## Precalculus (6th Edition) Blitzer

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.