Mathematics Advanced • Year 11 • Module 1 • Lesson 8

Working with Functions — Synthesis

Build fluency in two-step problems that combine composition with inverses, domains, and symmetry tests.

Build · Skill Drill

1. Quick recall — the toolkit

Answer each question in the space provided. 1 mark each

Q1.1 Complete the domain reminders:

For y = √(expression): need expression ________ 0.

For y = 1/(expression): need expression ____ 0.

For y = ln(expression): need expression ____ 0.

Q1.2 Three notation traps. True (T) or False (F)?

(a) f⁻¹(x) means 1/f(x). T / F

(b) (f ∘ g)⁻¹(x) and (f⁻¹ ∘ g⁻¹)(x) are always equal. T / F

Q1.3 Write the two-step strategy for finding (f ∘ g)⁻¹(x):

Step 1: ________________________________________________________________

Step 2: ________________________________________________________________

Stuck? Revisit lesson § Misconceptions and § Copy Into Your Books.

2. Worked example — (f ∘ g)⁻¹(x) for f(x) = 2x + 3, g(x) = x − 1

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

Problem. Find (f ∘ g)⁻¹(x) where f(x) = 2x + 3 and g(x) = x − 1.

Step 1 — Form the composite (f ∘ g)(x).

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

Reason: substitute g(x) into every x in f(x); use brackets to avoid sign errors.

Step 2 — Find the inverse of h(x) = 2x + 1 by swap-and-solve.

y = 2x + 1 ⇒ swap: x = 2y + 1 ⇒ solve: y = (x − 1)/2

Reason: the inverse swaps the roles of input and output.

Step 3 — Write the final answer in inverse-function notation.

(f ∘ g)⁻¹(x) = (x − 1)/2

Step 4 — Verification check.

(f ∘ g)((x − 1)/2) = 2 · ((x − 1)/2) + 1 = (x − 1) + 1 = x ✓

Conclusion. (f ∘ g)⁻¹(x) = (x − 1)/2.

3. Faded example — domain of a composite with a rational outer function

Let f(x) = 1/x and g(x) = x² − 4. 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) = x² − 4 is defined for ____________.

Step 3 — Domain condition 2 (outer): f(t) = 1/t requires t ________ 0, so we need x² − 4 ____ 0.

x² − 4 ____ 0 ⇒ (x − ____)(x + ____) ≠ 0 ⇒ x ≠ ____ and x ≠ ____.

Step 4 — Combine: Domain (in interval notation) = ________________________.

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

Stuck? Revisit lesson § Worked Example 2 — Domain of a Composite with a Rational Function.

4. Graduated practice — multi-step toolkit

Show all working. Two questions ask for verification — keep the check line short and clear.

Foundation — single-skill recall (4 questions)

Q	Task	Working / answer
4.1 1	State the domain of f(x) = √(x − 4) in interval notation.
4.2 1	State the domain of g(x) = 1/(x + 3) in interval notation.
4.3 1	Find the inverse of f(x) = 4x − 8.
4.4 1	Classify h(x) = x³ − 3x as even, odd, or neither (show the test in one line).

Standard — two-skill synthesis (6 questions)

Show the composite step, then the inverse / domain / verification step on a new line.

4.5 f(x) = 3x − 2, g(x) = x + 4. Find (f ∘ g)⁻¹(x). 2 marks

4.6 f(x) = √x, g(x) = x − 6. Find (f ∘ g)(x) and its domain in interval notation. 2 marks

4.7 f(x) = 2x + 5, g(x) = x − 3. Find (f ∘ g)(x), then (f ∘ g)⁻¹(x). 2 marks

4.8 C(w) = 10 + 2w (dollars to ship a parcel of w kg). Find C⁻¹(x) and interpret C⁻¹(30) in one sentence. 2 marks

4.9 Verify the identity f(f⁻¹(x)) = x for f(x) = 5x + 1 by computing f(f⁻¹(x)) directly. 2 marks

4.10 f(x) = √(x − 1), g(x) = 2/x. Find (f ∘ g)(x) (in surd form). State the two domain conditions separately. 2 marks

Extension — combine three skills (2 questions)

4.11 p(x) = x² + 4, with the restriction x ≥ 0. (a) Find p⁻¹(x). (b) State the domain of p⁻¹ in interval notation, and explain in one sentence why the restriction x ≥ 0 was needed. 3 marks

4.12 A scientist has T(t) = 80 − 5t (°C as a function of time t in minutes) and F(C) = (9/5)C + 32 (Fahrenheit as a function of Celsius). (a) Find a single formula for temperature in °F as a function of t, simplified. (b) Use your formula to find the temperature in °F at t = 4 minutes. 3 marks

Stuck on 4.12? It's a composite — (F ∘ T)(t) = F(T(t)).

5. Self-check the easy 3

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

For 4.1 I gave the domain as [4, ∞) (square bracket on the left because √0 is defined).

