The document discusses using the second derivative to identify extrema and classify flat points on a graph of y=f(x). It defines terms for the second derivative, explaining that if f''(x)>0, the slope f'(x) is increasing, meaning a downhill point is getting less steep and an uphill point is getting more steep. For a maximum point M, the curve must flatten out with f'(x) approaching 0+ and f'(x) becoming increasingly negative after M, resulting in f''(M)<0.

1. Extrema and the Second Derivative

2. In this section, we take a closer look at the extrema.
We also examine the geometric information offered by
the second derivative y'' about the graph of y = f(x).
Extrema and the Second Derivative

3. In this section, we take a closer look at the extrema.
We also examine the geometric information offered by
the second derivative y'' about the graph of y = f(x).
Extrema and the Second Derivative
c d eba
A
We summarize the geometric information obtained
from y' about the graph of y = f(x). We may view the
curve as a roller coaster track of and we’re going
from left to right.
y = f(x)

4. In this section, we take a closer look at the extrema.
We also examine the geometric information offered by
the second derivative y'' about the graph of y = f(x).
Extrema and the Second Derivative
C
c
D
E
d eba
A
y'=0
y'=0
y'=0
y'=0
B y = f(x)
We summarize the geometric information obtained
from y' about the graph of y = f(x). We may view the
curve as a roller coaster track of and we’re going
from left to right.

5. In this section, we take a closer look at the extrema.
We also examine the geometric information offered by
the second derivative y'' about the graph of y = f(x).
Extrema and the Second Derivative
C
c
D
E
d e
increasing
increasing
increasing
ba
A
y'>0
y'>0
y'=0
y'=0
y'=0
y'>0
y'=0
B y = f(x)
We summarize the geometric information obtained
from y' about the graph of y = f(x). We may view the
curve as a roller coaster track of and we’re going
from left to right.

6. In this section, we take a closer look at the extrema.
We also examine the geometric information offered by
the second derivative y'' about the graph of y = f(x).
Extrema and the Second Derivative
decreasing
decreasing
C
c
D
E
d e
increasing
increasing
increasing
ba
A
y'>0
y'>0
y'=0
y'=0
y'<0
y'<0
y'=0
y'>0
y'=0
B y = f(x)
We summarize the geometric information obtained
from y' about the graph of y = f(x). We may view the
curve as a roller coaster track of and we’re going
from left to right.

7. Extrema and the Second Derivative
Note that C is a maximum because we go uphill
(y' >0) from the left to C and downhill (y' < 0) after
passing through C.
decreasing
decreasing
C
c
D
E
d e
increasing
increasing
increasing
ba
A
y'>0
y'>0
y'=0
y'=0
y'<0
y'<0
y'=0
y'>0
y'=0
B y = f(x)

8. Extrema and the Second Derivative
Note that C is a maximum because we go uphill
(y' >0) from the left to C and downhill (y' < 0) after
passing through C.
Similarly it’s a minimum at E because it’s downhill
(y' < 0) to E and uphill (y' >0) after E. Following is a
summary of the information.
decreasing
decreasing
C
c
D
E
d e
increasing
increasing
increasing
ba
A
y'>0
y'>0
y'=0
y'=0
y'<0
y'<0
y'=0
y'>0
y'=0
B y = f(x)

9. Extrema and the Second Derivative
decreasing
decreasing
C
c
D
E
d e
increasing
increasing
increasing
ba
A
y'>0
y'>0
y'=0
y'=0
y'<0
y'<0
y'=0
y'>0
y'=0
B y = f(x)
y' > 0, the graph is going uphill.
y' < 0, the graph is going downhill.
y' = 0
maximum, if y' > 0 to the left, y' < 0 to the right
minimum, if y' < 0 to the left, y' > 0 to the right
uphill flat point, if y' > 0 on both sides
downhill flat point, if y' < 0 on both sides

10. Extrema and the Second Derivative
For the discussion below we assume y = f(x) is
infinitely differentiable everywhere in the domain.

