Calculus talks about functions constantly, and a little more carefully than College Algebra usually needed to. "Domain" and "range" stop being vocabulary-quiz words. They become the first thing you check before any calculus operation touches a function, because every one of those operations can fail at a point where the function itself was never defined.
Think of a function as a vending machine, not a calculator. You press a button (the input), and it drops exactly one item (the output). It never drops two items, and it never drops a random one each time. The domain is just the list of buttons that exist on this machine. Press a button that isn't there, like x=0 on a machine wired as 1/x, and nothing happens. The machine isn't broken. You just asked for something it was never built to give you. The range is just the pile of items you'd actually get if you pressed every working button once, nothing more.
A function f is a rule that pairs each allowed input x with exactly one output, written f(x). "Exactly one output" is the entire rule. A relation that sometimes hands back two different values for the same input isn't a function, no matter how naturally it arises. For example, y^2 = x fails this test: x=4 pairs with both y=2 and y=-2. The domain of f is the full set of inputs the rule is actually allowed to run on. The range is the set of outputs that actually occur, every value f(x) takes for some x in the domain, no more and no less. A function can fail to be defined at an input for three concrete reasons, and calculus runs into all three constantly. One is a denominator that would hit zero, as in \frac{1}{x}, undefined at x=0. Another is a quantity under a square root or other even-index radical that would go negative, as in \sqrt{x}, undefined for x<0. The third, starting a few sections from now, is a logarithm's argument that would be zero or negative. Finding a function's domain means ruling out every input where one of those three failures happens. Everything left over is the domain.
Theorem
Vertical Line Test
Recalled from College Algebra · Function fundamentals — first taught there.
A curve in the plane is the graph of a function of x if and only if no vertical line intersects the curve more than once.
The vertical line x = a consists of exactly the points whose first coordinate is a — so the curve's intersections with that line are exactly the points (a, y) the curve assigns to input a.
If the curve is the graph of a function, input a has exactly one output, so x = a meets it in at most one point — for every a. Conversely, if some vertical line x = a met the curve twice, at (a, y_1) and (a, y_2) with y_1 \neq y_2, that would mean input a produced two different outputs, which is precisely what the definition of a function forbids. So the curve fails to be a function exactly when some vertical line crosses it more than once.
Formula
Domain Restrictions of a Rational Expression
Recalled from College Algebra · Simplifying rational expressions — first taught there.
For a rational expression \dfrac{p(x)}{q(x)}, the domain excludes every value of x for which q(x) = 0, read off the original (unfactored, uncancelled) denominator.
Theorem
Domain of a Radical Function
Recalled from College Algebra · Square and cube root functions — first taught there.
For f(x) = \sqrt[n]{g(x)}:
- n even, bare radical: domain is \{x : g(x) \ge 0\}.
- n even, radical in a denominator: domain is \{x : g(x) > 0\}; the boundary point is excluded, since it would make the denominator 0.
- n odd, bare radical: domain is all of \mathbb{R}, with no condition to solve.
- n odd, radical in a denominator: domain is \{x : g(x) \neq 0\}, the only forbidden input being the one that zeroes the denominator.
A real nth root \sqrt[n]{x} is, by definition, a real number y with y^n = x. When n is even, y^n \ge 0 for every real y — an even power is never negative — so x must be \ge 0 for such a y to exist. That's the bare-radical, n-even case. When n is odd, y^n ranges over all of \mathbb{R} as y does, since odd powers preserve sign and grow without bound in both directions, so every real x has a real nth root — the bare-radical, n-odd case.
The two denominator cases layer the ordinary "a denominator can't be zero" restriction on top of that: the boundary g(x)=0 is excluded whenever the radical sits under a fraction, regardless of n's parity, since \sqrt[n]{0}=0 either way.
Where does this machine break?
Drag h — the break in the domain moves with it, always sitting right where the denominator hits 0.
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![Graph of the rational function with numerator coefficients [-2, 0, 8] and denominator coefficients [1, -4, 4], showing its asymptotes and holes](/assets/images/gen/rational-inspector-1ffe22317e.svg?v=1792993961)
A radical's domain
Drag h — the curve only ever exists to the right of x = h, never to the left.
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Which restriction actually applies?
Drag each expression into the bin naming the domain restriction that actually applies to it — one bin is a trap, since a squared term in a denominator can never actually hit zero.
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The range is the list of outputs the machine can actually drop. To find it, don't guess. Use one of three tricks. One: the formula may be a familiar one that's been nudged, like a square root with a number added on the end. Start from the familiar version, then move its outputs by that number. Two: if it's a parabola (an x^2 formula), find its lowest or highest point, and everything on one side of that point is fair game. Three: try to "run the machine backwards." Pick an output and ask whether some input would produce it. Any output that no input could produce is missing from the range.
