WEBVTT
1
00:00:00.000 --> 00:00:07.000
Concavity and second derivative test.
2
00:00:07.000 --> 00:00:10.000
Concavity.
3
00:00:10.000 --> 00:00:15.000
Let F be differentiable at each x and an open interval i.
4
00:00:15.000 --> 00:00:19.000
The graph of F is concave up on i, if F prime,
5
00:00:19.000 --> 00:00:24.000
that is the slope of the graph, is increasing on i.
6
00:00:24.000 --> 00:00:30.000
The graph of F is concave down on i, if F prime is decreasing on i.
7
00:00:30.000 --> 00:00:33.000
Let's look at a couple of pictures that illustrate these definitions.
8
00:00:33.000 --> 00:00:36.000
Here the slope is increasing.
9
00:00:36.000 --> 00:00:39.000
Here the left end of the graph, there's a large negative slope,
10
00:00:39.000 --> 00:00:42.000
which becomes smaller as we move to the right,
11
00:00:42.000 --> 00:00:47.000
eventually becoming positive.
12
00:00:47.000 --> 00:00:53.000
Increasing slope results in a graph that is concave up or convex.
13
00:00:53.000 --> 00:00:55.000
Here the slope is decreasing.
14
00:00:55.000 --> 00:00:58.000
Here the left end of the graph, there's a large positive slope,
15
00:00:58.000 --> 00:01:01.000
which becomes smaller as we move to the right,
16
00:01:01.000 --> 00:01:06.000
eventually becoming negative.
17
00:01:06.000 --> 00:01:12.000
Increasing slope results in a graph that is concave down or simply concave.
18
00:01:12.000 --> 00:01:15.000
Even though both of these pictures indicate a local extreme value,
19
00:01:15.000 --> 00:01:18.000
we point out that that need not be the case.
20
00:01:18.000 --> 00:01:22.000
Here's the graph that is concave up, where the function is always decreasing.
21
00:01:22.000 --> 00:01:28.000
Here's a graph that is concave up, where the function is always increasing.
22
00:01:28.000 --> 00:01:33.000
This graph is concave down, and the function is always increasing.
23
00:01:33.000 --> 00:01:40.000
And this graph is concave down, while the function is always decreasing.
24
00:01:40.000 --> 00:01:46.000
A slopeless interpretation of concavity.
25
00:01:46.000 --> 00:01:50.000
Let i be an open interval in the domain of F.
26
00:01:50.000 --> 00:01:55.000
If the graph of F is concave up on i,
27
00:01:55.000 --> 00:02:01.000
then f of x is less than f of x plus h plus f of x minus h over 2,
28
00:02:01.000 --> 00:02:05.000
whenever x plus and minus h are in i.
29
00:02:05.000 --> 00:02:14.000
If the graph is concave down on i, then f of x is greater than f of x plus h plus f of x minus h over 2,
30
00:02:14.000 --> 00:02:18.000
whenever x plus and minus h are in i.
31
00:02:18.000 --> 00:02:21.000
It basically says that if the graph is concave up,
32
00:02:21.000 --> 00:02:29.000
given a closed sub-interval of i, the value of f at the midpoint is always less than the average of the endpoint values of f.
33
00:02:29.000 --> 00:02:36.000
And when the graph is concave down, the value of f at the midpoint is always greater than the average of the endpoint values.
34
00:02:36.000 --> 00:02:39.000
This has a simple geometric interpretation.
35
00:02:39.000 --> 00:02:44.000
In the concave up case, it says that the point x comma f of x on the graph of f
36
00:02:44.000 --> 00:02:51.000
always lies below the point x comma y that lies on the segment joining x comma f of x minus h
37
00:02:51.000 --> 00:02:57.000
and x comma f of x plus h, such as segment is called a chord.
38
00:02:57.000 --> 00:03:03.000
So roughly speaking, the graph is concave up, wherever it lies underneath every chord.
39
00:03:03.000 --> 00:03:08.000
Here's another picture of the Illustrates the idea.
40
00:03:08.000 --> 00:03:13.000
Likewise, when the graph of f is concave down, it lies above every chord.
41
00:03:13.000 --> 00:03:20.000
Now, to see why the first part of this is true,
42
00:03:20.000 --> 00:03:24.000
suppose that f prime is increasing on i.
43
00:03:24.000 --> 00:03:33.000
That is, the graph of f is concave up on i, and let x plus and minus h be in i, where h is positive.
