Answer
$f^{-1}(x)=\sqrt{4-x^{2}}$
Work Step by Step
$f(x)=\sqrt{4-x^{2}}$ $,$ $0\le x\le2$
Rewrite this expression as $y=\sqrt{4-x^{2}}$ and solve for $x$:
$y=\sqrt{4-x^{2}}$
Square both sides:
$y^{2}=(\sqrt{4-x^{2}})^{2}$
$y^{2}=4-x^{2}$
Take $-x^{2}$ to the left side:
$x^{2}+y^{2}=4$
Take $y^{2}$ to the right side:
$x^{2}=4-y^{2}$
Take the square root of both sides:
$\sqrt{x^{2}}=\sqrt{4-y^{2}}$
$x=\sqrt{4-y^{2}}$
Interchange $x$ and $y$:
$y=\sqrt{4-x^{2}}$
The inverse of the original function is $f^{-1}(x)=\sqrt{4-x^{2}}$