Explanation Given information: The given curve is f ( x ) = 18 ( x − 3 ) ( x − 1 ) 2 3 . The given first and second order derivative of the curve are below: f ' ( x ) = 30 ( x − 9 5 ) ( x − 1 ) 1 3 .......... ( 1 ) f ' ' ( x ) = 20 ( x − 3 5 ) ( x − 1 ) 4 3 ......... ( 2 ) Calculation: In order to find the behavior of the curve, it is required to find the critical points, strictly increasing and strictly decreasing interval of the function, Maxima and Minima if exists. For critical points, substitute first order derivative to be equal to zero. Therefore from (1), f ' ( x ) = 0 30 ( x − 9 5 ) ( x − 1 ) 1 3 = 0 30 ( x − 9 5 ) = 0 x − 9 5 = 0 x = 9 5 x = 1.8 x = 1.8 is the critical point of thecurve. For x < 1 , consider the point x = 0 and check the value of first order derivative at that point. For the interval ( − ∞ , 1 ) ( f ' ( x ) ) x = 0 = 30 ( 0 − 9 5 ) ( 0 − 1 ) 1 3 ( f ' ( x ) ) x = 0 = 30 ( − 1.8 ) ( − 1 ) 1 3 ( f ' ( x ) ) x = 0 = − 5.4 ( − 1 ) ( f ' ( x ) ) x = 0 = 5.4 > 0 Since, the value of first order derivative at x = 0 is positive therefore, the behavior of curve is strictly increasing for the interval − ∞ < x < 1 . For 1 < x < 1.8 , consider the point x = 1.5 and check the value of first order derivative at that point. For the interval ( 1 , 1.8 ) ( f ' ( x ) ) x = 1.5 = 30 ( 1.5 − 9 5 ) ( 1.5 − 1 ) 1 3 ( f ' ( x ) ) x = 1.5 = 30 ( 1.5 − 1.8 ) ( 0.5 ) 1 3 ( f ' ( x ) ) x = 1

4th Edition

ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Question

Chapter 4.6, Problem 37E

To determine

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The behavior of the curve f(x)=18(x−3)(x−1)23.

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Chapter 4.6, Problem 36E

Chapter 4.6, Problem 38E

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Chapter 4 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

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