Answer
The function $f$ is not continuous at $x=0$.
For $y=0$ in $(f(-2),f(2))$ there is no corresponding $x$ in $(-2,2)$
Work Step by Step
The Intermediate Value Theorem "asks" for a function $f$ 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 $[-2,2]$, so $a=-2,b=2$.
$f(a)=f(-2)=-(-2)^2=-4$
$f(b)=f(2)=2$
Therefore $f(-2)\not=f(2)$.
Now let's check the continuity.
The function $f$ is continuous on each of the intervals $[-2,0), (0,2]$. We have to check the continuity at $x=0$.
Calculate the left and right limits at $x=0$:
$$\begin{align*}
\lim_{x\rightarrow0^-}f(x)&=\lim_{x\rightarrow0^-} (-x^2)=0\\
\lim_{x\rightarrow0^+}f(x)&=\lim_{x\rightarrow0^+} x=0\\
f(0)&=1
\end{align*}$$
As $\lim_{x\rightarrow0^-}f(x)=\lim_{x\rightarrow0^+}f(x)\not=f(0)$, it means the function is not continuous at $x=0$.
We conclude that the Intermediate Value Theorem cannot be applied.
A value between $f(-2)$ and $f(2)$ for which there is no $x$ in $(-2,2)$ is $0$.
