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miL2872x ch10 689-78410/6/0610:50 AMPage 689IACONFIRMING PAGES10Exponential andLogarithmicFunctions10.1 Algebra and Composition of Functions10.2 Inverse Functions10.3 Exponential Functions10.4 Logarithmic Functions10.5 Properties of Logarithms10.6 The Irrational Number eProblem Recognition Exercises—Logarithmic andExponential Forms10.7 Logarithmic and Exponential EquationsChapter 10 is devoted to the study exponential and logarithmic functions.These functions are used to study many naturally occurring phenomena suchas population growth, exponential decay of radioactive matter, and growth ofinvestments.The following is a Sudoku puzzle. As you work through this chapter, tryto simplify the expressions or solve the equations in the clues given below.Use the clues to fill in the boxes labeled a–n. Then fill in the remaining part ofthe grid so that every row, every column, and every 2 3 box contains thedigits 1 through 6.Cluesaa.c.d.e.g.h.j.l.n.b. In elog2 8Solution to 32x 10 9Solution to ln 1x 32 ln 8f. log2 16x log2 x1 log 1Solution to log12 1x 42 1 log12 xln 22i. f1f 1 1622ln 2k. log7 75ln e41 xm. a b 16e02log 100bcedfghijklmn689

miL2872x ch10 689-78410/4/0611:40 PM690IAPage 690CONFIRMING PAGESChapter 10 Exponential and Logarithmic FunctionsSection 10.1Algebra and Composition of Functions1. Algebra of FunctionsConcepts1. Algebra of Functions2. Composition of Functions3. Multiple Operations onFunctionsAddition, subtraction, multiplication, and division can be used to create a newfunction from two or more functions. The domain of the new function will bethe intersection of the domains of the original functions. Finding domains offunctions was first introduced in Section 4.2.Sum, Difference, Product, and Quotient of FunctionsfGiven two functions f and g, the functions f g, f g, f g, and g are definedas1f g21x2 f 1x2 g1x21f g2 1x2 f 1x2 g1x21f g2 1x2 f 1x2 g1x2f 1x2fa b1x2 gg1x2( f g)(x) x 3f (x) x y7654f (2) g(2)g(x) 321 5 4 3 2 1 1 2 31 2 3 4 5Figure 10-1provided g1x2 0For example, suppose f 1x2 ƒ x ƒ and g1x2 3. Taking the sum of the functions produces a new function denoted by 1 f g2. In this case, 1 f g21x2 ƒ x ƒ 3. Graphically, the y-values of the function 1 f g2 are given by the sum of the correspondingy-values of f and g. This is depicted in Figure 10-1. The function 1 f g2 appears inred. In particular, notice that 1 f g2122 f 122 g122 2 3 5.xExample 1Given: g1x2 4xAdding, Subtracting, and Multiplying Functionsh1x2 x2 3xk1x2 2x 2a. Find 1g h2 1x2 and write the domain of 1g h2 in interval notation.b. Find 1h g2 1x2 and write the domain of 1h g2 in interval notation.c. Find 1g k2 1x2 and write the domain of 1g k2 in interval notation.Solution:a. 1g h2 1x2 g1x2 h1x2 14x2 1x 2 3x2 4x x 2 3x x2 xb. 1h g2 1x2 h1x2 g1x2The domain is all real numbers 1 , 2 . 1x2 3x2 14x2 x2 3x 4x x 2 7xc. 1g k2 1x2 g1x2 k1x2 14x21 1x 22 4x 1x 2The domain is all real numbers 1 , 2 .The domain is 32, 2 because x 2 0for x 2.

