`x^3 = tan^-1 (x)` Use Newton's method to find all roots of the equation correct to six decimal places.
To solve using Newton's method apply the formula,
Plug in f(x) and f'(x) in the formula
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,
Now let's solve for the second zero , x_1=-0.9
So the roots of the equation are 0.902025 and -0.902025