College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 5 - Section 5.4 - Polynomial and Rational Inequalities - 5.4 Assess Your Understanding - Page 373: 34

Answer

The inequality is valid for values less than -1 and values more than 3 (not including them) i.e. $(-\infty,-1)\cap (3,\infty)$

Work Step by Step

First, we are going to find the x-intercepts by equating to zero: $\dfrac{x-3}{x+1}=0$ $\dfrac{x-3}{x+1}\cdot(x+1)=0(x+1)$ $x-3=0$ $x=3$ It is important to note that even though we have eliminated the expression $x+1$, the equation has the restriction $x\ne-1$; therefore we are going to count it as a critical point. $x_1=-1$ $x_2=3$ These are the critical points. We are going to take three values: one less than -1; one between -1 and 3; and one more than 3 to test in the original equation and check if the inequality is true or not: First test with a value less than -1: $\dfrac{-2-3}{-2+1}>0$ $\dfrac{-5}{-1}>0$ $5>0 \rightarrow \text{ TRUE}$ Second test with a value between -1 and 3: $\dfrac{0-3}{0+1}>0$ $\dfrac{-3}{1}>0$ $-3>0 \rightarrow \text{ FALSE}$ Third test with a value more than 3: $\dfrac{4-3}{4+1}>0$ $\dfrac{1}{5}>0$ $\dfrac{1}{5}>0 \rightarrow \text{ TRUE}$ These tests show that the inequality $\dfrac{x-3}{x+1}>0$ is valid for values less than -1 and values more than 3 (not including them) i.e. $(-\infty,-1)\cap (3,\infty)$
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