Lesson 20: The Mean Value Theorem

The Mean Value Theorem is the most important theorem in calculus! It allows us to infer information about a function from information about its derivative. Such as: a function whose derivative is zero must be a constant function.

1. . . . . . .
Section 4.2
The Mean Value Theorem
V63.0121.027, Calculus I
November 10, 2009
Announcements
IQuiz this week on §§3.1–3.5
.
.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
Supposefis a continuous function on the closed, bounded
interval[a;b], andcis a global maximum point.
.
.
start
.
Iscan
endpoint?
.
c=aor
c=b
.
cis a
local max
.
Isfdiff’ble
atc?
.
fis not
diff atc
.
f
′
(c) =0
.no.yes.no.yes

4. . . . . . .
The Closed Interval Method
This means to find the maximum value offon[a;b], we need to:
IEvaluatefat theendpointsaandb
IEvaluatefat thecritical pointsxwhere eitherf
′
(x) =0 orfis
not differentiable atx.
IThe points with the largest function value are the global
maximum points
IThe 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..c

8. . . . . . .
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

9. . . . . . .
Proof of Rolle’s Theorem
Proof.
IBy the Extreme Value Theoremfmust achieve its maximum
value at a pointcin[a;b].
IIfcis in(a;b), great: it’s a local maximum and so by
Fermat’s Theoremf
′
(c) =0.
IOn the other hand, ifc=aorc=b, try with the minimum.
The minimum offon[a;b]must be achieved at a pointdin
[a;b].
IIfdis in(a;b), great: it’s a local minimum and so by
Fermat’s Theoremf
′
(d) =0. If not,d=aord=b.
IIf we still haven’t found a point in the interior, we have that
the maximum and minimum values offon[a;b]occur at
both endpoints.But we already know thatf(a) =f(b).If
these are the maximum and minimum values,fisconstant
on[a;b]and any pointxin(a;b)will havef
′
(x) =0.

10. . . . . . .
Proof of Rolle’s Theorem
Proof.
IBy the Extreme Value Theoremfmust achieve its maximum
value at a pointcin[a;b].
IIfcis in(a;b), great: it’s a local maximum and so by
Fermat’s Theoremf
′
(c) =0.
IOn the other hand, ifc=aorc=b, try with the minimum.
The minimum offon[a;b]must be achieved at a pointdin
[a;b].
IIfdis in(a;b), great: it’s a local minimum and so by
Fermat’s Theoremf
′
(d) =0. If not,d=aord=b.
IIf we still haven’t found a point in the interior, we have that
the maximum and minimum values offon[a;b]occur at
both endpoints.But we already know thatf(a) =f(b).If
these are the maximum and minimum values,fisconstant
on[a;b]and any pointxin(a;b)will havef
′
(x) =0.

11. . . . . . .
Proof of Rolle’s Theorem
Proof.
IBy the Extreme Value Theoremfmust achieve its maximum
value at a pointcin[a;b].
IIfcis in(a;b), great: it’s a local maximum and so by
Fermat’s Theoremf
′
(c) =0.
IOn the other hand, ifc=aorc=b, try with the minimum.
The minimum offon[a;b]must be achieved at a pointdin
[a;b].
IIfdis in(a;b), great: it’s a local minimum and so by
Fermat’s Theoremf
′
(d) =0. If not,d=aord=b.
IIf we still haven’t found a point in the interior, we have that
the maximum and minimum values offon[a;b]occur at
both endpoints.But we already know thatf(a) =f(b).If
these are the maximum and minimum values,fisconstant
on[a;b]and any pointxin(a;b)will havef
′
(x) =0.

12. . . . . . .
Proof of Rolle’s Theorem
Proof.
IBy the Extreme Value Theoremfmust achieve its maximum
value at a pointcin[a;b].
IIfcis in(a;b), great: it’s a local maximum and so by
Fermat’s Theoremf
′
(c) =0.
IOn the other hand, ifc=aorc=b, try with the minimum.
The minimum offon[a;b]must be achieved at a pointdin
[a;b].
IIfdis in(a;b), great: it’s a local minimum and so by
Fermat’s Theoremf
′
(d) =0. If not,d=aord=b.
IIf we still haven’t found a point in the interior, we have that
the maximum and minimum values offon[a;b]occur at
both endpoints.But we already know thatf(a) =f(b).If
these are the maximum and minimum values,fisconstant
on[a;b]and any pointxin(a;b)will havef
′
(x) =0.

