Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.1 Four Ways to Represent a Function - 1.1 Exercises - Page 21: 40

Answer

The Domain is $t \ne -1 $ the Range is: $(-\infty, -2) and (-2, \infty)$ The sketch is:

Work Step by Step

$g(t) = \frac{t^2 - 1}{t + 1}$ Lets find the Domain of the function $t+1\ne0$ $t \ne -1 $ The Domain is $t \ne -1 $ Lets find the Range: $g(t) = \frac{t^2 - 1}{t + 1} = \frac{(t-1)(t+1)}{t + 1}$ when $t \ne -1 $, $g(t) =(t-1)$ So the Range is: $(-\infty, -2) and (-2, \infty)$ The sketch is:
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