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Math Help - local extremes

  1. #1
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    local extremes

    Find function y=\sqrt[3]{x^3+3x^2} local extremes.

    I start:

     y'= \frac{1}{3}(x^3+3x^2)^{\frac{-2}{3}}*(x^3+3x^2)'=\frac{1}{3}(x^3+3x^2)^{\frac{-2}{3}}(3x^2+6x)

     y'= \frac{1}{3(\sqrt[3]{(x^3+3x^2)^2})}(3x^2+6x)=0
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Bernice View Post
    Find function y=\sqrt[3]{x^3+3x^2} local extremes.

    I start:

     y'= \frac{1}{3}(x^3+3x^2)^{\frac{-2}{3}}*(x^3+3x^2)'=\frac{1}{3}(x^3+3x^2)^{\frac{-2}{3}}(3x^2+6x)

     y'= \frac{1}{3(\sqrt[3]{(x^3+3x^2)^2})}(3x^2+6x)=0
    now find the critical values of y' the root of the denominator and the numerator

     y'= \frac{1}{3(\sqrt[3]{(x^2(x+3))^2})}(3x(x+2))=0

    after you find the critical points study y' signs around critical points
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Easier to consider the function

    f(x)=x^3+3x^2
    f'(x)=3x^2+6x
    f'(x)=3x(x+2)
    f'(x)=0 at x=0 and x=-2
    f(x) has its local extremes at x=0 and x=-2.

    Hence, y has its local extremes at x=0 (minima) and x=-2 (maxima).

    [Use 2nd derivative test to figure if it's a minima or maxima.
    f''(x)=6(x+1)
    f''(0)=6>0 (indicates minima)
    f''(-2)=-6<0 (indicates maxima)
    ]

    The local extremes are y=0 (minima) and \sqrt[3]{4} (maxima).
    Last edited by alexmahone; June 11th 2009 at 07:27 AM.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by alexmahone View Post
    Easier to consider the function

    f(x)=x^3+3x^2
    f'(x)=3x^2+6x
    f'(x)=3x(x+2)
    f'(x)=0 at x=0 and x=-2
    f(x) has its local extremes (minima) at x=0 and x=-2
    what you said is true for this function but if the power of the denominator is odd number (always positive) you can't use you method
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