Answer
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.
Work Step by Step
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.
