1. 4.1 Extreme Values of Functions
Absolute extreme values are either maximum or minimum points on a curve.
They are sometimes called global extremes.
They are also sometimes called absolute extrema.
(Extrema is the plural of the Latin extremum.)

2. 4.1 Extreme Values of Functions
Definition Absolute Extreme Values
Let f be a function with domain D. Then f (c) is the
absolute minimum value on D if and only
if f(x) < f (c) for all x in D.
absolute maximum value on D if and only
if f(x) > f (c) for all x in D.

3. 4.1 Extreme Values of Functions
Extreme values can be in the interior or the end points of a function.
No Absolute
Maximum
Absolute
Minimum

4. 4.1 Extreme Values of Functions
Absolute
Maximum
Absolute Minimum

5. 4.1 Extreme Values of Functions
Absolute
Maximum
No Minimum

6. 4.1 Extreme Values of FunctionsNo
Maximum
No Minimum

7. 4.1 Extreme Values of Functions
Extreme Value Theorem:
If f is continuous over a closed interval, [a,b] then f has a maximum and minimum value over that interval.
Maximum & minimum
at interior points
Maximum & minimum
at endpoints
Maximum at interior point, minimum at endpoint

8. 4.1 Extreme Values of Functions
Local Extreme Values:
A local maximum is the maximum value within some open interval.
A local minimum is the minimum value within some open interval.

9. 4.1 Extreme Values of Functions
Absolute maximum
(also local maximum)
Local maximum
Local minimum
Local minimum
Local extremes are also called relative extremes.
Absolute minimum
(also local minimum)

10. 4.1 Extreme Values of Functions
Absolute maximum
(also local maximum)
Local maximum
Local minimum
Notice that local extremes in the interior of the function occur where is zero or is undefined.

11. 4.1 Extreme Values of Functions
Local Extreme Values:
If a function f has a local maximum value or a local minimum value at an interior point c of its domain, and if exists at c, then

12. 4.1 Extreme Values of Functions
Critical Point:
A point in the domain of a function f at which
or does not exist is a critical point of f .
Note:
Maximum and minimum points in the interior of a function always occur at critical points, but critical points are not always maximum or minimum values.

13. 4.1 Extreme Values of Functions
EXAMPLE 3 FINDING ABSOLUTE EXTREMA
Find the absolute maximum and minimum values of
on the interval
There are no values of x that will make
the first derivative equal to zero.
The first derivative is undefined at x=0,
so (0,0) is a critical point.
Because the function is defined over a
closed interval, we also must check the
endpoints.

14. 4.1 Extreme Values of Functions
To determine if this critical point is
actually a maximum or minimum, we
try points on either side, without
passing other critical points.
At:
Since 0<1, this must be at least a local minimum, and possibly a global minimum.
At:
At:

15. 4.1 Extreme Values of Functions
Absolute
minimum:
maximum:
At:
At:

16. 4.1 Extreme Values of Functions
y = x2/3

17. 4.1 Extreme Values of Functions
Finding Maximums and Minimums Analytically: 1
Find the derivative of the function, and determine where the derivative is zero or undefined. These are the critical points. 2
Find the value of the function at each critical point. 3
Find values or slopes for points between the critical points to determine if the critical points are maximums or minimums. 4
For closed intervals, check the end points as well.

18. 4.1 Extreme Values of Functions
Find the absolute maximum and minimum of the function
Find the critical numbers

19. 4.1 Extreme Values of Functions
Find the absolute maximum and minimum of the function
Check endpoints and critical numbers
The absolute maximum is 2 when x = -2
The absolute minimum is -13 when x = -1

20. 4.1 Extreme Values of Functions
Find the absolute maximum and minimum of the function
Find the critical numbers

21. 4.1 Extreme Values of Functions
Find the absolute maximum and minimum of the function
Check endpoints and critical numbers
The absolute maximum is 3 when x = 0, 3
The absolute minimum is 2 when x = 1

22. 4.1 Extreme Values of Functions
Find the absolute maximum and minimum of the function
Find the critical numbers

23. 4.1 Extreme Values of Functions
Find the absolute maximum and
minimum of the function
The absolute maximum is 1/4 when x = /6, 5/6
The absolute minimum is –2 when x =3/2

24. 4.1 Extreme Values of Functions
Critical points are not always extremes!
(not an extreme)

25. 4.1 Extreme Values of Functions
(not an extreme)

26. 4.2 Mean Value Theorem Mean Value Theorem for Derivatives
If f (x) is a differentiable function over [a,b], then at some point between a and b:
Mean Value Theorem for Derivatives

27. 4.2 Mean Value Theorem Mean Value Theorem for Derivatives
If f (x) is a differentiable function over [a,b], then at some point between a and b:
Mean Value Theorem for Derivatives
Differentiable implies that the function is also continuous.

