Explanation Given information : Given equation x 5 − x 4 + 3 x 2 − 3 x − 2 = 0 , x ∈ [ 1 , 2 ] , let f ( x ) = x 5 − x 4 + 3 x 2 − 3 x − 2 and f | ( x ) = 5 x 4 − 4 x 3 + 6 x − 3 let f ( x ) = x 5 − x 4 + 3 x 2 − 3 x − 2 and f | ( x ) = 5 x 4 − 4 x 3 + 6 x − 3 ⇒ f ( 1 ) = 1 5 − 1 4 + 3 ( 1 2 ) − 3 ( 1 ) − 2 ⇒ f ( 1 ) = − 2 and ⇒ f ( 2 ) = 2 5 − 2 4 + 3 ( 2 2 ) − 3 ( 2 ) − 2 ⇒ f ( 2 ) = 20 Here f ( 1 ) = − 2 is closure to 0 . Thus, x 0 = 1 would be the initial approximation. From Newton’s method we have x n + 1 = x n − f ( x n ) f | ( x n )

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Author: Stewart, JAMES, Day, Troy

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Chapter 3, Problem 87E

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The root of the equation x5−x4+3x2−3x−2=0 in the interval [1,2] correct to six decimal places.

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Chapter 3, Problem 86E

Chapter 3, Problem 88E

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Chapter 3 Solutions

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The root of the equation x 5 − x 4 + 3 x 2 − 3 x − 2 = 0 in the interval [ 1 , 2 ] correct to six decimal places.

Solution Summary: The author explains that the root of the equation x5-X4+32-3x-2=0 is correct to six decimal places.

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The root of the equation x5−x4+3x2−3x−2=0 in the interval [1,2] correct to six decimal places.

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Chapter 3 Solutions