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

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 2.3, Problem 22E

(a)

To determine

To find: The points of discontinuity of f(x)={1−x2, x≠−12, x=−1 .

(a)

Expert Solution

Answer to Problem 22E

The point of discontinuity of f(x)={1−x2, x≠−12, x=−1 is x=−1 .

Explanation of Solution

Given information:

The function is f(x)={1−x2, x≠−12, x=−1 .

Calculation:

Check the points of discontinuity of the function f(x)={1−x2, x≠−12, x=−1 with the help of graph.

To graph a function f(x)={1−x2, x≠−12, x=−1 , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press Y= key and enter right hand side of the equation f(x)=1−x2 after the symbol Y1 . Enter the keystrokes (1−X^2)÷(X2ndMATH2(−)1) . Now press the ENTER button and enter right hand side of the equation f(x)=2 after the symbol Y2 . Enter the keystrokes (2)÷(X2ndMATH1(−)1) .

The display will show the equation,

Y1=(1−X^2)/(X≠−1)Y2=(2)/(X=−1)

Now, press the GRAPH key and TRACE key to produce the graph of given function in standard window as shown in Figure (1).

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 2.3, Problem 22E

Figure (1)

As observed from the graph, the function f(x) is continuous for all the points x<−1 and also for all the points x>−1 .

Now check the continuity for x=−1 .

As observed from graph, the value of f(x) approaches to 0 as x approaches to −1 from the left side. So, the left hand limit of f(x) at x=−1 is 0 .

As observed from graph, the value of f(x) approaches to 0 as x approaches to −1 from the right side. So, the right hand limit of f(x) at x=−1 is 0 .

As the left hand limit of f(x) is equal to the right hand limit at x=−1 . So, the value of limx→−1f(x) is equal to 0 .

As the value of f(x) at the point x=−1 is equal to 2 . So, the value of f(−1)=2 .

As the value of limx→−1f(x) is not equal to the value of f(−1) . So, the function is discontinuous at x=−1 .

Therefore, the point of discontinuity of f(x)={1−x2, x≠−12, x=−1 is x=−1 .

(b)

To determine

To check: Whether the discontinuity of f(x)={1−x2, x≠−12, x=−1 is removable or not, explain.

(b)

Expert Solution

Answer to Problem 22E

The discontinuity of f(x)={1−x2, x≠−12, x=−1 at x=−1 is removable as the limit value of f(x) at x=−1 and f(−1) are not equal.

Explanation of Solution

Given information:

The function is f(x)={1−x2, x≠−12, x=−1 .

Calculation:

As calculated in part(a), the value of limx→−1f(x) is equal to 0 and the value of f(−1) is equal to 2 .

So, the function f(x) can be made continuous be setting the value of f(−1) .

Therefore, the discontinuity of f(x)={1−x2, x≠−12, x=−1 at x=−1 is removable.

Chapter 2.3, Problem 21E

Chapter 2.3, Problem 23E

Chapter 2 Solutions

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

Ch. 2.1 - Prob. 1QR Ch. 2.1 - Prob. 2QR Ch. 2.1 - Prob. 3QR Ch. 2.1 - Prob. 4QR Ch. 2.1 - Prob. 5QR Ch. 2.1 - Prob. 6QR Ch. 2.1 - Prob. 7QR Ch. 2.1 - Prob. 8QR Ch. 2.1 - Prob. 9QR Ch. 2.1 - Prob. 10QR

Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E Ch. 2.1 - Prob. 11E Ch. 2.1 - Prob. 12E Ch. 2.1 - Prob. 13E Ch. 2.1 - Prob. 14E Ch. 2.1 - Prob. 15E Ch. 2.1 - Prob. 16E Ch. 2.1 - Prob. 17E Ch. 2.1 - Prob. 18E Ch. 2.1 - Prob. 19E Ch. 2.1 - Prob. 20E Ch. 2.1 - Prob. 21E Ch. 2.1 - Prob. 22E Ch. 2.1 - Prob. 23E Ch. 2.1 - Prob. 24E Ch. 2.1 - Prob. 25E Ch. 2.1 - Prob. 26E Ch. 2.1 - Prob. 27E Ch. 2.1 - Prob. 28E Ch. 2.1 - Prob. 29E Ch. 2.1 - Prob. 30E Ch. 2.1 - Prob. 31E Ch. 2.1 - Prob. 32E Ch. 2.1 - Prob. 33E Ch. 2.1 - Prob. 34E Ch. 2.1 - Prob. 35E Ch. 2.1 - Prob. 36E Ch. 2.1 - Prob. 37E Ch. 2.1 - Prob. 38E Ch. 2.1 - Prob. 39E Ch. 2.1 - Prob. 40E Ch. 2.1 - Prob. 41E Ch. 2.1 - Prob. 42E Ch. 2.1 - Prob. 43E Ch. 2.1 - Prob. 44E Ch. 2.1 - Prob. 45E Ch. 2.1 - Prob. 46E Ch. 2.1 - Prob. 47E Ch. 2.1 - Prob. 48E Ch. 2.1 - Prob. 49E Ch. 2.1 - Prob. 50E Ch. 2.1 - 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Prob. 12E Ch. 2.2 - Prob. 13E Ch. 2.2 - Prob. 14E Ch. 2.2 - Prob. 15E Ch. 2.2 - Prob. 16E Ch. 2.2 - Prob. 17E Ch. 2.2 - Prob. 18E Ch. 2.2 - Prob. 19E Ch. 2.2 - Prob. 20E Ch. 2.2 - Prob. 21E Ch. 2.2 - Prob. 22E Ch. 2.2 - Prob. 23E Ch. 2.2 - Prob. 24E Ch. 2.2 - Prob. 25E Ch. 2.2 - Prob. 26E Ch. 2.2 - Prob. 27E Ch. 2.2 - Prob. 28E Ch. 2.2 - Prob. 29E Ch. 2.2 - Prob. 30E Ch. 2.2 - Prob. 31E Ch. 2.2 - Prob. 32E Ch. 2.2 - Prob. 33E Ch. 2.2 - Prob. 34E Ch. 2.2 - Prob. 35E Ch. 2.2 - Prob. 36E Ch. 2.2 - Prob. 37E Ch. 2.2 - Prob. 38E Ch. 2.2 - Prob. 39E Ch. 2.2 - Prob. 40E Ch. 2.2 - Prob. 41E Ch. 2.2 - Prob. 42E Ch. 2.2 - Prob. 43E Ch. 2.2 - Prob. 44E Ch. 2.2 - Prob. 45E Ch. 2.2 - Prob. 46E Ch. 2.2 - Prob. 47E Ch. 2.2 - Prob. 48E Ch. 2.2 - Prob. 49E Ch. 2.2 - Prob. 50E Ch. 2.2 - Prob. 51E Ch. 2.2 - Prob. 52E Ch. 2.2 - Prob. 53E Ch. 2.2 - Prob. 54E Ch. 2.2 - Prob. 55E Ch. 2.2 - Prob. 56E Ch. 2.2 - Prob. 57E Ch. 2.2 - Prob. 58E Ch. 2.2 - Prob. 59E Ch. 2.2 - Prob. 60E Ch. 2.2 - Prob. 61E Ch. 2.2 - Prob. 62E Ch. 2.2 - Prob. 63E Ch. 2.2 - Prob. 64E Ch. 2.2 - Prob. 65E Ch. 2.2 - Prob. 66E Ch. 2.2 - Prob. 67E Ch. 2.2 - Prob. 68E Ch. 2.2 - Prob. 69E Ch. 2.2 - Prob. 70E Ch. 2.2 - Prob. 71E Ch. 2.2 - Prob. 1QQ Ch. 2.2 - Prob. 2QQ Ch. 2.2 - Prob. 3QQ Ch. 2.2 - Prob. 4QQ Ch. 2.3 - Prob. 1QR Ch. 2.3 - Prob. 2QR Ch. 2.3 - Prob. 3QR Ch. 2.3 - Prob. 4QR Ch. 2.3 - Prob. 5QR Ch. 2.3 - Prob. 6QR Ch. 2.3 - Prob. 7QR Ch. 2.3 - Prob. 8QR Ch. 2.3 - Prob. 9QR Ch. 2.3 - Prob. 10QR Ch. 2.3 - Prob. 1E Ch. 2.3 - Prob. 2E Ch. 2.3 - Prob. 3E Ch. 2.3 - Prob. 4E Ch. 2.3 - Prob. 5E Ch. 2.3 - Prob. 6E Ch. 2.3 - Prob. 7E Ch. 2.3 - Prob. 8E Ch. 2.3 - Prob. 9E Ch. 2.3 - Prob. 10E Ch. 2.3 - Prob. 11E Ch. 2.3 - Prob. 12E Ch. 2.3 - Prob. 13E Ch. 2.3 - Prob. 14E Ch. 2.3 - Prob. 15E Ch. 2.3 - Prob. 16E Ch. 2.3 - Prob. 17E Ch. 2.3 - Prob. 18E Ch. 2.3 - Prob. 19E Ch. 2.3 - Prob. 20E Ch. 2.3 - Prob. 21E Ch. 2.3 - Prob. 22E Ch. 2.3 - Prob. 23E Ch. 2.3 - Prob. 24E Ch. 2.3 - Prob. 25E Ch. 2.3 - Prob. 26E Ch. 2.3 - Prob. 27E Ch. 2.3 - 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Prob. 5E Ch. 2.4 - Prob. 6E Ch. 2.4 - Prob. 7E Ch. 2.4 - Prob. 8E Ch. 2.4 - Prob. 9E Ch. 2.4 - Prob. 10E Ch. 2.4 - Prob. 11E Ch. 2.4 - Prob. 12E Ch. 2.4 - Prob. 13E Ch. 2.4 - Prob. 14E Ch. 2.4 - Prob. 15E Ch. 2.4 - Prob. 16E Ch. 2.4 - Prob. 17E Ch. 2.4 - Prob. 18E Ch. 2.4 - Prob. 19E Ch. 2.4 - Prob. 20E Ch. 2.4 - Prob. 21E Ch. 2.4 - Prob. 22E Ch. 2.4 - Prob. 23E Ch. 2.4 - Prob. 24E Ch. 2.4 - Prob. 25E Ch. 2.4 - Prob. 26E Ch. 2.4 - Prob. 27E Ch. 2.4 - Prob. 28E Ch. 2.4 - Prob. 29E Ch. 2.4 - Prob. 30E Ch. 2.4 - Prob. 31E Ch. 2.4 - Prob. 32E Ch. 2.4 - Prob. 33E Ch. 2.4 - Prob. 34E Ch. 2.4 - Prob. 35E Ch. 2.4 - Prob. 36E Ch. 2.4 - Prob. 37E Ch. 2.4 - Prob. 38E Ch. 2.4 - Prob. 39E Ch. 2.4 - Prob. 40E Ch. 2.4 - Prob. 41E Ch. 2.4 - Prob. 42E Ch. 2.4 - Prob. 43E Ch. 2.4 - Prob. 44E Ch. 2.4 - Prob. 45E Ch. 2.4 - Prob. 46E Ch. 2.4 - Prob. 47E Ch. 2.4 - Prob. 48E Ch. 2.4 - Prob. 49E Ch. 2.4 - Prob. 50E Ch. 2.4 - Prob. 51E Ch. 2.4 - Prob. 52E Ch. 2.4 - Prob. 53E Ch. 2.4 - Prob. 54E Ch. 2.4 - Prob. 55E Ch. 2.4 - Prob. 1QQ Ch. 2.4 - Prob. 2QQ Ch. 2.4 - Prob. 3QQ Ch. 2.4 - Prob. 4QQ Ch. 2 - Prob. 1RE Ch. 2 - Prob. 2RE Ch. 2 - Prob. 3RE Ch. 2 - Prob. 4RE Ch. 2 - Prob. 5RE Ch. 2 - Prob. 6RE Ch. 2 - Prob. 7RE Ch. 2 - Prob. 8RE Ch. 2 - Prob. 9RE Ch. 2 - Prob. 10RE Ch. 2 - Prob. 11RE Ch. 2 - Prob. 12RE Ch. 2 - Prob. 13RE Ch. 2 - Prob. 14RE Ch. 2 - Prob. 15RE Ch. 2 - Prob. 16RE Ch. 2 - Prob. 17RE Ch. 2 - Prob. 18RE Ch. 2 - Prob. 19RE Ch. 2 - Prob. 20RE Ch. 2 - Prob. 21RE Ch. 2 - Prob. 22RE Ch. 2 - Prob. 23RE Ch. 2 - Prob. 24RE Ch. 2 - Prob. 25RE Ch. 2 - Prob. 26RE Ch. 2 - Prob. 27RE Ch. 2 - Prob. 28RE Ch. 2 - Prob. 29RE Ch. 2 - Prob. 30RE Ch. 2 - Prob. 31RE Ch. 2 - Prob. 32RE Ch. 2 - Prob. 33RE Ch. 2 - Prob. 34RE Ch. 2 - Prob. 35RE Ch. 2 - Prob. 36RE Ch. 2 - Prob. 37RE Ch. 2 - Prob. 38RE Ch. 2 - Prob. 39RE Ch. 2 - Prob. 40RE Ch. 2 - Prob. 41RE Ch. 2 - Prob. 42RE Ch. 2 - Prob. 43RE Ch. 2 - Prob. 44RE Ch. 2 - Prob. 45RE Ch. 2 - Prob. 46RE Ch. 2 - Prob. 47RE Ch. 2 - Prob. 48RE Ch. 2 - Prob. 49RE Ch. 2 - Prob. 50RE Ch. 2 - Prob. 51RE Ch. 2 - Prob. 52RE Ch. 2 - Prob. 53RE Ch. 2 - Prob. 54RE Ch. 2 - Prob. 55RE

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The points of discontinuity of f ( x ) = { 1 − x 2 , x ≠ − 1 2 , x = − 1 .

Concept explainers

Videos

To find: The points of discontinuity of f(x)={1−x2, x≠−12, x=−1 .

Answer to Problem 22E

Explanation of Solution

To check: Whether the discontinuity of f(x)={1−x2, x≠−12, x=−1 is removable or not, explain.

Answer to Problem 22E

Explanation of Solution

Chapter 2 Solutions

Additional Math Textbook Solutions