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: 73


$fog$ is a one-to-one function.

Work Step by Step

- First, since $g(x)$ is one-to-one, according to definition: $$g(x_1)\ne g(x_2)\hspace{1cm}\text{whenever}\hspace{1cm} x_1\ne x_2$$ - And since $g(x_1)\ne g(x_2)$, we can deduce $$f(g(x_1))\ne f(g(x_2))$$ as $f$ is also one-to-one. This, however, means that $fog(x_1)\ne fog(x_2)$ whenever $x_1\ne x_2$. Therefore, $fog$ is a one-to-one function.
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