11. Extrema and the Second Derivative
For the discussion below we assume y = f(x) is
infinitely differentiable everywhere in the domain.
Another way to identify the different types of flat
points (y' = 0) is to use the 2nd derivative y'' at that
point.

12. Extrema and the Second Derivative
For the discussion below we assume y = f(x) is
infinitely differentiable everywhere in the domain.
Another way to identify the different types of flat
points (y' = 0) is to use the 2nd derivative y'' at that
point.
M
y = f(x)
x

13. Extrema and the Second Derivative
For the discussion below we assume y = f(x) is
infinitely differentiable everywhere in the domain.
Another way to identify the different types of flat
points (y' = 0) is to use the 2nd derivative y'' at that
point. To go uphill to a maximum M the curve
must flatten out, i.e. y' 0+.
M
y = f(x)
x

14. Extrema and the Second Derivative
y' 0+
M
For the discussion below we assume y = f(x) is
infinitely differentiable everywhere in the domain.
Another way to identify the different types of flat
points (y' = 0) is to use the 2nd derivative y'' at that
point. To go uphill to a maximum M the curve
must flatten out, i.e. y' 0+.
y = f(x)
x

15. Extrema and the Second Derivative
y' 0+
M
For the discussion below we assume y = f(x) is
infinitely differentiable everywhere in the domain.
Another way to identify the different types of flat
points (y' = 0) is to use the 2nd derivative y'' at that
point. To go uphill to a maximum M the curve
must flatten out, i.e. y' 0+.
y = f(x)
x

16. Extrema and the Second Derivative
y' 0+
M
For the discussion below we assume y = f(x) is
infinitely differentiable everywhere in the domain.
Another way to identify the different types of flat
points (y' = 0) is to use the 2nd derivative y'' at that
point. To go uphill to a maximum M the curve
must flatten out, i.e. y' 0+.
y = f(x)
x

17. Extrema and the Second Derivative
y' 0+
M
For the discussion below we assume y = f(x) is
infinitely differentiable everywhere in the domain.
Another way to identify the different types of flat
points (y' = 0) is to use the 2nd derivative y'' at that
point. To go uphill to a maximum M the curve
must flatten out, i.e. y' 0+.
After the curve passes the
maximum it must go
downhill steadily steeper,
i.e. the derivative y'
gets more negative.
y = f(x)
x

18. Extrema and the Second Derivative
y' 0+
y' becomes
increasingly
negative
M
For the discussion below we assume y = f(x) is
infinitely differentiable everywhere in the domain.
Another way to identify the different types of flat
points (y' = 0) is to use the 2nd derivative y'' at that
point. To go uphill to a maximum M the curve
must flatten out, i.e. y' 0+.
After the curve passes the
maximum it must go
downhill steadily steeper,
i.e. the derivative y'
gets more negative.
y = f(x)
x

19. Extrema and the Second Derivative
y' 0+
y' becomes
increasingly
negative
M
For the discussion below we assume y = f(x) is
infinitely differentiable everywhere in the domain.
Another way to identify the different types of flat
points (y' = 0) is to use the 2nd derivative y'' at that
point. To go uphill to a maximum M the curve
must flatten out, i.e. y' 0+.
After the curve passes the
maximum it must go
downhill steadily steeper,
i.e. the derivative y'
gets more negative.
y = f(x)
x

20. Extrema and the Second Derivative
y' 0+
y' becomes
increasingly
negative
M
For the discussion below we assume y = f(x) is
infinitely differentiable everywhere in the domain.
Another way to identify the different types of flat
points (y' = 0) is to use the 2nd derivative y'' at that
point. To go uphill to a maximum M the curve
must flatten out, i.e. y' 0+.
After the curve passes the
maximum it must go
downhill steadily steeper,
i.e. the derivative y'
gets more negative.
y = f(x)
x