28. 4.2 Mean Value Theorem Mean Value Theorem for Derivatives
If f (x) is a differentiable function over [a,b], then at some point between a and b:
Mean Value Theorem for Derivatives
Differentiable implies that the function is also continuous.
The Mean Value Theorem only applies over a closed interval.

29. 4.2 Mean Value Theorem Mean Value Theorem for Derivatives
If f (x) is a differentiable function over [a,b], then at some point between a and b:
Mean Value Theorem for Derivatives
The Mean Value Theorem says that at some point in the closed interval, the actual slope equals the average slope.

30. 4.2 Mean Value Theorem Tangent parallel to chord. Slope of tangent:
Slope of chord:

31. 4.2 Mean Value Theorem Rolle’s Theorem
If f (x) is a differentiable function over [a,b], and if f(a) = f(b) = 0, then there is at least one point c between a and b such that f’(c)=0:
(a,0)
(b,0)

32. 4.2 Mean Value Theorem Show the function satisfies the hypothesis of
the Mean Value Theorem
The function is continuous on [0,/3] and differentiable on
(0,/3). Since f(0) = 1 and f(/3) = 1/2, the Mean Value
Theorem guarantees a point c in the interval (0,/3) for
which
c = .498

33. 4.2 Mean Value Theorem at x = .498, the slope of the tangent line is
(0,1)
at x = .498, the slope
of the tangent line is
equal to the slope of
the chord.
(/3,1/2)

34. 4.2 Mean Value Theorem
Definitions Increasing Functions, Decreasing Functions
Let f be a function defined on an interval I and let x1 and x2
be any two points in I.
f increases on I if x1 < x2  f(x1) < f(x2).
f decreases on I if x1 > x2  f(x1) > f(x2).

35. 4.2 Mean Value Theorem
Corollary Increasing Functions, Decreasing Functions
Let f be continuous on [a,b] and differentiable on (a,b).
If f’ > 0 at each point of (a,b), then f increases on [a,b].
If f’ < 0 at each point of (a,b), then f decreases on [a,b].
A couple of somewhat obvious definitions:
A function is increasing over an interval if the derivative is always positive.
A function is decreasing over an interval if the derivative is always negative.

36. 4.2 Mean Value Theorem Find where the function
is increasing and decreasing and find the local extrema. + - 2 4
f’(x)
x = 2, local maximum
x = 4, local minimum

37. 4.2 Mean Value Theorem
(2,20) local max
(4,16) local min

38. 4.2 Mean Value Theorem
These two functions have the same slope at any value of x.
Functions with the same derivative differ by a constant.

39. 4.2 Mean Value Theorem
Find the function whose derivative is and whose graph passes through
so:

40. 4.2 Mean Value Theorem
Find the function f(x) whose derivative is sin(x) and whose graph passes through (0,2).
so:
Notice that we had to have initial values to determine the value of C.

41. 4.2 Mean Value Theorem
The process of finding the original function from the derivative is so important that it has a name:
Antiderivative
A function is an antiderivative of a function
if for all x in the domain of f. The process of finding an antiderivative is antidifferentiation.
You will hear much more about antiderivatives in the future.
This section is just an introduction.

42. 4.2 Mean Value Theorem
Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward.
(We let down be positive.)
Since acceleration is the derivative of velocity, velocity must be the antiderivative of acceleration.

43. 4.2 Mean Value Theorem
The power rule in reverse: Increase the exponent by one and multiply by the reciprocal of the new exponent.
Since velocity is the derivative of position, position must be the antiderivative of velocity.

44. 4.2 Mean Value Theorem
The initial position is zero at time zero.

45. 4.3 Connecting f’ and f’’ with the Graph of f
In the past, one of the important uses of derivatives was as an aid in curve sketching. We usually use a calculator of computer to draw complicated graphs, it is still important to understand the relationships between derivatives and graphs.

