Chapter 6 Curve Sketching
Introduction…….
6.1 The First Derivative
At this point we should be comfortable calculating the derivative of a given function which in turn tells us
about the instantaneous rate of change of the given function. Now we want to reverse the process. That
is, if we are given information about the first derivative of some function what can we find out about the
original function? To investigate this, let’s look at the graph of f ( x)  x3  2x2  x  1 shown in figure
6.1 below.
Figure 6.1 Graph of f ( x)  x3  2x2  x  1
As we look at the graph of f ( x)  x3  2x2  x  1 from left to right, notice that the graph increases from
negative infinity to somewhere between x = 0 and x = 0.5. Then, the graph begins to decrease until it
nears x = 1 and finally the graph increases into infinity. Figure 6.1 below shows the graph of some
tangent lines for f ( x)  x3  2x2  x  1.
Figure 6.2 Graph of f ( x)  x3  2x2  x  1 and tangent lines at various x values.
(a) the slope of tangent line at
x = 0.5 is positive.
(c) the slope of tangent line at
x = 0.5 is negative.
(b) the slope of tangent line at
x = 0.33 is zero.
(d) the slope of tangent line at
x = 1 is zero.
(e) the slope of tangent line at
x = 1.5 is zero.
Notice that when the tangent line has a positive slope, the graph of f ( x)  x3  2x2  x  1 is increasing as
shown in figures (a) and (e) above. When the tangent line has a negative slope, the graph of
f ( x)  x3  2x2  x  1 is decreasing as shown in figure (c) above. Lastly, when the tangent line has a
slope of zero, the graph of f ( x)  x3  2x2  x  1 is neither increasing nor decreasing, we say the graph
is constant, as shown in figures (b) and (d) above. Will this always be true? As it turns out this
relationship between slopes of tangent lines and original functions will always be true. Consequently, we
can conclude the following about a given function f ( x) and its first derivative f '( x) . For a formal proof
of this concept see your Calculus textbook.
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Increasing/Decreasing/Constant Intervals
Given a function, f ( x) , that is differentiable and continuous on an open interval (a, b)
1) if f '( x)  0 for all x on (a, b) , then f ( x) is increasing on (a, b) .
2) if f '( x)  0 for all x on (a, b) , then f ( x) is decreasing on (a, b) .
3) if f '( x)  0 for all x on (a, b) , then f ( x) is constant on (a, b) .
Example 6.1
Use the graph of f ( x) below to find the intervals where the function is increasing,
decreasing, and constant.
Solution
First we need to find where the graph of f ( x) appears to be constant, that is we need to
find where f '( x)  0 . If we approximate where the tangent lines have a slope of zero we see that f ( x)
appears to be constant at x  3,  1, 2 as shown in the figure below.
The values of x where the graph is constant are the values where the graph changes from either increasing
to decreasing or decreasing to increasing. These values, called critical values, will help us determine the
endpoints of intervals. The arrows in the figure below show where the graph of f ( x) increases and
decreases.
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(a) The arrows show where f ( x) is increasing
(b) The arrows show where f ( x) is decreasing
Thus, the graph of f ( x) increases on  , 3   1, 2  and decreases on  3,  1   2,   .

When a function changes from increasing to decreasing at a critical value x = c, a relative maximum is
formed. That is, at x = c, f ( x)  f (c) for all x in an open interval that contains c. When a function
changes from decreasing to increasing at a critical value x = c, a relative minimum is formed. That is, at
x = c, f ( x)  f (c) for all x in an open interval that contains c. It is important to note here that the term
relative extrema is often used to refer to a relative minimum, relative maximum, or both. Figure 6.3
below shows an examples of relative extrema at x = 3.
Figure 6.3
(a) a relative minimum exists at x = 3.
(b) a relative maximum exists at x = 3.
Relative minimum
at x =3.
Relative minimum
at x =3.
Once the critical value(s) have been found, we can then determine the intervals where the function
increases and decreases. Critical values are found using the following definition.
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Definition Critical Value
The value x = c is a critical value if the following two criteria are upheld.
1) f '(c)  0 or f '(c) does not exist and
2) x = c is in the domain of f ( x)
Example 6.2 Use the graph of f ( x) below and locate any the critical values.
2
–4
–2
2
Solution
At x = –3, f '( x)  0 and at x = –1 f '( x) does not exist. Thus the critical values are at x = –1, – 3.

Example 6.3 Find the critical values for f ( x)  x3  2x2  x  1.
Solution
As discussed at the beginning of this chapter, it appears that f ( x) has critical values somewhere on the
interval (0, 0.5) and at x = 1. To find the exact location of the critical values, we must use the definition
of critical value and find where f '( x)  0 .
f '( x)  3 x 2  4 x  1
0  (3 x  1)( x  1)
1
x  ,1
3
Since x  13 and x  1 are in the domain of f ( x) , we can accept both answers as critical values. (Note:
we did not try to find where f '( x) was undefined because f '( x) is a polynomial function which is
defined for all values of x.)

