Find all points of discontinuity of $f,$ where $f$ is defined by $f(x) = \begin{cases} x^3 - 3, & \text{if } x \le 2 \\ x^2 + 1, & \text{if } x > 2 \end{cases}$

(NONE) The given function $f$ is $f(x) = \begin{cases} x^3 - 3, & \text{if } x \le 2 \\ x^2 + 1, & \text{if } x > 2 \end{cases}$
The function $f$ is defined for all real numbers.
Case $I$: If $c < 2,$ then $f(c) = c^3 - 3.$
$\lim_{x \to c} f(x) = \lim_{x \to c} (x^3 - 3) = c^3 - 3 = f(c).$
Thus,$f$ is continuous for all $x < 2.$
Case $II$: If $c = 2,$ then $f(2) = 2^3 - 3 = 5.$
Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^3 - 3) = 2^3 - 3 = 5.$
Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 + 1) = 2^2 + 1 = 5.$
Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 5,$ the function is continuous at $x = 2.$
Case $III$: If $c > 2,$ then $f(c) = c^2 + 1.$
$\lim_{x \to c} f(x) = \lim_{x \to c} (x^2 + 1) = c^2 + 1 = f(c).$
Thus,$f$ is continuous for all $x > 2.$
Conclusion: The function $f$ is continuous at every point on the real line. Therefore,there are no points of discontinuity.