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Math Help - relative extrema and C.P

  1. #1
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    relative extrema and C.P

    Find the critical points of:

    (3x^2+3x+3)e^{-x}

    I get x=0, x=1 am I correct?
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  2. #2
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    Re: relative extrema and C.P

    Quote Originally Posted by Dobby View Post
    Find the critical points of:

    (3x^2+3x+3)e^{-x}

    I get x=0, x=1 am I correct?
    And how did you get this answer?
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  3. #3
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    Re: relative extrema and C.P

    found the derivative and set it=0

    Let f(x) = (3x^2 + 3x + 3) * e^(-x)

    Using the product rule, we have:
    f'(x) = (6x + 3) * e^(-x) + (3x^2 + 3x + 3) * (-e^(-x))
    f'(x) = (6x + 3) * e^(-x) - (3x^2 + 3x + 3) * e^(-x)
    f'(x) = ((6x + 3) - (3x^2 + 3x + 3)) * e^(-x)
    f'(x) = (6x + 3 - 3x^2 - 3x - 3) * e^(-x)
    f'(x) = (-3x^2 + 3x) * e^(-x)

    Setting f'(x) to zero, we have:
    f'(x) = 0
    (-3x^2 + 3x) * e^(-x) = 0
    (-3x)(x - 1) * e^(-x) = 0
    -3x = 0 or x - 1 = 0 or e^(-x) = 0
    x = 0 or x = 1
    Last edited by Dobby; April 25th 2012 at 05:11 AM.
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  4. #4
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    Re: relative extrema and C.P

    Looks good.
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  5. #5
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    Re: relative extrema and C.P

    Prove It could have just said "yes" but it would still be possible that you got the right answer by making two mistakes that happened to cancel- and so you would have thought you had done it right when, perhaps, you hadn't. It is always better, when turning in homework or posting on a forum, to show all of your work.
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