1 Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. x5 - x - 6 = 0, x1 = 2 Part 1 of 3 If f( x) = x5 - x - 6, then f ' ( x ) = 5 0 8 5 x4 - 1 0 8 1 . Part 2 of 3 We have Xn + 1 = Xn - Fix f( xn ) = Xn - Xn - Xn - 6. Therefore, 5X4 - 1 X2 = 2 - L 79 (rounded to six decimal places). Submit Skip_(you cannot come back) 2 Use Newton's method with initial approximation X1 = -2 to find x2, the second approximation to the root of the equation x + x + 7 = 0. (Round your answer to four decimal places.) X2 = 3 Use Newton's method to approximate the given number correct to eight decimal places. V20 Part 1 of 8 Note that x = \\20 is a root of f(x) = x5 - 20. We need to find f'(x). f ' ( x ) = 5x4 Part 2 of 8 We know that Xn + 1 = Xn - F'(Xn) f( xn) -. Therefore, Xn5 - 20 20 Xn + 1= Xn - 5 8 5 X 4 Part 3 of 8 Since 32 = 2, and 32 is reasonably close to 20, we'll use x1 = 2. This gives us X2 = 2 - -1(2) f ' (2 ) = 2 - 12 12 80 = 1.85 1.850000000 (rounded to nine decimal places). Part 4 of 8 Now we can find X3. 1.8500000005 - 20 20 X3 = 1.850000000 - 51.85 1.850000000 1.821486137 1.821486137 (rounded to nine decimal places) Part 5 of 8 Since X2 and x3 don't agree to eight decimal places, we must find another iteration. x35- 20 X4 = X3 - 5 5 0 5 5 X 3 4 = 1.820565135 1.820565136 (rounded to nine decimal places) Part 6 of 8 Since x3 and x4 don't agree to eight decimal places, we must find yet another iteration. X5 = X4 - 24 - 5x4 4 (rounded to nine decimal places) 4 EXAMPLE 1 Starting with X1 = 3, find the third approximation x3 to the root of the equation x3 - 2x - 16 = 0. SOLUTION We apply Newton's method with f(x) = x3 - 2x - 16 and f'(x) = 3x2- 2 So the iterative equation becomes x3 - 2xn - 16 Xn+1 = Xn - 3 ( x , )2 - 2 With n = 1, we have X2 = X1 - Ta x13 - 2x1 - 16 x12 - 2 = 3 - 3 - 2(3) - 16 25 = 2.8 Then with n = 2, we obtain *23- 2X2 - 16 X3 = X2 - 3 V X 2 2 - 2 2.8 3 - 2 2.8 - 16 2.7836 . (Round to four decimal places.) It turns out that this third approximation X3 ~ 2.784 is accurate to three decimal places