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Derivative Proof

  1. Jan 17, 2009 #1
    1. The problem statement, all variables and given/known data
    if f' is continuous, show that:
    [tex]\mathop{\lim}\limits_{x \to 0}(\frac{f(x+h)-f(x-h)}{2h})=f'(x)[/tex]
    be sure to explain why f' must be continuous

    2. Relevant equations
    not really any equations, this is for AP Calc BC and we've just done L'Hospital's theorem and the derivatives/integrals of logs and inverse trig functions.

    3. The attempt at a solution
    I know that as x-->0 it becomes [tex]\frac{f(h)-f(-h)}{2h}[/tex] . I thought about proving that the difference quotient can be manipulated into the above formula, but haven't had any success.

    Any pointers?
     
  2. jcsd
  3. Jan 17, 2009 #2

    jgens

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    Gold Member

    Are you certain that it's supposed to be as x -> 0 and not h? Aside from that, my only suggestion is to add and subtract f(x)/2h and apply some limit lemmas.
     
  4. Jan 17, 2009 #3
    yeah it says x-->0
    what is a 'limit lemma'? o_O
     
  5. Jan 17, 2009 #4
    I think jgens is right, it's likely h -> 0.

    Hmmm, I'm surprised this is Calc BC. I think I've seen this problem in Spivak before.

    The trick is to consider the difference quotient and replace h with -h, and then take the limit as h -> 0. Then you should be able to work with the expression given.

    EDIT: Ok, this is a slightly different problem. Hmmm, see if you can make use of the definition of continuity, and the equivalence of the statements lim x -> a f(x) and lim h -> 0 f(a+h).
     
    Last edited: Jan 17, 2009
  6. Jan 17, 2009 #5

    jgens

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    Gold Member

    I think this approach may work:

    Derivative of a function: y = f(x) than y - dy = f(x - h), hence dy/dx = lim h -> 0 (f(x) - f(x - h))/h

    Given (f(x + h) - f(x - h))/2h = (f(x +h) - f(x) + f(x) - f(x - h))/2h. Take the limit as h -> 0 and I think that should yield a solution.

    A limit lemma would be lim x-> c kf(x) = klim x -> c f(x) where k is a constant.
     
  7. Jan 17, 2009 #6
    thanks i'll give that a try--my only question is how do i get it from x-->0 to h-->0?
     
  8. Jan 19, 2009 #7
    So it turns out my math teacher actually did write the question wrong... it really is h-->0. it's solved with a simple l'hospital. thanks guys
     
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