Example 6.4 Find the critical values for f ( x)  ln( x  0.5)  x2 .
Solution
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We begin by taking the first derivative and find the values of x that make f '( x)  0 .
1
 2x
x  0.5
1
0
 2x
x  0.5
1
2x 
x  0.5
2
2x  x  1
f '( x) 
2 x2  x  1  0
(2 x  1)( x  1)  0
x  0.5,1
Since x  0.5 is not in the domain of f ( x) it can not be a critical value. In addition, f '( x) is undefined
when x = 0.5 but once again, this is not in the domain of f ( x) therefore it is not a critical value. Thus,
the only critical value is x  1 .

Example 6.5 Find the critical values for f ( x)  x .
2
3
Solution
Once again, we will find where the first derivative is equal to zero and/or where the first derivative does
not exist.
f '( x) 
2  13
2
x  1
3
3x 3
If we try to set the first derivative equal to zero we will obtain the following false equation.
2
1
3x 3
02
0
This tells us that the first derivative will never equal zero because 0  2 . However, the first derivative is
1
undefined where 3 x 3  0 which happens when x  0 . Thus, x  0 is the only critical value.

Now that we know how to find critical values, we will further our analysis by finding the intervals where
functions increase and decrease. To make this process as organized as possible, we will always construct
a first derivative chart to help us perform the analysis. Suppose we want to find the intervals where
f ( x)  x3  2x2  x  1 increases and decreases. We found in example 6.3 that this function has critical
values at x  13 and x  1 . To find the intervals where this function increases and decreases we first plot
the critical values on a number line as shown in figure 6.4.
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Figure 6.4 First derivative chart for f ( x)  x3  2x2  x  1.
Behavior of f ( x)
test point
Sign of
f '( x )
x
1
3
x 1
To determine the behavior of f ( x) on these intervals, we need to determine the sign (whether it be
positive or negative) of f '( x ) on these intervals. To do this, we must choose a test point for each interval,
substitute this test point into f '( x) , and then record the sign of our answer. A test point is any number
that lies on the interval. For simplicity reasons let’s choose the values x = 0, x = 0.5, and x = 2 as test
points for our example. The table 6.1 below shows the results of evaluating f '( x) at each test and the
resulting sign of each answer.
Table 6.1
f '( x)  3x2  4 x  1 evaluated at each chosen test point.
Test Point
f '(test point)
Sign of f '(test point)
x=0
f '(0)  3(0)2  4(0)  1  1
+
x = 0.5
f '(0.5)  3(0.5)2  4(0.5)  1  0.25
–
x=2
f '(2)  3(2)2  4(2)  1  5
+
To finish the first derivative chart, we need to record the sign of f '(test point) and the resulting behavior
of f ( x) . Since f ( x) increases when f '( x)  0 and decreases when f '( x)  0 we complete the first
derivative chart as shown in figure 6.5 below.
Figure 6.5 Completed first derivative chart for f ( x)  x3  2x2  x  1.
Behavior of f ( x)
test point
0
0.5
2
Sign of f '( x )
+
–
+
x
1
3
x 1
Thus f ( x) increases on (, 13 )  (1, ) and decreases on  1 3 ,1 . We can also see that a relative
maximum exists at x  13 and a relative minimum exists at x  1 .
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Example 6.6
Find the intervals on which f ( x)  ln( x  0.5)  x2 increases and decreases then find
any relative extrema.
Solution
We found in Example 6.4 that the only critical value was at x = 1. In addition, the domain of the function
is  0.5,   , so the first derivative chart will start at x = 0.5 and continue infinitely to the right. Noting
1
 2 x and choosing test points of x = 0.75 and x = 2 we can complete table 6.2 as
x  0.5
shown below.
that f '( x) 

Table 6.2
f '( x) 
1
 2 x evaluated at each chosen test point.
x  0.5
Test Point
x = 0.75
x=2
Sign of f '(test point)
f '(test point)
f '(0.75) 
f '(2) 
1
 2(0.75)  2.5
0.75  0.5
+
1
10
 2(2)  
2  0.5
3
–
The table allows us to construct the first derivative chart shown in figure 6.6 below.
Figure 6.6 Completed first derivative chart for f ( x)  ln( x  0.5)  x2 .
Behavior of f ( x)
test point
Sign of
0.75
2
+
–
f '( x )
x 1
x  0.5
Thus, f ( x) increases on  0.5,1 and decreases on 1,  . In addition f ( x) has a relative maximum at
x 1.

Example 6.7
Find the intervals on which f ( x)  x
2
3
increases and decreases then find any relative
extrema.
Solution
We found in example 6.5 that f ( x)  x 3 has a critical value at x = 0. Recall that the derivative did not
exist at x = 0 which means that the graph of f ( x) has a cusp, a vertical tangent, or discontinuity at x = 0.
2
Since f ( x)  x
2
3
is continuous everywhere we rule out the possibility of discontinuity at x = 0. Thus, we
must keep in mind that a vertical tangent or a cusp exists at x =0. Since the domain of f ( x)  x
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is
2
and
1
3x 3
making our test points x = –1 and x = 1 we obtain the following table shown in table 6.3 below.
 ,  
our first derivative chart will encompass the entire number line. Noting that f '( x) 
Table 6.3
f '( x) 
2
evaluated at each chosen test point.
1
3x 3
Test Point
x = –1
x=1
Sign of f '(test point)
f '(test point)
f '(1) 
f '(1) 
2
2

1
3
3
3(1)
–
2
2

1
3
3
3(1)
+
The table allows us to construct the first derivative chart shown in figure 6.7 below.
Figure 6.7 Completed first derivative chart for f ( x)  x .
2
3
Behavior of f ( x)
–1
–
test point
Sign of
f '( x )
1
+
x0
Thus the graph of f ( x) decreases on (,0) and increases on (0, ) . The graph also has a relative
minimum at x = 0. As discussed earlier a cusp or a vertical tangent must exist x = 0. Since the graph
changes from decreasing to increasing at x = 0, a vertical tangent can not exist. For a vertical tangent to
exist, the graph must increase on both sides of x = 0 or decrease on both sides of x = 0. Thus, a cusp
must exist at x = 0.

