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Math Help - Intergral- Max and min of function problem

  1. #1
    Junior Member madmax29's Avatar
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    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?
    Last edited by madmax29; December 14th 2010 at 08:32 AM. Reason: added math tags
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  2. #2
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    Quote Originally Posted by madmax29 View Post
    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
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  3. #3
    Junior Member madmax29's Avatar
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    How would i solve the problem from this? to find the local maximum and local minimum?

    f'x=e^x (x^2-1x-6)
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  4. #4
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    Quote Originally Posted by madmax29 View Post
    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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