The monotony criterion is quite intuitive: if the derivative of a function (i.e. the slope) is positive, it goes up, if the derivative is negative, it goes down. Mathematically, if the derivative
of a differentiable function
is non-negative or (non-positive) on an interval
, then
is monotonously increasing (or decreasing) on
. If
is even strictly positive (or negative)
, then
is strictly monotonously increasing (or decreasing).
In the first case, even inversion of the statement is true: If a differentiable function is monotonously increasing on
, then
and if the function is monotonously decreasing on
, then then
. However, the inversion does not hold true in the strict case, monotone functions do not always have
or
. For instance,
is strictly monotonous, but
.
Theorem (Monotony criterion for differentiable functions)
Let
be continuous and differentiable on
. Then, there is
on
monotonously increasing on ![[a,b]](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
on
monotonously decreasing on ![[a,b]](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
on
strictly monotonously increasing on ![[a,b]](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
on
strictly monotonously decreasing on ![[a,b]](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
The four directions "
" follow from the mean value theorem. The two directions "
" follow by differentiability of the function:
Proof (monotony criterion for differentiable functions)
We first show the four directions "
" and then the two "
".
1.
: From
on
we get that
in monotonously increasing on
.
Let
for all
and let
with
. We need to show
. By assumption,
is continuous on
and differentiable on
. By the mean-value theorem, there is a
with
By assumption,
, and hence
. Since
we have in the enumerator
. This is equivalent to
, i.e.
is monotonously increasing.
2.
: From
on
we get that
in monotonously decreasing on
.
Let
for all
and let
with
. We need to show
. By assumption,
is continuous on
and differentiable on
. By the mean-value theorem, there is a
with
Now,
, and hence
. Since
we have
. This is equivalent to
, i.e.
is monotonously decreasing.
3.
:
on
implies that
is strictly monotonously increasing on ![[a,b]](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
We prove this by contradiction: Let
be not strictly monotonously increasing. That means, we have some
with
and
. We need to find a
with
. Now,
is continuous on
and differentiable on
. So by the mean value theorem, we can find a
with
Since
, the enumerator of the quotient is non-positive, and because of
the denominator is positive. Thus the whole fraction is non-positive, and therefore
.
4.
:
on
implies that
is strictly monotonously increasing on ![[a,b]](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
Another proof by contradiction: Let
be not strictly monotonously decreasing. That means, we have some
with
and
. We need to find a
with
. Now,
is continuous on
and differentiable on
. So by the mean value theorem, we can find a
with
Since
, the enumerator of the quotient is non-positive, and because of
the denominator is positive. Thus the whole fraction is non-positive, and therefore
.
Now, the two directions "
" follow:
1.
:
being monotonously increasing on
implies
on 
Let
with
. By monotony,
. Further, let
with
. Then we have for the difference quotient
If
, then
. The enumerator and denominator of the difference quotient are thus non-negative, and so is the total quotient. Similarly in the case of
and
enumerator and denominator are non-positive. Thus the whole fraction is again non-negative. Now we form the differential quotient by taking the limit
. This limit exists because
is differentiable on
. Furthermore, the inequality remains valid because of the monotony rule for limit values. Thus we have
Since
and
have been arbitrary, we get
on all of
.
2.
:
being monotonously decreasing on
implies
on 
Let again
with
. By monotony,
. Further, let
with
. Then we have for the difference quotient
If
, then
and thus the total quotient is non-positive. An analogous statement holds in the case
and
. By forming the differential quotient we now obtain
Since
and
have been arbitrary, we get
on all of
.
Examples: monotony criterion[Bearbeiten]
Quadratic and cubic functions[Bearbeiten]
Example (Monotony of quadratic and cubic functions)
Graphs of the functions

and

Graphs of the functions

and

For the quadratic power function
there is
So
is strictly monotonously decreasing by the monotony criterion on
and strictly monotonously increasing on
.
For the cubic power function
there is
So by the monotony criterion,
is monotonously increasing on
and strictly monotonously increasing on
and
. The cubic power function
is even strictly monotonously increasing on all of
.
The fact that
with
is strictly' monotonously increasing, although only
and not
, stems from its derivative being zero at only a single point (namely 0). In the end of this article, we will treat a criterion, which tells us when a function is strictly monotonous, even if there is not everywhere
.
Question: Why is
strictly monotonously increasing on
?
Exponential and logarithm function[Bearbeiten]
Question: What is the monotonicity behaviour of the logarithm function extended to
, i.e.
?
Trigonometric functions[Bearbeiten]
Example (Monotony of the sine function)
For the sine function
there is
So for all
, the
is strictly monotonously increasing on the intervals
and strictly monotonously decreasing on the intervals
.
Question: Where does the cosine function
show monotonous behaviour?
Here,
.
So for all
, the
is strictly monotonously increasing on the intervals
and strictly monotonously decreasing on the intervals
.
Question: Where does the cotangent function
show monotonous behaviour?
Monotony intervals and existence of a zero[Bearbeiten]
Exercise (Monotony intervals and existence of a zero)
Where is the following polynomial function monotonous?
Prove that
has exactly one zero.
Solution (Monotony intervals and existence of a zero)
Graph of the function

Monotony intervals:
The function
is differentiable on all of
, with
So
According to the monotony criterion,
is strictly monotonously increasing on
and on
. Further,
According to the monotony criterion,
is strictly monotonously decreasing on
.
has exactly one zero:
For
, we have the following table of values:
Based on the monotonicity properties and the continuity of
that we have previously investigated, we can read off that:
- On
is
strictly monotonously increasing. Because of
there is
for all
.
- On
is
then strictly monotonously decreasing. So there is also
for all
.
- Subsequently
increases on
again strictly monotonously. Because of
and
, there must be an
with
by the mean value theorem. Because of the strict monotony of
on
, there cannot be any further zeros.
Necessary and sufficient criterion for strict monotony[Bearbeiten]
Proof (Necessary and sufficient criterion for strict monotony)
From the monotony criterion we already know that
is monotonously increasing exactly when
. So we only have to show that
is strictly monotonously increasing exactly when the second condition is additionally fulfilled.
:
strictly monotonously increasing
the set of zeros of
does not contain an open interval.
We perform a proof by contradiction. In other words, we show: If the set of zeros of
contains an open interval,
is not strictly monotonously increasing. Assume there is
with
for all
. Then, by the mean value theorem there is a
with
So
. If now
, then since
is monotonously increasing, there is
So there is
for all
. Hence,
is not strictly monotonously increasing.
: the set of zeros of
does not contain an open interval
strictly monotonously increasing
We perform a proof by contradiction. In other words, we show: if
is monotonously, but not strictly monotonously increasing, then the zero set of
contains an open interval. Assume there is
with
with
. Because of the monotony of
there is
So
for all
. That means
is constant on
. Hence there is for all
:
so the set of zeros of
does contain an open interval and we get a contradiction.
Exercise:
is strictly monotonously increasing