21. Extrema and the Second Derivative
y' 0+
y' becomes
increasingly
negative
The fact the derivative y' turns from positive to 0
then negative means y' is decreasing at M,
that translates to (y')' = y'' < 0 at the maximum M.
M
For the discussion below we assume y = f(x) is
infinitely differentiable everywhere in the domain.
Another way to identify the different types of flat
points (y' = 0) is to use the 2nd derivative y'' at that
point. To go uphill to a maximum M the curve
must flatten out, i.e. y' 0+.
After the curve passes the
maximum it must go
downhill steadily steeper,
i.e. the derivative y'
gets more negative.
y = f(x)
x

22. Extrema and the Second Derivative
y = f(x)
x
We define the following terms based on the 2nd
derivative.

23. Extrema and the Second Derivative
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing.
y = f(x)
x

24. Extrema and the Second Derivative
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing. So if we are at a downhill
point, the downhill is getting less steep,
y = f(x)
x

25. Extrema and the Second Derivative
(a, f(a))
f ''(a) > 0
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing. So if we are at a downhill
point, the downhill is getting less steep,
y = f(x)
x

26. Extrema and the Second Derivative
(a, f(a))
f ''(a) > 0
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing. So if we are at a downhill
point, the downhill is getting less steep,
y = f(x)
x

27. Extrema and the Second Derivative
(a, f(a))
f ''(a) > 0
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing. So if we are at a downhill
point, the downhill is getting less steep,
y = f(x)
x

28. Extrema and the Second Derivative
(a, f(a))
f ''(a) > 0
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing. So if we are at a downhill
point, the downhill is getting less steep,
y = f(x)
x

29. but if we are at a uphill
point, the uphill is getting
more steep.
Extrema and the Second Derivative
(a, f(a)) f ''(b) > 0
(b, f(b))
f ''(a) > 0
y = f(x)
x
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing. So if we are at a downhill
point, the downhill is getting less steep,

30. but if we are at a uphill
point, the uphill is getting
more steep.
Extrema and the Second Derivative
(a, f(a)) f ''(b) > 0
(b, f(b))
f ''(a) > 0
y = f(x)
x
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing. So if we are at a downhill
point, the downhill is getting less steep,

31. but if we are at a uphill
point, the uphill is getting
more steep.
Extrema and the Second Derivative
(a, f(a)) f ''(b) > 0
(b, f(b))
f ''(a) > 0
y = f(x)
x
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing. So if we are at a downhill
point, the downhill is getting less steep,

32. but if we are at a uphill
point, the uphill is getting
more steep.
Extrema and the Second Derivative
(a, f(a)) f ''(b) > 0
(b, f(b))
f ''(a) > 0
y = f(x)
x
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing. So if we are at a downhill
point, the downhill is getting less steep,

33. but if we are at a uphill
point, the uphill is getting
more steep.
Extrema and the Second Derivative
(a, f(a)) f ''(b) > 0
(b, f(b))
f ''(a) > 0
y = f(x)
x
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing. So if we are at a downhill
point, the downhill is getting less steep,

34. but if we are at a uphill
point, the uphill is getting
more steep.
Extrema and the Second Derivative
(a, f(a)) f ''(b) > 0
(b, f(b))
f ''(a) > 0
y = f(x)
Another useful way to
identify where f ''(x) > 0
is to view the curve as the
trail of an ant crawling from left to right.
x
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing. So if we are at a downhill
point, the downhill is getting less steep,

35. but if we are at a uphill
point, the uphill is getting
more steep.
Extrema and the Second Derivative
(a, f(a)) f ''(b) > 0
(b, f(b))
f ''(a) > 0
y = f(x)
Another useful way to
identify where f ''(x) > 0
is to view the curve as the
trail of an ant crawling from left to right,
We say that y = f(x) is concave up at x if f ''(x) > 0
(turning left).
x
f ''(x) > 0 corresponds to where the ant is turning left.
We define the following terms based on the 2nd
derivative. At a generic x, f ''(x) > 0 means the “slope
at x” or f '(x) is increasing. So if we are at a downhill
point, the downhill is getting less steep,

