Answer
Proved below.
Work Step by Step
$h\left( x \right)=\left( f\circ g \right)\left( x \right)$
The function $I\left( x \right)$ is defined as $I\left( x \right)=x$; then, $\left( f\circ I \right)\left( x \right)=f\left( I\left( x \right) \right)$ for any function f.
To prove that ${{h}^{-1}}\left( x \right)=\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( x \right)$ it is enough to show that $\left( \left( {{g}^{-1}}\circ {{f}^{-1}} \right)\circ h \right)\left( x \right)=\left( h\circ \left( {{g}^{-1}}\circ {{f}^{-1}} \right) \right)\left( x \right)=x$.
Evaluate $\left( \left( {{g}^{-1}}\circ {{f}^{-1}} \right)\circ h \right)\left( x \right)$ as follows.
$\left( \left( {{g}^{-1}}\circ {{f}^{-1}} \right)\circ h \right)\left( x \right)=\left( \left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( f\circ g \right) \right)\left( x \right)$
Apply the associative property,
$\left( \left( {{g}^{-1}}\circ {{f}^{-1}} \right)\circ h \right)\left( x \right)=\left( \left( {{g}^{-1}}\circ \left( {{f}^{-1}}\circ f \right) \right)\circ g \right)\left( x \right)$
Use the definition of composition of inverse functions,
$\begin{align}
& \left( \left( {{g}^{-1}}\circ {{f}^{-1}} \right)\circ h \right)\left( x \right)=\left( \left( {{g}^{-1}}\circ I \right)\circ g \right)\left( x \right) \\
& =\left( {{g}^{-1}}\circ g \right)\left( x \right) \\
& =x
\end{align}$
Therefore, $\left( \left( {{g}^{-1}}\circ {{f}^{-1}} \right)\circ h \right)\left( x \right)=x$.
Evaluate $\left( h\circ \left( {{g}^{-1}}\circ {{f}^{-1}} \right) \right)\left( x \right)$ as follows.
$\left( h\circ \left( {{g}^{-1}}\circ {{f}^{-1}} \right) \right)\left( x \right)=\left( \left( f\circ g \right)\circ \left( {{g}^{-1}}\circ {{f}^{-1}} \right) \right)\left( x \right)$
Apply the associative property,
$\left( h\circ \left( {{g}^{-1}}\circ {{f}^{-1}} \right) \right)\left( x \right)=\left( \left( f\circ \left( g\circ {{g}^{-1}} \right) \right)\circ {{f}^{-1}} \right)\left( x \right)$
Use the definition of compositions of inverse functions,
$\begin{align}
& \left( h\circ \left( {{g}^{-1}}\circ {{f}^{-1}} \right) \right)\left( x \right)=\left( \left( f\circ I \right)\circ {{f}^{-1}} \right)\left( x \right) \\
& =\left( f\circ {{f}^{-1}} \right)\left( x \right) \\
& =x
\end{align}$
Therefore, $\left( h\circ \left( {{g}^{-1}}\circ {{f}^{-1}} \right) \right)\left( x \right)=x$.
So, $\left( \left( {{g}^{-1}}\circ {{f}^{-1}} \right)\circ h \right)\left( x \right)=\left( h\circ \left( {{g}^{-1}}\circ {{f}^{-1}} \right) \right)\left( x \right)=x$.
From the composition of inverse functions, it can be concluded that ${{h}^{-1}}\left( x \right)=\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( x \right)$.
Hence, the required result is proved.
