1. Chapter 3 Additional Applications of the Derivative
CALCULUS For Business, Economics, and the Social and life Sciences Hoffmann, L.D. & Bradley, G.L.
Chapter 3 Additional Applications of the Derivative

2. Additional Applications of the Derivative
In this Chapter, we will encounter some important concepts.
Increasing and Decreasing
Functions; Relative Extrema（相对极值）
Concavity（凹凸性） and Points of Inflection（拐点）.
Optimization（最优化）

3. Section 3.1 Increasing and Decreasing Functions
Increasing and Decreasing Function Let f(x) be a function
defined on the interval a<x<b, and let x1 and x2 be two numbers
in the interval, Then
f(x) is increasing on the interval if f(x2)>f(x1) whenever x2>x1
f(x) is decreasing on the interval if f(x2)<f(x1) whenever x2 >x1
Monotonic Increasing
Monotonic
Decreasing

4. Tangent line with positive slope
f(x) will be increasing
Tangent line with negative slope
f(x) will be decreasing

5. If for every x on some interval I, then f(x) is increasing on the interval.
If for every x on some interval I, then f(x) is decreasing on the interval.
If for every x on some interval I, then f(x) is constant on the interval.
How to determine all intervals of increase and decrease
for a function ? How to find all intervals on which the sign of the derivative does not change.
Intermediate value property
A continuous function cannot change sign without first becoming 0.

6. Procedure for using the derivative to determine intervals of increase and decrease for a function of f.
Step 1. Find all values of x for which or is
not continuous, and mark these numbers on a number line. This divides the line into a number of open intervals.
Step 2. Choose a test number c from each interval a<x<b
determined in the step 1 and evaluate Then
If the function f(x) is increasing on a<x<b.
If the function f(x) is decreasing on a<x<b.

7. Rising f is increasing 2 x>1 Falling f is deceasing -2<x<1
Example 1
Find the intervals of increase and decrease for the function
Solution:
The derivative of f(x) is
which is continuous everywhere, with where x=1 and x=-2. The number -2 and 1 divide x axis into three open intervals.
x<-2, -2<x<1 and x>1.
Rising
f is increasing 2
x>1
Falling
f is deceasing
-2<x<1
- 3
x<-2
Direction
of graph
Conclusion
Test
Number
Interval

8. Find the interval of increase and decrease for the function
Example 2
Find the interval of increase and decrease for the function
Solution:
The function is defined for x≠2, and its derivative is
which is discontinuous at x=2 and has f’(x)=0 at x=0 and x=4. Thus, there are four intervals on which the sign of f’(x) does not change: namely, x<0, 0<x<2, 2<x<4 and x>4. Choosing test numbers in these intervals (say, -2,1,3, and 5, respectively), we find that
To be continued

9. We conclude that f(x) is increasing for x<0 and for x>4 and that it is decreasing for 0<x<2 and for 2<x<4.

10. Relative (Local) Extrema The Graph of the function f(x) is said to be have a relative maximum at x=c if f(c)≥f(x) for all x in interval a<x<b containing c. Similarly the graph has a relative Minimum at x=c if f(c)≤ f(x) on such an interval. Collectively, the relative maxima and minima of f are called its relative extrema（相对极值或局部极值）.
Peaks: C,E,
(Relative maxima)
Valleys: B, D, G (Relative minima)

11. Since a function f(x) is increasing when f’(x)>0 and decreasing when f’(x)<0, the only points where f(x) can have a relative extremum are where f’(x)=0 or f’(x) is not continuous. Such points are so important that we give them a special name.
Critical Numbers and Critical Points（关键点）: A number c in the domain of f(x) is called a critical number if either or is not continuous. The corresponding point (c,f(c)) on the graph of f(x) is called a critical point for f(x).
Relative extrema can only occur at critical points!

12. Not all critical points correspond to relative extrema!
Figure. Three critical points where f’(x) = 0: (a) relative maximum, (b) relative minimum (c) not a relative extremum.

13. Not all critical points correspond to relative extrema!
Figure Three critical points where f’(x) is undefined:(a) relative maximum, (b) relative minimum (c) not a relative extremum.