36. In a similar manner,
y = g(x) is concave down
at x if g''(x) < 0 ( the trail is
turning right).
Extrema and the Second Derivative
(d, g(d))
g''(c) < 0
(c, g(c))
g''(d) < 0
y = g(x)

37. In a similar manner,
y = g(x) is concave down
at x if g''(x) < 0 ( the trail is
turning right).
Extrema and the Second Derivative
(d, g(d))
g''(c) < 0
(c, g(c))
g''(d) < 0
y = g(x)
y = g(x) shown here is
concave down at x = c, d.

38. In a similar manner,
y = g(x) is concave down
at x if g''(x) < 0 ( the trail is
turning right).
Extrema and the Second Derivative
(d, g(d))
g''(c) < 0
(c, g(c))
g''(d) < 0
y = g(x)
The point where the
concavity changes is
called an inflection point.
x
y = g(x) shown here is
concave down at x = c, d.
f ''(e) = 0
f ''(x) > 0
f ''(x) < 0
e
An inflection point at x = e

39. In a similar manner,
y = g(x) is concave down
at x if g''(x) < 0 ( the trail is
turning right).
Extrema and the Second Derivative
(d, g(d))
g''(c) < 0
(c, g(c))
g''(d) < 0
y = g(x)
The point where the
concavity changes is
called an inflection point.
x
y = g(x) shown here is
concave down at x = c, d.
f ''(e) = 0
f ''(x) > 0
f ''(x) < 0
Specifically,
x = e is an inflection point
if f''(e) = 0 and f''(x) changes
signs at e (as shown).
e
An inflection point at x = e

40. Extrema and the Second Derivative
The other possibility of an
inflection point is also
shown here.
x
f ''(e) = 0
f ''(x) > 0
f ''(x) < 0
e
An inflection point at x = e

41. Extrema and the Second Derivative
The other possibility of an
inflection point is also
shown here.
x
f ''(e) = 0
f ''(x) > 0
f ''(x) < 0
e
An inflection point at x = e
The fact that
f ''(x) = 0 does not make
x an reflection point.

42. Extrema and the Second Derivative
The other possibility of an
inflection point is also
shown here.
x
f ''(e) = 0
f ''(x) > 0
f ''(x) < 0
e
An inflection point at x = e
y = x2y = x4y = x6
The fact that
f ''(x) = 0 does not make
x an reflection point.
Let y = xEven,
then y'' = #xEven,
so that y''(0) = 0 at x = 0.
(0, 0)
x

43. Extrema and the Second Derivative
The other possibility of an
inflection point is also
shown here.
x
f ''(e) = 0
f ''(x) > 0
f ''(x) < 0
e
An inflection point at x = e
y = x2y = x4y = x6
The fact that
f ''(x) = 0 does not make
x an reflection point.
Let y = xEven,
then y'' = #xEven,
so that y''(0) = 0 at x = 0.
However (0, 0) is not an
inflection point because
there is no change in the
concavity. The point (0, 0)
is the minimum.
(0, 0)
x

44. Extrema and the Second Derivative
y = x3
y = x5
y = x7
(0, 0)
For y = x3, 5, 7., y'' = #x0dd,
the 2nd derivative changes
signs at (0, 0), hence it’s an
inflection point.

45. Extrema and the Second Derivative
y = x3
y = x5
y = x7
(0, 0)
Let’s summarize the techniques
for graphing.
For y = x3, 5, 7., y'' = #x0dd,
the 2nd derivative changes
signs at (0, 0), hence it’s an
inflection point.

