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Domain and Range Calculator

Find the domain and range of a function with each restriction identified separately — zero denominators, negative radicands, non-positive log arguments — and then combined correctly.

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How to use it

Type the function and say whether you want the domain, the range, or both — they take different work, and the range is usually the harder half. Write roots as sqrt(...) and logs as log(...) or ln(...). If the question restricts the domain in advance, include that condition, because it changes the range. Functions that appear inside a wider precalculus question belong in the precalculus solver.

The method behind it

The domain is the set of inputs the formula can legally accept. The range is the set of outputs it actually produces. Finding a domain is a search for what breaks; finding a range is a question about what the function can reach.

Only three structures create domain restrictions in the functions a first course uses, and each has a fixed condition attached.

\frac{p ( x )}{q ( x )} \Rightarrow q (x) \neq = 0, 2 n u (x) \Rightarrow u (x) \geq 0, lo g_{b} (u (x)) \Rightarrow u (x) > 0

Odd roots and polynomials contribute nothing, because they accept every real number. When several restrictions appear together, solve each one separately and then take the intersection — every condition must hold at the same time for a single input to be legal. Solving those conditions is usually an inequality problem, and finding the zeros of a denominator is usually a factoring problem.

Range takes a different tool for each family. A linear function with non-zero slope reaches everything. A quadratic reaches everything on one side of its vertex value, which completing the square exposes directly.

f (x) = a (x - h)^{2} + k \Rightarrow range = [k, \infty) if a > 0, (- \infty, k] if a < 0

An even root outputs only non-negative values before any shifting, so the range starts at the vertical shift. An exponential is always strictly positive before shifting. For a rational function the general technique is to swap the roles of the variables: solve $y = f (x)$ for $x$ , and any $y$ that makes that impossible is missing from the range.

That missing value is usually the horizontal asymptote, which is the same number you get by comparing the degrees of numerator and denominator. The language of approaching a value without attaining it belongs to limits and continuity, and rewriting radicals as fractional powers using the exponent and radical rules often makes a restriction easier to see.

Worked examples

Worked example

f (x) = \frac{2 x + 1}{x - 4}

1
Only one structure can break: the denominator. Set it to zero and solve to find the single excluded input.
$x - 4 = 0 \Rightarrow x = 4$
2
The domain is every real number except that one. Note the round brackets around 4 on both sides — a single point is being removed.
$(- \infty, 4) \cup (4, \infty)$
3
For the range, set y equal to the function and solve for x. Multiply out and gather the x terms.
$y (x - 4) = 2 x + 1 \Rightarrow y x - 4 y = 2 x + 1$
4
Factor x out of the left-hand side and divide. That division is only legal when the bracket is non-zero.
$x (y - 2) = 4 y + 1 \Rightarrow x = \frac{4 y + 1}{y - 2}$
5
y = 2 makes the inverse undefined, so no input produces an output of 2. That value is the horizontal asymptote, which matches the ratio of the leading coefficients.
$y \neq = 2$

Answer

Domain: x \neq = 4; Range: y \neq = 2

Worked example

h (x) = - 2 x^{2} + 8 x - 3

1
A polynomial has no restrictions at all, so the domain question is finished immediately.
$Domain = (- \infty, \infty)$
2
For the range, complete the square. Factor the leading coefficient out of the first two terms only.
$- 2 (x^{2} - 4 x) - 3$
3
Half of −4 is −2 and its square is 4. Add and subtract inside the bracket, so nothing changes in value.
$- 2 [(x - 2)^{2} - 4] - 3$
4
Distribute the −2 across both terms in the bracket. The −4 becomes +8, which is where the sign error usually happens.
$- 2 (x - 2)^{2} + 8 - 3 = - 2 (x - 2)^{2} + 5$
5
The squared term is never negative, so multiplying by −2 makes it never positive. The largest possible output is 5, reached at x = 2, and confirmed by substitution: −8 + 16 − 3 = 5.
$Range = (- \infty, 5]$

Answer

Domain: (- \infty, \infty); Range: (- \infty, 5]

Which restriction applies?

Domain questions are a checklist, not an insight. Run down the structures in the function, collect one condition from each, and then combine them. The tree below is that checklist made explicit.

Find the restriction on the domain

Scan the formula. Which of these structures does it contain?

Common mistakes

Taking the union instead of the intersection

For $\frac{x + 2}{x - 3}$ the conditions are $x \geq - 2$ and $x \neq = 3$ . Both must hold at once, so the domain is $[- 2, 3) \cup (3, \infty)$ . Combining them with or would readmit values the function cannot accept.

Cancelling away a restriction

$\frac{x ^{2} - 4}{x - 2}$ simplifies to $x + 2$ , but the original function is still undefined at $x = 2$ . The simplified formula describes a line with a hole in it. Determine the domain from the original expression, always before simplifying.

Assuming the range follows the domain

$f (x) = x^{2}$ accepts every real number but produces only non-negative ones, so the range is $[0, \infty)$ . An unrestricted domain says nothing about the range; that has to be worked out from the shape of the function.

Practice

Answers are checked here — nothing is sent anywhere.

1
$Which two values are excluded from the domain of \frac{1}{x ^{2} - 9} ?$
2
$Domain of 5 - x$
3
$Range of f (x) = x^{2} + 4$

Frequently asked questions

Interval notation or set notation?

Both are given. Interval notation is more compact for continuous stretches; set-builder notation is clearer when isolated points are removed, as in all real x with x not equal to 4.

How do I find a range without drawing the graph?

Swap the roles of the variables: solve y = f(x) for x, and whatever values of y make that impossible are the ones missing from the range. For quadratics, completing the square is faster, because the vertex is the extreme value.

Does it handle piecewise functions?

Yes. The domain is the union of the pieces, and the range is the union of what each piece produces on its own sub-domain. Endpoints need checking individually because a piece may approach a value without reaching it.

What if a square root sits in the denominator?

Then two restrictions collide and the strict one wins: the radicand must be greater than zero, not merely greater than or equal to it, because zero in the denominator is still forbidden.

What does it cost?

The first solution is free with no account. Free accounts include three solutions a day, and Gauth Plus lifts that for $11.99 a month with a three-day free trial.

Keep going

Precalculus solverFunctions, logarithms, and conics worked through with the reasoning shown.Exponent and radical rulesRewrite roots as fractional powers and the restrictions become visible.Limits and continuity, intuition firstHoles, asymptotes, and the difference between approaching a value and reaching it.Inequality calculatorSolve the radicand and denominator conditions a domain question produces.