Chapter 3
Applications of Derivatives
3.1
Extrema of Functions on Intervals
• Maximum and Minimum Values of a Function • Relative Extrema and Critical Numbers • Finding the Extrema on a Closed Interval
Maximum and Minimum Values of a Function
In Chapter 2, we studied several rules for finding the derivative of a function. In this
section, we use derivatives to find the maximum and minimum values of a differentiable
function. Such values have important consequences. The information of knowing how to
maximize returns or minimize costs is important to know.
Definition 1 Maximum and Minimum Values of a Function
What you have been obliged
to discover by yourself leaves a
path in your mind which you
can use again when the need
arises. - G.C. Lichtenberg
y
5
Let f be a function that is defined on an interval I containing c.
1. f (c) is the minimum value of f on I if f (c) ≤ f (x) for all x in I.
2. f (c) is the maximum value of f on I if f (c) ≥ f (x) for all x in I.
The maximum and minimum values of f on I are called the extreme values or extrema
of f on I.
1
1 2 3 4
x
Figure 1a
In [1, 4], the maximum
value is f (4) = 5 and the
minimum value is f (2) = 1.
In the next example, we use the graph to identify any extrema. It is possible for a
function not to have a maximum or minimum value.
Example 1 Finding the Extrema of a Function
Find the extrema of f (x) = (x − 2)2 + 1 in the indicated interval.
a) [1, 4]
y
5
b) [1, 4)
Solution
a)
b)
The graph of y = f (x) in [1, 4] is given in Figure 1a. The ‘highest point’ is (4, 5)
and the ‘lowest point’ is (2, 1). Then the maximum value of f is f (4) = 5 and the
minimum value is f (2) = 1.
In Figure 1b, we see the graph of y = f (x) in [1, 4). The point (4, 5) is not the
‘highest point’ since (4, 5) does not belong to the graph of y = f (x) in [1, 4).
137
1
1 2 3 4
Figure 1b
In [1, 4), f has
no maximum value and the
minimum value is f (2) = 1.
x
138
CHAPTER 3. APPLICATIONS OF DERIVATIVES
There is no highest point in [1, 4) because given any point we can always find a
higher point. In other words, none of the following values of f (x)
f (3.9)
=
f (3.99)
=
..
.
f (x)
=
(3.9 − 2)2 + 1 = 4.61
(3.99 − 2)2 + 1 = 4.9601
(x − 2)2 + 1, if x < 4
is the maximum value of f (x) on [1, 4). Then f has no maximum value on [1, 4).
However, the point (2, 1) is the ‘lowest point’ on the graph. Hence, the minimum
value of f on [1, 4) is f (2) = 1.
✷
y
Try This 1
Find the extrema of f (x) = x2 + 1 in the indicated interval, see Figure 1c.
2
a)
�1
1
[−1, 0]
b)
(−1, 1)
c)
(0, 1)
d)
(0, 1]
x
Example 2 Finding the Extrema of a Function
Figure 1c
The graph of f (x) = x2 + 1.
Find the extrema of
g(x) =
in the interval [1, 4]
y
�
(x − 2)2 + 1
3
if
if
x �= 2
x=2
Solution In Figure 2, the point (4, 5) is the ‘highest point’ on the graph of y = g(x).
Then the maximum value of g on [1, 4] is g(4) = 5.
Note, the point (2, 1) does not lie on the graph of y = g(x) and cannot be the ‘lowest
point ” on the graph. We cannot claim that a certain point on the graph is the lowest
point since there will be another point on the graph that is ‘lower’. Hence, g(x) has no
minimum value on [1, 4].
5
3
1
1 2 3 4
✷
x
Figure 2
The graph has no minimum
value in [1, 4], but the
maximum value is 5.
Try This 2
Find the extrema of g(x) = (x − 3)2 + 4 where x lies in [1, 4] and x �= 3. In particular,
g(3) is undefined.
The next theorem states a condition when a function is guaranteed to have extreme
values. A proof of the theorem is omitted since it is beyond the scope of this book.
Theorem 3.1 The Extreme Value Theorem
A continuous function on a closed interval [a, b] is guaranteed to have a maximum value
and a minimum value on [a, b].
3.1. EXTREMA OF FUNCTIONS ON INTERVALS
139
Relative Extrema and Critical Numbers
A function f has a relative maximum value at point P (c, f (c)) if P is ‘higher than nearby
points’, or geometrically P is at the ‘top of a hill’. Similarly, f has a relative minimum
value at P if P is ‘lower than nearby points’, or P is at the ‘bottom of a valley’.
As seen in Figure 3, the graph of y = f (x) has a relative maximum value at the
point (1, 5), or f (1) = 5 is a relative maximum value of f . Similarly, f has relative
minimum values at (0, 0) and (3, −27). That is, f (0) = 0 and f (3) = −27 are relative
minimum values of f .
y
y� f �x�
5
�1
1
2
3
4
x
�27
Definition 2 Definition of Relative Extrema
Let f be a function.
1. f (c) is a relative maximum value of f if there is an open interval I containing c for
which f (c) is the maximum value of f on I.
Figure 3
Relative minimum values
are f (0) = 0 and
f (3) = −27, and the relative
maximum value is f (1) = 5.
2. f (c) is a relative minimum value of f if there is an open interval I containing c for
which f (c) is the minimum value of f on I.
The relative maximum values and relative minimum values of f are called the relative
extreme values, or relative extrema of f . The relative maximum values and relative
minimum values are also called relative maxima and relative minima, respectively.
y
Clearly, an extreme value of a function is a relative extreme value. However, a relative
extreme value is not necessarily an extreme value. In Figure 3, f (0) = 0 and f (1) = 5 are
relative extreme values of f but they are not the minimum and maximum values of f .
In Example 3, we will see a particular case of a general phenomena. That is, if f (c) is
a relative extreme value then f � (c) is either zero or undefined.
Example 3 The Derivative at a Relative Extrema
Determine the relative extreme values f (c) of f , and evaluate f � (c).
a)
f (x) = x3 − 3x2 + 2
b)
f (x) = 1 − x2/3
See the graphs in Figures 4a and 4b.
Solution
a)
In Figure 4a, we see that the point (0, 2) is a ‘relative maximum point’. Then
f (0) = 2 is a relative maximum value of f . Similarly, the point (2, −2) is a ‘local
minimum point’. Thus, f (2) = −2 is a relative minimum value of f .
The derivative of f is
f � (x) = 3x2 − 6x = 3x(x − 2).
f �x��x3�3x2�2
2
�1
1 2 3 4
�2
Figure 4a
f (0) = 2 is the relative
maximum value, and
f (2) = −2 is the relative
minimum value of f .
y
Substituting x = 0 and x = 2 into f � (x), we obtain
�
1
�
f (0) = 0 and f (2) = 0.
b)
x
f �x��1�x2�3
Note, the derivative of f is
2
2
f � (x) = − x−1/3 = − √
.
3
33x
Using Figure 4b, we see that f (0) = 1 is the maximum value of f .
2
Then f � (0) is undefined since it evaluates to the expression − 3 √
3 .
0
Finally, from the graph we see that there is no relative minimum value as seen in
Figure 4b.
✷
�1
Figure 4b
f (0) = 1 is the
maximum value of f .
1
x
140
y
f �x��4x�x2
Try This 3
For each relative extreme value f (c), evaluate f � (c). See Figures 5a and 5b.
4
a)
f (x) = 4x − x2
�
�3/2
f (x) = 1 − x2/3
b)
x
2
In Example 3, we have seen that the values of the derivative at the relative extrema are
either zero or undefined. These x-values are called critical numbers.
Figure 5a
f (x) = 4x − x2
Definition 3 The Definition of a Critical Number
y
1
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Let f be a function which is defined at a number c. If f � (c) = 0 or f � (c) does not exist,
then c is a critical number of f .
f �x���1�x 2�3�3�2
Example 4 Finding the Critical Numbers
�1
1
x
Find the critical numbers of each function.
a)
Figure 5b
�
�3/2
f (x) = 1 − x2/3
f (x) = 2x3 − 3x2 − 36x + 1
b)
g(x) = x2/3 (x + 4)
Solution
a)
Evaluate and factor the derivative f � as follows:
f � (x)
=
=
=
�
6x2 − 6x − 36
6(x2 − x − 6)
6(x − 3)(x + 2).
Then f (x) = 0 exactly when x = −2 or x = 3. Thus, the critical numbers are
x = −2, 3; see local extrema in Figure 6a.
y
f �x��
2x3�3x2�36x�1
45
�2
3
x
�80
b)
Figure 6a: The critical numbers of f are −2 and 3.
Apply the product rule and factor g � (x).
g(x)
=
g � (x)
=
g � (x)
=
g � (x)
=
x2/3 (x + 4)
2 −1/3
x
(x + 4) + x2/3
3
x−1/3
[2(x + 4) + 3x]
3
5x + 8
√
33x
3.1. EXTREMA OF FUNCTIONS ON INTERVALS
141
Note, g � (x) = 0 if and only if
5x + 8 = 0.
− 85
is a critical number.
Then x =
Moreover, the derivative g � (0) is undefined since it evaluates to an undefined expres8
sion 3 √
3 . Thus, x = 0 is also a critical number.
0
Hence, the critical numbers are − 85 and 0; see the relative extrema in Figure 6b.
y
g�x��x2�3�x�4�
4
x
� 85
Figure 6b: The critical numbers of g are 0 and −8/5.
✷
Try This 4
Find the critical numbers of each function.
a)
f (x) = x3 + 3x2 − 24x
b)
f (x) =
√
x
c)
f (x) = 3x + 1
A first step in finding the extreme values of a differentiable function is to find the
critical numbers. To illustrate this point, in Figure 7 we see that f has relative minimum
and maximum values at C1 and C2 , respectively. By Theorem 3.2, the abcissas of C1 and
C2 are critical numbers of f . Thus, the relative extrema can be found from the list of
critical numbers. However, we caution the student that there are critical numbers that do
not provide extreme values. In any case, the set of critical numbers is a short list that we
could use to find the relative extrema.
y
C2
y� f �x�
x
C1
Figure 7 The graph of f has
relative extreme values at the
critical points C1 and C2 .
142
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Theorem 3.2 The Relative Extrema Occur Exactly at Critical Points
Let f be a function. If f (c) is a relative extreme value, then c is a critical number of f .
Proof If f � (c) does not exist, then the theorem is true and there would be nothing else
to prove. So, we assume f � (c) exists and we have to prove f � (c) = 0. If f � (c) �= 0, then
either f � (c) > 0 or f � (c) < 0. Suppose
f � (c) = lim
h→0
f (c + h) − f (c)
> 0.
h
By Exercise 47 in Chapter 1, there exists an open interval (−δ, δ) with δ > 0 such that
f (c + h) − f (c)
>0
h
whenever −δ < h < δ with h �= 0. If we multiply both sides of the previous inequality
with a positive h that is less than δ, then we obtain
h·
f (c + h) − f (c)
h
f (c + h) − f (c)
f (c + h)
>
h·0
>
0
>
f (c).
This implies f (c) is not a relative maximum value of f . On the other hand if we multiply
the same inequality with a negative h with −δ < h < 0, then we reverse the direction of
the inequality and find
h·
f (c + h) − f (c)
h
f (c + h) − f (c)
f (c + h)
<
h·0
<
0
<
f (c).
This shows f (c) is not a relative minimum value of f . Since f (c) is a relative extreme
value of f , we have a contradiction.
Similarly, if f � (c) < 0 then we would reach the same contradiction. Hence, we conclude
�
f (c) = 0.
✷
Finding the Extrema on a Closed Interval
Recall, the extrema of a continuous function f defined on a closed interval [a, b] are guaranteed to exist because of Theorem 3.1. As a consequence of Theorem 3.2, the following
guidelines can be used to find the extrema of f on [a, b].
Guidelines for Finding the Extrema of a
Continuous Function f on a Closed Interval [a, b]
1.
Find the critical numbers c of f in (a, b).
2.
Evaluate f (c) at the critical numbers c in (a, b).
3.
Evaluate f (a) and f (b).
4.
The largest value in steps 2-3 is the maximum value of f on [a, b].
The smallest value in steps 2-3 is the minimum value of f on [a, b].
3.1. EXTREMA OF FUNCTIONS ON INTERVALS
143
y
Example 5 Finding the Extrema on a Closed Interval
y�x3�3x�12
6
Find the extrema of f (x) = x3 − 3x − 12 on [0, 3], as in Figure 8
Solution The derivative of f is
x
3
f � (x) = 3x2 − 3.
The critical numbers of f are found as follows:
2
�12
�14
3x − 3
=
0
3x2
=
3
2
=
1
x
=
±1.
x
Figure 8
On [0, 3], the maximum,
value of f is f (3) = 6,
and the minimum value
is f (1) = −14.
The only critical number in the open interval (0, 3) is x = 1, and
f (1) = 13 − 3(1) − 12 = −14.
The values of f at the endpoints of [0, 3] are
f (0) = −12 and f (3) = 33 − 3(3) − 12 = 6.
From the list below
x
1, critical number
0, endpoint
3, endpoint
f (x)
−14
−12
6
we conclude the maximum value of f is 6, and the minimum value is −14.
✷
Try This 5
Find the extrema of the given function on [0, 3].
f (x) = x3 + 3x2 − 24x + 18.
Example 6 Find the Extrema on [a, b]
Find the extrema of g(x) = x2 − x2/3 on the interval [−2, 1], see Figure 9.
Solution
y
The derivative of g is
2
6x4/3 − 2
√
g � (x) = 2x − √
=
.
3
3 x
33x
Note, x = 0 is a critical number of g. If g � (x) = 0, then
6x
4/3
−2
x
=
4/3
=
x
=
0
1
3
1
±√
≈ ±0.44
4
27
√
Then x = ±1/ 4 27 are critical numbers on (−2, 1). Moreover, we find
� �
�2 �
�2/3
�
1
1
1
1
2
1
√
√
=
−
= √ −√ =− √ .
g ±√
4
4
4
27
27
27
3 3
3
3 3
y�x2�x2�3
�2
1
Figure 9
On [−2, 1], the
maximum value of g is
achieved at x = −2, and
the minimum value at√the
critical numbers ±1/ 4 27.
x
144
CHAPTER 3. APPLICATIONS OF DERIVATIVES
In addition, at the endpoints of [−2, 1] we have
√
3
g(−2) = 4 − 4 and g(1) = 0.
Now, identify the largest and smallest values of g in the table
x
0, critical
number
√
±1/ 4 27, critical numbers
−2, endpoint
1, endpoint
g(x)
0
√
−2/(3
3) ≈ −0.38 .
√
4 − 3 4 ≈ 2.4
0
Hence, the maximum value of g is
g(−2) = 4 −
√
3
4
and the minimum value is
√
√
2
2 3
4
g(±1/ 27) = − √ = −
.
9
3 3
✷
Try This 6
Find the extrema of g(x) = x − 3x1/3 on [0, 8].
Example 7 Finding the Extrema on a Closed Interval
y
Find the extrema of h(x) = 2 cos x + sin 2x on [−π, π], see Figure 10.
Solution
y�2cos�x��sin�2x�
3
3
h� (x)
2
�Π
� Π2
The derivative of h is
Π
6
5Π
Π
6
=
x
=
�
h (x)
� 323
Figure 10
The maximum
√ value is
h (π/6) = 3 3/2 and
the minimum value
is
√
h (5π/6) = −3 3/2.
=
=
−2 sin x + 2 cos 2x
−2 sin x + 2(1 − 2 sin2 x)
2
−2(2 sin x + sin x − 1)
Since cos 2x = 1 − 2 sin2 x
−2(2 sin x − 1)(sin x + 1)
�
If h (x) = 0 for x on the interval (−π, π), then
sin x
=
1
2
or
sin x
=
−1
x
=
π 5π
,
6 6
or
x
=
π
− .
2
Moreover, we find
�π�
h
6
� �
5π
h
6
� π�
h −
2
=
=
=
√
√
√
π
π
3
3 3
+ sin = 3 +
=
,
6
3
2
2
√
√
√
5π
5π
3
3 3
2 cos
+ sin
=− 3−
=−
, and
6
3
2
2
� π�
2 cos −
+ sin (−π) = 0.
2
2 cos
The values of h at the endpoints of [−π, π] are
h (−π)
=
h (π)
=
2 cos (−π) + sin (−2π) = −2 + 0 = −2
2 cos π + sin 2π = −2 + 0 = −2.
3.1. EXTREMA OF FUNCTIONS ON INTERVALS
145
The values of h at the critical numbers and endpoints are listed below:
x
π/6, critical number
5π/6, critical number
−π/2, critical number
−π, endpoint
π, endpoint
h(x)
√
3
3/2
≈
2.6
√
−3 3/2 ≈ −2.6
0
−2
−2
Identify the largest
and smallest values of h in the table. Hence,
the maximum value of h
√
√
is h(π/6) = 3 2 3 , and the minimum value is h(5π/6) = − 3 2 3 .
✷
Try This 7
Find the extrema of f (x) = cos 2x + 2 sin x on [0, π].
3.1 Check-It Out
√
1. Find the critical numbers of f (x) = 2 x − x
2. Find the critical numbers of g(t) = 2 sin t − t on the open interval (0, 2π).
3. Find the extrema of f (x) = x4 − 4x on [0, 2].
True or False. If false, explain or show an example that shows it is false.
1. The number f (1) is the minimum value of f (x) = x2 − 2x.
2. If f (c) is a relative extreme value of f , then f � (c) = 0.
3. If c is a critical number of a function f , then f (c) is a relative extreme value of f .
√
4. The critical numbers of f (x) = x3 − x are x = ± 3.
5. The critical number of f (x) = (x − a)(x − b) is x = (a + b)/2.
6. The critical numbers of y = sin(2πx) on the open interval (0, 1) are x =
1 3
, .
4 4
1
7. g(x) = √ has a critical number.
x
8. The extrema of f (x) = x3 − 12x on [0, 3] are 0 and −16.
√
9. The extrema of f (x) = 2 sin x − x on [0, π] are 3 − π/3 and −π.
10. The critical numbers of f (t) =
t2
are t = 0, −2.
t+1
Exercises for Section 3.1
In Exercises 1-4, use the graph to find the extrema of the function in the indicated interval.
1. f (x) = −(x − 3)2 + 5, [1, 4]
y
2. f (x) = −(x − 3)2 + 5, (1, 4)
y
5
5
1
1
1
For No. 1
3
4
x
1
For No. 2
3
4
x
146
CHAPTER 3. APPLICATIONS OF DERIVATIVES
3. g(x) =
�
−(x − 2)2 + 5
3
if
if
x �= 2
,
x=2
[0, 3]
y
y
5
4
4
3
3
2
1
1
2
x
3
2
For No. 3
4. g(x) =
�
3
x
4
For No. 4
(x − 3)3 + 2
4
if
if
x is in (2, 3) ∪ (3, 4)
,
x=3
[2, 4]
In Exercises 5-8, find the value of the derivative at each relative extremum.
5. f (x) = x3 − 12x
6. f (x) = 2x3 − 9x2 + 12x − 2
y
y
16
3
�2
x
2
2
�16
1
For No. 5
For No. 6
7. f (x) = 2 − 3x2/3
x
2
8. f (x) = 18x2/3 − x4/3
y
y
2
�1
1
x
�27
For No. 7
27
For No. 8
In Exercises 9-18, find the critical numbers of the function.
9.
f (x) = x3 − 48x + 2
10.
g(x) = x5/3 − x2/3
12.
x2
13.
f (x) =
14.
x+1
15. g(x) = 4 cos x + 2x − 1, 0 < x < 2π
11.
16. g(x) = 3x − 6 sin x + 2, 0 < x < 2π
17. f (x) = 2 sin x − cos 2x, 0 < x < 2π
1
18. f (x) = sin 2x − sin x, 0 < x < 2π
2
f (x) = x3 + 3x2 − 72x + 4
g(x) = x4/3 (x − 3)
x2 + 8
f (x) =
1−x
x
3.1. EXTREMA OF FUNCTIONS ON INTERVALS
147
In Exercises 19-34, find the extrema of the function in the indicated closed interval.
19.
f (x) = 2x(x − 4), [1, 4]
20.
f (x) = −2x(5 + x), [−3, 0]
21.
f (x) = x3 − 27x + 5, [−4, 4]
22.
f (x) = 4x3 − 3x + 2, [−1, 1]
23.
g(x) = x3 + 6x2 − 15x + 10, [−6, 2]
24.
g(x) = x3 + x2 − x + 1, [−1.5, 0.5]
25.
g(x) = (x − 1)2 (x + 1)2 , [−1/2, 2]
√
f (x) = (x − 2) x, [0, 1]
26.
g(x) = (x + 1)3 (2x − 1), [−1, 1/4]
√
f (x) = x 1 − x, [−1, 1]
27.
29.
31.
33.
f (x) = cos x, [0, 2π]
t2 − 1
, [−1, 1]
t2 + 1
k(t) = 6 sin t − 3t, [0, π]
f (t) =
28.
30.
f (x) = 1 − sin 2πx, [0, 1]
32.
h(t) =
34.
t
, [0, 3]
t2 + 4
M (s) = 2 sin s + cos 2s, [0, π/3]
Applications
35. The Path of Least Cost A plumbing project involves installing pvc pipes from A
to C to B, see figure below. Along the horizontal, the installation costs $4 per foot.
Along the diagonal, the cost is $12 per foot due to extra labor. Find the minimum
cost of the plumbing project.
y
y
30 A
75
50
C
B
40
x
P
For No. 35
100
x
For No. 36
36. Staking Two Antennas Two antennas are 100 feet apart, and their heights from
the ground are 50 feet and 75 feet. Suppose a cable is connected from point P to
the top of each antenna, see above figure. Where should P be located so that the
least amount of cable is used?
37. Generating a Right Circular Cylinder The perimeter of a rectangle is 6 feet. If
the rectangle is revolved about one of it sides, a right circular cylinder is generated.
Find the maximum volume of the cylinder. See figure below.
y
A�0,1�
B�x,0�
2
x
C�2,�1�
For No. 37
For No. 38
38. Bring out a Calculator A line segment has endpoints A(0, 1) and B(x, 0) where
0 ≤ x ≤ 2. A second line segment has endpoints B(x, 0) and C(2, −1). Find x if
two times the length of the first segment plus the length of the second segment is
the minimum. See figure above.
148
3.2
CHAPTER 3. APPLICATIONS OF DERIVATIVES
The Mean Value Theorem
• Rolle’s Theorem • The Mean Value Theorem
Michel Rolle (1652-1719),
a French mathematician,
published Rolle’s Theorem
in 1691. His theorem plays
an important role in the proofs
of several calculus theorems.
Rolle’s Theorem
The above-named theorem is a basic result that makes possible the application of derivatives to finding the extreme values of a differentiable function. Also, Rolle’s Theorem
establishes a connection between critical numbers and the functional values of a differentiable function, see Figure 1.
y
�c1, f �c1��
a
b
x
�c2, f �c2��
Figure 1
Rolle’s Theorem: If f (a) = f (b),
then f � (c) = 0 for some c in (a, b).
Theorem 3.3 Rolle’s Theorem
Let f be a function that is continuous on [a, b] and differentiable on (a, b). If f (a) = f (b),
then there exists a number c in (a, b) satisfying f � (c) = 0.
Proof We consider three cases.
