4.1 What is it a function?

For the purposes of this text, the term „function“ is understood as follows:

Definition 4.1 (Real function of a real variable)

Let us have a non-empty set of real numbers $A \subset \mathbb{R}$. By Real function of a real variable (abbreviated function) $f$ we mean an unambiguous rule that assigns to each number of $A$ a unique real number. We denote such a function as $f: A \to \mathbb{R}$. If the function $f$ assings to $a \in A$ the number $b$, we write $f(a) = b$. The number $a$ is said pre-image of the number $b$ and $b$ is said image of $a$ through the function $f$. We also say that $f(a)$ is the value of the function $f$ at the point $a$.

Example 4.1

Let us consider the set $A = [ -1,1 ]$. Let us try to construct the function $g$ such that „to each $x$ in the set $A$ it corresponds the real number $y$ satisfying $x^2 + y^2 = 1$“. Can $g$ be uniquely specified as a function $g: A \to \mathbb{R}$? Let us take $x \in A$. We ask if $y \in \mathbb{R}$ such that $x^2 + y^2 = 1$ can be found unambiguously. This is equivalent to solve the equality

\begin{equation}\label{eq-ex-kruh}\tag{4.1} y^2 = 1 - x^2. \end{equation}

Since $x \in A$, $1 - x^2 \geq 0$ and therefore the equation (4.1) has two solutions (for $x \neq \pm 1$)

\begin{equation*} y = \pm \sqrt{1-x^2}. \end{equation*}

Which $y$ should we take? This is not an anumbiguous way to assign a number $y$ to every $x$ in $A$, as required in the definition of function . Therefore, we cannot construct a function as asked in this example. We need to slightly adjust the entry.

Example 4.2

Consider the set $A =[ -1,1 ]$. Let us try to construct a function $g$ as follows: „to every $x$ in the set $A$ we assign a non-negative real $y$ such that $x^2 + y^2 = 1$“. Can $g$ be uniquely specified as a function $g: A \to \mathbb{R}$ in this way? We can start as in the previous example. Thus we solve (4.1) with respect to $y$ for given $x \in A$. However now we realize that the equation has just one non-negative solution

\begin{equation*} y = \sqrt{1-x^2}. \end{equation*}

This $y$ is the image of a given $x \in A$. Therefore, after this consideration, we see that $g$ is now well defined aas a function. We can write $g$ more explicitly as

\begin{equation*} g(x) = \sqrt{1-x^2}. \end{equation*}

Speaking of functions it is very often necessary to talk about displayed objects and possible functional values.

Definition 4.2

Let us take a function $f: A \to \mathbb{R}$ as defined in 4.1. We say that the set $A$ is the domain of $f$ and we denote it as $D_f$. The set

\begin{equation}\label{eq-obor-hodnot}\tag{4.2} H_f \href{Thus, it is the set of all real numbers \(b\) for which there exists \(a\) in the domain of the function \(f\) such that \(f(a) = b\).}{\class{mathpopup bg-info-subtle}{:=}} \{ b \in \mathbb{R} \mid (\exists a \in D_f)(f(a) = b) \} \end{equation}

is said image set (or range) of the function $f$.

In the previous example the domain of the fuction $g$ id $D_g=[-1,1]$. Remember the meaning of the symbols used in equation (4.2). The set $H_f$ contains all the real numbers $b$ for which there exists $a$ in the domain of $f$ such that $f(a) = b$. We often denote the domain of the function $f$ also without index, i.e. $D(f)$.

Here let us point out a frequent mistake. When a function $f: A \to \mathbb{R}$ is given, $\mathbb{R}$ is not necessarely its image set. For example, the function $\sin$, that we will describe later, is in the usual notation denoted as 4.1 funkce $\sin: \mathbb{R} \to \mathbb{R}$. However its image set is $H_{\sin} = [ -1,1 ]$, which certainly it is not the whole real axis.

The reader is certainly used to introduce a function as $f (x)$, using an explicit formula indicating what operations need to be done with the (real) $x$ to get its image $f (x)$. This is not the only (nor most common) way to denote a function $f$. You will see other notations in  BIE-ZMA. If a function is given thorugh a formula, without any further comment, then the set of all real $x$, for which $f (x)$ has meaning as a real number, is called the natural domain of the function $f$.

Example 4.3

Suppose a function is specified by the formula

\begin{equation*} h(z) = \sqrt{z^2 - 3z + 2}, \end{equation*}

without any comment on the domain. Its domain is then the above mentioned natural domain. We have to find it. For the square root to make sense, we need that its argument is non-negative, thus $z$ belonging to the natural domain of $h$ must satisfy

\begin{equation*} 0 \leq z^2 - 3z + 2 \href{To find the roots of the second degree polynomial and decompose it into factors is easy.}{\class{mathpopup bg-info-subtle}{=}} (z-2)(z-1) \end{equation*}

The product of two real numbers is non-negative, if the given numbers are both non-negative or both non-positive. Thus $z$ belong to $D_h$ if it satisfies $z \geq 2$ and simultaneously $z \geq 1$ (i.e. $z \geq 2$) or $z \leq 2$ and simultaneously $z \leq 1$ (tj. $z \leq 1$). Therefore the natural domain of our function is

\begin{equation*} D_h = (-\infty, 1] \cup [ 2, +\infty). \end{equation*}

Example 4.4

Not every formula specifies a function. For example, the expressions

\begin{equation*} \sqrt{-1-x^2}, \quad \ln \ln \sin(x), \end{equation*}

have no sense in the set of real numbers $x$.

