Home / IB DP Maths 2026, 2027 & 2028 / Application and Interpretation HL / IB Mathematics AI SL Concept of a function, domain, range and graph MAI Study Notes

IB Mathematics AI SL Concept of a function, domain, range and graph MAI Study Notes - New Syllabus

IB Mathematics AI SL Concept of a function, domain, range and graph MAI Study Notes

LEARNING OBJECTIVE

  • Concept of a function, domain, range and graph.

Key Concepts: 

  • Functions

MAI HL and SL Notes – All topics

 FUNCTIONS IN GENERAL

A function from a set $X$ to a set $Y$, written as:

$f: X \rightarrow Y$

assigns to each element $x \in X$ exactly one element $y \in Y$.

This relationship is written as:

$
\begin{aligned}
f(x) &= y \\
f : x &\mapsto y
\end{aligned}
$

We say that:

“$f$ maps $x$ to $y$” or “$y$ is the image of $x$ under $f$”

For example: $X = {a, b, c}$ and $Y = {4, 5}.$ A function from X to Y can be represented

Example

Let $X = \{1, 2, 3\}$ and $Y = \{a, b, c, d\}$. A relation $f: X \rightarrow Y$ is defined as follows:

 $f(1) = a$
 $f(2) = b$
 $f(3) = d$

Is this relation a function from $X$ to $Y$? Justify your answer.

▶️Answer/Explanation

Solution:

Indeed, each element of \( X \) has a unique image in \( Y \).
We say
\( f \) maps 1 to \( a \) or \( a \) is the image of 1
2 to \( b \) or\( b \) is the image of 2
3 to \( d \) or \( d \) is the image of 3

We write

\( \begin{aligned}
f(1) &= a, \quad f(2) = b, \quad f(3) = d \\
\text{or } f: 1 &\mapsto a \quad f: 2 \mapsto b \quad f: 3 \mapsto d
\end{aligned} \) 

Example

Let $X = \{1, 2, 3\}$ and $Y = \{a, b, c, d\}$. A relation $f: X \rightarrow Y$ is defined as follows:

 $f(1) = a$
 $f(2) = b$
 $f(3) = b$

Is this relation a function from $X$ to $Y$? Justify your answer.

▶️Answer/Explanation

Solution:

 The following is a function(we do not mind if two elements of \( X \) have the same image)

Note:


(we said “each \( x \) of \( X \),” but here 3 has no image) so this is not function.


(we said “unique \( y \) of \( Y \),” but 2 has two images) for f(2) , so this is not function.

DOMAIN AND RANGE

For a function \( f: X \rightarrow Y \),

The set of all \( x \)’s involved is called DOMAIN
The set of all \( y \)’s involved (only the images) is called RANGE

Consider again the function \( f: X \rightarrow Y \) given by

Then

DOMAIN: \( x \in X = \{1, 2, 3\} \)
RANGE: \( y \in \{a, b, d\} \)

We usually denote the domain by \( D_f \) and the range by \( R_f \).

GRAPH


DOMAIN:

Projection on the \( x \)-axis, i.e. \( D_f: x \in [1, 8] \)

RANGE:

Projection on the \( y \)-axis, i.e. \( R_f: y \in [2, 6] \)

Example

Consider the function

 \( f(x) = \begin{cases} x^2, & -2 \leq x \leq 0 \\ x, & 1 \leq x \leq 5 \end{cases} \)

Sketch the graph.

Also mention domain and range.

▶️Answer/Explanation

Solution:

$D_f : x \in [-2, 0] \cup [1, 5] \quad \text{and} \quad R_f : y \in [0, 5]$

NOTE:

The graph also shows if we have a function or not

This is not a function, since \( f(3) \) for example is not unique!

Vertical line test:

Any vertical line intersects the graph at most once.

AN “AGREEMENT” FOR THE DOMAIN

Usually, a function is simply given as a formula of the form \( y = f(x) \), where \( x \) and \( y \) are real variables.

If the domain of the function is not given, we agree that

\( D_f \text{ is } \mathbb{R} \)

or \( D_f \) is the largest possible subset of \( \mathbb{R} \)

For example,

 if \( f \) is given by \( f(x) = 2x \), we assume that \( x \in \mathbb{R} \)
 if \( f \) is given by \( f(x) = \frac{2}{x} \), we assume that \( x \in \mathbb{R} – \{0\} = \mathbb{R} \)
(we may also write \( D_f: x \neq 0 \))

1. \( f(x) \) is a function with no restrictions on \( x \),

for example a polynomial [say \( f(x) = 2x^3 + 3x^2 + 1 \)], then

\( D_f = \mathbb{R} \)

2. \( f(x) = \frac{A}{B} \), then \( B \) cannot be 0, thus

\( D_f = \mathbb{R} – \{\text{roots of the equation } B = 0\} \)

3. \( f(x) = \sqrt{A} \), then \( A \geq 0 \).

