Because f(0)=f( 1) and f(x) is continuously differentiable on [0, 1], according to Rolle's theorem, there exists k∈(0, 1), so f'(k)=0.

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

Because G (k) = f' (k) (1-k) 2 = 0 and g( 1)=0, according to Rolle's theorem, there exists ξ∈(k, 1)? (0, 1), so g'(ξ)=0.

f''(ξ)( 1-ξ)^2-f'(ξ)*2( 1-ξ)=0

f''(ξ)=2f'(ξ)/( 1-ξ)

Certificate of completion