IB Maths AA SL
Paper 1 & 2
13 min read
Composite Functions
A composite function means putting one function inside another — using the output of one as the input of the next. The single most important rule: read inside-out and start with the function closest to x.
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What you need to know
- (f ∘ g)(x) means f(g(x)) — apply g first, then f
- Three equivalent notations: (f ∘ g)(x), fg(x), f(g(x))
- Always read inside-out: start with the function closest to x
- Order matters! (f ∘ g)(x) is usually NOT the same as (g ∘ f)(x)
- ff(x) means apply f twice — it’s not the same as [f(x)]2
- For the domain of (f ∘ g): the outputs of g must be valid inputs of f
What is a Composite Function?
A composite function feeds the output of one function straight into another. You take an input, apply one function to it, then apply a second function to the result. There are three equivalent notations — all mean the same thing:
(f ∘ g)(x)
“f composed with g of x”
f(g(x))
expanded form (clearest)
The Inside-Out Rule
The trickiest part of composite functions is figuring out which function to apply first. The rule is simple: start with the function closest to x.
How (f ∘ g)(x) Works
Read inside-out: the function next to x goes first.
Memory hook: in (f ∘ g)(x), the letter g is closer to x, so g goes first. Same with fg(x) — the rightmost function is touching x directly, so it acts first.
Order Matters
Swapping the order of composition usually changes the answer:
f ∘ g
(f ∘ g)(x) = f(g(x))
g first, then f
g ∘ f
(g ∘ f)(x) = g(f(x))
f first, then g
(f ∘ g)(x) ≠ (g ∘ f)(x) (in general)
Important: ff(x) = (f ∘ f)(x) means apply f twice — first to x, then to the result. It is not the same as squaring [f(x)]2. Don’t confuse them!
How to Build a Composite Function Expression
To find an expression for (f ∘ g)(x), follow these steps:
- Start with g(x) — the function closest to x.
- Substitute g(x) into f: wherever you see x in the formula for f, replace it with the entire expression for g(x).
- Simplify the result.
Domain and Range of a Composite Function
For (f ∘ g)(x), the input x first goes into g, and the output g(x) then becomes the input of f. So:
- The inputs have to be valid for g
- The outputs of g must also be valid as inputs of f
This means the domain of (f ∘ g) might be smaller than the domain of g on its own.
- Start with the domain of g.
- Find the outputs of g on that domain.
- Restrict so those outputs fit inside the domain of f. This may shrink the original domain.
- For the range, push the (now correct) domain through g, then through f in order.
Worked Examples
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Example 1 — Numerical evaluation
Given f(x) = 3x and g(x) = x + 2, find (f ∘ g)(5).
Answer:
Apply g first (closest to input).
g(5) = 5 + 2 = 7
Then apply f to that result.
f(7) = 3 × 7 = 21
(f ∘ g)(5) = 21
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Example 2 — Two-step evaluation
Given f(x) = √(x + 4) and g(x) = 3 + 2x, find the value of (g ∘ f)(12).
Answer:
Apply f first (closest to the input 12).
f(12) = √(12 + 4) = √16 = 4
Then apply g to that result.
g(4) = 3 + 2(4) = 11
(g ∘ f)(12) = 11
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Example 3 — Building an expression
Given f(x) = √(x + 4) and g(x) = 3 + 2x, find an expression for (f ∘ g)(x).
Answer:
Apply g first — write down g(x).
g(x) = 3 + 2x
Substitute g(x) into f. Wherever f has “x”, put “(3 + 2x)” instead.
f(g(x)) = f(3 + 2x)
= √((3 + 2x) + 4)
= √(7 + 2x)
(f ∘ g)(x) = √(7 + 2x)
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Example 4 — Same function twice
Given g(x) = 3 + 2x, find an expression for (g ∘ g)(x).
Answer:
Apply g once, then apply g to that result.
g(g(x)) = g(3 + 2x)
Substitute (3 + 2x) wherever g has “x”.
= 3 + 2(3 + 2x)
= 3 + 6 + 4x
(g ∘ g)(x) = 9 + 4x
Note: this is NOT the same as [g(x)]² = (3 + 2x)² = 9 + 12x + 4x².
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Example 5 — Order matters: (f ∘ g) vs (g ∘ f)
Given f(x) = x2 and g(x) = x + 1, find expressions for both (f ∘ g)(x) and (g ∘ f)(x).
Answer:
(f ∘ g)(x): apply g first, then f.
f(g(x)) = f(x + 1)
= (x + 1)²
= x² + 2x + 1
(f ∘ g)(x) = x² + 2x + 1
(g ∘ f)(x): apply f first, then g.
g(f(x)) = g(x²)
= x² + 1
(g ∘ f)(x) = x² + 1
Same f and g — different order — different answers.
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Tips
- Always read inside-out. The function next to x acts first. This is the single rule that prevents most mistakes.
- Use brackets generously when substituting. (x + 1)2 is very different from x + 12.
- Check by substituting a number. If you find an expression for (f ∘ g)(x), pick a value like x = 2 and check both methods give the same answer.
- For the domain of (f ∘ g), keep only the inputs whose g-output is allowed by f‘s domain.
- Use the GDC on Paper 2 to plot the composite directly — saves time and confirms your algebra.
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Common mistakes
- Doing the wrong function first. In (f ∘ g)(x), g goes first — not f. The rightmost (closest to x) acts first.
- Confusing ff(x) with [f(x)]2. ff(x) is f applied twice; [f(x)]2 is f(x) squared. Different operations entirely.
- Forgetting brackets when substituting. If f(x) = x2 and g(x) = 2x + 1, then f(g(x)) = (2x + 1)2, NOT 2x + 12.
- Assuming (f ∘ g) = (g ∘ f). They’re usually different. Compute both separately if asked.
- Not restricting the domain. If f can’t take certain inputs, the domain of f ∘ g may need to shrink.
- Skipping the simplification step. Always simplify your final expression — examiners expect a clean answer.
Final word: Composite functions are about order and substitution. Read inside-out, work step by step, and brackets are your best friend.
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