Answer
$ f^{-1}(x)= \sqrt{\frac{1}{x}}$
Domain of $ f^{-1}$ is $(0,\infty)$
Range of $ f^{-1}$ is $(0,\infty)$
Work Step by Step
Given:
$ f(x) =\frac{1}{x^{2}}$
Find the inverse:
$y=\frac{1}{x^{2}}$
$x^{2} =\frac{1}{y}$
$x=\sqrt{\frac{1}{y}}$
Switch $x$ and $y$:
$y= \sqrt{\frac{1}{x}} =f^{-1}(x) $
We see that:
Domain of $ f^{-1}$ is $(0,\infty)$
and
Range of $ f^{-1}$ is $(0,\infty)$
Confirm the inverse:
\begin{aligned} f\left(f^{-1}(x)\right) &=f\left(\frac{1}{\sqrt{x}}\right) \\ &=\frac{1}{\left(\frac{1}{\sqrt{x}}\right)^{2}} \\ &=\frac{1}{1/x} \\ &=x \\f^{-1}(f(x)) &=f^{-1}\left(\frac{1}{x^{2}}\right) \\ &=\frac{1}{\sqrt{\frac{1}{x^{2}}}} \\ &=\frac{1}{\frac{1}{x}} \\ &=x \end{aligned}
