Introduction to Extension 1 Functions
You already know what a function is, an input, a rule, an output. But Extension 1 asks a deeper question: what happens when we flip that rule around? What does it mean for two expressions to be constrained by inequality rather than equality? This module unlocks four powerful function tools, inequalities, inverses, graphical transformations, and parametric equations, that will reappear in every calculus and proof topic to come.
What is the difference between a relation and a function? Write down the definition of domain and range, and explain how you would use the vertical line test. Don't look anything up, write what you already know.
Everything in Extension 1 Functions flows from two core ideas. First: test whether a relation is a function using the vertical line test. Second: test whether a function is one-to-one using the horizontal line testbecause only one-to-one functions have inverses.
The vertical line test asks: does any vertical line cross the graph more than once? If yes, it's not a function. The horizontal line test asks: does any horizontal line cross the graph more than once? If yes, the function is many-to-one and has no inverse function over that domain.
Key facts
- The four key areas of this module: inequalities, inverse functions, graphical relationships, parametric equations
- Domain and range of common functions: $x^2$, $\sqrt{x}$, $\frac{1}{x}$, $e^x$, $\ln x$
- A function has an inverse only when it is one-to-one
Concepts
- How Extension 1 extends Advanced function concepts into inequalities and inverses
- Why the horizontal line test determines whether an inverse function exists
- How domain restrictions allow non-one-to-one functions to have inverses
Skills
- State the domain and range of standard functions
- Apply vertical and horizontal line tests to graphs
- Identify appropriate domain restrictions for inverse functions
In Mathematics Advanced, you studied functions and their transformations: translations, reflections and dilations. In Extension 1, the module Further Work with Functions goes much deeper into four interconnected areas:
- Inequalities solving linear, quadratic, rational and absolute value inequalities algebraically and graphically.
- Inverse functions finding inverses algebraically, understanding when domains must be restricted, and sketching inverse graphs.
- Graphical relationships sketching $y = |f(x)|$, $y = f(|x|)$, $y = \dfrac{1}{f(x)}$, $y = \sqrt{f(x)}$ and $y = [f(x)]^2$ from the graph of $y = f(x)$.
- Parametric equations describing curves using a parameter, eliminating the parameter, and sketching parametric curves.
All four areas are essential for the HSC Extension 1 examination and form the foundation for calculus and proof in later modules.
Four key areas: (1) Inequalities, (2) Inverse functions, (3) Graphical relationships, (4) Parametric equations; A function has an inverse function only if it is one-to-one (passes the horizontal line test)
Pause, copy the four key areas of Module 1 into your book, inequalities, inverse functions, graphical transformations, parametric equations, and the horizontal line test condition for inverse functions.
Quick check: Which of the following is NOT one of the four key areas in the Extension 1 Functions module?
We just saw that Module 1 covers four areas: inequalities, inverse functions, graphical transformations, and parametric equations. That raises a question: what prerequisite function knowledge, domains, ranges, function notation, do you need before tackling any of these four areas? This card answers it → by reviewing the essential toolkit: domain, range, the horizontal line test, and the key functions $sqrt{x}$ and $1/x$.
Before diving into new material, recall the essential function concepts from Advanced:
- A function is a relation where every input ($x$-value) has exactly one output ($y$-value).
- The domain is the set of all possible inputs.
- The range is the set of all possible outputs.
- A function is one-to-one if every output comes from exactly one input (passes the horizontal line test).
Key domains and ranges to memorise:
| Function | Domain | Range |
|---|---|---|
| $f(x) = x^2$ | All real $x$ | $y \ge 0$ |
| $f(x) = \sqrt{x}$ | $x \ge 0$ | $y \ge 0$ |
| $f(x) = \dfrac{1}{x}$ | $x \ne 0$ | $y \ne 0$ |
| $f(x) = e^x$ | All real $x$ | $y > 0$ |
| $f(x) = \ln x$ | $x > 0$ | All real $y$ |
Memory rule for domains: square roots need $\ge 0$ under the root; denominators cannot be zero; logarithms need a positive argument.
Domain: set of allowed inputs. Range: set of possible outputs.; x: domain x 0, range y 0. 1{x}: domain x 0, range y 0.