46. 4.3 Connecting f’ and f’’ with the Graph of f
First Derivative Test for Local Extrema at a critical point c
If f ‘ changes sign from positive to
negative at c, then f has a local
maximum at c.
local max
f’>0
f’<0
2. If f ‘ changes sign from negative to
positive at c, then f has a local
minimum at c.
local min
f’<0
f’>0
If f ‘ changes does not change sign
at c, then f has no local extrema.
no extreme
f’>0

47. 4.3 Connecting f’ and f’’ with the Graph of f
First derivative:
is positive
Curve is rising.
is negative
Curve is falling.
is zero
Possible local maximum or minimum.

48. 4.3 Connecting f’ and f’’ with the Graph of f
concave up
Definition Concavity
The graph of a differentiable
function y = f(x) is
concave up on an open interval
I if y’ is increasing on I. (y’’>0)
concave down on an open interval
I if y’ is decreasing on I. (y’’<0)
concave down

49. 4.3 Connecting f’ and f’’ with the Graph of f
Second Derivative Test for Local Extrema at a critical point c
If f’(c) = 0 and f’’(c) < 0, then f has a local maximum at x = c.
If f’(c) = 0 and f’’(c) > 0, then f has a local minimum at x = c.
+ +

50. 4.3 Connecting f’ and f’’ with the Graph of f
Second derivative:
is positive
Curve is concave up.
is negative
Curve is concave down.
is zero
Possible inflection point
(where concavity changes).

51. 4.3 Connecting f’ and f’’ with the Graph of f
Definition Point of Inflection
A point where the graph of a function has a tangent line and
where the concavity changes is called a point of inflection.
inflection point

52. 4.3 Connecting f’ and f’’ with the Graph of f
Sketch the graph
Possible extreme at
zeros at x = -1, x = 2
First derivative test:
Set
negative
positive
positive

53. 4.3 Connecting f’ and f’’ with the Graph of f
Possible extreme at
Set
First derivative test:
maximum at
minimum at

54. 4.3 Connecting f’ and f’’ with the Graph of f
Possible extreme at
Set
Or you could use the second derivative test:
negative
concave down
local maximum
positive
concave up
local minimum
maximum at
minimum at

55. 4.3 Connecting f’ and f’’ with the Graph of f
We then look for inflection points by setting the second derivative equal to zero.
Possible inflection point at
negative
positive
inflection point at

56. 4.3 Connecting f’ and f’’ with the Graph of f
Make a summary table:
rising, concave down
local max
falling, inflection point
local min
rising, concave up

57. 4.4 Modeling and Optimization
A Classic Problem
You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?

58. 4.4 Modeling and Optimization

59. 4.4 Modeling and Optimization
To find the maximum (or minimum) value of a function:
Understand the Problem.
Develop a Mathematical Model.
Graph the Function.
Identify Critical Points and Endpoints.
Solve the Mathematical Model.
Interpret the Solution.

60. 4.4 Modeling and Optimization
What dimensions for a one liter cylindrical can will use the least amount of material?
Motor
Oil
We can minimize the material by minimizing the area.
We need another equation that relates r and h:
area of
ends
lateral
area

61. 4.4 Modeling and Optimization
area of
ends
lateral
area

62. 4.4 Modeling and Optimization
Find the radius and height of
the right-circular cylinder of
largest volume that can be
inscribed in a right-circular
cone with radius 6 in. and
height 10 in.
10 in r h
6 in

63. 4.4 Modeling and Optimization
The formula for the volume of
the cylinder is h r
10 in
6 in
To eliminate one variable, we
need to find a relationship
between r and h. 6 h
10-h r 10

64. 4.4 Modeling and Optimizationh r
10 in
6 in

65. 4.4 Modeling and Optimizationh r
10 in
6 in
Check critical points and endpoints.
r = 0, V = 0
r = 4 V = 160/3
r = 6 V = 0
The cylinder will have a
maximum volume when
r = 4 in. and h = 10/3 in.

66. 4.4 Modeling and Optimization
Determine the point on the
curve y = x2 that is closest to
the point (18, 0).                
Substitute for x

67. 4.4 Modeling and Optimization
Determine the point on the
curve y = x2 that is closest to
the point (18, 0).                

68. 4.4 Modeling and Optimization
Determine the point on the
curve y = x2 that is closest to
the point (18, 0).                 2

69. 4.4 Modeling and Optimization
Notes:
If the function that you want to optimize has more than one variable, use substitution to rewrite the function.
If you are not sure that the extreme you’ve found is a maximum or a minimum, you have to check.
If the end points could be the maximum or minimum, you have to check.

70. 4.5 Linearization and Newton’s Method
For any function f (x), the tangent is a close approximation of the function for some small distance from the tangent point.
We call the equation of the tangent the linearization of the function.

