Functions: Domain, Range and Mappings

Understanding mappings, function notation, and how to find domain and range

SPEC 2.8 • ALGEBRA AND FUNCTIONS

What is a Mapping?

Definition

A mapping is a rule that connects each element from one set (the input) to one or more elements in another set (the output).

Think of a mapping like a machine: you put something in, and something comes out based on a rule.

Input (x)

→→→

Output

Example Rule

The mapping "double the input" takes 1 → 2, 2 → 4, 3 → 6

Types of Mappings

Mappings can be classified into four types based on how inputs connect to outputs:

One-to-One

Each input gives a unique output, and each output comes from exactly one input.

Example: f(x) = 2x + 1

✓ IS a function

Many-to-One

Multiple inputs can give the same output, but each input gives only one output.

Example: f(x) = x²

✓ IS a function

One-to-Many

A single input gives multiple outputs.

Example: y² = x (gives ±√x)

✗ NOT a function

Many-to-Many

Multiple inputs give multiple outputs.

Example: x² + y² = 9

✗ NOT a function

What Makes a Function?

Key Rule

A function is a mapping where each input has EXACTLY ONE output.

This means functions can be:

One-to-one: f(x) = 3x − 2 (each x gives unique y)

Many-to-one: f(x) = x² (both x = 2 and x = −2 give y = 4)

Important

One-to-many mappings are NOT functions because a single input would give multiple outputs.

Example: y = ±√x is NOT a function because x = 4 gives both y = 2 and y = −2

Vertical Line Test

On a graph, if any vertical line crosses the curve more than once, it is NOT a function.

Domain

Definition

The domain of a function is the set of all permitted input values (x-values).

The domain tells us which values we are allowed to substitute into the function.

Common Restrictions

SituationRestrictionExampleDivision by zeroDenominator ≠ 0f(x) = 1/(x−2) → x ≠ 2Square root of negativeExpression ≥ 0f(x) = √(x−3) → x ≥ 3Logarithm of non-positiveExpression > 0f(x) = ln(x) → x > 0

Notation

Domain is written as: x ∈ ℝ, x > 0, x ≠ 2, {x : x ≥ 3}, or [3, ∞)

Range

Definition

The range of a function is the set of all possible output values (y-values or f(x)-values).

The range tells us which values the function can actually produce.

From a Graph

Look at how high and low the curve goes.

Minimum/maximum points

Asymptotes

End behaviour

Algebraically

Use techniques like:

Completing the square (quadratics)

Finding turning points

Considering limiting values

Examples

f(x) = x² has domain x ∈ ℝ and range f(x) ≥ 0

f(x) = √x has domain x ≥ 0 and range f(x) ≥ 0

f(x) = 1/x has domain x ≠ 0 and range f(x) ≠ 0

Function Notation

Functions can be written in several equivalent ways:

Notation Styles

Standard notation: f(x) = 2x + 3

Mapping notation: f : x ↦ 2x + 3

Full formal notation: f : x ↦ 2x + 3, x ∈ ℝ

Reading Function Notation

f(x) is read as "f of x" and means the output when x is the input

f(3) means "substitute x = 3 into the function"

f : x ↦ 2x + 3 is read as "f maps x to 2x + 3"

Evaluating Functions

If f(x) = x² − 4x + 1, then:

f(2) = (2)² − 4(2) + 1 = 4 − 8 + 1 = −3

f(−1) = (−1)² − 4(−1) + 1 = 1 + 4 + 1 = 6

Standard Functions

Know these common function types and their typical domains and ranges:

Function TypeExampleDomainRange

Linearf(x) = 2x + 1x ∈ ℝf(x) ∈ ℝ

Quadraticf(x) = x²x ∈ ℝf(x) ≥ 0

Square rootf(x) = √xx ≥ 0f(x) ≥ 0

Reciprocalf(x) = 1/xx ≠ 0f(x) ≠ 0

Rationalf(x) = 1/(x−a)x ≠ af(x) ≠ 0

Key Point

For quadratics, use completing the square to find the exact range by identifying the minimum or maximum value.

