Find the domain and the range of $ \arcsin \frac{2x-y}{x+y}$? Well, the domain of arcsin is $-1 \leq \arcsin x \leq 1$, so I can write that $-1 \leq  \frac{2x-y}{x+y} \leq 1$
 A: Not well expresed the domain. That's the way the range is wrote, but the bounds being $-\pi/2\leq\arcsin f\leq\pi/2$, but, yes, you can write $-1 \leq  \dfrac{2x-y}{x+y} \leq 1$
The domain of $\sin z$ is the real line and its range is $[-1,1]$. Then, the domain of $\arcsin f(x,y)$ is then at most $[-1,1]$.
Domain
$-1\le\dfrac{2x-y}{x+y}\le 1$ We have to consider two separate cases, denominator positive and in the negative.
$-1\le\dfrac{2x-y}{x+y}\le1\begin{cases}
  x+y\gt0;&-x-y\le2x-y\le x+y;&(y\gt-x\land0\le x\land x\le2y)\\
  x+y\lt0;&-x-y\ge2x-y\ge x+y;&(y\lt-x\land0\ge x\land x\ge2y)
\end{cases}$
$D=\{(x,y)\in\mathbb R^2\vert(y\gt-x\land0\le x\land x\le2y)\lor(y\lt-x\land0\ge x\land x\ge2y)\}$
Range
$\dfrac{2x-y}{x+y}=1\;;2x-y=x+y\;;x=2y$ So, it reaches the value $\pi/2$ at, e.g. $(2,1)$
Consider the he path $x=2,y\ge1$, so is, the vertical line strting at $(2,1)$.We have that $\displaystyle\lim_{y\to+\infty}\frac{4-y}{2+y}=-1$. Because the quotient is continuous, it reaches every value in $[-1,1]$ and because $\arcsin f$ is continuous for $f\in[-1,1]$, it reaches every value in $[-\pi/2,\pi/2]$
$R=\{z\in\mathbb R\,\vert-\pi/2\le z\le\pi/2\}$
