Issuu

CONTENTS PREFACE...................................................................................................................................................................

HINTS FOR TEACHING CALCULUS WITH APPLICATIONS ............................................................................

PRETESTS ................................................................................................................................................................. xix ANSWERS TO PRETESTS ....................................................................................................................................... xxvii FINAL EXAMINATIONS ......................................................................................................................................... xxix ANSWERS TO FINAL EXAMINATIONS............................................................................................................... xliii

SOLUTIONS TO ALL EXERCISES CHAPTER R ALGEBRA REFERENCE R.1 Polynomials ..................................................................................................................................................

R.2 Factoring .......................................................................................................................................................

R.3 Rational Expressions ....................................................................................................................................

R.4 Equations ......................................................................................................................................................

R.5 Inequalities ...................................................................................................................................................

R.6 Exponents .....................................................................................................................................................

R.7 Radicals ........................................................................................................................................................

CHAPTER 1 LINEAR FUNCTIONS 1.1 Slopes and Equations of Lines ......................................................................................................................

1.2 Linear Functions and Applications................................................................................................................

1.3 The Least Squares Line .................................................................................................................................

Chapter 1 Review Exercises..................................................................................................................................

Extended Application: Predicting Life Expectancy ..............................................................................................

iii

CHAPTER 2 NONLINEAR FUNCTIONS 2.1 Properties of Functions..................................................................................................................................

2.2 Quadratic Functions; Translation and Reflection ..........................................................................................

2.3 Polynomial and Rational Functions .............................................................................................................. 102 2.4 Exponential Functions ................................................................................................................................... 115 2.5 Logarithmic Functions .................................................................................................................................. 125 2.6 Applications: Growth and Decay; Mathematics of Finance .......................................................................... 137 Chapter 2 Review Exercises.................................................................................................................................. 146 Extended Application: Power Functions ............................................................................................................... 162

CHAPTER 3 THE DERIVATIVE 3.1 Limits ............................................................................................................................................................ 163 3.2 Continuity...................................................................................................................................................... 177 3.3 Rates of Change ............................................................................................................................................ 184 3.4 Definition of the Derivative .......................................................................................................................... 195 3.5 Graphical Differentiation .............................................................................................................................. 219 Chapter 3 Review Exercises.................................................................................................................................. 223 Extended Application: A Model for Drugs Administered Intravenously .............................................................. 235

CHAPTER 4 CALCULATING THE DERIVATIVE 4.1 Techniques for Finding Derivatives .............................................................................................................. 237 4.2 Derivatives of Products and Quotients .......................................................................................................... 247 4.3 The Chain Rule ............................................................................................................................................. 258 4.4 Derivatives of Exponential Functions ........................................................................................................... 269 4.5 Derivatives of Logarithmic Functions ........................................................................................................... 283 Chapter 4 Review Exercises.................................................................................................................................. 296 Extended Application: Electric Potential and Electric Field ................................................................................. 308

CONTENTS

CHAPTER 5 GRAPHS AND THE DERIVATIVE 5.1 Increasing and Decreasing Functions ............................................................................................................ 309 5.2 Relative Extrema ........................................................................................................................................... 324 5.3 Higher Derivatives, Concavity, and the Second Derivative Test .................................................................. 339 5.4 Curve Sketching

............................................................................................................................... 362

Chapter 5 Review Exercises.................................................................................................................................. 383 Extended Application: A Drug Concentration Model for Orally Administered Medications ............................... 400

CHAPTER 6 APPLICATIONS OF THE DERIVATIVE 6.1 Absolute Extrema .......................................................................................................................................... 401 6.2 Applications of Extrema................................................................................................................................ 411 6.3 Further Business Applications....................................................................................................................... 432 6.4 Implicit Differentiation ................................................................................................................................. 440 6.5 Related Rates ................................................................................................................................................. 457 6.6 Differentials: Linear Approximation ............................................................................................................. 466 Chapter 6 Review Exercises.................................................................................................................................. 472 Extended Application: A Total Cost Model for a Training Program .................................................................... 483

CHAPTER 7 INTEGRATION 7.1 Antiderivatives .............................................................................................................................................. 485 7.2 Substitution ................................................................................................................................................... 497 7.3 Area and the Definite Integral ....................................................................................................................... 507 7.4 The Fundamental Theorem of Calculus ........................................................................................................ 521 7.5 The Area Between Two Curves .................................................................................................................... 538 7.6 Numerical Integration ................................................................................................................................... 556 Chapter 7 Review Exercises.................................................................................................................................. 570 Extended Application: Estimating Depletion Dates for Minerals ......................................................................... 588

CONTENTS

CHAPTER 8 FURTHER TECHNIQUES AND APPLICATIONS OF INTEGRATION 8.1 Integration by Parts ....................................................................................................................................... 589 8.2 Volume and Average Value .......................................................................................................................... 601 8.3 Continuous Money Flow ............................................................................................................................... 612 8.4 Improper Integrals ......................................................................................................................................... 619 Chapter 8 Review Exercises.................................................................................................................................. 633 Extended Application: Estimating Learning Curves in Manufacturing with Integrals ......................................... 643

CHAPTER 9 MULTIVARIABLE CALCULUS 9.1 Functions of Several Variables...................................................................................................................... 645 9.2 Partial Derivatives ........................................................................................................................................ 655 9.3 Maxima and Minima ..................................................................................................................................... 671 9.4 Lagrange Multipliers ..................................................................................................................................... 686 9.5 Total Differentials and Approximations ........................................................................................................ 702 9.6 Double Integrals ............................................................................................................................................ 711 Chapter 9 Review Exercises.................................................................................................................................. 730 Extended Application: Using Multivariable Fitting to Create a Response Surface Design .................................. 749

CHAPTER 10 DIFFERENTIAL EQUATIONS 10.1 Solutions of Elementary and Separable Differential Equations .................................................................. 751 10.2 Linear First-Order Differential Equations ................................................................................................... 769 10.3 Euler’s Method ............................................................................................................................................ 778 10.4 Applications of Differential Equations........................................................................................................ 788 Chapter 10 Review Exercises ............................................................................................................................... 800 Extended Application: Pollution of a Lake ........................................................................................................... 813

CHAPTER 11 PROBABILITY AND CALCULUS 11.1 Continuous Probability Models ................................................................................................................... 815 11.2 Expected Value and Variance of Continuous Random Variables ............................................................... 828 11.3 Special Probability Density Functions ........................................................................................................ 845 Chapter 11 Review Exercises ................................................................................................................................ 857 Extended Application: Exponential Waiting Times .............................................................................................. 871 vi

CONTENTS

CHAPTER 12 SEQUENCES AND SERIES 12.1 Geometric Sequences ................................................................................................................................. 873 12.2 Annuities: An Application of Sequences .................................................................................................... 882 12.3 Taylor Polynomials at 0 .............................................................................................................................. 896 12.4 Infinite Series .............................................................................................................................................. 914 12.5 Taylor Series at 0 ........................................................................................................................................ 923 12.6 Newton’s Method ........................................................................................................................................ 940 12.7 L’Hospital’s Rule ........................................................................................................................................ 951 Chapter 12 Review Exercises ................................................................................................................................ 962 Extended Application: Living Assistance and Subsidized Housing ...................................................................... 979

CHAPTER 13 THE TRIGONOMETRIC FUNCTIONS 13.1 Definitions of the Trigonometric Functions ................................................................................................ 981 13.2 Derivatives of Trigonometric Functions ..................................................................................................... 993 13.3 Integrals of Trigonometric Functions .......................................................................................................... 1010 Chapter 13 Review Exercises ................................................................................................................................ 1019 Extended Application: The Shortest Time and the Cheapest Path ........................................................................ 1031

CONTENTS

vii

PREFACE This book provides several resources for instructors using Calculus with Applications, Twelfth Edition, by Margaret L. Lial, Raymond N. Greenwell, and Nathan P. Ritchey, with Katherine Ritchey, Sarah Ritchey Patterson, and Blain Patterson. ·

Hints for teaching Calculus with Applications are provided as a resource for faculty.

One open-response form and one multiple-choice form of a pretest are provided. These tests are an aid to instructors in identifying students who may need assistance.

One open-response form and one multiple-choice form of a final examination are provided.

Solutions for nearly all of the exercises in the textbook are included. Solutions are usually not provided for exercises with open-response answers.

PREFACE

HINTS FOR TEACHING CALCULUS WITH APPLICATIONS Algebra Reference Some instructors obtain best results by going through this chapter carefully at the beginning of the semester. Others find it better to refer to it as needed throughout the course. Use whichever method works best for your students. As in the previous edition, we refer to the chapter as a “Reference” rather than a “Review,” and the regular page numbers don’t begin until Chapter 1. We hope this will make your students less anxious if you don’t cover this material.

Chapter 1 Instructors sometimes go to either of two extremes in this chapter and the next. Some feel that their students have already covered enough precalculus in high school or in previous courses, and consequently begin with Chapter 3. Unfortunately, if they are wrong, their students may do poorly. Other instructors spend at least half a semester on Chapters 1 and 2 and the algebra reference chapter, and subsequently have little time for calculus. Such a course should not be labeled as calculus. We recommend trying to strike a balance, which may still not make all your students happy. A few may complain that the review of algebra, functions, and graphs is too quick; such students should be sent to a more basic course. Those students who are familiar with this material may become lazy and develop habits that will hurt them later in the course. You may wish to assign a few challenging exercises to keep these students on their toes. Chapter 1 of Calculus with Applications is identical to Chapter 1 of Finite Mathematics and may be skipped by students who have already taken a course using that text. In this edition, we have streamlined the chapter from four to three sections, allowing instructors to reach the calculus material more quickly.

Section 1.1 This section and the next may seem fairly basic to students who covered linear functions in high school. Nevertheless, many students who have graphed hundreds of lines in their lifetime still lack a thorough understanding of slope, which hampers their understanding of the derivative. Such students could benefit from doing dozens of exercises similar to 43–46. Perpendicular lines are not used in future chapters and could be skipped if you are in a hurry.

Section 1.2 Much recent research has been devoted to students’ misunderstandings of the function concept. Such misunderstandings are among the major impediments to learning calculus. One way to help students is to study a simple class of functions first, as we do in this section. In this edition, even more of the general material on functions, including domain and range, is postponed until Chapter 2. Supply and demand provides the students’ first experience with a mathematical model. Spend time developing both the economics and the mathematics involved. Stress that for cost, revenue, and profit functions, x represents the number of units. For supply and demand functions, we use the economists’ notation of q to represent the number of units. Emphasize the difference between the profit earned on 100 units sold as opposed to the number of units that must be sold to produce a profit of $100.

Section 1.3 The statistical functions on a calculator can greatly simplify these calculations, allowing more time for discussion and further examples. In this edition, we use “parallel presentation” to allow the instructor a choice on the extent technology is used. This section may be skipped if you are in a hurry, but your students can benefit from the realistic model and the additional work with equations of lines.

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Section 2.1 After learning about linear functions in the previous chapter, students now learn about functions in general. This concept is critical for success in calculus. Unless sufficient time is devoted to this section, the results will become apparent later when students don’t understand the derivative. One device that helps students distinguish f ( x + h) from f ( x) + h is to use a box in place of the letter x, as we do in this section before Example 6.

Section 2.2 This section combines the topics of quadratic functions and translation and reflection, with a minimal amount of material on completing the square. Our experience is that students graph quadratics most easily by first finding the y-intercept, then finding the x-intercepts when they exist (using factoring or the quadratic formula), and finding the vertex last by locating the point midway between the x-intercepts or, if the quadratic formula was used, by letting x = -b /(2a). Quadratics are among a small group of functions that can be analyzed completely with ease, so they are used throughout the text. On the other hand, the advent of graphing calculators has made ease of graphing less important, so we rely on quadratics less than in previous editions. Some instructors pressed for time may choose to skip translations and reflections. But we have found that students who understand that the graph of f ( x) = 5 - 4 - x is essentially the same as the graph of f ( x) = x , just shifted and reflected, will have an easier time when using the derivative to graph functions. Since students are familiar with very few classes of functions at this point, it helps to work with functions defined solely by their graphs, such as Exercises 35–38. Exercises 43–50 cover stretching and shrinking of graphs in the vertical and horizontal directions. Covering these exercises carefully will not only give students a better grasp of functions, but will help them later to interpret the chain rule.

Section 2.3 Graphing calculators have made point plotting of functions less important than before. Plotting points by hand should not be entirely neglected, however, because a small amount is helpful when using the derivative to graph functions. The two main goals of this section are to have an understanding of what an n-th degree polynomial looks like, and to be able to find the asymptotes of a rational function. Students who master these ideas will be better prepared for the chapter on curve sketching. Exercise 59 is the first of several in this chapter asking students to find what type of function best fits a set of data. (See also Section 2.4, Exercises 57, 58 and 59, and Review Exercise 111.) The class can easily get bogged down in these exercises, particularly if you decide to explore the regression features in a calculator such as the TI-84 Plus C. But there is a powerful payoff in terms of mastery of functions for the student who succeeds at these exercises.

Section 2.4 Some instructors may prefer at this point to continue with Chapter 3 and to postpone discussion of the exponential and logarithmic functions until later. The overwhelming preference of instructors we surveyed, however, was to cover exponential and logarithmic functions early and then to use these functions throughout the rest of the course. Instructors who wish to postpone this material will also need to omit for now those examples and exercises in Sections 3.1–4.3 that refer to exponential and logarithmic functions. Students typically have no problem with f ( x) = 2 x , but the number e often remains a mystery. Like π, the number e is a transcendental number, but students have had years of schooling to get used to π. Have your students approximate e with a calculator, as the textbook does before the definition of e. Notice how we use compound interest to help students get a handle on this number.

Hints for Teaching Calculus with Applications

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Section 2.5 Logarithms are a very difficult topic for many students. It’s easy to say that a logarithm is just an exponent, but the fact that it is the exponent to which one must raise the base to get the number whose logarithm we are calculating is a rather obtuse concept. Therefore, spend lots of time going over examples that can be done without a calculator, such as log 2 8. Students will also tend to come up with many incorrect pseudoproperties of logarithms, similar in form to the properties of logarithms given in this section. Take as much time and patience as necessary in gently correcting the many errors students inevitably will make at first. Even after receiving a thorough treatment of logarithms, some students will still be stumped when solving a problem such as Example 8. Some of these students can get the correct answer using trial-and-error. The instructor should take consolation in the fact that at least such a student understands exponentials better than the one who uses logarithms incorrectly to solve Example 8 and comes up with the nonsensical answer t = -7.51 without questioning whether this makes sense. Be sure to teach your students to question the reasonableness of their answers; this will help them catch their errors.

Section 2.6 This section gives students much needed practice with exponentials and logarithms, and the applications keep students interested and motivated. Instructors should keep this in mind and not worry about having students memorize formulas. We have removed the formulas for present value in this edition, having decided that it’s better for students to just solve the compound amount formula for P. This reduces by two the number of formulas that students need to remember. There is a summary of graphs of basic functions in the end-of-chapter review.

Section 3.1 This is the first section on calculus, and perhaps the most important, since limits are what really distinguish calculus from algebra. Students will have the best understanding of limits if they have studied them graphically (as in Exercises 11–18), numerically (as in Exercises 21–26), and analytically (as in Exercises 35–62). The graphing calculator is a powerful tool for studying limits. Notice in Example 12 (c) and (d) that we have modified the method of finding limits at infinity by dividing by the highest power of x in the denominator, which avoid the problem of division by 0.

Section 3.2 The section on continuity should be straightforward if students have mastered limits from the previous section.

Section 3.3 This section introduces the derivative, even though that term doesn’t appear until the next section. In a class full of business and social science majors, an instructor may wish to place less emphasis on velocity, an approach more suited to physics majors. But we have found velocity to be the manifestation of the derivative that is most intuitive to all students, regardless of their major. Instructors in a hurry can skip the material on estimating the instantaneous rate of change from a set of data, but it helps solidify students’ understanding of the derivative by giving them one more point of view.

Section 3.4 Students who have learned differentiation formulas in high school usually want (and deserve) some explanation of why they need to learn to take derivatives using the definition. You might try explaining to your students that getting the right formula is not the only goal; graphing calculators can give derivatives numerically. The most important thing for students to learn is the concept of the derivative, which they don’t learn if they only memorize differentiation formulas. Zooming in on a function with a graphing calculator until the graph appears to be a straight line gives students a very concrete image of what the derivative means. After students have learned the differentiation formulas, they may forget about the definition of derivative. We have found that if we want them to use the definition on a test, it is important to say so clearly and emphatically, or they will simply use the shortcut formulas.

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Section 3.5 One way to get students to focus on the concept of the derivative, rather than the mechanics, is to emphasize graphical differentiation. We have therefore devoted an entire section to this topic. Graphical differentiation is difficult for many students because there are no formulas to rely on. One must thoroughly understand what’s going on to do anything. On the other hand, we have seen students who are weak in algebra but who possess a good intuitive grasp of geometry find this topic quite simple.

Section 4.1 Students tend to learn these differentiation formulas fairly quickly. These and the formulas in the next few sections are included in a summary at the end of the chapter.

Section 4.2 The product and quotient rules are more difficult for students to keep straight than those of the previous section. People seem to remember these rules better if they use an incantation such as “The first times the derivative of the second, plus the second times the derivative of the first.” Some instructors have argued that this formulation of the product rule doesn’t generalize well to products of three or more functions, but that’s not important at this level. Some instructors allow their students to bring cards with formulas to the tests. This does not eliminate the need for students to understand the use of the formulas, but it does eliminate the anxiety students may have about forgetting a key formula under the pressure of an exam.

Section 4.3 No matter how many times an instructor cries out to his or her students, “Remember the chain rule!”, many will still forget this rule at some time later in the course. But if a few more students remember the rule because the instructor reminds them so often, such reminders are worthwhile.

Section 4.4 and 4.5 In going through these sections, you may need to frequently refer to the rules of differentiation in the previous sections. You may also need to review the last three sections of Chapter 2.

Section 5.1 and 5.2 If students have understood Chapter 3, then the connection between the derivative, increasing and decreasing functions, and relative extrema should be obvious, and these sections should go quickly and smoothly.

Section 5.3 Students often confuse concave downward and upward with increasing and decreasing; carefully go over Figure 31 or the equivalent with your class.

Section 5.4 Graphing calculators have made curve sketching techniques less essential, but curve sketching is still one of the best ways to unify the various concepts introduced in this and the previous two chapters. Students should use graphing calculators to check their work. Because this section is the culmination of many ideas, students often find it difficult and start to forget things they previously knew. For example, a student might state that a function is increasing on an interval and then draw it decreasing. The best solution seems to be lots of practice with immediate help and feedback from the instructor. Students sometimes stumble over this topic because they use the rules for differentiation incorrectly, or because they make mistakes in algebra when simplifying. Exercises such as 29–34 are excellent for testing whether students really grasp the concepts.

Section 6.1 This section should not be conceptually difficult, but students need constant reminders to check the endpoints of an interval when finding the absolute maxima and minima.

Hints for Teaching Calculus with Applications

Section 6.2 This section is one of the high points of the course. Some of the best applications of calculus involve maxima and minima. Notice that some exercises have a maximum or minimum at the endpoint of an interval, so students cannot ignore checking endpoints. Almost everyone finds this material difficult because most people are not skilled at word problems. Remind your students that if they ever wonder whether mathematics is of any use, this section will show them. Why are word problems so difficult? One theory is that word problems require the use of two different modes of thinking, which students are using simultaneously for the first time. People use words in daily life without difficulty, but when they study mathematics, they often turn off that part of their brain and begin thinking in a very formal, mechanical way. In word problems, both modes of thinking must be active. If and when the NCTM Standards become widely accepted in the schools, children will get more practice at an early age in such ways of thinking. Meanwhile, the steps for solving applied problems given in this book might make the process a little more straightforward, and hence achievable by the average student.

Section 6.3 This section continues the ideas of the previous one. The point of studying economic lot size should not be to apply Equation (3), but to learn how to apply calculus to solve such problems. We therefore urge you to cover Exercises 17–20, in which we vary the assumptions, so Equation (3) does not necessarily apply. In this edition, we have changed the presentation to be consistent with that of business texts. The material on elasticity can be skipped, but it is an important application that should interest students who have studied even a little economics.

Section 6.4 There are two main reasons for covering implicit differentiation. First, it reinforces the chain rule. Second, it is needed for doing related rate problems. If you skip related rates due to lack of time, it is not essential to cover implicit differentiation, either.

Section 6.5 Related rate applications are less important than applied extrema problems, but they use some of the same skills in setting up word problems, and for that reason are worth covering. The best application exercises are under the heading “Physical Sciences,” because those are the exercises in which no formula is given to the student; the student must construct a formula from the words. The geometrical formulas needed are kept to a minimum: the Pythagorean theorem, the area of a circle, the volume of a sphere, the volume of a cone, and the volume of a cylinder with a triangular cross section. Some instructors allow their students to use a card with such formulas on the exam. These formulas are summarized in a table in the back of the book.

Section 6.6 Differentials may be skipped by instructors in a hurry; you need not fear that this omission will hamper your students in the chapter on integration. The differentials used there are not the same as those used here, and the required techniques are easily picked up when integration by substitution is covered. The exception is for instructors who intend to cover Section 10.3 on Euler’s Method, since differentials are used to motivate and derive that method. As in the previous edition, our presentation of differentials emphasizes linear approximation, a topic of considerable importance in mathematics.

Section 7.1 Students sometimes start to get differentiation and antidifferentiation confused when they reach this section. Some believe the antiderivative of x-2 is (-1/3) x-3; after all, if n is −2, isn’t n + 1 = -3? Carefully clarify this point.

Section 7.2 The main difficulty here is teaching students what to choose as u. The advice given before Example 3 should be helpful.

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Section 7.3 Some instructors who are pressed for time go lightly over the topic of the area as the limit of the sum of rectangular areas. This is possible, but care should be taken that students don’t lose track of what the definite integral represents. Also, a light treatment here lessens the excitement of the Fundamental Theorem of Calculus. We have continued the trend of the previous edition in covering three ways of approximating a definite integral with rectangles: the right endpoint, the left endpoint, and the midpoint. The trapezoidal rule is briefly introduced here as the average of the left sum and the right sum.

Section 7.4 The Fundamental Theorem of Calculus should be one of the high points of the course. Make a big deal about how the theorem unifies these two separate topics of area as a limit of sums and the antiderivative. When using substitution on a definite integral, the text recommends changing the limits and the variable of integration. (See Example 4 and the Caution which follows the example.) Some instructors prefer instead to have their students solve an indefinite integral, and then to evaluate the integral using the limits on x. One advantage of this method is that students don’t have to remember to change the limits. This method also has two disadvantages. The first is that it takes slightly longer, since it requires changing the integral to u and then back to x. Second, it prevents students at this stage from solving problems such as ò

1/2

that the integral ò

x 1 - 16 x 4 dx, which can be solved using the substitution u = 4 x 2 and the fact

1 - u 2 du represents the area of a quarter circle. This is one section in which we deliberately did not

use more than one method of presentation, because this would inevitably lead to confusion in the minds of some students.

Section 7.5 This section gives more motivation to the topic of integration. Consumers’ and producers’ surplus are important, realistic applications. We have downplayed sketching the curves that bound the area under consideration. Such sketches take time and are not necessary in solving these problems. But they clarify what is happening and make it possible to avoid memorizing formulas. Using a graphing calculator to sketch the curves can be helpful.

Section 7.6 The ubiquity of computers and graphing calculators has made numerical integration more important. Rather than computing a definite integral by an integration technique, one can just as easily enter the function into a calculator and press the integration key. Point out to students that this is valuable when the function to be integrated is complicated. On the other hand, using the antiderivative makes it easier to see the effect of changing the upper limit, say, from 2 to 3, or for working with the definite integral when one or both limits are parameters, such as a and b, rather than numbers. Simpson’s rule is the most accurate of the simple integration formulas. To achieve greater accuracy, a more complicated method must be used. This is why, unlike the trapezoidal rule, Simpson’s rule is actually used by mathematicians and engineers. You may wish to give your students the programs for the trapezoidal rule and Simpson’s rule in The Graphing Calculator Manual that is available with the text.

Section 8.1 Students usually find column integration simpler than traditional integration by parts. We show both methods to give instructors a choice, and also to emphasize that the two methods are equivalent. Column integration makes organizing the details simpler, but is no more mechanical than the traditional method, as some have mistakenly claimed. At Hofstra University, students even use this method when neither the instructor nor the book discuss it. They find out about it from other students, and so it has become an underground method. Some instructors feel that students will lose any theoretical understanding of what they are doing if they use this method. Our experience is that almost no students at this level have a theoretical understanding of what integration by parts is about, but the better students can at least master the mechanics. With column integration, almost all of the students master the mechanics.

Section 8.2 and 8.3 These two sections give more applications of integration. Coverage of either section is optional.

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xvii

Section 8.4 Improper integration is not really an application of integration, but it makes further applications of integration possible. Many mathematicians use shorthand notation such as the following: ò

¥ -x

dx = -e-x

¥ 0

= 0 - (-1) = 1.

For students at this level, it may be best to avoid the shorthand notation.

Section 9.1 The major difficulty students have with this section, and indeed with this entire chapter, is that they cannot visualize surfaces in 3-dimensional space, even though they live there. Fortunately, such visualization is not really necessary for doing the exercises in this chapter. A student who wants to explore what various surfaces look like can use any of the commercial or public domain computer programs available.

Section 9.2 Students who have mastered the differentiation techniques should have no difficulty with this section.

Section 9.3 This section corresponds to the section on applied extrema problems in the chapter on applications of the derivative, but with less emphasis on word problems. In Exercise 36, we revisit the topic of the least squares line, first covered in Section 1.3.

Section 9.4 Lagrange multipliers are an important application of calculus to economics. At some colleges, the business school is very insistent that the mathematics department cover this material.

Section 9.5 This section corresponds to the section on differentials in the chapter on applications of the derivative.

Section 9.6 Students who have trouble visualizing surfaces in 3-dimensional space are sometimes bothered by double integrals over variable regions. Instructors should assure such students that all they need to do is draw a good sketch of the region in the xy-plane, and not try to draw the volume in three dimensions.

Section 10.1 Differential equations of the form dy /dx = f ( x) are treated lightly in this section because they were already covered in the chapter on integration, although the terminology and notation were different then. Remind students that solving such differential equations is the same as antidifferentiation. The rest of the section is on separable differential equations. Students sometimes have trouble with this section because they have forgotten how to find an antiderivative, particularly one requiring substitution.

Section 10.2 If you get this far, your students have covered most of the techniques for solving first-order differential equations. You can find further techniques in differential equations texts, but most first-order equations that come up in real applications are separable, linear, or not solvable by any exact method.

Section 10.3 This is one of the most calculator/computer-intensive sections of the book. In practice, more accurate methods than Euler’s method are almost always used, but Euler’s method introduces students to a way of solving problems that would otherwise be beyond their grasp. You may wish to give your students the program for Euler’s method in The Graphing Calculator Manual that is available with the text.

Section 10.4 This is a fun section of assorted applications, showing students that the techniques they have learned were not in vain. You can pick and choose those applications of greatest interest to yourself and your students. You can also supplement

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the text with applications from other sources, such as those published by the Consortium for Mathematics and Its Applications Project (COMAP).

Chapter 11 Probability is one of the best applications of calculus around. In fact, statistics instructors sometimes feel the temptation to start discussing the definite integral even when their students know no calculus. This chapter is just a brief introduction, but it covers some of the most important concepts, such as mean, variance, standard deviation, expected value, and probability as the area under the curve. The third section covers three of the most important continuous probability distributions: uniform, exponential, and normal.

Chapter 12 Except for geometric series, most of the material in this chapter is of greater interest to the professional mathematician or engineer than to the student in business, management, economics, life sciences or social sciences, so many instructors may choose to skip this chapter. Many students confuse sequences with series, and often have trouble with the various tests for convergence. To avoid these sources of confusion, we emphasize summation as a unifying concept, whether of a few terms of a sequence or of an infinite series. Except for convergence of geometric series and a few words on the interval of convergence, we devote little coverage to the topic of convergence. We believe this approach is appropriate for a course at this level.

Section 12.1 Geometric sequences are the most important type of sequences for applications. Even instructors who wish to skip most of this chapter may want to cover the first two sections.

Section 12.2 This material is similar to the material in the Mathematics of Finance chapter in our textbook Finite Mathematics. The section shows how geometric sequences are critical for an understanding of annuities, mortgages, and amortization. Don’t be put off by the strange notation for the amount or the present value of an annuity; this notation is not at all strange in the world of finance.

Section 12.3 Taylor polynomials are introduced here as an approximation method, with no hint of infinite series.

Section 12.4 The emphasis here is on geometric series, the most important and simplest type of infinite series to understand.

Section 12.5 Some students find Taylor series a strange and abstract concept. To help make this concept more concrete, cover Example 4 on the normal curve, as well as the derivation of the rule of 70 and the rule of 72, introduced without proof in Chapter 2.

Section 12.6 Students with a “zero” feature on their calculator may lack motivation to learn Newton’s method unless you can interest them in how one might develop techniques for finding a zero. Newton’s method is not, however, the method typically used by calculators; calculator manufacturers are usually reluctant to discuss the actual algorithms.

Section 12.7 We have no applications for this sections, but students and instructors who enjoy symbol manipulation may still find this section satisfying.

Chapter 13 This chapter is a brief introduction to trigonometry and its uses in calculus. Students who need a more thorough treatment of this subject would be better served by a calculus book designed for mathematics majors. The presentation is brief, with a limited number of examples. As a result, students may find some of these exercises challenging. Therefore, tread carefully through this chapter.

