(a) Use Newton’s method with x1 = 1 to find the root of the equation x3 – x = 1 correct to six decimal places.
(b) Solve the equation in part (a) using x1 = 0.6 as the initial approximation.
(c) Solve the equation in part (a) using x1 = 0.57. (You definitely need a programmable calculator for this part.)
(d) Graph f(x) = x3 – x – 1 and its tangent lines at x1 = 1. 0.6 And 0.57 to explain why Newton’s method is so sensitive to the value of the initial approximation.