求证：已知f(x)在[a,b]存在二阶导数,f'(a)=f'(b)=0,则在存在c∈[a,b],有|f''(c)|≥2|f(b)-f(a)|/(a-b)^2

Question

求证：已知f(x)在[a,b]存在二阶导数,f'(a)=f'(b)=0,则在存在c∈[a,b],有|f''(c)|≥2|f(b)-f(a)|/(a-b)^2

Answer 1

由泰勒公式得
f(x)=f(a)+f'(a)(x-a)+f''(c1)/2!(x-a)^2=f(a)+f''(c1)/2!(x-a)^2 (c1介于a和x之间) (1)
f(x)=f(b)+f'(b)(x-b)+f''(c2)/2!(x-b)^2=f(b)+f''(c2)/2!(x-b)^2 (c2介于a和x之间) (2)
两式相减并整理得
f(b)-f(a)=f''(c1)/2!(x-a)^2 -f''(c2)/2!(x-b)^2
于是|f(b)-f(a)|= | f''(c1)/2!(x-a)^2 -f''(c2)/2!(x-b)^2|