miL2872x ch10 689-78410/4/0611:41 PMIAPage 691Section 10.1Skill PracticeCONFIRMING PAGES691Algebra and Composition of FunctionsGiven:f 1x2 x 1g 1x2 5x2 xh1x2 25 xPerform the indicated operations. Write the domain of the resulting function in interval notation.2. 1g f 21x21. 1f g21x2Example 23. 1f h21x2Dividing FunctionsGiven the functions defined by h1x2 x2 3x and k1x2 1x 2, find Qkh R 1x2 andwrite the domain of Qkh R in interval notation.Solution:k1x 2a b1x2 2hx 3xTo find the domain, we must consider the restrictions on x imposed by the squareroot and by the fraction. From the numerator we have x 2 0 or, equivalently, x 2. From the denominator we have x2 3x 0 or, equivalently, x1x 32 0.Hence, x 3 and x 0.Thus, the domain of kh is the set of real numbersgreater than or equal to 2, but not equal to 3 or 0.This is shown graphically in Figure 10-2.[ 5 4 3 2 1 0 1 2 3 4 5Figure 10-2The domain is [2, 3) (3, ).Skill PracticeGiven:f 1x2 1x 1g 1x2 x2 2xf4. Find a b1x2 and write the domain interval notation.g2. Composition of FunctionsComposition of FunctionsThe composition of f and g, denoted f g, is defined by the rule1 f g21x2 f 1g1x2 2provided that g1x2 is in the domain of fThe composition of g and f, denoted g f, is defined by the rule1g f 21x2 g1 f 1x2 2provided that f 1x2 is in the domain of gNote: f g is also read as “f compose g,” and g f is also read as “g compose f.”Skill Practice Answers1. 5x 2 2x 1; domain: 1 , 22. 5x 2 1; domain: 1 , 23. 1x 12 15 x ; domain: 1 , 5 41x 14. 2; domain: 3 1,02 10, 2x 2x

miL2872x ch10 689-78410/4/0669211:41 PMIAPage 692CONFIRMING PAGESChapter 10 Exponential and Logarithmic FunctionsFor example, given f 1x2 2x 3 and g1x2 x 5, we have1f g2 1x2 f 1g1x22 f 1x 52Substitute g1x2 x 5 into the function f. 21x 52 3 2x 10 3 2x 7In this composition, the function g is the innermost operation and acts on x first.Then the output value of function g becomes the domain element of the functionf, as shown in the figure.gxExample 3fg(x)f(g(x))Composing FunctionsGiven: f 1x2 x 5, g1x2 x2, and n1x2 1x 2a. Find 1 f g21x2 and write the domain of 1 f g2 in interval notation.b. Find 1g f 21x2 and write the domain of 1g f 2 in interval notation.c. Find 1n f 2 1x2 and write the domain of 1n f 2 in interval notation.Solution:a. 1 f g21x2 f 1g1x2 2 f 1x 2 2Evaluate the function f at x2. 1x 2 2 5TIP: Examples 3(a) and3(b) illustrate that theorder in which twofunctions are composedmay result in differentfunctions. That is, f gdoes not necessarilyequal g f. x2 5b. 1g f 21x2 g1 f 1x2 2 g1x 52 1x 522Evaluate the function g at 1x 52.Replace x by 1x 52 in function g.The domain is all real numbers 1 , 2 . n1x 52Evaluate the function n at x 5.c. 1n f 21x2 n1 f 1x22 1x 3Skill Practice5. 2x 2 12x 18; domain: 1 , 26. 2x 2 3; domain: 1 , 27. 1x 2; domain: 3 2, 2The domain is all real numbers 1 , 2 . x2 10x 25 11x 52 2Skill Practice AnswersReplace x by x2 in the function f.Replace x by the quantity 1x 52 in function n.The domain is 33, 2 .Given f 1x2 2x2, g 1x2 x 3, and h 1x2 1x 1,5. Find 1 f g2 1x2 . Write the domain of 1 f g2 in interval notation.6. Find 1g f 2 1x2 . Write the domain of 1g f 2 in interval notation.7. Find 1h g2 1x2 . Write the domain of 1h g2 in interval notation.

miL2872x ch10 689-78410/4/0611:41 PMIAPage 693Section 10.1CONFIRMING PAGES693Algebra and Composition of Functions3. Multiple Operations on FunctionsExample 4Combining FunctionsGiven the functions defined by f 1x2 x 7 and h1x2 2x3, find the functionvalues, if possible.a. 1 f h2132hb. a b 172fSolution:a. 1 f h2132 f 132 h132 13 72 2132 3c. 1h f 21221f h2 (3) is a product (not a composition). 1 42 21272 216b. The functionhhas restrictions on its domain.fh1x2h2x3a b 1x2 ff 1x2x 7h172hTherefore, x 7 is not in the domain, and a b 172 is undefined.ff 172Avoiding Mistakes:If you had tried evaluating the function hf at x 7, the denominator would be zero and the functionundefined.h 1722172 3 f 1727 7c. 1h f 2122 h1 f 1222 h1 52 21 52Evaluate f(2) first. f 122 2 7 5.Substitute the result into function h.3 21 1252 250Skill PracticeGiven:h 1x2 x 4k1x2 x2 38. Find 1h k2 1 22.h9. Find a b142.k10. Find 1k h2 112 .Skill Practice Answers8. 29.81310. 22

