One machine feeding another — and the order changes the answer.
Given f(x) = 2x + 3 and g(x) = x² − 1
Find (f ∘ g)(2).
(f ∘ g) means g runs first: g(2) = 2² − 1 = 3, then f(3) = 2(3) + 3 = 9. B does them in the wrong order — f(2) = 7 then g(7) = 48, which is (g ∘ f)(2). Read the ring from the inside out.
f(x) is a rule. Whatever goes in the bracket replaces every x.
f(2) = 2(2) + 3 = 7
a number in
f(a) = 2a + 3
a letter in
f(x + 1) = 2(x + 1) + 3 = 2x + 5
a whole expression in
The bracket is a slot. Keeping the substituted expression in its own brackets protects it from the operations around it.
start
2
g →
then
3
f →
end
9
(f ∘ g)(2) = f(g(2))
g runs first
g(2) = 4 − 1 = 3
the inner one
f(3) = 6 + 3 = 9
feed it into f
f ∘ g reads “f of g”, so g is the one touching the input. It runs first even though it is written second.
(g ∘ f)(2) = g(f(2))
now f runs first
f(2) = 7
the inner one
g(7) = 49 − 1 = 48
very different from 9
Composition is not commutative. Socks then shoes is not shoes then socks.
(f ∘ g)(x) = f(x² − 1)
put g into f's slot
= 2(x² − 1) + 3
f doubles and adds 3
= 2x² + 1
expand and tidy
(g ∘ f)(x) = (2x + 3)² − 1
the other way
= 4x² + 12x + 8
expand the square first
The middle term is 2(2x)(3) = 12x. Squaring a binomial always leaves three terms.
When do the two orders agree? With f(x) = 5x + m and g(x) = 2x + 3: f(g(x)) = 5(2x + 3) + m = 10x + 15 + m, and g(f(x)) = 2(5x + m) + 3 = 10x + 2m + 3. The x terms already match, so set the constants equal: 15 + m = 2m + 3, m = 12.
For f(x) = 1x + 2, simplify f(f(x)).
f(f(x)) = 1 ÷ (1x + 2 + 2)
the whole of f(x) goes in the slot
1x + 2 + 2 = 1 + 2x + 4x + 2 = 2x + 5x + 2
one fraction on the bottom
1 ÷ 2x + 5x + 2 = x + 22x + 5
dividing by a fraction flips it
1(x + 2)² multiplies f by itself. Check at x = 0: f(0) = ½, then f(½) = 1 ÷ 2½ = 25, and 0 + 20 + 5 = 25 ✓.
Three layers work the same way, one at a time. For f(x) = 3x + 2: f(f(x)) = 3(3x + 2) + 2 = 9x + 8, then f(f(f(x))) = 3(9x + 8) + 2 = 27x + 26.
(a) f(x) = 3x − 2 and f(g(x)) = 12x + 7. Find g(x). (b) h(x) = x − 3 and g(h(x)) = x² − 6x + 14. Find g(x).
(a) 3·g(x) − 2 = 12x + 7
outer known: treat g(x) as the unknown
3·g(x) = 12x + 9 → g(x) = 4x + 3
solve like any equation
(b) let u = x − 3, so x = u + 3
inner known: name it
g(u) = (u + 3)² − 6(u + 3) + 14
rewrite every x as u + 3
= u² + 6u + 9 − 6u − 18 + 14 = u² + 5
the u terms cancel
g(x) = x² + 5
rename the letter
Always check by composing forward: 3(4x + 3) − 2 = 12x + 7 ✓, and (x − 3)² + 5 = x² − 6x + 14 ✓. In (b), (x − 3)² + 5 is g(x − 3); the answer must be written with plain x.
The inner value comes out negative. f(x) = x² − 3x and g(x) = 1 − 2x. Find (f ∘ g)(2).
g(2) = 1 − 4 = −3
inner first
f(−3) = (−3)² − 3(−3)
−3 goes in with its brackets
= 9 + 9 = 18
minus times minus is plus
What is different: the value passed along is negative, so every sign in the outer rule matters.
Writing −3² − 3 × 3 = −9 − 9, or 9 − 9 = 0, loses a sign. Substitute (−3) with its brackets every time.
Solving a composite equation. f(x) = 2x − 1 and g(x) = x². Solve (g ∘ f)(x) = 25.
(2x − 1)² = 25
f goes inside g
2x − 1 = 5 or 2x − 1 = −5
a square has two roots
x = 3 or x = −2
check: f(−2) = −5, and (−5)² = 25 ✓
What is different: the unknown is the starting input. Taking only +5 loses half the answer.
An input the outside function refuses. f(x) = √x and g(x) = x − 6. Find (f ∘ g)(x), its domain, and (f ∘ g)(2).
(f ∘ g)(x) = √x − 6
g goes inside the root
x − 6 ≥ 0, so x ≥ 6
the root needs a non-negative inside
g(2) = −4, f(−4) = √−4
undefined — 2 is outside the domain
What is different: g accepts 2 happily, but its output must also be a legal input for f.
A function and its inverse cancel each other out.
f(f⁻¹(x)) = x and f⁻¹(f(x)) = x
f(x) = 2x + 3
our function
f⁻¹(x) = (x − 3) ÷ 2
undo, in reverse order
f(f⁻¹(9)) = f(3) = 9
back where we started
That is the standard way to test a claimed inverse — compose the two and see whether x comes back unchanged.
Composition comes as a number (often read from a table — look up the inner value, then the outer), as a formula, or backwards with g(x) unknown. Whatever the form, run the right-hand function first, and check a formula answer by plugging in x = 0 or 1 on both sides.
Given f(x) = 3x − 1 and g(x) = x² + 2
Find (f ∘ g)(3).
Inside first: g(3) = 3² + 2 = 11, then f(11) = 3(11) − 1 = 32. C is (g ∘ f)(3): f(3) = 8, then g(8) = 66. The two compositions are almost never equal.
