We define what it means for a function to have a maximum or minimum value, and explain the Extreme Value Theorem, which indicates these maxima and minima must be there under certain conditions. Fermat's Theorem says that at differentiable extreme points, the derivative should be zero, and thus we arrive at a technique for finding extrema: look among the endpoints of the domain of definition and the critical points of the function. There's also a little digression on Fermat's Last theorem, which is not related to calculus but is a big deal in recent mathematical history.

1. Section 4.1
Maximum and Minimum Values
V63.0121, Calculus I
March 24, 2009
Announcements
Homework due Thursday
Quiz April 2, on Sections 2.5–3.5
Final Exam Friday, May 8, 2:00–3:50pm
.
.
Image: Flickr user Karen with a K
. . . . . .

2. Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
. . . . . .

3. Optimize
. . . . . .

4. Why go to the extremes?
Rationally speaking, it is
advantageous to find the
extreme values of a
function (maximize profit,
minimize costs, etc.)
Pierre-Louis Maupertuis
(1698–1759)
. . . . . .

5. Design
.
.
Image credit: Jason Tromm
. . . . . .

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

7. Optics
.
.
Image credit: jacreative
. . . . . .

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

9. Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
. . . . . .

10. Extreme points and values
Definition
Let f have domain D.
The function f has an absolute
maximum (or global maximum)
(respectively, absolute minimum)
at c if f(c) ≥ f(x) (respectively,
f(c) ≤ f(x)) for all x in D
.
.
Image credit: Patrick Q
. . . . . .

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

12. Extreme points and values
Definition
Let f have domain D.
The function f has an absolute
maximum (or global maximum)
(respectively, absolute minimum)
at c if f(c) ≥ f(x) (respectively,
f(c) ≤ f(x)) for all x in D
The number f(c) is called the
maximum value (respectively,
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
. . . . . .

13. Theorem (The Extreme Value Theorem)
Let f be a function which is continuous on the closed interval [a, b]. Then f
attains an absolute maximum value f(c) and an absolute minimum value
f(d) at numbers c and d in [a, b].
. . . . . .

14. Theorem (The Extreme Value Theorem)
Let f be a function which is continuous on the closed interval [a, b]. Then f
attains an absolute maximum value f(c) and an absolute minimum value
f(d) at numbers c and d in [a, b].
.
.
. .
a
. b
.
. . . . . .

15. Theorem (The Extreme Value Theorem)
Let f be a function which is continuous on the closed interval [a, b]. Then f
attains an absolute maximum value f(c) and an absolute minimum value
f(d) at numbers c and d in [a, b].
.
maximum . (c)
f
.
value
.
.
minimum . (d)
f
.
value
. ..
. c
d
a
. b
.
maximum
minimum
. . . . . .

16. No proof of EVT forthcoming
This theorem is very hard to prove without using technical facts
about continuous functions and closed intervals.
But we can show the importance of each of the hypotheses.
. . . . . .

17. Bad Example #1
Example
Consider the function
{
0≤x<1
x
f(x) =
x−2 1 ≤ x ≤ 2.
. . . . . .

18. Bad Example #1
Example
Consider the function
{
0≤x<1
x
f(x) =
x−2 1 ≤ x ≤ 2.
.
. .
.
|
1
.
.
. . . . . .

19. Bad Example #1
Example
Consider the function
{
0≤x<1
x
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.
. . . . . .

20. Bad Example #2
Example
The function f(x) = x restricted to the interval [0, 1) still has no
maximum value.
. . . . . .

21. Bad Example #2
Example
The function f(x) = x restricted to the interval [0, 1) still has no
maximum value.
.
. .
|
1
.
. . . . . .

22. Final Bad Example
Example
1
The function f(x) = is continuous on the closed interval [1, ∞) but
x
has no minimum value.
. . . . . .

23. Final Bad Example
Example
1
The function f(x) = is continuous on the closed interval [1, ∞) but
x
has no minimum value.
.
. .
1
.
. . . . . .

24. Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
. . . . . .

25. Local extrema
Definition
A function f has a local maximum or relative 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.
Similarly, f has a local minimum at c if f(c) ≤ f(x) when x is
near c.
. . . . . .

