# Rolle's theorem – Serlo

## Motivation

We already know from the extreme value theorem that a continuous function ${\displaystyle f}$ attains a maximum and a minimum on a closed interval ${\displaystyle [a,b]}$:

This is of course also true, if ${\displaystyle f(a)=f(b)}$. 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 ${\displaystyle [a,b]}$, i.e. within the open interval ${\displaystyle (a,b)}$:

Let us now additionally assume that ${\displaystyle f}$ is differentiable on ${\displaystyle (a,b)}$. Let ${\displaystyle \xi }$ be a maximum or minimum. If ${\displaystyle \xi }$ is inside the domain of definition, i.e. if ${\displaystyle \xi is\in (a,b)}$, then ${\displaystyle f'(\xi )=0}$ according to the main criterion for extremes values of a differentiable function. This means that the tangent to ${\displaystyle f}$ at ${\displaystyle \xi }$ is horizontal. This is exactly what Rolle's theorem says: For every continuous function ${\displaystyle f:[a,b]\to \mathbb {R} }$ with ${\displaystyle f(a)=f(b)}$, which is differentiable at ${\displaystyle (a,b)}$, there is an argument ${\displaystyle \xi \in (a,b)}$ with ${\displaystyle f'(\xi )=0}$.

Of course, ${\displaystyle f}$ can also assume several (partly local) maxima and minima on ${\displaystyle (a,b)}$ . Furthermore, it is possible that ${\displaystyle f}$ attains only one maximum (and no minimum) or one minimum (and no maximum) on ${\displaystyle (a,b)}$:

A special case is ${\displaystyle f}$ being constant on ${\displaystyle [a,b]}$. In this case there is ${\displaystyle f'(x)=0}$ for all ${\displaystyle x\in (a,b)}$:

This may also happen on a finite sub-interval of ${\displaystyle [a,b]}$, 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 ${\displaystyle f:[a,b]\to \mathbb {R} }$ be a continuous function with ${\displaystyle a and ${\displaystyle f(a)=f(b)}$. Furthermore, ${\displaystyle f}$ is assumed to be differentiable on the open interval ${\displaystyle (a,b)}$. Then there exists a ${\displaystyle \xi \in (a,b)}$ with ${\displaystyle f'(\xi )=0}$.

If ${\displaystyle f}$ is differentiable on ${\displaystyle (a,b)}$, then ${\displaystyle f}$ is continuous on ${\displaystyle (a,b)}$. Therefore, it is sufficient to prove the continuity of ${\displaystyle f}$ at the boundary points ${\displaystyle a}$ and ${\displaystyle b}$ in order to check the requirements.

Example (Rolle's theorem)

Let us consider the function ${\displaystyle f:[0,2]\to \mathbb {R} }$ with ${\displaystyle f(x)=x^{2}-2x-3}$. There is

• ${\displaystyle f}$ continuous as a polynomial on ${\displaystyle [0,2]}$
• ${\displaystyle f(0)=-3=f(2)}$
• ${\displaystyle f}$ continuous as a polynomial on ${\displaystyle (0,2)}$

Rolle's theorem now asserts: there is at least one ${\displaystyle \xi \in (0,2)}$ with ${\displaystyle f'(\xi )=0}$.

Question: What is a value ${\displaystyle \xi }$ where the derivative of ${\displaystyle f}$ in the above example is zero?

The derivative of ${\displaystyle f}$ is ${\displaystyle f'(x)=2x-2}$. We set ${\displaystyle f}$ equal to 0 and get:

{\displaystyle {\begin{aligned}&f'(\xi )=2\xi -2=0\\\iff \ &2\xi =2\\\iff \ &\xi =1\end{aligned}}}

At the position ${\displaystyle \xi =1}$ the derivative of ${\displaystyle f}$ is zero. This value lies within the domain of definition ${\displaystyle [0,2]}$ of ${\displaystyle f}$ and is the only zero of the derivative. Thus ${\displaystyle \xi =1}$ 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: ${\displaystyle f}$ is continuous on ${\displaystyle [a,b]}$

Exercise (Condition: continuity)

Find a function ${\displaystyle f:[a,b]\to \mathbb {R} }$, which is differentiable only on ${\displaystyle (a,b)}$ and for which ${\displaystyle f(a)=f(b)}$ holds, but for which the implication of Rolle's theorem does not hold.

