Answer
The function $f\left( x \right)={{\left( x \right)}^{9}}$ and $g\left( x \right)={{x}^{2}}-3x+7$.
Work Step by Step
Take ${{x}^{2}}-3x+7$ as $g\left( x \right)$ from the function $h\left( x \right)={{\left( {{x}^{2}}-3x+7 \right)}^{9}}$ and consider ${{x}^{2}}-3x+7$ as x for the another function $f\left( x \right)$.
So, the functions are $f\left( x \right)={{\left( x \right)}^{9}}$ and $g\left( x \right)={{x}^{2}}-3x+7$ .
Check by composition of function $f\left( x \right)$ and $g\left( x \right)$ to find $\left( f\circ g \right)\left( x \right)$. This should give the original function,
$h\left( x \right)={{\left( {{x}^{2}}-3x+7 \right)}^{9}}$
$\begin{align}
& \left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right) \\
& =f\left( {{x}^{2}}-3x+7 \right) \\
& ={{\left( {{x}^{2}}-3x+7 \right)}^{9}}
\end{align}$
Since, $\left( f\circ g \right)\left( x \right)=h\left( x \right)$
Hence, the functions are $f\left( x \right)={{\left( x \right)}^{9}}$ and $g\left( x \right)={{x}^{2}}-3x+7$.
