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Math Help - Critical points

  1. #1
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    Critical points

    Hello,

    Can someone show me how to find the critical points for the function f(x)=x^2*e^(-5x^2)?

    I appreciate your help! Thank you!
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  2. #2
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    Re: Critical points

    Find f'(x) using the product rule, then solve f'(x) = 0
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  3. #3
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    Re: Critical points

    Yes, I understand that I have to do that. But I'm getting confused when I'm differentiating. Could you help me with that part? Thank you!
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  4. #4
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    Re: Critical points

    \displaystyle f'(x) = (x^2)'e^{-5x^2} +(e^{-5x^2})'x^2

    Recall if \displaystyle y= e^{f(x)} \implies \frac{dy}{dx} = f'(x)e^{f(x)}
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  5. #5
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    Re: Critical points

    Right, so this is what I get from using the product rule and chain rule:

    f'(x)=[2x]e^(-5x^2)+x^2[e^(-5x^2)*(-10x)]

    Then I factored it (hopefully correctly) and got:

    2xe^(-5x^2)[1-5x^2*e^(-5x^2)]

    I set that equal to 0. But since e can never equal zero, I threw out 2xe^(-5x^2). So I'm looking for (1-5x^2*e^(-5x^2))=0. This is where I got confused. I'm not quite sure how to solve for x. Any ideas?
    Last edited by juicysharpie; December 7th 2011 at 11:58 PM.
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  6. #6
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    Re: Critical points

    True, that should've been 0 = 2x- 10x^3 \implies x = 0,\frac{1}{\sqrt{5}},\frac{-1}{\sqrt{5}}
    Last edited by pickslides; December 8th 2011 at 12:31 AM.
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  7. #7
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    Re: Critical points

    Ok, so I did factor it wrong. The answer given to me on my worksheet is x=0, root(1/5), -root(1/5).

    After factoring it correctly i got:

    0=2xe^(-5x^2)[1-5x^2]
    0=[1-5x^2]
    x= +/- root(1/5) and x also equals 0 because of the 2xe^(-5x^2)

    Thanks for the help!
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