Inverse, Exponential, and. Logarithmic Functions NOT FOR SALE

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1 Inverse, Eponential, and Logarithmic Functions The magnitudes of earthquakes, the loudness of sounds, and the growth or deca of some populations are eamples of quantities that are described b eponential functions and their inverses, logarithmic functions.... Inverse Functions Eponential Functions Logarithmic Functions Summar Eercises on Inverse, Eponential, and Logarithmic Functions. Evaluating Logarithms and the Change-of-Base Theorem Chapter Quiz.5.6 Eponential and Logarithmic Equations Applications and Models of Eponential Growth and Deca Summar Eercises on Functions: Domains and Defining Equations Copright Pearson. All Rights Reserved. 5

2 6 Chapter Inverse, Eponential, and Logarithmic Functions. Inverse Functions One-to-One Functions Inverse Functions Equations of Inverses An Application of Inverse Functions to Crptograph Domain 5 f Range 7 9 Not One-to-One Figure 6 8 One-to-One Functions Suppose we define the following function F. F = 5-,, -,,,,,,, 56 (We have defined F so that each second component is used onl once.) We can form another set of ordered pairs from F b interchanging the - and -values of each pair in F. We call this set G. G = 5, -,, -,,,,, 5, 6 G is the inverse of F. Function F was defined with each second component used onl once, so set G will also be a function. (Each first component must be used onl once.) In order for a function to have an inverse that is also a function, it must ehibit this one-to-one relationship. In a one-to-one function, each -value corresponds to onl one -value, and each -value corresponds to onl one -value. The function ƒ shown in Figure is not one-to-one because the -value 7 corresponds to two -values, and. That is, the ordered pairs, 7 and, 7 both belong to the function. The function ƒ in Figure is one-to-one. Domain Range f One-to-One Figure One-to-One Function A function ƒ is a one-to-one function if, for elements a and b in the domain of ƒ, a b implies ƒa ƒb. That is, different values of the domain correspond to different values of the range. Using the concept of the contrapositive from the stud of logic, the boldface statement in the preceding bo is equivalent to ƒa = ƒb implies a = b. This means that if two range values are equal, then their corresponding domain values are equal. We use this statement to show that a function ƒ is one-to-one in Eample (a). Classroom Eample Determine whether each function is one-to-one. (a) ƒ = (b) ƒ = 9 - Answers: (a) It is one-to-one. (b) It is not one-to-one. Eample Deciding Whether Functions Are One-to-One Determine whether each function is one-to-one. (a) ƒ = - + (b) ƒ = 5 - Solution (a) We can determine that the function ƒ = - + is one-to-one b showing that ƒa = ƒb leads to the result a = b. ƒa = ƒb -a + = -b + ƒ = - + -a = -b Subtract. a = b Divide b -. B the definition, ƒ = - + is one-to-one. Copright Pearson. All Rights Reserved.

3 . Inverse Functions 7 (b) We can determine that the function ƒ = 5 - is not one-to-one b showing that different values of the domain correspond to the same value of the range. If we choose a = and b = -, then -, but ƒ = 5 - = 5-9 = 6 = and ƒ- = = 5-9 =. Here, even though -, ƒ = ƒ- =. B the definition, ƒ is not a one-to-one function. Now Tr Eercises 7 and 9. As illustrated in Eample (b), a wa to show that a function is not oneto-one is to produce a pair of different domain elements that lead to the same function value. There is a useful graphical test for this, the horizontal line test. 5 f() = Ë5 (, ) (, ) Horizontal Line Test A function is one-to-one if ever horizontal line intersects the graph of the function at most once. 5 5 Figure Note In Eample (b), the graph of the function is a semicircle, as shown in Figure. Because there is at least one horizontal line that intersects the graph in more than one point, this function is not one-to-one. Classroom Eample Determine whether each graph is the graph of a one-to-one function. (a) Eample Using the Horizontal Line Test Determine whether each graph is the graph of a one-to-one function. (a) (b) (, ) (, ) (, ) (b) Answers: (a) It is one-to-one. (b) It is not one-to-one. Solution (a) Each point where the horizontal line intersects the graph has the same value of but a different value of. Because more than one different value of (here three) lead to the same value of, the function is not one-to-one. (b) Ever horizontal line will intersect the graph at eactl one point, so this function is one-to-one. Now Tr Eercises and. The function graphed in Eample (b) decreases on its entire domain. In general, a function that is either increasing or decreasing on its entire domain, such as ƒ =, g =, and h =, must be one-to-one. Copright Pearson. All Rights Reserved.

4 8 Chapter Inverse, Eponential, and Logarithmic Functions Tests to Determine Whether a Function Is One-to-One. Show that ƒa = ƒb implies a = b. This means that ƒ is one-to-one. (See Eample (a).). In a one-to-one function, ever -value corresponds to no more than one -value. To show that a function is not one-to-one, find at least two -values that produce the same -value. (See Eample (b).). Sketch the graph and use the horizontal line test. (See Eample.). If the function either increases or decreases on its entire domain, then it is one-to-one. A sketch is helpful here, too. (See Eample (b).) Teaching Tip This is a good place to review composition of functions. Inverse Functions Certain pairs of one-to-one functions undo each other. For eample, consider the functions g = and ƒ = We choose an arbitrar element from the domain of g, sa. Evaluate g. g = Given function g = 8 # + 5 Let =. g = 85 Multipl and then add. Now, we evaluate ƒ85. ƒ = Given function ƒ85 = ƒ85 = Let = 85. Multipl. ƒ85 = Subtract and then divide. Starting with, we applied function g and then applied function ƒ to the result, which returned the number. See Figure. Function 85 g() = f() = Function These functions contain inverse operations that undo each other. Figure As further eamples, confirm the following. g = 9 and ƒ9 = g-5 = -5 and ƒ-5 = -5 g = and ƒ = ƒ = - 8 and ga - 8 b = Copright Pearson. All Rights Reserved.

5 . Inverse Functions 9 In particular, for the pair of functions g = and ƒ = 8-5 8, In fact, for an value of, ƒg = and gƒ =. ƒg = and gƒ =. Using the notation for composition of functions, these two equations can be written as follows. ƒ g = and g ƒ = The result is the identit function. Because the compositions of ƒ and g ield the identit function, the are inverses of each other. Inverse Function Let ƒ be a one-to-one function. Then g is the inverse function of ƒ if ƒ g = for ever in the domain of g, and g ƒ = for ever in the domain of ƒ. Classroom Eample Let functions ƒ and g be defined respectivel b and ƒ = + 5 g = - 5. Is g the inverse function of ƒ? The condition that f is one-to-one in the definition of inverse function is essential. Otherwise, g will not define a function. Eample Determining Whether Two Functions Are Inverses Let functions ƒ and g be defined respectivel b Is g the inverse function of ƒ? ƒ = - and g = +. Answer: no Solution As shown in Figure 5, the horizontal line test applied to the graph indicates that ƒ is one-to-one, so the function has an inverse. Because it is oneto-one, we now find ƒ g and g ƒ. ƒ g g ƒ ƒ() = = ƒg = A + B - = + - = = gƒ = - + = = Figure 5 Since ƒ g = and g ƒ =, function g is the inverse of function ƒ. Now Tr Eercise. A special notation is used for inverse functions: If g is the inverse of a function ƒ, then g is written as ƒ (read ƒ-inverse ). ƒ = - has inverse ƒ - = +. See Eample. Copright Pearson. All Rights Reserved.

