Mean value theorem · Selected prerequisite proofs

L4.2.2: Scalar Fermat theorem at an interior extremum.

Lemma 4.2.2.

Suppose

f \colon (a,b) \to \R

is differentiable at

c \in (a,b)\text{,}

and

f

has a relative minimum or a relative maximum at

c\text{.}

Then

f'(c) = 0\text{.}

Proof.

Suppose

c

is a relative maximum of

f\text{.}

That is, there is a

\delta > 0

such that for every

x \in (a,b)

where

\sabs{x-c} < \delta\text{,}

we have

f(x)-f(c) \leq 0\text{.}

Consider the difference quotient. If

c < x < c+\delta\text{,}

then

\begin{equation*} \frac{f(x)-f(c)}{x-c} \leq 0 , \end{equation*}

and if

c-\delta < y < c\text{,}

then

\begin{equation*} \frac{f(y)-f(c)}{y-c} \geq 0 . \end{equation*}

See Figure 4.3 for an illustration.

A graph of a function f is shown. Three points are marked on the horizontal axis, from left to right, y, c, and x. The function has a maximum at the point c. The points corresponding to x, c, and y are marked on the graph. Through the points corresponding to y and c is a secant line that slopes upwards and is marked with slope equals f of y minus f of c the whole thing over quantity y-c is greater than or equal to 0. Through the points corresponding to c and x is a secant line that slopes downwards and is marked with slope equals f of x minus f of c the whole thing over quantity x-c is less than or equal to 0. — Figure 4.3. Slopes of secants at a relative maximum.

As

a < c < b\text{,}

there exist sequences

\{ x_n \}_{n=1}^\infty

and

\{ y_n \}_{n=1}^\infty

in

(a,b)

such that

c < x_n < c+\delta

and

c-\delta < y_n < c

for all

n \in \N\text{,}

and such that

\lim_{n\to\infty} x_n = \lim_{n\to\infty} y_n = c\text{.}

Since

f

is differentiable at

c\text{,}

\begin{equation*} 0 \geq \lim_{n\to\infty} \frac{f(x_n)-f(c)}{x_n-c} = f'(c) = \lim_{n\to\infty} \frac{f(y_n)-f(c)}{y_n-c} \geq 0. \end{equation*}

We are done with a maximum. For a minimum, consider the function

-f\text{.}

L4.2.3: Rolle theorem on a nondegenerate closed interval.

Theorem 4.2.3. Rolle.

Let

f \colon [a,b] \to \R

be a continuous function differentiable on

(a,b)

such that

f(a) = f(b)\text{.}

Then there exists a

c \in (a,b)

such that

f'(c) = 0\text{.}

Proof.

As

f

is continuous on

[a,b]\text{,}

it attains an absolute minimum and an absolute maximum in

[a,b]\text{.}

We wish to apply Lemma 4.2.2, and so we need to find some

c \in (a,b)

where

f

attains a minimum or a maximum. Write

K \coloneqq f(a) = f(b)\text{.}

If there exists an

x

such that

f(x) > K\text{,}

then the absolute maximum is larger than

K

and hence occurs at some

c \in (a,b)\text{,}

and therefore

f'(c) = 0\text{.}

On the other hand, if there exists an

x

such that

f(x) < K\text{,}

then the absolute minimum occurs at some

c \in (a,b)\text{,}

and so

f'(c) = 0\text{.}

If there is no

x

such that

f(x) > K

or

f(x) < K\text{,}

then

f(x) = K

for all

x

and then

f'(x) = 0

for all

x \in [a,b]\text{,}

so any

c \in (a,b)

works.

L4.2.4: Scalar mean-value theorem by subtraction of the affine secant.

Theorem 4.2.4. Mean value theorem.

Let

f \colon [a,b] \to \R

be a continuous function differentiable on

(a,b)\text{.}

Then there exists a point

c \in (a,b)

such that

\begin{equation*} f(b)-f(a) = f'(c)(b-a) . \end{equation*}

Proof.

Define the function

g \colon [a,b] \to \R

by

\begin{equation*} g(x) \coloneqq f(x)- \left( f(b)+\frac{f(b)-f(a)}{b-a}(x-b) \right) = f(x)- f(b)-\frac{f(b)-f(a)}{b-a}(x-b) . \end{equation*}

The function

g

is differentiable on

(a,b)\text{,}

continuous on

[a,b]\text{,}

such that

g(a) = 0

and

g(b) = 0\text{.}

Thus there exists a

c \in (a,b)

such that

g'(c) = 0\text{,}

that is,

\begin{equation*} 0 = g'(c) = f'(c)-\frac{f(b)-f(a)}{b-a} . \end{equation*}

In other words,

f(b)-f(a) = f'(c)(b-a)\text{.}

L4.2.6: Zero derivative implies constant on an interval.

Proposition 4.2.6.

Let

I

be an interval and let

f \colon I \to \R

be a differentiable function such that

f'(x) = 0

for all

x \in I\text{.}

Then

f

is constant.

Proof.

Take arbitrary

x,y \in I

with

x < y\text{.}

As

I

is an interval,

[x,y] \subset I\text{.}

Then

f

restricted to

[x,y]

satisfies the hypotheses of the mean value theorem. Therefore, there is a

c \in (x,y)

such that

\begin{equation*} f(y)-f(x) = f'(c)(y-x). \end{equation*}

As

f'(c) = 0\text{,}

we have

f(y) = f(x)\text{.}

Hence, the function is constant.