Composition runs one function on the output of another: (f ∘ g)(x) means f(g(x)), with g first. The order is not a convention that can be swapped, and the composite's domain can be smaller than either function's.
Core ideas
“Transformations of Graphs”'s horizontal shifts were compositions written informally. Here the operation gets its own symbol and its own rules.
In (f ∘ g)(x) the function nearest the input is g, and it goes first. The notation reads right to left, which is the reverse of English reading order, and that mismatch is the whole difficulty. f(g(x)) is the same statement with the order visible.
With f(x) = 2x + 3 and g(x) = x − 4, (f ∘ g)(x) = 2x − 5 and (g ∘ f)(x) = 2x − 1. At x = 2 they give −1 and 3. Composition is not commutative even when both functions and both composites are lines.
An input must first be acceptable to g, and then g(x) must be acceptable to f. Both tests happen on the original expression, before any simplification, because simplifying can erase the evidence of the first stage.
The material
The definition, why order matters, how the domain is assembled, and why simplification cannot enlarge it.
Given functions f and g, the composite f ∘ g is defined by (f ∘ g)(x) = f(g(x)). Its domain is the set of inputs x in the domain of g for which g(x) lies in the domain of f. The symbol is read "f circle g" or "f after g".
"After" is the reading worth adopting, because it says the order out loud. f ∘ g is f after g: g happens, then f happens to the result. The alternative reading, "f of g", is also correct but is easier to say without thinking about which one moves first.
Substitution again, exactly as in “Function Notation”, with a whole function in the slot. Take f(x) = 2x + 3 and g(x) = x − 4.
Both composites are lines and they are still different functions. The check at x = 2 is worth doing every time a composite is built: evaluate step by step through the two functions, then evaluate the combined formula, and confirm they agree. −1 against −1 catches an order or substitution error immediately.
Two conditions, applied in order. The input must be legal for the inner function, and the inner function's output must be legal for the outer one.
Same two functions, two different domains, two different outputs. In the first, the subtraction happens before the root, so the root restricts which inputs survive. In the second the root goes first and the subtraction cannot cause a failure, so only the root's own condition applies.
Take f(x) = x² and g(x) = √x. The composite simplifies to a formula whose appearance has forgotten the inner function's domain.
The simplified formula x accepts negative inputs, but the composite does not: g(−1) never produces a real number, so f is never reached. This needs no cancellation or rational-function technique. It follows directly from the two stages: find the domain before simplifying, and carry it beside the simplified rule.
Interactive
f(g(x)) and g(f(x)), computed separately
Method
Four steps. The domain step comes before the simplifying step, and that ordering is the point.
f(g(x)) first, on its own line, before any substitution. For (f ∘ g) with g(x) = x − 4, that is f(x − 4). Seeing the inner expression sitting in the slot prevents the order from flipping.
Same discipline as “Function Notation”: brackets around the whole inner expression, in every slot. 2(x − 4) + 3, not 2x − 4 + 3.
Ask which inputs g rejects, then which of g's outputs f rejects. For f(x) = x² after g(x) = √x, the inner root requires x ≥ 0; the outer square adds no restriction.
Now expand and collect. Then pick an input, run it through g and f separately, and compare with the simplified formula. Disagreement means the expansion is wrong; agreement at one point is not a proof, but it catches nearly every algebra slip.
The first pair shows that order matters. The second shows that the simplified formula cannot be trusted for a domain.
The table is the honest way to show that order matters. Both results are lines with the same slope, so a quick glance could mistake them; the intermediate values reveal which constant was applied first.
In the second example the simplified formula x is perfectly happy at x = −1, but the composite is not defined there because √(−1) is not real and the outer square is never reached. This is why stage 1 of the domain test is not optional.
Both checks were done by running an input through the two functions in sequence. That is slower than trusting the algebra and it is the only verification that tests the order as well as the simplification.
For a faded composite, take u(x) = 3x − 1 and v(x) = x + 2. The first line (u ∘ v)(x) = u(x + 2) is supplied. Finish both orders, find each domain, and run x = −1 through each chain before checking the two simplified formulas.
Given two functions you should produce f ∘ g and g ∘ f, show they differ by evaluating both at a specific input, and give each domain with the excluded values justified by which stage rejected them. One independent pair should be linear; a second should include a square root whose restriction survives simplification.
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