6 Chapter Inverse, Eponential, and Logarithmic Functions Caution Do not confuse the in ƒ with a negative eponent. The smbol ƒ - represents the inverse function of ƒ, not ƒ. Teaching Tip Remind students that composition of functions is not commutative. Therefore, ƒ g and g ƒ must both be checked. Use an eample such as ƒ = and g = to illustrate the necessit of checking both compositions if ou do not verif the functions are one-to-one. B the definition of inverse function, the domain of ƒ is the range of ƒ, and the range of ƒ is the domain of ƒ. See Figure 6. Domain of f Range of f X f f Figure 6 Y Range of f Domain of f Number of Year Hurricanes 9 7 Source: Classroom Eample Find the inverse of each function that is one-to-one. (a) F = 5-, -8, -, -,,,,,, 86 (b) G = 5-, 5, -,,,,,,, 56 (c) Let h be the function defined b the table in Eample (c) if the number of hurricanes for 9 is decreased b. Answers: (a) F - = 5-8, -, -, -,,,,, 8, 6 (b) G is not one-to-one. (c) h is not one-to-one. Eample Finding Inverses of One-to-One Functions Find the inverse of each function that is one-to-one. (a) F = 5-,, -,,,,,,, 6 (b) G = 5,,,,,,, 6 (c) The table in the margin shows the number of hurricanes recorded in the North Atlantic during the ears 9. Let ƒ be the function defined in the table, with the ears forming the domain and the numbers of hurricanes forming the range. Solution (a) Each -value in F corresponds to just one -value. However, the -value corresponds to two -values, and. Also, the -value corresponds to both - and. Because at least one -value corresponds to more than one -value, F is not one-to-one and does not have an inverse. (b) Ever -value in G corresponds to onl one -value, and ever -value corresponds to onl one -value, so G is a one-to-one function. The inverse function is found b interchanging the - and -values in each ordered pair. G - = 5,,,,,,, 6 Notice how the domain and range of G become the range and domain, respectivel, of G -. (c) Each -value in ƒ corresponds to onl one -value, and each -value corresponds to onl one -value, so ƒ is a one-to-one function. The inverse function is found b interchanging the - and -values in the table. ƒ - = 5, 9,,, 7,,,,, 6 The domain and range of ƒ become the range and domain of ƒ -. Now Tr Eercises 7, 5, and 5. Equations of Inverses The inverse of a one-to-one function is found b interchanging the - and -values of each of its ordered pairs. The equation of the inverse of a function defined b = ƒ is found in the same wa. Copright Pearson. All Rights Reserved.

7 (+)+* NOT FOR SALE. Inverse Functions Finding the Equation of the Inverse of = f For a one-to-one function ƒ defined b an equation = ƒ, find the defining equation of the inverse as follows. (If necessar, replace ƒ with first. An restrictions on and should be considered.) Step Interchange and. Step Solve for. Step Replace with ƒ -. Classroom Eample 5 Determine whether each function defines a one-to-one function. If so, find the equation of the inverse. (a) ƒ = (b) g = - 7 (c) h = + 5 Answers: (a) ƒ is not one-to-one. (b) g is one-to-one; g - = + 7. (c) h is one-to-one; h - = - 5. Teaching Tip To emphasize the relationship between inverse functions, for ƒ = + 5 and =, find = # + 5 =. Now, for ƒ - = - 5 and =, find = - 5 =. This shows numericall that when we substitute the original -value in ƒ -, we obtain the original -value. Eample 5 Finding Equations of Inverses Determine whether each equation defines a one-to-one function. If so, find the equation of the inverse. (a) ƒ = + 5 (b) = + (c) ƒ = - Solution (a) The graph of = + 5 is a nonhorizontal line, so b the horizontal line test, ƒ is a one-to-one function. Find the equation of the inverse as follows. ƒ = + 5 Given function = + 5 Let = ƒ. Step = + 5 Interchange and. Step - 5 = Subtract 5. = - 5 Divide b. Solve for. Rewrite. Step ƒ - = - 5 Replace with ƒ -. a - b c = A c B a - b c Thus, the equation ƒ - = - 5 = - 5 represents a linear function. In the function = + 5, the value of is found b starting with a value of, multipling b, and adding 5. The equation ƒ - = - 5 for the inverse subtracts 5 and then divides b. An inverse is used to undo what a function does to the variable. (b) The equation = + has a parabola opening up as its graph, so some horizontal lines will intersect the graph at two points. For eample, both = and = - correspond to =. Because of the presence of the -term, there are man pairs of -values that correspond to the same -value. This means that the function defined b = + is not one-toone and does not have an inverse. Proceeding with the steps for finding the equation of an inverse leads to Remember both roots. = + = + Interchange and. - = Solve for. { - =. Square root propert The last equation shows that there are two -values for each choice of greater than, indicating that this is not a function. Copright Pearson. All Rights Reserved.

8 (++)++* (++++)++++* NOT FOR SALE Chapter Inverse, Eponential, and Logarithmic Functions 8 f() = ( ) (c) Figure 7 shows that the horizontal line test assures us that this horizontal translation of the graph of the cubing function is one-to-one. ƒ = - Given function = - Replace ƒ with. Step = - Interchange and. Step = - Take the cube root on each side. = - a = a Solve for. 8 This graph passes the horizontal line test. + = Add. Step ƒ - = + Replace with ƒ -. Rewrite. Now Tr Eercises 59(a), 6(a), and 65(a). Figure 7 Eample 6 Finding the Equation of the Inverse of a Rational Function Classroom Eample 6 The following rational function is one-to-one. Find its inverse. ƒ = , 5 Answer: ƒ - = 5 + +, - The following rational function is one-to-one. Find its inverse. ƒ = + -, Solution ƒ = + -, = + - Given function Replace ƒ with. Step = +, Interchange and. - Step - = + Multipl b -. Pa close attention here. - = + Distributive propert - = + Add and -. - = + Factor out. Solve for. = +, Divide b -. - In the final line, we give the condition. (Note that is not in the range of ƒ, so it is not in the domain of ƒ -.) Step ƒ - = + -, Replace with ƒ-. Now Tr Eercise 7(a). One wa to graph the inverse of a function ƒ whose equation is known follows. Step Find some ordered pairs that are on the graph of ƒ. Step Interchange and to find ordered pairs that are on the graph of ƒ -. Step Plot those points, and sketch the graph of ƒ - through them. Another wa is to select points on the graph of ƒ and use smmetr to find corresponding points on the graph of ƒ -. Copright Pearson. All Rights Reserved.

