Suppose that your first guess with Newton's method is lucky, in the sense that X, is a root of ((x) =0. Assuming that "' (0) is defined and not 0, what happens to X, and later approximations? O A. X and later approximations will move away from the root. O B. X, and later approximations will move to other solutions of the equation f(x)=0. O C. X and later approximations will all be equal to 0. O D. X, and later approximations will all be equal to Xg