a. Answer The function will be one-to-one because no two value of x produce the same value of f ( x ) . Explanation Given information : The function is f ( x ) = 9 5 x + 32 . from the function it can be observed that there is no possibility to produce the same value of f ( x ) for two different value of x . The function will be one-to-one because no two value of x produce the same value of f ( x ) . b. To determine To state : the representation of f − 1 ( 50 ) . Answer The f − 1 ( 50 ) will represent the value of 50 degree Fahrenheit in degree Celsius. Explanation Given information : The function is f ( x ) = 9 5 x + 32 . from the function it can be observed that the in the function x represent the degree Celsius and f ( x ) represent the corresponding temperature in degree Fahrenheit. Thus, the f − 1 ( 50 ) will represent the value of 50 degree Fahrenheit in degree Celsius.

8th Edition

ISBN: 9781337904285

Author: Larson

Publisher: CENGAGE L

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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 1.6, Problem 135E

To determine

To make : the conceptual argument to show that f has an inverse function.

Expert Solution

Answer to Problem 135E

The function will be one-to-one because no two value of x produce the same value of f(x).

Explanation of Solution

Given information :

The function is f(x)=95x+32.

from the function it can be observed that there is no possibility to produce the same value of f(x) for two different value of x.

The function will be one-to-one because no two value of x produce the same value of f(x).

To determine

To state : the representation of f−1(50).

Expert Solution

Answer to Problem 135E

The f−1(50) will represent the value of 50 degree Fahrenheit in degree Celsius.

Explanation of Solution

Given information :

The function is f(x)=95x+32.

from the function it can be observed that the in the function x represent the degree Celsius and f(x) represent the corresponding temperature in degree Fahrenheit.

Thus, the f−1(50) will represent the value of 50 degree Fahrenheit in degree Celsius.

