Explanation Given information : Given equation sin x = x 2 − 3 x + 1 , let f ( x ) = x 2 − 3 x + 1 − sin x and f | ( x ) = 2 x − 3 − cos x Let us find the initial approximation by the graph of f ( x ) = x 2 − 3 x + 1 − sin x as follows Graph: Interpretation: from above graph we observe that the roots are closure to 0 and 3 , Thus, we use Newton’s method to find the solutions with initial approximation as x 0 = 0 and 3. let f ( x ) = x 2 − 3 x + 1 − sin x and f | ( x ) = 2 x − 3 − cos x ⇒ f ( 0 ) = 1 ⇒ f ( 3 ) = 9 − 9 + 1 − sin ( 3 ) ⇒ f ( 3 ) = 0.858880 and ⇒ f | ( 0 ) = 2 ( 0 ) − 3 − cos ( 0 ) = − 4 ⇒ f ( 3 ) = 2 ( 3 ) − 3 − cos ( 3 ) = 3.9899925 From Newton’s method we have x n + 1 = x n − f ( x n ) f | ( x n ) ......... ( 1 ) For x 0 = 0 , then First iteration: by replacing n = 0 in equation ( 1 ) , we get x 1 = x 0 − f ( x 0 ) f | ( x 0 ) ⇒ x 1 = 0 − f ( 0 ) f | ( 0 ) ⇒ x 1 = 0 − ( 1 − 4 ) ⇒ x 1 = 0 + 0

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ISBN: 9781305420878

Author: Stewart

Publisher: CENGAGE L

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Chapter 3, Problem 88E

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The root of the equation sinx=x2−3x+1 , correct to six decimal places using Newton’s method.

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Chapter 3, Problem 87E

Chapter 3, Problem 89E

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The root of the equation sin x = x 2 − 3 x + 1 , correct to six decimal places using Newton’s method.

Solution Summary: The author explains that the root of the equation mathrmsinx is correct to six decimal places using Newton’s method.

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The root of the equation sinx=x2−3x+1 , correct to six decimal places using Newton’s method.

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