1. The problem statement, all variables and given/known data Assume that f is twice differentiable on the entire real line. Show that f(-1) + f(1) - 2f(0) = f"(c) for some c in [-1,1] 2. Relevant equations I'm thinking the mean value theorem will be helpful here -- the MVT states that, given a function f differentiable on [a,b], there is some point c in (a,b) s.t. ( f(b) - f(a) ) / (b-a) = f'(c). 3. The attempt at a solution By applying to MVT to ( f(-1) - f(0) ) and ( f(1) - f(0) ) and then adding the results, I've managed to show that f(-1) + f(1) - 2f(0) = f'(d) - f'(e) for some d,e in (-1, 1). But then I am stuck. How to prove that f'(d) - f'(e) = f"(c) for some c in [-1,1]? Or am I just completely on the wrong track? Thanks for your help!