Lesson 20: The Mean Value Theorem

Section 4.2
The Mean Value Theorem

V63.0121.027, Calculus I

November 10, 2009

Announcements
Quiz this week on §§3.1–3.5

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.
Image credit: Jimmywayne22
. . . . . .

Outline

Review: The Closed Interval Method

Rolle’s Theorem

Applications

Why the MVT is the MITC

. . . . . .

Flowchart for placing extrema
Thanks to Fermat
Suppose f is a continuous function on the closed, bounded
interval [a, b], and c is a global maximum point.
.
. . c is a
start
local max

. . .
Is c an Is f diff’ble f is not
n
.o n
.o
endpoint? at c? diff at c

y
. es y
. es
. c = a or .
f′ (c) = 0
c = b

. . . . . .

The Closed Interval Method

This means to ﬁnd the maximum value of f on [a, b], we need to:
Evaluate f at the endpoints a and b
Evaluate f at the critical points x where either f′ (x) = 0 or f is
not differentiable at x.
The points with the largest function value are the global
maximum points
The points with the smallest or most negative function value
are the global minimum points.

. . . . . .

Heuristic Motivation for Rolle’s Theorem

If you bike up a hill, then back down, at some point your
elevation was stationary.

.

.
Image credit: SpringSun
. . . . . .

Mathematical Statement of Rolle’s Theorem

Theorem (Rolle’s Theorem)
Let f be continuous on [a, b]
and differentiable on (a, b).
Suppose f(a) = f(b). Then
there exists a point c in (a, b)
such that f′ (c) = 0. . . .
a
. b
.

. . . . . .

Mathematical Statement of Rolle’s Theorem

c
..
Theorem (Rolle’s Theorem)
Suppose f(a) = f(b). Then
there exists a point c in (a, b)
such that f′ (c) = 0. . . .
a
. b
.

. . . . . .

Proof of Rolle’s Theorem
Proof.
By the Extreme Value Theorem f must achieve its maximum
value at a point c in [a, b].

. . . . . .

Proof.
If c is in (a, b), great: it’s a local maximum and so by
Fermat’s Theorem f′ (c) = 0.

. . . . . .

Proof.
On the other hand, if c = a or c = b, try with the minimum.
The minimum of f on [a, b] must be achieved at a point d in
[a, b].

. . . . . .

Proof.
[a, b].
If d is in (a, b), great: it’s a local minimum and so by
Fermat’s Theorem f′ (d) = 0. If not, d = a or d = b.

. . . . . .

Proof.
[a, b].
If we still haven’t found a point in the interior, we have that
the maximum and minimum values of f on [a, b] occur at
both endpoints.

. . . . . .

Proof.
[a, b].
both endpoints. But we already know that f(a) = f(b).

. . . . . .

Proof.
[a, b].
both endpoints. But we already know that f(a) = f(b). If
these are the maximum and minimum values, f is constant
on [a, b] and any point x in (a, b) will have f′ (x) = 0.
. . . . . .

Flowchart proof of Rolle’s Theorem

. . .
Let c be Let d be endpoints
. . .
are max
the max pt the min pt
and min

.
. . f is
is c. an
y
. es is d. an. y
. es .
constant
endpoint? endpoint?
on [a, b]

n
.o n
.o
.
. . f′ (x) .≡ 0
. f (c) .= 0
′
f (d) .= 0
′
on (a, b)

. . . . . .

Heuristic Motivation for The Mean Value Theorem
If you drive between points A and B, at some time your
speedometer reading was the same as your average speed over
the drive.

.
.
Image credit: ClintJCL
. . . . . .


Theorem (The Mean Value
Theorem)
Then there exists a point c in
(a, b) such that .
b
.
f(b) − f(a) . .
= f′ (c). a
.
b−a

. . . . . .


Theorem (The Mean Value
Theorem) c
.
Then there exists a point c in
(a, b) such that .
b
.
f(b) − f(a) . .
= f′ (c). a
.
b−a

. . . . . .

Rolle vs. MVT

f(b) − f(a)
f′ (c) = 0 = f′ (c)
b−a

c
.. c
..

.
b
.
. . . . .
a
. b
. a
.

. . . . . .

Rolle vs. MVT

f(b) − f(a)
f′ (c) = 0 = f′ (c)
b−a

c
.. c
..

.
b
.
. . . . .
a
. b
. a
.

If the x-axis is skewed the pictures look the same.

. . . . . .

