1185 seats 2 61. 210 beige and 190 blue 63. {Tn} = { - 5.5n + 67}; T5 = 39.5°F 65. The amphitheater has 1647 seats. 67. 8 yr 70. 16.42% 71. v = 4i - 6j 72. Center: (0, 0); Vertices: (0, - 5), (0, 5); Foci: 10, - 2212, 10, 2212 y 6 (0, 5) (0, Ï21) 2 (22, 0) 26 24
(0, 2Ï21)
22
(2, 0) 4 6 x
26 (0, 25)
1
73. c 23 2
0 d -1
ANSWERS Section 11.4
AN95
Historical Problems (page 875) 1. 1
2 5 1 1 loaves, 10 loaves, 20 loaves, 29 loaves, 38 loaves 3 6 6 3
2. (a) 1 person (b) 2401 kittens
(c) 2800
11.3 Assess Your Understanding (page 876) a 1 - r 1 3 11. r = ; a1 = - , a2 2 2 3 1 17. r = ; t 1 = , t 2 = 2 2 3. geometric 4.
5. divergent series
6. True 7. False
8. True 9. r = 3; s1 = 3, s2 = 9, s3 = 27, s4 = 81
3 3 3 1 1 = - , a3 = - , a4 = 13. r = 2; c1 = , c2 = , c3 = 1, c4 = 2 15. r = 21/3; e1 = 21/3 , e2 = 22/3 , e3 = 2, e4 = 24/3 4 8 16 4 2 3 9 27 n-1 # , t = , t4 = 19. a5 = 162; an = 2 3 21. a5 = 5; an = 5 # ( - 1)n - 1 23. a5 = 0; an = 0 4 3 8 16 1 1 n-1 1 n-2 29. a9 = 1 31. a8 = 0.00000004 33. an = 7 # 2n - 1 35. an = - 3 # a - b 25. a5 = 4 22; an = ( 22)n 27. a7 = = a- b 64 3 3 n 7 1 2 n-1 n-1 n 2 n n 39. an = (15) = 7 # 15 41. - (1 - 2 ) 43. 2 c 1 - a b d 45. 1 - 2 37. an = - ( - 3) 15 4 3 51. 47. 49.
53. Converges;
3 2
55. Converges; 16
69. Arithmetic; d = 1; 1375
57. Converges;
71. Neither
8 5
59. Diverges 61. Converges;
2 73. Arithmetic; d = - ; - 700 3
1 79. Geometric; r = - 2; - 3 1 - 1 - 22 50 4 3 23 1/2 81. Geometric; r = 3 ; 11 + 232 11 - 325 2 2
83. - 4
75. Neither
20 3
63. Diverges
65. Converges;
77. Geometric; r =
18 5
67. Converges; 6
50
2 2 ; 2c 1 - a b d 3 3
85. $39,392.81
87. (a) 0.775 ft (b) 8th (c) 15.88 ft (d) 20 ft 89. $349,496.41 91. $96,885.98 93. $305.10 95. 1.845 * 1019 97. 10 99. $72.67 per share 101. December 20, 2014; $9999.92 103. Option A results in a higher salary in 5 years ($25,250 versus $24,761); option B results in a higher 5-year total ($116,801 versus $112,742). 105. Option 2 results in the most: $16,038,304; option 1 results in the least: $14,700,000. 107. Yes. A constant sequence is both arithmetic and geometric. For example, 3, 3, 3, . . . is an arithmetic sequence with a1 = 3 and d = 0 and is a y2 8 15 x2 = 1 114. 54 geometric sequence with a = 3 and r = 1. 111. 2.121 112. i j 113. 17 17 4 12
11.4 Assess Your Understanding (page 882) 1. (I) n = 1: 2(1) = 2 and 1(1 + 1) = 2 (II) If 2 + 4 + 6 + g + 2k = k(k + 1), then 2 + 4 + 6 + g + 2k + 2(k + 1) = (2 + 4 + 6 + g + 2k) + 2(k + 1) = k(k + 1) + 2(k + 1) = k 2 + 3k + 2 = (k + 1)(k + 2) = (k + 1)[(k + 1) + 1]. 1 1 3. (I) n = 1: 1 + 2 = 3 and (1)(1 + 5) = (6) = 3 2 2 1 (II) If 3 + 4 + 5 + g + (k + 2) = k(k + 5), then 3 + 4 + 5 + g + (k + 2) + [(k + 1) + 2] 2 1 1 1 1 = [3 + 4 + 5 + g + (k + 2)] + (k + 3) = k(k + 5) + k + 3 = (k 2 + 7k + 6) = (k + 1)(k + 6) = (k + 1)[(k + 1) + 5]. 2 2 2 2 1 1 5. (I) n = 1: 3(1) - 1 = 2 and (1)[3(1) + 1] = (4) = 2 2 2 1 (II) If 2 + 5 + 8 + g + (3k - 1) = k(3k + 1), then 2 + 5 + 8 + g + (3k - 1) + [3(k + 1) - 1] 2 1 1 1 = [2 + 5 + 8 + g + (3k - 1)] + (3k + 2) = k(3k + 1) + (3k + 2) = (3k 2 + 7k + 4) = (k + 1)(3k + 4) 2 2 2 1 = (k + 1)[3(k + 1) + 1]. 2 7. (I) n = 1: 21 - 1 = 1 and 21 - 1 = 1 (II) If 1 + 2 + 22 + g + 2k - 1 = 2k - 1, then 1 + 2 + 22 + g + 2k - 1 + 2(k + 1) - 1 = (1 + 2 + 22 + g + 2k - 1) + 2k = 2k - 1 + 2k = 2(2k) - 1 = 2k + 1 - 1. 1 1 9. (I) n = 1: 41 - 1 = 1 and (41 - 1) = (3) = 1 3 3 1 k 2 k-1 = (4 - 1), then 1 + 4 + 42 + g + 4k - 1 + 4(k + 1) - 1 = (1 + 4 + 42 + g + 4k - 1) + 4k (II) If 1 + 4 + 4 + g + 4 3 1 k 1 k 1 1 k = (4 - 1) + 4 = [4 - 1 + 3(4k)] = [4(4k) - 1] = (4k + 1 - 1). 3 3 3 3
AN96
ANSWERS Section 11.4
1 1 1 1 = and = 1#2 2 1 + 1 2 1 1 1 k 1 1 1 1 1 1 = , then # + # + # + g + + (II) If # + # + # + g + 1 2 2 3 3 4 k(k + 1) k + 1 1 2 2 3 3 4 k(k + 1) (k + 1)[(k + 1) + 1] k(k + 2) + 1 1 1 1 1 k 1 1 = c # + # + # + g + d + = + = 1 2 2 3 3 4 k(k + 1) (k + 1)(k + 2) k + 1 (k + 1)(k + 2) (k + 1)(k + 2) (k + 1)2 k + 1 k + 1 k 2 + 2k + 1 = = = . = (k + 1)(k + 2) (k + 1)(k + 2) k + 2 (k + 1) + 1
11. (I) n = 1:
1# # # 1 2 3 = 1 6 1 2 2 2 (II) If 1 + 2 + 3 + g + k 2 = k(k + 1)(2k + 1), then 12 + 22 + 32 + g + k 2 + (k + 1)2 6 1 1 = (12 + 22 + 32 + g + k 2) + (k + 1)2 = k(k + 1)(2k + 1) + (k + 1)2 = (2k 3 + 9k 2 + 13k + 6) 6 6 1 1 = (k + 1)(k + 2)(2k + 3) = (k + 1)[(k + 1) + 1][2(k + 1) + 1]. 6 6
13. (I) n = 1: 12 = 1 and
1 1 15. (I) n = 1: 5 - 1 = 4 and (1)(9 - 1) = # 8 = 4 2 2 1 (II) If 4 + 3 + 2 + g + (5 - k) = k(9 - k), then 4 + 3 + 2 + g + (5 - k) + [5 - (k + 1)] 2 1 1 1 = [4 + 3 + 2 + g + (5 - k)] + 4 - k = k(9 - k) + 4 - k = (9k - k 2 + 8 - 2k) = ( - k 2 + 7k + 8) 2 2 2 1 1 = (k + 1)(8 - k) = (k + 1)[9 - (k + 1)]. 2 2 1 17. (I) n = 1: 1 # (1 + 1) = 2 and # 1 # 2 # 3 = 2 3 1 (II) If 1 # 2 + 2 # 3 + 3 # 4 + g + k(k + 1) = k(k + 1)(k + 2), then 1 # 2 + 2 # 3 + 3 # 4 + g + k(k + 1) 3 + (k + 1)[(k + 1) + 1] = [1 # 2 + 2 # 3 + 3 # 4 + g + k(k + 1)] + (k + 1)(k + 2) 1 1 1 1 = k(k + 1)(k + 2) + # 3(k + 1)(k + 2) = (k + 1)(k + 2)(k + 3) = (k + 1)[(k + 1) + 1][(k + 1) + 2]. 3 3 3 3 19. (I) n = 1: 12 + 1 = 2, which is divisible by 2. (II) If k 2 + k is divisible by 2, then (k + 1)2 + (k + 1) = k 2 + 2k + 1 + k + 1 = (k 2 + k) + 2k + 2. Since k 2 + k is divisible by 2, and 2k + 2 is divisible by 2, then (k + 1)2 + (k + 1) is divisible by 2. 21. (I) n = 1: 12 - 1 + 2 = 2, which is divisible by 2. (II) If k 2 - k + 2 is divisible by 2, then (k + 1)2 - (k + 1) + 2 = k 2 + 2k + 1 - k - 1 + 2 = (k 2 - k + 2) + 2k. Since k 2 - k + 2 is divisible by 2, and 2k is divisible by 2, then (k + 1)2 - (k + 1) + 2 is divisible by 2. 23. (I) n = 1: If x 7 1, then x1 = x 7 1. (II) Assume, for an arbitrary natural number k, that if x 7 1 then xk 7 1. Multiply both sides of the inequality xk 7 1 by x. If x 7 1, then xk + 1 7 x 7 1. 25. (I) n = 1: a - b is a factor of a1 - b1 = a - b. (II) If a - b is a factor of ak - bk, then ak + 1 - bk + 1 = a(ak - bk) + bk(a - b). Since a - b is a factor of ak - bk and a - b is a factor of a - b, then a - b is a factor of ak + 1 - bk + 1.
27. (I) n = 1: (1 + a)1 = 1 + a Ú 1 + 1 # a (II) Assume that there is an integer k for which the inequality holds. So (1 + a)k Ú 1 + ka. We need to show that (1 + a)k + 1 Ú 1 + (k + 1)a. (1 + a)k + 1 = (1 + a)k (1 + a) Ú (1 + ka)(1 + a) = 1 + ka2 + a + ka = 1 + (k + 1)a + ka2 Ú 1 + (k + 1)a.
6 + g + 2k = k 2 + k + 2, then 2 + 4 + 6 + g + 2k + 2(k + 1) + 4 + 6 + g + 2k) + 2k + 2 = k 2 + k + 2 + 2k + 2 = k 2 + 3k + 4 = (k 2 + 2k + 1) + (k + 1) + 2 + 1)2 + (k + 1) + 2. 2 and 12 + 1 + 2 = 4. The fact is that 2 + 4 + 6 + g + 2n = n2 + n, not n2 + n + 2 (Problem 1). 1 # (1 - 1) 31. (I) n = 1: [a + (1 - 1)d] = a and 1 # a + d = a. 2 k(k - 1) (II) If a + (a + d) + (a + 2d) + g + [a + (k - 1)d] = ka + d , then 2 k(k - 1) a + (a + d) + (a + 2d) + g + [a + (k - 1)d] + [a + ((k + 1) - 1)d] = ka + d + a + kd 2 k(k - 1) + 2k (k + 1)(k) (k + 1)[(k + 1) - 1] = (k + 1)a + d = (k + 1)a + d = (k + 1)a + d . 2 2 2 35. {251} 36. Left: 448.3 kg; right: 366.0 kg 33. (I) n = 3: The sum of the angles of a triangle is (3 - 2) # 180° = 180°. 1 1 (II) Assume for some k Ú 3 that the sum of the angles of a 37. x = , y = - 3; a , - 3 b 2 2 convex polygon of k sides is (k - 2) # 180°. A convex polygon k sides 7 -3 of k + 1 sides consists of a convex polygon of k sides plus 38. c d -7 8 a triangle (see the illustration). The sum of the angles is k ⴙ 1 sides 29. If 2 + 4 + = (2 = (k But 2 # 1 =
(k - 2) # 180° + 180° = (k - 1) # 180° = [(k + 1) - 2] # 180°.
ANSWERS Chapter Test
AN97
11.5 Assess Your Understanding (page 888) 1. Pascal triangle 2. 1; n 3. False 4. Binomial Theorem 5. 10 7. 21 9. 50 11. 1 13. ≈1.8664 * 1015 15. ≈1.4834 * 1013 17. x5 + 5x4 + 10x3 + 10x2 + 5x + 1 19. x6 - 12x5 + 60x4 - 160x3 + 240x2 - 192x + 64 21. 81x4 + 108x3 + 54x2 + 12x + 1 23. x10 + 5x8y2 + 10x6y4 + 10x4y6 + 5x2y8 + y10 25. x3 + 6 22x5/2 + 30x2 + 40 22x3/2 + 60x + 24 22x1/2 + 8 27. a5x5 + 5a4bx4y + 10a3b2x3y2 + 10a2b3x2y3 + 5ab4xy4 + b5y5 29. 17,010 31. - 101,376 33. 41,472 35. 2835x3 37. 314,928x7 39. 495 41. 3360 43. 1.00501 n # (n - 1)! n n! n! n n! n! n! 45. a b = = = = n; a b = = = = 1 n - 1 (n - 1)! [n - (n - 1)]! (n - 1)! 1! (n - 1)! n n! (n - n)! n!0! n! n n n n n n 47. 2n = (1 + 1)n = a b 1n + a b (1)n - 1(1) + g + a b 1n = a b + a b + g + a b 0 1 n 0 1 n 5 51. e f ≈ 58.827 6 52. (a) 0 (b) 90° (c) Orthogonal 2n 6 - 2n 5 53. x = 1, y = 3, z = - 2; (1, 3, - 2) y 54. Bounded
49. 1
(0, 6) (4, 2) (0, 0) (5, 0)
x
Review Exercises (page 891) 4 5 6 7 8 1. a1 = - , a2 = , a3 = - , a4 = , a5 = 3 4 5 6 7
4 8 16 , a = , a5 = 3 4 9 27 13 n 1 4. a1 = 8, a2 = 19, a3 = 41, a4 = 85, a5 = 173 5. 6 + 10 + 14 + 18 = 48 6. a 1 - 12 k + 1 7. Arithmetic; d = 1; Sn = 1n + 112 8. Neither k 2 k=1 8 n 1 1 n 9. Geometric; r = 8; Sn = 18 - 12 10. Arithmetic; d = 4; Sn = 2n 1n - 12 11. Geometric; r = ; Sn = 6 c 1 - a b d 12. Neither 13. 3825 7 2 2 1 1 1 9 14. 9515 15. - 1320 16. a1 - 10 b ≈ 0.49999 17. 35 18. 12 19. 9 22 20. 5an 6 = 53n + 1 6 21. 5an 6 = 5n - 10 6 22. Converges; 2 2 5 3 4 23. Converges; 24. Diverges 25. Converges; 8 3 3#1 26. (I) n = 1: 3 # 1 = 3 and (1 + 12 = 3 2 3k 1k + 12, then 3 + 6 + 9 + g + 3k + 3 1k + 12 = 13 + 6 + 9 + g + 3k2 + 13k + 32 (II) If 3 + 6 + 9 + g + 3k = 2 3 1k + 12 3k 3k 2 3 3k 6k 6 3 = 1k + 12 + 13k + 32 = + + + = 1k 2 + 3k + 22 = 1k + 12 1k + 22 = 3 1k + 12 + 1 4 . 2 2 2 2 2 2 2 2 2. c1 = 2, c2 = 1, c3 =
8 32 , c = 1, c5 = 9 4 25
3. a1 = 3, a2 = 2, a3 =
27. (I) n = 1: 2 # 31 - 1 = 2 and 31 - 1 = 2 (II) If 2 + 6 + 18 + g + 2 # 3k - 1 = 3k - 1, then 2 + 6 + 18 + g + 2 # 3k - 1 + 2 # 31k + 12 - 1 = 12 + 6 + 18 + g + 2 # 3k - 1 2 + 2 # 3k = 3k - 1 + 2 # 3k = 3 # 3k - 1 = 3k + 1 - 1. 1 28. (I) n = 1: 13 # 1 - 22 2 = 1 and # 1 # 3 6 112 2 - 3 112 - 1 4 = 1 2 1 2 2 2 (II) If 1 + 4 + 7 + g + 13k - 22 2 = k 16k 2 - 3k - 12, then 12 + 42 + 72 + g + 13k - 22 2 + 3 3 1k + 12 - 2 4 2 2 1 1 = 312 + 42 + 72 + g + 13k - 22 2 4 + 13k + 12 2 = k 16k 2 - 3k - 12 + 13k + 12 2 = 16k 3 - 3k 2 - k2 + 19k 2 + 6k + 12 2 2 1 1 1 3 2 2 2 = 16k + 15k + 11k + 22 = 1k + 12 16k + 9k + 22 = 1k + 12 3 6 1k + 12 - 3 1k + 12 - 1 4 . 2 2 2
29. 10 30. 32x5 + 240x4 + 720x3 + 1080x2 + 810x + 243 31. 81x4 - 432x3 + 864x2 - 768x + 256 32. 144 33. 6048 34. (a) 8 bricks (b) 1100 bricks 3 3 135 3 n 35. 360 tiles 36. (a) 20 a b = ft (b) 20 a b ft (c) 13 times (d) 140 ft 37. $151,873.77 4 16 4 38. $23,397.17
Chapter Test (page 892) 1. 0,
3 8 5 24 , , , 10 11 4 13
2. 4, 14, 44, 134, 404
7. Geometric; r = 4; Sn =
2 (1 - 4n) 3
3. 2 -
3 4 61 + = 4 9 36
4. -
10 k + 1 5. a ( - 1)k a b 6. Neither k + 4 k=1 1 n 9. Arithmetic; d = - ; Sn = (27 - n) 2 4
1 14 73 308 680 = 3 9 27 81 81
8. Arithmetic: d = - 8; Sn = n(2 - 4n)
AN98
ANSWERS Chapter Test
10. Geometric; r =
2 125 2 n ;S = c1 - a b d 5 n 3 5
11. Neither
14. First show that the statement holds for n = 1. a1 +
12. Converges;
1024 5
13. 243m5 + 810m4 + 1080m3 + 720m2 + 240m + 32
1 b = 1 + 1 = 2. The equality is true for n = 1, so Condition I holds. Next, assume that 1
a1 +
1 1 1 1 b a1 + b a1 + b g a1 + b = n + 1 is true for some k, and determine whether the formula then holds for k + 1. Assume that 1 2 3 n 1 1 1 1 1 1 1 1 1 a1 + b a1 + b a1 + b g a1 + b = k + 1. Now show that a1 + b a1 + b a1 + b g a1 + b a1 + b 1 2 3 k 1 2 3 k k + 1 = (k + 1) + 1 = k + 2. Do this as follows: a1 +
1 1 1 1 1 1 1 1 1 1 b a1 + b a1 + b g a1 + b a1 + b = c a1 + b a1 + b a1 + b g a1 + b d a1 + b 1 2 3 k k + 1 1 2 3 k k + 1 1 1 b (induction assumption) = (k + 1) # 1 + (k + 1) # = k + 1 + 1 = k + 2 = (k + 1) a1 + k + 1 k + 1 Condition II also holds. Thus, the formula holds true for all natural numbers. 15. After 10 years, the car will be worth $6103.11. 16. The weightlifter will have lifted a total of 8000 pounds after 5 sets.
