f(0)=f(x)+f'(x)(0-x)+f''(d)/2(0-x)^2,
f(1)=f(x)+f'(x)(1-x)+f"'(e)/2(1-x)^2,两个式子相减,取绝对值得|f'(x)|=|f(1)-f(0)-f"'(e)/2(1-x)^2+f''(d)/2(0-x)^2|小于等于|f(1)-f(0)|+|f"'(e)/2(1-x)^2+f''(d)/2(0-x)^2|小于等于2a+b/2(x^2+(1-x)^2)小于等于2a+b/2（二次函数x^2+(1-x)^2在【0,1】上的最大值为1）