# How do you find the domain and range of f(x)=(2x-1)/(3-x)?

May 30, 2017

Perform polynomial division on $f \left(x\right)$ to put it into partial fraction form. From this, you can determine asymptotes which help you to determine domain and range.

#### Explanation:

For a rational function of the form $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$ such as the one above, $f \left(x\right) = \frac{2 x - 1}{3 - x}$, since the degree of the polynomial is the same in the denominator as it is in the numerator, you must divide through. Doing so, we get $f \left(x\right) = - 2 + \frac{5}{3 - x}$.

From this it is evident that this is a rectangular hyperbola with asymptotes at $x = 3$ and $y = - 2$, so neither of these are included in the domain or range respectively.

Therefore we get, $\mathrm{do} m f \in \left(- \infty , 3\right) \cup \left(3 , \infty\right)$ and $r a n f \in \left(- \infty , - 2\right) \cup \left(- 2 , \infty\right)$.

May 30, 2017

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

#### Explanation:

$\text{f(x) is defined for all real values of x apart from values that }$
$\text{make the denominator zero}$

$\text{Equating the denominator to zero and solving gives the value}$
$\text{that x cannot be}$

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

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

$\text{to find any excluded values in the range rearrange y = f(x)}$
$\text{making x the subject}$

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

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

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

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

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

$\text{the denominator cannot equal zero}$

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

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