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 39E

To determine

To find: The reason for the continuity of the function y=|x2−4x| , check the domain of the function, also mention the theorems used in the continuity of function with the functions assumed as continuous.

Expert Solution & Answer

Answer to Problem 39E

The domain of the function y=|x2−4x| is equal to interval (−∞,∞) and the difference, product, constant multiples and composition theorems are used and the functions x and |x| are assumed as continuous.

Explanation of Solution

Given information:

The function is y=|x2−4x| .

Calculation:

The function y=|x2−4x| is the composition of two functions f(x)=|x| and g(x)=x2−4x .

The function y=|x2−4x| is equal to the function (fog)(x) .

Assume the functions x and |x| are continuous functions.

The function g(x)=x2−4x is the subtraction of two functions h(x)=x2 and j(x)=4x .

The function h(x)=x2 is the product of function x twice. So, by the product theorem the function x2 is also a continuous function.

The function j(x)=4x is the constant multiple of function x . So, by the constant multiple theorem the function 4x is also a continuous function.

As the functions h(x)=x2 and j(x)=4x are continuous functions. So, by the subtraction theorem the function g(x)=x2−4x is also a continuous function.

As the functions f(x)=|x| and g(x)=x2−4x are continuous functions. So, by the composition theorem y=(fog)(x) is also continuous function.

The function y=|x2−4x| is defined for all the real values of x . So, the domain of the function y=|x2−4x| is equal to interval (−∞,∞) .

Therefore, the domain of the function y=|x2−4x| is equal to interval (−∞,∞) and the difference, product, constant multiples and composition theorems are used and the functions x and |x| are assumed as continuous.

Chapter 2.3, Problem 38E

Chapter 2.3, Problem 40E

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. 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 reason for the continuity of the function y = | x 2 − 4 x | , check the domain of the function, also mention the theorems used in the continuity of function with the functions assumed as continuous.

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To find: The reason for the continuity of the function y=|x2−4x| , check the domain of the function, also mention the theorems used in the continuity of function with the functions assumed as continuous.

Answer to Problem 39E

Explanation of Solution

Chapter 2 Solutions

Additional Math Textbook Solutions