miL2872x ch10 689-78410/4/0611:41 PM694IAPage 694CONFIRMING PAGESChapter 10 Exponential and Logarithmic FunctionsFinding Function Values from a GraphExample 5For the functions f and g pictured, find the function values if possible.ya. g(2)b. 1 f g21 325y g(x) 432gc. a b 152f 5 4 3 2 1 1 2y f(x) 3 4 5d. 1 f g 21421 2 3 4 5xSolution:a. g122 1b. 1 f g21 32 f 1 32 g1 32 2 132The value g(2) represents the y-value ofy g1x2 (the red graph) when x 2.Because the point (2, 1) lies on the graph,g122 1.Evaluate the difference of f 1 32 andg1 32 .Estimate function values from the graph. 5g152gc. a b 152 ff 152 Evaluate the quotient of g(5) and f(5).1(undefined)0gThe function f is undefined at 5 because the denominator is zero.d. 1 f g 2142 f 1g142 2From the red graph, find the value of g142first. f 102From the blue graph, find the value of f atx 0. 1Skill PracticeFind the values from the graph.y54321y f(x)y g(x) 5 4 3 2 1 1 2 3 4 511. g132Skill Practice Answers11. 212. 013. 414. 212. 1f g21 421 2 3 4 5f13. a b122gx14. 1g f 21 22

miL2872x ch10 689-78410/4/0611:41 PMIAPage 695Section 10.1Section 10.1CONFIRMING PAGESAlgebra and Composition of Functions695Practice ExercisesBoost your GRADE atmathzone.com! Practice Problems Self-Tests NetTutor e-Professors VideosStudy Skills Exercise1. Define the key term composition of functions.Concept 1: Algebra of FunctionsFor Exercises 2–13, refer to the functions defined below.f 1x2 x 4g1x2 2x2 4xh1x2 1x 1k1x2 1xFind the indicated functions. Write the domain in interval notation.2. 1 f g21x23. 1 f g2 1x24. 1g f 21x25. 1f h21x26. 1f h21x27. 1h k2 1x28. 1g f 21x29. 1 f k21x2g11. a b1x2ff12. a b1x2gh10. a b 1x2ff13. a b 1x2hConcept 2: Composition of FunctionsFor Exercises 14–22, find the indicated functions and their domains. Use f, g, h, and k as defined in Exercises 2–13.14. 1f g 21x215. 1f k21x216. 1g f 21x217. 1k f 21x218. 1k h 2 1x219. 1h k 2 1x220. 1k g 2 1x221. 1g k 2 1x222. 1f h 21x223. Based on your answers to Exercises 15 and 17 is it true in general that 1 f k 2 1x2 1k f 2 1x2 ?24. Based on your answers to Exercises 14 and 16 is it true in general that 1 f g2 1x2 1g f 2 1x2 ?For Exercises 25–30, find 1f g2 1x2 and 1g f 21x2.25. f 1x2 x2 3x 1, g1x2 5x28. f 1x2 1, g1x2 x 2 x 226. f 1x2 3x2 8, g1x2 2x 427. f 1x2 x , g1x2 x3 129. For h1x2 5x 4,find 1h h2 1x2.30. For k1x2 x2 1,find 1k k2 1x2.

