Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.1 - Four Ways to Represent a Function - 1.1 Exercises - Page 21: 36


all real $u$, $u\ne-1,-2$

Work Step by Step

$\because$ the denominator of $\frac{1}{u+1}\ne0,$ $u+1\ne0, u\ne-1$ $f(u)=\frac{(u+1)(u+1)}{(u+1)+1} =\frac{u^{2}+2u+1}{u+2}$ $\because$ the denominator of $f(u)\ne0$, $u+2\ne0, u\ne-2$ $\therefore$ the domain is all real $u$, excluding $u=-1,-2$
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