1. **State the problem:** We have a piecewise function $$f(x) = \begin{cases} 26 - cx^2 & \text{if } x < 3 \\ d & \text{if } x = 3 \\ cx + 2 & \text{if } x > 3 \end{cases}$$ We need to find values of $c$ and $d$ such that $f(x)$ is continuous at $x=3$. 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. **Calculate the left-hand limit as $x \to 3^-$:** $$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (26 - cx^2) = 26 - c \cdot 3^2 = 26 - 9c$$ 4. **Calculate the right-hand limit as $x \to 3^+$:** $$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (cx + 2) = c \cdot 3 + 2 = 3c + 2$$ 5. **Set the limits equal to $f(3) = d$ for continuity:** $$26 - 9c = d = 3c + 2$$ 6. **Solve the system:** From $26 - 9c = 3c + 2$, we get $$26 - 9c = 3c + 2$$ $$26 - 2 = 3c + 9c$$ $$24 = 12c$$ $$c = \frac{24}{12} = 2$$ 7. **Find $d$ using $d = 3c + 2$:** $$d = 3 \times 2 + 2 = 6 + 2 = 8$$ 8. **Find the limit at $x=3$:** $$\lim_{x \to 3} f(x) = d = 8$$ **Final answers:** $$c = 2, \quad d = 8, \quad \lim_{x \to 3} f(x) = 8$$