1. **State the problem:** We want to find an estimated root of the nonlinear equation $$e^{\cos(x+2)} = \ln(2 - x^2) + 2$$ near the initial guess $$x_0 = -0.7$$ using Newton's method with three iterations. 2. **Rewrite the equation:** Define the function $$f(x) = e^{\cos(x+2)} - \ln(2 - x^2) - 2$$. We want to find $$x$$ such that $$f(x) = 0$$. 3. **Newton's method formula:** $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ We need to compute $$f'(x)$$. 4. **Compute the derivative:** - Derivative of $$e^{\cos(x+2)}$$ is $$e^{\cos(x+2)} \cdot (-\sin(x+2))$$ by chain rule. - Derivative of $$-\ln(2 - x^2)$$ is $$- \frac{-2x}{2 - x^2} = \frac{2x}{2 - x^2}$$. - Derivative of constant $$-2$$ is 0. So, $$f'(x) = e^{\cos(x+2)} (-\sin(x+2)) + \frac{2x}{2 - x^2}$$ 5. **Perform iterations:** - **Iteration 1:** - Calculate $$f(x_0)$$ and $$f'(x_0)$$ at $$x_0 = -0.7$$. - $$f(-0.7) = e^{\cos(-0.7+2)} - \ln(2 - (-0.7)^2) - 2 = e^{\cos(1.3)} - \ln(2 - 0.49) - 2$$ - $$\cos(1.3) \approx 0.2675$$, so $$e^{0.2675} \approx 1.306$$. - $$\ln(1.51) \approx 0.413$$. - So, $$f(-0.7) \approx 1.306 - 0.413 - 2 = -1.107$$. - $$f'(-0.7) = e^{0.2675} (-\sin(1.3)) + \frac{2(-0.7)}{1.51}$$. - $$\sin(1.3) \approx 0.9636$$, so $$-\sin(1.3) = -0.9636$$. - $$f'(-0.7) \approx 1.306 \times (-0.9636) + \frac{-1.4}{1.51} = -1.259 - 0.927 = -2.186$$. - Update: $$x_1 = -0.7 - \frac{-1.107}{-2.186} = -0.7 - 0.506 = -1.206$$. - **Iteration 2:** - Calculate $$f(x_1)$$ and $$f'(x_1)$$ at $$x_1 = -1.206$$. - $$f(-1.206) = e^{\cos(-1.206+2)} - \ln(2 - (-1.206)^2) - 2 = e^{\cos(0.794)} - \ln(2 - 1.454) - 2$$ - $$\cos(0.794) \approx 0.700$$, so $$e^{0.700} \approx 2.014$$. - $$2 - 1.454 = 0.546$$, $$\ln(0.546) \approx -0.605$$. - So, $$f(-1.206) \approx 2.014 - (-0.605) - 2 = 0.619$$. - $$f'(-1.206) = e^{0.700} (-\sin(0.794)) + \frac{2(-1.206)}{0.546}$$. - $$\sin(0.794) \approx 0.713$$, so $$-\sin(0.794) = -0.713$$. - $$f'(-1.206) \approx 2.014 \times (-0.713) + \frac{-2.412}{0.546} = -1.436 - 4.417 = -5.853$$. - Update: $$x_2 = -1.206 - \frac{0.619}{-5.853} = -1.206 + 0.106 = -1.100$$. - **Iteration 3:** - Calculate $$f(x_2)$$ and $$f'(x_2)$$ at $$x_2 = -1.100$$. - $$f(-1.100) = e^{\cos(-1.100+2)} - \ln(2 - (-1.100)^2) - 2 = e^{\cos(0.900)} - \ln(2 - 1.21) - 2$$ - $$\cos(0.900) \approx 0.621$$, so $$e^{0.621} \approx 1.861$$. - $$2 - 1.21 = 0.79$$, $$\ln(0.79) \approx -0.236$$. - So, $$f(-1.100) \approx 1.861 - (-0.236) - 2 = 0.097$$. - $$f'(-1.100) = e^{0.621} (-\sin(0.900)) + \frac{2(-1.100)}{0.79}$$. - $$\sin(0.900) \approx 0.783$$, so $$-\sin(0.900) = -0.783$$. - $$f'(-1.100) \approx 1.861 \times (-0.783) + \frac{-2.2}{0.79} = -1.457 - 2.785 = -4.242$$. - Update: $$x_3 = -1.100 - \frac{0.097}{-4.242} = -1.100 + 0.023 = -1.077$$. 6. **Final answer:** After three iterations, the estimated root is approximately $$\boxed{-1.077}$$.