Cumulative Review (page 892) 1. { - 3, 3, - 3i, 3i}
2. (a)
(b) e a
y
B
- 1 + 13601 - 1 + 23601 - 1 + 13601 - 1 + 23601 , b, a , bf 18 6 B 18 6
(c) The circle and the parabola intersect at ⴚ4
6
x
a
B
- 1 + 13601 - 1 + 23601 - 1 + 13601 - 1 + 23601 b, a b. , , 18 6 B 18 6
5 6x + 3 1 (d) e x ` x ≠ f 3. e ln a b f 4. y = 5x - 10 5. 1x + 12 2 + 1y - 22 2 = 25 6. (a) 5 (b) 13 (c) 2 2x - 1 2 y2 1 2x x2 -1 -1 2 (g) g (x) = (x - 1); all reals (h) f (x) = ; {x x ≠ 3} 7. + = 1 8. (x + 1) = 4(y - 2) 2 x - 3 7 16
(e)
7x - 2 x - 2
(f) {x x ≠ 2}
CHAPTER 12 Counting and Probability 12.1 Assess Your Understanding (page 899) 5. subset; ⊆ 6. finite 7. n 1A2 + n 1B2 - n 1AxB2 8. True 9. ∅, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {b, c, d}, {a, c, d}, {a, b, d}, {a, b, c, d} 11. 25 13. 40 15. 25 17. 37 19. 18 21. 5 23. 15 different arrangements 25. 9000 numbers 27. 175; 125 29. (a) 15 (b) 15 (c) 15 (d) 25 (e) 40 31. (a) 13.6 million (b) 78.9 million 33. 480 portfolios y 37. 2, 5, - 2 38. 163.9 ft 39. - 14i + 11j + 13k 36. 6
(x22)2 1 (y11)2 2 9
26
6 x
26
12.2 Assess Your Understanding (page 906) n! n! 6. 7. 30 9. 24 11. 1 13. 1680 15. 28 17. 35 19. 1 21. 10,400,600 1n - r2! 1n - r2!r! 23. 5abc, abd, abe, acb, acd, ace, adb, adc, ade, aeb, aec, aed, bac, bad, bae, bca, bcd, bce, bda, bdc, bde, bea, bec, bed, cab, cad, cae, cba, cbd, cbe, cda, cdb, cde, cea, ceb, ced, dab, dac, dae, dba, dbc, dbe, dca, dcb, dce, dea, deb, dec, eab, eac, ead, eba, ebc, ebd, eca, ecb, ecd, eda, edb, edc 6; 60 25. 5123, 124, 132, 134, 142, 143, 213, 214, 231, 234, 241, 243, 312, 314, 321, 324, 341, 342, 412, 413, 421, 423, 431, 432 6; 24 27. 5abc, abd, abe, acd, ace,ade, bcd, bce, bde, cde 6; 10 29. 5123, 124, 134, 234 6; 4 31. 16 33. 8 35. 24 37. 60 39. 18,278 41. 35 43. 1024 45. 120 47. 132,860 49. 336 51. 90,720 53. (a) 63 (b) 35 (c) 1 55. 1.157 * 1076 57. 362,880 59. 660 61. 15 63. (a) 125,000; 117,600 (b) A better name for a combination lock would be a permutation lock, because the order of the numbers matters. 67. 10 sq. units 3. permutation
4. combination
68. (g ∘ f )(x) = 4x2 - 2x - 2
5.
69. sin 75° =
Historical Problem (page 916)
22 + 26 22 + 26 1 ; cos 15° = 22 + 13 or cos 15° = 4 2 4
70. a5 = 80
1. (a) 5AAAA, AAAB, AABA, AABB, ABAA, ABAB, ABBA, ABBB, BAAA, BAAB, BABA, BABB, BBAA, BBAB, BBBA, BBBB6; 11 5 P(A wins) = ; P(B wins) = 16 16 C14, 22 + C14, 32 + C14, 42 C14, 32 + C14, 42 6 + 4 + 1 11 4 + 1 5 (b) P1A wins2 = = = ; P1B wins2 = = = ; the results are the same. 16 16 16 16 24 24
ANSWERS Cumulative Review
AN99
12.3 Assess Your Understanding (page 916) 1. equally likely 2. complement
3. F 4. T 5. 0, 0.01, 0.35, 1 7. Probability model 9. Not a probability model 1 1 1 1 11. S = 5HH, HT, TH, TT 6; P1HH2 = , P1HT2 = , P1TH2 = , P1TT2 = 4 4 4 4 13. S = 5HH1, HH2, HH3, HH4, HH5, HH6, HT1, HT2, HT3, HT4, HT5, HT6, TH1, TH2, TH3, TH4, TH5, TH6, TT1, TT2, TT3, TT4, TT5, TT6 6; 1 each outcome has the probability of . 24 1 15. S = 5HHH, HHT, HTH, HTT, THH, THT, TTH, TTT6; each outcome has the probability of . 8 17. S = 5 1 Yellow, 1 Red, 1 Green, 2 Yellow, 2 Red, 2 Green, 3 Yellow, 3 Red, 3 Green, 4 Yellow, 4 Red, 4 Green 6; each outcome has the probability 1 1 1 1 of ; thus, P12 Red2 + P14 Red2 = + = . 12 12 12 6 19. S = 5 1 Yellow Forward, 1 Yellow Backward, 1 Red Forward, 1 Red Backward, 1 Green Forward, 1 Green Backward, 2 Yellow Forward, 2 Yellow Backward, 2 Red Forward, 2 Red Backward, 2 Green Forward, 2 Green Backward, 3 Yellow Forward, 3 Yellow Backward, 3 Red Forward, 3 Red Backward, 3 Green Forward, 3 Green Backward, 4 Yellow Forward, 4 Yellow Backward, 4 Red Forward, 4 Red Backward, 4 Green Forward, 4 1 1 1 1 Green Backward 6 ; each outcome has the probability of ; thus, P11 Red Backward2 + P11 Green Backward2 = + = . 24 24 24 12 21. S = 511 Red, 11 Yellow, 11 Green, 12 Red, 12 Yellow, 12 Green, 13 Red, 13 Yellow, 13 Green, 14 Red, 14 Yellow, 14 Green, 21 Red, 21 Yellow, 21 Green, 22 Red, 22 Yellow, 22 Green, 23 Red, 23 Yellow, 23 Green, 24 Red, 24 Yellow, 24 Green, 31 Red, 31 Yellow, 31 Green, 32 Red, 32 Yellow, 32 Green, 33 Red, 33 Yellow, 33 Green, 34 Red, 34 Yellow, 34 Green, 41 Red, 41 Yellow, 41 Green, 42 Red, 42 Yellow, 42 Green, 43 Red, 43 Yellow, 43 Green, n 1E2 4 1 1 = = . 44 Red, 44 Yellow, 44 Green6; each outcome has the probability of ; thus, E = 522 Red, 22 Green, 24 Red, 24 Green 6; P1E2 = 48 n 1S2 48 12 4 1 1 2 3 1 1 1 23. A, B, C, F 25. B 27. P1H2 = ; P1T2 = 29. P112 = P132 = P152 = ; P122 = P142 = P162 = 31. 33. 35. 37. 5 5 9 9 10 2 6 8 1 1 1 17 11 1 3 2 41. 43. 45. 0.55 47. 0.70 49. 0.30 51. 0.88 53. 0.67 55. 0.936 57. 59. 61. 63. 65. 39. 4 6 18 20 20 2 10 5 25 25 67. (a) 0.57 (b) 0.95 (c) 0.83 (d) 0.38 (e) 0.29 (f) 0.05 (g) 0.78 (h) 0.71 69. (a) (b) 71. 0.167 73. 0.000033069 33 33 74. 2; left; 3; down 75. ( - 3, 3 23) 76. {22} 77. (2, - 3, - 1)
Review Exercises (page 920) 1. ∅, 5Dave 6, 5Joanne 6, 5Erica 6, 5Dave, Joanne 6, 5Dave, Erica 6, 5Joanne, Erica 6, 5Dave, Joanne, Erica 6 2. 17 3. 24 4. 29 5. 34 6. 7 7. 45 8. 25 9. 7 10. 336 11. 56 12. 48 13. 128 14. 3024 15. 1680 16. 1820 17. 1,600,000 18. 216,000 19. 256 (allowing numbers with initial zeros, such as 011) 20. 2522520 21. (a) 381,024 (b) 1260 22. (a) 8.634628387 * 1045 (b) 0.6531 (c) 0.3469 23. (a) 0.081 (b) 0.919 24. 0.13 25. 0.5; 0.24 26. (a) 0.68 (b) 0.58 (c) 0.32
Chapter Test (page 921) 1. 22 2. 3 3. 8 4. 45 5. 5040 6. 151,200 7. 462 8. There are 54,264 ways to choose 6 different colors from the 21 available colors. 9. There are 840 distinct arrangements of the letters in the word REDEEMED. 10. There are 56 different exacta bets for an 8-horse race. 11. There are 155,480,000 possible license plates using the new format. 12. (a) 0.95 (b) 0.30 13. (a) 0.25 (b) 0.55 1 1250 14. 0.19 15. P1win on $1 play2 = ≈ 0.0000000083 16. P1exactly 2 fours2 = ≈ 0.1608 120,526,770 7776
Cumulative Review (page 922) 1. e
1 1 22 22 i, + if 3 3 3 3
3.
y 10
2. (ⴚ5, 0)
(1, 0) 10 x (0, ⴚ5)
(ⴚ2, ⴚ9) x ⴝ ⴚ2
5. b -
1 27 1 27 1 + i, - i, - , 3 r 2 2 2 2 5
6.
y 10 (1, 6)
yⴝ5
5 x
Domain: all real numbers Range: 5y y 7 5 6 Horizontal asymptote: y = 5
4. 5x 3.99 … x … 4.01 6 or 33.99, 4.01 4
y 5
(ⴚ2, ⴚ2)
5 x (0, ⴚ2)
(ⴚ1, ⴚ4)
7. 2
8 8. e f 3
11.
y
9. x = 2, y = - 5, z = 3
3 P x
10. 125; 700
12. a ≈ 6.09, B ≈ 31.9°, C ≈ 108.1°; area ≈ 14.46 square units 1 13. ≈ 0.00000000571; 175,223,510 The new format has decreased the probability of winning the jackpot.
AN100
ANSWERS Section 13.1
CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function 13.1 Assess Your Understanding (page 927) 3. lim f 1x2 xSc
23.
4. does not exist
5. True
6. False
25.
y 15
7. 32
9. 1
11. 4
13. 2
27.
y 5
15. 0
17. 3
19. 4 29.
y 8
21. Does not exist y 1.25
31. P
5 x
y 2.5
x
2.5 x
5 x 5 x
lim f 1x2 = 13
lim f 1x2 = - 3
xS4
33.
35.
y 5
37.
lim f 1x2 = - 1
lim f 1x2 = 0
xS0
41.
y 5
y 5
5 x
5 x
lim f 1x2 does not exist.
xS1
47. 0 49. d = 4 25; M = (4, - 7)
45. 1.6
39.
5 x
xS0
lim f 1x2 = 1
x S p/2
y 5
5 x
x S -1
lim f 1x2 = 1
x S -3
y 5
5 x
43. 0.67
lim f 1x2 = 6
xS2
lim f 1x2 = 0
lim f 1x2 = 0
xS0
xS0
50. Center: (2, - 1); foci: (2, - 3), (2, 1); vertices: (2, 213 - 1), (2, - 213 - 1)
2p 52. ¢ 4, ≤ 3
51. $7288.48
13.2 Assess Your Understanding (page 934) 3. c
1. product 2. b 33. 3
35. 0
37.
57.
39.
8 5
41. 0
43. 5
58. g - 1(x) =
y
12 3
2 3
4. True 5. False
2
f (x) 5 x 1x 11
6. False 45. 6
7. 5
47. 0
3 - x x - 2
11. - 10
9. 4
51. - 1
49. 0
59. 60° or
p red 3
13. 80
53. 1
15. 8
55.
17. 8
19. - 1
21. 8
23. 3
25. - 1
27. 32
29. 2
31.
7 6
3 4
60. x4 + 8x3 + 24x2 + 32x + 16
3x
23 212
13.3 Assess Your Understanding (page 940) 7. one-sided 8. lim+ f 1x2 = R xSc
9. continuous; c 10. False
13. 5x - 8 … x 6 - 6 or - 6 6 x 6 4 or 4 6 x … 6 6
11. True 12. True
2 15. - 8, - 5, - 3 17. f 1 - 82 = 0; f 1 - 42 = 2 19. q 21. 2 23. 1 25. Limit exists; 0 27. No 29. Yes 31. No 33. 5 35. 7 37. 1 39. 4 41. 3 3 43. 45. Continuous 47. Continuous 49. Not continuous 51. Not continuous 53. Not continuous 55. Continuous 57. Not continuous 2 59. Continuous 61. Continuous for all real numbers 63. Continuous for all real numbers 65. Continuous for all real numbers kp 67. Continuous for all real numbers except x = , where k is an odd integer 69. Continuous for all real numbers except x = - 2 and x = 2 2 71. Continuous for all positive real numbers except x = 1 73. Discontinuous at x = - 1 and x = 1; 1 1 lim R1x2 = : hole at a1, b xS1 2 2 lim -R1x2 = - q ; lim + R1x2 = q ;
75. Discontinuous at x = - 1 and x = 1; 1 1 lim R1x2 = : hole at a - 1, b x S -1 2 2 lim-R1x2 = - q ; lim+ R1x2 = q ;
y 5
5 x
y0
x S -1
x S -1
vertical asymptote at x = - 1
79. x = - 3: asymptote; x = 2: hole 85. 20
−20
5
20
−3
91. Vertical: x = 4; horizontal: y = 3
−20
92. 60
y1 5 x
xS1
vertical asymptote at x = 1
x 1
3 77. x = - 2 2 : asymptote; x = 1: hole 3 83.
−3
xS1
y 5
93. ln ¢
x 5y 2 z4
≤
3 94. C 1 0
x1
3 81. x = - 2 2 : asymptote; x = - 1: hole 87. 3
−3
1
−2
1 0 1
2 4 23 5S -3 -2
AN101
ANSWERS Section 13.5
13.4 Assess Your Understanding (page 948) 3. tangent line
4. derivative
5. velocity
9. mtan = 3
6. True 7. True 8. True
11. mtan = - 2
13. mtan = 12
y 10 (1, 8)
f (x) 3x 2 f (x) x 2 2
(1, 3)
(2, 12) f (x) 2x 2 x
17. mtan = - 4
y 18
10
y 13x 16 (1, 6)
(2, 10) f(x) x 2 2x 3 5 x
f (x) x 3 x
5 x
52. (2, 6), ( - 1, 3)
53. 10
5 x y 5x 2
5 x
19. mtan = 13
y
(1, 3)
y 12x 12
5 x
5 x
7 51. Vertex: (1, 3); focus: ¢ 1, ≤ 2
y 8
y 15
y 2x 1
f(x) 3x 5
y 4x 2
15. mtan = 5
y 10
21. - 4 23. 0 25. 7 27. 7 29. 3 31. 1 33. 60 35. - 0.8587776956 37. 1.389623659 39. 2.362110222 41. 3.643914112 43. 18p ft 3/ft 45. 16p ft 3/ft 47. (a) 6 sec (b) 64 ft/sec (c) 1 - 32t + 962 ft/sec (d) 32 ft/sec (e) 3 sec (f) 144 ft (g) - 96 ft/sec 1 49. (a) - 23 ft/sec (b) - 21 ft/sec (c) - 18 ft/sec 3 (d) s 1t2 = - 2.631t 2 - 10.269t + 999.933 (e) Approximately - 15.531 ft/sec
54. 23.66 sq. units
13.5 Assess Your Understanding (page 955) b
3.
La
9. (a)
f 1x2 dx
b
4.
La
f 1x2 dx
5. 3
7. 56 11. (a)
y 24
13. (a)
y 10
15. (a)
y 20
y 90
8 x
(b) 36 (d) 45
17. (a)
(c) 72 (e) 63
(f) 54
(b) 18 (c) 9 63 45 (d) (e) 4 4 19. (a)
y 1.5
5 x
5 x
4 x
51 (b) 22 (c) 2
27 (f) 2
4
(d) 21. (a)
y 20
L0
(b) 36
1x2 + 22 dx
(e)
88 3
y 1.2
(d)
L0
x3 dx
(e) 64
23. (a) Area under the graph of f 1x2 = 3x + 1 from 0 to 4 (b) y P x
5 x
(c) 49 4
16
4 x
(b)
25 12
(c)
(b) 11.475
4609 2520
(c) 15.197
(b) 1.896 p
3
(c) 1.974
5 x
x
(d) e dx (e) 19.718 (d) sin x dx (e) 2 (c) 28 L-1 L0 1 dx (e) 1.609 (d) x L1 27. (a) Area under the graph of f 1x2 = sin x 29. (a) Area under the graph of f 1x2 = e x from 25. (a) Area under the graph of f 1x2 = x2 - 1 p 0 to 2 from 0 to from 2 to 5 2 (b) y (b) y (b) y 8 5
30
1.5
P x
4 x
5 x
(c) 36
(c) 1
31. Using left endpoints: n = 2: 0 + 0.5 = 0.5; n = 4: 0 + 0.125 + 0.25 + 0.375 = 0.75; 10 10 + 0.182 = 0.9; 2 n = 100: 0 + 0.0002 + 0.0004 + 0.0006 + g + 0.0198 100 = 10 + 0.01982 = 0.99; 2 n = 10: 0 + 0.02 + 0.04 + 0.06 + g + 0.18 =
(c) 6.389 Using right endpoints: n = 2: 0.5 + 1 = 1.5; n = 4: 0.125 + 0.25 + 0.375 + 0.5 = 1.25; 10 10.02 + 0.202 = 1.1; n = 10: 0.02 + 0.04 + 0.06 + g + 0.20 = 2 n = 100: 0.0002 + 0.0004 + 0.0006 + g + 0.02 100 10.0002 + 0.022 = 1.01 = 2
AN102 y
33.
