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Copyright © 2007 Pearson Education, Inc. Slide 5-1 Chapter 5: Inverse, Exponential and Logarithmic Functions 5.1 5.2 5.3 5.4 5.5 Inverse Functions Exponential Functions Logarithms and Their Properties Logarithmic Functions Exponential and Logarithmic Equations and Inequalities 5.6 Further Applications and Modeling with Exponential and Logarithmic Functions Copyright © 2007 Pearson Education, Inc. Slide 5-2 5.1 Inverse Functions 1 Example Let f ( x) 8 x and g ( x) x. 8 f (12) 8 12 96 1 g (96) 96 12 i.e. g[ f (12)] 12 8 Also, f [g(12)] = 12. For these functions, it can be shown that f [ g ( x)] x and g[ f ( x)] x for any value of x. These functions are inverse functions of each other. Copyright © 2007 Pearson Education, Inc. Slide 5-3 5.1 One-to-One Functions A function f is a one-to-one function if, for elements a and b from the domain of f, ab • implies f (a) f (b). Only functions that are one-to-one have inverses. Copyright © 2007 Pearson Education, Inc. Slide 5-4 5.1 One-to-One Functions Example Decide whether the function is one-to-one. (a) f ( x) 4 x 12 (b) f ( x) 25 x 2 Solution (a) For this function, two different x-values produce two different y-values. Suppose that a b, then 4a 4b and 4a 12 4b 12. Since f (a ) f (b), f is one - to - one. (b) If we choose a = 3 and b = –3, then 3 –3, but f (3) 25 32 4 and f (3) 25 (3) 2 4, so f (3) f (3), therefore f is not one - to - one. Copyright © 2007 Pearson Education, Inc. Slide 5-5 5.1 The Horizontal Line Test If every horizontal line intersects the graph of a function at no more than one point, then the function is one-to-one. Example Use the horizontal line test to determine whether the graphs are graphs of one-to-one functions. (a) (b) Not one-to-one Copyright © 2007 Pearson Education, Inc. One-to-one Slide 5-6 5.1 Inverse Functions Let f be a one-to-one function. Then, g is the inverse function of f and f is the inverse of g if ( f g )( x) x for every x in the domain of g , and ( g f )( x) x for every x in the domain of f . Example Show that f ( x) x3 1 and g ( x) 3 x 1 are inverse functions of each other. ( f g )( x) f [ g ( x)] x 1 1 x 1 1 x 3 3 ( g f )( x) g[ f ( x)] 3 x3 1 1 3 x3 x Copyright © 2007 Pearson Education, Inc. Slide 5-7 5.1 Finding an Equation for the Inverse Function • Notation for the inverse function f -1 is read “f-inverse” Finding the Equation of the Inverse of y = f(x) 1. Interchange x and y. 2. Solve for y. 3. Replace y with f -1(x). Any restrictions on x and y should be considered. Copyright © 2007 Pearson Education, Inc. Slide 5-8 5.1 Example of Finding f -1(x) Example Find the inverse, if it exists, of 4x 6 f ( x) . 5 4x 6 Write f (x) = y. y Solution 5 4y 6 Interchange x and y. x 5 Solve for y. 5x 4 y 6 5x 6 y 4 5x 6 1 Replace y with f -1(x). f ( x) 4 Copyright © 2007 Pearson Education, Inc. Slide 5-9 5.1 The Graph of f -1(x) • f and f -1(x) are inverse functions, and f (a) = b for real numbers a and b. Then f -1(b) = a. • If the point (a,b) is on the graph of f, then the point (b,a) is on the graph of f -1. If a function is one-to-one, the graph of its inverse f -1(x) is a reflection of the graph of f across the line y = x. Copyright © 2007 Pearson Education, Inc. Slide 5-10 5.1 Finding the Inverse of a Function with a Restricted Domain Example Let f ( x) x 5. Find f 1 ( x). Solution Notice that the domain of f is restricted to [–5,), and its range is [0, ). It is one-to-one and thus has an inverse. y x5 x y5 x2 y 5 2 y x 5 The range of f is the domain of f -1, so its inverse is f 1 ( x) x 2 5, x 0. Copyright © 2007 Pearson Education, Inc. Slide 5-11 5.1 Important Facts About Inverses 1. If f is one-to-one, then f -1 exists. 2. The domain of f is the range of f -1, and the range of f is the domain of f -1. 3. If the point (a,b) is on the graph of f, then the point (b,a) is on the graph of f -1, so the graphs of f and f -1 are reflections of each other across the line y = x. Copyright © 2007 Pearson Education, Inc. Slide 5-12 5.1 Application of Inverse Functions Example Use the one-to-one function f(x) = 3x + 1 and the numerical values in the table to code the message BE VERY CAREFUL. A 1 F 6 K 11 P 16 U 21 B 2 G 7 L 12 Q 17 V 22 C 3 H 8 M 13 R 18 W 23 D 4 I 9 N 14 S 19 X 24 E 5 J 10 O 15 T 20 Y 25 Z 26 Solution BE VERY CAREFUL would be encoded as 7 16 67 16 55 76 10 4 55 16 19 64 37 because B corresponds to 2, and f(2) = 3(2) + 1 = 7, and so on. Copyright © 2007 Pearson Education, Inc. Slide 5-13