## Calculus: Early Transcendentals (2nd Edition)

There is a local maximum of $f(2) = 2$ which is also an absolute maximum. There is no absolute minimum, since the function is unbounded in the negative direction as $x→−∞$.
Note that f is continuous on $(−∞, 3)$. $f'(x) = \frac{−x}{2\sqrt{3-x}}+\sqrt{{3}-x} = \frac{−3x+6} {2\sqrt{3-x}}$ , which is $0$ only for $x = 2$, so there is only one critical point on the stated interval. Note that $f > 0$ for $x < 2$ and $f < 0$ on $(2, 3)$. Thus there is a local maximum of $f(2) = 2$ which is also an absolute maximum. There is no absolute minimum, since the function is unbounded in the negative direction as $x→−∞$.