miL2872x ch10 689-78469610/4/0611:42 PMIAPage 696CONFIRMING PAGESChapter 10 Exponential and Logarithmic FunctionsConcept 3: Multiple Operations on FunctionsFor Exercises 31–44, refer to the functions defined below.m 1x2 x3r1x2 1x 4Find the function values if possible.n1x2 x 3p1x2 1x 231. 1m r210232. 1n p2 10233. 1m r21 4234. 1n m214235. 1r n 2 13236. 1n r 2 15237. 1p m 21 1238. 1m n 215239. 1m p 212240. 1r m 212241. 1r p2 1 3242. 1n p2 1 2243. 1m p 21 2244. 1r m 21 22For Exercises 45–62, approximate the function values from the graph, if possible.45. f 1 4246. f(1)47. g1 2248. g(3)49. 1 f g2 12250. 1g f 2 13251. 1 f g21 1252. 1g f 2 1 42g53. a b 102ff54. a b 1 22gf55. a b 102gg56. a b 1 22f57. 1g f 21 1258. 1 f g 2 10259. 1 f g 2 1 4260. 1g f 21 4261. 1g g 212262. 1 f f 21 22y54y g(x) 321 5 4 3 2 1 1 2 3 4 564. a(1)65. b1 1266. b(3)67. 1a b21 1268. 1a b210269. 1b a2 11270. 1a b212271. 1b a210272. 1a b2 1 2273. 1a b21 4274. 1b a21 32b75. a b132aa76. a b142b77. 1a a2 1 221 2 3 4 554321 5 4 3 2 1 1 2 3a(x)1 2 3 4 5b(x) 4 578. 1b b2 11279. The cost in dollars of producing x toy cars is C1x2 2.2x 1. The revenue received is R1x2 5.98x. Tocalculate profit, subtract the cost from the revenue.a. Write and simplify a function P that represents profit in terms of x.b. Find the profit of producing 50 toy cars.xyFor Exercises 63–78, approximate the function values from the graph, if possible.63. a1 32y f(x)x

miL2872x ch10 689-78410/4/0611:42 PMIAPage 697Section 10.1CONFIRMING PAGES697Algebra and Composition of Functions80. The cost in dollars of producing x lawn chairs is C1x2 2.5x 10.1. The revenue for selling x chairs isR1x2 6.99x. To calculate profit, subtract the cost from the revenue.a. Write and simplify a function P that represents profit in terms of x.b. Find the profit in producing 100 lawn chairs.a. Find the function F defined by F1t2 D1t2 R1t2. What doesF represent in the context of this problem?b. Find F(0), F(2), and F(4). What do these function valuesrepresent in the context of this problem?Difference Between Child Support Due andChild Support Paid, United States, 2000–2006Amount ( billions)81. The functions defined by D1t2 0.925t 26.958 andR1t2 0.725t 20.558 approximate the amount of child support(in billions of dollars) that was due and the amount of childsupport actually received in the United States between the years2000 and 2006. In each case, t 0 corresponds to the year 2000.35302520151050D(t) 0.925t 26.958.558R(t) 0.725t 20F(t) 0.2t 6.4012345Year (t 0 corresponds to 2000)6(Source: U.S. Bureau of the Census)r1t2 3.497t2 266.2t 20,220where t 0 corresponds to the year 1900 and r(t)represents the rural population in thousands.u1t2 0.0566t 3 0.952t 2 177.8t 4593where t 0 corresponds to the year 1900 and u(t)represents the urban population in thousands.Rural and Urban Populations in the South,United States, 1900–1970Population(thousands)82. If t represents the number of years after 1900, thenthe rural and urban populations in the South(United States) between the years 1900 and 1970can be approximated by45,00040,00035,000230,000 r(t) 3.497t 266.2t 20,22025,00020,00015,00010,000u(t) 0.0566t 3 0.952t2 177.8t 45935000t0010203040506070Year (t 0 corresponds to 1900)(Source: Historical Abstract of the United States)a. Find the function T defined by T(t) r(t) u(t). What does the function T represent in the context ofthis problem?b. Use the function T to approximate the total population in the South for the year 1940.83. Joe rides a bicycle and his wheels revolve at 80 revolutions per minute (rpm). Therefore, the total numberof revolutions, r, is given by r1t2 80t, where t is the time in minutes. For each revolution of the wheels ofthe bike, he travels approximately 7 ft. Therefore, the total distance he travels, D, depends on the totalnumber of revolutions, r, according to the function D1r2 7r.a. Find 1D r 2 1t2 and interpret its meaning in the context of this problem.b. Find Joe’s total distance in feet after 10 min.84. The area, A, of a square is given by the function a1x2 x2, where x is the length of the sides of the square.If carpeting costs 9.95 per square yard, then the cost, C, to carpet a square room is given by C1a2 9.95a,where a is the area of the room in square yards.a. Find 1C a 21x2 and interpret its meaning in the context of this problem.b. Find the cost to carpet a square room if its floor dimensions are 15 yd by 15 yd.

