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

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thanks :)

Jun 12, 2021

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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!

Jun 12, 2021