# What is the domain and range of f(x) =(x-3)/(7x+4)?

##### 2 Answers
Jun 4, 2017

$\mathrm{do} m f \in \mathbb{R} \setminus \setminus \setminus \left\{- \frac{4}{7}\right\}$ and $r a n f \in \mathbb{R} \setminus \setminus \setminus \left\{\frac{1}{7}\right\}$.

#### Explanation:

So this is a rational function, so to get $f \left(x\right)$ into hyperbolic form (from which you can read off your asymptotes) perform polynomial division.

$\frac{x}{7 x} = \frac{1}{7}$

$\frac{1}{7} \cdot 7 x = x$ and $\frac{1}{7} \cdot 4 = \frac{4}{7}$

$x - x = 0$ and $- 3 - \frac{4}{7} = - 3 \frac{4}{7} = - \frac{25}{7}$.

So your quotient, $Q \left(x\right) = \frac{1}{7}$ and your remainder $R \left(x\right) = - \frac{25}{7}$.

Therefore, $f \left(x\right) = \frac{1}{7} + - \frac{25}{7} \div 7 x + 4$

Rearrange this to make your solution of the form $f \left(x\right) = \frac{a}{x - h} + k$

Therefore, $f \left(x\right) = - \frac{25}{49 \left(x + \frac{4}{7}\right)} + \frac{1}{7}$

This means that you have asymptotes at $x = - \frac{4}{7}$ and $y = \frac{1}{7}$. Since hyperbolas have infinite domains and ranges excluding any point on the asymptote, you have:

$\mathrm{do} m f \in \mathbb{R} \setminus \setminus \setminus \left\{- \frac{4}{7}\right\}$ and $r a n f \in \mathbb{R} \setminus \setminus \setminus \left\{\frac{1}{7}\right\}$.

Jun 4, 2017

$x \in \mathbb{R} , x \ne - \frac{4}{7}$
$y \in \mathbb{R} , y \ne \frac{1}{7}$

#### Explanation:

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

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

$\text{solve " 7x+4=0rArrx=-4/7larrcolor(red)" excluded value}$

$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne - \frac{4}{7}$

$\text{to find any excluded value in the range, rearrange }$

$y = f \left(x\right) \text{ making x the subject}$

$\Rightarrow y \left(7 x + 4\right) = x - 3 \leftarrow \textcolor{b l u e}{\text{ cross-multiplying}}$

$\Rightarrow 7 x y + 4 y = x - 3$

$\Rightarrow 7 x y - x = - 3 - 4 y$

$\Rightarrow x \left(7 y - 1\right) = - \left(3 + 4 y\right)$

$\Rightarrow x = - \frac{3 + 4 y}{7 y - 1}$

$\text{the denominator cannot equal zero}$

$\text{solve " 7y-1=0rArry=1/7larrcolor(red)" excluded value}$

$\Rightarrow \text{range is } y \in \mathbb{R} , y \ne \frac{1}{7}$