PRETESTS AND ANSWERS

CALCULUS WITH APPLICATIONS

Name: Pretest, Form A

Evaluate the expression -x 2 + 3 y - z 3 when x = -2, y = 3, and z = -1.

1. ____________________

Perform the indicated operations. 2.

(2 x 2 - 3x + 7) - (5 x 2 - 8x - 9)

2. ____________________

( y - 4) 2

3. ____________________

(2 x - 1)( x 2 + 3x - 4)

4. ____________________

(5a + 2b)(5a - 2b)

5. ____________________

Factor each polynomial completely. 6.

3x 2 y3 - 6 x3 y 2 + 15 x 2 y 2

6. ____________________

2ac - 3ad + 8bc - 12bd

7. ____________________

Solve each equation. 8.

6( x - 1) - 4(2 x + 8) = 2 - 5( x + 2)

8. ____________________

1 2x 1 x-7 = + 4 3 2

9. ____________________

10.

x -1 3 - 2x = 5 4

10. ____________________

11.

( x - 2)( x + 1) = 4

11. ____________________

12.

Solve the inequality 6 x - 8( x + 2) £ 5 - ( x + 14). Write your answer in interval notation.

12. ____________________

13.

Solve the equation 2 x + 4 y = ay - bx for y.

13. ____________________

14.

Solve the equation a5 + 2cb = 37a for a.

14. ____________________

PRETEST, FORM A

xxi

xxii 15.

PRETEST, FORM A Solve the following system of equations. 3x - 2 y = 12 -4 x + 5 y = -23

15. ____________________

Perform the indicated operation. Write answers in lowest terms. 16.

18a 2b 2

10 xy 2

27 xy

8ab 2

⋅ 3

17.

2x + 8 x 2 + 7 x + 12 ¸ 2 3 y - 15 y - 10 y + 25

18.

2 + 2 x -1 x + 3x + 2

16. ____________________

17. ____________________

18. ____________________

Simplify each expression. Write answers with only positive exponents. 19.

20.

(2 x 2 y-3 )-3

19. ____________________

6 x-2 y 5

20. ____________________

15 x 4 y-2

Simplify each radical. Assume all variables represent nonnegative real numbers. 21.

22.

72x3 y 4

21. ____________________

72x3 y 4

22. ____________________

Write an equation in the form ax + by = c for each line. 23.

The line through the points with coordinates (-3, -14) and (5, 2)

23. ____________________

24.

The line through the points with coordinates (3, - 7) and (3, 5)

24. ____________________

25.

Graph the solution of the inequality 2 x - 3 y £ 12.

25.

CALCULUS WITH APPLICATIONS

Name: Pretest, Form B

Choose the best answer. 1.

Evaluate the expression (a)

12 5

(b) 4

y2 - z when x = -2, y = 4, and z = -8. -x 2 + 6

(d)

1. ____________________

4 5

Perform the indicated operations. 2.

( x 2 - 3x + 7) + (4 x3 - 6 x 2 - 5x + 4)

(a) 4 x3 - 6 x 4 - 8 x 2 + 11

(b) 4 x3 - 5 x 2 - 8 x + 11

(d) 4 x3 - 7 x 2 - 8 x + 11

(5 x - 4w)2

2. ____________________

3. ____________________

(a) 25 x 2 - 16w2

(b) 25 x 2 + 16w2

(d) 25 x 2 - 20wx + 16w2

( x - 3) ( x 2 + x - 2)

4. ____________________

(a) x3 - 2 x 2 - 5x + 6

(b) x3 - 3x 2 - 5 x + 6

(d) x3 - 2 x 2 + 5x + 6

(4 x + 9) (4 x - 9)

5. ____________________

(a) 16 x 2 + 72 x - 81y 2

(b) 16 x 2 - 72 xy - 81y 2

(d) 16 x 2 - 72 xy + 81y 2

Factor each polynomial completely. 6.

14a 2b3 - 7a 2b 2 + 35ab 4

6. ____________________

(a) 7ab(2ab 2 - ab + 5b3 )

(b) a 2b3 (14 - 7b + 35b 2 )

(d) 7ab 2 (2ab - a + 5b 2 )

6 xt + 8x - 3 yt - 4 y

7. ____________________

(a) (2 x - y) (3t + 4)

(b) (2 x + y) (3t - 4)

(d) (2 x + y) (3t + 4)

PRETEST, FORM B

xxiii

xxiv

PRETEST, FORM B

Solve each equation. 8.

4m - (7m - 6) = -m

(a)

3 2

(b) - 32

12.

43 7

(b)

(d) -6

27 7

10. ____________________ (c)

27 17

43 17

(d)

11. ____________________

(a) -4, 3

(b) 4, - 3

(d) 4, - 72

Solve the inequality -6 y + 2 ³ 4 y - 7. Write your answer in interval notation.

( -¥, -109 ùúû

9 ,¥ (c) éê 10 ë

14.

23 11

9. ____________________

2 z 2 + z = 28

(a)

13.

(d) 3

3y + 2 7- y = 5 4

(a) 11.

5 2 1 + = 3 r -1 6

(a) - 13 (b) 10.

8. ____________________

)

Solve the equation x =

(b) éê 12 , ¥ ë (d) 4( y - z ) 3k

12. ____________________

)

( -¥, 109 ùúû for y.

13. ____________________

(a)

y = 3kx 4+ 4 z

(b)

y = kx 4- z

(c)

y = - 3kxz 4

(d)

y =

3x + 4 z 4

Solve the equation 4x + ay = 23 for x.

14. ____________________

(a) x =

2 xy - 12 y 3s

(b) x = 2 y - 3a

12 y + 3ax 2y

(d) x =

12 y

3a - 2 y 12 y

PRETEST, FORM B 15.

xxv

Solve the following system of equations and then determine the value of x + y for the solution.

15. ____________________

3x + 2 y = 0 x - 5 y = 17

(a) -1

(b) 1

(d) -3

Perform the indicated operation. Write answer in lowest terms. 16.

17.

6 yz 3

15a3b3 ⋅ 30a 2b 4 12 y 2 z 3

(a)

3a 2 12aby

(b)

(c)

a 4by

(d)

x2 - y 2 4a5b7 (a) (c)

18.

16. ____________________

4 y3 z 6 25a5b7 b 4ay

x 2 + 3xy + 2 y 2

17. ____________________

12a3b12

3b5 2a 2

(b)

3b5 ( x - y )

(d)

a ( x + 2 y)

3b5 ( x + y ) a2 ( x - 2 y)

3b3 ( x + y ) ( x - 2 y)

1 2 5 + 2x 3y 6 xy

(a)

18. ____________________

2 3xy

(b)

3y - 4x + 5 6 xy

(d)

4x - 3y + 5 6 xy

Simplify each expression. Write answers with only positive exponents. 19.

(3a-2b3 )-4

19. ____________________ 8

(b) - 1212a

(a) - 12 6

a b

(c)

20.

a8 12b12

(d)

a8 81b12

æ r 2m-1 ö÷-2 çç ÷ çç 3 2 ÷÷ è r m ø÷

(a) r 2m2

20. ____________________ (b)

1 r 2m2

(d)

1 r 2 m6

xxvi

PRETEST, FORM B

Simplify each radical. Assume all variables represent nonnegative real numbers. 108a 4b3

21.

22.

21. ____________________

(a) 3a 2b 12b

(b) 6a 2b 3b

(d) 54a 2b b

108a 4b3

22. ____________________

(a) 3ab 3 4a

(b) 6a 2b 3 3b

(d) 9a 2b 3 12a

Write an equation in the form ax + by = c for each line. 23.

24.

25.

The line through the points with coordinates (1, -1) and (-1, -2) (a) x - 2 y = 3

(b) x + 2 y = 3

(d) x - 2 y = -3

The line through the points with coordinates (-5, - 2) and (-3, - 2) (a) x + y = -2

(b) x = -2

(d)

23. ____________________

24. ____________________

y = -2

Graph the solution of the inequality 5 x - 3 y ³ 15.

25. ____________________

ANSWERS TO PRETESTS

xxvii

ANSWERS TO PRETESTS PRETEST, FORM A 2 y7

11.

-2, 3

20.

-3x 2 + 5 x + 16

12.

[-7, ¥)

21.

6 xy 2 2 x

y 2 - 8 y + 16

13.

y =

22.

2 xy 3 9 y

2 x3 + 5x 2 - 11x + 4

14.

b a = - 16 3c

23.

2x - y = 8

25a 2 - 4b 2

15.

(2, - 3)

24.

x =3

3x 2 y 2 ( y - 2 x + 5)

16.

5a 6y

25.

(b + 2) x a-4

5 x6

PRETEST, FORM B 1. (c)

6. (d)

11. (c)

16. (c)

21. (b)

2. (b)

7. (a)

12. (d)

17. (c)

22. (a)

3. (c)

8. (d)

13. (d)

18. (b)

23. (a)

4. (a)

9. (b)

14. (b)

19. (d)

24. (d)

5. (c)

10. (c)

15. (a)

20. (c)

25. (b)

FINAL EXAMINATIONS AND ANSWERS

CALCULUS WITH APPLICATIONS

Name: Final Examination, Form A

1. Find the coordinates of the vertex of the graph of f ( x) = -2 x 2 + 30 x + 45.

1. ________________________

2. Write the equation 2a = d using logarithms.

2. ________________________

3. Graph the function y = log 2 ( x - 3) + 4.

4. Evaluate lim

x4

x-4 . x -2

4. ________________________

x-2

5. Find all values of x at which g ( x) = x - 5 is not continuous.

5. ________________________

6. Find the average rate of change of y = between x = 4 and x = 12.

6. ________________________

2x + 1

7. Find the derivative of y = 7 x 2e3x .

7. ________________________

ln (3 x ) . x +1

8. ________________________

8. Find the derivative of y =

9. Find the instantaneous rate of change of s(t ) = 3t 2 -

8 at t = 2. t

9. ________________________

10. Find an equation of the tangent line to the graph of y = ( x 2 + 3x + 3)8 at the point (-1, 1). 11. Find h¢(2) if h( x) =

6 . 4 x +17

10. ________________________ 11. ________________________

12. Find the largest open intervals where f is increasing or decreasing if f ( x) = 2 x3 - 24 x 2 + 72 x.

FINAL EXAM, FORM A

12. ________________________

xxxi

xxxii

FINAL EXAM, FORM A

13. Find the locations and values of all relative extrema x2 + 5x + 3 of g if g ( x) = . x -1 14.

13. ________________________

For the function f ( x) = x 4 - 4 x3 - 5, find (a) all intervals where f is concave upward;

14. (a) _____________________

(b) all intervals where f is concave downward;

(b) _____________________

15. Find the fourth derivative of h( x) = xe x .

15. ________________________

16. Find the locations and values of all absolute extrema of f ( x) = x3 - 6 x 2 on the interval [-1, 2]. 17. For a particular commodity, its price per unit in dollars x2 , is given by P( x) = 120 10 where x, measured in thousands, is the number of units sold. This function is valid on the interval [0, 34]. How many units must be sold to maximize revenue? dy

18. Find dx if 3 x - 5 y 3 = 7 xy.

16. ________________________

17. ________________________ 18. ________________________

19. If xy = x - 9, find dt if dx = 12, x = -3, and y = 4. dt

19. ________________________

20. If y = -3(2 + x 2 )3 , find dy.

20. ________________________

21. Using differentials, approximate the volume of coating on a sphere of radius 5 inches, if the coating is 0.03 inch thick.

21. ________________________

22. Find

ò (3x - 7 x + 2) dx. 2

22. ________________________

23. Find

8 dx. x+9

23. ________________________

24. Evaluate

ò0 x 3x2 + 4 dx.

24. ________________________

25. Find the area of the region between the x-axis and the 2

graph of f ( x) = xe x on the interval [0, 2].

25. ________________________

FINAL EXAM, FORM A

xxxiii

26. Find the area of the region enclosed by the graphs of f ( x) = x 2 - 4 and g (x) = 1 - x 2.

27. Find

26. ________________________

ò x e dx.

28. Evaluate

2 x

27. ________________________

ò 3x In x dx.

28. ________________________

29. Find the average value of f ( x) = x( x 2 + 1)3 over the interval [0, 2].

29. ________________________

30. Find the volume of the solid of revolution formed by rotating the region bounded by the graphs of f ( x) = y = 0, and x = 12 about the x-axis.

30. ________________________

31. Determine whether the improper integral

-1

-¥

x + 4,

x-2 dx

converges or diverges. If it converges, find its value.

31. ________________________

32. If f ( x, y) = 3x 2 - 4 xy + y3, find f xy (1, -3). 33.

32. ________________________

For f ( x, y) = 2 x 2 - 2 xy + 2 y 2 + 4 x + 4 y + 5, find all of the following: (a) relative maxima;

33. (a) _____________________

(b) relative minima;

(b) _____________________

34. Maximize f ( x, y ) = x 2 y, subject to the constraint y + 4 x = 84.

34. ________________________

35. Find dz, given z = 3x 2 - 4 xy + y 2 , where x = 0, y = 2, dx = 0.02, and dy = 0.01.

35. ________________________

36. Evaluate

ò1 ò0 e2 x dy dx.

36. ________________________

37. Find the general solution of the differential equation dx =

1 + ex . 2y

37. ________________________

38. Find the particular solution of the differential dy

equation dx = 3x 2 - 7 x + 2; y = 4 when x = 0.

38. ________________________

xxxiv

FINAL EXAM, FORM A

39. The marginal cost function for a particular commodity dy

is dx = 8 x - 9 x 2. The fixed cost is $20. Find the cost function.

39. ________________________

40. The probability density function of a random variable is defined by f ( x) = 14 for x in the interval [12, 16]. Find P( X £ 14). 41.

40. ________________________

For the probability density function f ( x) = 3x-4 on [1, ¥), find (a) the expected value;

41. (a) _____________________

(b) the variance;

(b) _____________________

42. The average height of a particular type of tree is 20 feet, with a standard deviation of 2 feet. Assuming a normal distribution, what is the probability that a tree of this kind will be shorter than 17 feet?

42. ________________________

43. For the function f ( x) = e-3x , find (a) the Taylor polynomial of degree 4 at x = 0;

43. (a) _____________________

(b) an approximation, rounded to four decimal places, of e-0.03 using the Taylor polynomial from part (a). 44. Find the sum of the convergent geometric series 5 +

5 5 5 + + + . 3 9 27

(b) _____________________ 44. ________________________

45. Use Newton’s method to find a solution to the nearest hundredth of 3x3 - 2 x 2 + x - 1 = 0 in the interval [0, 1].

45. ________________________

46. Use l’Hospital’s rule to find lim

3e2 x - 3 . x x ¥

46. ________________________

47. Find the derivative of f ( x) = tan ( x 2 + 1).

47. ________________________

48. Find the derivative of y = x3 sin 2 x.

48. ________________________

ò sec2 (3x + 1) dx.

49. ________________________

49. Find

50. Evaluate

 /6

ò0 sin3 x cos x dx.

50. ________________________

CALCULUS WITH APPLICATIONS

Name: Final Examination, Form B

Choose the best answer. 1. Find the coordinates of the vertex of the graph of f ( x) = 2 x 2 + 10 x - 17. æ 5 ö çç - , - 59 ÷÷ (b) çè 2 2 ÷ø

(a)

æ5 ö çç , - 59 ÷÷ çè 2 2 ÷ø

(-5, -42) (d) (5, -42)

(c)

2. Write the equation m P = q using logarithms. (a) log m p = q

(b) log q p = m

(d) log m q = p

3. Solve

( 1e )

2. _______________

= 14. (Round to nearest thousandth.)

(a) -1.320 4. Evaluate lim

x9

3. _______________

(b) 2.639

(d) 1.320

x-9 . x-3

(a) 0

4. _______________

(b) 3

5. Find all values of x at which f ( x) = (a) 3

1. _______________

(b) 4

x2 - 5x + 6 x 2 - 7 x + 12

(d) The limit does not exist.

is not continuous.

5. _______________

(d) The function is continuous everywhere. 6. Find the average rate of change of y = x 2 + x between x = 4 (a) 9

(b) 10

(d) 25

7. Find the derivative of y = -3x3e 2 x .

7. _______________

(a) -27 x 2e2 x - 6 x3e2 x

(b) -18x 2e3x

(d) -6 x 2e2 x - 6 x3e2 x 2

8. Find the derivative of y = Inx 3x . (a) (c)

2x2 (ln 3x )

2 x ln 3x - x (ln 3x )

8. _______________ (b)

FINAL EXAM, FORM B

6. _______________

(d)

6 x ln 3x - x 3(ln 3x )2 6 x ln 3x + x 3(ln 3x )2

xxxv

xxxvi

FINAL EXAM, FORM B

9. Find the instantaneous rate of change of s(t ) =

(a)

3 2

(b)

5 2

(c)

t2 2 - 2 at t = 2. 2 t

7 2

(d)

9. _______________

9 2

(

)

1 10. Find an equation of the tangent line to the curve f ( x) = - x + at the point 2, - 14 . 2

(a) x - 16 y = 6 (c)

f ¢( x ) =

(b) x - 4 y = 3

1 16

(d)

y =

1 ( x + 2)2

11. Find h¢(3) if h( x) = - 400 .

11. _______________

3x + 7

(a) 100

10. _______________

(b)

25 8

(c)

75 8

(d) 200

12. Find the largest open interval(s) over which the function g ( x) = x3 - 12 x 2 + 36 x + 1 is increasing.

(a) (2, 6)

(b) (-¥, 2)

12. _______________

13. Find the location and value of all relative extrema for x2 + 5x + 3 . the function f ( x) = x -1

13. _______________

(a) Relative minimum of 1 at -2; relative maximum of 13 at 4 (b) Relative maximum of 1 at -2; relative minimum of 13 at 4 (c) Relative minimum of 1 at 4; relative minimum of 13 at -2 (d) No relative extrema 14. Find the coordinates of all points of inflection of the function g ( x) = x3 - 6 x 2.

(a) (2, -16)

(b) (2, -12)

15. Find the third derivative of f ( x) =

(a)

3 (2 x + 1)-5/2 8

14. _______________

(d) (-2, 16)

2 x + 1.

(b) 3(2 x + 1)-5/2 1 (d) - (2 x + 1)-3/2 4

15. _______________

FINAL EXAM, FORM B

xxxvii

16. Find the absolute minimum of f ( x) = x 4 - 4 x3 - 5 on the interval [-1, 2].

(a) -30

(b) 0

16. _______________

(d) No absolute minimum

17. Botanists, Inc., a consulting firm, monitors the monthly growth of an unusual plant. They determine that the growth (in inches) is given by g ( x) = 4 x - x 2 ,

where x represents the average daily number of ounces of water the plant receives. Find the maximum monthly growth of the plant. (a) 2 inches

(b) 8 inches

(d) 6 inches

18. Find dx , given 4 x 2 - 7 = 3 y + 23 .

18. _______________

(a)

(c)

4 y (4 x5 - 3)

(b)

3x 4 4 y (4 x3 - 3)

(d)

3x 2

4 y (4 x5 + 3) 3x 4 4 y (4 x3 + 3) 3x 2

19. If xy = y + 6, find dt if dx = -3, x = 4, and y = 2. dt

(a) 2

(b) 0

19. _______________ (d) -4

20. Evaluate dy, given y = 45 - 2 x - 3x 2 , with x = 3 and  x = 0.02.

(a) -0.4

(b) 0.4

17. _______________

(d) -0.32

21. Using differentials, approximate the volume of coating on a cube with 3-cm sides, if the coating is 0.02 cm thick.

(a) 0.54 cm3

(b) 27.54 cm3

(d) 81.54 cm3

22. Find

ò (4x + 3x - 5) dx.

(a)

4 3 3 x + x2 - 5 + C 3 2

(b)

(c)

3 3 3 x + x2 - 5 + C 4 2

(d) 8x + 3

20. _______________

21. _______________

22. _______________ 4 3 3 x + x 2 - 5x 3 2

xxxviii 23. Find

FINAL EXAM, FORM B

ò x

-3 dx. (2 + 5)

23. _______________

3 (a) - ln |2x + 5| + C 2

(b) -3 ln |2x + 5| + C

1 (c) - ln |2 x + 5| + C 6

(d)

24. Evaluate

1 ln |2x + 5| + C 2

ò 2x 4x + 5 dx. 2

24. _______________

(a)

3 8

(b)

27 - 5 5 6

(c)

27 - 5 5 8

(c)

27 - 5 5 4

25. Find the area of the region between the x-axis and the graph of f ( x) = x + 1 on the interval [0,3].

(a)

14 3

(b) 7

(c)

1 4

25. _______________ (d)

21 2

26. Find the area of the region enclosed by the

graphs of f ( x) = 9 - x 2 and g(x) = x 2 - 9. (a) -36 27. Find

(b) 36

26. _______________

(d) 72

ò 3x e x dx. 2 2

27. _______________ 3 2 2x 3 3 x e - xe2 x + e 2 x + C 2 2 4

(a) 3e x + C

(b)

(d) 3x 2e 2 x - 6 xe2 x + 6e 2 x + C

28. Evaluate

ò 9x ln x dx. 2

28. _______________

(a) 192 ln 4 - 45

(b) 192 ln 4 - 65

(d) 192 ln 4 - 63

29. Find the average value of the function f ( x) = interval [0, 8].

(a)

31 3

(b)

248 9

(c)

31 9

3x + 1 over the

29. _______________ (d)

248 3

FINAL EXAM, FORM B

xxxix

30. Find the volume of the solid of revolution formed by rotating the region bounded by f ( x) = x - 1, y = 0, and x = 10 about the x-axis.

(a)

40.5

31. Evaluate

(a) -

(b)

-2

-¥ x

81 2

(d) 40

31. _______________

dx.

1 8

(b) Diverges

(c)

1 64

(d) -

1 64

32. Given z = f ( x, y ) = x 2 - 3xy + 2 y 3, find f yx (0, -2).

(a) -16

30. _______________

(b) 2

32. _______________ (d) -3

33. Which of the following applies to the

function f ( x, y ) = 2 x 2 + 8 y 3 - 12 xy + 7 ? (a) f has a relative maximum at

33. _______________

( 92 , 23 )

(b) f has a relative minimum at (0, 0) (c) f has a relative minimum at (d) f has a saddle point at

( 92 , 23 )

34. Maximize f ( x, y) = x 2 y, subject to the constraint y + 4 x = 84.

(a) 0

(b) 14

(d) 3642

x 2 + y 2 , with x = 3, y = 4, dx = -0.01,

35. Find dz, given z =

and dy = 0.02.

35. _______________

(a) 0.022

36. Evaluate

(b) 0.01 2

(a)

112 3

(d) 0.05

ò ò x y dy dx. 0

34. _______________

36. _______________

(b) 16

(d) 12

FINAL EXAM, FORM B dy

37. Find the general solution of the differential equation dx =

(a)

æ3 ö1/3 y = çç e 2 x - 12 x + C ÷÷÷ çè 2 ø

(b)

y =

(c)

æ1 ö1/3 y = çç e 2 x - 4 x + C ÷÷÷ çè 2 ø

(d)

y 2 = e2 x - 4

e2 x - 4 y2

37. _______________

1 2x e - 4x + C 2

38. Find the particular solution of the differential equation dy = 7 - 2 x + 3x 2 ; y = -2 when x = 0. dx

38. _______________

(a)

y = 7 x - x 2 + x3 + C

(b)

y = 7 x - x 2 + x3 - 2

(c)

y = -x 2 + x 3 + C

(d)

y = 7 x - x 2 + x3 + 2 + C dy

39. The marginal cost function for a particular commodity is dx = 10 x - 7 x 2.

The fixed cost is $200. Find the cost function. (a)

y = 200 + 5 x 2 - 14 x

(b)

y = 5x 2 - 14 x

(c)

y = 5x2 -

7 x3 + 200 3

(c)

y = 10 - 14 x

39. _______________

40. The probability density function of a random variable is defined by f ( x) = 15 for [20, 25]. Find P( X £ 22).

(a) 0.2

(b) 0.8

40. _______________ (d) 0.6

41. Find the standard deviation for the probability

density function f ( x ) = 3x-4 on [1, ¥). (a) 1.5

(b) 0.866

41. _______________ (d) 0.67

(a) 0.0668

(b) 0.9332

(d) 0.1977

43. Find the monthly house payment necessary to amortize a $150,000 loan at 7.2% for 30 years.

(a) $1018.18

(b) $975.34

42. _______________

(d) $5478.44

43. _______________

FINAL EXAM, FORM B

xli

44. Find the sum of the convergent geometric series 2 +

(a) 2

(b) 8

1 1 1 + + + . 2 8 32

(d)

44. _______________

8 3

45. The nth term of the Taylor series expansion of ln(1 + 2x ) is

given by which of the following? (a)

(-1)n x n +1 (n + 1) ⋅ 2n +1

(b)

45. _______________

x n +1 n +1

(c)

x n +1 (n + 1) ⋅ 2n +1

(d)

(-1) n x n +1 n +1

-2e 4 x + 2 . x x 0

46. Evaluate lim

(a) -2

46. _______________

(b) -8

(d) The limit does not exist.

47. Find the derivative of f ( x) = tan 2 x + 1.

(a)

sec 2 2 x + 1 2x + 1

(b)

sec 2 2 x + 1 2

(d) 2 x sec2 2 x + 1

48. Find the derivative of y = x 2 cos3 x.

(a) x cos2 x (2 cos x + 3x)

(b) x cos2 x (2 cos x - 3x sin x)

(d) x cos2 x (2 cos x + 3x sin x)

49. Find

ò tan (2x) sec (2x) dx. 2

1 tan 2 (2 x) + C 4

(b) tan 2 (2 x) + C

(c)

1 sec (2 x) + C 4

(d) sec (2 x) + C

 /2

48. _______________

49. _______________

(a)

50. Evaluate

47. _______________

(10 - 10sin 2 x cos x) dx.

(a)

20 3

(b) 5 - 10

(c)

30 - 10 3

(d)

15 - 10 3

50. _______________

ANSWERS TO FINAL EXAMINATIONS

xliii

ANSWERS TO FINAL EXAMINATIONS FINAL EXAMINATION, FORM A 1.

(7.5,157.5)

log 2 d = a

15.

( x + 4)e x

16.

Absolute maximum of 0 at 0; absolute minimum of -16 at 2

17.

20, 000

18.

3 - 14 x1/2 y dy = dx 14 x 3/2 + 30 x1/2 y 2

19.

20.

dy = -18 x(2 + x 2 ) 2 dx

21.

3 in.3

1 4

22.

x3 - 72 x 2 + 2 x + C

y ¢ = 7 xe3x (2 + 3x)

23.

1 ln | x + 9 | + C 8

y¢ =

24.

56 9

25.

1 (e4 - 1) 2

10.

y = 8x + 9 26.

10 10 3

27.

e x ( x 2 - 2 x + 2) + C

on (2, 6)

28.

3 4

13.

Relative maximum of 1 at -2; relative minimum of 13 at 4

29.

14.

(a) (-¥, 0) and (2, ¥)

30.

120

11. 12.

x + 1 - x ln (3 x ) x ( x + 1) 2

12 - 125

» 10.54

Increasing on (-¥, 2) and (6, ¥); decreasing

(b) (0, 2)

( 3e4 + 1) » 123.60

xliv

ANSWERS TO FINAL EXAMINATIONS

31.

Converges; 1

42.

0.0668

32.

-4

43.

(a) 1 - 3x + 92x - 92x + 278x

33.

(a) None

34.

5488 when x = 14 and y = 28

35.

-0.12

36.

(b) -3 at ( - 2, - 2)

(b) 0.9704 44.

15 2

45.

0.78

3 e4 - 3 e2 » 70.81 2 2

46.

37.

y2 = x + ex + C

47.

2 x sec2 x 2 + 1

38.

y = x3 - 72 x 2 + 2 x + 4

48.

x 2 sin x (3sin x + 2 x cos x)

39.

y = 4 x 2 - 3x3 + 20

49.

1 tan (3x + 1) + C 3

40.

0.5

50.

1 64

41.

(a)

3 2

(b) 0.75

(

)

FINAL EXAMINATION, FORM B 1.

(a)

11.

(c)

21.

(a)

31.

(a)

41.

(b)

(d)

12.

(d)

22.

(c)

32.

(d)

42.

(b)

(a)

13.

(b)

23.

(a)

33.

(c)

43.

(a)

(c)

14.

(a)

24.

(b)

34.

(c)

44.

(d)

(b)

15.

(b)

25.

(a)

35.

(b)

45.

(a)

(c)

16.

(c)

26.

(d)

36.

(b)

46.

(b)

(c)

17.

(c)

27.

(b)

37.

(a)

47.

(a)

(c)

18.

(b)

28.

(d)

38.

(b)

48.

(b)

19.

(a)

29.

(c)

39.

(c)

49.

(a)

10.

(a)

20.

(a)

30.

(b)

40.

(c)

50.

