Find the absolute max and min values of the function, if they exist, over the indicated interval.
F(x) = x (square root of x+3) ; [-3,6]
$\displaystyle f(x)=x\sqrt{x+3}$
$\displaystyle f'(x)=\frac{x}{2\sqrt{x+3}}+\sqrt{x+3}$
$\displaystyle =\frac{3x+6}{2\sqrt{x+3}}$
The critical numbers are given by
$\displaystyle 3x+6=0$
$\displaystyle x=-2$
So $\displaystyle f(-2)$ should be an extreme point.
So to determine whether it's a max or min you can just examine points write around $\displaystyle f(-2)$ and see if they are less than or great than $\displaystyle f(-2)$.