In this chapter we will use derivatives to derive necessary and sufficient criteria for the existence of extrema. In calculus, one often uses the theorem that a function
must necessarily satisfy
so that
has a (local) extremum in
. If the derivative function
in addition changes the sign at
, then we have found an extremum. The sign change of the derivative is therefore a sufficient criterion for the extremum. We will now derive this statement and other consequences mathematically and illustrate them with the help of numerous examples. First, however, we will clearly define what an extremum is and what kinds of extrema there are.
A function
can have two types of an extremum: A maximum or a minimum. This in turn can be local or global. As the names already suggest, a local minimum is for example a value
, which is "locally minimal", i.e. In a neighbourhood of
there is also
. Mathematically: There is an interval
around
, so that
for all arguments
which lie in
. A global maximum on the other hand is a value
, which is "everywhere maximal". That means, for all arguments
from the domain of definition, we must have
. This intuitive idea is illustrated in the following figure:
For local extrema, a distinction is also made between strict and non-strict extrema. A strict local minimum, for example, is one that is only "strictly" attained at a single point. A non-strict extremum can be attained on an entire subinterval.
We now define the intuitively explained terms formally:
A local maximum/minimum is also sometimes referred to in the literature as relative maximum/minimum, and a strict maximum/minimum as isolated maximum/minimum. With this definition it is also clear that every global extremum is also a local one. Similarly, every strict local extremum is also a local extremum in the usual sense. In the following we want to determine some necessary and sufficient conditions for (strict) local extrema, using the derivative. Unfortunately our criteria are not sufficient to characterise global extrema. Those are a bit harder to catch!
Question: Consider the functions
Are the following statements true or false?
has a local minimum at
.
has a strict local minimum at
.
has a global minimum at
.
has a global maximum at
.
has a local maximum at
.
has a strict local maximum at
.
has a local maximum at
.
has a global minimum at
.
has a local minimum at
.
Graph of the function

with extrema
Graph of the function

with extrema
Solution:
- True. Since for
there is
for all
with
.
- True. Since for
there is
for all
with
.
- True. Since for all
there is
.
- False. Since e.g. for
there is
.
- True. Since for all
there is
. So
has at
a global and hence also a local maximum.
- False. Since for all
and all
there is
.
- True. Since for
there is
for all
with
.
- True. Since for all
there is
.
- False. Since for all
there is an
with
.
Necessary condition for extrema[Bearbeiten]
In order for a function to have a local extremum at a position within its domain of definition, the function must have a horizontal tangent there. This means that the derivative at this point must be zero. This is exactly what the following theorem says:
Example (Since function)
Graph of the sine function with extrema
Of course, the condition
can also be fulfilled at an infinite number of places in a function. An example is the sine function
. There is
Further there is
,
and
Therefore
has local maxima at
with
, and local minima at
with
.
Example (Exponential function)
The exponential function has no extrema
Finally, the case
for all
can occur. Let us consider the exponential function
. Then, there is for all
:
Since extrema at boundaries are also not possible, it follows from our criterion that the exponential function has no (local) extrema.
Exercise (Local extrema of functions)
Check if the following functions have local extrema:



Solution (Local extrema of functions)
Solution sub-exercise 2:
Here, there is for all
Hence,
has no extrema.
Solution sub-exercise 3:
Finally there is
Further there is
, as well as
and
Therefore
has local maxima at
,
, and local minima at
,
.
Application: Intermediate values for derivatives[Bearbeiten]
We have already stated in the previous sections that the derivative function of a differentiable function does not necessarily have to be continuous. An example for this is the following function, which we have learned about in the chapter "derivatives of higher order":
However, it can be shown that the derivative function always fulfils the intermediate value property. The reason why this is not a contradiction is that continuity is a stronger property than the intermediate value property. To prove this, we will use our necessary criterion from the previous theorem. This result is also known in the literature as "Darboux's theorem":
Proof (Darboux's theorem)
We define the auxiliary function
This function is differentiable with
So there is
and
. Hence
Thus there is an
with
Since the denominator
is positive,
follows. Similarly, there is a
with
. By the extreme value theorem,
attains a minimum on
. Since we have shown that
with
and
, the minimum must be in
. Let
be the minimum. According to our necessary criterion for an extremum, the following must now hold
From this follows
.
Necessary condition: sign change[Bearbeiten]
For many functions it can be tedious to determine only with the necessary condition
whether
has an extremum in
: There is a proof necessary that the function actually does not get greater or lower within the environment. Therefore we are now looking for sufficient conditions for an extremum, which saves us the extra proof work. One possibility is to investigate
in the surroundings of the possible extremum
. If the function increases on the left of
and decreases on the right, then there is a maximum. If the function first decreases and then increases, there is a minimum.
Theorem (Necessary condition for extrema by sign change of the derivative)
Let
and
be a differentiable function. And let
for some
. Then, there is
has a strict maximum at
, if there is an
, such that for all
there is
and for all
there is
.
has a strict Minimum at
, if there is an
, such that for all
there is
and for all
there is
.
Proof (Necessary condition for extrema by sign change of the derivative)
For the proof we use the mean value theorem:
Proof step:
has a strict maximum at 
Let
be arbitrary. By the mean value theorem there is a
, such that
. Since according to our assumption
and
, there is
or
.
Furthermore, for all
there is a
, such that
. We know that
and
. So we get
or
.
If
, then we have for all
with
that
. Thus
in
has a strict maximum.
Proof step:
has a strict minimum at 
The proof is analogous to case 1: For all
there is according to the mean value theorem a
with
. But there is
and
and thus we get
or
.
There is also for all
that we can find a
such that
. But because there is
and
, we also have
or
.
If
, then it holds for all
with
that
. So
has a strict minimum at
.
Alternative proof (Necessary condition for extrema by sign change of the derivative)
Alternatively, the proposition can be proved with the monotony criterion. We show this only for the first statement. The second can be proved analogously. Because of
for all
,
is strictly monotonously increasing according to the monotony criterion on
. For all
there is hence
.
In the same way it follows from
for all
that
is strictly monotonously decreasing on
. For all
there is hence
. With
we obtain
for all
. Thus
is a strict local maximum of
.
Hint
If in the previous theorem only
or
applies, the statements are still valid. The only difference is that the extrema no longer have to be necessarily strict.
Warning
With the sufficient criterion only local extrema can be found. Whether these are also global, or whether there are global extrema at other places, must be examined separately.
Example (Where are zeros of the following polynomial functions)
Graph of the function

