IB Maths AI HL Functions Toolkit Paper 1 & 2 f(g(x)), domain, range ~8 min read

Composite Functions

A composite function chains two functions together — the output of one becomes the input of the next. (f ∘ g)(x) = f(g(x)) means “do g first, then f“. Order matters: swapping it usually gives a different function.

📘 What you need to know

Notation: (f ∘ g)(x) = fg(x) = f(g(x)) — all the same thing.
Apply the inner function first: start with the one closest to x.
Order matters: (f ∘ g)(x) and (g ∘ f)(x) are generally different.
(f ∘ f)(x) is NOT [f(x)]²: the first composes f with itself, the second squares the output.
Domain: the outputs of the inner function must be valid inputs to the outer. Restrict the original domain accordingly.
Range: substitute the (restricted) domain through both functions in order.

How composition works

Think of each function as a machine. In (f ∘ g)(x), the input x is fed into the inner machine g first, producing g(x); that output then enters f, giving f(g(x)). To evaluate at a number, work inside-out: compute g(x), then apply f to that result. To find an algebraic expression, substitute the inner function’s formula wherever x appears in the outer formula, then simplify.

The same input x = 2 gives different outputs depending on which function acts first — (p ∘ q) and (q ∘ p) are different functions.

Composite function at a glance (f ∘ g)(x) = f(g(x)) apply inner first · (f ∘ g) ≠ (g ∘ f) in general · (f ∘ f) ≠ [f(x)]²

Domain & range of a composite

For (f ∘ g)(x) to be defined at x, two things must hold: x must be in the domain of g, and g(x) must be in the domain of f. Start with g‘s domain, find the range of g on that domain, then intersect that range with f‘s domain. If the intersection is smaller, work backwards through g to restrict the original domain. The range of (f ∘ g) is then whatever values come out when you push the restricted domain through both functions in order.

Sketching helps. On the GDC, graph f(g(x)) directly — the visible portion of the curve immediately shows you the domain and range.

🧭 Recipe — composite functions

Identify which function is inner: the one closest to x in the notation.
Apply inner first: compute g(x) numerically, or substitute g‘s formula wherever x appears in f.
Apply outer: feed the result into f, then simplify.
Domain: x must be in g‘s domain AND g(x) must lie in f‘s domain — restrict if needed.
Range: push the restricted domain through g then through f; the resulting interval is the range of the composite.

Worked examples

WE 1

Order matters at a number

Let p(x) = 3x − 5 and q(x) = x² + 1. Find (a) (p ∘ q)(2) and (b) (q ∘ p)(2). What do you notice?

(a) inner q first q(2) = 2² + 1 = 5 p(5) = 3(5) − 5 = 10 (p ∘ q)(2) = 10 (b) inner p first p(2) = 3(2) − 5 = 1 q(1) = 1² + 1 = 2 (q ∘ p)(2) = 2 order matters: 10 ≠ 2.

WE 2

Composite expressions both ways

Let h(x) = 2x − 3 and k(x) = x² + 4. Find an expression for (a) (h ∘ k)(x) and (b) (k ∘ h)(x).

(a) substitute k inside h (h ∘ k)(x) = h(x² + 4) = 2(x² + 4) − 3 = 2x² + 8 − 3 (h ∘ k)(x) = 2x² + 5 (b) substitute h inside k (k ∘ h)(x) = k(2x − 3) = (2x − 3)² + 4 = 4x² − 12x + 9 + 4 (k ∘ h)(x) = 4x² − 12x + 13

WE 3

(f ∘ f)(x) is not [f(x)]²

Let f(x) = 2x + 1. (a) Find (f ∘ f)(x). (b) Find [f(x)]². (c) Compare.

(a) compose f with itself (f ∘ f)(x) = f(2x + 1) = 2(2x + 1) + 1 = 4x + 2 + 1 (f ∘ f)(x) = 4x + 3 (b) square the output [f(x)]² = (2x + 1)² [f(x)]² = 4x² + 4x + 1 (c) compare linear (4x + 3) vs quadratic (4x² + 4x + 1) — very different functions.

WE 4

Domain and range with restrictions

Let f(x) = 3x + 4 with −8 ≤ x ≤ 8, and g(x) = √x with 0 ≤ x ≤ 81. Find the domain and range of (f ∘ g)(x).

expression first (f ∘ g)(x) = 3√x + 4 g’s domain: 0 ≤ x ≤ 81 g outputs: 0 ≤ √x ≤ 9 f needs input in [−8, 8] intersect [0, 9] ∩ [−8, 8] = [0, 8] need 0 ≤ √x ≤ 8 square: 0 ≤ x ≤ 64 domain: 0 ≤ x ≤ 64 range: push through both at x = 0: 3(0) + 4 = 4 at x = 64: 3(8) + 4 = 28 range: 4 ≤ (f ∘ g)(x) ≤ 28

WE 5

Composite of rational functions

Let f(x) = 1x − 3 and g(x) = 2x + 5. (a) Find (f ∘ g)(x) and state the value of x for which it is undefined. (b) Find (g ∘ f)(x).

(a) sub g into f (f ∘ g)(x) = 1/((2x + 5) − 3) = 1/(2x + 2) = 1/[2(x + 1)] (f ∘ g)(x) = 12x + 2 denominator zero when 2x + 2 = 0 ⇒ x = −1 undefined at x = −1 (b) sub f into g (g ∘ f)(x) = 2·(1/(x − 3)) + 5 = (2 + 5(x − 3))/(x − 3) (g ∘ f)(x) = 5x − 13x − 3 (g ∘ f) is undefined at x = 3.

WE 6

Applied: rocket pressure

A rocket’s altitude above the launch site after t minutes is h(t) = 2t² + 5 km. The atmospheric pressure at altitude h km is P(h) = 100e^−0.12h kPa. (a) Find (P ∘ h)(t). (b) Find the pressure 3 minutes after launch (2 d.p.).

(a) sub h into P (P ∘ h)(t) = 100 e⁻⁰·¹²(2t² + 5) (P ∘ h)(t) = 100 e⁻⁰·²⁴t² ⁻ ⁰·⁶ (b) sub t = 3 exponent: −0.24(9) − 0.6 = −2.76 P = 100 e⁻²·⁷⁶ ≈ 100(0.06329) P ≈ 6.33 kPa at altitude h(3) = 23 km the air is very thin.

💡 Top tips

Read inside-out: the function next to x is applied first.
Use brackets when substituting — the entire inner expression goes wherever x appears in the outer.
Domain check before answering: outputs of inner must be valid inputs to outer.
For domain restrictions, work backwards through the inner function to translate the constraint into a restriction on the original x.
On the GDC, graph the composite directly to see domain/range visually.

⚠ Common mistakes

Applying the outer function first: (f ∘ g)(x) means g first, not f first.
Confusing (f ∘ f)(x) with [f(x)]² — the first composes, the second multiplies.
Forgetting brackets when substituting an expression — turns f(2x + 1) into the wrong thing.
Reporting the inner function’s domain as the composite’s domain — you may need to restrict further so the inner output fits the outer’s domain.
Dividing by zero in a composite of rational functions — check where the denominator vanishes.

Next up: Inverse Functions — functions that undo each other, including how to find them, when they exist, and how restricting the domain rescues many-to-one functions.

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