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What is Domain?

The domain of a function is all the possible $x$-values, or inputs, that that function can take. This includes positive numbers and negative numbers.

If a function is a vending machine you can think of the domain as the different forms of payment you can use in order to get a candy bar. Maybe you use dollars, maybe quarters, or maybe you just aggressively shake the machine and hope for the best.

The best way to understand this, though, is to look at a graph.

Finding Domain Graphically

Suppose we wanted to find the domain for the following graphed function.

What are the $x$-values that this function takes? We basically want to identify the inputs.

Our graph starts at $-2$ on the $x$-axis and goes to $4$ on the $x$-axis. Since there's an open circle on $-2,$ we don't include it in the domain.

Because the graph is only defined between $-2$ and $4$ on the $x$-axis, our domain is only between $-2$ and $4$. More precisely, the domain is: $$D:\,\,-2<x\le4$$

Notice we didn't say $$D:\,\,-2\,\,{\color{#8925ae}{\le}}\,\,x\le4$$ because $-2$ is not included in the domain!

Interval Notation

One way to write the domain of a graph is by using interval notation.

We start from the left and write the intervals that $x$ is defined on. Use brackets, $\left[\quad\right]$, when the endpoints are included and parentheses, $\left(\quad\right)$, when the endpoints are excluded.

Using the graph above, the domain would be: $$D:\,\,(-2,4]$$ Because $-2$ is not included we use a parentheses. $4$ is included so we use a bracket.

There are three more points that our example hasn't covered.

1. All real numbers. If our domain is any real number value, we can say it spans from negative infinity to infinity, $(-\infty, \infty ).$ But more concisely, we use the symbol $\mathbb{R}$ to say the domain is all real numbers.
2. Infinities. If our domain spans continuously from a point to the left or right, we say it goes to negative infinity or positive infinity respectively. When using interval notation for infinity we always use parentheses, since infinity isn't a point. For example, we might write $(-\infty, 5)$ or $[24, \infty).$
3. Multiple intervals. If a graph has multiple segments you might need to use multiple intervals. We can use the union symbol, $\cup$, to join them together. So if a graph's domain is defined from $-2$ to $1$ and then from $4$ to infinity, we can write: $D:\,\,[-2,1]\cup[4,\infty)$.

Finding Domain Algebraically

Sometimes you'll be given an equation and asked to find its domain. You can graph it but that might not always be easy.

When looking at an equation, we just have to find the $x$-values that would "break" the equation, or cause it to be undefined.

For example, if you were given a linear number line like $y=x$, you can say the domain is all real numbers. No $x$-value you plug in causes the equation to be undefined.

Now, suppose you have $y=\frac1x$. If we plug in $x=0$, we get a zero in the denominator which we can't have. This "breaks" the equation. So we say the domain is all real numbers except $0$.

Luckily, there are only a few cases to keep an eye out for.

1. Fractions. As explained above, fractions "break" if their denominator equals $0$.
2. Square roots. Square roots break if there is a negative value under the root.

Example 1

We want to find the domain for $$f(x)=\frac{\sqrt{x+1}}{3-x}$$ Notice there's both a fraction and a square root to tackle.

Let's start with the fraction. We know the denominator cannot be equal to $0$ (you can't divide by $0$!). $$\begin{align} \text{denominator}&\ne0\\ 3-x&\ne0 \end{align}$$ Solving for $x$ will show us a value of $x$ that is not in the domain. $$\begin{align} 3-x&\ne0\\ 3&\ne x \end{align}$$ When $x=3$ we get a $0$ in the denominator and our function "breaks", so we can exclude that from the domain.

Next, we have to look at the square root. The function expression under the root cannot be negative. $$\begin{align} \text{expression under root}&\ge0\\ x+1&\ge0 \end{align}$$ Solving for $x$ will give us that values that give us nonnegative terms. $$\begin{align} x+1&\ge0\\ x&\ge-1 \end{align}$$ As long as $x$ is greater than or equal to $-1$, our function is defined. In other words, an $x$ values less than $-1$ gives us a negative square root which "breaks" our function.

Finally, we write the domain. We know $x$ has to be a number greater than or equal to $-1$ and it cannot be equal to $3$. We have two intervals: a portion from $-1$ to $3$ (not included) and a portion from $3$ (not included) to infinity. $$D:\,\,[-1,3)\cup(3,\infty)$$

Graphing this, we see we were correct.

