# How do you find the domain and range ofy = 3/ (x + 6) ?

Mar 2, 2018

Domain: $\left(- \infty , - 6\right) \cup \left(- 6 , \infty\right)$

Range: $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$

#### Explanation:

The domain of a function is all possible values of $x$where $f \left(x\right)$ is defined. Here, when the denominator is equal to 0, the function is undefined. In this case:

$x + 6 = 0$

$x = - 6$

So $y$ is only undefined at $x = - 6$. In interval notation, we write the domain as $\left(- \infty , - 6\right) \cup \left(- 6 , \infty\right)$.

The range of a function is all possible values for $y$. Another way to solve for the range is to find ${y}^{-} 1$, as in, the inverse function of $y$, and find its domain. Here, we can find ${y}^{-} 1$:

$y = \frac{3}{x + 6}$

Switch the variables and solve for $y$:

$x = \frac{3}{y + 6}$

$\frac{1}{x} = \frac{y + 6}{3}$

$\frac{3}{x} = y + 6$

$y = \frac{3}{x} - 6$

This is ${y}^{-} 1$. It is also defined when its denominator is equal to $0$.

So ${y}^{-} 1$ will be undefined when $x = 0$, and therefore $y$ will be undefined when $y = 0$. The range in interval notation is:

$\left(- \infty , 0\right) \cup \left(0 , \infty\right)$