Recipe

Compute a function composition

Use when: you're given f and g and asked to evaluate or simplify g \circ f or f \circ g, at a number or as a general formula

Goal: the correct output or formula for the composite, with its domain checked, not just the algebra carried out

Taught in Function composition

1. 1

   In g \circ f, apply f first and g second, even though g is written first. Read the notation right to left.

   Example

   (g \circ f)(4) means: apply f to 4 first, then apply g to that result.

   Why: composition is generally not commutative, so applying the two functions in the wrong order gives the wrong composite entirely.

2. 2

   Plug the given input into whichever function is applied first, and simplify.

   Example

   For f(x) = x+3: f(4) = 4 + 3 = 7.

   Why: this intermediate output is exactly what the outer function receives next, so it has to be computed before anything else.

3. 3

   Take the value found in the previous step and plug it into the other function to get the final output.

   Example

   For g(x) = 2x: g(7) = 2(7) = 14, so (g \circ f)(4) = 14.

   Why: this is the literal meaning of (g\circ f)(a) = g(f(a)), the outer function applied to the inner function's output.

4. 4

   Confirm that every output the inner function can produce is something the outer function is allowed to accept.

   Example

   For f(x) = x-9 into g(x) = \sqrt{x} with domain [0,\infty): f(3) = -6, which is outside [0,\infty), so g(f(x)) is only defined where x \ge 9.

   • the inner function's outputs always land inside the outer function's stated domain — the composition is defined everywhere the inner function is.
   • some outputs of the inner function fall outside the outer function's domain — the composition's true domain is only the inputs that avoid that problem.

   Why: a formula for the composite can be written down whether or not the composition is actually defined, so this check is the only thing that catches a domain mismatch.