   Chapter 4.8, Problem 23E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

#### Solutions

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

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

# Use Newton’s method to find all the solutions of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations.−2x7 − 5x4 + 9x3 + 5 = 0

To determine

To find: The solutions of f(x)=2x75x4+9x3+5 correct to eight decimal places.

Explanation

Formula used:

Newton’s method

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

Given:

Calculation:

f(x)=2x75x4+9x3+5

Differentiating with respect to x ,

dfdx=14x620x3+27x2

In Figure 1,

It is clear that the 1st zero is x1=1.8 .

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

Substitute, n=1 in xn+1=xnf(xn)f(xn),

x2=x1f(x1)f(x1)=1.82(1.8)75(1.8)4+9(1.8)3+514(1.8)620(1.8)3+27(1.8)2=1.7174126

Substitute n=2 in xn+1=xnf(xn)f(xn),

x3=x2f(x2)f(x2)=1.71741262(1.7174126)75(1.7174126)4+9(1.7174126)3+514(1.7174126)620(1.7174126)3+27(1.7174126)2=1.6947062

Substitute n=3 in xn+1=xnf(xn)f(xn),

x4=x3f(x3)f(x3)=1.69470622(1.6947062)75(1.6947062)4+9(1.6947062)3+514(1.6947062)620(1.6947062)3+27(1.6947062)2=1.6931276

Substitute n=4 in xn+1=xnf(xn)f(xn),

x5=x4f(x4)f(x4)=1.69312762(1.6931276)75(1.6931276)4+9(1.6931276)3+514(1.6931276)620(1.6931276)3+27(1.6931276)2=1.6931203

Substitute n=5 in xn+1=xnf(xn)f(xn),

x6=x5f(x5)f(x5)=1.69312032(1.6931203)75(1.6931203)4+9(1.6931203)3+514(1.6931203)620(1.6931203)3+27(1.6931203)2=1.6931203

The first 8th decimal places repeat at the 6th iteration.

So, one root is x=1.6931203

In Figure 1,

It is clear that the 2nd zero is x1=0.8

Substitute n=1 in xn+1=xnf(xn)f(xn),

x2=x1f(x1)f(x1)=0.82(0.8)75(0.8)4+9(0.8)3+514(0.8)620(0.8)3+27(0.8)2=0.74815218

Substitute n=2 in xn+1=xnf(xn)f(xn),

x3=x2f(x2)f(x2)=0

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