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?