Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.2 One-to-One Functions; Inverse Functions - 4.2 Assess Your Understanding - Page 293: 85

Answer

Quadrant I

Work Step by Step

Since $f^{-1} (x)$ is the reflection of $f(x)$ about the line $y=x$, the graph will remain in Quadrant I. This is because when we reflect the graph of $f(x)$ about the line $y=x$, the inverse will remain in Quadrant I. In other words, consider the pair of coordinates $(a,b)$ of $f(x)$. These coordinates will both be positive since we are in Quadrant I. The inverse coordinates will then be $(b,a)$, which are also positive and also in Quadrant I.
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