The searched function ${\displaystyle f}$ 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 ${\displaystyle f}$ may then diverge to infinity at the end points.

Solution (Condition: continuity)

${\displaystyle f:[0,1]\to \mathbb {R} }$ with

${\displaystyle f(x)={\begin{cases}x&{\text{ if }}x\neq 1,\\0&{\text{ if }}x=1,\end{cases}}}$

is differentiable on ${\displaystyle (0,1)}$ and there is ${\displaystyle f(0)=0=f(1)}$. But since ${\displaystyle f'(x)=1}$ for all ${\displaystyle x\in (0,1)}$, there is no ${\displaystyle \xi \in (0,1)}$ with ${\displaystyle f'(\xi )=0}$.

### Condition 2: ${\displaystyle f(a)=f(b)}$:

Exercise (Equality of function values)

Find a continuous function ${\displaystyle f:[a,b]\to \mathbb {R} }$ which is differentiable on ${\displaystyle (a,b)}$ for which the implication of Rolle's theorem does not hold.

This task shows that the condition ${\displaystyle f(a)=f(b)}$ 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)

Such a function is for example ${\displaystyle f:[0,1]\to \mathbb {R} }$ with ${\displaystyle f(x)=x}$. This function is continuous at ${\displaystyle [0,1]}$ and also differentiable at ${\displaystyle (0,1)}$. There is however ${\displaystyle f(0)=0\neq 1=f(1)}$.

For this function, there is ${\displaystyle f'(x)=1}$ for all ${\displaystyle x\in (0,1)}$. So there is no ${\displaystyle \xi \in (0,1)}$ with ${\displaystyle f'(\xi )=0}$.

### Condition 3: ${\displaystyle f}$ is differentiable on ${\displaystyle (a,b)}$:

Exercise (Condition: differentiability)

Find a continuous function ${\displaystyle f:[a,b]\to \mathbb {R} }$ with ${\displaystyle f(a)=f(b)}$, for which the implication of Rolle's theorem does not hold.

Solution (Condition: differentiability)

The function ${\displaystyle f:[0,1]\to \mathbb {R} }$ with

${\displaystyle f(x)={\begin{cases}x&{\text{ if }}x\leq {\tfrac {1}{2}}\\1-x&{\text{ if }}x>{\tfrac {1}{2}}\end{cases}}}$

is continuous and there is ${\displaystyle f(0)=0=f(1)}$. This function is only differentiable on the intervals ${\displaystyle \left[0,{\tfrac {1}{2}}\right)}$ and ${\displaystyle \left({\tfrac {1}{2}},1\right]}$. The derivative function ${\displaystyle g:[a,b]\to \mathbb {R} }$ has the assignment rule:

${\displaystyle g(x)={\begin{cases}1&x<{\tfrac {1}{2}}\\-1&x>{\tfrac {1}{2}}\end{cases}}}$

Hence, there is no ${\displaystyle \xi \in (0,1)}$ with ${\displaystyle f'(\xi )=0}$.

## Proof

Summary of proof (Rolle's theorem)

We first consider the special case that ${\displaystyle f}$ is a constant function. Here the derivative is zero everywhere. If ${\displaystyle f}$ 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 ${\displaystyle f:[a,b]\to \mathbb {R} }$ be continuous function with ${\displaystyle a, which is differentiable on ${\displaystyle (a,b)}$ . Let further ${\displaystyle f(a)=f(b)}$.

Fall 1: ${\displaystyle f}$ is constant.

Let ${\displaystyle f}$ be constant. Then, there is ${\displaystyle f'(\xi )=0}$ for all ${\displaystyle \xi \in (a,b)}$. So there is at least one ${\displaystyle \xi \in (a,b)}$ with ${\displaystyle f'(\xi )=0}$ (any ${\displaystyle \xi }$ can be chosen from ${\displaystyle (a,b)}$). Role's theorem is fulfilled in that simple case.

Fall 2: ${\displaystyle f}$ is not constant.

Let ${\displaystyle f}$ now be non-constant. By the extreme value theorem, ${\displaystyle f}$ attains both maximum and minimum on the compact interval ${\displaystyle [a,b]}$. The maximum or minimum of ${\displaystyle f}$ must be different from ${\displaystyle f(a)=f(b)}$, otherwise ${\displaystyle f}$ would be constant. Thus (at least) one extremum is attained at a position ${\displaystyle \xi \in (a,b)}$.