9 (+)* NOT FOR SALE. Inverse Functions b a a (a, b) = (b, a) b For eample, suppose the point a, b shown in Figure 8 is on the graph of a one-to-one function ƒ. Then the point b, a is on the graph of ƒ -. The line segment connecting a, b and b, a is perpendicular to, and cut in half b, the line =. The points a, b and b, a are mirror images of each other with respect to =. Thus, we can find the graph of ƒ from the graph of ƒ b locating the mirror image of each point in ƒ with respect to the line =. Figure 8 Classroom Eample 7 Determine whether functions ƒ and g graphed are inverses of each other. = g() = f() = Answer: The are not inverses. Eample 7 Graphing f Given the Graph of f In each set of aes in Figure 9, the graph of a one-to-one function ƒ is shown in blue. Graph ƒ - in red. Solution In Figure 9, the graphs of two functions ƒ shown in blue are given with their inverses shown in red. In each case, the graph of ƒ - is a reflection of the graph of ƒ with respect to the line =. (, ) 5 f (, ) (, ) f 5 (, ) = 5 f (, ) (, ) (, ) 5 = f Figure 9 Now Tr Eercises 77 and 8. Classroom Eample 8 Let ƒ = +,. Find ƒ -. Answer: ƒ - = - -, Ú Eample 8 Finding the Inverse of a Function (Restricted Domain) Let ƒ = + 5, Ú -5. Find ƒ -. Solution The domain of ƒ is restricted to the interval -5,. Function ƒ is one-to-one because it is an increasing function and thus has an inverse function. Now we find the equation of the inverse. ƒ = + 5, Ú -5 Given function = + 5, Ú -5 Replace ƒ with. Step = + 5, Ú -5 Interchange and. Step = A + 5 B Square each side. = + 5 A a B = a for a Ú Solve for. = - 5 Subtract 5. Rewrite. However, we cannot define ƒ - as - 5. The domain of ƒ is -5,, and its range is,. The range of ƒ is the domain of ƒ -, so ƒ - must be defined as follows. Step ƒ - = - 5, Ú As a check, the range of ƒ -, -5,, is the domain of ƒ. Copright Pearson. All Rights Reserved.

10 Chapter Inverse, Eponential, and Logarithmic Functions Graphs of ƒ and ƒ - are shown in Figures and. The line = is included on the graphs to show that the graphs of ƒ and ƒ - are mirror images with respect to this line. 5 f() = + 5, 5 f() = +5, 5 = 5 = f () = 5, Figure f () = 5, Figure Now Tr Eercise 75. Important Facts about Inverses =. If ƒ is one-to-one, then ƒ - eists.. The domain of ƒ is the range of ƒ -, and the range of ƒ is the domain of ƒ -... If the point a, b lies on the graph of ƒ, then b, a lies on the graph of ƒ -. The graphs of ƒ and ƒ - are reflections of each other across the line = To find the equation for ƒ -, replace ƒ with, interchange and, and solve for. This gives ƒ -.. = Despite the fact that = is not one-to-one, the calculator will draw its inverse, =. Figure Some graphing calculators have the capabilit of drawing the reflection of a graph across the line =. This feature does not require that the function be one-to-one, however, so the resulting figure ma not be the graph of a function. See Figure. It is necessar to understand the mathematics to interpret results correctl. Classroom Eample 9 The function ƒ = - was used to encode a message as Find the inverse function and decode the message. (Use the same numerical values for the letters as in Eample 9.) Answer: ƒ - = + ; DO YOU TWEET An Application of Inverse Functions to Crptograph A one-to-one function and its inverse can be used to make information secure. The function is used to encode a message, and its inverse is used to decode the coded message. In practice, complicated functions are used. Eample 9 Using Functions to Encode and Decode a Message Use the one-to-one function ƒ = + and the following numerical values assigned to each letter of the alphabet to encode and decode the message BE MY FACEBOOK FRIEND. A B C D E 5 F 6 G 7 H 8 I 9 J K L M N Copright Pearson. All Rights Reserved. O 5 P 6 Q 7 R 8 S 9 T U V W X Y 5 Z 6

11 . Inverse Functions 5 A B C D E 5 F 6 G 7 H 8 I 9 J K L M N O 5 P 6 Q 7 R 8 S 9 T U V W X Y 5 Z 6 Solution because The message BE MY FACEBOOK FRIEND would be encoded as B corresponds to and ƒ = + = 7, E corresponds to 5 and Using the inverse ƒ - = - to decode ields ƒ5 = 5 + = 6, and so on. ƒ - 7 = 7 - =, which corresponds to B, ƒ - 6 = 6 - = 5, which corresponds to E, and so on. Now Tr Eercise 97.. Eercises. one-to-one. not one-to-one. one-to-one. ; g ƒ 5. range; domain 6. b, a = does not; it is not one-to-one. one-to-one. one-to-one. not one-to-one. not one-to-one 5. one-to-one 6. one-to-one 7. one-to-one 8. one-to-one 9. not one-to-one. not one-to-one. one-to-one. one-to-one. one-to-one. one-to-one 5. not one-to-one 6. not one-to-one 7. one-to-one 8. one-to-one 9. no. no. unting our shoelaces. stopping a car. leaving a room. descending the stairs 5. unscrewing a light bulb 6. empting a cup 7. inverses 8. not inverses 9. not inverses. inverses. inverses. inverses. not inverses. not inverses 5. inverses 6. inverses 7. not inverses 8. not inverses 9. inverses 5. inverses 5. 56, -,,, 8, E -,,, 5, 5,, A, B F Concept Preview Determine whether the function represented in each table is one-to-one.. The table shows the number of registered passenger cars in the United States for the ears 8. Year Registered Passenger Cars (in thousands) 8 7,8 9,88 9,89 5,657,9 Source: U.S. Federal Highwa Administration.. The table gives the number of representatives currentl in Congress from each of five New England states. Number of State Representatives Connecticut 5 Maine Massachusetts 9 New Hampshire Vermont Source: Concept Preview Fill in the blank(s) to correctl complete each sentence.. For a function to have an inverse, it must be.. If two functions ƒ and g are inverses, then ƒ g = and =. 5. The domain of ƒ is equal to the of ƒ -, and the range of ƒ is equal to the of ƒ If the point a, b lies on the graph of ƒ, and ƒ has an inverse, then the point lies on the graph of ƒ -. Copright Pearson. All Rights Reserved.

