Calculus: Graphical, Numerical, Algebraic

3rd Edition

ISBN: 9780133688399

Author: Finney, Ross L.

Publisher: Pearson/Prentice Hall

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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: Graphical, Numerical, Algebraic

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The local extrema of the function k ( x ) = x x 2 − 4 .

Videos

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

Answer to Problem 26E

Explanation of Solution

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

Answer to Problem 26E

Explanation of Solution

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

Answer to Problem 26E

Explanation of Solution

Chapter 5 Solutions

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