14. The First Derivative Test for Relative Extrema
Let c be a critical number for f(x) [that is, f(c) is defined
and either or is not continuous]. Then the critical point (c,f(c)) is c
A relative maximum if to the left of c and to the right of c c
A relative minimum if to the left of c and to the right of c c
Not a relative extremum if
has the same sign on both sides of c c

15. Choose a test number in each of these intervals
Example 3
Find all critical numbers of the function and classify each critical point as a relative maximum, a relative minimum, or neither.
Solution:
The derivative exists for all x, the only critical numbers are where that is, x=0,x=-1,x=1. These numbers divide that x axis into four intervals, x<-1, <x<0, 0<x<1, x>1.
Choose a test number in each of these intervals
To be continued

16. Thus the graph of f falls for x<-1 and for
0<x<1, and rises for -1<x<0 and for x>1, so x=0 is relative maximum, x=1 and x=-1 are relative minimum.
0 max
1 min
-1 min

17. A Procedure for Sketching the Graph of a Continuous
Function f(x) Using the Derivative
Step 1. Determine the domain of f(x).
Step 2. Find and each critical number, analyze the sign of
derivative to determine intervals of increase and decrease for f(x).
Step 3. Plot the critical point P(c,f(c)) on a coordinate plane,
with a “cap” at P if it is a relative maximum or a “cup”
if P is a relative minimum. Plot intercepts and other key points that
can be easily found.
Step 4 Sketch the graph of f as a smooth curve joining the critical
points in such way that it rise where , falls where
and has a horizontal tangent where

18. Sketch the graph of the function
Example 4
Sketch the graph of the function
Solution:
The derivative exists for all x, the only critical numbers are where that is, x=0, x=-3. These numbers divide that x axis into three intervals, x<-3, -3<x<0, x>0.
Choose test number in each interval (say, -5, -1 and 1 respectively)
Thus the graph of f has a horizontal tangents where x is and 0, and it is falling in the interval x<-3 and -3<x<0 and is rising for x>0.
To be continued

19. Plot a “cup” at the critical point (0,-8)-3
neither
min
f(-3)=19 f(0)=-8
Plot a “cup” at the critical point (0,-8)
Plot a “twist” at (-3,19) to indicate a galling graph with
a horizontal tangent at this point .
Complete the sketch by passing a smooth curve through the
Critical point in the directions indicated by arrow

20. Example 5
The revenue derived from the sale of a new kind of motorized skateboard t weeks after its introduction is given by
million dollars. When does maximum revenue occur? What is the maximum revenue?

21. Differentiating R(t) by the quotient rule, we get
Solution:
Differentiating R(t) by the quotient rule, we get
By setting the numerator in this expression for R’(t) equal to 0, we find that t=7 is the only solution in the interval 0≤t≤63, and hence is the only critical number of R(t) in its domain. The critical number divides the domain into two intervals: 0≤t<7 and 7<t≤63.
++++++ t 7
Max 63

22. Section 3.2 Concavity（凹凸性） and Points of Inflection （拐点）
Increase and decrease of the slopes are our concern!
Figure: The output Q(t) of a factory worker t hours after coming to work.

23. Concavity: If the function f(x) is differentiable on the interval a<x<b then the graph of f is
Concave upward（凹的） on a<x<b if is increasing on the interval
Concave downward （凸的）on a<x<b if is decreasing on the interval

24. A graph is concave upward on the interval if it lies above all its tangent lines on the interval and concave downward on an Interval where it lies below all its tangent lines.

