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Math Help - Differentiability at a point

  1. #1
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    Differentiability at a point

    For
    f(x)= \begin{cases}x^2,&\quad \textrm{x is rational}\\x^3,&\quad \textrm{x irrational}\end{cases}

    I want to know where f is differentiable/not differentiable. I know it is differentiable at x=0.. but I kind of cheated and used the Squeeze Theorem to prove that.

    I now want to that to show that f is not differentiable at the only other point where it is continuous, i.e. at x=1. I don't know how to show it explicitly in limit form, however.

    Any help or tips are appreciated - thanks in advance!
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Differentiability at a point

    The limit form requires that for any \epsilon>0 there is a \delta>0, such that |f'(x)-{f(x+h)-f(x) \over h}|<\epsilon if 0<h<\delta.
    However, if \epsilon<|x^3-x^2|/2, there is always an h, for which this is not true (for any \delta and any choice of f'(x)).

    This is the case, since between any 2 different rational numbers, there is always an irrational number.
    And between any 2 different irrational numbers, there is always a rational number.
    Thanks from jsndacruz
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  3. #3
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    Re: Differentiability at a point

    Worded very well - thanks so much. I didn't think about using that form of limit.
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