13. . . . . . .
Proof of Rolle’s Theorem
Proof.
IBy the Extreme Value Theoremfmust achieve its maximum
value at a pointcin[a;b].
IIfcis in(a;b), great: it’s a local maximum and so by
Fermat’s Theoremf
′
(c) =0.
IOn the other hand, ifc=aorc=b, try with the minimum.
The minimum offon[a;b]must be achieved at a pointdin
[a;b].
IIfdis in(a;b), great: it’s a local minimum and so by
Fermat’s Theoremf
′
(d) =0. If not,d=aord=b.
IIf we still haven’t found a point in the interior, we have that
the maximum and minimum values offon[a;b]occur at
both endpoints.
But we already know thatf(a) =f(b).If
these are the maximum and minimum values,fisconstant
on[a;b]and any pointxin(a;b)will havef
′
(x) =0.

14. . . . . . .
Proof of Rolle’s Theorem
Proof.
IBy the Extreme Value Theoremfmust achieve its maximum
value at a pointcin[a;b].
IIfcis in(a;b), great: it’s a local maximum and so by
Fermat’s Theoremf
′
(c) =0.
IOn the other hand, ifc=aorc=b, try with the minimum.
The minimum offon[a;b]must be achieved at a pointdin
[a;b].
IIfdis in(a;b), great: it’s a local minimum and so by
Fermat’s Theoremf
′
(d) =0. If not,d=aord=b.
IIf we still haven’t found a point in the interior, we have that
the maximum and minimum values offon[a;b]occur at
both endpoints.But we already know thatf(a) =f(b).
If
these are the maximum and minimum values,fisconstant
on[a;b]and any pointxin(a;b)will havef
′
(x) =0.

15. . . . . . .
Proof of Rolle’s Theorem
Proof.
IBy the Extreme Value Theoremfmust achieve its maximum
value at a pointcin[a;b].
IIfcis in(a;b), great: it’s a local maximum and so by
Fermat’s Theoremf
′
(c) =0.
IOn the other hand, ifc=aorc=b, try with the minimum.
The minimum offon[a;b]must be achieved at a pointdin
[a;b].
IIfdis in(a;b), great: it’s a local minimum and so by
Fermat’s Theoremf
′
(d) =0. If not,d=aord=b.
IIf we still haven’t found a point in the interior, we have that
the maximum and minimum values offon[a;b]occur at
both endpoints.But we already know thatf(a) =f(b).If
these are the maximum and minimum values,fisconstant
on[a;b]and any pointxin(a;b)will havef
′
(x) =0.

16. . . . . . .
Flowchart proof of Rolle’s Theorem
.
.
..
Letcbe
the max pt
..
Letdbe
the min pt
..
endpoints
are max
andmin
.
..
iscan
endpoint?
..
isdan
endpoint?
..
fis
constant
on[a;b]
..
f
′
(c) =0
..
f
′
(d) =0
..
f
′
(x)≡0
on(a;b)
.no.no.yes.yes

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 pointsAandB, 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
f(b)−f(a)
b−a
=f
′
(c):
.
..a..b.c

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
f(b)−f(a)
b−a
=f
′
(c):
.
..a..b.c

21. . . . . . .
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

22. . . . . . .
Rolle vs. MVT
f
′
(c) =0
f(b)−f(a)
b−a
=f
′
(c)
.
..a..b..c
.
..a..b..c
If thex-axis is skewed the pictures look the same.

23. . . . . . .
Rolle vs. MVT
f
′
(c) =0
f(b)−f(a)
b−a
=f
′
(c)
.
..a..b..c
.
..a..b..c
If thex-axis is skewed the pictures look the same.

