1. **Problem 1: Solve $f(x) = x^4 - 4x - 9 = 0$ using Newton-Raphson and Secant Methods with given initial guesses.** 2. **Newton-Raphson Method:** - Formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ - Derivative: $$f'(x) = 4x^3 - 4$$ - Initial guess: $x_0 = 2$ 3. **Iterations for Newton-Raphson:** - $x_0 = 2$ - Calculate $f(2) = 2^4 - 4(2) - 9 = 16 - 8 - 9 = -1$ - Calculate $f'(2) = 4(2)^3 - 4 = 4(8) - 4 = 32 - 4 = 28$ - Update: $$x_1 = 2 - \frac{-1}{28} = 2 + \frac{1}{28} = 2 + 0.0357 = 2.0357$$ 4. Next iteration: - $f(2.0357) = (2.0357)^4 - 4(2.0357) - 9 \approx 17.15 - 8.14 - 9 = 0.01$ - $f'(2.0357) = 4(2.0357)^3 - 4 \approx 4(8.43) - 4 = 33.72 - 4 = 29.72$ - Update: $$x_2 = 2.0357 - \frac{0.01}{29.72} = 2.0357 - 0.00034 = 2.03536$$ 5. Next iteration: - $f(2.03536) \approx 0.00001$ - $f'(2.03536) \approx 29.7$ - Update: $$x_3 = 2.03536 - \frac{0.00001}{29.7} \approx 2.03536$$ - Converged to 3 decimal places: $x \approx 2.035$ 6. **Secant Method:** - Formula: $$x_{n+1} = x_n - f(x_n) \times \frac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})}$$ - Initial guesses: $x_0 = 2$, $x_1 = 3$ 7. Calculate: - $f(2) = -1$ - $f(3) = 3^4 - 4(3) - 9 = 81 - 12 - 9 = 60$ - Update: $$x_2 = 3 - 60 \times \frac{3 - 2}{60 - (-1)} = 3 - 60 \times \frac{1}{61} = 3 - 0.9836 = 2.0164$$ 8. Next iteration: - $f(2.0164) \approx -0.27$ - $f(3) = 60$ - Update: $$x_3 = 2.0164 - (-0.27) \times \frac{2.0164 - 3}{-0.27 - 60} = 2.0164 + 0.27 \times \frac{-0.9836}{-60.27} = 2.0164 + 0.0044 = 2.0208$$ 9. Next iteration: - $f(2.0208) \approx -0.05$ - $f(2.0164) \approx -0.27$ - Update: $$x_4 = 2.0208 - (-0.05) \times \frac{2.0208 - 2.0164}{-0.05 - (-0.27)} = 2.0208 + 0.05 \times \frac{0.0044}{0.22} = 2.0208 + 0.001 = 2.0218$$ 10. Continue until convergence to 3 decimals: $x \approx 2.036$ --- 11. **Problem 2: Solve system using Jacobi Method:** $$\begin{cases} 5x - y + z = 10 \\ 2x + 8y - z = 12 \\ x - y + 4z = 6 \end{cases}$$ 12. Rearrange for Jacobi iteration: $$x = \frac{10 + y - z}{5}$$ $$y = \frac{12 - 2x + z}{8}$$ $$z = \frac{6 - x + y}{4}$$ 13. Initial guess: $x^{(0)}=0$, $y^{(0)}=0$, $z^{(0)}=0$ 14. Iteration 1: - $x^{(1)} = \frac{10 + 0 - 0}{5} = 2$ - $y^{(1)} = \frac{12 - 2(0) + 0}{8} = 1.5$ - $z^{(1)} = \frac{6 - 0 + 0}{4} = 1.5$ 15. Iteration 2: - $x^{(2)} = \frac{10 + 1.5 - 1.5}{5} = 2$ - $y^{(2)} = \frac{12 - 2(2) + 1.5}{8} = \frac{12 - 4 + 1.5}{8} = 1.1875$ - $z^{(2)} = \frac{6 - 2 + 1.5}{4} = 1.375$ 16. Iteration 3: - $x^{(3)} = \frac{10 + 1.1875 - 1.375}{5} = 1.9625$ - $y^{(3)} = \frac{12 - 2(2) + 1.375}{8} = 1.1719$ - $z^{(3)} = \frac{6 - 1.9625 + 1.1719}{4} = 1.3023$ 17. Iteration 4: - $x^{(4)} = \frac{10 + 1.1719 - 1.3023}{5} = 1.9747$ - $y^{(4)} = \frac{12 - 2(1.9625) + 1.3023}{8} = 1.1762$ - $z^{(4)} = \frac{6 - 1.9747 + 1.1762}{4} = 1.3004$ 18. Continue iterations until values converge to 3 decimal places: - Final approximate solution: $x = 1.975$, $y = 1.176$, $z = 1.300$