# Math Help - Absolute Extrema Problem Stuck again

1. ## Absolute Extrema Problem Stuck again

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.

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

Heres what ive done so far:
$f'(x)=\frac{(x^2+2)(1)-(x)(2x)}{(x^2+2)^2}$
$F'(x)=\frac{(x^2+2)+(2x^2)}{(x^2+2)^2}$
$F'(x)=\frac{3x^2+2}{(x^2+2)^2}$

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

2. You suddenly switched signs in the numerator. This is what you should have:

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

$f'(x)=0$ when $2-x^{2}=0$ and that happens when $x=\pm \sqrt{2}$. However, $x=-\sqrt{2}$ is not in the interval $[0,4]$, so the only critical point to be concerned with is $x=\sqrt{2}$. Evaluate $f(0),f(\sqrt{2}),f(4)$ and determine which are your max and min.

3. Originally Posted by Jim Marnell
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.

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

Heres what ive done so far:
$f'(x)=\frac{(x^2+2)(1)-(x)(2x)}{(x^2+2)^2}$
$F'(x)=\frac{(x^2+2)+(2x^2)}{(x^2+2)^2}$
What happened to the "-"?
That should be
$f'(x)= \frac{(x^2+ 2)- (2x^2)}{x^2+ 2)^2}$

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

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

4. f(0)=0
f( $\sqrt{2}$=0
f(4)= $\frac{-7}{162}$

Did i do this wrong, cause those numbers seem weird. If not would max be 0,0 and min be $(4,\frac{-7}{162})$

5. You have to plug those values into $f(x)$, not $f'(x)$

$f(0)=0$
$f(\sqrt{2})=\frac{\sqrt{2}}{4}$
$f(4)=\frac{2}{9}$

Your maximum is at $(\sqrt{2},\frac{\sqrt{2}}{4})$ and your minimum is at $(0,0)$.