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

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Concept explainers

Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

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Question

Chapter 5.2, Problem 27E

To determine

To find : The local extrema of the function f(x)=x3−2x−2cosx .

Expert Solution

Answer to Problem 27E

The maximum is at x=−1.126 and the maximum value is −0.036 and the minimum is at 0.559 and the minimum value is −2.639 .

Explanation of Solution

Given information :

The given function is f(x)=x3−2x−2cosx

Calculation :

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

f(x)=x3−2x−2cosxf'(x)=3x2−2−2(−sinx)f'(x)=3x2−2+2sinx

Set f'(x)=0 and solve for x to using calculator:

3x2−2+2sinx=0x=−1.126, 0.559

Substitute x=−1.126 in the given equation and solve:

f(x)=x3−2x−2cosxf(−1.126)=(−1.126)3−2(−1.126)−2cos(−1.126)f(−1.126)=−0.036 Maximum

Also, substitute x=0.559 in the given equation and solve:

f(x)=x3−2x−2cosxf(0.559)=(0.559)3−2(0.559)−2cos(0.559)f(0.559)=−2.639 Minimum

Therefore, the maximum is at x=−1.126 and the maximum value is −0.036 and the minimum is at 0.559 and the minimum value is −2.639 .

Hence,

The maximum is at x=−1.126 and the maximum value is −0.036 and the minimum is at 0.559 and the minimum value is −2.639 .

To determine

To find : The interval on which the function f(x)=x3−2x−2cosx is increasing.

Expert Solution

Answer to Problem 27E

The function is increasing in the interval (−∞,−1.126]∪[0.559, ∞) .

Explanation of Solution

Given information :

The given function is f(x)=x3−2x−2cosx

Calculation :

To find the interval on which the function is increasng first find the derivative of f(x) . Therefore,

f(x)=x3−2x−2cosxf'(x)=3x2−2−2(−sinx)f'(x)=3x2−2+2sinx

Set f'(x)=0 and solve for x to using calculator:

3x2−2+2sinx=0x=−1.126, 0.559

Use sign chart to get the interval in which the function is increasing

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.2, Problem 27E , additional homework tip 1

Therefore, the function is increasing in the interval (−∞,−1.126]∪[0.559, ∞) .

Hence,

The function is increasing in the interval (−∞,−1.126]∪[0.559, ∞) .

To determine

To find : The interval on which the function f(x)=x3−2x−2cosx is decreasing.

Expert Solution

Answer to Problem 27E

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

Explanation of Solution

Given information :

The given function is f(x)=x3−2x−2cosx

Calculation :

To find the interval on which the function is decreasing first find the derivative of f(x) . Therefore,

f(x)=x3−2x−2cosxf'(x)=3x2−2−2(−sinx)f'(x)=3x2−2+2sinx

Set f'(x)=0 and solve for x to using calculator:

3x2−2+2sinx=0x=−1.126, 0.559

Use sign chart to get the interval in which the function is decreasing.

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.2, Problem 27E , additional homework tip 2

Therefore, the function is decreasing on the interval [−1.126, 0.599] .

Hence,

The function is decreasing in the interval [−1.126, 0.599] .

Chapter 5.2, Problem 26E

Chapter 5.2, Problem 28E

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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Differential Equation | MIT 18.01SC Single Variable Calculus, Fall 2010; Author: MIT OpenCourseWare;https://www.youtube.com/watch?v=HaOHUfymsuk;License: Standard YouTube License, CC-BY

The local extrema of the function f ( x ) = x 3 − 2 x − 2 cos x .

Concept explainers

Videos

To find : The local extrema of the function f(x)=x3−2x−2cosx .

Answer to Problem 27E

Explanation of Solution

To find : The interval on which the function f(x)=x3−2x−2cosx is increasing.

Answer to Problem 27E

Explanation of Solution

To find : The interval on which the function f(x)=x3−2x−2cosx is decreasing.

Answer to Problem 27E

Explanation of Solution

Chapter 5 Solutions

Additional Math Textbook Solutions