Answer
The function is not continuous at $x=0$, but it takes all the values between $g(-1)$ and $g(1)$.
Work Step by Step
The Intermediate Value Theorem "asks" for a function $g$ to be continuous on an interval $[a,b]$ so that $f(a)\not=f(b)$.
Let's check these conditions for the given function.
The interval $[a,b]$ is $[-1,1]$, so $a=-1,b=1$.
$g(a)=g(-1)=-(-1)=1$
$g(b)=g(1)=1^3+1=2$
Therefore $g(-1)\not=g(1)$.
Now let's check the continuity.
The function $g$ is continuous on each of the intervals $[-1,0)$ and $[0,1]$. We have to check the continuity at $x=0$.
Calculate the left and right limits at $x=0$:
$$\begin{align*}
\lim_{x\rightarrow0^-}g(x)&=\lim_{x\rightarrow0^-} (-x)=0\\
\lim_{x\rightarrow0^+}g(x)&=\lim_{x\rightarrow0^+} (x^3+1)=0^3+1=1\\
g(0)&=1
\end{align*}$$
As $\lim_{x\rightarrow0^-}g(x)\not=\lim_{x\rightarrow0^+}g(x)$, it means the function is not continuous at $x=0$.
Since $g(-1)=1$ and $g(1)=2$ we notice that even if the conditions are not met, the function takes all the values between $g(-1)$ and $g(1)$.