4
ANSWERS Section 13.5
34. J
f(x) 5 log2x
22
19 43
22 50
2 R 4
35. 4x + 2h + 3
36.
1 1 4 x - 2 x + 2 (x + 2)2
x
8
Review Exercises (page 957) 1 1 3 3 7. 4 8. 9. 0 10. 11. 12. Continuous 13. Not continuous 14. Continuous 15. Not continuous 4 2 2 2 16. 5x - 6 … x 6 2 or 2 6 x 6 5 or 5 6 x … 6 6 17. All real numbers 18. 1, 6 19. 4 20. f 1 - 62 = 2; f 1 - 42 = 1 21. 4 22. - 2 23. - q 24. q 25. Does not exist 26. No 27. Yes 1. 8 2. - 343 3. 4 4. 25 5. 27 6. -
28. R is discontinuous at x = - 4 and x = 4. 1 1 lim R1x2 = - : hole at a - 4, - b x S -4 8 8 lim-R1x2 = - q ; lim+ R1x2 = q :
29. Undefined at x = 2 and x = 9; R has a hole at x = 2 and a vertical asymptote at x = 9.
y 5
5 x
y0
xS4
xS4
The graph of R has a vertical asymptote at x = 4.
x4
31. mtan = 0
30. mtan = 12 y 21 f (x) 2x 2 8x
y 5
32. mtan = 16 y 21
f (x) x 2 2x 3
(2, 12)
(1, 10) 5 x 10 x (1, 4)
y 12x 2
5 x y 16x 20
f (x) x 3 x 2
y 4
33. - 24 34. - 3 35. 7 36. - 158 37. 0.6662517653 38. (a) 7 sec (b) 6 sec (c) 64 ft/sec (d) 1 - 32t + 962 ft/sec (e) 32 ft/sec (f) At t = 3 sec (g) - 96 ft/sec (h) - 128 ft/sec 39. (a) $61.29/watch (b) $71.31/watch (c) $81.40/watch (d) R1x2 = - 0.25x2 + 100.01x - 1.24 (e) Approximately $87.51/watch 40. (a)
41. (a)
y 14
42. (a)
y 5
43. (a) Area under the graph of f 1x2 = 9 - x2 from - 1 to 3 (b) y
y 2
10 5 x
4 x 3 x
(b) 24 (d) 26
(c) 32 (e) 30
(f) 28
77 (c) 8
(b) 10 2
(d)
L-1
49 ≈ 1.36 (c) 1.02 36 4 1 (d) dx (e) 0.75 L1 x2 (b)
14 - x2 2 dx
44. (a) Area under the graph of f 1x2 = e x from - 1 to 1
(b)
(e) 9
4 x
(c)
(c) 2.35
y 4
2.5 x
Chapter Test (page 959) 1 2 3. 5 4. - 2 5. 135 6. 7. - 1 8. - 3 9. 5 10. 2 11. Limit exists; 2 3 3 12. (a) Yes (b) No; lim f 1x2 ≠ f 112 (c) No; lim- f 1x2 ≠ f 132 (d) Yes 13. x = - 7: asymptote; x = 2: hole 1. - 5
2.
xS1
xS3
4
14. (a) 5 (b) y = 5x - 19 (c) f(x) 4x 2 11x 3 y
15. (a)
16.
y 5
5 x
30 y 5x 19 5 x (2, 9)
(b) 13.359 (c) 4p ≈ 12.566
L1
1 - x2 + 5x + 32 dx
1 17. 35 ft/sec 3
80 3
AN103
ANSWERS Section A.4
APPENDIX A Review A.1 Assess Your Understanding (page A10) 1. variable 2. origin 3. strict 4. base; exponent or power 5. True 6. True 7. False 8. False 9. 51, 2, 3, 4, 5, 6, 7, 8, 9 6 11. 54 6 13. 51, 3, 4, 6 6 15. 50, 2, 6, 7, 8 6
17. 50, 1, 2, 3, 5, 6, 7, 8, 9 6
19. 50, 1, 2, 3, 5, 6, 7, 8, 9 6
21.
– 2.5
–1
31 4
0
23. 7
5 2
25. 7
0.25
27. 7 51. - 28
29. = 53.
31. 6
4 5
55. 0
77. 5x x ≠ - 4 6 105. - 4
57. 1
79. 0°C
107. 5
131. C = pd
33. x 7 0
109. 4
59. 5
35. x 6 2 61. 1
81. 25°C 111. 2
37. x … 1
63. 22
83. 16
113. 15
115.
39.
41.
67. x = 0
65. 2
85.
1 16
1 2
117. 10; 0
87.
1 9
69. x = 3
89. 9
91. 5
119. 81
47. 6
49. 4
95. 64x6
121. 304,006.671
75. 5x x ≠ 5 6
73. x = 0, x = 1, x = - 1
71. None
93. 4
23 2 4 x 135. V = pr 3 137. V = x3 139. (a) $6000 4 3 1 147. No; is larger; 0.000333 c 149. No 3
(b) No
45. 2
–1
133. A =
145. (a) Yes
43. 1
–2
97.
123. 0.004
x
4
y
2
99.
x y
101. -
125. 481.890
141. x - 4 Ú 6
(b) $8000
8x3z 9y
127. 0.000
103.
16x2 9y2
129. A = lw
143. (a) 2 … 5
(b) 6 7 5
A.2 Assess Your Understanding (page A19) 1 bh 3. C = 2pr 4. similar 5. True 6. True 7. False 8. True 9. True 10. False 11. 13 13. 26 15. 25 2 17. Right triangle; 5 19. Not a right triangle 21. Right triangle; 25 23. Not a right triangle 25. 8 in2 27. 4 in2 29. A = 25p m2; C = 10p m 256 31. V = 224 ft 3; S = 232 ft 2 33. V = p cm3; S = 64p cm2 35. V = 648p in3; S = 306p in2 37. p square units 39. 2p square units 3 41. x = 4 units; A = 90°; B = 60°; C = 30° 43. x = 67.5 units; A = 60°; B = 95°; C = 25° 45. About 16.8 ft 47. 64 ft 2 1. right; hypotenuse
2. A =
49. 24 + 2p ≈ 30.28 ft 2; 16 + 2p ≈ 22.28 ft
51. 160 paces
53. About 5.477 mi
55. From 100 ft: 12.2 mi; From 150 ft: 15.0 mi
A.3 Assess Your Understanding (page A30) 1. 4; 3 2. x4 - 16 3. x3 - 8 4. False 5. True 6. False 7. Monomial; variable: x; coefficient: 2; degree: 3 9. Not a monomial; the exponent of the variable is not a nonnegative integer 11. Monomial; variables: x, y; coefficient: - 2; degree: 3 13. Not a monomial; the exponent of one of the variables is not a nonnegative integer 15. Not a monomial; it has more than one term 17. Yes; 2 19. Yes; 0 21. No; the exponent of the variable of one of the terms is not a nonnegative integer 23. Yes; 3 25. No; the polynomial of the denominator has a degree greater than 0 27. x2 + 7x + 2 3 2 5 4 2 29. x - 4x + 9x + 7 31. 6x + 5x + 3x + x 33. 7x2 - x - 7 35. - 2x3 + 18x2 - 18 37. 2x2 - 4x + 6 39. 15y2 - 27y + 30 3 2 5 2 3 2 2 2 2 41. x + x - 4x 43. - 8x - 10x 45. x + 3x - 2x - 4 47. x + 6x + 8 49. 2x + 9x + 10 51. x - 2x - 8 53. x2 - 5x + 6 55. 2x2 - x - 6 57. - 2x2 + 11x - 12 59. 2x2 + 8x + 8 61. x2 - xy - 2y2 63. - 6x2 - 13xy - 6y2 65. x2 - 49 67. 4x2 - 9 69. x2 + 8x + 16 71. x2 - 8x + 16 73. 9x2 - 16 75. 4x2 - 12x + 9 77. x2 - y2 79. 9x2 - y2 81. x2 + 2xy + y2 83. x2 - 4xy + 4y2 85. x3 - 6x2 + 12x - 8 87. 8x3 + 12x2 + 6x + 1 89. 4x2 - 11x + 23; remainder - 45 91. 4x - 3; remainder x + 1 1 5 1 2 2 2 2 99. - 4x2 - 3x - 3; remainder - 7 93. 5x - 13; remainder x + 27 95. 2x ; remainder - x + x + 1 97. x - 2x + ; remainder x + 2 2 2 101. x2 - x - 1; remainder 2x + 2 103. x2 + ax + a2; remainder 0
A.4 Assess Your Understanding (page A39) 1. 3x(x - 2)(x + 2) 15. (x + 1)(x - 1) 29. (2x + 1)
2
2. prime 2
79. (x + 1)(x + 10)
81. (x - 7)(x - 3)
95. 2(3x + 1)(x + 1)
123. 2(3x + 4)(9x + 13)
35. (x + 3)(x - 3x + 9)
69. 25; (x + 5)2 83. 4(x2 - 2x + 8)
47. (x - 8)(x + 1)
71. 9; ( y - 3)2 85. Prime
37. (2x + 3)(4x - 6x + 9) 2
49. (x + 8)(x - 1)
27. (x - 5)2 39. (x + 2)(x + 3)
51. (x + 2)(2x + 3)
61. (x + 2)(3x - 4)
63. (x - 2)(3x + 4)
1 1 ; ax - b 75. (x + 6)(x - 6) 77. 2(1 + 2x)(1 - 2x) 16 4 87. - (x - 5)(x + 3) 89. 3(x + 2)(x - 6) 91. y2( y + 5)( y + 6) 73.
99. (x - 1)2(x2 + x + 1)2
119. (x - 1)(x + 1)(x + 2)
127. 5(x + 3)(x - 2)2(x + 1)
25. (x + 2)
23. (x + 1)
13. 3xy(x - 2y + 4)
2
2
109. - (3x - 1)(3x + 1)(x2 + 1)
117. (x + 5)(3x + 11)
11. 2x(x - 1) 2
59. (z + 1)(2z + 3)
97. (x - 3)(x + 3)(x + 9) 2
125. 2x(3x + 5)
9. x(x2 + x + 1)
21. (5x + 2)(5x - 2)
57. (3x + 1)(x + 1)
107. (2y - 5)(2y - 3)
115. 13x - 52 19x - 3x + 72 2
7. a(x2 + 1) 2
45. (x - 8)(x - 2)
55. (2x + 3)(3x + 2) 67. (x + 4)(3x - 2)
105. - (4x - 5)(4x + 1)
19. (x + 4)(x - 4)
33. (x - 3)(x + 3x + 9)
65. (x + 4)(3x + 2)
93. (2x + 3)
5. 3(x + 2)
2
43. (x + 5)(x + 2)
53. (x - 2)(2x + 1)
2
4. False
17. (2x + 1)(2x - 1)
31. (4x + 1)
41. (x + 6)(x + 1)
3. True
101. x5(x - 1)(x + 1)
111. (x + 3)(x - 6)
113. (x + 2)(x - 3)
121. (x - 1)(x + 1)(x - x + 1)
129. 3(4x - 3)(4x - 1)
2
131. 6(3x - 5)(2x + 1)2(5x - 4)
133. The possibilities are (x { 1)(x { 4) = x { 5x + 4 or (x { 2)(x { 2) = x { 4x + 4, none of which equals x2 + 4. 2
2
103. (4x + 3)2
AN104
ANSWERS Section A.5
A.5 Assess Your Understanding (page A44) 3. True 4. True 5. x2 + x + 4; remainder 12 11. 4x5 + 4x4 + x3 + x2 + 2x + 2; remainder 7 19. Yes 21. Yes 23. No 25. Yes 27. - 9
1. quotient; divisor; remainder 2. - 3) 2 0 - 5 1 9. x4 - 3x3 + 5x2 - 15x + 46; remainder - 138 15. x4 + x3 + x2 + x + 1; remainder 0 17. No
A.6 Assess Your Understanding (page A53) 1. lowest terms 2. least common multiple 3. True 4. False 19.
2x(x2 + 4x + 16) x + 4
21.
8 3x
23.
x - 3 x + 7
25.
5.
3 x - 3
4x (x - 2)(x - 3)
7.
(x - 2)(x + 2) x + 5 3x - 2 x + 9 4 - x 37. 39. 41. 43. 2 2x - 3 x - 3 2x - 1 x - 2 2(x2 - 2) 51. 53. (x - 2)(x + 2)(x + 1) 55. x(x - 1)(x + 1) x(x - 2)(x + 2) 5x (x - 6)(x - 1)(x + 4)
71.
-1 x(x + h)
87.
73.
x + 1 x - 1
(x + 1)(x - 1) (x + 1) 2
63.
2
89.
75.
2(2x2 + 5x - 2)
65.
(x - 2)(x + 2)(x + 3)
(x - 1)(x + 1)
77.
2x(2x + 1)
x(3x + 2) (3x + 1)
(x - 1) (x + 1) 79.
2
93. f =
2
y + 5
4x 47.
(x - 1)(x + 2)
67.
2
13.
2( y + 1)
(x - 4)2
57. x3(2x - 1)2
2
(x - 2)(x + 1)
(x + 1)
29. -
11.
2(x + 5)
45.
2(5x - 1)
2
4x 2x - 1
5x + 1
(x + 3)(3x - 1)
91. -
2
9.
4 5(x - 1)
27.
35.
61.
x 3
7. 3x2 + 11x + 32; remainder 99 13. 0.1x2 - 0.11x + 0.321; remainder - 0.3531
31.
x + 5 x - 1
15. - (x + 7)
(x + 3)2
33.
(x - 3)2
3x2 - 2x - 3 (x + 1)(x - 1)
17.
3 5x(x - 2)
(x - 4)(x + 3) (x - 1)(2x + 1)
49.
- (11x + 2) (x + 2)(x - 2)
59. x(x - 1)2(x + 1)(x2 + x + 1)
- x2 + 3x + 13 (x - 2)(x + 1)(x + 4)
- 2x(x2 - 2)
81.
(x + 2)(x2 - x - 3)
R1 # R2 2 ; m (n - 1)(R1 + R2) 15
69.
-1 x - 1
x3 - 2x2 + 4x + 3 x2(x + 1)(x - 1)
83.
3x - 1 2x + 1
85.
19 (3x - 5)2
A.7 Assess Your Understanding (page A60) 3. index
4. True
25. 12 23 43.
22 2
65.
27 22 32
81.
27. 7 22 45. -
67. x7>12
2 + x 2(1 + x)
215 5
3>2
83.
5. cube root 29. 22 47.
15
93. (x2 + 4)1>3(11x2 + 12) 107. (a) 15,660.4 gal
49.
23 71. x2>3y
4 - x 3>2
85.
7. 3
73.
8x5>4 y
x (x - 1) 2
1>2
87.
15. x3y2
17. x2y
19. 6 1x
3 52 4 2
2x + h - 2 2x2 + xh h
55. 4
x(3x2 + 2)
22x + 5
51.
53.
3x + 2 (1 + x)
77.
1>2
1 - 3x2 2 1x(1 + x2)2
95. (3x + 5)1>3(2x + 3)1>2(17x + 27)
(b) 390.7 gal
3 13. - 2x 1 x
3 37. (2x - 1) 2 2x
75.
3>4
11. 2 22
35. x - 2 1x + 1
8 25 - 19 41
1 2
9. - 2
3 33. - 2 2
31. 2 23
+ 22 2 23
69. xy2
(x + 4)
6. False
109. 2 22p ≈ 8.89 sec
111.
97.
(x2 + 1)1>2
79.
39. (2x - 15) 22x
2x1>2
99. 1.41
3 41. - (x + 5y) 2 2xy
59. 64
61.
1 27
63.
27 22 32
10 2 1x - 52 14x + 32
1 89. (5x + 2)(x + 1)1>2 2 3(x + 2)
57. - 3
3 23. 15 2 3
21. 6x 1x
91. 2x1>2(3x - 4)(x + 1)
101. 1.59
103. 4.89
105. 2.15
p 23 ≈ 0.91 sec 6
A.8 Assess Your Understanding (page A70) 5 5. equivalent equations 6. identity 7. extraneous 8. True 9. {4} 11. e f 13. 5 - 1 6 15. 5 - 18 6 17. 5 - 3 6 19. 5 - 16 6 21. 50.5 6 23. {2} 4 1 25. {2} 27. {3} 29. {0, 9} 31. 5 - 3, 0, 3 6 33. {21} 35. {6} 37. ∅ or { } 39. 5 - 5, 0, 4 6 41. 5 - 1, 1 6 43. e - 2, , 2 f 45. {1} 47. No real solution 2 49. 5 - 13 6
51. {3}
53. {8}
55. 5 - 1, 3 6
57. {1, 5}
59. {5}
61. {2}
63. e -
11 f 6
65. 5 - 6 6
67. R =
R1R2 R1 + R2
69. R =
mv2 F
71. r =
S - a S
73. The distance is approximately 229.94 ft. 75. Approx. 221 ft 79. The apparent solution - 1 is extraneous. 81. When multiplying both sides by x + 3, we are actually multiplying both sides by 0 when x = - 3. This violates the Multiplication Property of Equality.
