# Math for Non-Geeks/ Rolle's theorem
## Motivation

We already know from the extreme value theorem that a continuous function $f$ attains a maximum and a minimum on a closed interval $[a,b]$:
This is of course also true, if $f(a)=f(b)$. 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 $[a,b]$, i.e. within the open interval $(a,b)$:
Let us now additionally assume that $f$ is differentiable on $(a,b)$. Let $\xi$ be a maximum or minimum. If $\xi$ is inside the domain of definition, i.e. if $\xi is\in (a,b)$, then $f'(\xi )=0$ according to the main criterion for extremes values of a differentiable function. This means that the tangent to $f$ at $\xi$ is horizontal. This is exactly what Rolle's theorem says: For every continuous function $f:[a,b]\to \mathbb {R}$ with $f(a)=f(b)$, which is differentiable at $(a,b)$, there is an argument $\xi \in (a,b)$ with $f'(\xi )=0$.
* The derivative at the maximum of $f$ is zero.
* The derivative at the minimum of $f$ is zero.
Of course, $f$ can also assume several (partly local) maxima and minima on $(a,b)$ . Furthermore, it is possible that $f$ attains only one maximum (and no minimum) or one minimum (and no maximum) on $(a,b)$:
* The function $f$ attains one maximum and no minimum within its domain of definition. At that point, the derivative is zero.
* The function $f$ attains one minimum and no maximum within its domain of definition. At that point, the derivative is zero.
A special case is $f$ being constant on $[a,b]$. In this case there is $f'(x)=0$ for all $x\in (a,b)$:
This may also happen on a finite sub-interval of $[a,b]$, 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.

## Rolle's theorem

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 $f:[a,b]\to \mathbb {R}$ be a continuous function with $a<b$ and $f(a)=f(b)$. Furthermore, $f$ is assumed to be differentiable on the open interval $(a,b)$. Then there exists a $\xi \in (a,b)$ with $f'(\xi )=0$.
If $f$ is differentiable on $(a,b)$, then $f$ is continuous on $(a,b)$. Therefore, it is sufficient to prove the continuity of $f$ at the boundary points $a$ and $b$ in order to check the requirements.
Example (Rolle's theorem)
Let us consider the function $f:[0,2]\to \mathbb {R}$ with $f(x)=x^{2}-2x-3$. There is
* $f$ continuous as a polynomial on $[0,2]$
* $f(0)=-3=f(2)$
* $f$ continuous as a polynomial on $(0,2)$
Rolle's theorem now asserts: there is at least one $\xi \in (0,2)$ with $f'(\xi )=0$.
Question: What is a value $\xi$ where the derivative of $f$ in the above example is zero?
The derivative of $f$ is $f'(x)=2x-2$. We set $f$ equal to 0 and get:
${\begin{aligned}&f'(\xi )=2\xi -2=0\\\iff \ &2\xi =2\\\iff \ &\xi =1\end{aligned}}$
At the position $\xi =1$ the derivative of $f$ is zero. This value lies within the domain of definition $[0,2]$ of $f$ and is the only zero of the derivative. Thus $\xi =1$ is the value sought.

## About conditions used in the theorem

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: $f$ is continuous on $[a,b]$

Math for Non-Geeks: Template:Aufgabe

### Condition 2: $f(a)=f(b)$:

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### Condition 3: $f$ is differentiable on $(a,b)$:

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## Proof

Summary of proof (Rolle's theorem)
We first consider the special case that $f$ is a constant function. Here the derivative is zero everywhere. If $f$ 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 $f:[a,b]\to \mathbb {R}$ be continuous function with $a<b$, which is differentiable on $(a,b)$ . Let further $f(a)=f(b)$.
Case 1: $f$ is constant.
Let $f$ be constant. Then, there is $f'(\xi )=0$ for all $\xi \in (a,b)$. So there is at least one $\xi \in (a,b)$ with $f'(\xi )=0$ (any $\xi$ can be chosen from $(a,b)$). Role's theorem is fulfilled in that simple case.
Case 2: $f$ is not constant.
Let $f$ now be non-constant. By the extreme value theorem, $f$ attains both maximum and minimum on the compact interval $[a,b]$. The maximum or minimum of $f$ must be different from $f(a)=f(b)$, otherwise $f$ would be constant. Thus (at least) one extremum is attained at a position $\xi \in (a,b)$.
Since $f$ is differentiable at $(a,b)$, $f$ is also differentiable at the extremum $\xi$. Here, according to the necessary criterion for extrema, there is $f'(\xi )=0$. Thus there exists at least one $\xi \in (a,b)$ where the derivative is zero. So the theorem of Rolle also gives the right implication in this case.

## Exercise

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## Application: Zeros of functions

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 $p(x)=x^{3}+x+1$ on the interval $[-1,0]$. Now,
* $p$ is continuous on $[-1,0]$. Furthermore, $p(-1)=-1<0$ and $p(0)=1>0$. According to the intermediate value theorem, the polynomial has at least one zero at $[-1,0]$.
* $p$ is differentiable at $(-1,0)$ with $p'(x)=3x^{2}+1$. We now assume that $p$ had two zeros on $[-1,0]$ $x_{1}$ and $x_{2}$. Let without loss of generality be $x_{1}<x_{2}$. There is also $p(x_{1})=0=p(x_{2})$. Since $p$ is continuous on $[x_{1},x_{2}]\subseteq [-1,0]$ and differentiable on $(x_{1},x_{2})\subseteq (-1,0)$, Rolle's theorem can be applied. Hence, there is a $\xi \in (x_{1},x_{2})$ with $p'(\xi )=3\xi ^{2}+1=0$. But now $p'$ has no zeros because of $p'(x)=\underbrace {3x^{2}} _{\geq 0}+1\geq 1$. So on $[-1,0]$, the polynomial $p$ cannot have more than one zero.
From both points we get that $p$ has exactly one zero on $[-1,0]$.

## Further exercise

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## Outlook: Rolle's theorem and the mean value theorem

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.