VCE Mathematical Methods

Relations and functions

A relation is a set of ordered pairs $(x, y)$. A function is a relation in which each $x$-value is paired with exactly one $y$-value. On a graph this is the vertical line test: a function is cut at most once by any vertical line. The graph of $y = x^2$ passes the test, while the graph of $x = y^2$ does not, because $x = 4$ gives both $y = 2$ and $y = -2$.

A function is one-to-one if each $y$-value also comes from exactly one $x$-value, which on a graph means every horizontal line cuts it at most once. The function $y = x^3$ is one-to-one and $y = x^2$ is not. This matters later, because only a one-to-one function has an inverse function.

Function notation

A function is specified by its rule, its domain and its codomain, and is written

The domain is the set of inputs, the codomain is the set the outputs are taken from, and the range is the set of outputs that actually occur. The range is always a subset of the codomain. For $f: \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$, the codomain is $\mathbb{R}$ but the range is $[0, \infty)$.

Notation for sets of real numbers:

A square bracket includes the end-point and a round bracket excludes it, and $\infty$ always takes a round bracket. Sets are combined with $\cup$ (union), so $\mathbb{R} \setminus \{2\}$ can also be written $(-\infty, 2) \cup (2, \infty)$.

Function notation also works for substitution. If $f(x) = 2x^2 - 3x$, then $f(3) = 18 - 9 = 9$, $f(-1) = 2 + 3 = 5$ and $f(a + 1) = 2(a + 1)^2 - 3(a + 1) = 2a^2 + a - 1$.

Maximal domain

When a rule is given without a domain, it is taken on its maximal domain (also called the natural or implied domain): every real number for which the rule gives a real value. Three situations restrict it.

• A denominator cannot be zero, so $f(x) = \dfrac{1}{x - 3}$ has maximal domain $\mathbb{R} \setminus \{3\}$.
• An even root needs a non-negative number under it, so $f(x) = \sqrt{5 - 2x}$ needs $5 - 2x \ge 0$, giving $x \le \frac{5}{2}$, and the maximal domain is $\left(-\infty, \frac{5}{2}\right]$.
• A logarithm needs a positive argument, so $f(x) = \log_e(x + 4)$ has maximal domain $(-4, \infty)$.

Where two conditions apply, both must hold. For $f(x) = \dfrac{\sqrt{x + 1}}{x - 2}$ the root needs $x \ge -1$ and the denominator needs $x \ne 2$, so the domain is $[-1, 2) \cup (2, \infty)$.

Finding the range

The range is found by reading the graph, or by working out the smallest and largest outputs the domain allows.

For $f(x) = x^2 - 4x + 7$ on $\mathbb{R}$, completing the square gives $f(x) = (x - 2)^2 + 3$, so the minimum is $3$ at $x = 2$ and the range is $[3, \infty)$. If the domain is restricted to $[0, 5]$, the minimum is still $3$ (since $2$ lies in the domain) and the largest value occurs at the end-point farthest from $x = 2$, which is $x = 5$, giving $f(5) = 12$. The range is $[3, 12]$.

For $f(x) = \dfrac{1}{x - 3}$ the output is never zero, so the range is $\mathbb{R} \setminus \{0\}$. For $f(x) = \sqrt{x}$ the range is $[0, \infty)$.

A restricted domain changes the function. The functions $f: \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$ and $g: [0, \infty) \to \mathbb{R}$, $g(x) = x^2$ have the same rule but are different functions, since $g$ is one-to-one and $f$ is not.

Ways of representing a function

A function can be given by a rule, a graph or a table. A table gives a function if no input appears twice with different outputs. A graph of real data, such as temperature against time, is read qualitatively: where the function is increasing or decreasing, where it is steepest, and the approximate position of intercepts, stationary points and points of inflection.

Specifying a function from information

The rule of a function can be found from conditions. A quadratic with $x$-intercepts $-1$ and $5$ has the form $f(x) = a(x + 1)(x - 5)$, and if its graph also passes through $(0, 10)$, then $-5a = 10$, so $a = -2$ and $f(x) = -2(x + 1)(x - 5)$. A parameter such as $a$ represents a whole family of functions sharing a feature, here all quadratics with those intercepts.

Relations that are not functions

Some relations fail the vertical line test. The circle with centre $(h, k)$ and radius $r$ has equation

For example $(x - 2)^2 + (y + 1)^2 = 9$ has centre $(2, -1)$ and radius $3$. Its domain is $[-1, 5]$ and its range is $[-4, 2]$. Solving for $y$ gives $y = -1 \pm \sqrt{9 - (x - 2)^2}$. The upper sign gives the upper semicircle, which is a function with domain $[-1, 5]$ and range $[-1, 2]$, and the lower sign gives the lower semicircle, a function with range $[-4, -1]$.

The relation $y^2 = x$ is a sideways parabola. It is not a function, but $y = \sqrt{x}$ and $y = -\sqrt{x}$ are each functions, being its upper and lower halves.