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Math Help - local maxima and minima

  1. #1
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    local maxima and minima

    How would you find the local maxima and minima for y= xe^-x

    Im not sure how to do it and would appreciate very much if someone could help me.
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  2. #2
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    Generally, you would examine it for continuity, then find the first derivative. Where the first derivative exists and is zero, this is likely to be a local max or min. That's only part of the story, of course. You tell me the rest.

    Can you find dy/dx? It will take the Product Rule.
    Last edited by TKHunny; October 18th 2009 at 05:51 PM.
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  3. #3
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    so y'= (e^-x)(-x e^-x)
    = e^-x(1-x)

    so do u let y'= 0?

    if so, i put e^-x(1-x)=0 and thats when i got stuck

    so what do i do next?
    Last edited by mr fantastic; October 16th 2009 at 09:16 PM. Reason: Merged posts
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  4. #4
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    Quote Originally Posted by fvaras89 View Post
    so y'= (e^-x)(-x e^-x)
    = e^-x(1-x)

    so do u let y'= 0?

    if so, i put e^-x(1-x)=0 and thats when i got stuck

    so what do i do next?
    Well, obviously either e^{-x} = 0 or (1 - x) = 0 .... (Hint: e^{-x} = 0 has no real solution ....)
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  5. #5
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    Personally, I'm a little curious how how got from here

    Quote Originally Posted by fvaras89 View Post
    so y'= (e^-x)(-x e^-x)
    to here

    = e^-x(1-x)
    Seems to be some notational difficulty requiring magic to get to the right result. Please be more careful. You WILL unconfuse yourself.
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  6. #6
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    I just thought u could take out the common factor like i did before or would there be another way to present it?
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  7. #7
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    I think it would help if the addition actually appeared in the first equation. Thus my encouragement to be more careful. I have little doubt that you had the correct thing in your mind, you just didn't write it.
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