1. **Problem 3a:** Solve the inequality $$\frac{x^2-9}{x+3} < \frac{2x}{x-1}$$ algebraically. 2. **Rewrite the inequality:** Note that $$x^2-9 = (x-3)(x+3)$$, so $$\frac{(x-3)(x+3)}{x+3} < \frac{2x}{x-1}$$. 3. **Simplify the left side:** For $$x \neq -3$$, $$\frac{(x-3)\cancel{(x+3)}}{\cancel{x+3}} = x-3$$. 4. **Rewrite inequality:** $$x-3 < \frac{2x}{x-1}$$ with restrictions $$x \neq -3$$ and $$x \neq 1$$ (denominator zeroes). 5. **Bring all terms to one side:** $$x-3 - \frac{2x}{x-1} < 0$$. 6. **Find common denominator:** $$\frac{(x-3)(x-1)}{x-1} - \frac{2x}{x-1} < 0$$ 7. **Expand numerator:** $$(x-3)(x-1) = x^2 - x - 3x + 3 = x^2 - 4x + 3$$ 8. **Combine numerator:** $$\frac{x^2 - 4x + 3 - 2x}{x-1} < 0 \implies \frac{x^2 - 6x + 3}{x-1} < 0$$ 9. **Find zeros of numerator:** Solve $$x^2 - 6x + 3 = 0$$ Use quadratic formula: $$x = \frac{6 \pm \sqrt{36 - 12}}{2} = \frac{6 \pm \sqrt{24}}{2} = 3 \pm \sqrt{6}$$ 10. **Critical points:** - Numerator zeros: $$x = 3 - \sqrt{6}$$ and $$x = 3 + \sqrt{6}$$ - Denominator zero: $$x = 1$$ - Also exclude $$x = -3$$ from original domain 11. **Intervals to test:** $$(-\infty, -3), (-3, 1), (1, 3-\sqrt{6}), (3-\sqrt{6}, 3+\sqrt{6}), (3+\sqrt{6}, \infty)$$ 12. **Test points:** - For $$(-\infty, -3)$$, test $$x = -4$$: $$\frac{(-4)^2 - 6(-4) + 3}{-4 - 1} = \frac{16 + 24 + 3}{-5} = \frac{43}{-5} < 0$$ (True) - For $$(-3, 1)$$, test $$x = 0$$: $$\frac{0 - 0 + 3}{0 - 1} = \frac{3}{-1} = -3 < 0$$ (True) - For $$(1, 3-\sqrt{6})$$, note $$3 - \sqrt{6} \approx 0.55$$ so this interval is empty (since 1 > 0.55), skip. - For $$(3-\sqrt{6}, 3+\sqrt{6})$$, test $$x = 3$$: $$\frac{9 - 18 + 3}{3 - 1} = \frac{-6}{2} = -3 < 0$$ (True) - For $$(3+\sqrt{6}, \infty)$$, test $$x = 5$$: $$\frac{25 - 30 + 3}{5 - 1} = \frac{-2}{4} = -0.5 < 0$$ (True) 13. **Check domain restrictions:** Exclude $$x = -3$$ and $$x = 1$$ where denominator is zero. 14. **Solution:** $$(-\infty, -3) \cup (-3, 1) \cup (3-\sqrt{6}, 3+\sqrt{6}) \cup (3+\sqrt{6}, \infty)$$ Since $$3 - \sqrt{6} < 1$$ is false, intervals overlap incorrectly; correct intervals are: - $$(-\infty, -3)$$ - $$(-3, 1)$$ - $$ (1, 3 - \sqrt{6})$$ is empty (since $$3 - \sqrt{6} \approx 0.55 < 1$$) - $$ (3 - \sqrt{6}, 3 + \sqrt{6})$$ - $$ (3 + \sqrt{6}, \infty)$$ But since $$3 - \sqrt{6} < 1$$, the interval $$(1, 3 - \sqrt{6})$$ is empty, so the intervals are: $$(-\infty, -3) \cup (-3, 1) \cup (3 - \sqrt{6}, 3 + \sqrt{6}) \cup (3 + \sqrt{6}, \infty)$$ 15. **Final answer for 3a:** $$x \in (-\infty, -3) \cup (-3, 1) \cup (3 - \sqrt{6}, 3 + \sqrt{6}) \cup (3 + \sqrt{6}, \infty)$$ --- 16. **Problem 3b:** Sketch $$f(x) = \frac{x^2 - 9}{x+3}$$ and $$g(x) = \frac{2x}{x-1}$$ on the same grid. 17. **Simplify $$f(x)$$:** $$f(x) = \frac{(x-3)(x+3)}{x+3} = x - 3, \quad x \neq -3$$ 18. **Asymptotes:** - $$f(x)$$ has a vertical asymptote at $$x = -3$$ (denominator zero). - $$g(x)$$ has a vertical asymptote at $$x = 1$$. 19. **Intercepts:** - $$f(x)$$ zero at $$x=3$$ (numerator zero). - $$g(x)$$ zero at $$x=0$$. 20. **Plot key points and asymptotes:** - Draw vertical dashed lines at $$x=-3$$ and $$x=1$$. - Plot $$f(x) = x-3$$ except undefined at $$x=-3$$. - Plot $$g(x) = \frac{2x}{x-1}$$. 21. **Shade regions where $$f(x) < g(x)$$:** Use test points from 3a intervals to determine where $$f(x) < g(x)$$. --- 22. **Problem 4:** Given $$\sin(x) = -\frac{5}{13}$$ and $$0 \leq x \leq \frac{3\pi}{2}$$, find exact $$\sin(2x)$$. 23. **Determine quadrant:** Since $$\sin(x) < 0$$ and $$x \leq \frac{3\pi}{2}$$, $$x$$ is in quadrant III. 24. **Find $$\cos(x)$$:** Use Pythagorean identity: $$\sin^2(x) + \cos^2(x) = 1$$ $$\cos^2(x) = 1 - \sin^2(x) = 1 - \left(-\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}$$ Since $$x$$ is in quadrant III, $$\cos(x) < 0$$, so $$\cos(x) = -\frac{12}{13}$$ 25. **Use double angle formula:** $$\sin(2x) = 2 \sin(x) \cos(x) = 2 \times \left(-\frac{5}{13}\right) \times \left(-\frac{12}{13}\right) = 2 \times \frac{60}{169} = \frac{120}{169}$$ 26. **Final answer for 4:** $$\sin(2x) = \frac{120}{169}$$