(a) Explanation Given: Consider the concentration of a drug in the bloodstream t hours after the drug is administered, K ( t ) = 4 t 3 t 2 + 27 for t ≥ 0 . Definition used: Critical point: A point whose x- coordinate is the critical number c such that either f ′ ( c ) = 0 or f ′ ( c ) does not exists and y -coordinate is f ( c ) . Calculation: Obtain the derivative of the function as follows. K ′ ( t ) = ( 3 t 2 + 27 ) ( 4 t ) ′ − ( 4 t ) ( 3 t 2 + 27 ) ′ ( 3 t 2 + 27 ) 2 = 4 ( 3 t 2 + 27 ) − 4 t ( 3 ( 2 t ) + 0 ) ( 3 t 2 + 27 ) 2 = 12 t 2 + 108 − 24 t 2 ( 3 t 2 + 27 ) 2 = 108 − 12 t 2 ( 3 t 2 + 27 ) 2 Assume c as a critical number such that K ′ ( c ) = 0 or K ′ ( c ) does not exist and obtain its value. 108 − 12 c 2 ( 3 c 2 + 27 ) 2 = 0 There is no critical number such that the derivative does not exist. Thus, only 108 − 12 c 2 = 0 such that c = ± 3 . Thus, the values of c on computation are − 3 and 3. There is only one critical number c that lies in the range t ≥ 0 which is c = 3 (b) To determine To find: The intervals where the function K ( t ) decreases.

Finite Mathematics and Calculus with Applications (10th Edition)

10th Edition

ISBN: 9780321979407

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 13.1, Problem 52E

(a)

To determine

To find: The intervals where the function K(t) increases.

(b)

To determine

To find: The intervals where the function K(t) decreases.

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Chapter 13.1, Problem 51E

Chapter 13.1, Problem 53E

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Finite Mathematics and Calculus with Applications (10th Edition)

Ch. 13.1 - Find where the function is increasing and...Ch. 13.1 - Prob. 2YT Ch. 13.1 - Prob. 3YT Ch. 13.1 - Prob. 4YT Ch. 13.1 - Prob. 1WE Ch. 13.1 - Prob. 2WE Ch. 13.1 - Prob. 3WE Ch. 13.1 - Prob. 4WE Ch. 13.1 - Prob. 5WE Ch. 13.1 - Prob. 6WE

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The intervals where the function K ( t ) increases.

Solution Summary: The author explains how the function K(t) increases in the interval 'left'.