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haha asked in 科學數學 · 1 decade ago

兩題 M.V.T. 證明題

兩題 M.V.T. 證明題

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  • 1 decade ago
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    (1) f(4) = 1, f(1) = 1/4

    f'(x) = -2/(x - 3)3

    Hence we have:

    -6/(c - 3)3 = 1/4 - 1

    -6/(c - 3)3 = -3/4

    (c - 3)3 = 8

    c - 3 = 2

    c = 5 which is beyond 1 < x < 4

    It does not contradict to the Mean Value Thm since f(x) is NOT differentiable at x = 3.

    (2) f(3) = -3 and f(0) = 1

    When x < 1/2, f(x) = 1 + 2x, f'(x) = 2

    When x >= 1/2, f(x) = 3 - 2x, f'(x) = -2

    So we have:

    f(3) - f(0) = -4 which is not equal to 2 nor -2

    It does not contradict to the Mean Value Thm since f(x) is NOT differentiable at x = 1/2

    Source(s): Myself
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