1. ## Critical Points

Fine the absolute minimum and absolute maximum of the function on the interval [0,6]

$\displaystyle F(x) = \frac{x}{x^2+16}$

I did the minimum already, it's just the maximum.

f(x) = x
f'(x) = 1
$\displaystyle g(x) = x^2+16$
g'(x) = 2x

$\displaystyle F'(x) = \frac{1*(x^2+16) - 2x * x}{(x^2+16)^2}$
$\displaystyle F'(x) = \frac{-x^2+16}{(x^2+16)^2}$

Bottom is never 0, so do the quadratic formula on the top:

$\displaystyle r = \frac{0\pm\sqrt{0-(4)(-1)(16)}}{-2}$

Which in the end gave me 4 and -4, which when plugging into F(x) gave me $\displaystyle \frac{1}{8}$ and $\displaystyle \frac{-1}{8}$

But plugging 6 in gave me $\displaystyle \frac{3}{26}$

Is this valid?

2. Originally Posted by CFem
Fine the absolute minimum and absolute maximum of the function on the interval [0,6]

$\displaystyle F(x) = \frac{x}{x^2+16}$

I did the minimum already, it's just the maximum.

f(x) = x
f'(x) = 1
$\displaystyle g(x) = x^2+16$
g'(x) = 2x

$\displaystyle F'(x) = \frac{1*(x^2+16) - 2x * x}{(x^2+16)^2}$
$\displaystyle F'(x) = \frac{-x^2+16}{(x^2+16)^2}$

Bottom is never 0, so do the quadratic formula on the top:

$\displaystyle r = \frac{0\pm\sqrt{0-(4)(-1)(16)}}{-2}$

Which in the end gave me 4 and -4, which when plugging into F(x) gave me $\displaystyle \frac{1}{8}$ and $\displaystyle \frac{-1}{8}$

But plugging 6 in gave me $\displaystyle \frac{3}{26}$

Is this valid?
I'm not sure what your point is about the 3/26. 3/26 is LESS then 1/8.

3. 3/26 gets spit back.

If it's correct then I'll figure it out. If not, then I need to know where I went wrong.

4. Originally Posted by CFem
3/26 gets spit back.

If it's correct then I'll figure it out. If not, then I need to know where I went wrong.
You were asked to find the minimum and maximum of the function. Those are -1/8 and 1/8 respectively, and they result when you plug in values of 4 and -4 respectively.

Why are you worried about plugging in 6 at all? It's irrelevant.

5. Because the absolute minimum and maximum are the smallest and largest points (respectively) when you plug in the critical points and the end points of an interval?

Unless I'm mistaken.

6. Originally Posted by CFem
Because the absolute minimum and maximum are the smallest and largest points (respectively) when you plug in the critical points and the end points of an interval?

Unless I'm mistaken.
Well, when you're asked to find the absolute minima/maxima, you use the $\displaystyle f'(x) = 0$ condition to find some minima/maxima, HOWEVER, these are not necessarily the ABSOLUTE... they may only be the relative or local minima/maxima. So we check whether or not we have a global (absolute) maxima/minima by comparing our values from the $\displaystyle f'(x) = 0$ condition with the values we get at the end points. If the values we get at the end points are LARGER or SMALLER than the maxima or minima that we got using that $\displaystyle f'(x) = 0$ condition respectively, then we know that the condition didn't give us the global extrema, but in fact, only local.

In your case, we use the $\displaystyle f'(x) = 0$, and we find that the minima/maxima we get from that is -1/8 and 1/8 respectively. Now, to make sure these are global, and not just local, we compare them to the end points. At the end points we get 0 and 3/26 respectively. Now, -1/8 is LESS than 0, and 1/8 is MORE than 3/26. Therefore, we can confirm that the GLOBAL maxima is 1/8, and the GLOBAL minima is -1/8.