Let

$f(x)=\begin{cases} |x|+3 & \text{if } x\le -3 \\ -2x & \text{if } -3<x<3 \\ 6x+2 & \text{if } x\ge 3 \end{cases}$

Then, which of the following is true?

A function is continuous at a point if the left-hand limit, right-hand limit, and the actual function value are all equal.

The piecewise function has potential discontinuities at $x = -3$ and $x = 3$.

---

At $x = -3$, using the condition $x \leq -3$:

$f(-3) = |-3| + 3$

$f(-3) = 3 + 3$

$f(-3) = 6$

For the left-hand limit (when $x < -3$), using $f(x) = |x| + 3$:

$\lim_{x \to -3^-} f(x) = |-3| + 3 = 6$

For the right-hand limit (when $-3 < x < 3$), using $f(x) = -2x$:

$\lim_{x \to -3^+} f(x) = -2(-3) = 6$

Since $f(-3) = 6$, left limit $= 6$, and right limit $= 6$, the function is continuous at $x = -3$.

---

At $x = 3$, using the condition $x \geq 3$:

$f(3) = 6(3) + 2$

$f(3) = 18 + 2$

$f(3) = 20$

For the left-hand limit (when $-3 < x < 3$), using $f(x) = -2x$:

$\lim_{x \to 3^-} f(x) = -2(3) = -6$

For the right-hand limit (when $x \geq 3$), using $f(x) = 6x + 2$:

$\lim_{x \to 3^+} f(x) = 6(3) + 2 = 20$

Since left limit $= -6$ and right limit $= 20$, the limits are not equal. The function is discontinuous at $x = 3$.

---

The function is continuous at $x = -3$ and discontinuous at $x = 3$.

Therefore, $f(x)$ is discontinuous at $x = 3$.