International Baccalaureate · MYP 5

Functions

⌚ ~202 min ▢ 23 blocks ✓ 97 exercises ➣ Prerequisites : Coordinate Geometry, Quadratic Functions, Quadratic Equations

I Functions and Function Notation

Definition — Relation

A relation describes how two variables (quantities) are connected.

A Relation as a Set of Ordered Pairs

A relation can also be written as a set of ordered pairs $(x, y)$: each pair links a value $x$ of the first variable with a value $y$ of the second variable that it is connected to.

The table

$x$ $1$ $2$ $3$ $4$

$y$ $3$ $5$ $7$ $9$

gives the relation $\{(1, 3), (2, 5), (3, 7), (4, 9)\}$. Each ordered pair gives the coordinates of a point that can be plotted: together, these points form the graph of the relation.
The relation “is a factor of”, from the numbers $2$ and $3$ to the numbers $4$, $6$ and $9$, is the set $\{(2, 4), (2, 6), (3, 6), (3, 9)\}$. Here the same first value $2$ appears in two pairs: a relation may connect one value with several values.

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Discover

A function is like a machine that follows a specific rule. Each number put in gives exactly one number out.
Consider a machine whose rule is “multiply by 2”.

If we put in a 3, we get out a 6. If we put in a 5, we get out a 10. A table of values helps us organise these pairs:

Input	3	5	8	10
Output	6	10	16	20

To work with these rules more efficiently, mathematicians developed a special notation.
To represent this machine, we write $\textcolor{olive}{f}(\textcolor{colordef}{\text{input}}) = \textcolor{colorprop}{\text{output}}$. The brackets $($ $)$ (also called parentheses) indicate the action of the function $\textcolor{olive}{f}$ on the input.
We use function notation to name functions and their variables, replacing “$\textcolor{colordef}{\text{input}}$” by “$\textcolor{colordef}{x}$” and “$\textcolor{colorprop}{\text{output}}$” by “$\textcolor{colorprop}{f(x)}$”.
For example, if the rule is “twice the input”:

we have $\textcolor{olive}{f}(\textcolor{colordef}{x}) = \textcolor{colorprop}{2x}$.

When the input is $\textcolor{colordef}{x} = \textcolor{colordef}{1}$, we get: $$\begin{aligned} \textcolor{olive}{f}(\textcolor{colordef}{1}) &= 2 \times \textcolor{colordef}{(1)}\\ &= \textcolor{colorprop}{2} \end{aligned}$$

Definition — Function

A function, $f$, is a rule that assigns to each input value exactly one output value.

$f$ is the name of the function (the rule).
$x$ is the input variable.
$f(x)$ is the output value when the input is $x$. It is read as “$f$ of $x$”.
$f(x)$ is called the image of $x$ under $f$, and $x$ is called an antecedent (also called a preimage) of $f(x)$.

Example

The function $f$ is defined by the rule $f(x)=2x-1$. Find the value of $f(5)$.

Answer

To find $f(5)$, we substitute the input value $x=5$ into the function's rule: $$ \begin{aligned}[t] f(x) &= 2x - 1 \\ f(5) &= 2(5) - 1 \\ &= 10 - 1 \\ &= \boldsymbol{9} \end{aligned} $$

Example

Consider the following calculation program:

Choose a number.
Multiply the chosen number by $3$.
Add $1$ to the result.

Let $x$ be the number chosen initially. Determine the function $f$ that corresponds to the result obtained with this program.

Answer

Run the program on the chosen number $x$, step by step: $$\textcolor{colordef}{x} \;\xrightarrow{\;\times\, 3\;}\; 3 \times \textcolor{colordef}{x} \;\xrightarrow{\;+\, 1\;}\; 3 \times \textcolor{colordef}{x} + 1.$$ The result is always $3 \times x + 1$, so the function is $$\textcolor{colorprop}{f(x) = 3x + 1}.$$ Check with a number: for $x = 2$, the program gives $2 \to 6 \to 7$, and $f(2) = 3 \times 2 + 1 = 7$.

Skills to practice

Deciding Whether a Set of Ordered Pairs Is a Function
Calculating Images
Evaluating a Function at an Expression

II Tables of Values

Definition — Table of Values

A table of values is a table that organises the relationship between the input values ($x$) and their corresponding output values ($f(x)$) for a function.

