The graph of the function is cut off in the vertical direction because the function escapes to negative infinity.
I know [2,-1)U(-1,1)U(1,3)U(3,4], but the system is telling me that my second set is incorrect.
$\endgroup$1 Answer
$\begingroup$Recall that the definition of $f'(0)$ is $$ \lim_{x \to 0} \frac{f(0 + h) - f(0)}{h}$$ From the graph you have shown, the function isn't even defined at $0$ thus the expression in the limit doesn't even make sense thus f can't be differentiable at $0$. So the second set should be $(-1,0) \cup (0,1)$.
Note that even if the function was defined arbitrarily at $0$, it still can't be differentiable at $0$. Because if it was differentiable at $0$, it would be continuous at $0$, and thus there is some interval around $0$ on which the function is bounded (this should be intuitively clear, but a rigorous proof of this is a straight forward exercise using the $\varepsilon - \delta$ definition of continuity... if you are unfamiliar with this, just take this statement for granted). However the bold statement is clearly false because from the graph, it is clear that $f$ is unbounded in every interval around $0$. Thus our initial assumption that $f$ is differentiable at $0$ is false.
$\endgroup$
