University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 50: 74


$g(x)$ is one-to-one.

Work Step by Step

$g(x)$ is one-to-one. We will prove this by contradiction. Suppose $g(x)$ is not one-to-one, thus there exist $x_1\ne x_2$ such that $g(x_1)=g(x_2)$, but then $f(g(x_1))=f(g(x_2))$, which contradicts the fact that $f \circ g$ is one-to-one.
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