g(x)=f(x)-x²[f(1)-f(0)]g(0)=g(1)=f(0)所以存在rr∈(0,1)使得g'(r)=0g'(x)=f'(x)-2x [f(1)-f(0)]g'(r)=f'(r)-2r[f(1)-f(0)]=0f'(r)=2r[f(1)-f(0)]
