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

. . . . . .
Flowchart proof of Rolle's Theorem
.
.
.
.
Let c be
the max pt
.
.
Let d be
the min pt
.
.
endpoints
are max
and min
.
.
.
is c an
endpoint?
.
.
is d an
endpoint?
.
.
f is
constant
on [a, b]
.
.
f′
(c) = 0 .
.
f′
(d) = 0 .
.
f′
(x) ≡ 0
on (a, b)
.no .no
.yes .yes

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

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

. . . . . .
Rolle vs. MVT
f′
(c) = 0
f(b) − f(a)
b − a
= f′
(c)
. .
.a
.
.b
..c
. .
.a
.
.b
..c

. . . . . .
Rolle vs. MVT
f′
(c) = 0
f(b) − f(a)
b − a
= f′
(c)
. .
.a
.
.b
..c
. .
.a
.
.b
..c
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
y − f(a) =
f(b) − f(a)
b − a
(x − a)

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

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

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

. . . . . .
Proof.
y − f(a) =
f(b) − f(a)
b − a
(x − a)
g(x) = f(x) − f(a) −
f(b) − f(a)
b − a
(x − 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
0 = g′
(c) = f′
(c) −
f(b) − f(a)
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
− x = 100 in the
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
− x = 100 in the
interval [4, 5].
Solution
− x
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
− x = 100 in the
interval [4, 5].
Solution
− x
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)
4 − 1
= f′
(c) < 2
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
f(4) − f(1)
4 − 1
= f′
(c) < 2
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.
. .x
.y
.
.(1, 3)
..(4, 9)
.
.(4, f(4))

. . . . . .
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
(x) = 0 on (a, b).
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
(x) = 0 on (a, b).
Question
If f′
(x) = 0 is f necessarily a constant function?

. . . . . .
Fact
(x) = 0 on (a, b).
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)
y − x
= f′
(z) = 0.
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
= g′
.
Proof.

. . . . . .
Theorem
= g′
.
Proof.
Let h(x) = f(x) − g(x)

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

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

. . . . . .
Theorem
= g′
.
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
f(x) =
{
−x if x ≤ 0
x2
if x ≥ 0
Is f differentiable at 0?

. . . . . .
Example
Let
f(x) =
{
−x if x ≤ 0
x2
if x ≥ 0
Solution (from the definition)
We have
lim
x→0−
f(x) − f(0)
x − 0
= lim
x→0−
−x
x
= −1
lim
x→0+
f(x) − f(0)
x − 0
= lim
x→0+
x2
x
= lim
x→0+
x = 0
Since these limits disagree, f is not differentiable at 0.

. . . . . .
Example
Let
f(x) =
{
−x if x ≤ 0
x2
if x ≥ 0
Solution (Sort of)
If x < 0, then f′
(x) = −1. If x > 0, then f′
(x) = 2x. Since
lim
x→0+
f′
(x) = 0 and lim
x→0−
f′
(x) = −1,
the limit lim
x→0
f′
(x) does not exist and so f is not differentiable at 0.

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

. . . . . .
Differentiable with discontinuous derivative
It is possible for a function f to be differentiable at a even if lim
x→a
f′
(x)
does not exist.
Example
Let f′
(x) =
{
x2
sin(1/x) if x ̸= 0
0 if x = 0
. Then when 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′
(0) = lim
x→0
f(x) − f(0)
x − 0
= lim
x→0
x2 sin(1/x)
x
= lim
x→0
x sin(1/x) = 0
So f′
(0) = 0. Hence f is differentiable for all x, but f′
is not continuous
at 0!

. . . . . .
Differentiability FAIL
. .x
.f(x)
This function is differentiable at
0.
. .x
.f′
(x)
.
But the derivative is not
continuous at 0!

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

. . . . . .
MVT to the rescue
Lemma
Suppose f is continuous on [a, b] and lim
x→a+
f′
(x) = m. Then
lim
x→a+
f(x) − f(a)
x − a
= m.
Proof.
Choose x near a and greater than a. Then
f(x) − f(a)
x − a
= f′
(cx)
for some cx where a < cx < x. As x → a, cx → a as well, so:
lim
x→a+
f(x) − f(a)
x − a
= lim
x→a+
f′
(cx) = lim
x→a+
f′
(x) = m.

. . . . . .
Theorem
Suppose
lim
x→a−
f′
(x) = m1 and lim
x→a+
f′
(x) = m2
If m1 = m2, then f is differentiable at a. If m1 ̸= m2, then f is not

. . . . . .
Theorem
Suppose
lim
x→a−
f′
(x) = m1 and lim
x→a+
f′
(x) = m2
If m1 = m2, then f is differentiable at a. If m1 ̸= m2, then f is not
Proof.
We know by the lemma that
lim
x→a−
f(x) − f(a)
x − a
= lim
x→a−
f′
(x)
lim
x→a+
f(x) − f(a)
x − a
= lim
x→a+
f′
(x)
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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