# How do you find the domain and range of p(x) = 1/(x+5) -9?

Dec 22, 2016

"Domain: " {x|x!="-5"};
"Range: " {y|y!="-9"}.

## Domain:

The domain of a function is simply the set of all possible $x$-values for which the function produces an output. For any number $x$, as long as $p \left(x\right)$ is also a number, that $x$ is in the domain of $p \left(x\right)$.

Conversely, any number $x$ which $p \left(x\right)$ does not know what to do with is not in the domain of $p \left(x\right)$. This alternate approach is often easier.

In this example, $p \left(x\right) = \frac{1}{x + 5} - 9$, and we can immediately see where a potential problem is. Since $x$ appears in a denominator, $x$ cannot take on any value that makes the denominator 0; that would create division by 0, which is undefined. So we simply need to keep $x$ from making $x + 5 = 0$. (All other $x$ values are okay.)

The only $x$ that would make that true is $x = \text{-5}$ (which we get by solving $x + 5 = 0$ for $x$). That means, our domain is all numbers other than -5, or:

"Domain: " {x|x!="-5"}
("all $x$ such that $x$ is not $\text{-5}$")

## Range:

Similar to domain, the range of a function is all the possible output values that function can take on. If $x$ is in the domain, and $p \left(x\right) = y$, then that $y$ is in the range of $p$.

And, conversely, if there is no $x$ such that $p \left(x\right) = y$, then that $y$ is not in the range of $p$. So, can we find some values that are impossible for the function to be?

Yes we can, since it is not possible for $\frac{1}{x + 5}$ to be 0. No matter how large (positive or negative) $x$ is, we cannot divide 1 by a real number and get 0.

(Side note: we can, however, make $\frac{1}{x + 5}$ as close to $\pm \infty$ as we like, by taking $x$ close to $\text{-5}$. So $\frac{1}{x + 5}$ can be anything else, just not 0.)

Since $\frac{1}{x + 5}$ cannot be 0, we know that $\frac{1}{x + 5} - 9$ cannot be -9. Thus, our range is all $y$-values except -9:

"Range: " {y|y!="-9"}
("all $y$ such that $y$ is not $\text{-9}$")

graph{1/(x+5)-9 [-22.47, 6, -12.71, 1.53]}