1. **State the problem:** Find the root of the equation $f(x) = x^3 - 5x - 1$ using the Bracketing method and Newton-Raphson method with error tolerance $\varepsilon_{step} = 0.001$. 2. **Bracketing Method (Bisection):** - Formula: If $f(a)$ and $f(b)$ have opposite signs, root lies in $[a,b]$. - Midpoint $c = \frac{a+b}{2}$. - Check sign of $f(c)$ and replace $a$ or $b$ accordingly. - Repeat until $|b - a| < \varepsilon_{step}$. 3. **Newton-Raphson Method:** - Formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ - Requires derivative $f'(x) = 3x^2 - 5$. - Iterate until $|x_{n+1} - x_n| < \varepsilon_{step}$. --- ### Bracketing Method Steps: 4. Choose initial interval where $f(a)f(b) < 0$. - Test $f(1) = 1 - 5 - 1 = -5$ (negative) - Test $f(2) = 8 - 10 - 1 = -3$ (negative) - Test $f(3) = 27 - 15 - 1 = 11$ (positive) So root lies in $[2,3]$. 5. Compute midpoint $c = \frac{2+3}{2} = 2.5$. - $f(2.5) = 15.625 - 12.5 - 1 = 2.125$ (positive) - Since $f(2)$ negative and $f(2.5)$ positive, new interval is $[2,2.5]$. 6. Next midpoint $c = \frac{2+2.5}{2} = 2.25$. - $f(2.25) = 11.39 - 11.25 - 1 = -0.86$ (negative) - Interval is $[2.25, 2.5]$. 7. Next midpoint $c = 2.375$. - $f(2.375) = 13.39 - 11.88 - 1 = 0.51$ (positive) - Interval $[2.25, 2.375]$. 8. Next midpoint $c = 2.3125$. - $f(2.3125) = 12.35 - 11.56 - 1 = -0.18$ (negative) - Interval $[2.3125, 2.375]$. 9. Next midpoint $c = 2.34375$. - $f(2.34375) = 12.87 - 11.72 - 1 = 0.16$ (positive) - Interval $[2.3125, 2.34375]$. 10. Interval length $2.34375 - 2.3125 = 0.03125 > 0.001$, continue. 11. Next midpoint $c = 2.328125$. - $f(2.328125) = 12.61 - 11.64 - 1 = -0.01$ (negative) - Interval $[2.328125, 2.34375]$. 12. Next midpoint $c = 2.3359375$. - $f(2.3359375) = 12.74 - 11.68 - 1 = 0.07$ (positive) - Interval $[2.328125, 2.3359375]$. 13. Interval length $2.3359375 - 2.328125 = 0.0078125 > 0.001$, continue. 14. Next midpoint $c = 2.33203125$. - $f(2.33203125) = 12.67 - 11.66 - 1 = 0.03$ (positive) - Interval $[2.328125, 2.33203125]$. 15. Interval length $2.33203125 - 2.328125 = 0.00390625 > 0.001$, continue. 16. Next midpoint $c = 2.330078125$. - $f(2.330078125) = 12.64 - 11.65 - 1 = 0.01$ (positive) - Interval $[2.328125, 2.330078125]$. 17. Interval length $2.330078125 - 2.328125 = 0.001953125 > 0.001$, continue. 18. Next midpoint $c = 2.3291015625$. - $f(2.3291015625) = 12.63 - 11.64 - 1 = 0.00$ (approx zero) - Interval $[2.328125, 2.3291015625]$. 19. Interval length $2.3291015625 - 2.328125 = 0.0009765625 < 0.001$, stop. **Root by Bracketing method:** approximately $2.329$. --- ### Newton-Raphson Method Steps: 20. Start with initial guess $x_0 = 2.5$ (from bracketing interval). 21. Compute $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$: - $f(2.5) = 2.125$ - $f'(2.5) = 3(2.5)^2 - 5 = 18.75 - 5 = 13.75$ - $$x_1 = 2.5 - \frac{2.125}{13.75} = 2.5 - 0.1545 = 2.3455$$ 22. Compute $x_2$: - $f(2.3455) = (2.3455)^3 - 5(2.3455) - 1 \approx 0.07$ - $f'(2.3455) = 3(2.3455)^2 - 5 \approx 11.52$ - $$x_2 = 2.3455 - \frac{0.07}{11.52} = 2.3455 - 0.0061 = 2.3394$$ 23. Compute $x_3$: - $f(2.3394) \approx 0.002$ - $f'(2.3394) \approx 11.43$ - $$x_3 = 2.3394 - \frac{0.002}{11.43} = 2.3394 - 0.000175 = 2.3392$$ 24. Compute $|x_3 - x_2| = |2.3392 - 2.3394| = 0.0002 < 0.001$, stop. **Root by Newton-Raphson method:** approximately $2.339$. --- **Final answers:** - Root by Bracketing method: $\boxed{2.329}$ - Root by Newton-Raphson method: $\boxed{2.339}$