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How to prove differentiability?

  1. Dec 12, 2007 #1
    1. The problem statement, all variables and given/known data
    Given an interval I. The function f goes from I to the real line. Define f as f(x)=x^2 if x rational or 0 if x is irrational. show that f i differentiablle at x=0 and find its derivative at this point, i.e. x=0.

    2. Relevant equations
    I have a given lead on this. That |f(x)/x|<|x|. But I'm not sure how this should help me. I have started with the definition and I have tried to determine a certain epsilon for which this is true. But I don't know how.

    3. The attempt at a solution
    I have shown that the function is continuous at this point and therefore know that the derivative might exist. But when it comes to the actual differentiability I have that (according to the definition),
    given a positive epsilon there exists a delta. For x satisfying the inequality
    0<|x-c|<delta, the inequality |(f(x)-f(c))/(x-c)-L|<epsilon holds. L then is called f's derivative in the point c.

    I'm not sure if this counts as an attempt. But this problem is really bugging me.

    Hopes for an answer!
  2. jcsd
  3. Dec 12, 2007 #2


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    the derivative is [tex]lim_{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} = lim_{x \rightarrow 0} \frac{f(x)}{x}[/tex] as f(0)=0 by definition. But notice f(x)=x2 or 0, so f(x)/x is either 0 or x. See what you can do with that
  4. Dec 12, 2007 #3


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    Okay, so your function f is defined like as follow:

    [tex]f(x) : = \left\{ \begin{array}{ll} x ^ 2 \quad & , \mbox{if } x \in \mathbb{Q} \\ 0 \quad & , \mbox{if } x \notin \mathbb{Q}\end{array} \right.[/tex]

    So that means, for all x:
    [tex]0 \leq |f(x)| \leq x ^ 2[/tex]

    Divide all by |x| > 0, (since |x| > 0, the signs won't change) we have:

    [tex]0 \leq \left| \frac{f(x)}{x} \right| \leq \left| \frac{x ^ 2}{x} \right|[/tex]

    [tex]\Rightarrow 0 \leq \left| \frac{f(x)}{x} \right| \leq | x |[/tex] (1) (this is true for x <> 0)

    Now, the derivative at x = 0, of the above function is:

    [tex]f'(0) = \lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x - 0}[/tex]

    [tex]= \lim_{x \rightarrow 0} \frac{f(x) - 0}{x - 0}[/tex] (0 is rational, so f(0) = 0)

    [tex]= \lim_{x \rightarrow 0} \frac{f(x)}{x}[/tex]

    Now, using the inequality (1) above can you show that the limit exists? And what is that limit? Note that:

    [tex]|a| \rightarrow 0 \Leftrightarrow a \rightarrow 0[/tex], i.e a tends to 0, if and only if |a| tends to 0.

    So, can you go from here? :)
    Last edited: Dec 12, 2007
  5. Dec 13, 2007 #4
    Yes, I know where to go from there.

    You are amazing! Thank you! :)
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