For a given

continuous function on a

closed interval, if there is a point in this closed interval where the

derivative of

f(x) = 0, then this point is either the low point (minima) or high point (maxima) for the continuous function in this interval.

This very basic idea of calculus requires a few lemmas before we are able to prove it.

If you are not familiar with the concept of a

function,

continuous function, or a

closed interval, start

here.

A

function is said to be

bounded if there is a value

L such that for all

x ∈ [a,b], f(x) ≤ L.

Lemma 1: Nested Interval Property for the real numbersSuppose that

I_{1}, I_{2}, ..., I_{n} is a sequence of nested, closed intervals where:

(a) Each

I_{i+1} is contained with

I_{i}(b) Each

I_{i} interval is of the form

[a_{i},b_{i}](c) The

lim (i → inf) (b_{i} - a_{i}) = 0 (See

here for review of

lim notation and the concept of limit)

Then:

There exists

1 and only

1 point

c such that

{c} = I_{1} ∩ I_{2} ∩ ... ∩ I_{n}Proof:

(1) We know that there can be at most

1 number since

lim(i → inf) (b_{i} - a_{i}) = 0 since:

If there were more than

1 number, the

lim(i → inf) (b_{i} - a_{i}) would be greater than

0.

The only way that it can be

0 is if

b_{i} = c, a_{i} = c, and

c-c=0.

(2) We know that there is at least

1 number that is common to all intervals since:

(a)

a_{i} has a limit

a_{n} that fits somewhere in the interval. For all intervals, it is clear that

a_{i} ≤ a_{n} ≤ b

_{i}(b) Likewise,

b_{i} has a limit

b_{n} such that

a_{i} ≤ b_{n} ≤ b_{i}(c) So we see that

a_{n},b_{n} are both elements of all intervals.

(d) We further note that

a_{n} = b_{n} since the

lim(i → inf)(b_{i} - a_{i})= 0.

(e) So, if we let

c = a_{n} = b_{n}, then we are done.

QED

Lemma 2: If a function f is continuous on a closed interval [a,b], then f is bounded there.(1) Assume that a function

f is not bounded on

[a,b](2) We can bisect the interval

[a,b] into two halves which I will label

I_{1} and

I_{2}.

(3) We can now pick an interval

I which is unbounded. If both intervals are unbounded, then we can pick either one.

(4) We can repeat this process and create a sequence of nested, closed intervals which we can call

I_{i} where each interval selects a subset which is unbounded.

(5) From Lemma 1, we know that there exists a point

c which is common to all the intervals in #4.

(6) Because

f is continuous, we know that there is a number

ε such that

f is bounded on the interval

c - ε and

c + ε [See

here for definition of Continuous Functions]

(7) But one of the unbounded values in

I_{n} must lie within

(c - ε, c + ε)(8) And this is a contradiction since from (#6), it must be bounded.

(9) Therefore, we reject our assumption.

QED

Lemma 3: Maximum value property of continuous functionsIf a function

f is continuous on the closed interval

[a,b], then there exists a number

c in

[a,b] such that

f(x) ≤ f(c) for all

x in

[a,b](1) Let

I be the the closed interval

[a,b](2) From Lemma 2, we know that

I is bounded.

(3) Let

λ be its least upper bound.

(4) We can divide

I in half.

(5) At least one of these halves will have a least upper bound

= λ (although it is possible that both have this least upper bound). Let

I_{1} be the division of

I which contains

λ as the least upper bound.

(6) We can keep dividing up

I_{1} in the same way until we have

I_{n} which has

λ as its upper bound and

b_{n} - a_{n} = 0.

(7) From Lemma 1, we know that there exists a point

c which is common to all these intervals.

(8) It follows from (6) that

f(c)=λ since:

(a) There exists a positive value

δ such that if

x-c is in between

-δ and

+δ, then

f(x)-f(c) is between

-ε and

ε. [From the definition of a continuous function, see

here]

(b) From (a), we have that

f(c) - ε is less than

f(x) which is less than

f(c) + ε(c) Since

ε can be arbitrarily small, we can have

f(c) ≤ f(x) ≤ f(c) which means that

f(x)=f(c) at some point.

(d) In this case,

f(c) cannot be more than

λ since

λ is an upper bound. [See

here for the definition of an upper bound]

(e) Likewise,

f(c) cannot be less than

λ since

λ is the least upper bound. [See

here for the definition of a least upper bound]

(f) Therefore,

f(c) = λQED

Definition 1: Right Hand Limit: lim (x → a+) f(x)lim (x → a+) f(x) = L if and only if:

if

x is between

a and

a + δ, then

f(x) - f(a) is between

-ε and

εDefinition 2: Left Hand Limit: lim(x → a-) f(x)lim(x → a-) f(x) = L if and only if:

if

x is between

a - δ and

a, then

f(x) - f(a) is between

-ε and

εLemma 4: One-sided and two-sided limits

The limit

lim (x → a) for

f(x) exists and is equal to the number

L if and only if the one-sided liimits

lim (x → a+) f(x) and

lim (x → a-) f(x) both exist and both are equal to the number

L.

