## Calculus 10th Edition

Published by Brooks Cole

# Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.3 Exercises: 8

#### Answer

$\text{a) The functions are inverses of each other.}$ $\text{b) Please see "step by step" for graph and explanation.}$

#### Work Step by Step

If f(x) and g(x) are inverses (of each other, then a. $f(g(x))=x$ and $g(f(x))=x$ for x in respective domains (of g and f). (definition, p.337) b. The graphs of f and g are reflections of each other across the line x=y. (Theorem 5.6, figure 5.12) ------------------ a. $f(g(x))=\displaystyle \frac{1}{1+g(x)}=\frac{1}{1+\frac{1-x}{x}}$ $=\displaystyle \frac{1}{\frac{x+1-x}{x}}=\frac{1}{\frac{1}{x}}=x$ $g(f(x)) =\displaystyle \frac{1-f(x)}{f(x)}=\frac{1-\frac{1}{1+x}}{\frac{1}{1+x}}$ $=\displaystyle \frac{\frac{(1+x)-1}{1+x}}{\frac{1}{1+x}}=\frac{\frac{x}{1+x}}{\frac{1}{1+x}}=x$ b. Using a graphing device, (see below) we see that the blue graph (f(x) ) and the red graph (g(x)) are reflections of each other about the line x=y: .

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