1. The problem asks for the expression of $x_{n+1}$ in Newton's Method to find the root of the function $$f(x) = \cos x (x^2 + 2x + 2).$$ 2. Newton's Method formula is: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ 3. First, find the derivative $f'(x)$ using the product rule: $$f(x) = u(x)v(x) \text{ where } u(x) = \cos x, v(x) = x^2 + 2x + 2$$ 4. Derivative of $u(x)$: $$u'(x) = -\sin x$$ 5. Derivative of $v(x)$: $$v'(x) = 2x + 2$$ 6. Using product rule: $$f'(x) = u'(x)v(x) + u(x)v'(x) = -\sin x (x^2 + 2x + 2) + \cos x (2x + 2)$$ 7. Substitute $f(x)$ and $f'(x)$ into Newton's formula: $$x_{n+1} = x_n - \frac{\cos x_n (x_n^2 + 2x_n + 2)}{-\sin x_n (x_n^2 + 2x_n + 2) + \cos x_n (2x_n + 2)}$$ 8. Simplify the denominator sign: $$x_{n+1} = x_n - \frac{\cos x_n (x_n^2 + 2x_n + 2)}{\cos x_n (2x_n + 2) - \sin x_n (x_n^2 + 2x_n + 2)}$$ 9. This matches option C: $$x_{n+1} = x_n - \frac{\cos x (x^2 + 2x + 2)}{\cos x (2x + 2) - \sin x (x^2 + 2x + 2)}$$ Final answer: Option C.