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 5.2, Problem 26E

To determine

To find : The local extrema of the function k(x)=xx2−4 .

Expert Solution

Answer to Problem 26E

k(x) does not have local extrema.

Explanation of Solution

Given information :

The given function is k(x)=xx2−4

Calculation :

To find the local extrema first find the derivative of k(x) . Therefore,

k(x)=xx2−4k'(x)=1(x2−4)2−x(2x)(x2−4)2k'(x)=x2−4−2x2(x2−4)2k'(x)=−4−x2(x2−4)2

Set k'(x)=0 and solve for x to find the critical value.

−4−x2(x2−4)2=0−4−x2=0x2=−4

Since, there is no real solution, so there is no critical values, therefore k(x) does not have a local extrema.

Hence,

k(x) does not have local extrema.

To determine

To find : The interval on which the function k(x)=xx2−4 is increasing.

Expert Solution

Answer to Problem 26E

The function increases nowhere.

Explanation of Solution

Given information :

The given function is k(x)=xx2−4

Calculation :

From (a) k'(x)=−4−x2(x2−4)2 and there is no critical value, so the graph will be either only increasing or only decreasing.

Substitute any point into k'(x) to check its only increasing or only decreasing. Therefore,

k'(0)=−14

So, the function is only decreasing.

Hence,

The function increases nowhere.

To determine

To find : The interval on which the function k(x)=xx2−4 is decreasing.

Expert Solution

Answer to Problem 26E

The interval on which the function is decreasing is (−∞,∞)−{2,−2} .

Explanation of Solution

Given information :

The given function is k(x)=xx2−4

Calculation :

From (a) k'(x)=−4−x2(x2−4)2 and there is no critical value, so the graph will be either only increasing or only decreasing.

Substitute any point into k'(x) to check its only increasing or only decreasing. Therefore,

k'(0)=−14

So, the function is only decreasing.

Also, for x2−4=0

k'(x) is undefined. Therefore,

x2−4=0x2=4x=±2

So, the interval on which the function is decreasing is (−∞,∞)−{2,−2} .

Hence,

The interval on which the function is decreasing is (−∞,∞)−{2,−2} .

Chapter 5.2, Problem 25E

Chapter 5.2, Problem 27E

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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