Chapter 3: 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）
1
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
2
Tangent line with positive slope
f(x) will be increasing
f ( x)  0
f ( x )  0
Tangent line with negative slope
f(x) will be decreasing
3
If f ( x)  0 for every x on some interval I, then f(x) is increasing
on the interval.
If f ( x)  0 for every x on some interval I, then f(x) is decreasing
on the interval.
If f ( x)  0 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.
4
Using the derivative to determine intervals of
increase and decrease for a function of f.
Step 1. Find all values of x for which f ( x)  0 or f ( x ) 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 f (c ) . Then
If f (c)  0 the function f(x) is increasing on a<x<b.
If f (c)  0 the function f(x) is decreasing on a<x<b.
5
Example 1
Find the intervals of increase and decrease for the function
f ( x)  2 x3  3x 2  12 x  7
Solution:
The derivative of f(x) is f ( x)  6 x 2  6 x  12  6( x  2)( x  1)
which is continuous everywhere, with f ( x)  0 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.
Interval
Test
Number
x<-2
-3
-2<x<1
0
x>1
2
f (c )
Conclusion
Direction
of graph
f ( 3)  0 f is increasing
Rising
f (0)  0
f ( 2)  0
f is deceasing
Falling
f is increasing
Rising
6
Example 2
Find the interval of increase and decrease for the function
x2
f ( x) 
x2
Solution:
The function is defined for x≠2, and its derivative is
( x  2)(2 x)  x 2 (1) x( x  4)
f ( x) 

2
( x  2)
( x  2) 2
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
7
f (2) 
3
0
4
f (1)  3  0
f (3)  3  0
f (5) 
5
0
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.
8
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)
9
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 f (c)  0 or f (c ) 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!
10
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.
11
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.
12
The First Derivative Test for Relative Extrema
Let c be a critical number for f(x) [that is, f(c) is defined
and either f ( x)  0 or f (c ) is not continuous]. Then the
critical point (c,f(c)) is
A relative maximum if f ( x)  0 to the
left of c and f ( x)  0 to the right of c
A relative minimum if f ( x)  0 to the
left of c and f ( x)  0 to the right of c
Not a relative extremum if f ( x )
has the same sign on both sides of c
f0 c
f0
f0 c
f0
f0 c f0
f0
c
f   013
Example 3
Find all critical numbers of the function f ( x)  2 x 4  4 x 2  3
and classify each critical point as a relative maximum, a
relative minimum, or neither.
Solution:
f ( x)  8x3  8x  8x( x  1)( x  1)
The derivative exists for all x, the only critical numbers
are where f ( x)  0 that is, x=0,x=-1,x=1. These
numbers divide that x axis into four intervals, x<-1,
-1<x<0, 0<x<1, x>1.
Choose a test number in each of these intervals
f (5)  960  0
1

f ( )  3  0
2
1
15

f ( ) 0
4
8
f (2)  48  0
To be continued
14
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.
--------
++++++
-1 min
--------
0 max
++++++
1 min
15
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 f ( x) 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 f ( x)  0 , falls where f ( x)  0
and has a horizontal tangent where f ( x)  0 .
16
Example 4
4
3
2
Sketch the graph of the function f ( x)  x  8x  18x  8
Solution:
f ( x)  4 x3  24 x 2  36 x  4 x( x  3)2
The derivative exists for all x, the only critical numbers
are where f ( x)  0 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) f (5)  80  0 f (1)  16  0 f (1)  64  0
Thus the graph of f has a horizontal tangents where x is
-3 and 0, and it is falling in the interval x<-3 and -3<x<0
and is rising for x>0.
To be continued
17
--------
--------
-3
neither
f(-3)=19 f(0)=-8 Plot a “cup”
++++++
0
min
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
18
Example 5
The revenue derived from the sale of a new kind of
motorized skateboard t weeks after its introduction is
given by
63t  t
R(t )  2
t  63
2
million dollars. When does maximum revenue occur?
What is the maximum revenue?
19
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
2
63(7)

