1. **Problem 9.1: Calculate the coordinates of E (turning point).** Given the function $$f(x) = x^3 - 8x^2 + 5x + 14$$ - Turning points occur where the first derivative $$f'(x)$$ equals zero. Calculate $$f'(x)$$: $$f'(x) = 3x^2 - 16x + 5$$ Set $$f'(x) = 0$$ to find critical points: $$3x^2 - 16x + 5 = 0$$ Use the quadratic formula: $$x = \frac{16 \pm \sqrt{(-16)^2 - 4 \times 3 \times 5}}{2 \times 3} = \frac{16 \pm \sqrt{256 - 60}}{6} = \frac{16 \pm \sqrt{196}}{6} = \frac{16 \pm 14}{6}$$ So, $$x_1 = \frac{16 + 14}{6} = 5$$ $$x_2 = \frac{16 - 14}{6} = \frac{2}{6} = \frac{1}{3}$$ Point E is the turning point with the smaller x-value (local minimum), so $$x = \frac{1}{3}$$. Calculate $$f\left(\frac{1}{3}\right)$$: $$f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - 8\left(\frac{1}{3}\right)^2 + 5\left(\frac{1}{3}\right) + 14 = \frac{1}{27} - 8 \times \frac{1}{9} + \frac{5}{3} + 14$$ Simplify: $$= \frac{1}{27} - \frac{8}{9} + \frac{5}{3} + 14 = \frac{1}{27} - \frac{24}{27} + \frac{45}{27} + \frac{378}{27} = \frac{1 - 24 + 45 + 378}{27} = \frac{400}{27} \approx 14.81$$ So, coordinates of E are $$\left(\frac{1}{3}, \frac{400}{27}\right)$$. 2. **Problem 9.2: For which values of x is $$f$$ concave down?** Concavity is determined by the second derivative $$f''(x)$$: $$f''(x) = 6x - 16$$ Concave down means $$f''(x) < 0$$: $$6x - 16 < 0 \implies 6x < 16 \implies x < \frac{16}{6} = \frac{8}{3}$$ So, $$f$$ is concave down for $$x < \frac{8}{3}$$. 3. **Problem 9.3: For which values of x is $$f(x) \cdot f''(x) < 0$$?** Given $$B = (2,0)$$ is an x-intercept. We want to find $$x$$ such that: $$f(x) \cdot f''(x) < 0$$ Recall: - $$f''(x) = 6x - 16$$ Analyze sign of $$f(x)$$ and $$f''(x)$$ around the roots and turning points. From the graph and given points: - At $$x=2$$, $$f(2) = 0$$. - The x-intercepts are at points A, B, C. Check intervals: - For $$x < 2$$, say at $$x=1$$: - $$f(1) = 1 - 8 + 5 + 14 = 12 > 0$$ - $$f''(1) = 6 - 16 = -10 < 0$$ - Product $$> 0 \times < 0 = < 0$$ (true) - For $$2 < x < \frac{8}{3}$$ (since $$f''(x) < 0$$ for $$x < \frac{8}{3}$$), say at $$x=2.5$$: - $$f(2.5)$$ is negative (from graph, since it dips below x-axis after B) - $$f''(2.5) = 15 - 16 = -1 < 0$$ - Product $$< 0 \times < 0 = > 0$$ (false) - For $$x > \frac{8}{3}$$, say at $$x=3$$: - $$f(3)$$ is positive (graph rises above x-axis after E) - $$f''(3) = 18 - 16 = 2 > 0$$ - Product $$> 0 \times > 0 = > 0$$ (false) Therefore, $$f(x) \cdot f''(x) < 0$$ for $$x < 2$$. 4. **Problem 9.4: For which values of $$t$$ will $$y = -11x + t$$ intersect $$f$$ at 3 distinct points?** Set: $$f(x) = -11x + t$$ Rewrite: $$x^3 - 8x^2 + 5x + 14 = -11x + t$$ Bring all terms to one side: $$x^3 - 8x^2 + 5x + 14 + 11x - t = 0$$ Simplify: $$x^3 - 8x^2 + 16x + (14 - t) = 0$$ This cubic equation will have 3 distinct real roots if its discriminant $$\Delta > 0$$. Calculate the discriminant of cubic $$x^3 + ax^2 + bx + c$$ where: - $$a = -8$$ - $$b = 16$$ - $$c = 14 - t$$ Discriminant formula: $$\Delta = 18abc - 4b^3 + b^2a^2 - 4a^3c - 27c^2$$ Calculate each term: - $$18abc = 18 \times (-8) \times 16 \times (14 - t) = -2304 (14 - t)$$ - $$-4b^3 = -4 \times 16^3 = -4 \times 4096 = -16384$$ - $$b^2a^2 = 16^2 \times (-8)^2 = 256 \times 64 = 16384$$ - $$-4a^3c = -4 \times (-8)^3 \times (14 - t) = -4 \times (-512) \times (14 - t) = 2048 (14 - t)$$ - $$-27c^2 = -27 (14 - t)^2$$ Sum: $$\Delta = -2304(14 - t) - 16384 + 16384 + 2048(14 - t) - 27(14 - t)^2$$ Simplify constants: $$-16384 + 16384 = 0$$ Combine terms: $$\Delta = (-2304 + 2048)(14 - t) - 27(14 - t)^2 = -256(14 - t) - 27(14 - t)^2$$ Let $$u = 14 - t$$: $$\Delta = -256u - 27u^2 = -u(256 + 27u)$$ For 3 distinct real roots, $$\Delta > 0$$: $$-u(256 + 27u) > 0$$ Analyze sign: - If $$u > 0$$, then $$256 + 27u > 256 > 0$$, so $$-u(positive) < 0$$ (false) - If $$u < 0$$, then $$256 + 27u$$ may be positive or negative. Find roots of $$256 + 27u = 0$$: $$27u = -256 \implies u = -\frac{256}{27} \approx -9.48$$ Test intervals: - For $$u < -9.48$$, $$256 + 27u < 0$$, so $$-u(negative) > 0$$ if $$-u > 0$$ (i.e., $$u < 0$$), so product positive. - For $$-9.48 < u < 0$$, $$256 + 27u > 0$$, so $$-u(positive) > 0$$ if $$-u > 0$$ (i.e., $$u < 0$$), so product negative. Therefore, $$\Delta > 0$$ when: $$u < -9.48$$ Recall $$u = 14 - t$$: $$14 - t < -9.48 \implies t > 23.48$$ So, the line intersects $$f$$ at 3 distinct points when $$t > 23.48$$. --- 5. **Problem 11.1.1: Show that $$e = 84$$ given male and preferring juice are independent.