International Baccalaureate · DP 1: Applications and Interpretation SL

Functions

⌚ ~114 min ▢ 12 blocks ✓ 48 exercises ➣ Prerequisites : Sets and Venn Diagrams, Equations

I Functions and Their Graphs

I.1 What Is a Function?

Recappreviously studied — click to reread

A function is like a machine that follows a specific rule. For every number you put in, you get 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 organize 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 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 "$\textcolor{colorprop}{f(x)}$".
A function maps every element from a set of inputs, the domain $X$, to an element in a set of possible outputs, the codomain $Y$. This is written as $f: X \to Y$. The specific rule that maps an individual element $x$ from the domain to an element $f(x)$ in the codomain is written as $x \mapsto f(x)$.
In general we write $$\Function{\textcolor{olive}{f}}{\textcolor{colordef}{X}}{\textcolor{colorprop}{Y}}{\textcolor{colordef}{x}}{\textcolor{colorprop}{f(x)}}$$ For example, if the rule is "twice the input" and we are mapping real numbers to real numbers, $f: \mathbb{R} \to \mathbb{R}$:

we have $\textcolor{olive}{f}(\textcolor{colordef}{x}) = \textcolor{colorprop}{2x}$, which is written in full as:$$\Function{\textcolor{olive}{f}}{\textcolor{colordef}{\mathbb{R}}}{\textcolor{colorprop}{\mathbb{R}}}{\textcolor{colordef}{x}}{\textcolor{colorprop}{2x}}$$ 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 $f: X \to Y$ is a rule that assigns to each element $x$ in a set $X$ exactly one element $f(x)$ in a set $Y$.

The set $X$ of all possible inputs is called the domain of $f$.
The set $Y$ is called the codomain of $f$.

We write $f:x \mapsto f(x)$ to indicate the rule that maps an element $x$ to its corresponding image $f(x)$.$$\Function{\textcolor{olive}{f}}{\textcolor{colordef}{X}}{\textcolor{colorprop}{Y}}{\textcolor{colordef}{x}}{\textcolor{colorprop}{f(x)}}$$

$f$ is the name of the function.
$x$ is the input variable, an element from the domain.
$f(x)$ is the output value in the codomain when the input is $x$. It is read as "$f$ of $x$".
$f(x)$ is the image of $x$ under $f$.
$x$ is a preimage of $y=f(x)$.

Example

Let the function $\Function{f}{\mathbb{R}}{\mathbb{R}}{x}{2x-1}$. Find the image of $5$ under $f$.

Answer

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

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 Preimages

I.2 Domain and Range

Recappreviously studied — click to reread

Definition — Range

The range of a function $f:X\to Y$ is the set of all actual output values produced by the function. It is the set of all images of the elements in the domain.$$ \text{Range} = f(X)=\{f(x):x\in X\} $$ The range is a subset of the codomain ($f(X) \subseteq Y$). While the codomain is the set of potential outputs, the range is the set of actual outputs.

Example

For the function $\Function{f}{\mathbb{R}}{\mathbb{R}}{x}{x^2}$:

The domain is $\mathbb{R}$ (all real numbers).
The codomain is also $\mathbb{R}$.
However, since $x^2\geq 0$, the range is $[0, \infty)$, which is a subset of the codomain.

Definition — Natural Domain

When a function is given by a formula and the domain is not explicitly specified, the natural domain is the largest set of real numbers for which the formula yields a defined real number.

Note

When a function's rule is given without explicitly specifying the domain and codomain, we assume the natural domain as the domain and the set of all real numbers, $\mathbb{R}$, as the codomain.
For example, when we write $f:x\mapsto \sqrt{x}$, we implicitly assume the domain is the set of non-negative real numbers, $[0, \infty)$, and the codomain is $\mathbb{R}$. Thus, it is shorthand for $\Function{f}{[0, \infty)}{\mathbb{R}}{x}{\sqrt{x}}$.
Sometimes, we simply refer to "the function $\sqrt{x}$". This is also a shorthand for the function defined on its natural domain.

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Method — Finding the Natural Domain

To find the natural domain of a function, we assume the domain is all real numbers ($\mathbb{R}$) and then exclude any values of $x$ that would lead to an undefined mathematical operation. The restrictions looked for are caused by:

Rational Functions: The denominator of a fraction cannot be zero. We solve denominator = 0 to find values to exclude.
Even Roots: The expression inside an even root (like a square root, $\sqrt{\cdot}$, or fourth root, $\sqrt{\cdot}$) must be non-negative ($\ge 0$).
Logarithms: The argument of a logarithm must be strictly positive ($> 0$).

Example

Find the natural domain of the function $f:x\mapsto\frac{1}{x-2}$.