Example

Complete the table of values for the function $f(x)=x^2$.

$x$	$-2$	$-1$	$0$	$1$	$2$
$f(x)$

Answer

We substitute each value of $x$ into the function $f(x)=x^2$:

$\begin{aligned}[t] f(-2) &= (-2)^2 \\ &= 4 \end{aligned}$
$\begin{aligned}[t] f(-1) &= (-1)^2\\ &= 1 \end{aligned}$
$\begin{aligned}[t] f(0) &= (0)^2 \\ &= 0 \end{aligned}$
$\begin{aligned}[t] f(1) &= (1)^2 \\ &= 1 \end{aligned}$
$\begin{aligned}[t] f(2) &= (2)^2 \\ &= 4 \end{aligned}$

The completed table is:

$x$	$-2$	$-1$	$0$	$1$	$2$
$f(x)$	$4$	$1$	$0$	$1$	$4$

Skills to practice

Using Tables of Values

III Graphs of Functions

While a table of values is useful for listing some input–output pairs of a function, a graph is a powerful tool for visualising how the output changes when the input changes. A graph gives us a picture of the function's behaviour.

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Definition — Graph of a Function

The graph of a function $f$ is the set of all points with coordinates $(\textcolor{colordef}{x}, \textcolor{colorprop}{f(x)})$ in a coordinate plane. The input value, $x$, is plotted on the horizontal axis (the x-axis), and the output value, $f(x)$, is plotted on the vertical axis (the y-axis). When the function is defined for all $x$ in an interval, we can connect these points to form the curve of the function.

Proposition — Vertical Line Test

A drawing in the plane is the graph of a function exactly when every vertical line crosses it at most once. Indeed, the points of a vertical line all share the same input $x$ — so two crossings would give one input two different outputs, which a function never does.

Method — Plotting a Graph from a Table

To plot the graph of a function from its table of values:

Draw a coordinate plane with a suitable scale on each axis and label the axes.
For each pair $(x, f(x))$ in the table, plot the corresponding point on the coordinate plane.
If the function is defined for all $x$ in the interval shown, connect the points with a straight line or a smooth curve.

Example

Plot the graph of the function $f(x) = x - 1$ using its table of values.

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$f(x)$	$-3$	$-2$	$-1$	$0$	$1$	$2$

Answer

We plot the points $(-2, -3)$, $(-1, -2)$, $(0, -1)$, $(1, 0)$, $(2, 1)$, and $(3, 2)$ from the table. These points lie on the same straight line, so we connect them to draw the graph of $f(x)=x-1$.

Method — Finding the Value of $f(x)$ from a Graph

To find the output $f(x)$ for a given input $x$ using a graph:

Locate the input value on the horizontal x-axis.
Move vertically from that point to the curve of the function.
Move horizontally from the intersection point to the vertical y-axis and read the corresponding value. This y-value is the output $f(x)$.

Example

Using the graph of the function $f$ below, find the value of $f(2)$.

Answer

We follow the graphical method:

Start at $x=2$ on the horizontal axis.
Move up to meet the curve.
Move horizontally to the vertical axis and read the value, which is $3$.

Therefore, $\boldsymbol{f(2) = 3}$.

Skills to practice

Reading Images on a Graph
Deciding Whether a Relation Is a Function
Plotting Graphs

IV Finding Inputs from Outputs

Taking an input ($x$) gives its output ($f(x)$). Here we work backwards: if the output is known, can the input(s) that produced it be found? This process is called finding the preimage(s), or antecedent(s), of a given value.

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Method — Finding Inputs from Outputs on a Graph

To find the preimage(s) of a value $y$ (i.e., find all $x$ such that $f(x)=y$):

Locate the output value $y$ on the vertical y-axis.
Draw a horizontal line from this value across the graph.
Find the intersection point(s) where the horizontal line crosses the function's curve.
Move vertically to the x-axis from each intersection point to read the corresponding input value(s). These are the preimages.

Example

Using the graph of the function $f$ below, find all values of $x$ such that $f(x)= 3$.