44
00:03:33.000 --> 00:03:45.000
We will show that f of x plus h minus 2 times f of x plus f of x minus h is greater than 0, which is equivalent to the desired inequality.
45
00:03:45.000 --> 00:03:50.000
Now, let's divide the left side of that inequality by a positive h.
46
00:03:50.000 --> 00:03:56.000
And split the fraction into two, and notice that each has the form of a difference quotient of f.
47
00:03:56.000 --> 00:04:01.000
So by the mean value theorem, each of these is equal to some value of f prime.
48
00:04:01.000 --> 00:04:11.000
So we have f prime of t minus f prime of s for some number s between x minus h and x, and some number t between x and x plus h.
49
00:04:11.000 --> 00:04:16.000
This difference is positive, because f prime is increasing.
50
00:04:16.000 --> 00:04:25.000
Therefore, f of x plus h minus 2 times f of x plus f of x minus h is greater than 0.
51
00:04:25.000 --> 00:04:38.000
The analogous statement about downward concavity may be proved by simply reversing inequalities.
52
00:04:38.000 --> 00:04:43.000
Concavity in the second derivative.
53
00:04:43.000 --> 00:04:51.000
Let i be an open interval in the domain of f, and suppose that f double prime of x exists for all x in that interval.
54
00:04:51.000 --> 00:05:01.000
Then, if the graph of f is concave up on i, then f double prime of x is greater than or equal to 0 for all x in i.
55
00:05:01.000 --> 00:05:09.000
If f double prime of x is greater than 0 for all x in i, then the graph of f must be concave up on i.
56
00:05:09.000 --> 00:05:26.000
If the graph of f is concave down on i, then f double prime of x is less than or equal to 0 for all x in i.
57
00:05:26.000 --> 00:05:30.000
For example, consider f of x equals x squared.
58
00:05:30.000 --> 00:05:44.000
If f prime of x is 2x, and f double prime of x is equal to 2, since that's positive, we're guaranteed the graph of f is concave up on any interval.
59
00:05:44.000 --> 00:05:48.000
Now consider f of x equals cosine x.
60
00:05:48.000 --> 00:06:01.000
f prime of x is minus sine x, and f double prime of x is minus cosine x. Now notice that f double prime of x is just the negative of f of x.
61
00:06:01.000 --> 00:06:09.000
So wherever the function is positive, the second derivative is negative, and the graph is concave down.
62
00:06:09.000 --> 00:06:27.000
Now let's look at f of x equals x to the fourth. f prime of x is 4x cubed, and f double prime of x is 12x squared.
63
00:06:27.000 --> 00:06:36.000
Now notice that the graph is indeed concave up, but f double prime of 0 is equal to 0.
64
00:06:36.000 --> 00:06:47.000
This is why we can only conclude that f double prime of x is greater than or equal to 0 when the graph of f is concave up.
65
00:06:47.000 --> 00:06:52.000
Now let's look at f of x equals x to the 8 fifths power.
66
00:06:52.000 --> 00:07:03.000
The first derivative is 8 fifths times x to the 3 fifths, and the second derivative is 24 over 25 times x to the minus 2 fifths power.
67
00:07:03.000 --> 00:07:21.000
The graph of f is always concave up, because f prime of x is always increasing, yet f double prime of 0 does not exist.
68
00:07:21.000 --> 00:07:38.000
Example, let f of x equal 2 times x to the 6th minus 9x to the 5th plus 10x to the 4th. Find the intervals on which the graph of f is concave up, and the intervals on which the graph of f is concave down.
69
00:07:38.000 --> 00:07:49.000
So let's compute derivatives. f prime of x is 12x to the 5th minus 45x to the 4th plus 40x cubed.
70
00:07:49.000 --> 00:08:01.000
So the second derivative is 60x to the 4th minus 180 times x cubed plus 120 times x squared.
71
00:08:01.000 --> 00:08:05.000
We can factor f double prime of x as follows.
72
00:08:05.000 --> 00:08:15.000
It equals 60x squared times x squared minus 3x plus 2, which factors further to give us 60x squared times x minus 1 times x minus 2.
73
00:08:15.000 --> 00:08:25.000
So we notice that f double prime of x is equal to 0 at x equals 0, x equals 1, and x equals 2.
74
00:08:25.000 --> 00:08:40.000
So f double prime of x is greater than 0 when x is less than 0, because x minus 1 and x minus 2 are both negative when x is less than 0.