A.9 Assess Your Understanding (page A78) 1. mathematical modeling 2. interest 3. uniform motion 4. False 5. True 6. 100 - x 7. A = pr 2; r = radius, A = area 9. A = s2; A = area, s = length of a side 11. F = ma; F = force, m = mass, a = acceleration 13. W = Fd; W = work, F = force, d = distance 15. C = 150x; C = total variable cost, x = number of dishwashers 17. Invest $31,250 in bonds and $18,750 in CDs. 19. $11,600 was loaned out at 8%. 21. Mix 75 lb of Earl Grey tea with 25 lb of Orange Pekoe tea. 23. Mix 160 lb of cashews with the almonds. 25. The speed of the current is 2.286 mi/hr. 27. The speed of the current is 5 mi/hr. 29. Karen walked at 4.05 ft/sec. 31. A doubles tennis court is 78 feet long and 36 feet wide. 33. Working together, it takes 12 min. 35. (a) The dimensions are 10 ft by 5 ft. (b) The area is 50 sq ft. (c) The dimensions will be 7.5 ft by 7.5 ft. (d) The area will be 56.25 sq ft. 2 2 37. The defensive back catches up to the tight end at the tight end’s 45-yd line. 39. Add gal of water. 41. Evaporate 10 oz of water. 3 1 3 43. 40 g of 12-karat gold should be mixed with 20 g of pure gold. 45. Mike passes Dan mile from the start, 2 min from the time Mike started to race. 3 47. The latest the auxiliary pump can be started is 9:45 am. 49. The tub will fill in 1 hr. 51. Run: 12 miles; bicycle: 75 miles 53. Bolt would beat Burke by 19.75 m. 55. Set the original price at $40. At 50% off, there will be no profit.
59. The tail wind was 91.47 knots.
AN105
ANSWERS Section A.11
A.10 Assess Your Understanding (page A86) 3. negative
4. closed interval
13. 32, q 2; x Ú 2 (c) 12 7 - 9
15. 3 0, 32; 0 … x 6 3
(d) - 8 6 6
17. (a) 6 6 8
21. (a) 2x + 4 6 5
23. [0, 4]
6. True
(b) 2x - 4 6 - 3
4
4
31. 2 … x … 5
3
5
41. 7
43. Ú
45. 6
53. {x x 6 4} or 1 - q , 42
47. …
49. 7
5
77. 5x x 6 - 5 6 or 1 - q ,- 52
10 10 r or a , q b 3 3
37. x 6 - 3 3
4
57. {x x 7 3} or 13, q 2
2 3
59. {x x Ú 2} or [2, q 2 2
65. {x x 6 - 20} or ( - q , - 20)
67. b x ` x Ú
4 4 r or c , q b 3 3
20
2 2 … x … 3 r or c , 3 d 3 3 3
79. 5x x Ú - 1 6 or 3 - 1, q 2 1
73. b x `
81. b x `
4 3
11 1 11 1 6 x 6 r or a - , b 2 2 2 2
11 2
75. 5x - 6 6 x 6 0 6 or 1 - 6, 02 6
1 2
1 5 1 5 … x 6 r or c , b 2 4 2 4 1 2
87. 5x x 7 3 6 or 13, q 2
10 3
4
3
2 2 f or a - q , d 3 3
5
29. 1 - q , - 42
4
2 3
71. b x `
(b) - 1 7 - 8
51. Ú
7
69. 5x 3 … x … 5 6 or 3 3, 5 4
19. (a) 7 7 0
(d) - 4x - 2 7 - 4
27. 3 4, q 2
11. [0, 2]; 0 … x … 2
10. True
35. x Ú 4
55. {x x Ú - 1} or [ - 1, q 2
63. e x x …
9. False
(d) - 6 7 - 10
(c) 6x + 3 6 6
−1
61. {x x 7 - 7} or 1 - 7, q 2
85. b x ` x 7
(c) 9 6 15
2
4
3
8. True
6
33. - 3 6 x 6 - 2
2
7. True
(b) - 2 6 0
25. [4, 6) 0
39. 6
5. Multiplication Property
1 1 83. b x ` x 6 - r or a - q , - b 2 2
5 4
89. 5x x Ú - 2 6
0
1 2
91. 21 6 Age 6 30
93. (a) Male Ú 77.3 years (b) Female Ú 81.6 years (c) A female can expect to live 4.3 years longer.
3
95. The agent’s commission ranges from $45,000 to $95,000, inclusive. As a percent of selling price, the commission ranges from 5% to 8.6%, inclusive. 97. The amount withheld varies from $111.20 to $161.20, inclusive. 99. The usage varies from 650 kW # hr to 2500 kW # hr, inclusive. 101. The dealer’s cost varies from $15,254.24 to $16,071.43, inclusive. 103. (a) You need at least a 74 on the fifth test. (b) You need at least a 77 on the fifth test. a + b a + b - 2a b - a a + b a + b 2b - a - b b - a a + b - a = = 7 0; therefore, a 6 .b = = 7 0; therefore, b 7 . 105. 2 2 2 2 2 2 2 2 2 2 2 2 2 107. 1 1ab2 - a = ab - a = a 1b - a2 7 0; thus 1 1ab2 7 a and 1ab 7 a. b2 - 1 2ab2 2 = b2 - ab = b 1b - a2 7 0; thus b2 7 1 2ab2 2 and b 7 2ab.
a 1b - a2 b 1b - a2 ab - a2 2ab b2 - ab 2ab - a = = 7 0; thus h 7 a. b - h = b = = 7 0; thus h 6 b. a + b a + b a + b a + b a + b a + b a - b a b 1 1 1 1 1 111. Since 0 6 a 6 b, then a - b 6 0 and 6 0. So 6 0, or 6 0. Therefore, 6 . And 0 6 because b 7 0. ab ab ab b a b a b 113. x2 + 1 Ú 1 for all real numbers x.
109. h - a =
A.11 Assess Your Understanding (page A94) 4. real; imaginary; imaginary unit 5. - 1; - i; 1
6. False
7. True 8. True 9. 8 + 5i
11. - 7 + 6i
13. - 6 - 11i
15. 6 - 18i
6 8 5 7 1 23 19. 10 - 5i 21. 37 23. + i 25. 1 - 2i 27. - i 29. - + i 31. 2i 33. - i 35. i 37. - 6 39. - 10i 41. - 2 + 2i 5 5 2 2 2 2 47. 2i 49. 5i 51. 5i 53. 6 55. 25 57. 2 + 3i 59. z + z = 1a + bi2 + 1a - bi2 = 2a; z - z = 1a + bi2 - 1a - bi2 = 2bi 61. z + w = 1a + bi2 + 1c + di2 = 1a + c2 + 1b + d2i = 1a + c2 - 1b + d2i = 1a - bi2 + 1c - di2 = z + w
67. 2a # 2b = 2ab only when 2a and 2b are real numbers.
17. 6 + 4i 43. 0
45. 0
AN106
ANSWERS Section B.1
APPENDIX B Graphing Utilities B.1 Exercises (page B2) 1. ( - 1, 4); II 3. (3, 1); I 5. Xmin = - 6, Xmax = 6, Xscl = 2, Ymin = - 4, Ymax = 4, Yscl = 2 7. Xmin = - 6, Xmax = 6, Xscl = 2, Ymin = - 1, Ymax = 3, Yscl = 1 9. Xmin = 3, Xmax = 9, Xscl = 1, Ymin = 2, Ymax = 10, Yscl = 2 11. Xmin = - 11, Xmax = 5, Xscl = 1, Ymin = - 3, Ymax = 6, Yscl = 1 13. Xmin = - 30, Xmax = 50, Xscl = 10, Ymin = - 90, Ymax = 50, Yscl = 10 15. Xmin = - 10, Xmax = 110, Xscl = 10, Ymin = - 10, Ymax = 160, Yscl = 10
B.2 Exercises (page B4) 1. (a)
(b)
4
5
10
5
5
4
5
8
10
5
4
5
8
10
5
10
5
8
8
10
5
4
19.
21.
23.
25.
27.
29.
31.
B.3 Exercises (page B6) 3. - 1.71
5. - 0.28
7. 3.00
9. 4.50
11. 1.00, 23.00
B.5 Exercises (page B8) 1. Yes
3. Yes
5. No
7. Yes
10
8
17.
1. - 3.41
10
8
(b)
4
5
10
4
8
4
15. (a)
10
8
(b)
4
8
(b)
10
5
5
10
4
13. (a)
8
4
11. (a)
10
8
(b)
4
8
(b)
10
5
5
10
4
9. (a)
7. (a)
8
10
5
8
4
8
(b)
4
(b)
4
5
10
4
5. (a)
3. (a)
8
9. Answers may vary. A possible answer is Ymin = 4, Ymax = 12, and Yscl = 1.
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C1
Subject Index Δ (change in), 52, 104 ± (plus or minus), 67 Abel, Niels, 255, 888 Abscissa, 34 Absolute maximum and absolute minimum of function, 102–3 Absolute value, 641, A5 Absolute value equations, 210–11 Absolute value function, 111–12, 114 Absolute value inequalities, 211–13 Acute angles, 564–66 Addition. See also Sum of complex numbers, A90 of polynomials, A24–A25 of rational expressions, A47–A48 least common multiple (LCM) method for, A48–A50 triangular, 888 of vectors, 649, 652–53 in space, 672 Addition principle of counting, 896–97 Addition property of inequalities, A83 Additive inverse, A47 Adjacency matrix, 811 Ahmes (Egyptian scribe), 875 Airfoil, 615 Airplane wings, 615 Algebra essentials, A1–A13 distance on the real number line, A5–A6 domain of variable, A7 evaluating algebraic expressions, A6–A7 evaluating exponents, A10 graphing inequalities, A4–A5 Laws of Exponents, A7–A9 real number line, A4 sets, A1–A4 to simplify trigonometric expressions, 525 square roots, A9–A10 Algebraic functions, 305 Algebraic vector, 651–52 Algorithm, 245n Alpha particles, 723 Altitude of triangle, A15 Ambiguous case, 578–79 Amount of annuity, 874–75 Amplitude of simple harmonic motion, 600 of sinusoidal functions, 456–58, 460, 476–79 Analytic trigonometry, 495–62 algebra to simplify trigonometric expressions, 525 Double-angle Formulas, 543–53 to establish identities, 544–47 to find exact values, 544 Half-angle Formulas, 547–50 to find exact values, 547–50 for tangent, 549–50 Product-to-Sum Formulas, 553–55 Sum and Difference Formulas, 531–43 for cosines, 531–32 defined, 531 to establish identities, 533–36 to find exact values, 532, 534–35 involving inverse trigonometric function, 537 for sines, 534 for tangents, 536 Sum-to-Product Formulas, 555 trigonometric equations, 514–23 calculator for solving, 517 graphing utility to solve, 519 identities to solve, 518–19 involving single trigonometric function, 514–17 linear, 515–16
linear in sine and cosine, 538–39 quadratic in from, 517–18 solutions of, defined, 514 trigonometric identities, 523–30 basic, 524 establishing, 525–28, 533–36, 544–47 Even-Odd, 524 Pythagorean, 524 Quotient, 524 Reciprocal, 524 Angle(s), 408–21 acute, 564–66 central, 411 converting between decimals and degrees, minutes, seconds forms for, 410–11 of depression, 568, 569 direction, 655, 675–77 drawing, 409–10 of elevation, 568 elongation, 586 of incidence, 522 of inclination, 438 initial side of, 408, 409 negative, 408 optical (scanning), 552 positive, 408 quadrantal, 409, 425–27 of refraction, 522 of repose, 513 right, 409, A13 in standard position, 408 straight, 409 terminal side of, 408, 409 trigonometric functions of, 424–27 between vectors, 663–64, 674–75 viewing, 507 Angle–angle case of similar triangle, A16, A17 Angle–side–angle case of congruent triangle, A16 Angular speed, 416 Annuity(ies), 874–75 amount of, 874–75 defined, 874 formula for, 874 ordinary, 874–75 Aphelion, 710, 737, 753 Apollonius of Perga, 691 Applied (word) problems, A72–A80 constant rate job problems, A77 interest problems, A73–A74 mixture problems, A74–A75 steps for solving, A73 translating verbal descriptions into mathematical expressions, A72 uniform motion problems, A75–A77 Approaches infinity, 265–66 Approximate decimals, A3 Approximating area, 951–54 Araybhatta the Elder, 433 Arc length, 411–12 Area definition of, 954 formulas for, A15 under graph of function, 951–54 of parallelogram, 683 of triangle, 593–99, A15 SAS triangles, 594–95 SSS triangles, 595–96 Argument of complex number, 641 of function, 79 Arithmetic calculator, A10 Arithmetic mean, A89 Arithmetic progression. See Arithmetic sequences
Arithmetic sequences, 862–68 common difference in, 862 defined, 862 determining, 862–63 formula for, 863–64 nth term of, 863 recursive formula for, 863–64 sum of, 864–66 Ars Conjectandi (Bernoulli), 916 Ars Magna (Cardano), 255 ASA triangles, 578 Associative property of matrix addition, 797 of matrix multiplication, 802 of vector addition, 649 Asymptote(s), 266–72, 716–18 horizontal (oblique), 267, 269–72, 321 vertical, 267, 268–69, 321 Atomic systems, 808 Augmented matrix, 770–71, 772 in row echelon form, B9–B10 Average cost function, 92 Average rate of change, 52, 329–30 of function, 104–6 defined, 104 finding, 104–5 secant line and, 105–6 limit of, 934 of linear functions, 151–54 Axis/axes of complex plane, 640–41 of cone, 692 coordinate, 34 of ellipse, 702, 732 of hyperbola conjugate, 712 transverse, 712, 732 polar, 616–17 of quadratic function, 183–84 rotation of, 725–27 analyzing equation using, 727–29 formulas for, 726 identifying conic without, 729–30 of symmetry of parabola, 182, 693 of quadratic function, 183–84 Azimuth, 571n Babylonians, ancient, 875 Back substitution, 758 Barry, Rick, 95 Base of exponent, A8 Basic trigonometric identities, 524 Bearing (direction), 571 Bernoulli, Jakob, 637, 916 Bernoulli, Johann, 746 Bessel, Friedrich, 573 Bessel’s Star, 573 Best fit line of, 164–65 polynomial function of, 238–39 quadratic functions of, 201–2 Beta of stock, 150, 221–22 Bezout, Étienne, 826 Binomial(s), A23 cubing, A27 squares of (perfect squares), A26 Binomial coefficient, 886 Binomial Theorem, 883–89 n to evaluate a b , 884–85 j expanding a binomial, 886–88 historical feature on, 888
I1
I2
Subject Index
Binomial Theorem (Continued) proof of, 888 using, 886–88 Bisection Method, 254, 258 Blood alcohol concentration (BAC), 351 Bode Law, 861 Bonds, zero-coupon, 380 Book value, 156–57 Boole, George, 916 Bounded graphs, 835 Bounding curves, 603 Bounds on zeros, 252–53, 254 Box, volume and surface area of, A15 Brachistochrone, 745–46 Brancazio, Peter, 95 Branches of hyperbola, 712 Break-even point, 160 Brewster’s Law, 523 Briggs, Henry, 363 Bürgi, Joost, 363 Calculator(s), A10. See also Graphing utility(ies) approximating roots on, A56 converting from polar coordinates to rectangular coordinates, 619 to evaluate powers of 34, 327 factorial key on, 855 functions on, 80–81 inverse sine on, 499n kinds of, A10 logarithms on, 360–61 nCr key on, 884 trigonometric equations solved using, 517 Calculus angle measurement in, 411 approximating f(x) = ex, 861 area under curve, 133, 507 area under graph, 108 complicated functions in, 84 composite functions in, 310 derivative, 274 difference quotient in, 80, 109, 342, 364, 542 double-angle formulas in, 546 e in, 334 end behavior of graphs, 233 exponential equations in, 368 Extreme Value Theorem, 103 factoring problems occurring in, A36 functions and approximated by polynomial functions, 243 exponential, 327 increasing, decreasing, or constant, 100, 387 local maxima and local minima in, 101 graph of polynomial functions in, 225 independent variable in, 453 integral, 547, 954–55 area under graph, 951–54 graphing utility to approximate, 955 Intermediate Value Theorem, 253–54 limit notation, 872 limits, 233 logarithms and, 359, 368 partial fraction decomposition and, 814 polar equations and, 637 projectile motion, 741 removable discontinuity in, 285 secant line and, 105, 109 simplifying expressions with rational exponents in, A59 Simpson’s Rule, 204 Snell’s Law and, 522 tangent line, 71 trigonometric functions and equations in, 519, 522, 546, 552, 553 trigonometric identities useful in, 547 turning points and, 231, 233 of variations, 746 Carbon dating, 383–84
Cardano, Girolamo, 255, 646, 915 Cardioid, 630, 636 Carlson, Tor, 394 Carrying capacity, 386 Cartesian (rectangular) coordinates, 33, 34–35 Cartesian (rectangular) form of complex number, 641–42 Catenary, 200n, 701 Cayley, Arthur, 754, 808 Ceilometer, 569 Cell division, 381–82, 386 Cell phones, 74 Center of circle, 66 of hyperbolas, 712 of sphere, 678 Central angle, 411 Change-of-Base Formula, 361 Chebyshëv, P.I., 545n Chebyshëv polynomials, 545n Chu Shih-chieh, 888 Circle(s), 66–72, 692 arc length of, 411–12 area of, A15 area of a sector of, 415 center of, 66 circumference of, A15 defined, 66 general form of equation of, 68–69 graphing, 67, 635 inscribed, 599 intercepts of, 67–68 polar equation of, 626, 628 radius of, 66 standard form of equation of, 66–67 unit, 66, 422–38 Circular functions, 423 Circular motion, simple harmonic motion and, 600–1 Circumference, A15 Clark, William, 563 Clock, Huygens’s, 746 Closed interval, A81, A82 Coefficient, 224, A22, A23 binomial, 886 correlation, 165 damping, 603 of friction, 661 leading, 224, 259, A23 Coefficient matrix, 771 Cofactors, 788–89 Cofunctions, 433, 566 names of, 433 Coincident lines, 757 Column index, 770, 795 Column vector, 799–800 Combinations, 903–5 defined, 903 listing, 903–4 of n distinct objects taken r at a time, 904 Combinatorics, 895, 916 Combined variation, 139–40 Common difference, 862 Common logarithms (log), 348, 361, 363 Common ratio, 868 Commutative property of dot products, 663, 674 of vector addition, 649 Commutative property of matrix addition, 797 Complementary angle theorem, 566 Complement of event, 914 Complement of set, A2–A3 Complement Rule, 914–15 Complete graph, 43 Completing the square, 172, A38–A39 identifying conics without, 724–25 Complex fraction, A50n Complex function, 76n
Complex number(s), 658, 667 argument of, 641 conjugates of, 641 De Moivre’s Theorem and, 643–44 geometric interpretation of, 640 magnitude (modulus) of, 641 in polar form converting from rectangular form to, 641–42 converting to rectangular form, 641–42 products and quotients of, 642–43 product of, 642–43 quotient of, 642–43 Complex number system, A90 Complex numbers, 76n, A89–A95 addition, subtraction, and multiplication of, A90–A94 conjugates of, A91–A93 definition of, A90 equality of, A90 imaginary part of, A90 real part of, A90 in standard form, A90 power of, A93 reciprocal of, A92 Complex plane, 640–42 defined, 640 imaginary axis of, 640 plotting points in, 640–41 real axis of, 640 Complex polynomial function, 259 Complex rational expressions, A50–A52 Complex roots, 644–46 Complex variable, 259 Complex zeros, 173n of polynomials, 258–64 Conjugate Pairs Theorem, 259–60 defined, 259 finding, 261–62 polynomial function with specified zeros, 260–61 of quadratic function, 207–9 Components of vectors, 651, 652, 671 Composite functions, 306–13 calculus application of, 310 components of, 310 defined, 306 domain of, 307–10 equal, 309–10 evaluating, 307 finding, 307–10 forming, 306–7 Compound interest, 372–75 computing, 372–73 continuous, 374–75 defined, 372 doubling or tripling time for money, 377 effective rates of return, 375 formula, 373 present value of lump sum of money, 376 Compound probabilities, 912 Compressions, 124–25, 127 Conditional equation, 524 Cone axis of, 692 generators of, 692 right circular, 692 vertex of, 692 Congruent triangles, A16–A19 Conics, 692 defined, 732 degenerate, 692 directrix of, 732 eccentricity of, 732 ellipse, 692, 701–11 with center at (h, k), 706–7 with center at the origin, 702–6 with center not at origin, 706–7 center of, 702 defined, 702, 732