Case 1 If f (x) = f (a) for all x in [a, b], then f is a constant function. Thus, f � (c) = 0
for all c in (a, b).
Case 2 Suppose f (x0 ) > f (a) for some x0 in (a, b). Recall, the Extreme Value Theorem
assures that f has a maximum value f (c) where c is some number c in [a, b]. Since
f (x0 ) > f (a) = f (b), we find c �= a, b. Then f (c) is a relative extreme value of f . By
Theorem 3.2, c is a critical number of f in (a, b). Since f is differentiable in (a, b), we have
f � (c) = 0.
Case 3 Suppose f (x0 ) < f (a) for some x0 in (a, b). We consider the minimum value f (c)
of f . Similarly, as in Case 2 we conclude that c lies in (a, b) and f � (c) = 0.
✷
Example 1 Illustrating Rolle’s Theorem
Show Rolle’s Theorem applies to the function on the indicated interval:
f (x) = x1/3 − x4/3 , [0, 1]
Then find all the numbers c that satisfy Rolle’s Theorem.
Solution To show Rolle’s Theorem applies we have to verify the following: a) f is
continuous on [0, 1], b) f is differentiable on (0, 1), and c) f (0) = f (1). We establish the
validity of statements a), b), and c) as follows.
3.2. THE MEAN VALUE THEOREM
a)
b)
149
√
√ 4
The radical function y = 3 x and the composite function y = ( 3 x) are continuous
everywhere, see page 44 and Theorem 1.10. Then the difference function f (x) =
x1/3 − x4/3 is continuous on [0, 1].
Applying the power rule, we find
f � (x)
=
0.47
1 −2/3 4 1/3
x
− x
3
3
1
(1 − 4x)
3x2/3
�
Since f (x) is undefined only when x = 0, f is differentiable on (0, 1).
Clearly, f (1) = 11/3 − 14/3 = 0 and f (0) = 0.
=
c)
y
1
4
(1)
Thus, Rolle’s Theorem applies. Using (1), we find that f � (c) = 0 implies
1 − 4c = 0.
Hence, c = 1/4 is the only number in (0, 1) that satisfies Rolle’s Theorem, see Figure 2.
✷
1
x
Figure 2
The values of f at the
endpoints of [0, 1] are
f (0) = f (1) = 0. Rolle’s
Theorem guarantees the
existence of c = 1/4 in the
interval (0, 1) such that
f � (c) = 0.
Try This 1
Find all numbers c satisfying Rolle’s Theorem.
a) f (x) = 9x − x3 , [−3, 0]
b) f (x) = sin 2x, [π/4, 5π/4]
Many functions satisfy the hypothesis of Rolle’s Theorem. These include polynomial,
rational, and trigonometric functions provided these functions are defined on the interval
[a, b] in Rolle’s Theorem. Recall, polynomial, rational, and trigonometric functions are
differentiable in their domains of definition, see Sections 1.3
Example 2 Applying Rolle’s Theorem
Apply Rolle’s Theorem and the Intermediate Value Theorem to show that the graph of
f (x) = x5 + 2x + 2
has exactly one x-intercept.
Solution
y
We find two functional values of f with opposite signs:
f (1) = 15 + 2(1) + 2 = 5 and f (−1) = (−1)5 + 2(−1) + 2 = −1
2
By the Intermediate Value Theorem (see page 45), we can find a number x0 in (−1, 1)
such that f (x0 ) = 0. Then (x0 , 0) is an x-intercept of the graph of f .
Suppose (x1 , 0) is another x-intercept of the graph f and x1 �= x0 . Then f (x1 ) =
f (x0 ) = 0. Applying Rolle’s Theorem, there exists a real number c between x1 and x0
such that f � (c) = 0. But this is impossible for the derivative has no zero, i.e.,
�1
f � (x) = 5x4 + 2 �= 0 for all x.
Hence, the graph of f has exactly one x-intercept, namely, (x0 , 0).
✷
Try This 2
Show the graph of
f (x) = 4 − 9x − x3
has exactly one x-intercept using Rolle’s Theorem and the Intermediate Value Theorem.
Figure 3
A sketch of
the graph of
f (x) = x5 + 2x + 2.
1
x
150
The Mean Value Theorem was
first proved by Joseph-Louis
Lagrange (1736-1813). At the
young age of 19, Lagrange became a professor in Turin.
CHAPTER 3. APPLICATIONS OF DERIVATIVES
The Mean Value Theorem
The following theorem is a generalization of Rolle’s Theorem. The Mean Value Theorem
can be applied to classify ‘increasing’ or ‘decreasing’ functions in terms of the derivative,
as we will see in Section 4.3.
Theorem 3.4 The Mean Value Theorem
y
�b, f �b��
Let f be a function that is continuous on [a, b] and differentiable on (a, b). Then there
exists a number c in (a, b) satisfying
�a, f �a��
f � (c) =
Proof
a
c
b
Figure 4
Mean Value Theorem:
The line joining
(a, f (a)) to (b, f (b))
is parallel to a
tangent line at some
point (c, f (c)) where
a < c < b.
f (b) − f (a)
.
b−a
The slope of the secant line joining A(a, f (a)) to B(b, f (b)) is
x
m=
f (b) − f (a)
.
b−a
An equation of the secant line is
s(x) = m(x − a) + f (a).
Let g(x) be the difference f (x) and s(x) as defined below:
g(x) = f (x) − (m(x − a) + f (a)) .
Then g(a) = 0 and g(b) = 0 because of the definition of m. Observe,
g � (x)
=
d
d
[f (x)] −
[m(x − a) + f (a)]
dx
dx
=
f � (x) − m.
Applying Rolle’s Theorem, there exists a number c in (a, b) satisfying
g � (c)
=
0
f (c) − m
=
0.
�
Hence, we obtain
f (b) − f (a)
.
b−a
This completes the proof of the Mean value Theorem.
f � (c) = m =
✷
In other words, the Mean Value Theorem implies that the average rate of change of a
function f over [a, b] is equal to the rate of change of f at some number c in (a, b).
Geometrically, the line joining the end points (a, f (a)) and (b, f (b)) is parallel to a tangent
line at some point (c, f (c)) where a < c < b, see Figure 4.
Example 3 Illustrating the Mean Value Theorem
Find the values of c that satisfy the Mean Value Theorem for the given function in the
indicated interval.
f (x) = x3 − 2x2 − 3x + 4, [−3, 5]
3.2. THE MEAN VALUE THEOREM
Solution
151
y
If c satisfies the Mean Value Theorem, then
f � (c)
=
=
=
�
f (c)
=
f (5) − f (−3)
5+3
64 − (−32)
8
96
8
12.
Next, solve the equation f � (x) = 12.
2
3x − 4x − 3
2
=
12
3x − 4x − 15
=
0
(3x + 5)(x − 3)
=
0
x
=
5
− ,3
3
Note, c must lie in the open interval (−3, 5). Thus, the values of c satisfying the Mean
Value Theorem are c = 3 and c = −5/3, as shown in Figure 5.
✷
Try This 3
Find the values of c that satisfy the Mean Value Theorem for the function
√
h(x) = 2 x + x
in the interval [0, 1].
Note, the values of c in the Mean Value Theorem belong to the open interval (a, b). The
theorem does not indicate how many such c’s there are. The theorem simply states the
existence of such a number.
Example 4 An Application of the Mean Value Theorem
A police officer clocks the speed of a certain car at 65 mph. Five minutes later another
police officer finds the same car going at 60 mph. If the police officers are eight miles
apart, explain why the car was speeding at 96 mph at some point between the two police
officers.
Solution Let t be a fraction of an hour after the first officer clocked the car’s speed.
Denote by s(t) the corresponding distance in miles between the first officer and the car at
time t.
Note, five minutes is equivalent to 1/12 of an hour. Then
� �
1
s(0) = 0 and s
= 8.
12
Suppose s(t) is a differentiable function of t. Then the Mean Value Theorem applies to
1
1
s(t) on [0, 12
]. Thus, there exists a number c in (0, 12
) such that
s� (c) =
s(1/12) − s(0)
8−0
=
= 96 mph.
1/12 − 0
1/12
Hence, at some point between the two police officers the car’s speed was exactly 96 mph.
✷
�3 � 53
�5,64�
3
5
��3,�32�
Figure 5
The tangent lines at
the points where x = −5/3
and x = 3 are parallel
to the secant line that
contains the endpoints
(−3, −32) and (5, 64).
x
152
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Try This 4
Johnny made a 300-mile trip in 4 hours. Show that at some point in the trip Johnny was
driving at 75 mph.
3.2 Check-It Out
1. Find the values of c that satisfy Rolle’s Theorem for f (x) = x3 − 3x + 2 in [−1, 2].
√
2. Find the values of c that satisfy the Mean Value Theorem for f (x) = x in [0, 4].
3. Use Rolle’s Theorem to explain why f (x) = cos x + 2x − 1 has only one x-intercept.
True or False. If false, explain or show an example that shows it is false.
1. For the function f (x) = x3 − 6x2 + 9x − 2 in [0, 3], the value of c that satisfies Rolle’s
Theorem is c = 1.
2. Rolle’s Theorem applies to f (x) = 3(x − 1)2/3 in the interval [0, 2].
√
3. For the function h(t) = t − 4t in [1, 4], the value of c that satisfies the Mean Value
Theorem is c = 3/2.
4. If y = f (x) is a differentiable function and f (a) = f (b), then there is exactly one c
in (a, b) satisfying f � (c) = 0.
5. Suppose a differentiable function y = f (x) has x-intercepts (a, 0) and (b, 0), a �= b.
Then y = f (x) has a critical number c in (a, b).
6. A police officer spotted a car traveling at 55 mph. After 3 minutes, a state trooper
clocked the same car at 60 mph. If the officer and trooper are 5 miles apart, then
the car’s speed at some point in between was exactly 100 mph.
7. The function f (x) = sin(2πx) + 2x2 − x satisfies Rolle’s Theorem in the interval
[0, 1/2].
8. For f (x) = 1/x2 in [−2, 1], there exists a number c in (−2, 1) such that
f � (c) =
f (1) − f (−2)
.
1 − (−2)
9. For f (x) = 1/x in [−1, 1], there exists a number c in (−1, 1) such that
f � (c) =
f (1) − f (−1)
.
1 − (−1)
10. If a car can accelerate from 60 mph to 70 mph in 1 minute, then the car’s acceleration
at some instant is exactly 600 mph per hour.
Exercises for Section 3.2
In Exercises 1-8, determine if Rolle’s theorem applies to the function in the indicated
interval. If it does, find the values of c that satisfy Rolle’s Theorem.
1.
f (x) = 2 + 6x − x2 , [0, 6]
2.
f (x) = 3x2 − x − 2, [0, 1/3]
3.
s(t) = t3 − t2 + 4, [0, 1]
4.
g(t) = t3 − 6t2 + 11t − 2, [1, 2]
5.
T (θ) = 2 sin θ − 1, [π/6, 5π/6]
6.
T (θ) = tan θ + cot θ, [π/6, π/3]
8.
h(α) = cos[(α2 − 4α + 5)π], [1, 3]
7.
2
p(w) = sin w , [−2, 2]
3.2. THE MEAN VALUE THEOREM
153
In Exercises 9-16, find the values of c that satisfy the Mean Value Theorem.
9.
g(t) = t3 − t2 + t − 1, [0, 1]
10.
f (x) = 2x2 − x3 − 3x + 1, [−3, 0]
11.
R(s) = s2 − s4 + 3s, [−1, 1]
12.
C(w) = w(w − 3)2 + w, [0, 2]
13.
15.
x+1
, [2, 3]
x−1
√
A(h) = h − h, [0, 4]
f (x) =
14.
16.
x−2
, [0, 3]
x+2
√
p(t) = 2t + t, [0, 1]
f (x) =
Applications of the Mean Value Theorem
17. Let y = f (x) be a function such that f � (a) �= 1 for any a. Prove that the equation
f (w) = w has at most one solution w.
18. Let y = g(t) be a function such that for some nonzero constant k we have g � (t) �= k
for any t. Prove g(x) = kx has at most one solution x.
19. Arithmetic Mean Verify that the value of c that satisfies the Mean Value Theorem
for f (x) = x2 on the interval [a, b] is c = (a + b)/2.
20. Geometric Mean Let a, b > 0 be positive numbers. Show that the value of √
c that
satisfies the Mean Value Theorem for r(x) = 1/x on the interval [a, b] is c = ab.
21. Let f be a differentiable function such that f (1) = 4 and 0 ≤ f � (x) ≤ 2 for all x.
Find a maximum possible value for f (6).
22. Let p be a differentiable function such that p(10) = 2 and p� (x) ≥ 3 for all x. Find
a minimum possible value for p(12).
23. If a �= b, show that
24. If x �= y, show that
�
�
� sin b − sin a �
�
� ≤ 1.
�
�
b−a
�
�
� tan x − tan y �
�
� ≥ 1.
�
�
x−y
25. Let y = f (x) be differentiable function that satisfies f (1) = 1 and f (2) = 2. Show
that there exists a number c in the open interval (1, 2) such that the tangent line to
the graph of f at the point (c, f (c)) passes through the origin.
√
3 3
26. Show that y = x3 − Ax2 + 1 has three distinct zeros if A >
, and has exactly
2
√
3 3
one zero if A <
.
2
27. Driving in an interstate Jimmie went from one exit to another exit in 80 seconds.
If the exits are 2 miles apart, explain why Jimmie’s speed was exactly 90 mph at
some point between the two exits.
28. Let y = f (x) be continuous on [a, b] and differentiable on (a, b). If f � (x) = 0 for all
x in (a, b), then f (s) = f (t) for all s, t in [a, b].
154
3.3
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Increasing and Decreasing Functions
• Increasing and Decreasing Functions • The First Derivative Test
Increasing and Decreasing Functions
In this section, we will use the derivative to identify the relative extrema of a differentiable
function. First, we need a few preliminaries.
Definition 4 Increasing, Decreasing, and Constant Functions
Let f be a function on an interval I.
1. f is increasing on I if
y
f (x1 ) < f (x2 ) whenever x1 < x2 and x1 , x2 belong to I.
2. f is decreasing on I if
f (x1 ) > f (x2 ) whenever x1 < x2 and x1 , x2 belong to I.
3. f is constant on I if f (x1 ) = f (x2 ) for all x1 , x2 in I.
a
x
b
Geometrically, a function f is increasing on an interval I if its graph is rising as x
moves to the right in I. In Figure 1a, f is increasing on an open interval (a, b). Likewise,
f is decreasing on I if its graph is falling as x moves to the right in I. In Figure 1b, f is
decreasing on (b, c).
The derivative can tell us if a differentiable function is increasing or decreasing. In
Figure 1c, we see f is increasing on (a, b), the tangent lines have positive slopes for x in
(a, b), and consequently f � (x) > 0. Similarly, f is decreasing on (b, c), the tangent lines
have negative slopes, and f � (x) < 0 for x in (b, c). The next theorem summarizes these
results.
Figure 1a
The function f is
increasing on the
open interval (a, b).
y
y
b
c
Figure 1b
The function f is
decreasing on (b, c).
x
a
b
c
d
x
Figure 1c
A function is increasing, decreasing, or constant
according to whether its derivative is positive,
negative, or zero, respectively.
Theorem 3.5 Increasing and Decreasing Test
Suppose f is a continuous function on [a, b], and differentiable on (a, b).
1. If f � (x) > 0 for all x in (a, b), then f is increasing on [a, b].
2. If f � (x) < 0 for all x in (a, b), then f is decreasing on [a, b].
3 If f � (x) = 0 for all x in (a, b), then f is constant on [a, b].
3.3. INCREASING AND DECREASING FUNCTIONS
155
Proof Suppose f � (x) > 0 for all x in (a, b). If a ≤ x1 < x2 ≤ b, then by the Mean Value
Theorem there exists a number c such that x1 < c < x2 and
f � (c) =
f (x2 ) − f (x1 )
.
x2 − x1
Since f � (c) and x2 − x1 are positive, f (x2 ) − f (x1 ) is also positive. Thus,
f (x2 ) > f (x1 )
and consequently f is increasing on [a, b]. The remaining two cases are proved similarly,
see Exercises 59 and 61 at the end of the section.
✷
The above theorem implies that we have to solve inequalities such as f � (x) > or
f (x) < 0 to determine where a function is increasing or decreasing. In the next example,
we use a standard method to solve such inequalities. The method involves finding the
critical numbers and determining the signs of f � (x) off the critical numbers. This method
is described in more details after Try This 1.
�
Example 1 Identify Where f is Increasing or Decreasing
Find the open intervals where
f (x) = 4x3 − 11x2 + 6x + 15
is increasing or decreasing.
Solution
Since the derivative is given by
f � (x) = 12x2 − 22x + 6 = 2(3x − 1)(2x − 3)
the critical numbers are x = 13 , 32 .
If we delete the critical numbers from the number line R, we obtain a union of open
intervals:
�
� �
� �
� �
�
1 3
1
1 3
3
R−
,
= −∞,
∪
,
∪
,∞ .
3 2
3
3 2
2
�
�
1
Choose a number in I1 = −∞, 3 , say x1 = −1. Determine the sign of f � (x1 ); in fact
f � (−1) > 0. Then f � (x) > 0 for all x in I1 .1 Thus, f is increasing on I1 by Theorem 3.5.
Repeat this procedure and choose test values x2 and x3 from
�
�
�
�
1 3
3
I2 =
,
and I3 =
,∞ .
3 2
2
Then determine the sign of f � (xi ) which is the sign of f � on the test open interval Ii . The
following table summarizes the results.
�
�
�1 3�
�3
�
Test Open Interval
−∞, 13
,
,∞
3 2
2
Test Value xi
x1 = −1
x2 = 1
x3 = 2
Sign of f � (xi )
f � (−1) > 0 f � (1) < 0
f � (2) > 0
Apply Theorem 3.5 Increasing Decreasing Increasing
Hence, f is increasing on (−∞, 13 ), increasing on ( 32 , ∞), and decreasing on ( 13 , 32 ). The
graph of f is shown in Figure 2.
✷
1
f � (−1)
f � (x)
The sign of
will be the same as the sign of
for any other test value x in I1 . For
if f � (x1 ) and f � (x2 ) have opposite signs with x1 and x2 in I1 , then by the Intermediuate Value
Theorem there exist a number x0 in I1 for which f � (x0 ) = 0; but this is a contradiction since I1
does not contain any critical number of f .
y
�1�3, 430�27�
�3�2, 51�4�
1
3
3
2
Figure 2
f is increasing on
(−∞, 13 ) and ( 32 , ∞),
and decreasing on ( 13 , 32 ).
x
156
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Try This 1
Find the open intervals on which the function is increasing or decreasing.
a) s(t) = −16t2 + 96t + 10
b) p(x) = x3 − 6x2 + 4
The guidelines below are for students to help them develop a strategy for finding open
intervals on which a function is increasing or decreasing.
Guidelines for finding open intervals
where a function is increasing or decreasing
Let f be a differentiable function on an open interval (a, b).
1. Find the critical numbers xi in (a, b) and arrange in ascending order:
a < x1 < x2 < · · · < xn < b.
2. Form the open intervals I1 = (a, x1 ), I2 = (x1 , x2 ), ..., In+1 = (xn , b)
3. Select any test value xi in Ii .
4. Apply Theorem 3.5:
• If f � (xi ) > 0 then f is increasing on Ii .
• If f � (xi ) < 0 then f is decreasing on Ii .
In the above, we allow for a = −∞ or b = ∞.
The First Derivative Test
The sign of the slope of a tangent line is an indicator of whether a function is locally
increasing or decreasing. This observation is the basis of the First Derivative Test.
However, we need to introduce some terminologies. For our purposes, think of the function
g in the next theorem as the first or second derivative of a function f .
Definition 5 A Function Changing its Sign at a Number
y
Let g be a function, and let c be a number.
1. g changes from positive to negative at c if there are open intervals (b, c) and (c, d)
such that g is positive on (b, c) and g is negative on (c, d).
�1�3, 430�27�
�3�2, 51�4�
1
3
3
2
Figure 3
The derivative changes
sign at a local extrema.
2. g changes from negative to positive at c if there are open intervals (b, c) and (c, d)
such that g is negative on (b, c) and g is positive on (c, d).
x
3. The sign of g stays constant about c if there are open intervals (b, c) and (c, d)
such that either g is negative on (b, c) and (c, d), or g is positive on (b, c) and (c, d).
For instance, see Figure 3, the sign of the derivative f � changes from positive to negative
at x = 13 . Also, f � changes its sign from negative to positive at x = 32 . Moreover, we see
that f has relative maxima f ( 13 ) and relative minima f ( 32 ). The First Derivative Test
states that sign changes of f � are indicators of the local extrema of a function.
3.3. INCREASING AND DECREASING FUNCTIONS
157
Theorem 3.6 The First Derivative Test
Let f be a differentiable function, and let c be a critical number of f .
1. If f � changes from positive to negative at c, then f (c) is a relative maximum of f .
2. If f � changes from negative to positive at c, then f (c) is a relative minimum of f .
3. If the sign of f � stays constant about c, then f (c) is not a relative extremum of f .
Proof
Suppose f � changes from positive to negative at c. Choose b and d such that
f � (x) > 0 for all x in (b, c)
and
f � (x) < 0 for all x in (c, d).
Then f is increasing on (b, c) and f is decreasing on (c, d) by Theorem 3.5. Since f is
continuous at c, f is increasing on (b, c] and f is decreasing on [c, d). Thus, f (c) is the
maximum value of f on the (b, d). Hence, f (c) is a relative maximum value of f .
The proof of the second part is similar. We leave the proof as an exercise for the
student to prove, see Exercise 62.
If the sign of f � stays constant at c, then there exist numbers c1 , c2 such that either f �
is positive on (c1 , c) and (c, c2 ), or f � is negative on (c1 , c) and (c, c2 ). Applying Theorem
3.5 and the continuity of f at c, we obtain that f is either increasing or decreasing on
(c1 , c2 ). Hence, f (c) is neither a relative maximum nor relative minimum value of f .
✷
Example 2 Applying the First Derivative Test
Find the relative extrema of g(x) = x1/3 − x2/3 .
Solution
y
The derivative of g is
g � (x)
=
1 −2/3 2 −1/3
1 − 2x1/3
x
− x
=
.
3
3
3x2/3
1
Then x = 0 is a critical number of g since g � (0) is undefined and g(0) is defined. To find
another critical number, suppose g � (x) = 0. Then
1 − 2x
1/3
=
x
=
0
� �3
1
1
= .
2
8
Thus, the critical numbers are x = 0, 18 . Consider the open intervals
�
�
�
�
1
1
I1 = (−∞, 0) , I2 = 0,
, I3 =
,∞
8
8
following the guidelines after Example 1. Find test values xi in Ii , and evaluate g � (xi ) as
follows:
�
�
�
�
Test Open Interval I1 = (−∞, 0) I2 = 0, 18
I3 = 18 , ∞
Test Value xi
x1 = −1
x2� =� 1/9
x3 = 1
Sign of g � (xi )
g � (−1) > 0
g � 19 > 0
g � (1) < 0
Positive
Positive
Negative
Note, g � changes from positive to negative at 18 . Applying the First Derivative Test, g has
a relative maximum value at x = 18 . Since the sign of g � stays constant at x = 0, g does
not have a relative extreme value at x = 0. Hence, the relative maximum value of g is
� � � �1/3 � �2/3
1
1
1
1
1
1
g
=
−
= − = .