Example 6.8 Find the intervals on which
f ( x) 
1
x 1
2
increases and decreases then find any relative extrema.
Solution
We begin by finding the first derivative.
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f '( x)  2 x( x 2  1)2 


2 x
( x 2  1)2
2
Notice that f '( x) does not exist where x2  1  0 which is when x  1,1 . The domain of f ( x) ,
however, does not contain x  1,1 . Hence x  1,1 are not critical values. Critical values, however,
may also exist where f '( x)  0 .
2 x
( x 2  1) 2
0  2 x
0x
0
Since x = 0 is in the domain of f ( x) we accept x = 0 as the only critical value. We can now construct the
first derivative chart for f ( x) . Since f ( x) is a rational function with vertical asymptotes at x  1 and
x  1 , its first derivative chart does not only contain its critical values but it must also contain the vertical
asymptote values. Thus, we need to plot x  1 , x = 0 , and x  1 on the first derivative chart and choose
test points that lie on these intervals. Simple test points to choose are x  2 , x  0.5 . x  0.5 , and
x  2 . The completed table is shown in table 6.4 below.
Table 6.4
f '( x) 
2 x
evaluated at each chosen test point.
( x 2  1)2
Test Point
x = –2
x = –0.5
x = 0.5
x=2
Sign of f '(test point)
f '(test point)
f '(2) 
2(2)
 (2)  1
2
f '(0.5) 
f '(0.5) 
f '(2) 
2

2(0.5)
(0.5)  1
2
2
4
9

+
16
9
2(0.5)
16

2
2
(0.5  1)
9
2(2)
(2) 2  1
2

4
9
+
–
–
The table allows us to construct the first derivative chart shown in figure 6.8 below.

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Figure 6.8 Completed first derivative chart for f ( x) 
1
.
x 1
2
Behavior of f ( x)
test point
Sign of
f '( x )
–2
–0.5
0.5
2
+
+
–
–
x  1
x 1
x0
Thus f ( x) increases on  ,  1   1, 0  and decreases on  0,1  1,   and has a relative maximum
at x = 0.

Example 6.9
A company has found that their revenue function, in thousands of dollars, is closely
modeled by the function R( x)  0.05x2  10 x  350 for 0  x  185 where x is the number of individual
units sold. How many units must the company sell to maximize their revenue? What is the maximum
revenue?
Solution To find the maximum revenue we need to find the critical values of R( x) .
R '( x)  0.1x  10
0  0.1x  10
 10  0.1x
100  x
Setting up a first derivative chart and choosing test points x = 1 and x = 101 we can construct the first
derivative chart shown in figure 6.9 below.
Figure 6.9 Completed first derivative chart for R( x)  0.05x2  10x  350 .
Behavior of f ( x)
test point
1
+
Sign of f ( x)
101
–
x  100
The chart reveals a relative maximum at x = 100. Therefore the company must sell 100 units to maximize
their revenue. To find the maximum revenue when 100 units are sold we compute the following
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R (100)  0.05(100) 2  10(100)  350
 150
Hence, the company will reach a maximum revenue of $150,000 when 100 units are sold.

Example 6.10
The same company from example 6.4 has also determined that their cost function
can be modeled by C ( x)  0.8x  20 where C ( x) is in thousands of dollars and x is the number of
individual units sold. How many units must the company sell in order to maximize their profit? What is
the company’s maximum profit?
Solution To find profit we need to use P( x)  R( x)  C ( x) . Thus, the profit equation is
P( x)  0.05 x 2  10 x  350   0.8 x  20 
 0.05 x 2  10 x  350  0.8 x  20
 0.05 x 2  9.2 x  370
To find the maximum profit, we need to find the critical value(s) of P( x) .
P '( x)  0.1x  9.2
0  0.1x  9.2
 9.2  0.1x
92  x
Setting up a first derivative chart and choosing test points of x = 90 and x = 95 we obtain the chart shown
in figure 6.10 below.
Figure 6.10 Completed first derivative chart for. P( x)  0.05x2  9.2x  370
Behavior of f ( x)
test point
90
+
Sign of f ( x)
95
–
x  92
The chart reveals a relative maximum at x = 92. Therefore the company must sell 92 units to maximize
their profit. The maximum profit when 92 units are sold is
P(92)  0.05(92)2  9.2(92)  370
 53.2
Hence, the company will reach maximum revenue of $53,200 when 92 units are sold.

