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

Mar 14, 2017

The domain of is $= \mathbb{R} - \left\{- 3\right\}$
The range is $= \mathbb{R} - \left\{0\right\}$

#### Explanation:

As you cannot divide by $0$, $x \ne - 3$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 3\right\}$

To find the range, we need the domain of ${f}^{-} 1 \left(x\right)$

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

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

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

Therefore,

${f}^{-} 1 \left(x\right) = \frac{1 - 3 x}{x}$

The domain of ${f}^{-} 1 \left(x\right)$ is ${D}_{{f}^{- 1}} \left(x\right) = \mathbb{R} - \left\{0\right\}$

So, the range is $= \mathbb{R} - \left\{0\right\}$