Answer
See graph and explanations below.
Work Step by Step
Step 1. See graph for the piecewise function: $f(x)=\begin{cases} 1\hspace1,3cm x\leq-1 \\ -x\hspace1cm -1\lt x\lt0 \\ 1\hspace1.3cm x=0 \\-x\hspace1cm 0\lt x\lt1 \\1\hspace1.3cm x\geq1 \\ \end{cases}$
Step 2. For $x=-1$, $\lim_{x\to-1^-}f(x)=1$, $\lim_{x\to-1^+}f(x)=-(-1)=1$, thus $\lim_{x\to-1}f(x)=1=f(-1)$, and the function is continuous at this point.
Step 3. For $x=0$, $\lim_{x\to^-}f(x)=-0=0$, $\lim_{x\to0^+}f(x)=-(0)=9$, thus $\lim_{x\to0}f(x)=0$, however, the function is not continuous at this point because $\lim_{x\to0}f(x)=0\ne f(0)=1$. This discontinuity is removable by redefining the function with $f(0)=0$
Step 4. For $x=1$, $\lim_{x\to1^-}f(x)=-(1)=-1$, $\lim_{x\to1^+}f(x)=1\ne \lim_{x\to1^-}f(x)$, thus $\lim_{x\to-1}f(x)$ does not exist. The function is not continuous at this point and the discontinuity can not be easily removed.
