# Thread: Intergral- Max and min of function problem

1. ## Intergral- Max and min of function problem

Hi there

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

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

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

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

2. Originally Posted by madmax29
Hi there

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

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

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

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

It is correct.

Tonio

3. How would i solve the problem from this? to find the local maximum and local minimum?

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

4. Originally Posted by madmax29
How would i solve the problem from this? to find the local maximum and local minimum?

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

First derivative equal to zero, then second derivative, signs and etc....as usual!

tonio