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Year 12 Maths - Methods (Unit 3 & Unit 4) Functions, relations and graphs

Implied domains and types of functions

20 practice questions 0 video lessons Theory + worked examples
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Theory

The implied (maximal) domain of a function is the largest set of real \(x\)-values for which its rule gives a real number. It is found by excluding division by zero, square roots of negatives and logarithms of non-positive numbers. Functions can also be classified by symmetry as even \((f(-x)=f(x))\) or odd \((f(-x)=-f(x))\), and by whether they are increasing or decreasing, with domain and range stated in interval notation.

Unless a domain is written down, a function is taken to have its implied (maximal) domain: the set of all real numbers \(x\) for which the rule produces a real \(y\)-value. Finding it means starting with all of \(\mathbb{R}\) and removing every \(x\) that would make the rule undefined.

There are three things a rule can never do, and each one restricts the domain. A denominator cannot be zero, so for \(\dfrac{p(x)}{q(x)}\) we require \(q(x)\neq 0\). A square root (or any even root) needs a non-negative radicand, so for \(\sqrt{g(x)}\) we require \(g(x)\ge 0\). A logarithm needs a strictly positive argument, so for \(\log_a\!\big(h(x)\big)\) we require \(h(x)>0\). When several of these appear at once, the domain is the set of \(x\) satisfying all of the conditions.

Functions are also described by their type. A function is even if \(f(-x)=f(x)\), giving a graph symmetric about the \(y\)-axis; it is odd if \(f(-x)=-f(x)\), giving rotational symmetry about the origin. Separately, a function is increasing on an interval where its graph rises left-to-right (positive gradient) and decreasing where it falls (negative gradient). Domain and range are then reported in interval notation.

Key idea. Implied domain = all real \(x\) that keep the rule defined: \(q(x)\neq 0\), \(g(x)\ge 0\) under an even root, \(h(x)>0\) inside a log. Even means \(f(-x)=f(x)\); odd means \(f(-x)=-f(x)\); most functions are neither.
An even function is symmetric about the y-axisThe parabola y equals x squared with the symmetric points (-1.6, 2.56) and (1.6, 2.56) marked, showing f(-x)=f(x): the left half is the mirror image of the right half across the y-axis. x y y = x²
Even: \(f(-x)=f(x)\) — the graph is its own mirror image in the \(y\)-axis
An odd function has rotational symmetry about the originThe cubic y equals x cubed with the symmetric points (1.2, 1.728) and (-1.2, -1.728) marked, showing f(-x)=-f(x): a half-turn about the origin maps the curve onto itself. x y y = x³
Odd: \(f(-x)=-f(x)\) — a half-turn about the origin maps the graph onto itself

The three restrictions that shrink an implied domain:

\[\frac{p(x)}{q(x)}:\ q(x)\neq 0 \qquad \sqrt{g(x)}:\ g(x)\ge 0 \qquad \log_a\!\big(h(x)\big):\ h(x)>0\]
q(x)0,g(x)0,h(x)>0

Symmetry test — replace \(x\) with \(-x\):

\[\text{even}:\ f(-x)=f(x) \qquad \text{odd}:\ f(-x)=-f(x)\]
f(x)=f(x)f(x)=f(x)

Increasing and decreasing on an interval (for all \(x_1

\[\text{increasing}:\ f(x_1)f(x_2)\]
f(x1)<f(x2)f(x1)>f(x2)
Interval notation. \([\,a,b\,]\) includes both ends, \((a,b)\) excludes them; \(\infty\) always takes a round bracket. Disjoint pieces are joined with \(\cup\), e.g. \([1,3)\cup(3,\infty)\).

