286 CHAPTER 5 Exponential and Logarithmic Functions 4 Verify that a Function Defined by an Equation Is an Inverse Function Suppose f is a one-to-one function. Then f has an inverse function −f .1 Figure 14 shows the relationship between the domain and range of f and the domain and range of −f .1 Based on the results of Examples 3 and 4 and Figure 14, two facts are now apparent about a one-to-one function f and its inverse −f .1 Figure 14 f f 21 x Domain of f Range of f Domain of f 21 Range of f 21 • = − f f Domainof Rangeof 1 • = − f f Rangeof Domainof 1 CAUTION Be careful! f 1− is a symbol for the inverse function of f. The 1− used in f 1− is not an exponent. That is, f 1− does not mean the reciprocal of f ; f x 1( ) − is not equal to f x 1 . ( ) j Look again at Figure 14 to visualize the relationship. Starting with x , applying f, and then applying −f 1 gets x back again. Starting with x , applying −f ,1 and then applying f gets the number x back again. To put it simply, what f does, −f 1 undoes, and vice versa. See the illustration that follows. x f Input from domain of x f Input from domain of 1− f x( ) f x 1( ) − f f x x 1( ) ( ) = − f f x x 1 ( ) ( ) = − f Apply f Apply 1− f Apply 1− f Apply In other words, Consider the function ( ) = f x x2 , which multiplies the argument x by 2. The inverse function −f 1 undoes whatever f does. So the inverse function of f is ( ) = −f x x 1 2 , 1 which divides the argument by 2. For example, ( ) = ⋅ = f 3 2 3 6 and ( ) = ⋅ = −f 6 1 2 6 3, 1 so −f 1 undoes f. This is verified by showing that ( ) ( ) ( ) ( ) ( ) ( ) = = ⋅ = = = ⋅ = − − − f f x f x x x f f x f x x x 2 1 2 2 and 1 2 2 1 2 . 1 1 1 See Figure 15. Figure 15 x f f 21 ∙2x = x f21(2x) = f(x) = 2x 1 – 2 • ( ) ( ) = −f f x x 1 where x is in the domain of f • ( ) ( ) = − f f x x 1 where x is in the domain of −f 1 Verifying Inverse Functions (a) Verify that the inverse of ( ) = g x x3 is ( ) = −g x x. 1 3 (b) Verify that the inverse of ( ) = + f x x2 3 is f x x 1 2 3 . 1( ) ( ) = − − Solution EXAMPLE 6 (a) g g x g x x x 1 1 3 3 3 ( ) ( ) ( ) = = = − − for all x in the domain of g g g x g x x x 1 3 3 3 ( ) ( ) ( ) ( ) = = = − for all x in the domain of −g 1 (b) f f x f x x x x 2 3 1 2 2 3 3 1 2 2 1 1 ( ) ( ) ( ) ( ) [ ] = + = + − = ⋅ = − − for all x in the domain of f f f x f x x x 1 2 3 2 1 2 3 3 1 ( ) ( ) ( ) ( ) ( ) = − = ⋅ − + = − for all x in the domain of −f 1
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