Lesson Objectives
- Learn how to find the domain of a function
- Learn how to find the range of a function
How to Find the Domain and Range of a Function
In this lesson, we want to review the concept of domain and range. In our last lesson, we discussed the basics of relations and functions. Specifically, we reviewed how to determine if a relation, which we defined as a set of ordered pairs, was a function. Additionally, we reviewed the concepts of domain and range. The domain of a relation or function is the set of allowable inputs for our independent variable x. A more official definition is that the domain of a relation is the set of all real numbers that produce real numbers when substituted in for the independent variable x. Additionally, the range is the set of all outputs for our dependent variable y. In the case where we are just given a simple set of ordered pairs, finding the domain and range is pretty easy. Let’s suppose we had some function F: $$F=\{(6,9), (-2,3), (5,-1)\}$$ What’s the domain? Again the domain is just the set of x-values in our function. $$\text{domain:}\hspace{.1em}\{6, -2, 5\}$$ What’s the range? Again the range is just the set of y-values in our function. $$\text{range:}\hspace{.1em}\{9, 3, -1\}$$ What happens when we are asked to find the domain and range of a relation or function that is in the format of an equation or a graph? Let’s suppose we wanted to find the domain and range for the following function: $$y=2x - 1$$ 
When thinking about the domain, the first thing you would want to consider is if there are any restrictions on what can be plugged in for the independent variable x. Well, all we are doing is multiplying x by 2 and then subtracting away 1 from the result. So there’s no restriction on x. x can be any real number. Additionally, since we can vary x as much as we want, we can make y as small or as large as we want. Therefore, y can also be any real number. When we work with most linear functions, the domain and range are both all real numbers. We of course have special case scenarios such as horizontal and vertical lines.
domain: {all real numbers}
range: {all real numbers}
In most cases, we will want to figure out just the domain from the equation alone. When solving a domain problem, think about the following:
Example 1: Find the domain for each.
$$y=\frac{1}{x - 9}$$ Since we are not allowed to divide by zero, think about the denominator here: $$x - 9$$ We can set this equal to zero and solve: $$x - 9=0$$ $$x=9$$ This means x can't be 9. If we let x be 9, our denominator will be zero, and division by zero is not defined. $$\text{domain:}\hspace{.1em}\{x | x ≠ 9\}$$
Example 2: Find the domain for each.
$$y=\sqrt{x - 12}$$ Since we can only take the square root of a non-negative number and end up with a real number, we think about what is under the square root symbol: $$x - 12$$ We know that whatever is plugged in for x, the result of subtracting away 12 has to be 0 or larger: $$x - 12 ≥ 0$$ $$x ≥ 12$$ This means x can be 12 or any larger value. If we plug in a value that is less than 12, we end up with the square root of a negative. $$\text{domain:}\hspace{.1em}\{x | x ≥ 12\}$$
In some cases, we need to consider a combination of restrictions. Let's look at an example.
Example 3: Find the domain for each. $$y=\frac{\sqrt{x - 1}}{3x - 5}$$ Consider the fact that you have a square root with a variable and a variable in the denominator. Consider each part separately and then combine the two restrictions to build your domain. $$\sqrt{x - 1}$$ $$x - 1 ≥ 0$$ $$x ≥ 1$$ $$3x - 5=0$$ $$3x=5$$ $$x=\frac{5}{3}$$ We know that x needs to be greater than or equal to 1 and it can't be equal to 5/3. We can now combine these restrictions and create our domain.
$$\text{domain:}\hspace{.1em}\left\{x|x ≥ 1, x ≠ \frac{5}{3}\right\}$$
Example 4: Find the domain for each. $$y=\frac{7x - 1}{|2x +1| - 1}$$ Consider the fact that you have a variable in the denominator and the absolute value operation is involved. Let's set the denominator equal to zero and solve the resulting absolute value equation. $$|2x + 1| - 1=0$$ $$|2x + 1|=1$$ Split into a compound equation with "or": $$2x + 1=1$$ $$\text{or}$$ $$2x + 1=-1$$ First Part: $$2x + 1=1$$ $$2x=0$$ $$x=0$$ Second Part: $$2x + 1=-1$$ $$2x=-2$$ $$x=-1$$ We know that x can't be -1 or 0 as this results in a denominator of zero and division by zero is not defined. $$\text{domain:}\hspace{.1em}\{x | x \ne 0, -1\}$$ 
When thinking about the domain, the first thing you would want to consider is if there are any restrictions on what can be plugged in for the independent variable x. Well, all we are doing is multiplying x by 2 and then subtracting away 1 from the result. So there’s no restriction on x. x can be any real number. Additionally, since we can vary x as much as we want, we can make y as small or as large as we want. Therefore, y can also be any real number. When we work with most linear functions, the domain and range are both all real numbers. We of course have special case scenarios such as horizontal and vertical lines. domain: {all real numbers}
range: {all real numbers}
In most cases, we will want to figure out just the domain from the equation alone. When solving a domain problem, think about the following:
- We can't divide by zero
- We can't take the square root of a negative number and get a real number
- When we square a number or take the absolute value of a number, the result is non-negative
Example 1: Find the domain for each.
