设f(x)在[a,b]有二阶导数,f✀(a)=f✀(b)=0 证明,在(a,b)内至少存在一点ξ

2024-12-14 05:33:57
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回答1:

设 c=(a+b)/2
f(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