C o l l College Algebra & Trigonometry e g e A Unit Circle Approach A l Mark Dugopolski g e b Fifth Edition r a & T r i g o n o m e t r y D u g o p o l s k i F i f t h E d i t i o n ISBN 978-1-29202-381-6 9 781292 023816 College Algebra & Trigonometry A Unit Circle Approach Mark Dugopolski Fifth Edition ISBN 10: 1-292-02381-3 ISBN 13: 978-1-292-02381-6 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.pearsoned.co.uk © Pearson Education Limited 2014 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 licence 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 affi liation with or endorsement of this book by such owners. ISBN 10: 1-292-02381-3 ISBN 13: 978-1-292-02381-6 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America 123445667788976217417183318574792333135 P E A R S O N C U S T O M L I B R AR Y Table of Contents Chapter 1. Equations, Inequalities, and Modeling Mark Dugopolski 1 Chapter 2. Functions and Graphs Mark Dugopolski 95 Chapter 3. Polynomial and Rational Functions Mark Dugopolski 178 Chapter 4. Exponential and Logarithmic Functions Mark Dugopolski 265 Chapter 5. The Trigonometric Functions Mark Dugopolski 327 Chapter 6. Trigonometric Identities and Conditional Equations Mark Dugopolski 414 Chapter 7. Applications of Trigonometry Mark Dugopolski 477 Chapter 8. Systems of Equations and Inequalities Mark Dugopolski 549 Chapter 9. Matrices and Determinants Mark Dugopolski 612 Chapter 10. The Conic Sections Mark Dugopolski 673 Chapter 11. Sequences, Series, and Probability Mark Dugopolski 713 Answers to Selected Exercises Mark Dugopolski 783 Credits Mark Dugopolski 831 Index 833 I II 1 Equations, Inequalities, and Modeling Even infamous Heartbreak Hill couldn’t break the winning spirit of Ethiopian runner Dire Tune as she focused on first place. It was the 112th running of the Boston Marathon, a 26.2-mile ordeal that one runner called “14 miles of fun, 8 miles of sweat, and 4 miles of hell!” Sporting events like the 2008 Marathon have come a long way since the first Olympic Games were held over 2500 years ago. Today, sports and sci- ence go hand in hand. Modern athletes often use mathematics to analyze the variables that help them increase aerobic capacity, reduce air resistance, or strengthen muscles. Learn... WHAT YOU WILL In this chapter you will see examples of sports applications while you learn to solve equations and inequalities. By the time you reach the finish line, you will be using algebra to model and solve problems. 1.1 Equations in One Variable 1.2 Constructing Models to Solve Problems 1.3 Equations and Graphs in Two Variables 1.4 Linear Equations in Two Variables 1.5 Scatter Diagrams and Curve Fitting 1.6 Quadratic Equations 1.7 Linear and Absolute Value Inequalities From Chapter 1 of College Algebra and Trigonometry: A Unit Circle Approach, Fifth Edition. Mark Dugopolski. Copyright © 2011 by Pearson Education, Inc. All rights reserved. 1 80 Chapter 1 Equations, Inequalities, and Modeling 1.1 Equations in One Variable One of our main goals in algebra is to develop techniques for solving a wide variety of equations. In this section we will solve linear equations and other similar equations. Definitions An equation is a statement (or sentence) indicating that two algebraic expressions are equal. The verb in an equation is the equali#ty symbol. For example, 2x + 8 = 0 is an equation. If we replace xby -4,we get 2 -4 + 8 = 0,a true statement. So 1 2 we say that -4is a solution# or rootto the equation or -4satisfiesthe equation. If we replace xby 3, we get 2 3 + 8 = 0,a false statement. So 3 is not a solution. Whether the equation 2x + 8 = 0is true or false depends on the value of x, and so it is called an open sentence.The equation is neither true nor false until we choose a value for x.The set of all solutions to an equation is called the solution setto the equation. To solve an equation means to find the solution set. The solution set for 2x + 8 = 0is {-4}.The equation 