Functions

The idea of a function

Definition: Let $A$ and $B$ be two sets. By a function defined on the set $A$, with values in $B$ we mean any rule (procedure or convention) $f$ which associates to every element $a\in A$ a single element, written $f(a)$, in $B$. The set $A$ is called the domain, and the set $B$ is called the codomain of the function, or the set of values of the function.

Let $f:A\to B$ be a function. By the graph of this function we mean the subset ${{G}_{f}}$ of the Cartesian product $A\times B$ formed from the pairs $(a,f(a)), a\in A$. So ${{G}_{f}}=\{(a,f(a))|a\in A\}$.

Definition: A numerical function is a function $f:A\to B$ for which both the domain $A$ and the set of values $B$ are subsets of the set of real numbers.

Injective, surjective and bijective functions

Definition: Let $f:A\to B$ be a function. We say $f$ is an injective function (one-to-one) if for any two elements $x\text{ i }y$ of $A$, $x\ne y$, we have $f(x)\ne f(y).$ The fact that $f$ is injective can also be written:

$(\forall )x,y\in A;\text{ }f(x)=f(y)\Rightarrow x=y.$

Let $y=m, m\in \mathbb{R}$ be a family of lines parallel to the $Ox$ axis. If these lines meet the graph of the function $f$ in at most one point (one or none), then $f$ is injective.

Definition: Let $f:A\to B$ be a function. We say $f$ is a surjective function (onto) if for every $b\in B$ there is at least one element $a\in A$ such that $f(a)=b$. So $f:A\to B$ is not surjective if $\exists b\in B$ we have $f(a)\ne b$, $(\forall )a\in A.$

Let $y=m, m\in \mathbb{R}$ be a family of lines parallel to the $Ox$ axis. If these lines meet the graph of the function $f$ in at least one point, then $f$ is surjective.

Definition: A function $f:A\to B$ which is both injective and surjective is called a bijective function.

Let $y=m, m\in \mathbb{R}$ be a family of lines parallel to the $Ox$ axis. If these lines meet the graph of the function $f$ in exactly one point, then $f$ is bijective.

Every continuous, strictly monotonic function is bijective.

Composition of functions

Definition: Let the functions be $f:A\to B$ and $g:B\to C$ (the codomain of $f$ coincides with the domain of $g$). Let $a\in A$; then $f(a)\in B$, so its image under $g$ exists, namely $g(f(a))\in C$. We can therefore define a function $h:A\to C$ where $h(a)=g(f(a))$ for $(\forall ) a\in A$. The function $h$ defined this way is written $g\circ f$ and is called the composition of the function $g$ with the function $f$.

1. If $f:A\to B$ and $g:C\to D$ are two functions, it only makes sense to speak of the composition of $g$ with $f$ when $B=C$.

2. If $f:A\to B$ and $g:B\to A$ are two functions, both $g\circ f:A\to A$ and $f\circ g:B\to B$ make sense. In general $f\circ g\ne g\circ f$.

3. Let $f:A\to B$, $g:B\to C$ and $h:C\to D$ be three functions. Then each of the functions $h\circ (g\circ f)$, $(h\circ g)\circ f$ makes sense, and the following equality holds: $h\circ (g\circ f)=(h\circ g)\circ f$.

The inverse of a function

Definition: Let $A$ be any set. Write ${{1}_{A}}:A\to A$ for the function defined by ${{1}_{A}}(a)=a$ for $(\forall ) a\in A$. ${{1}_{A}}$ is called the identity function of the set $A$.

Let $A$ be a set and ${{1}_{A}}$ its identity function. Then:

1. For any set $B$ and any function $f:A\to B$, we have $f\circ {{1}_{A}}=f$.

2. For any set $C$ and any function $g:C\to A$, we have ${{1}_{A}}\circ g=g$.

Definition: A function $f:A\to B$ is called invertible if there is a function $g:B\to A$ such that $g\circ f={{1}_{A}}$ and $f\circ g={{1}_{B}}$.

$f(x)=y\Leftrightarrow {{f}^{-1}}(y)=x, \forall x\in A$

The graphs of the functions $f$ and ${{f}^{-1}}$ are symmetric about the first bisector of the coordinate system — the line y = x.

Theorem: A function is invertible if and only if it is bijective.

Examples of invertible functions

1. If $f:\mathbb{R}\to \mathbb{R}$ is a linear function, $f(x)=ax+b, a\ne 0,$ then there is a function $g:\mathbb{R}\to \mathbb{R}$, $g(x)=\frac{1}{a}x-\frac{b}{a}, a\ne 0,$ such that $f\circ g=g\circ f=x$:

$g\circ f=(g\circ f)(x)=\frac{1}{a}(ax+b)-\frac{b}{a}=x, \forall x\in \mathbb{R}$

$f\circ g=(f\circ g)(x)=a(\frac{1}{a}x-\frac{b}{a})+b=x, \forall x\in \mathbb{R}$

So the function $f(x)=ax+b$ is invertible.

