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I am asked to find a function $f$ that is differentiable at $x=0$ but not differentiable at any other point.

I am looking at the function $f(x)=x^2$ when $x\in \mathbb{Q}$ and $f(x)=0$ when $x\notin \mathbb{Q}$.

I think it is easy to see that it is not differentiable whenever $x\neq 0$ but I do not know how to go about proving that the function is differentiable at $x=0$. I am thinking about using the idea of convergence (that we can create a sequence of rational numbers close to $x=0$ that approach $0$ or are within the $\epsilon$-neighborhood of $0$) but I am not sure if I need to worry about the irrational numbers? Specifically, how do I guarantee that this function is continuous at $0$?

MathStudent1324
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1 Answers1

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Hint: You're on the right track with that function, but you're overthinking it. You have $|f(x)| \le x^2$ for all $x.$ Use the definition of the derivative at $0.$

zhw.
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  • Using the definition, I have $\lim_{h\to 0} \frac{(x+h)^2-x^2}{h} = \lim_{h\to 0} \frac{2xh+h^2}{h} = \lim_{h\to 0} 2x+h = 2x$. But I am unsure of how the bound you provided helps me? – MathStudent1324 Feb 28 '17 at 01:59
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    What you want to consider is

    $$\frac{f(x) - f(0)}{x-0}.$$

     – zhw. Feb 28 '17 at 02:06
  • So I have $\lim_{x\to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x\to 0} \frac{x^2}{x} = \lim_{x\to 0} x = 0$. Does this alone prove that the derivative at $0$ exists? – MathStudent1324 Feb 28 '17 at 02:31
  • Well $f(x) = x^2$ some of the time. Youll want to consider the difference quotient separately for rational and irrational points. – zhw. Mar 01 '17 at 18:09
  • Thanks, I think I was overthinking it again, I figured everything out! – MathStudent1324 Mar 02 '17 at 05:13