Answer
$$
f(x)=\left\{\begin{array}{ll}{1-x} & {\text { if } \quad x \lt 1} \\
{2} & {\text { if } \quad 1\leq x\leq 2}\\
{4-x} & {\text { if }\quad x>2}\end{array}\right.
$$
(a) graphed the function.
(b) The graph is discontinuous at $x=1.$
(c)
$$
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(1-x \right)=0
$$
From the right, where $x$-values are greater than 1,
$$
\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(2\right)=2
$$
so, $ \lim _{x \rightarrow 1} f(x) $ does not exist since
$$
\lim _{x \rightarrow 1^{+}} f(x) \ne \lim _{x \rightarrow 1^{-}}f(x)
$$
Work Step by Step
$$
f(x)=\left\{\begin{array}{ll}{1-x} & {\text { if } \quad x \lt 1} \\
{2} & {\text { if } \quad 1\leq x\leq 2}\\
{4-x} & {\text { if }\quad x>2}\end{array}\right.
$$
(a) graphed the function.
(b) The graph is discontinuous at $x=1.$
(c)
Since each piece of this function is a polynomial, the only $x$-values where
$f$ might be discontinuous here are 1 and 2. We investigate at $x=1$ first. From the left, where $x$-values are less than $1$,
$$
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(1-x \right)=0
$$
From the right, where $x$-values are greater than 1,
$$
\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(2\right)=2
$$
so, $ \lim _{x \rightarrow 1} f(x) $ does not exist since
$$
\lim _{x \rightarrow 1^{+}} f(x) \ne \lim _{x \rightarrow 1^{-}}f(x)
$$
Now, we investigate at $x=2$ first. From the left, where $x$-values are less than $2$,
$$
\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}\left(2 \right)=2
$$
From the right, where $x$-values are greater than 2,
$$
\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}}\left(4-x\right)=2
$$
then we have
$$
f(2)=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{-}}f(x)
$$
so, $f(x)$ is continuous at $x=2$.
