Differentiability at a point

For

$\displaystyle f(x)= \begin{cases}x^2,&\quad \textrm{x is rational}\\x^3,&\quad \textrm{x irrational}\end{cases}$

I want to know where $\displaystyle f$ is differentiable/not differentiable. I know it is differentiable at $\displaystyle x=0$.. but I kind of cheated and used the Squeeze Theorem to prove that.

I now want to that to show that $\displaystyle f$ is *not* differentiable at the only other point where it is continuous, i.e. at $\displaystyle x=1$. I don't know how to show it explicitly in limit form, however.

Any help or tips are appreciated - thanks in advance!

Re: Differentiability at a point

The limit form requires that for any $\displaystyle \epsilon>0$ there is a $\displaystyle \delta>0$, such that $\displaystyle |f'(x)-{f(x+h)-f(x) \over h}|<\epsilon$ if $\displaystyle 0<h<\delta$.

However, if $\displaystyle \epsilon<|x^3-x^2|/2$, there is always an h, for which this is not true (for any $\displaystyle \delta$ and any choice of f'(x)).

This is the case, since between any 2 different rational numbers, there is always an irrational number.

And between any 2 different irrational numbers, there is always a rational number.

Re: Differentiability at a point

Worded very well - thanks so much. I didn't think about using that form of limit.