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:
Let and be a function. Then, at has a
local maximum or minimum, if there is an , such that (or ) for all with holds.
strict local maximum or minimum, if there is an such that (or ) for all with holds.
global maximum or minimum, if (or ) for all holds.
Extremum is the umbrella term for a maximum or minimum. is then called maximum or minimum.
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 .
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.
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:
Theorem (Necessary condition for extrema)
Let with . Let and be differentiable at . Let further be a local minimum (or maximum). Then, there is .
Proof (Necessary condition for extrema)
We consider the case where has a local minimum at . The proof in the case of a local maximum is analogous. We want to show that
Since is differentiable at , there is
Since has a local minimum at , there is an , such that for all there is
From the limit value rules follows
On the other hand there is an , such that for all there is
The function has a local (and even global) minimum at . Since , the necessary criterion yields .
If there is an extremum at a boundary point, the condition at this point does not have to be fulfilled! For instance, is a local maximum. But there is .
Example (Cubic power functions)
The necessary condition is not not sufficient. That means, it does not necessarily follow from that has an extremum at . An example for this is the function with . There is and therefore . But has no extremum at , because for all there is and for all there is . The zero point in this case is also called terrace or saddle point.
Example (Since function)
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 ,
Therefore has local maxima at with , and local minima at with .
Example (Exponential function)
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.
Check if the following functions have local extrema:
Solution (Local extrema of functions)
Solution sub-exercise 1:
Since furthermore for all , the point is a local (even global) minimum of .
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
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":
Theorem (Darboux's theorem)
Let be differentiable. Further let and . Then there is an with .
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
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 .
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.
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)
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?
Again there is
The derivative has in the zero . In the domain of definition, is always positive. On there is , and therefore . On however , and therefore . Hence has a strictly local minimum at .
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
is differentiable on according to the product rule with
According to the necessary criterion for the existence of a maximum , we need . Now
So is the only candidate for our local maximum on . Furthermore
So there is for all and for all . According to the sufficient criterion is therefore a (strict) local maximum of .
Proof step: has a global maximum at
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
There is and for all . So a global and therefore also a local minimum of .
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.
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]
We again look at the polynomial function with . As we already know there is
Hence there is to . Further
and therefore . So has a strict local maximum at .
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
The following monotony table ("smi" = strictly monotonously increasing, "smd" = strictly monotonously decreasing) we get the first step of the proof for :
Further there is . Thus is a global minimum of . Finally
So is unbounded from above and therefore has no global maximum.
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:
Theorem (sufficient criterion 2b for local extrema)
Let be an -times differentiable function (), where is continuous on . Further let
Then, there is:
If is even, then in the case of in has a strict local maximum. If , then has a strict local minimum.
If is odd, then has a saddle point in .
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.
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