Answer
The IVT can be applied
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 $[-1,1]$, so $a=-1,b=1$.
$f(a)=f(-1)=-1$
$f(b)=f(1)=1^2=1$
Therefore $f(-1)\not=f(1)$.
Now let's check the continuity.
The function $f$ 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 $x=0$:
$$\begin{align*}
\lim_{x\rightarrow0^-}f(x)&=\lim_{x\rightarrow0^-} x=0\\
\lim_{x\rightarrow0^+}f(x)&=\lim_{x\rightarrow0^+} (x^2)=0^2=0\\
f(0)&=0
\end{align*}$$
As $\lim_{x\rightarrow0^-}f(x)=\lim_{x\rightarrow0^+}f(x)=f(0)$, it means the function is continuous at $x=0$.
We conclude that the Intermediate Value Theorem can be applied.
