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: 33


$(f\circ g)(t)=\dfrac{1}{\sqrt{-t^2}}$; $D_{f\circ g}=\phi$ $(g\circ f)(t)=-\dfrac{1}{t}$; $D_{g\circ f}=(0,\infty)$

Work Step by Step

We are given the functions: $f(t)=\dfrac{1}{\sqrt t}$ $g(x)=-t^2$ Determine the domains of $f$ and $g$: $D_f=(0,\infty)$ $D_g=(-\infty,\infty)$ Determine $(f\circ g)$: $(f\circ g)(t)=f(g(t))=f(-t^2)=\dfrac{1}{\sqrt{-t^2}}$ As $-t^2\leq 0$ for any real $t$ and for $t=0$ the denominator $\sqrt{-t^2}$ is not defined, it means that $(f\circ g)$ is not defined at all. $D_{f\circ g}=\phi$ Determine $(g\circ f)$: $(g\circ f)(x)=g(f(x))=g\left(\dfrac{1}{\sqrt t}\right)=-\left(\dfrac{1}{\sqrt t}\right)^2=-\dfrac{1}{t}$ The domain of $(g\circ f)$ is: $D_{g\circ f}=(0,\infty)$
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.