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hscscience Maths Adv · Y11
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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 composite" or "state the domain." 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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Orient and recall

Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.

Worksheets

Practise this lesson

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

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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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Formula reference · Synthesis
core notation

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

$(f \circ g)(x)$  = composite: inner → outer
$(g \circ f)(x)$  = the other order: $g(f(x))$, usually different
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.

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What you'll master
Know

Key facts

  • How to combine evaluation, domain, composite and piecewise 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 carry through a composite
  • Why checking your answer prevents careless errors
Can do

Skills

  • Solve multi-step problems involving functions, composites and piecewise rules
  • Determine domains for combined operations
  • Sketch and interpret graphs using odd and even symmetry and absolute value
  • Translate between algebraic rules and real-world contexts
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Key terms
Function
A rule giving exactly one output for each input. Like this: $f(x) = 2x$ sends 3 to 6 and to nothing else.
Domain
The inputs a rule is allowed to take. Like this: $f(x) = \sqrt{x - 1}$ has domain $x \geq 1$, because a square root needs a non-negative inside.
Range
The outputs the rule actually produces. Like this: $f(x) = x^2$ has range $y \geq 0$, because squaring never gives a negative.
Composite function
Feeding one rule’s output straight into another. Like this: if $g(x) = x + 1$ and $f(x) = x^2$, then $f(g(x)) = (x + 1)^2$.
Order of composition
Which function is on the outside, because swapping changes the answer. Like this: $f(g(x)) = (x + 1)^2$ but $g(f(x)) = x^2 + 1$.
Domain of a composite
The inner rule’s domain, narrowed so its output suits the outer rule. Like this: for $f(x) = \sqrt{x}$ and $g(x) = x - 4$, $f(g(x))$ needs $x \geq 4$.
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Bringing it all together

Work through the core explanation before applying it.

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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
  • Absolute value the piecewise definition and graphing $y = |ax + b|$
  • 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 the domain of $(f \circ g)(x)$, you first need $f \circ g$, then you check both restrictions. 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:

  • Piecewise + domain: Reading which rule applies on which part of the domain, then testing whether the pieces join
  • 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 whether $f(g(x))$ and $g(f(x))$ really differ, or confirming a claimed property
Misconception to fix, $f(g(x))$ does not mean $f(x) \times g(x)$. A composite substitutes one function into the other; a product multiplies them. With $f(x) = x^2$ and $g(x) = x + 1$, the composite is $(x+1)^2$ but the product is $x^2(x+1)$.

Composite order: $f(g(x))$ substitutes $g$ into $f$, and $g(f(x))$ is usually different; Domain checklist: domain of inner $g$, then $g(x)$ must lie in domain of $f$, combine with intersection

Pause, copy the composite order rule 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: $f(g(x))$ always gives the same result as $g(f(x))$.

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

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Common connection patterns
core concept

We just saw the full procedure: form a composite in the stated order, 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: composition in both orders, 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 for both $(f \circ g)(x)$ and $(g \circ f)(x)$. Do NOT assume they are equal, because composition is not commutative. The correct approach is:

  1. Form $(f \circ g)(x) = f(g(x))$, substituting $g$ into $f$
  2. Form $(g \circ f)(x) = g(f(x))$ separately, substituting $f$ into $g$

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 chains two real-world rules, for example cost per item $C(n)$ and items per hour $n(t)$:

  • The composite $C(n(t))$ answers "what is the cost after $t$ hours?", feeding the output of one rule into the next
  • Always state the units at each stage: $t$ in hours gives $n$ in items, which gives $C$ in dollars

Pattern 1: $f(g(x))$ and $g(f(x))$ are different, substitute in the stated order every time; Pattern 2: Set up two inequalities and intersect, draw a number line if needed

Pause, copy both exam patterns: Pattern 1 (substitute in the stated order, the two composites differ) 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 the inner function must sit inside the ___ of the outer one. For a composite, first check the ___ function's domain, then check the ___ function's requirements.

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Work examples end to end

Follow the reasoning through complete worked solutions.

Worked example 1 · composition in both orders +5 XP on full reveal

Let $f(x) = 2x + 3$ and $g(x) = x - 1$. Find $(f \circ g)(x)$ and $(g \circ f)(x)$, and state whether they are equal.

