Answer
$f^{-1}(x)=\sqrt{25-x^2}, x\le25$.
Work Step by Step
Let us find the inverse function.Replace $f(x)$ with y.
$y=\sqrt{25-x^2}$
Interchange x and y.
$x=\sqrt{25-y^2}$
Solve for y and square each side.
$x^2=25-y^2$
Subtract 25 from each side.
$x^2+25=-y^2$
Divide each side by -1.
$25=x^2-y^2$
Now take the square root of both sides.
$y=\sqrt{25=x^2}$
Replace y with $f^{-1}(x)$
$f^{-1}(x)=\sqrt{25-x^2}$
The range of f is $y\le0$. This means that the domain of $f^{-1}$ has to be $x\le25$.
The inverse function is $f^{-1}(x)=\sqrt{25-x^2}, x\le25$.