Calculus: Early Transcendentals, Chapter 4, 4.8, Section 4.8, Problem 20

1/x=1+x^3
Set the left side equal to zero.
0=1+x^3 -1/x
To solve this using Newton's method, apply the formula:
x_(n+1) = x_n - (f(x_n))/(f'(x_n))
Let the function be:
f(x) = 1+x^3-1/x
Take the derivative of f(x).
f'(x) = 3x^2 +1/x^2
Plug-in f(x) and f'(x) to the formula of Newton's method.
x_(n+1)=x_n-(1+x_n^3-1/x_n)/(3x_n^2+1/x_n^2)
This simplifies to:
x_(n+1) = x_n- (x_n^5+x_n^2-x_n)/(3x_n^4+1)
Then, refer to the graph of the function to get the initial values x when f(x) =0. (See attached figure.)
Notice that the function has two zeros. These two zeros are near x=-1.2 and x=0.8.
Let's solve for the approximate values of zeros of f(x) up to six decimal places.
For our first zero, let the initial value be -1.2.
x_=-1.2
x_2= x_1- (x_1^5+x_1^2-x_1)/(3x_1^4+1)=-1.221005982
x_3= x_2- (x_2^5+x_2^2-x_1)/(3x_2^4+1)=-1.220744126
x_4= x_3- (x_1^5+x_1^2-x_1)/(3x_1^4+1)=-1.220744084
Now there are two approximations that agree to six decimal places. So one of the approximate solution of the equation is x=-1.220744 .
Next, let's solve for second zero. Let its initial value be 0.8.
x_1=0.8
x_2= x_1- (x_1^5+x_1^2-x_1)/(3x_1^4+1)=0.7247666905
x_3= x_2- (x_2^5+x_2^2-x_1)/(3x_2^4+1)=0.7244919491
x_4= x_3- (x_1^5+x_1^2-x_1)/(3x_1^4+1)=0.7244919590
There are already two approximations that have exact six decimal places. So, one of the solution is x=0.724492 .

Therefore, the approximate solution to the equation 1/x=1+x^3 are x={-1.220744,0.724492} .