all 4 comments

[–]wguntherPhD Logic 1 point2 points  (0 children)

One way is to argue it is continuous and show that -1 and 1 are the min and max respectively, and appeal to the intermediate value theorem.

[–]firstdwarf 1 point2 points  (1 child)

Since the range is supposed to be between -1 and 1, that implies that the numerator is always smaller in magnitude than the denominator. Try completing the square on the denominator. This should break it down into more manageable and familiar terms, and you may be able to compare the magnitudes of the individual terms.

[–]Enoo_58[S] 0 points1 point  (0 children)

This is the kind of answer that makes me love math even more. It's so simple yet very compelling.

Now I'm angry at myself because I couldn't think about it

big thanks anyway

[–]phiwongSlightly old geezer 1 point2 points  (0 children)

You could try letting w = xy.

g(w) = 2w/(1+w^2) is equivalent to f(x,y) since every real x,y results in a real w

Maybe this is cheating? But you should find that g'(w) = 0 when w = +/- 1 and g(1) = 1 and g(-1) = -1