46. Extrema and the Second Derivative
y = x3
y = x5
y = x7
(0, 0)
Let’s summarize the techniques
for graphing.
1. Get as much information as possible from y with
the sign chart. Find the limits of the boundary values.
Use these to get the general shape of the graph.
For y = x3, 5, 7., y'' = #x0dd,
the 2nd derivative changes
signs at (0, 0), hence it’s an
inflection point.

47. Extrema and the Second Derivative
y = x3
y = x5
y = x7
(0, 0)
Let’s summarize the techniques
for graphing.
2. Find the critical points, extrema, intervals of
increasing and decreasing using signs of y‘. Confirm
these information graphically and refine our graph.
1. Get as much information as possible from y with
the sign chart. Find the limits of the boundary values.
Use these to get the general shape of the graph.
For y = x3, 5, 7., y'' = #x0dd,
the 2nd derivative changes
signs at (0, 0), hence it’s an
inflection point.

48. Extrema and the Second Derivative
y = x3
y = x5
y = x7
(0, 0)
Let’s summarize the techniques
for graphing.
1. Get as much information as possible from y with
the sign chart. Find the limits of the boundary values.
Use these to get the general shape of the graph.
2. Find the critical points, extrema, intervals of
increasing and decreasing using signs of y‘. Confirm
these information graphically and refine our graph.
3. Find the concavity and the inflection point using
y'' (sign–chart). Fill in more details of the graph.
For y = x3, 5, 7., y'' = #x0dd,
the 2nd derivative changes
signs at (0, 0), hence it’s an
inflection point.

49. Example A. Graph y = 3x5 – 5x3.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative

50. Example A. Graph y = 3x5 – 5x3.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
Set y = 3x5 – 5x3 = 0

51. Example A. Graph y = 3x5 – 5x3.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
Set y = 3x5 – 5x3 = 0
x3(3x2 – 5) = 0
so x = 0, ±√5/3

52. Example A. Graph y = 3x5 – 5x3.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
x
y = 3x5 – 5x3
Set y = 3x5 – 5x3 = 0
x3(3x2 – 5) = 0
so x = 0, ±√5/3
– √5/3 √5/30The sign–chart of y is ++– –

53. Example A. Graph y = 3x5 – 5x3.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
x
y = 3x5 – 5x3
Set y = 3x5 – 5x3 = 0
x3(3x2 – 5) = 0
so x = 0, ±√5/3
– √5/3 √5/30The sign–chart of y is ++– –
y' = 15x4 – 15x2 = 0

54. Example A. Graph y = 3x5 – 5x3.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
x
y = 3x5 – 5x3
Set y = 3x5 – 5x3 = 0
x3(3x2 – 5) = 0
so x = 0, ±√5/3
– √5/3 √5/30The sign–chart of y is ++– –
y' = 15x4 – 15x2 = 0
15x2(x2 – 1) = 0 or x = 0, ±1

55. Example A. Graph y = 3x5 – 5x3.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
x
y = 3x5 – 5x3
Set y = 3x5 – 5x3 = 0
x3(3x2 – 5) = 0
so x = 0, ±√5/3
– √5/3 √5/30The sign–chart of y is ++– –
y' = 15x4 – 15x2 = 0
15x2(x2 – 1) = 0 or x = 0, ±1
x
– 1 1
0The sign–chart of y' is ++ ––
y' = 15x4 – 15x2

56. Extrema and the Second Derivative
x
x
y = 3x5 – 5x3
– √5/3 √5/30
The sign–chart of y
++– – y = 3x5 – 5x3
√5/3– √5/3

57. Extrema and the Second Derivative
The sign–chart of y
x
x
– √5/3 √5/30
++– – y = 3x5 – 5x3
– √5/3 √5/3
y = 3x5 – 5x3

58. Extrema and the Second Derivative
x
–
(0, 0)
x
y = 3x5 – 5x3
– √5/3 √5/30
The sign–chart of y
++– –
x
– 1 1
0
The sign–chart of y'
++ –– y' = 15x4 – 15x2
y = 3x5 – 5x3
(– 1, – 2)
– √5/3 √5/3
(– 1, 2)

