Precalculus (6th Edition) Blitzer

The required values are $f\left( g\left( x \right) \right)=x$ and $g\left( f\left( x \right) \right)=x$. And the functions $f\left( x \right)=\sqrt[3]{x-4}$ and $g\left( x \right)={{x}^{3}}+4$ are inverses of each other.
Consider the functions: $f\left( x \right)=\sqrt[3]{x-4}$ and $g\left( x \right)={{x}^{3}}+4$ The equation for $f$ is given as: $f\left( x \right)=\sqrt[3]{x-4}$ Replace $x$ with $g\left( x \right)$ \begin{align} & f\left( g\left( x \right) \right)=\sqrt[3]{g\left( x \right)-4} \\ & =\sqrt[3]{\left( {{x}^{3}}+4 \right)-4} \\ & =\sqrt[3]{{{x}^{3}}} \\ & =x \end{align} Now, to find $g\left( f\left( x \right) \right)$ Consider the function $g\left( x \right)$: $g\left( x \right)={{x}^{3}}+4$ Replace $x$ with $f\left( x \right)$ \begin{align} & g\left( f\left( x \right) \right)={{\left( f\left( x \right) \right)}^{3}}+4 \\ & ={{\left( \sqrt[3]{x-4} \right)}^{3}}+4 \\ & =\left( x-4 \right)+4 \\ & =x \end{align} Because $g$ is inverse of $f$ (and vice-versa), the inverse notation can be used: $f\left( x \right)=\sqrt[3]{x-4}$ and $\text{ }{{f}^{-1}}\left( x \right)={{x}^{3}}+4$ It can be easily observed that ${{f}^{-1}}$ can be expressed in $f$ if cube root property is applied. \begin{align} & \text{ }{{f}^{-1}}\left( x \right)={{x}^{3}}+4 \\ & =g\left( x \right) \end{align} Hence, the required values are $f\left( g\left( x \right) \right)=x$ and $g\left( f\left( x \right) \right)=x$. And the functions $f\left( x \right)=\sqrt[3]{x-4}$ and $g\left( x \right)={{x}^{3}}+4$ are inverses of each other.