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6.2 The Second Derivative
Since the derivate of a function is a function itself it is possible to take its derivative. The derivative of
the first derivative is called the second derivative and it is typically denote using a double prime, f ''( x) .
The second derivative of a function is a function itself and therefore has its own derivative. The
derivative of the second derivative is called the third derivative and can be denoted with a triple prime,
f '''( x) . We can continue this process and take numerous derivatives of functions; however, derivatives
higher than the second do not serve any real purpose for our function analysis so we will stick to taking
just the first and second derivative of a given function.
Example 6.11 Find the second derivative of f ( x)   x3  4x2  x  3 .
Solution To solve this, we must find f '( x) and then take its derivative.
f '( x)  3 x 2  8 x  1
f ''( x)  6 x  8

What does the second derivative of a function tell us about the original function? To investigate this, let’s
look at the graph of the function f ( x)   x3  4x2  x  3 on (, 4 3 ) and some values for f '( x) as
shown in figure 6.11 below.
Figure 6.11 Graph of f ( x)   x3  4x2  x  3 on (, 4 3 ) and a table of values for f '( x) .
X
f '( x)
-1.5
-1
-.5
0
.5
1
1.1
1.2
1.3
-19.75
-12
-5.75
-1
2.25
4
4.17
4.28
4.33
If we look at the table of values we notice that the values of f '( x) are increasing as x approaches 1.3 from
its left side and the graph of f ( x) is concave up.
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Now let’s looks at the graph of f ( x)   x3  4x2  x  3 on (, 4 3 ) and some values for f '( x) as shown
in figure 6.12 below.
Figure 6.12 Graph of f ( x)   x3  4x2  x  3 on ( 4 3 , ) and a table of values for f '( x) .
x
f '( x)
1.3
1.4
1.5
1.6
1.8
2
2.5
3
3.5
4.33
4.32
4.25
4.12
3.68
3
0.25
-4
-9.75
If we look at the table of values we notice that the values of f '( x) are decreasing as x goes from 1.3
toward positive infinity and the graph of f ( x) is concave down.
Thus, we see that
3
2
4
4
f ( x)   x  4x  x  3 is concave up on (, 3 ) and concave down on ( 3 , ) . Since the concavity
changes and a tangent line exists at x  4 3 , f ( x) has a point of inflection at  4 3 , 11 27   1.33,0.41 .
Figure 6.13 below organizes the information we acquired about f ( x)   x3  4x2  x  3 in the form of a
chart.
Figure 6.13 Chart displaying information about f ( x)   x3  4x2  x  3 .
Behavior of
f ( x)
Know about
f '( x)
x
4
3
Recall that when a function is increasing the slopes of its tangent lines are all positive and when a
function is decreasing the slopes of its tangent lines are all negative. Thus, when f '( x) is increasing,
f ''( x) is positive and when f '( x) is decreasing, f ''( x) is negative. If we add this information to the
chart from figure 6.13 we obtain a chart as shown in figure 6.14 below.
Figure 6.14 Chart displaying information about f ( x)   x3  4x2  x  3 .
Behavior of
f ( x)
Know about
f '( x)
Sign of
f ''( x)
–
+
x
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We can see from the figure 6.14 that when the second derivative is positive ( f ''( x)  0 ) the graph f ( x) is
concave up and when the second derivative is negative ( f ''( x)  0 ) the graph of f ( x) is concave down.
When the second derivative is zero ( f ''( x)  0 ) the graph of f ( x) has a point of inflection. Although a
proof is not shown in this book, the conclusions drawn from our single example have been proven to be
true for all functions an you may find the proof in your Applied Calculus textbook. The conclusions
about the second derivative of a function are stated below.
Intervals of Concavity and Points of Inflection
Given a function f ( x) that is twice differentiable and continuous on an open
interval (a, b)
• if f ''( x)  0 for all x on (a, b) , then f ( x) is concave up on (a, b) .
• if f ''( x)  0 for all x on (a, b) , then f ( x) is concave down on (a, b) .
• if f ''( x)  0 or f ''( x) does not exist, a point of infection exists at  x, f ( x) 
provided f ( x) changes concavity and a tangent line exists at  x, f ( x)  ,
Recall from the example above the value x  43 was the value at which concavity changed and thus
 4 3 , 11 27   1.33,0.41
was a point of inflection. Use the definition of point of inflection from above we
can now show the calculation used to find the point of inflection.
f '( x)  3x2  8x  1 so f ''( x)  6 x  8
0  6 x  8
8  6 x
4
x
3
Example 6.12 Given f ( x) below, find the intervals where it is concave up and concave down then
determine all points of inflection.
f ( x) 
1 4 1 3
x  x  x2
12
6
Solution We begin by finding the second derivative of f ( x) .
1
1
f '( x)  x3  x 2  2 x so f ''( x)  x2  x  2
3
2
Possible points of inflection exist where f ''( x)  0 and/or where f ''( x) does not exist. Since the second
derivative is a quadratic and is defined for all values of x, we just need to find where the second derivative
equal to zero.
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0  x2  x  2
0  ( x  2)( x  1)
x  2,1
To determine the intervals of concavity we need to plot these values of x on a second derivative chart and
choose test points for each interval as shown in figure 6.15 below. Notice that the set up of the second
derivative chart is very similar to the set up of the first derivative chart.
Figure 6.15 Second derivative chart for f ( x)  1 x 4  1 x3  x 2 .
12
Behavior of
f ( x)
test point
Sign of
6
-3
0
2
f ''( x)
x  2
x 1
We now substitute the test points into the second derivative and determine the resulting sign as shown in
table 6.5 below.
Table 6.5
f ''( x)  x2  x  2 evaluated at each chosen test point.
Test Point
f ''(test point)
Sign of f ''(test point)
x = –3
f ''( x)  (3)2  3  2  4
+
x=0
f ''( x)  02  0  2  2
–
x=2
f ''( x)  22  2  2  4
+
To complete the second derivative chart we need to transfer the sign of f ''(test point) to the chart and
then determine the behavior of f ( x) . Since f ( x) is concave up when f ''( x)  0 and concave down
when f ''( x)  0 we can complete the second derivative chart as shown in figure 6.16 below.
Figure 6.16 Completed second derivative chart for f ( x)  1 x 4  1 x3  x 2 .
12
Behavior of
f ( x)
test point
-3
0
2
f ''( x)
+
–
+
Sign of
x  2
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x 1
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Thus f ( x) is concave up on (, 2)  (1, ) and concave down on (2,1) . Since there is a change in
concavity and a tangent line exists at x  2 and x  1 , the two points of inflection are
(2, f (2))  (2,  4) and (1, f (1))  (1, 0.75)

