Chapter 4.8, Problem 23E

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

8th Edition
James Stewart
ISBN: 9781305270336

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:

Newtonâ€™s method

xn+1=xnâˆ’f(xn)fâ€²(xn)

Given:

Calculation:

f(x)=âˆ’2x7âˆ’5x4+9x3+5

Differentiating with respect to x ,

dfdx=âˆ’14x6âˆ’20x3+27x2

In Figure 1,

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

xn+1=xnâˆ’f(xn)fâ€²(xn)

Substitute, n=1 in xn+1=xnâˆ’f(xn)fâ€²(xn),

x2=x1âˆ’f(x1)fâ€²(x1)=âˆ’1.8âˆ’âˆ’2(âˆ’1.8)7âˆ’5(âˆ’1.8)4+9(âˆ’1.8)3+5âˆ’14(âˆ’1.8)6âˆ’20(âˆ’1.8)3+27(âˆ’1.8)2=âˆ’1.7174126

Substitute n=2 in xn+1=xnâˆ’f(xn)fâ€²(xn),

x3=x2âˆ’f(x2)fâ€²(x2)=âˆ’1.7174126âˆ’âˆ’2(âˆ’1.7174126)7âˆ’5(âˆ’1.7174126)4+9(âˆ’1.7174126)3+5âˆ’14(âˆ’1.7174126)6âˆ’20(âˆ’1.7174126)3+27(âˆ’1.7174126)2=âˆ’1.6947062

Substitute n=3 in xn+1=xnâˆ’f(xn)fâ€²(xn),

x4=x3âˆ’f(x3)fâ€²(x3)=âˆ’1.6947062âˆ’âˆ’2(âˆ’1.6947062)7âˆ’5(âˆ’1.6947062)4+9(âˆ’1.6947062)3+5âˆ’14(âˆ’1.6947062)6âˆ’20(âˆ’1.6947062)3+27(âˆ’1.6947062)2=âˆ’1.6931276

Substitute n=4 in xn+1=xnâˆ’f(xn)fâ€²(xn),

x5=x4âˆ’f(x4)fâ€²(x4)=âˆ’1.6931276âˆ’âˆ’2(âˆ’1.6931276)7âˆ’5(âˆ’1.6931276)4+9(âˆ’1.6931276)3+5âˆ’14(âˆ’1.6931276)6âˆ’20(âˆ’1.6931276)3+27(âˆ’1.6931276)2=âˆ’1.6931203

Substitute n=5 in xn+1=xnâˆ’f(xn)fâ€²(xn),

x6=x5âˆ’f(x5)fâ€²(x5)=âˆ’1.6931203âˆ’âˆ’2(âˆ’1.6931203)7âˆ’5(âˆ’1.6931203)4+9(âˆ’1.6931203)3+5âˆ’14(âˆ’1.6931203)6âˆ’20(âˆ’1.6931203)3+27(âˆ’1.6931203)2=âˆ’1.6931203

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

So, one root is x=âˆ’1.6931203

In Figure 1,

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

Substitute n=1 in xn+1=xnâˆ’f(xn)fâ€²(xn),

x2=x1âˆ’f(x1)fâ€²(x1)=âˆ’0.8âˆ’âˆ’2(âˆ’0.8)7âˆ’5(âˆ’0.8)4+9(âˆ’0.8)3+5âˆ’14(âˆ’0.8)6âˆ’20(âˆ’0.8)3+27(âˆ’0.8)2=âˆ’0.74815218

Substitute n=2 in xn+1=xnâˆ’f(xn)fâ€²(xn),

x3=x2âˆ’f(x2)fâ€²(x2)=âˆ’0

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