1.1 Function Language and Notation

What is a Function?

A function is a rule that maps every element in a starting set (domain) to exactly one element in a target set (range). Think of it as a machine: for every valid input, you get exactly one output.

Types of Mappings

One-to-OneEvery input gives a unique output. (e.g., $f (x) = 2 x$ ).
Function? YES. Invertible? YES.
Many-to-OneMultiple inputs can give the same output. (e.g., $f (x) = x^{2}$ since $(- 2)^{2} = 2^{2}$ ).
Function? YES. Invertible? NO.
One-to-ManyOne input produces multiple outputs. (e.g., $y = \pm \sqrt{x}$ ).
Function? NO.

Visualizing Mappings

Arrows show how inputs (x) travel to outputs (y).

The Vertical Line Test

A graph represents a function if any vertical line drawn passes through the curve at most once. If it hits twice, it's not a function.

Notation, Domain & Range

Function Notation

We can write functions in two main ways:

f(x) = 2x + 1Standard notation

f : x ↦ 2x + 1Mapping notation ("f maps x to...")

Definitions

Domain: The set of all possible input values (x-values) for which the function is defined.
Range: The set of all possible output values (y-values) produced by the function.

Common Restrictions

Function Type	Problem	Restriction (Domain)
$f (x) = \frac{1}{x - a}$	Division by zero	$x \neq a$
$f (x) = \sqrt{x - a}$	Square root of negative	$x \geq a$
$f (x) = \ln (x - a)$	Log of non-positive	$x > a$

Worked Examples

Find the domain and range of the function

f (x) = x^{2} - 4 x + 3

for

x \in R

1. Determine the shape.

This is a quadratic function opening upwards (coefficient of

x^{2}

is positive).

The domain of a standard polynomial is all real numbers.

Domain:

x \in R

(or

- \infty < x < \infty

2. Find the minimum point (vertex).

Complete the square or use

x = - b / 2 a

x = - (- 4) / 2 (1) = 2

f (2) = (2)^{2} - 4 (2) + 3 = 4 - 8 + 3 = - 1

So the vertex is at

(2, - 1)

3. State the range.

Since the graph opens upwards from a minimum y-value of -1:

Range:

f (x) \geq - 1

A function is defined by

g (x) = \sqrt{2 x - 5}

. Find the largest possible domain for

g (x)

and the corresponding range.

1. Condition for square roots.

The term inside the square root must be non-negative for real values.

2 x - 5 \geq 0

2 x \geq 5 ⟹ x \geq 2.5

Domain:

x \geq 2.5

2. Find the range.

The square root function

\sqrt{u}

always produces non-negative results (

\geq 0

When

x = 2.5

g (2.5) = \sqrt{0} = 0

x

increases,

g (x)

increases.

Range:

g (x) \geq 0

The function

h : x \mapsto \frac{3}{x - 2}

is defined for

x \in R, x \neq 2

. Find an expression for

h^{- 1} (x)

1. Set $ y = h(x) $.

y = \frac{3}{x - 2}

2. Rearrange to make x the subject.

Multiply by

(x - 2)

y (x - 2) = 3

x y - 2 y = 3

x y = 3 + 2 y

x = \frac{3 + 2 y}{y}

3. Swap x and y notation.

h^{- 1} (x) = \frac{3 + 2 x}{x}

(or

\frac{3}{x} + 2

Given

f (x) = 2 x + 1

and

g (x) = x^{2} - 3

, find the values of

x

for which

f g (x) = g f (x)

1. Find $ fg(x) $.

f g (x) = f (g (x)) = f (x^{2} - 3)

= 2 (x^{2} - 3) + 1 = 2 x^{2} - 6 + 1 = 2 x^{2} - 5

2. Find $ gf(x) $.

g f (x) = g (f (x)) = g (2 x + 1)

= (2 x + 1)^{2} - 3

= (4 x^{2} + 4 x + 1) - 3 = 4 x^{2} + 4 x - 2

3. Equate and solve.

2 x^{2} - 5 = 4 x^{2} + 4 x - 2

0 = 2 x^{2} + 4 x + 3

4. Check discriminant.

b^{2} - 4 a c = 4^{2} - 4 (2) (3) = 16 - 24 = - 8

Since the discriminant is negative, there are no real solutions.

Function

f (x) = 3 x - 5

for

x \in R

. Sketch the graph of

y = | f (x) |

and solve

| 3 x - 5 | = 4

1. Table of Values.

$x$	0	$1.67 (5 / 3)$	3
$y = \| 3 x - 5 \|$	5	0	4

2. Sketching.

Draw

y = 3 x - 5

. It crosses y-axis at -5 and x-axis at

5 / 3

Reflect the negative part (below x-axis) to positive.

The graph is a 'V' shape touching the x-axis at

x = 5 / 3

and crossing y-axis at 5.

3. Solve equation.

Case 1:

3 x - 5 = 4 ⟹ 3 x = 9 ⟹ x = 3

Case 2:

3 x - 5 = - 4 ⟹ 3 x = 1 ⟹ x = 1 / 3

Solutions:

x = 3, x = 1 / 3

State whether the function

k (x) = x^{2} + 1

for

x \in R

has an inverse. If not, suggest a restriction on the domain to make it invertible.

1. Check One-to-One.

A horizontal line test shows that

y = x^{2} + 1

(a parabola opening up) is hit twice for any

y > 1

For example,

k (1) = 2

and

k (- 1) = 2

Since it is a Many-to-One function, it does not have an inverse.

2. Suggest restriction.

To make it One-to-One, we can restrict the domain to one side of the vertex (x=0).

Suggestion: Consider domain

x \geq 0

(or

x \leq 0