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

Aug 4, 2017

The domain is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{\frac{1}{2}\right\}$
The range is $y \in \mathbb{R}$

#### Explanation:

Our function is

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

The denominator cannot be $= 0$

So, $2 x - 1 \ne 0$, $x \ne \frac{1}{2}$

Therefore,

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{\frac{1}{2}\right\}$

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

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

$2 {x}^{2} - 1 = 2 y x - y$

$2 {x}^{2} - 2 y x + \left(y - 1\right) = 0$

In order for this quadratic equation in ${x}^{2}$ to have solutions, the discriminant is $\ge 0$

$\Delta = {b}^{2} - 4 a c = {\left(- 2 y\right)}^{2} - 4 \cdot \left(2\right) \cdot \left(y - 1\right) \ge 0$

$4 {y}^{2} - 8 \left(y - 1\right) \ge 0$

${y}^{2} - 2 y + 1 \ge 0$

${\left(y - 1\right)}^{2} \ge 0$

$\forall y \in \mathbb{R}$, ${\left(y - 1\right)}^{2} \ge 0$

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

graph{(2x^2-1)/(2x-1) [-8.89, 8.89, -4.444, 4.445]}