miL2872x ch10 689-78410/4/0669811:42 PMIAPage 698CONFIRMING PAGESChapter 10 Exponential and Logarithmic FunctionsSection 10.2Concepts1. Introduction to InverseFunctions2. Definition of a One-to-OneFunction3. Finding an Equation of theInverse of a Function4. Definition of the Inverse of aFunctionInverse Functions1. Introduction to Inverse FunctionsIn Section 4.2, we defined a function as a set of ordered pairs (x, y) such thatfor every element x in the domain, there corresponds exactly one element y inthe range. For example, the function f relates the price, x (in dollars), of a USBFlash drive to the amount of memory that it holds, y (in megabytes).price, xmegabytes, y f {(52, 512), (29, 256), (25, 128)}That is, the amount of memory depends on how much money a person has tospend. Now suppose we create a new function in which the values of x and yare interchanged. The new function is called the inverse of f and is denoted byf 1. This relates the amount of memory, x, to the cost, y.megabytes, xAvoiding Mistakes: 1f denotes the inverse of afunction. The 1 does not represent an exponent.price, y f 1 {(521, 52), (256, 29), (128, 25)}Notice that interchanging the x- and y-values has the following outcome. Thedomain of f is the same as the range of f 1, and the range of f is the domainof f 1.2. Definition of a One-to-One FunctionA necessary condition for a function f to have an inverse function is that no twoordered pairs in f have different x-coordinates and the same y-coordinate. Afunction that satisfies this condition is called a one-to-one function. The functionrelating the price of a USB Flash drive to its memory is a one-to-one function.However, consider the function g defined byg {(1, 4), (2, 3), ( 2, 4)}same ydifferent xThis function is not one-to-one because the range element 4 has two differentx-coordinates, 1 and 2. Interchanging the x- and y-values produces a relationthat violates the definition of a function.{(4, 1), (3, 2), (4, 2)}same xdifferent yThis relation is not afunction because forx 4 there are twodifferent y-values, y 1and y 2.In Section 4.2, you learned the vertical line test to determine visually if a graphrepresents a function. We use a horizontal line test to determine whether a function is one-to-one.

miL2872x ch10 689-78410/4/0611:42 PMIAPage 699CONFIRMING PAGESSection 10.2Inverse FunctionsHorizontal Line TestConsider a function defined by a set of points (x, y) in a rectangular coordinate system. The graph of the ordered pairs defines y as a one-to-one functionof x if no horizontal line intersects the graph in more than one point.To understand the horizontal line test, consider the functions f and g.f {(1, 2.99), (1.5, 4.49), (4, 11.96)}g {(1, 4), (2, 3), ( 2, 4)}y12y10865432142 12 10 8 6 4 2 2 4 6 8 10 122 4 6 8 10 12x 5 4 3 2 1 1 2 3 4 5This function is one-to-one.No horizontal line intersects morethan once.Example 11 2 3 4 5xThis function is not one-to-one.A horizontal line intersects morethan once.Identifying One-to-One FunctionsDetermine whether the function is one-to-one.a.b.yyxxSolution:a. Function is not one-to-one.A horizontal line intersectsin more than one point.b. Function is one-to-one.No horizontal line intersects morethan once.yyxx699

miL2872x ch10 689-78410/4/0670011:42 PMIAPage 700CONFIRMING PAGESChapter 10 Exponential and Logarithmic FunctionsSkill PracticeUse the horizontal line test to determine if the functions are oneto-one.y1.y2.5432154321 5 4 3 2 1 1 2 3 4 51 2 3 4 5x 5 4 3 2 1 1 2 3 4 51 2 3 4 5x3. Finding an Equation of the Inverseof a FunctionAnother way to view the construction of the inverse of a function is to find afunction that performs the inverse operations in the reverse order. For example,the function defined by f(x) 2x 1 multiplies x by 2 and then adds 1.Therefore, the inverse function must subtract 1 from x and divide by 2. We havef 1 1x2 x 12To facilitate the process of finding an equation of the inverse of a one-to-onefunction, we offer the following steps.Finding an Equation of an Inverse of a FunctionFor a one-to-one function defined by y f(x), the equation of the inversecan be found as follows:1. Replace f (x) by y.2. Interchange x and y.3. Solve for y.4. Replace y by f 1(x).Example 2Find the inverse.Finding an Equation of the Inverseof a Functionf (x) 2x 1Solution:We know the graph of f is a nonvertical line. Therefore, f (x) 2x 1 definesa one-to-one function. To find the inverse we haveSkill Practice Answers1. Not one-to-one2. One-to-oney 2x 1Step 1: Replace f(x) by y.x 2y 1Step 2: Interchange x and y.

