# Differentiable functions/Mean value theorem/Section

## Theorem

Let ${\displaystyle {}a, and let

${\displaystyle f\colon [a,b]\longrightarrow \mathbb {R} }$

be a continuous function, which is differentiable on ${\displaystyle {}]a,b[}$, and such that ${\displaystyle {}f(a)=f(b)}$. Then there exists some ${\displaystyle {}c\in {]a,b[}}$, such that

${\displaystyle {}f'(c)=0\,.}$

### Proof

The statement is true if ${\displaystyle {}f}$ is constant. So suppose that ${\displaystyle {}f}$ is not constant. Then there exists some ${\displaystyle {}x\in {]a,b[}}$, such that ${\displaystyle {}f(x)\neq f(a)=f(b)}$. Let's say that ${\displaystyle {}f(x)}$ has a larger value. Due to fact, there exists some ${\displaystyle {}c\in [a,b]}$, where the function attains its maximum. This point is not on the border. For this ${\displaystyle {}c}$, we have ${\displaystyle {}f'(c)=0}$, due to fact.

${\displaystyle \Box }$

This theorem is called Theorem of Rolle.

The following theorem is called Mean value theorem. It says that if a function describes a differentiable one-dimensional movement, then the average velocity is obtained at least once as the instantaneous velocity.

## Theorem

Let ${\displaystyle {}a, and let

${\displaystyle f\colon [a,b]\longrightarrow \mathbb {R} }$

be a continuous function which is differentiable on ${\displaystyle {}]a,b[}$. Then there exists some ${\displaystyle {}c\in {]a,b[}}$, such that

${\displaystyle {}f'(c)={\frac {f(b)-f(a)}{b-a}}\,.}$

### Proof

We consider the auxiliary function

${\displaystyle g\colon [a,b]\longrightarrow \mathbb {R} ,x\longmapsto g(x):=f(x)-{\frac {f(b)-f(a)}{b-a}}(x-a).}$

This function is also continuous and differentiable in ${\displaystyle {}]a,b[}$. Moreover, we have ${\displaystyle {}g(a)=f(a)}$ and

${\displaystyle {}g(b)=f(b)-(f(b)-f(a))=f(a)\,.}$

Hence, ${\displaystyle {}g}$ fulfills the conditions of fact, and therefore there exists some ${\displaystyle {}c\in {]a,b[}}$, such that ${\displaystyle {}g'(c)=0}$. Because of the rules for derivatives, we obtain

${\displaystyle {}f'(c)={\frac {f(b)-f(a)}{b-a}}\,.}$
${\displaystyle \Box }$

## Corollary

Let

${\displaystyle f\colon {]a,b[}\longrightarrow \mathbb {R} }$

be a differentiable function such that ${\displaystyle {}f'(x)=0}$ for all ${\displaystyle {}x\in {]a,b[}}$. Then ${\displaystyle {}f}$ is constant.

### Proof

If ${\displaystyle {}f}$ is not constant, then there exists some ${\displaystyle {}x such that ${\displaystyle {}f(x)\neq f(x')}$. Then there exists, due to the mean value theorem, some ${\displaystyle {}c}$, ${\displaystyle {}x, such that ${\displaystyle {}f'(c)={\frac {f(x')-f(x)}{x'-x}}\neq 0}$, which contradicts the assumption.

${\displaystyle \Box }$

## Theorem

Let ${\displaystyle {}I\subseteq \mathbb {R} }$ be an open interval, and let

${\displaystyle f\colon I\longrightarrow \mathbb {R} }$

be a

differentiable function. Then the following statements hold.
1. The function ${\displaystyle {}f}$ is increasing (decreasing) on ${\displaystyle {}I}$, if and only if ${\displaystyle {}f'(x)\geq 0}$ (${\displaystyle {}f'(x)\leq 0}$) holds for all ${\displaystyle {}x\in I}$.
2. If ${\displaystyle {}f'(x)\geq 0}$ holds for all ${\displaystyle {}x\in I}$, and ${\displaystyle {}f'}$ has only finitely many zeroes, then ${\displaystyle {}f}$ is strictly increasing.
3. If ${\displaystyle {}f'(x)\leq 0}$ holds for all ${\displaystyle {}x\in I}$, and ${\displaystyle {}f'}$ has only finitely many zeroes, then ${\displaystyle {}f}$ is strictly decreasing.

### Proof

(1). It is enough to prove the statements for increasing functions. If ${\displaystyle {}f}$ is increasing and ${\displaystyle {}x\in I}$, then the difference quotient fulfills

${\displaystyle {}{\frac {f(x+h)-f(x)}{h}}\geq 0\,}$

for every ${\displaystyle {}h}$ with ${\displaystyle {}x+h\in I}$. This estimate carries over to the limit as ${\displaystyle {}h\rightarrow 0}$, and this limit is ${\displaystyle {}f'(x)}$.
Suppose now that the derivative is ${\displaystyle {}\geq 0}$. We assume, in order to obtain a contradiction, that there exist two points ${\displaystyle {}x in ${\displaystyle {}I}$ with ${\displaystyle {}f(x)>f(x')}$. Due to the mean value theorem, there exists some ${\displaystyle {}c}$ with ${\displaystyle {}x and

${\displaystyle {}f'(c)={\frac {f(x')-f(x)}{x'-x}}<0\,,}$

(2). Suppose now that ${\displaystyle {}f'(x)>0}$ holds with finitely many exceptions. We assume that ${\displaystyle {}f(x)=f(x')}$ holds for two points ${\displaystyle {}x. Since ${\displaystyle {}f}$ is increasing, due to the first part, it follows that ${\displaystyle {}f}$ is constant on the interval ${\displaystyle {}[x,x']}$. But then ${\displaystyle {}f'=0}$ on this interval, which contradicts the condition that ${\displaystyle {}f'}$ has only finitely many zeroes.

${\displaystyle \Box }$

## Corollary

A real polynomial function

${\displaystyle f\colon \mathbb {R} \longrightarrow \mathbb {R} }$

of degree ${\displaystyle {}d\geq 1}$ has at most ${\displaystyle {}d-1}$ local extrema, and one can partition the real numbers into at most ${\displaystyle {}d}$ intervals, on which ${\displaystyle {}f}$ is alternatingly strictly increasing or strictly decreasing.

### Proof

${\displaystyle \Box }$

## Corollary

Let ${\displaystyle {}I}$ denote a real interval,

${\displaystyle f\colon I\longrightarrow \mathbb {R} }$

a twice continuously differentiable function, and ${\displaystyle {}a\in I}$ an inner point of the interval. Suppose that ${\displaystyle {}f'(a)=0}$

holds. Then the following statements hold.
1. If ${\displaystyle {}f^{\prime \prime }(a)>0}$ holds, then ${\displaystyle {}f}$ has an isolated local minimum in ${\displaystyle {}a}$.
2. If ${\displaystyle {}f^{\prime \prime }(a)<0}$ holds, then ${\displaystyle {}f}$ has an isolated local maximum in ${\displaystyle {}a}$.

### Proof

${\displaystyle \Box }$

We will encounter a more general statement in fact.