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Rolle's Theorem

  1. Jul 29, 2009 #1
    If we consider Rolle's Theorem:
    "If f is continuous on [a, b], differentiable in
    (a,b), and f (a) = f (b), then there exists a point c in (a, b) where f'(c) = 0."

    Why do we need to state continuity of f in interval and differentiability of f in open segment? Why can't we say f differentiable on [a,b]?


  2. jcsd
  3. Jul 29, 2009 #2


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    If you said f is differentiable on [a,b], the result would still be true, just slightly less useful. As stated the theorem shows that the result is true even if f is not differentiable at a or b. If it was restated to require f to be differentiable on [a,b], it could not be used in these cases.
  4. Jul 29, 2009 #3
    for example... [itex]f(x) = \sqrt{(1-x^2)}[/itex] is continuous on [itex][-1,1][/itex] and differentiable on [itex](-1,1)[/itex] but not differentiable on [itex][-1,1][/itex].
  5. Jul 29, 2009 #4


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    Similarly, f(x)= |1- x^2| is continous on [-1, 1], differentiable on (-1, 1) but not differentiable on [-1, 1].
  6. Jul 30, 2009 #5
    A possible definition of "differentiable on [-1,1]" requires only one-sided derivatives at the two endpoints.
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