Recipe

Find the domain of a function

Use when: the formula is built from pieces that each carry their own restriction — a square root, a fraction, a logarithm — and you need every input that survives all of them at once

Goal: the domain written as one interval or union of intervals, guaranteed to satisfy every restriction the formula imposes, simultaneously

Taught in Functions: domain, codomain, and composition

Quick recipe

Identify every piece that restricts the domain
Write and solve each restriction on its own
Intersect every restriction — never union them
Write the final domain and check a boundary point

1
Scan the formula for the three classic restriction sources: an even root (needs a non-negative radicand), a denominator (can't equal zero), a logarithm (needs a positive argument). List every one that appears — a formula can have zero, one, or several at once.

Example

$h(x) = \dfrac{\sqrt{x+9}}{x-1}$ has two restriction sources: a square root (an even root, on $x+9$ ) and a denominator (on $x-1$ ). There's no logarithm here.

Why: missing even one restriction source produces a domain that's too big — an $x$ that clears the restrictions you found gets reported as legal even though a restriction you missed would have rejected it.
2
For each source, write its condition and solve it by itself, without combining it with any other restriction yet: an even root needs $(\text{inside}) \ge 0$ ; a denominator needs $(\text{denominator}) \ne 0$ ; a logarithm needs $(\text{argument}) > 0$ .

Example

Square root: $x+9\ge0 \Longrightarrow x\ge-9$ . Denominator: $x-1\ne0 \Longrightarrow x\ne1$ .

Why: each restriction is an independent condition on its own piece of the formula — solving two of them together too early is exactly how a boundary value or an excluded point gets lost.
3
The domain is every $x$ that satisfies ALL of the restrictions at once, not any $x$ that satisfies at least one of them.

Example

$x\ge-9$ intersected with $x\ne1$ : since $1$ lies inside $[-9,\infty)$ , it has to be explicitly removed, giving $[-9,1)\cup(1,\infty)$ .
- an excluded point from a $\ne$ condition already falls outside the interval from a $\ge$ or $\le$ condition — that exclusion never mattered; no extra carve-out needed in the final answer.
- an excluded point from a $\ne$ condition falls inside the interval from another condition — carve that single point out of the interval explicitly.
Why: a function only produces an output when it clears every one of its restrictions simultaneously — one restriction failing jams the whole formula, so combining restrictions can only shrink the domain, never grow it.
4
State the domain in interval notation, then substitute one of the boundary numbers back into the original formula to double-check whether it's correctly included or excluded.

Example

Check $x=-9$ : $\sqrt{-9+9}=0$ is defined, and $-9-1=-10\ne0$ , so $x=-9$ is correctly included. Check $x=1$ : the denominator $1-1=0$ , so $x=1$ is correctly excluded — matching $[-9,1)\cup(1,\infty)$ .

Why: boundary points are exactly where a " $\ge$ " gets mistyped as a strict " $>$ ," or an excluded point gets left sitting inside the interval by accident — substituting back in catches that whole class of error directly.

Common mistakes

Taking the union of the restrictions instead of their intersection, which reports a domain that's too generous.
Solving only the more visually obvious restriction (a square root) and missing a plain denominator elsewhere in the formula.
Getting a boundary's inclusion or exclusion backwards — writing $\ge$ where a strict $>$ or $\ne$ was needed, or the reverse.
For a composed function, applying only the outer function's restriction and forgetting the inner function carries its own restriction too — the domain of $g\circ f$ has to satisfy both $f$ 's domain and $g$ 's.