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.

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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. : )Attachment 27328

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).

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.

Can you tell what its range is?

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!!

Re: What is the range of this function?

For all we have and . Hence ( ).

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.