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, Problem 5RE

To determine

Tofindthe interval on which the function y=1+x−x2−x4 is increasing.

Expert Solution

Answer to Problem 5RE

The function is then increasing for (−∞,0.385] .

Explanation of Solution

Given information:

The given function is y=1+x−x2−x4 .

Formula:

Chain rule:

ddx(uv)=u′v+uv′

Consider the function y=1+x−x2−x4 ,

Using product and chain rule: ddx(uv)=u′v+uv′ find the derivative,

y′=1−2x−4x3

y″=−2−12x2

Here the second derivative is always negative, so the function is concave down for all values of x .

y′=1−2x−4x3=0x≈ 0.38545…

Here, the function is positive until one point which is at zero and then negative.

Therefore, the function is increasing for (−∞,0.385] .

To determine

To find the interval on which the function y=1+x−x2−x4 is decreasing.

Expert Solution

Answer to Problem 5RE

The function is then decreasing for x>0.385 .

Explanation of Solution

Given information:

The given function is y=1+x−x2−x4 .

Formula:

Chain rule:

ddx(uv)=u′v+uv′

Consider the function y=1+x−x2−x4 ,

Using product and chain rule: ddx(uv)=u′v+uv′ find the derivative,

y′=1−2x−4x3

y″=−2−12x2

Here the second derivative is always negative, so the function is concave down for all values of x .

y′=1−2x−4x3=0x≈ 0.38545…

Here, the function is positive until one point which is at zero and then negative.

Therefore, the function is decreasing for x>0.385 .

To determine

To find the interval on which the function y=1+x−x2−x4 is concave up.

Expert Solution

Answer to Problem 5RE

The function is never concave up.

Explanation of Solution

Given information:

The given function is y=1+x−x2−x4 .

Formula:

Chain rule:

ddx(uv)=u′v+uv′

Consider the function y=1+x−x2−x4 ,

Using product and chain rule: ddx(uv)=u′v+uv′ find the derivative,

y′=1−2x−4x3

y″=−2−12x2

Here the second derivative is always negative, so the function is concave down for all values of x and there are no inflection points.

y″=−2−12x2=0−12x2=2x2=−16

This has no solution so there are no values of x that makes y″=0 .

−2−12x2<0∀x

Therefore, the function is never concave up.

To determine

To find the interval on which the function y=1+x−x2−x4 is concave down.

Expert Solution

Answer to Problem 5RE

The function is always concave down.

Explanation of Solution

Given information:

The given function is y=1+x−x2−x4 .

Formula:

Chain rule:

ddx(uv)=u′v+uv′

Consider the function y=1+x−x2−x4 ,

Using product and chain rule: ddx(uv)=u′v+uv′ find the derivative,

y′=1−2x−4x3

y″=−2−12x2

Here the second derivative is always negative, so the function is concave down for all values of x and there are no inflection points.

y″=−2−12x2=0−12x2=2x2=−16

This has no solution so there are no values of x that makes y″=0 .

−2−12x2<0∀x

Therefore, the function is always concave down.

To determine

To find the interval on which the function y=1+x−x2−x4 has local extreme values.

Expert Solution

Answer to Problem 5RE

The function haslocal maximum at (0.385,1.215) .

Explanation of Solution

Given information:

The given function is y=1+x−x2−x4 .

Formula:

Chain rule:

ddx(uv)=u′v+uv′

Consider the function y=1+x−x2−x4 ,

Using product and chain rule: ddx(uv)=u′v+uv′ find the derivative,

y′=1−2x−4x3

Equating the equation to zero,

y′=1−2x−4x3=0x≈0.385

Now to determine if the critical value x≈0.385 is a maximum or minimum,

y′(0)=1−2(0)−4(0)3=1>0y′(1)=1−2(1)−4(1)3=−5<0

Since the derivative switches from positive to negative, the critical point is a local maximum. It is also an absolute maximum since there are no endpoints or other critical points.

y(0.385)=1+0.385−0.3752−0.3854≈1.215

Therefore, the function has local maximum at (0.385,1.215) .

To determine

To find the interval on which the function y=1+x−x2−x4 has inflection points.

Expert Solution

Answer to Problem 5RE

The function has noinflection point.

Explanation of Solution

Given information:

The given function is y=1+x−x2−x4 .

Formula:

Chain rule:

ddx(uv)=u′v+uv′

Consider the function y=1+x−x2−x4 ,

Using product and chain rule: ddx(uv)=u′v+uv′ find the derivative,

y′=1−2x−4x3

y″=−2−12x2

The point of inflection occurs at the values x that make the second derivative undefined or equal to zero.

y″=−2−12x2=0−12x2=2x2=−16x=±−16

Since the square root of a negative number is not a real number, there are no values of x that make y″=0 .

Since there are no values of x that make y″=0 or be undefined, the second derivative is strictly positive or strictly negative.

y″(0)=−2−12(0)2=−2<0−12x2=2x2=−16

y″=−2−12x2 is always negative, so the function is concave down for all values of x and there are no inflection points.

Therefore, the function has no inflection point.

Chapter 5, Problem 4RE

Chapter 5, Problem 6RE

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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