Explanation Given information: The function M ( x ) = ( x + 2 ) ln ( x ) and domain 1 ≤ x ≤ 3 . Graph: Consider the provided function M ( x ) = ( x + 2 ) ln ( x ) . Evaluate the first derivative of the function.Apply the product rule, d d x ( f ( x ) g ( x ) ) = g ( x ) d d x f ( x ) + f ( x ) d d x g ( x ) . d d x M ( x ) = d d x ( ( x + 2 ) ln ( x ) ) = ( x + 2 ) d d x ( ln ( x ) ) + ln ( x ) d d x ( x + 2 ) = ( x + 2 ) 1 x + 1 ⋅ ln ( x ) = x + 2 x + ln ( x ) Recall if first derivative of the function is greater than zero that is g ' ( x ) > 0 for every x in ( c , d ) , then the function g is increasing on interval [ c , d ] . If first derivative of the function is less than zero that is g ' ( x ) < 0 for every x in ( c , d ) , then the function g is decreasing on interval [ c , d ] . The first derivative contains the term ln ( x ) which only expect values of x greater than or equal to 1. So, the entire expression L ' ( x ) = x + 2 x + ln ( x ) is positive for x > 1 . Therefore, the function is increasing when x > 1 . Evaluate the second derivative of the function, differentiate first derivative again with respect to x . Apply the quotient rule of differentiation, d d x ( f ( x ) g ( x ) ) = g ( x ) d d x f ( x ) − f ( x ) d d x g ( x ) [ g ( x ) ] 2

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists

3rd Edition

ISBN: 9780840064189

Author: Frederick R. Adler

Publisher: Cengage Learning

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Chapter 2.8, Problem 24E

To determine

To graph:The function M(x)=(x+2)ln(x) for 1≤x≤3 with help of first and second derivative of the function and identify the regions in which function is increasing and concave up.

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Chapter 2.8, Problem 23E

Chapter 2.8, Problem 25E

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Chapter 2 Solutions

Modeling the Dynamics of Life: Calculus and Probability for Life Scientists

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E

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To graph: The function M ( x ) = ( x + 2 ) ln ( x ) for 1 ≤ x ≤ 3 with help of first and second derivative of the function and identify the regions in which function is increasing and concave up.

Solution Summary: The author analyzes the function M(x) for 1le x-le 3 with help of first and second derivatives.

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To graph:The function M(x)=(x+2)ln(x) for 1≤x≤3 with help of first and second derivative of the function and identify the regions in which function is increasing and concave up.

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Chapter 2 Solutions