6.3 Application - Point of Diminishing Returns
The law of diminishing returns is a concept in economics that has been around since the late 1700’s. It
states that when one of the factors of production is held fixed in supply, successive additions of the other
factors will lead to an increase in returns up to a point, known as the point of diminishing returns, and
beyond this point, returns will decrease. (http://www.cr1.dircon.co.uk/TB/2/dreturns.htm) To get a better
understanding of this concept let’s look at the following scenario.
Suppose a crop of wheat is ready to be harvested. Due to factors like weather and insects, the crop must
be harvested as quickly as possible if the farmer expects his crop to give him the best returns. At this
point, the farmer must make some important decisions about hiring laborers to help with the harvest. He
may choose to harvest the crop by himself, however, this is a big job for one man and though he will
harvest some bushels of wheat, it is likely he could get better returns if he hires some helpers. So suppose
he hires one other person to help him. The work load would be cut in half and each person could
specialize in a task thus increasing the overall number of wheat bushels harvested, but due to lunch breaks
and exhaustion, there would still be some time in which no work was getting done. So the farmer may
decide to hire one more person. This third person would lessen the work load of the other two workers,
decrease the need for workers to break and increase the overall number of bushels of wheat harvested. At
this point, one might think that having more than three workers would increase overall number of bushels
of wheat harvested even more. But, it is possible that hiring a fourth person may lead to a decrease in
marginal production of the wheat. That is, the gain in production of adding the fourth worker was not a
great as the gain in production of adding the third worker. As a matter of fact, it is likely that hiring more
than four workers will lead to an overall decrease in total production because workers may be waiting
around for work to do. So how do you figure out the ideal number of workers to hire? You find the point
of diminishing returns, which is also known as the point of inflection.
Suppose table 6.___ below represents the total number of bushels of wheat harvested when different
number of workers are hired.

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Table 6.__
Table representing total number of bushels of wheat
Total Production
(in bushels)
TP
10
30
90
120
130
120
Number of
Workers
1
2
3
4
5
6
Marginal Production
MP
10
30 – 10 = 20
90 – 30 = 60
120 – 90 = 30
130 – 120 = 10
120 – 130 = –10
These data points have been plotted and fitted with a smooth curve as shown in figure 6.___ below.
Figure 6.__ Graph of TP and MP.
140
Number of Bushels
120
TP (x)
point of inflection
100
relative
maximum
80
60
MP(x)
40
20
0
-20
1
2
3
4
5
6
Num ber of Workers
Notice that marginal production MP(x) is maximized at x = 3 which means at this point the returns for
additional workers begin to decrease, or diminish. Consequently, the point x = 3 is known as the point of
diminishing returns. Recall that MP(x) = TP’(x) so MP’(x) = TP’’(x). Since MP(x) is maximized at x = 3,
MP’(x) = 0 at x = 3 so we can conclude that TP’’(x) = 0 at x =3 as well. By examining the graph in figure
6.___ above, we see that TP(x) changes concavity and has a tangent line at x = 3, thus TP(x) has a point of
inflection at x = 3. Since x = 3 is also the point of diminishing returns, we conclude that the point of
diminishing returns is the same point as the point of inflection. Though this is not a proof, it does show
that the point of inflection for a model is found by locating the point of inflection.
Example 6.____
Use table 6.___ above and find a cubic function that fits the total production
data. Then find the point of diminishing returns for total production.
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Solution
Using the ACOW modeling applet we find total production function to be
TP( x)  2.5x3  20x2  10x , 1  x  6 , with R 2  0.98423 .
To find the point of diminishing returns we need to find the point of inflection for
TP( x)  2.5x3  20x2  10x which means we need to find where TP "( x)  0 or does not exist.
TP '( x)  7.5x2  40x  10 so TP "( x)  15x  40
0  15 x  40
40  15 x
8
x
3
2.67  x
Now we construct a second derivative chart to find intervals of concavity for
TP( x)  2.5x3  20x2  10x .
Behavior of
Sign of
TP( x)
test point
2
4
TP ''( x)
+
–
x = 2.67
The second derivative chart reveals that a point of inflection, and hence a point of diminishing returns,
exists at x = 2.67 . Since x represents the number of workers, we would round x = 2.67 to the nearest
whole number x = 3. Thus, the farmer should have 3 workers harvesting the crop.
Recall MP '( x)  TP ''( x) , so x = 2.67 is also a possible point of inflection. Constructing a second
derivative chart would show x = 2.67 is a point of inflection, therefore it is the point of diminishing
returns for TP( x) . This value would also round to x = 3 as explained above.

