The closed interval method tells us how to find the extreme values of a continuous function defined on a closed, bounded interval: we check the end points and the critical points.

1. Section 4.1
Maximum and Minimum Values
V63.0121.034, Calculus I
November 4, 2009
Announcements
Quiz next week on §§3.1–3.5
Final Exam Friday, December 18, 2:00–3:50pm
.
.
Image credit: 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
. . . . . .

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.
. . . . . .

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
. . ..
a
. d c
b
.
minimum maximum
. . . . . .

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
{
x 0≤x<1
f (x ) =
x − 2 1 ≤ x ≤ 2.
. . . . . .

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

19. Bad Example #1
Example
Consider the function .
{
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.
. . . . . .

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, ∞)
x
but has no minimum value.
. . . . . .

23. Final Bad Example
Example
1
The function f(x) = is continuous on the closed interval [1, ∞)
x
but 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 . .
|.
b
local and global . global
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→0 h
. . . . . .

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→0 h
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→0 h
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→0 − h
. . . . . .

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→0 h
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→0 − h
f(c + h) − f(c)
Since the limit f′ (c) = lim exists, it must be 0.
h→0 h
. . . . . .

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. Flowchart for placing extrema
Thanks to Fermat
Suppose f is a continuous function on the closed, bounded
interval [a, b], and c is a global maximum point.
.
. . c is a
start
local max
. . .
Is c an Is f diff’ble f is not
n
.o n
.o
endpoint? at c? diff at c
y
. es y
. es
. c = a or .
f′ (c) = 0
c = b
. . . . . .

42. 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 critical points or critical numbers x where
either f′ (x) = 0 or f is not differentiable at x.
The points with the largest function value are the global
maximum points
The points with the smallest or most negative function value
are the global minimum points.
. . . . . .

43. 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
. . . . . .

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

45. 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
. . . . . .

46. 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.
. . . . . .

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

48. 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.
. . . . . .

49. 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) =
. . . . . .

50. 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) =
. . . . . .

51. 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) =
. . . . . .

52. 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
. . . . . .

53. 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
. . . . . .

54. 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)
. . . . . .

55. Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
. . . . . .

56. Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at
x = 0 and x = 1.
. . . . . .

57. Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at
x = 0 and x = 1. The values to check are
f(−1) =
f(0) =
f(1) =
f(2) =
. . . . . .

58. Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at
x = 0 and x = 1. The values to check are
f(−1) = − 4
f(0) =
f(1) =
f(2) =
. . . . . .

59. Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at
x = 0 and x = 1. The values to check are
f(−1) = − 4
f(0) = 1
f(1) =
f(2) =
. . . . . .

60. Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at
x = 0 and x = 1. The values to check are
f(−1) = − 4
f(0) = 1
f(1) = 0
f(2) =
. . . . . .

61. Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at
x = 0 and x = 1. The values to check are
f(−1) = − 4
f(0) = 1
f(1) = 0
f(2) = 5
. . . . . .

62. Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at
x = 0 and x = 1. The values to check are
f(−1) = − 4 (absolute min)
f(0) = 1
f(1) = 0
f(2) = 5
. . . . . .

63. Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at
x = 0 and x = 1. The values to check are
f(−1) = − 4 (absolute min)
f(0) = 1
f(1) = 0
f(2) = 5 (absolute max)
. . . . . .

64. Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at
x = 0 and x = 1. The values to check are
f(−1) = − 4 (absolute min)
f(0) = 1 (local max)
f(1) = 0
f(2) = 5 (absolute max)
. . . . . .

65. Example
Find the extreme values of f(x) = 2x3 − 3x2 + 1 on [−1, 2].
Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at
x = 0 and x = 1. The values to check are
f(−1) = − 4 (absolute min)
f(0) = 1 (local max)
f(1) = 0 (local min)
f(2) = 5 (absolute max)
. . . . . .

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

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) =
f(−4/5) =
f(0) =
f(2) =
. . . . . .

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) =
f(0) =
f(2) =
. . . . . .

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) =
f(2) =
. . . . . .

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
f(0) = 0
f(2) =
. . . . . .

72. 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
. . . . . .

73. 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
. . . . . .

74. 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)
. . . . . .

75. 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)
. . . . . .

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

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

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

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

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

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

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

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

84. 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
. . . . . .

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

86. 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:
. . . . . .

87. 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.
. . . . . .

88. 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.
. . . . . .

89. 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.
. . . . . .

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