J Jason R New member Mar 31, 2010 #1 Use Newton's method with the specified initial approximation x1 and find x3, the third approximation to the root of the given equation. (Give your answer to two decimal places.) x^3 - x^2 - 1 = 0 x1 = 1
Use Newton's method with the specified initial approximation x1 and find x3, the third approximation to the root of the given equation. (Give your answer to two decimal places.) x^3 - x^2 - 1 = 0 x1 = 1