Answer

The function involves division. Division is undefined when the denominator is zero. Therefore, we must exclude any value of $x$ that makes the denominator $x-2$ equal to zero.
Set the denominator to zero and solve for $x$:$$x-2 = 0 \Leftrightarrow x=2$$ The natural domain is the set of all real numbers except $2$. Using set notation, we write:$$\text{Domain} = \{x \in \mathbb{R} \mid x \neq 2\} \quad \text{or} \quad \mathbb{R} \setminus \{2\}$$ In interval notation, this is $(-\infty, 2) \cup (2, \infty)$.

Skills to practice

Recognising a Natural Domain
Finding the Natural Domain and the Range

I.3 Graphs of Functions

Recappreviously studied — click to reread

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 $(x, f(x))$ plotted on a coordinate plane. The input, $x$, is plotted on the horizontal axis (the x-axis), and the output, $f(x)$, is plotted on the vertical axis (the y-axis). When we connect these points, we form the curve of the function.$$\text{Graph of } f = \{(x,f(x)): x \in X\}$$

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.

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

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 $\boldsymbol{2}$.

Skills to practice

Finding Inputs from Outputs on a Graph

I.4 The Sign of a Function

Recappreviously studied — click to reread

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

The Sign of a Function

I.5 Graphs with Technology

Method — Reading the Key Features of a Graph on the Screen

To study the graph of a function $f$ with a graphing calculator:

Enter $f(x)$ as the first function ($Y_1$) of the function editor.
Choose the window: the least and greatest values of $x$ and of $y$ shown on the screen. Start from the domain the question gives, or from a table of values of $f$, and adjust until every feature asked for is visible. A window that is too narrow hides an intercept or a turning point; a window that is too wide flattens the curve.
Read each feature with the analysis tools of the graph menu, never by eye:
- the $y$-intercept: the value of $f$ at $x = 0$;
- each $x$-intercept: the tool zero (or root);
- a local maximum or minimum: the tool maximum or minimum, which gives both coordinates of the turning point.
For the tools zero, maximum and minimum, bound the search: give a left bound and a right bound on either side of the feature, or a first guess close to it, so that the calculator finds that feature and not a neighbouring one.

The mock-up shows the maximum of $f(x) = -x^2 + x + 5$, in the window $-3 \le x \le 3$, $-5 \le y \le 10$.

Example

Let $f(x) = x^3 - 5x + 3$.

Choose a window that shows the intercepts and the turning points of the graph of $f$.
Find the $y$-intercept and the $x$-intercepts of the graph.
Find the local maximum point and the local minimum point.

Answer

A table of values of $f(x) = x^3 - 5x + 3$ at the integers guides the choice:

$x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$

$f(x)$ $-9$ $5$ $7$ $3$ $-1$ $1$ $15$

The sign of $f(x)$ changes between $-3$ and $-2$, between $0$ and $1$, and between $1$ and $2$: three $x$-intercepts, all between $-3$ and $3$. Since $f(-1) = 7$ is greater than both $f(-2) = 5$ and $f(0) = 3$, the curve has a highest point strictly between $-2$ and $0$: a local maximum lies there. Likewise $f(1) = -1$ is less than both $f(0) = 3$ and $f(2) = 1$: a local minimum lies between $0$ and $2$. The window $-3 \le x \le 3$, $-5 \le y \le 10$ shows them all; the two ends of the curve leave the screen, where there is nothing to read.
The $y$-intercept is $f(0) = 3$. The zero tool, bounded between $-3$ and $-2$, then between $0$ and $1$, then between $1$ and $2$, displays $-2.4908636$, $0.6566204$ and $1.8342432$: the $x$-intercepts are $x \approx -2.49$, $x \approx 0.657$ and $x \approx 1.83$. An equation of degree $3$ has at most three real roots, so none is missing.
The maximum tool, bounded between $-2$ and $0$, displays $X = -1.2909944$ and $Y = 7.3033148$; the minimum tool, bounded between $0$ and $2$, displays $X = 1.2909944$ and $Y = -1.3033148$. The local maximum point is $(-1.29, 7.30)$ and the local minimum point is $(1.29, -1.30)$.

Example

Let $g(x) = \dfrac{2x - 1}{x + 1}$, for $x \neq -1$. Graph $g$ with technology. Give the intercepts of the graph and the equations of the asymptotes that the screen suggests, then confirm each asymptote with the formula.

Answer

The graph of $g(x) = \dfrac{2x - 1}{x + 1}$, in the window $-6 \le x \le 4$, $-4 \le y \le 8$, shows two branches, one on each side of the vertical line $x = -1$; both flatten towards the horizontal line $y = 2$.

