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.
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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.
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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.
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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
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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.
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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.
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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.
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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 = x2), 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
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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.
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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.)
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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.
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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
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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.
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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).
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Maxima and Minima
Quick Examples
1. Stationary Points: Let f(x) = x3 – 12x. Then to locate
the stationary points, set f(x) = 0 and solve for x. This
gives 3x2 – 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).
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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.
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Example 1 – Maxima and Minima
Find the relative and absolute maxima and minima of
f(x) = x2 – 2x
on the interval [0, 4].
Solution:
We first calculate f(x) = 2x – 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 2x – 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.
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Example 1 – Solution
cont’d
Singular Points: We look for interior points where the
derivative is not defined. However, the derivative is 2x – 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 ):
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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
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Example 1 – Solution
cont’d
Connecting these points must give us a graph something
like that in Figure 10.
Figure 10
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Example 1 – Solution
cont’d
From Figure 10 we can see that f has the following
extrema:
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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:
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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).
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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.
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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.
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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:
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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.)
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Example 2 – Unbounded Interval
Find all extrema of f(x) = 3x4 – 4x3 on [–1,
).
Solution:
We first calculate f(x) = 12x3 – 12x2.
Stationary points:
We solve the equation f(x) = 0, which is
12x3 – 12x2 = 0 or
12x2(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.
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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:
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Example 2 – Solution
cont’d
We will illustrate two methods we can use to determine
which are minima, which are maxima, and which are
neither:
1. Plot these points and sketch the graph by hand.
2. Use the First Derivative Test.
Use the method you find most convenient.
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Example 2 – Solution
cont’d
Using a Hand Plot :
If we plot these points by hand, we obtain Figure 12(a),
which suggests Figure 12(b).
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?
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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
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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.
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Example 2 – Solution
cont’d
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.
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.
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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.
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First Derivative Test
Quick Example
The function f(x) = 3x – x3 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.
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