Approximate the fixed point of the function to two decimal places. [A fixed point xo of a function f is a value of x such that f(xo) = xo.] (x) = cos x Part 1 of 4 A fixed point xo of a function f is a value of x such that (xo) = x0. To approximate the fixed point of the function x) - cos x, let g(x) = r(x) – x = COs(x) X. COS r Differentiate g(x) with respect to x. 9 '(x) =-sinx 1 sin r Part 2 of 4 To approximate the fixed point of function f, approximate a zero of g(x) = cos x - x. The iterative process of Newton's Method is given by the formula, cos Xn - Xn 9(xn) 9 '(Xn) Xn + 1 = Xn - = Xn - x, +1 - 1 sin an = Xn + COS Xn - Xp COs I. sin x, + 1 sin a Part 3 of 4 For n- 1, let x = 1.0. Now calculate the iterations and enter the values in the table below. (Round your answers to four decimal places.) g(xn) 9 '(Xn) 9(xn) g(xn) Xn - 9 (Xp) ) ("x), 6 1 1.0000 -0.4597 -1.8415 0.2496 0.7504 2 0.7504 -0.0190-1.6819 0.0113 3 0.0000 -1.6736 0.0000

Calculus: Early Transcendentals
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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# Approximating the Fixed Point of a Function Using Newton's Method

## Understanding Fixed Points

A **fixed point** \( x_0 \) of a function \( f \) is a value of \( x \) such that:
\[ f(x_0) = x_0 \]

Consider the function:
\[ f(x) = \cos x \]

## Part 1 of 4

To approximate the fixed point \( x_0 \) of the function \( f(x) = \cos x \), let:
\[ g(x) = f(x) - x = \cos(x) - x \]

Next, differentiate \( g(x) \) with respect to \( x \):
\[ g' (x) = -\sin x - 1 \]

## Part 2 of 4

To approximate the fixed point of function \( f \), approximate a zero of \( g(x) = \cos x - x \).

The iterative process of Newton's Method is given by the formula:
\[ x_{n+1} = x_n - \frac{g(x_n)}{g' (x_n)} = x_n - \frac{\cos x_n - x_n}{-\sin x_n - 1} \]

## Part 3 of 4

For \( n = 1 \), let \( x_1 = 1.0 \). Now calculate the iterations and enter the values in the table below. (Round your answers to four decimal places.)

| \( n \) | \( x_n \)  | \( g(x_n) \)  | \( g' (x_n) \) | \( \frac{g(x_n)}{g' (x_n)} \) | \( x_n - \frac{g(x_n)}{g' (x_n)} \) |
|---------|------------|---------------|----------------|-------------------------------|--------------------------------------|
| 1       | 1.0000     | -0.4597       | -1.8415        | 0.2496                        | 0.7504                               |
| 2       | 0.7504     | -0.0190       | -1.6819        | 0.0113                        |                                      |
| 3       |            | 0.0000        | -1.6736        | 0.0000                        |                                      |

##
Transcribed Image Text:--- # Approximating the Fixed Point of a Function Using Newton's Method ## Understanding Fixed Points A **fixed point** \( x_0 \) of a function \( f \) is a value of \( x \) such that: \[ f(x_0) = x_0 \] Consider the function: \[ f(x) = \cos x \] ## Part 1 of 4 To approximate the fixed point \( x_0 \) of the function \( f(x) = \cos x \), let: \[ g(x) = f(x) - x = \cos(x) - x \] Next, differentiate \( g(x) \) with respect to \( x \): \[ g' (x) = -\sin x - 1 \] ## Part 2 of 4 To approximate the fixed point of function \( f \), approximate a zero of \( g(x) = \cos x - x \). The iterative process of Newton's Method is given by the formula: \[ x_{n+1} = x_n - \frac{g(x_n)}{g' (x_n)} = x_n - \frac{\cos x_n - x_n}{-\sin x_n - 1} \] ## Part 3 of 4 For \( n = 1 \), let \( x_1 = 1.0 \). Now calculate the iterations and enter the values in the table below. (Round your answers to four decimal places.) | \( n \) | \( x_n \) | \( g(x_n) \) | \( g' (x_n) \) | \( \frac{g(x_n)}{g' (x_n)} \) | \( x_n - \frac{g(x_n)}{g' (x_n)} \) | |---------|------------|---------------|----------------|-------------------------------|--------------------------------------| | 1 | 1.0000 | -0.4597 | -1.8415 | 0.2496 | 0.7504 | | 2 | 0.7504 | -0.0190 | -1.6819 | 0.0113 | | | 3 | | 0.0000 | -1.6736 | 0.0000 | | ##
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