1. **Problem 3.1:** Graph the circle given by $$x^2 + y^2 = 7$$ and determine its domain and range. 2. The equation represents a circle centered at the origin with radius $$r = \sqrt{7}$$. 3. The domain is all $$x$$ values for which $$y$$ is real. Since $$y^2 = 7 - x^2$$, $$y$$ is real if $$7 - x^2 \geq 0$$, so $$-\sqrt{7} \leq x \leq \sqrt{7}$$. 4. The range is similarly $$-\sqrt{7} \leq y \leq \sqrt{7}$$. --- 5. **Problem 3.2:** Sketch the graph of $$y = x^3 - 4x^2 - 21x$$. 6. Find critical points by computing the derivative: $$y' = 3x^2 - 8x - 21$$. 7. Solve $$3x^2 - 8x - 21 = 0$$ using quadratic formula: $$x = \frac{8 \pm \sqrt{64 + 252}}{6} = \frac{8 \pm \sqrt{316}}{6}$$. 8. Approximate $$\sqrt{316} \approx 17.78$$, so critical points at $$x_1 \approx \frac{8 - 17.78}{6} = -1.63$$ and $$x_2 \approx \frac{8 + 17.78}{6} = 4.63$$. 9. Evaluate $$y$$ at critical points: $$y(-1.63) \approx (-1.63)^3 - 4(-1.63)^2 - 21(-1.63) \approx -4.32 - 10.62 + 34.23 = 19.29$$ $$y(4.63) \approx 4.63^3 - 4(4.63)^2 - 21(4.63) \approx 99.3 - 85.7 - 97.2 = -83.6$$ 10. These points are local max and min respectively. --- 11. **Problem 3.3:** Plot $$f(x) = (\frac{1}{8})^x$$ and $$g(x) = \log_{\frac{1}{8}} x$$ on the same axes. 12. Note that $$f(x)$$ is an exponential decay function since base $$\frac{1}{8} < 1$$. 13. The function $$g(x)$$ is the inverse of $$f(x)$$, so its domain is $$x > 0$$ and range is all real numbers. 14. Key points for $$f(x)$$: $$f(0) = 1$$, $$f(1) = \frac{1}{8}$$, $$f(-1) = 8$$. 15. Key points for $$g(x)$$: $$g(1) = 0$$, $$g(\frac{1}{8}) = 1$$, $$g(8) = -1$$. --- 16. **Problem 4.1:** Find derivative of $$f(x) = 3x^2 - 5x - 9$$ from first principles. 17. Use definition: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 18. Compute numerator: $$f(x+h) = 3(x+h)^2 - 5(x+h) - 9 = 3(x^2 + 2xh + h^2) - 5x - 5h - 9 = 3x^2 + 6xh + 3h^2 - 5x - 5h - 9$$ 19. Subtract $$f(x)$$: $$f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 5x - 5h - 9) - (3x^2 - 5x - 9) = 6xh + 3h^2 - 5h$$ 20. Divide by $$h$$: $$\frac{6xh + 3h^2 - 5h}{h} = 6x + 3h - 5$$ 21. Take limit as $$h \to 0$$: $$f'(x) = 6x - 5$$ --- 22. **Problem 4.2:** Find tangent line to $$f(x) = 5x^2 + 4x - 1$$ at $$x=3$$. 23. Compute derivative: $$f'(x) = 10x + 4$$ 24. Slope at $$x=3$$: $$m = f'(3) = 10(3) + 4 = 34$$ 25. Find point on curve: $$y = 5(3)^2 + 4(3) - 1 = 45 + 12 - 1 = 56$$ 26. Equation of tangent line: $$y - 56 = 34(x - 3)$$ or $$y = 34x - 102 + 56 = 34x - 46$$ --- 27. **Problem 4.3:** Given total cost $$TC = \frac{1}{8}Q^3 + 7Q^2 - 9Q + 12$$, find marginal cost and check if it varies with output. 28. Marginal cost is derivative of total cost with respect to $$Q$$: $$MC = \frac{dTC}{dQ} = \frac{3}{8}Q^2 + 14Q - 9$$ 29. Since $$MC$$ depends on $$Q$$, it varies with output. --- 30. **Problem 4.4:** Find $$\frac{dy}{dx}$$ for $$y = \frac{\frac{7}{x} - 5x + 3x^4}{\sqrt{x}}$$. 31. Rewrite $$y$$ as: $$y = \left(7x^{-1} - 5x + 3x^4\right) x^{-\frac{1}{2}} = 7x^{-\frac{3}{2}} - 5x^{\frac{1}{2}} + 3x^{\frac{7}{2}}$$ 32. Differentiate term-by-term: $$\frac{dy}{dx} = 7 \cdot \left(-\frac{3}{2}\right) x^{-\frac{5}{2}} - 5 \cdot \frac{1}{2} x^{-\frac{1}{2}} + 3 \cdot \frac{7}{2} x^{\frac{5}{2}}$$ 33. Simplify: $$\frac{dy}{dx} = -\frac{21}{2} x^{-\frac{5}{2}} - \frac{5}{2} x^{-\frac{1}{2}} + \frac{21}{2} x^{\frac{5}{2}}$$