Question

sketch the domain of f
f(x,y)= arcsin(2x-y)/(x+y)

Answer 1

let me rewrite the equation
\[f(x,y)=\arcsin \frac{ 2x-y }{ x+y }\]

Answer 2

@SolomonZelman help please

Answer 3

I think that the domain of the function above, is given by the solution of these inequalities:
\[ - 1 \leqslant \frac{{2x - y}}{{x + y}} \leqslant 1,\quad x + y \ne 0\]
wherein the first inequality, comes from the natural limits of \(\sin \) function whereas the second inequality, comes from the impossibility to divide by zero

Answer 4

so, first, we have to subtract the line \(y=-x\) from the \((x,y)-\)plane:

Answer 5

@Michele_Laino does the line of 2x-y need to draw too?

Answer 6

no, I don't think. Here we have to solve these two inequalities:
\[\frac{{2x - y}}{{x + y}} \leqslant 1\]
and:
\[\frac{{2x - y}}{{x + y}} \geqslant - 1\]

Answer 7

the first inequality, is equivalent to these two systems of inequalities:
\[\left\{ \begin{gathered}
2x - y \leqslant x + y \hfill \\
x + y > 0 \hfill \\
\end{gathered} \right. \cup \left\{ {\begin{array}{*{20}{c}}
{2x - y \geqslant x + y} \\
{x + y < 0}
\end{array}} \right.\]

Answer 8

is this equation valid?
2x-y=0

Answer 9

the solutions of those systems are:

Answer 10

how can i know which part to shade??

Answer 11

since you have to solve the two systems above

Answer 12

for example, I solve the first system:
\[\left\{ \begin{gathered}
2x - y \leqslant x + y \hfill \\
x + y > 0 \hfill \\
\end{gathered} \right.\]

Answer 13

here is the next step:
\[\left\{ \begin{gathered}
2x - y \leqslant x + y \hfill \\
x + y > 0 \hfill \\
\end{gathered} \right. \Rightarrow \left\{ \begin{gathered}
x \leqslant 2y \hfill \\
y > - x \hfill \\
\end{gathered} \right. \Rightarrow \left\{ \begin{gathered}
y \geqslant x/2 \hfill \\
y > - x \hfill \\
\end{gathered} \right.\]

Answer 14

the first inequality is represented by all points \((x,y)\) such that \(y \geqslant x/2\), whereas the second inequality is represented by all points \((x,y)\) such that \(y>-x\).
The solution is given by the intersection between both regions

Answer 15

ohh.. i get it, thank you so much

Answer 16

ok! Now we have to solve the second system:
\[\left\{ {\begin{array}{*{20}{c}}
{2x - y \geqslant x + y} \\
{x + y < 0}
\end{array}} \right.\]

Answer 17

I got this:
\[\left\{ {\begin{array}{*{20}{c}}
{2x - y \geqslant x + y} \\
{x + y < 0}
\end{array}} \right. \Rightarrow \left\{ \begin{gathered}
x \geqslant 2y \hfill \\
y < - x \hfill \\
\end{gathered} \right. \Rightarrow \left\{ \begin{gathered}
y \leqslant x/2 \hfill \\
y < - x \hfill \\
\end{gathered} \right.\]

Answer 18

he first inequality selects all points (x,y) such that \(y \leqslant x/2\), whereas the second inequality is represented by all points \((x,y)\) such that \(y<−x\).
The solution is given by the intersection between both regions

the solution of the first inequlity, is given by the union of the two solution of those two system.
Such solution is represented in my drawing above

Answer 19

oops...the* first...

Answer 20

another typo: inequality*

Answer 21

now, in order to get the requested domain, you have to add, using the same procedure above, the solution of this second inequality:
\[\frac{{2x - y}}{{x + y}} \geqslant - 1\]

Answer 22

i dont get this part, why \[2x-y \] become\[y \le x/2\] shouldnt it be \[x \le y/2\]

Answer 23

if we start from this inequality:
\[2x - y \leqslant x + y\]
then I add \(y\) to both sides, so I can write:
\[\begin{gathered}
2x - y + y \leqslant x + y + y \hfill \\
\hfill \\
2x \leqslant x + 2y \hfill \\
\end{gathered} \]
finally, I subtract \(x\) from both sides:
:\[\begin{gathered}
2x - x \leqslant x + 2y - x \hfill \\
\hfill \\
x \leqslant 2y \hfill \\
\end{gathered} \]

Answer 24

with the same procedure, if I start with this inequality:
\[{2x - y \geqslant x + y}\]
I get:
\[{x \geqslant 2 y}\]

Answer 25

ohhh.. i see my mistake now

Answer 26

next step:
the solution of the subsequent inequality:
\[\frac{{2x - y}}{{x + y}} \geqslant - 1\]
is given by the union of the solutions of these two systems of inequalities:
\[\left\{ \begin{gathered}
2x - y \geqslant - x - y \hfill \\
x + y > 0 \hfill \\
\end{gathered} \right. \cup \left\{ {\begin{array}{*{20}{c}}
{2x - y \leqslant - x - y} \\
{x + y < 0}
\end{array}} \right.\]

Answer 27

please try to solve such systems of inequalities

Answer 28

ok

Answer 29

we need to do both in order to obtain the graph?

Answer 30

yes! That's right!

Answer 31

ok, thanks for the help :)