miL2872x ch10 689-78410/4/0611:42 PMIAPage 701Section 10.2x 1 2y701Inverse FunctionsStep 3: Solve for y. Subtract 1 from both sides.x 1 y2f 1 1x2 CONFIRMING PAGESDivide both sides by 2.x 12Step 4: Replace y by f 1(x).Skill Practice3. Find the inverse of f (x) 4x 6.The key step in determining the equation of the inverse of a function is to interchange x and y. By so doing, a point (a, b) on f corresponds to a point (b, a) onf 1. For this reason, the graphs of f and f 1 are symmetric with respect to the liney x (Figure 10-3). Notice that the point ( 3, 5) of the function f corresponds tothe point ( 5, 3) of f 1. Likewise, (1, 3) of f corresponds to (3, 1) of f 1.y54321 5 4 3 2 1 1 2( 5, 3) 3 4 5f(x) 2x 1y x(1, 3)1f 1(x) x 2(3, 1)1 2 3 4 5x( 3, 5)Figure 10-3Finding an Equation of the Inverseof a FunctionExample 3Find the inverse of the one-to-one function.3g1x2 15x 4Solution:3y 15x 4Step 1: Replace g(x) by y.3x 1 5y 4Step 2: Interchange x and y.3x 4 15y1x 42 1 1 5y2331x 42 3 5y1x 42 y5Step 3: Solve for y. Add 4 to both sides.3To eliminate the cube root, cube both sides.Simplify the right side.3g 1 1x2 1x 42 35Divide both sides by 5.Step 4: Replace y by g 1(x).Skill Practice34. Find the inverse of h 1x2 12x 1.Skill Practice Answersx 643x 14. h 1 1x2 23. f 1 1x2

miL2872x ch10 689-78470210/4/0611:42 PMIAPage 702CONFIRMING PAGESChapter 10 Exponential and Logarithmic FunctionsThe graphs of g and g 1 from Example 3 are shown in Figure 10-4. Onceagain we see that the graphs of a function and its inverse are symmetric withrespect to the line y x.(x 4)3 y5108642g 1(x) 10 8 6 4 2 2 4 6 8 10y x3g(x) 5x 42 4 6 8 10xFigure 10-4For a function that is not one-to-one, sometimes we can restrict its domain tocreate a new function that is one-to-one. This is demonstrated in Example 4.Example 4Finding the Equation of an Inverse of a Functionwith a Restricted DomainGiven the function defined by m(x) x2 4 for x 0, find an equationdefining m 1.Solution:From Section 8.4, we know that y x2 4 is a parabola with vertex at (0, 4)(Figure 10-5). The graph represents a function that is not one-to-one. However,with the restriction on the domain x 0, the graph of m(x) x2 4, x 0,consists of only the “right” branch of the parabola (Figure 10-6). This is aone-to-one function.y108642 10 8 6 4 2 2 4 6 8 10yy x2 42 4 6 8 10Figure 10-5108642x 10 8 6 4 2 2 4 6 8 10m(x) x2 4; x 02 4 6 8 10xFigure 10-6To find the inverse, we havey x2 4x 0Step 1: Replace m(x) by y.x y 4y 0Step 2: Interchange x and y. Notice that therestriction x 0 becomes y 0.2x 4 y21x 4 yy 0y 0Step 3: Solve for y. Subtract 4 from both sides.Apply the square root property. Noticethat we obtain the positive square rootof x 4 because of the restriction y 0.