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6.4 Function Analysis using Derivatives
At this point you may be feeling a bit overwhelmed with all the information concerning function analysis
using the first and second derivatives. So instead of tossing this ACOW book out the window, let’s try to
ease your pain by organizing this information in the form of a flow chart as shown in the flow chart below
in figure 6.___.
Figure 6.__ Flow chart for function analysis.
Step 1
Find Domain of f ( x)
CRITICAL
VALUES
Step 2
Perform a First
Derivative Test
Step 3
Perform a Second
Derivative Test
Step 4
Graph Function
Step 2a
Find f '( x )
Step 3a
Find f ''( x)
Step 4a
Graph all asymptotes.
Step 2b
Find values of x, in the
domain, such that
f '( x)  0 .
Step 3b
Find values of x, in the
domain, such that
f ''( x)  0 .
Step 4b
Let x =c be a critical
value. Plot extrema at
(c, f (c)) .
Step 2c
Find values of x, in the
domain, such that
f '( x ) does not exist.
A vertical tangent,
cusp, or discontinuity
exists at these values.
Step 3c
Find values of x, in the
domain, such that
f ''( x) does not exist.
Step 4c
Let x = p be a point of
inflection. Plot it at
( p, f ( p)) .
Step 3c
Complete a second
derivative chart
Step 4d
Construct a curve that
fits the plotted
information. Be sure to
include any cusps or
points of discontinuity.
Step 2d
Complete a first
derivative chart
Step 2e
Use the first derivative
chart to find intervals
where f ( x) increases
and decreases.
Step 2f
Use first derivative
chart to find relative
extrema.
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Step 3d
Use the second
derivative chart to find
f ( x ) ’s intervals of
concavity.
Step 3e
Use second derivative
chart to find points of
inflection.
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Example 6.__ Given f ( x)  x find the intervals where f ( x) increases and decreases, all relative
1
3
extrema, the intervals where f ( x) is concave up and down, and all points of inflection. Then construct a
graph of f ( x).
Solution Following the flow chart in figure 6.___ above we need to first find the domain. Since
f ( x)  x 3 is a power function, its domain (, ) . Next we perform a first derivative test and start by
finding f '( x) .
1
1 2
1
f '( x)  x  3  2
3
3x 3
Now we find critical values by first finding where f '( x)  0 and then find where f '( x) does not exist.
The first derivative we found above, however, will never equal zero. If we tried to solve f '( x)  0 we
would obtain the false equation shown below.
1
2
3x 3
0 1
0
Since the equation is false, we can conclude that no values of x make this first derivative equal to zero.
2
This first derivative, however, does not exist when 3 x 3  0 which is when x  0 . Since x = 0 is in the
1
domain of f(x) it is the only critical value for f ( x)  x 3 . Recall when the first derivative does not exist
the function either has discontinuity, a vertical tangent, or a cusp at the value of x that makes the first
1
derivative undefined. Since f ( x)  x 3 is everywhere continuous a vertical tangent or a cusp must exist
at x = 0. We will decide which of the two it is after we construct the derivative charts. We now use this
critical value to complete a first derivative chart as shown figure 6.__ below.
Figure 6.__ First derivative chart for f ( x)  x .
1
Test
Point
x = –1
x=1
f '(test point)
f '(1) 
f '(1) 
Behavior of
3
Sign of
f '(test point)
1
1

2
3
3
3(1)
+
1
1

2
3
3
3(1)
+
f ( x)
test point
–1
1
Sign of f '( x)
+
+
x =0
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We can see from the first derivative chart that there are no relative extrema and that f ( x) is increasing on
(,0)   0,   . To find the concavity and point(s) of inflection we need to first find f ''( x) .
2 5
2
f ''( x)   x 3   5
9
9x 3
Possible point(s) of inflection will exists where f ''( x)  0 or where f ''( x) does not exist. But, f ''( x)
will never be zero (for the same reasons the first derivative did not equal zero), so we just need to find
5
where f ''( x) does not exist. f ''( x) does not exist when 9 x 3  0 which will happen when x  0 . Thus,
the only possible point of inflection is at x = 0. To make the final decision about points of inflection and
intervals of concavity we need to complete a second derivative chart as shown in figure 6.__ below.
Figure 6.__ Second derivative chart for f ( x)  x .
1
Test
Point
3
f ''(test point)
x = –1
f ''(1) 
x=1
1
1

