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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)
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- Anonymous7 years agoFavorite 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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