1. **State the problem:** Determine the value of $k$ such that the piecewise function $$f(x) = \begin{cases} kx + 3, & x < 1 \\ 8 - x^2, & x \geq 1 \end{cases}$$ is continuous at $x = 1$. 2. **Recall the conditions for continuity at $x = a$:** 1. $f(a)$ is defined. 2. $\lim_{x \to a} f(x)$ exists. 3. $\lim_{x \to a} f(x) = f(a)$. 3. **Apply these conditions at $x = 1$:** - Calculate $f(1)$ from the second piece since $x \geq 1$: $$f(1) = 8 - 1^2 = 7$$ - Calculate the left-hand limit as $x \to 1^-$: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (kx + 3) = k(1) + 3 = k + 3$$ - Calculate the right-hand limit as $x \to 1^+$: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (8 - x^2) = 8 - 1 = 7$$ 4. **Set the left-hand limit equal to $f(1)$ for continuity:** $$k + 3 = 7$$ 5. **Solve for $k$:** $$k = 7 - 3 = 4$$ **Final answer:** $$\boxed{4}$$