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Newton's Method : If xn is an approximation a solution of f ( x ) = 0 and if given by, f ′ ( xn ) ≠ 0 the next approximation is given by
xn+1 = xn - f(xn)/f'(xn)
It has to lead to the question of while do we stop? How several times do we go through this procedure? One of the more common stopping points in the procedure is to continue till two successive approximations agree upon a given number of decimal places.
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A pair of pants costs $24. The cost was decreased by 8%. What is the new cost of the pants? If the cost of the pants is decreased by 8%, the cost of the pants is 92 percent of
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