# an absolute max and min problem involving e

• Oct 26th 2009, 05:50 PM
hazecraze
an absolute max and min problem involving e
Find the absolute max and mins of $f(x)$ on the given interval.

$f(x)= xe^\frac{-x^2}{8}, [-1,4]$
$f'(x)= e^\frac{-x^2}{8}+\frac{-x^2}{4}e^\frac{-x^2}{8}$

$e^\frac{-x^2}{8}(1-\frac{x^2}{4})^4$

$e^\frac{-x^2}{8}=0$

$e^\frac{-x^2}{8}=0$
=?

$(1-\frac{x^2}{4})^4=0$
$(1-\frac{x^2}{4})=0$
$(\frac{x^2}{4})=1$

$(x^2)=4
$

2,-2
Help?
• Oct 26th 2009, 06:02 PM
Debsta
e^(whatever)=0 doesn't have any solutions.

2 is the only solution in [-1, 4]

So now find f(2) and also f(-1) and f(4) at the limits of the function, and compare them.