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: 32


all real x, $x\ne-3,2$

Work Step by Step

$f(x)$ is defined for all real x where the denominator $x^{2}+x-6\ne0$. When $x^{2}+x-6=(x+3)(x-2)=0, x=-3,2$ Therefore, $x=-3,2$ are excluded from the domain.
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.