Question

f(x)在[0,1]上二阶可导,且f(0)=f(1)=0,试证：存在ξ∈（0,1）,使f``(ξ)=2f`(ξ)/1-ξ

Answer 1

由条件,存在η∈(0,1),满足f'(η)=0.

令G(x) = (1-x)²f'(x),则G(η) = G(1) = 0

所以,存在ξ∈(η,1),使G'(ξ)=0,即(1-ξ)²f''(ξ)-2(1-ξ)f'(ξ)=0

由于ξ