12 6 Chapter Inverse, Eponential, and Logarithmic Functions 5. not one-to-one 5. not one-to-one 55. inverses 56. inverses 57. not inverses 58. not inverses 59. (a) ƒ - = + (b) ƒ ƒ (c) Domains and ranges of both ƒ and ƒ - are -,. 7. If ƒ =, then ƒ - =. 8. If a function ƒ has an inverse, then the graph of ƒ - ma be obtained b reflecting the graph of ƒ across the line with equation. 9. If a function ƒ has an inverse and ƒ- = 6, then ƒ - 6 =.. If ƒ- = 6 and ƒ = 6, then ƒ have an inverse because. (does/does not) Determine whether each function graphed or defined is one-to-one. See Eamples and (a) ƒ - = + 5 (b) f 5 f.. 5 (c) Domains and ranges of both ƒ and ƒ - are -,. 6. (a) ƒ - = - + (b) f f (c) Domains and ranges of both ƒ and ƒ - are -,. 6. (a) ƒ - = (b) 8 ƒ ƒ 8 (c) Domains and ranges of both ƒ and ƒ - are -,. 6. (a) ƒ - = - (b) ƒ ƒ (c) Domains and ranges of both ƒ and ƒ - are -,. 6. (a) ƒ - = - - (b) f f (c) Domains and ranges of both ƒ and ƒ - are -,. 7. = = + 9. = 6 -. = - -. = -. = - 6. = - +. = Concept Check Answer each question Can a constant function, such as ƒ =, defined over the set of real numbers, be one-to-one?. Can a polnomial function of even degree defined over the set of real numbers have an inverse? Concept Check An everda activit is described. Keeping in mind that an inverse operation undoes what an operation does, describe each inverse activit.. ting our shoelaces. starting a car. entering a room. climbing the stairs 5. screwing in a light bulb 6. filling a cup 5. = = = = Copright Pearson. All Rights Reserved.

13 . Inverse Functions not one-to-one 66. not one-to-one 67. (a) ƒ - =, (b) ƒ = ƒ (c) Domains and ranges of both ƒ and ƒ - are -,,. 68. (a) ƒ - =, (b) f = f (c) Domains and ranges of both ƒ and ƒ - are -,,. 69. (a) ƒ - = +, (b) ƒ ƒ ƒ ƒ (c) Domain of ƒ = range of ƒ - = -,,. Domain of ƒ - = range of ƒ = -,,. 7. (a) ƒ - = -, (b) ƒ ƒ ƒ ƒ (c) Domain of ƒ = range of ƒ - = -, - -,. Domain of ƒ - = range of ƒ = -,,. 7. (a) ƒ - = + -, (b) Determine whether the given functions are inverses. See Eample. 7. ƒ g ƒ = 5, 5,, 5,, 56; g = 55, 6 8. ƒ ƒ = 5,,,, 5, 56; g = 5,,,, 5, 56 g Use the definition of inverses to determine whether f and g are inverses. See Eample.. ƒ = +, g = -. ƒ = + 9, g = -. ƒ = - +, g = - -. ƒ = - +, g = ƒ = + +, g = ƒ =, g = ƒ = +, Ú ; g = -, Ú 5. ƒ = + 8, Ú -8; g = - 8, Ú 6. ƒ = - +, g = ƒ = - +, g = - Find the inverse of each function that is one-to-one. See Eample , 6,,, 5, e, -, 5,,, 5, a, b f 5. 5, -,, -7,, -, 5, , -8,, -,, -8, 5, -6 Determine whether each pair of functions graphed are inverses. See Eample = 56. = 5 ƒ ƒ ƒ ƒ (c) Domain of ƒ = range of ƒ - = -,,. Domain of ƒ - = range of ƒ = -,,. = = Copright Pearson. All Rights Reserved.

14 8 Chapter Inverse, Eponential, and Logarithmic Functions 7. (a) ƒ - = + -, (b) ƒ = ƒ ƒ = ƒ (c) Domain of ƒ = range of ƒ - = -,,. Domain of ƒ - = range of ƒ = -,,. 7. (a) ƒ - = + 6 -, (b) ƒ ƒ ƒ For each function that is one-to-one, (a) write an equation for the inverse function, (b) graph ƒ and ƒ - on the same aes, and (c) give the domain and range of both f and ƒ -. If the function is not one-to-one, sa so. See Eamples ƒ = - 6. ƒ = ƒ = ƒ = ƒ = + 6. ƒ = ƒ = ƒ = ƒ =, 68. ƒ =, 69. ƒ = -, 7. ƒ = +, - 7. ƒ = + +, 7. ƒ = - -, 7. ƒ = , 7. ƒ =, ƒ = + 6, Ú ƒ = - - 6, Ú ƒ (c) Domain of ƒ = range of ƒ - = -,,. Domain of ƒ - = range of ƒ = -,,. Graph the inverse of each one-to-one function. See Eample (a) ƒ - = (b) ƒ ƒ 6 + +, ƒ ƒ (c) Domain of ƒ = range of ƒ - = -, 6 6,. Domain of ƒ - = range of ƒ = -, - -,. 75. (a) ƒ - = - 6, Ú (b) ƒ 6 6 (c) Domain of ƒ = range of ƒ - = -6,. Domain of ƒ - = range of ƒ =,. 76. (a) ƒ - = + 6, (b) f (c) Domain of ƒ = range of ƒ - =,. Domain of ƒ - = range of ƒ = -,. ƒ f Concept Check The graph of a function ƒ is shown in the figure. Use the graph to find each value. 8. ƒ - 8. ƒ ƒ ƒ ƒ ƒ - - Concept Check Answer each of the following. 89. Suppose ƒ is the number of cars that can be built for dollars. What does ƒ - represent? 9. Suppose ƒr is the volume (in cubic inches) of a sphere of radius r inches. What does ƒ - 5 represent? 9. If a line has slope a, what is the slope of its reflection across the line =? 9. For a one-to-one function ƒ, find ƒ - ƒ, where ƒ =. Copright Pearson. All Rights Reserved.

