`x^6 - x^5 - 6x^4 - x^2 + x + 10 = 0` Use Newton's method to find all roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations.

Textbook Question

Chapter 4, 4.8 - Problem 23 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`f(x)=x^6-x^5-6x^4-x^2+x+10`

`f'(x)=6x^5-5x^4-24x^3-2x+1`

`x_(n+1)=x_n-((x_n)^6-(x_n)^5-6(x_n)^4-(x_n)^2+x_n+10)/(6(x_n)^5-5(x_n)^4-24(x_n)^3-2(x_n)+1)`

See the attached graph. From the graph the roots of f are approximately -1.9 , -1.2 , 1.1 and 3.

Approximate until the eight decimal places are same.

For x_1=-1.9

`x_2=(-1.9)-((-1.9)^6-(-1.9)^5-6(-1.9)^4-(-1.9)^2+(-1.9)+10)/(6(-1.9)^5-5(-1.9)^4-24(-1.9)^3-2(-1.9)+1)`

`x_2~~-1.94278290`

`x_3~~-1.93828380`

`x_4~~-1.93822884`

`x_5~~-1.93822883`

`x_6~~-1.93822883`

Now for x_1=-1.2

`x_2=(-1.2)-((-1.2)^6-(-1.2)^5-6(-1.2)^4-(-1.2)^2+(-1.2)+10)/(6(-1.2)^5-5(-1.2)^4-24(-1.2)^3-2(-1.2)+1)`

`x_2~~-1.22006245`

`x_3~~-1.21997997`

`x_4~~-1.21997997`

Now for x_1=1.1

`x_2=(1.1)-(1.1^6-1.1^5-6(1.1)^4-(1.1)^2+1.1+10)/(6(1.1)^5-5(1.1)^4-24(1.1)^3-2(1.1)+1)`

`x_2~~1.14111662`

`x_3~~1.13929741`

`x_4~~1.13929375`

`x^5~~1.13929375`

Now for x_1=3

`x_2=3-(3^6-3^5-6(3)^4-3^2+3+10)/(6(3)^5-5(3)^4-24(3)^3-2(3)+1)`

`x_2~~2.99`

`x_3~~2.98984106`

`x_4~~2.98984102`

`x_5~~2.98984102`

To eight decimal places the roots of the equation are,

-1.93822883 , -1.21997997 , 1.13929375 , 2.98984102

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