Explanation Given information: The equation: x 4 = 1 + x Calculation: We are given x 4 = x + 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 4 − x − 1 . From the diagram it can be seen that the solutions are close to x = - 1 and x = 1.5. Thus we use Newton’s method to find the solutions with the first approximations as x = - 1 and x = 1.5. 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 − 0.8 x 3 − 0.731233595801 x 4 − 0.724548480011 x 5 − 0

Bundle: Biocalculus: Calculus, Probability, And Statistics For The Life Sciences + Webassign Printed Access Card For Stewart/day's Biocalculus: Calculus For Life Sciences, 1st Edition, Multi-term

1st Edition

ISBN: 9781305779037

Author: James Stewart, Troy Day

Publisher: Cengage Learning

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Chapter 3.8, Problem 23E

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Chapter 3.8, Problem 22E

Chapter 3.8, Problem 24E

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Bundle: Biocalculus: Calculus, Probability, And Statistics For The Life Sciences + Webassign Printed Access Card For Stewart/day's Biocalculus: Calculus For Life Sciences, 1st Edition, Multi-term

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all roots of x 4 = 1 + x correct to six decimal places, using Newton’s method .

Solution Summary: The author analyzes Newton's method to find the roots of x4=1+x correct to six decimal places.

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To find: all roots of x4=1+x correct to six decimal places, using Newton’s method

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all roots of x 4 = 1 + x correct to six decimal places, using Newton’s method .

Solution Summary: The author analyzes Newton's method to find the roots of x4=1+x correct to six decimal places.

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To find: all roots of x4=1+x correct to six decimal places, using Newton’s method .

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Chapter 3 Solutions

To find: all roots of x4=1+x correct to six decimal places, using Newton’s method

.