a. Explanation Given information : The given statement is that there is almost no leeway in the locations of the inflection point and extrema of f ( x ) = a x 3 + b x 2 + c x + d , a ≠ 0 , because the one inflection point occurs at x = − b 3 a and extrema, if any, must be located symmetrically about this value of x . Proof : The graph of the cubic y = 4 x 3 + 21 x 2 + 36 x − 20 The inflection point of the graph is at x = − 1.75 and its extrema are at x = − 2 and x = − 1.5 . The extrema are symmetric about the inflection pint since they are both 0.25 unit from the inflection point. b. To determine To check the statement by examining for cubic with equation y = − 2 x 3 + 6 x 2 − 20 . Explanation Given information : The given statement is that there is almost no leeway in the locations of the inflection point and extrema of f ( x ) = a x 3 + b x 2 + c x + d , a ≠ 0 , because the one inflection point occurs at x = − b 3 a and extrema, if any, must be located symmetrically about this value of x . Proof : The graph of the cubic y = − 2 x 3 + 6 x 2 − 20 The inflection point of the graph is at x = 1 and its extrema are at x = 2 and x = 0 . The extrema are symmetric about the inflection pint since they are both 1 unit from the inflection point. c. To determine To prove the statement for the general case. Explanation Given information: The given statement is that there is almost no leeway in the locations of the inflection point and extrema of f ( x ) = a x 3 + b x 2 + c x + d , a ≠ 0 , because the one inflection point occurs at x = − b 3 a and extrema, if any, must be located symmetrically about this value of x . Proof: Take 1 s t and 2 n d derivatives f ′ ( x ) = 3 a x 2 + 2 b x + c f ″ ( x ) = 6 a x + 2 b Apply quadratic formula to get extrema point for which f ′ ( x ) = 0 3 a x 2 + 2 b x + c = 0 x = − 2 b ± 4 b 2 − 12 a c 6 a At an inflection point f ″ ( x ) changes sign, which means you need to find points for which f ″ ( x ) = 0 6 a x + 2 b = 0 x = − 2 b 6 a = − b 3 a Hence, − b 3 a is an unique inflection point. If b 2 − 3 a c ≥ 0 there are two extremum point positioned symmetrically around the inflection point at a distance of b 2 − 3 a c 3 a

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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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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Question

Chapter 5.3, Problem 61E

To determine

To check the statement by examining for cubic with equation y=4x3+21x2+36x−20 .

Expert Solution

Explanation of Solution

Given information:

The given statement is that there is almost no leeway in the locations of the inflection point and extrema of f(x)=ax3+bx2+cx+d,a≠0 , because the one inflection point occurs at x=−b3a and extrema, if any, must be located symmetrically about this value of x .

Proof:

The graph of the cubic y=4x3+21x2+36x−20

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.3, Problem 61E , additional homework tip 1

The inflection point of the graph is at x=−1.75 and its extrema are at x=−2 and x=−1.5 . The extrema are symmetric about the inflection pint since they are both 0.25 unit from the inflection point.

To determine

To check the statement by examining for cubic with equation y=−2x3+6x2−20 .

Expert Solution

Explanation of Solution

Given information:

Proof:

The graph of the cubic y=−2x3+6x2−20

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.3, Problem 61E , additional homework tip 2

The inflection point of the graph is at x=1 and its extrema are at x=2 and x=0 . The extrema are symmetric about the inflection pint since they are both 1 unit from the inflection point.

To determine

To prove the statement for the general case.

Expert Solution

Explanation of Solution

Given information:

Proof:

Take 1^s^t and 2ⁿ^d derivatives

f′(x)=3ax2+2bx+cf″(x)=6ax+2b

Apply quadratic formula to get extrema point for which f′(x)=0

3ax2+2bx+c=0x=−2b±4b2−12ac6a

At an inflection point f″(x) changes sign, which means you need to find points for which f″(x)=0

6ax+2b=0x=−2b6a =−b3a

Hence, −b3a is an unique inflection point.

If b2−3ac≥0 there are two extremum point positioned symmetrically around the inflection point at a distance of b2−3ac3a

Chapter 5.3, Problem 60E

Chapter 5.3, Problem 62E

Chapter 5 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 5.1 - Prob. 1QR Ch. 5.1 - Prob. 2QR Ch. 5.1 - Prob. 3QR Ch. 5.1 - Prob. 4QR Ch. 5.1 - Prob. 5QR Ch. 5.1 - Prob. 6QR Ch. 5.1 - Prob. 7QR Ch. 5.1 - Prob. 8QR Ch. 5.1 - Prob. 9QR Ch. 5.1 - Prob. 10QR

Ch. 5.1 - Prob. 11QR Ch. 5.1 - Prob. 12QR Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E Ch. 5.1 - Prob. 11E Ch. 5.1 - Prob. 12E Ch. 5.1 - Prob. 13E Ch. 5.1 - Prob. 14E Ch. 5.1 - Prob. 15E Ch. 5.1 - Prob. 16E Ch. 5.1 - Prob. 17E Ch. 5.1 - Prob. 18E Ch. 5.1 - Prob. 19E Ch. 5.1 - Prob. 20E Ch. 5.1 - Prob. 21E Ch. 5.1 - Prob. 22E Ch. 5.1 - Prob. 23E Ch. 5.1 - Prob. 24E Ch. 5.1 - Prob. 25E Ch. 5.1 - Prob. 26E Ch. 5.1 - Prob. 27E Ch. 5.1 - Prob. 28E Ch. 5.1 - Prob. 29E Ch. 5.1 - Prob. 30E Ch. 5.1 - Prob. 31E Ch. 5.1 - Prob. 32E Ch. 5.1 - Prob. 33E Ch. 5.1 - Prob. 34E Ch. 5.1 - Prob. 35E Ch. 5.1 - Prob. 36E Ch. 5.1 - Prob. 37E Ch. 5.1 - Prob. 38E Ch. 5.1 - Prob. 39E Ch. 5.1 - Prob. 40E Ch. 5.1 - Prob. 41E Ch. 5.1 - Prob. 42E Ch. 5.1 - Prob. 43E Ch. 5.1 - Prob. 44E Ch. 5.1 - Prob. 45E Ch. 5.1 - Prob. 46E Ch. 5.1 - Prob. 47E Ch. 5.1 - Prob. 48E Ch. 5.1 - 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Finding The Focus and Directrix of a Parabola - Conic Sections; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=KYgmOTLbuqE;License: Standard YouTube License, CC-BY

To check the statement by examining for cubic with equation y = 4 x 3 + 21 x 2 + 36 x − 20 .

Concept explainers

Videos

To check the statement by examining for cubic with equation y=4x3+21x2+36x−20 .

Explanation of Solution

To check the statement by examining for cubic with equation y=−2x3+6x2−20 .

Explanation of Solution

To prove the statement for the general case.

Explanation of Solution

Chapter 5 Solutions

Additional Math Textbook Solutions