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Math Help - f differentiable implies f' continuous.

  1. #1
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    f differentiable implies f' continuous.

    Suppose that f is differentiable on (a, b) and that f ' is increasing on (a, b). Prove that f ' is continuous on (a, b).


    I'm actually at a lost on how to do this problem. The method I keep finding myself trying is to assume that some point on f ' is not continuous. Then show this a contradiction. The problem is I keep finding myself assuming that the rest of f ' is continuous and I have no clue how f ' increasing contributes to this proof.

    It seems to me that any differentiable function would have f ' has continuous, otherwise, the original function would have a 'sharp corner' causing it to not be differentiable.

    Thanks in advance!
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  2. #2
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    You need to use the fact that f' is increasing - the statement is not true otherwise. The proof probably would make use of the fact that derivative attains intermediate values.
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  3. #3
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    The derivative of a function is not necessarily continuous but it does satisfy the "intermediate value" property: if f is differentiable on [a, b] and c is a number between f'(a) and f'(b), then there exist an x_0 such that f'(x_0) is between f'(a) and f'(b). A consequence of that is that if \lim_{x\to x_0^+} f'(x) and \lim_{x\to x_0^-} f'(x) exist, then they are equal and the derivative is continuous there. So if f' is NOT continuous at least one of those one-sided limits does not exist. Use "increasing" to show that cannot happen.
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  4. #4
    Member kalyanram's Avatar
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    As suggested f^' satisfies the intermediate value property now for any arbitrary c \in (a,b), given an \epsilon >0 consider (f^{'}(c) - \epsilon , f^{'}(c)+\epsilon) \subset (f^{'}(a) , f^{'}(b)) by intermediate value property \exists (\alpha , \beta) \subset (a,b) such that f^{'}(\alpha, \beta)=(f^{'}(c) - \epsilon , f^{'}(c)+\epsilon), \alpha < c < \beta now lets choose  \delta = min((c-\alpha),(\beta-c)) we always have |f^{'}(x) -f^{'}(c)| < \epsilon, \forall  |x-c| < \delta. Hence f^{'} is continuous on (a,b).
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