Solution

Again, we have both a square root and a fraction.

The square root portion must be nonnegative. $$x-1\ge0$$

But, because it's in the denominator of a fraction, it can't be equal to $0$. $$\sqrt{x-1}\ne0$$

Combining these two, we can exclude $0$ and just look for: $$x-1>0$$ Solve this gives us: $$x>1$$ This tells us that the domain is all the $x$ values greater than $1$. Because $1$ itself is not included, we use parentheses when writing the domain. $$D:\,\,(1,\infty)$$

An extension of this concept...is to find the range! Range consists of all $y$ values of a function. We can also explore domain and range with complex numbers. We'll learn about that a little later.

(Video) Introduction to Functions

by mathman1024

In this video, mathman1024 explains the basics of functions. He goes over what the domain is and how you find it.

$\large{\text{Summary}}$

Relations can be a set of ordered pairs: $$\{(2,3),(4,1),(0,1),(2,-2),(-1,5)\}$$ Or they can be a continuous graph:

Domain

When we talk about relations or functions, we want to talk about the domain and range. The domain is the set of $x$-values used in the relation.

You can describe the domain of a set by just listing the $x$ values of all the ordered pairs in the set.

In the example set above, $\{(2,3),(4,1),(0,1),(2,-2),(-1,5)\}$, the domain is $D=\{-1,0,2,4\}$ (notice that there are multiple $2$'s in the relation but only one was written down for the domain).

What about the graph above? The domain is all real numbers, since the $x$ values span from negative infinity to positive infinity. This can be written in a couple different ways: $D=(-\infty,\infty)$ or $D=\mathbb{R}$. The $\mathbb{R}$ symbol is used to represent the set of all real numbers.

Domain of a Function

The domain of a function is the set of all valid inputs for the function. A valid input yields a valid output. For example, say your function is $$f(x)=x+1$$

You can plug any real number into this function and get a valid output. For example: $$f(1)=1+1=2\\ f(5.5) = 5.5+1=6.5\\ f(\pi) = \pi+1$$

Since any real number works, we say that the domain of this function is the set of all real numbers. In other words, the domain of $f(x)$ is $\mathbb{R}.$

However, sometimes the domain is more restricted. For example, let's look at the following function: $$g(x)=\frac{1}{x}$$

You can definitely plug in most numbers and get a valid output. For example: $$g(1)=\frac{1}{1}=1\\ g(5.5)=\frac{1}{5.5} = 0.\overline{18}\\ g(\pi)= \frac{1}{\pi}\approx 0.31830988618$$

These outputs may not always be integers or rational numbers (like $g(\pi)$) but they are still valid real numbers.

But what happens if we have an input of $0?$ $$g(0)=\frac{1}{0}$$

But wait...we learned earlier that dividing by zero is undefined — it will not yield a valid output! So, $0$ is not a valid input, and we cannot include $0$ in the domain for this function. All other values are valid, so they can be included in the domain. Because of this, our domain is the set of all real numbers, besides $0.$ In other words, the domain of $g(x)$ is $\mathbb{R}\setminus \{0\}.$

(Note: We did not write $``\mathbb{R}\setminus 0"$ for the domain because the domain is a set, and to do a set minus operation (which is represented by the $\setminus$ symbol meaning "exclude") you need two sets. $0$ is not a set, while $\{0\}$ is a set.)

Let's do another example!

Solution

Let's find the domain of $h(x)=\frac{1}{x-2}.$

You can plug in most numbers and get a valid output. For example: $$g(1)=\frac{1}{1-2}=\frac{1}{-1}=-1\\ g(6)=\frac{1}{6-2} = \frac{1}{4} = 0.25\\ g(-2)= \frac{1}{-2-2}=\frac{1}{-4} = -0.25$$

However, remember in the last example when we said that we could not divide by $0?$ Let's figure out which value will yield a denominator of $0.$ This will be the only value we have to exclude from the domain.

Let's set $x-2$ equal to $0$ in order to solve for the value of $x$ that makes the denominator $0:$

$$x-2=0 \iff x=2$$

So, the only input that yields an invalid output is $2,$ because $h(2)=\frac{1}{2-2}=\frac{1}{0},$ which is undefined. This means that the domain of $h(x)$ is $\boxed{\mathbb{R}\setminus \{2\}}.$