Intergral- Max and min of function problem

**madmax29** · December 14th 2010, 09:31 AM

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?

**tonio** · December 14th 2010, 10:40 AM

Originally Posted by madmax29

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

**madmax29** · December 14th 2010, 01:12 PM

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

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

**tonio** · December 14th 2010, 04:37 PM

Originally Posted by madmax29

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

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

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

tonio

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