Explanation Test used: A function f with its derivatives f ′ and f ′ ′ exists at all points in an interval ( a , b ) . Then f is concave upward on ( a , b ) if f ′ ′ ( x ) > 0 for all x in ( a , b ) and concave downward on ( a , b ) if f ′ ′ ( x ) < 0 for all x . Definition used: A function f has an inflection point, if the second derivative is 0 or does not exist. Calculation: The given function is f ( x ) = sin ( 2 x ) . Obtain the first derivative f ′ ( x ) as follows. f ′ ( x ) = d d x ( sin ( 2 x ) ) = cos ( 2 x ) ⋅ 2 = 2 cos ( 2 x ) That is, the first derivative f ′ ( x ) is 2 cos ( 2 x ) . Obtain the second derivative f ′ ′ ( x ) as follows. f ′ ′ ( x ) = d d x ( f ′ ( x ) ) = d d x ( 2 cos ( 2 x ) ) = 2 ( − sin ( 2 x ) 2 ) = − 4 sin ( 2 x ) That is, f ′ ′ ( x ) = − 4 sin ( 2 x ) . Here, the second derivative is less than zero, the function sin ( 2 x ) must satisfy sin ( 2 x ) > 0 for 0 < x < π 2 , π < x < 3 π 2 , and so on. The second derivative is greater than zero, the function sin ( 2 x ) must satisfy sin ( 2 x ) < 0 for π 2 < x < π , 3 π 2 < x < 2 π , and so on

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ISBN: 2818440028625

Author: Lial

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Chapter 13.2, Problem 45E

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The open intervals where the functions are concave upward or concave downward, and the inflection points.

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Chapter 13.2, Problem 44E

Chapter 13.2, Problem 46E

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