15 . Eponential Functions ƒ ƒ ƒ = = ƒ ƒ ƒ = ƒ = ƒ Use a graphing calculator to graph each function defined as follows, using the given viewing window. Use the graph to decide which functions are one-to-one. If a function is one-to-one, give the equation of its inverse. 9. ƒ = ; -, b -, 95. ƒ = - 5 +, -; -8, 8 b -6, 8 9. ƒ = - 5 ; -, b -8, ƒ = - -, ; -, 8 b -6, = ƒ ƒ ƒ It represents the cost, in dollars, of building cars. 9. It represents the radius of a sphere with volume 5 in.. 9. a not one-to-one 9. not one-to-one 95. one-to-one; ƒ - = , 96. one-to-one; ƒ - = +, ƒ - = + ; MIGUEL HAS ARRIVED 98. ƒ - = + 9 ; BIG GIRLS DONT CRY ; ƒ - = ; ƒ - = - ƒ = Use the following alphabet coding assignment to work each problem. See Eample 9. A B C D E 5 F 6 G 7 H 8 I 9 J K L M N 97. The function ƒ = - was used to encode a message as Find the inverse function and determine the message. 98. The function ƒ = - 9 was used to encode a message as Find the inverse function and determine the message. 99. Encode the message SEND HELP, using the one-to-one function ƒ = -. O 5 P 6 Q 7 R 8 S 9 T U Give the inverse function that the decoder will need when the message is received.. Encode the message SAILOR BEWARE, using the one-to-one function ƒ = +. V W X Y 5 Z 6 Give the inverse function that the decoder will need when the message is received.. Eponential Functions Eponents and Properties Eponential Functions Eponential Equations Compound Interest The Number e and Continuous Compounding Eponential Models Eponents and Properties Recall the definition of a m/n : If a is a real number, m is an integer, n is a positive integer, and n a is a real number, then a m/n = A! n a B m. For eample, 6 / = A 6 B = = 8, 7 -/ = 7 / = 7 =, and 6-/ = 6 / = 6 = 8. Copright Pearson. All Rights Reserved.

16 Chapter Inverse, Eponential, and Logarithmic Functions Teaching Tip Use this opportunit to emphasize the distinction between eact and approimate values. In this section, we etend the definition of a r to include all real (not just rational) values of the eponent r. Consider the graphs of = for different domains in Figure = ; integers as domain = ; selected rational numbers as domain Figure = ; real numbers as domain The equations that use just integers or selected rational numbers as domain in Figure leave holes in the graphs. In order for the graph to be continuous, we must etend the domain to include irrational numbers such as. We might evaluate b approimating the eponent with the rational numbers.7,.7,.7, and so on. Because these values approach the value of more and more closel, it is reasonable that should be approimated more and more closel b the numbers.7,.7,.7, and so on. These epressions can be evaluated using rational eponents as follows..7 = 7/ = Q R Because an irrational number ma be approimated more and more closel using rational numbers, we can etend the definition of a r to include all real number eponents and appl all previous theorems for eponents. In addition to the rules for eponents presented earlier, we use several new properties in this chapter. Additional Properties of Eponents For an real number a 7, a, the following statements hold. Propert (a) a is a unique real number for all real numbers. (b) a b = a c if and onl if b = c. (c) if a + and m * n, then a m * a n. Description = a can be considered a function ƒ = a with domain -,. The function ƒ = a is one-to-one. Eample: 6 a 7 Increasing the eponent leads to a greater number. The function ƒ = is an increasing function. (d) if * a * and m * n, then a m + a n. Eample: A B 7 A B 6 a 6 Increasing the eponent leads to a lesser number. The function ƒ = A B is a decreasing function. Copright Pearson. All Rights Reserved.

17 . Eponential Functions Eponential Functions We now define a function ƒ = a whose domain is the set of all real numbers. Notice how the independent variable appears in the eponent in this function. In earlier chapters, this was not the case. Eponential Function If a 7 and a, then the eponential function with base a is ƒ = a. Note The restrictions on a in the definition of an eponential function are important. Consider the outcome of breaking each restriction. If a 6, sa a = -, and we let =, then ƒ A B = - / = -, which is not a real number. If a =, then the function becomes the constant function ƒ = =, which is not an eponential function. Classroom Eample For ƒ =, find each of the following. (a) ƒ- (b) ƒ5 (c) ƒa b (d) ƒ.5 Answers: (a) 6 (b) (c) (d) Eample Evaluating an Eponential Function For ƒ =, find each of the following. (a) ƒ- (b) ƒ (c) ƒa 5 b (d) ƒ.9 Solution (a) ƒ- = - = Replace with -. (b) ƒ = = 8 (c) ƒ a 5 b = 5/ = 5 / = / = = 6 # = (d) ƒ.9 = Use a calculator. Now Tr Eercises, 9, and. We repeat the final graph of = (with real numbers as domain) from Figure and summarize important details of the function ƒ = here. The -intercept is,. Because 7 for all and S as S -, the -ais is a horizontal asmptote. As the graph suggests, the domain of the function is -, and the range is,. The function is increasing on its entire domain. Therefore, it is one-to-one. These observations lead to the following generalizations about the graphs of eponential functions. Q, R 8 6 f() = (, ) (, ) Figure (repeated) (, ) Graph of ƒ = with domain H, H Copright Pearson. All Rights Reserved.

18 Chapter Inverse, Eponential, and Logarithmic Functions Eponential Function For ƒ = : ƒ Domain: -, (, a) f = a f() = a, a > Figure Range:, ƒ = a, for a +, is increasing and continuous on its entire domain, -,. The -ais is a horizontal asmptote as S -. (, ) (, a) The graph passes through the points A -, a B,,, and, a. f() = a, a > This is the general behavior seen on a calculator graph for an base a, for a +. For ƒ = A B : ƒ f() = a, < a < (, a) (, ) f() = a, < a < (, a) Figure 5 ƒ = a, for * a *, is decreasing and continuous on its entire domain, -,. The -ais is a horizontal asmptote as S. This is the general behavior seen on a calculator graph for an base a, for * a *. The graph passes through the points A -, a B,,, and, a. Recall that the graph of = ƒ- is the graph of = ƒ reflected across the -ais. Thus, we have the following. If ƒ =, then ƒ- = - = - # = - = a b. This is supported b the graphs in Figures and 5. The graph of ƒ = is tpical of graphs of ƒ = a where a 7. For larger values of a, the graphs rise more steepl, but the general shape is similar to the graph in Figure. When 6 a 6, the graph decreases in a manner similar to the graph of ƒ = A B in Figure 5. Copright Pearson. All Rights Reserved.

