Explanation Given: The function f ( x ) = cos x + x 2 . And its graph for the period [ 0 , π ] : Key concepts used: In order to decide the sign of the first derivative, we look where the graph is increasing and where it is decreasing. The sign of f ' is positive when graph is increasing. The sign of f ' is negative when graph is decreasing. In order to decide the sign of the double derivative, we look where the graph is concave up and where it is concave down. The sign of f ' ' is positive when graph is concave up. The sign of f ' ' is negative when graph is concave down. Maximum of a function occurs when derivative changes its sign from positive to negative. Minimum of a function occurs when derivative changes its sign from negative to positive. Calculation: Let us first find critical values of the given function. For that, first we find the derivative of the function and set it equal to zero: ⇒ f ( x ) = cos x + x 2 ⇒ f ' ( x ) = − sin x + 1 2 ⇒ f ' ( x ) = 0 ⇒ − sin x + 1 2 = 0 ⇒ sin x = 1 2

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.5, Problem 13E

To determine

To graph:

The given function on the period [0,5π]

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Chapter 4.5, Problem 12E

Chapter 4.5, Problem 14E

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Chapter 4 Solutions

CALCULUS (CLOTH)

Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E

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To graph: The given function on the period [ 0 , 5 π ]

Solution Summary: The author explains that the graph for the period [0,5pi] is the function f(x)=mathrmcosx+

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To graph:

The given function on the period [0,5π]

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Chapter 4 Solutions

To graph: The given function on the period [ 0 , 5 π ]

Solution Summary: The author explains that the graph for the period [0,5pi] is the function f(x)=mathrmcosx+

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To graph: The given function on the period [0,5π]

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Chapter 4 Solutions

To graph:

The given function on the period [0,5π]