# What is the domain and range of  f(x)= (3x) /(x^2-1)?

Jul 2, 2018

The domain is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , 1\right) \cup \left(1 , \infty\right)$. The range is $y \in \mathbb{R}$.

#### Explanation:

As you cannot divide by $0$, the denominator is $\ne 0$

Therefore,

${x}^{2} - 1 \ne 0$

$\implies$, $\left(x - 1\right) \left(x + 1\right) \ne 0$

So,

$x \ne 1$ and $x \ne - 1$

The domain is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , 1\right) \cup \left(1 , \infty\right)$

To calculate the range, let

$y = \frac{3 x}{{x}^{2} - 1}$

$\implies$, $y \left({x}^{2} - 1\right) = 3 x$

$\implies$, $y {x}^{2} - y = 3 x$

$\implies$. $y {x}^{2} - 3 x - y = 0$

This ia a quadratic equation in $x$ and in order to have solutions, the discriminant must be $\ge 0$

Therefore,

$\Delta = {\left(- 3\right)}^{2} - 4 \left(y\right) \left(- y\right) \ge 0$

$9 + 4 {y}^{2} \ge 0$

So,

$\forall y \in \mathbb{R}$, $9 + 4 {y}^{2} \ge 0$

The range is $y \in \mathbb{R}$

graph{3x/(x^2-1) [-18.02, 18.02, -9.01, 9.02]}