Answer
The inverse of the function is $f\left( x \right)={{f}^{-1}}\left( x \right)={y}'=\frac{1+x}{1-x}$.
Work Step by Step
We have to compute the inverse as follows:
Let $f\left( x \right)=y$
So,
$\begin{align}
& y=\frac{x-1}{x+1} \\
& y\left( x+1 \right)=x-1 \\
& yx+y=x-1
\end{align}$
Then, solving the equation:
$\begin{align}
& xy-x=-1-y \\
& x\left( y-1 \right)=-1-y \\
& x\left( y-1 \right)=-\left( 1+y \right) \\
& x=\frac{-\left( 1+y \right)}{\left( y-1 \right)}
\end{align}$
And the value of x is,
$x=\frac{1+y}{1-y}$
Now, replacing $x$ by ${y}'$ and $y$ by $x$ to get,
$\begin{align}
& {y}'=\frac{1+x}{1-x} \\
& \text{inverse of }f\left( x \right)={{f}^{-1}}\left( x \right) \\
& ={y}' \\
& =\frac{1+x}{1-x}
\end{align}$
Thus, the result is $f\left( x \right)={{f}^{-1}}\left( x \right)={y}'=\frac{1+x}{1-x}$.
