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F(x) = x^(1/3), [0,1]?

You are supposed to use the Mean Value Theorem and come up with the answer 3^(1/2)/9, but I got (1/3) and -(1/3)

1 Answer

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  • Anonymous
    7 years ago
    Favorite Answer

    You need to find c in [0,1] such that f'(c) = [f(1) - f(0)] / (1-0).

    Since f'(x) = (1/3)x^(-2/3), we get

    (1/3)c^(-2/3) = (1 - 0) / (1 - 0) = 1, so

    c^(-2/3) = 3

    c = 3^(-3/2)

    c = 1 / 3^(3/2)

    c = 1 / sqrt(27)

    c = sqrt(27) / 27

    c = 3sqrt(3) / 27

    c = sqrt(3) / 9, which is what you wanted.

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