#### Answer

$g$ is left-continuous at $c_1=1$
$g(1)=3$

#### Work Step by Step

A function has a jump discontinuity if both one-sided limits exist, but they are not equal.
$\displaystyle\lim_{x\rightarrow 1^{-}} g(x)=2$
$\displaystyle\lim_{x\rightarrow 1^{+}} g(x)=3$
As $\displaystyle\lim_{x\rightarrow 1^{-}} g(x)$ and $\displaystyle\lim_{x\rightarrow 1^{+}} g(x)$, but they are not equal, the function $g$ has a discontinuity at point $x=1$.
We also have:
$\displaystyle\lim_{x\rightarrow 1^{-}} g(x)=g(1)$
therefore $g$ is left-continuous at $c_1=1$.
In order to make $g$ right-continuous in $x=c_1=1$, we must have:
$\displaystyle\lim_{x\rightarrow 1^{+}} g(x)=g(1)$
$3=g(1)$
$g(c_1)=g(1)=3$