Answer
$f^{-1}(x)=\sqrt{x+4}$
Work Step by Step
We are given the function:
$f(x)=x^2-4,x\geq 0$
Determine the inverse of $f$:
$y=x^2-4$
Interchange $x$ and $y$:
$x=y^2-4$
$y^2=x+4$
$y=\sqrt {x+4}$
$f^{-1}(x)=\sqrt{x+4}$
Compute $f\circ f^{-1}$ and $f^{-1}\circ f$:
$(f\circ f^{-1})(x)=f(f^{-1}(x))=f(\sqrt{x+4})=(\sqrt{x+4})^2-4=x+4-4=x$
$(f^{-1}\circ f)(x)=f^{-1}\left(x^2-4\right)=\sqrt{x^2-4+4}=\sqrt{x^2}=x$
Thus, the functions are inverses.
