Answer
$(-\infty,-1-\sqrt 3) \cup[0, -1+\sqrt 3)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a Rational function.
$\displaystyle \frac{x}{x^2+2x-1} \leq 0$
$f(x)=\displaystyle \frac{x}{(x+1+\sqrt 3)(x+1-\sqrt 3)} \leq 0$
2. The cut points are:
$\displaystyle \displaystyle \frac{x}{(x+1+\sqrt 3)(x+1-\sqrt 3)} = 0$
$x=0$ or $x=-1-\sqrt 3$ or $x=-1+\sqrt 3$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \leq 0 ? \\
& & \displaystyle \displaystyle \frac{a}{(a+1-\sqrt 3)(a+1+\sqrt 3)} & \\
(-\infty,-1-\sqrt 3) & -5 & \frac{(-)}{(-)(-)}=(-) & T\\
(-1-\sqrt 3,0) & -1 & \frac{(-)}{(-)(+)}=(+) & F\\
(0, -1+\sqrt 3) & \frac{1}{2} & \frac{(+)}{(-)(+)}=(-) & T\\
(-1+\sqrt 3,\infty) & 10 & \frac{(+)}{(+)(+)}=(+) & F\\
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $(-\infty,-1-\sqrt 3) \cup[0, -1+\sqrt 3)$
