1. **Problem 1: Continuity at $x=3$ for the function** $$f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & x \neq 3 \\ 7, & x = 3 \end{cases}$$ Step 1: Identify if the function is defined at $x=3$. Here, $f(3) = 7$ is given. Step 2: Simplify the expression for $x \neq 3$: $$\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3 \quad \text{for} \quad x \neq 3$$ Step 3: Find the limit as $x$ approaches 3: $$\lim_{x \to 3} f(x) = \lim_{x \to 3} (x + 3) = 3 + 3 = 6$$ Step 4: Compare the limit with $f(3)$: Since $\lim_{x \to 3} f(x) = 6 \ne f(3) = 7$, the function is **not continuous** at $x = 3$. --- 2. **Problem 2: Derivative of** $$y = (5x^2 - 1)(6x^3 + x)$$ Step 1: Identify the two functions for the product rule: Let $u = 5x^2 - 1$ and $v = 6x^3 + x$. Step 2: Find derivatives: $$u' = \frac{d}{dx}(5x^2 - 1) = 10x$$ $$v' = \frac{d}{dx}(6x^3 + x) = 18x^2 + 1$$ Step 3: Apply the product rule: $$\frac{dy}{dx} = u'v + uv' = (10x)(6x^3 + x) + (5x^2 - 1)(18x^2 + 1)$$ Step 4: Expand terms: $$(10x)(6x^3 + x) = 60x^4 + 10x^2$$ $$(5x^2 - 1)(18x^2 + 1) = 5x^2 \cdot 18x^2 + 5x^2 \cdot 1 - 1 \cdot 18x^2 - 1 \cdot 1 = 90x^4 + 5x^2 - 18x^2 - 1 = 90x^4 - 13x^2 - 1$$ Step 5: Combine all terms: $$\frac{dy}{dx} = 60x^4 + 10x^2 + 90x^4 - 13x^2 - 1 = (60x^4 + 90x^4) + (10x^2 - 13x^2) - 1 = 150x^4 - 3x^2 - 1$$ **Final answers:** - The function is not continuous at $x=3$. - The derivative is $$\boxed{150x^4 - 3x^2 - 1}$$.