To illustrate a function we can use its graph. If we introduce two orthogonal coordinate axes, denoted by default $x$ (horizontal axis, independent variable) and $y$ (vertical axis, dependent variable), we call graph of the function $f$ the set which contains the pairs $(x,y) \in \mathbb{R}\times\mathbb{R}$ such $y = f(x)$. It holds then

\begin{equation*} \mathrm{graph}\, f = \{ (x,f(x)) \in \mathbb{R} \href{The Cartesian product of the real axis with itself, i.e. the set of all ordered pairs of real numbers.}{\class{mathpopup bg-info-subtle}{\times}} \mathbb{R} \mid x\in D_f \}. \end{equation*}

Now we will deal with several kinds of known functions. An overview of the properties of many functions can be found, for example, in (NIST Digital Library of Mathematical Functions, b.r.).

4.1.1 Properties of functions

In order to easily talk about the behavior of functions, it is worthwhile introducing some useful terms. By slope, we distinguish the following types of functions:

Definition 4.3

A function $f$ with domain $D_f \subset \mathbb R$ is said to be, on the set $A \subset D_f$,

  • increasing, if $(\forall x,y \in A)\big(x < y \Rightarrow f(x) \leq f(y)\big)$,

  • stricly increasing, if $(\forall x,y \in A)\big(x < y \Rightarrow f(x) < f(y)\big)$,

  • decreasing, if $(\forall x,y \in A)\big(x < y \Rightarrow f(x) \geq f(y)\big)$,

  • strictly decreasing, if $(\forall x,y \in A)\big(x < y \Rightarrow f(x) > f(y)\big)$,

  • monotonic, if it is increasing or decreasing,

  • strictly monotonic, if it is strictly increasing or strictly decreasing.

Here again, the reader is reminded that our nomenclature is not widely used in every country, it is used in Anglo-Saxon literature. It is therefore more likely that the reader will encounter it when searching the Internet and studying English literature.

In terms of symmetry, we distinguish between odd, even and periodic functions.

Definition 4.4

A function $f$ is said

  • even, if $(\forall x \in D_f)((-x \in D_f) \ \text{and} \ (f(-x) = f(x)))$,

  • odd, if $(\forall x \in D_f)((-x \in D_f) \ \text{and} \ (f(-x) = -f(x)))$,

  • periodic with period $T > 0$, if $(\forall x \in D_f)((x + T \in D_f) \ \text{and} \ (f(x) = f(x + T)))$.

The graph of an even function is axially symmetrical with respect to the $y$ axis. The graph of an odd function is symmetric with respct to the origin of the coordinates axis. The function value of an aperiodic function at a point $x$ does not change with a shift to the point $x + T$, $T$ being the period.

Finally, let us recall the notion of injective function here.

Definition 4.5

We call a function $f: D_f \to \R$ injective, when for very different number $a$ and $b$ from the domain of the function $f$ also the functional values $f(a)$ and $f(b)$ are different. Equivalently, in symbols

\begin{equation*} (\forall a,b \in D_f)(a \neq b \Rightarrow f(a) \neq f(b)). \end{equation*}

Alternatively, the requirement in the definition can be reformulated as follows: a function $f$ is injective, if for every $a,b \in D_f$ such that $f(a) = f(b)$, ti holds that $a = b$.

Example 4.5

For example the function $f(x) = x^2$ defined on the whole $\R$ is not injective. The requirement in the definition is not met: it is enough to choose two different numbers, as for example $a = 1$ a $b = -1$, for which obviously $f(1) = f(-1)$.In contrast, the function $f(x) = x^3$ defined on the whole $\R$ is injective. Indeed, let us take two $a,b \in \R$ such that $f(a) = a^3 = b^3 = f(b)$. Does it follow from this that $a = b$? The use of a known algebraic formula results in equality

\begin{equation}\label{eq-prostota}\tag{4.3} 0 = a^3 - b^3 = (a-b) (a^2 + ab + b^2). \end{equation}

The expression in the second bracket is zero only if $a = b = 0$. We can make sure of this by adjusting the square:

\begin{equation*} a^2 + ab + b^2 = a^2 + 2 a \frac{b}{2} + \frac{b^2}{4} + \frac{3}{4} b^2 = \left(a + \frac{b}{2} \right)^2 + \frac{3}{4} b^2 \end{equation*}

If at least one of $a, b$ is non-zero, then from (4.3) it necessarily follows $a = b$.

Remark 4.1

Frequent student myths include the statement: The function $f$ is injective when every element in the domain has just one image. This statement applies to every function (it is in the definition of function)! It does not express the injectivity of the function.