(d)

SOLUTIONS TO ALL EXERCISES

Chapter R

ALGEBRA REFERENCE R.1 Polynomials

(-4 y 2 - 3 y + 8) - (2 y 2 - 6 y - 2)

Your Turn 1

= (-4 y 2 - 3 y + 8) + (-2 y 2 + 6 y + 2)

3( x 2 - 4 x - 5) - 4(3x 2 - 5x - 7)

= -4 y 2 - 3 y + 8 - 2 y 2 + 6 y + 2

= 3x 2 - 12 x - 15 - 12 x 2 + 20 x + 28

= (-4 y 2 - 2 y 2 ) + (-3 y + 6 y ) + (8 + 2)

= -9 x 2 + 8 x + 13

= -6 y 2 + 3 y + 10

Your Turn 2

(2 x + 7)(3x - 1)

æ1 ö æ1 1 ö æ2 4ö = ççç z 2 + z 2 ÷÷÷ + ççç z + z ÷÷÷ + ççç + ÷÷÷ è2 ø è3 ø è 2 3 3ø

= (2 x )(3x ) + (2 x )(-1) + (7)(3x ) + (7)(-1) = 6 x 2 - 2 x + 21x - 7

= 6 x 2 + 19 x - 7

Your Turn 3

(3 y + 2)(4 y - 2 y - 5) = (3 y )(4 y 2 - 2 y - 5) + (2)(4 y 2 - 2 y - 5) = 12 y 3 - 6 y 2 - 15 y + 8 y 2 - 4 y - 10

Your Turn 4

æ ö æ ö çç 2 t 2 + 3 t + 5 ÷÷ - çç 1 t 2 + t + 1 ÷÷ ÷ çè 3 ç 2 6ø è6 6 ø÷ æ2 3 5ö æ 1 1ö = ççç t 2 + t + ÷÷÷ + ççç - t 2 - t - ÷÷÷ è3 ø è 2 6 6 6ø

(3x + 2 y )3

= (3x + 2 y )(3x + 2 y )(3x + 2 y )

- 6(2q 2 + 4q - 3) + 4(-q 2 + 7q - 3) + (-4q 2 + 28q - 12)

= (9 x 2 + 12 xy + 4 y 2 )(3x + 2 y )

= (-12q 2 - 4q 2 ) + (-24q + 28q) + (18 - 12)

= 9 x 2 (3x + 2 y ) + 12 xy (3x + 2 y ) + 4 y 2 (3x + 2 y ) = 27 x 3 + 18 x 2 y + 36 x 2 y + 24 xy 2 + 12 xy 2 + 8 y 3

= -16q 2 + 4q + 6

= 27 x 3 + 54 x 2 y + 36 xy 2 + 8 y 3

R.1 Exercises

2(3r 2 + 4r + 2) - 3(-r 2 + 4r - 5) = (6r 2 + 8r + 4) + (3r 2 - 12r + 15)

(2 x - 6 x + 11) + (-3x + 7 x - 2)

= (6r 2 + 3r 2 ) + (8r - 12r ) + (4 + 15)

= 2 x 2 - 6 x + 11 - 3x 2 + 7 x - 2

= 9r 2 - 4r + 19

= (2 - 3) x + (7 - 6) x + (11 - 2) = -x 2 + x + 9

1 2 1 2 t + t+ 2 2 3

= (-12q 2 - 24q + 18)

= (9 x 2 + 6 xy + 6 xy + 4 y 2 )(3x + 2 y )

3 2 5 z + z+2 2 6

æ2 ö æ5 1 ö æ3 1ö = ççç t 2 - t 2 ÷÷÷ + ççç t - t ÷÷÷ + ççç - ÷÷÷ è ø è3 ø è 6 2 6 6ø

= 12 y 3 + 2 y 2 - 19 y - 10

æ1 2 ö æ ö çç z + 1 z + 2 ÷÷ + çç z 2 + 1 z + 4 ÷÷ ÷ ç èç 2 ø è 3 3 2 3 ø÷

(0.613x 2 - 4.215x + 0.892) - 0.47(2 x 2 - 3x + 5) = 0.613x 2 - 4.215x + 0.892 - 0.94 x 2 + 1.41x - 2.35 = -0.327 x 2 - 2.805x - 1.458

2 8.

Chapter R ALGEBRA REFERENCE 0.5(5r 2 + 3.2r - 6) - (1.7r 2 - 2r - 1.5)

16.

= (2.5r 2 + 1.6r - 3) + (-1.7r 2 + 2r + 1.5)

æ 3 öæ 5 ö æ 3 öæ 1 ö æ 2 öæ 5 ö = çç r ÷÷÷çç r ÷÷÷ + çç r ÷÷÷çç s ÷÷÷ + çç - s ÷÷÷çç r ÷÷÷ èç 4 øèç 4 ø èç 4 øèç 3 ø èç 3 øèç 4 ø

= (2.5r 2 - 1.7r 2 ) + (1.6r + 2r ) + (-3 + 1.5) = 0.8r 2 + 3.6r - 1.5

æ 2 öæ 1 ö + çç - s ÷÷÷çç s ÷÷÷ çè 3 øèç 3 ø 15 2 1 5 2 = r + rs - rs - s 2 16 4 6 9 15 2 7 2 2 = r rs - s 16 12 9

-9m(2m 2 + 3m - 1) = -9m(2m2 ) - 9m(3m) - 9m(-1) = -18m3 - 27m 2 + 9m

10.

6 x(-2 x3 + 5x + 6) = -12 x 4 + 30 x 2 + 36 x

11.

(3t - 2 y )(3t + 5 y )

17.

= 3 p(9 p 2 ) - 1(9 p 2 ) + 3 p(3 p )

= 9t 2 + 15ty - 6ty - 10 y 2

- 1(3 p) + 3 p(1) - 1(1)

= 9t 2 + 9ty - 10 y 2

= 27 p 3 - 9 p 2 + 9 p 2 - 3 p + 3 p - 1

(9k + q)(2k - q)

= 27 p 3 - 1

= (9k )(2k ) + (9k )(-q) + (q)(2k ) + (q)(-q)

18.

= 18k 2 - 9kq + 2kq - q2

+ (2)(5 p 2 ) + (2)( p ) + (2)(-4)

(2 - 3x )(2 + 3x )

= 15 p 3 + 3 p 2 - 12 p + 10 p 2 + 2 p - 8

= (2)(2) + (2)(3x ) + (-3x )(2) + (-3x )(3x ) = 4 + 6x - 6x - 9x

= 15 p 3 + 13 p 2 - 10 p - 8

19.

= 4 - 9x2

14.

(3 p + 2)(5 p 2 + p - 4) = (3 p )(5 p 2 ) + (3 p )( p) + (3 p )(-4)

= 18k 2 - 7kq - q2

13.

(3 p - 1)(9 p 2 + 3 p + 1) = (3 p - 1)(9 p 2 ) + (3 p - 1)(3 p) + (3 p - 1)(1)

= (3t )(3t ) + (3t )(5 y ) + (-2 y )(3t ) + (-2 y )(5 y )

12.

æ3 öæ ö çç r - 2 s ÷÷çç 5 r + 1 s ÷÷ ÷ ÷ çè 4 ç 3 øè 4 3 ø

(2m + 1) (4m2 - 2m + 1) = 2m(4m 2 - 2m + 1) + 1(4m2 - 2m + 1) = 8m3 - 4m 2 + 2m + 4m2 - 2m + 1

(6m + 5)(6m - 5) = (6m)6m) + (6m)(-5) + (5)(6m) + (5)(-5)

= 8m3 + 1

= 36m - 30m + 30m - 25

20.

= 36m2 - 25

15.

(k + 2)(12k 3 - 3k 2 + k + 1) = k (12k 3 ) + k (-3k 2 ) + k (k ) + k (1) + 2(12k 3 ) + 2(-3k 2 ) + 2(k ) + 2(1)

æ2 öæ ö çç y + 1 z ÷÷çç 3 y + 1 z ÷÷ ÷ çè 5 ç 8 øè 5 2 ÷ø

= 12k 4 - 3k 3 + k 2 + k + 24k 3 - 6k 2 + 2k + 2

æ 2 öæ 3 ö æ 2 öæ 1 ö æ 1 öæ 3 ö = çç y ÷÷÷çç y ÷÷÷ + çç y ÷÷÷çç z ÷÷÷ + çç z ÷÷÷çç y ÷÷÷ èç 5 øèç 5 ø èç 5 øèç 2 ø èç 8 øèç 5 ø æ 1 öæ 1 ö + çç z ÷÷÷çç z ÷÷÷ çè 8 øèç 2 ø

6 2 1 3 1 2 y + yz + yz + z 25 5 40 16 6 2 æç 8 3 ö÷ 1 2 = y +ç + z ÷÷ yz + ç è 40 25 40 ø 16 6 2 11 1 2 = y + yz + z 25 40 16

= 12k 4 + 21k 3 - 5k 2 + 3k + 2

21.

( x + y + z )(3x - 2 y - z ) = x(3x ) + x(-2 y ) + x(-z ) + y(3x ) + y(-2 y )

+ y(-z ) + z(3x ) + z(-2 y ) + z(-z ) = 3x 2 - 2 xy - xz + 3xy - 2 y 2 - yz + 3xz - 2 yz - z 2 = 3x 2 + xy + 2 xz - 2 y 2 - 3 yz - z 2

Section R.2 22.

( r + 2s - 3t )(2r - 2s + t ) = r(2r ) + r(-2s ) + r(t ) + 2s(2r ) + 2s(-2s )

28.

(3x + y )3 = (3x + y )(3x + y )2 = (3x + y )(9 x 2 + 6 xy + y 2 )

+ 2s(t ) - 3t (2r ) - 3t (-2s ) - 3t (t )

= 27 x 3 + 18 x 2 y + 3xy 2 + 9 x 2 y

= 2r - 2rs + rt + 4rs + 2st - 6rt + 6st 2

- 3t + 2rt - st + t

+ 6 xy 2 + y 3

= 27 x 3 + 27 x 2 y + 9 xy 2 + y 3

= 2r + 2rs - 5rt - 4s 2 + 8st - 3t 2

23.

R.2 Factoring

( x + 1)( x + 2)( x + 3) = [ x( x + 2) + 1( x + 2)]( x + 3)

Your Turn 1

= [ x 2 + 2 x + x + 2]( x + 3)

Factor 4 z 4 + 4 z 3 + 18z 2 .

= [ x 2 + 3x + 2]( x + 3)

4 z 4 + 4 z 3 + 18z 2

= x 2 ( x + 3) + 3x( x + 3) + 2( x + 3)

= (2 z 2 ) ⋅ 2 z 2 + (2 z 2 ) ⋅ 2 z + (2 z 2 ) ⋅ 9

= x 3 + 3x 2 + 3x 2 + 9 x + 2 x + 6

= (2 z 2 )(2 z 2 + 2 z + 9)

= x 3 + 6 x 2 + 11x + 6

24.

Your Turn 2

( x - 1)( x + 2)( x - 3) = [ x( x + 2) + (-1)( x + 2)]( x - 3) = ( x 2 + 2 x - x - 2)( x - 3)

Your Turn 3

= ( x 2 + x - 2)( x - 3) = x 2 ( x - 3) + x( x - 3) + (-2)( x - 3) = x 3 - 3x 2 + x 2 - 3x - 2 x + 6 = x 3 - 2 x 2 - 5x + 6

25.

x 2 - 3x - 10 = ( x + 2)( x - 5) since (2)(-5) = -10 and -5 + 2 = -3.

Factor 6a 2 + 5ab - 4b2 . 6a 2 + 5ab - 4b2 = (2a - b)(3a + 4b)

R.2 Exercises 1.

( x + 2) 2 = ( x + 2)( x + 2) = x( x + 2) + 2( x + 2)

= 7a 2 (a + 2)

= x2 + 2x + 2x + 4

7a3 + 14a 2 = 7a 2 ⋅ a + 7a 2 ⋅ 2

3 y 3 + 24 y 2 + 9 y = 3 y ⋅ y 2 + 3 y ⋅ 8 y + 3 y ⋅ 3 = 3 y( y 2 + 8 y + 3)

= x2 + 4x + 4

26.

(2a - 4b) 2 = (2a - 4b)(2a - 4b) = 2a(2a - 4b) - 4b(2a - 4b)

= 13 p 2q ⋅ p 2q - 13 p 2q ⋅ 3 p + 13 p 2q ⋅ 2q = 13 p 2q( p 2q - 3 p + 2q)

= 4a 2 - 8ab - 8ab + 16b 2 = 4a 2 - 16ab + 16b 2

27.

( x - 2 y )3 = [( x - 2 y )( x - 2 y )]( x - 2 y ) 2

60m4 - 120m3n + 50m2n 2 = 10m2 ⋅ 6m2 - 10m2 ⋅ 12mn + 10m2 ⋅ 5n 2 = 10m2 (6m2 - 12mn + 5n 2 )

= ( x - 2 xy - 2 xy + 4 y )( x - 2 y ) = ( x 2 - 4 xy + 4 y 2 )( x - 2 y )

m2 - 5m - 14 = (m - 7)(m + 2) since (-7)(2) = -14 and -7 + 2 = -5.

x 2 + 4 x - 5 = ( x + 5)( x - 1) since 5(-1) = -5 and -1 + 5 = 4.

z 2 + 9 z + 20 = ( z + 4)( z + 5) since 4 ⋅ 5 = 20 and 4 + 5 = 9.

= ( x 2 - 4 xy + 4 y 2 ) x + ( x 2 - 4 xy + 4 y 2 )(-2 y ) = x 3 - 4 x 2 y + 4 xy 2 - 2 x 2 y + 8 xy 2 - 8 y 3 = x 3 - 6 x 2 y + 12 xy 2 - 8 y 3

13 p 4q 2 - 39 p3q + 26 p 2q 2

4 8.

Chapter R ALGEBRA REFERENCE b 2 - 8b + 7 = (b - 7)(b - 1) since (-7)(-1) = 7 and -7 + (-1) = -8.

s 2 + 2st - 35t 2 = (s - 5t )(s + 7t )

since (-5t )(7t ) = -35t 2 and 7t + (-5t ) = 2t. 11.

y 2 - 4 yz - 21z 2 = ( y + 3z )( y - 7 z )

since (3z )(-7 z ) = -21z 2 and 3z + (-7 z ) = -4 z. 12.

13.

= 10( x 2 - 42 ) = 10( x + 4)( x - 4)

a 2 - 6ab + 5b 2 = (a - b)(a - 5b)

since (-b)(-5b) = 5b 2 and -b + (-5b) = -6b. 10.

23. 10 x 2 - 160 = 10( x 2 - 16)

3x 2 + 4 x - 7

24.

9 x 2 + 64 is the sum of two perfect squares. It cannot be factored. It is prime.

25.

z 2 + 14 zy + 49 y 2 = z 2 + 2 ⋅ 7 zy + 72 y 2 = ( z + 7 y )2

s 2 - 10st + 25t 2 = s 2 - 2 ⋅ 5st + (5t )2

26.

= ( s - 5t )2

The possible factors of 3x2 are 3x and x and the possible factors of –7 are –7 and 1, or 7 and –1. Try various combinations until one works.

27.

3x 2 + 4 x - 7 = (3x + 7)( x - 1)

28.

9 p 2 - 24 p + 16 = (3 p)2 - 2 ⋅ 3 p ⋅ 4 + 42 = (3 p - 4)2

a 3 - 216 = a 3 - 63 = (a - 6)[(a)2 + (a)(6) + (6)2 ]

3a + 10a + 7

The possible factors of 3a2 are 3a and a and the possible factors of 7 are 7 and 1. Try various combinations until one works.

= (a - 6)(a 2 + 6a + 36)

27r 3 - 64s 3 = (3r )3 - (4s)3

29.

= (3r - 4s )(9r 2 + 12rs + 16s 2 )

3a + 10a + 7 = (a + 1)(3a + 7) 2

14. 15 y + y - 2 = (5 y + 2)(3 y - 1) 2

21m + 13mn + 2n = (7m + 2n)(3m + n)

16.

6a 2 - 48a - 120 = 6(a 2 - 8a - 20) = 6(a - 10)(a + 2)

18.

19.

3m3 + 12m 2 + 9m = 3m(m 2 + 4m + 3) = 3m(m + 1)(m + 3)

= 3(m + 5)(m2 - 5m + 25)

x 4 - y 4 = ( x 2 )2 - ( y 2 )2

31.

= ( x 2 + y 2 )( x 2 - y 2 ) = ( x 2 + y 2 )( x + y )( x - y )

32. 16a 4 - 81b4 = (4a 2 )2 - (9b2 )2

4a 2 + 10a + 6 = 2(2a 2 + 5a + 3) = 2(2a + 3)(a + 1)

= (4a 2 + 9b2 )(4a 2 - 9b2 )

24a 4 + 10a3b - 4a 2b 2

= (4a 2 + 9b2 )(2a + 3b)(2a - 3b)

= 2a 2 (12a 2 + 5ab - 2b 2 ) = 2a 2 (4a - b)(3a + 2b)

20.

= 3(m3 + 53 )

15.

17.

3m3 + 375 = 3(m3 + 125)

30.

R.3 Rational Expressions Your Turn 1

2 2

24 x + 36 x y - 60 x y 2

= (4a 2 + 9b2 )[(2a )2 - (3b)2 ]

= 12 x (2 x + 3xy - 5 y ) = 12 x 2 ( x - y)(2 x + 5 y)

21.

x 2 - 64 = x 2 - 82 = ( x + 8)( x - 8)

22.

9m2 - 25 = (3m)2 - (5)2 = (3m + 5)(3m - 5)

Write in lowest terms z 2 + 5z + 6

z 2 + 5z + 6 2z 2 + 7z + 3

( z + 3)( z + 2) ( z + 3)(2 z + 1) 2z + 7z + 3 z+2 = 2z + 1 2

Section R.3

Your Turn 2 Perform each of the following operations. (a)

2z 2 - z - 1 ⋅ 2 z 2 - 5z - 3 z 2 + 2 z - 3 (z + 2)( z + 3) (2 z + 1)( z - 1) = ⋅ (2 z + 1)( z - 3) ( z + 3)( z - 1)

z 2 + 5z + 6

( z + 2) ( z + 3) (2 z + 1) ( z - 1) (2 z + 1) ( z - 3) ( z + 3) ( z - 1)

z+2 z-3 a-3

(b)

5a + 2 a + 3a + 2 a -4 5a a-3 = + (a + 2)(a + 1) (a - 2)(a + 2) (a - 2) a-3 = ⋅ (a + 2)(a + 1) (a - 2) 5a (a + 1) + ⋅ (a - 2)(a + 2) (a + 1)

(a 2 - 5a + 6) + (5a 2 + 5a) (a - 2)(a + 2)(a + 1)

6a 2 + 6 (a - 2)(a + 2)(a + 1)

6(a 2 + 1) (a - 2)(a + 2)(a + 1)

10.

11.

m 2 - 4m + 4

(m - 2)(m - 2) (m - 2)(m + 3) m +m-6 m-2 = m+3 2

r2 - r - 6

( r - 3)( r + 2) ( r + 4)( r - 3) r + r - 12 r+2 = r+4 =

3x 2 + 3x - 6 2

x -4

z 2 - 5z + 6 2

z -4 m 4 - 16 4m2 - 16

( z - 3)( z - 2) z-3 = ( z + 2)( z - 2) z+2

(m2 + 4)(m + 2)(m - 2) 4(m + 2)(m - 2)

m2 + 4 4

12.

6 y 2 + 11 y + 4 (3 y + 4)(2 y + 1) 2 y +1 = = 2 3y + 7 y + 4 (3 y + 4)( y + 1) y +1

13.

9k 2 5 3 ⋅ 3 ⋅ 5k 2 3k 2 3k ⋅ = = = 25 3k 5 ⋅ 5 ⋅ 3k 5k 5

14.

R.3 Exercises

15 p3

10 p

¸ 2

= 2

5v 2 5⋅v⋅v v = = 35v 5⋅7⋅v 7

25 p 3

5⋅5⋅ p⋅ p⋅ p 5p = = 2 2⋅5⋅ p⋅ p 2 10 p

8k + 16 8(k + 2) 8 = = 9k + 18 9(k + 2) 9

2(t - 15) 2 = (t - 15)(t + 2) t+2

4 x3 - 8 x 2 4x2

4 x 2 ( x - 2) 4x2

15.

= x-2

36 y 2 + 72 y 36 y( y + 2) = 9y 9y 9 ⋅ 4 ⋅ y( y + 2) = 9⋅ y = 4( y + 2)

3( x + 2)( x - 1) 3( x - 1) = ( x + 2)( x - 2) x-2

16.

15 p3 10 p 2 ⋅ 6p 9 p2 150 p5 54 p3 25 ⋅ 6 p5 9 ⋅ 6 p3 25 p 2 9

3a + 3b 12 3(a + b) 3⋅ 4 ⋅ = ⋅ 4c 5(a + b) 4c 5(a + b) 3⋅3 = c⋅5 9 = 5c a-3 a-3 a-3 32 ¸ = ⋅ 16 32 16 a-3 a - 3 16 ⋅ 2 = ⋅ 16 a-3 2 = = 2 1

6 17.

18.

19.

20.

Chapter R ALGEBRA REFERENCE 2k - 16 4k - 32 2k - 16 3 ¸ = ⋅ 6 3 6 4k - 32 2(k - 8) 3 = ⋅ 6 4(k - 8) 1 = 4

4a + 12 a2 - 9 ¸ 2 2a - 10 a - a - 20 4(a + 3) (a - 5)(a + 4) = ⋅ 2(a - 5) (a - 3)(a + 3) 2(a + 4) = a-3

25.

6r - 18

12r - 16 9r + 6r - 24 4r - 12 6(r - 3) 4(3r - 4) = ⋅ 2 3(3r + 2r - 8) 4(r - 3) ⋅

6(r - 3) 4(3r - 4) ⋅ 3(3r - 4)(r + 2) 4(r - 3) 6 2 = = 3(r + 2) r+2 2

22.

k + 4k - 12 2

⋅

m 2 + 3m + 2 m 2 + 5m + 4 =

26.

27. 28.

m2 + 10m + 24 m 2 + 10m + 24

m 2 + 5m + 4 m2 + 5m + 6 (m + 1)(m + 2) (m + 6)(m + 4) = ⋅ (m + 4)(m + 1) (m + 3)(m + 2) m+6 = m+3

2m2 - 5m - 12

⋅

m2 - 9m + 18

4n 2 + 4n - 3

⋅

8n 2 + 32n + 30

6n 2 - n - 15 4n 2 + 16n + 15 (2n + 3)(2n - 1) 2(2n + 3)(2n + 5) = ⋅ (2n + 3)(3n - 5) (2n + 3)(2n + 5) 2(2n - 1) = 3n - 5 (a + 1) - (a - 1) a +1 a -1 = 2 2 2 a +1- a +1 = 2 2 = =1 2

3 1 + p 2 2⋅3 p ⋅1 6 p 6+ p + = + = 2⋅ p p⋅2 2p 2p 2p

m2 + 5m + 6

⋅

m2 - 9m + 18

k + k - 12

m2 + 3m + 2

4m2 - 9

Multiply the first term by 22 and the second by p .

k + 10k + 24 k2 - 9 (k + 6)(k - 2) (k + 4)(k - 3) = ⋅ (k + 6)(k + 4) (k + 3)(k - 3) k-2 = k +3

m2 - 10m + 24

m2 - 10m + 24 4m 2 - 9 (2m + 3)(m - 4)(m - 6)(m - 3) = (m - 6)(m - 4)(2m - 3)(2m + 3) m-3 = 2m - 3

24.

2m2 - 5m - 12

9 y - 18 3 y + 6 9( y - 2) 3( y + 2) ⋅ = ⋅ 6 y + 12 15 y - 30 6( y + 2) 15( y - 2) 27 3⋅3 3 = = = 90 10 ⋅ 3 10

21.

23.

29.

6 3 6⋅2 3⋅5y 12 -15 y - = = 5y 2 5y ⋅ 2 2⋅5y 10 y 1 2 4 5 ⋅1 6⋅2 30 ⋅ 4 + + = + + 6m 5m 5 ⋅ 6m 6 ⋅ 5m 30 ⋅ m m 5 12 120 = + + 30m 30m 30m 5 + 12 + 120 = 30m 137 = 30m 1 2 m æ 1 ö÷ m -1 çæ 2 ÷ö + = çç ÷+ ç ÷ m -1 m m èç m -1ø÷ m -1 çè m ÷ø m + 2m - 2 = m( m -1) 3m - 2 = m( m -1)

Section R.3 30.

31.

32.

33.

5 2 5r 2(2r + 3) - = 2r + 3 r r (2r + 3) r (2r + 3) 5r - 2(2r + 3) 5r - 4r - 6 = = r (2r + 3) r (2r + 3) r -6 = r (2r + 3) 8 2 8 3 æ 2 ö÷ + = + çç ÷ 3( a -1) 3( a -1) 3 çè a -1ø÷ a -1 8+6 = 3( a -1) 14 = 3( a -1)

4 2

4( x - 2) ( x - 2)( x + 3)( x + 1)

3( x + 3) ( x - 2)( x + 3)( x + 1) 4( x - 2) + 3( x + 3) = ( x - 2)( x + 3)( x + 1) 4 x - 8 + 3x + 9 = ( x - 2)( x + 3)( x + 1) 7x + 1 = ( x - 2)( x + 3)( x + 1) +

y2 + y - y + 1 ( y + 3)( y + 1)( y - 1)

y2 + 1 ( y + 3)( y + 1)( y - 1) 3k

1 2

y + 2y - 3 y + 4y + 3 y 1 = ( y + 3)( y - 1) ( y + 3)( y + 1) y( y + 1) = ( y + 3)( y + 1)( y - 1) 1( y - 1) ( y + 3)( y + 1)( y - 1)

2k 2

2k + 3k - 2 2k - 7k + 3 3k 2k = (2k - 1)(k + 2) (2k - 1)(k - 3) æ k - 3 ÷ö 3k = çç çè k - 3 ÷÷ø (2k - 1)(k + 2)

(3k 2 - 9k ) - (2k 2 + 4k ) (2k - 1)(k + 2)(k - 3)

k 2 - 13k (2k - 1)(k + 2)(k - 3) k (k - 13) = (2k - 1)(k + 2)(k - 3) =

3 2

x + 4x + 3 x -x-2 4 3 = + ( x + 3)( x + 1) ( x - 2)( x + 1)

y( y + 1) - ( y - 1) ( y + 3)( y + 1)( y - 1)

æ k + 2 ö÷ 2k - çç çè k + 2 ÷÷ø (2k - 1)(k - 3)

2 3 4⋅2 5⋅ 3 + = + 5( k - 2) 4( k - 2) 4 ⋅ 5( k - 2) 5 ⋅ 4( k - 2) 8 15 = + 20( k - 2) 20( k - 2) 8 + 15 = 20( k - 2) 23 = 20( k - 2)

34.

35.

36.

4m 2

m 2

3m + 7m - 6 3m - 14m + 8 m 4m = (3m - 2)(m + 3) (3m - 2)(m - 4) 4m(m - 4) = (3m - 2)(m + 3)(m - 4) m(m + 3) (3m - 2)(m - 4)(m + 3) 4m(m - 4) - m(m + 3) = (3m - 2)(m - 4)(m + 3) =

4m2 - 16m - m2 - 3m (3m - 2)(m + 3)(m - 4)

3m 2 - 19m (3m - 2)(m + 3)(m - 4) m(3m - 19) = (3m - 2)(m + 3)(m - 4) =

Chapter R ALGEBRA REFERENCE

37.

a -1 2 1 + + 2 a+2 a a + 2a a -1 2 1 = + + a+2 a a(a + 2)

Your Turn 3

Solve z 2 + 6 = 8z. z 2 - 8z + 6 = 0

æaö 2 æ a + 2 ö÷ 1 a -1 = çç ÷÷÷ + çç + çè a ø a + 2 çè a + 2 ø÷÷ a a(a + 2)

2a + a + 2 + a - 1 a(a + 2) 4a + 1 = a(a + 2) =

38.

5x + 2 2

x -1

Use the quadratic formula with a = 1, b = -8, and c = 6. -(-8)  (-8)2 - 4(1)(6) 2(1)

8  64 - 24 2 8  40 = 2 =

8  4 ⋅ 10 2 8  2 10 = 2 = 4  10

x +x x -x 5x + 2 3 1 = + x( x + 1) x( x - 1) ( x + 1)( x - 1) ö ö÷ æ x ö÷æç æ 5x + 2 3 ÷÷ + ç x - 1 ö÷÷æçç ÷ = çç ÷÷ç ç ÷ ÷ çè x øçè ( x + 1)( x - 1) ø÷ çè x - 1 øçè x( x + 1) ÷ø÷ ö÷ æ x + 1 ö÷æç 1 ÷ - çç ÷ç ÷ çè x + 1 øçè x( x - 1) ÷÷ø =

x(5 x + 2) + ( x - 1)(3) - ( x + 1)(1) x( x + 1)( x - 1)

5 x 2 + 2 x + 3x - 3 - x - 1 x( x + 1)( x - 1)

5x 2 + 4 x - 4 x( x + 1)( x - 1)

Your Turn 4 1 2 1 Solve 2 + = . x-2 x x -4

1 2 1 + = x-2 x ( x - 2)( x + 2) 1 ( x - 2)( x + 2)( x ) ⋅ ( x - 2)( x + 2) 2 + ( x - 2)( x + 2)( x ) ⋅ x-2 = ( x - 2)( x + 2)( x ) ⋅

R.4 Equations

x + 2x2 + 4x = x 2 - 4

Your Turn 1 Solve 3x - 7 = 4(5x + 2) - 7 x.

x 2 + 5x + 4 = 0 ( x + 1)( x + 4) = 0

3x - 7 = 20 x + 8 - 7 x 3x - 7 = 13x + 8

x = -1 or x = -4

-10 x = 15 15 x=10 3 x=2

Neither of these values makes a denominator equal to zero, so both are solutions.

R.4 Exercises 1.

x = -12

Your Turn 2

The solution is –12.

Solve 2m2 + 7m = 15.

2m - 3 = 0

2x + 8 = x - 4 x + 8 = -4

2m2 + 7m - 15 = 0 (2m - 3)(m + 5) = 0 3 m= or 2

5 x + 2 = 8 - 3x 8x + 2 = 8

or m + 5 = 0 m = -5

1 x

8x = 6 x =

3 4

Section R.4 3.