8
8
8
2
4
4
The graph of g is shown in Figure 4.
✷
Rel. max.
�1�8,1�4�
1
x
Figure 4
g has a relative maxima at
x = 18 since g � changes from
positive to negative at x = 18 .
158
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Try This 2
Find the relative extrema of
g(t) =
1 4/3 1 5/3
t
− t .
4
5
Example 3 Using the First Derivative Test
Find the relative extrema of
h(x) =
√
3 sin x + cos x
in the open interval (0, 2π), see Figure 5.
y
Solution
The derivative is
h� (x) =
1
Π
3
4Π
3
x
√
3 cos x − sin x.
To find the critical numbers, we find the zeros
√
3 cos x − sin x
√
3 cos x
√
3
of the derivative.
=
0
=
sin x
=
tan x
=
π 4π
,
.
3 3
�2
Figure 5
The relative extrema
of h are y = ±2.
x
Then we find test values xi in the open intervals Ii where
�
�
�
�
� π�
π 4π
4π
I1 = 0,
, I2 =
,
, I3 =
, 2π .
3
3 3
3
following the guidelines after Example 1.
Test Open Interval
�
Test Value xi
x1 =
Sign of h� (xi )
0,
π�
3
π
6
h� (x1 ) = 1
Positive
�
π 4π
,
3 3
x2 =
�
π
2
h� (x2 ) = −1
Negative
�
4π
, 2π
3
x3 =
�
3π
2
h� (x3 ) = 1
Positive
Hence, by the First Derivative Test, the relative maximum value of h is
�π� √
π
π
h
= 3 sin + cos = 2
3
3
3
and the relative minimum value of h is
� �
√
4π
4π
4π
h
= 3 sin
+ cos
= −2.
3
3
3
✷
Try This 3
Find the relative extrema of
k(x) = 2 cos x − sin 2x where 0 < x < 2π.
3.3. INCREASING AND DECREASING FUNCTIONS
159
Example 4 Using the First Derivative Test
m(x) = 23 x3 +
Find the relative extrema of
Solution
32
.
x
y
Evaluate m� (x) as follows:
m� (x)
32
x2
=
2x2 −
=
2(x4 − 16)
x2
��2,�64�3�
�2
2
=
2(x − 2)(x + 2)(x + 4)
x2
I1 = (−∞, −2) , I2 = (−2, 0) , I3 = (0, 2) , I4 = (2, ∞).
Select certain test values xi in Ii and evaluate m� (xi ).
(−∞, −2)
x1 = −3
m� (x1 ) > 0
Positive
(−2, 0)
x2 = −1
m� (x2 ) < 0
Negative
2
Factoring
The critical numbers of m are x = ±2. Note, x = 0 is not a critical number for m(0) is
undefined. Form the open intervals using the critical numbers:
Test Open Interval
Test Value xi
Value of m� (xi )
�2,64�3�
20
(0, 2)
x3 = 1
m� (x3 ) < 0
Negative
Figure 6
Relative maxima and
relative minima
(2, ∞)
x4 = 3
h� (x4 ) > 0
Positive
We apply the First Derivative Test. Since m� changes from positive to negative at
x = −2, the relative maximum value of m is
m (−2) =
2
32
16
64
(−2)3 +
=−
− 16 = −
3
−2
3
3
Similarly, m� changes from negative to positive at x = 2. Hence, the relative minimum
value of m is
16
64
m (2) =
+ 16 =
.
3
3
The graph of m is shown in Figure 6.
✷
Try This 4
Find the relative extrema of R(x) = x2 +
25
.
x2 +4
Example 5 The Longest Rod Through a Corner
Find the length of the longest rod that can be carried horizontally from one hall onto the
other hall, as in Figure 7. Suppose the halls are 4 feet wide and they meet at a right angle.
Solution The length of the longest rod that can be carried horizontally through the halls
is the minimum value of L1 + L2 , as seen Figure 7. Using right triangle trigonometry, we
obtain
π
L = L1 + L2 = 4 sec α + 4 csc α, 0 < α < .
2
The derivative of L with respect to α is
L� (α)
=
4 sec α tan α − 4 csc α cot α
=
4
�
=
4
�
sin α
cos α
−
cos2 α
sin2 α
sin3 α − cos3 α
sin2 α cos2 α
�
�
.
L1
Α
4
4
L2
Α
Figure 7
L1 and L2 are the
hypotenuse of right
triangles with angle α.
x
160
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Then the critical numbers of L in (0, π2 ) must satisfy
sin3 α − cos3 α = 0
Thus, the critical number is α =
or
tan α = 1.
π
.
4
Consider the open intervals
�π π�
π�
I1 = 0,
, I2 =
,
.
4
4 2
�
The sign of L� (α) in the above open intervals are described below.
�
�
�
�
Test Open Interval
I1 = 0, π4
I2 = π4 , π2
π
π
Test Value αi
� α�1 = 6
�α2�= 3
Sign of L� (αi )
L� π6 ≈ −11.2 L� π3 ≈ 11.2
Negative
Positive
L1
Α
3 3
1
L2
Α
Note, L is decreasing on (0, π/4) and L is increasing on (π/4, π/2) by
Theorem 3.5. Thus, the minimum value of L is
�π�
�π�
�π�
√
L
= 4 sec
+ 4 csc
= 8 2 ≈ 11.3 feet.
4
4
4
√
Hence, the length of the longest possible rod is 8 2 feet.
✷
Figure 8
Find the minimum sum of
L1 + L2 given√the widths
of 1 yd and 3 3 yd.
Try This 5
√
As in Example 5 but suppose the widths of the halls are 1 yard and 3 3 yards, see Figure
8. Find the length of the longest rod that can be carried horizontally from one hall onto
the other hall.
3.3 Check-It Out
1. Find the open intervals on which f (x) = 12x − x3 + 10 is increasing or decreasing.
2. Find the relative extrema of s(t) = 2t3 + 3t2 − 12t + 7 by using the First Derivative
Test.
3. Find the relative extrema of y = sin x + cos x where 0 < x < 2π.
4. Find the maximum product of two positive numbers x and y where x + y = 1.
True or False. If false, explain or show an example that shows it is false.
1. If f � (x) > 0 for x in (−10, 10), then f is increasing on (−10, 10).
2. If f � (x) < 0 for x in (−1, 1), then f is decreasing on [−1, 1].
3. If f (x) = (x − 2)5 , then the sign of f � (x) stays constant about 2.
4. If y = f (x) is continuous on (0, 6) and the sign of f � (x) changes from negative to
positive at 4, then f (4) is a relative minimum of f .
5. If f � (x) > 0 for x in (−∞, 0) and f � (x) < 0 for x in (0, ∞), then f (0) is a relative
maximum of f .
6. If f � (x) = (x − 1)2 (x + 1)3 , then f is increasing on (−1, ∞).
7. If f � (x) = −3(x − 2)5 , then f � (x) > 0 for x in (2, ∞).
√
8. If g(t) = t(t − 1), then g(1/3) is a relative minimum of g.
9. The function s(t) = 2 cos t + t − 15 has a relative maximum at t = π/6.
10. The minimum sum x + y of two positive numbers x and y for which xy = 1 is two.
3.3. INCREASING AND DECREASING FUNCTIONS
161
Exercises for Section 3.3
In Exercises 1-20, determine the open intervals on which the function is increasing or
decreasing.
1.
f (x) = 5(4 − x)(x + 2)
2.
g(x) = −3x(x + 2)
3.
y = 3x(x − 4) − x(4 − x)
4.
y = (x − 1)(x + 5) − 10(x + 5)
5.
p(x) = 4x3 − x2 − 2x − 5
6.
r(x) = 8x3 − 3x2 − 9x + 4
7.
S(t) = −4t3 − t2 + 2t − 6
8.
g(t) = −4t3 + 11t2 − 6t + 7
9.
y = t3 + 6t2 + 12t − 3
10.
y = 6t2 − 3t3 − 4t + 2
x+1
2x − 3
12.
P (x) =
3x − 1
x+4
14.
N (x) =
1 − x2
x2 − 4
11.
R(x) =
13.
m(x) =
15.
s(θ) = sin2 θ, 0 < θ < 2π
16.
t(θ) = cos2 θ, 0 < θ < 2π
17.
y = sin x − cos x, 0 < x < 2π
18.
y = cos x + sin x, 0 < x < 2π
19.
r = 2 sin α − cos 2α, 0 < α < 2π
20.
r = cos α −
x2 + 1
x2 − 9
sin 2α
, 0 < α < 2π
2
In Exercises 21-46, determine all the relative extreme values of the function. Apply the
First Derivative Test.
21.
f (x) = 4x(x − 4)
22.
g(x) = 3(x − 1)(x + 3)
23.
h(x) = x3 + 3x2 − 9x + 15
24.
y(t) = (t + 1)2 (t − 3)
25.
y(t) = (3t − 1)2 (t + 1)
26.
y = 27x − 4x3 − 7
27.
f (x) = 3x4 − 14x3 + 9x2 + 2
28.
M (x) =
29.
L(x) =
31.
33.
x2 − 9
x2 + 1
x
k(x) =
8 − x3
√
f (x) = x(2 − x)2
�
2+x
22 − x2
30.
if
if
35.
f (x) =
37.
f (x) = (x + 1)2/3 − 2x
39.
41.
43.
45.
x≤4
x>4
32.
f (x) = x1/3 + x−1/3
34.
f (x) = x4 − 18x2
36.
f (x) =
38.
f (x) = x2/3 (3 − x)1/3
40.
2
x2 + 1
x2 − 4
x
N (x) = 3
x +4
�
1 − 2x
x2 + 1
if
if
x ≤ −2
x > −2
1 3 16
x +
3
x
√
f (x) = x(x − 1)3
f (x) =
g(t) = 4 + sin t, 0 < t < 2π
√
v(θ) = sin θ + 3 cos θ, 0 < θ < 2π
42.
g(t) = cos2 t − 3, −π < t < π
44.
A(w) = 2 sin w + w −
f (α) = cos2 α − sin α,
46.
0 < α < 2π
π
, −π < w < π
3
g(β) = sin2 β + cos β, 0 < β < 2π
Applications
47. The difference between two numbers is one. Find the minimum product of two such
numbers.
48. What is the minimum sum of two positive numbers whose product is one?
49. Find the maximum area of a rectangle that has a perimeter of 12 feet. What if the
perimeter is p feet?
162
CHAPTER 3. APPLICATIONS OF DERIVATIVES
50. Find the minimum perimeter of a rectangle that has an area of 16 square inches.
What if the area is A square inches?
51. Let f (x) be the square of the distance between (0, 1) and a point (x, 4 − x2 ) on the
parabola y = 4 − x2 .
a) Find open intervals on which y = f (x) is increasing or decreasing.
b) Which point on the parabola is closest to (0, 1)?
y
52. Let D(x) be the square of the distance between (1, 0) and a point (x,
√
parabola y = x.
6
√
x) on the
a) Find open intervals on which y = D(x) is increasing or decreasing.
√
b) Which point on the graph of y = x is nearest to (0, 1)?
y
53. A rectangle is bounded by the x-and y-axis and the line 3x + 2y = 12, see figure
below. Find the dimensions of the rectangle that has the maximum area.
x
4
x
Figure for No.53
54. A triangle is bounded by the x-and y-axis and a line that passes through the point
(1, 2). Find the dimensions of the triangle of minimum area.
Theory and Proofs
55. Find and sketch the graph of a function y = p(x) that satisfies
y
a) p(2) = 1, p� (2) = 0, p� (x) > 0 if x > 2, and p� (x) < 0 when x < 2
b) p(1) = −3, p� (1) = 0, p� (x) > 0 if x �= 1
y
56. Sketch the graph of a differentiable function m = v(t) that satisfies
�1,2�
x
Figure for No.54
a) v(2) = 5, v(−2) = −5, v � (−2) = v � (2) = 0, v � (t) > 0 if |t| < 2,
and v � (t) < 0 when |t| > 2
x
b) v(1) = v(−1) = 1, v � (1) = v � (−1) = 0, v � (t) > 0 if t < −1, v � (t) < 0
if −1 < t < 0, v � (t) > 0 if 0 < t < 1, and v � (t) < 0 if t > 1
57. Let f (x) = x(x − a)(x − b). Prove y = f (x) is decreasing on the open
√
�
�
interval a+b−c
, a+b+c
where c = a2 − ab + b2
3
3
58. Let p(x) = (x − a)2 (x − b)2 , a < b. Prove y = p(x) is increasing
�
�
on a, a+b
and (b, ∞).
2
59. Let f be continuous on [a, b] and differentiable on (a, b). Suppose f � (x) < 0 for all
x in (a, b). Prove f is decreasing on [a, b]. Hint: Mean Value Theorem.
60. Suppose f is continuous on [a, b]. If f � (x) > 0 for x in (a, b), prove f is increasing
on [a, b].
61. Let f be continuous on [a, b] and differentiable on (a, b). If f � (x) = 0 for all x in
(a, b), prove f is a constant function on [a, b].
62. Let c be a critical number of a differentiable function f . If f � changes from negative
to positive at c, prove f (c) is a local minimum of f .
Odd Ball Problems
63. Find a 3rd degree polynomial that has a relative maximum point at (−1, 2) and a
relative minimum point at (1, 1).
64. Find a 3rd degree polynomial that has a relative maximum point at (4, 5) and a
relative minimum point at (1, 3).
�x�
π
≤ tan2 x.
65. If 0 ≤ x < , prove tan2
2
2
�x�
√
π
66. If 0 ≤ x ≤ , prove 3 cos x ≤ cos
.
2
3
3.4. CONCAVITY OF THE GRAPHS OF FUNCTIONS
3.4
163
Concavity of the Graphs of Functions
• Concavity • Points of Inflection • The Second Derivative Test
Concavity
In Section 3.3, we discussed how the sign of the derivative f � determines whether f is
increasing or decreasing. Also, we studied the First Derivative Test which is a criterion for
finding the relative extrema of f . In this section, we will see how the sign of the second
derivative f �� determines whether the graph of f is curving upward or curving downward.
In addition, we discuss the Second Derivative Test which is another criterion for finding
the relative extrema of f .
Definition 6 Concavity of a Graph
Let f be a differentiable function on an open interval I. The graph of f is concave
upward if the derivative f � is an increasing function on I. Likewise, the graph of f is
concave downward if f � is decreasing on I.
For simplicity, we say f is concave upward or downward on an interval I if the graph
of f is concave or downward on I, respectively.
Consider the graphs of f and p in Figures 1a and 2a, respectively. Note, f � and p� are
increasing since the slopes of the tangent lines to the graphs of f and p are increasing. By
definition, f and p are concave upward. Similarly, we see the graphs of g and q in Figures
1b and 2b. Since g � and q � are decreasing functions, g and q are concave downward.
y
y
The graph of g is
concave downward
The graph of f is
concave upward
x
x
Figure 1a. f � is increasing.
y
Figure 1b. g � is decreasing.
y
The graph of q is
The graph of p is
concave downward
concave upward
x
�
Figure 2a. p is increasing.
x
�
Figure 2b. q is decreasing.
Moreover, we claim f is concave upward if and only if the graph of f lies above all its
tangent lines. Also, f is concave downward if and only if the graph of f lies below all its
tangent lines. For a proof, see Theorem A.6, page 413.
164
CHAPTER 3. APPLICATIONS OF DERIVATIVES
The next theorem is a straightforward way to determine concavity. The proof follows
directly from Definition 6 and Theorem 3.5 in Section 3.3.
y
f �x� � x3�3x�3
Theorem 3.7 Concavity Test
concave
downward
concave
upward
Let f be twice differentiable on an open interval I.
x
1. If f �� (x) > 0 for all x in I, then the graph of f is concave upward on I.
2. If f �� (x) < 0 for all x in I, the graph of f is concave downward on I.
In Figure 3, we see the graphs of f (x) = x3 − 2x + 3 and f �� (x) = 6x. Since f �� (x) > 0
on (0, ∞), f is concave upward on (0, ∞) by Theorem 3.7. Likewise, since f �� (x) < 0 on
(−∞, 0), f is concave downward on (−∞, 0).
y
Example 1 Determining Concavity
f ''�x��6x
Find the open intervals where the graph of
f '' is positive
f (x) = x4 − 2x3 + 2
x
is concave upward or concave downward.
f '' is negative
Solution
Figure 3
The concavity of the
graph of f is determined
by the signs of f �� .
We obtain the derivatives as follows:
f � (x)
=
f �� (x)
=
4x3 − 6x2
12x2 − 12x = 12x(x − 1).
Then x = 0, 1 are the solutions to f �� (x) = 0. We test the signs of f �� (x) on the open
intervals I0 = (−∞, 0), I1 = (0, 1), and I2 = (1, ∞).
Test open interval
Test value xi
Value of f �� (xi )
Sign of f �� on Ii
y
(−∞, 0)
x1 = −1
f �� (x1 ) = 24
Positive
(0, 1)
x2 = 12
f �� (x2 ) = −3
Negative
(1, ∞)
x3 = 2
f �� (x3 ) = 24
Positive
Applying Theorem 3.7, f is concave upward on (−∞, 0) and (1, ∞), and concave downward
on (0, 1). The graph of f is shown in Figure 4.
✷
3
Try This 1
1
x
Find the open intervals on which the graph of y = x4 + 2x3 − 1 is concave upward or
concave downward.
Figure 4
A sketch of the graph of
f (x) = x4 − 2x3 + 2
Example 2 Determining Concavity on Open Intervals
Discuss the concavity of the graph of f (x) =
Solution
24
.
x2 + 12
Applying the General Power Rule,we obtain
f � (x)
=
−24(x2 + 12)−2 (2x)
=
−48x
.
(x2 + 12)2
3.4. CONCAVITY OF THE GRAPHS OF FUNCTIONS
165
Then by the Quotient Rule, we find
f �� (x)
(x2 + 12)2 (−48) + 48x(2(x2 + 12)(2x))
(x2 + 12)4
�
�
2
48(x + 12) −(x2 + 12) + 4x2
Factor out 48(x2 + 12)
(x2 + 12)4
�
�
48(x2 + 12) 3x2 − 12
(x2 + 12)4
144(x − 2)(x + 2)
.
Simplify
(x2 + 12)3
=
=
=
f �� (x)
=
y
��2,1.5�
Note, f �� (x) = 0 precisely when x = ±2. We test the signs of f �� (x) on (−∞, −2), (−2, 2),
and (2, ∞) as follows:
Test open interval
Test value xi
Value of f �� (xi )
Sign of f �� on test
open interval
(−∞, −2)
x0 = −3
f �� (x1 ) ≈ 0.1
(−2, 2)
x0 = 0
f �� (x2 ) = −1/3
(2, ∞)
x0 = 3
f �� (x3 ) ≈ 0.1
Positive
Negative
Positive
�2,1.5�
�2
2
x
Figure 5.
The graph of f is concave
upward on (−∞, −2) and
(2, ∞), and concave downward on (−2, 2).
By Theorem 3.7, f is concave upward (−∞, −2) and (2, ∞), and concave downward on
(−2, 2). The graph of f is shown in Figure 5.
✷
Try This 2
Find the open intervals on which the graph of y =
downward.
1
is concave upward or concave
x2 + 3
Example 3 Determining Concavity on Open Intervals
Discuss the concavity of the graph of g(x) =
Solution
The first derivative is
g � (x)
x
, as shown in Figure 6.
x2 − 1
=
(x2 − 1) − x(2x)
(x2 − 1)2
=
−
Then
y
Quotient rule
1 + x2
.
(x2 − 1)2
3
Simplify
�1
g �� (x)
2
=
−
2
2
2
(x − 1) (2x) − (1 + x )[2(x − 1)(2x)]
Quotient rule
(x2 − 1)4
�
2x(x2 − 1) (x2 − 1) − 2(1 + x2 )
(x2 − 1)4
=
−
=
2x(x2 + 3)
.
(x2 − 1)3
�
Factor out 2x(x2 − 1)
Simplify
Then x = 0 is the only solution to g �� (x) = 0, Note, g �� (x) is undefined when x = ±1.
Next, we obtain the sign of g �� (x) on the open intervals determined by x = 0, ±1.
Test open interval
Test value xi
Value of g �� (xi )
Sign of g �� (x)
(−∞, −1)
x1 = −2
g �� (x1 ) ≈ −1.0
Negative
(−1, 0)
x2 = −0.5
g �� (x2 ) ≈ 7.7
Positive
(0, 1)
x3 = 0.5
g �� (x3 ) ≈ −7.7
Negative
(1, ∞)
x4 = 2
g �� (x4 ) ≈ 1.0
Positive
1
x
�3
Figure 6
Apply the Concavity Test to
determine the concavity of
the graph of g.
166
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Hence, by Theorem 3.7, g is concave upward on (−1, 0) and (1, ∞), and concave downward
on (−∞, −1) and (0, 1).
✷
Try This 3
Find the open intervals on which the graph of
y=
1
1−x
is concave upward or concave downward.
Points of Inflection
A point on a graph where concavity changes is called a point of inflection. In Figure 7,
we see three points of inflection P , Q, and R. Observe, the concavity to the nearby left
side is different from the concavity to the nearby right side of a point of inflection.
y
3
Q
2
1
P
1
R
2
3
x
Figure 7
Points of Inflection at P, Q, and R.
Definition 7 Point of Inflection
Let f be continuous at c. Then (c, f (c)) is a point of inflection of the graph of f if there
are open intervals (a, c) and (c, b) such that either
a) f is concave upward on (a, c) and concave downward on (c, b), or
b) f is concave downward on (a, c) and concave upward on (c, b).
For brevity, we may write inflection point of f . A point of inflection of f is not
necessarily a critical point of f . However, an inflection point is a critical point of the
derivative f � as the next theorem shows.
3.4. CONCAVITY OF THE GRAPHS OF FUNCTIONS
167
Theorem 3.8 Necessary Condition for a Point of Inflection
If (c, f (c)) is a point of inflection of f , then f �� (c) = 0 or f �� (c) does not exist.
Proof If f �� (c) does not exist, then the theorem is true. Suppose f �� (c) exists. Then we
apply the continuity of f � at c and Definition 7. That is, there exist intervals (a, c] and
[c, b) such that
a) f � is increasing on (a, c] and decreasing on [c, b), or
b) f � is decreasing on (a, c] and increasing on [c, b).
Thus, f � (c) is a relative extreme value of f � . Hence, f �� (c) = 0 by Theorem 3.2 in page
142.
✷
Example 4 Finding Points of Inflection
Find the points of inflection and discuss the concavity of the graph of
f (x) = x4 + 2x3 − 1.
Solution
y
The first two derivatives of f are given by
f � (x)
=
4x3 + 6x2
f �� (x)
=
12x2 + 12x = 12x(x + 1).
�4
Then f �� (x) = 0 only if x = 0 or x = −1. We determine the signs of f �� on
I1 = (−∞, −1) , I2 = (−1, 0) , I3 = (0, ∞) .
Test open interval
Test value xi
Value of f �� (xi )
Sign of f ��
I1 = (−∞, −1)
x1 = −2
f �� (−2) = 24
Positive
I2 = (−1, 0)
x2 = −1/2
f �� (−1/2) = −3
Negative
Figure 8
A sketch of the graph of
Applying Theorem 3.7, f is concave upward on (−∞, −1) and (0, ∞), and concave downward on (−1, 0). Hence, by Definition 7 the points of inflection of f are (0, f (−1)) = (0, −2)
and (−1, f (−1)) = (−1, −2). See the graph of f in Figure 8.
