Calculus, 10th Edition (Anton)

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