Function Domain Calculator
Calculate function domain instantly with our math tool. Shows detailed work, formulas used, and multiple solution methods.
Reviewed by Manoj Kumar, Mathematics Educator
Formula
Domain = {x | f(x) is defined}The domain is found by identifying all x-values where the function produces a valid output. For rational functions, exclude values making the denominator zero. For square roots, require the radicand to be non-negative. For logarithms, require the argument to be positive.
Worked Examples
Example 1: Domain of a Rational Function
Problem:Find the domain of f(x) = (x + 2) / (3x - 6).
Solution:Set denominator equal to zero: 3x - 6 = 0\n3x = 6\nx = 2\nThe function is undefined at x = 2.\nDomain: All real numbers except x = 2\nInterval notation: (-Infinity, 2) U (2, Infinity)\nSet-builder: {x | x is not equal to 2}
Result:Domain: (-Infinity, 2) U (2, Infinity)
Example 2: Domain of a Square Root Function
Problem:Find the domain of f(x) = sqrt(3x - 9).
Solution:The radicand must be >= 0:\n3x - 9 >= 0\n3x >= 9\nx >= 3\nDomain: x >= 3\nInterval notation: [3, Infinity)\nSet-builder: {x | x >= 3}
Result:Domain: [3, Infinity)
Frequently Asked Questions
What is the domain of a function?
The domain of a function is the complete set of all possible input values (typically x-values) for which the function produces a valid, defined output. Think of it as the collection of all numbers you can legally plug into the function without causing mathematical problems like division by zero, taking the square root of a negative number, or computing the logarithm of a non-positive number. For example, the function f(x) = 1/x has a domain of all real numbers except zero, because plugging in x = 0 would require dividing by zero. Understanding the domain is fundamental because it tells you the complete set of inputs where the function actually works and produces meaningful results.
How do you find the domain of a rational function?
To find the domain of a rational function (a fraction where both numerator and denominator are polynomials), you must identify all x-values that make the denominator equal to zero, then exclude those values from the set of all real numbers. Start by setting the denominator equal to zero and solving the resulting equation. For example, for f(x) = (x + 3)/(x^2 - 9), the denominator x^2 - 9 = 0 gives x = 3 and x = -3. So the domain is all real numbers except 3 and -3, written in interval notation as (-Infinity, -3) U (-3, 3) U (3, Infinity). Always check if the denominator can be factored, as this might reveal multiple restrictions that are not immediately obvious.
How do you find the domain of a square root function?
For a square root function f(x) = sqrt(expression), the expression inside the square root (called the radicand) must be greater than or equal to zero, since the square root of a negative number is not a real number. Set the radicand greater than or equal to zero and solve the resulting inequality. For example, for f(x) = sqrt(2x - 6), solve 2x - 6 >= 0, which gives x >= 3, so the domain is [3, Infinity). If the square root appears in a denominator, the radicand must be strictly greater than zero (not equal to zero) to avoid division by zero. For cube roots and other odd-indexed roots, there are no domain restrictions since odd roots of negative numbers are defined.
What is the difference between domain and range?
Domain and range are complementary concepts that together describe the complete behavior of a function. The domain is the set of all valid input values (x-values) that you can plug into the function, while the range is the set of all possible output values (y-values) that the function can produce. For example, for f(x) = x^2, the domain is all real numbers (any x works), but the range is y >= 0 (the output is always non-negative). For f(x) = sqrt(x), the domain is x >= 0 (cannot take square root of negative) and the range is also y >= 0. Finding the domain focuses on input restrictions, while finding the range requires analyzing what outputs are actually achievable.
References
Reviewed by Manoj Kumar, Mathematics Educator · Editorial policy