Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.7 - Polynomial and Rational Inequalities - Exercise Set - Page 413: 68


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Work Step by Step

To calculate the inequality, first express the inequality in the form of $f(x)\,>0$ $\begin{align} & \frac{1}{x+1}>\frac{2}{x-1}\, \\ & \frac{1}{x+1}-\frac{2}{x-1}>\frac{2}{x-1}\,-\frac{2}{x-1} \\ & \frac{1}{x+1}-\frac{2}{x-1}>0 \\ & \frac{(x-1)-2\left( x+1 \right)}{\left( x+1 \right)\left( x-1 \right)}>0 \end{align}$ Then simplify the inequality, $\begin{align} & \frac{(x-1)-2\left( x+1 \right)}{\left( x+1 \right)\left( x-1 \right)}>0 \\ & \frac{x-1-2x-2}{\left( x+1 \right)\left( x-1 \right)}>0 \\ & \frac{-x-3}{\left( x+1 \right)\left( x-1 \right)}>0 \end{align}$ Second, solve the equation $f(x)=0$ $\frac{-x-3}{\left( x+1 \right)\left( x-1 \right)}=0$ Evaluate the values of $x$ that make the numerator and denominator of the inequality equal to $0$. $\begin{align} & -x-3=0 \\ & x=-3 \\ \end{align}$ And $\begin{align} & x+1=0 \\ & x=-1 \\ & x-1=0 \\ & x=1 \end{align}$ So, the required values of $x$ are $-3,-1,1$. Hence, the required intervals for the provided inequality are $\left( -\infty ,-3 \right)\left( -3,-1 \right)\left( -1,1 \right)\text{ and }\left( 1,\infty \right)$ Choose one test value within each interval and evaluate f at that number. The provided function is greater than zero.
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