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Langrange's Mean Value Theorem question

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Find a function f on [-1,1] such that :-

    (a) there exists c [itex]\in[/itex] (-1,1) such that f'(c) = 0 and
    (b) f(a) [itex]\neq[/itex] f(b) for any a[itex]\neq[/itex] b [itex]\in[/itex] [-1,1]

    2. Relevant equations

    Lagrange's Mean Value Theorem (LMVT), which states that if f:[a,b]-->ℝ is a function which is continuous in [a,b] and differentiable in (a,b), there exists some c[itex]\in[/itex] (a,b) such that f'(c) = (f(b) - f(a))/(b-a).

    3. The attempt at a solution

    If f was said to be differentiable in (-1,1), then it's quite easy to prove that such a function cannot exist, using LMVT. But here, nothing is said about the continuity or differentiability of f in (-1,1). The function could just be differentiable at the point x = c.

    I'm quite sure that such a function cannot exist. However, this is purely intuitive, and I'm unable to prove it. Can someone please hint on a proof, or an example if there's any?

    Thank you.
     
  2. jcsd
  3. Nov 9, 2012 #2

    haruspex

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    I don't see that. In fact, I don't see the relevance of LMVT here. Were you told to use it somehow? Unless I've misunderstood something, there's quite an easy (and differentiable) solution.
     
  4. Nov 10, 2012 #3
    Thank you, haruspex for your reply. No, I wasn't told to use LMVT, but that was the only way I could think of.

    Well, what I had in mind was assuming that such a function f exists, which is continuous in [-1,1] and differentiable in (-1,1), and f(a) not equals f(b), for any a not equals b, in [-1,1] (Sorry for my abstinence from symbols, I'm typing from my phone). Then by LMVT, there would exist some c such that

    f'(c) = (f(1) - f(-1))/(1-(-1))

    Now, if f'(c) were to be 0, then essentially f(-1) = f(1).

    But we assumed that, f(a) not equals f(b), for any a not equals b, yet we got f(-1) = f(1). Hence, our assumption is wrong, and therefore such a function cannot exist.

    But the above proof relies on the assumption that f is continuous in [-1,1] and differentiable in (-1,1). But this need not be the case. From the question, what I understand is that f could very well be differentiable ONLY at x = c.

    Could you share your thoughts on this, haruspex? Please don't give out the solution, as I'd like to have the pleasure of solving it on my own. Thanks.
     
  5. Nov 10, 2012 #4

    haruspex

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    Sure, there exists some c s.t. f'(c) = (f(1) - f(-1))/(1-(-1)), but why should f'(c)=0 for that c? f could quite easily be 0 somewhere else.
    Think of some point at which f' is zero. Now extend f left and right satisfying f(a)≠f(b).
     
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