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?

Here it is another approach for the sake of curiosity.
I would start with squaring both sides as it is done next:
\begin{align*}
\left|\frac{x-1}{2x}\right| \leq 1 & \Longleftrightarrow \left(\frac{x-1}{2x}\right)^{2} \leq 1\\\\
& \Longleftrightarrow \frac{x^{2} - 2x + 1 - 4x^{2}}{4x^{2}} \leq 0\\\\
& \Longleftrightarrow \frac{3x^{2} + 2x - 1}{4x^{2}} \geq 0
\end{align*}
We may now study the behavior of the function at the numerator.
To begin with, let us rearrange its expression:
\begin{align*}
3x^{2} + 2x - 1 & = \left(3x^{2} + 2x + \frac{1}{3}\right) - \frac{4}{3}\\\\
& = \left(\sqrt{3}x + \frac{1}{\sqrt{3}}\right)^{2} - \frac{4}{3}
\end{align*}
Consequently, the solution set to problem proposed is given by
\begin{align*}
3x^{2} + 2 x - 1 \geq 0 & \Longleftrightarrow \left(\sqrt{3}x + \frac{1}{\sqrt{3}}\right)^{2} \geq \frac{4}{3}\\\\
& \Longleftrightarrow \left(\sqrt{3}x + \frac{1}{\sqrt{3}} \geq \frac{2}{\sqrt{3}}\right)\vee\left(\sqrt{3}x + \frac{1}{\sqrt{3}} \leq -\frac{2}{\sqrt{3}}\right)\\\\
& \Longleftrightarrow \left(x \geq \frac{1}{3}\right)\vee\left(x \leq -1\right)
\end{align*}
Since both intervals does not contain the value $x = 0$, we conclude the sought domain is given by
\begin{align*}
D_{f} = (-\infty,-1]\cup\left[\frac{1}{3},+\infty\right)
\end{align*}
Hopefully this helps!
