Domain of a function quick question $$f(x)=\arcsin\left(\frac{x-1}{2x}\right)$$
We need to find the domain of this function.
My try:
$$-1\leq \frac{x-1}{2x} \leq 1 $$
We can split this into
$$-1\leq \frac{x-1}{2x} \quad\text{and} \quad\frac{x-1}{2x} \leq 1$$
My idea is to solve for this two inequalities and then take the intersection of them both.
$$-2x\leq x-1\\
-3x\leq -1\\
x\geq\frac{1}{3}$$
$$x-1 \leq 2x\\
-1 \leq x$$
By doing the intersection I get the wrong answer which is: $x\geq\frac{1}{3}$
What have I done wrong?

The two inequalities can be handled at once as following without inviting mistakes :
For $x \neq 0$
$$\frac{x-1}{2x} \in [-1,1]$$
$$\frac{1}{2} - \frac{1}{2x} \in [-1,1]$$
$$-\frac{1}{2x} \in [\frac{-3}{2},\frac{1}{2}]$$
$$\frac{1}{x} \in [-1,3]$$
Taking reciprocal (and changing the direction),
$$x \in \; ]-1,\frac{1}{3}[$$
That is $$x \le -1 \vee x \ge 1/3$$