Answer

We apply the graphical method:

We locate $y=3$ on the vertical axis.
We draw a horizontal line at $y=3$.
The line intersects the curve at two points.
We move vertically from these points to the x-axis to read the values, which are $-2$ and $2$.

The preimages of $3$ are $\boldsymbol{-2}$ and $\boldsymbol{2}$.

Method — Finding Inputs from Outputs Algebraically

To find the preimage(s) of a value $y$ for a function $f(x)$:

Set the function's formula equal to the output value: $\boldsymbol{f(x) = y}$.
Solve the resulting equation for $x$.

Example

Let $f(x) = 3x + 12$. Find all values of $x$ such that $f(x)=0$.

Answer

We need to find the value of $x$ such that $f(x)=0$. We set up the equation and solve: $$\begin{aligned} f(x) &= 0 \\ 3x + 12 &= 0 \\ 3x &= -12 && \color{gray}\text{(subtract 12 from both sides)} \\ x &= \frac{-12}{3} &&\color{gray}\text{(divide both sides by 3)} \\ x &= -4 \end{aligned}$$ The preimage of $0$ is $\boldsymbol{x = -4}$.
Check: $f(-4) = 3(-4) + 12 = -12 + 12 = 0$. The answer is correct.

Skills to practice

Finding Inputs on a Graph
Finding Inputs Algebraically
Finding Inputs of Various Functions

V Domain and Range

Definition — Domain and Range

The domain of a function is the set of all input values $x$ for which the expression $f(x)$ is defined (i.e., for which the rule of the function gives a real output).
The range of a function is the set of all output values $f(x)$ actually obtained when $x$ runs through the domain.

Unless otherwise stated, we work with real-valued functions, so the domain and the range are subsets of the real numbers.
When a function is given with its domain (for example “$f$ defined on $[0, 5]$”), that set is its domain; when only the expression $f(x)$ is given, the domain is the set of all real numbers $x$ for which $f(x)$ is defined.

Method — Finding the Domain

To find the domain of a function (over the real numbers), we identify any values of $x$ for which the expression $f(x)$ would not make sense and exclude them. Two main restrictions are looked for:

Division by zero: The denominator of a fraction cannot be zero.
Square roots of negative numbers: For a real square root, the expression inside the root (the radicand) must be greater than or equal to zero.

In word problems, the context (for example, a time or a length that cannot be negative) may impose additional restrictions on the domain.

Example

Find the domain of the function $f(x)=\dfrac{1}{x-2}$.

Answer

The function is undefined when the denominator is zero. We set the denominator equal to zero and solve for $x$: $$x-2 = 0 \implies x=2.$$ So $x=2$ is not allowed.
The function is defined for all real numbers except $x=2$, that is, for all $x$ in the set $\{x \mid x \neq 2\}$.

Skills to practice

Recognising the Domain of a Function
Finding the Domain and the Range

VI Reading the Sign on a Curve

Definition — Sign of a Function

Let $f$ be a function defined on a set $D$ and $I$ an interval contained in $D$.

$f$ is non-negative on $I$ if $f(x) \geq 0$ for every $x$ in $I$: the curve is on or above the $x$-axis.
$f$ is non-positive on $I$ if $f(x) \leq 0$ for every $x$ in $I$: the curve is on or below the $x$-axis.

The sign table of $f$ summarises, with the symbols $+$ and $-$, the sign of $f(x)$ on the intervals separated by the values where $f$ vanishes ($f(x) = 0$: the curve meets the $x$-axis) or is not defined.

Method — Reading the Sign of a Function from its Curve

Mark the points where the curve meets the $x$-axis: at these values, $f$ vanishes ($f(x) = 0$).
Where the curve is above the axis, $f(x) \gt 0$; where it is below, $f(x) \lt 0$.
Summarise in a sign table: the values where $f$ vanishes in the top row, the symbols $+$, $0$ and $-$ in the bottom row.

Example

The function $f$ of the definition above is drawn on $[-3, 4]$, and vanishes at $-1$ and $2$. Construct the sign table of $f$.

Answer

The curve is above the axis before $-1$, below between $-1$ and $2$, above after $2$:

Skills to practice

Reading the Sign of a Function on a Curve
Building a Sign Table from a Curve

VII The Sign of an Affine Expression

Proposition — Sign of an Affine Expression

Let $a$ and $b$ be real numbers with $a \neq 0$. The expression $ax + b$ vanishes at $x_0 = -\dfrac{b}{a}$; it has the sign of $a$ for $x \gt -\dfrac{b}{a}$, and the opposite sign for $x \lt -\dfrac{b}{a}$.

