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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Use Newton’s method to find the root of the equation

x 5 x 4 + 3 x 2 3 x 2 = 0

in the interval [1, 2] correct to six decimal places.

To determine

To find: The roots of equation x5x4+3x23x2=0 using Newton’s method.

Explanation

Newton’s formula: xn+1=xnf(xn)f(xn) .

Calculation:

Consider the function f(x)=x5x4+3x23x2 (1)

Obtain the derivative of f(x)

f(x)=ddx(f(x))=ddx(x5x4+3x23x2)=5x44x3+6x3

Thus, f(x)=5x44x3+6x3 . (2)

Let x1=1.5 and f(x) calculate the value of f(x)

f(x1)=(1.5)5(1.5)4+3(1.5)23(1.5)2=(7.59375)(5.0625)+6.754.52=2.78125

Calculate the value of f'(x1) at x1=1.5

f(x1)=5(1.5)44(1.5)3+6(1.5)3=5(5.0625)4(3.375)+93=25.312513.5+6=17.8125

Calculate x2 by using Newton method, xn+1=xnf(xn)f(xn) .

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

Substitute x1=1.5,f(x1)=2.78125 and f'(x1)=17.8125 ,

x2=1.52.7812517.8125x2=1.50.15614035x21.343859

Substitute x2=1.343859 in f(x2) ,

f(x2)=(1.343859)5(1.343859)4+3(1.343859)23(1.343859)2=(4.382970)(3.261480)+3(1.805957)4.0315772=0.507784

Substitute x2=1.34386 in f'(x2) ,

f(x2)=5(1.343859)44(1.343859)3+6(1.343859)3=5(3.261480)4(2.426951)+8.0631543=16.30749.707806+8.0631543=11.662748

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

Substitute x2=1.343859,f(x2)=0.507784 and f'(x2)=11.662748 and find the value of x3 ,

x3=1.343859(0.507784)11.662748x3=1.3438590.043538x31.300320

Substitute x3=1.300320 in f(x3)

f(x3)=(1.300320)5(1.300320)4+3(1.300320)23(1.300320)2=(3.717502)(2.858913)+3(1.690832)3

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