Subject Index eccentricity of, 711, 733, 734 foci of, 702 graphing of, 704 intercepts of, 704 length of major axis, 702 major axis of, 702, 732 minor axis of, 702 solving applied problems involving, 707–8 vertices of, 702 focus of, 732 general form of, 724–31 hyperbolas, 691, 692, 711–23 asymptotes of, 716–18 branches of, 712 with center at (h, k), 718–19 with center at the origin, 712–16 with center not at the origin, 718–19 center of, 712 conjugate, 723 conjugate axis of, 712 defined, 712, 732 eccentricity of, 723, 734 equilateral, 723 foci of, 712 graphing equation of, 713–14 solving applied problems involving, 719–20 transverse axis of, 712, 732 vertices of, 712 identifying, 724–25 without a rotation of axes, 729–30 names of, 692 parabola, 182–88, 692, 693–701 axis of symmetry of, 182, 693 defined, 693, 732 directrix of, 693 focus of, 693 graphing equation of, 694 solving applied problems involving, 697–98 with vertex at (h, k), 696–97 with vertex at the origin, 693–96 vertex of, 182, 693 paraboloids of revolution, 691, 698 parametric equations, 737–49 applications to mechanics, 745–46 for curves defined by rectangular equations, 743–45 cycloid, 745 defined, 737 describing, 740 graphing, using graphing utility, 738 rectangular equation for curve defined parametrically, 738–40 time as parameter in, 740–43 polar equations of, 731–37 analyzing and graphing, 731–35 converting to rectangular equation, 735 focus at pole; eccentricity e, 733–35 rotation of axes to transform equations of, 725–27 analyzing equation using, 727–29 formulas for, 726 Conjugate of complex number, A91–A93 of conjugate of complex number, A93 of product of two complex numbers, A93 of real number, A92–A93 of sum of two complex numbers, A93 Conjugate axis, 712 Conjugate golden ratio, 860 Conjugate hyperbola, 723 Conjugate of complex numbers, 641 Conjugate Pairs Theorem, 259–60 Connected mode, 115 Consistent systems of equations, 756, 757, 761–62 Constant(s), A6, A22 limit of, 929 of proportionality, 138 Constant functions, 100–1, 112 Constant linear functions, 154–55
Constant rate job problems, A77 Constant term, 224 Constraints, 839 Consumer Price Index (CPI), 380 Continued fractions, A55 Continuous compounding, 374–75 Continuous function, 253, 938–40 Continuous graph, 225 Convergent geometric series, 871–73 Cooling, Newton’s Law of, 384–86 Coordinates, 34, 35 of point on number line, A4 rectangular (Cartesian), 33, 34–35 Copernicus, 417 Corner points, 835 Correlation coefficient, 165 Correspondence between two sets, 75 Cosecant, 425, 426, 440 graph of, 471–73 periodic properties of, 442 Cosecant function, 423 continuous, 939, 940 domain of, 440, 441 inverse, 510–11 approximate value of, 511 calculator to evaluate, 511 definition of, 510 exact value of, 510–11 range of, 440, 441 Cosine(s), 425, 426, 439 direction, 676–77 exact value of, 532 Law of, 587–93 in applied problems, 589–90 defined, 587–88 historical feature on, 590 proof of, 588 Pythagorean Theorem as special case of, 588 SAS triangles solved using, 588–89 SSS triangles solved using, 589 periodic properties of, 441, 442 Sum and Difference Formula for, 531–32 trigonometric equations linear in, 538–39 Cosine function, 423 continuous, 940 domain of, 439, 441, 455 graphs of, 452–67 amplitude and period, 456–58 equation for, 462 key points for, 458–61 inverse, 500–2 defined, 500 exact value of, 501–2, 509–10 implicit form of, 500 properties of, 455 range of, 439, 440, 441, 455 Cosine function hyperbolic, 342 Cost(s) fixed, 63 marginal, 191 variable, 63 Cotangent, 425, 426, 440 periodic properties of, 442 Cotangent function, 423 continuous, 939, 940 domain of, 440, 441 graph of, 470–71 inverse, 510–11 approximating the value of, 511 calculator to evaluate, 511 definition of, 510 range of, 440, 441 Counting, 895–900 addition principle of, 896–97 combinations, 903–5 defined, 903 listing, 903–4 of n distinct objects taken r at a time, 904
I3
formula, 896 multiplication principle of, 897–98 permutations, 900–3 computing, 903 defined, 901 distinct objects with repetition, 901 distinct objects without repetition, 901–2 involving n nondistinct objects, 905–6 Counting numbers (natural numbers), 879n, 881–82, A3 Cramer, Gabriel, 754 Cramer’s Rule, 754, 784–91 for three equations containing three variables, 789–91 for two equations containing two variables, 785–87 Cross (vector) product, 658, 679–85 algebraic properties of, 681 area of parallelogram using, 683 defined, 679 geometric properties of, 682 of two vectors, 679–81 vector orthogonal to two given vectors, 682 Cube(s) of binomials (perfect cubes), A27 difference of two, A27, A33 sum of two, A27, A33 Cube function, 79, 113 Cube root, 110–11, 113, A55 complex, 644–46 Cubic function of best fit, 239 Cubic models from data, 238–39 Curve(s) bounding, 603 defined by rectangular equations, 743–45 defined parametrically, 738–40 graphing utility to graph parametrically-defined, B12 of quickest descent, 745–46 sawtooth, 608 Curve fitting, 765 sinusoidal, 479–83 hours of daylight, 482–83 sine function of best fit, 483 temperature data, 479–81 Curvilinear motion, 740 Cycle of sinusoidal graph, 453, 457 Cycloid, 745 Cygni, 573 Damped motion, 600, 603–4 Damping factor (damping coefficient), 603 Dantzig, George, 838n Data arrangement in matrix, 795 fitting exponential functions to, 392–93 sinusoidal models from, 479–83 Day length, 223, 303–4, 407, 493 “Deal or No Deal” (TV show), 894, 922 Decay, Law of, 381, 383–84. See also Exponential growth and decay Decimals, A3 approximate, A3 repeating, 872–73, A3 Declination of the Sun, 506 Decomposition, 665–66 Decreasing functions, 100–2 Decreasing linear functions, 154–55 Deflection, force of, 615 Degenerate conics, 692 Degree(s), 409–11 converting between decimals and, 410–11 converting between radians and, 412–15 Degree of monomial, A22, A30 Degree of polynomial, 224–28, A23, A30 odd, 252, 260 Demand equation, 197 De Moivre, Abraham, 643
I4
Subject Index
De Moivre’s Theorem, 643–44 Denominator, A45 rationalizing the, A57–A58 Dependent systems of equations, 757 containing three variables, 764–65 containing two variables, 761 matrices to solve, 777–78 Dependent variable, 79 Depressed equation, 249–50 Depression, angle of, 568, 569 Derivative, 945–46 Descartes, René, 33, 34n, 74, 522n Descartes’ Rule of Signs, 247–48 Descartes’s Law, 522n Determinants, 667, 680–81, 754, 784–94 cofactors, 788–89 Cramer’s Rule to solve a system of three equations containing three variables, 789–91 Cramer’s Rule to solve a system of two equations containing two variables, 785–87 evaluating, 680 expanding across a row or column, 789 to find cross products, 680–81 minors of, 788–89 properties of, 791–92 3 by 3, 788–89 2 by 2, 785, 791–92 Diagonal entries, 802 Difference(s). See also Subtraction common, 862 first, 475 limits of, 930 of logarithms, 359 of two cubes, A27, A33 of two functions, 83–84 of two matrices, 796–98 of two squares, A26, A33 of vectors, 649 Difference quotient, 80, 109, 342, 364, 542 Diophantus, 875 Directed line segment, 648, 650–51 Direction angles, 655, 675–77 Direction (bearing), 571 Direction cosines, 676–77 Direction of vectors, 648, 654–55 Directrix, 732 of parabola, 693 Direct variation, 138, 139–40 Dirichlet, Lejeune, 74, 76n Discontinuity, 114–15 nonremovable, 285 removable, 285 Discontinuous function, 938–39 Discrete mathematics, 851 Discriminant, 174, 209 Disjoint sets, A3 Distance, mean, 710, 753 Distance formula, 35–37 proof of, 35–36 in space, 671 Distributive Property of dot products, 663, 674 of matrix multiplication, 802 of real numbers, A4 Divergent geometric series, 871–73 Dividend, 245, A28 Division. See also Quotient(s) of complex numbers, A92 of polynomials, A27–A30 algorithm for, 245 synthetic, A41–A44 of rational expressions, A46 of two integers, A28 Divisor, 245, A28 Domain, 76, 81–83 of absolute value function, 114 of composite function, 307–10 of constant function, 112
of cosecant function, 440, 441 of cosine function, 439, 441, 455 of cotangent function, 440, 441 of cube function, 113 of cube root function, 113 of difference function, 83–84 of greatest integer function, 114 of identity function, 112 of inverse function, 317 of logarithmic function, 345 of logistic models, 386 of one-to-one function, 314–15 of product function, 83–84 of quotient function, 84 as range of inverse, 497 of rational function, 265–67 of reciprocal function, 113 of secant function, 439, 440, 441 of sine function, 439, 441, 454 of square function, 113 of square root function, 113 of sum function, 83–84 of tangent function, 439, 441, 469 of trigonometric functions, 439–40 unspecified, 85 of variable, A7 Domain-restricted function, 321–22 Doppler, Christian, 289 Doppler effect, 289 Dot mode, 115 Dot product, 658, 662–69, 673–74 angle between vectors using, 663–64 to compute work, 667 defined, 662, 673 finding, 662–63 historical feature on, 667 orthogonal vectors and, 664–65 parallel vectors and, 664 properties of, 663, 674 of two vectors, 662–63 Double-angle Formulas, 543–53 to establish identities, 544–47 to find exact values, 544 Double root (root of multiplicity 34), 170 Drag, 615 Dry adiabatic lapse rate, 867 e, 333–35, 342 Earthquakes, magnitude of, 355 Eccentricity, 732 of ellipse, 711, 733, 734 of hyperbola, 723, 734 Eddin, Nasir, 417, 590 Effective rates of return, 375 Egyptians, ancient, 875 Elements (Euclid), 590, 875 Elements of sets, 895-658, A1 Elevation, angle of, 568 Elimination, Gauss-Jordan, 777 Elimination method, 754, 759, 761 systems of nonlinear equations solved using, 822–25 Ellipse, 692, 701–11 with center at (h, k), 706–7 with center at the origin, 702–6 major axis along x-axis, 703–5 major axis along y-axis, 705–6 with center not at origin, 706–7 center of, 702 defined, 702, 732 eccentricity of, 711, 733, 734 foci of, 702 graphing of, 704 intercepts of, 704 major axis of, 702, 732 length of, 702 minor axis of, 702
solving applied problems involving, 707–8 vertices of, 702 Ellipsis, A3 Elliptical orbits, 691 Elongation angle, 586 Empty (null) sets, 895, A1 End behavior, 233–35, 267 Engels, Friedrich, 923 Entries of matrix, 770, 795 diagonal, 802 Epicycloid, 749 Equality of complex numbers, A90 of sets, 895, A2 of vectors, 649, 652 Equally likely outcomes, 911–12 Equation(s) conditional, 524 demand, 197 depressed, 249–50 domain of a function defined by, 82 equivalent, A63–A64, B3 even and odd functions identified from, 99–100 exponential, 335–36, 350–51, 367–68 quadratic in form, 368 as function, 78 graphing utility to graph, B3–B5 intercepts from, 44 inverse function defined by, 319–22 linear. See Linear equation(s) polar. See Polar equations quadratic in form, 176–77 solving, 176–77 satisfying the, 41, A63 sides of, 41, A63 solution set of, A63 solving, A63–A71 algebraically, A64 by factoring, A67–A68 with graphing calculator, B6–B7 systems of. See Systems of equations in two variables, graphs of, 41–43 intercepts from, 43 by plotting points, 41–43 symmetry test using, 44–46 x = y2, 47 y = 1 , x, 47–48 y = x3, 46–47 Equilateral hyperbola, 723 Equilateral triangle, 39 Equilibrium, static, 656–57 Equilibrium price, 157 Equilibrium quantity, 157 Equilibrium (rest) position, 600 Equivalent equations, A63–A64, B3 Equivalent systems of equations, 759 Error triangle, 40 Euclid, 590, 875, 888 Euler, Leonhard, 74, 417, 916 Even functions defined, 98 determining from graph, 98–99 identifying from equation, 99–100 Evenness ratio, 354 Even-Odd identity, 448–49, 524 Events, 911 complement of, 914 mutually exclusive, 913–14 probabilities of union of two, 913–15 Explicit form of function, 81 Exponent(s), A8–A9 Laws of, 327, 335, 336, A7–A9 logarithms related to, 343 Exponential equations, 335–36 defined, 335 solving, 335–36, 350–51, 367–68 equations quadratic in form, 368 using graphing utility, 368–69
Subject Index Exponential expressions, changing between logarithmic expressions and, 344 Exponential functions, 326–42, 495, 940 continuous, 940 defined, 328 e, 333–35, 342 evaluating, 326–30 fitting to data, 392–93 graph of, 330–33 using transformations, 333 identifying, 328–30 power function vs., 328 properties of, 331, 333, 337 Exponential growth and decay, 327–30, 381–84 law of decay, 381, 383–84 logistic models, 386–88 defined, 386 domain and range of, 386 graph of, 386 properties of, 386 uninhibited growth, 381–83 Exponential law, 381 Extended Principle of Mathematical Induction, 883 Extraneous solutions, A67, A68 Extreme values of functions, 102–3 Extreme Value Theorem, 103 Factored completely, A32 Factorial symbol, 854–55 Factoring defined, A32 equations solved by, A67–A68 of expression containing rational exponents, A60 over the integers, A32 polynomials, A32–A41 Ax2 + Bx + C, A37–A38 difference of two squares and the sum and the difference of two cubes, A33 by grouping, A36–A37 perfect squares, A34 x2 + Bx + C, A34–A36 quadratic equations, 170 Factors, A32 linear, 814–17 nonrepeated, 814–15 repeated, 816–17 quadratic, 252, 817–19 synthetic division to verify, A43–A44 Factor Theorem, 245–47 Family of lines, 64 of parabolas, 131 Feasible point, 839, 840 Fermat, Pierre de, 33, 342, 915 Ferrari, Lodovico, 255 Ferris, George W., 71, 521 Fertility, population growth and, 851 Fibonacci, 875 Fibonacci numbers, 856 Fibonacci sequences, 856 Financial models, 371–81 effective rates of return, 375 future value of a lump sum of money, 372–75 present value of a lump sum of money, 376 rate of interest or time required to double a lump sum of money, 377 Finck, Thomas, 417, 433 Finite sets, 895 First-degree equation. See Linear equation(s) First differences, 475 Fixed costs, 63 Focus/foci, 732 of ellipse, 702 of hyperbola, 712 of parabola, 693 FOIL method, A26 Foot-pounds, 667
Force, 600–1 of deflection, 615 resultant, 655, 656 Force vector, 654–55 Formulas, geometry, A15–A16 Foucault, Jean-Bernard-Leon, A71 Fractions complex, A50n continued, A55 partial, 814 Frequency, 466 in simple harmonic motion, 601 Friction, coefficient of, 661 Frobenius, Georg, 808 Function(s), 74–149. See also Composite functions; Exponential functions; Inverse functions; Linear functions; Polynomial functions; Trigonometric functions absolute maximum and absolute minimum (extreme values) of, 102–3 absolute value, 111–12, 114 area under graph of, 951–54 argument of, 79 average cost, 92 average rate of change of, 104–6 defined, 104 finding, 104–5 secant line and, 105–6 building and analyzing, 133–35 on calculators, 80–81 circular, 423 complex, 76n constant, 100–1, 112 continuous, 253, 938–40 cube, 79, 113 cube root, 110–11, 113 decreasing, 100–2 defined, 76 derivative of, 945–46 difference of two, 83–84 difference quotient of, 80 discontinuous, 938–39 domain of, 76, 81–83 unspecified, 85 domain-restricted, 321–22 equation as, 78 even and odd determining from graph, 98–99 identifying from equation, 99–100 explicit form of, 81 graph of, 88–97, 121–33 combining procedures, 123–24, 127–29 determining odd and even functions from, 98–99 determining properties from, 100–1 extreme values, 102–3 identifying, 89–90 information from or about, 90–92 using compressions and stretches, 124–25, 127 using reflections about the x-axis or y-axis, 126 using vertical and horizontal shifts, 121–24, 127 greatest integer, 114 identically equal, 524 identity, 112 implicit form of, 81 important facts about, 81 increasing, 100–2 intersection of two, 175–76 library of, 110–20 graphing, 110–17 local maxima and local minima of, 101–2, 231 with no limit at 33, 927 nonlinear, 151, 153 objective, 839–40 one-to-one, 314–16 periodic, 441 piecewise-defined, 115–17, 939–40 power, 225–28 graph of, 226–27
I5
of odd degree, 227–28 properties of, 227–28 product of two, 83–84 properties of, 98–110 quotient of two, 83–84 range of, 76 real zeros of, 176–77 reciprocal, 113, 471, 473. See also Cosecant function; Secant function relation as, 75–78 square, 112–13 square root, 110, 113 step, 115 sum of two, 83–84 graph of, 604–6 value (image) of, 76, 78–81 zeros of, 91–92 Bisection Method for approximating, 258 Function keys, A10 Function notation, 85 Fundamental identities of trigonometric functions, 443–45 quotient, 444 reciprocal, 443–44 Fundamental period, 441 Fundamental Theorem of Algebra, 259 Conjugate Pairs Theorem and, 259–60 proof of, 259 Future value, 372–75 Galois, Evariste, 255 Gauss, Karl Friedrich, 259, 646, 754 Gauss-Jordan method, 777 General addition principle of counting, 897 General form of conics, 724–31 of equation of circle, 68–69 linear equation in, 58–59 General term, 853 Generators of cone, 692 Geometric mean, A89 Geometric progression. See Geometric sequences Geometric sequences, 868–79 common ratio of, 868 defined, 868 determining, 868–69 formula for, 869–70 nth term of, 869–70 sum of, 870–71 Geometric series, 871–73 infinite, 871–73 Geometric vectors, 648–49 Geometry essentials, A13–A22 congruent and similar triangles, A16–A19 formulas, A15–A16 Pythagorean Theorem and its converse, A13–A15 Geometry problems, algebra to solve, 36–37 Gibbs, Josiah, 658 Golden ratio, 860 conjugate, 860 Grade, 65 Graph(s)/graphing area under, 951–54 bounded, 835 of circles, 67–68, 635 complete, 43 of cosecant function, 471–73 using transformations, 472 of the cosine function, 455 of cotangent function, 470–71 of ellipse, 704 of equations in two variables, 41–50 intercepts from, 43 by plotting points, 41–43 symmetry test using, 44–46 x = y2, 47 y = 1 , x, 47–48 y = x3, 46–47
