The Mean Value Theorem gives us tests for determining the shape of curves between critical points.

1. Sec on 4.3
Deriva ves and the Shapes of Curves
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University
April 11, 2011
.

2. Announcements
Quiz 4 on Sec ons 3.3,
3.4, 3.5, and 3.7 this
week (April 14/15)
Quiz 5 on Sec ons
4.1–4.4 April 28/29
Final Exam Thursday May
12, 2:00–3:50pm

3. Objectives
Use the deriva ve of a func on
to determine the intervals along
which the func on is increasing
or decreasing (The
Increasing/Decreasing Test)
Use the First Deriva ve Test to
classify cri cal points of a
func on as local maxima, local
minima, or neither.

4. Objectives
Use the second deriva ve of a
func on to determine the
intervals along which the graph
of the func on is concave up or
concave down (The Concavity
Test)
Use the first and second
deriva ve of a func on to
classify cri cal points as local
maxima or local minima, when
applicable (The Second
Deriva ve Test)

5. Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Deriva ve Test
Concavity
Defini ons
Tes ng for Concavity
The Second Deriva ve Test

6. Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be con nuous on [a, b]
and differen able on (a, b).
Then there exists a point c in
(a, b) such that
b
f(b) − f(a) .
= f′ (c). a
b−a

7. Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be con nuous on [a, b]
and differen able on (a, b).
Then there exists a point c in
(a, b) such that
b
f(b) − f(a) .
= f′ (c). a
b−a

8. Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
c
Let f be con nuous on [a, b]
and differen able on (a, b).
Then there exists a point c in
(a, b) such that
b
f(b) − f(a) .
= f′ (c). a
b−a

9. Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
c
Let f be con nuous on [a, b]
and differen able on (a, b).
Then there exists a point c in
(a, b) such that
b
f(b) − f(a) .
= f′ (c). a
b−a
Another way to put this is that there exists a point c such that
f(b) = f(a) + f′ (c)(b − a)

10. 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 con nuous on
[x, y] and differen able on (x, y). By MVT there exists a point z in
(x, y) such that
f(y) = f(x) + f′ (z)(y − x)
So f(y) = f(x). Since this is true for all x and y in (a, b), then f is
constant.

11. Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Deriva ve Test
Concavity
Defini ons
Tes ng for Concavity
The Second Deriva ve Test

12. Increasing Functions
Defini on
A func on f is increasing on the interval I if
f(x) < f(y)
whenever x and y are two points in I with x < y.

13. Increasing Functions
Defini on
A func on f is increasing on the interval I if
f(x) < f(y)
whenever x and y are two points in I with x < y.
An increasing func on “preserves order.”
I could be bounded or infinite, open, closed, or
half-open/half-closed.
Write your own defini on (muta s mutandis) of decreasing,
nonincreasing, nondecreasing

14. The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0
on an interval, then f is decreasing on that interval.

15. The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0
on an interval, then f is decreasing on that interval.
Proof.
It works the same as the last theorem. Assume f′ (x) > 0 on an
interval I. Pick two points x and y in I with x < y. We must show
f(x) < f(y).

16. The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0
on an interval, then f is decreasing on that interval.
Proof.
It works the same as the last theorem. Assume f′ (x) > 0 on an
interval I. Pick two points x and y in I with x < y. We must show
f(x) < f(y). By MVT there exists a point c in (x, y) such that
f(y) − f(x) = f′ (c)(y − x) > 0.

17. The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0
on an interval, then f is decreasing on that interval.
Proof.
It works the same as the last theorem. Assume f′ (x) > 0 on an
interval I. Pick two points x and y in I with x < y. We must show
f(x) < f(y). By MVT there exists a point c in (x, y) such that
f(y) − f(x) = f′ (c)(y − x) > 0.
So f(y) > f(x).

18. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.

19. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solu on
f′ (x) = 2 is always posi ve, so f is increasing on (−∞, ∞).

20. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solu on
f′ (x) = 2 is always posi ve, so f is increasing on (−∞, ∞).
Example
Describe the monotonicity of f(x) = arctan(x).

21. Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solu on
f′ (x) = 2 is always posi ve, so f is increasing on (−∞, ∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solu on
1
Since f′ (x) = is always posi ve, f(x) is always increasing.
1 + x2

22. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
.

23. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
f′ (x) = 2x, which is posi ve when x > 0 and nega ve when x is.
.

24. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
f′ (x) = 2x, which is posi ve when x > 0 and nega ve when x is.
We can draw a number line:
− 0 + f′
.
0

25. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
f′ (x) = 2x, which is posi ve when x > 0 and nega ve when x is.
We can draw a number line:
− 0 + f′
.
↘ 0 ↗ f

26. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
− 0 + f′
.
↘ 0 ↗ f
So f is decreasing on (−∞, 0) and increasing on (0, ∞).

27. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
− 0 + f′
.
↘ 0 ↗ f
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing on
[0, ∞)

28. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).

29. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
.
f′ (x) = 2 x−1/3 (x + 2) + x2/3
3
= 3 x−1/3 (5x + 4)
1

30. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
.
f′ (x) = 2 x−1/3 (x + 2) + x2/3
3
= 3 x−1/3 (5x + 4)
1
The cri cal points are 0 and
and −4/5.

31. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
. x−1/3
f′ (x) = 2 x−1/3 (x + 2) + x2/3
3
= 3 x−1/3 (5x + 4)
1
The cri cal points are 0 and
and −4/5.

32. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
×.
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3
= 3x (5x + 4)
The cri cal points are 0 and
and −4/5.

33. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×.
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3
= 3x (5x + 4)
The cri cal points are 0 and
and −4/5.

34. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3
= 3x (5x + 4)
The cri cal points are 0 and
and −4/5.

35. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3
= 3x (5x + 4) 5x + 4
The cri cal points are 0 and
and −4/5.

36. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 0
= 3x (5x + 4) 5x + 4
−4/5
The cri cal points are 0 and
and −4/5.

37. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0
= 3x (5x + 4) 5x + 4
−4/5
The cri cal points are 0 and
and −4/5.

38. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
−4/5
The cri cal points are 0 and
and −4/5.

39. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
−4/5 f′ (x)
The cri cal points are 0 and
and −4/5. f(x)

40. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
−4/5 f′ (x)
0
The cri cal points are 0 and
and −4/5. −4/5 f(x)

41. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
−4/5 × f′ (x)
0
The cri cal points are 0 and
and −4/5. −4/5 0 f(x)

42. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
+ −0/5 f′ (x)
4
×
The cri cal points are 0 and
and −4/5. −4/5 0 f(x)

43. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
+ −0/5− × f′ (x)
4
The cri cal points are 0 and
and −4/5. −4/5 0 f(x)

44. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
+ −0/5− × f′ (x)
4
+
The cri cal points are 0 and
and −4/5. −4/5 0 f(x)

45. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
+ −0/5− × f′ (x)
4
+
The cri cal points are 0 and
and −4/5. ↗ −4/5 0 f(x)

46. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
+ −0/5− × f′ (x)
4
+
The cri cal points are 0 and
and −4/5. ↗ −4/5 0
↘ f(x)

47. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
+ −0/5− × f′ (x)
4
+
The cri cal points are 0 and
and −4/5. ↗ −4/5 0
↘ ↗ f(x)

48. The First Derivative Test
Theorem (The First Deriva ve Test)
Let f be con nuous on [a, b] and c a cri cal point of f in (a, b).
If f′ changes from posi ve to nega ve at c, then c is a local
maximum.
If f′ changes from nega ve to posi ve at c, then c is a local
minimum.
If f′ (x) has the same sign on either side of c, then c is not a local
extremum.

49. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
− 0 + f′
.
↘ 0 ↗ f
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing on
[0, ∞)

50. Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
− 0 + f′
.
↘ 0 ↗ f
min
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing on
[0, ∞)

51. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
+ −0/5− × f′ (x)
4
+
The cri cal points are 0 and
and −4/5. ↗ −4/5 0
↘ ↗ f(x)

52. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
+ −0/5− × f′ (x)
4
+
The cri cal points are 0 and
and −4/5. ↗ −4/5 0
↘ ↗ f(x)
max

53. Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
− ×. +
′
f (x) = 2 −1/3
3x (x + 2) + x2/3 x−1/3
0
1 −1/3 − 0 +
= 3x (5x + 4) 5x + 4
+ −0/5− × f′ (x)
4
+
The cri cal points are 0 and
and −4/5. ↗ −4/5 0
↘ ↗ f(x)
max min

54. Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Deriva ve Test
Concavity
Defini ons
Tes ng for Concavity
The Second Deriva ve Test

55. Concavity
Defini on
The graph of f is called concave upwards on an interval if it lies
above all its tangents on that interval. The graph of f is called
concave downwards on an interval if it lies below all its tangents on
that interval.

