(a) Answer The value of f ( − 1 ) exists. Explanation Given information: The given function is f ( x ) = { x 2 − 1 , − 1 ≤ x < 0 2 x , 0 < x < 1 1 , x = 1 − 2 x + 4 , 1 < x < 2 0 , 2 < x < 3 and the graph of the function is given below: Calculation: As observed from the graph of the function, the value of the function f ( x ) at x = − 1 is 0 . ⇒ f ( − 1 ) = 0 Therefore, the value of f ( − 1 ) exists. (b) To determine To check: Whether the value of lim x → − 1 + f ( x ) exists or not. Answer The value of lim x → − 1 + f ( x ) exists. Explanation Given information: The given function is f ( x ) = { x 2 − 1 , − 1 ≤ x < 0 2 x , 0 < x < 1 1 , x = 1 − 2 x + 4 , 1 < x < 2 0 , 2 < x < 3 and the graph of the function is given below: Calculation: As observed from the graph of the function, the function f ( x ) approaches to 0 as x approaches to − 1 from the right side. lim x → − 1 + f ( x ) = 0 Therefore, the value of lim x → − 1 + f ( x ) exists. (c) To determine To prove: Whether the value of lim x → − 1 + f ( x ) is equal to f ( − 1 ) or not. Explanation Given information: The given function is f ( x ) = { x 2 − 1 , − 1 ≤ x < 0 2 x , 0 < x < 1 1 , x = 1 − 2 x + 4 , 1 < x < 2 0 , 2 < x < 3 and the graph of the function is given below: Calculation: As calculated in part(a) and part (b), the value of f ( − 1 ) and lim x → − 1 + f ( x ) both are 0. ⇒ lim x → − 1 + f ( x ) = f ( − 1 ) Hence, it is proved that the value of lim x → − 1 + f ( x ) is equal to f ( − 1 ) . (d) To determine To check: Whether the function f is continuous at x = − 1 or not. Answer The function f is continuous at x = − 1 . Explanation Given information: The given function is f ( x ) = { x 2 − 1 , − 1 ≤ x < 0 2 x , 0 < x < 1 1 , x = 1 − 2 x + 4 , 1 < x < 2 0 , 2 < x < 3 and the graph of the function is given below: Calculation: As calculated in part(c), the value of both lim x → − 1 + f ( x ) and f ( − 1 ) is equal to 0 . From the graph it can be observed that x = − 1 is an endpoint and lim x → − 1 + f ( x ) = f ( − 1 ) . So, the function f ( x ) is continuous at x = − 1 . Therefore, the function f is continuous at x = − 1 .

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 11E

(a)

To determine

To check:Whetherthe value of f(−1) exists or not.

(a)

Expert Solution

Answer to Problem 11E

The value of f(−1) exists.

Explanation of Solution

Given information:

The given function is f(x)={x2−1, −1≤x<02x, 0<x<11, x=1−2x+4, 1<x<20, 2<x<3 and the graph of the function is given below:

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 2.3, Problem 11E , additional homework tip 1

Calculation:

As observed from the graph of the function, the value of the function f(x) at x=−1 is 0 .

⇒f(−1)=0

Therefore, the value of f(−1) exists.

(b)

To determine

To check:Whether the value of limx→−1+f(x) exists or not.

(b)

Expert Solution

Answer to Problem 11E

The value of limx→−1+f(x) exists.

Explanation of Solution

Given information:

The given function is f(x)={x2−1, −1≤x<02x, 0<x<11, x=1−2x+4, 1<x<20, 2<x<3

and the graph of the function is given below:

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 2.3, Problem 11E , additional homework tip 2

Calculation:

As observed from the graph of the function, the function f(x) approaches to 0 as x approaches to −1 from the right side.

limx→−1+f(x)=0

Therefore, the value of limx→−1+f(x) exists.

(c)

To determine

To prove: Whether the value of limx→−1+f(x) is equal to f(−1) or not.

(c)

Expert Solution

Explanation of Solution

Given information:

The given function is f(x)={x2−1, −1≤x<02x, 0<x<11, x=1−2x+4, 1<x<20, 2<x<3

and the graph of the function is given below:

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 2.3, Problem 11E , additional homework tip 3

Calculation:

As calculated in part(a) and part (b), the value of f(−1) and limx→−1+f(x) both are 0.

⇒limx→−1+f(x)=f(−1)

Hence, it is proved that the value of limx→−1+f(x) is equal to f(−1) .

(d)

To determine

To check: Whether the function f is continuous at x=−1 or not.

(d)

Expert Solution

Answer to Problem 11E

The function f is continuous at x=−1 .

Explanation of Solution

Given information:

The given function is f(x)={x2−1, −1≤x<02x, 0<x<11, x=1−2x+4, 1<x<20, 2<x<3

and the graph of the function is given below:

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 2.3, Problem 11E , additional homework tip 4

Calculation:

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

From the graph it can be observed that x=−1 is an endpoint and limx→−1+f(x)=f(−1) . So, the function f(x) is continuous at x=−1 .

Therefore, the function f is continuous at x=−1 .

Chapter 2.3, Problem 10E

Chapter 2.3, Problem 12E

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. 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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Finding Local Maxima and Minima by Differentiation; Author: Professor Dave Explains;https://www.youtube.com/watch?v=pvLj1s7SOtk;License: Standard YouTube License, CC-BY

To check: Whetherthe value of f ( − 1 ) exists or not.

Concept explainers

Videos

To check:Whetherthe value of f(−1) exists or not.

Answer to Problem 11E

Explanation of Solution

To check:Whether the value of limx→−1+f(x) exists or not.

Answer to Problem 11E

Explanation of Solution

To prove: Whether the value of limx→−1+f(x) is equal to f(−1) or not.

Explanation of Solution

To check: Whether the function f is continuous at x=−1 or not.

Answer to Problem 11E

Explanation of Solution

Chapter 2 Solutions

Additional Math Textbook Solutions