Find the range of $x$, given $y_{min} \leq y(x) \leq y_{max}$, where $y(x) $ can be any function ( Updated) I have a series of inequalities:
$$y_{1min} \leq y_{1}(x) \leq y_{1max}$$
$$y_{2min} \leq y_{2}(x) \leq y_{2max}$$
$$..$$
$$y_{nmin} \leq y_{n}(x) \leq y_{nmax}$$
Note that $x\in\mathbb{R}$
The question is, is there a general method that allows me to find the range of $x$ that satisfies the above inequalities?
Things are easy if there is just one inequality, and $y_1(x)$ is at most a quadratic function. But I am looking for a general solution here. 
Edit: My $y$ function can roughly be written as
$$y=\Biggl[\frac{(x^2+bx+c)^f}{(dx+e)^g}\Biggr]$$
where $c$ and $e$ are small comparatively.$(dx+e)^g$ must be positive. 
In one of the $y_i(x)$, $f=5/3, g=2/3$, but in another, $f=1, g=1$. It is safe to assume that both $f,g\in\mathbb{Q}$, that is they are rational numbers. 
$b,c,d,e\in\mathbb{R}$ , and they vary from one $y_i(x)$ to another. 
Hope this helps.
Edit:Let's make this question a little general; instead of letting $y(x)$ depends on one variable, what if $y$ depending on multiple variables? That is, how to find the range for $(x_1,x_2,x_3,..., x_i)$, given that  
$$y_{1min} \leq y_{1}(x_1,x_2,x_3,..., x_i) \leq y_{1max}$$
$$y_{2min} \leq y_{2}(x_1,x_2,x_3,..., x_i) \leq y_{2max}$$
$$..$$
$$y_{nmin} \leq y_{n}(x_1,x_2,x_3,..., x_i) \leq y_{nmax}$$

 A: If $f$ and $g$ are rational the problem is very easy.  Multiply out by the denominator, take the $d^{th}$ power of both sides where $d$ is the least common denominator of $f$ and $g$, and you've reduced to the polynomial case, which is easy (check the extrema and so forth).  This is why it is important to provide details.
A: For a start, suppose we can find the interval $I_i\subseteq\mathbb{R}$ such that $y_{imin}\leq y_i(x)\leq y_{imax}$. Then the interval $I$ which satisfies all the constraints is 
$I=\bigcap_{i=1}^n I_i$
Thus we can concentrate on finding $I_i$ separately for each $1\leq i\leq n$. Here I'm sure we will need some conditions on $y_i(x)$. Do your functions have compact support? Are they continuous? Do they oscillate wildly in any region?
In the general case - finding a feasible point
You mentioned that $y_i(x)$ are continuous. Lets suppose that in addition they are continuously differentiable. I'll do the problem with $1$ constraint for easy of notion but the general case is exactly the same though more difficult to write down.  
Rewrite the constraints as 
$y_1(x)\geq y_{1min}$
$-y_1(x)\geq -y_{1max}$
Now let $A(x) = \left(\nabla y_1(x),\nabla(-y_1(x))\right)^T$ (the jacobian matrix of the constraints). 
Also let $b=\left(y_{1min},-y_{1max}\right)^T$. 
Then we need to find a $x_0$ such that $Ax_0\geq b$. We do this by solving the following (phase-1) problem. 
let $x_{guess}\in\mathbb{R}^n$. Let $r=\min(b-Ax_{guess},0)$ and solve the problem
$\min_{x\in\mathbb{R}^n,\mu\in\mathbb{R}}$  $\mu$ subject to $Ax + \mu r\geq b$ and $\mu\geq 0$
Now, you may not think that we have come very far but because $(x,\mu)=(x_{guess},1)$ is an initial feasible point for this minimisation problem it can be solved by standard software. If the optimal point gives an optimal value $>0$ then the orignal problem is not feasible. If the optimal value is $0$ then the orignal problem is feasible and the optimal point is a feasible point for the orignal problem. 
