Let $$f(x) = \begin{cases} x^2(x^2-1),&x \in\mathbb{Q} \\ 0,&x \not\in\mathbb{Q} \end{cases}$$
A. When is this function continuous? when is it differentiable?
I solved these kind of excercises, but never when I have $x\in$\ $\not\in \mathbb{Q}$
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ The function is continuous at 3 points, which we can find by considering $x^{2}(x^{2}-1)=0$, giving us: $x=\{0,\pm 1\}$. We have that it is differentiable at $x=0$ only. $\endgroup$– Thomas RussellJun 23, 2014 at 10:33
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Hints: (a) By definition, a function is continuous at $x_0$ iff $\lim_{x\to x_0} f(x) = f(x_0)$, for your function, you need hence $$ 0 = \lim_{x\to x_0, x\not\in \mathbb Q} f(x) \stackrel != \lim_{x\to x_0, x\in \mathbb Q} f(x) = x_0^2(x_0^2 - 1) $$ (b) If $f$ is not continous, it is not differentiable at $x_0$. To check differentiability at $x_0$ argue as in (a), but for $$ \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0} $$ instead of $\lim f(x)$.
-
$\begingroup$ Didn't get the part $$\lim_{x\to x_0, x\not\in \mathbb Q} f(x) \stackrel != \lim_{x\to x_0, x\in \mathbb Q} f(x)$$ and why doesn't it differentiable at $\pm 1$ $\endgroup$ Jun 23, 2014 at 11:40
-
$\begingroup$ If this holds, the limit exists ... (as the limits on and off $\mathbb Q$ are equal) ... $\endgroup$– martiniJun 23, 2014 at 11:43