✷
Example 5 Finding Points of Inflection
Discuss the concavity and find the points of inflection of
f (x) = 4x5/3 − x8/3 + 1.
4
�4
I3 = (0, ∞)
x3 = 1
f �� (1) = 24
Positive
Try This 4
Find the points of inflection and discuss the concavity of y = 2x3 − x4 − 2.
�1
f (x) = x4 + 2x3 − 1.
x
168
y
�0,1�
Solution
CHAPTER 3. APPLICATIONS OF DERIVATIVES
The first two derivatives are
�1,4�
1 2 3
x
Figure 9
At a point of inflection
(c, f (c)), either f �� (c) = 0
or f �� (c) is undefined.
f � (x)
=
20 2/3 8 5/3
x
− x
3
3
f �� (x)
=
40 −1/3 40 2/3
x
−
x
9
9
=
40(1 − x)
√
.
93x
Note, f �� (x) = 0 if x = 1, and f �� (x) is undefined if x = 0. Next, determine the signs of
f �� (x) on the open intervals
I1 = (−∞, 0) , I2 = (0, 1) , I3 = (1, ∞) .
The results are as follows:
Test open interval
Test value xi
Value of f �� (xi )
Sign of f ��
I1 = (−∞, 0)
x1 = −1
f �� (x1 ) ≈ −8.8
Negative
I2 = (0, 1)
x2 = 0.5
f �� (x2 ) ≈ 2.8
Positive
I3 = (1, ∞)
x3 = 2
f �� (x3 ) ≈ −3.5
Negative
By Theorem 3.7, f is concave upward on (0, 1), and concave downward on (−∞, 0) and
(1, ∞). Since the direction of concavity changes at x = 0 and x = 1, the points of inflection
of f are (0, f (0)) = (0, 1) and (1, f (1)) = (1, 4). See the graph of f in Figure 9.
✷
Try This 5
Find the points of inflection of the graph of
y
Concave Upward
y = x5/3 − 5x2/3 .
Include a discussion of the concavity of the graph.
c
x
Figure 10a
If f � (c) = 0 and f �� (c) > 0,
then f (c) is a relative
minimum value of f .
9
x
The Second Derivative Test
The above named-test uses the second derivative to identify certain relative extreme values
of a function. We describe a geometric interpretation of the test. If f � (c) = 0 and f is
concave upward on an open interval I containing c as in Figure 10a, then f (c) is a relative
minima of f . Similarly, if f is concave downward on I, then f (c) is a relative maxima of
f , see Figure 10b.
y
Concave Downward
c
Observe, if f �� (c) = 0 or f �� (c) is undefined, we cannot conclude that (c, f (c)) is a
point of inflection. For example, let g(x) = x4 and h(x) = x4/3 . Since g � (x) = 4x3
√
and h� (x) = 43 3 x are increasing functions, the graphs of g and h are concave upward on
(−∞, ∞) by Definition 6. Thus, g and h do not have points of inflections even though
4
g �� (0) = 0 and h�� (0) is undefined for g �� (x) = 12x2 and h�� (x) = √
3 2.
x
Theorem 3.9 The Second Derivative Test
Figure 10b
If f � (c) = 0 and f �� (c) < 0,
then f (c) is a relative
maximum value of f .
Let f be a function and suppose f � (c) = 0.
a) If f �� (c) > 0, then f (c) is a relative minimum value of f .
b) If f �� (c) < 0, then f (c) is a relative maximum value of f .
3.4. CONCAVITY OF THE GRAPHS OF FUNCTIONS
Proof
169
Suppose f �� (c) > 0.
f �� (c) = lim
x→c
f � (x) − f � (c)
f � (x)
= lim
>0
x→c
x−c
x−c
Then there exists an open interval (a, b) containing c such that
f � (x)
> 0 whenever x is in (a, b) and x �= c.
x−c
Thus, f � (x) < 0 for all x in (a, c), and f � (x) > 0 for all x in (c, b). Note, f is continuous
at c since f � (c) exists. Then f is decreasing on (a, c] and decreasing on [c, b). Hence, f (c)
is a relative minimum of f by the First Derivative Test in page 154. The proof of part b)
is left as an exercise.
✷
Note, the Second Derivative Test is inconclusive if f �� (c) = 0. In such a case, an
alternative is to apply the First Derivative Test to identify any relative extrema.
Example 6 Applying the Second Derivative Test
Find the relative extrema of
f (x) = x3 −
Solution
3 5
x − 1.
20
y
First, we determine the derivative of f .
f � (x)
=
3x2 −
3 4
x
4
3
2
=
3x
(4 − x2 )
4
(2)
�
Then the solutions to f (x) = 0 are x = 0, ±2. The second derivative of f is
�
�
3
d
f �� (x) =
3x2 − x4
dx
4
=
f �� (x)
=
6x − 3x3
3x(2 − x2 ).
We apply the Second Derivative Test with c = 0, ±2.
c
Sign of f �� (c)
Conclusion
−2
f �� (−2) > 0
Relative minimum
0
f �� (0) = 0
Test fails
2
f �� (2) < 0
Relative maximum
Hence, a local maximum value of f is
f (2) = 8 −
3
(32) − 1 = 2.2
20
and a local minimum value is
f (−2) = −8 −
3
(−32) − 1 = −4.2
20
Note, the Second Derivative Test fails when x = 0. However, using (2) we claim that the
sign of f � (x) stays constant at x = 0. Thus, f (0) is not a relative extrema of f by the
First Derivative Test, see page 157. The graph of f is given in Figure 11
✷
Try This 6
Apply the Second Derivative Test to find the relative extrema of
s(t) = 8t2 − t4 + 5.
�2
2
�3
Figure 11
The point (2, 2.2) is a
relative maximum, and
(−2, −4.2) is a relative
minimum of f .
x
170
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Example 7 Applying the Second Derivative Test
Find the relative extrema of
g(x) = 2 sin x + cos(2x), 0 < x < 2π.
y
Solution
The derivative is
g � (x)
3
2
Π
6
Π
2
5Π
6
3Π
2
2Π
x
=
=
=
2 cos x − 2 sin(2x)
2 cos x − 4 sin x cos x
Since sin 2x = 2 sin x cos x
2 cos x(1 − 2 sin x).
Then the critical numbers x in (0, 2π) satisfy either
�3
Figure 12
The graph of
g(x) = 2 sin x + cos(2x)
1
.
2
cos x = 0 or sin x =
Thus, the critical numbers are
in 0 < x < 2π.
x=
π 3π π 5π
,
, ,
.
2 2 6 6
=
d
[2 cos x − 2 sin(2x)]
dx
=
−2 sin x − 4 cos(2x).
The second derivative is
g �� (x)
We apply the Second Derivative Test as follows:
c
Value of g �� (c)
Conclusion
π
6
�� π
g (6) =
−3
Rel max
π
2
�� π
g (2)
5π
6
�� 5π
g ( 6 )=
−3
Rel max
=2
Rel min
3π
2
�� 3π
g (2)
=6
Rel min
Hence, the relative maximum value of g(x) is
g
�π�
6
=
3
=g
2
�
5π
6
�
.
and the relative minimum values are
� �
�π�
3π
g
= 1 and g
= −3.
2
2
The graph of g is in Figure 12.
✷
Try This 7
Use the Second Derivative Test to find the relative extrema of
f (x) = cos(2x) − 2 cos x,
π
5π
<x<
.
4
4
3.4. CONCAVITY OF THE GRAPHS OF FUNCTIONS
3.4 Check-It Out
1. Find the open intervals on which the graph of f (x) = x4 + 4x3 + 12 is concave upward
or concave downward.
2. Find the points of inflection of the graph of p(x) = 3x5 − 10x4 + 10x3 + 2.
1
3. Use the Second Derivative Test to find the relative extrema of s(t) = t4 − 2t2 .
4
True or False. If false, explain or show an example that shows it is false.
1. The graph of p(x) = x3 − 6x2 + 9x + 1 is concave upward on (−∞, 2).
2. The graph of T (x) = sin x is concave downward on (0, π).
3. A point of inflection of the graph of y = x4 is (0, 0).
4. A point of inflection of the graph of y =
√
3
x is (0, 0).
5. If (c, f (c)) is a point of inflection of y = f (x), then f �� (c) = 0 or f �� (c) is undefined.
6. If g �� (c) = 0, then (c, g(c)) is a point of inflection of y = g(x).
7. If R� (c) = 0 and R�� (c) > 0, then R(c) is relative minimum value of y = R(x).
8. If y = M (x) is concave upward on (a, b) and concave downward on (b, c),
then (b, M (b)) is a point of inflection of the graph of M .
9. If y = p(x) is a polynomial such that p�� (x) < 0 on (a, b) and p�� (x) > 0 on (b, c),
then (b, p(b)) is a point of inflection of y = p(x).
10. It is possible that a point of inflection of y = f (x) is also a relative extrema of f .
Exercises for Section 3.4
In Exercises 1-4, find the open intervals where the graph is concave upward or concave downward.
Also, identify the inflection points.
1. y = x2 − 4x
2.
y = −x2 − 2x
y
y
1
1
2
3
1
x
�2 �1
�2
�1
�2
�4
Figure for No. 1
1
Figure for No. 2
x
171
172
CHAPTER 3. APPLICATIONS OF DERIVATIVES
2. p(t) = t3 − 3t2 + 3t + 1
3.
y = 2 sin t, 0 ≤ t ≤ 2π
y
y
2
2
Π
2
1
1
Figure for No. 3
t
2Π
�2
t
3
3Π
2
Π
Figure for No. 4
In Exercises 5-22, find the open intervals where the graph of the function is concave upward or downward.
Then determine the points of inflection.
5.
p(x) = x4 − 8x3 + 18x2 + 3
6.
p(x) = x5 − 5x + 2
7.
g(x) = x4/3 − 4x1/3
8.
h(x) = x4/3 + 2x1/3
9.
f (x) = x5/3 − 5x4/3
10.
g(x) = x7/3 − 14x4/3
11.
y = 1 + cos2 x, 0 < x < π
12.
y = x + sin2 x, 0 < x < π
13.
y = sin x + cos x, 0 < x < π
14.
y = cos x − sin x, 0 < x < 2π
15.
f (x) =
64
x2 + 12
16.
f (x) =
17.
P (t) =
32t
3t2 + 16
18.
R(w) =
19.
m(x) = √
20.
n(x) = √
21.
s(t) = √
22.
T (w) = √
x
x2
+1
t
1−t
50
x2 + 75
196w
21w2 + 343
x
x+1
w
1 − w2
In Exercises 23-38, find the relative extrema by using the Second Derivative Test when applicable.
23.
f (x) = x3 − 3x2 + 5
24.
f (x) = 3x2 − 2x3 + 4
25.
g(x) = x3 + x2 − 8x + 20
26.
f (x) = 2x3 + 5x2 − 4x + 15
27.
C(w) = (w2 + 64)(100 − w2 )
28.
A(w) = (w2 + 25)(169 − w2 )
29.
g(x) = 3 sin 2x, 0 ≤ x ≤ π
30.
f (x) = 4 cos 3x + 1,
31.
L(α) = tan α − 4α, − π2 < α <
32.
R(β) = 4β + 3 cot β, 0 < β < π
33.
y=
34.
Q(p) =
−10p
p2 + 25
35.
f (x) =
36.
g(x) =
x3
9 − x2
37.
K(x) = 4x1/3 − x4/3
38.
M (c) = 8c2/3 − c4/3
8t
t2 + 16
x2 + x − 2
x−2
π
2
π
6
≤x≤
5π
6
3.4. CONCAVITY OF THE GRAPHS OF FUNCTIONS
In Exercises 39-42, sketch a graph of a function with the given properties. There are several possible answers.
Then find a formula y = f (x) for such a function.
39. f (1) = 2, f � (1) = 0, and f �� (x) > 0 for −∞ < x < ∞
40. f (3) = −1, f � (3) = 0, and f �� (x) < 0 for −∞ < x < ∞
41. f (0) = 2, f � (0) = 0, f �� (x) < 0 for x < 1, and f �� (x) > 0 for x > 1
42. f (1) = 5, f (−1) = −5, f � (±1) = 0, f �� (x) > 0 for x < 0, and f �� (x) < 0 for x > 0
Applications
43. Find the points on the graph of y = 1/x that are closest to the origin.
44. Find the shortest distance between the origin and the graph of y =
1
.
x2
45. Find the shortest distance between the parabola y = x2 and the point (3, 0).
46. Find the point on the graph of y = sin x that is nearest to the point (π/2, 0).
47. A rectangle is bounded by the x and y-axes and the graph of y = 1/x3 in the first quadrant,
see figure below. Find the smallest possible perimeter of such a rectangle.
y
y
y�1�x3
y�1�x
x
Figure for No. 47
x
Figure for No. 48
48. Find the minimum length of the line segment that is tangent to the graph of y = 1/x
and whose endpoints lie on the positive x- and y-axes, see figure above.
49. The diagonals of a rectangle contain the origin and two vertices of the rectangle lie on the
branches of the graph of y = 1/x3 . Find the minimum perimeter of such a rectangle.
50. Find the points on the graph of y = 1/x3 that are nearest the origin.
173
174
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Prove, or disprove by finding a counter example
51. If f �� (x) is a twice differentiable concave upward function, prove or disprove that f �� (x) > 0 for all x.
52. If f (x) is a concave upward function, prove or disprove that y = (f (x))2 is concave upward.
53. If f � and f �� are positive functions on the interval (0, ∞), prove that y = f (x2 ) is concave upward on (0, ∞).
Is the converse true?
54. Find conditions on f and g so that y = f (g(x)) is concave upward.
3.5. LIMITS AT INFINITY
3.5
175
Limits at Infinity
• Limits at Infinity • Horizontal Asymptotes • Infinite Limits at Infinity
Limits at Infinity
In the section, we study the ‘end behavior’ of a function. Specifically, we analyze the limit
of a function f (x) as x increases or decreases without bound. To illustrate, we list several
values of a function for large |x|:
√
4 + x2
f (x) =
x
x
f (x)
−1000
−1.00001
−100
−1.0002
−10
−1.0198
10
1.0198
100
1.0002
1000
1.00001
The above values suggest that f (x) approaches 1 as x increases without bound, and
we write
lim f (x) = 1.
Limit at Positive Infinity
y
x→∞
Similarly, as x decreases without bound the table of values indicate f (x) approaches −1,
and we write
lim f (x) = −1.
Limit at Negative Infinity
x→−∞
1
�4
�1
4
In Figure 1, we see the graph of f is getting closer to the horizontal line y = 1 as x → ∞.
Also, the graph of f is getting nearer to the line y = −1 as x → −∞. For clarity, we point
out that ∞ does not represent a number.
Example 1 Finding a Limit at Infinity
Complete the table below where f (x) =
x
f (x)
−1000
Figure 1
As x → ∞, we find
f (x) approaches 1.
As x → −∞, we see
f (x) approaches −1.
2x
.
x+1
−100
−10
10
100
1000
Then estimate the limits at ∞ and −∞:
a) lim
x→∞
2x
x+1
Solution
b) lim
x→−∞
2x
x+1
Using a calculator, we find
x
f (x)
−1000
2.002
−100
2.020
−10
2.222
10
1.818
100
1.980
1000
1.998
Observe, f (x) approaches 2 as x → ∞ or x → −∞. Then we find
lim f (x) = 2 and
x→∞
lim f (x) = 2.
x→−∞
✷
Try This 1
Estimate the limits
a) lim
x→∞
2x
x2 + 1
b) lim
x→−∞
by constructing a table of values for f (x) =
2x
x2 + 1
2x
.
x2 +1
x
176
y
CHAPTER 3. APPLICATIONS OF DERIVATIVES
f �x� approaches L
x approaches �
For exactness, we will need the definition of a limit at infinity.
L�Ε
L
L�Ε
x
M
Definition 8 Limits at Infinity
Let L be a real number and let f be a function.
Figure 2a
For x > M , the graph
of y = f (x) lies
between the lines
y = L + � and y = L − �.
1. The statement lim f (x) = L means that for each ε > 0 there exists a positive number
x→∞
M > 0 such that
|f (x) − L| < ε whenever x > M.
2. The statement
lim f (x) = L means that for each ε > 0 there exists a negative
x→−∞
number N < 0 such that
|f (x) − L| < ε whenever x < N.
y
f �x� approaches L
x approaches ��
x
N
L�Ε
L
L�Ε
Figure 2b
For x < N , the graph
of y = f (x) lies
between the lines
y = L + � and y = L − �.
In words, the definition of a limit at infinity implies that f (x) is within ε units of L
for all x > M (where M depends on ε). Geometrically, the graph of y = f (x) for x > M
lies between the horizontal lines y = L + ε and y = L − ε, as shown in Figure 2a. Similar
statements can be made for a limit at minus infinity (or negative infinity), see Figure 2b.
Not all functions have limits at ±∞. The graph of y = sin x oscillates infinitely many
times between the lines y = −1 and y = 1, see Figure 2c. Then it is impossible to find
a positive number M > 0 such that the graph of y = sin x for x > M lies between the
horizontal lines Y = L ± 21 . Thus, the infinite limit lim sin x does not exist. Similarly,
lim sin x
x→∞
does not exist.
x→−∞
Example 2 Proving a Limit at Infinity
Prove the following limit by using Definition 8:
y
lim
x→∞
1
�2Π
�Π
Π
2Π
x
�1
Solution
analyze
Let f (x) =
4
,
x2
4
=0
x2
ε > 0, and L = 0. Applying the notation in Definition 8, we
�
�
�4
�
4
|f (x) − L| = �� 2 − 0�� = 2 < ε
x
x
The above inequality is equivalent to
2
√ < |x|.
ε
Figure 2c
lim sin x does not exist
Let M = √2ε . Thus, |f (x) − L| < ε whenever x > M .
Hence, by Definition 8 we have proved
x→∞
lim sin x does not exist
lim
x→−∞
x→∞
4
= 0.
x2
✷
Try This 2
Use Definition 8 to prove
lim
x→∞
1
= 0.
x
3.5. LIMITS AT INFINITY
177
Several properties of limits extend to limits at infinity. Such properties of limits are
discussed in Section 1.3. For instance, if lim f (x) and lim g(x) both exist then
x→∞
x→∞
1. lim (f (x) + g(x)) = lim f (x) + lim g(x)
x→∞
x→∞
x→∞
2. lim (f (x) − g(x)) = lim f (x) − lim g(x)
x→∞
x→∞
x→∞
3. lim (f (x)g(x)) = lim f (x) lim g(x)
x→∞
x→∞
x→∞
lim f (x)
4. lim
x→∞
f (x)
x→∞
=
g(x)
lim g(x)
lim g(x) �= 0, and
provided
x→∞
x→∞
5. lim (cf (x)) = c lim f (x) where c is any real number.
x→∞
x→∞
The above formulas remain true when x → ∞ is replaced by x → −∞. The next
theorem is useful for evaluating limits at infinity of rational functions. A proof of the
theorem is given in Appendix A.1, page 414.
Theorem 3.10 Basic Rules for Limits at Infinity
If r > 0 is a positive rational number and c is a real number, then
c
1. lim r = 0, and
x→∞ x
c
2. lim
= 0 provided xr is well-defined for all x < 0.
x→−∞ xr
Example 3 Finding Limits at Infinity
Evaluate the limits:
�
�
10
a) lim 3 − 2
x→∞
x
b)
lim
x→−∞
�
4+
5
x
�
Solution
a)
Applying the limit of a difference, we find
�
�
10
10
lim 3 − 2
= lim 3 − lim 2
x→∞
x→∞
x→∞ x
x
3−0
=
=
b)
By Theorem 3.10
3.
Likewise, applying the limit of a sum we obtain
�
�
5
5
lim
4+
=
lim 4 + lim
x→−∞
x→−∞
x→−∞ x
x
=
4+0
=
4.
Theorem 3.10
✷
Try This 3
Evaluate the limits.
10
a) lim √
x→∞
x
b) lim
x→−∞
120
x2
1
c) lim √
x→−∞
x
178
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Horizontal Asymptotes
When they exist, the limits at infinity are useful in describing the ‘end behavior’ of the
graph of a function.
Definition 9 Horizontal Asymptote
A line y = L is a horizontal asymptote of the graph of a function y = f (x) if either
lim f (x) = L
or
x→∞
lim f (x) = L.
x→−∞
Since the limits in Definition 9 involve x → ∞ or x → −∞, a function may have at
most two distinct horizontal asymptotes.
Example 4 Finding the Horizontal Asymptotes
y
Evaluate the limits
a) lim
3x�5
f �x��
2x�9
3
2
x→∞
3x − 5
2x + 9
b) lim
x→∞
3x − 4
2x2 + 1
Then find the horizontal asymptotes of the graph of f (x) =
x
Solution
3x − 5
.
2x + 9
Evaluate the limit as follows:
lim
x→∞
Figure 3
3x − 5
2x + 9
=
3x − 5
·
2x + 9
lim
3−
2+
x→∞
=
3
The line y = is a
2
horizontal asymptote.
lim
x→∞
1
x
1
x
5
x
9
x
3−0
3
=
2+0
2
=
By Theorem 3.10
Similarly, as x → −∞ we obtain
lim
x→−∞
Thus, the line y =
Finally, we have
3
2
3−
3x − 5
= lim
x→−∞ 2 +
2x + 9
5
x
9
x
=
3
.
2
is the horizontal asymptote of the graph of f as shown in Figure 3.
lim
x→∞
3
− x42
3x − 4
= lim x
= 0.
2
x→∞ 2 + 12
2x + 1
x
✷
Try This 4
Evaluate the limits
a) lim
x→∞
4x
3 − 8x
b)
lim
x→−∞
4x
5 − 8x2
Then find the horizontal asymptotes of y =
4x
.
3 − 8x
3.5. LIMITS AT INFINITY
179
For the rational functions, the limit at infinity is determined by the leading coefficients
in the numerator and denominator, as discussed below.
Summary for Finding Limits at Infinity of Rational Functions
Let p(x) and q(x) be polynomials of degree n and m, respectively. Suppose a and b are
the leading coefficients of p(x) and q(x), respectively.
1. If n = m, then
lim
p(x)
p(x)
a
= = lim
.
x→−∞ q(x)
q(x)
b
lim
p(x)
p(x)
= 0 = lim
.
x→−∞ q(x)
q(x)
x→∞
2. If n < m, then
x→∞
3. If n > m, the limit does not exist as x → ∞ or x → −∞.
In any case, a rational function has at most one horizontal asymptote.
The proof of the above summary is left as an exercise at the end of this section. For
example, the polynomials in the numerator and denominator in the rational function
R(x) =
3x2 + 10x
5x2 + 12
are of both of degree two. Thus, as x → ∞ or x → −∞, we obtain that R(x) → 3/5,
which is the quotient of the leading coefficients. In Figure 4, the line y = 35
y
r�x��
3 x2 � 10 x
5 x2 � 12
3
5
x
Figure 4:
R(x) =
3x2 + 10x
5x2 + 12
is the horizontal asymptote of the graph of R. However, for a non-rational function its
graph may possibly have two distinct horizontal asymptotes.
