Answer
$y=f^{-1}(x)=\dfrac{1}{\sqrt{x}}$
Work Step by Step
$f(x)=\dfrac{1}{x^{2}}$, for $x\gt0$
Substitute $f(x)$ by $y$:
$y=\dfrac{1}{x^{2}}$
Take $x^{2}$ to multiply the left side:
$x^{2}y=1$
Take $y$ to divide the right side:
$x^{2}=\dfrac{1}{y}$
Take the square root of both sides:
$\sqrt{x^{2}}=\sqrt{\dfrac{1}{y}}$
$x=\dfrac{1}{\sqrt{y}}$
Interchange $x$ and $y$:
$y=\dfrac{1}{\sqrt{x}}$
Express in $y=f^{-1}(x)$ form:
$y=f^{-1}(x)=\dfrac{1}{\sqrt{x}}$
The graph of the function and its inverse is shown in the answer section.
