Finding range of $\frac{x}{(1-x)^2}$ I'm working though some exercises. One of them is asking to find the range of $f(x) = \frac{x}{(1-x)^2}$. The chapter this exercise belongs to is after the one where the differentiation power rule is introduced, but before the one where the quotient rule is introduced. My approach was to draw a table of values, but this was getting cumbersome and I got frustrated with the approach before finding the answer. My question is, are there simple analytical ways of finding the range that doesn't rely upon the quotient rule? I suspect there must be because of where the question was located. All the existing fraction-related examples in the book have been easily manipulated so that they are $\frac{number}{f(x)}$. This one is $\frac{f(x)}{g(x)}$ and my attempts at making it simpler are failing.
 A: Inferring what you were trying to do, you could have written $f(x) = x (1-x)^{-2}$.
As an aside, you could just solve the equation $y = x / (1-x)^2$. It might even be worth solving the problem both ways!
A: Let $$y=\frac{x}{(1-x)^2}\implies yx^2-(2y+1)x+y=0$$
This is a quadratic equation in $x$ and as $x$ is real, the discriminant  $(2y+1)^2-4\cdot y\cdot y=4y+1$ must be $\ge 0$ 
So, $-\frac14\le y<\infty $
A: Assume $x/(1-x)^2=k$. Now this can be rewriiten as $k(1-x)^2-x=0$, which is a quadratic equation ($k$ is constant). Now find the condition on $k$ for which this equation has real roots. That will give you the range.Also, you need to care of the case when $x=1$ since it is not in the domain.But in this case, $x=1$ can never be the solution of the above equation.
A: $f$ is defined in $(-1,1)$ and continuous there. In addition
$$f^{\prime}(x)=\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\frac{1-x+2x}{(1-x)^3}=\frac{1+x}{(1-x)^3}>0$$
as $-1<x<1$. 
Also note that if $g$ is increasing in $(a,b)$ and $g$ is continuous there then $$R_g=(\lim_{x\to a^+}g(x),\lim_{x\to b^-}g(x))$$
Even if the quotient rule is not introduced, you can use the product rule to prove it. $f$ can't be manipulated the way you want. The best you can do is the following:
$$\frac{x}{(1-x)^2}=\frac{x-1+1}{(x-1)^2}=\frac{1}{x-1}+\frac{1}{(x-1)^2}$$
but I don't think this helps too much. Of course you could solve the equation
$$y=f(x)$$
and demand that it has a solution for $x$ but this is hardly analytical