Chapter 1.6, Problem 134E

Chapter 1.6, Problem 136E

Chapter 1 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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Prob. 79E Ch. 1.6 - Prob. 80E Ch. 1.6 - Prob. 81E Ch. 1.6 - Prob. 82E Ch. 1.6 - Prob. 83E Ch. 1.6 - Prob. 84E Ch. 1.6 - Prob. 85E Ch. 1.6 - Prob. 86E Ch. 1.6 - Prob. 87E Ch. 1.6 - Prob. 88E Ch. 1.6 - Prob. 89E Ch. 1.6 - Prob. 90E Ch. 1.6 - Prob. 91E Ch. 1.6 - Prob. 92E Ch. 1.6 - Prob. 93E Ch. 1.6 - Prob. 94E Ch. 1.6 - Prob. 95E Ch. 1.6 - Prob. 96E Ch. 1.6 - Prob. 97E Ch. 1.6 - Prob. 98E Ch. 1.6 - Prob. 99E Ch. 1.6 - Prob. 100E Ch. 1.6 - Prob. 101E Ch. 1.6 - Prob. 102E Ch. 1.6 - Prob. 103E Ch. 1.6 - Prob. 104E Ch. 1.6 - Prob. 105E Ch. 1.6 - Prob. 106E Ch. 1.6 - Prob. 107E Ch. 1.6 - Prob. 108E Ch. 1.6 - Prob. 109E Ch. 1.6 - Prob. 110E Ch. 1.6 - Prob. 111E Ch. 1.6 - Prob. 112E Ch. 1.6 - Prob. 113E Ch. 1.6 - Prob. 114E Ch. 1.6 - Prob. 115E Ch. 1.6 - Prob. 116E Ch. 1.6 - Prob. 117E Ch. 1.6 - Prob. 118E Ch. 1.6 - Prob. 119E Ch. 1.6 - Prob. 120E Ch. 1.6 - Prob. 121E Ch. 1.6 - Prob. 122E Ch. 1.6 - Prob. 123E Ch. 1.6 - Prob. 124E Ch. 1.6 - Prob. 125E Ch. 1.6 - Prob. 126E Ch. 1.6 - Prob. 127E Ch. 1.6 - Prob. 128E Ch. 1.6 - Prob. 129E Ch. 1.6 - Prob. 130E Ch. 1.6 - Prob. 131E Ch. 1.6 - Prob. 132E Ch. 1.6 - Prob. 133E Ch. 1.6 - Prob. 134E Ch. 1.6 - Prob. 135E Ch. 1.6 - Prob. 136E Ch. 1.6 - Prob. 137E Ch. 1.6 - Prob. 138E Ch. 1.6 - Prob. 139E Ch. 1.6 - Prob. 140E Ch. 1.6 - Prob. 141E Ch. 1.6 - Prob. 142E Ch. 1.6 - Prob. 143E Ch. 1.6 - Prob. 144E Ch. 1.7 - Prob. 1E Ch. 1.7 - Prob. 2E Ch. 1.7 - Prob. 3E Ch. 1.7 - Prob. 4E Ch. 1.7 - Prob. 5E Ch. 1.7 - Prob. 6E Ch. 1.7 - Prob. 7E Ch. 1.7 - Prob. 8E Ch. 1.7 - Prob. 9E Ch. 1.7 - Prob. 10E Ch. 1.7 - Prob. 11E Ch. 1.7 - Prob. 12E Ch. 1.7 - Prob. 13E Ch. 1.7 - Prob. 14E Ch. 1.7 - Prob. 15E Ch. 1.7 - Prob. 16E Ch. 1.7 - Prob. 17E Ch. 1.7 - Prob. 18E Ch. 1.7 - Prob. 19E Ch. 1.7 - Prob. 20E Ch. 1.7 - Prob. 21E Ch. 1.7 - Prob. 22E Ch. 1.7 - Prob. 23E Ch. 1.7 - Prob. 24E Ch. 1.7 - Prob. 25E Ch. 1.7 - Prob. 26E Ch. 1.7 - Prob. 27E Ch. 1.7 - Prob. 28E Ch. 1.7 - Prob. 29E Ch. 1.7 - Prob. 30E Ch. 1.7 - Prob. 31E Ch. 1.7 - Prob. 32E Ch. 1.7 - Prob. 33E Ch. 1.7 - Prob. 34E Ch. 1 - Prob. 1CR Ch. 1 - Prob. 2CR Ch. 1 - Prob. 3CR Ch. 1 - Prob. 4CR Ch. 1 - Prob. 5CR Ch. 1 - Prob. 6CR Ch. 1 - Prob. 7CR Ch. 1 - Prob. 8CR Ch. 1 - Prob. 9CR Ch. 1 - Prob. 10CR Ch. 1 - Prob. 11CR Ch. 1 - Prob. 12CR Ch. 1 - Prob. 13CR Ch. 1 - Prob. 14CR Ch. 1 - Prob. 15CR Ch. 1 - Prob. 16CR Ch. 1 - Prob. 17CR Ch. 1 - Prob. 18CR Ch. 1 - Prob. 19CR Ch. 1 - Prob. 20CR Ch. 1 - Prob. 21CR Ch. 1 - Prob. 22CR Ch. 1 - Prob. 23CR Ch. 1 - Prob. 24CR Ch. 1 - Prob. 25CR Ch. 1 - Prob. 26CR Ch. 1 - Prob. 27CR Ch. 1 - Prob. 28CR Ch. 1 - Prob. 29CR Ch. 1 - Prob. 30CR Ch. 1 - Prob. 31CR Ch. 1 - Prob. 32CR Ch. 1 - Prob. 33CR Ch. 1 - Prob. 34CR Ch. 1 - Prob. 35CR Ch. 1 - Prob. 36CR Ch. 1 - Prob. 37CR Ch. 1 - Prob. 38CR Ch. 1 - Prob. 39CR Ch. 1 - Prob. 40CR Ch. 1 - Prob. 41CR Ch. 1 - Prob. 42CR Ch. 1 - Prob. 43CR Ch. 1 - Prob. 44CR Ch. 1 - Prob. 45CR Ch. 1 - Prob. 46CR Ch. 1 - Prob. 47CR Ch. 1 - Prob. 48CR Ch. 1 - Prob. 49CR Ch. 1 - Prob. 50CR Ch. 1 - Prob. 51CR Ch. 1 - Prob. 52CR Ch. 1 - Prob. 53CR Ch. 1 - Prob. 54CR Ch. 1 - Prob. 55CR Ch. 1 - Prob. 56CR Ch. 1 - Prob. 57CR Ch. 1 - Prob. 58CR Ch. 1 - Prob. 59CR Ch. 1 - Prob. 60CR Ch. 1 - Prob. 61CR Ch. 1 - Prob. 62CR Ch. 1 - Prob. 63CR Ch. 1 - Prob. 64CR Ch. 1 - Prob. 65CR Ch. 1 - Prob. 66CR Ch. 1 - Prob. 67CR Ch. 1 - Prob. 68CR Ch. 1 - Prob. 69CR Ch. 1 - Prob. 70CR Ch. 1 - Prob. 71CR Ch. 1 - Prob. 72CR Ch. 1 - Prob. 73CR Ch. 1 - Prob. 74CR Ch. 1 - Prob. 75CR Ch. 1 - Prob. 76CR Ch. 1 - Prob. 77CR Ch. 1 - Prob. 78CR Ch. 1 - Prob. 79CR Ch. 1 - Prob. 80CR Ch. 1 - Prob. 81CR Ch. 1 - Prob. 82CR Ch. 1 - Prob. 83CR Ch. 1 - Prob. 84CR Ch. 1 - Prob. 85CR Ch. 1 - Prob. 86CR Ch. 1 - Prob. 87CR Ch. 1 - Prob. 88CR Ch. 1 - Prob. 89CR Ch. 1 - Prob. 90CR Ch. 1 - Prob. 91CR Ch. 1 - Prob. 92CR Ch. 1 - Prob. 93CR Ch. 1 - Prob. 94CR Ch. 1 - Prob. 95CR Ch. 1 - Prob. 96CR Ch. 1 - Prob. 97CR Ch. 1 - Prob. 98CR Ch. 1 - Prob. 99CR Ch. 1 - Prob. 100CR Ch. 1 - Prob. 101CR Ch. 1 - Prob. 102CR Ch. 1 - Prob. 103CR Ch. 1 - Prob. 104CR Ch. 1 - Prob. 105CR Ch. 1 - Prob. 106CR Ch. 1 - Prob. 107CR Ch. 1 - Prob. 108CR Ch. 1 - Prob. 109CR Ch. 1 - Prob. 110CR Ch. 1 - Prob. 1CT Ch. 1 - Prob. 2CT Ch. 1 - Prob. 3CT Ch. 1 - Prob. 4CT Ch. 1 - Prob. 5CT Ch. 1 - Prob. 6CT Ch. 1 - Prob. 7CT Ch. 1 - Prob. 8CT Ch. 1 - Prob. 9CT Ch. 1 - Prob. 10CT Ch. 1 - Prob. 11CT Ch. 1 - Prob. 12CT Ch. 1 - Prob. 13CT Ch. 1 - Prob. 14CT Ch. 1 - Prob. 15CT Ch. 1 - Prob. 16CT Ch. 1 - Prob. 17CT Ch. 1 - Prob. 18CT Ch. 1 - Prob. 19CT Ch. 1 - Prob. 20CT

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