Explanation Given information: The given curve is y = 12 − 5 x − 2 x 2 . Calculation: The first and second order derivative of the curve are below: y ' = − 5 − 4 x ................... ( 1 ) y ' ' = − 4........................... ( 2 ) 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. ∴ y ' = 0 − 4 x − 5 = 0............... ( Using (1) ) 4 x = − 5 x = − 5 4 x = − 1.25 x = − 1.25 is the critical point of the curve. Since the given curve is a polynomial function of x, therefore its domain is real. For x > − 1.25 consider the point x = 0 and check the value of first order derivative at point x = 0 . For the interval ( − 1.25 , ∞ ) [ ( y ) ' ] x = 0 = − 5 − 4 ( 0 ) [ ( y ) ' ] x = 0 = − 5 − 0 [ ( y ) ' ] x = 0 = − 5 < 0 Since, the value of first order derivative at x = 0 is negative therefore, the behavior of curve is strictly decreasing for all x > − 1

4th Edition

ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Maximization

Have you ever seen a bridge in the shape of a parabola? Where is its maximum on the bridge? The maximum of that bridge is the highest point on that bridge. The process of finding the maximum of the bridge here is called the maximization.

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Chapter 4.6, Problem 10E

To determine

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The behavior of the curve y=12−5x−2x2 and sketch it.

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Chapter 4.6, Problem 9E

Chapter 4.6, Problem 11E

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Chapter 4 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

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