Question: by showing all the possible steps included in the solution of each Use stated methods to find solutions accurate to within $10^{-4}$ for the following problems. a. $x^3 + 3x^2 - 1 = 0$, $[-3, -2]$ (Newton's method) b. $x - 0.8 - 0.2 \sin x = 0$, $[0, \frac{\pi}{2}]$ (Secant method)
1. **Problem a:** Solve $x^3 + 3x^2 - 1 = 0$ on $[-3, -2]$ using Newton's method accurate to within $10^{-4}$.
2. **Newton's Method Formula:**
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
where $f(x) = x^3 + 3x^2 - 1$ and $f'(x) = 3x^2 + 6x$.
3. **Initial guess:** Choose $x_0 = -3$ (start of interval).
4. **Iteration 1:**
Calculate $f(-3) = (-3)^3 + 3(-3)^2 - 1 = -27 + 27 - 1 = -1$
Calculate $f'(-3) = 3(-3)^2 + 6(-3) = 27 - 18 = 9$
Apply Newton's formula:
$$x_1 = -3 - \frac{-1}{9} = -3 + \frac{1}{9} = -2.8889$$
5. **Iteration 2:**
Calculate $f(-2.8889) = (-2.8889)^3 + 3(-2.8889)^2 - 1 \approx -24.1 + 25.0 - 1 = -0.1$
Calculate $f'(-2.8889) = 3(-2.8889)^2 + 6(-2.8889) \approx 25.0 - 17.3 = 7.7$
$$x_2 = -2.8889 - \frac{-0.1}{7.7} = -2.8889 + 0.0130 = -2.8759$$
6. **Iteration 3:**
Calculate $f(-2.8759) \approx -0.002$
Calculate $f'(-2.8759) \approx 7.5$
$$x_3 = -2.8759 - \frac{-0.002}{7.5} = -2.8759 + 0.00027 = -2.8756$$
7. **Iteration 4:**
Calculate $f(-2.8756) \approx -0.00001$
Calculate $f'(-2.8756) \approx 7.5$
$$x_4 = -2.8756 - \frac{-0.00001}{7.5} = -2.8756 + 0.0000013 = -2.8756$$
8. Since $|x_4 - x_3| < 10^{-4}$, stop. The root is approximately $-2.8756$.
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9. **Problem b:** Solve $x - 0.8 - 0.2 \sin x = 0$ on $[0, \frac{\pi}{2}]$ using the Secant method accurate to within $10^{-4}$.
10. **Secant Method Formula:**
$$x_{n+1} = x_n - f(x_n) \cdot \frac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})}$$
where $f(x) = x - 0.8 - 0.2 \sin x$.
11. **Initial guesses:** $x_0 = 0$, $x_1 = \frac{\pi}{2} \approx 1.5708$.
12. Calculate $f(0) = 0 - 0.8 - 0 = -0.8$
Calculate $f(1.5708) = 1.5708 - 0.8 - 0.2 \times 1 = 1.5708 - 0.8 - 0.2 = 0.5708$
13. **Iteration 1:**
$$x_2 = 1.5708 - 0.5708 \times \frac{1.5708 - 0}{0.5708 - (-0.8)} = 1.5708 - 0.5708 \times \frac{1.5708}{1.3708} = 1.5708 - 0.6539 = 0.9169$$
14. Calculate $f(0.9169) = 0.9169 - 0.8 - 0.2 \sin(0.9169) \approx 0.9169 - 0.8 - 0.2 \times 0.793 = 0.9169 - 0.8 - 0.1586 = -0.0417$
15. **Iteration 2:**
$$x_3 = 0.9169 - (-0.0417) \times \frac{0.9169 - 1.5708}{-0.0417 - 0.5708} = 0.9169 + 0.0417 \times \frac{-0.6539}{-0.6125} = 0.9169 + 0.0445 = 0.9614$$
16. Calculate $f(0.9614) = 0.9614 - 0.8 - 0.2 \sin(0.9614) \approx 0.9614 - 0.8 - 0.2 \times 0.820 = 0.9614 - 0.8 - 0.164 = -0.0026$
17. **Iteration 3:**
$$x_4 = 0.9614 - (-0.0026) \times \frac{0.9614 - 0.9169}{-0.0026 - (-0.0417)} = 0.9614 + 0.0026 \times \frac{0.0445}{0.0391} = 0.9614 + 0.0030 = 0.9644$$
18. Calculate $f(0.9644) \approx -0.0001$
19. **Iteration 4:**
$$x_5 = 0.9644 - (-0.0001) \times \frac{0.9644 - 0.9614}{-0.0001 - (-0.0026)} = 0.9644 + 0.0001 \times \frac{0.003}{0.0025} = 0.9644 + 0.00012 = 0.9645$$
20. Since $|x_5 - x_4| < 10^{-4}$, stop. The root is approximately $0.9645$.
**Final answers:**
- a) Root $\, \approx -2.8756$
- b) Root $\, \approx 0.9645$