24. . . . . . .
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):
Thengis continuous on[a;b]and differentiable on(a;b)sincef
is.Alsog(a) =0 andg(b) =0 (check both)So by Rolle’s
Theorem there exists a pointcin(a;b)such that
0=g
′
(c) =f
′
(c)−
f(b)−f(a)
b−a
:

25. . . . . . .
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):
Thengis continuous on[a;b]and differentiable on(a;b)sincef
is.Alsog(a) =0 andg(b) =0 (check both)So by Rolle’s
Theorem there exists a pointcin(a;b)such that
0=g
′
(c) =f
′
(c)−
f(b)−f(a)
b−a
:

26. . . . . . .
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):
Thengis continuous on[a;b]and differentiable on(a;b)sincef
is.
Alsog(a) =0 andg(b) =0 (check both)So by Rolle’s
Theorem there exists a pointcin(a;b)such that
0=g
′
(c) =f
′
(c)−
f(b)−f(a)
b−a
:

27. . . . . . .
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):
Thengis continuous on[a;b]and differentiable on(a;b)sincef
is.Alsog(a) =0 andg(b) =0 (check both)
So by Rolle’s
Theorem there exists a pointcin(a;b)such that
0=g
′
(c) =f
′
(c)−
f(b)−f(a)
b−a
:

28. . . . . . .
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):
Thengis continuous on[a;b]and differentiable on(a;b)sincef
is.Alsog(a) =0 andg(b) =0 (check both)So by Rolle’s
Theorem there exists a pointcin(a;b)such that
0=g
′
(c) =f
′
(c)−
f(b)−f(a)
b−a
:

29. . . . . . .
Using the MVT to count solutions
Example
Show that there is a unique solution to the equationx
3
−x=100
in the interval[4;5].
Solution
IBy the Intermediate Value Theorem, the function
f(x) =x
3
−x must take the value100at some point on c in
(4;5).
IIf there were two points c1and c2with f(c1) =f(c2) =100,
then somewhere between them would be a point c3
between them with f
′
(c3) =0.
IHowever, f
′
(x) =3x
2
−1, which is positive all along(4;5).
So this is impossible.

30. . . . . . .
Using the MVT to count solutions
Example
Show that there is a unique solution to the equationx
3
−x=100
in the interval[4;5].
Solution
IBy the Intermediate Value Theorem, the function
f(x) =x
3
−x must take the value100at some point on c in
(4;5).
IIf there were two points c1and c2with f(c1) =f(c2) =100,
then somewhere between them would be a point c3
between them with f
′
(c3) =0.
IHowever, f
′
(x) =3x
2
−1, which is positive all along(4;5).
So this is impossible.

31. . . . . . .
Using the MVT to count solutions
Example
Show that there is a unique solution to the equationx
3
−x=100
in the interval[4;5].
Solution
IBy the Intermediate Value Theorem, the function
f(x) =x
3
−x must take the value100at some point on c in
(4;5).
IIf there were two points c1and c2with f(c1) =f(c2) =100,
then somewhere between them would be a point c3
between them with f
′
(c3) =0.
IHowever, f
′
(x) =3x
2
−1, which is positive all along(4;5).
So this is impossible.

32. . . . . . .
Using the MVT to count solutions
Example
Show that there is a unique solution to the equationx
3
−x=100
in the interval[4;5].
Solution
IBy the Intermediate Value Theorem, the function
f(x) =x
3
−x must take the value100at some point on c in
(4;5).
IIf there were two points c1and c2with f(c1) =f(c2) =100,
then somewhere between them would be a point c3
between them with f
′
(c3) =0.
IHowever, f
′
(x) =3x
2
−1, which is positive all along(4;5).
So this is impossible.

