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I saw an interesting question:

Let $D(x)=\begin{cases} 1 & \text{if } x \in \mathbb{Q}, \\ 0 & \text{if } x \in \mathbb{R} \setminus \mathbb{Q}\end{cases}$

Let $f(x) = x \cdot D(x)$.

Which is true:

  1. $f$ is not differentiable in $x=0$
  2. $f'(0)=0$
  3. $f'(0)=\frac{1}{2}$
  4. $f'(0)=1$

So, what's the right answer? I know that in every neighborhood of $x$ there are points where $D(x)=1$ but also points where $D(x)=0$, so how do I go about calculating this?

Thanks.

  • 3
    Despite your title, $f$ is *not* nowhere continuous.2012-05-19

1 Answers 1