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: 32


$(f\circ g)(x)=\dfrac{x^4}{1+x^4}$; $D_{f\circ g}=(-\infty,0)\cup(0,\infty)$ $(g\circ f)(x)=(x^2+1)^2$; $D_{g\circ f}=(-\infty,\infty)$

Work Step by Step

We are given the functions: $f(x)=\dfrac{1}{x^2+1}$ $g(x)=x^{-2}=\dfrac{1}{x^2}$ Determine the domains of $f$ and $g$: $D_f=(-\infty,\infty)$ $D_g=(-\infty,0)\cup(0.\infty)$ Determine $(f\circ g)$: $(f\circ g)(x)=f(g(x))=f\left(\dfrac{1}{x^2}\right)=\dfrac{1}{\left(\dfrac{1}{x^2}\right)^2+1}=\dfrac{1}{\dfrac{1}{x^4}+1}=\dfrac{x^4}{1+x^4}$ The domain of $(f\circ g)$ is: $D_{f\circ g}=(-\infty,0)\cup(0,\infty)$ Determine $(g\circ f)$: $(g\circ f)(x)=g(f(x))=g\left(\dfrac{1}{x^2+1}\right)=\dfrac{1}{\left(\dfrac{1}{x^2+1}\right)^2}=(x^2+1)^2$ The domain of $(g\circ f)$ is: $D_{g\circ f}=(-\infty,\infty)$
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