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

Aug 20, 2017

$x \in \mathbb{R} , x \ne 3$
$y \in \mathbb{R} , y \ne - 2$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.

$\text{solve "3-x=0rArrx=3larrcolor(red)" excluded value}$

$\text{domain is } x \in \mathbb{R} , x \ne 3$

To find any excluded values in the range rearrange f(x) making x the subject.

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

$\Rightarrow y \left(3 - x\right) = 2 x - 1 \leftarrow \textcolor{b l u e}{\text{ cross-multiplying}}$

$\Rightarrow 3 y - x y = 2 x - 1$

$\Rightarrow - x y - 2 x = - 3 y - 1 \leftarrow \textcolor{b l u e}{\text{ collecting terms in x together}}$

$\Rightarrow x \left(- y - 2\right) = - \left(3 y + 1\right)$

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

$\text{the denominator cannot equal zero}$

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

$\Rightarrow \text{range is } y \in \mathbb{R} , y \ne - 2$

Aug 20, 2017

The domain is $x \in \left(- \infty , 3\right) \cup \left(3 , + \infty\right)$. The range is $y \in \left(- \infty , - 1\right) \cup \left(- 1 , + \infty\right)$

#### Explanation:

The function is $f \left(x\right) = \frac{2 x - 1}{3 - x}$

The denominator must be $\ne 0$

So,

$3 - x \ne 0$, $\implies$, $x \ne 3$

The domain is $x \in \left(- \infty , 3\right) \cup \left(3 , + \infty\right)$

Let,

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

$y \left(3 - x\right) = 2 x - 1$

$3 y - y x = 2 x - 1$

$2 x + y x = 1 + 3 y$

$x = \frac{1 + 3 y}{2 + y}$

$2 + y \ne 0$

$y \ne - 1$

The range is $y \in \left(- \infty , - 1\right) \cup \left(- 1 , + \infty\right)$

graph{(y-(2x-1)/(3-x))=0 [-58.53, 58.54, -29.26, 29.24]}