Since ${\displaystyle f}$ is differentiable at ${\displaystyle (a,b)}$, ${\displaystyle f}$ is also differentiable at the extremum ${\displaystyle \xi }$. Here, according to the necessary criterion for extrema, there is ${\displaystyle f'(\xi )=0}$. Thus there exists at least one ${\displaystyle \xi \in (a,b)}$ where the derivative is zero. So the theorem of Rolle also gives the right implication in this case.

## Exercise

Exercise (Exercise)

Let ${\displaystyle k\in \mathbb {N} }$. Show with Rolle's theorem that the derivative function ${\displaystyle f'(x)}$ of the function ${\displaystyle f:[0,k\pi ]\to [-1,1]}$ with ${\displaystyle f(x)=\sin(x)}$ has at least ${\displaystyle k}$ zeros.

Solution (Exercise)

The sine function is differentiable on all of ${\displaystyle \mathbb {R} }$, i.e. also continuous. Furthermore there is ${\displaystyle \sin(l\pi )=0}$ for all ${\displaystyle l\in \mathbb {Z} }$. By Rolle's theorem, there is a ${\displaystyle \xi \in (l\pi ,(l+1)\pi )}$ with ${\displaystyle f'(\xi )=0}$. For every ${\displaystyle l\in \mathbb {N} }$ with ${\displaystyle 0\leq l we find a ${\displaystyle \xi }$ where the derivative is zero. Since there are ${\displaystyle k}$ different natural numbers for ${\displaystyle l}$ with ${\displaystyle 0\leq l, we can also find ${\displaystyle k}$ different zeroes ${\displaystyle \xi }$ of the derivative function. The derivative of ${\displaystyle f}$ must therefore have at least ${\displaystyle k}$ distinct zeros.

## 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 ${\displaystyle p(x)=x^{3}+x+1}$ on the interval ${\displaystyle [-1,0]}$. Now,

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

From both points we get that ${\displaystyle p}$ has exactly one zero on ${\displaystyle [-1,0]}$.

## Further exercise

Exercise (Finding a unique zero)

Show that

${\displaystyle f:[1,2]\to \mathbb {R} :x\mapsto {\frac {2}{x^{4}}}-x+1}$

has exactly one zero.

Summary of proof (Finding a unique zero)

First, use the intermediate value theorem to show that ${\displaystyle f}$ has at least one zero. Then we show by Rolle's theorem that ${\displaystyle f}$ has at most one zero. From both steps the assertion follows.

Proof (Finding a unique zero)

Proof step: ${\displaystyle f}$ has at least one zero.

${\displaystyle f}$ is continuous as a composition of continuous functions. Further, there is ${\displaystyle f(1)={\tfrac {2}{1}}-1+1=2>0}$ and ${\displaystyle f(2)={\tfrac {2}{16}}-2+1=-{\tfrac {7}{8}}<0}$. By the intermediate value theorem the function has therefore at least one zero.

Proof step: ${\displaystyle f}$ has at most one zero.

${\displaystyle f}$ is differentiable on ${\displaystyle (1,2)}$ as it is a composition of differentiable functions. In particular, ${\displaystyle f'(x)=-4{\tfrac {2}{x^{5}}}-1=-{\tfrac {8}{x^{5}}}-1}$. We now assume that on ${\displaystyle [1,2]}$, the function ${\displaystyle f}$ has two distinct zeros ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$. Let us assume that ${\displaystyle x_{1}. There is therefore ${\displaystyle f(x_{1})=0=f(x_{2})}$.

Now, ${\displaystyle f}$ is continuous on ${\displaystyle [x_{1},x_{2}]\subseteq [1,2]}$ and differentiable on ${\displaystyle (x_{1},x_{2})\subseteq (1,2)}$. According to Rolle's theorem, there hence is a ${\displaystyle \xi \in (x_{1},x_{2})}$ with ${\displaystyle f'(\xi )=-{\tfrac {8}{\xi ^{5}}}-1=0}$. But since

${\displaystyle f'(x)=\underbrace {-{\tfrac {8}{x^{5}}}} _{\leq -{\frac {8}{2^{5}}}=-{\tfrac {1}{4}}}-1\leq -{\tfrac {5}{4}}<0}$

${\displaystyle f'}$ has no zeros. So ${\displaystyle f}$ has at most one zero.

It follows from both steps of proof that ${\displaystyle f}$ has exactly one zero.

## 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.