Example 5 A Graph With Two Horizontal Asymptotes
Evaluate the limits:
a)
lim √
x→∞
4x
x2 + 9
b)
lim √
x→−∞
4x
x2 + 9
180
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Solution
a) We rewrite as follows:
y
y�
4
√
4x
x2 � 9
1
x
1
x
4x
·
x2 + 9
= √
4
= �
x2 +9
√
x2
Applying Theorem 3.10, we obtain
x
lim √
x→∞
�4
4x
4
= lim �
x→∞
x2 + 9
1+
4
1+
= √
9
x2
9
x2
4
= 4.
1+0
√
b) Since x = − x2 whenever x < 0, we rewrite
√
Figure 5
The lines y = ±4 are
horizontal asymptotes.
4x
·
x2 + 9
1
x
1
x
= √
4
x2 +9
√
− x2
4
�
1+
=
4
√
= −4.
− 1+0
=
−
9
x2
Thus, by Theorem 3.10 we obtain
lim √
x→−∞
4x
4
�
= lim
x→−∞
x2 + 9
− 1+
9
x2
In particular, the lines y = ±4 are the horizontal asymptotes of the graph of y = √ 4x
2
x +9
as seen in Figure 5.
✷
Try This 5
Evaluate each limit.
a) lim √
x→∞
x
x2 + 1
b)
lim √
x→−∞
x
9x2 + 4
In evaluating limits at infinity involving trigonometric functions, the Squeeze Theorem
as explained in Exercise 54 will be helpful to know.
Example 6 Limits at Infinity with Trigonometric Functions
Evaluate each limit:
a)
lim
x→∞
Solution
x + cos x
x
b)
lim
x→−∞
x + cos x
x
For all x, we recall
Adding x, we obtain
If we divide by x > 0, then
−1 ≤ cos x ≤ 1.
x − 1 ≤ x + cos x ≤ x + 1.
x−1
x + cos x
x+1
≤
≤
.
x
x
x
However, if x < 0 then
x−1
x + cos x
x+1
≥
≥
.
x
x
x
(3)
3.5. LIMITS AT INFINITY
181
y
Using the leading coefficients, we find
lim
x→∞
x+1
x−1
= 1 = lim
x→∞
x
x
y�
1
and
x−1
x+1
lim
= 1 = lim
.
x→−∞
x→−∞
x
x
Hence, by the Squeeze Theorem we obtain
x
�20
x + cos x
x + cos x
lim
= 1 = lim
.
x→∞
x→−∞
x
x
As seen in Figure 6, the line y = 1 is the horizontal asymptote of y =
x+cos x
.
x
✷
Figure 6
The graph of
y=
Try This 6
Evaluate the following limits.
a)
x � cos �x�
x
x + cos x
x
and the horizontal
asymptote y = 1.
2x + sin x
lim √
x2 + 1
b)
x→∞
lim
x→−∞
2x + sin x
√
x2 + 1
Example 7 The Limit of a Population of Bacteria
A nutrient is needed to help grow a population of bacteria. The number of bacteria is
modeled by the function
30000t2
N (t) = 5000 +
100 + t2
where t ≥ 0 is in hours. Determine the number of bacteria when t = 24, t = 48 and t = 72
hr. Then find the limit of N (t) as t → ∞.
Solution
N�t�
N�t��5000�
30000 t2
100 � t2
Using a calculator, the number of bacteria are
N (24)
=
30000(24)2
5000 +
= 30, 562
100 + (24)2
N (48)
=
5000 +
30000(48)2
= 33, 752
100 + (48)2
N (72)
=
5000 +
30000(72)2
= 34, 432.
100 + (72)2
We evaluate the limit as follows:
�
30000t2
lim 5000 +
·
t→∞
100 + t2
1
t2
1
t2
�
=
=
lim
t→∞
�
30000
5000 + 100
+1
t2
5000 +
24
48
72
Figure 7
As t increases without
bound, N (t) approaches
35,000.
�
30000
= 35, 000
0+1
Thus, N (t) → 35, 000 bacteria as t → ∞. In Figure 7, we see a graph of the population
versus time.
✷
t
182
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Try This 7
�
The population of a bacteria is approximated by N (t) = 10, 000 1 +
�
4t2
.
100 + t2
Find the number of bacteria after t = 24 hours. Also, find the limit of N (t) as t → ∞.
Infinite Limits at Infinity
If f (x) increases without bound as x → ∞, we say f (x) has an infinite limit at infinity
and write lim f (x) = ∞. Similarly, if f (x) decreases without bound as x → ∞, then
x→∞
f (x) has a negative infinite limit at infinity and we write lim f (x) = −∞. The definition
below describes the notion of an infinite limit at infinity.
x→∞
Definition 10 Infinite Limits at Infinity
Let y = f (x) be a function.
1. We write
lim f (x) = ∞
x→∞
if for each positive number M > 0 there exists a positive number N > 0 such that
f (x) > M whenever x > N.
2. The statement
lim f (x) = −∞
x→∞
means that for each negative number M < 0 there exists a positive number N > 0
such that
f (x) < M whenever x > N.
We have similar statements for lim f (x) = ∞ and
x→−∞
lim f (x) = −∞.
x→−∞
In evaluating infinite limits, the following observations are useful.
If L > 0 is a positive number, lim f (x) = L, and lim g(x) = ∞, then
x→∞
x→∞
lim f (x)g(x) = ∞
x→∞
(4)
and
g(x)
= ∞.
f (x)
The proofs of these observations are left as exercises at the end of the section.
lim
x→∞
(5)
Example 8 Evaluating Infinite Limits at Infinity
Evaluate the limits at infinity.
a)
lim x3
x→∞
Solution
�
�
b) lim 4x3 − 10x2 − 5
x→∞
a) Since x3 > x when x > 1, it follows that x3 increases without bound as x → ∞. Then
lim x3 = ∞.
x→∞
3.5. LIMITS AT INFINITY
183
b) Factor the monomial with the highest exponent:
�
�
5
5
4x3 − 10x2 − 5 = 4x3 1 −
− 3 .
2x
4x
y
3�106
By part a), 4x3 → ∞ as x → ∞. Moreover, we find
�
�
5
5
lim 1 −
− 3 = 1 − 0 − 0 = 1.
x→∞
2x
4x
2�106
106
Applying observation (4) in the previous page, we obtain
�
�
�
�
5
5
lim 4x3 − 10x2 − 5
= lim 4x3 lim 1 −
− 3
x→∞
x→∞
x→∞
2x
4x
= ∞·1
=
25
25
56, 245
50
474, 995
x
50
Figure 8
As x increases without
bound, f (x) increases
without bound.
∞
Note, the few values of f (x) = 4x3 − 10x2 − 5 shown below
x
f (x)
f �x�� 4x3�10x2�5
100
3, 899, 995
support the fact that f (x) increases without bound as x → ∞. The ‘right-end’
behavior of the graph of f is seen in Figure 8.
✷
Try This 8
Evaluate the limits at infinity.
a)
lim x3
x→−∞
b) lim
x→−∞
� 4
�
x + x3
y
Example 9 Limits at Infinity of a Rational Function
15
Find the limits at infinity.
a)
lim
x→∞
x2 + 1
x−2
b) lim
x→−∞
x5 + 1
2x3 + 1
�4
7
�10
Solution
x2 + 1
a) The degree of the denominator in
is smaller than the degree of the numerator.
x−2
In such a case, multiply and divide by 1/xd where d is the degree of the denominator.
x + 1/x
x2 + 1 1/x
·
=
.
x − 2 1/x
1 − 2/x
Applying Theorem 3.10 and observation (5), we find
lim
x→∞
x2 + 1
x−2
=
=
=
2
In Figure 9a, we see the graph of y =
lim
x→∞
x + 1/x
1 − 2/x
∞+0
∞
=
1−0
1
∞
x +1
.
x−2
Figure 9a
The graph of
y=
x2 + 1
.
x−2
x
184
CHAPTER 3. APPLICATIONS OF DERIVATIVES
y
x5 + 1
is smaller than the degree of the numerator.
2x3 + 1
We follows the strategy applied in part a). Then we multiply and divide by 1/x3 .
b) The degree of the denominator in
3
�3
3
x2 + 1/x3
x5 + 1 1/x3
·
=
3
3
2x + 1 1/x
2 + 1/x3
x
Then
�3
lim
x→−∞
x5 + 1
2x3 + 1
Figure 9b
The graph of
lim
x→−∞
x2 + 1/x3
2 + 1/x3
=
∞+0
∞
=
2+0
2
=
∞
x5 + 1
The graph of y =
is shown in Figure 9b.
2x3 + 1
5
y=
=
x +1
.
2x3 + 1
Try This 9
Evaluate the limits at infinity.
a)
x4
+1
lim
b) lim
x→∞ x2
x3
+1
x→−∞ x2
3.5 Check-It Out
1. Evaluate the limits.
a)
c)
lim
x→∞
lim
100x
x2 + 1
b)
x2
x+1
d)
x→−∞
3x
4 − 6x
√
4x2 + 1
lim
x→−∞
x
lim
x→∞
1 − x2
.
x2 − 4
1
3. Find a positive number N > 0 such that 2 < 10−800 whenever x > N .
x
2. Find the horizontal asymptote of f (x) =
True or False. If false, revise the statement to make it true.
f (x)
1. If lim f (x) = lim g(x) = ∞, then lim
= 1.
x→∞
x→∞
x→∞ g(x)
2. If lim f (x) = lim g(x) = ∞, then lim f (x) + lim g(x) = ∞.
x→∞
x→∞
x→∞
x→∞
3. If lim f (x) = 0.1 and lim g(x) = ∞, then lim (f (x)g(x)) = ∞.
x→∞
x→∞
x→∞
sin x
4. lim
does not exist
x→∞
x
5.
7.
1
=∞
x
6.
lim (2 − x) = ∞
8.
lim x2 sin
x→∞
x→−∞
lim
x→∞
lim
x − cos x
=1
x
3x3 + 1
=∞
+x+3
x→∞ x4
9. The graph of a function may have two horizontal asymptotes.
x4 + 3x
10. The graph of f (x) =
has no horizontal asymptote.
2x4 + 1
✷
3.5. LIMITS AT INFINITY
185
Exercises for Section 3.5
In Exercises 1-8, use the given graph to evaluate the limit.
a)
lim f (x)
b)
x→∞
1. f (x) =
2x + 1
1−x
2.
lim f (x)
x→−∞
f (x) =
y
3x + 1
x+1
3.
f (x) =
y
2x
+1
4.
x2
f (x) =
y
10x2
x4 + 1
y
2
3
2
�4
x
4
�2
5
�4
x
4
�3
x
3
x
�5
For No. 1
5. f (x) =
√
For No. 2
4x2 + 1
x
6. f (x) =
y
For No. 3
√
x4 + 1
x2
For No. 4
7. f (x) =
y
sin x
x
8. f (x) =
y
y
2
2
2
�4
4
x
2
1
�4
x
4
�7
�2
10
x
7
�1
For No. 5
For No. 6
For No. 7
For No. 8
In Exercises 9-34, evaluate the limit. If applicable, indicate if the limit is ∞ or −∞.
9.
12.
15.
18.
21.
24.
27.
30.
�
10.
x2 − 2x + 7
x2 − 4x + 3
13.
lim
x→−∞
lim
x→∞
�
2+
3
x2
lim
4 − 5x3
x2 + 1
16.
lim
x4
x2 + 1
19.
x→−∞
x→−∞
lim √
x→−∞
lim
√
x→∞
lim
x→−∞
x2
22.
16x2 + 5x + 2
3x + 11
25.
1
2x + sin x
28.
1
x
31.
lim x cos
x→∞
4x
+ 3x + 1
lim
x→∞
�
1−
2
x
�
3x3 − 1
+ x2 + 3
lim
x→−∞ 2x3
x−2
x2 + 4
√
4x2 − 3x + 1
lim
x→∞
x
lim
x→∞
5 − 2x
lim �
(x − 4)2
x→−∞
11.
14.
17.
20.
23.
lim
cos 2πx
x
26.
lim
3x
x + cos 4x
29.
1
x
32.
x→∞
x→∞
lim x sin
x→∞
2 + cos x
x
x2 − 6
−x−2
lim
x→∞ x2
lim
x→−∞
lim
x→∞
x3 + x2 + 3x + 3
2x3 + 2x2 + 5x + 5
x2 − 1
x3 + 2x2
x
9x2 + 5
√
3
8x3 + x2 − 5
lim
x→∞
3x − 1
lim √
x→∞
lim
x→−∞
lim
x→∞
sin 3x
x
sin(πx) − 1
x
lim x sin
x→∞
3
x
x
186
33.
lim
x→∞
cos x
x
CHAPTER 3. APPLICATIONS OF DERIVATIVES
34.
lim
x→∞
sin x
x−π
Encounters with the indeterminate form ∞ − ∞.
In Exercises 35-40, evaluate the infinite limits.
35.
38.
�
x2 + x − x
x→∞
√
�√
�
lim
x+9− x+4
lim
��
36.
39.
x→∞
��
�
x2 + 6x − x
x→∞
��
�
3
lim
x3 + 8x2 − x
lim
37.
40.
x→∞
lim
x→∞
lim
x→−∞
In Exercises 41-44, determine the horizontal asymptotes of the function.
41.
f (x) =
8 − 27x3
9x3 + 2
42.
h(t) =
√
x2 + 1
x
43.
s(t) =
�
4t
t2 + 3
44.
x−
��
3
�
x2 + 4x
�
x3 + 2x2 − x
R(x) = 2 +
�
sin x
x
In Exercises 45-48, sketch the graph of a function that satisfies the given statements.
45. f (0) = 0, lim f (x) = 3, lim f (x) = 3
x→∞
x→−∞
46. f (0) = 1, lim f (x) = ∞, lim f (x) = 0
x→∞
x→−∞
47. f (0) = 0, lim f (x) = π/2, lim f (x) = −π/2
x→∞
x→−∞
48. f (0) = 2, f (1) = 4, f (−1) = 0, lim f (x) = 2, lim f (x) = 2
x→∞
x→−∞
Applications
t2
where
4t2 + 1
t is the number of weeks past March 21, the first day of spring. Determine the limit of P (t) as t → ∞. What
is a practical interpretation of the limit?
√
x
50. The probability that a certain injury will heal in 12 hours is given by P (x) = √
where x is the number
2 x+1
milligrams of an ointment that should be administered right after the injury. Evaluate lim P (x).
49. During the spring season the proportion of adults with sinus problems due to pollen is P (t) =
x→∞
51. A company has to spend $40,000 to set-up a manufacturing equipment. After the set-up, it will cost the company
$5 to manufacture each item.
a) Find the total cost C(x) of manufacturing x items with the set-up cost included.
b) Find the limit at infinity of the average cost per item, i.e., the limit of C(x)
as x → ∞.
x
52. In a given day, the amount of nitrogen dioxide in the air is modeled by
A(t) =
50t2 + 15(242 )
2t2 + 242
where t is the number of hours past midnight, and A(t) is measured using the pollutant standard index (PSI ).
a) Find the amount of nitrogen dioxide in the air at 11 AM.
b) Find the limit of A(t) as t → ∞.
Theory and Writing Proofs
53. Let p(x) and q(x) be nonzero polynomials of degree n and m, respectively. Suppose a and b are the leading
coefficients of p(x) and q(x), respectively.
p(x)
p(x)
p(x)
p(x)
a
a) If n = m, prove lim
= = lim
.
b) If n < m, show lim
= 0 = lim
.
x→∞ q(x)
x→−∞ q(x)
x→∞ q(x)
x→−∞ q(x)
b
54. Suppose h(x) ≤ f (x) ≤ g(x) for all x in (a, ∞), and lim h(x) = lim g(x) = L. Prove lim f (x) = L.
x→∞
x→∞
x→∞
55. Let L be a nonzero real number, lim f (x) = L, and lim g(x) = ∞. Prove lim f (x)g(x) = sign(L) · ∞ where
x→∞
sign(L) denotes the sign of L.
x→∞
x→∞
3.6. SUMMARY OF CURVE SKETCHING
3.6
187
Summary of Curve Sketching
Curve Sketching Techniques
We use the following features to sketch a detailed graph of a function. For a review of the
features, we list the pertinent sections in parentheses.
y
• Domain and range
10
• Symmetry
f �x��
• x-intercepts and y-intercepts
• Continuity (Section 1.4)
• Vertical and horizontal asymptotes (Sections 1.5 and 3.5)
x3
16 � x2
�10
10
x
10
x
• Differentiability (Section 2.1)
• Relative extrema (Section 3.1)
�10
• Concavity (Section 3.4)
Figure 1a
• Points of inflection (Section 3.4)
• Limits at infinity (Section 3.5)
A graphing utility is helpful in obtaining different parts or sections of the graph of a
function. For example, the viewing windows in Figure 1a and 1b show certain parts of the
graph of
x3
f (x) =
.
16 − x2
However, the second viewing window shows a holistic view of the graph.
The guidelines listed below should help the student determine important features of
the graph. Consequently help her or him to choose a good viewing window. Not every
feature is relevant to every function. For instance, a graph may have no asymptotes and
the range may not be determined before the relative extrema are identified.
Guidelines For Sketching the Graph of a Function
1. Determine the domain, and if possible the range.
2. Find the intercepts, symmetry, and horizontal and vertical asymptotes.
y
25
f �x��
x3
16 � x2
�10
�25
Figure 1b
Viewing windows
for the graph of
3. Find the x-values for which f � (x) and f �� (x) are either zero or undefined.
f (x) =
4. Determine the relative extrema and points of inflection from step 3.
y
Example 1 Sketching the Graph of a Polynomial Function
Sketch the graph of p(x) = x4 − 8x3 + 18x2 + 3.
Solution
�3,30�
30
The first three features of the graph follow directly.
Domain:
x-intercept:
Infinite limits:
(−∞, ∞)
(0, 3)
lim p(x) = ∞, lim p(x) = ∞
x→∞
x→−∞
x3
.
16 − x2
�1,14�
�0,3�
�3
1
3
x
Then we find
First derivative: p� (x) = 4x3 − 24x2 + 36x = 4x(x − 3)2
and by the product rule we obtain
Second derivative: p�� (x)
=
=
=
4 · (x − 3)2 + 4x · 2(x − 3)
4(x − 3)((x − 3) + 2x)
12(x − 3)(x − 1)
Figure 2
The graph of
p(x) = x4 − 8x3 + 18x2 + 3.
188
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Then the critical numbers and the zeros of p�� (x) determine the following test open intervals:
Critical numbers:
x = 0, 3
Test open intervals:
(−∞, 0), (0, 1), (1, 3), (3, ∞)
We list the signs of p� (x) and p�� (x) on the test open intervals, determine whether the
graph of f is increasing or decreasing, and discuss concavity. Also, we list all relative
extrema and points of inflection.
(−∞, 0)
(0, 1)
(1, 3)
(3, ∞)
x=0
x=1
x=3
p� (x)
−
+
+
+
0
+
0
p�� (x)
+
+
−
+
+
0
0
Property of the Graph
Decreasing, concave upward
Increasing, concave upward
Increasing, concave downward
Increasing, concave upward
Relative Minimum
Point of Inflection
Point of inflection
Applying the Second Derivative Test, the relative minimum value of p is p(0) = 3. Since
the sign of p�� (x) changes at x = 1 and x = 3, then (1, 14) and (3, 30) are inflection points.
In Figure 2, the graph implies that the range is [3, ∞).
✷
Try This 1
Sketch the graph of F (w) = w5 − 5w4 . Apply the sketching guidelines in page 187.
Example 2 The Graph of a Radical Function
Sketch the graph of f (x) = √
y
Solution
x
.
x−1
The first few features of the graph are
Domain:
4
2
�4,4� 3 �
�2,2�
2
Limit at infinity:
Vertical asymptote:
[1, ∞)
lim f (x) = ∞
x→∞
x=1
Applying the quotient rule, we find
4
8
x
�
First derivative: f (x)
x−1−x·
2
√1
x−1
x−1
=
x−2
2(x − 1)3/2
Similarly, we obtain
Figure 3
The graph of
f (x) = √
=
√
Second derivative: f �� (x)
x
.
x−1
=
√
2(x − 1)3/2 − (x − 2) · 3 x − 1
4(x − 1)3
=
2(x − 1) − 3(x − 2)
4(x − 1)5/2
=
4−x
4(x − 1)5/2
Then the critical numbers and the test open intervals are as follows:
Critical number:
Test open interval:
x=2
(1, 2), (2, 4), (4, ∞)
Note f �� (4) = 0
3.6. SUMMARY OF CURVE SKETCHING
189
We summarize the signs of f � and f �� on the test open intervals. From which, we find
where f is increasing, decreasing, and concave upward or downward. Also, we identify the
relative extrema by the Second Derivative Test and find the points of inflection.
(1, 2)
(2, 4)
(4, ∞)
x=2
x=4
f � (x)
−
+
+
0
+
f �� (x)
+
+
−
+
0
Property of the Graph
Decreasing, concave upward
Increasing, concave upward
Increasing, concave downward
Relative Minimum
Point of inflection
The relative minimum value is f (2) = 2. Since the sign of f �� (x) changes at x = 4, the
point (4, √43 ) is an inflection point. From the graph in Figure 3, we conclude the range of
f is [2, ∞).
✷
Try This 2
x+1
Sketch the graph of R(x) = √ . Apply the guidelines in page 187.
x
Example 3 The Graph of a Rational Function
Sketch the graph of f (x) =
Solution
x2 − 1
.
9 − x2
The following features can be computed directly:
Domain:
Symmetry:
Intercepts:
Vertical asymptotes:
Limits at infinity:
Horizontal asymptote:
{x : x �= ±3}
With respect
to� the y-axis
�
(±1, 0) , 0, − 19
x = ±3
lim f (x) = −1, lim f (x) = −1
x→∞
x→−∞
y = −1
Applying the quotient rule, we obtain
1st derivative: f � (x)
=
(9 − x2 )(2x) − (x2 − 1)(−2x)
16x
=
(9 − x2 )2
(9 − x2 )2
Differentiating again, we find
2nd derivative: f �� (x)
=
(9 − x2 )2 (16) − (16x)(−4x(9 − x2 ))
(9 − x2 )4
=
16(9 − x2 )((9 − x2 ) + 4x2 )
(9 − x2 )4
=
48(x2 + 3)
(9 − x2 )3
The following test open intervals are determined from the critical numbers and the domain.
Critical number:
Test open intervals:
x=0
(−∞, −3), (−3, 0), (0, 3), (3, ∞)
190
y
�3
�1
�1
CHAPTER 3. APPLICATIONS OF DERIVATIVES
We determine the signs of f and f �� , and where f is increasing or decreasing. Also, we
discuss concavity and identify the relative extrema by using the Second Derivative Test.
1
3
(−∞, −3)
(−3, 0)
(0, 3)
(3, ∞)
x=0
x
f � (x)
−
−
+
+
0
f �� (x)
−
+
+
−
+
Property of the Graph
Decreasing, concave downward
Decreasing, concave upward
Increasing, concave upward
Increasing, concave downward
Relative Minimum
�
�
The relative minimum value of f is f (0) = − 19 . Finally, the range is (−∞, −1) ∪ − 19 , ∞
as seen from the graph of f in Figure 4.
✷
Figure 4
Important features of
a graph are determined
by using calculus.