Pause, copy the domain and range of the key functions into your book: $\sqrt{x}$ has domain $x \ge 0$, range $y \ge 0$; $1/x$ has domain $x \ne 0$, range $y \ne 0$.
Did you get this? True or false: the function $f(x) = x^3$ is one-to-one over all real $x$.
Worked examples · 3 in a row, reveal as you go
State the domain and range of $f(x) = \sqrt{x - 3}$.
Show that $f(x) = x^3$ is one-to-one, but $g(x) = x^2$ is not.
The 15 lessons in this module are grouped into four areas. Describe each area and the lesson range it covers.
Fill the gap: The domain of $f(x) = \sqrt{x - 5}$ is $x \ge $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $f(x) = x^2$ over the domain $x \ge 0$ has an inverse function.
Odd one out: Which of the following is the odd one out? Select the function with a different property from the rest.
Activities · practice with the ideas
State the domain and range of $f(x) = \dfrac{1}{x - 2}$.
Is $f(x) = \ln x$ one-to-one? Explain your reasoning and state whether it has an inverse.
Explain why $f(x) = x^2$ does not have an inverse function over its natural domain. State a restricted domain for which an inverse does exist.
List the four key topic areas studied in this module. Which area do you think will be most challenging? Give a reason.
Determine whether each of these is a function and, if so, whether it is one-to-one: (a) $y = x^2 - 4$, (b) $x = y^2$, (c) $y = e^x$.
Earlier you were asked: if $f(x) = x^2$, what is $f^{-1}(9)$?
Over the natural domain (all real $x$), $f(x) = x^2$ is not one-to-one, both $x = 3$ and $x = -3$ give $f(x) = 9$. So $f^{-1}$ does not exist without a domain restriction. If we restrict to $x \ge 0$, then $f^{-1}(9) = 3$. If we restrict to $x \le 0$, then $f^{-1}(9) = -3$. The domain restriction determines which inverse you get.
Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. State the domain and range of $f(x) = \dfrac{1}{x - 2}$. (2 marks)
Q2. Explain why $f(x) = x^2$ does not have an inverse function over its natural domain, and state a restricted domain for which an inverse does exist. (2 marks)
Q3. List the four key topic areas studied in the Extension 1 Functions module, and briefly describe what is covered in each. (4 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. Domain $x \ne 2$, range $y \ne 0$. · 2. $\ln x$ is one-to-one (always increasing, horizontal line test passes); inverse is $e^x$. · 3. $x^2$ fails horizontal line test ($g(2) = g(-2) = 4$); restrict to $x \ge 0$ or $x \le 0$. · 4. (1) Inequalities, (2) Inverse functions, (3) Graphical relationships, (4) Parametric equations. · 5. (a) $y = x^2 - 4$: function, not one-to-one. (b) $x = y^2$: not a function (vertical line test fails). (c) $y = e^x$: function, one-to-one.
Q1 (2 marks): Domain: $x \ne 2$ [or $(-\infty, 2) \cup (2, \infty)$] [1 mark]. Range: $y \ne 0$ [or $(-\infty, 0) \cup (0, \infty)$] [1 mark].
Q2 (2 marks): $f(x) = x^2$ fails the horizontal line test, for example $f(2) = f(-2) = 4$, so two inputs give the same output [1 mark]. A restricted domain such as $x \ge 0$ makes $f$ one-to-one and its inverse $f^{-1}(x) = \sqrt{x}$ exists [1 mark].
Q3 (4 marks): (1) Inequalities, solving $<$, $>$, $\le$, $\ge$ statements algebraically and graphically [0.5+0.5]. (2) Inverse functions, finding $f^{-1}$, restricting domains, using $f^{-1}(f(x)) = x$ [0.5+0.5]. (3) Graphical relationships, sketching $|f(x)|$, $f(|x|)$, $1/f(x)$, $\sqrt{f(x)}$, $[f(x)]^2$ [0.5+0.5]. (4) Parametric equations, expressing $x$ and $y$ in terms of parameter $t$, eliminating the parameter [0.5+0.5].
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⚔ Enter the arenaClimb platforms by answering Extension 1 functions questions. Lighter alternative to the boss.
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