The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.
Lesson 20: Derivatives and the Shapes of Curves (handout)
Lesson 19: The Mean Value Theorem (Section 041 handout)
1. Section 4.2
The Mean Value Theorem
V63.0121.041, Calculus I
New York University
November 10, 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
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 3 / 29
Notes
Notes
Notes
1
Section 4.2 : The Mean Value TheoremV63.0121.041, Calculus I November 10, 2010
2. Outline
Rolle’s Theorem
The Mean Value Theorem
Applications
Why the MVT is the MITC
Functions with derivatives that are zero
MVT and differentiability
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 4 / 29
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 5 / 29
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
c
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 6 / 29
Notes
Notes
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2
Section 4.2 : The Mean Value TheoremV63.0121.041, Calculus I November 10, 2010
3. 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
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 8 / 29
Outline
Rolle’s Theorem
The Mean Value Theorem
Applications
Why the MVT is the MITC
Functions with derivatives that are zero
MVT and differentiability
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 9 / 29
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 10 / 29
Notes
Notes
Notes
3
Section 4.2 : The Mean Value TheoremV63.0121.041, Calculus I November 10, 2010
4. The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on [a, b]
and differentiable on (a, b).
Then there exists a point c in
(a, b) such that
f (b) − f (a)
b − a
= f (c).
a
b
c
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 11 / 29
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 12 / 29
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)
Apply Rolle’s Theorem to the function
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. 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
.
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 13 / 29
Notes
Notes
Notes
4
Section 4.2 : The Mean Value TheoremV63.0121.041, Calculus I November 10, 2010
5. 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].
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).
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 14 / 29
Using the MVT to estimate
Example
We know that |sin x| ≤ 1 for all x. Show that |sin x| ≤ |x|.
Solution
Apply the MVT to the function f (t) = sin t on [0, x]. We get
sin x − sin 0
x − 0
= cos(c)
for some c in (0, x). Since |cos(c)| ≤ 1, we get
sin x
x
≤ 1 =⇒ |sin x| ≤ |x|
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 15 / 29
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 16 / 29
Notes
Notes
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Section 4.2 : The Mean Value TheoremV63.0121.041, Calculus I November 10, 2010
6. 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).
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 17 / 29
Outline
Rolle’s Theorem
The Mean Value Theorem
Applications
Why the MVT is the MITC
Functions with derivatives that are zero
MVT and differentiability
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 18 / 29
Functions with derivatives that are zero
Fact
If f is constant on (a, b), then f (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
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 19 / 29
Notes
Notes
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Section 4.2 : The Mean Value TheoremV63.0121.041, Calculus I November 10, 2010
7. Why the MVT is the MITC
Most Important Theorem In Calculus!
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 20 / 29
Functions with the same derivative
Theorem
Suppose f and g are two differentiable functions on (a, b) with f = g .
Proof.
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 21 / 29
MVT and differentiability
Example
Let
f (x) =
−x if x ≤ 0
x2
if x ≥ 0
Is f differentiable at 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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 22 / 29
Notes
Notes
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Section 4.2 : The Mean Value TheoremV63.0121.041, Calculus I November 10, 2010
8. MVT and differentiability
Example
Let
f (x) =
−x if x ≤ 0
x2
if x ≥ 0
Is f differentiable at 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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 23 / 29
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)
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 24 / 29
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!
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 25 / 29
Notes
Notes
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Section 4.2 : The Mean Value TheoremV63.0121.041, Calculus I November 10, 2010
9. Differentiability FAIL
x
f (x)
This function is differentiable at
0.
x
f (x)
But the derivative is not
continuous at 0!
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 26 / 29
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 27 / 29
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
differentiable at a.
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 28 / 29
Notes
Notes
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Section 4.2 : The Mean Value TheoremV63.0121.041, Calculus I November 10, 2010
10. 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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Mean Value Theorem November 10, 2010 29 / 29
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Section 4.2 : The Mean Value TheoremV63.0121.041, Calculus I November 10, 2010