Find the root of the equation. Give your answers correct to six decimal places?

We have:
# x^3-x=2 => x^3-x-2 = 0#

Let # f(x) = x^3-x-2 # then # f'(x)=3x^2-1 # and we can use Newton's method using the iterative formula;

# x_(n+1) = x_n - f(x_n) / (f'(x_n)) #
# :. x_(n+1) = x_n - (x_n^3-x_n-2) / (3x_n^2-1) #

(a) If we start with # x_0=1 #, then we can tabulate the results as follows (in this case using Excel working to 8dp);

So we see that very rapidly the Newton-Rhapson method converges to the solution # x=1.521380 # (6dp)
.

(b) If we start with # x_0=0.6= #, then we can tabulate the results as follows (in this case using Excel working to 8dp);

So again we see that the Newton-Rhapson method converges to the solution # x=1.521380 # (6dp), but this time it takes a few more steps.

(c) If we start with # x_0=0.58= #, then we can tabulate the results as follows (in this case using Excel working to 8dp);

So again we see that the Newton-Rhapson method converges to the solution # x=1.521380 # (6dp), but this time it takes many more steps.