5
3
3
3(1)
f ''(1) 
Behavior of
Sign of
f ''(test point)
–
1
1

5
3
3
3(1)
+
f ( x)
test point
–1
1
Sign of f ''( x)
–
+
x =0
Since concavity changes at x = 0 and a tangent line exists at x = 0 , we can conclude that (0, f (0))  (0,0)
is a point of inflection and that f ( x) is concave down on (,0) and concave up on (0, ) . We
discussed earlier that a vertical tangent or a cusp must exist at x = 0. Since (0, 0) is a point of inflection,
we conclude that a vertical tangent exists at x = 0.
Before we construct a graph of f ( x) let’s summarize the information we found.
• the domain of f ( x) is (, )
• f '( x)  0 on (,0)   0,  
• f ( x) has no relative extrema
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• a vertical tangent exists at x = 0.
• f ''( x)  0 on (0, ) and f ''( x)  0 on (,0)
• f ( x) has a point of inflection at (0,0)
Now we can construct a graph of f '( x) .
Figure 6.__ Graph of f ( x)  x
1
3

3
Example 6.__ Given f '( x)  ( x  2) and the domain of f ( x) is (, ) , find the intervals where
f ( x) increases and decreases, all relative extrema, intervals where f ( x) is concave up and down, and all
points of inflection. Then sketch a possible graph of f(x).
Solution Since we are given the first derivative and the domain of f ( x) , we will start the first
derivative test by finding where f '( x)  0 and where f '( x) is undefined.
0  ( x  2)3
0 x2
2x
Since f '( x) is a polynomial function, it is defined for all values of x, thus the only critical value is at x =
2. We now use this critical value to complete a first derivative chart as shown figure 6.21 below.
Figure 6.__ First derivative chart for f '( x)  ( x  2)3 .
ACOW Group
Test
Point
f '(test point)
Sign of
f '(test point)
x=1
f '(1)  (1  2)3  1
–
x=3
f '(3)  (3  2)3  1
+
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Behavior of
f ( x)
test point
1
3
Sign of f '( x)
–
+
x=2
We can see from the first derivative chart that a relative minimum exists at x = 2 and f ( x) is decreasing
on (, 2) and increasing on (2, ) . To find the concavity and point(s) of inflection we need to find
f ''( x) .
f ''( x)  3( x  2)2
Possible point(s) of inflection will exists where f ''( x)  0 or where f ''( x) does not exist. First we find
where the second derivative is equal to zero.
0  3( x  2) 2
0 x2
2x
Since f ''( x) is a polynomial it is defined for all values of x. Thus, the only possible point of inflection is
at x = 2. To make the final decision about points of inflection and intervals of concavity we need to
complete a second derivative chart as shown in figure 6.__ below.
Figure 6.__ Seocond derivative chart for f '( x)  ( x  2)2 .
Test
Point
f ''(test point)
Sign of
f ''(test point)
x=1
f ''(1)  3(1  2)2  3
+
x=3
f ''(3)  3(3  2)2  3
+
Behavior of
f ( x)
test point
1
3
Sign of f ''( x)
+
+
x=2
Since concavity does not change, f ( x) does not have any points of inflection and f ( x) is concave up on
(,2)  (2, ) .
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Before we construct a graph of f ( x) let’s summarize the information we found.
• the domain of f ( x) is (, )
• f '( x)  0 on (, 2) and f '( x)  0 on  2, 
• f ( x) has a relative minimum at x = 2.
• f ''( x)  0 on (,2)  (2, )
• f ( x) does not have any points of inflection.
Now we can construct a graph of f '( x) as shown in figure 6.__ below.
Figure 6.__ Possible graph of f ( x)
Example 6.__ Sketch a graph of a function that fulfils the following criteria.
•
•
•
•
•
Domain (, 4) and Range (,3)
f '( x)  0 on (, 2)  (0,2) and f '( x)  0 on (2,0)  (2,4)
f '( x)  0 at x  2 and x  2 and f '( x) does not exist at x  0
f ''( x)  0 on (,0)  (0,4) and f ''( x)  0 on (,0)  (0,4)
f (2)  3, f (0)  0, f (2)  2, f (4)  1
Solution
Using the information given about f ‘ (x) to construct a first derivative chart (as shown in
figure 6.___) to locate any relative extrema and increasing/decreasing intervals. We will also note that
the first derivative does not exist at x = 0.
Figure 6.__ First derivative chart using information given about f ‘(x).
Behavior of
f ( x)
test point
Sign of
f '( x)
n/a
n/a
n/a
n/a
+
–
+
–
x = –2
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x=0
DNE
x=2
x=4
5/28/2016
The first derivative chart reveals that f (x) has two relative maximums, one at x = –2 and the other at x=
2, and one relative minimum at x = 0. Now we will use the information given about f “(x) to construct a
second derivative chart (as shown in figure 6.___ below) to locate any points of inflection and intervals of
concavity.
Figure 6.__ Second derivative chart using information given about f “(x).
Behavior of
f ( x)
test point
Sign of
n/a
n/a
–
–
f ''( x)
x=4
x=0
The second derivative chart reveals that f(x) does not have any points of inflection. We also conclude that
a cusp must exist at x = 0. There can not be a vertical tangent at x = 0 because there is not a change in
concavity and there can not be any discontinuity since f(x) is everywhere continuous.
We are now ready to begin constructing a possible graph of f(x). First plot the given points
f (2)  3, f (0)  0, f (2)  2, f (4)  1 , label them as maximum or minimum, and then shade any part of
the graph that is not in the domain or range of f(x). This step is shown in figure 6.___ below.
Figure 6.__ Stage 1 of constructing a graph of f ( x) .
max
4
max
2
–6
–4
–2
min
2
4
6
–2
Next we’ll draw arrows to represent the increasing/decreasing intervals and use
up and
to represent concave down as shown in figure 6.__ below.
to represent concave
Figure 6.__ Stage 2 of constructing a graph of f ( x) .
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max
4
max
2
–2
–4
–6
min
2
4
6
–2
We can easily see that a cusp must exist at x = 0. So let’s finish the graph by drawing a smooth curve
following the direction arrows in figure 6.__ and adding the cusp. The final graph is shown in figure 6.__
below.
Figure 6.__
Final graph of f ( x) .
4
2
–6
–4
–2
2
4
6
–2

