For the function $f(x,y)=(xy+x+y,y-x)$ on $\mathbb R^2$, how would I go about finding the image of the region $\{(x,y):1<x,y<2\}$?
I am not looking for a solution, just a hint.
EDIT:
If I parameterize the sides, I get
$f(1,y)=(2y+1,y-1)$
$f(2,y)=(3y+2,y-2)$
$f(x,1)=(2x+1,1-x)$
$f(x,2)=(3x+2,2-x)$
Where do I go from here?
Consider the boundary of your rectangle. Parameterize one side, then put it through your transformation. Rinse and repeat for the other 3 sides.
Edit: So I'll use $t$ as the parameter, to avoid confusion between the new and old coordinates.
Boundaries: $$\{(1,t) : 1 < t < 2\} \mapsto \{ (2t + 1, t - 1) : 1 < t < 2 \}$$ $$\{(2,t) : 1 < t < 2\} \mapsto \{ (3t + 2, t - 2) : 1 < t < 2 \}$$ $$\{(t,1) : 1 < t < 2\} \mapsto \{ (2t + 1, 1 - t) : 1 < t < 2 \}$$ $$\{(t,2) : 1 < t < 2\} \mapsto \{ (3t + 2, 2 - t) : 1 < t < 2 \}$$
What do these look like? Can you express them in terms of $x$ and $y$? Note the restrictions on $t$!
Edit: Proof that the interior does the right thing (I'm really not satisfied by this, if someone can find a better proof, please let me know):
Since $x < 2$ and $y > 0$, $(x-2)(y+4) < 0$.
We mess around a bit: $xy + 4x - 2y - 8 < 0 \implies xy + x + y < 3y - 3x + 8 \implies x^\prime < 3y^\prime + 8$.
Similarly, $(x+4)(y-2) < 0 \implies x^\prime < -3y^\prime + 8$.
Since $x > 1$ and $y > 0$, $(x - 1)(y + 3) > 0$.
Again, we expand: $xy + 3x - y - 3 > 0 \implies xy + x + y > -2x + 2y + 3 \implies x^\prime > 2y^\prime + 3$
Similarly, $(x+3)(y-1) > 0 \implies x^\prime > -2y^\prime + 3$.
We know there must be points near these lines, because they are the images of the boundaries, and the map is continuous. Since there are points arbitrarily close to the boundary of the original rectangle, there must be points arbitrarily close to the image of the boundary.
Unfortunately, all I have is intuition to say that those two pieces come together to mean "the boundary of the image is the image of the boundary"
