Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Additional and Advanced Exercises - Page 38: 5

Answer

See graph and explanations.

Work Step by Step

To graph this function, we need to consider several different cases: Case 1. $x\geq0$, we have $x+|y|=1+x$ or $|y|=1$ which gives $y=\pm1$, or two lines as shown in the figure. Case 2. $x\lt0$, we have $-x+|y|=1+x$ or $|y|=2x+1$, which leads to two more cases: (i) $y\geq0$, we have $y=2x+1$ shown as a line segment in the figure. (ii) $y\lt0$, we have $-y=2x+1$ or $y=-2x-1$ shown as another line segment in the figure.
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.