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Math Help - differentiability question

  1. #1
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    differentiability question

    My question is twofold:
    The first one is this:
    I need a function which is continuous at only one point and differentiable at only one point.
    I claim f(x)={x^2 if x is rational, 0 if x is irrational} is such a function. Is this true?
    The second part is, I need a function which is differentiable at only two points and continuous at only one point. This seems impossible to me, unless I am missing something. I have been operating under the assumption that a function must be continuous at a point to be differentiable at that point... any suggestions?
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  2. #2
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    Quote Originally Posted by dannyboycurtis View Post
    My question is twofold:
    The first one is this:
    I need a function which is continuous at only one point and differentiable at only one point.
    I claim f(x)={x^2 if x is rational, 0 if x is irrational} is such a function. Is this true?
    The second part is, I need a function which is differentiable at only two points and continuous at only one point. This seems impossible to me, unless I am missing something. I have been operating under the assumption that a function must be continuous at a point to be differentiable at that point... any suggestions?
    The function you have supplied is fine.

    Another alternative is

    f(x) = x if x is rational and 0 if x is irrational.


    As for the second, logic would say that it would be impossible, as a function needs to be continuous at a point for it to be differentiable at that point...
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Prove It View Post
    Another alternative is

    f(x) = x if x is rational and 0 if x is irrational.
    That function is not differentiable at zero.

    \lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{f(x)}{x}=

    1.) \lim_{x\to0}\frac{x}{x}=1

    or

    2.) \lim_{x\to0}\frac{0}{x}=0

    So the derivative doesn't exist because the subsequential limits are not the same.

    You need the x^2 to make it work.
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  4. #4
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    can you be more clear
    Last edited by Jhevon; November 15th 2009 at 12:09 PM.
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  5. #5
    Super Member redsoxfan325's Avatar
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    About what?
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