# How do you find the domain and range of  f(x)= 2- 1/(x+6)^2?

May 1, 2018

Domain: $\left(- \infty , - 6\right) \cup \left(- 6 , \infty\right)$
Range: $\left(- \infty , 2\right)$

#### Explanation:

The domain is the set of all possible inputs $x$ to a function that give a valid result. Ie, all of the numbers you can replace $x$ with and not break any algebraic rules is the domain.

Here, we note that we have the term $\frac{1}{x + 6} ^ 2$ in our function. We know that we cannot divide by zero, so it follows that ${\left(x + 6\right)}^{2} \ne 0$. This implies that $x \ne - 6$. Thus, $x = - 6$ is not included in our domain. All other values of $x$ provide a legal answer, however, so our domain is $\left(- \infty , - 6\right) \cup \left(- 6 , \infty\right)$.

The range is the set of all possible outputs of a function. That is, if you plug in every $x$ from the domain into $f \left(x\right)$, all the results you get consist of the range.

We note that when $x$ gets very large, the denominator of $\frac{1}{x + 6} ^ 2$ gets very large, making the term itself quite small. Thus, $f \left(x\right)$ will get very near $2$ with large values of $x$.

With values of $x$ very close to $- 6$, we find that the denominator of $\frac{1}{x + 6} ^ 2$ gets very small, making the term evaluate to a large number. For closer and closer values of $x$ to $- 6$, $f \left(x\right) = 2 - \frac{1}{x + 6} ^ 2$ begins to get very negative, tending toward negative infinity.

Thus, our function covers all values between $- \infty$ and $2$, though it never quite reaches $2$. (And it cannot reach $- \infty$.) So our range is $\left(\infty , 2\right)$.

A graph of this function reveals this behavior.

graph{2 - 1/(x+6)^2 [-16, 16, -8, 8]}