# Differentiability at a point

• Apr 23rd 2012, 11:33 AM
jsndacruz
Differentiability at a point
For
$f(x)= \begin{cases}x^2,&\quad \textrm{x is rational}\\x^3,&\quad \textrm{x irrational}\end{cases}$

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

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

Any help or tips are appreciated - thanks in advance!
• Apr 23rd 2012, 02:47 PM
ILikeSerena
Re: Differentiability at a point
The limit form requires that for any $\epsilon>0$ there is a $\delta>0$, such that $|f'(x)-{f(x+h)-f(x) \over h}|<\epsilon$ if $0.
However, if $\epsilon<|x^3-x^2|/2$, there is always an h, for which this is not true (for any $\delta$ 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.
• Apr 27th 2012, 02:28 PM
jsndacruz
Re: Differentiability at a point
Worded very well - thanks so much. I didn't think about using that form of limit.