The problem in the image involves using Newton's Method to approximate the root of the function $f(x) = \cos(x)$ with an initial guess of $x_1 = 1.1$.

Newton's Method formula is given by: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ where:

• $f(x_n) = \cos(x_n)$
• $f'(x_n) = -\sin(x_n)$

First iteration $n = 1$:

Given:

• $x_1 = 1.1$

1. Compute $f(x_1) = \cos(1.1)$: $f(x_1) = \cos(1.1) \approx 0.4536$ (This value is correct as per the image.)

2. Compute $f'(x_1) = -\sin(1.1)$: $f'(x_1) = -\sin(1.1) \approx -0.8912$ (This value is correct as well.)

3. Now, calculate the Newton's Method formula: $x_2 = 1.1 - \frac{0.4536}{-0.8912} \approx 1.6095$ This step also matches the image.

Second iteration $n = 2$:

Now, using $x_2 = 1.6095$:

1. Compute $f(x_2) = \cos(1.6095)$: $f(x_2) = \cos(1.6095) \approx -0.0001$ (This value is rounded correctly.)

2. Compute $f'(x_2) = -\sin(1.6095)$: $f'(x_2) = -\sin(1.6095) \approx -0.9999$ (This value is also correct.)

3. Now, apply Newton's Method again: $x_3 = 1.6095 - \frac{-0.0001}{-0.9999} \approx 1.6096$ This matches the calculation in the table as well.

Corrections:

The values calculated for $f(x_n)$, $f'(x_n)$, and $\frac{f(x_n)}{f'(x_n)}$ in the image are mostly correct, but some entries like the second $x_n$ values (for iteration 2) are slightly off due to rounding.

Let me know if you'd like further clarifications or have any questions. Here are five questions that follow:

1. How does Newton's Method converge to a root?
2. What is the significance of the derivative $f'(x)$ in Newton's Method?
3. What would happen if $f'(x_n) = 0$ during an iteration?
4. Why is the initial guess important in Newton's Method?
5. Can Newton's Method fail to converge? Under what conditions?

Tip: Newton's Method generally converges faster when the initial guess is closer to the actual root.