59. Extrema and the Second Derivative
x
– √5/3 √5/3,0)
–
(0, 0)
x
y = 3x5 – 5x3
– √5/3 √5/30
The sign–chart of y
++– –
x
– 1 1
0
The sign–chart of y'
++ –– y' = 15x4 – 15x2
y = 3x5 – 5x3
(– 1, 2)
(– 1, – 2)

60. Extrema and the Second Derivative
y '' = 60x3 – 30x = 0

61. Extrema and the Second Derivative
y '' = 60x3 – 30x = 0
30x(2x2 – 1) = 0
x = 0, ±√1/2

62. Extrema and the Second Derivative
y '' = 60x3 – 30x = 0
– √1/2 0
The sign–chart of y'' is
++ ––
30x(2x2 – 1) = 0
x = 0, ±√1/2
√1/2
x

63. Extrema and the Second Derivative
y '' = 60x3 – 30x = 0
All three points are inflection points because the
concavity changes at each one of them.
– √1/2 0
The sign–chart of y'' is
++ ––
30x(2x2 – 1) = 0
x = 0, ±√1/2
√1/2
Increasing: (–∞,–1), (1, ∞).
Decreasing: (–1, 0), (0, –1).
Concave up: (–√1/2, 0), (√1/2, ∞).
Concave down: (–√1/2, 0), (0, √1/2).
x

64. Extrema and the Second Derivative
Increasing: (–∞,–1), (1, ∞)
Decreasing: (–1, 0), (0, –1)
Concave up: (–√1/2, 0), (√1/2, ∞)
Concave down: (–√1/2, 0), (0, √1/2)
inflection points
Your turn: Find the coordinate of the inflection points.
0
Intervals:

65. Example B. Graph y =
Extrema and the Second Derivative
x
x2 + 4 .
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.

66. Example B. Graph y =
Extrema and the Second Derivative
x
x2 + 4 .
Note that the domain is the set of all real numbers
because the denominator is always positive.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.

67. Example B. Graph y =
Extrema and the Second Derivative
x
x2 + 4 .
Note that the domain is the set of all real numbers
because the denominator is always positive.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
x
Set y = 0, we have that x = 0
0
The sign–chart of y is +–
x
x2 + 4y =

68. Example B. Graph y =
Extrema and the Second Derivative
x
x2 + 4 .
x
Set y = 0, we have that x = 0
0
The sign–chart of y is +–
y' =
set y' = 0, (4 – x2 ) = 0 or x = ±2
x
x2 + 4y =
Note that the domain is the set of all real numbers
because the denominator is always positive.
4 – x2
(x2 + 4)2
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
The derivative is ,

69. Example B. Graph y =
Extrema and the Second Derivative
x
x2 + 4 .
x
Set y = 0, we have that x = 0
0
The sign–chart of y is +–
y' =
set y' = 0, (4 – x2 ) = 0 or x = ±2
– 2
The sign–chart of y' is +
x
x2 + 4y =
Note that the domain is the set of all real numbers
because the denominator is always positive.
4 – x2
(x2 + 4)2
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
The derivative is ,
0
–
2
–

70. As x  ∞, y goes to 0+, as x  –∞, y goes to 0– .
Extrema and the Second Derivative

71. x
x
x2 + 4y =
As x  ∞, y goes to 0+, as x  –∞, y goes to 0– .
y goes to 0+
y goes to 0–
x
– 2
+–
0
– 4 – x2
(x2 + 4)2y' =
Extrema and the Second Derivative
2

72. x
x
x2 + 4y =
(2, 1/4)
(2, – 1/4)
As x  ∞, y goes to 0+, as x  –∞, y goes to 0– .
y goes to 0+
y goes to 0–
x
– 2
+–
0
– 4 – x2
(x2 + 4)2y' =
Extrema and the Second Derivative
2