75
00:08:40.000 --> 00:08:50.000
When x is between 0 and 1, f double prime of x is positive, because x minus 1 and x minus 2 are still both negative.
76
00:08:50.000 --> 00:09:00.000
When x is between 1 and 2, f double prime of x is negative, because x minus 1 is positive while x minus 2 is negative.
77
00:09:00.000 --> 00:09:09.000
And when x is greater than 2, x minus 1 and x minus 2 are both positive and so f double prime of x is positive.
78
00:09:09.000 --> 00:09:25.000
So we can say that the graph of f is concave up on the open intervals minus infinity to 0, 0 to 1, and 2 to infinity, and the graph of f is concave down on the open interval from 1 to 2.
79
00:09:25.000 --> 00:09:32.000
Now let's have a look at the graph of this function.
80
00:09:32.000 --> 00:09:50.000
So the graph is concave down between 1 and 2 and concave up everywhere else. In fact, we can combine the two intervals minus infinity to 0 and 0 to 1, and say that the graph of f is concave up on the interval from minus infinity to 1.
81
00:09:50.000 --> 00:10:09.000
The two points on the graph, where the concavity changes from up to down or down to up, are called inflection points. In this graph, they're the points 1,3 and 2,0.
82
00:10:09.000 --> 00:10:19.000
Critical points in local extrema. Recall the following definition and theorem from a preceding lecture.
83
00:10:19.000 --> 00:10:31.000
A number C in the interior of the domain of f is called a critical number of f if either f prime of C equals 0 or f prime of C does not exist.
84
00:10:31.000 --> 00:10:38.000
If f of C is a local extreme value, then C is a critical number of f.
85
00:10:38.000 --> 00:10:52.000
Now suppose that C is an isolated critical number of f, where f prime of C equals 0. That is, there's an interval surrounding C that contains no other critical numbers of f.
86
00:10:52.000 --> 00:11:08.000
F of C is a local minimum if and only if the graph of f is concave up on an open interval containing C. And F of C is a local maximum if and only if the graph of f is concave down on an open interval containing C.
87
00:11:08.000 --> 00:11:26.000
These facts suggest that concavity may be used as a way of classifying critical numbers. In fact, a simplified version of these ideas takes the form of a test called the second derivative test for classifying critical numbers.
88
00:11:26.000 --> 00:11:42.000
The second derivative test, let C be a critical number of f where f prime of C equals 0 and f double prime of C exists. If f double prime of C is positive, then f of C is a local minimum.
89
00:11:42.000 --> 00:11:59.000
If f double prime of C is less than 0, then f of C is a local maximum. If f double prime of C equals 0, then f of C may be a local minimum, a local maximum, neither, or even both.
90
00:11:59.000 --> 00:12:03.000
That is, the test fails.
91
00:12:03.000 --> 00:12:17.000
For example, consider f of x equals x to the fourth. Here f prime of 0 is 0 and 0 is the only critical number. And f double prime of 0 is also 0.
92
00:12:17.000 --> 00:12:36.000
And the graph of the function is always concave up and f of 0 equals 0 is a local minimum. In fact, it's a global minimum. On the other hand, if f of x equals minus x to the fourth, then again, f prime of 0 equals 0 and 0 is the only critical number.
93
00:12:36.000 --> 00:12:52.000
And f double prime of 0 is also 0. This time, the graph is always concave down. And f of 0 is a local maximum. In fact, a global maximum.
94
00:12:52.000 --> 00:13:06.000
If f of x equals x cubed, then f prime of 0 equals 0. And f double prime of 0 is also 0. But in this case, f of 0 is neither a local maximum nor a local minimum.
95
00:13:06.000 --> 00:13:15.000
Notice that the graph is concave down to the left and concave up to the right. So 0, 0 is an inflection point.
96
00:13:15.000 --> 00:13:29.000
For an example where f of c is both a local maximum and a local minimum, we only need to consider any constant function.
97
00:13:29.000 --> 00:13:43.000
Example, that f of x equals 3x to the fifth minus 5x cubed. Let's find the critical numbers of f and classify the value of f at each one as a local minimum, local maximum or neither.
98
00:13:43.000 --> 00:13:53.000
Let's compute the first derivative and find the critical numbers. Using the power rule, f prime of x is 15x to the fourth minus 15x squared.
99
00:13:53.000 --> 00:14:03.000
Factoring gives us 15x squared times x squared minus 1. And further factoring gives us 15x squared times x plus 1 times x minus 1.