How to find and state an implied domain

  1. Scan for trouble spots. Look for denominators, even roots (square roots) and logarithms in the rule — these are the only things that restrict the domain.
  2. Write a condition for each. Set every denominator \(\neq 0\), every radicand \(\ge 0\), and every log argument \(>0\).
  3. Solve each condition for \(x\) as an equation or inequality.
  4. Combine. The domain is the set of \(x\) that satisfies all the conditions at once (their intersection).
  5. State it. Write the domain in interval notation, using round brackets for excluded values, and read the range from the graph or key values.
Tip. A square root includes its endpoint (radicand \(=0\) is allowed, closed bracket), but a value that makes a denominator zero is always excluded (open bracket) — that single difference decides most bracket choices.
Example 1 — a denominator restriction
State the implied domain of \(f(x)=\dfrac{1}{x-3}\).
Solution
The only trouble spot is the denominator — it cannot be zero:
\(x-3\)\(\neq\)\(0\)
\(x\)\(\neq\)\(3\)
Every other real number is allowed, so remove only \(x=3\):
\(\therefore\) domain \(=(-\infty,3)\cup(3,\infty)\), i.e. \(\mathbb{R}\setminus\{3\}\)
(,3)(3,)
Example 2 — a square-root restriction
Find the implied domain and range of \(f(x)=\sqrt{x-2}\).
Solution
A square root needs a non-negative radicand:
\(x-2\)\(\ge\)\(0\)
\(x\)\(\ge\)\(2\)
The output starts at \(0\) (when \(x=2\)) and grows without bound:
\(\therefore\) domain \(=[2,\infty)\); range \(=[0,\infty)\)
Graph of y = sqrt(x-2)The square-root curve y equals sqrt(x-2) beginning at the endpoint (2,0) and rising to the right; it exists only for x greater than or equal to 2. x y y = √(x−2) (2, 0)
[2,)
Example 3 — even or odd?
Classify \(f(x)=x^{4}-3x^{2}\) and \(g(x)=x^{3}-x\) as even, odd or neither.
Solution
\(f(x)=x^{4}-3x^{2}\)
Replace \(x\) with \(-x\) and simplify:
\(f(-x)\)\(=\)\((-x)^{4}-3(-x)^{2}\)
\(=\)\(x^{4}-3x^{2}=f(x)\)
\(\therefore\) \(f(-x)=f(x)\): \(f\) is even (symmetric about the \(y\)-axis)
\(g(x)=x^{3}-x\)
Replace \(x\) with \(-x\) and factor out \(-1\):
\(g(-x)\)\(=\)\((-x)^{3}-(-x)\)
\(=\)\(-x^{3}+x=-(x^{3}-x)\)
\(\therefore\) \(g(-x)=-g(x)\): \(g\) is odd (symmetric about the origin)
f(x)=f(x),g(x)=g(x)
Example 4 — two restrictions combined
State the implied domain of \(f(x)=\dfrac{\sqrt{x-1}}{x-3}\).
Solution
There are two trouble spots — a square root and a denominator:
\(x-1\)\(\ge\)\(0 \ \Rightarrow\ x\ge 1\)
\(x-3\)\(\neq\)\(0 \ \Rightarrow\ x\neq 3\)
Both conditions must hold, so take \(x\ge 1\) and exclude \(x=3\):
domain\(=\)\(\{x:\ x\ge 1,\ x\neq 3\}\)
\(\therefore\) domain \(=[1,3)\cup(3,\infty)\)
[1,3)(3,)

Common pitfalls

A root includes its endpoint; a denominator never does. For \(\sqrt{x-2}\) the value \(x=2\) is allowed, so the domain is \([2,\infty)\); but for \(\dfrac{1}{x-2}\) the value \(x=2\) is excluded, giving an open bracket. Mixing these up flips a bracket.
A logarithm needs a strictly positive argument. The domain of \(\log_a(x-4)\) is \((4,\infty)\) with a round bracket — \(x=4\) gives \(\log_a 0\), which is undefined, so it is never included as \([4,\infty)\).
Even and odd are special, not the default. Always test \(f(-x)\): a function such as \(x^{2}+x\) is neither even nor odd, because \(f(-x)=x^{2}-x\) equals neither \(f(x)\) nor \(-f(x)\). Do not assume "even powers make an even function" once other terms appear.

Frequently asked questions

What is the implied domain of a function?

It is the largest set of real \(x\)-values for which the rule gives a real number. Unless a domain is stated, this is assumed. You find it by starting with all real numbers and excluding every \(x\) that would make the rule undefined.

What restrictions determine an implied domain?

Three situations must be avoided: a denominator cannot equal zero, the expression under a square root (or any even root) must satisfy \(g(x)\ge 0\), and the argument of a logarithm must satisfy \(h(x)>0\). Each gives a condition you solve for \(x\).

What is an even function?

A function is even if \(f(-x)=f(x)\) for every \(x\) in its domain. Its graph is symmetric about the \(y\)-axis, so the left half is the mirror image of the right half. Examples include \(x^{2}\), \(x^{4}\) and \(\cos x\).

What is an odd function?

A function is odd if \(f(-x)=-f(x)\) for every \(x\) in its domain. Its graph has rotational symmetry about the origin, so a half-turn about the origin leaves it unchanged. Examples include \(x\), \(x^{3}\) and \(\sin x\).

What does it mean for a function to be increasing or decreasing?

A function is increasing on an interval where its output rises as \(x\) increases, so the graph slopes upward and the gradient is positive. It is decreasing where the output falls as \(x\) increases, so the graph slopes downward and the gradient is negative.

How do you state a domain or range in interval notation?

Use a square bracket to include an endpoint and a round bracket to exclude one, with \(\infty\) always taking a round bracket. Join separate pieces with \(\cup\); for example the domain of \(\dfrac{\sqrt{x-1}}{x-3}\) is \([1,3)\cup(3,\infty)\).