# How do you find its inverse and check your answer and state the domain and the range of f(x) = 4 /( x+2) and f^-1?

Dec 8, 2017

Explained below

#### Explanation:

The domain: $D \left(f\right) = \mathbb{R} / \left\{- 2\right\}$
$x + 2 \ne 0$
$x \ne - 2$

In order to find inverse function the following statement must be true:
For every ${x}_{1} / {x}_{2} \in D \left(f\right) : {x}_{1} \ne {x}_{2} \implies f \left({x}_{1}\right) \ne f \left({x}_{2}\right)$

$\frac{4}{{x}_{1} + 2} = \frac{4}{{x}_{2} + 2}$
$\cancel{4} \left({x}_{2} + \cancel{2}\right) = \cancel{4} \left({x}_{1} + \cancel{2}\right)$
${x}_{2} = {x}_{1}$
So now we know that this function is simple and has an inverse function. Let's find it.

$y = \frac{4}{x + 2}$

$x = \frac{4}{y + 2}$

$x \left(y + 2\right) = 4$

$y + 2 = \frac{4}{x}$

${f}^{-} 1 : y = \frac{4}{x} - 2$

If we want to find $H \left(f\right)$ we have to find $D \left({f}^{-} 1\right)$. It's:
$x \ne 0$
so $H \left(f\right) = \mathbb{R} / \left\{0\right\}$

(english is not my native language which means you may use different way to solve it)