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# 兩題 M.V.T. 證明題

兩題 M.V.T. 證明題

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- 六呎將軍Lv 71 decade agoFavorite Answer
(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

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