a. Answer The graph of our given function would be: Explanation Given: A function: f ( x ) = ln ( x x 2 + 1 ) Calculation: Upon graphing the given functionusing a graphing utility, we will get our required graph as shown below: b. To determine To find: The domain of the given function. Answer The domain of our given logarithmic function would be x > 0 or ( 0 , ∞ ) . Explanation Given: A function: f ( x ) = ln ( x x 2 + 1 ) Calculation: We know that logarithmic functions are not defined for negative values. To find domain of our given function, we will set argument of log function greater than 0 as: x x 2 + 1 > 0 x x 2 + 1 ⋅ ( x 2 + 1 ) > 0 ⋅ ( x 2 + 1 ) x > 0 Therefore, the domain of our given logarithmic function would be x > 0 or ( 0 , ∞ ) . c. To determine To find: Open intervals on which the given function is increasing and decreasing using the graph of function. Answer The function is increasing on interval ( 0 , 1 ) and decreasing on ( 1 , ∞ ) . Explanation Given: A function: f ( x ) = ln ( x x 2 + 1 ) Calculation: Upon looking at graph of our given function, we can see that function is increasing from 0 to 1 and decreasing from 1 to positive infinity. Therefore, the function is increasingon interval ( 0 , 1 ) and decreasing on ( 1 , ∞ ) . d. To determine To approximate: The relative maximum or minimum values of the function. Round to three decimal places. Answer The relative maximum of the given function is ( 1 , − 0.693 ) . Explanation Given: A function: f ( x ) = ln ( x x 2 + 1 ) Calculation: We can see from our given graph that there is no minimum value for our given function. We can see that the maximum value of our given function occurs at x = 1 that is − 0.693 . Therefore, the relative maximum of the given function is ( 1 , − 0.693 ) .

8th Edition

ISBN: 9781337904285

Author: Larson

Publisher: CENGAGE L

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Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topic…

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Question

Chapter 3.2, Problem 106E

To determine

To graph:

The given function using a graphing utility.

Expert Solution

Answer to Problem 106E

The graph of our given function would be:

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 3.2, Problem 106E , additional homework tip 1

Explanation of Solution

Given:

A function: f(x)=ln(xx2+1)

Calculation:

Upon graphing the given functionusing a graphing utility, we will get our required graph as shown below:

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 3.2, Problem 106E , additional homework tip 2

To determine

To find:

The domain of the given function.

Expert Solution

Answer to Problem 106E

The domain of our given logarithmic function would be x>0 or (0,∞).

Explanation of Solution

Given:

A function: f(x)=ln(xx2+1)

Calculation:

We know that logarithmic functions are not defined for negative values. To find domain of our given function, we will set argument of log function greater than 0 as:

xx2+1>0xx2+1⋅(x2+1)>0⋅(x2+1)x>0

Therefore, the domain of our given logarithmic function would be x>0 or (0,∞).

To determine

To find:

Open intervals on which the given function is increasing and decreasing using the graph of function.

Expert Solution

Answer to Problem 106E

The function is increasing on interval (0,1) and decreasing on (1,∞).

Explanation of Solution

Given:

A function: f(x)=ln(xx2+1)

Calculation:

Upon looking at graph of our given function, we can see that function is increasing from 0 to 1 and decreasing from 1 to positive infinity.

Therefore, the function is increasingon interval (0,1) and decreasing on (1,∞).

To determine

To approximate:

The relative maximum or minimum values of the function. Round to three decimal places.

Expert Solution

Answer to Problem 106E

The relative maximum of the given function is (1,−0.693).

Explanation of Solution

Given:

A function: f(x)=ln(xx2+1)

Calculation:

We can see from our given graph that there is no minimum value for our given function.

We can see that the maximum value of our given function occurs at x=1 that is −0.693.

Therefore, the relative maximum of the given function is (1,−0.693) .

