Answer
$[-2, 0) \cup (1, 3]$
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.
$\displaystyle \frac{6}{x-1} -\frac{6}{x} \geq 1$,
$\displaystyle\frac{6}{x-1} -\frac{6}{x} -1\geq 0$ ,
$\displaystyle\frac{6x-6x+6-x^2+x}{(x-1)(x)}\geq 0$ ,
$\displaystyle \frac{(-x+3)(x+2)}{(x-1)(x)}\geq 0$ ,
$f(x)=\displaystyle \frac{(-x+3)(x+2)}{(x-1)(x)}\geq 0$,
2.The cut points are:
$\displaystyle \frac{(-x+3)(x+2)}{(x-1)(x)}=0$
$x=-2$ or $x=0$ 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,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \geq 0 ? \\
& & \displaystyle \displaystyle \frac{(-x+3)(x+2)}{(x-1)(x)} & \\
(-\infty,-2) & -5 & \frac{(+)(-)}{(-)(-)}=(-) & F\\
(-2, 0) & -1 & \frac{(+)(+)}{(-)(-)}=(+) & T\\
(0, 1) & \frac{1}{2} & \frac{(+)(+)}{(-)(+)}=(-) & F\\
(1, 3) & 2 & \frac{(+)(+)}{(+)(+)}=(+) & T\\
(3,\infty) & 5 & \frac{(-)(+)}{(+)(+)}=(-) & F\\
\end{array}$
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 unless they belong to the denominator.
Solution set: $[-2, 0) \cup (1, 3]$
