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: 20

Answer

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Work Step by Step

(a) Consider the function: $f\left( x \right)={{\left( x-1 \right)}^{3}}$ Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$. $y={{\left( x-1 \right)}^{3}}$ Step 2: Interchange the variables x and y. $x={{\left( y-1 \right)}^{3}}$ Step 3: Solve the equation for y. The variable y has to be isolated. As $\sqrt[3]{{{y}^{3}}}=y$ , take the cube root of both sides of the equation. So, the equation becomes, ${{\left( x \right)}^{{1}/{3}\;}}={{\left( {{\left( y-1 \right)}^{3}} \right)}^{{1}/{3}\;}}$ Simplify the powers. So, $\sqrt[3]{x}=y-1$ Add $1$ to both sides of the equation and simplify. $\begin{align} & \sqrt[3]{x}+1=y-1+1 \\ & \sqrt[3]{x}+1=y \\ \end{align}$ Because the obtained equation defines y as a function of x, then the inverse function exists for the function f. Step 4: Replace y by ${{f}^{-1}}\left( x \right)$ in the equation obtained in step 3. Thus, the inverse function ${{f}^{-1}}\left( x \right)$ is obtained as, ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}+1$ Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}+1$. (b) Consider the function: $f\left( x \right)={{\left( x-1 \right)}^{3}}$. The inverse of the function is obtained as ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}+1$ in part (a). The inverse of an equation can be verified by showing that $f\left( {{f}^{-1}}\left( x \right) \right)=x\text{ and }{{f}^{-1}}\left( f\left( x \right) \right)=x\text{ }$. First calculate $f\left( {{f}^{-1}}\left( x \right) \right)$. Replace ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}+1$ and evaluate the function f. $\begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \sqrt[3]{x}+1 \right) \\ & ={{\left( \sqrt[3]{x}+1-1 \right)}^{3}} \\ & ={{\left( \sqrt[3]{x} \right)}^{3}} \\ & =x \end{align}$ Now, calculate ${{f}^{-1}}\left( f\left( x \right) \right)$. Replace $f\left( x \right)={{\left( x-1 \right)}^{3}}$ and evaluate the inverse function ${{f}^{-1}}$. $\begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( {{\left( x-1 \right)}^{3}} \right) \\ & =\sqrt[3]{{{\left( x-1 \right)}^{3}}}+1 \\ & =x-1+1 \\ & =x \end{align}$ Thus, it verifies that $f\left( {{f}^{-1}}\left( x \right) \right)=x\text{ and }{{f}^{-1}}\left( f\left( x \right) \right)=x\text{ }$. Therefore, the obtained inverse function ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}+1$ is correct. Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.
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