71. 4.5 Linearization and Newton’s Method
Start with the point/slope equation:
linearization of f at a
is the standard linear approximation of f at a.
The linearization is the equation of the tangent line, and you can use the old formulas if you like.

72. 4.5 Linearization and Newton’s Method
Find the linearization of f(x) = x4 + 2x at x = 2
f’(x) = 4x3 + 2
L (x) = f(3) + f’(3)(x - 3)
L (x) = (x - 3)
L (x) = 110x - 243

73. 4.5 Linearization and Newton’s Method
Important linearizations for x near zero:
This formula also leads to non-linear approximations:

74. 4.5 Linearization and Newton’s Method
Estimate using local linearization.

75. 4.5 Linearization and Newton’s Method
Estimate sin 31º using local linearization.
Need to
be in radians

76. 4.5 Linearization and Newton’s Method
Differentials:
When we first started to talk about derivatives, we said that becomes when the change in x and change in y become very small.
dy can be considered a very small change in y.
dx can be considered a very small change in x.

77. 4.5 Linearization and Newton’s Method
Let y = f(x) be a differentiable function. The differential dx is an independent variable.
The differential dy is: dy = f ’(x)dx

78. 4.5 Linearization and Newton’s Method
Example: Consider a circle of radius 10. If the radius increases by 0.1, approximately how much will the area change?
very small change in r
very small change in A
(approximate change in area)

79. 4.5 Linearization and Newton’s Method
Compare to actual change:
New area:
Old area:
Absolute
error
percent
error

80. 4.5 Linearization and Newton’s Method
True
Estimated
Absolute Change
Relative Change
Percent Change

81. 4.5 Linearization and Newton’s Methodx y
y = f(x)
Root
sought x1
First
(x1,f(x1))
(x2,f(x2))
(x3,f(x3)) x3
Third x2
Second

82. 4.5 Linearization and Newton’s Method
This is Newton’s Method of finding roots. It is an example of an algorithm (a specific set of computational steps.)
This is a recursive algorithm because a set of steps are repeated with the previous answer put in the next repetition. Each repetition is called an iteration.

83. 4.5 Linearization and Newton’s Method
Finding a root for:
We will use Newton’s Method to find the root between 2 and 3.

84. 4.5 Linearization and Newton’s Method
Guess x1 = 2

85. 4.5 Linearization and Newton’s Method
Guess x2 = 2.5

86. 4.5 Linearization and Newton’s Method
Find where crosses

87. 4.5 Linearization and Newton’s Method

88. 4.5 Linearization and Newton’s Method
There are some limitations to Newton’s Method:
Looking for this root.
Bad guess.
Wrong root found
Failure to converge

89. 4.6 Related Rates First, a review problem:
Consider a sphere of radius 10 cm.
If the radius changes 0.1 cm (a very small amount) how much does the volume change?
The volume would change by approximately 40 cm

90. 4.6 Related Rates Now, suppose that the radius is
changing at an instantaneous rate
of 0.1 cm/sec.
The sphere is growing at a rate of 40 cm3/sec
Note: This is an exact answer, not an approximation like
we got with the differential problems.

91. 4.6 Related Rates
Water is draining from a cylindrical tank at 3 liters/second. How fast is the surface dropping?
(We need a formula to relate V and h. )
Find
(r is a constant.)

92. 4.6 Related Rates Steps for Related Rates Problems:
1. Draw a picture (sketch).
2. Write down known information.
3. Write down what you are looking for.
4. Write an equation to relate the variables.
5. Differentiate both sides with respect to t.
6. Evaluate.

93. 4.6 Related Rates Hot Air Balloon Problem: Given:
How fast is the balloon rising?
Find

94. 4.6 Related Rates Hot Air Balloon Problem: Given:
How fast is the balloon rising?
Find

95. 4.6 Related Rates Truck Problem: Truck A travels east at 40 mi/hr.
Truck B travels north at 30 mi/hr.
How fast is the distance between the trucks changing 6 minutes later? B A

96. 4.6 Related Rates Truck Problem: Truck A travels east at 40 mi/hr.
Truck B travels north at 30 mi/hr.
How fast is the distance between the trucks changing 6 minutes later? B A

97. 4.6 Related Rates Truck Problem: Truck A travels east at 40 mi/hr.
Truck B travels north at 30 mi/hr.
How fast is the distance between the trucks changing 6 minutes later?