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

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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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 2.3, Problem 52E

(a)

To determine

To show: The equation that represent salary of a person is given by y=36,500(1.035)int t .

(a)

Expert Solution

Explanation of Solution

Given information: The initial salary of a person is $36,500 and a contract promises to increase salary each year for 4 years at 3.5% rate.

Proof:

The general formula for an amount is given by,

A=P(1+r)t

To find the first year salary (0≤t<1) , substitute 0 for t , 0.035 for r and 36500 for P in the above formula.

A=36500(1+0.035)0=36500

To find the second year salary (1≤t<2) , substitute 1 for t , 0.035 for r and 36500 for P in the above formula.

A=36500(1+0.035)1=36500(1.035)

To find the third year salary (2≤t<3) , substitute 2 for t , 0.035 for r and 36500 for P in the above formula.

A=36500(1+0.035)2=36500(1.035)2

To find the fourth year salary (3≤t<4) , substitute 3 for t , 0.035 for r and 36500 for P in the above formula.

A=36500(1+0.035)3=36500(1.035)3

So, the above function corresponds to y=36500(1.035)int t .

Hence, it is proved that the Salary increment is given by y=36500(1.035)int t .

(b)

To determine

To find: The graph of the function that represents salary and the values of t for which the function is continuous.

(b)

Expert Solution

Answer to Problem 52E

The graph of the function y=36500(1.035)int t is shown in Figure (1) and it is continuous at all points of domain [0, 5) except t=1, 2, 3, 4 .

Explanation of Solution

Given information: The initial salary of a person is $36,500 and a contract promises to increase salary each year for 4 years at 3.5% rate.

Calculation:

To draw the graph of the function y=36500(1.035)int t , follow the steps given below.

Press “ON” button to open the calculator. Press Y= key and enter the right hand side of the equation y=36500(1.035)int t after the symbol Y1 . Enter the keystrokes as given below:

36500(1.035)∧MATH⇒5(X)

Set the window at [0, 5] by [35000, 45000] . Press GRAPH key to produce the graph of the function as shown below.

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 2.3, Problem 52E

Figure (1)

From the graph it can be seen that the function is not continuous at points t=1, 2, 3, 4 .

Therefore, the graph of the function y=36500(1.035)int t is shown in Figure (1) and it is continuous at all points of domain [0, 5) except t=1, 2, 3, 4 .

Chapter 2.3, Problem 51E

Chapter 2.3, Problem 53E

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 equation that represent salary of a person is given by y = 36 , 500 ( 1.035 ) int t .

Concept explainers

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To show: The equation that represent salary of a person is given by y=36,500(1.035)int t .

Explanation of Solution

To find: The graph of the function that represents salary and the values of t for which the function is continuous.

Answer to Problem 52E

Explanation of Solution

Chapter 2 Solutions

Additional Math Textbook Solutions