Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.2 One-to-One Functions; Inverse Functions - 4.2 Assess Your Understanding - Page 292: 39

Answer

$f[g(x)]=g[f(x)]=x$. This means that $f(x)$ and $g(x)$ are inverses of each other. We know that $x=0$ has to be excluded from both domains because the denominator cannot be $0$ for the function to be undefined.

Work Step by Step

We wish to plug $f(x)$ into $g(x)$ to obtain: $$\displaystyle g[f(x)]=g (\dfrac{1}{x}) \\ =\dfrac{1}{(1/x)} \\=x$$ We wish to plug $g(x)$ into $f(x)$ to obtain: $$f[(g(x)]=f (\dfrac{1}{x}) \\ =\dfrac{1}{(1/x)}\\=x$$ We see that $f[g(x)]=g[f(x)]=x$. This means that $f(x)$ and $g(x)$ are inverses of each other. We know that $x=0$ has to be excluded from both domains because the denominator cannot be $0$ for the function to be undefined.
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