19 . Eponential Functions In Figure 6, the graphs of several tpical eponential functions illustrate these facts. Teaching Tip The point, lies on the graph of ƒ = a for all a. We can informall consider it home plate when graphing translations and reflections, as seen later in Eample. For < a <, the function is decreasing. f() = ( ) f() = ( ) f() = ( ) 5 f() = For a >, f() = f() = f() = a Domain: (, ); Range: (, ) the function is increasing. The -ais is a horizontal asmptote. Figure 6 In summar, the graph of a function of the form ƒ = a has the following features. Teaching Tip Encourage students to learn the characteristics of the graph of ƒ = a. Ask them how the coordinates, and, a are affected b different variations of the graph of ƒ = -a -h + k. Characteristics of the Graph of f = a. The points A -, a B,,, and, a are on the graph.. If a 7, then ƒ is an increasing function. If 6 a 6, then ƒ is a decreasing function.. The -ais is a horizontal asmptote.. The domain is -,, and the range is,. Classroom Eample Eample Graphing an Eponential Function Graph ƒ = a. b Give the domain and range. Answer: Graph ƒ = A 5 B. Give the domain and range. Solution The -intercept is,, and the -ais is a horizontal asmptote. Plot a few ordered pairs, and draw a smooth curve through them as shown in Figure 7. f() ( ) ƒ -, ;, (, 5) 5 f() = ( ) 5 5 (, 5) 5 (, 5) (, ) Figure 7 This function has domain -,, range,, and is one-to-one. It is decreasing on its entire domain. Copright Pearson. All Rights Reserved. Now Tr Eercise 9.

20 Chapter Inverse, Eponential, and Logarithmic Functions Classroom Eample Graph each function. Give the domain and range. (a) ƒ = - (b) ƒ = - (c) ƒ = + - Answers: (a) 5 f() = Eample Graphing Reflections and Translations Graph each function. Show the graph of = for comparison. Give the domain and range. (a) ƒ = - (b) ƒ = + (c) ƒ = - - Solution In each graph, we show in particular how the point, on the graph of = has been translated. (a) The graph of ƒ = - is that of ƒ = reflected across the -ais. See Figure 8. The domain is -,, and the range is -,. (b) The graph of ƒ = + is the graph of ƒ = translated units to the left, as shown in Figure 9. The domain is -,, and the range is,. (b) -, ; -, f() = (c) The graph of ƒ = - - is that of ƒ = translated units to the right and unit down. See Figure. The domain is -,, and the range is -,. = f() = (c) -, ;, f() = + 7 (, ) = (, ) f() = f() = + (, ) = (, ) (, ) (, ) = -, ; -, Figure 8 Figure 9 Figure Now Tr Eercises 9,, and 7. Eponential Equations Because the graph of ƒ = a is that of a one-to-one function, to solve a = a, we need onl show that =. This propert is used to solve an eponential equation, which is an equation with a variable as eponent. Classroom Eample Solve 5 = 5. Answer: 5-6 Eample Solving an Eponential Equation Solve A B = 8. Solution Write each side of the equation using a common base. a b = 8 - = 8 Definition of negative eponent - = 8 a m n = a mn 5 The -intercept of the graph of = A B - 8 can be used to verif the solution in Eample. - = Write 8 as a power of. - = Set eponents equal (Propert (b) given earlier). = - Multipl b -. Check b substituting - for in the original equation. The solution set is 5-6. Now Tr Eercise 7. Copright Pearson. All Rights Reserved.

21 . Eponential Functions 5 Classroom Eample 5 Solve + = 9 -. Answer: 576 Teaching Tip Now is a good time to review the properties of eponents. Teaching Tip Give an eample such as 6 - = 9 - that cannot be solved using the methods of this section. Show how a graphing calculator can be used to find the approimate solution.558. Eample 5 Solve + = 8-6. Solving an Eponential Equation Solution Write each side of the equation using a common base. + = = -6 Write 8 as a power of. + = -8 a m n = a mn + = - 8 Set eponents equal (Propert (b)). - = - Subtract and. = Divide b -. Check b substituting for in the original equation. The solution set is 56. Now Tr Eercise 8. Later in this chapter, we describe a general method for solving eponential equations where the approach used in Eamples and 5 is not possible. For instance, the above method could not be used to solve an equation like 7 = because it is not eas to epress both sides as eponential epressions with the same base. In Eample 6, we solve an equation that has the variable as the base of an eponential epression. Classroom Eample 6 Solve / = 5. Answer: 5{ 56 Teaching Tip Caution students that in solving equations like the one in Eample 6, the must consider both even roots. Eample 6 Solve / = 8. Solving an Equation with a Fractional Eponent Solution Notice that the variable is in the base rather than in the eponent. / = 8 A B = 8 = { = {7 Radical notation for a m/n Take fourth roots on each side. Remember to use {. Cube each side. Check both solutions in the original equation. Both check, so the solution set is 5 {76. Alternative Method There ma be more than one wa to solve an eponential equation, as shown here. / = 8 / = 8 Cube each side. = Write 8 as. = a m n = a mn = { Take fourth roots on each side. = { Simplif the radical. = { 7 Appl the eponent. The same solution set, 5{76, results. Now Tr Eercise 8. Copright Pearson. All Rights Reserved.

22 6 Chapter Inverse, Eponential, and Logarithmic Functions Compound Interest Recall the formula for simple interest, I = Prt, where P is principal (amount deposited), r is annual rate of interest epressed as a decimal, and t is time in ears that the principal earns interest. Suppose t = r. Then at the end of the ear, the amount has grown to the following. P + Pr = P + r Original principal plus interest If this balance earns interest at the same interest rate for another ear, the balance at the end of that ear will increase as follows. P + r + P + rr = P + r + r Factor. = P + r a # a = a After the third ear, the balance will grow in a similar pattern. P + r + P + r r = P + r + r Factor. = P + r a # a = a Continuing in this wa produces a formula for interest compounded annuall. A = P + r t The general formula for compound interest can be derived in the same wa. Compound Interest If P dollars are deposited in an account paing an annual rate of interest r compounded (paid) n times per ear, then after t ears the account will contain A dollars, according to the following formula. A = P a + r n b tn Classroom Eample 7 Repeat Eample 7 if $5 is deposited in an account paing % per ear compounded semi-annuall (twice per ear). Answers: (a) $67. (b) $867. Eample 7 Using the Compound Interest Formula Suppose $ is deposited in an account paing % interest per ear compounded quarterl (four times per ear). (a) Find the amount in the account after r with no withdrawals. (b) How much interest is earned over the -r period? Solution (a) A = P a + r n b t n Compound interest formula A = a +. b Let P =, r =., n =, and t =. A = +. Simplif. A = Round to the nearest cent. Thus, $88.86 is in the account after r. (b) The interest earned for that period is $ $ = $ Now Tr Eercise 97(a). Copright Pearson. All Rights Reserved.

