Explanation Given information:- The given derivative of f ( x ) = x 3 − x x 2 + 1 . Calculations:- The given derivative of f ( x ) = x 3 − x x 2 + 1 . Find the inflection point by using the first and second derivative of the given equation as shown below: f ( x ) = x 3 − x x 2 + 1 [The given derivative of the equation] y ' = d d x [ x 3 − x x 2 + 1 ] [Differentiate with respect to x ] Apply the quotient rule [ u ( x ) v ( x ) ] ' = u ' ( x ) ⋅ v ( x ) − u ( x ) ⋅ v ' ( x ) v ( x ) 2 ] to simplify y ' = ( d d x [ x 3 − x ] ⋅ ( x 2 + 1 ) − ( x 3 − x ) d d x [ x 2 + 1 ] ) ( x 2 + 1 ) 2 y ' = ( 3 x 2 − 1 ) ( x 2 + 1 ) − ( 2 x + 0 ) ( x 3 − x ) ( x 2 + 1 ) 2 [ ∴ , d d x [ x 3 ] = 3 x , d d x [ x ] = 1 ] y ' = x 4 + 4 x 2 − 1 ( x 2 + 1 ) 2 Find the second derivative by using first derivative are as shown below: y ' = x 4 + 4 x 2 − 1 ( x 2 + 1 ) 2 [Find the second derivative] y ' ' = d d x [ x 4 + 4 x 2 − 1 ( x 2 + 1 ) 2 ] [Differentiate by the quotient rule [ u ( x ) v ( x ) ] ' = u ' ( x ) ⋅ v ( x ) − u ( x ) ⋅ v '

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4, Problem 43CRE

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The solution of point of inflection.

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Chapter 4, Problem 42CRE

Chapter 4, Problem 44CRE

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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 solution of point of inflection.

Solution Summary: The author calculates the inflection point by using the first and second derivatives of the given equation as shown below.

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Chapter 4 Solutions