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$(f \circ g)(x) = f(g(x)) = f(x - 1)$
Substitute $g(x) = x - 1$ into $f$
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$= 2(x - 1) + 3 = 2x - 2 + 3 = 2x + 1$
Expand and simplify: $(f \circ g)(x) = 2x + 1$
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$(g \circ f)(x) = g(f(x)) = g(2x + 3) = (2x + 3) - 1$
Now substitute in the other order, $f$ into $g$
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$= 2x + 2$, and $2x + 1 \neq 2x + 2$ ✓
The two composites differ, so order matters
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 $g(x) = x^2 - 4$.

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

A factory makes $n(t) = 12t$ items in $t$ hours, and producing $n$ items costs $C(n) = 40 + 3n$ dollars. (a) Find $C(n(t))$ and explain what it represents. (b) Find the cost after 5 hours.

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$C(n(t)) = C(12t) = 40 + 3(12t)$
Substitute the inner rule $n(t)$ into the cost rule $C$
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$C(n(t)) = 40 + 36t$
The composite gives the cost directly in terms of hours
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$40 + 36(5) = 40 + 180 = 220$ ✓
After 5 hours the cost is $\$220$, and checking directly, $n(5) = 60$ items at $C(60) = 220$
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Dodge the traps, then apply

Meet the mistakes that cost marks, then do it yourself.

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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 state its domain), complete the first part before moving to the second.

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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.

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Composing in the wrong order

$f(g(x))$ means substitute $g$ into $f$. Writing $g(f(x))$ instead changes the answer: with $f(x) = x^2$ and $g(x) = x + 1$, $f(g(x)) = (x+1)^2$ but $g(f(x)) = x^2 + 1$.

✓ Fix: Read which function is on the outside, then substitute the other one into it.

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.

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

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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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A courier charges $C(w) = 10 + 2w$ dollars for a parcel of weight $w$ kg, and packing gives $w(n) = 0.4n$ kg for $n$ items. Find $C(w(n))$ and explain what $C(w(25))$ means in this context.

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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 $g(x) = x - 6$?

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Drill it, then lock it in

Run the quick drill and copy the summary into your book.

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

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State the domain of $f(g(x))$ if $g(x) = x - 6$ and $f(x) = \sqrt{x}$.

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

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

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If $f(x) = 3x - 9$ and $g(x) = x + 2$, find $f(g(4))$.

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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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Multiple choice

Answer the drill bank and rate your confidence.

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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.

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Short answer

Write full responses, then check them against the model answers.

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Short answer
ApplyBand 44 marks

Q8. Let $f(x) = 2x + 5$ and $g(x) = x - 3$. (a) Find $(f \circ g)(x)$. (b) Find $(g \circ f)(x)$. (c) Show that $(f \circ g)(3) \neq (g \circ f)(3)$ by substituting into each. (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 $f(x) = x^2$ and $g(x) = x + 3$ and writes: "$f(g(x))$ and $g(f(x))$ must be the same, because both involve squaring and adding 3." Evaluate this claim. Is the student correct? If not, find both composites and explain in one sentence why the order changes the result. (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. $(f \circ g)(x) = f(x + 4) = 3(x + 4) - 2 = 3x + 10$. $(g \circ f)(x) = g(3x - 2) = (3x - 2) + 4 = 3x + 2$. The two differ, so order matters.

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

3. $C(w(n)) = 10 + 2(0.4n) = 10 + 0.8n$. $C(w(25)) = 10 + 0.8(25) = 30$. This means 25 items cost $\$30$ to send.

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) $(g \circ f)(x) = g(2x + 5) = (2x + 5) - 3 = 2x + 2$ [1]
(c) At $x = 3$: $(f \circ g)(3) = 2(3) - 1 = 5$ and $(g \circ f)(3) = 2(3) + 2 = 8$, so they differ [1 for both substitutions, 1 for the 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 domposites are not equal [1]. $f(g(x)) = (x + 3)^2 = x^2 + 6x + 9$ [1]. $g(f(x)) = x^2 + 3$ [1]. The order matters because in $f(g(x))$ the addition happens before the squaring, so the $+3$ is squared too [1].