Composition of functions

With f(x) = 2x + 1 and g(x) = 3 − x, f(g(x)) = 7 − 2x but g(f(x)) = 2 − 2x. The order changes the answer.

A discount and a coupon: which goes first?

Two ways to pay for a $50 item. The shop takes 10% off, d(x) = 0.9x, and you also hold a $5 coupon, c(x) = x − 5. Discount first, then coupon: c(d(50)) = 45 − 5 = 40 dollars. Coupon first, then discount: d(c(50)) = 0.9 × 45 = 40.50 dollars. Fifty cents apart. In both cases the output of one function went straight into the other. Only the order changed, and the order changed the bill.

Sit-ups to calories: the inside function acts first

Sometimes the output of one function is the input of the next. OpenStax describes a person doing sit-ups: s(t) is the number of sit-ups done in t minutes, and c(s) is the number of calories burned by s sit-ups. To get calories from minutes you first find the sit-ups, then the calories: c(s(t)). This chaining is called composition, written (c ∘ s)(t) = c(s(t)) and read "c composed with s". The function written last, on the inside, acts first.

f(g(x)) = 7 − 2x, but g(f(x)) = 2 − 2x

To evaluate f(g(1)) with f(x) = 2x + 1 and g(x) = 3 − x, work from the inside out. First g(1) = 3 − 1 = 2. Then f(2) = 2 × 2 + 1 = 5. To get a formula, substitute the whole of g(x) for x in f: f(g(x)) = 2(3 − x) + 1 = 6 − 2x + 1 = 7 − 2x. Check with x = 1: 7 − 2 = 5, the same as before. The other order is g(f(x)) = 3 − (2x + 1) = 2 − 2x, and g(f(1)) = g(3) = 0. The two compositions are different functions, so composition is not commutative.

f(g(2)) is not f(2) · g(2)

Do not confuse composition with multiplication. f(g(x)) means "put g(x) into f", while f(x)g(x) means "multiply the two outputs". With f(x) = x² and g(x) = x + 1, f(g(2)) = f(3) = 9, but f(2) · g(2) = 4 × 3 = 12. And the order matters again: g(f(2)) = g(4) = 5. Some pairs do give the same function both ways, such as f(x) = 4x and g(x) = x/2, where both orders give 2x, but that is the exception.

Look up the inside value first, and check the domain from the inside too

Composition works with tables too. If a table says g(2) = 5 and another says f(5) = 1, then f(g(2)) = f(5) = 1; you only need to look up the inner value first. Domains need one extra thought: x must be allowed in g, and g(x) must be allowed in f. With f(x) = 1/x and g(x) = x − 3, f(g(x)) = 1/(x − 3), which fails at x = 3 because g(3) = 0 cannot go into f. Going the other way, you can take a complicated function apart: h(x) = (x − 3)² is f(g(x)) with g(x) = x − 3 inside and f(x) = x² outside.

The lab shows both orders as two chains of machines. Choose f and g from the lists and change the input x. The top chain sends x through g and then f; the bottom chain sends it through f and then g. The panel writes out both compositions as formulas, so you can see whether the order matters for the pair you picked.