©Tomáš Madaras 2016
The notion of function
Definition
Let A, B be nonempty sets. We say that, on the set A, is defined a
function f ∶ A → B, if there is a rule which assigns to each element x
of A exactly one element f (x) from B; we write f ∶ y = f (x) or just
y = f (x). The symbol x in this notation is called the argument of the
function (or the input variable), the symbol f (x) is the value of
function f in the point x.
The set A is called domain of f (denoted by D(f )). The image of a
function f is the set H(f ) = {y ∈ B ∶ (∃x ∈ A) y = f (x)}.
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Definition
The set {[x, f (x)] ∶ x ∈ D(f )} is called the graph of the function f .
If f is a function of real variable, then its graph is visualized as a set of
points in the plane with fixed Cartesian coordinate system.
A set of points in the plane is the graph of a function f ∶ R → R if
and only if each line parallel with the y-axis has exactly one common
point with that set.
If D(f ) is a finite set, then the graph of f consists of isolated points.
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The function can be specified in several ways:
1 by equation y = f (x),
2 by table of arguments and values (if its domain is a finite set),
3 by description in natural language,
4 by graph.
Example
The function π ∶ N → N is given by description "π(n) is the number
of primes which are less or equal n". One of possible ways to specify
the function π by an equation (for n > 3) is
n
π(n) = −1 + ∑ ((j − 2)! − j ⌊
j=3
(j − 2)!
⌋)
j
A function is always specified both by its description and domain. If,
in the specification of a function, the domain is not explicitly stated,
we will assume that the domain is the set of all x for which is the
expression f (x) defined (or the natural language specification has
sense).
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Relations and operations on functions
Definition
The functions f1 and f2 are equal (we write f1 = f2 ) if the following
holds:
1
D(f1 ) = D(f2 ),
2
(∀x ∈ D(f1 )) f1 (x) = f2 (x).
Warning
The checking of equality f1 (x) = f2 (x) in formal way (without
checking for which arguments x the equality holds) is not enough to
claim that the functions f1 , f2 are equal!
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Definition
Let f, g be functions and their images be subsets of a field F with
binary operations ⊕ and ⊙.
The sum of functions f, g is the function h = f + g such that
D(h) = D(f ) ∩ D(g) and
(∀x ∈ D(h)) h(x) = (f + g)(x) = f (x) ⊕ g(x)
The difference of functions f, g is the function h = f − g such that
D(h) = D(f ) ∩ D(g) and
(∀x ∈ D(h)) h(x) = (f − g)(x) = f (x) ⊕ (−g(x))
(where −g(x) is the negative element of g(x))
The product of functions f, g is the function h = f ⋅ g such that
D(h) = D(f ) ∩ D(g) and
(∀x ∈ D(h)) h(x) = (f ⋅ g)(x) = f (x) ⊙ g(x)
The quotient of functions f, g i sthe function h = fg such that
D(h) = D(f ) ∩ D(g) ∩ {x ∶ g(x) ≠ 0} and
(∀x ∈ D(h)) h(x) = ( fg ) (x) = f (x) ⊙ (g(x))−1
(where (g(x))−1 is the inverse element of g(x))
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Definition
Let f, g be functions. The function h = f ○ g such that
1
D(h) = {x ∈ D(f ) ∶ f (x) ∈ D(g)},
2
(∀x ∈ D(h)) (f ○ g)(x) = g(f (x)).
is called the composition of functions f, g.
The function g is called outer function, the function f is called inner
function of the composition f ○ g.
Warning
The composition of functions is not commutative, that is, in general
it does not hold that f ○ g = g ○ f .
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Example
Given functions f ∶ y =
f ○ f, f ○ g, g ○ f, g ○ g.
√
x, g ∶ y = x2 , determine
We have D(f ) = ⟨0, ∞), D(g) = R. Then
√√
√
(f ○ f )(x) = f (f (x)) =
x = 4 x; D(f ○ f ) = ⟨0, ∞)
√
(f ○ g)(x) = g(f (x)) = ( x)2 = x; D(f ○ g) = ⟨0, ∞)
√
(g ○ f )(x) = f (g(x)) = x2 = ∣x∣; D(g ○ f ) = R
(g ○ g)(x) = g(g(x)) = (x2 )2 = x4 ; D(g ○ g) = R
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Definition
Let f be a function with domain M and image N. A function g = f −1
is called the inverse function of f , if
1
D(g) = N,
2
(∀a ∈ N) g(a) = b ⇔ a = f (b).