0.2m - 0.5 = 0.1m + 0.7

10(0.2m - 0.5) = 10(0.1m + 0.7)

4[2 p - (3 - p) + 5] = -7 p - 2 4[2 p - 3 + p + 5] = -7 p - 2

2m - 5 = m + 7

4[3 p + 2] = -7 p - 2

m-5 = 7

12 p + 8 = -7 p - 2

m = 12

19 p + 8 = -2

The solution is 12.

19 p = -10

2 3 1 k-k + = 3 8 2

p =-

Multiply both sides of the equation by 24. æ2 ö æ3ö æ1ö 24 çç k ÷÷ - 24(k ) + 24 çç ÷÷ = 24 çç ÷÷ çè 2 ÷ø èç 3 ø÷ èç 8 ÷ø 16k - 24k + 9 = 12 -8k + 9 = 12 -8k = 3 3 k =8

The solution is - 10 . 19 9.

x+3= 0

10.

x - 3 = 0 or

-7 = 8r

11.

7 = r 8

(m - 7)2 = 0 m-7 = 0 m=7 The solution is 7.

5a + 15 + 4a - 5 = - 2a + 4 9a + 10 = - 2a + 4 11a + 10 = 4

12.

11a = -6 6 a =11 6. The solution is - 11

2[3m - 2(3 - m) - 4] = 6m - 4 2[3m - 6 + 2m - 4] = 6m - 4 2[5m - 10] = 6m - 4 10m - 20 = 6m - 4

m= 4

m2 = 14m - 49 (m) 2 - 2(7m) + (7)2 = 0

5(a + 3) + 4a - 5 = - (2a - 4)

4m = 16

x = -1

m 2 - 14m + 49 = 0

The solution is - 78 .

4m - 20 = -4

x +1 = 0

x = 3 or The solutions are 3 and -1.

-3 = 8r + 4

The solution is 4.

x2 = 3 + 2x ( x - 3)( x + 1) = 0

-3 - 2r = 6r + 4

x = -2

x2 - 2x - 3 = 0

3r + 2 - 5r - 5 = 6r + 4

x+2 =0

x = -3 or The solutions are -3 and -2.

3r + 2 - 5(r + 1) = 6r + 4

x 2 + 5x + 6 = 0 ( x + 3)( x + 2) = 0

The solution is - 83 . 5.

10 19

2k 2 - k = 10 2k 2 - k - 10 = 0 (2k - 5)(k + 2) = 0 2k - 5 = 0 or k + 2 = 0 5 k = k = -2 or 2 The solutions are 52 and -2.

13. 12 x 2 - 5 x = 2 12 x 2 - 5 x - 2 = 0 (4 x + 1)(3x - 2) = 0 4x + 1 = 0 or 3x - 2 = 0 4 x = -1 or 3x = 2 1 2 x = - or x = 4 3 1 2 The solutions are - 4 and 3 .

10 14.

Chapter R ALGEBRA REFERENCE m(m - 7) = -10

19.

m 2 - 7m + 10 = 0 (m - 5)(m - 2) = 0 m - 5 = 0 or m - 2 = 0 m = 5 or m= 2 The solutions are 5 and 2.

15.

2m 2 - 4m = 3 2m 2 - 4m - 3 = 0 m= =

4 x 2 - 36 = 0 x2 - 9 = 0 ( x + 3)( x - 3) = 0 x+3=0 or x - 3 = 0 x = -3 or x =3 The solutions are -3 and 3. z (2 z + 7) = 4 2

2z + 7 z - 4 = 0 (2 z - 1)( z + 4) = 0 2 z - 1 = 0 or z + 4 = 0 1 z = z = -4 or 2 The solutions are 12 and -4.

The solutions are 2 - 10 2

20.

18.

y-4=0 y = 4

Use the quadratic formula.

5

25 - 12 6 5 + 13 5 - 13 x = or x = 6 6 » 1.4343 » 0.2324 5 + 13 » 1.4343 and 6

5 - 13 » 0.2324. 6

4 ⋅ 10 4

4

2 + 10 2

» 2.5811 and

-1  12 - 4(1)(-1) 2(1) -1  5 2

The solutions are

-1 + 5 » 0.6180 and 2

21.

k 2 - 10k = -20 k 2 - 10k + 20 = 0 - (-10) 

(-10)2 - 4(1)(20) 2(1)

10  100 - 80 2 10  20 k = 2 10  4 5 k = 2 10  2 5 k = 2 2(5  5) k = 2 k =5 5 k =

-(-5)  (-5) 2 - 4(3)(1) 2(3)

The solutions are

4

» -0.5811.

k =

3x - 5 x + 1 = 0

-1 - 5 » -1.6180. 2

x =

40 4

p2 + p - 1 = 0

p =

17. 12 y - 48 y = 0 12 y( y) - 12 y(4) = 0 12 y( y - 4) = 0 12 y = 0 or y = 0 or The solutions are 0 and 4.

4

4 10 4 4  2 10 2  10 = = 4 2 =

Divide both sides of the equation by 4.

16.

(-4) 2 - 4(2)(-3) 2(2)

-(-4) 

The solutions are 5 + 5 » 7.2361 and 5 - 5 » 2.7639.

Section R.4 22.

5x 2 - 8x + 2 = 0 x = = = = =

-(-8)  (-8)2 - 4(5)(2) 2(5)

23.

5(3x - 2) = 7( x + 2) 15x - 10 = 7 x + 14

8

4⋅6 10

8

4 6 82 6 = 10 10

8 x = 24

4 6 5 4+ 6 5

3x - 2 x+2 = 7 5 æ 3x - 2 ö÷ æ x + 2 ÷ö 35çç ÷ = 35ççç ÷ èç 7 ÷ø è 2 ÷ø

8  24 10

The solutions are 4- 6 5

27.

x =3 The solution is x = 3.

28.

x 3x -7 = 63 4

Multiply both sides by 12, the least common denominator of 3 and 4. æx ö æ 3x ö÷ 12 çç - 7 ÷÷ = 12 çç 6 ÷ ÷ø çè 3 çè 4 ø÷ æxö æ 3x ö 12 çç ÷÷÷ - (12)(7) = (12)(6) - 12 çç ÷÷÷ èç 3 ø èç 4 ø

» 1.2899 and

» 0.3101.

2r 2 - 7r + 5 = 0 (2r - 5)( r - 1) = 0 2r - 5 = 0 or r - 1 = 0

4 x - 84 = 72 - 9 x

2r = 5 5 or r = 2 The solutions are 52 and 1.

24.

13x - 84 = 72 x = 12

The solution is 12.

2 x 2 - 7 x + 30 = 0

29.

(-7)2 - 4(2)(30) 2(2) 7  49 - 240 x = 4 7  -191 x = 4 Since there is a negative number under the radical sign, -191 is not a real number. Thus, there are no real number solutions. x =

25.

- (-7) 

3k 2 + k = 6

4 8 3 + =0 x-3 2x + 5 x-3 4 3 8 + =0 x-3 x-3 2x + 5 7 8 =0 x-3 2x + 5

Multiply both sides by ( x - 3)(2 x + 5). Note that x ¹ 3 and x ¹ 52 . æ 7 8 ö÷ ( x - 3)(2 x + 5) çç ÷ çè x - 3 2 x + 5 ÷ø = ( x - 3)(2 x + 5)(0)

3k 2 + k - 6 = 0 k =

13x = 156

r =1

7(2 x + 5) - 8( x - 3) = 0

-1  1 - 4(3)(-6) 2(3)

14 x + 35 - 8x + 24 = 0 6 x + 59 = 0

-1  73 = 6

6 x = -59

-1 + 73 The solutions are » 1.2573 and 6 -1 - 73 » -1.5907. 6

26.

5m 2 + 5m = 0 5m(m + 1) = 0

x =-

59 6

Note: It is especially important to check solutions of equations that involve rational expressions. Here, a check shows that - 59 is a solution. 6

5m = 0 or m + 1 = 0 m = 0 or m = -1 The solutions are 0 and –1.

12 30.

Chapter R ALGEBRA REFERENCE 5 7 12 = 2 p-2 p+2 p -4

32.

2y 5 10 - 8 y = + 2 y -1 y y -y

5 7 12 = p-2 p+2 ( p - 2)( p + 2)

2y 5 10 - 8 y = + y -1 y y( y - 1)

Multiply both sides by ( p - 2)( p + 2). Note that p ¹ 2 and p ¹ -2.

Multiply both sides by y( y - 1). Note that y ¹ 0 and y ¹ 1.

æ 5 7 ö÷ ÷ ( p - 2)( p + 2) çç çè p - 2 p + 2 ÷÷ø æ ö÷ 12 ÷ = ( p - 2)( p + 2) çç ÷ çè ( p - 2)( p + 2) ÷ø

æ 2 y ö÷ é5 10 - 8 y úù ÷ = y( y - 1) ê + y( y - 1) çç êy çè y - 1 ÷÷ø y( y - 1) úû ë æ 2 y ÷ö æ5ö ÷ = y( y - 1) çç ÷÷÷ y( y - 1) çç çè y - 1 ÷÷ø çè y ÷ø

æ 5 ö÷ æ 7 ÷ö ÷÷ - ( p - 2)( p + 2) çç ÷ ( p - 2)( p + 2) çç çè p + 2 ÷÷ø èç p - 2 ø÷

é 10 - 8 y ù ú + y( y - 1) ê ê y( y - 1) ú ë û y(2 y) = ( y - 1)(5) + (10 - 8 y)

æ ö÷ 12 ÷ = ( p - 2)( p + 2) çç ÷ èç ( p - 2)( p + 2) ÷ø

2 y 2 = 5 y - 5 + 10 - 8 y

( p + 2)(5) - ( p - 2)(7) = 12 5 p + 10 - 7 p + 14 = 12 -2 p + 24 = 12 -2 p = -12 p =6

2 y2 = 5 - 3y 2 y2 + 3y - 5 = 0 (2 y + 5)( y - 1) = 0 2y + 5 = 0 5 y =2

The solution is 6. 31.

2m 6 12 = 2 m-2 m m - 2m 2m 6 12 = m-2 m m(m - 2)

y -1 = 0

y =1

Since y ¹ 1, 1 is not a solution. The solution is - 52 .

Multiply both sides by m(m - 2). Note that m ¹ 0 and m ¹ 2. æ 2m 6ö m(m - 2) çç - ÷÷÷ çè m - 2 mø æ ö÷ 12 ÷ = m(m - 2) çç çè m(m - 2) ÷÷ø

33.

1 3x 2x + 1 = 2 x-2 x -1 x - 3x + 2 1 3x 2x + 1 = x-2 x -1 ( x - 2)( x - 1)

Multiply both sides by ( x - 2)( x - 1). Note that x ¹ 2 and x ¹ 1. æ 1 3x ö÷ ( x - 2)( x - 1) çç ÷ çè x - 2 x - 1 ÷ø

m(2m) - 6(m - 2) = 12 2m2 - 6m + 12 = 12 2m 2 - 6m = 0 2m(m - 3) = 0 2m = 0 or m - 3 = 0 m = 0 or

m=3

Since m ¹ 0, 0 is not a solution. The solution is 3.

é ù 2x + 1 ú = ( x - 2)( x - 1) ⋅ ê ê ( x - 2)( x - 1) ú ë û æ 1 ö÷ æ 3x ÷ö ( x - 2)( x - 1) çç ÷ - ( x - 2)( x - 1) ⋅ ççç ÷ èç x - 2 ø÷ è x - 1 ÷ø ( x - 2)( x - 1)(2 x + 1) = ( x - 2)( x - 1)

Section R.4

( x - 1) - ( x - 2)(3x) = 2 x + 1

5b(b + 2) - 4(b + 5) = 6b

x - 1 - 3x + 6 x = 2 x + 1

5b 2 + 10b - 4b - 20 = 6b

-3 x 2 + 7 x - 1 = 2 x + 1

5b 2 - 20 = 0 b2 - 4 = 0

-3 x 2 + 5 x - 2 = 0

(b + 2)(b - 2) = 0 b+2= 0 or b-2= 0

3x 2 - 5 x + 2 = 0 (3x - 2)( x - 1) = 0 3x - 2 = 0 or x =

2 3

b = -2

x -1 = 0

Since b ¹ -2, - 2 is not a solution. The solution is 2.

x =1

1 is not a solution since x ¹ 1. The solution is 23 . 36.

34.

5 a - 2a + 4 -7 + = a a +1 a2 + a æ5 -7 ÷ö a(a + 1) çç + ÷ çè a a + 1 ø÷ æ a 2 - 2a + 4 ö÷ ç ÷÷ = a(a + 1) çç çè a 2 + a ÷ø÷

Note that a ¹ 0 and a ¹ -1. 5a + 5 - 7a = a 2 - 2a + 4

5 + 2 x - 2x - 3 x -x-6 1 = 2 x + 3x + 2 2 5 + ( x - 3)( x + 1) ( x - 3)( x + 2) 1 = ( x + 2)( x + 1)

æ ö÷ 2 ÷ ( x - 3)( x + 1)( x + 2) çç çè ( x - 3)( x + 1) ÷ø÷

5 - 2a = a - 2a + 4

æ ö÷ 5 ÷ + ( x - 3)( x + 1)( x + 2) çç çè ( x - 3)( x + 2) ÷÷ø

5 = a2 + 4 0 = a2 - 1

æ 1 ÷÷ö = ( x - 3)( x + 1)( x + 2) çç çè ( x + 2)( x + 1) ÷÷ø 2( x + 2) + 5( x + 1) = x - 3

0 = (a + 1)(a - 1) a = -1

Multiply both sides by ( x - 3)( x + 1)( x + 2). Note that x ¹ 3, x ¹ -1, and x ¹ -2.

5(a + 1) + (-7)(a) = a 2 - 2a + 4

a +1= 0

a -1 = 0

a =1

2 x + 4 + 5x + 5 = x - 3

Since 1 would make two denominators zero, 1 is the only solution. 35.

b= 2

5 4 6 - 2 = 2 b+5 b + 2b b + 7b + 10 5 4 6 = b+5 b(b + 2) (b + 5)(b + 2)

Multiply both sides by b(b  5)(b  2). Note that b  0, b  5, and b  2. æ 5 ö÷ 4 ÷ b(b + 5)(b + 2) çç çè b + 5 b(b + 2) ø÷÷ æ ö÷ 6 ÷ = b(b + 5)(b + 2) çç çè (b + 5)(b + 2) ÷÷ø

7x + 9 = x - 3 6 x + 9 = -3 6 x = -12 x = -2 However, x ¹ -2. Therefore there is no solution.

37.

4 2

2 x + 3x - 9

2 2

2x - x - 3 =

3 2

x + 4x + 3 4 2 + (2 x - 3)( x + 3) (2 x - 3)( x + 1) 3 = ( x + 3)( x + 1)

Multiply both sides by (2 x - 3)( x + 3)( x + 1). Note that x ¹ 32 , x ¹ -3, and x ¹ -1.

Chapter R ALGEBRA REFERENCE (2 x - 3)( x + 3)( x + 1) æ 4 2 ÷÷ö ⋅ çç + çè (2 x - 3)( x + 3) (2 x - 3)( x + 1) ÷÷ø æ ö÷ 3 ÷ = (2 x - 3)( x + 3)( x + 1) çç çè ( x + 3)( x + 1) ø÷÷ 4( x + 1) + 2( x + 3) = 3(2 x - 3)

The numbers in interval B satisfy the inequality, and since the sign was less than or equal to, the boundary points of interval B are also part of the solution. The solution is [-2/3, 6]. Your Turn 3

Solve

4x + 4 + 2x + 6 = 6x - 9 6 x + 10 = 6 x - 9

k 2 - 35 ³ 2. k

First solve the corresponding equation

10 = -9

k 2 - 35 =2 k

This is a false statement. Therefore, there is no solution.

k 2 - 35 = 2k k 2 - 2k - 35 = 0 (k - 7)(k + 5) = 0

R.5 Inequalities

k = 7 or k = -5

Your Turn 1 Solve 3z - 2 > 5z + 7. 3z - 2 > 5z + 7

The denominator is 0 when k = 0, so there are four intervals to consider: (-¥, -5), (-5, 0), (0, 7), and (7, ¥).

3z - 2 + 2 > 5z + 7 + 2 3z > 5z + 9

Choose a test point in each interval.

3z - 5z > 5z - 5z + 9 -2 z > 9 -2 z 9 < -2 -2 9 z<2

k = -8;

(-8)2 - 35 <2 -8

k = -1;

(-1)2 - 35 >2 -1

k = 5;

(-5)2 - 35 <2 5

k = 10;

(10)2 - 35 >2 10

Your Turn 2

Solve 3 y 2 £ 16 y + 12. 3 y 2 £ 16 y + 12 2

3 y - 16 y - 12 £ 0

First solve the equation 3 y 2 - 16 y - 12 = 0. 3 y 2 - 16 y - 12 = 0 (3 y + 2)( y - 6) = 0 3y + 2 = 0 y =-

The second and fourth intervals are part of the solution. Since the inequality is greater than or equal to, we can include the endpoints -5 and 7 but not the endpoint 0, which makes the denominator 0. The solution is [-5, 0) È [7, ¥).

R.5 Exercises 1. x  4

or y - 6 = 0 2 3

k 2 - 35 = 2. k

y =6

Because the inequality symbol means “less than,” the endpoint at 4 is not included. This inequality is written in interval notation as (-¥, 4). To graph this interval on a number line, place an open circle at 4 and draw a heavy arrow pointing to the left.

Determine three intervals on the number line and choose a test point in each interval. B

A 2 3

C 6

Choose -1 from interval A: 3(-1)2 - 16(-1) - 12 > 0 Choose 0 from interval B: 3(0)2 - 16(0) - 12 < 0 Choose 7 from interval C: 3(7)2 - 16(7) - 12 > 0

x ³ -3

Because the inequality sign means “greater than or equal to,” the endpoint at -3 is included. This inequality is written in interval notation as [-3, ¥).

Section R.5

To graph this interval on a number line, place a closed circle at -3 and draw a heavy arrow pointing to the right. –3

1£ x < 2

The endpoints at -2 and 3 are both included. This inequality is written in interval notation as [-2, 3]. To graph this interval, place an open circle at -2 and another at 3 and draw a heavy line segment between them.

11. Notice that the endpoint -2 is included, but 6 is not. The interval shown in the graph can be written as the inequality -2 £ x < 6. 12. Notice that neither endpoint is included. The interval shown in the graph can be written as 0 < x < 8. 13. Notice that both endpoints are included. The interval shown in the graph can be written as x £ -4 or x ³ 4.

-2 £ x £ 3

Ð2

14. Notice that the endpoint 0 is not included, but 3 is included. The interval shown in the graph can be written as x < 0 or x ³ 3. 15.

p£2

This inequality may be rewritten as x < -9, and is written in interval notation as (-¥, -9). Note that the endpoint at -9 is not included. To graph this interval, place an open circle at -9 and draw a heavy arrow pointing to the left.

The solution in interval notation is (-¥, 2].

16. 6.

6k < 3k + 3 3k < 3

This inequality may be written as x ³ 6, and is written in interval notation as [6, ¥). Note that the endpoint at 6 is included. To graph this interval, place a closed circle at 6 and draw a heavy arrow pointing to the right.

k <1

m - (3m - 2) + 6 < 7m - 19 m - 3m + 2 + 6 < 7m - 19 -2m + 8 < 7m - 19

[-7, -3]

[4, 10)

This represents all the numbers between 4 and 10, including 4 but not including 10. This interval can be written as the inequality 4 £ x < 10. 9.

The solution in interval notation is (-¥,1).

17.

This represents all the numbers between -7 and -3, including both endpoints. This interval can be written as the inequality -7 £ x £ -3. 8.

6k - 4 < 3k - 1

6£ x

6 p + 7 £ 19 6 p £ 12 æ 1 ö÷ æ ö çç ÷ (6 p) £ çç 1 ÷÷ (12) çè 6 ÷ø èç 6 ÷ø

-9 > x

–9

(3, ¥)

This represents all the numbers to the right of 3, and does not include the endpoint. This interval can be written as the inequality x > 3.

The endpoint at 1 is included, but the endpoint at 2 is not. This inequality is written in interval notation as [1, 2). To graph this interval, place a closed circle at 1 and an open circle at 2; then draw a heavy line segment between them.

10.

-9m + 8 < -19 -9m < -27

1 1 - (-9m) > - (-27) 9 9 m>3

The solution is (3, ¥).

(-¥, -1]

This represents all the numbers to the left of -1 on the number line and includes the endpoint. This interval can be written as the inequality x £ -1.

16 18.

Chapter R ALGEBRA REFERENCE 22.

-2(3 y - 8) ³ 5(4 y - 2)

8 £ 3r + 1 £ 13 8 + (-1) £ 3r + 1 + (-1) £ 13 + (-1)

-6 y + 16 ³ 20 y - 10

7 £ 3r £ 12

-6 y + 16 + (-16) ³ 20 y - 10 + (-16)

1 1 1 (7) £ (3r ) £ (12) 3 3 3 7 £r £4 3 The solution is éê 73 , 4 ùú . ë û

-6 y ³ 20 y - 26 -6 y + (-20 y) ³ 20 y + (-20 y) - 26 -26 y ³ -26 -

1 1 (-26) y £ - (-26) 26 26 y £1

19.

23.

3 p - 1 < 6 p + 2( p - 1) 3p - 1 < 6 p + 2 p - 2

1 1 1 - (-9) > - (-3k ) ³ - (15) 3 3 3 Rewrite the inequalities in the proper order. -5 £ k < 3

1 1 - (-5 p) > - (-1) 5 5 1 p> 5

The solution is [ -5, 3 ).

( 15 , ¥ ). 01

Ð5

24.

x + 5( x + 1) > 4 (2 - x) + x 6 x + 5 > 8 - 3x

5y + 2 £4 3

-5 £ 5 y £ 10

9x > 3

-1 £ y £ 2

1 x> 3

The solution is [ -1, 2 ].

( 13 , ¥ ). 0 1

–1

25.

-11 < y - 7 < -1 -11 + 7 < y - 7 + 7 < -1 + 7 -4 < y < 6

3 1 (2 p + 3) ³ (5 p + 1) 5 10

æ3ö æ1 ö 10 çç ÷÷ (2 p + 3) ³ 10 çç ÷÷ (5 p + 1) ÷ èç 5 ø èç 10 ÷ø 6(2 p + 3) ³ 5 p + 1 12 p + 18 ³ 5 p + 1

The solution is (-4, 6). Ð4

-3 £ 5 y + 2 £ 12

6 x > 3 - 3x

21.

-1 £

æ 5 y + 2 ÷ö 3(-1) £ 3çç £ 3(4) çè 3 ÷÷ø

x + 5x + 5 > 8 - 4 x + x

The solution is

1 - 3k £4 4 æ 1 - 3k ö÷ 4(-2) < 4 çç £ 4(4) çè 4 ÷÷ø

-2 <

-9 < -3k £ 15

- 5 p - 1 < -2 -5 p < -1

20.

-8 < 1 - 3k £ 16

3p - 1 < 8p - 2

The solution is

7 3

The solution is (-¥, 1].

7 p ³ -17 17 p³7

The solution is éê - 17 , ¥ ). ë 7 – 17 7

Section R.5 26.

8 2 ( z - 4) £ (3z + 2) 3 9 8 2 (9) ( z - 4) £ (9) (3z + 2) 3 9 24( z - 4) £ 2(3z + 2)

29.

y2 - 3y + 2 < 0 ( y - 2)( y - 1) < 0

Solve ( y - 2)( y - 1) = 0. y = 2

24 z - 96 £ 6 z + 4

Intervals: (-¥, 1), (1, 2), (2, ¥) For (-¥, 1), choose y = 0.

24 z £ 6 z + 100 18z £ 100

02 - 3(0) + 2 = 2 < / 0

100 8 50 z £ 9

For (1, 2), choose y = 32 .

z £

æ 3 ÷ö2 æ ö çç ÷ - 3çç 3 ÷÷ + 2 = 9 - 9 + 2 çè 2 ÷ø çè 2 ÷ø 4 2

(

ù. The solution is -¥, 50 9 úû

9 - 18 + 8 4 1 =- <0 4 =

50 9

27.

(m - 3)(m + 5) < 0

For (2, ¥), choose 3.

Solve (m - 3)(m + 5) = 0. (m - 3)(m + 5) = 0 m=3 or m = -5

32 - 3(3) + 2 = 2 < / 0

The solution is (1, 2).

Intervals: (-¥, - 5), (-5, 3), (3, ¥) For (-¥, - 5), choose -6 to test for m. (-6 - 3)(-6 + 5) = -9(-1) = 9 < / 0 For (-5, 3), choose 0. (0 - 3)(0 + 5) = -3(5) = -15 < 0 For (3, ¥), choose 4. (4 - 3)(4 + 5) = 1(9) = 9 < / 0

30.

28.

2k 2 + 7k - 4 > 0 Solve 2k 2 + 7k - 4 = 0. 2k 2 + 7 k - 4 = 0 (2k - 1)(k + 4) = 0 1 k = or k = -4 2

The solution is (-5, 3). Ð5

y =1

Intervals: (-¥, -4), ( -4, 12 ),( 12 , ¥ )

For (-¥, -4), choose -5.

(t + 6)(t - 1) ³ 0

2(-5)2 + 7(-5) - 4 = 11 > 0

Solve (t + 6)(t - 1) = 0. (t + 6)(t - 1) = 0 t = -6 or t =1

For -4, 12 , choose 0.

Intervals: (-¥, -6), (-6, 1), (1, ¥) For (-¥, -6), choose -7 to test for t. (-7 + 6)(-7 - 1) = (-1)(-8) = 8 ³ 0 For (-6, 1), choose 0. (0 + 6)(0 - 1) = (6)(-1) = -6 ³ / 0 For (1, ¥), choose 2. (2 + 6)(2 - 1) = (8)(1) = 8 ³ 0 Because the symbol ³ is used, the endpoints -6 and 1 are included in the solution. The solution is (-¥, -6] È [1, ¥).

For

–6

(

)

2(0)2 + 7(0) - 4 = -4 > / 0

0 1

( 12 , ¥ ), choose 1.

2(1) 2 + 7(1) - 4 = 5 > 0

The solution is (-¥, -4) È ( 12 , ¥ ). –4

01 2

31.

x 2 - 16 > 0

Solve x 2 - 16 = 0. x 2 - 16 = 0 ( x + 4)( x - 4) = 0 x = -4 or x = 4

Chapter R ALGEBRA REFERENCE Intervals: (-¥, -4), (-4, 4), (4, ¥)

For (5, ¥), choose 6.

For (-¥, -4), choose - 5.

(6)2 - 4(6) = 12 ³ 5 The solution is (-¥, -1] È [5, ¥).

(-5) 2 - 16 = 9 > 0

For (-4, 4), choose 0. 02 - 16 = -16 > / 0

Ð1 0

34. 10r + r £ 2

For (4, ¥), choose 5.

Solve 10r 2 + r = 2.

5 - 16 = 9 > 0

10r 2 + r = 2

The solution is (-¥, -4) È (4, ¥).

10r 2 + r - 2 = 0

Ð4

32.

(5r - 2)(2r + 1) = 0 2 1 or r = r = 5 2

2k 2 - 7k - 15 £ 0

(

(2k + 3)(k - 5) = 0 3 2

(

)

For - 12 , 52 , choose 0. 10(0)2 + 0 = 0 £ 2

Intervals: ( -¥, - 23 ), ( - 23 , 5 ), (5, ¥)

For

For ( -¥, - 23 ), choose -2.

)

For - 32 , 5 , choose 0.

–1

2(0)2 - 7(0) - 15 = -15 £ 0

2 5

35.

For (5, ¥), choose 6.

( 52 , ¥ ), choose 1.

10(1) 2 + 1 = 11 £ / 2 The solution is éê - 12 , 52 ùú . ë û

2(-2) 2 - 7(-2) - 15 = 7 £ / 0

(

)

10(-1) 2 + (-1) = 9 £ / 2

k =5

3x 2 + 2 x > 1

Solve 3x 2 + 2 x = 1.

2(6)2 - 7(6) - 15 £ / 0

3x 2 + 2 x = 1

The solution is éê - 23 , 5 ùú . ë û

3x 2 + 2 x - 1 = 0

x2 - 4x ³ 5

(3x - 1)( x + 1) = 0 1 or x = -1 x = 3

Solve x 2 - 4 x = 5.

Intervals: (-¥, -1), -1, 13 ,

–3

(

3(-2)2 + 2(-2) = 8 > 1

x2 - 4x - 5 = 0

(

)

For -1, 13 , choose 0.

( x + 1)( x - 5) = 0 or

) ( 13 , ¥ )

For (-¥, -1), choose -2.

x2 - 4x = 5

x +1= 0

) ( 52 , ¥ )

For -¥, - 12 , choose -1.

2k 2 - 7k - 15 = 0

k =-

)(

(

Intervals: -¥, - 12 , - 12 , 52 ,

Solve 2k 2 - 7k - 15 = 0.

33.

3(0)2 + 2(0) = 0 > / 1

x-5 = 0

x = -1 or x =5 Intervals: (-¥, -1), (-1, 5), (5, ¥) For (-¥, -1), choose -2. (-2)2 - 4(-2) = 12 ³ 5

For

( 13 , ¥ ), choose 1. 3(1)2 + 2(1) = 5 > 1

The solution is (-¥, -1) È ( 13 , ¥ ).

For (-1, 5), choose 0.

–1

02 - 4(0) = 0 ³ / 5

01 3

Section R.5 36.

19 For (-¥, 0), choose -1.

3a 2 + a > 10

(-1)2 - 16(-1) = 17 > 0

Solve 3a 2 + a = 10. 3a 2 + a = 10

For (0, 16), choose 1.

