Solution:
The function provided is $f(x)=\left\{\begin{array}{lll}2 x+3, & \text { if } & x \leq 2 \\ 2 x-3, & \text { if } & x>2\end{array}\right.$
Given here the $f(x)$ is defined for $x \leq 2$ or $(-\infty, 2)$ and even for $x>2$ or $(2, \infty)$.
As a result, Domain of $f$ is $(-\infty, 2) \cup(2, \infty)=(-\infty, \infty)=R$
For all $x<2, f(x)=2 x+3$ is a polynomial and hence continuous and for all $x>2, f(x)=2 x-3$ is a continuous and hence it is also continuous on $R-\{2\}$.
Now
Left Hand limit = $\lim _{x \rightarrow 2} f(x)=\lim _{x \rightarrow 2^{2}}(2 x+3)=2 \times 2+3=7$
Right Hand Limit = $\lim _{x \rightarrow 2+} f(x)=\lim _{x \rightarrow 2^{-}}(2 x-3)=2 \times 2-3=1$
Since, $\lim _{x \rightarrow 2^{-}} f(x) \neq \lim _{x \rightarrow 2^{+}} f(x)$
As a result, $\lim _{x \rightarrow 2} f(x)$ does not exist and therefore $f(x)$ is discontinuous at only $x=2$.