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 4.1 Composite Functions - 4.1 Assess Your Understanding - Page 279: 21

Answer

$\left\{x| x\ne0, x\ne2\right\}$

Work Step by Step

The domain of $f\circ g$ or $f(g(x))$ is the set of all $x$ such that $x$ is in the domain of $g$ and $g(x)$ is in the domain of $f$. Since the denominator can't be equal to $0$, then $x=0$ is not in the domain of $g$. For $f(x)$, $1x=$ is not in the domain of $f$ as it will make the denominator equal to zero.. This means that $g(x)\ne1$. Hence, \begin{align*} g(x)&\ne1\\ \frac{2}{x}&\ne1\\ 2&\ne x \end{align*} Therefore, the domain of $f\circ g$ is all real numbers except $2$ and $0$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.