   Chapter 4.4, Problem 71E ### 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 2 x − 1 g ( x ) = 1 2 + ln x

To determine

To prove: The functions f(x)=e2x1 and g(x)=12+ln(x) are inverse functions and also verify this by using technology.

Explanation

Given Information:

The provided functions are f(x)=e2x1 and g(x)=12+ln(x).

Formula used:

The property of logarithms for the expression lnxn is lnxn=nlnx where x and y are real numbers greater than 0.

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)=e2x1 and g(x)=12+ln(x)

Apply the property of lnxn in the function g(x) as,

g(x)=12+ln(x)=12+12lnx

The value of f(g(x)) is computed by substituting x=12+12lnx in the function f(x) as,

f(g(x))=f(12+12lnx)=e2(12+12lnx)1

Solve further as,

f(g(x))=e22+2lnx21=e1+lnx1=elnx

Now, apply the property of elnx as,

f(g(x))=elnx=x

Now, the value of g(f(x))

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