1. **Problem statement:** Use Newton's method to calculate $x_4$ starting from $x_0=1$ for the function $$f(x) = \cos(x)(x^2 + 2x + 2).$$ 2. **Newton's method formula:** $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ 3. **Find the derivative $f'(x)$:** Using the product rule: $$f'(x) = -\sin(x)(x^2 + 2x + 2) + \cos(x)(2x + 2)$$ 4. **Calculate iterations:** Start with $x_0 = 1$. - Calculate $f(x_0)$: $$f(1) = \cos(1)(1^2 + 2\cdot1 + 2) = \cos(1) \times 5 \approx 0.5403 \times 5 = 2.7015$$ - Calculate $f'(x_0)$: $$f'(1) = -\sin(1) \times 5 + \cos(1) \times 4 = -0.8415 \times 5 + 0.5403 \times 4 = -4.2075 + 2.1612 = -2.0463$$ - Compute $x_1$: $$x_1 = 1 - \frac{2.7015}{-2.0463} = 1 + 1.320 = 2.320$$ Repeat for $x_2$: - $f(2.320) \approx \cos(2.320)(2.320^2 + 2\times2.320 + 2) \approx -0.680 \times 11.382 = -7.737$ - $f'(2.320) \approx -\sin(2.320) \times 11.382 + \cos(2.320) \times (2\times2.320 + 2) \approx -0.729 \times 11.382 - 0.680 \times 6.64 = -8.298 - 4.515 = -12.813$ - $x_2 = 2.320 - \frac{-7.737}{-12.813} = 2.320 - 0.604 = 1.716$ Repeat for $x_3$: - $f(1.716) \approx \cos(1.716)(1.716^2 + 2\times1.716 + 2) \approx -0.145 \times 7.618 = -1.104$ - $f'(1.716) \approx -\sin(1.716) \times 7.618 + \cos(1.716) \times (2\times1.716 + 2) \approx -0.989 \times 7.618 - 0.145 \times 5.432 = -7.533 - 0.788 = -8.321$ - $x_3 = 1.716 - \frac{-1.104}{-8.321} = 1.716 - 0.133 = 1.583$ Repeat for $x_4$: - $f(1.583) \approx \cos(1.583)(1.583^2 + 2\times1.583 + 2) \approx -0.009 \times 6.665 = -0.060$ - $f'(1.583) \approx -\sin(1.583) \times 6.665 + \cos(1.583) \times (2\times1.583 + 2) \approx -0.9999 \times 6.665 - 0.009 \times 5.166 = -6.664 - 0.046 = -6.710$ - $x_4 = 1.583 - \frac{-0.060}{-6.710} = 1.583 - 0.009 = 1.574$ 5. **Conclusion:** The value of $x_4$ is approximately $1.574$, which is closest to option D (None of the above choices).