Proof:

(1) Assume

lim(x → a) f(x) = L(2) Then if

x - a is between

δ and

- δ, then

f(x) - f(a) is between

-ε and

+ε [ By the definition of continuous functions, see

here]

(3) Now

x - a is between

δ and

-δ implies that:

x is between

a - δ and

a + δ.

(4) Since

a is greater than

a - δ (since

δ is a positive value), this implies that:

if

x is between

a and

a + δ, then

x is necessarily between

a - δ and

a + δ.

(5) (#4) combined with (#2) gives us:

lim (x → a+) f(x) = L. [See definition 1 above]

(6) Since

a is less than

a + δ, this implies that:

if

x is between

a - δ and

a, then

x is between

a - δ and

a + δ.

(7) (#6) combined with (#2) gives us:

lim (x → a-) f(x)= L. [See definition 2 above]

(8) So, that proves the first half of the above theorem.

(9) Assume that

lim (x → a+) f(x) = L and

lim(x → a-) f(x) = L(10) From definition 1, we have if

x is between

a and

a + δ, then

f(x) - f(a) is between

-ε and

ε(11) From definition 2, we have if

x is between

a-δ and

a, then

f(x) - f(a) is between

-ε and

ε(12) Combining (10) and (11) gives us:

if

x is between

a-δ and

a+δ then

f(x) - f(a) is between

-ε and

ε [For either condition (10) applies or condition (11) applies]

(13) Then, applying the definition of limit (see

here), we get:

lim(x → a) f(x) = LQED

Clarification: Local Maxima and MinimaA

maximum or

minimum is

local if it is true for a given interval. A

maximum or

minimum is

absolute if it is true across a domain. A maximum is value that is greater than all other points in for the range in question. A minimum is a value that is smaller or equal to all other points in the range in question.

Theorem: Local MaximaIf a function

f(x) is differentiable at

c and is defined as an open interval containing

c and if

f(c) is a local maximum value of

f(x), then

f'(c)=0.

Proof:

(1) Assume that

f(c) is a local maximum value for

f(x) on the open interval

(a,b).(2) Since

c is differentiable, it means that right-hand and left-hand limits both exist and are equal to

f'(c). [See Lemma 4 above]

lim (Δx → 0+) [f(c + Δx) - f(c)]/Δx = f'(c)lim (Δx → 0-) [f(c + Δx) - f(c)]/Δx = f'(c)(3) If

Δx is greater than

0, then:

[f(c + Δx) - f(c)]/Δ x ≤ 0This is true since

f(c) ≥ f(c +Δx) for all small positive values of

Δx since

f(c) is a local maximum by assumption.

(4) If

Δx is less than

0, then:

[f(c + Δx) - f(c)]/Δ x ≥ 0This is true since in this case, we have a negative value

f(c + Δx) - f(c) over another negative value

Δx.

(5) So, we have from (#2):

f'(c) = lim(Δx → 0+) [f(c + Δx) - f(c)]/Δ x ≤ 0(6) But we also have from (#2):

f'(c) = lim(Δx → 0-) [f(c + Δx) - f(c)]/Δx ≥ 0(7) Combining (#5) and (#6), we can conclude that

f'(c) = 0 [since

f'(c) ≥ 0 and

f'(c) ≤ 0 ]

QED

Corollary: Local MinimaIf a function

f(x) is differentiable at

c and is defined as an open interval containing

c and if

f(c) is a local minimum value of

f(x), then

f'(c)=0.

Proof:

(1) Assume that

f(c) is a local minimum value for

f(x) on the open interval

(a,b).

(2) Since

c is differentiable, it means that right-hand and left-hand limits both exist and are equal to

f'(c). [See Lemma 4 above]

lim (Δx → 0+) [f(c + Δx) - f(c)]/Δx = f'(c)lim (Δx → 0-) [f(c + Δx) - f(c)]/Δx = f'(c)(3) If

Δx is greater than

0, then:

[f(c + Δx) - f(c)]/Δ x ≥ 0This is true since

f(c) ≤ f(c +Δx) for all small positive values of

Δx since

f(c) is a local minimum by assumption.

(4) If

Δx is less than

0, then:

[f(c + Δx) - f(c)]/Δ x ≤ 0This is true since in this case, we have a positive value

f(c + Δx) - f(c) over a negative value

Δ x.

(5) So, we have from (#2):

f'(c) = lim(Δx → 0+) [f(c + Δx) - f(c)]/Δ x ≥ 0(6) But we also have from (#2):

f'(c) = lim(Δx → 0-) [f(c + Δx) - f(c)]/Δx ≤ 0(7) Combining (#5) and (#6), we can conclude that

f'(c) = 0 [since

f'(c) ≥ 0 and

f'(c) ≤ 0 ]

QED

References