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Module 1 · L8 of 15 ~45 min ⚡ +50 XP in Learn · +25 to complete

Working with Functions — Synthesis

Real problems do not arrive labelled "find the inverse" or "evaluate the composite." They demand that you choose the right tool, combine several ideas, and work step by step. This lesson brings together everything you have learned about functions into integrated, exam-style problems.

Today's hook — A scientist models cooling with $T(t) = 80 - 5t$ and then converts to Fahrenheit with $F(C) = \frac{9}{5}C + 32$. How do you combine these into a single formula? What kind of function operation is this?
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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

01
Recall — your gut answer first
+5 XP warm-up

A scientist models the temperature of a cooling liquid with $T(t) = 80 - 5t$, where $T$ is in degrees Celsius and $t$ is in minutes. She then converts Celsius to Fahrenheit using $F(C) = \frac{9}{5}C + 32$. Without doing any detailed algebra, describe in words how you would find a single formula for temperature in Fahrenheit as a function of time. What kind of function operation is this?

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02
Formula reference · Synthesis
core notation

$$(f \circ g)(x) = f(g(x))$$

$(f \circ g)(x)$  = composite: inner → outer
$f^{-1}$  = inverse: swap $x$ and $y$, then solve
Domain  = denominator $\neq 0$; radicand $\geq 0$; log arg $> 0$

Key insight: Multi-step problems are just a sequence of single-step problems. Break them down, solve each part, and check your final answer.

03
What you'll master
Know

Key facts

  • How to combine evaluation, domain, inverse, and composite skills
  • The connections between algebraic and graphical representations
  • Common exam question structures that mix multiple concepts
Understand

Concepts

  • Why breaking a complex problem into smaller steps is essential
  • How domain restrictions affect inverses and composites
  • Why checking your answer prevents careless errors
Can do

Skills

  • Solve multi-step problems involving functions, inverses, and composites
  • Determine domains for combined operations
  • Sketch and interpret graphs using symmetry and reflection properties
  • Translate between algebraic rules and real-world contexts
04
Key terms
Function
A relation where each input has exactly one output.
Domain
The set of all possible input values for a function.
Range
The set of all possible output values for a function.
Inverse Function
A function that reverses the effect of the original function.
Quadratic
A polynomial of degree 2, in the form $ax^2 + bx + c$.
Discriminant
The expression $b^2 - 4ac$ that determines the nature of quadratic roots.
05
Bringing it all together
core concept · +3 XP at end

By now you have mastered the individual building blocks of functions:

  • Notation & evaluation — interpreting $f(x)$ and finding outputs
  • Domain & range — identifying where a function is defined and what values it can produce
  • Piecewise & absolute value — handling functions that change rules at boundaries
  • Odd & even — recognising symmetry algebraically and graphically
  • Inverse functions — undoing operations and reflecting in $y = x$
  • Composite functions — chaining processes together

This lesson focuses on integration. HSC exam questions rarely test these skills in isolation. Instead, they present scenarios where you must:

  1. Read and interpret a real-world or abstract problem
  2. Choose the appropriate function tool
  3. Work through multiple algebraic steps carefully
  4. Check that domain restrictions and logical conditions are satisfied
Strategy for multi-step problems: Before you write any algebra, ask yourself: "What is the final thing I need to find?" Then work backwards. If you need $(f \circ g)^{-1}(x)$, you first need $f \circ g$, then you find its inverse. If you need the domain of a composite involving a square root and a rational function, check both the inner and outer function restrictions.

Here are the most common ways these concepts connect in exam questions:

  • Inverse + domain: Finding an inverse and then stating its domain (which is the range of the original)
  • Composite + domain: Forming a composite and identifying where it is undefined
  • Context + composite: Modelling a two-stage real-world process as $(f \circ g)(x)$
  • Symmetry + sketching: Using even/odd properties to complete a graph faster
  • Verification: Checking that $f(f^{-1}(x)) = x$ or confirming a claimed property
Misconception to fix — $f^{-1}(x)$ does not mean $\frac{1}{f(x)}$. $f^{-1}(x)$ denotes the inverse function, not the reciprocal. The reciprocal would be written $[f(x)]^{-1}$ or $\frac{1}{f(x)}$.

