Explanation Given: The function is f ( x ) = 2 x 4 − 3 x 2 + 2 . 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 ) = 2 x 4 − 3 x 2 + 2 . First, find the derivative. f ′ ( x ) = 2 ( 4 ) x 4 − 1 − 3 ( 2 x 2 − 1 ) = 8 x 3 − 6 x Now, find the second derivative. f ″ ( x ) = 8 ( 3 x 3 − 1 ) − 6 = 24 x 2 − 6 Find the value of x by substitute f ″ ( x ) = 0 . 24 x 2 − 6 = 0 24 x 2 = 6 x 2 = 6 24 x = 1 4 x = ± 1 2 Thus, the zero of the f ″ ( x ) are x = ± 1 2 . This divides the x − axis into two intervals: ( − ∞ , − 1 2 ) , ( − 1 2 , 1 2 ) , and ( 1 2 , ∞ ) . Interval Point Function Sign of f ″ Behavior of f ( x ) ( − ∞ , − 1 2 ) − 1 24 x 2 − 6 = 24 ( − 1 ) 2 − 6 = 24 − 6 = 18 + v e Concave up ( − 1 2 , 1 2 ) 0 24 x 2 − 6 = 24 ( 0 ) 2 − 6 = 0 − 6 = − 6 − v e Concave down ( 1 2 , ∞ ) 1 24 x 2 − 6 = 24 ( 1 ) 2 − 6 = 24 − 6 = 18 + v e Concave up Thus, f is concave up for interval ( − ∞ , − 1 2 ) and ( 1 2 , ∞ ) and f is concave down for interval ( − 1 2 , 1 2 ) . When c = ± 1 2 is the point of inflection then f ″ ( x ) changes sign at ± 1 2 . Now, find the critical points by substitute f ′ ( x ) = 0 . 8 x 3 − 6 x = 0 x ( 8 x 2 − 6 ) = 0 That is, x = 0 or ( 8 x 2 − 6 ) = 0 x = 0 or 8 x 2 = 6 x = 0 or x 2 = 6 8 x = 0 or x = ± 3 2 Thus, the critical points are 0 , 3 2 and − 3 2 . This divides the x − axis into three intervals: ( − ∞ , − 3 2 ) , ( − 3 2 , 0 ) , ( 0 , 3 2 ) , and ( 3 2 , ∞ )

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.4, Problem 42E

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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 41E

Chapter 4.4, Problem 43E

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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