33. . . . . . .
Example
We know that|sinx| ≤1 for allx. Show that|sinx| ≤ |x|.
Solution
Apply the MVT to the function f(t) =sint on[0;x]. We get
sinx−sin 0
x−0
=cos(c)
for some c in(0;x). Since|cos(c)| ≤1, we get




sinx
x




≤1=⇒ |sinx| ≤ |x|

34. . . . . . .
Example
We know that|sinx| ≤1 for allx. Show that|sinx| ≤ |x|.
Solution
Apply the MVT to the function f(t) =sint on[0;x]. We get
sinx−sin 0
x−0
=cos(c)
for some c in(0;x). Since|cos(c)| ≤1, we get




sinx
x




≤1=⇒ |sinx| ≤ |x|

35. . . . . . .
Example
Letfbe a differentiable function withf(1) =3 andf
′
(x)<2 for
allxin[0;5]. Couldf(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))

36. . . . . . .
Example
Letfbe a differentiable function withf(1) =3 andf
′
(x)<2 for
allxin[0;5]. Couldf(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))

37. . . . . . .
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.

38. . . . . . .
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.

39. . . . . . .
Outline
Review: The Closed Interval Method
Rolle’s Theorem
The Mean Value Theorem
Applications
Why the MVT is the MITC

40. . . . . . .
Fact
If f is constant on(a;b), then f
′
(x) =0on(a;b).
IThe limit of difference quotients must be 0
IThe tangent line to a line is that line, and a constant
function’s graph is a horizontal line, which has slope 0.
IImplied by the power rule sincec=cx
0
Question
Iff
′
(x) =0 isfnecessarily a constant function?
IIt seems true
IBut so far no theorem (that we have proven) uses information
about the derivative of a function to determine information
about the function itself

41. . . . . . .
Fact
If f is constant on(a;b), then f
′
(x) =0on(a;b).
IThe limit of difference quotients must be 0
IThe tangent line to a line is that line, and a constant
function’s graph is a horizontal line, which has slope 0.
IImplied by the power rule sincec=cx
0
Question
Iff
′
(x) =0 isfnecessarily a constant function?
IIt seems true
IBut so far no theorem (that we have proven) uses information
about the derivative of a function to determine information
about the function itself

42. . . . . . .
Fact
If f is constant on(a;b), then f
′
(x) =0on(a;b).
IThe limit of difference quotients must be 0
IThe tangent line to a line is that line, and a constant
function’s graph is a horizontal line, which has slope 0.
IImplied by the power rule sincec=cx
0
Question
Iff
′
(x) =0 isfnecessarily a constant function?
IIt seems true
IBut so far no theorem (that we have proven) uses information
about the derivative of a function to determine information
about the function itself

43. . . . . . .
Fact
If f is constant on(a;b), then f
′
(x) =0on(a;b).
IThe limit of difference quotients must be 0
IThe tangent line to a line is that line, and a constant
function’s graph is a horizontal line, which has slope 0.
IImplied by the power rule sincec=cx
0
Question
Iff
′
(x) =0 isfnecessarily a constant function?
IIt seems true
IBut so far no theorem (that we have proven) uses information
about the derivative of a function to determine information
about the function itself

44. . . . . . .
Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f
′
=0on an interval(a;b).
Then f is constant on(a;b).
Proof.
Pick any pointsxandyin(a;b)withx<y. Thenfis continuous
on[x;y]and differentiable on(x;y). By MVT there exists a point
zin(x;y)such that
f(y)−f(x)
y−x
=f
′
(z) =0:
Sof(y) =f(x). Since this is true for allxandyin(a;b), thenfis
constant.

45. . . . . . .
Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f
′
=0on an interval(a;b).Then f is constant on(a;b).
Proof.
Pick any pointsxandyin(a;b)withx<y. Thenfis continuous
on[x;y]and differentiable on(x;y). By MVT there exists a point
zin(x;y)such that
f(y)−f(x)
y−x
=f
′
(z) =0:
Sof(y) =f(x). Since this is true for allxandyin(a;b), thenfis
constant.

46. . . . . . .
Why the MVT is the MITC
Most Important Theorem In Calculus!
Theorem
Let f
′
=0on an interval(a;b).Then f is constant on(a;b).
Proof.
Pick any pointsxandyin(a;b)withx<y. Thenfis continuous
on[x;y]and differentiable on(x;y). By MVT there exists a point
zin(x;y)such that
f(y)−f(x)
y−x
=f
′
(z) =0:
Sof(y) =f(x). Since this is true for allxandyin(a;b), thenfis
constant.

