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.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
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?
$$(f \circ g)(x) = f(g(x))$$
Key insight: Multi-step problems are just a sequence of single-step problems. Break them down, solve each part, and check your final answer.
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
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
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
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:
- Read and interpret a real-world or abstract problem
- Choose the appropriate function tool
- Work through multiple algebraic steps carefully
- Check that domain restrictions and logical conditions are satisfied
The connections web
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
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?
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.
Pattern 1 — Composite then inverse
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:
- Form $h(x) = (f \circ g)(x) = f(g(x))$
- Find $h^{-1}(x)$ by swapping $x$ and $y$ and solving
Pattern 2 — Domain of a composite with restrictions
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.
Pattern 3 — Context with inverse
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.
The ___ of $f^{-1}$ equals the ___ of $f$. For a composite, first check the ___ function's domain, then check the ___ function's requirements.
Worked examples · reveal as you go
Let $f(x) = 2x + 3$ and $g(x) = x - 1$. Find $(f \circ g)^{-1}(x)$.
Find the domain of $(f \circ g)(x)$ where $f(x) = \dfrac{1}{x}$ and $g(x) = x^2 - 4$.
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.
Common mistakes · the 4 traps that cost marks
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.
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.
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.
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.
Activity 1 — Synthesis drills
Solve each problem carefully. Show all steps and state any domain restrictions.
If $f(x) = 3x - 2$ and $g(x) = x + 4$, find $(f \circ g)^{-1}(x)$.
Find the domain of $(f \circ g)(x)$ where $f(x) = \sqrt{x}$ and $g(x) = x - 6$. Write your answer in interval notation.
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.
Determine whether $h(x) = x^3 - 3x$ is even, odd, or neither. Show the algebraic test.
Odd one out: Which answer is INCORRECT for the domain of $(f \circ g)(x)$ where $f(x) = \sqrt{x}$ and $g(x) = x - 6$?
Quick-fire practice · 5 reps +2 XP per reveal
If $f(x) = 2x + 5$ and $g(x) = x - 3$, find $(f \circ g)(x)$.
State the domain of $f^{-1}$ if $f$ has range $[2, \infty)$.
Find the domain of $(f \circ g)(x)$ where $f(x) = \sqrt{x + 1}$ and $g(x) = x^2 - 2$.
Test whether $p(x) = x^4 - 2x^2$ is even, odd, or neither.
If $f(x) = 3x - 9$, find $f^{-1}(7)$.
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.
Pick your answer, then rate your confidence — that tells the system what to drill next.
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)
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)
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)
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].
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