3a 2 + a - 10 = 0

(1)2 - 16(1) = -15 > / 0

(3a - 5)(a + 2) = 0 5 or a = -2 a = 3

For (16, ¥), choose 17. (17)2 - 16(17) = 17 > 0

The solution is (-¥, 0) È (16, ¥).

æ5 ö Intervals: (-¥, -2), ( -2, 53 ), çç , ¥ ÷÷÷ çè 3 ø For (-¥, -2), choose -3. 3(-3)2 + (-3) = 24 > 10

(

39.

x3 - 4x = 0

3(0) 2 + 0 = 0 ³ / 10

x( x 2 - 4) = 0 x( x + 2)( x - 2) = 0 x = 0, or x = -2, or

For ( 53 , ¥ ), choose 2. 3(2) 2 + 2 = 14 > 10

The solution is (-¥, - 2) È ( 53 , ¥ ).

37.

x3 - 4 x ³ 0

Solve x3 - 4 x = 0 .

)

For -2, 53 , choose 0.

–2

x = 2

Intervals: (-¥, -2), (-2, 0), (0, 2), (2, ¥) For (-¥, -2), choose -3. (-3)3 - 4(-3) = -15 ³ / 0

5 3

For (-2, 0), choose 1.

9 - x2 £ 0

(-1)3 - 4(-1) = 3 ³ 0

Solve 9 - x 2 = 0.

For (0, 2), choose 1.

9 - x2 = 0

(1)3 - 4(1) = -3 ³ / 0

(3 + x )(3 - x ) = 0 x = -3 or x = 3 Intervals: (-¥, -3), (-3, 3), (3, ¥) For (-¥, -3), choose -4.

For (2, ¥), choose 3. (3)3 - 4(3) = 15 ³ 0

The solution is [-2, 0] È [2, ¥).

9 - (-4) 2 = -7 £ 0

For (-3, 3), choose 0.

2

9 - (0)2 = 9 £ / 0

For (3, ¥), choose 4. 9 - (4)2 = -7 £ 0

40.

Ð3

x3 + 7 x 2 + 12 x £ 0

Solve x 3 + 7 x 2 + 12 x = 0. x3 + 7 x 2 + 12 x = 0

The solution is (-¥, -3] È [3, ¥).

38.

x( x 2 + 7 x + 12) = 0 x( x + 3)( x + 4) = 0 or x = -3, or x = -4

p 2 - 16 p > 0

x = 0,

Solve p 2 - 16 p = 0.

Intervals: (-¥, -4), (-4, -3), (-3, 0), (0, ¥) For (-¥, -4), choose -5.

p 2 - 16 p = 0

(-5)3 + 7(-5) 2 + 12(-5) = -10 £ 0

p( p - 16) = 0 p = 0 or p = 16

For (- 4, - 3), choose - 72 . 3

Intervals: (-¥, 0), (0, 16), (16, ¥)

( - 72 ) + 7 ( - 72 ) + 12( - 72 ) = 78 £/ 0 For (-3, 0), choose -1.

Chapter R ALGEBRA REFERENCE For (0, 4), choose 1.

(-1)3 + 7(-1)2 + 12(-1) = -6 £ 0

3(1)3 - 9(1)2 - 12(1) = -18 > / 0

For (0, ¥), choose 1.

For (4, ¥), choose 5.

(1)3 + 7(1)2 + 12(1) = 20 £ / 0

3(5)3 - 9(5) 2 - 12(5) = 90 > 0

The solution is (-¥, -4] È [-3, 0]. −4 −3

41.

The solution is (-1, 0) È (4, ¥).

 0

2 x3 - 14 x 2 + 12 x < 0 Solve 2 x3 - 14 x 2 + 12 x = 0 .

43.

2 x3 - 14 x 2 + 12 x = 0

(m + 5)

x =6

Intervals: (-¥, 0), (0, 1), (1, 6), (6, ¥) For (-¥, 0), choose -1.

For (0, 1), choose 12 . æ 1 ö3 æ 1 ö2 æ1ö 11 2 çç ÷÷ - 14 çç ÷÷ + 12 çç ÷÷ = < / 0 çè 2 ÷ø çè 2 ø÷ çè 2 ø÷ 4 2(2)3 - 14(2)2 + 12(2) = -16 < 0

For (-5, 3), choose 0.

For (6, ¥), choose 7.

0-3 3 =- £0 0+5 5

2(7)3 - 14(7)2 + 12(7) = 84 < / 0

For (3, ¥), choose 4.

The solution is (-¥, 0) È (1, 6).

42.

4-3 1 = £ / 0 4+5 9 Although the £ symbol is used, including -5 in the solution would cause the denominator to be zero.

3x3 - 9 x 2 - 12 x > 0 Solve 3x3 - 9 x 2 - 12 x = 0 .

The solution is (-5, 3].

3x3 - 9 x 2 - 12 x = 0 3x( x 2 - 3x - 4) = 0 3x( x - 4)( x + 1) = 0 x = 0, or x = 4, or x = -1

Intervals: (-¥, -1), (-1, 0), (0, 4), (4, ¥) For (-¥, -1), choose -2. 3(-2)3 - 9(-2)2 - 12(-2) = -36 > / 0

For (-1, 0), choose - 12 . 3

( ) - 9( - 12 ) - 12 ( - 12 )

3 - 12

m-3 = (m + 5)(0) m+5 m-3= 0 m=3

-6 - 3 =9£ / 0 -6 + 5

For (1, 6), choose 2.

m-3 = 0. m+5

Set the denominator equal to 0 and solve. m+5=0 m = -5 Intervals: (-¥, -5), (-5, 3), (3, ¥) For (-¥, -5), choose - 6.

2(-1)3 - 14(-1) 2 + 12(-1) = -28 < 0

m-3 £0 m+5

Solve

2 x( x 2 - 7 x + 6) = 0 2 x( x - 1)( x - 6) = 0 x = 0, or x = 1, or

44.

r +1 >0 r -1

Solve the equation

r +1 = 0. r -1

r +1 =0 r -1 r +1 (r - 1) = (r - 1)(0) r -1 r + 1 = 0 r = -1

Find the value for which the denominator equals zero. 27 = >0 8

r -1 = 0 r =1

Intervals: (-¥, -1), (-1, 1), (1, ¥)

Section R.5

(

For (-¥, -1), choose -2.

)

For -2, 32 , choose 0.

-2 + 1 -1 1 = = >0 -2 - 1 -3 3

0-5 -5 5 = = - < -1 0+2 2 2

For (-1, 1), choose 0.

For

0 +1 1 = = -1 > / 0 0 -1 -1

( 32 , ¥ ), choose 2. 2-5 -3 3 = =- < / -1 2+2 4 4

For (1, ¥), choose 2. 2 +1 3 = =3>0 2 -1 1 The solution is (-¥, -1) È (1, ¥).

45.

(

47.

k -1 >1 k +2

Solve

2y + 3 £1 y-5

Solve

k -1 = 1. k +2

Solve y - 5 = 0. y =5

Intervals: (-¥, - 8), (-8, 5), (5, ¥) For (-¥, -8), choose y = -10. 2(-10) + 3 17 = £ / 1 -10 - 5 15

-3 - 1 = 4>1 -3 + 2

For (-8, 5), choose y = 0.

For (-2, ¥), choose 0.

2(0) + 3 3 =- £1 0-5 5

0 -1 1 =- > / 1 0+2 2 The solution is (-¥, -2).

For (5, ¥), choose y = 6. 2(6) + 3 15 = £ / 1 6-5 1

a-5 < -1 a+2

Solve the equation

2y + 3 = 1. y-5 2y + 3 = y - 5 y = -8

k -1 = k + 2 -1 ¹ 2 The equation has no solution. Solve k + 2 = 0. k = -2 Intervals: (-¥, -2), (-2, ¥) For (-¥, -2), choose -3.

46.

)

The solution is -2, 32 .

The solution is [-8, 5). a-5 = -1. a+2

48.

a-5 = -1 a+2 a - 5 = -1(a + 2) a - 5 = -a - 2 2a = 3 3 a = 2 Set the denominator equal to zero and solve for a. a+2=0 a = -2

Intervals: (-¥, -2), ( -2, 23 ), ( 23 , ¥ )

a+2 £5 3 + 2a a+2

Solve the equation 3 + 2a = 5. a+2 =5 3 + 2a a + 2 = 5(3 + 2a) a + 2 = 15 + 10a -9a = 13 13 a =9 Set the denominator equal to zero and solve for a.

For (-¥, -2), choose -3. -3 - 5 -8 = =8< / -1 -3 + 2 -1

3 + 2a = 0 2a = -3 3 a =2

Chapter R ALGEBRA REFERENCE Intervals: ( -¥, - 23 ), ( - 23 , - 13 , - 13 , ¥ ) 9 ) ( 9

(

For (3, ¥), choose 4. 2(4) = 8 and 4-3 2(4) 4 . £ / 4-3 4-3

)

For -¥, - 32 , choose -2. -2 + 2 0 = =0£5 3 + 2(-2) -1

(

)

The solution is [2, 3).

, choose -1.46. For - 32 , - 13 9 -1.46 + 2 0.54 = = 6.75 £ / 5 3 + 2(-1.46) 0.08

(

50.

)

, ¥ , choose 0. For - 13 9

5 12 > p +1 p +1

Solve the equation

0+2 2 = £5 3 + 2(0) 3

since it would make the denominator zero. The ,¥ . solution is -¥, - 32 È éê - 13 ë 9 49.

)

The equation has no solution. Set the denominator equal to zero and solve for p.

)

p +1= 0 p = -1

2k 4 £ k -3 k -3 2k 4 Solve = . k -3 k -3

Intervals: (-¥, -1), (-1, ¥) For (-¥, -1), choose -2. 5 = -5 and -2 + 1 5 12 > . -2 + 1 -2 + 1

2k 4 = k -3 k -3 2k 4 =0 k -3 k -3 2k - 4 =0 k -3 2k - 4 = 0 k = 2

For (-1, ¥), choose 0. 5 = 5 and 0 +1 5 12 > / . 0 +1 0 +1

Set the denominator equal to 0 and solve for k. k -3= 0 k =3

51.

For (-¥, 2), choose 0. 4 4 = - , so 0-3 3

For (2, 3), choose 52 .

( ) = 5 = -10

2 52

5 -3 2

and

4 5 -3 2

2 ( 52 ) 5 -3 2

- 12

4 = -8, so - 12 4 . 3 2

£ 5

12 = -12, so -2 + 1

12 = 12, so 0 +1

The solution is (-¥, -1).

Intervals: (-¥, 2), (2, 3), (3, ¥)

2(0) = 0 and 0-3 2(0) 4 £ / . 0-3 0-3

5 12 = . p +1 p +1

5 12 = p +1 p +1 5 = 12

The value - 32 cannot be included in the solution

(

4 = 4, so 4-3

2x 2

x -x-6

Solve

³0

2x 2

x -x-6

= 0.

=0 x -x-6 2x = 0 x =0 2

Set the denominator equal to 0 and solve for x. x2 - x - 6 = 0 ( x + 2)( x - 3) = 0 x+2 = 0 or x - 3 = 0 x = -2 or x =3

Intervals: (-¥, -2), (-2, 0), (0, 3), (3, ¥)

Section R.5

For (-¥, -2), choose -3.

For (2, ¥), choose 3.

2(-3)

= -1 ³ / 0

(-3) - (-3) - 6

(3) + (2)(3)

2(-1) (-1) - (-1) - 6

1 ³0 2

53.

For (0, 3), choose 2. 2(2) 2

2 -2-6 For (3, ¥), choose 4. 2(4)

8 2

p + 2p

z2 - 1

8 2

p + 2p

= 3.

z 2 + z = 3z 2 - 3 4 ³0 3

-2 z 2 + z + 3 = 0 -1(2 z 2 - z - 3) = 0 -1( z + 1)(2 z - 3) = 0 3 z = -1 or z = 2

Solve the equation

³3

z2 - 1 Solve

z2 + z

4 -4-6 The solution is (-2, 0] È (3, ¥).

52.

z2 + z

= -1 ³ / 0

Set z 2 - 1 = 0.

= 1.

p + 2p

z2 = 1 z = -1 or z = 1

Intervals: (-¥, -1), (-1, 1), (1, 23 ),( 23 , ¥ )

8 = p2 + 2 p

For (-¥, -1), choose x = -2.

0 = p2 + 2 p - 8 0 = ( p + 4)( p - 2) p+4 =0 or p - 2 = 0 p = -4 or p = 2

For (-1, 1), choose x = 0.

Set the denominator equal to zero and solve for p.

02 + 3

p2 + 2 p = 0 p( p + 2) = 0 p = 0 or p + 2 = 0 p = -2 Intervals: (-¥, -4), (-4, -2), (-2, 0), (0, 2), (2, ¥) For (-¥, -4), choose -5.

02 - 1

8 2

(-5) + 2(-5)

(-2)2 + 3 2

(-2) - 1

22 + 3 2 -1

8 8 = = >1 2 9-6 3 (-3) + 2(-3) 8 = = -8 > / 1 2 -1 (-1) + 2(-1)

7 ³ / 3 3

The solution is 1, 32 ùú . û 54.

a 2 + 2a a2 - 4

£2 2

Solve the equation a 2+ 2a = 2. a -4

For (0, 2), choose 1. (1) + 2(1)

(

= -3 ³ / 3

( 32 ) + 3 = 21 ³ 3 2 ( 32 ) - 1 5 For ( 32 , ¥ ) , choose x = 2. 2

7 ³ / 3 3

( )

8 > / 1 15

For (-2, 0), choose -1.

For 1, 32 , choose x = 32 .

For (-4, -2), choose -3.

8 > / 1 15

The solution is (-4, -2) È (0, 2).

For (-2, 0), choose -1. 2

8 >1 3

Chapter R ALGEBRA REFERENCE a 2 + 2a a2 - 4

= 2

a 2 + 2a = 2(a 2 - 4)

-2

( (

) )

æ y 2 z-4 ö÷-2 y 2 z -4 y (2)(-2) z (-4)(-2) çç ÷÷ = = çç -3 4 ÷÷ -2 y (-3)(-2) z (4)(-2) èy z ø y -3 z 4

a 2 + 2a = 2a 2 - 8 0 = a 2 - 2a - 8 0 = (a - 4)(a + 2) a - 4 = 0 or a + 2 = 0

a = 4 or

a = -2

But -2 is not a possible solution. Set the denominator equal to zero and solve for a. a2 - 4 = 0 (a + 2)(a - 2) = 0 a + 2 = 0 or a - 2 = 0 a = -2 or a = 2

Intervals: (-¥, -2), (-2, 2), (2, 4), (4, ¥) For (-¥, -2), choose -3. (-3) 2 + 2(-3) 2

(-3) - 4

9-6 3 = £2 9-4 5

For (-2, 2), choose 0. (0) 2 + 2(0) 0 = = 0£2 -4 0-4

(3)2 - 4

9+6 15 = = £ / 2 9-5 4

For (4, ¥), choose 5. (5)2 + 2(5) (5)2 - 4

25 + 10 35 = = £2 25 - 4 21

The value 4 will satisfy the original inequality, but the values -2 and 2 will not since they make the denominator zero. The solution is (-¥, -2) È (-2, 2) È [4, ¥).

R.6 Exponents Your Turn 1

æ 2 ö÷-3 çç ÷ = çè 3 ø÷

1 27 = 8 = 3 8 2 27

1251/ 3 = 5, since 53 = 125.

Your Turn 4

16-3/ 4 = (161/ 4 )-3 = 2-3 =

1 8

æ x1/ 2 x 4 ö÷1/ 3 æ (1/ 2) + 4 ÷ö1/ 3 æ 9 / 2 ÷ö1/ 3 çç ÷÷ = çç x ÷÷ = çç x ÷ çç 3/ 2 ÷÷ çç çç 3/ 2 ÷÷÷ 3/ 2 ÷÷ è x ø è x ø èx ø

(

= x (9 / 2) -(3/ 2)

1/ 3

)

1/ 3

( )

= x3

= x 3(1/ 3) = x1 = x

Your Turn 6

Factor 5z1/3 + 4 z-2/3.

(

5z1/3 + 4 z-2/3 = z-2/3 5z (1/3)+(2/3) + 4 z (-2/3)+(2/3) = z-2/3(5z + 4)

R.6 Exercises 1.

1 1 8-2 = 2 = 64 8

1 1 3-4 = 4 = 81 3

50 = 1, by definition.

æ 3 ö÷0 çç - ÷ = 1, by definition. çè 4 ÷ø

-(-3)-2 = -

æ 1 ö æ 1ö 1 -(-3-2 ) = -çç - 2 ÷÷ = -çç - ÷÷÷ = çè 9 ø çè 3 ÷ø 9

æ 1 ö-2 ççç ÷÷÷ = è6ø

Your Turn 2 æ y 2 z-4 ö÷-2 Simplify ççç -3 4 ÷÷÷ . çè y z ÷ø

Your Turn 5

(3)

z8-(-8) z16 = = y 6 z -8 y 6-(-4) y10

Your Turn 3

For (2, 4), choose 3. (3) 2 + 2(3)

y -4 z 8

1 (-3)

1 2

( 16 )

1 9

1 = 1 = 36 36

)

Section R.6 8.

æ 4 ö÷-3 çç ÷ = çè 3 ÷ø

11.

1 27 = 64 = 3 64 4

(3)

18.

-7

8 ⋅8

5-2 m 2 y-2 = ⋅ ⋅ 52 m-1 y-2 52 m-1 y-2 = 5-2- 2 m2-(-1) y-2-(-2) = 5-2- 2 m2 +1 y-2 + 2 = 5-4 m3 y 0 =

= 89 + (-7)-(-3) = 89-7 +3 = 85

8-3

104 ⋅ 102 108+ (-10) -2 - 6

106

= 10

1 108

20. -1

æ 7-12 ⋅ 73 ö÷ çç ÷÷ çç ÷÷ -8 7 è ø

14.

15.

x 4 ⋅ x3 x5

y10 ⋅ y-4 y6

(4k -1)2 2k -5

x 4 +3

(-1)(-1)

c (3)(-2) c-6 = -2 (-2)(-2) = -2 4 7 d 7 d 72 49 = 6 4 = 6 4 c d c d

=7 =7

x7 x

= x 7 -5 = x 2 5

21.

1 1 + a b æ b öæ ö æ a öæ 1 ö 1 = çç ÷÷çç ÷÷ + çç ÷÷çç ÷÷ ÷ç b ÷ø çè b ÷øèç a ø÷ çè a øè

a-1 + b-1 =

= y10- 4-6 = y 0 = 1

b a + ab ab b+a = ab a+b = ab =

4 2 k -2 2 k -5

16k -2 - (-5) 2

= 8k -2 + 5 = 8k 3

16.

(3z 2 )-1 z5

= =

3-1( z 2 )-1

z5 -1 -2

3-1 ⋅ x ⋅ y 2 x-4 ⋅ y 5

-1

3-1 z 2(-1) z5

1 1 1 ⋅ 7 = 7 3 z 3z 1-(-4)

⋅x

⋅ y

= 3-1 ⋅ x1+ 4 ⋅ y-3 = =

22.

b-2 - a =

= 3-1 z-2-5

= 3-1 z-7 =

17.

æ c3 ö÷-2 (c 3 )-2 çç ÷÷ = çç -2 ÷÷ 7-2 (d -2 )-2 è 7d ø

= (7-12 +3+8 )-1 = (7-1)-1

13.

⋅ m3 ⋅ 1

æ a-1 ö÷-3 (a-1)-3 a (-1)(-3) çç ÷ = çç 2 ÷÷÷ = 2 -3 (b ) b2(-3) è b ø

= (7-12 +3-(-8) )-1

a3 = -6 = a 3b6 b

-8

= 10 =

10-2

104 + 2

19.

108 ⋅ 10-10

12.

5-2 m2 y-2

4-2 1 1 = 4-2-1 = 4-3 = 3 = 4 64 4 9

10.

= =

2-5

1 5 1 ⋅x ⋅ 3 3 y x5 3 y3

1 b2

-a

æ b 2 ö÷ çç ÷ a çç 2 ÷÷ 2 b è b ø÷ 1 1 b

ab 2 b2

1 - ab 2 b2

26 23.

Chapter R ALGEBRA REFERENCE 2

2n-1 - 2m-1

æ xy - 1 ö÷-2 ÷÷ = ççç çè y 2 ÷÷ø

- m = n m + n2

m + n2

2 ⋅ m - 2 ⋅ n m m n 2

= n =

or -1

24.

æ m ö÷ çç ÷ èç 3 ø÷

-2

(m + n ) 2m - 2n

mn(m + n 2 ) 2(m - n)

mn(m + n 2 ) 1

ænö + çç ÷÷ èç 2 ø÷

æ y 2 ö÷2 ç ÷÷ = çç çè xy - 1 ÷÷ø

= 2

æ3ö æ2ö = çç ÷÷ + çç ÷÷ èç m ø÷ èç n ø÷

3 4 + 2 m n æ æ 3 öç n 2 ö÷ æ 4 öæ m ö = çç ÷÷çç 2 ÷÷÷ + çç 2 ÷÷çç ÷÷ èç m ø÷çè n ø÷ çè n ÷øçè m ø÷ 3n

4m mn 2

3n 2 + 4m

25.

29.

322/5 = (321/5 )2 = 22 = 4

30.

-1252 / 3 = -(1251/ 3 )2 = -52 = -25

31.

æ 36 ö÷1/ 2 361/ 2 6 1 çç = = = ÷ 1/ 2 çè 144 ÷ø 12 2 144

mn 2

This can also be solved by reducing the fraction first. æ 36 ö÷1/2 æ 1 ö1/ 2 11/2 1 çç = çç ÷÷ = 1/2 = ÷ çè 144 ÷ø çè 4 ÷ø 2 4

1 1 ⋅ y - 1 ⋅ x x y y x

x - xy xy

1 = y- x

32.

1/3 æ 64 ö÷1/3 çç ÷ = 64 = 4 ÷ çè 27 ø 3 271/3

33.

8-4 / 3 = (81/3 )-4 = 2-4 =

34.

625-1/4 =

35.

æ 27 ö÷-1/3 27-1/3 641/3 4 çç ÷ = = = çè 64 ÷ø 3 64-1/3 271/3

26.

( xy - 1) 2

271/ 3 = 3 27 = 3

1 - 1y x

= y

28.

( x-1 - y-1)-1 = 1 =

( xy - 1) 2

27. 1211/2 = (112 )1/2 = 112(1/2) = 111 = 11

( y 2 )2

xy y-x

-2 æx 1 ö÷ ( x ⋅ y-1 - y-2 )-2 = ççç - 2 ÷÷ çè y y ÷÷ø

36.

-2 é æ x öæ y ö 1 ùú ÷ ÷ ç ç ê = ê ç ÷÷ç ÷÷ - 2 ú y úû êë çè y ÷øèç y ÷ø -2 æ xy 1 ö÷ = ççç 2 - 2 ÷÷ çè y y ÷÷ø

1/4

625

1 2

1 16

1 5

æ 121 ö÷-3/2 1 1 çç = = 3/2 çè 100 ÷÷ø 121 é 121 1/2 ù 3 ê 100 ú 100 ëê ûú =

( )

1 1000 = 1331 = 1331 1000

( ) 11 10

37.

32/3 ⋅ 34 / 3 = 3(2/3) + (4/3) = 36/3 = 32 = 9

38.

272/3 ⋅ 27-1/3 = 27(2/3) + (-1/3) = 272 / 3-1/ 3 = 271/ 3 = 3

Section R.6 39.

49/4 ⋅ 4-7/4

= 49/4-7/4-(-10/4)

-10/4

45.

= 412/4 = 43 = 64

40.

3-5/2 ⋅ 33/2 37/2 ⋅ 3-9/2

41/2 ⋅ k 7/2 =

= 3(-5/2) + (3/ 2)-(7/2)-(-9/2)

= 30 = 1

æ x6 y-3 ö÷1/ 2 çç ÷ = ( x6-(-2) y-3-5 )1/ 2 çç -2 5 ÷÷ è x y ø÷

3k 2 ⋅ 4-1k 3 2 ⋅ k 7/2

= 3 ⋅ 2-1 ⋅ 4-1k 2 + 3-(7/2) 3 = ⋅ k 3/2 8

= 3-5/2 + 3/2-7/2 + 9 / 2

41.

3k 2 ⋅ (4k -3 )-1

46.

3k 3/2 8

8 p-3 (4 p 2 )-2 p-5

= ( x8 y-8 )1/ 2 = ( x8 )1/ 2 ( y-8 )1/ 2

= x 4 y -4 = æ a-7b-1 ö÷ çç ÷ çç -4 2 ÷÷÷ è b a ø

= (a-7- 2b-1-(-4) )1/ 3

= a-3b1 b = 3 a

47.

7-1/3 ⋅ 7r -3

7-1/3+1r -3 = 2/3 -4 2/3 -2 2 7 ⋅ (r ) 7 ⋅r = 70 r -3+ 4 = 1 ⋅ r1 = r

48.

128/4 ⋅ y-2

12-1 y 6

= 122-(-1) ⋅ y-2-6 = 123 y-8 =

⋅

b1/2 -3/2

1 2 p2

= a 4/3- 2/3b1/2-(-3/2)

x3/2 ⋅ y 4/5 ⋅ z-3/4 x5/3 ⋅ y-6/5 ⋅ z1/2

12-1 ⋅ y 6

122 ⋅ y-2

123

2/3

16 p

= x-1/6 ⋅ y 2 ⋅ z-5/4

12-1 ⋅ ( y-3 )-2 12-1 ⋅ y (-3)(-2)

a 4/3

= x3/2-(5/3) ⋅ y 4/5-(-6/5) ⋅ z-3/4-(1/2)

123/4 ⋅ 125/4 ⋅ y-2

p-5

= a 2/3b 2

= 7-1/3+3/3- 2/3 r -3-(-4)

123/4 + 5/4 ⋅ y-2

8 p-3 4-2 p-4

= 8 ⋅ 4-2 p-2 1 1 = 8⋅ 2 ⋅ 2 p 4 1 1 = 8⋅ ⋅ 2 16 p

= (a-9 )1/ 3 (b3 )1/ 3

44.

p -5

= 8 ⋅ 4-2 p-3-4 + 5

= (a-9b3 )1/ 3

43.

8 p-3 ⋅ 4-2 p (2) (-2)

= 8 ⋅ 4-2 p (-3) + (-4)-(-5)

1/ 3

42.

49.

y2 x1/6 z 5/4

k -3/5 ⋅ h-1/3 ⋅ t 2/5 k -1/5 ⋅ h-2/3 ⋅ t1/5 = k -3/5-(-1/5) h-1/3-(-2/3)t 2/5-1/5 = k -3/5+1/5h-1/3+ 2/3t 2/5-1/5 = k -2/5h1/3t1/5 =

h1/3t1/5 k 2/5

28 50.

Chapter R ALGEBRA REFERENCE m7/3 ⋅ n-2/5 ⋅ p3/8

56.

(4 x 2 + 1)2 (2 x - 1)-1/2

m-2/3 ⋅ n3/5 ⋅ p-5/8

+ 16 x(4 x 2 + 1)(2 x - 1)1/ 2

= m7/3-(-2/3) n-2/5-(3/5) p3/8-(-5/8)

= (4 x 2 + 1)(4 x 2 + 1)(2 x - 1)-1/2

= m7/3+ 2/3n-2/5-3/5 p3/8+5/8

+ 16 x(4 x 2 + 1)(2 x - 1)-1/2 (2 x - 1)

= m9/3n-5/5 p8/8

= (4 x 2 + 1)(2 x - 1)-1/2

= m3n-1 p1 =

51.

⋅ [(4 x 2 + 1) + 16 x(2 x - 1)]

m3 p n

= (4 x 2 + 1)(2 x - 1)-1/2

⋅ (4 x 2 + 1 + 32 x 2 - 16 x)

3x 3 ( x 2 + 3x )2 - 15x( x 2 + 3x )2

= (4 x 2 + 1)(2 x - 1)-1/2 (36 x 2 - 16 x + 1)

= 3x ⋅ x 2 ( x 2 + 3x )2 - 3x ⋅ 5( x 2 + 3x )2 = 3x( x 2 + 3x )2 ( x 2 - 5)

52.

6 x( x 3 + 7)2 - 6 x 2 (3x 2 + 5)( x 3 + 7) = 6 x( x 3 + 7)( x 3 + 7) - 6 x( x )(3x 2 + 5)( x 3 + 7) = 6 x( x 3 + 7)[( x 3 + 7) - x(3x 2 + 5)] = 6 x( x 3 + 7)( x 3 + 7 - 3x 3 - 5x )

R.7 Radicals Your Turn 1

28 x 9 y 5 .

Simplify

28 x 9 y 5 =

= 2 x 4 y 2 7 xy

= 6 x( x 3 + 7)(-2 x 3 - 5x + 7)

53. 10 x 3 ( x 2 - 1)-1/2 - 5x( x 2 - 1)1/ 2 = 5x ⋅ 2 x 2 ( x 2 - 1)-1/2 - 5x( x 2 - 1)-1/2 ( x 2 - 1)1 2

-1/2

= 5x( x - 1)

Your Turn 2

Rationalize the denominator in

[2 x - ( x - 1)]

= 5x( x 2 - 1)-1/ 2 ( x 2 + 1)

54.