47. . . . . . .
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.
ILeth(x) =f(x)−g(x)
IThenh
′
(x) =f
′
(x)−g
′
(x) =0 on(a;b)
ISoh(x) =C, a constant
IThis meansf(x)−g(x) =Con(a;b)

48. . . . . . .
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.
ILeth(x) =f(x)−g(x)
IThenh
′
(x) =f
′
(x)−g
′
(x) =0 on(a;b)
ISoh(x) =C, a constant
IThis meansf(x)−g(x) =Con(a;b)

49. . . . . . .
MVT and differentiability
Example
Let
f(x) =
{
−xifx≤0
x
2
ifx≥0
Isfdifferentiable at 0?

50. . . . . . .
MVT and differentiability
Example
Let
f(x) =
{
−xifx≤0
x
2
ifx≥0
Isfdifferentiable 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
+
x
2
x
=lim
x→0
+
x=0
Since these limits disagree, f is not differentiable at0.

51. . . . . . .
MVT and differentiability
Example
Let
f(x) =
{
−xifx≤0
x
2
ifx≥0
Isfdifferentiable 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) =0andlim
x→0
−
f
′
(x) =−1;
the limitlim
x→0
f
′
(x)does not exist and so f is not differentiable at0.

52. . . . . . .
IThis solution is valid but less direct.
IWe seem to be using the following fact:Iflim
x→a
f
′
(x)does not
exist, then f is not differentiable at a.
Iequivalently:If f is differentiable at a, thenlim
x→a
f
′
(x)exists.
IBut this “fact” is not true!

53. . . . . . .
Differentiable with discontinuous derivative
It is possible for a functionfto be differentiable ataeven if
lim
x→a
f
′
(x)does not exist.
Example
Letf
′
(x) =
{
x
2
sin(1=x)ifx̸=0
0 ifx=0
:Then whenx̸=0,
f
′
(x) =2xsin(1=x)+x
2
cos(1=x)(−1=x
2
) =2xsin(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
x
2
sin(1=x)
x
=lim
x→0
xsin(1=x) =0
Sof
′
(0) =0. Hencefis differentiable for allx, butf
′
is not
continuous at 0!

54. . . . . . .
MVT to the rescue
Lemma
Suppose f is continuous on[a;b]andlim
x→a
+
f
′
(x) =m. Then
lim
x→a
+
f(x)−f(a)
x−a
=m:
Proof.
Choosexnearaand greater thana. Then
f(x)−f(a)
x−a
=f
′
(cx)
for somecxwherea<cx<x. Asx→a,cx→aas well, so:
lim
x→a
+
f(x)−f(a)
x−a
=lim
x→a
+
f
′
(cx) =lim
x→a
+
f
′
(x) =m:

55. . . . . . .
MVT to the rescue
Lemma
Suppose f is continuous on[a;b]andlim
x→a
+
f
′
(x) =m. Then
lim
x→a
+
f(x)−f(a)
x−a
=m:
Proof.
Choosexnearaand greater thana. Then
f(x)−f(a)
x−a
=f
′
(cx)
for somecxwherea<cx<x. Asx→a,cx→aas well, so:
lim
x→a
+
f(x)−f(a)
x−a
=lim
x→a
+
f
′
(cx) =lim
x→a
+
f
′
(x) =m:

56. . . . . . .
Theorem
Suppose
lim
x→a
−
f
′
(x) =m1andlim
x→a
+
f
′
(x) =m2
If m1=m2, then f is differentiable at a. If m1̸=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.

57. . . . . . .
Theorem
Suppose
lim
x→a
−
f
′
(x) =m1andlim
x→a
+
f
′
(x) =m2
If m1=m2, then f is differentiable at a. If m1̸=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.

58. . . . . . .
What have we learned today?
IRolle’s Theorem: there is a stationary point
IMean Value Theorem: at some point the instantaneous rate
of change equals the average rate of change (The Most
Important Theorem in Calculus)
IOnly constant functions have a derivative of zero.