\( D_f = \text{the solution set of the inequality } A \geq 0 \)

4. \( f(x) = \log A \) or \( f(x) = \ln A \), then \( A > 0 \).¹

\( D_f = \text{the solution set of the inequality } A > 0 \)

Example

Find Domain of following

a) \( f(x) = 3x – 9 \)

b) \( f(x) = \dfrac{5}{3x – 9} \)

c) \( f(x) = \sqrt{3x – 9} \)

d) \( f(x) = \ln(3x – 9) \)

e) \( f(x) = \dfrac{x + 2}{x^2 – 3x + 2} \)

f) \( f(x) = \sqrt{x – 1} + \sqrt[3]{2 – x} \)

g) \( f(x) = \dfrac{\sqrt{1 – x^2}}{x} \)

▶️Answer/Explanation

Solution:

a) $D_f : x \in \mathbb{R}$

b) $D_f : x \in \mathbb{R} \setminus \{3\} \quad$ or $\quad D_f : x \neq 3$

c) $D_f : x \in [3, +\infty) \quad$ or $\quad D_f : x \geq 3$

d) $D_f : x \in (3, +\infty) \quad$ or $\quad D_f : x > 3$

e) $D_f : x \in \mathbb{R} \setminus \{1, 2\}$

f) $D_f : x \in [1, 2] \quad$ or $\quad D_f : 1 \leq x \leq 2$

g) $D_f : x \in [-1, 0) \cup (0, 1]$

 AN INFORMAL IDEA OF THE INVERSE FUNCTION: \( f^{-1} \)

The inverse function of \( f \), that is \( g \), will be denoted by \( f^{-1} \)

\( \begin{aligned}
f(x) &= x + 10 \\
f^{-1}(x) &= x – 10
\end{aligned} \)

Mathematically

\( \text{If } f(x) = y \text{ then } f^{-1}(y) = x . \)

In fact, \( f \) and \( f^{-1} \) are inverse to each other.

The graph of \( f^{-1} \) is a reflection of \( f \) about the line \( y = x \)

 

Example

If 

\( f(x) = x^2 \), for \( x \geq 0 \), then

 \( f^{-1}(x) = \sqrt{x} \).

Discuss it true or not? 

▶️Answer/Explanation

Solution:

 \( f \) and \( f^{-1} \) intersect on the line \( y = x \).
 The points of intersection are \( (0, 0) \) and \( (1, 1) \)
(they are on the line \( y = x \))
\( f(2) = 4 \) and thus \( f^{-1}(4) = 2 \).

THE INVERSE FUNCTION IN REAL LIFE PROBLEMS

In real life problems instead of \( x \) and \( y \) we may have other parameters.

For example, a square of side \( a \) has

Perimeter \( P = 4a \)
Area \( A = a^2 \)

(it is the function \( y = 4x \))
(it is the function \( y = x^2 \))

If they give us the perimeter \( P \), then

\( P = 4a \Leftrightarrow a = \frac{P}{4} \)

Now the side is given in terms of the perimeter. This is in fact the inverse function of \( P = 4a \)

If they give us the area \( A \), then

\( A = a^2 \Leftrightarrow a = \sqrt{A} \)

Now the side is given in terms of the area. This is in fact the inverse function of \( A = a^2 \). 

Example

Let \( F \) denote the temperature in Fahrenheit degrees & \( C \) denote the temperature in Celsius degrees

The conversion from Celsius to Fahrenheit is given by

\( F = 1.8C + 32 \)

\( F(30) = 86 \) implies that \( 30^\circ \) Celsius is equal to \( 86^\circ \) Fahrenheit

▶️Answer/Explanation

Solution:

If we solve for \( C \) then

\( C = \frac{F – 32}{1.8} \)

\( C(86) = 30 \) implies that \( 86^\circ \) Fahrenheit is equal to \( 30^\circ \) Celsius

We can say that

\( \begin{aligned}
F(30) &= 86 \quad \text{(Celsius to Fahrenheit)} \\
\text{and} \\
F^{-1}(86) &= 30 \quad \text{(Fahrenheit to Celsius)}
\end{aligned} \)

Scroll to Top