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Proof Using Rolle's Theorem

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f be continuous on [a,b] and differentiable on (a,b) Suppose that:

    f2(b) f2(a) = b2 - a2:

    Prove (using Rolle's theorem) that:

    ( exists x belonging to (a, b) ) ( f'(x)f(x) = x )

    I just don't know where to start i've done basic proofs with the theorem but only when f(a) = f(b). Any help would be greatly appreciated.

  2. jcsd
  3. Mar 5, 2009 #2


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    I think you mean f(b)^2-f(a)^2=b^2-a^2. Define g(x)=f(x)^2-x^2. Can you apply Rolle's theorem to g(x) on the interval [a,b]?
  4. Mar 5, 2009 #3
    No the notation i used is what is displayed in the question. I've never seen it before. Is it just the second derivative?
  5. Mar 5, 2009 #4


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    No. '^2' just means squared. E.g. f(b)^2=f(b)*f(b). Which is the same as [itex]f^2(b)[/itex]. The problem I'm having is that what you wrote looks like f(b)^2*f(a)^2=b^2-a^2. I really think it should be f(b)^2-f(a)^2. With a minus sign between the two squares, not a product of them.
  6. Mar 5, 2009 #5
    Yep. So sorry your spot on =]
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