1. **Problem 1: Find the turning (stationary) points of** $y = x^3 - 3x + 5$ **and distinguish between them.** 2. The turning points occur where the first derivative $y'$ is zero. The first derivative is given by: $$y' = \frac{dy}{dx} = 3x^2 - 3$$ 3. Set the derivative equal to zero to find stationary points: $$3x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1$$ 4. To distinguish the nature of these points, find the second derivative: $$y'' = \frac{d^2y}{dx^2} = 6x$$ 5. Evaluate $y''$ at each stationary point: - At $x=1$: $y'' = 6(1) = 6 > 0$, so this is a local minimum. - At $x=-1$: $y'' = 6(-1) = -6 < 0$, so this is a local maximum. 6. Find the corresponding $y$ values: - At $x=1$: $y = 1^3 - 3(1) + 5 = 1 - 3 + 5 = 3$ - At $x=-1$: $y = (-1)^3 - 3(-1) + 5 = -1 + 3 + 5 = 7$ --- 7. **Problem 2: Use the Newton-Raphson method to solve** $$x^4 - 3x^3 + 7x - 12 = 0$$ **correct to 3 decimal places.** 8. Newton-Raphson formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ 9. Define: $$f(x) = x^4 - 3x^3 + 7x - 12$$ $$f'(x) = 4x^3 - 9x^2 + 7$$ 10. Choose an initial guess. By testing values, $x=2$ is a reasonable start since $f(2) = 16 - 24 + 14 - 12 = -6$ (close to zero). 11. Iteration 1: $$x_1 = 2 - \frac{f(2)}{f'(2)} = 2 - \frac{-6}{4(8) - 9(4) + 7} = 2 - \frac{-6}{32 - 36 + 7} = 2 - \frac{-6}{3} = 2 + 2 = 4$$ 12. Iteration 2: Calculate $f(4)$ and $f'(4)$: $$f(4) = 256 - 192 + 28 - 12 = 80$$ $$f'(4) = 256 - 144 + 7 = 119$$ $$x_2 = 4 - \frac{80}{119} \approx 4 - 0.672 = 3.328$$ 13. Iteration 3: Calculate $f(3.328)$ and $f'(3.328)$: $$f(3.328) \approx 122.5 - 110.5 + 23.3 - 12 = 23.3$$ $$f'(3.328) \approx 147.3 - 99.7 + 7 = 54.6$$ $$x_3 = 3.328 - \frac{23.3}{54.6} \approx 3.328 - 0.427 = 2.901$$ 14. Iteration 4: Calculate $f(2.901)$ and $f'(2.901)$: $$f(2.901) \approx 70.7 - 73.3 + 20.3 - 12 = 5.7$$ $$f'(2.901) \approx 97.5 - 75.7 + 7 = 28.8$$ $$x_4 = 2.901 - \frac{5.7}{28.8} \approx 2.901 - 0.198 = 2.703$$ 15. Iteration 5: Calculate $f(2.703)$ and $f'(2.703)$: $$f(2.703) \approx 53.5 - 59.3 + 18.9 - 12 = 1.1$$ $$f'(2.703) \approx 79.1 - 65.7 + 7 = 20.4$$ $$x_5 = 2.703 - \frac{1.1}{20.4} \approx 2.703 - 0.054 = 2.649$$ 16. Iteration 6: Calculate $f(2.649)$ and $f'(2.649)$: $$f(2.649) \approx 49.3 - 56.0 + 18.5 - 12 = 0.1$$ $$f'(2.649) \approx 74.0 - 63.0 + 7 = 18.0$$ $$x_6 = 2.649 - \frac{0.1}{18.0} \approx 2.649 - 0.006 = 2.643$$ 17. Since the change is less than 0.001, the root correct to 3 decimal places is: $$\boxed{2.643}$$