26. Local extrema
Definition
A function f has a local maximum or relative 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.
Similarly, f has a local minimum at c if f(c) ≤ f(x) when x is
near c.
.
.
.
.
....
| |
.
.
a local local b
.
maximum minimum
. . . . . .

27. So a local extremum must be inside the domain of f (not on the
end).
A global extremum that is inside the domain is a local extremum.
.
.
.
.
....
| |.
.
.
a local local and global . global
b
max min max
. . . . . .

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

29. Sketch of proof of Fermat’s Theorem
Suppose that f has a local maximum at c.
. . . . . .

30. Sketch of proof of Fermat’s Theorem
Suppose that f has a local maximum at c.
If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This
means
f(c + h) − f(c)
≤0
h
. . . . . .

31. Sketch of proof of Fermat’s Theorem
Suppose that f has a local maximum at c.
If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This
means
f(c + h) − f(c) f(c + h) − f(c)
≤ 0 =⇒ lim+ ≤0
h h
h→0
. . . . . .

32. Sketch of proof of Fermat’s Theorem
Suppose that f has a local maximum at c.
If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This
means
f(c + h) − f(c) f(c + h) − f(c)
≤ 0 =⇒ lim+ ≤0
h h
h→0
The same will be true on the other end: if h is close enough to 0
but less than 0, f(c + h) ≤ f(c). This means
f(c + h) − f(c)
≥0
h
. . . . . .

33. Sketch of proof of Fermat’s Theorem
Suppose that f has a local maximum at c.
If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This
means
f(c + h) − f(c) f(c + h) − f(c)
≤ 0 =⇒ lim+ ≤0
h h
h→0
The same will be true on the other end: if h is close enough to 0
but less than 0, f(c + h) ≤ f(c). This means
f(c + h) − f(c) f(c + h) − f(c)
≥ 0 =⇒ lim ≥0
h − h
h→0
. . . . . .

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

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

36. Tangent: Fermat’s Last Theorem
Plenty of solutions to
x2 + y2 = z2 among
positive whole numbers
(e.g., x = 3, y = 4, z = 5)
. . . . . .

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

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

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

40. Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
. . . . . .

41. The Closed Interval Method
Let’s put this together logically. Let f be a continuous function
defined on a closed interval [a, b]. We are in search of its global
maximum, call it c. Then:
. . . . . .

42. The Closed Interval Method
Let’s put this together logically. Let f be a continuous function
defined on a closed interval [a, b]. We are in search of its global
maximum, call it c. Then:
Either the maximum
occurs at an endpoint of
the interval, i.e., c = a or
c = b,
. . . . . .

43. The Closed Interval Method
Let’s put this together logically. Let f be a continuous function
defined on a closed interval [a, b]. We are in search of its global
maximum, call it c. Then:
Either the maximum
occurs at an endpoint of
the interval, i.e., c = a or
c = b,
Or the maximum occurs
inside (a, b). In this case, c
is also a local maximum.
. . . . . .

44. The Closed Interval Method
Let’s put this together logically. Let f be a continuous function
defined on a closed interval [a, b]. We are in search of its global
maximum, call it c. Then:
Either the maximum
occurs at an endpoint of
the interval, i.e., c = a or
c = b,
Or the maximum occurs
inside (a, b). In this case, c
is also a local maximum.
Either f is differentiable
at c, in which case
f′ (c) = 0 by Fermat’s
Theorem.
. . . . . .

45. The Closed Interval Method
Let’s put this together logically. Let f be a continuous function
defined on a closed interval [a, b]. We are in search of its global
maximum, call it c. Then:
Either the maximum
occurs at an endpoint of
the interval, i.e., c = a or
c = b,
Or the maximum occurs
inside (a, b). In this case, c
is also a local maximum.
Either f is differentiable
at c, in which case
f′ (c) = 0 by Fermat’s
Theorem.
Or f is not
differentiable at c.
. . . . . .

46. The Closed Interval Method
Let’s put this together logically. Let f be a continuous function
defined on a closed interval [a, b]. We are in search of its global
maximum, call it c. Then:
This means to find the
Either the maximum
maximum value of f on [a, b],
occurs at an endpoint of
we need to check:
the interval, i.e., c = a or
c = b,
Or the maximum occurs
inside (a, b). In this case, c
is also a local maximum.
Either f is differentiable
at c, in which case
f′ (c) = 0 by Fermat’s
Theorem.
Or f is not
differentiable at c.
. . . . . .

