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

May 15, 2017

The domain is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 1 , 1\right\}$
The range is $f \left(x\right) \in \left(- \infty , - 1\right) \cup \left[0 , + \infty\right)$

#### Explanation:

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

As we cannot divide by $O$, $x \ne 1$ and $x \ne - 1$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 1 , 1\right\}$

To calculate the range, we need to calculate ${f}^{-} 1 \left(x\right)$

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

We interchange $y$ and $x$

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

Now, we calculate $y$ in terms of $x$

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

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

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

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

$y = \sqrt{\frac{x}{x + 1}}$

The domain of $y$ is the range of $f \left(x\right)$

What is underneath the sqrt sign is $\ge 0$

Therefore,

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

We build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$${f}^{-} 1 \left(x\right)$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

${f}^{-} 1 \left(x\right) \ge 0$ when $x \in \left(- \infty , - 1\right) \cup \left[0 , + \infty\right)$