Transcript Slide 1
5
Further Applications of the Derivative
Copyright © Cengage Learning. All rights reserved.
5.1
Maxima and Minima
Copyright © Cengage Learning. All rights reserved.
Maxima and Minima
Figure 1 shows the graph of a function
f
whose domain is the closed interval [
a
,
b
].
Figure 1
A mathematician sees lots of interesting things going on here. There are hills and valleys, and even a small chasm (called a
cusp
) near the center. For many purposes, the important features of this curve are the highs and lows.
3
Maxima and Minima
Figure 2 shows the graph once again with the highs and lows marked.
Figure 2
Mathematicians have names for these points: the highs (at the
x
-values
p
,
r
, and
b
) are referred to as
relative maxima,
and the lows (at the
x
-values
a
,
q
, and
s
) are referred to as
relative minima
. 4
Maxima and Minima
Collectively, these highs and lows are referred to as
relative extrema
. (A point of language: The singular forms of the plurals
minima, maxima,
and
extrema
are
minimum, maximum,
and
extremum
.) Why do we refer to these points as relative extrema? Take a look at the point corresponding to
x
=
r
. It is the highest point of the graph
compared to other points nearby
.
If you were an extremely nearsighted mountaineer standing at the point where
x
=
r
, you would
think
that you were at the highest point of the graph, not being able to see the distant peaks at
x
=
p
and
x
=
b
.
5
Maxima and Minima
Let’s translate into mathematical terms. We are talking about the heights of various points on the curve. The height of the curve at
x
=
r
is
f
(
r
), so we are saying that
f
(
r
) is greater than or equal to
f
(
x
) for every
x
near
r
.
In other words,
f
(
r
)
is the greatest value that f
(
x
)
has for all choices of x between r
–
h and r
+
h
for some (possibly small)
h
. (See Figure 3.)
Figure 3
6
Maxima and Minima
We can phrase the formal definition as follows.
Relative Extrema
f
has a
relative maximum
at
x
=
r
if there is some interval (
r
–
h
,
r
+
h
) (even a very small one) for which
f
(
r
) ≥
f
(
x
) for all
x
in (
r
–
h
,
r
+
h
) for which
f
(
x
) is defined.
f
has a
relative minimum
at
x
=
r
if there is some interval (
r
–
h
,
r
+
h
) (even a very small one) for which
f
(
r
) ≤
f
(
x
) for all
x
in (
r
–
h
,
r
+
h
) for which
f
(
x
) is defined.
7
Maxima and Minima
Quick Example In Figure 2,
f
has the following relative extrema: Relative maxima at
p
and
r
.
Looking carefully at Figure 2, we can see that the lowest point on the whole graph is where
x
=
s
and the highest point is where
x
=
b
.
Figure 2
This means that
f
(
s
) is the least value of
f
on the whole domain of
f
(the interval [
a
,
b
]) and
f
(
b
) is the greatest value. We call these the
absolute
minimum and maximum.
8
Maxima and Minima
Absolute Extrema
f
has an
absolute maximum
at
x
=
r
if
f
(
r
) ≥
f
(
x
) for every
x
in the domain of
f
.
f
has an
absolute minimum
at
x
=
r
if
f
(
r
) ≤
f
(
x
) for every
x
in the domain of
f
.
Quick Example In Figure 2,
f
has an absolute maximum at
b
and an absolute minimum at
s
.
9
Maxima and Minima
Some graphs have no absolute extrema at all (think of the graph of
y
=
x
), while others might have an absolute minimum but no absolute maximum (like
y
=
x
2 ), or vice versa.
When
f
does have an absolute maximum, there is only one absolute maximum
value
of
f
, but this value may occur at different values of
x
, and similarly for absolute minima. (See Figure 6.) Absolute maxima at
x
=
a
and
x
=
b
Figure 6
10
Maxima and Minima
In Figure 7 we see the graph from Figure 1 once more, but we have labeled each extreme point as one of three types.
Figure 1 Figure 7
Notice that two extrema occur at endpoints and the others at
interior points;
that is, points other than endpoints. At the points labeled “Stationary,” the tangent lines to the graph are horizontal, and so have slope 0, so
f
(which gives the slope) is 0. 11
Maxima and Minima
Any time
f
(
x
) = 0, we say that
f
has a
stationary point
at
x
because the rate of change of
f
is zero there.
We call an extremum that occurs at a stationary point a
stationary extremum
. In general, to find the exact location of each stationary point, we need to solve the equation
f
(
x
) = 0.
There is a relative minimum in Figure 7 at
x
=
q
, but there is no horizontal tangent there. In fact, there is no tangent line at all;
f
(
q
) is not defined. (We know a similar situation with the graph of
f
(
x
) = |
x
| at
x
= 0.) 12
Maxima and Minima
When
f
(
x
) does not exist for some interior point
x
in the domain of
f
, we say that
f
has a
singular point
at
x
.
