f(0)=f(x)+f'(x)(0-x)+0.5f''(a)(0-x)^2
f(1)=f(x)+f'(x)(1-x)+0.5f''(b)(1-x)^2
两式相减,移项,取绝对值得|f'(x)|=|f(1)-f(0)+0.5f''(a)x^2-0.5f''(b)(1-x)^2|