miL2872x ch10 689-78410/4/0611:42 PMIAPage 703Section 10.2m 1 1x2 1x 4CONFIRMING PAGES703Inverse FunctionsStep 4: Replace y by m 1(x). Notice that thedomain of m 1 has the same valuesas the range of m.Figure 10-7 shows the graphs of mand m 1. Compare the domain andrange of m and m 1.y m(x) x2 4; x 0y x108642Domain of m: 30, 2Range of m: 34, 2 10 8 6 4 2 2 4 6 8 10Domain of m 1: 34, 2Range of m 1: 30, 2m 1(x) x 42 4 6 8 10xFigure 10-7Skill Practice5. Find the inverse. g(x) x 2 2x 04. Definition of the Inverse of a FunctionDefinition of an Inverse FunctionIf f is a one-to-one function represented by ordered pairs of the form (x, y),then the inverse function, denoted f 1, is the set of ordered pairs denotedby ordered pairs of the form (y, x).An important relationship between a function and its inverse is shown inFigure 10-8.Domain of fRange of ffyxf 1Range of f 1Domain of f 1Figure 10-8Recall that the domain of f is the range of f 1 and the range of f is the domainof f 1. The operations performed by f are reversed by f 1. This leads to theinverse function property.Skill Practice Answers5. g 1 1x2 1x 2

miL2872x ch10 689-78410/4/0611:42 PM704IAPage 704CONFIRMING PAGESChapter 10 Exponential and Logarithmic FunctionsInverse Function PropertyIf f is a one-to-one function, then g is the inverse of f if and only if(f g)(x) xfor all x in the domain of g(g f )(x) xfor all x in the domain of fandExample 5Composing a Function with Its InverseShow that the functions are inverses.h1x2 2x 1k1x2 andx 12Solution:To show that the functions h and k are inverses, we need to confirm that(h k)(x) x and 1k h2 1x2 x.1h k21x2 h1k1x2 2 h a 2ax 1b2x 1b 12 x 1 1 x 1h k2 1x2 x as desired.1k h21x2 k1h1x2 2 k12x 12 12x 12 12 2x 1 12 2x2 x 1k h2 1x2 x as desired.The functions h and k are inverses because (h k)(x) x and (k h)(x) xfor all real numbers x.Skill Practice6. Show that the functions are inverses.f 1x2 3x 2Skill Practice Answers6. 1f g2 1x2 f 1g 1x2 2x 2 3ab 2 x31g f 2 1x2 g 1f 1x2 23x 2 2 x3andg 1x2 x 23

miL2872x ch10 689-78410/4/0611:42 PMIAPage 705Section 10.2Section 10.2CONFIRMING PAGES705Inverse FunctionsPractice ExercisesBoost your GRADE atmathzone.com! Practice Problems Self-Tests NetTutor e-Professors VideosStudy Skills Exercise1. Define the key terms.a. Inverse functionb. One-to-one functionc. Horizontal line testReview Exercises2. Write the domain and range of the relation {(3, 4), (5, 2), (6, 1), (3, 0)}.For Exercises 3–8, determine if the relation is a function by using the vertical line test. (See Section 4.2.)3.4.y5.yx6.yxx7.yyy8.xxxConcept 1: Introduction to Inverse FunctionsFor Exercises 9–12, write the inverse function for each function.9. g {(3, 5), (8, 1), ( 3, 9), (0, 2)}11. r {(a, 3), (b, 6), (c, 9)}10. f {( 6, 2), ( 9, 0), ( 2, 1), (3, 4)}12. s {( 1, x), ( 2, y), ( 3, z)}Concept 2: Definition of a One-to-One Function13. The table relates a state, x, to the number of representatives in theHouse of Representatives, y, in the year 2006. Does this relationdefine a one-to-one function? If so, write a function defining theinverse as a set of ordered pairs.StatexNumber sianaPennsylvania719

miL2872x ch10 689-78410/4/0611:42 PM706IAPage 706CONFIRMING PAGESChapter 10 Exponential and Logarithmic Functions14. The table relates a city x to its average January temperature y. Does this relation define a one-to-onefunction? If so, write a function defining the inverse as a set of ordered pairs.CityxTemperaturey ( C)Gainesville, Florida13.6Keene, New Hampshire 6.0Wooster, Ohio 4.0Rock Springs, Wyoming 6.0Lafayette, Louisiana10.9For Exercises 15–20, determine if the function is one-to-one by using the horizontal line test.15.y16.17.yx18.yyx19.x20.yxyxxConcept 3: Finding an Equation of the Inverse of a FunctionFor Exercises 21–30, write an equation of the inverse for each one-to-one function as defined.21. h(x) x 422. k(x) x 325. p(x) x 1026. q1x2 x 329. g1x2 22x 130. f 1x2 x3 423123. m1x2 x 2324. n(x) 4x 227. f(x) x3328. g1x2 1x31. The function defined by f (x) 0.3048x converts a length of x feet into f (x) meters.a. Find the equivalent length in meters for a 4-ft board and a 50-ft wire.b. Find an equation defining y f 1(x).c. Use the inverse function from part (b) to find the equivalent length in feet for a 1500-m race in trackand field. Round to the nearest tenth of a foot.

