In a supply chain, raw materials go to a factory, and the factory's output goes to a distribution centre. The final product depends on two connected processes — one after the other. Composite functions work exactly the same way: one function feeds its output directly into another.
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Imagine two machines. Machine A doubles its input. Machine B adds 3 to its input. If you put a number into Machine A, then take the output and feed it into Machine B, what is the overall rule? Does it matter whether Machine A or Machine B goes first?
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Wrong: The domain of a function is always all real numbers.
Right: The domain depends on the function; rational functions exclude values that make the denominator zero.
📚 Core Content
A composite function is formed when the output of one function becomes the input of another. We write this as $(f \circ g)(x)$, which means $f(g(x))$: first apply $g$, then apply $f$ to the result.
The notation can be confusing at first. Remember:
To find $(f \circ g)(x)$ algebraically, substitute the entire expression for $g(x)$ into every $x$ in $f(x)$. Use brackets to avoid errors.
The domain of $f \circ g$ has two requirements:
Both conditions must be satisfied. If either fails, the composite is undefined at that point.
In most cases, $f(g(x)) \neq g(f(x))$. Composition is not commutative. The order in which you apply the functions usually changes the result.
For example, if $f(x) = x + 1$ and $g(x) = x^2$:
These are clearly different expressions. In real-world terms, it matters whether you double a price and then add tax, or add tax and then double the price.
🧮 Worked Examples
🧪 Activities
1 $(f \circ g)(1)$
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2 $(g \circ f)(1)$
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3 $(f \circ g)(x)$ in simplified form
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4 $(g \circ f)(x)$ in simplified form
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1 $(f \circ g)(x)$ where $f(x) = \sqrt{x}$ and $g(x) = x - 7$
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2 $(f \circ g)(x)$ where $f(x) = \dfrac{1}{x}$ and $g(x) = x + 2$
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3 A manufacturing process has two stages. Stage 1 transforms raw material according to $g(t) = t + 5$, and Stage 2 processes the output according to $f(s) = \sqrt{s}$. Explain why the composite process $(f \circ g)(t)$ cannot begin with $t = -10$.
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Earlier you were asked: If Machine A doubles its input and Machine B adds 3, what is the overall rule? Does it matter which machine goes first?
If A goes first and then B, the overall rule is $B(A(x)) = 2x + 3$. If B goes first and then A, the overall rule is $A(B(x)) = 2(x + 3) = 2x + 6$. These are different results, so yes — the order matters. This is the fundamental nature of composite functions: $(f \circ g)(x)$ is not the same as $(g \circ f)(x)$ unless the functions have a special relationship (like being inverses). In supply chains, manufacturing, and even cooking, the order in which you apply processes almost always changes the final outcome.
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Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. Let $f(x) = 2x - 1$ and $g(x) = x^2 + 3$. (a) Find $(f \circ g)(x)$ in expanded form. (b) Find $(g \circ f)(x)$ in expanded form. (c) Show that $(f \circ g)(x) \neq (g \circ f)(x)$. 4 MARKS
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9. Find the domain of $(f \circ g)(x)$ where $f(x) = \sqrt{x + 2}$ and $g(x) = x - 3$. Show the inequality you solve and write your final answer in interval notation. 3 MARKS
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10. A student claims that $(f \circ g)(x)$ and $(g \circ f)(x)$ are always different. Evaluate this claim by considering the functions $f(x) = x + 2$ and $g(x) = x - 2$. Calculate both composites and explain what this example tells you about the student's claim. 3 MARKS
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1. $g(1) = 1^2 - 1 = 0$, so $(f \circ g)(1) = f(0) = 2(0) + 3 = 3$.
2. $f(1) = 2(1) + 3 = 5$, so $(g \circ f)(1) = g(5) = 5^2 - 1 = 24$.
3. $(f \circ g)(x) = f(x^2 - 1) = 2(x^2 - 1) + 3 = 2x^2 + 1$.
4. $(g \circ f)(x) = g(2x + 3) = (2x + 3)^2 - 1 = 4x^2 + 12x + 9 - 1 = 4x^2 + 12x + 8$.
1. $(f \circ g)(x) = \sqrt{x - 7}$. Domain: $[7, \infty)$ because $x - 7 \geq 0$.
2. $(f \circ g)(x) = \frac{1}{x + 2}$. Domain: $(-\infty, -2) \cup (-2, \infty)$ because $x + 2 \neq 0$.
3. $(f \circ g)(t) = \sqrt{t + 5}$. If $t = -10$, then $g(-10) = -5$, and $f(-5) = \sqrt{-5}$ is undefined in the real numbers. The composite process cannot start at $t = -10$ because the output of Stage 1 would be a negative input for Stage 2.
1. B — $(f \circ g)(x) = f(g(x)) = f(x + 3) = 2(x + 3) = 2x + 6$.
2. B — $(g \circ f)(2) = g(f(2)) = g(4) = 4 + 1 = 5$.
3. B — $(f \circ g)(x) = f(g(x))$.
4. C — $\sqrt{x - 4}$ requires $x \geq 4$.
5. A — $(f \circ g)(x) = f(x^2) = 3x^2 - 1$.
Q8 (4 marks):
(a) $(f \circ g)(x) = f(x^2 + 3) = 2(x^2 + 3) - 1 = 2x^2 + 5$ (b) $(g \circ f)(x) = g(2x - 1) = (2x - 1)^2 + 3 = 4x^2 - 4x + 1 + 3 = 4x^2 - 4x + 4$(c) $2x^2 + 5 \neq 4x^2 - 4x + 4$ for most values of $x$ [1]. Therefore $(f \circ g)(x) \neq (g \circ f)(x)$ [1].
Q9 (3 marks): $(f \circ g)(x) = f(x - 3) = \sqrt{(x - 3) + 2} = \sqrt{x - 1}$ [1]. We need $x - 1 \geq 0$ [1]. Domain: $[1, \infty)$ [1].
Q10 (3 marks): The student's claim is false [1]. $(f \circ g)(x) = f(x - 2) = (x - 2) + 2 = x$ [0.5]. $(g \circ f)(x) = g(x + 2) = (x + 2) - 2 = x$ [0.5]. In this case, both composites equal $x$ because $f$ and $g$ are inverses of each other [1]. This shows that while composition is not generally commutative, there are special cases where the two composites are the same.
Answer questions on composing and evaluating composite functions. Pool: lessons 1–7.
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