Answer
$f$ is right-continuous in $x=1$
$f$ is continuous in $x=2$
Work Step by Step
We are given the function:
$f(x)=\begin{cases}
x^2+3,\text{ for }x<1\\
10-x,\text{ for }1\leq x\leq 2\\
6x-x^2,\text{ for }x>2
\end{cases}$
The functions $x^2+3$, $10-x$ and $6x-x^2$ are continuous on each of the subintervals as they are polynomial functions.
Compute the left hand and right hand limits at $x=1$ and $x=2$:
$\displaystyle\lim_{x\rightarrow 1^{-}} (x^2+3)=1^2+3=4$
$f(1)=10-1=9$
$\displaystyle\lim_{x\rightarrow 1^{+}} (10-x)=10-1=9$
$\displaystyle\lim_{x\rightarrow 2^{-}} (10-x)=10-2=8$
$f(2)=10-2=8$
$\displaystyle\lim_{x\rightarrow 2^{+}} (6x-x^2)=6(2)-2^2=8$
Therefore we got:
$\displaystyle\lim_{x\rightarrow 1^{+}} f(x)=f(1)=9$
$\Rightarrow f$ is right-continuous in $x=1$
$\displaystyle\lim_{x\rightarrow 2^{-}} f(x)=f(2)=\displaystyle\lim_{x\rightarrow 2^{+}} f(x)=8$
$\Rightarrow f$ is continuous in $x=2$
