(a) Explanation Given: Consider the approximate blood alcohol concentration function, A ( x ) = 0.003631 x 3 − 0.03746 x 2 + 0.1012 x + 0.009 where 0 ≤ x ≤ 5 is the number of hours after a lady weighing 170-lb drank 2 oz of alcohol on an empty stomach. 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. A ′ ( x ) = 0.003631 ( 3 x 2 ) − 0.03746 ( 2 x ) + 0.1012 ( 1 ) + 0 = 0.010893 t 2 − 0.07492 t + 0.1012 Assume c as a critical number such that A ′ ( c ) = 0 or A ′ ( c ) does not exist and obtain its value. 0.010893 c 2 − 0.07492 c + 0.1012 = 0 Use quadratic formula to find the values of c as follows. c = 0.07492 ± ( − 0.07492 ) 2 − 4 ( 0.010893 ) ( 0.1012 ) 2 × 0.003631 Thus, the values of c on computation are 1.85 and 5.03. There is only one critical number c that lies in the range 0 ≤ x ≤ 5 . Hence, the only critical number c = 1.85 . These points determine the interval ( − ∞ , 0 ) , ( 0 , 1 (b) To determine To find: The intervals where the function A ( x ) decreases.

Finite Mathematics and Calculus with Applications

1st Edition

ISBN: 9781323188361

Author: Margaret Lial

Publisher: Pearson Education

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Chapter 13.1, Problem 51E

(a)

To determine

To find: The intervals where the function A(x) increases.

(b)

To determine

To find: The intervals where the function A(x) decreases.

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Chapter 13.1, Problem 50E

Chapter 13.1, Problem 52E

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Finite Mathematics and Calculus with Applications

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 A ( x ) increases.

Solution Summary: The author explains that the function A(x) increases in the interval 'left'.