I6
Subject Index
Graph(s)/graphing (Continued) of exponential functions, 330–33 using transformations, 333 of function, 88–97, 121–33 combining procedures, 123–24, 127–29 determining odd and even functions from, 98–99 determining properties from, 100–1 extreme values, 102–3 identifying, 89–90 information from or about, 90–92 using compressions and stretches, 124–25, 127 using reflections about the x-axis or y-axis, 126 using vertical and horizontal shifts, 121–24, 127 functions listed in the library of functions, 110–17 of inequalities, 830–35, A4–A5 linear inequalities, 831–32 steps for, 831 of inverse functions, 318–19 of linear functions, 151 of lines given a point and the slope, 54 using intercepts, 58–59 of logarithmic functions, 345–49 base not 42 or e, 362 inverse, 347–48 of logistic models, 386 of parabola, 694 of parametric equations, 738, B11–B12 of piecewise-defined functions, 115–17 of polar equations, 625–40 cardioid, 630, 636 circles, 635 of conics, 733–35 by converting to rectangular coordinates, 626–29 defined, 626 lemniscate, 634, 636 limaçon with inner loop, 632, 636 limaçon without inner loop, 631, 636 by plotting points, 630–35 polar grids for, 625 rose, 633, 636 sketching, 636–37 spiral, 634–35 using graphing utility, 627, B11 of polynomial functions analyzing, 236–38 smooth and continuous, 225 using bounds on zeros, 254 using transformations, 228 of polynomial inequalities, 291 of quadratic functions properties of, 183 steps for, 187 using its vertex, axis, and intercepts, 184–87 using transformations, 181–83 of rational functions, 275–90 analyzing, 275–85 constructing rational function from, 285–86 using transformations, 266 of rational inequalities, 292–93 of secant function, 471–72 using transformations, 472 of sequences, 852, 853 of sine and cosine functions, 452–67, 476–79, 604–5 amplitude and period, 456–58 equation for, 462 key points for, 458–61 y = sin x, 453 to solve systems of equations, 757 of systems of nonlinear inequalities, 834 of vectors, 650–51 of y = 33/x2, 265–66 Graphing calculator(s), A10 composite functions on, 307 scale setting, 35 TABLE feature, 254
Graphing utility(ies), B1–B12 to analyze graph of polynomial function, 237–38 connected mode, 115 converting between degrees, minutes, seconds, and decimal forms on, 410 coordinates of point shown on, B2 derivative of function using, 945 dot mode, 115 eVALUEate feature, 246, B5 to find limit, 927 to find sum of arithmetic sequence, 864–66 to fit exponential function to data, 391–92 to fit logistic function to data, 393–94 functions on, 103–4 geometric sequences using, 870, 871 to graph a circle, 69 to graph equations, 55, B3–B5 to graph inequalities, B9 to graph parametric equations, B11–B12 to graph polar equations, 627, B11 identity established with, 526 intercepts found using, 44 INTERSECT feature, B6–B7 line of best fit from, 164–65 to locate intercepts and check for symmetry, B5–B6 MAXIMUM and MINIMUM features, 104, 200 PARametric mode, 742 polar equations using, 627 REGression options, 392 row echelon form on, 780 RREF command, B10 sine function of best fit on, 483 to solve equations, B6–B7 to solve logarithmic and exponential equations, 368–69 to solve systems of linear equations, B9–B10 square screens, B8 TABLE feature, 853 tables on, B4 TRACE feature, 853 trigonometric equations solved using, 519 turning points in, 232 viewing rectangle, B1–B3 setting, B1–B2 ZERO (or ROOT) feature, 200, B5, B6 ZOOM-STANDARD feature, B3n ZSQR function on, B8n Grassmann, Hermann, 658, 667 Greatest integer function, 114–15 Grouping, factoring by, A36–A37 Growth, exponential, 327–30 Growth, uninhibited, 381–83 Growth factor, 328 Hale–Bopp comet, 691, 753 Half-angle Formulas, 547–50 to find exact values, 547–50 for tangent, 549–50 Half-life, 383–84 Half-line (ray), 408 Half-open/half-closed intervals, A81, A82 Half-planes, 832 Hamilton, William Rowan, 658 Harmonic mean, A89 Heron of Alexandria, 595, 596, 875 Heron’s Formula, 595–96 historical feature on, 596 proof of, 595–96 Horizontal component of vector, 652 Horizontal compression or stretches, 125 Horizontal lines, 55–56, 627, 635 Horizontal-line test, 315–16 Horizontal (oblique) asymptote, 267, 269–72, 321 Horizontal shifts, 121–24, 127 to the left, 122–23 to the right, 122 HP 48G, B8n
Huygens, Christiaan, 746, 915 Huygens’s clock, 746 Hyperbolas, 691, 692, 711–23 asymptotes of, 716–18 branches of, 712 with center at (h, k), 718–19 with center at the origin, 712–16 transverse axis along x-axis, 713–14 transverse axis along y-axis, 715–16 with center not at the origin, 718–19 center of, 712 conjugate, 723 conjugate axis of, 712 defined, 712, 732 eccentricity of, 723, 734 equilateral, 723 foci of, 712 graphing equation of, 713–14 solving applied problems involving, 719–20 transverse axis of, 712, 732 vertices of, 712 Hyperbolic cosine function, 342 Hyperbolic sine function, 342 Hyperboloid, 722 Hypocycloid, 749 Hypotenuse, A13 i, A93 Identically equal functions, 524 Identity(ies), A63 definition of, 524 polarization, 669 Pythagorean, 445, 524 trigonometric, 523–30 basic, 524 establishing, 525–28, 533–36, 544–47 Even-Odd, 524 Pythagorean, 524 Quotient, 524 Reciprocal, 443–44, 524 trigonometric equations solved using, 518–19 Identity function, 112 Identity matrix, 802–3 Identity Properties, 803 Image (value) of function, 76, 78–81 Imaginary axis of complex plane, 640 Imaginary unit, A89 Implicit form of function, 81 Improper rational expression, 814 Improper rational function, 270 Incidence, angle of, 522 Inclination, angle of, 438 Inconsistent systems of equations, 756, 757, 761–62, 763–64 containing three variables, 763–64 containing two variables, 760 matrices to solve, 778–79 Increasing functions, 100–2 Increasing linear functions, 154–55 Independent systems of equations, 757 Independent variable, 79 in calculus, 453 Index/indices of radical, A55, A60 of refraction, 522 row and column, 770, 795 of sum, 856 Induction, mathematical, 879–83 Extended Principle of, 883 principle of, 880, 882 proving statements using, 879–82 Inequality(ies) absolute value, 211–13 combined, A85–A86 graphing, 830–35 on graphing utility, B9 linear inequalities, 831–32 steps for, 831
Subject Index interval notation to write, A82 involving quadratic functions, 192–96 nonstrict, A5 in one variable, A84 polynomial, 291–92 algebraically and graphically solving, 291–92 multiplicity in, 292 steps for solving, 294 properties of, A82–A84, A86 rational algebraically and graphically solving, 292–94 multiplicity in, 294 steps for solving, 294 satisfying, 830 sides of, A5 solving, A84–A86 strict, A5 systems of, 830–37 graphing, 832–35 in two variables, 830 Inequality symbols, A5 Inertia moment of, 557 product of, 552 Infinite geometric series, 871–73 Infinite limit, 234 Infinite sets, 895 Infinity, A81 approaches, 265–66 limits at, 234 Inflation, 379 Inflection point, 386 Initial point of directed line segment, 648 Initial side of angle, 408, 409 Initial value, 328 Input to relation, 75 Inscribed circle, 599 Instantaneous rate of change, 946 Integers, A3 dividing, A28 factoring over the, A32 positive, 879n Integrals, 547, 955 area under graph, 951–54 graphing utility to approximate, 955 Intercept(s) of circle, 67–68 of the ellipse, 704 from an equation, 44 from a graph, 43 graphing an equation in general form using, 58–59 graphing utility to find, B5–B6 graph of lines using, 58–59 of quadratic function, 184–87 Interest compound computing, 372–73 continuous, 374–75 defined, 372 doubling or tripling time for money, 377 effective rates of return, 375 formula, 373 present value of lump sum of money, 376 problems involving, A73–A74 rate of, 372, A73 simple, 372, A73 Intermediate Value Theorem, 253–54 Intersection of sets, A2 of two functions, 175–76 Intervals, A81–A82 closed, A81, A82 endpoints of, A81 half-open, or half-closed, A81, A82 open, A81, A82 writing, using inequality notation, A82 Invariance, 731
Inverse additive, A47 of matrix, 803–8 finding, 805–6 multiplying matrix by, 804–5 solving system of linear equations using, 807–8 Inverse functions, 316–22, 495, 496–528. See also Logarithmic functions cosine, 500–2 defined, 500 exact value of, 501–2 exact value of expressions involving, 509–10 implicit form of, 500 defined by a map or set of ordered pairs, 316–18 domain of, 317 of domain-restricted function, 321–22 finding, 316–18, 504–5 defined by an equation, 319–22 graph of, 318–19 range of, 317, 322 secant, cosecant, and cotangent, 510–11 approximating the value of, 511 calculator to evaluate, 511 definition of, 510 sine, 496–500 approximate value of, 498–99 defined, 497 exact value of, 497–98, 509–10, 537 implicit form of, 497 properties of, 499–500 solving equations involving, 505 Sum and Difference Formulas involving, 537 tangent, 502–4 defined, 503 exact value of, 503–4 exact value of expressions involving, 509–10 implicit form of, 503 verifying, 318 written algebraically, 511–12 Inverse trigonometric equations, 505 Inverse variation, 139 Irrational numbers, A3, A90 decimal representation of, A3 Irreducible quadratic factor, 252, 817–19 Isosceles triangle, 39 Jiba, 433 Jiva, 433 Joint variation, 139–40 Jordan, Camille, 754 Joules (newton-meters), 667 Karmarkar, Narendra, 838n Kepler, Johannes, 140, 143 Khayyám, Omar, 888 Kirchhoff’s Rules, 768, 783 Koukounas, Teddy, 598n Kôwa, Takakazu Seki, 754 Latus rectum, 694 Law of Cosines, 587–93 in applied problems, 589–90 defined, 587–88 historical feature on, 590 proof of, 588 Pythagorean Theorem as special case of, 588 SAS triangles solved using, 588–89 SSS triangles solved using, 589 Law of Decay, 381, 383–84. See also Exponential growth and decay Law of Sines, 576–77, 588n in applied problems, 581–83 defined, 577 historical feature on, 590 proof of, 582–83 SAA or ASA triangles solved using, 578 SSA triangles solved using, 578–81 Law of Tangents, 587, 590
Laws of Exponents, 327, 335, 336, A7–A9 Leading coefficient, 224, 259, A23 Leading term, 224 Least common multiple (LCM) to add rational expressions, A48–A50 Left endpoint of interval, A81 Left limit, 936 Left stochastic transition matrix, 811 Legs of triangle, A13–A14 Leibniz, Gottfried Wilhelm, 74, 754 Lemniscate, 634, 636 Lensmaker’s equation, A54 Lewis, Meriwether, 563 Lift, 615 Light detector, 569 Light projector, 569 Like radicals, A57 Like terms, A23 Limaçon with inner loop, 632, 636 without inner loop, 631, 636 Limits, 233–34, 924–42 of average rate of change, 934 of constant, 929 of difference, 930 finding, 924–29 algebra techniques for, 929–35 by graphing, 926–27 using a table, 924–26 at infinity, 234 of monomial, 931 one-sided, 936–37 of polynomial, 931–32 of power or root, 932 of product, 930 of quotient, 933–34 of sum, 930 of x, 929 Line(s), 51–65 of best fit, 164–65 coincident, 757 equation of secant, 105–6. See also Linear equation(s); Systems of linear equations equations of perpendicular lines, 664n family of, 64 graphing given a point and the slope, 54 using intercepts, 58–59 horizontal, 55–56, 627, 635 point-slope form of, 55–56 polar equation of, 626–27, 635 slope of, 51–54, 57 containing same point, 52–53 containing two points, 52 from linear equation, 57 tangent, 71 vertical, 51, 627, 635 y-intercept of, 57 Linear algebra, 795 Linear equation(s), A64–A66. See also Line(s); Systems of linear equations defined, 59 in general form, 58–59 given two points, 56 for horizontal line, 55–56 in one variable, A63, A64 for parallel line, 59–60 for perpendicular line, 60–61 slope from, 57 in slope-intercept form, 56–57 solving equations that lead to, A65–A66 for vertical line, 54–55 Linear factors, 814–17 nonrepeated, 814–15 repeated, 816–17 Linear functions, 151–62 average rate of change of, 151–54 defined, 151
I7
I8
Subject Index
Linear functions (Continued) graphing utility to find the line of best fit, 164–65 graph of, 151 identifying, 328–30 increasing, decreasing, or constant, 154–55 models from data, 162–68 models from verbal descriptions, 156–58 scatter diagrams, 162–63 zero of a, 155 Linear programming problems, 754, 838–44 definition of, 839 maximum, 841–42 minimum, 840–41 setting up, 838–39 solution to, 840 location of, 840 solving, 839–42 in two variables, 839 Linear relations, nonlinear relations vs., 163–64 Linear speed, 416–17 Linear trigonometric equation, 515–16 LINE command, 55 Line segment, 648, 650–51 midpoint of, 38 Local maxima and local minima of functions, 101–2, 231 Logarithmic equations, 365–66 defined, 350 solving, 350–51, 365–66 Logarithmic functions, 343–56 changing between logarithmic expressions and exponential expressions, 344 continuous, 940 defined, 343 domain of, 345 evaluating, 344 fitting to data, 393–94 graph of, 345–49 properties of, 346, 352 range of, 345 Logarithmic spiral, 635 Logarithms, 356–64 with bases other than 42 or e evaluating, 360–62 graphing, 362 on calculators, 360–61 common (log), 348, 361, 363 historical feature on, 363 logarithmic expression as single, 359–60 logarithmic expression as sum or difference of, 359 natural (ln), 346, 360–61, 363 properties of, 356–58 establishing, 357 proofs of, 357 summary of, 362 relating to exponents, 343 Logistic models, 386–88 defined, 386 domain and range of, 386 graph of, 386 growth models, 394–95 properties of, 386 Loudness, 355 Lowest terms rational expressions in, A45–A46 rational function in, 265, 268 Magnitude of vectors, 648, 650–51, 652, 654–55, 676–77 in space, 673 Magnitude (modulus), 641, 642 Magnitude of earthquake, 355 Major axis, 732 Malthus, Thomas Robert, 923, 960 Mandel, Howie, 894 Mandelbrot sets, 647 Mapping, 75
Marginal cost, 191 Marginal propensity to consume, 878 Markov chains, 850 Marx, Karl, 923 Mathematical induction, 879–83 Extended Principle of, 883 principle of, 880, 882 proving statements using, 879–82 Mathematical modeling, A72 Matrix/matrices, 754, 770–84, 794–813 adjacency, 811 arranging data in, 795 augmented, 770–71, 772 in row echelon form, B9–B10 coefficient, 771 defined, 770, 795 entries of, 770, 795, 802 equal, 796 examples of, 795–96 historical feature on, 808 identity, 802–3 inverse of, 803–8 finding, 805–6 multiplying matrix by, 804–5 solving system of linear equations using, 807–8 left stochastic transition, 811 m by n, 795 nonsingular, 803, 805–6 product of two, 799–803 in reduced row echelon form, 776–77 row and column indices of, 770, 795 in row echelon form, 773–80 row operations on, 772–73 scalar multiples of, 798–99 singular, 803 to solve system of linear equations, 773–80 square, 795–96 sum and difference of two, 796–98 transition, 850 zero, 798 Maxima of functions, local, 101–2, 231 Mean arithmetic, A89 geometric, A89 harmonic, A89 Mean distance, 710, 753 Mechanics, parametric equations applied to, 745–46 Medians of triangle, 39 Menelaus of Alexandria, 417 Metrica (Heron), 596 Midpoint formula, 37–38 Midpoint of triangle, 39 Mindomo software, 562 Minima of functions, local, 101–2, 231 Minors, 788–89 Minutes, 410–11 Miranda, Kathleen, 598n Mixture problems, A74–A75 Model(s), A72 conic sections in, 692 financial, 371–81 effective rates of return, 375 future value of a lump sum of money, 372–75 present value of a lump sum of money, 376 rate of interest or time required to double a lump sum of money, 377 probability, 909–11 quadratic, 196–206 from data, 201–2 from verbal descriptions, 197–201 using direct variation, 138, 139–40 using inverse variation, 139 using joint variation or combined variation, 139–40 with vectors, 655–57 from verbal descriptions, 156–58 Modulus (magnitude), 641, 642 Mollweide, Karl, 586
