Inversing a function

Question

I'm having some problems calculating the inverse of this function:

$f(u,v)=(u+v,v-u^2)$, its domain is $D=\{(u,v)$ in $\Bbb R^2 : u>0\}$

Thanks in advance.

Answer 1

If you consider the inverse function $f^{-1}(a,b)$, you have to solve $$\left\{\begin{array}{l} u+v = a\\ v-u^2 = b \end{array}\right.$$ $$\begin{align} (u +v) - (v-u^2) =& a-b\\ u^2 + u - a + b =& 0\\ u =& \frac{-1+\sqrt{1+4(a-b)}}2\\ v =& a-\frac{-1+\sqrt{1+4(a-b)}}2\\ f^{-1}(a,b) =& \left(\frac{-1+\sqrt{1+4(a-b)}}2,a-\frac{-1+\sqrt{1+4(a-b)}}2\right) \end{align}$$

Answer 2

Hint: Try to solve the simultaneous equations of $(u, v)$ $$ x=u+v \qquad y=v-u^2 $$

Answer 3

Setting $x=u+v$ and $y=v-u^2$, we find that

$$u=\frac{-1\pm\sqrt{1+4(x-y)}}{2}$$

Note that the constraint $u \gt0$ implies that we take the plus sign above and constrain $x \gt y$.