23 . Eponential Functions 7 In the formula for compound interest A = P a + r n b t n, A is sometimes called the future value and P the present value. A is also called the compound amount and is the balance after interest has been earned. Classroom Eample 8 Brendan must pa a lump sum of $5, in 8 r. (a) What amount deposited toda (present value) at.% compounded annuall will give $5, in 8 r? (b) If onl $,5 is available to deposit now, what annual interest rate is necessar for the mone to increase to $5, in 8 r? Answers: (a) $,7.7 (b).8% Eample 8 Finding Present Value Beck must pa a lump sum of $6 in 5 r. (a) What amount deposited toda (present value) at.% compounded annuall will grow to $6 in 5 r? (b) If onl $5 is available to deposit now, what annual interest rate is necessar for the mone to increase to $6 in 5 r? Solution (a) A = P a + r n b t n 6 = P a +. 5 b 6 = P. 5 Simplif. Compound interest formula Let A = 6, r =., n =, and t = 5. P = 6. 5 Divide b. 5 to solve for P. P 55.6 Use a calculator. If Beck leaves $55.6 for 5 r in an account paing.% compounded annuall, she will have $6 when she needs it. Thus, $55.6 is the present value of $6 if interest of.% is compounded annuall for 5 r. (b) A = P a + r n b t n Compound interest formula 6 = 5 + r 5 Let A = 6, P = 5, n =, and t = = + r5 Divide b 5. a 6 /5 5 b = + r Take the fifth root on each side. a 6 /5 5 b - = r Subtract. r.7 Use a calculator. An interest rate of.7% will produce enough interest to increase the $5 to $6 b the end of 5 r. Now Tr Eercises 99 and. Caution When performing the computations in problems like those in Eamples 7 and 8, do not round off during intermediate steps. Keep all calculator digits and round at the end of the process. Copright Pearson. All Rights Reserved.

24 8 Chapter Inverse, Eponential, and Logarithmic Functions n a + n b n (rounded) ,.785,,.788 The Number e and Continuous Compounding The more often interest is compounded within a given time period, the more interest will be earned. Surprisingl, however, there is a limit on the amount of interest, no matter how often it is compounded. Suppose that $ is invested at % interest per ear, compounded n times per ear. Then the interest rate (in decimal form) is., and the interest rate per period is n. According to the formula (with P = ), the compound amount at the end of r will be A = a + n b n. A calculator gives the results in the margin for various values of n. The table suggests that as n increases, the value of A + n B n gets closer and closer to some fied number. This is indeed the case. This fied number is called e. (In mathematics, e is a real number and not a variable.) Figure = = e = Value of e e? Figure shows graphs of the functions =, =, and = e. Because 6 e 6, the graph of = e lies between the other two graphs. As mentioned above, the amount of interest earned increases with the frequenc of compounding, but the value of the epression A + n B n approaches e as n gets larger. Consequentl, the formula for compound interest approaches a limit as well, called the compound amount from continuous compounding. Teaching Tip Tell students that as n increases without bound, A + n B n approaches e. Continuous Compounding If P dollars are deposited at a rate of interest r compounded continuousl for t ears, then the compound amount A in dollars on deposit is given b the following formula. A = Pe rt Classroom Eample 9 Suppose $8 is deposited in an account paing 5% interest compounded continuousl for 6 r. Find the total amount on deposit at the end of 6 r. Answer: $, Eample 9 Solving a Continuous Compounding Problem Suppose $5 is deposited in an account paing % interest compounded continuousl for 5 r. Find the total amount on deposit at the end of 5 r. Solution A = Pe rt Continuous compounding formula A = 5e.5 Let P = 5, r =., and t = 5. A = 5e.5 Multipl eponents. A or $589.7 Use a calculator. Check that dail compounding would have produced a compound amount about $. less. Now Tr Eercise 97(b). Copright Pearson. All Rights Reserved.

25 . Eponential Functions 9 Classroom Eample In Classroom Eample 7, we found the total amount on deposit after r in an account paing % semiannuall in which $5 was invested. Find the amounts from the same investment for interest compounded quarterl, monthl, dail, and continuousl. Answer: quarterl: $7.87; monthl: $7.8; dail: $7.6; continuousl: $7.65 Eample Comparing Interest Earned as Compounding Is More Frequent In Eample 7, we found that $ invested at % compounded quarterl for r grew to $ Compare this same investment compounded annuall, semiannuall, monthl, dail, and continuousl. Solution Substitute. for r, for t, and the appropriate number of compounding periods for n into the formulas A = P a + r n b t n Compound interest formula and A = Pe rt. Continuous compounding formula The results for amounts of $ and $ are given in the table. Compounded $ $ Annuall +..8 $8. Semiannuall a +. b.8595 $85.95 Quarterl a +. b.8886 $88.86 Monthl a +. b.98 $9.8 Dail a b $9.79 Continuousl e..98 $9.8 Comparing the results for a $ investment, we notice the following. Looking Ahead To Calculus In calculus, the derivative allows us to determine the slope of a tangent line to the graph of a function. For the function ƒ = e, the derivative is the function ƒ itself: ƒ = e. Therefore, in calculus the eponential function with base e is much easier to work with than eponential functions having other bases. Compounding semiannuall rather than annuall increases the value of the account after r b $5.7. Quarterl compounding grows to $.9 more than semiannual compounding after r. Dail compounding ields onl $.96 more than monthl compounding. Continuous compounding ields onl $. more than dail compounding. Each increase in compounding frequenc earns less additional interest. Now Tr Eercise 5. Eponential Models The number e is important as the base of an eponential function in man practical applications. In situations involving growth or deca of a quantit, the amount or number present at time t often can be closel modeled b a function of the form = e kt, where is the amount or number present at time t = and k is a constant. Eponential functions are used to model the growth of microorganisms in a culture, the growth of certain populations, and the deca of radioactive material. Copright Pearson. All Rights Reserved.

26 Chapter Inverse, Eponential, and Logarithmic Functions Classroom Eample Refer to the model given in Eample. (a) What will be the atmospheric carbon dioide level in parts per million in 5? (b) Use a graph of this model to estimate when the carbon dioide level will be double the level that it was in. Answers: (a) 5 ppm (b) b Eample Using Data to Model Eponential Growth Data from recent ears indicate that future amounts of carbon dioide in the atmosphere ma grow according to the table. Amounts are given in parts per million. (a) Make a scatter diagram of the data. Do Year Carbon Dioide (ppm) the carbon dioide levels appear to grow 99 5 eponentiall? 75 (b) One model for the data is the function =.9e.69, 75 9 where is the ear and Use a graph of this model to estimate when future levels of carbon dioide will double and triple over the preindustrial level of 8 ppm. 75 Source: International Panel on Climate Change (IPCC). 975 (a) =.9e.69 Solution (a) We show a calculator graph for the data in Figure (a). The data appear to resemble the graph of an increasing eponential function. (b) A graph of =.9e.69 in Figure (b) shows that it is ver close to the data points. We graph = # 8 = 56 in Figure (a) and = # 8 = 8 in Figure (b) on the same coordinate aes as the given function, and we use the calculator to find the intersection points. =.9e.69 =.9e (b) Figure = 56 = 8 (a) (b) Figure (a) =.9(.69) The graph of the function intersects the horizontal lines at -values of approimatel 6. and.9. According to this model, carbon dioide levels will have doubled during 6 and tripled b. Now Tr Eercise 7. Graphing calculators are capable of fitting eponential curves to scatter diagrams like the one found in Eample. The TI-8 Plus displas another (different) equation in Figure (a) for the atmospheric carbon dioide eample, approimated as follows. 975 =.9.69 (b) Figure This calculator form differs from the model in Eample. Figure (b) shows the data points and the graph of this eponential regression equation. Copright Pearson. All Rights Reserved.

