Hint:

To identify points of discontinuity, check the limits of $f(x)$ as $x$ approaches the boundaries where the function's definition changes. In this case, consider the limit as $x$ approaches 1 from both the left and the right.

Step-by-Step Solution:

We will analyze the function f at each of the two pieces that define it: x ≥ 1 and x < 1.

Case 1: x ≥ 1

At any point x ≥ 1, f(x) = x + 1. To check if f is continuous at x, we need to check if the following three conditions hold:

f(x) exists. Yes, f(x) = x + 1 exists for all x ≥ 1.

$lim_{x \to c} f(x)$ exists. Let's find the one-sided limits of f at x = c:

$lim_{x \to c^{-}} f(x) = lim_{x \to c^{-}} (x + 1) = c + 1$

$lim_{x \to c^{+}} f(x) = lim_{x \to c^{+}} (x + 1) = c + 1$

Since the one-sided limits agree, $lim_{x \to c} f(x)$ exists and is equal to $c +1$.

$lim_{x \to c} f(x)=f(c)$. Since $lim_{x \to c} f(x) = c+1$ and $f(c) = c + 1$, this condition is met.

Therefore, f is continuous at all points x ≥ 1.

Case 2: x<1

At any point x < 1, $f(x) = x^2 + 1$. To check if f is continuous at x, we need to check if the following three conditions hold:

f(x) exists. Yes, $f(x) = x^2 + 1$ exists for all x < 1.

$lim_{x \to c} f(x)$ exists. Let's find the one-sided limits of f at x = c:

$lim_{x \to c^{-}}f(x)= lim_{x \to c^{-}}(x^2 +1)= c^2+1$

$lim_{x \to c^{+}}f(x)= lim_{x \to c^{+}}(x^2 +1)= c^2+1$

Since the one-sided limits agree, $lim_{x \to c} f(x)$ exists and is equal to $c^2 +1$.

$lim_{x \to c} f(x) = f(c)$. Since $lim_{x \to c} f(x) = c^2 +1$ and $f(c) = c^2 +1$, this condition is met.

Therefore, f is continuous at all points x < 1.

Note:

This function seamlessly transitions between $x^2 + 1$ and $x + 1$ at $x = 1$. It's a demonstration of how piecewise functions can be crafted to be continuous across their entire domain.