What is finding the range?

To find the range of f on a stated domain:

The function f is defined by f(x) = (x + 1)^2 - 5 for x ≥ -1. Find the range and state whether f is one-to-one. [3 marks]

SingaporeMathsSyllabus dot point

What makes a relation a function, and how do its domain and range determine its behaviour?

Define a function and its domain and range, decide whether a relation is a function or one-to-one, and find the range of a given function over a stated domain

A focused answer to the H2 Mathematics outcome on functions. The definition of a function, the vertical and horizontal line tests, one-to-one functions, and finding the range from a stated domain.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to state precisely what a function is, identify its domain and range, decide whether a given relation is a function and whether it is one-to-one, and find the range of a function over a stated domain. This underpins composite functions, inverses and all later graphing work.

The answer

What a function is

A function $\mathrm{f}$ from a set (the domain) to the real numbers assigns to each element $x$ of the domain exactly one value $\mathrm{f}(x)$ . The set of all output values is the range.

Two ideas matter from the start:

The domain is part of the definition of the function, not an afterthought. The same rule on different domains gives different functions.
The range is determined by the rule together with the domain.

The vertical line test

A relation in the $x$ - $y$ plane defines $y$ as a function of $x$ exactly when every vertical line $x = c$ meets the graph at most once. If some vertical line meets it twice, one input has two outputs and the relation is not a function. A circle and the sideways parabola $y^2 = x$ both fail this test.

One-to-one functions and the horizontal line test

A function is one-to-one (injective) if different inputs give different outputs: $\mathrm{f}(a) = \mathrm{f}(b) \implies a = b$ . Graphically, every horizontal line meets the graph at most once. A function that is strictly increasing or strictly decreasing on its whole domain is automatically one-to-one. Being one-to-one is the condition for an inverse function to exist.

Finding the range

To find the range of $\mathrm{f}$ on a stated domain:

Identify the shape (sketch it).
Find any maximum or minimum (often by completing the square for a quadratic, or by calculus).
Check the values at the endpoints of the domain.
Read off the set of achievable outputs.

Worked example

The function $\mathrm{g}$ is defined by $\mathrm{g}(x) = \dfrac{1}{x - 3}$ for $x \in \mathbb{R}$ , $x > 3$ . Find the range of $\mathrm{g}$ .

Step 1: Understand the domain restriction

For $x > 3$ the denominator $x - 3$ is positive and can be any value in $(0, \infty)$ .

Step 2: Track the outputs

As $x \to 3^+$ , $x - 3 \to 0^+$ , so $\mathrm{g}(x) \to +\infty$ . As $x \to \infty$ , $x - 3 \to \infty$ , so $\mathrm{g}(x) \to 0^+$ .

Step 3: Confirm monotonicity

On $x > 3$ the function is strictly decreasing (larger $x$ gives a larger positive denominator and hence a smaller positive output), so every value between $0$ and $\infty$ is achieved exactly once.

Step 4: State the range

The range is $\mathrm{g} > 0$ , that is $(0, \infty)$ . The value $0$ is never reached because $\dfrac{1}{x-3}$ is never zero.

Examples in context

Example 1. A restricted quadratic. The function $\mathrm{f}(x) = 9 - x^2$ for $-1 \leq x \leq 2$ has a maximum of $9$ at $x = 0$ and endpoint values $\mathrm{f}(-1) = 8$ , $\mathrm{f}(2) = 5$ . The lowest value on the domain is $5$ , so the range is $[5, 9]$ . The function is not one-to-one here because it rises then falls.

Example 2. A naturally one-to-one model. A savings balance $B(t) = 1000(1.03)^t$ for $t \geq 0$ is strictly increasing, so it is one-to-one: any balance corresponds to a unique time. Its range is $[1000, \infty)$ , beginning at the initial deposit.

Try this

Q1. State what is meant by a function and by a one-to-one function. [2 marks]

Cue. A function gives exactly one output per input; one-to-one means different inputs give different outputs (no horizontal line crosses the graph twice).

Q2. The function $\mathrm{f}$ is defined by $\mathrm{f}(x) = (x + 1)^2 - 5$ for $x \geq -1$ . Find the range and state whether $\mathrm{f}$ is one-to-one. [3 marks]

Cue. Minimum $-5$ at $x = -1$ ; increasing for $x \geq -1$ , so range $[-5, \infty)$ and one-to-one.

Q3. Explain why the relation $x^2 + y^2 = 4$ does not define $y$ as a function of $x$ . [2 marks]

Cue. A vertical line such as $x = 0$ meets the circle at $y = 2$ and $y = -2$ , two outputs for one input, failing the vertical line test.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksThe function

\mathrm{f}

is defined by

\mathrm{f}(x) = x^2 - 4x + 7

for

x \in \mathbb{R}

x \geq 2

. Find the range of

\mathrm{f}

and explain why

\mathrm{f}

is one-to-one on this domain.

Show worked answer →

Complete the square: $\mathrm{f}(x) = (x - 2)^2 + 3$ .

The vertex is at $x = 2$ , where $\mathrm{f}(2) = 3$ . For $x \geq 2$ the curve increases, so the minimum value is $3$ and the range is $\mathrm{f} \geq 3$ , that is $[3, \infty)$ .

On $x \geq 2$ the function is strictly increasing, so each output corresponds to exactly one input: any horizontal line meets the graph at most once. Hence $\mathrm{f}$ is one-to-one.

Markers reward completing the square, identifying the minimum at the domain endpoint, the correct range in interval or inequality form, and a clear one-to-one justification.

Original3 marksA relation is given by

y^2 = x

for

x \geq 0

. Determine, with reasons, whether this relation defines

y

as a function of

x

Show worked answer →

For a given $x > 0$ there are two values $y = +\sqrt{x}$ and $y = -\sqrt{x}$ .

A function assigns exactly one output to each input, so a vertical line $x = c$ (for $c > 0$ ) meets the graph twice. The relation therefore fails the vertical line test.

Hence $y^2 = x$ does not define $y$ as a function of $x$ . Restricting to $y \geq 0$ (taking $y = +\sqrt{x}$ ) would make it a function.

Markers reward identifying the two outputs, citing the vertical line test, and a correct conclusion with the restriction that fixes it.