## Calculus: Early Transcendentals (2nd Edition)

$y=f^{-1}(x)=\dfrac{1}{\sqrt{x}}$
$f(x)=\dfrac{1}{x^{2}}$, 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.