Writing the answer down is its own small skill. The answer is a stretch of the number line, written as its two ends inside brackets. A square bracket [ means "this end counts." A round bracket ( means "get as close as you like, but this end itself is left out." The symbol \infty just means "keeps going forever," which is never a real stopping point, so it always gets a round bracket. The symbol \cup, called "union," means "and also." It glues two stretches together.
Finding a range takes more work than finding a domain, because nothing forbids an output outright. You have to work out which outputs actually happen. Three moves cover almost every readiness-level function. First, start from a parent whose range you know and track the shifts: \sqrt{x} has range [0,\infty), so \sqrt{x-2}+1 has range [1,\infty). The -2 inside only slides inputs, and the +1 outside lifts every output. Second, for a quadratic, find the vertex (completing the square if you must). The vertex's y-value is the top or bottom of the range. Third, solve y = f(x) for x and ask which y-values leave a legal x. For y = \frac{1}{x}+3 you get x = \frac{1}{y-3}, which fails only at y=3.
Write the answer in interval notation. A square bracket [\,] means the endpoint belongs to the set, and a round bracket (\,) means it doesn't. \infty is never a number the function reaches, so it always takes a round bracket. \cup ("union") glues separate pieces together: (-\infty,3)\cup(3,\infty) is every real number except 3. The same sets in inequality notation read y \ge 1 and y \ne 3. Both notations describe the same set; a question will say which one it wants.
Where this goes next
Every calculus technique that follows
Calculus is built on two questions: what a function's outputs approach near a point, and how fast the function is changing at a point. Both questions are meaningless the instant you've stepped outside the domain. That is why the first line of so many calculus solutions is "note the domain." It isn't habit: that line catches half of this course's most common mistakes before they happen.
Check your understanding (counts toward your progress)
For f(x)=-5x^2+14x+6, find f(11).
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
f(11)=-5(11)^2+14(11)+6=-445.
Correct answer
Also accepted:
Why
Substitute the input directly for x and simplify.
A sideways parabola x = y^2 fails the vertical line test at every x>0. What does that mean for y as a function of x?
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Shape alone never disqualifies a graph -- the vertical line test failing is exactly what disqualifies it, since it means one input has two outputs.
The failure happens at every x>0, not just at the single point x=0 -- x=0 is actually the one point with a single y-value, y=0.
The vertical-line-test failure is the same for every positive x regardless of which quadrant the resulting y-values land in.
Why
For any x>0, both y=\sqrt x and y=-\sqrt x satisfy x=y^2 -- one input, two outputs -- which is exactly the condition a function is never allowed to have.
A vertical line crossing the graph twice means one x-value produces two different y-values -- check what that does to the function definition.
Practice: Functions, domain, and range -- one more pass (7 problems · doesn’t affect your progress)
Rewrite the inequality -1 < x \le 4 in interval notation.
Answer in interval notation
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Interval notation: [3, ∞), (-∞, 2) ∪ (2, ∞) — square bracket means the endpoint is included, round means it isn't, and ∞ always gets a round one.
Correct answer
Also accepted:
Why
Why
The left end is strict (-1 < x), so -1 is excluded and gets a round bracket. The right end allows equality (x \le 4), so 4 is included and gets a square bracket. Result: (-1, 4]. The smaller number always goes on the left, matching the number line.
Each end gets its own bracket. A strict < means that endpoint is left out (round bracket). A \le means it's included (square bracket).
What is the domain of f(x) = \sqrt{2x-10}?
Answer in interval notation
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Interval notation: [3, ∞), (-∞, 2) ∪ (2, ∞) — square bracket means the endpoint is included, round means it isn't, and ∞ always gets a round one.
Correct answer
Also accepted:
Why
Why
An even radical needs a nonnegative radicand. 2x-10\ge0 gives x\ge5. x=5 itself is included, since \sqrt0=0 is defined, so the bracket at 5 is square: [5, \infty). Two common slips: writing (5, \infty), which throws away the perfectly good input 5, and flipping the inequality to x \le 5. Larger inputs make the radicand bigger, so those are the inputs that work.
Set the radicand 2x-10 greater than or equal to 0, then solve for x. An even root needs a nonnegative argument.
Which function excludes fewer real numbers from its domain, f(x)=\dfrac{1}{x^2-9} or g(x)=\dfrac{1}{x^2+9}?
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x^2-9=0 at x=\pm3, so f is the one with two exclusions, not zero -- this has the two functions swapped.
x^2+9 is a sum of a square and a positive number, so it's always positive and never zero -- g has no exclusions, unlike f.
x^2+9 never equals 0 for any real x, since x^2\ge0 makes the sum at least 9 -- g has no exclusions at all.
