Explanation Given: f ( x ) = x cos x , 0 ≤ x ≤ π , Choose initial approximation x 1 = π 4 . Result used : (Newton’s method). x n + 1 = x n − f ′ ( x n ) f ′ ′ ( x n ) Where f ( x ) is function defined above. Calculation: The given equation is f ( x ) = x cos x , 0 ≤ x ≤ π . Calculate the derivative of the above function f ( x ) . Let the derivative of f ( x ) is f ′ ( x ) = d f ( x ) d x . f ′ ( x ) = d ( x cos ( x ) ) d x = cos ( x ) d ( x ) d x + x d ( cos ( x ) ) d x = cos ( x ) − x sin ( x ) To find the absolute maximum value at first start with critical point. For that take f ′ ( x ) = 0 and solve for critical point. From the above problem f ′ ( x ) exist for all x . Calculate the second derivative of the f ( x ) . Let f ′ ′ ( x ) = d f ′ ( x ) d x ,Where f ′ ( x ) is function mentioned above. f ′ ′ ( x ) = d ( cos ( x ) − x sin ( x ) ) d x = d ( cos ( x ) ) d x − x d ( sin ( x ) ) d x − sin ( x ) d x d = − sin ( x ) − x cos ( x ) − sin ( x ) = − 2 sin ( x ) − x cos ( x ) The value of f ′ ( x ) at the point x 1 = π 4 is, f ′ ( π 4 ) = cos ( π 4 ) − π 4 sin ( π 4 ) = 1 2 − π 4 × 1 2 = 0.707106 − 0.785398 × 0.707106 = 0.707106 − 0.555359 = 0.151746 Calculate the value of f ′ ′ ( x 1 ) at the point x 1 = π 4 . f ′ ′ ( π 4 ) = − 2 sin ( π 4 ) − π 4 cos ( π 4 ) = − 2 × 1 2 − π 4 × 1 2 = − 2 × 0.707106 − 0.785398 × 0.707106 = − 1.414212 − 0.555359 Further simplification of the above problem is, − 1.414212 − 0.555359 = − 1.969571 Calculate the value of x 2 by using above result and set n = 1 to obtain x 2 = x 1 − f ′ ( x 1 ) f ′ ′ ( x 1 ) Substitute, x 1 = 0.785398 , f ′ ( x 1 ) = 0.151746 and f ′ ′ ( x 1 ) = − 1.969571 . x 2 = 0.785398 − 0.151746 ( − 1.969571 ) = 0.785398 + 0.151746 1.969571 = 0.785398 + 0

Essential Calculus: Early Transcendentals

2nd Edition

ISBN: 9781133112280

Author: James Stewart

Publisher: Cengage Learning

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Chapter 4.6, Problem 28E

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To Find: The absolute maximum value of the given function correct to six decimal places.by Newton’s method.

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Chapter 4.6, Problem 27E

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Essential Calculus: Early Transcendentals

Ch. 4.1 - Explain the difference between an absolute minimum...Ch. 4.1 - Suppose f is a continuous function defined on a...Ch. 4.1 - For each of the numbers a, b, c, d, r, and s,...Ch. 4.1 - For each of the numbers a, b, c, d, r, and s,...Ch. 4.1 - Use the graph to state the absolute and local...Ch. 4.1 - Use the graph to state the absolute and local...Ch. 4.1 - Sketch the graph of a function f that is...Ch. 4.1 - 710 Sketch the graph of a function f that is...Ch. 4.1 - 710 Sketch the graph of a function f that is...Ch. 4.1 - 710 Sketch the graph of a function f that is...

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The absolute maximum value of the given function correct to six decimal places.by Newton’s method.

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