Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.8 - Inverse Functions - Exercise Set - Page 269: 10

Answer

The required values are $f\left( g\left( x \right) \right)=x$ and $g\left( f\left( x \right) \right)=x$. And the functions $f\left( x \right)=\sqrt[3]{x-4}$ and $g\left( x \right)={{x}^{3}}+4$ are inverses of each other.

Work Step by Step

Consider the functions: $f\left( x \right)=\sqrt[3]{x-4}$ and $g\left( x \right)={{x}^{3}}+4$ The equation for $f$ is given as: $f\left( x \right)=\sqrt[3]{x-4}$ Replace $x$ with $g\left( x \right)$ $\begin{align} & f\left( g\left( x \right) \right)=\sqrt[3]{g\left( x \right)-4} \\ & =\sqrt[3]{\left( {{x}^{3}}+4 \right)-4} \\ & =\sqrt[3]{{{x}^{3}}} \\ & =x \end{align}$ Now, to find $g\left( f\left( x \right) \right)$ Consider the function $g\left( x \right)$: $g\left( x \right)={{x}^{3}}+4$ Replace $x$ with $f\left( x \right)$ $\begin{align} & g\left( f\left( x \right) \right)={{\left( f\left( x \right) \right)}^{3}}+4 \\ & ={{\left( \sqrt[3]{x-4} \right)}^{3}}+4 \\ & =\left( x-4 \right)+4 \\ & =x \end{align}$ Because $g$ is inverse of $f$ (and vice-versa), the inverse notation can be used: $f\left( x \right)=\sqrt[3]{x-4}$ and $\text{ }{{f}^{-1}}\left( x \right)={{x}^{3}}+4$ It can be easily observed that ${{f}^{-1}}$ can be expressed in $f$ if cube root property is applied. $\begin{align} & \text{ }{{f}^{-1}}\left( x \right)={{x}^{3}}+4 \\ & =g\left( x \right) \end{align}$ Hence, the required values are $f\left( g\left( x \right) \right)=x$ and $g\left( f\left( x \right) \right)=x$. And the functions $f\left( x \right)=\sqrt[3]{x-4}$ and $g\left( x \right)={{x}^{3}}+4$ are inverses of each other.
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