Let $f(x) = x^3 - 3x^2 + 6$ and $$g(x) = \begin{cases} {\max \{f(t) : x + 1 \le t \le x+2 \}} & -3\le x\lt 0\\ 1-x &x\ge 0 \end{cases}$$
I need to find the points of discontinuity of $g(x)$ in $[-3, 1]$.
Since $f(x)$ is increasing in $(-\infty, 0) \cup (2, \infty)$,
$g(x) = {\max \{f(t) : x + 1 \le t \le x+2 \}} -3\le x\lt 0$
$$g(x) = \begin{cases} f(x+2) & -3\le x\lt -2\\ 6 &-2\le x\lt-1 \\f(x+1) &-1\le x \lt0 \\1-x &x\ge 0 \end{cases}$$
$g(x)$ is discontinuous at $x = 0$ because $\lim_{x \to 0^-} g(x) = f(1) = 4 \ne \lim_{x \to 0^+} g(x) = 1$.
The answer given is that $g(x)$ is continuous in $[-3, 1]$.
I don't see how $g$ is continuous at $x =0$, any hints on how to solve this problems ?