   Chapter 4.8, Problem 16E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use Newton’s method to approximate the indicated root of the equation correct to six decimal places.The positive root of 3 sin x = x

To determine

To find: The positive root of the equation correct to six decimal places.

Explanation

Formula used:

The newton’s formula is xn+1=xnf(xn)f(xn).

Given:

The initial approximation x1=2, and 3sin(x)=x.

Calculation:

Rewrite the given equation as f(x)=3sin(x)x.

Calculate the derivative of f(x).

f(x)=ddx(3sin(x)x)=ddx(3sin(x))ddx(x)=3cos(x)1

The value of f(x) at x1=2 is,

f(2)=3cos(2)1=3×(0.416146)1=1.2484401=2.248440

Calculate the value of f(x) at x1=2.

f(2)=3sin(2)2=3×0.9092972=2.7278922=0.727892

Use Newton’s method to calculate x2.

xn+1=xnf(xn)f(xn).

Set n=1 and obtain the value of x2.

x2=x1f(x1)f(x1)

Substitute x1=2,f(x1)=0.727892 and f(x1)=2.248440,

x2=2(0.727892)(2.248440)=2+(0.727892)(2.248440)=2+0.3237322.323732

Calculate the value of f(x) at x2=2.323732.

f(2.323732)=3sin(2.323732)2.323732=3×0.7296842.323732=2.1890532.323732=0.134679

Calculate the value of f(x) at x2=2.323732.

f(2.323732)=3cos(2.323732)1=3×(0.683783)1=2.0513511=3.051351

Use Newton’s method to calculate x3.

Set n=2 and obtain the value of x3.

x3=x2f(x2)f(x2).

Substitute x2=2.323732, f(x)=0.134679 and f(x2)=3

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Study Guide for Stewart's Multivariable Calculus, 8th 