# Thread: What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

1. ## What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

the function is f(x,y)=2xy/(x^2+y^2)
the domain is R-{(x,y)=(0,0)}
I managed to know that the lower bound of the function is -1 because:
2xy=(x+y)^2-x^2-y^2
so z=((x+y)^2+(x^2-y^2))/(x^2+y^2)
which simplifies to (x+y)^2/(x^2+y^2) -1
(x+y)^2/(x^2+y^2) >=0 so (x+y)^2/(x^2+y^2) -1>=-1
but I didn't manage to know the upper bound of the domain. Thank you for helping me.

2. ## Re: What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

This is a cool question. I think the range is within [-1, 1]. But I haven't found a good reason why. Have you graphed it? Check it out. : )

3. ## Re: What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

Yes you're right that the range is [-1,1], and to prove that I was trying to prove why (x+y)^2/(x^2+y^2)<=2 (you can try with different number combinations and you will always see that the results are less or equal to 2 but there's no way to prove why I think).

4. ## Re: What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

Hi mohomad!

The trick is to switch to polar coordinates.

$\frac {2xy} {x^2+y^2} = \frac {2 \cdot r\cos\theta \cdot r\sin \theta} { r^2} = 2\cos\theta\sin\theta$

Can you tell what its range is?

5. ## Re: What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

I tried switching it to polar but got stuck on 2cos(O)sin(O) but now after seeing it here I remembered that it's equal to sin(2O). Thank you very much!!

6. ## Re: What is the range of this function?

For all $x,y\in\mathbb R$ we have $(x+y)^2\geqslant0$ $\implies$ $2xy\geqslant-(x^2+y^2)$ and $(x-y)^2\geqslant0$ $\implies$ $x^2+y^2\geqslant2xy$. Hence $-(x^2+y^2)\leqslant2xy\leqslant x^2+y^2$ $\implies$ $-1\leqslant\frac{2xy}{x^2+y^2}\leqslant1$ ( $(x,y)\ne(0,0)$).

7. ## Re: What is the range of this function? the function is f(x,y)=2xy/(x^2+y^2)

Thank you Nehushtan that what I was looking for.