2x + 8 = 0is an example of a linear equation. Definition: Linear Equation A linear equation in one variable is an equation of the form ax + b = 0, in One Variable where aand bare real numbers, with a Z 0. Note that other letters can be used in place of x. For example, 3t + 5 = 0, 2w - 6 = 0,and -2u + 7 = 0are linear equations. Solving Linear Equations The equations 2x + 8 = 0 and 2x = -8 both have the solution set {-4}. Two equations with the same solution set are called equivalent equations. Adding the same real number to or subtracting the same real number from each side of an equa- tion results in an equivalent equation. Multiplying or dividing each side of an equa- tion by the same nonzero real number also results in an equivalent equation. These properties of equalityare stated in symbols in the following box. Properties of Equality If Aand Bare algebraic expressions and Cis a real number, then the following equations are equivalent to A = B: A + C = B + C Addition property of equality A - C = B - C Subtraction property of equality CA = CB C Z 0 Multiplication property of equality 1 2 A B = C Z 0 Division property of equality C C 1 2 We can use an algebraic expression for Cin the properties of equality, because the value of an algebraic expression is a real number. However, this can produce nonequivalent equations. For example, 1 1 x = 0 and x + = 0 + x x appear to be equivalent by the addition property of equality. But the first is satisfied by 0 and the second is not. When an equation contains expressions that are unde- fined for some real number(s) then we must check all solutions carefully. 2 1.1 Equations in One Variable 81 Any linear equation, ax + b = 0, can be solved in two steps. Subtract b from each side and then divide each side by a a Z 0 , to get x = -b a.Although the 1 2 > equations in our first example are not exactly in the form ax + b = 0,they are often called linear equations because they are equivalent to linear equations. 1 EXAMPLE Using the properties of equality Solve each equation. 1 3 a. 3x - 4 = 8 b. x - 6 = x - 9 c. 3 4x - 1 = 4 - 6 x - 3 2 4 1 2 1 2 Solution a. 3x - 4 = 8 3x - 4 + 4 = 8 + 4 Add 4 to each side. 3x = 12 Simplify. 3x 12 = Divide each side by 3. 3 3 x = 4 Simplify. Since the last equation is equivalent to the original, the solution set to the original eq#uation is {4}. We can check by replacing x by 4 in 3x - 4 = 8. Since 3 4 - 4 = 8is correct, we are confident that the solution set is {4}. b. Multiplying each side of the equation by the least common denominator, LCD, will eliminate all of the fractions: 1 3 x - 6 = x - 9 2 4 1 3 4a2 x - 6b = 4a4 x - 9b Multiply each side by 4, the LCD. 2x - 24 = 3x - 36 Distributive property 2x - 24 - 3x = 3x - 36 - 3x Subtract 3xfrom each side. -x - 24 = -36 Simplify. -x = -12 Add 24 to each side. -1 -x = -1 -12 Multiply each side by -1. 1 21 2 1 21 2 x = 12 Simplify. Check 12 in the original equation. The solution set is {12}. c. 3 4x - 1 = 4 - 6 x - 3 1 2 1 2 12x - 3 = 4 - 6x + 18 Distributive property 12x - 3 = 22 - 6x Simplify. 18x - 3 = 22 Add 6xto each side. 18x = 25 Add 3 to each side. 25 x = Divide each side by 18. 18 25 Check 25 18in the original equation. The solution set is . > E18F You can use a graphing calculator to calculate the value of each side of the Figure 1.1 equation when xis 25 18as shown in Fig. 1.1. > (cid:2)TRY THIS. Solve 5 3x - 2 = 5 - 7 x - 1 . ■ 1 2 1 2 3 82 Chapter 1 Equations, Inequalities, and Modeling Note that checking the equation in Example 1(c) with a calculator did not prove that 25 18 is the correct solution. The properties of equality that were > applied correctly in each step guarantee that we have the correct solution. The val- ues of the two sides of the equation could agree for the 10 digits shown on the calculator and disagree for the digits not shown. Since that possibility is extremely unlikely, the calculator check does support our belief that we have the correct