a. Answer The interval is- Explanation Given Information: The given function is- Calculation : The intermediate value theorem states that “If a function is continuous on a closed interval and c is any number between and inclusive, then there is at least one number in the closed interval such that It can be known that the equation given to us are continuous. So, if it can be found out that two numbers such that the value at these two points changes its sign then we could say that the graph should have passed 0 in that interval. So, for the function if substitute 0 the result is obtained -1 as and if 5 is substituted, -1 as a resultant. Likewise, it can be found a range by substituting 1 and 2. So the values at both these points changed it its sign. Therefore, in this interval there must be a zero. So, the answer is b. To determine To plot the graph of the above function. Explanation If the graphing utility is used that could be found, the values at which these zeros occur. So, the value obtained by using computer aid is Using the graphing utility to draw the graph and see that there is a zero in this region. So, the graph is,

4th Edition

ISBN: 9781337516853

Author: Larson

Publisher: CENGAGE CO

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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, Problem 22RE

To determine

The find out the interval in which the zero of the polynomial lie.

Expert Solution

Answer to Problem 22RE

The interval is-

EBK PRECALCULUS W/LIMITS, Chapter 2, Problem 22RE , additional homework tip 1

Explanation of Solution

Given Information:

The given function is-

EBK PRECALCULUS W/LIMITS, Chapter 2, Problem 22RE , additional homework tip 2

Calculation:

The intermediate value theorem states that “If a function EBK PRECALCULUS W/LIMITS, Chapter 2, Problem 22RE , additional homework tip 3 is continuous on a closed interval and c is any number between and inclusive, then there is at least one number in the closed interval such that

It can be known that the equation given to us are continuous. So, if it can be found out that two numbers such that EBK PRECALCULUS W/LIMITS, Chapter 2, Problem 22RE , additional homework tip 9 the value at these two points changes its sign then we could say that the graph should have passed 0 in that interval.

So, for the function EBK PRECALCULUS W/LIMITS, Chapter 2, Problem 22RE , additional homework tip 10 if substitute 0 the result is obtained -1 as and if 5 is substituted, -1 as a resultant. Likewise, it can be found a range by substituting 1 and 2. So the values at both these points changed it its sign. Therefore, in this interval there must be a zero.

So, the answer is EBK PRECALCULUS W/LIMITS, Chapter 2, Problem 22RE , additional homework tip 11

To determine

To plot the graph of the above function.

Expert Solution

Explanation of Solution

If the graphing utility is used that could be found, the values at which these zeros occur. So, the value obtained by using computer aid is EBK PRECALCULUS W/LIMITS, Chapter 2, Problem 22RE , additional homework tip 12 Using the graphing utility to draw the graph and see that there is a zero in this region. So, the graph is,

EBK PRECALCULUS W/LIMITS, Chapter 2, Problem 22RE , additional homework tip 13

Chapter 2, Problem 21RE

Chapter 2, Problem 23RE

Chapter 2 Solutions

EBK PRECALCULUS W/LIMITS

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Prob. 33E Ch. 2.7 - Prob. 34E Ch. 2.7 - Prob. 35E Ch. 2.7 - Prob. 36E Ch. 2.7 - Prob. 37E Ch. 2.7 - Prob. 38E Ch. 2.7 - Prob. 39E Ch. 2.7 - Prob. 40E Ch. 2.7 - Prob. 41E Ch. 2.7 - Prob. 42E Ch. 2.7 - Prob. 43E Ch. 2.7 - Prob. 44E Ch. 2.7 - Prob. 45E Ch. 2.7 - Prob. 46E Ch. 2.7 - Prob. 47E Ch. 2.7 - Prob. 48E Ch. 2.7 - Prob. 49E Ch. 2.7 - Prob. 50E Ch. 2.7 - Prob. 51E Ch. 2.7 - Prob. 52E Ch. 2.7 - Prob. 53E Ch. 2.7 - Prob. 54E Ch. 2.7 - Prob. 55E Ch. 2.7 - Prob. 56E Ch. 2.7 - Prob. 57E Ch. 2.7 - Prob. 58E Ch. 2.7 - Prob. 59E Ch. 2.7 - Prob. 60E Ch. 2.7 - Prob. 61E Ch. 2.7 - Prob. 62E Ch. 2.7 - Prob. 63E Ch. 2.7 - Prob. 64E Ch. 2.7 - Prob. 65E Ch. 2.7 - Prob. 66E Ch. 2.7 - Prob. 67E Ch. 2.7 - Prob. 68E Ch. 2.7 - Prob. 69E Ch. 2.7 - Prob. 70E Ch. 2.7 - Prob. 71E Ch. 2.7 - Prob. 72E Ch. 2.7 - Prob. 73E Ch. 2.7 - Prob. 74E Ch. 2.7 - Prob. 75E Ch. 2.7 - Prob. 76E Ch. 2.7 - Prob. 77E Ch. 2.7 - Prob. 78E Ch. 2.7 - Prob. 79E Ch. 2.7 - Prob. 80E Ch. 2.7 - Prob. 81E Ch. 2.7 - Prob. 82E Ch. 2.7 - Prob. 83E Ch. 2.7 - Prob. 84E Ch. 2.7 - Prob. 85E Ch. 2.7 - Prob. 86E Ch. 2.7 - Prob. 87E Ch. 2.7 - Prob. 88E Ch. 2.7 - Prob. 89E Ch. 2.7 - Prob. 90E 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 Ch. 2 - Prob. 56RE Ch. 2 - Prob. 57RE Ch. 2 - Prob. 58RE Ch. 2 - Prob. 59RE Ch. 2 - Prob. 60RE Ch. 2 - Prob. 61RE Ch. 2 - Prob. 62RE Ch. 2 - Prob. 63RE Ch. 2 - Prob. 64RE Ch. 2 - Prob. 65RE Ch. 2 - Prob. 66RE Ch. 2 - Prob. 67RE Ch. 2 - Prob. 68RE Ch. 2 - Prob. 1CT Ch. 2 - Prob. 2CT Ch. 2 - Prob. 3CT Ch. 2 - Prob. 4CT Ch. 2 - Prob. 5CT Ch. 2 - Prob. 6CT Ch. 2 - Prob. 7CT Ch. 2 - Prob. 8CT Ch. 2 - Prob. 9CT Ch. 2 - Prob. 10CT Ch. 2 - Prob. 11CT Ch. 2 - Prob. 12CT Ch. 2 - Prob. 13CT Ch. 2 - Prob. 14CT Ch. 2 - Prob. 15CT Ch. 2 - Prob. 16CT Ch. 2 - Prob. 17CT Ch. 2 - Prob. 18CT Ch. 2 - Prob. 1PS Ch. 2 - Prob. 2PS Ch. 2 - Prob. 3PS Ch. 2 - Prob. 4PS Ch. 2 - Prob. 5PS Ch. 2 - Prob. 6PS Ch. 2 - Prob. 7PS Ch. 2 - Prob. 8PS Ch. 2 - Prob. 9PS Ch. 2 - Prob. 10PS Ch. 2 - Prob. 11PS Ch. 2 - Prob. 12PS Ch. 2 - Prob. 13PS

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