Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - 12.1 Composite Functions and Inverse Functions - 12.1 Exercise Set - Page 788: 66

Answer

The graph is shown in below.

Work Step by Step

$g\left( x \right)=\frac{1}{3}{{x}^{3}}$ Evaluate the inverse of the function $g\left( x \right)=\frac{1}{3}{{x}^{3}}$ as follows. Replace the function $g\left( x \right)$ with y. $y=\frac{1}{3}{{x}^{3}}$ Interchange the variables x and y. $x=\frac{1}{3}{{y}^{3}}$ Solve for y value. $\begin{align} & 3x={{y}^{3}} \\ & y=\sqrt[3]{3x} \\ \end{align}$ Replace y with ${{g}^{-1}}\left( x \right)$ as follows. ${{g}^{-1}}\left( x \right)=\sqrt[3]{3x}$ Thus, the inverse function is ${{g}^{-1}}\left( x \right)=\sqrt[3]{3x}$. Consider the function. $g\left( x \right)=\frac{1}{3}{{x}^{3}}$ Substitute $x=-1.732,0,1.732$ in the function $g\left( x \right)=\frac{1}{3}{{x}^{3}}$. For $x=-1.732$, the value of $g\left( x \right)$ is, $\begin{align} & g\left( -1.732 \right)=\frac{1}{3}{{\left( -1.732 \right)}^{3}} \\ & =-1.732 \end{align}$ For $x=0$, the value of $g\left( x \right)$ is, $\begin{align} & g\left( 0 \right)=\frac{1}{3}{{\left( 0 \right)}^{3}} \\ & =0 \end{align}$ For $x=1.732$, the value of $g\left( x \right)$ is, $\begin{align} & g\left( 1.732 \right)=\frac{1}{3}{{\left( 1.732 \right)}^{3}} \\ & =1.732 \end{align}$ Tabulate for obtained values as shown below. $\begin{matrix} x & g\left( x \right)=\frac{1}{3}{{x}^{3}} \\ -1.732 & -1.732 \\ 0 & 0 \\ 1.732 & 1.732 \\ \end{matrix}$ Consider the function. ${{g}^{-1}}\left( x \right)=\sqrt[3]{3x}$ Substitute $x=-1.732,0,1.732$ in the function ${{g}^{-1}}\left( x \right)=\sqrt[3]{3x}$. For $x=-1.732$, the value of ${{g}^{-1}}\left( x \right)$ is, $\begin{align} & {{g}^{-1}}\left( -1.732 \right)=\sqrt[3]{3\left( -1.732 \right)} \\ & =\sqrt[3]{-5.196} \\ & =-1.732 \end{align}$ For $x=0$, the value of ${{g}^{-1}}\left( x \right)$ is, $\begin{align} & {{g}^{-1}}\left( 0 \right)=\sqrt[3]{3\left( 0 \right)} \\ & =\sqrt[3]{0} \\ & =0 \end{align}$ For $x=1.732$, the value of ${{g}^{-1}}\left( x \right)$ is, $\begin{align} & {{g}^{-1}}\left( 0 \right)=\sqrt[3]{3\left( 1.732 \right)} \\ & =\sqrt[3]{5.196} \\ & =1.732 \end{align}$ Tabulate for obtained values as shown below. $\begin{matrix} x & {{g}^{-1}}\left( x \right)=\sqrt[3]{2x} \\ -1.732 & -1.732 \\ 0 & 0 \\ 1.732 & 1.732 \\ \end{matrix}$ Plot these points and sketch the graphs of the functions $g\left( x \right)=\frac{1}{3}{{x}^{3}}$ and ${{g}^{-1}}\left( x \right)=\sqrt[3]{3x}$ as shown in the figure below.
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