Answer
$[-1,0)\cup(0,1)\cup(1,2)\cup(2,3)$
Work Step by Step
In Exercise $5$, we concluded that f is right-continuous at $x=-1$.
Polynomials (all piecewise component functions are polynomials) are continuous on their domains.
So, f is continuous on $[-1,0)$.
f is not continuous at $c=0,2,3$, because $f(c)$ is not defined.
f is not continuous at $1$, as we found in exercise $6\ \ (\displaystyle \lim_{x\rightarrow 1}f(x)\neq f(1))$.
The piecewise component functions are continuous on the intervals between them,
that is, on $(0,1)\cup(1,2)\cup(2,3).$
Taking the first half-closed interval into account, f is continuous on
$[-1,0)\cup(0,1)\cup(1,2)\cup(2,3)$
