Explanation Given: The function is f ( x ) = x x 2 + 27 . 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 ) = x x 2 + 27 . First, find the derivative. f ′ ( x ) = ( x 2 + 27 ) d d x x − x d d x ( x 2 + 27 ) ( x 2 + 27 ) 2 = ( x 2 + 27 ) − x ( 2 x ) ( x 2 + 27 ) 2 = x 2 + 27 − 2 x 2 ( x 2 + 27 ) 2 = 27 − x 2 ( x 2 + 27 ) 2 Now, find the second derivative. f ″ ( x ) = ( x 2 + 27 ) 2 d d x ( 27 − x 2 ) − ( 27 − x 2 ) d d x ( x 2 + 27 ) 2 ( ( x 2 + 27 ) 2 ) 2 = ( x 2 + 27 ) 2 ( − 2 x ) − ( 27 − x 2 ) 2 ( x 2 + 27 ) ( 2 x ) ( ( x 2 + 27 ) 2 ) 2 = ( x 2 + 27 ) [ ( x 2 + 27 ) ( − 2 x ) − ( 27 − x 2 ) 4 x ] ( ( x 2 + 27 ) 2 ) 2 = [ − 2 x 3 − 54 x − 108 x + 4 x 3 ] ( x 2 + 27 ) 3 = 2 x 3 − 162 x ( x 2 + 27 ) 3 Find the value of x by substitute f ″ ( x ) = 0 . 2 x 3 − 162 x ( x 2 + 27 ) 3 = 0 x ( 2 x 2 − 162 ) = 0 That is, x = 0 or ( 2 x 2 − 162 ) = 0 x = 0 or 2 x 2 = 162 x = 0 or x 2 = 81 x = 0 or x = ± 9 Thus, the zeroes of the f ″ ( x ) are x = 0 , ± 9 . This divides the x − axis into four intervals: ( − ∞ , − 9 ) , ( − 9 , 0 ) , ( 0 , 9 ) , and ( 9 , ∞ ) . Interval Point Function Sign of f ″ Behavior of f ( x ) ( − ∞ , − 9 ) − 10 2 x 3 − 162 x ( x 2 + 27 ) 3 = 2 ( − 10 ) 3 − 162 ( − 10 ) ( ( − 10 ) 2 + 27 ) 3 = − 2000 + 1620 ( 127 ) 3 = − 380 ( 127 ) 3 − v e Concave down ( − 9 , 0 ) − 1 2 x 3 − 162 x ( x 2 + 27 ) 3 = 2 ( − 1 ) 3 − 162 ( − 1 ) ( ( − 1 ) 2 + 27 ) 3 = − 2 + 162 ( 28 ) 3 = 160 ( 28 ) 3 + v e Concave up ( 0 , 9 ) 1 2 x 3 − 162 x ( <

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.4, Problem 45E

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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 44E

Chapter 4.4, Problem 46E

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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