Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.3 The Basic Classes of Functions - Exercises - Page 23: 27


$(f \circ g)(x)=f(g(x))=f(x+1)=\sqrt {x+1}$ $(g \circ f)(x)=g(f(x))=g(\sqrt{x})=\sqrt{x} + 1$ $D_{f \circ g} = [-1, \infty)$ $D_{g \circ f} = [0, \infty)$

Work Step by Step

The domain $D_f$ of $f(x)$ is $[0,\infty)$ and the domain $D_g$ of $g(x)$ is $\mathbb{R}$. The composite functions $f \circ g$ and $g \circ f$ are defined by $(f \circ g)(x)=f(g(x))=f(x+1)=\sqrt {x+1}$ and $(g \circ f)(x)=g(f(x))=g(\sqrt{x})=\sqrt{x} + 1$ The domain of $f \circ g$ is defined by: $D_{f \circ g} = \{x \in D_g : g(x) \in D_f\} = \{x \in \mathbb{R} : x+1 \in [0, \infty)\}$ that is, $D_{f \circ g} = \{x \in \mathbb{R} : x \ge -1\}$ Hence, $D_{f \circ g} = [-1, \infty)$. The domain of $g \circ f$ is defined by: $D_{g \circ f} = \{x \in D_f : f(x) \in D_g\} = \{x \in [0, \infty) : \sqrt{x} \in \mathbb{R}\}$ Hence, $D_{g \circ f} = [0, \infty)$.
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.