Mathematics Advanced • Year 11 • Module 1 • Lesson 7

Composite Functions

Build procedural fluency in evaluating composites, finding the algebraic expression, and stating the correct order.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the notation:

(f ∘ g)(x) = ____________________

In the symbol (f ∘ g), the function applied first is ____ and the function applied second is ____.

Q1.2 State the two domain conditions for (f ∘ g)(x):

Condition 1: ________________________________________________________________

Condition 2: ________________________________________________________________

Q1.3 True or false: For all functions, f(g(x)) = g(f(x)). Circle: T / F (Give a one-line reason.)

Stuck? Revisit lesson § Formula Reference and § Order Matters.

2. Worked example — finding (f ∘ g)(x) when f(x) = 3x − 2 and g(x) = x + 4

Follow each line. The reason is given on the right.

Problem. Find (f ∘ g)(x) and (g ∘ f)(x) for f(x) = 3x − 2, g(x) = x + 4.

Step 1 — Identify INNER and OUTER for (f ∘ g)(x).

INNER = g(x) = x + 4 OUTER = f( · ) = 3( · ) − 2

Reason: in (f ∘ g)(x), the function on the right is applied first.

Step 2 — Substitute the whole inner expression into the outer rule, in brackets.

f(g(x)) = 3(x + 4) − 2

Reason: brackets ensure the entire inner expression is multiplied by 3.

Step 3 — Simplify.

= 3x + 12 − 2 = 3x + 10

Step 4 — Now reverse the order for (g ∘ f)(x).

INNER = f(x) = 3x − 2 OUTER = g( · ) = ( · ) + 4

g(f(x)) = (3x − 2) + 4 = 3x + 2

Conclusion. (f ∘ g)(x) = 3x + 10 and (g ∘ f)(x) = 3x + 2. They differ — order matters.

3. Faded example — fill in the missing steps

Let f(x) = √x and g(x) = x − 5. Find (f ∘ g)(x) and its domain. 4 marks

Step 1 — Write the composite:

(f ∘ g)(x) = f(g(x)) = f( ________ ) = ____________________

Step 2 — Domain condition 1 (inner): g(x) = x − 5 is defined for ____________.

Step 3 — Domain condition 2 (outer): f requires its input to be ____________, so we need g(x) ________ 0.

x − 5 ____ 0 ⇒ x ____ ________

Step 4 — Combine both conditions: Domain = ________________ (interval notation).

Conclusion. (f ∘ g)(x) = ______________ with domain ______________.

Stuck? Revisit lesson § Worked Example 3 — Domain of a Composite Function.

4. Graduated practice

Use f and g as stated under each question. Show the substitution line (with brackets) and a simplified final expression.

Foundation — numerical evaluation (4 questions)

Q	With f(x) = 2x + 1, g(x) = x² − 3, evaluate:	Working line	Value
4.1 1	(f ∘ g)(2)
4.2 1	(g ∘ f)(2)
4.3 1	(f ∘ g)(0)
4.4 1	(g ∘ f)(−1)

Standard — algebraic expressions (6 questions)

Show the substitution step in brackets before simplifying.

4.5 f(x) = 2x + 3, g(x) = x² − 1. Find (f ∘ g)(x) in simplified form. 2 marks

4.6 f(x) = 2x + 3, g(x) = x² − 1. Find (g ∘ f)(x) in expanded form. 2 marks

4.7 f(x) = 3x − 1, g(x) = x². Find (f ∘ g)(x). 2 marks

4.8 f(x) = √x, g(x) = x − 7. Find (f ∘ g)(x) and state its domain in interval notation. 2 marks

4.9 f(x) = 1/x, g(x) = x + 2. Find (f ∘ g)(x) and state its domain. 2 marks

4.10 f(x) = x² + 3, g(x) = 2x − 1. Find (f ∘ g)(x) in expanded form. 2 marks

Extension — combine concepts (2 questions)

4.11 Let f(x) = √(x + 2) and g(x) = x − 3. Find (f ∘ g)(x), state the inequality you must solve for the domain, and write the domain in interval notation. 3 marks

4.12 Let f(x) = x + 2 and g(x) = x − 2. Find (f ∘ g)(x) and (g ∘ f)(x). Hence explain in one line why these are an example of the special case where (f ∘ g)(x) = (g ∘ f)(x). 3 marks

Stuck on 4.12? Think about what f and g do to a number when applied in sequence — does the net effect leave x unchanged?

