There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.

1. Sec on 4.1
Maximum and Minimum Values
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University
April 4, 2011
.

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

3. Objectives
Understand and be able to
explain the statement of the
Extreme Value Theorem.
Understand and be able to
explain the statement of
Fermat’s Theorem.
Use the Closed Interval Method
to find the extreme values of a
func on defined on a closed
interval.

4. Outline
Introduc on
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples

5. Optimize
.

6. Why go to the extremes?
Ra onally speaking, it is
advantageous to find the
extreme values of a func on
(maximize profit, minimize costs,
etc.)
Pierre-Louis Maupertuis
(1698–1759)

7. Design
.

8. Why go to the extremes?
Ra onally speaking, it is
advantageous to find the
extreme values of a func on
(maximize profit, minimize costs,
etc.)
Many laws of science are
derived from minimizing
principles.
Pierre-Louis Maupertuis
(1698–1759)

9. Optics
.

10. Why go to the extremes?
Ra onally speaking, it is
advantageous to find the
extreme values of a func on
(maximize profit, minimize costs,
etc.)
Many laws of science are
derived from minimizing
principles.
Maupertuis’ principle: “Ac on is
minimized through the wisdom Pierre-Louis Maupertuis
of God.” (1698–1759)

11. Outline
Introduc on
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples

12. Extreme points and values
Defini on
Let f have domain D.
.
Image credit: Patrick Q

13. Extreme points and values
Defini on
Let f have domain D.
The func on f has an absolute maximum
(or global maximum) (respec vely,
absolute minimum) at c if f(c) ≥ f(x)
(respec vely, f(c) ≤ f(x)) for all x in D
.
Image credit: Patrick Q

14. Extreme points and values
Defini on
Let f have domain D.
The func on f has an absolute maximum
(or global maximum) (respec vely,
absolute minimum) at c if f(c) ≥ f(x)
(respec vely, f(c) ≤ f(x)) for all x in D
The number f(c) is called the maximum
value (respec vely, minimum value) of f
on D.
.
Image credit: Patrick Q

15. Extreme points and values
Defini on
Let f have domain D.
The func on f has an absolute maximum
(or global maximum) (respec vely,
absolute minimum) at c if f(c) ≥ f(x)
(respec vely, f(c) ≤ f(x)) for all x in D
The number f(c) is called the maximum
value (respec vely, minimum value) of f
on D.
An extremum is either a maximum or a .
minimum. An extreme value is either a
maximum value or minimum value.
Image credit: Patrick Q

16. The Extreme Value Theorem
Theorem (The Extreme Value
Theorem)
Let f be a func on which is
con nuous on the closed
interval [a, b]. Then f a ains
an absolute maximum value
f(c) and an absolute minimum
value f(d) at numbers c and d
in [a, b].

17. The Extreme Value Theorem
Theorem (The Extreme Value
Theorem)
Let f be a func on which is
con nuous on the closed
interval [a, b]. Then f a ains
an absolute maximum value
f(c) and an absolute minimum
value f(d) at numbers c and d .
in [a, b]. a b

18. The Extreme Value Theorem
Theorem (The Extreme Value
Theorem) maximum
value
Let f be a func on which is f(c)
con nuous on the closed
interval [a, b]. Then f a ains
an absolute maximum value
f(c) and an absolute minimum
value f(d) at numbers c and d .
a c
in [a, b]. b
maximum

19. The Extreme Value Theorem
Theorem (The Extreme Value
Theorem) maximum
value
Let f be a func on which is f(c)
con nuous on the closed
interval [a, b]. Then f a ains minimum
an absolute maximum value value
f(c) and an absolute minimum f(d)
value f(d) at numbers c and d .
a d c
in [a, b]. b
minimum maximum

20. No proof of EVT forthcoming
This theorem is very hard to prove without using technical facts
about con nuous func ons and closed intervals.
But we can show the importance of each of the hypotheses.

21. Bad Example #1
Example
Consider the func on
{
x 0≤x<1
f(x) =
x − 2 1 ≤ x ≤ 2.

22. Bad Example #1
Example
Consider the func on
{
x 0≤x<1 .
f(x) = |
x − 2 1 ≤ x ≤ 2. 1

23. Bad Example #1
Example
Consider the func on
{
x 0≤x<1 .
f(x) = |
x − 2 1 ≤ x ≤ 2. 1
Then although values of f(x) get arbitrarily close to 1 and never
bigger than 1, 1 is not the maximum value of f on [0, 1] because it is
never achieved.

