Answer
For $x\in(-\infty,0]$ the inverse and its domain are:
$f^{-1}(x)=-\sqrt{\dfrac{2-2x}{x}}$
$D_{f^{-1}}=(0,1]$
For $x\in[0,\infty)$ the inverse and its domain are:
$f^{-1}(x)=\sqrt{\dfrac{2-2x}{x}}$
$D_{f^{-1}}=(0,1]$
Work Step by Step
We are given the function:
$f(x)=\dfrac{2}{x^2+2}$
The domain of and range of $f$ are:
$D_f=(-\infty,\infty)$
$R_f=(0,1]$
Determine the inverse $f^{-1}$:
$y=\dfrac{2}{x^2+2}$
$x=\dfrac{2}{y^2+2}$ (we switched $x$ and $y$)
$x(y^2+2)=2$
$y^2+2=\dfrac{2}{x}$
$y^2=\dfrac{2}{x}-2$
$y^2=\dfrac{2-2x}{x}$
$y=\pm\sqrt{\dfrac{2-2x}{x}}$
There are two inverses:
For $x\in(-\infty,0]$ the inverse and its domain are:
$f^{-1}(x)=-\sqrt{\dfrac{2-2x}{x}}$
$D_{f^{-1}}=(0,1]$
For $x\in[0,\infty)$ the inverse and its domain are:
$f^{-1}(x)=\sqrt{\dfrac{2-2x}{x}}$
$D_{f^{-1}}=(0,1]$