We now consider the polynomial function
with
. To find the extreme points, we first differentiate
. There is
So the derivative on the interval
is zero only at the position
. In our domain of definition
the factor
is always negative. In the interval
there is
. So we have
. In the interval
there is
and so we get
.
According to our theorem
has a strict local maximum at
.
Question: Does the polynomial function
have an extremum?
Exercise (Extremum of a function)
Show that for
the function
has a local maximum and minimum, which is a global maximum and minimum respectively
Solution (Extremum of a function)
Proof step:
has a local maximum at 
Proof step:
has a global maximum at 
Graphs of the functions

Since
has no boundary points, according to part 1 only
can be considered for a global maximum of
. We have to show
for all
. Because of
for all
, the function
is strictly monotonously increasing according to the monotony criterion on
. Therefore there is for all
Analogously it follows from
for all
that
is strictly monotonously decreasing on
. So for all
In total, there is
for all
. Thus
is a global maximum of
. Just like in the first part, we can also justify that
is also a global minimum of
.
Proof step:
has a global minimum at 
Conditions are not necessary[Bearbeiten]
The condition in the previous theorem is a sufficient condition for the existence of an extreme point. There is however no necessary condition. We do not have that an extreme position exists exactly when one of the conditions in the previous sentence is fulfilled. The following example illustrates this.
Example
We consider the function
We have already seen that the function
is differentiable and
For all
there is
. Consequently,
is differentiable with the derivative function
For all
there is
and
. Thus
. Hence
There is
and therefore the function
has a (global) minimum at the position
. Next we show that there is no
, such for all
the inequality
is fulfilled. For this we construct a sequence
in
, which converges towards
and has the property that for all
we have
. We define for all
Let
. Then, there is
Necessary condition: presign of the second derivative[Bearbeiten]
If
is twice differentiable, we can also use the following sufficient criterion:
Proof (Necessary condition for extrema via second derivative)
1st statement:
,
has a strict maximum at 
There is
Therefore there is an
such that for all
there is:
If now
, then because of
we immediately get
. If, on the other hand,
, then because of
it follows that
. According to the first sufficient criterion,
is therefore a strict local maximum of
.
Warning
Graph of the function

This sufficient criterion is also not necessary. Since we had deduced it from the first criterion, it is even weaker than this one. An example is given by the function
As we considered above,
has a strict local minimum at
. However, the second sufficient criterion is not applicable. There is in fact
This can be remedied by extending the second sufficient criterion, which we will discuss later.
Example and Exercise[Bearbeiten]
Exercise (Determining extrema of a function)
Consider the function
Determine all local and global extrema of
.
Solution (Determining extrema of a function)
Proof step: Determining local extrema of 
is differentiable on
with
For local extrema in
there must necessarily be
. Now
This equation is fulfilled on
for
and
. So
and
are candidates for local extrema.
is also twice differentiable on
with
Hence there is
According to our second criterion,
has a strict local maximum at
. Furthermore
So
has a strict local minimum at
. Now we still have to examine the boundary point
, because our criteria do not apply there! Since
has a local maximum at
, and on
there are no further zeros of
, the function
is strictly monotonously decreasing on
. So there is
for all
. Therefore
v at
.
Proof step: Determining global extrema of 
Extended sufficient crietrion[Bearbeiten]
The problem with functions like
is that
and so the second derivative vanishes. We cannot decide just by the second derivative whether and what kind of extrema are present. If we now differentiate
two more times, we get
. The question now is whether we can conclude from this, analogous to the second criterion, that
in
has a strict local minimum.
The answer is "yes" - but there is something we need to take care of: Let us look at the example
. This has, in contrast to
no extremum at
, but a saddle point. And this although for the third derivative is also
. The difference is that here the smallest derivative order, which is not equal to zero, is equal to
and therefore odd. With
on the other hand, the smallest order is
, so it is even. We can generalize this to the following criterion:
Summary of proof (sufficient criterion 2b for local extrema)
For the proof we need the Taylor formula for
up to the order
with the Lagrange residuals
Proof (sufficient criterion 2b for local extrema)
Proof step:
and
even
has a strict local minimum at 
Since
is continuous at
there is a
, so that
for
. According to Taylor's theorem, there is now for every m
some
(or
) with
Since
it follows that
If
, then
, and so there is even
for all
. So
has a strict local maximum at
. The proof that
has a strict local minimum at
, if
is analogous.
Proof step:
and
odd
has a saddle point at 
As in the proof of part 1, since
and by Taylor's theorem there is :
for some
(or
). But since now
is odd, there is
if
, and
if
. If now
, then there is
for
and
for
. Conversely, if
, the inequalities apply in the opposite way. In either case
is a saddle point.