Proof of the Mean Value Theorem
Proof.
The line connecting (a, f(a)) and (b, f(b)) has equation

f(b) − f(a)
y − f(a) = (x − a)
b−a

. . . . . .

Proof.

f(b) − f(a)
y − f(a) = (x − a)
b−a
Apply Rolle’s Theorem to the function

f(b) − f(a)
g(x) = f(x) − f(a) − (x − a).
b−a

. . . . . .

Proof.

f(b) − f(a)
y − f(a) = (x − a)
b−a

f(b) − f(a)
g(x) = f(x) − f(a) − (x − a).
b−a
Then g is continuous on [a, b] and differentiable on (a, b) since f
is.

. . . . . .

Proof.

f(b) − f(a)
y − f(a) = (x − a)
b−a

f(b) − f(a)
g(x) = f(x) − f(a) − (x − a).
b−a
is. Also g(a) = 0 and g(b) = 0 (check both)

. . . . . .

Proof.

f(b) − f(a)
y − f(a) = (x − a)
b−a

f(b) − f(a)
g(x) = f(x) − f(a) − (x − a).
b−a
is. Also g(a) = 0 and g(b) = 0 (check both) So by Rolle’s
Theorem there exists a point c in (a, b) such that

f(b) − f(a)
0 = g′ (c) = f′ (c) − .
b−a

. . . . . .

Using the MVT to count solutions

Example
Show that there is a unique solution to the equation x3 − x = 100
in the interval [4, 5].

. . . . . .


Example

Solution
By the Intermediate Value Theorem, the function
f(x) = x3 − x must take the value 100 at some point on c in
(4, 5).

. . . . . .


Example

Solution
(4, 5).
If there were two points c1 and c2 with f(c1 ) = f(c2 ) = 100,
then somewhere between them would be a point c3
between them with f′ (c3 ) = 0.

. . . . . .


Example

Solution
(4, 5).
If there were two points c1 and c2 with f(c1 ) = f(c2 ) = 100,
then somewhere between them would be a point c3
between them with f′ (c3 ) = 0.
However, f′ (x) = 3x2 − 1, which is positive all along (4, 5).
So this is impossible.

. . . . . .

Example
We know that |sin x| ≤ 1 for all x. Show that |sin x| ≤ |x|.

. . . . . .

Example
Let f be a differentiable function with f(1) = 3 and f′ (x) < 2 for
all x in [0, 5]. Could f(4) ≥ 9?

. . . . . .

Example
Let f be a differentiable function with f(1) = 3 and f′ (x) < 2 for
all x in [0, 5]. Could f(4) ≥ 9?

Solution . 4, 9 )
(
y
. .
By MVT

f(4) − f(1) . 4, f(4))
(
= f′ (c) < 2 .
4−1
for some c in (1, 4). Therefore
. 1, 3 )
(
.
f(4) = f(1) + f′ (c)(3) < 3 + 2 · 3 = 9.

So no, it is impossible that f(4) ≥ 9.
. x
.

. . . . . .

Question
A driver travels along the New Jersey Turnpike using EZ-Pass. The
system takes note of the time and place the driver enters and exits
the Turnpike. A week after his trip, the driver gets a speeding
ticket in the mail. Which of the following best describes the
situation?
(a) EZ-Pass cannot prove that the driver was speeding
(b) EZ-Pass can prove that the driver was speeding
(c) The driver’s actual maximum speed exceeds his ticketed
speed
(d) Both (b) and (c).
Be prepared to justify your answer.

. . . . . .

Fact
If f is constant on (a, b), then f′ (x) = 0 on (a, b).

. . . . . .

Fact

The limit of difference quotients must be 0
The tangent line to a line is that line, and a constant
function’s graph is a horizontal line, which has slope 0.
Implied by the power rule since c = cx0

. . . . . .

Fact


Question
If f′ (x) = 0 is f necessarily a constant function?

. . . . . .

Fact


Question
If f′ (x) = 0 is f necessarily a constant function?

It seems true
But so far no theorem (that we have proven) uses information
about the derivative of a function to determine information
about the function itself

. . . . . .

Most Important Theorem In Calculus!

Theorem
Let f′ = 0 on an interval (a, b).

. . . . . .


Theorem
Let f′ = 0 on an interval (a, b). Then f is constant on (a, b).

. . . . . .


Theorem
Let f′ = 0 on an interval (a, b). Then f is constant on (a, b).

