Differentiable functions/Mean value theorem/Section

Theorem

Let ${}a<b$, and let

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

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

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

Proof

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

$\Box$

This theorem is called Theorem of Rolle.

The mean value theorem means that, for every secant, there exists a parallel tangent.

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 ${}a<b$, and let

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

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

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

Proof

We consider the auxiliary function

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

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

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

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

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

$\Box$

Corollary

Let

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

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

Proof

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

$\Box$

Theorem

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

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

be a

differentiable function. Then the following statements hold.

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

Proof

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

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

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

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

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

$\Box$

Corollary

A real polynomial function

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

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

Proof

See exercise.

$\Box$

Corollary

Let ${}I$ denote a real interval,

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

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

holds. Then the following statements hold.

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

Proof

See exercise.

$\Box$

We will encounter a more general statement in fact.