Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Cumulative Review Exercises - Page 1127: 17


The value of the expression $\frac{x-1}{x+3}\le 0$ is $(-3,1]$

Work Step by Step

We know that for the inequality $\frac{x-1}{x+3}\le 0$, two conditions are possible: $\left( x-1 \right)\le 0$ And $\left( x+3 \right)\gt0$ Or $\left( x-1 \right)\ge 0$ And $\left( x+3 \right)\lt0$ There is no equal sign with the expression $ x+3$ since at $ x=-3$ the function does not exist. Now, from the first condition, we have $ x\le 1$ and $ x>-3$. From the second condition, we have $ x\ge 1$ and $ x-3$. Thus, the solution for x is $(-3,1]$.
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