Answer
Solution set: $(-\infty, -2) \cup \left[-\frac{1}{3}, 1\right) \cup(3, \infty) $
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.
$\displaystyle \frac{1}{x-3} +\frac{1}{x+2} \geq \frac{2x}{x^2+x-2}$,
$\displaystyle \frac{1}{x-3} +\frac{1}{x+2}- \frac{2x}{(x+2)(x-1) }\geq 0$,
$\displaystyle \frac{3x+1}{(x-3)(x+2)(x-1)}\geq 0$,
$f(x)=\displaystyle \frac{3x+1}{(x-3)(x+2)(x-1)}\geq 0$,
2.The cut points are:
$\displaystyle \frac{3x+1}{(x-3)(x+2)(x-1)}=$
$x=-2$ or $x=-\frac{1}{3}$ or $x=1$ or $x=3$
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,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \geq 0 ? \\
& & \displaystyle \displaystyle \frac{3a+1}{(a-3)(a+2)(a-1)} & \\
(-\infty,-2) & -5 & \frac{(-)}{(-)(-)(-)}=(+) & T\\
(-2, -\frac{1}{3}) & -1 & \frac{(-)}{(-)(+)(-)}=(-) & F\\
(-\frac{1}{3}, 1) & 0 & \frac{(+)}{(-)(+)(-)}=(+) & T\\
(1, 3) & 2 & \frac{(+)}{(-)(+)(+)}=(-) & F\\
(3,\infty) & 5 & \frac{(+)}{(+)(+)(+)}=(+) & T\\
\end{array}$
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, -2) \cup \left[-\frac{1}{3}, 1\right) \cup(3, \infty) $
