Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 0 - Before Calculus - 0.2 New Functions From Old - Exercises Set 0.2 - Page 25: 32


$$(f \circ g)(x)=\sqrt{\sqrt{x^2+3}-3}$$ $$(g \circ f)(x)=\sqrt{x}$$ $$D_{f \circ g}= (- \infty , \sqrt{6}] \cup [\sqrt{6} , + \infty )$$ $$D_{g \circ f}= [3, + \infty )$$

Work Step by Step

$$(f \circ g)(x)=f(g(x))=\sqrt{\sqrt{x^2+3}-3}$$ $$(g \circ f)(x)=g(f(x))=\sqrt{(\sqrt{x-3})^2+3}=\sqrt{x}$$ The domain of $f \circ g$ consists of all $x$ in the domain of $g$ for which $g(x)$ is in the domain of $f$. So we have $$D_{f \circ g}= \{ x \in \mathbb{R} \mid \sqrt{x^2+3} \ge 3 \}=(- \infty , \sqrt{6}] \cup [\sqrt{6} , + \infty )$$ $$D_{g \circ f}= \{ x \ge 3 \mid \sqrt{x-3} \in \mathbb{R} \}=[3, + \infty ).$$
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.