Intercepts. The value of $g$ at $0$ is $g(0) = \dfrac{-1}{1} = -1$: the $y$-intercept is $-1$. The zero tool displays $X = 0.5$, an exact value, since $2x - 1 = 0$ for $x = \dfrac{1}{2}$: the $x$-intercept is $0.5$.
Vertical asymptote. The denominator $x + 1$ is zero for $x = -1$, while the numerator equals $-3 \neq 0$ there: the line $x = -1$ is a vertical asymptote.
Horizontal asymptote. The numerator is rewritten around the denominator: $$\begin{aligned}[t] g(x) &= \frac{2(x + 1) - 3}{x + 1} &&\text{(}2x - 1 = 2(x + 1) - 3\text{)} \\ &= 2 - \frac{3}{x + 1}. &&\text{(divide each term by } x + 1\text{)} \end{aligned}$$ When $x$ is large, positive or negative, the fraction $\dfrac{3}{x + 1}$ is as small as desired, so $g(x)$ is as close to $2$ as desired: the line $y = 2$ is a horizontal asymptote.

Skills to practice

Reading Key Features with Technology
Studying Graphs with Technology

I.6 Injective, Surjective and Bijective Functions

For a function to be perfectly reversible (i.e., to have an inverse), it must create a perfect pairing between the elements of its domain and its codomain. This leads to the concept of a bijection, which is a function that is both injective and surjective.

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Definition — Injective, Surjective, and Bijective Functions

Let $f: X \to Y$ be a function.

$f$ is injective (one-to-one) if different inputs produce different outputs. For any $x_1, x_2 \in X$, if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.
$f$ is surjective (onto) if its range is equal to its codomain ($f(X)=Y$). This means every element in the codomain $Y$ is the image of at least one element from the domain $X$.
$f$ is bijective if it is both injective and surjective. This means every element in the codomain is the image of exactly one element from the domain.

Example

Consider two versions of the squaring function:

$\Function{f}{\mathbb{R}}{\mathbb{R}}{x}{x^2}$
$\Function{g}{[0,\infty)}{[0,\infty)}{x}{x^2}$

The function $f$ is not bijective. It is not injective because $f(-2)=f(2)$. It is not surjective because its range, $[0, \infty)$, is not equal to its codomain, $\mathbb{R}$.
The function $g$ is bijective. It is injective because its domain is restricted to non-negative numbers. It is surjective because its range, $[0, \infty)$, is equal to its specified codomain, $[0, \infty)$.

Method — The Horizontal Line Test for Bijectivity

To determine visually if a function $f: X \to Y$ is bijective:

Draw the graph of the function $f(x)$.
Imagine drawing horizontal lines across the graph for every possible $y$-value in the codomain $Y$.
Conclusion:
- The function is bijective if and only if every horizontal line (for all $y \in Y$) intersects the graph exactly once.
  - Intersecting at most once shows it is injective.
  - Intersecting at least once shows it is surjective.

Example

Use the horizontal line test to determine if the function $\Function{f}{\mathbb{R}}{\mathbb{R}}{x}{x^2}$ is a bijective function.

Answer

First, we graph the function $f(x) = x^2$. The specified codomain is $\mathbb{R}$.

We test for both conditions:

Injectivity: The horizontal line $y=4$ intersects the graph at two distinct points, $(-2,4)$ and $(2,4)$. Since a horizontal line intersects the graph more than once, the function is not injective.
Surjectivity: The horizontal line $y=-1$ is in the codomain ($\mathbb{R}$) but does not intersect the graph at all. This means that $-1$ has no preimage. Since there is an element in the codomain that is not in the range, the function is not surjective.

Since the function is neither injective nor surjective, it is not bijective.

Skills to practice

Deciding Whether a Function Is Injective, Surjective or Bijective

II Operations on Functions

II.1 Sums, Differences and Products of Functions

Discover

Let $\textcolor{colordef}{f(x)=2x+1}$ and $\textcolor{colorprop}{g(x)=x^2}$.
$\textcolor{olive}{(f+g)(x)}$ means that, for any input $x$ that belongs to the domains of both functions, $\textcolor{colordef}{f(x)}$ and $\textcolor{colorprop}{g(x)}$ are calculated separately, then the results are added. In other words, $(f+g)$ is a new function defined by $(f+g)(x)=f(x)+g(x)$.
This is illustrated by the function machine below:

So, $$\begin{aligned}[t]\textcolor{olive}{(f+g)(x)} &= \textcolor{colordef}{f(x)} + \textcolor{colorprop}{g(x)} \\ &= \textcolor{colordef}{2x+1} + \textcolor{colorprop}{x^2}.\end{aligned}$$

Definition — Operations on Functions

Given two functions $f$ and $g$, we can define new functions by performing arithmetic operations on their outputs, for each $x$ where both are defined:

Sum: $(f+g)(x) = f(x) + g(x)$
Difference: $(f-g)(x) = f(x) - g(x)$
Product: $(fg)(x) = f(x) \times g(x)$
Quotient: $\displaystyle\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$, provided that $g(x) \neq 0$.

