According to Rolle theorem, a∈(0, 1) exists, so f'(a)=0.

Let g(x) = f' (x) * (x- 1) 2, then g(x) is derivable on [0, 1].

Because g(a)=g( 1)=0, then according to Rolle's theorem, there is ξ∈(a, 1)? (0, 1), so g'(ξ)=0.

f''(ξ)*(ξ- 1)^2+f'(ξ)*2(ξ- 1)=0

f''(ξ)=2f'(ξ)/( 1-ξ)

Certificate of completion