Mollweide’s Formula, 586 Moment of inertia, 557 Monomial(s), A22–A23 common factors, A32 degree of, A22, A30 examples of, A22–A23 limit of, 931 recognizing, A22–A23 in two variables, A30 Monter, 65 Monty Hall Game, 922 Motion circular, 600–1 curvilinear, 740 damped, 600, 603–4 Newton’s second law of, 650 projectile, 741–42 simple harmonic, 600–2 uniform, A75–A77 Multiple root, 230n Multiplication. See also Product(s) of complex numbers, A91 of polynomials, A25 of rational expressions, A46 scalar, 798–99 of vectors, by numbers, 649–50 Multiplication principle of counting, 897–98 Multiplication properties for inequalities, A83 Multiplicity of polynomial functions, 229–35 roots of, 170, 230n in solving polynomial inequalities, 292 in solving rational inequalities, 294 vertical asymptote and, 268–69 Multiplier, 878 Mutually exclusive events, 913–14 Napier, John, 363 Nappes, 692 Natural logarithms (ln), 346, 360–61, 363 Natural numbers (counting numbers), 879n, 881–82, A3 Nautical miles, 420 Negative numbers real, A4 square roots of, A94 Newton, Isaac, 384n Newton-meters (joules), 667 Newton’s Law of Cooling, 384–86, 389 Newton’s Law of Heating, 389 Newton’s Law of universal gravitation, 297 Newton’s Method, 274 Newton’s Second Law of Motion, 600, 650 Niccolo of Brescia (Tartaglia), 255, 646 Nonlinear equations, systems of, 821–29 elimination method for solving, 822–25 historical feature on, 826 substitution method for solving, 821–22 Nonlinear functions, 151, 153 Nonlinear inequalities, systems of, 834 Nonlinear relations, 163–64 Nonnegative property of inequalities, A82 Nonremovable discontinuity, 285 Nonsingular matrix, 803, 805–6 Nonstrict inequalities, A5 Normal plane, 664n nth roots, A55–A56 historical feature, A60 rationalizing the denominator, A57–A58 simplifying, A55 simplifying radicals, A56–A57 Null (empty) sets, 895, A1 Numbers Fibonacci, 856 irrational, A3 natural (counting), 879n, 881–82, A3 rational, A3 Numerator, A45
Subject Index Objective function, 839–40 Oblique (horizontal) asymptote, 267, 269–72, 321 Oblique triangle, 577 Odd functions defined, 98 determining from graph, 98–99 identifying from equation, 99–100 One-sided limits, 936–37 One-to-one functions, 314–16 defined, 314 horizontal-line test for, 315–16 inverse of, 320 Open interval, A81, A82 Opens down, 182 Opens up, 182 Optical (scanning) angle, 552 Optimization, quadratic functions and, 197 Orbits elliptical, 691 planetary, 710 Ordered pair(s), 34 inverse function defined by, 316–18 as relations, 75–76 Ordinary annuity, 874–75 Ordinary (statute) miles, 420 Ordinate (y-coordinate), 34 Orientation, 738 Origin, 34 distance from point to, 133–34 of real number line, A4 in rectangular coordinates, 670 symmetry with respect to, 44–45 Orthogonal vectors, 664–67 Outcome of probability, 909 equally likely, 911–12 Output of relation, 75 Parabola, 182–88, 692, 693–701 axis of symmetry of, 182, 693 defined, 693, 732 directrix of, 693 family of, 131 focus of, 693 graphing equation of, 694 solving applied problems involving, 697–98 with vertex at (h, k), 696–97 with vertex at the origin, 693–96 finding equation of, 694–95 focus at (a, 0), a > 0, 694 vertex of, 182, 693 Paraboloids of revolution, 691, 698 Parallax, 573 Parallel lines, 59–60 Parallel vectors, 664 Parallelepiped, 684 Parallelogram, area of, 683 Parameter, 737 time as, 740–43 Parametric equations, 737–49 for curves defined by rectangular equations, 743–45 applications to mechanics, 745–46 cycloid, 745 defined, 737 describing, 740 graphing, 738 using graphing utility, B11 rectangular equation for curve defined parametrically, 738–40 time as parameter in, 740–43 Partial fraction decomposition, 754, 813–20 defined, 814 where denominator has nonrepeated irreducible quadratic factor, 818–19 where denominator has only nonrepeated linear factors, 814–15
where denominator has repeated irreducible quadratic factors, 819 where denominator has repeated linear factors, 816–17 Partial fractions, 814 Participation rate, 88 Partitioning, 951–54 Pascal, Blaise, 746, 885, 915 Pascal triangle, 885, 888 Payment period, 372 Peano, Giuseppe, 916 Pendulum period of, 141, A62 simple, 141 Perfect cubes, A27 Perfect roots, A56 Perfect squares, A26, A34 Perfect triangle, 598 Perihelion, 710, 737, 753 Perimeter, formulas for, A15 Period fundamental, 441 of simple harmonic motion, 600 of sinusoidal functions, 456–58, 460, 476–79 Periodic functions, 441 Periodic properties, 441–42 Period of pendulum, 141, A62 Permutations, 900–3 computing, 903 defined, 901 distinct objects with repetition, 901 distinct objects without repetition, 901–2 involving n nondistinct objects, 905–6 Perpendicular line, 664n Perpendicular lines, equations of, 60–61 Phase shift, 475–86 to graph y = A sin(vx - w), 475–79 Phones, cellular, 74 Physics, vectors in, 648 Piecewise-defined functions, 115–17 continuous, 940 Pitch, 65 Pixels, B1 Plane(s) complex defined, 640 imaginary axis of, 640 plotting points in, 640–41 real axis of, 640 normal, 664n Plane curve, 737 Planets, orbit of, 710 PLOT command, 55 Plotting points, 34, 616–18 graph equations by, 37–39 Point(s) coordinates of, 35 on graphing utility, B2 on number line, A4 corner, 835 distance between two, 35 in space, 671 distance from the origin to, 133–34 feasible, 839, 840 inflection, 386 initial, 648 of intersection of two functions, 175–76 plotting, 34, 616–18 graph equations by, 37–39 polar coordinates of, 617–18 of tangency, 71 terminal, 648 turning, 231–32 Point-slope form of equation of line, 55–56 Polar axis, 616 Polar coordinates, 616–25 conversion from rectangular coordinates to, 620–22
I9
conversion to rectangular coordinates, 618–19 defined, 616 plotting points using, 616–18 of a point, 617–18 polar axis of, 616 pole of, 616 rectangular coordinates vs., 616 Polar equations calculus and, 636 classification of, 635–36 of conics, 731–37 analyzing and graphing, 733–35 converting to rectangular equation, 735 focus at pole; eccentricity e, 733–35 defined, 626 graph of, 625–40 cardioid, 630, 636 circles, 635 by converting to rectangular coordinates, 626–29 defined, 626 lemniscate, 634, 636 limaçon with inner loop, 632, 636 limaçon without inner loop, 631, 636 by plotting points, 630–35 polar grids for, 625 rose, 633, 636 sketching, 636–37 spiral, 634–35 using graphing utility, 627, B11 historical feature on, 637 identifying, 626–28 testing for symmetry, 629 transforming rectangular form to, 622–23 transforming to rectangular form, 622–23 Polar form of complex number, 641–42 Polar grids, 625 Polarization identity, 669 Pole, 616 Polynomial(s), A22–A44 adding, A24–A25 Chebyshëv, 545n degree of, 224–28, A23, A30 odd, 252, 260 second-degree, A34–A36 dividing, 245–47, A27–A30 synthetic division, A41–A44 examples of, A23–A24 factoring, A32–A41 Ax2 + Bx + C, A37–A38 difference of two squares and the sum and the difference of two cubes, A33 by grouping, A36–A37 perfect squares, A34 x2 + Bx + C, A34–A36 limit of, 931–32 monomials, 931 multiplying, A25 prime, A32, A36 recognizing, A23–A24 solving, 251–52 special products formulas, A26–A27 in standard form, A23 subtracting, A24–A25 terms of, A23 in two variables, A30 zero, A23 Polynomial functions, 224–44, 938 of best fit, 238–39 complex, 259 complex zeros of, 258–64 Conjugate Pairs Theorem, 259–60 defined, 259 finding, 261–62 polynomial function with specified zeros, 260–61 continuous, 938 defined, 224 end behavior of, 233–35
I10
Subject Index
Polynomial functions (Continued) graph of analyzing, 236–38 smooth and continuous, 225 using bounds on zeros, 254 using transformations, 228 historical feature on, 255 identifying, 224–28 multiplicity of, 229–35 real zeros (roots) of, 229–35, 244–58 approximating, 254 Descartes’ Rule of Signs for, 247–48 finding, 250–51 Intermediate Value Theorem, 253–54 number of, 247–48 Rational Zeros Theorem, 248–49, 262 Remainder Theorem and Factor Theorem, 245–47 repeated, 230 theorem for bounds on, 252–53 solving, 250–51 in standard form, 224 unbounded in the negative direction, 233–34 writing, 235 Polynomial inequalities, 291–92 algebraically and graphically solving, 291–92 multiplicity in solving, 292 steps for solving, 294 Polynomial in one variable, second-degree, quadratic function as, 169 Population, world, 851 Population increases, 923, 960 Position vector, 651–52, 671–72 Positive integers, 879n Positive real numbers, A4 Power(s). See also Exponent(s) of i, A93 limit of, 932 log of, 358 Power (electrical), 466 Power functions, 225–28 exponential function vs., 328 graph of, 226–27 of odd degree, 227–28 properties of, 227–28 Present value, 373, 376 Price, equilibrium, 157 Prime polynomials, A32, A36 Principal, 372, A73 Principal nth root of real number, A55–A56 Principal square root, A9, A94 Probability(ies), 850, 909–19 Complement Rule to find, 914–15 compound, 912 constructing models, 909–11 defined, 909 of equally likely outcomes, 911–12 of event, 911 mutually exclusive, 913–14 historical feature on, 915–16 outcome of, 909 sample space, 909 of union of two events, 913–15 Product(s). See also Dot product; Multiplication of complex numbers, A91 in polar form, 642–43 of inertia, 552 limits of, 930–31 log of, 358 special, A26–A27, A30 of two functions, 83–84 of two matrices, 799–803 vector (cross), 658 Product function, 83 Product-to-Sum Formulas, 553–55 Projectile motion, 741–42 Projection, vector, 665–66 Projection of P on the x -axis, 601
Projection of P on the y -axis, 601 Prolate spheroid, 710 Proper rational expressions, 814 Proper rational function, 270 Proper subsets, 895 Proportionality, constant of, 138 Ptolemy, 523, 590 Pure imaginary number, A90 Pythagorean Identities, 445, 524 Pythagorean Theorem, A13–A15 applying, A14–A15 converse of, A13–A15, A18–A19 proof of, A18–A19 as special case of Law of Cosines, 588 Pythagorean triples, A21 Quadrantal angles, 409, 425–27 Quadrants, 34 Quadratic equations, 169–76 character of solutions of, 209 completing the square to solve, 172 defined, 169 discriminant of, 174, 209 factoring, 170 quadratic formula for, 173–75 solutions in complex number system, 208 Square Root Method for solving, 170–71 in standard form, 169 Quadratic factors, irreducible, 252, 817–19 Quadratic formula, 173–75 Quadratic functions, 169–80 of best fit to data, 201–2 defined, 169 graph of properties of, 183 steps for, 187 using its vertex, axis, and intercepts, 184–87 using transformations, 181–83 inequalities involving, 192–96 maximum or minimum value of, 188 optimizations and, 197 properties of, 180–92 vertex and axis of symmetry of, 184–87 zeros of completing the square to find, 172 complex, 207–9 factoring to find, 170 quadratic formula to find, 173–75 Square Root Method to find, 170–71 Quadratic models, 196–206 from data, 201–2 from verbal descriptions, 197–201 Quantity, equilibrium, 157 Quantity demanded, 157–58 Quantity supplied, 157–58 Quaternions, 658 Quotient(s), 245, A28. See also Division of complex numbers in polar form, 642–43 difference, 80, 109, 342, 364, 542 limit of, 933–34 log of, 358 synthetic division to find, A42–A43 of two functions, 83–84 Quotient identity(ies), 524 of trigonometric functions, 444 Radians, 411 converting between degrees and, 412–15 Radical equations, A68–A69 defined, A68 graphing utility to solve, B7 solving, A68–A69 Radicals, A55 fractional exponents as, A58–A59 index of, A55, A60 like, A57 properties of, A56
rational exponents defined using, A58 simplifying, A56–A57 Radical sign, A9, A60 Radicand, A55 Radioactive decay, 383–84 Radius, 66 of sphere, 678 Range, 76 of absolute value function, 114 of constant function, 112 of cosecant function, 440, 441 of cosine function, 439, 440, 441, 455 of cotangent function, 440, 441 of cube function, 113 of cube root function, 113 of greatest integer function, 114 of identity function, 112 of inverse function, 317, 322 inverse of, domain as, 497 of logarithmic function, 345 of logistic models, 386 of one-to-one function, 314–15 of projectile, 547 of reciprocal function, 113 of secant function, 440, 441 of sine function, 439, 440, 441, 454 of square function, 113 of square root function, 113 of tangent function, 440, 441, 469 of trigonometric functions, 439–40 Rate of change, average, 52, 329–30 of function, 104–6 defined, 104 finding, 104–5 secant line and, 105–6 limit of, 934 of linear functions, 151–54 Rate of change, instantaneous, 946 Rate of interest, 372, A73 Rates of return, effective, 375 Ratio common, 868 evenness, 354 Rational equations, A66–A67 defined, A66 with no solution, A67 solving, A66–A67 Rational exponents, A58–A60 Rational expressions, A45–A55 adding and subtracting, A47–A48 least common multiple (LCM) method for, A48–A50 application of, A52 complex, A50–A52 decomposing. See Partial fraction decomposition defined, A45 improper, 814 multiplying and dividing, A46 proper, 814 reducing to lowest terms, A45–A46 Rational functions, 223, 264–90 applied problems involving, 286–87 asymptotes of, 266–72 horizontal or oblique, 267 vertical, 267 continuous, 938–39 defined, 264 domain of, 264–67 examples of, 264 graph of, 275–90 analyzing, 275–85 constructing rational function from, 285–86 using transformations, 266 with a hole, 283–85 improper, 270 in lowest terms, 265, 268 proper, 270
Subject Index unbounded in positive direction, 265 zeros of denominator of, 285 Rational inequalities algebraically and graphically solving, 292–94 multiplicity in solving, 294 steps for solving, 294 Rationalizing the denominator, A57–A58 Rational numbers, 264, A3, A89–A90 Rational Zeros Theorem, 248–49, 262 Ray (half-line), 408 Real axis of complex plane, 640 Real number(s), A3–A6, A89–A90 approximate decimals, A3 conjugate of, A92–A93 defined, A3 principal nth root of, A55–A56 Real number line, A4 Real part of complex numbers, A90 Real zeros (roots) of function, 176–77 of polynomial functions, 244–58 approximating, 254 Descartes’ Rule of Signs for, 247–48 finding, 250–51 Intermediate Value Theorem, 253–54 number of, 247–48 Rational Zeros Theorem, 248–49, 262 Remainder Theorem and Factor Theorem, 245–47 repeated, 230 theorem for bounds on, 252–53 Reciprocal function, 113, 471, 473. See also Cosecant function; Secant function Reciprocal identities, 443–44, 524 Reciprocal property for inequalities, A83–A84, A86 Rectangle, area and perimeter of, A15 Rectangular (Cartesian) coordinates converted to polar coordinates, 620–22 polar coordinates converted to, 618–19 polar coordinates vs., 616 polar equations graphed by converting to, 626–29 in space, 670–71 Rectangular (Cartesian) coordinate system, 33, 34–35 Rectangular (Cartesian) form of complex number, 641–42 Rectangular equations for curve defined parametrically, 738–40 polar equations converted to, 622–23, 735 transforming to polar equation, 622–23 Rectangular grid, 625 Recursive formula, 855–56 for arithmetic sequences, 863–64 terms of sequences defined by, 855–56 Reduced row echelon form, 776–77 Reflections about x-axis or y-axis, 126 Refraction, 522 Regiomontanus, 417, 590 Relation(s), 75–78. See also Function(s) defined, 75 as function, 75–78 input to, 75 linear vs. nonlinear, 163–64 ordered pairs as, 75–76 output of, 75 Remainder, 245, A28 synthetic division to find, A42–A43 Remainder Theorem, 245–47 Removable discontinuity, 285 Repeated zeros (solutions), 170, 230 Repeating decimals, 872–73, A3 Repose, angle of, 513 Rest (equilibrium) position, 600–2 Resultant force, 655 Review, A1–A95 of algebra, A1–A13 distance on the real number line, A5–A6 domain of variable, A7 evaluating algebraic expressions, A6–A7 evaluating exponents, A10
graphing inequalities, A4–A5 Laws of Exponents, A7–A9 sets, A1–A4 square roots, A9–A10 complex numbers, A89–A95 of geometry, A13–A22 congruent and similar triangles, A16–A19 formulas, A15–A16 Pythagorean theorem and its converse, A13–A15 inequalities combined, A85–A86 properties of, A82–A84 solving, A84–A86 interval notation, A81–A82 of nth roots, A55–A56 historical feature, A60 rationalizing the denominator, A57–A58 simplifying, A55 simplifying radicals, A56–A57 of polynomials, A22–A44 adding, A24–A25 dividing, A27–A30 factoring, A32–A41 monomials, A22–A23 multiplying, A25 recognizing, A23–A24 special products formulas, A26–A27 subtracting, A24–A25 synthetic division of, A41–A44 in two variables, A30 of rational exponents, A58–A60 of rational expressions, A45–A55 adding and subtracting, A47–A48 application of, A52 complex, A50–A52 multiplying and dividing, A46 reducing to lowest terms, A45–A46 Revolutions, 416 Rhaeticus, 417 Rhind papyrus, 875 Richter scale, 355 Right angle, 409, A13 Right circular cone, 692 Right circular cylinder, volume and surface area of, A15 Right endpoint of interval, A81 Right-hand rule, 670 Right limit, 936–37 Right triangles, 564–76, A13 applications of, 567–71 solving, 566–67 Right triangle trigonometry, 564–76 Rise, 51 Root(s), A63. See also Solution(s); Zeros complex, 644–46 limit of, 932 of multiplicity 34 (double root), 170 of multiplicity m, 230n perfect, A56 Rose, 633, 636 Roster method, A1 Rotation of axes, 725–27 analyzing equation using, 727–29 formulas for, 726 identifying conic without, 729–30 Rounding, A10 Round-off errors, 567 Row echelon form, 773–80 augmented matrix in, B9–B10 reduced, 776–77 Row index, 770, 795 Row operations, 772–73 Row vector, 799–800 Rudolff, Christoff, A60 Ruffini, P., 255 Rule of Signs, Descartes’, 247–48 Run, 51 Rutherford, Ernest, 723
