Explanation It is known that, sin x < x ∀ x 1 sin x > 1 x 1 sin x − 1 x > 0 f ( x ) = 1 sin x − 1 x : 0 < x ≤ π 2 is positive Now, f ( x ) = 1 sin x − 1 x : 0 < x ≤ π 2 f ' ( x ) = 1 x 2 − cos x sin 2 x It is

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 4.3, Problem 53PS

To determine

The proof that f(x)=1sinx−1x : 0<x≤π2 is positive and increasing

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Chapter 4.3, Problem 52PS

Chapter 4.3, Problem 54PS

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Chapter 4 Solutions

Calculus

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The proof that f ( x ) = 1 sin x − 1 x : 0 < x ≤ π 2 is positive and increasing

Solution Summary: The author explains the proof that f(x)=1mathrmsinx forallarrayl

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The proof that f(x)=1sinx−1x : 0<x≤π2 is positive and increasing

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The proof that f ( x ) = 1 sin x − 1 x : 0 &lt; x ≤ π 2 is positive and increasing

Solution Summary: The author explains the proof that f(x)=1mathrmsinx forallarrayl

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The proof that f(x)=1sinx−1x : 0<x≤π2 is positive and increasing

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Chapter 4 Solutions

The proof that f ( x ) = 1 sin x − 1 x : 0 < x ≤ π 2 is positive and increasing