** From the table: - Total learners: 210 - Male preferring juice: 36 - Male total: $$f$$ - Female preferring juice: $$a$$ - Female total: $$c$$ - Total preferring juice: $$e$$ Independence means: $$P(\text{male} \cap \text{juice}) = P(\text{male}) \times P(\text{juice})$$ Calculate probabilities: $$P(\text{male} \cap \text{juice}) = \frac{36}{210}$$ $$P(\text{male}) = \frac{f}{210}$$ $$P(\text{juice}) = \frac{e}{210}$$ Set equality: $$\frac{36}{210} = \frac{f}{210} \times \frac{e}{210}$$ Multiply both sides by $$210^2$$: $$36 \times 210 = f \times e$$ From total rows and columns: $$f = 36 + 54 = 90$$ Substitute $$f = 90$$: $$36 \times 210 = 90 \times e \implies e = \frac{36 \times 210}{90} = \frac{7560}{90} = 84$$ Hence, $$e = 84$$. 6. **Problem 11.1.2: Calculate probability a female learner likes energy drinks.** From table: - Female total: $$c$$ - Female energy drinks: $$b$$ - Total learners: 210 We know: $$c = a + b$$ $$e = a + 36 = 84 \implies a = 84 - 36 = 48$$ Total juice: $$e = 84$$ Total energy drinks: $$d = 210 - 84 = 126$$ Male energy drinks: 54 So female energy drinks: $$b = d - 54 = 126 - 54 = 72$$ Female total: $$c = a + b = 48 + 72 = 120$$ Probability female likes energy drinks: $$P = \frac{b}{c} = \frac{72}{120} = \frac{3}{5} = 0.6$$ 7. **Problem 11.2: Calculate number of cups of coffee sold on a non-rainy day.** Given: - Total people: 120 - Probability of rain: 0.75 - Probability of coffee on rainy day: 3 times probability of coffee on non-rainy day - Overall probability of coffee: $$\frac{7}{12}$$ Let: - $$p =$$ probability of coffee on non-rainy day - Then coffee on rainy day = $$3p$$ Use total probability: $$P(\text{coffee}) = P(\text{rain}) \times P(\text{coffee} | \text{rain}) + P(\text{no rain}) \times P(\text{coffee} | \text{no rain})$$ Substitute: $$\frac{7}{12} = 0.75 \times 3p + 0.25 \times p = 2.25p + 0.25p = 2.5p$$ Solve for $$p$$: $$p = \frac{7}{12} \div 2.5 = \frac{7}{12} \times \frac{1}{2.5} = \frac{7}{12} \times \frac{2}{5} = \frac{14}{60} = \frac{7}{30}$$ Number of people on non-rainy day: $$120 \times 0.25 = 30$$ Number of coffee cups sold on non-rainy day: $$30 \times \frac{7}{30} = 7$$ 8. **Problem 11.3.1: Number of ways 8 runners finish with Bongi immediately after Andrew.** Treat Andrew and Bongi as a single block with order fixed (Andrew then Bongi). Number of entities to arrange: $$7$$ (the block + 6 other runners) Number of permutations: $$7! = 5040$$ 9. **Problem 11.3.2: Probability that two or more runners finish after Andrew and before Bongi.** Positions of Andrew (A) and Bongi (B) must be such that B is immediately after A for 11.3.1, but here we consider any order. Total permutations of 8 runners: $$8! = 40320$$ Number of permutations where 0 or 1 runner is between A and B: - 0 runners between means A and B are adjacent: $$2 \times 7! = 2 \times 5040 = 10080$$ (A before B or B before A) - 1 runner between A and B: Choose 1 runner from 6 others: $$6$$ Positions of A and B with 1 runner between: 2 ways (A _ B or B _ A) Arrange remaining 5 runners: $$5! = 120$$ Total: $$6 \times 2 \times 120 = 1440$$ Sum for 0 or 1 runner between: $$10080 + 1440 = 11520$$ Number of permutations with 2 or more runners between A and B: $$40320 - 11520 = 28800$$ Probability: $$\frac{28800}{40320} = \frac{60}{84} = \frac{5}{7} \approx 0.714$$ --- Final answers summary: - 9.1: $$E = \left(\frac{1}{3}, \frac{400}{27}\right)$$ - 9.2: $$f$$ concave down for $$x < \frac{8}{3}$$ - 9.3: $$f(x) \cdot f''(x) < 0$$ for $$x < 2$$ - 9.4: 3 distinct intersections when $$t > 23.48$$ - 11.1.1: $$e = 84$$ - 11.1.2: Probability female likes energy drinks = 0.6 - 11.2: Coffee cups sold on non-rainy day = 7 - 11.3.1: Number of ways with Bongi immediately after Andrew = 5040 - 11.3.2: Probability two or more runners finish between Andrew and Bongi = $$\frac{5}{7}$$