25. Determining Intervals of Concavity Using the Sign of f’’
Step 1. Find all values of x for which or is not continuous, and mark these numbers on a number line. This divides the line into a number of open intervals.
Step 2. Choose a test number c from each interval a<x<b
determined in the step 1 and evaluate Then
If , the graph of f(x) is concave upward on
a<x<b.
If , the graph of f(x) is concave downward on

26. Determine intervals of concavity for the function
Example 6
Determine intervals of concavity for the function
Solution:
We find that and
The second derivative f’’(x) is continuous for all x and f’’(x)=0 for x=0, x=1, and x=-1. These numbers divide the x axis into four intervals on which f’’(x) does not change sign; namely, x<1, -1<x<0, 0<x<1, and x>1. Evaluating f’’(x) at test numbers in each of these intervals (say, at x=-2, x=-1/2, x=1/2, and x=5, respectively), we find that
to be continued

27. Thus, the graph of f(x) is concave up for x<-1 and for x>1 and concave down for -1<x<0 and for 0<x<1, as indicated in this concavity diagram.
Type of concavity
Sign of x -1 1

28. Note Don’t confuse the concavity of a graph with its “direction” (rising or falling). A function may be increasing or decreasing on an interval regardless of whether its graph is concave upward or concave downward on the interval.

29. Inflection Point （拐点） An inflection point is a point (c,f(c)) on the graph of a function f where the concavity changes. At such a point, either f’’(c)=0 or f’’(c) is not continuous.
is not continuous.

30. In each case, find all inflection points of the given function.
Example 7
In each case, find all inflection points of the given function.
Solution:
to be continued

31. Type of concavity Sign of No inflection 1 inflection - - - - - -
No inflection 1
inflection
to be continued

32. Type of concavity
Sign of
inflection

33. Note: A function can have an inflection point only where it is continuous!!
For example, if f(x)=1/x, then
, so f’’(x)<0 if x<0 and f’’(x)>0 if x>0. The concavity changes from downward to upward at x=0 but there is no inflection point at x=0 since f(0) is not defined,

34. For instance, if , then f(0)=0 and so f’’(0)=0
For instance, if , then f(0)=0 and so f’’(0)=0. However, f’’(x)>0 for any number x≠0 , so the graph of f is always concave upward, and there is no inflection point at (0,0).

35. Behavior of Graph f(x) at an inflection point P(c,f(c))

36. A graph showing concavity and inflection points

37. Determine where the function
Example 8
Determine where the function
Is increasing and decreasing, and where its graph is concave up and concave down. Find all relative extrema and points of inflection, and sketch the graph.
Solution:
First, note that since f(x) is a polynomial, it is continuous for all x, as are the derivatives f’(x) and f’’(x). The first derivative of f(x) is
and f’(x)=0 only when x=1 and x=-1.5. The sign of f’(x) does not change for x<-1.5, nor in the interval -1.5<x<1, nor for x>1.
to be continued

38. Evaluating f’(x) at test numbers in each interval (say, at -2,0, and 3), you obtain the arrow diagram shown. Note that there is a relative minimum at x=-1.5 but no extremum at x=1.
-1.5
min 1
Neither
Sign of f’(x) x
The second derivative is and f’’(x)=0 only when x=1 and x=-2/3. The sign of f’’(x) does not change on each of the intervals x<-2/3, -2/3<x<1, and x>1. Evaluating f’’(x) at test numbers in each interval, we obtain the concavity diagram.
to be continued

39. Type of concavity
Sign of 1
inflection
-2/3
inflection
The patterns in these two diagrams suggest that there is a relative minimum at x=-1.5 and inflection points at x=-2/3 and x=1 (since the concavity changes at both points).
to be continued

43. Example 9
Solution:
to be continued

45. Example 10
Solution:
to be continued

46. t 3
Max 4

47. Section 3.3 Curve Sketching
limits involving infinity: Limits at infinity and infinite limits
Our goal is to see how limits involving infinity may be interpreted as graphical features!!

48. 垂直渐进线

49. Determine all vertical asymptotes of the graph of
Example 11
Determine all vertical asymptotes of the graph of
Solution:
Let and be the numerator and denominator of f(x), respectively. The q(x)=0 when x=-3 and when x=0. However, for x=-3, we also have p(-3)=0 and
This means that the graph of f(x) has a “hole” at the point (-3,2) and x=-3 is not a vertical asymptote of the graph.
to be continued

50. On the other hand, for x=0 we have q(0)=0 but p(0)≠0, which suggests that the y axis is a vertical asymptote for the graph of f(x). This asymptote behavior is verified by noting that

51. 水平渐进线

52. Determine all horizontal asymptotes of the graph of
Example 12
Determine all horizontal asymptotes of the graph of
Solution:
Dividing each term in the rational function f(x) by (the highest power of x in the denominator), we find that
and similarly,
to be continued

53. Thus, the graph of f(x) has y=1 as a horizontal asymptote.
NOTE: The graph of a function f(x) can never cross a vertical asymptote x=c because at least one of the one-sided limits must be infinite. However, it is possible for a graph to cross its horizontal asymptotes.