Chapter 3.2, Problem 105E

Chapter 3.2, Problem 107E

Chapter 3 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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Prob. 78E Ch. 3.6 - Prob. 1E Ch. 3.6 - Prob. 2E Ch. 3.6 - Prob. 3E Ch. 3.6 - Prob. 4E Ch. 3.6 - Prob. 5E Ch. 3.6 - Prob. 6E Ch. 3.6 - Prob. 7E Ch. 3.6 - Prob. 8E Ch. 3.6 - Prob. 9E Ch. 3.6 - Prob. 10E Ch. 3.6 - Prob. 11E Ch. 3.6 - Prob. 12E Ch. 3.6 - Prob. 13E Ch. 3.6 - Prob. 14E Ch. 3.6 - Prob. 15E Ch. 3.6 - Prob. 16E Ch. 3.6 - Prob. 17E Ch. 3.6 - Prob. 18E Ch. 3.6 - Prob. 19E Ch. 3.6 - Prob. 20E Ch. 3.6 - Prob. 21E Ch. 3.6 - Prob. 22E Ch. 3.6 - Prob. 23E Ch. 3.6 - Prob. 24E Ch. 3.6 - Prob. 25E Ch. 3.6 - Prob. 26E Ch. 3.6 - Prob. 27E Ch. 3.6 - Prob. 28E Ch. 3.6 - Prob. 29E Ch. 3.6 - Prob. 30E Ch. 3.6 - Prob. 31E Ch. 3.6 - Prob. 32E Ch. 3.6 - Prob. 33E Ch. 3.6 - Prob. 34E Ch. 3.6 - Prob. 35E Ch. 3.6 - Prob. 36E Ch. 3.6 - Prob. 37E Ch. 3.6 - Prob. 38E Ch. 3.6 - Prob. 39E Ch. 3.6 - Prob. 40E Ch. 3.6 - Prob. 41E Ch. 3.6 - Prob. 42E Ch. 3.6 - Prob. 43E Ch. 3.6 - Prob. 44E Ch. 3.6 - Prob. 45E Ch. 3.6 - Prob. 46E Ch. 3 - Prob. 1CR Ch. 3 - Prob. 2CR Ch. 3 - Prob. 3CR Ch. 3 - Prob. 4CR Ch. 3 - 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Prob. 58CR Ch. 3 - Prob. 59CR Ch. 3 - Prob. 60CR Ch. 3 - Prob. 61CR Ch. 3 - Prob. 62CR Ch. 3 - Prob. 63CR Ch. 3 - Prob. 64CR Ch. 3 - Prob. 65CR Ch. 3 - Prob. 66CR Ch. 3 - Prob. 67CR Ch. 3 - Prob. 68CR Ch. 3 - Prob. 69CR Ch. 3 - Prob. 70CR Ch. 3 - Prob. 71CR Ch. 3 - Prob. 72CR Ch. 3 - Prob. 73CR Ch. 3 - Prob. 74CR Ch. 3 - Prob. 75CR Ch. 3 - Prob. 76CR Ch. 3 - Prob. 77CR Ch. 3 - Prob. 78CR Ch. 3 - Prob. 79CR Ch. 3 - Prob. 80CR Ch. 3 - Prob. 81CR Ch. 3 - Prob. 82CR Ch. 3 - Prob. 83CR Ch. 3 - Prob. 84CR Ch. 3 - Prob. 85CR Ch. 3 - Prob. 86CR Ch. 3 - Prob. 87CR Ch. 3 - Prob. 88CR Ch. 3 - Prob. 89CR Ch. 3 - Prob. 90CR Ch. 3 - Prob. 91CR Ch. 3 - Prob. 92CR Ch. 3 - Prob. 93CR Ch. 3 - Prob. 94CR Ch. 3 - Prob. 95CR Ch. 3 - Prob. 96CR Ch. 3 - Prob. 97CR Ch. 3 - Prob. 98CR Ch. 3 - Prob. 99CR Ch. 3 - Prob. 100CR Ch. 3 - Prob. 101CR Ch. 3 - Prob. 102CR Ch. 3 - Prob. 103CR Ch. 3 - Prob. 104CR Ch. 3 - Prob. 105CR Ch. 3 - Prob. 106CR Ch. 3 - Prob. 107CR Ch. 3 - Prob. 108CR Ch. 3 - Prob. 109CR Ch. 3 - Prob. 110CR Ch. 3 - Prob. 111CR Ch. 3 - Prob. 112CR Ch. 3 - Prob. 113CR Ch. 3 - Prob. 114CR Ch. 3 - Prob. 115CR Ch. 3 - Prob. 116CR Ch. 3 - Prob. 117CR Ch. 3 - Prob. 118CR Ch. 3 - Prob. 119CR Ch. 3 - Prob. 120CR Ch. 3 - Prob. 121CR Ch. 3 - Prob. 122CR Ch. 3 - Prob. 123CR Ch. 3 - Prob. 124CR Ch. 3 - Prob. 125CR Ch. 3 - Prob. 126CR Ch. 3 - Prob. 1CT Ch. 3 - Prob. 2CT Ch. 3 - Prob. 3CT Ch. 3 - Prob. 4CT Ch. 3 - Prob. 5CT Ch. 3 - Prob. 6CT Ch. 3 - Prob. 7CT Ch. 3 - Prob. 8CT Ch. 3 - Prob. 9CT Ch. 3 - Prob. 10CT Ch. 3 - Prob. 11CT Ch. 3 - Prob. 12CT Ch. 3 - Prob. 13CT Ch. 3 - Prob. 14CT Ch. 3 - Prob. 15CT Ch. 3 - Prob. 16CT Ch. 3 - Prob. 17CT Ch. 3 - Prob. 18CT Ch. 3 - Prob. 19CT Ch. 3 - Prob. 20CT Ch. 3 - Prob. 21CT Ch. 3 - Prob. 22CT Ch. 3 - Prob. 23CT Ch. 3 - Prob. 24CT Ch. 3 - Prob. 25CT Ch. 3 - Prob. 26CT Ch. 3 - Prob. 27CT Ch. 3 - Prob. 28CT Ch. 3 - Prob. 29CT Ch. 3 - Prob. 30CT Ch. 3 - Prob. 1STP Ch. 3 - Prob. 2STP Ch. 3 - Prob. 3STP Ch. 3 - Prob. 4STP Ch. 3 - Prob. 5STP Ch. 3 - Prob. 6STP Ch. 3 - Prob. 7STP Ch. 3 - Prob. 8STP Ch. 3 - Prob. 9STP Ch. 3 - Prob. 10STP Ch. 3 - Prob. 11STP Ch. 3 - Prob. 12STP Ch. 3 - Prob. 13STP Ch. 3 - Prob. 14STP Ch. 3 - Prob. 15STP Ch. 3 - Prob. 16STP Ch. 3 - Prob. 17STP Ch. 3 - Prob. 18STP Ch. 3 - Prob. 19STP Ch. 3 - Prob. 20STP Ch. 3 - Prob. 21STP Ch. 3 - Prob. 22STP Ch. 3 - Prob. 23STP Ch. 3 - Prob. 24STP Ch. 3 - Prob. 25STP Ch. 3 - Prob. 26STP

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