We shall call an extremum that occurs at a singular point a
singular extremum
. The points that are either stationary or singular we call collectively the
critical points
of
f
.
The remaining two extrema are at the
endpoints
of the domain.
13
Maxima and Minima
As we see in the figure, they are (almost) always either relative maxima or relative minima.
We bring all the information together in Figure 8:
Figure 8
14
Maxima and Minima
Locating Candidates for Extrema If
f
is a real-valued function, then its extrema occur among the following types of points: 1. Stationary Points: f has a stationary point at
x
if
x
is in the interior of the domain and
f
(
x
) = 0. To locate stationary points, set
f
(
x
) = 0 and solve for
x
.
2. Singular Points: f has a singular point at
x
if
x
is in the interior of the domain and
f
(
x
) is not defined. To locate singular points, find values of
x
where
f
(
x
) is
not
defined, but
f
(
x
)
is
defined.
15
Maxima and Minima
3. Endpoints:
These are the endpoints, if any, of the domain. We know that closed intervals contain endpoints, but open intervals do not. If the domain of
f
is an open interval or the whole real line, then there are no endpoints.
Once we have a candidate for an extremum of
f
, we find the corresponding point (
x
,
y
) on the graph of
f
using
y
=
f
(
x
).
16
Maxima and Minima
Quick Examples
1. Stationary Points:
Let
f
(
x
) =
x
3 – 12
x
. Then to locate the stationary points, set
f
(
x
) = 0 and solve for
x
. This gives 3
x
2 – 12 = 0, so
f
has stationary points at
x
= ±2. The corresponding points on the graph are ( –2,
f
( –2)) = (–2, 16) and (2,
f
(2)) = (2, –16).
2. Singular Points:
Let
f
(
x
) = 3(
x
– 1) 1
/
3 . Then
f
(
x
) = (
x
– 1) −2
/
3 = 1
/
(
x
– 1) 2
/
3 .
f
(1) is not defined, although
f
(1)
is
defined. Thus, the (only) singular point occurs at
x
= 1. The corresponding point on the graph is (1,
f
(1)) = (1, 0).
17
Maxima and Minima
3. Endpoints:
Let
f
(
x
) = 1/
x
, with domain ( , 0) [1, ). The corresponding point on the graph is (1, 1).The natural domain of 1/
x
, on the other hand, has no endpoints.
18
Example 1 –
Maxima and Minima
Find the relative and absolute maxima and minima of
f
(
x
) =
x
2 – 2
x
on the interval [0, 4].
Solution: We first calculate
f
(
x
) = 2
x
– 2. We use this derivative to locate the critical points (stationary and singular points).
Stationary Points:
To locate the stationary points, we solve the equation
f
(
x
) = 0, or 2
x
– 2 = 0, getting
x
= 1. The domain of the function is [0, 4], so
x
= 1 is in the interior of the domain. Thus, the only candidate for a stationary relative extremum occurs when
x
= 1.
19
Example 1 –
Solution
cont’d
Singular Points:
We look for interior points where the derivative is not defined. However, the derivative is 2
x
– 2, which is defined for every
x
. Thus, there are no singular points and hence no candidates for singular relative extrema.
Endpoints:
The domain is [0, 4], so the endpoints occur when
x
= 0 and
x
= 4.
We record these values of
x
in a table, together with the corresponding
y
-coordinates (values of
f
): 20
Example 1 –
Solution
cont’d This gives us three points on the graph, (0, 0), (1, –1), and (4, 8), which we plot in Figure 9.
We remind ourselves that the point (1, –1) is a stationary point of the graph by drawing in a part of the horizontal tangent line.
Figure 9
21
Example 1 –
Solution
Connecting these points must give us a graph something like that in Figure 10.
cont’d
Figure 10
22
Example 1 –
Solution
From Figure 10 we can see that
f
has the following extrema: cont’d 23
First Derivative Test
24
First Derivative Test
The
first derivative test
gives another, very systematic, way of checking whether a critical point is a maximum or minimum.
To motivate the first derivative test, consider again the critical point
x
= 1.
If we look at some values of
f
(
x
) to the left and right of the critical point, we obtain the information shown in the following table: 25
First Derivative Test
At
x
= 0.5 (to the left of the critical point) we see that
f
(0.5) = –1
<
0, so the graph has negative slope and
f
is decreasing. We note this with the downward pointing arrow. At
x
= 2 (to the right of the critical point), we find
f
(2) = 2
>
0, so the graph has positive slope and
f
is increasing.
