Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 9 - Sections 9.1-9.6 - Integrated Review - Functions and Properties of Logarithms - Page 579: 7

Answer

One-to-one, $f^{-1}={(6,-2), (8,4), (-6,2), (3,3)}$

Work Step by Step

We are given that $f={(-2,6), (4,8), (2,-6), (3,3)}$. This function is one-to-one, because each x-value corresponds to only one y-value and each y-value corresponds to only one x-value. We can find the inverse by interchanging the coordinates of each point. For example, the point (-2,6) will become (6,-2). Therefore, $f^{-1}={(6,-2), (8,4), (-6,2), (3,3)}$.
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