Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} x + 1, & \text{if } x \ge 1 \\ x^2 + 1, & \text{if } x < 1 \end{cases}$. Is $f$ a continuous function?

(NONE) The given function is $f(x) = \begin{cases} x + 1, & \text{if } x \ge 1 \\ x^2 + 1, & \text{if } x < 1 \end{cases}$.
The function $f$ is defined for all real numbers.
Case $I$: If $c < 1$,then $f(c) = c^2 + 1$. The limit is $\lim_{x \to c} f(x) = \lim_{x \to c} (x^2 + 1) = c^2 + 1$. Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous for all $x < 1$.
Case $II$: If $c = 1$,then $f(1) = 1 + 1 = 2$.
The left-hand limit is $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 1) = 1^2 + 1 = 2$.
The right-hand limit is $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + 1) = 1 + 1 = 2$.
Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 2$,$f$ is continuous at $x = 1$.
Case $III$: If $c > 1$,then $f(c) = c + 1$. The limit is $\lim_{x \to c} f(x) = \lim_{x \to c} (x + 1) = c + 1$. Since $\lim_{x \to c} f(x) = f(c)$,$f$ is continuous for all $x > 1$.
Conclusion: The function $f$ has no points of discontinuity and is a continuous function.