Answer
$f^{-1} (x)=x^2+2 $; Domain: $x \geq 0$.
Work Step by Step
When we apply the horizontal test, it has been noticed that the function is one-to-one and verifies the horizontal test.
Therefore, the function has an inverse function.
To compute the inverse, we will have to interchange $x$ and $y$.
$x=\sqrt {y-2} \implies x^2=y-2$
or, $y=x^2+2 $
Replace $y$ with $f^{-1} (x)$.
so, $f^{-1} (x)=x^2+2 $
Since the range of the function is $x \geq 0$, the domain for the function is $x \geq 0$.
