Lesson 19: The Mean Value Theorem (Section 021 slides)

Section 4.2
The Mean Value Theorem

V63.0121.021, Calculus I

New York University

November 11, 2010

Announcements
Quiz 4 next week (November 16, 18, 19) on Sections 3.3, 3.4, 3.5,
3.7

. . . . . .

Announcements

Quiz 4 next week
(November 16, 18, 19) on
Sections 3.3, 3.4, 3.5, 3.7

. . . . . .

V63.0121.021, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 11, 2010 2 / 29

Objectives

Understand and be able to
explain the statement of
Rolle’s Theorem.
Understand and be able to
explain the statement of
the Mean Value Theorem.

. . . . . .


Outline

Rolle’s Theorem

Applications

Why the MVT is the MITC
Functions with derivatives that are zero
MVT and differentiability

. . . . . .


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

Theorem (Rolle’s Theorem)

c
..

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

. . . . . .


Flowchart proof of Rolle's Theorem

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

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

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


Outline

Rolle’s Theorem

Applications


. . . . . .


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

. . . . . .



Theorem (The Mean Value Theorem)

c
.
Then there exists a point c
in (a, b) such that
.
f(b) − f(a) b
.
= f′ (c).
b−a . .
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
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
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
interval [4, 5].

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
interval [4, 5].

Solution

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
interval [4, 5].

Solution

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.

. . . . . .


Using the MVT to estimate II
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
[0, 5]. Could f(4) ≥ 9?

Solution

By MVT

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

f(4) = f(1) + f′ (c)(3) < 3 + 2 · 3 = 9.

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


Example
[0, 5]. Could f(4) ≥ 9?

Solution

By MVT
y
. . 4, 9) .
(

f(4) − f(1) . 4, f(4))
(
= f′ (c) < 2 .
4−1
for some c in (1, 4). Therefore
.
f(4) = f(1) + f′ (c)(3) < 3 + 2 · 3 = 9.
. 1, 3)
(
So no, it is impossible that f(4) ≥ 9.
. x
.
. . . . . .


Food for Thought

Question
A driver travels along the New Jersey Turnpike using E-ZPass. 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) E-ZPass cannot prove that the driver was speeding
(b) E-ZPass can prove that the driver was speeding
(c) The driver’s actual maximum speed exceeds his ticketed speed
(d) Both (b) and (c).

. . . . . .


Outline

Rolle’s Theorem

Applications


. . . . . .


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.

. . . . . .


Functions with the same derivative

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

Proof.

. . . . . .



Theorem

Proof.

Let h(x) = f(x) − g(x)

. . . . . .



Theorem

Proof.

Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b)

. . . . . .



Theorem

Proof.

Then h′ (x) = f′ (x) − g′ (x) = 0 on (a, b)
So h(x) = C, a constant

. . . . . .



Theorem

Proof.

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)

. . . . . .


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

. . . . . .


Why only “sort of"?

.′ (x)
f
This solution is valid but y
. f
.(x)
less direct.
We seem to be using the
following fact: If lim f′ (x)
x→a
does not exist, then f is not
. x
.
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)
x→a
does not exist.
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!
. . . . . .


Differentiability FAIL

f
.(x) .′ (x)
f

. x
. . x
.

This function is differentiable at But the derivative is not
0. 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′ (cx )
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

. . . . . .


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

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.
. . . . . .


Summary

Rolle’s Theorem: under suitable conditions, functions must have
critical points.
Mean Value Theorem: under suitable conditions, functions must
have an instantaneous rate of change equal to the average rate of
change.
A function whose derivative is identically zero on an interval must
be constant on that interval.
E-ZPass is kinder than we realized.

. . . . . .


Lesson 19: The Mean Value Theorem (Section 021 slides)

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