https://file.aiquickdraw.com/mathbot/file/1727104209959zwx7rimz.jpg

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.

Calculate two iterations of Newton's Method to approximate a zero of the function f(x) = cos(x) using the initial guess x1 = 1.1. Round your answers to four decimal places.

This page provides a detailed solution for calculating two iterations of Newton's Method to approximate a zero of f(x) = cos(x) starting with the initial guess x1 = 1.1. Step-by-step calculations include the derivative of cos(x) and the application of Newton's formula. Ideal for university students studying calculus or numerical methods.

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