Let $p > 0$, and let $f:\mathbb{R} \rightarrow \mathbb{R}$ be defined piecewise by $f(x)= |x|^p$ if $x \in \mathbb{Q}$ and $f(x)=0$ if $x \in \mathbb{R} \setminus \mathbb{Q}$. For what values of $p$ is $f$ differentiable?
Do I have to show the left/right hand limits exist and agree using the definition of absolute value (i.e. $|x|=x$ if $x \geq 0$ and $|x|=-x$ if $x < 0$)? But how do I proceed? By fixing $p$ and then solving for it somehow?
Also, since both $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ are dense, we can pick a sequence $(x_n) $of rationals and $(y_n)$ of irrationals that both converge to 0. Then, by the sequential definition of continuity, we must have $(f(x_n)) \rightarrow 0$ and $(f(y_n)) \rightarrow 0$ (since before being differentiable, a function has to be continuous).
That means that $f$ will be differentiable only for $p=0$?
Thanks!
As you have observed the function is not continous at $x$ if $x \neq 0$ so it is not differentiable either. For differentiability at $x=0$ consider $\lim_{x \to 0} \frac {|x|^{p}} x$. If $p>1$ this limit is $0$ because $|\frac {|x|^{p}} x|=|x|^{p-1} \to 0$. For $p<1$ it is clear that $|\frac {|x|^{p}} x|=|x|^{p-1} \to \infty$. For $p=1$ consider right hand and left hand limits to see that the function is not differentiable