Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.8 - Modeling Using Variation - Exercise Set - Page 426: 60


${{f}^{-1}}\left( x \right)=\sqrt[3]{x-2}$

Work Step by Step

Inverse of a function $f\left( x \right)$ is defined by a function ${{f}^{-1}}\left( x \right)$ such that the domain $f\left( x \right)$ is equal to the range of ${{f}^{-1}}\left( x \right)$. This means that if the function $f$ is the set of ordered pairs $\left( x,y \right)$ , then ${{f}^{-1}}$ is the set of ordered pairs $\left( y,x \right)$. And if we interchange the values of x and y in the equation function $f$ , it will give the equation for function ${{f}^{-1}}$. The provided equation is $f\left( x \right)={{x}^{3}}+2$. Now replace $f\left( x \right)$ with y: $y={{x}^{3}}+2$ Interchange x and y to provide the inverse function as follows: $x={{y}^{3}}+2$ Subtract 2 from both sides of the equation. $x-2={{y}^{3}}$ Take the cubic root of y. $\begin{align} & \sqrt[3]{x-2}=\sqrt[3]{{{y}^{3}}} \\ & \sqrt[3]{x-2}=y \\ \end{align}$ Now, $y={{f}^{-1}}\left( x \right)$ ${{f}^{-1}}\left( x \right)=\sqrt[3]{x-2}$ Thus, the inverse of $f\left( x \right)={{x}^{3}}+2$ is ${{f}^{-1}}\left( x \right)=\sqrt[3]{x-2}$
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