Explanation Given: The function is f ( x ) = cos 2 x , [ 0 , π ] . Formula used: Test for concavity: Let f ″ ( x ) exist for all x ∈ ( a , b ) , then · y = f ″ ( x ) > 0 for all x ∈ ( a , b ) , here the function f is concave upward on ( a , b ) . · y = f ″ ( x ) < 0 for all x ∈ ( a , b ) , here the function f is concave downward on ( a , b ) . y = f ( x ) has a point of inflection if f ″ ( x ) = 0 . 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 ) = cos 2 x , [ 0 , π ] . First, find the derivative. f ′ ( x ) = − 2 cos x sin x = − sin 2 x Now, find the second derivative. f ″ ( x ) = − 2 cos 2 x Find the value of x by substitute f ″ ( x ) = 0 . − 2 cos 2 x = 0 cos 2 x = 0 cos 2 x = cos ( 2 × π 4 ) or cos 2 x = cos ( 2 × 3 π 4 ) cos 2 x = cos π 2 or cos 2 x = cos 3 π 2 x = π 4 or x = 3 π 4 Thus, the zeroes of the f ″ ( x ) are x = π 4 , 3 π 4 . This divides the x − axis into three intervals: ( 0 , π 4 ) , ( π 4 , 3 π 4 ) , and ( 3 π 4 , π ) . Interval Point Function Sign of f ″ Behavior of f ( x ) ( 0 , π 4 ) π 6 − 2 cos 2 x = − 2 cos ( 2 × π 6 ) = − 2 cos π 3 = − 2 × 1 2 = − 1 − v e Concave down ( π 4 , 3 π 4 ) π 2 − 2 cos 2 x = − 2 cos ( 2 × π 2 ) = − 2 cos π = − 2 × ( − 1 ) = 2 +

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.4, Problem 50E

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Find the intervals on which f is concave up or down, the points of inflection, the critical points and the local minima and maxima.

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Chapter 4.4, Problem 49E

Chapter 4.4, Problem 51E

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Chapter 4 Solutions

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Find the intervals on which f is concave up or down, the points of inflection, the critical points and the local minima and maxima .

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Find the intervals on which f is concave up or down, the points of inflection, the critical points and the local minima and maxima.

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Chapter 4 Solutions