27 . Eponential Functions. Eercises. 6; 6. rises. falls. -, ;, 5. 8 ; ; 8 6. ; ; { f() = f() =. Concept Preview Fill in the blank(s) to correctl complete each sentence.. If ƒ =, then ƒ = and ƒ- =.. If a 7, then the graph of ƒ = a from left to right. (rises/falls). If 6 a 6, then the graph of ƒ = a from left to right. (rises/falls). The domain of ƒ = is and the range is. 5. The graph of ƒ = 8 passes through the points (-, ), (, ), and (, ). 6. The graph of ƒ = - A B is that of ƒ = A B reflected across the -ais, translated units to the left and units down. Concept Preview Solve each equation. Round answers to the nearest hundredth as needed. 7. a b = 6 8. / = 6 9. A = a +. b 8., = 5 + r 5 f() = ( ) f() = ( ) For ƒ = and g = A B, find each of the following. Round answers to the nearest thousandth as needed. See Eample.. ƒ. ƒ. ƒ-. ƒ g 6. g 7. g- 8. g- 5 f() = ( ) f() = ( 5) 9. ƒa b. ƒa - 5 b. ga b. ga - 5 b. ƒ.. ƒ g g.. f() = ( ) f() = f() = f() = ( ) 6 f() = 8. f() = Graph each function. See Eample. 7. ƒ = 8. ƒ = 9. ƒ = a b. ƒ = a b. ƒ = a b. ƒ = a 5 b. ƒ = a - b. ƒ = a - 6 b 5. ƒ = - 6. ƒ = - 7. ƒ = 8. ƒ = - Graph each function. Give the domain and range. See Eample. 9. ƒ = +. ƒ = -. ƒ = +. ƒ = -. ƒ = - +. ƒ = ƒ = - 6. ƒ = ƒ = ƒ = ƒ = ƒ = - - Copright Pearson. All Rights Reserved.

28 Chapter Inverse, Eponential, and Logarithmic Functions 9. -, ;,. 8 f() = + -, ;,. 5. f() = + 8 = f() = + -, ; -, f() = -, ;, 7. f() =. Graph each function. Give the domain and range. See Eample. 5. ƒ = a b - 5. ƒ = a b + 5. ƒ = a + b 5. ƒ = a b - -, ; -, 57. ƒ = a -. b 8 f() = -, ;,. 6. f() = f() = -, ; -, = f() = -, ; -, ƒ = a b ƒ = - a b ƒ = a b ƒ = a - b + 6. ƒ = a - b + 6. ƒ = a + b - 6. ƒ = a + b - Connecting Graphs with Equations Write an equation for the graph given. Each represents an eponential function ƒ with base or, translated and/or reflected (, ) (, 7) (, ) (, ) = (, 7) (, 5) = (, ) (, ) (, ) (, 9) (, ) (, ) = (, ) (, ) (, ) (, ) = (, 7) = (, ) -, ;, 9. = -, ;, 5. = f() = + f() = 69. (, ) ( = (, ), ) 7. = 5 (, ) (, ) (, ) = f() = + -, ; -, = Solve each equation. See Eamples = 7. 5 = 5 7. a 5 b = 7. a 5 b = 9 -, ; -, = = e - = e 78. e - = e - 5. = 9 f() = ( ) 5. = -, ; -, -, ;, 5 f() = ( ) = = = = / = 8. /5 = / = 86. / = = = / = /5 = a - e b = a + e b 9. e - = a + e b 9. A B + = 9. A 5 B - = a 5 b = = -5 Copright Pearson. All Rights Reserved.

29 . Eponential Functions 5. f() = ( ) + -, ;, f() = ( ) + -, ;, , ;, -, ;, 6. 9 = f() = ( ) f() = ( ) 5. -, ;, 9 6 f() = ( ) -, ;, f() = ( ) 9 -, ; -, = -, ;, f() = ( ) f() = ( ) Solve each problem. See Eamples Future Value Find the future value and interest earned if $896.5 is invested for 9 r at % compounded (a) semiannuall (b) continuousl. 98. Future Value Find the future value and interest earned if $56,78 is invested at.8% compounded (a) quarterl for quarters (b) continuousl for 5 r. 99. Present Value Find the present value that will grow to $5, if interest is.% compounded quarterl for quarters.. Present Value Find the present value that will grow to $5, if interest is.6% compounded monthl for r.. Present Value Find the present value that will grow to $5 if interest is.5% compounded quarterl for r.. Interest Rate Find the required annual interest rate to the nearest tenth of a percent for $65, to grow to $65,78.9 if interest is compounded monthl for 6 months.. Interest Rate Find the required annual interest rate to the nearest tenth of a percent for $ to grow to $5 if interest is compounded quarterl for 9 r.. Interest Rate Find the required annual interest rate to the nearest tenth of a percent for $5 to grow to $6 if interest is compounded quarterl for 8 r. Solve each problem. See Eample. 5. Comparing Loans Bank A is lending mone at 6.% interest compounded annuall. The rate at Bank B is 6.% compounded monthl, and the rate at Bank C is 6.5% compounded quarterl. At which bank will we pa the least interest? 6. Future Value Suppose $, is invested at an annual rate of.% for r. Find the future value if interest is compounded as follows. (a) annuall (b) quarterl (c) monthl (d) dail (65 das) 8 f() = ( ) + = -, ; -, , ; -, f() = ( ) + = 6. ƒ = - 6. ƒ = ƒ = ƒ = ƒ = ƒ = ƒ = ƒ = (Modeling) Solve each problem. See Eample. 7. Atmospheric Pressure The atmospheric pressure (in millibars) at a given altitude (in meters) is shown in the table. Altitude Pressure Altitude Pressure , Source: Miller, A. and J. Thompson, Elements of Meteorolog, Fourth Edition, Charles E. Merrill Publishing Compan, Columbus, Ohio. (a) Use a graphing calculator to make a scatter diagram of the data for atmospheric pressure P at altitude. (b) Would a linear or an eponential function fit the data better? (c) The following function approimates the data. P = e -. Use a graphing calculator to graph P and the data on the same coordinate aes. (d) Use P to predict the pressures at 5 m and, m, and compare them to the actual values of 86 millibars and 7 millibars, respectivel. Copright Pearson. All Rights Reserved.

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