In fact, because
f
(
x
) = 0 only at
x
= 1, we know that
f
(
x
)
<
0 for all
x
in (0, 1), and we can say that
f
is decreasing on the interval (0, 1).
Similarly,
f
is increasing on (1, 4).
26
First Derivative Test
So, starting at
x
= 0, the graph of
f
goes down until we reach
x
= 1 and then it goes back up, telling us that
x
= 1 must be a minimum. Notice how the minimum is suggested by the arrows to the left and right.
27
First Derivative Test
First Derivative Test for Extrema Suppose that
c
is a critical point of the continuous function
f
, and that its derivative is defined for
x
close to, and on both sides of,
x
=
c
. Then, determine the sign of the derivative to the left and right of
x
=
c
.
1.
If
f
(
x
) is positive to the left of
x
=
c
and negative to the right, then
f
has a maximum at
x
=
c
.
2.
If
f
(
x
) is negative to the left of
x
=
c
and positive to the right, then
f
has a minimum at
x
=
c
.
3.
If
f
(
x
) has the same sign on both sides of
x
=
c
, then
f
has neither a maximum nor a minimum at
x
=
c
.
28
First Derivative Test
Quick Example Here is a graph showing a function
f
with a singular point at
x
= 1: The graph gives us the information shown in the table: 29
First Derivative Test
Since
f
(
x
) is positive to the left of
x
= 1 and negative to the right, we see that
f
has a maximum at
x
= 1. (Notice again how this is suggested by the direction of the arrows.) 30
Example 2 –
Unbounded Interval
Find all extrema of
f
(
x
) = 3
x
4 – 4
x
3 on [ –1, ).
Solution: We first calculate
f
(
x
) = 12
x
3 – 12
x
2 .
Stationary points:
We solve the equation
f
(
x
) = 0, which is 12
x
3 – 12
x
2 = 0 or 12
x
2 (
x
– 1) = 0.
There are two solutions,
x
= 0 and
x
= 1, and both are in the domain. These are our candidates for the
x
-coordinates of stationary extrema.
31
Example 2 –
Solution
cont’d
Singular points:
There are no points where
f
(
x
) is not defined, so there are no singular points.
Endpoints:
The domain is [ –1, ), so there is one endpoint, at
x
= –1.
We record these points in a table with the corresponding
y
-coordinates: 32
Example 2 –
Solution
We will illustrate two methods we can use to determine which are minima, which are maxima, and which are neither: cont’d
1.
Plot these points and sketch the graph by hand.
2.
Use the First Derivative Test.
Use the method you find most convenient.
33
Example 2 –
Solution
Using a Hand Plot :
If we plot these points by hand, we obtain Figure 12(a), which suggests Figure 12(b).
cont’d
Figure 12
We can’t be sure what happens to the right of
x
= 1. Does the curve go up, or does it go down?
34
Example 2 –
Solution
cont’d To find out, let’s plot a “test point” to the right of
x
= 1. Choosing
x
= 2, we obtain
y
= 3(2) 4 – 4(2) 3 = 16, so (2, 16) is another point on the graph. Thus, it must turn upward to the right of
x
= 1, as shown in Figure 13.
From the graph, we find that
f
has the following extrema: A relative (endpoint) maximum at ( –1, 7) An absolute (stationary) minimum at (1, –1)
Figure 13
35
Example 2 –
Solution
cont’d
Using the First Derivative Test :
List the critical and endpoints in a table, and add additional points as necessary so that each critical point has a noncritical point on either side.
Then compute the derivative at each of these points, and draw an arrow to indicate the direction of the graph.
Notice that the arrows now suggest the shape of the curve in Figure 13.
36
Example 2 –
Solution
The first derivative test tells us that the function has a relative maximum at
x
= –1, neither a maximum nor a minimum at
x
= 0, and a relative minimum at
x
= 1. cont’d Deciding which of these extrema are absolute and which are relative requires us to compute
y
-coordinates and plot the corresponding points on the graph by hand, as we did in the first method.
37
First Derivative Test
Extreme Value Theorem If
f
is
continuous
on a
closed interval
[
a
,
b
], then it will have an absolute maximum and an absolute minimum value on that interval. Each absolute extremum must occur at either an endpoint or a critical point. Therefore, the absolute maximum is the largest value in a table of the values of
f
at the endpoints and critical points, and the absolute minimum is the smallest value.
38
First Derivative Test
Quick Example The function
f
(
x
) = 3
x
–
x
3 on the interval [0, 2] has one critical point at
x
= 1. The values of
f
at the critical point and the endpoints of the interval are given in the following table: From this table we can say that the absolute maximum value of
f
on [0, 2] is 2, which occurs at
x
= 1, and the absolute minimum value of
f
is –2, which occurs at
x
= 2.
39