4 ⋅ x8 ⋅ y 4 ⋅ 7 xy

5 x-

9(6 x + 2)1/2 + 3(9 x - 1)(6 x + 2)-1/2

5 x-

= 3(6 x + 2)-1/2[3(6 x + 2) + (9 x - 1)] = 3(6 x + 2)-1/2 (18 x + 6 + 9 x - 1) = 3(6 x + 2)-1/2 (27 x + 5)

55.

x(2 x + 5)2 ( x 2 - 4)-1/2 + 2( x 2 - 4)1/2 (2 x + 5) = (2 x + 5)2 ( x 2 - 4)-1/2 ( x )

y y

Your Turn 3

4+ x 4+ x 4- x = ⋅ 16 - x 16 - x 4 - x 16 - x 1 = = (16 - x)(4 - x ) 4- x

R.7 Exercises 125 = 5 because 53 = 125.

= (2 x + 5)( x 2 - 4)-1/2

4 1296 = 4 64 = 6

⋅ [(2 x + 5)( x ) + ( x 2 - 4)(2)] = (2 x + 5)( x 2 - 4)-1/2 ⋅ (2 x 2 + 5x + 2 x 2 - 8)

= (2 x + 5)( x 2 - 4)-1/ 2 (4 x 2 + 5x - 8)

x-y

+ ( x 2 - 4)1( x 2 - 4)-1/2 (2)(2 x + 5)

( x + y)

= 3 ⋅ 3(6 x + 2)-1/2 (6 x + 2)1 + 3(9 x - 1)(6 x + 2)-1/ 2

x+ x+

⋅

5 x-

-3125 = -5 because (- 5)5 = -3125.

50 =

25 ⋅ 2 =

2000 =

4 ⋅ 100 ⋅ 5

= 2 ⋅ 10 5 = 20 5

25 2 = 5 2

Section R.7

32 y 5 =

(16 y 4 )(2 y)

14.

23 5 - 43 40 + 33 135 = 23 5 - 43 8 ⋅ 5 + 33 27 ⋅ 5

= 16 y 4 2 y

= 23 5 - 4(2) 3 5 + 3(3) 3 5

= 4 y2 2 y

= 2 3 5 - 83 5 + 9 3 5

27 ⋅ 3 =

81 = 9

2 ⋅ 32 =

64 = 8

7 2 - 8 18 + 4 72 = 7 2 - 8 9 ⋅ 2 + 4 36 ⋅ 2

= 33 5

15.

2 x3 y 2 z 4 =

16.

160r 7 s 9t12

= 7 2 - 8(3) 2 + 4(6) 2

(16 ⋅ 10)( r 6 ⋅ r )( s8 ⋅ s )(t12 )

= 7 2 - 24 2 + 24 2

(16r 6 s8t12 )(10rs)

16r 6 s8t12 10rs

=7 2

10.

= 4r 3s 4t 6 10rs

4 3 - 5 12 + 3 75 = 4 3 - 5( 4 3) + 3( 25 3) = 4 3 - 5(2 3) + 3(5 3)

11.

17.

128 x3 y8 z 9 = 3 64 x3 y 6 z 9 ⋅ 2 y 2

= 4 3 - 10 3 + 15 3

= 3 64 x3 y 6 z 9 3 2 y 2

= (4 - 10 + 15) 3 = 9 3

= 4 xy 2 z 3 3 2 y 2

4 7 - 28 + 343 = 4 7 - 4 7 +

18.

4 x8 y 7 z11 = 4 ( x8 )( y 4 ⋅ y 3 )( z 8 z 3 )

49 7

= 4 ( x8 y 4 z 8 )( y3 z 3 )

= 4 7 -2 7 +7 7

12.

= (4 - 2 + 7) 7

= 4 x8 y 4 z 8 4 y 3 z 3

=9 7

= x 2 yz 2 4 y 3 z 3

3 28 - 4 63 + 112

19.

= 3( 4 7) - 4( 9 7) + ( 16 7)

a 2b 4ab - 2 a 6b 2ab +

a 2b8ab

= (ab 2 - 2a3b + ab 4 ) ab

= (6 - 12 + 4) 7

= ab ab (b - 2a 2 + b3 )

= -2 7 2 - 3 16 + 23 54

a3b9

= ab 2 ab - 2a3b ab + ab 4 ab

= 6 7 - 12 7 + 4 7

a3b5 - 2 a 7b3 + =

= 3(2 7) - 4(3 7) + (4 7)

13.

x 2 y 2 z 4 ⋅ 2 x = xyz 2 2 x

20.

p 7 q3 -

p5q 9 +

p9 q

= 3 2 - ( 3 8 ⋅ 2) + 2( 3 27 ⋅ 2)

( p 6 p)(q 2q) - ( p 4 p)(q8q) + ( p8 p)q

= 3 2 - 3 8 3 2 + 2( 3 27 3 2)

( p 6q 2 )( pq) - ( p 4q8 ) ( pq) + ( p8 ) pq

= 3 2 - 23 2 + 2(33 2)

p 6q 2 pq -

= 3 2 - 23 2 + 63 2

= p3q pq - p 2q 4 pq + p 4 pq

= 53 2

= p 2 pq pq - p 2q 4 pq + p 2 p 2 pq

p 4q8 pq +

= p 2 pq ( pq - q 4 + p 2 )

21.

a ⋅ 3 a = a1/2 ⋅ a1/3

= a1/2 + (1/3) = a5/6 =

6 5

p8 pq

30 22.

Chapter R ALGEBRA REFERENCE b3 ⋅ 4 b3 = b3/2 ⋅ b3/4

32.

= b3/2 + (3/4) = b9/4 = 4 b9 = 4 b8 ⋅ b

( )

5(2 + 6) 4 + 2 6 - 2 6 - 36

5(2 + 6) 4 - 36

= 4 b8 4 b = b2 4 b

23.

16 - 8 x + x 2 =

(4 - x)

5(2 + 6) 4-6 5(2 + 6) = -2 5(2 + 6) =2 =

=| 4 - x|

24.

9 y 2 + 30 y + 25 =

25.

4 - 25z 2 =

(3 y + 5) 2 = |3 y + 5|

(2 + 5z )(2 - 5z )

This factorization does not produce a perfect square, so the expression 26.

4 - 25 z

cannot be simplified.

33.

and cannot be factored. Therefore, cannot be simplified. 27.

5 5 = ⋅ 7 7

28.

5 = 10

6 6 2- 2 = ⋅ 2+ 2 2+ 2 2- 2 6(2 - 2) 4-2 2 +2 2 - 4

9k 2 + h 2 The expression 9k 2 + h 2 is the sum of two squares

29.

5 5 2+ 6 = ⋅ 2- 6 2- 6 2+ 6

9k + h

7 5 7 = 7 7

6(2 - 2) 6(2 - 2) = 4-2 2 = 3(2 - 2) =

34.

5 = 5 + 2

5 10 5 10 ⋅ = 10 10 100

5 10 5 10 = 10 100

10 2

-3 = 12

-3 4⋅3

-3 ⋅ 2 3

3 -3 3 3 = =6 2 3

4 4 = ⋅ 8 8

31.

3 3 1+ = ⋅ 1- 2 1- 2 1+

5- 2 5- 2

5( 5 - 2) 25 - 10 + 10 - 4

5 - 10 5-2 5 - 10 = 3

35.

30.

5 ⋅ 5 + 2

2 4 2 4 2 = = = 4 2 16

1 = r - 3 =

36.

5 = m- 5 =

2 2

3(1 + 2) 1- 2 = -3(1 + 2) =

1 ⋅ r - 3

r + 3 r + 3

r + 3 r-3

5 ⋅ m- 5

m + 5 m + 5

5( m + 5) 2

( m ) + 5m - 5m - 25 5( m + 5) 2

( m ) - 25 5( m + 5) m-5

Section R.7 37.

y-5 = y - 5

42.

3- 3 3- 3 3+ 3 = ⋅ 6 6 3+ 3

( y - 5)( y + 5) y-5

38.

y + 5 y + 5

y-5 ⋅ y - 5

y + 5

z -1 = z - 5

z -1 ⋅ z - 5

= =

z + 5 z + 5

( z ) + 5z - z - 5 = 2 ( z ) + 5z - 5z - 25 z + 5z - z - 5 z-5

39.

x + x -

x +1 = x +1

x + x -

x +1 ⋅ x +1

43. x + x +

x +1 x +1

2 x + 2 x( x + 1) + 1 -1 = -2 x - 2 x( x + 1) - 1 p +

p2 - 1

p -

p2 - 1

44.

p2 - 1

p -

p2 - 1

⋅

p +

p2 - 1

p +

p2 - 1

( p) + 2

= =

41.

p -1 -

p + 2 p p 2 - 1 + ( p 2 - 1)

p -

(

p -1

)

= =

p 2 + p + 2 p( p 2 -1) - 1

- p2 + p + 1

(

)(

(

1- 2

2 1- 2

(

)

2 1- 2

)

1 3+ 3

p-2 ⋅ p p

)

p + p +

p-2 p-2

p -2 -

( p) +

p -2

p - ( p - 2) p + p( p - 2)

p - ( p 2 - 1)

1+ 2 1- 2 1+ 2 = 2 2 1- 2

(

6 3+ 3

p-2 p

)

x +1 x +1

6 3+ 3

( p) + =

( p ) + 2 p p - 1 + ( p - 1) 2

p -

p +

(

9-3

)

x +1 x - x +1 ⋅ x +1 x - x +1 x - ( x + 1) = x - 2 x ⋅ x + 1 + ( x + 1) -1 = 2 x - 2 x( x + 1) + 1

40.

(

6 3+ 3

x + x -

x + 2 x( x + 1) + ( x + 1) x - ( x + 1)

x + x -

9+3 3-3 3- 9

2 p( p - 2)

( p - 2)

Chapter 1

LINEAR FUNCTIONS 1.1 Slopes and Equations of Lines Your Turn 1 Find the slope of the line through (1, 5) and (4, 6). Let ( x1, y1) = (1, 5) and ( x2 , y2 ) = (4, 6). m=

6-5 1 = 4 -1 3

First find the slope of the line 3x - 6 y = 7 by solving this equation for y. 3x - 6 y = 7

Your Turn 2 Find the equation of the line with x-intercept -4 and y-intercept 6. We know that b = 6 and that the line crosses the axes at (-4, 0) and (0, 6). Use these two intercepts to find the slope m. m=

Your Turn 5 Find (in slope-intercept form) the equation of the line that passes through the point (4, 5) and is parallel to the line 3x - 6 y = 7.

6-0 6 3 = = 0 - (-4) 4 2

6 y = 3x - 7 y = 3x- 7 6 6 1 y = x-7 2 6 Since the line we are to find is parallel to this line, it will also have slope 1/2. Use the point-slope form with ( x1, y1) = (4, 5).

y - y1 = m( x - x1)

Thus the equation for the line in slope-intercept form is 3 y = x + 6. 2

y - 5 = 1 ( x - 4) 2 y-5 = 1x-2 2 1 y = x+3 2

Your Turn 3 Find the slope of the line whose equation is 8 x + 3 y = 5. Solve the equation for y.

Your Turn 6 Find (in slope-intercept form) the equation of the line that passes through the point (3, 2) and is perpendicular to the line 2 x + 3 y = 4.

8x + 3 y = 5 3 y = -8 x + 5 8 5 y =- x+ 3 3

First find the slope of the line 2 x + 3 y = 4 by solving this equation for y.

The slope is -8/3. Your Turn 4 Find the equation (in slope-intercept form) of the line through (2, 9) and (5, 3). First find the slope. 3-9 -6 = = -2 5-2 3 Now use the point-slope form, with ( x1, y1) = (5, 3).

y - y1 = m( x - x1) y - 3 = -2( x - 5) y - 3 = -2 x + 10 y = -2 x + 13

2x + 3 y = 4 3 y = -2 x + 4 y = -2 x + 4 3 3 Since the line we are to find is perpendicular to a line with slope –2/3, it will have slope 3/2. (Note that (-2/3)(3/2) = -1.)

Use the point-slope form with ( x1, y1) = (3, 2). y - y1 = m( x - x1) y - 2 = 3 ( x - 3) 2 y-2= 3x-9 2 2 3 5 y = x2 2

Chapter 1 LINEAR FUNCTIONS

1.1 Warmup Exercises W1.

15 - (-3) 18 = = -3 -2 - 4 -6

W2. y - (-3) = -2( x + 5) y + 3 = -2 x - 10 y = -2 x - 13

1 2æ 1ö = çç x + ÷÷÷ ç 2 5è 3ø 1 2 2 y- = x+ 2 5 15 2 4 15 + y = x+ 5 30 30 2 19 y = x+ 5 30

10.

11.

y = 3x - 2

5x - 9 y = 11

Rewrite the equation in slope-intercept form. 9 y = 5x - 11 5 11 y = x9 9

-3 y = -2 x + 7 2 7 y = x3 3

The slope is 95 . 12.

1.1

Exercises

False. Any nonvertical line can be represented in the form y = mx + b.

True

False. A vertical line has an undefined slope. A horizontal line has a slope of 0.

False. If two lines are perpendicular, the product of their slopes is –1.

Find the slope of the line through (4, 5) and (-1, 2).

4x + 7 y = 1

Rewrite the equation in slope-intercept form. 7 y = 1 - 4x 1 1 1 (7 y ) = (1) - (4 x ) 7 7 7 1 4 y = - x 7 7 4 1 y =- x+ 7 7 The slope is - 74 .

5-2 4 - (-1) 3 = 5

13.

Find the slope of the line through (5, -4) and

15.

x 5

This is a vertical line. The slope is undefined. 14. The x-axis is the horizontal line y = 0. Horizontal lines have a slope of 0.

(1, 3).

y =8

This is a horizontal line, which has a slope of 0.

3 - (-4) m= 1- 5 3+4 = -4 7 =4

16.

y = -6

By rewriting this equation in the slope-intercept form, y = mx + b, we get y = 0 x - 6, with the slope, m, being 0. 17. Find the slope of a line parallel to 6 x - 3 y = 12.

Find the slope of the line through (8, 4) and (8, -7). 4 - (-7) 11 = 8-8 0 The slope is undefined; the line is vertical. m=

y = x

This equation is in slope-intercept form, y = mx + b. Thus, the coefficient of the x-term, 3, is the slope.

W4. 2 x - 3 y = 7

5-5 0 = =0 -2 - 1 -3

Using the slope-intercept form, y = mx + b, we see that the slope is 1.

W3. y -

Find the slope of the line through (1, 5) and (-2, 5).

Rewrite the equation in slope-intercept form. -3 y = -6 x + 12 y = 2x - 4 The slope is 2, so a parallel line will also have slope 2.

Section 1.1

18. Find the slope of a line perpendicular to 8 x = 2 y - 5.

First, rewrite the given equation in slope-intercept form. 8x = 2 y - 5 8x + 5 = 2 y 5 5 4 x + = y or y = 4 x + 2 2 Let m be the slope of any line perpendicular to the given line. Then. 4 ⋅ m = -1 1 m =- . 4 19. The line goes through (1, 3), with slope m = -2. Use point-slope form. y - 3 = -2( x - 1) y = -2 x + 2 + 3 y = -2 x + 5

20. The line goes through (2, 4), with slope m = -1. Use point-slope form.

24. The line goes through (8, -1) and (4, 3). Find the slope, then use point-slope form with either of the two given points. 3 - (-1) 4-8 3+1 = -4 4 = = -1 -4 y - (-1) = -1( x - 8) m=

y + 1 = -x + 8 y = -x + 7

25. The line goes through m =

3 2 3 y = 6x - - 2 2 3 4 y = 6x - 2 2 7 y = 6x 2

(

)

26. The line goes through -2, 34 and

( 23 , 52 ).

5 - 3 10 - 3 4 4 m = 22 = 24 6 ( 2) + 3 3 3

22. The line goes through (-8, 1), with undefined slope. Since the slope is undefined, the line is vertical. The equation of the vertical line passing through (-8, 1) is x = -8.

3-2 1 =1- 4 3 1 y - 3 = - ( x - 1) 3 1 1 y =- x+ +3 3 3 1 10 y =- x+ 3 3

- 42 - 12

y + 2 = 6x -

y = -x + 6

æ 1ö y - (-2) = 6 çç x - ÷÷÷ çè 4ø

y - 4 = -x + 2

23. The line goes through (4, 2) and (1, 3). Find the slope, then use point-slope form with either of the two given points.

-2 - 12

3 - 8 1 - 2 4 3 12 12 -5 60 m = 25 = =6 - 12 10

y - 4 = -1( x - 2)

21. The line goes through (-5, - 7) with slope m = 0. Use point-slope form. y - (-7) = 0[ x - (-5)] y+7 = 0 y = -7

( 23 , 12 ) and ( 14 , -2 ).

= 84 = 3

21 32

3 21 = [ x - (-2)] 4 32 3 21 42 y- = x+ 4 32 32 21 42 3 y = x+ + 32 32 4 21 21 12 y = x+ + 32 16 16 21 33 y = x+ 32 16 y-

Chapter 1 LINEAR FUNCTIONS

27. The line goes through (-8, 4) and (-8, 6). m=

4-6 -2 = ; -8 - (-8) 0

Use m = - 32 and the point (-4, 6) in the pointslope form. 3 y - 6 = - [ x - (-4)] 2 3 y = - ( x + 4) + 6 2 3 y =- x-6+6 2 3 y =- x 2 2 y = -3 x

which is undefined. This is a vertical line; the value of x is always -8. The equation of this line is x = -8. 28. The line goes through (-1, 3) and (0, 3). m=

3-3 0 = =0 -1 - 0 -1

This is a horizontal line; the value of y is always 3. The equation of this line is y = 3. 29. The line has x-intercept -6 and y-intercept -3. Two points on the line are (-6, 0) and (0, -3). Find the slope; then use slope-intercept form. 1 -3 - 0 -3 = =0 - (-6) 6 2 b = -3 1 y = - x-3 2 2 y = -x - 6

m =

x + 2 y = -6

30. The line has x-intercept –2 and y-intercept 4. Two points on the line are (-2, 0) and (0, 4). Find the slope; then use slope-intercept form. 4-0 4 m= = = 2 0 - (-2) 2 y = mx + b y = 2x + 4 2 x - y = -4 31. The vertical line through (-6, 5) goes through the point (-6, 0), so the equation is x = -6. 32. The line is horizontal, through (8, 7). The line has an equation of the form y = k where k is the y-coordinate of the point. In this case, k = 7, so the equation is y = 7. 33. Write an equation of the line through (-4, 6), parallel to 3x + 2 y = 13.

Rewrite the equation of the given line in slopeintercept form. 3x + 2 y = 13 2 y = -3x + 13 3 13 y =- x+ 2 2 The slope is - 32 .

3x + 2 y = 0

34. Write the equation of the line through (2, -5), parallel to y - 4 = 2 x. Rewrite the equation in slope-intercept form. y - 4 = 2x y = 2x + 4

The slope of this line is 2. Use m = 2 and the point (2, -5) in the pointslope form. y - (-5) = 2( x - 2) y + 5 = 2x - 4 y = 2x - 9 2x - y = 9 35. Write an equation of the line through (3, -4), perpendicular to x + y = 4.

Rewrite the equation of the given line as y = -x + 4. The slope of this line is -1. To find the slope of a perpendicular line, solve -1m = -1. m=1 Use m = 1 and (3, -4) in the point-slope form. y - (-4) = 1( x - 3) y = x-3-4 y = x-7 x- y =7

36. Write the equation of the line through (-2, 6), perpendicular to 2 x - 3 y = 5.

Rewrite the equation in slope-intercept form. 2x - 3 y = 5 -3 y = -2 x + 5 2 5 y = x3 3

Section 1.1

37 æ 2 öù 1é y - 0 = - ê x - çç - ÷÷÷ ú èç 3 ø úû 2 êë 1æ 2ö y = - çç x + ÷÷÷ ç 2è 3ø

The slope of this line is 23 . To find the slope of a perpendicular line, solve 2 m = -1. 3 3 m=2

1 1 y =- x2 3 6 y = -3 x - 2 3 x + 6 y = -2

Use m = - 32 and (-2, 6) in the point-slope form. 3 y - 6 = - [ x - (-2)] 2 3 y - 6 = - ( x + 2) 2 3 y-6 = - x-3 2 3 y =- x+3 2 2 y = -3 x + 6 3x + 2 y = 6

37. Write an equation of the line with y-intercept 4, perpendicular to x + 5 y = 7.

39. Do the points (4, 3), (2, 0), and (-18, -12) lie on the same line? Find the slope between (4, 3) and (2, 0). m =

Find the slope between (4, 3) and (-18, -12). -12 - 3 -15 15 = = -18 - 4 -22 22 Since these slopes are not the same, the points do not lie on the same line. m =

40. (a) Write the given line in slope-intercept form.

Find the slope of the given line. x + 5y = 7 5 y = -x + 7 1 7 y =- x+ 5 5

2x + 3 y = 6 3 y = -2 x + 6 2 y =- x+2 3

This line has a slope of - 23 . The desired line

The slope is - 15 , so the slope of the perpendicular

has a slope of - 23 since it is parallel to the

line will be 5. If the y-intercept is 4, then using the slope-intercept form we have y = mx + b y = 5 x + 4, or 5x - y = -4

given line. Use the definition of slope. m =

2 2 - (-1) = k-4 3 2 3 - = k-4 3 -2(k - 4) = (3)(3)

Find the slope of the given line. 2x - y = 4 2x - 4 = y

-2k + 8 = 9 -2 k = 1

The slope of this line is 2. Since the lines are perpendicular, the slope of the needed line is - 12 . The line also has an x-intercept of - 23 . Thus, it

)

passes through the point - 23 , 0 .

y2 - y1 x2 - x1

38. Write the equation of the line with x-intercept - 23 , perpendicular to 2 x - y = 4.

(

0-3 -3 3 = = 2-4 -2 2

1 2 Write the given line in slope-intercept form. 5 x - 2 y = -1 k =-

(b)

2 y = 5x + 1

Using the point-slope form, we have

y =

5 1 x+ 2 2

This line has a slope of 52 . The desired line has a slope of - 52 since it is perpendicular to the given line. Use the definition of slope.

Chapter 1 LINEAR FUNCTIONS m =

y2 - y1 x2 - x1

2 - (-1) k-4 2 2 +1 - = k-4 5 3 -2 = k-4 5 -2(k - 4) = (3)(5)

The product of the slopes is (1)(-1) = -1, so the diagonals are perpendicular.

43. The line goes through (0, 2) and (-2, 0) m=

41. A parallelogram has 4 sides, with opposite sides parallel. The slope of the line through (1, 3) and (2, 1) is 3-1 m= 1- 2 2 = -1 = -2.

)

(

The slope of the line through - 52 , 2 and - 72 , 4 is m=

2-4

- 52 -

( ) - 72

-2 = -2. 1

Since these slopes are equal, these two sides are parallel.

(

)

The slope of the line through - 72 , 4 and (1, 3) is

)

44. The line goes through (1, 3) and (2, 0). 3-0 3 m= = = -3 1- 2 -1 The correct choice is (f). 45. The line appears to go through (0, 0) and (-1, 4). m=

Slope of the line through m =

(

) and (2, 1) is

2 -1 1 2 = 9 =- . - 52 - 2 -2 9

Since these slopes are equal, these two sides are parallel. Since both pairs of opposite sides are parallel, the quadrilateral is a parallelogram. 42. Two lines are perpendicular if the product of their slopes is -1.

The slope of the diagonal containing (4, 5) and (-2, -1) is 5 - (-1) 6 = = 1. m= 4 - (-2) 6

The slope of the diagonal containing (-2, 5) and (4, -1) is

4-0 4 = = -4 -1 - 0 -1

46. The line goes through (-2, 0) and (0, 1). 1- 0 1 m= = 0 - (-2) 2 47. (a) See the figure in the textbook. Segment MN is drawn perpendicular to segment PQ. Recall that MQ is the length of segment MQ.

m1 =

4-3 1 2 m= 7 = 9 =- . 9 -2 -1 -2 - 52 , 2

2-0 2 = =1 0 - (-2) 2

The correct choice is (a).

-2k + 8 = 15 -2k = 7 7 k =2

(

5 - (-1) 6 = = -1. -2 - 4 -6

y MQ = x PQ

From the diagram, we know that PQ = 1. MQ

Thus, m1 = 1 , so MQ has length m1. y -QN -QN (b) m2 = = = PQ x 1 QN = -m2 (c)

Triangles MPQ, PNQ, and MNP are right triangles by construction. In triangles MPQ and MNP, angle M = angle M , and in the right triangles PNQ and MNP, angle N = angle N .

Since all right angles are equal, and since triangles with two equal angles are similar, triangle MPQ is similar to triangle MNP and triangle PNQ is similar to triangle MNP. Therefore, triangles MNQ and PNQ are similar to each other.

Section 1.1

(d) Since corresponding sides in similar triangles are proportional, MQ = k ⋅ PQ and

x y + =1 a b y 0+ =1 b y =1 b y = b

PQ = k ⋅ QN .

MQ k ⋅ PQ = PQ k ⋅ QN MQ PQ = PQ QN

From the diagram, we know that PQ = 1. MQ =

(c)

1 QN

From (a) and (b), m1 = MQ and -m2 = QN .

49.

Substituting, we get

The y-intercept is b. If the equation of a line is written as x y + = 1 , we immediately know the a b intercepts of the line, which are a and b.

y = x -1

Three ordered pairs that satisfy this equation are (0, -1), (1, 0), and (4, 3). Plot these points and draw a line through them.

1 m1 = . -m2

Multiplying both sides by m2 , we have m1m2 = -1. 48. (a) Multiplying both sides of the equation x y + = 1 by ab, we have a b æxö æ yö ab çç ÷÷÷ + ab çç ÷÷÷ = ab (1) èç a ø èç b ø

50.

bx + ay = ab.

Three ordered pairs that satisfy this equation are (-2, -3), (-1, 1), and (0, 5). Plot these points and draw a line through them.

Solve this equation for y. bx + ay = ab ay = ab - bx ab - bx y = a b y =- x+b a b If we let m = - , then the equation a becomes y = mx + b.

(b)

Let y = 0 .

y = 4x + 5

y y = 4x + 5

1 –1

51.

y = -4 x + 9

Three ordered pairs that satisfy this equation are (0, 9), (1, 5), and (2, 1). Plot these points and draw a line through them.

x y + =1 a b x +0 =1 a x =1 a x = a The x-intercept is a. Let x = 0 .

40 52.

Chapter 1 LINEAR FUNCTIONS y = -6 x + 12

Plot the ordered pairs (-3, 0) and (0, 9) and draw a line through these points. (A third point may be used as a check.)

There ordered pairs that satisfy this equation are (0, 12), (1, 6), and (2, 0). Plot these points and draw a line through them.

y 3x – y = –9

y = –6x + 12

55. 53.

2 x - 3 y = 12

3 y - 7 x = -21

Find the intercepts. If y = 0, then

3(0) + 7 x = -21 -7 x = -21 x =3

2 x - 3(0) = 12 2 x = 12 x =6

so the x-intercept is 3. If x = 0, then

so the x-intercept is 6. If x = 0, then

3 y - 7(0) = -21 3 y = -21 y = -7

2(0) - 3 y = 12 -3 y = 12 y = -4

So the y-intercepts is -7.

so the y-intercept is -4.

Plot the ordered pairs (3, 0) and (0, -7) and draw a line through these points. (A third point may be used as a check.)

Plot the ordered pairs (6, 0) and (0, -4) and draw a line through these points. (A third point may be used as a check.)

3y – 7x = –21 –7

54.

56.

3 x - y = -9

Find the intercepts. If y = 0, then 3 x - 0 = -9 3 x = -9 x = -3

If x = 0, then 3(0) - y = -9 - y = -9 y =9

5 y + 6 x = 11

Find the intercepts. If y = 0, then 5(0) + 6 x = 11 6 x = 11 11 x = 6 11 so the x-intercept is 6 .

If x = 0, then 5 y + 6(0) = 11 5 y = 11 11 y = 5

so the y-intercept is 9.

Section 1.1

( 116 , 0 ) and ( 0, 115 ) and

Plot the ordered pairs

for any value of x. The graph is the horizontal line with y-intercept -8.

draw a line through these points. (A third point may be used as a check.)

y x

5y + 6x = 11

61. 57.

y = 2x

Three ordered pairs that satisfy this equation are (0, 0), (-2, -4), and (2, 4). Use these points to draw the graph.

y = -2

The equation y = -2, or, equivalently, y = 0 x - 2, always gives the same y-value, -2, for any value of x. The graph of this equation is the horizontal line with y-intercept -2.

62.

58.

y = -5 x

Three ordered pairs that satisfy this equation are (0, 0), (-1, 5), and (1, -5). Use these points to draw the graph.

x = 4

For any value of y, the x-value is 4. Because all ordered pairs that satisfy this equation have the same first number, this equation does not represent a function. The graph is the vertical line with x-intercept 4.

y y = –5x 1 0

63.

x=4

59.

–5

y+8=0

–8

4 x

x+5=0

This equation may be rewritten as x = -5. For any value of y, the x-value is -5. Because all ordered pairs that satisfy this equation have the same first number, this equation does not represent a function. The graph is the vertical line with x-intercept -5.

x + 4y = 0

If y = 0, then x = 0, so the x-intercept is 0. If x = 0, then y = 0, so the y-intercept is 0. Both intercepts give the same ordered pair, (0, 0). To get a second point, choose some other value of x (or y ). For example if x = 4, then x + 4y = 0 4 + 4y = 0 4 y = -4 y = -1, giving the ordered pair (4, -1). Graph the line through (0, 0) and (4, -1).

60.

y+8= 0

42 64.

Chapter 1 LINEAR FUNCTIONS 3x - 5 y = 0

(c)

If y = 0, then x = 0, so the x-intercept is 0. If x = 0, then y = 0, so the y-intercept is 0. Both intercepts give the same ordered pair (0, 0). To get a second point, choose some other value of x (or y ). For example, if x = 5, then

y = 0.9 x + 36 y = 0.9(180) + 36 = 198 180 gourmet cupcakes would cost $198.