For 4.2 I wrote the domain as (−∞, −3) ∪ (−3, ∞), excluding only x = −3.

For 4.3 I used swap-and-solve: y = 4x − 8 → x = 4y − 8 → y = (x + 8)/4.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Domain reminders

√: expression ≥ 0. 1/: expression ≠ 0. ln: expression > 0.

Q1.2 — Notation traps

(a) F — f⁻¹ is the inverse function; the reciprocal would be [f(x)]⁻¹ or 1/f(x). (b) F — in general (f ∘ g)⁻¹ = g⁻¹ ∘ f⁻¹ (note the order is reversed); they are not equal in general. (c) T.

Q1.3 — Two-step strategy

Step 1: Form the composite (f ∘ g)(x) and simplify. Step 2: Find the inverse of the result by swap-and-solve.

Q3 — Faded example: f(x) = 1/x, g(x) = x² − 4

Step 1: (f ∘ g)(x) = f(x² − 4) = 1/(x² − 4).
Step 2: g(x) = x² − 4 is defined for all real x.
Step 3: f requires its input to be ≠ 0, so x² − 4 ≠ 0 ⇒ (x − 2)(x + 2) ≠ 0 ⇒ x ≠ 2 and x ≠ −2.
Step 4: Domain = (−∞, −2) ∪ (−2, 2) ∪ (2, ∞).
Conclusion: (f ∘ g)(x) = 1/(x² − 4) with domain (−∞, −2) ∪ (−2, 2) ∪ (2, ∞).

Q4.1 — Domain of √(x − 4)

Need x − 4 ≥ 0 ⇒ x ≥ 4. Domain: [4, ∞).

Q4.2 — Domain of 1/(x + 3)

Need x + 3 ≠ 0 ⇒ x ≠ −3. Domain: (−∞, −3) ∪ (−3, ∞).

Q4.3 — Inverse of f(x) = 4x − 8

y = 4x − 8 ⇒ swap: x = 4y − 8 ⇒ 4y = x + 8 ⇒ f⁻¹(x) = (x + 8)/4.

Q4.4 — Classify h(x) = x³ − 3x

h(−x) = (−x)³ − 3(−x) = −x³ + 3x = −(x³ − 3x) = −h(x). Odd.

Q4.5 — (f ∘ g)⁻¹(x) for f(x) = 3x − 2, g(x) = x + 4

(f ∘ g)(x) = 3(x + 4) − 2 = 3x + 10. Inverse: y = 3x + 10 → x = 3y + 10 → y = (x − 10)/3. (f ∘ g)⁻¹(x) = (x − 10)/3.

Q4.6 — (f ∘ g)(x) for f(x) = √x, g(x) = x − 6

(f ∘ g)(x) = √(x − 6). Need x − 6 ≥ 0 ⇒ x ≥ 6. Domain: [6, ∞).

Q4.7 — Composite then inverse

(f ∘ g)(x) = 2(x − 3) + 5 = 2x − 1. Inverse: y = 2x − 1 → x = 2y − 1 → (f ∘ g)⁻¹(x) = (x + 1)/2.

Q4.8 — C(w) = 10 + 2w

y = 10 + 2w → swap: x = 10 + 2y → C⁻¹(x) = (x − 10)/2. C⁻¹(30) = (30 − 10)/2 = 10, meaning a $30 fee corresponds to a 10 kg parcel.

Q4.9 — Verify f(f⁻¹(x)) = x for f(x) = 5x + 1

f⁻¹(x) = (x − 1)/5. f(f⁻¹(x)) = 5 · ((x − 1)/5) + 1 = (x − 1) + 1 = x. ✓

Q4.10 — f(x) = √(x − 1), g(x) = 2/x

(f ∘ g)(x) = √((2/x) − 1) = √((2 − x)/x). Domain conditions: (i) x ≠ 0 (from g); (ii) (2/x) − 1 ≥ 0 (from f). (Combined domain — solved fully in W2/W3 — is (0, 2].)

Q4.11 — p(x) = x² + 4, x ≥ 0

(a) y = x² + 4 → swap: x = y² + 4 → y² = x − 4 → y = √(x − 4) (positive root, because original was restricted to x ≥ 0 so the inverse outputs are ≥ 0). p⁻¹(x) = √(x − 4).
(b) Domain of p⁻¹ = range of p = [4, ∞). The restriction x ≥ 0 on p was needed because without it, p is not one-to-one (p(2) = p(−2) = 8), so no inverse function exists on the full real line.

Q4.12 — (F ∘ T)(t) for T(t) = 80 − 5t, F(C) = (9/5)C + 32

(a) (F ∘ T)(t) = F(80 − 5t) = (9/5)(80 − 5t) + 32 = (9/5)(80) − (9/5)(5t) + 32 = 144 − 9t + 32 = 176 − 9t (°F).
(b) At t = 4: 176 − 9(4) = 176 − 36 = 140 °F.