Explanation Given information: Given function is S ( t ) = A t e − k t where A, k > 0 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 : S ( t ) = A t e − k t S ' ( t ) = A t ( e − k t ( − k ) ) + ( A ) e − k t S ' ( t ) = − A k t e − k t + A e − k t S ' ' ( t ) = − A k t ( e − k t ( − k ) ) + ( − A k ) e − k t + A ( e − k t ( − k ) ) S ' ' ( t ) = A k 2 t e − k t − 2 A k e − k t The function is concave up when S ' ' ( t ) > 0 and concave down when S ' ' ( t ) < 0 The domain of the function can be written as the intervals: − ∞ < t < 2 k , 2 k < t < ∞ (As k is a positive number) Since the second derivative has zeroes at t = 2 k . Checking the value of the second derivative in each of the intervals: For − ∞ < t < 2 k , let t = 0 S ' ' ( t ) = A k 2 t e − k t − 2 A k e − k t S ' ' ( 0 ) = A k 2 ( 0 ) e − k ( 0 ) − 2 A k e − k ( 0 ) S ' ' ( 0 ) = 0 − 2 A k ( 1 ) S ' ' ( 0 ) = − 2 A k Hence the function is concave down in the interval − ∞ < t < 2 k

Loose-leaf Version For Calculus: Early Transcendentals

4th Edition

ISBN: 9781319055912

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Chapter 4.4, Problem 24E

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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 4.4, Problem 23E

Chapter 4.4, Problem 25E

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Chapter 4 Solutions

Loose-leaf Version For Calculus: Early Transcendentals

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

Solution Summary: The author calculates 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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