Calculus: Early Transcendentals, Chapter 4, 4.8, Section 4.8, Problem 23

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