Answer
See below:
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.
