Explanation Given information: The equation: 1 x = 1 + x 3 Calculation: We are given 1 x = x 3 + 1 . We have to find the solution of this equation. We use Newton’s method to solve this and thus we consider f ( x ) = x 3 − 1 x + 1 . From the diagram it can be seen that the solutions are close to x = - 1 and x = 1. Thus we use Newton’s method to find the solutions with the first approximations as x = - 1 and x = 1. We know from Newton’s method that for a function f ( x ) and the initial approximation: x n + 1 = x n + f ( x n ) f ' ( x n ) So when we substitute x = − 1 we get: [ x 2 − 1.25 x 3 − 1

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ISBN: 9781305420878

Author: Stewart

Publisher: CENGAGE L

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Chapter 3.8, Problem 26E

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To find: all roots of 1x=1+x3 correct to six decimal places, using Newton’s method.

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Chapter 3.8, Problem 25E

Chapter 3.8, Problem 27E

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Chapter 3 Solutions

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all roots of 1 x = 1 + x 3 correct to six decimal places, using Newton’s method.

Solution Summary: The author analyzes Newton's method to find the solution of the equation 1x=1+X3.

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To find: all roots of 1x=1+x3 correct to six decimal places, using Newton’s method.

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Chapter 3 Solutions