Thomas' Calculus 13th Edition

$f^{-1}(x)= \sqrt{\frac{1}{x}}$ Domain of $f^{-1}$ is $(0,\infty)$ Range of $f^{-1}$ is $(0,\infty)$
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}