# What is the domain and range of  y =1/(2x-4)?

May 2, 2017

The domain of $y$ is $= \mathbb{R} - \left\{2\right\}$
The range of $y$, $= \mathbb{R} - \left\{0\right\}$

#### Explanation:

As you cannot divide by $0$,

$2 x - 4 \ne 0$

$x \ne 2$

Therefore, the domain of $y$ is ${D}_{y} = \mathbb{R} - \left\{2\right\}$

To determine the range, we calculate ${y}^{-} 1$

$y = \frac{1}{2 x - 4}$

$\left(2 x - 4\right) = \frac{1}{y}$

$2 x = \frac{1}{y} + 4 = \frac{1 + 4 y}{y}$

$x = \frac{1 + 4 y}{2 y}$

So,

${y}^{-} 1 = \frac{1 + 4 x}{2 x}$

The domain of ${y}^{-} 1$ is ${D}_{{y}^{-} 1} = \mathbb{R} - \left\{0\right\}$

This is the range of $y$, ${R}_{y} = \mathbb{R} - \left\{0\right\}$
graph{1/(2x-4) [-11.25, 11.25, -5.625, 5.625]}

May 2, 2017

$\text{domain } x \in \mathbb{R} , x \ne 2$

$\text{range } y \in \mathbb{R} , y \ne 0$

#### Explanation:

The denominator of y cannot be zero as this would make y $\textcolor{b l u e}{\text{undefined}} .$Equating the denominator to zero and solving gives the value that x cannot be.

$\text{solve " 2x-4=0rArrx=2larrcolor(red)" excluded value}$

$\text{domain } x \in \mathbb{R} , x \ne 2$

$\text{to find excluded value/s in the range}$

$\text{Rearrange the function making x the subject}$

$\Rightarrow y \left(2 x - 4\right) = 1$

$\Rightarrow 2 x y - 4 y = 1$

$\Rightarrow 2 x y = 1 + 4 y$

$\Rightarrow x = \frac{1 + 4 y}{2 y}$

$\text{the denominator cannot be zero}$

$\text{solve " 2y=0rArry=0larrcolor(red)" excluded value}$

$\text{range } y \in \mathbb{R} , y \ne 0$
graph{1/(2x-4) [-10, 10, -5, 5]}