We already know from the extreme value theorem that a continuous function attains a maximum and a minimum on a closed interval :
This is of course also true, if . In this case (if the function is not constant) there must be a maximum or minimum inside the domain of definition. In the following figure, both the maximum and the minimum are inside , i.e. within the open interval :
Let us now additionally assume that is differentiable on . Let be a maximum or minimum. If is inside the domain of definition, i.e. if , then according to the main criterion for extremes values of a differentiable function. This means that the tangent to at is horizontal. This is exactly what Rolle's theorem says: For every continuous function with , which is differentiable at , there is an argument with .
The derivative at the maximum of is zero.
The derivative at the minimum of is zero.
Of course, can also assume several (partly local) maxima and minima on . Furthermore, it is possible that attains only one maximum (and no minimum) or one minimum (and no maximum) on :
The function attains one maximum and no minimum within its domain of definition. At that point, the derivative is zero.
The function attains one minimum and no maximum within its domain of definition. At that point, the derivative is zero.
A special case is being constant on . In this case there is for all :
This may also happen on a finite sub-interval of , i.e. on a "horizontal plateau".
No matter which case we looked at, there was always at least one point inside the domain of definition where the derivative of the function is zero.
The theorem named after Michel Rolle (1652-1719) represents a special case of the mean value theorem of differential calculus and reads as follows:
Theorem (Rolle's theorem)
Let be a continuous function with and . Furthermore, is assumed to be differentiable on the open interval . Then there exists a with .
If is differentiable on , then is continuous on . Therefore, it is sufficient to prove the continuity of at the boundary points and in order to check the requirements.
Example (Rolle's theorem)
Let us consider the function with . There is
- continuous as a polynomial on
- continuous as a polynomial on
Rolle's theorem now asserts: there is at least one with .
Question: What is a value where the derivative of in the above example is zero?
and its derivative
The derivative of is . We set equal to 0 and get:
At the position the derivative of is zero. This value lies within the domain of definition of and is the only zero of the derivative. Thus is the value sought.
About conditions used in the theorem[Bearbeiten]
There are several necessary requirements in Rolle's theorem. We will show now that if we drop any one of them, the theorem is no longer true.
Condition 1: is continuous on [Bearbeiten]
Exercise (Condition: continuity)
Find a function , which is differentiable only on and for which holds, but for which the implication of Rolle's theorem does not hold.
The searched function fulfils all requirements of the set of roles except continuity in the complete domain of definition. If we drop continuity on the endpoints, we run into trouble! The reason is that may then diverge to infinity at the end points.
Solution (Condition: continuity)
Graph of the function
is differentiable on and there is . But since for all , there is no with .
Condition 2: :[Bearbeiten]
Exercise (Equality of function values)
Find a continuous function which is differentiable on for which the implication of Rolle's theorem does not hold.
This task shows that the condition is necessary for Rolle's theorem. Otherwise, we may "build a slight slope" between the end points, which has no maximum or minimum.
Solution (Equality of function values)
The identity function
Such a function is for example with . This function is continuous at and also differentiable at . There is however .
For this function, there is for all . So there is no with .
Condition 3: is differentiable on :[Bearbeiten]
Exercise (Condition: differentiability)
Find a continuous function with , for which the implication of Rolle's theorem does not hold.
Solution (Condition: differentiability)
Plot of the function
The function with
is continuous and there is . This function is only differentiable on the intervals and . The derivative function has the assignment rule:
Hence, there is no with .
Summary of proof (Rolle's theorem)
We first consider the special case that is a constant function. Here the derivative is zero everywhere. If is not constant, we use the extreme value theorem to find a maximum or minimum within the domain of definition. At this extremum, the derivative vanishes according to the criterion for the existence of an extremum.
Proof (Rolle's theorem)
Let be continuous function with , which is differentiable on . Let further .
Fall 1: is constant.
Let be constant. Then, there is for all . So there is at least one with (any can be chosen from ). Role's theorem is fulfilled in that simple case.
Fall 2: is not constant.
Let now be non-constant. By the extreme value theorem, attains both maximum and minimum on the compact interval . The maximum or minimum of must be different from , otherwise would be constant. Thus (at least) one extremum is attained at a position .
Since is differentiable at , is also differentiable at the extremum . Here, according to the necessary criterion for extrema, there is . Thus there exists at least one where the derivative is zero. So the theorem of Rolle also gives the right implication in this case.
Let . Show with Rolle's theorem that the derivative function of the function with has at least zeros.
The sine function is differentiable on all of , i.e. also continuous. Furthermore there is for all . By Rolle's theorem, there is a with . For every with we find a where the derivative is zero. Since there are different natural numbers for with , we can also find different zeroes of the derivative function. The derivative of must therefore have at least distinct zeros.
Application: Zeros of functions[Bearbeiten]
Rolle's theorem can also be used in proofs of existence of zeros. And it can be used to show that a function has at most one zero on an interval. On the other hand, the intermediate value theorem can be used to show that a function has at least one zero on an interval. Together the existence of exactly one zero can be implied.
Example (Zeros of a polynomial)
Let us consider the polynomial on the interval . Now,
- is continuous on . Furthermore, and . According to the intermediate value theorem, the polynomial has at least one zero at .
- is differentiable at with . We now assume that had two zeros on and . Let without loss of generality be . There is also . Since is continuous on and differentiable on , Rolle's theorem can be applied. Hence, there is a with . But now has no zeros because of . So on , the polynomial cannot have more than one zero.
From both points we get that has exactly one zero on .
Exercise (Finding a unique zero)
has exactly one zero.
Summary of proof (Finding a unique zero)
First, use the intermediate value theorem to show that has at least one zero. Then we show by Rolle's theorem that has at most one zero. From both steps the assertion follows.
Proof (Finding a unique zero)
Proof step: has at least one zero.
is continuous as a composition of continuous functions. Further, there is and . By the intermediate value theorem the function has therefore at least one zero.
Proof step: has at most one zero.
is differentiable on as it is a composition of differentiable functions. In particular, . We now assume that on , the function has two distinct zeros and . Let us assume that . There is therefore .
Now, is continuous on and differentiable on . According to Rolle's theorem, there hence is a with . But since
has no zeros. So has at most one zero.
It follows from both steps of proof that has exactly one zero.
Outlook: Rolle's theorem and the mean value theorem[Bearbeiten]
As mentioned above, Rolle's theorem is a special case of the mean value theorem. This is one of the most important theorems from real Analysis, as many other useful results can be derived from it. Conversely, we will show that the mean value theorem follows from Rolle's theorem. Both theorems are thus equivalent.