Explanation Method used: Newton’s Method: 1. Pick an initial estimate x 0 of the root c. 2. Find a new estimate using the iterative formula x n + 1 = x n − f ( x n ) f ′ ( x n ) , n = 0 , 1 , 2 , … . 3. Compute | x n − x n + 1 | . If the number is less than a prescribed positive number, then stop the iteration. The required approximation to the root x = c is x = x n + 1 . Calculation: The function is given by f ( x ) = x 3 − 14 . Find the derivative of f ( x ) = x 3 − 14 as follows. f ′ ( x ) = 3 x 2 The required iterative formula is, x n + 1 = x n − x n 3 − 14 3 x n 2 = 3 x n 3 − x n 3 + 14 3 x n 2 = 2 x n 3 + 14 3 x n 2 By using the above mentioned Newton’s method, find the three iterations for the function f ( x ) = x 3 − 14 at x 0 = 2.5

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Author: Tan

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Chapter 11.7, Problem 5E

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To estimate: The value of a radical 143 to five decimal places by using three iterations of Newton’s method with the indicated initial guess x0=2.5 for the function f(x)=x3−14.

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Chapter 11.7, Problem 4E

Chapter 11.7, Problem 6E

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To estimate: The value of a radical 14 3 to five decimal places by using three iterations of Newton’s method with the indicated initial guess x 0 = 2.5 for the function f ( x ) = x 3 − 14 .

Solution Summary: The author calculates the value of a radical sqrt[3]14 to five decimal places by using three iterations of Newton's method.

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To estimate: The value of a radical 143 to five decimal places by using three iterations of Newton’s method with the indicated initial guess x0=2.5 for the function f(x)=x3−14.

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Chapter 11 Solutions