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

Apr 15, 2017

$x \in \mathbb{R} , x \ne 0$
$y \in \mathbb{R} , y \ne 3$

#### Explanation:

You may wish to consider f(x) as a single rational function.

$f \left(x\right) = \frac{1}{x} + 3 = \frac{1}{x} + \left(\frac{3}{1} \times \frac{x}{x}\right) = \frac{1}{x} + \frac{3 x}{x}$

$\Rightarrow y = f \left(x\right) = \frac{1 + 3 x}{x}$

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.

$\Rightarrow x = 0 \leftarrow \textcolor{red}{\text{ excluded value in domain}}$

$\text{domain is } x \in \mathbb{R} , x \ne 0$

$\text{Rearrange f(x) to make x the subject}$

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

$\Rightarrow x y = 1 + 3 x$

$\Rightarrow x y - 3 x = 1$

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

$\Rightarrow x = \frac{1}{y - 3} \to \left(y \ne 3\right) \textcolor{red}{\text{ excluded value in range}}$

$\text{range is } y \in \mathbb{R} , y \ne 3$