## Calculus 10th Edition

f is one-to-one, $f^{-1}(x)=x^{2}+2, \quad x \geq 0$
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 $f^{-1}(x):$ $f^{-1}(x)=x^{2}+2, \quad x \geq 0$