54. to be continued

56. Sketch the graph of the function
Example 13
Sketch the graph of the function
Solution:
to be continued

57. -1
Asymptote 1
Max
to be continued

58. Type of concavity
Sign of -1
No inflection 2
inflection
to be continued

59. 7. The vertical asymptote (dashed line) breaks the graph into two parts. join the features in each separate part by a smooth curve to obtain the completed graph.
Exercise
Sketch the graph of

60. Section 3.4 Optimization（最优化）
Absolute Maxima and Minima （极大值和极小值）of a Function Let f be a Function defined on an interval I that contains the number c. Then
f(c) is the absolute maximum of f on I if f(c) ≥ f(x) for all x in I
f(c) is the absolute minimum of f on I if f(c) ≤ f(x) for all x in I
Absolute extrema often
coincide with relative
extrema but not always!

61. The Extreme Value Property（极值定理） A function f(x) that is continuous on the closed Interval a≤x≤b attains its absolute extrema on the interval either at an endpoint of the interval (a or b) or at a critical number c such that a<c<b.

63. Example 14
Solution:
to be continued

65. Example 14
Solution:
to be continued

68. Example 15

69. Solution:
to be continued

70. to be continued

71. to be continued

73. Two General Principle of Marginal Analysis

75. Explanation in Economics The marginal cost (MC) is approximately the same as the cost of producing one additional unit. If the additional unit costs less to produce than the average cost (AC) of the existing units (If MC<AC), then this less-expensive unit will cause the average cost per unit to decrease. On the other hand, if the additional unit costs more than the average cost of the existing units (if MC>AC), then this more-expensive unit will cause the average cost per unit to increase. However (if MC=AC), then the average cost will neither increase nor decrease, which means (AC)’=0

76. Price Elasticity of Demand（需求的价格弹性）
In general, an increase in the unit price of a commodity will result in decreased demand, but the sensitivity or responsiveness of demand to a change in price varies from one product to another. For some products, such as soap, flashlight batteries, and salt, small percentage changes in price have little effect on demand. For other products, such as airline tickets, designer furniture, and home loans, small percentage changes in price can affect demand dramatically.

79. Example 16
Suppose the demand q and price p for a certain commodity are related by the linear equation q=240-2p (for 0≤p≤120).

80. Solution:
to be continued

86. Example 16
Solution:
to be continued

87. to be continued

88. b. The total revenue, R=pq, increases when demand is inelastic; that is, when 0≤ p<10. For this range of prices, a specified percentage increase in price results in a smaller percentage decrease in demand, so the bookstore will take in more money for each increase in price up to $10 per copy.
to be continued

89. However, for the price range 10<p≤√300 , the demand is elastic, so the revenue is decreasing. If the book is priced in this range, a specified percentage increase in price results in a larger percentage decrease in demand. Thus, if the bookstore increases the price beyond $10 per copy, it will lose revenue.
This means that the optimal price is $10 per copy, which corresponds to unit elasticity.

90. Summary
Find the intervals of increase and decrease for the function f(x); Critical numbers and critical points; First derivative test for relative extrema; Sketch the graph of a continuous function.
Concavity, concave upward and concave downward; Using the second derivative to find intervals of concavity; The points of inflection; The second derivative test for relative extrema. Sketch the graph of a continuous function.

91. Limits involving infinity: Vertical Asymptotes and
Horizontal Asymptotes (rational functions)
Optimization: Absolute maximum and absolute
minimum, marginal analysis criterion for maximum,
profit and minimal average cost, elasticity of demand