$$y=\frac{1}{x - 9}$$ Since we are not allowed to divide by zero, think about the denominator here: $$x - 9$$ We can set this equal to zero and solve: $$x - 9=0$$ $$x=9$$ This means x can't be 9. If we let x be 9, our denominator will be zero, and division by zero is not defined. $$\text{domain:}\hspace{.1em}\{x | x ≠ 9\}$$
Example 2: Find the domain for each. $$y=\sqrt{x - 12}$$ Since we can only take the square root of a non-negative number and end up with a real number, we think about what is under the square root symbol: $$x - 12$$ We know that whatever is plugged in for x, the result of subtracting away 12 has to be 0 or larger: $$x - 12 ≥ 0$$ $$x ≥ 12$$ This means x can be 12 or any larger value. If we plug in a value that is less than 12, we end up with the square root of a negative. $$\text{domain:}\hspace{.1em}\{x | x ≥ 12\}$$
In some cases, we need to consider a combination of restrictions. Let's look at an example. Example 3: Find the domain for each. $$y=\frac{\sqrt{x - 1}}{3x - 5}$$ Consider the fact that you have a square root with a variable and a variable in the denominator. Consider each part separately and then combine the two restrictions to build your domain. $$\sqrt{x - 1}$$ $$x - 1 ≥ 0$$ $$x ≥ 1$$ $$3x - 5=0$$ $$3x=5$$ $$x=\frac{5}{3}$$ We know that x needs to be greater than or equal to 1 and it can't be equal to 5/3. We can now combine these restrictions and create our domain.
$$\text{domain:}\hspace{.1em}\left\{x|x ≥ 1, x ≠ \frac{5}{3}\right\}$$
Example 4: Find the domain for each. $$y=\frac{7x - 1}{|2x +1| - 1}$$ Consider the fact that you have a variable in the denominator and the absolute value operation is involved. Let's set the denominator equal to zero and solve the resulting absolute value equation. $$|2x + 1| - 1=0$$ $$|2x + 1|=1$$ Split into a compound equation with "or": $$2x + 1=1$$ $$\text{or}$$ $$2x + 1=-1$$ First Part: $$2x + 1=1$$ $$2x=0$$ $$x=0$$ Second Part: $$2x + 1=-1$$ $$2x=-2$$ $$x=-1$$ We know that x can't be -1 or 0 as this results in a denominator of zero and division by zero is not defined. $$\text{domain:}\hspace{.1em}\{x | x \ne 0, -1\}$$ Skills Check:
Example #1
Find the domain. $$y=9x - \frac{1}{4}$$
Please choose the best answer.
A
$$(-\infty, 9)$$
B
$$\left(-\frac{1}{4}, 9\right)$$
C
$$\left(-\frac{1}{4}, 9\right)$$
D
$$(-9, \infty)$$
E
$$(-\infty, \infty)$$
Example #2
Find the domain. $$y=\frac{\sqrt{x - 5}}{x}$$
Please choose the best answer.
A
$$(-5, \infty)$$
B
$$(-5, 0) ∪ (0, \infty)$$
C
$$(-\infty, 0) ∪ (0, 5)$$
D
$$[5, \infty)$$
E
$$(-\infty, \infty)$$
Example #3
Find the domain. $$y=\frac{2x - 1}{\sqrt{x - 3}}$$
Please choose the best answer.
A
$$(-2, 3) ∪ (3, \infty)$$
B
$$(-3, \infty)$$
C
$$[3, \infty)$$
D
$$(3, \infty)$$
E
$$(-\infty, \infty)$$
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