Explanation Given: f ( x ) = x 1 − x 2 Concept Used: Invertible function: the function fis called an inverse function if for every value of y in the defined range there exist unique value of x in the defined domain i.e. y = f ( x ) . Graph: Graph 1 To determine (b) To find: h ( − 10 ) , h ( − 2 ) , h ( 0 ) , h ( 3 ) , h ( 9 ) To determine (c) To find: the expression for h ( x )

4th Edition

ISBN: 9781319309671

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 1.5, Problem 58E

To determine

(a)

To graph:

f(x)=x1−x2

To explain:

f is not invertible. Furthermore, explain why f is invertible if we restrict the domain to (−1,1)

To determine

(b)

To find:

h(−10),h(−2),h(0),h(3),h(9)

To determine

(c)

To find:

the expression for h(x)

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Chapter 1.5, Problem 57E

Chapter 1.5, Problem 59E

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Chapter 1 Solutions

CALCULUS: EARLY TRANS 4TH ED W/ ACCESS

Ch. 1.1 - Prob. 1PQ Ch. 1.1 - Prob. 2PQ Ch. 1.1 - Prob. 3PQ Ch. 1.1 - Prob. 4PQ Ch. 1.1 - Prob. 5PQ Ch. 1.1 - Prob. 6PQ Ch. 1.1 - Prob. 7PQ Ch. 1.1 - Prob. 8PQ Ch. 1.1 - Prob. 9PQ Ch. 1.1 - Prob. 10PQ

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - 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Prob. 5E Ch. 1.5 - Prob. 6E Ch. 1.5 - Prob. 7E Ch. 1.5 - Prob. 8E Ch. 1.5 - Prob. 9E Ch. 1.5 - Prob. 10E Ch. 1.5 - Prob. 11E Ch. 1.5 - Prob. 12E Ch. 1.5 - Prob. 13E Ch. 1.5 - Prob. 14E Ch. 1.5 - Prob. 15E Ch. 1.5 - Prob. 16E Ch. 1.5 - Prob. 17E Ch. 1.5 - Prob. 18E Ch. 1.5 - Prob. 19E Ch. 1.5 - Prob. 20E Ch. 1.5 - Prob. 21E Ch. 1.5 - Prob. 22E Ch. 1.5 - Prob. 23E Ch. 1.5 - Prob. 24E Ch. 1.5 - Prob. 25E Ch. 1.5 - Prob. 26E Ch. 1.5 - Prob. 27E Ch. 1.5 - Prob. 28E Ch. 1.5 - Prob. 29E Ch. 1.5 - Prob. 30E Ch. 1.5 - Prob. 31E Ch. 1.5 - Prob. 32E Ch. 1.5 - Prob. 33E Ch. 1.5 - Prob. 34E Ch. 1.5 - Prob. 35E Ch. 1.5 - Prob. 36E Ch. 1.5 - Prob. 37E Ch. 1.5 - Prob. 38E Ch. 1.5 - Prob. 39E Ch. 1.5 - Prob. 40E Ch. 1.5 - Prob. 41E Ch. 1.5 - Prob. 42E Ch. 1.5 - Prob. 43E Ch. 1.5 - Prob. 44E Ch. 1.5 - Prob. 45E Ch. 1.5 - Prob. 46E Ch. 1.5 - Prob. 47E Ch. 1.5 - Prob. 48E Ch. 1.5 - Prob. 49E Ch. 1.5 - Prob. 50E Ch. 1.5 - Prob. 51E Ch. 1.5 - Prob. 52E Ch. 1.5 - Prob. 53E Ch. 1.5 - Prob. 54E Ch. 1.5 - Prob. 55E Ch. 1.5 - Prob. 56E Ch. 1.5 - Prob. 57E Ch. 1.5 - Prob. 58E Ch. 1.5 - Prob. 59E Ch. 1.5 - Prob. 60E Ch. 1.6 - Prob. 1PQ Ch. 1.6 - Prob. 2PQ Ch. 1.6 - Prob. 3PQ Ch. 1.6 - Prob. 4PQ Ch. 1.6 - Prob. 5PQ Ch. 1.6 - Prob. 6PQ Ch. 1.6 - Prob. 7PQ Ch. 1.6 - Prob. 8PQ Ch. 1.6 - Prob. 9PQ Ch. 1.6 - Prob. 1E Ch. 1.6 - Prob. 2E Ch. 1.6 - Prob. 3E Ch. 1.6 - Prob. 4E Ch. 1.6 - Prob. 5E Ch. 1.6 - Prob. 6E Ch. 1.6 - Prob. 7E Ch. 1.6 - Prob. 8E Ch. 1.6 - Prob. 9E Ch. 1.6 - Prob. 10E Ch. 1.6 - Prob. 11E Ch. 1.6 - Prob. 12E Ch. 1.6 - Prob. 13E Ch. 1.6 - Prob. 14E Ch. 1.6 - Prob. 15E Ch. 1.6 - Prob. 16E Ch. 1.6 - Prob. 17E Ch. 1.6 - Prob. 18E Ch. 1.6 - Prob. 19E Ch. 1.6 - Prob. 20E Ch. 1.6 - Prob. 21E Ch. 1.6 - Prob. 22E Ch. 1.6 - Prob. 23E Ch. 1.6 - Prob. 24E Ch. 1.6 - Prob. 25E Ch. 1.6 - Prob. 26E Ch. 1.6 - Prob. 27E Ch. 1.6 - Prob. 28E Ch. 1.6 - Prob. 29E Ch. 1.6 - Prob. 30E Ch. 1.6 - Prob. 31E Ch. 1.6 - Prob. 32E Ch. 1.6 - Prob. 33E Ch. 1.6 - Prob. 34E Ch. 1.6 - Prob. 35E Ch. 1.6 - Prob. 36E Ch. 1.6 - Prob. 37E Ch. 1.6 - 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(a) To graph: f ( x ) = x 1 − x 2 To explain: f is not invertible. Furthermore, explain why f is invertible if we restrict the domain to ( − 1 , 1 )

Solution Summary: The author explains that f is not invertible if we restrict the domain to (-1,1)

Concept explainers

Videos

(a)

To graph:

f(x)=x1−x2

To explain:

f is not invertible. Furthermore, explain why f is invertible if we restrict the domain to (−1,1)

(b)

To find:

h(−10),h(−2),h(0),h(3),h(9)

(c)

To find:

the expression for h(x)

Want to see the full answer?

Chapter 1 Solutions

(a) To graph: f ( x ) = x 1 − x 2 To explain: f is not invertible. Furthermore, explain why f is invertible if we restrict the domain to ( − 1 , 1 )

Solution Summary: The author explains that f is not invertible if we restrict the domain to (-1,1)

Concept explainers

Videos

(a) To graph: f(x)=x1−x2 To explain: f is not invertible. Furthermore, explain why f is invertible if we restrict the domain to (−1,1)

(b) To find: h(−10),h(−2),h(0),h(3),h(9)

(c) To find: the expression for h(x)

Want to see the full answer?

Chapter 1 Solutions

(a)

To graph:

f(x)=x1−x2

To explain:

f is not invertible. Furthermore, explain why f is invertible if we restrict the domain to (−1,1)

(b)

To find:

h(−10),h(−2),h(0),h(3),h(9)

(c)

To find:

the expression for h(x)