Hence, for a function and its inverse, it holds
(∀b ∈ M) (f ○ f −1 )(b) = b
(∀a ∈ N) (f −1 ○ f )(a) = a
The graph of the inverse of a real function is mirror-symmetric to the
graph of original function according to the symmetry axis being the
line y = x.
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Example
x3
.
x3 + 1
We have D(f ) = R − {−1} (as x3 ≠ −1, t.j. x ≠ −1). Swapping the
symbols x and y in the formula for specification of f , we obtain
y3
x =
y3 + 1
3
x(y + 1) = y 3
xy 3 + x = y 3
x = y 3 − xy 3
x = y 3 (1 − x)
x
= y3
1
−
x
√
x
3
= y
1−x
√
x
Thus f −1 ∶ y = 3
. From this we obtain D(f −1 ) = R − {1}.
1−x
Hence, H(f ) = R − {1}, H(f −1 ) = R − {−1}.
Find the inverse function of f ∶ y =
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Properties of functions
Definition
A function f is called even, if
(∀x ∈ D(f )) (−x ∈ D(f ) ∧ f (−x) = f (x)).
A function f is called odd, if
(∀x ∈ D(f )) (−x ∈ D(f ) ∧ f (−x) = −f (x)).
To hold the evenness / oddness property of a function, the domain
must be (as the subset of real axis) symmetric with respect to 0.
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The graph of an even function is mirror-symmetric with respect to
y-axis, the graph of an odd function is centrally-symmetric with
respect to the point [0, 0]:
The graph of a function which is neither even nor odd need not be
symmetric at all, or can be symmetric but not with respect to y-axis or
the point [0, 0].
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Example
Determine whether the function f ∶ y = ln
neither even nor odd.
5−x
is even, odd or
5+x
First, we explore the domain of f :
5−x
>0⇔
5+x
(5 − x > 0
∧
5 + x > 0)
∨
(5 − x < 0
∧
5 + x < 0)
(5 > x
x
∧
∈
x > −5)
(−5, 5)
∨
∨
(5 < x
x
∧
∈
x < −5)
∅
Thus D(f ) = (−5, 5), which is symmetric set with respect to 0.
5 − (−x)
5+x
1
5−x
= ln
= ln
= ln 1 − ln
=
5
−
x
5 + (−x)
5−x
5+x
5+x
5−x
= −f (x). Thus f is odd function.
0 − ln
5+x
Next, f (−x) = ln
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Definition
Let f be a function and M ⊆ D(f ). Then f is called:
strictly increasing on M, if
(∀x1 , x2 ∈ M) x1 < x2 Ô⇒ f (x1 ) < f (x2 ),
strictly decreasing on M, if
(∀x1 , x2 ∈ M) x1 < x2 Ô⇒ f (x1 ) > f (x2 ),
nondecreasing on M, if
(∀x1 , x2 ∈ M) x1 < x2 Ô⇒ f (x1 ) ≤ f (x2 ),
nonincreasing on M, if
(∀x1 , x2 ∈ M) x1 < x2 Ô⇒ f (x1 ) ≥ f (x2 ).
If f has some of these properties, then it is called monotonic; if it is
strictly increasing or strictly decreasing, then it is called strictly
monotonic.
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Example
Show that the function f ∶ y =
interval ⟨−5, 5⟩.
10x
is strictly increasing at the
x2 + 25
Let x1 , x2 ∈ ⟨−5, 5⟩, x1 < x2 . Then x2 − x1 > 0 a x1 x2 ≤ 25, hence
x1 x2 (x2 − x1 )
x1 x22 − x21 x2
x1 x22 + 25x1
<
<
<
25(x2 − x1 )
25x2 − 25x1
x21 x2 + 25x2
x1 (x22 + 25)
<
x2 (x21 + 25)
x1
+ 25
<
10x1
x21 + 25
f (x1 )
<
x21
<
x22
x2
+ 25
10x2
x22 + 25
f (x2 ).
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Definition
A function f is called injective on a set M, if
(∀x1 , x2 ∈ M) x1 ≠ x2 Ô⇒ f (x1 ) ≠ f (x2 ).
Lemma
If a function is strictly monotonic on a set, then it is injective on that
set.