Try This 3
Sketch the graph of g(t) =
t2 − 4
. Use the guidelines in page 187.
t2 − 1
Example 4 Sketching the Graph of a Radical Function
Sketch the graph of f (x) = x2/3 (x + 5). Use the guidelines in page 187.
y
Solution
50
��2, 3 4 �
3
�5
�10
�1,6�
�0,0�
Figure 5
The graph of
f (x) = x2/3 (x + 5)
The following features of the graph follow directly.
Domain:
Intercepts:
Limits at infinity:
x
(−∞, ∞)
(0, 0), (−5, 0)
lim f (x) = ∞, lim f (x) = −∞
x→∞
x→−∞
Applying the product rule, we obtain
5(x + 2)
2
√
First derivative: f � (x) = x2/3 + √
(x + 5) =
.
33x
33x
Differentiating again, by the quotient rule we find
√
15 3 x − 5(x + 2) ·
√
Second derivative: f �� (x) =
3
9 x2
1
√
3 2
x
=
10(x − 1)
.
9x4/3
We list the test open intervals determined by the critical numbers and the the zero of
f �� (x), i.e., x = 1.
Critical number:
Test open intervals:
x = −2, 0
(−∞, −2), (−2, 0), (0, 1), (1, ∞)
We tabulate the signs of f � and f �� on the test open intervals. We determine where f
is increasing or decreasing, and discuss concavity. Also, we identify the relative extrema
and the points of inflection.
(−∞, −2)
(−2, 0)
(0, 1)
(1, ∞)
x = −2
x=0
x=1
f � (x)
+
−
+
+
0
undefined
+
f �� (x)
−
−
−
+
−
undefined
0
Property of the Graph
Increasing, concave downward
Decreasing, concave downward
Increasing, concave downward
Increasing, concave upward
Relative Maximum
Relative Minimum
Point of inflection
3.6. SUMMARY OF CURVE SKETCHING
191
We have a local maximum
√value at x = −2 by the Second Derivative Test. A relative
maximum value is f (−2) = 3 3 4. Note, the sign of f � (x) changes from negative to positive
at x = 0. Thus, by the First Derivative Test a relative minimum value is f (0) = 0.
Since the sign of f �� (x) changes at x = 1 the point (1, 6) is an inflection point. In
Figure 5, we see the graph of f and the range is (−∞, ∞).
✷
Try This 4
√
Sketch the graph of H(t) = 6 3 t − 3t2/3 . Apply the sketching guidelines in page 187.
Example 5 The Graph of a Trigonometric Functions
Sketch the graph of T (x) =
sin x
.
2 + cos x
y
Solution First, we sketch the graph for x in [0, 2π]. Since −1 ≤ cos x ≤ 1 and sin(kπ) = 0
where k is an integer, we obtain
Domain:
x-intercepts:
�
1
(−∞, ∞)
(0, 0), (π, 0), (2π, 0)
2
2Π
3
,
�
3 3
x
�Π
2Π
2
Applying the quotient rule and the identity cos x + sin x = 1, we find
First derivative: T � (x)
=
(2 + cos x) cos x − (sin x)(− sin x)
(2 + cos x)2
=
2 cos x + 1
(2 + cos x)2
�
Figure 6a
y
Second derivative:
T �� (x)
4Π
3
,�
�
3
3
=
(2 + cos x)2 (−2 sin x) − (2 cos x + 1)(−2 sin x(2 + cos x))
(2 + cos x)4
=
2 sin x(2 + cos x)(−(2 + cos x) + (2 cos x + 1))
(2 + cos x)4
=
2 sin x(cos x − 1)
(2 + cos x)3
��
4Π
3
,
�
3 3
1
�
2Π
3
,
�
3 3
x
�2Π
2Π
��
2Π
3
,�
�
3
3
�
4Π
3
,�
�
3
3
Note, x = 0, π, 2π satisfy T �� (x) = 0. Together with the critical numbers, we list the test
open intervals:
Critical numbers:
Test open intervals:
2π 4π
x=
,
3 3
�
� �
� �
� �
�
2π
2π
4π
4π
0,
,
, π , π,
,
, 2π
3
3
3
3
We tabulate the signs of T � (x) and T �� (x) on the test open intervals. Then determine
where T is increasing or decreasing, and discuss concavity.
(0, 2π/3)
(2π/3, π)
(π, 4π/3)
(4π/3, 2π)
x = 2π/3
x=π
x = 4π/3
T � (x)
+
−
−
+
0
−
0
T �� (x)
−
−
+
+
−
0
+
Property of the Graph
Increasing, concave downward
Decreasing, concave downward
Decreasing, concave upward
Increasing, concave upward
Relative Maximum
Point of Inflection
Relative Minimum
Figure 6b
Apply the periodicity
to extend the graph of
T (x) =
sin x
.
2 + cos x
192
CHAPTER 3. APPLICATIONS OF DERIVATIVES
√
By the Second Derivative Test, a local maximum value is T ( 2π
) = 33 , and a relative
3
√
��
4π
3
minimum is T ( 3 ) = − 3 . Since the sign of T (x) changes at x = π, (π, 0) is a point of
inflection. The graph of T in [0, 2π] is given in Figure 6a.
Since T (x + 2π) = T (x), we extend √
the √
graph to [−2π, 2π], see Figure 6b. Using the
extrema, we obtain that the range is [− 33 , 33 ].
✷
Try This 5
Sketch the graph of N (x) =
cos x
. Apply the guidelines in page 187.
2 + sin x
Example 6 Sketching the Graph of a Rational Function
Sketch the graph of f (x) =
�
�
3 3
,
2
3
16
y
�
�
1
2
Solution
1
2
x
�
3
2
,�
3
3
16
�
x3
.
1 − 4x2
The first few features of the graph can be obtained directly.
Domain:
Vertical asymptotes:
Symmetry:
Intercept:
Limits at infinity:
{x : x �= ± 12 }
x = ± 12
with respect to the origin
(0, 0)
lim f (x) = −∞, lim f (x) = ∞
x→∞
x→−∞
Applying the quotient rule, we find
=
(1 − 4x2 )(3x2 ) − x3 (−8x)
(1 − 4x2 )2
=
x2 (3(1 − 4x2 ) + 8x2 )
(1 − 4x2 )2
f � (x)
=
x2 (3 − 4x2 )
3x2 − 4x4
=
(1 − 4x2 )2
(1 − 4x2 )2
Critical numbers: x
=
0, ±
1st derivative: f � (x)
Figure 7
The graph of
x3
f (x) =
.
1 − 4x2
√
3
2
Differentiating again, we obtain
2nd derivative: f �� (x)
f �� (x)
=
(1 − 4x2 )2 (6x − 16x3 ) − (3x2 − 4x4 )(−16x(1 − 4x2 ))
(1 − 4x2 )4
=
(1 − 4x2 )(6x − 16x3 ) + 16x(3x2 − 4x4 )
(1 − 4x2 )3
=
6x − 16x3 − 24x3 + 64x5 + 48x3 − 64x5
(1 − 4x2 )3
=
2x(4x2 + 3)
8x3 + 6x
=
2
3
(1 − 4x )
(1 − 4x2 )3
We consider the test open intervals determined by the critical numbers, the zeros of f �� (x),
and the domain:
�
� �
√ � � √
�
Test open intervals:
−∞, − 23 , − 23 , − 12 , − 12 , 0 ,
� 1� �1
0, 2 , 2 ,
�
√ � �√
3
3
,
,
∞
2
2
3.6. SUMMARY OF CURVE SKETCHING
193
Below, we list the signs of f � and f �� on the test open intervals. Then we identify any
relative extrema by the Second Derivative Test. We describe concavity and the points of
inflection, too.
f �� (x)
Property of the Graph
+
Decreasing, concave upward
+
Increasing, concave upward
−
Increasing, concave downward
+
Increasing, concave upward
−
Increasing, concave downward
−
Decreasing, concave downward
−
Relative maximum
x = 0√
0
Point of inflection
x = − 23
+
Relative minimum
�√ �
√
The relative maximum value of f is f 23 = − 3163 and the relative minimum value is
� √ �
√
f − 23 = 3163 . Note, the sign of f �� changes at x = 0. Thus, (0, 0) is an inflection point.
√
(−∞
− 23 )
√
3
(− 2 , − 12 )
(− 12 , 0)
(0, √12 )
(√12 , 23 )
( 23 , ∞)
√
x = 23
f � (x)
−
+
+
+
+
−
0
0
0
Finally, the range of f is (−∞, ∞) as seen in Figure 7.
Try This 6
Sketch the graph of h(t) =
✷
t2
using the guidelines in page 187.
t−3
y
The graph in Example 6 has a slant asymptote or oblique asymptote. Precisely,
a line y = mx + b where m �= 0 is a slant asymptote of the graph of a function f if either
1
1
y�� x
4
lim [f (x) − (mx + b)] = 0
x→∞
x
or
lim [f (x) − (mx + b)] = 0.
x→−∞
If R(x) = p(x)/q(x) is a rational function and the degree of p(x) is one more than the
degree of q(x), then the graph of y = R(x) has a slant asymptote. To find the slant
asymptote, use long division to express R(x) as a sum of linear function and another
rational function. For example, using long division we find
f (x) =
Then we obtain
lim
x→∞
�
x/4
1
x3
=− x+
.
1 − 4x2
4
1 − 4x2
�
1
f (x) − − x
4
��
=
lim
x→∞
x/4
= 0
1 − 4x2
Thus, y = − x4 is a slant asymptote of the graph of f , see Figure 8.
3.6 Check-It Out
x
.
x2 − 1
x2
2. Find the domain, range, and open intervals where the graph of f (x) =
x+1
is increasing, decreasing, concave upward, or concave downward.
1. Find the domain, asymptotes, and the symmetry of the graph of R(x) =
3. Find the critical numbers and open intervals in [0, 2π] where g(t) = sin t + cos t is
increasing or decreasing.
�1
Figure 8
1
The line y = − x
4
is a slant asymptote
of the graph of
x3
f (x) =
.
1 − 4x2
194
CHAPTER 3. APPLICATIONS OF DERIVATIVES
True or False. If false, explain or show an example that shows it is false.
In Exercises 1-6, use the graph below to answer the problems.
y
4
2
�2
1 2 3 4
x
�2
Figure for nos 1-6: The graph of f (x) =
1.
3.
5.
x2
x−1
f �� (x) > 0 on (0, ∞)
2.
f is increasing on (−∞, 0)
lim f (x) = ∞
4.
The range of f is (−∞, ∞).
The graph has no slant asymptote.
6.
(0, 0) is a local maximum point.
x→1
7. The domain of y =
sin x
is {x : x �= kπ where k is an integer }.
1 + cos x
8. The graph of y = x2 sin x is symmetric about the y-axis.
9. The graph of y = tan x is symmetric about the origin.
10. If f � (x0 ) = 0, then f (x0 ) is a relative extreme value of f .
Exercises for Section 3.6
In Exercises 1-4, an approximate sketch of the graph of a function is given. Find the
relative extrema, open intervals where the function is increasing or decreasing, limits at
infinity, and range.
1. f (x) = 6x2 − x4 + 2
2.
g(x) =
y
x3
x2 − 1
y
4
12
6
�3 �2
�2
�1
1
2
2
x
�4
For No. 1
For No. 2
3
x
3.6. SUMMARY OF CURVE SKETCHING
3. p(t) = t5/3 − 5t2/3
195
C(t) = sin t + cos t, 0 ≤ t ≤ 2π
4.
y
y
1
�2
1 2 3
1
t
Π
2Π
t
�1
For No. 3
For No. 4
In Exercises 5-14, sketch the graph of the function. Find the intercepts, relative extrema,
points of inflection, asymptotes, and the range.
t2
8t
5. g(t) = 2
6. h(t) =
t −1
4 − t2
7.
9.
2x3
x2 − 9
w
r(w) = √
w−2
8.
f (x) =
10.
11.
p(x) = 8x1/3 − x4/3
12.
13.
y=
x4 − 24x2
24
14.
x4
x2 − 1
w
C(w) = √
w4 + 1
�
1�
F (x) =
36x1/3 − x4/3
9
R(x) =
y=
x5
2x3
−
20
3
In Exercises 15-18, sketch the graph of the function over the indicated interval. Find the
coordinates of the local extrema and points of inflection.
15.
17.
4 sin x
, [0, 2π]
2 − cos x
� π π�
R(x) = 4x − tan x, − ,
2 2
T (x) =
16.
18.
5 cos θ
, [0, 2π]
2 − sin θ
� π π�
f (θ) = 3 tan 2θ − 8θ, − ,
4 4
S(θ) =
In Exercises 19-22, find a function with the given asymptotes. There are several possible
answers.
19.
20.
21.
22.
Vertical
Vertical
Vertical
Vertical
asymptote x = 3, horizontal asymptote y = 1
asymptotes x = ±1, no horizontal asymptote
asymptote x = 0, slant asymptote y = x + 1
asymptote x = −1, slant asymptotes y = x, y = −x
In Exercises 23-26, the graphs of y = f � (x) and y = f �� (x) are given. Sketch the graph of
f if the graph passes through the indicated point P .
23. P (0, 0)
y� f ''�x�
24.
y
P (0, 0)
y� f '�x�
3
y
2
1
1
1
y� f '�x�
For No. 23
x
1
2
y� f ''�x�
For No. 24
x
196
CHAPTER 3. APPLICATIONS OF DERIVATIVES
25. P (0, 1)
26.
P (1, 0)
y
y� f ''�x�
y
2
y� f '�x�
1
1
3
�1
x
x
�3
y� f '�x�
y� f ''�x�
For No. 25
For No. 26
Applications
v
v � s'�t�
3
6.25
t
27. Free-falling Object
A ball is thrown from a certain height s0 above the ground. The graph shown is
that of the velocity function v(t) in ft/sec over [0 sec, 6.25 sec], which is the time
interval when the ball is above the ground. If v(3) = 0, find s0 and the maximum
height reached by the ball.
28. FM Radio
Sketch the graph and find all relative extrema of F (θ) = cos (θ + sin 2θ) over the
interval [0, π]. A function such as F is used in frequency modulation FM synthesis.
29. Fourier sine series
Figure for 27
1
1
Find all the relative extrema of S(θ) = sin θ + sin 3θ over the interval [0, π],
2
6
and sketch the graph of S. The function S is called a Fourier sine series and has
applications in signal processing.
30. Harmonic oscillator
3
cos(4t) + cos(3t) over the interval
4
[0, π]. Moreover, show that the function satisfies the differential equation y �� + 16y =
39 cos 3t. Such a differential equation is said to model the motion of a weight that
is attached to a spring.
Find the critical numbers of the function y =
Theory and Writing Proofs
31. Let y = ax2 + bx + c be a quadratic function where a �= 0. Use calculus to determine
the coordinates of the vertex of the parabola. When is the graph concave upward?
Concave downward?
32. Determine the local extrema and points of inflection of the cubic polynomial c(x) =
x3 + kx2 where k is a constant.
33. The signs of the first two derivatives of y = G(t) on certain open intervals are listed
below. If the graph of G passes through the points A(−1, 3), B(1, 1), and (3, −1),
then sketch a possible graph of G.
Open interval
(−∞, −1)
(−1, 1)
(1, 3)
(3, ∞)
G� (t)
+
−
−
+
G�� (t)
−
−
+
+
34. The derivative of a polynomial y = p(x) is p� (x) = (x − 1)2 (2 − x). Find the relative
extrema and points of inflection of the graph of p.
35. Let p be a polynomial with an even degree n ≥ 2. Prove that p has at least one
relative extreme value.
3.7. OPTIMIZATION PROBLEMS
3.7
197
Optimization Problems
Solving Optimization Problems
The applications of derivatives in finding minimum and maximum values are useful in
solving optimization problems. For instance, we may want to know the maximum area,
shortest distance, maximum profit, and least cost among others. In the process, we have
to convert word problems into equations as the following examples illustrate.
Example 1 Maximizing the Area of a Rectangle
y
A homeowner plans to build a rectangular enclosure along the side of her house using 500
feet of fencing material, see Figure 1a. Find the dimensions of the enclosure that has the
largest area assuming no fencing material is needed along the side of the house.
Solution
x
The area A of the rectangle is
A = xy.
Wall
Primary equation
Since there are 500 feet of fencing material and only three sides of the rectangle require
fencing, we have
2x + y = 500.
Secondary equation
Substituting
into the primary equation, we obtain
x
y = 500 − 2x.
(6)
Figure 1a
A rectangular enclosure
where the dotted side represents the side that uses no
fencing material.
A
A = xy = x(500 − 2x) = 500x − 2x2 .
�125, 31250�
Since x and y must be nonnegative, using (6) we find that x ≤ 250. Then the feasible
domain of A as a function of x is [0, 250]. The graph of A as a function of x is seen in
Figure 1b.
Now, we find the critical numbers of A in (0, 250).
dA
= 500 − 4x
dx
x
=
0
Set the derivative to zero.
=
125
Critical number
x
Figure 1b
A sketch of
the graph of
A = 500x − 2x2 in the
feasible domain [0, 250].
Next, evaluate A at the critical number and at the endpoints of [0, 125].
x
A
0
0
125
31, 250
150
0
Thus, the maximum value of A occurs when x = 125 ft. Correspondingly, we get y = 250
ft by using (6). Therefore, the dimensions of the rectangular enclosure with the largest
possible area are
125 ft by 250 ft.
✷
In Example 1, with the given 500 feet of fencing we must have realized that there
were infinitely many ways to construct a rectangular enclosure. For instance, we list a few
possible dimensions of such rectangular enclosures with their corresponding areas.
Dimensions
50 ft by 50 ft by 400 ft
80 ft by 80 ft by 340 ft
150 ft by 150 ft by 200 ft
Area
20, 000 ft2
27, 200 ft2
30, 000 ft2
198
CHAPTER 3. APPLICATIONS OF DERIVATIVES
400
340
200
50
50
80
80
150
Area � 20,000 ft 2
Area � 27,200 ft2
150
Area � 30,000 ft2
Figure 2
Rectangular enclosures with the same perimeters but different areas.
Try This 1
Find the dimensions of the rectangle with the maximum possible area that has a perimeter
of 100 feet.
The following guidelines are useful in solving optimization problems, as illustrated in
Example 1.
Guidelines for Solving Optimization Problems
1. Identify all quantities and draw a picture of the problem. A quantity could be a variable
or constant such as area, perimeter, etc.
2. Write a primary equation that involves the variable to be optimized, i.e., maximized
or minimized.
3. Express the quantity to be optimized as a function of one variable. A substitution and
a secondary equation may be needed to do this.
4. Find the feasible domain, i.e., the set of numbers for which the optimization problem
makes sense.
5. Find the maximum or minimum value of the quantity to be optimized using the calculus
techniques in Sections 3.1 through 3.6.
y
f �x�� x
�x,y�
Example 2 Minimizing the Distance from a Point to a Graph
�1,0�
Figure 3a
Find the point (x, y) on
√
the graph of f (x) = x
that is nearest to (1, 0).
x
√
Find the point on the graph of f (x) = x that is closest to the point (1, 0). What is the
minimum distance between the graph of f and (1, 0)?
Solution Let d be the distance between (1, 0) and a point (x, y) on the graph of f , see
Figure 3a.
�
d =
(x − 1)2 + (y − 0)2
�
=
(x2 − 2x + 1) + y 2
Primary equation
√
Since (x, y) lies on the graph of f (x) = x, we can substitute
√
y= x
Secondary equation
into the primary equation. Then we write
�
�
√
d(x) = (x2 − 2x + 1) + ( x)2 = x2 − x + 1.
3.7. OPTIMIZATION PROBLEMS
199
√
Using the domain of f (x) = x, we find that the feasible domain of d(x) is [0, ∞).
Applying the Chain Rule, we obtain
�
d �� 2
2x − 1
1
d� (x) =
x − x + 1 = (x2 − x + 1)−1/2 (2x − 1) = √
.
dx
2
2 x2 − x + 1
y
d�x��
� 12 , 23 �
Now, if d� (x) = 0 then
2x − 1
=
x
=
0
1
2
x2 � x � 1
x
Critical number
Using the First Derivative Test, we find that the minimum value of d occurs when x = 1/2
as seen in Figure 3b. Hence, the minimum distance between the graph of f and (1, 0) is
�� �
√
� �
2
1
1
3
d(1/2) =
−
+1=
.
2
2
2
√
Finally, the point on the graph of f (x) = x that is closest to (1, 0) is
�
� � � �
� √ �
1
1
1
1
2
, f (1/2) =
,
, or
,
.
2
2
2
2 2
Figure 3b
The shortest distance
between (1, 0) and the
graph of f is
√
d(1/2) = 3/2.
✷
Try This 2
√
Find the point on the graph of f (x) = x that is closest to the point (4, 0).
y
Example 3 Maximizing the Area of an Inscribed Rectangle
√
A rectangle is inscribed in the region bounded by the semicircle y = 1 − x2 and the
x-axis, see Figure 4a. Find the length and width of the rectangle with the maximum
possible area.
Solution Let P (x, y) be a vertex of a rectangle such that P lies on the semicircle in the
first quadrant, see Figure 4a. Since 2x and y are the length and width of the rectangle,
respectively, the area of the rectangle is
A
Since
y=
we obtain
�
=
2xy.
A = 2x
Secondary equation
�
1 − x2 .
A feasible domain of A is [0, 1]. Using the Product Rule, we obtain
dA
dx
=
=
=
dA
dx
=
�
−2x
2x · √
+ 2 1 − x2
2
2 1−x
�
−2x2
√
+ 2 1 − x2
1 − x2
� 2
�
2
√
−x + (1 − x2 )
1 − x2
2(1 − 2x2 )
√
1 − x2
Factor √
1 � x2
P�x,y�
�1
1
x
Figure 4a
Inscribing a rectangle in a
semicircle with radius 1.
Primary equation
1 − x2 ,
y�
2
1 − x2
Simplify
200
y
CHAPTER 3. APPLICATIONS OF DERIVATIVES
If dA/dx = 0, then
A�2x 1 � x
2
� 2 �2,1�
2
1 − 2x = 0 or x =
x
1
Figure 4b
The maximum possible
value of A is 1 and
√ this
occurs when x = 2/2.
√
2
2
Critical number
The values of A at the critical number and the endpoints of [0, 1] are
√
A(0) = 0, A( 2/2) = 1, and A(1) = 0.
√
Thus, the maximum value of A occurs when x = 2/2, see Figure 4b. Hence, the length
and width of the rectangle with the maximum area are
�√ �
√
2
length = 2x = 2
= 2
2
and
width = y =
�
1−
x2
=
�
√
� √ �2 �
2
1
2
1−
= 1− =
.
2
2
2
✷
Try This 3
√
A rectangle is inscribed in a region bounded by the semicircle y = 16 − x2 and the x-axis.
Find the length and width of the rectangle that has the maximum possible area.
Example 4 Minimizing the Total Length
A 3-feet high post and a 5-feet high post are 10 feet apart, see Figure 5a. The posts are to
be supported by two wires staked at a common point C in the horizontal ground between
the posts. Where should the wires be staked to minimize the sum of the lengths of the
wires?
E
A
3 ft
5 ft
B
C
BD�10 ft
D
Solution Let x = BC and y = CD be the lengths of the line segments shown in Figure
5a. By the Pythagorean Theorem, we obtain
�
�
AC = 9 + x2 and CE = 25 + y 2 .
Then the sum L of the lengths of the wires is
�
�
L = AC + CE = 9 + x2 + 25 + y 2 .
Primary equation
Since the posts are 10 feet apart, we have
Figure 5a
A 3-ft post and a 5-ft post
are supported by wires that
are staked at point C.
y = 10 − x.
