MISSION · Algebra
Domain and Range
What you are allowed to put in, and what can possibly come out
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LEVEL 2 · LEARN
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The idea in one sentence
The domain is every input the function will accept, and the range is every output it can actually produce — one is a question about the \(x\)-axis, the other about the \(y\)-axis.
Three things that break a function
Most functions accept every real number. Finding a domain is really about hunting for the few that don’t, and in school mathematics there are only three culprits.
Division by zero. \(f(x) = \frac{1}{x - 5}\) takes anything except \(x = 5\). Set the denominator to zero and solve; whatever you get is excluded.
Even roots of negatives. \(f(x) = \sqrt{x - 4}\) needs \(x - 4 \ge 0\), so \(x \ge 4\). Set the inside \(\ge 0\) and solve. Odd roots are fine — \(\sqrt[3]{-8} = -2\) exists.
Logs of anything not positive. \(f(x) = \log(x + 2)\) needs \(x + 2 > 0\), so \(x > -2\). Strictly greater: \(\log 0\) is undefined too.
If a function has several of these, every restriction applies at once and you keep only the overlap.
Context can restrict it further
A function modelling something real inherits the limits of the thing it models. If \(A(s) = s^2\) is the area of a square of side \(s\), the algebra is happy with \(s = -3\) but the square is not, so the domain is \(s > 0\).
If \(h(t)\) is the height of a ball, the domain runs from launch to landing and nothing past it. Exam questions call this a “sensible domain”, and the algebra will never tell you what it is — the situation does.
Range is the harder one
There is no checklist for range. It is asking what heights the graph reaches, and the reliable approach is to sketch or reason about the shape.
- \(f(x) = x^2 + 3\): a square is never negative, so the smallest output is \(3\) at \(x = 0\). Range \(y \ge 3\).
- \(f(x) = \sqrt{x - 4}\): a square root is never negative and grows without bound. Range \(y \ge 0\).
- \(f(x) = \frac{1}{x - 5}\): a fraction with 1 on top is never zero. Range: all \(y \ne 0\).
- \(f(x) = \sin x\): it oscillates between \(-1\) and \(1\). Range \(-1 \le y \le 1\).
Notice the pattern in the middle two: an excluded input usually leaves a vertical asymptote, and an impossible output usually leaves a horizontal one. The graph shows both as lines the curve refuses to cross.
If the function is invertible, there is a shortcut worth knowing: the range of \(f\) is the domain of \(f^{-1}\), and the two swap when you find the inverse.
Writing the answer down
Three notations say the same thing, and different courses prefer different ones.
| In words | Inequality | Interval |
|---|---|---|
| 4 and above | \(x \ge 4\) | \([4, \infty)\) |
| bigger than \(-2\) | \(x > -2\) | \((-2, \infty)\) |
| everything but 5 | \(x \ne 5\) | \((-\infty, 5) \cup (5, \infty)\) |
A square bracket includes the endpoint, a round one excludes it. Infinity always gets a round bracket — you never reach it.
Common mix-ups
- Swapping them. Domain is inputs, \(x\). Range is outputs, \(y\). Alphabetical order matches: d before r, x before y.
- Simplifying away a restriction. \(f(x) = \frac{x^2 - 9}{x - 3}\) tidies to \(x + 3\), but \(x = 3\) is still excluded — the graph has a hole there. The domain comes from the original expression.
- Assuming the range is everything because the domain is. Any parabola disproves it.
- Using \(\ge\) for a log or a denominator. Those need strictly greater; only even roots allow the endpoint.
Check yourself
Which value does the domain of \(\frac{1}{x-5}\) exclude, what is the domain of \(\sqrt{x-4}\), and what is the range of \(x^2 + 3\)?
\(x - 5 = 0\) at \(x = 5\), so 5 is excluded. \(x - 4 \ge 0\) gives \(x \ge 4\). And \(x^2\) bottoms out at 0, so \(x^2 + 3\) bottoms out at 3: \(y \ge 3\).
LEVEL 3 · REFLECT
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