I11
SAA triangles, 578 Sample space, 909 SAS triangles, 577, 588–89, 594–95 Satisfying equations, 41, A63 Satisfying inequalities, 830 Sawtooth curve, 608 Scalar, 649–50, 798 Scalar multiples of matrix, 798–99 Scalar product. See Dot product Scale of number line, A4 Scanning (optical) angle, 552 Scatter diagrams, 162–63, 479–81 Schroeder, E., 916 Scientific calculators, A10 converting between degrees, minutes, seconds, and decimal forms on, 410 Secant, 425, 426, 439 graph of, 471–72 periodic properties of, 442 Secant function, 423 continuous, 939, 940 domain of, 439, 441 inverse, 510–11 approximating the value of, 511 calculator to evaluate, 511 definition of, 510 range of, 440, 441 Secant line, 105–6 Second-degree equation. See Quadratic equations Second-degree polynomials, A34–A36 Seconds, 410–11 Seed, 647 Sequences, 852–61 annuity problems, 874–75 arithmetic, 862–68 common difference in, 862 defined, 862 determining, 862–63 formula for, 863–64 nth term of, 863 recursive formula for, 863–64 sum of, 864–66 defined, 852 factorial symbol, 854–55 Fibonacci, 856 geometric, 868–79 common ratio of, 868 defined, 868 determining, 868–69 formula for, 869–70 nth term of, 869–70 sum of, 870–71 graph of, 852, 853 historical feature on, 875 from a pattern, 854 properties of, 857 summation notation, 856–57 sum of, 857–58 terms of, 852–54, 855–56 alternating, 854 defined by a recursive formula, 855–56 general, 853 Set(s), A1–A4 complement of, A2 correspondence between two, 75 disjoint, A3 elements of, 895–97, A1 empty (null), 895, A1 equal, 895, A2 finite, 895 infinite, 895 intersection of, A2 Mandelbrot, 647 of numbers, A1–A4 subsets of, 895 proper, 895
I12
Subject Index
Set(s) (Continued) union of, A2 universal, 896, A2 as well-defined, A1 Set-builder notation, A1–A2 Shannon’s diversity index, 354 Shifts, graphing functions using vertical and horizontal, 121–24, 127 Side–angle–side case of congruent triangle, A16 Side–angle–side case of similar triangle, A16, A17 Sides of equation, 41, A63 of inequality, A5 Side–side–side case of similar triangle, A16, A17 Similar triangles, 577n, A16–A19 Simple harmonic motion, 600–2 amplitude of, 600 analyzing, 602 circular motion and, 600–1 defined, 600 equilibrium (rest) position, 600 frequency of object in, 601 model for an object in, 600–2 period of, 600 Simple interest, 372, A73 Simple pendulum, 141 Simplex method, 838n Simplifying complex rational expressions, A50–A52 expressions with rational exponents, A58–A60 nth roots, A55 radicals, A56–A57 rational expressions, A45–A46 Simpson’s rule, 204 Sine, 425, 426, 439 historical feature on, 433 Law of, 576–77, 588n in applied problems, 581–83 defined, 577 historical feature on, 590 proof of, 582–83 SAA or ASA triangles solved using, 578 SSA triangles solved using, 578–81 periodic properties of, 441, 442 Sum and Difference Formula for, 534 trigonometric equations linear in, 538–39 Sine function, 423 of best fit, 483 continuous, 940 domain of, 439, 441, 454 graphs of, 452–67 amplitude and period, 456–58 equation for, 462 key points for, 458–61 y = sin x, 453 inverse, 496–500 approximate value of, 498–99 defined, 497 exact value of, 497–98, 509–10 implicit form of, 497 properties of, 499–500 properties of, 454 range of, 439, 440, 441, 454 Sine function, hyperbolic, 342 Singular matrix, 803 Sinusoidal curve fitting, 479–83 hours of daylight, 482–83 sine function of best fit, 483 temperature data, 479–81 Sinusoidal graphs, 452–67, 604–5 amplitude and period, 456–58 defined, 456 equation for, 462 key points for, 458–61 steps for, 479 y = sin x, 453 Six trigonometric functions of u, 425, 447 of t, 423
Slope, 51–54, 57 containing same point, 52–53 containing two points, 52 graphing lines given, 54 from linear equation, 57 of secant line, 105–6 Slope-intercept form of equation of line, 56–57 Smooth graph, 225 Snell, Willebrord, 522 Snell’s Law of Refraction, 522 Solution(s), A63. See also Zeros extraneous, A67, A68 of inequality, A84 of linear programming problems, 840 location of, 840 repeated, 170, 230 of systems of equations, 756, 761 of trigonometric equations, 514 Solution set of equation, A63 Special products, A26–A27, A30 Speed angular, 416 instantaneous, 946–48 linear, 416–17 Sphere, 678 volume and surface area of, A15 Spheroid, prolate, 710 Spiral, 634–35 Square(s) of binomials (perfect squares), A26, A33 difference of two, A26, A33 perfect, A26, A34 Square function, 112–13 Square matrix, 795–96 Square root(s), A9–A10, A55 complex, 644 of negative number, A94 principal, A9, A94 Square root function, 110, 113 Square Root Method, 170–71 SSA triangles, 578–81 SSS triangles, 577, 589, 595–96 Standard deviation, A88 Standard form complex number in, A90 power of, A93 quotient of two, A92 reciprocal of, A92 of equation of circle, 66–67 linear equation in, 58–59 polynomial function in, 224 polynomials in, A23 quadratic equations on, 169 Standard position, angle in, 408 Static equilibrium, 656–57 Statute (ordinary) miles, 420 Step function, 115 Stirling’s formula, 889 Stock risk, beta to measure, 150 Stock valuation, 150, 221–22 Straight angle, 409 Stretches, graphing functions using, 124–25, 127 vertical, 124 Strict inequalities, A5 Subscript, A23n Subscripted letters, 852 Subsets, 895, A2 proper, 895 Substitution method, 754, 757–58 systems of nonlinear equations solved using, 821–22 Subtraction. See also Difference(s) of complex numbers, A90 of polynomials, A24–A25 of rational expressions, A47–A48 least common multiple (LCM) method for, A48–A50
of vectors, 652–53 in space, 672 Sum. See also Addition of arithmetic sequences, 864–66 of geometric sequences, 870–71 index of, 856 of infinite geometric series, 872 limits of, 930 of logarithms, 358 of sequences, 857–58 of two cubes, A27, A33 of two functions, 83–84 graph of, 604–6 of two matrices, 796–98 Sum and Difference Formulas, 531–43 for cosines, 531–32 defined, 531 to establish identities, 533–36 to find exact values, 532, 534–35 involving inverse trigonometric function, 537 for sines, 534 for tangents, 536 Sum function, 83 Sum-to-Product Formulas, 555 Summation notation, 856–57 Sun, declination of, 506 Surface area, formulas for, A15 Sylvester, James J., 808 Symmetry, 44–46 axis of of parabola, 182, 693 of quadratic function, 183–84 graphing utility to check for, B5–B6 of polar equations, 629 with respect to origin, 44–45 p (y-axis), 629 with respect to the line u = 2 with respect to the polar axis (x-axis), 629 with respect to the pole (origin), 629 with respect to the x-axis, 44–45 with respect to the y-axis, 44–45 Synthetic division, A41–A44 Systems of equations consistent, 756, 761–62 dependent, 757 containing three variables, 764–65 containing two variables, 761 equivalent, 759 graphing, 757 inconsistent, 756, 761–62, 763–64 containing three variables, 763–64 containing two variables, 760 independent, 757 solutions of, 756, 761 Systems of inequalities, 830–37 graphing, 832–35 bounded and unbounded graphs, 835 vertices or corner points, 835 Systems of linear equations, 755–94 consistent, 757, 761–62 defined, 755 dependent, 757 containing three variables, 764–65 containing two variables, 761 matrices to solve, 777–78 determinants, 784–94 cofactors, 788–89 Cramer’s Rule to solve a system of three equations containing three variables, 789–91 Cramer’s Rule to solve a system of two equations containing two variables, 785–87 minors of, 788–89 properties of, 791–92 3 by 3, 788–89 2 by 2, 785, 791–92 elimination method of solving, 759, 761 equivalent, 759 examples of, 755–56
Subject Index graphing, 757 inconsistent, 757, 761–62, 763–64 containing three variables, 763–64 containing two variables, 760 matrices to solve, 778–79 independent, 757 matrices. See Matrix/matrices partial fraction decomposition, 813–20 defined, 814 where denominator has a nonrepeated irreducible quadratic factor, 818–19 where denominator has only nonrepeated linear factors, 814–15 where denominator has repeated irreducible quadratic factors, 819 where denominator has repeated linear factors, 816–17 solution of, 756, 761 solving, 756 with graphing utility, B9–B10 substitution method of, 757–58 three equations containing three variables, 761–63 Systems of nonlinear equations, 821–29 elimination method for solving, 822–25 historical feature on, 826 substitution method for solving, 821–22 Systems of nonlinear inequalities, graphing, 834 Tables, on graphing utility, B4 Tangency, point of, 71 Tangent(s), 425, 426, 439 graph of, 468–70 Half-angle Formulas for, 549–50 historical feature on, 433 Law of, 587, 590 periodic properties of, 442 Sum and Difference Formulas for, 536 Tangent function, 423 continuous, 939, 940 domain of, 439, 441, 469 inverse, 502–4 defined, 503 exact value of, 503–4, 509–10 implicit form of, 503 properties of, 469 range of, 440, 441, 469 Tangent line, 71 to the graph of a function, 943–44 Greek method for finding, 71 Tangent problem, 943 Tartaglia (Niccolo of Brescia), 255, 646 Tautochrone, 746 Terminal point of directed line segment, 648 Terminal side of angle, 408, 409 Terminating decimals, A3 Terms constant, 224 leading, 224 like, A23 of polynomial, A23 of sequences, 852–54, 855–56 alternating, 854 defined by a recursive formula, 855–56 general, 853 3 by 3 determinants, 680, 788–89 Thrust, 615 TI-84 Plus, B3, B9, B10 TI-85, B8n Time, as parameter, 740–43 Transcendental functions, 305 Transformations combining, 123–24, 127–29 compressions and stretches, 124–25, 127 cosecant and secant graphs using, 472 of cosine function, 455–56 defined, 121 graphs using of exponential functions, 333
of polynomial functions, 228 of quadratic functions, 181–83 of rational functions, 266 reflections about the x-axis or y-axis, 126 of sine function, 454 vertical and horizontal shifts, 121–24, 127 Transition matrix, 850 Transverse axis, 712, 732 Tree diagram, 898 Triangle(s). See also Law of Sines altitude of, A15 area of, 593–99, A15 ASA, 578 congruent, A16–A19 equilateral, 39 error, 40 isosceles, 39 legs of, A13–A14 medians of, 39 midpoint of, 39 oblique, 576–77 Pascal, 885, 888 perfect, 598 right, 564–76, A13 applied problems involving, 567–71 solving, 566–67 SAA, 578 SAS, 577, 588–89, 594–95 similar, 577n, A16–A19 SSA, 578–81 SSS, 577, 589, 595–96 Triangular addition, 888 Trigonometric equations, 514–23 calculator for solving, 517 graphing utility to solve, 519 identities to solve, 518–19 involving single trigonometric function, 514–17 linear, 515–16 linear in sine and cosine, 538–39 quadratic in from, 517–18 solutions of, defined, 514 Trigonometric expressions, written algebraically, 511–12, 537 Trigonometric functions, 407–94 of acute angles, 564–66 angles and their measure, 408–21 arc length of a circle, 411–12 area of a sector of a circle, 415 degrees, 409–11, 412–15 radians, 411, 412–15 applications of, 563–614 damped motion, 600, 603–4 graphing sum of two functions, 604–6 involving right triangles, 567–71 Law of Cosines, 587–93 Law of Sines, 576–87, 589–90 Law of Tangents, 587, 589–90 simple harmonic motion, 600–2 calculator to approximate values of, 432 circle of radius r to evaluate, 432–33 cosecant and secant graphs, 471–73 exact values, 425–27 given one of the functions and the quadrant of the angle, 445–48 p p = 30°, = 45°, and for integer multiples of 6 4 p = 60°, 428–30 3 p = 45°, 427 4 p p = 30° and = 60°, 428–30 6 3 using a point on the unit circle, 424–25 using periodic properties, 442 fundamental identities of, 443–45 quotient, 444 reciprocal, 443–44 historical feature on, 417 phase shift, 475–86 to graph y = A sin(vx - w), 475–79
I13
properties of, 439–52 domain and range, 439–40 even-odd, 448–49 periodic, 441–42 of quadrantal angles, 425–27 right triangle trigonometry, 564–66 signs of, in a given quadrant, 443 sine and cosine graphs, 452–67 amplitude and period, 456–58, 460, 476–79 equation for, 462 key points for, 458–61 sinusoidal curve fitting, 479–83 of t, 423 tangent and cotangent graphs, 468–71 of u, 425, 447 unit circle approach to finding exact values of, 422–38 Trigonometric identities, 523–30 basic, 524 establishing, 525–28 Double-angle Formulas for, 544–47 Sum and Difference Formulas for, 533–36 Even-Odd, 524 Pythagorean, 524 Quotient, 524 Reciprocal, 524 Trinomials, A23 factoring, A35, A37–A38 Truncation, A10 Turning points, 231–32 2 by 2 determinants, 680, 785 proof for, 791 Umbra versa, 433 Unbounded graphs, 835 Unbounded in positive direction, 265 Unbounded in the negative direction, 233–34 Uniform motion, A75–A77 Uninhibited growth, 381–83 Union of sets, A2 of two events, probabilities of, 913–15 Unit circle, 66, 422–38 Unit vector, 651, 654, 671 in direction of v, 673 Universal sets, 896, A2 Value (image) of function, 76, 78–81 Variable(s), A6, A22 complex, 259 dependent, 79 domain of, A7 independent, 79 in calculus, 453 Variable costs, 63 Variation, 138–43 combined, 139–40 direct, 138, 139–40 inverse, 139 joint, 139–40 Vector(s), 648–62 adding, 649, 652–53 algebraic, 651–52 angle between, 663–64 column, 799–800 components of, 651, 652, 671 horizontal and vertical, 652 cross product of, 679–81 decomposing, 665–66 defined, 648 difference of, 649 direction of, 648, 654–55 dot product of two, 662–63 equality of, 649, 652 finding, 654–55 force, 654–55 geometric, 648–49 graphing, 650–51
I14
Subject Index
Vector(s) (Continued) historical feature on, 658 magnitudes of, 648, 650–51, 652, 654–55 modeling with, 655–57 multiplying by numbers, 649–50 objects in static equilibrium, 656–57 orthogonal, 664–67 parallel, 664 in physics, 648 position, 651–52 row, 799–800 scalar multiples of, 650, 653, 663 in space, 670–79 angle between two vectors, 674–75 direction angles of, 675–77 distance between two points in space, 671 dot product, 673–74 operations on, 672–73 position vectors, 671–72 subtracting, 652–53 unit, 651, 654, 671 in direction of v, 673 velocity, 654–55 writing, 655 zero, 649 Vector product. See Cross (vector) product Vector projection, 665–66 Velocity, instantaneous, 946–48 Velocity vector, 654–55 Venn diagrams, A2 VERT command, 55 Vertex/vertices, 835 of cone, 692 of ellipse, 702 of hyperbola, 712 of parabola, 182, 693 of quadratic function, 183–87 of ray, 408 Vertical asymptote, 267–69, 321 multiplicity and, 268–69
Vertical component of vector, 652 Vertical line, 51, 627, 635 equation of, 54–55 Vertical-line test, 89 Vertically compressed or stretched graphs, 124 Vertical shifts, 121–24, 127 Viète, François, 590 Viewing angle, 507 Viewing rectangle, 35, B1–B3 setting, B1–B2 Vinculum, A60 Volume, formulas for, A15 Wallis, John, 646 Waves, traveling speeds of, 522 Weight, 615 Whispering galleries, 708 Wings, airplane, 615 Work, 678 dot product to compute, 667 World population, 851, 923, 960 x-axis, 34 projection of P on the, 601 reflections about, 126 symmetry with respect to, 44–45 x-coordinate, 34 x-intercept(s), 43 polynomial graphed using, 229, 230–31 of quadratic function, 175, 184. See also Zeros: of quadratic function xy-plane, 34, 670–71 xz-plane, 670–71 Yang Hui, 888 y-axis, 34 projection of P on the, 601 reflections about, 126 symmetry with respect to, 44–45
y-coordinate (ordinate), 34 y-intercept, 43, 57 from linear equation, 57 yz-plane, 670–71 Zero-coupon bonds, 380 Zero-level earthquake, 355 Zero matrix, 798 Zero polynomial, A23 Zero-Product Property, A4 Zeros bounds on, 252–53 complex, of polynomials, 258–64 Conjugate Pairs Theorem, 259–60 defined, 259 finding, 261–62 polynomial function with specified zeros, 260–61 complex, of quadratic function, 207–9 of function, 91–92 real, 176–77 of linear function, 155 of quadratic function completing the square to find, 172 factoring to find, 170 quadratic formula to find, 173–75 Square Root Method to find, 170–71 real, of polynomials, 229–35, 244–58 approximating, 254 Descartes’ Rule of Signs for, 247–48 finding, 250–51 Intermediate Value Theorem, 253–54 number of, 247–48 Rational Zeros Theorem, 248–49, 262 Remainder Theorem and Factor Theorem, 245–47 repeated, 230 theorem for bounds on, 252–53 Zero vector, 649