Domain & Range
Find the domain of a polynomial, rational, root or log function — and the range where it can be stated exactly.
f(x) = √(x − 3)
- 1The radicand cannot be negative.A real square root needs x − 3 ≥ 0.
- 2Solve that inequality.It holds on [3, ∞).
- 3A square root returns non-negative values.Over a half-line the radicand takes every non-negative value, so the output does too.
The domain is every input a function accepts; the range is every output it produces. Only three things restrict a domain: division by zero, an even root of a negative, and the logarithm of a non-positive. For √(x − 3) the domain is [3, ∞) and the range is [0, ∞).
Start by assuming everything is allowed
The efficient way to find a domain is to begin with all real numbers and then remove what breaks. A polynomial never breaks — there is no division and no root — so its domain is always every real number, however complicated it looks. Everything else is a matter of spotting which of three forbidden operations appears and solving the small inequality it implies.
A denominator must not be zero, so set it equal to zero and exclude the solutions. An even root must not sit over a negative, so set the radicand ≥ 0 and solve. A logarithm needs a strictly positive argument, so set it > 0. When a function combines these, the domain is the overlap of all the conditions, not their union.
Range is the harder half
Domain is mechanical; range often is not. For a parabola the range follows from the vertex, and for a logarithm it is every real number, but for most other families finding it means locating turning points — which is a calculus question. The tool above states a range only when it can be derived exactly, and says so plainly when it cannot, rather than offering a number it cannot justify.
- 1 Assume every real number works. Start from (−∞, ∞) and remove only what an operation forbids.
- 2 Exclude anything that makes a denominator zero. For (x + 1)/(x² − 4), solving x² − 4 = 0 gives x = −2 and x = 2, both excluded.
- 3 Require every even radicand to be non-negative. For √(x − 3) solve x − 3 ≥ 0, which gives x ≥ 3.
- 4 Require every logarithm argument to be positive. For ln(x + 2) solve x + 2 > 0, which gives x > −2.
- 5 Combine the conditions and write it as intervals. A root in a denominator needs both at once, so 1/√(x − 5) needs x − 5 > 0 and the domain is (5, ∞).
What restricts a domain, and what does not
Only these three operations can remove values. Everything else is defined for every real input.
| Feature | Condition | Example domain |
|---|---|---|
| Polynomial | No restriction | x² − 4x + 7 → all real numbers |
| Denominator | Denominator ≠ 0 | 1/(x − 2) → all reals except 2 |
| Even root | Radicand ≥ 0 | √(x − 3) → [3, ∞) |
| Even root in a denominator | Radicand > 0 | 1/√(x − 5) → (5, ∞) |
| Logarithm | Argument > 0 | ln(x + 2) → (−2, ∞) |
| Odd root | No restriction | ∛(x − 1) → all real numbers |
The distinctions that cost marks
Even and odd roots behave differently. √ needs a non-negative radicand because no real number squares to a negative, but ∛ accepts anything — the cube root of −8 is −2. Writing a restriction for an odd root is a common wasted step.
The strict-versus-inclusive boundary matters too. A square root allows its radicand to equal zero, so √(x − 3) includes x = 3 and the interval is closed at that end. Put that root in a denominator and zero is no longer allowed, so 1/√(x − 5) excludes x = 5 and the bracket opens. And a hole from a cancelled factor still leaves the point out of the domain: (x² − 4)/(x − 2) simplifies to x + 2, but x = 2 was never allowed and stays excluded.