(a) Explanation Given Information: The provided function is f ( x ) = x 3 + 1 . Consider the provided function f ( x ) = x 3 + 1 , Calculate the value of the function f ( x ) for different values of x . Substitute 0 for x in f ( x ) = x 3 + 1 . f ( 0 ) = 0 3 + 1 = 1 Thus, f ( 0 ) = 1 . Substitute 1 for x in f ( x ) = x 3 + 1 . f ( 1 ) = 1 3 + 1 = 2 Thus, f ( 1 ) = 2 . Substitute 2 for x in f ( x ) = x 3 + 1 . f ( 2 ) = 2 3 + 1 = 9 Thus, f ( 2 ) = 9 . Substitute 3 for x in f ( x ) = x 3 + 1 . f ( 3 ) = 3 3 + 1 = 28 Thus, f ( 3 ) = 28 . Substitute − 1 for x in f ( x ) = x 3 + 1 . f ( − 1 ) = ( − 1 ) 3 + 1 = − 1 + 1 = 0 Thus, f ( − 1 ) = 0 . Substitute − 2 for x in f ( x ) = x 3 + 1 . f ( − 2 ) = ( − 2 ) 3 + 1 = − 8 + 1 = − 7 Thus, f ( − 2 ) = − 7 . Substitute − 3 for x in f ( x ) = x 3 + 1 . f ( − 3 ) = ( − 3 ) 3 + 1 = − 27 + 1 = − 26 Thus, f ( − 3 ) = − 26 (b) To determine To calculate: The formula of the inverse of the function f ( x ) = x 3 + 1 . (c) To determine To graph: The function defined by f ( x ) = x 3 + 1 and its inverse on same set of axes.

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Chapter 8.2, Problem 14DE

(a)

To determine

Whether the function f(x)=x3+1 is one-to-one or not.

(b)

To determine

To calculate: The formula of the inverse of the function f(x)=x3+1.

(c)

To determine

To graph: The function defined by f(x)=x3+1 and its inverse on same set of axes.

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Chapter 8.2, Problem 13DE

Chapter 8.2, Problem 15DE

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Chapter 8 Solutions

EBK INTERMEDIATE ALGEBRA

Ch. 8.1 - Graph. Ch. 8.1 - Prob. 2SR Ch. 8.1 - Prob. 1DE Ch. 8.1 - Prob. 2DE Ch. 8.1 - Prob. 3DE Ch. 8.1 - Prob. 4DE Ch. 8.1 - Prob. 5DE Ch. 8.1 - Prob. 6DE Ch. 8.1 - Prob. 1CCE Ch. 8.1 - Prob. 7DE

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Whether the function f ( x ) = x 3 + 1 is one-to-one or not.

Solution Summary: The author explains how to determine whether the function f(x) is one-to-one or not.

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Whether the function f(x)=x3+1 is one-to-one or not.

To calculate: The formula of the inverse of the function f(x)=x3+1.

To graph: The function defined by f(x)=x3+1 and its inverse on same set of axes.

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Chapter 8 Solutions