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Math Help - Simple Prime Question

  1. #1
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    Thumbs down Simple Prime Question

    F(X) = (2/3)(X-2)^(-1/3)

    F'(X)= (-2/9)(X-2)^(-4/3)

    F''(X) = (8/27)(X-2)^(-5/3) Is F''(X) correct???
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  2. #2
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    Hi.

    Quote Originally Posted by Brazuca View Post
    F(X) = (2/3)(X-2)^(-1/3)

    F'(X)= (-2/9)(X-2)^(-4/3)

    F''(X) = (8/27)(X-2)^(-5/3) Is F''(X) correct???
    No!

    I think it is a typo, but it is

    F''(X) = (8/27)(X-2)^{\displaystyle-(4/3) - 1}= (8/27)(X-2)^{\displaystyle-4/3 - 3/3} = (8/27)(X-2)^{\displaystyle-7/3}


    (F'(X) is correct!)

    Yours
    Rapha
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  3. #3
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    Wasn't a typo. I just had it incorrect.

    Thanks now I understand how it works.
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  4. #4
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    I have two more questions involving this problem.

    F(X) = (2/3)(X-2)^(-1/3) when X = 1 why is the final answer negative?

    F'(X)= (-2/9)(X-2)^(-4/3) when X = 1 why is the final answer negative?
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  5. #5
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    Quote Originally Posted by Brazuca View Post
    I have two more questions involving this problem.

    F(X) = (2/3)(X-2)^(-1/3) when X = 1 why is the final answer negative?
    Just substitute X=1, then

    F(1) = \frac{2}{3}*(1-2)^{-1/3} = \frac{2}{3}*(-1)^{-1/3}

    We know that a^{-1} = \frac{1}{a}

    Therefor

    F(1) = \frac{2}{3}*(-1)^{-1/3} = \frac{2}{3}*\frac{1}{\displaystyle(-1)^{1/3}}

    Furthermore we know that x^{1/2} = \sqrt{x} and x^{1/3} = \sqrt[3]{x}

    Thus

    F(1) = \frac{2}{3}*\frac{1}{\displaystyle(-1)^{\displaystyle1/3}} = \frac{2}{3}*\frac{1}{\sqrt[3]{-1}} = \frac{2}{3}*(-1) = - 2/3 < 0

    Quote Originally Posted by Brazuca View Post
    F'(X)= (-2/9)(X-2)^(-4/3) when X = 1 why is the final answer negative?
    Same here

    F'(1) = (-2/9) (1-2)^{-4/3} = -2/9*(-1)^{\displaystyle-4/3} = -2/9 * \frac{1}{(-1)^{\displaystyle4/3}}

    = -2/9 * \frac{1}{\displaystyle((-1)^4)^{\displaystyle1/3}} = -2/9*\frac{1}{1^{\displaystyle1/3}} = -2/9 * \frac{1}{\sqrt[3]{1}} = -2/9*1 = -2/9 <0
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