Unit 10 · Lesson 2 of 4
Domain and range
You'll be able to: find the domain of a function from its formula, and read the domain and range from a graph, in interval notation.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first. Pulling it out of your head is what makes it stick.
\(f(x) = x^2 - 4x\). Find \(f(-1)\).
\(f(-1) = (-1)^2 - 4(-1)\).
\(= 1 + 4 = 5\).
Solve \(2x - 3 \ge 5\). Give the answer in interval notation.
Add 3: \(2x \ge 8\). Divide by 2: \(x \ge 4\).
4 is included, so use a bracket. Infinity always gets a parenthesis. Answer: \([4, \infty)\).
Step 2 of 5The idea
The domain of a function is every input you are allowed to use. The range is every output the function actually gives.
From a formula, start with all real numbers and remove the inputs that break the math. On this test, only two things break it:
- a denominator equal to 0 (you can't divide by 0);
- a negative number under a square root.
A polynomial, like \(x^2 - 5x\), has neither, so its domain is all real numbers, \((-\infty, \infty)\).
From a graph, read the domain left to right along the x-axis, and the range bottom to top along the y-axis. A solid dot means that endpoint is included; an open dot means it is not.
Denominator: set the denominator \(= 0\) and solve. Those x-values are removed from the domain.
Square root: the radicand must be \(\ge 0\). Solve that inequality.
Graph: domain = leftmost x to rightmost x. Range = lowest y to highest y.
Writing it: use interval notation. Solid dot or \(\le\): [ ]. Open dot, removed value, or \(\infty\): ( ). Join pieces with ∪.
Why this shows up on the placement test: "find the domain" questions are quick to ask and catch people who forget either the denominator rule or the square-root rule.
Step 3 of 5Worked examples
Example 1: domains from formulas
Find the domain of \(f(x) = \dfrac{x + 1}{x^2 - 9}\), then of \(g(x) = \sqrt{6 - 2x}\).
- For \(f\), set the denominator to 0: \(x^2 - 9 = 0\), so \(x^2 = 9\) and \(x = 3\) or \(x = -3\).
- Remove both from the number line. That leaves three pieces: \((-\infty, -3) \cup (-3, 3) \cup (3, \infty)\).
- For \(g\), the radicand must be at least 0: \(6 - 2x \ge 0\).
- Subtract 6: \(-2x \ge -6\). Divide by \(-2\) and flip the sign: \(x \le 3\). Dividing by a negative flips the inequality (Lesson 12).
- 3 is allowed (\(\sqrt{0} = 0\)), so the domain is \((-\infty, 3]\).
Example 2: domain and range from a graph
- Domain: the graph runs from \(x = -3\) on the left to \(x = 2\) on the right.
- \(x = -3\) has a solid dot (included); \(x = 2\) has an open dot (not included). Domain: \([-3, 2)\).
- Range: the lowest point is at \(y = -1\) (solid) and the highest is at \(y = 4\) (open). Range: \([-1, 4)\).
Step 4 of 5Practice
Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went.
Find the domain of \(f(x) = 3x^2 - 5x + 1\).
This is a polynomial: no denominator with \(x\), no square root.
Every real number works. Domain: all real numbers, \((-\infty, \infty)\).
Find the domain of \(g(x) = \dfrac{5}{x - 4}\).
Set the denominator to 0: \(x - 4 = 0\), so \(x = 4\) must be removed.
Domain: \((-\infty, 4) \cup (4, \infty)\).
Find the domain of \(h(x) = \sqrt{x + 6}\).
The radicand must be at least 0: \(x + 6 \ge 0\), so \(x \ge -6\).
Domain: \([-6, \infty)\).
Find the domain of \(f(x) = \dfrac{x + 2}{x^2 - x - 12}\).
Set the denominator to 0 and factor: \(x^2 - x - 12 = (x - 4)(x + 3) = 0\).
So \(x = 4\) or \(x = -3\). Remove both.
Domain: \((-\infty, -3) \cup (-3, 4) \cup (4, \infty)\).
Find the domain of \(g(x) = \sqrt{10 - 2x}\).
Radicand at least 0: \(10 - 2x \ge 0\).
Subtract 10: \(-2x \ge -10\). Divide by \(-2\) and flip: \(x \le 5\).
Domain: \((-\infty, 5]\).
Use the graph. What is the domain of this function?
Read the x-values from left to right: from \(x = -4\) to \(x = 2\).
\(-4\) has a solid dot (bracket); \(2\) has an open dot (parenthesis).
Domain: \([-4, 2)\).
Use the graph. What is the range of this function?
Range runs from the lowest y to the highest y on the graph.
Lowest: the bottom of the U, at \(y = -3\). It is not an endpoint, so don't use the left endpoint's \(y = 1\).
Highest: the right endpoint, \(y = 6\) (solid dot). Range: \([-3, 6]\).
Find the domain of \(f(x) = \dfrac{\sqrt{x - 1}}{x - 5}\).
Two rules at once. The square root needs \(x - 1 \ge 0\), so \(x \ge 1\).
The denominator needs \(x - 5 \ne 0\), so remove \(x = 5\).
Start at 1 (included) and skip over 5: \([1, 5) \cup (5, \infty)\).
Step 5 of 5Wrap-up
You can now find a domain by throwing out zero denominators and negative radicands, and read the domain and range of a graph using solid and open dots.
Read or watch next: OpenStax College Algebra 2e, 3.2 Domain and Range (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Algebra 1: Introduction to the domain and range of a function (short videos). Quick rules: Functions cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "Why does the domain of a square root use greater than or equal to, but a denominator uses not equal?" It knows this course and can explain things another way.