Solve the inequality (x - 1) / ((x - 3)(x - 2)) <= 0 and give the answer in interval notation.

Step-by-step solution

1. 1

   Find the critical points, not a common denominator. The expression is already a single factored fraction, so the sign can only change where a factor vanishes: the numerator zero $x = 1$ and the denominator zeros $x = 2$ and $x = 3$. Never multiply both sides by $(x-3)(x-2)$ — its sign is unknown, so the inequality direction would be unreliable.

2. 2

   Split the line into four intervals. The three critical points cut $\mathbb{R}$ into $(-\infty,1)$, $(1,2)$, $(2,3)$ and $(3,\infty)$. Within each, all three factors keep a fixed sign, so the whole quotient does too — one test value per interval is enough.

3. 3

   Test one point in each interval.

   interval	test $x$	value	sign
   $(-\infty,1)$	$0$	$\tfrac{-1}{(-3)(-2)} = -\tfrac16$	negative
   $(1,2)$	$1.5$	$\tfrac{0.5}{(-1.5)(-0.5)} = +\tfrac23$	positive
   $(2,3)$	$2.5$	$\tfrac{1.5}{(-0.5)(0.5)} = -6$	negative
   $(3,\infty)$	$3.5$	$\tfrac{2.5}{(0.5)(1.5)} = +\tfrac{10}{3}$	positive

4. 4

   Decide each endpoint separately. The inequality is $\le$, so zeros of the numerator are allowed: at $x = 1$ the value is $0$, which satisfies $\le 0$, so $x=1$ is included. Zeros of the denominator are never allowed — at $x = 2$ and $x = 3$ the expression is undefined, so both are excluded, no matter which way the inequality points.

5. 5

   Assemble the solution. Collect the negative intervals and attach the closed endpoint:

   $(-\infty,\,1] \;\cup\; (2,\,3)$

6. 6

   Cross-check against the usual distractors. A symbolic solver returns exactly $(-\infty,1] \cup (2,3)$ ✓. The option $[1,2) \cup [3,\infty)$ takes the positive intervals and wrongly closes a pole; $[1,2] \cup (3,\infty)$ closes $x=2$, where the expression does not even exist.