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I evaluated g(2) first, then put the result into f.

For 4.2 I evaluated f(2) first, then put the result into g — and got a different value than in 4.1.

For 4.3 I wrote g(0) = (0)² − 3 = −3 in brackets, then f(−3) = 2(−3) + 1 = −5.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Notation

(f ∘ g)(x) = f(g(x)). First: g. Second: f.

Q1.2 — Domain conditions

(1) x must be in the domain of g. (2) g(x) must be in the domain of f. Both must hold.

Q1.3 — Order

False. Composition is not commutative; in general f(g(x)) ≠ g(f(x)). They are equal only in special cases (e.g. when f and g are inverses).

Q3 — Faded example: f(x) = √x, g(x) = x − 5

Step 1: (f ∘ g)(x) = f(x − 5) = √(x − 5).
Step 2: g(x) = x − 5 is defined for all real x.
Step 3: f requires its input to be ≥ 0, so g(x) ≥ 0 ⇒ x − 5 ≥ 0 ⇒ x ≥ 5.
Step 4: Domain = [5, ∞).
Conclusion: (f ∘ g)(x) = √(x − 5) with domain [5, ∞).

Q4.1 — (f ∘ g)(2) with f(x)=2x+1, g(x)=x²−3

g(2) = 4 − 3 = 1. f(1) = 2(1) + 1 = 3.

Q4.2 — (g ∘ f)(2)

f(2) = 2(2) + 1 = 5. g(5) = 5² − 3 = 22.

Q4.3 — (f ∘ g)(0)

g(0) = (0)² − 3 = −3. f(−3) = 2(−3) + 1 = −5.

Q4.4 — (g ∘ f)(−1)

f(−1) = 2(−1) + 1 = −1. g(−1) = (−1)² − 3 = −2.

Q4.5 — (f ∘ g)(x) with f(x)=2x+3, g(x)=x²−1

(f ∘ g)(x) = f(x² − 1) = 2(x² − 1) + 3 = 2x² − 2 + 3 = 2x² + 1.

Q4.6 — (g ∘ f)(x)

(g ∘ f)(x) = g(2x + 3) = (2x + 3)² − 1 = 4x² + 12x + 9 − 1 = 4x² + 12x + 8.

Q4.7 — (f ∘ g)(x) with f(x)=3x−1, g(x)=x²

(f ∘ g)(x) = f(x²) = 3(x²) − 1 = 3x² − 1.

Q4.8 — (f ∘ g)(x) with f(x)=√x, g(x)=x−7

(f ∘ g)(x) = √(x − 7). Need x − 7 ≥ 0 ⇒ x ≥ 7. Domain: [7, ∞).

Q4.9 — (f ∘ g)(x) with f(x)=1/x, g(x)=x+2

(f ∘ g)(x) = 1/(x + 2). Need x + 2 ≠ 0 ⇒ x ≠ −2. Domain: (−∞, −2) ∪ (−2, ∞).

Q4.10 — (f ∘ g)(x) with f(x)=x²+3, g(x)=2x−1

(f ∘ g)(x) = (2x − 1)² + 3 = 4x² − 4x + 1 + 3 = 4x² − 4x + 4.

Q4.11 — f(x)=√(x+2), g(x)=x−3

(f ∘ g)(x) = f(x − 3) = √((x − 3) + 2) = √(x − 1). Need x − 1 ≥ 0 ⇒ x ≥ 1. Domain: [1, ∞).

Q4.12 — f(x)=x+2, g(x)=x−2

(f ∘ g)(x) = f(x − 2) = (x − 2) + 2 = x. (g ∘ f)(x) = g(x + 2) = (x + 2) − 2 = x. Both composites equal x because f and g are inverses of each other — applying one then the other returns the original input. (Inverses are the prototypical case where composition does commute.)