$f(x) = \left\{ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{2x + 3,}&{if}&{x \le 2}\\{2x - 3,}&{if}&{x > 2}\end{array}} \right.$ }

For x $< 2$, function f (x)$=$ 2x + 3 is polynomial and hence, continuous.
For $x > 2$, function f(x) $=$ 2x$-$3 is polynomial and hence, continuous.

For continuity at x $=$ 2,

$\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} (2x + 3) = \mathop {\lim }\limits_{\scriptstyle x \to 2 - h\atop\scriptstyle h \to 0} [2(2 - h) + 3]$

$= \mathop {\lim }\limits_{\scriptstyle x \to 2 - h\atop\scriptstyle h \to 0} (4 - 2h + 3) = \mathop {\lim }\limits_{\scriptstyle x \to 2 - h\atop\scriptstyle h \to 0} (7 - 2h) = 7$

$\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} (2x - 3) = \mathop {\lim }\limits_{\scriptstyle x \to 2 + h\atop\scriptstyle h \to 0} [2(2 + h) - 3]$

$= \mathop {\lim }\limits_{\scriptstyle x \to 2 + h\atop\scriptstyle h \to 0} (4 + 2h - 3) = \mathop {\lim }\limits_{\scriptstyle x \to 2 + h\atop\scriptstyle h \to 0} (1 + 2h) = 1$

Thus, $\mathop {\lim }\limits_{x \to {2^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {2^ + }} f(x)$

Hence we can say that f(x) is not continuous at x $=$ 2.

So, the only point of discontinuity of the given function f is 2.