2. Let $f:[0,\infty )\to [0,\infty )$, $f(x)={{x}^{2}}$ and ${{f}^{-1}}:[0,\infty )\to [0,\infty )$, ${{f}^{-1}}(x)=\sqrt{x}$

$(f\circ {{f}^{-1}})(x)=f({{f}^{-1}}(x))=f(\sqrt{x})={{(\sqrt{x})}^{2}}=x, \forall x\ge 0$

$({{f}^{-1}}\circ f)(x)={{f}^{-1}}(f(x))={{f}^{-1}}({{x}^{2}})=\sqrt{{{x}^{2}}}=\left| x \right|=x, \forall x>0$

3. Let $f:(0,\infty )\to (0,\infty )$, $f(x)=\frac{1}{x}$ have ${{f}^{-1}}(x)=\frac{1}{x}$, so: ${{f}^{-1}}(x)=f(x), \forall x\in (0,+\infty )$

4. Let $\sin :\left[ -\frac{\pi }{2},\frac{\pi }{2} \right]\to \left[ -1,1 \right]$, a bijective function on this domain with inverse $\arcsin :\left[ -1,1 \right]\to \left[ -\frac{\pi }{2},\frac{\pi }{2} \right]$. The graphs of these functions are shown in the figure below, so that the symmetry about the line y = x can be seen.

5. Let $\cos :\left[ 0,\pi \right]\to \left[ -1,1 \right]$, a bijective function on this domain with inverse $\arccos :\left[ -1,1 \right]\to \left[ 0,\pi \right]$. The graphs of these functions are shown in the adjacent figure, so that the symmetry about the line y = x can be seen.

6. Let $f:[-1,2]\to [0,4]$, defined by

$f(x)=\left\{ \begin{aligned} & \frac{1}{2}x+\frac{1}{2},\text{ dac }\!\!\breve{\mathrm{a}}\!\!\text{ }x\in [-1,1] \\ & {{x}^{2}},\text{ dac }\!\!\breve{\mathrm{a}}\!\!\text{ }x\in (1,2] \\ \end{aligned} \right.$

The function $f$ is invertible and its inverse is: ${{f}^{-1}}:[0,4]\to [-1,2]$

${{f}^{-1}}(x)=\left\{ \begin{aligned} & 2x-1,\text{ dac }\!\!\breve{\mathrm{a}}\!\!\text{ }x\in [0,1] \\ & \sqrt{x},\text{ dac }\!\!\breve{\mathrm{a}}\!\!\text{ }x\in (1,4] \\ \end{aligned} \right.$

Geometric transformations of graphs

Translating graphs along the coordinate axes. Suppose the function $y=f(x)$ and the numbers $a,b>0$. Then:

1. the graph of the function $y=f(x-a)$ is obtained by translating the graph of $y=f(x)$ by $a$ units to the right;

2. the graph of the function $y=f(x+a)$ is obtained by translating the graph of $y=f(x)$ by $a$ units to the right;

3. the graph of the function $y-b=f(x)$ or $y=f(x)+b$ is obtained by translating the graph of $y=f(x)$ by $b$ units in the positive direction of the $Oy$ axis (upwards);

4. the graph of the function $y+b=f(x)$ or $y=f(x)-b$ is obtained by translating the graph of $y=f(x)$ by $b$ units in the negative direction of the $Oy$ axis (downwards).

The graph of $y-3={{(x-2)}^{2}}$ is obtained from the graph of $y={{x}^{2}}$ by two translations.

Compressing and stretching graphs. Suppose the function $y=f(x)$ and the number $a>1$. Then:

1. the graph of the function $y=f(ax)$ is obtained by compressing the graph of $y=f(x)$ horizontally by a factor of $a$;

The graph of $g(x)=\sin (2\cdot x)$ is obtained from the graph of $f(x)=\sin (x)$ by a contraction along the x-axis.

2. the graph of the function $y=f(\frac{x}{a})$ is obtained by stretching the graph of $y=f(x)$ horizontally by a factor of $a$;

3. the graph of the function $y=af(x)$ is obtained by stretching the graph of $y=f(x)$ vertically by a factor of $a$;

The graphs of the functions $f(x)=\sin (x)$ and $g(x)=3\cdot \sin (x)$

The graph of the function $f(x)=a{{x}^{2}}$ for the following values of a = 1, 2, 3.

4. the graph of the function $y=\frac{f(x)}{a}$ is obtained by compressing the graph of $y=f(x)$ vertically by a factor of $a$;

Reflecting graphs in the coordinate axes. Suppose the function $y=f(x)$. Then:

1. the graph of the function $y=f(-x)$ is obtained by reflecting the graph of $y=f(x)$ in the $Oy$ axis;

2. the graph of the function $y=-f(x)$ is obtained by reflecting the graph of $y=f(x)$ in the $Ox$ axis.

In physics one meets the function: $f:\mathbb{R}\to \mathbb{R},\text{ }f(x)=A\cdot \cos (\omega x+b), \omega \ne 0$

$A=amplitudine,$ $\omega =pulsatie,$ $b=faza iniial\breve{a}.$ The fundamental period of the function is $T=\frac{2\pi }{\left| \omega \right|}.$

The graph of the function $f(x)=\sin (x-b)$ for $b=0, \frac{\pi }{4}, \pi , \frac{3\pi }{2}$