Composite + inverse: find $(f \circ g)(x)$ first, then find the inverse of the result; Domain checklist: domain of inner $g$, then $g(x)$ must lie in domain of $f$ — combine with intersection

Pause — copy the composite-then-inverse procedure and the domain checklist (check inner domain, then restrict so $g(x)$ lies in domain of $f$) into your book.

Did you get this? True or false: $f^{-1}(x)$ means the same as $\dfrac{1}{f(x)}$.

Quick check: When finding the domain of $(f \circ g)(x)$, which conditions must you check?

06
Common connection patterns
core concept

We just saw the full procedure: form a composite, find its inverse, then track the domain through both steps. That raises a question: are there recurring question structures in exams that require this exact sequence? This card answers it → two high-frequency HSC patterns: $(f \circ g)^{-1}$ and intersecting domain inequalities.

Multi-step problems in the HSC always involve at least one of these patterns. Recognising them quickly saves time in the exam.

The question asks you to find $(f \circ g)^{-1}(x)$. Do NOT try to find $f^{-1}$ and $g^{-1}$ separately. The correct approach is:

  1. Form $h(x) = (f \circ g)(x) = f(g(x))$
  2. Find $h^{-1}(x)$ by swapping $x$ and $y$ and solving

For $(f \circ g)(x)$, the domain is the set of $x$-values satisfying both:

  • $x$ is in the domain of $g$
  • $g(x)$ is in the domain of $f$

Find each condition as an inequality, then intersect the results.

When a problem gives a real-world function $C(h)$ and asks for the inverse $C^{-1}(x)$:

  • The inverse reverses the input and output — it answers "given the output, what was the input?"
  • Always interpret the inverse in context: "$C^{-1}(250)$ = 10 hours" means $250 pays for 10 hours

Pattern 1: For $(f \circ g)^{-1}$, first form the composite, then invert the whole thing; Pattern 2: Set up two inequalities and intersect — draw a number line if needed

Pause — copy both exam patterns: Pattern 1 (form composite first, then invert the whole) and Pattern 2 (set up two domain inequalities and intersect on a number line) into your book.

Fill the blanks: drag each token to the correct blank.

range domain inner outer

The ___ of $f^{-1}$ equals the ___ of $f$. For a composite, first check the ___ function's domain, then check the ___ function's requirements.

Worked example 1 · composite, then inverse +5 XP on full reveal

Let $f(x) = 2x + 3$ and $g(x) = x - 1$. Find $(f \circ g)^{-1}(x)$.

1
$(f \circ g)(x) = f(g(x)) = f(x - 1)$
Substitute $g(x) = x - 1$ into $f$
2
$= 2(x - 1) + 3 = 2x - 2 + 3 = 2x + 1$
Expand and simplify — call this $h(x) = 2x + 1$
3
$y = 2x + 1 \Rightarrow x = 2y + 1$
Swap $x$ and $y$ to find the inverse
4
$(f \circ g)^{-1}(x) = \dfrac{x - 1}{2}$ ✓
Solve for $y$
Worked example 2 · domain of a composite with a rational function +5 XP on full reveal

Find the domain of $(f \circ g)(x)$ where $f(x) = \dfrac{1}{x}$ and $g(x) = x^2 - 4$.

