Solve the inequality |x - 1| / (x^2 - 4) >= 0.

Step-by-step solution

1. 1

   Start with the domain. The denominator must not be zero:

   $x^2-4=(x-2)(x+2)\neq 0\ \Longrightarrow\ x\neq 2,\ x\neq -2$

   These two points can never appear in the answer, no matter what the rest of the analysis says.

2. 2

   Use the fact that the numerator has a fixed sign. For every real $x$, $|x-1|\ge 0$, and it equals $0$ only at $x=1$. So the numerator can never make the fraction negative — the sign of the quotient is decided entirely by the denominator, except at the one point where the numerator is zero.

3. 3

   Case A: the fraction is strictly positive. This needs $|x-1|>0$ (so $x\neq 1$) together with $x^2-4>0$. Factoring, $(x-2)(x+2)>0$ holds when both factors are negative or both positive:

   $x<-2\quad\text{or}\quad x>2$

   Since $x=1$ lies in neither region, the whole of $(-\infty,-2)\cup(2,\infty)$ qualifies.

4. 4

   Case B: the fraction is exactly zero. A quotient is zero when its numerator is zero and its denominator is not:

   $|x-1|=0\Rightarrow x=1,\qquad 1^2-4=-3\neq 0\ \checkmark$

   Because the inequality is $\ge 0$ rather than $>0$, this single point is a solution, even though it sits inside the interval $(-2,2)$ where the fraction is otherwise negative.

5. 5

   Rule out the rest of $(-2,2)$. There $x^2-4<0$ while $|x-1|>0$, so the fraction is strictly negative — for example at $x=0$ it is $\tfrac{1}{-4}=-0.25$. Every point of $(-2,2)$ except $x=1$ fails.

6. 6

   Combine the cases.

   $(-\infty,-2)\cup\{1\}\cup(2,\infty)$

   The lone brace around $1$ is not a typo: an isolated point in a solution set is exactly what a "$\ge$" produces when a non-negative numerator has a zero inside a negative region.