1. **Problem 1: Determine continuity of** $$f(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & x \neq 3 \\ 7 & x = 3 \end{cases}$$ at $$x=3$$. 2. To check continuity at $$x=3$$, evaluate: $$\lim_{x \to 3} f(x)$$ and compare with $$f(3)$$. 3. Factor numerator for $$x \neq 3$$: $$x^2 - 9 = (x - 3)(x + 3)$$. 4. Simplify for $$x \neq 3$$: $$\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x + 3$$. 5. Find $$\lim_{x \to 3} f(x) = \lim_{x \to 3} (x+3) = 6$$. 6. Given $$f(3) = 7$$, since $$\lim_{x \to 3} f(x) \neq f(3)$$, **function is not continuous at** $$x=3$$. --- 7. **Problem 2: Find derivative $$\frac{dy}{dx}$$ for** $$y = (5x^2 - 1)(6x^3 + x)$$. 8. Use product rule: $$\frac{d}{dx}[uv] = u'v + uv'$$. 9. Define $$u = 5x^2 - 1$$, $$v = 6x^3 + x$$. 10. Compute derivatives: - $$u' = \frac{d}{dx}(5x^2 - 1) = 10x$$ - $$v' = \frac{d}{dx}(6x^3 + x) = 18x^2 + 1$$. 11. Apply product rule: $$\frac{dy}{dx} = u'v + uv' = 10x(6x^3 + x) + (5x^2 - 1)(18x^2 + 1)$$. 12. Expand terms: $$10x(6x^3 + x) = 60x^4 + 10x^2$$ $$(5x^2 - 1)(18x^2 + 1) = 5x^2 \times 18x^2 + 5x^2 \times 1 - 1 \times 18x^2 - 1 \times 1 = 90x^4 + 5x^2 - 18x^2 -1$$ 13. Simplify: $$90x^4 + 5x^2 - 18x^2 - 1 = 90x^4 - 13x^2 -1$$ 14. Combine all terms: $$\frac{dy}{dx} = (60x^4 + 10x^2) + (90x^4 - 13x^2 -1) = 150x^4 - 3x^2 - 1$$. **Final answers:** - Function $$f$$ is **not continuous** at $$x=3$$. - Derivative $$\frac{dy}{dx} = 150x^4 - 3x^2 -1$$.