1
$(f \circ g)(x) = f(x^2 - 4) = \dfrac{1}{x^2 - 4}$
Write the composite
2
$g(x) = x^2 - 4$ is defined for all real $x$
Check domain of inner function $g$ — no restrictions
3
$x^2 - 4 \neq 0 \Rightarrow (x-2)(x+2) \neq 0 \Rightarrow x \neq \pm 2$
$f$ requires its input non-zero — solve for excluded $x$-values
4
Domain: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$ ✓
Combine results — exclude $x = \pm 2$
Worked example 3 · context problem with inverse +5 XP on full reveal

A company charges $C(h) = 50 + 20h$ for $h$ hours of consulting. A client has a budget of $250. (a) Find $C^{-1}(x)$ and explain what it represents. (b) Find the maximum number of whole hours the client can afford.

1
$y = 50 + 20h \Rightarrow x = 50 + 20y \Rightarrow 20y = x - 50$
Swap variables and isolate $y$
2
$C^{-1}(x) = \dfrac{x - 50}{20}$
The inverse gives the hours purchasable for $x
3
$C^{-1}(250) = \dfrac{250 - 50}{20} = \dfrac{200}{20} = 10$ ✓
The client can afford exactly 10 hours at a $250 budget
1

Trying to do too many steps at once

In synthesis problems, students often skip intermediate working and try to jump straight to the final answer. This increases the chance of algebraic slips and makes it harder for markers to award partial credit.

✓ Fix: Show every distinct step on a new line. If a question has two parts (find the composite, then find its inverse), complete the first part before moving to the second.

2

Forgetting to check the domain of the inner function

When finding the domain of a composite, it is easy to focus only on the final simplified expression and forget that the inner function may have its own restrictions.

✓ Fix: Always state: "$x$ must be in the domain of $g$, and $g(x)$ must be in the domain of $f$." Then solve both conditions.

3

Giving the domain of the original function instead of the inverse

When asked for the domain of $f^{-1}$, students sometimes give the domain of $f$. Remember: the domain of $f^{-1}$ is the range of $f$.

✓ Fix: After finding an inverse, explicitly state its domain by considering what outputs the original function could produce.

4

Not interpreting answers in context

Context questions often ask you to explain what your answer means. A purely numerical answer without interpretation will not score full marks.

✓ Fix: End every context question with a sentence that explains what the number means in the real-world situation.

Solve each problem carefully. Show all steps and state any domain restrictions.

1

If $f(x) = 3x - 2$ and $g(x) = x + 4$, find $(f \circ g)^{-1}(x)$.

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2

Find the domain of $(f \circ g)(x)$ where $f(x) = \sqrt{x}$ and $g(x) = x - 6$. Write your answer in interval notation.

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3

A shipping company charges $C(w) = 10 + 2w$ dollars to ship a package of weight $w$ kg. Find the inverse function and explain what $C^{-1}(30)$ means in this context.

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4

Determine whether $h(x) = x^3 - 3x$ is even, odd, or neither. Show the algebraic test.

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Odd one out: Which answer is INCORRECT for the domain of $(f \circ g)(x)$ where $f(x) = \sqrt{x}$ and $g(x) = x - 6$?

Work mode · how are you completing this lesson?
1

If $f(x) = 2x + 5$ and $g(x) = x - 3$, find $(f \circ g)(x)$.

2

State the domain of $f^{-1}$ if $f$ has range $[2, \infty)$.

3

Find the domain of $(f \circ g)(x)$ where $f(x) = \sqrt{x + 1}$ and $g(x) = x^2 - 2$.

4

Test whether $p(x) = x^4 - 2x^2$ is even, odd, or neither.

5

If $f(x) = 3x - 9$, find $f^{-1}(7)$.

13
Revisit your thinking

Earlier you were asked: A scientist has $T(t) = 80 - 5t$ and $F(C) = \frac{9}{5}C + 32$. How would you find temperature in Fahrenheit as a function of time?

You would form the composite function $(F \circ T)(t) = F(T(t)) = F(80 - 5t) = \frac{9}{5}(80 - 5t) + 32$. This single formula takes time in minutes as input and directly produces temperature in Fahrenheit. The first function $T$ converts time to Celsius, and the second $F$ converts Celsius to Fahrenheit. Chaining them together — a composite function — is exactly how scientists and engineers build complex models from simpler pieces.

