Answer
See the explanation below.
Work Step by Step
(a)
Consider the function,
$f\left( x \right)=\frac{x-7}{x+2}$
Step 1: Replace $f\left( x \right)$ with $y$:
$y=\frac{x-7}{x+2}$
Step 2: Interchange $x$ and $y$:
$x=\frac{y-7}{y+2}$
Step 3: Now solve for the value of $y$:
$xy+2x=y-7$
Or
$2x+7=y-xy$
Or
$2x+7=y\left( 1-x \right)$
That is,
$y=\frac{2x+7}{1-x}$
Step 4: Replace $y$ with ${{f}^{-1}}\left( x \right)$:
${{f}^{-1}}\left( x \right)=\frac{2x+7}{1-x}$
Hence, the inverse function ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)=\frac{x-7}{x+2}$ is ${{f}^{-1}}\left( x \right)=\frac{2x+7}{1-x}$.
(b)
Consider the function, $f\left( {{f}^{-1}}\left( x \right) \right)$
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \frac{2x+7}{1-x} \right) \\
& =\frac{\left( \frac{2x+7}{1-x} \right)-7}{\left( \frac{2x+7}{1-x} \right)+2} \\
& =\frac{2x+7-7+7x}{2x+7+2-2x} \\
& =x
\end{align}$
Next consider the function, ${{f}^{-1}}\left( f\left( x \right) \right)$
$\begin{align}
& {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( \frac{x-7}{x+2} \right) \\
& =\frac{2\left( \frac{x-7}{x+2} \right)+7}{1-\left( \frac{x-7}{x+2} \right)} \\
& =\frac{2x-14+7x+14}{x+2-x+7} \\
& =x
\end{align}$
Hence, $f\left( {{f}^{-1}}\left( x \right) \right)=x$ and ${{f}^{-1}}\left( f\left( x \right) \right)=x$ for the function $f\left( x \right)=\frac{x-7}{x+2}$.
