# How do you find the domain and range of 1/(x+6)?

Jul 6, 2017

Domain: $\text{ } x \in \mathbb{R} , x \ne 6$

Range: $\text{ } y \in \mathbb{R} , y \ne 0$

#### Explanation:

The domain is all possible values of $x$ for which $\frac{1}{x + 6}$ is defined.

We see that the function is only undefined if the denominator is 0, meaning that $x + 6 = 0$

This tells us that $x$ cannot be $- 6$.

So we can say our domain is: $\text{ } x \in \mathbb{R} , x \ne - 6$

(This is just a fancy way of saying "$x$ can be all real numbers except for -6")

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Range is a little harder to find. We need to find all possible values that $\frac{1}{x + 6}$ could be.

Let's think about it this way: what does the graph of $y = \frac{1}{x + 6}$ look like? It will be the graph of $y = \frac{1}{x}$, but translated 6 units to the left, like this:
graph{y = 1/(x+6) [-13.71, 6.29, -4.76, 5.24]}

We need to find all possible $y$ values that can be generated from this graph.

When $x < - 6$, we can see that the function $\frac{1}{x + 6}$ will be negative, since $x + 6$ will be negative.

As we approach $x = - 6$ from the left side, the function flies downwards towards $- \infty$, hitting every possible negative value.

As we approach $x = - \infty$, the function tends towards zero, but never actually reaches it. This is because the denominator is getting bigger and bigger, so the fraction is getting closer and closer to 0 without ever reaching it.

Therefore, from $x = - \infty$ to $x = - 6$, we can say that we will hit all possible negative values of $y$.

The same logic can be used for the positive side of the graph.

As we approach $x = - 6$ from the right side, the function flies upwards towards $\infty$, hitting every possible positive value.

As we approach $x = \infty$, the function tends towards zero but never actually reaches it.

Therefore, from $x = - 6$ to $x = \infty$, we can say that we will hit all possible positive values of $y$.

We've checked every possible $x$ value now, and our possible $y$ values are:

$y$ is a real number
$y$ is positive OR negative

In other words:

$y \in \mathbb{R} , y \ne 0$

This is the range of 1/(x+6