What is it called when a function is not continuous but still can have a derivative? Consider the following function (I think it has a name, but I don't remember it):
$$
f(x) = \cases{-1 & $x < 0$ \\
0 & $x = 0$ \\
1 & $x > 0$}
$$
$f'(x)$ is zero everywhere except at $x=0$, where $f$ is not continuous. But suppose we ignore the right half of the real line and define $f(0)$ to be $-1$. Then $f$ has a left derivative at $x=0$, and it is zero. We can do the same thing from the right, so in a way it could make a little bit of sense to say that $f'(0) =0$.
Of course, I understand that going by the definition $f$ isn't differentiable at $x=0$. But one could imagine an alternative definition of derivative for discontinous functions, in which one calculates lateral derivatives by redefining the function to be continuous, and then we see if the lateral derivatives match. This doesn't always work; for example it's hard to meaningfully assign a derivative to $x \mapsto |x|$ at $x=0$.
Are there other functions with this property? Does it have a name?
 A: This "lateral derivative" approach is already taken in many analysis texts. Royden's analysis text does this, for example. (I though that I have seen them referred to as left and right "derivates", but I can't seem to come up with a source...) 
By the way, differentiability of a function on its domain always implies it is continuous there. This works for "left differentiability implies left continuity" and on the right hand too. That is not contradicting your example, however!
If you "ignore the right half of the plane", you are effectively changing the domain to be only the half you are focused on, and indeed, you defined it to be continuous there.
A: Interestingly, you can assign a derivative of the function $\operatorname{abs}$ at $0$ by using the following definition:
$$\frac{\mathrm df(x)}{\mathrm dx}=\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}.$$
Thus, taking the limit,
$$\operatorname{abs}'(x)=\frac{\mathrm d}{\mathrm dx}|x|=\operatorname{sgn}(x)=\cases{-1, & $x < 0;$ \\ \phantom{-}0, & $x = 0;$ \\\phantom{-}1, & $x > 0.$}$$.