Why
x^2-9=0 at x=\pm3, so f excludes two values; x^2+9\ge9>0 for every real x, so g excludes nothing at all.
Check whether each denominator can actually equal 0 for some real x -- one of them, x^2+9, is a sum of a square and a positive constant.
g excludes no real numbers at all, while f excludes two
For f(x) = x^2 - 3x, write f(a+1) as a simplified expression in a.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Substituting a+1 for x gives (a+1)^2 - 3(a+1), which expands to a^2-a-2 -- either form is a correct evaluation of f at a+1.
Correct answer
Also accepted:
Why
Replace every x in x^2 - 3x with (a+1), exactly as you would with any other input.
What is the range of f(x) = -2(x+3)^2 + 5?
Answer in interval notation
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Interval notation: [3, ∞), (-∞, 2) ∪ (2, ∞) — square bracket means the endpoint is included, round means it isn't, and ∞ always gets a round one.
Correct answer
Also accepted:
Why
Why
The vertex is (-3, 5), and the leading -2 makes the parabola open downward. So 5 is the highest output, reached at x=-3, and the outputs fall without limit from there. The 2 makes the parabola narrower but doesn't change its peak. Range: (-\infty, 5]. The bracket at 5 is square because 5 is actually reached.
Read the vertex straight off the vertex form. Then check the sign in front: does this parabola open up or down?
What is the range of g(x) = \sqrt{9 - x^2}?
Answer in interval notation
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Interval notation: [3, ∞), (-∞, 2) ∪ (2, ∞) — square bracket means the endpoint is included, round means it isn't, and ∞ always gets a round one.
Correct answer
Also accepted:
Why
Why
The domain is -3 \le x \le 3, where 9-x^2 \ge 0. On that domain, 9-x^2 runs from 0 (at x=\pm3) up to 9 (at x=0), so its square root runs from 0 up to 3. The graph is the top half of a circle of radius 3. Range: [0, 3], with both brackets square because both 0 and 3 are reached. Here the domain is what keeps the range bounded.
Find the domain first: 9 - x^2 \ge 0. That limits which inputs you get to try.
Over that domain, what's the largest value 9 - x^2 reaches, and what's the smallest?
What is the range of h(x) = \dfrac{1}{x^2 + 1}?
Answer in interval notation
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Interval notation: [3, ∞), (-∞, 2) ∪ (2, ∞) — square bracket means the endpoint is included, round means it isn't, and ∞ always gets a round one.
Correct answer
Also accepted:
Why
Why
x^2+1 \ge 1, with equality at x=0, so the fraction is at most \frac{1}{1}=1, and 1 is reached. As |x| grows, the denominator grows without limit and the fraction shrinks toward 0, but a fraction with numerator 1 is never exactly 0. The outputs are every number above 0 up to and including 1: (0, 1]. It's round at 0 (approached, never reached) and square at 1 (reached).
The denominator x^2+1 is never smaller than 1. What does that make the largest possible output?
As x grows, the denominator grows without limit. Does the fraction ever actually reach 0?
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
Determine the domain of f(x) = \sqrt{5 - x}.
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That flips the inequality. Test x=0: 5-0=5\geq 0 works fine, so small x values ARE allowed — the restriction points the other way.
x=5 itself is fine: \sqrt{5-5}=\sqrt{0}=0 is a real number, so 5 should be included, not excluded.
A square root does restrict the domain — the inside can't be negative.
Why
The inside of the root must be \geq 0: 5 - x \geq 0 gives -x \geq -5, and flipping the inequality (divide by -1) gives x \leq 5.
Set the expression under the square root greater than or equal to zero and solve for x.
From College Algebra
What is the domain of f(x) = x^4 - 3x + 7?
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There's no denominator here to jam at zero — polynomials never have division trouble.
There's no square root here either — nothing requires x to stay nonnegative.
The 3 is just a coefficient being multiplied, not a value the input is forbidden from equaling.
Why
A polynomial has no denominators and no roots — nothing can jam. Its domain is all real numbers.
Quick review
Practice from earlier coursework, pulled in for this topic.
From College Algebra
If f(x) = x^2 - 4x, find f(-2). (four and negative two, spelled out.)
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Use parentheses around the input: f(-2) = (-2)^2 - 4(-2) = 4 + 8 = 12. Both signs matter: (-2)^2 is +4, and subtracting 4 \cdot (-2) adds 8. Getting -4 or -12 usually means a dropped parenthesis.
Correct answer
Also accepted:
Why
Put parentheses around the input everywhere x appears before you simplify.