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Mean value theorem, closed intervals

  1. Jun 26, 2015 #1
    Mean Value Theorem

    Suppose that ##f## is a function that is continuous on ##[a,b]## and differentiable on ##(a,b)##. Then there is at least one ##c## in ##(a,b)## such that:
    $$f'(c) = \frac{f(b) - f(a)}{b - a}$$

    My question is: wouldn't it be better to state that ##c## is in ##[a,b]## rather than ##(a,b)##? For example, if ##f(x) = 2## for ##1 \leq x \leq 3##, then:
    $$f'(x) = 0 = \frac{f(3) - f(1)}{2}$$
    For all ##x##, including 3, which is one of the endpoints of the interval.
     
  2. jcsd
  3. Jun 26, 2015 #2

    micromass

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    If ##c## is in ##(a,b)##, then ##c## is in ##[a,b]##. Besides, ##f## might not be differentiable at ##a## or ##b##.
     
  4. Jun 26, 2015 #3

    verty

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    Because the theorem is a stronger result.
     
  5. Jun 26, 2015 #4
    Yeah, I think that's it. ##f## is differentiable on ##(a,b)##, not ##[a,b]##.
     
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