Precalculus (6th Edition) Blitzer

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

Chapter 1 - Review Exercises - Page 303: 109


See the explanation below.

Work Step by Step

(a) Consider $f\left( x \right)=8{{x}^{3}}+1$ Step 1: Replace $f\left( x \right)$ with $y$: $y=8{{x}^{3}}+1$ Step 2: Interchange $x$ and $y$: $x=8{{y}^{3}}+1$ Step 3: Now solve for the value of $y$: $x-1=8{{y}^{3}}$ Or ${{y}^{3}}=\frac{x-1}{8}$ That is, $y=\sqrt[3]{\frac{x-1}{8}}$ Step 4: Replace $y$ with ${{f}^{-1}}\left( x \right)$: ${{f}^{-1}}\left( x \right)=\sqrt[3]{\frac{x-1}{8}}$ Hence, the inverse function ${{f}^{-1}}\left( x \right)$ of the $f\left( x \right)=8{{x}^{3}}+1$ is ${{f}^{-1}}\left( x \right)=\sqrt[3]{\frac{x-1}{8}}$. (b) Consider the function, $f\left( {{f}^{-1}}\left( x \right) \right)$ $\begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \sqrt[3]{\frac{x-1}{8}} \right) \\ & =8{{\left( \sqrt[3]{\frac{x-1}{8}} \right)}^{3}}+1 \\ & =8\left( \frac{x-1}{8} \right)+1 \\ & =x \end{align}$ Next consider the function, ${{f}^{-1}}\left( f\left( x \right) \right)$ $\begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( 8{{x}^{3}}+1 \right) \\ & =\sqrt[3]{\frac{\left( 8{{x}^{3}}+1 \right)-1}{8}} \\ & =\sqrt[3]{{{x}^{3}}} \\ & =x \end{align}$ Hence, $f\left( {{f}^{-1}}\left( x \right) \right)=x$ and ${{f}^{-1}}\left( f\left( x \right) \right)=x$ for the function $f\left( x \right)=8{{x}^{3}}+1$.
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