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Tutorial Exercise Use Newton's method with the specified initial approximation x, to find X3, the third approximation to the root of the given equation. x5 - x - 2 =0, x1 = 2 Part 1 of 3 If f (x) = x - x - 2, then f ' ( x ) = 5 0 5 * 4 1 1 Part 2 of 3 f(xn ) 5 - Xn - 2 We have Xn + 1 = Xn- = Xn Therefore, f ' ( x ) 5x4 - 1 X2 = 2 - 79 (rounded to six decimal places). Submit Skip (you cannot come back)6. [-/1 Points] DETAILS SCALCET7 4.8.009. MY NOTES ASK Use Newton's method with initial approximation x1 = -1 to find x2, the second approximation to the root of the equation x5 + x + 6 = 0. (Round your answer to four decimal places.) *2 = Need Help? Watch It 7. [1/14 Points] DETAILS PREVIOUS ANSWERS SCALCET7 4.8.011.MI.SA. MY NOTES ASK This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise Use Newton's method to approximate the given number correct to eight decimal places. $48 Part 1 of 8 Note that x = \\48 is a root of f(x) = x6 - 48. We need to find f '(x). f ' ( x ) = 675 Part 2 of 8 We know that Xn + 1 = Xn F (xn ) Therefore, f ' ( xn ) Xn+ 1 = Xn Xn Submit Skip (you cannot come back)