73. x
x
x2 + 4y =
(2, 1/4)
(2, – 1/4)
As x  ∞, y goes to 0+, as x  –∞, y goes to 0– .
y goes to 0+
y goes to 0–
x
– 2
+–
0
– 4 – x2
(x2 + 4)2y' =
Extrema and the Second Derivative
2

74.  x = 0, ±√12 all are inflection points.
x
x
x2 + 4y =
(2, 1/4)
(2, – 1/4)
As x  ∞, y goes to 0+, as x  –∞, y goes to 0– .
y goes to 0+
y goes to 0–
y'' =
2x(x2 – 12)
(x2 + 4)3
Find the 2nd derivative
x
– 2 2
+–
0
– 4 – x2
(x2 + 4)2y' =
Extrema and the Second Derivative

75.  x = 0, ±√12 all are inflection points.
x
x
x2 + 4y =
(2, 1/4)
(2, – 1/4)
As x  ∞, y goes to 0+, as x  –∞, y goes to 0– .
y goes to 0+
y goes to 0–
y'' =
2x(x2 – 12)
(x2 + 4)3
Find the 2nd derivative
x
– √12
––
(x2 + 4)3y '' =
2x(x2 – 129)
++
0
x
– 2 2
+–
0
– 4 – x2
(x2 + 4)2y' =
Extrema and the Second Derivative
√12

76.  x = 0, ±√12 all are inflection points.
x
x
x2 + 4y =
(2, 1/4)
(2, – 1/4)
As x  ∞, y goes to 0+, as x  –∞, y goes to 0– .
y goes to 0+
y goes to 0–
y'' =
2x(x2 – 12)
(x2 + 4)3
Find the 2nd derivative
x––
(x2 + 4)3y '' =
2x(x2 – 12)
++
0
x
– 2 2
+–
0
– 4 – x2
(x2 + 4)2y' =
Extrema and the Second Derivative
– √12 √12
(√12, (√3)/8)
(√12, (– √3)/8)

77.  x = 0, ±√12 all are inflection points.
x
x
x2 + 4y =
(2, 1/4)
(2, – 1/4)
As x  ∞, y goes to 0+, as x  –∞, y goes to 0– .
y goes to 0+
y goes to 0–
y'' =
2x(x2 – 12)
(x2 + 4)3
Find the 2nd derivative
x––
(x2 + 4)3y '' =
2x(x2 – 12)
++
0
(You list the intervals asked in the question.)
x
– 2 2
+–
0
– 4 – x2
(x2 + 4)2y' =
Extrema and the Second Derivative
– √12 √12
(√12, (– √3)/8)
(√12, (√3)/8)

78. Example C. Graph y = 5x2/5 – 2x.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative

79. Example C. Graph y = 5x2/5 – 2x.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
y = 5x2/5 – 2x = 0

80. Example C. Graph y = 5x2/5 – 2x.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
y = 5x2/5 – 2x = 0 pull out the lowest degree of x
x2/5(5 – 2x3/5) = 0

81. Example C. Graph y = 5x2/5 – 2x.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
y = 5x2/5 – 2x = 0 pull out the lowest degree of x
x2/5(5 – 2x3/5) = 0  x = 0 or
5 – 2x3/5 = 0

82. Example C. Graph y = 5x2/5 – 2x.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
y = 5x2/5 – 2x = 0 pull out the lowest degree of x
x2/5(5 – 2x3/5) = 0  x = 0 or
5 – 2x3/5 = 0  5 = 2x3/5 so
(5/2)5/3 = x ≈ 4.61

83. Example C. Graph y = 5x2/5 – 2x.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
y = 5x2/5 – 2x = 0 pull out the lowest degree of x
x2/5(5 – 2x3/5) = 0  x = 0 or
5 – 2x3/5 = 0  5 = 2x3/5 so
(5/2)5/3 = x ≈ 4.61
The sign–chart of y is x+ –
0
–
(5/2)5/3

