I though i had a grasp of these but this one is a little different from the other ones.

Find the locations (ordered pairs) of all absolute extrema (min max) for the following function with the specified domain.

$\displaystyle F(x)=\frac{x}{x^2+2}$ Domain [0,4]

Heres what ive done so far:

$\displaystyle f'(x)=\frac{(x^2+2)(1)-(x)(2x)}{(x^2+2)^2}$

$\displaystyle F'(x)=\frac{(x^2+2)+(2x^2)}{(x^2+2)^2}$

$\displaystyle F'(x)=\frac{3x^2+2}{(x^2+2)^2}$

The bottom can never equal 0 so how do i test to find the absolute max and absolute min? Thanks for any help!