First of all, my understanding of the term "continuously differentiable" is that you can continue to take derivatives of your function ad infinitum. Does this sound right? 

The function in question is the following: 

$f(x) = (2x-1)^{\frac{1}{3}}$

The function $f$ is differentiable on all of $\mathbb{R}$ but its derivative is not, since $f'(x) = \frac{2}{3}(2x-1)^{-\frac{2}{3}}$ is not differentiable at $x = \frac{1}{2}$. So would I conclude that $f$ is not cont. diff. or that $f$ is cont. diff. on $\mathbb{R} \setminus \{\frac{1}{2}\}$?