Explanation Given information: Given function is f ( t ) = t e − t 2 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 t : f ( t ) = t e − t 2 f ' ( t ) = t ( e − t 2 ( − 2 t ) ) + ( 1 ) e − t 2 f ' ( t ) = − 2 t 2 e − t 2 + e − t 2 f ' ' ( t ) = − 2 t 2 ( e − t 2 ( − 2 t ) ) + ( − 2 ( 2 t ) ) e − t 2 + ( e − t 2 ( − 2 t ) ) f ' ' ( t ) = 4 t 3 e − t 2 − 4 t e − t 2 − 2 t e − t 2 f ' ' ( t ) = 4 t 3 e − t 2 − 6 t e − t 2 The function is concave up when f ' ' ( t ) > 0 and concave down when f ' ' ( t ) < 0 The domain of the function can be written as the intervals: − ∞ < t < − 3 2 , − 3 2 < t < 0 , 0 < t < 3 2 , 3 2 < t < ∞ Since the second derivative has zeroes at t = − 3 2 , 0 , 3 2 . Checking the value of the second derivative in each of the intervals: For − ∞ < t < − 3 2 , let t = − 2 f ' ' ( t ) = 4 t 3 e − t 2 − 6 t e − t 2 f ' ' ( − 2 ) = 4 ( − 2 ) 3 e − ( − 2 ) 2 − 6 ( − 2 ) e − ( − 2 ) 2 f ' ' ( − 2 ) = − 32 e − 4 + 12 e − 4 f ' ' ( − 2 ) = − 20 e − 4 Hence the function is concave down in the interval − ∞ < t < − 3 2 . For − 3 2 < t < 0 , let t = − 1 f ' ' ( t ) = 4 t 3 e − t 2 − 6 t e − t 2 f ' ' ( − 1 ) = 4 ( − 1 ) 3 e − ( − 1 ) 2 − 6 ( − 1 ) e − ( − 1 ) 2 f ' ' ( − 1 ) = − 4 e − 1 + 6 e − 1 f ' ' ( − 1 ) = 2 e − 1 Hence the function is concave up in the interval − 3 2 < t < 0

4th Edition

ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Question

Chapter 4.4, Problem 23E

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 22E

Chapter 4.4, Problem 24E

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Chapter 4 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

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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