设 c=(a+b)/2f(c)=f(a)+f'(a)(c-a)+1/2 f''(t1)(c-a)^2, af(c)=f(b)+f'(b)(c-b)+1/2 f''(t2)(c-b)^2, c相减,利用f'(a)=f'(b)=0,得:f(a)-f(b)=1/2 (c-a)^2(f''(t2)-f''(t1))取 ξ为t1,t2之一, 使得 |f''(ξ)|=max{|f''(t1)|,|f''(t2)|}于是 |f''(ξ)|>=|f''(t2)-f''(t1)|/2 =|f(a)-f(b)|/ (c-a)^2=4×|f(b)-f(a)|/(b-a)^2