Explanation Given: y = x x 3 + 1 Formula Used: Calculation: We know that the inflection points are the points at which the sign of the second derivatives of the function changes. Here, the given function is y = x x 3 + 1 Differentiating the above function, we get d y d x = − 2 x 3 + 1 ( x 3 + 1 ) 2 To find out the critical points, we equate the above equation to zero. i.e. − 2 x 3 + 1 ( x 3 + 1 ) 2 = 0 On solving this, we get x = 2 3 and x = − 1 Therefore, the critical points of the given functionoccur at x = 2 3 and x = − 1 . Now we second derivative to find the inflection points. d 2 y d x 2 = − 6 x 2 ( − x 3 + 2 ) ( x 3 + 1 ) 3 To find out the inflection points, we equate the above equation to zero

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ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4, Problem 49CRE

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To find the points of inflection for the function, y=xx3+1.

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Chapter 4, Problem 48CRE

Chapter 4, Problem 50CRE

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Chapter 4 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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To find the points of inflection for the function, y = x x 3 + 1 .

Solution Summary: The author explains that the inflection points are the points at which the sign of the second derivatives of a function changes.

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Chapter 4 Solutions