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Math Help - stationary point

  1. #1
    Senior Member furor celtica's Avatar
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    stationary point

    locate the stationary point(s) on this curve:
    y=x^3 - 2lnx^3
    i keep on finding (1,1), which apparently is so wrong. helpful hints?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by furor celtica View Post
    locate the stationary point(s) on this curve:
    y=x^3 - 2lnx^3
    i keep on finding (1,1), which apparently is so wrong. helpful hints?
    y=x^3-2lnx^3

    y=x^3-6lnx

    \frac{dy}{dx}=3x^2-\frac{6}{x}

    At a stationary point,

    \frac{dy}{dx}=0

    3x^2-\frac{6}{x}=0

    x^2=\frac{2}{x}

    x^3=2

    x=2^\frac{1}{3}=1.26

    y=x^3-2lnx^3

    y=2-2ln2=0.61

    So the stationary point is (1.26, 0.61).
    Last edited by alexmahone; August 2nd 2009 at 10:35 PM.
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