# How do you find the range of f(x)= x^2/(1-x^2)?

Mar 21, 2018

$\left(- \infty , - 1\right) \cup \left[0 , \infty\right)$

#### Explanation:

Given:

$f \left(x\right) = {x}^{2} / \left(1 - {x}^{2}\right)$

Let $y = f \left(x\right)$ and attempt to solve for $x$...

$y = f \left(x\right) = {x}^{2} / \left(1 - {x}^{2}\right) = \frac{1 - \left(1 - {x}^{2}\right)}{1 - {x}^{2}} = \frac{1}{1 - {x}^{2}} - 1$

Add $1$ to both ends to get:

$y + 1 = \frac{1}{1 - {x}^{2}}$

Multiply both sides by $\frac{1 - {x}^{2}}{y + 1}$ to get:

$1 - {x}^{2} = \frac{1}{y + 1}$

Add ${x}^{2} - \frac{1}{y + 1}$ to both sides to get:

$1 - \frac{1}{y + 1} = {x}^{2}$

In order for this to have a real valued solution, we require:

$1 - \frac{1}{y + 1} \ge 0$

That is:

$\frac{y}{y + 1} \ge 0$

Hence we require one of:

$\left(y \ge 0 \wedge y + 1 > 0\right) \rightarrow y \in \left[0 , \infty\right)$

$\left(y < 0 \wedge y + 1 < 0\right) \rightarrow y \in \left(- \infty , - 1\right)$

So the range of $f \left(x\right)$ is $\left(- \infty , - 1\right) \cup \left[0 , \infty\right)$

graph{x^2/(1-x^2) [-10, 10, -5, 5]}