The domain of the function $r(x) = \frac{x^2}{1-x}$ is $(-\infty,1)\cup(1,\infty)$. What is the range?

In order to find the range, 

Let \(y=r(x)\)

      \(y={x^2 \over 1-x^2}\)

⇒  \({1\over y}={1\over x^2}-1\)

⇒  \({1\over x^2}={1+y\over y}\)

⇒  \(x^2={y\over 1+y}\)                          ...(1)

 

Eq (1) has solutions if and only if, 

\({y\over 1+y} ≥0\)

i.e. if either \(y≥0\)  and  \(y+1>0 \)

or  \(y≤0\)  and  \(y+1<0\)

 

Thus range of r(x) is \( ( − ∞ , − 1 ) ∪ [ 0 , ∞ )\).

 

 

~Thank You!