# What is the domain and range of  f(x) = 4/(9-x) ?

Feb 3, 2016

domain: $\left\{x \in \mathbb{R} | x \ne 9\right\}$

range: $\left\{x \in \mathbb{R} | x \ne 0\right\}$

#### Explanation:

The domain of a function is the set of possible values you can input into it. In this case, the only value that cannot be entered into $f \left(x\right)$ is $9$, as that would result in $f \left(9\right) - \frac{4}{9 - 9} = \frac{4}{0}$. Thus the domain of $f \left(x\right)$ is $\left\{x \in \mathbb{R} | x \ne 9\right\}$

The range of $f \left(x\right)$ is the set of all possible outputs of the function. That is, it is the set of all values which can be obtained by inputting something from the domain into $f \left(x\right)$. In this case, the range consists of all real numbers besides $0$, as for any nonzero real number $y \in \mathbb{R}$, we can input $\frac{9 y - 4}{y}$ into $f$ and obtain

$f \left(\frac{9 y - 4}{y}\right) = \frac{4}{9 - \frac{9 y - 4}{y}} = \frac{4 y}{9 y - 9 y + 4} = \frac{4 y}{4} = y$

The fact that this works shows that ${f}^{- 1} \left(y\right) = \frac{9 y - 4}{y}$ is actually the inverse function of $f \left(x\right)$. It turns out that the domain of the inverse function is the same as the range of the original function, meaning that the range of $f \left(x\right)$ is the set of possible values you can input into ${f}^{- 1} \left(y\right) = \frac{9 y - 4}{y}$. As the only value that cannot be entered into this is zero, we have the desired range as
$\left\{x \in \mathbb{R} | x \ne 0\right\}$