Secondary equation
Substituting into the primary equation, we find
�
�
L =
9 + x2 + 25 + (10 − x)2
�
�
=
9 + x2 + x2 − 20x + 125.
See Figure 5b. Since x and 10 − x are nonnegative, let [0, 10] be the feasible domain of
L. Using the Chain Rule, we find
dL
dx
=
√
x
x − 10
+√
.
9 + x2
x2 − 20x + 125
3.7. OPTIMIZATION PROBLEMS
201
y
If dL/dx = 0, then
�
√
x
9 + x2
�2
�
=
2
−√
x − 10
x2 − 20x + 125
�2
15
10
2
x
9 + x2
L�
x − 20x + 100
x2 − 20x + 125
=
5
3.75
Then cross-multiply and set one side of the equation to zero.
16x2 + 180x − 900
4(4x − 15)(x + 15)
x
=
0
=
0
=
15/4
Critical number in [0, 10]
9 � x2 � 25 � �10 � x�2
10
x
Figure 5b
The minimum total length L
occurs when x = 3.75 ft.
The values of L at the endpoints of [0, 10] and the critical number are
L(0) ≈ 14.2, L(3.75) ≈ 12.8, and L(10) ≈ 15.4.
Hence, the wires must be staked at a point in the ground that is x = 3.75 ft from the 3-ft
post.
✷
Try This 4
If the posts in Example 4 are 16 feet apart, then where should the wires be staked to
minimize the sum of the lengths of the wires?
Example 5 Using the Least Material
Find the radius and height of a 1-liter can with the least surface area. Assume the can is
a right circular cylinder that has a cover at the top and bottom.
Solution Let r and h be the radius and height of the can, respectively. Since the can
has a top and bottom, its surface area A is
A = 2πr2 + 2πrh.
Primary equation
Since the volume is 1 liter or equivalently 1000 cm3 , we find
πr2 h = 1000.
Secondary equation
Substituting
h=
1000
πr2
A
into the primary equation, we obtain
A = 2πr2 + 2πr
�
1000
πr2
�
Figure 6a
A right circular cylinder with
a top and bottom. Let r be
the radius and h the height.
A�2Πr2�2000�r
= 2πr2 +
2000
.
r
Since r must be positive, the feasible domain of A is (0, ∞) as in Figure 6b. Then
dA
2000
4πr3 − 2000
= 4πr − 2 =
.
dr
r
r2
600
300
3 500
If dA/dr = 0, then
r
Π
4πr3 − 2000
=
r
=
0
�
3
√
3
500
5 4π 2
=
π
π
Critical number
Figure 6b
The graph of the surface
area A as a function
of its radius r.
202
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Applying the First Derivative Test, we find that the minimum surface area A occurs at
the critical number. Hence, the radius and height of the can with the least possible surface
area are
√
3
5 4π 2
radius = r =
≈ 5.4 cm, and
π
√
3
1000
1000
10 4π 2
height = h =
=
=
≈ 10.8 cm.
�
�
√
2
2
3
πr
π
5 4π 2
π
π
✷
Try This 5
Find the radius and height of a 1-liter can with the least surface area. Assume the can is
a right circular cylinder that has a cover at the bottom but is open at the top.
3.7 Check-It Out
1. Gary wants to build a fence on two adjacent lots, as shown in Figure 1. No fencing
is needed along the boundary represented by the dotted lines. If 1200 feet of fencing
material is available, then find x and y as indicated in the figure if the area enclosed
is the maximum.
y
x
Figure for No. 1. The length from the left-most side to the
right-most side is y, and the width is x.
2. Find the minimum value of p if the primary equation is p = 2x + y,
secondary equation is xy = 32, and the feasible domain is x > 0.
True or False. If false, explain or show an example that shows it is false.
1. The distance between (a, b) and a point (x, y) on the parabola y = x2 is
�
(x + a)2 + (x2 + b)2 .
2. If a right circular cylinder has a cover at the top and bottom, then its surface area
is S = 2πrh + 2πr2 where r and h are the radius and height, respectively.
3. If the perimeter of a rectangle is 4 feet, then the area of such a rectangle is 1 ft2 .
4. If the sum of two numbers is 10, then the maximum possible product of two such
numbers is 100.
5. If the area of a rectangle is 12 square inches, then the dimensions of such a rectangle
are 4 inches by 3 inches.
3.7. OPTIMIZATION PROBLEMS
6. The two positive numbers that have a product of 15 and the minimum possible sum are
203
√
15 and
√
15.
7. If two negative numbers differ by 1, then the minimum product of two such numbers does not exist.
8. A box with a square base and open top has a surface area of 1200 square inches. Then the shape of the box
that has the maximum volume is a cube, i.e., the height of the box is the same as side of the base.
9. A rectangle is inscribed in a semicircle with radius 1, see figure below. If the indicated radial line makes an
angle θ with the positive x-axis where 0 < θ < π/2, then the area of the rectangle is sin 2θ.
y
�1
Θ
1
x
Figure for No. 9 and 10
10. In the above figure, the minimum perimeter of the inscribed rectangle is 2.
Exercises for Section 3.7
In Exercises 1-6, optimize the indicated variable given the primary and
secondary equations, and feasible domain.
1. Minimize D, primary equation D = (y − 1.5)2 + x2 , secondary equation y = x2 , feasible domain −∞ < x < ∞.
2. Minimize D, primary equation D =
feasible domain −∞ < x < ∞.
�
(x − 1)2 + y 2 , secondary equation y = x + 1,
3. Maximize A, primary equation A = l · w, secondary equation 2l + 2w = 20, feasible domain 0 ≤ w ≤ 10.
4. Maximize A, primary equation A =
√
1
bh, secondary equation b2 + h2 = 27, feasible domain 0 ≤ b ≤ 27.
2
5. Minimize S, primary equation S = 2πrh + πr2 , secondary equation r2 h = π, feasible domain 0 < r < ∞.
6. Maximize V , primary equation V = πr2 h, secondary equation 2πrh + 2πr2 = 24π, feasible domain 0 < r < ∞.
Optimization Problems
√
7. Find the point on the graph of f (x) = x that is nearest the given point.
1
a) (9, 0)
b) ( 2 , 0)
√
8. Find the point on the graph of f (x) = 2 x that is closest to the indicated point.
a) (4, 0)
b) (1, 0)
9. Find the number which exceeds its square by the greatest value.
10. Find the positive number that exceeds its cube by the greatest amount.
204
CHAPTER 3. APPLICATIONS OF DERIVATIVES
11. A homeowner plans to fence a rectangular region that must have an area of 30,000 square feet. In addition,
the homeowner plans to subdivide the region into three equal rectangles by inserting two smaller fences,
as seen below. Find the dimensions of the rectangular region that will use the least amount of fence.
y
h
x
2r
Figure for No. 11.
Figure for No. 12.
12. A window consists of a rectangle and a semicircle that is mounted on top of the rectangle, see figure above.
Let h be the height from the base of the rectangle to the base of the semicircle. If the perimeter of the
window is 20 feet, find the radius of the semicircle so that the window has the maximum area.
13. From an aluminum square sheet, 12 inches on each side, a box with an open top is to be made by cutting
off small squares from each corner and bending up the sides. How large a square should be cut from each
corner so that the box has the maximum possible volume?
Figure for No. 13
14. The sum of the perimeters of a square and an equilateral triangle is 4 feet. Find the length of the side
of the square that minimizes the sum of the areas of the square and triangle.
15. A right triangle in the first quadrant is bounded by coordinate axes, and a line through point (4, 2), as seen
below. Find the coordinates of the vertices if the triangle has the minimum area. Does the triangle have
the minimum perimeter?
y
�4,2�
x
Figure for No. 15
Figure for No. 16
16. Find the area of the largest isosceles triangle that can be inscribed in a circle with radius 2 units.
17. Find the dimensions of the right circular cylinder with a volume of 30 cubic inches and that has the
minimum surface area. Assume the cylinder has a top and bottom.
3.7. OPTIMIZATION PROBLEMS
205
18. A track and field has the shape of a rectangle and two semicircles attached to the sides of the rectangle. The
perimeter of the track and field is 400 meters. Find the length of the rectangle and the radius of the semicircle
if the rectangular region has the maximum possible area.
√
19. Two towns A and B that are 37 miles from each other plan to build a recreation center C for their residents
along a certain highway, see figure below. Suppose the perpendicular distances from A and B to the highway
are 3 miles and 4 miles, respectively. Where on the highway should the recreation center be built so that the
sum of the distances from the recreation center to the towns is minimized?
C
B
3
P
C
4
3 km
A
A
37
B
Figure for No. 19
Figure for No. 20
20. A forest ranger must travel from point A to point C through some point P , see figure above. There is a road
from P to C that runs from west to east. The point B on the road nearest A is 3 km away, and the distance
from B to C is 10 km. The forest ranger can travel diagonally from A to P at the speed of 15 kph, and drive
faster from P to C at 30 kph. How far from B should P be located so that the forest ranger travels the least
amount of time from A to P to B?
21. Mark plans to make a rectangular garden using x feet of fencing material. He figures that only three sides of
the garden will be fenced and the fourth side will be left open without a fence. What is the maximum area of
the garden that that he can enclose?
22. If the base b and perimeter p of a triangle are fixed, then determine the remaining two sides of the triangle with
the maximum area.
23. A crew is needed to service a manufacturing plant. It has been found that the number of hours needed to do
the job is 15 + 64/x where x is the size of the crew. If each worker is paid $16 per hour and the use of an
equipment is $60 per hour, find the most economical size of the crew.
24. A field trip will cost each student $60 if 200 students join the trip. If more than 200 students join, then the cost
of the trip per student will be reduced by 25 cents times the number of students in excess of 200. How many
students should join the trip to realize the largest gross income?
25. A contractor wishes to find the dimensions of a rectangular piece of land where she can build a house that will
have an area of 22,000 square feet. Moreover, the land must have 50 feet of space from the exterior of the house
to the front and back property lines, and 27.5 feet of space from the exterior to the left and right property
boundaries. Find the dimensions of the land with the least area on which the house can be built.
26. What are the dimensions of the rectangle with the greatest area that can be inscribed in a triangle with base
40 ft and height 15 ft, if one side of the rectangle lies on the base of the triangle? See figure below.
x
y
Figure for No. 26
27. Find the minimum distance between a point (p, q) and a line y = mx + b.
206
CHAPTER 3. APPLICATIONS OF DERIVATIVES
28. A right circular cone is made by cutting a sector from a unit circle, see figure below. Let θ be the central
angle of the sector. Find θ if the cone has the maximum volume.
!
Figure for No. 28
Figure for No. 29
29. A rain gutter is made by folding up the edges of an 18-inch metal sheet, see figure above. Suppose each of the
left and right edges of the gutter are 6 inches long. Find the angle that the edges must be folded so that the
gutter holds the maximum volume.
30. Find the volume of the largest circular cone that can be inscribed in a sphere of radius r.
31. A 4-by-6 rectangular card is folded in such a way that vertex A is placed at some point B on the opposite
longer side, see figure below. Let c be the length of the crease that is formed. Find the minimum value of c.
y
B
1
x
c
x
A
Figure for No. 31
R
1
x
Figure for No. 32
32. A rectangle region R in the first quadrant has a vertex at the origin, the opposite vertex is a point in the
parabola y = (x − 1)2 , and two sides that lie in the coordinate axes, see figure above. Find the maximum
area of such a rectangle.
33. A rope of length L is formed into a sector of a circle. Find the central angle of the sector that gives the
maximum area of the sector.
Odd Ball Problems
34. Determine the constant C if G(x) = x2 +
C
has a relative minimum at x = 3.
x
35. Determine the constant K if the function H(t) = t3 +
K
has points of inflection at t = ±1
t
36. Find the maximum and minimum values of f (x) = A cos(2x) + B cos x assuming B ≥ 4|A| > 0.
37. Find the maximum and minimum values of f (x) = A cos(2x) + B cos x assuming 4|A| > B > 0.
38. A circle of radius 2 is centered at (0, 2). A second circle of radius 3 lies in the first quadrant, and is tangent
to the first circle and x-axis. In the figure above, we see a third circle that is tangent to the first and second
circles. If the center of the third circle lies in the y-axis, find its radius. What is the limit of the common point
between the first and third circles as the radius of the third circle increases to ∞?
3.8. NEWTON’S METHOD
3.8
207
Newton’s Method
• Aproximating a Real Zero of a Function • A Sufficient Condition for Convergence
Aproximating a Real Zero of a Function
The solutions of f (x) = 0 are not easy to solve. In fact not all solutions are necessarily
real numbers. If c is a real zero of f , we can approximate c to any degree of precision
under certain conditions. We will discuss and apply Newton’s method to approximate
c.
The first step is to choose an arbitrary number x1 that is approximately c, as in Figure
1a. The graph of f will help you choose x1 . The tangent line at (x1 , f (x1 )) is given by
y − f (x1 )
f � (x1 )(x − x1 )
=
y
y
c
x1
x
x1
x
f (x1 )(x − x1 ) + f (x1 ).
=
If (x2 , 0) is the x-intercept of the tangent line, then
0
=
f � (x1 )(x2 − x1 ) + f (x1 )
x2
=
x1 −
f (x1 )
f � (x1 )
if f � (x1 ) �= 0
Figure 1a
y
Now, we have two approximations for c, namely, x1 and x2 . We continue the process.
Similarly, we can find the x-intercept (x3 , 0) of the tangent line at (x2 , f (x2 )). In general,
the nth approximation to c is
xn+1 = xn −
x2
�
f (xn )
f � (xn )
�x1, f �x1��
�x2, f �x2��
if f � (xn ) �= 0
c
The repeated applications of the above procedure is called Newton’s Method.
x2
A Summary of Newton’s Method
Let f be a differentiable function such that f (c) = 0.
Step 1. Find an initial estimate x1 of c.
Step 2. The (n + 1)st approximation is
xn+1 = xn −
f (xn )
.
f � (xn )
Step 3. If |xn+1 − xn | is less than the desired accuracy, let xn+1 be the final estimate of
c. Otherwise, return to Step 2 and find a new approximation.
Under certain conditions, the approximations become more precise as n increases.
Example 1 Approximating
√
4
y
f �x��x4�2
2
Apply Newton’s Method to approximate the real root of x4 − 2 = 0 as follows: use x1 = 1
and estimate x4 .
Solution
formula
Figure 1b
Newton’s Method finds
sequential approximations
x1 , x2 , x3 , ... to an
x-intercept of the graph
of a function.
x1
x2
Let f (x) = x4 − 2. Since f � (x) = 4x3 , the iterative process is given by the
xn+1
=
xn −
f (xn )
f � (xn )
=
xn −
x4n − 2
.
4x3n
Figure 2
The tangent line at the
point where x1 = 1 has
x-intercept x2 = 1.25.
x
208
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Since x1 = 1, we find
x2
=
x1 −
=
1.25.
x41 − 2
14 − 2
= 1−
3
4x1
4 · 13
Then the third estimate is
x3
(1.25)4 − 2
x42 − 2
= 1.25 −
3
4x2
4(1.25)3
=
x2 −
≈
1.1935
Continuing the process, we obtain
x4
x43 − 2
4x33
=
x3 −
≈
1.18923.
The first iteration of Newton’s Method is shown in Figure 2. Using a calculator, we find
√
4
2 ≈ 1.189207 which agrees with x4 up to four decimal places
✷
Try This 1
Repeat Example 1 but approximate the real root of x3 − 2 = 0.
Example 2 Finding a Zero of a Function
The function
has a zero in [1, 2], see Figure 3. Apply Newton’s Method with x1 = 1.5 to find xn+1 if
|xn+1 − xn | < 0.001. Then find the value of f (xn+1 ).
y
3
f (x) = x5 − x4 + x − 3
Solution
�1.5, f �1.5��
Since f � (x) = 5x4 − 4x3 + 1, the iterative process is given by the formula
xn+1 = xn −
1
2
x
Using a calculator with x1 = 1.5, we obtain the following estimates
x2
�3
x3
x4
Figure 3
The tangent line to the
graph of f at the point
with x1 = 1.5 has
x-intercept x2 = 1.41951.
f (xn )
x5 − x4 + xn − 3
= xn − n 4 n 3
.
�
f (xn )
5xn − 4xn + 1
≈
1.41951
≈
1.40678
≈
1.40705
|x3 − x2 | ≈ 0.01246
|x4 − x3 | ≈ 0.00027
We stop iterating since |x4 − x3 | < 0.001. Then the zero of f (x) in [1, 2] is approximately
x4 ≈ 1.40678. Finally, x4 satisfies
f (x4 ) ≈ 1.406785 − 1.406784 + 1.40678 − 3 ≈ 0.000001.
✷
Try This 2
The function g(x) = cos x − x has a zero in [0, 1]. Apply Newton’s Method to find xn+1 if
x1 = 1 and |xn+1 − xn | < 0.0001.
3.8. NEWTON’S METHOD
209
Newton’s Method does not always guarantee that the values of
xn+1 = xn −
f (xn )
f � (xn )
will approximate a zero of f (x) to any accuracy desired. For instance, let f (x) = x3 −2x+2
whose graph is given in Figure 4. The tangent line at (1, 1) has x-intercept (0, 0), and
the tangent line at (0, 2) has x-intercept (1, 0). Applying Newton’s Method with initial
estimate x1 = 1 will generate the following values: x1 = x3 = x5 = · · · 1 and x2 = x4 =
x6 = · · · 0.
A Sufficient Condition for Convergence
The next theorem will give a condition when Newton’s Method is guaranteed to work.
That is, the values of xn ’s generated by the method will approximate a given zero c of a
function to any desired precision. In such a case, we write
lim xn = c.
n→∞
Also, we say xn converges to c as n → ∞. The proof of the theorem relies on the Mean
Value Theorem. Before stating the theorem, if a < b, we observe that the interval [a, b] is
the middle third of [2a − b, 2b − a].
Theorem 3.11
Let f (c) = 0 and suppose
�
�
� f (x)f �� (x) �
�
�
� (f � (x))2 � ≤ α < 1
(7)
for all x in (c − β, c + β) for some positive constants α, β. If x1 lies in [c − β, c + β] and
n)
xn+1 = xn − ff�(x
, then lim xn = c.
(xn )
n→∞
Proof
Let N (x) = x −
f (x)
. Observe N (c) = c and
f � (x)
N � (x) = 1 −
(f � (x))2 − f (x)f �� (x)
f (x)f �� (x)
=
.
(f � (x))2
(f � (x))2
In particular, |N � (x)| ≤ α for all x in (c − β, c + β) . Applying the Mean Value Theorem,
we find
|x2 − c| = |N (x1 ) − N (c)| ≤ α |x1 − c| .
Since 0 < α < 1 and |x1 − c| ≤ β, it follows that |x2 − c| < β. Thus, x2 lies in (c − β, c + β).
Similarly, by the Mean Value Theorem we obtain
|x3 − c| = |N (x2 ) − N (c)| ≤ α |x2 − c| ≤ α2 |x1 − c|.
In general, for any integer n ≥ 1 we have
|xn+1 − c| ≤ αn |x1 − c|
and xn+1 lies in (c − β, c + β).
Let ε > 0 be a given degree of accuracy. Then |xn+1 − c| < ε whenever n is sufficiently
large . Hence, lim xn = c.
n→∞
✷
210
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Example 3 Convergent Approximations Using Newton’s Method
Let f (x) = x3 − 3 and x1 =
satisfy
4
.
3
Prove the values of xn generated by Newton’s Method
√
3
lim xn =
n→∞
3
√
Solution The graph of f is seen in Figure 4a, and 3 3 is the real zero of f (x). The first
two derivatives are
f � (x) = 3x2 and f �� (x) = 6x.
Then inequality (7) in Theorem 3.11 reduces to
�
� �
�
�
�
� f (x)f �� (x) � � (x3 − 3)(6x) �
�
�
�
�=�
� = 2 �1 − 3 �
� [f � (x)]2 � �
�
9x4
3�
x3 �
�
�
The graph of g(x) = 23 �1 − x33 � is shown in Figure 4b. We find g(2) =
�
and g( 3 65 ) = 1.
y
5
,
12
g( 43 ) < g(2),
y
1
1
1
4
3
x
5
3
5
12
3
6
5
4
3
5
3
x
2
�
�
2 ��
3 ��
Figure 4b: g(x) = �1 − 3 �.
3
x
3
Figure 4a: f (x) = x − 3.
√
5
From Figure 4b, we see that g(x) < 12
for all x in [ 43 , 2]. Let β = 3 3 − 43 > 0. Then
√
√
5
g(x) < 12
for all x in ( 3 3 − β, 3 3 + β). Thus, we can apply Theorem 3.11 with x1 = 43 .
√
3
Hence, the values of xn generated by Newton’s Method satisfy lim xn = 3.
n→∞
✷
Try This 3
Let f (x) = x2 − 60
√ and x1 = 8. Prove the values of xn generated by Newton’s Method
satisfy lim xn = 60.
n→∞
Example 4 Convergent Approximations Using Newton’s Method
Let f (x) = cos(x) − x and x1 = 1. If f (c) = 0, prove the values of xn generated by
Newton’s Method satisfy lim xn = c. Then find x2 to the nearest thousandth.
n→∞
Solution The graph of f is seen in Figure 5, and its zero c is near
derivatives are
f � (x) = − sin(x) − 1 and f �� (x) = − cos x.
Then inequality (7) in Theorem 3.11 reduces to
�
�
� f (x)f �� (x) �
�
g(x) = ��
[f � (x)]2 �
=
�
�
� (cos(x) − x) cos x �
�
�
� (1 + sin x)2
�
3
.
4
The first two
3.8. NEWTON’S METHOD
In Figure 5, we find g(x) ≤
1
2
211
for all x in [ 12 , 1].
y
1
2
y�g�x�
y� f �x�
1
2
�
3
4
1
x
1
2
�
�
� f (x)f �� (x) �
�.
Figure 5: f (x) = cos(x) − x and g(x) = ��
(f � (x))2 �
Then Theorem 3.11 applies with x1 = 1 or for any x1 in [ 12 , 1] . Thus, the values of xn generated by Newton’s Method
will satisfy lim xn = c. Finally, since x1 = 1, we find
n→∞
x2 = x1 −
f (x1 )
cos(1) − 1
=1−
≈ 0.739
f � (x1 )
− sin(1) − 1
✷
Try This 4
As in Example 4, but f (x) = sin x + x − 2. As done earlier, let x1 = 1.
An equation f (x) = 0 is said to be solvable by radicals if its solutions can be determined in a finite number of
√
steps using addition, subtraction, multiplication, division, and the radicals n x. For example, ax2 + bx + c = 0 with
a �= 0 is solvable by radicals for the solutions are
√
−b ± b2 − 4ac
x=
.
2a
Moreover, 3rd and 4th degree polynomials are solvable by radicals as proved by Geronimo Cardano (1501-1576). For
instance, the real solution to x3 + px + q = 0 where p > 0 is
��
��
3
3
q2
p3
q
q2
p3
q
x=
+
− −
+
+
4
27
2
4
27
2
The verification of this cubic solution is an exercise at the end of this section. However, the zeros of 5th and higher
degree polynomials are not solvable by radicals. The latter fact was established by Evariste Galois (1811-1832).
Newton’s Method is a robust numerical algorithm for solving equations whether solvable by radicals or not.
