## Precalculus (6th Edition)

$x^{1/3}$ or $\sqrt[3]{x}$
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:$(y^{3})^{1/3}=y^{3\cdot\frac{1}{3}}=y^{1}=y)$ $y=f^{-1}(x)=x^{1/3}$