Explanation Given: The function is f ( x ) = sin 2 x + cos x , [ 0 , π ] . Formula used: To find the critical points substitute f ′ ( x ) = 0 . Second derivative test: If f ′ ( c ) = 0 and f ″ ( c ) exists, then f ( c ) is a local maximum if f ″ ( c ) < 0 . f ( c ) is a local minimum if f ″ ( c ) > 0 . The test fails if f ″ ( c ) = 0 . When test fails then use the first derivative test. Calculation: The function is f ( x ) = sin 2 x + cos x . We have to find the critical points and apply the second derivative test. First, find the derivative. f ′ ( x ) = 2 sin x cos x − sin x = sin 2 x − sin x Now, find the second derivative. f ″ ( x ) = 2 cos 2 x − cos x Now, find the critical points by substitute f ′ ( x ) = 0 . sin 2 x − sin x = 0 2 sin x cos x − sin x = 0 sin x ( 2 cos x − 1 ) = 0 That is, sin x = 0 or 2 cos x − 1 = 0 sin x = sin 0 or sin x = sin π or cos x = 1 2 sin x = sin 0 or sin x = sin π or cos x = cos π 3 x = 0 or x = π or x = π 3 Thus, the critical points are 0 , π and π 3 . Now, find the sign of the second derivative at the critical points. f ″ ( 0 ) = 2 cos 2 ( 0 ) − cos ( 0 ) = 2 − 1 = 1 Thus, f ″ ( 0 ) = 1 > 0

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 4.4, Problem 31E

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Chapter 4.4, Problem 30E

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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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