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