Taylor展式：对任意的x, f(0)＝f(x)+f'(x)(0－x)+f''(c1)(0－x)^2/2, f(1)＝f(x)+f'(x)(1－x)+f''(c2)(1－x)^2/2. 两式相减,得 f'(x)＝f''(c1)x^2/2－f''(c2)(1－x)^2/2, 取绝对值并利用条件得 |f'(x)|,6,
