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Thread: a differential funtion

  1. #1
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    a differential funtion

    Assume f:R \rightarrow R has a finite derivative everywhere on (a,b) except possibly at a point c \in (a,b). If we have lim_{x \rightarrow c} f'(x)=A, then prove f'(c) exists and is equal to A.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,

    Hint : use the mean value theorem.
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    mean value theorem

    i thought about it but is it possible to even use it? f has to be ontinuous on the endpoints and has to be differentiable on (a,b). but we don know if it is continuoust at a and b or it is diffentiable at c.
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  4. #4
    Member alunw's Avatar
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    You cannot hope to prove this unless you are given that f is continuous at c. Otherwise you can have a function that is continuous and differentiable everywhere in the interval apart from c but which takes an arbitrary value at c itself, or you could have a function that looks like tan(x) but with a small section round pi/2 removed. Then there is a huge discontinuity at c, and yet the derivative is the same both sides of c.
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by Kat-M View Post
    i thought about it but is it possible to even use it? f has to be ontinuous on the endpoints and has to be differentiable on (a,b). but we don know if it is continuoust at a and b or it is diffentiable at c.
    We must be allowed to assume that f is continuous on (a,b), otherwise the propostion would be clearly false (because in that case f(c) could be anything).

    So, assuming f is continuous on (a,b), I suggest that you use the mean value theorem, applied to (a,c] and [c,b), respectively, to show that the left and the right derivative of f at c both exist and are both equal to A.
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  6. #6
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    Quote Originally Posted by Kat-M View Post
    Assume f:R \rightarrow R has a finite derivative everywhere on (a,b) except possibly at a point c \in (a,b). If we have lim_{x \rightarrow c} f'(x)=A, then prove f'(c) exists and is equal to A.
    The question does not actually say that f is continuous at c. But it better had be, as noted by alunw and Failure. With that assumption, the result is true, but it seems to be quite a tricky piece of analysis, requiring a very careful proof.

    We are told that \lim_{x\to c}f'(x) = A. So given \varepsilon>0 there exists \delta>0 such that |f'(\xi)-A|<\varepsilon whenever |\xi-c|<\delta.

    Now suppose that |x-c|<\delta and |y-c|<\delta. By the mean value theorem (as hinted by flyingsquirrel ) \frac{f(x) - f(y)}{x-y} = f'(\xi) for some \xi between x and y. But if x and y are both within distance \delta from c then so is \xi. It follows that \left|\frac{f(x) - f(y)}{x-y} - A\right|<\varepsilon.

    Next, let y\to c in that last inequality. By continuity of f at c, this gives \left|\frac{f(x) - f(c)}{x-c} - A\right|\leqslant\varepsilon. Since this holds whenever |x-c|<\delta, and \varepsilon is an arbitrary positive number, it follows that f is differentiable at c, with derivative A.
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  7. #7
    Super Member Failure's Avatar
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    Quote Originally Posted by Opalg View Post

    Now suppose that |x-c|<\delta and |y-c|<\delta. By the mean value theorem (as hinted by flyingsquirrel ) \frac{f(x) - f(y)}{x-y} = f'(\xi) for some \xi between x and y.
    How can you apply the mean value theorem here if you don't know (yet) that f is differentiable in the entire intervall between x and y?
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  8. #8
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    Quote Originally Posted by Failure View Post
    How can you apply the mean value theorem here if you don't know (yet) that f is differentiable in the entire interval between x and y?
    Good point. I should have said that y must be on the same side of c as x. Then I think the proof should work. But in fact your comment #5 above gives a shorter proof: once you know that f is continuous on the closed interval [c,x], you can apply the MVT on that interval and the result comes out straight away.
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