1. **Problem Statement:** Find the turning (stationary) points of the curve $y = x^3 - 3x + 5$ and determine their nature. 2. **Step 1: Find the first derivative $\frac{dy}{dx}$:** $$\frac{dy}{dx} = 3x^2 - 3$$ 3. **Step 2: Set $\frac{dy}{dx} = 0$ to find stationary points:** $$3x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1$$ 4. **Step 3: Find the corresponding $y$ values by substituting $x$ into $y = x^3 - 3x + 5$:** - For $x = 1$: $$y = 1^3 - 3(1) + 5 = 1 - 3 + 5 = 3$$ - For $x = -1$: $$y = (-1)^3 - 3(-1) + 5 = -1 + 3 + 5 = 7$$ So the stationary points are $(1, 3)$ and $(-1, 7)$. 5. **Step 4: Find the second derivative $\frac{d^2y}{dx^2}$:** $$\frac{d^2y}{dx^2} = 6x$$ 6. **Step 5: Determine the nature of each stationary point by substituting $x$ into $\frac{d^2y}{dx^2}$:** - At $x = 1$: $$\frac{d^2y}{dx^2} = 6(1) = 6 > 0$$ This means $(1, 3)$ is a minimum turning point. - At $x = -1$: $$\frac{d^2y}{dx^2} = 6(-1) = -6 < 0$$ This means $(-1, 7)$ is a maximum turning point. --- 7. **Problem Statement:** Use the Newton-Raphson method to solve $x^4 - 3x^3 + 7x - 12 = 0$ correct to 3 decimal places. 8. **Step 1: Define the function and its derivative:** $$f(x) = x^4 - 3x^3 + 7x - 12$$ $$f'(x) = 4x^3 - 9x^2 + 7$$ 9. **Step 2: Choose an initial guess $x_0$.** By inspection or trial, let's choose $x_0 = 2$. 10. **Step 3: Apply the Newton-Raphson iteration formula:** $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ 11. **Step 4: Perform iterations until desired accuracy:** - Iteration 1: $$f(2) = 16 - 24 + 14 - 12 = -6$$ $$f'(2) = 32 - 36 + 7 = 3$$ $$x_1 = 2 - \frac{-6}{3} = 2 + 2 = 4$$ - Iteration 2: $$f(4) = 256 - 192 + 28 - 12 = 80$$ $$f'(4) = 256 - 144 + 7 = 119$$ $$x_2 = 4 - \frac{80}{119} \approx 4 - 0.672 = 3.328$$ - Iteration 3: $$f(3.328) \approx 122.7 - 110.5 + 23.3 - 12 = 23.5$$ $$f'(3.328) \approx 147.3 - 99.7 + 7 = 54.6$$ $$x_3 = 3.328 - \frac{23.5}{54.6} \approx 3.328 - 0.430 = 2.898$$ - Iteration 4: $$f(2.898) \approx 70.5 - 65.0 + 20.3 - 12 = 13.8$$ $$f'(2.898) \approx 97.0 - 75.5 + 7 = 28.5$$ $$x_4 = 2.898 - \frac{13.8}{28.5} \approx 2.898 - 0.484 = 2.414$$ - Iteration 5: $$f(2.414) \approx 33.9 - 42.3 + 16.9 - 12 = -3.5$$ $$f'(2.414) \approx 56.3 - 52.6 + 7 = 10.7$$ $$x_5 = 2.414 - \frac{-3.5}{10.7} \approx 2.414 + 0.327 = 2.741$$ - Iteration 6: $$f(2.741) \approx 56.3 - 61.7 + 19.2 - 12 = 1.8$$ $$f'(2.741) \approx 82.3 - 67.5 + 7 = 21.8$$ $$x_6 = 2.741 - \frac{1.8}{21.8} \approx 2.741 - 0.083 = 2.658$$ - Iteration 7: $$f(2.658) \approx 47.3 - 56.3 + 18.6 - 12 = 0.6$$ $$f'(2.658) \approx 75.0 - 63.3 + 7 = 18.7$$ $$x_7 = 2.658 - \frac{0.6}{18.7} \approx 2.658 - 0.032 = 2.626$$ - Iteration 8: $$f(2.626) \approx 44.3 - 54.0 + 18.4 - 12 = 0.1$$ $$f'(2.626) \approx 71.9 - 61.5 + 7 = 17.4$$ $$x_8 = 2.626 - \frac{0.1}{17.4} \approx 2.626 - 0.006 = 2.620$$ - Iteration 9: $$f(2.620) \approx 43.7 - 53.6 + 18.3 - 12 = 0.01$$ $$f'(2.620) \approx 71.3 - 61.2 + 7 = 17.1$$ $$x_9 = 2.620 - \frac{0.01}{17.1} \approx 2.620 - 0.001 = 2.619$$ 12. **Final answer:** The root correct to 3 decimal places is approximately $x = 2.619$.