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

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 5, Problem 17RE

To determine

To find points at which f has a local maxima using the derivative of function.

Expert Solution

Answer to Problem 17RE

f does not have local maximum value because there does not exist any value of x for which y″(x)<0

Explanation of Solution

Given:

The derivative of function is y′=6(x+1)(x−2)2 .

Calculation:

Since, the derivative of function is y′=6(x+1)(x−2)2 put y′=0 to find the critical points

6(x+1)(x−2)2=0∴x=−1,2

Now , check the value of second derivative at these critical points so that if y″(x)<0 then function has maximum value at that point and if y″(x)>0 then function has minimum value at that point.

y″(x)=18x(x−2)y″(−1)=18(−1)(−1−2)=54>0y″(2)=18(2)(2−2)=0

Therefore, f does not have local maximum value because there does not exist any value of x for which y″(x)<0

To determine

To find points at which f has a local minima, using the derivative of function.

Expert Solution

Answer to Problem 17RE

f have local minimum value at x=−1

Explanation of Solution

Given:

The derivative of function is y′=6(x+1)(x−2)2 .

Calculation:

Since, the derivative of function is y′=6(x+1)(x−2)2 put y′=0 to find the critical points

6(x+1)(x−2)2=0∴x=−1,2

Now , check the value of second derivative at these critical points so that if y″(x)<0 then function has maximum value at that point and if y″(x)>0 then function has minimum value at that point.

y″(x)=18x(x−2)y″(−1)=18(−1)(−1−2)=54>0y″(2)=18(2)(2−2)=0

Therefore, f have local minimum value at x=−1

To determine

To find inflections points.

Expert Solution

Answer to Problem 17RE

Inflection points are x=0 and x=2

Explanation of Solution

Given:

The derivative of function is y′=6(x+1)(x−2)2 .

Calculation:

Inflection point of any function is a point where the graph of function has a tangent line and where the concavity changes.

Below is the graph of y′=6(x+1)(x−2)2

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5, Problem 17RE

From graph of y′=6(x+1)(x−2)2 it is clear that concavity changes at points (0,24) and (2,0) therefore inflection points are x=0 and x=2

Chapter 5, Problem 16RE

Chapter 5, Problem 18RE

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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