Proof.
Pick any points x and y in (a, b) with x < y. Then f is continuous
on [x, y] and differentiable on (x, y). By MVT there exists a point
z in (x, y) such that

f(y) − f(x)
= f′ (z) = 0.
y−x

So f(y) = f(x). Since this is true for all x and y in (a, b), then f is
constant.

. . . . . .

Theorem
Suppose f and g are two differentiable functions on (a, b) with
f′ = g′ . Then f and g differ by a constant. That is, there exists a
constant C such that f(x) = g(x) + C.

. . . . . .

Theorem
Suppose f and g are two differentiable functions on (a, b) with
f′ = g′ . Then f and g differ by a constant. That is, there exists a
constant C such that f(x) = g(x) + C.

Proof.
Let h(x) = f(x) − g(x)
Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b)
So h(x) = C, a constant
This means f(x) − g(x) = C on (a, b)

. . . . . .

MVT and differentiability

Example
Let {
−x if x ≤ 0
f(x) =
x2 if x ≥ 0
Is f differentiable at 0?

. . . . . .


Example
Let {
−x if x ≤ 0
f(x) =
x2 if x ≥ 0

Solution (from the deﬁnition)
We have
f(x) − f(0) −x
lim = lim = −1
x→0− x−0 x→0− x
f(x) − f(0) x2
lim = lim+ = lim+ x = 0
x→0+ x−0 x→0 x x→0

Since these limits disagree, f is not differentiable at 0.
. . . . . .


Example
Let {
−x if x ≤ 0
f(x) =
x2 if x ≥ 0

Solution (Sort of)
If x < 0, then f′ (x) = −1. If x > 0, then f′ (x) = 2x. Since

lim f′ (x) = 0 and lim f′ (x) = −1,
x→0+ x→0−

the limit lim f′ (x) does not exist and so f is not differentiable at 0.
x→0

. . . . . .

This solution is valid but less direct.
We seem to be using the following fact: If lim f′ (x) does not
x→a
exist, then f is not differentiable at a.
equivalently: If f is differentiable at a, then lim f′ (x) exists.
x→a
But this “fact” is not true!

. . . . . .

Differentiable with discontinuous derivative

It is possible for a function f to be differentiable at a even if
lim f′ (x) does not exist.
x→a

Example {
x2 sin(1/x) if x ̸= 0
Let f′ (x) = . Then when x ̸= 0,
0 if x = 0

f′ (x) = 2x sin(1/x) + x2 cos(1/x)(−1/x2 ) = 2x sin(1/x) − cos(1/x),

which has no limit at 0. However,

f(x) − f(0) x2 sin(1/x)
f′ (0) = lim = lim = lim x sin(1/x) = 0
x→0 x−0 x→0 x x→0

So f′ (0) = 0. Hence f is differentiable for all x, but f′ is not
continuous at 0!

. . . . . .

MVT to the rescue
Lemma
Suppose f is continuous on [a, b] and lim f′ (x) = m. Then
x→a+

f(x) − f(a)
lim = m.
x→a+ x−a

. . . . . .

MVT to the rescue
Lemma
Suppose f is continuous on [a, b] and lim f′ (x) = m. Then
x→a+

f(x) − f(a)
lim = m.
x→a+ x−a

Proof.
Choose x near a and greater than a. Then

f(x) − f(a)
= f ′ (c x )
x−a
for some cx where a < cx < x. As x → a, cx → a as well, so:

f(x) − f(a)
lim = lim f′ (cx ) = lim f′ (x) = m.
x→a+ x−a x→a+ x→a+

. . . . . .

Theorem
Suppose
lim f′ (x) = m1 and lim f′ (x) = m2
x→a− x→a+

If m1 = m2 , then f is differentiable at a. If m1 ̸= m2 , then f is not
differentiable at a.

. . . . . .

Theorem
Suppose
lim f′ (x) = m1 and lim f′ (x) = m2
x→a− x→a+

If m1 = m2 , then f is differentiable at a. If m1 ̸= m2 , then f is not
differentiable at a.

Proof.
We know by the lemma that

f(x) − f(a)
lim = lim f′ (x)
x→a− x−a x→a−
f(x) − f(a)
lim = lim f′ (x)
x→a+ x−a x→a+

The two-sided limit exists if (and only if) the two right-hand sides
agree.

. . . . . .

What have we learned today?

Rolle’s Theorem: there is a stationary point
Mean Value Theorem: at some point the instantaneous rate
of change equals the average rate of change (The Most
Important Theorem in Calculus)
Only constant functions have a derivative of zero.

. . . . . .

Lesson 20: The Mean Value Theorem

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