## Calculus: Early Transcendentals (2nd Edition)

(a). $f'(x) = \frac{1}{2\sqrt{x−2}}$ , which is never zero. Also, $f(x)$ exists on $(2,∞)$, so there are no critical points. (b). $f(2) = 0$ and $f(6) = 2$, so the absolute maximum of this function on the given interval is $2$ and the absolute minimum is $0$.