56. Concavity
Defini on
The graph of f is called concave upwards on an interval if it lies
above all its tangents on that interval. The graph of f is called
concave downwards on an interval if it lies below all its tangents on
that interval.
. .
concave up concave down

57. Concavity
Defini on
. .
concave up concave down
We some mes say a concave up graph “holds water” and a concave
down graph “spills water”.

58. Synonyms for concavity
Remark
“concave up” = “concave upwards” = “convex”
“concave down” = “concave downwards” = “concave”

59. Inflection points mean change in concavity
Defini on
A point P on a curve y = f(x) is called an inflec on point if f is
con nuous at P and the curve changes from concave upward to
concave downward at P (or vice versa).
concave
up
inflec on point
.
concave
down

60. Testing for Concavity
Theorem (Concavity Test)
If f′′ (x) > 0 for all x in an interval, then the graph of f is concave
upward on that interval.
If f′′ (x) < 0 for all x in an interval, then the graph of f is concave
downward on that interval.

61. Testing for Concavity
Proof.
Suppose f′′ (x) > 0 on the interval I (which could be infinite). This
means f′ is increasing on I.

62. Testing for Concavity
Proof.
Suppose f′′ (x) > 0 on the interval I (which could be infinite). This
means f′ is increasing on I. Let a and x be in I. The tangent line
through (a, f(a)) is the graph of
L(x) = f(a) + f′ (a)(x − a)

63. Testing for Concavity
Proof.
Suppose f′′ (x) > 0 on the interval I (which could be infinite). This
means f′ is increasing on I. Let a and x be in I. The tangent line
through (a, f(a)) is the graph of
L(x) = f(a) + f′ (a)(x − a)
By MVT, there exists a c between a and x with
f(x) = f(a) + f′ (c)(x − a)

64. Testing for Concavity
Proof.
Suppose f′′ (x) > 0 on the interval I (which could be infinite). This
means f′ is increasing on I. Let a and x be in I. The tangent line
through (a, f(a)) is the graph of
L(x) = f(a) + f′ (a)(x − a)
By MVT, there exists a c between a and x with
f(x) = f(a) + f′ (c)(x − a)
Since f′ is increasing, f(x) > L(x).

65. Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .

66. Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .
Solu on
We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2.

67. Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .
Solu on
We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2.
This is nega ve when x < −1/3, posi ve when x > −1/3, and 0
when x = −1/3

68. Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2 .
Solu on
We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2.
This is nega ve when x < −1/3, posi ve when x > −1/3, and 0
when x = −1/3
So f is concave down on the open interval (−∞, −1/3), concave
up on the open interval (−1/3, ∞), and has an inflec on point
at the point (−1/3, 2/27)

69. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).

70. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solu on
We have + ×. +
x−4/3
10 −1/3 4 −4/3 0
f′′ (x) = x − x − 0 +
9 9 5x − 2
2/5
2 −4/3
= x (5x − 2)
9

71. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solu on
We have + ×. +
x−4/3
10 −1/3 4 −4/3 0
f′′ (x) = x − x − 0 +
9 9 5x − 2
2 −4/3 ×
2/5
0 f′′ (x)
= x (5x − 2)
9 0 2/5 f(x)

72. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solu on
We have + ×. +
x−4/3
10 −1/3 4 −4/3 0
f′′ (x) = x − x − 0 +
9 9 5x − 2
2 −4/3 −− ×
2/5
0 f′′ (x)
= x (5x − 2)
9 0 2/5 f(x)

73. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solu on
We have + ×. +
x−4/3
10 −1/3 4 −4/3 0
f′′ (x) = x − x − 0 +
9 9 5x − 2
2 −4/3 −−
2
× /5
−− 0 f′′ (x)
= x (5x − 2)
9 0 2/5 f(x)

74. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solu on
We have + ×. +
x−4/3
10 −1/3 4 −4/3 0
f′′ (x) = x − x − 0 +
9 9 5x − 2
2 −4/3 −−
2
× /5
−− 0 ++ f′′ (x)
= x (5x − 2)
9 0 2/5 f(x)

75. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solu on
We have + ×. +
x−4/3
10 −1/3 4 −4/3 0
f′′ (x) = x − x − 0 +
9 9 5x − 2
2 −4/3 −−
2
× /5
−− 0 ++ f′′ (x)
= x (5x − 2) ⌢
9 0 2/5 f(x)

76. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solu on
We have + ×. +
x−4/3
10 −1/3 4 −4/3 0
f′′ (x) = x − x − 0 +
9 9 5x − 2
2 −4/3 −−
2
× /5
−− 0 ++ f′′ (x)
= x (5x − 2) ⌢
9 0⌢ 2/5 f(x)

77. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solu on
We have + ×. +
x−4/3
10 −1/3 4 −4/3 0
f′′ (x) = x − x − 0 +
9 9 5x − 2
2 −4/3 −−
2
× /5
−− 0 ++ f′′ (x)
= x (5x − 2) ⌢
9 0⌢ 2/5 ⌣ f(x)

78. Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solu on
We have + ×. +
x−4/3
10 −1/3 4 −4/3 0
f′′ (x) = x − x − 0 +
9 9 5x − 2
2 −4/3 −−
2
× /5
−− 0 ++ f′′ (x)
= x (5x − 2) ⌢
9 0⌢ 2/5 ⌣ f(x)
IP

79. The Second Derivative Test
Theorem (The Second Deriva ve Test)
Let f, f′ , and f′′ be con nuous on [a, b]. Let c be be a point in (a, b)
with f′ (c) = 0.
If f′′ (c) < 0, then c is a local maximum of f.
If f′′ (c) > 0, then c is a local minimum of f.

80. The Second Derivative Test
Theorem (The Second Deriva ve Test)
Let f, f′ , and f′′ be con nuous on [a, b]. Let c be be a point in (a, b)
with f′ (c) = 0.
If f′′ (c) < 0, then c is a local maximum of f.
If f′′ (c) > 0, then c is a local minimum of f.
Remarks
If f′′ (c) = 0, the second deriva ve test is inconclusive
We look for zeroes of f′ and plug them into f′′ to determine if
their f values are local extreme values.

81. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
+ f′′ = (f′ )′
.
c f′
0 f′
c f

82. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is con nuous,
f′′ (x) > 0 for all x + f′′ = (f′ )′
sufficiently close to c. .
c f′
0 f′
c f

83. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is con nuous,
f′′ (x) > 0 for all x + + f′′ = (f′ )′
sufficiently close to c. .
c f′
0 f′
c f

84. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is con nuous,
f′′ (x) > 0 for all x + + + f′′ = (f′ )′
sufficiently close to c. .
c f′
0 f′
c f

85. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is con nuous,
f′′ (x) > 0 for all x + + + f′′ = (f′ )′
sufficiently close to c. .
c f′
Since f′′ = (f′ )′ , we know
0 f′
f′ is increasing near c.
c f

86. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is con nuous,
f′′ (x) > 0 for all x + + + f′′ = (f′ )′
sufficiently close to c. .
↗ c f′
Since f′′ = (f′ )′ , we know
0 f′
f′ is increasing near c.
c f

87. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is con nuous,
f′′ (x) > 0 for all x + + + f′′ = (f′ )′
sufficiently close to c. .
↗ c ↗ f′
Since f′′ = (f′ )′ , we know
0 f′
f′ is increasing near c.
c f

88. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′ (c) = 0 and f′ is
increasing, f′ (x) < 0 for x + + + f′′ = (f′ )′
.
close to c and less than c, c
↗ ↗ f′
and f′ (x) > 0 for x close
0 f′
to c and more than c.
c f

89. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′ (c) = 0 and f′ is
increasing, f′ (x) < 0 for x + + + f′′ = (f′ )′
.
close to c and less than c, c
↗ ↗ f′
and f′ (x) > 0 for x close
− 0 f′
to c and more than c.
c f

90. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′ (c) = 0 and f′ is
increasing, f′ (x) < 0 for x + + + f′′ = (f′ )′
.
close to c and less than c, c
↗ ↗ f′
and f′ (x) > 0 for x close
− 0 + f′
to c and more than c.
c f

91. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
This means f′ changes
sign from nega ve to + + + f′′ = (f′ )′
.
posi ve at c, which c
↗ ↗ f′
means (by the First
− 0 + f′
Deriva ve Test) that f has
a local minimum at c. c f

92. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
This means f′ changes
sign from nega ve to + + + f′′ = (f′ )′
.
posi ve at c, which c
↗ ↗ f′
means (by the First
− 0 + f′
Deriva ve Test) that f has
a local minimum at c. ↘ c f

93. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
This means f′ changes
sign from nega ve to + + + f′′ = (f′ )′
.
posi ve at c, which c
↗ ↗ f′
means (by the First
− 0 + f′
Deriva ve Test) that f has
a local minimum at c. ↘ c ↗ f

94. Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
This means f′ changes
sign from nega ve to + + + f′′ = (f′ )′
.
posi ve at c, which c
↗ ↗ f′
means (by the First
− 0 + f′
Deriva ve Test) that f has
a local minimum at c. c
↘ min ↗ f

95. Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2 .

96. Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2 .
Solu on
f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.

97. Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2 .
Solu on
f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′ (x) = 6x + 2

98. Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2 .
Solu on
f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′ (x) = 6x + 2
Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum.

99. Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2 .
Solu on
f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.
Remember f′′ (x) = 6x + 2
Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum.
Since f′′ (0) = 2 > 0, 0 is a local minimum.

100. Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3 (x + 2)

101. Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solu on
1
Remember f′ (x) = x−1/3 (5x + 4) which is zero when x = −4/5
3

102. Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solu on
1
Remember f′ (x) = x−1/3 (5x + 4) which is zero when x = −4/5
3
10
Remember f′′ (x) = x−4/3 (5x − 2), which is nega ve when
9
x = −4/5

103. Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solu on
1
Remember f′ (x) = x−1/3 (5x + 4) which is zero when x = −4/5
3
10
Remember f′′ (x) = x−4/3 (5x − 2), which is nega ve when
9
x = −4/5
So x = −4/5 is a local maximum.

104. Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3 (x + 2)
Solu on
1
Remember f′ (x) = x−1/3 (5x + 4) which is zero when x = −4/5
3
10
Remember f′′ (x) = x−4/3 (5x − 2), which is nega ve when
9
x = −4/5
So x = −4/5 is a local maximum.
No ce the Second Deriva ve Test doesn’t catch the local
minimum x = 0 since f is not differen able there.

105. Using the Second Derivative Test II
Graph
Graph of f(x) = x2/3 (x + 2):
y
(−4/5, 1.03413)
(2/5, 1.30292)
. x
(−2, 0) (0, 0)

106. When the second derivative is zero
Remark
At inflec on points c, if f′ is differen able at c, then f′′ (c) = 0
If f′′ (c) = 0, must f have an inflec on point at c?

107. When the second derivative is zero
Remark
At inflec on points c, if f′ is differen able at c, then f′′ (c) = 0
If f′′ (c) = 0, must f have an inflec on point at c?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3

108. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = , f′ (0) =
f(x) = x
f′′ (x) = , f′′ (0) =
g′ (x) = , g′ (0) =
g(x) = −x 4
g′′ (x) = , g′′ (0) =
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

109. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) =
f(x) = x
f′′ (x) = , f′′ (0) =
g′ (x) = , g′ (0) =
g(x) = −x 4
g′′ (x) = , g′′ (0) =
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

110. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x
f′′ (x) = , f′′ (0) =
g′ (x) = , g′ (0) =
g(x) = −x 4
g′′ (x) = , g′′ (0) =
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

111. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x
f′′ (x) = 12x2 , f′′ (0) =
g′ (x) = , g′ (0) =
g(x) = −x 4
g′′ (x) = , g′′ (0) =
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

112. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x
f′′ (x) = 12x2 , f′′ (0) = 0
g′ (x) = , g′ (0) =
g(x) = −x 4
g′′ (x) = , g′′ (0) =
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

113. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x .
f′′ (x) = 12x2 , f′′ (0) = 0
g′ (x) = , g′ (0) =
g(x) = −x 4
g′′ (x) = , g′′ (0) =
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

114. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
g′ (x) = , g′ (0) =
g(x) = −x 4
g′′ (x) = , g′′ (0) =
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

115. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
g′ (x) = − 4x3 , g′ (0) =
g(x) = −x 4
g′′ (x) = , g′′ (0) =
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

116. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
g′ (x) = − 4x3 , g′ (0) = 0
g(x) = −x 4
g′′ (x) = , g′′ (0) =
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

117. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
g′ (x) = − 4x3 , g′ (0) = 0
g(x) = −x 4
g′′ (x) = − 12x2 , g′′ (0) =
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

118. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
g′ (x) = − 4x3 , g′ (0) = 0
g(x) = −x 4
g′′ (x) = − 12x2 , g′′ (0) = 0
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

119. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
.
g′ (x) = − 4x3 , g′ (0) = 0
g(x) = −x4
g′′ (x) = − 12x2 , g′′ (0) = 0
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

120. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
.
g′ (x) = − 4x3 , g′ (0) = 0
g(x) = −x4 max
′′ ′′
g (x) = − 12x , g (0) = 0
2
3
h′ (x) = , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

121. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
.
g′ (x) = − 4x3 , g′ (0) = 0
g(x) = −x4 max
′′ ′′
g (x) = − 12x , g (0) = 0
2
3
h′ (x) = 3x2 , h′ (0) =
h(x) = x
h′′ (x) = , h′′ (0) =

122. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
.
g′ (x) = − 4x3 , g′ (0) = 0
g(x) = −x4 max
′′ ′′
g (x) = − 12x , g (0) = 0
2
3
h′ (x) = 3x2 , h′ (0) = 0
h(x) = x
h′′ (x) = , h′′ (0) =

123. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
.
g′ (x) = − 4x3 , g′ (0) = 0
g(x) = −x4 max
′′ ′′
g (x) = − 12x , g (0) = 0
2
3
h′ (x) = 3x2 , h′ (0) = 0
h(x) = x
h′′ (x) = 6x, h′′ (0) =

124. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
.
g′ (x) = − 4x3 , g′ (0) = 0
g(x) = −x4 max
′′ ′′
g (x) = − 12x , g (0) = 0
2
3
h′ (x) = 3x2 , h′ (0) = 0
h(x) = x
h′′ (x) = 6x, h′′ (0) = 0

125. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
.
g′ (x) = − 4x3 , g′ (0) = 0
g(x) = −x4 max
′′ ′′
g (x) = − 12x , g (0) = 0
2
3
h′ (x) = 3x2 , h′ (0) = 0
h(x) = x .
h′′ (x) = 6x, h′′ (0) = 0

126. When first and second derivative are zero
func on deriva ves graph type
4
f′ (x) = 4x3 , f′ (0) = 0
f(x) = x . min
′′ 2 ′′
f (x) = 12x , f (0) = 0
.
g′ (x) = − 4x3 , g′ (0) = 0
g(x) = −x4 max
′′ ′′
g (x) = − 12x , g (0) = 02
3
h′ (x) = 3x2 , h′ (0) = 0
h(x) = x . infl.
′′ ′′
h (x) = 6x, h (0) = 0

127. When the second derivative is zero
Remark
At inflec on points c, if f′ is differen able at c, then f′′ (c) = 0
If f′′ (c) = 0, must f have an inflec on point at c?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
All of them have cri cal points at zero with a second deriva ve of
zero. But the first has a local min at 0, the second has a local max at
0, and the third has an inflec on point at 0. This is why we say 2DT
has nothing to say when f′′ (c) = 0.

128. Summary
Concepts: Mean Value Theorem, monotonicity, concavity
Facts: deriva ves can detect monotonicity and concavity
Techniques for drawing curves: the Increasing/Decreasing Test
and the Concavity Test
Techniques for finding extrema: the First Deriva ve Test and
the Second Deriva ve Test