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Creative Application of Mean Value Theorem

  1. Dec 14, 2007 #1
    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!
     
  2. jcsd
  3. Dec 14, 2007 #2

    morphism

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    I would use Taylor's theorem: f(x) = f(0) + f'(0)x + remainder.
     
  4. Dec 14, 2007 #3
    you would use taylor's for the whole thing, or just for f'(d) - f'(e) = f"(c)?

    thanks for replying, by the way.
     
  5. Dec 14, 2007 #4

    morphism

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