miL2872x ch10 689-78410/4/0611:42 PMIAPage 707Section 10.2CONFIRMING PAGES707Inverse Functions32. The function defined by s(x) 1.47x converts a speed of x mph to s(x) ft/sec.a. Find the equivalent speed in feet per second for a car traveling 60 mph.b. Find an equation defining y s 1(x).c. Use the inverse function from part (b) to find the equivalent speed in miles per hour for a traintraveling 132 ft/sec. Round to the nearest tenth.For Exercises 33–39, answer true or false.33. The function defined by y 2 has an inverse function defined by x 2.34. The domain of any one-to-one function is the same as the domain of its inverse.35. All linear functions with a nonzero slope have an inverse function.36. The function defined by g1x2 0 x 0 is one-to-one.37. The function defined by k(x) x2 is one-to-one.38. The function defined by h1x2 0 x 0 for x 0 is one-to-one.39. The function defined by L(x) x2 for x 0 is one-to-one.40. Explain how the domain and range of a one-to-one function and its inverse are related.41. If (0, b) is the y-intercept of a one-to-one function, what is the x-intercept of its inverse?42. If (a, 0) is the x-intercept of a one-to-one function, what is the y-intercept of its inverse?43. Can you think of any function that is its own inverse?44. a. Find the domain and range of the function defined by f 1x2 1x 1.b. Find the domain and range of the function defined by f 1(x) x2 1, x 0.45. a. Find the domain and range of the function defined by g(x) x2 4, x0.b. Find the domain and range of the function defined by g 1x2 1x 4. 1For Exercises 46–49, the graph of y f(x) is given.a. State the domain of f.c. State the domain of f 1b. State the range of f.d. State the range of f 1.e. Graph the function defined by y f 1(x). The line y x is provided for your reference.46.y f(x)47.y54321 5 4 3 2 1 1 2y x1 2 3 4 5y f(x)xy54321 5 4 3 2 1 1 2 3 3 4 5 4 5y x1 2 3 4 5x

miL2872x ch10 689-78410/4/0611:43 PM708IAPage 708CONFIRMING PAGESChapter 10 Exponential and Logarithmic Functions48.49.y54321y54321y xy f(x) 5 4 3 2 1 1 21 2 3 4 5x 5 4 3 2 1 1 2 3 3 4 5 4 5y xy f(x)1 2 3 4 5xConcept 4: Definition of the Inverse of a FunctionFor Exercises 50–55, verify that f and g are inverse functions by showing thata. 1 f g21x2 xb. 1g f 2 1x2 x50. f 1x2 6x 1 and g1x2 52. f 1x2 x 1651. f 1x2 5x 2 and g1x2 32xand g1x2 8x3253. f 1x2 54. f 1x2 x2 1, x 0, and g1x2 1x 1, x 1x 2532xand g1x2 27x3355. f 1x2 x2 3, x 0, and g1x2 1x 3, x 3Expanding Your SkillsFor Exercises 56–67, write an equation of the inverse of the one-to-one function.56. f 1x2 x 1x 157. p1x2 59. w1x2 4x 260. g1x2 x2 9x 061. m1x2 x2 163. g1x2 x2 1x64. q1x2 1x 462. n1x2 x2 9x065. v1x2 1x 163 xx 358. t1x2 066. z1x2 1x 42x 1x 067. u1x2 1x 16Graphing Calculator ExercisesFor Exercises 68–71, use a graphing calculator to graph each function on the standard viewing window definedby 10x10 and 10y10. Use the graph of the function to determine if the function is one-to-oneon the inter

Exponential and Logarithmic Functions 689 Chapter 10is devoted to the study exponential and logarithmic functions. These functions are used to study many naturally occurring phenomena such as population growth, exponential decay of radioactive matter, and growth of investments. The following is a Sudoku puzzle. As you work through this chapter, try

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