(7)
7<t≤63.
R(7) 
 3.5
(7)2  63
++++++ - - - - -
0
7
Max
t
63
20
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.
21
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 f ( x ) is
increasing on the interval
Concave downward （凸的）on a<x<b if f ( x ) is
decreasing on the interval
22
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.
23
Determining Intervals of Concavity Using the
Sign of f’’
Step 1. Find all values of x for which f ( x)  0 or f (x)
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 f (c) . Then
If f (c)  0 , the graph of f(x) is concave upward on
a<x<b.
If f (c)  0 , the graph of f(x) is concave downward on
a<x<b.
24
Example 6
Determine intervals of concavity for the function
f ( x)  2 x 6  5 x 4  7 x  3
Solution:
We find that
f ( x)  12 x 5  20 x 3  7
and
f ( x)  60 x 4  60 x 2  60 x 2 ( x 2  1)  60 x 2 ( x  1)( x  1)
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
25
f ( 2)  720  0
 45
 1 
f 
0

2
4


 45
1
f   
0
2
4
 
f (5)  36,000  0
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
0
1
26
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.
27
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.
28
Example 7
In each case, find all inflection points of the given
function.
1
a. f ( x)  3x5  5x 4  1
b. g ( x)  x 3
Solution:
to be continued
29
Type of concavity
Sign of
------
0
No inflection
-----
+++++
1
inflection
to be continued
30
Type of concavity
Sign of
++++
------
0
inflection
31
Note: A function can have an inflection point only
where it is continuous!!
For example, if f(x)=1/x, then
3

f ( x)  2 / x , 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,
32
For instance, if f ( x)  x 4 ,
2


f
(
x
)

12
x
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).
33
Behavior of Graph f(x) at an inflection point P(c,f(c))
34
A graph showing concavity and inflection points
35
Example 8
Determine where the function
f ( x)  3x  2 x  12 x  18x  15
4
3
2
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
f ( x)  12 x3  6 x 2  24 x  18  6( x  1) 2 (2 x  3)
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
36
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.
Sign of f’(x) - - - - - - - -
-1.5
min
++++++
++++++
x
1
Neither
The second derivative is f ( x)  36 x 2  12 x  24  12( x  1)(3x  2) 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
37
Type of concavity
Sign of
+++++
-----
-2/3
inflection
+++++
1
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
38
39
40
41
Example 9
Solution:
to be continued
42
43
Example 10
Solution:
to be continued
44
++++
0
------
3
Max
t
4
45
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!!
46
垂直渐进线
47
Example 11
Determine all vertical asymptotes of the graph of
x2  9
f ( x)  2
x  3x
Solution:
Let p( x)  x 2  9 and q( x)  x 2  3x 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
x2  9
x 3
lim
 lim
2
x  3 x 2  3 x
x  3
x
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
48
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
x2  9
x2  9
lim 2
   and lim 2

x 0 x  3 x
x 0 x  3 x
49
水平渐进线
50
Example 12
Determine all horizontal asymptotes of the graph of
x2
f ( x)  2
x  x 1
Solution:
2
Dividing each term in the rational function f(x) by x (the
highest power of x in the denominator), we find that
x2
x2 / x2
lim f ( x)  lim 2
 lim 2 2
x 
x  x  x  1
x  x / x  x / x 2  1 / x 2
1
 lim
1
reciprocal power rule
2
x  1  1 / x  1 / x
and similarly,
x2
lim f ( x)  lim
1
x  
x   x 2  x  1
to be continued
51
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.
52
to be continued
53
54
Example 13
Sketch the graph of the function
x
f ( x) 
( x  1) 2
Solution:
to be continued
55
------
++++
-1
Asymptote
------
1
Max
to be continued
56
Type of concavity
Sign of
------
-----
-1
No inflection
+++++
2
inflection
to be continued
57
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
3x 2
Sketch the graph of f ( x)  2
x  2 x  15
58
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!
59
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.
60
61
Example 14
Solution:
to be continued
62
63
Example 14
Solution:
to be continued
64
65
66
Example 15
67
Solution:
to be continued
68
to be continued
69
to be continued
70
71
Two General Principle of Marginal Analysis
72
73
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
74
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.
75
76
77
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).
78
Solution:
to be continued
79
80
81
82
83
84
Example 16
Solution:
to be continued85
to be continued
86
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
87
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.
88
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.
89
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
90