If we have $f(x)=|x|$ this is not differentiable at $x=0$ because the limits from the right and left are different. This makes me think that a function is never differentiable at a cusp.
This piecewise function is differentiable at $0$, does that mean there is no cusp at $x=0$?
$f(x) = \begin{cases} x^{2}-3, & x < 0 \\ -3, & x \ge 0 \end{cases}$
If I had the following piecewise function, where it is discontinuous and I am asked what is $f'(1)$
$f(x) = \begin{cases} 4x+2, & x \le 1 \\ x+1, & x > 1 \end{cases}$
Here I would say that the $f'(1)$ does not exist because it is not continuous at $x=1$?
For this example, if the piecewise function is continuous but has a corner (would this be considered a cusp?) is it differentiable at $x=1$?
$f(x) = \begin{cases} 4x+2, & x \le 1 \\ x+5, & x > 1 \end{cases}$