3.8 Check-It Out
Use Newton’s Method to find xn given n, x1 , and f (x) = 0.
1. n = 2, x1 = 1, x3 + x − 1 = 0
2.
n = 3, x1 = 1, x2 − 2 = 0
True or False. If false, explain or show an example that shows it is false.
1. In Newton’s Method, the (n + 1)st estimate is xn+1 = xn −
f (xn )
f � (xn )
2. If f (x) = mx + b, m �= 0, and x1 is any real number, then the second value generated by Newton’s Method
satisfies f (x2 ) = 0.
3. If p(x) is a 3rd degree polynomial and x1 = 0, then the values xn generated by Newton’s Method converge to a
zero of p(x).
4. Let x1 be a first estimate of the zero of f (x) =
Method is x2 = x1 (2 − ax1 ).
1
x
− a. Then the second estimate of the zero of f (x) by Newton’s
212
CHAPTER 3. APPLICATIONS OF DERIVATIVES
5. Let x1 = 1 be a first estimate of the zero of f (x) = x2 − a. By Newton’s Method the second estimate of the
zero is x2 = 1−a
.
2
�
�
� (c)f �� (c) �
6. Let f (c) = 0 and suppose f �� (x) is continuous. If � f(f
� (c))2 � < 1, then there exists x1 satisfying lim xn = c.
n→∞
Exercises for Section 3.8
In Exercises 1-8, apply Newton’s method using the given x1 to find the next two estimates x2 and x3 that solves the
indicated equation f (x) = 0. Round the answers to the nearest thousandth.
1.
x1 = 2, x2 − x − 5 = 0
2.
x1 = −1, x2 − 2x − 10 = 0
3.
x1 = 2, x3 − 2 = 0
4.
x1 = 1, x5 − 2 = 0
5.
x1 = 2, cos(πx) − x = 0
6.
x1 = 3, 2 sin(x) − x + 2 = 0
7.
x1 = 3, sin(x) + cos(x) + 1 = 0
8.
x1 =
1
,
4
cos(2πx) − πx = 0
In Exercises 9-12, explain why Newton’s Method fails to find a zero of f (x) given the initial estimate x1 .
9.
12.
3
x1 = −1, f (x) = x − 5x
10.
x1 = −3, f (x) = 2x3 + 9x2 − 1
13.
x1 = 1, f (x) =
√
3
x1 = 5, f (x) =
1
x
x
11.
−1
14.
x1 = 1, f (x) =
�
x1 = 2.5, f (x) =
√
x
if
x≥0
− −x
if
x<0
√
1
x2
−1
In Exercises 15-18, you are given an equation f (x) = 0 and an initial estimate x1 of a zero of f (x). Apply Newton’s
Method to find xn+1 that satisfies |xn+1 − xn | < 0.001. Round xn+1 to four decimal places.
15.
x1 = 2, x3 − 5 = 0
17.
x1 = 3, sin( x6 ) −
1
2
=0
16.
x1 = 4, x5 + 15x − 1000 = 0
18.
x1 = 3, tan( x3 ) −
√
3=0
In Exercises 19-22, given x1 show that the values xn generated by Newton’s Method converges to a zero of the given
function f . Then approximate x4 to the nearest thousandth. Apply Theorem 3.11 and see Examples 3 and 4.
19.
x1 = 3, f (x) = x2 − 10
20.
x1 = −1, f (x) = x3 + 5x + 10
21.
x1 = 0.35, f (x) = cos(πx2 )
22.
x1 = 7, f (x) = x2 sin(x) − 2x
Theory and Proofs
23. Let f (x) = x3 + x + 1 and let x1 be any real number. Prove lim f (xn ) = 0 where xn+1 = xn −
n→∞
f (xn )
.
f � (xn )
24. Cardano’s Formulas Let x3 + px + q = 0 where p > 0.
a) Verify the identity: (a − b)3 + 3ab(a − b) − (a3 − b3 ) = 0
��
��
b) If a =
3
q2
4
+
p3
27
−
q
2
and b =
3
q2
4
+
p3
27
+ 2q , prove 3ab = p and a3 − b3 = −q
c) Using the notation in part b), show x = a − b is the only real solution of x3 + px + q = 0.
b
25. Let f (x) = ax2 + bx + c, a > 0, and b2 − 4ac ≥ 0. If x1 �= − 2a
and xn+1 = xn −
26. Let f (c) = 0 where a < c < b. Suppose for some α we have
�
�
� f (x)f �� (x) �
�
�
� (f � (x))2 � ≤ α < 1
for all x in (2a − b, 2b − a). If x1 lies in [a, b] and xn+1 = xn −
f (xn )
,
f � (xn )
f (xn )
,
f � (xn )
prove lim f (xn ) = 0.
prove lim xn = c.
n→∞
n→∞
3.9. LINEAR APPROXIMATIONS AND DIFFERENTIALS
3.9
213
Linear Approximations and Differentials
• Linear Approximations • Differentials • Approximating a Functional Value
• Propagated Error
Linear Aproximations
If f is differentiable at c, the tangent line to the graph of f at (c, f (c)) is
L(x) = f � (c)(x − c) + f (c).
(8)
The linear function in (8) is called the linear approximation, linearization, or tangent
line approximation of f at c. Clearly, L(x) → f (c) as x → c. Also, f (x) → f (c) as
x → c by continuity. Combining, we obtain
f (x) ≈ f � (c)(x − c) + f (c).
(9)
whenever x is near c.
Example 1 Finding a Linear Approximation
Find the tangent line approximation of f (x) = sin x at (0, 0).
Then approximate the values of sin(0.1) and sin(0.05).
Solution Note, f � (x) = cos x and f � (0) = cos 0 = 1. Using (8) with c = 0, the tangent
line to f (x) = sin x at (0, 0) is given by
L(x)
=
=
L(x)
=
y
y�x
0.5
f � (0)(x − 0) + f (0)
1(x − 0) + 0
f �x�� sin �x�
0.5
x.
x
Thus, the tangent line approximation to f (x) = sin x at (0, 0) is
sin x ≈ x provided x is near 0.
Hence, sin(0.1) ≈ 0.1 and sin(0.05) ≈ 0.05. These are good approximations for sin(0.1) ≈
0.09983 and sin(0.05) ≈ 0.04998 by using a calculator.
✷
Try This 1
Apply (8) to find the linearization of f (x) = tan x at (0, 0).
π
Then apply (9) to approximate the value of tan( 24
) using c = 0 and x =
Differentials
Definition 11 The Differential of a Function
Let y = f (x) be differentiable on an open interval containing c.
a) The differential dx of x is an independent variable.
b) The differential dy of y is a function of x and dx, and defined by
dy = f � (x)dx.
π
.
24
Figure 1
The tangent line y = x
approximates f (x) = sin x
near x = 0.
214
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Also, the independent variable dx is denoted by �x and is called the change in x. In
the definition of f � (x) in Section 2.1 we encountered the expression
�y = f (x + �x) − f (x)
We call �y the change in y. If �x = x − c, then the linearization in (9) may be rewritten
as follows:
f � (c)(x − c) + f (c)
≈
f (x)
f (c + �x)
f (c + �x) − f (c)
for x near c
f � (c)�x + f (c)
≈
�
≈
f (c)�x
for �x ≈ 0
Since c is an arbitrary number, we may replace c by x:
f (x + �x) − f (x)
f � (x)�x
≈
�y
≈
for �x ≈ 0
dy
(10)
In Figure 2, we see a (dotted) tangent line at x to the graph of f .
y
Tangent line at x
y� f �x�
�y
dy
x
x��x
x
Figure 2: �y ≈ dy if �x ≈ 0.
The quantity �y represents a change in the functional values of f . While dy represents
a change in the functional values of the tangent line.
Example 2 Comparing �y With dy
If y =
y
�1
Let f (x) = (1 + x)1/2 . By the Chain Rule we find
1
f � (x) = √
.
2 1+x
f �x�� 1 � x
3
1 + x, evaluate dy and �y if x = 3 and �x = 0.1.
Solution
Tangent line at 3
2
√
Then we obtain
dx
dy = f � (x)dx = √
2 1+x
x
and
�y
f (x + �x) − f (x)
=
�
=
Figure 3
dy is the change in the values
of the tangent line, and �y
is the change in the values of
y = f (x).
1 + x + �x −
√
1 + x.
Substitute x = 3 and dx = �x = 0.1. Then we find
0.1
dy = √ = 0.025
2 4
and
�y =
√
4.1 −
√
4 ≈ 0.0248.
Finally, note dy ≈ �y since dx = 0.1 is approximately zero. In Figure 3, we see
the tangent line at x = 3.
✷
3.9. LINEAR APPROXIMATIONS AND DIFFERENTIALS
215
Try This 2
1
If y = √ , evaluate dy and �y given x = 1 and �x = 0.1.
x
Approximating a Functional Value
Differentials may be used to approximate the values of a function. We rewrite (10) as
follows:
f (x + �x) ≈ f (x) + f � (x)�x
(11)
where �x ≈ 0. The above approximation shows how to estimate f (x + �x) if we know
the values of f (x), f � (x), and �x.
Example 3 Approximating a Square Root
Approximate
√
10 by applying (11).
√
In (11), let f (x) = x, x = 9, and �x = 1. Then
√
10 = f (10) = f (9) + f � (9)�x
1
1
= 3 + (1)
for f � (x) = √
6
2 x
Solution
Thus,
√
19
= 2.375
8
The latter decimal is obtained by using a calculator.
10 ≈
✷
Try This 3
Estimate
√
15 by applying (11) with x = 16 and �x = −1.
Example 4 Approximating a Functional Value
Approximate tan 47◦ by applying (11).
Solution
We recall
tan(45◦ ) = tan
�π�
4
y
= 1 and 2◦ = 2
� π �
π
=
.
180
90
π
Let x = π4 , dx = 90
, and f (x) = tan x. Then f � (x) = sec2 x .
Applying (11) we find
�π
�π�
�π� π
π�
f
+
≈ f
+ f�
4
90
4
4 90
�π�
�π� π
= tan
+ sec2
4
4 90
�π�
= 1+2
90
�π
�
π
π
f
+
= 1+
4
90
45
�
�
π
Since tan(47◦ ) = f π4 + 90
, we obtain
π
tan(47◦ ) ≈ 1 +
≈ 1.0698
45
The latter decimal is obtained by using a calculator.
1
f �x��tan �x�
Π
4
x
Figure 4
The linear approximation
at π4 underestimates
f (x) = tan x since the
tangent line lies below
the graph of f locally.
✷
216
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Try This 4
Estimate the value of sin 29◦ by applying (11) with x =
π
6
π
and �x = − 180
.
Propagated Error
Many scientific experiments involve approximations of position, rate, etc. Suppose x is an
approximation, and x + �x is the true value of a quantity. We say �x is the error in the
approximation.
Let y = f (x) be a function of x. The propagated error
�
� is defined as |�f | =
�
�
|f (x + �x) − f (x)|, the relative error in f (x) is given by � f�f
and the percentage
(x) �,
�
�
� �f �
relative error in f (x) is � f (x) � · 100%.
Example 5 Approximating a Propagated Error
The approximate radius of a sphere is 6 in. with an error of at most ±0.05 in. of the true
radius. Use differentials to approximate the propagated error and the percentage relative
error in calculating the surface area of the sphere.
Solution
The surface area of a sphere of radius x is
f (x) = 4πx2 .
If the true radius is 6 + �x, then the propagated error is
|�f | = |f (6 + �x) − f (6)|
Figure 5
A sphere with a
true radius of 6 + �x.
where −0.05 ≤ �x ≤ 0.05.
Note f � (x) = 8πx. Applying the differential in (10), we obtain
�
�
|f (6 + �x) − f (6)| ≈ �f � (6)�x�
=
≈
≈
|48π�x|
48π(0.05)
7.5 sq. in.
Thus, the propagated error is |�f | ≈ 7.5 sq. in. Finally, the percentage
relative error is
�
�
� �f �
48π(0.05)
�
�
� f (6) � 100 ≈ 144π · 100% ≈ 1.7%.
✷
Try This 5
A rectangular box has a square base with an exact area of 16 sq. in. The height of the
box is 6 in. with an error of at most ±0.1 in. Find the propagated error and percentage
relative error in the volume of the box.
3.9. LINEAR APPROXIMATIONS AND DIFFERENTIALS
217
3.9 Check-It Out
1. Find the linear approximation of f (x) =
2. Let y = tan x. Evaluate dy if x =
π
4
√
x at (4, 2). Then use it to estimate
√
4.1.
and dx = 0.005.
1
3. The side of a cube is measured to be 12 inches with an error of at most ± 16
inch. Use differentials to find the
propagated error in the volume of the cube.
True or False. If false, explain or show an example that shows it is false.
1. If L(x) is the linear approximation of y = f (x) at x = c,
then L(c) = f (c) and L� (c) = f � (c).
2. f (c + �x) ≈ f (c) + f � (c)�x whenever �x ≈ 0
3. �f = f (x + �x) + f (x)
�
�
4. If y = f (x), then dy = f � (x)dx.
√
√
5. If x > 0 and �x ≈ 0, then x + �x ≈ x +
�x
√ .
2 x
6. For any x, (x + �x)5 − x5 ≈ (�x)5 if �x ≈ 0
7. If x is the measurement
� of the side of� a cube with a possible error of �x, then the propagated error in the
volume of the cube is �(x + �x)3 − x3 �.
8. If L(x) is the linear approximation of y = f (x) at x = c, then f (x) ≈ L(x) for values of x near c.
9. The linear approximation of f (x) = mx + b at x = 0 is L(x) = mx.
10. The tangent line approximation of f (x) = cos(x +
π
)
2
at x = 0 is L(x) = −x.
Exercises for Section 3.9
In Exercises 1-10, find the linear approximation of f (x) at the given point. Then use it to estimate the indicated
number. See Example 1.
1.
f (x) = x3 , (2, 8); estimate (2.1)3
3.
f (x) =
5.
f (x) = tan x, ( π4 , 1); estimate tan( π4 + 0.1)
7.
f (x) =
9.
f (x) = cos x, ( π3 , 12 ); estimate cos(61◦ )
√
x, (289, 17); estimate
√
3
√
8 − x, (0, 2); estimate
4.
1
1
, (2, 14 ); estimate (1.9)
2
x2
√
√
f (x) = 3 x, (64, 4); estimate 3 70
6.
f (x) = sec x, ( π6 ,
8.
f (x) = √
2.
285
√
3
7
10.
f (x) =
√2 );
3
estimate sec( π6 + 0.2)
2
, (0, 12 ); estimate
x + 16
f (x) = sin x, ( π4 ,
√
2
);
2
√2
17
estimate sin(44◦ )
In Exercises 11-16, evaluate �y and dy for the given function and constants x and �x = dx. See Example 2.
11.
y = x2 , x = 5, �x = 0.1
13.
y = cot x, x =
15.
y = |x|3 , x = −2, �x = 0.01
π
,
4
dx =
π
180
12.
1
y = √ , x = 4, dx =
x
14.
y = csc x, x =
16.
y=
1
x
π
,
6
1
10
�x =
π
90
− [[ x1 ]], x = 34 , �x = 0.01
In Exercises 17-22, use differentials to approximate the following. See Examples 3 and 4.
17.
√
20.
sec 59.5◦
26
18.
√
3
21.
(2.01)5
7
19.
sin 121◦
22.
(27.5)2/3
218
CHAPTER 3. APPLICATIONS OF DERIVATIVES
23. The side of a square is measured to be 12 inches with a possible error of at most 0.125 inches. Use differentials
to approximate the propagated error in the area of the square. Approximate to the nearest square inch.
24. The radius of a circle is measured to be 2 in. with a possible error of at most 0.0625 in. Use differentials to
approximate the propagated error in the area of the circle. Approximate to the nearest thousandth of a sq. in.
25. A sphere has a radius that is measured to be 20 in. with an error of at most 0.01 in. Use differentials to
approximate the propagated error in the volume of the cube. Round to the nearest cubic inch.
26. A cube has six faces. If an edge is measured to be 5 inches with an error of at most 0.05 inch, use differentials to
approximate the propagated error in the surface area of the cube. Round to the nearest hundredth of a square
inch.
27. A projectile is fired at an angle of elevation of 60◦ with an error of at most ±1◦ . Let v0 = 800 ft/sec be the initial
velocity of the firing. Use differentials to approximate the propagated change in the range of the projectile.
v2
Assume the range is R = 320 sin(2θ) where θ is the angle of elevation. Round to the nearest foot.
28. A circular cylinder has a height of 6 inches. The radius is 2 inches with a possible error of �x. If the propagated
error in the volume is 1 cubic inch, use differentials to approximate �x. Round to the nearest thousandth of
an inch.
29. A sphere has a radius of 5 inches with a possible error of �r. If the propagated error in the surface area of the
sphere is 10 square inches, use differentials to approximate �r. Round to the nearest hundredth of an inch.
30. A homeowner is standing 10 ft from a tree. She knows that the true angle of elevation θ of the top of the tree
is between 45◦ and 60◦ . She likes to approximate the height of the tree to the nearest foot. Use differentials
to estimate the maximum error �θ for measuring the true angle of elevation? Round to the nearest tenth of a
degree.
Chapter 3 Multiple Choice Test
Choose the best answer.
1. If f (c) = 10 and f � (c) does not exist, then
A. c is a critical number of f
C. f (c) is a relative extremum of f
2. The graph y =
A. x = −2
B. x = c is an asymptote of the graph of f
D. The point (c, f (c)) is an inflection point of the graph of f
2x
has an asymptote, namely
x+2
B. y = 1
C. x = 0
D. None of the above
3. The function f (x) = x2 increases on the interval
A. (−∞, ∞)
B. (−∞, 0)
C. (0, ∞)
D. Nowhere
3
4. The value of the second derivative of f (x) = x at x = −1 is
A. −3
B. 3
C. −6
D. 6
2x2
equals
+3
A. ∞
B. 2
5. lim
x→∞ x2
3x3 − 2
6. lim
equals
x→−∞ 4x3 + 1
C. −2
D. 3
3
D. −2
4
7. The horizontal tangent line to the graph of y = x3 − 1 passes through the point
A. (0, 0)
B. (0, −1)
C. (0, 1)
D. (1, 1)
A. ∞
B. −∞
C.
1
is
x
B. Decreasing
8. In (−∞, 0) ∪ (0, ∞), the function y =
A. Increasing
√
3
C. Constant
9. The graph of y = x has a point of inflection at x equals
A. −1
B. 1
C. 0
D. ∞
D. Increasing and decreasing
3.9. LINEAR APPROXIMATIONS AND DIFFERENTIALS
10. If f (x) =
A. 0
219
x−2
, then f can be made continuous everywhere by defining f (2) equal to
2−x
B. 1
C. −1
D. 2
11. The graph of y = x(x2 − 1) is concave downward on the interval
A. (−∞, 0)
B. (0, 1)
C. (0, ∞)
D. (−1, 1)
dy
π
12. If y = sec x(1 + sin x) then
at x =
equals
dx
4
√
A. 0
B. 1 + 2
C. 2
√
2
dy
13. If xy + y 2 = 1 defines y as an implicit function of x, then
equals
dx
−1
−2y
−y
A.
B.
C.
D. None of the above
x + 2y
x
x + 2y
14. If y = u1/3 and u = tan x, then
A. 1
B.
√
2
C.
2
3
dy
π
at x =
equals
dx
4
D. ∞
15. If f (x) = x|x|, then f � (x) equals
A. 2x
B. −2x
C. 2|x|
16. If f (x) = x7 , then lim
x→1
A. 1
B. 7
D. 2 +
D. −2|x|
f (x) − f (1)
equals
x−1
C. −1
D. Does not exist
17. The slope of the curve xy 2 − x − 3 = 0 at the point (1, 2) is
3
3
A. 0
B. −1
C. −
D.
4
4
�π
�
cos 2 + h
18. lim
equals
h→0
h
A. −1
B. 1
C. 0
D. Does not exist
19. The graphs of 2x + y = 1 and x − 2y = 1 are
A. Parallel
B. Perpendicular
20.
C. Coincident
D. None of the above
The shortest distance between the graph of y = 12 − x2 and the origin is
√
√
√
√
47
23
A. 47
B.
C. 23
D.
2
2
Investigation Projects
Inventory: Delivery in One Batch per Order
We store goods in an inventory in order to meet a demand. If a company has a large inventory, the cost of holding the
goods is high. However, with a low inventory the demand may not be satisfied, customers can go to the competition,
and there is a loss in revenue.
Leonardo, Inc., is a bookstore with an annual demand of D units for a certain children’s book. The bookstore submits
several orders for the book in a year to a third party publisher. The size of an order is Q units. Given an order the
publisher delivers the books to the bookstore in one batch.
Then the ratio D/Q represents the number of orders. Let Co be the cost to the bookstore for preparing an order.
The annual ordering cost is
O1 (Q)
=
(Number of orders per year) × (Ordering cost)
=
D
× Co
Q
220
CHAPTER 3. APPLICATIONS OF DERIVATIVES
Since the inventory level fluctuates due to the demand, we assume the average annual inventory is one-half the highest
inventory. That is, the average annual inventory is Q/2 units. Let Ch denote the annual holding cost per unit. Then
the annual holding cost is
Q
× Ch
2
To minimize the annual inventory cost we do not consider the cost of the books since the demand is constant. The
annual inventory cost is
H1 (Q)
(Average inventory ) × (Holding cost) =
=
I1 (Q)
O1 (Q) + H1 (Q)
=
Let
be the order quantity that minimizes I1 (Q). Note, Q∗1 is a critical number of I(Q). In Management Science,
∗
Q1 is called the Economic Order Quantity.
Q∗1
Exercises
�
2DCo
Ch
3. For the bookstore Leonardo, Inc., suppose the annual demand is 10, 000 units, ordering cost is $50 per order,
and holding cost is $0.25 per book/year. Determine the economic order quantity Q∗1 .
1. Prove O1 (Q∗1 ) = H1 (Q∗1 ), i.e., annual ordering cost equals holding cost.
2.
Show that Q∗1 =
The Production Run Model
A production run will be a period when goods are produced and sold simultaneously. There is a setup cost Cs
associated to a production run. The setup cost replaces the ordering cost in the above inventory model.
Let Q be the number of units produced during a production run. Denote the production rate by p units/day, and
the demand rate by d units/day. If t is the length of a production run, then Q = pt. We declare the variables:
(Annual demand)
=
D units
(Number of units per production run)
=
Q units
(Setup cost)
=
Cs per production run
(Carrying cost)
=
Ch per unit per year
(Daily production rate)
=
p units/day
(Daily demand rate)
=
d units/day
(Length of a production run)
=
t days
Then the annual setup cost is
Sp (Q)
=
(Number of production runs) × Cs
=
D
× Cs
Q
Similarly, the annual holding cost depends on one-half of the maximum inventory. The annual holding is calculated
as follows:
Hp (Q)
=
=
(Maximum inventory)
pt − dt
× Ch =
× Ch
2
2
�
�
Q
d
Q
1−
× Ch
since t =
2
p
p
The annual inventory cost is
Ip (Q) = Sp (Q) + Hp (Q).
Let Q∗p be the optimal value of Q which minimizes Ip (Q).
Exercises
∗
∗
4. Prove Sp (Q ) = Hp (Q )
5. Show that
Q∗p
�
�
=�
�
�
2DC
� s �.
d
Ch 1 −
p