24. Bad Example #1
Example
Consider the func on
{
x 0≤x<1 .
f(x) = |
x − 2 1 ≤ x ≤ 2. 1
Then although values of f(x) get arbitrarily close to 1 and never
bigger than 1, 1 is not the maximum value of f on [0, 1] because it is
never achieved. This does not violate EVT because f is not
con nuous.

25. Bad Example #2
Example
Consider the func on f(x) = x restricted to the interval [0, 1).
. |
1

26. Bad Example #2
Example
Consider the func on f(x) = x restricted to the interval [0, 1).
There is s ll no maximum
value (values get
arbitrarily close to 1 but
do not achieve it).
. |
1

27. Bad Example #2
Example
Consider the func on f(x) = x restricted to the interval [0, 1).
There is s ll no maximum
value (values get
arbitrarily close to 1 but
do not achieve it).
This does not violate EVT
. |
because the domain is 1
not closed.

28. Final Bad Example
Example
1
The func on f(x) = is con nuous on the closed interval [1, ∞).
x
.
1

29. Final Bad Example
Example
1
The func on f(x) = is con nuous on the closed interval [1, ∞).
x
.
1
There is no minimum value (values get arbitrarily close to 0 but do
not achieve it).

30. Final Bad Example
Example
1
The func on f(x) = is con nuous on the closed interval [1, ∞).
x
.
1
There is no minimum value (values get arbitrarily close to 0 but do
not achieve it). This does not violate EVT because the domain is not
bounded.

31. Outline
Introduc on
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples

32. Local extrema
Defini on
A func on f has a local
maximum or rela ve maximum
at c if f(c) ≥ f(x) when x is near
c. This means that f(c) ≥ f(x)
for all x in some open interval
containing c. |. |
a b
Similarly, f has a local minimum
at c if f(c) ≤ f(x) when x is near
c.

33. Local extrema
Defini on
A func on f has a local
maximum or rela ve maximum
at c if f(c) ≥ f(x) when x is near
c. This means that f(c) ≥ f(x)
for all x in some open interval
containing c. |. |
a
local b
Similarly, f has a local minimum maximum
at c if f(c) ≤ f(x) when x is near
c.

34. Local extrema
Defini on
A func on f has a local
maximum or rela ve maximum
at c if f(c) ≥ f(x) when x is near
c. This means that f(c) ≥ f(x)
for all x in some open interval
containing c. |. |
local local b
a
Similarly, f has a local minimum maximum minimum
at c if f(c) ≤ f(x) when x is near
c.

35. Local extrema
A local extremum could be a
global extremum, but not if
there are more extreme values
elsewhere.
A global extremum could be a
local extremum, but not if it is
an endpoint.
|. |
a b
local local and global
maximum global max
min

36. Fermat’s Theorem
Theorem (Fermat’s Theorem)
Suppose f has a
local extremum at c
and f is
differen able at c.
Then f′ (c) = 0. |. |
a local local b
maximum minimum

37. Fermat’s Theorem
Theorem (Fermat’s Theorem)
Suppose f has a
local extremum at c
and f is
differen able at c.
Then f′ (c) = 0. |. |
a local local b
maximum minimum

38. Proof of Fermat’s Theorem
Suppose that f has a local maximum at c.

39. Proof of Fermat’s Theorem
Suppose that f has a local maximum at c.
If x is slightly greater than c, f(x) ≤ f(c). This means
f(x) − f(c)
≤0
x−c

40. Proof of Fermat’s Theorem
Suppose that f has a local maximum at c.
If x is slightly greater than c, f(x) ≤ f(c). This means
f(x) − f(c) f(x) − f(c)
≤ 0 =⇒ lim ≤0
x−c x→c+ x−c

41. Proof of Fermat’s Theorem
Suppose that f has a local maximum at c.
If x is slightly greater than c, f(x) ≤ f(c). This means
f(x) − f(c) f(x) − f(c)
≤ 0 =⇒ lim ≤0
x−c x→c+ x−c
The same will be true on the other end: if x is slightly less than
c, f(x) ≤ f(c). This means
f(x) − f(c)
≥0
x−c

42. Proof of Fermat’s Theorem
Suppose that f has a local maximum at c.
If x is slightly greater than c, f(x) ≤ f(c). This means
f(x) − f(c) f(x) − f(c)
≤ 0 =⇒ lim ≤0
x−c x→c+ x−c
The same will be true on the other end: if x is slightly less than
c, f(x) ≤ f(c). This means
f(x) − f(c) f(x) − f(c)
≥ 0 =⇒ lim ≥0
x−c x→c− x−c