solution. (cid:3) Identities, Conditional Equations, and Inconsistent Equations An equation that is satisfied by every real number for which both sides are defined is an identity.Some examples of identities are x 3x - 1 = 3x - 1, 2x + 5x = 7x, and = 1. x The solution set to the first two identities is the set of all real numbers, R.Since 0 0 > is undefined, the solution set to x x = 1 is the set of nonzero real numbers, > {x|x Z 0}. A conditional equationis an equation that is satisfied by at least one real num- ber but is not an identity. The equation 3x - 4 = 8 is true only on condition that x = 4,and it is a conditional equation. The equations of Example 1 are conditional equations. An inconsistent equation is an equation that has no solution. Some inconsis- tent equations are # 0 x + 1 = 2, x + 3 = x + 5, and 9x - 9x = 8. Note that each of these inconsistent equations is equivalent to a false statement: 1 = 2, 3 = 5,and 0 = 8,respectively. 2 EXAMPLE Classifying an equation Determine whether the equation 3 x - 1 - 2x 4 - x = 2x + 1 x - 3 is an 1 2 1 2 1 21 2 identity, an inconsistent equation, or a conditional equation. Solution 3 x - 1 - 2x 4 - x = 2x + 1 x - 3 1 2 1 2 1 21 2 3x - 3 - 8x + 2x2 = 2x2 - 5x - 3 Simplify each side. 2x2 - 5x - 3 = 2x2 - 5x - 3 Since the last equation is equivalent to the original and the last equation is an iden- tity, the original equation is an identity. (cid:2)TRY THIS. Determine whether x x - 1 - 6 = x - 3 x + 2 is an identity, 1 2 1 21 2 an inconsistent equation, or a conditional equation. ■ Equations Involving Rational Expressions Recall that division by zero is undefined and we can’t have zero in the denomina- tor of a fraction. Since the rational expressions in the next example have variables in their denominators, these variables can’t be replaced by any numbers that would cause zero to appear in a denominator. Our first step in solving these equa- tions is to multiply by the LCD and eliminate the denominators. But we must check our solutions in the original equations and discard any that cause unde- fined expressions. 4 1.1 Equations in One Variable 83 3 EXAMPLE Equations involving rational expressions Solve each equation and identify each as an identity, an inconsistent equation, or a conditional equation. y 3 1 1 2 1 1 a. + 3 = b. - = c. + = 1 y - 3 y - 3 x - 1 x + 1 x2 - 1 2 x - 1 Solution a. Since y - 3 is the denominator in each rational expression, y - 3 is the LCD. Note that using 3 in place of yin the equation would cause 0 to appear in the de- nominators. So we know up front that 3 is not a solution to this equation. y 3 1 y - 32a y - 3 + 3b = 1 y - 32 y - 3 Multiply each side by the LCD. y 1 y - 32 y - 3 + 1 y - 323 = 3 Distributive property y + 3y - 9 = 3 4y - 9 = 3 4y = 12 Add 9 to each side. y = 3 Divide each side by 4. If we replace yby 3 in the original equation, then we get two undefined expres- sions. So 3 is not a solution to the original equation. The original equation has no solution. The equation is inconsistent. b. Since x2 - 1 = x - 1 x + 1 , the LCD is x - 1 x + 1 . Note that using 1 21 2 1 21 2 1 or -1for xin the equation would cause 0 to appear in a denominator. 1 1 2 - = x - 1 x + 1 x2 - 1 1 1 2 x - 1 x + 1 - = x - 1 x + 1 Multiple by the LCD. 1 21 2ax - 1 x + 1b 1 21 2x2 - 1 1 1 1x - 121x + 12 x - 1 - 1x - 121x + 12 x + 1 = 2 Distributive property x + 1 - x - 1 = 2 1 2 2 = 2 Since the last equation is an identity, the original equation is also an identity. The solution set is x|x Z 1and x Z -1 ,because 1 and -1cannot be used for xin 5 6 the original equation. c. Note that we cannot use 1 for x in the original equation. To solve the equation multiply each side by the LCD: 1 1 + = 1 2 x - 1 1 1 2 x - 1 + = 2 x - 1 1 Multiply by the LCD. 1 2a2 x - 1b 1 2 x - 1 + 2 = 2x - 2 x + 1 = 2x - 2 3 = x Check 3 in the original equation. The solution set is 3 , and the equation is a 5 6 conditional equation. 2 3 4 (cid:2)TRY THIS. Solve - = . ■ x - 3 x + 3 x2 - 9 5