The Mean Value Theorem is the Most Important Theorem in Calculus. It allows us to relate information about the derivative of a function to information about the function itself.

1. Section 4.2
The Mean Value Theorem
V63.0121, Calculus I
March 25–26, 2009
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Quiz 4 next week: Sections 2.5–3.5
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Image credit: Jimmywayne22
. . . . . .

2. Outline
Review: The Closed Interval Method
Rolle’s Theorem
The Mean Value Theorem
Applications
Why the MVT is the MITC
. . . . . .

3. 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
. . . . . .

4. 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.
. . . . . .

5. Outline
Review: The Closed Interval Method
Rolle’s Theorem
The Mean Value Theorem
Applications
Why the MVT is the MITC
. . . . . .

6. 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
. . . . . .

7. 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
.
. . . . . .

8. Mathematical Statement of Rolle’s Theorem
c
.
.
•
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
.
. . . . . .

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

10. Proof of Rolle’s Theorem
Proof.
By the Extreme Value Theorem f must achieve its maximum
value at a point c in [a, b].
If c is in (a, b), great: it’s a local maximum and so by Fermat’s
Theorem f′ (c) = 0.
. . . . . .

11. Proof of Rolle’s Theorem
Proof.
By the Extreme Value Theorem f must achieve its maximum
value at a point c in [a, b].
If c is in (a, b), great: it’s a local maximum and so by Fermat’s
Theorem f′ (c) = 0.
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].
. . . . . .

12. Proof of Rolle’s Theorem
Proof.
By the Extreme Value Theorem f must achieve its maximum
value at a point c in [a, b].
If c is in (a, b), great: it’s a local maximum and so by Fermat’s
Theorem f′ (c) = 0.
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].
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.
. . . . . .

13. Proof of Rolle’s Theorem
Proof.
By the Extreme Value Theorem f must achieve its maximum
value at a point c in [a, b].
If c is in (a, b), great: it’s a local maximum and so by Fermat’s
Theorem f′ (c) = 0.
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].
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.
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.
. . . . . .

14. Proof of Rolle’s Theorem
Proof.
By the Extreme Value Theorem f must achieve its maximum
value at a point c in [a, b].
If c is in (a, b), great: it’s a local maximum and so by Fermat’s
Theorem f′ (c) = 0.
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].
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.
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. But we already know that f(a) = f(b).
. . . . . .

15. Proof of Rolle’s Theorem
Proof.
By the Extreme Value Theorem f must achieve its maximum
value at a point c in [a, b].
If c is in (a, b), great: it’s a local maximum and so by Fermat’s
Theorem f′ (c) = 0.
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].
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.
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. 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.
. . . . . .

16. 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 d. an .
is c. 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)
. . . . . .

17. Outline
Review: The Closed Interval Method
Rolle’s Theorem
The Mean Value Theorem
Applications
Why the MVT is the MITC
. . . . . .

18. 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
. . . . . .

19. 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
b
.
f(b) − f(a) .
= f′ (c). .
•
a
.
b−a
. . . . . .

20. 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
b
.
f(b) − f(a) .
= f′ (c). .
•
a
.
b−a
. . . . . .

21. The Mean Value Theorem
Theorem (The Mean Value c
.
.
•
Theorem)
Let f be continuous on [a, b] and
differentiable on (a, b). Then
there exists a point c in (a, b)
.
•
such that
b
.
f(b) − f(a) .
= f′ (c). .
•
a
.
b−a
. . . . . .

22. 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
. . . . . .

23. 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
Apply Rolle’s Theorem to the function
f(b) − f(a)
g(x) = f(x) − f(a) − (x − a).
b−a
. . . . . .

24. 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
Apply Rolle’s Theorem to the function
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.
. . . . . .

25. 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
Apply Rolle’s Theorem to the function
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).
. . . . . .

26. 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
Apply Rolle’s Theorem to the function
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
. . . . . .

27. 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].
. . . . . .

28. 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).
. . . . . .

29. 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.
. . . . . .

30. 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.
. . . . . .

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

32. 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
= cos(c)
x−0
for some c in (0, x). Since |cos(c)| ≤ 1, we get
sin x
≤ 1 =⇒ |sin x| ≤ |x|
x
. . . . . .

33. 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.
. . . . . .

34. 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.
. . . . . .

35. Outline
Review: The Closed Interval Method
Rolle’s Theorem
The Mean Value Theorem
Applications
Why the MVT is the MITC
. . . . . .

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

37. 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
. . . . . .

38. 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?
. . . . . .

39. 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
. . . . . .

40. Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f′ = 0 on an interval (a, b).
. . . . . .

41. 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).
. . . . . .

42. 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)
= 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.
. . . . . .

43. 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.
. . . . . .

44. 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)
. . . . . .