43. Proof of Fermat’s Theorem
Suppose that f has a local maximum at c.
If x is slightly greater than c, f(x) ≤ f(c). This means
f(x) − f(c) f(x) − f(c)
≤ 0 =⇒ lim ≤0
x−c x→c+ x−c
The same will be true on the other end: if x is slightly less than
c, f(x) ≤ f(c). This means
f(x) − f(c) f(x) − f(c)
≥ 0 =⇒ lim ≥0
x−c x→c− x−c
f(x) − f(c)
Since the limit f′ (c) = lim exists, it must be 0.
x→c x−c

44. Meet the Mathematician: Pierre de Fermat
1601–1665
Lawyer and number
theorist
Proved many theorems,
didn’t quite prove his last
one

45. Tangent: Fermat’s Last Theorem
Plenty of solu ons to
x2 + y2 = z2 among posi ve
whole numbers (e.g., x = 3,
y = 4, z = 5)

46. Tangent: Fermat’s Last Theorem
Plenty of solu ons to
x2 + y2 = z2 among posi ve
whole numbers (e.g., x = 3,
y = 4, z = 5)
No solu ons to x3 + y3 = z3
among posi ve whole numbers

47. Tangent: Fermat’s Last Theorem
Plenty of solu ons to
x2 + y2 = z2 among posi ve
whole numbers (e.g., x = 3,
y = 4, z = 5)
No solu ons to x3 + y3 = z3
among posi ve whole numbers
Fermat claimed no solu ons to
xn + yn = zn but didn’t write
down his proof

48. Tangent: Fermat’s Last Theorem
Plenty of solu ons to
x2 + y2 = z2 among posi ve
whole numbers (e.g., x = 3,
y = 4, z = 5)
No solu ons to x3 + y3 = z3
among posi ve whole numbers
Fermat claimed no solu ons to
xn + yn = zn but didn’t write
down his proof
Not solved un l 1998!
(Taylor–Wiles)

49. Outline
Introduc on
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples

50. Flowchart for placing extrema
Thanks to Fermat
Suppose f is a c is a
.
start
con nuous local max
func on on
the closed,
bounded Is c an
no Is f diff’ble no f is not
interval endpoint? at c? diff at c
[a, b], and c is
a global
yes yes
maximum
c = a or
point. f′ (c) = 0
c = b

51. 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 cri cal points or cri cal numbers x where
either f′ (x) = 0 or f is not differen able at x.
The points with the largest func on value are the global
maximum points
The points with the smallest or most nega ve func on value
are the global minimum points.

52. Outline
Introduc on
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples

53. Extreme values of a linear function
Example
Find the extreme values of f(x) = 2x − 5 on [−1, 2].

54. Extreme values of a linear function
Example
Find the extreme values of f(x) = 2x − 5 on [−1, 2].
Solu on
Since f′ (x) = 2, which is never
zero, we have no cri cal points
and we need only inves gate
the endpoints:
f(−1) = 2(−1) − 5 = −7
f(2) = 2(2) − 5 = −1

55. Extreme values of a linear function
Example
Find the extreme values of f(x) = 2x − 5 on [−1, 2].
Solu on
So
Since f′ (x) = 2, which is never
zero, we have no cri cal points The absolute minimum
and we need only inves gate (point) is at −1; the
the endpoints: minimum value is −7.
f(−1) = 2(−1) − 5 = −7 The absolute maximum
(point) is at 2; the
f(2) = 2(2) − 5 = −1
maximum value is −1.

56. Extreme values of a quadratic
function
Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].

57. Extreme values of a quadratic
function
Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solu on
We have f′ (x) = 2x, which is zero when x = 0.

58. Extreme values of a quadratic
function
Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solu on
We have f′ (x) = 2x, which is zero when x = 0. So our points to
check are:
f(−1) =
f(0) =
f(2) =

59. Extreme values of a quadratic
function
Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solu on
We have f′ (x) = 2x, which is zero when x = 0. So our points to
check are:
f(−1) = 0
f(0) =
f(2) =

60. Extreme values of a quadratic
function
Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solu on
We have f′ (x) = 2x, which is zero when x = 0. So our points to
check are:
f(−1) = 0
f(0) = − 1
f(2) =

61. Extreme values of a quadratic
function
Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solu on
We have f′ (x) = 2x, which is zero when x = 0. So our points to
check are:
f(−1) = 0
f(0) = − 1
f(2) = 3

62. Extreme values of a quadratic
function
Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solu on
We have f′ (x) = 2x, which is zero when x = 0. So our points to
check are:
f(−1) = 0
f(0) = − 1 (absolute min)
f(2) = 3

63. Extreme values of a quadratic
function
Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solu on
We have f′ (x) = 2x, which is zero when x = 0. So our points to
check are:
f(−1) = 0
f(0) = − 1 (absolute min)
f(2) = 3 (absolute max)

64. Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].

65. Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solu on
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
and x = 1.

66. Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solu on
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
and x = 1. The values to check are

67. Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solu on
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
and x = 1. The values to check are
f(−1) = − 4

68. Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solu on
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
and x = 1. The values to check are
f(−1) = − 4
f(0) = 1

69. Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solu on
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
and x = 1. The values to check are
f(−1) = − 4
f(0) = 1
f(1) = 0

70. Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solu on
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
and x = 1. The values to check are
f(−1) = − 4
f(0) = 1
f(1) = 0
f(2) = 5

71. Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solu on
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
and x = 1. The values to check are
f(−1) = − 4 (global min)
f(0) = 1
f(1) = 0
f(2) = 5

72. Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solu on
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
and x = 1. The values to check are
f(−1) = − 4 (global min)
f(0) = 1
f(1) = 0
f(2) = 5 (global max)

73. Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solu on
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
and x = 1. The values to check are
f(−1) = − 4 (global min)
f(0) = 1 (local max)
f(1) = 0
f(2) = 5 (global max)

74. Extreme values of a cubic function
Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solu on
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have cri cal points at x = 0
and x = 1. The values to check are
f(−1) = − 4 (global min)
f(0) = 1 (local max)
f(1) = 0 (local min)
f(2) = 5 (global max)

75. Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].

76. Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solu on
Write f(x) = x5/3 + 2x2/3 .

77. Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solu on
Write f(x) = x5/3 + 2x2/3 . Then
5 4 1
f′ (x) = x2/3 + x−1/3 = x−1/3 (5x + 4)
3 3 3

78. Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solu on
Write f(x) = x5/3 + 2x2/3 . Then
5 4 1
f′ (x) = x2/3 + x−1/3 = x−1/3 (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not differen able at 0. Thus there are two
cri cal points.

79. Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) =
f(−4/5) =
f(0) =
f(2) =

80. Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) =
f(0) =
f(2) =

81. Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) = 1.0341
f(0) =
f(2) =

82. Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0
f(2) =

83. Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0
f(2) = 6.3496

84. Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0 (absolute min)
f(2) = 6.3496

85. Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0 (absolute min)
f(2) = 6.3496 (absolute max)

86. Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solu on
Write f(x) = x5/3 + 2x2/3 .
f(−1) = 1
f(−4/5) = 1.0341 (rela ve max)
f(0) = 0 (absolute min)
f(2) = 6.3496 (absolute max)

87. Extreme values of another algebraic function
Example
√
Find the extreme values of f(x) = 4 − x2 on [−2, 1].

88. Extreme values of another algebraic function
Example
√
Find the extreme values of f(x) = 4 − x2 on [−2, 1].
Solu on
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differen able at ±2 as well.)

89. Extreme values of another algebraic function
Example
√
Find the extreme values of f(x) = 4 − x2 on [−2, 1].
Solu on
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differen able at ±2 as well.) So our points to check are:
f(−2) =

90. Extreme values of another algebraic function
Example
√
Find the extreme values of f(x) = 4 − x2 on [−2, 1].
Solu on
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differen able at ±2 as well.) So our points to check are:
f(−2) = 0
f(0) =

91. Extreme values of another algebraic function
Example
√
Find the extreme values of f(x) = 4 − x2 on [−2, 1].
Solu on
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differen able at ±2 as well.) So our points to check are:
f(−2) = 0
f(0) = 2
f(1) =

92. Extreme values of another algebraic function
Example
√
Find the extreme values of f(x) = 4 − x2 on [−2, 1].
Solu on
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differen able at ±2 as well.) So our points to check are:
f(−2) = 0
f(0) = 2
√
f(1) = 3

93. Extreme values of another algebraic function
Example
√
Find the extreme values of f(x) = 4 − x2 on [−2, 1].
Solu on
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differen able at ±2 as well.) So our points to check are:
f(−2) = 0 (absolute min)
f(0) = 2
√
f(1) = 3

94. Extreme values of another algebraic function
Example
√
Find the extreme values of f(x) = 4 − x2 on [−2, 1].
Solu on
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differen able at ±2 as well.) So our points to check are:
f(−2) = 0 (absolute min)
f(0) = 2 (absolute max)
√
f(1) = 3

95. Summary
The Extreme Value Theorem: a con nuous func on on a closed
interval must achieve its max and min
Fermat’s Theorem: local extrema are cri cal points
The Closed Interval Method: an algorithm for finding global
extrema
Show your work unless you want to end up like Fermat!