100
00:14:03.000 --> 00:14:19.000
So we conclude that the critical numbers of f are 0, negative 1, and plus 1. Now let's compute the second derivative. Have double prime of x, again using the power rule, is 60x cubed minus 30x.
101
00:14:19.000 --> 00:14:28.000
And factoring gives us 30x times 2x squared minus 1. Now we'll apply the second derivative test to each of the three critical numbers.
102
00:14:28.000 --> 00:14:44.000
Let's start with negative 1. The value of f double prime at negative 1 turns out to be negative 30. And since that's negative, we conclude that f of minus 1, which is 2, is a local maximum value.
103
00:14:44.000 --> 00:14:58.000
Now let's look at the critical number 1. f double prime of 1 is 30. And since that's positive, we conclude that f of 1, which is negative 2, is a local minimum value.
104
00:14:58.000 --> 00:15:06.000
Now let's consider 0. The value of f double prime at 0 is 0, so the second derivative test fails.
105
00:15:06.000 --> 00:15:26.000
So now let's apply the first derivative test. Since f prime of x does not change sine at x equals 0, this is because of the x squared factor in f prime of x, the first derivative test tells us that f of 0, which is 0, is neither a local minimum nor a local maximum.
106
00:15:26.000 --> 00:15:38.000
Let's start by computing the derivative and finding the critical numbers.
107
00:15:38.000 --> 00:15:59.000
f prime of x is 1 plus 2 times cosine x. And that's equal to 0 when cosine x has a value minus 1 half.
108
00:15:59.000 --> 00:16:17.000
So let's think about the unit circle and what angles between 0 and 2 pi give a cosine value of minus 1 half.
109
00:16:17.000 --> 00:16:32.000
Those angles are our critical numbers and they are 2 thirds pi and 4 thirds pi. Now let's compute the second derivative and apply the second derivative test to each of those critical points.
110
00:16:32.000 --> 00:16:40.000
We take the derivative of f prime of x to obtain f double prime of x equal to minus 2 times sine x.
111
00:16:40.000 --> 00:16:53.000
Now let's evaluate f double prime at each of the critical points. f double prime at 2 thirds pi is equal to minus 2 times the sine of 2 thirds pi, which is root 3 over 2.
112
00:16:53.000 --> 00:17:14.000
So f double prime of 2 thirds pi is negative root 3, which is negative. And so we conclude that the value of f at 2 thirds pi, which is 2 thirds pi, plus 2 times the sine of 2 thirds pi or root 3, is a local maximum value.
113
00:17:14.000 --> 00:17:26.000
f double prime at the other critical number, 4 thirds pi is minus 2 times the sine of 4 thirds pi, which is negative root 3 over 2.
114
00:17:26.000 --> 00:17:53.000
So f double prime at 4 thirds pi is equal to positive root 3. And therefore we conclude that the value of f at 4 thirds pi, which is 4 thirds pi, plus 2 times the sine of 4 thirds pi or negative root 3, is a local minimum value.
115
00:17:53.000 --> 00:18:07.000
Now let's take another example. Let f of x equal x plus 1 over x. Find the critical numbers of f and classify the value of f at each 1 as a local minimum, local maximum or neither.
116
00:18:07.000 --> 00:18:20.000
Let's compute the first derivative and find the critical numbers. f prime of x equals 1 plus the derivative of x to the minus 1 power, which is minus x to the minus 2 power.
117
00:18:20.000 --> 00:18:29.000
So f prime of x equals 1 minus 1 over x squared, or x squared minus 1 over x squared.
118
00:18:29.000 --> 00:18:43.000
Therefore we conclude that the critical numbers are plus and minus 1. Let's note that even though f prime of 0 is undefined, it's not a critical number because it's not in the domain of f.
119
00:18:43.000 --> 00:18:58.000
So now let's apply the second derivative test to each of the critical numbers minus 1 and 1. We first compute f double prime of x, which is 2 times x to the minus 3 power, or 2 divided by x cubed.
120
00:18:58.000 --> 00:19:14.000
And evaluating f double prime at minus 1 gives us negative 2. Therefore we conclude that the value of f at minus 1, which is negative 3 halves, is a local maximum.
121
00:19:14.000 --> 00:19:27.000
The value of f double prime at 1 is positive 2, and therefore the value of f at 1, which is positive 3 halves, is a local minimum value.
122
00:19:45.000 --> 00:19:51.000
So now let's look at the value of f at 1, which is negative 3 halves.