# How to find the range of x^2/(1-x^2)?

## I thought that the answer is $R : \left(- \infty , - 1\right) \cup \left(- 1 , + \infty\right)$but the book says that the correct answer is $R : \left(- \infty , - 1\right) \cup \left[0 , + \infty\right)$ Can anyone help please?

Aug 10, 2018

The range of ${x}^{2} / \left(1 - {x}^{2}\right)$ is $\left(- \infty , - 1\right) \cup \left[0 , \infty\right)$

#### Explanation:

Let:

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

and solve for $x$...

Multiplying both sides by $1 - {x}^{2}$, we get:

$y - y {x}^{2} = {x}^{2}$

Adding $y {x}^{2}$ to both sides, this becomes:

$y = \left(y + 1\right) {x}^{2}$

Then dividing both sides by $\left(y + 1\right)$ we get:

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

This has solutions if and only if:

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

That is, if either of the following:

• $y \ge 0 \text{ }$ and $\text{ } y + 1 > 0$. That is $y \ge 0$

• $y \le 0 \text{ }$ and $\text{ } y + 1 < 0$. That is $y < - 1$

So the range of ${x}^{2} / \left(1 - {x}^{2}\right)$ is $\left(- \infty , - 1\right) \cup \left[0 , \infty\right)$

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