   Chapter 4.8, Problem 12E ### 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 given number correct to eight decimal places. 500 8

To determine

To approximate: The given number 754 correct to eight decimal places by using Newton’s method.

Explanation

Given:

The function is f(x)=x8500.

The initial condition is x1=2.2.

Observation: The number 5008 is equivalent to finding the positive root of x8500=0.

Formula used:

The newton’s formula is, xn+1=xnf(xn)f(xn), where f(x) is the given function.

Calculation:

Calculate the value of above equation at x1=2.2.

f(2.2)=(2.2)8500=548.75873536500=48.75873536

Calculate the derivative of f(x).

f(x)=dydxd(x8)dx=8x7

Calculate the value of above equation at x1=2.2.

f'(2.2)=8×(2.2)7=8×249.4357888=1995.4863104

Calculate x2 by using newton method.

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

Set n=1, and obtain x2=x1f(x1)f'(x1).

Substitute x1=2.2,f(x1)=48.75873536 and f'(x1)=1995.4863104,

x2=2.248.758735361995.4863104=2.20.024434512.17556548

Substitute x2=2.17556548 in f(x2),

f(2.17556548)=(2.17556548)8500=501.85386534500=1.85386543

Substitute x2=2.17556548 in f'(x2),

f'(2.17556548)=8×(2.17556548)7=8×230.67743534=1845.41948274

Set n=3, and obtain x3=x2f(x2)f'(x2).

Substitute x2=2.17556548,f(x2)=1

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