Calculus: Early Transcendentals, Chapter 4, 4.8, Section 4.8, Problem 21

x^3=tan^-1(x)
f(x)=x^3-tan^-1(x)=0
f'(x)=3x^2-(1/(x^2+1))
To solve using Newton's method apply the formula,
x_(n+1)=x_n-f(x_n)/(f'(x_n))
Plug in f(x) and f'(x) in the formula
x_(n+1)=x_n-((x_n)^3-tan^-1(x_n))/(3(x_n)^2-(1/((x_n)^2+1)))
See the attached graph for finding the initial values of x_1.
The curve of the function intersects the x axis at ~~ 0.9 and -0.9
Let's solve for the first zero x_1=0.9,
x_2=0.9-(0.9^3-tan^-1(0.9))/(3(0.9)^2-(1/(0.9^2+1)))
x_2~~0.90203199
x_3~~0.90202549
x_4~~0.90202549
Now let's solve for the second zero , x_1=-0.9
x_2=-0.9-((-0.9)^3-tan^-1(-0.9))/(3(-0.9)^2-(1/((-0.9)^2+1)))
x_2~~-0.9020443393
x_3~~-0.902025494
x_4~~-0.9020254924
x_5~~-0.9020254924
So the roots of the equation are 0.902025 and -0.902025