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01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next.

02
Short answer
ApplyBand 44 marks

Q8. Let $f(x) = 2x + 5$ and $g(x) = x - 3$. (a) Find $(f \circ g)(x)$. (b) Find $(f \circ g)^{-1}(x)$. (c) Verify that $(f \circ g)^{-1}(11) = 6$ by showing your substitution. (4 marks)

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AnalyseBand 54 marks

Q9. Consider $f(x) = \sqrt{x - 1}$ and $g(x) = \dfrac{2}{x}$. (a) Find $(f \circ g)(x)$ in simplified form. (b) Find the domain of $(f \circ g)(x)$. Show the inequalities you solve and write the final domain in interval notation. (4 marks)

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EvaluateBand 54 marks

Q10. A student is given $f(x) = x^2 + 2x$ and is asked to find $f^{-1}(x)$. They write: "This function does not have an inverse because it is a quadratic." Evaluate this claim. Is the student correct? If not, explain what additional step would allow an inverse to be found, and find that inverse with the appropriate restriction. (4 marks)

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Comprehensive answers (click to reveal)

Multiple choice — drill bank

MC answers and feedback are shown inline as you complete each question. Use the retry button to attempt a fresh set.

Activity 1 — Synthesis Drills model answers

1. $(f \circ g)(x) = f(x + 4) = 3(x + 4) - 2 = 3x + 10$. Inverse: $y = 3x + 10 \Rightarrow x = 3y + 10 \Rightarrow y = \frac{x - 10}{3}$. So $(f \circ g)^{-1}(x) = \frac{x - 10}{3}$.

2. $(f \circ g)(x) = \sqrt{x - 6}$. Domain: $[6, \infty)$.

3. $C^{-1}(x) = \frac{x - 10}{2}$. $C^{-1}(30) = \frac{20}{2} = 10$. This means a $30 shipping fee corresponds to a 10 kg package.

4. $h(-x) = (-x)^3 - 3(-x) = -x^3 + 3x = -(x^3 - 3x) = -h(x)$. Therefore $h$ is odd.

Short answer model answers

Q8 (4 marks):

(a) $(f \circ g)(x) = f(x - 3) = 2(x - 3) + 5 = 2x - 1$ [1]
(b) $y = 2x - 1 \Rightarrow x = 2y - 1 \Rightarrow y = \frac{x + 1}{2}$. So $(f \circ g)^{-1}(x) = \frac{x + 1}{2}$ [1]
(c) $(f \circ g)^{-1}(11) = \frac{11 + 1}{2} = \frac{12}{2} = 6$ ✓ [1 for correct substitution, 1 for conclusion]

Q9 (4 marks):

(a) $(f \circ g)(x) = f\!\left(\frac{2}{x}\right) = \sqrt{\frac{2}{x} - 1}$ [1]
(b) Need $x \neq 0$ (domain of $g$) and $\frac{2}{x} - 1 \geq 0$ [1].
$\frac{2 - x}{x} \geq 0$. Critical values at $x = 0$ and $x = 2$. Testing: positive on $(0, 2]$ [1]. Domain: $(0, 2]$ [1].

Q10 (4 marks): The student's claim is incomplete, not fully correct [1]. While $f(x) = x^2 + 2x$ is a quadratic and is not one-to-one over $(-\infty, \infty)$, we can restrict the domain to make it one-to-one [1]. Completing the square: $f(x) = (x + 1)^2 - 1$, vertex at $(-1, -1)$. A suitable restriction is $x \geq -1$ [1]. With this restriction, $f^{-1}(x) = \sqrt{x + 1} - 1$ with domain $x \geq -1$ [1].

01
Boss battle
earn bronze · silver · gold

Five timed questions on working with functions — synthesis. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).

Enter the arena
02
Science Jump · working with functions
arcade practice

Climb platforms, hit checkpoints, and answer synthesis questions. Quick recall, lighter than the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

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