ACOW Group
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Example 6.__ Given the graph of f '( x) below, sketch a possible graph of f ( x) that is everywhere
continuous and has a domain and range of (, ) .
6
4
2
–6
–4
–2
2
4
6
–2
Solution
First we need to notice that f '( x) does not exist at x = 2. Since f ( x) is everywhere
continuous, a cusp or vertical tangent must exist at x = 2. A little more analysis will help us determine
what happens at x = 2. Let’s start the analysis by finding where f '( x) is positive and where f '( x) is
negative. f ( x) is positive when its graph lies above the x-axis and is negative when it lies below the xaxis. Thus f '( x)  0 on (,2)  (4, ) , and f '( x)  0 on (4, ) . We will constructe a first derivative
chart help us organize this information as shown in figure 6.__ below.
Figure 6.__ First derivative chart for graph given in example 6.___
Behavior of f ( x)
test point
Sign of f '( x )
n/a
n/a
n/a
+
+
–
x=2
x=4
The first derivative chart reveals that f(x) has a relative minimum at x = 4 but no relative minimums.
Now we need to figure out where f ''( x) is positive, negative, and zero. We do this by finding where
f '( x) increases and decreases. Looking at the graph, it appears that f '( x) is increasing on
(,2)  (6, ) and f '( x) is decreasing on (2,6) . Recall that when f '( x) is increasing f ''( x)  0 and
when f '( x) is decreasing f ''( x)  0 . We can construct a second derivative chart as shown in figure
6.___ below to help us organize this information.
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Figure 6.__ Second derivative chart for graph given in example 6.___
Behavior of
f ( x)
test point
Sign of f ''( x)
n/a
n/a
n/a
+
–
+
x=6
x=2
The second derivative chart reveals that f ( x) has two points of inflection, one at x = 2 and the other at x
= 6. We can not conclude that a vertical tangent exist at x = 2 because the graph continues to increase and
the concavity changes. Now we’ll plot the relative maximum and the points of inflection. We do not
have enough information about f ( x) to determine the to determine the y values for these points so we’ll
plot the relative maximum somewhere on the line x = 4 and the points of inflection somewhere on the
line s x = 2 and x = 6. In addition we will draw arrows representing increasing/decreasing intervals and
draw the concavity icons where needed as shown in figure 6.___ below.
Figure 6.__ Information plotted from derivative charts for example 6.___
Now we’ll draw a smooth curve that follows the information displayed in figure 6.__ above. The final
graph is drawn in figure 6.___ below.
Figure 6.__ Final graph for example 6.___
6
4
2
–6
–4
–2
2
4

6
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Sample Quiz
Question 6.1 Find all relative extrema for f ( x) 
1 4 1 3 3 2
x  x  x .
2
3
2
1
Question 6.2 Find the interval(s) where f ( x)  e2 x  3e x  2 x decreases. (If this problem is too hard
2
3 8
2
because of the exponential function then use the function f ( x)  x 3  6 x 3 )
8
Question 6.3 A company’s cost function, in thousands of dollars, is modeled by
C( x)  0.005x2  0.87 x  63 , where x represents the number of items produced. Determine the number
of items the company must produce to minimize their cost.
Question 6.4 If the company from exercise 6.3 has a revenue function, in hundreds of dollars, of
R( x)  0.5x , find the interval(s) where the company’s marginal profit is negative.
Question 6.5 Given f '( x) 
2 3 1 2
x  x  x  1 , find interval(s) where f ( x) is concave up.
3
2
Question 6.6 Find the point of inflection for f ( x)  ex  3x2
Question 6.7 The table below shows the total number of hockey sticks produced for varying numbers of
workers. Find a cubic model that fits the data and then determine the point of diminishing returns.
Number of Workers
1
3
6
7
8
10
Total Number of
Hockey Sticks Produced
6
33
84
96 101 87
Question 6.8 Use the graph of g ( x) below and find
a) interval(s) where f ( x)  0 .
b) the critical values of f ( x) .
c) interval(s) where f '( x)  0 .
d) interval(s) where f ''( x)  0 .
g ( x)
a
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b
c
d
e
f g h
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Question 6.9 Use the information given below to construct a possible graph of f ( x) .
lim f ( x)  3
f '( x)  0 on (0,1)  (2, )
f ''( x)  0 on (0.5,2)
lim f ( x)  2
f '( x)  0 on (,0)  (1,2)
f ''( x)  0 on (,0.5)
Domain (, )
f '( x)  0 at x  0,1
f ''( x) does not exist on (2, )
Range (3, )
f '( x)  1 on (2, )
x  2
x  2
Question 6.10 Sketch a possible graph of f ( x) given the graph of f '( x) below.
–2
2
4
6
8
10
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