# Find Range of $x_1,x_2$, where $y_{min} \leq y(x_1, x_2) \leq y_{max}$

This is a similar question with this one. But with a bit of twist.

I have two inequalities:

$$y_{1min} \leq y_{1}(x_1,x_2) \leq y_{1max}$$ $$y_{2min} \leq y_{2}(x_1,x_2) \leq y_{2max}$$

where

$$y_i(x_1,x_2)=\Biggl[\frac{(x_{1}^2x_{2}+b_{i}x_{1}x_{2}+c_{i})^{f_i}}{(x_{1}(2+x_{2})+e_{i})^{g_i}}\Biggr]$$

where $f_1=g_1=1, f_2=5, g_2=2.$

$b_{i},c_{i},e_{i},x_1,x_2\in\mathbb{R}$

Question: How to find the range of $x_{1},x_{2}$ that satisfy the above inequalities?

The range of $x_{1},x_{2}$ is defined as all the $x_1,x_2$ pairs that satisfy the above inequalities.

Edit: Maybe I'll relax the question a bit: let's assume that $b_{1}=b_{2}, c_{1}=c_{2}, e_{1}=e_{2}$.

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What do you mean by "range"? In general you will get a region much more complicated than a finite union of rectangles. –  Qiaochu Yuan Sep 23 '10 at 7:52
@Qiaochu, the set of ${x_1,x_2}$ that satisfy the above inequalities. –  Graviton Sep 23 '10 at 7:54
Are the brackets supposed to mean "integer part"? –  Mariano Suárez-Alvarez Sep 23 '10 at 8:12
@Mariano, they are not. Sorry for the confusion; I'm not that well-versed in standard mathematical notation. –  Graviton Sep 23 '10 at 8:32
If you can remove them without changing the sense of the formula, then remove them :) –  Mariano Suárez-Alvarez Sep 23 '10 at 8:34