84. Example C. Graph y = 5x2/5 – 2x.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
y = 5x2/5 – 2x = 0 pull out the lowest degree of x
x2/5(5 – 2x3/5) = 0  x = 0 or
5 – 2x3/5 = 0  5 = 2x3/5 so
(5/2)5/3 = x ≈ 4.61
The sign–chart of y is x+ –
0
–
(5/2)5/3
The derivative is y' = 2x–3/5 – 2
set y' = 0,

85. Example C. Graph y = 5x2/5 – 2x.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
y = 5x2/5 – 2x = 0 pull out the lowest degree of x
x2/5(5 – 2x3/5) = 0  x = 0 or
5 – 2x3/5 = 0  5 = 2x3/5 so
(5/2)5/3 = x ≈ 4.61
The sign–chart of y is x+ –
0
–
(5/2)5/3
The derivative is y' = 2x–3/5 – 2
set y' = 0, 2x–3/5(1 – x3/5) = 0 pull out the lowest
degree of x

86. Example C. Graph y = 5x2/5 – 2x.
Find all critical points, extrema, inflection points,
intervals of increasing and decreasing, intervals of
concave–up and intervals of concave-down.
Extrema and the Second Derivative
y = 5x2/5 – 2x = 0 pull out the lowest degree of x
x2/5(5 – 2x3/5) = 0  x = 0 or
5 – 2x3/5 = 0  5 = 2x3/5 so
(5/2)5/3 = x ≈ 4.61
The sign–chart of y is x+ –
0
–
(5/2)5/3
The derivative is y' = 2x–3/5 – 2
set y' = 0, 2x–3/5(1 – x3/5) = 0
or 1 – x3/5 = 0  x = 1.
pull out the lowest
degree of x

87. Extrema and the Second Derivative
x
(0, 0)
(5/2)5/3
Roots at x=0, (5/2)5/3
y = 5x2/5 – 2x

88. Extrema and the Second Derivative
The sign–chart of y' is x+ –
0
–
1
y' = 2x–3/5 – 2
x
(0, 0)
(5/2)5/3
Roots at x=0, (5/2)5/3
y = 5x2/5 – 2x

89. Extrema and the Second Derivative
The sign–chart of y' is x+ –
0
–
1
y' = 2x–3/5 – 2
x
(1, 3)
y = 5x2/5 – 2x
(0, 0)
(5/2)5/3
Roots at x=0, (5/2)5/3

90. Extrema and the Second Derivative
The sign–chart of y' is x+ –
0
–
1
y' = 2x–3/5 – 2
As x  ∞, y goes to –∞, as x  –∞, y goes to ∞.
x
(1, 3)
y = 5x2/5 – 2x
(0, 0)
(5/2)5/3
Roots at x=0, (5/2)5/3

91. Extrema and the Second Derivative
The sign–chart of y' is x+ –
0
–
1
y' = 2x–3/5 – 2
As x  ∞, y goes to –∞, as x  –∞, y goes to ∞.
x
(1, 3)
y = 5x2/5 – 2x
y goes to –∞
y goes to ∞
(0, 0)
(5/2)5/3
Roots at x=0, (5/2)5/3

92. Extrema and the Second Derivative
The sign–chart of y' is x+ –
0
–
1
y' = 2x–3/5 – 2
As x  ∞, y goes to –∞, as x  –∞, y goes to ∞.
x
(1, 3)
y = 5x2/5 – 2x
y goes to –∞
y goes to ∞
(0, 0)
The graph is shown below.
(5/2)5/3
Roots at x=0, (5/2)5/3

93. Extrema and the Second Derivative
x
(1, 3)
y = 5x2/5 – 2x
(0, 0)
(5/2)5/3
The 2nd derivative is y'' = (–6/5)x–8/5 which is always
negative so y is always concave down when x ≠ 0.
At x = 0, we have a cusp (why?).
y = 5x2/5 – 2x
Roots at x=0, (5/2)5/3