# Creative Application of Mean Value Theorem

1. Dec 14, 2007

### michaelxavier

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?

2. Dec 14, 2007

### morphism

I would use Taylor's theorem: f(x) = f(0) + f'(0)x + remainder.

3. Dec 14, 2007

### michaelxavier

you would use taylor's for the whole thing, or just for f'(d) - f'(e) = f"(c)?

thanks for replying, by the way.

4. Dec 14, 2007

### morphism

For the whole thing - don't bother with the MVT. If you choose the right form for the remainder, it's pretty much done!