## Precalculus (6th Edition)

Published by Pearson

# Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.1 Inverse Functions - 4.1 Exercises - Page 416: 7

#### Answer

$x^{1/3}$ or $\sqrt[3]{x}$

#### Work Step by Step

To find the equation for $f^{-1}$, replace $f(x)$ with $y$, interchange $x$ and $y$, and solve for $y$. This gives $f^{-1}(x)$. ------------ $y=x^{3}\qquad$ ... swap x and y $x=y^{3}\qquad /(...)^{1/3}\quad$... solve for y $x^{1/3}=y$ (RHS:$(y^{3})^{1/3}=y^{3\cdot\frac{1}{3}}=y^{1}=y)$ $y=f^{-1}(x)=x^{1/3}$

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