Explanation Given information: g ( x ) = x x − 1 Formula used: Domain is the input value for the given function. Calculation: Consider the function g ( x ) = x x − 1 . To show that g ( x ) = g − 1 ( x ) , consider g ( g ( x ) ) . g ( g ( x ) ) = ( x x − 1 ) ( x x − 1 ) − 1 = ( x x − 1 )

4th Edition

ISBN: 9781319379421

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 1, Problem 62CRE

To determine

Show that g(x)=xx−1 is equal to its inverse on the domain [x:x≠1].

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Chapter 1, Problem 61CRE

Chapter 1, Problem 63CRE

Chapter 1 Solutions

CALCULUS 4E (HC) W/ ACHIEVE 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. 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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Show that g ( x ) = x x − 1 is equal to its inverse on the domain [ x : x ≠ 1 ] .

Solution Summary: The author explains that g(x)=xx-1 is equal to its inverse on the domain.

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Show that g(x)=xx−1 is equal to its inverse on the domain [x:x≠1].

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Chapter 1 Solutions