Question

Find the domain and range of f(x,y)= 2xy/x^2+y^2

Answer 1

Let us find the domain and range of $f(x,y)=\frac{2xy}{x^2+y^2}.$ Since $x^2+y^2=0$ implies $(x,y)=(0,0),$ we conclude that the domain is $\R^2\setminus\{(0,0)\}.$ Taking into account that $(x-y)^2\ge0,$ we conclude that $x^2-2xy+y^2\ge 0,$ and hence $x^2+y^2\ge 2xy.$ It follows that $\frac{2xy}{x^2+y^2}\le 1.$ On the other hand, $(x+y)^2\ge0$ implies $\frac{2xy}{x^2+y^2}\ge -1.$ It follows that the range of $f$ is $[-1,1].$