Explanation Given: The equation the curve is f ( x ) = − sin x . The infinite number of the line that are tangent to the curve ad passing through the origin. Result Used: ( Newton’s method). (a) x n + 1 = x n − g ( x n ) g ′ ( x n ) (b) The equation of the line between two point ( x 1 , y 1 ) and ( x 2 , y 2 ) is y − y 1 x − x 1 = y 2 − y 1 x 2 − x 1 . Calculation: Calculate the derivative of f ( x ) . Let f ′ ( x ) = d f ( x ) d x . f ′ ( x ) = d ( − sin x ) d x = − cos x The slope of the tangent line at the arbitrary point x = a is, f ′ ( a ) = − cos a . From result (b) with two points ( a , f ( a ) ) and ( 0 , 0 ) is, y − 0 x − 0 = − sin a − 0 a − 0 y = − sin a a x If the above line is tangent to the curve f ( x ) = − sin x at the point x = a then its slope is f ′ ( a ) . f ′ ( a ) = − cos a − sin a a = − cos a tan a = a To find the root of the above equation, let g ( x ) = tan x − x . Calculate the derivative of the above g ( x ) . Let g ′ ( x ) = d g ( x ) d x , g ′ ( x ) = d ( tan x − x ) d x = sec 2 ( x ) − 1 The value of g ( x 1 ) at the point x 1 = 4.5 . g ( 4.5 ) = tan ( 4.5 ) − 4.5 = 4.6373320 − 4.5 = 0.137332 The value of g ′ ( x 1 ) at the point x 1 = 4.5 . g ′ ( 4.5 ) = sec 2 ( 4.5 ) − 1 = 22.504848 − 1 = 21.504848 Set n = 1 in result (a) to obtain x 2 . x 2 = x 1 − g ( x 1 ) g ′ ( x 1 ) Substitute x 1 = 4.5 , g ( x 1 ) = 0

Essential Calculus: Early Transcendentals

2nd Edition

ISBN: 9781133112280

Author: James Stewart

Publisher: Cengage Learning

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Chapter 4.6, Problem 30E

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To Find: The largest slope of the tangent line correct to the six decimal places by Newton’s method

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Chapter 4.6, Problem 29E

Chapter 4.6, Problem 31E

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Essential Calculus: Early Transcendentals

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The largest slope of the tangent line correct to the six decimal places by Newton’s method

Solution Summary: The author calculates the largest slope of the tangent line correct to the six decimal places by Newton's method.

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To Find: The largest slope of the tangent line correct to the six decimal places by Newton’s method

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