Explanation Given information: w ( t ) = t 4 − 1 t Calculation: We have: w ( t ) = t 4 − 1 t . f ′ ( t ) = 4 t 3 ⋅ t − ( t 4 − 1 ) t 2 = 3 t 4 + 1 t 2 f ″ ( t ) = 2 ( 3 t 4 − 1 ) t 3 Solve, f ′ ( t ) = 0 . 3 t 4 + 1 t 2 = 0 3 t 4 + 1 = 0 3 t 4 = − 1 t = ± 1 3 4 f ( t ) is not defined at t = 0 , so t = ± 1 3 4 . So combine inflection point is t = 1 3 4 , − 1 3 4 . With domain to get f ″ ( t ) sign intervals: − ∞ < t < − 1 3 4 , − 1 3 4 < t < 0 , 0 < x < 1 3 4 , 1 3 4 < x < ∞

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 4.4, Problem 18E

To determine

To find: the interval on which the function is concave up or down and points of inflection.

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Chapter 4.4, Problem 17E

Chapter 4.4, Problem 19E

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Chapter 4 Solutions

CALCULUS (CLOTH)

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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the interval on which the function is concave up or down and points of inflection.

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To find: the interval on which the function is concave up or down and points of inflection.

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Chapter 4 Solutions