# 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.

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Thanks to everyone who answered. The speed and quality of the responses on this site is incredible. Happy Christmas! – PeteUK Dec 25 '12 at 21:58

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, $-\frac14\le y<\infty$

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Is being one-one function important for $y$ when using the inverse? – Babak S. Dec 24 '12 at 9:49
... and so the range is contained in the interval $[-1/4, \infty)$. Note that the argument needs to be reversed to show that every element of the interval $[-1/4, \infty)$ is in the range of $f(x)$ (which requires some care regarding the case of $x=1$). – Hurkyl Dec 24 '12 at 9:57
Just what I was looking for. Appreciated. – PeteUK Dec 25 '12 at 21:54

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!

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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.

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$f$ 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

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