## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 1 - Precalculus Review - 1.3 The Basic Classes of Functions - Exercises - Page 23: 30

#### Answer

$f \circ g (x) = |sin(x)|$ $g \circ f (x) =sin(|x|)$. Domains: $\mathbb{R}$

#### Work Step by Step

$f \circ g$ is defined by $f \circ g (x) = f(g(x)) = |sin(x)|$ and $g \circ f$ is defined by $g \circ f (x) = g(f(x)) = sin(|x|)$. Note: the domain of $f$ is $\mathbb{R}$ (all real numbers) and the domain of $g$ is $\mathbb{R}$. Hence, we have: The domain of $f \circ g$ is given by $D_{f \circ g} = \{x \in D_g : \, g(x) \in D_f\} = \{x \in \mathbb{R} : \, |x| \in \mathbb{R}\}$ and then the domain of $f \circ g$ is equal to $\mathbb{R}$. The domain of $g \circ f$ is given by $D_{g \circ f} = \{x \in D_f : \, f(x) \in D_g\} = \{x \in \mathbb{R} : \, sin(x) \in \mathbb{R}\}$ and then the domain of $g \circ f$ is equal to $\mathbb{R}$.

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