67. (a)

3x - 5 y = 0 3(5) - 5 y = 0 15 - 5 y = 0 -5 y = -15 y =3

giving the ordered pair (5, 3). Graph the line through (0, 0) and (5, 3).

(b)

Yes, the data appear to lie roughly along a straight line. The line goes through (10, 31,500) and (19, 36,880).

m= 5 x

36,880 - 31,500 » 597.8 19 - 10

Use (10, 31,500) and the point-slope form. y - y1 = m(t - t1)

3x – 5y = 0

y - 31,500 = 597.8(t - 10)

65. (a) The line goes through (2, 27,000) and (5, 63,000). 63, 000 - 27, 000 5-2 = 12, 000

y = 597.8t - 25,522

The slope 597.8 indicates that tuition and fees have increased approximately $598 per year.

(c)

y - 27, 000 = 12, 000( x - 2) y - 27, 000 = 12, 000 x - 24, 000 y = 12, 000 x + 3000

(b) Let y = 100, 000; find x.

The year 2035 is too far in the future to rely on this equation to predict costs; too many other factors may influence these costs by then.

68. (a)

100, 000 = 12, 000 x + 3000 97, 000 = 12, 000 x 8.08 = x

Sales would surpass $100,000 after 8 years, 1 month. 66. (a) The line goes through (100, 126) and (120, 144) . 144 - 126 m = 120 - 100 = 0.9 The average cost of producing gourmet cupcakes increases by $0.90 per cupcake. (b) Use the point slope form with the given points. y - 126 = 0.9( x - 100) y = 0.9 x - 90 + 126 y = 0.9 x + 36

(b)

The line goes through (6, 233.04) and (18, 421.79). 421.79 - 233.04 m= » 15.729 18 - 6 Use (10, 31,500) and the point-slope form. y - y1 = m(t - t1) y - 233.04 = 15.729(t - 6)

y = 15.729t + 138.67

Section 1.1 (c)

The line goes through (10, 296.29) and (18, 421.79). m = 421.79 - 296.29 » 15.688 18 - 10 Use (10, 296.29) and the point-slope form. y - 296.29 = 15.688(t - 10) y = 15.688t + 139.41

(d)

(e)

The data is approximately linear because all the data points do not fall on a straight line. So the lines between different pairs of points have different slopes that are close in value. y = 15.729t + 138.67 y = 15.729(16) + 138.67 y » 390.33 million subscribers y = 15.688t + 139.41

71. (a) Let x = age. u = 0.85(220 - x ) = 187 - 0.85x l = 0.7(200 - x ) = 154 - 0.7 x

(b)

u = 187 - 0.85(20) = 170 l = 154 - 0.7(20) = 140

The target heart rate zone is 140 to 170 beats per minute. (c)

u = 187 - 0.85(40) = 153 l = 154 - 0.7(40) = 126

The target heart rate zone is 126 to 153 beats per minute. (d)

154 - 0.7 x = 187 - 0.85( x + 36) 154 - 0.7 x = 187 - 0.85x - 30.6 154 - 0.7 x = 156.4 - 0.85 x

y = 15.688(16) + 139.41

0.15 x = 2.4 x = 16

y » 390.42 million subscribers

Both the estimated values are slightly less than the actual number of subscribers of 395.88 million.

The younger woman is 16; the older woman is 16 + 36 = 52. l = 0.7(220 - 16) » 143 beats per minute.

69. (a) The line goes through (3, 100) and (35, 251.1). m = 251.1 - 100 » 4.722 35 - 3 Use the point (3, 100) and the point-slope form. y - 100 = 4.722(t - 3) y = 4.722t - 14.166 + 100 y = 4.722t + 85.834

72. Let x represent the force and y represent the speed. The linear function contains the points (0.75, 2) and (0.93, 3).

(b)

(c)

The year 2000 corresponds to t = 2000 – 1980 = 20. y = 4.722(20) + 85.834 » 180.274 The predicted value is slightly more than the actual CPI of 172.2. The annual CPI is increasing at a rate of approximately 4.7 units per year.

70. (a) The line goes through (4, 0.17) and (7, 0.33). m = 0.33 - 0.17 » 0.053 7-4 y - 0.33 = 0.16 (t - 7) 3 y - 0.33 = 0.053t - 0.373 y » 0.053t - 0.043

(b)

Let y = 0.5; solve for t. 0.5 = 0.053t - 0.043 0.543 = 0.053t 10.2 = t

3-2 1 = 0.93 - 0.75 0.18 1 100 50 = 18 = = 18 9 100

Use point-slope form to write the equation. 50 ( x - 0.75) y-2= 9 50 50 (0.75) y-2= x9 9 50 75 +2 y = x9 18 50 13 y = x9 6 Now determine y, the speed, when x, the force, is 1.16. 50 13 y = (1.16) 9 6 58 13 77 = = » 4.3 9 6 18 The pony switches from a trot to a gallop at approximately 4.3 meters per second.

Chapter 1 LINEAR FUNCTIONS

73. Let x = 0 correspond to 1900. Then the “life expectancy from birth” line contains the points (0, 46) and (110, 78.7). 78.7 - 46 32.7 m = = » 0.297 110 - 0 110

Since (0, 46) is one of the points, the line is given by the equation. y = 0.297 x + 46. The “life expectancy from age 65” line contains the points (0, 76) and (110, 84.1). 84.1 - 76 8.1 m = = » 0.074 110 - 0 110 Since (0, 76) is one of the points, the line is given by the equation y = 0.074 x + 76. Set the two expressions for y equal to determine where the lines intersect. At this point, life expectancy should increase no further. 0.297 x + 46 = 0.074 x + 76 0.223x = 30 x » 135 Determine the y-value when x = 129. Use the first equation. y = 0.297(135) + 46 = 40.095 + 46 = 86.095

75. (a) The line goes through (50, 249,187) and (117, 1,127,167). 1,127,167 - 249,187 m= » 13,104.18 117 - 50

Use the point (50, 249,187) and the pointslope form. y - 249,187 = 13,104.18(t - 50) y = 13,104.18t - 655, 209 + 249,187 y = 13,104.18t - 406, 022

(b)

(c)

Use the point (9, 26.2) and the point-slope form. y - 26.2 = 0.12(t - 9) y = 0.12t + 25.12 (b)

y = -1.93t + 266.7

(c)

Since 0.14 > 0.12 , women seem to have the faster increase in median age at first marriage.

(d)

Let y = 32. 32 = 0.12t + 25.12 6.88 = 0.12t 57.33 » t

25 = -1.93t + 266.7 -241.7 = -1.93t 125.2 » t

If the trend continues, the goal of the mortality rate for children under 5 years of age being 25 per 1000 live births would be reached in 2026.

The line (for the data for women) goes through (9, 23.8) and (39, 28.0). 28.0 - 23.8 m= = 0.14 39 - 9 Use the point (9, 23.8) and the point-slope form. y - 23.8 = 0.14(t - 9) y = 0.14t + 22.54

y - 93 = -1.93(t - 90) y = -1.93t + 266.7

(b) Let y = 25 and solve for t.

The number of immigrants admitted to the United States in 2020 will be about 1,166,480. The equation y = 13,104.18t - 406, 022 has –406,022 for the y-intercept, indicating that the number of immigrants admitted in the year 1900 was –406,022. Realistically, the number of immigrants cannot be a negative value, so the equation cannot be used for valid predicted values.

76. (a) The line (for the data for men) goes through (9, 26.2) and (39, 29.8). 29.8 - 26.2 m= = 0.12 39 - 9

Thus, the maximum life expectancy for humans is about 86 years. 74. (a) Let t = 0 correspond to 1900. Then the “mortality rate for children under 5 years of age” line contains the points (90, 93) and (118, 39). 39 - 93 m= » -1.93 118 - 90 Use the point (90, 93) and the point-slope form.

The year 2020 corresponds to t = 120. y = 13,104.18(120) - 406, 022 y » 1,166, 480

The median age at first marriage for men will reach 32 in the year 1980 + 58 = 2038.

Section 1.1 (e)

45 So, 1970 + 133 = 2103.

Let t = 57.33. y = 0.14(57.33) + 22.54 y » 30.6

The temperature will rise to 19°C in about the year 2103.

The median age at first marriage for women will reach be 30.6 when the median age for men is 30. 77. (a) Plot the points (15, 1600), (200,15, 000), (290, 24, 000), and (520, 40, 000).

(c)

From part (a) If the temperature rises 0.2C° per decade, it rises 0.02C° per year. m = 0.02 b = 15, since a point is (0, 15). T = 0.02t + 15

Velocity

5 10

2.5 10

From part (b) Let T = 19; find t. 19 = 0.02t + 15

4 = 0.02t t = 200 0 100 200 300 400 500 600 Distance

The points lie approximately on a line, so there appears to be a linear relationship between distance and time. (b) The graph of any equation of the form y = mx goes through the origin, so the line goes through (520, 40, 000) and (0, 0). 40, 000 - 0 » 76.9 520 - 0 b=0

So, 1970 + 200 = 2170 The temperature will rise to 19°C in about the year 2170. 79. (a) Let t = 0 correspond to 2000. Then the line representing the percentage of Americans 12 or older who had listened to online radio contains the points (10, 27) and (18, 64). m=

Use the point (10, 27) and the point-slope form.

y = 76.9 x + 0

yo - 27 = 4.625(t - 10)

y = 76.9 x

(c)

Let y = 60, 000; solve for x. 60, 000 = 76.9 x 780.23 » x Hydra is about 780 megaparsecs from earth.

(d)

yo = 4.625t - 19.25

(b) The line representing the percent of U.S. cellphone users who ever listened to online radio in a car using a phone contains the points (10, 6) and (18, 44).

9.5 ´ 1011 , m = 76.9 m

9.5 ´ 1011 76.9 = 12.4 billion years

44 - 6 = 4.75 18 - 10

Use the point (10, 6) and the point-slope form.

yc  6  4.75(t  10)

78. (a) If the temperature rises 0.3C° per decade, it rises 0.03C° per year. m = 0.03 b = 15, since a point is (0,15).

T = 0.03t + 15

(b)

64 - 27 = 4.625 18 - 10

yc  4.75t  41.5

(c) The percent of Americans who had listened to online radio in the previous month increased by 4.625% per year, while the percent of U.S. cellphone users who had ever listened to online radio in the car using a phone increased by 4.75% per year.

Let T = 19; find t. 19 = 0.03t + 15 4 = 0.03t t = 133.3 » 133

Chapter 1 LINEAR FUNCTIONS

1.2 Linear Functions and Applications Your Turn 1 For g ( x ) = -4 x + 5, calculate g (-5).

g ( x ) = -4 x + 5 g (-5) = -4(-5) + 5 = 20 + 5 = 25 Your Turn 2 For the demand and supply functions given in Example 2, find the quantity of watermelon demanded and supplied at a price of $3.30 per watermelon. p = D(q) = 9 - 0.75q 3.30 = 9 - 0.75q 0.75q = 5.7 5.7 q= = 7.6 0.75

Since the quantity is in thousands, 7600 watermelon are demanded at a price of $3.30. p = S (q) = 0.75q 3.30 = 0.75q

Your Turn 5 The fixed cost is b, this function has the form C ( x)  mx  7145. To find m, use the fact that producing 100 items costs $7965. C ( x) = mx + 7145 C (100) = 100m + 7145 7965 = 100m + 7145 820 = 100m m = 8.2

Thus the cost function is C ( x)  8.2 x  7145. Your Turn 6 The cost function is C ( x ) = 35x + 250 and the revenue function is R( x) = 59 x. Thus the profit function is P ( x) = R ( x ) - C ( x ) = 59 x - (35x + 250) = 24 x - 250

The profit is to be $8030. P( x) = 24 x - 250

3.3 q= = 4.4 0.75

8030 = 24 x - 250 24 x = 8280

Since the quantity is in thousands, 4400 watermelon are supplied at a price of $3.30. Your Turn 3 Set the two price expressions equal and solve for the equilibrium quantity q. 10 - 0.85q = 0.4q 10 = 1.25q 10 =8 q= 1.25

8280 = 345 24

Sale of 345 units will produce $8030 profit.

1.2 Warmup Exercises W1. 3( x - 2)2 + 6( x + 4) - 5x + 4 3(5 - 2)2 + 6(5 + 4) - 5(5) + 4 = 3(3)2 + 6(9) - 5(5) + 4 = 27 + 54 - 25 + 4 = 60

The equilibrium quantity is 8000 watermelon. Use either price expression to find the equilibrium price p. W2.

p = 0.4q p = 0.4(8) = 3.2 The equilibrium price is $3.20 per watermelon. Your Turn 4 The marginal cost is the slope of the cost function C ( x ), so this function has the form C ( x ) = 15x + b. To find b, use the fact that producing 80 batches costs $1930.

C ( x ) = 15x + b C (80) = 15(80) + b 1930 = 1200 + b b = 730

p 8 6 4 2 0

y = 7 – 2.5x

4 q

1.2

Exercises

This statement is true. When we solve y = f ( x) = 0, we are finding the value of x when y = 0, which is the x-intercept. When we evaluate f (0), we are finding the value of y when x = 0, which is the y-intercept.

Thus the cost function is C ( x ) = 15x + 730.

This statement is false. The graph of f ( x) = -5 is a horizontal line.

Section 1.2

This statement is true. Only a vertical line has an undefined slope, but a vertical line is not the graph of a function. Therefore, the slope of a linear function cannot be undefined.

This statement is true. For any value of a, f (0) = a ⋅ 0 = 0,

so the point (0, 0), which is the origin, lies on the line.

24. $10 is the fixed cost and $0.99 is the cost per downloaded song—the marginal cost. Let x = the number of downloaded songs and C ( x) = cost of downloading x songs. Then, C ( x ) = (marginal cost) ⋅ (number of downloaded songs) + fixed cost C ( x ) = 0.99 x + 10.

25. $2 is the fixed cost and $0.75 is the cost per halfhour.

Let x = the number of half-hours;

False. The variable x in the linear function f ( x) = 3x + 4 is the independent variable.

True

f (2) = 7 - 5(2) = 7 - 10 = -3

10.

f (4) = 7 - 5(4) = 7 - 20 = -13

11.

f (-3) = 7 - 5(-3) = 7 + 15 = 22

12.

f (-1) = 7 - 5(-1) = 7 + 5 = 12

13.

g (1.5) = 2(1.5) - 3 = 3 - 3 = 0

14.

g (2.5) = (2.5) - 3 = 5 - 3 = 2

15.

æ 1ö æ 1ö g çç - ÷÷ = 2 çç - ÷÷ - 3 = -1 - 3 = -4 çè 2 ÷ø çè 2 ø÷

C ( x) = mx + b, where b is the fixed cost.

16.

æ 3ö æ 3ö 3 9 g çç - ÷÷ = 2 çç - ÷÷ - 3 = - - 3 = èç 4 ø÷ èç 4 ø÷ 2 2

Now,

17.

f (t ) = 7 - 5(t ) = 7 - 5t

18.

g (k 2 ) = 2(k 2 ) - 3 = 2k 2 - 3

C ( x) = the cost of parking a car for x half-hours.

Thus,

C ( x) = 2 + 0.75x = 0.75x + 2

26. $44 is the fixed cost and $0.28 is the cost per mile. Let x = the number of miles; R( x) = the cost of renting for x miles.

Thus, R( x ) = fixed cost + (cost per mile) ⋅ (number of miles) R( x ) = 44 + 0.28 x. 27. Fixed cost, $100; 50 items cost $1600 to produce.

Let C ( x) = cost of producing x items. C ( x) = mx + 100 C ( x) = 1600 when x = 50, so 1600 = m(50) + 100 1500 = 50m 30 = m.

19. The fixed cost is constant for a particular product and does not change as more items are made. The marginal cost is the rate of change of cost at a specific level of production and is equal to the slope of the cost function at that specific value; it approximates the cost of producing one additional item. 23. $10 is the fixed cost and $2.25 is the cost per hour. Let x = number of hours;

Thus, C ( x) = 30 x + 100. 28. Fixed cost: $35; 8 items cost $395.

Let C ( x) = cost of x items C ( x) = mx + b, where b is the fixed cost C ( x) = mx + 35

Now, C ( x) = 395 when x = 8, so 395 = m(8) + 35 360 = 8m 45 = m.

R( x) = cost of renting a snowboard for x hours.

Thus, R( x ) = fixed cost + (cost per hour) ⋅ (number of hours) R( x ) = 10 + (2.25)( x ) = 2.25x + 10

Thus, C ( x) = 45x + 35.

Chapter 1 LINEAR FUNCTIONS

29. Marginal cost: $75; 50 items cost $4300.

(g)

C ( x) = 75 x + b

Now, C ( x) = 4300 when x = 50. 4300 = 75(50) + b 4300 = 3750 + b 550 = b

Thus, C ( x) = 75 x + 550.

(h)

p = 16 – 1.25q

2 4 6 8 10 12 14 q

S (q) = 0.75q

Let S (q) = 0. Find q.

30. Marginal cost, $120; 700 items cost $96,500 to produce. C ( x) = 120 x + b Now, C ( x) = 96,500 when x = 700. 96,500 = 120(700) + b 96,500 = 84, 000 + b 12,500 = b Thus, C ( x ) = 120 x + 12,500. 31.

p 16 14 12 10 8 6 4 2

0 = 0.75q 0 = q

When the price is $0, the number of watches supplied is 0. (i)

Let S (q) = 10. Find q. 10 = 0.75q 40 = q 3 q = 13.3

D(q) = 16 - 1.25q (a) D(0) = 16 - 1.25(0) = 16 - 0 = 16

When 0 watches are demanded, the price is $16. (b)

(c)

When the price is $10, The number of watches supplied is about 1333.

D(4) = 16 - 1.25(4) = 16 - 5 = 11 When 400 watches are demanded, the price is $11.

(j) Let S (q) = 20. Find q. 20 = 0.75q 80 = q 3 q = 26.6

D(8) = 16 - 1.25(8) = 16 - 10 = 6 When 800 watches are demanded, the price is $6.

When the price is $20, the number of watches demanded is about 2667.

(d) Let D(q) = 8. Find q. 8 = 16 - 1.25q 5 q =8 4 q = 6.4

(k)

When the price is $8, the number of watches demanded is 640. (e)

Let D(q) = 10. Find q. 10 = 16 - 1.25q 5 q =6 4 q = 4.8

(l)

When the price is $10, the number of watches demanded is 480. (f)

Let D(q) = 12. Find q. 12 = 16 - 1.25q 5 q = 4 4 q = 3.2 When the price is $12, the number of watches demanded is 320.

p 16 14 12 10 8 6 4 2

p = 16 – 1.25q

(8, 6) p = 0.75q 2 4 6 8 10 12 14 q

D( q ) = S ( q ) 16 - 1.25q = 0.75q 16 = 2q 8= q S (8) = 0.75(8) = 6

The equilibrium quantity is 800 watches, and the equilibrium price is $6. 32.

D(q) = 6 - 0.25q

(a)

D(0) = 6 - 0.25(0) = 6 - 0 = 6

When 0 quarts are demanded, the price is $6.

Section 1.2 (b)

49 When the price is $6.30, 1800 quarts are supplied.

D(4) = 6 - 0.25(4) = 6 - 1 = 5

When 400 quarts are demanded, the price is $5. (c)

(k)

D(8.4) = 6 - 0.25(8.4) = 6 - 2.1 = 3.9

When 840 quarts are demanded, the price is $3.90. (d) Let D(q) = 4.5. Find q. 4.5 = 6 - 0.25q 0.25q = 1.5 q =6

(l)

When the price is $4.50, 600 quarts are demanded. (e)

Let D(q) = 3.25. Find q.

S (10) = 0.35(10) = 3.5

3.25 = 6 - 0.25q 0.25q = 2.75 q = 11

When the price is $3.25, 1100 quarts are demanded. (f)

D( q ) = S ( q) 6 - 0.25q = 0.35q 6 = 0.6q 10 = q

33.

The equilibrium quantity is 1000 quarts and the equilibrium price is $3.50. 2 4 p = S (q) = q; p = D(q) = 120 - q 5 5 (a)

Let D(q) = 2.4. Find q. 2.4 = 6 - 0.25q 0.25q = 6.6 q = 14.4

When the price is $2.40, 1440 quarts are demanded. (g)

(h)

(b)

S (q) = 0.35q

The equilibrium quantity is 100, the equilibrium price is $40.

Let S (q) = 0. Find q. 0 = 0.35q q =0

34. (a)

Let S (q) = 2.80. Find q.

p = 1.4q – 0.6

1 0

When the price is $2.80, 800 quarts are supplied. Let S (q) = 6.30. Find q.

p = –2q + 3.2

2.80 = 0.35q q =8

(j)

p 3

When the price is $0, 0 quarts are supplied. (i)

S (q ) = D(q ) 2 4 q = 120 - q 5 5 6 q = 120 5 q = 100 2 S (100) = (100) = 40 5

(b)

S (q) = p = 1.4q - 0.6 D(q) = p = -2q + 3.2

Set supply equal to demand and solve for q.

6.30 = 0.35q q = 18

Chapter 1 LINEAR FUNCTIONS 1.4q - 0.6 = -2q + 3.2

(b)

1.4q + 2q = 0.6 + 3.2

P( x ) = 15x - (5x + 20)

3.4q = 3.8 3.8 q = » 1.12 3.4 S (1.12) = 1.4(1.12) - 0.6 = 0.968

P(100) = 15(100) - (5 ⋅ 100 + 20) = 1500 - 520 = 980

The equilibrium quantity is about 1120 pounds; the equilibrium price is about $0.96 35. Use the supply function to find the equilibrium quantity that corresponds to the given equilibrium price of $4.50. S (q) = p = 0.3q + 2.7 4.50 = 0.3q + 2.7 1.8 = 0.3q 6= q The line that represents the demand function goes through the given point (2, 6.10) and the equilibrium point (6, 4.50). 4.50 - 6.10 m= = -0.4 6-2

Use point-slope form and the point (2, 6.10) . D(q) - 6.10 = -0.4(q - 2) D(q) = -0.4q + 0.8 + 6.10 D(q) = -0.4q + 6.9

36. Use the supply function to find the equilibrium quantity that corresponds to the given equilibrium price of $5.85. p = S ( q) 5.85 = 0.25q + 3.6 2.25 = 0.25q 9 = q

(c)

5.85 - 7.60 9-4 = -0.35 Use point-slope form and the point (4, 7.60) .

The profit from 100 units is $980. P( x ) = 500 15x - (5x + 20) = 500 10 x - 20 = 500 10 x = 520 x = 52 For a profit of $500, 52 units must be produced.

38.

C ( x) = 12 x + 39; R( x) = 25 x

(a)

C ( x ) = R( x ) 12 x + 39 = 25x 39 = 13x 3= x

The break-even quantity is 3 units. (b) P( x ) = R( x ) P( x ) = 25x - (12 x + 39) P( x ) = 13x - 39 P(250) = 13(250) - 39 = 3250 - 39 = 3211 The profit from 250 units is $3211. (c)

P( x) = $130; find x. 130 = 13x - 39 169 = 13x 13 = x

The line that represents the demand function goes through the given point (4, 7.60) and the equilibrium point (9, 5.85) . m=

P( x ) = R( x ) - C ( x )

For a profit of $130, 13 units must be produced. 39. (a)

C ( x) = mx + b ; m = 3.50; C (60) = 300 C ( x) = 3.50 x + b

Find b.

D(q) - 7.60 = -0.35(q - 4)

300 = 210 + b

D(q) = -0.35q + 1.4 + 7.60

90 = b C ( x ) = 3.50 x + 90

D(q) = -0.35q + 9

37.

C ( x) = 5x + 20; R( x) = 15x

(a)

300 = 3.50(60) + b

(b)

R( x) = 9 x C ( x) = R( x)

C ( x ) = R( x ) 5x + 20 = 15x

3.50 x + 90 = 9 x

20 = 10 x

90 = 5.5x

2 = x

16.36 = x

The break-even quantity is 2 units.

Joanne must produce and sell 17 shirts.

Section 1.2 (c)

51 P( x) = R( x) - C ( x); P( x) = 500

(c)

500 = 9 x - (3.50 x + 90)

= 97 + 1.32

500 = 5.5x - 90

= 98.32

The total cost of producing 1000 cups is $98.32.

590 = 5.5x 107.27 = x

(d)

To make a profit of $500, Joanne must produce and sell 108 shirts. 40. (a)

C (1000) = 2675; b = 525

(e)

2675 = m(1000) + 525 2150 = 1000m

(f)

2.15 = m C ( x ) = 2.15x + 525

(b)

C (1001) = 0.097(1001) + 1.32 = 97.097 + 1.32 = 98.417

C ( x) = mx + b

Find m.

C (1000) = 0.097(1000) + 1.32

R( x) = 4.95x

42.

C ( x) = R( x)

The total cost of producing 10001 cups is $98.42. Marginal cost = 98.417 - 98.32 = $0.097 or 9.7¢ The marginal cost for any cup is the slope, $0.097 or 9.7¢. This means the cost of producing one additional cup of coffee would be 9.7¢.

C (10, 000) = 547,500; C (50, 000) = 737,500

(a)

C ( x) = mx + b

2.15 x + 525 = 4.95 x

737,500 - 547,500 50, 000 - 10, 000 190, 000 = 40, 000 = 4.75

525 = 2.80 x 187.5 = x

In order to break even, he must produce and sell 188 books. (c)

P( x) = R( x) - C ( x); P( x) = 1000

y - 547,500 = 4.75( x - 10, 000) y - 547,500 = 4.75 x - 47,500 y = 4.75 x + 500, 000 C ( x) = 4.75 x + 500, 000

1000 = 4.95 x - (2.15 x + 525) 1000 = 4.95 x - 2.15 x - 525 1000 = 2.80 x - 525 1525 = 2.80 x

(b)

The fixed cost is $500,000.

544.6 = x

(c)

C (100, 000) = 4.75(100, 000) + 500, 000 = 475, 000 + 500, 000 = 975, 000

In order to make a profit of $1000, he must produce and sell 545 books. 41. (a) Using the points (100, 11.02) and (400, 40.12), 40.12 - 11.02 m = 400 - 100 29.1 = = 0.097. 300 y - 11.02 = 0.097( x - 100) y - 11.02 = 0.097 x - 9.7 y = 0.097 x + 1.32 C ( x ) = 0.097 x + 1.32

(b) The fixed cost is given by the constant in C ( x). It is $1.32.

(d)

43.

The total cost to produce 100,000 items is $975,000. Since the slope of the cost function is 4.75, the marginal cost is $4.75. This means that the cost of producing one additional item at this production level is $4.75.

C ( x) = 85 x + 900 R( x) = 105x

Set C ( x) = R( x) to find the break-even quantity. 85 x + 900 = 105x 900 = 20 x 45 = x The break-even quantity is 45 units. You should decide not to produce since no more than 38 units can be sold.

Chapter 1 LINEAR FUNCTIONS P( x) = R( x) - C ( x) = 105 x - (85 x + 900) = 20 x - 900

105 x + 6000 = 250 x 6000 = 145 x 41.38 » x

The profit P( x) = R( x) - C ( x) is 0 at the given break-even quantity of 80. P( x ) = px - (mx + 400) P( x ) = px - mx - 400 P( x ) = Mx - 400 (Let M = p - m.) P(80) = M ⋅ 80 - 400 0 = 80 M - 400 400 = 80 M 5= M

The break-even quantity is about 41 units, so you should decide to produce.

So, the linear profit function is P( x) = 5 x - 400, and the marginal profit is 5.

The profit function is P( x) = 20 x - 900. 44.

C ( x) = 105 x + 6000 R( x) = 250 x

Set C ( x) = R( x) to find the break-even quantity.

P( x) = R( x) - C ( x) = 250 x - (105 x + 6000) = 145x - 6000

The profit function is P( x) = 145x - 6000. 45.

C ( x) = 70 x + 500 R( x) = 60 x 70 x + 500 = 60 x 10 x = -500 x = -50

This represents a break-even quantity of -50 units. It is impossible to make a profit when the break-even quantity is negative. Cost will always be greater than revenue. P( x) = R( x) - C ( x) = 60 x - (70 x + 500) = -10 x - 500

The profit function is P( x) = -10 x - 500. 46.

The revenue function is R( x) = px, where p is the price per unit.

C ( x) = 1000 x + 5000 R( x) = 900 x 900 x = 1000 x + 5000 -5000 = 100 x -50 = x

It is impossible to make a profit when the breakeven quantity is negative. Cost will always be greater than revenue. P ( x ) = R ( x ) - C ( x) = 900 x - (1000 x + 5000) = -100 x - 5000

The profit function is P( x) = -100 x - 5000 (always a loss). 47. Since the fixed cost is $400, the cost function is C ( x) = mx + 100 , where m is the cost per unit.

48. Since the fixed cost is $650, the cost function is C ( x) = mx + 650 , where m is the cost per unit. The revenue function is R( x) = px, where p is the price per unit.

The profit P( x) = R( x) - C ( x) is 0 at the given break-even quantity of 25. P( x ) = px - (mx + 650) P( x ) = px - mx - 650 P( x ) = Mx - 650 (Let M = p - m.) P(25) = M ⋅ 25 - 650 0 = 25M - 650 650 = 25M 26 = M So, the linear profit function is P( x) = 26 x - 650, and the marginal profit is 26. 49. (a)

f ( x) = 34 x + 230 1000 = 34 x + 230 770 = 34 x x = 22.647

Approximately 23 acorns per square meter would produce 1000 deer tick larvae per 400 square meters. (b) The slope is 34, which indicates that the number of deer tick larvae per 400 square meters in the spring will increase by 34 for each additional acorn per square meter in the fall. 50. (a) Let t correspond to the number of years since 1990. Then for the cause of death due to tobacco, we have at t  0, m  2.8 since the quantity was rising at the rate of 28 million years per decade and b  35 since 35 million years of healthy life were lost. The linear function is f1 (t )  2.8t  35.

