Answer
$f^{-1}(x)=-\dfrac{2x+2}{x-1}$
Work Step by Step
$f(x)=\dfrac{x-2}{x+2}$
Rewrite this expression as $y=\dfrac{x-2}{x+2}$ and solve for $x$:
$y=\dfrac{x-2}{x+2}$
Take $x+2$ to multiply the left side:
$(x+2)y=x-2$
$xy+2y=x-2$
Take $x$ to the left side and $2y$ to the right side:
$xy-x=-2-2y$
Take out common factor $x$ from the left side:
$x(y-1)=-2-2y$
Take $y-1$ to divide the right side:
$x=\dfrac{-2-2y}{y-1}$
$x=-\dfrac{2y+2}{y-1}$
Interchange $x$ and $y$:
$y=-\dfrac{2x+2}{x-1}$
The inverse of the initial function is $f^{-1}(x)=-\dfrac{2x+2}{x-1}$
