## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 0 - Before Calculus - 0.4 Inverse Functions - Exercises Set 0.4 - Page 45: 11

#### Answer

$f^{-1}(x)$ = $\sqrt[3] \frac{x + 5}{3}$.

#### Work Step by Step

1. Switch $f^{}(x)$ and $x^{}$. x = 3*$(y)^3$ - 5 *Lets call f(x) by "y" to simplificate. 2. Isolate "y". x + 5 = 3$y^{3}$ $y^{3}$ = $\frac{x + 5}{3}$ y = $\sqrt[3] \frac{x + 5}{3}$. 3. Rename y as $f^{-1}(x)$ $f^{-1}(x)$ = $\sqrt[3] \frac{x + 5}{3}$.

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