Find all points of discontinuity of $f,$ where $f$ is defined by
$f(x) = \begin{cases} |x| + 3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \ge 3 \end{cases}$

(D) The given function is $f(x) = \begin{cases} |x| + 3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \ge 3 \end{cases}$.
Case $I$: If $c < -3$,then $f(c) = -c + 3$. The limit $\lim_{x \to c} f(x) = \lim_{x \to c} (-x + 3) = -c + 3 = f(c)$. Thus,$f$ is continuous for all $x < -3$.
Case $II$: If $c = -3$,then $f(-3) = |-3| + 3 = 6$. The left-hand limit is $\lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} (-x + 3) = -(-3) + 3 = 6$. The right-hand limit is $\lim_{x \to -3^+} f(x) = \lim_{x \to -3^+} (-2x) = -2(-3) = 6$. Since $\lim_{x \to -3^-} f(x) = \lim_{x \to -3^+} f(x) = f(-3)$,$f$ is continuous at $x = -3$.
Case $III$: If $-3 < c < 3$,then $f(c) = -2c$. The limit $\lim_{x \to c} f(x) = \lim_{x \to c} (-2x) = -2c = f(c)$. Thus,$f$ is continuous for all $x \in (-3, 3)$.
Case $IV$: If $c = 3$,then $f(3) = 6(3) + 2 = 20$. The left-hand limit is $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (-2x) = -2(3) = -6$. The right-hand limit is $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (6x + 2) = 6(3) + 2 = 20$. Since the left-hand limit $\neq$ right-hand limit,$f$ is discontinuous at $x = 3$.
Case $V$: If $c > 3$,then $f(c) = 6c + 2$. The limit $\lim_{x \to c} f(x) = \lim_{x \to c} (6x + 2) = 6c + 2 = f(c)$. Thus,$f$ is continuous for all $x > 3$.
Therefore,the only point of discontinuity is $x = 3$.