Answer
The graph is enclosed below.
1) At $x=-1$:
$\lim_{x\to-1}f(x)=\lim_{x\to-1^+}f(x)=\lim_{x\to-1^-}f(x)=1$
$f(x)$ is left-continuous, right-continuous, and overall, continuous at $x=-1$.
2) At $x=0$:
$\lim_{x\to0}f(x)=\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)=0$
$f(x)$ is discontinuous at $x=0$. It is also not left-continuous or right-continuous at $x=0$. This discontinuity is removable.
3) At $x=1$:
$\lim_{x\to1^+}f(x)=1$, $\lim_{x\to1^-}f(x)=-1$, $\lim_{x\to1}f(x)$ does not exist.
$f(x)$ is discontinuous at $x=1$. But it is right-continuous here.
This discontinuity is non-removable.
Work Step by Step
The graph is enclosed below.
1) At $x=-1$:
- $\lim_{x\to-1^-}f(x)=\lim_{x\to-1^-}1=1$
- $\lim_{x\to-1^+}f(x)=\lim_{x\to-1^+}-x=-(-1)=1$
So $\lim_{x\to-1}f(x)=\lim_{x\to-1^+}f(x)=\lim_{x\to-1^-}f(x)=1$
- $f(-1)=1$
So $\lim_{x\to-1}f(x)=\lim_{x\to-1^+}f(x)=\lim_{x\to-1^-}f(x)=f(-1)=1$
Therefore, $f(x)$ is left-continuous, right-continuous, and overall, continuous at $x=-1$.
2) At $x=0$:
- $\lim_{x\to0^-}f(x)=\lim_{x\to0^-}-x=-0=0$
- $\lim_{x\to0^+}f(x)=\lim_{x\to0^+}-x=-0=0$
So $\lim_{x\to0}f(x)=\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)=0$
- $f(0)=1$
So $\lim_{x\to0}f(x)\ne f(0)$. Neither is $f(0)$ equal with $\lim_{x\to0^+}f(x)$ or $\lim_{x\to0^-}f(x)$
Therefore, $f(x)$ is discontinuous at $x=0$. It is also not left-continuous or right-continuous at $x=0$.
This discontinuity is removable because only $\lim_{x\to0}f(x)\ne f(0)$, so we can extend the function $f(x)$ to cover $f(0)=0$ so that $\lim_{x\to0}f(x)=f(0)$, thus removing the discontinuity.
3) At $x=1$:
- $\lim_{x\to1^-}f(x)=\lim_{x\to1^-}-x=-1$
- $\lim_{x\to1^+}f(x)=\lim_{x\to1^+}1=1$
So $\lim_{x\to1^+}f(x)\ne\lim_{x\to1^-}f(x)$. $\lim_{x\to1}f(x)$, as a result, does not exist.
- $f(1)=1$
Since $\lim_{x\to1}f(x)$ does not exist, $f(x)$ is discontinuous at $x=1$.
However, since $\lim_{x\to1^+}f(x)=f(1)=1$, $f(x)$ is right-continuous at $x=1$.
This discontinuity is not removable because $\lim_{x\to1}f(x)$ does not exist. We cannot remove the discontinuity by making $f(x)$ include any values so that $\lim_{x\to1}f(x)=f(1)$.
