Answer
$f^{-1}(x)=\dfrac{2x}{2-x}$
$D_{f^{-1}}=(-\infty,2)\cup(2,\infty)$
Work Step by Step
We are given the function:
$f(x)=\dfrac{2x}{x+2}$
The domain of $f$ is:
$D_f=(-\infty,-2)\cup(-2,\infty)$
Determine the inverse $f^{-1}$:
$y=\dfrac{2x}{x+2}$
$x=\dfrac{2y}{y+2}$ (we switched $x$ and $y$)
$x(y+2)=2y$
$xy+2x=2y$
$2x=2y-xy$
$2x=y(2-x)$
$y=\dfrac{2x}{2-x}$
$f^{-1}(x)=\dfrac{2x}{2-x}$
The domain of the inverse is:
$D_{f^{-1}}=(-\infty,2)\cup(2,\infty)$
