(a) Use Newton's method with $ x_1 = 1 $ to find the roof of the equation $ x^3 - x = 1 $ correct to six decimal places.
(b) Solve the equation in part (a) using $ x_1 == 0.6 $ as the initial approximation.
(c) Solve the equation in part (a) using $ x_1 = 0.57 $.
(You definitely need a programmable calculator for this part.)
(d) Graph $ f(x) = x^3 - x - 1 $ and its tangent lines at $ x_1 = 1 $, $ 0.6 $, and $ 0.57 $ to explain why Newton's method is so sensitive to the value of the initial approximation.
d. see graph
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