transformations of $\mathbb R^2$ Consider the transformation $(u,v)=f(x,y)=(x-y,xy)$. Demonstrate the effect of this transformation on the lines $x-y=\text{constant}$, $x+y=0$, and the curves $xy=\text{constant}$. In particular demonstrate the effect of this transformation of the region of the region of the $x,y$-plane bounded by the curves $y=1/x$, $y=2/x$, $x-y=1$ and $x-y=2$.
Could any point of the $xy$ plane be mapped to the point $(2,-2)$ of the $u,v$ plane? Try to determine the range of the function $f$.
I know I am supposed to give some of my work first, but I have spent hours on reading the book about transformation and I have no idea at all about this question :( can anyone at least give me some hints or explain the question ?
 A: I believe you should simply calculate the images of the sets specified. For example:
Let $c$ be a constant, and consider the line $x-y=c$. In other words, this line is simply the set $A=\left\{(x,y)\in\mathbb{R}^2:x-y=c\right\}$. By the "effect" of this transformation, you're supposed to calculate $f(A)=\left\{f(x,y):(x,y)\in A\right\}$.
Notice that if $(x,y)\in A$, then $x=y+c$, and $f(x,y)=(c,xy)=(c,y^2+yc)$. The possibles values for $y^2+yc$ are all the numbers $\geq -\dfrac{c^2}{4}$, so the "effect" of $f$ over $A$ is simple given by $\left\{(c,t):t\geq -\dfrac{c^2}{4}\right\}$. The other cases should be similar.
Now, suppose there exists a point $(x,y)$ such that $f(x,y)=(2,-2)$. then $x=2+y$ and
$-2=xy=(2+y)y=2y+y^2$, so $-1=y^2+2y+1=(y+1)^2>0$, an absurd. Therefore, no point of the xy plane is mapped to $(2,-2)$.
Now, let's calculate the range of $f$. Let $(u,v)=(x-y,xy)$ be in the range of $f$. Then $x=u+y$ and $v=xy=(u+y)y=y^2+uy$, so the equation $y^2+uy-v=0$ has a root. Hence, $0\leq u^2+4v$, that is, $u^2\geq -4v$. This shows that the range $f(\mathbb{R}^2)$ is contained in the set $\left\{(u,v):u^2\geq-4v\right\}$. On the other hand, if $u^2\geq -4v$, take any root $y$ of the polynomial $y^2+uy-v=0$ and take $x=u+y$. Then $f(x,y)=(u,v)\in f(\mathbb{R}^2)$. Therefore, the range of $f$ is given by
$$f(\mathbb{R}^2)=\left\{(u,v):u^2\geq-4v\right\}.$$
(notice that $(2,-2)\not\in f(\mathbb{R}^2)$, since $2^2=4<8=-4(-2)$.)
