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

May 24, 2017

$x \in \mathbb{R} , x \ne - 1$

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

Explanation:

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

$\text{expressing y as a single rational function}$

$y = \frac{1}{x + 1} + \frac{3 \left(x + 1\right)}{x + 1} = \frac{3 x + 4}{x + 1}$

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

$\text{solve "x+1=0rArrx=-1larrcolor(red)" excluded value}$

$\Rightarrow \text{domain is } x \in \mathbb{R} , x \ne - 1$

To find any excluded value in the range, rearrange the function making x the subject.

$y \left(x + 1\right) = 3 x + 4 \leftarrow \text{ cross-multiplying}$

$x y + y = 3 x + 4$

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

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

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

$\text{The denominator cannot be zero.}$

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

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