Lesson 19: The Mean Value Theorem

Section 4.2
The Mean Value Theorem

V63.0121.002.2010Su, Calculus I

New York University

June 8, 2010

Announcements
Exams not graded yet
Assignment 4 is on the website
Quiz 3 on Thursday covering 3.3, 3.4, 3.5, 3.7

Announcements

Exams not graded yet
Assignment 4 is on the
website
Quiz 3 on Thursday covering
3.3, 3.4, 3.5, 3.7

V63.0121.002.2010Su, Calculus I (NYU) Section 4.2 The Mean Value Theorem June 8, 2010 2 / 28

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
yes yes constant
endpoint? endpoint?
on [a, b]

no no

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
Then g is continuous on [a, b] and diﬀerentiable on (a, b) since f 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
Then g is continuous on [a, b] and diﬀerentiable on (a, b) since f 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 x 3 − x = 100 in the
interval [4, 5].



Example
interval [4, 5].

Solution

By the Intermediate Value Theorem, the function f (x) = x 3 − 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) = 3x 2 − 1, which is positive all along (4, 5). So this is
impossible.


Using the MVT to estimate

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


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?


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?

Solution
y (4, 9)
By MVT

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 = cx 0


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 .


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 .

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)


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”?

f (x)
This solution is valid but less y f (x)
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
x 2 sin(1/x) if x = 0
Let f (x) = . Then when x = 0,
0 if x = 0

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

which has no limit at 0. However,

f (x) − f (0) x 2 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) f (x)

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

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