Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} 2x + 3, & \text{if } x \le 2 \\ 2x - 3, & \text{if } x > 2 \end{cases}$

(C) The given function is $f(x) = \begin{cases} 2x + 3, & \text{if } x \le 2 \\ 2x - 3, & \text{if } x > 2 \end{cases}$.
It is evident that the function $f$ is defined for all real numbers. Let $c$ be any real number. We consider three cases:
Case $I$: $c < 2$. Here $f(x) = 2x + 3$. $\lim_{x \to c} f(x) = 2c + 3 = f(c)$. Thus,$f$ is continuous for all $x < 2$.
Case $II$: $c > 2$. Here $f(x) = 2x - 3$. $\lim_{x \to c} f(x) = 2c - 3 = f(c)$. Thus,$f$ is continuous for all $x > 2$.
Case $III$: $c = 2$. We check the limits at $x = 2$.
Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x + 3) = 2(2) + 3 = 7$.
Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x - 3) = 2(2) - 3 = 1$.
Since the left-hand limit $(7)$ is not equal to the right-hand limit $(1)$,the function is discontinuous at $x = 2$.
Therefore,$x = 2$ is the only point of discontinuity.