# Intergral- Max and min of function problem

• Dec 14th 2010, 08:31 AM
Intergral- Max and min of function problem
Hi there

I have a problem to find the maximum and minimum functions of the following,
$\int e^x (x-3) (x+2)$

but i am not sure what to do with the terms in brackets in order to find $(dy/dx)$ and then the second derivative; $(d^2y/dx^2)$

in order to determine which values are local; $y''<0= MAX$? or $y'' >0 = MIN$?

would $f'x=e^x (x^2-1x-6)$ be correct, however I do belive this is wrong could someone explain why?
• Dec 14th 2010, 09:40 AM
tonio
Quote:

Hi there

I have a problem to find the maximum and minimum functions of the following,
$\int e^x (x-3) (x+2)$

but i am not sure what to do with the terms in brackets in order to find $(dy/dx)$ and then the second derivative; $(d^2y/dx^2)$

in order to determine which values are local; $y''<0= MAX$? or $y'' >0 = MIN$?

would $f'x=e^x (x^2-1x-6)$ be correct, however I do belive this is wrong could someone explain why?

It is correct.

Tonio
• Dec 14th 2010, 12:12 PM
How would i solve the problem from this? to find the local maximum and local minimum?

$f'x=e^x (x^2-1x-6)$
• Dec 14th 2010, 03:37 PM
tonio
Quote:

$f'x=e^x (x^2-1x-6)$