f(x)在(0,1)上三阶可导,f(1)=f(0)=0 F(x)=x²f(x) 求证有一点z使得F(z)的三阶导=0 z属于(0,1)
追答F(x)=x²f(x),
F'(x)=2xf(x)+x^2*f'(x),
F''(x)=2f(x)+4xf'(x)+x^2*f''(x),
F'''(x)=6f'(x)+6xf''(x)+x^2*f'''(x),
f(0)=f(1)=0,
所以存在x0,满足0<x0<1,f'(x0)=0,
F'(0)=0,F'(1)=f'(1),F'(x0)=2x0f(x0),
F''(0)=2f(0)=0,待续