(a) Use Newton’s method with x1 − 1 to find the root of the equation x3 2 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.)

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(d) Graph f sxd − x3 2 x 2 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.