47. The Closed Interval Method
Let’s put this together logically. Let f be a continuous function
defined on a closed interval [a, b]. We are in search of its global
maximum, call it c. Then:
This means to find the
Either the maximum
maximum value of f on [a, b],
occurs at an endpoint of
we need to check:
the interval, i.e., c = a or
a and b
c = b,
Or the maximum occurs
inside (a, b). In this case, c
is also a local maximum.
Either f is differentiable
at c, in which case
f′ (c) = 0 by Fermat’s
Theorem.
Or f is not
differentiable at c.
. . . . . .

48. The Closed Interval Method
Let’s put this together logically. Let f be a continuous function
defined on a closed interval [a, b]. We are in search of its global
maximum, call it c. Then:
This means to find the
Either the maximum
maximum value of f on [a, b],
occurs at an endpoint of
we need to check:
the interval, i.e., c = a or
a and b
c = b,
Points x where f′ (x) = 0
Or the maximum occurs
inside (a, b). In this case, c
is also a local maximum.
Either f is differentiable
at c, in which case
f′ (c) = 0 by Fermat’s
Theorem.
Or f is not
differentiable at c.
. . . . . .

49. The Closed Interval Method
Let’s put this together logically. Let f be a continuous function
defined on a closed interval [a, b]. We are in search of its global
maximum, call it c. Then:
This means to find the
Either the maximum
maximum value of f on [a, b],
occurs at an endpoint of
we need to check:
the interval, i.e., c = a or
a and b
c = b,
Points x where f′ (x) = 0
Or the maximum occurs
inside (a, b). In this case, c Points x where f is not
is also a local maximum. differentiable.
Either f is differentiable
at c, in which case
f′ (c) = 0 by Fermat’s
Theorem.
Or f is not
differentiable at c.
. . . . . .

50. The Closed Interval Method
Let’s put this together logically. Let f be a continuous function
defined on a closed interval [a, b]. We are in search of its global
maximum, call it c. Then:
This means to find the
Either the maximum
maximum value of f on [a, b],
occurs at an endpoint of
we need to check:
the interval, i.e., c = a or
a and b
c = b,
Points x where f′ (x) = 0
Or the maximum occurs
inside (a, b). In this case, c Points x where f is not
is also a local maximum. differentiable.
Either f is differentiable
The latter two are both called
at c, in which case
f′ (c) = 0 by Fermat’s critical points of f. This
technique is called the Closed
Theorem.
Interval Method.
Or f is not
differentiable at c.
. . . . . .

51. Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
. . . . . .

52. Example
Find the extreme values of f(x) = 2x − 5 on [−1, 2].
. . . . . .

53. Example
Find the extreme values of f(x) = 2x − 5 on [−1, 2].
Solution
Since f′ (x) = 2, which is never zero, we have no critical points and we
need only investigate the endpoints:
f(−1) = 2(−1) − 5 = −7
f(2) = 2(2) − 5 = −1
So
The absolute minimum (point) is at −1; the minimum value is −7.
The absolute maximum (point) is at 2; the maximum value is −1.
. . . . . .

54. Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
. . . . . .

55. Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solution
We have f′ (x) = 2x, which is zero when x = 0.
. . . . . .

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

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

58. Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solution
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) =
. . . . . .

59. Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solution
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
. . . . . .

60. Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solution
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
. . . . . .

61. Example
Find the extreme values of f(x) = x2 − 1 on [−1, 2].
Solution
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)
. . . . . .

62. Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
. . . . . .

63. Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solution
Write f(x) = x5/3 + 2x2/3 , then
5 2/3 4 −1/3 1 −1/3
f′ (x) = x +x =x (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not differentiable at 0.
. . . . . .

64. Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solution
Write f(x) = x5/3 + 2x2/3 , then
5 2/3 4 −1/3 1 −1/3
f′ (x) = x +x =x (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check
are:
f(−1) =
f(−4/5) =
f(0) =
f(2) =
. . . . . .

65. Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solution
Write f(x) = x5/3 + 2x2/3 , then
5 2/3 4 −1/3 1 −1/3
f′ (x) = x +x =x (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check
are:
f(−1) = 1
f(−4/5) =
f(0) =
f(2) =
. . . . . .

66. Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solution
Write f(x) = x5/3 + 2x2/3 , then
5 2/3 4 −1/3 1 −1/3
f′ (x) = x +x =x (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check
are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) =
f(2) =
. . . . . .

67. Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solution
Write f(x) = x5/3 + 2x2/3 , then
5 2/3 4 −1/3 1 −1/3
f′ (x) = x +x =x (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check
are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0
f(2) =
. . . . . .

68. Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solution
Write f(x) = x5/3 + 2x2/3 , then
5 2/3 4 −1/3 1 −1/3
f′ (x) = x +x =x (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check
are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0
f(2) = 6.3496
. . . . . .

69. Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solution
Write f(x) = x5/3 + 2x2/3 , then
5 2/3 4 −1/3 1 −1/3
f′ (x) = x +x =x (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check
are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0 (absolute min)
f(2) = 6.3496
. . . . . .

70. Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solution
Write f(x) = x5/3 + 2x2/3 , then
5 2/3 4 −1/3 1 −1/3
f′ (x) = x +x =x (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check
are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0 (absolute min)
f(2) = 6.3496 (absolute max)
. . . . . .

71. Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].
Solution
Write f(x) = x5/3 + 2x2/3 , then
5 2/3 4 −1/3 1 −1/3
f′ (x) = x +x =x (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to check
are:
f(−1) = 1
f(−4/5) = 1.0341 (relative max)
f(0) = 0 (absolute min)
f(2) = 6.3496 (absolute max)
. . . . . .

72. Example √
4 − x2 on [−2, 1].
Find the extreme values of f(x) =
. . . . . .

73. Example √
4 − x2 on [−2, 1].
Find the extreme values of f(x) =
Solution
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differentiable at ±2 as well.)
. . . . . .

74. Example √
4 − x2 on [−2, 1].
Find the extreme values of f(x) =
Solution
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differentiable at ±2 as well.) So our points to check are:
f(−2) =
f(0) =
f(1) =
. . . . . .

75. Example √
4 − x2 on [−2, 1].
Find the extreme values of f(x) =
Solution
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differentiable at ±2 as well.) So our points to check are:
f(−2) = 0
f(0) =
f(1) =
. . . . . .

76. Example √
4 − x2 on [−2, 1].
Find the extreme values of f(x) =
Solution
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differentiable at ±2 as well.) So our points to check are:
f(−2) = 0
f(0) = 2
f(1) =
. . . . . .

77. Example √
4 − x2 on [−2, 1].
Find the extreme values of f(x) =
Solution
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differentiable at ±2 as well.) So our points to check are:
f(−2) = 0
f(0) = 2
√
f(1) = 3
. . . . . .

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

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

80. Outline
Introduction
The Extreme Value Theorem
Fermat’s Theorem (not the last one)
Tangent: Fermat’s Last Theorem
The Closed Interval Method
Examples
Challenge: Cubic functions
. . . . . .

81. Challenge: Cubic functions
Example
How many critical points can a cubic function
f(x) = ax3 + bx2 + cx + d
have?
. . . . . .

82. Solution
If f′ (x) = 0, we have
3ax2 + 2bx + c = 0,
and so
√ √
−2b ± 4b2 − 12ac −b ± b2 − 3ac
x= = ,
6a 3a
and so we have three possibilities:
b2 − 3ac > 0, in which case there are two distinct critical points. An
example would be f(x) = x3 + x2 , where a = 1, b = 1, and c = 0.
b2 − 3ac < 0, in which case there are no real roots to the quadratic,
hence no critical points. An example would be f(x) = x3 + x2 + x,
where a = b = c = 1.
b2 − 3ac = 0, in which case there is a single critical point. Example:
x3 , where a = 1 and b = c = 0.
. . . . . .

83. Review
Concept: absolute (global) and relative (local) maxima/minima
Fact: Fermat’s theorem: f′ (x) = 0 at local extrema
Technique: the Closed Interval Method
. . . . . .