use newtons method to approximate the x - coordinates where the two functions intersect. you may use the…

Explanation:

Step1: Define the function

Let (f(x)=\cos x - x^{2}). Then (f'(x)=-\sin x - 2x).

Step2: Choose an initial - guess

By looking at the graph of (y = \cos x) and (y=x^{2}), we can see that the intersection points are symmetric about the (y) - axis. For the positive (x) - value, we can choose an initial guess (x_0 = 0.8).

Step3: Apply Newton's Method formula

Newton's Method formula is (x_{n + 1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}). For (n = 0): [ \begin{align*} x_1&=x_0-\frac{f(x_0)}{f'(x_0)}\ f(0.8)&=\cos(0.8)-0.8^{2}\approx0.6967 - 0.64=0.0567\ f'(0.8)&=-\sin(0.8)-2\times0.8\approx - 0.7174-1.6=-2.3174\ x_1&=0.8-\frac{0.0567}{-2.3174}\approx0.8 + 0.0245=0.8245 \end{align*} ] For (n = 1): [ \begin{align*} f(0.8245)&=\cos(0.8245)-0.8245^{2}\approx0.687 - 0.6798 = 0.0072\ f'(0.8245)&=-\sin(0.8245)-2\times0.8245\approx-0.726 - 1.649=-2.375\ x_2&=0.8245-\frac{0.0072}{-2.375}\approx0.8245 + 0.003 = 0.8275 \end{align*} ] For (n = 2): [ \begin{align*} f(0.8275)&=\cos(0.8275)-0.8275^{2}\approx0.685 - 0.6847=0.0003\ f'(0.8275)&=-\sin(0.8275)-2\times0.8275\approx - 0.728-1.655=-2.383\ x_3&=0.8275-\frac{0.0003}{-2.383}\approx0.8275+0.0001 = 0.8276 \end{align*} ] Since the functions (y = \cos x) and (y = x^{2}) are symmetric about the (y) - axis, the intersection points are (x\approx\pm0.8276).

Answer:

(x\approx\pm0.8276)