设函数f(x)在[0,1]上连续且不恒为零,在(0,1)内可导,且f(0)=0,证明:存在ξ∈(0,1),使得f(ξ)f‘(ξ)>0

问题描述:

设函数f(x)在[0,1]上连续且不恒为零,在(0,1)内可导,且f(0)=0,证明:存在ξ∈(0,1),使得f(ξ)f‘(ξ)>0

f(x) = f(0) +f'(ξ)x ( Taylor expansion)
put x=ξ
f(ξ) = f'(ξ)ξ
f'(ξ)f(ξ) = [f(ξ)]^2/ξ >0