For the sum, difference and product, the domain of the new function is the intersection of the domains of $f$ and $g$ (all $x$ for which both $f(x)$ and $g(x)$ are defined).
For the quotient, the domain is the intersection of the domains of $f$ and $g$, excluding any $x$ such that $g(x)=0$.

Example

Let $f(x)=2x+1$ and $g(x)=x^4-1$. Find $(f+g)(x)$.

Answer

$\begin{aligned} (f+g)(x) &= f(x) + g(x) \\ &= (2x+1) + (x^4-1) \\ &= x^4 + 2x. \end{aligned}$
Since both $f$ and $g$ are polynomials, $(f+g)(x)$ is defined for all real numbers $x$.

Skills to practice

Adding, Subtracting and Multiplying Functions

II.2 Inverse Functions

In arithmetic, we are familiar with inverse operations. For example, subtraction is the inverse of addition because it “undoes” the addition. Starting with 5, adding 3 gives 8, and subtracting 3 then returns to 5: $$(5+3) - 3 = 5.$$ Similarly, division is the inverse of multiplication: $$(5\times 3) \div 3 = 5.$$ The concept of an inverse function follows the same idea. An inverse function, denoted $\boldsymbol{f^{-1}}$, is a function that “undoes” or reverses the action of another function, $f$.
If a function $f$ takes an input $x$ to an output $y$, the inverse function $f^{-1}$ takes that output $y$ back to the original input $x$. This creates a perfect loop:

However, not every function has an inverse function. For an inverse to exist, each output $y$ must come from exactly one input $x$ (the function must never take the same value twice on its domain).

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Definition — Inverse Function

Let $f: X \to Y$ be a bijective function.
The inverse function, denoted $f^{-1}$, is the function $\boldsymbol{f^{-1}: Y \to X}$ that reverses the action of $f$.
This means that if $f$ maps an input $x$ to an output $y$, then $f^{-1}$ maps that output $y$ back to the original input $x$. This relationship is defined by:$$ f(x) = y \iff f^{-1}(y) = x $$

The composition of a function and its inverse results in the identity function (which returns the original input):

$\boldsymbol{(f^{-1} \circ f)(x) = x}$ for all $x$ in the domain of $f$ ($X$).
$\boldsymbol{(f \circ f^{-1})(y) = y}$ for all $y$ in the domain of $f^{-1}$ ($Y$).

Method — Finding the Inverse Function

To find the inverse of a function $f$ (when it exists):

Set $y = f(x)$.
Solve the equation for $x$ in terms of $y$. This gives an expression of the form $x = f^{-1}(y)$.
Swap the variables $x$ and $y$ to write the inverse in terms of $x$. The result is $y = f^{-1}(x)$.

This procedure defines an inverse function only if each output corresponds to exactly one input (i.e., if $f$ is invertible on its domain).

Example

Find the inverse of the function $\Function{f}{[0,\infty)}{[0,\infty)}{x}{\sqrt{x}}$.

Answer

The function is $f(x) = \sqrt{x}$ with Domain: $[0, \infty)$ and Range: $[0, \infty)$.

Set $y = \sqrt{x}$.
Solve for $x$: Since the domain of $f$ is non-negative, $x \ge 0$, and the range is non-negative, $y \ge 0$. We can square both sides: $$ y^2 = x \Leftrightarrow x = y^2 $$
Swap variables to get the rule: $y = x^2$.
The domain of $f^{-1}$ is the range of $f$, which is $[0, \infty)$. The range of $f^{-1}$ is the domain of $f$, which is $[0, \infty)$.

So, the inverse function is $\boldsymbol{\Function{f^{-1}}{[0,\infty)}{[0,\infty)}{x}{x^2}}$.

Proposition — Symmetry of Inverse Functions

The graph of a function $f$ and its inverse $f^{-1}$ are reflections of each other across the line $\boldsymbol{y = x}$.

Skills to practice

Graphing and Finding Inverse Functions

III Going further

Skills to practice

Reasoning and Problem-Solving
Solving Problems on Functions
Quiz

\(x\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)
\(f(x)\)	\(-9\)	\(5\)	\(7\)	\(3\)	\(-1\)	\(1\)	\(15\)