Answer
f is one-to-one,
$f^{-1}(x)=x^{2}+2, \quad x \geq 0$
Work Step by Step
We apply Theorem 5.7. here. Test whether f is strictly monotonic.
If we presume that $x_{1} < x_{2}$, both from the domain of f,
$x_{1} \geq 2$ and $x_{2} \geq 2, $
then
$x_{1}-2 < x_{2} - 2$, and,
since $\sqrt{x}$ is a rising function, (strictly monotonic), it follows that
$\sqrt{x_{1}-2} < \sqrt{x_{2}-2}\qquad$that is,
$f(x_{1}) < f(x_{2})$
So, f is strictly monotonic.
By Th.5.7, it is also one-to-one, and has an inverse.
To find the inverse,
1. swap f(x)=y and x:
$y=\sqrt{x-2} \qquad , x \geq 2\Rightarrow \mathrm{y} \geq 0$
$x=\sqrt{y-2} \qquad , y \geq 2\Rightarrow x \geq 0$
square both sides (both are positive)
2. Solve for y:
$x^{2}=y-2\qquad+2$
$y=x^{2}+2, \quad x \geq 0$
3. Replace $y $ with $f^{-1}(x):$
$f^{-1}(x)=x^{2}+2, \quad x \geq 0$
