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 17E

To determine

To find the local extrema.

Expert Solution

Answer to Problem 17E

The local extrema at none.

Explanation of Solution

Given information:

The given function is h(x)=2x

Concept used:

Extreme values occur only at critical points and end points.

A point in the interior of the domain of a function f at which f′=0 or f′ does not exist is called a critical point of f .

Calculation :

Take derivative of the function

h′(x)=−2x2

Equate the derivative to 0.

−2x2=0

The value of x is not defined

Therefore,

The local extrema at none.

To determine

To find the intervals on which the function is increasing.

Expert Solution

Answer to Problem 17E

The function increases at none.

Explanation of Solution

Given information:

The given function is h(x)=2x

Concept used:

If f′>0 at each point of (a,b) , then f increases on [a,b]

It is assumed that f be continuous on [a,b] and differentiable on (a,b) .

Calculation :

For increasing function h′(x) must be greater than 0.

−2x2>0

h′(x) Is never positive because a perfect square is multiplied by a negative sign and the value of a perfect square is always non-negative.

Therefore,

The function increases at none.

To determine

To find the intervals on which the function is decreasing.

Expert Solution

Answer to Problem 17E

The function decreases in the interval (−∞,0) and (0,∞)

Explanation of Solution

Given information:

The given function is h(x)=2x

Concept used:

If f′<0 at each point of (a,b) , then f decreases on [a,b]

It is assumed that f be continuous on [a,b] and differentiable on (a,b) .

Calculation :

For decreasing function h′(x) must be less than 0.

−2x2<0

h′(x) Is never positive because a perfect square is multiplied by a negative sign and the value of a perfect square is always non-negative. Hence, h′(x) is always negative except at x=0 because h′(x) is undefined at x=0 .

Therefore,

The function decreases in the interval (−∞,0] and [0,∞)

Chapter 5.2, Problem 16E

Chapter 5.2, Problem 18E

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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