Warning
If a function is injective on a set, then it need not be monotonic; for
1
example, f ∶ y = is injective on (−∞, 0) ∪ (0, ∞), but not
x
monotonic.
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Theorem
The inverse function of a function f exists if and only if f is injective
on D(f ).
Proof. Let f have the inverse f −1 . Suppose that f is not injective on
D(f ). Then there exist x1 , x2 ∈ D(f ), x1 ≠ x2 such that
f (x1 ) = f (x2 ). We have f (x1 ), f (x2 ) ∈ H(f ), hence
(f ○ f −1 )(x1 ) = (f ○ f −1 )(x2 ), t.j.
x1 = f −1 (f (x1 )) = f −1 (f (x2 )) = x2 , a contradiction.
Conversely, suppose that f is injective on D(f ). For each x ∈ D(f ),
there exists unique y ∈ H(f ) such that f (x) = y; define the function
g ∶ H(f ) → D(f ) by assignment g(y) = x. Then
(f ○ g)(x) = g(f (x)) = g(y) = x,
hence g is inverse of f .
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Definition
A function f is called lower bounded on the set M, if
(∃k ∈ R)(∀x ∈ M) f (x) ≥ k.
f is called upper bounded on M, if
(∃l ∈ R)(∀x ∈ M) f (x) ≤ l.
If a function f is both upper and lower bounded on M, then it is called
bounded on M.
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Lemma
A function f is bounded on M if and only if
(∃c ∈ R)(∀x ∈ M) ∣f (x)∣ ≤ c.
Proof. Let f be bounded on M. Then, by definition, there exist
k, l ∈ R such that, for all x ∈ M,
k ≤ f (x) ≤ l.
Put c = max{∣k∣, ∣l∣}. Then, for all x ∈ M, we have
−c ≤ k ≤ f (x) ≤ l ≤ c,
hence ∣f (x)∣ ≤ c.
Assume now that there exists c ∈ R such that, for all x ∈ M, ∣f (x)∣ ≤ c
holds. Then for each x ∈ M, −c ≤ f (x) ≤ c, hence the function f is
both lower and upper bounded, so it is bounded on M.
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Thus, the graph of a bounded function can be fully covered by an
infinite horizontal strip (which is, in addition, symmetric with respect
to x-axis):
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Definition
Let f ∶ M → N be a function. We say that a point x0 ∈ D(f ) is, for f ,
global minimum, if (∀x ∈ D(f )) f (x0 ) ≤ f (x)
strict minimum, if (∀x ∈ D(f )) f (x0 ) < f (x)
local minimum, if (∃ε > 0)(∀x ∈ (x0 − ε, x0 + ε)) f (x0 ) ≤ f (x)
strict local minimum, if
(∃ε > 0)(∀x ∈ (x0 − ε, x0 + ε)) f (x0 ) < f (x)
A point having some of the mentioned properties is global (or local)
extremum of function.
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Examples of global/local extrema:
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Definition
A function f is called periodic, if
(∃p ≠ 0)(∀x ∈ D(f )) (x + p ∈ D(f ) ∧ f (x + p) = f (x)).
If, for a periodic function f there exist the smallest number p0 among
the above mentioned p ≠ 0, then p0 is called the basic period of f .
According to this definition, the constant function f ∶ y = a is
periodic, but has no basic period.
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Example
Prove that the function f ∶ y = 2 + (−1)⌊x⌋ (where ⌊x⌋ is the largest
integer which is less or equal to x - so called "floor of x") is periodic.
We have D(f ) = R. We show that, for each x ∈ R, f (x + 2) = f (x)
holds. Clearly, for each x ∈ R, ⌊x⌋ ≤ x < ⌊x⌋ + 1, thus (after adding 2 to
these inequalities)
⌊x⌋ + 2 ≤ x + 2 < (⌊x⌋ + 2) + 1.
This means (by the definition of the floor function) that ⌊x⌋ + 2 is the
largest integer which is not greater thatn x + 2, and so
⌊x + 2⌋ = ⌊x⌋ + 2. By subsequent transformations, we get
f (x + 2) = 2 + (−1)⌊x+2⌋ = 2 + (−1)⌊x⌋+2 =
2 + (−1)⌊x⌋ (−1)2 = 2 + (−1)⌊x⌋ = f (x).
Thus f is periodic.
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