Proof

We solve $ax + b \gt 0$, that is $ax \gt -b$.

If $a \gt 0$, dividing by $a$ keeps the direction: $x \gt -\dfrac{b}{a}$. So $ax+b$ is strictly positive exactly for $x \gt -\dfrac{b}{a}$ — the sign of $a$.
If $a \lt 0$, dividing by $a$ reverses the direction: $x \lt -\dfrac{b}{a}$. So $ax+b$ is strictly positive exactly for $x \lt -\dfrac{b}{a}$ — again on one side of $x_0$, with the sign of $a$ on the other side.

In both cases $ax + b = 0$ exactly when $x = -\dfrac{b}{a}$.

Example

Construct the sign table of $3 - 2x$ on $\R$.

Answer

Here $a = -2$ and $b = 3$: the expression vanishes at $x_0 = -\dfrac{3}{-2} = \dfrac{3}{2}$. Since $a = -2 \lt 0$, the expression is positive before $\dfrac{3}{2}$ and negative after:

Check: $3 - 2 \times 0 = 3 \gt 0$ and $3 - 2 \times 2 = -1 \lt 0$.

Skills to practice

Completing Sign Tables for Affine Expressions

VIII Sign Tables of Products

When an inequality involves a product of factors, such as $(x-2)(x-1) > 0$, we cannot simply isolate $x$. We must instead determine where the expression is positive or negative. The sign table (or sign chart) is a tool used to organise this analysis.

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Proposition — Where an Expression Can Change Sign

A product of linear factors can change sign only at a critical value — a value where one of the factors is zero. Between two consecutive critical values, every factor keeps a constant sign, so the whole expression keeps a constant sign too.

Proof

A linear factor $x - a$ is negative for $x < a$ and positive for $x > a$: it changes sign only at its own zero $a$; a general factor $ax+b = a\left(x + \frac{b}{a}\right)$ behaves the same way around its zero, the constant $a$ never changing sign. On an interval containing no critical value, no factor changes sign; a product of factors of fixed signs has a fixed sign. This is why, in the table, each row keeps a single sign on each interval between two consecutive critical values — checking a single test value per interval would give the same result.

Method — Constructing a Sign Table

To create a sign table for a factored expression:

Find the critical values: Find the values of $x$ that make each factor equal to zero. These are the points where the expression can change sign.
Create the table: Write the critical values in increasing order in the top row of the table. List each factor in a separate row on the left, with the full expression in the last row.
Determine the sign of each factor in the intervals created by the critical values. A linear factor $(x-a)$ is negative for $xa$, and equal to $0$ when $x=a$. For a general affine factor $ax+b$, use the proposition above: it has the sign of $a$ after its zero $-\frac{b}{a}$, the opposite sign before.
Determine the sign of the full expression in each interval by multiplying the signs of the factors in that column.

Example

Solve the inequality $(x-2)(x-1) \ge 0$.

Answer

Critical values: The factors are $(x-1)$ and $(x-2)$.
- $x-1=0 \implies x=1$
- $x-2=0 \implies x=2$
Construct the sign table:
Conclusion: The inequality asks for $(x-1)(x-2) \ge 0$, which means where the product is positive or zero. From the last row of the table, this occurs when $x \le 1$ or $x \ge 2$. In interval notation: $x \in (-\infty,1] \cup [2,+\infty)$.

Skills to practice

Identifying the Sign Table of an Expression
Reading a Sign Table
Building Sign Tables for Products

IX Combining Transformations

Definition — Vertical Translation by Adding a Constant

A vertical translation shifts the graph of a function up or down. The transformation is defined by:$$g(x) = f(x) + k$$ This transformation maps a point $(x, y)$ on the graph of $f$ to a new point $(x, y+k)$ on the graph of $g$.

If $k>0$, the graph is translated $\boldsymbol{k}$ units up.
If $k<0$, the graph is translated $\boldsymbol{|k|}$ units down.