01Worked Example

Determine whether each of the following is a function. If it is, state whether it is one-to-one or many-to-one.

(a) f(x) = 3x − 5(b) g(x) = x² + 1(c) y² = x

a)

f(x) = 3x − 5 is a function (each x gives exactly one output)

•

It is one-to-one because different inputs give different outputs.

b)

g(x) = x² + 1 is a function (each x gives exactly one output)

•

It is many-to-one because g(2) = g(−2) = 5 (different inputs, same output).

c)

y² = x means y = ±√x, so each positive x gives TWO values of y.

•

This is NOT a function (one-to-many mapping).

Worked Example02

Find the domain of $f (x) = \frac{3}{x - 4}$

1

Identify restriction: Cannot divide by zero

2

Set denominator ≠ 0: x − 4 ≠ 0

3

Solve: x ≠ 4

Domain: x ∈ ℝ, x ≠ 4

Alternative notation: {x : x ∈ ℝ, x ≠ 4} or (−∞, 4) ∪ (4, ∞)

Worked Example03

Find the domain and range of $f (x) = x - 5$

1

Domain: Cannot take square root of negative number

2

Need: x − 5 ≥ 0

3

Solve: x ≥ 5

4

Range: Square root always gives non-negative output

5

Minimum output: when x = 5, f(5) = √0 = 0

6

As x increases, f(x) increases without bound

Domain: x ≥ 5

Range: f(x) ≥ 0

Worked Example04

Find the range of $f (x) = x^{2} - 6 x + 11$ for x ∈ ℝ

1

Complete the square to find the minimum value

2

f(x) = x² − 6x + 11

3

f(x) = (x² − 6x + 9) + 11 − 9

4

f(x) = (x − 3)² + 2

5

Since (x − 3)² ≥ 0 for all x, the minimum value is when (x − 3)² = 0

6

Minimum value of f(x) = 0 + 2 = 2, occurring at x = 3

Range: f(x) ≥ 2

05Worked Example

The graph shows the function f. State the domain and range of f.

1

Domain: Read the x-values where the graph exists

2

The graph starts at x = 1 and ends at x = 7

3

Domain: 1 ≤ x ≤ 7

4

Range: Read the y-values the graph reaches

5

Lowest point: y = 1 (at x = 1 and x = 7)

6

Highest point: y = 4 (at x = 4)

Range: 1 ≤ f(x) ≤ 4

06Worked Example

The function f is defined by:

f(x) = { x² − 4, x < 0 { 2x + 1, x ≥ 0

(a) Find f(−3) and f(4)

(b) State the range of f

(a)

Find f(−3): Since −3 < 0, use f(x) = x² − 4

•

f(−3) = (−3)² − 4 = 9 − 4 = 5

•

Find f(4): Since 4 ≥ 0, use f(x) = 2x + 1

•

f(4) = 2(4) + 1 = 9

(b)

For x < 0: f(x) = x² − 4. As x → 0⁻, f(x) → −4. As x → −∞, f(x) → ∞

•

So this piece gives f(x) > −4 (but not equal to −4)

•

For x ≥ 0: f(x) = 2x + 1. When x = 0, f(0) = 1. As x → ∞, f(x) → ∞

•

So this piece gives f(x) ≥ 1

Range: f(x) > −4 or f(x) ≥ 1

Which simplifies to: f(x) > −4

Worked Example07

The function f(x) = x² − 6x + 5 has domain 0 ≤ x ≤ 5. Find the range of f.

1

Complete the square: f(x) = x² − 6x + 5 = (x − 3)² − 9 + 5 = (x − 3)² − 4

2

The vertex is at x = 3, giving minimum f(3) = −4

3

Check: x = 3 is within the domain [0, 5] ✓

4

Find values at domain endpoints:

•

f(0) = 0² − 6(0) + 5 = 5

•

f(5) = 5² − 6(5) + 5 = 25 − 30 + 5 = 0

5

Compare all critical values: minimum = −4, f(0) = 5, f(5) = 0

6

Maximum value is 5 (at x = 0), minimum is −4 (at x = 3)

Range: −4 ≤ f(x) ≤ 5

Worked Example08

The function h(x) = x² − 8x + 20 has domain x ≥ a. Given that h(x) is a one-to-one function, find the smallest possible value of the constant a.

