Explanation Given: Suppose f and g are functions. If f ( 0 ) = g ( 0 ) and f ′ ( x ) ≤ g ′ ( x ) for x ≥ 0 , then f ( x ) ≤ g ( x ) for all x ≥ 0 . Calculation: Assume the function f ( x ) as shown below. f ( x ) = x Assume the function g ( x ) as shown below. g ( x ) = tan x Substitute 0 for x in equation f ( x ) = x as shown below. f ( 0 ) = 0 Substitute 0 for x in equation g ( x ) = tan x as shown below. g ( 0 ) = tan 0 = 0 Thus, the equality f ( 0 ) = g ( 0 ) = 0 holds true. Differentiate the function f ( x ) with respect to x as shown below. f ′ ( x ) = d ( x ) d x = 1 Differentiate the function g ( x ) with respect to x as shown below. g ′ ( x ) = d ( tan x ) d x = sec 2 x Here, 1 ≤ sec 2 x for 0 ≤ x < π 2 . For 0 ≤ x < π 2 , the inequality f ′ ( x ) ≤ g ′ ( x ) holds true. Thus, by the above result both the conditions f ( 0 ) = g ( 0 ) and f ′ ( x ) ≤ g ′ ( x ) for 0 ≤ x < π 2 holds true

4th Edition

ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Question

Chapter 4.3, Problem 74E

To determine

To show:

The inequality x≤tanx for 0≤x<π2 and sinx≤x for x≥0 is true.

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Chapter 4.3, Problem 73E

Chapter 4.3, Problem 75E

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Chapter 4 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

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The inequality x ≤ tan x for 0 ≤ x < π 2 and sin x ≤ x for x ≥ 0 is true.

Solution Summary: The author explains how the inequality xle mathrmtanx is true.

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The inequality x≤tanx for 0≤x<π2 and sinx≤x for x≥0 is true.

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The inequality x ≤ tan x for 0 ≤ x &lt; π 2 and sin x ≤ x for x ≥ 0 is true.

Solution Summary: The author explains how the inequality xle mathrmtanx is true.

Concept explainers

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To show: The inequality x≤tanx for 0≤x<π2 and sinx≤x for x≥0 is true.

Want to see the full answer?

Chapter 4 Solutions

The inequality x ≤ tan x for 0 ≤ x < π 2 and sin x ≤ x for x ≥ 0 is true.

To show:

The inequality x≤tanx for 0≤x<π2 and sinx≤x for x≥0 is true.