Section 1.2

(b) For the cause of death due to diarrhea, we have at t = 0, m = -2.2 since the quantity was falling at the rate of 22 million years per decade and b = 100 since 100 million years of healthy life were lost. The linear function is f d (t ) = -2.2t + 100.

2.8t + 35 = -2.2t + 100 5.0t = 65 t = 13

51. Use the formula derived in Example 8 in this section of the textbook. 9 5 C + 32 or C = ( F - 32) 5 9 F = 58; find C.

(a)

5 (58 - 32) 9 5 C = (26) = 14.4 9 The temperature is 14.4°C.

9 (37) + 32 5 333 F = + 32 = 98.6 5 The Fahrenheit equivalent of 37°C is 98.6°F. C = 36.5; find F. 9 F = (36.5) + 32 5 F = 65.7 + 32 = 97.7

C = 37.5; find F. 9 F = (37.5) + 32 5 = 67.5 + 32 = 99.5

The amount of healthy life lost to tobacco was expected to equal that lost to diarrhea in 2003.

F =

C = 37; find F. F =

(b)

f t (t ) = f d ( t )

(c)

(a)

The range is between 97.7°F and 99.5°F. 53. If the temperatures are numerically equal, then F = C. 9 C + 32 5 9 C = C + 32 5 F =

C =

(b)

F = -20; find C. 5 ( F - 32) 9 5 C = (-20 - 32) 9 5 C = (-52) = -28.9 9 C =

4 - C = 32 5 C = -40 The Celsius and Fahrenheit temperatures are numerically equal at -40.

54. (a) m = 7.17 b = 3.308 ´ 109 C ( x) = mx + b C ( x) = 7.17 x + 3.308 ´ 109

The temperature is -28.9C. (c)

(b) C (120) = 7.17 ⋅ 120, 000, 000 + 3.308 ´ 109 = 4,168, 400, 000

C = 50; find F.

9 C + 32 5 9 F = (50) + 32 5 F = 90 + 32 = 122

The total cost to generate 120 million MWh will be about $4.168 billion.

F =

The temperature is 122°F.

(c)

Let C ( x) = 4,500, 000, 000 . 4.5 ´ 109 = 7.17 x + 3.308 ´ 109 1.192 ´ 109 = 7.17 x

52. Use the formula derived in Example 8 in this section of the textbook. 9 5 F = C + 32 or C = ( F - 32) 5 9

166, 248, 256.6 = x

Spending $4.5 billion would result in generating about 166 million MWh of energy.

Chapter 1 LINEAR FUNCTIONS

1.3

The Least Squares Line

Your Turn 1

True

(a)

y 6 5

2 3

0.5 1

1 3

4 9

0.25 1

4 5

2 2.5

8 12.5

16 25

4 6.25

6 7

3 3

18 21

36 49

9 9

8 9

4 4.5

32 40.5

64 81

16 20.25

10 55

5 25.5

50 186

100 385

25 90.75

3 2 1

The number of data points is n  6. Putting the column totals into the formula for the slope m, we get

n ( å xy ) - ( å x )( å y )

(

)

n å x2 - (å x) 6(132) - (21)(33)

6(91) - (21) m » 0.9429

r =

 y - m ( x ) n 33 - (0.9429)(21) = » 2.2 6

x = 21 y = 33 xy = 132 x 2 = 91 y 2 = 199

(b)

(

(c)

n(å x ) - (å x )2 ⋅ n(å y 2 ) - (å y )2 10(186) - (55)(25.5) 10(385) - (55)2 ⋅ 10(90.75) - (25.5)2

The least squares line is of the form Y = mx + b. First solve for m. m =

)

n å x 2 - (å x) ⋅ n å y 2 - (å y ) 6(132) - (21)(33) 6(91) - (21)2 ⋅ 6(199) - (33)2

» 0.9429

n(å xy ) - (å x )(å y ) n(å x 2 ) - (å x )2 10(186) - (55)(25.5)

10(385) - (55)2 = 0.5545454545 » 0.555

Now find b. å y - m(å x) b= n 25.5 - 0.5545454545(55) = 10 = -0.5

1.3

Exercises

True

y 6

False. Additionally, å y 2 must be known to

calculate the correlation coefficient.

n(å xy ) - (å x )(å y )

n ( å xy ) - ( å x )( å y )

)

Your Turn 2 Put the column totals computed in Your Turn 1 into the formula for the correlation r.

(

10 x

» 0.993

The least square line is Y  0.9429 x  2.2.

Thus, Y = 0.555x - 0.5.

False. A positive correlation coefficient between two variables means that the line has a positive slope.

4 2 1

10 x

Section 1.3 (d)

Let x = 11. Find Y.

(b)

Y = 0.55(11) - 0.5 = 5.6

y2 0.64 1.44

5.44 8.4

x2 46.24 49.0

0.9 0.9 1.5

6.39 6.48 11.1

50.41 51.84 54.76

0.81 0.81 2.25

5.3

37.81 252.25

5.95

6.8 7.0

0.8 1.2

7.1 7.2 7.4 35.5

n= 4 n ( å xy ) - ( å x )( å y )

m =

(

5(37.81) - (35.5)(5.3)

r =

4(10) - (6)2

» 0.6985

r = (0.6985)2 » 0.5

The answer is choice (c). 9.

å y - m( å x ) 6 - (0)(6) = = 1.5 n 4

b =

4(9) - (6)(6)

5(252.25) - (35.5)2 ⋅ 5(5.95) - (5.3)2

)

n å x2 - ( å x )

Thus, Y = 0 x + 1.5, or Y = 1.5 .

n ( å xy ) - ( å x )( å y )

r =

(

)

n å x2 - ( å x )

(

)

⋅ n å y 2 - ( å y )2

4(9) - (6)(6) 4(10) - (6)2 ⋅ 4(10) - (6)2

(c)

y 10 8 6

n=5

(a)

m = =

4 2

n ( å xy ) - ( å x )( å y )

(

n åx

) - (å x)

10.

å y - m( å x ) n 15 - (0.9782608)(15) = » 0.0652 5 Thus, Y = 0.9783x + 0.0652 . b =

n ( å xy ) - ( å x )( å y )

(

)

n å x2 - ( å x )

(

10 x

The point (9, 9) is an outlier that has a strong effect on the least squares line and the correlation coefficient.

5(90) - (15)(15)

5(91) - (15)2 = 0.9782608 » 0.9783

r =

)

⋅ n å y 2 - ( å y )2

-20

-180

400

-10

-150

111

430

5(90) - (15)(15) » 0.9783 2 2 5(91) - (15) ⋅ 5(91) - (15)

Chapter 1 LINEAR FUNCTIONS (a)

n=5 m= =

n ( å xy ) - ( å x )( å y )

(

n åx

) - (å x)

(

)

n å x2 - ( å x )

(

)

⋅ n å y 2 - ( å y )2

4(30) - (10)(10)

4(30) - (10)2 ⋅ 4(30) - 102

5(-150) - (19)(-10)

5(111) - (19)2 = -2.886597 » -2.887 b=

n ( å xy ) - ( å x )( å y )

r =

(c)

å y - m( å x )

y 10 5 0 –5 –10 –15 –20 –25

n -10 - (-2.886597)(19) = 5 = 8.969069 » 8.969

10 x

Thus, Y = -2.887 x + 8.969 . The point (9, -20) is an outlier that has a strong effect on the least squares line and the correlation coefficient.

n (å xy ) - (å x)(å y )

r =

(

n åx

) - (å x)2 ⋅ n (å y 2 ) - (å y)2

5(-150) - (19)(-10)

5(111) - (19)2 ⋅ 5(430) - (-10)2

11.

= -0.887994 » -0.8880

(b)

1.1

4.4

1.21

4.1

10.4

4.21

(a)

n=4

(

n ( å xy ) - ( å x )( å y )

(

n åx

) - (å x)

4(30) - (10)(10) 4(30) - (10)2

)

n å x 2 - (å x )

(

)

⋅ n å y 2 - (å y )2

4(10.4) - (10)(4.1)

n=4 m=

n (å xy ) - (å x )(å y )

4(30) - (10)2 ⋅ 4(4.21) - (4.1)2

= 0.7745966 » 0.7746

(b)

y 2

=1 1

å y - m( å x ) n 10 - (1)(10) = =0 4

b =

(c)

Thus, Y = 1x + 0, or Y = x .

Yes; because the data points are either on or very close to the horizontal line y = 1 , it seems that the data should have a strong linear relationship. The correlation coefficient does not describe well a linear relationship if the data points fit a horizontal line.

Section 1.3 12.

(å x)[(å y) - m(å x)] + nm(å x 2 ) = n(å xy) (å x)(å y) - m(å x) 2 + nm(å x 2 ) = n(å xy)

nm(å x 2 ) - m(å x )2 = n(å xy ) - (å x )(å y ) m[n(å x 2 ) - (å x )2 ] = n(å xy ) - (å x )(å y ) n(å xy ) - (å x )(å y ) m= n(å x 2 ) - (å x )2

(a) n = 5 m = =

æ å y - m(å x) ö÷ 2 (å x) çç ÷÷ + (å x )m = å xy çè ø n

14. (a)

(

)

n å x2 - ( å x ) 5(20) - (10)(10) 5(30) - (10)2

10(1645) - (125) 2 = 0.06814848 » 0.06815

(b)

Thus, Y  0.06815 x  1.844. .

n ( å xy ) - ( å x )( å y )

(

)

n å x2 - ( å x )

(b)

⋅ n ( å y 2 ) - ( å y )2

5(20) - (10)(10) =0 5(30) - (10)2 ⋅ 5(34) - (10)2

(c)

Let Y  6 and find x. 6  0.06815 x  1.844 4.156  0.06815 x 60.98  x The total value of consumer durable goods will reach at least $6 trillion in the year 2000 + 61 = 2061.

(d)

y 4 3 2

(c)

13.

No; a correlation coefficient of 0 means that there isn’t a linear relationship between the x and y values. A parabola (a quadratic relationship) seems to fit the given data points. nb + (å x)m = å y

The year 2025 corresponds to x = 25. Y  0.06815(25)  1.844  3.54775 The total value of consumer durable goods in 2025 will be about $3.548 trillion.

1 0

å y - m (å x)

n 26.9571 - (0.06815)(125) = 10 = 1.843835 » 1.844

Thus, Y = 0 x + 2, or Y = 2 .

n(å x 2 ) - (å x)2 10(342.586) - (125)(26.9571)

n ( å xy ) - ( å x )( å y )

å y - m( å x ) b = n 10 - (0)(10) = = 2 5

r =

n (å xy ) - (å x )( å y )

n ( å xy ) - ( å x)( å y ) n(å x 2 ) - (å x)2 ⋅ n(å y 2 ) - (å y)2 10(342.586) - (125)(26.9571) 10(1645) - 1252 ⋅ 10(73.2813) - 26.95712

» 0.7907

(å x)b + (å x 2 )m = å xy nb + (å x)m = å y nb = (å y) - (å x)m å y - m(å x) b = n

Since r is very close to 1, the data has a strong linear relationship, and the least squares line fits the data very well.

Chapter 1 LINEAR FUNCTIONS m=

15. (a)

n (å xy ) - ( å x )(å y ) 2

n 154 - 0.815(70) = 5 = 19.39 » 19.4

n(å x ) - (å x) 10(753.387) - (135)(57.232)

10(1905) - (135)2 = -0.2332727 » -0.2333 å y - m (å x)

Thus, Y = 0.815 x + 19.4 .

n 57.232 - (-0.2333)(135) = 10 = 8.87275 » 8.873 Thus, Y  0.2333 x  8.873.

(b)

(b) The number of subscriptions is growing at a rate of about 0.815 per year. (c) The year 2026 corresponds to x = 26. Y = 0.815(26) + 19.4 = 40.59 » 40.6

The year 2025 corresponds to x = 25.

If the trend continues linearly, the percent of subscriptions will be about 40.6 in 2026.

Y  0.2333(25)  8.873  3.0405

If the trend continues linearly, there will be about 3041 banks in 2025. (c)

å y - m( å x )

(d)

Let Y = 45 and find x. 45 = 0.815x + 19.4 25.6 = 0.815x

Let Y = 4 (since Y is the number of banks in thousands) and find x.

31.4 » x

1.800  0.2333 x  8.873

The number of subscriptions will exceed 45 in the year 2000 + 32 = 2032.

7.073  0.2333 x

30.32  x

The number of U.S. banks will drop below 1800 in the year 2000 + 31 = 2031. (d)

(e)

n ( å xy ) - ( å x )( å y )

n ( å x 2 ) - ( å x ) 2 ⋅ n( å y 2 ) - ( å y ) 2

(a)

27.3

273

100

745.29

29.4

352.8

144

864.36

30.8

431.2

196

948.64

32.7

523.2

256

1069.29

33.8

608.4

324

1142.44

154

2188.6

1020

4770.02

m= =

n (å xy ) - (å x )( å y ) 2

⋅ n (å y 2 ) - (å y)2

5(2188.6) - (70)(154) 5(1020) - 702 ⋅ 5(4770.02) - 1542

17.

(a)

58.1

755.3

169

3375.61

53.3

746.2

196

2840.89

49.3

739.5

225

2430.49

47.5

760

256

2256.25

43.8

744.6

289

1918.44

41.7

750.6

324

1738.89

293.7

4496.2

1459

14560.57

n(å x ) - (å x) 5(2188.6) - (70)(154)

5(1020) - (70) = 0.815

n ( å x 2 ) - ( å x)

This strong positive correlation means that the least squares line fits the data points extremely well.

» -0.9991

Since r is very close to –1, the data has a strong linear relationship, and the least squares line fits the data very well.

n (å xy ) - (å x)(å y)

» 0.9953

10(753.387) - (135)(57.232) 10(1905) - 1352 ⋅ 10(332.048) - 57.2322

16.

r =

n ( å xy ) - ( å x )( å y ) n( å x 2 ) - ( å x ) 2 6(4496.2) - (93)(293.7)

6(1459) - (93)2 » -3.20857 » -3.21

Section 1.3

59 b=

å y - m( å x )

n 293.7 - (-3.21)(93) = 6 » 98.7 Thus, Y = -3.21x + 98.7 .

å y - m( å x )

n 21,319.8 - 176.84(93) = 6 = 812.28

Thus, Y = 176.84 x + 812.28 .

(b)

The percent of households with landlines is decreasing at a rate of about 3.21% per year.

(b)

The consumer credit is growing at a rate of about $176.84 billion per year.

(c)

The year 2024 corresponds to x = 24.

(c)

The year 2030 corresponds to x = 30.

Y = -3.21(24) + 98.7 = 21.66 » 21.7 If the trend continues linearly, the percent of households will be about 21.7% in 2024.

Let Y = 20 and find x.

(d)

-78.7 = -3.21x 24.5 » x

The percent of households with landlines will dip below 20% in the year 2000 + 25 = 2025. r =

n (å xy) - (å x)(å y) n ( å x ) - ( å x)

⋅ n (å y ) - (å y ) 2

6(4496.2) - (93)(293.7)

6(1459) - 93

If the trend continues linearly, the consumer credit will be about $6117 in 2030. (d) Let Y  7000 and find x. 7000 = 176.84 x + 812.28 6187.72 = 176.84 x 35 » x

20 = -3.21x + 98.7

(e)

Y = 176.84(30) + 812.28 = 6117.48

⋅ 6(14,560.57) - 293.7

» -0.9896

The total debt will first exceed $7000 billion in the year 2000 + 35 = 2035. (e) r =

n (å xy) - (å x)(å y) 2

n( å x ) - ( å x )

⋅ n(å y 2 ) - (å y)2

6(333,551.6) - (93)(21,319.8) 6(1459) - 932 ⋅ 6(76,308,191.92) - 21,319.82

» 0.9952

This strong negative correlation means that the least squares line fits the data points extremely well. 18.

The strong positive correlation means that the least squares line fits the data points extremely well. 19. (a)

3078.7

40,023.1

169

9,478,393.69

3287.7

46,027.8

196

10,808,971.29

3527.0

52,905

225

12,439,729

3632.3

58,116.8

256

13,193,603.29

3814.9

64,853.3

289

14,553,462.01

3979.2

71,625.6

324

15,834,032.64

21,319.8

333,551.6

1459

76,308,191.92

(a)

n ( å xy ) - ( å x )( å y ) 2

Yes, the data points lie in a linear pattern. (b)

r  0.9516; there is a strong positive correlation among the data. (c)

The mean earnings of high school graduates are growing by about $615.16 per year.

(d)

Y  1071.91x  48,383.

r  0.9568; there is a strong positive correlation among the data.

n( å x ) - ( å x ) 6(333,551.6) - 93(21,319.8)

6(1459) - (93) 2 = 176.84

Y  615.16 x  26,104.

(e)

The mean earning of workers with a bachelor’s degree are growing by about $1072 per year.

Chapter 1 LINEAR FUNCTIONS 80, 000  615.16 x  26,104

(f)

53,896  615.16 x

å y - m( å x )

n 20.468 - 0.0721(126) = 9 » 1.2648 Thus, Y = -0.0721x + 1.2648 .

87.6  x The mean earnings for high school graduates will exceed $80,000 in the year 2000 + 88 = 2088. 80, 000  1071.91x  48,383

(b)

31, 617  1071.91x

n (å xy) - (å x)(å y)

r =

29.5  x The mean earning for workers with a bachelor’s degree will exceed $80,000 in the year 2000 + 30 = 2030.

n ( å x ) - ( å x)

⋅ n (å y 2 ) - (å y)2

9(290.878) - (126)(20.468)

9(1824) - 1262 ⋅ 9(48.237098) - 20.4682

» 0.4298

20. (a)

(c)

No, the data points do not lie in a linear pattern. r  0.2050; there is a negative correlation between price and distance. The cost of a ticket tends to decrease as the distance flown increases.

(b)

(c)

The outlier in the scatterplot is Philadelphia. Removing that data point results in r  0.0006244;

Y  0.00003964 x  191.16; the marginal cost per mile is about $0.00003964 per mile. (Much less than a penny.)

(d)

21.

(a)

22.

1.251

12.51

100

1.565001

2.387

26.257

121

5.697769

2.241

26.892

144

5.022081

2.369

30.797

169

5.612161

2.879

40.306

196

8.288641

2.574

38.61

225

6.625476

2.116

33.856

256

4.477456

2.068

35.156

289

4.276624

2.583

46.494

324

6.671889

126

20.468

290.878

1824

48.237098

n ( å xy ) - ( å x )( å y ) n( å x 2 ) - ( å x ) 2 9(290.878) - (126)(20.468)

(a)

9.8

100

96.04

9.1

100.1

121

82.81

8.3

99.6

144

68.89

8.0

104

169

6.6

92.4

196

43.56

5.7

85.5

225

32.49

4.9

78.4

256

24.01

4.8

81.6

289

23.04

4.1

73.8

324

16.81

126

61.3

813.4

1824

451.65

n ( å xy ) - ( å x )( å y )

n( å x 2 ) - ( å x ) 2 9(813.4) - (126)(61.3)

9(1824) - (126) 2 » -0.74667 » -0.7467

å y - m( å x )

n 61.3 - (-0.74667)(126) = 9 » 17.264 Thus, Y = -0.7467 x + 17.264 .

9(1824) - (126)2 = 0.0721

Section 1.3

(b) r =

n ( å x 2 ) - ( å x)

5(77.36) - (13.9)(19.6)

r =

(d)

n (å xy) - (å x)(å y)

⋅ n (å y 2 ) - (å y)2

5(53.07) - 13.92 ⋅ 5(113.18) - 19.62

» 0.999

9(813.4) - (126)(61.3) 9(1824) - 1262 ⋅ 9(451.65) - 61.32

24. (a)

» -0.9900

(c)

16 10

Yes, the points lie in a linear pattern. Using a calculator’s STAT feature, the correlation coefficient is found to be r » 0.959.

(b)

23. (a)

This indicates that the percentage of successful hunts does trend to increase with the size of the hunting party.

Length (cm)

y 8

Y = 3.98 x + 22.7

(c)

4 2 0

Yes, the data appear to be linear. 2

5.8

8.6

49.88

33.64

73.96

1.5

1.9

2.85

2.25

3.61

88.6

20.0

1772

7849.96

400.0

2.3

3.1

7.13

5.29

9.61

71.6

16.0

1145.6

5126.56

256.0

1.0

93.3

19.8

1847.34

8704.89

392.04

3.3

5.0

16.5

10.89

25.0

84.3

18.4

1551.12

7106.49

338.56

13.9

19.6

77.36

53.07

113.18

80.6

17.1

1378.26

6496.36

292.41

75.2

15.5

1165.6

5655.04

240.25

69.7

14.7

1024.59

4858.09

216.09

82.0

17.1

1402.2

6724

292.41

69.4

15.4

1068.76

4816.36

237.16

83.3

16.2

1349.46

3938.89

262.44

79.6

15.0

1194

6336.16

225

82.6

17.2

1420.72

6822.76

295.84

80.6

16.0

1289.6

6496.36

256.0

83.5

17.0

1419.5

6972.25

289.0

76.3

14.4

1098.72

5821.69

207.36

1200.6

249.8

20,127.47

96,725.86

4200.56

(b)

n(å xy) - (å x)(å y) 2

n(å x ) - (å x) 5(77.36) - (13.9)(19.6)

5(53.07) - 13.92 = 1.585250901 » 1.585 å y - m(å x ) b = n 19.6 - 1.585250901(13.9) = » -0.487 5 Y = 1.585x - 0.487 Length (cm)

y 8 6

25. (a)

m =

2 0

(c)

16 10

No, it gives negative values for small widths.

n(å xy) - (å x)(å y) n(å x 2 ) - (å x)2 15(20,127.47) - (1200.6)(249.8)

15(96,725.86) - 1200.62 = 0.211925009 » 0.212

Chapter 1 LINEAR FUNCTIONS å y - m(å x ) n 249.8 - 0.211925(1200.6) = » -0.309 15

b =

Y = 0.212 x - 0.309

Y = 0.212(73) - 0.309 » 15.2

If the temperature were 73° F, you would expect to hear 15.2 chirps per second. Let Y = 18; find x.

(c)

27.

86.4 » x

When the crickets are chirping 18 times per second, the temperature is 86.4°F. (d) r =

15(20,127) - (1200.6)(249.8) 15(96,725.86) - (1200.6)2 ⋅ 15(4200.56) - (249.8)2

= 0.835

26.

17.8

316.84

16.6

149.4

275.56

16.4

213.2

169

268.96

15.7

266.9

289

246.49

16.3

342.3

441

265.69

16.2

405

625

262.44

90 (a)

1465.8

1630

1635.98

6(1630) - (90)2 ⋅ 6(1635.98) - (99)2

13,359

178,462,881

15,569

77,845

24,2393,761

17,604

176,040

100

309,900,816

19,971

229,565

225

398,840,841

22,315

446,300

400

497,959,225

24,257

606,425

625

588,402,049

25,701

719,628

784

660,541,401

103

138,776

2,325,803

2159

2,876,500,974

(a)

Yes, the data appear to lie along a straight line. (b) r =

n(å xy ) - (å x)(å y ) 2 2 2 n( å x ) - ( å x ) ⋅ n( å y ) - ( å y ) 2

7(2, 325, 803) - (103)(138, 776) 2 2 7(2159) - 103 ⋅ 7(2, 876, 500, 974) - 138, 776

n ( å xy ) - ( å x )( å y )

n( å x 2 ) - ( å x ) 2 6(1465.8) - (90)(99)

= 0.9996

Yes, the value indicates a strong linear correlation.

6(1630) - (90) 2 » -0.06857

6(1465.8) - (90)(99)

18 = 0.212 x - 0.309 18.309 = 0.212 x

n( å x 2 ) - ( å x ) 2 ⋅ n( å y 2 ) - ( å y )

» -0.7286 The value indicates a strong negative linear correlation.

Let x = 73; find Y.

(b)

n ( å xy ) - ( å x )( å y )

å y - m( å x )

n 99 - (-0.06857)(90) = 6 » 17.53 Thus, Y  0.06857 x  17.53.

n ( å xy ) - ( å x )( å y ) n( å x 2 ) - ( å x ) 2 7(2,325,803) - (103)(138, 776)

7(2159) - (103) 2

» 441.1

(b) The year 2025 corresponds to x = 35. Y  0.06857(35)  17.53  15.1 If the trend continues linearly, the pupil-teacher ratio will be about 15.1 in 2035.

å y - m(å x) n 138, 776 - (441.1)(103) = » 13,335 7

Thus, Y = 441.1x + 13,335.

Section 1.3

The year 2030 corresponds to x = 40.

(d)

Y = 441.1(40) + 13,335 = 30,979

The predicted poverty level in the year 2030 is about $13,335.

3894

3481

4356

4402

3844

5041

4752

4356

5184

4964

4624

5329

5325

5041

5625

4221

4489

3969

4410

4900

3969

4757

5041

4489

4818

5329

4356

4950

5625

4356

682

46,493

46,730

46,674

28.

m =

(a)

)

(

n åx =

n ( å xy ) - ( å x )( å y )

(

) - ( å x )2 ⋅ n ( å y 2 ) - ( å y )2

» 0.9459

Data for male students:

» -0.08915

4221

4489

3969

4419

4900

3969

4757

5041

4489

4818

5329

4356

4950

5625

4356

356

325

23,156

25,384

21,139

m =

n ( å xy ) - ( å x )( å y )

(

)

n å x2 - ( å x )

5(23,156) - (356)(325)

5(23, 337) - (326)(357) 2 2 5(21, 346) - (326) ⋅ 5(25,535) - (357)

) - ( å x) ⋅ n ( å y ) - ( å y ) 2

» 0.6674

5(21,346) - (326)2

n åx

n (å xy) - (å x)(å y) 2

5(23,337) - (326)(357)

r =

å y - m( å x ) b = n 682 - (-0.08915)(682) = » 74.28 10 Thus, Y = -0.08915x + 74.28 . r =

å y - m( å x ) n 357 - (0.6674)(326) = » 27.89 5 Thus, Y = 0.6674 x + 27.89 .

10(46,730) - (682)2

)

b =

n å x2 - ( å x )

10(46,493) - (682)(682)

(

n å x2 - ( å x )

n ( å xy ) - ( å x )( å y )

(

n ( å xy ) - ( å x )( å y )

m =

5(25,384) - (356)2

» 0.4348

å y - m( å x ) n 325 - (0.4348)(356) = » 34.04 5 Thus, Y = 0.4348 x + 34.04 . b =

10(46, 493) - (682)(682) 10(46,730) - (682)2 ⋅ 10(46,674) - (682)2

» -0.1035

The taller the student, the shorter the ideal partner’s height is. (b) Data for female students:

3894

3481

4356

4402

3844

5041

4752

4356

5184

4964

4624

5329

5325

5041

5625

326

357

23,337

21,346

25,535

n ( å xy ) - ( å x )( å y )

r =

(

n åx

) - ( å x )2 ⋅ n ( å y 2 ) - ( å y )2

5(23,156) - (356)(325) 2 2 5(25, 384) - (356) ⋅ 5(21,139) - (325)

» 0.7049

(c)

80 60

Chapter 1 LINEAR FUNCTIONS There is no linear relationship among all 10 data pairs. However, there is a linear relationship among the first five data pairs (female students) and a separate linear relationship among the second five data pairs (male students).

(c)

Let x = 620; find Y. Y = -0.0067(620) + 14.75 = 10.596 » 11

(d) 19(115, 400) - (10, 490)(210) 2 2 19(5,872,500) - (10, 490) ⋅ 19(2522) - 210

r =

29.

(a)

540

10,800

291,600

400

510

8160

260,100

256

490

4900

240,100

100

560

4480

313,600

470

5640

220,900

144

600

6600

360,000

121

540

5400

291,600

100

580

4640

336,400

680

10,200

462,400

560

4480

560

500

» -0.13

30. (a)

y 3 2 1 0 1

(b)

225

1.0

1.11

1.2321

313,600

1.5

1.36

2.04

2.25

1.8496

7280

313,600

169

2.0

1.57

3.14

2.4649

7000

250,000

196

2.5

1.76

4.4

6.25

3.0976

470

4700

220,900

100

3.0

1.92

5.76

3.6864

440

4400

193,600

100

3.5

2.08

7.28

12.25

4.3264

520

5720

270,400

121

4.0

2.22

8.88

4.9844

620

6820

384,400

121

17.5

12.02

32.61

50.75

21.5854

680

5440

462,400

550

4400

302,500

620

4340

384,400

10,490

210

115,400

5,872,500

2522

m= m=

n(å xy) - (å x)(å y) n(å x 2 ) - (å x)2 19(115, 400) - (10, 490)(210)

19(5,872,500) - 10, 4902 m = -0.0066996227 » -0.0067 å y - m(å x ) n 210 - (-0.0066996227)(10, 490) b = » 14.75 19 b =

Y = -0.0067 x + 14.75

(b)

(e) There is no linear relationship between a student’s math SAT and mathematics placement test scores.

m = m =

n(å xy ) - (å x )(å y ) n(å x 2 ) - (å x )2 7(32.61) - (17.5)(12.02)

7(50.75) - 17.52 m = 0.3657142857 » 0.366 å T - m(å L) n 12.02 - 0.3657142857(17.5) b = 7 » 0.803 b =

Y = 0.366 x + 0.803

The line seems to fit the data. y

Let x = 420; find Y.

Y = -0.0067(420) + 14.75 = 11.936 » 12

2 1 0 1