   Chapter 4.4, Problem 70E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Inverse Functions In Exercises 69-72, analytically show that the functions are inverse functions. Then use a graphing utility to show this graphically. f ( x ) = e x − 1 g ( x ) = ln ( x + 1 )

To determine

To prove: The functions f(x)=ex1 and g(x)=ln(x+1) are inverse functions and also verify this by using technology.

Explanation

Given Information:

The provided functions are f(x)=ex1 and g(x)=ln(x+1).

Formula used:

The function g is the inverse function of the function f if the value of f(g(x)) and g(f(x)) is equal to x for every x in the domain of f and g.

The inverse property of exponents for the expression elnx is elnx=x.

The inverse property of logarithms for the expression lnex is lnex=x.

Proof:

Consider the functions, f(x)=ex1 and g(x)=ln(x+1)

The value of f(g(x)) is computed by substituting x=ln(x+1) in the function f(x) as,

f(g(x))=f(ln(x+1))=eln(x+1)1

Now, apply the property of elnx as,

f(g(x))=eln(x+1)1=x+11=x

Now, the value of g(f(x)) is computed by substituting x=ex1 in the function g(x) as,

g(f(x)

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