Explanation Given: The function f ( x ) = sin ( 2 x ) . Approach: A function f with derivatives f ′ and f ″ existing at all points in an interval ( a , b ) is concave upward on ( a , b ) if f ″ > 0 for all x in ( a , b ) and concave downward on ( a , b ) f ″ < 0 for all x in ( a , b ) . The points at which f ″ is 0 or doesn’t exist are the inflection points. Calculation: f ( x ) = sin ( 2 x ) f ′ ( x ) = 2 cos ( 2 x ) f ″ ( x ) = − 4 sin ( 2 x ) Solve f ″ ( x ) = 0 . f ″ ( x ) = 0 − 4 sin ( 2 x ) = 0 sin ( 2 x ) = 0 x = n π 2 ,

2nd Edition

ISBN: 9780321964038

Author: GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher: Pearson Addison Wesley,

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Chapter 5.3, Problem 55E

To determine

To find:

The open intervals on which the function f(x)=sin(2x) is concave upward or concave downward and inflection points, if any.

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Chapter 5.3, Problem 54E

Chapter 5.3, Problem 56E

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The open intervals on which the function f ( x ) = sin ( 2 x ) is concave upward or concave downward and inflection points, if any.

Solution Summary: The author explains how the function f(x)=mathrmsin

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To find:

The open intervals on which the function f(x)=sin(2x) is concave upward or concave downward and inflection points, if any.

Want to see the full answer?

Chapter 5 Solutions

The open intervals on which the function f ( x ) = sin ( 2 x ) is concave upward or concave downward and inflection points, if any.

Solution Summary: The author explains how the function f(x)=mathrmsin

Concept explainers

Videos

To find: The open intervals on which the function f(x)=sin(2x) is concave upward or concave downward and inflection points, if any.

Want to see the full answer?

Chapter 5 Solutions

To find:

The open intervals on which the function f(x)=sin(2x) is concave upward or concave downward and inflection points, if any.