1

Complete the square: h(x) = x² − 8x + 20

2

h(x) = (x² − 8x + 16) + 20 − 16 = (x − 4)² + 4

3

The vertex (minimum point) is at x = 4

4

For h to be one-to-one, we need only ONE half of the parabola

5

Since domain is x ≥ a, we're taking the right half

6

We must include only x-values from the vertex onwards (x ≥ 4)

7

The smallest value of a that makes h one-to-one is a = 4

a = 4

(With domain x ≥ 4, the range would be h(x) ≥ 4)

09Worked Example

The function g is defined by g(x) = (2x + 1)/(x − 3), x ∈ ℝ, x ≠ 3.

(a) Find the value of a such that g(a) = 5

(b) State the range of g

(a)

Set g(a) = 5: (2a + 1)/(a − 3) = 5

•

Multiply both sides by (a − 3): 2a + 1 = 5(a − 3)

•

Expand: 2a + 1 = 5a − 15

•

Rearrange: 1 + 15 = 5a − 2a

•

16 = 3a, so a = 16/3

(b)

Let y = (2x + 1)/(x − 3) and rearrange for x:

•

y(x − 3) = 2x + 1 → xy − 3y = 2x + 1

•

xy − 2x = 3y + 1 → x(y − 2) = 3y + 1

•

x = (3y + 1)/(y − 2)

•

This is undefined when y = 2

(a) a = 16/3

(b) Range: g(x) ∈ ℝ, g(x) ≠ 2

Practice

Question 1: The function f(x) = x² − 4x + 7 has domain x ≥ 1. Find the range of f.

Complete the square:

f(x) = x² − 4x + 7 = (x − 2)² − 4 + 7 = (x − 2)² + 3

Vertex at x = 2, minimum value f(2) = 3

Check domain:

Domain is x ≥ 1, and vertex x = 2 is within this domain ✓

At endpoint: f(1) = 1 − 4 + 7 = 4

Since x = 2 is in the domain, minimum is f(2) = 3

As x → ∞, f(x) → ∞

Range: f(x) ≥ 3

Practice

Question 2: The function g(x) = x² + 2x − 8 has domain x ≥ k. Find the smallest value of k for which g is a one-to-one function.

Complete the square:

g(x) = x² + 2x − 8 = (x + 1)² − 1 − 8 = (x + 1)² − 9

Vertex (minimum) is at x = −1

For one-to-one:

The parabola must be restricted to one side of the vertex

Since domain is x ≥ k, we need the right half of the parabola

The smallest k that gives the entire right half is k = −1

k = −1

Practice

Question 3: The function f is defined by f(x) = (3x − 2)/(x + 1), x ≠ −1.

(a) Find the value of a such that f(a) = 4

(b) State the range of f

(a) Find a such that f(a) = 4:

(3a − 2)/(a + 1) = 4

3a − 2 = 4(a + 1)

3a − 2 = 4a + 4

−2 − 4 = 4a − 3a

−6 = a, so a = −6

(b) Range:

Let y = (3x − 2)/(x + 1) and rearrange:

y(x + 1) = 3x − 2

xy + y = 3x − 2

xy − 3x = −2 − y

x(y − 3) = −2 − y

x = (−2 − y)/(y − 3)

Undefined when y = 3

(a) a = −6

(b) Range: f(x) ∈ ℝ, f(x) ≠ 3

Summary

Key Definitions

TermDefinition

FunctionA mapping where each input gives exactly one output

DomainThe set of all permitted input values

RangeThe set of all possible output values

One-to-oneEach output comes from exactly one input

Many-to-oneSome outputs come from multiple inputs

Key Exam Skills

Find range of quadratics on restricted domains (check endpoints AND vertex)

Find smallest value of a for one-to-one functions (vertex of parabola)

Evaluate piecewise functions and find their range

Solve f(a) = k by substituting and rearranging

Find range of rational functions by rearranging for x