Explanation Given information: Given function is f ( x ) = x 3 1 + x . Calculation: A function f ( x ) is said to be concave up when f ' ' ( x ) > 0 and is said to be concave down when f ' ' ( x ) < 0 Applying derivative to the given function twice with respect to x : f ( x ) = x 3 1 + x f ' ( x ) = ( 1 + x ) ( 3 x 2 ) − ( x 3 ) ( 0 + 1 ) ( 1 + x ) 2 f ' ( x ) = 3 x 2 + 3 x 3 − x 3 ( 1 + x ) 2 f ' ( x ) = 3 x 2 + 2 x 3 ( 1 + x ) 2 f ' ' ( x ) = ( 1 + x ) 2 ( 3 ( 2 x ) + 2 ( 3 x 2 ) ) − ( 3 x 2 + 2 x 3 ) ( 2 ( 1 + x ) ) ( ( 1 + x ) 2 ) 2 f ' ' ( x ) = ( 1 + x ) 2 ( 6 x + 6 x 2 ) − ( 3 x 2 + 2 x 3 ) ( 2 ( 1 + x ) ) ( 1 + x ) 4 f ' ' ( x ) = ( 1 + x ) [ ( 1 + x ) 2 ( 6 x ) − ( 6 x 2 + 4 x 3 ) ] ( 1 + x ) 4 f ' ' ( x ) = 6 x + 12 x 2 + 6 x 3 − 6 x 2 − 4 x 3 ( 1 + x ) 3 f ' ' ( x ) = 2 x 3 + 6 x 2 + 6 x ( 1 + x ) 3 f ' ' ( x ) = 2 x ( x 2 + 3 x + 3 ) ( 1 + x ) 3 The function is concave up when f ' ' ( x ) > 0 and concave down when f ' ' ( x ) < 0 The domain of the function can be written as the intervals: − ∞ < x < − 1 , − 1 < x < 0 , 0 < x < ∞ Since the function is not defined at x = -1 and the function has zero at x = 0

ACHIEVE FOR CALCULUS 1 TERM >CSI CUSTOM<

4th Edition

ISBN: 9781319411657

Author: Rogawski

Publisher: MAC HIGHER

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Concept explainers

Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

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Question

Chapter 4.4, Problem 17E

To determine

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

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Chapter 4.4, Problem 16E

Chapter 4.4, Problem 18E

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Chapter 4 Solutions

ACHIEVE FOR CALCULUS 1 TERM >CSI CUSTOM<

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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Prob. 63CRE Ch. 4 - Prob. 64CRE Ch. 4 - Prob. 65CRE Ch. 4 - Prob. 66CRE Ch. 4 - Prob. 67CRE Ch. 4 - Prob. 68CRE Ch. 4 - Prob. 69CRE Ch. 4 - Prob. 70CRE Ch. 4 - Prob. 71CRE Ch. 4 - Prob. 72CRE Ch. 4 - Prob. 73CRE Ch. 4 - Prob. 74CRE Ch. 4 - Prob. 75CRE Ch. 4 - Prob. 76CRE Ch. 4 - Prob. 77CRE Ch. 4 - Prob. 78CRE Ch. 4 - Prob. 79CRE Ch. 4 - Prob. 80CRE Ch. 4 - Prob. 81CRE Ch. 4 - Prob. 82CRE Ch. 4 - Prob. 83CRE Ch. 4 - Prob. 84CRE Ch. 4 - Prob. 85CRE Ch. 4 - Prob. 86CRE Ch. 4 - Prob. 87CRE Ch. 4 - Prob. 88CRE Ch. 4 - Prob. 89CRE

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

Solution Summary: The author explains how to find the intervals on which the function is concave up or down and the points of inflection.

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

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Chapter 4 Solutions