Definition — Horizontal Translation

A horizontal translation shifts the graph of a function left or right. The transformation is defined by:$$g(x) = f(x-h)$$ This transformation maps a point $(x, y)$ on the graph of $f$ to a new point $(x+h, y)$ on the graph of $g$.

If $h>0$, the graph is translated $\boldsymbol{h}$ units to the right.
If $h<0$, the graph is translated $\boldsymbol{|h|}$ units to the left.

Definition — Vertical Stretch

A vertical stretch stretches or compresses the graph of a function vertically. It is defined, for a number $a \neq 0$, by:$$g(x) = a \cdot f(x)$$ This transformation maps a point $(x, y)$ to $(x, ay)$.

If $|a|>1$, the graph is stretched vertically by a factor of $|a|$.
If $0 < |a| < 1$, the graph is compressed vertically by a factor of $|a|$.
If $a < 0$, the stretch or compression is followed by a reflection in the $x$-axis ($a = -1$ gives the reflection alone).

Definition — Horizontal Stretch

A horizontal stretch stretches or compresses the graph of a function horizontally. It is defined, for a number $b \neq 0$, by:$$g(x) = f(bx)$$ This transformation maps a point $(x, y)$ to a new point $(\frac{x}{b}, y)$.

If $|b|>1$, the graph is compressed horizontally by a factor of $\frac{1}{|b|}$.
If $0 < |b| < 1$, the graph is stretched horizontally by a factor of $\frac{1}{|b|}$.
If $b < 0$, the stretch or compression is followed by a reflection in the $y$-axis ($b = -1$ gives the reflection alone).

When multiple transformations are applied to a function, the order in which they are performed is crucial. The standard form for a transformed function is:$$ g(x) = a \cdot f(b(x-c)) + d $$ To graph this function from the parent function $f(x)$, apply the transformations in the following order:

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Method — Order of Transformations

Horizontal Transformations (inside the brackets):
- Apply the horizontal stretch/compression by the factor $\frac{1}{|b|}$ ($b \neq 0$), followed by a reflection in the $y$-axis if $b < 0$.
- Apply the horizontal translation (shift) by $c$ units.
Vertical Transformations (outside the brackets):
- Apply the vertical stretch/compression by the factor $|a|$, followed by a reflection in the $x$-axis if $a < 0$.
- Apply the vertical translation (shift) by $d$ units.

Note: Always take out the coefficient $b$ as a factor of the term inside the function to correctly identify the horizontal shift $c$. For example, transform $f(2x-6)$ as $f(2(x-3))$. This shows a compression by $\frac{1}{2}$ followed by a shift of 3 units right, not 6.

Example

Describe the sequence of transformations that maps the graph of $f(x)=\sqrt{x}$ onto the graph of $g(x) = 3\sqrt{-x+2} - 4$.

Answer

First, we rewrite the function $g(x)$ in the standard form $a \cdot f(b(x-c)) + d$:$$ g(x) = 3\sqrt{-(x-2)} - 4 $$ Comparing this to $f(x)=\sqrt{x}$, we have $a=3$, $b=-1$, $c=2$, and $d=-4$. The transformations are applied as follows:

Horizontal Transformations on $f(x)=\sqrt{x}$:
- Reflection: Since $b=-1$, reflect the graph in the y-axis ($y = \sqrt{-x}$).
- Translation: Since $c=2$, translate the graph 2 units to the right ($y = \sqrt{-(x-2)}$).
Vertical Transformations on the result:
- Stretch: Since $a=3$, stretch the graph vertically by a factor of 3 ($y = 3\sqrt{-(x-2)}$).
- Translation: Since $d=-4$, translate the graph 4 units down ($y = 3\sqrt{-(x-2)} - 4$).

Skills to practice

Sketching Combined Transformations
Finding Equations from a Sequence of Transformations

X Going further

Skills to practice

Reasoning about Functions
Solving Problems with Functions
Investigating Transformations of Graphs
Quiz

\(x\)	\(1\)	\(2\)	\(3\)	\(4\)
\(y\)	\(3\)	\(5\)	\(7\)	\(9\)

\(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)
\(f(x)\)

\(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)
\(f(x)\)	\(4\)	\(1\)	\(0\)	\(1\)	\(4\)

\(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)
\(f(x)\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)