1. **State the problem:** Determine if the piecewise function $$f(x) = \begin{cases} x^3 + 1, & x \leq 1 \\ x + 1, & x > 1 \end{cases}$$ is continuous at the point $a = 1$. 2. **Recall the continuity condition:** A function $f$ is continuous at $x = a$ if $$\lim_{x \to a^-} f(x) = f(a) = \lim_{x \to a^+} f(x)$$ 3. **Evaluate $f(1)$:** Since $1 \leq 1$, use the first piece: $$f(1) = 1^3 + 1 = 1 + 1 = 2$$ 4. **Find the left-hand limit as $x \to 1^-$:** Use the first piece again: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^3 + 1) = 1^3 + 1 = 2$$ 5. **Find the right-hand limit as $x \to 1^+$:** Use the second piece: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + 1) = 1 + 1 = 2$$ 6. **Compare the values:** $$\lim_{x \to 1^-} f(x) = 2, \quad f(1) = 2, \quad \lim_{x \to 1^+} f(x) = 2$$ Since all three are equal, the function is continuous at $x = 1$. **Final answer:** The function is continuous at $a = 1$.