1. **Problem 3.1:** Graph the circle given by the equation $$x^2 + y^2 = 7$$ 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 all $$y$$ values for which $$x$$ is real. 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 differentiating: $$y' = 3x^2 - 8x - 21$$. 7. Solve $$3x^2 - 8x - 21 = 0$$ using quadratic formula: $$x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-21)}}{2 \cdot 3} = \frac{8 \pm \sqrt{64 + 252}}{6} = \frac{8 \pm \sqrt{316}}{6}$$. 8. Approximate $$\sqrt{316} \approx 17.78$$, so critical points at $$x \approx \frac{8 + 17.78}{6} = 4.63$$ and $$x \approx \frac{8 - 17.78}{6} = -1.63$$. 9. Evaluate $$y$$ at critical points: - At $$x=4.63$$, $$y \approx 4.63^3 - 4(4.63)^2 - 21(4.63) \approx 99.3 - 85.7 - 97.2 = -83.6$$. - At $$x=-1.63$$, $$y \approx (-1.63)^3 - 4(-1.63)^2 - 21(-1.63) \approx -4.3 - 10.6 + 34.2 = 19.3$$. 10. Sketch the cubic curve showing these turning points and label axes. --- 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 logarithm base $$\frac{1}{8}$$, which is the inverse of $$f(x)$$. 14. Calculate some points: - For $$f(x)$$: $$f(0) = 1$$, $$f(1) = \frac{1}{8}$$, $$f(-1) = 8$$. - For $$g(x)$$: $$g(1) = 0$$, $$g(\frac{1}{8}) = 1$$, $$g(8) = -1$$. 15. Plot both curves on the same axes, label points and axes. --- 16. **Problem 4.1:** Find the 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: $$f(x+h) = 3(x+h)^2 - 5(x+h) - 9 = 3(x^2 + 2xh + h^2) - 5x - 5h - 9 = 3x^2 + 6xh + 3h^2 - 5x - 5h - 9$$. 19. Then: $$\frac{f(x+h) - f(x)}{h} = \frac{3x^2 + 6xh + 3h^2 - 5x - 5h - 9 - (3x^2 - 5x - 9)}{h} = \frac{6xh + 3h^2 - 5h}{h} = 6x + 3h - 5$$. 20. Taking limit as $$h \to 0$$: $$f'(x) = 6x - 5$$. --- 21. **Problem 4.2:** Find equation of tangent line to $$f(x) = 5x^2 + 4x - 1$$ at $$x=3$$. 22. Derivative: $$f'(x) = 10x + 4$$. 23. Slope at $$x=3$$: $$m = f'(3) = 10(3) + 4 = 34$$. 24. Point on curve: $$y = 5(3)^2 + 4(3) - 1 = 45 + 12 - 1 = 56$$. 25. Equation of tangent line: $$y - 56 = 34(x - 3) \implies y = 34x - 102 + 56 = 34x - 46$$. --- 26. **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. 27. Marginal cost is derivative of total cost with respect to $$Q$$: $$MC = \frac{dTC}{dQ} = \frac{3}{8}Q^2 + 14Q - 9$$. 28. Since $$MC$$ depends on $$Q$$, it varies with output. --- 29. **Problem 4.4:** Find $$\frac{dy}{dx}$$ for $$y = \frac{\frac{7}{x} - 5x + 3x^4}{\sqrt{x}}$$. 30. Rewrite $$y$$ as: $$y = \left(7x^{-1} - 5x + 3x^4\right) x^{-\frac{1}{2}} = 7x^{-\frac{3}{2}} - 5x^{\frac{1}{2}} + 3x^{\frac{7}{2}}$$. 31. Differentiate term-by-term: $$\frac{dy}{dx} = 7 \cdot \left(-\frac{3}{2}\right) x^{-\frac{5}{2}} - 5 \cdot \frac{1}{2} x^{-\frac{1}{2}} + 3 \cdot \frac{7}{2} x^{\frac{5}{2}} = -\frac{21}{2} x^{-\frac{5}{2}} - \frac{5}{2} x^{-\frac{1}{2}} + \frac{21}{2} x^{\frac{5}{2}}$$. **Final answers:** - 3.1 Domain and range: $$[-\sqrt{7}, \sqrt{7}]$$. - 3.2 Critical points at $$x \approx -1.63, 4.63$$ with corresponding $$y$$ values. - 3.3 Functions plotted with points as above. - 4.1 $$f'(x) = 6x - 5$$. - 4.2 Tangent line: $$y = 34x - 46$$. - 4.3 Marginal cost: $$MC = \frac{3}{8}Q^2 + 14Q - 9$$ varies with $$Q$$. - 4.4 $$\frac{dy}{dx} = -\frac{21}{2} x^{-\frac{5}{2}} - \frac{5}{2} x^{-\frac{1}{2}} + \frac{21}{2} x^{\frac{5}{2}}$$.