Explanation Given: The functions are g ( x ) = x 2 − 2 x and f ( x ) = 3 x + 2 . Formula used: Quotient of function ( g f ) ( x ) = g ( x ) f ( x ) , f ( x ) ≠ 0 Calculation: The given expression is ( g f ) ( − 1 ) . Use the formula for quotient of function, ( g f ) ( x ) = g ( x ) f ( x ) , f ( x ) ≠ 0 to determine the solution of expression and put the value of f ( x ) and g ( x ) in the expression. ( g f ) ( x ) = g ( x ) f ( x ) , f ( x ) ≠ 0 = x 2 − 2 x 3 x + 2 The denominator of the quotient can never be zero

8th Edition

ISBN: 9780134178967

Author: John Tobey Jr., Jeffrey Slater, Jamie Blair, Jenny Crawford

Publisher: PEARSON

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Chapter 10.3, Problem 29E

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To calculate: The value of the expression, (gf)(−1) for the functions, g(x)=x2−2x and f(x)=3x+2.

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Chapter 10.3, Problem 28E

Chapter 10.3, Problem 30E

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Chapter 10 Solutions

Intermediate Algebra (8th Edition)

Ch. 10.1 - For the function f ( x ) = 3 x − 5 , find the...Ch. 10.1 - For the function , find the following. 2. Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - For the function , find the following. 7. Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E

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The value of the expression, ( g f ) ( − 1 ) for the functions, g ( x ) = x 2 − 2 x and f ( x ) = 3 x + 2 .

Solution Summary: The author explains how to calculate the value of the expression, (gf) (-1 ), using the formula for quotient of function.

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To calculate: The value of the expression, (gf)(−1) for the functions, g(x)=x2−2x and f(x)=3x+2.

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Chapter 10 Solutions